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1100 | http://hyperphysics.phy-astr.gsu.edu/hbase/gpot.html | Gravitational Potential Energy
Gravitational Potential Energy
Gravitational potential energy is energy an object possesses because of its position in a gravitational field. The most common use of gravitational potential energy is for an object near the surface of the Earth where the gravitational acceleration can be assumed to be constant at about 9.8 m/s 2. Since the zero of gravitational potential energy can be chosen at any point (like the choice of the zero of a coordinate system), the potential energy at a height h above that point is equal to the work which would be required to lift the object to that height with no net change in kinetic energy. Since the force required to lift it is equal to its weight, it follows that the gravitational potential energy is equal to its weight times the height to which it is lifted.
PE = kg x 9.8 m/s 2 x m = joules.
PE = lbs x ft = ft lb.Index
Energy concepts
HyperPhysics MechanicsR NaveGo Back
Gravitational Potential Energy
The general expression for gravitational potential energy arises from the law of gravity and is equal to the work done against gravity to bring a mass to a given point in space. Because of the inverse square nature of the gravity force, the force approaches zero for large distances, and it makes sense to choose the zero of gravitational potential energy at an infinite distance away. The gravitational potential energy near a planet is then negative, since gravity does positive work as the mass approaches. This negative potential is indicative of a "bound state"; once a mass is near a large body, it is trapped until something can provide enough energy to allow it to escape. The general form of the gravitational potential energy of mass m is:
where G is the gravitation constant, M is the mass of the attracting body, and r is the distance between their centers.
This is the form for the gravitational potential energy which is most useful for calculating the escape velocity from the earth's gravity.Index
Energy concepts
HyperPhysics MechanicsR NaveGo Back
Gravitational Potential Energy
From the work done against the gravity force in bringing a mass in from infinity where the potential energy is assigned the value zero, the expression for gravitational potential energy is
This expression is useful for the calculation of escape velocity, energy to remove from orbit, etc. However, for objects near the earth the acceleration of gravity g can be considered to be approximately constant and the expression for potential energy relative to the Earth's surface becomes
where h is the height above the surface and g is the surface value of the acceleration of gravity.Index
Energy concepts
Gravity concepts
HyperPhysics MechanicsR NaveGo Back
Gravitational Potential
Index
Energy concepts
HyperPhysics MechanicsR NaveGo Back
Gravitational Potential Integral
Index
Energy concepts
HyperPhysics MechanicsR NaveGo Back |
1101 | https://dn790005.ca.archive.org/0/items/arrlacv3/ARRL_AC_v1.pdf | The ARRL Antenna Compendium Volume 1 Editors Gerald L. Hall, K1TD Paul Rinaldo, W4RI Maureen Thompson, KA1DYZ Contributors Those 31 authors whose material is published herein Production Maureen Thompson, KA1DYZ Cover Design Sue Fagan Published by the American Radio Relay League Newington, CT USA 06111 THE COVER: A fall sunrise silhouettes the antennas of station W1AW. (Photo by Jerry Hall, K1TD) Copyright 1985 by The American Radio Relay League, Inc. Copyright secured under the Pan-American Convention International Copyright secured This work is Publication No. 58 of the Radio Amateur's Library, published by the League. ISBN: 0-87259-019-4 Foreword The topic of antennas is one of the most popular in Amateur Radio literature. The material that has been written for amateurs on antennas would fill several volumes, and, indeed, several volumes are already available. So popular is the topic that ARRL Headquarters receives many more antenna manu-scripts for consideration as feature article mate-rial for the journal, QST, than can be published. In the past we have had to turn away good material because of space limitations in the journal. Instead of returning that material to t h e authors unpublished, why not collect it and publish it in a single volume? With this thought, The ARRL Antenna Compendium was born. Additional material was solicited for Volume 1, and you have the result in your hands. Early in the planning stages, we chose to use the format of QEX in preparing material for publication. The authors have provided their own camera-ready art work, and "typesetting" has been done with Apple lie computers and an NEC-3550 letter-quality printer. This approach was an experiment; we hope you will like the results. There is a wealth of material between these covers, and on a variety of antenna subjects. If you have a serious interest in antenna design or construction, you'll likely find something here that is "right up your alley." At this writing we anticipate that the Compendium will become quite popular, and that Volume 2 will be a l o g i c a l follow-on. We'd appreciate receiving your comments and suggestions for the next volume. David Sumner, K1ZZ Executive Vice President Newington, CT Contents Quad and Loop Antennas 8 The Sloping Diamond, A Full-Wave Loop for Four Bands the Easy Way Duane R. Sanderson, W0TID 11 Optimum Gain Boomless Quad Harold T. Mitchell, NJ0ARQ 18 Bicycle Wheel Quad Dave Guimont, WB6LL0 20 Quad Antennas for 80 and 160 Meters William M. Kelsey, N8ET 24 Gossamer Quad Update R. F. Thompson, W30DJ/ZF2CD 29 The PV4 Quad — A New Twist John DeWitt, AI9P 41 Cubical-Quad Antenna Design Norris G. Boucher, W3GNR Log Periodic Arrays 48 A Wide-Band, Low-Z Antenna — New Thoughts on Small Antennas Ken Heitner, WB4AKK 50 Development of the W8JF Waveram: A Planar Log-Periodic Quad Array Jim Fisher, W8JF 55 A Second-Generation Spiderweb Antenna Dick A. Mack, W6PGL Other Beam Antennas 62 A Simple Log-Yag Array for 50 MHz John J. Meyer, N5JM 64 Designing X-Beams Brice Anderson, W9PNE 67 An HF Phased Array Using Twisted-Wire Hybrid Directional Couplers James V. Melody, WA2NPJ/KX6JM 72 LARAE — Line Array of Rotary Antennas in Echelon A. J. F. Clement, W6KPC Multiband Antennas 84 A Great 10 Through 40 Portable Antenna Edward L. Henry, K0GPD 86 The G5RV Multiband Antenna . . . Up-to-Date Louis Varney, G5RV Vertical Antennas 92 Wiring Up the Old Spruce Kris Merschrod, KA2I0G 94 A Triband Parasitic Vertical Directional Array Walter J. Schulz, K30QF 101 The 5/8-Wavelength Antenna Mystique Donald K. Reynolds, K7DBA Antennas of Reduced Size 108 Optimum Design of Short Coil Loaded High-Frequency Mobile Antennas Bruce P. Brown, W6TWW 116 Short Loaded Half-Wave Dipole Design — The Easy Way H. L. Ley, Jr., N3CDR Miscellaneous Antennas 124 Dielectric Antennas for the 10-GHz and Higher Amateur Bands David Andersen, KK9W 127 A Crossed-Loop/Goniometer DF Antenna for 160 Meters Charles P. W. Anderson, N4KF 133 Subsurface Antennas and the Amateur Richard Silberstein, W0YBF Antenna Construction and Installation 142 A New Approach to the Construction of Large Yagi Beams Arie Bles, VK2AVA 144 Raising Beam Antennas Lawson Young, N4LY General Antenna and Transmission-Line Information 148 The Horizontal Dipole Over Lossy Ground Robert B. Sandell, W9RXC 152 Antenna Polarization Gerd Schrick, WB8IPM 157 Baluns: What They Do and How They Do It Roy W. Lewallen, W7EL 165 Available Power, SWR and Loading David T. Geiser, WA2ANU 170 Mr. Smith's "Other" Chart and Broadband Rigs Roger K. Ghormley, W0KK Quad and Loop Antennas The Sloping Diamond, A Full-Wave Loop for Four Bands the Easy Way Duane R. Sanderson, W0TID •3735 S. E. Stanley Rd., Tecumseh, KS 66542 The full-wave loop antenna has become the sub-ject of increasing interest among antenna exper-imenters in recent years. Judging by the volume of questions and comments I receive on the air, the loop is gaining popularity as an above-average performer. After 34 years as an active licensed amateur in pursuit of "that one great antenna," I think I have tried just about all applications of the standard quarter and half-wave antenna. My experience has shown that folded-dipole antennas have always performed better for me than standard dipoles. The folded dipole is basically a folded full-wave loop. An argument as to which an-tenna is best is not the point here. I only mention this to identify the experimenting trail that led me to an antenna I currently use. I decided my search for a better antenna would be in the full-wave loop category. Research Before Construction A review of published antenna articles, day-dreaming and mental pictures was necessary before I began to string wire. Published data on loops stated that some of their advantages are to provide gain over a dipole, and they are usually quiet even in a noisy environment. More broadbanded than half-wave types, they will work on more than one band. These good points are a match for my anteima requirements. My QTH is in a rural setting with leaky high tension power lines at the front of the property. I work both SSB and CW, and wanted an antenna that would work on 40 meters with 15-meter capability, and be coaxial line fed. I decided to put up a full-wave loop. Now, what's next? Loop Shape Different handbooks state that loops can be circular, square, rectangular, delta or diamond. My use of this term "diamond" differs from the square in that the two inside distances of a diamond, between the sets of opposite points, are not equal. It is longer than it is wide. From a construction standpoint, I concluded the delta was the most practical shape. Only one mast is needed to support the apex of a vertical delta, and metal fence posts can be driven into the yard at ap-propriate places to tie off the two bottom points. It just so happened that I had a 45-foot mast near the garage with a TV antenna on top. The center of the 40-meter dipole was attached about one foot below the TV antenna. The dipole was fed with a balun and a 52-ohrn mini-foam coaxial cable. I lowered the ends of the dipole and connected the appropriate half-wavelength of wire to them to form the loop. The formula I used for the loop length is the standard L = 1005/f (MHz). I used 142 feet of no. 12 gauge copper wire for my loop that I stretched into a vertical delta shape, tying the bottom corners to appropriately spaced 4-foot metal fence posts. I could just reach the lower horizontal part of the delta while standing on the ground. The appropriate prune-and-tune period took many trips from the rig to the antenna and back again with the best SWR figure of 1.6:1. Performance was a bit disappointing. Orienta-tion was broadside east and west. Forty-meter daytime short skip was not as good as on the old dipole, and signal reports often included heavy "QSB" and "fading" comments. Long skip and nighttime I Fig, 1 — The sloping-diamond loop antenna. performance was somewhat improved, but not signif-icantly better than the dipole. The delta did not perform well on 20, 15, and 10 meters. A f t e r s e v e r a l weeks of delta operation, I elected to try another loop shape. Again, the new shape had to work using only one mast for primary support, so I pulled the delta into the shape of a diamond and tied the two side points and the lower point to nearby objects. After pruning and tuning, I found the SWR was 1.3:1. 1 was making progress in obtaining a match, at least. Vertical vs Horizontal At this point my quad loop was a vertical diamond, broadside to the southeast and northwest. It was fed at the top, using a 1:1 balun and a 52-ohm RG-58 mini-foam coaxial cable. The total length of the loop wire was approximately 142 feet. Daytime reports were fairly good but not spec-tacular. However, nighttime reports were signifi-cantly better than any wire antenna I have had at this QTH. The diamond outperformed my half slopers, which are mounted on a 50-foot tower, by at least one S unit or more. The next few days and evenings were spent on 40-meter CW enjoying the short-skip daytime 579 reports, and the nighttime 589 to 599 reports from the West Coast. The power input was 60 watts. Operation of the loop on 20 CW resulted in rather poor performance compared with a 20-meter dipole on the tower. I concluded my diamond loop to be a good 40-meter nighttime long skip and DX antenna. As time progressed, I encountered a few sta-tions in nearby states who were using horizontal quad loops. These stations were consistently loud on the band in the daytime, and seemed to be more fade-free than others. They seemed to be using an antenna with high-angle radiation with rein-forcement from nearby ground reflection, and had the equivalent of a two-element quad pointing straight up. In choosing this antenna for use at my QTH, I recognized that most of my operating is in the day-time or early evening hours, with late-night DXing an occasional event. This is probably true for the majority of amateurs on 40 meters. I found that, still using the 45-foot TV mast as the top of the diamond, I could pull the loop away from the vertical plane to nearly horizontal, but with some slope. I placed three metal fence posts in appropriate locations in the yard as tie points for the diamond shape. My horizontal quad became a loop sloper with an approximate 30-degree slope off horizontal. See Fig. 1. Once I used the prune-and-tune method on my new antenna, these were the results I obtained. At 7.1 MHz no reflected power was measured (1:1 SWR). The horizontal-loop impedance was a perfect match to the RG-58 coaxial cable and the broadband character-istics are the best I have ever seen in any of my antenna projects. A check of the SWR on each band from 40 through 10 revealed the following: 7.0 MHz 1.2:1 14.0 MHz 1.3:1 7.1 MHz 1:1 14.1 MHz 1.1:1 7.2 MHz 1.2:1 7.3 MHz 1.5:1 14.2 MHz 14.3 MHz 1.2:1 1.6:1 21.0 MHz 21.1 MHz 21.2 MHz 21.3 MHz 1.8:1 1.7:1 1.9:1 2.0:1 28.0 MHz 28.1 MHz 28.2 MHz 28.3 MHz 1.9:1 1.9:1 2.3:1 2.2:1 Does It Work? You bet it does, and it works well on more than one band. Daytime 40-meter operation is great. Reports received while using 60 watts input are usually 599, and often there are reinforcing com-ments or questions from the other stations about my antenna. The receiving performance of the sloping diamond is equally good and demonstrates the quiet nature of loops, with low noise and loud signals. Nighttime operation is similar to daytime operation until late evening when West Coast stations overtake the band. At that hour, my half slopers are about two S units better than the loop. This is because of their low-angle characteristics. The bonus came when the loop was used on 20 meters. The SWR was almost as good as 40 meters, and performance on 20 is generally better than my dipole on the tower. Per-formance on 15 is almost as good as the 3/4-wave slopers on the tower. Performance on 10 has not been evaluated as of this writing because the band is generally dead, a victim of the sunspot cycle. My approach to operating with this antenna is to use a line tuner on 21 and 28 MHz to tune out the small amount of reactance present on those bands, and to provide an ideal match to the rig. The Clincher The sloping diamond was put through a good test a few weeks after it was erected. The world-wide QRP contest arrived on the calendar and about mid-morn-ing of the first day, I decided to hook my Argonaut to the loop and have a go at it. I spent about 2-1/2 hours working the 40-, 20-, and 15-meter CW bands. I worked 50 stations without difficulty. Among my con-tacts were G4, KH6, KL7, VE, and a good spread of stations across the U. S. A. This antenna appears to be omnidirectional. At present, I am unable to find holes or blind spots in its performance. Some Construction Comments A rectangular horizontal loop has four corners to support, but the use of one mast as the primary support keeps things simple. A point on an existing tower, building or other structure can serve equally well. The mast I use is made of steel conduit, 1-3/4-inch diameter at the bottom. Successively smaller diameter sections are telescoped together a few inches at each joint with bolts and sheet metal screws through the overlapping points. I have found that sheet metal screws will help maintain a good electrical bond between each mast section. The base of the mast is hinged and set in a small poured concrete base for stability. My need for mast hinging stems from a lot of antenna experimentation at my location and with this setup, I can work alone to raise or lower the antenna. The mast hinge is made of two 16-inch pieces of 1-1/2-inch iron pipe set vertically with 10 inches in the concrete and 6 inches exposed. The two pipes are about 1-3/4 inches apart with sufficient space for the mast to set between them. All three pipes are then drilled so that a long bolt can pass through all three, as shown in Fig. 2. The bolt is the hinge or pivot point. My mast has one top guy on the side opposite the antenna, and the diamond shape of the antenna forms the equivalent of two other guys. The end product is a 3-guy arrangement, and the stability of the mast is surprisingly good. A house bracket about midway up the mast ties it to the eaves of the roof. This makes the mast rigid and flexproof. Summary The unique propagation advantages of a hori-zontal loop seem to combine, to some degree, with the propagation advantages of a vertical loop when the plane of the loop is sloping. The amount of ground space required for the fence-post anchors is determined by the degree to which the constructor pulls the diamond sides apart. My diamond is nearly twice as long as it is wide. The ARRL Antenna Book states that next to a circular loop, a square loop is the most efficient, and rectangular loops or unequal diamonds have some energy cancellation. My loop is more broadbanded and provides the best match to 52-ohm coax, when it is shaped as an unequal diamond. The use of a balun is optional. My use of balanced feed stems from my preference for a symmetrical pattern. My experience with this antenna as a sloping loop has produced reports that are considerably better than when the loop is hung vertically. The exception is for late-night long skip where a ver-tical loop would perform best. My home is surrounded by hilly terrain that is 40 feet plus higher than my yard elevation. Ama-teurs with low lying antenna sites, or at locations surrounded by tall objects, may find the horizon-tal or sloping loop to be the best choice. Those with large lots may want to try a really big loop and go for an 80-meter loop or one for 160 meters. Soon I will have an 80-meter diamond sloper strung from a mast on top of one of those hills I mentioned. Listen for me on the bands and help me find out how well it works! Pig. 2 — A mast hinge is made of two lengths of 1-1/2-in iron pipe set in concrete. Their 1-3/4-in separation accommodates the steel conduit of the homemade telescoping mast. Optimum Gain Boomless Quad By Harold T. Mitchell, N0ARQ •2403 Inca Lane, New Brighton, MN 55112 This 2-element beam can be precut for 10, 15, and 20 meters, with provision for 12 and 17 meters. It can also be adjusted for maximum front-to-back ratio using mini-stubs. Perhaps you are interested in a quiet beam antenna, superior performance at low heights, single coaxial cable feed without an antenna tuner, high quad strength through cross ties, and a one-man installation of a 3-band quad. This article tells how those were obtained, plus optimum gain spacing for each band, and maximum front-to-back ratio ad-justability. The antenna material cost a total of $75, and a guyed 13-foot roof-top mast was purchased for $45. History and Background The common 2-element 3-band concentric quad was patented by James C. McCaig of England in 1960.\1 It used bamboo canes and had an 8-foot boom for 10, 15, and 20 meters. Eighteen years earlier, Clarence C. Moore, W9LZX, had invented the cubical quad anten-na to solve corona problems encountered at the 9,500 foot altitude site of HCJB in Quito, Ecuador.\i A beautiful picture of four 6-element HCJB quads against the Andes was shown on the cover of a recent broadcasting magazine.Y& Quads have withstood the test of time. The large thin planar elements of either a sin-gle or multiband quad have trouble resisting high bending loads. R. Michael Doherty invented a quad with spreader-reinforced crossarms, stating in his patent, "The chief disadvantage to cubical quad antennas is their lack of strength and vulnerability to high winds and icing conditions.'M In New Brighton, MN, we are faced with severe weather, plus a restrictive antenna height ordi-nance. I turned to the quad for its superior per-formance when installed at low heights, and picked up the challenge of designing a strong quad. At my location, the 20-m loop bottom wire is only 1/4 wavelength above ground, yet the vertical angle of radiation for 20 m e t e r s is s t i l l b e l o w 40 d e -grees.VS. I designed this quad with not only performance and strength in mind, but also simplicity, one-man i n s t a l l a t i o n , and l o w c o s t . P a r a l l e l invention, simultaneous development, or redevelopment are not uncommon in history. When almost done with the detail design based on a local quad, I investigated the literature and found the Gem Quad antenna in-vented by Emerson G. Partridge, VE4RA.\fuZ The Partridge patent describes a boomless quad with unique fiberglass construction having inherently low wind resistance. My design uses Sitka spruce for spreaders. For the ham willing to construct a beam antenna, this design offers several new advantages, plus a rock bottom price well below that for even a bamboo quad (Fig. 1). Why Boomless? Quad wire loops can be square, circular or some intermediate shape, but they have one thing in com-mon: All are planar. In the case of a 3-band con-centric quad, the three loops of each element are also in the same plane. Wind and ice loads produce high bending s t r e s s e s in the spreaders; high-strength fiberglass tubes, and even vaulting poles, are resorted to.\£ Boomless construction and 12 cross ties turn the quad into a structure with rigidity. Loads can be shared and mechanical oscillations eliminated. In addition, the spreaders are stabilized against side loads and high bending moments. The boomless quad is strong through s t r u c t u r a l d e s i g n , and spruce spreaders can be used without undue concern. The quad contains 36 triangles, which are cross-coupled in three dimensions at each vertex. Spiders were designed with 18-degree angles to obtain optimum gain 0.12-wave spacing for each band.\£ A concentric quad only can provide optimum gain for one band, since the boom length determines common spacing for all bands. COPffiP.W6.UD - . 8.33' _ AHTENHA 6-7dBd GAIN lO1 TURNING RAS>\OS 1 11 SQUARE. FETET WIND MVEA 5 4 POUMDS M A S T SCHEt>Ul_e H-O PVPE t h r e e s e e m cms &H- P o u n d s Fig. 1 — The end view of the roof-mounted 10-, 15-, and 20-m boomless quad. Each band has 0.12 wavelength optimum gain spacing. Precut or Tunable? Table 1 gives precut wire loops and spreader attachment points for five bands. The emerging 12-and 17-meter bands are covered in addition to 10, 15, and 20 metersAlfl There are a number of precut loop designs with booms for single band or concen-tric quads.\11JL2. Precut antennas are convenient because construction and tuning steps are elimi-nated, which may be intimidating or burdensome. However, a considerable performance penalty is paid for this convenience. R. J. Eckersley, G4FTJ, has stated an excellent case for antenna tuning, "For working DX it is clear that good front-to-back ratio is more important than forward gain — often the limiting factor in copying a weak signal is interference coming from the op-posite d i r e c t i o n . . B i l l Orr, W6SAI, adds weight to the need for quad tuning as he writes, "...the front-to-back ratio of the array is quite critical as to stub placement."\14 There are problems with most tuning stubs as they are relatively long and flimsy affairs, with 34- to 38-inch stub length for a 20-m reflector loop.\l£ I have added 3 percent to the driver length to obtain precut reflector loops for the boomless quad. This is the accepted relationship that goes back to the work of Lee Bergren, W0AIW, in 1963.\16r17r18 The reflector loop lengths of Table 1 are also used for the tuned reflector quad. This permits small 12-inch x 3-inch mini-stubs rather than long stubs used where driver and reflector loops are initially of equal length. A tuned boomless quad is said to give an excellent 25-dB front-to-back ratio on 10 and 20 meters, and a very good 20 dB on 15 me-ters.Ua Relays, Baluns, Line Transformers, or Gammas? A boomless quad requires solution of the basic antenna problem — how best to maximize radiated power and minimize SWR at the transmitter. Multiband quads have been matched to coaxial feed line by several different methods.\2J1 A tri-gamma matching system has been devised by Jack A. McCullough, W6CHE, and applied to a precut 4-element concentric quad with a 30-foot boom.\21 I have adapted the tri-gamma match (Fig. 2) to the optimum-gain boomless quad. A reactance capacitor balances gammas and the open-wire transmission line, which is the key to the system. The open-wire line connects all driver loops and the single coaxial cable from the transmitter. Detail Design and Construction The heart of the boomless quad consists of two short steel pipe spider hubs and a pipe axle mast T (Fig. 3). Welded to the hubs are steel angle stock spider arms to support the spreaders. With this hub and axle construction, quad elements can be in-stalled one at a time (Fig. 4). The pair of ele-ments then rotate easily in the vertical plane for complete cross tying (Fig. 5) prior to bolting to the hub axle. If a diamond quad is preferred instead of a horizontal quad, it is only a matter of changing hole locations at the hub axle. For a diamond quad, the signal will be polarized at 45 degrees and contain both vertical and horizontal components unless the feed point is shifted to an apex. To obtain a vertically polarized signal, the quad is rotated to obtain feed on a vertical side. CE.NTE.W. OF DRIVER _OOPS INPUT COA 50-72. r H \oivi 50 f (-Z0 1" SPAC\N6 X-2-O" 00 "2.8 MWz. GMNANIA I5M Jfc -"X -3D" 75>F (57) "Z-l G^MMA 3/V SbPAONG, "TRAusMisstoH l i n e "20 M X 4-0" (."33) 1 H - MHz. GPMtM tarJlOOpKH-5) 3 S O (>2_) REACTANCE. CPAC\TOR Fig. 2 — The tri-gamma matching system provides impedance matching and reduces interaction between loops. Open-wire transmission line and gamma rods are 12 AWG solid. Back-to-back alligator clips are used for gamma rod length adjustment during match-ing. Values in parentheses are for RG-8 coaxial cable and N0ARQ low elevation. Precision assemblies follow from large simple welding fixtures glued together from 3/4-inch plywood. Spider accuracy results from a square hub face in contact with the fixture surface (Fig. 6A) and equally spaced 18 degree (5-3/16 inch x 16 inch) ramps to guide the steel angle. Axle T accuracy is obtained from a 3/4-inch slot, which lifts and aligns the mast pipe for welding to the larger hub axle pipe resting in a 1-1/2-inch slot at right angles (Fig. 6B). FOUR HOLES FOR 3/8" X 1" BOLTS HUB'AXLE. . ^ ^ S . SCHEDULE H O P\PE SraE. OD ID l"P\PEXIO» 238" 2 07" =,TFF\ ANGLE. 1 rlrL X IU -\Jz. 1.90 I ' d e r ^ m s m w t s t u b \%z 1.0s |-1A"X l-lA" V^" 1 8' 1 Fig. 3 — The spider hub and axle mast T are the key to this boomless quad. Construction allows separate installation of each element as well as rotation together in the vertical plane for cross tieing. Sitka spruce was selected for spreader material for its strength and low cost. Clear boards without knots come from these very tall trees that grow along the coast from Alaska to California. Sitka spruce is still used for propellers and framing small aircraft. Vertical grain boards are twice as expensive as flat grain and are unnecessary as the original grain direction is lost when ripped into a square cross section. Ten 7/8-inch x 7/8-inch spreaders were obtained from two 1-inch x 6-inch x 14-foot boards, with several thinner strips left over. Each spreader cost $1.76 and weighed 2.4 Fig. 4 — N0ARQ slips the reflector element onto the mast T. A 63-inch spruce tie bar supports 12-inch x 3-inch mini-stubs in the bottom center of each reflector loop. Fig. 5 — N0ART stands on the roof to add 12 polyester cross ties by rotating the quad in the vertical plane. Polyester is strong and stretches very little. It is also highly resistant to sun-light. Fig. 6 — At A, the spider welding fixture is constructed on a 2 foot x 2 foot x 3/4-inch plywood base. Evenly spaced 5 3/16-inch x 16-inch ramps provide the 18-degree spreader angles. At B, a hub axle and mast T fixture lifts the 1-8 inch mast pipe for welding at right angles to the 1-1/2-inch axle pipe. pounds. A sharp saw blade left slightly rough surfaces, which were then primed and painted with light blue latex house paint to blend with the sky. Sanding or shaping of spreaders is not necessary. Number 12 AWG solid, steel-core copperweld was used for strength and resistance to stretching. If all copper wire were chosen, stranded would be used for added resistance to breaking.\22 Spreaders were measured according to Table 1 and drilled on the diagonal for the 0.08-inch OD wire. It was nice working with solid wood of constant cross section rather than thin tapered bamboo or f i b e r g l a s s cylinders. After threading and positioning the wires at each spreader, wrap wires were added and soldered to p r e v e n t s h i f t i n g of the s p r e a d e r s on wire loops.\2j£ Silicone rubber sealant was applied at drilled holes to prevent water entry and give electrical insulation. Each spreader was designed to butt against its spider hub and not require adjustment. Spreaders were fastened to spider arms with glass filament shipping tape instead of the usual 16 stainless steel hose clamps (more costly). The shipping tape was then over-wrapped with black vinyl tape to protect it from sunlight. It was necessary to plan ahead for placing various gamma-rod spacing insulators on driver loops before closure. To avoid breakage, the copperweld should be spliced and soldered without sharply bending w i r e s . H e a v y scissors were used to cut out the open-wire line spacers from 1/16-inch Lexan polycarbonate sheet. Holes can be drilled or punched in this excellent electrical insulation, which is also tough and resiliant. The spacers were laced in place with nylon cord.\2£ This is a very rapid, effective and economical method of building open-wire line (Fig. 7). It proved especially useful since five different spacings were required. As the first spacer of each reflector loop stub, an egg insulator was used to carry loop tension under load.\2£ Fig. 7 — Coaxial cable input is at the 15-m point to the tri-gamma matching system. Open-wire line polycarbonate spacers were laced in place with nylon twine. Variable capacitors used during matching were removed, measured, and replaced with silver mica fixed capacitors. Coaxial braid and ends were coated with silicone rubber sealant to prevent moisture wicking, corrosion and change. Final Assembly The boomless quad was assembled on a home built, roof-top mast designed to withstand a 40 pound per square foot (100 mi/h) wind load on the system.V21 A mast height of 13 feet was chosen as a minimum to be consistent with the 12.8-foot vertical turning radius of the quad. The basic mast was fab-ricated from two 1-1/2-inch schedule 40 steel pipes. A 10-foot length of pipe weighing 27 pounds was the heaviest item lifted during installation of the quad and mast. A 6-foot sleeve of 2-inch schedule 40 pipe was centered and fastened with 5/16-inch eyebolts in the middle of the mast to pin the two 1-1/2-inch pipe sections. The main function of the sleeve is to pro-vide additional bending strength in that critical length. The mast, rotator, and tee were preassembled, for f i t and then finished with R u s t - O l e u rrr1 primer and paint before taking sections to the roof for reassembly. Each of three evenly spaced 3/16-inch EHS galvanized steel guy wires was brought down at a 45-degree angle from the eyebolts and terminated in a thimble.\2& A bridle through the thimble (Fig. 8) divided the load of each main cable between two 5/16-inch eyebolts screwed into the house frame. All cable ends were secured with 3/16-inch cable clamps. Turnbuckles are unnecessary in this application, and can even be undesirable.\22 Fig. 8 — A bridle running through a thimble in each guy wire divides the load between two eye boits screwed into the house frame. Two cable clamps are used at each end of the 3/16-inch EHS guy cable. One clamp is sufficient for each end of the 3/16-inch bridle cable. From inside the roof, an entrance point for the 1.90-inch OD pipe was selected so that the pipe would rest against the roof center beam and a rafter (Fig. 9A). A center-line hole was drilled up through the roof and a plumb bob was dropped from a nail in the hole to locate the exact center of the mast base. A 2 x 8 was fastened to the joists above the ceiling and a circumferential pattern drilled for 8 nails, which form a funnel socket (Fig. 9B) for the mast. This may appear unorthodox, but it is very simple, and the shearing strength of three nails for any wind load direction is certainly sufficient. A corner-notched 2 x 8 was used to wedge the mast against the center beam and rafter. A roof-top mast clamp to prevent rotation was made from a V cut 2 x 4 (Fig. 10), and it was installed after sealing the roof entry. There are other ways to roof mount an antenna, but this mast proved very strong, easy to build, and economical. It provided a precisely vertical mast without adjustment.\Mwll After the mast was as-sembled and guyed, each quad element was lifted into place. Then the elements were rotated together vertically and cross tied with 1/4-inch braided polyester cord at the 20-m loops and 3/16 inch for the 10- and 15-m loops. Polyester (not poly-propylene) was usee for its superior UV resistance and minimal stretch when compared with nylon. Although this antenna and mast were value engineered to reduce cost and overdesign, higher than normal quality materials were used where system strength and stability could be enhanced. After bolting spider hubs to the axle (Fig. 11), the antenna and rotator were aligned with true north, and final electrical and ground connections were made to the antenna, rotator and mast. (A) (B) Fig. 9 — At A, the mast rests against the roof center beam and a rafter. A notched 2 x 8 completes a tight box below the roof. At B, a funnel pattern of nails in a 2 x 8 serves as roast socket and base. Fig. 10 — A V-cut 2 x 4 and screw-adjustable block make a roof-mounted mast clamp to prevent rotation. The roof was sealed before clamp in-stallation. A heavy ground wire is attached to the mast for lightning protection. Dressed along the mast are coaxial and rotator control cable, which has been wrapped with grounded copper mesh to prevent RF pickup from feeding the shack. Tuning the Booraless Quad 1. Set the gamma system broadcast type temporary variable capacitors and rod l e n g t h s approximately to the values found by N0ARQ (Fig. 2). Since loops and adjustments i n t e r a c t , it is recommended that these initial values provide a starting point and that the following steps be taken Fig. 11 — An end view of the quad shows pre-cut spreaders against the hub. The hub and axle pipe are self-aligning. The quad is symmetrical with low rotational inertia and presents modest loads to the rotator. Coaxial cable was brought up over the mast T for strain relief. A 7 foot 2 x 2 stiffener was fastened to each upper spreader with shipping tape overwrapped with black vinyl tape for protection from the sunlight. c a r e f u l l y and sequentially. If tuning stubs are used, adjust each reflector loop stub for maximum f r o n t - t o - b a c k r a t i o . This can be done with a receiver S meter at the antenna, or with earphones on a long cord. Listen for a distant signal with the back (reflector) of the quad aimed broadside at the source. Move the alligator clip shorting bar until a minimum response is heard. Tf the 20-m reflector loop is too long with the stub length entirely shorted, then bridge the loop bottom corners an equal amount and fine tune with the stub. Because of local reflections, it may be desirable to try sig-nals from more than one direction. The stubs are adjusted first, as these reflector adjustments will affect driver loop matching. 2. Check that the shield of the input coaxial cable is connected at the 15-m level to the side of the open-wire transmission line that connects the center of each driver loop, and check that the reactance capacitor is at the 20-m level. Apply low power on 10 meters and adjust the 10-m gamma capacitor and rod length for lowest SWR as the transmitter frequency is varied, to find the band resonant frequency. 3. Switch to 15 meters and tune the 15-m gamma for lowest SWR at resonance. 4. Switch to 20 meters and tune similarly. Now also vary the reactance capacitor as well as the gamma capacitor and rod to further reduce the 20-m SWR. 5. Recheck 15 meters, then 10 meters, and readjust for lowest SWR. If adjustments and results are not adding up, it will seem as if the SWR im-proves as gamma lengths are shortened. If the gamma lengths are much shorter than noted in Fig. 2, it is a good indication that the reactance capacitor set-ting is incorrect. If band c o n d i t i o n s are not cooperating to deliver distant signaLs when front-to-baok adjust-ments are to be made, it may be desirable to enlist the services of a local ham. Your amateur friend should transmit a horizontally polarized signal from more than 500 feet to avoid confusing results. This adjustment sequence produced excellent matching as measured by a Drake MN7 on the end of 35 feet of RG-8. SWR below 2.0 fop two bands and the CW portion of 10 meters is possible because of the low Q characteristic of a cubical quad. From a loss standpoint, an SWR below 2.0 is considered fully acceptable.\33f34 It also permits full output of s o l i d - s t a t e transmitters that automatically reduce output for SWR in excess of 2.0. The four variable capacitors were removed and measured, then substituted, with 500-WVDC silver micas. Jumper wires were soldered in place of alligator-clip shorting bars. This quad is easy to work on, as all ties and adjustments are within arm's reach from either the roof or only the mast. From the data (Fig. 12), very broadband per-formance is documented. With such low SWR, the boomless quad can be fed directly without an antenna tuner. An often overlooked benefit is also realized. The deleterious effect of antenna tuner Q on band-width is also eliminated.^ If the transmitter fre-quency is changed without retuning an antenna tuner, then the tuner Q reduces power transfer to the antenna. Other things being equal, a d i r e c t l y matched antenna will have the greatest bandwidth. Conclusion The excellent performance of this antenna on the bands means that we have joined the big leagues. The entry fee was a modest dollar expenditure, plus time. For this investment, there has been the satisfaction of building a multiband quad that is the equal of 2-element commercial quads. I also believe that it is superior to the 3-element Yagi. It clearly has better bandwidth and front-to-back ratio, and is within tenths of a dB on forward gain. It wins on gain, too, if the Yagi is tuned for maximum front-to-back r a t i o . A t low heights there is no contest as the quad wins hands down for low-angle power radiation. The 11 square feet of wind area causes little concern. Rotators are rated for unsupported antennas within 24 inches. The center of the boomless quad is within 6 inches. It is also a symmetrical structure with relatively low rotational inertia. Rotator drive and stopping requirements are modest. With this roof mast and guying design, a total system with high strength was obtained. The 13-foot mast should handle practically any antenna and is recommended for a neat roof-top installation. Structural design and 12 AWG copperweld, together with quality materials and assembly, should solve the old strength problems of 2-element quads. The lace-like appearance of this antenna and mast will also draw lots of attention. One said it looks like art. Appendix A massive storm with 100 mi/h+ winds ripped through the Minneapolis — St. Paul area during the summer of 1983. It produced the most extensive local power failure in history. The mast stood but one spreader and several spider angles failed. The heavier spider steel angle stock of Fig. 3 gives a factor of 6 increase in bending strength. The top spreaders have been stiffened from the hub to above the 10-m wire, using 1-1/2 inch x 1-1/2 inch x 7 feet 2 x 2s. These were taped to the spreaders to give a factor of 5 increase in bending strength. As a result of the storm, an even stronger antenna was rebuilt and is reflected in this article. With heavier steel and new top stiffeners, I now recommend against one person lifting elements into place. Moment arms magnify actual weight. Although possible, it becomes difficult for one person to align an element to slip its hub on the T. With a gin pole it would be an easy one-man job. Acknowledgement I wish to thank my son, NjOART, for getting me started on beam antennas. Todd found a new HAM-M rotator (we have added our own brake variable delay modification) for sale at his Junior High School and pointed out that the quad is an unusually effective low-angle radiator when installed at low heights. Thanks also to Todd's brothers Tom, N0BSY, for improving this article, and Tore for taking and printing the photographs. I have appreciated the patience and tolerance of my wife during the period S W R , 3.0 Z.O l.o urn/ / I I I N ^ A R q BOOMLE.SS Q U A D — —) urn/ t — —) t r r - - - - " ISN\ X e>/\ N D E-o^e. 1 0 M 1 5 M 20M 2 8 . 0 21.0 14.0 " 28.2 21.05 14.05 28.4 21.1 14.1 28.6 21.15 14. 15 2 8 . 8 2 9 . 0 2 9 . 2 2 9 . 4 2 9 . 6 2 9 . 7 2 9 . 8 MHz 2 1 . 2 2 1 . 2 5 2 1 . 3 2 1 . 3 5 2 1 . 4 2 1 . 4 5 1 4 . 2 1 4 . 2 5 1 4 . 3 1 4 . 3 5 1 4 . 4 1 4 . 4 5 SWR = (i + /ftf^ j (i - rrpr^ l'r = the power in the reflected wave P e = the power in the forward wave 10 M 20 M 15 M Band Edge Fig. 12 — SWR plots of N0ARQ boomless quad. of intense activity required to design, build, and publish something of worth. R. P. Gruetzman, K0KLH, W. W. Karjalahti, WA0ODW, and B. W. Schultz, WAjOOOU, are credited with developing the basic quad structure. References \1 McCaig, J. S., "Improvements Relating to Composite Aerials," U.K. Patent No. 850,974, Oct. 12, I960. \Z Orr, W. I., and Cowan, S. D., All About Cubical Quad Antennas, 3rd edition, (Wilton, CT: Radio Publications, Inc., 1982), p. 6. \2. Woram, J. M., "Broadcasting and Recording in Ecuador," dB Sound Engineering, April 1982, pp. cover, 47. Doherty, R. M., "Cubical Quad Antennas With Spreader-Reinforced Crossarms," U.S. Patent No., 4,138,682, Feb. 6, 1979. See note 2, p. 33. "Pointers on the Gem Quad Antenna," QST, Feb. 1979, p. 50, and "Product Review," QST, Jan. 1978, p. 38. \Z Partridge, E. G., "Antenna Wire Support Struc-ture," U.S. Patent No. 3,532,315, Oct. 6, 1970. "Fiberglass Poles for Antenna Construc-tion," Hints and Kinks, QST, March 19 83, p. 41. \9_ See note 2, p. 30. "FCC Takes Big Step Toward Putting WARC-79 Into U.S. Law," QST, March 1983, p. 55. \11 Hall, G., ed., Hie ARRL Antenna Book, 14th edition (Newington: ARRL, 1982), p. 9-7. \12 See note 2, pp. 40, 46, 52. M l Eckersley, R. J., ed., "Working DX Stations," Amateur Radio Operating Manual, 2nd edition, (London: RSGB, 1982), p. 36. \14_See note 2, p. 89 and note 11, p. 6-16. \UL Orr, W. I., and Cowan, S. D., All About Cubical Quad Antennas, 2nd edition, (Wilton, CT: Radio Publications, Inc., 1970), p. 40. \16_ See note 11, p. 6-23. \12 Lindsay, J. E., "Quads and Yagis," QST, May 1968, pp. 11-19, 150. \1& Bergren, L., "The Multielement Quad," QST, May 1963, pp. 11-16. \19 See note 2, p. 50 and note 6, p. 50. \20_ See note 2, pp. 60-74 and note 11, pp. 5-14, 9-9. \21_ See note 2, pp. 75-80, 107. \22_ See note 11, pp. 8-21, 9-5. \23 See note 11, p. 9-8. \M. See note 11, pp. 8-22, 8-24. Woodward, G., ed., The Radio Amateur's Hand-book, 60th edition, (Newington:ARRL, 1983), p. 17-6. \26 See note 11, p. 8-23. Structural Standards for Steel Antenna Towers and Antenna Supporting Structures, EIA Standard RS-222-C, Electronic Industries Assoc., March 1976, available from EIA, 2001 Eye St., N.W., Washington, D.C. 20006. \28_, See note 11, p. 9-19. \29 See note 11, p. 8-29. Mil Ellis, C. J., "A N o v e l Way to Mount a Rotary-Beam Antenna," QST, May 1979, pp. 32-33. \31 See note 11, p. 9-24. \32. See note 2, p. 37. \33 See note 11, pp. 3-12, 3-14, 3-20. \M Gibilisco, S., "What Does Your SWR Cost You?," QST, January 1979, pp. 19-20. \35 See note 11, p. 5-22. \M See note 11, pp. 6-16, 6-18. Table 1 Side and Spreader Lengths Band f S D S.A. S„ S.A (meters) (MHz) (ft) (ft)u (ffi) (ft)K 20 14.17 17.57 12.90 18.14 13.32 17 18.12 13.74 10.05 14.18 10.38 15 21.22 11.73 8.56 12.11 8.84 12 24.94 9.98 7.25 10.30 7.49 10 28.85 3.63 6.25 8.91 6.46 Spacing (ft) 8.33 6.51 5.56 4.73 4.09 All length dimensions are feet f = design resonant frequency in MHz driver side length S„ = 257/f = reflector side length = spreader attach length SD = 249/f R S.A. S.A. = [side/(2sin45° x cosl8 0)] - hub clearance S.A. = 0.7435 x side - 0.167 Free space wave = 984/f Spacing = 0.12 wavelength = 118/f •Measured from hub end of spreader Bicycle Wheel Quad By Dave Guimont, WB6LLO •5030 July St., San Diego, CA 92110 The prototype of this antenna was originally developed for OSCAR 8 uplink modes A and J. Per-formance in those modes was highly satisfactory, and uplink comparison checks were made against a 16-element crossed Vagi operating in the following configurations: vertical, horizontal, axial, LHCP and RHCP. The Bicycle Wheel Quad signals equaled or bettered those configurations 90% of the time. The "bicycle wheel" construction is not my design, but is patterned after an article in Ham Radio.M Aluminum clothesline wire (no. 9), drilled at the spoke junctions, was used in the original antenna. Two years of exposure to a mild, but corrosive, atmosphere finally took its toll. The 1/4-inch aluminum tubing and split-ring anchoring method used here should prove more durable (Figs. 1 through 3). j 2 f X Fig. 1 — Element joint. Various electro-mechanical polarity rotation systems were considered. The fishline lanyard works quickly and efficiently, other devices may be re-quired at your location. Fig. 4 shows the rotation arm support used in my setup. Gain figures are unknown. Again, on-the-air performance comparisons are the only methods avail-able to me. Brief testing on OSCAR 10 indicates reception is equal to or better than a 6-turn RHCP helix or a 16-element crossed Yagi in all config-urations. Element spacing was determined empiri-cally. Prototype spacings (no. 9 wire) were within 1/2 inch of these dimensions (Fig. 5). A field strength meter and SWR bridge were used to set tho elements and gamma for maximum gain. Minimum SWR and maximum gain did not coincide. These dimensions provided maximum gain at a SWR of 1.3:1 measured at the antenna. Photos and the five illus-trations provide construction details. Mount the antenna so a pull on the fishline provides vertical polarization (feed point at the side). Rig a small diameter bungee so its tension plus the weight of the coexial feed line returns the feed point to the bottom (horizontal polarization). My Bicycle Wheel Quad is shown in Fig. 6. I would be interested in comments and improvements from builders. Information regarding actual gain would be particularly appreciated. CU on OSCAR 10! C O -l b . n y l o n flan l.ne " s p o k e s " P l i o b c n d C e a e n t Split ring (3 p e r e l e m e n t ) Fig. 2 — Spoke attachnerit. Fig. 3 — Hub detail. The materials are inexpensive. Ordinary hand tools will do the j o b . One f e e d line provides vertical, horizontal or any degree of polarization in between. The turning radius is less than 30 inches, and my antenna weighs 2 pounds, 12 ounces. References \1 Kennicott, J. W., W40V0, "Dream Beam," Ham Radio, p. 12, May 1980. fiberglass S o d 1 2 -i M O M l o n g Fig. 4 — Rotation arm support. ?
7 C T e e — split a n d s p r e a d to fi CS-) J-V2 3 -1 / 2 -Fig. 5 — Element spacing. MATERIALS OTY 1 4 ft, 6 inch length 1/2-inch schedule 40 PVC (boom) 4 6 inch length 1 inch schedule 40 PVC (element hubs) 30 ft 1/4-inch aluminum tubing 80 ft 60-]b test nylon fishline (spokes) 32 3/8-inch split rings (perimeter spoke supports) 1 3/4 inch x 3/4 inch x 1/2 inch T schedule 40 PVC (rotation arm support) 1 1 ft x 3/8 inch fiberglass rod (rotation arm) GAMMA MATCH BNC connector 20-pF variable capacitor 6 inch, no. 10 copper wire 3/8 inch x 2 inch stainless strap 8-32 x 1/2 inch stainless hardware ELEMEm1 DIMENSIONS Reflector Driven 1st Director 2nd Director 27 inch 25 3/4 inch 24 13/16 inch 23 7/8 inch 84 3/4 inch 8C 7/8 inch 78 inches 75 inches (I added 6 inches to the circumference for overlap, and flattened and secured it with 8-32 stainless hardware.) SPACING Reflector to driven 15 1/2 inch Driven to 1st director 13 1/2 inch 1st director to 2nd director 18 1/2 inches Fig. 6 — A view of the completed Bicycle Wheel Quad. Quad Antennas for 80 and 160 Meters By William M. Kelsey, N8ET 2716 C.R. 26, Mt. Cory, OH 45868 During 1977 and 1978, I constructed and used a 3-element quad antenna for 40 meters.\l It was the best 40-meter antenna I ever used. European stations could be heard over most of the east coast stations, and a simple switching unit, constructed with the quad, enabled the quad's direction to be reversed so I could work ZL/VK stations. I wanted to try the same arrangement on 80 meters, but was transferred to England for f i v e years. It was difficult to erect even a small tri-bander where I lived, so the 80-meter quad was out of the question. Returning to Ohio in 1983, I again had room to construct the 80-meter quad antenna and found that KS8S had just erected a 180-ft tower. I began to think about a quad for 160 meters, also. General Design A QST a r t i c l e I was reading at the t i m e indicated that a quad-loop reflector could be created by cutting the loop the same size as the driven loop, and adding approximately 150 ohms of inductive reactance.\2. A director could be con-structed in the same manner, except that approx-imately 150 ohms of capacitive, not inductive, reactance was to be added. See Fig. 1. By applying some basic transmission line theory and making use of a Smith Chart, 150 ohms of reactance can be created at one end of the feed line. This is done by placing a capacitor at the other end of the line. The line length transforms any capacitive reactance at one end to 150 ohms of capacitive reactance at the other end; or, if the line is a bit longer, to 150 ohms of inductive reactance. In fact, any value of reactance can be created using the correct line length. As an example, Fig. 2 shows how 150 ohms of inductive reactance can be created by placing a capacitor of 900 ohms reactance at the end of a length of a 450-ohm line, 0.226 wavelength long. HiR^cro^ " L X Z I U J -M R<cC=<-tCToe. Fig. 1 — A basic quad with 150-ohm capacitive reactance added to the director. This is the distance shown on the outer edge of the Smith Chart between point A (150-ohms inductive) and point B (900-ohms capacitive). Note that if 300-ohm line were used, the line length would be different. Fig. 3 shows how a quad reflector would be con-structed using 900 ohms of capacitive reactance and a length of 450 line 0.226 wavelength long. For an additional explanation of how the Smith Chart works and how to use it, see The ARRL Antenna Book.\i By taking the above ideas one step further, and using a fixed length of line, either a director or a reflector can be created by changing the value of capacitance at the end of the line. For example, if 0.16 wavelength of 450-ohm line were used between the loop and the capacitor, the loop would act as a reflector, if the capacitor were 373 ohms, (point A, Fig. 4). If the capacitor were 1755 ohms (point B, Fig. 4), it would act as a director. This can be done by switching the two capacitors at the end of the line, or by adjusting the value of a variable capacitor. Study the Smith Chart for a few minutes. You will also note that with an appropriate length of open (or shorted) line you can create either a director or reflector from a one-wavelength loop. Fig. 2 — By placing a capacitor at one end of your feed line, an inductive reactance can be created at the other end of the line. Transmission Lines By performing a few hypothetical design prob-lems on the Smith Chart (transforming reactances along a transmission line), and using a higher impedance line, it will be much easier to create a reflector and director by switching the two capac-itors and using a fixed length of line. Because of this, and because a loop is balanced, I have always used open-wire line between the loop and the capac-itors. Open wire is available commercially, or can be constructed at home as shown in The ARRL An-tenna Book. The characteristic impedance of the open wire depends primarily on the wire size used and the spacing between the wires. Again, see The ARRL Antenna Book for information.\A y> 2 n. it 9 o o -n_ Co-f>a-c. t.Me. Fig. 3 — This shows how a quad r e f l e c t o r would be constructed using 900 ohms of capacitive reactance and a length of 450 l i n e 0.226 wavelength long. = ISO A. F i g . 4 — Using a f i x e d length of l i n e and by changing the value of capacitance at the end of the l i n e , a d i r e c t o r or r e f l e c t o r can be crcated. Antenna Dimensions Spacing of the quad elements is not critical. The ARRL Antenna Book states that spacings of 0.14 to 0.2 wavelength are common. I used about a 40 ft spacing for a 2-element 80-meter quad because I had a 40-ft boom on hand to take the elements from. Spacings between 25 ft and 65 ft should be ac-ceptable. Loop dimensions are c a l c u l a t e d using t h e standard formula for the driven element of a quad: i _ 1005 1 ~ f I constructed my open wire two different ways. The first method I tried used thin wall PVC tubing. This resulted in a rather heavy line which was dif-ficult to handle and the spacers did not stay on as well as I had hoped. I would not recommend this method of constructing open-wire line. The second method I tried was much more suc-c e s s f u l . 1 cut several plastic milk bottles into strips about 1 inch x 4 inches. Using an awl, I poked two holes in each end of each strip. The wire was fed through both holes, and the plastic was springy enough that the spacers maintained their position on the line (see Fig. 5). Amazingly, this open-wire line survived several Ohio blizzards. As a side benefit, I found that triangular pieces of the sa.ne plastic milk bottles made excellent strain-relief insulators for the corners of the quad ele-ments, and I constructed an excellent center insu-lator out of the same material. This line was easier to construct and is light; the plastic can be cut with scissors. It now seems as if I have a never-ending supply of spacer material. where 1 = length in f e e t f = frequency in megahertz Loop dimensions for various frequencies in the 80- and 160-meter bands are shown in Tables 1 and 2, respectively. Practical Design Example - 80 Meters A 3-element quad for 3.5 MHz could be con-structed as follows: Loop size from Table 1: 287 ft for each of the three elements. Line length: A line length of 0.14 wavelength of 450-ohm line was chosen to go between the parasitic loops and the capacitors. This allows capacitors of a reasonable value to be used at the ends of the line to make either a director or reflector. Using the formula below, 0.14 wavelength was calculated to be approximately 38 ft of line: , _ nk ( 9 8 4 ' ) 1 ~ f where 1 = length in feet, n = number of wavelengths, k = velocity factor of the line (about 0.96 for open-wire line) f = frequency in MHz. P b "Plo-sV.C "STV'.p a p p r o x I ' x V Fig. 5 — Plastic strips for use as spacers were cut from milk bottles. With the help of an awl, two holes were poked in each end of each strip to allow the wire to be fed through. The plastic is springy and light enough that the spacers maintain their position on the line. Capacitor value - R e f l e c t o r : The reactance required at the end of the 0.14-wavelength 450-ohm line, to make a reflector, is approximately 293 ohms. This was determined by rotating the Smith Chart 0.14 wavelength toward the load from the 150 ohm (inductive) point, and reading the reactance from the chart. The capacitance is calculated using the formula: C 2 ' f X c where C = capacitance in farads, Xc = capacitive reactance in ohms, f = frequency in hertz. The capacitor should, therefore, be about 155 pF. A variable capacitor of about 250 pF maximum will work nicely. Capacitor value - Director: To make a director, the capacitor required at tlie end of the 0.14 wave-length of 450-ohm line is approximately 1125 ohms. This works out to be about 40 pF. A 100 pF, or another 250 pF, variable would work well here. A diagram of this quad is shown in Fig. 6. The quad could be suspended from a large boom, or a line between two tall supports. The quad I built had the apex at about 90 ft and the bottom was about 20 ft above the ground. The corners were pulled out and supported by small nylon cord. Element shape is not critical. The direction the main lobe is headed is changed by switching the relays shown in Fig. 6. Practical Design Example - 160 Meters The 160-meter quad could be designed in the same manner as the 80-meter quad, however I only had to' > — > vc '-c, " muat- be Cj » G -' S j-r.l--.ble +o I^pfrc to . SO Met-cv C 3 o .
o -< L S xe^plc Fig. 6 — A three-element quad for 3.5 MHz is constructed using the capacitor values listed. CI and C4 form the reflector, while the director consists of C2 and C3. one support available which was high enough for a 160 quad. The 160 quad was therefore set up using only one switchable parasitic element. The peaks of both elements were hung from the same support and the bottoms were spread out as far as was physically p o s s i b l e . T h e top was at 170 ft and the bottom was about 40 ft above the ground. Dimensions for a quad designed for 1.8 MHz using 0.14 wavelength line are as follows. Loop size: 558 ft Line length: 73 ft Capacitor size - reflector: pF (450-pF variable) Capacitor size - director: 79 pF (150-pF variable) Tuning the Quad The elements should be in place before per-forming tuning adjustments. The parasitics will affect the resonant frequency of the driven element. First, adjust the resonant frequency of the driven element using an SWR or noise bridge, and trim the length of the driven element. Then adjust the parasitic elements by adjusting the capacitors, using a receiver for the best rejection of signals off the back of the antenna. This will correspond closely to the point of maximum forward gain. It is much easier to see a dip in signal strength off the back than to adjust for a peak In a signal off the front. The peak is very broad, while the dip off the back is sharper. Use a horizontally polarized signal when doing these adjustments, preferably a long distance away. If that cannot be done, use a hori-zontal dipole several wavelengths away to provide a signal. Do not use a vertical or a short-wire antenna as a source or you will receive strange results. Final Comments I hope this article has provided ideas and enough information to encourage other amateurs to build and design a few monster quads. The article presents very few specific dimensions because it would be difficult to reproduce the quads exactly as I built them. In any event, there are several variables that can be adjusted to fit a particular physical situation; spacing, height, line length, and capacitor value are just a few examples. Note that the lines can be extended by adding multiples of half wavelengths of line, and the capacitors could be brought into the operating position. The capacitors on my quads have all been mounted in plastic ice cream containers; they were receiving type capacitors, and the relays were small DPDT types. The driven element was fed directly with 52-ohm coaxial cable. I would be happy to correspond with anyone having questions about these quad antennas. References \1 Kelsey, William M., "A Three-Element Switchable Quad for 40 Meters," Ham Radio, p. 26, Oct. 1980. \2 Kaufmann, John E., and Kopec, Gary E., "A Convenient Stub-Tuning System f o r Quad Antennas," QST, p. 18, May 1975. U The ARRL Antenna Book, 14th edition, Chapter 3. \ 4 _ See Reference 3. Pfaff, William, "The K2GNC Giza Beam," Ham Radio, p. 52, May 1981. Table 1. Loop Lengths — 80 Meters FREQ (MHz) LOOP LENGTH (FT) 3.5 287 3.6 279 3.7 272 3.8 265 3.9 258 4.0 251 Tiable 2. Loop Lengths — 160 Meters FREQ (MHz) LOOP LENGTH (FT) 1.800 1.850 1.900 1.950 2.000 558 543 529 515 503 Gossamer Quad Update By R. F. Thompson, W30DJ/ZF2CD •Highway 5, Box 31, Waldorf, MD 20601 It takes less than you think to keep quad loops in the air. This feather weight 20-meter beam has been flying for four years. What is a Gossamer? The gossamer quad is electrically similar to most conventional two-element 20-meter quad an-tennas. The essential difference between gossamer and conventional construction is in the mechanical support and steering structures that keep the one-pound radiating and reflecting loops in the air, properly spaced, spread and oriented. All the parts of a conventional quad are pushed up from below by a massive, compressive structure that includes an expensive metal tower. The essential parts of the Gossamer quad, however, hang from an overhead ten-sile support system that is both lightweight and inexpensive. Fig. 1 is a perspective sketch of the gossamer quad, and Figs. 2 through 5 magnify some of the assembly details. The "sky line" suspension cable is simply a nylon rope stretched tautly over the crowns of two tall trees, and it supports the two quad loops and their lightweight framework. The mechan-ical parts of the antenna, such as the rotator, spindle and steering pole, sit atop a ground based vertical cedar pole. There is nothing new in di-viding the load between antenna parts that will be high above the ground, and those parts that are lower in height. It has been tried and proved at Arecibo in Puerto Rico where the largest radio telescope in the world uses suspension cables to support its antenna high above a large reflector resting on the ground. The framework for the two gossamer quad loops consists only of four spacer poles and one spreader pole, and weighs about eleven pounds. The supporting ropes weigh less than ten pounds, so only about 20 pounds of material are needed to keep the one-pound loops near tree top height and properly positioned. Conventional quads use several hundred pounds of tower, mast, boom, spiders and spreaders to do the same thing. Towers do have one advantage, however. They can raise antennas higher than natural sup-ports. The gossamer quad takes advantage of the over-head support that the sky line provides , by uti-lizing the tension in the loop wires to supply most of the support for the loops and their framework. Thus, the loop wires are not only the most important electrical component of the antenna, they are very important mechanical components as well. In short, the essential difference between conventional and gossamer construction is that conventional quads use massive, compressive support, while the gossamer uses lightweight, mostly tensile support. The First Four Years The original gossamer quad was described in QST.\1 With some minor modifications on my part, it has been actively used for over four years at my ROTATOR -SO OHM COAX>- ' ill I CEDAR POLE - J ? / Fig. 1 — A perspective sketch of the support system looking up from the ground. The sky line, pulley and mast rope support the spacers, spreader, loops and the 75-ohm matching section. The corners of the loop wires are shown as dashed lines at the ends of the four spacer poles. The cedar pole supports the rotator motor, steering pole, spindle, and the 50-ohm feed line. The steering lines are tied to the ends of the spreader. location. Only one important weak spot has been found in the original design, and this has been strengthened considerably by a simple fix described below. In addition, the original lower rigging has been replaced by a self-adjusting rigging that uses a spindle pole to keep the bottom of the antenna centered over the rotator motor. Neither of these improvements add any significant weight to the load on the sky line. The original lop rigging on the gossamer quad was proved to be highly satisfactory. The sky line, pulley and mast rope easily survive the weather's worst tricks. As mentioned, the sky line is simply a nylon suspension rope, and the pulley "sky hook" is fixed on the sky line directly above the rotator motor. The quad loops and their framework hang from the mast rope which passes up through the pulley and The rotatable gossamer quad is a two-element 20-meter beam which has a minimum of heavy, self-supporting components. The loops are mechanical parts of a lightweight framework that hangs from an overhead line. Fig. 2 — Spreader-sleeve assembly. The two halves of the spreader pole are joined by slipping the large ends inside the spreader sleeve. Hose clamps prevent either half from pushing the other half out of the sleeve. The assembly hangs by the harness rope from a metal rope clip on the mast rope. The mast rope passes freely behind the sleeve. Fig. 3 — Spacer-spreader mounting. The spacer pole is fastened to the mounting board by two U-shaped aluminum straps which are bolted to the board. The mounting tube passes through a snug hole in the board. The small end of the spreader pole is inserted into the tube and pushes against the end cap. The hose clamp transfers the push to the board and spacer, and this spreads the loops into a diamond shape. then goes over the crown of a nearby tree and down to a cleat where it is secured after it is used to raise the quad loops to operating height, or to lower them for repairs. During the summer of 1982 the mast rope was sabotaged by squirrels! They apparently enjoyed chewing through the nylon strands. Luckily, the damage was spotted before the rope broke, and the badly frayed portion was cut out. The spliced rope was then rerouted away from the squirrel's playground. The framework for the loops, consisting of the spreader and spacer poles, has also been satis-factory. The 27-ft spreader pole is the only com-pressive part of the framework, and it weighs only 2.6 pounds. The two halves of the spreader pole were originally aligned and held together by thin alu-minum L-angle stock and hose clamps. Unfortunately, this design subjected the thin L-angle to forces Fig. 4 — Feed-point assembly. The ends of the driven loop are attached to short standoff insulators mounted on a transparent Plexiglas plate. Short wires connect the loop to an SO-239 coaxial cable jack. A square hole in the Plexiglas provides a snug f i t for the bottom spacer, and the plate is held in place between two hose clamps attached on the spacer. Fig. 5 — Lower support structure. The mast rope passes through the spindle and exits the bottom of the mast pipe. The bottom spacer is tethered to the spindle so that it is always centered above the rotator, and i t is free to move vertically. The spindle slips down over the mast pipe and rests on a hose clamp tightened around the mast pipe. U-bolts through the mounting plate attach the steering pole to the mast. The steering lines pass through metal rings hose clamped to the ends of the steering pole< and are kept taut by 8-ounce weights. which easily distorted the aluminum, and in April 1982 the L-angle ruptured during a week-long attack of severe winds. The torsional weakness of the L-angle has been the only significant design flaw discovered during more than four years of continuous service. The L-angle has been replaced by a round aluminum tube, the spreader sleeve (Fig. 2). The tube is a half-pound heavier, but much stronger structurally. The original lower rigging used four guy lines to limit motion of the lower spacer and keep the spacer in position above the rotator motor. The guy ropes, however, needed to be adjusted during times when the rain or ice changed the height of the sky line. When the guys were too loose, the lower spacer was poorly constrained; when they were too tight, they would pull down on the sky line and lower the height of the antenna. The original rigging has been replaced by a new design that is self adjusting, and it does not add any load on the sky line. The lower rigging keeps the bottom part of the antenna exactly centered on a spindle pole which slips over the mast pipe mounted on the rotator motor. A loose loop of nylon rope tethers the center of the spacer to the spindle and allows the spacer to ride up and down the spindle whenever changes in the weather change the height of the sky line. After more than a year in service, it seems that little could be done to improve on this design. Parts and Materials Table 1 catalogs the major components of the gossamer quad in two categories: parts supported by the sky line, and parts sitting atop the cedar pole. Most of the material was purchased from local elec-tronics, lumber and hardware stores, but the two halves of the spreader pole were obtained through mail order from Skylane Products.\2 These tough, but light weight poles are called Dynaflex fiberglass spreaders. Each consists of three telescoping sec-tions that extend to a maximum length of 15 f t . These two poles are joined in the spreader sleeve and the lengths of the sections are adjusted and locked by small hose clamps to give the spreader pole a total length of 27 f t . The original sky line is still in service and shows no sign of wear. It is made of 0.25-inch-diameter Goldline nylon rope purchased from REI.\2 REI has discontinued selling Goldline in 1982, but they now o f f e r a superior rope called Skyline II. A 1984 catalog lists 1/4-inch-diameter Skyline II at 15 cents per foot, obtainable in any length up to 600 feet. Compared to Goldline, the new Skyline II is 100% more abrasion resistant when dry, 50% more abrasion resistant when wet, and 20% more resistant to the degrading effects of ultraviolet rays. Any rigid pole having a length of 13 feet or more can be used for the steering pole. The gossamer quad uses a fiberglass pole that has a honeycomb cellular cross section. This pole is a four-pound spreader designed for use in conventional quads.(A conventional quad would use eight of these, 32 pounds of spreader in all. The gossamer uses less than three pounds.) The cedar pole was once a tree in my back yard, and it is the only eye-level object in the support system, blending naturally with its surrounding. It elevates the rotator about 13 ft above ground, and supports the weight of the rotator, steering pole, and spindle pole. Since the spindle pole is a fair guide for the lower mast rope, the slender cedar pole also absorbs some of the shocks of wind loading on the gossamer quad. The original cedar pole is still being used, and shows no sign of weakening. Untreated cedar fence posts have a life expectancy of between 15 and 20 years.Vi Conclusion This article presents observations made during the gossamer quad's first four years of operation, and it has given details of changes made since the first gossamer quad article was published in QST. The QST article contains additional information not repeated here, which describes how to attach the loops to the spreaders, how to launch the sky line using a sling shot, formulas for the lengths of the quad loops, and use of a quarter-wave transformer to feed the driven loop. O t h e r a r t i c l e s c o n t a i n i n g useful ideas for gossamer construction have been written. VE3CUI de-scribed his gossamer Yagi in CQ and KC7M described some useful knots in a recent issue of 73.\5,6 Finally, the caveat must be repeated that the gossamer quad is not a rigid structure, and it may be a poor design for use in locations that are subject to frequent strong winds. But the design has proved itself to be rugged, durable and well suited to the usually calm weather in Maryland. In my backyard, a many-hundred-pound metal tower is not needed to keep one-pound loops near t r e e - t o p height. Acknowledgement: I would like tc thank Les Salter, K A 3 D C W , who assisted me with various computer tasks such as transferring files from my Apple II floppy disk to an IBM PC diskette. References \1 Thompson, R. F., W30DJ, "Build A Gossamer Quad," QST, December 1981. \2 Skylane Products, 359 Glenwood Ave., Satellite Beach, Florida, 32937 (305-773-1342). \3 REI, P. O. Box C-88125, Seattle, WA 98188. Back To Basics, Reader's Digest Association, 1981. \5. Swynar, E. P., VE3CUI,"A Tree-Mounted Vertical Yagi Array," CQ, August 1982. \£ Jacobs, G., "All Tied Up In Knots?" 73, Oct. 1982. TABLE 1. GOSSAMER QUAD COMPONENTS COMPONENTS SUPPORTED BY THE SKY LINE WEI6HT COMPONENT MATERIAL DESCRIPTION SIZE pounds TOP SPACER WOOD ROUND CLOSET POLE 8 ft 7 in LONG 1.25 in DIA 3.0 LEFT SPACER WOOD SQUARE THIN POLE 8 ft 7 in LONG 0.75 in SQUARE 1.0 RIGHT SPACER WOOD SQUARE THIN POLE 8 ft 7 in LONG 0.75 in SQUARE 1.0 BOTTOM SPACER WOOD SQUARE THIN POLE 8 ft 7 in LONG 0.75 in SQUARE 1.0 SPREADER POLE FIBERGLAS ROUND POLE TELESCOPING 6 SECTIONS 27 ft LONG Max 1.06 in DIA Min 0.50 in DIA 2.6 SLEEVE ALUMINUM ROUND TUBING 6ft LONG 1.25 in OD 1.12 in ID 1.3 SPACER/SPREADER MOUNTING BOARDS WOOD RECTANGULAR (2 needed) 6 i n v . 3 i n 0.75 in THICK 0.25 SPACER/SPREADER MOUNTING BOARDS WOOD RECTANGULAR (2 needed) 6 i n v . 3 i n 0.75 in THICK 0.25 MOUNTING TUBES COPPER ROUND PIPE (2 needed) 10 in LONG 0.63 in OD 0.25 MOUNTING TUBES COPPER ROUND PIPE (2 needed) 10 in LONG 0.63 in OD 0.25 FEED POINT PLEXIGLAS (clear) RECTANGULAR PLATE 6 in x 2.25 i n 0.25 in THICK 0. 13 DRIVEN LOOP COPPER WIRE SQUARE LOOPS 14AWG 7—strand 17.9 ft/SIDE 1.0 REFLECTOR COPPER WIRE SQUARE LOOPS 14AWG 7—strand 18.4 ft/SIDE 1.0 QUARTER WAVE TRANSFORMER COAXIAL CABLE RG-ll/U 75 ohm Approx. 11.6 ft (Tuned to 14 MHz) 1.0 COMPONENTS SUPPORTED BY THE ROTATOR COMPONENT MATERIAL DESCRIPTION SIZE SPINDLE PVC ROUND PIPE 10 ft LONG 1.88 in OD, 1.56 in ID STEERING POLE FIBERGLAS ROUND POLE VERY RIGID 13 ft LONG 1.12 in DIA MOUNTING PLATE ALUMINUM RECTANGULAR PLATE 7 in x 4.5 in 0.125 in THICK MAST ALUMINUM ROUND PIPE 6 ft LONG 1.5 in OD, 1.25 in ID (in=inch=2.54 cm, f t=f oot=0. 3048 m) 15' - 4 5 0 - _ n _ o p e n w i r e ' A X M a t c h e r P .
G , 8 Treed Line to-to-gamma, matcher Fig. 1 — An overall view of the array. •driven ELEMENT BoTTon VERTICAL T43l£- 1 CAficiroR VALUCS C I C e i K P t i j r r t T o R 2s-o CP i/ABiasle FWAL 14KUS ISO F { X E D (i oippeo »ic ft i) C 2 2o mrr /O O Pfr l / A R PHO/JE SCcTltM StTTMZ BO CP CI J? »rg Table I Tri-Gamma Hatch Dimensions Band Length " Space " 20 35 2 15 27 1.5 10 18 1 Capacitor pF 100 50 75 Set pF 80 30 60 S~o PF i / f t R PHOA/E ICTT/rJc 3orr CI- / o <hra 7 S PF MR PKowt lerridc 6opr Compensator capacitor 250 pF variable; set to 150 pF. Fig. 2 — Details of the Tri-Gairma Match. a larger 2-inch ID PVC tube. The sections of the boom between the center of the driven element and the reflector or director are hollow 2-inch ID tubes. Aluminum mesh around the spreaders acts as a cushion for the boom laminate. The mesh allows for some floppy movement of the horizontal and vertical spreaders. Note that the wooden core of the hub sections must be varnished completely before encasing it with the PVC tubes. Stainless steel screws hold the spreaders in place at the hubs. A cork washer, fabricated from linoleum, cushions the boom structure where the element hubs meet the hollow boom tubes. The hub section that telescopes inside the boom tube may be 3 or 4 feet long. This provides a "handle" for carrying the fully assembled element around the yard during tuning tests. Fig. 7 shows a view of a typical hub. Vertical spreaders of telescoping aluminum are 1 inch OD at the boom, then 7/8 inch OD, with 3/4 inch OD at the ends. Lightning protection is pro-vided by a ground strap system from each vertical spreader, across the boom and down the tower to ground. Unique PVC tubing wire holders are attached to the spreaders by galvanized no. 16 gauge straps. The wire loops are firmly anchored by a 3-step lock-in loop through the PVC holder (Fig. 8). Dress the wire through the wire holders such that the wire is clear of the aluminum vertical spacers. Six years of Chicago weather didn't wear through the house wire insulation. Note that the wire holders on the bottom vertical spreader of each element must be movable to provide slack in the wire loops for tuning. In Phase III, short nylon string holders are used along the horizontal spreaders for less wear and tear on the helix tape windings. Each is pro-tected against abrasion by a layer of 3M "strapping" tape on top of the copper or aluminum tape winding. Once the array is in the air, it is also at the mercy of curious birds and squirrels. The varnish coat and strapping tape help preserve the helix tapes. There is a mounting problem for the 32-ft long 40-m, 1/4-wave matching section. The 450-ohm open wire runs from the helix feed point, up to the top of the vertical spreader and back down to the tri-gamma match coaxial cable feed point. TV type screw-eye insulators along the vertical spreader hold the ladder line in place. Boom support in Phase II was by suspension from a T bar at the mast. The Phase in boom uses a truss system support. Assembly, Tuning and Testing Each of the three elements measure 23 feet across the spreaders and 18 feet for each side of a quad loop. A critical series of holes drilled in the tETAIL A .
' . Fig. 3 — Construction details of a typical element. ALUfn. VERTICAL SPREAPEHS 8' / "o-P. ALU 1. 3ft Rop 6' 1/r'O.V. 4' y -f ' f o . H o r i z o n t a l S-PREAPERS „ I r4-TU«NS/FooT HELIX 3/s" ~«Vcx\ : ——-4= = - = ^ = : -1,005 1 Wl"o. Dou£. (. I o.o.ilLur<\ \ A t" TTfW 14 / X.D. J7 r u ^ t flfl v w At + 0 Fig, 6 — Boom section details. Boom SccrtoAfS PV+ OVflX> Fig. 7 — Typical hub. Note the loading coil tuned reflector helix. Fig. 8 — The wire loops are firmly anchored by a three-step lock-in loop through the FVC holder. Fig. 9 — Each element is tuned in a horizontal attitude and mounted on temporary poles 8 feet above the ground. the feed point at the 15-m driven element loop bot-tom. The sequence of tuning steps is as follows: 1. Set the variable 250-pF compensator capacitor to 200 pF. 2. Adjust the 10-m tap point distance from the bottom apex of the loop, and the variable capacitor (both for minimum SWR). 3. Do the same for the 15-m tap point and tuning capacitor. 4. Then do the same for the 20-m tap point and tuning capacitor. 5. Finally, the compensator capacitor is used to recheck each band for minimum S W R . 6. Then start over again with the 10-m adjust-ments. Each of the variable capacitors is mounted in a plastic freezer jer and weatherproofed with duct tape and varnish as shown in Fig. 10. Adjustments of the 40-m helix involved peeling off turns and resoldering turns at the outer ends. A tuning coil at the center of the reflector and director helix windings provided additional adjust-ment. Most of the 40-m tuning was accomplished with the individual elements mounted at 4 to 8 feet above ground in a horizontal attitude. Performance Evaluation The PV4-Quad was used for five years atop a 34-ft Heights aluminum tower at my Illinois loca-t i o n ( F i g . 11). P e r f o r m a n c e data consists of hundreds of worldwide contacts on all four bands. Numerous QSOs involved tests using an HP 7035 XY plotter. The plotter X scale was driven from the o f f s e t v o l t a g e of the beam rotator controller circuit. The receiver "S" meter circuit fed the Y scale. See Fig. 12. Fig. 13 shows the X-Y plotter interface. The XY data was transposed to coordinates of the f a m i l i a r cardioid pattern. Typical SWR measurements for a 20-, 15-, 10-, and 40-meter band are shown in Fig. 14. Compare this data to that shown in Fig. 15 for the PV4-Quad. Fig. 16 displays a radiation pattern for the PV4-Quad. Fig. 10 — Freezer jars weatherproDf the ganma capacitors. Fig. 11 — PV4-Quad on the author's former 34-foot tower. Fig. 12 — X-Y plotter test setup. Note the S-meter magnifier. X-V P L O T T E R i N T f R T A C E Fig. 13 — The X-Y plotter interface. /If). 4 0 RoToR PoTEWT loMETEfc I ISVAt.r H. tf.U I I plott eR "X" flfAPtl F | £ R ^ — W -O V A.C. - W O R T H —•-Oo CoNTftOLL ER TERlMfl M_S RECEIVER S-fAETElZ. T JL . i ^ i . F L o A T PloTTE R 2 0 I J S M E T E R S £ 1 i . = — 1 5 1JS M E I E R S i ' V -X t •i :t4 Fig. 14 — Typical S W R measurements for the 20-, 15-, 10-, and 40-meter bands. 10 U i METERS — ( — 4o METERS H E L I X 212 . 2IJ 2 8 .
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-. Fig. 15 — SWR measurements for the PV4-Quad. 38 Summary In the move from Illinois to New Jersey, the helix tapes were damaged. New tape windings are now being evaluated. The present mounting on a test tower beside a former chicken coop serves tuning and pruning very well. A nearby inverted V antenna is used for comparison measurements. Note the overhanging oak trees in the Dlinois photos. Once again the quad design low angle of radiation has proved to work well at "low alti-tude." During the summer of 1984, the PV4-Quad was again reassembled atop a new tower. For many of the parts, it was the sixth time of disassembly and reassembly. The PVC sections have proved to be much more serviceable than I ever planned at the start of the project. References \1 Meyers, R., W1FBY, "The HW-40 Micro Beam," QST, Feb. 1974. \Z Orr, Bill, W6SAI, All About Cubical Quad An-tennas. U Simon, B., W2UNN, "Surface Area of Quad An-tennas," CQ, July 1967. Y4 Lindsay, J. E., W0HTH, "Quads and Yagis," QST, May 1969. Bergren, L., W0AIW, "The Multi-Element Quad," QST, May 1963. \£ Schultz, John, K3EZ, "A New Look at Helical-ly-Loaded Antennas," CQ, April 1976. \1 Gooch, et. al., "The Hairpin Match," QST, April 1962. Y2 Grammer, George, W1DF, "Simplified Design of Impedance Matching Networks," QST, March 1957. Perrott, J. H., W4FRU, "Quad vs Triband Yagi," QST, February 1971. Table III Tape Helix Test Windings 0.3-inch wide copper tape on 1.25-inch CD PVC Test # Tape Length Turns Helix Length Frequency Tape Length (Feet) (MHz) 1 50' 1 TPinch 12.5 18.5 2 50' 0.8 TPinch 13 17.6 3 50' 16 TPfoot 9 19.5 4 50' 6 TPfoot 14.5 20.5 5 50' 6 TPfoot 20 15.5 6 75' 16 TPfoot 14 14.0 l/2 40-meter helix El 35 6 TPfoot 24 14 E2 35 15 TPfoot 6 28 2 inch ID 1 1/2 inch ID 1-3/8 inch List of Materials Aluminum tubing: 6 and 8 ft lengths of 1-7/8 inch, and 3/4 inch OD. PVC pipe (must be "Schedule 40" grade): 6 20 foot lengths 1 inch ID 3 12 foot lengths 4 12 foot lengths Wood clothes closet pole dia., 12 foot length 500 feet no. 14 stranded copper house wiring (7 strands, no. 20 gauge) 33 feet of 450-ohm open^wire ladder line 6 feet of no. 8 copper rod (hairpin) RG8 coax, aluminum ground wire, steel guy wire Miscellaneous bolts, clamps, and so on. A Heath AR-40 rotor Poly clothesline and pulleys for sky-hooks Note: Copper foil tape with CONDUCTIVE pressure sensitive adhesive. Two suppliers are: Prehler Electrical, Inc. 2300 N. Kilbourn Chicago, ILL 60639 tel. (312) 384-6100 (3M Type 1181 Electrical Tape) Chomerics, Inc. 77 Dragon Court Woburn, MA 01801 tel. (617) 935-4850 Weight of Materials 500 feet of copper wire 20-foot boom, PVC and wood 3 26-foot aluminum verticals 3 PVC horizontals Boom center vertical yoke Coax and misc. clamps, boom guy Rotor Steel tubing, center mast and T bar 5 lbs 25 lbs 19 lbs 26 lbs 15 lbs . 10 lbs 100 lbs 15 lbs 10 lbs 125 lbs Fig. 16 — Radiation pattern for the FV4-Quad. Helix Antenna Data 6 test windings on 20 foot 1.25-inch OD PVC pipe (Schedule 40). Copper tape was 0.3-inch wide (3M stickytape, 1-inch wide, cut into three pieces). Stock reel, 1" x 18 yards. Two dip meters were used: General Radio (vacuum tube) and a Heath solid state. 1) 50 feet of tape, 1 turn per inch, 48" tape per foot, 12.5 foot long coil 2) 50 feet of tape, widened spacing 1/2 coil length, 6' coil, 1 turn per inch and 7' coil 0.8 turn per inch, both a 13' coil 3) 50 feet of tape, close spaced, 9' coil length, 16 turns per inch 4) 50 feet of tape (29' excess coiled), 14.5' coil length, 6 turns per inch, wide spaced 5) 50 feet of tape, 20' coil length, 6 turns per inch 6) 75 feet of tape, 14' coil length, 16-17 turns per foot Estimates: 35' tape (0.3" wide), 6 turns per inch, 24' long — 14 MHz 35' tape (0.3" wide), 15 turns per foot, 6 long — 28 MHz G. R. 18.6 MHz 17.6 MHz 19.5 MHz 20.5 MHz 15.5 MHz 14.0 MHz Heath 18.5 MHz 17.7 MHz 20.0 MHz 21.0 MHz 15.5 MHz 14.0 MHz Cubical-Quad Antenna Design By Norris G. Boucher, W3GNR 913 Chestnut Avenue, DuBois, PA 15801 D i a m o n d s , d e l t a s , Swiss and squares: These various types of quad antennas have provoked much speculation until direct gain comparisons between cubical quad and Yagi antennas settled crucial arguments in favor of the Yagi antenna.\l Many prospective antenna builders were so disillusioned by the report that the cubical quad is seldom consid-ered for directive antenna applications. This is unfortunate because the cubical quad can be an inexpensive alternative to the mono-band Yagi or specialty antennas, especially at frequencies that allow construction from locally available mate-rials. Unlike for the Yagi antenna, the selection of quad antenna designs is limited and often unre-liable. Attempting to define satisfactory cubical quad antenna models for various spacings and numbers of elements, I developed the information presented in this article by experiments, which I will de-scribe. Project Goals My first attempts to design and build quads of various types generated baffling problems. Frequency scaled cubical-quad antennas usually worked poorly, as did successive duplicate models at other sites. Taking reliable antenna gain measurements on loca-tion also proved to be discouragingly difficult. The practical goal for this series of e x p e r -iments was to produce a model set of low SVVR, fre-quency scaleable quad antennas with three to eight elements and maximized gain. For simplification, only models with one reflector and equal-length directive parasitic elements were planned (several different models were tested). Other curious effects required explanations, and I hoped to discover their cause as my work progressed. Element Materials I h a v e constructed dozens of antennas with little regard for the type of insulation, if any, on the wire. So it was interesting to learn that the resonant length of insulated quad elements is sig-nificantly d i f f e r e n t from their uninsulated coun-terparts. All information found in this a r t i c l e applies to uninsulated element quad antenna de-signs. I have not measured the extent of this effect relative to resonant frequency, gain or quality of insulation (surface corrosion or thickness). Some trials i n d i c a t e d that i n s u l a t e d - w i r e e l e m e n t s lowered the operating frequency of the quad antenna perhaps by two percent. So note carefully that in-sulated-quad-element designs may not exhibit other effects reported here. Element Diameter Scaling Experiment The cubical-quad antenna has a relatively broad operating bandwidth. This is one of its better qualities. Unfortunately, that quality makes it dif-ficult to determine optimum quad-element design k factors ( f t - MHz). While trying to tablulate k factors for different elements from published de-signs, it became obvious that many experimenters, like myself, were floundering for lack of consistent design information. Applying frequency scaling techniques added to the confusion. Researchers have found it necessary to consider the thickness of an element (expressed as an element diameter-to-wavelength ratio, d/1, in consistent units) to successfully frequency scale model antenna designs.\2 Furthermore, the infor-mation I found on the quad-element d/1 ratio effect was virtually a sidebar in some quad antenna lit— erature.\l An experiment to define the extent of the d/1 effect on the circumference of cubical-quad elements was performed in this manner. Two open circular quad elements of equal length were placed in parallel, physically and clcetrieaUy, and connected to a 50-ohm coaxial line. One loop was axiaily rotated 90 degrees from the other to roughly define a spherical cage. That configuration presented the proper load impedance to the feed line at resonance, decoupled the elements to minimize mutual impedance effects and increased the Q factor of the antenna. This concept is sometimes found in other ref-erences as a wavelength-to-diameter ratio, 1/d. This is the inverse of d/1. Increased Q was imperative since a single driven quad element exhibited a very shallow SWR curve at 52 +/- 2 MHz, the frequencies of mea-surement. When fed at the bottom or top of the loops, this quad loop cage makes an excellent, broadband, omnidirectional, horizontally polarized antenna. Crossed-loop quad antennas were built and resonated, using no. 6 to 14 AWG bare copper wire. For each gauge wire, SWR measurements were made at numerous frequencies, and the values plotted to generate a characteristic SWR curve. From those curves, the frequency of the lowest SWR was selected, the loops measured and a k factor was calculated for each gauge element. As a group, the resonant k factors (Kr) in-crease with an increase in element diameters, al-though the specific values are not sufficiently accurate for design purposes. The measurement error of my equipment exceeded the percentage differences of the k factors. It was confirmed that the k factor approached Kr = 1005. a value given for resonant quad elements in many references, as the d/1 ratio decreases (thinner elements).\4 At least two adequately accurate resonant element k factor values resulted from further work in this series of experiments. It was found that eighth-inch bare aluminum wire has a Kr = 1020 at 144 MHz; 0.064 inch (no. 14 AWG) bare copper wire has a Kr = 1005 at 50.125 MHz. This, and information from another source, make it possible to construct a graph (Fig. l).\ji This graph may be used as a guide, until more data is available. 1 0 3 0 ^ _ + 1 0 2 5 ' n ! ffi ! a """ T i ; •M ! ( D i w 1 0 1 5 + 1010 1 0 0 5 ! 1 0 0 3 +• 0 Pig. 1 — Resonant element k factor vs. element diameter-to-wavelength ratio. This curve was generated from limited data to serve as a guide, until more accurate values are available. Note that the circumference of quad elements increases with increasing element diameter. Element Diameter Scaling To use Fig. 1, calculate the d/1 ratio of the element diameter selected. For example, if you choose no. 14 AWG bare copper wire for use at 50.125 MHz, as I did: d = diameter (inches) x f (MHz) (Eq. 1) 1 11803 (in-MHz) = 0.064 x 50.125 11803 d. = 0.00027 1 Then, referring to the graph, locate the k factor for that d/1 ratio. It should be Kr = 1005. Using two k factors, a model correction factor (Ck) can be calculated. For example, Kr = 1020 (model) at 144 MHz and Kr =1005 (new) at 50.125 MHz, then calculate: C = K .
( n e w ) . (Eq. 2) k(model) = 1005 1020 Ck = 0.985 To complete the sealing process, multiply the individual model element k factors by Ck, then divide by the selected operating frequency. The result is the length of the selected quad element. Quad Antenna Model Experiment Two computer-designed, five dBd, two-element Yagi reference antennas were deployed about 100 ft apart.\£ One was excited by a 144-MHz crystal con-trolled exciter. The other, used as a reference, was placed two wavelengths to one side of the antenna test area for the quad models. Each measurement was taken relative to the fixed reference antenna to compensate for system and signal path changes. Test signals are routed through feed lines of equal length from the reference and test antennas, a two-port coaxial line switch, a precision attenuator or an SWR/power meter. This depends on the measure-ment being taken. Signals are finally fed to the 144-MHz receiver where a variable scale, slide-back voltmeter is attached to the receiver for mea-surements. I used an antenna tuner in the model antenna feed line during gain measurements. The quad antenna test jig consisted of two vertically separated and parallel wooden strips. They were drilled at many places along their length to allow various trial element spacings. U-shaped portions of the quad elements, made from eighth-inch solid alu minim wire, were slipped into those loca-tions. The other side of the elements, shorter U-shaped portions, were attached by slip couplings to close the loops. The driven element was fixed at a circumference determined from previous tests. Accuracy of Measurements Although all nearby structures were wooden, the area set aside for modeling the quad antennas was selected more for convenience than theoretic suit-ability. Therefore, in my opinion, the gain of these models cannot be compared to other antennas within 3.0 dB. The meter I used to make these relative gain measurements had enough resolution to guarantee measurements within the accuracy of my standard at-tenuator (estimated at +/- 0.25 dB). Relative gain measurements were routinely repeatable to within +/-0.2 dB, averaging around +/- 0.1 dB. Selecting a Design Model A f t e r researching available sources, I decided to model three different element spacings of 0.130, 0.165 and 0.195 wavelength.\l These values are grouped around spacings that are reported to produce maximum gain and low SWR. The results are tabulated in Tables no. 1 through 3, respectively (the element k factors must be corrected for other d/1 ratios). Being a controversial subject, quad antenna gain was the first quantity plotted from the ex-perimental data. Looking at Fig. 2, you will see how the relative gain of my quad models varied with boom length for each of three spacings. Except for one point on the 0.195-wavelength element-spacing curve (attributed to a noisy attenuator switch), a smooth curve was easily drawn through the data points. There is a design consideration for these spac-ings that tends to make the selection of the number of elements less arbitrary. An oscillation in the parasitic element k factors is apparent in the tabulated data for even and odd numbers of direc-tors. To analyze this phenomenon in more detail, graphs of model director (Fig. 3) and model re-flector (Fig. 4) k factors were constructed. 40 8 0 1 2 0 1 6 0 2 0 0 240 2 8 0 3 2 0 D/L X 1 0 " 3 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 B X M L Q C 1 H 1 K A V E U M 3 T K S ) Pig. 2 — Model quad antenna relative gain vs. boom length. The slope of these collateral curves indicates the marginal gain of constant length director designs is approximately nine-tenths that of an optimized Yagi antenna for boom lengths in the region between 0.7 and 1.2 wavelengths. Optimized quad antennas should be expected to yield about the same marginal gain in that region. = C U ^ D A N T E O T n D E S I G N = • locond: . - 0.130 spaci.ig • - 0.165 spacirc • 5 - 0.195 spacing N X to a resonant element k factor. In this case, erring by making the element shorter would quickly reduce gain at the design frequency, but not as quickly as a lengthening error in the director elements. This effect makes even numbered element quad antenna construction relatively critical. If you are willing to accept the increased accuracy require-ments of this design, there is yet another con-sideration. When changing operating frequencies, the resonant frequency of the parasitic elements is more quickly approached. Fig. 5 graphically displays the different ele-ment tuning for both even and odd numbered director element cases. Obviously, the gain and SWR curves of the even numbered director element quad will be sharper than the odd number designs. Given this data, the odd numbered director element quad antenna designs are a better choice. < i > < p K 1041 ! 1 0 3 5 ! 1029 -1023 t K(rj-> 1 0 1 7 — = Q U A D W I E N N A D E S I Q J = • legend: . - 0.130 spacing • - 0.165 spacing • # - 0.195 spacing 2 3 4 5 6 N T J M B 5 1 O F C U A D D I R E C T O R E L E M E O T S Fig. 3 — Director k factor vs. number of director elements. This graph clearly shows an oscillation in equal length director tuning for the first five directors. Notice the even numbered element K-factor groupings near the model resonant element k factor, Kr. Note: The 0.165-wavelength spaced model produced two gain peaks at the sixth director. The lower k factor produced more forward gain (+0.6 dB). Clearly, these even and odd numbered element quad antenna models are quite different. Referring back to Fig. 3, notice that the four- and six-ele-ment quads have director k factors in the area of Kd = 1008, while three-, five- and seven-element de-signs have much lower k factors, grouped in the Kd = 984 region. Knowing the correct design k fac-tors, you may reason that either odd or even num-bered director quads could be built with equal ease. Theoretically that is true, but the resonant element k factor in thase examples is Kr = 1020. An error in director length measurement of several tenths of one percent longer will noticeably decrease antenna gain at the design frequency. This problem with an even number director ele-ment design is compounded by an accompanying oscil-lation in the quad reflector k factors. Looking at Fig. 4, notice that these designs are grouped closer 1 2 3 4 5 6 N U X 8 E R O F Q U f f i D I R E C T O R E L E M E N T S Fig. 4 — Reflector k factor vs. number of director elements. Although not as well defined, this graph shows a characteristic model k factor o s c i l l a t i o n , as seen in the corresponding model quad director k factors (Fig. 3 ) . Notice the even numbered element design k factors are grouped closer to a resonant element k factor, Kr. Design Example I used the information in this article to design and build five- and seven-element, 0.165-wavelength spaced, 50.125-MHz quad antennas. My choice of element material was no. 14 AWG solid copper wire stripped of insulation. Use Eq. 1 to compute the d/1 ratio for this wire, d/1 = 0.00027, as before. Then go to Fig. 1. Finding that d/1 ratio on the horizontal axis, tracing that point upward to the curve, then left to the vertical axis locates Kr = 1005 for this size wire. Next, use Eq. 2 to com-pute the correction factor, Ck = 0.985, also as before. Consulting Table 2 for five-element model k factor values, the reflector k factor (Kr) is 1083, and the k factor of the directors (Kd) is 984. Now, Element length = C(k) x k(model) (ft) f(MHz) So, Directors = 0.985 x 984 (ft), or 50.125 (Eq. 3) ( | 3 o a k relative gain} 0 1 2 2 3 3 4 5 5 6 6 7 8 8 9 9 0 1 1 2 2 3 4 4 5 5 6 7 7 8 8 9 0 0 1 2 9 5 1 7 3 9 5 1 7 3 9 5 1 7 3 9 5 1 7 3 9 5 1 7 3 9 5 1 7 3 9 5 1 7 3 0 D I R E C T O R (ly aild R E F L E C T O R I K ^ ) M O D E L K -F A C T O R S Fig. 5 — Relative gain vs. quad element length. These curves demonstrate the difference in lengths (for 4/1 = 0.00027) of quad directors and reflectors, which produce maximum gain for even and odd numbered director model quad antenna designs of 0.165-wavelength element spacing. Directors = 19.34 ft (19' 4"). And, Reflector = 0.985 x 1083 (ft), or 50.125 Reflector = 21.28 ft (21' 3-3/8"). Finally, Driven element = 0.985 x 1020 (ft), or 50.125 Driven element = 20.04 ft (20' 1/2"). Now that the element lengths have been de-termined, the element spacing, boom length and spreader length are calculated. Thus, Spaces = Spacing(waves) x 983.58 (ft), f (MHz) = 0.165 x 983.58 (ft), or 50.125 Spaces = 3.24 ft (38-7/8"). Next, Boom = (no. of elements - 1) x spaces (ft) = (5-1) x 3.24 (ft), or Boom = 12.96 ft (121 11-1/2"). And finally, Spreader length = Element length (ft), 2.828 For example, the driven-element spreader length = 2SLM ( f t ) , or 2.828 Spreader length = 7.09 ft (7 ft 1-1/32 in) between wire holes for the driven element. Calculate the reflector and director spreader lengths in the same manner. An interesting feature of this model is that it can be modified to make a five- or seven-element convertible quad antenna. If you look at Table 2 again, notice the relatively minor difference be-tween the director k factors of the five- and seven-element model quads. Using either k factor will produce a satisfactory antenna. So this design can be changed by adding two director elements to accommodate changing circumstances, like portable and fixed operation with attendant differences in gain. Sample Construction Materials The spreaders used to build the 50.1-MHz quad test antenna were lengths of 1/2 x 3/4 inch, rec-tangular, wooden doorstop molding. Each was gener-ously coated with several layers of high grade, oil base (not latex) paint. This wooden material is inexpensive and available in 15-ft lengths. If you wish to use it for spreaders greater than eight but less than twelve ft, cut the strip in half. Overlap the halves to the proper length and glue the pieces together to inoreasc their strength. The spreaders were individually held to the boom with U bolts which, when disassembled, allow the elements to be folded for transportation. Also, the spreaders were aligned vertically and hori-zontally to produce diamond-shaped elements. With the feed point for horizontal polarization occurring at the end of the lower driven element spreader, the feed line was conveniently dressed along that arm, back the boom and down the mast. The boom material was sections of steel TV mast, and a metal, square electrical box cover was drilled and fitted with U bolts to hold the boom to the mast. The mast was a 10-ft length of 1-1/4 inch thinwall EMT conduit. So constructed, the antenna was light in weight, and two of these quad antennas were easily turned by a TV antenna rotator. The Obligatory Comparison Experimenters find it difficult to justify a number of persistently reported quad-antenna char-acteristics. To explain those, I believe it is necessary to look beyond the physical quad antenna and to integrate fact, theory, observation, circum-stance and environment. Unfortunately, that is be-yond the scope of this article. However, I have no argument with the claim that the horizontally polar-ized Yagi antenna is superior to a quad antenna for competitive communications. There is one quad antenna characteristic I would like to discuss. Consider that the closed quad loop has no physically constrained or inherent polarization. It follows, then, that an array of quad loop parasites will have equal gain with any angle linear polarization. Therefore, a quad antenna is dependent on a single element, the driven ele-ment, to fix its majority polarization. In compar-ison to the Yagi, the quad antenna lacks the cas-caded polarization selectivity of Yagi parasites and does not competitively reject off-horizontal wave components. These facts are experimentally observable as an anti-fade characteristic of the quad antenna. Over a period of months, I observed reduced signal fading on a 150-mile VHF communications path from a station using a high-gain quad antenna array. Numerous sta-tions using Yagi antennas were available for com-parison during the same time period. Then for fur-ther verification, the local Yagi was replaced by a quad antenna. Signal fading from other stations was reduced. All other things being equal, it is rea-sonable to assume that the observed signal fading was caused, in part, by polarization rotation and that the anti-face characteristic of the quad an-tenna was because of its parasitic element response to nonhorizontally polarized wave components. This discussion and earlier remarks define areas of communication where the quad antenna may be expected to perform well. For casual or low-budget communications, construction of a quad antenna yields economy of resources, reduced fade characteristic, increased compatibility with other antenna types, respectable gain and broader oper-ating bandwidth than the Yagi antenna. The quad antenna may even have its niche in competition with single polarization Yagi antennas. Since quad antenna polarization is dependent on the point where the driven element is excited, moving the feed point in various ways allows the user to take advantage of the prevailing optimum polar-ization with no loss in antenna gain. This cha-meleon-like quality, on balance, may make the quad antenna very competitive in some circumstances. Conclusion The practical goals of these experiments were met. Using the information presented in this arti-cle, a quad antenna that required no tuning was designed and built for 50.125 MHz. In use, the pat-tern appears clean, the SVVR is excellent and the gain appears to be in line with the model. By taking care to duplicate the models, using bare elements and nonmetallic spreaders and correcting for dif-fering element d/1 ratios, good results can be expected. The narrowest element spacing may be of greater interest for quad antenna designs in the upper end of the HF spectrum. Translating the boom length of the longest 0.130 wavelength spaced model into feet for example, a 17.8-ft boom is necessary to accom-modate five elements at 28.7 MHz. For economy of elements, rather than boom length, the wider element spacings are attractive at VHF. There for example, a 19.4-ft boom will hold seven either 0.165- or six 0.195-wavelength spaced elements for 50.1-MHz quad designs. Do not be concerned with SWR, unless it differs greatly from that of the model at your design fre-quency. SWR is not an indication of gain, nor an indication of the frequency where maximum gain occurs. No attempt was made to match the driven element to the feed line, when modeling, so the listed values of SWR are intrinsic to these specific models. These quad antenna models were gain maximized for equal length quad directors. Front-to-back (f/b) ratio was never a factor in these experiments. While it may be impressive and sometimes desirable to ob-serve large f/b ratios in use, the factor is not a valid indicator of forward gain. Indeed, some for-ward gain must be sacrificed to optimize the f/b ratio. Several dual reflector quad antenna models were modeled, but the gain improvements were not thor-oughly investigated (see notes in Tables 1 through 3). However, a second quad reflector looks promising instead of a fourth quad director for increased gain. There it also preserves the desirable odd numbered director element design format. References \1 W. Overbeck, N6NB, "Quad vs Yagi Revisited," Ham Radio, May 1979, p. 12. \2. P. Viezbicke, "Yagi Antenna Design," NBS Technical Note 688, U. S. Department of Commerce, Washington, DC, December 1976. The ARRL Antenna Handbook, 13th edition, Chapter 4, "The Quad Antenna," p. 159, second paragraph. \4_ See Ref. 3. See Ref. 3. J. L. Lawson, W2PV, A Series: "Yagi Antenna Design," Ham Radio, 1980. Y Z . William I. Orr, W6SAI, AH About Cubical Quad Antennas, 2nd edition, 1970. Table 1. quad model: 0.130 wavelength-element spacing. Number of elements (including driven), element design k factors (ref: Kr = 1020ft-MHz), boom length (wavelengths), gain (relative to a two-element Yagi), and SWR (:1). Number quad Model k factors of Els Reflector All Dir' ' s BoomL G(dB) SWR 3 1059 975 0.260 2.3 1.3 4 1077 1011 0.390 4.4 1.1 4 1101 981 0.390 3.0 1.3 5 1083 993 0.520 5.0 1.3 Note: Two reflectors, one driven, plus directors. Table 2. quad models: 0.165 wavelength-element spacing. Number of elements (including driven), element design k factors (ref: Kr = 1020ft-MHz), boom length (wavelengths), gain (relative to a two element yagi), and SWR (:1). Number quad Model k factors Of Els Reflector All Dir's BoomL G(dB) SWR 3 1065 975 0.330 3.5 1.4 4 1059 1011 0.495 4.4 1.2 4 1065 987 0.495 3.7 1.3 5 1083 984 0.660 5.2 1.2 6 1059 1008 0.825 5.7 1.1 7 1083 981 0.990 5.9 1.3 8 1068 969 1.155 6.1 1.1 8# 1068 1005 1.155 5.5 -8 1053 981 1.155 6.3 1.3 Note: #As above, directors at another gain peak. Two reflectors, one driven, plus directors. Table 3. quad models: 0.195 wavelength-element spacing. Number of elements (including driven), element design k factors (ref: Kr - 1020 ft-MIIz), boom length (wavelengths), gain (relative to a two-element Yagi), and SWR (:1). Nuntoer quad Model k factors of Els Reflector All Dir's BoomL G(dB) SWR 3 1071 975 0.390 3.7 1.2 4 1053 1011 0.585 4.4 -4 1059 987 0.585 3.6 1.4 5 1065 978 0.780 5.3 1.2 6 1047 999 0.975 5.8 1.4 7 1059 999 1.170 6.2 1.1 Note: Two reflectors, one driven, plus directors. Log Periodic Arrays A Wide-Band, Low-Z Antenna — New Thoughts on Small Antennas By Ken Heitner, WB4AKK 2410 Garnett Court, Vienna, VA 22180 Today's radio amateur has the opportunity to select flexible equipment that will operate on a number of amateur frequencies. Confronted by the dilemma of living in crowded urban areas, these same amateurs don't have the acreage to install separate antennas for each band. This clearly creates a need for a small, v e r s a t i l e antenna — one that is physically compact, but having the bandwidth to cover several amateur bands. If such antennas pro-vided directivity and were inexpensive, they would also be attractive. Most amateur thinking on an-tennas, however, is along "traditional" lines. An-tennas are typically quarter- or half-wavelength resonant structures, and matched directly to 50- or 72-ohm feed lines. A few amateurs have thoughtfully worked on solving the problems of small antennas, indicating that their size can be reduced. The developed de-signs, however, are for resonant antennas with limited band width.\1 Characteristics of Small Antennas With a reduction in size, the antenna radiation resistance decreases and the reactance increases. Fig. 1 shows these radiation resistance changes for small dipoles and loops. Lower radiation resistance presents two prob-lems. First, the lower resistance must be coupled to the nominal 50-ohm transmission line. Second, as the radiation resistance becomes very low, resistive losses in the antenna become significant and antenna efficiency goes down. Resonant antennas have pre-viously been built with radiation resistances in the order of 3 ohms with good performance, and mobile whip antennas for 80 and 160 meters routinely use lower radiation resistance levels. These antennas, h o w e v e r , o p e r a t e with s i g n i f i c a n t r e s i s t i v e losses. When a dipole is less than its resonant length, its impedance has an increasing capacitive component (see Fig. 2). To resonante a short dipole, inductive loading must be added. This introduces additional resistive losses. Small loop antennas exhibit an inductive im-pedance characteristic, but can be made resonant with a parallel capacitor. These antennas are desirable f o r this size category because large losses do not occur in the tuning capacitor. 60 r 0 i'f3o\ < 15 O Dipole Length or Loop Width — Electrical Degrees Fig. 1 — Radiation resistance for small dipoles and loops. Antenna Length (Degrees) O 5D lOO O w - S O O s & o ( u o C 5 - IOOO o C O t u os 1SOO \SO - 1 For Length/Diameter of 1000 Fig. 2 — Capacitive reactance for small dipoles. Broadband Antenna Concepts Various broadband antenna types have been de-veloped with designs based on the fact that antenna structures defined by angles have inherently broad-band characteristics. Examples include the discone and conical monipole.\2. Other designs are based on the concept of a repeating geometrically similar structure. The log periodic dipole array (LPDA) falls into this group. It consists of a series of dipoles that are geometrically related and connected to a common feed line. Elements resonant at or near the operating frequency actively radiate the sig-nal. For a small broadband antenna design, the LPDA offers a potential approach. Antenna size can be reduced by modifying the individual elements. If the geometrical similarity and the element's elec-trical properties are preserved, the achievement of creating a smaller antenna can become reality. LPDA Design A description of conventional LPDA design is provided in Hie ARRL Antenna Book (1974 edition, pages 160 to 164). This design consists of tubular elements at right angles to the boom. The resulting antennas are large and heavy, requiring large rota-tors and sturdy towers for support. The need for a lighter, more compact LPDA led to an alternative design shown in Fig. 3.\£ The design approach shown in Fig. 3 is notable for several reasons. The dipole elements are raked forward, reducing the turning radius of the array, and the elements are wires held in tension by the array structure. Because the array structure is a boom and a single cross piece, it is lighter and has a low cross-sectional area. Also, the number of wire elements can be easily increased if desired. Since the array geometry for the LPDA departs from the published data in The ARRL Antenna Book, I cannot evaluate it's performance. Thoughts on Compact LPDA Designs Several considerations come to mind when trying to evolve a compact LPDA for amateur use. Although the basic dipole element can be physically short-ened, it requires retuning to some new frequency. On the other hand, if the dipole element was reduced in size (by folding it into a half-wave loop), no re-tuning would be necessary. With the loop element having directivity, it could offset gains loss by the use of close element spacing. Low radiation resistance of the half-wavelength loop can be raised by folding the loop element. This procedure is analagous to a folded dipole. The nom-inal impedance of the half-wave loop is about 10 ohms and it would be raised to about 40 ohms with a second folding. The second folding could be accom-plished by constructing the wire elements out of 300-ohm twin lead in the manner of the classic folded dipole. If the elements are constructed as a series of triangular loops, an array would appear as shown in Fig. 4. Although the loop elements overlap, the in-dividual wires are insulated. There is no contact between elements at the crossover points. They are connected at their feed points by a parallel line feeder that has alternating "phasing" to each ele-ment set, and a balun would be used to assure a balanced feed impedance with respect to ground. The triangular element shape also appears reasonable from a mechanical and structural view-point. The beam can be constructed with cross pieces for the smallest and largest elements. The remaining elements can be constructed of wires in tension, as is the design shown in Fig. 3. Maximum Radiation Pig. 3 — Granger LPDA design. Conclusion It is possible to c o n c e i v e of an antenna structure that possesses both wide bandwidth and compact qualities. In addition, it has directivity, lends itself to amateur construction techniques and though it has inherertly low radiation resistance, the impedance level at the feed point can be trans-formed to the 50-ohm level. Radiation 2.5m (98") 3.75m (1l(.8») Fig. 4 — Compact LPDA-triangular loops (20-30 MHz). References \1 The ARRL Antenna Anthology, 1S78, pp. 22-29. Hie ARRL Antenna Book, 1974, pp. 298 and 299. Y2. Solutions for Telecommunications, Granger A s s o c i a t e s , Santa C l a r a , CA ( R e c e n t ) . Development of the W8JF Waveram: A Planar Log-Periodic Quad Array By Jim Fisher, W8JF Director,Compensation & Benefits, Air Products & Chemicals, Inc., P. 0 . Box 538, Allentown, PA 18105 Learn to use log-periodie principles to develop broadband antennas for the new post-W ARC age, and discover some practical log-periodic quad arrays. This array won third place as a five-band design in the ARRL Antenna Competition (1984). Radio amateurs have made extensive use of log-periodie dipole arrays (LPDAs), including the hybrid log-periodic Yagi (LPY) in which parasitic Yagi elements are excited by a small LPDA. However, most amateurs are not familiar with log-periodic approaches that differ from LPDAs. This article describes the development of one such alternative approach — log-periodic quad arrays (LPQAs), cul-minating in the W8JF Waveram. Several antenna types useful in Amateur Radio are also presented and different ways to use log-periodic principles in non-dipole applications are illustrated. With the advent of the new WARC bands, interest in the LPA's broadband capabilities should grow. Characteristics of a Basic LP Fig. 1 shows a conventional log-periodic array. It is sometimes seen in the form of a TV antenna. Note several characteristics of the LPA that are used in later discussions: The arrey is balanced; elements on one bay are offset by equally long opposing elements on the other. The two bays are fed 180 degrees out of phase through the booms, which are single-conductor feed lines. To provide uniform gain over a wide frequency range, adjacent elements and the spacing between them are scaled down progressively by a constant multiplier such as 0.9. Frequency coverage in the fundamental mode is from the low frequency (the longest element is a quarter wave), to the high frequency (the shortest element is a quarter wave). By virtue of the phase reversals and increasing element lengths along the feed line moving away from the feed point, the array has gain along its central plane (a horizontal plane as in the case of Fig. 1). The array has a front-to-back ratio in the direction of the smaller elements because the two bays "aim forward," as opposed to being in the same plane. Deriving the LPDA from the Basic LP Let's start to play with these concepts. If the two bays are moved together, the opposing elements look like dipoles. The booms may either be very close or replaced with a two-wire balanced trans-mission line twisted between each set of elements. The result, shown in Fig. 2, is an LPDA. It has gain in the direction of the smaller elements, giv-ing it front-to-back ratio. Deriving the LPQA from the LPDA — a Misdirec-tion? A full-wave loop has a gain of several dB over a dipole. Why not replacc cach dipole with a loop (quad element) resonant at the same frequency? I built such an antenna (represented in Fig. 3) for 6 to 10 MHz in the mid-1970s and described it in an article appearing in CQ Magazine.Vl The same con-Gain Front-to-Back Ratio Fig. 1 — The conventional log-periodic array. The array is fed at the apex with balanced line, and exhibits gain and front-to-back ratio in the direction of the apex. p Gain Front-to-Back Ratio Fig. 2 — The log-periodic dipole array (LPDA). cept was described by W4AEO in QST.\2. As I studied the matter further, I began to realize that such an antenna, while performing well over a limited frequency range, has deficiencies that can be identified either through quad theory or log-periodic theory. Starting with quad theory, consider a full-wave quad loop as two half-waves in phase. We identify the gain as being derived from current maxima separated by a stacking distance. Normally the loop is fed only at one point, but the two current maxima are positioned by the half-wave distance from the feed point around the loop to the opposite point. What happens to the LPQA constructed as shown in Fig. 3? At most, one loop is exactly a full wave; the others are progressively larger or smaller. The resulting current maxima in the other loops are not distributed in a symmetrical manner about a single central plane. Even viewed from a distant point, the current patterns are not contributing efficiently toward a single pattern of forward gain. Furthermore, I found that the resistance and reactance patterns at the feed point (listed in my article) showed signif-icant excursions even within the 6- to 10-MHz range. There are,also potential problems, hinted at in my article, from nonfundamental resonances that can occur within the fundamental frequency range. Try Again Let's now reconsider the problem in light of log-periodic theory. Rather than thinking of an LPQA as an expanded LPDA, let us think of it as being derived from a log-periodic array resembling Fig. 1, but having four bays instead of two. The top two bays are collapsed into one LPDA and the bottom two bays are rearranged to form an identical LPDA. The four booms become not one, but two balanced trans-mission lines. The ends of the LPDA elements are folded to couch each other, and voila! An LPQA as shown in Fig. 4, with symmetrical current dis-tribution and a concerted f o r w a r d p a t t e r n , is formed. My 6- to 10-MHz LPQA was converted to the improved version simply by cutting the top of each loop and inserting a second transmission line. Checks of resistance and reactance patterns showed significant improvements, with excellent SWR over a wide frequency range, while feeding coaxial cable with a 4:1 balun. I thought the idea was original, but discovered that Collins had already obtained a U.S. Patent number on it (no. 3,279,159). I constructed several such antennas for SWL and Amateur Radio pur-poses and described them in the 1978 World Radio TV Handbook (WRTH).! Dimensions given in Tabic 1 arc for two an-tennas that are most likely to be of interest to Amateur Radio operators. A 6- to 18-MHz fixed LPQA wire antenna — e f f e c t i v e l y , a unidirectional 40-, 30-, 20- and 17-meter antenna fed with a coaxial cable through a 4:1 balun. • A 9- to 18-MHz rotatable four-element LPQA — think of it as a 30-, 20- and 17-meter tribander based on standard ham radio quad hardware. Fig. 3 — First attempt at a log-periodic quad array (LPQA). Don't build it this way! The fixed array can be constructed between two tall supports. If additional supports are available, two or more arrays can be constructed to cover var-ious directions. Continuing the array to progres-sively smaller elements and shorter spacing or, at the bigger end, to iarger elements and longer spacing, could broaden the already creditable 3:1 frequency range. At one time, I was using two of these arrays aimed about 60 degrees apart. Switching Wires to lower resonance (ro-tating model reversed between elements, but top in phase with bottom. Fig. i j . — An improved LPQA feeds each loop in phase at its top and bottom. from one to another with a coaxial-cable switch yielded a satisfying instant demonstration of pat-tern lobes. This antenna is effective and inexpen-sive to build, and could be used to cover 5:1 or greater frequency ranges anywhere from 160 meters on. The W4AEO article and my two articles contain some practical suggestions for methods and mate-rials. With proper spacing, the array may function well on odd multiples of its fundamental resonances, yielding extremely broad continuous frequency coverage. The rotatable array o f f e r e d fair gain and front-to-back ratio, which is not easy to realize in a compact antenna over that continuous a frequency range. It was taken down shortly thereafter to make room for further experiments, some of which are described below. Derivation of the W8JF Waveram Let's derive another antenna two different ways to see if our results are consistent. As an objec-tive, let's seek LPQA-type performance with a min-imum boom length. First, it occurred to me to reduce the boom length of an LPQA to zero, creating a planar log-periodic quad array which I call a W8JF Waveram (see Fig. 5). This is a broadband bidirectional array. Several versions have been tested at VHF, all dis-playing gain over a reference dipole. The first full-scale version, built while I lived in Michigan, was used to make two-way Amateur Radio contacts with hams on six continents while using an output power of two watts on 20 and 15 meters. For this article, I constructed and tested a ten-loop version (dimensions in Table II) in VHF scale model and full-sized HF versions. I used it with five watts output to make intercontinental QSOs on 10, 15, 20, and 30 meters. (Thirty-meter operation is below the fundamental frequency and is in another mode. This is not uncommon with LPs.) Without a balun, SWR was no more than 2:1 for the 14-, 18-, 21-, 24-, and 28-MHz bands with a small capacitance across the feed points. Devel-opment for lower SWR should be possible, measuring SWR or R and X while varying spacing, element thickness and transmission-line characteristics, and possibly using a tapped toroidal RF transformer. It should be noted that the SWR will change if the W8JF Waveram is used as part of a larger antenna array. To understand how this works and to confirm the approach, we will review the original LP model and try to develop a similar concept. First, the bays in Fig. 1 are straightened into a single plane. The front-to-back ratio drops off to zero; the array becomes bidirectional, but continues to have gain in tlie plane perpendicular to the ar-ray. Sweep the tips of the elements to the center line and we have Fig. 6, with offset elements not meeting as loops. The feed lines are still single conductors. This is a regular LP, one of the traditional possibilities, although the sweep is more severe than we would normally see. (One with no sweep is called 90 degrees from the transmission line. A mild sweep of 60 degrees from the trans-mission line is more common in LPs than this 45-degree sweep.) An advantage of this design is that it can be constructed with standard 20-meter quad hardware; 13-foot spreaders provide a fundamental resonance just below 20 meters. I constructed VHF scale models and a full-scale HF version. Gain of the VHF scale model was on the order of that of the equivalent fully balanced version. Reception was good and some DX was worked on the HF version, but the SWR was not as low as with the balanced transmission line. Again, there are ways to influence the SWR, and the Fig. 6 version may be better for some purposes. periodic array. Try Again To derive the fully balanced W8JF Waveram, we have to start with a four-bay LP. For instance, the bays should be spaced 90 degrees apart in an "X." We move the top bays together and the bottom bays to-gether and we have opposing LPDAs. Sweep the element tips together to form continuous loops and the fully balanced version is formed. Summary and Further Challenges We have used log-periodic principles to derive several different practical broadband arrays for amateur use. All could use some further experi-mentation to optimize gain and SWR through element shape and sweep, scaling factors, shorting loops at the end of the transmission lines, and so on. Ele-ments could be center loaded or linearly loaded with inductance or capacitance to lower or raise their resonant frequencies. All appear suitable for single coaxial feed through a balancing device, though some development work is needed before typical solid-state finals will run flat out into the load. At one extreme (Fig. 4), there are fixed arrays that could cover wide expanses of the MF and HF spectrum, and are inexpensive to build provided two high supports are available (one support can be lower than the one supporting the biggest elements). The array in Fig. 4 varies from huge to fairly compact, depending on spacing and frequency cover-age. A rotatable version of the Fig. 4 array has demonstrated gain and front-to-back ratio down to 9 MHz. The W8JF Waveram is usable by itself as a very compact antenna. It offers gain over a dipole and broadband coverage with a single feed line. I would certainly prefer it to trap dipoles for most ap-plications. Different versions have been tried, with help from my family. For example, the scaling factor has been varied from 0.7 to 0.8366, the array has been configured in a rectangle with the transmis-sion line running parallel to the longer side, and a version has been built with the elements swept even more than the 45-degree model. Each version dis-played good performance, to the extent that I v/ould recommend the builder choose what is mechanically convenient. For example, the highly swept, flat-diamond model, mounted with the transmission line horizontal, yields a lower quad that is easier to handle when raising and lowering than the 26-foot diagonal model — but it is vertically polarized! (The polarization of a Fig. 1, Fig. 5, or Fig. 6 array is perpendicular to the run of the transmis-sion line; the polarization of a Fig. 2 LPDA is in the plane of the elements; the polarization of the Fig. 3 (don't build it!) or Fig. 4 arrays is hori-zontal with the transmission line(s) at the top and bottom and vertical with them at the sides.) One purpose of building the W8JF Waveram was to create a compact, broadband building block and investigate whether it could be a basis for some more complex arrays. Actually, I first thought of it as a substitution for the forward element of a Fig. 4 array. Time has not permitted much work on these possibilities, but initial results indicate that the cookbook dimensions of parasitic dipole and quad elements may not be optimum for use with a W8JF Waveram as a driver, possibly because of the E and H-plane patterns of the array in the near field. I have also acquired a second parallel W8JF Waveram to show some activity as a parasitic element, with and without a tunable reactance across the feed point. In addition, I have obtained some hint of activity as a broadband sink for absorbing waves from the back as a second reflector and improving front-to-back ratio over a broad frequency range. The latter was attempted with a fixed resistance of 200 ohms, which may be too high. A more elegant solution would be attenuating open-wire line such as is used in non-resonant rhombics. Broadband reflectors such as corner reflectors or parabolic dishes have not yet been tried. For those who heve decided that these quad versions of the LP are not for them, I hope this has been an interesting exercise in LP approaches. It may be of assistance should you decide to devise your own arrays. If you experiment with my ideas, I would appreciate hearing of your results. References \1 Fisher, J., "Log Periodic Quad Array," CQ, February 1977. \2 Smith, G., "Quad Log-Periodic Fixed Beam An-tennas," QST, April 1977. Y 2 . Fisher, J., "Build Your Own Log-Periodic Quad Antenna," World Kadio TV Handbook, 1978 edi-tion. T a b l e 1 Dimensions for 6 to 18 MHz Fixed LPQA Length of One Side of Loop Spacing Element Feet (Meters) Feet (Meters) 1 8 3 i g n ( 2 5 . 5 4 ) lO'O" ( 3 . 0 5 ) 2 7 1 1 2 " ( 2 1 . 7 1 ) 8 ' 6 " ( 2 . 5 9 ) 3 6 0 ' 6 n ( 1 8 . 4 6 ) 7 ' 3 " ( 2 . 2 0 ) 4 5 1 1 5 " ( 1 5 . 6 9 ) 6 1 2 " ( 1 . 8 7 ) 5 4 3 ' 9 n ( 1 3 . 3 3 ) 5 1 3 n ( 1 . 5 9 ) 6 3 7 ' 2 " ( 1 1 . 3 3 ) 4 1 5 " ( 1 . 3 5 ) 7 3 1 ' 7 n ( 9 . 6 3 ) 3 ' 9 " ( 1 . 1 5 ) 8 26 ' l l n (8.19) Note: The above lengths are for one-half of the loop (from transmission line to transmission line). Dimensions for Rotatable 9 to 18-MHz Array Length of One Side of Loop Spacing Element Feet (Meters) Feet (Meters) 1 36'9" (11.21) 9'9" (2.97) 2 3619" (11.21) 7'10" (2.39) 3 3418" (10.56) 614" (1.93) 4 27'11" (8.52) Note: To reduce the resonant frequency of the lowest loop, 16'7n (5.05 m) of wire is connected to each of the two outside corners, run along the horizontal spreader and doubled back on itself. Similarly, for element 2, 6'2" (1.89 m) of wire is connected to each of the outside corners and run along the horizontal spreaders. Table 2 Dimensions for Planar Arrays Distance along spreader from center for each connection Distance to Center Element Feet (Meters) 1 1 2 ' l g n ( 3 . 8 8 ) 2 1 0 1 1 8 " ( 3 . 2 5 ) 3 8 ' ' 1 1 " ( 2 . 7 2 ) 4 7 1 ' 6 » ( 2 . 2 7 ) 5 6 ' i ^ n ( 1 . 9 0 ) 6 5 ' l 3 n ( 1 . 5 9 ) 7 4 ' r 5 " ( 1 . 3 3 ) 8 3 ' i g n ( 1 . 1 1 ) 9 3 ' •l" ( 0 . 9 3 ) 1 0 2 l 7 n ( 0 . 7 8 ) Note: In the fully balanced W8JF Waveram, wires are connected on both sides of each transmission line at each of the above points. In the W8JF Planar Log Periodic (PLP), wires are connected to alternating sides of each point as shown in Fig. 6. However, to balance the stress of the wires, it is suggested that nylon cord be used for the runs without wires. A Second-Generation Spiderweb Antenna Diek A. Mack, W6PGL 600 Loekewood Lane, Santa Cruz, CA 95066 T h e o r i g i n a l d e s c r i p t i o n of the Telerana (Spanish for Spiderweb) log-periodic antenna ap-peared the same time my son-in-law, Jim Nelson, N6EWP/VS6JA, departed as a career missionary to Hong Kong. Among our plans were skeds for frequent QSOs across the Pacific, Depending on the propagation, we expected to use 10, 15 and 20 meters. The Telerana appeared to be an almost ideal antenna for this service. It was light weight, rotatable, and pre-sented low resistance to the wind. Electrically, log-periodic arrays have good forward gain and acceptable front-to-back ratio characteristics. The idea of switching from 10 to 15 and then to 20 meters without an antenna tuner was most intriguing. Also, the future 18- and 25-MHz bands could be covered without additional elements. Now, three years later, I am convinced that the log-periodic antenna can rightly take its place among other high-performance beam antennas. Over this period, Jim and I have had numerous SSB QSOs, usually several times a week, with good results. A second array was built for my friend K7JIO two years ago. The construction described here is based on the length and spacing of the antenna radiators de-scriDed by Ansyl Eckols, YV5DLT, but with a dif-ferent supporting structure.\1 Several improvements were made while building the second unit and are included here. The beam uses 13 wire elements with a 90& taper and a 0.0596 wavelength spacing. As in all log-periodic dipole arrays, each pair of radiators is connected to the feed line in a transposed fash-ion with respect to the pair ahead and behind it (Fig. 1). The array is fed from the forward end of the feed line (where the elements are the shortest), with a 4:1 balun and RG-8 coaxial cable running back to the shack. A review of an earlier article by YV5DLT is helpful. When I began assembling materials for con-struction of the antenna, fiberglass vaulting poles were hard to find and expensive. I considered plas-tic PVC water pipe, but that thought was soon aban-doned as it was too limber. A local sporting goods store had a stock of fishing rod blanks, but none were strong or long enough (at least 15 feet). Tapered rods are also hard to splice together. Fi-nally, I located a plastics supplier who stocked solid fiberglass rods 1/2 inch, 3/4 inch, and 1 inch in diameter and 6 feet long.\2 The 3/4-inch diameter rods were selected as the best trade-off between stiffness, weight and cost. These rods came without surface covering and after a few hours of handling, my hands itched from tiny glass slivers. The sup-plier advised that an aerosol spray of clear plastic should eliminate this problem and minimize the weathering that I have noted with time. I chose a design that would use five fiberglass t Direction af Forward Loke Coa io Xtnjr Element/$ ^ V ^ ; / Balun Element" Distance Along feed Line. Mtasured from this Point Fig. 1 — Connection of radiators. Alternate pairs are transposed as they connect to the feed line. spreaders fastened at a common hub for the support construction, rather than the four vaulting poles and crossbow employed by Eckols. See Fig. 2. The hub was made from a 6-inch square, 1/4-ineh thick steel plate to which a local welder attached five pieces of 3/4-inch pipe at the appropriate angles as shown in Fig. 3. A 1-1/4-inch pipe flange was mounted on the underside of the plate with four no. 1/4-20 screws. To assemble the antenna, I found a flat spot in the yard where I could position a section of 1-1/4-inch pipe on the ground to place the hub about waist height. It was easy to assemble the various parts this way, and I could cut and stretch the antenna elements close to the ground. After com-pletion, the array was tested and operated on the air, before being hoisted into position on the roof. The ends of the first lengths of fiberglass were wrapped with plastic electrical tape for about 8 inches and pushed as far into the pipes at the hub as they would go. Two holes were then drilled about four inches apart and at right angles to each other for placement of the screws that secure the rods to 'feed Line., 30'6g",3,3rrt Long • . f-^—Coq^to Xrntr ; ' rDccron t ' y4 Rope / , > Catenary. / , x f^'ic, 5J3m , 6s" ' Long. — i - V W ^ o . ' 90- - / / fish/me Distances \ o/cng Catemn/ < Was. / " 5 1.25 Antennas fed with Kenwood TS-820 and MFJ Versa Tuner II, Model 949 B in Direct Coax position. Other Beam Antennas A Simple Log-Yag Array for 50 MHz By John J. Meyer, N5JM •112 Sherwood Forest Drive, New Orleans, LA 70119 The 50-MHz band is a playground of sorts for the antenna builder. It offers many interesting possibilities such as high gain, small physical size and the potential of making DX contacts with a homemade antenna and low power. Interestingly, at 50 MHz and higher, antenna construction becomes simpler and within the realm of the novice builder. Hand tools can be employed in constructing many antenna designs for this frequency range and since most construction materials for VHF antennas can be pur-chased at a hardware store, the builders job is easier and inexpensive. This antenna makes an ideal weekend project for the amateur who would like to build a beem at a fraction of the cost of a "store bought" Yagi. Most 50-MHz antennas are Yagis. They offer high gain for their size, yet there are lesser known an-tenna variations offering benefits such as greater bandwidth, ease of matching and higher gain for the same boom length. The Log-Yag Array is such an an-tenna. The Log-Yag Array, popularized by Oliver Swan in the 1960s, combines the bandwidth of the log-periodic antenna with the high gain of a Yagi.\l Those interested in the theory and design should reference the articles by K4EWG, W4BBP, W4AEO and WGSAI for an in-depth explanation on how they work.\2,3,4 In essence, the Log-Yag has two or more dipole-driven elements. Each is fed 180 degrees apart, and consists of one or more parasitic Yagi elements to both sharpen the beam pattern and increase the front-to-baek ratio. Element spacing tends to be closer than a Yagi, and element lengths are also slightly longer. This article describes a basic design that can be duplicated easily. The boom is a 10-ft section of TV mast from Radio Shack (or aluminum, if avail-able) and the elements are 1/2-inch EMT (electri-cians metallic tubing) from the local hardware store. Dimensions and spacing are not critical, and good construction techniques should be followed, just as you would with any antenna. Moisture and corrosion are serious problems so care should be taken to paint and protect the driven-element supports made of wood and all the associated hardware. Element-to-boom mounting is best ac-complished as illustrated in The ARRL Handbook.^ The e l e m e n t s a r e cut to the f o l l o w i n g lengths: Reflector: 116 " Rear Driven Element: 54 " each length (111 n tip to tip, 3 " center) Front Driven Element: 53 " each length (109 " tip to tip, 3 " center) First Director: 107 " Second Director: 104 " A complete diagram of the 50-MHz Log-Yag Array can be seen in Fig. 1. The driven elements have a 3-inch center spac-ing. The elements should be mounted on the 10-ft boom as follows: Reflector to Rear-Driven Element: 23 " KDE to Front-Driven Element: 26 " EDE to First Director: 19 " FD to Second Director: 50 " The driven elements are insulated from the boom with wood, while the other parasitic elements are mounted directly to the boom. A copper or aluminum feed line is transposed between the two driven ele-ments to achieve the 180 degree phase shift. The Log-Yag Array is fed at the front driven element with a 52-ohm balanced feed and exhibits an imped-ance of about 60 ohms. The 1:1 balun in Hie ARRL Handbook works quite well for this purpose. It should be noted that the balun must hang free from the antenna and should not be taped or otherwise attached to the antenna, or the results will be poor radiation patterns and strange SWR ratios. There is little else in the construction that will affect the SWR as much as an improperly built or placed balun. Moving the first director about on the boom and changing its length affects both the SWR and im-pedance. Should the builder wish to vary these values, this would be the place to start. To a lesser extent, the rcflector also controls these values. Experimentation can be done by placing the reflector element on a wood or other non-metallic support and pointing the antenna straight up in the air (away from any wires!) so that it can be worked on at ground level. This Log-Yag Array is flat across the first megahertz of the 6-meter band with an SWR of less than 1.3:1 and is usable up to 52 MHz. The front-to-back ratio is better than 25 dB and it has a side null in excess of 35 dB. The front lobe is broad and the antenna can be rotated 45 degrees off a station without any appreciable loss of signal. There are no side lobes. Increasing the boom length and adding additional directors will sharpen the front lobe for those who want greater directivity. Although no comparisons have been made with larger antennas, I believe this antenna will outperform a Yagi of the same boom length. Good DX! References \1 "The Swan Multidrive 2-Meter Antenna," QST, p. 42, Oct. 1969. Rhodes and Painter, "The Log-Yag Array," Hie ARRL Antenna Anthology, pp. 49 to 52. Smith, numerous articles in Ham Radio since 1974. U Orr, Radio Handbook, 22nd edition, pp. 30-15 to 30-17. \5. The ARRL Handbook for the Radio Amateur, 62nd edition, (ARRL: Newington), 1984. struction details are found in the text. Designing X-Beams By Brice Anderson, W9PNE P. O. Box 14, Lancaster, IL 62855 The X-Beam is a high-performance broadband antenna that is ideal for radio operators with limited space. If you are searching for an under-standing of this compact but powerful antenna, the explanation and formulas found in this article should put it on solid ground for you. Description The X-Beam is a compact version of a 2-element Yagi whose performance equals or exceeds that of a Yagi. It has good bandwidth, a 6-dB forward gain over a dipole, and a 15- to 18-dB front-to-baek ratio. The formulas presented will enable you to calculate X-Beam arm and tail lengths for any frequency. The X-Beam name is derived from its shape, shown in Fig. 1. It consists of four arms often built from aluminum tubing and four loading wire tails. As there is no boom, the antenna is usually mounted in a horizontal plane, though a vertical-plane position can be used. It is considerably smaller and lighter than a 2-element Yagi for the same band. Construction A square piece of heavy plywood, treated for weather protection, makes a suitable hub for 15- and 20-meter X-Beams. The arms are secured to the hub with pipe brackets, and a nylon cord is strung tightly around the beam from tip to tip. This pro-cedure strengthens the assembly and supports the tails. You could use an alternative method of con-struction with two pieces of aluminum angle stock bolted together as the X hub. The arms may then be clamped to the hub with screw type hose clamps, using plastic tubing to insulate the driven arms. Versions Formulas are given for full-size and minimum-size X-Beams. The full-size beam provides a close match to 50-ohm coax and the greatest possible bandwidth. It is recommended for 10, 15 and 20 meters. The minimum-size beam has similar per-formance, but the SWR is approximately 1.4:1. Because of the shorter arms, the ends of the tails nearly touch. It is used for 30 and 40 meters. Many measurements on a large number of experimental X-Beams were used to develop the formulas. The arms were made of tubing with a length-to-dla meter ratio of about 200:1. Stranded no. 19 vinyl-covered wire was used for the tails. The use of larger tubing or larger wire size will require slightly shorter tails than given by the formulas, and vice versa. Arms may be made of wire supported by nonconductive materials such as wood or bamboo. In this case, the tails will have to be made considerably longer. Formulas For Full-Size: Arm length (ft) = 195/f (MHz) (each arm, about 82% of half a Yagi driven element) Total driven element (ft) = 603/f (MHz) (about 11% greater than a Yagi driven element) Driven element tail (ft) = 106.5/f (MHz) (each tail) Total director element (ft) = 575/f (MHz) (4.65SS shorter than driven element) Director tail (ft) = 92.5/f (MHz) (each tail) Do not use arms much longer than calculated by the full-size formula. As the arms get longer and the tails get shorter, the pattern changes gradually to four broad lobes. For Minimum-Size: Arm length (ft) = 177/f (MHz) (each arm, about 75% of half a Yagi driven element) Total driven element (ft) = 603/f (MHz) (same as for full size) Driven element tail (ft) = 125/f (MHz) (each tail) Total director element (ft) = 575/f (MHz) (4.65« shorter than driven element) Director tail (ft) = 110.5/f (MHz) (each tail) General A purchased beam is shipped unassembled, but it comes with a table of the exact element lengths and spacings for CW or phone frequencies. Since it is a replica of the manufacturer's model, it will perform well once it is constructed. A homemade beam, on the other hand, will not be like the one described here. It will probably have to be tuned to resonance. Luckily, this procedure is simple. Tuning and Adjustment The tails should be made a few inches longer than the calculated lengths and the director tails shorter than the driven element tails as shown in the tables. For tuning, the beam should be mounted on a pole at least 10 ft in height. To determine resonance, an SWR bridge is placed in the coax feed line. An alternative method is to couple a dipper to the director element. Using 5 W of power or less, quickly make some SWR measurements starting at the low-frequency end Pig. 1 — Top view of the X-Beam. The arms are made of tubing. Nylon cord is strung tightly around the beam from tip to tip, to strengthen the assembly and support the wire tails. of the band. Do not cause QRM! The lowest SWR occurs at the resonant frequency. If the resonant frequency is too low, cut one inch from each of the four tails (1/2 inch on 10 meters). Again determine resonance. Repeat this process until the antenna resonates at the desired frequency. The difference in tail length must be as shown in the tables. The resonant frequency will increase somewhat when the antenna is mounted on the tower. Baluns The radiation pattern is slightly unsym metrical because of currents on the coax shield. While this scarcely effects the antenna performance, purists may connect a 1:1 balun to the driven element. For a balun, the coax can be coiled into a 6-in inside diameter coil, taped together and attached to the mast under the beam. The coil should be 12 turns for 15 and 20 meters and 6 turns for 10 meters. Conclusion The actual peformance of the simple X-Beam is amazingly similar to that of a 3-element Yagi. Put one up and challenge the "big guns." Table of Dimensions for Full-Size X-Beams Driven El. Director Tail Arm Tail Tail Diff MHz Ft In Ft In Ft In Ft : 7.050 27 8 15 1 1/2 13 1 1/2 2 10.125 19 3 10 6 9 1 1/2 1 -14.100 13 10 7 6 1/2 6 6 1/2 1 21.100 9 3 5 1/2 4 4 1/2 8 28.200 6 11 3 9 1/2 3 3 1/2 6 Table of Dimensions for Minimm-Size X-Beams Driven El. Director Tail Arm Tail Tail Diff. MHz Ft In Ft In Ft In Ft In 7.050 25 1 1/2 17 9 15 9 2 10.125 17 6 12 4 10 11 1 5 14.100 12 6 1/2 8 10 7 10 1 21.100 8 4 1/2 5 11 5 3 8 28.200 6 3 1/2 4 5 3 11 6 An HF Phased Array Using Twisted-Wire Hybrid Directional Couplers By James V. Melody, WA2NPJ/KX6JM •145 South Zurich Ave., Egg Harbor City, NJ 08215 Many articles have been published on phased arrays for 40 and 80 metersAl The advantage of a phased array over a beam is that it can be turned electronically, with no need for large, expensive eleotro-mechanical rotators. In addition, the beam can be electronically aimed at a station during transmit, with an interferring station switched to a null on receive. Several combinations of antennas at various phases have been simulated and tested by experi-menters. From their data, it seems that the most practical phased array for 40 and 80 meters consists of four vertical elements arranged in a square and spaced one-quarter wavelength apart. These four elements are fed with equal power at phases shown in Fig. 1. Such a configuration will give a gain of 5.3 dB, a front-to-back ratio of 25 dB, and a front-to-side ratio of 12 dB.\2, Previous designs of HF phased arrays have used Wilkinson power dividers or phasing lines, both of which employ long lengths of coaxial cable to obtain the necessary phase change and power split.\3.4 In t h i s d e s i g n , a s e r i e s o f t w i s t e d - w i r e quadrature hybrid directional couplers are used to obtain the power split and phase shift. A circuit diagram of a single twisted-wire hybrid is shown in Fig. 2. A single hybrid divides the power input equally between ports 2 and 4 at port 1, with no power entering port 3 if all Ztf are equal. The hybrid also shifts the phase a total of 90 degrees between ports 2 and 4. If these hybrids are cascaded as shown in the block diagram in Fig. 3, four out-puts are obtained. Each of the four outputs are of equal power and two have zero degrees phase, one -90 degrees, and one +90 degrees. This is the exact power split and phase shift needed for our antenna array. The complete circuit is shown in Fig. 4. The makeup of such a network begins with the design of the twisted-wire hybrid.Vi^fi The coil, LI, shown in Fig. 2 is a bifilar winding on a toroidal core capable of handling the expected power from the transmitter. In Fig. 5A, the inductance of this twisted pair is given by: Zfi L = 2 T £f& The frequency of interest is the middle of the 40-meter band and the impedance is 50 ohms. 50 L = 2 w x 7.15 MHz = 1.12 pa By an examination of catalogs and application notes, I decided to use an iron-powder toroid.\1 I picked the "mix 2" with permeability 10 because it covers the 7-MHz band well, and it is readily available. The number of turns required for an iron-powder toroid is given by: 1/2 T = 100 (Eq. 1) where AL is a parameter of the toroid given by the distributor. Since the coil will be dealing only with an ac voltage, a core large enough to handle the maximum ac flux density is required. The maximum ac flux density gauss (G) is given by: EDMS x 108 Broax = 4.44 AeNf/rf (Eq. 2) where Ae = effective cross-sectional area (cm) EHMS = the applied voltage £s& = the frequency of interest N = the number of turns ANT 1 +90° A 4 4 ANT 3 n f V v A N T 4 T T V ANT 2 -90° Fig. 1 — Top view of 4-element square array showing phases and relative power at each antenna. L 0 DB -3 DB 223 x 108 Bmax = 4.44 x 1.33 x 9.6 x 7.15 x 10 6 Bmax = 54.6 gauss From the manufacturer's specification sheet, type 2 material will support 14 kG. The ten turns of no. 18 wire will easily fit on the core. Therefore, the T-200-2 is an acceptable choice. If similar calculations are performed for a T-130-2 core at 500 W RMS (3 dB down), ten turns are required and the maximum gauss is 53.5. Thus, the T-130-2 core is a good selection for the second and third hybrid. Having chosen the core and the turns for LI, the capacitors must now be determined. In Fig. 5B, the capacitance required is given by: 1 CReq = 2 7 t fj&Z0 Fig. 2 — Circuit diagram of a single twisted-, wire hybrid. A l l Z^ must be equal, and CI equals C2. In addition to the above Eqs. 1 and 2, there is a general rule to follow for choosing the largest core: Be consistent with the price and space available. Based on this data, a T-200-2 core was selected. The following are the parameters for a T-200-2 supplied by the manufacturer: AL = 120 jiH/100 turns 2 Ae = 1.33 cm = effective area Substituting the required 1.12 jjH inductance and the value for AL into Eq. 1, the number of turns is: F = 100(1.12/120) 1/2 = 9.6 turns Assuming that we now have 1000 W RMS into a 50-ohm load, Bmax from Eq. 2 is: Z«S = 50 ohms f(6 = 7.15 MHz Therefore, CReq is 445 pF. This value includes both CI and C2 as well as the capacitance between the twisted bifilar turns of LI. To implement the three hybrids, first wind the toroids with the twisted pair. No. 18 wire with Teflon covering is a good choice. Twist the wires together with about six twists to the inch. Wind all three toroids with about ten turns. Here, it is convenient to hold the wires in place with two plastic tie wraps. Connect the wires as shown in Fig. 5A and measure the inductance of the coil. (See The ARRL Handbook for the Radio Amateur.) It should be 1.12 ^iH. If not, remove or add turns to obtain a value as close as possible to this number. Once this is done, tighten the tie wraps to hold the wires firmly in place. Recheck the inductance and spread or compress the winding's to decrease or increase the inductance, respectively, to get it to the exact value desired. In actual tests, the cores shown in Fig. 3 — A block diagram of three twisted-wire hybrids stacked to give four outputs of equal power and correct phase f o r a 4-element phased array. Fig. 6 required 7.5 turns spaced over about 80% of the core. Although you purchase parts that look identical, you will still need to measure the inductance because of variations in manufacturing. Now measure the capacitance (Cbifilar) between the wires of the twisted pair. This should be very small — about 30 pF. Since this capacitance is part of the capacitance between the halves of the hybrid (Fig. 5B), each capacitor in the hybrid must be: a = a = C R e q 2 t > l f i i a r For hybrid 1, C b i f i l a r is 32 pF, and for hybrids 2 and 3, Cbifilar is 28 pF. These values must be measured for your unit if other than no. 18 Teflon-covered wire is used. The capacitors required for each hybrid are therefore: 445 - 32 Hybrid Is 2 = 206 pF = CL = C2 Hybrids 445 - 23 2 and 3s 2 = 2 0 8 p F = C 3 = C 4 = C 5 = C 6 Select a parallel set of capacitors to obtain values close to this range. This, of course, must be done by measuring individual capacitors since they can vary from marked values. V/e now have the three hybrids of our phase shifter. Looking at Fig. 4, we can see how the three hybrids are connected into the final circuit for a switchable phased array for 40 meters. The last major design problem is the dummy load at port 3 of each hybrid. As stated earlier, the power into port 3 is zero if all Z < t > are equal. This is not the case in actual practice since no antenna will maintain a lsl SWR across the entire band. By experiment, it was found that less than 10 W of power went into port 3 of hybrid 1 if all four verticals had an SWR of less than 1.5:1. These tests were run with an SB-200 transmitter which was put-ting out about 600 W. It was therefore determined nnrr~\ "'l —p '""l ~ — ^ nonrT | Fig. 5 — fit A, measurement connections f o r hybrid connected as an inductor. At B, measure-ment connections f o r hybrid connected as a capacitor. Fig. 6 — The completed phase s h i f t e r . (Photo by Jim Wedge) +<jo° Fig. 4 — Complete circuit diagram for a 4-element switchable phase shifter for a 40-meter phased array. that a 20-W noninduetive resistor was needed for Rl, and 10-W resistors for R2 and R3. These were con-structed of parallel 2-W resistors. The value should be measured and resistors added or deleted to get as close to 50 ohms as possible. For Rl it was found that eleven 470-ohm, 2-W, 5% resistors gave almost 50 ohms exactly, while R2 and R3 consist of six 300-ohm and one 470-ohm resistors in parallel. Check your values before you solder them together. The finished product is shown in Fig. 6. Since toroids are used, the system can be extremely compact and internal shielding can be kept to a minimum. One switch position is used to feed one vertical directly. This was to give a "nondirec-tional" position for testing and listening purposes. If you plan to use one of the larger kW rigs or will not be able to get all your verticals close to 1.5:1 over the band, you should use higher wattage resis-tors for Rl, R2, and R3. There are some important points to remember. All lines from the phase shifter to the antennas must be the same length and have the same velocity factor. All antennas must be fed the same way. Different loading schemes at each antenna will, of course, change the phase. Four one-quarter wavelength verticals were spaced at four sides of my home. Two are wires running up trees and the other two are the remains of a TH6DXX. See Fig. 7 for an example of the antenna. I installed an 8-ft ground rod at each antenna since I have no room for radials. Also, because of space limitations, the distance between antennas is not exactly one-quarter wavelength nor exactly equal. Still, the gain closely approximates the theoretical value, as do the front-to-back and front-to-side ratios. Look in The Handbook, The ARRL Antenna Anthology, and The ARRL Antenna Book for a complete discussion of verticals and ground systems which could be used at your location. [Editor's Note: Fig. 8 shows the calculated radiation pattern of this array.] References \1 Anderson, Marian, WB1FSB, The ARRL Antenna Anthology, 1978, pp. 114-122. \2 Atchley, D. W., Stinehelfer, H. E., White, J. F., "360 Steerable Vertical Phased Arrays," QST, April 1976, p. 27. Wilkinson, Ernest J., "An N-Way Hybrid Power Divider," IRE Transactions on Microwave TTieory and Techniques, January 1960, p. 116. \4 Atchley, D. W., "A Switchable Four-Element 80-Meter Phased Array," QST, March 1965, p. 48. Fisher, R. E., "Broad-band Twisted-Wire Quadrature Hybrids," IEEE Transactions on Microwave Theory and Techniques, May 1973, p. 355. Fisher, Reed, "Twisted-Wire Quadrature Hybrid Directional Couplers," QST, January 1978, p. 21. \1 Application Notes, Amidon Associates. 50 Ohm 3d T \ 4 1 Coax Fig. 7 — Possible antenna and feed for each element of phased array. Fig. 8 — Calculated radiation pattern of the array shown in Fig. 1, for a zero-degree radiation angle. A perfect earth and equal currents in the four elements are assumed. Parts List LI: T-200-2 Amidon iron-powder toroid L2 and L3: T-130-2 Amidon iron-powder toroid Rl: Ten 470-ohm, 2-W carbon resistors R2 and R3: Six 300-ohm, 2-W carbon resistors a thru 06: See text. Switch: 6 pole, 5 position (need only 5 pole) Oentralab no. PA-2021 Metal Utility Cabinet: 5-1/4 x 3 x 5-7/8 inch, Radio Shack no. 270-253 Two views of the phase-shifter-network box. In the nondirectional switch position (ND), only Ant. 4 is fed. (Photos by Jim Wedge) LARAE—Line Array of Rotary Antennas in Echelon By A. J. F. Clement , W6KPC •Western Science Foundation, Rt. 1, Box 116, McFarland, CA 93250 As an ardent and devoted DXer, I have built, tested and operated four large antenna arrays. Each was completed near the peak of successive sunspot c y c l e s , i.e., 1948, 1958, 1989 and 1979.X1.2.3.4 These high-performance antenna arrays were always directed toward better performance on the 10-, 15-and 20-meter bands. During 1975 I decided to design a high-performance array-type antenna that could be adapted easily to both the low-frequency and high-frequency bands. As this article is being prepared, LARAE is being constructed on my antenna farm for the 75-meter band. The scheme offered here is intended to offer a much larger antenna aperture than can be realized from even a long-boom standard Yagi antenna. This array will also reduce QRM and QRN considerably, since it will have a very narrow beam width in azi-muth. LARAE's angle of take-off (vertical angle of radiation) will depend, principally, on its height above the earth, but not entirely so. Height con-sideration alone would indicate a vertical angle of radiation of 20 degrees (for a dipole antenna 3/4 wavelength high) but, since LARAE's individual components show considerable d i r e c t i v i t y , this 20-degree angle is depressed somewhat to approxi-mately 18 degrees above the horizon for this 75-meter application.NJL DX performance on 75 meters will generally be excellent with a wave angle of 18 degrees. See Fig, 1. How does LARAE work? Let's first review how gain is obtained with a collinear array of directive antennas (such as Yagis). The gain in dB over a dipole at the same height, for well-separated Yagis lined up in a row (see Fig. 2), will approximate the antenna gain of one Yagi, with roughly 3 dB to be added each time the number of Yagis in the array is doubled.VS_ To illustrate, consider a simple Yagi to have 7-dB gain; then two Yagis, suitably separated, will show a gain of 7 + 3 = 10 dB as a two-unit collinear array of Yagis. Another 3 dB can now be had by doubling the size of the two-yagi array. That is, we now place four Yagis in a line. Consequently, we have a 13-dB—gain array! Offhand, it would seem that such a four-unit system would act as an array in only two directions, since there are only two broadsides to a fixed straight line. A more careful examination of such an array will prove otherwise, and also reveal some very startling possibilities. (All gain figures are with reference to a dipole of the same height as the center height of the 3/4-wavelength Yagi array.) Let's return to the simple case of two 7-dB Yagis, each at 3/4-wavelength above ground. Also, let's make these Yagis into rotatable beam antennas. To start with, let's also place these two towers exactly one wavelength apart on a north-south line and point them 60 degrees to the east of north (toward Africa, from California). The Geometry of LARAE This arrangement of the two Yagis is shown in both elevation and plan view in Fig. 3A. Our next assumption is that we feed these two Yagis precisely 180 degrees out of phase; this is done easily by selecting proper lengths of feed lines from the final amplifier. We can readily visualize how far out from antenna B a wave will progress in exactly 1/2 cycle if part of a circle is drawn around the antenna (with circle radius equal to half a wave-length). It will travel a half wavelength through the air, as shown by the arrow BC in Fig. 3B. Let me now draw your attention to the line drawn through antenna A and tangent to the circle around antenna B. We will call this line the wave-3-5 7-0 14 0 - 210— FK£QU£HCY - MH 3 : ..; Fig. 1 — Vertical angle of arrival of sky-wave signals. Data below 7 MHz is extrapolated. Curve A, angle above which signals arrive 99% of the time. Curve B, angle above which signals arrive 5 0 f c of the time. Curve C, angle below which signals arrive 99% of the time. .75 A M ) Pig. 2 — Line-array of Yagi antennas in echelon. (B) .75. / K \R h t -a -A--\ 1 I R.F. FEED .15X A — TO .15 \ =<SA X W Cos © - _ = o . r Fig. 3 — TWo Yagi antennas spaced 1 wavelength apart and fed 180 degrees out of phase, pointed in a direction other than broadside. front line, since both Yagis are firing exactly perpendicular to this line AC. My next point is crucial, so follow me carefully. Were the two an-tennas to be fed in phase, then by the time the wave has traveled from antenna B to point C (1/2 cycle of time), the antenna at A is ready to reverse the current in its elements. The radiation field from the two antennas would now cancel each other in the forward direction of the two Yagis. Fortunately, we can cure this ambiguity simply by transposing the feed wires to antenna B, as shown in Fig. 3A. This will assure that a wave arriving at point C (from antenna B) will have the same phase as the wave just starting to leave antenna A. We now have the two antennas working together to produce an additive, common wavefront described by line AC. This composite wave will now progress outward in a direction perpendicular to line AC. Sometimes, knowing the answer beforehand helps to illustrate the solution to a geometry problem. This was the case when I chose 60 degrees as the common firing angle for our two antennas. Had we not known what it would turn out to be, we could have found out as follows: c o s i n e 0 = x /2 = x/2 = = 1 = 0 . 5 0 A + i X 2 A 2 2 2 (Eq. 1) The angle whose cosine = 0.50 is 60 degrees, or cos -0 = 60 degrees. What is the meaning of all of this geometry? Simply stated, we have now discovered that two high-performance rotary antennas can be positioned for complete array-effect additive gain for two separate angles from line AC, namely 60 degrees and 90 degrees. Also, when these two anten-nas are both set at zero degrees in azimuth, they will be additive if they are fed in phase. This gives us direction no. 3. Not bad, so far! (I will discuss this third case later on.) By now you are asking yourself, "What if we line up three or even four Yagis on a single straight line, all one wavelength apart?" The answer is, "More gain at the same three angles as before, that is, 90, 60 and 0 degrees." For instance, four Yagis (each with 7.0 dBd gain) would show an approx-imate gain of 7.0 + 3.0 + 3.0 = 13 dB over a dipole. At 3.8 MHz, this would be fantastic! The next question a DXer might ask is, "How can we get these separate rotaries to work together at angles besides 60 degrees?" (We will ignore 0 and 90 degrees for the time being). Well, let's try spacing our two Yagis out even more, to 1-1/2 wavelengths apart. See Fig. 4. cox .u = 48-iq D AUT'I Fig. 4 — Two Yagis spaced 1.5 wavelengths apart. Two "angles of joy" exist. (See Text.) Now our geometry has changed a bit, but it still looks familiar. We have two "angles of joy," 61 and 02. 01 has a vector angle of less than 60 degrees, and 02 seems somewhat greater than 60 degrees. Again, let's fall back on our high school geometry: c o s i n e 01 01 c o s i n e 62 1 3 0.667 (Eq. 2) -1 cos 0.667 = 48.19 degrees ±12, = _L = 3/2X -1 I = 0.333 3 (Eq. 3) 02 = cos A 0.333 = 70.53 degrees Now let's examine Fig. 5A. A look at this draw-ing should definitely raise the blood pressure a bit and rev up the old ticker of a true DXer! As antici-pated, we now have three angles of joy. Without de-lay, let's calculate them: c o s i n e 01 = 3 x 1 " = 3_L = 1 = 0.75 2. 4 4 4 2 (Eq. 4) 01 = cos-1 0.75 = 41.41 degrees c o s i n e 02 = 2 x x / 2 = 4 x A A\ 2 2 4 0.5 02 = cos-1 0.5 (Eq. 5) 60.00 degrees c o s i n e 03 = -Z2 = _ . = 1 4 x A 2 4?. 0.25 (Eq. 6) 03 = cos"1 0.25 = 75.52 degrees I expanded the spacing between Yagis, all the way out to three wavelengths. A general rule devel-oped: Not counting 0 degrees or 90 degrees, the number of discrete, useful angles of joy are related to the spacing between antennas by the equation: N = S - 1 (Eq. 7) where S is the number of 1/2-wavelength spaces between antennas. (See Figs. 2 and 6.) An extension of our geometry will produce Figs. 6 and 7, along with two more sets of angles. For Fig. 6 we have: For Fig. 7 we have: 36.87 degrees 53.13 degrees 66.42 degrees 78.46 degrees 33.56 degrees 48.19 degrees 60.00 degrees 70.53 degrees 80.41 degrees S =5 S =6 Remember, S always equals the number of 1/2-wavelength spacings between antennas. LARAE's Radiation Patterns At about this point in LARAE's development, I began to wonder what the azimuthal pattern would look like. I had to know the half-power beam width, to see if the array would cover the required DX headings in a given azimuthal quadrant, such as 0 degrees to 90 degrees. Would this beam width change substantially as I went from one angle of joy to the next? Professor Simon Lheto, 0H80S, is a true DXer and one of the world's best antenna men. Simon agreed to program LARAE's array characteristics on the big computer at the University of Oulu as free time became available. After a few months Simon sent me beautiful pattern printouts, accurate in every detail. The results were both good and bad. (See Figs. 8 and 9.) The broadside pattern was only 8 degrees wide at the half-power points, and there were two somewhat sig-nificant sidelobes at 22 degrees either side of the main lobe. This was for a LARAE of three 7-dBd Yagis at 3-wavelength antenna spacing. Our "cannon" had turned into a "laser-gun," with large gaps in the azimuth not covered! The good news was approx-imately 12.0 dB of gain over a dipole. Also, there were no significant lobes to the rear half of the pattern. About this time Simon was able to visit me, and we discussed how to improve LARAE. We finally de-cided that we would increase the width of the main lobe by decreasing the spacing of LARAE. This meant a decrease in antenna spacing from 3 toward 2 wavelengths. The final spacing chosen was 2.103 wavelengths. (I'll explain this odd value later.) Theory also indicated that the sidelobes at plus and minus 22 degrees would be attenuated, as a further benefit. Simon returned to the computer in northern Finland, and I returned to the drawing board. Our last few hours together resulted in our agreeing to incorporate delay-line phasing into the LARAE. The main idea was to wobble each of the above calculated angles (vector directions) of Fig. 5A a couple of steps, both minus then plus, as we Pig. 5 — Antennas spaced 2 wavelengths apart. c o s «-7S us".IS' C3S 4 as~' -5 ,M-j Cej"' .41 66.^2° -a.i - COS'' -Z-- 7lHb° 5 Graph of Phase Angle of Weights The graph in Fig. 10A was prepared by Simon's computer crew. This graph is the result of an X-, Y-and Z- coordinate plot of the advancing unified wave front when, and only when, the three antenna's phases have been adjusted to provide maximum field intensity at a far-away place in a given direction. Weight, as used here, defines the combination of phase and power at each antenna. The power part is simple; we chose to put 1/3 the RF power into each antenna. We also chose to vary the phase in incre-ments of 45 degrees to step from one angle of joy to the next. Now is the time that the reader fix in his or her mind how LARAE is steered. First, the phase angle on the end antennas is advanced on one and retarded on the other by 22-1/2 degrees. This gives a total change, between the two end antennas, of 45 degrees (1/8 wavelength). Once this phasing change has been made, then all three antennas are rotated to a common angle where the distant-point field strength is maximized. This specific angle can be computed from an expanded version of Fig. 5A, as shown in Fig. 5B. Basically, we insert new partial circles about point B showing wave progressions in increments of 1/8 wavelength. We can then solve arccos x for every total change of 1/8 wavelength. Since we have the antennas spaced by 2 wavelengths, we have 16 angles that we can solve for as separate angles of joy. This, of course, totals to sixteen firing angles for a 90-degree quadrant on a great-circle map. The complete success of any project depends on close attention to important details. One such detail has to do with how LARAE will work when all the Yagis are firing precisely down the line of the towers. In this case the phasing should be zero degrees (firing line = tower line). Fig. 11 shows a side view of the three towers and depicts the take-off angle of main-lobe radiation at 18 degrees from each of the three antennas, all being fed in phase. Simple geometry shows that the three towers must be spaced 2.103 wavelengths apart in order for the advancing wave front from each antenna to combine precisely with those from the other two and produce an overall wave front with a take-off angle of 18 degrees above the earth. At this unique angle, LARAE is working as a stacked array, for all practical purposes. The vertical distance between the energy beams is 0.65 wavelength, as shown in Fig. 11. As LARAE turns away from this unique zero-degree angle, it grad-ually goes from a stacked array to a collinear array. If you ponder this awhile, you will conclude that the graph of phase angle of weights (Fig. 10A) must be solved as a problem in three-dimensional geometry in order that each f i r i n g angle be precisely computed. Using the plane of the earth's surface only would introduce serious errors for all angles near 0 = zero. Simon's crew did use the solid geometry! (The actual math and solid geometry used to compute these sixteen solid angles are not shown in this article.) The horizontal components of each of these solid angles are shown on the graph in Fig. 10A as the abscissa, while phase-angle differ-ences between end antennas are shown as ordinates. Intercepts of the five horizontal lines on the graph delineate the sixteen firing angles (optimum angles) for the array, and also show the required phase dif-ference between the end antennas for each angle. Engineering Implementation The towers used at W6KPC are the 180-ft Tri-Ex "Clementower," which are rotated from the bottom with computer-controlled stepping motors. The amount (angle) of rotation is controlled by an Intel 8085-2 central processing unit, which is part of an Intel ICS580main-frame computer equipped with ROM, RAM, and analog-to-digital converters (and vice-versa) as well as clocks, printers, keyboard, video monitor, and so on. This industrial control computer is pro-grammed to (1) rotate all Yagis, in unison, to the c t s 8' " J cos'-833 ll^se.' ccs 0, . i eos-'M'iAJl' / !£«.»»; coi".S COS«+ <|is . 333 ; ear'.i) j •!(•!• 1 C o S m '.IWio- t Pig. 8 — Radiation pattern of 3-wavelength, 3-Yagi LARAE firing broadside. Fig. 9 — Radiation patterns 3-wavelength, 3-Yagi LARAE. Only major lobes are shown in one quad-rant for different amounts of beam steering and Yagi rotation. preprogrammed DX angle heading (as entered on the keyboard in real time or as called for by the clock); (2) set all relays on all feed systems to give the proper weights (phases) to the feed lines of the two end Yagis for the heading indicated; (3) manage all interlocks to prevent hot switching of RF relays; and (4) measure the signal strength (fre-quently) while in the receive mode, and then in-stantly (and automatically) wobble the main lobe to see if signals are perhaps arriving on a skewed path. If skewed path is detected, the computer will then rotate all Yagis to the correct heading for maximum incoming signal strength. { Consideration is being given to a voice-recognition command unit, coupled to a foot switch, which would instruct the computer to change the LARAE heading on voice command. Thus, when the foot switch is closed and the word "Finland" is spoken, the whole array would obediently rifle in on a bearing of 20 degrees from central California). Fig. 10 — At A, phase-angle of weights for an antenna spacing of 2.1 wavelengths. At B, radi-ation patterns. ^ ^ z - J .
J .
A I Z.MjJj Z70' 0/:Ly K'ajoz (B) Antennas and Feeds The LARAE at W6KPC was designed for 3.8 MHZ as a center frequency. The three Yagi antennas are KLM four-element Yagis constructed on a 76-foot boom with a gain of 7.25 dB over a dipole. The drive-point impedance is nonreactive at 200 ohms, using a two-element log-periodic cell as the double driven dipole. Each Yagi antenna is equipped with a 4:1 balun to step the antenna's balanced 200-ohm feed impedance point down to 50 ohms unbalanced. This allows the use of 50-ohm coaxial cable between the three towers and the transmitter. Fig. 12 shows how the three antennas are fed as an array. Note that the two end antennas are fed through the two relay boxes, while the center anten-na is fed directly as the pivot antenna. The same system that was developed to feed the collinear Yagi Sextet's three separate bays (levels)YL is utilized here to feed LARAE's three separate antennas. Notice the manner in which K6SSJ and W6KPC devised the quarter-wave matching sections so the transmitter constantly looks at a 50-ohm feed-line impedance.\8_ Table 1 is a tabulation of phase-angle weights and relay settings. Refer to Fig. 13 to see how the relay activations shown in the table insert the various line sections, causing the desired delays in the end antennas, no. 1 and no. 3. Notice, partic-ularly, that line sections are used two at a time to give a total phasing step of 45 degrees, each of the two sections representing 22-1/2 degrees. LARAE Radiation Pattern and Gain The KLM four-element Yagi used as the unit antenna in LARAE has somewhat shortened (folded) elements approximately 90 feet long, and a boom length of 76 feet. It uses a broadbanded feed system made famous by KLM, the double driven dipole, or two-element log cell. KLM specifies a power gain of 7.25 dB over a dipole for this antenna. The hori-zontal-plane pattern for this antenna is shown in Fig. 14. I developed an enormous respect for this Yagi three winters ago. I listened all winter long to Arnold, W2HCW, work the world with one, receiving incredible reports. W2HCW dominated the 75-meter band from the U. S. A. at that time. OF L 1 ' I •SCC. FIG. XiL/ir-SeX FOX INC :£fjrAL L , fairj.a rvfc X, SO^-S£S. SO-^i To Xnr/rrt/;. (actually 47-n-) Fig. 12 — Feed arrangement of a 3-Yagi LARAE. Fig. 8 shows the typical interferometer grating pattern of three widely spaced Yagi antennas when used in a broadside array.\9 This is the broadside pattern (0 = 90 degrees) for the antenna diagrammed in Fig. 7. Fig. 9 shows the main lobes of 01 through €5 as calculated from Fig. 7. Also, there is a fat north lobe and a skinny east lobe for the stacked (north) and broadside special conditions. This completes the family of seven lobes shown in Fig. 9. Remember, however, that the five angles cal-Fig. 11 — LARAE firing down the tower line. This is the "stacked" node of operation. FROM XMIRREN Fig. 13 — Diagram of r e l a y box f o r a 3-Yagi LARAE. Two boxes are used, one a t each end Yagi. Fig. 14 — Approximate horizontal radiation pattern of a 3.8-MHz Yagi ( s i n g l e antenna). Performance data i s suirmarized below. Manufacturer: KLM Measured gain: 7.25 dBd Half-power beamwidth: 70° Front-to-back r a t i o : 25 dB Front-to-side r a t i o : Greater than 30 dB culated in Fig. 7 are not the result of three-dimensional solid geometry, but of only plane geometry, for the sake of simplifying the theory of how we get natural phase coincidence at certain angles. When Simon ran these computations on the computer we came up with considerable variance in these angles when solid geometry was applied. Summary Sixteen angles, it turns out, give excellent pattern overlapping such that there is never a drop in far-point field strength versus beam heading of more than 1 dB. This is shown by how the patterns intersect each other at the bottom of Fig. 10A. As a net result, we have a practical antenna array that will show a real-world gain of approximately 12 dB over a dipole at the same antenna height. Also, the antenna array is steerable over 360 degrees of TO ANTSNNA azimuth in a step-wise manner that approaches the "all azimuths available" advantage of a single Yagi rotary-beam antenna. Fig. 1 shows that obtaining a radiation angle /FYLC A DSLA Y- below 15 degrees on 75 meters is not cost effective, -L/NZ SUE T/OAJS. since signals arrive above 14.5 degrees for 99& of the time.YLQ. A good height for a horizontally polarized 75-meter LARAE would appear to be 0.75 wavelength, which is 180 feet at 3.8 MHz. This height will give a vertical radiation angle of 18 degrees.\11.12 This 18-degree angle will come as no surprise to veteran DXers who already know that one does not need as low a radiation angle on 75 as is necessary on 10, 15, and 20 meters. I want to emphasize that the points on the graph of Fig. 1 that fall on the 3.5 MHz ordinate are extrapolated, as shown by the dotted lines. There is l i t t l e reason, however, to challenge the validity of these 3.5-MHz points. Earlier in this article reference was made to the fact that the LARAE would reduce both QRM and QRN. Some contest operators will argue for a broad azimuth pattern accompanied by a narrow vertical pattern.\13_ This may be okay on the 20- to 10-meter bands, but is patently undesirable on 75 meters. The reasons are simple: (1) Static crashes are much worse on 75 meters than on the higher frequency bands, and this kind of QRN comes from many direc-tions. (2) Nonamateur QRM and QRN may come from any direction. (3) Omnidirectional amateur QRM is often fierce in the relatively narrow DX portion of 75 meters. (4) High-angle radiation should not be suppressed unduly since we have previously noted the likelihood of high-angle arrival of DX signals on 3.5 MHz. (This 3/4-wavelength-high array does have a high-angle, vestigal, minor lobe, which is useful for close-in DX QSO's). Epilog As this article was being readied for mailing to ARRL Headquarters, Simon, 0H80S and I decided to make one more computer run. This run was for a Yagi antenna spacing of only 1.05 wavelengths and using three antennas, as before. The results were wonder-ful! The reduction in array gain is negligible, the minor lobes all but disappear, and each directional lobe is fattened to the extent that eight beams fill a quadrant with excellent overlap. This is all illustrated in Figs. 15 (which shows the geometrical math) and 16. Fig. 16A shows a polar plot of the angles calculated and shown by Fig. 15. Curves are added to this plot for 0 and 90 degrees, plus the first lobe in each neighboring quadrant, to illus-trate symmetry for a whole circle. Fig. 16B is included to show how minimal the worst case really is (90 degree, broadside firing), as far as inter-ferometer-type grating lobes are concerned. Fig. 17 shows a graph of phase angle of weights for 1.05 wavelengths. Acknowledgements The basic idea for the LARAE occurred to me in the early 1970s, when I made a few sketches. Calcu-lations did not start until about 1975, when I tried the idea with a couple of professional antenna-array engineers. These engineers were intrigued with the LARAE concept and urged me to pursue the idea further. When Simon, 0H80S, visited me in the mid-703, I showed him the drawings, charts and calcu-lations. He became enthused and volunteered to run some computer calculations and pattern plots. Simon has been a great help to me in properly sorting out some of the finer details and concepts that were involved in understanding the LARAE. I wish to thank one of Simon's associates, Rauno Suikola, for his excellent work on the computer programming associated with this project. References \1 Clement, A. J. F., "The Yagi Dagi", QST, Sep-tember 1951. \1 Clement, A. J. F., "The Driven Beast" QST, May 1958. Clement, A. J. F., "The Collinear Yagi Quartet" QST, November 1969. \4 Clement, A. J. F., "The Collinear Yagi Sextet" QST, June 1980. Bachelor, W. B., "Combined Vertical Direc-tivity" QST, February 1981. \£ The The ARRL Antenna Book, 13th Ed., p. 135, Fig. 4-9. \1 See Ref. 4. Kraus, John D., Antennas (McGraw-Hill), p. 437. Jasik, H., Antenna Engineering Handbook (Mc-Graw-Hill), p. 29-20. Vlil See R e f . 6, pages 18-19 and Table 1-1. \11 See Ref. 6, p. 49, Fig. 2-42. \12. See Ref. 5. \1£ Lawson, James L., "Yagi Antenna Design," Ham Radio p. 23, November 1980. C^fc®, • = 0.17 , -m, eoi''o.g1s> 0 »'«.7f. toS'1 0< 7 > 0 " 4 / . 4" ' i " ..!. . ; , : CM e,v£i-.;0.M£ M, C°i~'0.i2S CMal 'if io/V.o --, an.'1 e.Sio Ut.o' ; . i • ' 1 '• C M .
® < u £»A."'I>.)7R. W . O " Ctaat,' S i , c.tso -tr 7s.S' tK J Cat-o, = i - -n> C^r'a.nf' SZ.S' 7 s-a ' Fig. 15 — LARAE with 1.05-wavelength spacing and phasing such that radio energy arriving at anten-na A lags energy arriving at antenna B by mul-tiples of 1/8 to 7/8 wavelength. 18' oJt peuer p°/n4s. W Fig. 16 — Radiation patterns of LARAE with 1.05-wavelength spacing and increments of 1/8-wavelength phasing. Height is 3/4 wave-length above ground. (ilOA "Si D £ F i H'VCr) (B) Fig. 17 — Phase-angle of weights for an antenna spacing of 1.05 wavelengths. TABLE 1 TABLE OF PilASE-ANGLES AND RELAY SETTINGS (Antennas Spaced At 2.10 A ) Horiz. Component Azimuth -- Angle, 9 Ant. "A" Phase, (Degrees) Ant. "B" Phase, (Degiees) Relays - Settings, Ant. "A" 1 2 3 4 5 6 7 Relays - Settings, Total Ant. "D" (A + D) 1 2 3 4 5 6 7 Phase 616 (East) = 90 00° -0--0-1 1 2 2 2 2 1 2 2 -0-015 - £6 4 2° - 22 5 + 22 5 2 1 --- - 1 2 2 2 1 - 2 2 4 5° 914 = 82 80° - 4 5 0 + 4 5 0 2 2 1 - - - 1 2 2 1 -- 2 2 90° 813 = 79 19° - 67 5 + 6 7 5 2 2 2 1 - - 1 2 1 --- 2 135° 912 = 75 50° - 90 0 + 90 0 1 ---- 2 2 2 2 2 2 1 -1 180° Oil = 71 79° -112 5 + 112 0 2 1 -- - 2 2 2 2 2 1 - -1 225° 610 -68 00° -135 0 + 135 0 2 2 1 - - 2 2 2 2 1 - - -1 270° 6 9 = 64 06° -1 57 5 + 157 5 2 2 2 1 - 2 2 2 1 ----1 315° e 8 = 60 00° 180 0 180 0 2 2 2 2 1 2 2 1 1 360° 8 7 = 55 77° + 22 5 - 22 5 2 2 2 1 - 2 2 2 1 -- - -1 405° e 6 = 51. 30° • 45 0 - 4 T > 0 2 2 1 - - 2 2 2 2 1 - --1 450° 8 5 = 46 57° + 67 5 - 67 5 2 1 --- 2 2 2 2 2 1 --1 495° 9 4 41 40° + 90 0 - 90 0 2 2 2 2 1 - 1 1 --- - 2 2 540° 8 3 = 35 66° + 112 5 -112 5 2 2 2 1 - - 1 2 1 - - - 2 2 585° 9 2 = 29 00 + 1 35 0 -1 35 0 2 2 1 - - - 1 2 2 1 - -2 2 630° 9 1 = 20 36° 1 57 5 -157 5 2 1 - -- -1 2 2 2 1 - 2 2 675° 9 0 (Nor th) = -0-180 0 18C 0 1 1 2 2 2 2 1 2 2 720° NOTF.: Tionsmission line to 3ntenna "A" is 1/4 A longer then to middle antenna. Also, line to antenna "B" is 1/4 \ shorter than to middle antenna. Multiband Antennas L A Great 10 Through 40 Portable Antenna By Edward L. Henry, K0GPD 25 Gold Run Drive, St. Peters, MO 63376 For years I have seen local elubs construct Field Day antennas. Two or more vehicles were usually necessary to transport the array of antenna equipment to the site. Then a group of people would labor for several hours to erect the antennas. It can be a big, tiring job. Obviously, this is not the way to do it. In a real emergency, you do not want to be delayed. You need the antenna now! I designed a light-weight 10- through 40-meter inverted-V antenna that can be transported easily in the trunk of a small car. The elements are full-length dipoles (no traps) and can be erected by one man and ready to operate in 15 minutes. You do not need trees or existing structures to set it up, and it handles full legal power. The antenna performs so well that I use it for other occasions such as demonstrating Amateur Radio to schools and scout meetings. A park on a Sunday afternoon is a pleasant place to set up and talk to the world. Depending on where you are, spectators will be attracted and it is a good way to promote our hobby. It can even be set up on concrete parking lots, where nails can be driven into the cracks for anchors. An antenna tuner is not necessary, but using one will give you a perfect match. All construction information is given on the drawings. Most materials are inexpensive and available at hardware stores. The antenna pole consists of six standard Radio Shack 1-1/4 inch diameter TV mast sections. The mast sections are 4 to 5 feet long and arc made of aluminum or steel. Aluminum masts are preferred because of their light weight. Insulated wire is used for the antenna elements in the event one element touches another when the wind blows. I have experimented with many b a t t e r i e s (lead-acid and NiCcl) for emergency power. The marine type battery used by motorboaters outperforms them all. When you use a 100- to 2U0-W mobile type transceiver, a 12-V marine battery can operate 6 or more hours with 50% talk time. A 3-A charger is great for recharging the battery. The charger and a specific gravity tester can be taken on vacations. Remember to keep sparks away from a charging bat-tery! (Using a newspaper, always fan the air away from the battery and have equipment turned off before making any electrical connections to or near the battery.) No matter how many antennas you own, this portable multiband antenna will be a welcome SOLDER 4 WIRES TO ONE COMMON LUG (ATTACH TO BALUN WITH WING-NUT) J ~ ONE HALF OF INVERTED VEE ANTENNA USE NO. 14 INSULATED BRAIDED ANTENNA WIRE (NOT DRAWN TO SCALE) 1/4 IN. 1 3/4 IN. (TYPICAL) CUT AND FILE OFF PROTRUDING NAIL 1/2 IN. WIRE 8RAD NAIL THROUGH ANT. WIRE (PREDRILL TIGHT HOLE FOR NAIL) 1/8 IN. DIA. NYLON ROPE addition. Except for the pole sections, the antenna and hardware fit into a one-cubic-foot cardboard box. Keep it stored in the corner of your garage and it will be ready to go on a moment's notice. 1 TO 1 8ALJN (I KW) 0 0 NOT PUT METAL AROUND BALUN (USE PLASTIC CLAMPS) AOO STOP SCREW ON EACH POLE SECTION TO PREVENT BINDING WHEN ASSEMBLED ADD 3 CLIPS AT 2 0 FOOT HEIGHT "SMALL QUICK DISCONNECT HOOKS (3 REQUIRED) ONE MAN SET-UP PROCEDURE WARNING: DO NOT 0 0 THIS NEAR STORMY WEATHE R SET ANT. GUYS A F T E R POLE IS UP COAX L E A O - I N "yT. The G5RV Multiband Antenna ... Up-to-Date# By Louis Varney, G5RV 82 Folders Lane, Burgess Hill, W. Sussex RH15 J D D X , United Kingdom # Adapted from an article of the same title in Radio Communication, July 1984, pp. 572-575. The G5RV antenna, with its s p e c i a l f e e d e r arrangement, is a multiband center-fed antenna capable of efficient operation on all HF bands from 3.5 to 28 MHz. Its dimensions are specifically designed so it can be installed in areas of limited space, but which can accommodate a reasonably straight run of about 102 ft f o r the f l a t - t o p . Because the most useful radiation from a horizontal or inverted-V resonant antenna takes place from the center two-thirds of its total length, up to one-sixth of this total length at each end of the antenna may be dropped vertically, semi-vertically, or bent at a convenient angle to the main body of the antenna without significant loss of effective radiation e f f i c i e n c y . For installation in very limited areas, the dimensions of both the flat-top and the matching section can be divided by a factor of two to form the half-size G5RV, which is an ef-ficient antenna from 7 to 28 MHz. The full-size G5RV will also function on the 1.8-MHz band if the sta-tion end of the feeder (either balanced or coaxial type) is strapped and fed by a suitable matching network using a good earth connection or a coun-terpoise wire. Similarly, the half-size version may be used on the 3.5- and 1.8-MHz bands. In contradistinction to multiband antennas in general, the full-size G5RV antenna was not designed as a A/2 dipole on the lowest frequency of opera-tion, but as a 3 A / 2 center-fed long-wire antenna on 14 MHz, where the 34 ft open-wire matching section functions as a l t l impcdanee transformer. This enables the 75-ohm twin lead, or 50/80-ohm coaxial cable feeder, to see a close impedance match on that band with a consequently low SWR on the feeder. How-ever, on all the other HF bands, the function of this section is to act as a "make-up" section to accommodate that part of the standing wave (current and voltage components) which, on certain operating frequencies, cannot be completely accommodated on the flat-top (or inverted-V) radiating portion. The design center frequency of the full-size version is 14.150 MHz, and the dimension of 102 ft is derived from the formula for long-wire antennas which is: Length (ft) = 492(n - Q.Q5) MHz = 492 x 2.95 14.15 = 102.57 ft (31.27 m) where n = the number of half wavelengths of the wire (flat-top) Because the whole system will be brought to resonance by the use of a matching network in practice, the antenna is cut to 102 ft. As the antenna does not make use of traps or ferrite beads, the dipole portion bceomcs pro-gressively longer in electrical length with increas-ing frequency. This effect confers certain advan-tages over a trap or ferrite-bead loaded dipole because, with increasing electrical length, the major lobes of the vertical component of the polar diagram tend to be lowered as the operating fre-quency is increased. Thus, from 14 MHz up, most of the energy radiated in the vertical plane is at angles suitable for working DX. Furthermore, the polar diagram changes with increasing frequency from a typical A/2 dipole pattern at 3.5 MHz and a two A / 2 in-phase pattern at 7 and 10 MHz to that of a long-wire antenna at 14, 18, 21, 24 and 28 MHz. Although the impedance match for 75-ohm twin lead or 80-ohm coaxial cable at the base of the matching section is good on 14 MHz, and even the use of 50-ohm coaxial cable results in only about a 1.8:1 SWR on this band, the use of a suitable matching network is necessary on all the other HF bands. This is because the antenna plus the match-ing section will present a reactive load to the feeder on those bands. Thus, the use of the correct type of matching network is essential in order to ensure the maximum transfer of power to the antenna from a typical transceiver having a 50-ohm coaxial (unbalanced) output. This means unbalanced input to balanced output if twin-wire feeder is used, or unbalanced to unbalanced if coaxial feeder is used. A matching network is also employed to satisfy the stringent load conditions demanded by such modern equipment that has an automatic level control system. The system senses the SWR condition present at the solid state transmitter output stage to protect it from damage, which could be caused by a reactive load having an SWR of more than about 2:1.\1 The above reasoning does not apply to the use of the full-size G5RV antenna on 1.8 MHz, or to the use of the half-size version on 3.5 and 1.8 MHz. In these cases, the station end of the feeder conduc-tors should be "strapped" and the system tuned to resonance by a suitable series-connected inductance and capacitance circuit connected to a good earth or counterpoise wire. Alternatively, an unbalanced-to-unbalanced type of matching network such as a T or L matching circuit can be used.\2 Under these conditions the flat-top (or inverted-V) portion of the antenna, plus the matching section and feeder, function as a Marconi or T antenna, with most of the effective radiation taking place from the vertical, or near vertical, portion of the system; the flat-top acts as a top-capacitance loading element. How-ever, with the system fed as described above, very effective radiation on these two bands is obtainable even when the flat-top is as low as 25 ft above ground. Theory of Operation The general theory of operation has been ex-plained above. The detailed theory of operation on each band from 3.5 to 28 MHz follows, aided by figures showing the current standing wave conditions on the flat-top, and the matching (or make-up) section. The relevant theoretical horizontal plane polar diagrams for each band may be found in any of the specialized antenna handbooks. However, it must be borne in mind that: (a) the polar diagrams gen-erally shown in two dimensional form are, in fact, three dimensional (i.e., solid) figures around the plane of the antenna; and (b) all theoretical polar diagrams are modified by reflection and absorption effects of nearby conducting objects such as wire fences, metal house guttering, electric wiring systems, and even large trees. Also, the local earth conductivity will materially affect the actual polar radiation pattern produced by an antenna. Theoret-ical polar diagrams are based on the assumptions that an antenna is supported in "free space" above a perfectly conducting ground. Such conditions are obviously impossible of attainment in the case of typical amateur installations. What this means in practice is that the reader should not be surprised if any particular antenna in a typical amateur location produces contacts in directions where a null is indicated in the theoretical polar diagram, and perhaps not such effective radiation in the directions of the major lobes as theory would indicate. 3.5 MHz: On this band each half of the flat-top, plus about 17 ft of each leg of the matching section, forms a foreshortened or slightly folded up A / 2 dipole. The remainder of the matching section acts as an unwanted, but unavoidable reactance between the electrical center of the dipole and the feeder to the matching network. The polar diagram is effectively that of a A/2 antenna. See Fig. 1. despite this, by using a suitable matching network, the system loads well and radiates very effectively on this band. See Fig. 2. 102ft (31 1m) Center of -|-dipole_ on 3.5MHz Reactive load ( R ± j X ) --Current standing wave • 3 A f t (10.34m) o/w matching section Pig. 1 — Current standing-wave distribution on the G5KV antenna and matching section at 3.5 MHz. The antenna functions as a 1/2 dipole partially folded up at the center. 7 MHz: The flat-top, plus 16 ft of the matching section, now functions as a partially folded up two half waves in phase antenna producing a polar dia-gram with a somewhat sharper lobe pattern than a A /2 dipole because of its collinear characteristics. Again, the matching to a 75-ohm twin-lead or 50/80-ohm coaxial feeder at the base of the matching section is degraded somewhat by the unwanted react-ance of the lower half of the matching section, but, "Current standing wave Center of a n t e n n a 2 y in phase Reactive l o a d ( R ± j X ) ^ Fig. 2 — Current distribution on the antenna and matching section at 7 MHz. The antenna now func-tions as a collinear array with two half waves fed in phase. 10 MHz: On this band the antenna functions as a two half-wave in-phase collinear array, producing a polar diagram virtually the same as on 7 MHz. A reactive load is presented to the feeder at the base of the matching section but, as for 7 MHz, the per-formance is very effective. See Fig. 3. •Current standing wave Reactive load <R± Fig. 3 — Current standing-wave distribution on the antenna and matching section at 10 MHz. The antenna functions as a collinear array with two half waves fed in phase. 14 MHz: At this frequency the conditions are ideal. The flat-top forms a 3A/2 long center-fed antenna which produces a multilobe polar diagram with most of its radiated energy in the vertical plane at an angle of about 14 degrees, which is effective for working DX. Since the radiation re-sistance at the center of a 3A/2 long-wire antenna supported at a height of A / 2 above ground of average conductivity is about 90 ohms, and the 34-ft match-ing section now functions as a 1:1 impedance trans-former, a feeder of anything between 75 and 80 ohms characteristic impedance will see a nonreactive (i.e., resistive) load of about this value at the base of the matching section, so that the SWR on the feeder will be near 1:1. Even the use of 50-ohm coaxial feeder will result in an SWR of only about 1.8:1. It is assumed here that 34 ft is a reasonable average antenna height in amateur installations. See Fig. 4. 18 MHz: The antenna functions as two full-wave antennas fed in phase; it combines the broadside gain of a two-element collinear array with a some-what lower zenithal angle radiation than a A/2 dipole because of its long-wire characteristic. See Fig. 5. 21 MHz: On this band the antenna works as a long wire of five halfwaves, producing a multilobe polar diagram with effective low zenithal angle radiation. Although a high resistive load is pre-sented to the feeder at the base of the make-up I -1 -^ ^ C u r r e n t standing wave Resistive load approx 90J2' High 2 load (resistive) Fig. 4 — Current standing-wave distribution on the antenna and matching section at 14 MHz. In this case the antenna functions as a center-fed long wire of three half waves out of phase. The matching section now functions as a 1:1 impedance transformer, presenting a resistive load of ap-proximately 90 ohms at the lower end. Pig. 6 —Current standing-wave distribution on the antenna and matching section at 21 MHz. On this band the antenna works as a long wire of five half waves. The base of the matching section presents a virtually nonreactive high impedance load to the feeder. i i < i » / / \ / / \ / / H i g h Z load (slightly reactive) Fig. 5 — Current standing-wave distribution on the antenna and matching section at 18 MHz. The antenna functions as two full-wave antennas, slightly folded up at the center, fed in phase. section, the system loads well when used in con-junction with a suitable matching network and radiates effectively for DX contacts. See Fig. 6. 24 MHz: The antenna again functions effectively as a 5 A/2 long wire, but because of the shift in the positions of the current antinodes on the flat-top and the matching section (Fig. 7), the matching or make-up section now presents a much lower resistive load condition to the feeder connected to its lower end than it does on 21 MHz. Again, the polar diagram is multilobed with low zenithal angle radiation. 28 MHz: On this band, the antenna functions as two long-wire antennas, each of three half waves, fed in phase. The polar diagram is similar to that of a 3A/2 long-wire, but with even more gain over a A/2 dipole because of the collinear effect obtained by feeding two 3 A/2 antennas, in line and in close proximity, in phase. See Fig. 8. Construction The Antenna The dimensions of the antenna and its matching section are shown in Fig. 9. If possible, the flat-top should be horizontal and run in a straight line, and should be erected as high as can be above ground. In describing the theory of operation, it has been assumed that it is generally possible to erect the antenna at an average height of about 34 ft, which happens to be the optimum height for the antenna at 14 MHz. Although this is too low for optimum radiation efficiency on 1.8, 3.5, and 7 MHz for any horizontal type of antenna, in practice few amateurs can install masts of the optimum height of \ / A I i ' X Resistive load approx 90|l0QJ3 Fig. 7 — Current standing-wave distribution on the antenna and matching section at 24 MHz. The antenna functions as a long wire of five half waves. H i g h Z load (slig'htly reactive) Fig. 8 — Current standing-wave distribution on the antenna and matching section at 28 MHz. The antenna functions as two long-wire antennas each of three half waves length, fed in phase. A very effective form of antenna giving good multilobe, low zenithal angle, radiation. half a wavelength at 3.5 or 7 MHz, and certainly not at 1.8 MHz. If it is not possible to accommodate the 102-ft top in a straight line because of space limitations, up to about 10 ft of the antenna wire at each end may be allowed to hang vertically or at some conven-ient angle, or be bent in the horizontal plane, with little practical effect on performance. This is because, for any resonant dipole antenna, most of the effective radiation takes place from the center two-thirds of its length where the current antinodes are situated. Near each end of such an antenna, the amplitude of the current standing wave falls rapidly to zero at the outer extremities; consequently, the effective radiation from these parts of the antenna is minimal. The antenna may also be used in the form of an inverted V. However, it should be remembered that for sueh a configuration to radiate at maximum efficiency, the included angle at the apex of the V should not be less than 120 degrees.\2 The use of 14 AWG enameled copper wire is recommended for the flat-top or V, although thinner gauges such as 16 or even 18 AWG can be used. 2 (5cm) 5111 (15.5tm) 12 AWQ • 31ft (10.36m) 14 AWG 51 ft (15.54m) 12 AWQ -Matching section Fig. 9 — Construction dimensions of the G5RV antenna and matching section. The Matching Section This should be, preferably, of open-wire feeder construction for minimum loss. Since this section always carries a standing wave of current (and voltage), its actual impedance is unimportant. A typical, and satisfactory, form of construction is shown in Fig. 10. The feeder spreaders may be made of any high-grade plastic strips or tubing; the clear plastic tubing sold for beer or wine syphoning is ideal. If you decide to use 300-ohm ribbon type feeder for this section, it is strongly recommended that the type with "windows'1 be used. It has lower loss than a feeder with solid insulation throughout its length, and it possesses relative freedom from the detuning effect caused by rain or snow. If this feeder is used for the matching section, allowance must be made for its velocity factor (VF) in calcu-lating the mechanical length required to resonate as a half-wave section electrically at 14.15 MHz. Since the VF of standard 300-ohm ribbon feeder is 0.82, the mechanical length should be 28 ft. However, if 300-ohm ribbon with windows is used, its VF will be almost that of open-wire feeder, say 0.90, so its mechanical length should be 30.6 ft. This section should hang vertically from the center of the antenna for at least 20 ft or more if possible. It can then be bent and tied off to a suitable post with a length of nylon or terylene cord at an above-head height. Supported by a second post, its lower end is connected to the feeder. "Rie Feeder The antenna can be fed by any convenient type of feeder provided always that a suitable type of matching network is used. In the original article describing the G5RV antenna, published in the RSGB Bulletin for November 1966, it was suggested that if a coaxial cable feeder was used, a balun might be employed to provide the necessary unbalanced-to-balanced transformation at the base of the matching section. This was because the antenna and its matching section constitute a balanced system, whereas a coaxial cable is an unbalanced type of feeder. However, later experiments and a better understanding of the theory of operation of the balun indicated that such a device was unsuitable 3/eI (1cm) > E 5 H Notch at " each end j ^ f o r 14 AWU c x 7 Holes tor binding wire Matching section spreader Approx 1cm wide plastic strip (or 8mm dia plastic tube) copper wire of, matching : section 14 AWG - c o p p e r -^Binding wire^ Detail of matching section 1 1 2 ' (30cm) approx •J Fig. 10 — Constructional details of the matching section. Also suitable for open-wire feeder construction. because of the highly reactive load it would see at the base of the matching or make-up section on most HF bands. If a balun is connected to a reactive load with an SWR of more than 2:1, its internal losses in-crease. The result is heating of the windings and saturation of its core, if one is used. In extreme cases with relatively high power operation, the heat generated in the device can cause it to burn out. The main reason for not employing a balun in the G5RV antenna, however, is that unlike a matching network, which employs a tuned circuit, the balun cannot compensate for the reactive load condition presented to it by the antenna on most of the HF bands, whereas a suitable type of matching network can do this most effectively and efficiently. Experiments were conducted to determine the importance, or otherwise, of unbalance effects caused by the direct connection of a coaxial feeder to the base of the matching section. There was a rather surprising result. The research showed that the HF currents measured at the junction of the inner conductor of the coaxial cable with one side of the (balanced) matching section, and at the junction of the outer coaxial conductor (the sheath) with the other side of this section, are virtually identical on all bands up to 28 MHz, where a slight, but inconsequential difference in these currents has been observed. There is, therfore, no need to pro-vide an unbalanced-to-balanced device at this junction when using a coaxial feeder. The use of an unbalanced-to-unbalanced type of matching network between the coaxial output of a modern transmitter (or transceiver) and the coaxial feeder is essential. This is because of the reactive condition presented at the station end of this feeder, which on all but the 14-MHz band, will have a fairly high to high SWR on it. The SWR, however, will result in insignificant losses on a good-quality coaxial feeder of reasonable length; say, up to about 70 ft. Either 50- or 80-ohm coaxial cable can be used. Because it will have standing waves on it, the actual characteristic impedance of the cable is unimportant. Another convenient feeder type that can be employed is 75-ohm twin lead. It exhibits a rela-tively high loss at frequencies above 7 MHz, how-ever, especially when a high SWR is present. I recommend that not more than 50 to 60 ft of this type be used between the base of the matching section and the matching network. The 75-ohm twin lead available in the United Kingdom is of the r e c e i v e r t y p e ; less lossy transmitter type is available in the United States. By far the most efficient feeder is the open wire type. A suitable length of such can be con-structed in the same manner as that described for the open-wire matching section. If this form is employed, almost any length may be used from the center of the antenna to the matching network (balanced) output terminals. In this case, the matching section becomes an integral part of the feeder. A convenient length of open-wire feeder is 84 ft. It permits parallel tuning of the matching network circuit on all bands from 3.5 to 28 MHz, and with conveniently located coil taps in the matching network coils for each band, or where the altern-ative form of a matching network employing a three-gang 500 pF/section variable coupling capacitor is used, the optimum loading condition can be obtained for each band.\4 This is not a rigid feeder-length requirement, and almost any mechanically con-venient length may be used. Since this type of f e e d e r will always carry a standing wave, its characteristic impedance is unimportant. Sharp bends, if necessary, may be used without detriment to its efficiency. It is only when this type of feeder is correctly terminated by a resistive load equal to its characteristic impedance that such bends must be avoided. Coaxial Cable HF Choke Under certain conditions a current may flow on the outside of the coaxial outer conductor. This is because of inherent unbalanced-to-balanced effect caused by the direct connection of a coaxial feeder to the base of the (balanced) matching section, or to pickup of energy radiated by the antenna. It is an undesirable condition and may increase the chances of TVI [from fundamental overload, if the feeder is routed near a TV receiving antenna — Ed.]. This effect may be reduced or eliminated by winding the coaxial cable feeder into a coil of 8 to 10 turns about 6 inches in diameter immediately below the point of connection of the coaxial cable to the base of the matching section. The turns may be taped together or secured by nylon cord. It is important that the junction of the coax-ial cable to the matching section be made thoroughly waterproof by any of the accepted methods. Binding with several layers of plastic insulating tape or self-amalgemating tape and then applying two or three coats of polyurethane varnish, or totally enclosing the end of the coaxial cable and the con-nections to the base of the matching section in a sealant sueh as epoxy resin are a few methods used. References \1 Varney, L., "ATU or astu?," Radio Communica-tion, August 1983. See Ref. 1. Y2. Varney, L., "HF Antennas in Theory and Prac-tice — A Philosophical Approach," Radio Communication, Sept. 1981. U See Ref. 1. Vertical Antennas Wiring Up the Old Spruce By Kris Mersehrod, KA201G 123 Warren Road, Ithaca, NY 14850 One serious p r o b l e m an aspiring 8 0 - m e t e r operator may have is acquiring an antenna that will offer broadband performance and a low angle of radiation. I've been a Novice for two years, but early on discovered that the challenge of "getting out" on 80 meters had turned me into an antenna enthusiast. Friends suggested that only by going to a vertically polarized antenna could I expect to make it on 80 meters. That meant a vertical antenna. One characteristic of vertical antennas in articles I referenced is the need for an extensive radial system.VL A second characteristic is the need for a matching system if the height is significantly less than 1/4 wave. If height was necessary to avoid using a matching system, the idea of a vertical was out of the question. The zoning regulations in my town were written to prohibit commercial radio antennas by establishing a 35-foot maximum antenna height. According to the articles in QST, this is a constitutional issue, and I'd rather spend my time on antennas than at the Supreme Court. Although I'm currently limited to the Novice section of the band, broad bandwidth was not an immediate concern. Why put so much effort into an antenna of limited future? I read the articles on cage dipoles, but decided one of those "structures" would only invite comments from neighbors without solving the problem of the angle of radiation.\24. 2. F i n a l l y , Wilson published an inspiring and straightforward article on a vertically polarized antenna with only two radials!\i I built it im-mediately and hung it from dipole supports up 35 f e e t . It provided excellent results; I could copy stations that couldn't be heard with my former dipole. It was not designed for local work, however, and I had complaints from the XYL (KA20XG) because it was hanging in front of a main doorway. I carefully rolled up the "K4YF Special" for other locations; it's a great field day antenna. I was hooked on vertical polarization, though, and resumed reading about this type of antenna with a little more skepticism concerning the number of radials really needed. One side remark in The ARRL Antenna Anthology on 80-meter phased arrays follows the usual admonition to keep the antenna free of obstacles and in the clear.YSi The comment was in relation to the swamp site of the array and an advertisement of the system by the COL-ATCH-CO.\£ Their ad stated that, "Foliage absorption at 40 and 80 meters is negligible." Their video tape available for loan was an opportunity to see the foliage in the swamp. Except to tie one end of the dipole, I had been avoiding the windbreak of tall spruces on the south side of my house for two years. The idea of a "hidden" antenna began to jell. At the local scrap recycling yard, I obtained coils of cable TV feeder lines 3/4 inch in diameter. They are made of a fairly stiff aluminum-walled tube, filled with foam and a single copper con-ductor. At 50 cents a pound, I purchased them for a dime a foot. This chain of events convinced me to wire a 60-ft spruce in the following manner. The Antenna Specifications Spruce trees are like ladders, they have nicely spaced branches. A 60-ft length of TV feeder cable was laid on the ground (234/f (MHz) = 62.4 ft; I rounded the figure). This was pushed up the tree along the trunk until it began to bend over. Then I climbed the branches and pulled the cable along with me. Once at the 50-ft level, I slid the remainder of the "element" up ahead and lashed the cable to the tree. On the way down, I straightened the cable. After testing, the antenna was lashed at three levels to keep the lower end about 6 inches above the end of the central grounding rod. The Radial System An 8-ft ground rod was driven at the foot of the tree. Two 65-ft radials from the K4YF Special plus four 60-ft lengths of no. 8 aluminum were secured to the clamp. All of the radials are under the spread of the spruce trees so the "system" is about 65-ft east and west of the base, and 6 ft north and south. The base feed point of the spruced-up vertical system for 80 meters. Results The noise bridge showed the impedance to be 80 ohms at the base of the antenna and a bit on the capaeitive side (it could be shortened). A ratio of 1.2:1 from 3.5 MHz to 4.0 MHz was measured on the SWR meter. The feed line is a 52-ohm coaxial cable which is not a perfect match, but the SWR is pretty good. Thus, without a matching network, no capac-itance hats, and a limited radial system, a 1/4-wave vertical put me on the air. Using this antenna in the Novice band, I find that, like the K4YF Special, I can hear more than I can work. The stations that I can hear in the west or in the deep south probably can't hear me — at least they do not respond. Texas is as far as I have worked. One major problem is that most operators on the Novice segment of 80 meters work from dipoles, and distant signals are overcome by local stations. With my vertical, local stations are much weaker than they were on my dipole; thus, the distant stations are audible in comparison. My rig is a Drake TR-4C which puts in about 250 watts to the final, and I am consistently given 579s or better. DXing has eluded me so far, but I think that the problem is with the Novice band allocation. Devoldere's booklet on 80-meter DXing shows that only Argentina, Chile, Canada, the Marshall Islands and the U.S. territories overlap CW sections with the U.S. Novice section (1978: p. 4-1).\1 These countries are a distance away, plus they may avoid the Novice segment because of the activity. Outside of the Novice band, reception is great. It is fun to listen in Hie Extra-Class segment in the so-called DX window. I've heard some European and Japanese stations in pileups never experienced on the Novice segment. Conclusion What would hams do without trees? Sounds like a good opener for a joke, but on 80 meters and below, trees for antenna supports may be the easiest method to get on the air. The neighborhood looks the same and the zoning board isn't bothered. As luck would have it, there are other spruces a quarter wave away east and west. I'm planning on sprucing up the next phase of my antenna experiments. The vertical radiator (left) snakes its way up the trunk of the spruce tree, where it is lashed at the top. References \1 Sevick, Jerry, W2FMI, "The Ground-Image Verti-cal Antenna," Hie ARRL Antenna Anthology, 1978, (ARRL: Newington, CT), p. 22. \2 Hall, Jerry, K1TD, 'The Search for a Single Broadband 80-Meter Dipole," QST, April 1983, pp. 22-27. \1 Wolfgang, Larry, WA3VIL, "West Jersey Com-munication Products 80-Meter 'BN Cage' Antenna," QST, Sept. 1983, pp. 43-44. Wilson, John J., K4YF, "The 'K4YF Special' Antenna," QST, Sept. 1978, pp. 26-27. Atchley, Dana W., "Switchable 4-Element 80-Meter Phased Array," QST, March 1965; The ARRL Antenna Anthology, 1978, pp. 119-122. \ £ COL-ATCH-CO, advertising folder for "Instan-tarry Systems," P. O. Box 230, Carlisle, MA 01741, 1983. \I Devoldere, John, ON4UN, 80-Meter DXing, Com-munications Technology, Inc., Greenville, NH 03048, 1978. A Triband Parasitic Vertical Directional Array By Walter J. Schulz, K30QF •3617 Nanton Terrace, Philadelphia, PA 19154 Vertical directional arrays are employed in AM broadcast stations to protect other radio stations that share the same frequency from interference. In other instances, they are used to direct radiation to a specific target area. It is with that purpose in mind that as radio amateurs we are interested in this type of array. Much of this work was completed during the 1930s on the broadcast band in the United States, based on the research of Dr. George H. Brown and Carl E. SmithAl^. We have occasionally seen directional arrays in Amateur Radio articles with all the vertical ele-ments driven. Sometimes the array is constructed with one driven element and one parasitic element. This type of array is a much simpler system to em-ploy. However, when the number of array vertical elements is increased beyond two, the difficulty increases. I was perplexed on how to solve this problem in a simple way and felt that by using an array where each element was driven would prove a difficult task to maintain in adjustment, especially with an inadequate ground radial system. One of the other associated problems is sustaining the correct phase relationships and current ratios between array elements. Therefore, something less complicated was needed to overcome the problems noted. The answer occurred to me in the form of a triband Yagi-Uda dipole array conforming to the configuration of a vertical triband parasitic array placed on the earth's surface. Directional Array with Parasitic Elements A quarter-wave element can be constructed by dividing a Yagi-Uda array element in half. By em-ploying the earth's surface with a good ground radial system, an image quarter-wave antenna is pro-duced under the earth's surface. Using our imagi-nation, we can visualize that the image element exists along with the real element above the earth's surface. The image and real element produce a half-wave element whose feed point is at the earth's sur-face. Such a Yagi-Uda imaginary configuration would be vertically polarized. Constructing such a direc-tional array in real life would be a good compro-mise, offering good forward gain, possessing a fairly good front-to-back ratio to discriminate against unwanted signals, and having the capability to operate over the 20-, 15-, and 10-m radio amateur bands. The goal then is to design a triband parasitic vertical array consisting of one driven element, one reflector, and one director. The antenna system must endfire to the target area and be fed with a coaxial transmission line having a 52-ohm characteristic impedance. Judicious selection of spacing is re-quired to maintain an inductive or capacitive reac-tance at the feed point over the three bands of operation. I decided to keep the reactance capac-itive and employ a T-matching network using vari-able inductors in its arms. In this manner, you can tune for an impedance match in your station, pro-vided the resonant traps on the driven element (dielectric) do not break down from the voltage standing waves. To find the right values for these electrical parameters, a certain amount of per-turbation was necessary. The electrical parometers were obtained with a pocket calculator such as the Hewlett-Packard HP-15C. By proper selection of self-and mutual-impedance values from known charted values, current magnitudes and phases on each ele-ment were found. Then, from these electrical current values, the drive-point impedance and radiation pattern over the three bands were found. Algorithm for Finding Currents in Array Elements Before we discuss a keystroke sequence that will yield the current value for each individual element in the array, let us look at some diagrams. We know from observing a directional array that by making the observation directly above the system, an analogy to isotropic point sources can be seen.\l It becomes useful to represent the triband parasitic directional array in this manner for each band of operation. Notice for each illustration shown in Fig. 1A, B and C, that the electrical spacing is represented in electrical degrees. Between elements, the spacing becomes larger electrically as the wavelength becomes shorter, although the physical distance between the elements remains the same. This occurs because the wavelength changes for each band of operation. We begin by illustrating the problem graph-ically, finding the mutual impedance representative of the physical spacings for each band of operation. The driven element height is approximately 0.236 wavelength, the reflector element is 0.247 wave-length, and the director element is 0.225 wave-length. The antenna uses a 1-inch OD aluminum tub-ing. The self-impedance value is based on these forementioned heights and the outside diameter throughout the design. Our next step, once the self-and mutual impedances are obtained, is to calculate the currents flowing in the array. It is helpful to think of each vertical element in terms of self-impedance, and one that forms a network. When we think of a network, the mutual im-pedances must be considered also. Our goal is to solve for the current magnitude and phase of each mesh that represents a vertical element in the array or network.\l We may start by writing the simul-taneous equation describing the problem: 1 = x Z n + l 2 x Z12 + I 3 x Z13 0 = x Z ^ + I 2 x Z 2 2 + I 3 x Z23 0 = x Z31 + l 2 x Z32 + i 3 x Z33 Notice that the parasitic elements have no voltage flowing in the mesh, but the driving voltage is generated in the first equation. This driving voltage will induce current flow in the other meshes. We must determine the current magnitude and phase for II, 12, and 1 3 before we can determine the drive-point impedance and the radiation patterns. There are a number of systems that will yield these current values. Determinants or Guassian Successive Elimination methods are commonly used. However, with the advent of the pocket calculator, and particular the HP-15C, these methods are already programmed in the calculator. This reduces the design work to just plugging in the values. The two mathematical systems used to solve for currents in the array are included in the appendix. Hie Matrix Let's consider the problem described above for 20 meters and use the electrical parameters below for this given band of operation. Self-Impedances: Z11 31 -j8 driven element Z22 3 6 reflector element Z33 27 -j37 director element Mutual Impedances: Matrix A Matrix B Z12' Z21 = 36 + j7 Z13' Z31 - 36 + j7 Z23' Z32 = 29 - j6 We can then write the complex impedances in terms of individual entries in a matrix along with the voltages. ~Z11 Z12 Zl5" " h " r r Z21 Z22 Z23 X J2 = 0 31 Z32 Z33" LlJ _0-Matrix A Unknown Matrix B Next, enter the complex values into the individual entries. Matrix A Matrix B "(31 - j8) (36 + j7) (36 + j7)~ _11 ~1 + jo" (36 + j7) (36 + j21)(29 - j6) X X2 = 0 + jo .(36 + 37) (29 - j26)(27 - j37). -0 + jo-36 36 36 36 36 29 36 29 27 -8 7 7 7 21 -6 7 -6 -37 Because the impedance values are complex (rec-tangular form), we must determine a method to write the impedance matrix in such a way that the HP-15C calculator can handle the matrix. This is accom-plished by partitioning the matrix into two sub-matrices, one having a real-number entry, while the other submatrix has the imaginary (j-operator) value entered. We write the new matrix ass Notice that the submatrix entry positions are identical; the associated imaginary number entries are identical with the entry positions in the upper submatrix. It is important to keep the entries in the correct positions in both submatrices. Next to be considered is Matrix B. This matrix must be written for the voltage values in the same manner. Remember that this matrix is also partitioned and it may prove helpful for you to write one column at a time with its real and associated imaginary num-bers. We are now ready to proceed with the actual keystroke procedure as described in the HP-15C owner's manual.\5j£ Keystroke Sequence 0 f DIM (i) f MATRIX 1 6 Enter 3 f DIM A (size of Matrix A i s 6 x 3) f MATRIX 1 f USER 31 SEO A 1,1 36 STO A 1,2 36 SK) A 1,3 36 STO A 2,1 36 STO A 2,2 29 STO A 2,3 36 STO A 3,1 29 STO A 3,2 27 STO A 3,3 -8 STO A 4,1 7 STO A 4,2 7 STO A 4,3 7 SID A 5,1 21 STO A 5,2 -6 STO A 5,3 7 STO A 6,1 -6 STO A 6,2 -37 STO A 6,3 6 ENTER 1 f DIM B 1 ENTER 1 EHTER STO g (B) RCL MATRIX B 6 1 RCL MATRIX A 6 3 f MATRIX 2 A 6 6 f RESULT C % 06 1 g Cy,x (converts partition to coirplex form -In = RCL C Re 1,1 0.038 rectangular) x RCL C j 1,2 0.015 = RCL C Re 2,1 -0.0.36 1 RCL C j 2,2 0.027 I- = RCL C Re 3,1 0.027 J RCL C j 3,2 -0.030 The Drive-point Impedance Using the above method, we are able to derive the current magnitude and phase for each element in the antenna array. Next, we must determine two elec-trical values: the drive-point impedance and the radiation pattern. The equation shown below gives the drive-point impedance for the array, provided we know the current ratios between the driven element to the reflector and the director elements. Zin = Z11 + V 1 ! (Z12» + V 1 ! (Z13> Z. = (31 - j8) + (-0.036 + j0.027)/0.038 + j0.015 (36 + j7) 111 + (0.027 - j0.030)/0.038 + j0.015 (36 + 7) Z ^ = (31 - j8) + (-27.338 + j29.729) + (18.902 - j30.921) Z ^ = 22.564 - j9.192 ohms The procedure for finding the element currents and drive-point impedances must be completed for each band of operation. This must be done because the mutual impedances will change between elements with different electrical spacings. The Radiation Pattern Refer to Figs. 1A, B and C. The radiation pattern is determined from the current ratios between the elements. Take a closer look at the drive-point impedance equation and observe the two ratios of interest: 12/11 and 13/11. When the drive-point impedance is divided by itself, it will be equal to 1 + jO; when converted to polar form it is 1.0/ 0 • Now we must divide the current in complex rectangular form, II, into 1 2 and 13, and then convert the quotient to polar form for mani-pulation. It becomes evident that there are now three polar form values that describe the magnitude and phase between elements. Further, by considering the physical spacing between elements, the radiation pattern field strength intensity is determined for each azimuth angle. The easiest method to write the vectors is in the following form: Magnitude /spacing cos Q - 91 + electrical phase where 0 = azimuth angle ©^ = azimuth position of vertical element The vectors are written using the chain method, which is simply the addition of all vectors graph-ically.VZJl Once all the vectors have been added, the resultant magnitude is obtained for a specific azimuth angle that is under consideration. The chain, also known as the head to tail method, is repeated again until all azimuth angles in the pattern have been computed. Once the horizontal radiation pattern has been obtained (Figs. 2A, B and C), the gain figure over the isotropic point source is derived by finding the integral of the calculated radiation pattern. There are a number of methods that can be employed, but the simplest is that of the solid angle concept, borrowed from optics. To use this concept, we must know the beam width in both the horizontal and vertical planes. The 0.707 root-mean-square value is multiplied by the maximum field strength intensity in both planes, 'l'he product is the half-power point in the radiation pattern. This value is found along the radiation pattern plot. Once it is located, note the azimuth angle. There will always be two half power points in the horizontal plane, while there will be only one at a specific elevation angle in the vertical plane. Once we know the beam width, the equation below is used to find directivity and power gain in decibels. FIGURE IA 20MTR PARASITIC DIRECTIONAL ARRAY -WD 1.I0l/QC03g O M g r t.O/^0'CQS«-90a-0 o 222 Z|| Z33 MUTUAL IMPEDANCES 30"/Z| 2 j 2t,l3,3l 60°/Z23,32 =29_J6 t L t M t N T CURRENTS I,- 0.038 + JO.015 l 2 - 0 . 0 3 6 +J0.027 |j- 0.027-J0.030 0RIVEP0INT Zin =22 -J9 ohms FIGURE IS I5METER PARASITIC DIRECTIONAL ARRAY _ " 4 0 ° _ -K3 O O O Q .
5 7 / O O O S 0 t O -H I 3 ° 1 .
0 / 4 0 ° C O S B -9 0 ° + !) O RII/a0°r.r>SB-90°-9B° ^22 Hi 33 DRIVEPOINT Zin s 20"jl7ohms MUTUAL IMPEDANCES aOVZ23,32 -23-JI2 ELEMENT CURRENTS |,= 0.029 + J 0 . 0 2 4 l2=-O.OI9 «5.0l0 I 3=0,0I6-J0.026 Fig. 1 — Self- and mutual-impedance values taken from known charted values were employed to calculate current magnitudes and p h a s e s 0n each element. From these electrical current values, the drive-point impedance and radiation pattern for each was found. F I G U R E I C I O M E T E R P A R A S I T I C DIRECTIONAL ARRAY ->•0 R E F L E C T O R D R I V E N D I R E C T O R O O O 0.42/0 coseto-Mi9 i.n/snTO-0.0ia +J0.004 I,<0.014 -J0.030 D = 41, 253/(9V X e ^ and dB gain = 10 l o g ^ D In our particular case the system under design will yield a higher value than can actually be expected in the real world. However, the technique serves the purpose of allowing us to obtain a better understanding of system gain for the array. It helps to evaluate our design goals — and to see whether we are on the right path. Traps and Loading We have designed a 3-element triband parasitic directional array, but now we must calculate the effect that resonant LC trap loading will have on each element length. Not only does a resonant trap on its operation frequency isolate the antenna element beyond it, but when the trap is operated lower in frequency, it will effectively load the antenna element inductively. This loading effect makes the antenna element longer electrically, but also makes it possible for shorter element length (the actual physical dimension). It is evident from the preceding discussion that we must consider trap inductive loading in our design. In recent years, Joseph Boyer, W6UYH, wrote a series of articles describing how to calculate and design trap verticals. This knowledge is applied here to design the reflector, driven and director elements.\9r10rllf12 However, there is one slight difference. Rather than assume each element is of quarter-wave height, note in Fig. 3A that each element height is different. This is important because each length-to-diameter ratio is different; that means the shape factor is different. All values are used with this shape factor in our calculations on each element. From the shape factor, it is possible to consider how element and trap series inductive reactance adds to the element length band of operation we must describe the length-to-diameter ratio of each element height. The length-to-diameter ratio has already been used in the calculation of element self impedance, but for those who are not familiar with the procedure, I use the equation by Schelkunoff: Z_ = 60 (log 2h/a - 1) (7 6 where a = radius h = height Not only does this shape-factor equation deter-mine the electrical height, but will enable us to determine the voltage potential at each trap. c s 1 5 ' T 2 .
7 1 7 r . 1 .
0 7 5 e t 0.3 1 \ = = 3 T , 2 .
8 5 6 1 .
5 1 6 X dtivepoinl lis 6 9 15 TRAP r 1 .
1 4 7 W -t J iow t r a p r 7 .
7 4 I Fig. 2 — The horizontal radiation pat-terns for each band are shown. At A, we have the 20-m pattern, B depicts the 15-m band, and C is the 10-m band. electrically. The net result of this is that the element physical height is shortened. Our procedure is to arbitrarily select a fixed capacitance for the trap and then use its reactance value to calculate the inductance value to make the trap resonant. Refer to Fig. 3B. Following step by step, we must determine the loss resistance of the inductance, (RL = XL/Q), and the impedance of the LC constants in the trap at its resonant frequency: (Zp = V H ) Finally, we must find the value of series reactance that the trap presents as an inductive loading when operated on frequencies other than resonance. Xs = (1/Q M (1/M2 - 1)) (Zp) M(1q) = 21.225/28.6 = 0.742 M(15) = 14-175/28.6 = 0.496 M(20) = 14«175/21.225 + 0.668 Shape Factor Let's briefly return to the subject of shape factor. Dr. S. A. Schelkunoff gave a reliable equation for shape factor, which is sometimes referred to as Ka or ZQVli He equated an antenna to a transmission line in its behavior and this enabled him to find the antenna feed point complex impedance. What interests us here is that on each FIGURE 3A TRIBAND PARASITIC DIRECTIONAL ARRAY Fig. 3A — Note that each element height varies. This is important because each length-to-diameter ratio is different, thus the shape factor also varies. TRAP VALUES CAPACITANCE INDUCTANCE BAND 25 PF 1.24 UH 10 MTR 2 8 . 6 MHZ 25 PF 2 . 2 5 UH 15 MTR 2I.225MHZ ASSUME Q = IOO FOR DESIGN r L FIGURE 3B TRAP LC VALU£S Fig. 3B — Trap LC values for the 10- and 15-meter band. Voltage Potential on the Trap Practical experience has shown that a large amount of voltage is present on the upper element portion of a vertical antenna. For short vertical heights less than one quarter wave, the maximum cur-rent flows at the antenna base while high voltage potential is a cosine function in regard to antenna height. What concerns us the most in the config-uration described is the voltage potential across the traps. The trap components must have significant dielectric strength to prevent breakdown from either tuning standing wave voltages or just daily usage. We begin by finding the quality factor of the driven element.\H °(20) = Z9 ~ 3 / 4 6 ' 5 = 2 9 5 , 5 9 3 ~ 3/46.5 = 6.3 From calculations performed, we know that the array exhibits a radiation resistance of 22 ohms on the 20-m band. We assume the reactive component will be canceled by the series capacitance arm of the T network. Finding the current at the array drive point for an output power of one kilowatt: I = (p/Rr) 0 , 5 = (1000/22) 0' 5 = 6.7 arrperes The drive-point voltage is determined next: Voltage at drive point = I Rr = (6.7)(22) = 148.3 V and from the product of voltage times the quality factor (Q) we find: Voltage at trap = Q V/2 = (6.3)(148.3)/2 = 467.2 V It is advisable to use a safety f a c t o r of approximately four times the calculated voltage value. This is satisfactory with materials available today. Also, I suggest the use of half-wavelength multiples of coaxial transmission line to feed the array. The traps may be constructed from other than lumped-constant component parts and it is probably cheaper to employ coaxial LC traps. Not only is it less expensive, but the dielectric strength will be greater depending on the coaxial cable chosen. Some last words about trap construction. It may be easier to construct coaxial traps first, and then measure their quality factor with a Q meter. Also, the inductive reactance and coil loss resistance can be calculated. Once this is accomplished, parallel resonant impedance may be determined. Finally, from those values, the series reactive loading of the trap is found. By changing the procedure, we are still able to calculate the correct linear antenna height in each element. This article is a brief outline of a design for a triband parasitic directional array. It has sig-nificant advantages in that by doubling the element identical we have a blueprint for a Yadi-Uda dipole array. T n this case, the drive-point complex imped-ance will double in value and a matching network will have to be employed for matching the trans-mission line to the antenna. Using the array as a directional array, employ a ground radial system to prevent loss resistance and maintain a close resemblance to the theoretical radiation pattern shown in Figs. 2A, B and C for each band. References \1 Brown, G. H., "Directional Antennas," Pro-ceedings of the Institute of Radio Engineers, Jan. 1937, p. 78-145, vol. 25, no. 1, part 1. \2 Smith, Carl, Directional Antenna Patterns, Cleveland: Cleveland Institute of Radio Electronics, 1958. U Sehulz, W., "Vertical Array Analysis," QST, Feb. 1981, pp. 22-25. U Schulz, W., "Key to Three Element Yagi Design," Ham Radio, March 1984, pp. 48-51. \ J L Hewlett-Packard HP-15C Owner's Handbook, Aug. 1982, pp. 160-173. \£ Hewlett-Packard HP-15C Advanced Functions Handbook, Aug. 1982, pp. 128-131. \Z Schulz, W., "HP-41C Tweaks Vertical Antenna Arrays," Electronics, J a n . 17, 1980, pp. 141-143. \8_ See Reference 3. \2. Boyer, Joseph, "The Multiband Trap Antenna, Part 1," CQ Magazine, Feb. 1977, pp. 26-30 and 73-74. \lfl. Boyer, Joseph, "The Multiband Trap Antenna, Part 2," CQ Magazine, March 1977, pp. 51-55 and 72. M l Boyer, Joseph, "The Multiband Trap Antenna, Part 3," CQ Magazine, April 1977, pp. 46-50 and 72. \12 Boyer, Joseph, "The Multiband Trap Antenna, Part 4," CQ Magazine, May 1977, pp. 22-27. Schelkunoff, S.A., "Theory of Antennas of Arbitrary Size and Shape," Proceedings of the Listitute of Radio Engineers, Sept. 1941, pp. 493-520. \14 Schelkunoff, S.A., and Friis, Harold, Antennas Theory and Practice, New York: John Wiley & Sons, Inc., 1952, pp. 435-436. Kurosh, A., Higher Algebra, Moscow: Mir Publishers, 1980, pp. 15-22. \16 See Reference 4. Appendix The Algorithm for Successive Elimination of Three Simultaneous Equations r i z n + V 1 2 + hzn = V 1 hhl + Z2Z22 + hhl = V 2 X1Z31 + X2Z32 + ^ 3 3 = V 3 The first step is to eliminate I ^ i and I j Z ^ . I1 Z11 + I2 Z12 + I 3 Z 1 3 = V 1 + I2 Z22' + W = V 2 ' + W + hh3 = V 3 ' V ' = (Z22 ~ Z12Z21/Z11) Z23' ' = <Z23 - Y l W Z33' ' = <Z33 ~ Z13Z31/Z11) V ' 2 = < V 2 " V1Z21/Z11> V • 3 = < V3~ W 2 ! ! ' Z32 ' = <Z32 - z 1 2 z 3 i / z n ) Gaussian pivot is Z^/Z. f o r t h e f i r s t - s t e p elimination method. Second Step In the second s t e p , the first simultaneous equation is no longer involved. We only manipulate the second and third equation using the following pivot Z 3 2VZ 2 2\ Z33" = <Z33' - Z23' Z32'/ Z22') V3n = (V3 - V2'Z 3 2'/Z 2 2') then: h h l + h h z + I 3 Z 1 3 = V 1 + I2Z22' + I3 Z23' = V 2 ' + I3Z33n = V 3" finding in the last equation that 13 = V3"/Z33" and using this value for 13, we then back fill the second equation to find 12 = V21 - I3Z23'/Z22. Again using the values for 13 and 12, we back fill the first equation to find II. It is in this manner that the solution to all unknown current values are found using t h e Gaussian s u c c e s s i v e e l i m i n a t i o n method.Ua The Second Method Called Determinants A far more simpler method to find unknown current values is the use of determinants. This method is useful so long as the simultaneous equations are non-singular, or you can arrive at a determinant for the system. = " (Z33 Z21)/DE7r X3 = (Z21 Z32> " (Z31 Z22 ) / D E T ^ = (Z11 Z22 Z33 + Z12 Z23 Z31 + Z13 Z21 Z32 " Z31 Z22 Z13 ~ Z32 Z23 Z11 Z33 Z21 Z12} The 5/8-Wavelength Antenna Mystique By Donald K. Reynolds, K7DBA 230 B St., No. 301, Friday Harbor, WA 98250 The 5/8-wave monopole antenna is well known to amateurs operating in the VHF/UHF mobile commun-ications area. Often the operator is not an expert in electromagnetic theory or wave propagation. Be-cause of this, many amateurs have the misconcep-tion that there must be something mysterious and superior about this particular length. What is so magical about the 5/8-wave monopole? Actually, it is an old idea. The earliest publi-cation on the subject (to my knowledge) appeared in the Proceedings of the Institute of Radio Engineers in 1924.\1^_ The purpose of this article is to review the operation of the 5/8-wave monopole. I will point out its desirable properties, as well as certain ap-plications in which it is not only undesirable, but can considerably degrade the performance of a com-munications system. Radiation Pattern and Gain AM broadcasting was in its infancy in the early 1920s. Broadcast engineers were interested in de-termining the height of a base-excited vertical-tower radiator that would produce the maximum ground wave field strength for a given constant input power to che antenna. A technical paper published in 1924 showed, t h e o r e t i c a l l y , that subject to certain assumptions, the tower height for maximum ground-wave field strength was 5/8 w a v e l e n g t h . T h e s e are the assumptions: 1. Consider the tower to be a thin filament of perfectly conducting wire, mounted perpendicular to a plane of infinite extent and conductivity. 2. Consider the current a m p l i t u d e t o be a sinusoidal function of the distance down from the upper extremity of the antenna, as given by: I(z) = I sinjZJLtizzL (Eq. 1) degrees, the input impedance is a c a p a c i t i v e reactance. Actual current distributions on wires and rods of finite diameter differ from this assumed distri-bution in certain details. The current on the wire may be broken down into two components; one in phase with the driving voltage, the other 90 degrees out of phase. The resultant total current is the phasor sum of the two components, and is plotted in dashed lines for a typical case in Figs. IB and 1C. The phase of the total current no longer shifts abruptly from -90 to +90 degrees at the point 1/2 wavelength down from the upper end, but varies more gradually. Also, the magnitude of the current does not go to zero at the point 1/2 wavelength down from the top, but has a minimum value a little less than 1/2 wavelength down. The smaller the diameter of the antenna, the lower is this current minimum, but the diameter must be extremely small compared to the wavelength for the current minimum to approach zero. For example, if the antenna consists of a uniform cylindrical wire whose radius is 0.0001 wavelength, the current minimum is about 209s of the maximum value. For a radius of 0.01 wavelength, the minimum current is nearly 50% of the maximum value. The amplitude and phase of the current along cylindrical monopoles of different lengths and diameters have been c a l c u l a t e d using digital computers with what are called "moment methods."V2. Accurately measured current distributions reported by T. Morita show close correspondence with the more recently calculated distributions. where I m is the maximum current amplitude, z is the distance from the base to the point where I is calculated. J . is the total length of the antenna. A is the wavelength, in the same units as z and z . The standing-wave current distribution we have assumed is shown in Fig. 1 for a 5/8-wave antenna. The relative phase of the assumed current, with rcspect to the phase of a voltage generator at the base, is also plotted. Note that the current in the upper 1/2 wavelength is of constant phase — 180 degrees out of phase with the current in the lower 1/8 wavelength. Also note that the amplitude of the current at the base (feed point) is 0.707 I m . Since the input current leads the applied voltage by 90 B a s e Driven Monopole (A) Magnitude Rel.t0V0lta.9e (B) (C) Fig. 1 — Current distribution on a 5/8-wave monopole mounted on an infinite conducting plane. Solid lines are for assumed current distribution (Eq. 2). Dashed lines are typical of actual cur-rent distributions. Even though the actual current distribution on real 5/8-wave monopoles differ considerably from the ideal standing wave distribution given by Eq. 1, the radiation pattern computed on the basis of Eq. 1 is quite close to measured patterns. The expression for the radiation pattern, calculated from the current distribution of Eq. 1, is to be found in many textbooks, and is given by:\£ E = 0.5858 Gcos(225° x cos 91 + 0.707l3 (Eq. 2) sin e where E is the relative electric field strength at a large radial distance from the antenna. 9 is the vertical angle measured with re-spect to the axis of the antenna, as shown in Fig. 1A. The maximum value of E occurs at e = 9 0 de-grees, and the factor of 0.5858 adjusts this value to unity, making Eq. 2 the "normalized E-plane pattern." Fig. 2 shows the pattern of Eq. 2, expressed in decibels. I have also plotted the patterns of a 1/4-and a 1/2-wave monopole. All three patterns corre-spond to the same radiated power, with the maximum value of the 1/4-wave monopole taken as /) dB. Note that the maximum radiated power occurs at 0 = 90 degrees for all three monopoles. The 5/8-wave mono-pole is seen to give almost exactly 3 dB of gain over the 1/4-wave monopole, a fact which was ob-served by Ballentine.\£ Fig. 3 shows the calculated theoretical power gains of vertical monopoles of different lengths mounted above an infinite conducting plane, with respect to the gain of a 1/4-wave monopole. All gains are calculated for 0 = 90 degrees, that is, toward the horizon. This curve is derived from Terman.\l Note that the gain of a very short monopole compared to the wavelength is only about 0.4 dB below that of 1/4-wave monopole. For lengths less than 1/2 wavelength, the radi-ation pattern has a single lobe centered on the horizon. Above 1/2 wave, a minor lobe appears at a high angle, an effect which tends to reduce the gain toward the horizon. However, the main lobe at 8 =90 degrees continues to narrow, tending to increase the gain, with the result that maximum gain at 0 . = 90 degrees occurs for the well known length of 5/8 wavelength, with a value of almost exactly 3.0 dB with respect to the 1/4 wavelength. For lengths greater than 5/8 wavelength, the minor lobe rapidly increases in amplitude, and the gain in the direc-tion e = 90 degrees falls off, reaching zero (- ~ dB) for a length of 1 wavelength. The foregoing predicted radiation patterns and gains occur only when the following two require-ments are satisfied: 1) The base of the antenna is located on a flat ground plane of infinite extent and infinite con-ductivity. 2) The observer is on the ground plane when measuring the maximum radiated field. The ground plane extends beyond the observer (theoretically to infinity). These requirements are approximated reasonably well for vertical AM broadcast antenna towers (albeit with some degradation because of finite ground conductivity), since the receiving stations 180 5 / 8 A M O N O P O L E 1 / 2 X M O N O P O L E 1 / 1 I X K 0 N 0 P 0 L E Fig. 2 — Theoretically calculated radiation patterns in vertical plane for monopoles on an infinite conducting plane. Zero dB corresponds to maximum radiation intensity of 1/4-wave mono-pole. Fig. 3 — Power gain toward the horizon of a vertical monopole on an infinite conducting plane plotted vs length of monopole. are usually at ground level, and the earth continues beyond the receiver. These requirements are not satisfied in the five common installations of 5/8-wave monopoles described below, with the result that the signal level radiated toward the horizon can be much less than the theoretical value. Improper Uses of 5/8-Wave Monopoles The following section describes the unhappy consequences of five common installations of 5/8-wave monopoles. 1) An antenna mounted at the end of a length of coaxial cable, with no ground plane of any kind. In a rush to get the 2-meter FM base station on the air, operators sometimes screw on a commercial 5/8-wuve whip, with a coaxial fitting on the base, to the end of a length of coaxial cable. The antenna is erected, possibly attached to a broom handle. What we have now is actually a dipole; the upper element is the 5/8-wave whip, and the lower element is the outer conductor of the coaxial cable, extending back to the transceiver. The RF current on the center conductor of the coaxial line excites the whip. The current on the outer conductor, equal in magnitude to the current on the center conductor, simply folds over and excites the outside of the outer conductor below the feed point (the other element of the dipole). The radiation pattern for the system varies with each installation, depending on such factors as the length of the coaxial cable, how it is dressed, proximity to other objects, and so on. The radiation intensity in the desired direction (toward the hor-izon) will almost certainly be severely degraded, compared to that of a properly designed vertical base-station antenna. Furthermore, there will probably be RF currents on the outside of the transceiver, power supply leads, and possibly the house wiring. To illustrate these points, vertical-plane radiation patterns were made on a 5/'8-wave com-mercial whip antenna for two feed distances: 1 wavelength and 1-1/2 wavelengths. The transceiver was connected to a length of coaxial cable that extended below the feed point in a straight line with the whip. These measurements were taken using the technique described in the last section of this article. The resulting patterns are plotted in dB vs vertical angle in Fig. 4, along with the pattern of a theoretical 5/8-wave monopole. Note the almost total lack of correspondence between the measured and theoretical patterns. Gains cannot be compared on these superimposed patterns, since all of them are normalized to ti dB in the direction of maximum radiation for each antenna. However, it can be positively stated that since gain in a desired direction can only be obtained by reducing the radiation intensity in undesired directions, and redirecting it in the desired direction, the gains toward the horizon in the two measured eases must be much lower than in the theoretical case. Note par-ticularly that in the case of the whip with one wavelength of coaxial line below, maximum radiation intensity occurs at an angle of about 25 degrees above the horizon, and drops 6 dB below that level in the desired direction of 6 = 90 degrees. 2) An antenna mounted on a finite ground plane, and elevated to some height above ground. There may be a misconception among some oper-ators that a small ground plane at the base of a vertical whip antenna will make the radiated field spread out into space as it would above an infinite ground plane (radiate zero power at angles below the ground plane). This idea is false. Regardless of the diameter of the ground plane, even if it were 100 wavelengths in diameter, as long as it is hoisted up in the air, and the observer is beyond the edge of the ground plane, there will be radi-ation below the "horizon" plane. The principal cause is "edge diffraction." It causes the angle of max-imum radiation to be lifted above the horizon, and I—8-K o 5/8MONOPOLE ON INFINITE GROUND PLANE 5/8TvMONOPOLE ON 1 K 0 F COAXIAL CABLE 5 / S M M P O L E ON 1 1 / 2 OF COAXIAL CABLE Fig. 4 — Measured vertical patterns of a 5/8-wave whip connected to a length of coaxial line with no ground plane. strong radiation to be directed below the horizon. For vertical antennas mounted at the center of ground planes several wavelengths in diameter, the loss of radiation intensity toward the horizon is about 6 dB, compared to the same antenna on an infinite ground plane. For the case of a 5/8-wave monopole mounted on a circular ground plane 1/2 wavelength in diameter, the shape of the radiation pattern is greatly changed compared to the theoretical pattern with an infinite ground plane. Measured patterns of both a 5/8-wave and a 1/4-wave whip mounted on a circular ground plane 1/2 wavelength in diameter are pre-sented in Fig. 5, along with the theoretical pattern of the 5/8-wave whip on an infinite ground plane. There appears to be very little to choose between the 1/4-wave and 5/8-wave whips on the 1/2-wave-length ground plane. Both patterns are very "fat" compared to the theoretical pattern, such that the 3-dB gain factor has been entirely lost. The substitution of four 1/4-wave radials for the solid metal ground plane can be expected to give substantially the same performance. It might be a little more degraded by the fact that the four radials cannot intercept all of the antenna current flowing out from the outer conductor of the coaxial cable at the feed point. Some of the current still flows along the outside of the outer conductor below the radials, resulting in pattern distortion. 3) An antenna mounted on a hand-held trans-ceiver. Hand-held transceivers used in the VHF/XJHF bands typically use helically loaded flexible whip antennas (rubber duckies) for compactness and convenience. These antennas, however, are notably 3 0 0 2 4 0 5 / 8 \ M Q N O P D L E ON INFINITE GROUND PLANE 5/8AH0N0P0LE OH V 2 DIAMETER GROUND PLANE 1 / 4 MONOPOLE ON ty2 DIAMETER GROUND PLANE Fig. 5 — Measured vertical patterns of 1/4-wave and 5/8^wave monopoles mounted on a circular ground plane 1/2-wave in diameter. For comparison, the theoretical vertical pattern of 5/8-wave monopole on an infinite conducting plane is shown. quite inefficient. As a result, pull-out antennas of various designs are offered to provide higher radiation efficiency. The HT and its antenna constitute a dipole. One side consists of the whip antenna; the other in-cludes all the components in the unit which are directly or capacitively connected to the outer conductor of the coaxial connector to which the antenna is attached. At the 2-meter band, the length of this lower half of the dipole is typically less than 0.1 wavelength. The transmitter's role is to place an RF voltage between the two halves of the dipole. The current distribution along both sides of the dipole is sketched in Fig. 6 for whip lengths of 1/2 and 5/8 wavelengths. Note that the feed point is at a current minimum in the case of the 1/2-wave whip, and the current flowing down onto the lower half of the dipole is consequently low. With the 5/8-wave whip, the current at the feed point is much larger than for the 1/2-wave whip. Furthermore, there is a current phase reversal in the lower part with relation to the phase in the upper portion. The radiation intensity toward the horizon for the case of the 5/8-wave whip should, therefore, be weaker for the same power delivered to the antenna ter-minals than for the 1/2-wave whip. The stronger excitation of the lower part of the dipole when the 5/8-wave whip is used results in greater coupling of RF energy into the operator's hand and face, com-pared to when the 1/2-wave whip is used. Experimentally measured radiation patterns of a 2-meter HT with a helically loaded whip, a 1/2-wave whip, and a 5/8-wave whip are shown superimposed in Fig. 7. The three measured patterns were all made at the same power level. The first pattern measurement with the helical antenna was repeated at the end in order to ensure that the radiated power had not changed during the series of measurements. This pattern agreed with the first to within 0.1 dB. Note that the radiation intensity toward the horizon is less for the 5/8-wave whip (about 2 dB) than for the 1/2-wave whip, as predicted. 4) An antenna mounted atop the mast of a sail-boat. This is essentially the same situation as in Case 1. Monopole whips mounted atop sailboat masts are frequently seen. They are principally used in the 156- to 162-MHz marine VHF band. If the coaxial cable comes through a hollow metal mast, as is often the case, the dipole is constructed with the whip on one side and the entire mast on the other, along with various attached stays and shrouds. The radiation pattern (which is difficult to measure) must certainly consist of a mass of thin lobes resulting from radiation from the various conductors which are long compared to the wavelength. This structure of lobes will rock with the boat. The 5/8-weve whip has a large amount of current at the feed point and performs more poorly than the 1/2-wave whip. The 1/2-wave whip should also be avoided. The proper solution is to use a base-excited vertical whip with an integral, properly designed decoupling system. This guarantees that virtually all antenna currents are confined to the radiator, and little current is excited on the structure below.\S. 5) The antenna is mounted on the metal body of an automobile. In spite of the many years during which VHF and UHF whips have been mounted on vehicles, all of the data regarding the relative performances of the 1/4-, 1/2-, and 5/8-wave whip antennas have not been collected. A ear top represents a finite, elevated ground plane; thus the famous 3-dB gain of a 5/8-wave monopole over a 1/4-wave monopole cannot be claimed. The measured patterns of 1/4-wave and 5/8-wave monopoles over a circular ground plane (Fig. 5) testify to this statement. A whip antenna mounted on a car can also be regarded as a dipole, one side being the whip, the other the car. Again, continuity of current at the feed point demands that the total current flowing into the base of the whip be exactly equal to the total current flowing out over the car body from the outer conductor of the coaxial feed cable, at the base of the whip. Here, there may be some advantage of a 1/2-wave whip over 5/8 wavelength, because the smaller feed point current of the half wave results in lower current spreading out from the feed point over the car body. As a result, the radiated field from the 1/2-wave whip is probably less sensitive to the mounting location on the vehicle than the field from the 5/8-wave whip. Asymmetry in the mounting location may favor one over the other. For example, trunk-lid mounting on a sedan may favor the 5/8 over the 1/2 wavelength in the forward direction, because the reversed current phase at the base of the 5/8 wave is shielded by the rising car top, in the forward direction. In the reverse direction (with trunk-lid mount) the 1/2 wavelength would probably Current Diaiributton i i transceivers with 1/2-wave and 5/8-wave whip antennas. Where 5/8-Wavelength Pays Off It seems that up to this point, except for its use as a broadcast antenna, the G/8-wave moitopole has l i t t l e to o f f e r . There is one application, however, for which there is no controversy. This is the 1-1/4-wave center-driven dipole — a dipole in free space in which the length of each leg is 5/8 wavelength. Such an antenna can and does deliver the 3-dB gain figure over a center driven 1/2-wave dipole. This is a well known antenna type, sometimes called the "extended double Zepp." No infinite ground plane is needed to make this antenna perform; the antenna itself has the same geometry and current distribution as a 5/8-wave monopole plus its "image" in an infinite conducting plane. The 1-1/4-wave center-driven dipole has been used either as a stand-alone antenna, or as an element in an array, driven by a balanced 2-wire open transmission line or a coaxial line with a balun. It may also be used vertically, fed by a coaxial line led up through the lower 5/8-wave element, which must be hollow! Resonant decoupling sleeves must be used in this case to inhibit coupling between the lower radiating element of the dipole and the metallic structure below.YlH Impedance Considerations So far, nothing has been said about the input impedance of the various lengths of whip antennas. There is reason to suspect that some antenna enthusiasts believe that if an SWR meter reveals no reflected power from an antenna, the antenna must be radiating properly. This assumption is not correct. Lack of rcflcctcd power indicates only that the antenna input impedance has been matched to the impedance of the coaxial line. It does not tell how the radiated power is distributed in space. However, even though this article is concerned exclusively with the radiation properties of various antennas, a few words on antenna input impedance are in order. The input impedance of a 5/8-wave whip above a ground plane 1/2 wavelength or more in diameter has a resistive component close to 50 ohms, and a capacitive reactance that depends considerably on the whip diameter, typically in the range of 50 to 150 ohms. To match the antenna to 50-ohm coaxial cable, an inductor in series with the base should be provided to tune out the capacitive reactance. The simplicity of accomplishing the match is probably one reason for the popularity of the 5/8-wave whip (especially from the manufacturer's standpoint!). A 1/2-wave whip above a ground plane has an input resistance at its resonant frequency that depends strongly on the whip diameter. Typically this resistance is in the range of 1000 to 2000 ohms. Matching to 50 ohms is readily accomplished with an L-network at the base, which may consist of an inductive reactance shunting the antenna termi-nals, and a series capacitance to tune out the residual inductive reactance. Practical realizations of such a network can be built in about the same physical volume as the inductor for the 5/8-wave whip. Measurement Techniques Free-space radiation-pattern measurements of antennas that eliminate the effects of reflected signals from the earth require expensive experi-mental facilities. This includes tall, widely spaced towers, and directive antennas. A much simpler method that applies to the vertical pattern of whip antennas designed to radiate o m n i d i r e c t i o n a l , v e r t i c a l l y polarized signals, is to support the antenna horizontally, about 1 wavelength above the earth on a turntable that is mounted flush with the earth. A test signal is transmitted from the ro-tating antenna to a fixed, horizontally polarized receiving antenna at a distant point, over level ground. The receiving antenna is mounted at the same height above ground as the rotating antenna. The received signal is amplified and filtered from adjacent-channel i n t e r f e r e n c e using a tunable receiver, whose detected output is measured on a chart recorder that is connected by servos to the turntable. The r e c e i v e d signal inevitably contains a directly radiated component and a ground-reflected component. As long as the antenna being measured is rotated in the horizontal plane, and the spacing between the transmitting and receiving antennas is much greater than the heights of both antennas above the earth, the way that the direct and ground-reflected signals phase together at the receiving entenna will be essentially independent of the rotation angle. The measured pattern will then be that of the vertical pattern of the antenna, when it is mounted in its normal, vertical orientation. The validity of this method can be checked experimen-tally by measuring the radiation pattern of a well constructed reference dipole, whose free-space radiation pattern is accurately known. This pro-cedure was followed, using antenna measurement facilities at the University of Washington. References \1 Ballantine, Stuart, "On the Optimum Wavelength for a Vertical Antenna Over Perfect Earth," Proc. I.R.E., Vol. 12, p. 833, December 1924. \2. See Ref. 1. U Elliott, Robert S.,Antenna Theory and Design, Prentice Hall, 1981, Chapter 7. VI Morita, T., "Current Distributions on Transmitting and Receiving Antennas," Proc. I.R.E., Vol. 38, 1950, pp. 898 to 904. \Ji Stutzman, W. L., and G. A. Thiele, Antenna Theory and Design, John Wiley and Sons, 1981, Chapter 5. \£ See Ref. 1. Y L Terman, F. E., Radio Engineer's Handbook, Me-Gi-aw Hill, 1943, p. 843. \£ Balanis, C. A., Antenna Theory — Analysis and Design, Harper and Row, 1982, p. 515. \£ Reynolds, D. K., Facts About Proper VHF Ver-tical Antenna Design, AEA, Inc., P. O. Box 2160, Lynnwood, WA 98036. Vlfl. See Ref. 9. 1/2 W H I P 5 / 8 M W I P " R U B B E R D U C K I E " Fig. 7 — Measured vertical radiation patterns of hand-held transceiver with "rubber duckie," 1/2-, and 5/8-wave antennas. Antennas of Reduced Size Optimum Design of Short Coil-Loaded High-Frequency Mobile Antennas By Bruee F. Brown, W6TWW 1241 Arroyo Seeo Dr., Campbell, CA 95008 Optimum design of short high frequency mobile antennas results from a carefully considered balance among loading-eoil Q factor, loading-coil position in the antenna, ground loss resistance, and the length-to-diameter ratio of the antenna. To realize this optimum balance, a thorough understanding of how these antenna parameters interact is required. This paper presents a mathematical approach to designing mobile antennas for maximum radiation efficiency, and is an analysis of the parameters affecting optimum placement of a loading coil. The optimum loading-coil location could be found experimentally, but it would require many hours of design, construction, and measurements. A faster and more reliable way of determining optimum coil location is to write a mathematical program, using a personal computer. With the aid of the computer, all parameters except for one can be held constant, allowing exploration of the effects of a varying parameter. When plotted graphically, the data reveals that the placement of the loading coil is critical if we are to obtain maximum radiation efficiency. Radiation Resistance To determine antenna radiation e f f i c i e n c y , power losses caused by resistive losses in the antenna system and radiation loss must be defined. Radiation loss is expressed in terms' of radiation resistance. Radiation resistance is defined as the value of resistance that accounts for the portion of power input to the antenna system that is radiated, rather than dissipated as I R loss. In the interest of brevity, the variables used in the equations that follow are defined once in the text, and summarized in Table 1 for easy reference. R a d i a t i o n r e s i s t a n c e of v e r t i c a l a n t e n n a s shorter than 45 electrical degrees (1/8 wavelength) is approximatelyM o R r _ J l L (Eq. 1) ^ - 312 where Rr is the radiation resistance in ohms, and h is t h e height of the antenna in e l e c t r i c a l degrees. Antenna height in electrical d e g r e e s is e x -pressed by h = x f ^ 2 ) 360 (BJ- 2) 984 where L is the antenna height in feet, and f(MHz) is the operating frequency in megahertz. End effect is purposely omitted, to ensure that an antenna is electrically long. Thus, resonance at the design frequency can be obtained easily by removing a turn or two from the loading coil. Eq. 1 is valid only f o r antennas having a sinusoidal current profile and no reactive loading. It can be used, however, as a starting place to derive an equation that is useful for short antennas not having a sinusoidal current profile. From Fig. 1, it can be seen that the current profile on an antenna 90 electrical degrees high (1/4 wavelength) varies as the cosine of the height in electrical degrees. It is also apparent that the current profile of the top 30 degrees of the antenna is essentially linear. It is this l i n e a r i t y that allows derivation of a simpler, more useful equation for radiation resistance. The radiation resistance of an electrically short base-loaded vertical antenna can be con-veniently defined in terms of a geometric figure, a t r i a n g l e , as shown in Fig. 2.\2, The radiation resistance is Rr = KA (Eq. 3) where K is a constant that will be derived shortly, and A is the area of the triangular current profile in degree-amperes. Degree-amperes is expressed by A = i A 2 h X X , base (Eq. 4) By equating Eqs. 1 and 3 and solving for K, we S e t „ h 2 K = 312 x A (Eq. 5) By substituting the parametric values from Fig. 2 into Eq. 5 we get (30)' 312 x (0.5 x 30 x 1)" 0.012820513 and by substituting the derived value of K into Eq. 3 we get Rr = 0.012820513 X Az (Eq. 6) Eq. 6 is useful for determining the radiation resistance of coil-loaded vertical antennas of 30 degrees or less in height. The derived constant differs slightly from that presented by Laport because of the use of a different equation for 30 " " ' T O P = 0 A m P s E S S E N T I A L L Y L I N E A R C U R R E N T P R O F I L E 0.61279 0.76604 0.86603 0 - L - 1 . 0 LU, l B A S E = 1 A m p e r e Feed Points /7777777T777 C round Plane Pig. 1 — Relative current profile in a vertical antenna of height h equal to 90 electrical degrees. radiation resistance (Eq. 1). I did not round the constant for less decimal places. When the loading coil is moved up the antenna, the current profile is modified as shown in Fig. 3. The current varies as the cosine of the height in electrical degrees at any point in the base section. Thus, the current flowing into the bottom of the loading coil is less than the current flowing at the base of the antenna. But what about the current in the top section of the antenna? Well, the loading coil acts as the lumped constant that it is, and maintains the same current flow throughout. Therefore, the current exiting the top of the coil is the same as that entering the bottom of the coil. (This is true for conventional coils. However, radiation from long skinny coils allows coil current to decrease, as in helically wound antennas.) This is easily verified by installing RF ammeters immediately above and below the loading coil in a test antenna. Thus, the coil forces a much higher current into the top sec-tion than would flow in the equivalent part of a full 90-degree-high antenna. The cross-hatched area in Fig. 3 shows the current that would flow in the equivalent part of a 90-degree-high antenna, and reveals that the degree-ampere area of the whip section of the short antenna is greatly increased as a result of the modified current distribution. The extremely high voltage that appears at the top of the loading coil forces this high current flow into the top section. The result is more radiation from the top section than would occur from the equivalent part of a quarter-wave antenna. The current flow in the top section decreases almost linearly to zero at the top. This can also be seen in Fig. 3. h 'a 3 0 — — 0.0 2 5 -- 0.167 15 - - 0.5 1 0 -- 0.667 0 1 — 1 .
0 Feed Point h - 3 0 ° -h2 hi 0 — 1 — 1 .
0 . Feed Point < Z7777^7777 Cround Plane Fig. 2 — Relative current distribution in a base-loaded vertical antenna of height h equal to 30 electrical degrees (linearized). A base loading coil is omitted. BASE /77777T77777 Cround Plane Pig. 3 — Relative current distribution in a center-loaded antenna with base and top sections each equal to 15 electrical degrees in length. The cross-hatched area shows the current profile that would exist in the top 15 of a 90 -high vertical fed with 1 ampere at the base. The degree-ampere area of Fig. 3 is the sum of the triangular area represented by the current pro-file in the top section, and the nearly trapezoidal current profile in the base section. Radiation from the coil is not included in the degree-ampere area because it is small and difficult to define. What-ever radiation occurs can be considered a bonus. The degree-ampere area is expressed by A = ^ [hi x ( 1 + cos hi) + h2 x (cos hi)] 2 (Eq. 7) where hi is the electrical height in degrees of the base section, and h2 is the electrical height in degrees of the top section. The degree-ampere area (calculated by substi-tuting Eq. 7 into Eq. 6) determines the radiation resistance when the loading coil is at any position other than the base of the antenna. Using these equations, radiation resistance has been calculated and plotted against loading-coil positions for 8-and 11-foot antennas. Eight feet is a typical length for commercial antennas, and 11-foot antennas are about the maximum practical length that can be in-stalled on a vehicle. Calculations were made for three different frequencies. With reference to Fig. 4, the curves reveal that the radiation resistance rises almost linearly as the loading coil is moved up the antenna. They also show that the radiation resistance rises rapidly as the frequency is in-creased. If the analysis were stopped at this point, an erroneous conclusion that the loading coil should be at the top of the antenna could be reached. That is not so, and will be apparent shortly. Loading Inductance Required Calculations of the loading-coil inductance needed to resonate a short antenna can be done easily and a c c u r a t e l y by using the antenna transmission-line analog described in an article by Boyer.M For a base-loaded antenna, Fig. 2, the loading-coil reactance required to resonate the antenna is 9 8 7 ui s I O 6 z uj u z 5 < 1-— t / 1 o Q < O S 2 1 0 0 1 2 3 4 5 6 7 3 9 10 11 12 L O A D I N G C O I L P O S I T I O N IN F E E T A B O V E A N T E N N A B A S E Fig. 4 — Radiation resistance plotted as a function of loading coil position. If the loading coil is moved from its base position to the antenna midsection, the antenna is divided into a base and top section, as depicted in Fig. 3. The loading-coil inductive reactance re-quired to resonate the antenna when the coil is above the base position is expressed by XL = j Km2 x (cot h2) - j Kml x (tan hi) (Eq. 10) XL = -j Km x cot h (Eq. 8) where XL is the inductive reactance required, and - j indicates that the antenna presents capacitive reactance at the feed point. (This reactance must be canceled by a loading coil.) Km is the mean characteristic impedance, to be defined in Eq. 9. The mean characteristic impedance of an antenna is expressed by K m = 60 x [ J l o g e - TJ ( E q ' 9 ) where H is the height of the antenna (excluding length of the loading coil) in English or metric units, and a is the radius of the antenna in like units of measurement. From inspection of Eq. 9, it can be seen that decreasing the height-to-diameter ratio of an an-tenna by increasing the radius results in a decrease in mean characteristic impedance, Km. With reference to Eq. 8, a decrease in Km decreases the inductive reactance required to resonate an antenna. As will be shown later, this will increase radiation effi-ciency. Of course, in mobile applications we quickly run into wind-loading problems if we attempt to use an antenna with too large a diameter. In mobile-antenna design and construction, the top section of a whip usually has a much smaller diameter than the base section. Because of this, it is necessary to compute separate values of Km for the top and base sections. Kml and Km2 are the mean characteristic impedances of the base and top sections, respectively. Loading-coil reactance curves for the 75-meter antennas of Fig. 4 have been calculated and plotted in Fig. 5 to show the influence of the loading-coil position in the antenna on the reactance required for resonance. The curves in Fig. 5 show that the required reactance decreases with longer antennas. S i g n i f i c a n t l y , they reveal that the required loading-coil reactance grows at an increasingly rapid rate after the coil passes the center of the antenna. Since the highest possible loading-coil Q factor is needed, and since optimum Q is attained when the loading-coil diameter is twice the loading coil length, the coil would grow like a smoke ring above the center of the antenna, quickly reaching an intolerable size.\l It is for this reason that in all my computations the highest loading-coil posi-tion is limited to one foot from the top of the antenna. A3 3. AH 3. 9 Mhz, »' 1 9 Mhz, 1 1 ' j B A S E S E C T O P S E C T T I O N D I A . 1 ION D I A . 1/ 1/16" 4" A 3 1 A t j 1 / E = Pi X 1 0 0 P ^ 0 6 ^ (Eq. 12) 2 « 6 8 10 L O A D I N G C O I L L O C A T I O N IN F E E T A B O V E B A S E where E is radiation efficiency in percent, Pr is the power radiated, and Pi is the power fed to the antenna at the feed point. In a short coil-loaded mobile antenna, a large portion of the power fed to the antenna is dissi-pated in ground and coil resistances. A relatively insignificant amount of power is also dissipated in the antenna conductor resistance and in leakage resistance of the base insulator. Because these latter two losses are both very small and difficult to define, they are neglected in calculating radiation efficiency. Another loss, worthy of note in passing, is coupling-network loss. Since we are concerned only with power fed to the antenna in the determination of radiation efficiency, coupling-network loss is not considered in any of the equations. Suffice it to say that coupling networks should be designed for minimum loss in order to maximize the transmitter power available to the antenna. The radiation efficiency equation may be rewritten and expanded as follows: I X Rr X 1 Q Q I 2 x Rr + I 2 x Rg + (I cos h i ) 2 x Rc Fig. 5 — Loading c o i l reactance required for resonance, plotted as a function of c o i l height above the antenna base. The resonant frequency is 3900 kHz. Loading Coil-Resistance Loading-coil resistance is one of the resistive losses that consumes power which could otherwise be radiated. Heat loss in the loading coil is not of any benefit, so it should be minimized by using the highest Q possible. Loading-coil loss resistance is a function of the coil Q factor anc is expressed b y r c = XL (Eq. 11) Q where Rc is the loading-coil loss resistance, XL is the loading-coil inductive reactance, and Q is the coil figure of merit. From inspection of Eq. 11, we can see that, for a specific value of inductive reactance, the loss resistance will be lower for higher-Q coils. High Q can be obtained by going to l a r g e -diameter coils having a diameter-to-length ratio of two, by using larger diameter wire, by using more spacing between turns, and by using low-loss poly-styrene supporting and enclosure materials. Load-ing-coil turns should not be shorted for tuning purposes because shorted turns degrade Q. The highest possible Q is needed. Pruning to resonance should be performed by removing turns from the coil instead. Radiation Efficiency The ratio of power radiated to power fed to an antenna determines the radiation efficiency. It is expressed by (Eq. 13) where I is the antenna base current, Hg is the ground loss resistance, and Rc is the coil loss resistance. Each term of Eq. 13 represents the power dissipated in its associated resistance. By in-spection it can be seen that all the I values cancel out, simplifying the equation to E = x 100 Rr + Rg + Rc x cos 2 hi (Eq. 14) 2 For base-loaded antennas the term cos hi reduces to unity and may be omitted. Ground Loss Eq. 14 shows that the total resistive losses in the antenna system are 2 Rt = Rr + Rg + Rc x cos hi (Eq. 15) where Rt is the total resistive loss. Ground loss resistance can be determined by rearranging Eq. 15 as follows: 2 Rg = Rt - Rr - Rc x cos hi (Eq. 16) Rt may be measured in a test antenna instal-lation on a vehicle using an antenna noise bridge or an RX bridge. Rr and Rc can be calculated. Ground loss is a function of vehicle size, placement of the antenna on the vehicle, and con-ductivity of the ground over which the vehicle is traveling. Only the first two variables can be controlled. While I don't discourage anyone from going lo a small vehicle to reduce fuel consumption, be aware that larger vehicles provide a better ground plane than smaller ones. In any event, the vehicle ground plane is only partial, so the result is considerable RF current flow (and ground loss) in the ground around and under the vehicle.V5. By raising the antenna base as high as possible on the vehicle, the ground losses are decreased. This results from a decrease in antenna capacitance to ground, which increases the capacitive reactance to ground. This, in turn, reduces ground currents and ground losses. This effect has been verified by installing the same antenna at three different locations on two different vehicles, and by determining the ground loss from Eq. 16. In the first test, the antenna was mounted 6 inches below the top of my large station wagon, just behind the left rear window. This placed the antenna base 4 feet 2 inches above the roadway, and resulted in a measured ground loss resistance of 2.5 ohms. The second test used the same antenna mounted on the left rear fender of a mid-sized sedan, just to the left of the trunk lid. In this test, the measured ground loss resistance was 4 ohms. The third test used the same mid-sized sedan, but with the antenna mounted on the rear bumper. In this last test, the measured ground loss resistance was 6 ohms. So we have the same antenna seeing three different ground loss resistances purely as a result of the antenna mounting location and size of the vchicle. It is important to note that the measured ground loss increased as the antenna base was nearer the ground. The importance of minimizing ground losses in mobile antenna installations will become apparent shortly. Efficiency Curves With the previously defined equations, a computer program was written in BASIC and used to calculate the radiation-efficiency curves depicted in Figs. 6 through 9. These curves were calculated for 75- and 40-meter antennas cf 8- and 11-foot lengths. Several values of loading-coil Q factor were used, for both 2 and 10 ohms of ground loss resistance. For the calculations, the base section is half-inch-diameter electrical metallic tubing (EMT), which has an outside diameter of 11/16 inch. The top section is fiberglass bicycle-whip material covered with Belden braid. These are readily avail-able materials which can be used by the average amateur to construct an inexpensive but rugged antenna. Upon inspection, these r a d i a t i o n - e f f i c i e n c y curves reveal some significant information: (1) higher coil Q produces higher radiation e f f i c -iencies; (2) longer antennas produce higher radia-tion efficiencies; (3) higher frequencies produce higher radiation efficiencies; (4) lower ground loss resistances produce higher radiation efficiencies; (5) higher ground loss resistances force the loading coil above the antenna center to reach a crest in the radiation-effieicncy curve; and (6) higher coil Q sharpens the radiation-efficiency curves, result-ing in the coil position being more critical for optimum radiation efficiency. The reason that the radiation-efficiency curves reach a crest and then begin to decline as the load-ing coil is raised farther is because of the rapid 9 3 7 " 6 > U z uj y 5 u. u. u i z O 1 ) I-< Q < 3 a: J 2 1 ° 0 2 "I 6 8 LOADING COIL POSITION IN FEET ABOVE BASE Fig. 6 — Radiation e f f i c i e n c y of 8 - f t antennas at 3900 kHz. 18 16 1 4 U z U J u. L U I-< a 1 2 0 0 2 It 6 8 10 12 LOADINC COIL POSITION IN FEET ABOVE BASE Pig. 7 — Radiation e f f i c i e n c y of 1 1 - f t antennas at 3900 kHz. increase in loading-coil reactance required above the antenna center. Refer to Fig. 5. The rapid increase in coil size required for resonance results \ V C IF ? i 1 0 1 Q : " I a ar / C O I L Q CURVE A N T E N N A D I A M E T E R 150 300 150 A , D B , E C , F B A S E S E C T I O N = 1 1 / 1 6 " T O P S E C T I O N = 1 / 1 " 0 2 1 6 8 L O A D I N C C O I L P O S I T I O N IN FEET A B O V E T H E BASE Pig. 8 — Radiation efficiency of 8-ft antennas at 7225 kHz. ANTENNA CIAMETER BASE SECTION =11/16" TOP SECTION = 1/1" 2 4 6 8 10 LOADINC COIL POSITION IN FEET ABOVE THE BASE Fig. 9 — Radiation efficiency of 11-ft antennas at 7225 kHz. in the coil loss resistance increasing much more rapidly than the radiation resistance. See Fig. 4. This results in deercesing radiation efficiency. A slight reverse curvature in the curves be-tween the base-loaded position and the one-foot coil-height position is visible. This results from a shift in the curve because of insertion of a base section of larger diameter than the whip when the coil moves above the base. The curves in Figs. 6 through 9 were calculated with constant, but not equal-diameter base and whip sections. Because of wind loading, it is not desir-able to increase the diameter of the whip section. However, the base-section diameter can be increased within reason to further improve radiation e f f i c -iency. Fig. 10 was calculated for base-section diameters ranging from 11/16 of an inch to three inches. The curves reveal that a small increase in radiation efficiency results from larger diameter base sections. The curves in Figs. 6 through 9 show that radiation efficiencies can be quite low in the 75-meter band compared to the 40-meter band. They would be lower still in the 160-meter band. To gain some perspective on what these low efficiencies mean in terms of signal strength, Fig. 11 was calculated using the following equation: dB = log (Eq. 17) where dB is the signal loss in decibels, and E is efficiency in percent. The curve in Fig. 11 reveals that an antenna having 25% efficiency has a signal loss of 6 dB (one S unit) below a quarter-wave vertical reference antenna operated over perfect ground. An antenna efficiency in the neighborhood of 6 S & will produce a signal strength on the order of two S units or 12 dB below a quarter wave reference vertical. By careful optimization of mobile-antenna design, it can be seen that signal strengths from mobiles can be fairly competitive with those from fixed stations of comparable power. Impedance Matching The input impedance of short high-Q coil-loaded antennas is low. As an example, an optimized 8-foot antenna for 3,900 kHz having a coil Q of 300 and a ground loss resistance of two ohms has a base input impedance of about 13 ohms. This low value of imped-ance causes a standing wave ratio in 50-ohm coaxial line of 3.85:1 at resonance. This high SWR is not compatible with the low SWR requirements of solid-state finals. In addition, the bandwidth of short vertical antennas is very narrow. This severely limits the capability to maintain transmitter load-ing with a small frequency change. Impedance matching can be accomplished by means of L networks or impedance-matching transformers, but the narrow bandwidth limitation remains. A more elegant solution to the impedance matching and narrow bandwidth problem is to install en automatic tuner at the antenna base.\£ Such a device matches the antenna and coaxial line properly, and permits operation over a wide frequency range. Conclusion The mathematical modeling technique employing a personal computer revealed that loading-coil Q fac-tor and ground loss resistance greatly influence the optimum loading-coil position in a short vertical < 5 6 < C U R V E Q HEIGHT HEIC - A 300 11' 5' B 150 11' 5' C 300 8' D 150 8' V 3 1 1 .5 2 2.3 BASE SECTION DIAMETER IN INCHES Fig. 10 — Radiation efficiency plotted as a function of base section diameter. Frequency — 3900 kHz. Ground loss resistance — 2 ohms. Whip section — 1/4-in diameter. j i O S I C N A L LOSS IN '5' UNITS (S db PER 'S' UNIT) Fig. 11 — Mobile-antenna signal loss as a function of radiation efficiency, compared to a quarter-wave vertical antenna over perfect ground. antenna. It also shows that longer antennas, higher coil Q, and higher operating frequencies produce higher radiation efficiencies. The tools are now available to tailor a mobile antenna design to produce maximum radiation effic-iency.YZ. A missing ingredient is the ready availa-bility of very high-Q commercial coils. It is hoped that this paper will stimulate both the demand for and the production of such coils. End effect has not been included in any of the equations to assure that the loading coil will be slightly larger than actually required. Antenna pruning to resonance should be done by removing coil turns, rather than by shorting turns or shortening the whip section. Shortening the whip section reduces radiation efficiency, by both shortening the antenna and moving the optimum coil position. Shorting turns degrades the coil Q factor. Shortened Dipoles The mathematical modeling technique can be applied to shortened dipoles by using zero ground loss resistance and by doubling the computed values of radiation resistance and feed-point impedance. However, radiation efficency does not double. The reason is that to complete the other half of the shortened dipole requires a second loading coil. The doubling of radiation resistance is offset by the coil resistance of two coils, leaving the radiation efficiency unchanged. There is a gain in radiation efficiency over a vertical antenna worked against ground because the dipole configuration permits ground loss resistance to be zeroed out of the model. Acknowledgem ents I wish to acknowledge and express my apprec-iation to Rudy Severns, N6LF, for patiently urging me to prepare this paper, his numerous suggestions, experimental verification of the current flow into and out of the loading coil, and assistance in proofreading the manuscript. I also wish to express my appreciation to Syd Padrick, VVB6HXJ, for his assistance in making the ground loss measurements, and to John McCollum, WB6LVD, for his word-processing assistance. Table I A = area in degree-amperes a = antenna radius in English or metric units dB = signal loss in decibels E = efficiency in percent f(MHz) = frequency in megahertz H = height in English or metric units h = height in electrical degrees hi = height of base section in electrical degrees h2 = height of top section in electrical degrees I = I[jase = 1 ampere base current K = 0.012820513 Km = mean characteristic impedance Kml = mean characteristic impedance of base sec-tion Kin2 =mean characteristic impedance of top section L = length or height of the antenna in feet Pi = power fed to the antenna Pr = power radiated Q = coil figure of merit Rc = coil loss resistance in ohms Rg = ground loss resistance in ohms Rr = radiation resistance in ohms XL = loading-coil inductive reactance References \1 Smith, Carl E. and Earl M. Johnson, "Per-formance of Short Antennas," Proceedings of the IRE, October 1947. \2 Laport, Edmund A., Radio Antenna Engineering, 1952, p. 23. \2 Boyer, Joseph M., "Antenna-transmission line analog," Ham Radio, May 1977. Terman, Frederick E., Radio Engineering Hand-book, Third Edition, 1943, p. 74. V 5 . Belrose, J. S., "Short Antennas for Mobile Operation," QST, Sept. 1953. \£ Brown, Bruce F., "Tennamatic: An Auto-Tuning Mobile Antenna System", 73, July 1979. \1 A computer program is available from B Square Enterprises, P.O. Box 71, Campbell, CA 95008. Short Loaded Dipole Design Easy Way By H. L. Ley, Jr., N3CDR •Herbert L. Ley Associates, Inc., P. O. Box 2047, Rockville, MD 20852 An important but often neglected element in a station is the antenna. The present-day ham operates at a disadvantage with respect to his colleagues of 25 to 50 years ago. Available real estate for an-tenna installation is elusive, and antennas are now being placed in limited quarters such as an apart-ment or boat. Hie ARRL Handbook and Hie ARRL Antenna Book offer useful, but limited information on problems like these. In the past, I've inserted a loading inductor at an appropriate point in the antenna. By measuring the resonant frequency with a grid-dip meter or noise bridge, I would prune the coil until I brought the shortened, loaded antenna to resonance. Enter, the Home Computer With the availability of home computers, it is now possible to approach short loaded half-wave dipole design in a rational manner. The basic for-mulas have been known for some time, but they are complex and long combinations of calculations are sometimes required to arrive at a single solution. Most hams are unwilling to spend laborious hours evolving the best design for a given physical layout. I have brought the necessary equations together into a cassette tape computer program for both the Texas Instruments 99/4A home computer and the Sin-clair ZX-81/Timex-Sinclair 1000 home computer with 16 K added memory. The programs are convenient to use and permit anyone to explore more than a half-dozen options for a short-loaded dipole in a given physical layout during a 15-30 minute period. If sufficient demand develops, I will also provide the program in 5 inch single-sided, single-density or double-density disk format for the Heath/Zenith 89 computer in Microsoft BASIC running on a CPM operating system. Persons desiring full documen-tation of the program are invited to write me directly. Basics of Loaded Dipole Design Before I provide examples of the programs, the basic factors involved in the design of loaded dipoles should be reviewed. This will help us to understand how the computer programs work. Fig. 1A presents the voltage and current distributions in a conventional, full-length center-fed half-wave di-pole. Current is maximum at the center feed point where the voltage is lowest. Conversely, voltage is maximum and current is zero at the ends of the antenna. Half-Wave — The Figure IB presents the same curves for a resonant dipole loaded in the center of each ele-ment. It is half the length of the antenna shown in the first view and its voltage and current in the first and last half of each element is about the same as in Fig. 1A. At the location of the loading coils, there is an abrupt change in both voltage and current distribution along the antenna. The loading coils replace the missing portions of the original, full length antenna, and carry a significant voltage and current drop over the length of the coils. VOLTAGE SHORTENED LOADED DIPOLE Fig. 1 — A comparison of currents and voltages in a full length and shortened loaded dipole. The short half-wavelength loaded dipole ex-hibits two distinctive features in comparison with a full length antenna. The harmonic resonant fre-quencies cannot be predicted as easily as they can in a full-length dipole.M For this reason, short loaded dipoles are not recommended for harmonic operation. If this antenna is used for more than one frequency, arrangements must be made to short out unused inductors or have a separate plug-in loading inductor for each band of operation. The second difference is antenna bandwidth, which is decreased in comparison with a conventional dipole.\2, Careful adjustment of the loading-coil values used to center the resonant frequency at the desired operating frequency is required if you are to obtain a low SWR on the feed line. If any antenna is resonant (exhibits only resistance and no reactance at the resonant fre-quency), it is possible to calculate the voltage at the feed point by the conventional power equa-tions: E r a s - J f T I f Design Examples Irms = H T Problem 1: Design a short-loaded dipole, 16 ft R long, to fit a balcony space in an apartment. wnere, P = power transmitted to antenna (1:1 &1R) in watts. R = radiation resistance of antenna in ohms (as measured with an impcdance or noise b r i d g e ) . At a power level of 100 W and a radiation resistance of 50 ohms, the appropriate values are E = 70.7 V and I = 1.4 A. These values are important because they indicate the insulation and current-carrying capacity required at the feed point. Maximum radiation from the antenna occurs in the high central current portions. This establishes rule no. 1 of a loaded antenna design: Make the distance from the feed point to the loading coil as long as possible, consistent with your space re-quirements. Given a short antenna of a fixed length, the inductance of the loading coils required to bring it to resonance increases In value as they are moved from the center of the antenna toward the ends. This is important when working with loading coils. In-deed, at the very tips of the short antenna, a loading coil would require infinite inductance to resonate it. Rule no. 2 therefore states: Place the loading coil at a position in a short antenna such that it does not require an unreasonable value of inductance to resonate the antenna. Because rules 1 and 2 conflict with each other, a compromise is necessary. This can best be achieved by developing a series of options for possible an-tennas to fit a given space in terms of the distance of the loading coils from the feed point, and the inductance values required in the various positions to resonate the antenna. As a general rule, placing the loading coils at the feed point should be avoided for two reasons. First, the high-current portion of the antenna that is responsible for most radiation is replaced by a coil. Second, the high current at that location results in the highest I R heat losses in the in-ductor. The best position then is somewhere along the antenna element away from the feed point. This location depends on physical constraints and the size of the loading coil you have. Previous Articles on Loaded Dipole Design The basic mathematics for computation of the inductance required for loading short dipoles to resonance was developed by J. Hall, K1TD.\1 His article was acquired by Sander who wrote an ab-breviated computer program in BASIC for the Apple II Plus Computer, with Orr and Rasor later revising the program for the TRS-80 c o m p u t e r . B o t h programs are highly abbreviated and do not include sub-programs for calculating the matching impedance at the feed point by the method of Gooch et al,\fL or for calculating the winding data for both the loading and matching inductances. The consolidated program I wrote includes all of these features. To illustrate the use of the program, an example of an actual design is presented. After loading the cassette into the computer, you will receive a prompt for the operating frequency. I picked 7.125 MHz, the center of the 40-meter novice band (see Fig. 2). I specified a total length of 16 ft, and a distance of 2 ft from the center for the loading coils. My next step was to decide if I would use wire or aluminum tubing for the antenna. For the first run, I chose no. 18 wire. The computer cal-culated a value of 33.0 yuH for each loading coil. Those are big coils! What would have been the value if I chose Reynolds (TM) 0.75-in diameter with 0.047-in wall thickness tubing? It takes only a minute to obtain the new results. 6 fc ene scction, 0.~5 in tubing • I fr sprrior d Ft section 8BCSS& L4SJ2U IS.8 pH 365 pf ^ 3 6 5 pf iS.S ^ l H 52 ohm coax i iu feedlinc Length, 40 in Short, 55 in Pig. 2 — Configuration of indoor loaded dipole for 7.10-7.15 MHz. ENTER OPERATING ?REQ IN MHZ = 7.125 ENTER TOTAL LENGTH IN FEET = 16 ENTER DISTANCE F30M CENTER TO LOADING CDIL IN FEET = 2 DO YOU WANT TO USE THE WIRE TABLE? ( Y OR N) ENTER WIRE GAUGE (10 TO 30) = 18 LOADING CDIL IS 33.000237 UHY DO YOU WISH TO CALCULATE WINDING DATA FOR COIL? (Y OR N) C o r r p u t e r printout of calculation for 16 ft no. 18 wire antenna. When tubing is used for the antenna, a much smaller inductor is needed for resonance (see Fig. 3). The calculated value is 19.5 jiH — a consid-erable decrease in inductance and coil size. Matching Stub No. 12 copper wire c -I i — ENTER OPERATING FREQ IN MHZ = 7.125 ENTER TOTAL LENGTH IN FEET = 16 ENTER DISTANCE FROM CENTER TO LORDING OOIL IN FEET = 2 DO YOU WANT TO USE THE WIRE TABLE? (Y OR N) WHAT IS ELEMENT DIAMETER IN DECIMAL INCHES = 0.75 LOADING CDIL IS 19.469348 UHY DO YOU WISH 1 X 3 CALCULATE WINDING DATA FOR CDIL? (Y OR N) INPUT COIL DIAMETER, DECIMAL INCHES = 2 INPUT CDIL LENGTH, DECIMAL INCHES = 4 NUMBER OF TURNS REQUIRED = 30.886859 DO YOU WISH TO CALCULATE ANOTHER VERSION OF CDIL? (Y OR N) Fig. 3 — Computer printout of calculation for 16 ft 0.75-in tubing antenna. The computer program next prompted me to decide if I wanted to calculate the winding data for the inductance. If I answered "Y" for yes, the computer would perform the necessary mathematics, giving an answer of 31 turns of wire, 2 in diameter and 4 in long. The ARRL L/C/F Slide Rule Calculator may used as an alternative. It is easier to use if you're working with commercial coil stock where turns per inch is the critical parameter. In that case, answer "N" for no to the computer query. The program provides an opportunity to calcu-late the center matching inductance, connected to the feed point, to match a specified feed line impedance (see Fig. 4). The program calculates the antenna impedance (unless you have a measured value to substitute), the matching inductive reactance and the required inductance at the operating frequency. A word of caution is appropriate here. If the an-tenna length originally chosen is longer than a half wavelength at some frequency, the program will return a negative value for the inductance. The necessary matching element is then capacitive, and its value can be calculated by hand from the reactance at chosen the frequency.VZ. These calculations can be continued for the same 16-ft antenna, varying the distance from the center for the loading coils, and for other bands. Results from such calculations are shown in Table 1. This data clearly shows the increase in value of the loading coils as they are moved away from the feed point. It also shows the rising antenna impedance as frequency is increased. ANTENNA IMPEDANCE IS ESTIMATED BY MULTIPLYING 52 OHMS BY RATIO OF TOTAL LENGTH TO FULL DIPOLE LENGTH = 12.664889 OHMS DO YOU HAVE A MEASURED VALUE TO SUBSTITUTE? (Y OR NO) INDUCTIVE REACTANCE (XL) = 29.506238 OHMS MATCHING INDUCTANCE = 0.65918917 UHY THIS ODIL WILL HAVE TO BE TRIMMED TO LOWEST SWR DO YOU WISH TO CALCULATE WINDING DATA FOR CDIL? (Y OR N) Fig. 4 — Computer printout of calculation for center matching inductor. Problem 2: Design a short, loaded dipole, 20 ft long, to fit on the ceiling of a room measuring 8 by 10 ft. In this problem, the antenna is to be installed indoors suspended from the ceiling of a room. The second part of Table 1 gives the results of a series of calculations, for 0.75-in aluminum tubing, based on inserting the loading inductances 2, 3 and 4 ft from the feed point for a resonant frequency of 7.125 MHz. The most desirable physical arrangement for this antenna turns out to be a "U" shape with the loading coils placed 4 ft from the feed point on an 8 ft side. Each element is then extended with a 6-ft length of tubing at right angles to the 8-ft center portion. With this configuration, a loading coil of 18.78 ^iH should bring the antenna to resonance for 7.125 MHz. For 14.2 and 21.15 MHz, the appropriate values are 3.29 and 0.43 jjH. This computer-designed antenna was fabricated and installed in a room on the fourth floor of a multi-unit apartment building. It was suspended 10 in below the ceiling from hooks installed in the ceiling. Nylon cord was used as insulators. A Plexiglas corner fitting held the tubing sections at right angles and provided a mounting for the loading coils installed at the corners. When initially checked for resonance with a Palomar noise bridge and a general-coverage receiver, the antenna was found to be resonant approximately 10% below the design frequencies. Because of capacitive loading by house wiring and heating ductwork, the antenna impedance was estimated to be about 15 ohms at 7.125 MHz and 40 ohms at 21.15 MHz. Although the loading coils could have been pruned to bring the antenna to resonance, it was elected to use a simpler method — series capacitive correction. The configuration worked out for 40 m is shown in Fig. 2. This design permits the resonance ad-justment to be made at the antenna feed point. The receiver and noise bridge are connected to the antenna with a half wavelength of coax cable to determine the proper settings of the series ca-pacitors. When tuned to resonance with the matching stub shown, the antenna impedance at 7.125 MHz was found to be approximately 50 ohms. At one point, the capacitors were placed in series with each of the loading coils at the corners of the antenna on the feed-point side. The electrical characteristics of the antenna appeared identical to those it exhibited with the capacitors at the center, but the incon-venience of setting two capacitors located at opposite ends of the room was bothersome. There-after, the feed point location was used! To simplify the adjustment further, the two variable capacitors were mounted on a piece of Plexiglas and ganged by three gears. One was placed on each capacitor shaft and the third, an idler gear, was mounted between the two capacitors. A tuning knob was attached to the shaft of the idler gear to permit one-hand tuning. The logic behind this method of tuning the antenna to resonance is simple. The antenna appears to be longer than it actually is when loaded near building structures. It therefore exhibits inductive reactance. By adding an equal capacitive reactance to cancel the inductive reactance, the antenna is adjusted to resonance. Using this antenna over a period of one month on the 7.10-7.15 MHz band, I confirmed contacts with both New York and Wisconsin using a Heath HW-8 QRP transceiver. Maine, Massachusetts, New York, Penn-sylvania, Ohio, Indiana, Illinois, Wisconsin and West Virginia were contacted with a 30-year-old crystal-controlled Johnson Adventurer transmitter (30 W measured output) and a Collins 51J-4 receiver. Other contacts with New Hampshire and Vermont have not yet been confirmed. The antenna favored the north and west, probably because the center section was hung in a general east-west direction. Little effort was put into 21-MHz contacts, other than local. I had to prove to my satisfaction that it could be adjusted to resonance on that band as well. Most of those I contacted were surprised, to put it mildly, to learn that I was working them on a 20-ft, indoor loaded dipole! Myth or Reality? When the 16-ft loaded tubing dipole shown in Table 1 was tested in a balcony installation, the antenna resonated at a lower frequency than 7.125 MHz for which it was designed. Inductor trimming was necessary and this observation caused doubts about the validity of the equations used for the design work. It must be stated that the antenna was located parallel to and about 3 ft distant from both a rain gutter and a steel balcony railing. These obstacles could be expected to load the antenna capacitively, thereby lowering its resonant fre-quency and its efficiency as a radiator. A 1:1 balun was installed at the feed point of the same antenna. With the aid of KA3FLA, the loading coils were trimmed for 7.125 MHz in the balcony installation. The antenna was erected in open space about 7 ft above moist ground. A Palomar R-X noise bridge was connected to the antenna with 50 ft of RG-58 foam coax cable. The cable's measured velocity factor was 0.770, making the cable one-half electrical wavelength at 7.57 MHz. Several addi-tional lengths of cable were available for coupling to the 50—ft length to make total cable lengths corresponding to half wavelengths at 7.050, 7.125, 7.200 and 7.300 MHz. A Heath HD-1250 dip meter was used as a radiating extrapolation oscillator coupled with a small pickup coil to a Heath IM-4100 fre-quency counter. It gave an accurate frequency mea-surement of the noise null in a general-coverage SWL-type receiver to determine the resonant fre-quency of the antenna with the noise bridge. Much to my surprise, the resonant frequency of the antenna installed in an open area was about 7.5 MHz. A series of four frequency measurements of the reso-nant frequency of the antenna was made with the 50-ft coax section alone as the best compromise length. These were averaged to a value of 7.47 MHz. The radiation resistance of the antenna was esti-mated with the use of resistance standards to be 15 ohms (the resistance scale of the Palomar noise bridge is not suitable for accurate measurements). When the resonant frequency of 7.47 MHz was entered into the computer program, a calculated value for the loading coil placed 2 ft from the feed point was 17.59 uH. The ARRL L/C/F Slide Rule Cal-culator yielded a calculated value for the coil installed in the antenna of 17.7 jiH. The two values differed by only 0.6X, validating rather well the equations used in the computer program. No effort was made to accurately measure the resonant frequency of the antenna with smaller loading coils pruned to produce resonance in the 20-and 15-meter bands in the balcony installation be-cause of the limited choices of length of the coax feed line. We were fortunate that the 50-ft length of coax approximated in its electrical length a half wavelength at the resonant frequency of the antenna errected in open space. This experience proved to me that the formulas used in the computer program were adequate to pre-dict values of loading coils used to resonate the short antenna. It also provided evidence that the detuning effects of parallel conductors in proximity to the antenna could be major, as much as 5& in my case. Matching the Antenna and Feed line The 40-meter, short loaded dipole presents a major problem of matching to a 52-ohm coax feed line. The radiation resistance of the 16-ft antenna approximates 13 ohms. Delrin rod (0.625-in diameter) is recommended for the center joint of the antenna, for spacers for the break at the loading coils, and for the end insulators. It possesses excellent electrical, mechanical and machining character-istics. If the 0.75-in aluminum tubing is slit for approximately 1.5 In at the ends, a no. 10 bolt through the tubing and Delrin rod compresses the tubing around the rod to make a stable joint. The bolt, preferably aluminum to minimize corrosion problems, should be placed about 0.75 in from the end of the tubing.\& This matching problem has been approached with a gamma match in many beam antennas, and a similar approach could be used with the short loaded dipole. However, one reason for the popularity of the gamma match in beams is that it permits the use of an unbroken radiator, giving greater structural strength to the antenna. I have not found that to be a necessary feature in apart-ment antennas, so I have resorted to an older matching network called the "hairpin match," de-scribed by Gooch.Vfl The "hairpin" is installed at the feed point between two electrically isolated halves of the antenna. If made of no. 8 copper wire spaced 2 in apart, it can be adjusted easily by means of a shorting bar that can be tightened on the wires to provide good electrical contact. A useful approximation of the inductance of such a hairpin is provided by the following formula: L (uH) = Length to shorting bar, in 26 To calculate the approximate length of the hairpin to the shorting bar, the formula can be rearranged, as follows: Length to shorting bar, in = L^iU) x 26 These relationships are approximately correct forlengths of 6 to 24 in. The graph in Gooch's article is not a linear function, so these formulas are not exact. Nevertheless, because the shorting bar has to be adjusted empirically for lowest SWR, they do provide a starting point for adjustments. The computer provides the value of the inductance required at the feed point to convert the low im-pedance of the loaded dipole to any specified feed line impedance. As with the loading coils, if the computer returns the value of a negative inductance, the matching element is capacitive, not inductive. Obviously, the matching element could be a typical helically wound eoil rather than the hairpin. The hairpin is so much easier to adjust that I have found it preferable to a conventional coil. Summary The availability of the home computer permits rapid, reasonably accurate calculations of a number of options for installation of shortened, loaded dipoles in limited space. My experience Willi the use of my program developed from Amateur Radio related literature has been convincing with its predict-ability of the resonant frequency of the short loaded dipole in open space. It is also clear from the results of installation of apartment antennas that the detuning effect of conductors in the vicinity of the antenna is not easily predictable. In brief, the computer program will provide a starting point for an antenna design in a dwelling area, but it does not substitute for the "cut and try" pruning method used to compensate for the detuning effects of the environment on the an-tenna. I would be interested to correspond with other amateurs using the basic formulas employed in my computer program for antenna design. Please enclose an s.a.s.e. for a prompt reply. Good luck, and good radiation! References \1 National Bureau of Standards, Circular C74, Radio Instruments and Measurements (Wash-ington: U.S. Government Printing Office, 1937), pp. 75-78. \2. ARRL Staff, Hie ARRL Handbook, 55th Ed., (Newington: ARRL, 1978), p. 608. \1 J. Hall, "Off-Center-Loaded Dipole Antennas", QST, Sept. 1974, p. 28. D. Sander, "A Computer Designed Loaded Dipole Antenna", CQ, Dec. 1981, p. 44. B. Orr, "Ham Radio Techniques", Ham Radio, Apr. 1983, p. 52. \£ J. D. Gooch, O. E. Gardner and G. L. Roberts, "The Hairpin Match", QST, Apr. 1962, p. 11. \X Formula for calculating capacitance from capacitive reactance: C(F) = ] 2 x pi x f (H) x Xa (ohms) Delrin rod may be purchased from plastic suppliers or from Read Plastics, Inc., 12331 Wilkins Ave., Rockville, MD 20852, tel. (301) 881-7900. \2. See Ref. 6. •Cable 1 Antenna Design Exaiqples Total Load Coil Load Coil Antenna Match Match Length Freq. Ft From Induct Imped React Induct Efc/MafcL M H z Genter ^JH Ohms Ohms uH 16/#18 7.125 2 33.0 12.7 29.5 r 0.66 wire 16/0.75 7.125 2 19.46 12.7 29.5 0.66 tubing n 3 22.36 n n ft n 4 26.49 n n n 16/0.75 14.2 2 3.95 25.2 50.5 0.57 tubing 21.15 2 1.10 37.6 84.0 0.63 20/0.75 7.125 2 15.02 15.8 34.4 0.77 tubing n 3 16.65 n n tt it 4 18.78 0 Tt n 20/0.75 14.2 4 3.29 31.6 64.6 0.72 tubing 21.15 4 0.43 47.0 159.4 1.20 16/118 7.125 2 33.0 12.7 29.5 0.66 wire 16/0.75 7.125 2 19.46 12.7 29.5 0.66 tubing 3 22.36 n n rt n 4 26.49 n tt ti 16/0.75 14.2 2 3.95 2 5 . 2 50.5 0.57 tubing 21.15 2 1.10 37.6 84.0 0.63 20/0.75 7.125 2 15.02 15.8 34.4 0.77 tubing n 3 16.65 n n n n 4 18.78 n n tt 20/0.75 14.2 4 3.29 31.6 64.6 0.72 tubing 21.15 4 0.43 47.0 159.4 1.20 Computer program for ZX-81/Times-Sinclair 1000 computer +16k RAM 1000 CLEAR 1010 CLS 1020 PRINT "PROGRAM TO DESIGN LOADED DIPOLB" 1030 PRINT "q/R 1983, HERBERT L. LEY ASSOC., INC." 1040 PRINT 1050 PRINT "ENTER OPERATING FREQ IN MHZ" 1060 INPUT F 1070 PRINT "= ";F 1080 PRINT 1090 PRINT "ENTER TOTAL LENGTH IN FEET "; 1100 INPUT A 1110 PRINT "= ";A 1120 PRINT 1130 PRINT "ENTER DISTANCE FROM CENTER TO LOADING COIL IN FEET"; 1140 INPUT B 1150 PRINT "= ";B 1160 PRINT 1170 LET X=((234/F)-B) 1180 LET Y=((A/2)-B) 1190 PRINT "DO YOU WANT TO USE THE WIRE TABLE? (Y OR N)" 1200 INPUT Y$ 1210 IF Y$="Y" THEN GOSUB 1520 1220 IF Y$^"Nn THEN GOSUB 1680 1230 LET S1=1E6/(68PI2F2) 1240 LET S2=1/X(LN (24Ji/D)-l) 1250 LET S3=(l-FB/234)2-l 1260 LET S4=l/Y (LN (24Y/D)-1) 1270 LET S5=(YF/234)2-l 1280 LET L=S1(S2S3-S4S5) 1290 PRINT 1300 PRINT "LOADING COIL IS "; 1310 PRINT L;n JUHY" 1320 PRINT 1330 PRINT "DO YOU WISH TO CALCULATE WINDING DATA FOR CDIL? (Y OR N)" 1340 INPUT Y$ 1350 IF Y$="Y" THEN GOSUB 2090 1360 CLS 1370 PRINT 1380 PRINT "DO YOU WISH TO CALCULATE VALUE OF CENTER MATCHING CDIL? (Y OR N)n 1390 INPUT Y$ 1400 CLS 1410 IF Y$="Y" THEN GOSUB 1730 1420 CLS 1430 PRINT 1440 PRTNT "DO YOU WISH TO CALCULATE ANOTHER ANTENNA? (Y OR N) " 1450 INPUT Y$ 1460 IF Y$="Y" THEN GOTO 1000 1470 PRINT 1480 PRINT "PROGRAM TERMINATED, TO RUN AGAIN ENTER ""R""" 1490 INPUT R$ 1500 IF R$="R" THEN GOTO 1000 1510 STOP 1520 PRINT 1530 PRINT "ENTER WIRE GAUGE 1540 INPUT D 1550 PRINT " = = ";D 1560 IF D=10 THEN LET D=.101 1570 IF D=12 THEN LET D=.081 1580 IF D=14 THEN LET D=.064 1590 IF D=16 THEN LET D=.051 1600 IF D=18 THEN LET D=.040 1610 IF D=20 THEN LET D=.032 1620 IF D=22 THEN LET D=.025 1630 IF D=24 THEN LET D=.020 1640 IF D=26 THEN LET D=.016 1650 IF D=28 THEN LET D=.013 1660 IF D=30 THEN LET D=.010 1670 RETURN 1680 PRINT 1690 PRINT "WHAT IS ELEMENT DIAMETER IN DECIMAL INCHES"; 1700 INPUT D 1710 PRINT "= ";D 1720 RETURN 1730 CLS 1740 PRINT "EOTER FEED LINE IMPEDANCE IN OHMS"; 1750 INPUT ZT 1760 PRINT "= n;ZT 1770 PRINT 1780 PRINT "ANTENNA IMPEDANCE IS ESTIMATED BY MULTIPLYING 52 OHMS BY RATIO OF TOTAL LENGTH TO FULL DIPOLE LENGTH "; 1790 LET ZA=52(A/(468/F)) 1800 PRINT "= ";ZA;" OHMS" 1810 PRINT 1820 PRINT "DO YOU HAVE A MEASURED VALUE TO SUBSTITUTE? ( Y OR N)" 1830 INPUT Y$ 1840 IF Y$="N" THEN GOTO 1880 1850 PRINT "INPUT VALUE IN OHMS"; 1860 INPUT ZA 1870 PRINT "= ";ZA 1880 LET XL=ZT(SQR (ZV(ZT-ZA))) 1890 PRINT "INDUCTIVE REACTANCE (XL) = 1900 PRINT XL;" OHMS" 1910 PRINT 1920 LET L=XI/(2PIF) 1930 PRINT "MATCHING INDUCTANCE = "; 1940 PRINT L;" UHY" 1950 PRINT 1960 PRINT "THIS COIL WILL HAVE TO BE TRIMMED 10 LOWEST SWR" 1970 PRINT 1980 PRINT "DO YOU WISH TO CALCULATE WINDING DATA FOR OOIL? (Y OR N) " 1990 INPUT Y$ 2000 IF Y$="Y" THEN GOSUB 2090 2010 RETURN 2020 PRINT "DO YOU WANT TO CALCULATE ANOTHER ANTENNA? ( Y OR N)" 2030 INPUT Y$ 2040 IF Y$="Y" THEN GOTO 1000 2050 PRINT 2060 PRINT "PROGRAM TERMINATED, TO RUN ENTER ""R""" 2070 INPUT R$ 2080 IF R$="R" THEN GOTO 1000 2090 CLS 2100 PRINT "INPUT OOIL DIAMETER, DECIMAL INCHES "; 2110 INPUT D 2120 PRINT "= ";D 2130 PRINT 2140 PRINT "INPUT CDIL LENGTH, DECIMAL INCHES"; 2150 INPUT B 2160 PRINT "= ";B 2170 LET R=D/2 2180 LET N=SQR ((L(9R+10B))/R2) 2190 PRINT 2200 PRINT "NUMBER OF TURNS REQUIRED = "; 2210 PRINT N 2220 PRINT 2230 PRINT "DO YOU WISH TO CALCULATE ANOTHER VERSION OF OOIL? ( Y OR N)" 2240 INPUT Y$ 2250 IF Y$="Y" THEN GOTO 2090 2260 RETURN Miscellaneous Antennas L Dielectric Antennas for the 10-GHz and Higher Amateur Bands By David Andersen, KK9W •3240 MeCormick Kd., W. Lafayette, IN 47906 Dielectric antennas are useful devices for coupling power from metal waveguides into free space. They are used primarily in the microwave and millimeter regions of the radio frequency spectrum, an area where the dimensions of familiar wire an-tennas become too small for practical implemen-tation. This paper presents a qualitative descrip-tion of the electric field configuration in the antenna, and provides a set of design rules for implementing dielectric antennas. The rectangular waveguide is a popular trans-mission line in the microwave region of the RF spectrum. In most systems, the guide is operated in what is known as the TE „ mode. TE stands for trans-verse electric, and means that there is no component of the electric field in the direction of propaga-tion in the waveguide. The only component of the electric field present is that which points from one long side to the other long side of the rectangle for cross section. In addition, the field undergoes a half sinusoidal variation as an object moves along the long side of the cross section. At one edge, it starts at zero, moves through a maximum, and back to zero at the other edge. This type of field variation is specified by the subscript 10 in the mode name. To efficiently radiate energy from a rectan-gular waveguide, a system must be used to match the impedance between the guide and free space. A horn antenna, constructed from conductive material, is one such method used. The cross section of the horn becomes progressively larger as the wave propagates through it. Finally, the horn is terminated, after the cross section is sufficiently large that the effects of the metal boundaries are minimized. The major drawback of using only a horn antenna is that it does not exhibit high-gain character-istics. As a result, it does not have a narrow beam width. This problem can be overcome in several ways. The use of a horn and a second reflector, such as a parabolic dish, can be employed to increase the gain of the horn antenna. The horn illuminates the parabolic dish and reradiates, or reflects, the energy with a much narrower beam width and higher gain. A second method involves the use of a die-lectric antenna. Dielectric material allows radio waves to pro-pagate through it without loss. It also decreases the free space velocity of the radio wave from c = 3 x 108meters per second. The insulating material be-tween capacitor plates, the plastic material used in coaxial transmission lines, and the material between the traces of a microstrip transmission line and its groundplane are examples of dielectrics. Dielectric material is characterized by a num-ber called the relative dielectric constant. This is a unitless number that indicates the amount a radio wave is slowed by the dielectric medium. The rela-tive dielectric constant is denoted E r and is de-fined as: Er = where V is the velocity of the radio wave in the dielectric medium. A number sometimes used instead of the relative dielectric constant is called the index of refraction. It is denoted n and is simply the square root of the relative dielectric con-stant. The dielectric antenna is known as a leaky-wave antenna. It allows some radiation to leak into free space, as the input or feed energy propagates along the structure. The amount and phase shift of the energy which is leaked or radiated can be adjusted by varying the dimensions of the antenna. If the antenna is designed correctly, the input signal, by the time it reaches the end of the antenna struc-ture, should have been completely radiated. Ad-justing the phase of the energy radiated also ad-justs the sidelobe levels and beam width. There are three general configurations for dielectric antennas. These are specified as the E-plane taper, the H-plane taper, and a combination of these two, the EH taper. As their names suggest, the antennas contain tapers in the respective po-larization planes. Experimental work has been done to demonstrate that all of the advantages of di-electric antennas can be realized by using an E-plane taper exclusively.Xl Thus, the type of an-tenna discussed here will be the E-plane tapered configuration shown in Fig. 1. The antenna of Fig. 1 consists of several sub-parts. The first is the metal horn. This can be thought of as the launching mechanism for the radio wave, which is then focused by the dielectric ma-terial. Thus, by exciting the dielectric antenna, the horn plays a role analogous to its role in ex-citing the parabolic dish antenna. The metal horn excites the dielectric antenna, but the excitation is not 100% efficient. This ef-ficiency can be increased by adjusting the feed taper, denoted L, in Fig. 1. The active part of the antenna, which does most of the radiating, is the part denoted L . This part of the antenna must be long enough tor allow a large amount of the exci-tation energy to be reradiated; however, it must be short enough that construction is possible. Finally, the terminal taper, denoted L , is used to terminate any small amount of energy nSt radiated so that the energy does not reflect off the end of the structure and degrade the radiation pattern. -L-t" Fig. 1 — Configuration of E-plane tapered antenna. A qualitative description of the fields inside the dielectric can be obtained by assuming that the fields do not change radically for the first few wavelengths of propagation in the dielectric. The T E J Q mode of the metal waveguide has fields per-pendicular to the direction of propagation. As the wave propagates along the dielectric rod, a com-ponent of the electric field parallel to the surface of the antenna must exist in order for the antenna to radiate. This results in a slight curving of the electric field as it propagates. The qualitative changes the electric field undergoes are shown in Fig. 2. Fig. 2 — Change in E-field along dielectric antenna. Design principles for leaky-wave antennas have been established by Zucker, and some experimental work has been performed by Kobayashi, et.al., to demonstrate the feasibility of applying the design principles to dielectric antennas.Vi The result of this work has been a set of design rules for di-electric antennas. It is the goal of this paper to demonstrate the use of these design rules for the 10-GHz amateur band. Several things must be specified in order to begin the design: the frequency of operation, the dimensions of the metal waveguide to be used, and the dielectric constant of the dielectric material. Typical numbers for amateur use might be the fol-lowing: an X-band frequency of 10.250 GHz, a die-lectric constant Er = 2.25, and metal waveguide of type WR-90, which has dimensions of approximately A = 2.5 cm and B = 1.25 cm. Additionally, the free-spacc wavelength at 10.250 GHz is 2.927 cm. The next step in the design is to obtain the ratio of the wavelength in free space to the wave-length at the metal guide/dielectric interface. This ratio, denoted Ro, is equivalent to the square root of the effective dielectric constant and can be obtained from propagation curves of the E^j mode of the dielectric waveguide. These plots are avail-able in Ref. 3. For our particular problem, the plot used was Fig. 21 of Ref. 3, and the value obtained for Ro was 1.167. The next step involves selecting the beam width and gain of the desired antenna. Zucker has plotted two experimental curves (see Ref. 2, Figs. 16-10 and 16-11) that can be used. Initially, select the length of the antenna, L, which gives the desired beamwidth and/or gain from Fig. 16-11. For our problem, the value of L selected was L = 17.56 cm or 6 wavelengths. This results in a gain of approxi-mately 18 dB and a beamwidth of 21 degrees. Gain can be increased by increasing L. When L has been selected, the designer must then determine the average ratio of the wavelength in free space to the wavelength in the dielectric medium along the length of the dielectric rod. This is obtained by using Fig. 16-10 of Ref. 2 and is found, for our example, to be Ra = 1.05. Now that the numbers Ro, Ra, and L have been specified, the length of the feed taper can be calculated. This is done using the following for-mulas: where: and: Lf =~T |a - |/Q2 - 4 a j a =\o + 6(Ro + Ra - 2)L 6(So - 1) C2 = \ o L 3 (Ro - 1) Additionally, the effective dielectric constant at the end of the feed taper can be calculated as: Rr =~Xo - Ro + 2 3Lf For our example problem, these numbers are: a = 25.74 cnu C2 = 102.6 cm Lf = 4.93 cm Rr = 1.031 Using the value of Rr, and referring back to the propagation curves of Ref. 3, the width of the radiating section of the dielectric antenna, B', can be obtained. Here, an approximation is made by as-suming the propagation curve for the A/B = 2 sit-uation still applies. The ratio A/B' is actually greater than 2. This is a relatively good approx-imation, however, because of the weak coupling of the fields to the dielectric at this point. For the example problem, B' = 0.7078 cm. The last two items to be specified are the horn dimensions, L, , and the terminal taper, L . Exper-iment has shown that L = 1.1 L, and L = one-half wavelength are good values to use. Some of the considerations involved in de-signing dielectric antennas have been discussed. In addition, a design for an 18-dB gain, 21 degree beamwidth antenna at X-band was formulated. The design rules presented are also applicable to the amateur bands above 10 GHz. I hope that this information is of use to amateurs and will promote greater use of this unique antenna. References \1 Kobayashi, S., Mittra, R., Lampe, R., IEEE IVans. Ant. and Prop., V. AP-30, no. 1, p. 54 (1982). \2. Zucker, F. J., Antenna Engineering Handbook, H. Jasik, Ed., New York: McGraw-Hill, 1961, chap. 16. U. Goell, J. E., Bell System Technical Journal, V. 48, p. 2133 (1969). K i e l y , D. G., Dielectric Aerials, London: Methuen and Co. (1953). M L Watson, R. B., and Horton Jr., C. W., Applied Physics, V. 19, no. 7, p. 661 (1948). A Crossed-Loop/Goniometer DF Antenna for 160 Meters Charles F. W. Anderson, N4KF •1716 Reppard Rd., Orlando, FL 32803 What would low-band enthusiasts say to a DF antenna with performance comparable to WIFB's 1.5-meter square loop? \1 It measuresonly one meter on a side and its pattern is rotatable by nothing more than a hand-operated control right on your desk. If this interests you, read on! The crossed-loop/goniometer direction finder dates back to the early days of wireless. Developed by two Italian engineers, Bellini and Tosi, it con-sists of two fixed loops, arranged at 90-degree angles. The outputs of these loops are routed through transmission lines leading to two sets of coils. Called primary coils, they are also at right angles to each other. Inside these coils, or between the halves of each, another coil, called the second-ary coil, is rotated. The output of the secondary, or pickup coil, feeds the receiver input. These inductors form the goniometer. Fig. 1 shows a typical setup. It is based on descriptions and diagrams in Terman and Henney's radio engineering handbooks, and from Chapter 10 of Electronic Countermeasures.\2jl^l W2IMB also describes a crossed-loop/goniometer in The Low and Medium Frequency Radio Scrapbook,\5. Preliminaries Refer to the Bill of Materials. Most of the items listed, other than the coaxial cable and shielded, twisted-pair line, should be available at any building supply store. First, cut the pipe using information from the following table. Chamfer the inside edge of each end of all the pieces. Identification A pieces B pieces C pieces Stanchion length 35 in 25 in 1 1/2 in 54 in Number 8 4 16 1 Two of the 10-ft pieces of pipe will make six of the A pieces, plus the C pieces. A third piece of the pipe should be cut to make two more A pieces and two of the B pieces. The remaining piece of pipe should have the stanchion cut from it, plus the other two B pieces. Cut two 40-ft pieces of the RG-174 coaxial cable. At the exact center of each of these, remove about 3/8-inch of the jacket and outer conductor, using great care not to cut into the dielectric. Wrap the areas with insulating tape. J\Io-r£- Loops SHOWN W/O SHIZL.DS FOP CJ-ARJTr OUTPUT Fig. 1 — Crossed-loop/goniometer. Loop Housing Assembly The housing for the loops is about one meter on a side and stands 1.4 meters high. Refer to Fig. 2. If you follow the steps below with reasonable care, you should have no difficulty in constructing the housing. At the exact center of each of the cross fit-tings, cut a 3/4-inch diameter hole on each side. Of the methods I tried, the carpenter's brace-and-bit seemed to work best. Save at least one of the cir-cular pieces you've removed for use later. About 1/2 inch from each of the cross ends, drill a small hole into which you can thread a self-tapping screw. Com-parable holes at each port of the Ts and the 45-degree elbows must also be drilled. Slide one of the cross fittings onto the stan-chion piece and position it exactly at the middle Hold the fitting in place with a wrapping of tape on each side. Three-eighths inch from each end, wrap several turns of tape around the stanchion. These wrappings will keep the top and bottom cross fit-tings from sliding too far onto the stanchion while the rest of the housing is assembled. Insert the four B p i e c e s into the c e n t e r cross-fitting ports and hold them in place with self-tapping screws. Now put the other two cross fittings on the ends of the stanchion. Hold them in place by running a length of monofilament line through the stanchion and tying the ends together. At this point, you'll need to make an arrangement to hold the framework in a vertical position while completing the assembly. I suspended mine from a ceiling air duct in the shack, using heavy cord. Mate a C piece with each of the 45-degree elbows, and secure with set screws. Insert one of these into each of the ports of the bottom fitting. Align them accurately, and hold in place with set screws. Insert four of the A pieces into the other side of the elbows, and then put an elbow/C-piece on the ends of each of the A pieces. Mate each of the Ts with the four B and C pieces. Recheck alignment, loosening and tightening the set screws as neces-sary. Put four additional elbow/C-pieces on the upper ports of the Ts and the remaining four on the top cross fitting. Now place the last four A pieces in position, make final alignment adjustments to get the sides at right angles to each other, then tighten the set screws just enough to keep every-thing together. This completes the assembly of the loop housing. If you have room to mount the loop assembly indoors, (attic or air space, for example), you could eliminate the A pieces, the C pieces, and the 45-degree elbows. This would simplify assembly and reduce the cost. Note my remarks on this later in the article. Installing the Loops Spray some silicone lubricant into each of the four sides of the housing. This will help you through the following steps! Squeeze a split shot onto the end of a piece of monofilament line about 15 ft long. Start the shot down one side of the housing and get it down to the bottom cross fitting by shaking, bumping or any other means of persuasion you might think of. See Fig. 3A. When the end of the monofilament exits the bottom of the housing assembly, lash the end of one piece of coaxial cable along the monofilament at the top of the assembly. Use a short piece of monofil-ament for this purpose. See Fig. 3B. Start pulling on the end of the monofilament to bring the cable down through the housing. There might be a little trouble when the end approaches the first elbow. A little jiggling of the monofil-ament and cable should get it around the corner. If you run into trouble here, loosen some of the set screws and fish the coaxial-cable end through corner. Pull the monofilament until the end of the coaxial cable comes out at the bottom. Untie the Fig. 2 — Loop housing. cable from the monofilament, and secure the upper end of the monofilament so that it won't follow along with the cable during the next step (Fig. 3C). The cable should be pulled through the housing until the center is just inside of the top cross fitting. This is shown in Fig 3D. Repeat these steps using another piece of mono-filament and the other end of the coaxial cable. Care should be taken to keep the kinks out of the cable while you're pulling it into the top fitting. The middle of the cable should be centered in the top fitting, with the two ends coming out the bot-tom. Attach one of the cable ends to the monofila-ment which is in the opposite side of the housing. See Fig. 3E. Pull the monofilament out of the top cross fitting until the end emerges, then remove the lashing, and pull the cable through. At this point, be careful about kinks in the cable at the low cross fitting (Fig. 3F). Attach the other end of the cable to the bottom of the monofilament which is in the side of the housing opposite to the one you've just pulled through, and repeat the previous step. You should now have both ends of the cable at the top of the housing, with monofilament lines in both sides with which to get the cable back to the bottom. One end of the cable should be connected to the monofilament in the correct side of the housing. Pull it down and out of the bottom cross fitting. Repeat this step with the other end of the cable. Tie the ends of the coaxial cable out of the way for the time being. Remove the monofilament line used to retain the top and bottom cross fittings. Repeat this procedure from the start for the other length of cable. Close the hole in the top fitting with the circular piece you saved (step one of the housing assembly) and seal it with silicone adhesive. Loosen each of the set screws slightly and put a bead of silicone around each joint to make a good rain-tight seal. Push the joint back together and retighten the set screw. Assembly of the crossed loops is now complete. Interfacing the Loops to the Goniometer The loop outputs in my setup are run directly to the primary coils. Resonating the loops and primary coils with a variable capacitor in series gives higher input levels at the receiver. The addition of preamps at the loop outputs should give even better results. See WIFB's article in April 1984 QST for information on preampsAfi Whichever route you decide to take, an enclosure is needed at the base of the loop assembly in which to make connections to the transmission lines, and to house preamps, if used. Selection of the proper Minibox should be based on what it will contain. For connecting the loop outputs to the transmission lines, I mounted three standoff insulators where each of the sets of loop conductors and transmission lines are connected together. One of each group is used as the tie point for the two ends of the loop shield (the coaxial cable outer conductor) and the shield of the transmission line. The other two are used to connect the inner conductors of the loop to the twisted pair of the associated line. Trim the coax ends as necessary to make these connections. Mounting the Crossed-Loop Assembly My loop assembly is currently secured in a tree outside the shack. Reroofing of the house is immi-nent, which accounts for the unconventional "roost." Later, the assembly will be clamped to a 20-ft PVC pipe (Schedule 80, 2-inch dia.), to be supported at the eaves by a mast bracket. You might want to con-sider roof-mounting the loop assembly. One approach is to use a TV mast base to support the bottom, and substitute cross fittings for the Ts at the side corners. These could be used in combination with 90-degree elbows to provide support. The four legs made of Ts or elbows should be of the right length for the pitch of the roof and could rest on the surface. Some kind of tie-down should be provided. Adequate guying, using nonconducting lines, might preclude the need for the legs. Transmission Lines If the distance from the location of the loop to your operating table doesn't exceed 20 ft, almost any kind of shielded, twisted pair should be ade-quate. If the distance is greater, I suggest using one of the RF-grade lines. I'm using Raychem 10602, which is designed for the 1-megabit data bus used in military airborne installations. It has a nominal loss of 1.4 dB per 100 ft at 1 MHz. Gore and others make similar cables. RG-22 or RG-108 would also be suitable. The two transmission lines are routed into my shack using grommeted holes in the Plexiglas panel that has replaced the conventional windowlight and serves as the feedthrough for antenna cables. From there, they are run to the chassis in which the goniometer is installed. Goniometer This part of the setup consists of the primary and the secondary coils. A good method for fabri-cating this was the subject of much head-scratching. Part of the answer was found on the household-items shelf of a store. A plastic container, with cover, measuring about 9 x 9 x 3 inches, proved to be the right form for the primary coils. It included room for the rotatable pickup coil, as shown in Fig. 4. Preparing the plastic box for winding the coils requires a little work. A raised lip at the top of the box mates with the groove in the lid. Mark the exact center of the lip and of the ridge that runs around the outside of the box on each side. At each of the marked spots on the lip, remove 1 1/4 inches of the lip (5/8 inch on each side of the center marks) down to the level of the ridge. Now, starting 3/16 inch from the center of each side of the ridge, cut a shallow, flat-bottomed notch 1/8-inch wide. These notches (two on each side of the box) provide winding spaces for the primary coils. On two adjacent edges of the box, just out-side of the notches, drill holes to mount the stand-off insulators. These will be the terminals for the primary windings and the transmission lines. This is also shown in Fig. 4. Mark the exact center of the bottom of the box, and drill a hole to install the shaft bushing. To prevent the sides of the box from being pulled inwards by the primary windings, a block of styrofoam about the same thickness as the depth of the box is cut and shaped to fit snuggly inside. Here's a tip to spare yourself some extra work. Cut the styrofoam into eight pieces, corner-to-corner and side-to-side. It will be easy to remove these pieces after the windings are in place. After the bushing was installed, I had another problem. How do I space the windings away from the bushing and above it? My wife provided the solution in the form of a champagne cork, which she found in one of_^he kitchen drawers. A little work with an X-acto knife converted the cork into what was needed. Remove the shank of the cork so that about 1/8 inch is left. Drill a 3/8-inch hole through the center of the cork. This spacer holds the windings away from the shaft. For the primary windings, solder the end of a no. 26 enamel-insulated wire to one of the standoffs. Wind six turns around the box, using the notches to hold the wire, cross over on the top side of the box, and wind six more turns. Cut the wire, making certain that the windings are neither too loose nor so tight that the lid won't fit. Solder the end of the wire to the other standoff. Repeat these instructions for the other primary windings. Place the spacer over the shaft (bottom side of box), spread the windings over it to clear the shaft, and remove the styofoam. The coils measured about 39 yH. Finding a suitable form for the pickup coil also presented a problem. Here again, styrofoam was employed. The coil was wound on a block measuring 5-1/4 x 1-3/4 x 1-1/2 inches. It was calculated that the coil should have an inductance of about 50 uH (reactance of about 600 ohms at 1.8 MHz) to match the balanced input of the ancient SX-110 receiver used for 160 meters. To keep the number of turns within reason and to improve the Q, a broadcast-band ferrite loop core, 5-1/8 x 5/16-inch diameter was inserted into a hole cut lengthwise through the block. The hole was offset to allow the end of the coil-rotating shaft to be fastened in place. The coil was wound with no. 26 enamel-insulated wire; 16 turns spaced about the diameter of the wire on each side of the block, with a space between the halves of the winding sufficient for clearance for the FI/OTCHZSIN K/2><?£W (TYP a pi) \ IS — CUT-OUT— X IN RIM (TYP 4- PL) R.ON a, W/R.E TUHNS IN THESE NOTCHES NOTE: FOR. CLAZJTY, ONLY PRIMARY CO/L. SHOWM STANDOFF ON OPPOSRRF SW£~ OF E/B<I£(TYP 4- PL) ( < 9 i 'PLASTIC RUN MORE TURNS IN THESE NOTCHES CJZOSS-OVER —^-P/CKUP COIL SC£ FIGURES M/U< JVG, TOP CTYP/CAL 4-PQ s ^-VLASRRC BOX (SER TEXT & BILL OF MATERIALS) CHASSIS -(ZEF) Fig. 4 — Goniometer box with one primary c o i l . shaft. A 1/4-inch hole was drilled part way through the block and the shaft was inserted and bonded with silicone adhesive. See Fig. 5. To make the output line for the secondary coil, cover a length of fine twisted pair with shielding braid. The twisted pair I used came from a defunct in-the-ear headphone unit. The shielding braid used was removed from a piece of subminiature coaxial cable. Connect the twisted pair to the secondary coil leads. Then make one turn or more of the output line around the coil shaft so that it can be rotated easily. Connect the shield to the chassis at both ends. The goniometer assembly is now ready to be mounted. The four milk jug tops are used to space it away from the chassis. Install the plastic lids after the box has been secured in place. At a convenient point on the chassis, two standoffs are mounted. The twisted-pair leads are terminated at this spot. The shield should be connected to the chassis near the standoffs. Another length of the shielded transmission line makes the connections from this point to the receiver. Also at this time, the loop output lines should be connected to the primary coils. The shields of these lines are c o n n e c t e d to the chassis close to the coil terminals. The shield should be bonded to the chassis at all entry and exit points to minimize stray pickup of RFI and BC station harmonies. The chassis bottom p l a t e i s p a r t o f t h i s o v e r a l l s h i e l d i n g implementation. Install a knob on the pickup coil shaft, and the goniometer is complete. Receiver Input Connections As mentioned above, my receiver has a balanced 600-ohm input. Connecting the goniometer output line only involved attaching the twisted-pair leads to the antenna terminals and the shield to the ground terminal. Coaxial cable was tried earlier for the connection from the pickup coil to the receiver, but didn't work well in my case. If your receiver has a 50-ohm input, consider using a balun. Operation There is nothing complicated about using the setup. Rotate the pickup coil to null out noise or QRM. Some DF Dope and Test Results DF antenna patterns are affected by any c o n d u c t i n g o b j e c t in the v i c i n i t y . W h i l e flight-testing radio installations during W W II, I spent many hours making calibrations of what is called "quadrantal error." For a medium-sized aircraft (wingspan of approximately 50 ft), this error can run from about 15 degrees at 200 kHz to over 20 degrees in the BC band (it goes higher as the frequency increases). Quandrantal error curves usually show little or no deviation at the fh, 90-, 180- and 270-degree points relative to the station being received. How-ever, the error increases to levels of the order mentioned above at or near the intercardinal points. In addition, at those points, the nulls of the pat-tern will not be as sharp. The major factor con-tributing to the deviations is the "static dipole" field (also known as the induction electric field), a component of the complex near-field geometry in the vicinity of any antenna. This field varies in-versely as the cube of the distance from radiating or reradiating objects, and is therefore the pre-dominant field in the immediate vicinity. Also, its intrinsic impedance is quite high, compared with the 377 ohms of the far field. To minimize some of the e f f e c t s of this field, loops are provided with electrostatic shields; hence, the RG-59 which Doug DeMaw used for his big loop, the RG-58 of his "W1FB 4T-ES" loop and the RG-174 in the unit described herein. Don't expect that your crossed-loop/goni-ometer setup will give accurate bearings on local stations or i n t e r f e r e n c e sources. The wiring, piping, ducting, metal flashing and rain gutters and spouts of your house will all contribute to errors, so think twice before installing your loop in-doors! To d e m o n s t r a t e the importance of loop shielding, I made some checks. Using a lash-up version of the W1FB 1.5-meter square loop, I ran a test on a local station near the high end of the BC band. With the loop shield ends not connected to the shield of the line running to the receiver input, hardly any change in S-meter reading was seen as the loop was rotated through 180 degrees. With the loop shield connected properly, a sharp null of at least 30 dB was observed. Similar results were obtained r£ND OF JfT { TUR.N PVSHE. M- -rUD/i D THRU BLOCK T i o m / I / J I L 5~J/AFT n£ld in place W/SJ LICO^IE: s£Ale"R Ebtv OF LAST ^L-OGK I ' [ f l l k f f l M K/6 TURN^ •F£RR/TE: Pig. 5 — Secondary c o i l . using the crossed-loop/goniometer setup. These tests were made in my living room, so there were a lot of conducting objects in the vicinity of the loops. Some further investigations are in order, obviously. I'd like to find out if preamps work as well at the goniometer position compared with how they work at the loops. Another item is that of providing an electrostatic shield between primary coils and the secondary coil. This is mentioned in Terman's description of the goniometer. An 80-meter version about 0.75 meter on a side would be worth looking into. I'm hoping to find time to investigate these and and some other ideas. Meanwhile, here's hoping to BCNU on 160! References \1 DeMaw, Doug, W1FB, "A Receiving Loop for 160 Meters"; QST, March 1974. (See also Hie Antenna Anthology, 1978, Chap-ter 4; and The Antenna Book, 14th edi-tion.) \2 Terman, Radio Engineers' Handbook; New York, McGraw-Hill, 1943; pp. 878-879. Henney, The Radio Engineering Handbook, 3rd Edition; New York, McGraw-Hill, 1941; p. 581. \4 de Rosa, Electronic Counter measures, Chapter 10; Los Altos, CA, Peninsula Publishing, 1978; pp. 10-2 to 10-5, 10-30 to 10-31 and 10-80 to 10-81. Y5. Cornell, The Low and Medium Frequency Radio Scrapbook, 3rd Edition; published by the Ham Radio Publishing Group, 1977; pp. 22, 44, and 48. \£ DeMaw, "Receiver Preamps and How to Use Them"; QST, April 1984. (See also The An-tenna Anthology and The Antenna Book 14th Edition.) Bill of Materials Item Quantity FVC pipe, 1/2 in x 10 ft, 4 pes lawn irrigation type, not Schedule 40 FVC cross fittings, 1/2 in 3 ea FVC T fittings, 1/2 in 4 ea FVC 45-degree elbows, 1/2 in 16 ea No. 4 self-tapping screws 44 ea (min) RG-174 coaxial cable 80 ft Aluminum chassis, 12 x 17 x 3 in, 1 ea w/bottom plate (see text) Plastic refrigerator box (see text) 1 ea Ferrite rod (see text) 1 ea Styrofoam block, 12 x 12 x 3 in 1 pc Plastic champagne cork 1 ea Plastic milk jug tops 4 ea No. 26 enanel-insulated wire 75 ft Flexible twisted pair (see text) 2 ft Very fine shielding braid 2 ft (to cover above; see text) Small standoff insulators 12 ea (type w/no. 4 mounting studs) 5-lb. test nylon monofilament line 50 ft Shielde<Vtwisted pair See text Split-shot sinkers, no. 7 2 ea Silicone sealer 1 tube Silicone spray lubricant 1 can Shaft bushing, 1/4-in dia. shaft 1 ea Subsurface Antennas and the Amateur By Richard Silberstein, W0YBF •3915 Pleasant Ridge Rd., Boulder, CO 80301 The term "subsurface" applies to buried or submerged antennas. Because of its general d e f i -nition, it could also apply to a microwave antenna hidden in a meatloaf. There are two reasons for placing an antenna under ground or in the water: (1) concealment, and (2) hardening (protection against a nuclear blast). Why would anyone want to conceal an antenna? An Amateur Radio operator with available attic space would be reluctant to uproot his yard. Subsurface antennas are used mainly for military and Civil Defense purposes, but amateurs should find them an exciting topic for e x p e r i m e n t a t i o n and public service. Scientists were thinking about buried antennas. as early as 1912, and a loop antenna for submarines was proposed around 1919. Two instances of Amateur Radio experiments using this method were reported during 1948 to 1952. In the mid-50s, interest in buried antennas flourished, and many published papers appeared in the 1960s. A f t e r this period, interest died and the subject of subsurface antennas was put aside. R. W. P . King and G. S. Smith recently pub-lished a book on subsurface antennas.\l However, each paper I researched, each page I thumbed through, had one thing in common: They expound on antenna theory, but fail to give construction de-tails. Why? Because nearly all the building, per-formance testing, and evaluation is done by con-tractors. This includes universities and "in-house" government groups. Much of this work goes astray, never reaching the journals. Either the results become proprietary (belong to the company that performed the research because of c o m m e r c i a l potential), or they remain classified for security reasons. Although theoretical literature on subsurface antennas is incomprehensible to most people, there is no reason to despair. The simplest antenna could never have been built if it had been necessary to understand the exact theory. L u c k i l y , there is always some communication between the builders and researchers. Thinking About Subsurface Attenuation To help us relate to the theory of subsurface antennas, we must understand that much of what happens takes place near a boundary between two surfaces. Also, waves in a lossy medium weaken rapidly. Near-field wavefronts are curved, and so is the earth's surface. This causes the wavefront to tilt forward as it travels along the surface, radi-ating energy into the ground or the sea. To illus-trate the result of applied theory used later in this article, calculations are made using the con-cept of a plane wave rapidly losing energy as it goes through ground or sea water. To express attenuation loss, the decibel (dB) is used, where: 1 dB of voltage change = 20 log 1 Q E J / E J 1 dB of power change = 10 log 1 Q Some receivers are calibrated with S units of 5 dB each up to S9. The term "skin depth" is often used when work-ing with buried antennas. The skin depth of any soil is the depth at which the electric field is atten-uated 1 neper (8.68 dB). Skywave to Ground Amateurs who work in the HF band are most familiar with skywave signals reflected from the ionosphere. At each receiving site, the pattern of the elevated antenna shown in Fig. 1 at P results partly from the combination of an incident wave A and a wave B' reflected from the ground. There is also wave C, the refracted wave, going into the ground. This is the one of interest for buried antennas such as the one at P1. In this case, the reflected wave is wasted. Note that the refracted wave is heavily attenuated. To transmit from a buried antenna, start at P' in Fig. 1. The signal becomes skywave and travels toward the ionosphere along B so there is no energy loss to the reflected wave B\ At some angle, how-ever, total internal reflection begins to prevail. Fig. l — Reception above and below ground via skywave. Incident- and Ground-Reflected Waves Fig. 2 shows reception for short distances from any antenna. From a transmitter, T, there is a di-rect wave A and reflected wave B' combining at the receiver, R. For underground reception, the heavily attenuated wave, C, is still received at R'. The behavior for Fig. 2 is qualitatively valid at all frequencies (except for some high-angle sky-waves at lower frequencies), but at distances not exceeding the line of sight. The received wave is called the "ground wave." The ground (also water), has a dielectric constant and conductivity that, acting together, cause its reflective properties to change with frequency. Its attenuation is also a function of frequency of the wave, and particularly of the distance traveled in the medium. The Surface Wave As distance increases, curvature of the earth becomes important. This curvature and the wave front give rise to a "surface wave"that becomes the sole groundwave beyond line of sight (see Fig. 3).\2r3,4r5 The surface wave is much stronger for vertical electric polarization (horizontal magnetic polar-ization) and attenuates less with distance at lower frequencies. At standard broadcast frequencies over average soil, a station may have a service range of 50 miles, with more for "good" soil. Low and very-low frequencies propagated as surface waves over sea water may be detected beyond a thousand miles. Sea water has a dielectric constant of 81 and possesses high conductivity. It is vastly better than soil for surface propagation, but correspondingly worse for subsurface propagation at any depth. On the lower frequencies at great distances, it is the ionospherically reflected wave penetrating sea water that is the important mode for undersea reception. As noted earlier, the surface wave in Fig. 3 tilts forward as it moves, radiating part of its energy into the ground or water. This same energy reaches a subsurface antenna. In transmission from below the surface, energy goes upward and contri-butes to the tilted wave. Beyond the surface-wave range, which may be relatively short distances in the HF band, energy penetrating the soil or sea must come first via sky waves. Fig. 2 — Reception above and below ground from a source within line of sight. Fig. 3 — Surface wave. T and R are a surface transmitter and receiver. S, and S„ are sub-marines transmitting and receiving via the sur-face wave. Estimates of Subsurface Attenuation of the Re-fracted Wave It is difficult to estimate the attenuation of a surface wave in terms of depth below the surface, but easier to see what happens to the refracted wave from a downcoming skywave (Fig. 1). When working with skywave, important sources of field-strength loss are reflection at the point of entry, atten-uation in going through the medium (in the refracted wave), and direct ohmic loss caused by currents going from the antenna structure through the earth or water. Ohmic loss is also evident in broadcast-antenna ground screens. Insulation of ground-plane wires (if the ground plane lies on the ground or roof) was recognized by manufacturers of amateur verticals before 1960. H. A. Wheeler describes these losses in a more sophisticated manner and his papers should be studied for a better understanding.^ Using the "propagation constant" of Appendix I, specifically the attenuation constant, a , for a simple plane wave, calculations were made to show qualitatively how attenuation loss varies with frequency and distance under different subsurface conditions. In plane-wave theory, it is assumed that a traveling wave in the earth or water will be attenuated exponentially. This means that starting at any point, the proportion of energy lost per unit distance is always constant. Results show that in penetrating average soil at 5 MHz, a wave will be attenuated about 1.87 dB per meter of travel. The reflection loss must be added to this. With an average loss of 4 dB in the reflected wave, 40% is reflected and 60% must be refracted, or it is said that the refracted wave is down by 2 dB. Think about it. In sea water the dB attenuation per meter is proportional to the square root of the frequency, at frequencies up into the lower VHF region. Lower frequencies, then, should show less loss in going through salt water. At 20 kHz, the calculated subsurface propa-gation loss is only 5.4 dB per meter. However, for an arriving skywave, most of the power has already been lost upon reflection. Since salt water is a good reflector, perhaps there is a loss of 50 dB to the refracted wave. Now the total signal loss at 1 meter in depth is 5.4 dB + 50 dB = 55.4 dB. At a depth of 10 meters, it is 54 dB + 50 dB = 104 dB. At a frequency 100 times as great or 2 MHz, the atten-uation would be 10 times as great or 54 dB at one meter. It would have a slightly smaller reflection loss, giving a total loss of about 100 dB at 1 meter in depth. This does not include antenna circuit losses. Two MHz doesn't look good for undersea communication. In any subsurface reception, atmospheric noise and interference will be attenuated almost as much as the signal. The receiver noise figure limits reception, and a sensitive one can be used advan-tageously. Spectrum space is limited at low frequencies. With weak signals, narrow bandwidths must be used for reasons of signal-to-noise ratio and spectrum space available. This reduces the speed of com-munications. Some Practical Antennas Practical Antennas for Very Low Frequencies The VLF frequency range (3-30 kHz) and the LF range (30-300 kHz) could support reception of European broadcast stations, submarine communi-cations, reception of standard-frequency and nav-igation-aid signals. Loop antennas, possibly in radomes, are logical antennas for submarines. On land, large conventional antennas or buried ones similar to those described under Medium Frequenices would be used. As we have seen, direct propagation through salt water is highly attenuated. The predominant source of signal energy is a surface wave up to about 1,000 miles, with energy propagating to or from submarines to the surface-wave region. At long distances, skywave becomes the superior mode above the surface, with downcoming waves penetrating the ocean. The skywave is guided between the earth and the lower ionosphere. At the time of this writing, a project is undergoing construction in Michigan using VLF. It consists of burying miles of underground wire in a forest area and will enable submarine communication to naval bases. Medium Frequencies, MF (300-3000 kHz) These are short-distance surface-wave c o m -munication frequencies, free of ionospheric dis-turbances. They are also used in transmitting and receiving information in the broadcast bands to and from emergency shelters. Some skywave communica-tions are possible at night in the 160-meter band during t i m e s of minimum a t m o s p h e r i c n o i s e . Ionospheric absorption on these frequency ranges are too great in the daytime. Antennas for the MF are horizontal, insulated, buried wires fed in the center or at one end. Fig. 4 shows such wires that are grounded or insulated at the ends. The end-grounded wire has a high-loss return path to the feed point and resembles a cable with a lossy sheath. When the wire is near the surface, the largest return currents are below the surface so that the wire acts like an elongated vertical loop. It generates or receives a maximum surface wave off the ends of the antenna, and provides vertically propagated energy for feeding the forward tilt component of the surface wave. The lossiness produces a low-Q antenna, but with resonance points visible as f r e q u e n c y is varied. This makes the antenna broadband, but inefficient. The ends can also be free, however, earth losses will be fairly large. Low-loss loop antennas are available that tune to the upper part of the MF band.VL Wheeler suggests a more efficient buried an-tenna where the the straight long wire is replaced by a narrow insulated loop of the same length (Fig. Fig. 4 — Buried, insulated horizontal dipoles. At A, the dipole is center fed with the ends grounded. B displays a center-fed dipole with the ends free, and C shows one end fed, with the other free. 5).\& This eliminates the return-current ground loss and would give the antenna a higher Q, allowing it to work best near resonant frequencies. Refs. 6 and 9 give total losses for MF test antennas that differ markedly. Wheeler shows -28.5 dB for an antenna 1000 ft long and 3 ft deep. Fen-wick and Weeks show as much as 10 dB more for an antenna nearer the surface, but only 1/10 as long.\2 The tests were performed on soil with different losses. Fig. 5 — An insulated, elongated loop as a buried antenna. High Frequencies, HF (3-30 MHz) Since skywaves are associated with high fre-quencies, buried antennas here would generally be used for fairly long-distance communications. Buried single dipoles, dipole arrays, loops and various other antenna configurations can be used in this band. In 1965, I attempted to test the feasibility of using buried wires for communication when the ob-jective was to conceal a tactical position. There was no doubt that this could be done; the question was whether something simpler would work as well or better. The experiment was devised for reception of high-angle skywave from standard-frequency station WWV, then at Beltsville, MD on 5 MHz, at a distance of about 200 miles. A half-wave center-fed dipole (a doublet), insulated and free at the ends, was cho-sen. It was known that the effect of the ground is to reduce the physical length of the wire for res-onance, because the velocity of propagation is reduced. To get some idea of the maximum required length for 5 MHz, the resonant frequency of a measured length of wire lying on the ground was noted. Now the length of a half wavelength of a wave in free space at 5 MHz is 492/5.0 = 98.4 ft. Because the velocity of propagation along a high thin wire is 95% of the free-space velocity C, a half wave-length of this wire at 5 MHz is determined by the well-known 468/5.0 procedure, yielding 93.6 ft. For a wire lying on our particular ground, using a grid dipper and measuring resonance, the velocity was found to be 55.3% of C. Taking 55.336 of the above free-space half wavelength for 5.0 MHz, 98.4 ft, yielded 54.4 ft for the wire on the sur-face. When the antenna was finally insulated and buried in a trench 8 inches below the ground, the total antenna length for a 5-MHz half-wave resonance was 46.6 ft for a velocity factor of 47.4%. Some of the references show velocity factors as low as 25% for deeper antennas. Both the buried antenna and a comparison doublet were placed across the path to Beltsville. Fig. 6 is a diagram of the buried antenna. Since the experiment was done before the time of PVC tubing and joints, the wire was put into vinyl hose for insulation (not an original idea); the wire and the hose were cut into two equal lengths with the wire shorter than the hose. Each length of hose was ce-mented at one end around a hole in a vinyl trash container with a lid, which was located in a sump at the middle of the trench. Inside the container there was a smaller metal can containing the balanced matching network. Feedthrough insulators were pro-vided at each side of the metal can for connection to the antenna wires as well as a standard socket for the 50-ohm output cable, near the top. Nylon strings were attached to the ends of the no. 12 soft-drawn copper wires and brought up to the sur-face through the bent-up ends of the hose. Because we started with longer wires than necessary, this allowed us to shorten them to the proper length, keeping them taut. Precautions were taken to see that all joints were waterproof and that there was some room in the sump for water accumulation. A cable (RG-8) was brought out the top of the sump and the area nearby was covered with a piece of galvanized sheet metal. Fig. 6 — The half-wave dipole and matching unit in 5-MHz tests. The first version used v i n y l hose for insulation. Better results should be available using plastic tubes, spacers and water-tight fittings. Comparison of reception was made using a half-wavelength horizontal doublet 0.3 wavelength above ground. Both antennas were positioned across the path to Beltsville, and not too close to each other. The balanced and matched cables from each antenna went to a test shelter. A stable high-quality communications receiver was calibrated in decibels relative to one microvolt, using a good-quality signal generator and a strip-chart recorder fed by the AGC voltage. The recorder was an old Esterline-Angus galvanometric type. Sometimes a bridge-balancing or digital recorder should not be used with a narrow-band receiver, especially if the receiver and signal generator are unstable and require manual tuning. Some recorders will not respond quickly enough to show when the receiver is resonated. The test consisted of switching antennas every 5 minutes for two hours with the recorder running. The most constant difference was about 16 dB of loss in the buried antenna, both in the signal and in the noise during station-off periods. To ascertain that the signal was not being contaminated by pickup through the cable shielding, the cable was unplugged at the buried antenna and terminated in 50 ohms: The signal strength fell 41 dB. With these results, it was concluded that communications with a site using a buried antenna was certainly possible. If this method is used for concealment, something simpler and easier to install might be desirable such as a loop or a disguised loaded whip. I recently noted the existence of a report describing experimental work done several years later on buried antennas at the Rome Air Development Center (RADC).\m The report gives theoretical background and describes some careful, detailed research on buried antennas. The buried wires in the RADC experiment were laid in trenches much as described for the 5-MHz tests above. The insulated wires were actually RG-19 cable without the outside shield. One difference was that several wires were buried so that it was pos-sible to obtain some small gain by paralleling pairs separated by enough distance to make mutual coupling insignificant. Separation of 13.8 ft gave a distance of 1.4 "skin depths" near 7 MHz for their soil con-stants. One advantage of the RADC construction meth-od was that solid insulation and an above-ground matching network eliminated water-accumulation problems, although insulation by wide-diameter air-filled tubes might have reduced ground losses. Pulse and CW signals with a dipole pair buried 3 ft afforded surface-wave signals at 7 MHz out to 40 miles under favorable conditions of low noise. Power was probably only 22 W. The signal strength was 18.2 dB worse than for a 35-ft whip on the surface. Similarly, under favorable absorption and noise conditions, sky waves were seen out to 500 miles and were shown to be 24 dB weaker than for a half-wave dipole 1/4-wavelength above ground. Results and theory were combined at RADC to determine the performance of an array of buried dipoles, one half free-space wavelength long and 3 ft deep. With a transmitter power of 10 kW, received signal conditions were as follows: 1. For surface waves at 2 and 5 MHz under bad atmospheric noise conditions at night, the signal was down to about 10 dB above noise at around 100 mi. Bandwidth was not stated. 2. A similar antenna at 7 MHz for skywave under winter daylight conditions at a sunspot number of 100 had a signal equal to about 10 dB above noise at 1000 mi. Bandwidth was not stated. 3. Winter-night skywave at 2 MHz had a range over 2500 mi for 10 dB signal to noise. Bandwidth was not stated. HF Dipole Antenna Characteristics Plots of antenna resistance and reactance in Refs. 9 and 10 show resonant frequencies which in-dicate a velocity of propagation of only 25 to 30% of C. In Ref. 10, consider E < j > , the horizontal broadside electric field (TE mode), and E 0 , the off-end "vertical" electric field (TM mode) pat-tern. For a buried dipole at a given frequency, these patterns are similar in shape to those for a half-wave doublet a quarter wavelength above ground. With a buried dipole, the TM-mode pattern shows a stronger response than does the TE-mode pattern as the angle above the horizon decreases. The current along a lengthy, buried dipole, may exhibit resonances. Because of losses, however, it shows much less of a standing-wave pattern than does an above-ground antenna. Also, because of the slow velocity of propagation in the medium, these standing waves are crowded close together. Therefore, not much of a lobe structure above ground is created. This apparently means that for a velo-city factor on a wire in soil of 25%, a given dipole should be able to operate on frequencies up to values corresponding to four half wavelengths in soil without a bad antenna pattern. Of course, the antenna must be properly matched to accept power. In Fig. 3-46 of Ref. 10, there are some dips in the field strength-versus-frequency curve at 3 X /2 and 3 A which are not easily explained. There is obviously never any vertical lobe structure caused by height above ground! Other HF Antennas An intriguing antenna for concealment and partial hardening is a high-Q tuned loop. Those made by one manufacturer have about an 8-to-l tuning ratio and can be used for transmitting.Yll A recent compact model, the MLH-2/D is a half-loop on a ground plane. The manufacturer feels that these loops would not function well if buried in the ground, but could be used in a metal-lined conical cavity with protective dielectric lid or, less expensively, in a wire-mesh lined crater covered with such a lid. Considering a claim made many years ago for an early version of this antenna, only 3 dB worse than an average outdoor doublet at 4 MHz, I wonder if it might not perform well in a large subterranean cavity. For obtaining gain, the usual Yagi or log-periodic antenna would have to be installed in an enormous subsurface room. However, it is much simpler to obtain gain in buried antennas by re-ducing loss. This is done by paralleling dipoles as noted above. Another feasible way to obtain gain would be to design a buried V or rhombic antenna. Perhaps inserting capacitors in the wire period-ically could increase the phase velocity to simulate a wire above ground. Wheeler has mentioned this technique for acquiring a uniform current in buried wires; others have suggested it for above ground antennas.\12 It is suggested that experimenters try dipoles with more insulation than has been used. Wide-dia-meter PVC water pipes and even plastic sewer pipes suggest themselves. Internal supports would be needed and waterproofing is essential. VHF Antennas VHF antennas can be employed just below ground, the main purpose being concealment. A study made near 2 meters and above was concerned with com-munication at low angles to aircraft and nearby fixed stations.\l£ As expected, the TM mode was the most efficient in generating this surface wave. A f u l l , i n s u l a t e d horizontal dipole and a variety of miniature tuned loops and dipoles mounted a few centimeters deep produced TM fields at angles of about 10 degrees above the surface that were 14 or 15 dB weaker than TM fields from a test quarter-wave monopole on the ground. Considering cosmic noise as the only competition, it was estimated that with such an antenna and only one watt of power, communication with aircraft was possible at 50,000 ft to distances of 70 to 90 mi. An antenna not covered in the references is the slot antenna. It is not very difficult to construct at VHF and might be mounted just below a plastic lid on the ground. Beneath the antenna a mesh-lined cavity could exist. Arrays of resonant slots could offer substantial gain. It is worth noting that reception has been possible with a 2-meter hand-held radio in a plastic bag under salt water to a depth of 20 ft.\M Buried Antennas in a Nuclear Emergency Although the peril of a nuclear attack is "un-thinkable," most people would want to survive for as long as possible. The topic of survival communica-tion then is something many will want to consider. Because of the great attenuation of radio waves associated with a deeply buried antenna, Civil Defense (FEMA) personnel apparently now lean toward pop-up monopole antennas capable of pushing aside a ton of debris. These would be used only after an attack had ceased. This policy raises the question of how survivors would learn, in a deep shelter, that the attack had ceased. Some of the loop an-tennas described above are installed at Civil Defense locations in the recommended manner that would be of assistance in this situation. One problem that has not been mentioned is that of EMP; the short, powerful electromagnetic pulse from a nuclear b l a s t . U n d e r certain high-altitude conditions, explosions can disable a receiver a thousand miles away. It would be wishful thinking to hope there is no solution to EMP. Unfortunately, such a solution must exist, otherwise one could relax in the comforting thought that if missile-guidance systems would fail because of EMP, there could be no sustained nuclear war. Maybe some day everybody will be able to buy a kit which will " E M P - p r o o f " c o m p u t e r s , televisions, and radio equipment — a poor compromise, but possibly realistic. Consider that only a small area of a country is in the direct-hit category. It becomes obvious that the great difficulty of total hardening of a commun-ications antenna without destroying its e f f e c t i v e -ness should not deter e f f o r t to install antennas which, though only partially hardened, would provide some useful communications in the vast majority of cases. Hie Role of the Amateur Many years ago, I noted that it was possible to hear 20-meter daytime DX signals with a two-tube regenerative receiver connected to a short wire in the basement. Signals were received, even if they might have been reradiated from power lines, pipes, and heat ducts. For most people in a nuclear emer-gency, the basement would be the logical shelter. A few others would have bomb shelters. In either case, an amateur should be able to communicate with, say 10 watts of battery power, using narrow receiver band widths and hand-keyed CW. The antenna could be a very high-quality loop, a wire strung in the base-ment, or a center-fed long wire, insulated, ends not grounded, and buried just under the ground. An ad-justable matching network might be desirable, espe-cially for a wire buried in the soil, because im-pedance changes with frequency as well as with soil moisture. Amateurs could possibly assist Civil Defense units by conducting experiments and tests. The idea of a loop in a large underground room is intriguing; one would have to consider the possible need in some cases of filtering out strong broadcast- or power-frequency fields causing front-end overload and inter modulation. Common-mode pickup in certain types of balanced loops could be a problem. APPENDIX I Attenuation of Plane Waves in Subsurface Media An understanding of the material presented here is not necessary for a general application of the problems of subsurface radio communication. However, for those whose science education has taken them through Maxwell's Equations, use of a plane-wave analytical approach should make it possible to have a more quantitative understanding of the effect of soil or water on wave attenuation as distance, frequency and constants of the media are varied. For exact methods of solving the problems of subsurface antennas and propagation, the equivalent of a graduate science background and knowledge of complex electromagnetic problems is required. Many textbooks show that in the propagation of a plane electromagnetic wave in a homogeneous me-dium along the x axis, Maxwell's Equations yield the following for the electric field: E = E e o + rx (Eq. 1) where, (in the MKS system of units) x is the distance in meters E is the electric field in volts per meter at x = 0 E is the electric field at any distance along the x axis e is the Napierian base r is the propagation constant For attenuation and phase retardation, the negative sign in (1) is taken so that: E = E e o r x (Eq. 2) where, i u l f o e (1 + ) (BJ 3 ) w = 2 ' £ £ = K c 0 f is frequency in Hz k is relative dielectric constant ^ = p for non-magnetic material r = 8.855 x 10~12 1 o yo = 1.2566 x 10-6 o is soil conductivity, mhos/meter Since r is complex, it can be written: r = a + j ( : ! so Eq. 2 becomes: E = Eoe - ( a X + j S X) - a X - j S X = E e x e J o (Eq. 4) < is now attenuation in nepers per meter.(1 neper = 8.68 dB) and e represents phase retardation per meter. In free space a is zero and 8 = 2 t t / \ . However, we will be interested only in a . There is a problem in reducing (3) to obtain the value of a because one runs into a 4th-degree equation. A solution appears in Ref. 16, p. 271, Eq. 3. However, using De Moivre's Equation yields the identical solution. \1I Adapting that equation to the square root of a complex quantity: /A + jB = (cos e/2 + j sin s/2) (Eq. 5) where • + B , cos 6 = P i / o sin 9 = B / , o For convenience we now write (Eq. 3) as: r = j " n j r > f >•• 1 Fig. 1 — How to calculate anchor-point distances. Two 1/4-in eyebolts or hooks can be installed at the top of the mast to secure the pilot-catenary and back guy. These are permanent and will be out of the way. If you use a temporary gin pole, install the hooks at the top. Using 3/16- or 1/4-in stranded steel cable, attach the cables from the hooks at the top of the mast (or temporary gin pole) to the anchor points previously selected. The anchor points can be a tree, secure post or a 1-1/2-in pipe or rod driven 3 to 4 ft in the ground. The cables should be attached and drawn as tight as possible by hand. Cable clamps should be used to secure the cable. If a hoist or "come-along" is used to tighten the cable, be sure it is not too tight as the mast might be bent once the pilot catenary and back guys are in place. Temporarily drop the top set of tower guy wires, if in place, as the PCBG holds the tower very steady. Usually for a large beam, the set of guy wires at the top of the tower will be in the way. Disconnect them; dropping one guy wire will probably suffice. At most, two of the guys will have to be dropped. A pulley that will fit the pilot-catenary cable should be installed before the cable is tightened to its anchor point. The pulley should have an eyelet or hook and it will be upside down because the an-tenna assembly hangs from it. This can be seen in Fig. 3. A rope, 3/8 or 1/2 in, should be tied to this pulley eyelet, run to the top of the mast on the tower, through a second pulley and then to the base of the tower. For a hundred-foot tower, this will take approximately 250 to 275 ft of rope. This rope is used to pull the beam assembly to the top of the tower. The next step is to get your beam ready to be pulled up. The beam should be assembled and tested as would normally be done prior to installing it on any tower. A short temporary mast is placed in the normal mast position of the antenna, with one ex-ception. The mast can be any convenient length, 1 to 3 ft. If your antenna has boom guying, then the temporary mast should be long enough to allow the boom guys to be attached as they would ordinarily. Fig. 2 — Arrangement of antenna mast with hooks and pulley. Apx. 3" pulley the top of the temporary mast. This distance should be adjusted to match the distance from the top of the mast on the tower to the point above the tower top where the antenna will be attached. This is not critical, but should be considered. To secure the temporary mast to the antenna boom-to-mast plate is simple, but slightly dif-ferent. The mast is normally mounted to the boom-to-mast plate, on the side opposite from the boom. Naturally, they could not be mounted on the same side. The mast is usually held to this plate by one or two bolts above the boom and one or two below. To mount the temporary mast, put it on the same side as the boom, using only the top U bolts to fasten it. These top bolts are sufficient to hold the antenna while it is being hauled to the top of the tower. See Fig. 4. This leaves the opposite side of the boom-to-mast plate clear and ready to attach to the regular mast, using the bottom U bolts. The bolts will hold the antenna in place while the temporary mast is removed and the top U bolts installed on the regular mast. When you have the temporary mast attached to the antenna, and connected to the pulley on the pilot catenary, you are ready to haul it to the top. You should be sure all the cables, clamps and con-nections are tight and secure. The antenna can float free as it is ascending, but guidelines can be con-nected to turn the beam if necessary. Be sure you have a way to disconnect these guidelines if used because the end of your antenna is difficult to reach from the ground. Care should be used in pull-ing. For safety, a second person should snub the rope around a post or tree in case the rope slips out of your hand. The antenna can be raised and lowered with one person, but for safety's sake, two or more people should be employed. A person on top of the tower while the antenna is being raised is recommended because he or she can watch for any obstructions the antenna might encounter. The an-tenna suffers no ill effects from this technique. It is subjected to less force than it will encounter during strong winter winds. See Fig. 5. Once the antenna is intalled at the top, the lowered guy wires can be restored and the pilot catenary and back guy cables removed. The pre-paration for this technique takes a little time and thought, but the actual raising and installation of the antenna can be done in less than an hour. Rope Co temporary mast Fig. 3 — Arrangement of inverted pulley or trolley on pilot-catenary cable with pull rope. Provisions should be made at the top of this mast to secure a line or hook so that it can be connected to the pulley that is on the pilot catenary. A short piece of rope works fine, as this allows some flex-ibility in adjusting the distance from the pulley to Fig. 4 — Temporary mast installation. Fig. 5 — Thirty-two-ft boom antenna, in transit, with temporary mast in place and suspended from pilot catenary. General Antenna and Transmission-Line Information The Horizontal Dipole Over Lossy Ground By Robert B. Sandell, W9RXC 26 G. H. Baker Dr., Urbana, IL 61801 Technical material available to radio amateurs view antennas as wires hanging in free space, or above a ground of infinite conductivity. While these idealized conditions are necessary for the teaching of radio physics, they are of little help to the amateur trying to understand, or evaluate, his own real-world antenna. This paper, in a qualitative fashion, examines some of the effects of operating a horizontal dipole over a ground of finite conduc-tivity. The structure of the near field and the far field are discussed briefly, and then four parame-ters of the horizontal dipole are examined as the antenna is moved from an environment of free space to one of close proximity with lossy ground. Driv-ing-point impedance is first examined and a cor-rected curve is developed for radiation resistance versus h e i g h t . Efficiency is explored next and graphical presentations exhibit heat losses incurred as the near field contacts the lossy media. In con-nection with efficiency, changes in bandwidth are noted. Finally, the vertical angle of radiation is calculated as a function of height and the reflec-tion c o e f f i c i e n t calculated as a function of soil constants. Comments such as, "I put up a new antenna but it doesn't work like the Hie ARRL Handbook said it should," are frequently heard on the air. This is usually followed by a condemnation of antenna theory as worthless gibberish because it seems to deal with horizontal antennas in free space or vertical anten-nas above a perfectly conducting ground plane. What these well-meaning enthusiasts fail to realize is that antenna physics is divided into two parts. The first deals with antennas located in idealized en-vironments (free space and above ground planes of infinite conductivity), while the second part deals with the effects that occur when the antenna is moved into a hostile environment (lossy ground). It is regrettable that the subject has been given such cursory treatment in amateur literature since real-world antennas can, at times, be profoundly effected [sic] by ground presence. T h e f o l l o w i n g discussion deals, in l a r g e l y qualitative fashion, with the e f f e c t of a lossy (real life) ground on four important parameters of horizontal antennas. Detailed calculations of ground effects are, as we shall see, largely impossible to make. An understanding of the basic ground proximity mechanisms, however, should enhance the ability to design antennas or to interpret their results. Ground effects are not the same for vertical and horizontal polarization .\i Although the remarks and data contained herein refer to single horizontal dipoles, they apply equally well to all horizontal antennas that are dipole derived. Uda-Yagi arrays, c o l l i n e a r beams, inverted V's and horizontally polarized cubical quads and delta loops fall into this category. Earths and Fields Soil or earth appears almost purely resistive at very long wavelengths. At very short wavelengths, it presents itself as an almost pure dielectric. In the so-called high-frequency range of 10 to 160 meters, earth takes on a combination of the two, or more simply, appears as a lossy d i e l e c t r i c . T h i s is unfortunate because it greatly increases the complexity of mathematical calculation. Earth can be categorized by parameters of permittivity and con-ductivity or the ability to store and conduct charge. Permittivity is generally specified in rel-ative units, and conductivity in millisiemens per meter. Some typical values are: Good Soil 15 Average Soil 10 Poor Soil 2 Residential & Industrial 1 20 15 10 4 Maps containing regional conductivity figures are readily available; however, they are based on averages made from wide ranging data and thus are totally ineffective in pinpointing the conductivity at a s p e c i f i c location.\3,4 In addition, conduc-tivity can vary both daily and seasonally. Maps give no hint as to dielectric qualities. Conductivity may be measured with some difficulty, but the measure-ment of soil dielectric qualities is beyond the scope of Amateur Radio.Y5. It is for these reasons that most operators have only scant knowledge of their own soil parameters. The above table will serve as a fair estimate to use for evaluation pur-poses. In all probability, a majority of Amateur Radio sites will approach the residential-industrial category. While this may seem unduly stringent, it must be remembered that the original prairie con-tained no roads, sidewalks, or a host of other ob-stacles to good ground performance. Also, most urban and suburban areas usually have been subject to filling, scraping and leveling, all of which serve to alter ground characteristics. Fig. 1 describes the fundamental geometry of a dipole operating over lossy ground. Point A repre-sents the end of the antenna (the wire is perpen-dicular to the page). Region A is air, Region B is lossy ground, and the x axis is the air-earth in-t e r f a c e . H is the height of the dipole above ground. The area immediately surrounding an antenna contains not only the radiating field, but also strong reactive or s t o r a g e f i e l d s . This s t o r e d energy is returned to the antenna at a later part of the c y c l e . T h e strength of the radiating field falls o f f inversely to the distance from the an-tenna, while the stored energy falls off at rates inverse to the square and cube of the distance. At a distance of one-half wavelength, the stored energy has fallen to an insignificant value. At one-sixth of a wavelength, the radiation and storage fields are of equal strength.XZ In Fig. 1, r^ has a length of one-sixth wavelength and r, h a s a l e n g t h o f one-half wavelength. The area within the circle whose radius is r„ is known as the near field. When a resonant half-wave dipole is operated in the environment of free space, its current and volt-age will have a fixed ratio which is, by definition, an impedance. Since the current loop is coincident with the terminal points, it is also the radiation resistance. Traditionally, this ratio has been given as 73.13 ohms. (Later research indicates that at true resonance the figure should be 68 ohms, but the old value is still most commonly used.\14) A change occurs when the dipole is moved into the earth's environment. A portion of the radiated energy is reflected by the ground and in turn is intercepted by the antenna. This intercepted signal causes a second current to flow in the antenna and thus modifies the original voltage/current ratio. This new ratio is known as the driving-point im-pedance. It is this impedance, which is effected by ground proximity, that we are concerned with in feeding and matching the antanna.\8r9 The solid curve in Fig. 2 is a commonly published repre-sentation of the driving point (radiation) r e -sistance of a horizontal half-wave dipole at various heights above a perfectly conducting ground. It is obvious from examination of the curve that the greater the height, the less profound the ground effect on impedance. If the curve were extended to about ten wavelengths, it would be difficult to distinguish from a straight line. Waves incident on a perfectly conducting sur-face are 100% reflected. This is not true for a lossy medium, such as earth, where a portion of the energy is absorbed. When the dipole is a long way from earth, the effect of its reflected signal is insignificant. As it moves closer to earth, the impedance variation becomes greater and greater. Eventually the near field will come into contact with the lossy earth. Stored energy is then absorbed by the earth and converted to heat instead of being returned to the antenna for later use. The dashed-portion curve in Fig. 2 represents the impedance variation when the near field comes into contact with a poor-quality ground.\lflJLl This phenomenon is known as near-field coupling. As the quality of the soil improves, the curve approaches the solid-line value. This effect of height versus impedance is especially important when evaluating pre-cut or pre-tuned antennas or in tuning arrays at ground level. Efficiency The efficiency problem versus height al^pve ground is closely related to the impedance problem; both involve the concept of near-field coupling. When the antenna is high above ground, the amount of power lost to ground heating is insignificant. As the near field, with its large quantity of stored energy, comes into contact with lossy earth, power is lost that would normally end up as radiated power.\12. Fig. 3 illustrates the estimated fall-off of efficiency as a dipole, operating at 18 MHz, is moved close to a poor-quality earth. Note that below — The basic geometry. Ohm a I O O 90 8o ro 6o So 4o 30 lo to / A / i / / / / \ 1 ' / » ' 1 •..11 / / / i. H Fig. 2 — The driving-point resistance of a half-wave dipole over a ground of infinite conductivity and over lossy ground. one-fifth wavelength, the curve falls off mono-tonically toward zero. While good electrical height is easily attained at high frequencies, it comes at a high cost below about 5 MHz. Those who contemplate antennas for the 80- and 160-meter bands should certainly consider the trade-off between efficiency and the cost of vertical support structures. Bandwidth Bandwidth is defined as the frequency range in which an antenna parameter will hold to within a given limitAli The' parameter may be impedance, pattern or gain. The most frequently used parameter is impedance, as indicated by the standing wave ratio. Thus, the impedance bandwidth would be the band of frequencies where the antenna exhibits an SWR of less than, or equal to, some fixed limit (often 2:1). Bandwidth is an inverse function of Q (the figure of merit) which, in the ease of an antenna, is the ratio of the ability to store power to the ability to radiate power.\14 Expressed in more definitive form: Bandwidth 1 « Power Radiated Q Power Stored While plural lobes may occur at some heights, it is generally the lowest lobe that is of interest. Its location may be approximated, with good accur-acy, by a simple formula in which /a is the angle of radiation, expressed in degrees above the horizon-tal, and H is the height of the dipole above ground. A portion of the storage field is dissipated when it interfaces with lossy earth. As stored power is reduced, Q is reduced, resulting in increased bandwidth. While wide bandwidth is a desirable characteristic, in this case it is accomplished at the high price of efficiency. Some misleading band-width figures are sometimes published because an experimental or test antenna was located too close to a lossy ground. J ? c >. c 9o So To bo So 4o J o .1 .2 .3 .4 .5 Heiqht in wovalcnqths Fig. 3 — Efficiency of a half-wave dipole over poor quality soil. Angle of Radiation The fourth parameter effected by height is the angle of radiation or the angle of signal departure. Most amateurs are familiar with the concept of height versus radiation angle, having seen it il-lustrated using an image antenna that is hypoth-esized symmetrically below the air-earth interface. While image theory is a powerful tool in electro-statics, it can be misleading when used in antenna work because it implies a ground of infinite con-ductivity.VLi Fig. 4 presents a more realistic pic-ture in which a dipole is located at point A; its axis is perpendicular to the page at a height H above the x axis, which represents lossy ground. In order to obtain maximum signal strength at remote point P, the direct . ray (AD) and the ground re-flected ray (BC) must arrive at point P in phase. Since earth r e f l e c t i o n shifts the phase of a horizontally polarized wave by 180 degrees, the phase relationship of the two signals arriving at point P is controlled by the length of ray AB. When AB is an odd number of half wavelengths in length, the two components will add in phase and create a lobe. Conversely, when AB is an even number of half wavelengths long, a null will be created. Fig. 5 illustrates lobes at 30 degrees. Fig. 4 — Ray geometry of a dipole above lossy earth. Fig. 5 — The lobes of a dipole located one-half wavelength above a perfect conductor (solid line) and above lossy earth (dashed line). Za_ 14.33 H (H is in wavelengths) Our main concern herein is to determine what effect lossy ground has on the lobes. If the earth in Fig. 4 were a perfect electrical conductor, all of the energy in the incident ray AB would be re-flected, and energy BC will equal that of AB. We know from geometric optics that the angle of in-cidence, Za^in Fig. 4, will equal the angle of reflection, Zfllx When the earth is not a perfect electrical conductor, part of the incident energy will be transmitted into the ground to be dissipated as heat, and the remainder reflected as ray BC. As before, equals lal. The relationship: Reflected Power Incident Power is known as the power reflection coefficient, often designated as jr|2 , and when multiplied by 100, indicates the percentage of incident power re-flected. The formalism of this calculation is contained in my note at the end of this article. Table 1 illustrates the percentage of power re-flected for various angles of incident and soil conditions. Note that the Joss increases both as the angle of radiation (incidence) increases and as the soil quality decreases. Thus, the long-distance enthusiast who has raised his antenna to attain a favorable low angle of radiation will suffer less transmission loss than the amateur whose antenna is at a low height with its resulting high angle of radiation. In Fig. 5, the solid line indicates the lobes of a dipole one-half wavelength above a per-fectly conducting ground, while the dashed curve is the same antenna above poor ground.\l£ Note not only the difference in lobe amplitude, but also the fact that the complete null between the symmetrical lobes is partially filled when the earth is lossy. Conclusion The driving-point resistance of a horizontal dipole antenna is altered as ground proximity in-creases. Below about one fifth wavelength, the variation is controlled not only by height but also by soil characteristics. Both the bandwidth and the efficiency are affected when the near field comes into earth contact. While the bandwidth may be broadened in a desirable fashion, it is done at the expense of efficiency. High antennas, with their predominantly low angle of radiation, suffer less transmission loss over poor quality ground than do low antennas with their high angle of radiation. Author's Note When the Fresnel reflection coefficient, Rh, is squared, it will yield a power reflection coeffi-cient. For horizontal polarization: sin /a-A ' - cos /a R^ = sin /a+/r' - cos Z& where e ' is the complex permittivity of the medium. Refs. 1 and 17 contain good discussions about Fresnel coefficients. References \1 Burrows & Atwood, Radio Wave Propagation, Academic Press, New York City, 1949. \2. Terman, Radio Engineers Handbook, McGraw-Hill, New York, 1943. Fine, An Effective Ground Conductivity Map for the Continental United States, Proceed-ings of the IRE, September 1954. V4 Jordan & Balmain, Electromagnetic Waves & Radiating Systems, Prentice-Hall, Englewood Cliffs, New Jersey, 1968. IEEE, Recommended Guide for Measuring Ground Resistance and Potential Gradients in the Earth, Std. 81-1962, IEEE, New York, 1962. V S . Blake, Antennas, John Wiley & Sons, New York, 1966. \1 Ramsey & Dreisbach, Radiation and Induction, Proceedings of the IRE, New York, 1966. \£ Williams, Antenna Theory and Design, Vol. II, Sir Isaac Pitman, London, 1950. \2. Schelkunoff & Friis, Antenna Theory and Practice, John Wiley & Sons, New York, 1952. \lfl Ma, Theory and Application of Antenna Ar-rays, John Wiley & Sons, New York, 1974. Ml Sommerfield & Renner, Radiation and Earth Absorbtion for Dipole Antennae, Wireless Engineer, London, September 1942. \12. Hansen, "Hie Radiation Efficiency of a Dipole Antenna Located Above an Imperfectly Con-ducting Ground, IEEE Transactions, APS, New York, November 1972. \li IEEE, Standard Definitions of Terms for Antennas, IEEE, New York, 1983. Kraus, Antennas, McGraw-Hill, New York, 1950. \15 Rao, Elements of Engineering Electro-magnetics, Prentice-Hall, Englewood Cliffs, New Jersey, 1977. \1£ Ladner & Stoner, Short Wave Wireless Com-munication, Chapman & Hall, London, 1942. \11 Radford (contributor), The Handbook of An-tenna Design, Peter Peregrinius Ltd., London, 1983. Additional Reading: Bhattacharya, Input Resistance of Dipoles Over Homogeneous Ground, IEEE, Transactions, APS, New York, 1963. Miller, Poggio, Burke & Selden, Analysis of Wire Antennas in the Presence of a Conducting Half Space, Part II, Canadian Journal of Physics, Vol. no. 50, Montreal, 1972. Karwowski, Low-Frequency Approach to the Problem of a Horizontal Wire Antenna Above an Imperfect Ground, IEE, Proceedings, Vol. 131, Part H, London, 1984. 10° 40° Good soil 9 3 % 8 7 % 7 6 % 6 0 % Average soil 9 2 % 8 5 % 7 3 % 5 5 % Poor soil 8 9 % 8 0 % 6 4 % 4 3 % Residential & Industrial 8 2 % 6 8 % 4 6 % 2 5 % Table 1. Percentage of power reflected for various angles of radiation and soil conditions. Biography Robert B. Sandell began his Amateur Radio ac-tivity in 1937 and currently holds the call letters W9RXC. He served with the United States Navy in World War II working with the technical aspects of radar. He is a graduate of the University of Min-nesota and has studied antenna and propagation theory at The Georgia Institute of Technology, George Washington University, the Southeastern Cen-ter for Electrical Engineering Education, and the University of Dlinois. He is a member of the Institute of Electrical and Electronic Engineers, the Antenna and Propa-gation Society, the Microwave Theory and Techniques Society, the Antenna Measurement Techniques Assoc., and is an affiliate member of the American Institute of Physics. He was a participant at the 1981 and 1983 International Conference on Antennas and Propa-gation, sponsored by the Institution of Electrical Engineers, (London), at the University of York, England, and the University of East Anglia, England, respectively. Antenna Polarization Gerd Schrick, WB8IFM •4741 Harlou Drive, Dayton, Ohio 45432 In point-to-point communication using elec-tromagnetic waves, the antennas are usually posi-tioned and oriented in such a way as to maximize the transfer of energy. This requires pointing the an-tennas in the direction of maximum gain and using the proper polarization. A problem arises when one or both of the sta-tions are mobile and antenna rotation becomes im-practical. Traditionally, this problem was solved by using vertical antennas. In aeronautical mobile op-eration, however, this problem is further aggravated by requiring elevated radiation. Finally, in space communication, the spacecraft moves overhead and rotates (spins) at the same time, so the question of polarization becomes a major concern. See Fig. 1. Electromagnetic radiation is best understood by looking back at the basic radiator: a short loaded dipole (Fig. 2). By alternately charging the end spheres of this dipole with positive and negative electricity at a high rate, a propagating electro-magnetic field is generated with preferred direc-tions in regard to the orientation of the dipole. It should be understood here that positive means a lack of electrons and negative means a surplus. ^ h Fig. 1 — Conmunication with moving platforms. O—.r—O Fig. 2 — The short dipole (Hertz dipole). The radiation of a dipole is the f a m i l i a r doughnut shape directional pattern with omnidirec-tivity around the axis and no radiation o f f both ends of the dipole (the hole of the doughnut). Refer to Fig. 3. In this pattern the polarization of the electric field vector at some distance (far field) at the point of maximum radiation is in line with the axis of the dipole. As a general rule then, the polarization of an antenna is parallel to the cur-rent carrying elements. As one deviates from the maximum of the radiation pattern, the field vector, and therefore the optimum polarization of the re-ceiving antenna, follow the doughnut pattern (Fig. 4). It is helpful to keep these simple facts in mind for the following discussion. In practice, antennas for HF are usually hor-izontally polarized. This helps to minimize ground conductivity problems. Vertical antennas, in order to have the desirable low-angle radiation, require good ground conductivity or radials. On the higher frequency bands, where line-of-sight propagation prevails, the question of polarization is less im-portant as long as it is compatible. Therefore, for 2-m FM, vertical polarization is used because it can compensate for moving objects. In the more serious DX work that takes place on VHF/UHF, horizontal polarization is used, taking advantage of the gain afforded by ground reflection. Fig. 3 — The familiar doughnut radiation pattern of a dipole. But what about c i r c u l a r polarization? It is often used by commercial stations, the military and in space communication. In general, all single-fed antennas generate a linear-polarized wave such as the elementary dipole described above. Examples are depicted in Fig. 5. As can be seen, the shape of the antenna does not result in any special polarization other than linear. The outer lengths of the antenna conductors are merely used to resonate the antenna or provide proper phasing (as in the case of the quad) to other elements for the same antenna struc-ture. The RSGB VHF/UHF Manual states that, "Being merely a dipole bent into c i r c u l a r shape, the popular halo antenna for 2 m is far from omni-directional. Most of the radiation from a half-wave S Fig. 4 — Sample electrical field of a transmitting dipole and optimum alignment of receiving dipoles 1...3. element is from the middle of the radiator, regard-less of what is done with the ends."\i. Now, when considering circular-polarization waves or antennas, contemplate one more dimension: time. It is e a s i e s t t o v i s u a l i z e a c i r c u l a r -polarized wave by imagining a normal transmitting dipole rotating at the frequency of the wave. Take a 146-MHz signal for example. The dipole rotates 146 million times per second. If this dipole is opposed by a receiving dipole in any arbitrary linear fixed position, it would intercept part of the transmitted wave, but not all. As a matter of fact, it will re-ceive 5 O S S of the power that it could if both dipoles were perfectly aligned. On the other hand, if both dipoles are fixed and poorly aligned, or even at a right angle, less than 50% or nothing at all would be received. Finally, if the receiving antenna is capable of receiving the rotating wave, power transfer would again be 10085. This is shown in Fig. 6 and Box 1. 1 H f -tk J L r TX. noh rotating rotating hof«f»/15 hot) rotating ft aligned. ahjj position poorly alifl/iul 100 % So % <50 7o 0 Fig. 6 — The rotating transmitter dipole vs. the stationary receiving dipole. (Effects of polarization mismatch.) It is clearly impossible to rotate the antenna physically. So what are the methods to generate the circular polarized wave? There are two methods known and used. The first consists of feeding two or more independent linear-polarized antennas so as to gen-erate a circular-polarized wave. The second method uses a wave-structure antenna where the electric wave generated is guided along in a circular way. Both methods are widely used, but the second method is used mostly for the higher frequencies with the 70-cm band forming the dividing line. So far we have explained the function of these antennas in terms of transmitting. These same an-tennas, of course, function as well in a similar manner as receiving antennas. \ T \ — I t O — 0 c O 3 3 1 I oO 0 3 3 0 3 3 0 00 p 3 3 oO 0 Fig. 5 — Different shaped antennas generate the same linear polarization. Polarization mismatch is a term used to de-scribe the less than 100% power transfer. Circular polarization then is the best choice when dealing with unknown or changing polarization. The event of the first ham in space (the Columbia space shuttle mission no. 9, Dec. 1983) is one example of how Amateur Radio operators showed concern about using proper polarization. The interaction of various polarizations is summarized in Fig. 7. Fig. 7 — Interaction of various polarizations (extra attenuation in dB). Before going into details about the actual antennas used, a few more terms should be explained. Two rotational directions are possible with the circular-polarized antenna. They are labeled right hand circular polarized (RHCP) and left hand cir-cular polarized (LHCP). By convention, the polar-ization is viewed from the transmitting antenna in the direction of the radiation (Fig. 8A). Elliptical polarization is a term often heard. It refers to a two-feed system antenna and means that one of the Pig. 8 — The generation of a circular polarized wave (Lissajous figure on the scope, cross dipole). in which ease the feed lines to the radiators should be equal. By adding a half wave to the quarter wave delay or to one of the equal lines, the sense of rotation is reversed. Switching and inserting lengths of coaxial cable will switch the antenna to a number of linear polarizations as well as to LHCP or RHCP. Note an interesting effect: From the back of the antenna a reverse circular-polarized wave attenuated by the front-to-back ratio eminates. These features are, however, difficult to implement at UHF and higher. Fig 9. suggests some feed-line and matching approaches depending on the quarter-wave transformer in use. Helical Antenna Because of its physical size, the axial helical beam is practical from about 435 MHz and up.\2 It uses only one feed and generates a circular polar-ization in space by guiding the wave along the he-lical structure. Here it is easy to determine the rotational sense just by looking at the helix. The helical antenna has one polarization and cannot be switched. The antenna design is not critical and is rather broadband. Bandwidth ratios of 2:1 are not uncommon. This feature makes it possible to use the same helix on 144 and 220, or 220 and 435 MHz. Fig. 10, courtesy of RSGB Handbook, offers some basic dimension and gain figures.\2 Pig. 8A — Right hand circular-polarized wave; electric field vector. field vectors generated is stronger than the other, leading to a deformation of the circle into an ellipse. Most practical antennas exhibit some amount of ellipticity. If more than two antennas are used, the resulting pattern could be distorted more from the circle than a simple ellipse. Such is the case with the 144-MHz downlink signal of OSCAR 10; ap-parently one of the three antennas, excited for RHCP, was damaged when it was deployed. Crossed Dipoles/Yagis To generate a circular-polarized wave, which we have described as a time varying linear-polarized wave, two crossed dipoles are fed with the same sig-nal, but out of phase. The phase difference deter-mines the polarization, as does the orientation of the dipoles. See Box 2. The two dipoles are at right angles to each other and in the same plane, and the signal to one of the dipoles is delayed by a quarter wave. The way this works can be compared to gener-ating a circle on an oscilloscope (Lissajous fig-ures). If the same sine wave is fed to the x and y plates of the scope, a diagonal line is obtained. If we now delay (or shift the phase of) the signal to one of the plates, an ellipse will be formed. When the delay is right, a circle is formed as in Fig. 8. What can be performed with the dipoles can also be done with Yagi antennas. Sometimes the antenna may be shifted on the boom by a quarter wavelength, xt'Oh : Pig. 9A — Quarter-wave transformer. Fig. 9B — Methods of feeding crossed dipole/yagis for circular polarization using quarterwave delay lines and transformers. Pig. 10A — Helical antenna, courtesy RSGB. OSCAR 10 Refer to Fig. 10. Aside from the fact that the 2-m-gain antenna structure (Fig. 11) has been dam-aged, the elliptical orbit naturally varies the distance to earth. However, the attitude relative to space remains the same, and the radiation pattern is designed to keep the signal strength constant. This does not take polarization into consideration, and many observers with the capability to switch the polarization type have found it superior for dif-ferent OSCAR 10 positions. At or near apogee (the highest position), the intended RHCP is definitely optimum. The exclusive use of this polarization, although not always optimum, is definitely suffi-cient for operation. Fig. 12 shows the OSCAR 10 beacon signal at 145.810 MHz recorded by Guenter Schwarz, DL1BU, using a switchable crossed Yagi arrangement which permitted four linear polarizations and RHCP and LHCP. The recording took place at apogee and is most indicative of the intended and achieved link energy transfer. The rapid fluctuations of the signal (re-ferred to as spin fading or spin modulation) most noticeable on all the linear polarizations is caused by the damaged antenna in conjunction with the spinning (about 2 revolutions per second) of the s a t e l l i t e . The f a c t that the circular-polarized antenna does not pick up this "spin fading" does reveal that the real cause is related to polar-ization mismatch. The circular-polarized antenna (RHCP) takes care of this effect nicely. The LHCP is received at about 12 dB weaker and is akin to the front-to-back ratio. The linear polarizations are all expected to produce a 3-dB weaker signal which is generally confirmed by the recordings. Fig. 11 — OSCAR 10, 2-m antenna feed system for circular polarization. Lengths of coaxial feeds: dl = i + 1/3 ; d2 = J L + 2/3 X . Final Comments Space communication is the wave of the future. The great capacity of the line-of-sight waves used, as well as their low distortion, have already proved its superiority to the HF band. For example, world-wide television could hardly be realized by HF. Hams must become more involved in the high frequencies. Understanding wave propagation and polarization is part of it. OSCAR 10 ORBIT 171 M$tpl1983] 20:00 UTC ipojti ( j ? 0 0 0 „,;lr<.^ Fig. 12 — Plot of the OSCAR 10 beacon signal. Box 1 Gain of Circular Polarized Antennas To consider the gain of a circular-polarized antenna, it is best to consider the crossed Yagis. If the Yagis were side by side or stacked, the gain of the arrangement would obviously be 3 dB higher than a single Yagi. In the cross arrangement, this 3 dB is not available and is considered polarization loss. However, if the intercepted wave were circular polarized, there would be no polarization loss for the crossed Yagi and 3-dB polarization loss for the side-by-side Yagi, thus making the two antenna arrangements equally efficient. Box 2 How to Determine Polarization of the Crossed Dipole/Yagi Locate the dipole halves or gamma match rods that are connected to the center conductor of the coaxial lines. Rotate the tip of the dipole half directly connected to the feed line on the short way to the tip of the dipole half connected through the quarter wave delay line. You now have the sense of rotation. If you then move out into the direction of the radiation (forward), the screw formed will either be right hand or left hand corresponding to RHCP or LHCP. If you move to the opposite side of the antenna (backwards), the polarization is op-posite. Some delay lines are 3/4 wavelength; in that case the polarization is also opposite (reverse) from the one at 1/4 wavelength. References \1 Jessop, G. R., VHF/UHF Manual, RSGB, 1971. \1 Kraus, John D., Antennas, McGraw-Hill, Inc., New York, 1950, Chapter 7, p. 173. \1 Radio Society of Great Britain, RSGB Handbook, 4th edition, 3rd printing, Feb. 1971, pp. 14-20 to 14-21, 9-28. Table 1 The helical antenna — dimensions and performance. Courtesy of RSGB. Dimensions Band D R P a d General 0 . 3 2 X 0 . 8 X 0 . 2 2 X 0 . 1 2 X 4 3 2 8 - 3 / 4 " 2 2 " 6 " 3 " 1 / 4 " 1 2 5 0 3 " 7 n 2 " 1 - 1 / 8 " 1 / 4 to Turns 6 8 1 0 1 2 2 0 Gain 1 2 dB 1 4 dB 1 5 dB 1 6 dB 1 7 dB Beamwidth 47 o 4 1 ° 3 6 ° 3 1 ° 2 4 ° Baluns: What They Do And How They Do It By Roy W. Lewallen, W7EL •5470 SW 152nd Ave., Beaverton, OR 97007 I've always been a bit bothered by baluns, since I was never sure what they are supposed to do, let alone how they might go about doing it. The majority of articles deal with various ways of building and testing baluns, or the advantages of one type over another, but almost never a word about when or why a balun is necessary, if at all. Like most amateurs, there have been few occasions when I have been able to tell if a balun has any effect on an antenna system, and when it has, the effect hasn't always been good! The turning point came when I was trying to measure the resonant frequency of a folded dipole through a one-wavelength coaxial line. The bridge null varied a great deal as I moved my hand around the coaxial cable, or if the line or bridge was moved. A hastily constructed balun in-stalled at the center of the dipole eliminated the problem. But why? I found a brief, but clear explanation of one phenomenon involved in a paper by Maxwell, W2DU.U However, many questions remained. This led me to an investigation of just how baluns are supposed to work, and what problems they are supposed to cure. One surprising conclusion I found from my research is that one popular type of balun, when properly designed and used in an antenna system, may not solve the problems that baluns are expected to solve. Other results indicated that the type of feed line (balanced or unbalanced) has little to do with how well a system is balanced. In order to verify, or r e f u t e , the t h e o r e t i c a l r e s u l t s , s e v e r a l experiments were carefully set up and run, and the data analyzed. The result is a much clearer view of the operation of baluns in antenna systems, and some definite "dos" and "don'ts" regarding their use. What Problems are Baluns Supposed to Solve? Baluns usually solve problems caused by an imbalance. An imbalance of what? To answer this question, we need to look at current flow in transmission lines. In a coaxial cable, the currents on the inner conductor and the inside of the shield are equal and opposite. This is because the fields from the two currents are confined to the same space.\2 With the presence of skin effect, a different current flows on the outside of the shield than on the inside. The current on the outside, if significant, causes the feed line to act like an antenna, radiating a field that is proportional to this current. A twin-lead feed line has similar properties, despite its different physical nature. Since it is physically symmetrical, if the currents flowing through the conductors are equal and opposite, the radiation from the line is minimal (assuming that the conductor spacing is very small relative to a wavelength). However, several factors may cause the currents in the two conductors to be imbalanced, that is, other than equal and opposite. If this happens, the balanced feed line will radiate like a coaxial cable that has current on the outside of the shield. This occurs because the components of the currents on the two conductors that are equal and opposite create fields which cancel. But the field from any remaining component on either conductor (called a common-mode, secondary-mode, or antenna current) will cause radiation.\4,5,6 In this ar-ticle, the current on the outside of the coaxial shield, or the antenna current on the twin lead, will be called the imbalance current: They are caused by the same things and produce the same effects. Imbalance current, on either kind of line, is the cause of a number of undesirable effects: o pattern distortion (caused by the feed-line radiation adding to the antenna-radiated field, or by unequal currents in the antenna halves) o TVI (radiation from a feed line coupling into nearby television sets, house wiring, and so on) o RF in the shack (caused by a "hot" radiator — the feed line — residing in the shack) If you have read other articles on baluns, you'll recognize these as the problems baluns are supposed to solve. What isn't usually too clear is that they are all caused by current imbalance, on either coaxial or twin-lead feed line. Of course, if the imbalance current is sufficiently small to begin with, a balun is not necessary at all. Or it can be said that a properly designed balun will not solve the problem being experienced. What Causes System Imbalance? The first cause of imbalance currents was explained by Maxwell. It will be repeated here for completeness. When a balanced antenna is fed with coaxial cable (Fig. l), the outside of the shield appears as an extra, separate conductor connected to the right side of the antenna at the feed point. The current in the cable's center conductor flows into the left half of the dipole. The equal and opposite current on the inside of the shield flows partly into the right half of the dipole, and partly along the outside of the shield. The proportion of current which flows each way is determined by the relative impedances of the two paths. The current on the outside is the greatest when the total effective length of the path along the outside of the coaxial cable from the antenna to ground is an integral number of half wavelengths, since this makes the impedance presented by the undesirable path relatively low. If the rig is effectively an odd number of quarter wavelengths from actual ground, it is at a voltage maximum and can be hot. On the other hand, there are other combinations of lengths for which the imbalance current will be negligible — cases where a balun does not make any noticeable difference. The obvious solution to this problem is to feed F i g . 1 — Imbalance caused by another path to ground from only one side of the dipole. the balanced antenna with a balanced feed line — twin lead. This solves the problem neatly, until you encounter the problem which most of us have today, illustrated in Fig. 2. Suppose that we went ahead and connected the line as shown schematically in Fig. 3. If the rig could be totally isolated from ground, the feed-line conductor currents would be equal and opposite, just as they would be if coaxial cable were used; the imbalance current would be zero, and the feed line would not r a d i a t e . \ 1 However, when we connect the rig to ground, as shown by the broken line, we've again provided a third conductor in parallel with the right side of the feed line, and the same problem occurs as with the coaxial cable (see Fig. 4). So either type of line is unbalanced if a direct path to ground is provided from one side, and both can be a balanced, non-radiating line if the imbalance current is elim-inated. Imbalance current can be caused also by situations where the two sides of the antenna are not precisely symmetrical: Coupling to nearby ob-jects, the tilt relative to ground, or slight dif-ferences in lengths of the two antenna halves.\& Another cause of imbalance currents is induction. If the feed line is not exactly placed at a right angle to the antenna, a net current is induced into it by the antenna field. This current appears as an im-balance current. At UHF, where the diameter of c o a x i a l cable is a substantial fraction of the length of the antenna elements, coaxial line is more difficult to place symmetrically relative to the antenna than twin lead is (this is sometimes given as the only reason for using a balun!). The problem is negligible at VHF (except perhaps with very large diameter coaxial cable) or below. What Baluns Do Let's recall what we want a balun to do: cause the currents in the feed-line conductors to be equal in magnitude and opposite in phase, resulting in a zero imbalance current. How well do the popular balun types do this? One type of balun is known as a transformer-type balun or balun with a tertiary winding (Fig. 5).\a This type is commonly used for providing single-ended to differential conversion for driving balanced mixers, push-pull amplifiers, and so on. It seems to be suitable for our purpose. An analysis of its operation (see Appendix 1) shows that it does indeed perform an unbalanced-to-balanced c o n -version. The voltages at the balanced port are caused to be equal, and opposite, in phase relative to the cold side of the unbalanced port. Thus, the use of this sort of balun will eliminate the problem of current flow on the outside of a line only if the antenna is perfectly balanced. There is nothing gained by forcing the voltages of the two antenna halves, whether balanced or not, to be equal and opposite relative to the cold side of the balun input (usually connected to the shield of a coaxial feed line), since the antenna field is proportional to the currents in the elements, not the voltages at the feed point. I will call this type of balun a voltage balun to emphasize that it balances the output voltages regardless of load impedances. F i g . 2 — One solution t o the imbalance problem i s t o feed the balanced antenna with a balanced feed l i n e — twin lead. r i g -— r — rh Fig. 3 — If the rig were isolated from ground, the feed-line conductor currents would be equal and opposite, the imbalance current would be zero and the feed line would not radiate. • y i Imbalance Current G-Current to Ground cable, it can be visualized as an RF choke acting only on the outside of the coaxial-cable shield, reducing the current to a very small value. This is the exact function a balun needs to accomplish when used in an antenna system. A current balun can be constructed by winding coaxial feed line into a coil, winding either type of feed line onto a core, or by stringing ferrite cores along either type of line.Ml Even if the balun is mediocre, there will be no effect on the desired properties of the line itself (impedance, e l e c t r i c a l length, SWR, and so on). A less-than-perfect voltage balun can have a profound effect on the impedance seen at its input because of the tertiary winding. Impedance-transforming (4:1) baluns are discussed in Appendix 3. "Hot" Side Unbalanced Port "VAJUUfT-" u u j j - r BoloncecS Port | UXUJT Pig. 5 — Voltage balun (see text). The bottom winding is sometimes referred to as a tertiary winding. All windings are closely coupled. UJJJJT IN OUT " U M A T 7 Fig. 4 — When the rig is connected to ground, a third conductor in parallel with the right side of the feed line is introduced. This causes an imbalance in both coaxial cable and twin lead feed lines. Another type of balun that appears in the literature has been called a choke-type balun (Fig. 6).\1Q It resembles the voltage balun, except that the tertiary winding is missing. The analysis of both types of balun in Appendix 1 shows that the effect of a tertiary winding is not a minor one. The two types of baluns produce fundamentally different results. The voltage balun causes equal and opposite voltages to appear at the balanced port regardless of load impedances, but the second type of balun causes equal and opposite currents on the conductors at both ports for any load impedances. For this reason, I will call this type of balun a current balun. Intuitively, the current balun produces the sort of effect we would expect. When wound with twisted pair or twin lead, it is nothing but a bifilar RF choke that impedes any net current which tries to flow through it. When wound with coaxial Fig. 6 — Current balun (see text). Both windings are closely coupled. Experiments A series of experiments was designed to test the validity of the results of the theoretical investigation. A 10-meter dipole was set about 12 feet above the ground, and about five feet above the edge of an elevated wooden deck (Fig. 7). One-half wavelength from the center of the dipole, a 4-foot rod was driven into the ground, which was completely saturated with water at the time the experiments were run (during November, in Oregon). To further lower ground-system impedance, six radials were placed on the ground around the ground rod. Two feed lines were cut to a half wavelength: one of RG-59/U coaxial cable, and one of 72-ohm transmitting twin lead. The velocity factors of the cables were not taken into account, since the intent was to have the outside of the coaxial cable, or the two parallel conductors of the twin lead, be an electrical half-wavelength long. A low-power 10-meter transmitter located at, and connected to, the ground system was used as a signal source. Current probes and baluns were built as de-scribed in Appendix 2. Two of the current probes were permanently wired into each side of the dipole near the feed point, and a third was used for all feed-line measurements. A single detector was used for all measurements, and it was calibrated over the range of encountered output levels by using a signal source and precision attenuator. The results of the experiment have been corrected to account for the measured nonlinearity of the detector. No atteir.pt was made to keep the power level or impedance match constant from one test to another. When running an experiment with no balun, a current balun, and a voltage balun, the only variation in the system was to change the balun. Initially, the intent was to use the antenna current probe readings as a measure of current balance in the antenna halves. However, a case was encountered in which the antenna halves showed equal currents, but a large imbalance current was measured in the feed line at the antenna feed point — a seemingly impossible combination! (The equal antenna currents were even more suspicious because no balun was being used, and the antenna had intentionally been made nonsym-metrical for that test.) A bit of thought provided the answer. The imbalance current is measured by placing the feed line through the current-probe toroid. In conjunction with the detector, it measures the magnitude of the vector sum of all currents flowing through the toroid. Each antenna current probe, with the detector, measures the magnitude of the current in each half of the an-tenna, at the feed point. What must be happening is that the currents in the dipole halves are equal in magnitude, but not 180 degress out of phase. A check of the current-probe outputs with a good-quality dual-channel oscilloscope confirmed the hypothesis: The currents were 230 degrees, rather than 180 degrees, apart, although equal in magnitude. What an interesting pattern that dipole would have! But this illustrates how misleading the magnitudes of element currents can be when judging balance. Measuring the imbalance currcnt in the feed line at the feed point does, however, provide a good indication of the balance of the currents in the antenna halves. If the imbalance current is very small, the currents in the sides of the antenna must be nearly equal in magnitude and opposite in phase. A significant im-balance current, on the other hand, indicates that one or both conditions have not been met. Measurement of the imbalance current on the feed line also indicates how much the feed line will radiate. The imbalance current at the rig provides a measure of RF in the shack. In the following tests, the magnitude of the current was measured in each conductor, then the magnitude of the imbalance current was measured by placing the complete feed line through the current-probe toroid. A single figure of merit, balance, was calculated as: balance (dB) = 20 log (average of magnitudes of currents in each conductor^ (magnitude of imbal-__ance current) . Experiments 1 through 4 were done using a nominally symmetrical dipole, although results indicate that some asymmetry was present. For ex-periments 5 through 7, the dipole was intentionally made nonsymmetrical by lengthening one side by five inches, and shortening the other side by the same amount. Experiments 1 and 5: See Fig. 8. The dipole was symmetrical for experiment 1, nonsymmetrical for experiment 5. Fig. 7 — Experimental setup. [ " n o b a l u n — 1 voltage b a l u n j _ c u r r e n c b a l u n [ n o b a l u n volcage b a l u n current b a l u n F v l G -X b a l u n t a p .
e b a l u n current b a l u n 13.-16.5 27.1 Pig. 8 — Setup and results of experiments 1 and 5. Numbers are measured balance in dB. Discussion If the dipole balance (symmetry) were indeed perfect for experiment 1, we would expect the currents in the sides of the dipole to be unbal-anced, resulting in imbalance current on the feed line. This is because the outside of the coaxial shield appears as a conductor in parallel with half of the dipole. Also, either a current or voltage balun should reduce the imbalance current to zero. Since the feed line is placed symmetrically relative to the antenna, no additional current should be induced into the feed line, so the imbalance should also be quite small at the rig end of the line when either type of balun is used. With the nonsymmetrical dipole (experiment 5), we would expect the voltage balun to do worse than in experiment 1. We would also expect the current balun to do about the same, and the no-balun case to be considerably worse. Results In experiment 1, the voltage balun did not perform as well as the current balun, indicating some asymmetry in the dipole. At the frequency chosen, the small differences in connections and a slight tilt of the antenna could easily account for what happened. When no balun is used, a curious result is the much better balance at the rig end than at the antenna end of the feed line. This may be because the feed lines weren't exactly an ef-fective half wavelength long, because there was a wire of about six inches in length connecting the rig to the ground system, or because the feed line was doubled back on itself for a short distance near the rig to provide strain relief. Perhaps the doubling back generated enough inductance to cause a current balun, or RF choking effect. The better balance at the rig end can be seen in the results of all experiments. The no-balun result was worse with the non-symmetrical dipole than the symmetrical one, as expected, and the current balun did about the same in both cases. The voltage balun, although slightly worse with the nonsymmetrical antenna, was better than expected, but still definitely inferior to the current balun. Experiments 2 and 6: See Fig. 9. The dipole was symmetrical for experiment 2, nonsymmetrical for experiment 6. Discussion The results of these experiments should dupli-cate those of the previous pair, since the feed line is placed symmetrically relative to the antenna to avoid induced current. The only difference is that the balun is placed farther down in a symmetrical or nonsymmetrical system. Results The trend is clearly the same as in experiments 1 and 5; the current balun provides the best bal-ance, the voltage balun is second best, and a feed line with no balun is Hie worst case. The balance with no balun was better in this experiment, however, (except at the rig end with the non-symmetrical antenna, which was about the same), and the balance at the rig end was substantially better when using the current balun. Time did not permit me to run additional experiments to explain these dif-ferences, but the ability of the current balun to achieve superior balance was again illustrated. fly.:-. 2 exp. 6 RIG pio balun I voltage balun I current balun e o balun oltage balun urrent balun [ ho balun voltage balun current balur. 9 . 1 1 3 . 2 2 1 . 3 1 7 . 3 1 7 . 6 4 1 . 6 17.3 18.6 3 4 . 6 2 . 7 7 . 2 1 9 . 5 9 . 8 1 4 . 4 3 3 . 4 9 . 8 1 7 . 2 32.6 Fig. 9 — Setup and results of experiments 2 and 6. Numbers are measured balanced in dB. Experiments 3 and 7: See Fig. 10. The dipole was symmetrical for experiment 3, nonsymmetrical for experiment 7. The voltage balun was connected with the balanced port toward the antenna. Discussion: These experiments, and experiment 4, were conducted to test the idea that coaxial cable and twin-lead feed lines would behave in the same fashion, as theorized earlier. If so, the results of fno balun JvoltAr.e bnlu [current balu 3.7 19.3 0 . 3 10.5 Iww I R i g fno balun 13. .4 10, .4 -jvoltaee balun 12. .5 — — [current balun 41, .1 32 .3 fno balun 13. ,4 10, .4 [voltage balun 15, .3 — — [current balun 42. ,0 33, ,7 Fig. 10 — Setup and results of experiments 3 and 7. Numbers are measured balance in dB. these experiments should be similar to those of the previous pair. Results: With no balun, the results were those of ex-periments 1 and 5 (the test with no balun was not rerun). With the current balun, the results were similar to those of experiments 2 and 6, indicating that coaxial cable can be used as a balanced feed line (in the sense discussed earlier) with a bal-anced or somewhat-unbalanced load. This data also points to the possibility that a current balun could be added to an existing antenna system at the rig end of the line, with results similar to those obtained by placing it at the antenna, in some eases at least. This would certainly be worth a try in systems where the symptoms indicate the need for a balun, but the antenna itself is difficult to get to. With the symmetrical antenna, the voltage balun made balance worse at both ends of the feed line than no balun at all. The balanced port of the voltage balun sees two unequal impedances to ground: the coaxial center conductor, ending in one dipole half, and the coaxial shield terminating in the other. The shield is capable of radiating but the inner conductor isn't, and the two are of different diameters, accounting for the different impedances. The voltage balun predictably generates unequal currents in the different impedances, causing additional current imbalance. A voltage balun was not evaluated in this application with a non-symmetrical dipole, having shown distinctly inferior results even with a symmetrical one. Experiment 4: See Fig. 11. The dipole was symmetrical. The voltage balun was connected with the balanced port toward the antenna. Discussion: Like experiments 3 and 7, this was intended to test the similarity between performance of the two kinds of feed line, provided that all other con-ditions and connections are the same. If the two feed lines act the same, the results should dupli-cate those of experiment 1. Results: The current balun again causes the predicted results, except it shows improved performance at the rig over experiment 1. In contrast, the voltage balun gave strikingly poorer balance at the antenna, and markedly poorer performance at the antenna end of the feed line, compared to no balun at all. If the antenna were completely symmetrical, there should be no current imbalance at the input end of no balun — voltage balun current balun no balun voltage balun current balun Jno balun -fvoltaee balun X current talun exp. ' 9.1 0.1 28.8 9 . 1 3.5 26.4 17.3 17.6 41.6 Pig. 11 — Setup and results of experiment 4. Numbers are measured balance in dB. the voltage balun, but with the moderate (unin-tentional) imbalance presented by the actual an-tenna, the current balance on the feed line was seriously degraded. This configuration isn't likely to be used in actual practice, but helps illustrate the operation of the baluns and feed lines. Conclusions Although some aspects of the experimental results remain to be explained (as they always will be unless performed under extremely controlled conditions), they certainly support the theoretical analysis. The current balun gave superior balance at every measured point in each experiment. The voltage balun improved balance in most cases, explaining its acceptance in spite of the theoretically and exper-imentally demonstrated superiority of the current balun to cure the problems we have discussed. As always, finding the answers to questions generates yet more questions. Lack of time did not permit experiments with the feed line placed non-symmetrical^ with respect to the antenna, to induce imbalance current into the feed line. The results of such an experiment should be interesting and en-lightening. Is there an optimum point in the feed line to place a balun? Suppose the effective distance along the feed line/ground wire from the antenna to ground is an integral number of half wavelengths, and the balun is placed a quarter wavelength below the an-tenna, as sometimes recommended. Wouldn't the im-balance current be conducted as before? Would in-duced current, if present, be reduced? What's the effect of poor coaxial shield coverage? More work needs to be done in evaluating the various styles of current baluns (such as coaxial cable wound into a choke, coaxial cable wound on a ferrite or powdered iron core, insertion of the feed line through one or more ferrite cores, and so on) for their primary characteristic: causing currents to be equal in magnitude and out of phase. The method I've used is briefly described in Appendix 2, but how good is good enough? The basic investigation reported here does answer some of the major questions regarding baluns. I now know what symptoms I can expect a balun to cure, why it will (or won't!) cure them, how to predict and measure the balun's success in doing so, and what type of balun to use. I hope you do, too! Appendix 1: A Brief Analysis of Balun Operation Analysis of both balun types assumes "ideal" operation: All flux is linked to all windings (coefficient of coupling is one), and each winding has sufficient self-impedance to make the mag-netizing currcnt negligible. "Cold" Side The Voltage Balun Because of transformer action, VI - V3 = V2 -V J 8 = V0 - VI. The third term comes about because of the "tertiary" winding shown at the bottom. Rearranging the last two terms, V2 - Vfi = -(VI -VP). So relative to V0 (the voltage at the cold side of the unbalanced port), the voltages at the balanced port are equal and opposite. The current It flowing in the tertiary winding is, by inspection, -(Ii + 10 - 12) and also (Ic -II), so -(Ii + 10 - 12) = (Ic - II). Because of the property of coaxial cable discussed in the body of the article, Ii = -Ic, so Io - 10 + 12 = Ic - II; thus 10 = II + 12. So for the current on the out-side of the shield, 10, to be zero, load currents II and 12 must be equal and opposite. Since VI and V2 are forced to be equal and opposite relative to V0, the only way for II and 12 to fulfill this requirement is for the impedances from each side of the balanced port to the cold side of the unbalanced port to be equal. Thus, only a perfectly balanced load will cause no current on the outside of the coaxial cable. Ironically, if this does occur, current It = 0, and the tertiary winding accom-plishes no function. The Current Balun In an ideal transformer of two windings having an equal number of turns, the currents in the windings are forced to be equal and opposite. So Ic = -(Ii + Io). Again, Ii = -Ic, so Ic = Ic - b, resulting in Io = 0. This result is independent of the load impedances. And, since the load currents are the winding currents, they are also equal in magnitude and opposite in phase. If the balun is constructed by winding coaxial cable on a core or into an air-core coil, or by stringing ferrite beads on the outside, the oper-ation can be understood by observing that the inside of the coaxial cable "can't tell" what's going on outside. The currents on the inside — equal and opposite — happen regardless of the outside en-vironment, but the construction causes a high impedance to current flow on the outside, acting like a choke to the imbalance current (hence the appropriate name choke balun). When constructed of twisted-pair line, the effect on imbalance current is the same and for the same reasons, but operation is more difficult to visualize. The current balun IS p , iy -hXo "UJUUU" Appendix 2: Construction and Test of Baluns, Current Probes and Detector Voltage Balun The voltage balun was constructed using the method described in Kef. 8. A piece of no. 26 wire was laid along a length of RG-178/U cable (small-diameter Teflon-insulated coaxial cable), and heat-shrinkable tubing was applied over the assem-bly. The modified cable was wound on an FT82-61 core using ten turns. This construction method was decided on after trying to wind a balun with two pieces of coaxial cable in bifilar fashion, the shield of the second being connected as the tertiary winding. The latter construction method was much poorer in providing good voltage balance. Voltage balance was evaluated by connecting the cold side of the unbalanced port to a ground plane and the balanced port to two resistors of unequal value, the other ends of which were connected to the same point on the ground plane. Using resistors of 27 and 54 ohms, the ratios of voltages appearing at the two resistors were measured as about 3/4 and 1-1/2 dB, depending on which resistor was connected to which lead of the balanced output. Current Balun The current balun consisted of 15 turns of RG-178/U coaxial cable on an FT82-61 core. Per-formance was evaluated by connecting the output end to 27- and 54-ohm resistors to ground, and measuring the voltages across them. A properly working current balun should generate twice the voltage across the 54-ohm resistor than across the 27-ohm resistor, regardless of which lead is connected to each resistor. The results were within 0.2 dB of theo-retical, with either lead connected to either resistor. Current Probe The current probes were constructed as shown in Fig. 12. The output voltage equals ten times the current, in amperes, being measured. Insertion resistance is one ohm. Detector The detector is shown in Fig. 13. It was calibrated using a signal source and precision attenuators, at the operating frequency. Calibration using a dc source was found to be inaccurate. FT82-^3 :err:te core " current-carrying conductor to be measured - 5 bifilar fjrrs on c;37-72 ferrite i 100 ohr.s 3KC tenale to detector Pig. 12 — Current probe. 3!»C r»a\e to current proDe \ K — Gen=aniurs diode L .001 ^iT to DVM CIO Megohm input resistance) — , / Appendix 3: Impedance-Transforming (4:1) Baluns The common 4:1 balun, shown schematically in Fig. A3-1, is a voltage balun. If used with a cur-rent balun as in Fig. A3-2, the combination acts like a 4:1 current balun. Or it can be converted to a 4:1 transforming current balun by adding a third winding, as shown in Fig. A3-3. A 1:1 voltage balun could be converted to a 4:1 current balun by re-connecting the existing windings. The difficulty with using this configuration is that, like the 1:1 voltage balun, all windings must be closely coupled, and rather severe impedance changes can occur be-cause of transformer imperfections. A better approach is shown in Fig. A3-4. Old-timers will recognize this as the configuration used by the balun coils commonly used some years ago. This balun does force equal and opposite currents at the input and output, so it is a true current balun, and it performs a 4:1 impedance transformation. Al-though it does require two cores which must not be coupled, it has several advantages: It's much easier to tightly couple two conductors than three, it's much more forgiving than the other config-urations, and it lends itself to easy construction. One method is simply to wind coaxial cable on two cores, with the center conductors being the con-ductors shown on the outsides in the figure. This balun can also be used in all-coaxial-cable systems. Besides effecting a 4:1 impedance transformation, it will greatly reduce any current flowing on the out-sides of the lines. If ferrite rods or air-core coils are used, don't place them end to end. Place them side by side and spaced a fair distance, or, better yet, at right angles. Less care needs to be taken with toroidal coils. Low-Z Unbalanced High-Z Unbalanced T1 Fig. A3-1 — The 4:1 voltage balun. "UJUJT^ 3 High-Z Pig. A3-2 — The 4:1 voltage balun used with 1:1 current balun. Low-Z High-Z No other connections to this point. Fig. 13 — Detector. Fig. A3-3 — A 4:1 current balun. References \1 Maxwell, Walter, W2DU, "Some Aspects of the Balun Problem," QST, March 1983, p. 38. \2. If a perfect shield is assumed (a reasonable approximation for this analysis), the result follows directly from Ampere's Law. For a more detailed explanation, see Electromagnetic Energy Transmission and Radiation, by Richard B. Adler, Lan Jen Chu, and Robert M. Fano (Wiley, 1960). A very clear development of the phenomenon of skin effect may be found in Chapter 7 of Elec-tric Transmission Lines by Hugh H. Skilling (McGraw-Hill, 1951). Winningstad, C. Norman, "Nanosecond Pulse Transformers," IRE Transactions on Nuclear Science, March 1959. V 5 . Matick, Richard E., "Transmission Line Pulse Transformers — Theory and Application," Pro-ceedings of the IEEE, Vol. 56, No. 1, Jan. 1968. Hall, Gerald L., K1TD, ed., The ARRL Antenna Book, 14th ed., Chapter 5, (ARRL, 1982). Y Z . In practice, there is always an RF path from the rig to ground, and its impedance should be made us low as possible. The rig should always be dc grounded for safety. \S See Ref. 6. \2 Nagle, John J., K4KJ, "High-Performance Broadband Balun," Ham Radio, Feb. 1980, p. 28. \lfl See Ref. 1. \11 Reisert, Joe, W1JR, "Simple and Efficient Broadband Balun," Ham Radio, Sept. 1978, p. 12. Available Power, SWR and Loading By David T. Geiser, WA2ANU RR 2, Box 787, Snowden Hill Rd., New Hartford, NY 13413 The Power Transfer Circuit Simple reasoning and easy arithmetic eliminate much of the mystery from SWR and loading. This ap-proach can spark new thoughts in the novice and old-timer alike. Any RF power-transfer circuit can be repre-sented by a series connection made up of a gener-ator, internal generator resistance, load resistance and reactance (Fig. 1). Remembering that the source may include an RF transmission line from a trans-mitter, the internal generator resistance may re-present generator or source losses. The power actually delivered through the output terminals goes to the load resistance. Notice that while power is delivered to the load, some of it is also dissipated to the source. Any circuit that is not intentionally tuned has an overwhelming chance of showing inductive or ca-pacitive reactance. This reactance (X) can be shown in series with the generator internal resistance (Ro) and the load resistance (RL). Power actually delivered to RL is: PL = (RL + RO)2 + X2 (Eq. 1) If we assume for the moment that we tune out the reactance (X = 0), delivered power PL is: v 2 (RL +So)' (Eq. 2) Let us plot resistance and reactance on a common graph, such that: R^ Ro = S W R (Eq. 3) Taking a tip from Table I, let us choose R1 and R2 with values of 25 and 100 ohms, respectively, and Ro equal to 50 ohms. Taking 25 and 100 as the end-points of a diameter (R2 - Rl) of a circle (Appendix B), we can plot that circle (Fig. 3). If we cheek several values of RL and X that lie on the circle's circumference, we find that the cir-cle is a plot of all possible values of RL and X for a constant delivered power, calculated with Eq. 1. (Some check values to try for an SWR equal to 2 are: RL = Ro and X = 0.707Ro, or RL = 2Ro and X = 0). SWR is the relation that links the portion of the available power that can be delivered to an unmodified load to the relation between the actual and ideal loads. XCVR LOAD We can work out an example to see what happens to the delivered power as we change the load, RL. Table I gives the values for E = 100 volts rms and Ro = 50 ohms. When we plot the results on semi-log graph paper (Fig. 2), we see a symmetrical curve with the power maximum at RL = 50 ohms. The power delivered is always a maximum when the generator and load resistances are equal and there is no net reactance. This amount of power is called available power. It would seem that this is an ideal condi-tion; however, two other cases exist and are dis-cussed in Appendix A. Circuit Reactance We must consider reactance because it is dif-ficult to adjust X, shown in Fig. 1, to exactly zero.\l Though a few ohms of reactance (compared to 50 ohms) do not cause much apparent difference, highly selective loads may display considerable reactance. Fig. 1 — Ideal RF generator with a partially reactive load. Maximum Reactance The impedance chart in Fig. 3 shows that for a specified SWR there is a maximum limit on capacitive and inductive reactance, 37.5 ohms for the SWR = 2 example. This inspires the thought that the reac-tance can be reduced to zero by series insertion of equal reactance of the opposite type. In the SWR = 2 example, reactance cancellation at X = 37.5 would reduce the SWR from 2 to 1.25. If we find the SWR of a load with an SWR bridge, we can calculate the maximum possible reactance (Appendix B). The actual reactance type and magnitude will not be known, but a compensating circuit can be made (point A in Fig. 4). The values given are one of the many choices that would satisfy the SWR = 2 example. 50 4 o 30 WATTS TO 20 Rl 10 o A 10 2.5 100 250 5 0 , CR0) R E L O A D R E S I S T A N C E ) + jX o • i X r -5 W R -2 (CIF ?
c le) u Ui i R o ^ \ F 'o 1 Fig. 2 — Power delivered from a 50-ohm source to various pure resistances. Use of a series circuit near resonance makes the choice of a low-loss inductor (XL)desirable. Even then, an increase in circuit resistance (from inductor losses) may result in less power being delivered to the actual load. A mobile Amateur Radio operator can minimize his narrow-band antenna problems by loading the body of the antenna so it is slightly capacitive at the high-frequency band edge. A manual or motor-driven variable inductor at the feed point allows the antenna to resonate anywhere in the band. While the mobile ham sees the resonated antenna as a low resistance, the base-station ham has many choices that allow the opportunity for minimizing the complexity of an antenna-matching circuit (Transmatch). When we look at the characteristics of a vertical antenna, we observe that the resistance at slightly more than a quarter wave (90 degrees) is close to 50 ohms, with different inductive reac-Fig. 3 — Impedance circle for all SWR = 2 in a 50-ohm system. (See text and Appendix B.) Fig. 4 — Simple series compensator for SWR = 2 in a 50-ohm system. There is no improvement for RL of 25 or 100 ohms. tances (Table II). \2 Let's assume we chose a 95-degree length for 3.5 MHz — this gives us about 110 degrees at 4.05 MHz. Series capacitance compensation reduces the SWR to 1.3 or less across the whole band. The experienced ham usually reminds me that you can also use parallel compensation for reactance, but the equivalent parallel resistance is not the same as series resistance (Appendix C). The lowest SWR would probably call for a length of 82 degrees at 3.5 MHz. In this case, the SWR is not as low as in the series example, but would be 2 or less between 3.5 and 4 MHz. The parallel choice also requires both inductive and capacitive reactance to cover the range. Exact and Approximate Hatching The ham who wants to make an exact match will have to use a Transmatch (Appendix C), remembering that the L network is the most efficient. Unfor-tunately, when high ratios of RP/RS are encoun-tered, the bandwidth of the L network can become very small. This is the reason the pi-L network is useful. I believe the broadest bandwidth is found when the Qs of each of its two L networks are equal, or, paraphrased, the ratio of Rp to the intermediate resistance is equal to the ratio of the intermediate resistance to RS. An approximate match may be all that is neces-sary. Most transmitters behave well if the load SWR is 1.5 or less, seen at the transmitter antenna terminals. Three one-eighth-wavelength sections of coaxial cable and four T connectors are very useful tools.Insertion of each of these sections between the load and the measuring point moves the measuring point on the impedance circle; four of them returns you to the starting point on the circle. Alone, they do not change- the SWR significantly. The procedure I follow is to cascade the sec-tions (Fig. 5A). I start by connecting a variable capacitor to (1) and adjust it to lower the SVVR. If Similarly, many generators deliver a better-the SWR changes, I leave it there and try a variable quality (less distortion, harmonics, or noise) capacitor on (2), (3) and (4), respectively. If the signal to a chosen mismatched load. The previous capacitor at (1) did not reduce the SWR, I try it at example choices (10-ohm source and a 50-ohm load) the other locations. One position will reduce the are not particularly unusual. Because third- and SWR. Sometimes a second capacitor only increases the fifth-order distortion hurts SSB intelligibility and SWR. This means an inductance is necessary. A harmonics cause interference, a designed mismatch at quarter-wave stub (Fig. 5B) will make a variable the transmitter is often made. Similarly, the best capacitor at (2) look like a variable inductor at noise-figure adjustment at the receiver input is (3), and losses are much less than in the conven- often a power mismatch. tional inductor. Once the SWR is reduced to an A mismatch at the load will usually appear acceptable value (it is surprising how often it can different at the source if there is any distance be reduced to the ideal value of 1.0) any excess between the two. Imagine the mismatch (in the 10-ohm cable between the last used tap and the SWR bridge example) changed to a "matched" 10 ohms. Now 90 can be removed. watts are dissipated in the source and the source There is nothing unique about this form of has to generate 180 watts instead of 60. Parts will compensation. Ideally, any load may be matched with often burn out. Yet this was a "matched" load. This two reactances spaced 1/8 wavelength on a trans- sort of reaction (distortion, harmonics, noise, and mission line. This method only minimizes precision damage) is dangerous when a high SWR is present, measurements and cable-cutting. We will call the 50 ohms the design load (not a matched load) and do our best to keep it at the resistive 50 ohms, or other design load value. f A ) The radius, r, of the circle is: r _ Ro(SWR - 1) SS5R (Eq. 5) < B ) Fig. 5 — Using 1/8-wave sections for impedance transformation (A) shows a cascade arrangement of shunt-capacitor taps. At B i s a transforming shunt capacitance to a low-loss shunt induc-tance. (See text.) Appendix A — Intentionally Mismatched Loads Efficiency and low distortion are two common reasons that a user may not want to feed a source into a matched load. Refreshing our memory of Fig. 1 and Table I, we note that the generator and load resistances were equal. The same current flowed through each, so each must, in the example, dis-sipate 50 watts. This is 50% efficiency. Consider a generator of 60 V rms and an in-ternal resistance of 10 ohms and zero reactance. A 50-ohm load still receives 50 watts, but the gen-erator dissipates only 10 watts. This is an ef-ficiency of 83%. Only 60 watts (instead of 100) has to be generated, and a smaller generator proves satisfactory. Note that this is also the maximum possible reactance. This is not the only kind of SWR circle that relates design loads to actual loads. The most common and useful is the Smith Chart, based on a logarithmic distribution of impedances. The Smith Chart places the design load at the center of the SWR circle (hinted at by the symmetry of Fig. 2) with phase angles proportional to angular rotation along the circle. It is more useful than the rectangular chart of this note, but harder to explain. Appendix C — Series-Parallel Equivalence and Matching At any single frequency, any two-terminal combinations of resistance and reactance may be expressed as either a series circuit or a parallel circuit. (Fig. 6.) We can make the circuit look purely resistive by scries or parallel reactance of the opposite type. Then only the equivalent series or parallel resistance will be seen at that fre-quency (Fig. 7). H e r e , t h e r e has been no change in the resistance. If, however, we use parallel compen-sation on the series circuit (Fig. 8) we see resistance RP instead of RS. Series compensation of the parallel circuit gives us RS instead of HP. The reactances, XP, XS, -XP, and -XS, can be, but do not have to be, part of an antenna. They can be ideal inductors and capacitors. Thus, with one capacitor and one inductor, we can transform any high resistance RP into a lower resistance RS. We first define a number, Q: From this simplification, Rp = Rg(Q 2 + 1) Xp = X sll + ®s p (Eq. 6) (Eq. 7) (Eq. 8) (Eq. 9) (Eq. 10) 1 r p X c = 1.732(50) = 86.6 ohms, o XP = 200 = 115.5 ohms, 1.732 The resulting circuit is called an L network, as shown in Fig. 9. It is called this because of the usual relative positions of the reactances in a circuit diagram. Pairs of L networks may be used for multiple impedance transformations; familiar types are the pi, T and pi-L. Relative impedances are indicated, and the usual harmonic-reducing configurations are shown in Fig. 10. K . (A) R« Fig. 6 — Series-parallel equivalence symbols. - x < R. (A) K i R , (B) Fig. 7 — Compensation. At A, series with series equivalence. At B, parallel with parallel equivalence. There is no change in equivalent resistance as seen looking into the circuit terminals. If, for example, we wanted to transform 200 ohms (RP) to 50 ohms (RS), Q = (200 - 50) (0.5) = (150) (0.5) 50 "55 = (3)(0.5) = 1.732 R, R r (B) Fig. 8 — Compensation. At A, parallel with series equivalence. RS transforms to RP. At B, series with parallel equivalence. RP transforms to RS. 8QX JTL O 501.L 115.5n? < 2.00 -O-(A) son zoo n Fig. 9 — Examples of L networks that transform 200 ohms to 50 ohms. At A, the series-inductance version attenuates higher frequencies more. The series-capacitance version at B attenuates lower frequencies more. References \1 Exact adjustment of reaetancc to zero (or another exact compensating value) may re-duce distortion or harmonics a dozen or so dB. I have seen reductions up to 40 dB. \2. The numbers in Table II apply to only one antenna, visually estimated from the graphs on page 2-24 of The ARRL Antenna Book. \j£ Remember that a wavelength of 50-ohm solid-polyethylene coaxial cable is only about 66% of a free-space wavelength. Other cables may have different velocity factors. (A) HIGH LOW HIGH oj-^rinr 1—o—•nnnr^-o -r Ll L2 t-O -I o 1 — o L, AND Lz MAY BE COMBINED LOW HIGH LOW o — — — — ° (B) lLI C, C1 AND C2 MAY BE COMBINED Table I—Power Delivered vs. Load Resistance VERY LOW LOW HIGH o—onf i o—onrwo CO -L 1 ^ T T O I o 4 -o LOW RL HIGH RL PL (WSTTS) SWR Fig. 10 — Transmatches. Pairs of L networks 50 50 50 1.00 are combined to make a pi or 1 network (A), a T 40 62.5 49.4 1.25 network (B), and a pi-L network (C). The "highs' 33.3 75 48 1.50 at the two ends of the network at A are not 25 100 44.4 2.00 required to be equal, and may differ by a ratio 16.67 150 37.5 3.00 of 100:1 or more. The same applies to the "lows' 10 250 27.8 5.00 at B. Table II—Reactance Compensation of Vertical Antenna Length Series Series SWR Series (Degrees) Resistance Reactance Compensated (RS) (XS) SWR 82 27 -35 2.96 1.85 85 30 -15 1.89 1.67 90 36 +17 1.67 1.39 95 40 +50 2.96 1.25 100 47 +80 4.51 1.06 105 56 +115 6.58 1.12 110 66 +150 8.78 1.32 Length Parallel Parallel Parallel (Degrees) Resistance Reactance Compensated (RP) (XP) SWR 82 72.73 -55.8 1.45 85 37.5 -75 1.33 90 44.0 +93.2 1.14 95 102.5 +82 2.05 Mr. Smith's "Other" Chart and Broadband Rigs By Roger K. Ghormley, W0KK •2115 South 24th St., Lincoln, NE 68502 Have you heard about the Smith Chart that is not the Smith Chart we know of today? This "other" ehart actually came before the well-known one, and can play an important role in the ham shack. The following paragraphs show how this lesser known chart can assist today's amateur. Mr. Smith During the early 1940s, Phillip H. Smith was involved in the Radio Development Department of the Bell Telephone Laboratories, Inc. His chart (Fig. 1) was published as part of an article in the January 1944 issue of Electronics.\l The other chart, actu-ally a series of charts, was published in the March 1942 Electronics article, "L-Type Impedance Trans-forming Circuits." It covered eight variations, but we will look at only two of them.\2. With the increasing use of no-tune or broadband rigs, Amateur Radio operators are faced with the need for an "impedance transforming circuit." To illustrate the factors involved, we'll take a look at John Q. Ham. He is active in RACES (Radio Amateur Civil Emergency Service) and wants to be able to function on RACES frequencies in the 80- and 75-meter band — 3500 to 3550 kHz and 3984 to 4000 kHz.\2 Until now, there's been no problem. The tune and load controls of John's rig have been able to transform the impedance at the shack end of the transmission line to something that "satisfies" the final amplifier of his rig. But with broadband rigs, a problem develops. Hie Problem John Q.'s new modern broadband rig does not have a tune or load control. Why is this a problem? His rig was designed to "see" a resistance of 50 ohms and wants it fairly "pure" (not much reac-t a n c e ) . If it does not see this, it will auto-matically cut back on output or blow the circuit breaker by going into oscillation! What's more, the shack end of John's transmission line (even though it is a 50-ohm RG-8 coaxial cable) shows an imped-ance that varies widely from 50 ohms, depending on the frequency. In f a c t , it is never a pure 50 ohms! Fig. 1 includes a plot of how the impedance of John's dipole (125 ft 6 in of no. 14 wire) varies with frequency.M Two additional plots show the input impedance at the shack end of 40-ft and 75-ft lengths of RG-8 coax if terminated in John's dipole. (For now, ignore the "backward S" divider; it is discussed later.) All notes are included in the dipole antenna (125.5 f e e t of no. 14 gauge wire) and of the input impedance f o r a 40-foot and a 75-foot RG-8 coaxial line terminated in the dipole over a frequency ranqe of 3.5 to 4.0 MHz. In addition, the backward "S" shows the impedance ranges matched by the two types of L networks discussed in the t e x t . appendix. Look at the "horseshoe" for John's 75-ft trans-mission line. The input impedance at 3.5 MHz is 1.91 + j2.42 ohms, normalized.\£ This is a pure resis-tance of 96 ohms in series with an inductive reac-tance of 121 ohms. That represents an SWR of 5.31:1 — a bit above the manufacturer's suggested limit of 2:1!\£ At a frequency of 3.75 MHz, the input impedance is 54 + jl7. Not bad. The SWR reading is 1.4:1. That should satisfy the rig. But at 4.0 MHz, the SWR is back up to 5.2:1. It is clear that John has to do something if the new rig is to work near the band edges. The L-Net Answer Having eliminated the tune and load knobs, John wonders what is the simplest workable solution to the problem — without adding more panel knobs. Mr. Smith's "L-Type Impedance Transforming Circuit," the L network, now enters the scene. The L net consists of two elements; a series and a shunt arm. This is about as simple as you can get! It transforms any complex impedance (resistance plus reactance) into a pure resistance of any de-sired value. As shown later, its tuning is quite broad, so it provides a workable match over a fairly good frequency range. Sound good? Let's see how John goes about making up such a net. We can now look at Mr. Smith's "other" chart, which we will call the Smith L Chart (Figs. 2 and 3). Fig. 2 applies to an L net when the shunt arm is next to the load. I have divided this figure into A and B for clarity. Fig. 3 applies when the shunt arm is next to the transmitter. Which figure we should use is covered in a later section. Using the Smith L Chart Locate the point on Fig. 2 or 3 that corre-sponds to the normalized load impedance. Read the reactance (in normalized ohms) directly from the chart for the shunt and series arms of the L net. For example, a few paragraphs earlier we learned the input impedance to John's 75-ft line at 3.5 MHz was 1.91 + J2.42 ohms. Find that spot on Fig. 2A (there is one there for your convenience), and read the required shunt arm reactance of 1.5 capacitive ohms. The same procedure on Fig. 2B gives a series arm reactance of 2.0 inductive ohms.\l Convert these normalized values to -75 ohms for the shunt capacitance and 100 ohms for the series inductance. Multiply by 50 to satisfy John's rig. Now convert these ohms into component values for the capacitor and the inductor, either by using the formulas: C = 1 and L = X L 2 t fX c 2 IT f or your favorite reactance chart in The ARRL Handbook for 3.5 MHz. The results at 3.5 MHz are summarized in Fig. 4. Note that the complex impedance 96 + jl21 ohms appears as a pure resistance of 50 ohms at the input to the L net. John's rig would like that! The Backward S An inspection of the Smith L Chart of Figs. 2 and 3 shows that Fig. 2 is used for all loads having a normalized resistive component greater than 1. For loads having a resistive component less than 1, the sign and size of the reactive component dictate whether Fig. 2 or 3 is used. The same information in a different form is shown by the "backward S" of Fig. 1. The area in the right half is served by Fig. 2, and the area in the left half by Fig. 3. Had John's transmission line been 40 ft long instead of 75 ft, we would have used Fig. 3 to determine the L net. In fact, let's use Fig. 3 and assume the transmission line is 40 ft long. Try to find a way to satisfy John's rig over the entire 80- and 75-meter band with a minimum of knob twisting. John's "Broadband" Dipole Fig. 5 shows a plot of SWR vs. frequency at the shack end of 40 ft of RG-8 that terminates in John's dipole. The SWR at the band edges is now up to 5.7:1. This is because the 40-ft line has less attenuation than the 75-ft line (0.16 dB vs. 0.30 dB). Since John's rig works reasonably well with an SWR at about 2:1, he needs no matching network over the frequency range of 3.66 to 3.81 MHz. Let's now see if we can find an L network to a a o .
» 1 .
0 Fig. 3 — Series reactancc X. and shunt reactance X c required to transform load impedance + jX^ to a pure resistance Rj + jjj ohms. H-.SS^. 10 O O U N & ITST SO + ,0 -75" ^ OHfli 9U> OHUi IZI O H U S SWR of less than 2:1 across the entire 80- and 75-meter band, John's rig has a nice broadband system. He does it with an ordinary dipole and a bit of matching. bOUpt Fig. 4 — The L network, which transforms a load of 96 + 121 ohms into 50 + jjJ ohms. Component values are for 3.5 MHz. keep the SVVR below 2:1 for the space below 3.66 MHz, and a second setting of the network for the space above 3.81 MHz. If so, then John can go from band edge to band edge with only two settings of the network, plus a "network out" position. Fig. 5 shows just that! The L network is set for 50 ohms input at 3.57 MHz with an SWR of 1:1 in Fig. 5.\£ As the frequency goes down to 3.5 MHz, everything changes. The anten-na resistive component goes down, the antenna reac-tive component gets bigger, the transmission-line effective length goes down, the transmission-line loss goes down, the L network shunt arm reactance gets bigger, and the series arm reactance gets smaller. As the frequency moves up to 3.66 MHz, all of these items change accordingly. Even so, the end result gives an SWR of 1.9:1 or less over the bottom portion of the band. This is shown in Fig. 5. Similarly, an L network is fashioned for the top portion of the band. With the L network designed for a perfect match of 50 ohms at 3.905 MHz, the SWR is 1.82:1 at 3.8 MHz, and 1.95:1 at 4.0 MHz. With an Fig. 5 — A plot of S W R vs. frequency for a dipole (125.5 f e e t of no. 14 wire) via 40 f e e t of B3-8 cable, showing taminq of S W R bv the use of two settings of an L network. Network values are in ohms for the frequency indicated. The input impedance of the line i s shown as part of Fig. 1. "However," You Say — Are you thinking that John must have had a good RF impedance bridge to measure all those input im-pedances? How about those of us who have only an SWR meter of questionable accuracy? Good question. Here's what we'll do. If we knew the largest values of shunt capac-itance and series inductance we might have to use, we could "breadboard" a test network with con-fidence. We would then be prepared for whatever actual values are required during a trial-and-error procedure. Fig. 6 shows the possible extremes in series and shunt arm values for an L network when only the SWR is knownAS The extreme for the shunt arm is the lowest value of capacitive reactance; this requires a large-value capacitor. The extreme for the series arm is the highest value of inductive reactance; this requires a large coil. These extremes are only possibilities, and they do not occur at the same time. They are required only if the transmission line length is such that the resistive component of the input impedance to the transmission line is either a minimum (requiring a large shunt capacitor) or a maximum (requiring a large series coil). A longer or shorter line reduces component values. A Practical Case Some years ago I acquired my first no-tune rig, a Triton IV. Appreciable operating was (and still is) done in the 80- and 75-meter band. (The same problem was also present for a tri-band beam, but the 80- and 75-meter case is the only one presented here.) The average of three dubious-quality SWR meters gave the SWR vs.frequency at the shack end of RG-8 transmission line as shown in Fig. 7. The simplicity of the L network was appealing, but no impedance bridge was at hand to help out. However, it looked probable that we would be liable for the largest component values for the L network down around 3.555 MHz, dealing with an SWR of about 4.5:1. Fig. 6 — Curves shewing the minimum possible shunt arm reactance, X_, and maximum possible series arm reactance, x., required for L networks terminated in lines of various SWRs. Fig. 7 — The plot of SWR vs. frequency for the W0KK dipole (123 feet of no. 12 wire) via 80 feet of RG-8 cable, showing taming of the SWR by the use of three settings of an L network. Network values are in ohms for the frequency indicated. Input impedance to the line is unknown. Entering Fig. 6 at an SWR of 4.5:1 yields a minimum liable X,, of 0.53 normalized ohm. That converts into a capacitance of 1690 pF at 3.55 MHz. Here was the first stumbling block. The l&rgest transmitting-type capacitor at hand was 1000 pF. Perhaps luck was also at hand, and the maximum would be less than 16S0. From Fig. 6, the maximum liable x l at an o f 4.5:1 is about 1.9 normalized ohm. That would be a maximum inductance of 4.3 uH at 3.55 MHz. No problem. An L network was breadboarded as diagrammed in Fig. 8, using the capacitor at hand and some Minidux with taps. The network was placed between the SWR meter and the transmission line. With low power applied to the system, the network was adjusted by trial and error for a low SWR at all frequencies of interest for each antenna via the line. A record was made of coil taps, capacitor settings, and whether the shunt arm was next to the load or transmitter. The 1000-pF capacitor proved to be large enough, though I had already decided to add transmission line if necessary to keep below this value for the shunt arm. By using the ARRL Type A L/C/F Calculator [Discontinued — Ed.], the inductance values found for the breadboard coil were translated into a heavier-duty transmitting version, with some extra for error. The transmitting version of the completed network is a duplicate of the circuit in Fig. 8. A perfectionist might want a roller-type in-ductor for the coil in order to get a "pure" match at all frequencies. The rest of us would be satis-fied with the quick and easy convenience of about four coil taps plus " s h o r t e d " and " f u l l - c o i l " positions. These fit nicely on the band switch taken from the network final of some defunct rig. The end result for my 80- and 75-meter dipole plus L network is seen in Fig. 7. The lower portion of the band is not quite covered with a single setting of the network, and a twist of only the capacitor is required. That's good enough for me. Some day I might add a few feet to the dipole, to make it look almost as good as John's. The Smith Chart is a powerful tool. Maybe all have described can help today's amateur benefit from Mr. Smith's "other" chart. It's one of the simplest, least expensive ways to get a broadband antenna system to match you broadband rig. Give it a try! experimentally determining component size. The transmitting v e r s i o n is the same except for heavier-duty components. Appendix \1 Phillip H. Smith retired from the Bell Labs in 1970 and presently operates the Analog Instruments Company of New Providence, NJ. One of the best, sources of i n f o r m a t i o n on t h e Smith C h a r l r and its use is his book, Electronic Applications of the Smith Chart, 1969, Robert E. Krieger Publishing House, Melbourne, FL. The book contains a number of overlays for use with a Smith Chart, including one that provides a single-chart method of determining the series and shunt arms of an L network for any impedance location on the chart. Y2 An L network may have either inductive or capacitive reactances in one or both arms. This results in eight possible configurations by moving the shunt arm next to the load or next to the power source. The series inductor/shunt capacitor styles used here were chosen because they include all load impedances and attenuate the higher harmonies be-tween them. Palm, FCC Rule Book, 1983, ARRL Publication No. 47, p. 6-11, Table 1, RACES Frequencies. \i The impedances for Fig. 1 were obtained as follows: a. Input impedance of the dipole itself was cal-culated from \UL Zd = R + j [Y - 120 (In 24I/d - 1) cot m] (Eq. 1) (Eq. 2) R = m 2 , 7 3 6/ 3048 Y = in2"234/ 549.7 (Eq. 3) (Eq. 4) m = 180 (f1/984 + 0.013) degrees L = antenna length in f t d = antenna diameter in inches f = frequency in MHz b. The input impedance of the transmission line terminat from \ll Z i (normalized) terminated in the impedance Z^ above was calculated cosh vx + Zr sinh vx Z = Z= = Dipole impedance r I s Trans, line Zn o ° v = a + jB the propagation constant (Eq. 5) (Eq. 6) (Eq. 9) From The Antenna Book, p. 3-19, Table 1 W = 0.66 224 f t for RG-8 coax f From Skilling\12 Eq. 1-48 w = 2 • " • B Hence, B = 2"f = 0.9675 f radians per 0.66 x 984 100 f t at f M H z ( E q > 1 0 ) \5 See Ref. 1. \ 6 _ The SWR for Figs. 1 and 6 was calculated fromVlH S W R = A + B A - B (Eq. 11) where, A = /(R + Z Q) 2 + x2 a = attenuation constant in nepers per unit length B = phase constant in radians per unit length (2 n radians = 360 degrees) x = line length in units From The ARRL Antenna Book, 14th edition, p. 3-20, Fig. 34, the formula to determine RG-8 cable in dB per 100 ft Loss at 1 MHz = 0.206 = a l x x Loss at 300 MHz = 3.9 = a 300 300 x = 18.932 (by dividing the second equa-tion by the first) x = 0.5156. From which, Loss at f MHz = 0.206 f 0- 5 1 5 6 dB per 100 ft Attenuation constant a = 0.02372 f 0 - 5 1 5 6 nepers per 100 ft (1 neper = 8.686 dB). (Eq. 7) From The Antenna Book, p. 2-1 w = 2S4_ f t 8 ) fMHz B = |/(R - Z ) 2 + X2 W Ql ZQ = Characteristic impedance of system R = Resistive component of impedance X = Reactive component of impedance \1 The charts drawn in Fig. 2 include all im-pedance combinations represented by SWR values up to 7:1, and most of them to an SWR = 10:1. (Fig. 3 is readily extrapolated by inspection for values of X, off the chart.) For loads involving higher SWR values, the L network values may be calculated directly. The equations are developed as outlined in Figs. 9 and 10 and as follows: a. In Fig. 9 (shunt arm next to the load as in Fig. 2), X is chosen to make R E Q = 1. Then X. is set equal eto the absolute valtfe^of the resulting XEQ -Xc = Xx - J&2+ (R^ - 1) (R, 2 + Xx2) (Eq. 12) w -c R l 2 + X l 2 + X l X c T V + ( X1 + X c r (Eq. 13) X is a negative number (X = -n) R^ = load resistive component (R 1 ? 1) X = load reactive component (-n for capacitive and +n for inductive, where n is the numerical value) Xc x - I Pig. 9 — The development of the L network equations with the shunt arm next to the load. (See Appendix, note 7a). R i - Rsq ~ I Pig. 10 — The development of the L network equations with a shunt arm next to the source. (See Appendix, note 7b). R j and X^ are normalized values If R^ is less than 1, X.. must be positive and equal to or greater than ^ (1 - R^) For the special ease where R. = 1. X L = x i For X. If X j is negative or zero, X e = ao If X j is positive, use R. = 1.00001 in Eq. 12 to get around trying to diviae by zero. Or, if you want X c exactly, use Eq. 14. Xc R1 | / R l ( 1 - Rl) X c is a negative number (nXc " (XQ depends only on R^) (Eq. 16) -n) R. = load resistive component and is less than l.A X. = load reactive component and must be either negative, or if positive, must be equal to or less than (1 - Rj_ ) c. Note that the semi-circle boundary between Figs. 2 and 3 is the locus of points where » 0 and x1 = ( 1 - , Figs. 9 and 10 hence become identical. In Fig. 5, the frequencies at which to set the L network for a perfect match were found by trial and error to hold the SWR below 2:1 as the frequency was varied above and below the design frequency over the required range. The process makes heavy use of a programmable calculator! \£ The reactance of the maximum-size components of Fig. S are given by the following: XCMin = T/3® o h m s XIMax CMin = /SWR ohms (Eq. 17) (Eg. 18) \lfi J. Hall, "The Search for a Simple, Broadband 80-Meter Dipole," QST, April, 1983. Eq. 1 and its terms are taken from the Appendix of Hall's article. Eq. 11 is also taken from the same article as is Eq. 6. \11 Skilling, Electric Transmission Lines (New York: McGraw-Hill Book Co.). Eq. 5-33, p. 88. Uilbid., p. 15. Xc = - Q X ^ + 1) ± /OX-,2 + l)2 - 8X1 2(X1 2 + 1) (Eq. 14) In the numerator, use a negative radical for Xj_ less than 1 and a positive radical for X greater than 1. (The radical is zero at Xx =1). (This unlikely looking Ea. 14 is derived from Eq. 13 by setting X L = XjJ b. In Fig. 10 (shunt arm next to the transmitter as in Fig. 3), X^ is chosen to make = l.Then X is set equal to the resulting X.,,,. y c XL = - X 1 ( 1 (Eq. 15) |
1102 | https://www.meracalculator.com/math/power-reduction-identity.php | Power Reducing Formula Calculator
Enter the angle in degree and press the buttons below to get the power reduced trigonometric identities using power reducing identities calculator.
Table of Contents:
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Formula
The below equations use cosine double angle and half angle to reduce the power of squared trigonometric identities.
sin2θ = [1 - cos(2θ) ]2
cos2θ = [1 + cos(2θ) ]2
tan2θ = [1 - cos(2θ) ][1 + cos(2θ) ]
Power reduction formula calculator uses the power reducing formulas to rewrite the expression.
Power reducing identity calculatoris an online trigonometric identity calculator that calculates the value for trigonometric quantities with powers. It proficiently reduces the power of sin2θ, cos2θ, and tan2θ and converts them to double angle.
Let’s find out formula of power reduction, and how to calculate power reducing trig function without using power reducing calculator.
What is power reducing?
Power reducing is the process of evaluating the squared value of the three basic trigonometric functions (sin, cos, tan) using a reducing power function.
The power reduction formulas are obtained by solving the second and third versions of the cosine double-angle and half-angle formulas. In power reduction formulas, a trigonometric function is raised to a power such as:
Sin2α or cos 2 α
How to reduce power of trigonometric identities?
To reduce the power of squared trig identities, follow the below steps:
Example:
Find the value of sin2θ, cos2θ, and tan2θ, if the given angle is 30 degree.
Solution:
Step 1: Write down the angle.
Θ = 30
Step 2: Place the angle value in the trig functions given above to calculate the value.
sin2 (30°) = [1 - cos (2(30°))]/2
sin2 (30°) = [1 - cos (60°)]/2
sin2 (30°) = [1 - cos (60°)]/2
sin2 (30°) = (1 – 0.5)/2
sin2 (30°) = 0.5/2
sin2 (30°) = 0.25
cos2 (30°) = [1 + cos (2(30°))]/2
cos2 (30°) = [1 + cos (60°)]/2
cos2 (30°) = (1 + 0.5)/2
cos2 (30°) = 1.5/2
cos2 (30°) = 0.75
tan2 (30°) = [1 - cos (2(30°)]/ [1 + cos (2(30°)]
tan2 (30°) = [1 - cos (60°)]/ [1 + cos (60°)]
tan2 (30°) = [1 – 0.5]/ [1 + 0.5]
tan2 (30°) = 0.5/ 1.5
tan2 (30°) = 0.33
Power reduction calculatorcan perform all of the above calculations in blink of an eye. Moreover, you can use this tool to verify the outcomes of your manual calculations.
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1103 | https://mindyourdecisions.com/blog/2013/05/06/what-are-the-chances-of-meeting-in-a-tournament/ | Skip to content
Mind Your Decisions
Math Videos, Math Puzzles, Game Theory. By Presh Talwalkar
What are the chances of meeting in a tournament?
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Posted May 6, 2013 By Presh Talwalkar. Read about me, or email me.
Alice and Bob enter into a rock, paper, and scissors tournament with 2n players.
What is the probability the two will face off in a match?
(Assume the initial tournament bracket pairs are chosen randomly and every player has an equal chance of winning a particular match.)
. .
"All will be well if you use your mind for your decisions, and mind only your decisions." It costs thousands of dollars to run a website and your support matters. If you like the posts and videos, please consider a monthly pledge on Patreon. You may also consider a one-time donation to support my work.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . METHOD 1: an elegant solution
I read about this problem at Math Stackexchange. There is a direct and simple combinatorial solution.
A 2n player knock-out tournament consists of 2n-1 matches with all pairings distinct. Since there are “2n choose 2″ pairings in total, the probability for a given pairing to occur among the pairings actually taking place, is
(2n-1)/(2n choose 2)=1/2n – 1
You will only truly appreciate the brevity of this solution after reading the longer proof presented below.
METHOD 2: work out smaller cases
Alas, I did not come across the quick solution. I cranked out the answer using induction, which is an instructive exercise.
When n = 1, the tournament is just 2 people and so Alice and Bob necessarily face off. The probability they meet is 1.
For n = 2, the tournament is 4 people. As illustrated, there are 3 possible ways in which they will meet.
One case they could meet is if they are paired in the first match (corresponding to the figure on the left). What are the odds this happens? Alice can be paired with any of the other three people equally, so there is a 1/3 chance she faces off against Bob in the first round.
A second case considers the 2/3 of the time they are not paired up (corresponding to the two figures on the right). In this case, they only meet if they both win and proceed to the next round. Since each has a 1/2 chance of winning, there is a 1/4 chance that both will win.
So the overall chance that Alice and Bob meet in a tournament of 4 people is:
Pr(meet) = Pr(meet first round) + Pr(not first round) Pr(meet second round)
Pr(meet) = 1/3 + (2/3)(1/4) = 1/2
What about n = 3, with a tournament of 8 people? We can count the number of ways in a similar manner.
The easy case is the chance they meet in the first round. Since Alice can face any of 7 opponents with equal chance, there is a 1/7 chance that Alice and Bob are paired up immediately.
What if they are not paired up, which happens with probability 6/7? There are a lot of possibilities to consider. They can only meet in the second round if they both win and by luck they get matched up. If they do not get matched up, they might still meet if they both win and make it to the third and final round. Rather than sort out this elaborate calculation, there is a trick to be used.
In the first round of the tournament, there are 8 people. Only half of them survive to the second round, leaving 4 people to face off against each other. There is a 1/4 chance that both Alice and Bob win the first round.
The key observation is this: if Alice and Bob both win in the first round of an 8-person tournament, the chances they will meet some time after that is the same as the probability they would meet in a 4-person tournament with random pairings. This suggests an inductive method to solve the problem.
What is that probability? It was calculated in the n = 2 case as 1/2. Putting this all together, the overall probability can be calculated.
Pr(meet) = Pr(meet first round) + Pr(not first round) Pr(win first round) Pr(meet 4-person tournament)
Pr(meet) = 1/7 + (6/7)(1/4)(1/2) = 1/4
Notice the results we have just calculated.
n = 1 , (21), probability = 1
n = 2 , (22), probability = 1/2
n = 3 , (23), probability = 1/4
…
n in general (2n), is the probability = (1/2)n – 1?
There seems to be a pattern: each time the tournament doubles in size, it halves the chances that Alice and Bob will meet. Can we prove that the probability will be 1/8 for the case n = 4, when the tournament has 16 people?
Let’s calculate and find out. The chance they meet in the first round is 1/15 since Alice can face any of 15 opponents with equal chance. If they do not meet, which happens with 14/15 chance, then there is a 1/4 chance they both win to advance to the next round. The probability they meet from the second round onward is the same as the chance they would meet in a 4-person tournament, which was calculated to be 1/4. The overall probability is therefore the following.
Pr(meet) = Pr(meet first round) + Pr(not first round) Pr(win first round) Pr(meet 8-person tournament)
Pr(meet) = 1/15 + (14/15)(1/4)(1/4) = 1/8
Will the pattern continue? Will the two have a chance of 1/2n-1 for meeting in a tournament with 2n people?
Let’s prove this by induction. Write C(n) for the probability that Alice and Bob will meet in a tournament with 2n people. Assuming that C(n) = 1/2n-1, we will then want to prove C(n+1) = 1/2n.
As argued above, the chance they meet can be split up into meeting in the first round, or, if they both win, the chance they meet in any subsequent round. The probability they meet will be:
C(n+1) = Pr(meet first round) + Pr(not first round) Pr(win first round) Pr(meet n-person tournament)
Since Pr(meet n-person tournament) = C(n) by the induction hypothesis, we have
C(n+1) = Pr(meet first round) + Pr(not first round) Pr(win first round) C(n)
C(n+1) = 1/(2n + 1 – 1) + (1 – 1/(2n + 1 – 1))(1/4)(1/2n-1)
(skipping some algebra)
C(n+1) = 1/2n
This proves the formula inductively. The chance of Alice and Bob meeting in a tournament of 2n people is (1/2)n – 1.
This is of course the same answer as the combinatorial proof, but it is still instructive because in a way it illustrates how the problem is inductive.
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2 thoughts on “What are the chances of meeting in a tournament?”
Brilliant!I remember having sent you this puzzle by email. I had solved it the rigorous way! Good to see the (much much) easier solution :)
Pingback: Monday Puzzle NCAA Tournament Edition. What are the odds of picking a perfect bracket? Should you enter more than 1 bracket? | Mind Your Decisions
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1104 | https://www.math.ntu.edu.tw/~hchen/teaching/StatInference/notes/lecture16.pdf | 3.2.5 Negative Binomial Distribution In a sequence of independent Bernoulli(p) trials, let the random variable X denote the trial at which the rth success occurs, where r is a fixed integer. Then P(X = x|r, p) = µx −1 r −1 ¶ pr(1 −p)x−r, x = r, r + 1, . . . , (1) and we say that X has a negative binomial(r, p) distribution.
The negative binomial distribution is sometimes defined in terms of the random variable Y =number of failures before rth success. This formulation is statistically equivalent to the one given above in terms of X =trial at which the rth success occurs, since Y = X −r. The alternative form of the negative binomial distribution is P(Y = y) = µr + y −1 y ¶ pr(1 −p)y, y = 0, 1, . . . .
The negative binomial distribution gets its name from the relationship µr + y −1 y ¶ = (−1)y µ−r y ¶ = (−1)y (−r)(−r −1) · · · (−r −y + 1) (y)(y −1) · · · (2)(1) , (2) which is the defining equation for binomial coefficient with negative integers. Along with (2), we have X y P ¡ Y = y ¢ = 1 from the negative binomial expansition which states that (1 + t)−r = X k µ−r k ¶ tk = X k (−1)k µr + k −1 k ¶ tk 1 EY = ∞ X y=0 y µr + y −1 y ¶ pr(1 −p)y = ∞ X y=1 (r + y −1)!
(y −1)!(r −1)!pr(1 −p)y = ∞ X y=1 r(1 −p) p µr + y −1 y −1 ¶ pr+1(1 −p)y−1 = r(1 −p) p ∞ X z=0 µr + 1 + z −1 z ¶ pr+1(1 −p)z = r1 −p p .
A similar calculation will show VarY = r(1 −p) p2 .
Example 3.2.6 (Inverse Binomial Sampling A technique known as an inverse binomial sampling is useful in sampling biological popula-tions. If the proportion of individuals possessing a certain characteristic is p and we sample until we see r such individuals, then the number of individuals sampled is a negative bnomial rndom variable.
0.1 Geometric distribution The geometric distribution is the simplest of the waiting time distributions and is a special case of the negative binomial distribution. Let r = 1 in (1) we have P(X = x|p) = p(1 −p)x−1, x = 1, 2, . . . , which defines the pmf of a geometric random variable X with success probability p.
X can be interpreted as the trial at which the first success occurs, so we are “waiting for a success”. The mean and variance of X can be calculated by using the negative binomial formulas and by writing X = Y + 1 to obtain EX = EY + 1 = 1 P and VarX = 1 −p p2 .
2 The geometric distribution has an interesting property, known as the “memoryless” property.
For integers s > t, it is the case that P(X > s|X > t) = P(X > s −t), (3) that is, the geometric distribution “forgets” what has occurred. The probability of getting an additional s −t failures, having already observed t failures, is the same as the probability of observing s −t failures at the start of the sequence.
To establish (3), we first note that for any integer n, P(X > n) = P(no success in n trials) = (1 −p)n, and hence, P(X > s|X > t) = P(X > s and X > t) P(X > t) = P(X > s) P(X > t) = (1 −p)s−t = P(X > s −t).
3 |
1105 | https://pmc.ncbi.nlm.nih.gov/articles/PMC5773993/ | Osteitis fibrosa cystica of mandible in hyperparathyroidism-jaw tumor syndrome: A rare presentation and review of literature - PMC
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Natl J Maxillofac Surg
. 2017 Jul-Dec;8(2):162–166. doi: 10.4103/njms.NJMS_48_17
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Osteitis fibrosa cystica of mandible in hyperparathyroidism-jaw tumor syndrome: A rare presentation and review of literature
Anupama Singh Satpathy
Anupama Singh Satpathy
1 Department of ENT and Head Neck Surgery, Medica Super Speciality Hospital, Kolkata, West Bengal, India
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1,✉, Arjun Dasgupta
Arjun Dasgupta
1 Department of ENT and Head Neck Surgery, Medica Super Speciality Hospital, Kolkata, West Bengal, India
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1, Chirajit Dutta
Chirajit Dutta
1 Department of ENT and Head Neck Surgery, Medica Super Speciality Hospital, Kolkata, West Bengal, India
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1, N V K Mohan
N V K Mohan
1 Department of ENT and Head Neck Surgery, Medica Super Speciality Hospital, Kolkata, West Bengal, India
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Shouvanik Satpathy
1 Department of ENT and Head Neck Surgery, Medica Super Speciality Hospital, Kolkata, West Bengal, India
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1 Department of ENT and Head Neck Surgery, Medica Super Speciality Hospital, Kolkata, West Bengal, India
✉
Address for correspondence: Dr. Anupama Singh Satpathy, Flat No. B-301, Mayfair Greens, 449 S. N. Ghosh Avenue, Narendrapur, Kolkata - 700 103, West Bengal, India. E-mail: dr.anupamasingh@gmail.com
Copyright: © 2017 National Journal of Maxillofacial Surgery
This is an open access article distributed under the terms of the Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License, which allows others to remix, tweak, and build upon the work non-commercially, as long as the author is credited and the new creations are licensed under the identical terms.
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PMCID: PMC5773993 PMID: 29386822
Abstract
Brown's tumor, also referred as osteitis fibrosa cystica is a rare nonneoplastic diagnostically challenging consequence of hyperparathyroidism (HPT) which occurs due to increased parathormone secretions in blood, causing excessive calcium resorption from kidneys, bone resorption, and phosphaturia. Brown's tumor is a misnomer, presenting as cystic expansile lesions in bone, often misdiagnosed as neoplastic lesion or granuloma or abscess in bones. It can affect long bones, clavicle, ribs, and pelvis. According to literature, skeletal manifestations of Brown tumor is relatively rare and occurs in <2% of the cases of HPT. We present a case of a female 15-year-old patient who presented with bleeding gums and an expansile lesion in mandible whose previous investigations elsewhere suggested a malignant lesion. However, further investigations revealed it to be Brown's tumor with primary HPT which is a rare genetic disorder, known as HPT-Jaw Tumor Syndrome (HPT-JT).
Keywords: Brown tumor, hyperparathyroidism jaw tumor, mandible, parathormone
INTRODUCTION
Parathyroids are four in number situated behind the thyroid gland. The function of Parathyroid hormone (PTH) is to maintain a proper calcium balance in the bloodstream and in tissues. PTH interacts with Vitamin D and its metabolites to regulate calcium absorption and excretion. Hyperparathyroidism (HPT) is a pathological condition characterised by elevated levels of parathormone in the blood due to overactivity of one or more parathyroid glands. Primary lesion in parathyroid with excessive PTH secretion resulting into high PTH and high calcium is called primary HPT. Secondary HPT is due to Vitamin D deficiency or chronic renal failure, with high PTH and low calcium. Tertiary is with autonomic PTH secretion leading to high PTH and high calcium, with a history of chronic renal failure.. HPT caused by parathyroid adenoma causing increased osteoclastic activity in skeletal system presents as osteitis fibrosa cystica also known as Von Recklinghausen's disease of bone. HPT can also present as a part of multiple endocrine neoplasia (MEN) I, IIa, and IIIb.
Most of the patients with HPT remain asymptomatic and presents as an incidental diagnosis on biochemical testing with hypophosphatemia, hypercalcemia, and increased alkaline phosphatase levels in the blood.
Brown tumor is a benign intraosseous, nonneoplastic lesion consisting of cellular fibrous tissue with multiple foci of mononuclear stromal cells mixed with hemorrhagic infiltrates, aggregations of multinucleated giant cells, hemosiderin deposits and occasionally trabeculae of woven bone. Radiologically, they present as solitary or multilocular soap bubble-like bone expanding radiolucencies. The reported prevalence of Brown tumor is 0.1% with male:female ratio of 1:3 in age group <30 years. Its rare in facial bones but more common in the mandible than the maxilla. They mainly occur in patients with secondary HPT with renal insufficiency, and calcium malabsorption in facial bones, clavicle, ribs, pelvis, and femur are affected. Primary HPT with ossifying fibroma of the jaw, which represents HPT-Jaw Tumor syndrome (HPT-JT) is a rare variant.[5,6]
We present a case report of a 15-year-old female with fibro-ossifying tumors of the mandible and multiple lytic lesions in skull, humerus, pelvis, femur, and other long bones with severe HPT.
CASE REPORT
A 15-year-old female patient referred from elsewhere was admitted in otorhinolaryngology department with slightly painful swelling in left side of the lower jaw, and bleeding from gums while brushing for the past 2 months. There was a history of abdominal pains with nausea, pain in limbs, and back every now and then for the last 3 years. She started limping in the past 2–3 months. There was no history of peptic ulcers, fractures, vitamin and calcium supplementation intake, urolithiasis, or any exposure to radiations.
On physical examination, there was a large 8 cm × 5 cm × 3 cm firm, slightly tender swelling in the left side of the mandible with the widening of the alveolar process. On palpation, no egg shell crackling was noted. On neck examination, no clinically palpable cervical lymph node detected. Rest of systemic examinations was normal [Figure 1].
Figure 1.
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Large swelling involving left side of the body of mandible with widening of alveolar process
Blood tests demonstrated haemoglobin (14.1), total count (8580), elevated intact S. PTH (2269 pg/ml), raised S. Calcium (12.3 mg/dl), raised S. Alkaline Phosphatase (3319U/L), low Vitamin D3 (10.63 ng/ml), normal S. Prolactin (12.86 ng/ml) and normal S. Phosphorus (19.4). Urine calcium concentration was normal, 24 h urine for Phosphorus was low (291 mg/24 h), creatinine in urine was also low (408 mg/24 h).
Fine-needle aspiration cytology showed: profuse Giant cells rich lesion-Giant cell reparative granuloma or Giant cell tumor of bone with scanty pus cells, no organism, acid-fast bacteria negative
On HPE: vascularized fibrous tissue in vague storiform pattern, bony spicules and numerous osteoclastic cells with areas of hemorrhage and extravasated RBCs consistent with Brown tumor of HPT
Contrast-enhanced computed tomography (CT) Face revealed: large destructive expansile lesion involving mandible-well defined, 7.0 cm × 5.2 cm × 5.0 cm with soap bubble appearance involving 2/3rd of the left side of body of mandible up to first molar tooth, to right side up to canine, medially involving the left geniohyoid/genioglossus muscle with early obscuration of sublingual space [Figure 2]. Diffuse ground glass opacity in skull vault with lytic and sclerotic foci in vertebrae and head of left humerus
X-ray skeletal survey: lytic with sclerotic lesions seen at the end of long bones with early deformity. Possible sclerosis seen on both SI joint. Bowing of both humeral neck regions
USG Neck: Cystic SOL in left side of mandible and a possible cervicle lymph node near inferior part of left lobe of thyroid
Sestamibi scan: left inferior parathyroid adenoma
USG whole abdomen showed: Normal study.
Figure 2.
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Axial section computed tomography scan showing large destructive expansile lesion with soap bubble appearance involving 2/3 rd of the left side of body of mandible
She was initially on conservative management for treatment of HPT, MEN I was ruled out after consultation with Endocrinologist and Orthopaedic surgeon. Left parathyroid adenoma which was 2 cm × 1 cm in size was excised [Figure 3]. Preexcision serum PTH level was 1472 pg/mL, whereas, intraoperatively, postexcision level came to be 248 pg/mL.
Figure 3.
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Resected parathyroid adenoma (2 cm × 1 cm)
After surgery, patient was on conservative management and advised for regular follow-up. Six months after initial management the patient had a minimal external deformity, no systemic manifestation and improved quality of life [Figures 4 and 5]. Repeat CT scan was done after 6 months and again after 1 year, [Figures 6 and 7] showing significant reduction of size of the lesion although complete bone remodeling will still take time.
Figure 4.
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Clinical picture six months postsurgery showing significant reduction in size of tumor
Figure 5.
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Clinical picture 6 months postsurgery showing correction of external facial deformity
Figure 6.
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Axial section computed tomography scan 6 months postsurgery
Figure 7.
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Axial section computed tomography scan 1 year postsurgery
DISCUSSION
Different bone expanding giant cell lesions that can arise in jaw bones include odontogenic cysts and tumors, i.e., periodontal cyst, radicular cyst, or ameloblastoma, infectious diseases such as osteomyelitis or bone abscess, primary bone tumors, and cysts such as giant cell reparative granuloma or myxoma or simple bone cysts or odontogenic fibromas, metabolic bone disease, i.e., HPT or metastasis from a known or unknown primary such as lung, breast, kidney, or prostate.
Excessive PTH secretion causes osteoclastic activity to exceed the osteoblastic activity resulting into bone resorption, intraosseus bleeding and tissue degeneration, with the formation of a cystic lesion filled with hemosiderin loaded macrophages, giant cells, and fibroblasts. Hemosiderin deposits, hemmorhages and vascularization results into color and name BROWN TUMOR. It is difficult to differentiate brown tumor from any other jaw bone expansile lesion on the basis of histopathology or radiology, but the clinical correlation with HPT favours the diagnosis. Symptoms caused by the lesion depends on their size and location.[8,9]
Histopathologically, it is very difficult to differentiate it from giant cell reparative granuloma, fibrous dysplasia or true giant cell tumors. Giant cell reparative granuloma is a localized lesion occurring mainly in young females but without HPT. True giant cell tumors will be more infiltrative and with some degree of cellular atypia histologically. Infectious diseases and other local cysts will also be localized and will lack the features of HPT.
Ossifying fibroma of the jaw (brown tumor) with primary HPT, as in our case, is a relatively rare autosomal dominant disorder with incomplete penetrance and variable expressions and is known as HPT-JT. Pathogenesis of HPT-JT involves an activation of HRPT2 gene (located on 1q25) which codes for a 531 amino acid protein called parafibromin. Tumor suppressor role is also suggested here for the allelic loss of 1q24–q32. There is reduced penetrance in females. HPTJT which is considered more aggressive as compared to sporadic HPT due to the risk of developing parathyroid carcinomas (10%–15% of affected individuals) and frequent multiglandular involvement. Single gland involvement occurs in 89% of cases. Treatment for HRPT2 related HPT is surgery. Radiologically, the unique parathyroid lesion is excised and so is a grossly enlarged gland. Recurrent cases need revision surgery. Single gland disease may also be confirmed by intraoperative PTH assay which should decrease by more than 50% in 10 min after excision. Long-term follow-ups in these patients is mandatory.
Brown tumors usually regress after normalization of S. PTH levels. Cystic brown tumors will not show radiographic ossification after parathyroidectomy. Bone lesion need not to be operated on. There are possibilities of remodelling after normocalcemia level is achieved, but in case of persistent bony lesion even after 6 months of metabolic control, curettage and enucleation is recommended.
CONCLUSION
Primary HPT with Vitamin D deficiency, hypercalcemia and brown tumor in mandible-HPT-JT syndrome is a relatively rare now, especially after recent improvements in analytical technologies which usually detects HPT in asymptomatic stage. At present, in Asian population, Primary HPT is three times more common than diabetes mellitus. It's presentation as osteitis fibrosa cystica is rarer, <2% of the population. Serum PTH level and levels of Vitamin D3, calcium, phosphorus, and alkaline phosphatase in blood is mandatory for planning management of any bone expansile lesion cases. Excision of parathyroid adenoma and achievement of normocalcemic level with long-term follow-up proves to be the treatment of choice in these cases. Curettage and enucleation should be considered in cases where primary bone lesion expands or does not regress even after 6 months of primary management.
Ethical approval
This article does not contain any studies with human participants or animals performed by any of the author.
Declaration of patient consent
The authors certify that they have obtained all appropriate patient consent forms. In the form the patient(s) has/have given his/her/their consent for his/her/their images and other clinical information to be reported in the journal. The patients understand that their names and initials will not be published and due efforts will be made to conceal their identity, but anonymity cannot be guaranteed.
Financial support and sponsorship
Nil.
Conflicts of interest
There are no conflicts of interest.
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Card Probability
Last Updated : 23 Jul, 2025
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Card Probability is the probability of the events involving a deck of playing cards. As we know, probability is one of the important topics of mathematics which deals with the calculation of the possibility of any event. In simple words, card probability is one part of probability in which we find the probability of drawing a card from the deck of cards.
Table of Content
What is Probability?
Probability Formula
What is Card Probability?
Deck of Cards in Probability
Types of Cards in a Deck
Playing Card Probability
How to Find the Probability of Cards?
Sample Questions on Card Probability
What is Probability?
Probability is the branch of mathematics that studies the possibilities of any event happening or not. Mathematically is nothing but the ratio of the number of favorable outcomes to the total number of outcomes (sample space) for an event.
Some of the real-life examples of probability are:
Playing card games, to find the probability of winning or losing the game.
Weather forecasting, to predict the rain.
Election results, to determine whether the candidate will win or lose.
Exam results, to identify whether the candidate will pass or fail.
Read More:Applications of Probability
Probability Formula
If E be an event with sample space S and the number of favorable outcomes are n(E) then the probability of event E i.e., P(E) is given by:
P(E) = n(E) / n(S)
Read More: Probability Formulas
What is Card Probability?
The probability of drawing a card or collection of cards from a deck is called Card Probability. In simple words, probability related to playing cards is called card probability. As this is the type of probability, it always lies between 0 and 1.
For example, if we have to find the probability of drawing an ace from the deck of cards i.e., 4/52 = 1/13 [As there are 4 aces in the deck of 52 cards].
Deck of Cards in Probability
Deck of Cards is a collection of 52 cards that have been around for thousands of years. Deck of Cards or playing cards are considered to have originated either from India or China, first documented proof of these cards is found in 9 th-century China during the Tang Dynasty. These cards were similar to modern-day cards and also divided into four suits but the names and symbols of those suits are different i.e., coins, strings of coins, myriads, and myriads of tens.
In the modern day, these cards come in various designs and are divided into four suits namely Spade (♠), Club (♣), Heart (♥\color{red}{\text{♥}}♥), and Diamond (⧫\color{red}{\blacklozenge}⧫). For a single chosen card, the sample space is 52 i.e., the total number of outcomes for a single chosen card from a deck is 52.
n(S) for deck of cards = 52
Types of Cards in a Deck
Any deck of cards can be classified in many ways, some of the parameters on which cards can be classified are:
Based on Colors
Based on Suits
Let's understand this classification in detail as follows:
Based on Colors
Based on colors a deck of cards can be classified into two categories,
Red Cards
Black Cards
A total of 52 cards are divided equally into red and black cards which means there are 26 red cards and 26 black cards in the deck.
Based on Suits
Four suits in the deck of cards are:
Hearts (♥\color{red}{\text{♥}}♥)
Diamonds (⧫\color{red}{\blacklozenge}⧫)
Clubs (♣)
Spades (♠)
Other than these there is one more classification of cards, based on the rank of cards:
Ace
Number Cards
Face Cards
Ace
Ace is one such card which either is the most important or least important based on the game. This card "A" written on it and each suit has one of such card i.e., four ace cards.
Number Cards
From 2 to 10, there are 9 cards per suit, thus there is a total of 36 such cards.
Face Cards
Face cards as the name suggests, contain a figure or face of the figure on the card. There are three cards of each suit i.e., Jack, Queen, King. Thus there are a total of 12 face cards.
All these classifications can be seen in the following table.
| Deck Of Cards (52 cards) |
| Colored Cards | Black Cards (26 cards) | Red Cards (26 cards) |
| Suits | Spade (13 cards) | Club (13 cards) | Heart (13 cards) | Diamond (13 cards) |
| Face Cards (12 cards in a deck and 3 cards in each suit) | K (King) | K (King) | K (King) | K (King) |
| Q (Queen) | Q (Queen) | Q (Queen) | Q (Queen) |
| J (Jack) | J (Jack) | J (Jack) | J (Jack) |
| Number Cards (36 cards in a deck and 9 cards in a suit) | 10 | 10 | 10 | 10 |
| 9 | 9 | 9 | 9 |
| 8 | 8 | 8 | 8 |
| 7 | 7 | 7 | 7 |
| 6 | 6 | 6 | 6 |
| 5 | 5 | 5 | 5 |
| 4 | 4 | 4 | 4 |
| 3 | 3 | 3 | 3 |
| 2 | 2 | 2 | 2 |
| Ace Cards (4 cards in a deck and 1 card in a suit) | A (Ace) | A (Ace) | A (Ace) | A (Ace) |
Deck of Cards Chart
The following chart represents the classification of the deck of playing cards:
Playing Card Probability
Some of the common events in card probabilities are discussed in the following table:
| Event E for drawing card | Probability P(E) |
--- |
| An Ace | P(E) = 4 / 52 = 1 / 13 |
| A King | P(E) = 4 / 52 = 1 / 13 |
| A Number Card | P(E) = 36 / 52 = 9 / 13 |
| A Face Card | P(E) = 12 / 52 = 3 / 13 |
| A Spade Card | P(E) = 13 / 52 = 1 / 4 |
| A Red Card | P(E) = 26 / 52 = 1 / 2 |
How to Find the Probability of Cards?
The steps to find the probability of events involving cards are the same as all the other probabilities, which are given as follows:
Step 1: First, find the number of favourable outcomes from the given question.
Step 2:Then, find the total number of outcomes.
Step 3:Apply the probability formula to find the card probability.
Example: What is the probability of drawing an ace from a deck of cards?
Answer:
Here, E is event of drawing an ace card
Total number of outcomes in a deck n(S) = 52
Number of favorable outcomes = n(E) = drawing an ace card from deck = 4 (There are 4 ace cards in 1 deck)
P(E) = n(E) / n(S) = 4 / 52
P(E) = 1 / 13
Probability of drawing an ace card = 1 / 13
Sample Questions on Card Probability
Problem 1: What is the probability of drawing the following cards from a deck of cards?
a spade
a black card
a number card
Solution:
(i) Here, E is event of drawing a spade card
Total number of outcomes in a deck n(S) = 52
Number of favorable outcomes = n(E) /n(s) = drawing a spade card from deck/total number of Outcomes from the deck
P(E) = n(E) / n(S) = 13 / 52
P(E) = 1 / 4
Probability of drawing a spade = 1 / 4
(ii) Here, E is event of drawing a black card
Total number of outcomes in a deck n(S) = 52
Number of favorable outcomes = n(E) = drawing a black card from deck = 26 (There are 26 black cards in 1 deck)
P(E) = n(E) / n(S) = 26 / 52
P(E) = 1 / 2
Probability of drawing a black card = 1 / 2
(iii) Here, E is event of drawing a number card
Total number of outcomes in a deck n(S) = 52
Number of favorable outcomes = n(E) = drawing a number card from deck = 36 (There are 36 number cards in 1 deck)
P(E) = n(E) / n(S) = 36 / 52
P(E) = 9 / 13
Probability of drawing a number card = 9 / 13
Problem 2: What is the probability of drawing the following cards from a deck of cards?
A king or a black card
A red and ace card
Solution:
(i) Here, E is event of drawing a king or a black card
Total number of outcomes in a deck n(S) = 52
Number of favorable outcomes = n(E) = drawing a king or a black card from deck = 26 + 2 = 28 (There are 26 black cards in which 2 are king and remaining 2 kings of black in 1 deck)
P(E) = n(E) / n(S) = 28 / 52
P(E) = 7 / 13
Probability of drawing a king or a black card = 7 / 13
(ii)Here, E is event of drawing a red and ace card
Total number of outcomes in a deck n(S) = 52
Number of favorable outcomes = n(E) = drawing a red and ace card from deck = 2 (There are 26 red cards in which 2 are ace cards)
According to question drawn card should be red and ace both. Therefore, n(E) = 2
P(E) = n(E) / n(S) = 2 / 52
P(E) = 1 / 26
Probability of drawing a red and ace card= 1 / 26
Problem 3: What is the probability of drawing the following cards from a deck of cards?
A non-club card
A non-face card
Solution:
(i) Here, E is event of drawing a non-club card
Total number of outcomes in a deck n(S) = 52
Number of favorable outcomes = n(E) = drawing a non-club card from deck = 39 (There are 13 clubs in 1 deck, non- deck = 52 - 13 = 39)
P(E) = n(E) / n(S) = 39 / 52
P(E) = 3 / 4
Probability of drawing a non-club card = 3 / 4
(ii) Here, E is event of drawing a non-face card
Total number of outcomes in a deck n(S) = 52
Number of favorable outcomes = n(E) = drawing a non-face card from deck = 40 (There are 12 face cards in 1 deck, non- deck = 52 - 12 = 40)
P(E) = n(E) / n(S) = 40 / 52
P(E) = 10 / 13
Probability of drawing a non-club card = 10 / 13
Problem 4: What is the probability of drawing a card that is neither red nor a face card?
Solution:
Here, E is event of drawing a neither red nor a face card
Total number of outcomes in a deck n(S) = 52
Number of favorable outcomes = n(E) = drawing neither red nor a face card from deck.
Total red cards = 26
There is total 12 face cards in a deck, but 6 red face cards are already removed. So remaining face cards = 12 - 6 = 6
Total back cards = 26
Black Face cards = 6
Black non-face cards = 26 - 6 = 20
Probability of drawing a neither red nor a face card= 20 / 52 = 5 / 13
Problem 5: What is the probability of drawing two cards from a deck of cards without replacement when the first card is a heart and the second card is a diamond?
Solution:
Probability of drawing first card as heart = 13 / 52
After drawing first card, the card is removed.
Probability of drawing second card as diamond = 13 / 51
Probability of drawing first card as heart and second as diamond = (13 / 52) × (13 / 51) = 169/2652
Probability of drawing first card as heart and second as diamond = 13/204
Related Reads:
Interesting Facts about Probability
Dice Probability
Probability Tricks
Practice Questions - Card Probability
Question 1.What is the probability of drawing an Ace from a standard 52-card deck?
Question 2. If you draw two cards without replacement, what is the probability of getting two Hearts?
Question 3. What is the probability of drawing a face card (Jack, Queen, or King)?
Question 4.If you're dealt a 5-card hand, what's the probability of getting a flush (all cards of the same suit)?
Question 5.What's the probability of drawing a red card (Hearts or Diamonds)?
Question 6.If you draw three cards without replacement, what's the probability of getting three different suits?
Question 7.What's the probability of drawing a card that is either a Spade or a King?
Question 8.If you're dealt two cards, what's the probability of getting a pair (two cards of the same rank)?
Question 9.What's the probability of drawing a card that is both red and even-numbered?
Question 10.If you draw five cards without replacement, what's the probability that at least one of them is an Ace?
Summary
Card probability involves calculating the likelihood of specific outcomes when drawing cards from a standard 52-card deck.
This deck consists of four suits (hearts, diamonds, clubs, spades) with 13 ranks each (Ace through King).
Probability is typically expressed as a fraction or percentage, calculated by dividing the number of favorable outcomes by the total number of possible outcomes.
Key concepts include independent events (where previous draws don't affect future probabilities) and dependent events (where probabilities change after each draw without replacement).
Common calculations involve finding the probability of drawing specific ranks, suits, or combinations of cards.
Understanding card probability is crucial for games of chance, strategic decision-making in card games, and as a foundation for more complex probability theory.
Suggested Quiz
10 Questions
In a standard 52-card deck, what is the probability of drawing a Queen or a Heart?
A 1/13
B 7/26
C 3/13
D 4/13
Explanation:
No of Heart Cards = 13
No of Queens = 4
Total no of cards that are either Queen or Hearts = 13 + 4 - 1 = 16 ( Since Queen of hearts is common to both hence 1 is subtracted )
then, p( Queen or Hearts) = 16/52 = 4/13
If two cards are drawn from a deck with replacement, what is the probability that both cards are Kings?
A 1/169
B 1/2652
C 1/221
D 2/52
Explanation:
p( drawing two successive kings ) = p(getting king in first draw) × p(getting king in second draw)
p(getting king in first draw) = 4/52 = 1/13
p(getting king in second draw) = 4/52 = 1/13
p( drawing two successive kings ) = 1/13 × 1/13 = 1/169
What is the probability of drawing a card that is either a red card or a face card from a standard deck?
A 1/2
B 8/13
C 3/13
D 7/13
Explanation:
No of Red card = 26
No of face cards = 12
total no of cards that are either red or face card = 26 + 12 - 6 = 32
p( getting either a red card or a face card ) = 32/52 = 8/13
In a deck of cards, what is the probability of drawing a card that is neither a Spade nor a Face card?
A 10/13
B 1/4
C 15/26
D 11/26
Explanation:
P(neither a Spade nor a Face card) = 1 - P( Spade or a Face card )
No of Spade or a Face card = 13 + 12 - 3 = 22
P( Spade or a Face card ) = 22/52 = 11/26
P(neither a Spade nor a Face card) = 1-(11/26) = 15/26
What is the probability of drawing a card that is both red and even-numbered from a standard deck?
A 5/52
B 5/13
C 5/26
D 1/2
Explanation:
No of cards that is both red and even = 5 ( 2,4,6,8,10 )
P( getting a card that is both red and even) = 5/52
If you draw three cards from a deck without replacement, what is the probability of drawing three different suits?
A [Tex]\frac{13 × 13 × 13 }{52 × 52 × 52}[/Tex]
B [Tex]\frac{13 × 13 × 13 }{52 × 51 × 50}[/Tex]
C [Tex]\frac{52 × 26 × 13 }{52 × 51 × 50}[/Tex]
D [Tex]\frac{52 × 26 × 13 }{52 × 52 × 52}[/Tex]
Explanation:
First Card: When you draw the first card, it can be of any suit. Therefore, there are no restrictions for the first card.
Sample Space for 1st card= 52
Total no of cards in first draw = 52
Second Card: The second card must be of a different suit from the first. Since there are 4 suits in total, and one is already drawn, there are 3 suits left for the second card.
Sample Space for 2nd card= 26
Total no of cards in 2nd draw= 51
Third Card: The third card must be of a different suit from the first two. This leaves 2 suits available for the third card.
Sample Space for 3rd card= 13
Total no of cards in 3rd draw= 50
Probability of drawing three different suits= [Tex]\frac{52 × 39 × 26 }{52 × 51 × 50}[/Tex] = 169/425 =
In a standard deck, what is the probability of drawing a card that is a King or a Red card?
A 1/4
B 7/13
C 5/13
D 3/13
Explanation:
No of red card = 26
No of kings == 4
No of cards that are red or king = 26 + 4 - 2 = 28
P(getting a King or a Red card) = 28/52 = 7/13
A random card is drawn from a shuffled deck. What is the probability of getting a king, when it is known that the card drawn is a red card?
A 1/2
B 2/26
C 2/13
D 1/26
Explanation:
By using conditional probability formula : P(A/B) = P(A ∩ B)/P(A)
A = Getting a red card
B = Getting a king
A ∩ B = getting a red king
P(A) = 26/52 = 1/2
P(B) = 4/52 = 1/13
P(A ∩ B) = 1/26
Probability of getting a king, when the card drawn is a red card = [Tex]\frac{\frac{2}{52}}{\frac{26}{52}}[/Tex] = 2/26 = 1/13
If you draw two cards without replacement from a deck, what is the probability that both cards are Kings?
A 2/169
B 1/221
C 2/26
D 1/169
Explanation:
Probability of drawing two kings without replacement = Probability of drawing a king in first draw × Probability of drawing a king in the second draw
No of kings available in the first draw = 4
Total No of cards in first draw = 52
Probability of drawing a king in first draw = 4/52 = 1/13
No of kings available in the first draw = 3
Total No of cards in first draw = 52
Probability of drawing a king in second draw = 3/51 = 1/17
Probability of drawing two kings without replacement = 1/13 × 1/17 = 1/221
One card is drawn from 10 different decks. What is the probability of drawing 4 aces?
A 0.005
B 0.045
C 0.0045
D 0.05
Explanation:
By using the formula for n successes
P( X=r) = n C r p r q(n-r)
n = number of trials = 10
r = number of successes = 4
p = probability of success = 4/52 = 1/13
q = probability of failure = 48/52 = 12/13
Probability of drawing a 4 aces from 10 different decks = 10 C 4 × (1/13)4 × (12/13)6 = \frac{10 × 9 × 8 × 7 × 6!}{6! × 4 × 3 × 2 × 1} × (\frac{1}{13})^4 × (\frac{12}{13})^6 = 0.00455
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1107 | https://ocw.mit.edu/courses/18-01sc-single-variable-calculus-fall-2010/713bd74e2658227fad9374a885208c23_MIT18_01SCF10_ex13sol.pdf | Implicit Differentiation and the Second Derivative Calculate y00 using implicit differentiation; simplify as much as possible. x 2 + 4y 2 = 1 Solution As with the direct method, we calculate the second derivative by differentiating twice. With implicit differentiation this leaves us with a formula for y00 that involves y and y0, and simplifying is a serious consideration. Recall that to take the derivative of 4y2 with respect to x we first take the derivative with respect to y and then multiply by y0; this is the “derivative of the inside function” mentioned in the chain rule, while the derivative of the outside function is 8y. So, differentiating both sides of: x 2 + 4y 2 = 1 gives us: 2x + 8yy0 = 0. We’re now faced with a choice. We could immediately perform implicit differentiation again, or we could solve for y0 and differentiate again. If we differentiate again we get: 2 + 8yy00 + 8(y0)2 = 0. In order to solve this for y00 we will need to solve the earlier equation for y0, so it seems most efficient to solve for y0 before taking a second derivative. 2x + 8yy0 = 0 8yy0 = −2x y0 = −2x 8y y0 = −x 4y Differentiating both sides of this expression (using the quotient rule and implicit differentiation), we get: y00 = (−1)4y (4 − y ( ) − 2 x) · 4y0 −4y + 4xy0 = 16y2 y00 = −y 4 + y2 xy0 1 We now substitute −x 4y for y0: y00 = −y + xy0 4y2 = −y + x −x 4y 4y2 = x −x 4y − y 4y 4y2 · 4y = −x2 − 4y2 16y3 y00 = − 1 16y3 (Don’t forget to use the relation x2 + 4y2 = 1 at the end!) How can we check our work? If we recognize x2 + 4y2 = 1 as the equation of an ellipse, we can test our equation y0 = −x/4y at the points (0, 1/2) and (1, 0). At (0, 1/2), y0 = −x/4y = 0 which agrees with the fact that the tangent line to the ellipse is horizontal at that point. At (1, 0) y0 is undefined, which agrees with the fact that the tangent line to the ellipse at (1, 0) is vertical. Once we have learned how the value of the second derivative is related to the shape of the graph, we can do a similar test of our expression for y00. 2 MIT OpenCourseWare 18.01SC Single Variable Calculus Fall 2010 For information about citing these materials or our Terms of Use, visit: |
1108 | https://openstax.org/books/organic-chemistry/pages/2-11-acids-and-bases-the-lewis-definition | Skip to Content
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Organic Chemistry
2.11 Acids and Bases: The Lewis Definition
Organic Chemistry
2.11
Acids and Bases: The Lewis Definition
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2.11 • Acids and Bases: The Lewis Definition
The Lewis definition of acids and bases is more encompassing than the Brønsted–Lowry definition because it’s not limited to substances that donate or accept just protons. A Lewis acid is a substance that accepts an electron pair, and a Lewis base is a substance that donates an electron pair. The donated electron pair is shared between the acid and the base in a covalent bond.
Lewis Acids and the Curved Arrow Formalism
The fact that a Lewis acid is able to accept an electron pair means that it must have either a vacant, low-energy orbital or a polar bond to hydrogen so that it can donate H+ (which has an empty 1s orbital). Thus, the Lewis definition of acidity includes many species in addition to H+. For example, various metal cations, such as Mg2+, are Lewis acids because they accept a pair of electrons when they form a bond to a base. We’ll also see in later chapters that certain metabolic reactions begin with an acid–base reaction between Mg2+ as a Lewis acid and an organic diphosphate or triphosphate ion as the Lewis base.
In the same way, compounds of group 3A elements, such as BF3 and AlCl3, are Lewis acids because they have unfilled valence orbitals and can accept electron pairs from Lewis bases, as shown in Figure 2.6. Similarly, many transition-metal compounds, such as TiCl4, FeCl3, ZnCl2, and SnCl4, are Lewis acids.
Figure
2.6
The reaction of boron trifluoride, a Lewis acid, with dimethyl ether, a Lewis base.
The Lewis acid accepts a pair of electrons, and the Lewis base donates a pair of nonbonding electrons. Note how the movement of electrons from the Lewis base to the Lewis acid is indicated by a curved arrow. Note also how, in electrostatic potential maps, the boron becomes more negative (red) after reaction because it has gained electrons and the oxygen atom becomes more positive (blue) because it has donated electrons.
Look closely at the acid–base reaction in Figure 2.6, and notice how it's shown. Dimethyl ether, the Lewis base, donates an electron pair to a vacant valence orbital of the boron atom in BF3, a Lewis acid. The direction of electron-pair flow from base to acid is shown using a curved arrow, just as the direction of electron flow from one resonance structure to another was shown using curved arrows in Section 2.5. We’ll use this curved-arrow notation throughout the remainder of this text to indicate electron flow during reactions, so get used to seeing it.
Some further examples of Lewis acids follow:
Lewis Bases
The Lewis definition of a base—a compound with a pair of nonbonding electrons that it can use to bond to a Lewis acid—is similar to the Brønsted–Lowry definition. Thus, H2O, with its two pairs of nonbonding electrons on oxygen, acts as a Lewis base by donating an electron pair to an H+ in forming the hydronium ion, H3O+.
In a more general sense, most oxygen- and nitrogen-containing organic compounds can act as Lewis bases because they have lone pairs of electrons. A divalent oxygen compound has two lone pairs of electrons, and a trivalent nitrogen compound has one lone pair. Note in the following examples that some compounds can act as both acids and bases, just as water can. Alcohols and carboxylic acids, for instance, act as acids when they donate an H+ but as bases when their oxygen atom accepts an H+.
Note also that some Lewis bases, such as carboxylic acids, esters, and amides, have more than one atom with a lone pair of electrons and can therefore react at more than one site. Acetic acid, for example, can be protonated either on the doubly bonded oxygen atom or on the singly bonded oxygen atom. Reaction normally occurs only once in such instances, and the more stable of the two possible protonation products is formed. For acetic acid, protonation by reaction with sulfuric acid occurs on the doubly bonded oxygen because that product is stabilized by two resonance forms.
Worked Example 2.6
Using Curved Arrows to Show Electron Flow
Using curved arrows, show how acetaldehyde, CH3CHO, can act as a Lewis base.
Strategy
A Lewis base donates an electron pair to a Lewis acid. We therefore need to locate the electron lone pairs on acetaldehyde and use a curved arrow to show the movement of a pair toward the H atom of the acid.
Solution
Problem
2-17
Using curved arrows, show how the species in part (a) can act as Lewis bases in their reactions with HCl, and show how the species in part (b) can act as Lewis acids in their reaction with OH–.
(a)
CH3CH2OH, HN(CH3)2, P(CH3)3
(b)
H3C+, B(CH3)3, MgBr2
Problem
2-18
Imidazole, which forms part of amino acid histidine, can act as both an acid and a base.
(a)
Look at the electrostatic potential map of imidazole, and identify the most acidic hydrogen atom and the most basic nitrogen atom.
(b)
Draw structures for the resonance forms of the products that result when imidazole is protonated by an acid and deprotonated by a base.
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1109 | https://www.youtube.com/watch?v=QWhsEep5ptI | Bouncing Ball Problem (Sum of an Infinite Geometric Series)
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In this video we use the sum of an infinite geometric series formula to find the total vertical distance travelled by a bouncing ball dropped from an initial height of 20 feet.
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in this video we're going to do a common problem referred to as the bouncing ball problem and it's using the sum of an infinite geometric Series so let's dive into this example and I'll show you how this works so it says here a ball is dropped from a height of 20 feet and then it rebounds 65% of its previous height on each successive bounce find the total vertical distance traveled by the ball before coming to rest so if you can kind of visualize this you're dropping it from a height of 20 ft it falls then it bounces back up but not quite as high just 65% of that previous height it falls again then it goes up but not quite as high as that that balce so it keeps getting smaller and smaller you can almost kind of visualize this kind of like this it drops then it comes up then it goes down then it comes up and it goes down now now the thing is it's just bouncing up and down like this right but it gradually less and less High each time until eventually settles what's interesting about this is two things one is that it's a geometric Series right so what you can think about is you start off with falling 20 ft then it goes up 20 time 65 then when it goes up the next time it's times 65 again the thing is is that when you first drop it you see you have this initial drop of 20 feet let's just diagram that out right here so 20 feet but then when it goes up on this next balance not only does it travel up but it also travels down travels UP Travels down but this one just has that one initial drop so what we're going to do is we're going to think about this okay let me just kind of show you on the calculator so 20 okay then 0 65 so this height right here okay that distance is 13 feet okay if we multiply by 65% again this one has a height of 8.45 and then if we do Again by okay 5 point I'm rounding here a little bit 49 except so basically what we're doing you see we're multiplying by 65 65 65 when you multiply by the same quantity each time that's a geometric series the dot dot dot this is infinite in order to find the sum we're going to use this infinite geometric sum formula A1 over 1 minus r where A1 is our first term R is our ratio what we're multiplying by H time okay so if we to look at that we're going to ignore the initial drop of 20 fet we're going to start with 13 and I'll show you y so 13 over 1 -65 that comes out to uh about 37.4 but that's just counting not the up and the down it's just see it's just 13 8.45 5.49 Etc we we're going to have to actually double this okay to account for the up and down okay so now we're looking at more like 74.2 right plus the initial drop of 20 ft so that's 94. 28 ft is the total distance traveled by this this ball until it finally comes to rest now is there another way to do this problem well you know one way to do it is you could treat this as two different uh series you could say well I'm going to start with 20 over 1 minus the ratio 65 Plus so that's the one doing all the down drops you had the down down down like that then we're going to do one for all on the up up you know going upwards directions that's going to be the first term would be 13 and we can add those together let's see what this comes out to so 20 divid .35 is about 57.14 and here we have 13 id35 which is 3714 add those together 94.28% follow me over to that video right there where I go into more details learning about sequences and series I'll see you there |
1110 | https://en.wiktionary.org/wiki/consensus_theorem | consensus theorem - Wiktionary, the free dictionary
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English
[edit]
English Wikipedia has an article on:
consensus theorem
Wikipedia
Noun
[edit]
consensustheorem
(logic) The following theorem of Boolean algebra: X Y+X′Z+Y Z=X Y+X′Z{\displaystyle XY+X'Z+YZ=XY+X'Z} where Y Z{\displaystyle YZ}, the algebraically redundant term, is called the "consensus term", or its dual form (X+Y)(X′+Z)(Y+Z)=(X+Y)(X′+Z){\displaystyle (X+Y)(X'+Z)(Y+Z)=(X+Y)(X'+Z)}, in which case Y+Z{\displaystyle Y+Z} is the consensus term. (Note: X+Y,X′+Z⊢Y+Z{\displaystyle X+Y,X'+Z\vdash Y+Z} is an example of the resolutioninference rule (replacing the +{\displaystyle +} with ∨{\displaystyle \vee } and the prime with prefix ¬{\displaystyle \neg } might make this more evident).)
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1111 | https://www.cryst.ehu.es/cryst/help/examples_lkvec.html | | |
| Bilbao Crystallographic Server Help |
Examples
The relation between the traditional and the reciprocal-space group descriptions of the wave-vector types is illustrated by the following examples. The included figures and tables form part of the output of the access tool LKVEC.
k-vector table and Brillouin zone for layer group p-62m (No. 78)
The k-vector table and the Brillouin-zone diagram of the hexagonal layer group p-6m2 (No. 78). The reciprocal lattice of a hexagonal p lattice is also a hexagonal p lattice and the Brillouin zone is a hexagon. The conventional basis for the reciprocal lattice has γ = 60° while the ITA description of hexagonal layer group is based on a basis aH, bH with γ = 120°. In the Brillouin zone diagrams, the axis kx is taken along aH while ky points out in the direction of aH+bH.
The k vectors of p-6m2 (No. 78) listed by L&W are distributed in four k-vector types: (i) the Wyckoff-position block 1a formed by the GM point, (ii) the block 2b formed by the K point, (iii) the k-vector point M and the k-vector lines SM and SN correspond to the block 3c, and (iv) the Wyckoff-position block 6d formed by the k-vector lines LD and T and the k-vector planes B and BB. The parameter description of a k-vector type is given in the last column of the table. Consider for example the line SM which, according to the ITA description, forms part of the k-vector type that is assigned to the Wyckoff position 3c with a site-symmetry group '..m'. Its parameter description x,0: -1/2 < x < 0 indicates that the independent segment of the line x,0 in the asymmetric unit is limited by the special k-vector points GM (x=0) and M(x=1/2) with x varying between 0 and 1/2. The parameter descriptions of the uni-arm regions (for a discussion of uni-arm description the reader is referred to ) of the k-vector types are shown in the last row of the corresponding Wyckoff position block. For example, in the block for position 3c, the k-vector line SN is equivalent (by a threefold rotation) to x,0 : -1/2 < x < 0 which in turn is equivalent (by a translation) to the line x,0: 1/2 < x < 1, denoted by SN1. This gives the uni-arm description M+SM+SN1 for the Wyckoff position 3c.
As the asymmetric unit and the representation domain do not coincide, their edges are coloured in pink and light blue, respectively. It has already been pointed out that k-vector points and lines are brought out in red only if they are special k-vector points and lines. For example, the lines T and LD (see figure of p-62m) are not coloured as special lines since they belong to the k-vector type of the Wyckoff-position block 6d, and their symmetry coincides with that of the neighbouring points of the symmetry plane B=[GM K M]. The point GM and K, however, are represented by red circles as they are special k-vector points. Likewise, the line SM is coloured in brown because it is an edge of the asymmetric unit and at the same time is a special k-vector symmetry line. The k-vector line SM is coloured in dark blue as it is a special symmetry line along the edge of the representation domain. As already indicated, the k-vector line SN together with SM and (see figure of p-62m) the point M belong to the special k-vector type of the Wyckoff-position block 3c, i.e. all these different wavevectors belong to the same k-vector type. Although M is explicity listed by L&W as a special k-vector point, it is represented by an open circle. In fact, it joins the symmetry lines SM and SN1 to a continuous line as its little co-group type coincides with those of the points on the lines.
Brillouin-zone diagrams of the layer group cm2m (No. 35)
The layer group cm2m (No. 35) is an example of orthorhombic layer groups with a c-centred lattice. It belongs to the arithmetic crystal class m2mc which also includes the layer group cm2e (No. 36). The k-vector tables and Brillouin-zone figures belong to the arithmetic crystal class m2mc.
The Brillouin-zone of the layer groups of the arithmetic crystal class m2mc is a non-regular hexagon. Depending on the relation between the lattice parameters a and b, two topologically different Brillouin zones are to be distinguished:
the acute case with a > b
2. the obtuse case with a < b
Due to the reflection -x,y (with normal kx) of the reciprocal-plane group (c1m1), the representation domain is only one half of the hexagon: for example, in the acute case it is the trapezium with vertices Δ0, F0, F2, Δ2 (light blue boundary). The asymmetric unit is different from the representation domain: it is the rectangle with vertices J2, J4, V4, V2 (pink boundary), i.e. the points x,y : 0<= x <= 1/2; -1/4 <= y <= 1/4. While the representation domain of the acute and obtuse with cells have the more complicated form of a trapezium, the asymmetric units in both cases have the topologically identical and relatively simple shape of a rectangle.
The k-vector tables show all special wave-vectors with their coefficients and layer little co-groups as specified in Tables 24 and 25 of L&W. The wave-vector coefficients with respect to the conventional reciprocal basis, i.e. dual to the conventional centred basis, are listed in the column under the heading 'Conventional' of the k-vector tables. For example, a k-vector point of the DT line with primitive coefficients (-1/4,1/4) is described as (0,1/2) with respect to a basis dual to the conventional basis cm2m.
The point GM, Y2 and S (acute case) and GM, Y and S (obtuse case) are not special k-vector points but form part of special lines and planes and in the diagrams they are represented by open circles. The line SM is not a symmetry line and is represented by a think black line because it is located inside the Brillouin zone. The lines DT and DA are coloured in brown because they are symmetry lines and at the same time are edges of the asymmetric unit. Part of DT and DA are also coloured in red because they correspond to flagpoles (for a discussion of flagpoles the reader is referred to ). The k-vector lines F and FA (acute case) are coloured in dark blue as they are symmetry lines along the edges of the representation domain.
Because of the special shape of the Brillouin zone and the representation domain for the acute case (a > b), the special k-vector line corresponding to the Wyckoff-position block 2a splits into several segments: the lines DT and DA, located inside the Brillouin zone, and the lines F and FA (coloured dark blue) at the border of the Brillouin zone. For the description of the end points of the segments, it is necessary to introduce additional parameters as dt0 and f0 whose values depend on the specific relations between the lattice parameters. The use of flagpoles enable the uni-arm description: the flagpole [J2 Y2] 0,y : 1/4 < y < 1/2 is equivalent to the segment [V4 Y] 1/2,y: -1/4 < y < 0 and the flagpole [V2 Y4] 0,y : -1/2 < y < -1/4 is equivalent to the segment [Y J4] 1/2,y : 0 < y < 1/4. The uni-arm description of the k-vector type of the Wyckoff position 2a is shown in the last row of the Wyckoff-position block and it is formed by the union of the points GM and Y2, the lines DT, DA, DT1 (∼ FA) and DA1 (∼ F). Its parameter description (0,y) with y varying in the range (-1/2,1/2) coincides with that of the acute case. The parameter description of the flagpole and its parameter range with respect to the basis of the reciprocal group are given below the k-vector table.
Litvin, D. B. & Wike, T. R. (1991). Character Tables and Compatibility Relations of the Eighty Layer Groups and Seventeen Plane Groups. New York: Plenum Press. [Abbreviated as L&W].
International Tables for Crystallography, Volume E: Subperiodic Groups (2010). Edited by V. Kopsky & D. Litvin, 2nd edition. Chichester: Wiley. [Abbreviated as ITE].
de la Flor, G., Souvignier, B., Madariaga, G. & M. I. Aroyo. Acta Cryst A (in review).
| | |
--- |
| Bilbao Crystallographic Server www.cryst.ehu.es | For comments, please mail to cryst@wm.lc.ehu.es |
| |
| Bilbao Crystallographic Server Help |
Examples
k-vector table and Brillouin zone for layer group p-62m (No. 78)
The k-vector table and the Brillouin-zone diagram of the hexagonal layer group p-6m2 (No. 78). The reciprocal lattice of a hexagonal p lattice is also a hexagonal p lattice and the Brillouin zone is a hexagon. The conventional basis for the reciprocal lattice has γ = 60° while the ITA description of hexagonal layer group is based on a basis aH, bH with γ = 120°. In the Brillouin zone diagrams, the axis kx is taken along aH while ky points out in the direction of aH+bH.
The k vectors of p-6m2 (No. 78) listed by L&W are distributed in four k-vector types: (i) the Wyckoff-position block 1a formed by the GM point, (ii) the block 2b formed by the K point, (iii) the k-vector point M and the k-vector lines SM and SN correspond to the block 3c, and (iv) the Wyckoff-position block 6d formed by the k-vector lines LD and T and the k-vector planes B and BB. The parameter description of a k-vector type is given in the last column of the table. Consider for example the line SM which, according to the ITA description, forms part of the k-vector type that is assigned to the Wyckoff position 3c with a site-symmetry group '..m'. Its parameter description x,0: -1/2 < x < 0 indicates that the independent segment of the line x,0 in the asymmetric unit is limited by the special k-vector points GM (x=0) and M(x=1/2) with x varying between 0 and 1/2. The parameter descriptions of the uni-arm regions (for a discussion of uni-arm description the reader is referred to ) of the k-vector types are shown in the last row of the corresponding Wyckoff position block. For example, in the block for position 3c, the k-vector line SN is equivalent (by a threefold rotation) to x,0 : -1/2 < x < 0 which in turn is equivalent (by a translation) to the line x,0: 1/2 < x < 1, denoted by SN1. This gives the uni-arm description M+SM+SN1 for the Wyckoff position 3c.
As the asymmetric unit and the representation domain do not coincide, their edges are coloured in pink and light blue, respectively. It has already been pointed out that k-vector points and lines are brought out in red only if they are special k-vector points and lines. For example, the lines T and LD (see figure of p-62m) are not coloured as special lines since they belong to the k-vector type of the Wyckoff-position block 6d, and their symmetry coincides with that of the neighbouring points of the symmetry plane B=[GM K M]. The point GM and K, however, are represented by red circles as they are special k-vector points. Likewise, the line SM is coloured in brown because it is an edge of the asymmetric unit and at the same time is a special k-vector symmetry line. The k-vector line SM is coloured in dark blue as it is a special symmetry line along the edge of the representation domain. As already indicated, the k-vector line SN together with SM and (see figure of p-62m) the point M belong to the special k-vector type of the Wyckoff-position block 3c, i.e. all these different wavevectors belong to the same k-vector type. Although M is explicity listed by L&W as a special k-vector point, it is represented by an open circle. In fact, it joins the symmetry lines SM and SN1 to a continuous line as its little co-group type coincides with those of the points on the lines.
Brillouin-zone diagrams of the layer group cm2m (No. 35)
The layer group cm2m (No. 35) is an example of orthorhombic layer groups with a c-centred lattice. It belongs to the arithmetic crystal class m2mc which also includes the layer group cm2e (No. 36). The k-vector tables and Brillouin-zone figures belong to the arithmetic crystal class m2mc.
The Brillouin-zone of the layer groups of the arithmetic crystal class m2mc is a non-regular hexagon. Depending on the relation between the lattice parameters a and b, two topologically different Brillouin zones are to be distinguished:
the acute case with a > b
2. the obtuse case with a < b
Due to the reflection -x,y (with normal kx) of the reciprocal-plane group (c1m1), the representation domain is only one half of the hexagon: for example, in the acute case it is the trapezium with vertices Δ0, F0, F2, Δ2 (light blue boundary). The asymmetric unit is different from the representation domain: it is the rectangle with vertices J2, J4, V4, V2 (pink boundary), i.e. the points x,y : 0<= x <= 1/2; -1/4 <= y <= 1/4. While the representation domain of the acute and obtuse with cells have the more complicated form of a trapezium, the asymmetric units in both cases have the topologically identical and relatively simple shape of a rectangle.
The k-vector tables show all special wave-vectors with their coefficients and layer little co-groups as specified in Tables 24 and 25 of L&W. The wave-vector coefficients with respect to the conventional reciprocal basis, i.e. dual to the conventional centred basis, are listed in the column under the heading 'Conventional' of the k-vector tables. For example, a k-vector point of the DT line with primitive coefficients (-1/4,1/4) is described as (0,1/2) with respect to a basis dual to the conventional basis cm2m.
The point GM, Y2 and S (acute case) and GM, Y and S (obtuse case) are not special k-vector points but form part of special lines and planes and in the diagrams they are represented by open circles. The line SM is not a symmetry line and is represented by a think black line because it is located inside the Brillouin zone. The lines DT and DA are coloured in brown because they are symmetry lines and at the same time are edges of the asymmetric unit. Part of DT and DA are also coloured in red because they correspond to flagpoles (for a discussion of flagpoles the reader is referred to ). The k-vector lines F and FA (acute case) are coloured in dark blue as they are symmetry lines along the edges of the representation domain.
Because of the special shape of the Brillouin zone and the representation domain for the acute case (a > b), the special k-vector line corresponding to the Wyckoff-position block 2a splits into several segments: the lines DT and DA, located inside the Brillouin zone, and the lines F and FA (coloured dark blue) at the border of the Brillouin zone. For the description of the end points of the segments, it is necessary to introduce additional parameters as dt0 and f0 whose values depend on the specific relations between the lattice parameters. The use of flagpoles enable the uni-arm description: the flagpole [J2 Y2] 0,y : 1/4 < y < 1/2 is equivalent to the segment [V4 Y] 1/2,y: -1/4 < y < 0 and the flagpole [V2 Y4] 0,y : -1/2 < y < -1/4 is equivalent to the segment [Y J4] 1/2,y : 0 < y < 1/4. The uni-arm description of the k-vector type of the Wyckoff position 2a is shown in the last row of the Wyckoff-position block and it is formed by the union of the points GM and Y2, the lines DT, DA, DT1 (∼ FA) and DA1 (∼ F). Its parameter description (0,y) with y varying in the range (-1/2,1/2) coincides with that of the acute case. The parameter description of the flagpole and its parameter range with respect to the basis of the reciprocal group are given below the k-vector table.
Litvin, D. B. & Wike, T. R. (1991). Character Tables and Compatibility Relations of the Eighty Layer Groups and Seventeen Plane Groups. New York: Plenum Press. [Abbreviated as L&W].
International Tables for Crystallography, Volume E: Subperiodic Groups (2010). Edited by V. Kopsky & D. Litvin, 2nd edition. Chichester: Wiley. [Abbreviated as ITE].
de la Flor, G., Souvignier, B., Madariaga, G. & M. I. Aroyo. Acta Cryst A (in review).
| | |
--- |
| Bilbao Crystallographic Server www.cryst.ehu.es | For comments, please mail to cryst@wm.lc.ehu.es |
k-vector table and Brillouin zone for layer group p-62m (No. 78)
The k vectors of p-6m2 (No. 78) listed by L&W are distributed in four k-vector types: (i) the Wyckoff-position block 1a formed by the GM point, (ii) the block 2b formed by the K point, (iii) the k-vector point M and the k-vector lines SM and SN correspond to the block 3c, and (iv) the Wyckoff-position block 6d formed by the k-vector lines LD and T and the k-vector planes B and BB. The parameter description of a k-vector type is given in the last column of the table. Consider for example the line SM which, according to the ITA description, forms part of the k-vector type that is assigned to the Wyckoff position 3c with a site-symmetry group '..m'. Its parameter description x,0: -1/2 < x < 0 indicates that the independent segment of the line x,0 in the asymmetric unit is limited by the special k-vector points GM (x=0) and M(x=1/2) with x varying between 0 and 1/2. The parameter descriptions of the uni-arm regions (for a discussion of uni-arm description the reader is referred to ) of the k-vector types are shown in the last row of the corresponding Wyckoff position block. For example, in the block for position 3c, the k-vector line SN is equivalent (by a threefold rotation) to x,0 : -1/2 < x < 0 which in turn is equivalent (by a translation) to the line x,0: 1/2 < x < 1, denoted by SN1. This gives the uni-arm description M+SM+SN1 for the Wyckoff position 3c.
As the asymmetric unit and the representation domain do not coincide, their edges are coloured in pink and light blue, respectively. It has already been pointed out that k-vector points and lines are brought out in red only if they are special k-vector points and lines. For example, the lines T and LD (see figure of p-62m) are not coloured as special lines since they belong to the k-vector type of the Wyckoff-position block 6d, and their symmetry coincides with that of the neighbouring points of the symmetry plane B=[GM K M]. The point GM and K, however, are represented by red circles as they are special k-vector points. Likewise, the line SM is coloured in brown because it is an edge of the asymmetric unit and at the same time is a special k-vector symmetry line. The k-vector line SM is coloured in dark blue as it is a special symmetry line along the edge of the representation domain. As already indicated, the k-vector line SN together with SM and (see figure of p-62m) the point M belong to the special k-vector type of the Wyckoff-position block 3c, i.e. all these different wavevectors belong to the same k-vector type. Although M is explicity listed by L&W as a special k-vector point, it is represented by an open circle. In fact, it joins the symmetry lines SM and SN1 to a continuous line as its little co-group type coincides with those of the points on the lines.
Brillouin-zone diagrams of the layer group cm2m (No. 35)
The layer group cm2m (No. 35) is an example of orthorhombic layer groups with a c-centred lattice. It belongs to the arithmetic crystal class m2mc which also includes the layer group cm2e (No. 36). The k-vector tables and Brillouin-zone figures belong to the arithmetic crystal class m2mc.
The Brillouin-zone of the layer groups of the arithmetic crystal class m2mc is a non-regular hexagon. Depending on the relation between the lattice parameters a and b, two topologically different Brillouin zones are to be distinguished:
the acute case with a > b
2. the obtuse case with a < b
Due to the reflection -x,y (with normal kx) of the reciprocal-plane group (c1m1), the representation domain is only one half of the hexagon: for example, in the acute case it is the trapezium with vertices Δ0, F0, F2, Δ2 (light blue boundary). The asymmetric unit is different from the representation domain: it is the rectangle with vertices J2, J4, V4, V2 (pink boundary), i.e. the points x,y : 0<= x <= 1/2; -1/4 <= y <= 1/4. While the representation domain of the acute and obtuse with cells have the more complicated form of a trapezium, the asymmetric units in both cases have the topologically identical and relatively simple shape of a rectangle.
The k-vector tables show all special wave-vectors with their coefficients and layer little co-groups as specified in Tables 24 and 25 of L&W. The wave-vector coefficients with respect to the conventional reciprocal basis, i.e. dual to the conventional centred basis, are listed in the column under the heading 'Conventional' of the k-vector tables. For example, a k-vector point of the DT line with primitive coefficients (-1/4,1/4) is described as (0,1/2) with respect to a basis dual to the conventional basis cm2m.
The point GM, Y2 and S (acute case) and GM, Y and S (obtuse case) are not special k-vector points but form part of special lines and planes and in the diagrams they are represented by open circles. The line SM is not a symmetry line and is represented by a think black line because it is located inside the Brillouin zone. The lines DT and DA are coloured in brown because they are symmetry lines and at the same time are edges of the asymmetric unit. Part of DT and DA are also coloured in red because they correspond to flagpoles (for a discussion of flagpoles the reader is referred to ). The k-vector lines F and FA (acute case) are coloured in dark blue as they are symmetry lines along the edges of the representation domain.
Because of the special shape of the Brillouin zone and the representation domain for the acute case (a > b), the special k-vector line corresponding to the Wyckoff-position block 2a splits into several segments: the lines DT and DA, located inside the Brillouin zone, and the lines F and FA (coloured dark blue) at the border of the Brillouin zone. For the description of the end points of the segments, it is necessary to introduce additional parameters as dt0 and f0 whose values depend on the specific relations between the lattice parameters. The use of flagpoles enable the uni-arm description: the flagpole [J2 Y2] 0,y : 1/4 < y < 1/2 is equivalent to the segment [V4 Y] 1/2,y: -1/4 < y < 0 and the flagpole [V2 Y4] 0,y : -1/2 < y < -1/4 is equivalent to the segment [Y J4] 1/2,y : 0 < y < 1/4. The uni-arm description of the k-vector type of the Wyckoff position 2a is shown in the last row of the Wyckoff-position block and it is formed by the union of the points GM and Y2, the lines DT, DA, DT1 (∼ FA) and DA1 (∼ F). Its parameter description (0,y) with y varying in the range (-1/2,1/2) coincides with that of the acute case. The parameter description of the flagpole and its parameter range with respect to the basis of the reciprocal group are given below the k-vector table.
Litvin, D. B. & Wike, T. R. (1991). Character Tables and Compatibility Relations of the Eighty Layer Groups and Seventeen Plane Groups. New York: Plenum Press. [Abbreviated as L&W].
International Tables for Crystallography, Volume E: Subperiodic Groups (2010). Edited by V. Kopsky & D. Litvin, 2nd edition. Chichester: Wiley. [Abbreviated as ITE].
de la Flor, G., Souvignier, B., Madariaga, G. & M. I. Aroyo. Acta Cryst A (in review).
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| Bilbao Crystallographic Server www.cryst.ehu.es | For comments, please mail to cryst@wm.lc.ehu.es |
As the asymmetric unit and the representation domain do not coincide, their edges are coloured in pink and light blue, respectively. It has already been pointed out that k-vector points and lines are brought out in red only if they are special k-vector points and lines. For example, the lines T and LD (see figure of p-62m) are not coloured as special lines since they belong to the k-vector type of the Wyckoff-position block 6d, and their symmetry coincides with that of the neighbouring points of the symmetry plane B=[GM K M]. The point GM and K, however, are represented by red circles as they are special k-vector points. Likewise, the line SM is coloured in brown because it is an edge of the asymmetric unit and at the same time is a special k-vector symmetry line. The k-vector line SM is coloured in dark blue as it is a special symmetry line along the edge of the representation domain. As already indicated, the k-vector line SN together with SM and (see figure of p-62m) the point M belong to the special k-vector type of the Wyckoff-position block 3c, i.e. all these different wavevectors belong to the same k-vector type. Although M is explicity listed by L&W as a special k-vector point, it is represented by an open circle. In fact, it joins the symmetry lines SM and SN1 to a continuous line as its little co-group type coincides with those of the points on the lines.
Brillouin-zone diagrams of the layer group cm2m (No. 35)
The layer group cm2m (No. 35) is an example of orthorhombic layer groups with a c-centred lattice. It belongs to the arithmetic crystal class m2mc which also includes the layer group cm2e (No. 36). The k-vector tables and Brillouin-zone figures belong to the arithmetic crystal class m2mc.
The Brillouin-zone of the layer groups of the arithmetic crystal class m2mc is a non-regular hexagon. Depending on the relation between the lattice parameters a and b, two topologically different Brillouin zones are to be distinguished:
the acute case with a > b
2. the obtuse case with a < b
Due to the reflection -x,y (with normal kx) of the reciprocal-plane group (c1m1), the representation domain is only one half of the hexagon: for example, in the acute case it is the trapezium with vertices Δ0, F0, F2, Δ2 (light blue boundary). The asymmetric unit is different from the representation domain: it is the rectangle with vertices J2, J4, V4, V2 (pink boundary), i.e. the points x,y : 0<= x <= 1/2; -1/4 <= y <= 1/4. While the representation domain of the acute and obtuse with cells have the more complicated form of a trapezium, the asymmetric units in both cases have the topologically identical and relatively simple shape of a rectangle.
The k-vector tables show all special wave-vectors with their coefficients and layer little co-groups as specified in Tables 24 and 25 of L&W. The wave-vector coefficients with respect to the conventional reciprocal basis, i.e. dual to the conventional centred basis, are listed in the column under the heading 'Conventional' of the k-vector tables. For example, a k-vector point of the DT line with primitive coefficients (-1/4,1/4) is described as (0,1/2) with respect to a basis dual to the conventional basis cm2m.
The point GM, Y2 and S (acute case) and GM, Y and S (obtuse case) are not special k-vector points but form part of special lines and planes and in the diagrams they are represented by open circles. The line SM is not a symmetry line and is represented by a think black line because it is located inside the Brillouin zone. The lines DT and DA are coloured in brown because they are symmetry lines and at the same time are edges of the asymmetric unit. Part of DT and DA are also coloured in red because they correspond to flagpoles (for a discussion of flagpoles the reader is referred to ). The k-vector lines F and FA (acute case) are coloured in dark blue as they are symmetry lines along the edges of the representation domain.
Because of the special shape of the Brillouin zone and the representation domain for the acute case (a > b), the special k-vector line corresponding to the Wyckoff-position block 2a splits into several segments: the lines DT and DA, located inside the Brillouin zone, and the lines F and FA (coloured dark blue) at the border of the Brillouin zone. For the description of the end points of the segments, it is necessary to introduce additional parameters as dt0 and f0 whose values depend on the specific relations between the lattice parameters. The use of flagpoles enable the uni-arm description: the flagpole [J2 Y2] 0,y : 1/4 < y < 1/2 is equivalent to the segment [V4 Y] 1/2,y: -1/4 < y < 0 and the flagpole [V2 Y4] 0,y : -1/2 < y < -1/4 is equivalent to the segment [Y J4] 1/2,y : 0 < y < 1/4. The uni-arm description of the k-vector type of the Wyckoff position 2a is shown in the last row of the Wyckoff-position block and it is formed by the union of the points GM and Y2, the lines DT, DA, DT1 (∼ FA) and DA1 (∼ F). Its parameter description (0,y) with y varying in the range (-1/2,1/2) coincides with that of the acute case. The parameter description of the flagpole and its parameter range with respect to the basis of the reciprocal group are given below the k-vector table.
Litvin, D. B. & Wike, T. R. (1991). Character Tables and Compatibility Relations of the Eighty Layer Groups and Seventeen Plane Groups. New York: Plenum Press. [Abbreviated as L&W].
International Tables for Crystallography, Volume E: Subperiodic Groups (2010). Edited by V. Kopsky & D. Litvin, 2nd edition. Chichester: Wiley. [Abbreviated as ITE].
de la Flor, G., Souvignier, B., Madariaga, G. & M. I. Aroyo. Acta Cryst A (in review).
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| Bilbao Crystallographic Server www.cryst.ehu.es | For comments, please mail to cryst@wm.lc.ehu.es |
As the asymmetric unit and the representation domain do not coincide, their edges are coloured in pink and light blue, respectively. It has already been pointed out that k-vector points and lines are brought out in red only if they are special k-vector points and lines. For example, the lines T and LD (see figure of p-62m) are not coloured as special lines since they belong to the k-vector type of the Wyckoff-position block 6d, and their symmetry coincides with that of the neighbouring points of the symmetry plane B=[GM K M]. The point GM and K, however, are represented by red circles as they are special k-vector points. Likewise, the line SM is coloured in brown because it is an edge of the asymmetric unit and at the same time is a special k-vector symmetry line. The k-vector line SM is coloured in dark blue as it is a special symmetry line along the edge of the representation domain. As already indicated, the k-vector line SN together with SM and (see figure of p-62m) the point M belong to the special k-vector type of the Wyckoff-position block 3c, i.e. all these different wavevectors belong to the same k-vector type. Although M is explicity listed by L&W as a special k-vector point, it is represented by an open circle. In fact, it joins the symmetry lines SM and SN1 to a continuous line as its little co-group type coincides with those of the points on the lines.
Brillouin-zone diagrams of the layer group cm2m (No. 35)
The layer group cm2m (No. 35) is an example of orthorhombic layer groups with a c-centred lattice. It belongs to the arithmetic crystal class m2mc which also includes the layer group cm2e (No. 36). The k-vector tables and Brillouin-zone figures belong to the arithmetic crystal class m2mc.
The Brillouin-zone of the layer groups of the arithmetic crystal class m2mc is a non-regular hexagon. Depending on the relation between the lattice parameters a and b, two topologically different Brillouin zones are to be distinguished:
the acute case with a > b
2. the obtuse case with a < b
Due to the reflection -x,y (with normal kx) of the reciprocal-plane group (c1m1), the representation domain is only one half of the hexagon: for example, in the acute case it is the trapezium with vertices Δ0, F0, F2, Δ2 (light blue boundary). The asymmetric unit is different from the representation domain: it is the rectangle with vertices J2, J4, V4, V2 (pink boundary), i.e. the points x,y : 0<= x <= 1/2; -1/4 <= y <= 1/4. While the representation domain of the acute and obtuse with cells have the more complicated form of a trapezium, the asymmetric units in both cases have the topologically identical and relatively simple shape of a rectangle.
The k-vector tables show all special wave-vectors with their coefficients and layer little co-groups as specified in Tables 24 and 25 of L&W. The wave-vector coefficients with respect to the conventional reciprocal basis, i.e. dual to the conventional centred basis, are listed in the column under the heading 'Conventional' of the k-vector tables. For example, a k-vector point of the DT line with primitive coefficients (-1/4,1/4) is described as (0,1/2) with respect to a basis dual to the conventional basis cm2m.
The point GM, Y2 and S (acute case) and GM, Y and S (obtuse case) are not special k-vector points but form part of special lines and planes and in the diagrams they are represented by open circles. The line SM is not a symmetry line and is represented by a think black line because it is located inside the Brillouin zone. The lines DT and DA are coloured in brown because they are symmetry lines and at the same time are edges of the asymmetric unit. Part of DT and DA are also coloured in red because they correspond to flagpoles (for a discussion of flagpoles the reader is referred to ). The k-vector lines F and FA (acute case) are coloured in dark blue as they are symmetry lines along the edges of the representation domain.
Because of the special shape of the Brillouin zone and the representation domain for the acute case (a > b), the special k-vector line corresponding to the Wyckoff-position block 2a splits into several segments: the lines DT and DA, located inside the Brillouin zone, and the lines F and FA (coloured dark blue) at the border of the Brillouin zone. For the description of the end points of the segments, it is necessary to introduce additional parameters as dt0 and f0 whose values depend on the specific relations between the lattice parameters. The use of flagpoles enable the uni-arm description: the flagpole [J2 Y2] 0,y : 1/4 < y < 1/2 is equivalent to the segment [V4 Y] 1/2,y: -1/4 < y < 0 and the flagpole [V2 Y4] 0,y : -1/2 < y < -1/4 is equivalent to the segment [Y J4] 1/2,y : 0 < y < 1/4. The uni-arm description of the k-vector type of the Wyckoff position 2a is shown in the last row of the Wyckoff-position block and it is formed by the union of the points GM and Y2, the lines DT, DA, DT1 (∼ FA) and DA1 (∼ F). Its parameter description (0,y) with y varying in the range (-1/2,1/2) coincides with that of the acute case. The parameter description of the flagpole and its parameter range with respect to the basis of the reciprocal group are given below the k-vector table.
Litvin, D. B. & Wike, T. R. (1991). Character Tables and Compatibility Relations of the Eighty Layer Groups and Seventeen Plane Groups. New York: Plenum Press. [Abbreviated as L&W].
International Tables for Crystallography, Volume E: Subperiodic Groups (2010). Edited by V. Kopsky & D. Litvin, 2nd edition. Chichester: Wiley. [Abbreviated as ITE].
de la Flor, G., Souvignier, B., Madariaga, G. & M. I. Aroyo. Acta Cryst A (in review).
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| Bilbao Crystallographic Server www.cryst.ehu.es | For comments, please mail to cryst@wm.lc.ehu.es |
Brillouin-zone diagrams of the layer group cm2m (No. 35)
The layer group cm2m (No. 35) is an example of orthorhombic layer groups with a c-centred lattice. It belongs to the arithmetic crystal class m2mc which also includes the layer group cm2e (No. 36). The k-vector tables and Brillouin-zone figures belong to the arithmetic crystal class m2mc.
The Brillouin-zone of the layer groups of the arithmetic crystal class m2mc is a non-regular hexagon. Depending on the relation between the lattice parameters a and b, two topologically different Brillouin zones are to be distinguished:
the acute case with a > b
2. the obtuse case with a < b
Due to the reflection -x,y (with normal kx) of the reciprocal-plane group (c1m1), the representation domain is only one half of the hexagon: for example, in the acute case it is the trapezium with vertices Δ0, F0, F2, Δ2 (light blue boundary). The asymmetric unit is different from the representation domain: it is the rectangle with vertices J2, J4, V4, V2 (pink boundary), i.e. the points x,y : 0<= x <= 1/2; -1/4 <= y <= 1/4. While the representation domain of the acute and obtuse with cells have the more complicated form of a trapezium, the asymmetric units in both cases have the topologically identical and relatively simple shape of a rectangle.
The k-vector tables show all special wave-vectors with their coefficients and layer little co-groups as specified in Tables 24 and 25 of L&W. The wave-vector coefficients with respect to the conventional reciprocal basis, i.e. dual to the conventional centred basis, are listed in the column under the heading 'Conventional' of the k-vector tables. For example, a k-vector point of the DT line with primitive coefficients (-1/4,1/4) is described as (0,1/2) with respect to a basis dual to the conventional basis cm2m.
The point GM, Y2 and S (acute case) and GM, Y and S (obtuse case) are not special k-vector points but form part of special lines and planes and in the diagrams they are represented by open circles. The line SM is not a symmetry line and is represented by a think black line because it is located inside the Brillouin zone. The lines DT and DA are coloured in brown because they are symmetry lines and at the same time are edges of the asymmetric unit. Part of DT and DA are also coloured in red because they correspond to flagpoles (for a discussion of flagpoles the reader is referred to ). The k-vector lines F and FA (acute case) are coloured in dark blue as they are symmetry lines along the edges of the representation domain.
Because of the special shape of the Brillouin zone and the representation domain for the acute case (a > b), the special k-vector line corresponding to the Wyckoff-position block 2a splits into several segments: the lines DT and DA, located inside the Brillouin zone, and the lines F and FA (coloured dark blue) at the border of the Brillouin zone. For the description of the end points of the segments, it is necessary to introduce additional parameters as dt0 and f0 whose values depend on the specific relations between the lattice parameters. The use of flagpoles enable the uni-arm description: the flagpole [J2 Y2] 0,y : 1/4 < y < 1/2 is equivalent to the segment [V4 Y] 1/2,y: -1/4 < y < 0 and the flagpole [V2 Y4] 0,y : -1/2 < y < -1/4 is equivalent to the segment [Y J4] 1/2,y : 0 < y < 1/4. The uni-arm description of the k-vector type of the Wyckoff position 2a is shown in the last row of the Wyckoff-position block and it is formed by the union of the points GM and Y2, the lines DT, DA, DT1 (∼ FA) and DA1 (∼ F). Its parameter description (0,y) with y varying in the range (-1/2,1/2) coincides with that of the acute case. The parameter description of the flagpole and its parameter range with respect to the basis of the reciprocal group are given below the k-vector table.
Litvin, D. B. & Wike, T. R. (1991). Character Tables and Compatibility Relations of the Eighty Layer Groups and Seventeen Plane Groups. New York: Plenum Press. [Abbreviated as L&W].
International Tables for Crystallography, Volume E: Subperiodic Groups (2010). Edited by V. Kopsky & D. Litvin, 2nd edition. Chichester: Wiley. [Abbreviated as ITE].
de la Flor, G., Souvignier, B., Madariaga, G. & M. I. Aroyo. Acta Cryst A (in review).
| | |
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| Bilbao Crystallographic Server www.cryst.ehu.es | For comments, please mail to cryst@wm.lc.ehu.es |
Brillouin-zone diagrams of the layer group cm2m (No. 35)
The layer group cm2m (No. 35) is an example of orthorhombic layer groups with a c-centred lattice. It belongs to the arithmetic crystal class m2mc which also includes the layer group cm2e (No. 36). The k-vector tables and Brillouin-zone figures belong to the arithmetic crystal class m2mc.
The Brillouin-zone of the layer groups of the arithmetic crystal class m2mc is a non-regular hexagon. Depending on the relation between the lattice parameters a and b, two topologically different Brillouin zones are to be distinguished:
the acute case with a > b
2. the obtuse case with a < b
Due to the reflection -x,y (with normal kx) of the reciprocal-plane group (c1m1), the representation domain is only one half of the hexagon: for example, in the acute case it is the trapezium with vertices Δ0, F0, F2, Δ2 (light blue boundary). The asymmetric unit is different from the representation domain: it is the rectangle with vertices J2, J4, V4, V2 (pink boundary), i.e. the points x,y : 0<= x <= 1/2; -1/4 <= y <= 1/4. While the representation domain of the acute and obtuse with cells have the more complicated form of a trapezium, the asymmetric units in both cases have the topologically identical and relatively simple shape of a rectangle.
The k-vector tables show all special wave-vectors with their coefficients and layer little co-groups as specified in Tables 24 and 25 of L&W. The wave-vector coefficients with respect to the conventional reciprocal basis, i.e. dual to the conventional centred basis, are listed in the column under the heading 'Conventional' of the k-vector tables. For example, a k-vector point of the DT line with primitive coefficients (-1/4,1/4) is described as (0,1/2) with respect to a basis dual to the conventional basis cm2m.
The point GM, Y2 and S (acute case) and GM, Y and S (obtuse case) are not special k-vector points but form part of special lines and planes and in the diagrams they are represented by open circles. The line SM is not a symmetry line and is represented by a think black line because it is located inside the Brillouin zone. The lines DT and DA are coloured in brown because they are symmetry lines and at the same time are edges of the asymmetric unit. Part of DT and DA are also coloured in red because they correspond to flagpoles (for a discussion of flagpoles the reader is referred to ). The k-vector lines F and FA (acute case) are coloured in dark blue as they are symmetry lines along the edges of the representation domain.
Because of the special shape of the Brillouin zone and the representation domain for the acute case (a > b), the special k-vector line corresponding to the Wyckoff-position block 2a splits into several segments: the lines DT and DA, located inside the Brillouin zone, and the lines F and FA (coloured dark blue) at the border of the Brillouin zone. For the description of the end points of the segments, it is necessary to introduce additional parameters as dt0 and f0 whose values depend on the specific relations between the lattice parameters. The use of flagpoles enable the uni-arm description: the flagpole [J2 Y2] 0,y : 1/4 < y < 1/2 is equivalent to the segment [V4 Y] 1/2,y: -1/4 < y < 0 and the flagpole [V2 Y4] 0,y : -1/2 < y < -1/4 is equivalent to the segment [Y J4] 1/2,y : 0 < y < 1/4. The uni-arm description of the k-vector type of the Wyckoff position 2a is shown in the last row of the Wyckoff-position block and it is formed by the union of the points GM and Y2, the lines DT, DA, DT1 (∼ FA) and DA1 (∼ F). Its parameter description (0,y) with y varying in the range (-1/2,1/2) coincides with that of the acute case. The parameter description of the flagpole and its parameter range with respect to the basis of the reciprocal group are given below the k-vector table.
Litvin, D. B. & Wike, T. R. (1991). Character Tables and Compatibility Relations of the Eighty Layer Groups and Seventeen Plane Groups. New York: Plenum Press. [Abbreviated as L&W].
International Tables for Crystallography, Volume E: Subperiodic Groups (2010). Edited by V. Kopsky & D. Litvin, 2nd edition. Chichester: Wiley. [Abbreviated as ITE].
de la Flor, G., Souvignier, B., Madariaga, G. & M. I. Aroyo. Acta Cryst A (in review).
| | |
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| Bilbao Crystallographic Server www.cryst.ehu.es | For comments, please mail to cryst@wm.lc.ehu.es |
Brillouin-zone diagrams of the layer group cm2m (No. 35)
The Brillouin-zone of the layer groups of the arithmetic crystal class m2mc is a non-regular hexagon. Depending on the relation between the lattice parameters a and b, two topologically different Brillouin zones are to be distinguished:
the acute case with a > b
2. the obtuse case with a < b
Due to the reflection -x,y (with normal kx) of the reciprocal-plane group (c1m1), the representation domain is only one half of the hexagon: for example, in the acute case it is the trapezium with vertices Δ0, F0, F2, Δ2 (light blue boundary). The asymmetric unit is different from the representation domain: it is the rectangle with vertices J2, J4, V4, V2 (pink boundary), i.e. the points x,y : 0<= x <= 1/2; -1/4 <= y <= 1/4. While the representation domain of the acute and obtuse with cells have the more complicated form of a trapezium, the asymmetric units in both cases have the topologically identical and relatively simple shape of a rectangle.
The k-vector tables show all special wave-vectors with their coefficients and layer little co-groups as specified in Tables 24 and 25 of L&W. The wave-vector coefficients with respect to the conventional reciprocal basis, i.e. dual to the conventional centred basis, are listed in the column under the heading 'Conventional' of the k-vector tables. For example, a k-vector point of the DT line with primitive coefficients (-1/4,1/4) is described as (0,1/2) with respect to a basis dual to the conventional basis cm2m.
The point GM, Y2 and S (acute case) and GM, Y and S (obtuse case) are not special k-vector points but form part of special lines and planes and in the diagrams they are represented by open circles. The line SM is not a symmetry line and is represented by a think black line because it is located inside the Brillouin zone. The lines DT and DA are coloured in brown because they are symmetry lines and at the same time are edges of the asymmetric unit. Part of DT and DA are also coloured in red because they correspond to flagpoles (for a discussion of flagpoles the reader is referred to ). The k-vector lines F and FA (acute case) are coloured in dark blue as they are symmetry lines along the edges of the representation domain.
Because of the special shape of the Brillouin zone and the representation domain for the acute case (a > b), the special k-vector line corresponding to the Wyckoff-position block 2a splits into several segments: the lines DT and DA, located inside the Brillouin zone, and the lines F and FA (coloured dark blue) at the border of the Brillouin zone. For the description of the end points of the segments, it is necessary to introduce additional parameters as dt0 and f0 whose values depend on the specific relations between the lattice parameters. The use of flagpoles enable the uni-arm description: the flagpole [J2 Y2] 0,y : 1/4 < y < 1/2 is equivalent to the segment [V4 Y] 1/2,y: -1/4 < y < 0 and the flagpole [V2 Y4] 0,y : -1/2 < y < -1/4 is equivalent to the segment [Y J4] 1/2,y : 0 < y < 1/4. The uni-arm description of the k-vector type of the Wyckoff position 2a is shown in the last row of the Wyckoff-position block and it is formed by the union of the points GM and Y2, the lines DT, DA, DT1 (∼ FA) and DA1 (∼ F). Its parameter description (0,y) with y varying in the range (-1/2,1/2) coincides with that of the acute case. The parameter description of the flagpole and its parameter range with respect to the basis of the reciprocal group are given below the k-vector table.
Litvin, D. B. & Wike, T. R. (1991). Character Tables and Compatibility Relations of the Eighty Layer Groups and Seventeen Plane Groups. New York: Plenum Press. [Abbreviated as L&W].
International Tables for Crystallography, Volume E: Subperiodic Groups (2010). Edited by V. Kopsky & D. Litvin, 2nd edition. Chichester: Wiley. [Abbreviated as ITE].
de la Flor, G., Souvignier, B., Madariaga, G. & M. I. Aroyo. Acta Cryst A (in review).
| | |
--- |
| Bilbao Crystallographic Server www.cryst.ehu.es | For comments, please mail to cryst@wm.lc.ehu.es |
The k-vector tables show all special wave-vectors with their coefficients and layer little co-groups as specified in Tables 24 and 25 of L&W. The wave-vector coefficients with respect to the conventional reciprocal basis, i.e. dual to the conventional centred basis, are listed in the column under the heading 'Conventional' of the k-vector tables. For example, a k-vector point of the DT line with primitive coefficients (-1/4,1/4) is described as (0,1/2) with respect to a basis dual to the conventional basis cm2m.
The point GM, Y2 and S (acute case) and GM, Y and S (obtuse case) are not special k-vector points but form part of special lines and planes and in the diagrams they are represented by open circles. The line SM is not a symmetry line and is represented by a think black line because it is located inside the Brillouin zone. The lines DT and DA are coloured in brown because they are symmetry lines and at the same time are edges of the asymmetric unit. Part of DT and DA are also coloured in red because they correspond to flagpoles (for a discussion of flagpoles the reader is referred to ). The k-vector lines F and FA (acute case) are coloured in dark blue as they are symmetry lines along the edges of the representation domain.
Because of the special shape of the Brillouin zone and the representation domain for the acute case (a > b), the special k-vector line corresponding to the Wyckoff-position block 2a splits into several segments: the lines DT and DA, located inside the Brillouin zone, and the lines F and FA (coloured dark blue) at the border of the Brillouin zone. For the description of the end points of the segments, it is necessary to introduce additional parameters as dt0 and f0 whose values depend on the specific relations between the lattice parameters. The use of flagpoles enable the uni-arm description: the flagpole [J2 Y2] 0,y : 1/4 < y < 1/2 is equivalent to the segment [V4 Y] 1/2,y: -1/4 < y < 0 and the flagpole [V2 Y4] 0,y : -1/2 < y < -1/4 is equivalent to the segment [Y J4] 1/2,y : 0 < y < 1/4. The uni-arm description of the k-vector type of the Wyckoff position 2a is shown in the last row of the Wyckoff-position block and it is formed by the union of the points GM and Y2, the lines DT, DA, DT1 (∼ FA) and DA1 (∼ F). Its parameter description (0,y) with y varying in the range (-1/2,1/2) coincides with that of the acute case. The parameter description of the flagpole and its parameter range with respect to the basis of the reciprocal group are given below the k-vector table.
Litvin, D. B. & Wike, T. R. (1991). Character Tables and Compatibility Relations of the Eighty Layer Groups and Seventeen Plane Groups. New York: Plenum Press. [Abbreviated as L&W].
International Tables for Crystallography, Volume E: Subperiodic Groups (2010). Edited by V. Kopsky & D. Litvin, 2nd edition. Chichester: Wiley. [Abbreviated as ITE].
de la Flor, G., Souvignier, B., Madariaga, G. & M. I. Aroyo. Acta Cryst A (in review).
| | |
--- |
| Bilbao Crystallographic Server www.cryst.ehu.es | For comments, please mail to cryst@wm.lc.ehu.es |
The point GM, Y2 and S (acute case) and GM, Y and S (obtuse case) are not special k-vector points but form part of special lines and planes and in the diagrams they are represented by open circles. The line SM is not a symmetry line and is represented by a think black line because it is located inside the Brillouin zone. The lines DT and DA are coloured in brown because they are symmetry lines and at the same time are edges of the asymmetric unit. Part of DT and DA are also coloured in red because they correspond to flagpoles (for a discussion of flagpoles the reader is referred to ). The k-vector lines F and FA (acute case) are coloured in dark blue as they are symmetry lines along the edges of the representation domain.
Because of the special shape of the Brillouin zone and the representation domain for the acute case (a > b), the special k-vector line corresponding to the Wyckoff-position block 2a splits into several segments: the lines DT and DA, located inside the Brillouin zone, and the lines F and FA (coloured dark blue) at the border of the Brillouin zone. For the description of the end points of the segments, it is necessary to introduce additional parameters as dt0 and f0 whose values depend on the specific relations between the lattice parameters. The use of flagpoles enable the uni-arm description: the flagpole [J2 Y2] 0,y : 1/4 < y < 1/2 is equivalent to the segment [V4 Y] 1/2,y: -1/4 < y < 0 and the flagpole [V2 Y4] 0,y : -1/2 < y < -1/4 is equivalent to the segment [Y J4] 1/2,y : 0 < y < 1/4. The uni-arm description of the k-vector type of the Wyckoff position 2a is shown in the last row of the Wyckoff-position block and it is formed by the union of the points GM and Y2, the lines DT, DA, DT1 (∼ FA) and DA1 (∼ F). Its parameter description (0,y) with y varying in the range (-1/2,1/2) coincides with that of the acute case. The parameter description of the flagpole and its parameter range with respect to the basis of the reciprocal group are given below the k-vector table.
Litvin, D. B. & Wike, T. R. (1991). Character Tables and Compatibility Relations of the Eighty Layer Groups and Seventeen Plane Groups. New York: Plenum Press. [Abbreviated as L&W].
International Tables for Crystallography, Volume E: Subperiodic Groups (2010). Edited by V. Kopsky & D. Litvin, 2nd edition. Chichester: Wiley. [Abbreviated as ITE].
de la Flor, G., Souvignier, B., Madariaga, G. & M. I. Aroyo. Acta Cryst A (in review).
| | |
--- |
| Bilbao Crystallographic Server www.cryst.ehu.es | For comments, please mail to cryst@wm.lc.ehu.es |
Because of the special shape of the Brillouin zone and the representation domain for the acute case (a > b), the special k-vector line corresponding to the Wyckoff-position block 2a splits into several segments: the lines DT and DA, located inside the Brillouin zone, and the lines F and FA (coloured dark blue) at the border of the Brillouin zone. For the description of the end points of the segments, it is necessary to introduce additional parameters as dt0 and f0 whose values depend on the specific relations between the lattice parameters. The use of flagpoles enable the uni-arm description: the flagpole [J2 Y2] 0,y : 1/4 < y < 1/2 is equivalent to the segment [V4 Y] 1/2,y: -1/4 < y < 0 and the flagpole [V2 Y4] 0,y : -1/2 < y < -1/4 is equivalent to the segment [Y J4] 1/2,y : 0 < y < 1/4. The uni-arm description of the k-vector type of the Wyckoff position 2a is shown in the last row of the Wyckoff-position block and it is formed by the union of the points GM and Y2, the lines DT, DA, DT1 (∼ FA) and DA1 (∼ F). Its parameter description (0,y) with y varying in the range (-1/2,1/2) coincides with that of the acute case. The parameter description of the flagpole and its parameter range with respect to the basis of the reciprocal group are given below the k-vector table.
Litvin, D. B. & Wike, T. R. (1991). Character Tables and Compatibility Relations of the Eighty Layer Groups and Seventeen Plane Groups. New York: Plenum Press. [Abbreviated as L&W].
International Tables for Crystallography, Volume E: Subperiodic Groups (2010). Edited by V. Kopsky & D. Litvin, 2nd edition. Chichester: Wiley. [Abbreviated as ITE].
de la Flor, G., Souvignier, B., Madariaga, G. & M. I. Aroyo. Acta Cryst A (in review).
| | |
--- |
| Bilbao Crystallographic Server www.cryst.ehu.es | For comments, please mail to cryst@wm.lc.ehu.es |
Litvin, D. B. & Wike, T. R. (1991). Character Tables and Compatibility Relations of the Eighty Layer Groups and Seventeen Plane Groups. New York: Plenum Press. [Abbreviated as L&W].
International Tables for Crystallography, Volume E: Subperiodic Groups (2010). Edited by V. Kopsky & D. Litvin, 2nd edition. Chichester: Wiley. [Abbreviated as ITE].
de la Flor, G., Souvignier, B., Madariaga, G. & M. I. Aroyo. Acta Cryst A (in review).
| | |
--- |
| Bilbao Crystallographic Server www.cryst.ehu.es | For comments, please mail to cryst@wm.lc.ehu.es |
| | |
--- |
| Bilbao Crystallographic Server www.cryst.ehu.es | For comments, please mail to cryst@wm.lc.ehu.es | |
1112 | https://faculty.uca.edu/kdooley/lab_titrations.pdf | Acid / Base Titrations 1 Acid / Base Titrations v051413_7pm Objectives: Determine the concentration of a base solution using an acid standard. Optional: Precipitate an ionic salt for percent yield determination using the standardized base or determine the concentration of an acid using the standardized base. Background: Carefully Read Sections 4.4 and 4.8 in Tro 2nd Edition The most common method for determining the concentration of a solution is via titration. A titration involves reacting a solution of unknown concentration against a known standard to reach an endpoint, often signaled by an abrupt color change in the reaction mixture. A standard is a compound whose composition is unchanged by light or heat and one that is not hygroscopic, meaning that it does not readily absorb water from the atmosphere. The titrant reagent is carefully added to the other reagent until an equivalence point is reached (p 158), the point where stoichiometric amounts of the reactants have combined and both are limiting. The endpoint of a titration typically occurs after a few drops of excess titrant are added, past the equivalence point. The majority of acid / base reactions involve colorless products and therefore the equivalence point and endpoint are not visually detectable. To allow for the identification of the endpoint, an indicator is added at the start of the reaction. An indicator (p 158) is a substance that changes color depending on the pH of the solution. pH is indicative of the amount of hydronium, H3O+, in solution. In this lab you will be standardizing (determining the concentration of) a sodium hydroxide solution by titrating it against an oxalic acid dihydrate standard according to the reaction: H2C2O4 · 2H2O (aq) + 2 NaOH (aq) à Na2C2O4 (aq) + 4 H2O (l) You will begin by weighing an amount of oxalic acid dihydrate and dissolving it in water. The amount of water need only be approximate as the amount of hydronium in solution will be determined by the mass of oxalic acid. To this solution you will add the indicator phenolphthalein. Phenolphthalein is colorless in the presence of excess hydronium but turns pink in the presence of excess hydroxide. Sodium hydroxide of approximately 0.1 M concentration will be the titrant added to the acid until your reach the endpoint. The mass of oxalic acid dihydrate, the volume of sodium hydroxide used, and the stoichiometric relationship of these two compounds will enable you to determine the concentration of the sodium hydroxide to an increased number of significant figures. Once you have determined the concentration of the sodium hydroxide solution, then you can use that sodium hydroxide for other tests. Your instructor might provide you with an additional procedure to perform additional titrations. Safety: Wear safety goggles. Wash your hands thoroughly before leaving the lab. Sodium hydroxide, Oxalic Acid Dihydrate, and Phenolphthalein • If ingested, immediately rinse mouth and drink plenty of water (200-300 mL) • If it comes in contact with your skin, remove contaminated clothing and wash before reuse. Wash the skin immediately with soap and water. • If it comes in contact with your eyes, rinse eyes for at least 15 minutes with water. Remove contact lenses if worn prior to rinsing. Disposal: Dispose of mixture in Erlenmeyer flasks down the drain. Dispose of excess NaOH down the drain with copious amounts of water. This is acceptable because the concentration is much less concentrated than that of Liquid Plumber Drain Cleaner. Acid / Base Titrations 2 Procedure I. Preparation of Oxalic Acid Dihydrate Solutions for Titration 1. Thoroughly wash 3 Erlenmeyer flasks with soap and tap water. 2. Rinse the flasks first using tap water a minimum of 3 times each. 3. Rinse the flasks second with distilled water a minimum of 3 times each. 4. Place a piece of weigh paper on the balance and hit the tare/zero button. 5. Using weigh paper, measure out a sample of oxalic acid dihydrate, approximately 0.2 – 0.25 g in mass. 6. Record the exact mass on the data sheet to the nearest 0.001 g. 7. Add the sample of oxalic acid dihydrate to the first Erlenmeyer flask and label that flask #1. Carefully rinse remaining residue from the weigh paper using distilled water from a water bottle. 8. Add approximately 50 mL of distilled water to the Erlenmeyer flask. 9. Mix until all of the oxalic acid dihydrate has dissolved. 10. Add 2 drops phenolphthalein to the Erlenmeyer flask. 11. Repeat steps 4-10 for the other two Erlenmeyer flasks, labeling the flasks #2 and #3 accordingly. II. Preparing the NaOH Solution for Titration 1. Thoroughly wash a buret and rinse it using distilled water. 2. Close the stopcock. 3. Add approximately 5 mL of NaOH to the buret. 4. Gently roll the buret back and forth to coat the sides of the buret. 5. Drain the rinse solution through the buret tip into a discard container. 6. Repeat steps 3-5. 7. Close the stopcock and add approximately 50 mL of NaOH solution to the buret. 8. Drain a small amount of NaOH solution through the buret tip into the discard container to remove all bubbles. 9. Record this volume of NaOH as your initial buret reading for the first experiment. III. Standardizing the NaOH Solution 1. Place Erlenmeyer flask #1 containing the oxalic acid dihydrate under the buret and lower the buret so that the buret tip is below the mouth of the flask. (see Figure 3.7 on page 90 of your text) 2. Add NaOH solution to the Erlenmeyer flask by swirling the flask with one hand and controlling the stopcock with the other. 3. As you approach the endpoint take particular care to add the NaOH solution dropwise. 4. Add NaOH solution to the flask until a pale pink color is obtained after one drop and persists for 30 seconds. 5. Have your instructor approve your result. 6. Record this volume as your final buret reading for the first experiment. 7. Discard the solution in the Erlenmeyer flask down the drain. 8. Refill the buret, using a funnel, to approximately 50 mL of NaOH. Record this volume as the initial reading for the 2nd experiment. 9. Repeat steps 1-8 for the other two Erlenmeyer flasks using flask #2 and #3 accordingly. Acid / Base Titrations 3 Acid / Base Titrations: Pre-Laboratory Assignment 1. Give at least two characteristics desirable in a primary standard. 2. Define equivalence point and titration end point. Do these two points in a titration occur when the same volume of titrant has been added? Explain. 3. A student standardized a NaOH solution using KHC8H4O4. Three different KHC8H4O4 samples were dissolved in water and titrated with the same NaOH solution, each to a phenolphthalein end point. The pertinent reaction is: KHC8H4O4(aq) + NaOH(aq) à NaKC8H4O4(aq) + H2O(l) Results from these three runs are summarized in the data table below. Complete the following summary table of calculated values using the given experimental measurements. 1 2 3 Mass of KHC8H4O4 titrated, g 0.452 0.470 0.442 Volume of NaOH(aq) required, mL 15.45 15.99 15.05 Number of moles of KHC8H4O4 titrated, mol _ _ _ Number of moles of NaOH required, mol _ _ _ Volume of NaOH solution required, L _ _ _ Molarity of NaOH solution, mol/L _ _ _ Mean molarity of NaOH solution, mol/L _ Show clear work for the following: a) Moles of KHC8H4O4 titrated for Run #1 b) Moles of NaOH required for Run #1 c) Molarity of NaOH as determined by Run #1 Acid / Base Titrations 4 3. Below is a set of pictorial diagrams representing five different stages of the reaction mixture during the titration of hydrochloric acid with sodium hydroxide as titrant. Water molecules, hydronium ions, chloride ions, and sodium ions in the various mixtures are shown. a) List the diagrams by letter in the order these mixtures would appear during the titration, beginning with the earliest and ending with the final mixture. earliest __ _ _ __ _ final mixture b) Answer the following questions about the pictures above. i) If phenolphthalein is used as the indicator, which of the solutions would be: Colorless _ Pink _ ii) Which diagram represents the endpoint of the titration? __ . Defend your choice. iii) Which diagram represents the equivalence point of the titration? __. Defend your choice. A B C D E Acid / Base Titrations 5 Data 1 2 3 4 Mass of H2C2O4 . 2H2O, g _ _ _ _ Initial buret reading, mL _ _ _ _ Final buret reading, mL _ _ _ _ Results 1 2 3 4 Moles of NaOH reacted, mol _ _ _ _ Volume of NaOH solution added, mL _ _ _ _ Volume of NaOH solution added, L _ _ _ _ Molarity of NaOH solution, mol/L _ _ _ _ Mean molarity of NaOH solution, mol/L _ Discuss your calculated molarities for Runs 1 – 3 with your instructor to determine if additional runs are necessary Acid / Base Titrations 6 Acid / Base Titrations: Post-Laboratory Questions 1. Sodium hydroxide solutions are best stored in tightly sealed containers to avoid exposure to atmospheric carbon dioxide. a) Sodium hydroxide readily absorbs carbon dioxide from the atmosphere. Write the balanced chemical reaction that occurs between carbon dioxide and sodium hydroxide (Hint p. 701 of Tro). b) Explain the effect on the concentration of your sodium hydroxide if you left it uncorked in an atmosphere where there was a significant carbon dioxide concentration. 2. Why was it acceptable to discard the final solution from your titration down the drain? 3. A student used a standardized 0.205 M NaOH solution to determine the mass percent of citric acid (H3C6H5O7) in a mixture of citric acid and inert potassium chloride. The pertinent reaction is H3C6H5O7(aq) + 3NaOH (aq) à Na3C6H5O7(aq) + 3H2O (l) Sample masses and titration data are given in the table below. Do the following calculations for each titration and enter your answers in this table. 1 2 3 Mass of mixture titrated, g 0.356 0.478 0.420 Volume of 0.205 M NaOH(aq) required, mL 19.52 26.18 23.20 Moles of citric acid titrated, mol _ _ _ Mass of citric acid in the sample, g _ _ _ Mass percent citric acid in the sample, % _ _ _ Mean percent citric acid, % _ Show clear calculations for Run #1 to receive full credit for your completed data table. This includes units and clear setup. |
1113 | https://stackoverflow.com/questions/37743021/finding-intersection-points-of-line-and-circle | geometry - Finding intersection points of line and circle - Stack Overflow
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Finding intersection points of line and circle
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Im trying to understand what this function does. It was given by my teacher and I just cant understands, whats logic behind the formulas finding x, and y coordinates. From my math class I know I my formulas for finding interception but its confusing translated in code. So I have some problems how they defined the formulas for a,b,c and for finding the coordinates x and y.
```
void Intersection::getIntersectionPoints(const Arc& arc, const Line& line) {
double a, b, c, mu, det;
std::pair xPoints;
std::pair yPoints;
std::pair zPoints;
//(m2+1)x2+2(mc−mq−p)x+(q2−r2+p2−2cq+c2)=0.
//a= m2;
//b= 2 (mc - mq - p);
//c= q2−r2+p2−2cq+c2
a = pow((line.end().x - line.start().x), 2) + pow((line.end().y - line.start().y), 2) + pow((line.end().z - line.start().z), 2);
b = 2 ((line.end().x - line.start().x)(line.start().x - arc.center().x)
+ (line.end().y - line.start().y)(line.start().y - arc.center().y)
+ (line.end().z - line.start().z)(line.start().z - arc.center().z));
c = pow((arc.center().x), 2) + pow((arc.center().y), 2) +
pow((arc.center().z), 2) + pow((line.start().x), 2) +
pow((line.start().y), 2) + pow((line.start().z), 2) -
2 (arc.center().x line.start().x + arc.center().y line.start().y +
arc.center().z line.start().z) - pow((arc.radius()), 2);
det = pow(b, 2) - 4 a c;
/ Tangenta na kružnicu /
if (Math::isEqual(det, 0.0, 0.00001)) {
if (!Math::isEqual(a, 0.0, 0.00001))
mu = -b / (2 a);
else
mu = 0.0;
// x = h + t ( p − h )
xPoints.second = xPoints.first = line.start().x + mu (line.end().x - line.start().x);
yPoints.second = yPoints.first = line.start().y + mu (line.end().y - line.start().y);
zPoints.second = zPoints.first = line.start().z + mu (line.end().z - line.start().z);
}
if (Math::isGreater(det, 0.0, 0.00001)) {
// first intersection
mu = (-b - sqrt(pow(b, 2) - 4 a c)) / (2 a);
xPoints.first = line.start().x + mu (line.end().x - line.start().x);
yPoints.first = line.start().y + mu (line.end().y - line.start().y);
zPoints.first = line.start().z + mu (line.end().z - line.start().z);
// second intersection
mu = (-b + sqrt(pow(b, 2) - 4 a c)) / (2 a);
xPoints.second = line.start().x + mu (line.end().x - line.start().x);
yPoints.second = line.start().y + mu (line.end().y - line.start().y);
zPoints.second = line.start().z + mu (line.end().z - line.start().z);
}
```
line
geometry
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asked Jun 10, 2016 at 8:02
dranecdranec
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Denoting the line's start point as A, end point as B, circle's center as C, circle's radius as r and the intersection point as P, then we can write P as
P=(1-t)A + tB = A+t(B-A) (1)
Point P will also locate on the circle, therefore
|P-C|^2 = r^2 (2)
Plugging equation (1) into equation (2), you will get
|B-A|^2t^2 + 2(B-A)\dot(A-C)t +(|A-C|^2 - r^2) = 0 (3)
This is how you get the formula for a, b and c in the program you posted. After solving for t, you shall obtain the intersection point(s) from equation (1). Since equation (3) is quadratic, you might get 0, 1 or 2 values for t, which correspond to the geometric configurations where the line might not intersect the circle, be exactly tangent to the circle or pass thru the circle at two locations.
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edited Jun 12, 2016 at 5:39
answered Jun 12, 2016 at 5:32
fangfang
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1114 | https://pubmed.ncbi.nlm.nih.gov/7000854/ | Histometric evaluation of periodontal surgery. II. Connective tissue attachment levels after four regenerative procedures - PubMed
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J Clin Periodontol
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. 1980 Jun;7(3):224-31.
doi: 10.1111/j.1600-051x.1980.tb01965.x.
Histometric evaluation of periodontal surgery. II. Connective tissue attachment levels after four regenerative procedures
J Caton,S Nyman,H Zander
PMID: 7000854
DOI: 10.1111/j.1600-051x.1980.tb01965.x
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Histometric evaluation of periodontal surgery. II. Connective tissue attachment levels after four regenerative procedures
J Caton et al. J Clin Periodontol.1980 Jun.
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The present study was undertaken to determine the effect of four periodontal regenerative procedures on the connective tissue attachment level. The procedures tested were: 1) the modified Widman flap procedure, 2) the modified Widman flap procedure combined with transplantation of previously frozen autogenous red marrow and cancellous bone, 3) the modified Widman flap procedure in combination with implantation of beta tricalcium phosphate, and 4) periodic root planing and soft tissue curettage. Eight adult Rhesus monkeys, divided into four equal groups, were used. Periodontal pockets were produced around contralateral teeth in a standardized manner. In each group of animals, the pockets on one side of the jaws were subjected to one of the above-mentioned surgical treatments, while the contralateral pockets remained as unoperated controls. Three weeks before surgery, a carefully designed plaque control program was instituted and continued until the animals were sacrificed 12 months after surgery. In histologic sections, linear measurements along the root surfaces were made from the cemento-enamel junction (CEJ) to the most apical cells of the junctional epithelium (JE). These measurements from operated and unoperated sites were then compared. The data revealed that healing following the four different regenerative procedures resulted in the reformation of an epithelial lining (long junctional epithelium) along the treated root surfaces, with no new connective tissue attachment.
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Histometric evaluation of periodontal surgery. I. The modified Widman flap procedure.Caton J, Nyman S.Caton J, et al.J Clin Periodontol. 1980 Jun;7(3):212-23. doi: 10.1111/j.1600-051x.1980.tb01964.x.J Clin Periodontol. 1980.PMID: 6933162
Effect of citric acid and lyophilized autologous plasma on healing following periodontal flap surgery in monkeys.Nasjleti CE, Caffesse RG, Castelli WA, Smith BA, Lopatin DE, Kowalski CJ.Nasjleti CE, et al.J Periodontol. 1987 Nov;58(11):770-9. doi: 10.1902/jop.1987.58.11.770.J Periodontol. 1987.PMID: 3480350
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Present status of the modified Widman flap procedure.Ramfjord SP.Ramfjord SP.J Periodontol. 1977 Sep;48(9):558-65. doi: 10.1902/jop.1977.48.9.558.J Periodontol. 1977.PMID: 409833 Review.No abstract available.
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Saxon Algebra 2 Homeschool: Online Textbook Help
Geometry Angle Relationships | Overview, Equations & Examples
Contributors: Rachel Mcconnell, Gerald Lemay
Author
RM
Author:
Rachel Mcconnell
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Instructor
Instructor:
Gerald Lemay
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Learn about angle relationships in geometry. Learn how to solve angle relationships and angle equations. Discover the angle terminologies with their equations. Updated: 11/21/2023
Table of Contents
Geometry Angle Relationships
How to Solve Angles
Angle Equations
Angle Equations Examples
Lesson Summary
Show
FAQ
How do you find the relationship of a geometric angle?
To find the angle relationship in geometry, first understand complementary, supplementary, and vertical angles. Complementary angles are two angles that add together to 90 degrees. Supplementary angles are two angles that add together to 180 degrees. Vertical angles are created by two lines intersecting and the vertical angles are the opposite angles that are equal to each other.
How do you find the degree of an angle?
Calculate the degree of an angle by using the angle relationships of complementary angles, supplementary angles, and vertical angles. Create an equation using the angle measurement and the angle relationship. Complementary angles add together to equal 90 degrees. Supplementary angles add together to equal 180 degrees. Vertical angles are set equal to each other.
What are angle relationships in geometry?
Angle relationships in geometry include the angle relationships for complementary angles, supplementary angles, and vertical angles. These angle relationships can help solve for missing angle measurements.
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Geometry Angle Relationships ----------------------------
Angles are critical in Geometry. They help us understand shapes, whether two or three-dimensional, and they have strong relationships within lines and other angles to solve for missing values. An angle in Geometry is created from two or more lines intersecting each other. Angles can be measured by hand using a protractor or the angle measurement can be solved by using angle relationships and equations. When using a protractor, an angle is measured by lining up one of the sides of the angles with the 0-degree mark of the protractor. From there, the angle can be measured by identifying where the other line of the angle matches up to. Notice, angle measurements range from 0 degrees to 360 degrees.
Angles are an important part of Geometry because they are used in every shape as well as helping shapes fit together. Today, we are discussing angle relationships with intersecting lines. Three important types of angles include supplementary, complementary, and vertical angles. Let's discover these angle relationships.
Supplementary Angles
Supplementary angles are two angles that have the sum of 180 degrees. This means that together they create a straight line because the angle measurement of a straight line is 180 degrees. An easy way to remember supplementary angles is the s in supplementary and the s in straight! These angles do not have to be next to each other as long as they add up to 180 degrees.
The two angles are supplementary angles because their sum is 180 degrees.
Complementary Angles
Complementary angles are two angles that have the sum of 90 degrees. These angles come together to make a right angle, or a corner. An easy way to remember supplementary angles is the c in complementary and the c in corner. These angles do not have to be next to each other as long as they add up to 90 degrees.
The two angles are complementary angles because their sum is 90 degrees.
Vertical Angles
Vertical angles are two angles that are equal to each other when two lines intersect. The vertical angles are created by the intersecting lines and they are opposite each other.
The vertical angle pairs are A and C, and B and D. They are opposite of each other and have equal angle measures.
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How to Solve Angles -------------------
When solving for missing angle measurements, the angle relationships of supplementary, complementary, and vertical angles are very helpful. For supplementary and complementary angles, the pair of angles add up to a specific number. For vertical angles, the angle measurements are set equal to each other since vertical angles are equal. Let's practice solving for the missing angle in each angle relationship.
Example One Find the missing angle measurement.
Solve for the missing angle in this right angle.
Solution
Step 1: Identify the angle relationship. The two angles are complementary angles because they make a right angle of 90 degrees.
Step 2: Solve for the missing angle. The given angle is 15 degrees. Subtract 15 from 90 because together these angles make a 90-degree right angle. 90−15=75.
Step 3: The measurement of the missing angle is 75 degrees.
Example Two Find the missing angle measurement.
Calculate the missing angle in this set of vertical angles.
Solution
Step 1: Identify the angle relationship. The two angles are vertical angles because the angles are created by two intersecting lines and the angles are opposite of each other.
Step 2: Since these angles are vertical angles, they are equal. The given angle is 60 degrees, which is equal to the missing angle. 60=x
Step 3: The measurement of the missing angle is 60 degrees.
Example Three Find the missing angle measurement.
Calculate the missing angle in this set of angles which create a straight line.
Solution
Step 1: Identify the angle relationship. The two angles are supplementary angles because they make a straight line of 180 degrees.
Step 2: The sum of both angles is 180 degrees. The given angle is 125 degrees, therefore subtract 125 from 180. 180−125=55.
Step 3: The measurement of the missing angle is 55 degrees.
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Angle Equations ---------------
Sometimes the problems get more difficult and it's not as easy as subtracting from 180 degrees. Setting up an angle equation using the angle relationships is an important skill when the difficulty increases. Angle equations can be used for the easy problems as well but they should definitely be used for the difficult problems in order to keep the problem organized.
Starting with supplementary angles, the idea of the angle equation is straightforward when thinking about the angle definition. The sum of two angles is equal to 180 degrees. Therefore, add the measurements of both angles together and set them equal to 180 degrees. Similar to complementary angles, add together the two complementary angle measurements and set them equal to 90 degrees. Last but not least, vertical angles are set equal to each other. Once the equation is set up, the problem can be solved like a normal equation. Let's try!
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Angle Equations Examples ------------------------
Example One Find the missing angle measurement.
Create an equation to find the value of x.
Solution
Step 1: Identify the angle relationship. The two angles are supplementary angles because they make a straight line of 180 degrees.
Step 2: Since the angles are supplementary angles, the sum of the angle measurements is 180 degrees. Add together the two angles to get the equation of (2 x−10)+30=180
Step 3: Solve for the equation. Start by adding together the like terms of -10 and 30. −10+30=20. Now the equation looks like 2 x+20=180. Next step, subtract 20 on both sides. 180−20=160. Now the equation looks like 2 x=160. Last step, divide both sides by 2. x=80
Step 4: The variable is solved but to solve for the angle plug in the value of x into the expression 2 x−10. 2(80)−10. To solve, multiply 2 and 80, 2⋅80=160, then subtract by 10, 160−10=150. The measurement of the missing angle is 150 degrees.
Example Two Find the missing angle measurement.
Create an equation to find the value of x.
Solution
Step 1: Identify the angle relationship. The two angles are complementary angles because they make a right angle of 90 degrees.
Step 2: Since the angles are complementary angles their sum is 90 degrees. Set up an equation by adding the two angles together and setting them equal to 90 degrees. (3 x+5)+40=90
Step 3: Solve the equation. First, combine the like terms: 5 and 40. 5+40=45. The new equation is 3 x+45=90. Next, subtract 45 on both sides. This removes the 45 to get x alone. 3 x=45. Last, divide by 3 on both sides. x=15
Step 4: The value of x=15. When putting this into the expression of the angle, 3 x+5 put 15 in for x. 3(15)+5. Solve by multiplying first then adding 5. 3⋅15=45, then 45+5=50. The missing angle is 50 degrees.
Example Three Find the missing angle measurement.
Create an equation to find the value of x.
Solution
Step 1: Identify the angle relationship. The two angles are vertical angles because the angles are created by two intersecting lines and the angles are opposite of each other.
Step 2: Set up an angle equation by setting the two angles equal to each other because they are vertical angles. The angle equation looks like, 6 x−30=120.
Step 3: To solve for the value of x, first add 30 on both sides. 120+30=150. The new equation looks like 6 x=150. To solve for x, divide by 6 on both sides. 150/6=25, therefore, the value of x is 25.
Step 4: The measurement of the missing angle is 120 degrees. This can be found because vertical angles are equal and the other angle is 120 degrees but it can also be found by plugging in the value of x into the expression, 6 x−30. This looks like 6(25)−30. To solve, first multiply 6 by 25. 6⋅25=150. Then subtract 150 by 30. 150−30=120. Therefore, the missing angle measurement is 120 degrees.
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Lesson Summary --------------
An angle in Geometry is created from two or more lines intersecting each other. Angles range from 0 to 360 degrees. There are a few special angle relationships in Geometry: supplementary angles, complementary angles, and vertical angles. Supplementary angles are two angles that have the sum of 180 degrees. Complementary angles are two angles that have the sum of 90 degrees. Vertical angles are two angles that are equal to each other when two lines intersect. An easy way to remember the difference between supplementary and complementary angles, because they both add up to a specific amount, is that the s in supplementary stands for a straight angle and the c in complementary stands for a corner angle, or a 90-degree right angle.
These angle relationships help to calculate the value of missing angles. Sometimes angle measurements can be represented by more complex expressions. With more complex angles, using angle equations is very useful. Angle equations are created by using the properties of angle relationships and setting up an equation. For complementary or supplementary angles, this would be adding together the two given angles and setting the sum equal to 90 for complementary angles and 180 for supplementary angles. For vertical angles, create an equation by setting the angles equal to each other because vertical angles are equal. Using these methods can make a complex problem into a simple problem.
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Video Transcript
Equations with Angle Relationships
There are some really delightful juice mixes, like orange and mango. How about peach and lemon? Okay, maybe not so delightful. Well, how about a mix of algebra and geometry?
In fact, a very nice mix of algebra and geometry happens when we identify geometric angle relationships and then express unknowns using algebraic equations. The angle relationships we will look at are supplementary angles, complementary angles, and vertical angles.
Supplementary Angles
To get this delightful mixture started, we'll look at a straight line and notice how it is a rotation of 180 o.
Supplementary angle: 180 degrees
Fact: Supplementary angles add to 180 o.
This means if we know one of two supplementary angles, we can easily calculate the other. In this example, the angles of 150 o and x are supplementary.
Find the angle
Then 150 + x = 180. Meaning, the angle x = 30 o.
Let's change the orientation of the straight line and have the unknown angle be some algebraic expression.
Finding x
The steps to solve for x:
Identify the angle relationship:
There's a straight line, and we see 150 o and 2 x are supplementary angles.
Write an equation:
150 + 2 x = 180; (supplementary angles add to 180 o)
Solve for the unknown
150 + 2 x - 150 = 180 - 150; (subtracting 150 from both sides)
2 x = 30; (simplifying)
2 x/2 = 30/2; (dividing both sides by 2)
x = 15; (simplifying)
Thus, x = 15 and the unknown angle is 2 x = 30 o.
What if the straight line is in a non-standard position and there are more lines?
Arbitrary orientation
The steps are still the same:
Identify the angle relationship:
The angles of 60 o and 4 x are supplementary.
Write an equation:
4 x + 60 = 180; (supplementary angles add to 180 o)
Solve for the unknown:
4 x + 60 - 60 = 180 -60; (subtract 60 from both sides)
4 x = 120; (simplify)
4 x/4 = 120/4; (divide both sides by 4)
x = 30; (simplify)
Thus, x = 30 and the unknown angle is 4 x = 120 o.
Continuing with our mix of geometry and algebra, we look at the relationship of complementary angles.
Complementary Angles
A little square in a figure indicates an angle of 90 o.
Fact: Complementary angles add to 90 o.
An angle of 90 degrees
Seeing a little square in a figure is a clue to look for complementary angles.
Finding the angle
We see x and 50 o are complementary. Thus, x = 40 o. Let's look at a problem that is a little more complicated.
Finding x
Using the steps:
Identify the angle relationship:
The angles of 35 o and 2 x - 5 are complementary.
Write an equation:
2 x - 5 + 35 = 90; (complementary angles add to 90 o)
Solve for the unknown:
2 x + 30 = 90; (-5 + 35 equals 30)
2 x + 30 - 30 = 90 - 30; (subtract 30 from both sides)
2 x = 60; (simplify)
2 x/2 = 60/2; (divide both sides by 2)
x = 30; (simplify)
Thus, x = 30 and the unknown angle is 2 x - 5 = 55 o.
There is one more juicy angle relationship to look at.
Vertical Angles
When two lines cross, the lines form pairs of equal angles. One pair looks like this:
Equal angles
The same crossing lines also form another equal pair:
The other equal angles
These types of angles are called vertical angles.
Fact: Vertical angles are the angles opposite each other when two lines cross. These angles are congruent. Angles that are equal are said to be congruent.
For example, find the angle labeled x:
Finding x
Two crossings lines create vertical angles. Thus, x = 35 o.
A Review Example: In the following figure, find the values for x, y, and z.
Finding x, y, and z
First of all, this figure seems to have letters everywhere. All places except where the lines meet at the origin. We will refer to this place as ''O''.
Remembering Step 1: Identify the angle relationship. We see a right-angle at AOE . This short-hand, ''AOE'', means the vertex is at O and the angle is formed by the line segments AO and OE. There may be a complementary angle relationship we can use. The angles x at EOF and 20 o are complementary. Thus, x + 20 = 90, and voila!
x = 70 o.
How about vertical angles? Look for lines that cross. Sure, the line AD crosses the line CF. The opposite angles are equal. Thus, y at COD equals 20 o.
See the line AD? The angle relation to identify is supplementary angles add to 180 o. Thus,
y + z + (5 y + 10) = 180, and now we substitute and solve.
20 + z + 5(20) + 10 = 180; having substituted 20 for y
130 + z = 180; having combined 20 + 5(20) + 10 to give 20 + 100 + 10, which is 130
z = 50 o; having subtracted 130 from both sides
See how a complicated looking geometrical figure can be solved using angle relationships and algebra?
Thus, the question of how to mix geometry and algebra has been answered. Now, we can go back to finding an alternate mix for peaches and lemons.
Lesson Summary
Using algebra and angle relationships, we can solve for unknown variables and unknown angles. First, we must identify the angle relationship. Next, we must write an equation. Then we solve for the unknown and simplify. The angle relationships include:
Supplementary angles are those which add to 180 o.
Complementary angles are those which add to 90 o.
Vertical angles are the angles opposite each other when two lines cross. These angles are equal. Congruent angles are equal.
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Saxon Algebra 2 Homeschool: Online Textbook Help
38 chapters
297 lessons
Chapter 1
Saxon Algebra 2: Decimal Numbers
Repeating Decimal Definition, Example & Symbol 4:54 min
Decimal to Fraction | Conversion & Examples 7:32 min
Scientific Notation | Definition, Conversion & Examples 6:49 min
Comparing Numbers Written in Scientific Notation 5:15 min
How to Multiply & Divide in Scientific Notation 7:31 min
Using Scientific Notation to Approximate
Using Scientific Notation on a Calculator | Overview & Steps 6:06 min
Using Scientific Notation in Ideal Gas Law Problems
Chapter 2
Saxon Algebra 2: Graphs on the Coordinate Plane
Distance Formula | Overview & Examples 5:27 min
Solving Linear Equations by Substitution
System of Equations in Algebra | Overview, Methods & Examples 8:39 min
Slope | Definition, Formula & Examples 7:10 min
Slope of a Line | Meaning, Calculation & Examples 6:19 min
Y-Intercept | Definition, Formula & Examples 2:57 min
Horizontal Line | Definition, Equations & Examples 5:10 min
Vertical Line | Slope & Equation 2:55 min
Parallel & Perpendicular Lines | Equation, Graph & Examples 6:07 min
Graphing Consistent, Inconsistent, Dependent & Independent System
How to Graph a Circle | Equation & Examples 8:32 min
Defining and Graphing Ellipses in Algebra 5:00 min
Hyperbola | Definition, Equation & Graphs 10:00 min
Parabola | Definition, Formula & Examples 8:33 min
Chapter 3
Saxon Algebra 2: Sets
Set Notation | Concept & Examples 6:44 min
Subsets of Real Numbers | Overview & Practice Problems
Venn Diagram | Uses, Sets & Symbols 6:01 min
Cardinality of a Set | Definition & Examples 4:13 min
Chapter 4
Saxon Algebra 2: Real Numbers
Types of Numbers & Its Classifications 6:56 min
What is Addition? | Definition, Operation & Examples 4:07 min
How to Perform Subtraction: Steps & Examples 3:46 min
How to Perform Multiplication: Steps & Examples 5:22 min
How to Perform Division: Steps & Examples 3:56 min
Absolute Value | Explanation & Examples 4:42 min
Division and Reciprocals of Rational Expressions 5:09 min
Commutative Property of Addition: Examples | What is the Commutative Property? 5:02 min
Commutative Property of Multiplication | Definition & Formula 4:40 min
Distributive Property in Math | Definition & Examples 6:20 min
Multiplication Property of Equality | Overview, Example & Formula 4:05 min
Additive Property of Inequality | Definition & Examples
Chapter 5
Saxon Algebra 2: Complex Numbers
Imaginary Numbers | Definition, History & Examples 8:40 min
Complex Numbers | Standard Form & Examples 4:54 min
How to Add, Subtract and Multiply Complex Numbers 5:59 min
Complex Numbers & Conjugates | Multiplication & Division 6:40 min
Complex Conjugate | Definition & Examples 7:45 min
Finding Complex Roots of Quadratic Equations 5:44 min
Chapter 6
Saxon Algebra 2: Conversion by Unit Multipliers
Metric Unit for Volume | Conversion & Examples 4:31 min
English System of Measurement | List, Advantages & Disadvantages 9:27 min
How to Convert Units in the English System of Measurement 5:34 min
How to Convert Rate Units in the English System of Measurement
Multiple Unit Multipliers: English System of Measurement
How to Convert Units in the Metric System 6:06 min
Multiple Unit Multipliers: Metric System of Measurement
Common Unit Conversions 4:13 min
Chapter 7
Saxon Algebra 2: Ratio & Proportion
How to Solve Ratio Word Problems 5:28 min
Solving Ratio Problems Involving Totals 5:55 min
Ideal Gas Equation | Derivations & Formula 4:26 min
Linear Scale Factor | Shapes, Area & Volume 7:33 min
How to Use Proportions With Chemical Compounds
Chapter 8
Saxon Algebra 2: Percent
Percentage | Definition & Calculation 4:20 min
Percent Change | Formula & Examples
Solve Problems Using Percents 7:50 min
Using Percentages in Chemical Weight Problems
How to Visualize Percents Using Diagrams
How to Use Percents Greater Than 100 5:12 min
Percentage Increase & Decrease | Formula & Examples 5:06 min
Chapter 9
Saxon Algebra 2: Rate
How to Change Rates Using Multipliers
How to Solve Uniform Motion Problems 6:13 min
Related Rates in Calculus | Problems, Formulas & Uses 9:11 min
Velocity and the Rate of Change 2:54 min
Chapter 10
Saxon Algebra 2: Exponents
Exponential Expressions & The Order of Operations 4:36 min
Simplifying Expressions with Exponents | Overview & Examples 4:52 min
Simplifying Powers of Fractions | Process, Rules & Examples 4:58 min
Product Theorem for Exponents: Definition & Examples
Exponents with Negative Bases | Overview, Formula & Examples 4:03 min
Negative Exponents: Writing Powers of Fractions and Decimals 3:55 min
Solving Equations With Exponents
Variables As Exponents: Practice Problems
Exponential Growth & Decay | Formula, Function & Graphs 8:41 min
Fractional Exponents | Definition, Rules & Examples 6:38 min
Simplifying a Sum Raised to a Power
Factoring Expressions With Exponents | Steps & Examples 6:09 min
Chapter 11
Saxon Algebra 2: Exponents on a Scientific Calculator
What Is a Scientific Calculator? 4:22 min
How to Use the y^x Calculator Key
Using Exponents on a Scientific Calculator 4:29 min
Chapter 12
Saxon Algebra 2: Roots
Negative Square Root | Definition & Examples 8:13 min
Product of Square Roots Rule: Definition & Example 4:57 min
Addition and Subtraction Using Radical Notation 3:08 min
How to Multiply Radical Expressions 6:35 min
Solving Radical Equations | Overview & Examples 6:48 min
Simplifying Square Roots | Overview & Examples 4:49 min
How to Rationalize the Denominator with a Radical Expression 3:52 min
How to Simplify Roots of Roots
How to Convert Roots to Fractional Exponents 3:48 min
Euler's Formula for Complex Numbers | Conversions & Examples
Evaluating Roots With a Scientific Calculator 4:37 min
Chapter 13
Saxon Algebra 2: Probability
Conditional Probability | Overview, Calculation & Examples 5:10 min
Independent & Dependent Events | Overview, Probability & Examples 12:06 min
Combinations in Probability | Equation, Formula & Calculation 11:00 min
Permutation Definition, Formula & Examples 6:58 min
Chapter 14
Saxon Algebra 2: Statistics
Creating & Reading Stem & Leaf Displays 4:27 min
Creating & Interpreting Histograms: Process & Examples 5:43 min
Box Plot | Definition, Uses & Examples 6:29 min
Measures of Central Tendency: Definitions & Practice 5:25 min
Normal Distribution | Curve, Table & Examples 11:40 min
Standard Deviation Equation, Formula & Examples 13:05 min
Chapter 15
Saxon Algebra 2: Simplifying Expressions
How to Simplify Expressions with Integers 5:12 min
Simplifying and Solving Exponential Expressions 7:27 min
Negative Exponents Definition, Rules & Examples 4:29 min
Negative Signs and Simplifying Algebraic Expressions 9:38 min
Practice Simplifying Algebraic Expressions 8:27 min
Fractional Exponents | Definition, Rules & Examples
Evaluating Expressions | Definition & Examples
Distributive Property & Algebraic Expressions | Rules & Examples 5:04 min
How to Simplify an Expression with Parentheses & Exponents 8:07 min
Combining Like Terms in Algebraic Expressions 7:04 min
Least Common Multiple | LCM Overview & Examples 10:56 min
How to Simplify Radical Expressions With Addition
Multiplying then Simplifying Radical Expressions 3:57 min
Conjugate in Math | Definition & Examples 4:51 min
Chapter 16
Saxon Algebra 2: Polynomials
Monomial | Definition, Components & Examples 6:04 min
Binomial Definition, Calculation & Examples 5:12 min
Difference of Two Squares | Definition & Factoring
Binomials: Sum and Difference of Two Cubes
Factoring Trinomials | Steps & Examples
Common Factors | Factoring & Examples 5:05 min
Trinomials: Lead Coefficients Greater Than One
Degree of a Polynomial | Definition, Function & Examples
Adding, Subtracting & Multiplying Polynomials | Steps & Examples 6:53 min
Polynomial Long Division | Overview & Examples 8:05 min
Polynomial Division: Missing Dividends
Dividing Polynomials | Calculation & Examples
Chapter 17
Saxon Algebra 2: Simplifying Rational Expressions
Simplifying Complex Rational Expressions | Steps & Examples 4:37 min
Adding & Subtracting Rational Expressions | Overview & Examples 8:02 min
Practice Adding and Subtracting Rational Expressions 9:12 min
How to Multiply and Divide Rational Expressions 8:07 min
Multiplying and Dividing Rational Expressions: Practice Problems 4:40 min
Simplifying Rational Expression | Overview, Steps & Examples
Rationalizing the Denominator | Overview & Examples 7:01 min
Complex Fractions | Definition, Simplification & Examples
Simplifying Complex Numbers: Addition of Like Terms
Simplifying Complex Numbers: Euler's Notation
Simplifying Imaginary & Complex Numbers | Definition & Examples 5:12 min
Simplifying Complex Numbers With Multiple Steps
How to Simplify Complex Numbers with Multiplication & Division
Chapter 18
Saxon Algebra 2: Simplifying and Solving Equations
Defining, Translating, & Solving One-Step Equations 6:15 min
Simplifying & Solving Equivalent Equations | Definition & Example 5:49 min
Solving Equations Using the Addition Principle 5:20 min
Multiplication Principle | Definition, Equations & Examples 4:03 min
How to Use the Fractional-Part-of-a-Number Equation
Abstract Equation: Definition & Concept
Simplifying & Solving Equations With Decimals & Mixed Numbers
Solving Equations Using the Least Common Multiple
Percent Equation | Definition, Formula & Calculation 7:21 min
Solving Addition Equations with Two or More Variables 7:57 min
Solving Subtraction Equations with Two or More Variables 6:28 min
Solving Multiple Step Equations | Explanation, Steps & Examples 5:44 min
How to Solve Multi-Step Equations | Definition, Steps & Examples 5:56 min
Solving Equations with Distributive Property | Formula & Examples
Translating Words into Algebraic Expressions | Phrases & Examples 6:31 min
Direct & Inverse Variation | Equations, Relationships & Problems 6:27 min
Solving Equations Involving Variation
Rational Equations | Definition, Formula & Examples 7:58 min
Solving Radical Equations with Two Radical Terms 6:00 min
Chapter 19
Saxon Algebra 2: Linear Equations
Finding Linear Equations to Fit Experimental Data
Find the Slope of a Line | Formula & Examples 9:27 min
Equation of a Line Using Point-Slope Formula 9:27 min
How to Find Equations of Lines With Given Slopes & Points
Horizontal & Vertical Line | Slope, Graph & Equation 5:10 min
Forms of a Linear Equation | Overview, Graphs & Conversion 6:38 min
Undefined & Zero Slope Graph | Definition & Examples 4:23 min
Rearranging Linear Equations Before Graphing
Chapter 20
Saxon Algebra 2: Solving Linear Equations
Solving Linear Equations With Substitution
How to Solve Problems with the Elimination in Algebra: Examples
Viewing now Geometry Angle Relationships | Overview, Equations & Examples 7:38 min
Up next Solving Simultaneous Linear Equations 5:33 min Watch next lesson
How to Solve Linear Equations By Graphing
Inconsistent and Dependent Systems: Using Gaussian Elimination 6:43 min
How to Solve a Linear System in Three Variables With a Solution 5:01 min
Chapter 21
Saxon Algebra 2: Quadratic Equations
Quadratic Equation | Definition, Formula & Examples 5:13 min
How to Solve a Quadratic Equation by Factoring 7:53 min
How to Use the Difference of Two Squares Theorem to Solve Quadratic Equations
How to Complete the Square | Method & Examples 8:43 min
Quadratic Function | Formula, Equations & Examples 9:20 min
How to Identify Lead Coefficients in Quadratic Equations
Discriminant Formula, Rules & Solutions 4:12 min
Chapter 22
Saxon Algebra 2: Other Types of Equations
Exponential Equations | Definition, Solutions & Examples 6:18 min
Exponential Population Growth | Formula, Calculation & Examples 9:50 min
Lead Coefficients of Completing the Square
Complex vs. Irrational Roots | Graph & Formula 7:14 min
Roots of a Quadratic Equation | Overview, Function & Formula 5:20 min
How to Solve Quadratics That Are Not in Standard Form 6:14 min
Simple Interest Definition, Formula & Examples 8:46 min
Compounding Interest | Formula, Types & Examples 7:45 min
Finding Compound Interest With a Calculator 7:51 min
Markup & Markdown | Purpose, Formulas & Examples 6:17 min
How to Solve Problems with Money 8:29 min
Using Equations to Solve Age Problems in Math 6:02 min
Using Equations to Solve Chemical Mixture Problems
Types of Parabolas | Overview, Graphs & Examples 6:15 min
Chapter 23
Saxon Algebra 2: Understanding Functions
Domain & Range of a Function | Definition, Equation & Examples 8:32 min
Function Notation Definition, Evaluation & Examples 9:26 min
Vertical Line Test | Definition, Function & Examples 5:42 min
Representing Functions as Ordered Pairs
Chapter 24
Saxon Algebra 2: Manipulating and Evaluating Functions
How to Add, Subtract, Multiply and Divide Functions 6:43 min
Exponential Function | Definition, Equation & Examples 7:24 min
Trigonometric Functions | Definition, Formula & Examples 6:40 min
Chapter 25
Saxon Algebra 2: Trigonometry
Trig Functions | Sine, Cosine & Tangent 7:26 min
Evaluating Trigonometric Functions With a Scientific Calculator
Solving Right Triangles | Overview, Methods & Examples
Using Trigonometry to Work With Vectors 5:38 min
Chapter 26
Saxon Algebra 2: Logarithms
Logarithmic Properties | Product, Power & Quotient Properties 5:11 min
Antilog | Definition & Scientific Calculator Procedures 6:21 min
Chapter 27
Saxon Algebra 2: Lines, Points, Segments, and Planes
Intersecting Lines | Definition, Properties & Examples 2:44 min
Parallel vs Perpendicular vs Transverse Lines Overview & Examples 6:06 min
Skew Lines | Definition, Conditions & Distance 3:41 min
Perpendicular Bisector Theorem | Converse & Examples 6:41 min
Distance Between Two Points | Formula, Calculation & Examples 4:37 min
Line Segments & Rays | Differences & Measurement 3:59 min
Identifying Proportional Line Segments
Line Segment Bisection & Midpoint Theorem: Geometric Construction 4:39 min
Lines & Planes in Space
Chapter 28
Saxon Algebra 2: Angles
How to Identify the Vertex of an Angle
Angles in Math | Definition & Types 2:56 min
Complementary Angles | Theorems & Examples 4:24 min
Supplementary Angle | Definition, Properties, Theorem & Examples 4:29 min
Adjacent Angles | Definition, Properties & Examples 3:21 min
Angle Relationships | Overview & Types 10:28 min
Reflex Angles | Definition, Types & Facts 5:45 min
Remote Interior vs. Exterior Angle | Definition & Examples 4:47 min
How to Find the Measure of an Inscribed Angle 5:09 min
Constructing an Angle Bisector in Geometry 3:36 min
Angles in a Polygon | Measurement, Formula & Examples 6:00 min
Using Angles With Vectors
Defining Negative Angles on the Coordinate Plane 4:41 min
Measure of an Arc: Process & Practice 4:51 min
Chapter 29
Saxon Algebra 2: Polygons
Characteristics of a Polygon | Overview & Examples 4:25 min
Equilateral vs. Equiangular Polygons | Definition & Shapes 5:05 min
Triangles in Geometry | Definition, Types & Formulas 4:30 min
Quadrilateral | Definition & Examples 4:45 min
What is a Square? | Definition, Properties & Examples 6:52 min
Trapezoid | Definition, Types & Attributes 4:24 min
Parallelograms | Definition, Properties & Theorems 5:20 min
Rhombus | Definition, Properties & Measurements 4:24 min
Rectangles: Definition, Properties & Construction 4:08 min
Pentagon | Overview, Area & Perimeter 3:24 min
Hexagon | Definition, Shape & Examples 3:28 min
Congruent Polygons | Definition, Tests & Examples 3:01 min
How to Translate, Rotate & Reflect Polygons
Polygon Symmetry | Definition, Shapes & Lines 4:06 min
Diagonals of a Polygon | Formula & Examples 4:49 min
Chapter 30
Saxon Algebra 2: Circles
Diameter of a Circle | Definition, Formula & Examples 3:12 min
Radius of a Circle Formula & Example | How to Find the Radius of a Circle 2:22 min
Chord of a Circle | Formula, Length & Examples 5:39 min
Arc Length & Sector Area | Definition, Formula & Examples 6:39 min
Central and Inscribed Angles: Definitions and Examples 6:32 min
Tangent of a Circle | Definition, Formula & Examples 3:52 min
Circumscribed & Inscribed Circles | Definition & Drawing 5:00 min
Measurements of Angles Involving Tangents, Chords & Secants 6:59 min
Measurements of Lengths Involving Tangents, Chords and Secants 5:44 min
Chapter 31
Saxon Algebra 2: Triangles
How to Classify Triangles | Overview & Examples 5:44 min
Special Right Triangles | Definition, Types & Examples 6:12 min
Congruent Triangles | Definition, Parts & Examples 5:19 min
Degrees in a Triangle | Measurement & Examples 5:14 min
Similar Triangles | Definition, Application Problems & Examples 6:23 min
Overlapping Triangles | Overview, Proofs & Examples
Chapter 32
Saxon Algebra 2: Geometric Solids
Volume & Surface Area of a Cylinder | Formula & Examples 5:09 min
Prisms: Definition, Area & Volume 6:12 min
Volume of a Cone | Definition, Formula & Examples 8:59 min
Volume & Surface Area of a Sphere | Formula & Examples 4:19 min
Chapter 33
Saxon Algebra 2: Perimeter and Circumference
Finding the Perimeter of Polygons 5:19 min
What is Pi? | Number & Examples 4:55 min
Finding the Area & Circumference of a Circle 7:24 min
Semicircle | Definition, Area & Formula 5:57 min
Chapter 34
Saxon Algebra 2: Area
Area of Triangles & Rectangles | Formula, Calculation & Examples 5:43 min
Area of a Regular Polygon | Formula & Examples 4:15 min
Area of Complex Figures 6:30 min
Chapter 35
Saxon Algebra 2: Surface Area and Volume
Surface Area of a Cylinder | Formula, Calculation & Examples 4:26 min
Surface Area of a Cube & Rectangular Prism | Definition & Formula 4:08 min
Volume of Prisms & Pyramid | Types, Formula & Calculation 6:15 min
Volume of Cylinders, Cones & Spheres | Formula & Examples 7:50 min
Finding Surface Area of Figures with Complex Shapes
Chapter 36
Saxon Algebra 2: Constructions
How to Construct Angles Using a Compass & Straight Edge
Perpendicular Bisectors | Overview, Construction & Theorem
Constructing Triangles: Types of Geometric Construction 5:59 min
Geometric Constructions Using Lines and Angles 4:32 min
Chapter 37
Saxon Algebra 2: Postulates & Pythagorean Theorem
Euclid's Axiomatic Geometry: Developments & Postulates 5:58 min
Pythagorean Theorem | Definition, Method & Examples 6:05 min
Pythagorean Theorem | Overview, Formula & Examples 7:33 min
Finding Distance with the Pythagorean Theorem 6:54 min
Chapter 38
Saxon Algebra 2: Proofs
Geometric Proof Types & Formats | What is a Proof in Geometry? 8:35 min
Proofs for Lines, Segments or Rays 3:37 min
Geometry Angle Proofs & Theorems 5:36 min
Proofs for Circles
Geometric Proofs for Polygons 6:34 min
Parallelogram | Proofs, Theorems & Formulas 6:20 min
Rhombus | Angles, Sides & Proofs 5:15 min
Proofs & Angles of an Isosceles Trapezoid | Overview & Diagram 5:26 min
Triangle Proofs | Congruence Theorems & Strategies 6:52 min
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Bring about Sentence Examples
He, however, protected the royal family against the violence of the mob and, on the 7th of August, even attempted to bring about a reconciliation, but his efforts were frustrated by Marie Antoinette.
Other variations in the mode of growth or budding bring about further differences in the building up of the colony, which are not in all cases properly understood and cannot be described in detail here.
The downward pull of gravity suffices to bring about the fall of such material, but the path it will follow and the distance it will travel before coming to rest depend upon the land form.
The proprietors struggled in vain to bring about a closer union.
The plea of the last named on behalf of Corsica served to enlist the sympathy of Napoleon in his wider speculations, and so helped to bring about that mental transformation which merged Buonaparte the Corsican in Bonaparte the Jacobin and Napoleon the First Consul and Emperor.
He knew from his English experiences that such a veto would be hardly ever used unless the king felt the people were on his side, and that if it were used unjustifiably the power of the purse possessed by the representatives of the people would, as in England in 1688, bring about a bloodless revolution.
A serious outbreak took place at Adrianople in 1804, where 20,000 of the new troops had been sent, ostensibly to put down the revolt in Servia, but really to try to bring about the reform of the European provinces.
Napoleon therefore came early to the conclusion that he must bring about a concentration of his seagoing fleet in the Channel, which would give him a temporary command of its waters.
He represented the antiFranco-Prussian portion of her council, and his object was to bring about an Anglo-Austro-Russian alliance which, at that time, was undoubtedly Russia's proper system, Hence the reiterated attempts of Frederick the Great and Louis XV.
The confederates, thereupon, appealed for help abroad and contributed to bring about a war between Russia and Turkey.
Further to strengthen their position, Pretorius and his party unsuccessfully endeavoured to bring about a union with the Orange Free State.
On paper the scheme had everything to recommend it as the expedient most likely to bring about the desired end.
The result was to bring about the deposition and banishment of the Monophysites from the latter city.
When the trigeminus nerve is divided (Majendie), or when its root is compressed injuriously, say Iby a tubercular tumour, the cornea begins to show points of ulceration, which, increasing in area, may bring about total disintegration of the eyeball.
Their principal function is to bring about the removal of foreign, dead or degenerating material.
It is believed also that they secrete bactericidal substances and ferments which bring about the liquefaction of the fibrin and the damaged tissues - histolysis - and thus assist the process of absorption.
Any of the abnormal conditions that bring about general or local defective nutrition is an important factor in producing fatty degeneration.
In the sphere of physiology and in the interpretation of associated arterial diseases much obscurity still remains; as, for instance, concerning the nature of the toxic substances which produce those bilateral changes in the kidneys which we call Bright's disease, and bring about the "uraemia" which is characteristic of it.
A study of the changes going on in the rif tvalley in which the lakes lie leads, however, to the belief that the Albert Edward and Albert Nyanzas are drying up, a process which the nature of the drainage areas is helping to bring about.
In 1815 the marquis endeavoured to bring about another Vendean rising for the king, and was shot in a skirmish with the Imperialist forces at the Pont des Marthes on the 4th of June 1815.
Following on his calculations from 1509, when the population may be supposed to have been about 50,000, Dr Creighton carries on his numbers to the Restoration The same causes that operated to bring about these changes in the whole kingdom were of course also at work in the case of the City of London.
The defenders employed mines drifting down with the current with striking success on this occasion, and ` the damage caused by them contributed largely to bring about the defeat of the naval force.
It was, therefore, the policy of Bestuzhev to bring about a quadruple alliance between Russia, Austria, Great Britain and Saxony, to counterpoise the Franco-Prussian league.
Was a compromise possible which would bring about a satisfactory settlement?
The oxygen of the air may also bring about chemical changes which result in the production of soluble substances removable by rain, the insoluble parts being left in a loosened state.
Among the factors, economic, geographic, political and social, which combined to bring about the decline of the Hanseatic League, none was probably more influential than the absence of a German political power comparable in unity and energy with those of France and England, which could quell particularism at home, and abroad maintain in its vigour the trade which these towns had developed and defended with their imperfect union.
It would thus seem that he was intriguing to bring about intervention by the United States with a view to annexation; and as the independence of the French Canadian race, which he professed to desire, could not have been achieved under the constitution of the American republic, it is inconsistent to regard his services to his fellow-countrymen as those of a true patriot.
At that date the peace of Peru was so seriously disturbed by internal troubles that the government was quite unable to take active steps to bring about any solution of the matter.
Petrarch had urgently pressed Urban V., Gregory's immediate predecessor, to accomplish the desired change; and Dante had at an earlier date laboured to bring about the same object.
After the outbreak of the Civil War many of the Democrats of the Middle West, who were opposed to the war policy of the Republicans, organized the Knights of the Golden Circle, pledging themselves to exert their influence to bring about peace.
Gates claimed precedence over Schuyler and, on failing to secure recognition, intrigued to bring about Schuyler's dismissal.
But certain forces were at work which were destined to bring about a great revival, viz.
Lenthall now turned his attention to bring about the Restoration.
Thus external and internal influences alike drove him into conflict with the Netherlands, France and England; with the first because political and religious discontent combined to bring about revolt, which he felt bound in duty to crush; with the second and third because they helped the Flemings and the Hollanders.
The dauphin and the duke of Alencon failed to bring about any sympathetic rising in Auvergne, and the Praguerie was over, except for some final pillaging and plundering in Saintonge and Poitou, which the royal army failed to prevent.
These results showed clearly that liquefied acetylene was far too dangerous for general introduction for domestic purposes, since, although the occasions would be rare in which the requisite temperature to bring about detonation would be reached, still, if this point were attained, the results would be of a most disastrous character.
A provincial synod, held at the instance of Wenceslaus in February 1413, broke up without having reached any practical result; and a commission appointed shortly afterwards also failed to bring about a reconciliation between Huss and his adversaries.
On being summoned by the commissioners of the allied powers at Copenhagen to bring about a union between Norway and Sweden in accordance with the terms of the treaty of Kiel, and then return to Denmark, he replied that, as a constitutional king, he could do nothing without the consent of the Storthing, to the convocation of which a suspension of hostilities on the part of Sweden was the condition precedent.
Later, when Herod's conduct aroused the suspicions of Augustus, Nicolaus was sent on a mission to bring about a reconciliation.
He aimed at improving relations with Austria, and also tried to bring about a reconciliation with France; it was in fact under his auspices that President Loubet visited Rome.
Basing themselves on St Gregory's counsel to St Augustine, Dunstan, lEthelwold and Oswald adopted from the observance of foreign monasteries, and notably Fleury and Ghent, what was suitable for the restoration of English monachism, and so produced the Concordia Regularis, interesting as the first serious attempt to bring about uniformity of observance among the monasteries of an entire nation.
Pious people were eager to bring about the conversion of the Indians, and were zealously served by missionaries.
When a proposal was set on foot to bring about a reconciliation between the Roman Church and the Christian Churches of the East, the Abbe Duchesne endeavoured to show that the union of those churches was possible under the Roman supremacy, because unity did not necessarily entail uniformity.
Efforts have been almost unceasingly made since 1872 by statistical experts in periodical conference to bring about a general understanding, first, as to the subjects which may be considered most likely to be ascertained with approximate accuracy at a census, and secondly - a point of scarcely less importance - as to the form in which the results of the inquiry should be compiled in order to render comparison possible between the facts recorded in the different areas.
Marthinus Pretorius, who had succeeded to his father's position as commandant general of Potchefstroom, wished to bring about a confederation between the two Boer states.
His active hostility to Seward did much to prevent the success of that statesman, and to bring about instead the nomination of Abraham Lincoln.
Having urged this view upon the country, when war was declared he felt that it would be inconsistent for him not to share personally in the perils of a conflict which he believed to be a just one, and which he had done as much as he could to bring about.
They regarded it as their principal task to bring about a compromise between the nationalities, and this again depended on the outcome of the GermanCzech negotiations which were always being started afresh.
Waitz was an adherent of the party who were eager to bring about a union of the German states under a German emperor; and when the king of Prussia declined the imperial crown the professor withdrew from the assembly in disappointment, and ended his active share in public life.'
In 1855 he supported Gladstone in the efforts to bring about peace with Russia before the capture of Sebastopol; in 1856 he opposed the opening of museums on Sunday; in the following year he supported Cobden in his disapproval of the second opium war with China.
What the Teutonic Order had Teutonic vainly endeavoured to bring about by fire and sword, Order.
This he calculated would bring about a retaliatory invasion of Poland by the Turks, which would justify him in taking the field against them also with all the forces of the Republic. In case of success he would be able to impose the will of a victorious king upon a discredited diet, and reform the constitution on an English or Swedish model.
In 1832 Lamennais, with his friends Lacordaire and Montalembert, visited Germany, and obtained considerable sympathy in their attempts to bring about a modification of the Roman Catholic attitude to modern problems. Dbllinger seems to have regarded favourably the removal, by the Bavarian government, in 1841, of Professor Kaiser from his chair, because he had taught the infallibility of the pope.
In the summer of 1827, through the persistent efforts of persons most interested in the woollen manufactures of Massachusetts and other New England states to secure legislative aid for that industry, a convention of about loo delegates - manufacturers, newspaper men and politicians - was held in Harrisburg, and the programme adopted by the convention did much to bring about the passage of the famous high tariff act of 1828.
The War of 1812, with the Embargo Acts (1807-1813), which were so destructive of New England's commerce, thoroughly aroused the Federalist leaders in this part of the country against the National government as administered by the Democrats, and in 1814, when the British were not only threatening a general invasion of their territory but had actually occupied a part of the Maine coast, and the National government promised no protection, the legislature of Massachusetts invited the other New England states to join with her in sending delegates to a convention which should meet at Hartford to consider their grievances, means of preserving their resources, measures of protection against the British, and the advisability of taking measures to bring about a convention of delegates from all the United States for the purpose of revising the Federal constitution.
Pleasure is strictly nothing more than the state of being pleased, and hedonism the theory that man's chief good consists in acting in such a way as to bring about a continuous succession of such states.
Cavalry could not bring about the decision in such country, and sought a field for its restless activity elsewhere.
Enlightened as he was, he fully recognized the intellectual inferiority of Russia as compared with the West, and did his utmost to bring about a better state of things.
The progress of navigation and the association of divinities of the sky with maritime affairs probably also assisted to bring about the change, although the memory of her earlier function as a goddess of childbirth survived till imperial times.
Gregory XIV., by the bull Ecclesiae Christi (July 28, 1591), again confirmed the Society, and granted that Jesuits might, for true cause, be expelled from the body without any form of trial or even documentary procedure, besides denouncing excommunications against every one, save the pope or his legates, who directly or indirectly infringed the constitutions of the Society or attempted to bring about any change therein.
The power to declare war formally belongs to Congress; but the executive may, without an act of Congress, virtually engage in hostilities and thus bring about a state of war, as happened in 184546, when war broke out with Mexico.
From Naples Armfelt communicated with Catherine II., urging her to bring about by means of a military demonstration a change in the Swedish government in favour of the Gustavians.
The government endeavoured to bring about an amalgamation of these rival companies, believing that the united energies and financial ability of the whole country were required for so vast an undertaking.
In 1838 the king began to suspect his heir of plotting with the Liberal party to bring about a change of ministry, or even his own abdication.
Daub was one of the leaders of a school which sought to reconcile theology and philosophy, and to bring about a speculative reconstruction of orthodox dogma.
The Polish Protestants hoped that he would take this course and thus bring about a breach with Rome at the very crisis of the confessional struggle in Poland, while the Habsburgs, who coveted the Polish throne, raised every obstacle to the childless king's remarriage.
These and other elaborate efforts, however, failed to bring about the return of the king to Hanover, though the Guelph party continued to agitate and to hope even after the Franco-German War had immensely increased the power and the prestige of Prussia.
It is composed of groups of the different parliaments of the world, who meet periodically to " bring about the acceptance in their respective countries, by votes in parliament and by means of arbitration treaties, of the principle that differences between nations should be submitted to arbitration and to consider other questions of international importance."
It is well known that he fed on inspirations, and expected each day the advent of some supernatural occurrence which should bring about the triumph of the Church.
Partly on this account, and in spite of the attempts of the Chinese authorities to bring about a settlement, there was some delay owing to the attitude of the lamas, but finally a treaty of peace was concluded on the 7th of September.
The temperature of the condenser is so regulated as to bring about the condensation of the nitric acid only, which runs out at the bottom of the pipe, whilst any uncondensed steam, nitrogen peroxide and other impurities pass into a Lunge tower, where they meet a descending stream of water and are condensed, giving rise to an impure acid.
But in no case does it appear that the modifications in shape and colour, which contribute to bring about a mimetic resemblance, are greater and more elaborate than those which result in the simpler examples of ordinary protective resemblance.
Instances of ant-mimicry, unique in the method employed to bring about the resemblance, are supplied by some insects of the Homopterous group of the Rhynchota, belonging to the family Membracidae.
He was totally opposed to the peace with Spain, and wished to bring about a speedy resumption of the war.
Among the enzymes already extracted from fungi are invertases (yeasts, moulds, &c.), which split cane-sugar and other complex sugars with hydrolysis into simpler sugars such as dextrose and levulose; diastases, which convert starches into sugars (Aspergillus, &c.); cytases, which dissolve cellulose similarly (Botrytis, &c.); peptases, using the term as a general one for all enzymes which convert proteids into peptones and other bodies (Penicillium, &c.); lipases, which break up fatty oils (Empusa, Phycomyces, &c.); oxydases, which bring about the oxidations and changes of colour observed in Boletus, and zymase, extracted by Buchner from yeast, which brings about the conversion of sugar into alcohol and carbondioxide.
As a result convection currents are produced in the air which are sufficient to catch the basidiospores in their fall and carry them, away from the regions of comparative atmospheric stillness near the ground, to the upper air where more powerful air-currents can bring about their wide distribution.
The gemmae formed on submerged Mucors may bud like a yeast, and even bring about alcoholic fermentation in a saccharine solution.
The invasion, however, failed, and Michael so far had his revenge in the "Sicilian Vespers," which he helped to bring about.
It is, however, only fair to add that the sultan was doubtless influenced by the desire to bring about a similar change in the succession to the Ottoman throne and to ensure the succession after him of his eldest son, Yussuf Izz-ed-din.
The Dutch were unprepared, and suffered severely through the loss of their carrying trade, and De Witt resolved to bring about peace as soon as possible.
He did his best to get at the real facts, and after a number of conferences with the leaders became so convinced that nothing but a separate administration of the two countries would restore tranquillity that he promised to use his influence with his father to bring about that object - on receiving assurances that the personal union under the house of Orange would be maintained.
After the Restoration in 1660 Baxter, who had helped to bring about that event, settled in London.
He helped to bring about the downfall of James II.
Ever since the conclusion of the Great Northern War, Danish statesmen had been occupied in harvesting its fruits, namely, the Gottorp portions of Schleswig definitely annexed to Denmark in 1721 by the treaty of Nystad, and endeavouring to bring about a definitive general understanding with the house of Gottorp as to their remaining possessions in Holstein.
It was through his initiative, too, that the convention of KlosterSeven was signed (loth of September 1757), and on the 4th of May 1758 he concluded a still more promising treaty with France, whereby, in consideration of Denmark's holding an army-corps of 24,000 men in Holstein till the end of the war, to secure Hamburg, Lubeck and the Gottorp part of Holstein from invasion, France, and ultimately Austria also, engaged to bring about an exchange between the king of Denmark and the cesarevitch, as regards Holstein.
The German king treated his foe generousli and was rewarded by receiving to the end of his reign the servic of a loyal vassal; he also gained the goodwill of the Poles by helping to bring about the return of their duke, Casimir I., who willingly did homage for his land.
Then being generally recognized as king he was able to do something to quell disturbances in various parts of the country, and, in April 1220, to bring about the election of his young son Henry as king of the Romans.
His plan was to bring about the meeting league of of a general council to make the necessary reforms in Schnial- the church, and then at whatever cost to compel the kalden...
A definite defeat of Prussia on an important question of policy must bring about a serious crisis; it is generally avoided because, as the meetings are secret, an arrangement or compromise can be made.
After Bismarcks retirement the emperor attempted to bring about a reconciliation with the duke and the Hanoverians.
The hope that this might bring about some agreement was frustrated by the sudden death of the cardinal, and his successor was more under the influence of the Jesuits and the more extreme party.
As these would inevitably bring about a large increase in the importation of corn from Rumania and Russia, a great agitation was begun in agricultural circles, and the whole influence of the Conservative party was opposed to the treaties.
It was used by the Nationalist parties, in Austria as well as in Germany, to spiead the conception of Pan-Germanism; the Boer3 as Low Germans were regarded as the representatives of Teutonic civilization, and it seemed possible that the conception might be used to bring about a closer friendship, and even alliance, with Holland.
He is a good representative of the type of the grands seigneurs holding advanced and liberal ideas, who helped to bring about the movement of 1789, and then tried in vain to arrest its course.
He had no previous experience of Austrian affairs, and was only anxious at once to bring about a settlement which would enable the empire to take a strong position in international politics.
Each measure had, therefore, to be considered not only on its own merits, but in relation to the general balance of advantage, and an amendment in one might bring about the rejection of all.
This was to bring about a reaction against the economic doctrines which had held the field for nearly twenty years; but the full effect of the change was not seen for some time.
In a country like Austria, in which a mistaken foreign policy or a serious quarrel with Hungary might bring about the disruption of the monarchy, parliamentary government was impossible unless the party which the government helped in internal matters were prepared to support it in foreign affairs and in the commercial policy bound up with the settlement with Hungary.
In 1890, however, instead of proceeding to the coronation as was expected, Taaffe attempted to bring about a reconciliation between the opposing parties.
It was Adam Gottlob Ohlenschldger (q.v.; 1 7791850), the greatest poet of Denmark, who was to bring about the new romantic movement.
This theory being accepted, it is evident that a small quantity of water, by successive dissolution and deposition of a substance capable of existing in a more soluble and in a less soluble form, is able to bring about the crystallization of an indefinitely large quantit y of material.
It contains but little alumina and oxide of iron, which are the constituents generally necessary to bring about the union of silica and lime to form a cement, but in spite of this the silica is so finely divided and so well distributed that it unites readily with the lime when the limestone is burned at a sufficiently high temperature.
At first he was willing to subordinate them to an attempt to win over Scotland to his anti-papal policy, and he made various efforts to bring about an interview with his nephew.
And similarly in the development of a complicated organism, the suppression or doubling of a single cell or group of cells may bring about striking differences in the symmetry of the adult, or the reduction or increase in the number of metameric organs.
In 1880 he became editor of Justice, and worked with success to bring about a revision of the sentences passed on the Communards.
In 1904 appeared the third volume, La Renaissance de Petal, in which the author describes the efforts of the Capetian kings to reconstruct the power of the Frankish kings over the whole of Gaul; and goes on to show how the clergy, the heirs of the imperial tradition, encouraged this ambition; how the great lords of the kingdom (the "princes," as Flach calls them), whether as allies or foes, pursued the same end; and how, before the close of the 12th century, the Capetian kings were in possession of the organs and the means of action which were to render them so powerful and bring about the early downfall of feudalism.
When the ecole normale was joined to the university of Paris, Lavisse was appointed director of the new organization, which he had helped more than any one to bring about.
He was entrusted with various missions in the interests of Catholic unity, the most important being to Constantinople, to endeavour to bring about a union of the Eastern and Western churches.
He wished to bring about the subjection of the church, and to this end sold bishoprics to the highest bidder, annulled the wills made in favour of the bishoprics and abbeys, and sought to impose upon his subjects a rationalistic conception of the Trinity.
In fact, while eager for the deliverance of Italy from Austria, his aim was to bring about a confederation of the states of the country, which was to be under the control of the pope.
The attempt of the Giovane Italia to bring about a general revolution in 1843 only resulted in a few sporadic outbreaks easily crushed.
The king refused to open parliament unless the barricades were removed, and while the moderate elements attempted to bring about conciliation, the ministry acted with great weakness.
Cavour's attempt to bring about the annexation of Sicily to Sardinia failed, for Garibaldi wished to use the island as a basis for an invasion of the mainland.
As the war which he had done so much to bring about did not eventually secure for Russia advantages commensurate with the sacrifices involved, he fell into disfavour, and retired from active service.
At the same time he endeavoured to bring about a union of Aragon with Navarre, by a contract of mutual adoption between himself and the Navarrese king, Sancho, who was old enough to be his grandfather.
In July 1909, General Teranchi, Japanese minister of war, became resident-general, with the mission to bring about annexation.
It was Liszt's aim to bring about a direct alliance or amalgamation of instrumental music with poetry.
So far from attempting to raise their standard of spiritual life, or even leaving it to ordinary intercourse to gradually bring about a certain community of intellectual culture and religious sentiment, they deliberately set up artificial barriers in order to prevent their own traditional modes of worship from being contaminated with the obnoxious practices of the servile race.
He saw that Poland, with her existing constitution, could not hope for a long future, and he determined to bring about a royalist reaction and a reform along with it by every means in his power.
In view of the fact that fresh grape juice contains innumerable bacteria and moulds, in addition to the yeast cells which bring about the alcoholic fermentation, and that the means which are adopted by the brewer and the distiller for checking the action of these undesirable organisms cannot be employed by the wine-maker, it is no doubt remarkable that the natural wine yeast so seldom fails to assert a preponderating action, particularly as the number of yeast cells at the beginning of fermentation is relatively small.
Until the following March, Washington's work was to bring about some semblance of military organization and discipline, to collect ammunition and military stores, to correspond with Congress and the colonial authorities, to guide military operations in widely separate parts of the country, to create a military system for a people entirely unaccustomed to such a thing and impatient and suspicious under it, and to bend the course of events steadily towards driving the British out of Boston.
Pillow, Polk's intimate friend, did much to bring about the nomination.
In 1169 Owen Gwynedd died and was buried in Bangor cathedral after a reign of 33 years, wherein he had successfully defended his own realm and had done much to bring about that union of all Wales which his grandson was destined to complete.
They have, for instance, attained a population of millions in such severe climates as Poland and Russia; in the towns of Algeria they have succeeded so conspicuously as to bring about an outburst of anti-semitism; and in Cochin-China and Aden they succeed in rearing children and forming permanent communities.
Their disunion, he argued, would open a door in the north to the Catholic league and so bring about the destruction of Denmark and Sweden alike.
The object of the party was to bring about a fusion between the representatives of the large landed proprietors and the regular peasant proprietors, to support the interests of landed proprietors in general against those of the town representatives, and to resist Crown interference in the administration of local affairs.
He contributed more than any other man to bring about the downfall and the third partition of Poland, for which he was magnificently recompensed.
On the other hand, when the ulceration is old and the circulation through it poor, the aim of the therapeutist is to reawaken the normal reparative process, to bring about increased circulation and increased tissue change, and thereby insure healing.
In short, its aim was to bring about the best conditions for an ideal civilization, reducing to a minimum the labour necessary for mere existence, and by this and by the simplicity of its social machinery saving the !maximum of time for mental and spiritual education and development.
The effect of marriage upon the property of the spouses is, by the Roman-Dutch law and in the absence of any ante-nuptial contract to the contrary, to bring about a complete community of property, virtually a universal partnership between husband and wife, subject to the sole and absolute control of the husband while the marriage lasts.
Fenelon was continued in his office, but he was recalled in 1575 when Catherine de' Medici wished to bring about a marriage between Elizabeth and the duke of Alencon, and thought that another ambassador would have a better chance of success in the negotiation.
The work of van Tieghem, van Senus, Fribes, Omeliansky and others has now shown that while certain anaerobic bacteria decompose the substance of the middle lamella - chiefly pectin compounds - and thus bring about the isolation of the cellulose fibres when, for instance, flax is steeped or " retted," they are unable to attack the cellulose itself.
It was natural, therefore, that in the series of stormy debates, protracted through several years, that ended in the downfall of Walpole, his eloquence should have been one of the strongest of the forces that combined to bring about the final result.
The brilliancy of the court and the triumph of the sense of unity in the German nation over the particularism of the smaller German states have conduced more than all else to bring about this result.
A year later he noticed that spongy platinum in presence of oxygen can bring about the ignition of hydrogen, and utilized this fact to construct his "hydrogen lamp," the prototype of numerous devices for the self-ignition of coal-gas burners.
The lords sent him to England to ask for assistance from Elizabeth, and his constant aim throughout his political career was to bring about a union between the two crowns.
Palaeologus, whom he accompanied to Italy in order to bring about a union between the Greek and Latin churches with the object of obtaining help from the West against the Turks.
Swift did not bring about the revolution with which, notwithstanding, he associated his name.
Its chemical properties are in general intermediate between those of chlorine and iodine; thus it requires the presence of a catalytic agent, or a fairly high temperature, to bring about its union with hydrogen.
Community of creed, ancient traditional influence, the entire absence of Russian merchants, and t the consequent avoidance of many small commercial rivalries, contributed to bring about a sort of passive preference for Russia, while the bitter disputes that had occurred with Germany on the question of railway finance had left a very hostile feeling.
The fall of meteoric matter into the sun must be a certain source of energy; if considerable, this external supply would retard the sun's contraction and so increase its estimated age, but to bring about a reconciliation with geological theory, very nearly the whole amount must be thus supplied.
In 1874 the 4th earl of Carnarvon, secretary of state for the colonies, who had been successful in aiding to bring about the federation of Canada, turned his attention to a similar scheme for the confederation of South Africa.
In the year 1888, in spite of the failure of statesmen and high commissioners to bring about political confederation, the members of the Cape parliament set about the establishment of a South African Customs Union.
The endeavour to bring about a customs union which would embrace the Transvaal was also little to the taste of President Kruger's Hollander advisers, interested as they were in the schemes of the Netherlands Railway Company, who owned the railways of the Transvaal.
Whilst premier of Cape Colony, by means of the customs union and in every other way, Rhodes endeavoured to bring about a friendly measure of at least commercial federation among the states and colonies of South Africa.
Many attempts have been made to reduce the chamber space by apparatus intended to bring about a better mixture of the gases, and to facilitate the interaction of the misty particles of nitrous vitriol and dilute acid floating in the chamber with each other and with the chamber atmosphere.
Charles had no difficulty in stirring up the commercial jealousy of England so as to bring about a second Dutch war (1672).
The fruitless intrigues carried on by Sophia Dorothea to bring about this match played a large part in Wilhelmina's early life.
At Berlin Jablonski worked hard to bring about a union between the followers of Luther and those of Calvin; the courts of Berlin, Hanover, Brunswick and Gotha were interested in his scheme, and his principal helper was the philosopher Leibnitz.
With the aid of Fialin and Eleonore Gordon, a singer, who is supposed to have been his mistress, and with the co-operation of certain officers, such as Colonel Vaudrey, an old soldier of the Empire, commanding the 4th regiment of artillery, and Lieutenant Laity, he tried to bring about a revolt of the garrison of Strassburg (October 30, 1836).
Avarice, luxury and the glaring inequality in the distribution of wealth, threatened to bring about the speedy fall of the state if no cure could be found.
He always hoped to bring about an honorable agreement between the two adversaries, and in his estimation the advantages of peace outweighed personal interest.
It is, however, easy to bring about an understanding between people in whom religious fury has been extinguished either by patriotism or by ambition, like that of the duke of Alencon, who had now escaped from the Louvre where he had been confined on account of his intrigues.
Necker was carried away in his turn by the reaction he had helped to bring about (1781).
If, therefore, one could correctly read and interpret the activity of these powers, one knew what the gods were aiming to bring about.
Prussia was bound by the treaty of London of 1852, which guaranteed the integrity of the Danish monarchy; to have disregarded this would have been to bring about a coalition against Germany similar to that of 1851.
During the next three years the Ultramontane party hoped to bring about an anti-Prussian revolution, and Napoleon was working for an alliance with Austria, where Beust, an old opponent of Bismarck's, was chancellor.
He also underwent much anxiety lest the efforts of Thiers to bring about an interference by the neutral powers might be successful.
Their alliance began as early as 1239, when Grosseteste exerted himself to bring about a reconciliation between the king and the earl.
Bonaparte brought hint back to France in the autumn of 1799, and it is known that the intervention of Eugene and Hortense helped to bring about the reconciliation which then took place between Bonaparte and Josephine.
The services rendered by Eugene at the time of the coup d'etat of Brumaire (1799) and during the Consulate (1799-1804) served to establish his fortunes, despite the efforts of some of the Bonapartes to destroy the influence of the Beauharnais and bring about the divorce of Josephine.
Other influences were at work to bring about their extinction.
As soon as the pump has sufficiently exhausted the air from the vessel containing the water, vapour is rapidly given off and is absorbed by the acid until sufficient heat has been abstracted to bring about the desired reduction in temperature, the acid becoming heated by the absorption of water vapour, while the water freezes.
In religious matters he interceded with the emperor and the diet for the Protestants, and sought, but without success, to bring about a reconciliation between Lutherans and Calvinists in Brandenburg.
Many circumstances assisted to bring about this change, among the chief of which were the want of harmonious action on the part of the estates, and the decline in the political power of the towns.
At the beginning of 1632, in order to bring about the general peace he so earnestly desired, he proposed to take the field with an overwhelming numerical majority.
The joke in fed circles was that the government could activate the Horsemen at will and bring about the destruction of the planet itself.
The two sides are ready to make continued efforts to bring about universal accession to and effective compliance with the Comprehensive Test Ban Treaty.
To what extent did economic problems in the communist bloc bring about the end of the Cold War?
The Reformation produced little bloodshed in London, with most higher class members co-operating to bring about gradual shifts to Protestantism.
The United Nations Security Council should act swiftly to bring about pressure on both sides to stop the carnage.
The movement, inspired by the late Father Edward Holloway, posits that God works through evolution to bring about an ordered cosmos.
The impact this is having on our education systems has been to bring about discord and an element of confusion.
This remedy is for excessive pride and arrogance and helps bring about humility.
We hope to help bring about a major rethink about the content of tactile graphics.
Do you think the visit of the Holy Father will bring about a new spiritual springtime for the Church in Germany?
Indeed the electron-phonon interactions that bring about low-temperature superconductivity is so weak in these metals that they have never been seen to superconduct.
He was suspected, too, of doing all in his power to bring about a revolt in Scotland.
He enjoyed considerable popularity in Belgium, as well as in Holland for his affability and moderation, and in 1830, on the outbreak of the Belgian revolution, he betook himself to Brussels, and did his utmost by personal conferences with the most influential men in the Belgian capital to bring about a peaceable settlement on the basis of the administrative autonomy of the southern provinces under the house of Orange.
The new parliament decided to adopt the procedure of again sending the premier, Mr Reid, into conference, armed with a series of resolutions affirming its desire to bring about the completion of federal union, but asking the other colonies to agree to the reconsideration of the provisions which were most generally objected to in New South Wales.
The symbolism employed by Mithraism finds its best illustration in the large central relief, which represents Mithras in the act of slaying the bull as a sacrifice to bring about terrestrial life, and thus portrays the concluding scenes in the legend of the sacred animal.
The efforts of the British authorities at this period (1882-1883) to bring about a satisfactory settlement were feeble and futile, and fighting continued until peace was made entirely on Boer lines.
Indignant protest in Cape Town and throughout South Africa, as well as England, led to the despatch g P in October 1884 of the Warren expedition, which was sent out by the British government to remove the filibusters, to bring about peace in the country, and to hold it until further measures were decided upon.
The varied sources of his work and its worthlessness as a transcript of actual Celtic poems do not alter the fact that he produced a work of art which by its deep appreciation of natural beauty and the melancholy tenderness of its treatment of the ancient legend did more than any single work to bring about the romantic movement in European, and especially in German, literature.
This latter complement may not suit the immune body, that is, may not be fixed to the bacterium by means of it, or if the latter event does occur, may fail to bring about the death of the bacteria.
Servia received financial assistance; a large consignment of arms was sent openly from St Petersburg to the prince of Montenegro; Prince Ferdinand of Bulgaria became ostensibly reconciled with the Russian emperor, and his son Boris was received into the Eastern Orthodox Church; the Russian embassy at Constantinople tried to bring about a reconciliation between the Bulgarian exarch and the oecumenical patriarch; Bulgarians and Servians professed, at the bidding of Russia, to lay aside their mutual hostility.
In this chapter, I offer forty-three developments, dynamics, and new realities I believe will work together to bring about an end to war.
Having covered the financial and political factors, let's look at thirteen ways communication and information will help bring about war's demise.
It helps us bring about our social ideals.
A term, "techno-utopian," is often applied to people who believe a technology will bring about a perfect world.
How can we bring about the closest possible rapprochement between philosophy and science?
The sex hormones are chemicals, which bring about gradual changes in the body.
In the case of purring, the part of the cat's brain responsible for vocalizations causes the muscles to move and bring about purring.
When the mediator steps in to help a divorcing couple, the mediator will help both spouses to work with through the issues and bring about a resolution that makes sense for everyone involved.
Air pollution is a real problem in many parts of the world, and it's up to us to reduce our emissions and bring about positive change.
This way you can nip a little twinge of stress in the bud before it spirals into a feeling that will ruin your day and maybe even bring about an anxiety attack.
People have claimed positive effects of meditation on the mind since the dawn of time, and for those who doubt its effect, recent scientific studies confirm that meditation can bring about emotional stability and a sense of peace.
Using a time management matrix will be helpful to you only if you commit to using it as a tool to bring about meaningful change.
Be forewarned, setting a wedding date may bring about tears as you learn your first choice for a wedding date won't work.
Even though an intervention is extremely difficult, it is one of the best ways to encourage and bring about change for an individual with a drug addiction.
Ennemond Boniface was a socialist nudist, who fervently believed that nudism was an alternative to bloody socialist revolution, and would bring about a new naturist era in which all would be equal under the sun (see sidebar).
It's often used to bring about good business deals.
That action has, at the very least, gotten the attention of fashion industry trendsetters and has prompted dialogue necessary as a first step to bring about needed change, hopefully in a not-too-distant future.
Along with these problems, dementia may bring about personality changes.
For instance, black cohosh may bring about nausea and gastric discomfort while reducing hot flashes.
In 1908, Henry Ford made the first Model T, and then in 1913 created a moving assembly line for mass producing automobiles which changed manufacturing and helped bring about the Industrial Revolution.
As an isolated abnormality, spina bifida is caused by the combination of genetic factors and environmental influences that bring about malformation of the spine and spinal column.
As with other porphyrias, the first line of defense is avoidance of factors, especially alcohol, that could bring about symptoms.
Acupuncture treatments can help bring about balance and facilitate sleep.
Vitamin D levels may be insufficient and bring about a softening of bones (osteomalacia), which produces pain and bony deformities, such as flattening or bending.
Hormones produced by the maturing body bring about physical changes that require greater attention when it comes to hygiene.
Illness with high fever may bring about delirium with frightening episodes of nighttime awakening.
However fatty foods or spicy meals that may bring about digestive distress at bedtime may trigger sleep disturbances and awaken a child out of an otherwise peaceful slumber.
However, few programs have the quality necessary to bring about the benefits promised.
Constitutional homeopathy can also work to quiet the symptoms of JA and bring about balance to the whole person.
Small physical additions of these colors to your workspace can bring about bigger changes in your career.
If a woman is close to her full term due date, using home methods to induce labor can bring about labor sooner and make the labor go more quickly, as well.
Whether they are simply ready to meet their baby, are tired of being pregnant, or want to avoid inducing labor medically, many women claim to have special secrets that purportedly bring about labor and delivery of their child.
The warm weather months also bring about outdoor activities, including picnics and bake sales.
If you are looking how to start a conversation with your boyfriend that will bring about a meaningful discussion, try catching him when you both have time to talk, limiting distractions and asking him questions.
Unfortunately, for many who are sensitive to xanthan gum, eating just a little can bring about gastrointestinal symptoms that persist for days.
Keep in mind that sometimes the things you hope for may also bring about certain fears.
Games help bring about social awareness and bonding which is why many parents opt to find games for preschool children.
On one hand, it can bring about a feeling of peacefulness or contentment to know that it's possible that one's spirit journeys on.
However, it took a disaster to bring about widespread acceptance of stringent standards.
If you are the parent of an autistic child who is allergic to gluten and casein, changing your child's diet can bring about dramatic changes within a few months.
Sprouting is the process of soaking seeds in order to bring about germination.
Induction is the strictest phase of the diet, but it is necessary to follow it to the letter so you can bring about ketosis.
Coupled with a sensible, non-starvation diet, this weekly routine should bring about a respectable set of male abs over time.
The biological changes induced by exercise bring about some important health benefits.
They use satin, lace, tulle…whatever it takes to bring about a woman's flirty, sexy side.
The other three saw their involvement in music as a way to bring about social change, while Chimes was in it to get rich and famous.
As another tick in the time line is added to the life of Harry Potter, J.K. Rowling begins the task of linking the elements of her previous novels to bring about the conclusion to her story.
These types of foods may exacerbate the virus and bring about outbreaks.
It was heart wrenching when sometimes our tip failed to bring about justice.
It is permanent in dry air, but in the finely divided state it rapidly combines with oxygen, the compact metal requiring a strong heating to bring about this combination.
The lime deposit or " fur " is a poor conductor of heat, and it is therefore most detrimental to the efficiency of the system to allow the interior of the boiler or any other portion to become furred up. Further, if not removed, the fur will in a short time bring about a fracture in the boiler.
The reforms which it was to bring about were eagerly and impatiently demanded by the public. This great operation had to be effected without interrupting the public service, and the department had immediately to reduce and to simplify the charges for transmission throughout the kingdom.
Seeing the hesitation of the Italian government, the Austrian and German semi-official press redoubled their efforts to bring about the visit.
The protoplasm appears to be able also to bring about thc change without secreting any enzyme.
The poison must not be strong enough to injure the roots, leaves, &c., of the host-plant, or allowed to act long enough to bring about such injury.
Mississippi has taken a leading part in the movement to bring about the removal of the common law disabilities of married women, the first statute for that purpose having been passed in 1839.
In March his illness was evidently gaining on him, to his great grief, because he knew that he alone could yet save France from the distrust of her monarch and the present reforms, and from the foreign interference, which would assuredly bring about catastrophes unparalleled in the history of the world.
Much the same had been the ultimate outcome of the spasmodic attempt of the British government to bring about the introduction of cotton to new districts, after it had been pressed to take some action a few years prior to the formation of the Cotton Supply Association.
Before a battle they often throw themselves between two armies to bring about peace.
He laboured much to bring about the reunion of the Oriental Churches with the see of Rome, establishing Catholic educational centres in Athens and in Constantinople with that end in view.
Turkey now sought for a rapprochement with France, and endeavoured to bring about her intervention in return for concessions as regards the holy places.
A conspiracy to bring about a change was hereupon formed by certain prominent statesmen, whose leaders were Midhat Pasha, Mehemed Rushdi Pasha and Mahmud Damad Pasha, the husband of a princess of the blood, sister to Prince Murad.
Changes of ministry at Constantinople were powerless to bring about an improvement, and early in 1896 Cretan affairs became so serious as to call for the intervention of the powers.
All efforts to bring about an understanding between the government and the opposition were fruitless.
If, in these circumstances, the food supply be also insufficient, the combination of influences is sure, in course of time, to bring about a physical deterioration of the race.
His great aim was to bring about peace, both international and internal.
In other lands things did not on the whole go so well, and many causes at work during the later middle ages tended to bring about relaxation in the Benedictine houses; above all the vicious system of commendatory abbots, rife everywhere except in England.
Repeated attempts of the grand-dukes to bring about a reform were stopped by the opposition of the Ritterschaft.
Its object was to support the attempts of the Pharisees to bring about a reform in the administration of the law courts.
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Related Articles
One of the tricky parts about the English language is that its vocabulary is vast. In a language with so many words, some are quite similar in definition. The trick is to understand the slight differences in the shades of meaning so you can use each word correctly and avoid confusion.
One set of similar words that are often confused are bring, take, fetch, and carry. These verbs all share the same basic definition: to move an object from one place to another. However, there are key differences in how each of these words is used.
In this case, the difference in meaning arises from the location of the subject in relation to the objects being moved. Let's break down the specific definition of each of these verbs so you can see exactly when to use bring, take, fetch and carry.
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1118 | https://pmc.ncbi.nlm.nih.gov/articles/PMC4360115/ | Every inch a finch: a commentary on Grant (1993) ‘Hybridization of Darwin's finches on Isla Daphne Major, Galapagos’ - PMC
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Philos Trans R Soc Lond B Biol Sci
. 2015 Apr 19;370(1666):20140287. doi: 10.1098/rstb.2014.0287
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Every inch a finch: a commentary on Grant (1993) ‘Hybridization of Darwin's finches on Isla Daphne Major, Galapagos’
Graham Bell
Graham Bell
1 Biology Department, McGill University, 1205 avenue Docteur Penfield, Montreal, Quebec, Canada H3A 1B1
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1 Biology Department, McGill University, 1205 avenue Docteur Penfield, Montreal, Quebec, Canada H3A 1B1
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e-mail: graham.bell@mcgill.ca
One contribution of 18 to a theme issue ‘Celebrating 350 years of Philosophical Transactions: life sciences papers’.
The featured article can be viewed at
© 2015 The Authors. Published by the Royal Society under the terms of the Creative Commons Attribution License which permits unrestricted use, provided the original author and source are credited.
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PMCID: PMC4360115 PMID: 25750230
Abstract
One of the most familiar features of the natural world is that most animals and plants fall into distinct categories known as species. The attempt to understand the nature of species and the origin of new species was the enterprise that drove the early development of evolutionary biology and has continued to be a major focus of research. Individuals belonging to the same species usually share a distinctive appearance and way of life, and they can mate together successfully and produce viable offspring. New species may evolve, therefore, either through ecological divergence or through sexual isolation. The balance between these processes will depend on the extent of hybridization, especially in the early stages of divergence. Detecting and measuring hybridization in natural populations, however, requires intensive, long-term field programmes that are seldom undertaken, leaving a gap in our understanding of species formation. The finch community of a small, isolated island in the Galapagos provided an opportunity to discover how frequently hybridization takes place between closely related species in a pristine location, and Peter Grant's paper, published in Philosophical Transactions B in 1993, reports the observations that he and his collaborators made during the first 20 years of what is now one of the classical studies of evolution in action. This commentary was written to celebrate the 350th anniversary of the journal Philosophical Transactions of the Royal Society.
Keywords: hybridization, species, ecological speciation, sexual isolation, mating barrier, Darwin's finches
1. Introduction
Daphne Major is a tiny island, not even 40 ha in area, of the Galapagos archipelago, which lies in the Pacific Ocean more than 1000 km off the coast of South America. Very few species have discovered a place so small and remote, and the only resident vertebrates apart from lizards are several hundred small birds in the genus Geospiza, which make their nests in the prickly pear cactus, Opuntia. These are the famous ‘Darwin's finches', which gave Darwin food for thought when he encountered them during the voyage of the Beagle. They are remarkable for their diversity of form, and in particular for variation in the shape of the beak, which is related in some degree to their diet. Two species are common on the island. One is the cactus finch, G. scandens, which has a rather long, slender beak and feeds mainly on the flowers and fruit of Opuntia. The other is the medium ground finch, G. fortis, which has a shorter, blunter beak and eats the small, soft seeds of shrubs such as Chamaesyce. There are also a few individuals of the small ground finch, G. fuliginosa, and of the large ground finch, G. magnirostris, which has a large, stout beak and is able to crack the hard seeds of plants such as Tribulus. Their diets overlap quite broadly, however, and vary over the year, and between years, in response to the availability of different seed crops. Their morphology is also variable; for example, the fortis of Daphne Major have beaks that are somewhat less robust than those of fortis individuals on other islands. Individuals that seem to be intermediate between species are also occasionally found. Nevertheless, the six species of Geospiza that have been recognized throughout the archipelago—together with eight other species of finch in closely related genera—vary consistently in morphology and ecology, even though their differences are far from absolute.
Darwin was impressed by the finches because all of them are endemic to the Galapagos. The existence of so many distinct forms, in a remote location where the familiar mainland birds were absent, seemed to contradict the universally accepted ideas of his time about the living world, and required a radically different interpretation. The idea that Darwin set out to overthrow, in part because of his observations on the Galapagos Islands, was the privileged status of the species. Before his time, a catalogue of species, building on the schemes of Linnaeus and Ray, was the main goal of natural history, because it would exhibit the plan of the world and thereby reveal the mind of God. This enterprise is credible only if species are fixed and immutable, which is what everyday experience seems to show: a cat is a cat, and has been since the time of the Pyramids. Darwin made the astonishing claim that, on the contrary, species are no more than strongly marked varieties, so that, just as one variety may admittedly give rise to another, so may any species arise by the transformation of an ancestor through a series of intermediate stages. At some point in this process, two lineages descending from a common ancestor become sufficiently distinct in morphology and ecology that they are given different names, but how this point is identified serves convenience rather than principle.
The extreme gradualism that is characteristic of Darwin's view of nature was widely accepted at the time with regard to the adaptation of species to new ways of life. Natural selection and sexual selection were thereby established as the principal agents of evolutionary modification, although they were viewed as acting only very weakly over very long periods of time (e.g. [1, p. 24]). (The emphasis on very slow change may have been necessary to gain the general acceptance of selection as the mechanism of evolution, but in many ways it impeded the development of evolutionary biology, especially by discouraging experimental studies of evolution, in the field or in the laboratory, for the best part of a century.) Darwin's views on speciation were less influential, and by the middle of the twentieth century they had been largely replaced by a new school of thought that emphasized sexual isolation rather than ecological distinctiveness as the criterion for the species boundary. The discrete nature of sexual isolation meant that it could be used, in principle at least, to define a unique set of individuals as constituting a species. In this way, the species re-emerged as the only natural category of classification: races and varieties were too loosely defined to have a consistent meaning, while the membership of genera, families and other Linnean categories was a matter of subjective judgement.
These two schools provided different interpretations of hybridization, and thereby of phylogeny. In the older, Darwinian tradition, hybrids were commonplace and expected: races and varieties often interbreed, and species will naturally do the same, although with lesser frequency as they become progressively more distinct. In the same way, hybrids are often completely viable and fertile, although they need not be, especially if the parents are dissimilar. The phylogenetic tree of a group of related species will show branching, caused by divergent natural section, but also anastomosis, caused by hybridization, at least until lineages become widely divergent. According to the newer view, hybrids are rare and regrettable instances in which the sexual barrier between species has been breached, perhaps as the consequence of some recent environmental change. Most hybrids are markedly inferior to either parent, and sexual isolation is thereby reinforced, through selection against inappropriate mating. The phylogenetic tree is strictly branching because hybrids are too rare or too feeble to make any appreciable contribution to it. This view is much clearer and more elegant that the rather vague notion of species inherent in the older tradition, and by the 1960s the newer had silently replaced the older.
This is a simplification, of course, perhaps an outrageous simplification, of the labyrinthine and occasionally acrimonious debates about the ‘species concept’ that have continued down to the present day (without, in my opinion, adding much of substance to our understanding of evolution). I think it is fair, however, to contrast these two views of species, and to conclude that sexual isolation and strictly branching phylogenies have been broadly accepted as the leading features of species formation, at least since the publication of Ernst Mayr's magisterial tome (as I think it must be described) in 1963 . Nevertheless, there were a few exceptional situations that seemed to support a more nuanced interpretation. One of these began to take shape when Peter and Rosemary Grant landed on Daphne Major in 1973 to begin a detailed study of its resident finches (figure 1).
Figure 1.
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Peter and Rosemary Grant. Photograph kindly supplied by Peter Grant.
2. The finch radiation on the Galapagos archipelago
There are 13 named species of finch endemic to the islands of the Galapagos: five species of tree finch, Camarhynchus; the warbler finch, Certhidea; the vegetarian finch, Platyspiza; and the six species of ground finch, Geospiza. The group is often cited as a classical example of adaptive radiation, with a very wide range of morphological variation. In particular, there is extensive variation in beak shape, from large, broad, stout beaks in seed-eating species to slender, pointed beaks in insect-eating species (figure 2). The association of morphology with diet immediately suggests a mechanism for the adaptive radiation of this group from an unspecialized ancestor arriving in this remote archipelago.
Figure 2.
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Gradation in beak size and shape of selected males of the six Geospiza species. Reprinted from Abbott et al. . Republished with permission from the Ecological Society of America; permission conveyed through Copyright Clearance Center, Inc.
The usual depiction of this radiation, however, as a phylogenetic tree branching into morphologically distinct species, is not wholly consistent with the rather limited DNA surveys that have been reported. Individual birds can indeed always be placed unequivocally in a genus, wherever they have been collected, and there is ample evidence that the genera are reciprocally monophyletic (i.e. all species within a genus have a common ancestor that is not the ancestor of any species in another genus). For species within a genus, and particularly for the species of Geospiza, the situation is different. Individuals which are assigned to different species often overlap in morphology and diet; individuals assigned to the same species may vary among islands; and it is often correspondingly difficult to assign an individual with confidence to any given species. Indeed, there may be greater genetic divergence between populations of a single species on different islands than between nominate species [4–7]. Consequently, the estimated phylogeny does not consistently recover the canonical species. The species assignment of an individual cannot be predicted from mitochondrial DNA data (e.g. ). Indeed, with respect to mitochondrial DNA sequences, a sample from all six Geospiza species is not very different from a sample from a single outcrossed population (; figure 3). Hence, morphological variation in this group has not yet condensed into fully discrete, permanently isolated lineages. Instead, it is kept in flux by a combination of four processes: ecological divergence, hybridization, geographical variation and dispersal.
Figure 3.
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Neighbour-joining phylogenetic tree of Geospiza derived from mitochondrial DNA sequences (fig. 1 of , which was redrawn from fig. 1 of ).
Ecological divergence evolves through selection for specialization on different food items. There is some evidence from detailed field studies that diet correlates with beak shape , although the correlation is not very strong or consistent, except that individuals with large stout beaks can handle large hard seeds and thereby have a broader diet . The most convincing evidence that beak size and shape constrains diet is the shift in average phenotype that occurs when the vegetation changes in response to rainfall. The seed supply is reduced by severe drought, and the limited amount of small, soft seeds is soon exhausted; consequently, selection favours birds with stout beaks able to crack the large, hard seeds that remain. Conversely, the profusion of small, soft seeds in years of heavy rainfall reverses the direction of selection to favour birds with smaller and more slender beaks. The observation of fluctuating selection over drier and wetter years [10–13] shows how beak morphology is related to diet and can shift in response to changes in seed quality.
Ecological divergence will be obstructed by hybridization. As hybrids are morphologically intermediate between their parents they may be ecologically intermediate too, and in consequence may actually survive better than either parental type when the environment changes . In this way, hybridization may actually facilitate the morphological response to shifts in seed supply , although it does not necessarily follow that hybridization is itself an adaptive response to environmental change.
There is also geographical variation in morphology among islands, as figure 2 illustrates. The fortis of Daphne Major, for example, have smaller, shallower beaks than fortis individuals on other islands. Some of this variation may be attributable to differences in vegetation among islands, as Grant & Grant suggest for G. conirostris. Whatever be the cause, it implies that dispersal may affect the amount of variation expressed by a particular population. The populations on Daphne, for example, might be disproportionately affected by immigration from the much larger nearby island of Santa Cruz. The phenotypic distinctiveness of the Daphne fortis, however, suggests that they are largely isolated from neighbouring populations on other islands, with only a low level of immigration. Direct observations of ringed birds have confirmed that there is only a low level of immigration (≤1 individual per generation) of fortis and scandens to Daphne .
The consequence of the contending processes of selection, hybridization and dispersal is that phylogenetic studies of Geospiza have not produced a simple, strictly nested tree. Instead, individuals assigned to the same species are not all grouped together in the same lineage but rather appear at different places in the tree, indicating that the species is not monophyletic (figure 3). This implies that characters such as ‘stout beak capable of cracking hard seeds' have not evolved once only, uniquely marking a single lineage, but have rather evolved repeatedly, in different lineages and often on different islands. On each occasion, the outcome is a morphologically distinct type that might in the course of time evolve into a permanently separate lineage (as exemplified by the genera), but that is usually prevented from doing so by dispersal and hybridization. On a larger scale, this results in a swarm of ecotypes representing incipient species, held apart by divergent selection but still united by occasional hybridization.
3. Hybridization on Daphne Major
Efforts to interpret the radiation of Darwin's finches on the Galapagos Islands mirrored the larger debate about the nature of species. Earlier studies attributed a good deal of the extensive morphological variation in the group to hybridization between rather loosely defined species (e.g. ). Later studies emphasized the role of ecological competition in generating divergent natural selection for specialized diets, resulting in distinctive morphology (e.g. ). It is easy to appreciate with hindsight that the issue could not be satisfactorily resolved without a detailed, long-term survey that would not only detect hybridization but would also be capable of estimating its frequency and its effects. This is what the Grants, working with an exceptionally talented group of students and collaborators, set out to do.
Daphne Major is no more than 500 m long in any direction, and supports a few hundred finches in most years. The populations of the four species of Geospiza found on the island are morphologically distinctive with respect to beak shape, and can be reliably diagnosed by a combination of beak depth and beak length. There are no predators, and the birds are quite tame. Almost every individual can be captured and uniquely tagged with leg bands. The nests of almost all breeding pairs can be visited, and the number, survival and subsequent fate of their progeny can be recorded. After two decades of work, enough information had accumulated to provide a complete picture of the demography, diet and behaviour of the finches, including the vexed question of hybridization, which was the subject of the landmark 1993 article in the Philosophical Transactions of the Royal Society .
The first result of the survey was that hybridization occurred sufficiently often for its frequency to be reliably estimated (figure 4). Indeed, hybridization was common at the level of nominate species: fortis individuals mated with both fuliginosa and scandens, and their hybrid offspring mated among themselves and with the parental types. At the same time, hybridization was rare at the level of individuals: only a few percent of all offspring are hybrids. A similar result had previously been obtained for a different community of finches on Isla Genovesa, where about 1% of all offspring were hybrids between G. conirostris and either G. magnirostris or G. difficilis . Hence, some low rate of hybridization seems to be widespread among species of Geospiza. This would explain why individuals can usually be assigned to one species or another on the basis of morphology, whereas related species are genetically more similar when on the same island than when on different islands .
Figure 4.
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Hybridization of Geospiza species on Daphne Major. (a) Hybridizing species and the F1 hybrids (fig. 7 of Grant , recreated using images supplied by Peter Grant). (b) Number of fledglings produced by interspecific crosses and hybrid pairs (fig. 8 of ).
The second striking result was that the success of matings between individuals of different species, gauged by the number of fledglings per clutch, was just as high as the success of matings between conspecific individuals. Moreover, matings between hybrids, or backcrosses of hybrids to the parental species, showed no consistent sign of reduced fitness. The admixture of fortis and scandens genomes, or fortis and fuliginosa genomes, in any proportion, had no detectable effect on the vigour of offspring. Hence, the ecological differences between species were not being maintained by selection against inviable intermediate types, at least after the El Niño event of 1983 that altered the vegetation of the island.
The pattern that the Grants and their collaborators discovered on Daphne Major, then, was a community of ecologically specialized groups that occasionally interbred to produce fully viable hybrids. Without divergent selection to maintain their distinctiveness, these groups should slowly coalesce, and this seems to be happening. Twenty years after the 1993 paper, the morphological differences between fortis and scandens have diminished appreciably, and a visit in mid-century might find only a single type, with no hint of the diversity that had once existed .
4. The rise of ‘ecological speciation’
There has been a long-running controversy in evolutionary biology about the possibility of ‘sympatric speciation’, which means the formation of two species from a single ancestral species in the same locality, where individuals can freely intermingle. It formed, so to speak, the left wing of speciation theory, in contrast to the more conventional right wing of allopatric speciation, which required geographical isolation between the diverging populations until they were completely sexually isolated. During the 1990s, sympatric speciation began to be supplemented, and to some degree supplanted, by the theory of ecological speciation, which is the view that ecological divergence is the primary stage in species formation, often but not necessarily occurring between sympatric populations. Ecological speciation was powerfully supported by new field studies, especially the extensive investigations of morphological divergence associated with diet in fish such as sticklebacks and whitefish in northern lakes. These frequently evolve into two sharply demarcated ecotypes, a smaller and more gracile type living in the open water and a stouter benthic type foraging in the littoral zone, that coexist in the same lake. This differentiation evolves rapidly and has been observed repeatedly in sticklebacks , whitefish and other postglacial fish populations (e.g. smelt, Osmerus ). Although hybrids are produced and are viable, they are often inferior to their parents because of their intermediate ecological attributes (in sticklebacks ) or partial genetic incompatibility during development (in whitefish ).
The theory of ecological specialization is opposed to the view that sexual isolation comes first, so that multilocus genetic divergence can occur. Without sexual isolation, the randomizing effect of recombination effaces divergent specialization ; this is neatly captured by the simple model of Kirkpatrick & Ravigné . This theory requires that hybridization is rare on an individual basis, but allows that it may be common at the level of ecotypes; that is, any two ecotypes may occasionally interbreed, provided that almost all individuals mate with their own ecotype. In this case, divergent specialization can be maintained, provided that, roughly speaking, the rate of selection acting against maladapted types is greater than the rate of hybridization.
The 1993 paper makes it clear that Darwin's finches are an instance of ecological speciation in the presence of hybridization. The species diverge morphologically, through natural selection for ecological specialization, in the same isolated archipelago, if not generally on the same island. Most individuals mate with another individual of the same species, but hybridization occurs occasionally. The hybrids may be fully viable despite—or sometimes because of—their intermediate morphology, or they may be much less fit than either parent, depending on the state of the environment. Divergent selection tends to maintain the morphological differences between nominate species, while hybridization tends to erode them. Depending on circumstances, species can be forced further apart or brought closer together, even to the point of losing their separate identities.
Building on these results, the research programme of the Grants has provided us with one of the classic accounts of evolution in action, revealing how natural selection routinely initiates the process of species formation. More than that, the 1993 paper marks the point at which interest in ecological speciation began to increase exponentially. At the time of its publication, only 5–10 papers per year referred to ecological speciation; by the end of the decade this number had increased to 30–40; at present about a thousand papers a year are published on this theme (figure 5). The painstaking, long-continued observations on a small, remote island have indeed borne abundant fruit.
Figure 5.
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Trend in number of articles referring to ‘ecological speciation’ published in the last 20 years. Source: Google Scholar.
The truth of the matter is likely to be, as usual, somewhere in between the extreme views of exclusively ecological and exclusively sexual divergence. We now understand that the phylogenies of sexual organisms are not the strictly branching trees of the classical school; but nor are they the free-for-all of reticulate evolution in bacteria exchanging genes by horizontal transfer. Hybridization is most frequent during the early stages of species formation, becomes less frequent as nascent species diverge and eventually ceases completely. The most general representation of a phylogeny, on this view, is a branching tree with sparse reticulation, largely confined to the diverging lineages close to the most recent common ancestor of sister taxa.
5. Saltation on Daphne
The contrast between strictly branching and basally reticulate phylogenies does not by any means exhaust the controversies about species formation; nor does it exhaust the lessons to be learned from the finches of Daphne Major. The suggestion that new species are most likely to arise suddenly in small populations, for example, those on the periphery of a species' range, has been called ‘peripatric speciation’ . This might involve some radical change in genome structure that could only be fixed in small populations , although this idea has been strongly criticized . The ecological idiosyncrasies and the constraints on mating that might occur in small populations, however, could provide a suitable arena for the initial stages of species formation—if only a convincing example could be found.
The most formidable obstacle faced by a sexual theory of speciation is the necessity for reciprocal change in sexual signal generation and signal reception between genders in order to produce a sexually compatible population isolated from its ancestor. If a mutant with an altered signal appears, it is unlikely to have a novel sexual identity; instead, it is almost certain to be incapable of mating. Even if by rare chance it encounters a reciprocally altered partner, their offspring will be rare types in the population as a whole, unlikely to meet and correspondingly unlikely to found a sexually distinct lineage. A sexually distinct lineage is likely to most emerge when mating occurs predominantly within small groups of close relatives, so that mating partners with novel sexual compatibility, created by a rare reciprocal shift of signal generation and reception, are kept together.
A reciprocal shift in sexual compatibility will readily arise, however, when mating preferences are governed by events during development that are shared by sibs. In many birds, including the ground finches, male nestlings learn the song of their father, and females learn to be attracted by this song. When the father's song differs from that sung by other males, brothers and sisters are predisposed to mate together, because they are likely to be rejected by, or are not attracted to, unrelated birds. If the population is restricted to a small area—such as Daphne Major—adult sibs may encounter one another often enough to form a sexually isolated lineage that is perpetuated by the heritability of male song, whether genetic or cultural.
The possibility of rapid speciation in birds through sexual isolation mediated by song was clearly recognized by the Grants , and by good fortune they actually observed such an event, and were able to trace its consequences. A somewhat aberrant fortis male, almost certainly with some scandens ancestry, arrived on Daphne Major in 1981, probably from the nearby, much larger island of Santa Cruz. It was an unusually stout individual with a broad, pointed beak, and it sang an unusual song. For the next 28 years, its descendants were traced generation by generation. In the fourth generation, the finch population was depleted by a severe drought, and the immigrant lineage was reduced to a single pair of brother and sister. From this point on, young birds heard only the song peculiar to their lineage, and in consequence mated when adult only among themselves, and not with individuals from the surrounding fortis population. They occupied clusters of territories, within auditory range of one another. They also retained the unusual beak morphology of their ancestor, suggesting some degree of ecological specialization. In short, ecological speciation and sexual speciation have in this case acted in concert; and in consequence this lineage had proceeded some distance down the path that eventually leads to distinctive ecological specialization and complete sexual isolation [22,34] (figure 6).
Figure 6.
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Pedigree of an immigrant G. fortis male (5110) with a line of descent to an exclusively inbreeding (endogamous) group (fig. 1 of ).
In all likelihood, it will proceed no further. Such a very small population is inevitably vulnerable to accidents and environmental fluctuations, and is likely to become extinct in the near future. It seems most unlikely, however, that it represents a very rare event. For such an event to be observed in one of the very few wild populations that have ever been scrutinized so thoroughly that it could be observed suggests rather strongly that similar events are occurring rather frequently in other populations where they would never be detected. This argument certainly falls short of proof, but it does raise the possibility that nascent species, far from arising only in exceptional circumstances at long intervals of time, may instead continually arise at a measurable frequency, at least in some kinds of organism, even though most become extinct before expanding enough to be noticed. ‘Speciation appears to be easy; the intermediate stages are all around us' [36, p. 2980].
I shall conclude with an historical footnote. To understand the phenotypic variation among the finches of Daphne Major, it was important to identify possible sources of immigrants. Santa Cruz was the obvious possibility; but there were others, including the very small nearby island of Daphne Minor. Unfortunately, this sheer-sided plug of lava rises abruptly from the ocean without offering any feasible landfall. The Grants therefore recruited a McGill colleague, Howard Bussey, as the leader of a small party to establish a fixed rope as a route to the interior of the island. The overhangs of soft, crumbling rock (‘hardish as cheeses go’) made this a hazardous enterprise , but after a series of adventures (including an unplanned descent, fortunately on a top rope) the interior was reached and the resident finches surveyed. As it turned out, only a few scandens had migrated there from Daphne Major, and no fortis; in the reverse direction, two fortis from Daphne Minor were seen on Daphne Major but did not breed [10,11]. The Grants and the climbing team were probably the first featherless bipeds to visit this inaccessible site; it was important to be sure.
Author profile
Graham Bell is James McGill Professor in the Department of Biology in McGill University. He is an evolutionary biologist and ecologist who has worked on a broad range of questions. One of the most striking results of his research has been the discovery that selection in laboratory microcosms over relatively short periods of time can produce radical changes to the structure or way of life of organisms. These studies show that experimental evolution offers an effective technique for studying radical changes in the ecological attributes of organisms. In the last five years, his laboratory has also pioneered the experimental study of evolutionary rescue, the evolutionary adaptation of a population through natural selection to stressful conditions that would otherwise lead to extinction.
In 1977, Peter Grant took a sabbatical year from McGill University to continue his systematic observations of the very birds that first aroused Darwin's interest in the origin of species. At that time, universities provided temporary replacements for professors on sabbatical leave, which is why Graham Bell arrived at McGill in 1976, just before Peter Grant set off for the Galapagos, to take over his teaching duties during his absence. Sixteen years later, Bell was still at McGill and Peter Grant published the results of his investigation of hybridization in Darwin's finches.
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连分数
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数学术语
连分数的意义是什么?为什么说连分数可能是无理数的最好表现形式?是因为存在规律还是精确度高
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本词条由“科普中国”科学百科词条编写与应用工作项目审核 。
连分数是数学中一种特殊的分式表示法，由整数或多项式序列构成，分为有限连分数和无限连分数两类，分别对应有理数和无理数的精确表达。其展开形式具有唯一性：有限连分数对应有理数，无限连分数对应无理数，而二次无理数的连分数项呈现周期性排列。连分数通过递归取整与倒数迭代生成，其截断收敛能提供最优有理数逼近。例如π的连分数展开[3;7,15,1,...]产生的逼近值355/113精度显著高于同分母的小数截断结果。有限连分数可扩展为两种等价形式，无限连分数则用于构造无理数的递进逼近 。该理论在密码学中被用于RSA算法的低解密指数攻击，并在动力学研究中应用于多项式系统的全局相图分析 。连分数计算可采用反向或正向算法，其中正向算法适用于处理无穷连分式 。
中文名 连分数
外文名 continued fraction
学 科数学
分类1 有限连分数
分类2 无限连分数
目录
1动机
2算法
3表示法
4分类
▪有限连分数
▪连分数的倒数
▪无限连分数
5定理
6半收敛
7逼近
动机
播报
图1 连分数
研究连分数的动机源于想要有实数在“数学上纯粹”的表示。
多数人熟悉实数的小数表示：（图1）
这里的 a 0 可以是任意整数，其它 a i 都是 {0, 1, 2, ..., 9} 的一个元素。在这种表示中，例如数 π 被表示为整数序列 {3, 1, 4, 1, 5, 9, 2, ...}。
这种小数表示有些问题。例如，在这种情况下使用常数 10 是因为我们使用了 10进制系统。我们还可以使用 8进制或 2 进制系统。另一个问题是很多有理数在这个系统内缺乏有限表示。例如，数 1/3 被表示为无限序列 {0, 3, 3, 3, 3, ....}。
连分数表示法是避免了实数表示的这两个问题。让我们考虑如何描述一个数如 415/93，约为 4.4624。近似为 4，而实际上比 4 多一点，约为 4 + 1/2。但是在分母中的 2 是不准确的；更准确的分母是比 2 多一点，约为 2 + 1/6，所以 415/93 近似为 4 + 1/(2 + 1/6)。但是在分母中的 6 是不准确的；更准确分母是比 6 多一点，实际是 6+1/7。所以 415/93 实际上是 4+1/(2+1/(6+1/7))。这样才准确 。
去掉表达式 4 + 1/(2 + 1/(6 + 1/7)) 中的冗余部分可得到简略记号 [4; 2, 6, 7]。
实数的连分数表示可以用这种方式定义。它有一些可取的性质：
一个数的连分数表示是有限的，当且仅当这个数是有理数。 “简单”有理数的连分数表示是简短的。 任何有理数的连分数表示是唯一的，如果它没有尾随的 1。（但是 [a0; a1, ... an, 1] = [a0; a1, ... an + 1]。） 无理数的连分数表示是唯一的。 连分数的项将会重复，当且仅当它是一个二次无理数（即整数系数的二次方程的实数解）的连分数表示 。 数 x 的截断连分数表示很早产生 x 的在特定意义上“最佳可能”的有理数逼近（参阅下述定理 5 推论 1）。 最后一个性质非常重要，且传统的小数点表示就不能如此。数的截断小数表示产生这个数的有理数逼近，但通常不是非常好的逼近。例如，截断 1/7 = 0.142857... 在各种位置上产生逼近比，如 142/1000、14/100 和 1/10。但是明显的最佳有理数逼近是“1/7”自身。π 的截断小数表示产生逼近比，如 31415/10000 和 314/100。π 的连分数表示开始于 [3; 7, 15, 1, 292, ...]。截断这个表示产生极佳的有理数逼近 3、22/7、333/106、355/113、103993/33102、...。 314/100 和 333/106 的分母相当接近，但近似值 314/100 的误差是远高于 333/106 的 19 倍。作为对π的逼近，[3; 7, 15, 1] 比 3.1416 精确 100 倍 。
算法
播报
考虑实数 r。设 i 是 r 的整数部分，而 f 是它的小数部分。则 r 的连分数表示是 [i; …]，这里的“…”是 1/f 的连分数表示。习惯上用分号取代 第一个 逗号。
要计算实数 r 的连分数表示，写下 r 的整数部分（技术上floor）。从 r 减去这个整数部分。如果差为 0 则停止；否则找到这个差的倒数并重复。这个过程将终止，当且仅当 r 是有理数。
连分数
数 3.245 还可以表示为连分数展开 [3; 4, 12, 3, 1]；参见下面的有限连分数。
这个算法适合于实数，但如果用浮点数实现的话，可能导致数值灾难。作为替代，任何浮点数是一个精确的有理数（在现代计算机上分母通常是 2 的幂，在电子计算器上通常是 10 的幂），所以 欧几里得GCD算法 的变体可以用来给出精确的结果......
表示法
播报
连分数的表示方法：
连分数图册(1张)
连分数图册(1张)
连分数图册(1张)
分类
播报
有限连分数
所有有限连分数都表示一个有理数，而所有有理数都可以按两种不同的方式表示为有限连分数。这两种表示除了最终项之外都是一致的。在较长的连分数表示，其最终项是 1；较短的表示去掉了最后的 1，而向新的终项加 1。在短表示中的最终项因此大于 1，如果短表示至少有两项的话。其符号表示：
=
连分数图册(1张)
连分数图册(1张)
连分数图册(1张)
连分数的倒数
有理数的连分数表示和它的倒数除了依据这个数小于或大于 1 而分别左移或右移一位以外是相同的。换句话说，图册2中的（左图）和（右图）互为倒数。这是因为如果 a是整数，接着如果x<1,则x=0+1/(a+1/b)且1/x=a+1/b,而且如果x>1,则x=a+1/b且1/x=0+1/(a+1/b)带有最后的数生成对x和它的倒数是同样的连数的余数。
图册2
连分数图册(1张)
连分数图册(1张)
连分数
无限连分数
所有无限连分数都是无理数，而所有无理数可用一种精确的方式表示为无限连分数。
无理数的无限连分数表示是非常有用的，因为它的初始段提供了对这个数的优异的有理数逼近。这些有理数可以叫做这个连分数的收敛（convergent，也译为“渐进”）。所有偶数编号的收敛都小于最初的数，而奇数编号的收敛都大于它。
定理
播报
如果 a 0,a 1,a 2, ... 是正整数的无限序列，递归的定义序列 hn 和 kn：
连分数图册(1张)
连分数图册(1张)
半收敛
播报
连分数
则如下形式的任何分数
这里的 a 是非负整数，而分子和分母在 n 和 n + 1 项（包含它们）之间，叫做“半收敛”、次收敛或中间分数。这个术语经常意味着排除了是收敛的可能性，而不是收敛是一种半收敛。
对实数 x 的连分数展开的半收敛包括了所有比有更小分母的任何逼近都好的有理数逼近。另一个有用的性质是连续的半收敛 a/b 和 c/d 有着 。
逼近
播报
最佳有理数逼近
设
是实数α的第 k 个渐进分数，则对任意分数
满足0<q<=
有
。
证明：若
则显然成立。否则不妨设
，若
则有
且
，所以
矛盾！若
则有
，由
有
矛盾！所以
此时由不等式性质显然成立，或者
即
所以定理得证 。
算法：对实数 x 的最佳有理数逼近是有理数 n/d（d > 0），它比带有更小分母的任何逼近都接近于 x。依据如下三个规则，从 x 的简单连分数生成所有对 x 的最佳有理数逼近：
截断连分数，并尽可能减小它的最后项。 减小的项不能小于它最初的值的一半。 如果最终项是偶数，则用特殊规则确定它的值是否可接受。（见后） 例如，0.84375 有连分数 [0;1,5,2,2]。下面是它的所有最佳有理数逼近。
[0;1] [0;1,3] [0;1,4] [0;1,5] [0;1,5,2] [0;1,5,2,1] [0;1,5,2,2] 13/4 4/5 5/6 11/13 16/19 27/32包含了分母严格单调递增的增补项允许在算法上施加限制，要么在分母的大小上，要么在逼近的接近性上。
要向有理数逼近并入新项，只需要两个前面的收敛。如果 a k+1 是新项，则新分子和分母是
nk+1 = nk−1 + ak+1 nk dk+1 = dk−1 + ak+1 dk 初始的收敛（要求前两项）是 0⁄1 和 1⁄0。例如，以下是对 [0;1,5,2,2] 的收敛。
k−2−101234ak01522nk010151127dk101161332减半规则的形式描述是减半的项 1/2 ak 是可接受的，当且仅当
[ak; ak−1, …, a1] > [ak; ak+1, …] 在实践中，经常使用类似欧几里得GCD的算法依序生成这些项，且它提供的辅助值可更方便的测试。例如，以下是为 0.84375 = 27⁄32 生成的项。
a0 = ⌊27⁄32⌋ = 0, f0 = 27 − 32a0 = 27a1 = ⌊32/27⌋ = 1, f1 = 32 − 27a1 = 5a2 = ⌊27/5⌋ = 5, f2 = 27 − 5a2 = 2a3 = ⌊5/2⌋ = 2, f3 = 5 − 2a3 = 1a4 = ⌊2/1⌋ = 2, f4 = 2 − 1a4 = 0使用以此生成的 f 值，1/2 ak 的可接受性测试是 dk−2 ⁄ dk−1 > fk ⁄ fk−1。对于例中的 a3，d1 ⁄ d2 = 1/6 且 f3 ⁄ f2 = 1/2，所以 1/2 a3 是不可接受的；在对 a4 的时候，d2 ⁄ d3 = 6⁄13 且 f4 ⁄ f3 = 0⁄1，所以 1/2 a4 是可接受的。
对 x 的收敛在更强的意义上是最佳逼近：n/d 是 x 的逼近，当且仅当 |dx−n| 是在所有逼近 m⁄c 带有 c ≤ d 中是最小的相对误差的；就是说，我们有 |dx−n| < |cx−m| 只要 c < d。（注意还有 |dkx−nk| → 0 当 k→∞。）
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参考资料
1 赵义超. 也谈连分数[J]. 文教资料,2005,(25):170-171. [2017-09-13]．中国知网 [引用日期2017-09-13]
2 何雅. 连分数及其基本性质[J]. 长江工程职业技术学院学报,2004,(01):50-52. [2017-09-13].．中国知网 [引用日期2017-09-13]
3 沈忠华,于秀源. 一类连分数的有理逼近[J]. 科技通报,2009,25(06):720-722. [2017-09-13].．中国知网 [引用日期2017-09-13]
4 21 函数逼近．北京大学主页．2020-02-04
5 王小云院士:公钥密码学的数学基础(第二版)．百家号．2023-02-06
7 数学与统计学院在基础数学研究领域连续取得重要进展性成果．中南大学．2024-04-10
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Power Mean Inequality
The Power Mean Inequality is a generalized form of the multi-variable Arithmetic Mean-Geometric Mean Inequality.
Inequality
For positive real numbers and positive real weights with sum , the power mean with exponent , where , is defined by
( is the weighted geometric mean.)
The Power Mean Inequality states that for all real numbers and , if . In particular, for nonzero and , and equal weights (i.e. ), if , then
Considering the limiting behavior, we also have , and .
The Power Mean Inequality follows from Jensen's Inequality.
Proof
We prove by cases:
for
for with
Case 1:
Note that As is concave, by Jensen's Inequality, the last inequality is true, proving . By replacing by , the last inequality implies as the inequality signs are flipped after multiplication by .
Case 2:
For , As the function is concave for all , by Jensen's Inequality, For , becomes convex as , so the inequality sign when applying Jensen's Inequality is flipped. Thus, the inequality sign in is flipped, but as , is a decreasing function, the inequality sign is flipped again after applying , resulting in as desired.
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Structure and randomness in the prime numbers
7 February, 2008 in math.NT, talk, travel | Tags: number theory, powerpoint, randomness, structure | by Terence Tao
This Thursday I was at the University of Sydney, Australia, giving a public lecture on a favourite topic of mine, “Structure and randomness in the prime numbers“. My slides here are a merge between my slides for a Royal Society meeting and the slides I gave for the UCLA Science Colloquium; now that I figured out to use Powerpoint a little bit better, I was able to make the latter a bit more colourful (and the former less abridged).
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76 comments
Comments feed for this article
9 January, 2011 at 3:37 pm
Pete Quinn
If anyone is interested, I’ve tried to explain the ideas more concisely in the form of a draft paper. Note it’s written by an engineer, not a mathematician, and therefore is written intuitively in the way the ideas developed, but not necessarily in a particularly elegant fashion.
I’d appreciate if anyone could tell me if the ideas have been published before, and please note it’s a rough draft, so references in particular need to be tightened up, and the formatting is a little icky.
10 January, 2011 at 8:01 am
Sorry, the attempt twin prime proof is here:
Reply
24 February, 2011 at 10:57 pm
mobiusfunction
Two recurrences:T(1,1)=1, n>1: T(n,1)=0, k>1: T(n,k) = (0 + (sum from i = 1 to k-1 of T(n-i,k-1))) mod 2.
T(1,1)=1, n>1: T(n,1)=0, k>1: T(n,k) = (1 + (sum from i = 1 to k-1 of T(n-i,k-0))) mod 2.
Mahonian numbers modulo 2
Mahonian numbers mod 2
Reply
21 March, 2011 at 8:02 am
petequinn
Hopefully this is not seen as using up unnecessary bandwidth.
Acknowledging again this is just a bit of a hobby project, for fun. We’ve been playing with this a little further, and can offer a couple of observations perhaps a little more concisely.
Within any primorial, there is a symmetric pattern of “non-primes” that repeats for all repetitions of that primorial to infinity. These numeric positions can never be prime, as they all represent multiples of primes (namely the factors of that primorial). There is also an asymmetric, non-repetitive component of non-primes within each primorial. This asymmetric component represents all additional “non-primes” that need to be deleted to leave behind only primes.
The repetitive pattern, whose symmetry turns out to be fairly easy (I think) to prove, involves only the primes that are factors of the primorial. The asymmetric, non-repetitive, component involves (prime) multiples of all primes higher than the prime factors but less than the square root of the primorial.
Each of the higher primes not involved in the symmetric pattern within a primorial has a repetitive, symmetric pattern that eventually emerges as the numbers get high enough. But this does not emerge until you reach the first primorial involving that prime as a factor.
Of interest (if perhaps only to me), is there are “long” sequences of non-primes leading up to and following every primorial, and every multiple of every primorial. If the nth primorial is Pn#, then (Pn# – 1) and (Pn# + 1) could be prime (common knowledge), but every other number between (xPn# – P(n+1)+1) and (xPn# + P(n+1)-1) must not be prime, where x is some arbitrary integer (also, I think, fairly simple to prove). So if for some reason you need an arbitrarily long string of sequential non-primes, simply go to the primorial corresponding to the closest higher prime (or any multiple thereof).
Reply
15 April, 2011 at 3:09 am
Mats Granvik
The prime zeta function (Wikipedia) is defined as:
It is known that:
We may consider:where zero is not included in the multiplication and .
Doctormatt rewrote as:
Can it be proven that:
Reply
23 July, 2011 at 8:07 pm
spaghetti monster
Ian Robinson: You said…..“The image of the physical makeup of the Whirlpool Galaxy looks identical to my drawing, I have deliberately withheld an important piece of the jigsaw until a later date.The truth is I require someone with a bit of mathematical clout to give these findings the validity and respect they deserve, no one else is aware of my findings.”—————————————-
You might find the link below more helpful than Terry Tao’s blog.
Reply
20 September, 2011 at 3:53 pm
petequinn
In case anyone is interested, I extended the previous bit of numerical experimentation to twin primes, and found (perhaps unsurprisingly) some similar patterns. Some details here:
Was just playing around a bit to try to understand the bounds of the twin prime conjecture problem. No breakthroughs to share (surprise!) but I find the patterns and trends to be interesting, and help me, as a visual thinker, to put the challenge in perspective.
Reply
31 October, 2011 at 8:05 am
petequinn
CRACKPOT ALERT :-)
More on primes and twin primes from a complete amateur. Looking for help answering a question…
What I’ve tried to describe in previous posts are some patterns that develop within primorials. For the n-th primorial, Pn#, the distribution of primes can obviously be fully developed by repetition of all primes less that sqrt(Pn#). The removal (or sieving) of all non-primes can be divided into two broad components: (1) a symmetric pattern of non-primes removed through repetition of all primes from 2 to Pn (symmetric since these all divide evenly into Pn#), and (2) a non-symmetric distribution of non-primes generated by all other primes from P(n+1) to the highest prime < sqrt(Pn#).
Numerical experimentation shows that the symmetric component of non-primes projects outward, presumably to infinity, for every subsequent repetition of Pn#, so that primes can never fall on those positions, and primes appear, at "random" (clearly not random, but apparently random) intervals at the other positions that were not "non-primes" within Pn# screened by the primes to Pn.
Interestingly, if only to me, is that the number of "possible primes" (i.e. not "non-primes" screened by the primes to Pn), which I will call Nn, within any given primorial, Pn# ( obviously defined as the product of all primes to Pn), is related to the primorial as follows: Nn = (product for all i=1 to n) [Pn – 1]
(sorry I don't have LaTEX)
In my last post I linked to some notes describing results of numerical experimentation that show that similar patterns develop for twin primes. Within any given primorial, all possible twin primes can be sieved via symmetric removal due to primes to Pn, and then non-symmetric removal by higher primes to sqrt(Pn#). The symmetric pattern of "non-twins" removed by, and lingering "possible twins" left alone by, the sieving of primes to Pn, then repeats to infinity for all repetitions of Pn#, so that higher twins will always only occur at these "possible twin" locations.
Even more interesting (probably still only to me, haha) is that the number of "possible twins" within any given primorial is ALSO related to the primorial, similar to the way the "possible primes" are, but with the following form, where Tn is the number of "possible twins" within any given repetition of a primorial:
Tn = (product for all i=2 to n) [Pn – 2]
Alternatively, one can describe the number of "possible twins" removed when going from Pn# to P(n+1)# as:
T(n+1) = P(n+1)Tn – 2 (product for all i=2 to n) [P(n-1) – 2]= P(n+1)Tn – 2 Tn= (Pn – 2)Tn
Using a couple of random examples to illustrate this observation:
consider P4 = 7, where P4# = 210.
The number of possible positions for primes within repetitions of 210 projecting out to infinity is 48, a result you can check for yourself fairly easily (work out Pi(mod 210) for as many primes as you like and you will find exactly 48 unique values). This is (7-1)(5-1)(3-1)(2-1)
Similarly, the number of possible twin prime pairs within repetitions of 210 is T4=15, which is (7-2)(5-2)(3-2). You can readily see that T3 would be 3, or (5-2)(3-2).
Note that T4 = 15 = (P4 – 2)T3 = 53
I can't work out the math to prove these results, but they seem far to simple to be accidental, and to not continue to infinity. Can anyone suggest a way to demonstrate these results algebraicly, or otherwise?
Thanks for humouring me.
Pete
Reply
31 October, 2011 at 8:08 am
petequinn
dangit, no editing function… there are a few silly typos that should be obvious by context, like some n’s in products should be i’s
Reply
1 November, 2011 at 12:10 pm
petequinn
Goodness I left numerous irritating little typos in that post, sorry. To correct a few:
“Possible primes” within the nth primorial, Pn#, is:
Nn = (product for all i=1 to n) [Pi – 1]
“Possible twins” within the nth primorial is:
Tn = (product for all i=2 to n) [Pi – 2]
or
T(n+1) = [P(n+1) – 2]Tn
For a few moments, let’s suspend disbelief and pretend these definitions for Nn and Tn are proven.
I now propose that the density of primes, within all natural numbers, is less than Nn/Pn#, and converges on Nn/Pn# as n approaches infinity. I will claim this without proof for the moment and maybe elaborate later. More disbelief to suspend, sorry. :-)
I also propose that the density of twin pairs, within all natural numbers, is less than Tn/Pn#, and converges on Tn/Pn# as n approaches infinity.
The ratio of the density of twin pairs to the density of primes, within all natural numbers, which is the same as the density of twin pairs within all primes, can be taken as (Tn/Pn#)/(Nn/Pn#) = Tn/Nn, n approaching infinity
Let’s now compare the density of twin pairs within the set of primes to the density of primes within the set of natural numbers, as:
[Tn/Nn]/[Nn/Pn#], as n approaches infinity.
I believe it is fairly straightforward to show that this ratio converges on about 1.320…, the twin prime constant.
This result tells us, I think, that the set of twin primes corresponding to the set of primes is 1.32 times larger than the set of primes corresponding to the set of natural numbers (with apologies, if this isn’t written correctly from a mathematical perspective). Since the set of primes is infinite, hence the set of twin primes is also infinite.
Does that make sense?
I’ve claimed a couple of things in here without proof – the first being the existence and definitions of Nn and Tn, which I’ve demonstrated (to myself anyway) through numerical experimentation but not proven mathematically. I would think that, given the simplicity of the relationships they should be easy to develop algebraically, but admit that they are beyond my ability. The second unproven claim is that the ratios converge to the values indicated as n approaches infinity. I think those claims are much easier to rationalize, and the fact that they lead to the twin prime constant would seem to suggest they must be correct.
Thoughts?
Thanks for reading. :-)
Pete
Reply
1 November, 2011 at 9:43 pm
petequinn
One more crackpot post, thanks for the indulgence Prof. Tao.
I’d appreciate if someone would comment, even if only to make fun of me. This is the internet after all. :-)
I know this is a famous problem and it’s bad form to hastily claim a proof (or “near-proof,” as in this case I think) of a famous problem, but since I am not a mathematician, I have no reputation to lose by exposing myself to ridicule for this, and I can easily slink back to the day job when someone pokes a big hole through this. :-)
Let me elaborate on two of the items I’ve claimed without proof in the previous post, and then also show why the ratio of twin pair/primes density to primes/natural numbers density is 1.32, or at least converges and is not less than 1. That will be in the next post to keep this one relatively short.
To find all prime numbers, we need to strip away all composite numbers. We can imagine doing this by progressively removing, or sieving, all prime multiples, working our way from the lowest prime through to infinity.
Let’s start with by removing all multiples of 2, the even numbers, from the set of natural numbers. This strips out half of all natural numbers as prime candidates, leaving only the odd numbers as possible primes. Then remove all multiples of 3, 5, 7 and so on through to Pn, the nth prime, with n approaching infinity.
Consider the second primorial, P2# = 6, after first removing the even numbers, so we have 1, 3 and 5 as prime candidates, and 2, 4 and 6 as confirmed “non-primes” between 1 and 6. Note that this pattern, which is symmetric within 6, repeats ad infinitum, or in other words, after sieving the even numbers, all remaining numbers (mod 6) are either 1, 3 or 5.
Now remove all multiples of 3 from the remaining set of “possible primes.” This removes one possible prime candidate between 1 and 6, leaving 1 and 5 as possible primes. Note again this leaves a symmetric pattern, which, since we have now only removed multiples of 2 and 3, repeats ad infinitum, removing 2/3 of all natural numbers as non-primes and leaving 1/3 for further consideration as possible primes. After having sieved the 2s and 3s, the remaining “possible prime” candidates (mod 6) are all either 1 or 5.
We can continue in the same fashion to remove all remaining multiples of 5, 7, 11, 13, …. to Pn. Each time we do this, we will be left with a set of non-primes and “possible primes” that is symmetric within the respective primorial, and which repeats ad infinitum. The density of “possible primes” within the set of natural numbers, after sieving all primes to some arbitrary Pn, is equal to the number of “possible primes” within the respective primorial divided by the value of the primorial.
The complete set of prime numbers is revealed when we have sieved all possible composite numbers by repeating to infinity each prime, and striking out all these repetitions as non-prime. Therefore, if we continue this process as described, ad infinitum, to remove all multiples of all primes to Pn, where n approaches infinity, then we will be left with only primes. Hence, if we call Nn the number of “possible primes” within Pn#, then Nn/Pn# converges on the density of primes within the set of natural numbers as n approaches infinity, since as n approaches infinity, Nn becomes the set of primes and Pn# becomes the set of natural numbers.
Now let’s examine how many natural numbers are removed as candidates for any given prime, Pn. Choose an arbitrary prime, Pn, and consider the number of multiples of Pn within Pn#. This is clearly P(n-1)#. If no multiples of primes smaller than Pn had yet been removed, then repetitions of Pn to infinity would eliminate Pn prime candidates out of every Pn# natural numbers. However, assuming Pn > 2, and following our sieving process sequentially with monotonically increasing n, by the time we are set to remove remaining candidates that are multiples of Pn, there are a number of multiples that have already been removed, namely all those that are multiples of lower primes < Pn. If you conduct a simple accounting exercise, starting with the smallest prime numbers and working up through a few, you should be able to convince yourself that of the P(n-1)# possible multiples of Pn within Pn#, [P(n-1)# – N(n-1)] have already been eliminated by the multiples of Pn with lower primes, where N(n-1) means the number of possible primes left within the lower primorial, P(n-1)#, after sieving of all primes up to P(n-1), and Nn is equal to, as claimed before:
Nn = (product for all i=1 to n) [Pi – 1] —– eqn
To demonstrate… removal of all even numbers eliminates half of all numbers between 1 and 2 (= P1#), and by extension removes half of all numbers within every multiple of P1# to infinity, or half of all natural numbers.
Removal of all remaining multiples of P2 = 3 between 1 and 6 (= P2#) eliminates the number 3 as a possible prime within subsequent repetitions of P2#. Note that 6 was already removed, since it is a multiple of 2. So there were 2 (= P1#) multiples of 3 within P2#, but 1 (i.e. {P1# – N1}, or {P1# – eqn for n-1, where n = 2} had already been removed by multiples of 3 by 2, leaving only 1 possible prime candidate to be removed at this sieving pass, to leave 2 remaining “possible prime” candidates (1 and 5) within P2#, or 2/6 = 1/3 “possible primes” within the set of all natural numbers after repeating this pattern an infinite number of times. Hence all primes (mod 6) greater than P2 must equal 1 or 5, a well known result.
Now if we remove the remaining multiples of P3 = 5 between 1 and 30, we find that of the 6 (i.e. P2#) possible multiples of 5, 4 had already been removed, leaving only 2 (i.e. {P2# – eqn for n-1 where n = 3}), namely 5 and 25, to be screened by repetitions of 5. Thus of the 10 possible primes within 30 existing after sieving by 2 and 3, two more candidates (5 and 25) have been removed, and all primes (mod 30) greater than P3 must equal 1, 7, 11, 13, 17, 19, 23 or 29 (i.e. exactly 8 possibilities).
Repeating the same logic now for multiples of P4 = 7 up to 210, of the 30 (= P3#) multiples of 7 to be removed, 22 ({P3# – product for i = 1 to 3 of [Pi – 1]}) were already screened by multiples of 7 with 2, 3 or 5, leaving 8 other multiples of 7 to be eliminated at this stage. Thus of 8 x 7 = 56 possible primes within 210 remaining after sieving by 2, 3 and 5, the further sieving by 7 removes 8 other possible prime candidates, leaving 48 “possible primes” within repetitions of 210, and therefore 48 unique values for all primes higher than 7 (mod 210), which I won’t list here.
Now let’s examine why Nn takes the form I’ve claimed, using P4 as a working example.
If we want to count the number of multiples of 7 with 2, 3 and 5, we should expect:
3 x 5 multiples of 2 x 7,2 x 5 multiples of 3 x 7,2 x 3 multiples of 5 x 7,2 multiples of 3 x 5 x 7,3 multiples of 2 x 5 x 7,5 multiples of 2 x 3 x 7, and1 multiple of 2 x 3 x 5 x 7
In order to identify unique multiples of 7 already removed by 2, 3 and 5, we obtain [(3 x 5) + (2 x 3) + (2 x 3)] – (2 + 3 + 5 + 1) = 22, which we can re-organize and re-write as {5 x 3 x 2 – [(5 – 1) x (3 – 1) x (2 – 1)]} = 22. Or, the number of unique multiples of 7 removed within multiples of 210 is 8 = (5 – 1) x (3 – 1). And hence the number of possible primes remaining within multiples of 210 is 8 (from P3#) x 7 – 8, or 48, or 8 x 6, or 8 x (P4 – 1), or (P4 – 1) x (P3 – 1) x (P2 – 1) x (P1 -1).
Hence, Nn = (product for all i=1 to n) [Pi – 1]
Phew. :-)
I think that is fairly straightforward, and I expect the logic to generate the product formula for Tn is very similar, but this came to me on today’s 9 mile run, and I think I need another good run or two to come around to figuring that one out. :-)
To wrap this post, I think I’ve defended the prior claimed definition of Nn, the number of “possible prime” candidates within the nth primorial (which by extension serves to define the density of “possible prime” candidates within the set of natural numbers after sieving by primes up to Pn). And, I think I’ve defended the claim that this density of “possible primes” within the nth primorial converges to the density of primes within the set of natural numbers as n approaches infinity.
OK, on to producing the twin prime constant, or something like it, in one more post…
Reply
1 November, 2011 at 10:32 pm
petequinn
So let’s start with the assumption that the claimed definitions of Nn and Tn are proven. I think I’ve proven the first, and can only say I believe the second from numerical experimentation, and will try to defend it, given more time and more miles. :-)
Let’s pick up from two posts back with the idea that the ratio of the density of twin pairs within the set of primes to the density of primes within the set of natural numbers can be expressed as:
D(t:p)/D(p:nat) = [Tn/Nn]/[Nn/Pn#] = Tn(Pn#)/(Nn)^2, with n approaching infinity
Where D(t:p) means the density of twin pairs within the set of primes, and D(p:nat) = density of primes within the set of natural numbers.
This can be re-written as:
D(t:p)/D(p:nat) = product for i = 1 to infinity {Pi x P(i – 2)/[P(i-1)]^2} ——- eqn
Where we include P0 = 1 to allow us to simplify the examination of the products with no change to the result.
This can be written as:
D(t:p)/D(p:nat) = limit as n approaches infinity {Pn x P(n – 2)/[P(n-1)]^2} x … x {13 x 11/12^2} x … x {3 x 1/2^2} x {2 x 1/1^2}
Numerical experimentation shows this series converges quite nicely to 1.320… as claimed. I won’t attempt to work out the arithmetic, but by inspection, as n increases, the nth term steadily converges on 1, being always < 1, and with n = 1, 2, 3 and 4 we get 2, 1.5, 1.406…, and 1.367…
Without knowing the actual value of this infinite product, we can obtain a lower bound by examining:
Product for i = 1 to infinity {i x (i – 2)/(i – 1)^2}
which converges on 1.
This latter product includes all the terms in eqn , but also includes additional terms less than 1 for values of i that are not prime. Hence it decreases more quickly than eqn , and is always lower than eqn , after the first two terms, when the same number of terms has been included in both products. Therefore in the limit, eqn converges on a positive number that is less than 1.367… (by inspection) and not less than 1.
So, if we believe the definition of Tn as claimed, I think we have shown that the set of twin prime pairs is infinite, since its density within the infinite set of primes is at least equal to the density of primes within the infinite set of natural numbers.
Of course the definition of Tn is not yet proven. A few more miles to be run first… :-)
Questions, comments, rotten tomatoes welcome.
Hopefully no fatal typos in this post, fingers crossed…
Reply
2 November, 2011 at 11:34 am
petequinn
And of course there were some critical typos. Corrections:
D(t:p)/D(p:nat) = product for i = 1 to infinity {Pi x (Pi – 2)/[(Pi-1)]^2} ——- eqn
Where we include (P1 – 1) = 1 (instead of 0)…
——
I think that’s the worst of it.
Reply
3 November, 2011 at 7:34 am
petequinn
Goodness even the corrections need corrections:
Where we include (P1 – 2) = 1…
The idea there is simply to keep 1 as a placeholder (instead of (2 – 2 = 0) for when we group the individual product terms. It’s not a necessary element, just a convenience to make the elaboration of the infinite product easier to manage and look at.
I see there are still some other minor typos in the preceding posts: a (2 x 3) instead of (2 x 5), and at least one Pn instead of the correct P(n-1)#. Sorry for those.
I think I’ve just about got the correct explanation for Tn, will try to elaborate after the end of the work day. Meanwhile, I should probably explain one thing I’ve glossed over that may have some readers stuck or rejecting my logic.
In the preceding posts I’ve made no specific mention of “actual primes,” but have rather focussed on a discussion of confirmed “non-primes” (as ruled out in any given sieving by a given Pn) and “possible primes” that remain after each sieving step. I’ve suggested that these values can be extrapolated to infinity to obtain the ultimate densities of primes and twin primes.
But what about the pesky ACTUAL primes that develop as we go, namely 2, 3, 5, …. Pn? These actual primes occupy positions I have called “non-primes.” Is this a fatal flaw in my logic? I’m fairly confident this has no bearing, and here is why…
If we take Nn and Tn to be the number of possible primes and possible twins within the nth primorial, where “possible” implies “still standing as candidates after sieving by primes to Pn,” then these values represent the numbers of possibilities within all primorials after the first one (accepting for the moment that Tn remains to be developed/defended).
Within the first primorial, we know that we have n ACTUAL primes, which are sitting on n “non-prime” positions. If we want to be precise, then, we should perhaps state that the number of “possible primes, “within the first Pn# only, is actually Nn + n, and thereafter remains Nn. We may also have some number, which must be less than n, of actual twin primes.
It is trivial to show to show that (Nn + n)/Pn#, the density of primes within the first primorial, converges to Nn/Pn# as n grows. It is also trivial to show that this addition of n “possible primes” within the first primorial has no bearing on the density of “possible primes” in the set of natural numbers after the nth sieving step, when we recall that the first primorial containing the extra n “possible primes” is only one of an infinite number, with the rest all containing exactly Nn “possible primes.”
Hence the density of “possible primes” within the set of natural numbers after n sieving steps remains Nn/Pn#, and the density of twin pairs Tn/Pn#.
Now if only we could prove the definition of Tn, we might have the problem licked. :-)
(or I could be completely wrong, and given the fact that I’ve never done anything elegant in math on my own, this latter outcome seems somehow more likely, yet this continues to be fun and intellectually stimulating to me, so thanks again for the continued indulgence)
Reply
3 November, 2011 at 8:30 am
Terence Tao
I think any further discussion of these matters should take place on your own web site, where you have the ability to edit your own posts.
Reply
3 November, 2011 at 8:32 am
petequinn
OK fair enough, thanks for letting it go on this far.
All the best!
Reply
6 November, 2011 at 12:28 am
petequinn
One final post to let anyone who may have been following this know the story continues to develop here:
Further numerical experimentation shows predictable numbers of candidate k-tuples, other than primes and twin primes, within the primorials.
Thanks again Prof. Tao, I won’t post here on this topic again, I promise.
Cheers,
Pete
Reply
11 February, 2012 at 5:27 am
jose javier garcia
Click to access 1111.0105v4.pdf
the Riemann Xi function can be written as $ \xi(s)=Cdet(H+s(s-1)+1/4) $
Here ‘H’ is a Hamiltonian operator whose potential is $ V^{-1}(x) = A \sqrt D n(x) $ where N(x) is the eigenvalue staircase :)
Reply
13 February, 2012 at 12:40 pm
jose javier garcia
my operator is correct but none put attention because i am no famous :( , you can check the energies and see that they agree with the square of the riemann zeros :D
Reply
16 July, 2014 at 3:14 am
latex \displaystyle {\mathbb P}\in \left(2,3,5,7,11\right)-prime\ numbers$To exhaust all combinations
The question is a mirror image of the entire sequence of numbers or through an extension of the product operator. Or the first and the second time. And most importantly, how it can help to understand the nature of prime numbers, I do not know
Reply
16 July, 2014 at 7:22 pm
@0_Lesh_a: Will you please stop your comments? They are obviously gibberish.
Reply
17 July, 2014 at 4:58 am
0_Lesh_a
It’s just your opinion
Reply
11 June, 2019 at 6:21 am
Otro valiente más (o una nueva "demostración" de la hipótesis de Riemann) - La Ciencia de la Mula Francis
[…] Fields en el ICM de Madrid de 2006, uno de los matemáticos más geniales vivos en la actualidad) afirma en su blog que ha encontrado ciertos “problemas” con una descomposición presentada en la página […]
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Revision Notes
Number
14 Topics · 43 Revision Notes
1.
Number Toolkit
1. ##### Types of Numbers
2. ##### Irrational Numbers
3. ##### Negative Numbers
4. ##### Mathematical Symbols
5. ##### Order of Operations (BIDMAS/BODMAS)
6. ##### Addition & Subtraction
7. ##### Multiplication & Division
8. ##### Operations with Decimals
2.
Set Notation & Venn Diagrams
1. ##### Set Notation & Venn Diagrams
3.
Prime Factors, HCF & LCM
1. ##### Prime Factor Decomposition
2. ##### Uses of Prime Factor Decomposition
3. ##### HCF & LCM
4.
Powers, Roots & Standard Form
1. ##### Powers & Roots
2. ##### Laws of Indices
3. ##### Converting to & from Standard Form
4. ##### Operations with Standard Form
5.
Fractions Toolkit
1. ##### Basic Fractions
2. ##### Mixed Numbers & Improper Fractions
3. ##### Adding & Subtracting Fractions
4. ##### Multiplying & Dividing Fractions
6.
Percentages Toolkit
1. ##### Basic Percentages
2. ##### Percentage Increases & Decreases
3. ##### Reverse Percentages
7.
Simple & Compound Interest, Growth & Decay
1. ##### Simple Interest
2. ##### Compound Interest
3. ##### Depreciation
4. ##### Exponential Growth & Decay
8.
Surds
1. ##### Simplifying Surds
2. ##### Rationalising Denominators
9.
Working with FDP
1. ##### Converting Fractions, Decimals & Percentages
2. ##### Recurring Decimals
3. ##### Ordering Fractions, Decimals & Percentages
10.
Working with Ratios
1. ##### Ratios
2. ##### Problem Solving with Ratios
3. ##### Working with Proportion
11.
Compound Measures (Speed, Density, Pressure)
1. ##### Compound Measures
2. ##### Speed, Density & Pressure
12.
Time, Currency & Conversions
1. ##### Time
2. ##### Money Calculations
3. ##### Exchange Rates
13.
Rounding, Estimation & Bounds
1. ##### Rounding & Estimation
2. ##### Upper & Lower Bounds
14.
Using a Calculator
1. ##### Using a Calculator
Algebra & Sequences
12 Topics · 41 Revision Notes
1.
Algebra Toolkit
1. ##### Algebraic Notation
2. ##### Algebraic Vocabulary
3. ##### Substitution
4. ##### Collecting Like Terms
2.
Algebraic Roots & Indices
1. ##### Algebraic Roots & Indices
3.
Expanding & Factorising Brackets
1. ##### Expanding & Simplifying Single Brackets
2. ##### Expanding Double Brackets
3. ##### Expanding Triple Brackets
4. ##### Factorising Out Terms
5. ##### Factorising by Grouping
6. ##### Factorising Simple Quadratics
7. ##### Factorising Harder Quadratics
8. ##### Difference of Two Squares
9. ##### Deciding the Factorisation Method
4.
Linear Equations & Inequalities
1. ##### Solving Linear Equations
2. ##### Solving Linear Inequalities
5.
Quadratic Equations
1. ##### Solving Quadratics by Factorising
2. ##### The Quadratic Formula
3. ##### Completing the Square
4. ##### Deciding the Quadratic Method
6.
Rearranging Formulas
1. ##### Formulas where Subject Appears Once
2. ##### Formulas where Subject Appears Twice
7.
Simultaneous Equations
1. ##### Linear Simultaneous Equations
2. ##### Quadratic Simultaneous Equations
8.
Algebraic Fractions
1. ##### Simplifying Algebraic Fractions
2. ##### Adding & Subtracting Algebraic Fractions
3. ##### Multiplying & Dividing Algebraic Fractions
4. ##### Solving Equations with Algebraic Fractions
9.
Forming & Solving Equations
1. ##### Forming Equations from Words
2. ##### Forming Equations from Shapes
3. ##### Problem Solving with Equations
10.
Functions
1. ##### Introduction to Functions
2. ##### Domain & Range
3. ##### Composite Functions
4. ##### Inverse Functions
11.
Sequences
1. ##### Introduction to Sequences
2. ##### nth Terms of Linear Sequences
3. ##### Quadratic Sequences
4. ##### Other Sequences
12.
Proportion
1. ##### Direct Proportion
2. ##### Inverse Proportion
Coordinate Geometry & Graphs
7 Topics · 23 Revision Notes
1.
Coordinate Geometry
1. ##### Coordinates
2. ##### Midpoint of a Line
3. ##### Gradient of a Line
4. ##### Length of a Line
2.
Linear Graphs
1. ##### Equations of Straight Lines (y = mx + c)
2. ##### Drawing Straight Line Graphs
3. ##### Parallel Lines
4. ##### Perpendicular Lines
3.
Quadratic Graphs
1. ##### Quadratic Graphs
4.
Further Graphs & Tangents
1. ##### Types of Graphs
2. ##### Drawing Graphs from Tables
3. ##### Solving Equations from Graphs
4. ##### Finding Gradients of Tangents
5.
Solving & Graphing Inequalities
1. ##### Representing Inequalities as Regions
2. ##### Finding Inequalities from Regions
6.
Real-Life Graphs
1. ##### Conversion Graphs
2. ##### Distance-Time Graphs
3. ##### Speed-Time Graphs
4. ##### Rates-of-Change Graphs
7.
Differentiation
1. ##### Differentiation
2. ##### Finding Stationary Points & Turning Points
3. ##### Classifying Stationary Points
4. ##### Problem Solving with Differentiation
Geometry
4 Topics · 19 Revision Notes
1.
Geometry Toolkit
1. ##### Rotational Symmetry
2. ##### Lines of Symmetry
3. ##### 2D Shapes
4. ##### 3D Shapes
5. ##### Planes of Symmetry
6. ##### Converting between Units
7. ##### Squared & Cubic Units
2.
Angles in Polygons & Parallel Lines
1. ##### Basic Angle Properties
2. ##### Angles in Polygons
3. ##### Angles in Parallel Lines
3.
Bearings, Constructions & Scale Drawings
1. ##### Bearings
2. ##### Scale
3. ##### Constructing Triangles
4.
Circle Theorems
1. ##### Angles at Centre & Circumference
2. ##### Angle in a Semicircle
3. ##### Theorems with Chords & Tangents
4. ##### Angles in Cyclic Quadrilaterals
5. ##### Angles in the Same Segment
6. ##### The Alternate Segment Theorem
Lengths, Areas & Volumes
4 Topics · 13 Revision Notes
1.
Area & Perimeter
1. ##### Perimeter
2. ##### Area
3. ##### Adding & Subtracting Areas
4. ##### Problem Solving with Areas
2.
Circles, Arcs & Sectors
1. ##### Area & Circumference of Circles
2. ##### Arc Lengths & Sector Areas
3.
Volume & Surface Area
1. ##### Volume
2. ##### Problem Solving with Volumes
3. ##### Surface Area
4.
Congruence & Similarity
1. ##### Congruence
2. ##### Similarity
3. ##### Similar Lengths
4. ##### Similar Areas & Volumes
Pythagoras & Trigonometry
4 Topics · 11 Revision Notes
1.
Right-Angled Triangles (Pythagoras & Trigonometry)
1. ##### Pythagoras Theorem
2. ##### SOHCAHTOA
3. ##### Angles of Elevation & Depression
4. ##### Exact Trig Values
2.
Sine, Cosine Rule & Area of Triangles
1. ##### The Sine Rule
2. ##### The Cosine Rule
3. ##### Area of a Triangle
4. ##### Deciding the Trig Rule
3.
3D Pythagoras & Trigonometry
1. ##### 3D Pythagoras & Trigonometry
4.
Trigonometric Graphs & Equations
1. ##### Trigonometric Graphs
2. ##### Solving Trig Equations
Vectors & Transformations
2 Topics · 10 Revision Notes
1.
Vectors
1. ##### Introduction to Column Vectors
2. ##### Representing Vectors as Diagrams
3. ##### Magnitude of a Vector
4. ##### Position & Displacement Vectors
5. ##### Finding Vector Paths
6. ##### Problem Solving with Vectors
2.
Transformations
1. ##### Translations
2. ##### Reflections
3. ##### Rotations
4. ##### Enlargements
Probability
3 Topics · 9 Revision Notes
1.
Basic Probability
1. ##### Basic Probability
2. ##### Possibility (Sample Space) Diagrams
3. ##### Relative & Expected Frequency
2.
Probability Diagrams (Tree & Venn Diagrams)
1. ##### Two-Way Tables
2. ##### Probabilities from Venn Diagrams
3. ##### Probability Tree Diagrams
4. ##### Combined Probabilities
3.
Conditional Probability
1. ##### Conditional Probability
2. ##### Combined Conditional Probabilities
Statistics
5 Topics · 20 Revision Notes
1.
Averages & Range
1. ##### Discrete & Continuous Data
2. ##### Mean, Median & Mode
3. ##### Calculations with the Mean
4. ##### Averages from Tables
5. ##### Averages from Grouped Data
6. ##### Range & Interquartile Range
7. ##### Comparing Data Sets
2.
Statistical Diagrams
1. ##### Stem & Leaf Diagrams
2. ##### Bar Charts & Pictograms
3. ##### Pie Charts
4. ##### Reading & Interpreting Statistical Diagrams
5. ##### Comparing Statistical Diagrams
3.
Histograms
1. ##### Frequency Density
2. ##### Drawing Histograms
3. ##### Interpreting Histograms
4.
Cumulative Frequency
1. ##### Cumulative Frequency
2. ##### Drawing Cumulative Frequency Diagrams
3. ##### Interpreting Cumulative Frequency Diagrams
5.
Scatter Graphs & Correlation
1. ##### Scatter Graphs & Correlation
2. ##### Lines of Best Fit
IGCSEMathsCambridge (CIE)ExtendedRevision NotesPythagoras & Trigonometry Right-Angled Triangles (Pythagoras & Trigonometry)Exact Trig Values
Exact Trig Values(Cambridge (CIE) IGCSE Maths):Revision Note
Exam code:0580 & 0980
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Written by:Amber
Reviewed by:Dan Finlay
Updated on 16 June 2025
Exam board:
Cambridge (CIE)
Cambridge (CIE) Maths: ExtendedCambridge (CIE) International Maths: Extended
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Exact trig values
What are exact values in trigonometry?
Forcertain angles the values of sin θ, cos θ and tan θ can be written exactly
This means using fractions and surds
You areexpected to know the exact values of sin, cos and tan for
0°, 30°, 45°, 60°, 90°, 180° and their multiples
| | 0º | 30º | 45º | 60º | 90º |
--- --- --- |
| sin | 0 | | | | 1 |
| cos | 1 | | | | 0 |
| tan | 0 | | 1 | | undefined |
How can I remember these exact trig values?
Look at patterns in the table
Note the values of sin θ from 0° to 90° match cos θ in reverse, from 90° to 0°
Some people remember sin θ using the trick which simplifies to
Two special right-angled trianglesbelow can help you to find the exact values for 30º, 45º and 60º
Remember that by rationalising the denominator,
The trig graphs can help you to remember the exact values for 0º and multiples of 90º
How do I use exact trig values?
You may come across trig questions in a non-calculator question
In trig calculations, substitutein theexact trig values and solve as usual
E.g. Solve the equation
Replace with to give
Then you can solve for
On trig graphs, you may be expected to find a coordinate
E.g. The coordinates lie on the graph , find
will be equal to
The exact value of is
Therefore
Examiner Tips and Tricks
Writing these out (or sketching the triangles/graphs) on your paper at the beginning of the exam means that you can use them as many times as you need to during the exam!
Worked Example
Find the value of in the diagram below.
Give your answer as an exact value.
Triangle ABC is a right-angled triangle, so use SOHCAHTOA
We know the hypotenuse (AC) and we want to calculate the opposite (BC), so use
Remember that
So,
Leave in exact (surd) form
cm
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1123 | https://en.wikipedia.org/wiki/Strain_(mechanics) | Jump to content
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From Wikipedia, the free encyclopedia
Relative deformation of a physical body
| | |
--- |
| | This article needs additional citations for verification. Please help improve this article by adding citations to reliable sources. Unsourced material may be challenged and removed. Find sources: "Strain" mechanics – news · newspapers · books · scholar · JSTOR (July 2025) (Learn how and when to remove this message) |
| Strain | |
--- |
| Other names | Strain tensor |
| SI unit | 1 |
| Other units | % |
| In SI base units | m/m |
| Behaviour under coord transformation | tensor |
| Dimension | |
| |
| Part of a series on |
| Continuum mechanics |
| Fick's laws of diffusion |
| Laws | Conservations | | Mass Momentum Energy | | Inequalities | | Clausius–Duhem (entropy) | |
| Solid mechanics Deformation Elasticity + linear Plasticity Hooke's law Stress Strain + Finite strain + Infinitesimal strain Compatibility Bending Contact mechanics + frictional Material failure theory Fracture mechanics |
| Fluid mechanics | Fluids | | Statics · Dynamics Archimedes' principle · Bernoulli's principle Navier–Stokes equations Poiseuille equation · Pascal's law Viscosity + (Newtonian · non-Newtonian) Buoyancy · Mixing · Pressure | | Liquids | | Adhesion Capillary action Chromatography Cohesion (chemistry) Surface tension | | Gases | | Atmosphere Boyle's law Charles's law Combined gas law Fick's law Gay-Lussac's law Graham's law | | Plasma | |
| Rheology | | | Viscoelasticity Rheometry Rheometer | | Smart fluids | | Electrorheological Magnetorheological Ferrofluids | |
| Scientists Bernoulli Boyle Cauchy Charles Euler Fick Gay-Lussac Graham Hooke Newton Navier Noll Pascal Stokes Truesdell |
| v t e |
In mechanics, strain is defined as relative deformation, compared to a reference position configuration. Different equivalent choices may be made for the expression of a strain field depending on whether it is defined with respect to the initial or the final configuration of the body and on whether the metric tensor or its dual is considered.
Strain has dimension of a length ratio, with SI base units of meter per meter (m/m).
Hence strains are dimensionless and are usually expressed as a decimal fraction or a percentage.
Parts-per notation is also used, e.g., parts per million or parts per billion (sometimes called "microstrains" and "nanostrains", respectively), corresponding to μm/m and nm/m.
Strain can be formulated as the spatial derivative of displacement:
where I is the identity tensor.
The displacement of a body may be expressed in the form x = F(X), where X is the reference position of material points of the body;
displacement has units of length and does not distinguish between rigid body motions (translations and rotations) and deformations (changes in shape and size) of the body.
The spatial derivative of a uniform translation is zero, thus strains measure how much a given displacement differs locally from a rigid-body motion.
A strain is in general a tensor quantity. Physical insight into strains can be gained by observing that a given strain can be decomposed into normal and shear components. The amount of stretch or compression along material line elements or fibers is the normal strain, and the amount of distortion associated with the sliding of plane layers over each other is the shear strain, within a deforming body. This could be applied by elongation, shortening, or volume changes, or angular distortion.
The state of strain at a material point of a continuum body is defined as the totality of all the changes in length of material lines or fibers, the normal strain, which pass through that point and also the totality of all the changes in the angle between pairs of lines initially perpendicular to each other, the shear strain, radiating from this point. However, it is sufficient to know the normal and shear components of strain on a set of three mutually perpendicular directions.
If there is an increase in length of the material line, the normal strain is called tensile strain; otherwise, if there is reduction or compression in the length of the material line, it is called compressive strain.
Strain regimes
[edit]
Depending on the amount of strain, or local deformation, the analysis of deformation is subdivided into three deformation theories:
Finite strain theory, also called large strain theory, large deformation theory, deals with deformations in which both rotations and strains are arbitrarily large. In this case, the undeformed and deformed configurations of the continuum are significantly different and a clear distinction has to be made between them. This is commonly the case with elastomers, plastically-deforming materials and other fluids and biological soft tissue.
Infinitesimal strain theory, also called small strain theory, small deformation theory, small displacement theory, or small displacement-gradient theory where strains and rotations are both small. In this case, the undeformed and deformed configurations of the body can be assumed identical. The infinitesimal strain theory is used in the analysis of deformations of materials exhibiting elastic behavior, such as materials found in mechanical and civil engineering applications, e.g. concrete and steel.
Large-displacement or large-rotation theory, which assumes small strains but large rotations and displacements.
Strain measures
[edit]
In each of these theories the strain is then defined differently. The engineering strain is the most common definition applied to materials used in mechanical and structural engineering, which are subjected to very small deformations. On the other hand, for some materials, e.g., elastomers and polymers, subjected to large deformations, the engineering definition of strain is not applicable, e.g. typical engineering strains greater than 1%; thus other more complex definitions of strain are required, such as stretch, logarithmic strain, Green strain, and Almansi strain.
Engineering strain
[edit]
Engineering strain, also known as Cauchy strain, is expressed as the ratio of total deformation to the initial dimension of the material body on which forces are applied. In the case of a material line element or fiber axially loaded, its elongation gives rise to an engineering normal strain or engineering extensional strain e, which equals the relative elongation or the change in length ΔL per unit of the original length L of the line element or fibers (in meters per meter). The normal strain is positive if the material fibers are stretched and negative if they are compressed. Thus, we have
,
where e is the engineering normal strain, L is the original length of the fiber and l is the final length of the fiber.
The true shear strain is defined as the change in the angle (in radians) between two material line elements initially perpendicular to each other in the undeformed or initial configuration. The engineering shear strain is defined as the tangent of that angle, and is equal to the length of deformation at its maximum divided by the perpendicular length in the plane of force application, which sometimes makes it easier to calculate.
Stretch ratio
[edit]
The stretch ratio or extension ratio (symbol λ) is an alternative measure related to the extensional or normal strain of an axially loaded differential line element. It is defined as the ratio between the final length l and the initial length L of the material line.
The extension ratio λ is related to the engineering strain e by
This equation implies that when the normal strain is zero, so that there is no deformation, the stretch ratio is equal to unity.
The stretch ratio is used in the analysis of materials that exhibit large deformations, such as elastomers, which can sustain stretch ratios of 3 or 4 before they fail. On the other hand, traditional engineering materials, such as concrete or steel, fail at much lower stretch ratios.
Logarithmic strain
[edit]
The logarithmic strain ε, also called, true strain or Hencky strain. Considering an incremental strain (Ludwik)
the logarithmic strain is obtained by integrating this incremental strain:
where e is the engineering strain. The logarithmic strain provides the correct measure of the final strain when deformation takes place in a series of increments, taking into account the influence of the strain path.
Green strain
[edit]
Main article: Finite strain theory
The Green strain is defined as:
Almansi strain
[edit]
Main article: Finite strain theory
The Euler-Almansi strain is defined as
Strain tensor
[edit]
Further information: Infinitesimal strain theory § Infinitesimal strain tensor
The (infinitesimal) strain tensor (symbol ) is defined in the International System of Quantities (ISQ), more specifically in ISO 80000-4 (Mechanics), as a "tensor quantity representing the deformation of matter caused by stress. Strain tensor is symmetric and has three linear strain and three shear strain (Cartesian) components."
ISO 80000-4 further defines linear strain as the "quotient of change in length of an object and its length" and shear strain as the "quotient of parallel displacement of two surfaces of a layer and the thickness of the layer".
Thus, strains are classified as either normal or shear. A normal strain is perpendicular to the face of an element, and a shear strain is parallel to it. These definitions are consistent with those of normal stress and shear stress.
The strain tensor can then be expressed in terms of normal and shear components as:
Geometric setting
[edit]
Consider a two-dimensional, infinitesimal, rectangular material element with dimensions dx × dy, which, after deformation, takes the form of a rhombus. The deformation is described by the displacement field u. From the geometry of the adjacent figure we have
and
For very small displacement gradients the squares of the derivative of and are negligible and we have
Normal strain
[edit]
For an isotropic material that obeys Hooke's law, a normal stress will cause a normal strain. Normal strains produce dilations.
The normal strain in the x-direction of the rectangular element is defined by
Similarly, the normal strain in the y- and z-directions becomes
Shear strain
[edit]
See also: Shear stress
| Shear strain | |
--- |
| Common symbols | γ or ε |
| SI unit | 1, or radian |
| Derivations from other quantities | γ = τ/G |
The engineering shear strain (γxy) is defined as the change in angle between lines AC and AB. Therefore,
From the geometry of the figure, we have
For small displacement gradients we have
For small rotations, i.e. α and β are ≪ 1 we have tan α ≈ α, tan β ≈ β. Therefore,
thus
By interchanging x and y and ux and uy, it can be shown that γxy = γyx.
Similarly, for the yz- and xz-planes, we have
Volume strain
[edit]
This section is an excerpt from Infinitesimal strain theory § Volumetric strain.[edit]
The volumetric strain, also called bulk strain, is the relative variation of the volume, as arising from dilation or compression; it is the first strain invariant or trace of the tensor:
Actually, if we consider a cube with an edge length a, it is a quasi-cube after the deformation (the variations of the angles do not change the volume) with the dimensions and V0 = a3, thus
as we consider small deformations,
therefore the formula.
In case of pure shear, we can see that there is no change of the volume.
Metric tensor
[edit]
Main article: Finite strain theory § Deformation tensors in curvilinear coordinates
A strain field associated with a displacement is defined, at any point, by the change in length of the tangent vectors representing the speeds of arbitrarily parametrized curves passing through that point. A basic geometric result, due to Fréchet, von Neumann and Jordan, states that, if the lengths of the tangent vectors fulfil the axioms of a norm and the parallelogram law, then the length of a vector is the square root of the value of the quadratic form associated, by the polarization formula, with a positive definite bilinear map called the metric tensor.
See also
[edit]
Stress measures
Strain rate
Strain tensor
References
[edit]
^ Lubliner, Jacob (2008). Plasticity Theory (PDF) (Revised ed.). Dover Publications. ISBN 978-0-486-46290-5. Archived from the original (PDF) on 2010-03-31.
^ Jump up to: a b Rees, David (2006). Basic Engineering Plasticity: An Introduction with Engineering and Manufacturing Applications. Butterworth-Heinemann. ISBN 0-7506-8025-3. Archived from the original on 2017-12-22.
^ "Earth."Encyclopædia Britannica from Encyclopædia Britannica 2006 Ultimate Reference Suite DVD ..
^ Rees, David (2006). Basic Engineering Plasticity: An Introduction with Engineering and Manufacturing Applications. Butterworth-Heinemann. p. 41. ISBN 0-7506-8025-3. Archived from the original on 2017-12-22.
^ Hencky, H. (1928). "Über die Form des Elastizitätsgesetzes bei ideal elastischen Stoffen". Zeitschrift für technische Physik. 9: 215–220.
^ Jump up to: a b "ISO 80000-4:2019". ISO. 2013-08-20. Retrieved 2023-08-28.
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Symmetry
A proof utilizes symmetry if the steps to prove one thing is identical to those steps of another. For example, to prove that in triangle ABC with all three sides congruent to each other that all three angles are equal, you only need to prove that if then the other cases hold by symmetry because the steps are the same.
Contents
[hide]
1 Hidden symmetry
2 Symmetry with respect angle bisectors
3 Symmetry with respect angle bisectors 1
4 Construction of triangle
5 Symmetry with respect angle bisectors 2
6 Symmetry of radical axes
7 Composition of symmetries
8 Composition of symmetries 1
9 Composition of symmetries 2
10 Symmetry and secant
11 Symmetry and incircle
12 Symmetry and incircle A
13 Symmetry for 60 degrees angle
14 See also
Hidden symmetry
Let the convex quadrilateral be given.
Prove that
Proof
Let be bisector
Let point be symmetric with respect
is isosceles.
Therefore vladimir.shelomovskii@gmail.com, vvsss
Symmetry with respect angle bisectors
Given the triangle is the incircle, is the incenter,
Points and are symmetrical to point with respect to the lines containing the bisectors and respectively.
Prove that is the midpoint
Proof Denote
The tangents from point to are equal
Point is symmetrical to point with respect is symmetrical to segment
Symilarly, vladimir.shelomovskii@gmail.com, vvsss
Symmetry with respect angle bisectors 1
The bisector intersect the incircle of the triangle at the point The point is symmetric to with respect to the point is symmetric to with respect to Prove that is the bisector of the segment
Proof
The point is symmetric to with respect to
The point is symmetric to with respect to
Similarly
vladimir.shelomovskii@gmail.com, vvsss
Construction of triangle
Given points and at which the segments of the bisectors and respectively intersect the incircle of centered at
Construct the triangle
Construction
We construct the incenter of as circumcenter of
If these points are collinear or if construction is impossible.
We construct bisectors and
We construct the points and symmetrical to point with respect to and respectively.
We construct the bisector and choose the point as the point intersection with the circle closest to the line
We construct a tangent to the the circle at the point It intersects the lines and at points and respectively.
We construct the tangents to which are symmetrical to sideline with respect to and
vladimir.shelomovskii@gmail.com, vvsss
Symmetry with respect angle bisectors 2
Given the triangle is the incircle, is the incenter,
Let be the point on sideline
Points and are symmetrical to point with respect to the lines and respectively. The line contains point
Prove that is the midpoint
Proof
The segment is symmetric to with respect to the segment is symmetric to with respect to So
Similarly at midpoint
or or We use the Law of Sines and get: vladimir.shelomovskii@gmail.com, vvsss
Symmetry of radical axes
Let triangle be given. The point and the circle are the incenter and the circumcircle of
Circle centered at has the radius and intersects at points and Line is the tangent for at the point
Prove that line is symmetry to the line with respect axis
Proof
circle centered at contain points and and is tangent for and
is the radical axis of and is the radical axis of and
is the radical axis of and and are concurrent (at point )
vladimir.shelomovskii@gmail.com, vvsss
Composition of symmetries
Let the inscribed convex hexagon be given, Prove that
Proof
Denote the circumcenter of
the common bisector the common bisector
the smaller angle between lines and
is the symmetry with respect axis is the symmetry with respect axis
It is known that the composition of two axial symmetries with non-parallel axes is a rotation centered at point of intersection of the axes at twice the angle from the axis of the first symmetry to the axis of the second symmetry.
Therefore vladimir.shelomovskii@gmail.com, vvsss
Composition of symmetries 1
Let the triangle be given.
is the incircle, is the incenter, is the circumcenter of The point is symmetric to with respect to is symmetric to with respect to is symmetric to with respect to
Prove: a)
b)
Proof
a) Denote the smaller angle between and
is the symmetry with respect axis is the symmetry with respect axis
counterclockwise direction.
clockwise direction.
Therefore is parallel to tangent line for at point
b) is homothetic to
is the circumcenter of
The center of the homothety lies on the line passing through the circumcenters of the triangles.
vladimir.shelomovskii@gmail.com, vvsss
Composition of symmetries 2
Let triangle be given. The point and the circle are the incenter and the incircle of
Let be the symmetry with respect axis be the symmetry with respect axis the symmetry with respect axis Find the composition of axial symmetries with respect and
Solution
It is known that the composition of three axial symmetries whose axes intersect at one point is an axial symmetry whose axis contains the same point
Consider the composition of axial symmetries for point
is a fixed point of transformation.
This means that the desired axis of symmetry contains points and , this is a straight line
vladimir.shelomovskii@gmail.com, vvsss
Symmetry and secant
The circle centered at and the point be given. Let and be the tangents, be the secant (
Segment intersects segment at point Prove that
Proof
Let be symmetric to with respect the line
It is known that We use symmetry and get It is known that
Triangles and have common side
Similar triangles and have the areas ratio
Therefore
According the Cross-ratio criterion the four points are a harmonic range (on the real projective line).
vladimir.shelomovskii@gmail.com, vvsss
Symmetry and incircle
Let with incircle be given.
Point
Let be the point in
Denote
Prove that
Let be the point in the segment
Let be the point in the ray such that
Denote
Prove that points and are collinear.
Proof
We use Menelaus theorem for a triangle and a transversal line and get:
Denote Then vladimir.shelomovskii@gmail.com, vvsss
Symmetry and incircle A
Denote is the arbitrary point.
Prove that and are concurrent.
Proof
Denote is the incenter of
Let us make the projective transformations mapping circle onto circle and point onto center of this circle.
Denote the result of transformation of point This transformation maps point to infinity.
Segment this transformation maps onto diameter onto
We use the cross-ratio which is fixed, equation the Claim, and get
so is the midpoint of
Point in infinity, so
Lines and are crossing at the line of symmetry therefore lines and are concurrent.
Claim
Let with incircle be given.
Point Prove
Proof
WLOG, Denote the point in such that Denote vladimir.shelomovskii@gmail.com, vvsss
Symmetry for 60 degrees angle
Let an isosceles triangle be given.
Let be the bisector of
a) Prove that
b) Prove that
Proof
a) One can find successively angles (see diagram).
b) Let
Let
Let
vladimir.shelomovskii@gmail.com, vvsss
See also
Symmedian
Antiparallel
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1125 | https://thirdspacelearning.com/gcse-maths/number/percentage-multipliers/ | What are percentage multipliers?
How to find a decimal multiplier from a percentage
Percentage multiplier worksheets
Percentage multiplier examples
Example 1: finding the decimal multiplier Example 2: finding the decimal multiplier Example 3: finding the decimal multiplier
How to use a percentage multiplier to calculate the percentage of an amount
↓
Example 4: finding the percentage of an amount Example 5: finding the percentage of an amount Example 6: finding the percentage of an amount Example 7: calculating a percentage change - a percentage increase Example 8: calculating a percentage change - a percentage decrease
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What are percentage multipliers?
How to find a decimal multiplier from a percentage
Percentage multiplier worksheets
Percentage multiplier examples
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Example 1: finding the decimal multiplier Example 2: finding the decimal multiplier Example 3: finding the decimal multiplier
How to use a percentage multiplier to calculate the percentage of an amount
↓
Example 4: finding the percentage of an amount Example 5: finding the percentage of an amount Example 6: finding the percentage of an amount Example 7: calculating a percentage change - a percentage increase Example 8: calculating a percentage change - a percentage decrease
Common misconceptions
Related lessons
Practice percentage multiplier questions
Percentage multiplier GCSE questions
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GCSE Maths Number Percentages
Percentage Multipliers
Percentage Multipliers
Here we will learn about about using a percentage multiplier including how to find the single multiplier from a percentage and use the single multiplier to answer percentage questions.There are also percentage multiplier worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.
What is a percentage multiplier?
A percentage multiplier is a number which is used to calculate a percentage of an amount or used to increase or decrease an amount by a percentage.
E.g.
In order to find 12% of a number we can multiply the number by a multiplier:
[12\% = \frac{12}{100} = 0.12]
So 0.12 is the multiplier.
What is a percentage multiplier?
These questions will often involve interest rates in financial situations such as simple interest or compound interest. It is sometimes referred to as the multiplier method.
How to find a decimal multiplier from a percentage
In order to write a decimal multiplier from a percentage:
Write down the percentage
Convert this percentage to a decimal by dividing by 100 – this is the multiplier
Multiply the original amount by the multiplier
Explain how to write a decimal multiplier from a percentage in 3 steps
Percentage multipliers worksheet
Get your free percentage multipliers worksheet of 20+ questions and answers. Includes reasoning and applied questions.
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Percentage multipliers worksheet
Get your free percentage multipliers worksheet of 20+ questions and answers. Includes reasoning and applied questions.
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Percentage multiplier examples
Example 1: finding the decimal multiplier
What is the decimal multiplier for 58 \%?
Write down the percentage required
[58 \%]
2Convert the percentage to a decimal by dividing by 100
[58\% = \frac{58}{100} = 0.58]
The decimal multiplier is 0.58.
Example 2: finding the decimal multiplier
What is the decimal multiplier for 26%?
Write down the percentage
[26 \%]
Convert the percentage to a decimal by dividing by 100
[26\% = \frac{26}{100} = 0.26]
The decimal multiplier is 0.26.
Example 3: finding the decimal multiplier
What is the decimal multiplier for 4.5%?
Write down the percentage.
[4.5 \%]
Convert the percentage to a decimal by dividing by 100
[4.5\% = \frac{4.5}{100} = 0.045]
The decimal multiplier is 0.045.
How to use a percentage multiplier to calculate the percentage of an amount
In order to use a percentage multiplier to calculate the percentage of an amount:
Write down what percentage you need
Convert this percentage to a decimal by dividing by 100; this is the decimal multiplier
Multiply the original amount in the question by the decimal multiplier
Example 4: finding the percentage of an amount
Work out 34% of £700
Write down the percentage
[34 \%]
Convert the percentage to a decimal by dividing by 100
[34\% = \frac{34}{100} = 0.34]
Multiply the original amount in the question by the decimal multiplier
What is the original amount? 700
What is the decimal multiplier? 0.34
[700\times0.34=238]
The answer is £238.
Example 5: finding the percentage of an amount
Work out 8% of £650
Write down the percentage
[8 \%]
Convert the percentage to a decimal by dividing by 100
[8\% = \frac{8}{100} = 0.08]
Multiply the original amount in the question by the decimal multiplier
What is the original amount? 650
What is the decimal multiplier? 0.08
[650\times0.08=52]
The answer is £52.
Example 6: finding the percentage of an amount
Work out 15.6% of 200 kg
Write down the percentage
[15.6 \%]
Convert the percentage to a decimal by dividing by 100
[15.6\% = \frac{15.6}{100} = 0.156]
Multiply the original amount in the question by the decimal multiplier
What is the original amount? 200
What is the decimal multiplier? 0.156
[200\times0.156=31.2]
The answer is 31.2 kg.
Example 7: calculating a percentage increase
Increase £320 by 25%
Write down the percentage. It is an increase so we need to add it to 100 \%
[100\%+25\%=125\%]
Convert the percentage to a decimal by dividing by 100
[125\% = \frac{125}{100} = 1.25]
Multiply the original amount in the question by the decimal multiplier
What is the original amount? 320
What is the decimal multiplier? 1.25
[320\times1.25=400]
The answer is £400.
Example 8: calculating a percentage decrease
Decrease £600 by 7%
Write down the percentage. It is a decrease so we need to subtract it from 100 \%
[100\%-7\%=93\%]
Convert the percentage to a decimal by dividing by 100
[93\% = \frac{93}{100} = 0.93]
Multiply the original amount in the question by the decimal multiplier
What is the original amount? 600
What is the decimal multiplier? 0.93
[600\times0.93=558]
The answer is £558.
Common misconceptions
Decimal multipliers greater than 1
If you need the decimal multiplier of 135% it would be 1.35, so multipliers can in fact be greater than 1.
[135\% = \frac{135}{100} = 1.35]
Take care to remember pence when you are working with money
An answer of £43.7 should be written as £43.70. You need two decimal places for the pence part of the answer.
Related lessons
Percentage multipliers is part of our series of lessons to support revision on percentages. You may find it helpful to start with the main percentages lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:
Percentages
Percentage of an amount
Percentage increase
Percentage decrease
Percentage change
Reverse percentages
One number as a percentage of another
Percentage profit
Practice percentage multiplier questions
What is the decimal multiplier of 61\% ?
0.61
6.10
610
0.061
61\% = \frac{61}{100} = 0.61
So 0.61 is the multiplier.
What is the decimal multiplier of 5\% ?
0.5
5.00
500
0.05
5\% = \frac{5}{100} = 0.05
So 0.05 is the multiplier.
What is the decimal multiplier of 18.3\% ?
0.183
1.83
18.3
0.0183
18.3\% = \frac{18.3}{100} = 0.183
So 0.183 is the multiplier.
Work out 29\% of £400
\pounds 371
\pounds 116
\pounds 29
\pounds 429
29\% = \frac{29}{100} = 0.29
So 0.29 is the multiplier.
0.29 \times 400 = 116
Work out 9\% of 450 km?
45.0 km
40.5 km
441 km
50.0 km
9\% = \frac{9}{100} = 0.09
So 0.09 is the multiplier.
0.09 \times 450 = 40.5
Work out 14.8\% of 560 kg?
8.288 kg
412 kg
148 kg
82.88 kg
14.8\% = \frac{14.8}{100} = 0.148
So 0.148 is the multiplier.
0.148 \times 560 = 82.88
Percentage multiplier GCSE questions
Write 61\% as a decimal
(1 mark)
Show answer
61\% = \frac{61}{100} = 0.61
0.61
(1)
Work out 38\% of 600 kg
(2 marks)
Show answer
0.38 × 600
(1)
= 228 kg
(1)
Fiona is booking a holiday.
The holiday costs £700 .
She pays a 15\% deposit.
Work out how much she has left to pay.
(2 marks)
Show answer
100\% \hspace{1mm}- 15\% = 85\%=0.85
0.85 × 700
(1)
\pounds 595
(1)
Learning checklist
You have now learned how to:
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Find the decimal multiplier of a percentage
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Interpret percentages and percentage changes as a decimal
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Use the decimal multiplier to work out the percentage of an amount
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Interpret percentages multiplicatively
Next lessons
Standard form
Straight line graphs
Compound interest
Simple interest
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Introduction
What are percentage multipliers?
How to find a decimal multiplier from a percentage
Percentage multiplier worksheets
Percentage multiplier examples
↓
Example 1: finding the decimal multiplier Example 2: finding the decimal multiplier Example 3: finding the decimal multiplier
How to use a percentage multiplier to calculate the percentage of an amount
↓
Example 4: finding the percentage of an amount Example 5: finding the percentage of an amount Example 6: finding the percentage of an amount Example 7: calculating a percentage change - a percentage increase Example 8: calculating a percentage change - a percentage decrease
Common misconceptions
Related lessons
Practice percentage multiplier questions
Percentage multiplier GCSE questions
Learning checklist
Next lessons
Still stuck?
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1126 | https://www.merckmanuals.com/professional/clinical-pharmacology/pharmacokinetics/overview-of-pharmacokinetics | honeypot link
skip to main content
Merck ManualProfessional Version
Overview of Pharmacokinetics
ByJennifer Le, PharmD, MAS, BCPS-ID, FIDSA, FCCP, FCSHP, Skaggs School of Pharmacy and Pharmaceutical Sciences, University of California San Diego
Reviewed ByEva M. Vivian, PharmD, MS, PhD, University of Wisconsin School of Pharmacy
Reviewed/Revised Modified Nov 2024
v1108791
View Patient Education
Pharmacokinetics, sometimes described as what the body does to a drug, refers to the movement of drug into, through, and out of the body—the time course of its absorption, bioavailability, distribution, metabolism, and excretion.
Pharmacodynamics, described as what a drug does to the body, involves receptor binding, postreceptor effects, and chemical interactions.
Drug pharmacokinetics determines the onset, duration, and intensity of a drug’s effect. Formulas relating these processes summarize the pharmacokinetic behavior of most drugs (see table Formulas Defining Basic Pharmacokinetic Parameters).
Table
Formulas Defining Basic Pharmacokinetic Parameters
Table
Formulas Defining Basic Pharmacokinetic Parameters
Formulas Defining Basic Pharmacokinetic Parameters
| Category | Parameter | Formula |
| Absorption | Absorption rate constant | Rate of drug absorption ÷ amount of drug remaining to be absorbed |
| Bioavailability | Amount of drug absorbed ÷ drug dose |
| Distribution | Apparent volume of distribution | Amount of drug in body ÷ plasma drug concentration |
| Unbound fraction | Plasma concentration of unbound drug ÷ total plasma drug concentration |
| Elimination (metabolism and excretion) | Rate of elimination | Renal excretion + extrarenal (usually metabolic) elimination |
| Clearance | Rate of drug elimination ÷ plasma drug concentration, or elimination rate constant × apparent volume of distribution |
| Renal clearance | Rate of renal excretion of drug ÷ plasma drug concentration |
| Metabolic clearance | Rate of drug metabolism ÷ plasma drug concentration |
| Fraction excreted unchanged | Rate of renal excretion of drug ÷ rate of drug elimination |
| Elimination rate constant | Rate of drug elimination ÷ amount of drug in body |
| Clearance ÷ volume of distribution |
| Biologic half-life | 0.693 ÷ elimination rate constant (for first-order elimination only―see Rate) |
Pharmacokinetics of a drug depends on patient-related factors as well as on the drug’s chemical properties. Some patient-related factors (eg, renal function, genetic makeup, sex, age) can be used to predict the pharmacokinetic parameters in populations. For example, the half-life of some drugs, especially those that require both metabolism and excretion, may be remarkably long in older adults (see figure Comparison of Pharmacokinetic Outcomes for Diazepam in a Younger Man [A] and an Older Man [B]Comparison of Pharmacokinetic Outcomes for Diazepam in a Younger Man [A] and an Older Man [B]). In fact, physiologic changes with aging affect many aspects of pharmacokinetics (see Pharmacokinetics in Older Adults and Pharmacokinetics in Children) (1, 2).
Other factors are related to individual physiology. The effects of some individual factors (eg, renal failure, obesity, hepatic failure, dehydration) can be reasonably predicted, but other factors are idiosyncratic and thus have unpredictable effects. Because of individual differences, drug administration must be based on each patient’s needs—traditionally, by empirically adjusting dosage until the therapeutic objective is met. This approach is frequently inadequate because it can delay optimal response or result in adverse effects.
Knowledge of pharmacokinetic principles helps prescribers adjust dosage more accurately and rapidly. Application of pharmacokinetic principles to individualize pharmacotherapy is termed therapeutic drug monitoring.
Comparison of Pharmacokinetic Outcomes for Diazepam in a Younger Man (A) and an Older Man (B)
| |
| Diazepam is metabolized in the liver to desmethyldiazepam through P-450 enzymes. Desmethyldiazepam is an active sedative, which is excreted by the kidneys. 0 = time of dosing. (Adapted from Greenblatt DJ, Allen MD, Harmatz JS, Shader RI: Diazepam disposition determinants. Clin Pharmacol Ther 27:301–312, 1980. doi: 10.1038/clpt.1980.40) |
Le J, Bradley JS. Optimizing Antibiotic Drug Therapy in Pediatrics: Current State and Future Needs. J Clin Pharmacol 2018;58 Suppl 10:S108-S122. doi:10.1002/jcph.1128
Mangoni AA, Jackson SH. Age-related changes in pharmacokinetics and pharmacodynamics: basic principles and practical applications. Br J Clin Pharmacol 2004;57(1):6-14. doi:10.1046/j.1365-2125.2003.02007.x
Drugs Mentioned In This Article
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1127 | https://thefactfactor.com/facts/pure_science/chemistry/physical-chemistry/gay-lussacs-law-of-combining-volumes/12444/ | Gay-Lussac's Law of Combining Volumes: Statement, Explanation
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Gay-Lussac’s Law of Combining Volumes
Post author By Hemant More
Post date May 20, 2020
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Science >Chemistry>Laws of Chemical Combinations> Gay-Lussac’s Law of Combining Volumes
In the previous article, we have studied the law of reciprocal proportions. In this article, we shall study Gay-Lussac’s Law of Combining Volumes. A French chemist Joseph L. Gay – Lussac in 1809,put forward this law.
Statement :
Whenever gases take part in a chemical reaction, either as reactants or as products, they do so in simple proportions by Volumes. Provided the volumes of gases are measured at the same temperature and pressure,
Illustration 1:
Consider following reaction
H 2(g)+ Cl 2(g) →2HCl
1vol 1vol 2 vol
Thus the simple ratio of volumes is 1 : 1 : 2
Illustration 2:
Consider following reaction
N 2(g)+ 3 H 2(g) →2NH 3(g)
1vol 3vol 2 vol
Thus the simple ratio of volumes is 1 : 3 : 2
Numerical Problems:
Example – 01:
Calculate the volume of oxygen required for the complete combustion of 0.25 dm 3 of methane at STP.
Solution:
CH 4(g) + 2O 2(g) → CO 2(g) + 2H 2 O(g)
1 vol 2 vol 1 vol 2 vol
By Gay-Lussac’s law of combining volumes of gases
1 vol of methane requires 2 vol of oxygen for complete combustion.
Hence 0.25 dm 3 of methane requires 2 x 0.25 = 0.50 dm 3 of oxygen.
Example – 02:
Calculate the volume of hydrogen required for the complete hydrogenation of 0.25 dm 3 of ethyne at STP.
Solution:
C 2 H 2(g) + 2H 2(g) →C 2 H 6(g)
1 vol 2vol 1 vol
By Gay-Lussac’s law of combining volumes of gases
1 vol of ethyne requires 2 vol of hydrogen for complete hydrogenation.
Thus 0.25 dm 3 of ethyne requires 2 x 0.25 = 0.50 dm 3 of hydrogen for complete hydrogenation.
Example – 03:
Calculate the volume of hydrogen required for the complete hydrogenation of 0.25 dm 3 of ethylene at STP.
Solution:
C 2 H 4(g) + H 2(g) →C 2 H 6(g)
1 vol 1vol 1 vol
By Gay-Lussac’s law of combining volumes of gases
1 vol of ethylene requires 1 vol of hydrogen for complete hydrogenation.
Thus 0.25 dm3 of ethylene requires 1 x 0.25 = 0.25 dm 3 of hydrogen for complete hydrogenation.
Example – 04:
Calculate the volume of oxygen required for the complete combustion of 0.25 mol of methane at STP.
Solution:
CH 4(g) + 2O 2(g) → CO 2(g) + 2H 2 O(g)
1 mol 2 mol 1 mol 2 mol
By Gay-Lussac’s law of combining volumes of gases
1 mol of methane requires 2 mol of oxygen for complete combustion.
Thus 0.25 mol of methane requires 2 x 0.25 = 0.50 mol of oxygen.
One mole of any gas occupies 22.4 dm 3 by volume at STP.
Volume of oxygen required = 22.4 x No. of moles= 22.4 x 0.5 = 11.2 dm 3
Example – 05:
Calculate the volume of oxygen required for the complete combustion of 0.5 dm 3 of H 2 S at STP.
Solution:
2H 2 S(g) + 3O 2(g) → 2SO 2(g) + 2H 2 O(g)
2 vol 3 vol 2 vol 2 vol
By Gay-Lussac’s law of combining volumes of gases
2 vol of H 2 S requires 3 vol of oxygen for complete combustion.
1 vol of H 2 S requires 3/2 = 1.5 vol of oxygen for complete combustion.
Thus 0.5 dm 3 of H 2 S requires 1.5 x 0.5 = 0.75 dm 3 of oxygen for complete combustion.
Example – 06:
Calculate the volume of oxygen required at STP for the complete combustion of 5.0 dm 3 of ethane at 295 K and 0.993 x 10 5 Nm-2
Solution:
2C 2 H 6(g) + 7O 2(g) → 4CO 2(g) + 6H 2 O(g)
2 vol 7 vol 4 vol 6 vol
By Gay-Lussac’s Law of Combining Volumes of gases
2 vol of C 2 H 6 requires 7 vol of oxygen for complete combustion.
1 vol of C 2 H 6 requires 7/2 = 3.5 vol of oxygen for complete combustion.
Thus 5 dm 3 of C 2 H 6 requires 3.5 x 5 = 17.5 dm 3 of oxygen for complete combustion.
P = 0.993 x 10 5 Nm-2, T = 295 K, V = 17.5 dm 3
P o = 1.013 x 10 5 Nm-2, T o= 273 K, V o= ?
Thus the volume of oxygen required at STP is 15.88 dm 3.
Example – 07:
6.0 dm 3 of hydrogen is reacted with 2.4 dm 3 of oxygen in a closed chamber. Calculate composition of resulting mixture.
Solution:
2H 2(g) + O 2(g) → 2 H 2 O(g)
2 vol 1 vol 2 vol
By Gay-Lussac’s law of combining volumes of gases.
The ratio of the volume of hydrogen to that of oxygen is 2 : 1.
In this case,oxygen is limiting reagent, and hydrogen is an excess reagent.
2.4 dm 3 of oxygen can combine with 2 x 2.4 = 4.8 dm 3 of hydrogen to form 2 x 2.4 = 4.8 dm 3 of water vapours.
Thus unreacted hydrogen = 6.0 – 4.8 = 1. 2 dm 3.
Thus resulting mixture contains 4.8 dm 3 of water vapours and 1.2 dm 3 of unreacted hydrogen.
Example – 08:
15 litres of nitrogen is made to react with 30 litres of hydrogen to prepare ammonia. Calculate composition of resulting mixture.
Solution:
N 2(g)+ 3 H 2(g) →2NH 3(g)
1vol 3vol 2 vol
By Gay-Lussac’s law of combining volumes of gases
The ratio of the volume of nitrogen to that of hydrogen is 1 : 3.
In this case, hydrogen is limiting reagent, and nitrogen is an excess reagent.
3 x 10 = 30 litres of hydrogen can combine with 1 x 10 = 10 litres of nitrogen to form 2 x 10 =20 litres of ammonia.
Thus unreacted nitrogen = 15.0 – 10.0 = 5 litres.
Thus resulting mixture contains 20 litres of ammonia and 5 litres of unreacted nitrogen.
Example – 09:
200 dm 3 of hydrogen gas is allowed to react with 250 dm 3 of chlorine gas. Calculate composition of resulting mixture.
Solution:
H 2(g)+ Cl 2(g) →2HCl
1vol 1vol 2 vol
By Gay-Lussac’s law of combining volumes of gases.
The ratio of the volume of hydrogen to that of chlorine is 1 : 1.
In this case, hydrogen is limiting reagent and chlorine is excess reagent.
200 dm 3 of hydrogen can combine with 200 dm 3 of chlorine to form 2 x 200 =400 dm 3 of hydrogen chloride.
Thus unreacted chlorine = 250 – 200 = 50 dm 3.
Thus resulting mixture contains 400 dm 3 of hydrogen chloride and 50 dm 3 of unreacted chlorine.
Example – 10:
10 dm 3 of hydrogen gas is allowed to react with 15 dm 3 of chlorine gas. Calculate composition of resulting mixture.
Solution:
H 2(g)+ Cl 2(g) →2HCl
1vol 1vol 2 vol
By Gay-Lussac’s law of combining volumes of gases.
The ratio of the volume of hydrogen to that of chlorine is 1 : 1
In this case, hydrogen is limiting reagent and chlorine is excess reagent.
10 dm 3 of hydrogen can combine with 10 dm 3 of chlorine to form 2 x 10 = 20 dm 3 of hydrogen chloride.
Thus unreacted chlorine = 15 – 10 = 5 dm 3.
Thus resulting mixture contains 20 dm 3 of hydrogen chloride and 5 dm3 of unreacted chlorine
In the next chaper we shall study the concept of atomic mass and equivalent mass.
Previous Topic: The Law of Reciprocal Proportions
Next Chapter: Concept of Atomic Mass and Equivalent Mass
Science >Chemistry>Laws of Chemical Combinations> Gay-Lussac’s Law of Combining Volumes
Tags Chemistry, Gay-Lussac, Gay-Lussac's law of combining volume, Laws of chemical combinations
←Law of Reciprocal Proportions→Concept of Atomic Mass
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A frog is at the bottom of a 10-meter well. Each day he climbs up 3 meters. Each night he slides down 1 meter. On what day will he reach the top of the well and escape? Which of the following problem-solving strategies would be the most appropriate to use to solve this problem? A. Work backwards B. Draw a diagram (picture) C. Set up an equation D. Identify a sub-goal Mathematics Proficiency Exam (ECAE)
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Brooklyn College
Mathematics Proficiency Exam
Question
A frog is at the bottom of a 10meter well Each day he climbs up 3
Brooklyn College
Mathematics Proficiency Exam
Question
Anonymous Student
3 years ago
A frog is at the bottom of a 10-meter well. Each day he climbs up 3 meters. Each night he slides down 1 meter. On what day will he reach the top of the well and escape? Which of the following problem-solving strategies would be the most appropriate to use to solve this problem? A. Work backwards B. Draw a diagram (picture) C. Set up an equation D. Identify a sub-goal
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Answer Created with AI
3 years ago
Solution: Let the distance climbed in a day be d and the total distance to be climbed be T.
d=3 m−1 m=2 m\begin{aligned} d&=3\,\text{m}-1\text{ m}&=2\,\text{m} \end{aligned}
Divide T by d to obtain the result.
T d=10 m 2 m/day=5 days\begin{aligned} \frac{T}{d}&=\frac{10\,\text{m}}{2\,\text{m}/\text{day}}&=5\,\text{days} \end{aligned}
Working backwards will not help in this situation. So, the problem cannot be solved in this way. Drawing a diagram is also a way to solve this problem but that will be a bit more tedious than forming equations. So, this is not the most appropriate way. The problem is easily solved by setting up equations. Identification of subgoals also involves setting up and solving equations. So, this is also not as easy as drawing a simple straight line diagram.
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1129 | https://fluentwords.net/en/collocations/original/en | Typical usage patterns for 'original' | FluentWords
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Top 10 collocations for original
| # | Collocations for original | Example Sentence |
---
| 1 | original version | These have been left as in the originalversion . |
| 2 | original building | In 1867 , a fire destroyed the originalbuilding . |
| 3 | original text | The following changes were made to the originaltext . |
| 4 | original series | The animation quality is far higher than the originalseries . |
| 5 | original member | Leaving only Pedro and Oscar as the originalmembers . |
| 6 | original plan | Many items now approved were not in the originalplans . |
| 7 | original name | The title refers to the band 's originalname . |
| 8 | original design | No motto had been included in the originaldesign . |
| 9 | original song | The rest of the tracks are all originalsongs . |
| 10 | original lineup | Saints of Los Angeles features the band 's originallineup . |
Click on the word cloud to see example sentences
original version original building original text original plan original name original design original song
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original + Noun
e.g. original version
version building text series member plan name design song lineup title owner production location material composition site idea form script concept settler film church release article book
original + Verb +(e)s
e.g. original release +(e)s
release read
Noun + Prep + original
e.g. copy of the original
copy of the
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1130 | https://www.freesoft.org/ModernIntegration/ModernIntegration.pdf | Modern Integration Brent Baccala Contents Table of Contents i Contents i 1 Introduction 1 1.1 Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3 1.2 Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10 1.3 Richardson’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14 1.4 Algorithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
16 1.5 Sage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17 1.5.1 Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17 1.5.2 Arrays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18 2 Commutative Algebra 21 2.1 Rings and Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
21 2.2 Fraction fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
23 2.3 Polynomial rings and rational function fields . . . . . . . . . . . . . . . .
25 2.4 Long Division . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
26 2.5 Greatest Common Divisors . . . . . . . . . . . . . . . . . . . . . . . . . .
27 2.6 Polynomial Diophantine Equations . . . . . . . . . . . . . . . . . . . . . .
32 2.7 Partial Fractions Expansion . . . . . . . . . . . . . . . . . . . . . . . . . .
36 2.8 Resultants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
41 2.9 Algebraic Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
44 2.10 Trace and Norm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
47 2.11 Hermite Normal Form . . . . . . . . . . . . . . . . . . . . . . . . . . . .
49 2.12 Factorization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
51 2.13 Primary Decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . .
55 2.14 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
57 3 Algebraic Geometry 59 3.1 Solving systems of equations . . . . . . . . . . . . . . . . . . . . . . . . .
59 4 Differential Algebra 61 4.1 Differential Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
61 4.2 Liouvillian Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
69 4.3 Liouville’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
70 5 Integration of Rational Functions 79 5.1 Logarithms and related functions . . . . . . . . . . . . . . . . . . . . . . .
79 5.2 Multi-valued logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . .
81 5.3 A Bit Of Perspective . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
90 6 The Logarithmic Extension 91 6.1 The Logarithmic Integration Theorem . . . . . . . . . . . . . . . . . . . .
93 6.2 Hermite Reduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 7 The Exponential Extension 113 7.1 The Exponential Integration Theorem . . . . . . . . . . . . . . . . . . . . 115 7.2 Risch Equations in C(x) . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 7.3 Risch Equations over Normal Polynomials . . . . . . . . . . . . . . . . . . 135 7.4 Normal Risch Equations in Sage . . . . . . . . . . . . . . . . . . . . . . . 138 7.5 Risch Equations over Special Polynomials . . . . . . . . . . . . . . . . . . 142 8 Algebraic Curves 151 8.1 The Topology of an Algebraic Curve . . . . . . . . . . . . . . . . . . . . . 153 8.2 Puiseux Expansions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161 8.3 Valuation Rings and Orders . . . . . . . . . . . . . . . . . . . . . . . . . . 168 8.4 Sage’s FunctionField code . . . . . . . . . . . . . . . . . . . . . . . . . . 178 8.5 Divisors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194 8.6 Riemann-Roch spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195 8.7 Mittag-Leffler Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . 199 8.8 Parallels with the Transcendental Cases . . . . . . . . . . . . . . . . . . . 201 9 Abelian Integrals 205 9.1 The Abelian Integration Theorem . . . . . . . . . . . . . . . . . . . . . . 206 9.2 Some Simple Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208 9.3 An integral Sage can’t solve . . . . . . . . . . . . . . . . . . . . . . . . . 217 9.4 Geddes’s example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222 9.5 Holliman’s Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225 9.6 Chebyshev’s Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228 9.7 The Risch Theorem: A First Look . . . . . . . . . . . . . . . . . . . . . . 234 9.8 Hermite reduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243 9.9 Señor Gonzalez, otra vez . . . . . . . . . . . . . . . . . . . . . . . . . . . 246 10 The Risch Theorem 247 10.1 The Riemann-Roch Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 248 10.2 Jacobian Varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250 10.3 Simple Algebraic Extensions over Finite Fields . . . . . . . . . . . . . . . 252 10.4 Endomorphism Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255 11 Algebraic Extensions 257 11.1 Exponential Extensions over Simple Algebraic Extensions . . . . . . . . . 257 11.2 Integral Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261 11.3 Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262 11.4 The K[θ]-module I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263 11.5 Basis for all Rational Functions . . . . . . . . . . . . . . . . . . . . . . . 266 11.6 Divisors and Integral Modules . . . . . . . . . . . . . . . . . . . . . . . . 269 12 Notes 277 12.1 Valuations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 280 12.2 Notes on Harris - Geometry of Alg Curves - Harvard 287 . . . . . . . . . . 281 12.3 Notes on [Fu08] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282 12.4 Notes on [Sh61] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285 12.5 Notes on Bronstein . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287 12.6 Function Fields (Stichtenoth) . . . . . . . . . . . . . . . . . . . . . . . . . 288 12.7 The Riemann-Roch Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 289 12.8 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289 12.9 arcsin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295 Bibliography 299 Preface In 1970, Robert Risch published [Ri70], which sketched in four pages how to bound the torsion of a divisor on an algebraic curve, and thus provided the “missing link” in a comprehensive algorithm that would either find an elementary form for a given integral, or prove that no such elementary form can exist. Risch’s method, suitably enhanced, is currently used in the symbolic integration routines of the most sophisticated computer algebra systems.
The goal of this book is to present the Risch integration algorithm in a manner suitable to be understood by undergraduate mathematics students, the prerequisites being calculus and abstract algebra, and the expected context being a senior-level university class.
Why, first of all, should math students study this subject, and why near the end of an undergraduate mathematics program?
First and foremost, for pedagogical reasons. Almost all modern college math curricula include higher algebra, yet this subject seems to be taught in a very abstract context. The integration problem puts this abstraction into concrete form. One of the best ways to learn a subject is to apply it in a specific and concrete way. The greatest difficulties I have encountered in math is when faced with abstract concepts lacking concrete examples.
Such, in my mind, is the primary benefit of studying Risch integration near the end of an undergraduate program. The student has no doubt been exposed to higher algebra, now we want to make sure we understand it by taking all those rings, fields, ideals, extensions and what not and applying them to a specific goal.
Secondly, there is a sense of both historical and educational completion to be obtained.
Not only has the integration problem challenged mathematicians since the development of the calculus, but there is a real danger of getting through an entire calculus sequence and be left thinking that if you really want to solve an integral, the best way is to use a computer! Due to the intricacy of the calculations involved, the best way probably is to use a computer, but without studying the Risch algorithm, the student is left with a vague sense that integration is nothing but a bag of tricks, and a real deficiency without understanding that the integration problem has been solved.
Third, an introduction to differential algebra may be quite appropriate at a point where students are starting to think about research interests. Though this field has profitably engaged the attentions of a number of late twentieth century mathematicians, it is still a young field that may turn out to be a major breakthrough in the solution of differential equations. It may also turn out to be a dead end (“interesting but not compelling” in the words of one commentator), which I why I hesitate to list this reason first on my list.
The big question, in my mind, is whether this theory can be suitably extended to handle partial differential equations, as both integrals and ordinary differential equations can now be adequately handled using numerical techniques. This question remains unanswered at this time, and that mystery has animated my own mathematical research for a number of years.
Furthermore, the available material on this subject is spread around among some terse research papers, some sparse lecture notes, and a single graduate level textbook ([Br05]), that while excellent, is unfortunately incomplete due to the untimely death of its author prior to completing an anticipated second volume. Having slowly assimilated this mate-rial over the course of years of study, and having given roughly a dozen lectures on Risch integration without the benefit of a textbook, the lack of a suitable text has become ob-vious. Although I began work on this textbook in 2006, I set it aside after a while and moved on to other interests. In the winter of 2016-17, I was once again preparing to lec-ture on Risch integration, and once again scrambling to pull everything together without a textbook.
Therefore, it seems appropriate to compile this knowledge together and offer it back to the mathematical community. Partly for the reasons I have listed above, and partly just to write something different from [Br05], I have decided to target this book at an undergrad-uate audience with some exposure to higher algebra.
I have liberally used the computer algebra system Sage in conjunction with the L A T EX package pythontex, which, suitable extended, allows Sage code in the L AT EX source to be automatically processed and typeset into the output. In keeping with my Christian religious principles, the book is freely available on the Internet, both in PDF form, and as L A T EX source in a github repository.
Since the book is still a work in progress, I can’t hope to properly conclude this preface at this time. I would, however, like to specifically thank my dear friend Bruce Caslow, whose support and encouragement has been invaluable in this, as well as many other pursuits.
Chapter 1 Introduction Who Wants to be a Mathematician?
$50,000 Question1 Which of the following integrals can not be expressed as an elementary function?
A.
R sin x dx B.
R e−x2 dx C.
R x{(x2e2x2−ln2(x+1))2+2xe3x2(x−(2x3+2x2+x+1) ln(x+1))} (x+1)(ln2(x+1)−x2e2x2)2 dx D.
R 2x6+4x5+7x4−3x3−x2−8x−8 (2x2−1)2√ x4+4x3+2x2+1 dx 1The instructor may not necessarily possess a $50,000 prize fund.
The answer to this “$50,000” question is, somewhat surprisingly, (B). Simplifying (A) as R sin dx = −cos x + C is an easy exercise from a first year calculus course. (C) and (D), while appearing more formidable, are solvable using the techniques of this book.
(C) is example 6.6, and can be written as: Z x{(x2e2x2 −ln2(x + 1))2 + 2xe3x2(x −(2x3 + 2x2 + x + 1) ln(x + 1))} (x + 1)(ln2(x + 1) −x2e2x2)2 dx = x −ln(x + 1) − xex2 ln(x + 1) ln2(x + 1) −x2e2x2 + 1 2 ln ln(x + 1) + xex2 ln(x + 1) −xex2 (D) is example 9.10: A(x) = 1023x8 + 4104x7 + 5048x6 + 2182x5 + 805x4 + 624x3 + 10x2 + 28x B(x) = 1025x10+6138x9+12307x8+10188x7+4503x6+3134x5+1598x4+140x3+176x2+2 C(x) = 32x10 −80x8 + 80x6 −40x4 + 10x2 −1 y = √ x4 + 4x3 + 2x2 + 1 Z (2x6 + 4x5 + 7x4 −3x3 −x2 −8x −8) (2x2 −1)2√ x4 + 4x3 + 2x2 + 1 dx = (x + 1 2)y 2x2 −1 + 1 2 ln A(x)y −B(x) C(x) Integral (B), on the other hand, can not be “solved” in this manner, and example 7.5 proves this claim of impossibility.
What does it mean to “solve” an integral?
Is there a formal procedure, an algorithm, that lets us solve any integral, or prove that such a solution is impossible?
These questions have puzzled mathematicians for over 300 years, since the invention of calculus, so much so that an introductory calculus sequence can start to seem like a series of puzzle problems, each chapter harder than the last.
This book aims to present our most sophisticated integration theory that provides defini-tive answers to these questions, but the existance of integrals like R e−x2 dx without any elementary form shows that any such theory has severe limitations. Futhermore, the devel-opment of the electronic computer, coupled with sophisticated numerical integration tech-niques, has provided us with powerful approximation methods that significantly reduce the importance of solving integrals. Nevertheless, more difficult differential equations continue to elude easy analysis, so perhaps the greatest benefit of studying integration is the insight it provides to solving differential equations in general.
1.1 Calculus Let’s consider again the integral R e−x2dx. We can trivially construct an anti-derivative as follows: E(x) = Z x 0 e−t2dt I claim that E(x) is an anti-derivative of e−x2. Let’s see...
First, is E(x) well defined? Let’s recall some material from a standard introductory cal-culus textbook, say, [BrCo10]: [BrCo10] Definition - Definite Integral (p. 324) A function f defined on [a, b] is integrable on [a, b] if lim∆→0 Pn k=1 f(¯ xk)∆xk exists and is unique over all partitions of [a, b] and all choices of ¯ xk on a partition. This limit is the definite integral of f from a to b, which we write Z b a f(x)dx = lim ∆→0 n X k=1 f(¯ xk)∆xk .
[BrCo10] Theorem 5.2 - Integrable Functions (p. 325) If f is continuous on [a, b] or bounded on [a, b] with a finite number of discontinuities, then f is integrable on [a, b].
So, E(x) = R x 0 e−t2dt is integrable on [0, x] if e−t2 is continuous on [0, x], and e−t2 is continuous everywhere on the real line. We can easily plot e−t2: -6 -4 -2 0 2 4 6 0 0.2 0.4 0.6 0.8 1 Figure 1.1: e−t2 It’s obviously continuous, so Theorem 5.2 tells us that E(x) is well defined for any real number x – the limit used to construct the Riemann sum exists and is unique. We can also plot E(x), using a numerical integration routine to approximate the integral at each point of the graph: -6 -4 -2 0 2 4 6 0 0.5 1 1.5 Figure 1.2: R x 0 e−t2dt We’re plotting the integral now... the height of each point on the graph was calculated by numerically approximating a Riemann sum.
Is this an anti-derivative of e−t2? Plotting various tangent lines suggests that it might be...
-6 -4 -2 0 2 4 6 0 0.5 1 1.5 2 Figure 1.3: R x 0 e−t2dt (with tangent lines at x0 = −1.5 and x0 = 1) The tangent lines in the graph were plotted using this formula: y(x) = E(x0) + e−x2 0(x −x0) i.e, the point-slope equation of a straight line, with point (x0, E(x0)) and slope e−x2 0.
I used the E(x) = R x 0 e−t2dt formula for the y-coordinate of the point, and the e−x2 formula for the slope. If E(x) is an anti-derivative of e−x2, then the derivative of E(x) is e−x2, and the formula will produce tangent lines for any value of x0. If E(x) were not an anti-derivative of e−x2, we’d get lines, but they probably wouldn’t be tangent lines.
Varying the value of x0 produces different lines (the two lines in the graph were generated using x0 = −1.5 and x0 = 1), and they appear to be tangent lines, so perhaps E(x) is an anti-derivative of e−x2.
In fact, we can do better than guess. Remember the Fundamental Theorem of Calculus?
[BrCo10] Theorem 5.3 (part 1) - Fundemental Theorem of Calculus (p. 338) If f is continuous on [a, b], then the area function A(x) = Z x a f(t) dt for a ≤x ≤b is continuous on [a, b] and differentiable on (a, b). The area function satisfies A′(x) = f(x); or, equivalently, A′(x) = d dx Z x a f(t) dt = f(x), which means that the area function of f is an antiderivative of f.
Pay particular attention to that last formula – it says that the derivative of an integral with respect to its upper bound of integration is just the integrand, with the name of the variable changed.
So, E(x), defined like this: E(x) = Z x 0 e−t2dt is trivially an anti-derivative of e−x2, because the Fundamental Theorem of Calculus tells us that: E′(x) = d dx Z x 0 e−t2dt = e−x2 [BrCo10] Theorem 5.2 tells us that E(x) exists (because e−x2 is continuous), and [BrCo10] Theorem 5.3 tells us that E(x) is an anti-derivative of e−x2.
Of course, we had something else in mind when we asked for an anti-derivative of e−x2.
We wanted a simplified form, something like this: Z x2 dx = 1 3x3 + C not some mathematical smart aleck telling us that the answer is R x2 dx!
The problem is that R e−x2dx doesn’t have a simplified form. It has an anti-derivative (we plotted it, remember?), and its anti-derivative is completely well-defined as a mathemati-cal function, but we can’t simplify it in the way that we can simplify R x2 dx.
Another example is R sin x x dx. It’s also continuous everywhere. The only point where that’s at all in question is x = 0, but L’Hospital’s Rule2 tells us that: 2Using L’Hospital’s Rule here is actually a circular argument, because we had to evaluate this limit to prove that sine’s derivative is cosine.
lim x→0 sin x x = lim x→0 cos x 1 = cos 0 1 = 1 which means that the division by zero in sin x x is a removable discontinuity. We can patch up our function like this: f(x) = ( sin x x x ̸= 0 1 x = 0 This is the cardinal sine, or sinc function, and it’s easy to plot: -20 -15 -10 -5 0 5 10 15 20 -0.2 0 0.2 0.4 0.6 0.8 1 Figure 1.4: sinc t = sin t t Since sinc is continuous everywhere, this integral is well defined everywhere: Si(x) = Z x 0 sin(t) t dt ...and we can plot it...
-20 -15 -10 -5 0 5 10 15 20 -1.5 -1 -0.5 0 0.5 1 1.5 Figure 1.5: R x 0 sin t t dt ...and we can check some of its tangent lines, using the formula: y(x) = Si(x0) + sin x0 x0 (x −x0) -20 -15 -10 -5 0 5 10 15 20 -1.5 -1 -0.5 0 0.5 1 1.5 2 Figure 1.6: R x 0 sin t t dt (with tangent lines at x0 = −9 and x0 = 5) Again, it’s trivial that Si(x): 1. exists, by [BrCo10] Theorem 5.2 and the continuity of sin x x , and 2. is an anti-derivative of sin x x , by [BrCo10] Theorem 5.3 and the definition of Si(x): Si(x) = Z x 0 sin(t) t dt = ⇒ Si′(x) = d dx Z x 0 sin(t) t dt = sin(x) x Yet, again, we have no simple closed form for Si(x).
Let’s see... how could we find simple expressions for R e−x2dx and R sin x x dx?
Could we try...
1. Integration by Parts 2. Trigonometric Substitution 3. Partial Fractions 4. ...some clever change of variables...
5. Google How about this instead – let’s develop a theory that lets us prove that these two integrals have no simple forms. One of the most surprising aspects of this theory is that it’s based not on analysis, but rather algebra.
1.2 Algebra In high school, we study what the Arabs called “al-jabr”, or what the Encyclopaedia Britannica calls “a generalization and extension of arithmetic”. “Elementary algebra," the encyclopedia goes on, “is concerned with properties of arbitrary numbers,” and cites the commutative law of addition (a+b = b+a) as an example of such a property. We use only a few others: the commutative law of multiplication; associative laws of both addition and multiplication; the distributive law. The key point is that all of these laws are valid for any numbers whatsoever, so we are justified in applying them to unknown numbers.
In addition to these basic laws, there is a language to be learned, as well as the more gen-eral Principle of Equality: given two identical quantities, the same operation applied to both must given identical results. This hold true no matter what the operation is, so long as it is deterministic (i.e, has no randomness). Thus, combining the Principle of Equal-ity with the commutative law of addition, I can conclude that sin(a + b) = sin(b + a), without any additional knowledge of what “sin” might be.
For example, consider the following sequence: (ax + b 2)2 = (ax + b 2)(ax + b 2) definition of square = ax(ax + b 2) + b 2(ax + b 2) distributive law = axax + ax b 2 + b 2(ax + b 2) distributive law = axax + ax b 2 + b 2ax + b 2 b 2 distributive law = aaxx + 1 2abx + 1 2abx + b 2 b 2 commutative law of multiplication (3 times) = a2x2 + 1 2abx + 1 2abx + b2 4 definition of square = a2x2 + ( 1 2 + 1 2)abx + b2 4 distributive law = a2x2 + abx + b2 4 basic arithmetic (ax + b 2)2 −b2 4 + ac = a2x2 + abx + b2 4 −b2 4 + ac principle of equality (ax + b 2)2 −b2 4 + ac = a2x2 + abx + ac definition of subtraction So, if ax2 + bx + c = 0, then ax2 + bx + c = 0 a(ax2 + bx + c) = 0a principle of equality a(ax2 + bx + c) = 0 zero theorem3 a2x2 + abx + ac = 0 distributive law (ax + b 2)2 −b2 4 + ac = 0 principle of equality4 (ax + b 2)2 −b2 4 + ac + b2 4 −ac = b2 4 −ac principle of equality (ax + b 2)2 = b2 4 −ac definition of subtraction 4(ax + b 2)2 = 4 b2 4 −4ac principle of equality 4(ax + b 2)2 = b2 −4ac definition of division 22(ax + b 2)2 = b2 −4ac definition of square (2(ax + b 2))2 = b2 −4ac commutative law of multiplication5 (2ax + 2 b 2)2 = b2 −4ac distributive law (2ax + b)2 = b2 −4ac definition of division p (2ax + b)2 = √ b2 −4ac principle of equality (2ax + b) = √ b2 −4ac !?!?!??!
(2ax + b) −b = √ b2 −4ac −b principle of equality 2ax = √ b2 −4ac −b definition of subtraction 1 2a2ax = 1 2a( √ b2 −4ac −b) principle of equality x = 1 2a( √ b2 −4ac −b) definition of division At each step in the sequence (except one), we’re just applying one of the basic rules above. The problem with the “mystery step” isn’t so much that we’re taking the square root, since the principle of equality tells us that we can perform the same operation on both sides of the equal sign, but rather that it cancels out the square in some undefined way. So, assuming that we can perform the mystery step, and noting that the division in the next to last step is only defined if a ̸= 0, we can legitimately conclude that the final result is true for any a, b, and c whatsoever.
The mystery step leads us to introduce complex numbers, typically when we want to use this equation to solve polynomials such as x2 + 1 = 0. At this point, the alert student, having been lured into a false sense of security by the encyclopedia’s “numbers”, and now finding himself facing a whole new type of number entirely, can rightly ask, “What is a number, anyway? Can we just make up new ones if the old ones weren’t good enough?” To which we wave our hands and reply, “A number is... you know, a number!” I am reminded of the time that I was asked to sub in a seventh grade pre-algebra class, and was promptly asked by one of the students to explain the difference between “3” and “2.9999999...” I think I mumbled something lame like “I don’t know, what do you think?” I certainly hadn’t come to class prepared to discuss Cauchy sequences!
In college we are no longer satisfied with this answer, and here is really the launching point for “higher” algebra. Our “numbers” become objects in a set, and our simple concepts of addition and multiplication morph into operations which map pairs of objects into other objects. When asked, “What is a number?”, we now confidently reply, “Anything whose operations obey the axioms!”, which really isn’t all that surprising an answer (anymore) because our entire theory had been built around those axioms to begin with.
The program of higher algebra (in fact much of modern mathematics) goes thus. We postulate the existance of one or more sets of objects and one or more operations, which are simply mappings defined on the objects of those sets. We write out a list of axioms that we assume those sets and operations obey. Which axioms are those? Whichever we find useful (or at least interesting). Then we develop as little or much of a theory as we can, reasoning always from the base axioms. Finally, we take some specific set of objects (like the integers), demonstrate that they obey our set of axioms, and conclude that the entire theory developed for those axioms must apply, therefore, to the integers. Sometimes we reverse the process by finding axioms obeyed by some specific set of objects that we wish to study, then developing a theory around them.6 The most important (i.e, repeatedly used) sets of axioms are given names, or more pre-cisely the sets and operators which obey them are given names. Thus, a “group” is any set and operator which obey three or four certain axioms. A “ring” is any set and pair of operators which obey about six axioms. Add another axiom or two and it becomes a “field”. If a different axiom is obeyed, it is a “Noetherian ring”.
It’s easy to get bogged down with terminology, especially in a classroom environment where you can’t raise your hand during a test and ask, “Excuse me, what’s a semigroup again?” Far more important, I think, is to grasp the central idea that any of these terms refers simultaneously to three things: a set of axioms, a theory logically developed from those axioms, and any particular object(s) that obeys those axioms, and therefore the theory. The ultimate goal is to develop far more sophisticated theories than are possible using the “numbers” of elementary algebra.
Our goal in this book is the development of an algebraic system that allows us to represent as a single object any expression written using elementary functions, putting √1 + sin x on par with 3 2, introducing the concept of a derivative so that we can write differential equations using these objects (it now becomes differential algebra), and equipping this system with a theory powerful enough to either integrate anything so expressed, or prove that it can’t be done, at least not using elementary functions. This is how computer pro-grams like Mathematica or AXIOM solve “crazy” integrals. Along the way, we will have 6How do we demonstrate that a certain set obeys certain axioms? By using more axioms, of course! Mathematics is probably the most self-contained of all major academic fields of study. Many other fields use its results, but math itself references nothing. It’s impossible to get started without assuming something, so the entire process becomes a bit of a chicken-and-egg operation, which leads you to wonder... which did come first?
cause to at least survey some of the deepest waters of modern mathematics. Differential algebra is very much a 20th century theory — the integration problem was not solved un-til roughly 1970; a really workable algorithm for the toughest cases wasn’t available until 1990; a key sub-problem (testing the equivalence of constants) remains unsolved still. Yet one thing is for sure. Three hundred years after the development of calculus, one of its most basic and elusive problems has finally yielded not to limits, sums, and series, but to rings, fields and polynomials. Quite a triumph for “al-jabr”.
1.3 Richardson’s Theorem The absolute value function (in real analysis) or modulus function (in complex analysis) presents a serious obstacle to any attempt to develop an comprehensive algorithm for symbolic integration, as has been known since Daniel Richardson’s proof of the following theorem in 1968: Richardson’s Theorem Let E be a set of expressions representing real, single valued, partially defined func-tions of one real variable. E∗will be the set of functions represented by expressions in E.
If A is an expression in E, A(x) is the function denoted by A.
It is assumed that E∗contains the identity function and the rational numbers as con-stant functions and that E∗is closed under addition, subtraction, multiplication and composition. In every case it is also supposed that given A and B in E there is an effective procedure for finding expressions in E to represent: A(x) + B(x) A(x) −B(x) A(x) · B(x) A(B(x)) A(x) ≡B(x) will mean that A(x) and B(x) are defined at the same points and are equal wherever they are defined.
The integration problem for (E, E∗) is the problem of deciding, given A in E, whether there is a function f(x) in E∗so that f ′(x) ≡A(x).
If E∗satisfies conditions 1, 2, and 3, the integration problem for (E, E∗) will be unsolvable.
1. E∗contains ln 2, π, ex, sin x.
2. There is a function, µ(x), in E∗so that µ(x) = |x| for x ̸= 0.
3. There is a totally defined function, B(x), in E∗so that for no function, f(x), in E∗and no interval I, is f ′(x) = B(x) on I.
Daniel Richardson, Some Undecidable Problems involving Elementary Functions of a Real Variable, The Journal of Symbolic Logic, Volume 33, Number 4, Dec. 1968 Obviously, any algorithm purporting to work over real or complex numbers as its coeffi-cient field will include π and ln 2. Furthermore, we will see later in this text that sin x and ex don’t present any serious problem. The presence of the absolute value function is the critical component that leads to the undecidability of the problem. Why?
Not only is the absolute value function non-differentiable at the origin, but its obvious generalization to complex numbers, the modulus function, fails to be analytic anywhere.
It also introduces an ordering on our field of constants, and allows the two square roots of a positive integer to be distinguished from each other.
Richardson’s proof strategy is not based on differential algebra, but rather is developed from the undecidability proof of Hilbert’s Tenth Problem, the so-called MRDP theorem, named after the surnames of its four principal contributors.
No attempt will be made to prove Richardson’s Theorem here, as the proof of the MRDP theorem is notoriously difficult. I don’t know it, and it would take us too far astray.
My approach in this textbook is simply to note the significance of Richardson’s Theorem, discard absolute value and complex modulus from our menu of elementary operations, and move on.
1.4 Algorithms Several algorithms are required, some of which are too complicated to discuss here in full detail. The difficult algorithms are: • Polynomial Factorization • Primary Decomposition • Constructing the basis for a Riemann-Roch space • Basic field operations on an algebraically closed field constants, including extension for polynomial factorization A complete implementation of the theory described in this book would require imple-mentations of all of these algorithms to be available, in both characteristic zero and prime characteristic.
1.5 Sage To demonstrate the power of this theory, it’s important to do some fairly difficult examples, such as integrals (C) and (D) in the “Who Wants to be a Mathematician?” question on the first page of this chapter. However, attempts to work such difficult integrals by hand quickly bogs down in a mess of algebra. Therefore, the judicious use of a computer algebra system is an essential learning tool in this subject. I’ve chosen the open source program Sage, and will use it liberally in the book’s exercises.
In the course of writing the book, I’ve encountered a number of computation that Sage couldn’t do, and have added code to Sage (taking advantage of its open source philosophy) to improve its functionality. While some of these features have been incorporated into the main Sage distribution, not all have, so the examples in the book are constructed using a customized version of Sage, the source code of while I maintain in a Github repository7.
To do the exercises yourself, it’s likely that you’ll need to download and compile this customized version of Sage.
In the section, I’ll collect several useful functions that I’ll use throughout the book.
1.5.1 Tables Sage has matrices where all entries share a common mathematical type, but lacks any native tool for formatting tables that contain entries with different mathematical types.
The following code creates a table function that accepts a list of lists and formats the output as a table: class LatexObject(SageObject): def init(self, str): self.str = str def latex(self): return self.str def table(listOfLists): return LatexObject('\begin{aligned}' + '\\'.join(['&\quad&'.
table1() is an attempt to create a table that contains series # expansions aligned so that terms of the same power line up under # each other.
We also have to deal with header cells that won't hav # any alignment chars at all.
It's got a lot of problems.
# I've modified LaurentSeries to accept a 'table' parameter that # causes TeX alignment characters to be inserted in the output.
We # assume that each series has the same absolute precision, so any # difference in the number of alignment characters is caused by the 7 # valuation of the series (i.e, the power of the leading term) # being different, and we patch this up by inserting extra # alignment characters at the beginning of the cell.
def table_latex(e): if isinstance(e, LaurentSeries): return e.latex(table=True) else: return latex(e) def fill_amps(target_amp_count, e): # If there are any TeX alignment characters in the cell, # this is (probably) a series with a higher valuation, so # pad it with enough &s to left-fill the cell.
Otherwise, # pad it with enough \omit\span's to span the cell, since # it's (probably) a header cell.
I don't use the suggested # practice of dropping the trailing \span, because that # seems to take us out of math mode and cause errors.
if e.count('&') > 0: return (target_amp_count-e.count('&'))'&' + e else: return (target_amp_count)'\omit\span' + e def table1(listOfLists): latex_listOfLists = [[table_latex(e) for e in l] for l in listOfList ampcount = map(max, zip([[e.count('&') for e in l] for l in latex_l latex_listOfLists_2 = [[fill_amps(ampcount[i], e) for i,e in enumera return LatexObject('\begin{alignedat}{' + str(sum(ampcount)) + "}" 1.5.2 Arrays This next function is convenient for displaying Sage arrays.
Here’s a Python trick8 to find out what variable name a caller uses: import inspect def varName(var): lcls = inspect.stack().f_locals for name in lcls: if id(var) == id(lcls[name]): return name return None def displayarray(b): 8 varname = varName(b) for k in sorted(b.keys()): print('$$', varname, '_{') if isinstance(k, tuple): print(','.join([latex(kk) for kk in k])) else: print(latex(k)) print('}=', latex(b[k]), '$$') Chapter 2 Commutative Algebra In this chapter I will outline the basic algebraic structures necessary to carry out the pro-gram sketched out in the previous chapter. This material is included mainly to provide a starting point for the rest of the book. Where I have omitted proofs, I have tried to provide references to [Go14], which is a good introductory algebra textbook that is freely available on-line.
2.1 Rings and Fields [van der Waerden], §3.1 [Go14], §1.11 We begin with two key definitions that we will use throughout: the ring and the field.
A ring is a mathematical system where addition, subtraction, and multiplication are defined.
A field is a mathematical system where addition, subtraction, multiplication, and division are defined.
Both concepts are associated with sets of axioms. Any algebraic system that obeys the ring axioms is called a ring; any algebraic system that obeys the field axioms is called a field.
A ring R obeys the axioms in figure 2.1.
Notice the commutative law of multiplication (CR1), along with the existence of a multi-plicative identity (RwU1). A substantial theory has been developed around non-commutative rings, probably because matrix multiplication (a critically important example) is non-commutative. Most of our rings are commutative ring, or abelian (the terms are synony-mous). Also, I require the existence of a multiplicative identity. Much of ring theory associative law of addition ∀a, b, c, (a + b) + c = a + (b + c) (R1) associative law of multiplication ∀a, b, c, (ab)c = a(bc) (R2) commutative law of addition ∀a, b, a + b = b + a (R3) distributive law ∀a, b, c, a(b + c) = ab + bc (R4) existence of an additive identity (zero) ∃0, ∀a, 0 + a = a (R5) invertibility of addition ∀a, ∃b, a + b = 0 (R6) commutative law of multiplication ∀a, b, ab = ba (CR1) existence of a multiplicative identity (unity) ∃1, ∀a, 1a = a (RwU1) Figure 2.1: Ring axioms All ring axioms, plus: invertibility of multiplication ∀a ̸= 0, ∃b, ab = 1 (F1) Figure 2.2: Field axioms can be developed without this axiom, but some theorems require it, and I don’t want to belabor the point, since all of our rings will have a unity element. Therefore, I will omit any additional terminology, adopt the (CR1) and (RwU1) axioms along with ring axioms (R1) through (R6) to obtain a commutative ring with unity, and call it a ring for the rest of the book.
A field F obeys all the ring axioms (thus all fields are also rings), as well as one additional axiom (Figure 2.1).
Informally, rings are mathematical systems in which addition and multiplication are cleanly defined. Subtraction is also defined due to (R6), the invertibility of addition, which allows a subtraction problem to be turned into an addition problem. Division, however, is not, since it requires (F1). Since the ring axioms do not require multiplication to be invertible, there is no guarantee that we can carry out division in a ring. A field, on the other hand, is a mathematical system in which all four elementary operations — addition, subtraction, multiplication, and division — are defined.
The simplest example of a ring is the set of integers, which I shall denote as Z (after the German word for number, zahl). A pair of integers can be added, subtracted or multiplied to form another integer. Note, however, that a pair of integers can not necessarily be divided to form another integer.
3 2 is not an integer, because multiplication (in Z) is not necessarily invertible — there is no integer that when multiplied by 2 forms 1. (F1) is not satisfied. Thus Z forms a ring but not a field.
2.2 Fraction fields The field Q [van der Waerden], §3.3 [Go14], §6.4 We can remedy our inability to divide using only integers by moving on the rational numbers, traditionally denoted Q (probably for quotient). This is the simplest example of a fraction field, in this case formed over the integers, Z. It is also our first example of a theme we’ll use repeatedly in this book, that of using a simple algebraic system to construct a more complex one.
To form a fraction field from a ring, we take pairs of elements from the ring (convention-ally arranged into fractions) and establish an equivalence relationship between them. We also require that the second element in the pair (the denominator) can not be zero. Two pairs (a, b) and (c, d) are equivalent if ad = bc, and we write them a b and c d. We group equivalent pairs together into equivalence classes and define our basic field operations as follows: a b + c d = ad + bc bd a b −c d = ad −bc bd a b c d = ac bd a b c d = ad bc Figure 2.3: Fraction field operations The additive identity element is 0 1 and the multiplicative identity is 1 1, using the original identities 0 and 1 from the base ring. Note that the division by zero is not defined, nor do our field axioms require it to be.
Notice that although we define all four field operations, we only use the three ring oper-ations to do it! I.e, when we divide a b by c d, we need only to form ad and bc in order to form the (ad, bc) pair, which we write as ad bc . We thus divide 1 2 by 2 3 to obtain 3 4 without ever having to divide the integers — only multiplying them (1 · 3 = 3 and 2 · 2 = 4).
In general, there is no guarantee that this kind of construction will work. We can’t just pair numbers up however we want and call it a field. Several other conditions have to be met.
First of all, we have to ensure that the equivalence relationship is well-defined. If x = y (in the sense of equivalence) and y = z, then we must have x = z, otherwise we can’t even cleanly establish the critical notion of an equivalence class (which says that 1 2 and 2 4 are basically the same thing). I emphasize here that the new field, and its new operations, are defined using the equivalence classes, although we muddle this distinction by using All ring axioms, plus: non-existence of zero divisors ∀a ̸= 0, b ̸= 0 ∈I, ab ̸= 0 (I1) Figure 2.4: Integral domain axioms the smallest fraction in a class to represent it. Strictly speaking, the multiplicative identity is not 1 1 but {1 1, 2 2, 3 3, . . .}, the additive identity is not 0 1 but {0 1, 0 2, 0 3, . . .}, and my example in the last paragraph should have read “we thus divide {1 2, 2 4, 3 6, . . .} by {2 3, 4 6, 6 9, . . .} to obtain {3 4, 6 8, 9 12, . . .}...” Having cleanly established equivalence classes, we have to make sure that our operations actually work consistently on them, since they are defined in terms of fractions within the classes. We need to verify that taking any fraction from one equivalence class and any fraction from other, then applying one of four operations to them, we always get an answer in a third equivalence class. The actual answer can (and will) vary depending on the choice of representative fractions, but it has to always be in the same class. In this way, we confirm that the operations are cleanly defined not just on the fractions, but on the classes. I’m not going to actually make this verification, but leave it as an exercise.
Which is why I excluded zero as a possible denominator. We do this because otherwise our operations aren’t cleanly defined on these equivalence classes. 1 0 is not equivalent to 0 1 (since 1 · 1 ̸= 0 · 0), so 1 0 must have a multiplicative inverse (by axiom F1); i.e, some fraction a b must exist which when multiplied by 1 0 produces 1 1, yet by the zero theorem, no such element b can exist in the base ring so that 1b = 0. Excluding zero as a possible denominator ensures that our field axioms are satisfied.
Yet in the fraction field operations, where we multiply two denominators together to get the result’s denominator, what would happen if two non-zero elements can be multiplied to form zero, producing a zero denominator? Nothing in the ring axioms prevents this from happening, so we add an additional axiom.
An integral domain I obeys all the ring axioms, plus one more (figure 2.2) that guarantees the non-existence of zero divisors. All of the rings in this book are integral domains.
The fraction field construction is only defined on integral domains, and I’ll leave it as an exercise to show that Z is an integral domain. The main point of this section is to recognize that the fraction field construction can be performed not only on the integers Z to obtain the rationals Q, but on any integral domain to obtain its fraction field.
2.3 Polynomial rings and rational function fields The ring F[x] and the field F(x) [van der Waerden], §3.4 Having built a field from a ring, can we build a ring from a field? The answer is yes, and the most important such construction is a polynomial ring, whose elements are poly-nomials in some variable with coefficients in the underlying field, all but a finite number of which must be zero.1 We write this ring using the underlying field, brackets and the variable, so F[x] is the ring of polynomials in x with coefficients from the field F.
F[x] is a ring but not a field. It is, however, an integral domain (left as an exercise), so we can form a fraction field from it, which we write using parenthesis instead of brackets: F(x). Elements in F(x) are fractions, both the numerator and denominator of which are polynomials in x. So, for example, x x−1 is a element of Q(x). Fractions of polynomials are called rational functions, so Q(x) is the field of rational functions in x over the rational numbers.
Now, you might ask, “Can’t (x −1) be zero? Say, if x is 1?”. The answer is no. x is not 1 or any other number. x is x, and in Q[x], (x −1) is as different from zero as 3 2 is. (x −1), and things like it, are completely distinct elements in the algebraic systems in which they are defined.
Now, obviously, we can set x to be 1. But now we are no longer working in Q[x] — for starters, there is no longer a distinct element x, since it’s equal to 1! Now we are working in Q. Setting x equal to 1 mapped everything from Q[x] into Q. This is a simple ex-ample of an evaluation homomorphism — a homomorphism (a mapping which preserves operations) from one system to another created by setting an independent variable equal to some constant value.
So, you ask, “what about 1 2 and 2 4? Are they distinct elements as well?” No. This time we are dealing with elements that are basically the same. This is where the technical details of the fraction field construction become significant. Strictly speaking, we are not working with elements like 1 2 at all. We are working with the equivalence classes defined above. 1 2 is a representative of an equivalence class that includes 2 4.
It’s convenient to select one unique representative of each equivalence class. In the case of fraction fields, we’ll use the fraction with no common factors between the numerator and the denominator, i.e, 1 2 instead of 2 4 and 1 x instead of x x2. To reduce any given fraction to its canonical form, we need to compute the greatest common divisor of the numerator and denominator, and divide it out. The simplest way to do this involves long division.
1If we relax the finiteness requirement and allow infinite “polynomials”, we obtain a ring of formal power series over the field, typically written F. We will have little use for formal power series in this book.
2.4 Long Division [van der Waerden], §3.4 As we all learned in grade school, polynomials can be divided using long division. To generalize this in our more abstract context, let’s consider a very simple calculation of this type: 1 2x + 3 4 2x + 1 x2 + 2x + 1 −(x2+ 1 2x ) 3 2x + 1 −( 3 2x+ 3 4 ) 1 4 Each step starts by dividing the leading terms, i.e, x2 is divided by 2x to form x 2. Actually, we can be a bit more precise. Each step starts by dividing the leading coefficients, since the variables are divided just by subtracting their powers. x2 divided by x is just x. We divide 1 by 2 to form 1 2 and in this manner obtain x · 1 2 = x 2.
Next, we multiply this value by the divisor to obtain a polynomial that we will subtract from the dividend (or what remains of it after prior steps). Again, let’s be more precise.
We multiply the polynomial variable just by adding its powers. What we really have to multiply are the coefficients. To multiple x 2 by 2x + 1 we multiply 1 2 by 2 to obtain 1, add the powers of x and x to obtain x2, and arrive at the first term 1 · x2 = x2. Next, we multiply 1 2 by 1, get 1 2, add the powers of 1 and x to obtain x, and have the second term 1 2 ·x = 1 2x. Adding these terms we get x2+ 1 2x — the first of the intermediate polynomials.
To perform the third step, we don’t have to do anything with the variables. We just subtract the coefficients. These three steps are repeated until we are left with a remainder of lower degree than the divisor.
So, to summarize, working with the polynomial variable is easy — we just add or subtract its integer powers. We perform polynomial long division by dividing, multiplying, and subtracting the coefficients. Now, these are three of the basic four operations provided by a field. It follows, therefore, that we can perform polynomial long division on polynomials whose coefficients lie in any field whatsoever. Given F[x], a polynomial ring over a field, we can use the field operations provided by F to divide any two elements from F[x] using polynomial long division and obtain a remainder and a quotient.
We can even say a bit more. Just like with grade school long division, we know that the degree of the quotient will be the difference in degrees of the dividend and the divisor, and that the degree of the remainder will be less than the degree of the divisor. We just need to keep in mind that these degrees are measured relative to the polynomial ring variable, not any other variable that might appear as part of the underlying field.
2.5 Greatest Common Divisors [van der Waerden], §3.7, §3.8, §5.4 (multivariate rings) [Ge92], Ch. 7 One of the most important uses of polynomial long division is to compute greatest com-mon divisors (GCDs), at least in theory. In practice, there are other, more efficient algo-rithms.2 However, because long division is a simple and straightforward way to compute GCDs, because it provides a theoretical underpinning for other methods, and because it leads us directly to solving polynomial diophantine equations, I’ll present it here in this section.
The first thing to observe is that the long division equation, D = qd + r (dividend equals quotient times divisor plus remainder), can be rearranged to read r = D −qd, which shows that any common divisor of the dividend and the divisor can be divided out from the right hand side of the equation, so must divide the left hand side also. Thus, common divisors of the dividend and divisor are preserved in the remainder.
Furthermore, since the remainder is always of lower degree than the divisor, we can re-peat the long division with the divisor as the new dividend and the remainder as the new divisor. The new remainder will also preserve common divisors of the original dividend and divisor, and will be of lower degree than the original remainder. This process can repeated, lowering the degree of the remainder at each step, until we are left with a zero remainder, i.e. D′ = q′d′, where I’ve used primes to emphasize that we are no longer deal-ing with the original dividend and divisor. Since common divisors have been preserved throughout by D′ and d′, it follows that d′, the divisor of the last step, must be a common divisor. It is, in fact, a greatest common divisor ([Go14] Theorem 1.8.16). This has been known since the time of Euclid, at least in the case of integers.
Nothing in our ring axioms guarantees the existence of GCDs. The problem is that there might be a lattice of divisors for a given element, instead of a strict ordering of them.
However, GCDs exist in any integral domain in which the procedure described in the last paragraph can be carried out ([Go14] Theorem 6.5.8 and Lemma 6.6.2).
A GCD domain is an integral domain in which any two elements have a greatest common divisor.
A unit is an invertable element.
There is usually more than one GCD. Any divisor can be transformed into another divisor by multiplying it by a unit, since if uu′ = 1, then ab = (ua)(u′b) for any a and b what-soever. In particular, a greatest common divisor can be transformed into another greatest common divisor by multiplying it by a unit.
2See [Ge92], for example Example 2.1. Compute a GCD of 4x4 + 13x3 + 15x2 + 7x + 1 and 2x3 + x2 −4x −3 in Q[x].
2x + 11 2 2x3 + x2 −4x −3 4x4 + 13x3 + 15x2 + 7x + 1 −(4x4+ 2x3 −8x2 −6x ) 11x3 + 23x2 + 13x + 1 −(11x3+ 11 2 x2 −22x −33 2 ) 35 2 x2 + 35x + 35 2 4 35x −6 35 35 2 x2 + 35x + 35 2 2x3 + x2 −4x −3 −(2x3+ 4x2 + 2x ) −3x2 −6x −3 −(−3x2 −6x −3 ) 0 The divisor of the last step, in this case 35 2 x2 + 35x + 35 2 , is a GCD, and multiplying it by any unit will produce a different GCD. In the case of a polynomial ring over a field, the units are the elements of the underlying field, so we can multiply by anything in Q (i.e, any rational number) and get another GCD. For this example, the obvious thing to multiply by is 2 35, which both clears the denominators and divides out the common factor in the numerators to produce x2 + 2x + 1. Both answers are acceptable.
The Sage function gcd compute GCDs.
sage: gcd(4x^4 + 13x^3 + 15x^2 + 7x + 1, 2x^3 + x^2 - 4x - 3) x2 + 2 x + 1 □ Multivariate GCDs A few words are in order here about GCDs of multivariate polynomials. A factorization in Q(x)[y] or Q(y)[x] (both of the form F[x]) is superficially so similar to a factorization in Q[x, y] (not of the form F[x]), that the distinction should be noted. In both of the first two cases, we form a fraction field with respect to one of the two variables and thus obtain a polynomial ring (in the other variable) over the fraction field. In the case of Q[x, y] we do not form a fraction field with respect to either variable; thus we have a polynomial ring over not a field, but over another polynomial ring.
Now a polynomial ring over a GCD domain is itself a GCD domain ([Go14] Lemma 6.6.2 and Theorem 6.6.7), so by induction any finite series of polynomial rings (like Q[x, y]) is also a GCD domain. This implies that GCDs exist in multivariate polynomial rings. The problem is finding them, since the procedure described above requires long division, and this only works cleanly in an F[x]-type system.
One solution is to pick one variable (say, x) to form coefficients (in the ring Q[x]), then use the second variable (say, y) to form polynomials in Q[x][y].
Then we factor out of each Q[x][y] polynomial the GCD of the coefficients (calculated in Q[x]), which we call the content of the polynomial, leaving a primitive polynomial. It can be shown ([Go14] Lemma 6.6.9) that if a primitive polynomial factors at all, then it factors into primitive polynomials. We thus can compute a GCD of the primitive parts and multiply this by the GCD of the contents to obtain a GCD in Q[x, y].
We will have little use for multivariate polynomial factorizations in this book, since in-variably we will calculate GCDs with respect to one variable, and form fraction fields from any others, and thus always be working in F[x] systems.
Example 2.2. Compute the GCD of 5xy −5y2 −7x + 7y and 2x2 −yx −y2 in Q[x, y].
Q[x, y] is a multivariate polynomial ring, but we’ll need to work in a F[x]-type system to perform the computation. Our choices are Q(x)[y] and Q(y)[x].
Let’s start with Q(x)[y], and rearrange the polynomials so that y is the polynomial variable and our coefficients are in Q[x]: −5y2 + (5x + 7)y −7x and −y2 −xy + 2x2 The first step is to compute the content (GCD of the coefficients) of each polynomial.
Clearly, the GCD of −5, (5x + 7), and −7x is 1 and the GCD of −1, −x, and 2x2 is also 1, so both polynomials are already primitive and we can just proceed with the GCD calculation in Q(x)[y]: 1 5 −5y2 + (5x + 7)y −7x −y2 − xy + 2x2 −(−y2 + (x + 7 5)y − 7x 5 ) −(2x + 7 5)y + (2x2 + 7x 5 ) 5 2x+1y − 7 2x+1 −(2x + 1)y + (2x2 + x) −5y2 + (5x + 7)y − 7x −(−5y2 + 5xy ) 7y − 7x −(7y− 7x ) 0 This leads us to conclude that the last divisor, −(2x+ 7 5)y+(2x2+ 7 5x) is a GCD in Q(x)[y].
Now we need to remove its content, which is the GCD of −(2x + 7 5) and (2x2 + 7 5x), or (2x + 7 5). Dividing through by this polynomial (a polynomial in Q[x], and thus a unit in Q(x)[y]) we obtain −y + x. We now multiply by the GCD of our original contents, but they were just 1, so we conclude that x −y is our GCD in Q[x, y].
Now let’s do all that again in Q(y)[x]. Our polynomials become: (5y −7)x −(5y2 −7y) and 2x2 −yx −y2 The second one has unit content (the GCD of 2, −y, and −y2), but the first one’s content is gcd(5y −7, 5y2 −7y) = 5y −7. Dividing this out, we obtain: x −y and 2x2 −yx −y2 and compute the GCD of these polynomials: 2x + y x −y 2x2 −yx −y2 −(2x2−2yx ) yx −y2 −(yx−y2 ) 0 Thus, x −y is the GCD of the primitive polynomials, and it has unit content gcd(1, −y).
The GCD of the original contents (1 and 5y −7) is 1, so the final result is again x −y.
sage: R. = QQ[] Q[x, y] sage: gcd(5xy - 5y^2 - 7x + 7y, 2x^2 - yx - y^2) −x + y □ 2.6 Polynomial Diophantine Equations The same long division procedure used for GCD computations can also be used solve a certain class of polynomial Diophantine equations. A Diophantine equation is one whose variables are restricted to be integers. The most famous example is Fermat’s equation, xn+yn = zn; Fermat’s theorem states that this equation has no solutions x, y, z, n ∈Z for n > 2. A generalized Diophantine equation is one whose variables are restricted to some algebraic system, not necessarily Z. A polynomial Diophantine equation is one whose variables are restricted to be polynomials of some form, and the one we will consider here is this: sa + tb = c; a, b, c ∈F[x] given; s, t ∈F[x] unknown Let’s begin by noting that any common divisor of a and b, and in particular gcd(a, b), can be divided out from the left hand side of the equation, and thus must also divide the right hand side, so c must be be a multiple of gcd(a, b), or the equation has no solution.
This necessary condition is also sufficient, and the simplest way to demonstrate this is to use the GCD computation in a constructive proof. Note that first step in computing gcd(a, b) is to solve a = qb + r. Rearranging this as r = a −qb we see how the remainder can be expressed in the Diophantine form sa + tb. More generally, at each step of the calculation, we solve D = qd+r, where D and d are each either a, b, or a remainder from a previous step, so using r = D −qd we can write each remainder in the form sa + tb. At the end of the calculation, we will have expressed gcd(a, b) in the form sa + tb.
We now use long division to divide c by gcd(a, b). Because of the necessity demonstrated above, the division must be exact (i.e, zero remainder) or the equation has no solution.
Having computed both gcd(a, b) = sa + tb and c = q gcd(a, b) we can now combine these expression to form c = (qs)a + (qt)b, which solves the original equation.
This solution is not unique. Given a solution to c = sa + tb, we can form any multiple of ab, say mab, and write another solution c = (s −mb)a + (t + ma)b. Note however, that (s −mb) has the form of a remainder after dividing s by b (m is the quotient). Since the degree of a remainder is always less than the degree of the divisor, it follows that if sa + tb = c can be solved, then we can always compute an s of lower degree than b, or a t of lower degree than a.
If deg(c) < deg(a) + deg(b), then these conditions are not exclusive; finding an s of lower degree than b implies a t of lower degree than a. To see this, simply note that if deg(s) < deg(b), then deg(sa) = deg(s)+deg(a) < deg(b)+deg(a). Since tb = c−sa, if deg(c) < deg(a) + deg(b) and deg(sa) < deg(a) + deg(b), then deg(tb) < deg(a) + deg(b), which implies that deg(t) < deg(a).
We will make repeated use of this polynomial Diophantine equation throughout the book.
Example 2.3. Solve: s(4x4 + 13x3 + 15x2 + 7x + 1) + t(2x3 + x2 −4x −3) = x3 + 5x2 + 7x + 3 for s, t ∈Q[x] satisfying minimal degree bounds.
Using the notation: a = 4x4 + 13x3 + 15x2 + 7x + 1; b = 2x3 + x2 −4x −3; c = x3 + 5x2 + 7x + 3 we see that we are trying to solve sa + tb = c. a and b are the same polynomials used for the first GCD example. Recalling that GCD calculation: 2x + 11 2 2x3 + x2 −4x −3 4x4 + 13x3 + 15x2 + 7x + 1 −(4x4+ 2x3 −8x2 −6x ) 11x3 + 23x2 + 13x + 1 −(11x3+ 11 2 x2 −22x −33 2 ) 35 2 x2 + 35x + 35 2 4 35x −6 35 35 2 x2 + 35x + 35 2 2x3 + x2 −4x −3 −(2x3+ 4x2 + 2x ) −3x2 −6x −3 −(−3x2 −6x −3 ) 0 The first division states that: a = (2x + 11 2 )b + (35 2 x2 + 35x + 35 2 ) Rearranging this equation we get: x2 + 2x + 1 = 2 35a −1 35(4x + 11)b The second division isn’t used in solving the polynomial Diophantine equation; it simply states that 35 2 x2 + 35x + 35 2 and x2 + 2x + 1 are GCDs.
Having concluded that x2 + 2x + 1 is a GCD of a and b, we divide it into c: x + 3 x2 + 2x + 1 x3 + 5x2 + 7x + 3 −(x3+ 2x2 + x ) 3x2 + 6x + 3 −(3x2+ 6x + 3 ) 0 The remainder is zero, so the original problem has a solution (remember that c had to be a multiple of gcd(a, b)).
We substitute our expansion for x2 + 2x + 1 above into c = (x + 3)(x2 + 2x + 1) and obtain: c = (x + 3)(x2 + 2x + 1) = 2 35(x + 3)a −1 35(4x2 + 23x + 33)b deg( 2 35(x + 3)) = 1 < deg(b) = 3 and deg( 1 35(4x2 + 23x + 33)) = 2 < deg(a) = 4, so the degree bounds are met and we have our solution: s = 2 35(x + 3) t = −1 35(4x2 + 23x + 33) Now let’s verify this solution using Sage: sage: R. = QQ[] Q[x] sage: a = 4x^4+13x^3+15x^2+7x+1; sage: b = 2x^3+x^2-4x-3; sage: c = x^3+5x^2+7x+3; sage: xgcd(a,b) x2 + 2x + 1, 2 35, −4 35x −11 35 def diophantine(a,b,c): (g,s,t) = xgcd(b,c) if not g.divides(a): raise ValueError("diophantine: a doesn't divide gcd(b,c)") s = s (a//g) t = t (a//g) # g = sb + ct # a = mg = (ms)b + (mc)t (q,r) = s.quo_rem(c) # s = r + qb, so r = s - qb return (r, t + qb) sage: (s,t) = diophantine(c,a,b) 2 35x + 6 35, −4 35x2 −23 35x −33 35 sage: c == as + bt True □ 2.7 Partial Fractions Expansion [van der Waerden], §5.10 As a first application of polynomial Diophantine equations, we use them to construct partial fractions expansions, which will be a major technical tool in the first half of the book.
Consider an element a from a polynomial fraction field F(x). We can write a = n d where n, d ∈F[x]. If we are now given a factorization of d = de1 1 de2 2 · · · dek k , where di ∈F[x] and gcdFx = 1 if i ̸= j, and assuming that a is a proper fraction (degx n < degx d), then we can construct a partial fractions expansion of a: a = n d = n X i=1 ei X j=1 ni,j di j degx(ni,j) < degx(di) (2.1) We begin by computing an expansion in the form: a = n d = n X i=1 ni di ei degx(ni) < ei degx(di) (2.2) i.e, each denominator factor is separated apart from the others, but there is only a single term for each denominator factor and the degree bounds are weaker.
ni is found by solving the following polynomial Diophantine equation for ni and ri: n = ni Y j̸=i dj ej + ri(d ei i ) The degree bounds from the previous section guarantee that degx(ni) < ei degx(di), and dividing through by d shows: n d = ni d ei i + ri Q j̸=i dj ej We can now either ignore ri and use this procedure to compute all of the ni’s, or we can note that the second term on the right is a fraction in the original form, but with one less factor in the denominator, so we can recurse and separate out all the di ei into seperate fractions.
Having computed an expansion of the form (2.2), a series of long divisions now suffices to seperate each of these fractions into a full partial fraction expansion in the form (2.1): qi,ei = ni qi,j = qi,j−1di + ni,j qi,1 = ni,1 The degree bounds on long division ensure that degx qi,j < j degx di and degx ni,j < degx di Assembling the ni,j’s together, we obtain the form required by equation (2.1): ni = ei X j=1 ni,j d ei−j i ni d ei i = ei X j=1 ni,j di j Theorem 2.4. Partial fractions expansions that meet the minimal degree bounds are unique.
Proof If two different partial fractions expansions can be constructed for the same proper frac-tion a, then subtracting them from each other would yield a non-trivial partial fractions expansion for 0: 0 = n X i=1 ei X j=1 ni,j di j degx(ni,j) < degx(di) (2.3) We clear the denominators: 0 = n X i=1 ei X j=1 ni,j d ei−j i Y k̸=i d ek k !
(2.4) Isolating the n1,ei term we obtain: −n1,e1 n Y k=2 d ek k = e1−1 X j=1 n1,j d e1−j 1 n Y k=2 d ek k !
+ n X i=2 ei X j=1 ni,j d ei−j i Y k̸=i d ek k !
(2.5) All of the terms on the right hand side include a d1 factor, which implies that d1 must also factor the left hand side. Since all of the di’s are relatively prime, this d1 factor can only come from n1,e1 itself. Yet degx(n1,e1) < degx(d1), so this is impossible, and no such expansion can exist.
□ Example 2.5. Compute the partial fractions expansion of 3x2 −4x + 2 x3 −3x2 + 4 We begin by factoring the denominator. While factoring can be quite complicated in practice, in this case we need only try some small integers to discover that either 2 or −1 solve the denominator, leading directly to a factorization: 3x2 −4x + 2 x3 −3x2 + 4 = 3x2 −4x + 2 (x + 1) · (x −2)2 Next, we solve the polynomial Diophantine equation: 3x2 −4x + 2 = s(x −2)2 + t(x + 1) = sa + tb First, we compute the GCD of a = (x −2)2 = x2 −4x + 4 and b = x + 1. The actual result is obvious, but now we’re interested in both the remainder and the quotient.
x −5 x + 1 x2 −4x + 4 −(x2+ x ) −5x + 4 −(−5x −5 ) 9 Since the remainder, 9, is a unit, a and b have no common factors and their GCD is 1. Of course, this result is hardly surprising since (x −2) and (x + 1) have no common factor between them.
a = (x −5)b + 9 9 = a −(x −5)b 1 = 1 9[a −(x −5)b] 3x2 −4x + 2 = 1 9(3x2 −4x + 2)[a −(x −5)b] 3x2 −4x + 2 = 1 9(3x2 −4x + 2)a −1 9(3x3 −19x2 + 22x −10)b Our degree bounds aren’t met yet, so we divide b = x+1 into a’s coefficient 3x2−4x+2: 3x −7 x + 1 3x2 −4x + 2 −(3x2+ 3x ) −7x + 2 −(−7x −7 ) 9 The remainder becomes our new a coefficient, and after subtracting (3x −7)a = 3x3 − 19x2 + 40x −28 from the b coefficient, we conclude that: x2 + 3x + 2 = a −(−2x + 2)b In other words (remember that a = (x −2)2 and b = x + 1), x2 + 3x + 2 (x −2)2(x + 1) = 1 x + 1 + 2x −2 (x −2)2 Now we need only divide (2x −2) by (x −2): 2 x −2 2x −2 −(2x−4 ) 2 so, x2 + 3x + 2 x3 −3x2 + 4 = 1 x + 1 + 2 (x −2)2 + 2 (x −2) Here’s Sage code to do a partial fractions expansion and save the results in an array: def partfrac(num, den): b = {} factorization = factor(den) for f in factorization: (t,s) = diophantine(num, f^f, den//(f^f)) for i in range(f,1,-1): (s, b[f,i]) = s.quo_rem(f) b[f,1] = s return(b) sage: b = partfrac(3x^2 - 4x + 2, x^3 - 3x^2 + 4); sage: displayarray(b); bx−2,1 = 2 bx−2,2 = 2 bx+1,1 = 1 □ 2.8 Resultants [van der Waerden], §5.8; [Lang], §IV.8 At times, we will want a simple way of testing two polynomials in F[x] to see if they have a GCD, without actually computing it. This is more than just a computational con-venience. The presence of the polynomial’s variable in the GCD often encumbers us. On the other hand, the resultant yields a simple element from the underlying field that is zero if the polynomials have a non-trivial GCD and non-zero otherwise. The GCD exists in F[x], while the resultant is in F.
For example, the polynomials tx + x + t + 1 and ty + y share t + 1 as a GCD, so their t-resultant is zero. On the other hand, their x-resultant is not zero, because in the ring C(y, t)[x], t + 1 is a unit, so the GCD is 1. The resultant is constructed in an F[x]-type system, so for multivariate polynomials, we always need to specify which variable is the ring variable; any remaining variables are implicitly field variables.
The resultant is defined3 as the determinant of the Sylvester matrix Sx(P, Q), which is the m + n × m + n matrix constructed from two polynomials (in F[x]) P and Q of degrees m and n (all the blanks are zeros): P = m X i=0 pi xi Q = n X i=0 qi xi Sx(P, Q) = pm pm−1 . . .
p0 pm pm−1 . . .
p0 . . .
. . .
pm pm−1 . . .
p0 .
.
.
.
.
.
.
.
.
qn qn−1 . . .
q0 qn qn−1 . . .
q0 . . .
. . .
qn qn−1 . . .
q0 In plain English, the matrix is constructed by forming the first row from the first polyno-mial coefficients, adding n −1 trailing zeros at the end of the row. The second row is formed by shifting the first row one position to the right. This shifting is repeated a total of m−1 times to obtain the first m rows. The last n rows are constructed in the same way from the second polynomial.
Now consider the following straightforward matrix identity: 3There are other equivalent ways of defining the resultant.
Sx(P, Q) xn+m−1 xn+m−2 .
.
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x 1 = Pxm−1 Pxm−2 .
.
.
Qx Q If det Sx(P, Q) is non-zero, then the Sylvester matrix is invertible, and we can form the following equation: xn+m−1 xn+m−2 .
.
.
x 1 = Sx(P, Q)−1 Pxm−1 Pxm−2 .
.
.
Qx Q Since the matrix is formed exclusively from the polynomials’ underlying field F, its in-verse must also be formed from F. Now consider the bottom element in the last equation.
It must have the following form (the fi’s are the bottom row of Sx(P, Q)−1): 1 = f0Pxm−1 + f1Pxm−2 + . . . + fn+m−1Qx + fn+mQ fi ∈F 1 = AP + BQ A, B ∈F[x] The only way this statement can be true is if P and Q (viewed as polynomials in x) have a trivial GCD, so a non-zero determinant of Sx(P, Q) imply that P and Q have only a trivial GCD.
Conversely, assume that gcdx(P, Q) = 1. Then we can solve a series of polynomial Diophantine equations to express 1, x, . . . , xn+m−1 as AP +BQ, where deg A < deg Q = n and deg B < deg P = m, which suffices to construct an inverse of the Sylvester matrix.
We have thus proved: Theorem 2.6. The resultant is zero iff the two polynomials have a non-trivial GCD.
□ Let me note two points. First, though we used a field construction for the proof, determi-nants are constructed using only ring operations, so resultants can be computed in any ring and the result will be a ring element. The proof depends on the ring having a well-defined fraction field, but since UFDs are integral domains, the GCD concept doesn’t make sense without a fraction field anyway.
Second, if the underlying ring involves multiple variables, the net effect of the resultant is to eliminate one of them. To see this, imagine arbitrary values being assigned to the other variables. The resultant yields a condition on the remaining variables for the original system to be solvable.
Example 2.7. Compute the t−resultant of t2 −1 −x and t3 −t −y.
sage: R. = QQ[] Q[x, y, t] sage: (t^2 - 1 - x).sylvester_matrix(t^3 - t - y, t) 1 0 −x −1 0 0 0 1 0 −x −1 0 0 0 1 0 −x −1 1 0 −1 −y 0 0 1 0 −1 −y sage: (t^2 - 1 - x).resultant(t^3 - t - y, t) −x3 −x2 + y2 This result implies that the polynomials t2 −1 −x and t3 −t −y have a common factor when y2 −x3 −x2 = 0. For example, this condition is satisfied when x = 3 and y = 6, and the two polynomials become t2 −4 and t3 −t −6, which have the common factor t −2: sage: (t^2-4).factor() (t −2) · (t + 2) sage: (t^3-t-6).factor() (t −2) · (t2 + 2t + 3) □ 2.9 Algebraic Extensions The field F[x] mod n(x) Both our fraction field and polynomial ring constructions are examples of extensions.
Simply put, when we use an algebraic system to construct a new algebraic system that includes the original system as a subset, then the new system is an extension of the origi-nal.4 So, Z[x] is an extension of Z because we can identify Z as a subset of Z[x] and, in fact, even homomorphicly map Z into Z[x]. Such an inclusion homomorphism should not be confused with an evaluation homomorphism, which would map the other way (from Z[x] into Z).
The only remaining type of extension that will be important to us is the algebraic exten-sion. It is another equivalence class construction that we build starting with a polynomial ring over a field, say F[x]. Our equivalence classes are all elements in F[x] whose dif-ferences are multiples of some distinguished irreducible polynomial in F[x], called the minimal polynomial of the field. And I do say field, because we don’t need to use the fraction field construction with an algebraic extension; the ring is already a field.
Algebraic extensions are used to model fields where some algebraic expression (the mini-mal polynomial) is zero. Since adding any multiple of zero to an expression doesn’t affect its value, all elements in an algebraic equivalence class are essentially the same value.
This isn’t exactly the same as the real number system. Real numbers are typically con-structed using Cauchy sequences of rational numbers. Thus, the square root of two is con-structed as a real number using the Cauchy sequence {1, 14 10, 141 100, 1414 1000, 14142 10000, · · ·}, which obeys the algebraic relationship γ2−2 = 0. The distinction is that the negative square root of two obeys the same algebraic relationship, but is constructed from a different Cauchy sequence: {−1, −14 10, −141 100, −1414 1000, −14142 10000, · · ·}. Thus γ, as a solution of γ −2 = 0, could be either the positive or the negative square root of two; we can’t distinguish them us-ing only the four field operations (addition, subtraction, multiplication, and division). To distinguish between + √ 2 and − √ 2, we’d have to introduce an additional relationship – greater than or less than – which would produce an ordered field, and that’s not part of our vocabulary. An ordered field can be treated as an unordered field by ignoring its ordering, which is fortunate for us, because it allows our integration theory to be applied to standard real numbers. For our purposes, γ, as a root of γ2 −2 = 0, is simply a field element that when squared equals two. It could be either + √ 2 or − √ 2; we can’t tell the difference, but it isn’t important because our results will be the same in either case.
Polynomial long division (section 2.4) can be used to divide any polynomial by the mini-mal polynomial, then we discard the quotient and use the remainder as the unique repre-sentative of our equivalence class. Thus, we can easily perform addition, subtraction, and multiplication in an algebraic extension field, by simply performing ordinary polynomial operations and then modulo reducing to a remainder, but how do we perform division?
The polynomial Diophantine equation algorithm described in section 2.6 can be used to solve the following equation: 4The key property is that the one system is a subset of the other, not the exact method of construction.
sx + tn = gcd(x, n) Now, if n is irreducible, its GCD with any polynomial of lower degree will be 1, so our equation becomes: sx + tn = 1 Reducing mod n, we obtain sx ≡1 mod n So s, when multiplied by x, yields 1, which means that s and x are multipicative inverses.
If n was not irreducible, this construction would not work for all values of x, and we would have a ring but not a field.
We need to specify the variable in order to distinguish between operations in C(x)[y] mod n(y) and C(y)[x] mod n(x). For example, reducing mod (y −x) effectively sets y equal to x, but it could also set x equal to y: sage: R. = QQ[] Q[x] sage: S. = Frac(R)[] Frac(Q[x])[y] sage: (x^3 + y) % (y-x) x3 + x sage: R. = QQ[] Q[y] sage: S. = Frac(R)[] Frac(Q[y])[x] sage: (x^3 + y) % (y-x) y3 + y 2.10 Trace and Norm [van der Waerden], §5.7 Given an algebraic extension E of a field F, two of the coefficients in the minimal poly-nomial have a special significance that often makes them particularly useful. Our only use of them will come in Chapter 4 in the proof of Liouville’s theorem.
First, let’s note that while we used a minimal polynomial to construct the extension field, in fact, every element in the extension field has a minimal polynomial associated with it, which can be constructed by raising the element to successive powers until one of the powers is a linear combination of the lower powers.
xn + cn−1 |{z} −Tr(x) xn−1 + · · · + c1x + c0 |{z} (−1)nN (x) = 0 The special significance of these elements becomes more obvious if we consider a split-ting field, which is a further field extension (an extension of the extension) in which the minimal polynomial factors completely into linear factors. The element x is itself one of the roots in these factors; the remaining roots are called the conjugates of x.
Let’s write x’s minimal polynomial in the form: Y σ (t −σx) = tn − X σ σxtn−1 + · · · + Y σ (−σx) where the sums and products are over all automorphisms that fix the base field.
Note that the n −1th coefficient in the polynomial is the sum of all the negatives of the conjugates, while the zeroth-order coefficient coefficient in the polynomial, is the product of all the negatives of the conjugates.
Definition 2.8. The trace of an element x in E, written Tr(x), is the sum of all the conju-gates of x.
Definition 2.9. The norm of an element x, written N(x), is the product of all the conju-gates of x.
The significance of these two functions is first, that they map elements from the extension field to the base field, and second, that trace commutes with addition and norm commutes with multiplication.
Example 2.10. Compute the norm and trace of x + 1 in Q[x] mod x2 −2.
The field Q[x] mod x2 −2 consists of the rational numbers, adjoined with x = √ 2.
Actually, x is any square root of 2.
(x + 1)2 = x2 + 2x + 1 mod x2 −2 = 2x + 3 = 2(x + 1) + 1 (x + 1)2 −2(x + 1) −1 = 0 Tr(x + 1) = 2 N (x + 1) = −1 Tr (x + (x + 1)) = Tr x + Tr(x + 1) = 0 + 2 = 2 = Tr 1 = 2 □ Example 2.11. Prove that (1 + 2i) is irreducible in Q[i] mod i2 + 1.
The field Q[i] mod i2 + 1 consists of the rational numbers, adjoined with i, the square root of −1. These form the Gaussian integers.
Since norm commutes with multiplication, any factorization of (1 + 2i) must lead to a factorization of its norm, so let’s compute the norm of (1 + 2i): (1 + 2i)2 = 1 + 4i + 4i2 mod i2 + 1 = −3 + 4i = 2(1 + 2i) −5 (1 + 2i)2 −2(1 + 2i) + 5 = 0 mod i2 + 1 So, N (1+2i) = 5, which is prime. Therefore, if (1+2i) factors, at least one of its factors must have norm 1 or −1. There are no elements in this field with norm −1, and the only elements with norm 1 are 1, −1, i, and −i, all of which are units.
□ 2.11 Hermite Normal Form If we’re given a ring R and a maximal ideal I, we know that R mod I will be a field, but how can we actually reduce an element e of R modulo I and find its residue, which is e’s value in R mod I, written e mod I?
The answer to that question depends heavily on the ring R and how it and the ideal I are represented.
In Chapter 8, we’ll need to perform this computation when R is a free K[x]-module (K is a field), which is to say that we’ll have a set of basis elements bi in a larger field, typically R’s fraction field. Multiplying these basis elements by a vector in K[x] will give us an element of R, and since the module is free, every vector in K[x] will correspond to a unique element of R.
Since an element of R is specified as a vector in K[x], we can form a square matrix corresponding to the element that, when multiplied by a vector v in K[x] will return a vector in K[x] that is exactly the multiplication of the element of R specified by the vector v by the element of R specified by the matrix. If we expand out the element horizontally (as a K[x] vector), and list out its products with bi vertically, we’ll get a matrix that when multiplied on the left by v (as a row vector) will return the product as a row vector.
I will be given by a set of generators, also in R. Multiplying any one of these vectors by bi will give us the corresponding generator. To form an element of I, we would then multiply each generator by an element of R and add the products.
In matrix notation, we can take the matrices that correspond to multiplying by each gen-erator and stack them vertically. Now we’ll left multiply by a row vector v in K[x] with length ng where n is the degree of the K[x]-module and g is the number of generators.
This corresponds to multiplying each generator by the corresponding element of R, of which there are g.
A matrix H (not necessarily square) with entries in K[x] is in Hermite Normal Form if: (wikipedia, modified slightly) 1. H is upper triangular 2. The leading coefficient of a row is strictly to the right of the leading coefficent of the row above it, and it is monic 3. The elements below pivots are zero and elements above pivots are nonnegative and have degree strictly smaller than the pivot A simple algorithm to convert a matrix M into HNF using elementary row operations works like this: 1. Let i run from 1 to the number of columns in M.
2. Use elementary row operations to move any rows with a zero in the i’th column, considering only rows i and lower, to the bottom of the matrix 3. Compute the monic GCD (use Section 2.5) of the entries in the i’th column, consid-ering only rows i and lower (i.e, the diagonal element and everything below it), but ignore the rows with zeros 4. Use elementary row operations to construct the monic GCD in element eii.
5. Element eii will now be a factor of all elements beneath it, so use elementary row operations to replace these elements with zeros 6. Divide eii into each of the elements above it, and use elementary row operations to replace these elements by their remainders mod eii.
We can form HNF by left multiplying by an invertable matrix. So, we were left multiply-ing by an arbitrary row vector v of length ng to get vM, which is an element of the ideal, specified as a vector of length n. Using HNF, we now get H = UM, or M = U −1H, or vU −1H as our result. Now the bottom (n −1)g rows of the HNF will be all zeros, so the rightmost (n −1)g elements of vU −1 will be ignored.
Given a HNF matrix H and an element e of R represented by a vector in K[x], we can reduce modulo the ideal and obtain a vector u in K[x] with each element of strictly lower degree than the corresponding diagonal element in H.
The algorithm to do this runs as follows: 1. Let i run from 1 to the number of columns in M.
HNF reduction will produce a row u with as many entries as rows in H, such that uH, dotted into bi (call it uHB), will be an element of R that differs from e by less than the degree of the diagonal elements (clarify this). Since vU −1 = u, padding this vector with zeros and multiplying by U gives us v = uU, which is a list of elements that when multiplied by the ideal generators, would produce the same result.
If the ideal is maximal, the residue field is, in fact, a field, and will be K. This implies that the matrix H will have at most a single linear polynomial along the diagonal, all other entries in the matrix will be constants, and vHB will differ from e by a constant.
The bottom row is somewhat special, in that it will correspond to a generator using only a single basis element (the last one), and that generator will be the GCD of all elements in the ideal formed using only that basis element.
We’ll usually want that final basis element to be 1, and the matrix element will be the GCD of all elements in the ideal formed with x and not y. In this case, we’ll end up with e −Hv (as vectors), being a vector with a single constant entry in the last element, and the rest of the vector zero.
2.12 Factorization Some integers, as we know, are prime numbers, meaning that they can not be factored into a product of smaller integers, and the primes play a central role in number theory. In a more abstract setting, they generalize into the concepts of irreducible elements and prime elements, which are similar, but not identical. First, let me first introduce the simpler concept of a unit: A unit is an invertable element.
Put another way, u is a unit if there exists v such that uv = 1. In a field, all elements except 0 are units, so units are only meaningful in the context of a ring.
An irreducible element is a non-zero, non-unit element that can not be written as the product of two non-units.
Any divisor can be transformed into another divisor by multiplying it by a unit, since if uv = 1, then ab = (ua)(vb) for any a and b whatsoever. Therefore, factors are only unique up to units. For example, x can be “factored” as 1·x or (−1)·(−1)·x, but we don’t regard these as factorizations for determining irreducibility, since 1 and −1 are units.
There exist rings in which indeterminates can be factored.
Example 2.12. In the ring F[x, y] mod (y2 −x), the element x is not irreducible.
x ≡y2 mod (y2 −x), so x can be factored as y · y.
Roughly speaking, y is the square root of x.
□ A prime element p is a non-zero, non-unit element with the property that when p divides ab, p divides either a or b.
∀a, b p|ab ⇒p|a ∪p|b There exist rings in which these two concepts do not coincide.
Example 2.13. In the ring F[x, y, z] mod (z2 −xy), the element z is irreducible but not prime. [Wikipedia] First, note that z is irreducible in this ring.
Then, consider the element xy. z divides xy since xy ≡z2 mod (z2 −xy), so xy z ≡z mod (z2 −xy). However, z divides neither x nor y. Therefore, z is not prime.
□ In particular, a greatest common divisor can be transformed into another greatest common divisor by multiplying it by a unit. I leave without proof the claims that in F[x], the units are all elements in F, and that all GCDs differ from each other by a unit multiple.
Unique Factorization A unique factorization domain U is an integral domain in which every non-zero element can be written as a product of a unit and prime elements. [Wikipedia] The Fundamental Theorem of Arithmetic states that Z is a unique factorization domain.
Theorem 2.14. For any field K, the ring K[x] is a unique factorization domain.
Proof Assume the contrary, that we’re working in a unique factorization domain that is not an integral domain. Pick two elements c and d which are divisors of zero, cd = 0. Obviously, we can pick a = 0 and b = 0 and have ab = 0 = cd. So, by U1, x exists so that 0x = c or 0x = d, which by the zero theorem implies that either c = 0 or d = 0. This proves I1.
Thus, a unique factorization domain is also an integral domain.
□ Also, F[x] is a unique factorization domain (proof omitted).
Not all rings satisfy U1. Consider, for example, Z[i]; i2 = −1, the Gaussian integers.
This ring differs from the polynomial ring Z[x] because polynomials of degree two and higher don’t exist since the square of i is -1; i is thus algebraic (see below) and this makes all the difference. The number 9 can be factored two different ways in this ring: 9 = 3 · 3 = (4 −i)(4 + i). It’s not too hard to see that 3 can’t be multiplied by any Gaussian integer to form either (4 −i) or (4 + i), so U1 is not satisfied. The Gaussian integers form an integral domain, but not a unique factorization domain.
Algebraic Closure [van der Waerden], §11 The most important way to characterize a field F is to extend it to a polynomial ring F[x] and then study how polynomials factor in F[x].
A field F is algebraically closed if all polynomials in F[x] can be completely factored into factors linear (i.e, first degree) in x.
Put another way, a field F is algebraically closed if the only irreducible polynomials in F[x] are linear in x. These irreducible polynomials do not have to be linear in any other indeterminate that might exist in F.
A polynomial is monic if its leading coefficient is 1. If F is algebrically closed, then the monic irreducible polynomials in F[x] are all of the form (x −λ), where λ is some element of F.
The Fundamental Theorem of Algebra states that the complex field C is algebraically closed.
There are several proof routes for this theorem. The most common involves complex analysis, and can be found in any standard complex analysis text, usually near Liouville’s Theorem on the behavior of bounded entire functions. I won’t go into it here.
I do want to note the difference between C and C(x). Both are fields. C is algebraically closed because any polynomial in C[x] can be completely factored into linear factors, i.e, x3 −3x2 + 4 = (x −2)2(x + 1). C(x) is not algebraically closed. If it were, then all polynomials in C(x)[y] could be completely factored, but in fact, there exist polynomials such as y2 −x that can not be factored, so C(x) is not algebraically closed.
In this book, the complex field C is the most common field that we’ll use for our constants, precisely because it’s algebraically closed and the irreducible polynomials of C[x] can be so easily characterized. It’s major disadvantage is that it isn’t computable, meaning that an arbitrary complex number can’t be expressed in a finite form. This makes C fine for theoretical proofs and for specific examples, but unsuitable for algorithms designed to work with arbitrary inputs, for which smaller, more complicated fields are required.
I use the algebraic closure of Q extensively. This option causes Sage to print elements of Q using radical notation, if possible: QQbar.options.display_format = 'radical'; Square-free factorization [Ge92], §8.2 A square-free polynomial is one with no repeated factors. x2 −1 is square-free because it factors as (x −1)(x + 1). x2 + 2x + 1 is not square-free because it factors as (x + 1)2.
Whether or not a polynomial is square-free is independent of the field in which the factor-ization occurs.
A square-free factorization of a polynomial is a factorization into square-free factors, each of which appears at a different power. It is much easy to compute than a full factorization into irreducible factors and for this reason will be quite useful to us.
Surprisingly, a polynomial’s square-free factorization is independent of its algebraic sys-tem! For example, x2 + 1 is irreducible in R[x], so its square-free factorization is simply x2 + 1. On the other hand, in C[x], x2 + 1 factors as (x + i)(x −i). Yet both of these factors combine together in the square-free factorization (since they both appear to the first power), so x2 + 1’s square-free factorization in C[x] is...x2 + 1.
To compute square-free factorizations, we’ll use an operation that, for lack of a better word, I’ll call “differentiation.” We “differentiate” a polynomial by multiplying each term by its power and then lowering the power by one.
This “differentiation” not to be confused with the field operation that I will define in the next chapter. “Differentiation” is simply a mechanical procedure of lowering powers and multiplying by constants. In particular, no attempt is made to “differentiate” the coefficients even if they are not constants.
To compute a square-free factorization, first we “differentiate” the polynomial. The result is a second polynomial with the degree of all factors reduced by one. Note in particular that any factors of unit degree (and only those factors) disappear completely. Dividing this into the original polynomial, we obtain a polynomial with no square factors — all factors now appear with unit degree. Computing the GCD of this polynomial with the original one also produces a polynomial with only factors of unit degree, except that the original unit degree factors are missing. We can divide this last two polynomials into each other to determine the original unit degree factors. Going back to the “differentiation” step, we can keep repeating the process until we have obtained all the square-free factors.
Full polynomial factorization (optional) [van der Waerden], §5.6 [Ge92], Chs. 5, 6, 8 Let’s conclude this chapter by taking a least a brief look at fully factoring a polynomial into its irreducible factors. There are several reasons to do this.
First of all, it’s easy to declare that “the Fundamental Theorem of Algebra tells us that any polynomial in C[x] can be factored into linear factors”, and maybe even prove it. That’s true, but when it comes time to actually do a computation, how do we proceed? How do we actually factor a polynomial? Call it the price of success. Differential algebra is solid enough to actually compute integrals, so existence theorems don’t cut it. We need constructive algorithms.
Second, it’s a surprisingly difficult problem. An appreciation of its difficulty now will motivate the discussion later when I show various techniques that have been developed to avoid full factorization whenever possible. Yet the fact remains that it is at times unavoid-able.
Finally, both techniques that I will discuss here work according to a basic principle that we’ll use again later in a more advanced context, so it makes sense to present it now in a simpler form. Specifically, we’ll solve a difficult problem in an algebraic system by using a homomorphism to map into a different algebraic system where we can solve the problem more easily, then find some way of “lifting” this answer back into the original system. This is one of the most powerful solution methods in algebra, and has been used to solve problems once thought impossible.
Let’s start simple. We want to factor a polynomial in Z[x], the ring of polynomials with integer coefficients. If such a polynomial has a factorization in Z[x], for example x2−1 = (x+1)(x−1), we want to find it. If it has no such factorization, for example x2+1 (which would require at least Z[i, x]; i2 = −1 to factor), we want to prove this.
Now consider what happens when we set x equal to some specific integer value, say a.
Any polynomial in Z[x] will be transformed into an integer. Thus, we have a mapping φx−a : Z[x] →Z from polynomials to integers. Not only is this a mapping, but it is a ho-momorphism, a mapping that preserves the operations, so φx−a(m+n) = φx−am ˆ + φx−an and φx−a(m · n) = φx−amˆ · φx−an, where I have used ˆ + and ˆ · to emphasize that these operations are operations in Z, and distinct from + and ·, which are operations in Z[x].5 I leave it an exercise to actually prove this is a homomorphism.
Since φx−a is a homomorphism, any factorization of a polynomial in Z[x] must map into a factorization of its image integer in Z. In other words, if a polynomial factors into smaller polynomials (all with integer coefficients), then setting the variable equal to some specific integer causes all the polynomials to evaluate into integers, which must themselves factor.
Consider x2 −1 = (x + 1)(x −1). If we set x = 2 (the evaluation homomorphism φx−2), then the equation becomes 3 = 3 · 1. This happens irregardless of our choice of integer.
Choosing x = 3 (φx−3) transforms x2 −1 = (x + 1)(x −1) into 8 = 4 · 2.
Thus, we have our homomorphism, which maps our problem from Z[x] into Z and trans-forms the factorization of polynomials into a factorization of integers. Although factoring integers is certainly not trivial (the security of the RSA cryptosystem depends on its near impossibility for large numbers), it is much easier than factoring polynomials. Not only easier, but finite. There are only a certain number of ways any given integer can factor, and for relatively small integers, we can enumerate them by computing a prime factorization and then listing the finite number of ways that the primes can be combined into factors.
The number 3, for example, can be split into two integer factors in only one of four ways: 3 · 1, 1 · 3, −3 · −1, and −1 · −3.
2.13 Primary Decomposition Primary decomposition generalizes polynomial factorization to systems of polynomial equations.
Ideals in rings generalize primes of integers. In the ring of integers, all ideals are princi-pal, meaning that they are generated by a single element, so the ideals are in one-to-one correspondence with the integers themselves. The prime ideals are exactly those ideals generated by prime integers.
In more general rings, ideals are not necessarily principal, and factorization, while well defined, is not as simple as factoring an integer, although the RSA algorithm would be useless if factoring an integer was trivial.
Ideal in general factor in the following manner. An ideal can be written as the intersection of primary ideals, each of which is contained within a single prime ideal. While the 5The symbols · and ˆ · represent multiplication, which we normally omit entirely, but I have written explicitly here to make this point.
primary ideals are not uniquely defined, the associated prime ideals are. The radical of a primary ideal is its associated prime ideal, and if the original ideal is radical, then the primary ideals are themselves both radical and prime. Thus, while the factorization of an arbitrary ideal is only defined up to its associated prime ideals, a radical ideal is exactly equal to the intersection of its prime ideals.
How, then, do we compute a primary decomposition of an ideal?
Let’s consider the easiest case first: a radical ideal in an polynomial ring over an alge-braically closed field of constants, like C[x, y, z].
First, consider all possible subsets of the variables. Which subsets u, when intersected with the ideal I ∪C[u], form (0)? Is all we need to check are its generators? A largest such subset is called a maximal independent set; it may not be unique, but all maximal independent sets have equal size, and their size equals the dimension of the ideal.
Pick a maximal independent set u, form the fraction field C(u), and consider the polyno-mial ring over that fraction field generated by the remaining elements C(u)[x\u]. In this ring, the ideal is zero-dimensional. We can find the components of a zero-dimensional ideal by computing a Grobner basis using an elimination order and then factoring polyno-mials.
Ideals, however, can have components of varying dimension. We’ve only found the com-ponents of largest dimension. Once we’ve found the d-dimensional components, we can use saturation to divide them out. If the resulting ideal is (1), we’re done, otherwise we continue going to find the remaining components, of smaller dimension. Obviously this process will eventually terminate.
Consider the ideal (xy, xz) in C(x, y, z). {y, z} is a maximal independent set, because setting y and z to zero annihilates both generators. Consider the ideal in the ring C(y, z)[x]; it’s generated by (x). Now we compute (xy, xz) : (x)inf = (y, z).
2.14 Exercises Factor the following polynomials in Z[x], Q[x], R[x], and C[x]: 1. x2 −1 (factors in all four rings) 2. 4x2 −1 (factors in all rings except Z[x]) 3. x2 −2 (factors in R[x] and C[x], but not in Z[x] or Q[x]) 4. x2 + 1 (factors only in C[x]) Write a computer program to factor the following polynomials: 5. x5 + 2x4 + 2x3 −x −1 factor( x^5+2x^4+2x^3-x-1 ) Ans: (x2 + x + 1) · (x3 + x2 −1) 6. 34x5 + 51x4 + 60x3 + 25x2 + 8x −1 factor( 34x^5+51x^4+60x^3+25x^2+8x-1 ) Ans: (x2 + x + 1) · (34x3 + 17x2 + 9x −1) 7. x5 + 2x4 + 2x3 −x −1 factor( x^5+2x^4+2x^3-x-1 ) Ans: (x2 + x + 1) · (x3 + x2 −1) Write a computer program to factor the following polynomials: (Hint: You’ll need a 2 dimensional grid) 8. 2x3 + 3x2y −7x2 + 14xy + 21y2 + 22x −16y −77 factor( 2x^3+3x^2y-7x^2+14xy+21y^2+22x-16y-77 ) Ans: (2x + 3y −7) · (x2 + 7y + 11) 9. x2y3 + 2x4 + y4 + 2x2y + y3 −x2 −3y −3 factor( x^2y^3+2x^4+y^4+2x^2y+y^3-x^2-3y-3 ) Ans: (x2 + y + 1) · (y3 + 2x2 −3) Chapter 3 Algebraic Geometry In pure algebra, such as we’ve studied in the last chapter, x is just x. It does not “take a value” and has no other interpretation. Everything we did in the last chapter is based solely on the axioms of a commutative ring or field.
We next want to consider what happens when we let x take a value, say a real number, and then conclude that the equation x2 −1 = 0 is true if x is either −1 or 1.
This is now algebraic geometry.
The roots of algebraic geometry lie in studying the zeros of polynomial equations. We began with a single polynomial in a single variable, and have learnt a great deal about it.
We know how to solve it (at least in terms of radicals) if its degree is less than 5. Galois proved that no such solution (in radicals) exists (in the general case) for larger degree, though abstract algebra provides us with a suitable general theory to handle this case.
Simple long division tells us that it can have no more roots than its degree, and Gauss showed that all of the roots exist as complex numbers — the Fundamental Theorem of Algebra.
The next logical step is to consider zeros of a single polynomial in two variables, and this equation has also received a great deal of attention from mathematicians. Like the univariate case, we have theories devoted to low-order special cases — linear equations (all terms first degree or constant), the conic sections (all terms second degree or less), and the elliptic curves (one term third degree; all others second degree or less). In the general case, P aijxiyj = 0 is called an algebraic curve.
3.1 Solving systems of equations Probably the most basic application of primary decomposition is solving systems of equa-tions.
An ideal in a polynomial ring has an associated variety, which is the set of points that zero all of the polynomials in the ideal. A variety can be expressed as the intersection of its irreducible components, each of which corresponds precisely with a primary ideal in the primary decomposition.
If the underlying field of constants has no zero divisors (as is the case with ordinary real or complex numbers), then we are justified in taking the radical of the ideal, and then the primary decomposition is just the prime decomposition because our primary ideals are all prime ideals.
Returning to the (xy, xz) example, we found that this ideal’s primary decomposition is (x)∩(y, z). In other words, the solution variety of the system of equations xy = 0, xz = 0 is formed by the z-axis y = 0 z = 0 and the x −y plane x = 0.
Chapter 4 Differential Algebra 4.1 Differential Fields The advent of the modern, axiomatized approach to mathematics at the turn of the twen-tieth century led directly to the development of abstract algebra, with its rings and fields, in the 1920s. By 1940, a Columbia University professor named J.F. Ritt had proposed the concepts of differential rings and differential fields. They are exactly analogous to ordi-nary rings and fields, except that they are equipped with a third basic operator (addition and multiplication are the first two), called derivation. A derivation is a unary operator (the other two are binary), which we shall typically denote by D. Since algebra is fun-damentally concerned with how operators commute with each other, the first question we are lead to ask are, “How does derivation commute with addition and multiplication?” The answer is to be found in two basic axioms: addition law of derivations ∀a, b ∈D, D(a + b) = Da + Db (D1) multiplication law of derivations ∀a, b ∈D, D(ab) = aDb + bDa (D2) Neither axiom should come as any great surprise. After all, these are just the basic addition and multiplication rules we learned in first year calculus. Yet note how they are being presented; not as results derived from some theorem involving fractions and limits, but as axioms that are assumed true from the start. One of the great themes of differential algebra is that we purge from the subject almost any mention of limits; for us, derivation is just a mapping in a field that carries an object a to another object b. Integration, then, is little more than the inversion of derivation: given an object b, can we find an object a which maps into b?
Yet the connection to calculus should be made clear. Since derivation (in the calculus sense) obeys these two axioms for derivation (in the algebra sense), the calculus derivation will always behave as an algebra derivation, so any theory we develop for the algebra derivation will apply immediately to the calculus version.
Let me immediately note the difference between differential algebra and algebraic ge-ometry. In algebraic geometry, the indeterminates generally take the value of constants, typically real or complex numbers. In differential algebra, the indeterminates are now functions with non-trivial derivatives. In particular, we should not immediately expect that we can use differential algebra to solve a system of differential equations in the same manner as we can use algebraic geometry to solve a system of polynomial equations.
Without introducing additional assumptions (which we absolutely can do), we’re only go-ing to find differential relationships between functions. The additional assumptions will typically take the form of assuming that the functions have some particular form, such as being elementary, and we’re not going to be able to “solve” differential equations without these additional assumptions. If this seems at all puzzling, ask yourself what form those solutions would take, if an indeterminate is not a real or complex number, but rather some function with no additional assumptions about its structure.
What can we determine from these two axioms? A surprising lot, in my opinion.
Theorem 4.1.
• D(0) = 0 • D(1) = 0 • D(1 a) = −1 a2D(a) • D(cx) = c D(x) if D(c) = 0 Proof D(0) = D(0) + D(0) −D(0) = D(0 + 0) −D(0) = D(0) −D(0) = 0 D(1) = D(1 · 1) = D(1) + D(1) D(1) = D(1) + D(1) −D(1) = D(1) −D(1) = 0 0 = D(1) = D(a · 1 a) = 1 aD(a) + aD(1 a) aD(1 a) = −1 aD(a) D(1 a) = −1 a2D(a) □ It follows immediately from this theorem that our entire prime subfield, as well as any purely algebraic extension thereof, must map to zero under derivation.
Theorem 4.2. The set of all elements in a differential field which map to zero under derivation forms a subfield.
□ The subfield which maps to zero is called the constant subfield. It necessarily includes the prime subfield and any elements algebraic over the prime subfield, but may include other transcendental elements as well. For example, consider R, the real numbers. 2 is in the prime subfield, so D(2) = 0; √ 2 is algebraic over the prime subfield, so D( √ 2) = 0; π is transcendental over the prime subfield, so doesn’t have to map to zero, but we will (obviously) set D(π) = 0. All three elements — 2, √ 2, π — are in the constant subfield.
Theorem 4.3. The derivation of an algebraic extension is determined uniquely by the derivation of its subfield.
Proof The derivation of an algebraic element is completely defined by the subfield’s derivation and the element’s minimal polynomial.
Given an element ξ, let its minimal polynomial be: X i aiξi = 0 Differentiating this polynomial (using the D1 and D2 axioms), we obtain: X i (a′ iξi + iaiξi−1ξ′) = 0 X i iaiξi−1ξ′ = − X i a′ iξi ξ′ = − P i a′ iξi P i iaiξi−1 □ The upshot of all this is that our basic D1 and D2 axioms completely define a derivation both for our prime subfield as well as any purely algebraic extensions. It therefore follows that we need only specify the behavior of a derivation on transcendental elements and we will have completely defined the derivation.
We will use four types of transcendental elements in our theory: 1. Constants. D(c) = 0 2. The distinguished variable of integration. D(x) = 1 Since this is an O.D.E. theory, and particularly an integration theory, we are always in-tegrating with respect to some variable of integration. There is no loss of generality in labeling it x. By setting D(x) = 1 we establish that our derivation is in fact a derivative and not a differential.
Incidently, Ritt had already conceived back in the 1940s of equipping a differential field with multiple derivations, one for each of a set of independent variables. This corresponds nicely to what is needed for a P.D.E. theory. Thus, given variables x, y and z, we could construct derivatives Dx, Dy and Dz so that Dx(x) = 1, Dx(y) = 0, Dx(z) = 0 and so on. Since our focus is on integration, I’ll have nothing more to say about fields with multiple derivations.
3. Logarithmic extensions. D(θ) = D(φ) φ 4. Exponential extensions. D(θ) = θD(φ) φ, in both of these cases, is some element in the underlying field.
These two extentions clearly correspond to θ = ln φ (in the logrithmic case) and θ = exp φ (in the exponential case). The key point I want to made immediately is that these are transcendental extensions...and not all logarithms and exponentials are transcendental!
Transcendental extensions are defined by exclusion — any extension that isn’t algebraic is transcendental. If we’re dealing with an algebraic extension, even if defined using logarithms and exponentials, we have to use our algebraic theory.
Example 4.4. Represent 4x + 1 2x + 1 in Liouvillian form There are three ways to do this — the easy way, the hard way, and the wrong way.
Let me first note that 4x = (22)x = (2x)2. The existence of this algebraic relationship between 4x and 2x means that we can not use two seperate transcendental extensions. So this: θ + 1 φ + 1; θ = exp(x ln 4); φ = exp(x ln 2) is the wrong way.
The easy way is to set up 2x first and then construct 4x as its square: φ2 + 1 φ + 1 ; φ = exp(x ln 2) You can also do this the hard way, setting up 4x first and then using an additional algebraic extension to get its square root, 2x: θ + 1 φ + 1; θ = exp(x ln 4); φ2 = θ See Example 7.7 for the actual integration.
□ That’s it! The basic two differential axioms, algebraic extensions, fraction fields, and these four types of transcendentals, round out the entire base algebraic structure we’ll need to construct our theory. We do need to be careful, though, as the last example illustrated. In these simple examples, figuring out which elements are algebraic and which are transcendental is easy, but in more complex expressions this may not be obvious. We’ll discuss in Chapter ?? how to test new elements for transcendence.
from sage.rings.fraction_field_element import is_FractionFieldElement from sage.rings.polynomial.polynomial_element import is_Polynomial from sage.rings.polynomial.multi_polynomial import is_MPolynomial from sage.misc.latex import str_function class Derivation(SageObject) : def init(self, parent, generator_map): self.generator_map = generator_map self.parent = parent def repr_defn(self): gm = self.generator_map return '\n'.join(['%s |--> %s'%(i, gm[i]) for i in gm]) def _repr(self): return 'Derivation of %s\n Defn: %s'%(self.parent, '\n '.join(self.repr_d def _latex_defn(self): gm = self.generator_map return '\\'.join(['%s & \rightarrow & %s'%(latex(i), latex(gm[i])) for i in gm]) def _latex(self): return '\begin{array}{rcl}\multicolumn{3}{c}{' + str_function('Derivation of ') + def call(self, x): gm = self.generator_map if is_FractionFieldElement(x): n = x.numerator() d = x.denominator() return (self(n)d - nself(d))/(d^2) elif is_Polynomial(x): var = x.args() result = 0 for power in range(0, x.degree()+1): coeff = x[power] if power != 0: result = result + power coeff var^(power-1) gm[var] result = result + self(coeff) var^power return self.parent(result) elif is_MPolynomial(x): vars = x.args() result = 0 for (monomial, coeff) in x.dict().items(): for i in range(len(monomial)): power = monomial[i] if power > 0: var = vars[i] Dvar = gm[var] result = result + power coeff x.parent().monomial(monomial) / var result = result + self(coeff) x.parent().monomial(monomial) return self.parent(result) elif x.parent() in [ZZ, QQ, AA, QQbar, RR, CC]: return 0 else: raise NotImplementedError sage: R. = QQ[] Q[x] sage: D = Derivation(R, {x: 1}) Derivation of Q[x] x → 1 sage: D(x^2) 2x sage: R. = QQ[] Q[x, θ, ψ] sage: D = Derivation(R, {x: 1, theta: 1/x, psi: psi}) Derivation of Q[x, θ, ψ] x → 1 θ → 1 x ψ → ψ sage: D(xthetapsi) xθψ + θψ + ψ Definition 4.5. An elementary extension of a differential field is a differential extension field constructed using a finite number of algebraic, logarithmic, and exponential exten-sions.
Definition 4.6. A elementary function of a single variable x over a specified field of constants K is a function in an elementary extension of the rational function field K(x).
What about sines and cosines, all those arc-functions, raising things to powers, and all that? Turns out we can express all those operations using just our basic extensions. The key here is Euler’s famous identity eiθ = i sin θ + cos θ.
Example 4.7. Express sin x in Liouvillian form Euler’s identity immediately gives: sin x = −i eix −e−ix 2 Therefore, starting from C(x), we add the exponential extension θ = exp(ix), and con-clude that sin x can be expressed as the rational function: θ2 −1 2iθ in the field C(x, θ); θ = exp(ix).
□ If trigonometric functions can be represented using complex exponentials, then it should come as no real surprise that inverse trigonometric functions can be represented with complex logarithms.
Example 4.8. Represent arcsin x in Liouvillian form Let’s start with Euler’s identity and take its logarithm: eiθ = i sin θ + cos θ iθ = ln(i sin θ + cos θ) Now, if θ = arcsin x, then x = sin θ, and we can use the basic sin2 θ + cos2 θ = 1 identity to compute cos θ = p 1 −sin2 θ = √ 1 −x2. Substituting above: iθ = ln ix + √ 1 −x2 θ = −i ln ix + √ 1 −x2 arcsin x = −i ln ix + √ 1 −x2 Thus, we need first an algebraic extension to construct φ = √ 1 −x2, followed by a logarithm extension to construct arcsin x = −i ln(ix + φ).
□ I think the details of further constructions along these lines are straightforward enough that I will simply summarize them in a table.
4.2 Liouvillian Forms Expression Liouvillian Form Expression Liouvillian Form f g e g ln f sin x −i eix −e−ix 2 sinh x ex −e−x 2 cos x eix + e−ix 2 cosh x ex + e−x 2 tan x −i eix −e−ix eix + e−ix tanh x ex −e−x ex + e−x sec x 2 eix + e−ix sech x 2 ex + e−x csc x 2i eix −e−ix csch x 2 ex −e−x cot x i eix + e−ix eix −e−ix coth x ex + e−x ex −e−x arcsin x −i ln ix + √ 1 −x2 sinh−1 x ln x + √ x2 + 1 arccos x −i ln x + i √ 1 −x2 cosh−1 x ln x + √ x2 −1 arctan x 1 2 i ln ix −1 ix + 1 tanh−1 x 1 2 ln 1 + x 1 −x sec−1 x −i ln 1 + i √ x2 −1 x sech−1x 1 2 ln 1 + √ 1 −x2 1 − √ 1 −x2 csc−1 x −i ln i + √ x2 −1 x csch−1x 1 2 ln √ 1 + x2 + 1 √ 1 + x2 −1 cot−1 x 1 2 i ln i + x i −x coth−1 x 1 2 ln x + 1 x −1 4.3 Liouville’s Theorem The next problem we must confront is to limit the number of possible fields in which we can find solutions to our problem. So far, we have seen how to construct an algebraic system to express any elementary function, but there are an infinity of such systems.
Searching them exhaustively for the solution to a given integral is out of the question.
Fortunately, it’s been known for almost 200 years that there are severe restrictions on what extensions can appear in an integral above and beyond those used in the original integrand.
For example, consider the expression ex. Differentiating it yields, well, ex. Now the key thing to note is that the exponential does not disappear after differentiation. This, in fact, is a general property of exponentials — differentiation never makes them disappear.
They can change around, to be sure, d dxe2x = 2e2x, but notice that the exponential is still present in the result. Therefore, since the solution to our integral must differentiate into the original integrand, we conclude that no new exponentials can appear in the integral beyond those in the integrand. If there were new exponentials in the result, then they would have to appear in the integrand as well, since they can never disappear under differentiation.
The same thing happens with roots. Differentiate √x and you get 1 2√x. This time the root moves from the numerator to the denominator, but again, it doesn’t completely disappear.
This is a general property of roots, algebraic extensions in general, in fact.
Logarithms are different, though. Differentiate ln x to get 1 x. The logarithm is gone. So new logarithms can appear in integrals, because they can disappear under differentiation to recover the original integrand. Even here, though, there are important restrictions. The logarithms have to appear with constant coefficients (because something like x ln x would differentiate into 1+ln x), can not appear in powers or in denominators ( d dx ln2 x = 2ln x x ), and can not be nested ( d dx ln(ln x) = 1 x ln x).
These examples are all special cases of Liouville’s Theorem — the only new extensions that can appear in an integral are simple logarithms with constant coefficients. Let’s begin by stating and proving some basic properties of our three basic types of extensions.
Theorem 4.9. Let E = K(θ) be a simple transcendental logarithmic extension of a differential field K with the same constant subfield as K, let p = P piθi be a polynomial in K[θ], and let r = a/b be a rational function in K(θ) (a, b ∈K[θ]). Then: 1. If p’s leading coefficient is constant (p′ n = 0), then Degθ p′ = Degθ p −1 2. If p’s leading coefficient is not constant (p′ n ̸= 0), then Degθ p′ = Degθ p 3. If p is monic and irreducible, then p′ ∤p 4. If an irreducible factor appears in r’s denominator with multiplicity m, then it ap-pears in r′’s denominator with multiplicity m + 1 5. r′ ∈K if and only if r has the form cθ + k, where c is a constant Proof The first two statements follow easily from considering p′: p′ = n X i=0 (p′ iθi + ipiθ′θi−1) = n X i=0 (p′ i + (i + 1) pi+1θ′) θi Note that since K(θ) is a logarithmic extension, θ′ ∈K, so for all i the entire expression (p′ i + (i + 1)pi+1θ′) is in K. In particular, since pn+1 is zero, the nth coefficient of p′ is just p′ n and the θ-degree of p′ will be n if p′ n is non-zero. On the other hand, if p′ n is zero, then the n −1th coefficient of p′ is (p′ n−1 + npnθ′) which would be zero only if θ′ = − p′ n−1 npn = (−pn−1 npn )′ (by Theorem 4.1 since pn is constant), which implies an algebraic relationship between θ and −pn−1 npn (specifically, they differ only by a constant, which must be in K), contradicting the transcendence of E over K.
Next, if p is monic and irreducible, then Degθ p′ = Degθ p −1, and no lower degree polynomial can divide an irreducible polynomial, establishing the third claim.
Now consider a rational function r = a(θ)/b(θ) in its normalized form, so gcd(a, b) = 1 and b is monic. Now we can factor b into irreducible factors (b = Q bi(θ)mi) and expand r using partial fractions (Section 2.7): r = a0(θ) + µ X i=1 mi X j=1 aij(θ) bi(θ)j where a0, aij, bi ∈K[θ] and Degθ aij < Degθ bi. Now let’s differentiate: r′ = a′ 0(θ) + µ X i=1 mi X j=1 a′ ij(θ) bi(θ)j −j aij(θ) b′ i(θ) bi(θ)j+1 aij does not divide bi (since Degθ aij < Degθ bi, and we proved above that b′ i does not divide bi (since bi is monic and irreducible), so there is exactly one term on the right hand side with bi(θ)mi+1 in its denominator and no other terms with higher powers. Therefore, r′ must have a bi(θ)mi+1 in its denominator, establishing the fourth claim.
Finally, since the hypothesis of the fifth claim states that r′ is in K, it can not have any θ terms in its denominator (or anywhere else), so there can not be any bi(θ) factors, and r must be a polynomial. Futhermore, our first two claims imply that if Degθ r′ = 0 (since r′ ∈K), then Degθ r can be at most 1, and its leading coefficient must be constant.
□ Theorem 4.10. Let E = K(θ) be a simple transcendental exponential extension of a differential field K with the same constant subfield as K, let p = P piθi be a polynomial in K[θ], and let r = a/b be a rational function in K(θ) (a, b ∈K[θ]). Then: 1. Degθ p′ = Degθ p 2. p′ | p if and only if p is monomial (i.e, has the form piθi) 3. If an irreducible factor other than θ appears in r’s denominator with multiplicity m, then it appears in r′’s denominator with multiplicity m + 1 4. r′ ∈K if and only if r ∈K Proof Again, p′ = n X i=0 (p′ iθi + ipiθ′θi−1) This time, however, θ′ = k′θ, so p′ = n X i=0 (p′ i + ipik′)θi Assume that one of these coefficients, say (p′ i + ipik′), was zero but pi was non-zero.
Then D(piθi) = (p′ i + ipik′)θi = 0, so piθi would be a constant, which must be in K, contradicting the transcendence of E. Therefore, none of these coefficients can be zero, establishing the first claim.
To establish the second claim, assume first that p′ | p. Since p′ has the same degree as p (by the first claim), it can only divide p if it has the form mp, where m ∈K. Equating coefficients of θ in the above sums leads us to conclude that m = (p′ i pi + ik′) If p was not monomial, then all of its coefficients must yield the same value for m, i.e, m = (p′ i pi + ik′) = (p′ j pj + jk′) p′ ipj −pip′ j + (i −j)k′pipj = 0 p′ ipj −pip′ j p2 j + (i −j)k′ pi pj = 0 pi pj ′ + (i −j)k′ pi pj = 0 Then D( pi pj θj−i) = pi pj ′ + (i −j) pi pj k′ = 0, again contradicting the transcendence of E over K. So p must be monomial.
Conversely, if p is monomial, say aθn, then p′ = (a′ + nak′)θn = a′+nak′ a p and p′ | p.
To prove the final two claims, we proceed as before, expanding r using partial fractions: r = a0(θ) + µ X i=1 mi X j=1 aij(θ) bi(θ)j and taking the derivative: r′ = a′ 0(θ) + µ X i=1 mi X j=1 a′ ij(θ) bi(θ)j −j aij(θ) b′ i(θ) bi(θ)j+1 θ is the only irreducible monomial, so if a bi is not θ, then it will not be canceled by b′ i, and again we’ll have a single term on the R.H.S. with bj+1 i in the denominator, so the L.H.S.
must also have a bj+1 i in its denominator.
This time, however, b′ i can divide bi if bi is monomial. When r′ is in K’, all other possi-bilities are excluded as before, so r must now have the form: r = n X i=−m riθi where ri ∈K. We’ve already established that if ri is non-zero, then the corresponding term in the derivative is also non-zero, so the only way for r′ to be in K is if r is in K.
□ Example 4.11. Let p = xex. Then d dxxex = ex + xex = (x + 1)ex = x+1 x ex.
We start with the rational function field K = C(x), and extend by the transcendental exponential θ = exp(x) to form the ring K[θ]. Both p and p′ are in K[θ]; p = xθ is monomial (in θ); note that x+1 x ∈K, so p′ | p in this ring.
□ Theorem 4.12. Let A be an algebraic extension of a differential field K with the same constant subfield as K, let σ be an automorphism of A/K, and let a be an element of A.
1. D(σx) = σ(Dx), 2. Tr(Dx) = D(Tr x), 3. Tr Dx x = N (x)′ N (x) 4. a′ ∈K ↔a ∈K.
Proof 1. Consider an automorphism σ of A/K, i.e, an automorphism of A that fixes the dif-ferential field K, so that σx = x for x ∈K. Writing the minimal polynomial of x as P i aixi = 0, applying σ to this equation, and remembering that automorphism commutes with multiplication and addition and that ai ∈K, we obtain P i ai(σx)i = 0, i.e, σx has the same minimal polynomial as x; we say that σx is a conjugate of x.
Theorem 4.3 now gives us the derivation of σx: D(σx) = − P i a′ i(σx)i P i iai(σx)i−1 Applying the operators in the other direction, however, and again using the fact that auto-morphism commutes with our field operators, we obtain: σ(Dx) = σ − P i a′ ixi P i iaixi−1 = − P i a′ i(σx)i P i iai(σx)i−1 i.e, D(σx) = σ(DX); automorphisms that fix the base field of an algebraic extension commute with derivation.
2. Now let’s consider how an automorphism σ of A/K interacts with Tr. Remember that trace, in a Galois extension, can be written as a sum over all automorphisms that fix the base field: Tr x = X σ σx We extend A, if necessary, into a Galois extension, and use the commutation relationship we just proved to establish that trace commutes with derivation: D(Tr x) = D X σ σx !
= X σ D(σx) = X σ σ(Dx) = Tr(Dx) 3. Using the commutation relationship we just proved, along with the definitions of Tr and N : Tr Dx x = X σ σ Dx x = X σ σDx σx = X σ Dσx σx = D Q σ σx Q σ σx = D(N (x)) N (x) 4. The right-to-left implication is obvious (since differential fields are closed under deriva-tion), so we need only to prove the left-to-right implication.
Consider a, with Da ∈K, so Tr(Da) = nDa, where n is the degree of the algebraic extension, by Theorem ??. It follows that Da = 1 n Tr(Da) = 1 nD(Tr a) Since Tr a ∈K, we have identified an element in K with the same derivation as a, which therefore can differ from a solely by an additive constant. Since A and K have the same constant subfield, all of our constants are in K, so a is therefore also in K.
□ Example 4.13. Explain the “disappearance” of the square root in: Z 1 √ 1 −x2 = arcsin x Finding arcsin x in the table, we see that: arcsin x = −i ln ix + √ 1 −x2 That’s where it went! It “disappeared” into the complex logarithm that arcsin x is formed from. New logarithms, of course, are acceptable. Notice that the new logarithm has a constant coefficient (−i), is not nested, and appears to the first power.
□ Notice the extra condition on the algebraic extension, that the extension has to preserve the constant subfield. The theorem would fail without this condition, as shown by numerous examples of roots appearing in integrals where only rational numbers were needed in the integrand. The simplest way to handle this situation is to use an algebraically closed constant subfield (like C), but this is not always practical.
Example 4.14.
Z 1 x2 −2dx = Z 1 2 √ 2 1 x − √ 2 − 1 x + √ 2 dx = 1 2 √ 2 h ln x − √ 2 −ln x + √ 2 i This integrand can be expressed in Q(x), but the integral requires Q(x, ξ, θ, ψ); ξ is alge-braic with minimal polynomial ξ2 −2 = 0; θ and ψ are logarithmic transcendental with θ′ = 1/(x −ξ) and ψ′ = 1/(x + ξ).
□ Finally, we want to prove the full Liouville theorem, establishing that the only new exten-sions that can appear in an integral are logarithmic ones.
Theorem 4.15. (Liouville) Let L be an elementary extension of a differential field K with the same constant subfield as K. Then ∀l ∈L, l′ ∈K iff l has the form: k + n X i=1 ciθi where k ∈K, K(θi) are simple logarithmic extensions of K and ci are constants. Proof By the definition of an elementary extension, L has the form K(t1, . . . , tn) where each ti is a simple elementary extension of K(t1, . . . , ti−1).
We’ll proceed by induction on the number of extensions n. Theorems 4.9(5), 4.10(4), and 4.12(4) establish the theorem for n = 1. So assume that the theorem is true for all i < n.
Let M = K(t1), so L = M(t2, . . . , tn), and the induction hypothesis implies that if l′ ∈M, then l has the form: l = m0 + X ciθi where m0 ∈M and M(θi) are simple logarithmic extensions of M.
l′ = m′ 0 + X ci m′ i mi l′ ∈K, and m0 and the various mi are rational functions in K(t1). We can use our basic logarithm identities: (ab)′ ab = a′ a + b′ b ( 1 a)′ 1 a = −a′ a2 1 a = −a′ a to reduce to the case where the mi (except m0) are all irreducible polynomials, so let’s consider our three cases: 1. K(t1) is logarithmic over K.
In this case, Theorem 4.9(4) states that if m0 has a non-trivial denominator with an irreducible factor p, then p appears in m′ 0’s denominator with multiplicity at least two. Since l′ ∈K, this implies that the sum must contribute a denominator with the same multiplicity in order to achieve cancellation, so p must be one of the mi’s.
However, since p is irreducible, Theorem 4.9(4) implies that the multiplicity of p in the sum’s denominator can be no more than one, so m0 must be a polynomial.
Futuremore, there is no cancellation between a normal irreducible polynomial and its derivative, so none of the mi’s can be polynomials in K[t1]; they must exist in K, since otherwise l′ would have a non-trivial denominator in K(t1), and by hypothesis, l′ ∈K.
Finally, Theorem 4.9(5) states that m0 must have the form k0 + c0t1. In other words, a simple logarithm extension can contribute a single term to the sum in the statement of the theorem.
2. K(t1) is exponential over K.
This case is similar to the logarithmic one, except that we must now consider the possibility of special polynomials in the mi’s. Actually, there is only one special irreducible polynomial, t1 itself. If one of the mi’s was t1, then it would cancel with its derivative as follows: t′ 1 t1 = k′ Since both l′ and k′ are in K, we can collect them together as follows: l′ −c1k′ = m′ 0 + X ci m′ i mi and proceed with the proof as before. This time, however, Theorem 4.10(4) tells us that m0 can only have the form k0, so exponential extensions contribute nothing to the sum in the statement of the theorem.
3. M = K(t1) is algebraic over K Applying the trace map to our induction equation: Tr(l′) = Tr(m′ 0) + X ci Tr m′ i mi Since l′ ∈K, Tr(l′) = nl′, where n is the degree of the algebraic extension K(t1) over K, and: nl′ = Tr(m0)′ + X ci N (mi)′ N (mi) l′ = Tr(m0) n ′ + X ci n N (mi)′ N (mi) Setting k0 = Tr(m0) n and ki = N (mi), we see that l′ can be written: l′ = k′ 0 + X ci d k′ i ki establishing that l has the form required by the theorem.
□ Chapter 5 Integration of Rational Functions Since our strategy will be to reduce integrals in complex fields by stripping away their extensions and obtaining integrals in the simpler underlying fields, it follows that we should start this discussion by describing how to integrate in C(x), the field that underlies all the others.
Perhaps this seems pedantic. After all, didn’t we go over all this in first year Calculus?
Don’t we already know everything we need to about integrating rational functions? We just factor the denominator, do a partial fractions expansion, plug in some simple known integrals, and we’re done, right?
Not so fast. To begin with, there’s that business of “just” factoring the denominator. As we’ve already seen, factoring a large polynomial can be quite a daunting undertaking.
Techniques have been developed to avoid it as much as possible. Also, if you’re seeing Liouville’s theorem for the first time, then a whole new dimension to things like arctan open up when you regard them as complex logarithms. And finally, the multi-valued nature of complex logarithms and related functions make them very slippery little beasts.
It’s easy to get nonsense answers from the simplest calculations if you’re not careful.
5.1 Logarithms and related functions Let’s start with a simple calculation, one we learned back in Calc I: Z 1 x2 −1dx = arctan x Now, we’ve already learned that the way to handle arctangents and the like is to convert them to E-L-R form, which for arctan is: arctan x = 1 2 i ln ix −1 ix + 1 Interesting, but not very illuminating. Nevertheless, as Sherlock Holmes was wont to say, “once you have eliminated the impossible...” Let’s examine the improbable remains in an attempt to find the truth: Z 1 x2 −1dx = Z 1 (x + i)(x −i)dx Applying one or another of our techniques for partial fractions expansion, we compute: 1 (x + i)(x −i) = 1 2 h i x + i − i x −i i = −1 2i h i −ix + 1 − i −ix −1 i On that last step, I multiplied through by 1 in the form −i −i for reasons that I’ll explain later. But now, since each numerator is just the negative of its denominator’s derivative, we proceed: Z 1 x2 −1dx = −1 2i Z h i −ix + 1 − i −ix −1 i dx = −1 2i h −ln(−ix + 1) + ln(−ix −1) i = −1 2i h ln(−ix −1) −ln(−ix + 1) i How do we evaluate something like ln(−ix −1)? Well, Euler’s identity is a good place to start: eiθ = i sin θ + cos θ iθ = ln[i sin θ + cos θ] We can always factor a complex number into its modulus and its angle: a + bi = √ a2 + b2 a √ a2 + b2 + b √ a2 + b2i Now, the expression in parenthesis on the right is just an x-y coordinate pair on the unit circle, which form the sine and cosine of an angle. What angle? Well, the tangent of the angle is going to be the ratio between the y (imaginary) and x (real) coordinates, so its b a, and therefore the angle must be arctan b a. Combining this logic with the last two equations and the addition law of logarithms lets us obtain a general expression for imaginary logarithms: ln(a + bi) = ln √ a2 + b2 + i arctan b a Plugging this back into our integral, and using the fact that arctan is an odd function, we conclude: Z 1 x2 −1dx = −1 2i h ln(−ix −1) −ln(−ix + 1) i = −1 2i h ( ln √ x2 + 1 + i arctan x) −(ln √ x2 + 1 + i arctan(−x)) i = −1 2i h 2i arctan x i = arctan x 5.2 Multi-valued logarithms The complex logarithm is a multi-valued function, since adding 2πi to any logarithm produces another power for the same value.
In Symbolic Integration I, Manuel Bronstein gave a detailed analysis, which was so en-lightening to me that I will repeat and expand it here, of the following definite integral: Z x4 −3x2 + 6 x6 −5x4 + 5x2 + 4dx sage: R. = QQ[] Q[x] sage: a = x^4-3x^2+6; sage: b = x^6-5x^4+5x^2+4; sage: gcd(b,b.differentiate()) 1 sage: S. = QQ[] Q[z] sage: T. = S[] Q[z][x] sage: trager = (T(a) - zT(b).differentiate()).resultant(b); sage: trager.factor() (2930944) · (z2 + 1 4)3 sage: SS. = QQbar[] Q[z] sage: SS(trager).roots() −1 2i, 3 , 1 2i, 3 These roots give us coefficients of the logarithmic terms.
sage: RR. = QQbar[] Q[x] sage: loglist = []; sage: for (r,m) in SS(trager).roots(): s = (T(a)-zT(b).differentiate()).map_coefficients(lambda v : v loglist.append(r log(gcd(RR(b), RR(s)))) sage: sum(loglist) 1 2i log x3 + i x2 −3 x −2i −1 2i log x3 −i x2 −3 x + 2i Z x4 −3x2 + 6 x6 −5x4 + 5x2 + 4dx = i 2 ln x3 + ix2 −3x −2i −i 2 ln x3 −ix2 −3x + 2i = tan−1(x3 −3x x2 −2 ) -4 -2 0 2 4 -4 -2 0 2 4 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 Figure 5.1: (x3 −3x) + (x2 −2)i Let us note briefly that the integrand is clearly positive over the entire real line. Now, using a straightforward application of the method of partial fractions, we conclude: Z x4 −3x2 + 6 x6 −5x4 + 5x2 + 4dx = X α|4α2+1=0 α log x3 + 2αx2 −3x −4α Since the zeros of 4α2 + 1 are α = ±i/2, we evaluate the definite integral using the indefinite integral, expand the complex logarithms, and obtain: Z 2 1 x4 −3x2 + 6 x6 −5x4 + 5x2 + 4dx = i 2 log(2 + 2i) −i 2 log(2 −2i) − i 2 log(−2 −i) −i 2 log(−2 + i) = −5π 4 + arctan 1 2 ≈−3.46 Since the integral was positive over the entire range of integration, this answer can not possibly be correct.
Alternately, we can apply the arctan identity from the last chapter and conclude: Z x4 −3x2 + 6 x6 −5x4 + 5x2 + 4dx = X α|4α2+1=0 α log x3 + 2αx2 −3x −4α = i 2 ln x3 + ix2 −3x −2i −i 2 ln x3 −ix2 −3x + 2i = i 2 ln x3 −3x + (x2 −2)i x3 −3x −(x2 −2)i = i 2 ln (x3 −3x)i −(x2 −2) (x3 −3x)i + (x2 −2) = arctan x3 −3x x2 −2 Z 2 1 x4 −3x2 + 6 x6 −5x4 + 5x2 + 4dx = arctan 1 −arctan 2 ≈−0.32 What went wrong? We gain a key insight, as is so often the case, by graphing first the integrand: 0 1 2 3 4 5 6 7 8 0 0.5 1 1.5 2 2.5 3 x4 −3x2 + 6 x6 −5x4 + 5x2 + 4 And now the (indefinite) integral: -4 -3 -2 -1 0 1 2 3 4 0 0.5 1 1.5 2 2.5 3 Z x4 −3x2 + 6 x6 −5x4 + 5x2 + 4dx = tan−1(x3 −3x x2 −2 ) A discontinuity has appeared, seemingly out of nowhere. A closer inspection reveals that the break occurs suspiciously close to √ 2 — exactly where a plot of (x3 −3x)+(x2 −2)i in the complex plane crosses the negative real axis: -4 -2 0 2 4 -4 -2 0 2 4 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 OK, so here’s what happened. When we evaluate a complex logarithm, we implicitly select one blade of ln’s Riemann surface. There’s no way to avoid this. ln is a multi-valued function, the different values corresponding to different blades of the Riemann surface, and if we want to actually obtain a numerical result from ln we need to pick one of those values. In the context of an indefinite integral, which is only defined within an unspecified additive constant, the choice is completely arbitrary.
So far, so good. Yet now we want a definite integral. Now we’re going to evaluate not just a single logarithm, but we’re going to trace out a curve along the Riemann surface.
As I noted above, if we assign a single fixed value to ln x, then there’s no way to avoid a discontinuity somewhere in the Riemann surface. In this specific case, we used the identity ln(a + bi) = ln √ a2 + b2 + i arctan b a, which just converts the discontinuity in ln to one expressed in terms of arctan, also a multi-valued function. See, there’s no way to completely avoid this.
Where’s the discontinuity in arctan? Typically, the function is defined so that it ranges from −π 2 to π 2 and is continuous over finite numbers, so its discontinuity is where its argument becomes infinite; i.e, where its denominator goes to zero. In the ln(a + bi) expansion, this occurs where a becomes zero; i.e, where a + bi crosses the negative real axis. This is easier to visualize if we reproduce that last graph in 3-D, superimposing (x3 −3x) + (x2 −2)i on arctan’s Riemann surface: -5 0 5 -5 0 5 The z-axis is now the value of arctan. Along the x-axis, it is zero, and it grows positive as we move counter-clockwise, and negative as we move clockwise. The resulting discon-tinuity along the negative real axis, where arctan jumps from π 2 to −π 2, clearly creates a matching discontinuity in the plot of (x3 −3x) + (x2 −2)i.
Of course, this isn’t really arctan’s Riemann surface, only a slice of it, and now we begin to find our solution. arctan’s complete Riemann surface looks like an infinite screw, with an infinite series of blades, each spaced 2π apart in the z-direction. So, when we plot our function, what we really want is something more like this, moving smoothly along the Riemann surface without any discontinuity.
Alas, while easy enough to graph, and easy enough to understand once you’re thinking about the continuity of a Riemann surface, this is just asking a bit much from poor old ln -1 0 1 -1 0 1 -10 0 10 20 30 40 (or arctan). There’s no way for the function to know which result is needed based solely on a single value as the argument.
However, there is a way out, at least in the case of rational functions. The method, due to R. Rioboo, is to take advantage of two things. First, the addition law of logarithms (ln ab = ln a + ln b), combined with Euler-?? interpretation of complex numbers, lets us split a complex logarithm into two logarithms, each of which require only half the range of angles of the original logarithm. Second, since a rational function only intersects the real axis in a finite number of points (the finite number of zeros of its real component), a finite (and easily computed) number of reductions converts the logarithm into a sum of logarithms, none of whose arguments cross the real axis.
-5 0 5 -5 0 5 Z x4 −3x2 + 6 x6 −5x4 + 5x2 + 4dx = i 2 ln x3 + ix2 −3x −2i −i 2 ln x3 −ix2 −3x + 2i sage: (s,t) = diophantine(1, x^3-3x, x^2-2) −1 2x, 1 2x2 −1 2 sage: ((x^3-3x)+(x^2-2)QQbar(I)) (t + sQQbar(I)) 2 x5 −3x3 + x + 2i ln h (x3 −3x) + (x2 −2)i i = ln h x5 −3x3 + x + 2i i −ln h (x2 −1) −xi i ln x3 + ix2 −3x −2i = ln x5 −3x3 + x + 2i −ln x3 −i + ln(x + i) ln x3 −ix2 −3x + 2i = ln x5 −3x3 + x −2i −ln x3 + i + ln(x −i) Z x4 −3x2 + 6 x6 −5x4 + 5x2 + 4dx = tan−1(x5 −3x2 + x 2 ) + tan−1(x3) + tan−1(x) 0 1 2 3 4 5 6 7 8 9 0 0.5 1 1.5 2 2.5 3 5.3 A Bit Of Perspective Now, something interesting happened during the course of this chapter. Something subtle enough that I didn’t notice it for the first year or so that I worked with this theory. Maybe you’re more cleaver than I am and have already seen it (or maybe I just organized this book well enough that it’s more obvious to you that it was to me).
How did we start with a problem that was pure algebra, and end up with a solution that involved topology?
Let’s review. We defined a differential field using mapping between elements as the derivation. No topology yet — the field was purely discrete, with no concept of “close-ness” between the elements. We went looking for an element that got mapped onto some specified element. Again, no topology. And finally, we extended to a logarithmic exten-sion, defined purely as a transcendental extension with a specific differential mapping.
Still no topology.
So why have we spent the last ? pages discussing topology?
The change happened when we went from regarding “ln x” as a transcendental element over the field C(x) to regarding it as ln(x), a function mapping one complex number to another...
In conclusion, we need to be careful when working with multi-valued functions like ln and arctan, but the concept of a Riemann surface provides an important conceptual tool for dealing with them. Liouville’s theorem tells us that additional logarithms can be in-troduced by integration, and the multi-valued nature of the complex logarithm leaves us with a choice as to which branch of the function (or blade of the Riemann surface) to use.
Yet fundamental principles of calculus tell us that an indefinite integral is only defined within a constant of integration, so it doesn’t matter exactly which value of the logarithm we choose to use. Having made that choice, however, we then need to remain consistent by preserving continuity on the Riemann surface during the evaluation of any definite integral. In the specific case of rational function integration, Rioboo’s method gives us an algorithm that automatically preserves this continuity. For more complicated integrals that introduce new logarithms, we apply the more general concept of continuity on the Riemann surface.
Chapter 6 The Logarithmic Extension When we express a function in Liouvillian form, we construct a tower of nested fields, starting with C(x) at the bottom, and building up to the extension required to express our integral.
For example, consider example 6.7: Z (ln(ln x))2 ln x − 2 ln x(ln(ln x) + 1) dx To express this integral, we need to construct ln(ln x), which requires ln x to be con-structed first. Thus, we obtain nested fields according to the following structure: C(x) C(x, θ) θ = ln x C(x, θ, ψ) ψ = ln θ Our integrand can now be expressed as a rational function in the top-most field: Z θ2ψ2 −2ψ −2 θ dx Our basic strategy for integration is to always work in the top most field extension, re-ducing to some kind of problem that must be solved in the next lower extension. Since we only have three basic types of field extension, our aim is to develop a theory to handle each type.
In this chapter, we’ll analyze the logarithmic extension.
For logarithmic extensions, the problem is particularly easy, since integration leads only to futher integration steps in the base field.
6.1 The Logarithmic Integration Theorem Theorem 6.1. Let K be a differential field with k ∈K, let K(θ = ln k) be a simple logarithmic extension of K, let ni(θ) be irreducible polynomials in K[θ], and let f be an element of K(θ) with partial fractions expansion: f = n X i=0 aiθi + ν X i=1 mi X j=1 bi,j(θ) ni(θ)j (6.1) ai ∈K bi,j(θ), ni(θ) ∈K[θ] If f has an elementary anti-derivative F, then F ∈K(θ, Ψ), where K(θ, Ψ) is a finite logarithmic extension of K(θ) and F has a partial fractions expansion of the form: F = cn+1θn+1 + n X i=1 [Ai + ci] θi + A0 + ν X i=1 mi−1 X j=1 Bi,j(θ) ni(θ)j + ν X i=1 Ci ln ni(θ) (6.2) A0 ∈K(Ψ) Ai ∈K Bi,j(θ), ni(θ) ∈K[θ] c′ i = C′ i = 0 i̸=0 and the following relationships hold: (An + (n + 1)cn+1θ)′ = an (6.3a) (Ai + (i + 1)ci+1θ)′ = ai −(i + 1)k′ k Ai+1 0≤i<n (6.3b) Ri,mi−1(θ) = bi,mi(θ) (6.4a) Ri,j(θ) = bi,j+1(θ) −B′ i,j+1(θ) −Qi,j+1(θ) 1≤j1 (6.7a) Ci = bi,1(θ) n′ i(θ) mi=1 (6.7b) Proof By Theorem 4.15, an elementary antiderivative of f can only exist in a finite logarithmic extension K(θ, Ψ) of K(θ) and therefore must have the form: F = R + η X i=1 CiΨi where R ∈K(θ), the Ci are constants, and the Ψi are logarithms.
We can perform a partial fractions expansion on R, then F becomes: F = N X i=0 Aiθi + ν X i=1 Mi X j=1 Bi,j(θ) ni(θ)j + η X i=1 CiΨi It will be convenient later to separate a constant from the Ai terms, so let’s do that now: F = N X i=0 (Ai + ci)θi + ν X i=1 Mi X j=1 Bi,j(θ) ni(θ)j + η X i=1 CiΨi Our basic logarithmic relationships: ln ab = ln a + ln b ln a b = ln a −ln b allow us to assume, without loss of generality, that the Ψi’s are logarithms of irreducible polynomials. Some of those irreducible polynomials will exist solely in our underlying field K, and those we collapse into A0, noting that this makes A0 unique among the Ai’s because it can include additional logarithms. The remaining irreducible polynomials (those involving θ) can be identified as ni(θ)’s, simply by increasing i and adding new ni(θ)’s if needed.
Now let’s differentiate, remembering that θ′ = k′ k : F ′ = N X i=0 A′ iθi + ik′ k (Ai + ci)θi−1 + ν X i=1 Mi X j=1 B′ i,j(θ)ni(θ) −jBi,j(θ)n′ i(θ) ni(θ)j+1 + ν X i=1 Ci n′ i(θ) ni(θ) F ′ = A′ NθN + N−1 X i=0 A′ i + (i + 1)k′ k (Ai+1 + ci+1) θi + ν X i=1 " −MiBi,Mi(θ)n′ i(θ) ni(θ)Mi+1 + Mi−1 X j=1 B′ i,j+1(θ) −jBi,j(θ)n′ i(θ) ni(θ)j+1 + Ci n′ i(θ) ni(θ) # F ′ has the form of a partial fractions decomposition, but it is not a partial fractions decom-position because the numerators in the B terms violate the partial fractions degree bounds.
The problem is that the product Bi,j(θ)n′ i(θ) might have degree greater than degθ ni(θ).
To fix this, let’s divide the −jBi,j(θ)n′ i(θ) terms by ni(θ) (think polynomial long division) and rewrite them as a quotient and a remainder: −jBi,j(θ)n′ i(θ) = Qi,j(θ)ni(θ) + Ri,j(θ) This fixes the B terms, since deg Qi,j(θ) < deg ni(θ) and deg Ri,j(θ) < deg ni(θ).
F ′ = A′ NθN + N−1 X i=0 A′ i + (i + 1)k′ k (Ai+1 + ci+1) θi + ν X i=1 " Ri,Mi(θ) ni(θ)Mi+1 + Mi−1 X j=1 B′ i,j+1(θ) + Qi,j+1(θ) + Ri,j(θ) ni(θ)j+1 + B′ i,1(θ) + Qi,1(θ) + Cin′ i(θ) ni(θ) # What is the degree of F ′? It’s N, if AN is not constant. If AN is constant and not zero, then the degree of F ′ is N −1, since otherwise the N −1 coefficient would be zero: A′ N−1 + N k′ k AN = 0 = ⇒ A′ N−1 = (−NAN)k′ k Since AN is constant, this could only be satisfied by AN−1 = −NANθ, contradicting the assumption that AN−1 ∈K.
Performing a partial fractions decomposition of f: f = n X i=0 aiθi + ν X i=1 mi X j=1 bi,j(θ) ni(θ)j setting F ′ = f and equating like terms, we establish the relationships listed in the state-ment of the theorem.
For the degree of F ′ to be n, either degθ F = n + 1 and An+1 is constant, or degθ F = n.
Since the highest order denominators in F ′ have order Mi + 1, and they must match with f’s denominators of order mi, we conclude that Mi = mi −1.
To establish the remaining relationships, let’s remember the definition of Ri,j and Qi,j: −jBi,j(θ)n′ i(θ) = Qi,j(θ)ni(θ) + Ri,j(θ) Reducing this equation modulo ni(θ), we obtain: −jBi,j(θ)n′ i(θ) ≡Ri,j(θ) mod ni(θ) Now we use the fact that ni(θ) is irreducible, and invoke Theorem ??, which states the quotient ring modulo a prime ideal is a field, so we can perform division: Bi,j(θ) ≡−Ri,j(θ) jn′ i(θ) mod ni(θ) This equation seems to identify Bi,j(θ) up to a multiple of ni(θ), but if we remember our degree bound on partial fractions expansions, degθ Bi,j(θ) < degθ ni(θ), we see that in fact we’ve completely determined Bi,j(θ) from Ri,j(θ).
□ A few comments are in order. First, let’s recall equations (6.3a) and (6.3b): [An + (n + 1)cn+1θ]′ = an (6.3a) [Ai + (i + 1)ci+1θ]′ = ai −(i + 1)k′ k Ai+1 (6.3b) These equations both require integration of the right hand side in the underlying field K. The resulting integral must take the form of a element of K (that’s the Ai), plus possibly a constant times a single additional logarithm, θ itself. Liouville’s theorem 4.15 tells us that integrals can generally include an arbitrary number of additional logarithms.
The integrations required by (6.3a) and (6.3b) are more restrictive; the only additional logarithm they can include is θ, which makes sense because θ is not part of the underlying field K, but could easily appear in the final result because it already appears in the original integrand.
The exception is A0, since it can include arbitrary additional logarithms, not just θ. It’s version of equation (6.3b) reads: [A0 + c1θ]′ = a0 −k′ k A1 (6.8) So, this integral can include a logarithm θ, just like the others, as well as additional log-arithms collapsed into the A0 term. The c0 term, by the way, does not appear in any of the theorem’s equations, but a moment’s thought shows that c0 is merely the constant of integration.
Example 6.2. Compute R 1 x ln xdx Operating in C(x, θ = ln x), we evaluate: Z 1 θx dx = Z 1 x θ dx This has the form of equation (6.1) with n1(θ) = θ, m1 = 1 and b1,1(θ) = 1 x.
C1 = b1,1(θ) n′ 1(θ) = 1 x 1 x = 1 Plugging C1 into equation (6.2) we get: Z 1 x ln xdx = ln n1(θ) = ln ln x □ Example 6.3. Compute R ln x dx Again we’ll use C(x, θ = ln x) Z θ dx This has the form of equation (6.1) with n = 1 and a1 = 1, so [A1 + 2c2θ]′ = a1 = 1 A1 + 2c2θ = x Since A1 ∈C(x), it can not involve θ, so A1 = x and c2 = 0.
[A0 + c1θ]′ = a0 −k′ k A1 = 0 −1 xx = −1 A0 + c1θ = −x So A0 = −x and c1 = 0. Plugging A0, A1, c1 and c2 into equation (6.2) we get: Z θ dx = xθ −x Z ln x dx = x ln x −x □ Example 6.4. Compute R tan−1 x dx Using the differential algebra identity tan−1 x = 1 2 i ln ix−1 ix+1 from Section 4.2, we use the differential field C(x, θ); θ = ln ix−1 ix+1; θ′ = − 2i x2+1 and compute Z 1 2 i θ dx This is in the form of equation (6.1) with n = 1 and a1 = 1 2i, so Theorem 6.1 tells us that the θ-degree of our integral is at most two.
[A1 + 2c2θ]′ = a1 = 1 2 i A1 + 2c2θ = 1 2 ix Since A1 ∈C(x), c2 is zero and A1 = 1 2ix.
[A0 + c1θ]′ = a0 −A1 k′ k = −1 2ix −2i x2 + 1 [A0 + c1θ]′ = − x x2 + 1 A0 + c1θ = −1 2 ln x2 + 1 Remembering that A0 can include new logarithmic extensions, we conclude that c1 = 0 A0 = −1 2 ln x2 + 1 and therefore our solution is: Z 1 2 i θ dx = 1 2 i xθ −1 2 ln x2 + 1 Z tan−1 x dx = x tan−1 x −1 2 ln x2 + 1 □ Example 6.5. Determine if P∞ n=0 1 n2xn is an elementary function.
We can differentiate the series term wise, and if the resulting series can be identified with an elementary function, then we need only to decide if that derivative can be integrated into an elementary function. Using a standard identity for the Taylor series of ln(1 −x), we determine that d dx " ∞ X n=0 1 n2xn # = ∞ X n=1 1 nxn−1 = 1 x ln(1 −x) so we need to determine if R 1 x ln(1 −x) dx is elementary. Working in the differential field C(x, θ = ln(1 −x)), we’re trying to integrate 1 xθ dx. Equation (6.3a) reads: [A1 + 2c2θ]′ = 1 x A1 + 2c2θ = ln x ln x is required to express A1, but new logarithms are not allowed at this point in the algorithm. Therefore, P∞ n=0 1 n2xn is not elementary.
□ Example 6.6. Compute1 Z x{(x2e2x2 −ln2(x + 1))2 + 2xe3x2(x −(2x3 + 2x2 + x + 1) ln(x + 1))} (x + 1)(ln2(x + 1) −x2e2x2)2 dx Due to the complexity of this integral, I’ll use Sage for the entire calculation.
sage: integrand = x \ ((x^2exp(2x^2)-log(x+1)^2)^2 \ +2xexp(3x^2)(x-(2x^3+2x^2+x+1)log(x+1))) \ / ((x+1)(log(x+1)^2 - x^2exp(2x^2))^2) − 2 2 x3 + 2 x2 + x + 1 log (x + 1) −x xe(3 x2) − x2e(2 x2) −log (x + 1)22 x x2e(2 x2) −log (x + 1)22 (x + 1) We begin by putting our integral into Liouvillian form, assigning ψ = exp(x2) and θ = ln(x + 1).
sage: theta = var('theta'); sage: psi = var('psi'); sage: lintegrand = integrand.subs( {log(x+1) : theta, exp(x^2) : psi, exp(2x^2) : psi^2, exp(3x^2) : psi^3}) − 2 ((2 x3 + 2 x2 + x + 1)θ −x)ψ3x −(ψ2x2 −θ2)2 x (ψ2x2 −θ2)2(x + 1) Since ψ and θ are both simple extensions of C(x), we can operate in either C(x, ψ, θ) or C(x, θ, ψ). Since C(x, θ, ψ) is an exponential extension of C(x, θ), working in C(x, θ, ψ) would require evaluating an integral in an exponential extension, which we won’t study until the next chapter. Therefore, we’ll work in C(x, ψ, θ), a logarithmic extension of C(x, ψ), and hope that we won’t have to do anything too complicated in C(x, ψ)!
Let’s declare this field to Sage and tell it how our variables differentiate: sage: F. = FractionField(ZZ['x', 'psi']); 1This integral [Ge92]’s Example 12.8.
sage: R. = F['theta'] Frac(Z[x, ψ])[θ] sage: D = Derivation(R, {x: 1, theta: 1/(x+1), psi: 2xpsi}) Derivation of Frac(Z[x, ψ])[θ] x → 1 θ → 1 x+1 ψ → 2xψ sage: num = R(lintegrand.numerator(False)) xθ4 −2x3ψ2θ2 + −4x5ψ3 −4x4ψ3 −2x3ψ3 −2x2ψ3 θ + x5ψ4 + 2x3ψ3 sage: den = R(lintegrand.denominator(False)) (x + 1) θ4 + −2x3ψ2 −2x2ψ2 θ2 + x5ψ4 + x4ψ4 We need to put our integrand into partial fractions form, so let’s begin by using Sage’s quo_rem function, which performs polynomial long division with respect to a specific variable, and obtain: sage: (a,N) = num.quo_rem(den) x x + 1, −4x5ψ3 −4x4ψ3 −2x3ψ3 −2x2ψ3 θ + 2x3ψ3 Z x x + 1 + (−4x5 −4x4 −2x3 −2x2)ψ3θ + 2x3ψ3 (x + 1)(θ2 −x2ψ2)2 dx The a0 = x x+1 term is easy.
sage: A = integrate(SR(a), x) x −log (x + 1) Let’s consider the fractional term. Polynomial factorization can be difficult, but this de-nominator is easy – it’s just a difference of squares: (θ2 −x2ψ2) = (θ −xψ)(θ + xψ).
We’ll factor the denominator into its irreducible factors and then perform a partial frac-tions expansion: sage: n = [f for f in factor(den)] [θ −xψ, θ + xψ] sage: b = partfrac(N, den); sage: displayarray(b); bθ−xψ,1 = 1 −2x −2 bθ−xψ,2 = 2x4ψ2 + 2x3ψ2 + x2ψ2 + xψ2 −xψ −2x −2 bθ+xψ,1 = 1 2x + 2 bθ+xψ,2 = 2x4ψ2 + 2x3ψ2 + x2ψ2 + xψ2 + xψ 2x + 2 Now we’re ready to apply Theorem 6.1. We’ll number our factors 0 and 1, since that’s how Python like to do it, and start with n0(θ). m0 = 2, so R0,1(θ) = b0,2(θ) = 2x4 + 2x3 + x2 + x 2(x + 1) ψ2 + x 2(x + 1)ψ and we wish to compute B0,1(θ) ≡−R0,1(θ) n′ 0(θ) mod n0(θ) As modulo calculations go, this one is easy.
sage: R = {}; sage: B = {}; sage: R[0,1] = b[n,2] 2x4ψ2 + 2x3ψ2 + x2ψ2 + xψ2 −xψ −2x −2 sage: B[0,1] = - R[0,1] / D(n) −xψ 2 Now we wish to compute Q0,1(θ): Q0,1(θ) = −R0,1(θ) + B0,1(θ)n′ 0(θ) n0(θ) sage: Q = {}; sage: Q[0,1] = -(R[0,1] + B[0,1] D(n)) / n 0 This division to obtain Q0,1 will always be exact. What might not be exact is the following division to obtain C0. If the division isn’t exact, or if C0 isn’t a constant, then the integral is not elementary.
C0 = b0,1(θ) −B′ 0,1(θ) −Q0,1(θ) n′ 0(θ) sage: C = {}; sage: C = (b[n,1] - D(B[0,1]) - Q[0,1]) / D(n) −1 2 A similar calculation handles the other irreducible factor: sage: R[1,1] = b[n,2] 2x4ψ2 + 2x3ψ2 + x2ψ2 + xψ2 + xψ 2x + 2 sage: B[1,1] = - R[1,1] / D(n) −xψ 2 sage: Q[1,1] = - (R[1,1] + B[1,1] D(n)) / n 0 sage: C = (b[n,1] - D(B[1,1]) - Q[1,1]) / D(n) 1 2 Plugging everything together, we conclude that our solution is: sage: lans = A + sum([B[i,1]/n[i] for i in range(2)]) \ + sum([2 C[i] log(n[i]) for i in range(2)]).simplify_log()/2 ψθx ψ2x2 −θ2 + x −log (x + 1) + 1 2 log −ψx + θ ψx −θ Converting back to our original form: sage: ans = lans.subs({theta : log(x+1), psi : exp(x^2)}) xe(x2) log (x + 1) x2e(2 x2) −log (x + 1)2 + x −log (x + 1) + 1 2 log −xe(x2) + log (x + 1) xe(x2) −log (x + 1) !
Finally, we verify that this is, in fact, an anti-derivative of the original integrand: sage: bool(diff(ans,x) == integrand) True Z x{(x2e2x2 −ln2(x + 1))2 + 2xe3x2(x −(2x3 + 2x2 + x + 1) ln(x + 1))} (x + 1)(ln2(x + 1) −x2e2x2)2 dx = x −ln(x + 1) − xex2 ln(x + 1) ln2(x + 1) −x2e2x2 + 1 2 ln ln(x + 1) + xex2 ln(x + 1) −xex2 □ Example 6.7. Compute Z (ln(ln x))2 ln x − 2 ln x(ln(ln x) + 1) dx We’ll use the extension C(x, θ, ψ) where θ = ln x and ψ = ln θ = ln ln x. Converting to Liouvillian form, our integral becomes Z ψ2θ −2 θ(ψ + 1) dx = Z θψ2 −2 θψ −2 θ dx Since we need θ to construct ψ, our extensions nest in only a single way: C(x) C(x, θ) θ = ln x C(x, θ, ψ) ψ = ln θ So, we’ll work in C(x, θ, ψ), recursing into C(x, θ) and C(x). In C(x, θ, ψ), n = 2, and we identify the coefficients of our integrand: a2 = θ a1 = −2 θ a0 = −2 θ So equation (6.3a) reads: [A2 + 3c3ψ]′ = a2 = θ We already performed this integration in example 6.3 with the result: A2 + 3c3ψ = xθ −x In this example, A2 ∈C(x, θ), so A2 = xθ −x, c3 = 0.
[A1 + 2c2ψ]′ = a1 −2 1 x θ [xθ −x] = −2 θ −2 xθ[xθ −x] = −2 A1 + 2c2ψ = −2x So A1 = −2x and c2 = 0. Finally, A′ 0 = a0 −1 xθ(−2x) = −2 θ + 2 θ = 0 So A0 = C and our result becomes: Z θψ2 −2 θ(ψ + 1) dx = (xθ −x)ψ2 −2xψ □ 6.2 Hermite Reduction Another, more efficient, approach to handling the polynomials in denominators is to re-duce their order until our denominator has only factors of multiplicity one. We’re attempt-ing to do this: Z N V n = A V n−1 + Z B V n−1 Differentiating: N V n = A′ V n−1 −(n −1)AV ′ V n + B V n−1 and multiplying through by V n: N = V A′ −(n −1)AV ′ + BV N = (A′ + B)V −(n −1)AV ′ This equation has the form of a polynomial Diophantine equation, and since we know N, V and V ′, we can use the extended Euclidian algorithm to find (n −1)A and (A′ + B), which easily gives us A and B. So long as V is square-free, we know that gcd(V, V ′) = 1 (EXPLAIN), so we’re guaranteed a solution (STATE THEOREM).
Example 6.8. Redo Example 6.6 using Hermite reduction.
Z x{(x2e2x2 −ln2(x + 1))2 + 2xe3x2(x −(2x3 + 2x2 + x + 1) ln(x + 1))} (x + 1)(ln2(x + 1) −x2e2x2)2 dx We proceed as before, putting the integral into Liouvillian form and dividing out x x+1 to obtain a proper fraction: Z x x + 1 + (−4x5 −4x4 −2x3 −2x2)ψ3θ + 2x3ψ3 (x + 1)(θ2 −x2ψ2)2 dx Now we apply the Hermite reduction, using: V = θ2 −x2ψ2 V ′ = 2 x + 1θ −(2x + 4x3)ψ2 N = −4x5 −4x4 −2x3 −2x2 x + 1 ψ3θ + 2x3 x + 1ψ3 Our polynomial Diophantine equation is: −4x5 −4x4 −2x3 −2x2 x + 1 ψ3θ + 2x3 x + 1ψ3 = sV + tV ′ sage: v = theta^2-x^2psi^2 θ2 −x2ψ2 sage: vtick = D(v) 2 x + 1 θ −4x3ψ2 −2xψ2 sage: (s,t) = diophantine(N/(x+1), v, vtick) −2xψ x + 1 , xψθ So, our solution is: −4x5 −4x4 −2x3 −2x2 x + 1 ψ3θ + 2x3 x + 1ψ3 = −2x x + 1ψV + xψθV ′ sage: A = -t −xψθ sage: B = s - D(A) 2x2ψ + ψ θ + −xψ x + 1 A = −xψθ A′ = −ψθ −x(2x)ψθ −xψ 1 x + 1 A′ + B = −2x x + 1ψ B = (2x2 + 1)ψθ − x x + 1ψ Which reduces our integral to: sage: integrate(SR(a),x) + A/v + integrate(SR(B/v), x, hold=True) ψθx ψ2x2 −θ2 + x + Z − (2 ψx2 + ψ)θ −ψx x+1 ψ2x2 −θ2 dx −log (x + 1) x −θ − xψθ θ2 −x2ψ2 + Z (2x2 + 1)ψθ − x x+1ψ θ2 −x2ψ2 Now we can compute a Rothstein-Trager resultant: sage: R1 = FractionField(ZZ['x','psi'])['z'] Frac(Z[x, ψ])[z] sage: R = FractionField(R1)['theta'] Frac(Frac(Z[x, ψ])[z])[θ] sage: r = R(v).resultant(R(B - R('z') D(v))) 16x8ψ4 + 32x7ψ4 + 32x6ψ4 + 32x5ψ4 + 20x4ψ4 + 8x3ψ4 + 4x2ψ4 −4x2ψ2 x2 + 2x + 1 z2 + −4x8ψ4 −8x7ψ4 −8x6ψ4 −8x5ψ4 −5x4ψ4 −2x3ψ4 −x2ψ4 + x2ψ2 x2 + 2x + 1 This result is in C(x, ψ)[z], so the first two factors are just content (EXPLAIN THIS TERM).
sage: R1(r) / R1(r).factor().unit() 4z2 −1 The result is really just 4z2 −1, which has two solutions: ± 1 2. Substituting in these two values for z, we obtain the corresponding logarithms: sage: gcd(B - (1/2)D(v), v) θ + xψ sage: gcd(B + (1/2)D(v), v) θ −xψ Z x(x + 1){(x2e2x2 −ln2(x + 1))2 + 2xe3x2(x −(2x3 + 2x2 + x + 1) ln(x + 1))} ((x + 1) ln2(x + 1) −(x3 + x2)e2x2)2 dx = x −θ − xψθ θ2 −x2ψ2 + 1 2 ln(θ + xψ) −1 2 ln(θ −xψ) = x −ln(x + 1) − xex2 ln(x + 1) ln2(x + 1) −x2e2x2 +1 2 ln h ln(x + 1) + xex2i −1 2 ln h ln(x + 1) −xex2i □ Chapter 7 The Exponential Extension The two distinctive features of integration in exponential extensions are the presence of special polynomials, which divide their own derivatives, and the appearance of the Risch differential equation.
Let’s recall our basic theorem on the behavior of exponential extensions: Theorem 4.10. Let E = K(θ) be a simple transcendental exponential extension of a differential field K with the same constant subfield as K, let p = P piθi be a polynomial in K[θ] (pi ∈K), and let r be a rational function in K(θ). Then: 1. Degθ p′ = Degθ p 2. If p is monic and irreducible, then p | p′ if and only if p = θ.
3. If an irreducible monic factor other than θ appears in r’s denominator with multi-plicity m, then it appears in r′’s denominator with multiplicity m + 1 4. r′ ∈K if and only if r ∈K Contrast this theorem with the behavior of the ordinary polynomials that we’re accus-tomed to. Ordinary irreducible polynomials never divide their own derivatives in the manner described in (2); polynomials that do are called special. Instead, ordinary poly-nomials always behave in the way described in (3); such polynomials are called normal.
Irreducible polynomials are characterized as either normal or special, depending on whether they divide their own their derivatives. Theorem 4.10 (2) states that in a exponential ex-tension, the only special irreducible polynomial is θ itself.
We attack integrands in exponential extensions in much the same way as we attack ordi-nary polynomials: we factor the denominator into irreducible factors and perform a partial fractions expansion. In this case, however, we have to classify the denominator factors as either normal or special. Normal factors can be handled in much the same way as we’re used to, but special factors are treated in a manner similar to polynomials.
For example, let p = P piθi be a polynomial in K[θ], with θ = exp k, and take its derivation: p′ = n X i=0 (p′ i + ipik′)θi Notice that, unlike the logarithm or rational cases, there is no interdependence between the various terms of the sum; each term is completely independent of the others. Instead, each coefficient of θi has the form p′ i + Api, and equating the p′ polynomial to the integrand’s polynomial produces a differential equation of the form: p′ i + Api = B A, B, pi ∈K This is called a Risch equation and is a primary object of our study. Solving Risch equa-tions in a differential extension is the principle problem that we need to solve in order to carry out our program of symbolic integration.
Special factors in the denominator behave in almost exactly the same way as polynomials.
They both give rise to Risch equations that need to be solved in the underlying field. On the other hand, partial fractions terms involving normal polynomials give rise to rational functions and logarithms in the result that can be solved using the extended Euclidean algorithm, again operating in the underlying field.
We’ve already studied, in Section 2.6, how to use the extended Euclidean algorithm over an arbitrary field, so the primary additional tool we need to develop is the ability to solve Risch equations in arbitrary differential fields, or at least in the differential fields that arise in the course of our study. Once we can do that, we can evaluate integrals in complicated extension fields by “peeling away” the extensions, and solving equations in successively simplier extensions until we’ve reached the rational function field C(x).
I’ll begin by presenting the basic integration theorem for exponential extensions, then we’ll consider how to solve Risch equations in C(x), which is a simplified case that lets us solve integrals in simple exponential extensions. Finally, we’ll study solving the Risch equation more generally.
7.1 The Exponential Integration Theorem Theorem 7.1. Let K be a differential field, let K(θ = exp k) be a simple exponential extension of K, let ni(θ) be normal irreducible polynomials in K[θ], and let f be an element of K(θ) with partial fractions expansion: f = n X i=0 aiθi + l X j=1 bj θj + ν X i=1 mi X j=1 ci,j(θ) ni(θ)j (7.1) ai, bj ∈K ci,j(θ), ni(θ) ∈K[θ] If f has an elementary anti-derivative F, then F ∈K(θ, Ψ), where K(θ, Ψ) is a finite logarithm extension of K(θ), F has a partial fractions expansion of the form: F = n X i=0 Aiθi + l X j=1 Bj θj + ν X i=1 mi−1 X j=1 Ci,j(θ) ni(θ)j + ν X i=1 Di ln ni(θ) (7.2) Ai, Bj ∈K Ci,j(θ), ni(θ) ∈K[θ] D′ i = 0 and the following relationships hold: A′ 0 = a0 − ν X i=1 Di lc n′ i(θ) lc ni(θ) (7.3) A′ i + iAik′ = ai B′ j −jk′Bj = bj (7.4) Ri,mi−1(θ) = ci,mi (7.5a) Ri,j−1(θ) = ci,j −C′ i,j(θ) −Qi,j(θ) 1<j≤mi (7.5b) Ci,j(θ) ≡−Ri,j(θ) jn′ i(θ) mod ni(θ) (7.6) Di = ci,1 −C′ i,1(θ) −Qi,1(θ) n′ i(θ) −lc n′ i(θ) lc ni(θ)ni(θ) (7.7) Proof By Theorem 4.15, an elementary antiderivative of f can only exist in a finite logarithm extension K(θ, Ψ) of K(θ) and therefore must have the form: F = R + η X i=1 DiΨi where R ∈K(θ), and the Di are constants.
Constructing a partial fractions expansion of R, separating the normal and special com-ponents of its denominator, and using the fact that s1 = θ is the only special irreducible polynomial (Theorem 4.10): F = N X i=0 Aiθi + L X j=1 Bj θj + ν X i=1 Mi X j=1 Ci,j(θ) ni(θ)j + η X i=1 DiΨi Now let’s differentiate, remembering that θ′ = k′θ: F ′ = N X i=0 (A′ i+iAik′)θi+ L X j=1 B′ j −jk′Bj θj + ν X i=1 Mi X j=1 C′ i,j(θ)ni(θ) −jCi,j(θ)n′ i(θ) ni(θ)j+1 + η X i=1 Di E′ i(θ) Ei(θ) Let’s examine the logarithmic elements Ei(θ). If an Ei(θ) doesn’t involve θ, i.e, Ei ∈K, then we can collapse DiΨi into A0, with the understanding that when we recurse into K to solve for A0, additional logarithm terms are allowed.
So now let’s consider what happens if Ei(θ) is a polynomial in K[θ]. If it’s reducible, then the basic properties of logarithms let us split it into multiple irreducible elements.
Otherwise, it’s irreducible and therefore either normal or special. If it’s special, then it must be θ itself and ln θ = ln exp k = k, which contracts the transcendence of the logarithm extension Ψ. So all of the Ei(θ)’s must be normal, and therefore F ′ must have the form: F ′ = A′ 0 + X Dk E′ k Ek + N X i=1 (A′ i + iAik′)θi + L X j=1 B′ j −jk′Bj θj + ν X i=1 Mi X j=1 C′ i,j(θ)ni(θ) −jCi,j(θ)n′ i(θ) ni(θ)j+1 + η X i=1 Di ni(θ)′ ni(θ) F ′ has the form of a partial fractions decomposition, but it is not a partial fractions decom-position because the numerators in the C and D terms violate the partial fractions degree bounds. To fix this, let’s divide the −jCi,j(θ)n′ i(θ) terms by ni(θ) (think polynomial long division) and rewrite them as a quotient and a remainder: −jCi,j(θ)n′ i(θ) = Qi,j(θ)ni(θ) + Ri,j(θ) This fixes the C terms, since deg Qi,j(θ) < deg ni(θ) and deg Ri,j(θ) < deg ni(θ).
We can fix the D terms by noting that deg n′ i(θ) = deg ni(θ), so by subtracting an appro-priate multiple of ni(θ) we can ensure the cancellation of the leading terms, achieving our degree bounds.
F ′ = A′ 0 + X Dk E′ k Ek + X Di lc n′ i(θ) lc ni(θ) + N X i=1 (A′ i + iAik′)θi + L X j=1 B′ j −jk′Bj θj + ν X i=1 " Ri,Mi(θ) ni(θ)Mi+1 + Mi X j=2 C′ i,j(θ) + Qi,j(θ) + Ri,j−1(θ) ni(θ)j + C′ i,1(θ) + Qi,1(θ) + Di h ni(θ)′ −lc n′ i(θ) lc ni(θ)ni(θ) i ni(θ) # Now F ′ is an actual partial fractions decomposition. It not only has the right form, but all of the other conditions, specifically the degree bounds, are met. Therefore, we can perform a partial fractions decomposition of f: f = n X i=0 aiθi + l X j=1 bj θj + ν X i=1 mi X j=1 cij(θ) ni(θ)j Setting F ′ = f and equating like terms, we establish that n = N, l = L, Mi + 1 = mi, and the relationships listed in the statement of the theorem. The zero-order term: A′ 0 + X Dk E′ k Ek + X Di lc n′ i(θ) lc ni(θ) = a0 Polynomial terms and special denominators give rise to Risch equations: A′ i + iAik′ = ai B′ j −jk′Bj = bj Normal denominators give rise to these terms: Ri,mi(θ) = ci,mi+1 C′ i,j(θ) + Qi,j(θ) + Ri,j−1(θ) = ci,j 1<j≤mi C′ i,1(θ) + Qi,1(θ) + Dini(θ)′ = ci,1 Remember the definition of Ri,j and Qi,j: −jCi,j(θ)n′ i(θ) = Qi,j(θ)ni(θ) + Ri,j(θ) Reducing this equation modulo ni(θ), we obtain: −jCi,j(θ)n′ i(θ) ≡Ri,j(θ) mod ni(θ) Now we use the fact that ni(θ) is irreducible, and invoke Theorem ??, which states the quotient ring modulo a prime ideal is a field, so we can perform division: Ci,j(θ) ≡−Ri,j(θ) jn′ i(θ) mod ni(θ) This equation seems to identify Ci,j(θ) up to a multiple of ni(θ), but if we remember our degree bound on partial fractions expansions, degθ Ci,j(θ) < degθ ni(θ), we see that in fact we’ve completely determined Ci,j(θ) from Ri,j(θ).
□ We’ll begin discussing the Risch equation in the next section, which is how we obtain the Ai’s and Bi’s.
How can we calculate the Ci,j(θ)’s?
The highest order term in the partial fractions expansion gives us an Ri,j directly. Then we use the extended Euclidean algorithm from Section 2.6, which is our major computational tool for calculating inverses in quotient rings, to calculate Ci,j(θ). A simple long division step then gives us the quotient Qi,j(θ), and we just move down the list, solving this system of equations from highest order terms to lowest. Once we get to the end, we need to see if the bottom equation can be solved using a constant Di. If not, then the equation has no solution.
Once all of the Di’s have been calculated, then we have all of the information needed to determine A′ 0, and an integration step in the underlying field yields A0 itself.
Example 7.2. Compute R sin x dx We’ll operate in C(x, θ = exp ix) and evaluate Z θ −θ−1 2i dx The first step is write the integrand in partial fractions form: Z 1 2iθ −1 2i 1 θ dx By Theorem 7.1, the integral must have the form a1θ + a−1 1 θ with a1, a−1 ∈C(x) and must satisfy the equations: 1 2i = a′ 1 + ia1 −1 2i = a′ −1 −ia−1 These are very simple Risch equations that can be solved by inspection to obtain a1 = a−1 = −1 2, so Z θ −θ−1 2i dx = −1 2(θ + θ−1) = −1 2(eix + e−ix) = −cos x □ Example 7.3. Compute R csc x dx We’ll operate in C(x, ψ = exp ix) and evaluate Z 2i ψ −ψ−1 dx = Z 2i ψ ψ2 −1 dx Now we want a partial fractions expansion. We could use a resultant, or the Euclidean G.C.D. algorithm, but it’s simpler to just note that the denominator’s a difference of squares and compute: c1 ψ −1 + c2 ψ + 1 = ψ ψ2 −1 c1(ψ + 1) + c2(ψ −1) = ψ c1 = c2 = 1 2 giving us Z i ψ −1 + i ψ + 1 dx Now we have an integral in the form of equation (7.1) with ν = 2, n1(ψ) = ψ −1, n2(ψ) = ψ + 1, m1 = m2 = 1 and c1,1 = c2,1 = i.
D1 = c1,1 n′ 1(θ) −lc n′ 1(θ) lc n1(θ)n1(θ) = i iψ −i(ψ −1) = 1 D2 = c2,1 n′ 2(θ) −lc n′ 2(θ) lc n2(θ)n2(θ) = i iψ −i(ψ + 1) = −1 so by Theorem 7.1 the integral can be written: Z i ψ −1 + i ψ + 1 dx = ln(ψ −1) −ln(ψ + 1) = ln ψ −1 ψ + 1 = ln eix −1 eix + 1 = ln eix/2 −e−ix/2 eix/2 + e−ix/2 = ln i sin x 2 cos x 2 = ln tan x 2 where I dropped the i at the end because, as a constant multiple inside a logarithm, it disappears into the constant of integration, and we conclude that Z csc x dx = ln tan x 2 □ Example 7.4. Compute R tan x dx We’ll operate in C(x, θ = exp ix) and evaluate −i Z θ −θ−1 θ + θ−1 dx The first step is write the integrand in partial fractions form: θ −θ−1 θ + θ−1 = θ2 −1 θ2 + 1 = 1 − 2 θ2 + 1 = 1 − 2 (θ + i)(θ −i) = 1 + i θ −i − i θ + i −i θ −θ−1 θ + θ−1 = −i + 1 θ −i − 1 θ + i This integrand is now in the form of equation (7.1) with n1(θ) = θ −i and n2(θ) = θ + i, c1,1 = 1, c2,1 = −1, a0 = 1 and k = ix, so plugging these values into equation (7.7), we obtain: D1 = c1,1 n1(θ)′ −lc n′ 1(θ) lc n1(θ)n1(θ) = 1 iθ −i(θ −i) = −1 D2 = c2,1 n2(θ)′ −lc n′ 2(θ) lc n2(θ)n2(θ) = −1 iθ −i(θ + i) = −1 Pluging this into equation (7.3), we obtain: A′ 0 = a0 − ν X i=1 Di lc n′ i(θ) lc ni(θ) A′ 0 = −i + 2i = i A0 = ix so our integral is: −i Z θ −θ−1 θ + θ−1 dx = ix −ln(θ + i) −ln(θ −i) = ix −ln [(θ + i)(θ −i)] = ix −ln θ2 + 1 We can convert this into a more familiar form by realizing that ix = ln exp ix = ln θ, so = ln θ −ln θ2 + 1 = ln θ θ2 + 1 = ln 1 θ−1 + θ Z tan x dx = ln csc x □ 7.2 Risch Equations in C(x) Risch equations in C(x) arise when our integrand exists in a simple exponential extension of C(x), i.e, an integrand formed as a rational function of x and a single exponential of a rational function of x. Theorem 7.1 then produces Risch equations in the underlying field; in this case, C(x).
Consider such a Risch equation: r′ + Sr = T S, T, r ∈C(x) (7.8) Recall that in C[x], all irreducible factors have the form (x −γ), since C is algebraically closed. Futhermore, all irreducible factors in r’s denominator increase in degree on differ-entiation, since all of C[x]’s factors are normal. Therefore, the factors in r’s denominator must appear in either S or T’s denominator, because otherwise r′ would have a factor in its denominator that could not cancel anything else in the equation. Thus, we can easily identify which factors can appear in r’s denominator, and we next wish to calculate the multiplicities with which they appear.
Let’s consider a single pole at γ, expand S, T, and r using partial fractions, and look at the highest powers of (x −γ) in the denominator: r = a (x −γ)j + · · · r′ = −ja (x −γ)j+1 + · · · S = b (x −γ)k + · · · T = c (x −γ)l + · · · Combining everything into the Risch equation (7.8), we find: −ja (x −γ)j+1 + · · · + ba (x −γ)j+k + · · · = c (x −γ)l + · · · We know k (the pole’s multiplicity in S’s denominator) and l (the pole’s multiplicity in T’s denominator), and we wish to determine j, the pole’s multiplicity in r’s denominator.
We can classify the equation into three basic cases, based on the value of k: 1. k = 0. In this case, the −ja (x−γ)j+1 term dominates the left hand side, and j = l −1 in order to match the right hand side.
2. k = 1. Here, the high order terms on the left are equal, so either j = l −1 in order to match the right hand side, or j = b and j > l −1 in order for the left hand terms to exactly cancel.
3. k > 1. Now the ba (x−γ)j+k term dominates the left hand side, so j = l −k in order to match the right hand side.
By checking all of S’s and T’s poles using this technique, we can identify all the poles in r’s denominator and determine the multiplicity with which they appear. This determines r’s denominator d completely.
Having done so, we can replace r with p/d, where p, d ∈C[x]: p′d −pd′ d2 + S p d = T (p′d −pd′) + Spd = Td2 We still might have lingering denominators, which can be cleared by multiplying through by D, the least common multiple of the denominators of Spd and Td2: D(p′d −pd′) + SDpd = TDd2 (Dd)p′ + (SDd −Dd′)p = TDd2 Setting A = Dd, B = D(Sd −d′) and C = TDd2, we now have a polynomial Risch equation that must be satisfied by the numerator p: Ap′ + Bp = C A, B, C, p ∈C[x] (7.9) Our next aim is to upper bound the degree of p, and there are again three cases.
First, the highest terms on the left can have higher degree than any term on the right, and so must cancel against each other. For this to occur, deg A = deg B + 1 (since deg p drops by one on differentiation), and we can determine deg p by looking at the leading coefficients in A and B: A = ajxj + · · · B = bj−1xj−1 + · · · p = pkxk + · · · Ap′ + Bp = (kajpk + bj−1pk)xj+k−1 · · · In order for this coefficient to be zero, k = −bj−1/aj. So, if these conditions are met: deg A = deg B + 1 k = −bj−1 aj = −lc B lc A then p may exist as a kth degree polynomial.
Otherwise, the leading terms of Ap′ + Bp do not cancel out, so they must match the leading term of C. This can only occur if deg p = deg C −max(deg A −1, deg B) The final case we need to consider is when p is a constant, which would solve the Risch equation if and only if C was a constant multiple of B, irregardless of A.
Summarizing, equation (7.9) can be solved only if one of these three conditions is met: 1. deg A = deg B + 1 and deg p = −lc B lc A 2. deg p = deg C −max(deg A −1, deg B) 3. p is a constant and pB = C They are not mutually exclusive. In particular, if both of the first two cases can be met, then the larger of the two values for deg p should be used.
Having determined the degree of p, we can now determine its coefficients by hypothesiz-ing p as a polynomial of the calculated degree and plugging it into equation (7.9).
Example 7.5. Prove that R e−x2 dx has no elementary form We’ll use C(x, ψ = exp −x2), so ψ′ = −2x and study Z ψ dx We know from Theorem 7.1 that our solution, if it exists, must have the form A1ψ, where A1 ∈C(x), and A1 must satisfy equation (7.4): A′ 1 −2xA1 = 1 This is already a polynomial Risch equation, and A′ 1 has only a constant coefficient, so A1 can not have a non-trivial denominator. Futhermore, identifying A as 1, B as −2x, and C as 1, we see that deg A = 0 and deg B = 1. Since deg A ̸= deg B + 1 (case 1), the leading terms on the left hand side can not cancel, so they must match the leading term on the right. We compute: deg C −max(deg A −1, deg B) = 0 −1 = −1 so that doesn’t work (case 2). Futhermore, C is not a constant multiple of B, so a constant can’t solve our equation (case 3).
We conclude that no solution to this Risch equation exists in C(x), so the integral can not be expressed in elementary form.
□ Example 7.6. Prove that R sin x x dx has no elementary form As we often do with trigonometric integrals, we’ll operate in C(x, ψ = exp ix), use Euler’s relationship eix = i sin x + cos x, and evaluate Z ψ −ψ−1 2ix dx Let’s begin by writing the integrand in the form of a rational function in C(x)(ψ), i.e, a ratio of ψ-polynomials, with coefficients in C(x): 1 2i Z 1 xψ −1 x 1 ψ dx Applying theorem 7.1, we see that the integral must have the form A1ψ + A−1 1 ψ with A1, A−1 ∈C(x) and must satisfy equations (7.4): A′ 1 + iA1 = 1 x A′ −1 −iA−1 = −1 x In both cases, we’ve got an equation of the form (7.8) with a single pole in the denominator of T, so k = 0, l = 1, and j = l −1 = 0, so there are no poles in the denominator of our solution. However, there is then no way to produce the denominator on the right, so the Risch equation has no solution in C(x).
Thus, the integral can not be expressed in elementary form.
□ Partial fractions terms involving normal polynomials are handled the same way as, well, normal polynomials. Terms with simple denominators give rise to logarithms in the solu-tion, while terms with higher powered denominators give rise to rational functions in the solution.
One unusual feature of exponential extensions is that the numerator of a derivative will have the same degree as the denominator, so a long division step is needed to make the fraction proper, and this will produce a constant that will modify the integrand. For this reason, it’s best to handle the denominator’s normal factors first, Example 7.7. Compute R 4x+1 2x+1dx We’ll use the field C(x, θ = exp(x ln 2)); θ′ = (ln 2)θ and the representation (see Exam-ple 4.4): θ2 + 1 θ + 1 = θ −1 + 2 θ + 1 This integrand has the form of equation (7.1) with a1 = 1, a0 = −1, n1(θ) = θ + 1, n′ 1(θ) = (ln 2)θ and c1,1 = 2.
We’ll start with equation (7.7): D1 = c1,1 n1(θ)′ −lc n′ 1(θ) lc n1(θ)n1(θ) = 2 (ln 2)θ −ln 2 1 (θ + 1) = −2 ln 2 Now, equation (7.4) yields: A′ 1 + (ln 2)A1 = 1 = ⇒ A1 = 1 ln 2 and equation (7.3) yields: A′ 0 = a0 −D1 lc n′ i(θ) lc ni(θ) = −1 + 2 ln 2 ln 2 = 1 A0 = x Plugging everything into equation (7.2), then substituting 2x for θ, we obtain our solution: Z 4x + 1 2x + 1dx = 1 ln 2θ + x − 2 ln 2 [ln(θ + 1)] = 2x ln 2 + x − 2 ln 2 ln(2x + 1) □ Example 7.8. Redo Example 6.6 using the exponential theory.
Z x{(x2e2x2 −ln2(x + 1))2 + 2xe3x2(x −(2x3 + 2x2 + x + 1) ln(x + 1))} (x + 1)(ln2(x + 1) −x2e2x2)2 dx We proceed as before, putting the integral into Liouvillian form and dividing out x x+1 to obtain a proper fraction: sage: F. = FractionField(ZZ['x', 'theta']); sage: Ring. = F['psi'] Frac(Z[x, θ])[ψ] sage: D = Derivation(Ring, {x: 1, theta: 1/(x+1), psi: 2xpsi}) Derivation of Frac(Z[x, θ])[ψ] x → 1 θ → 1 x+1 ψ → 2xψ sage: num = Ring(lintegrand.numerator(False)) x5ψ4 + −4x5θ −4x4θ −2x3θ + 2x3 −2x2θ ψ3 −2x3θ2ψ2 + xθ4 sage: den = Ring(lintegrand.denominator(False)) x5 + x4 ψ4 + −2x3θ2 −2x2θ2 ψ2 + xθ4 + θ4 sage: (a,N) = num.quo_rem(den) x x + 1, −4x5θ −4x4θ −2x3θ + 2x3 −2x2θ ψ3 This time, we’ll operate in C(x, θ = ln(x + 1), ψ = exp x2), treating this as an exponen-tial extension of C(x, θ). We’ll begin again by computing a partial fractions expansion, this time with respect to ψ: sage: n = [f for f in factor(den)] [xψ −θ, xψ + θ] sage: c = partfrac(N, den); sage: displayarray(c); cxψ−θ,1 = 2x3θ + 2x2θ + xθ −x + θ −x2 −x cxψ−θ,2 = −2x3θ2 −2x2θ2 −xθ2 + xθ −θ2 2x2 + 2x cxψ+θ,1 = −2x3θ −2x2θ −xθ + x −θ x2 + x cxψ+θ,2 = 2x3θ2 + 2x2θ2 + xθ2 −xθ + θ2 2x2 + 2x Now Theorem 7.1 tells us that sage: R = {}; sage: R[0,1] = c[n,2] −2x3θ2 −2x2θ2 −xθ2 + xθ −θ2 2x2 + 2x C0,1(ψ) = −R0,1(ψ) n′ 0(ψ) mod n0(ψ) This time, we do need to perform a modulo reduction, but as modulo reductions go, it’s trivial. We construct a quotient ring, map our operands into it, and perform the division there.
sage: F2 = Ring.quo(n); sage: C[0,1] = - F2(R[0,1]) / F2(D(n)) θ 2 Now we wish to compute Q0,1(ψ). There is no modulo reduction in this division, and it should always be exact. We lift the result of the modulo reduction back into its parent ring, and use // to perform exact division (as opposed to working in a fraction field).
Q0,1(ψ) = −R0,1(θ) + C0,1(θ)n′ 0(θ) n0(θ) sage: Q = {}; sage: Q[0,1] = - (R[0,1] + C[0,1].lift() D(n)) // n −2x2θ −θ 2x Having computed C0,1 and Q0,1, we are now able to compute D0: D0 = c0,1 −C′ 0,1(ψ) −Q0,1(ψ) n′ 0(ψ) −lc n′ 0(ψ) lc n0(ψ)n0(ψ) sage: myD = {}; sage: myD = (c[n,1] - D(C[0,1].lift()) - Q[0,1]) \ / (D(n)-D(n).lc()/n.lc()n) 1 −2 A similar calculation for n1(ψ) = xψ + θ yields yields sage: R[1,1] = c[n,2] 2x3θ2 + 2x2θ2 + xθ2 −xθ + θ2 2x2 + 2x sage: F2. = Ring.quo(n); sage: C[1,1] = - F2(R[1,1]) / F2(D(n)) −θ −2 sage: Q[1,1] = - (R[1,1] + C[1,1].lift() D(n)) // n 2x2θ + θ −2x sage: myD = (c[n,1] - D(C[1,1].lift()) - Q[1,1]) \ / (D(n)-D(n).lc()/n.lc()n) 1 2 In an exponential extension, our D coefficients can affect our A0 term...
A′ 0 = a0 − ν X i=1 Di lc n′ 1(ψ) lc n1(ψ) sage: a = a - sum([myD[i]D(n[i]).lc()/n[i].lc() for i in range(2)]) x x + 1 sage: A = integrate(a,x) x −log (x + 1) We end up with the same result that we obtained from the logarithmic theory: sage: lans = A + sum([C[i,1].lift()/n[i] for i in range(2)]) \ + sum([2 myD[i] log(n[i]) for i in range(2)]).simplify_log() x + ψθ ψ2 −θ2 x2 x −log (x + 1) + 1 2 log ψx + θ ψx −θ sage: ans = lans.subs({theta : log(x+1), psi : exp(x^2)}) x − e(x2) log (x + 1) x log(x+1)2 x2 −e(2 x2) −log (x + 1) + 1 2 log xe(x2) + log (x + 1) xe(x2) −log (x + 1) !
sage: bool(diff(ans,x) == integrand) True □ 7.3 Risch Equations over Normal Polynomials Now let’s expand our study to include Risch equations in more complicated differential fields, starting with normal polynomials, which will allow us to handle logarithmic exten-sions.
Consider again such a Risch equation, this time in a simple transcendental extension K(θ) of an arbitrary differential field K: r′ + Sr = T S, T, r ∈K(θ) (7.10) Once again, we can perform a partial fractions expansion on S, T, and r, except that this time our irreducible polynomials are more complicated that (x −γ). Consider one such normal irreducible polynomial n(θ): S = b(θ) n(θ)k + · · · T = c(θ) n(θ)l + · · · r = a(θ) n(θ)j + · · · r′ = a′(θ)n(θ) −ja(θ)n′(θ) n(θ)j+1 + · · · = −ja(θ)n′(θ) n(θ)j+1 + · · · Plugging this all into the Risch equation, we obtain: −ja(θ)n′(θ) n(θ)j+1 + · · · + a(θ)b(θ) n(θ)j+k + · · · = c(θ) n(θ)l + · · · Both of the numerators on the left hand side could have θ-degree greater than degθ n(θ), so we divide them by n(θ): R1(θ) = −ja(θ)n′(θ) mod n(θ) R2(θ) = a(θ)b(θ) mod n(θ) R1(θ) n(θ)j+1 + · · · + R2(θ) n(θ)j+k + · · · = c(θ) n(θ)l + · · · Since n(θ) is irreducible, K/(ni) is a field, so it has no zero divisors, and neither R1(θ) and R2(θ) are zero.
Our three cases are as before, except that when k = 1, our cancellation condition becomes: R1(θ) = −R2(θ) mod n(θ) ja(θ)n′(θ) = a(θ)b(θ) mod n(θ) Again, division is possible in a field, so j = b(θ) n′(θ) mod n(θ) Our three cases become: 1. k = 0 and j = l −1.
(7.11a) 2. k = 1 and either j = l −1 or j = b(θ) n′(θ) mod n(θ).
(7.11b) 3. k > 1 and j = l −k.
(7.11c) Once we have determined the factors and multiplicities in the denominator, then we can proceed as before, clearing out the denominator and obtaining a polynomial equation of the form: Ap′ + Bp = C A, B, C, p ∈K[θ] (7.12) In addition to the three cases that we considered for C[x], there is a fourth case that must be considered. p can be zeroth order in θ, yet not be a constant. So our cases become: 1. degθ A = degθ B + 1 and degθ p = −lc B lc A 2. degθ p = degθ C −max(degθ A −1, degθ B) 3. degθ p = 0 4. p is a constant and pB = C Example 7.9. Determine if R xxdx has an elementary form.
To handle this integral, we’ll rewrite it as R ex ln xdx and operate in the field C(x, θ, ψ), where θ = ln x and ψ = exp xθ. So, we’re trying to compute Z ψ dx Applying theorem 7.1, we obtain the following Risch equation in C(x, θ): A′ 1 + k′A1 = 1 A′ 1 + (θ + 1)A1 = 1 (7.13) A = 1 B = θ + 1 C = 1 Since deg A = 0 and deg B = 1, deg A ̸= deg B + 1, so we can’t have cancellation on the left hand side. This means that deg A1 = deg C −max(deg A −1, deg B) = 0 −max(−1, 1) = −1 which is impossible.
If degθ A1 is zero, then A1 ∈C(x) and equation (7.13) splits into two Risch equations in C(x), one for first degree terms in θ and one for zeroth degree terms in θ: A1 = 0 A′ 1 + A1 = 1 = ⇒ A1 = −1 These equations have no simultaneous solution.
We conclude that equation (7.13) has no solution in C(x, θ), and that the original integral is not elementary.
□ Example 7.10. Integrate R log(x)e−x2dx It’s noted in the Sage documentation that Sage can’t simplify this integral, though both Maple and Mathematica can (calculus/tests.html).
□ 7.4 Normal Risch Equations in Sage To facilitate use of Sage, I’ve written a Risch equation solver in Sage.
First, we want a partial fractions expansion that’s a little bit different from the standard function partfrac. If expr isn’t a fraction, or if its denominator doesn’t involve the variable var, we want an empty list returned, otherwise return a list of the fractions in the expansion. We just want the highest-powered term in the expansion.
def partfrac1(num, den): b = {} if den != 1: factorization = factor(den) for f in factorization: (t,s) = diophantine(num, f^f, den//(f^f)) (q,r) = s.quo_rem(f) b[f] = [r, f] return(b) Next, given a Risch equation in the form r′ + Sr = T normal_risch_factors returns a list of sublists, with one sublist for each irre-ducible factor in S or T’s denominator. Each sublist contains the factor along with a pair of lists, one for S and one for T, containing the highest power in that variable’s partial fractions expansion, along with the corresponding numerator, in the following format: [factor, [S-numerator, S-power], [T-numerator, T-power]] def normal_risch_factors(S, T): result = {} Sb = partfrac1(S.numerator(), S.denominator()) Tb = partfrac1(T.numerator(), T.denominator()) for f in set(Sb.keys()).union(Tb.keys()): result[f] = [Sb.get(f, [0,1]), Tb.get(f, [0,1])] return result normal_risch_denominator uses normal_risch_factors to compute the solutions’s denominator.
def is_integer(P): try: int(P) return True except (TypeError, ValueError): return False def normal_risch_denominator1(L, D): (f, ) = L if k == 0: return f^max(0,l-1) elif k == 1: R = f.parent().quo(f) j = R(Snum)/R(D(f)) if is_integer(j): return f^max(l-1, j, 0) else: return f^max(l-1, 0) else: return f^max(0,l-k) def normal_risch_denominator(S, T, D): return prod(map(lambda L: normal_risch_denominator1(L, D), normal_ The next function, normal_risch_polynomial, given S, T, and the denominator computed by normal_risch_denominator, returns the components of the polyno-mial equation satisfied by the solution’s numerator: An′ + Bn = C in the form [A, B, C].
def normal_risch_polynomial(S, T, D1, D): D2 = (SD1).denominator() (TD1).denominator() # XXX check to make sure that denominator is 1 return [D1D2, SD1D2 - D(D1)D2, (TD1^2D2).numerator()] Next, given A, B, and C, we wish to compute the maximum degree of the numerator, or None if we can determine at this point that there is no solution.
def is_constant(P, D): return (D(P) == 0) def normal_risch_degree(A, B, C, D): # XXX What if B is zero?
if C != 0: k = C.degree() - max(A.degree() - 1, B.degree()) else: k = -1 if is_constant(C/B, D): k = max(k,0) if A.degree() == B.degree() + 1: p = - B.lc() / A.lc() if is_integer(p): k = max(k,p) return k Finally, we combine all these pieces together into a function that either returns a solution to a normal Risch equation, or None if no such solution exists.
def normal_risch_equation(S, T, D): D1 = normal_risch_denominator(S, T, D) [A, B, C] = normal_risch_polynomial(S, T, D1, D) if is_constant(C/B, D): return C/B/D1 elif normal_risch_degree(A,B,C,D) == -1: return None else: raise NotImplementedError Let’s demonstrate the use of this code by using it to solve the Risch equations that we’ve already seen in this chapter’s examples so far.
sage: R. = QQbar[] Q[x] sage: I = R(sqrt(R(-1))) i sage: D = Derivation(R, {x: 1}) Derivation of Q[x] x → 1 sage: normal_risch_equation(I, 1/(2I), D) −1 2 sage: normal_risch_equation(-2x, R(1), D) None sage: normal_risch_equation(I, 1/x, D) None sage: C2. = QQbar[] Q[log2] sage: C2._latex_names = '{\ln 2}' {\ln 2} sage: R2. = C2[] Q[ln 2][x] sage: D = Derivation(R2, {x: 1, log2: 0}) Derivation of Q[ln 2][x] x → 1 ln 2 → 0 sage: normal_risch_equation(R2(log2), R2(1), D) 1 ln 2 sage: R2. = R[] Q[x][θ] sage: D = Derivation(R2, {x: 1, theta: 1/x}) Derivation of Q[x][θ] x → 1 θ → 1 x sage: normal_risch_equation(theta+1, R2(1), theta) None 7.5 Risch Equations over Special Polynomials Finally, let’s consider Risch equations over fields with special polynomials, i.e, exponen-tial extensions with θ = exp k and θ′ = k′θ.
r′ + Sr = T S, T, r ∈K(θ) (7.14) First, what happens when our partial fractions decomposition yields special polynomials in the denominators of S or T?
S = b θk + · · · T = c θl + · · · b, c ∈K r = a θj + · · · r′ = −jk′a + a′ θj + · · · a ∈K First, we should consider if the leading r′ term actually exists. Could the numerator actually be zero? If so, then jk′a = a′, but this could only happen if a were a constant multiple of θj (PROVE THIS), which contradicts the transcendance of θ over K.
Our Risch equation becomes: −jk′a + a′ θj + · · · + ab θk+j + · · · = c θl + · · · There are two cases: 1. k = 0 and either j = l or j = a′+ab ak′ 2. k > 0 and j = l −k.
Now we have computed j, the multiplicity of the special factor θ in the denominator, and our normal theory from the previous section gives us the denominator multiplicity of our normal factors, so we’ve computed d, the denominator of r, and thus can replace r with p/d, which will yield a polynomial Risch equation.
There are additional issues that arises with special polynomials when solving a polynomial Risch equation: Ar′ + Br = C A, B, C ∈K[θ] r ∈K(θ) (7.15) If K[θ] has only normal polynomials, then this equation can be solved as described be-fore, since r must be a polynomial. In the special case, however, r could have a special denominator. Expanding as before...
r = a θj + · · · r′ = −jk′a + a′ θj + · · · If A and B have no θ factors, then their zeroth order coefficients will produce j-th order fractions: A(0)−jk′a + a′ θj + · · · + B(0) a θj + · · · = C Since C is a polynomial, the fractions on the left must cancel, and we obtain: [−jak′ + a′] A(0) + aB(0) = 0 jk′ −a′ a = B(0) A(0) Integrating, we obtain: jk −ln a = Z B(0) A(0)dx (7.16) We don’t know a, but A, B, and k are all known, so solving this equation amounts to an integration step that must result in a constant multiple of k plus a possible logarithm.
This completes our determination of the denominator of r, and we are now reduced to a polynomial equation: Ap′ + Bp = C A, B, C, p ∈K[θ] (7.17) p = pnθn + · · · p′ = (p′ n + npnk′)θn + · · · So the leading term on the left hand side is: lc A(p′ n + npnk′) + lc Bpn = 0 nk′ + p′ n pn = −lc B lc A This equation has the same form as 7.16, so again, we integrate: nk + ln pn = − Z lc B lc Adx (7.18) If this integral has this desired form, then cancellation is possible between the terms on the left hand side of 7.17, and n is the θ-degree of the solution polynomial.
Otherwise, there is no cancellation and degθ p = degθ C −max(degθ A, degθ B) This degree can be negative, so long as it is no lower than the lower degree bound deter-mined earlier.
Example 7.11. ([Br05] examples 6.2.1, 6.3.3, 6.4.2) Integrate Z ex −x2 + 2x (ex + x)2x2 e(x2−1)/x+1/(ex+x)dx We’ll use the differential field C(x, θ, ψ) where θ = exp x and ψ = exp x2−1 x + 1 θ+x .
sage: var('x', 'theta', 'psi'); sage: integrand = \ (exp(x) - x^2 + 2x) \ / ((exp(x) + x)^2 x^2) \ exp((x^2-1)/x + 1/(exp(x)+x)) −(x2 −2 x −ex)e x2−1 x + 1 x+ex (x + ex)2x2 sage: exponent = (x^2-1)/x + 1/(theta+x) x2 −1 x + 1 θ + x sage: lintegrand = integrand.subs( {exp(x) : theta, exp(exponent) : psi}) −(x2 −θ −2 x)ψ (θ + x)2x2 sage: F. = FractionField(ZZ['x', 'theta']); sage: R. = F[] Frac(Z[x, θ])[ψ] sage: D1 = Derivation(R, {x: 1, theta: theta}) Derivation of Frac(Z[x, θ])[ψ] x → 1 θ → θ sage: D = Derivation(R, {x: 1, theta: theta, psi: D1(R(exponent))psi}) Derivation of Frac(Z[x, θ])[ψ] x → 1 θ → θ ψ → x4+2x3θ+x2θ2−x2θ+2xθ+θ2 x4+2x3θ+x2θ2 ψ The integrand has the form a1ψ, where a1 ∈C(x, θ), so we’re ready to apply Theorem 7.1. Equation (7.4) gives: sage: a1 = lintegrand/psi −x2 −θ −2 x (θ + x)2x2 sage: A1f = function('A1', nargs=1); sage: A1 = A1f(x) A1 (x) sage: eq = diff(A1,x) + D(psi)/psi A1 - a1 (θ2x2 + 2 θx3 + x4 −θx2 + θ2 + 2 θx)A1 (x) θ2x2 + 2 θx3 + x4 + x2 −θ −2 x (θ + x)2x2 + ∂ ∂xA1 (x) This is already in the form required by equation 7.14: r′ + Sr = T S, T, r ∈K(θ) with S and T expanded in partial fractions. None of the denominator factors are special (only factors of the form θn are special), and both (θ + x) and x are normal polynomials with k = 2 (the multiplicity in S’s denominator) and l = 2 (the multiplicity in T’s denominator). Case 3 in 7.11 tells us that j = l −k = 0, so neither of these (normal) factors can appear in the denominator of r, and r’s denominator can only involve special factors.
Next we can move on to equation 7.15: Ar′ + Br = C A, B, C ∈K[θ] r ∈K(θ) Let’s use numerator() to collapse the entire equation into a single fraction and discard the denominator, and then extract the coefficients of A1 and A′ 1.
sage: [BC, ], [A, ] = \ eq.numerator().coefficients(diff(A1,x)); sage: [C, ], [B, ] = \ BC.numerator().coefficients(A1); sage: A θ2x2 + 2 θx3 + x4 sage: B θ2x2 + 2 θx3 + x4 −θx2 + θ2 + 2 θx sage: C x2 −θ −2 x Now we want to integrate A(0)/B(0), to determine the highest power of θ that can appear in the denominator: sage: (A/B).subs(theta=0) 1 Since θ = ex, k = x, this solves equation 7.16 with j = 1 and a constant, which tells us that θ can appear in r’s denominator with multiplicity one.
What about the polynomial degree?
sage: R2 = ZZ['x']['theta'] Z[x][θ] sage: R2(B).lc() / R2(A).lc() x2 + 1 x2 sage: integrate(R2(B).lc() / R2(A).lc(), x) x −1 x This does not have the desired form of 7.18, specifically it is not an integer multiple of x plus a logarithm, so we lack cancellation. This means that A1’s θ-degree is upper bounded by degθ C −max(degθ A, degθ B) = −1. Since the lower bound is also −1, this gives us the following form for A1: A1 = A1,−1θ−1 A1,−1 ∈C(x) Sage doesn’t know how theta differentiates1, so I expanded it out by hand: sage: A1m1f = function('A1m1', latex_name='A_{1,-1}', nargs=1); sage: A1m1 = A1m1f(x); sage: eq2 = eq.subs({A1 : A1m1/theta, diff(A1,x) : (diff(A1m1,x)theta - A1m1D(theta))/theta^2}) (θ2x2 + 2 θx3 + x4 −θx2 + θ2 + 2 θx)A1,−1 (x) (θ2x2 + 2 θx3 + x4)θ −θA1,−1 (x) −θ ∂ ∂xA1,−1 (x) θ2 + x2 −θ −2 x (θ + x)2x2 sage: eq2.numerator() θ2x2 ∂ ∂xA1,−1 (x) + 2 θx3 ∂ ∂xA1,−1 (x) + x4 ∂ ∂xA1,−1 (x) −θx2A1,−1 (x) + θx2 + θ2A1,−1 (x) + 2 θxA1,−1 (x) −θ2 −2 θx This can also be done with more sophisticated tools. We create a differential ring involving A1, A1,−1 and θ, with x as its sole derivation. An elimination ordering is used (the default), with A1 appearing earlier in the list than A1,−1, thus eliminating A1 in favor of A1,−1.
Then we construct a differential ideal showing how θ differentiates and how A1 is related to A1,−1. Finally, we reduce our equation modulo the single differential chain in the differential ideal, producing a normal form with the desired result.2 A1,−1 is a rational functions in C(x). In particular, it does not involve θ, so each power of θ in this equation must be zero. Let’s extract the leading θ coefficient and set it equal to zero.
This Risch equation has an obvious solution. Now we’re solving in C(x), so if it were difficult, we could use the theory of the previous section: sage: R2. = QQbar[]; sage: D = Derivation(R2, {x: 1}); 1Maxima is somewhat better in this regard. It’s gradef command lets the user declare how variables differentiate.
2This is also somewhat painful with Sage, requiring us to convert θ into a “function” so that we can take its derivative.
sage: normal_risch_equation(1/x^2, 1/x^2, D) 1 Having solved for A1,−1, we’ve now solved for A1: sage: A1 = R(1/theta) 1 θ Since our integrand was of the form a1ψ, our integral is of the form A1ψ, so we’ve com-pleted our calculation. Is it correct?
sage: lans = psi/theta 1 θψ sage: ans = lans.subs({psi : exp(exponent), theta : exp(x)}) e −x+ x2−1 x + 1 x+ex sage: bool(diff(ans,x) == integrand) True □ Chapter 8 Algebraic Curves THIS CHAPTER IS INCOMPLETE.
Having addressed logarithmic and exponential extensions, we now turn to the algebraic extension, which at first appears to be completely different in character from the two transcendental cases. The differences stem largely from the lack of unique factorization in the algebraic case; algebraic extensions are not, in general, unique factorization domains.
There are four basic operations we perform on a rational function in order to integrate it: 1. We factor its numerator and denominator x3 + x2 −5x + 3 x2 −5x + 6 = (x + 3) · (x −1)2 (x −3) · (x −2) From the factorization, we can read off the locations of the function’s zeros and poles. In this example, our zeros are at -3 and 1 (multiplicity 2), and our poles are at 2 and 3.
2. We construct a partial fractions expansion x3 + x2 −5x + 3 x2 −5x + 6 = x + 6 + 24 x −3 − 5 x −2 = 1 1/x + 6 + 24 x −3 − 5 x −2 3. We reconstruct a function given a factorization of its numerator and denominator (or equivalently, a list of poles and zeros along with their multiplicities) 4. We reconstruct a function given a partial fractions expansion (or equivalently, a set of principal parts expansions at the function’s poles) In this chapter, we’ll develop a basic set of technical tools for working in the simplest kind of algebraic extension, an extension of C(x). This will prepare us for the next chapter, where we’ll study Abelian integrals, which are integrals whose integrands are formed from polynomials and roots of polynomials. In other words, integrands in an algebraic extension of C(x).
How might we handle an algebraic extension of C(x)? A crucial property of algebraic functions, as elements of an algebraic extension are called, is that they admit series ex-pansions everywhere, including infinity, so long as we allow a finite number of negative exponents. Such functions are called meromorphic. The logarithm function fails to be meromorphic at the origin, and the exponential function fails to be meromorphic at infin-ity, but algebraic functions are meromorphic everywhere, including infinity.
This means that around any specific point, we can construct a series expansion of the integrand and integrate termwise to obtain a series expansion for the integral. At first this doesn’t seem terribly useful, because series expansions are infinite and we’re trying to construct closed-form solutions, but it turns out that only a finite number of places will have negative exponents in their series expansions and that an algebraic function is completely specified, up to an additive constant, by the coefficients of the negative powers.
Thus, the basic strategy is first to identify the function’s poles, the places where its value becomes infinite, and compute the principal part of the series expansions there, which are the negative exponents and their coefficients. This is fairly straightforward, though there are issues of computational complexity that make it non-trivial. Then we integrate termwise, which is trivial, and obtain local series expansions at the poles of the solution.
Next, we need to reassemble this local information into a global function (if one exists), a Mittag-Leffler problem, for which I will present a basic algorithm in this chapter, although more efficient techniques have been developed.
What about the logarithmic terms? This turns out to be the most difficult part of the problem. We can begin to analyze them using the same techniques, by noting that the t−1 terms in the principal parts of the integrand lead directly to logarithms in the integral, and furthermore that the coefficients of these terms give us the locations and orders of the poles and zeros in the logarithms. This information specifies an algebraic function up to a multiplicative constant1, and our algorithm can be adapted without too much trouble to handle this case.
The problem is that no algebraic function might exist that match a given set of zeros and poles, but increasing the order of the zeros and poles might produce a solution. This corresponds to raising the logarithm term to powers, i.e, ln f is the same as 1 2 ln f 2, which is the same as 1 3 ln f 3, except that in our case the lower powers might not exist in our function field, even though higher powers do. What powers should we use? We could go on raising to higher and higher powers, hoping that something will work, but the only known algorithm to limit this search requires reducing modulo a prime, and that requires techniques that weren’t developed until the 1960s. Before heading into such modern algebraic geometry, however, let’s see how far we can get with the classical algebraic geometry of the nineteenth century.
1Of course. Due to the presence of a constant of integration, we expect to specify the main part of the integral up to an additive constand, and the logarithmic parts of the integral up to a multiplicative constant.
8.1 The Topology of an Algebraic Curve Recall that the graph of a bivariate polynomial, P aijxiyj = 0, is called an algebraic curve, and will be our main focus of attention in this chapter.
One of the first problems we face when dealing with algebraic curves is the multi-valued nature of their solutions. Consider the algebraic algebraic curve y2 = x2 −1. For almost any given x, there are two seperate y values that solve this equation. Conventionally, we express this by writing something like y = ±√x −1, but for higher degree curves this kind of notation becomes unsuitable. How, for example, do you express the three possible solutions to a cube root, and how do you deal with the general case where y appears multiple times in the curve’s defining polynomial, something like y3+x2y2−x+4y = 0?
Our solution to this problem is to regard the entire algebraic curve as a two-dimensional surface in a four-dimensional x-y space. Why four dimensions? Well, just as in the univariate case, we find it convenient to work with complex numbers because of their property of algebraic closure. Most algebraic geometers talk about dimension with re-spect to the coefficient field, so we can consistently say that an algebraic curve is a one dimensional curve in two dimensional space. Regarding both x and y as complex numbers (two dimensions each), and plotting them against each other, we obtain a four dimensional space. Just as in the real case, where an equation like x2 + y2 = 1 defines a circle, an algebraic curve defines a surface, the loci of x and y that satisfy the defining polynomial.
Once we have defined an algebraic curve, we will want to consider algebraic functions on the curve, which are simply rational functions (ratios of polynomials) in the two variables x and y. We will only be interested the values of these functions on the curve; the curve acts as a restriction on the possible values of x and y. If this seems at all puzzling, consider an integrand involving both x and √ x2 + 1. We’ll define y2 = x2 + 1 and replace all instances of the square root with y. Then we’ll have an integrand involving both x and y (the algebraic function), but y is always the square root of x2 + 1, so it makes no sense to consider values of the algebraic function where y2 = x2 + 1 is not satisfied.
Given an algebraic curve and an algebraic function, we’d like to construct a series expan-sion of the algebraic function at each point of the curve. How can we construct series expansions of rational functions that involve both x and y? At each point of the curve, we seek to find a single uniformizing variable that is suitable for constructing power series expansions that converge in an open neighborhood of the point. This is also called a local uniformizer, since no one function is a suitable uniformizing variable at all points of an algebraic curve.
Do local uniformizing variables always exist at every point of an algebraic curve?
The answer is a qualified “yes”. At most points on an algebraic curve, the answer is an unqualified “yes” as a result of the Implicit Function Theorem: Theorem 8.1. Implicit Function Theorem The two-dimensional complex analytic version: Let f(x, y) be an analytic mapping of an open set E ∈C2 into C, such that f(a, b) = 0 and d f dy(a, b) ̸= 0, then open sets U and V exist in C such that a ∈U, b ∈V , and an analytic function g(x) exists that maps U into V such that f(x, g(x)) = 0.
What does it mean for a C2 →C function to be analytic?
[Ru76] Theorem 9.28 is a real version of the theorem that is based on the inverse func-tion theorem, but only establishes that the implicit function is continuously differentiable, whereas we want to show that it’s analytic. Of course, for a complex version of this proof, differentiable would immediately imply analytic.
The Wikipedia article on the implicit function theorem includes a proof for the two-dimensional case (they only case we care about here) that derives a differential equation that the implicit function must satisfy and uses the Picard-Lindelöf theorem to show its existence, but this also only establishes continuous differentiability.
[Gu05] Lecture 7 starts with a complex version of the theorem.
See discusses the difficulty of moving along the transition R →Rn →C →Cn.
□ This theorem applies everywhere on the curve that d f dy ̸= 0, which is everywhere except a finite number of points. Where does it fail? Those points at which both f(x, y) = 0 and d f dy equals zero. In other words, points at which the defining function and its derivative with respect to one of its variables share a zero. At these points, the curve exhibits a behavior called ramification.
If the derivative with respect to one variable is zero, could we try to apply the theorem using the derivative with respect to the other variable? The answer is often yes, but not always. Ramification thus occurs with respect to a specific variable, more generally with respect to a specific mapping.
On some curves, however, there are points, called singularities, at which the derivative with respect to both variables is zero. One might hope to pick a uniformizing variable different from either coordinate variable that would allow the Implicit Function Theorem to be applied, but this turns out to be impossible. The derivative with respect to any alge-braic function could be expanded using partial derivatives with respect to the coordinate variables, and the resulting derivative would necessarily be zero: d f dz = d f dx dx dz + d f dy dy dz = 0dx dz + 0dy dz = 0 So there is no variable at a singularity that meets the requirements of the Implicit Function Theorem. Something more is required.
Since both f and d f dy share a zero at a given value of x, f’s zero must be at least second order, so singularities are necessarily multiple points of the curve.
Consider a specific multiple point where n sheets of the curve meet at a single point P.
A small circle at distance ϵ from P will map to n values of the curve. Without loss of generality, let’s assume that P is the origin.
[Silverman] Theorem 10.13: If a function is analytic on every point of a open disk K : |z −a|< R, then it has a power series expansion centered on a that converges everywhere in K.
By this theorem, we see that we can surround P with disks of radius ϵ; a finite number of them will circle around P. These disks don’t have to cover every point in a neighborhood of P; they need only circle P so that we cleanly identify a permutation of the sheets.
Label the r values of y at (x+ϵ) as y1(x), . . . , yn(x). As we trace along the circle defined by x = ϵeiθ these values deform continuously as θ goes from 0 to 2π. Once we reach 2π, we have come full circle and the y values match up, but permuted. Call the permutation σ, so y1(x), . . . , yn(x) map to yσ1(x), . . . , yσn(x). The permutation σ of the n sheets can decomposed into k disjoint cycles. We seek to show that a Riemann surface can be obtained by replacing the singular point with k distinct points.
Consider a single cycle of length r, the ramification index, and an open disk in the t-plane, ∆= {t : |t|< ϵ}. We wish to exhibit an bihomomorphism from ∆to the cycle. Set x(t) = tr, which is clearly holomorphic in ∆(in fact, everywhere on the t-plane). The function y(t) = yσ⌊(r arg t)/(2π)⌋(tr) where arg t is the complex argument, with range [0, 2π), ⌊·⌋is the integer floor function, and powering σ by an integer applies the permutation that many times. The permutation ensures continuity of the function at each transition between the various yn functions. y(t) is obviously holomorphic away from the origin, but is it also holomorphic at the origin?
Yes, according to Riemann’s theorem on removable singularities.
Riemann removable singularity theorem Theorem 8.2. Let D ⊂C be an open subset of the complex plane, a ∈D a point of D and f a holomorphic function defined on the set D \ {a}. The following are equivalent: 1. f is holomorphically extendable over a.
2. f is continuously extendable over a.
3. There exists a neighborhood of a on which f is bounded.
4. limz→a(z −a)f(z) = 0.
Proof The implications 1 ⇒2 ⇒3 ⇒4 are trivial. To prove 4 ⇒1, we first recall that the holomorphy of a function at a is equivalent to it being analytic at a, i.e. having a power series representation. Define h(z) = (z −a)2f(z) z ̸= a, 0 z = a.
Clearly, h is holomorphic on D{a}, and there exists h′(a) = limz→a (z−a)2f(z)−0 z−a = limz→a(z −a)f(z) = 0 by 4, hence h is holomorphic on D and has a Taylor series about a: h(z) = c0 + c1(z −a) + c2(z −a)2 + c3(z −a)3 + · · · .
We have c0 = h(a) = 0 and c1 = h′(a) = 0; therefore h(z) = c2(z −a)2 + c3(z −a)3 + · · · .
Hence, where z ̸= a, we have: f(z) = h(z) (z −a)2 = c2 + c3(z −a) + · · · .
However, g(z) = c2 + c3(z −a) + · · · .
is holomorphic on D, thus an extension of f.
□ Source: copied verbatim from Wikipedia To see that function y(t) is bounded on ∆, pick a δ so that f(0, δ) is G, then ensure that ϵ is small enough to ensure that f(x, δ) ̸= 0∀x ∈∆ϵ. Since y(t) is continuous on ∆ϵ, if it were not bounded, then f(tr, y(t)) = δ for some value t, contridicting our assumption.
Thus, y(t) is bounded on ∆, so it can be holomorphically extended to the origin by the previous theorem.
Lemma. Given a polynomial f(x, y), then at any given point (x0, y0) and any given real number δ > 0, there exists a real number ϵ such that f(x, y) ̸= f(x0, y0) ∀x, y; |x −x0|< ϵ; |y −y0|= δ Proof. Consider g(y) = f(x0, y) −f(x0, y0), a polynomial in y with a zero at y0. Pick a number a such that no other zero is within ±a of y0. Consider the complex circle of radius a centered at y0. g(y) on this circle must have a minimum value that is not zero; call this minimum value m. Now, at any point y0 + a, f(x0, y0 + a) by complex continuity will have a value ϵ such that x0 can be varied up to ϵ without changing the function by more than m. But do all of these values have a minimum greater than zero?
In short, to obtain a complex manifold, we need to modify our curve slightly by adding additional points at singularities. Theorem 8.2 tells us that no additional information is need to specify the behavior of holomorphic functions at those additional points – their behavior at an isolated point is completely determined by their behavior in an open neigh-borhood surrounding that point.
What about meromorphic functions? They can just be promoted to holomorphic functions by multiplying them by a suitable power of the uniformizing variable (t, in the above treatment), and Theorem 8.2 again tells us that their behavior is completely specified.
Interestingly enough, a rational function can have different values at the same points over a singularity.
Example 8.3. Example: y2 = x3 + x2 has a singularity at the origin, since f(x, y) = y2 −x3 −x2, d f dx = −3x2 −2x, d f dy = 2y, and d f dx(0, 0) = 0 and d f dy(0, 0) = 0.
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 Figure 8.1: y2 = x3 + x2 The function y/x has the value 1 on one branch and −1 on the other. It is also possible, straightforward even, to construct functions with a zero on only one branch (y/x −1) or the other (y/x+1), or a pole on only one branch (x/(y −x)), or a pole on one branch and zero on the other (x/(y −x) + 1/2).
□ Analytically, both partial derivatives of the curve’s polynomial are zero at a singular point, while at least one is non-zero at ordinary points. (PROOF) Resolving our singularities in this manner creates a complex manifold, but it lacks a cru-cial property: topological compactness. In order to apply Theorem 8.5, we need a com-pact manifold. We fix this problem by embedding our algebraic curve in projective space, using a standard construction.
Now, the coefficient of yn in the defining polynomial will be a polynomial in x, which has a finite number of roots at which it is zero, so there are only a finite number of points where the defining polynomial is of degree less than n in y. As x approaches one of these points, the value of the yn coefficient approaches zero, which causes at least one of the roots to approach infinity. We’ll deal with these points by introducing a line at infinity, forming projective space and creating a compact surface.
Definition 8.4. A topological space X is compact if any open covering of X admits a finite subset that covers X.
There are several crucial theorems that depend on the topological property of compact-ness. The complex plane is not compact; we remedy this by adding a point at infinity to obtain the Riemann sphere. Likewise, two-dimensional complex space is not compact, either; we remedy this by adding a line at infinity and obtaining projective space.
Theorem: Projective space is compact Munkres Theorem 26.2: Every closed subspace of a compact space is compact.
Theorem: The Zariski topology is the coursest topology in which singletons are closed.
Varieties are closed in the Zariski topology.
Standard topology is finer than Zariksi, so all open sets in Zariski are open in standard, and all Zariki-closed sets are closed in standard.
So, varieties are closed in the standard topology, and are therefore compact in projective space.
Projective space. Compactness.
Another highly desirable property is to be locally isomorphic to Euclidean space. A dif-ferentiable surface that is everywhere locally Euclidean is called a manifold.
By adding a line at infinity and resolving our singularities, we can coax our algebraic curve into a compact, connected, complex manifold. The primary utility of this construction is embodied in the following theorems.
Regular functions. Holomorphic functions. Identical on complex projective varieties.
Theorem 8.5. Every holomorphic function M →C on a compact, connected, complex manifold M is constant. 2 Proof [Gu05] Lecture 2 contains a proof of the Maximum Modulus Priciple.
□ Definition 8.6. The principal part of an algebraic function at a pole is the portion of its corresponding Laurent series with negative exponents.
Theorem 8.7. An algebraic function on an algebraic curve is completely characterized, up to an additive constant, by its principal parts.
Proof 2In fact, we don’t need the complex structure and can make the stronger statement that the only regular functions on any projective variety are the constants. See Proposition 4.2 in bertram/6030/Projective.pdf, or Hartshorne Theorem (I, 3.4a), or If the base field is not alge-braically closed, however, new constants may appear as algebraic functions. For example, the polynomial (x + y)2 + 1 is irreducible in the ring Q[x, y], so we can use it to construct an algebraic curve, but x + y is a constant on this curve, a square root of −1, in fact.
Consider two algebraic functions f and g with identical principal parts. Taking the differ-ence between them, we obtain a function f −g with no principal parts, i.e, a holomorphic function. By Theorem 8.5, f −g must be constant.
□ Theorem 8.8. An algebraic function on an algebraic curve is completely characterized, up to a multiplicative constant, by its divisor.
Proof Consider two algebraic functions f and g with identical divisors. Dividing f/g we obtain a function with no poles (or zeros), a holomorphic function. By Theorem 8.5, f/g must be constant.
□ Related: Hartshorne Corollary 6.10. A principle divisor on a complete nonsingular curve has degree zero.
Given the importance of an algebraic function’s principal parts, we will now develop tools to calculate them.
8.2 Puiseux Expansions The previous section showed that, in complex projective space, a covering surface can be constructed for an algebraic curve, which is isomorphic to the curve except at singularities (where additional points must be introduced), such that: • the covering surface is a complex manifold (a Riemann surface), • the curve’s rational functions are meromorphic functions on that manifold, and • the rational functions admit series expansions at every point of the manifold.
Our next task is to compute those series expansions.
Since the rational functions in the curve’s function field are formed as rational functions in x and y (or whatever our variables are named), our primary goal is to compute series expansions for x and y at arbitrary points on the curve. With such expansions in hand, it is straightforward to construct expansions for any algebraic function, simply by substituting in the x and y expansions.
At any point where the discriminant is non-zero and y obtains a finite value, a series expan-sion for y exists as a power series in (x −α), which, as we have seen, is a straightforward application of the Implicit Function Theorem. At these ordinary points, we need only pos-tulate a Taylor series for y in powers of (x −α), substitute this into the curve’s defining polynomial, and equate like powers to obtain a set of equations to be solved simultane-ously. Multiple solutions will typically be found, corresponding to multiple branches of the curve.
At ramification points, the series expansion exists in terms of fractional powers of (x−α), where the denominator of the fractions is the ramification index. Issac Newton, in 1676, first proposed a method of computing the ramification index using what are now called Newton polygons.
Let’s assume that we’re expanding around the point (0, 0), as this simplifies the analysis with no loss of generality. Consider factoring the defining polynomial of the algebraic curve: pnyn + pn−1yn−1 + · · · + p0 = (y −r1)(y −r2) · · · (y −rn) How might we do this, if the polynomial is irreducible? We need to extend to a larger field where the polynomial’s roots exist. The analysis above shows that Puiseux series form a suitable extension.
For each root ri, define its order as the lowest power of t that appears in its Puiseux ex-pansion, divided by its ramification index. Multiplying factors together adds their orders, so p0’s order will be the sum of all of the ri’s orders.
Now let’s consider increasing i by one. How does p0’s order change? p1 is formed by adding together all products of n −1 roots, so p1’s order will be lower than p0’s order by the largest of ri’s orders, unless there are multiple ri’s with the same order. In this case, cancellation between these multiple terms could result in p1 having a larger order than otherwise expected.
If there are j ri’s with the same largest order, increasing i by j will lower pi’s order by j times that largest order.
The Newton polygon is formed by plotting the orders of the pi coefficients, with i varying along the horizonal axis and the order plotted vertically. The easiest way to do this is to plot the powers of the monomials that appear in the equation, and construct the polygon’s lower convex hull.
Thus, a segment on the lower convex hull of the Newton polygon will correspond to as many solutions as the width of the line segment, each with order equal to the change in height divided by the width, i.e, the negative slope of the line segment. The denominator of the slope will be the ramification index, and the numerator of the slope will be the lowest exponent expected in the expansion of y.
Consider a Puiseux series corresponding to a single line segment of the Newton polygon.
Letting α be the x exponent and β be the y exponent, so the monomials in f have the form xαyβ, then the equation of the line segment is rα + sβ = p, where r, s, and p are integers and r and s are relatively prime. Making the substitution x = tr and y = tsu(t), we obtain: f(x, y) = X Aαβxαyβ = X Aαβtrαtsβu(t)β = tp X Aαβtrα+sβ−pu(t)β | {z } g(t,u) At least two of the (rα+sβ−p) exponents will be zero (those monomials corresponding to the endpoints of the line segment on the Newton polygon); all of the remaining exponents will be positive. This means that if we expand u(t) in a power series in t: u(t) = u0 + u1t + u2t2 + u3t3 + · · · then any power of u(t) will have the form: u(t)β = U0(u0) + U1(u0, u1)t + U2(u0, u1, u2)t2 + U3(u0, u1, u2, u3)t3 + · · · and g(t, u) will also have the form: g(t, u) = G0(u0) + G1(u0, u1)t + G2(u0, u1, u2)t2 + G3(u0, u1, u2, u3)t3 + · · · In order for g(t, u) = 0 at t = 0, G0(u0) must be zero, and since G0(u0) is a polynomial in u0, this gives us a finite number of values for u0 that can solve our equation.
Now, by setting g(t, u) = 0, can we obtain u(t) as a function of t?
The Implicit Function Theorem states that we can, if δg δu is not zero at the point we wish to expand around.
δg δu(0, u0) = δ δuG0(u0) In short, the roots of G0(u0) give us the starting values for our series expansion, and if the roots are simple, then the Implicit Function Theorem guarantees that we’ll have a unique series expansion for u(t) as a function of t. If any of the roots are not simple, then we can repeat this procedure for g(t, u). It can be shown ([Bl47] §15) that this procedure always terminates.
Example 8.9. Construct Puiseux expansions of y at the multiple points of the curve y2 = 1 −x2 We normalize the defining polynomial by writing it as y2 + x2 −1 = 0. Where does it have multiple points? We compute the discriminant: sage: R. = QQbar[]; sage: (y^2 + x^2 - 1).discriminant(y).factor() (−4) · (x −1) · (x + 1) The multiple points of y2 = 1 −x2 lie at the roots of the discriminant, which are x = ±1.
In both cases, y = 0 is the only solution, so the curve has multiple points at (x, y) = (±1, 0). The partial derivative of the defining polynomial with respect to x is 2x, which is non-zero, so neither of these multiple points are singular; we’ll get ramification instead.
The analysis is almost the same in both cases, so I’ll just do (1, 0).
First, construction of the Newton polygon requires recasting the curve’s polynomial into a form centered about the point being analyzed, i.e, y2 + (x −1)2 + 2(x −1) = 0. Next, we construct the Newton polygon by plotting the monomial powers, putting the y exponents on the horizontal axis and the (x −1) exponents on the vertical: 0 1 2 3 0 1 2 3 The only segment on the Newton polygon’s lower convex hull has slope −1/2 and width 2, telling us that two of our roots (the width of the segment) will require a single Puiseux series with ramification index 2 (the denominator of the slope): x = t2 + 1 We know that y can be expressed as a power series in t with initial exponent 1 (the nu-merator of the slope): y = a1t + a2t2 + a3t3 + · · · Now, substituting these expressions for x2 and y2 into the curve’s defining equation y2 + x2 −1 = 0 and setting all coefficients of t to zero, we find: sage: var('x, y, t, a1, a2, a3'); sage: f = y^2 + x^2 - 1; sage: exp = f.subs({x: t^2+1, y: a1t + a2t^2 + a3t^3}); sage: exp.collect(t) a2 3t6 + 2 a2a3t5 + 2 a1a2t3 + a2 2 + 2 a1a3 + 1 t4 + a2 1 + 2 t2 2 + a2 1 = 0 2a1a2 = 0 1 + 2a1a3 + a2 2 = 0 The first equation tells us that a1 = ± √ 2i, the second equation tells us that a2 = 0 and the third equation tells us that a3 = ± √ 2 4 i, so x = t2 + 1; y = ± " √ 2it + √ 2 4 it3 + · · · # (8.1) It would seem that we have two different series to chose from. This is not really the case, as they differ by only a 180◦rotation in the t-plane, as can been seen by substituting t = −t, which transforms one of the y-series into the other, while leaving the x-series unchanged.
Now, let’s analyze the point at infinity. We move infinity to a finite point (0) with the substitution x = u−1, then combine all of our terms over a common denominator and discard the denominator. Our curve becomes: y2u2 + 1 −u2 = 0 0 1 2 3 0 1 2 3 The Newton polygon’s lower convex hull has a single line segment, slope 1, length 2, telling us that we’ll have two separate poles, each with ramification index 1. Thus, u can be used directly as a uniformizing variable, and we postulate an expansion for y in the form: y = a−1 1 u + a0 + a1u + a2u2 + a3u3 + · · · Plugging this into y2u2 + 1 −u2 and setting all the resulting coefficients to zero, we conclude: sage: var('u, y, t, a0, a1, a2, a3'); sage: var('an1', latex_name='a_{-1}'); sage: f = y^2u^2 + 1 - u^2 u2y2 −u2 + 1 sage: exp = f.subs({y: an1(1/u) + a0 + a1u + a2u^2 + a3u^3}); sage: exp.collect(u) a2 3u8 + 2 a2a3u7 + a2 2 + 2 a1a3 u6 + 2 (a1a2 + a0a3)u5 + a2 1 + 2 a0a2 + 2 a3a−1 u4 + 2 (a0a1 + a2a−1)u3 + 2 a0a−1u + a2 0 + 2 a1a−1 −1 u2 + a−1 2 + 1 a−1 = ±i; a0 = 0; a1 = ∓1 2i; a2 = 0; a3 = ∓1 8i y = ±i1 u ∓1 2iu ∓1 8iu3 + · · · This time, there is no ramification, since u, and not a power of u, is 1 x. We actually have two distinct series that will yield two different values of y for each value of u. Geometri-cally, we have two sheets that approach each other and touch at a singular point where the curve is not locally Euclidean, in a manner somewhat like this: □ We’ve seen how to construct Puiseux expansions at arbitrary points of an algebraic curve, but some points have multiple expansions, corresponding to multiple cycles on their cor-responding surfaces. These points are the curve’s singularities, which are the only points where the curve is not locally Euclidean, and behaves instead like in the graphic above, where several Euclidean sheets touch at a single point.
We now seek some mechanisms for distinguishing between multiple cycles at a single point. To this end, we introduce the concept of a place. Intuitively speaking, a place is a cycle, and places are in one-to-one correspondence with cycles. Therefore, we can handle singularities by thinking in terms of places. Non-singular points have a unique place associated with them, while there are multiple places (and multiple cycles) associated with a singular point.
8.3 Valuation Rings and Orders Driven in no small part by the difficulty in visualizing higher dimensional geometric shapes, mathematicians have developed increasingly algebraic techniques to manipulate geometric objects. In particular, the techniques of the previous section were presented largely for educational purposes, as they are now considered obsolete. The current state of the art is to use the tools of abstract algebra developed in the early twentieth century, such as rings, ideals, and fields.
We can’t use ideals directly in a function field, or any field for that matter, because there are only two ideals in any field – the zero ideal, which is the ideal containing only the zero element, and the unit ideal, which is the entire field. This follows directly from the invertability of field elements. As all field element possess inverses, any non-zero ideal generator can be multiplied by its multiplicative inverse to generate 1, which then generates the entire field.
The obvious choice would be to use ideals of the coordinate ring, which is simply the polynomial ring modulo the defining polynomial of the curve, but these ideals are in one-to-one correspondence with the points of the curve, and we need some way to represent places.
To solve this problem, we introduce the concept of valuation rings and decompose the function field into orders, of which there are two of primary interest: the maximal finite order and the maximal infinite order. The ideals of the maximal finite order correspond to finite places, and the ideals of the maximal infinite order correspond to infinite places.
These maximal orders can be constructed using the algebraic process of normalization.
Using one or the other, we obtain ideals that represent all places in the function field.
Definition 8.10. The coordinate ring of the algebraic curve is the ring K[x, y] mod p(x, y).
If p(x, y) is irreducible, then the coordinate ring is an integral domain.
K is the curve’s field of constants. The easy case is to let K be the complex numbers C. Since C is algebraically closed, many of the theorems are simplified. Even if we’re looking for real solutions, it is often simplest to work with complex numbers because of the simplification of the theory, then restrict to real solutions only at the end. For example, the entire argument of the last section is based on the curve having n solutions at all but a finite number of values for x; this is only the case if we’re working over C, or some other algebraically closed field, such as Q, the algebraic number field, which is the algebraic closure of Q. For actually computations, however, C is often not the best choice because both the difficulty of expressing an arbitrary complex number, and due to the added complexity of having to fully factor a polynomial. See Example 8.21 for a problem where the results are simplified by using a different field of constants.
What do we do if p(x, y) is not irreducible? In this case, we’re trying something like computing an integral with a root of a polynomial that can be factored. We factor the polynomial, producing two separate roots. We then use the theorem of the primitive element (NEED TO PUT IT IN CHAPTER 2) to construct a single algebraic extension defined by a single irreducible polynomial in which we can construct both roots. So, for the remainder of this discussion, we need only consider the case where p(x, y) is irreducible.
Definition 8.11. The fraction field of the coordinate ring is called the curve’s function field, and its elements are called algebraic functions.
A curve’s function field contains all rational functions in x and y, grouped together into equivalence classes by the relation p(x, y) = 0.
Definition 8.12. A valuation ring of a function field F/K is a subring O ∈F such that K ⊂O ⊂F (both proper inclusions), and for every z ∈F, either z ∈O or z−1 ∈O Both ideals and valuation rings are subrings, but there are two different conditions that they must satisfy. Fields admit only two ideals (the unit ideal and the trivial ideal), but there are typically an infinite number of valuation rings within a given function field.
We now wish to show that valuation rings are in one-to-one correspondence with places.
The key observation is that an element z in is a valuation ring O only if z has non-negative valuation (i.e, is finite) at the corresponding place.
The condition that either z or z−1 must be in the valuation ring does not preclude the possibility that both will be in the valuation ring. In fact, these are precisely the elements of valuation zero, that have neither a zero nor a pole at the place in question. The elements of O whose inverses are not in O are the elements of positive valuation, that have a zero at the place in question. They form an ideal P of the valuation ring. O is, in fact, a local ring, and P is its unique maximal ideal.
Given a valuation ring O and its maximal prime ideal P, we define the residue class field of the valuation as FP := O/P. Given any element of O, its value at the corresponding place is its image in FP. The easy case in when the field of constants is algebraically closed. Otherwise, we can have non-rational places whose residue fields have degree (over the constants) greater than one.
A valuation ring might not be discrete valuation ring (DVR), which is a valuation ring with value group isomorphic to the integers under addition. All of our valuation rings are DVRs, mainly because we’re working on an algebraic curve, which has Krull dimension one.3 [St09] Theorem 1.1.6 shows that the valuation ring of a function field is a DVR by establishing one of a DVR’s key properties: it’s a principal ideal domain with a unique maximal ideal.
Theorem 8.13. The valuation rings of a curve’s function field are in one-to-one corre-spondence with its places.
Proof 1. Every cycle induces a valuation ring 3 contains a list of equivalent conditions that a DVR satisfies, including “R is an integrally closed Noetherian local ring with Krull dimension one”.
To do this, we just use the Puiseux expansion of a function field element to compute its valuation. If it is finite, it is in the valuation ring; if it is a pole, it is not in the valuation ring.
Show that valuation of the Puiseux expansion is independent of the choice of uni-formizing variable (PUT THIS IN THE PREVIOUS SECTION) Show compatibility of the Puiseux expansions – inverting one inverts the valuation (PUT THIS IN THE PREVIOUS SECTION) 2. Every valuation ring induces a cycle How to compute a Puiseux expansion from a valuation ring? Since we have a discrete valuation ring, which is a principal ideal domain, there exist elements which generate the DVR’s unique maximal ideal P. Every such element t ∈P such that P = tO is a uniformizing parameter.
Show that O mod P is isomorphic to the constants. x mod P and y mod P give us the coordinates of the point corresponding to the place.
Given an element in O, mod out by P to find its value. Subtract this value to get an element in P. Divide by a uniformizing parameter. Repeat to construct a Puiseux expansion.
Show that all elements have a well-defined valuation (P v contains them but P v+1 does not).
If the element is not in O, its inverse must be, show that its inverse respects valuation and we can multiply by t−v, where t is a uniformizing parameter, to move the inverse into O.
□ Since there are an infinite number of valuation rings, it is most convenient to separate them into two sets (the finite and the infinite), each of which can be represented as ideals of a particular subring (an order).
Definition 8.14. A valuation ring O is finite if both x ∈O and y ∈O. Otherwise, it is infinite. (or is it finite if only x ∈O?) Definition 8.15. An order O of a ring R (also called an R-order) is a subring of R such that 1. O is a finite-dimensional algebra over the field Q of rational numbers 2. O spans R over Q, and 3. O is a Z-lattice in R.
[wikipedia] Definition 8.16. The maximal finite order of a function field F is the intersection of all its finite valuation rings.
Show that it’s an order.
Show that valuation rings are maximal ideals of the order. (are there other maximal ide-als?) Definition 8.17. The maximal infinite order of a function field F is the intersection of all its infinite valuation rings.
Show that it’s an order.
All places correspond to a maximal ideal in one of these two orders. That ideal consists of all algebraic functions which are zero at that place and, in the finite case, have no finite poles. How can they be characterized in the infinite case? Equivalently, it consists of all integral elements (that satisify a monic polynomial) which are zero at that place. Two different places, even if they’re at the same point, correspond to two distinct maximal ideals of the order.
Definition 8.18. The integral closure of an integral domain R in its field of fractions F are the elements of F with no finite poles, or that admit monic defining polynomials.
Theorem: Every element of an R-order is integral over R.
Theorem: The integral closure of an integral domain R in its field of fractions is an R-order, and in fact is maximal (due to the last theorem).
Corollary: The maximal finite order is the integral closure of the coordinate ring.
We might need all this stuff to show that integral closure is the same as the intersection of all the finite valuation rings.
[SwHu06] Proposition 6.8.14 Let R be an integral domain. Then the integral closure of the ring R equals ∩V V , where V varies over all the valuation domains between R and its field of fractions. If R is Noetherian, all the V may be taken to be Noetherian.
[Ko07] Definition 1.23 Let S be an integral domain with quotient field Q(S). The normalization of S in Q(S), denoted by S, is the unique largest subring S ⊂Q(S) such that every homomorphism φ : S →R to a DVR extends to a homomorphism φ : S →R.
[Ko07] Lemma 1.24 A unique factorization domain is normal. In particular, any polynomial ring over a field is normal.
[Ko07] Lemma 1.25. Assume that t ∈Q(S) satisfies a monic equation with coeffi-cients in S (i.e, t is integral). Then t ∈S.
[Ko07] Defintion 1.27. Let S be an integral domain. The normalization of S is its integral closure in its quotient field [Ko07]: “The easy argument that every normal integral domain S is the intersection of all the valuation rings sitting between it and its quotient field is given in [AM69, 5.22].
Working only with discrete valuation rings is a bit harder. The strongest theorem in this direction is Serre’s condition for normality; see [Mat70, 17.1] or [Mat89, 23.8].” [Mat89]: Ri condition: Ap is regular for all P ∈SpecA with htP ≤i Si condition: depth Ap ≥min(htP, i) for all P ∈SpecA (S0) always holds. (S1) says that all the associated primes of A are minimal. (R0) + (S1) is n.a.s.c. for A to be reduced.
Theorem 23.8: (R1) + (S2) are n.a.s.c. for a Notherian ring A to be normal.
What do we need this for?
A Dedekind domain is a integral domain in which all ideals factor into a product of prime ideals.
Theorem: such a factorization is necessarily unique.
[Wiki Dedekind domain] Theorem: Let R be a Dedekind domain with fraction field K. Let L be a finite degree field extension of K and denote by S the integral closure of R in L. Then S is itself a Dedekind domain.
Theorem (Kummer): If {1, y, ..., yn−1} is a local integral basis at some prime polyno-mial p in x, then we can factor the ideal pe by factoring the field’s defining polynomial mod p. [Stichtenoch Theorem 3.3.7] An Artinian ring satisfied the descending chain condition on ideals.
Theorem: the modulo ring constructed from a Dedekind domain and a proper ideal is an Artinian ring.
Show that the maximal finite order is a finitely generated R[x]-module. See Section 11.4.
To work with the maximal finite order of an algebraic curve, or the integral closure of the coordinate ring (it’s same thing), we’d like to compute a basis for the order as an R[x]-module, called an integral basis. Computing an integral basis is not trivial, but algorithms to do this have been known for over a hundred years. Chapter 5 in [Al14] describes Trager’s algorithm, which was originally published in [Tr84]. If our algebraic curve is formed as the root of a polynomial, i.e, yn = p(x), then an integral basis has a particularly simple form, given in [Tr84] pp. 30-31.
Once we have an integral basis, we can represent places as ideals within the order using their ideal generators, of which there are also a finite number. This gives us the most convenient way of working with places on an algebraic curve: an integral basis to describe the maximal finite order, and a set of ideal generators to describe a place as an ideal in said order.
As a basic (and important) example of the technique, I’ll show how to evaluate an alge-braic function at a place, given a representation of the place as an ideal of an order.
Algorithm 8.19. Evaluation of an algebraic function at a place on a curve Given: a rational function f and a prime ideal I in a maximal order O.
Goal: compute the value of f mod I.
Step 0: Precompute a k[x]-module basis for I in Hermite normal form (doable since I is finitely generated over O and O is finitely generated over k[x]) and α, a rational function with a simple pole at the place corresponding to I and no other finite poles.
Step 1: Compute the valuation of f’s denominator (how?); call it ν. Multiply both f’s numerator and denominator by αν. Now both the numerator and denominator are in O, and the denominator is not in I (i.e, it has a finite value). So we’ve reduced to the case of computing the residue of an element of O, as we now can divide by the denominator’s residue, since we know it’s not zero.
Step 2: Use the techniques from Section 2.11 to compute residues of both the numer-ator and the denominator, and divide them to obtain the result.
Note: If the field k is not algebraically closed, then find a primitive element of the residue field and call it g. Compute g’s minimal polynomial and use it to construct the residue field as an algebraic extension of the constant base field. We can represent g and its powers w.r.t. the HNF basis, construct a matrix that converts from g-basis to HNF basis, then invert it to obtain a matrix that converts from HNF basis to g-basis.
Use reduction mod I’s HNF basis, then multiply by this inverse matrix to obtain an element in the residue field, expressed in g-basis.
Source: Sage’s FunctionFieldPlace_polymod._residue_field Example 8.20. Compute the valuation of y/x at the single place lying over the point (1, 1) in the function field y3 = x.
The answer is obviously 1, but let’s see how to compute this using the above algorithm.
First, we know that an integral basis for this function field is (1, y, y2) [Tr84] pp. 30-31.
We’ll use the ordering (y2, y, 1), because we want the 1 to be last.
Our first generator x −1 can be represented by the matrix x −1 0 0 0 x −1 0 0 0 x −1 Remember that the columns correspond to the integral basis (y2, y, 1), and the rows are the generators x −1, (x −1)y, and (x −1)y2.
Our second generator y −1 can be represented by the matrix 0 1 −1 1 −1 0 −1 0 x Putting this second matrix in HNF we obtain: 1 0 −1 0 1 −1 0 0 x −1 Combining the two matrices, we obtain: 1 0 −1 0 1 −1 0 0 x −1 x −1 0 0 0 x −1 0 0 0 x −1 Reducing further using HNF elementary row operations yields: 1 0 −1 0 1 −1 0 0 x −1 0 0 0 0 0 0 0 0 0 Since the two three rows are identical to those from the y −1 matrix, this implies that the ideal is principal and y −1 is the only generator required and our ideal’s HNF matrix is: 1 0 −1 0 1 −1 0 0 x −1 x is represented as (0, 0, x). Applying the HNF reduction algorithm to this vector, we reduce by left-multiplying the matrix by (0, 0, −1) to obtain (0, 0, 1 −x). Adding this vector to (0, 0, x) gives us (0, 0, 1), which, when dot-multiplied by (y2, y, 1), yields 1, our remainder.
sage: R. = FunctionField(QQbar) Rational function field in x over Algebraic Field sage: L. = R[] Rational function field in x over Algebraic Field[y] sage: F. = R.extension(y^3 - x) Function field in y defined by y^3 - x sage: F.maximal_order().ideal(x-1, y-1) (y −1) sage: F.maximal_order().ideal(y-1).hnf() x −1 0 0 −1 1 0 −1 0 1 □ Conceptually, the various rings, fields, and inclusion maps look like this: Rational Functions Algebraic Curve C(x) C(x, y) mod p(x, y) Fraction Fields I Maximal Finite Order / Integral Basis C[x] C[x, y] mod p(x, y) Coordinate Rings The coordinate rings are subsets of their fraction fields, and can be mapped into them.
The univariate coordinate rings and fields (the rational functions in x) are subsets of their counterparts on the algebraic curve, and can be mapped into them. Finally, the algebraic curve’s integral basis is a subset of its fraction field and a superset of its coordinate ring.
[St09] Corollary 1.2.3 shows that all places of C(x) (the rational function field) are in 1-to-1 correspondence with C ∪inf (as places).
We have the function field C(x) and its places, and also the function field defined by the curve, and its places. Since all the rational functions in C(x) are also rational functions in C(x, y) mod p(x, y) (the coordinate ring), it follows that the function field defined by the curve is a field extension of the rational function field.
The maximal finite order of an algebraic curve is an extension of the curve’s coordinate ring, which is an extension of the univariate polynomial ring. The maximal ideals in the univariate ring correspond to coordinates in a single variable x, the maximal ideals in the coordinate ring correspond to the points on the curve, and the maximal ideals in the maximal finite order correspond to the places of the curve. By computing how these ideals decompose under extension, we can compute the points that lie over a given x coordinate, as well as the places that lie over a given point, or a given x coordinate.
All points correspond to maximal ideals in the coordinate ring. For points, our ideal generators are just x −x0 and y −y0, and this ideal is maximal in the coordinate ring.
If (x0, y0) was not on the curve, then the ideal would be (x −x0, y −y0, p(x, y)), which would simplify to p(x, y) being a non-zero constant, and this would generate the unit ideal.
At a non-singular point, the ideal (x −x0, y −y0) is prime not only in the coordinate ring, but also in the maximal finite order. How can we show this?
At a singularity, the ideal (x −x0, y −y0) will be prime in the coordinate ring, but will decompose in the maximal finite order into multiple prime (and maximal) ideals, corre-sponding to the places that lie at that point.
1. All places of an extension are “over” a place of the underlying field 2. vP ′(x) = e · vP(x), so e is the ramification index (and is a finite integer) Do we need this discussion here? Only if we need to compute ramification indices, I think.
[Wiki Ideal] Theorem: if f : A →B is surjective and a ⊇ker f then: • aec = a and bce = b • a is a prime ideal in A ⇔ae is a prime ideal in B.
• a is a maximal ideal in A ⇔ae is a maximal ideal in B.
So, to factor p, we need to find all of the prime/maximal ideals in R mod p.
Constructing R mod p as a finite dimensional algebra, we have an algorithm to enu-merate all of the algebra’s maximal ideals. This is the same as the ideal decomposition algorithm.
We also want to find each ideal’s ramification index and relative degree. I wrote a paper on this.
The ramification indices are the powers of the prime ideals. Cohen Theorem 4.8.3: pZK = g Y i=1 pei i Stichnoch defines the ramification index as the integer such that νP ′(x) = eνP(x) for all x in the base field. (P ′ lies over P) We can find the ramification index in the modulo ring by raising each maximal ideal to successive powers and determining when it stabilizes. The Artinian condition guar-antees that it will eventually stabalize.
The relative (or residual) degree of p is defined (Cohen Definition 4.8.4): fi = [ZK/pi : Z/pZ] This seems to be the nullity of the matrix that defines the ideal in the algebra.
Why? The dimension of the algebra (over Z/pZ) is the dimension of ZK. The number of basis elements in the ideal is the dimension of pi. The dimension of ZK/pi is the difference of these two numbers (the dimension of the algebra minus the dimension of the ideal), which is the dimension of the vector space minus the dimension of the ideal’s basis matrix’s kernel, or the nullity of this matrix.
8.4 Sage’s FunctionField code Sage has built-in routines to compute Puiseux expansions without having to construct Newton polygons and substitute trial expansions. In fact, there are three different con-structions in Sage for algebraic curves – polynomial rings modulo an ideal, function fields, and curves in projective space. We’ll mostly use the FunctionField code.
Instead of constructing a polynomial ring the usual way (which is?), we’ll construct a univariate FunctionField over a single variable, specifying its constant field, then form its extension modulo the curve’s defining polynomial, which will yield another, bivariate, FunctionField.
A Sage FunctionField has methods to construct both the maximal finite order and the maximal infinite order. We can construct ideals of these orders, and from them construct the corresponding places. Given a place in the underlying univariate FunctionField, we have a method that will extend the corresponding ideal to the bivariate FunctionField, factor (decompose) it, and return a list of all places_above the original place.
The standard way to construct Puiseux expansions in Sage is the completion method, defined on FunctionField’s, but this method has some limitations. It doesn’t work on differentials (only on functions), doesn’t allow a uniformizing parameter to be spec-ified, and doesn’t allow absolute precision to be specified (only relative precision). To overcome this problems, I’ve written the following function that’s basically just a wrap-per around completion. Its default behavior is to compute either the principal part of a Puiseux expansion for poles, or a single term of the Puiseux expansion for finite values.
The main thing I don’t like about it is that when you pass it a differential, it returns a LaurentSeries, when it should actually return a LaurentSeries times ds, where s is the uniformizing variable.
If a uniformizing variable is specified, either as a function field element or as a LaurentSeries, we reverse the series, which only works if the series has valuation 1, giving us a new series that expresses s in powers of the uniformizing variable, then then substitute the reversed series into the original s-expansion of the function field element, which gives us an ex-pansion of the element in powers of the uniformizing variable.
from sage.rings.function_field.differential import FunctionFieldDifferential def puiseux(F, pl, uvar=None, absprec=0): def puiseux2(f): if isinstance(f, FunctionFieldDifferential): base_differential = f.parent()._gen_base_differential f = f._f is_differential = True valuation = f.valuation(pl) + base_differential.valuation(pl) - 1 else: is_differential = False valuation = f.valuation(pl) series = F.completion(pl, prec=max(1, absprec-valuation)) if uvar: if isinstance(uvar, LaurentSeries): uvar_series = uvar.reverse() else: uvar_series = series(uvar).reverse() f_series = series(f)(uvar_series) else: f_series = series(f) if is_differential: if uvar: f_series = series(base_differential).derivative()(uvar_series) else: f_series = series(base_differential).derivative() return f_series return puiseux2 Example 8.9 cont.
Construct Puiseux expansions of y at the multiple points of the curve y2 = 1 −x2 This is Example 8.9, redone now using Sage FunctionField.
First we create a function field in one variable with coefficients in Q: sage: R. = FunctionField(QQbar) Rational function field in x over Algebraic Field Next we create a ring of polynomials in y with coefficients in the function field, which allows us to write the minimal polynomial of the algebraic curve. We then create a new function field that is an extension of the rational function field: sage: L. = R[] Rational function field in x over Algebraic Field[y] sage: F. = R.extension(y^2 + x^2 - 1) Function field in y defined by y^2 + x^2 - 1 Recall from Example 8.9 that our multiple points lie at x = ±1. Since we wish to construct our series expansion at a point with finite coordinates (remember that this is projective space, so we also have points at infinity), we use the finite maximal order, construct the ideal corresponding the desired point, then construct the unique place at that point: sage: O = F.maximal_order() Maximal order of Function field in y defined by y^2 + x^2 - 1 sage: pl = O.ideal(x-1, y).place() (x −1, y) Finally, we construct the Puiseux expansion of y at the place, specifying the desired precision of the expansion. Notice that puiseux actually returns a function, to which we pass the function field element (y) that we wish to ex-pand: sage: puiseux(F, pl, absprec=4)(y) s + O(s4) This answer differs from the one we computed in Example 8.9 because the choice of uniformizing variable is not unique, and the computer made a differ-ent choice than we did. Our “t” variable in Example 8.9 was roughly √x −1.
Even though square roots don’t exist in this field, we can construct a series expansion of x −1, and construct the square root of the series expansion as another series expansion.
sage: xminus1 = puiseux(F, pl, absprec=5)(x-1) −1 2s2 −1 8s4 + O(s5) sage: t = xminus1^(1/2) 1 2i √ 2s + 1 8i r 1 2s3 + O(s4) Using this series as a uniformizing variables gives us y in powers of t, even though the computer still prints the results using s.
sage: puiseux(F, pl, absprec=4, uvar=t)(y) −i √ 2s −1 2i r 1 2s3 + O(s4) Comparing this to equation 8.1, we see that they’re the same.
Now let us turn to the point at infinity. We begin by constructing the maximal infinite order of the underlying rational function field C(x), because its struc-ture is quite simple and we know that it will only have a single point, and a single place, at infinity.
sage: ROinf = R.maximal_order_infinite(); sage: Rinf = ROinf.ideal(1/x).place() 1 x Now we’ll use the places_above method to obtain a list of all places in an extension field that lie over a given place in the underlying function field.
Recall from Example 8.9 that this curve has a singular point at infinity with two separate cycles and therefore two separate places.
sage: Pinf = F.places_above(Rinf) 1 x, 1 xy −i , 1 x, 1 xy + i Having obtained these places, represented as ideals in the maximal infinite or-der, it is straightforward to use the puiseux function to construct Puiseux series for x and y.
sage: [puiseux(F, pl, absprec=5)(x) for pl in Pinf] 1 s + O(s5), 1 s + O(s5) sage: [puiseux(F, pl, absprec=5)(y) for pl in Pinf] i s −1 2is −1 8is3 + O(s5), −i s + 1 2is + 1 8is3 + O(s5) Comparing this to Example 8.9, we see that the results are the same.
□ Example 8.21. [Bl47] §68 Compute expansions at all multiple points of y3 + x3y + x = 0 We begin by computing the discriminant of the equation, which gives us the locations of the multiple points.
sage: R. = QQbar[]; sage: f = y^3 + x^3y + x x3y + y3 + x sage: f.discriminant(y).factor() (−4) · (x + 1 4 · 27 1 74 6 7e(−6 7 i π)) · (x + 1 4 · 27 1 74 6 7e( 6 7 i π)) · (x + 1 4 · 27 1 74 6 7e(−4 7 i π)) · (x + 1 4 · 27 1 74 6 7e( 4 7 i π)) · (x + 1 4 · 27 1 74 6 7e(−2 7 i π)) · (x + 1 4 · 27 1 74 6 7e( 2 7 i π)) · (x + 1 4 · 27 1 74 6 7) · x2 That result is rather confusing. Let’s try factoring over Q instead of Q: sage: f.discriminant(y).change_ring(QQ).factor() (−1) · x2 · (4x7 + 27) The multiple points lie over the roots of this equation: x = 0 and the seven roots of 4x7 + 27 = 0. Infinity also needs to be examined. We begin with x = 0: sage: R. = FunctionField(QQbar) Rational function field in x over Algebraic Field sage: L. = R[] Rational function field in x over Algebraic Field[y] sage: F. = R.extension(y^3 + x^3y + x) Function field in y defined by y^3 + x^3y + x sage: O = F.maximal_order() Maximal order of Function field in y defined by y^3 + x^3y + x sage: F.places_above(R.maximal_order().ideal(x).place()) ((y)) sage: O.ideal(x).factor() ((1)) · (y)3 sage: pl = O.ideal(x, y).place() (y) sage: xseries = puiseux(F, pl)(x) −s3 + O(s4) sage: yseries = puiseux(F, pl)(y) s + O(s2) sage: yseries((xseries^(1/3)).reverse()) −1 2i √ 3 + 1 2 s + O(s2) This result shows that we have a single cycle at (x, y) = (0, 0) with three sheets. Now, let’s look at a specimen root of 4x7 + 27 = 0: sage: g = QQbar(-27/4)^(1/7) 1 4 · 27 1 74 6 7 (−1) 1 7 sage: pl = O.ideal(x-g, y+3/(2g^2)).place() x + 1 4 · 27 1 74 6 7e(−6 7 i π), y + 1 8 · 8 6 73 1 7e(−2 7 i π) sage: xseries = F.completion(pl, prec=3)(x) 1 4 · 27 1 74 6 7 (−1) 1 7 + 3 7 · 48 1 7 (−1) 1 7 e( 4 7 i π) s2 + O(s3) sage: yseries = F.completion(pl, prec=2)(y) 1 8 · 8 6 73 1 7 (−1) 1 7 e( 4 7 i π) + s + O(s2) sage: pl = O.ideal(x-g, y-3/g^2).place() x + 1 4 · 27 1 74 6 7e(−6 7 i π), y + 48 1 7 (−1) 1 7 e( 4 7 i π) sage: xseries = F.completion(pl, prec=3)(x) 1 4 · 27 1 74 6 7 (−1) 1 7 + s + O(s3) sage: yseries = F.completion(pl, prec=2)(y) 48 1 7e(−2 7 i π) + 10 81 · 64 1 79 6 7 (−1) 1 7 e(−4 7 i π) s + O(s2) sage: yseries((xseries-g).reverse()) 48 1 7e(−2 7 i π) + 10 81 · 64 1 79 6 7 (−1) 1 7 e(−4 7 i π) s + O(s2) sage: R1. = QQ[] Q[g] sage: S. = NumberField(4g^7+27) Q[g]/(4g7 + 27) sage: R. = FunctionField(S) Rational function field in x over Number Field in g with defining polyno sage: L. = R[] Rational function field in x over Number Field in g with defining polyno sage: F. = R.extension(y^3 + x^3y + x) Function field in y defined by y^3 + x^3y + x sage: O = F.maximal_order() Maximal order of Function field in y defined by y^3 + x^3y + x sage: O.ideal(x-g).factor() ((1)) · ( x −g, y + 4 9g5 ) · ( x −g, y −2 9g5 )2 sage: pl = O.ideal(x-g, y+3/(2g^2)).place() x −g, y −2 9g5 sage: xseries = F.completion(pl, prec=3)(x) g + 4 21g5s2 + O(s3) sage: yseries = F.completion(pl, prec=2)(y) 2 9g5 + s + O(s2) sage: xseries((yseries+3/(2g^2)).reverse()) g + 4 21g5s2 + O(s3) sage: pl = O.ideal(x-g, y-3/g^2).place() x −g, y + 4 9g5 sage: xseries = F.completion(pl, prec=2)(x) g + s + O(s2) sage: yseries = F.completion(pl, prec=2)(y) −4 9g5 −40 81g4s + O(s2) sage: yseries((xseries-g).reverse()) −4 9g5 −40 81g4s + O(s2) We have one sheet of two cycles at (g, −3/(2g2)) and an ordinary point at (g, 3/g2).
Finally, let’s look at what happens when x goes to infinity: sage: Rinf = R.maximal_order_infinite().ideal(1/x).place() 1 x sage: Pinf = F.places_above(Rinf) 1 x, 1 x3y2 , 1 x2y, 1 x3y2 + 1 sage: xseries = puiseux(F, Pinf)(x) 1 s + O(1) sage: yseries = puiseux(F, Pinf)(y) −s2 + O(s3) sage: xseries = puiseux(F, Pinf)(x) −1 s2 + O(1) sage: yseries = puiseux(F, Pinf)(y) 1 s3 + O(1) Here we have an ordinary point at (∞, 0) and a single cycle of two sheets at (∞, ∞).
We have examined all of this curve’s ramification points, including those at infinity (since we analyzed all of its points at infinity), and found that all of them admitted a single Puiseux expansion.
Therefore, this curve is non-singular, and according to the genus-degree formula (MORE INFO), its geometric and arithmetic genus are the same. Its arithmetic genus is 1 2(d − 1)(d −2) = 3, where d = 4 is the degree of the defining polynomial. Computing the geometric genus is more difficult4, but we can verify our information with Sage, being careful to work in projective space: sage: PP. = ProjectiveSpace(QQ, 2) P2 Q 4 sage: C = Curve(y^3z + x^3y + xz^3) x3y + y3z + xz3 sage: C.is_singular() False sage: C.arithmetic_genus() 3 sage: C.geometric_genus() 3 □ Example 8.22. Find the principal parts of 1 y on the curve y2 = 1 −x2 The principal part of an algebraic function is the part of its series expansion with negative exponents. Theorem 8.7 states that an algebraic function is completely determined, up to adding a constant, by its principal parts.
The first step is to locate the function’s poles, which in this case is simply the places where the denominator is zero, and that’s just x = ±1. Now, if we use puiseux, we can just request a series truncated at the −1 term: sage: R. = FunctionField(QQbar) Rational function field in x over Algebraic Field sage: L. = R[] Rational function field in x over Algebraic Field[y] sage: F. = R.extension(y^2 + x^2 - 1) Function field in y defined by y^2 + x^2 - 1 sage: D = (1/y).divisor() 1 x, 1 xy −i + 1 x, 1 xy + i −(x −1, y) −(x + 1, y) sage: table([[p, F.completion(p, prec=1)(1/y)] \ for p,m in D.list() if m < 0]) (x −1, y) 1 s + O(1) (x + 1, y) 1 s + O(1) □ Example 8.23. Find the principal parts of x y dx on the curve y2 = 1 −x2 Differential forms are not functions, and have different series expansions, due to the pres-ence of the differential, which must be adjusted at ramification points.
Let’s expand x y at x = 1: sage: R. = FunctionField(QQbar) Rational function field in x over Algebraic Field sage: L. = R[] Rational function field in x over Algebraic Field[y] sage: F. = R.extension(y^2 + x^2 - 1) Function field in y defined by y^2 + x^2 - 1 sage: O = F.maximal_order() Maximal order of Function field in y defined by y^2 + x^2 - 1 sage: pl = O.ideal(x-1, y).place() (x −1, y) sage: F.completion(pl, prec=6)(x) 1 −1 2s2 −1 8s4 + O(s6) sage: F.completion(pl, prec=6)(x/y) 1 s −1 2s −1 8s3 + O(s5) Now x = t2 + 1, so dx = 2t dt. Thus, multiplying x y by dx and changing our variable to t will multiply all of the terms in our expansion by 2t: sage: dx = F.completion(pl, prec=6)(x).derivative() −s −1 2s3 + O(s5) sage: F.completion(pl, prec=6)(x/y) dx −1 + O(s4) Even though x y has a pole at x = 1, x y dx does not!
Its behavior at infinity also requires analysis.
sage: F.maximal_order().ideal(x - sqrt(QQbar(-1))); sage: Rinf = R.maximal_order_infinite().ideal(1/x).place() 1 x sage: Pinf = F.places_above(Rinf) 1 x, 1 xy −i , 1 x, 1 xy + i sage: F.completion(Pinf, prec=2)(x) 1 s + O(s) sage: F.completion(Pinf, prec=2)(y) i s + O(s) sage: F.completion(Pinf, prec=2)(x/y) −i + O(s2) sage: dxinf = F.completion(Pinf, prec=2)(x).derivative() −1 s2 + O(1) sage: F.completion(Pinf, prec=2)(x/y) dxinf i s2 + O(1) x y has no poles at infinity, and approaches the limiting values ±i as x and y approach infinity. The differential x y dx, on the other hand, requires us to multiply by dx, and since x = 1 t, dx = −1 t2 dt.
In short, while x y has poles only at (±1, 0), x y dx has poles only at infinity.
sage: D = (x/y).divisor() + F(x).differential().divisor() −2 1 x, 1 xy −i −2 1 x, 1 xy + i + (x, y −1) + (x, y + 1) sage: table([[p, F.completion(p, prec=2)(x/y) F.completion(p)(x).deriv 1 x, 1 xy −i i s2 + O(1) 1 x, 1 xy + i −i s2 + O(1) □ Consider, for example the lemniscate of Bernoulli, defined by the equation (x2 + y2)2 −(x2 −y2) = 0 sage: R. = FunctionField(QQbar) Rational function field in x over Algebraic Field sage: L. = R[] Rational function field in x over Algebraic Field[y] sage: F. = R.extension((x^2+y^2)^2 - (x^2-y^2)) Function field in y defined by y^4 + (2x^2 + 1)y^2 + x^4 - x^2 sage: O = F.maximal_order() Maximal order of Function field in y defined by y^4 + (2x^2 + 1)y^ sage: I = O.ideal(x,y) (x, y) sage: I.factor() ((1)) · ( x, 1 xy3 + 1 xy −1 ) · ( x, 1 xy3 + 1 xy + 1 ) sage: O.basis() 1, y, y2, 1 xy3 + 1 xy One characterization of the maximal finite order O is that it contains all functions with no poles at finite places. The first three elements in the basis are obvious, but why is the fourth so complicated? Isn’t y x in O?
sage: y/x in O False Doesn’t y x approach either 1 or −1 as it approaches the origin? Remember that we’re working in complex space. We’ve got four roots, not two. Let’s look at some examples of limiting values using some numerical examples: sage: set_verbose(-1) None sage: R. = CC[] C[u, v] sage: ideal(((u^2+v^2)^2 - (u^2-v^2)), (u-.0001)) u4 + 2.00000000000000u2v2 + v4 −u2 + v2, u −0.000100000000000000 C[u, v] sage: ideal(((u^2+v^2)^2 - (u^2-v^2)), (u-.0001)).variety() [{u: 0.000100000000000000, v: 0.0000999999980000001}, {v: 1.00000001500 sage: [d[v] for d in ideal(((u^2+v^2)^2 - (u^2-v^2)), (u-.0001)).variety [0.0000999999980000001, 1.00000001500000i, −1.00000001500000i, −0.0000999999980000001] sage: [d[v]/.0001 for d in ideal(((u^2+v^2)^2 - (u^2-v^2)), (u-.0001)).v [0.999999980000001, 10000.0001500000i, −10000.0001500000i, −0.999999980000001] sage: [d[v]/.00001 for d in ideal(((u^2+v^2)^2 - (u^2-v^2)), (u-.00001)) [0.999999999800000, 100000.000015000i, −100000.000015000i, −0.999999999800000] sage: [(d[v]^3+d[v])/.00001 for d in ideal(((u^2+v^2)^2 - (u^2-v^2)), (u [0.999999999900000, −0.0000300000024822111i, 0.0000300000024822111i, −0.999999999900000] sage: Fs = O.ideal(x).factor() ((1)) · ((x, y −i)) · ( x, 1 xy3 + 1 xy −1 ) · ( x, 1 xy3 + 1 xy + 1 ) · ((x, y + i)) sage: F.completion(Fs.place(), prec=7)(y) i + 3 2is2 −25 8 is4 + 203 16 is6 + O(s7) sage: F.completion(Fs.place(), prec=7)(y) s −2s3 + 6s5 −28s7 + O(s8) sage: F.completion(Fs.place(), prec=7)(y) −s + 2s3 −6s5 + 28s7 + O(s8) sage: F.completion(Fs.place(), prec=7)(y) −i −3 2is2 + 25 8 is4 −203 16 is6 + O(s7) How does one of these ideal generators factor?
sage: A. = QQbar[] Q[a] sage: (1/2a^2+1/2sqrt(QQbar(-1))a+1).factor() 1 2 · (a −i) · (a + 2i) 8.5 Divisors Given a function on an algebraic curve, we can ask at which places it has poles and zeros.
The location and strengths of a function’s poles and zeros are called its divisor.
For non-singular curves, the points and places are in one-to-one correspondence, and a function’s divisor can be described in terms of the points where its poles and zeros lie.
Thus, one way of defining a divisor is to associate integers (positive for zeros, negative for poles) with each point of the curve, subject to the stipulation that all but a finite number of those integers is zero. Such a description is called a Weil divisor, and is most suitable for working in intersection theory.
?? Definition 1.5.6 defines a Weil differential.
For singular curves, the situation is more complicated. A divisor needs to be associated with places, not points. Such a divisor is called a Cartier divisor, and is more suitable for our purposes.
8.6 Riemann-Roch spaces A Riemann-Roch space is a subspace of an algebraic curve’s function field characterized by specifying a minimum order that the function must obtain at all of the curve’s points.
Aside from having great theoretical significance, Riemann-Roch spaces are practically useful because they are finite dimensional, and algorithms exist for constructing Riemann-Roch bases. Finding a basis for a Riemman-Roch space in a crucial first step in solving a Mittag-Leffler problem.
Numerous algorithms have been developed for computing bases of Riemann-Roch spaces.
Sage uses an implementation of Hess’s algorithm from [He02].
Here’s a simple example5 of a Riemann-Roch space calculation: sage: R. = FunctionField(QQbar) Rational function field in x over Algebraic Field sage: L. = R[] Rational function field in x over Algebraic Field[y] sage: F. = R.extension(y^2 - x^3 + x) Function field in y defined by y^2 - x^3 + x sage: O = F.maximal_order() Maximal order of Function field in y defined by y^2 - x^3 + x sage: P = O.ideal(x,y) (x, y) sage: D = P.divisor() (x, y) sage: D.basis_function_space() 5From sage: (2D).basis_function_space() 1, 1 x sage: (3D).basis_function_space() 1, 1 x, 1 x2y sage: (4D).basis_function_space() 1, 1 x, 1 x2, 1 x2y Here are the examples from [Al14] §6.3: sage: R. = FunctionField(QQbar) Rational function field in x over Algebraic Field sage: L. = R[] Rational function field in x over Algebraic Field[y] sage: F. = R.extension(y^2 - x^3 - 1) Function field in y defined by y^2 - x^3 - 1 sage: O = F.maximal_order() Maximal order of Function field in y defined by y^2 - x^3 - 1 sage: P1 = O.ideal(x-2,y-3) (x −2, y −3) sage: P2 = O.ideal(x-2,y+3) (x −2, y + 3) sage: Rinf = R.maximal_order_infinite().ideal(1/x).place() 1 x sage: Pinf = F.places_above(Rinf) 1 x2y sage: D1 = P1.divisor() (x −2, y −3) sage: D2 = P2.divisor() (x −2, y + 3) sage: Dinf = Pinf.divisor() 1 x2y sage: (Dinf-D1).basis_function_space() [] sage: (2Dinf-D1).basis_function_space() [x −2] sage: (3Dinf-D1).basis_function_space() [x −2, y −3] sage: (4Dinf-D1).basis_function_space() x2 −2x, x −2, y −3 sage: (Dinf).basis_function_space() sage: (2Dinf).basis_function_space() [x, 1] sage: (3Dinf).basis_function_space() [x, 1, y] 8.7 Mittag-Leffler Problems Theorem 8.7 tells us that a rational function on an algebraic curve is completely charac-terized, up to an additive constant, by the principal parts of the Puiseux expansions at its poles. Note that Theorem 8.7 does not guarantee the existence of a function with specified principal parts. It only shows that any two such functions, if they exist, differ by at most a constant.
A Mittag-Leffler problem is the practical application of this theorem – given a set of prin-cipal parts, find a function that matches them all, or prove that no such function exists.
The first step in solving a Mittag-Leffler problem is to identify the maximum strengths of the poles, and construct a basis for a Riemann-Roch space that includes all functions with poles of such strength. We now have a finite basis for a vector space that must include the function we are looking for. We construct Puiseux expansions for the basis functions, and use them to construct a matrix equation that, when solved, gives the coefficients needed to form the function we seek from the basis functions.
The input data is a set of principal parts or, alternately, a divisor combined with a vector of coefficients.
Let’s assume that we’ve got our data in the latter form, so we can run riemannroch on the divisor and obtain a set of basis functions. Now let’s construct a Sage function to extract the principal parts of the basis functions and form them into a matrix: def principal_parts_matrix(div, basis): F = div.parent().function_field() coeffs = [(puiseux(F,p), i) for p,m in div.list() for i in range return matrix( for c in coeffs] for b in basis]).
Given a vector b of coefficients, we now want to solve a matrix equation: m · v = b This will typically be an overspecified system – a non-square matrix that may or may not have a solution. That’s fine; since some integrals have no elementary form, this doesn’t represent a limitation in our theory. Failure to solve this matrix equation would only show that no function exists on this curve with the coefficients b.
Example 8.24. Let’s say that we’ve identified a divisor on an algebraic curve (example ??): We now compute its principal parts matrix: sage: R. = FunctionField(QQbar) Rational function field in x over Algebraic Field sage: L. = R[] Rational function field in x over Algebraic Field[y] sage: F. = R.extension(y^2 - x^8 - 1) Function field in y defined by y^2 - x^8 - 1 sage: y.divisor(); sage: O = F.maximal_order() Maximal order of Function field in y defined by y^2 - x^8 - 1 sage: Oinf = F.maximal_order_infinite() Maximal infinite order of Function field in y defined by y^2 - x^8 - 1 sage: Dfinite = add([O.ideal(x-aQQbar(-1).sqrt(), y-bQQbar(2).sqrt()) x −i, y − √ 2 + x −i, y + √ 2 + x + i, y − √ 2 + x + i, y + √ 2 sage: Rinf = R.maximal_order_infinite().ideal(1/x).place() 1 x sage: Dinf = add([pl.divisor() for pl in F.places_above(Rinf)]) 1 x, 1 x4y −1 + 1 x, 1 x4y + 1 sage: D1 = Dfinite + 2Dinf 2 1 x, 1 x4y −1 + 2 1 x, 1 x4y + 1 + x −i, y − √ 2 + x −i, y + √ 2 + x + i, y − √ 2 + x + i, y + √ 2 sage: basis = Dfinite.basis_function_space() x2 x2 + 1, x x2 + 1, 1 x2 + 1 sage: D1.basis_function_space() x4 x2 + 1, x3 x2 + 1, x2 x2 + 1, x x2 + 1, 1 x2 + 1, 1 x2 + 1 y sage: principal_parts_matrix(D1, basis) 0 0 0 0 0 0 0 0 0 0 0 0 1 2i 1 2 −1 2i 1 2i 1 2 −1 2i −1 2i 1 2 1 2i −1 2i 1 2 1 2i TODO Introduce a sample vector b and show how to proceed.
□ 8.8 Parallels with the Transcendental Cases At this point, it may seem that we’ve spent this entire chapter developing a suite of tech-nical tools that appear completely different from everything that came before them. Why should the algebraic case be so much different from the transcendental cases? What would happen if we used here the same kind of techniques from earlier in the book?
First, the key difference in the algebraic case is the lack of unique factorization. Algebraic extensions are not, in general, unique factorization domains, a classic example being the factorization of 6 into either 3 · 2 or (1 + √−5) · (1 −√−5) in the ring Z[√−5]. You can show that all four numbers 3, 2, (1 + √−5) and (1 −√−5) are all prime in Z[√−5], so we have two distinct factorizations in this ring.
Show an example in a function field.
The main problem with our earlier tools is the difficulty in defining factorization. How, for example, do you construct a partial fractions expansion? A review of Theorems 6.1 and 7.1 reveals that both depend not merely on the construction of a partial fractions ex-pansions, but also on its uniqueness. Without unique factorization, how can you possibly have a unique partial fractions expansion?
The primary goal of this chapter is to develop techniques to carry out the same kinds of operations we did earlier, but without relying on unique factorization.
For example, a principle parts expansion of a function on an algebraic curve is exactly analogous to a partial fractions expansion of a rational function.
Demonstrate Although we began our development using infinite series expansions, we ultimately con-cluded that we can completely specify a function (up to an additive constant), using only a finite number of constants – the principle parts coefficients, which turn out to align precisely with the coefficients in a partial fractions expansion.
Reassembling a partial fractions expansion into a rational function is easy – you just pro-mote all the fractions to a common denominator, add up the terms, and cancel any com-mon factors that remain between the numerator and the denominator. Solving a Mittag-Lefler problem is considerably more difficult, but is in principle the same operation – given the principle parts coefficients (resp. the partial fractions expansion), construct a single rational function that matches. The major caveat here is that, unlike reassembling a partial fractions expansion, there might be not solution. Not every principle parts expan-sion has a matching algebraic function.
Likewise, finding an algebraic function’s divisor is exactly analogous to factoring the nu-merator and denominator of a rational function. You get a finite set of poles and zeros with their locations and multiplicities. Again, in the algebraic case, it’s more complicated – you might have singularities with multiple places lying over a single point; the “coor-dinate” is more complicated that a simple (x, y) coordinate pair, but the principle is the same.
And finding a function with a specified set of poles and zeros is the same as taking a ra-tional function in factored form and multiplying the factors together again. Again, there’s a caveat – in the algebraic curve case there might be no solution.
So, if the tools we’ve developed in this chapter parallel neatly with the tools we used in Chapter 5 to solve integrals of rational functions, can we generalize these tools to handle more complicated transcendental fields, like we did in Chapters 6 and 7? And do we have anything like the Hermite reduction procedure we developed at the end of Chapter 5?
The answers to both of these questions is ’yes’. For the purpose of a clear exposition, I’ve developed this theory so far in its simplest form, and if you’re seeing it for the first time, I suspect that you already appreciate not having met it in its full generality! We can drop the assumption of an algebraically closed coefficient field and lose very little except simplicity; this will be the subject of Chapter 11. Barry Trager showed in [Tr84] how the Hermite reduction can be performed in an algebraic extension; it’s now called Hermite-Trager reduction and I’ll present it at the end of Chapter 9.
However, continuing with the intent of presenting the theory in its simplest form first, we’ll begin the next chapter by looking at how to use these tools to integrate Abelian integrals, much like we first met partial fractions expansion when learning to integrate in first year calculus, and only later generalized it into a form suitable for integrating in arbitrary transcendental extensions. We’ll find that completely solving integrals in even this simplest of algebraic extensions will require a significant excursion into modern algebraic geometry, so much so that the entirety of Chapter 10 will be devoted to proving the book’s most exotic theorem.
If there’s a lesson to be learned from Chapter 8, though, it’s this: Divisors, principle parts expansions, Riemann-Roch spaces, and Mittag Leffler problems are how we do factorization, partial fractions expansions, and their inverse operations in algebraic extensions where we’ve lost unique factorization.
Chapter 9 Abelian Integrals THIS CHAPTER IS INCOMPLETE.
We can now use the machinery developed in the previous chapter to solve Abelian inte-grals in general.
These techniques, taken together with Liouville’s Theorem, provide our method of attack for integration of Abelian integrals. We find all of the poles of the differential, construct Puiseux expansions of the differential there, and split the principal parts easily into two sets. The first order poles arise from logarithm components in the solution; all the higher order poles must come from the rational function. To find the rational function, we inte-grate term-wise to obtain its principal parts, and then solve a Mittag-Leffler problem to see if such a function exists. For the logarithmic components, it’s a little more complicated.
Once we’ve calculated the principal parts of an integrand, we can integrate the resulting series to obtain the principal parts of the integral. This is possible due to the simple but crucial observation that poles in the integral can only appear where there are poles in the integrand. Once we’ve determined the principal parts of the integral, we need to solve a Mittag-Leffler problem to find an algebraic function that matches the given principal parts. This can be done by finding a basis for a suitable Riemann-Roch space.
Thus, having identified the locations and orders of the integral’s principal parts, we can compute a Riemann-Roch basis for all functions with suitable poles at those locations.
Having done so, it then becomes a straightforward exercise in linear algebra to find a combination of those basis functions that match a specified set of principal parts.
9.1 The Abelian Integration Theorem Theorem 9.1. Let C be the complex field, let C(x) be the rational function field in x over C, let C(x, y) be an algebraic extension of C(x), and let f be an element of C(x, y) with pole divisor (f)∞and principle parts expansion: ∀P ∈(f)∞ f = X νP <i<0 aP,i ti + · · · If f has an elementary anti-derivative F, then F ∈C(x, y, Ψ), where C(x, y, Ψ) is a finite logarithm extension of C(x, y), F can be written as the sum of an algebraic function in A ∈C(x, y) and logarithms of algebraic functions Bj ∈C(x, y): F = A + X j cj ln Bj A’s pole divisor is a subset of f’s pole divisor, and its principle parts expansion has the form: ∀P ∈(f)∞ A = X (νP +1)<i<0 AP,i ti + · · · AP,i = 1 i aP,i−1 (what about ramification and infinity?) The residues aP,−1 can be written in terms of a Z-module in C, with basis {cj}: aP,−1 = X j bP,jcj bP,j ∈Z and the divisors of the Bj can be written: (Bj) = X P bP,jP (and any such basis is as good as any other) Proof By Theorem 4.15, an elementary antiderivative of f can only exist in a finite logarithm extension C(x, y, Ψ) of C(x, y) and therefore must have the form: F = A + X j cj ln Bj where A, Bj ∈C(x, y), and the cj are constants. Differentiating, we obtain: f = dF dx = dA dx + cj dBj dx Bj Consider a place P with uniformizing variable s.
f = dA ds ds dx + cj dBj ds ds dx Bj Let A’s Pusieux expansion at P be Aasa+Aa+1sa+1+· · ·, let Bj’s Pusieux expansions at P be Bj,kskj +Bj,k+1skj+1+· · ·, and let x’s Pusieux expansion at P be Ccsc+Cc+1sc+1+· · ·.
Substituting: f = aAasa−1 + (a + 1)Aa+1sa + · · · cCcsc−1 + (c + 1)Cc+1sc + · · · +kjBj,kskj−1 + (kj + 1)Bj,k+1skj + · · · cCcsc−1 + (c + 1)Cc+1sc + · · · 1 Bj,kskj + Bj,k+1skj+1 + · · · □ 9.2 Some Simple Examples Example 9.2. Evaluate R x √ x2+1 dx This is a simple integral that can be easily solved using first year calculus techniques, but let’s see how to attack it using the more sophisticated techniques of this chapter.
First, we convert the the integrand into a rational function on an algebraic curve: sage: R. = FunctionField(QQbar) Rational function field in x over Algebraic Field sage: L. = R[] Rational function field in x over Algebraic Field[y] sage: F. = R.extension(y^2 - (x^2+1)) Function field in y defined by y^2 - x^2 - 1 Next, we identify the poles of the integrand. The finite poles can only be located where the denominator is zero, which is where x2 + 1 = 0, over the points x = ±i.
Let’s compute the principal parts of x y at x = i. Calling puiseux with deg=-1 com-putes just the principal part of the differential: sage: D = (x/y).divisor() −(x −i, y) + (x, y −1) + (x, y + 1) −(x + i, y) sage: p = D.support() (x −i, y) sage: m = F.completion(p, prec=4) Completion map: From: Function field in y defined by y^2 - x^2 - 1 To: Laurent Series Ring in s over Algebraic Field sage: m(x/y) i s −1 2is + O(s3) There appears to be a pole here, but appearances are deceptive. We need to expand the differential, not the integrand: sage: m(x/y) m(x).derivative() 1 + O(s2) So, even though the integrand has a pole at (i, 0), the differential does not... and the differential is what matters!
The only other place we might have a pole is infinity.
sage: D2 = (x/y).divisor() + F(x).differential().divisor() −2 1 x, 1 xy −1 −2 1 x, 1 xy + 1 + (x, y −1) + (x, y + 1) sage: F.maximal_order_infinite().decomposition() 1 x, 1 xy −1 , 1, 1 , 1 x, 1 xy + 1 , 1, 1 Now it appears that we have two sheets with no poles, the expansions indicating simply that limx→∞ x √ x2+1 = ±1, depending on whether we use the positive or negative square root, but again we have to take the differential into account. Since x = 1 t, dx = −1 t2dt, and we actually have second order poles at infinity.
sage: m2 = F.completion(D2.support(), prec=2) Completion map: From: Function field in y defined by y^2 - x^2 - 1 To: Laurent Series Ring in s over Algebraic Field sage: m3 = F.completion(D2.support(), prec=2) Completion map: From: Function field in y defined by y^2 - x^2 - 1 To: Laurent Series Ring in s over Algebraic Field sage: m2(x/y) m2(x).derivative() −1 s2 + O(1) sage: m3(x/y) m2(x).derivative() 1 s2 + O(1) Integrating termwise, we see that since our differential has second order poles at infinity, our integral must have first order poles at infinity, and theorem ? states that this completely characterizes the integral.
What functions have first order poles at infinity and nowhere else?
sage: D3 = add([-(m+1)p for p,m in D2.list() if m < 0]) 1 x, 1 xy −1 + 1 x, 1 xy + 1 sage: D3.basis_function_space() [x, 1, y] Our solution, if it exists, is in the vector space spanned by these three basis elements.
1 is in the list, and we expect it to be there, because of the presence of the constant of integration, so we can always add a multiple of 1 to our solution and get another solution.
What about x and y? We want to combine them in such a way as to match the principal parts of our differential.
Let’s expand them: sage: [m2(x), m3(x)] 1 s + O(s), 1 s + O(s) sage: [m2(y), m3(y)] 1 s + O(s), −1 s + O(s) No linear algebra games are required to see that y will match the principal parts of the differential if it is itself differentiated. Therefore, y = √ x2 + 1 is our solution.
□ Example 9.3. Compute R 1 √ 1−x2 dx This is a familiar example from first year calculus, but let’s approach it using the tech-niques of this book. We’ll use the algebraic extension y2 = 1 −x2 and integrate 1 y dx.
sage: R. = FunctionField(QQbar) Rational function field in x over Algebraic Field sage: L. = R[] Rational function field in x over Algebraic Field[y] sage: F. = R.extension(y^2 - (1-x^2)) Function field in y defined by y^2 + x^2 - 1 sage: integrand = 1/y x.differential() −1 x2 −1 y dx sage: D = (1/y).divisor() + F(x).differential().divisor() − 1 x, 1 xy −i − 1 x, 1 xy + i sage: table([[pl, F.completion(pl)(1/y, prec=2) F.completion(pl)(x for pl,m in D.list() if m < 0]) 1 x, 1 xy −i i s + O(s) 1 x, 1 xy + i −i s + O(s) Our only poles are at infinity, and they’re first order poles, so this solution will be a logarithm.
The poles’ residues (coefficients) are −i and i. These exist in the field Q[i], which can be regarded as a vector field over Q with basis {1, i}, and we want to construct a func-tion whose poles and zeros match the i-component of the residues (the 1-component is uniformly zero), with the signs flipped due to the effect of differentiation of a negative power.
So, we have a singular point at infinity, and we want a function with a simple zero on one cycle and a simple pole on the other. The basis will have either one element or no elements, depending on whether an algebraic function exists with the desired properties.
sage: def logarithmic_divisor(differential, component): return add([- ZZ(differential.residue(pl) / component) pl.di sage: D = logarithmic_divisor(integrand, QQbar(-1).sqrt()) − 1 x, 1 xy −i + 1 x, 1 xy + i sage: D.basis_function_space() [y −ix] Yes, it does exist. Remembering that our residues came multiplied by a factor of i, we conclude that our solution is i ln(y −ix), or: Z 1 √ 1 −x2 dx = i ln √ 1 −x2 −ix = −i ln 1 √ 1 −x2 −ix = −i ln √ 1 −x2 + ix 1 −x2 + x2 = −i ln √ 1 −x2 + ix = arcsin x where I used the negative of a logarithm being the logarithm of the inverse, and the last transformation came from section 4.2.
□ Example 9.4. Compute R √ 4 −x2 dx A solution method from first year calculus might be to note that this integrand forms one leg of a right triangle with its other sides 2 and x, but we’ll attack this integral using the methods of this chapter.
First, transform the problem into an algebraic curve: sage: R. = FunctionField(QQbar) Rational function field in x over Algebraic Field sage: L. = R[] Rational function field in x over Algebraic Field[y] sage: F. = R.extension(y^2 - (4-x^2)) Function field in y defined by y^2 + x^2 - 4 With no denominator, there can be no poles at finite points, so we just need to check infinity: sage: integrand = y x.differential() (y) dx sage: D = sum([-m pl.divisor() for pl,m in integrand.divisor().lis 3 1 x, 1 xy −i + 3 1 x, 1 xy + i sage: table([[pl, F.completion(pl)(y, prec=2) F.completion(pl)(x) for pl in D.support()]) 1 x, 1 xy −i −i s3 + O(s−1) 1 x, 1 xy + i i s3 + O(s−1) Since these polar expansions have components at both t−3 and t−1, we’ll get a result with both algebraic and logarithmic components.
Let’s start with the algebraic part. The integration step lowers the order of the poles by one, so we want a divisor at the same places as the divisor of the poles, but with multiplicity one less: def reduced_divisor(D): return D - add([pl.divisor() for pl in D.support()]) sage: D2 = reduced_divisor(D) 2 1 x, 1 xy −i + 2 1 x, 1 xy + i We need a Riemann-Roch space with at most second order poles at infinity: sage: basis = D2.basis_function_space() x2, x, 1, xy, y Now let’s construct a matrix of principal parts: sage: m = principal_parts_matrix(D2, basis) 1 0 0 i 0 0 1 0 0 i 1 0 0 −i 0 0 1 0 0 −i Another Sage function will extract the principal parts of the differential, and divide them by the powers needed for term-wise integration: def solution_vector(div, differential): F = differential.parent().function_field() coeffs = [(puiseux(F, p), i) for p,m in div.list() for i in range(-m return matrix().t sage: b = solution_vector(D2, integrand) 1 2i 0 −1 2i 0 sage: v = m.solve_right(b) 0 0 0 1 2 0 sage: m v == b True sage: algebraic_solution = (matrix(basis) v)[0,0] 1 2xy This is the rational part of our answer. To find the logarithmic part, let’s return to the series expansion of the differential, note that the residues are −2i and 2i, which can be placed in the Z-module {2i}, and search for: sage: pls = D.support() 1 x, 1 xy −i , 1 x, 1 xy + i sage: basis2 = (pls.divisor() - pls.divisor()).basis_function_ [y + ix] sage: log_solution = basis2 y + ix The logarithmic component of our solution is thus: sage: var('x') x sage: F = FractionField(PolynomialRing(QQbar, x)) Frac(Q[x]) sage: log_solution.element().change_ring(F).subs({y:sqrt(4-x^2)}) i x + √ −x2 + 4 Remembering from the previous example that i ln ix − √ 1 −x2 = arcsin x, and that logarithmic components are only specified up to a multiplicative constant (it disappears into the constant of integration), we can rewrite this: 2i ln √ 4 −x2 −ix = 2i ln r 4 −4 x 2 2 −2i x 2 !
= 2i ln ix 2 − r 1 − x 2 2 !
= 2 arcsin x 2 Adding in the rational component we computed earlier, the final answer is: Z √ 4 −x2 dx = 2 arcsin x 2 + x √ 4 −x2 2 □ 9.3 An integral Sage can’t solve Example 9.5. Integrate R x9+2x7−x (x4+2x2+1) √ x8+1 dx When I say that Sage can’t solve this integral, I mean that its built-in integration routine can’t solve the integral: sage: y = sqrt(x^8+1); sage: integrand = (x^9+2x^7-x)/((x^4+2x^2+1)y) x9 + 2 x7 −x √ x8 + 1(x4 + 2 x2 + 1) sage: integrate(integrand,x) Z (x8 + 2 x6 −1)x √ x8 + 1(x2 + 1)2 dx Now let’s attack the problem using the techniques of this book.
sage: R. = FunctionField(QQbar) Rational function field in x over Algebraic Field sage: L. = R[] Rational function field in x over Algebraic Field[y] sage: F. = R.extension(y^2 - (x^8+1)) Function field in y defined by y^2 - x^8 - 1 sage: integrand = (x^9+2x^7-x)/((x^4+2x^2+1)y) x9 + 2x7 −x x12 + 2x10 + x8 + x4 + 2x2 + 1 y Where are our poles?
def differential_divisor_of_poles(differential): return sum([-m pl.divisor() for pl,m in differential.divisor() sage: D = differential_divisor_of_poles(integrand x.differential()) 3 1 x, 1 x4y −1 + 3 1 x, 1 x4y + 1 + 2 x −i, y − √ 2 + 2 x −i, y + √ 2 + 2 x + i, y − √ 2 + 2 x + i, y + √ 2 sage: table([[pl, puiseux(F, pl)(integrand x.differential())] for pl i 1 x, 1 x4y −1 −1 s3 + O(1) 1 x, 1 x4y + 1 1 s3 + O(1) x −i, y − √ 2 1 2i q 1 2 s2 + O(1) x −i, y + √ 2 −1 2i q 1 2 s2 + O(1) x + i, y − √ 2 −1 2i q 1 2 s2 + O(1) x + i, y + √ 2 1 2i q 1 2 s2 + O(1) We’ve found four second order poles at the ordinary points (±i, ± √ 2), as well as two third order poles at a singular point with two sheets at infinity.
Our next goal is to construct a basis for a suitable Riemann-Roch space. We invert the signs of the finite poles, since the convention for Riemann Roch spaces is their functions must have order greater than the negative of a divisor, remember that poles decrease in order by 1 when they are integrated, and conclude that the Riemann-Roch space that we’re interested in is: L(Z(i, √ 2)Z(−i, √ 2), Z(i, − √ 2)Z(−i, − √ 2)Z2(∞, ∞)) i.e, all the functions on our algebraic curve with at most first order poles at (±i, ± √ 2), no other finite poles, and at most second order poles at infinity.
sage: D = reduced_divisor(D) 2 1 x, 1 x4y −1 + 2 1 x, 1 x4y + 1 + x −i, y − √ 2 + x −i, y + √ 2 + x + i, y − √ 2 + x + i, y + √ 2 We now wish to see if a linear combination of these basis functions will match the poles in the differential. This problem is a bit more complicated than the last one, so let’s use the tools we developed in the last chapter for solving Mittag-Leffler problems: sage: basis = D.basis_function_space() x4 x2 + 1, x3 x2 + 1, x2 x2 + 1, x x2 + 1, 1 x2 + 1, 1 x2 + 1 y sage: m = principal_parts_matrix(D, basis) 1 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0 0 −1 0 1 0 0 0 0 −1 2i −1 2 1 2i 1 2 −1 2i −1 2i √ 2 −1 2i −1 2 1 2i 1 2 −1 2i 1 2i √ 2 1 2i −1 2 −1 2i 1 2 1 2i 1 2i √ 2 1 2i −1 2 −1 2i 1 2 1 2i −1 2i √ 2 sage: b = solution_vector(D, integrand x.differential()) 1 2 0 −1 2 0 −1 2i q 1 2 1 2i q 1 2 1 2i q 1 2 −1 2i q 1 2 Now we have a matrix equation that we want to solve: 1 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0 0 −1 0 1 0 0 0 0 −1 2i −1 2 1 2i 1 2 −1 2i −1 2i √ 2 −1 2i −1 2 1 2i 1 2 −1 2i 1 2i √ 2 1 2i −1 2 −1 2i 1 2 1 2i 1 2i √ 2 1 2i −1 2 −1 2i 1 2 1 2i −1 2i √ 2 · v = 1 2 0 −1 2 0 −1 2i q 1 2 1 2i q 1 2 1 2i q 1 2 −1 2i q 1 2 To find out if there actually is a solution, we simply have to check: sage: v = m.solve_right(b) 0 0 0 0 0 1 2 sage: m v == b True So, yes, this system does have a solution. Now let’s multiply our solution vector by the original basis: sage: solution = (matrix(basis) v)[0,0] 1 2 x2 + 1 y ...and convert back to our original form: sage: var('x') x sage: F = FractionField(PolynomialRing(QQbar, x)) Frac(Q[x]) sage: solution = solution.element().change_ring(F).subs({y:sqrt(x^8+ 1 2 √ x8 + 1 x2 + 1 Now we can verify the solution: sage: integrand = integrand.element().change_ring(F).subs({y:sqrt(x^ (x9 + 2 x7 −x) √ x8 + 1 x12 + 2 x10 + x8 + x4 + 2 x2 + 1 sage: bool(solution.diff(x) == integrand) True □ 9.4 Geddes’s example Example 9.6. Compute R 1 x √ x4+1 dx Two traditional solution techniques are either the substitution x2 = tan u, followed by the half angle formula for tan, or x4 + 1 = u2, which leads to a rational function and a partial fractions expansion.
We’ll use C(x, y); y2 = x4 + 1 and integrate 1 xy.
sage: R. = FunctionField(QQbar) Rational function field in x over Algebraic Field sage: L. = R[] Rational function field in x over Algebraic Field[y] sage: F. = R.extension(y^2 - (x^4+1)) Function field in y defined by y^2 - x^4 - 1 sage: integrand = 1/(xy) x.differential() 1 x5 + x y dx sage: D = differential_divisor_of_poles(integrand) (x, y −1) + (x, y + 1) sage: table([[pl, puiseux(F, pl)(integrand)] for pl in D.support()]) (x, y −1) 1 s + O(1) (x, y + 1) −1 s + O(1) Our only poles are a pair of first order poles at x = 0. This will give rise to a logarithmic term; can we find a function that matches?
sage: pls = D.support() [(x, y −1) , (x, y + 1)] sage: D2 = (pls.divisor() - pls.divisor()) (x, y −1) −(x, y + 1) sage: D2.basis_function_space() [] Since we can’t find a function that matches that divisor; let’s try doubling the strength of the poles.
sage: log_solution = (2D2).basis_function_space() 1 x2y + 1 x2 This is the function we are looking for inside our logarithm.
sage: var('x') x sage: F = FractionField(PolynomialRing(QQbar, x)) Frac(Q[x]) sage: 1/2log(log_solution.element().change_ring(F).subs({y:sqrt(x^4 1 2 log √ x4 + 1 x2 + 1 x2 !
Sage’s built-in integrator produces the result in a different form: sage: integrate(1/(xsqrt(x^4+1)),x) −1 4 log √ x4 + 1 + 1 + 1 4 log √ x4 + 1 −1 We can obtain this result from our algorithm by factoring 1 4 out of the residues and using fourth-order poles: sage: (4D2).basis_function_space() 1 x4y + 1 2x4 + 1 x4 which is another way of writing y−1 y+1 (remember that y2 = x4 + 1).
We conclude that: Z 1 x √ x4 + 1 dx = 1 2 ln √ x4 + 1 −1 x2 = 1 4 ln √ x4 + 1 −1 √ x4 + 1 + 1 □ 9.5 Holliman’s Integral Compute: Z 1 (x2 + 1)3/4 dx sage: R. = FunctionField(QQbar) Rational function field in x over Algebraic Field sage: L. = R[] Rational function field in x over Algebraic Field[y] sage: root = x^2+1 x2 + 1 sage: F. = R.extension(y^4 - root) Function field in y defined by y^4 - x^2 - 1 sage: integrand = y/(x^2+1) x.differential() 1 x2 + 1 y dx sage: D = integrand.divisor() 0 Any non-constant function will have poles that will increase in order under derivation, and any constant function will have a zero differential, which has no divisor.
Therefore, this integrand is not an exact differential (i.e, it is not the differential of a function), and the integral is non-elementary.
If we’re not just going to take the computer’s word that the integrand’s divisor is zero, how can we compute it by hand?
First, looking at the integrand, we see that it’s only possible finite poles are over the ramification points of the root, and we know that y is a uniformizing variable at those points. So, changing variables to y, we see...
y4 = (x2 + 1) = ⇒ 4y3 dy = 2x dx = ⇒ dx = 2y3 x dy Z y x2 + 1dx = Z y x2 + 1 2y3 x dy = Z 2 xdy The ramification points are at x = ±i, so the denominator (x) is finite and not zero, so there are no poles at the ramification points.
What about infinity? Here, 1 y is a uniformizing variable (WHY?), so let v = 1 y, and Z 2 x dy = Z 2 x −1 v2dv = − Z 2 xv2 dv y4 = x2 + 1 = ⇒ 1 v4 = x2 + 1 = ⇒ 1 x = v2 √ v4 −1 1 x = ± v2 i + i 2v4 + · · · = ⇒ 2 xv2 dv = ± 2i + iv4 + · · · dv so we see that the differential has no pole at infinity, just as the computer reported.
We’ve got a singularity at infinity with two places, 1 x being doubly ramified at both: sage: Rinf = R.maximal_order_infinite().ideal(1/x).place() 1 x sage: pls = [pl for pl in F.places_above(Rinf)] 1 xy, 1 xy2 −1 , 1 xy, 1 xy2 + 1 sage: Dinf = add([pl.divisor() for pl in F.places_above(Rinf)]) 1 xy, 1 xy2 −1 + 1 xy, 1 xy2 + 1 sage: F.completion(pls)(1/x) s2 −1 2s6 + 7 8s10 −33 16s14 + 715 128s18 + O(s22) sage: F.completion(pls)(1/x) −s2 + 1 2s6 −7 8s10 + 33 16s14 −715 128s18 + O(s22) Just checking that we’ve got the same result as from the kash code: sage: F.completion(pls)(1/x)(F.completion(pls)(1/y).reverse()) s2 + 1 2s6 + 3 8s10 + 5 16s14 + 35 128s18 + O(s22) sage: F.completion(pls)(1/x)(F.completion(pls)(1/y).reverse()) −s2 −1 2s6 −3 8s10 −5 16s14 −35 128s18 + O(s22) 9.6 Chebyshev’s Integral Example 9.7. Compute: Z √ x4 + 4 x3 + 2 x2 + 1(6 x2 + 5 x + 7) 2 x6 + 8 x5 + 3 x4 −4 x3 −1 dx sage: R. = FunctionField(QQbar) Rational function field in x over Algebraic Field sage: L. = R[] Rational function field in x over Algebraic Field[y] sage: root = x^4+4x^3+2x^2+1 x4 + 4x3 + 2x2 + 1 sage: F. = R.extension(y^2 - root) Function field in y defined by y^2 - x^4 - 4x^3 - 2x^2 - 1 sage: num = 6x^2 + 5x +7 6x2 + 5x + 7 sage: den = 2x^6 + 8x^5 + 3x^4 + - 4x^3 - 1 2x6 + 8x5 + 3x4 −4x3 −1 sage: integrand = ynum/den x.differential(); sage: D = differential_divisor_of_poles(integrand) x −1 2 √ 2, y − √ 2 −1 2 + x −1 2 √ 2, y + √ 2 + 1 2 + x + 1 2 √ 2, y − √ 2 + 1 2 + x + 1 2 √ 2, y + √ 2 −1 2 sage: table([[pl, puiseux(F,pl)(integrand)] for pl in D.support()]) x −1 2 √ 2, y − √ 2 −1 2 5 2 s + O(1) x −1 2 √ 2, y + √ 2 + 1 2 −5 2 s + O(1) x + 1 2 √ 2, y − √ 2 + 1 2 −5 2 s + O(1) x + 1 2 √ 2, y + √ 2 −1 2 5 2 s + O(1) We have poles at x = ± 1 √ 2 and nowhere else.
Now we can see that all of our residues are ± 5 2, so we need only a single logarithmic term. It must have two poles and two zeros, at the coordinates we just calculated, and the obvious choice is for each of these poles and zero to have degree one. Let’s use our Riemann-Roch basis space algorithm to see if such a function exists: sage: D2 = add([QQ(integrand.residue(pl)).sign() pl for pl in D.su x −1 2 √ 2, y − √ 2 −1 2 − x −1 2 √ 2, y + √ 2 + 1 2 − x + 1 2 √ 2, y − √ 2 + 1 2 + x + 1 2 √ 2, y + √ 2 −1 2 sage: D2.basis_function_space() [] No such function exists. However, a function may exist with higher degree poles and zero.
Let’s see... how about a function with second degree poles and zeros? Third degree?
sage: (2D2).basis_function_space() [] sage: (3D2).basis_function_space() [] sage: (4D2).basis_function_space() [] sage: res = (5D2).basis_function_space() 8832144405782038592248267177275666366163674962715995632074977306653940238509894494379110300209483153718734316011631386 54452646977767999887956444300007264946891217495505640729169989333675512121334073234070902649759444530540734772537783 + 30176493386421965190181579522358526751059222789279651742922839131067629148242139522461960192382400775205675579706407237 1856835261941888796179314750630247734688990516596742348864696636278334963337491897281817780356797058491439055743538406 Turns out that fifth degree poles and zeros give us our solution. Leaving aside for a moment the question of finding this degree in some other way than multiplying the degree of the divisor by every larger integers (and how would we ever know when to stop?), let’s continue with the problem by coaxing our result into a more reasonable form: sage: num2 = R.to_bivariate_polynomial(res) 30176493386421965190181579522358526751059222789279651742922839131067629148242139522 1856835261941888796179314750630247734688990516596742348864696636278334963337491897 + 8832144405782038592248267177275666366163674962715995632074977306653940238509894494 54452646977767999887956444300007264946891217495505640729169989333675512121334073 + 2649643321734611577674480153182699909849102488814798689622493191996182071552968348 27226323488883999943978222150003632473445608747752820364584994666837756060667036 + 1510296693388728599274453687314138948613988418624435253084821119437823780785191958 23210440774273609952241434382878096683612381457459279360808707953479187041718648 + 1168787109698489773707520689792813182455659653399416755311255330247204758229476038 5989791167554479987675208873000799144158033924505620480208698826704306333346748 + 1207059735456878607607263180894341070042368911571186069716913565242705165929685580 14974477918886199969188022182501997860395084811264051200521747066760765833366870 + 7498490600508950764818778833507040744872960043345880291631655733349195262494900425 46420881548547219904482868765756193367224762914918558721617415906958374083437297 + 321195651556940136804761983013591733516152312810771707819793341385314960007143163 92841763097094439808965737531512386734449525829837117443234831813916748166874594 + 1325704875307883992696464903309077521561167611903670944374454093728756429800335163 18568352619418887961793147506302477346889905165967423488646966362783349633374918 + 3385655355549781460361835751289005440362742069041131658962074634217343758095459556 26526218027741268516847353580432110495557007379953462126638523375404785190535598 + 4613323427953484824684344888930323065259492888858655051820496479842241451248334890 92841763097094439808965737531512386734449525829837117443234831813916748166874594 + 1148178772751665016992274733045836627601277745153079432169747049865012231006286284 11605220387136804976120717191439048341806190728729639680404353976739593520859324 + 6682989267041742534801188830805254217063847388455103361603399495368148113805820167 26526218027741268516847353580432110495557007379953462126638523375404785190535598 + 1472024067630339765374711196212611061027279160452665938679162884442323373084982415 928417630970944398089657375315123867344495258298371174432348318139167481668745948 + 1472024067630339765374711196212611061027279160452665938679162884442323373084982415 66315545069353171292118383951080276238892518449883655316596308438511962976338996 + 294404813526067953074942239242522212205455832090533187735832576888464674616996483 663155450693531712921183839510802762388925184498836553165963084385119629763389963 + 294404813526067953074942239242522212205455832090533187735832576888464674616996483 105502003519425499782915610831264075834601733897542178912766854333996304735084766 + 294404813526067953074942239242522212205455832090533187735832576888464674616996483 928417630970944398089657375315123867344495258298371174432348318139167481668745948 sage: den2 = R._to_bivariate_polynomial(res) x10 −5 2x8 + 5 2x6 −5 4x4 + 5 16x2 −1 32 sage: ld = lcm([QQ(c).denominator() for c in den2.coefficients()]); sage: ln = lcm([QQ(c).denominator() for c in num2.coefficients()]); sage: R2. = QQ[] Q[x] sage: F2 = Frac(R2) Frac(Q[x]) sage: R3. = F2[] Frac(Q[x])[y] sage: f = y^2 - (x^4 + 4x^3 + 2x^2 + 1) y2 −x4 −4x3 −2x2 −1 sage: Q = R3.quo(ideal(f)) Univariate Quotient Polynomial Ring in ybar over Fraction Field of Univa sage: R2. = ZZ[] Z[x] sage: F2 = Frac(R2) Frac(Z[x]) sage: R3. = F2[] Frac(Z[x])[y] sage: f = y^2 - (x^4 + 4x^3 + 2x^2 + 1) y2 −x4 −4x3 −2x2 −1 sage: Q2 = R3.quo(ideal(f)) Univariate Quotient Polynomial Ring in ybar over Fraction Field of Univa sage: Q2(Q(res.element()(ln/ld))) 3011761242371675159956659107451002230861813162286154510537567261568993621331874022 + 3017649338642196519018157952235852675105922278927965174292283913106762914824213952 This is a lot more complicated than what Geddes found: A(x) = 1023x8 + 4104x7 + 5048x6 + 2182x5 + 805x4 + 624x3 + 10x2 + 28x B(x) = 1025x10+6138x9+12307x8+10188x7+4503x6+3134x5+1598x4+140x3+176x2+2 C(x) = 32x10 −80x8 + 80x6 −40x4 + 10x2 −1 y = √ x4 + 4x3 + 2x2 + 1 Z (2x6 + 4x5 + 7x4 −3x3 −x2 −8x −8) (2x2 −1)2√ x4 + 4x3 + 2x2 + 1 dx = (x + 1 2)y 2x2 −1 + 1 2 ln A(x)y −B(x) C(x) □ 9.7 The Risch Theorem: A First Look At this point, there is only one major missing piece in our integration theory for Abelian integrals — how do we limit the multiples of a divisor to a testable set? We’ve seen how to repeatedly raise a divisor to higher and higher powers, but how do we know when to stop? At what point can we declare that a divisor has no multiple that is principal?
We’ll attack this problem the way Robert Risch discovered in 1970, by mapping into a finite field, solving the corresponding problem there, then lifting the result back to the original field. The finite field will be the integers modulo a prime. Reducing modulo some primes changes the genus of the algebraic curve, while other primes leave the genus unchanged. Those primes at which the genus of the curve remains unchanged yield good reduction The key theorems as stated by [Tr73] on page 67: Theorem 9.8. The homomorphism between divisor class groups under good reduction is an isomorphism when restricted to divisors whose orders are relatively prime to the characteristic of the reduced function field.
□ Let p be the characteristic of the reduced field, Theorem 9.9. if the divisor D has order pkn where gcd(n, p) = 1, then the reduction of D must have order pjn for some j ≤k.
Proof Let the order of the reduction be pjm. Since reduction is a group homomorphism, we must have m|n and j ≤k. Since Dpk has order exactly n, its reduction must have order exactly n. But the order of its reduction is a divisor of m and thus n|m and so finally we have n = m.
□ Since good reduction preserves the part of the divisor’s order relatively prime to the char-acteristic, by picking two different primes we can completely reconstruct the divisor’s order in characteristic zero.
A property of algebraic curves over finite fields is that all divisors of total degree zero have some multiple that is principal (proof in Chapter 10 when we construct the Jacobian variety). Thus, to carry out this program, we need to compute bases for Riemann-Roch spaces in prime characteristic. Then, we can keep raising a divisor to higher and higher powers until we find its order. Do this for two different primes, each exhibiting good reduction, and then we can find the order of the original divisor in characteristic zero.
I’ll prove these claims in the next chapter (not yet written), but for now let’s just see how to use these theorems to fix the degree of the divisor in the last section. First, let’s determine the genus of the curve over Q (what about Q?): sage: PP. = ProjectiveSpace(QQ, 2) P2 Q sage: root = x^4+4x^3+2x^2+1 x4 + 4x3 + 2x2 + 1 sage: C = Curve((y^2-root).homogenize(z)) −x4 −4x3z −2x2z2 + y2z2 −z4 sage: C.geometric_genus() 1 Now, let’s check reduction mod 2: sage: PP. = ProjectiveSpace(GF(2), 2) P2 F2 sage: root = x^4+4x^3+2x^2+1 x4 + 1 sage: C = Curve((y^2-root).homogenize(z)) x4 + y2z2 + z4 sage: C.geometric_genus() 0 The genus changed, so we don’t have good reduction mod 2.
What about mod 3 and mod 5?
sage: PP. = ProjectiveSpace(GF(3), 2) P2 F3 sage: root = x^4+4x^3+2x^2+1 x4 + x3 −x2 + 1 sage: C = Curve((y^2-root).homogenize(z)) −x4 −x3z + x2z2 + y2z2 −z4 sage: C.geometric_genus() 1 sage: PP. = ProjectiveSpace(GF(5), 2) P2 F5 sage: root = x^4+4x^3+2x^2+1 x4 −x3 + 2x2 + 1 sage: C = Curve((y^2-root).homogenize(z)) −x4 + x3z + 3x2z2 + y2z2 −z4 sage: C.geometric_genus() 1 Since the genus is unchanged, both of these primes exhibit good reduction. Let’s do the divisor calculation, first modulo 3: sage: A. = GF(3)[] F3[a] sage: B. = GF(3^2, modulus=a^2-2) F32 sage: R. = FunctionField(B) Rational function field in x over Finite Field in b of size 3^2 sage: L. = R[] Rational function field in x over Finite Field in b of size 3^2[y] sage: root = x^4+4x^3+2x^2+1 x4 + x3 + 2x2 + 1 sage: F. = R.extension(y^2 - root) Function field in y defined by y^2 + 2x^4 + 2x^3 + x^2 + 2 sage: num = 6x^2 + 5x +7 2x + 1 sage: den = 2x^6 + 8x^5 + 3x^4 + - 4x^3 - 1 2x6 + 2x5 + 2x3 + 2 sage: integrand = ynum/den x.differential() x + 2 x6 + x5 + x3 + 1 y dx sage: D = integrand.divisor() 1 x2y + 1 + 1 x2y + 2 + (x + 2, y + b) + (x + 2, y + 2b) −(x + b, y + b + 2) −(x + b, y + 2b + 1) −(x + 2b, y + b + 1) −(x + 2b, y + 2b + 2) sage: D2 = add([ZZ(1 if B(integrand.residue(pl))==1 else -1) pl fo −(x + b, y + b + 2) + (x + b, y + 2b + 1) + (x + 2b, y + b + 1) −(x + 2b, y + 2b + 2) sage: D2.basis_function_space() [] sage: (2D2).basis_function_space() [] sage: (3D2).basis_function_space() [] sage: (4D2).basis_function_space() [] sage: (5D2).basis_function_space() x6 + 2x5 + 2x4 + 2x2 + 2x x10 + 2x8 + x6 + x4 + 2x2 + 1 y + x10 + 2x8 + x5 + x4 + x3 + x2 + 1 x10 + 2x8 + x6 + x4 + 2x2 + 1 So, we see that the divisor has order 5, mod 3. This means that the original divisor’s order is 5 times some multiple of 3.
Again, let’s find the divisor’s order mod 5: sage: A. = GF(5)[] F5[a] sage: B. = GF(5^2, modulus=a^2-2) F52 sage: R. = FunctionField(B) Rational function field in x over Finite Field in b of size 5^2 sage: L. = R[] Rational function field in x over Finite Field in b of size 5^2[y] sage: root = x^4+4x^3+2x^2+1 x4 + 4x3 + 2x2 + 1 sage: F. = R.extension(y^2 - root) Function field in y defined by y^2 + 4x^4 + x^3 + 3x^2 + 4 sage: num = 6x^2 + 5x +7 x2 + 2 sage: den = 2x^6 + 8x^5 + 3x^4 + - 4x^3 - 1 2x6 + 3x5 + 3x4 + x3 + 4 sage: integrand = ynum/den x.differential() 3 x4 + 4x3 + 2x2 + 1 y dx sage: D = integrand.divisor() 0 What does this mean? Let’s take a look at how this function behaves over Q: sage: R. = FunctionField(QQ) Rational function field in x over Rational Field sage: L. = R[] Rational function field in x over Rational Field[y] sage: root = x^4+4x^3+2x^2+1 x4 + 4x3 + 2x2 + 1 sage: F. = R.extension(y^2 - root) Function field in y defined by y^2 - x^4 - 4x^3 - 2x^2 - 1 sage: num = 6x^2 + 5x +7 6x2 + 5x + 7 sage: den = 2x^6 + 8x^5 + 3x^4 + - 4x^3 - 1 2x6 + 8x5 + 3x4 −4x3 −1 sage: integrand = ynum/den x.differential() 3x2 + 5 2x + 7 2 x6 + 4x5 + 3 2x4 −2x3 −1 2 y dx sage: D = integrand.divisor() − x2 −1 2, y −2x −1 2 − x2 −1 2, y + 2x + 1 2 + x2 + 5 6x + 7 6 sage: ppoly = [pl.prime_ideal().gens() for pl,m in D.list() if m < 0] sage: zpoly = [pl.prime_ideal().gens() for pl,m in D.list() if m > 0] sage: R2. = QQ[] Q[x] sage: zpoly5 = R2(str(zpoly)).numerator().change_ring(GF(5)) x2 + 2 sage: ppoly5 = R2(str(ppoly)).numerator().change_ring(GF(5)) 2x2 + 4 Pay attention to how the ideals’ polynomials behaves when reduced mod 5. x2 + 5 6x + 7 6 mod 5 = x2 + 2, and x2 −1 2 mod 5 = 2x2 + 4, so the pole set and the zero set collapse together mod 5, and we get a zero divisor.
Since a zero divisor is principal (it matches any constant, non-zero function), we conclude that D is principal mod 5, and the original divisor’s order is 1 times some multiple of 5.
Combine this with the previous fact that the divisor’s order is 5 times some multiple of 3, and we conclude that the divisor’s order is exactly 5 – the only divisor we actually need to check over Q.
Let’s take a look at mod 7: sage: PP. = ProjectiveSpace(GF(7), 2) P2 F7 sage: root = x^4+4x^3+2x^2+1 x4 + 4x3 + 2x2 + 1 sage: C = Curve((y^2-root).homogenize(z)) −x4 + 3x3z + 5x2z2 + y2z2 −z4 sage: C.geometric_genus() 1 So, we have good reduction mod 7, as well as the interesting property that √ 2 already exists in this field, since 32 = 2 mod 7. This causes two of the places in our Q divisor to collapse together and cancel each other out, so we get a divisor supported over only two places.
sage: R. = FunctionField(GF(7)) Rational function field in x over Finite Field of size 7 sage: L. = R[] Rational function field in x over Finite Field of size 7[y] sage: root = x^4+4x^3+2x^2+1 x4 + 4x3 + 2x2 + 1 sage: F. = R.extension(y^2 - root) Function field in y defined by y^2 + 6x^4 + 3x^3 + 5x^2 + 6 sage: num = 6x^2 + 5x +7 6x2 + 5x sage: den = 2x^6 + 8x^5 + 3x^4 + - 4x^3 - 1 2x6 + x5 + 3x4 + 3x3 + 6 sage: integrand = ynum/den x.differential() 3x x5 + 2x4 + x3 + 3x2 + x + 5 y dx sage: D = integrand.divisor() (x, y + 1) + (x, y + 6) −(x + 5, y + 1) −(x + 5, y + 6) sage: [(integrand.residue(pl), pl) for pl,m in D.list() if m < 0] [(1, (x + 5, y + 1)) , (6, (x + 5, y + 6))] sage: D2 = add([ZZ(1 if (integrand.residue(pl))==1 else -1) pl for pl, (x + 5, y + 1) −(x + 5, y + 6) sage: D2.basis_function_space() [] sage: (2D2).basis_function_space() [] sage: (3D2).basis_function_space() [] sage: (4D2).basis_function_space() [] sage: (5D2).basis_function_space() 2x3 + 5x2 x5 + 4x4 + 5x3 + 4x2 + 3x + 3 y + x5 + 6x4 + 3x3 + x2 + 6x + 6 x5 + 4x4 + 5x3 + 4x2 + 3x + 3 The divisor also has order 5, mod 7. This means that the original divisor’s order is 5 times some multiple of 7. Combine this with the previous fact that the divisor’s order is 5 times some multiple of 3, and we conclude that the divisor’s order is exactly 5 – the only divisor we actually need to check over Q.
9.8 Hermite reduction The previous sections in this chapter have laid out the theoretical framework for the solu-tion of Abelian integrals and presented the simplest computational algorithm that I could formulate to acheive that goal. It is by no means the most efficient algorithm, and should not be used as a basis for a professional implementation. The remaining sections of this chapter showcase optimization techniques, and can be skipped without loss of continuity in the text. In particular, Puiseux expansions can be completely avoided, which offers significant savings in computational complexity.
[Tr84] shows how to extend Hermite reduction into the algebraic case. This offers a means of calculating the rational parts of the integrals without going through the calculations described in this chapter.
Example 9.10. Compute: Z 2x6 + 4x5 + 7x4 −3x3 −x2 −8x −8 (2x2 −1)2√ x4 + 4x3 + 2x2 + 1 dx The polynomial under the square root is square-free: sage: R. = QQ[] Q[x, y] sage: F=Frac(R) Frac(Q[x, y]) sage: root = x^4+4x^3+2x^2+1 x4 + 4x3 + 2x2 + 1 sage: num = 2x^6 + 4x^5 + 7x^4-3x^3-x^2-8x-8 2x6 + 4x5 + 7x4 −3x3 −x2 −8x −8 sage: den = 2x^2-1 2x2 −1 sage: root.factor() (x + 1) · (x3 + 3x2 −x + 1) . . . so y2 = x4 + 4x3 + 2x2 + 1; {1, y} is an integral basis; and our normal form for this integral is: Z (2x6 + 4x5 + 7x4 −3x3 −x2 −8x −8)y (2x2 −1)2(x4 + 4x3 + 2x2 + 1) dx Applying now Bronstein’s Hermite reduction from section 2.1 of his “Symbolic Integra-tion Tutorial” with v = 2x2 −1 to eliminate this square in the denominator: sage: D = Derivation(F, {x: 1, y: root.derivative(x)/2y/root}) Derivation of Frac(Q[x, y]) x → 1 y → 2x3y+6x2y+2xy x4+4x3+2x2+1 sage: D(y) 2x3y + 6x2y + 2xy x4 + 4x3 + 2x2 + 1 sage: U = root x4 + 4x3 + 2x2 + 1 sage: V = den 2x2 −1 sage: S2 = UV^2D(y/V) −4x4y −6x3y −6x2y −6xy Now we want to solve f2S2 = A2y where A2y is our numerator.
sage: f2 = numy/S2 2x6 + 4x5 + 7x4 −3x3 −x2 −8x −8 −4x4 −6x3 −6x2 −6x sage: T2 = f2.numerator() 2x6 + 4x5 + 7x4 −3x3 −x2 −8x −8 sage: Q = f2.denominator() −4x4 −6x3 −6x2 −6x sage: R2 = QQ['x'] Q[x] sage: (A,R) = diophantine(R2(1),R2(V),R2(Q)) −36 49x3 −38 49x2 −48 49x −1, −18 49x + 8 49 sage: (Q2,B2) = (T2R).quo_rem(V) −18 49x5 −4 7x4 −8 7x3 + 41 49x2 −31 49x + 177 98 , x + 1 2 sage: h = Anumy/(VU) - (D(V)Q2+D(B2))y/V + Q2D(y) 6x2y + 5xy + 7y 2x6 + 8x5 + 3x4 −4x3 −1 sage: F3 = Frac(ZZ['x']['y']) Frac(Z[x][y]) Z (2x6 + 4x5 + 7x4 −3x3 −x2 −8x −8) y 2x2 −1 dx = (2x + 1) y 4x2 −2 + Z (6x2 + 5x + 7) y 2x6 + 8x5 + 3x4 −4x3 −1 dx We solved this new integral in example 9.7. The final answer is: A(x) = 1023x8 + 4104x7 + 5048x6 + 2182x5 + 805x4 + 624x3 + 10x2 + 28x B(x) = 1025x10+6138x9+12307x8+10188x7+4503x6+3134x5+1598x4+140x3+176x2+2 C(x) = 32x10 −80x8 + 80x6 −40x4 + 10x2 −1 y = √ x4 + 4x3 + 2x2 + 1 Z (2x6 + 4x5 + 7x4 −3x3 −x2 −8x −8) (2x2 −1)2√ x4 + 4x3 + 2x2 + 1 dx = (x + 1 2)y 2x2 −1 + 1 2 ln A(x)y −B(x) C(x) □ 9.9 Señor Gonzalez, otra vez The Rothstein-Trager resultant allows us to compute all the residues at once. Trager, in his Ph.D. thesis, then showed how to construct a function that is zero at all poles with a given residue, and non-zero at all other poles, as well as at all places conjugate to a pole.
Chapter 10 The Risch Theorem THIS CHAPTER IS VERY VAGUE AND INCOMPLETE.
We now turn to the task of proving the Risch theorem.
There are three key steps in this proof.
1. Establish the existence of the Jacobian variety.
It’s not that hard to see that divisor classes on an algebraic curve can be composed and inverted, forming the divisor class group. It’s not nearly so obvious that these divisor classes can be identified with the points on an algebraic variety. The resulting Jacobian variety is thus both a group and an algebraic variety, and the group opera-tions can be represented by rational functions on the algebraic variety, thus forming an Abelian variety, which is an algebraic variety equipped with a group structure.
2. Prove that the following diagram commutes: C J(C) Cp J(Cp) The objects on the left are algebraic curves; the objects on the right are Abelian vari-eties. The horizontal arrows correspond to constructing a curve’s associated Jacobian variety; the vertical arrows correspond to reduction modulo a prime.
At the end of the previous chapter, we were working on the lower-left half of this diagram. We reduced a curve modulo a prime, then computed the order of various divisors on the reduced curve, which essentially computes orders of points on the reduced curve’s Jacobian variety. We now wish to show that those group orders are the same as if we had constructed the Jacobian of the original curve and then reduced the Jacobian modulo the prime.
3. Study the effects of reduction modulo a prime on an Abelian variety.
At this point, we no longer need any special properties of the Jacobian variety; it suffices to study how reduction modulo a prime affects the group structure on an Abelian variety. In particular, we will find that the factor of an element’s order coprime to the reduction prime is preserved.
10.1 The Riemann-Roch Theorem The Riemann-Roch Theorem is one of the most celebrated theorems in mathematics. Not only does it provide a crucial tool in understanding the structure of algebraic extensions, but it does so by tying together algebra, analysis, and geometry in one equation.
First, let’s review that equation: Theorem 10.1. (Riemann-Roch) For any divisor b, l(b) = −deg b + 1 −g + l(−b −c) where l(b) is the dimension of the vector space L(b) of multiples of b, g is the genus of the extension, and c is any divisor of the canonical class of differentials.
□ Interrelated by this theorem is the purely algebraic concept of the dimension of the vec-tor space of multiples of a divisor, the geometric concept of the genus, and the analytic concept of a differential.
However, this sophistication comes with a price. Specifically, we need a topology to define the genus, and we need a limit to define the differential.
André Weil showed how the Riemann-Roch theorem can be stripped of the analysis and the geometry, and proved as purely a result in algebra. The genus, instead of a topological invariant, now appears as merely a least upper bound on a divisor’s degree of specializa-tion, and a differential becomes an object in a dual space that maps a function into the field of constants. The advantage of this formulation is that does not require any topological structure, and is therefore well suited to use with finite fields. It is this formulation I will now adopt.
First, at any place in the function field, there is a local valuation ring with a maximal prime ideal. We can normalize the valuation (it is discrete) and pick a element of unit valuation to use as a uniformizing variable. By multiplying as necessary by some power of this element, we can adjust any field element to be a unit of the valuation ring and thus associate an order ordp with that element. The valuation ring’s units are a finite extension of the constant subfield; they are the constant subfield if it is algebraically closed. By subtracting out the remainder mod p, we get an element of higher order, which we can again subtract out, and so on, building a power series in the uniformizing variable. Each element of the function field thus has a power series associated with it at each place p.
A collection of such power series, one at each place, with arbitrary coefficients except that there are only a finite number of coefficients with negative powers, is called a vector.
Each individual power series is called a component of the vector. Clearly, every function has a vector associated with it; but the converse is not necessarily true. The mapping from functions to vectors is injective, though. Any two different functions will have a non-zero difference that must therefore have a finite value, of finite order, at some place p, and their vectors will differ at that point.
We also have a dual space of covectors. The coefficients of a covector component at a place are dual to the constant field at that place; if the constant field is algebraically closed, then the covector coefficients are in the constant field. Like vectors, covectors can only have a finite number of negative power coefficients.
We define a dot product between a vector v and a covector λ: v · λ = X p X i+j=−1 vp,iλp,j where vp,i is the coefficient of the ith power in v’s component at p, and likewise for λp,j.
Notice that the second summation requires at least one of i or j to be negative, so there will only be a finite number of places for the first sum at which the second sum contributes anything at all.
Weil also requires the Theorem of Independence, which states that, although an arbitrary (full) vector may not have a function associated with it, a function can always be found which matches a set of finite prefixes at a finite number of places. This can be demon-strated using Theorem 11.22, repeatedly applied a finite number of times. We also need to know that a function without a pole is constant.
With this setup, we can now prove a series of theorems that lead up the the Riemann-Roch Theorem.
Theorem 10.2.
l(p) ≤deg p + 1 i.e, l(p), the dimension (over the constants) of L(p), the multiples of −p, is no more than the degree of the divisor p, plus one.
Proof Since deg −p = −deg p, there are at least deg p poles (counting multiplicities) in −p, and at least deg p coefficients with negative powers in the vectors corresponding to the elements in L(p) . We can impose □ Theorem 10.3. If A is divisible by B, i.e, if AB−1 ⊆I, then n(A) −l(A) ≤n(B) −l(B) n(p) ≡degp Proof Consider C = AB−1. Now n(C) = n(A) −n(B) and since deg −C = −deg C, and C is integral (by supposition), there are exactly n(C) poles (counting multiplicities) in −C.
, and at least deg p coefficients with negative powers in the vectors corresponding to the elements in L(p) . We can impose □ It immediately follows (from b = 0) that l(c) = g, which can be taken as the definition of the genus.
10.2 Jacobian Varieties An algebraic extension is a simple example of what algebraic geometers term a variety, which is the zero locus of a set of polynomials defined over some field. Thus, for example, the unit circle is a variety (defined over the real numbers), because it is the zero locus of x2 + y2 = 1. But the points (1, 0) and (−1, 0) are also a variety, because they are the zero locus of the set of polynomials {x2 = 1; y = 0}.
An abelian variety is a variety accompanied by a commutative group structure on its elements, which typically includes picking an arbitrary zero point as the identity element.
The circle is an abelian variety, if we identify its points with their angles from the x-axis and make (1, 0) our identity element. Now any two points can be “added” or “subtracted” (by adding or subtracting their respective angles) to obtain a third point, and each point has an inverse associated with it (its mirror image across the x-axis). It should be obvious that the choice of a zero point was totally arbitrary. Likewise, the points (1, 0), (−1, 0) also form an abelian variety; their group structure is isomorphic to Z2 and the choice of one of them as the identity is, again, arbitrary.
Is every variety abelian? No, but any complete, non-singular variety can be homomor-phicly mapped into an associated abelian variety (typically of higher dimension), called its Jacobian variety. This fact, combined with the extensive body of literature on abelian varieties ([Mumford], [Birkenhake and Lange], [Lang], to mention a few), makes the Ja-cobian variety an important object of study (though David Mumford, in the preface to [Mumford], described it as a “crutch”).
We will be needing only a tiny bit of this theory here, so my goal in this section is only to demonstrate how the Riemann-Roch Theorem allows us to set up an abelian group structures on an algebraic extension.
The oldest construction of the Jacobian variety uses integrals and only works over the complex field C, and requires us to work on a non-singular model of the curve.
We can pick an arbitrary origin and 2g closed paths from that origin that form a basis for the surface’s holonomy group. We can also pick g independent holomorphic differentials and evaluate them over the 2g closed paths to get a lattice Λg in Cg. Given any point on the surface, we can now evaluate the differentials along a path from the origin to that point and thus map into the torus Cg/Λg.
Not all complex torii are algebraic varieties, but this one is (PROOF NEEDED).
Now, Abel’s Theorem and the Jacobi inversion theorem ([Griffiths and Harris], p. 235) shows that Pic0, the group of divisors of degree zero modulo linear equivalence is isomor-phic to Cg/Λg.
Alternately, ([Lang], II, §1, Theorem 3), we can factor a mapping of a product into an abelian variety into mappings on each factor.
Lang also characterizes Abel’s theorem as follows: Let ω1, ..., ωg be a basis for the differential forms on the first kind of V. If a = P niPi is a [divisor] of degree 0 on V, and P is a fixed point of V, then the map into C/Λg given by: a → X ni( Z Pi P ω1, ..., Z Pi P ωg) is well defined modulo the periods... the kernel consists of those divisors that are linearly equivalent to 0 (i.e, principal); this is Abel’s theorem.
Pulling this all together, we need to show that the group operation is defined by rational functions.
For our purposes, it isn’t enough to construct the Jacobian over the complex field C; we also need Jacobians for curves in prime characteristic. So the traditional construction using integrals isn’t available, we need something different.
Andre Weil, in Courbes Algébriques et Variétés Abéliennes (1948), constructs the Jaco-bian variety over arbitrary fields using the g-fold symmetric product of the curve. See Michel Raynaud, “André Weil and the Foundations of Algebraic Geometry” for an broad overview of this French text.
J.S. Milne’s “Jacobian Varieties” (jmilne.org/math) also uses the symmetric power of the variety.
Greg Anderson, “Abeliants and their application to an elementary construction of Jaco-bians” uses equivalence classes of matrices constructed using a Riemann-Roch space.
Equivalence of matrices defined over a vector space, as Joshua Grochow taught me1, is the same as simultaneous equivalence of matrices, a notoriously hard problem. Sergeichuk, “Canonical matrices for linear matrix problems” shows how to convert such a matrix into an canonical form, but there’s no obvious way to introduce the rank 1 condition required by Anderson’s construction, nor is the canonical form easily parametized.
10.3 Simple Algebraic Extensions over Finite Fields Let’s start with a simple but crucial observation: Theorem 10.4. In an algebraic extension over a finite field, the evaluation field is also finite.
Proof Consider a finite field of constants F, over which we’ll extend first into a rational function field F(x) and then add an algebraic extension F(x, y), where y satisfies some minimial polynomial f(x, y) = 0. Start with the constant field, which gives us a finite number of values for x. Plugging each of these values into the minimal polynomial gives a finite set of polynomials f(yi) = 0. By Theorem ?, we can extend F into a finite extension field F[γ] where all the roots of the polynomial exist. Since there a only a finite number of polynomials, we need at worst a finite set of extensions F[γ1, ..., γk] to construct a field in which all the roots of all the polynomials exist. Using the Theorem of the Primitive Element, we can collapse all of these into a single finite extension field F[φ]. Since all values of x exist in F, and all values of y exist in F[φ], an evaluation homomorphism carries any rational function in x and y into F[φ].
□ This theorem leads directly to the single more important difference (to us) between divi-sors in an infinite field versus those in a finite field. In a finite field, some multiple of every divisor is principal. The reason is that the multiplicative group of the evaluation field has finite order. The simplest way to demonstrate this is to construct theorems analogous to Theorems ? and ?: Theorem 10.5. In an algebraic extension of a finite field with characteristic greater than 2, a function can always be constructed with an mth-order zero at a specified place (α, β) 1 and zero order at all other finite places, where m is the multiplicative order of the evalu-ation field.
Proof The desired function is (x −α)m + (y −β)m .
Clearly, this function is zero at (α, β) and of mth order there (PROOF THIS). At all other places one of the two terms will be non-zero, and both exist in the evaluation field. By Theorem ?, any non-zero number raised to the multiplicative order of its field is one. Thus the value of this function will be either 1 + 0, 0 + 1, or 1 + 1 = 2, all finite and non-zero, and thus of zero order.
□ Theorem 10.6. In an algebraic extension of a finite field with characteristic greater than 2, a function can always be constructed with an mth-order pole at a specified place (α, β) and zero order at all other finite places, where m is the multiplicative order of the evalu-ation field.
Proof The desired function is f(α, y)m (x −α)m(y −β)m + 1 where the division by (y −β)m is exact. Clearly, this function has a pole at (α, β) and of mth order there (PROOF THIS). CONSIDER OTHER PLACES OVER α. At all other places the denominator term will be non-zero, and thus one, and the numerator will be either zero or one (by Theorem ?) Thus the value of this function at these places will be either 0 + 1 or 1 + 1 = 2, both finite and non-zero, and thus of zero order.
□ Example 10.7. Show that some multiple of Z(1, 1) is principal in Z5(x, y); y2 = x.
Let’s first construct a multiplication table for Z5: 0 1 2 3 4 0 0 0 0 0 0 1 0 1 2 3 4 2 0 2 4 1 3 3 0 3 1 4 2 4 0 4 3 2 1 Now, let’s list out the places on the Riemann surface for Z5(x, y); y2 = x.
x (x, y) 0 (0,0) 1 (1,1) (1,4) 2 (2, γ) (2, −γ); γ2 −2 = 0 3 (3, θ) (3, −θ); θ2 −3 = 0 4 (4,2) (4,3) It looks like we need Z5[γ, θ] to express these places. It’s simplest to collapse γ and θ into a single algebraic extension. We could use the Theorem of the Primitive Element to do this, but in this case just looking at the multiplication table and noting that 3 = 23 = γ6 shows that θ = ±γ3. So, in fact, we only need Z5[γ]: x (x, y) 0 (0,0) 1 (1,1) (1,4) 2 (2, γ) (2, −γ); γ2 −2 = 0 3 (3, γ3) (3, −γ3) 4 (4,2) (4,3) Since Z5[γ] has 52 = 25 elements, its multiplicative group has order one less than this.
We conclude that 24 is our “magic” multiple, and that Z24(1, 1) must be principal in this field. Its generator should be simply (x −1)24 + (y −1)24. Clearly this function is zero for (x, y) = (1, 1). Let’s verify that it’s non-zero for some other places on the Riemann surface: (0, 0) : (−1)24 + (−1)24 = 424 + 424 = 1 + 1 = 2 (1, 4) : 324 + 024 = 1 + 0 = 1 (2, γ) : (γ −1)24 + (2 −1)24 = 1 + 1 = 2, since : (γ −1)2 = (γ2 −2γ + 1) = 3 −2γ (γ −1)4 = (3 −2γ)2 = (9 −12γ + 4γ2) = 2 −2γ (γ −1)8 = (2 −2γ)2 = (4 −8γ + 4γ2) = 2 −3γ (γ −1)12 = (2 −2γ)(2 −3γ) = (4 −10γ + 6γ2) = 1 In the final series of calculations, I used γ2 = 2 and reduced mod 5 repeatedly. I think the pattern should be clear, and leave further verification as an exercise.
□ 10.4 Endomorphism Rings Any commutative group G induces a (non-commutative) ring structure on its endmor-phisms, defined as follows (remember that an endomorphism is a homomorphism from an object to itself): Two endmorphisms φ(g) : G →G and γ(g) : G →G are added using G’s group operation on the images: (φ + γ)(g) = φ(g) · γ(g), where · denotes the group operation.
The additive identify is the endmorphism that maps the entire group onto its identity element.
Two endmorphisms φ(g) : G →G and γ(g) : G →G are multiplied using composition of mappings: (φγ)(g) = φ(γ(g)). The multiplicative identity is the endmorphism that maps every element in the group onto itself.
Let us now verify that these operations define a ring, the endomorphism ring of G, which we shall denote End(G). The properties of the identity elements are fairly obvious, I think. Almost as obvious is that the associative and commutative properties of the un-derlying group translate directly into additive associative and commutative properties in the endmorphism ring. The multiplicative properties follow from composition of map-pings being associative, but not necessarily commutative. The distributive law follows from the easily verified identity φ(γ(g) · µ(g)) = φ(γ(g)) · φ(µ(g)), using the fact that φ is an endomorphism, and thus a homomorphism, and therefore maps the group operator through.
The ring of integers Z can be mapped homomorphicaly2 into any ring, and an endomor-phism ring is no exception. We’ll denote by [m] the endmorphism mapped to by the integer m. is clearly the additive identity mapping all elements to the group identity.
is, of course, the multiplicative identity mapping all elements to themselves. is + , the endmorphism that composes each element with itself (using the group oper-ator): : g →g · g. composes each element with itself thrice: : g →g · g · g, etc.
Because Z is commutative, the subring [m] it maps to is also commutative, even though End(G) may not be.
2An easy consequence of Z’s repelling universal property in the category of rings, see [Lang], p. ?
Chapter 11 Algebraic Extensions THIS CHAPTER IS VERY VAGUE AND INCOMPLETE.
We now turn to the algebraic extension in general. Theorem ? allows us to collapse any two adjacent algebraic extensions together, so we need only consider an algebraic exten-sion over a transcendental extension. The most basic case, the one that we’ve studied in the last two chapters, is when the integrand involves only polynomials and a single root, so we are integrating on an algebraic curve and our algebraic extension occurs directly over the variable of integration: C(x, y). However, many of these results are applicable in the more general case where we have a series of field extensions that end in a transcen-dental (exponential or logarithmic) extension followed by an algebraic extension. I’ll use the notation K(θ, y) to emphasize when this is the case.
The most important change that occurs when shifting from C(x, y) to K(θ, y) is the loss of algebraic closure. The complex field is algebraically closed, so the irreducible polynomi-als is C[x] are simply the linear factors (x −γ), and there’s a one-to-one correspondence between these linear factors and the complex numbers.
An arbitrary differential field K, however, is not likely to be algebraically closed, so we need to shift our viewpoint from thinking in terms of places over points to thinking in terms of places over irreducible polynomials. We’ve already looked (WHERE?) at factor-ing polynomials into their irreducible factors, and now we wish to reformulate everything we’ve done in the past three chapters to adapt.
11.1 Exponential Extensions over Simple Algebraic Extensions To analyze integrands formed from exponential extensions over algebraic extensions, we need to solve Risch equations in algebraic extensions. We haven’t yet developed the tools to handle this problem in general algebraic extensions, but we can use the tools of Chapter 8 to solve Risch equations in a simple algebraic extension of C(x). This will allow us to solve integrals involving exponentials of polynomial roots. While such integrals aren’t terribly common, this exercise will reinforce the principles of working over algebraic extensions.
Recall that a Risch equation has the form r′ + Sr = T S, T, r ∈K (11.1) K is a differential field, that in our case will be an algebraic extension of C(x): dR dx + SR = T S, T, r ∈C(x)[y]/p(x, y) (11.2) Our approach in Chapter 7 was to identify the poles in S and T and then use this informa-tion to restrict r enough to solve it.
This same approach works with algebraic extensions; we just need to work with places and principal parts expansions instead of denominator factors and partial fractions expansions.
Actually, working with respect to a local uniformizing variable simplies the theory, since we don’t have to treat infinity separately.
Multiplying through by dx, we obtain: dR + R(Sdx) = Tdx (11.3) Taking the differential of an algebraic function causes its poles to increase in order by exactly one, and introduces no new poles (Theorem ?). Therefore, we can’t have any poles (finite or infinite) at places where there aren’t poles in either S dx or T dx (the differentials, not the functions).
Once we find the poles of S dx and T dx, there are three cases, as before: 1. ν(S dx) > −1 = ⇒ ν(R) ≥ν(T dx) + 1.
(11.4a) 2. ν(S dx) = −1 = ⇒ ν(R) = ν(T dx) + 1 or ν(R) = −π(S dx).
(11.4b) 3. ν(S dx) < −1 = ⇒ ν(R) ≥ν(T dx) −ν(S dx).
(11.4c) Once we’ve figured out the maximum orders of all of R’s poles, it’s then a straightforward matter to construct a suitable divisor and compute a basis for its Riemann-Roch space.
Then we can use fairly straightfoward techniques of linear algebra to find the coefficients of the solutions, if they exist.
Example 11.1. Integrate R 5x4+2x−2 x2 1 + 1 √ x3+1 + x √ x3+1 ex √ x3+1dx From [Br91] The integrand can be expressed using the tower C(x) ⊂C(x, y) ⊂C(x, y, ψ), where y2 = x3 + 1, y′ = 3x2 2y , and ψ = exp xy: Integrate R 5x4+2x−2 x2 1 + 1 y + x y ψ dx Since the top-most extension is exponential, we apply Theorem 6.1, and see that k = xy, k′ = xy′ + y = ( 3x3 2y + y), and a1 = 5x4+2x−2 x2 1 + 1 y + x y , so equation 6.4 reads: A′ 1 + (3x3 2y + y)A1 = 5x4 + 2x −2 x2 1 + 1 y + x y This equation is in an algebraic form of equation 6.8, so we want to analyse the poles of the S and R functions.
sage: R. = FunctionField(QQbar) Rational function field in x over Algebraic Field sage: L. = R[] Rational function field in x over Algebraic Field[y] sage: F. = R.extension(y^2 - (x^3+1)) Function field in y defined by y^2 - x^3 - 1 sage: S = 3x^3/(2y) + y 5 2x3 + 1 x3 + 1 y sage: T = (5x^4+2x-2)/(x^2)(1+1/y) + x/y 5x3 −4x2 + 4x −2 x4 −x3 + x2 y + 5x4 + 2x −2 x2 sage: differential_divisor_of_poles(Sx.differential()) 6 1 x2y sage: Tdp = differential_divisor_of_poles(Tx.differential()) 7 1 x2y + 2 (x, y −1) We have two places at which either S dx or T dx have poles, so we need to consider them both.
At (0, 1), ν(S dx) = 0, so ν(R) >= ν(T dx) + 1 = −2 + 1 = −1.
At infinity, ν(S dx) = −6, so ν(R) >= ν(T dx) −ν(S dx) = −7 −(−6) = −1.
Our solution, if it exists, has at most first-order poles at (0, 1) and ∞.
sage: sum(Tdp.support()) 1 x2y + (x, y −1) sage: BFS = sum(Tdp.support()).basis_function_space() 1, 1 xy + 1 x Is there some linear combination of these basis functions that solves our Risch equation?
sage: R = 2 BFS 2 xy + 2 x sage: R.differential() + R S x.differential() == T x.differential True This solves the Risch equation (?), giving us the A1 term in our solution. Since there’s no term in the integrand of ψ-degree zero, the A1 term is the only term in our solution, and we conclude: sage: var('x,y') (x, y) sage: integrand = eval(preparse(str(T))).subs({y:sqrt(x^3+1)}) exp(xs (5 x3 −4 x2 + 4 x −2) √ x3 + 1 x4 −x3 + x2 + 5 x4 + 2 x −2 x2 !
e( √ x3+1x) sage: ans = eval(preparse(str(R))).subs({y:sqrt(x^3+1)}) exp(xsqr 2 √ x3 + 1 x + 1 x !
e( √ x3+1x) sage: bool(diff(ans,x) == integrand) True □ [Br91] gives a more sophisticated algorithm for attacking these kinds of problems.
11.2 Integral Elements An important component of Hess’s algorithm for computing bases of Riemann-Roch spaces is the construction of maximal orders. A moment’s thought suggests that the key characteristic of the finite maximal order is that it contains all of the functions with no finite poles. In fact, that is an important characterization of the finite maximal order, one that we now wish to define more concretely.
Definition 11.2. An element f ∈K(θ, y) is integral if it satisfies a monic polynomial with coefficients in K[θ].
Intuitively, an integral element is one with no finite poles. To see this, at least in the case where K = C, define z = 1 f and substitute this into f’s monic polynomial: f n + an−1f n−1 + · · · + a1f + a0 = 0 z−n + an−1z−n+1 + · · · + a1z−1 + a0 = 0 1 + an−1z + · · · + a1zn−1 + a0zn = 0 Now, if z is zero at a place p over x = x0, at least one of this polynomial’s roots must be zero at x = x0. Since all of the ai are finite at x = x0 (they are polynomials), multiplying any of them by zero yields zero, so substituting in z = 0 yields 1 = 0. We conclude that z can not be zero, and thus f can not have a pole over p.
What’s special about infinity? Why not exclude some other place? Well, nothing’s all that special about infinity. We’ve already seen how a birational transformation can be used to swap infinity with any finite point. Demanding that a field element have no poles anywhere is too restrictive, because Theorem ? tells us that such an element must be constant. So we want to relax this requirement slightly by allowing poles over a single point. We use infinity because it’s convenient.
It’s not always obvious from inspection which functions are integral. Something like y x, which appears to have a pole at x = 0, is actually integral if, for example, y2 = x3. Then we can consider squaring y x to obtain y2 x2 = x3 x2 = x. If the square is finite, then the original function had to be finite (you can’t square infinity and get a finite value), so we conclude that y x is, in fact, globally integral in C(x, y); y2 = x3, as it satisfies the monic polynomial f 2 −x = 0.
Unfortunately, we have no straightforward means to construct such a polynomial, or prove that one doesn’t exist, for any particular function f. To test a function to determine if it is integral, we’ll need a more advanced approach.
11.3 Modules We’ll resort now to modules, a fairly important algebra concept backed by a substantial body of theory, upon which I shall only draw as needed. General references include [Atiyah+McDonald] and [Lang].
Definition 11.3. An R-module over a ring R is an additive group M acted on by R (i.e, there is a mapping R × M →M) in a distributive manner: (r1 + r2)m = r1m + r2m r1, r2 ∈R; m ∈M where we have adopted the usual convention of writing R’s action on M as a multiplica-tion.
Definition 11.4. A free R-module is an R-module spanned by a linearly independent basis {b1, b2, ...bn}. It consists of all elements formed as follows: 1 a1b1 + a2b2 + ... + anbn; ai ∈R Not all modules have a finite set of generators, and not all those have a linearly indepen-dent set of generators. Elements formed from a basis can be added by using the module’s distributive property to factor out the coefficients from of each basis element and then performing the addition in the ring R: (a1b1 + a2b2 + ... + anbn) + (c1b1 + c2b2 + ... + cnbn) 1I’ll also note that a multiplication rule needs to be specified between the basis elements and the elements of the ring, and an addition rule between the elements of the module. Also, the expression has to be unique — you can’t be able to write an element two different ways. In our case, these rules are obvious, but that’s not always the case.
= (a1 + c1)b1 + (a2 + c2)b2 + ... + (an + cn)bn So the elements generated from a basis clearly form a module. R operates on them by multiplication by every coefficient.
Example 11.5.
An ideal I in a ring R is a R-module, but a subring S of R, in general, is not, because multiplication by an element of R might not produce a result in the subring. R, however, can always be viewed as an S-module.
□ Note that it is vitally important to specify the ring used for the coefficients. For example, consider the basis {1, y}. Treating this as a C(x)-module, I can form y x = 1 xy, since 1 x ∈C(x). However, y x does not belong to the C[x]-module generated by {1, y}. I would need to use polynomial coefficients to form a C[x]-module, not the rational functions coefficients allowed in a C(x)-module. We’ll be primarily interested in K[θ]-modules, K(θ)-modules, and I-modules, where I is the ring of integral elements in K(θ, y).
11.4 The K[θ]-module I Since polynomials have no finite poles, they are integral elements, and thus K[θ] ⊆I.
Thus, I (the ring of integral elements) is trivially a K[θ]-module (see Example 11.5), but what is not nearly so obvious is that it is also a free module, a fact which underlies a great deal of our theory. I’ll prove this first by showing that I is finitely generated as a K[θ]-module, then showing the existance of a linearly independent set of generators.
Let’s start with a preliminary theorem.
Theorem 11.6. If {w1, . . . , wn} is a basis for a finite separable field extension E/K, then a dual basis {u1, . . . , un} can be constructed such that Tr(wiuj) = δij. ([Lang] Corollary VI.5.3) Proof Consider the following matrix: M = Tr(w1w1) .
.
.
...
Tr(w1wn) · · · Tr(wnwn) Now take an element x ∈E, and represent it relative to the basis {w1, . . . , wn} as a row vector X = (xi). Multiplying XM produces a row vector whose jth element can be written: X i xiTr(wjwi) = Tr(wj X i xiwi) = Tr(wjx) = Trx(wj) where I used first the K-linearity and additive distributive properties of Tr, then wrote Trx : f(a) = Tr(ax) to emphasize that I’m regarding Trx as a linear form in HomE(E, K).
So, if M is singular, then there exists some non-zero element x such that Trx is zero for all of wi, which form a basis set, so Trx must therefore be the zero map. This can only happen if Tr is identically zero, which would be the case for an inseparable extension.
For the separable case, therefore, M must be invertible, and we can write: M −1M = 1 ...
1 A moment’s thought now shows that the rows of M −1 are the desired dual basis elements, written with respect to {w1, . . . , wn}.
□ Theorem 11.7. I is a finitely generated K[θ]-module. ([A+MacD] Proposition 5.17; [Lang] Exercise VII.3) Proof Regarding K(θ, y) as a vector space over K(θ), we can easily construct a basis of integral elements by starting with {1, y, . . . , yn−1} and multiplying each element (if needed) by a polynomial in θ which cancels all of its poles: K(θ, y) = K(θ){w1, . . . , wn} ∀i(wi ∈I) Using Theorem 11.6, construct a dual basis {u1, . . . , un} so that Tr(wiuj) = δij. Take any x ∈I and write it using the dual basis: x = X i aiui ai ∈K(x) Now consider Tr(xwj). Now, x and wj are both in I, so xwj is in I, and therefore has a monic minimum polynomial with coefficients in K[θ]. Since Tr equals some inte-ger multiple of the negative of the second coefficient in a monic minimum polynomial, Tr(xwj) ∈K[θ]. But also, Tr(xwj) = Tr( X i aiuiwj) = X i Tr(aiuiwj) = X i aiTr(uiwj) = ai which establishes that ∀i(ai ∈K[θ]), so I ⊆K[x]{u1, . . . , un} K[θ] is a Noetherian ring (Theorem ??), so K[θ]{u1, . . . , un} is a Noetherian module (The-orem ??), which means that I, as a submodule, is Noetherian and thus finitely generated (Theorem ??).
□ Theorem 11.8. Any submodule of a finite free module over a principal ideal ring is free.
([Lang] Theorem III.7.1) Proof Let F = R{w1, . . . , wn} be a free R-module (R a principal ideal ring) with a submodule M. Consider Fi = R{w1, . . . , wi}, the free R-module generated by the first i basis elements, and Mi = Fi ∩M. We will show inductively that all of the Mi are free R-modules, and since Fi = F and Mi = M, this will prove the theorem.
First, consider M1 = R{w1} ∩M. If M1 is not empty (and thus free), then any m ∈M1 can be written rw1. Since a module forms an additive group, and we can operate on the module using all the elements of R, it follows that all the r’s must form an ideal, and since R is principal, that ideal can be written with a single generator, say (r1), and M1 = R{r1w1} (or is empty).
Now, assume that Mj is free for all j < i. Consider all x ∈Mi, which can be written r1w1 + · · · + riwi. Either Mi = Mi−1 (and is therefore free), or at least some of the ri are non-zero. By the same rationale as the last paragraph, these ri form an ideal, which can be written (ri). Take any element x ∈Mi with its ith coefficient ri and add it to Mi−1’s basis set to form a basis set for Mi, since some multiple of this element can be used to cancel any ith coefficient from an element in Mi and leave an element in Mi−1 which can be formed using the remaining basis elements.
□ A torsion-free module has no “zero divisors”, in the sense that no non-zero element of its associated ring can operate on a non-zero element of the module and produce zero. Since fields are torsion-free, and all of our modules are subsets of the field K(θ, y), they are all torsion-free.
Theorem 11.9. Any finitely generated, torsion-free module M over a principal ideal ring R is free. ([Lang] Theorem III.7.3) Proof Take a maximal set of M’s linearly independent generators {w1, . . . , wn} and any remain-ing generators {yn+1, . . . , ym}. Every yi is therefore linearly dependant on {w1, . . . , wn}: aiyi + r1w1 + · · · + rnwn = 0 ai ̸= 0 Take the product of all the ai’s: a = an+1an+2 · · · am and consider the mapping x 7→ax which is injective, since a ̸= 0 and the module is torsion-free, so therefore maps M to aM, an isomorphic image which is a submodule of the free module R{w1, . . . , wn}. By Theorem 11.8, aM is therefore free, and since it is isomorphic to M, we can take a basis of aM, divide all of its basis elements by a (they are all multiples of a), and obtain a basis for M.
□ Now, since K[θ] is a principal ideal ring (Theorem ??), Theorems 11.7 and 11.9 demon-strate that I is a free K[θ]-module.
Definition 11.10. A basis for I will be called an integral basis.
Theorem 11.11. Any integral basis is also a basis for the K(θ, y) field as a K(θ)-module.
□ While the preceding theorems offer an existance proof for an integral basis, it is not im-mediately clear how to obtain one for any particular field, and in fact the calculation of an integral basis ultimately becomes one of the biggest computational barriers in this theory.
Therefore, I will defer a more detailed discussion until a later chapter, and instead present a simple construction for the special case of a simple radical extension.
11.5 Basis for all Rational Functions The first kind of basis we’re interested in, a basis for all rational functions, is one than spans the entire C(x, y) field as a C(x)-module. In other words, we’re looking for a basis {b1, b2, ...bn} so that everything in C(x, y) can be expressed in the form: a1b1 + a2b2 + ... + anbn; ai ∈C(x) Such a basis will always have n elements, where n is the degree of the C(x, y) extension over C(x), and can be most conveniently characterized using its conjugate matrix: Definition 11.12. The conjugates of a rational function η(x, y) in C(x, y) are the func-tions formed by replacing y with its conjugate values.
The trace of a rational function η(x, y) is the sum of its conjugates: T(η(x, y)) = X i η(x, yi) The norm of a rational function η(x, y) is the product of its conjugates: N(η(x, y)) = Y i η(x, yi) Both the trace and norm, as symmetric functions in y1, ..., yn, are functions in C(x).
The conjugate matrix Mω of n elements ωi in C(x, y), where n is the degree of C(x, y) over C(x), is the matrix whose each row consists of the n conjugate values of a single element, and whose n rows are formed in this way from the n elements.
A set of n elements ωi ∈C(x, y) form a rational function basis for C(x, y) if the deter-minant of their conjugate matrix is non-zero: |Mω|̸= 0 Definition 11.13. For any function η ∈C(x, y) and any rational function basis ωi, the trace vector Tη/ω = T(ηωi) of η relative to ω is formed from the traces of the n products of η with the n functions ωi.
The conjugate vector Cη = (η(x, yi)) is formed from the n conjugates of η.
Theorem 11.14. For any function η ∈C(x, y) and any rational function basis ωi, if Tη/ω is the zero vector, then η is zero.
Proof Tη/ω, Mω and Cη satisfy the matrix equation Tη/ω = MωCη since each row of this matrix equation has the form T(ηωi) = X j ωi(x, yj)η(x, yj) Since Mω is invertible (since its determinant is non-zero), if Tη/ω is identically zero, then so must be Cη, and η is the first element in Cη.
□ Theorem 11.15. A rational function basis ωi spans C(x, y) as a C(x)-module. ([Bliss], Theorem 19.1) Proof Note that when we multiply Mω by its transpose MT ω, the ijth element of MωMT ω is: X k ωi(x, yk)ωj(x, yk) = T(ωiωj) Since |Mω| is non-zero, |MT ω| is non-zero, and |MωMT ω| is non-zero, so given any func-tion η ∈C(x, y), we can solve the following equation for R: Tη = MωMT ωR each of row of which reads: T(ηωi) = X j T(ωiωj)rj Since both Tη and MωMT ω are composed of nothing but traces, they exist in C(x), so R must also exist in C(x) and its elements therefore commute with the trace: T(ηωi) = X j T(rjωjωi) Since the trace of a sum is the sum of the traces: T(ηωi) = T( X j rjωjωi) T((η − X j rjωj)ωi) = 0 which implies that η = P j rjωj, by Theorem 11.14, and since we’ve already shown that the rj are rational functions in C(x), this proves the theorem.
□ Let me illustrate with a simple example.
Example 11.16. Consider the basis {1, y} over the field C(x, y); y2 = x. The conjugate value of y is −y (PROVE THIS), so the conjugate matrix is: C = 1 1 y −y and its determinant: det C = 1 1 y −y = −2y Since −2y is not zero, we conclude that {1, y} is a basis for all rational functions over C(x, y); y2 = x.
□ Notice that I didn’t ask whether −2y was zero at some place in the field. The determinant of the conjugate matrix can be zero at certain places; in fact, often is. It just can’t be identically zero; i.e, it can’t be zero everywhere. If this isn’t clear, reread Theorems 11.14 and 11.15, noting that all the matrices are defined over the fields C(x) and C(x, y), where the only zero element is 0.
11.6 Divisors and Integral Modules In C(x), we were working with the quotient field of a principal ideal ring, so we could always find a single function to generate any finitely generated C[x]-module, simply by putting all the generators over a common denominator, then taking the G.C.D. of the numerators.
In K(θ, y), we are no longer working with a principal ideal ring, so we can’t guarantee that any particular ideal can be generated by a single function, but it turns out that every ideal can be generated by a pair of functions. Our course of attack is first to construct that pair of functions, then use them to determine if in fact the ideal is principal.
Definition 11.17. An integral module (or I-module) is a module formed over I, the ring of integral elements in K(θ, y).
Since I itself can be expressed as a K[θ]-module using an integral basis, any I-module is also a K[θ]-module. Not all K[θ]-modules are I-modules, however, since I is typically larger than K[x].
Some authors use the term fractional ideal to refer to an I-module. I have avoided use of this term for two reasons. First, I wish to emphasize the concept of a module. Second, I-modules are not ideals, either in the ring I (since they may contain elements not in I), nor in the field K(θ, y), since, as a field, K(θ, y) has only the trivial ideals. The term fractional ideal is used because an I-module can be regarded as a fraction of ideals in I.
Theorem 11.18. I is a Noetherian ring.
Proof Since K is a field, K[θ] is a Noetherian ring by the Hilbert basis theorem, K[θ] ⊆I, and I is finitely generated as a K[θ]-module, so I is a Noetherian ring by [Atiyah+McDonald] Proposition 7.2.
□ Theorem 11.19. The order of the norm of f at a point θ0 is the sum of the orders of f at all places over θ0.
□ Theorem 11.20. A function can always be constructed with a simple zero at a specified finite, ordinary place (α, β), zero order at an additional finite set of finite, ordinary places Σ, and non-negative order at all other finite places.
Proof Begin with the function (x −α), which is a uniformizing variable and thus has a simple zero at (α, β). If none of the other places in Σ have x-value α, then we are done, since (x −α) has no finite poles.
Otherwise, compute (x−α) (y−β) at all places in Σ that do not have y = β. Select a number γ different from all of these values. The function (x −α) −γ(y −β) has no finite poles and is non-zero at all places in Σ, but it may now have a zero of higher order at (α, β).
Consider a series expansion of y in terms of (x −α): y = β + c1(x −α) + c2(x −α)2 + · · · So long as γ is also selected different from c1, (x −α) −γ(y −β) will have a first order zero at (α, β) and meet all requirements of the theorem. The simplest way to do this is to pick a value for γ, use Theorem 11.19 to check if the function has a simple zero, and if not, choose a different value for γ.
□ Theorem 11.21. A function can always be constructed with a simple pole at a specified finite, ordinary place (α, β), zero order at an additional finite set of finite, ordinary places Σ, and non-negative order at all other places.
Proof Begin with the function: f(α, y) (x −α)(y −β) where f(x, y) is the minimum polynomial of the algebraic extension. Note that the divi-sion by (y −β) will always be exact, since f(α, β) = 0. So we have a rational function P(y) (x−α), where P(y) is a polynomial in y. It has a simple pole at (α, β), as can be seen from a series expansion in x −α (again, a uniformizing variable). Since the y −β factor has been divided out of f(α, y), the numerator is non-zero at (α, β), so the leading term in the series expansion involves (x −α)−1, and the pole is thus simple.
This function is finite at all other places, which is obvious except when x = α and y ̸= β, where it takes the form 0 0, so we can expand it using L’Hôpital’s rule: lim (x,y)→p P(y) (x −α) = lim (x,y)→p dP(y) dx d(x−α) dx = P ′(y) dy dx where ′ denotes differentiation with respect to the polynomial’s variable. P ′(y) is a poly-nomial, and is thus finite where y is finite, as is dy dx (consider a series expansion of y in terms of (x −α), since all places in Σ are finite and non-singular). It follows that the function is at least finite everywhere except at (α, β).
A more algebraic way to prove this is to note that f(x, y) has a simple zero at every place over x = α (assuming there are no multiple points over x = α), so P(y) will have a simple zero at every place over x = α except (α, β), which will exactly cancel the simple pole from (x −α).
Now, compute the value of the function at all other places in Σ, using either L’Hôpital’s rule or Puiseux expansion if some of these are over α. If the value of the function is non-zero at all of these places, then we are done. Otherwise, select a number γ different from all of these values. The function: f(α, y) (x −α)(y −β) −γ has the desired properties, since it still has a simple pole at (α, β), has no other poles, and is now non-zero at all places in Σ.
We can avoid computing any expansions by picking random values of γ, and using The-orem 11.19 to check for any extra zeros. Since only a finite number of γ values produce extra zeros, this process is guaranteed to terminate.
□ Theorem 11.22. A function can always be constructed with specified integer orders at a finite set of finite, non-singular places Σ and non-negative order at all other finite places.
Proof For each pole or zero, use Theorems 11.20 or 11.21 to construct a function with a simple pole or a simple zero at that place, zero order at all other places in Σ and non-negative order at all other finite places. Raise each of these function to the integer power that is the order of the corresponding pole or zero, then multiply them all together.
□ Definition 11.23. A finite multiple of a divisor is a function with order equal to or greater than that required by the divisor at all finite places.
For the remainder of this section, I’ll assume that our divisors involve only finite, ordinary places, which can always be guaranteed in the case of the integration theory.
Theorem 11.24. For any divisior D and any finite, non-singular place p, at least one finite multiple of D exists with order at p exactly that required by D.
Proof Use Theorem 11.22 with the zeros and poles required by D, adding p to Σ if necessary.
□ Theorem 11.25. There is a one-to-one relationship between finitely generated integral modules and divisors. Such a module consists of all finite multiples of its associated divisor, and the order of a module’s divisor at every finite place is the minimum of the orders of the module’s generators at that place.
Proof For a given divisor D, consider the set M(D) of all finite multiples of D. Now, adding two elements can not reduce their order at any finite place, nor can multiplying an element by an integral element i ∈I, so M(D) is clearly an I-module, but it might not be finitely generated.
Since D has only a finite number of poles, we can always construct a function with order equal or less than that of D at all finite places simply by taking the inverse of the polyno-mial r = (x−p1)m1 · · · (x−pn)mn where pi are the x-coordinates of the poles in D and mi are their multiplicities. For any m ∈M(D), mr is integral, so M(D) ⊆I{r−1}, where I{r−1} is the I-module generated by r−1. Now, since I{r−1} is a finitely generated module over a Noetherian ring (remember Theorem 11.18), I{r−1} is a Noetherian mod-ule by [Atiyah+McDonald] Proposition 6.5, and M(D) must also be a finitely generated I-module by [Atiyah+McDonald] Proposition 6.2.
Let (b1, ..., bn) be an I-module basis for M(D). Since there is no way to lower the orders of an element using I-module constructions, and by Theorem 11.24 for each place there is at least one function in M(D) with order exactly that required by D, it follows that for each place there must be at least one basis element with exactly the order required by D. Futhermore, no basis element can have an order less than required by D at any finite place, since that element would not be a finite multiple of D. Therefore, at each place p, the minimum of the orders of the basis elements must be exactly the order required by D.
Conversely, given a finitely generated I-module M, construct its associated divisor D by taking at every place the minimum of the orders of the module’s generators at that place.
Clearly, M ⊆M(D), but some finite multiple of D might not be in M.
To eliminate this possibility, take the module’s generators, say {b1, b2, b3} and expand them into a set where each additional generator beyond the first lowers the module’s order by one at a single place, say {b1, b′ 2, b2, b′′ 3, b′ 3, b3}. These additional generators can be constructed by multiplying the original generators by integral elements (constructed using Theorem 11.22) to remove any additional poles, so b′ 2 = i′ 2b2, where i′ 2 ∈I.
This new module M ′ clearly has the same associated divisor as M, and I’ll now show inductively that any f ∈M(D) can be found in M ′. Let Dn be the divisor associated with the first n basis elements of M ′. Clearly, any finite multiple of D1 can be constructed as integral element times b1, so let’s now assume that any finite multiple of Dn−1 can be constructed with the first n −1 generators, and consider the nth generator gn. It lowers the order by one at a single place, so any f ∈M(Dn) −M(Dn−1) must have exactly the same order as gn. Multiplying gn by a suitable constant (the ratio of coefficients in f and gn’s series expansion at this place) will exactly cancel this pole, so f −cg ∈M(Dn).
So any f ∈M(D) can be constructed using the integral module M ′. Writing this con-struction in matrix form shows how f can be constructed as an M-module element: a1 · · · am b1 b′ 2 b2 b′ 3 b3 = a1 · · · am 1 0 0 0 i′ 2 0 0 1 0 0 0 i′ 3 0 0 1 b1 b2 b3 = a′ 1 · · · a′ 3 b1 b2 b3 Consider such a finite multiple f. For every m ∈M (in particular, its basis elements), f must have lower order than m at at least one finite place p, since otherwise i = fm−1 would be integral and f would exist in M as mi. Yet D, by definition, is the minimum of the orders of M’s basis elements at every finite place. Therefore f can not have lower order than a basis element at any finite place and thus can not exist.
□ Theorem 11.25 shows that an I-module is associated with every divisor, but now we need a constructive procedure for forming an I-module basis for a given divisor.
Theorem 11.26. Given a divisor D, a pair of functions can always be constructed that generate the divisor’s associated integral module.
Proof Use Theorem 11.22 to construct a function f with the divisor’s required poles and zeros, zero order at all other places conjugate to those poles and zeros, and non-negative order elsewhere. Construct g, a polynomial in x with n-th order roots at all points under n-th order zeros. (f, g) is the required basis. The only finite poles are those of f and g has zero order everywhere except at f’s zeros and their conjugates, so by Theorem 11.25, (f, g) forms a basis for D’s associated I-module.
□ Of course, the whole point here is to actually find a function with a specified set of zeros and poles, so once we have constructed a basis for a divisor’s associated integral module, we need to determine if the module is principal. Since the total order of a field element is always zero, this only makes sense for divisors of zero order, since divisors of non-zero order can never be principal. Futhermore, since an integral module corresponds to finite multiples of a divisor, we can’t use this technique to find functions with poles or zeros at infinity, but that isn’t a serious limitation since if we need such a function, we can just transform into a field with a different point at infinity.
Since a finite multiple of a divisor differs from an exact multiple in that the finite multi-ple can have additional finite zeros, and thus additional infinite poles (since they always balance), a zero order I-module is principal iff it contains a function with no poles at infinity. We can determine this by expressing the I-module as a K[x]-module, simply by multiplying the I-module basis through by an integral basis (remember that an integral basis is simply a basis for I as a K[x]-module).
We now transform this K[x]-module basis to make it normal at infinity, i.e, to ensure that poles don’t cancel between terms. First, we use a series of row-equivalent transformations to reduce our 2n basis elements to n elements, then additional transformations to make it normal.
We then check these basis elements to see if one of them has no poles at infinity. The most straightforward way to do this is to invert the field using z = 1 x, which swaps zero with infinity. We can then construct an integral basis for the inverse field, and express each of the module’s basis elements (after inverting them) using this inverse basis. If there are any poles at infinity in the original field, they will appear as poles at zero in the inverse field, and can easily be detected by checking if z = 0 is a zero of the denominators.
Finally, let me note that since g in Theorem 11.26 is a polynomial, it always has poles at infinity (unless the divisor has no zeros, and is thus trivially constant), and can thus be excluded from consideration. We need only look at the function f, and perhaps not even all of its integral multiples (CHECK THIS).
Theorem 11.27. Let f dx be a differential with order greater than or equal to -1 at some place p with branching index r centered at x0. The residue of f dx at p is equal to the value of the function r(x −x0)f at p. ([Trager], p. 56, taken almost verbatim) Proof Let t be a uniformizing parameter at p. Since x −x0 has order r at p, it can be written as x −x0 = trg where g has order zero at p dx = (rtr−1g + tr dg dt )dt Since dg dt has non-negative order at p, dx has order r−1 at p and f must have order greater than or equal to −r at p f dx = rtr−1fg dt + trf(dg dt )dt the second term on the right side is holomorphic at p so the residue of f dx at p is the same as the residue of the first term on the right side. Since this term is expressed using the differential of a uniformizing paramter, its residue is the residue of rtr−1fg, which is the value of rtrfg = r(x −x0)f.
□ Chapter 12 Notes For a while I was thinking that the product of prime ideals is equal to their intersection.
This is true in principal ideal domains, but not in general.
For example, consider I = (x, z) and J = (x + z) in K[x, z]. Now, x + z ∈I ∩J, but x + z / ∈I · J.
Sage was useful in puzzling this out: sage: R. = QQ[]; sage: I = Ideal(x,z); sage: J = Ideal(x+z); sage: I.is_prime() True sage: J.is_prime() True sage: I.intersection(J) == IJ False Looking at the primary decomposition, we see that the product is smaller than the inter-section, because there’s an extra ideal that needs to be intersected (the original ideal is the intersection of the ideals in its primary decomposition). Sage’s comparison operator for ideals also shows us that the product is contained in the intersection.
sage: I.intersection(J).primary_decomposition() [(x + z) Q[x, z]] sage: (IJ).primary_decomposition() (x + z) Q[x, z], z, x2 Q[x, z] sage: I.intersection(J) > IJ True So now it’s a question of finding something in the intersection that isn’t in the product.
The ideal quotient isn’t useful for this, probably because of its Zariski closure property.
sage: I.intersection(J).quotient(IJ) (1) Q[x, z] sage: diff(sqrt(x^4+1),x) 2 x3 √ x4 + 1 sage: diff(sqrt(x^4+1),x,2) − 4 x6 (x4 + 1) 3 2 + 6 x2 √ x4 + 1 sage: diff(sqrt(x^4+1),x,3) 24 x9 (x4 + 1) 5 2 − 36 x5 (x4 + 1) 3 2 + 12 x √ x4 + 1 sage: diff(sqrt(x^4+1),x,4) −240 x12 (x4 + 1) 7 2 + 432 x8 (x4 + 1) 5 2 − 204 x4 (x4 + 1) 3 2 + 12 √ x4 + 1 sage: diff(sqrt(x^3+1),x) 3 x2 2 √ x3 + 1 sage: diff(sqrt(x^3+1),x,2) − 9 x4 4 (x3 + 1) 3 2 + 3 x √ x3 + 1 sage: diff(sqrt(x^3+1),x,3) 81 x6 8 (x3 + 1) 5 2 − 27 x3 2 (x3 + 1) 3 2 + 3 √ x3 + 1 12.1 Valuations [van der Waerden], §18.1 A valuation is a generalization of the absolute value. A valuation is a mapping φ from a field K to an ordered field R (typically the reals) obeying the following axioms: positivity ∀a ∈K, φ(a) ≥0 (V1) definiteness ∀a ∈K, φ(a) > 0 ⇐ ⇒a ̸= 0 (V2) homomorphism (on the multiplicative group) ∀a, b ∈K, φ(ab) = φ(a)φ(b) (V3) subadditivity (or triangle inequality) ∀a, b ∈K, φ(a + b) ≤φ(a) + φ(b) (V4) A moment’s thought will show that the standard absolute value on the reals obeys these axioms, as does the modulus on the complex field. Valuations are similar to norms, except that norms are defined on vector spaces, while valuations are defined on fields.
A valuation is said to be non-Archimedian if it also satisfies the following axiom, stronger than V4: non-Archimedian axiom ∀a, b ∈K, φ(a + b) ≤max(φ(a), φ(b)) (V4’) In this case, we can switch from a multiplicative to an additive notation and obtain expo-nential valuation by replacing φ(a) with w(a) = −ln φ(a): ∀a ∈K, w(a) ∈(−∞, ∞] (E1) ∀a ∈K, w(a) = ∞⇐ ⇒a = 0 (E2) ∀a, b ∈K, w(ab) = w(a) + w(b) (E3) ∀a, b ∈K, w(a + b) ≥min(w(a), w(b)) (E4) 12.2 Notes on Harris - Geometry of Alg Curves - Harvard 287 Abel’s theorem – p. 29 Classical Jacobian discussed on pp. 28-31 “As we’ve defined it, the Jacobian is only a complex torus so far. Note that a general complex torus is not embeddable in projective space. However, it turns out that the Jaco-bian has enough meromorphic functions to embed in projective space, so it is a projective variety.” 12.3 Notes on [Fu08] [Fu08] is a good, freely available introduction to algebraic geometry.
[Fu08] assumes an algebraically closed coefficient field throughout (except Chapter 1).
Riemann-Roch Theorem Let C be an algebraic curve, let X be its non-singular model, and let K be its function field.
Proposition 8.4.
Let x \in K, x \notin k. Let (x)_0 be the divisor of zeros of x and let n=[K:k(x)].
Then 1) (x)_0 is an effective divisor of degree n, 2) There is a constant \tau such that l(r(x)_0) \ge rn-\tau \forall r.
Proof Prop 6.9. K is an algebraic function field in one variable over k.
By definition, this means that exists some t such that K is algebraic over k(t).
So x \in K is algebraic over k(t), and \exists F \ in k[X,T] such that F[x,t] = 0.
x is not algebraic over k (1-48), so t must appear in F, so t is algebraic over k(x), and therefore k(x,t) is algebraic over k(x) (1-50), so K is algebraic over k(x) (1-46).
Problem 1-54: If R is a domain with quotient field K, and L is a finite algebraic extension of K, then there exists a basis for L over K such that each basis element is integral over R.
Proof 1-54: Let {w_1, ..., w_n} be any basis for L over K.
Since each basis element is algebraic over K, by clearing denominators we can write: a{i0} w_i^{n_i} + a_{i1} w_{n_i-1} + \cdots = 0 a_{ij} \in R We can pull a_{i0} into w_i, and thus adjust the w_i's to be integral over R by multiplying each one by something in R.
Since anything in L can be written l = \sum c_i w_i c_i \in K it can also be written l = \sum (c_i / r_i) w'i where the r_i adjust the w_i to be integral and c_i/r_i is still in K.
INTEGRAL ELEMENTS: w integral over k[x] means that w is finite everywhere x is, and has poles only where x does.
w integral over k[x^{-1}] means that w is finite everywhere x^{-1} is, and has poles only where x has zeros.
So, k[x^{-1}] is a domain with quotient field k(x), and K is a finite algebraic extension of k(x), so there exists a basis for K over k(x) such that each basis element is integral over k[x^{-1}].
Let {w_1, ..., w_n} be such a basis for K over k(x).
We will show that the poles of these functions must lie over the roots of x.
w_i^{n_i} + a{i1} w_{n_i-1} + \cdots = 0 a_{ij} \in k[x^{-1}] So, ord_P(a_ij) \ge 0 if P \ne S (zero set of x), since x^{-1} and thus anything in k[x^{-1}] is finite away from S.
Problem 2-29: if for some i, ord(a_i) < ord(a_j) \forall j \ne i, then a_1 + \cdots + a_n \ne 0 Proof of 2-29: Assume the contrary.
Then we can write a_i = \sum_{i\ne j} a_j.
Taking ord of both sides, and using ord(a+b) \ge min(ord(a), ord(b)), we see this is impossible.
Therefore, ord_P(w_i) \ge 0 if P \ne S, since otherwise ord_P(w_i^{n_i}) < ord_P(a_ij w^{n_i-j}) \forall j.
Therefore, the poles of w_i are isolated at the zeros of x, and since there are only a finite number of w_i and each has a finite number of poles, then for some t, div(w_i) + tZ > 0 \forall i.
So w_i \in L(tZ), and if j \le r, then w_i x^{-j} \in L((r+t)Z) Now w_i are independent over k(x) and 1, x^{-1}, ..., x^{-r} are independent over k, so l((r+t)Z) \ge n(r+1).
Now, l((r+t)Z) = l(rZ) + dim(L((r+t)Z) / L(rZ)) dim(L((r+t)Z) / L(rZ)) \le tm (Prop 3-1), where m is the degree of Z.
So, l(rZ) \ge n(r+1) - tm, so pick \tau = tm-n, and l(rZ) \ge nr - \tau, \forall r Riemman-Roch l(D) = deg(D) + 1 - g + l(W-D) deg(W) = 2g-2 l(W) = g l(0) = 1 l(D) = 0 if deg(D) < 0 If deg(D) > 2g-2, then l(D) = deg(D) + 1 - g If deg(D) = 2g-2, then deg(W-D) = 0, and l(W-D) = 1 iff D-W is principal, otherwise l(W-D) = Let D=W+X, where deg(X)=0, then l(D) = 2g-2 + 1 - g + (1/0) depending on whether X is principal l(D) = g (X is principal) or l(D) = g - 1 (X is not principal) Goal: an g-dimensional algebraic variety that represents Pic0 Consider (2g-2)-dimensional symmetric space.
Each point corresponds to an effective divisor of degree (2g-2).
Fix a (g-2)-tuple.
We're left with g free points.
Milne's construction Use r-dimensional symmetric space, with r > 2g-2.
Pick an (r-g) tuple.
12.4 Notes on [Sh61] A function (Y -> k) is regular at a point on a variety if there exists an open neighborhood of the point where the function is given by a rational function of polynomials, the denominator never zero.
(relation to coordinate ring?) A map (Y -> k) is regular on Y if it is regular at every point of Y.
A morphism (X -> Y) between varieties is a continuous map (in the Zariski topology) such that pullbacks of regular functions are regular.
A rational map is a morphism defined only on an open subset.
Given a rational map from affine varieties X to Y, if [x,y,z] is Y's coordinate system, then we pullback to a regular function, which is a rational function in X's function field.
So Y's coordinates are given by rational functions in X's coordinates.
A group variety is an algebraic variety equipped with a group structure, where the group operation and group inversion are rational maps.
If the group operation is commutative, it's an abelian variety.
A homomorphism between abelian varieties is a rational map that commutes with the group operation.
An endomorphism is a homomorphism from the variety to itself.
Endomorphisms of a group form a ring.
Addition is performed by mapping through both endomorphisms, then applying the group operation, which is commutative, so endomorphism addition is commutative.
Multiplication is performed by composition, and need not be commutative.
Using just the addition structure, we get an abelian group that can be structued as a Z-module.
Multiplication by an integer is just repeated application of the endomorphism.
We can promote the endomorphism ring into an algebra by tensoring with Q, call this EndQ(A).
Elements of EndQ(A) are basically endomorphisms with an associated 1/n denominator (numerators can be sucked into the endomorphism).
A lattice is a free Z-module of the same rank as the algebra over Q.
An order is a lattice that is also a subring and contains the identity.
The order of A, written t, is the image of the endomorphism ring in the endomorphism algebra.
a is a lattice in the endomorphism algebra contained in t, so it's a collection of actual endomorphisms.
g(a,A) is the set of points on A mapped to 0 by every element of a.
Prop 16. Let a be an integral ideal (p. 49) of F.
Reduction mod p defines a homomorphism of g(a,A) onto g(a,A^).
If a is prime to the characteristic of k^, this homomorphism is an isomorphism.
Div(C) is group of divisors Div0(C) is group of degree 0 divisors P(C) is group of principal divisors P(C) in Div0(C) in Div(C) define Pic(C) = Div(C)/P(C) define Pic0(C) = Div0(C)/P(C) Pic0(C) is isomorphic to the Jacobian variety with a point O.
Given a degree zero divisor in Div0(C), we can use the group law on the Jacobian to construct a single point corresponding to the divisor.
Asking if the divisor is principal is asking if this point is O.
Asking if any multiple of the divisor is principal is asking if any multiple of this point is O.
We can define an endomorphism to be addition by a point, using the group law.
NO - doesn't take identity to identity.
Let's consider [n], the endomorphism defined by applying the group operation n times (n is an integer).
This should generate a lattice contained in t, so it's an integral ideal.
g([n], A) is the set of points whose n-multiples are principal.
Given a divisor whose (k p^q)-multiple is principal, let's multiply by p^q and get a divisor whose k-multiple is principal.
Endomorphism ideal [k] is prime to p.
Then this divisor point will be in g(a,A) and g(a,A^), where a is [k].
We can determine that the divisor point is in g(a,A^) for a=[k] by determining that the divisor's k-multiple is principal on the module curve.
Since this is an isomorphism, the divisor point is also in g(a,A) for a=[k].
Given a non-singular algebraic curve C, we reduce mod p to get Cp.
Good reduction implies that Cp is non-singular with the same genus.
C has an associated Jacobian J.
Cp also has an associated Jacobian Jp.
PROBLEM: (hopefully) Show that J mod p is Jp.
Construct J as follows.
Pick r > 2g-2.
Symmetric group J^(r).
Pick r-g extra points and find an open covering of J^(r).
Within each open set, construct J locally.
12.5 Notes on Bronstein Definition 3.4.1. We say that t ∈K is a monomial over k (w.r.t. D), if 1. t is transcendental over k, 2. Dt ∈k[t].
Definition 3.4.3. We say that u ∈k is a logarithmic derivative of a k-radical if there exist v ∈k∗and an integer n ̸= 0 such that nu = Dv/v.
Definition 5.1.1. t ∈K is a primitive over k if Dt ∈k. t ∈K∗is an hyperexponential over k if Dt/t ∈k. t ∈K is Liouvillian over k if t is either algebraic, or a primitive or an hyperexponential over k. K is a Liouvillian extension of k if there are t1, ..., tn in K such that K = k(t1, ..., tn) and ti is Liouvillian over k(t1, ..., ti−1 for i in 1, ..., n.
Theorem 5.1.1. If t is a primitive over k and Dt is not the derivative of an element of k, then t is a monomial over k, Const(k(t)) = Const(k), and S = k (i.e. Sirr = Sirr 1 = 0).
Conversely, if t is transcendental and primitive over k, and Const(k(t)) = Const(k), then Dt is not the derivative of an element of k.
Theorem 5.1.2. If t is an hyperexponential over k and Dt/t is not a logarithmic derivative of a k-radical, then t is a monomial over k, Const(k(t)) = Const(k), and Sirr = Sirr 1 = t.
Conversely, if t is transcendental and hyperexponential over k, and Const(k(t)) = Const(k), then Dt/t is not a logarithmic derivative of a k-radical.
This next corollary (EK/C(x) and LK/C(x) are the exponential and logarithm elementary monomials in a tower of elementary extensions) leads directly to an algorithm for testing new logarithmic and exponential primitives to see if they are monomials.
Corollary 9.3.1. Let C, x, K, EK/C(x) and LK/C(x) be as in Theorem 9.3.1, a ∈K∗and b ∈K. Then, 1. Da/a is the derivative of an element of K if and only if there are ri ∈Q such that X i∈LK/C(x) riDti + X i∈EK/C(x) ri Dti ti = Da a 2. Db is the logarithmic derivative of a K-radical if and only if there are ri ∈Q such that X i∈LK/C(x) riDti + X i∈EK/C(x) ri Dti ti = Db 12.6 Function Fields (Stichtenoth) Stichtenoth, “Algebraic Function Fields and Codes”.
Let K be a field.
F/K is called rational if F = K(x) for some x ∈F A valuation ring of a function field F/K is a ring O ∈F such that K ∈O ∈F (both proper inclusions) for every z ∈F, either z ∈O or z−1 ∈O For the rational function field K(x), there’s a valuation ring for each irreducible polyno-mial A place of a function field F/K is the unique maximal ideal of a valuation ring O of F/K.
Every element of t ∈P such that P = tO is called a prime element, or local paramter, or uniformizing variable of P).
P determines O uniquely. OP is called the valuation ring of the place P Definition 1.4.1 - a divisor is a formal sum over the places of F/K 12.7 The Riemann-Roch Theorem Various proofs of the Riemann-Roch Theorem: Fulton (Chapter 8) - uses nonsingular model of curve Milne (Theorem 14.6) - assumes curve is nonsingular Stichtenoth - constructs divisor using places (Cartier divisor) Vakil (Eq 18.4.2.1) - unclear to me if he assumes regularity 12.8 Examples Example 12.1. Compute R √ 4 −x2 dx A solution method from first year calculus might be to note that this integrand forms one leg of a right triangle: √ 4 −x2 2 x θ x = 2 sin θ √ 4 −x2 = 2 cos θ dx = 2 cos θ dθ Z √ 4 −x2 dx = Z 4 cos2 θ dθ = Z (2 + 2 cos 2θ) dθ = 2θ + sin 2θ = 2θ + 2 sin θ cos θ = 2 arcsin x 2 + x √ 4 −x2 2 Now let’s attack this integral using the methods of this chapter. First, transform the prob-lem into an algebraic curve: Z y dx y2 = 4 −x2 Since limx→∞y = ∞, the integrand has poles at infinity. We want infinity to be an ordi-nary point of the curve (no ramification; no singularities) with no poles in the integrand.
The simplest transformation is to exchange zero with infinity, and in this case zero is an ordinary point with places (0, 2) and (0, −2), neither of which is a pole of the integrand.
So we’ll invert x and y into u and v: x = 1 u y = 1 v 1 v 2 = 4 − 1 u 2 = ⇒4u2v2 −v2 −u2 = 0 Z 1 v d 1 u = ⇒− Z 1 vu2 du The only poles in this integrand occur when either u = 0 or v = 0. Substituting these values into 4u2v2−v2−u2 = 0, we see that these condiutions only occur at (u, v) = (0, 0), so let’s analyze our curve at that point, starting with the Newton polygon: 4u2v2 −v2 −u2 = 0 x x x @ @ @ @ @ @ The Newton polygon has a single line segment of span 2 and slope -1, so we have two cycles, each with ramification index one: a singularity. Since there is no ramification, u is a uniformizing parameter and we expect to expand v as follows: v = c1u + c2u2 + c3u3 + · · · v2 = c2 1u2 + 2c1c2u3 + (2c1c3 + c2 2)u4 + · · · Substituting these expansions into 4u2v2 −v2 −u2 = 0, we obtain: 4c2 1u4 + 8c1c2u5 + (8c1c3 + 4c2 2)u6 + · · · −c2 1u2 −2c1c2u3 −(2c1c3 + c2 2)u4 + · · · −u2 = 0 Equating terms in u2, we see that c1 = ±i. Each of these two values corresponds to one branch of the singularity. There is only a single term in u3, which forces c2 to be zero, and equating terms in u4 produces c3 = 2c1, so v = ±(iu + 2iu3 + · · ·) @(0, 0) Inverting v and substituting into our 1-form, we obtain 1 v = ±(−i1 u + 2iu + · · ·) @(0, 0) 1 vu2 du = ± −i 1 u3 + 2i1 u + · · · du @(0, 0) The u−1 terms will integrate into logarithms, so let’s ignore them for the moment and concentrate on the u−3 terms, which will integrate into u−2 terms, so we’re looking for a function with second order poles at both places at the (0, 0) singularity.
Starting with our standard basis for all rational functions, {1, v}, we seek to modify it into a basis for P2(0, 0)aP2(0, 0)b. Note first that v has poles at u = ± 1 2. Using y = 1/u, we analyze at (± 1 2, ∞) as follows: y2 (u −1 2)2 + (u −1 2) + 1 4 −4(u −1 2)2 −4(u −1 2) x x x x x A A A A A A Our line segment has span 1 and slope -2, indicating a single place with ramification 2, and y as a uniformizing parameter. Setting (u −1 2) = c1y + c2y2 + · · · (u −1 2)2 = c2 1y2 + · · · Substituting, we find that c1 = 0 and c2 = 1 16, so (u −1 2) = 1 16y2 + · · · v = y−1 @(1 2, ∞) (u + 1 2) = 1 16y2 + · · · v = y−1 @(−1 2, ∞) In short, v has first order poles at (± 1 2, ∞) and (u ± 1 2) has second order zeros, so we can adjust our basis accordingly and obtain {1, (4u2 −1)v} for a basis with no finite poles.
We can also use a theorem of Trager to shortcut this calculation.
Returning to our analysis at (0, 0), we see that 1 has zero order (obviously) and (4u2−1)v has a first order zero at both sheets there, since 4u2 −1 = −1 is finite and v has first order zeros. We also know that u is a uniformizing parameter, so it’s easy to modify our basis and obtain 1 u2, 4u2 −1 u3 v is a C[x]−basis for P2(0, 0)aP2(0, 0)b Is this basis normal at infinity? Well, the representation order of 1 u2 is 2 and its u−2 coefficients at (∞, ± 1 2) are both 1, while the representation order of 4u2−1 u3 v is 1, and its u−1 coefficients are 2 and -2. Since det C = 1 2 1 −2 = −4 is non-zero, the basis is normal at infinity.
The Riemann-Roch theorem says that the dimension of l(D) is 5, 1 u2 can be multiplied by any polynomial up to second degree without introducing poles at infinity, and 4u2−1 u3 v can be multiplied by any polynomial up to first degree, so 1 u2, 1 u, 1, 4u2 −1 u3 v, 4u2 −1 u2 v is a C-module basis for l(D).
Any linear combination of these functions is a multiple of the divisor, but not all of them produce the correct residues. Looking at the residues, we see that only 4u2−1 u3 v = 1 uv has residues of ±i on the two sheets at the (0, 0) singularity. Dividing by 2 to correct for the 2 that will be introduced by the integration, we conclude that 1 2uv = xy 2 = x √ 4−x2 2 is the desired function.
Next, we have to deal with the logarithms. Going back to the series expansions of our 1-form, we see that we have residues of ±2i on our two sheets at (0, 0). The objective now is a bit different; we want a function with exactly the divisor Z(0, 0)aP(0, 0)b. Starting with an integral basis: {1, (4u2 −1)v} we want to modify these functions to make them multiples of Z(0, 0)aP(0, 0)b. The pole isn’t a problem for an integral basis, and looking at the series expansion for v at (0, 0) we see that it (and therefore (4u2 −1)v) has a simple zero there, but 1 needs to be replaced with u: {u, (4u2 −1)v} Now we construct a matrix with the coefficients in the series expansions: 1 −i 0 0 ←(0, 0)a ←(0, 0)b 1 −i 0 0 i 1 = 0 The solution shows us how to modify the basis: {u, iu + (4u2 −1)v u } = {u, i + (4u2 −1)v u } 1 0 0 0 ←(0, 0)a ←(0, 0)b 1 0 0 0 0 1 = 0 {u, i1 u + (4u2 −1)v u2 } 1 −2i 0 2i = 2i At the last step, the determinant is non-zero, which shows that we now have a basis for multiples of the divisor except at infinity. Is it normal at infinity? u’s expansion at both places at infinity is 1 u −1, so its representation order is -1, and the second element’s expansion at infinity starts ±2 + · · ·, so its representation order is 0 and: det C = 1 2 1 −2 = −4 So the basis is normal at infinity. If an exact multiple of the divisor exists, it is one of the basis elements. It’s not u, since u has a pole at infinity, but the second element is exact: i1 u + (4u2 −1)v u2 = i1 u −1 v = ix −y The desired residues are ±2i, so the function we want is 2i ln(ix −y) = 2i ln y 2 −ix 2 + 2i ln(−2) = 2i ln r 1 − x 2 2 −ix 2 !
= 2i(−i arcsin x 2) = 2 arcsin x 2 (the constant disappears into the constant of integration) and the final answer is: Z √ 4 −x2 dx = 2 arcsin x 2 + x √ 4 −x2 2 □ 12.9 arcsin Example 12.2. Compute R 1 √ 1−x2 dx The obvious attempt is to use the algebraic extension y2 = 1 −x2 and integrate 1 y dx.
But we first need to determine if this differential has any poles at infinity, by inverting the field and looking for poles at zero. Setting u = 1 x, we convert our minimal polynomial into u2y2 = u2 −1 (after multiplying through by u2), and using v = uy we obtain our inverse field C(u, v); v2 = u2 −1.
Since x = 1 u and y = v u, we convert our differential as follows: 1 y dx = u v (−1 u2 du) = −1 uv du Now, {1, v} is an integral basis for the inverse field, so we multiply through by v v to obtain: = −v uv2du = − 1 u(u2 −1)v du which is now in normal form and clearly has a pole at u = 0, or x = ∞. Note that 1 y = u v = uv v2 = u u2 −1v has no pole at u = 0, a clear example of a differential having a pole at a place where its constituent function has none.
In any event, we clearly can not use the original field to conduct the integration, since it would require constructing a function with a pole at infinity, and our algorithm can’t handle this. So we need to transform into a field where the differential has no pole at infinity.
Actually, we’ve already done this! Note that the integrand had no pole at zero in the original field: 1 y dx = y y2 dx = 1 1 −x2y dx Since the inverse field swapped zero with infinity, it follows that there is no pole at infinity in the inverse field, so we can proceed to integrate − 1 u(u2−1)v du in C(u, v); v2 = u2 −1.
Simple inspection of the integrand (already in normal form) shows that its poles are at (0, i), (0, −i), (1, 0), and (−1, 0). Remember that we’re now working on the Riemann surface of an algebraic extension, so we need to specify both u and v to specify a place.
The next step is to compute the residues at each of these places, using Theorem 11.27: (0, i) − 1 (u2 −1)v @ (0, i) = i (0, −i) − 1 (u2 −1)v @ (0, −i) = −i (1, 0) −2 1 u(u + 1)v @ (1, 0) = 0 (−1, 0) −2 1 u(u −1)v @ (−1, 0) = 0 The poles with zero residues can be ignored. We’re interested in the other two, which exist in Q[i], which can be regarded as a vector field over Q with basis {1, i}, and we want to construct a function whose poles and zeros match the i-component of the residues (the 1-component is uniformly zero).
We start by constructing an I-module generator set for the divisor with a simple zero at (0, i) and a simple pole at (0, −i). Theorem 11.21 shows that: f = v2 + 1 u(v + i) = v −i u has a simple pole at (0, −i). At (0, i), L’Hôpital’s rule gives: lim (u,v)→(0,i) v −i u = (v −i)′ u′ dv du = dv du = u v = 0 where the last transformation was accomplished by differentiating the mimimal polyno-mial. So f has a zero at (0, i), and I’ll note that we’ve just stumbled into the solution.
Theorem 11.21 already assures us that f has only a single finite simple pole, and we can see that its only zeros occur when v −i = 0, which, according to the minimum polyno-mial, can only occur at u = 0, thus (0, i) is its only finite zero, and it is simple, as we can verify by showing that the corresponding pole in its inverse is simple: 1 f = u v −i = u(v + i) v2 + 1 = u(v + i) u2 = 1 uv + i u So we’ve found the function we’re looking for by accident. Let’s save the general case for the next example, and convert back to our original field: v −i u = x(y x −i) = y −ix Remembering that our residues came multiplied by a factor of i, we conclude that our solution is i ln(y −ix), or: Z 1 √ 1 −x2 dx = i ln √ 1 −x2 −ix = −i ln 1 √ 1 −x2 −ix = −i ln √ 1 −x2 + ix 1 −x2 + x2 = −i ln √ 1 −x2 + ix = arcsin x where I used the negative of a logarithm being the logarithm of the inverse, and the last transformation came from section 4.2.
□ Bibliography [Al14] Paraskevas Alvanos; Riemann-Roch Spaces and Computation. De Gruyter Open, 2014. ISBN 978-3-11-042613-7, e-ISBN 978-3-11-042612-0 Free PDF download available from publisher’s web site [Bl47] Gilbert Ames Bliss; Algebraic Functions. American Mathematical Society, 1947 ISBN 0821874535, 9780821874530 [Br91] Bronstein, Manuel; The Risch Differential Equation on an Algebraic Curve. 1991.
[Br00] Bronstein, Manuel; Symbolic Integration Tutorial. 2000.
[Br05] Bronstein, Manuel; Symbolic Integration I. Springer, 2005.
[BrCo10] Briggs, Cochran, Calculus, 2nd Edition. Pearson, 2010. ISBN-10: 0321336119 [Co93] Henri Cohen, A Course in Computational Algebraic Number Theory. Graduate Texts in Mathematics 138. Springer, 1993.
[Co99] Henri Cohen, Advanced Topics in Computational Number Theory. Graduate Texts in Mathematics 193. Springer, 1999.
[DeWe82] Dedekind, R., Weber, H. (1882). Theorie der algebraischen Functionen einer Veanderlichen. J. reine angew. Math., 92:181-290.
[Fu08] Fulton, William, Algebraic Curves, An Introduction to Algebraic Geometry, 3rd Edition, 2008.
[Ge92] Geddes, Czapor, Labahn, Algorithms for Computer Algebra. Springer. ISBN: 978-0-585-33247-5 [Go14] Goodman, Algebra: Abstract and Concrete, 6th Edition. SemiSimple Press.
ISBN: 978-0-9799142-1-8 [Gu05] Victor Guillemin. 18.117 Topics in Several Complex Variables. Spring 2005.
Massachusetts Institute of Technology: MIT OpenCourseWare,
License: Creative Commons BY-NC-SA.
[He02] Hess, F. (2002). Computing Riemann-Roch Spaces in Algebraic Function Fields and Related Topics, Journal of Symbolic Computation, 33(4), 425-445.
[Ko07] Kollár, János (2007), Lectures on Resolution of Singularities, Princeton: Prince-ton University Press, ISBN 978-0-691-12923-5 Kollár’s Resolution of Singularities – Seattle Lecture, is a more advanced treat-ment of resolution of singularities on varieties in general, while his Princeton University Press book contains two introductory chapters on resolution for curves (the only one relevant here) and resolution for surfaces before it gets into the more general theory.
[Sh61] Shimura, Goro; Taniyama, Yutaka (1961), Complex multiplication of abelian vari-eties and its applications to number theory, Publications of the Mathematical Society of Japan, 6, Tokyo: The Mathematical Society of Japan, MR 0125113 Later expanded and published as Shimura (1997) [St09] Stichtenoth, Henning. Algebraic Function Fields and Codes, 2nd edition.
Springer-Verlag. Graduate Texts in Mathematics 254.
[SwHu06] Swanson, Irena; Huneke, Craig (2006), Integral Closure of Ideals, Rings, and Modules, Cambridge University Press, Cambridge.
Available on-line at iswanso/book/ [Ri70] Robert Risch; The Solution of the Problem of Integration in Finite Terms. Bulletin A.M.S, Vol 76, pp. 605-608.
[Ru76] Rudin, Walter (1976). Principles of mathematical analysis (Third ed.). New York.
ISBN 0-07-054235-X.
[Tr84] Trager, Barry; Integration of Algebraic Functions, Ph.D. thesis, Massachusetts Institute of Technology, 1984. |
1131 | https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/16%3A_Acids_and_Bases/16.10%3A_Acid_Strength_and_Molecular_Structure | Published Time: 2016-03-11T07:19:02Z
16.10: Acid Strength and Molecular Structure - Chemistry LibreTexts
16.10.1
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16: Acids and Bases
Map: A Molecular Approach (Tro)
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Fri, 14 Aug 2020 20:43:56 GMT
16.10: Acid Strength and Molecular Structure
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Contents
1. Home
2. Bookshelves
3. General Chemistry
4. Map: A Molecular Approach (Tro)
5. 16: Acids and Bases
6. 16.10: Acid Strength and Molecular Structure
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Map: A Molecular Approach (Tro)
Front Matter
1: Matter, Measurement, and Problem Solving
2: Atoms and Elements
3: Molecules, Compounds and Chemical Equations
4: Chemical Reactions and Aqueous Reactions
5: Gases
6: Thermochemistry
7: The Quantum-Mechanical Model of the Atom
8: Periodic Properties of the Elements
9: Chemical Bonding I- Lewis Structures and Determining Molecular Shapes
10: Chemical Bonding II- Valance Bond Theory and Molecular Orbital Theory
11: Liquids, Solids, and Intermolecular Forces
12: Solids and Modern Materials
13: Solutions
14: Chemical Kinetics
15: Chemical Equilibrium
16: Acids and Bases
17: Aqueous Ionic Equilibrium
18: Gibbs Energy and Thermodynamics
19: Electrochemistry
20: Radioactivity and Nuclear Chemistry
21: Organic Chemistry
22: Biochemistry
23: Chemistry of the Nonmetals
24: Metals and Metallurgy
25: Transition Metals and Coordination Compounds
Back Matter
16.10: Acid Strength and Molecular Structure
Last updated Aug 14, 2020
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16.9: Polyprotic Acids
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Table of contents
1. Learning Objectives
2. Bond Strengths
3. Stability of the Conjugate Base
1. the strongest acid Known: The hydrohelium Cation
Inductive Effects
Example 16.10.1
Exercise 16.10.1
Summary
Contributors and Attributions
Learning Objectives
To understand how molecular structure affects the strength of an acid or base .
We have seen that the strengths of acids and
bases
vary over many orders of magnitude. In this section, we explore some of the structural and electronic factors that control the acidity or basicity of a molecule.
Bond Strengths
In general, the stronger the A–H A–H or B–H+B–H+ bond, the less likely the bond is to break to form H+H+
ions
and thus the less acidic the substance. This effect can be illustrated using the hydrogen halides:
| Relative Acid Strength | HF | HCl | HBr | HI |
--- ---
| H–X Bond Energy (kJ/mol) | 570 | 432 | 366 | 298 |
| pKa | 3.20 | −6.1 | −8.9 | −9.3 |
The trend in bond energies is due to a steady decrease in overlap between the 1s orbital of hydrogen and the valence orbital of the halogen atom as the size of the halogen increases. The larger the atom to which H is bonded, the weaker the bond. Thus the bond between H and a large atom in a given family, such as I or Te, is weaker than the bond between H and a smaller atom in the same family, such as F or O. As a result, acid strengths of binary hydrides increase as we go down a column of the
periodic table
. For example, the order of acidity for the binary hydrides of Group 16 elements is as follows, with p K a p K a values in parentheses:
H 2 O(14.00=p K w)<H 2 S(7.05)<H 2 S e(3.89)<H 2 T e(2.6)
H 2 O(14.00=p K w)<H 2 S(7.05)<H 2 S e(3.89)<H 2 T e(2.6)(16.10.1)
Stability of the Conjugate
Base
Whether we write an acid–
base
reaction as A H⇌A−+H+A H⇌A−+H+ or as B H+⇌B+H+B H+⇌B+H+, the conjugate
base
(A−A− or B B) contains one more lone pair of electrons than the parent acid (A H A H or B H+B H+). Any factor that stabilizes the lone pair on the conjugate
base
favors dissociation of H+H+ and makes the parent acid a stronger acid. Let’s see how this explains the relative acidity of the binary hydrides of the elements in the second row of the
periodic table
. The observed order of increasing acidity is the following, with pKa values in parentheses:
C H 4(50)≪N H 3(36)<H 2 O(14.00)<H F(3.20)
C H 4(50)≪N H 3(36)<H 2 O(14.00)<H F(3.20)(16.10.2)
Consider, for example, the compounds at both ends of this series: methane and hydrogen fluoride. The conjugate
base
of C H 4 C H 4 is C H−3 C H−3, and the conjugate
base
of H F H F is F−F−. Because fluorine is much more electronegative than carbon, fluorine can better stabilize the negative charge in the F−F− ion than carbon can stabilize the negative charge in the CH3− ion. Consequently, HF HF has a greater tendency to dissociate to form H+H+ and F−F− than does methane to form H+H+ and C H−3 C H−3, making HF a much stronger acid than C H 4 C H 4.
The same trend is predicted by analyzing the properties of the conjugate acids. For a series of compounds of the general formula H E H E, as the
electronegativity
of E increases, the E–H bond becomes more polar, favoring dissociation to form E−E− and H+H+. Due to both the increasing stability of the conjugate
base
and the increasing polarization of the E–H bond in the conjugate acid, acid strengths of binary hydrides increase as we go from left to right across a row of the
periodic table
.
Acid strengths of binary hydrides increase as we go down a column or from left to right across a row of the
periodic table
.
the strongest acid Known: The hydrohelium Cation
The stornger acid, the weaker the
covalent bond
to a hydrogen atom. So the strongest acid possible is the molecule with the weakest bond. That is the hydrohelium (1+) cation, HeH+HeH+, which is a positively charged ion formed by the reaction of a proton with a helium atom in the gas phase. It was first produced in the laboratory in 1925 and is isoelectronic with molecular hydrogen (\ce{H2}}). It is the strongest known acid, with a proton affinity of 177.8 kJ/mol.
Ball and stick model of the hydrohelium ion. (CC BY-SA 3.0; CCoil).
HeH+HeH+ cannot be prepared in a condensed phase, as it would protonate any anion, molecule or atom with which it were associated. However it is possible to estimate a hypothetical aqueous acidity using Hess's law:
HHe+(g)→H+(g)+ He(g)+178 kJ/mol
HHe+(aq)→HHe+(g)+973 kJ/mol
H+(g)→H+(aq)−1530 kJ/mol
He(g)→He(aq)+19 kJ/mol
HHe+(aq)→H+(aq)+ He(aq)−360 kJ/mol
A free energy change of dissociation of −360 kJ/mol is equivalent to a p K a of −63.
It has been suggested that HeH+HeH+ should occur naturally in the interstellar medium, but it has not yet been detected.
Inductive Effects
Atoms
or groups of
atoms
in a molecule other than those to which H is bonded can induce a change in the distribution of electrons within the molecule. This is called an inductive effect, and, much like the coordination of water to a metal ion, it can have a major effect on the acidity or basicity of the molecule. For example, the hypohalous acids (general formula HOX, with X representing a halogen) all have a hydrogen atom bonded to an oxygen atom. In
aqueous solution
, they all produce the following
equilibrium
:
H O X(a q)⇌H+(a q)+O X−(a q)
H O X(a q)⇌H+(a q)+O X−(a q)(16.10.3)
The acidities of these acids vary by about three orders of magnitude, however, due to the difference in
electronegativity
of the halogen
atoms
:
| HOX | Electronegativity of X | pKa |
---
| HOCl | 3.0 | 7.40 |
| HOBr | 2.8 | 8.55 |
| HOI | 2.5 | 10.5 |
As the
electronegativity
of X X increases, the distribution of
electron density
within the molecule changes: the electrons are drawn more strongly toward the halogen atom and, in turn, away from the H in the O–H bond, thus weakening the O–H bond and allowing dissociation of hydrogen as H+H+.
The acidity of
oxoacids
, with the general formula H O X O n H O X O n (with n n = 0−3), depends strongly on the number of terminal oxygen
atoms
attached to the central atom X X. As shown in Figure 16.10.1 16.10.1, the K a K a values of the
oxoacids
of chlorine increase by a factor of about 10 4 10 4 to 10 6 10 6 with each oxygen as successive oxygen
atoms
are added. The increase in acid strength with increasing number of terminal oxygen
atoms
is due to both an inductive effect and increased stabilization of the conjugate
base
.
Any inductive effect that withdraws
electron density
from an O–H bond increases the acidity of the compound.
Because oxygen is the second most electronegative element, adding terminal oxygen
atoms
causes electrons to be drawn away from the O–H bond, making it weaker and thereby increasing the strength of the acid. The colors in Figure 16.10.1 16.10.1 show how the electrostatic potential, a measure of the strength of the interaction of a point charge at any place on the surface of the molecule, changes as the number of terminal oxygen
atoms
increases. In Figure 16.10.1 16.10.1 and Figure 16.10.2 16.10.2, blue corresponds to low electron densities, while red corresponds to high electron densities. The oxygen atom in the O–H unit becomes steadily less red from H C l O H C l O to H C l O 4 H C l O 4 (also written as H O C l O 3 H O C l O 3, while the H atom becomes steadily bluer, indicating that the
electron density
on the O–H unit decreases as the number of terminal oxygen
atoms
increases. The decrease in
electron density
in the O–H bond weakens it, making it easier to lose hydrogen as H+H+
ions
, thereby increasing the strength of the acid.
Figure 16.10.1 16.10.1: The Relationship between the Acid Strengths of the
Oxoacids
of Chlorine and the
Electron Density
on the O–H Unit. These electrostatic potential maps show how the
electron density
on the O–H unit decreases as the number of terminal oxygen
atoms
increases. Blue corresponds to low electron densities, whereas red corresponds to high electron densities. Source: Chlorine
oxoacids
pKa values from J.
R
. Bowser, Inorganic
Chemistry
(Pacific Grove, CA: Brooks-Cole,1993).
At least as important, however, is the effect of delocalization of the negative charge in the conjugate
base
. As shown in Figure 16.10.2 16.10.2, the number of
resonance structures
that can be written for the oxoanions of chlorine increases as the number of terminal oxygen
atoms
increases, allowing the single negative charge to be delocalized over successively more oxygen
atoms
.
Electron delocalization in the conjugate
base
increases acid strength.
The electrostatic potential plots in Figure 16.10.2 16.10.2 demonstrate that the
electron density
on the terminal oxygen
atoms
decreases steadily as their number increases. The oxygen atom in ClO− is red, indicating that it is electron rich, and the color of oxygen progressively changes to green in C l O+4 C l O+4, indicating that the oxygen
atoms
are becoming steadily less electron rich through the series. For example, in the perchlorate ion (C l O−4 C l O−4), the single negative charge is delocalized over all four oxygen
atoms
, whereas in the hypochlorite ion (O C l−O C l−), the negative charge is largely localized on a single oxygen atom (Figure 16.10.2 16.10.2). As a result, the perchlorate ion has no localized negative charge to which a proton can bind. Consequently, the perchlorate anion has a much lower affinity for a proton than does the hypochlorite ion, and perchloric acid is one of the strongest acids known.
As the number of terminal oxygen
atoms
increases, the number of
resonance structures
that can be written for the oxoanions of chlorine also increases, and the single negative charge is delocalized over more oxygen
atoms
. As these electrostatic potential plots demonstrate, the
electron density
on the terminal oxygen
atoms
decreases steadily as their number increases. As the
electron density
on the oxygen
atoms
decreases, so does their affinity for a proton, making the anion less basic. As a result, the parent oxoacid is more acidic.
Similar inductive effects are also responsible for the trend in the acidities of
oxoacids
that have the same number of oxygen
atoms
as we go across a row of the
periodic table
from left to right. For example, H 3 P O 4 H 3 P O 4 is a weak acid, H 2 S O 4 H 2 S O 4 is a strong acid, and H C l O 4 H C l O 4 is one of the strongest acids known. The number of terminal oxygen
atoms
increases steadily across the row, consistent with the observed increase in acidity. In addition, the
electronegativity
of the central atom increases steadily from P to S to C l C l, which causes electrons to be drawn from oxygen to the central atom, weakening the O–H O–H bond and increasing the strength of the oxoacid.
Careful inspection of the data in Table 16.10.1 16.10.1 shows two apparent anomalies: carbonic acid and phosphorous acid. If carbonic acid (H 2 C O 3(H 2 C O 3) were a discrete molecule with the structure (HO)2 C=O(HO)2 C=O, it would have a single terminal oxygen atom and should be comparable in acid strength to phosphoric acid (H 3 P O 4 H 3 P O 4), for which pKa1 = 2.16. Instead, the tabulated value of p K a 1 p K a 1 for carbonic acid is 6.35, making it about 10,000 times weaker than expected. As we shall see, however, H 2 C O 3 H 2 C O 3 is only a minor component of the aqueous
solutions
of C O 2 C O 2 that are referred to as carbonic acid. Similarly, if phosphorous acid (H 3 P O 3 H 3 P O 3) actually had the structure (H O)3 P(H O)3 P, it would have no terminal oxygen
atoms
attached to phosphorous. It would therefore be expected to be about as strong an acid as H O C l H O C l (pKa = 7.40). In fact, the p K a 1 p K a 1 for phosphorous acid is 1.30, and the structure of phosphorous acid is (HO)2 P(=O)H(HO)2 P(=O)H with one H atom directly bonded to P and one P=O P=O bond. Thus the pKa1 for phosphorous acid is similar to that of other
oxoacids
with one terminal oxygen atom, such as H 3 P O 4 H 3 P O 4. Fortunately, phosphorous acid is the only common oxoacid in which a hydrogen atom is bonded to the central atom rather than oxygen.
Table 16.10.1 16.10.1: Values of pKa for Selected Polyprotic Acids and
Bases
| H 2 C O 3 H 2 C O 3 and H 2 S O 3 H 2 S O 3 are at best minor components of aqueous solutions of C O 2(g)C O 2(g) and S O 2(g)S O 2(g), respectively, but such solutions are commonly referred to as containing carbonic acid and sulfurous acid, respectively. |
| Polyprotic Acids | Formula | p K a 1 | p K a 2 | p K a 3 |
| carbonic acid | “H 2 C O 3” | 6.35 | 10.33 | |
| citric acid | H O 2 C C H−2 C(O H)(C O 2 H)C H 2 C O 2 H | 3.13 | 4.76 | 6.40 |
| malonic acid | H O−2 C C H 2 C O 2 H | 2.85 | 5.70 | |
| oxalic acid | H O 2 C C O 2 H | 1.25 | 3.81 | |
| phosphoric acid | H 3 P O 4 | 2.16 | 7.21 | 12.32 |
| phosphorous acid | H 3 P O 3 | 1.3 | 6.70 | |
| succinic acid | H O 2 C C H 2 C H 2 C O 2 H | 4.21 | 5.64 | |
| sulfuric acid | H 2 S O 4 | −2.0 | 1.99 | |
| sulfurous acid | “H 2 S O 3” | 1.85 | 7.21 | |
| Polyprotic Bases | Formula | p K b 1 | p K b 2 | |
| ethylenediamine | H 2 N(C H 2)2 N H 2 | 4.08 | 7.14 | |
| piperazine | H N(C H 2 C H 2)2 N H | 4.27 | 8.67 | |
| propylenediamine | H 2 N(C H 2)3 N H 2 | 3.45 | 5.12 | |
Inductive effects are also observed in organic molecules that contain electronegative substituents. The magnitude of the electron-withdrawing effect depends on both the nature and the number of halogen substituents, as shown by the pKa values for several acetic acid derivatives:
p K a C H 3 C O 2 H 4.76<C H 2 C l C O 2 H 2.87<C H C l 2 C O 2 H 1.35<C C l 3 C O 2 H 0.66<C F 3 C O 2 H 0.52
As you might expect, fluorine, which is more electronegative than chlorine, causes a larger effect than chlorine, and the effect of three halogens is greater than the effect of two or one. Notice from these data that inductive effects can be quite large. For instance, replacing the –CH 3 group of acetic acid by a –CF 3 group results in about a 10,000-fold increase in acidity!
Example 16.10.1
Arrange the compounds of each series in order of increasing acid or
base
strength.
1. sulfuric acid [H 2 S O 4, or (H O)2 S O 2], fluorosulfonic acid (F S O 3 H, or F S O 2 O H), and sulfurous acid [H 2 S O 3, or (H O)2 S O]
2. ammonia (N H 3), trifluoramine (N F 3), and hydroxylamine (N H 2 O H)
The structures are shown here.
Given: series of compounds
Asked for: relative acid or
base
strengths
Strategy:
Use relative bond strengths, the stability of the conjugate
base
, and inductive effects to arrange the compounds in order of increasing tendency to ionize in
aqueous solution
.
Solution:
Although both sulfuric acid and sulfurous acid have two –OH groups, the sulfur atom in sulfuric acid is bonded to two terminal oxygen
atoms
versus one in sulfurous acid. Because oxygen is highly electronegative, sulfuric acid is the stronger acid because the negative charge on the anion is stabilized by the additional oxygen atom. In comparing sulfuric acid and fluorosulfonic acid, we note that fluorine is more electronegative than oxygen. Thus replacing an –OH by –F will remove more
electron density
from the central S atom, which will, in turn, remove
electron density
from the S–OH bond and the O–H bond. Because its O–H bond is weaker, F S O 3 H is a stronger acid than sulfuric acid. The predicted order of acid strengths given here is confirmed by the measured pKa values for these acids:
p K a H 2 S O 3 1.85<H 2 S O−2 4<F S O 3 H−10
The structures of both trifluoramine and hydroxylamine are similar to that of ammonia. In trifluoramine, all of the hydrogen
atoms
in NH3 are replaced by fluorine
atoms
, whereas in hydroxylamine, one hydrogen atom is replaced by OH. Replacing the three hydrogen
atoms
by fluorine will withdraw
electron density
from N, making the lone electron pair on N less available to bond to an H+ ion. Thus N F 3 is predicted to be a much weaker
base
than N H 3. Similarly, because oxygen is more electronegative than hydrogen, replacing one hydrogen atom in N H 3 by O H will make the
amine
less basic. Because oxygen is less electronegative than fluorine and only one hydrogen atom is replaced, however, the effect will be smaller. The predicted order of increasing
base
strength shown here is confirmed by the measured p K b values:
p K b N F 3—<<N H 2 O H 8.06<N H 3 4.75
Trifluoramine is such a weak
base
that it does not react with aqueous
solutions
of strong acids. Hence its
base ionization constant
has never been measured.
Exercise 16.10.1
Arrange the compounds of each series in order of
decreasing acid strength: H 3 P O 4, C H 3 P O 3 H 2, and H C l O 3.
increasing base strength: C H 3 S−, O H−, and C F 3 S−.
Answer a
H C l O−3>C H 3 P O 3 H 2>H 3 P O 4
Answer a
C F 3 S−<C H 3 S−<O H−
Summary
Inductive effects and charge delocalization significantly influence the acidity or basicity of a compound. The acid–
base
strength of a molecule depends strongly on its structure. The weaker the A–H or B–H+ bond, the more likely it is to dissociate to form an H+ ion. In addition, any factor that stabilizes the lone pair on the conjugate
base
favors the dissociation of H+, making the conjugate acid a stronger acid.
Atoms
or groups of
atoms
elsewhere in a molecule can also be important in determining acid or
base
strength through an inductive effect, which can weaken an O–H bond and allow hydrogen to be more easily lost as H+
ions
.
Contributors and Attributions
Anonymous
16.10: Acid Strength and Molecular Structure is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.
16.8: Molecular Structure and Acid-Base Behavior is licensed CC BY-NC-SA 4.0.
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1132 | https://kostochk.web.illinois.edu/docs/2012/british09ky.pdf | Extremal graph packing problems: Ore-type versus Dirac-type H. A. Kierstead, A. V. Kostochka, and Gexin Yu Abstract We discuss recent progress and unsolved problems concerning extremal graph packing, emphasizing connections between Dirac-type and Ore-type problems.
Extra attention is paid to coloring, and especially equitable coloring, of graphs.
1 Introduction An important instance of combinatorial packing problems is that of graph pack-ing. We say that n-vertex graphs G1, G2, . . . , Gk pack, if there exists an edge-disjoint placement of all these graphs onto the same set of n vertices. By definition, two graphs G1 and G2 pack, if G1 is a subgraph of the complement G2 of G2, or, equiv-alently, G2 is a subgraph of the complement G1 of G1. Many basic graph theory problems and concepts can be expressed in a unified (and sometimes more natural) form using the language of graph packing. Here are some relevant examples.
Example 1.1 The problem of existence of a spanning (hamiltonian) cycle in an n-vertex graph G (which is a close relative of the famous Travelling Salesman Problem) is equivalent to the question whether the n-cycle Cn packs with the complement G of G.
Example 1.2 The independence number α(G) of an n-vertex graph G is at least k if and only if G packs with the graph Kk + Kn−k consisting of the k-clique and n −k isolated vertices.
Example 1.3 An n-vertex graph G is k-colorable if and only if G packs with an n-vertex graph that is the union of k cliques.
A proper vertex coloring of a graph is equitable if the sizes of its color classes differ by at most one.
Example 1.4 An n-vertex graph G is equitably k-colorable if and only if G packs with the complement H(n, k) of the Tur´ an Graph T(n, k), i.e. with the n-vertex graph whose every component is a complete graph with either ⌊n/k⌋or ⌈n/k⌉ver-tices.
Tur´ an-type and Ramsey-type problems also can be naturally stated in the lan-guage of graph packing.
Since finding optimal solutions of many graph packing problems is NP-hard, corresponding extremal problems giving sufficient conditions for packing graphs are of great interest. Some well-known theorems of this type can be translated into the language of packing as follows.
Theorem 1.5 (Dirac ) If G is an n-vertex graph and ∆(G) ≤n/2 −1, then G packs with the cycle Cn of length n.
113 114 H. A. Kierstead, A. V. Kostochka, and G. Yu Theorem 1.6 (Ore ) If G is an n-vertex graph and d(x) + d(y) ≤n −2 for every edge xy in G, then G packs with the cycle Cn of length n.
Theorem 1.7 (Hajnal and Szemer´ edi ) Let G be an n-vertex graph with ∆(G) ≤r. Then G packs with the graph H(n, r+1), whose components are complete graphs with either ⌊n/(r + 1)⌋or ⌈n/(r + 1)⌉vertices.
Ore’s theorem motivates considering the notion of Ore-degree, θ(xy), of an edge xy in a graph G as the sum, d(x)+d(y), of the degrees of its ends in G. By definition, the Ore-degree of an edge xy is two greater than the degree of the vertex xy in the line graph of G, and coincides with the degree of xy in the total graph of G. We let the Ore-degree of a graph G be θ(G) = maxxy∈E(G) θ(xy). Thus, Ore’s theorem says that every n-vertex graph G with n ≥3 and θ(G) ≤n −2 packs with the cycle Cn.
Observe that θ(G) is closely related to the maximum degree of the line graph L(G): θ(G) = ∆(L(G)) + 2, and that any bound on θ(G) is in fact a bound on ∆(L(G)). Observe also that for every graph G, ∆(G) + δ∗(G) ≤θ(G) ≤2∆(G), (1.1) where δ∗(G) is the minimum positive degree of a vertex in G. The left inequality in (1.1) is obtained by considering an edge xy ∈E(G) incident with a vertex of the maximum degree.
In view of Dirac’s and Ore’s theorems, we call upper bounds in terms of maximum degree giving sufficient conditions for packing graphs Dirac-type bounds and those in terms of the Ore-degree Ore-type bounds.
In this survey, we compare recent progress in Dirac-type and Ore-type bounds for graph packing problems. We discuss the general problem in the next section, packing a graph with a power of a cycle or path in Section 3, graph coloring in Section 4, equitable coloring in Section 5, and equitable list coloring in Section 6.
We use the standard notation. In particular, for a graph G, |G| denotes the order and ||G|| denotes the size of G.
2 General packing results Some milestone results on extremal graph packing problems were obtained in the seventies. In the same issue of the Journal of Combinatorial Theory (B), funda-mental papers by Bollob´ as and Eldridge and Sauer and Spencer appeared.
The papers gave sufficient conditions for packing of two graphs with given average or maximum degree. Some of these results were also obtained by Catlin in his Ph.D.
Thesis and in . In particular, Sauer and Spencer proved the following Dirac-type bound.
Theorem 2.1 (Sauer and Spencer ) If G1 and G2 are n-vertex graphs and 2∆(G1)∆(G2) < n, then G1 and G2 pack.
The proof is simple (we will show it for a half-Ore version), but the bound is sharp for even n.
Graph packing problems 115 Example 2.2 Let G1 be a perfect matching on n vertices, and G2 contain K1+n/2 and have maximum degree n/2. Then 2∆(G1)∆(G2) = 2 · 1 · n/2 but G1 and G2 do not pack, since in any packing of G1 and G2 only one end of each edge of G1 can be placed onto a vertex in the copy of K1+n/2 in G2.
Example 2.3 Let n ≡2 (mod 4). Let G1 be a perfect matching on n vertices, and G2 = Kn/2,n/2. Again 2∆(G1)∆(G2) = 2 · 1 · n/2. Again, G1 and G2 do not pack, since in any packing of G1 and G2, the ends of each edge of G1 should be placed into the same partite set of Kn/2,n/2, but n/2 is odd.
Kaul and Kostochka proved that Examples 2.2 and 2.3 are the only examples where the bound of Sauer and Spencer is attained.
The half-Ore version of Theorem 2.1 mentioned above is as follows.
Theorem 2.4 If G1 and G2 are n-vertex graphs and θ(G1)∆(G2) < n, (2.1) then G1 and G2 pack.
Proof. For i ∈, set Gi = (Vi, Ei), di(v) = dGi(v), and ∆i = ∆(Gi). A packing of G2 with G1 will be viewed as a mapping f of V1 onto V2 such that if uv ∈E1 then f(u)f(v) / ∈E2. We argue by induction on ∥G1∥. The base step ∥G1∥= 0 is trivial, so consider the induction step. Let x be a vertex with the minimum positive degree δ∗in G1 and xy be any edge incident with x. By the induction hypothesis there exists a packing f of G1 −xy with G2. If f(x)f(y) / ∈E2, then we are done.
Otherwise, we will show that there is some good z ∈V −y such that the mapping gz obtained from f by switching the images of x and z is a packing of G1 with G2.
Suppose z ∈V −y is bad, i.e., not good. Then there exists an edge uv ∈E1 with gz(u)gz(v) ∈E2. Since f packs G1 −xy with G2, at least one, say v, of u, v is in {x, z}. Then either (1) xu ∈E1 and gz(x)gz(u) = f(z)f(u) ∈E2 or (2) zu ∈E1 and gz(z)gz(u) = f(x)f(u) ∈E2. The number of z for which (1) holds is at most d1(x)∆2: There are at most d1(x) choices for u and each choice witnesses at most ∆2 different z. Similarly, the number of z for which (2) holds is at most d2(f(x))∆1.
Noting that the choice z = x, u = y has been counted twice, and using both (1.1) and (2.1), the total number of such bad z is at most d1(x)∆2 + d2(f(x))∆1 −1 ≤(δ∗+ ∆1)∆2 −1 ≤θ(G1)∆2 −1 < n −1 = |V1 −y|.
Thus there exists a good z ∈V1 −y, and so gz is a packing of G1 with G2.
2 A sharpening of this result was proved in : Theorem 2.5 If two n-vertex graphs G1 and G2 satisfy θ(G1)∆(G2) ≤n, then G1 and G2 pack, with the following exceptions: 1. G1 is a perfect matching and G2 either is Kn/2,n/2 with n/2 odd or contains Kn/2+1; 2. G2 is a perfect matching, and G1 either is Kr,n−r with r odd or contains Kn/2+1.
116 H. A. Kierstead, A. V. Kostochka, and G. Yu Theorem 2.5 shows that Theorem 2.4 has extremal graphs other than Exam-ples 2.2 and 2.3, but not many. Observe also that the conditions in Theorems 2.5 and 2.4 involve ∆(G2). The following Ore-type analogue of Theorem 2.1 was con-jectured in .
Conjecture 2.6 If G1 and G2 are n-vertex graphs and θ(G1)θ(G2) < 2n, then G1 and G2 pack.
The conjecture looks natural and maybe has a simple proof, but so far we have failed to find any proof.
The main conjecture in the area (strengthening Theorem 2.1) is the following BEC-conjecture: Conjecture 2.7 If G1 and G2 are n-vertex graphs with maximum degrees ∆1 and ∆2, respectively, and (∆1 + 1)(∆2 + 1) ≤n + 1, then G1 and G2 pack.
This conjecture was posed by Bollob´ as and Eldridge (see also [8, 10]) and inde-pendently by Catlin .
The following examples show that the conjecture is sharp for all values of ∆(G1) and ∆(G2), if true.
Example 2.8 Let positive integers ∆1 and ∆2 be fixed. Let G1 be the disjoint union of ∆2 copies of K∆1+1 and one copy of K∆1−1. Let G2 be the disjoint union of ∆1 copies of K∆2+1 and one copy of K∆2−1. Then |V (G1)| = |V (G2)| = (∆1 + 1)(∆2 + 1) −2. Suppose that G1 and G2 pack. Then in this packing, each of the ∆1 copies of K∆2+1 in G2 should intersect all the ∆2 + 1 components of G1. But the K∆1−1-component of G1 cannot meet all these ∆1 copies of K∆2+1.
Example 2.9 Let a positive integer ∆1 and a positive odd integer ∆2 be fixed. Let G1 be the disjoint union of ∆2 copies of K∆1+1 and one copy of K∆1−1. Let G2 be the disjoint union of ∆1 −1 copies of K∆2+1 and one copy of K∆2,∆2. Again |V (G1)| = |V (G2)| = (∆1 + 1)(∆2 + 1) −2. Suppose that G1 and G2 pack. In this packing, each of the ∆1 −1 copies of K∆2+1 in G2 should intersect all the ∆2 + 1 components of G1. This leaves two vertices in each clique of size ∆1 + 1 in G1 and K∆2,∆2 in G2. So we come to Example 2.3 of graphs that do not pack.
Only very special cases of the BEC-conjecture have been proved. In particular, the Hajnal-Szemer´ edi Theorem (Theorem 1.7) on equitable colorings verifies the conjecture in the case when G2 is the disjoint union of cliques of the same size.
Aigner and Brandt and independently (for huge n) Alon and Fisher settled the conjecture in the case ∆1 ≤2 (this particular case was conjectured by Sauer and Spencer ).
Csaba, Shokoufandeh, and Szemer´ edi proved the BEC-conjecture for ∆1 = 3 and huge n. Bollob´ as, Kostochka and Nakprasit showed that although the BEC-conjecture is sharp, if one of the two graphs is sparse, to be precise, d-degenerate for a small d, then much weaker conditions on ∆1 and ∆2 imply the existence of a packing. Recall that a graph G is d-degenerate if every subgraph G′ of G has a vertex of degree (in G′) at most d. In this case, the vertices of G can be ordered so that each vertex has fewer than col(G) := d + 1 neighbors that precede it.
Graph packing problems 117 Theorem 2.10 (Bollob´ as, Kostochka and Nakprasit ) Let d ≥2. Let G1 be a d-degenerate graph of order n and maximum degree ∆1 and G2 be a graph of order n and maximum degree at most ∆2. If 40∆1 ln ∆2 < n and (2.2) 40d∆2 < n, (2.3) then there is a packing of G1 and G2.
If ∆2 ≥215, then ∆2/ ln ∆2 ≥40. Therefore, Theorem 2.10 yields that the BEC-conjecture holds if ∆2 ≥215 and ∆1 ≥40d. Adapting the proof of Theorem 2.10 to control the maximum degree of the union of the two packed graphs, implies the following result on simultaneous packing of many graphs.
Theorem 2.11 (Bollob´ as, Kostochka and Nakprasit ) Let n, d, ∆and q be positive integers such that d ≥2, q ≤ n 1500d2 , and 1000d∆< n ln n. Let F1, . . . , Fq be d-degenerate graphs of order n and maximum degree at most ∆. Then F1, . . . , Fq pack.
For a fixed d, Theorem 2.11 allows packing linearly many (in n) d-degenerate n-vertex graphs of moderate maximum degree.
The phenomenon here is that it is easier to pack graphs if the number of vertices is significantly greater than the maximum degrees of the graphs to be packed.
Clearly, restriction (2.3) cannot be weakened by more than 40 times. The fol-lowing result in shows that (2.2) is also weakest up to a constant factor.
Theorem 2.12 (Bollob´ as, Kostochka and Nakprasit ) Let k be a positive integer and q be a prime power.
Then for every n ≥q qk+1−1 q−1 , there are graphs G1(n, k) and G2(n, q, k) of order n that do not pack and have the following properties.
(a) G1(n, k) is a forest with n −k edges and maximum degree at most n/k; (b) G2(n, q, k) is a qk−1 q−1 -degenerate graph with maximum degree at most 2n/q.
So, for l ≥5, q = 3l, k ≥l, and n = 3 2(3k+1 −1), consider the graphs G1 = G1(n, k) and G2 = G2(n, 3l, k) of Theorem 2.12. Then (2.3) holds for l > 4 (we need d ≥2 in the theorem), so the bounding inequality is (2.2).
Furthermore, ∆(G1) ln ∆(G2) ≤ n k ln n < n k (1 + (k + 1) ln 3) < 2n, so restriction (2.2) can be possibly weakened only by a constant factor.
Motivated by Theorem 2.10, Theorem 2.1 has very recently been strengthened in terms of game coloring number. This parameter is defined by means of the marking game, which is played by two players Alice and Bob on a graph G. Bob decides who plays first, and then the players take turns choosing unchosen vertices until all vertices have been chosen. Let L be the order in which the vertices are chosen. The score of the game is the least integer s such that every vertex has fewer than s neighbors that precede it in L. The game coloring number gcol(G) is the least integer k such that Alice has a strategy for obtaining a score of at most k regardless of how Bob plays. The new result says: 118 H. A. Kierstead, A. V. Kostochka, and G. Yu Theorem 2.13 () If two graphs G1 and G2 satisfy (gcol(G1) −1)∆(G2) + (gcol(G2) −1)∆(G1) < n, then they pack.
Theorem 2.13 strengthens Theorem 2.1, since for every graph G, col(G) ≤gcol(G) ≤∆+ 1.
Moreover, for some important classes of graphs the upper bound on gcol(G) can be greatly improved. For instance, Zhu showed that every planar graph G satisfies gcol(G) ≤17.
In , the BEC-conjecture was attacked from a different direction: instead of proving the conjecture for another class of graphs, the following weaker bound for all graphs with high maximum degrees was proved.
Theorem 2.14 () Let G1 and G2 be n-vertex graphs with maximum degrees ∆1 and ∆2, respectively. If ∆1, ∆2 ≥300 and (∆1 + 1)(∆2 + 1) ≤0.6n + 1, then G1 and G2 pack.
This improves the bound of the Sauer–Spencer Theorem for large ∆1 and ∆2 and thus partially answers Problem 4.4 in . Helpful concepts used in the proof are those of critical pairs and of cyclic switchings. Cyclic switchings generalize the ordinary switchings used in the proofs of Theorem 2.4 and the original Sauer–Spencer Theorem.
In all extremal examples (G1, G2) for the BEC-conjecture that we know, G1 contains ∆(G1)-regular and G2 contains ∆(G2)-regular components. This fact and Theorem 2.10 suggest that maybe the following Ore-type analogue of the BEC-conjecture holds.
Conjecture 2.15 () If G1 and G2 are n-vertex graphs and (0.5θ(G1) + 1)(∆(G2) + 1) ≤n + 1, then G1 and G2 pack.
The case of Conjecture 2.15 when G2 is the union of vertex disjoint triangles (which is the analogue of Corr´ adi–Hajnal Theorem ) was proved independently by Enomoto and Wang . The following extension of the results of Enomoto and Wang is a small step towards Conjecture 2.15.
Theorem 2.16 () Each n-vertex graph G with θ(G) ≤4n−8 3 packs with any n-vertex graph H such that each component of H is either C3, or K2, or C5, or K4 −e, or a vertex.
Yet another way to attack the BEC-conjecture gives the concept of near packing introduced by Eaton . This is an extension of the notion of defective coloring.
In a near packing of degree d, the copies of the two graphs may overlap so that the maximum degree of the subgraph spanned by the edges common to both copies is at most d. Thus a near packing of degree 0 is an ordinary packing. The following result was proved in : Graph packing problems 119 Theorem 2.17 (Eaton ) Let 0 < a ≤1. If G1 and G2 are n-vertex graphs, ∆(G1) + ∆(G2) ≤n + a −1, and ∆(G1) · ∆(G2) < an, then there exists a near packing of G1 and G2 of degree d for some integer d < 2a. Furthermore, if (∆(G1)+ 1)(∆(G2) + 1) ≤n + 1, then there exists a near packing of G1 and G2 of degree 1.
Eaton also posed the conjecture below and showed that it is sharp if true: Conjecture 2.18 (Eaton ) Let G1 and G2 be n-vertex graphs and p be a pos-itive real number. If n ≥⌊∆(G1) p ⌋(∆(G2) + 1) + ⌊∆(G2) p ⌋(∆(G1) + 1 −p⌊∆(G1) p ⌋), then there exists a near packing of G1 and G2 of degree less than p.
It would be interesting to prove the Ore-version or half-Ore version of Theo-rem 2.17. For example, is it true that if (0.5θ(G1) + 1)(∆(G2) + 1) ≤n + 1, then there exists a near packing of G1 and G2 of degree 1?
Other interesting results on graph packing were obtained by Brandt , Csaba [19, 20], Fan and Kierstead [28, 29], Koml´ os , Koml´ os, S´ ark˝ ozy and Szemer´ edi [49, 50, 51], Sauer and Wang , Wozniak [37, 67], Yap [72, 73, 78] and others. One can look into the 75-page survey of the topic by Wozniak. Some of these results are discussed in the next section. After this paper was accepted, we learned about the following very interesting result by K¨ uhn, Osthus and Treglown : for every fixed graph H and large n, they found asymptotically exact Ore-type conditions for an n-vertex graph G of order divisible by |H| to contain a perfect H-packing of G, i.e., to contain vertex-disjoint copies of H that cover V (G).
3 Packing a graph with a power of a cycle Let H = v1v2 · · · vn be a path or a cycle. An r-chord is an edge of the form vivi+r, where addition is modulo n in the case that H is a cycle. The r-th power of H is the graph obtained by adding all i-chords with i ≤r. The second power of H is called the square of H. P´ osa (see ) conjectured that every graph on n vertices with minimum degree at least 2 3n contains the square of Cn. Seymour strengthened P´ osa’s Conjecture as follows.
Conjecture 3.1 (Seymour ) Every graph G on n vertices with δ(G) ≥ r r+1n contains the r-th power of Cn.
The case r = 1 is Dirac’s Theorem and the case r = 2 is P´ osa’s Conjecture.
Rephrased in terms of packing, Seymour’s Conjecture looks like this.
Conjecture 3.2 The r-th power of Cn packs with every graph G on n vertices with ∆(G) ≤ n r+1 −1.
Seymour’s conjecture implies the packing version of the Hajnal-Szemer´ edi Theorem, since any r+1 consecutive vertices of the r-th power of a cycle induce Kr+1. Indeed, Seymour’s original motivation for his conjecture was to find an understandable proof of the Hajnal-Szemer´ edi Theorem. Fan and Kierstead came close to proving 120 H. A. Kierstead, A. V. Kostochka, and G. Yu Posa’s conjecture, but had to settle for finding the square of a hamiltonian path, albeit with a slightly weaker hypothesis.
Theorem 3.3 (Fan and Kierstead ) Every graph G on n vertices which sat-isfies δ(G) ≥2n−1 3 contains the square of Pn.
The following example shows that the theorem is best possible for paths.
Example 3.4 Let G be the complete tripartite graph Kt−1,t+1,t+1.
Then n := |G| = 3t + 1 and δ(G) = 2t = 2n−2 3 . If H is a square path in G, then any three consecutive vertices must be in distinct parts. It follows that G does not contain the square of Pn, since when the last vertex in the small part is used there will still be four unused vertices. Notice also that adding edges inside the small part will not create the square of Pn.
Not only does the weakened hypothesis give the best possible result with respect to paths, but it also implies a strengthened version of the Aigner-Brandt Theorem.
Observe that every graph H on n vertices with ∆(H) ≤2 is contained in the square of Pn. In other words, the square of Pn is universal with respect to graphs H on n vertices with ∆(H) ≤2.
Theorem 3.5 (Fan and Kierstead ) There exists a universal graph U on n vertices with ∆(U) = 4 such that (1) U packs with every graph G on n vertices with ∆(G) ≤n−2 3 and (2) U contains every graph H on n vertices with ∆(H) ≤2.
Fan and Kierstead also proved: Theorem 3.6 (Fan and Kierstead ) If G is a graph on n vertices such that δ(G) ≥2 3n, then either (1) G contains the square of Cn or (2) G does not contain the square of Ck for k > 2 3n, but there exist integers i and j with i + j = n and i ≤j ≤2 3n such that G contains the squares of vertex disjoint copies of Ci and Cj.
P´ osa’s Conjecture remains open, but Koml´ os, S´ ark˝ ozy and Szemer´ edi [49, 50] have used Szemer´ edi’s Regularity Lemma and their own Blow-Up Lemma to prove it, and the more general conjecture of Seymour, for huge graphs.
Theorem 3.7 (Koml´ os, S´ ark˝ ozy and Szemer´ edi ) For every positive inte-ger r there exists an integer N such that for every n > N every graph G on n vertices with δ(G) ≥ r r+1n contains the r-th power of a hamiltonian cycle.
It is well known that every balanced bipartite graph G on n = 2s vertices with δ(G) ≥s+1 2 contains Cn. Wang conjectured that if δ(G) ≥s 2 + 1 then G is universal with respect to bipartite graphs H on n vertices with δ(H) ≤2.
He showed that if true, this is best possible.
Example 3.8 Let G be the balanced bipartite graph on n = 2s = 4t + 2 vertices formed by taking two copies of Kt,t+1 and joining the two larger parts by a matching with t + 1 edges. Then δ(G) = t + 1 = s+1 2 , but G does not contain Cs+1 + Cs−1.
Graph packing problems 121 Define the ladder Ln to be the balanced bipartite graph on n = 2s vertices a1, b1, · · · , as, bs such that aibj ∈E(Ln) if and only if |i −j| ≤1. It is easy to check that Ln is universal with respect to all bipartite graphs H on n vertices with ∆(H) ≤2. Czygrinow and Kierstead used the Regularity and Blow-Up lemmas to prove the following strengthening of Wang’s conjecture for huge graphs.
Theorem 3.9 (Czygrinow and Kierstead ) There exists an integer N such that every balanced bipartite graph G on n = 2s > N vertices with δ(G) ≥s 2 + 1 contains Ln.
Amar posed the following version of an Ore-type conjecture.
Conjecture 3.10 (Amar ) Every balanced A, B-bigraph G on n = 2s vertices satisfying d(a) + d(b) ≥s + k for all a ∈A and b ∈B contains a copy of every balanced bipartite graph H on n vertices with k components and ∆(H) ≤2.
Bondy and Chv´ atal proved the case k = 1 (prior to the conjecture) and Amar proved the case k = 2. Recently Czygrinow, DeBiasio and Kierstead proved the following two theorems. Together they imply Amar’s conjecture for sufficiently large graphs. Moreover, Theorem 3.12 shows that the degree bound can be relaxed to d(a) + d(b) ≥s + 2 as long as n is large in terms of k.
Theorem 3.11 (Czygrinow, DeBiasio and Kierstead ) For some integer K, every balanced A, B-bigraph G on n = 2s vertices satisfying d(a) + d(b) ≥s + K for all a ∈A and b ∈B contains Ln.
Theorem 3.12 (Czygrinow, DeBiasio and Kierstead ) For every integer k there exists an integer Nk such that every balanced A, B-bigraph G on n = 2s > Nk vertices satisfying d(a) + d(b) ≥s + 2 for all a ∈A and b ∈B contains a copy of every balanced bipartite graph H on n vertices with k components and ∆(H) ≤2.
Moreover if δ(G) ≥ 1 100ks, then G contains Ln.
4 Coloring As mentioned in the introduction, various coloring problems are important in-stances of packing problems.
An obvious (but sharp) Dirac-type bound on the (ordinary) chromatic number is χ(G) ≤∆(G) + 1, (4.1) where χ(G) is the chromatic number of G. Brooks’ Theorem below characterizes the graphs for which (4.1) holds with equality.
Theorem 4.1 (Brooks) If χ(G) = ∆(G) + 1, then either G contains the complete graph K∆(G)+1 or ∆(G) = 2 and G contains an odd cycle.
The counterpart of (4.1) for θ(G) is χ(G) ≤⌊θ(G)/2⌋+ 1.
(4.2) 122 H. A. Kierstead, A. V. Kostochka, and G. Yu x y x z u x y z’ u z’’ y Figure 1: Above: a graph with θ = 9 and χ = 5. Below: two graphs with θ = 7 and χ = 4.
The proof is also obvious and the bound is also attained at complete graphs.
However for small odd θ there are more connected graphs for which (4.2) holds with equality.
Example 4.2 Let G be the top graph in Figure 1, i.e., the 9-vertex graph with V (G) = Q ∪Q′, Q ∩Q′ = {x, y}, G[Q] = K5 −xy, and G[Q′] = K6 −T, where T is a tree with four leaves, xy ∈E(T) and d(x) = 3 = d(y). Since x and y are the only vertices with degree greater than 4 and are not adjacent to each other, θ(G) = 9. Every (proper) 4-coloring of G[Q] assigns x and y the same colors, since G[NQ(x) ∩NQ(y)] = K3. On the other hand, every 4-coloring of G[Q′] assigns x and y different colors, since G[NQ′(x) ∪NQ′(y)] = K4. Thus χ(G) > 4.
Example 4.3 Let G be a graph with θ(G) ≤7 and χ(G) = 4 (for example, G = K4).
We construct from G a (3 + |V (G)|)-vertex graph G′ with θ(G′) ≤7 and χ(G′) = 4 as follows. Choose a vertex v ∈V (G) that has no neighbors of degree 4 (each vertex of degree 4 has this property). Split v into two vertices v1 and v2 of degree at most two. Add two new vertices xv and yv that are adjacent to v1, v2, and to each other. By construction, θ(G′) = 7. Suppose that G′ has a 3-coloring f.
Since both, v1 and v2 are adjacent to xv and yv, we need f(v1) = f(v2). But then f yields a 3-coloring of G, contrary to our assumption. The two bottom graphs in Figure 1 illustrate the idea (with G on the left).
Iterating the idea of Example 4.3 yields infinitely many 2-connected graphs G with θ(G) ≤7 and χ(G) = 4. In contrast, for graphs with Ore-degree at least 12 (i.e., with chromatic number at least 7), the only extremal connected graphs are complete graphs.
Graph packing problems 123 Theorem 4.4 () If 7 ≤χ(G) = ⌊θ(G)/2⌋+ 1, then G contains the complete graph Kχ(G).
We think that the statement of Theorem 4.4 holds also for graphs G with Ore-degree at least 10 but cannot prove it. An even more challenging task would be to describe connected graphs G with θ(G) ≤7 and χ(G) = 4.
5 Equitable coloring In several applications of coloring as a partition problem there is an additional requirement that color classes be not too large or be of approximately the same size.
Examples are the mutual exclusion scheduling problem [6, 71], scheduling in communication systems , construction timetables , and round-the-clock scheduling . For other applications in scheduling, partitioning, and load balanc-ing problems, one can look into [7, 58, 71]. A model imposing such a requirement is equitable coloring—a proper coloring such that color classes differ in size by at most one. As mentioned in the introduction, equitable coloring is a particular case of the general graph packing problem. Alon and F¨ uredi , Pemmaraju and Janson, Luczak, and Ruci´ nski used equitable colorings to give new bounds on tails of distributions of sums of random variables. R¨ odl and Ruci´ nski used equitable colorings to give a new proof of the Blow-Up Lemma.
In contrast to ordinary coloring, a graph may have an equitable k-coloring (i.e., an equitable coloring with k colors) but have no equitable (k + 1)-coloring. Thus, it is natural to look for the minimum number, eq(G), such that for every k ≥eq(G), G has an equitable k-coloring.
Finding eq(G) even for planar graphs G is an NP-hard problem. This motivates a series of extremal problems on equitable colorings. The Dirac-type Theorem 1.7 by Hajnal and Szemer´ edi in the original language is as follows.
Theorem 5.1 (Hajnal and Szemer´ edi ) Every graph G with maximum de-gree at most r has an equitable (r + 1)-coloring.
The proof was long and sophisticated. A shorter proof appeared in . Then Kierstead, Kostochka, Mydlarz, and Szemer´ edi devised an algorithm that in time O(rn2) finds an equitable (r+1)-coloring for any n-vertex graph with maximum degree at most r. It is based on a modification of the proof of Theorem 5.1 in .
Here we present a yet shorter and simpler proof of Theorem 5.1.
Proof of Theorem 5.1. Let G be a graph with ∆(G) ≤r. We may assume that |G| is divisible by r+1: If |G| = s(r+1)−p, where p ∈[r], then set G′ = G+Kp.
By construction, |G′| is divisible by r + 1 and ∆(G′) ≤r. Moreover, the restriction of any equitable (r + 1)-coloring of G′ to G is an equitable (r + 1)-coloring of G. So we may assume |G| = (r + 1)s.
We argue by induction on ∥G∥. The base step ∥G∥= 0 is trivial, so consider the induction step. Let u be a non-isolated vertex. By the induction hypothesis, there exists an equitable (r +1)-coloring of G−E(u), where E(u) denotes the set of edges incident with u. We are done unless some color class V contains an edge uv. Since ∆(G) ≤r, some color class T contains no neighbors of u. Moving u to T yields an 124 H. A. Kierstead, A. V. Kostochka, and G. Yu (r +1)-coloring of G with all classes of size s, except for one small class V −= V −u of size s −1 and one large class V + = T + u of size s + 1. Such a coloring is called nearly equitable.
Given a nearly equitable (r + 1)-coloring, define an auxiliary digraph H, whose vertices are the color classes, so that UW is a directed edge if and only if some vertex y ∈U has no neighbors in W. In this case we say that y witnesses edge UW. Let A be the set of classes from which V −can be reached in H and B be the set of classes not in A. Set a = |A|, b = |B|, A = S A and B = S B. Then r + 1 = a + b. Since every vertex y ∈B has a neighbor in every class of A, dA(y) ≥a for all y ∈B.
() Case 0: V + ∈A.
Then there exists a V +, V −-path P = V1, . . . , Vk in H.
Moving each witness yj of VjVj+1 to Vj+1 yields an equitable (r + 1)-coloring of G.
We now argue by a secondary induction on b, whose base step b = 0 holds by Case 0. If V + / ∈A, then |A| = as −1 and |B| = bs + 1. Consider the secondary induction step.
A class W ∈A is terminal if every U ∈A −W can reach V −in H −W. Let A′ be the set of terminal classes, a′ = |A′| and A′ = S A′. An edge wz is solo if w ∈W ∈A′, z ∈B and NW (z) = {w}. Ends of solo edges are solo vertices and solo neighbors of each other.
If V −∈A′ then no other class can be in A. Thus in this case a = 1, b = r and by (), |E(A, B)| ≥|B| = rs+1, a contradiction to the fact that |E(A, B)| ≤r|A| = r(s −1). So, V −/ ∈A′.
Suppose that some solo vertex w ∈W ∈A′ witnesses an edge WX ∈E(H[A]).
Let y ∈B be a solo neighbor of w. Move w to X and y to W. This yields nearly equitable colorings of G[A + y] and G[B −y]. Since W is terminal, X + w can reach V −in H −W. Thus by Case 0, G[A + y] has an equitable a-coloring. By (), ∆(G[B −y]) ≤b −1. So by the primary induction hypothesis, G[B −y] has an equitable b-coloring.
Combining these equitable colorings yields an equitable (r + 1)-coloring of G. Thus, we may assume: Each solo vertex z ∈A′ witnesses no edges in H, and so dA(z) ≥a −1.
() Order A as X0, X1, . . . , Xa−1 so that X0 = V −and each Xi has a previous out-neighbor.
Case 1: For some a −b ≤i ≤a −1, class Xi is not terminal. This includes the case a ≤b. Then some Xj ∈A′ cannot reach V −in H −Xi. So j > i and Xj has no out-neighbors before Xi.
In particular, d+ A(Xj) < b.
Then for each w ∈Xj, dA(w) ≥a −b, and so dB(w) < 2b. Let S be the set of solo vertices in Xj, and D = Xj \ S.
By definition, each v ∈B either has a neighbor in S or at least two neighbors in D. Thus P w∈Xj dB(w) ≥2|B| −|NB(S)|. By (), P w∈S dA(w) ≥(a −1)|S| and hence |NB(S)| ≤b|S|. Since |S| + |D| = s, we have the following contradiction: rs ≥ X w∈Xj (dA(w) + dB(w)) ≥(a −b)|D| + (a −1)|S| + 2|B| −|NB(S)| ≥ ≥(a −b)|D| + (a −1)|S| + 2bs + 2 −b|S| = (a + b)|D| + 2 + (a + b −1)|S| > rs.
Graph packing problems 125 Case 2: All the last b classes Xa−b, . . . , Xa−1 are terminal. Then a′ ≥b. For y ∈B, let σ(y) be the number of solo neighbors of y. For each y ∈B, r ≥d(y) ≥a + dB(y) + (a′ −σ(y)) ≥r + 1 + dB(y) + a′ −b −σ(y).
So σ(y) ≥a′ −b+dB(y)+1. Let I be a maximal independent set with V + ⊆I ⊆B.
Then P y∈I(dB(y) + 1) ≥|B| = bs + 1. Since a′ ≥b, X y∈I σ(y) ≥ X y∈I (a′ −b + dB(y) + 1) ≥s(a′ −b) + bs + 1 > a′s = A′ .
So some vertex w ∈W ∈A′ has two solo neighbors y1 and y2 in the independent set I.
By the primary induction hypothesis, we can equitably b-color G[B −y1]. Let Y ′ be the class of y2 in this coloring. By (), dB−y1(w) ≤r −(a −1) −1 = b −1 and we can move w to some class U ⊆B −y1. Replacing w with y1 in W to get W ∗and moving w to U yields a new nearly equitable (r + 1)-coloring of G. Now at least a + 1 classes, W ∗, Y ′, and all X ∈A′ −W, can reach V −. In this case we are done by the secondary induction hypothesis.
2 As discussed in Example 2.3, if r is odd, then Kr,r has no equitable r-coloring.
Chen, Lih and Wu proposed the following common strengthening of Theorem 5.1 and Brooks’ theorem.
Conjecture 5.2 (Chen, Lih and Wu ) Let G be a connected graph such that ∆(G) = r. Then G has no equitable r-coloring if and only if either (1) G = Kr+1, or (2) r = 2 and G is an odd cycle, or (3) r is odd and G = Kr,r.
Some partial cases of Conjecture 5.2 were proved in [17, 52, 62, 79, 80].
In particular, Chen, Lih and Wu proved that the conjecture holds for r = 3: Theorem 5.3 (Chen, Lih and Wu ) Let G be a connected graph such that ∆(G) ≤3. Then G has no equitable 3-coloring if and only if G = K4 or G = K3,3.
Brooks’ theorem characterizes all graphs with maximum degree r that are r-colorable, since a graph is r-colorable if and only if each of its components is. This is not the case for equitable r-coloring. For example, for each odd r ≥3, the graph consisting of two disjoint copies of Kr,r has an equitable r-coloring, but the graph consisting of a copy of Kr,r and a copy of Kr does not. This construction can be generalized. Say that a graph H is r-equitable if |H| is divisible by r, H is r-colorable and every r-coloring of H is equitable. If G contains Kr,r and G−Kr,r is r-equitable, then G does not have an equitable r-coloring. This motivates the study of equitable graphs, i.e., graphs that are r-equitable for some r. It was proved in that there is a good description of the family of all r-equitable graphs; they can all be built from simple examples in a straightforward way.
If an r-colorable graph G has a spanning subgraph whose components are all r-equitable, then G is also r-equitable. We say that an r-equitable graph G is r-reducible if V (G) has a partition {V1, . . . Vt} into at least two parts such that G[Vi] is r-equitable for each i ∈[t]; otherwise G is r-irreducible. Clearly Kr is r-irreducible.
The reader can see one other 5-irreducible graph F1 and three other 4-irreducible 126 H. A. Kierstead, A. V. Kostochka, and G. Yu 3 2 F 1 F F 4 F Figure 2: One 5-equitable and three 4-equitable basic graphs.
F 8 F 5 6 F 7 F 9 F 10 F Figure 3: Six 3-equitable basic graphs.
Graph packing problems 127 graphs F2, F3, F4 in Figure 2 and six other 3-irreducible graphs F5, . . . , F10 in Fig-ure 3.
Together with Kr, the r-irreducible graphs from this list are the r-basic graphs.
An r-decomposition of G is a partition of V (G) into subsets V1, . . . , Vt such that each G[Vi] is r-basic. We say that G is r-decomposable if it has an r-decomposition. As was just mentioned, if r is odd and G is r-decomposable, then G ∪Kr,r has no equitable r-coloring. It was conjectured in that this is the only obstacle that prevents an r-colorable graph with ∆(G) ≤r from having an equitable r-coloring.
Conjecture 5.4 () Suppose that r ≥3 and G is an r-colorable graph with ∆(G) = r. Then G has no equitable r-coloring if and only if r is odd and there exists H ⊆G such that H = Kr,r and G −H is r-decomposable.
For r ≥6, this conjecture means that if an r-colorable graph G with ∆(G) ≤r has no equitable r-coloring, then r is odd and V (G) can be partitioned into sets V0, . . . , Vt such that G[V0] = Kr,r and G[Vi] = Kr for each i = 1, . . . , r.
A nearly equitable r-coloring of a graph G is defined in the above proof of Theo-rem 5.1. In this case |G| is divisible by r. If r is odd, G contains Kr,r and G −Kr,r has a nearly equitable r-coloring, then G has an equitable r-coloring, since the small class of one of the components can be combined with the large class of the other.
This explains our interest in nearly equitable r-colorings.
Let G(r) be the class of all graphs G with ∆(G) ≤r and χ(G) ≤r. Let G(r, n) be the set of graphs in G(r) with at most n vertices. The following result implies that Conjectures 5.4 and 5.2 are equivalent.
Theorem 5.5 () Let G ∈G(r) with |G| divisible by r. The following are equiv-alent: (A) G is r-decomposable; (B) G is r-equitable; (C) G has an equitable r-coloring, but does not have a nearly equitable r-coloring.
Corollary 5.6 () For all positive integers r and n > r, Conjecture 5.2 holds for all graphs in G(r, n) if and only if Conjecture 5.4 holds for all graphs in G(r, n).
Corollary 5.7 () Let G ∈G(r) be r-equitable. Then G has a unique r-decom-position.
Corollary 5.8 () There exists a polynomial time algorithm for deciding if a graph G ∈G(r) is r-equitable.
Very recently, the case r = 4 of Conjecture 5.2 was proved .
The following Ore-type analogue of Theorem 5.1 (slightly strengthening a con-jecture in [54, 56]) was proved in .
Theorem 5.9 () For every r ≥3, each graph G with θ(G) ≤2r + 1 has an equitable (r + 1)-coloring.
128 H. A. Kierstead, A. V. Kostochka, and G. Yu In particular, the theorem yields that Conjecture 5.2 holds for graphs in which vertices of maximum degree form an independent set. The bounding examples for this theorem (apart from Kr+1) are graphs Km,2r−m for every odd 0 < m ≤r. In the same paper, the following Ore-type analogue of Conjecture 5.2 was proposed.
Conjecture 5.10 () Let r ≥3 and G be a connected graph with θ(G) ≤2r.
If G is distinct from Kr+1 and from Km,2r−m for odd m, then G has an equitable r-coloring.
Conjecture 5.10 was proved to hold for r = 3. In the following common extension of Conjectures 5.4 and 5.10 was posed.
Conjecture 5.11 () Suppose that r ≥3. An r-colorable, n-vertex graph G with θ(G) ≤2r has no equitable r-coloring, if and only if n is divisible by r, and there exists W ⊆G such that W = Km,2r−m for some odd m and G−W is r-decomposable.
It was proved in that Conjecture 5.11 is equivalent to Conjecture 5.10 even for graphs with restricted number of vertices and restricted values of Ore-degree.
Theorem 5.12 () Assume that Conjecture 5.10 holds for all graphs with at most n vertices and Ore-degree at most 2r. Let G be an r-colorable n-vertex graph with θ(G) ≤2r. Then G has no equitable r-coloring if and only if n is divisible by r and there exists H ⊆G such that H = Km,2r−m for some odd m and G −H is r-decomposable.
It follows that Conjecture 5.11 holds for r = 3.
6 Equitable list coloring A list analogue of equitable coloring was introduced by Kostochka, Pelsmajer and West . A list assignment L for a graph G assigns to each vertex v ∈V (G) a set L(v) of allowable colors. An L-coloring of G is a proper vertex coloring such that for every v ∈V (G) the color on v belongs to L(v). Given a k-uniform list assignment L for an n-vertex graph G, the graph G is equitably L-colorable if G has an L-coloring where no color is used on more than ⌈n/k⌉vertices. A graph G is equitably k-choosable if G is equitably L-colorable whenever L is a k-uniform list assignment for G.
Because one cannot ensure the appearance of each color, the techniques previ-ously used for ordinary equitable colorings do not work well for equitable list color-ings. Nevertheless, Kostochka, Pelsmajer and West suggested that the following analogue of the Hajnal–Szemer´ edi Theorem holds.
Conjecture 6.1 (Kostochka, Pelsmajer and West ) Every graph G is eq-uitably (1 + ∆(G))-choosable.
Furthermore, Pelsmajer and independently Lih and Wang confirmed this conjecture for graphs with maximum degree at most 3: Theorem 6.2 (Pelsmajer , Lih and Wang ) If ∆(G) ≤3, then G is equitably 4-choosable.
Graph packing problems 129 Very recently, the conjecture was proved for graphs with maximum degree 4 .
We believe that the following Ore-type analogue of Conjecture 6.1 also holds.
Conjecture 6.3 Every graph G is equitably (1 + 0.5θ(G))-choosable.
The result below confirms the conjecture for graphs with small θ and somewhat extends Theorem 6.2.
Theorem 6.4 If θ(G) ≤6, then G is equitably 4-choosable.
Proof Let G be an edge-minimal counterexample to the theorem. In particular, |L(x)| = 4 for every x ∈V (G). Let v be a vertex of maximum degree in G and w1, . . . , wd(v) be the neighbors of v. Since θ(G) ≤6, d(v) ≤5. If d(v) ≤3, then we are done by Theorem 6.2. So, d(v) ∈{4, 5}.
If d(v) = 5, then the component of G containing v is K5,1. Let G1 = G−v−w1− w2 −w3. Since G1 is a proper subgraph of G, θ(G1) ≤6. By the minimality of G, the subgraph G1 admits some equitable L-coloring f. This means that each color is used on at most ⌈|V (G1)|/4⌉= ⌈|V (G)|/4⌉−1 vertices. We extend f to an equitable L-coloring of G as follows: Choose a color α0 ∈L(v) −f(w4) −f(w5) as f(v), and then for i = 1, 2, 3 choose a color αi ∈L(wi) −f(v) −{f(wj) : 1 ≤j ≤i −1} as f(wi). Since each color appears on our four “new” vertices only once, the resulting coloring is equitable.
Suppose now that d(v) = 4. Since θ(G) ≤6, d(wi) ≤2 for each i = 1, 2, 3, 4. For i = 1, 2, 3, 4, let ui be the neighbor of wi distinct from v, if it exists.
Case 1: u1 does not exist or u1 = w2. Consider G1 = G −v −w1 −w2 −w3. By the minimality of G, there exists an equitable L-coloring f of G1. Extend f to an L-coloring of G as follows: Choose a color α ∈L(v) −f(w4) as f(v), then choose f(w3) ∈L(w3) −f(v) −f(u3), f(w2) ∈L(w2) −f(v) −f(w3) −f(u2), and finally f(w1) ∈L(w1) −f(v) −f(w3) −f(w2). Again, since each color appears on our four “new” vertices only once, the resulting coloring is equitable.
So, in what follows, all ui exist and are distinct from all wj.
Case 2: |L(v) −L(w1)| ≥2. Consider G1 = G −v −w1 −w2 −w3 as in Case 1.
Let f be an equitable L-coloring of G1. Choose f(v) ∈L(v) −L(w1) −f(w4), then f(w3) ∈L(w3) −f(v) −f(u3), f(w2) ∈L(w2) −f(v) −f(w3) −f(u2), and finally f(w1) ∈L(w1) −f(u1) −f(w3) −f(w2). Since f(v) / ∈L(w1), the resulting coloring is proper.
From now on, |L(v) ∩L(wi)| ≥3 for all i ∈{1, 2, 3, 4}.
Case 3: |L(v) ∩L(w1) ∩L(w2)| ≥3. We may assume that |L(v) ∩L(w1)| ≥|L(v) ∩L(w2)|.
(6.1) Let G0 be obtained from G−v −w3 −w4 by identifying w1 and w2 into a new vertex w∗. Let L(w∗) = L(w2). By the minimality of G, the new graph G0 with the new list L has an equitable L-coloring f.
Subcase 3.1: f(w∗) ∈L(w1). We define f(w1) = f(w2) = f(w∗) and let f(w3) ∈ L(w3) −f(u3) −f(w∗), f(w4) ∈L(w4) −f(u4) −f(w∗) −f(w3), and f(v) ∈L(v) − f(w3) −f(w4) −f(w∗).
130 H. A. Kierstead, A. V. Kostochka, and G. Yu Subcase 3.2: f(w∗) / ∈L(w1). Under the conditions of Case 3, this means that |L(w1) ∩L(w2)| = |L(v) ∩L(w1) ∩L(w2)| = 3 and now (6.1) yields that f(w∗) / ∈ L(v). So, we can let f(w2) = f(w∗), f(w3) ∈L(w3) −f(u3) −f(w∗), f(w4) ∈ L(w4) −f(u4) −f(w∗) −f(w3), f(w1) ∈L(w1) −f(u1) −f(w3) −f(w4), and f(v) ∈ L(v) −f(w1) −f(w3) −f(w4). This finishes Case 3.
If none of Cases 1–3 takes place, then we may assume that L(v) = {1, 2, 3, 4} and for i = 1, 2, 3, 4, L(wi)∩L(v) = L(v)−i. Construct the graph G0 with the list L as in Case 3. Let f be an equitable L-coloring of G0. If f(w∗) ̸= 1, then we extend f to the whole of G as in Case 3. So, assume that f(w∗) = 1. If f(u3) ̸= 1, then let f(w3) = f(w2) = 1, f(w4) ∈L(w4) −f(u4) −1, f(w1) ∈L(w1) −f(u1) −f(w4), and f(v) ∈{2, 3, 4} −f(w4) −f(w1). Finally, suppose f(u3) = 1. In this case, let f(w2) = 1, f(w3) ∈L(w3) −L(v), f(w4) ∈L(w4) −f(u4) −1 −f(w3), f(w1) ∈ L(w1) −f(u1) −f(w3) −f(w4), and f(v) ∈{2, 3, 4} −f(w4) −f(w1).
□ Acknowledgements Research of the first author is supported in part by NSA grant H98230-08-1-0069.
Research of the second author is supported in part by NSF grant DMS-06-50784 and by grant 06-01-00694 of the Russian Foundation for Basic Research. Research of the third author is partially supported by NSF grant DMS-0852452.
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Department of Mathematics and Statistics, Arizona State University, Tempe, AZ 85287, USA kierstead@asu.edu Department of Mathematics, University of Illinois, Urbana, IL, 61801, USA and Sobolev Institute of Mathematics, Novosibirsk, 630090, Russia kostochk@math.uiuc.edu Department of Mathematics, College of William and Mary Williamsburg, VA 23187, USA gyu@wm.edu |
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24.2: Thermodynamics of Mixing
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: Jul 24, 2020
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24.1: Ideal Solutions - Raoult's Law
25: Extension 16 - Vapor-Solution Phase Diagrams
Page ID
: 238274
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When solids, liquids or gases are combined, the thermodynamic quantities of the system experience a change as a result of the mixing. This module will discuss the effect that mixing has on a solution’s Gibbs energy, enthalpy, and entropy, with a specific focus on the mixing of two gases.
Introduction
A solution is created when two or more components mix homogeneously to form a single phase. Studying solutions is important because most chemical and biological life processes occur in systems with multiple components. Understanding the thermodynamic behavior of mixtures is integral to the study of any system involving either ideal or non-ideal solutions because it provides valuable information on the molecular properties of the system.
Most real gases behave like ideal gases at standard temperature and pressure. This allows us to combine our knowledge of ideal systems and solutions with standard state thermodynamics in order to derive a set of equations that quantitatively describe the effect that mixing has on a given gas-phase solution’s thermodynamic quantities.
Gibbs Free Energy of Mixing
Unlike the extensive properties of a one-component system, which rely only on the amount of the system present, the extensive properties of a solution depend on its temperature, pressure and composition. This means that a mixture must be described in terms of the partial molar quantities of its components. The total Gibbs free energy of a two-component solution is given by the expression
where
is the total Gibbs energy of the system,
is the number of moles of component , and
is the partial molar Gibbs energy of component .
The molar Gibbs energy of an ideal gas can be found using the equation
where is the standard molar Gibbs energy of the gas at 1 bar, and P is the pressure of the system. In a mixture of ideal gases, we find that the system’s partial molar Gibbs energy is equivalent to its chemical potential, or that
This means that for a solution of ideal gases, Equation can become
where
is the chemical potential of the th component,
is the standard chemical potential of component at 1 bar, and
is the partial pressure of component .
Now pretend we have two gases at the same temperature and pressure, gas A and gas B. The Gibbs energy of the system before the gases are mixed is given by Equation , which can be combined with Equation to give the expression
If gas A and gas B are then mixed together, they will each exert a partial pressure on the total system, and , so that . This means that the final Gibbs energy of the final solution can be found using the equation
The Gibbs energy of mixing, , can then be found by subtracting from .
where
and is the mole fraction of gas . This equation can be simplified further by knowing that the mole fraction of a component is equal to the number of moles of that component over the total moles of the system, or
Equation then becomes
This expression gives us the effect that mixing has on the Gibbs free energy of a solution. Since and are mole fractions that range from 0 to 1, we can conclude that will be a negative number. This is consistent with the idea that gases mix spontaneously at constant pressure and temperature.
Entropy of mixing
Figure shows that when two gases mix, it can really be seen as two gases expanding into twice their original volume. This greatly increases the number of available microstates, and so we would therefore expect the entropy of the system to increase as well.
Thermodynamic studies of an ideal gas’s dependence of Gibbs free energy of temperature have shown that
This means that differentiating Equation at constant pressure with respect to temperature will give an expression for the effect that mixing has on the entropy of a solution. We see that
Since the mole fractions again lead to negative values for and , the negative sign in front of the equation makes positive, as expected. This agrees with the idea that mixing is a spontaneous process.
Enthalpy of mixing
We know that in an ideal system , but this equation can also be applied to the thermodynamics of mixing and solved for the enthalpy of mixing so that it reads
Plugging in our expressions for (Equation ) and (Equation ) , we get
This result makes sense when considering the system. The molecules of ideal gas are spread out enough that they do not interact with one another when mixed, which implies that no heat is absorbed or produced and results in a of zero. Figure illustrates how and change as a function of the mole fraction so that of a solution will always be equal to zero (this is for the mixing of two ideal gasses).
References
Chang, R. Physical Chemistry for the Biosciences, 1st Herdon, VA: University Science Books, 132-133. Print.
Meyer, E.F. (1987). Theromodynamics of “Mixing” of Ideal Gases. J. Chem. Educ. 64, 676-677.
Outside Links
Satter, S. (2000). Thermodynamics of Mixing Real Gases. J. Chem. Educ. 77, 1361-1365.
Brandani, V., Evangelista, F. (1987). Correlation and prediction of enthalpies of mixing for systems containing alcohols with UNIQUAC associated-solution theory. Ind. Eng. Chem. Res. 26 (12), 2423–2430.
Problems
Use Figure 2 to find the x1 that has the largest impact on the thermodynamic quantities of the final solution. Explain why this is true.
Calculate the effect that mixing 2 moles of nitrogen and 3 moles of oxygen has on the entropy of the final solution.
Another way to find the entropy of a system is using the equation ΔS = nRln(V2/V1). Use this equation and the fact that volume is directly proportional to the number of moles of gas at constant temperature and pressure to derive the final expression for . (Hint: Use the derivation of as a guide).
Answers
x1= 0.5
Increases the entropy of the system by 27.98 J/molK
Contributors
Elizabeth Billquist (Hope College)
24.1: Ideal Solutions - Raoult's Law
25: Extension 16 - Vapor-Solution Phase Diagrams |
1134 | https://www.aasld.org/sites/default/files/2022-07/Long-Term%20Management%20of%20the%20Successful%20Adult%20Liver%20Transplant.pdf | PRACTICE GUIDELINE Long-Term Management of the Successful Adult Liver Transplant: 2012 Practice Guideline by AASLD and the American Society of Transplantation Michael R. Lucey Norah Terrault Lolu Ojo J. Eileen Hay James Neuberger Emily Blumberg Lewis W. Teperman Jump to: CONTENTS RECOMMENDATIONS FULL TEXT REFERENCES FORWARD © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
2 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS CONTENTS Contents (click section title or page number) Recommendations and Rationales. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Full-text Guideline. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 Abbreviations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104 Preamble. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 LT as a Treatment for End-stage Liver Disease / Mortality After LT / Morbidity After LT. . . . 106 Complications of Portal Hypertension After LT. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 Liver Tests / Vascular Thrombosis. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 Immunosuppression. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 Late Rejection. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 Promoting Health After LT. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 Bone Health. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 Systemic Disease. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 Nutrition and Obesity. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 Oncology. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120 Reproductive Health . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 Infectious Disease. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 Immunizations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 Viral Hepatitis. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130 PBC / PSC. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132 Autoimmune Hepatitis (AIH) / Alcoholic Liver Disease (ALD). . . . . . . . . . . . . . . . . . . . 133 NASH and NAFLD / Late Surgical Complications. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 References. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 FORWARD BACK USING, SEARCHING, AND PRINTING GUIDELINES This document was designed for use on a variety of devices using Adobe Acrobat Reader.® Smaller screens should be held horizontally. You may search or print using your PDF viewer. Menu hyperlinks allow movement between sections and to the guidelines on the AASLD site. In Recommendations and Rationales, click on individual items to review specific rationales. Use the top menu to return to the list. This file reflects the most recently approved language of the published guideline. Your feedback is welcome on the design and usability and will help guide future publications. Please email your comments to adavisowino@aasld.org or visit our social media pages.
AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
3 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS Recommendations and Rationales 1. The frequency of monitoring with liver tests should be individualized by the transplant center according to the time from liver transplant (LT), the complications from LT, the stability of serial test results, and the underlying cause (grade 1, level A).
2. Depending on the pattern of liver tests, magnetic resonance imaging, computed tomography, endoscopic retrograde cholangiopancreatography, and sonography may be appropriate (grade 1, level A).
3. Liver histology should be obtained when parenchymal injury is suspected as the cause of abnormal liver tests (grade 1, level A).
4. Bilomas and biliary cast syndrome should be managed in a center with expertise in LT medicine, radiology, and biliary endoscopy (grade 1, level A).
5. Hepatic artery thrombosis or stenosis is most readily assessed initially by Doppler ultrasound, but angiography is usually required to confirm the diagnosis and plan therapy (grade 1, level B).
6. Immunosuppressive drugs for LT recipients should be prescribed and monitored only by those with knowledge and expertise in that area. The choice of agents will depend on many factors, and no one regimen can be recommended for any patient (grade 2, level A).
7. Every patient’s immunosuppressive regimen should be reviewed at least every 6 months and modified as required with the goal of minimizing long-term toxicities (grade 1, level B).
8. Rejection can be reliably diagnosed only on the basis of liver histology; a biopsy sample should be taken before treatment initiation and classified according to the Banff criteria (grade 1, level A).
9. Although the long-term withdrawal of all immunosuppression can be achieved in a small number of patients, this should be undertaken only with select recipients and under close supervision (grade 2, level C).
10. Frequent hand washing reduces the risk of infection with pathogens acquired by direct contact, including Clostridium difficile, community-acquired viral infections, and pathogens found in soil (grade 1, level A).
11. Shoes, socks, long-sleeve shirts, and long pants should be worn for activities that will involve soil exposure and tick exposure and also to avoid unnecessary sun exposure (grade 1, level A).
12. During periods of maximal immunosuppression, LT recipients should avoid crowds to minimize exposures to respiratory illnesses (grade 1, level A).
13. Work in high-risk areas, such as construction, animal care settings, gardening, landscaping, and farming, should be reviewed with the transplant team to develop appropriate strategies for the prevention of high-risk exposures (grade 2, level A).
14. LT recipients should avoid the consumption of water from lakes and rivers (grade 1, level A).
This guideline includes 93 specific recommendations. Please click on a recommendation to review the related rationale and supporting evidence. See Table 1 for an explanation of the grading system for recommendations.
FORWARD BACK AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
4 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS 15. LT recipients should avoid unpasteurized milk products and raw and undercooked eggs and meats (particularly uncooked pork, poultry, fish, and seafood; grade 1, level A).
16. LT recipients should avoid high-risk pets, which include rodents, reptiles, chicks, ducklings, and birds (grade 1, level A).
17. Travel by LT recipients, especially to developing countries, should be reviewed with the transplant team a minimum of 2 months before departure to determine optimal strategies for the reduction of travel-related risks (grade 1, level A).
18. LT recipients should take precautions to prevent vector (including mosquito) -borne diseases. These include avoiding going out during peak mosquito feeding times (dawn and dusk)and using N,N-diethyl-meta-toluamide–containing insect repellants (grade 1, level A).
19. LT recipients should undertake a thorough review of hobbies to assess potential infectious disease risks, particularly those associated with outdoor hobbies (grade 2, level A).
20. All LT recipients should be educated about the importance of sun avoidance and sun protection through the use of a sun block with a sun protection factor of at least 15 and protective clothing. They should be encouraged to examine their skin on a regular basis and report any suspicious or concerning lesions to their physicians (grade 1, level A).
21. Because of the strong association of lung, head, and neck cancers with smoking, the sustained cessation of smoking is the most important preventative intervention (grade 1, level A).
22. For female LT recipients of a child-bearing age, preconception counseling about contraception and the risks and outcomes of pregnancy should start in the pretransplant period and should be reinforced after transplantation (grade 1, level A).
23. In the first 5 years after transplantation, screening by bone mineral density (BMD) should be done yearly for osteopenic patients and every 2 to 3 years for patients with normal BMD; thereafter, screening depends on the progression of BMD and on risk factors (grade 2, level B).
24. If osteopenic bone disease is confirmed or if atraumatic fractures are present, then patients should be assessed for risk factors for bone loss; in particular, this should include an assessment of calcium intake and 25-hydroxy-vitamin D levels, an evaluation of gonadal and thyroid function, a full medication history, and thoracolumbar radiography (grade 1, level A).
25. The osteopenic LT recipient should perform regular weight-bearing exercise and receive calcium and vitamin D supplements (grade 1, level A).
26. Bisphosphonate therapy should be considered in LT recipients with osteoporosis or recent fractures (grade 1, level A).
27. Monitoring of renal function in LT recipients for the detection and management of chronic kidney disease should use an estimating equation to evaluate the glomerular filtration rate (grade 1, level B).
28. Urinary protein quantification using the concentration ratio of protein to creatinine in a spot urine specimen should be evaluated at least once yearly (grade 1, level B).
FORWARD BACK AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
5 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS 29. The reduction or withdrawal of calcineurin inhibitor (CNI) associated immunosuppression is an appropriate response to the development of chronic kidney disease in LT recipients (grade 1, level A).
30. Kidney transplantation from deceased or living donors is beneficial in improving survival and should be considered the optimal therapy for LT recipients who develop end-stage renal disease (grade 1, level A).
31. The treatment of diabetes mellitus after LT should aim for a hemoglobin A1c (HbA1c) target goal of <7.0% with a combination of lifestyle modifications and pharmacological agents as appropriate (grade 1, level B).
32. When high-dose corticosteroids are administered, insulin therapy is the most effective and safest agent with which to control hyperglycemia; however, as the interval from LT extends, patients with new-onset diabetes mellitus may experience a decline in insulin requirements, and oral hypoglycemic agents may be appropriate if allograft function is normal (grade 1, level C).
33. Metformin or sulfonylureas may be used in LT recipients with normal renal function, whereas sulfonylureas such as glipizide and glimepiride are preferable if there is any deterioration of renal function (grade 1, level C).
34. Consideration can be given to the conversion of immunosuppression from tacrolimus to cyclosporine in LT recipients with poor glycemic control (grade 2, level B).
35. The treatment of hypertension should aim for a target goal of 130/80 mm Hg with a combination of lifestyle modifications and pharmacological agents as appropriate (grade 1, level A).
36. Angiotensin-converting enzyme inhibitors, angiotensin receptor blockers, and direct renin inhibitors should be used as first-line antihypertensive therapy in LT recipients with diabetes mellitus, chronic kidney disease, and/or significant proteinuria (grade 1, level A).
37. The measurement of blood lipids after a 14-hour fast is recommended annually for healthy LT recipients. An elevated low-density lipoprotein (LDL) cholesterol level >100 mg/dL, with or without hypertriglyceridemia, requires therapy. If therapeutic lifestyle and dietary changes are not enough, statin therapy should be introduced. Suboptimal control with statins can be improved by the addition of ezetimibe (grade 2, level B).
38. Isolated hypertriglyceridemia is first treated with omega-3 fatty acids (up to 4 g daily if tolerated). If this is not sufficient for control, gemfibrozil or fenofibrate can be added, although patients must be followed carefully for side effects, especially with the concomitant use of statins and calcineurin inhibitors (grade 2, level C).
39. All LT patients require ongoing dietary counseling to avoid obesity (grade 1, level C).
40. Among LT recipients who become severely or morbidly obese and fail behavioral weight-loss programs, bariatric surgery may be considered, although the optimal procedure and its timing with respect to transplantation remain to be defined (grade 1, level C).
41. All LT recipients should see a dermatologist after transplantation to assess cutaneous lesions, with at least an annual evaluation by a dermatologist 5 years or more after transplantation (grade 1, level A).
FORWARD BACK AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
6 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS FORWARD BACK 42. Patients with primary sclerosing cholangitis and inflammatory bowel disease or other established risk factors for colorectal cancer should undergo an annual screening colonoscopy with biopsies. Colectomy, including continence-preserving pouch operations, should be considered when colonic biopsy reveals dysplasia. (grade 1, level B).
43. For patients without prior hepatocellular carcinoma (HCC) who develop recurrent cirrhosis of the allograft, surveillance for de novo HCC should be undertaken with abdominal imaging every 6 to 12 months (grade 1, level A).
44. An immunosuppressant regimen that includes sirolimus (started several weeks after transplantation) should be considered for patients undergoing transplantation for hepatocellular carcinoma (grade 2, level B).
45. Resection or ablation is usually the treatment of choice for a solitary extrahepatic metastasis or an intrahepatic recurrence of hepatocellular carcinoma (grade 1, level B).
46. Pregnancy in an LT recipient should be managed by a high-risk obstetrician in coordination with the transplant hepatologist (grade 1, level C).
47. Pregnancy should be delayed for 1 year after LT and occur at a time with good, stable allograft function, with maintenance immunosuppression, and with good control of any medical complications such as hypertension and diabetes (grade 1, level B).
48. The ideal immunosuppression for pregnancy is tacrolimus monotherapy, which should be maintained at therapeutic levels throughout pregnancy; cyclosporine, azathioprine, and prednisone may also be used if they are necessary (grade 1, level B).
49. Allograft function and calcineurin inhibitor serum levels are monitored every 4 weeks until 32 weeks, then every 2 weeks, and then weekly until delivery (grade 1, level B).
50. Contraception should begin before the resumption of sexual activity, although no particular form of contraception can be recommended over another (grade 2, level B).
51. An assessment for infections following LT should take into account the intensity of immunosuppression, the timing of the presentation, the environmental and donor exposures, the recipient’s history of both symptomatic and latent infections, and the utilization of prophylactic antimicrobials and immunizations (grade 1, level A).
52. Attention should be paid to potential drug interactions when new antimicrobial therapies are initiated (grade 1, level A).
53. High-risk recipients (cytomegalovirus-seronegative recipients of cytomegalovirus-seropositive donor organs) should receive prophylaxis with ganciclovir or valganciclovir for a minimum of 3 months after transplantation (grade 1, level B).
54. The treatment of LT recipients with cytomegalovirus should be maintained until viremia and all symptoms have resolved (grade 2, level B).
55. Prophylaxis against cytomegalovirus should be resumed whenever LT recipients receive anti-lymphocyte therapy for the treatment of rejection and should be continued for 1 to 3 months after the treatment of rejection (grade 2, level B).
AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
7 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS 56. The treatment of a cytomegalovirus infection should consist of the following: a. Consideration of immunosuppression reduction. b. High-dose intravenous ganciclovir or oral valganciclovir in individuals with mild to moderate disease without gastrointestinal involvement or a reduced capacity for absorption. c. A minimum of 2 weeks of treatment. Treatment should be continued to complete the resolution of all symptoms and viremia (grade 1, level A).
57. Resistant virus should be suspected in patients with a history of prolonged ganciclovir or valganciclovir exposure who have a persistent or progressive infection despite treatment with high-dose intravenous ganciclovir (grade 1, level A). In such instances, genotypic assays should be performed, and consideration should be given to the initiation of foscarnet with or in substitution for ganciclovir (grade 1, level B).
58. Posttransplant lymphoproliferative disorder (PTLD) should be considered in LT recipients (especially high-risk individuals) who present with unexplained fever, lymphadenopathy, or cytopenias (grade 1, level A).
59. Although Epstein-Barr virus (EBV) may be associated with the development of posttransplant lymphoproliferative disorder (PTLD), the detection of EBV viremia is not diagnostic for PTLD; a histopathological diagnosis is required (grade 1, level A).
60. The diagnosis of fungal infections may require diagnostic biopsy for pathological and microbiological confirmation (grade 1, level A). a. Blood cultures are most helpful for the diagnosis of Candida bloodstream infections (class 1, level B) and Blastomyces (grade 1, level B). b. Cryptococcal antigen testing of cerebrospinal fluid or blood is most helpful for the diagnosis of Cryptococcus (grade 1, level B). c. Urinary histoplasmosis and Blastomyces antigens are useful for the diagnosis of disseminated histoplasmosis and blastomycosis, respectively (grade 1, level B).
61. A cautious reduction of immunosuppression should be initiated to prevent immune reconstitution syndrome, especially for cryptococcal infections (grade 1, level B).
62. All LT recipients should receive prophylaxis against P. jirovecii with trimethoprim-sulphamethoxazole (single strength daily or double strength 3 times per week) for a minimum of 6 to 12 months after transplantation (grade 1, level A). Atovaquone and dapsone are the preferred alternatives for patients who are intolerant of trimethoprim-sulfamethoxazole (grade 1, level B).
63. Trimethoprim-sulphamethoxazole is the drug of choice for the treatment of P. jirovecii pneumonia. Intravenous pentamidine is the preferred alternative for patients intolerant of trimethoprim-sulphamethoxazole with more severe infections (grade 1, level A).
64. Patients with clinical signs and symptoms or radiological features suggestive of P. jirovecii pneumonia should undergo sputum sampling or bronchoalveolar lavage with a cytological examination using a silver or Giemsa stain, polymerase chain reaction, or a specific antibody stain to identify the organism (grade 1, level A).
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8 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS 65. The treatment of active tuberculosis (TB) should include the initiation of a 4-drug regimen using isoniazid, rifampin, pyrazinamide, and ethambutol (under the assumption of susceptible TB) with adjustments in accordance with subsequent culture results. This may be tapered to 2 drugs (isoniazid and rifampin) after 2 months (under the assumption of no resistance) and continued for a minimum of 4 additional months (grade 1, level B).
66. Close monitoring for rejection and hepatotoxicity is imperative while LT recipients receive anti-tuberculosis therapy (grade 1, level A).
67. HIV-infected LT recipients receiving highly active antiretroviral therapy require frequent monitoring of calcineurin inhibitor (CNI) levels because of the significant interaction between antiretrovirals and CNIs (grade 1, level A).
68. HIV-infected LT recipients receiving highly active antiretroviral therapy should be followed with scheduled HIV viral loads and T lymphocyte subset counts (grade 1, level A).
69. Standard prophylaxis for cytomegalovirus is recommended for HIV-infected LT recipients receiving highly active antiretroviral therapy, and lifelong Pneumocystis pneumonia prophylaxis is the norm (grade 1, level A).
70. Standard HIV-specific prophylaxis for low CD4 counts should be used (grade 1, level A).
71. All LT recipients should receive an annual influenza vaccination (grade 1, level B).
72. All LT recipients should avoid live virus vaccines (grade 1, level A).
73. Re-immunization is indicated for some vaccines, notably the influenza vaccine (annually) and the pneumococcal vaccine (every 3-5 years; no class or level provided). (grade 1, level A).
74. Long-term prophylactic therapy using a combination of antiviral agents and low-dose hepatitis B immune globulin on demand or at fixed intervals can effectively prevent HBV recurrence rates in ≥90% of transplant recipients (grade 1, level B).
75. In patients with low or undetectable hepatitis B virus (HBV) DNA levels before transplantation and an absence of high-risk factors for recurrence, hepatitis B immune globulin can be discontinued, and long-term treatment with antivirals (single or in combination) can be used as an alternative prophylactic strategy (grade 2, level B).
76. Lifelong antiviral therapy should be used to treat patients with recurrent HBV infections. Combination antiviral therapy is superior to monotherapy when drugs with a low genetic barrier to resistance are used, whereas the discontinuation of hepatitis B immune globulin is generally reserved for patients at low risk for HBV recurrence (grade 1, level B).
77. Retransplantation for recurrent HBV is appropriate when treatment strategies to prevent or treat recurrent HBV disease are available (grade1, level C).
78. Liver biopsy is useful in monitoring disease severity and progression and in distinguishing recurrent hepatitis C virus (HCV) disease from other causes of liver enzyme elevations (grade 1, level C).
79. Prophylactic antiviral therapy has no current role in the management of HCV disease (grade 1, level A).
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9 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS 80. Moderate acute rejection should be treated with increased maintenance immunosuppression and corticosteroid boluses, whereas lymphocyte-depleting drugs should be avoided (grade 1, level B).
81. Antiviral therapy is indicated for significant histological disease: grade 3 or higher inflammatory activity and/or stage 2 or higher fibrosis (on a scale of 4) or cholestatic hepatitis. Peginterferon and ribavirin are the current drugs of choice. The risks and benefits of triple therapy with protease inhibitors are to be determined. The goal of antiviral therapy is the achievement of a sustained virological response, and this confers a survival benefit (grade 1, class B).
82. Retransplantation for recurrent HCV disease should be considered selectively (grade 2, level B).
83. Primary biliary cirrhosis LT recipients should be routinely monitored for associated autoimmune diseases (eg, thyroid disease) and bone density (grade 2, level B).
84. For those with histological evidence of recurrent disease, treatment with ursodeoxycholic acid at 10 to 15 mg/kg/day (grade 2, level B) may be considered, and although its use is associated with the improvement of liver tests, no impact on graft survival has been documented (grade 2, level B). There is no indication for offering prophylaxis with ursodeoxycholic acid to patients with normal liver histology (grade 2, level B).
85. Although there are few data on prevention, it is recommended that those patients grafted for PSC in the presence of chronic ulcerative colitis (CUC) have an annual colonoscopy with mucosal biopsy (grade 2, level B).
86. Although there is no evidence for recommending a particular immunosuppressive regimen in patients undergoing transplantation for autoimmune hepatitis, it is prudent to maintain patients on long-term, low-dose corticosteroids in addition to routine immunosuppression (with attention to maintaining bone health; grade 2, level B).
87. All patients with a prior diagnosis of alcoholic liver disease should be encouraged to remain abstinent from alcohol (grade 1, level B).
88. Patients should be encouraged to enter therapy or counseling if they relapse into alcohol use (grade 1, level C).
89. All patients with a prior diagnosis of alcoholic liver disease who are users of tobacco should be encouraged to undertake smoking cessation (grade 1, level B).
90. Careful attention should be given to the risk of cardiovascular disease and/or new-onset cancers of the aerodigestive tract, especially in cigarette smokers (grade 1, level A).
91. The confirmation of recurrent or de novo nonalcoholic fatty liver disease, the recognition of fibrosis, and the exclusion of alternate causes of elevated liver chemistry tests require liver biopsy (grade 1, level B).
92. No specific recommendations regarding the prevention or treatment of nonalcoholic fatty liver disease or nonalcoholic steatohepatitis in LT recipients can be made other than general recommendations to avoid excessive gains in bodyweight and control hypertension and diabetes (grade 1, level C).
93. LT recipients with an incisional hernia should be instructed to recognize incarcerated hernias and advised to seek immediate medical assistance (grade 1, level B).
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10 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 1 The frequency of monitoring with liver tests should be individualized by the transplant center according to the time from LT, the complications from LT, the stability of serial test results, and the underlying cause (grade 1, level A).
RATIONALE 1 Liver tests are routinely monitored after LT. When liver tests are elevated for a healthy recipient, the course of action will depend on the severity and type of abnormality (cholestatic, hepatitic, or other). More than 1 cause may coexist in the same patient. When abnormal liver tests are recognized in a healthy, asymptomatic LT recipient, it is reasonable to repeat the tests in 1 to 2 weeks. A decision to investigate further should be based on the persistence and severity of the liver test abnormalities.
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11 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 2 Depending on the pattern of liver tests, magnetic resonance imaging, computed tomography, endoscopic retrograde cholangiopancreatography, and sonography may be appropriate (grade 1, level A).
RATIONALE 2 Elevated alkaline phosphatase, total bilirubin, and aminotransferase levels may arise from the late appearance of biliary anastomotic strictures due to thrombosis or stenosis of the hepatic artery or to recurrent PSC or PBC. Appropriate biliary imaging includes endoscopic retrograde cholangiopancreatography, magnetic resonance cholangiopancreatography, and/or ultrasound.
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12 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 3 Liver histology should be obtained when parenchymal injury is suspected as the cause of abnormal liver tests (grade 1, level A).
RATIONALE 3 More than 1 cause may coexist in the same patient. When abnormal liver tests are recognized in a healthy, asymptomatic LT recipient, it is reasonable to repeat the tests in 1 to 2 weeks. A decision to investigate further should be based on the persistence and severity of the liver test abnormalities. Investigations should include a thorough history and examination, appropriate laboratory tests, and Doppler ultrasound of the liver. It should not be assumed without appropriate histological confirmation that abnormal liver tests represent immune-mediated damage.
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13 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 4 Bilomas and biliary cast syndrome should be managed in a center with expertise in LT medicine, radiology, and biliary endoscopy (grade 1, level A).
RATIONALE 4 Biliary cast syndrome refers to a severe form of intrahepatic bile duct ischemic injury unique to post-LT patients,8 and it is associated with hepatic artery thrombosis and the use of a split liver, including partial grafts derived from living donors and, more commonly, from donation after cardiac death donors. Biliary cast syndrome may resolve with repeated clearance of bile duct debris either percutaneously or endoscopically.
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14 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 5 Hepatic artery thrombosis (HAT) or stenosis is most readily assessed initially by Doppler ultrasound, but angiography is usually required to confirm the diagnosis and plan therapy (grade 1, level B).
RATIONALE 5 Hepatic artery thrombosis (HAT) or stenosis may present clinically after 3 months, as: • Intrahepatic non-anastomotic strictures and/or sterile or infected fluid collections within the liver, sometimes referred to as bilomas, • Ischemic cholangiopathy or • Biliary cast syndrome.
The combination of hepatic artery thrombosis and biliary complications usually requires retransplantation.
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15 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 6 Immunosuppressive drugs for LT recipients should be prescribed and monitored only by those with knowledge and expertise in that area. The choice of agents will depend on many factors, and no one regimen can be recommended for any patient (grade 2, level A).
RATIONALE 6 The choice of immunosuppression depends on the following: • Indication for transplantation: the choice of immunosuppression may affect disease recurrence (eg, HCV, malignancy, or autoimmune disease).
• Comorbidities.
• Drug side effects: calcineurin inhibitors (CNIs) may cause renal impairment.
• Likelihood of pregnancy: mycophenolate and mammalian target of rapamycin (mTOR) inhibitors such as sirolimus are potential teratogens.
• History of severe or recurrent rejection.
• Prior experience with the various immunosuppressive agents.
• History of or risk for cancers.
• History of or risk for infections.
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16 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 7 Every patient’s immunosuppressive regimen should be reviewed at least every 6 months and modified as required with the goal of minimizing long-term toxicities (grade 1, level B).
RATIONALE 7 There is no reliable marker for determining the effective level of immunosuppression; therefore, the choice of the agent (or agents) and doses given will be determined by the clinical, laboratory, and histological response. The CNI dose is generally determined by the drug level; the target levels after 3 months are 5 to 10 ng/mL for tacrolimus and 100 to 150 ng/mL for cyclosporine (both are whole blood trough levels). The target whole blood trough level for sirolimus is 5 ng/mL. The need for therapeutic drug monitoring for mycophenolate is uncertain.
FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
17 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 8 Rejection can be reliably diagnosed only on the basis of liver histology; a biopsy sample should be taken before treatment initiation and classified according to the Banff criteria (grade 1, level A).
RATIONALE 8 Late rejection is defined as rejection that has its onset more than 90 days after transplantation. Traditionally, 2 forms have been recognized: cellular rejection (also known as acute cellular rejection and late-onset rejection) and ductopenic rejection (also known as vanishing bile duct syndrome). Both forms of rejection are, until the late stages, asymptomatic, and the diagnosis is made through the investigation of abnormal liver tests; the diagnosis can be confirmed only on the basis of histology.
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18 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 9 Although the long-term withdrawal of all immunosuppression can be achieved in a small number of patients, this should be undertaken only with select recipients and under close supervision (grade 2, level C).
RATIONALE 9 The majority of LT recipients need lifelong immunosuppression to maintain graft function. A very small number of LT recipients develop operational tolerance to the allograft and do not require long-term immunosuppression.5 FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
19 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 10 Frequent hand washing reduces the risk of infection with pathogens acquired by direct contact, including Clostridium difficile, community-acquired viral infections, and pathogens found in soil (grade 1, level A).
RATIONALE 10 (Please see full text.) FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
20 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 11 Shoes, socks, long-sleeve shirts, and long pants should be worn for activities that will involve soil exposure and tick exposure and also to avoid unnecessary sun exposure (grade 1, level A).
RATIONALE 11 (Please see full text.) FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
21 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 12 During periods of maximal immunosuppression, LT recipients should avoid crowds to minimize exposures to respiratory illnesses (grade 1, level A).
RATIONALE 12 (Please see full text.) FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
22 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 13 Work in high-risk areas, such as construction, animal care settings, gardening, landscaping, and farming, should be reviewed with the transplant team to develop appropriate strategies for the prevention of high-risk exposures (grade 2, level A).
RATIONALE 13 (Please see full text.) FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
23 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 14 LT recipients should avoid the consumption of water from lakes and rivers (grade 1, level A).
RATIONALE 14 (Please see full text.) FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
24 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 15 LT recipients should avoid unpasteurized milk products and raw and undercooked eggs and meats (particularly uncooked pork, poultry, fish, and seafood; grade 1, level A).
RATIONALE 15 (Please see full text.) FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
25 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 16 LT recipients should avoid high-risk pets, which include rodents, reptiles, chicks, ducklings, and birds (grade 1, level A).
RATIONALE 16 (Please see full text.) FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
26 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 17 Travel by LT recipients, especially to developing countries, should be reviewed with the transplant team a minimum of 2 months before departure to determine optimal strategies for the reduction of travel-related risks (grade 1, level A).
RATIONALE 17 (Please see full text.) FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
27 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 18 LT recipients should take precautions to prevent vector (including mosquito) -borne diseases. These include avoiding going out during peak mosquito feeding times (dawn and dusk) and using N,N-diethyl-meta-toluamide–containing insect repellants (grade 1, level A).
RATIONALE 18 (Please see full text.) FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
28 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 19 LT recipients should undertake a thorough review of hobbies to assess potential infectious disease risks, particularly those associated with outdoor hobbies (grade 2, level A).
RATIONALE 19 (Please see full text.) FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
29 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 20 All LT recipients should be educated about the importance of sun avoidance and sun protection through the use of a sun block with a sun protection factor of at least 15 and protective clothing. They should be encouraged to examine their skin on a regular basis and report any suspicious or concerning lesions to their physicians (grade 1, level A).
RATIONALE 20 (Please see full text.) FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
30 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 21 Because of the strong association of lung, head, and neck cancers with smoking, the sustained cessation of smoking is the most important preventative intervention (grade 1, level A).
RATIONALE 21 (Please see full text.) FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
31 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 22 For female LT recipients of a child-bearing age, preconception counseling about contraception and the risks and outcomes of pregnancy should start in the pretransplant period and should be reinforced after transplantation (grade 1, level A).
RATIONALE 22 (Please see full text.) FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
32 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 23 In the first 5 years after transplantation, screening by bone mineral density (BMD) should be done yearly for osteopenic patients and every 2 to 3 years for patients with normal BMD; thereafter, screening depends on the progression of BMD and on risk factors (grade 2, level B).
RATIONALE 23 Bone loss and fracturing are seen with 2 distinct phases after LT. In the first 4 postoperative months, there is accelerated bone loss in almost all liver recipients, regardless of the pretransplant bone mineral density (BMD), that is consistent with the effects of corticosteroids and possibly CNIs.12 After the first 4 postoperative months with normal allograft function, bone metabolism improves, and in the osteopenic patient, there will be a gain in bone mass over the next postoperative years with a gradual reduction in the incidence of fractures.12, 13 In patients with preexisting osteopenia or pretransplant fracturing, this early, rapid bone loss results in a high susceptibility to fracturing, mainly at sites of trabecular bone (vertebrae and ribs), especially in the first year after LT, but there is a smaller but steady cumulative increase in fracturing. In the early years after LT, BMD should be measured annually in osteopenic patients and every 2 to 3 years in patients with normal BMD. Later screening depends on risk factors.
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33 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 24 If osteopenic bone disease is confirmed or if atraumatic fractures are present, then patients should be assessed for risk factors for bone loss; in particular, this should include an assessment of calcium intake and 25-hydroxy-vitamin D levels, an evaluation of gonadal and thyroid function, a full medication history, and thoracolumbar radiography (grade 1, level A).
RATIONALE 24 (Please see full text.) FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
34 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 25 The osteopenic LT recipient should perform regular weight-bearing exercise and receive calcium and vitamin D supplements (grade 1, level A).
RATIONALE 25 In order to diminish factors that promote bone loss, glucocorticoids should be reduced or discontinued as soon as possible after LT. Calcium supplements are recommended for all LT recipients with (or at risk of) osteopenia, and all patients should receive 1000 to 1200 mg of elemental calcium daily to optimize bone remodeling and mineralization. Vitamin D levels should be maintained at a serum level of at least 30 ng/mL, and most LT patients will require supplementation (generally 400-1000 IU/day). The serum levels of 25-hydroxyvitamin D must be checked to assess the adequacy of replacement, even with supplementation, and should be rechecked annually or more frequently if a deficiency is diagnosed.
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35 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 26 Bisphosphonate therapy should be considered in LT recipients with osteoporosis or recent fractures (grade 1, level A).
RATIONALE 26 Many issues with regard to bisphosphonate therapy for LT recipients remain to be defined: the optimal duration of therapy, the optimal doses of bisphosphonates, whether oral or intravenous therapy is better, and the LT population most likely to benefit. Despite these caveats, we suggest that bisphosphonate therapy should be considered in the following circumstances: • T-score less than −2.5 or atraumatic fractures.
• T-score between −1.5 and −2.5 and other risk factors.
Oral alendronate at 70 mg weekly is an appropriate starting point, although other oral agents may, however, be equally efficacious.
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36 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 27 Monitoring of renal function in LT recipients for the detection and management of chronic kidney disease (CKD) should use an estimating equation to evaluate the glomerular filtration rate (grade 1, level B).
RATIONALE 27 The majority of LT recipients who survive the first 6 months develop chronic kidney disease (CKD).6 Predialysis CKD prevalence rates in this population range from 30% to 80%. A serum creatinine elevation is a late and insensitive indicator of CKD in this population. An estimating equation that has been shown to have reasonable precision should be routinely used. Both the 4-variable Modification of Diet in Renal Disease equation and the Chronic Kidney Disease Epidemiology Collaboration formula are superior to serum creatinine alone and 24-hour urinary creatinine clearance in estimating renal function.18, 19 FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
37 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 28 Urinary protein quantification using the concentration ratio of protein to creatinine in a spot urine specimen should be evaluated at least once yearly (grade 1, level B).
RATIONALE 28 Increased proteinuria (spot protein-to-creatinine ratio >0.3) may be absent even in the presence of advanced CKD because of the antiproteinuric effect of CNIs. Proteinuria is best assessed by the measurement of the concentration ratio of protein to creatinine in a spot urine specimen.20 FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
38 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 29 The reduction or withdrawal of calcineurin inhibitor (CNI)-associated immunosuppression is an appropriate response to the development of chronic kidney disease (CKD) in LT recipients (grade 1, level A).
RATIONALE 29 A reduction in the dosage or a complete withdrawal of CNIs several months to years after LT is a common practice aimed at ameliorating the progression of CKD. These renal-sparing maintenance protocols typically rely on sirolimus or everolimus, often in combination with mycophenolate, to prevent acute rejection; others use steroids and mycophenolate or azathioprine.21-24 Renal function is more likely to be preserved if CNI withdrawal is instituted when the estimated glomerular filtration rate is between 40 and 50 mL/minute/1.73 m2.25 FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
39 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 30 Kidney transplantation from deceased or living donors is beneficial in improving survival and should be considered the optimal therapy for LT recipients who develop end-stage renal disease (ESRD) (grade 1, level A).
RATIONALE 30 LT recipients with ESRD who subsequently receive a living or deceased kidney transplant have a 44% to 60% reduction in long-term mortality in comparison with their dialysis-treated counterparts.26, 27 FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
40 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 31 The treatment of diabetes mellitus (DM) after LT should aim for an hemoglobin A1c (HbA1c) target goal of <7.0% with a combination of lifestyle modifications and pharmacological agents as appropriate (grade 1, level B).
RATIONALE 31 Because stringent glycemic control significantly reduces morbidity and mortality in diabetic patients, it seems reasonable to assume that LT recipients would similarly benefit. The goals of the long-term management of diabetes after LT are not substantially different from the goals for nontransplant patients (Table 8). There is controversy regarding the appropriate target level of hemoglobin A1c (HbA1c), and consequently, our recommendation of a threshold of <7.0% rather than <6.0% reflects the view that the more demanding standard may confer no additional advantage.
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41 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 32 When high-dose corticosteroids are administered, insulin therapy is the most effective and safest agent with which to control hyperglycemia; however, as the interval from LT extends, patients with new-onset diabetes mellitus (NODM) may experience a decline in insulin requirements, and oral hypoglycemic agents may be appropriate if allograft function is normal (grade 1, level C).
RATIONALE 32 When insulin requirements are low, oral agents may be substituted if allograft function is normal. Metformin or a sulfonylurea may be used in LT recipients with normal renal function, whereas sulfonylureas such as glipizide and glimepiride are preferable if there is any deterioration in renal function.
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42 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 33 Metformin or sulfonylureas may be used in LT recipients with normal renal function, whereas sulfonylureas such as glipizide and glimepiride are preferable if there is any deterioration of renal function (grade 1, level C).
RATIONALE 33 Metformin or a sulfonylurea may be used in LT recipients with normal renal function, whereas sulfonylureas such as glipizide and glimepiride are preferable if there is any deterioration in renal function. The safety of thiazolidinediones in LT recipients is unproven.
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43 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RATIONALE 34 Because stringent glycemic control significantly reduces morbidity and mortality in diabetic patients, it seems reasonable to assume that LT recipients would similarly benefit. The goals of the long-term management of diabetes after LT are not substantially different from the goals for nontransplant patients (Table 8). There is controversy regarding the appropriate target level of hemoglobin A1c (HbA1c), and consequently, our recommendation of a threshold of <7.0% rather than <6.0% reflects the view that the more demanding standard may confer no additional advantage. When insulin requirements are low, oral agents may be substituted if allograft function is normal. Metformin or a sulfonylurea may be used in LT recipients with normal renal function, whereas sulfonylureas such as glipizide and glimepiride are preferable if there is any deterioration in renal function. The safety of thiazolidinediones in LT recipients is unproven. Retrospective data sets and a small prospective study suggest that the conversion of immunosuppression from tacrolimus to cyclosporine improves glycemic control in patients with established diabetes mellitus (DM) and new-onset diabetes mellitus (NODM).33 RECOMMENDATION 34 Consideration can be given to the conversion of immunosuppression from tacrolimus to cyclosporine in LT recipients with poor glycemic control (grade 2, level B).
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44 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 35 The treatment of hypertension should aim for a target goal of 130/80 mm Hg with a combination of lifestyle modifications and pharmacological agents as appropriate (grade 1, level A).
RATIONALE 35 Hypertension in LT recipients increases the risk of fatal and nonfatal cardiovascular disease events and chronic kidney disease.15 Although there are no clinical trials of antihypertensive therapy in LT recipients, it is prudent to target a blood pressure treatment goal of 130/80 mm Hg in LT recipients with systemic hypertension.34 FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
45 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 36 Angiotensin-converting enzyme inhibitors, angiotensin receptor blockers, and direct renin inhibitors should be used as first-line antihypertensive therapy in LT recipients with diabetes mellitus (DM), chronic kidney disease (CKD), and/or significant proteinuria (grade 1, level A).
RATIONALE 36 Angiotensin-converting enzyme inhibitors, angiotensin receptor blockers, and direct renin inhibitors should be used as first-line antihypertensive therapy in LT recipients with DM, CKD, and/or significant proteinuria. Monitoring of potassium levels is necessary when these drugs are used in conjunction with CNIs (particularly tacrolimus).
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46 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 37 The measurement of blood lipids after a 14-hour fast is recommended annually for healthy LT recipients. An elevated low-density lipoprotein cholesterol level >100 mg/dL, with or without hypertriglyceridemia, requires therapy. If therapeutic lifestyle and dietary changes are not enough, statin therapy should be introduced. Suboptimal control with statins can be improved by the addition of ezetimibe (grade 2, level B).
RATIONALE 37 Dyslipidemia occurs in up to 70% of LT recipients (a prevalence much higher than that before transplantation) and is a major risk factor for cardiovascular morbidity and mortality.37, 38 Although age, body weight, and genetics have some influence, medications—especially CNIs, mTOR inhibitors, and glucocorticoids—are the major influences on the high prevalence of dyslipidemia in LT recipients. Furthermore, 3-hydroxy-3-methyl-glutaryl-coenzyme A reductase inhibitors and CNIs share metabolic pathways and have significant drug-drug interactions. Measurement of blood lipids after a 14-hour fast is recommended annually for healthy LT recipients.
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47 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 38 Isolated hypertriglyceridemia is first treated with omega-3 fatty acids (up to 4 g daily if tolerated). If this is not sufficient for control, gemfibrozil or fenofibrate can be added, although patients must be followed carefully for side effects, especially with the concomitant use of statins and calcineurin inhibitors (CNI) (grade 2, level C).
RATIONALE 38 Dyslipidemia occurs in up to 70% of LT recipients (a prevalence much higher than that before transplantation) and is a major risk factor for cardiovascular morbidity and mortality.37, 38 Although age, body weight, and genetics have some influence, medications—especially CNIs, mTOR inhibitors, and glucocorticoids—are the major influences on the high prevalence of dyslipidemia in LT recipients.
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48 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 39 All LT patients require ongoing dietary counseling to avoid obesity (grade 1, level C).
RATIONALE 39 Weight accumulation is common after LT. In American and European cohorts, approximately 20% of lean patients become obese (body mass index >30 kg/m2) in the first 2 to 3 years after LT; this phenomenon is driven by the restoration of health and the stimulation of appetite by medicines such as corticosteroids.39, 40 FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
49 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 40 Among LT recipients who become severely or morbidly obese and fail behavioral weight-loss programs, bariatric surgery may be considered, although the optimal procedure and its timing with respect to transplantation remain to be defined (grade 1, level C).
RATIONALE 40 Weight accumulation is common after LT. In American and European cohorts, approximately 20% of lean patients become obese (body mass index >30 kg/m2) in the first 2 to 3 years after LT; this phenomenon is driven by the restoration of health and the stimulation of appetite by medicines such as corticosteroids.39, 40 FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
50 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 41 All LT recipients should see a dermatologist after transplantation to assess cutaneous lesions, with at least an annual evaluation by a dermatologist 5 years or more after transplantation (grade 1, level A).
RATIONALE 41 The incidence of de novo cancer is higher among LT recipients versus an age- and sex-matched nontransplant control population41 (Table 10). The cumulative incidence of de novo cancer after LT increases from 3% to 5% at 1 to 3 years to 11% to 20% at 10 years after LT.42, 43 Cutaneous malignancies are the most common form of malignancy in recipients of solid organ transplants, but cigarette smokers are at increased risk of developing lung cancer and oropharyngeal cancer, and the rate of colon cancer is increased in patients undergoing transplantation for PSC because of the comorbid risk from inflammatory bowel disease.42-45 FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
51 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 42 Patients with primary sclerosing cholangitis (PSC) and inflammatory bowel disease or other established risk factors for colorectal cancer should undergo an annual screening colonoscopy with biopsies. Colectomy, including continence-preserving pouch operations, should be considered when colonic biopsy reveals dysplasia. (grade 1, level B).
RATIONALE 42 Cutaneous malignancies are the most common form of malignancy in recipients of solid organ transplants, but cigarette smokers are at increased risk of developing lung cancer and oropharyngeal cancer, and the rate of colon cancer is increased in patients undergoing transplantation for PSC because of the comorbid risk from inflammatory bowel disease.42-45 FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
52 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 43 For patients without prior hepatocellular carcinoma (HCC) who develop recurrent cirrhosis of the allograft, surveillance for de novo HCC should be undertaken with abdominal imaging every 6 to 12 months (grade 1, level A).
RATIONALE 43 Guidelines for surveillance after LT, including the choice of the surveillance method, the intervals between surveillance tests, and the duration of surveillance, have not been established for patients undergoing transplantation for known HCC or for patients with incidental HCC found in the explanted liver.47 A reasonable plan is for the patient to undergo abdominal and chest computed tomography every 6 months for 3 years after LT.
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53 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 44 An immunosuppressant regimen that includes sirolimus (started several weeks after transplantation) should be considered for patients undergoing transplantation for hepatocellular carcinoma (HCC) (grade 2, level B).
RATIONALE 44 The proportion of patients undergoing LT for HCC has increased significantly in the past decade. Rates of recurrence at 4 years are 10% for patients with tumors within the Milan criteria and 40% to 60% for patients with tumors outside the Milan criteria.47 Tumor recurrence reduces long-term survival after LT for HCC. Accumulating data suggest that once postoperative healing is complete, the substitution of sirolimus for a CNI reduces the risk of recurrence of HCC.48 FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
54 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 45 Resection or ablation is usually the treatment of choice for a solitary extrahepatic metastasis or an intrahepatic recurrence of hepatocellular carcinoma (grade 1, level B).
RATIONALE 45 Guidelines for surveillance after LT, including the choice of the surveillance method, the intervals between surveillance tests, and the duration of surveillance, have not been established for patients undergoing transplantation for known HCC or for patients with incidental HCC found in the explanted liver.47 A reasonable plan is for the patient to undergo abdominal and chest computed tomography every 6 months for 3 years after LT. The serial measurement of alpha-fetoprotein is a useful adjunct for patients who had an elevated alpha-fetoprotein level before transplantation or ablation therapy. Any suspicious lesion discovered on surveillance should be characterized fully, and biopsy should be included when the diagnosis is in doubt. Ablation with radiofrequency is the best treatment for small solitary recurrences.
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55 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 46 Pregnancy in an LT recipient should be managed by a high-risk obstetrician in coordination with the transplant hepatologist (grade 1, level C).
RATIONALE 46 Pregnancy in the LT recipient has risks to both the mother and the fetus.51, 54 Although the numbers of pregnancies reported are relatively small, pregnancies completing the first trimester successfully generally proceed to a live birth, although there is a higher incidence of prematurity (29%-50%) and low birth weight (17%-57%).51, 54 Neonatal deaths or birth defects are not more frequent in comparison with the general population (except when the mother is on mTOR inhibitors).55, 56 The maternal risks include hypertension and pre-eclampsia, which occur more commonly in comparison with the general population.57 FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
56 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 47 Pregnancy should be delayed for 1 year after LT and occur at a time with good, stable allograft function, with maintenance immunosuppression, and with good control of any medical complications such as hypertension and diabetes (grade 1, level B).
RATIONALE 47 The National Transplant Pregnancy Registry guidelines51 recommend the female LT recipients postpone conception until • At least 1 year after LT.
• Allograft function is stable.
• Medical comorbidities such as diabetes and hypertension are well controlled.
• Immunosuppression is at a low maintenance level.
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57 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 48 The ideal immunosuppression for pregnancy is tacrolimus monotherapy, which should be maintained at therapeutic levels throughout pregnancy; cyclosporine, azathioprine, and prednisone may also be used if they are necessary (grade 1, level B).
RATIONALE 48 All immunosuppressive drugs cross the placenta and enter the fetal circulation with resulting concerns about teratogenicity and fetal loss. Table 11 shows the Food and Drug Administration (FDA) safety categories for drugs in pregnancy. Generally, CNIs (class C drugs), prednisone (class B), and azathioprine (class D) appear to be safe.54 The newer agents should be avoided if possible; in the National Transplant Pregnancy Registry,51 more structural abnormalities have been seen in babies born to mothers on mTOR inhibitors or mycophenolic acid, especially when they are used in early pregnancy.55, 56 FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
58 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 49 Allograft function and calcineurin inhibitor (CNI) serum levels are monitored every 4 weeks until 32 weeks, then every 2 weeks, and then weekly until delivery (grade 1, level B).
RATIONALE 49 Immunosuppression should be maintained during pregnancy to avoid rejection, and drug levels of CNIs should be monitored with dose adjustments for the increasing blood volume during the second half of pregnancy.59 Allograft function and CNI serum levels should be monitored frequently until delivery.
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59 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RATIONALE 50 Pregnancy in the LT recipient has risks to both the mother and the fetus.51, 54 Although the numbers of pregnancies reported are relatively small, pregnancies completing the first trimester successfully generally proceed to a live birth, although there is a higher incidence of prematurity (29%-50%) and low birth weight (17%-57%).51, 54 Neonatal deaths or birth defects are not more frequent in comparison with the general population (except when the mother is on mTOR inhibitors).55, 56 The maternal risks include hypertension and pre-eclampsia, which occur more commonly in comparison with the general population.57 Maternal deaths following pregnancy in LT recipients are rare and occur at a rate similar to that in the general population. The National Transplant Pregnancy Registry guidelines51 recommend the female LT recipients postpone conception until • At least 1 year after LT. • Allograft function is stable. • Medical comorbidities such as diabetes and hypertension are well controlled. • Immunosuppression is at a low maintenance level.
Contraception with whatever method is favored by the LT recipient should start before sexual activity is resumed.
RECOMMENDATION 50 Contraception should begin before the resumption of sexual activity, although no particular form of contraception can be recommended over another (grade 2, level B).
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60 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 51 An assessment for infections following LT should take into account the intensity of immunosuppression, the timing of the presentation, the environmental and donor exposures, the recipient’s history of both symptomatic and latent infections, and the utilization of prophylactic antimicrobials and immunizations (grade 1, level A). RATIONALE 51 The interval from the third to sixth month after LT is a high-risk period because of the occurrence of infections with opportunistic pathogens: herpes viruses (especially CMV, herpes zoster and simplex, and EBV), fungi (including Aspergillus and Cryptococcus), and unusual bacterial infections such as Nocardia, Listeria, and mycobacteria. The implementation of prophylactic antimicrobials, the avoidance of high-risk exposures, and the minimization of immunosuppression may reduce the occurrence of these pathogens.60 After the sixth posttransplant month, the risk of infection is lower, and this is related to the reduction of immunosuppression. From 3 to 24 months after LT, in the standard-risk LT recipient (ie, no augmented immunosuppression or specific environmental exposures), the most common infections are intra-abdominal or in the lower respiratory tract or infections by community-acquired pathogens such as enteric gram-negative infections, Streptococcus pneumonia, and respiratory viruses.
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61 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 52 Attention should be paid to potential drug interactions when new antimicrobial therapies are initiated (grade 1, level A).
RATIONALE 52 Table 4 outlines the drug-drug interactions involving anti-infectives and immunosuppressive agents.
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62 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 53 High-risk recipients (cytomegalovirus (CMV)-seronegative recipients of CMV-seropositive donor organs) should receive prophylaxis with ganciclovir or valganciclovir for a minimum of 3 months after transplantation (grade 1, level B).
RATIONALE 53 CMV remains the most significant opportunistic pathogen affecting LT recipients and produces diverse clinical manifestations and significant morbidity and mortality.61, 62 The most common clinical syndromes include viremia, bone marrow suppression, and involvement of the gastrointestinal tract and liver.
Risk factors for CMV61, 62 include the following: • CMV-seropositive donor organ (especially in the absence of prior immunity, ie, a CMV-seronegative recipient).
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63 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 54 The treatment of LT recipients with cytomegalovirus (CMV) should be maintained until viremia and all symptoms have resolved (grade 2, level B).
RATIONALE 54 The treatment of CMV should be started whenever recipients are symptomatic, have a tissue injury, or have persistent or increasing viremia.61, 63, 64 All LT recipients with a symptomatic CMV infection and/or end organ disease should receive antiviral therapy and have their immunosuppression reduced until viremia and all symptoms have resolved.
FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
64 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 55 Prophylaxis against cytomegalovirus (CMV) should be resumed whenever LT recipients receive anti-lymphocyte therapy for the treatment of rejection and should be continued for 1 to 3 months after the treatment of rejection (grade 2, level B).
RATIONALE 55 CMV remains the most significant opportunistic pathogen affecting LT recipients and produces diverse clinical manifestations and significant morbidity and mortality.61, 62 The most common clinical syndromes include viremia, bone marrow suppression, and involvement of the gastrointestinal tract and liver.
Risk factors for CMV61, 62 include the following: • CMV-seropositive donor organ (especially in the absence of prior immunity, ie, a CMV-seronegative recipient).
• Augmented immunosuppression (especially with the use of anti-lymphocyte antibodies or high-dose mycophenolate).
• Allograft rejection.
• Coinfection with other immunomodulating viruses (eg, human herpesviruses 6 and 7), bacteria, or fungi.
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65 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 56 The treatment of a cytomegalovirus infection should consist of the following: a. Consideration of immunosuppression reduction. b. High-dose intravenous ganciclovir or oral valganciclovir in individuals with mild to moderate disease without gastrointestinal involvement or a reduced capacity for absorption. c. A minimum of 2 weeks of treatment. Treatment should be continued to complete the resolution of all symptoms and viremia (grade 1, level A).
RATIONALE 56 Options for antiviral treatment include intravenous ganciclovir (5 mg/kg twice daily adjusted for renal impairment) and oral valganciclovir (900 mg twice daily adjusted for renal impairment) for mild to moderate disease if no significant gastrointestinal involvement is assumed (note: valganciclovir is not approved for use in LT). For those patients with more severe disease or gastrointestinal involvement, intravenous ganciclovir is preferable. A minimum of 2 weeks of treatment is recommended for those patients with rapid resolution of symptoms, but treatment should be continued until there is complete resolution of both symptoms and viremia. Whenever possible, a reduction in immunosuppression should be combined with antiviral therapy.
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66 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 57 Resistant virus should be suspected in patients with a history of prolonged ganciclovir or valganciclovir exposure who have a persistent or progressive infection despite treatment with high-dose intravenous ganciclovir (grade 1, level A). In such instances, genotypic assays should be performed, and consideration should be given to the initiation of foscarnet with or in substitution for ganciclovir (grade 1, level B).
RATIONALE 57 Options for antiviral treatment include intravenous ganciclovir (5 mg/kg twice daily adjusted for renal impairment) and oral valganciclovir (900 mg twice daily adjusted for renal impairment) for mild to moderate disease if no significant gastrointestinal involvement is assumed (note: valganciclovir is not approved for use in LT). For those patients with more severe disease or gastrointestinal involvement, intravenous ganciclovir is preferable. A minimum of 2 weeks of treatment is recommended for those patients with rapid resolution of symptoms, but treatment should be continued until there is complete resolution of both symptoms and viremia. Whenever possible, a reduction in immunosuppression should be combined with antiviral therapy. Ganciclovir resistance is uncommon in solid organ transplant recipients.
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67 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 58 Posttransplant lymphoproliferative disorder (PTLD) should be considered in LT recipients (especially high-risk individuals) who present with unexplained fever, lymphadenopathy, or cytopenias (grade 1, level A).
RATIONALE 58 Epstein-Barr virus (EBV)-associated PTLD is an uncommon but serious complication of LT with an incidence in adults of 0.9% to 2.9%.64, 65 Risk factors include a primary EBV infection, CMV donor-recipient mismatch or CMV disease, and augmented immunosuppression, especially with anti-lymphocyte antibodies.66 It is uncertain whether the etiology of liver disease influences the development of PTLD.67, 68 The association of PTLD with EBV infection is variable in adult LT recipients; later onset PTLD is less likely to be EBV-associated.64, 67, 68 Manifestations of PTLD include lymphadenopathy, cytopenias, unexplained fever, and disturbances of the gastrointestinal tract, lungs, spleen, and central nervous system.
FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
68 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 59 Although Epstein-Barr virus (EBV) may be associated with the development of posttransplant lymphoproliferative disorder (PTLD), the detection of EBV viremia is not diagnostic for PTLD; a histopathological diagnosis is required (grade 1, level A).
RATIONALE 59 The diagnosis of PTLD requires a high index of suspicion and should be considered in high-risk individuals who present with undiagnosed fever or unexplained lymphadenopathy or cytopenias.66–69 Radiographic studies can identify sites of involvement, especially when pulmonary or intra-abdominal sites are involved. The detection of EBV viremia with nucleic acid testing is not diagnostic of EBV-associated PTLD.
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69 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 60 The diagnosis of fungal infections may require diagnostic biopsy for pathological and microbiological confirmation (grade 1, level A).
a. Blood cultures are most helpful for the diagnosis of Candida bloodstream infections (class 1, level B) and Blastomyces (grade 1,level B).
b. Cryptococcal antigen testing of cerebrospinal fluid or blood is most helpful for the diagnosis of Cryptococcus (grade 1, level B).
c. Urinary histoplasmosis and Blastomyces antigens are useful for the diagnosis of disseminated histoplasmosis and blastomycosis, respectively (grade 1, level B).
RATIONALE 60 Risk factors for fungal infections after LT include preoperative fungal colonization, massive transfusion requirements (>40 U of blood products), choledochojejunostomy, reoperation, retransplantation, hepatic iron overload, renal replacement therapy, and extended intervals of intensive care immediately before LT.
The recognition of an invasive fungal infection after 90 days is challenging. Blood cultures are relatively insensitive for the diagnosis of many fungal infections, including Candida species, for which the (1,3)-β-D-glucan test is an inconsistent measure.71, 72 Aspergillus is especially difficult to diagnose with noninvasive testing.70 The sensitivity and specificity of galactomannan in either blood or bronchoalveolar lavage from LT recipients with presumed pulmonary aspergillosis are variable.72-74 Serum and cerebrospinal cryptococcal antigen testing is a sensitive tool for the rapid diagnosis of cryptococcal infections in organ transplant recipients.75 The isolation of Cryptococcus from a site other than cerebrospinal fluid should prompt lumbar puncture to rule out central nervous system involvement. Urinary histoplasmosis and Blastomyces antigens have been useful for the diagnosis of disseminated histoplasmosis and blastomycosis, respectively.76, 77 FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
70 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 61 A cautious reduction of immunosuppression should be initiated to prevent immune reconstitution syndrome, especially for cryptococcal infections (grade 1, level B).
RATIONALE 61 The treatment of fungal infections includes antifungal drug therapy as well as a reduction of immunosuppressive therapy.
The choice of antifungal agents varies with the pathogen and the site of involvement, as shown in Table 13.
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71 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 62 All LT recipients should receive prophylaxis against P. jirovecii with trimethoprim-sulphamethoxazole (single strength daily or double strength 3 times per week) for a minimum of 6 to12 months after transplantation (grade 1, level A). Atovaquone and dapsone are the preferred alternatives for patients who are intolerant of trimethoprim-sulfamethoxazole (grade 1, level B).
RATIONALE 62 P. jirovecii (formerly called P. carinii) is an uncommon pathogen in LT recipients, primarily because of the widespread use of antimicrobial prophylaxis after LT.78 Pneumocystis should be suspected in individuals presenting with respiratory symptoms, hypoxemia (often exacerbated by exercise), and fever.78 High-dose trimethoprim-sulfamethoxazole (administered orally or intravenously at 15-20 mg/kg/day in divided doses and adjusted for renal dysfunction) is the drug of choice.78 Corticosteroids (40-60 mg of prednisone or its equivalent) should be used in conjunction with antimicrobial therapy for patients with significant hypoxia (partial pressure of arterial oxygen <70 mm Hg on room air). The minimal duration of antimicrobial therapy is 14 days, but more severe infections may merit longer courses of treatment.
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72 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 63 Trimethoprim-sulphamethoxazole is the drug of choice for the treatment of P. jirovecii pneumonia. Intravenous pentamidine is the preferred alternative for patients intolerant of trimethoprim-sulphamethoxazole with more severe infections (grade 1, level A).
RATIONALE 63 High-dose trimethoprim-sulfamethoxazole (administered orally or intravenously at 15-20 mg/kg/day in divided doses and adjusted for renal dysfunction) is the drug of choice.78 Corticosteroids (40-60 mg of prednisone or its equivalent) should be used in conjunction with antimicrobial therapy for patients with significant hypoxia (partial pressure of arterial oxygen <70 mm Hg on room air). The minimal duration of antimicrobial therapy is 14 days, but more severe infections may merit longer courses of treatment.
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73 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 64 Patients with clinical signs and symptoms or radiological features suggestive of P. jirovecii pneumonia should undergo sputum sampling or bronchoalveolar lavage with a cytological examination using a silver or Giemsa stain, polymerase chain reaction, or a specific antibody stain to identify the organism (grade 1, level A).
RATIONALE 64 Pneumocystis should be suspected in individuals presenting with respiratory symptoms, hypoxemia (often exacerbated by exercise), and fever.78 Classic radiographic findings include bilateral interstitial infiltrates. The diagnosis is confirmed by the identification of the organism by a cytological examination of induced sputum or bronchoalveolar lavage fluid.
FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
74 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 65 The treatment of active tuberculosis (TB) should include the initiation of a 4-drug regimen using isoniazid, rifampin, pyrazinamide, and ethambutol (under the assumption of susceptible TB) with adjustments in accordance with subsequent culture results. This may be tapered to 2 drugs (isoniazid and rifampin) after 2 months (under the assumption of no resistance) and continued for a minimum of 4 additional months (grade 1, level B).
RATIONALE 65 Several risk factors for the development of symptomatic TB after LT have been identified: a prior infection with TB; intensified immunosuppression (especially anti–T lymphocyte therapies); DM; and coinfections with CMV, mycoses, P. jirovecii, and Nocardia.79 Donor-derived TB is rare. The diagnosis of TB may be confounded by an increased incidence of atypical presentations (especially extrapulmonary infections involving diverse organs and locations).80, 81 Standard antituberculous regimens for drug-susceptible isolates include 2 months of 4-drug therapy with isoniazid, rifampin, ethambutol, and pyrazinamide followed by an additional 4 months of isoniazid and rifampin.
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75 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 66 Close monitoring for rejection and hepatotoxicity is imperative while LT recipients receive anti-TB therapy (grade 1, level A).
RATIONALE 66 The use of anti-TB agents in LT recipients is complicated by hepatotoxicity and by the significant drug-drug interactions with immunosuppressive agents, which lead to the potential for hepatotoxicity associated with antitubercular chemotherapy. Because of the risk of a marked reduction in CNI and mTOR inhibitor levels with rifampin coadministration, the doses of CNIs will need to be increased 2- to 5-fold at the initiation of treatment.
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76 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 67 HIV-infected LT recipients receiving highly active antiretroviral therapy (HAART) require frequent monitoring of calcineurin inhibitor (CNI) levels because of the significant interaction between antiretrovirals and CNIs (grade 1, level A).
RATIONALE 67 The use of HAART in LT recipients is complicated by significant drug-drug interactions with immunosuppressive agents, which lead to a risk of cyclosporine or tacrolimus toxicity or inadequate immunosuppression.84 LT recipients with HCV-HIV coinfections have a higher frequency and severity of acute cellular rejection.85 The CNI doses and the frequency of their administration need to be reduced markedly in LT recipients receiving HAART containing protease inhibitors.83, 84 In contrast, those receiving nonnucleoside reverse transcriptase inhibitors (especially efavirenz) will require higher doses of CNIs.
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77 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 68 HIV-infected LT recipients receiving highly active antiretroviral therapy (HAART) should be followed with scheduled HIV viral loads and T lymphocyte subset counts (grade 1, level A).
RATIONALE 68 HIV-infected patients with well-controlled infections have undergone transplantation with success, although aggressive HCV recurrence has been problematic in LT recipients coinfected with HCV.83 HIV-infected patients maintained on highly active antiretroviral therapy (HAART) after transplantation do not experience an increase in opportunistic infections.
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78 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 69 Standard prophylaxis for cytomegalovirus (CMV) is recommended for HIV-infected LT recipients receiving highly active antiretroviral therapy, and lifelong Pneumocystis pneumonia prophylaxis is the norm (grade 1, level A).
RATIONALE 69 CMV remains the most significant opportunistic pathogen affecting LT recipients and produces diverse clinical manifestations and significant morbidity and mortality.61, 62 The most common clinical syndromes include viremia, bone marrow suppression, and involvement of the gastrointestinal tract and liver.
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79 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 70 Standard HIV-specific prophylaxis for low CD4 counts should be used (grade 1, level A).
RATIONALE 70 HIV-infected patients with well-controlled infections have undergone transplantation with success, although aggressive HCV recurrence has been problematic in LT recipients coinfected with HCV.83 HIV-infected patients maintained on highly active antiretroviral therapy (HAART) after transplantation do not experience an increase in opportunistic infections. The use of HAART in LT recipients is complicated by significant drug-drug interactions with immunosuppressive agents, which lead to a risk of cyclosporine or tacrolimus toxicity or inadequate immunosuppression.84 LT recipients with HCV-HIV coinfections have a higher frequency and severity of acute cellular rejection.85 The CNI doses and the frequency of their administration need to be reduced markedly in LT recipients receiving HAART containing protease inhibitors.83, 84 In contrast, those receiving nonnucleoside reverse transcriptase inhibitors (especially efavirenz) will require higher doses of CNIs.
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80 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 71 All LT recipients should receive an annual influenza vaccination (grade 1, level B).
RATIONALE 71 Appropriate advice regarding vaccination after LT has been reviewed by Danzinger-Isakov et al.86 and is also reviewed in Table 14. LT recipients should avoid live virus vaccines because of concerns about the dissemination of infections. Vaccine-related rejection has not been associated with immunization following LT.
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81 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 72 All LT recipients should avoid live virus vaccines (grade 1, level A).
RATIONALE 72 Appropriate advice regarding vaccination after LT has been reviewed by Danzinger-Isakov et al.86 and is also reviewed in Table 14. LT recipients should avoid live virus vaccines because of concerns about the dissemination of infections. Vaccine-related rejection has not been associated with immunization following LT.
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82 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 73 Re-immunization is indicated for some vaccines, notably the influenza vaccine (annually) and the pneumococcal vaccine (every 3-5 years; no class or level provided). (grade 1, level A).
RATIONALE 73 Appropriate advice regarding vaccination after LT has been reviewed by Danzinger-Isakov et al.86 and is also reviewed in Table 14. LT recipients should avoid live virus vaccines because of concerns about the dissemination of infections. Vaccine-related rejection has not been associated with immunization following LT.
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83 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 74 Long-term prophylactic therapy using a combination of antiviral agents and low-dose hepatitis B immune globulin (HBIG) on demand or at fixed intervals can effectively prevent hepatitis B virus (HBV) recurrence rates in ≥90% of transplant recipients (grade 1, level B).
RATIONALE 74 The survival for patients undergoing transplantation for HBV is excellent, and HBV ranks among the best of all indications for LT. The improvements in patient and graft survival evident over the past 10 to 15 years reflect the advances in therapeutics to prevent and control HBV infections after LT. The combination of hepatitis B immune globulin (HBIG) and nucleos(t)ide analogues can prevent recurrent infections in almost all HBV-infected LT patients.88-91 The combination of HBIG and nucleos(t)ide analogues is superior to HBIG alone. The individualization of prophylactic combination therapy can be undertaken on the basis of pretransplant clinical and virological characteristics.
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84 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 75 In patients with low or undetectable HBV DNA levels before transplantation and an absence of high-risk factors for recurrence, hepatitis B immune globulin (HBIG) can be discontinued, and long-term treatment with antivirals (single or in combination) can be used as an alternative prophylactic strategy (grade 2, level B).
RATIONALE 75 The individualization of prophylactic combination therapy can be undertaken on the basis of pretransplant clinical and virological characteristics. For example, low-dose intramuscular HBIG is much less expensive and avoids painful side effects associated with intravenous HBIG. The discontinuation of HBIG is generally reserved for patients at low risk for HBV recurrence.92 FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
85 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 76 Lifelong antiviral therapy should be used to treat patients with recurrent HBV infections. Combination antiviral therapy is superior to monotherapy when drugs with a low genetic barrier to resistance are used, whereas the discontinuation of hepatitis B immune globulin is generally reserved for patients at low risk for HBV recurrence (grade 1, level B).
RATIONALE 76 A recurrent infection is manifest with persistently detectable HBV DNA and hepatitis B surface antigen in serum and is usually due to a failure of prophylactic therapy. Liver biopsy is useful for assessing the severity of HBV recurrence and the progression of fibrosis.
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86 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 77 Retransplantation for recurrent HBV is appropriate when treatment strategies to prevent or treat recurrent HBV disease are available (grade1, level C).
RATIONALE 77 Fibrosing cholestatic HBV is a unique histological variant observed in LT recipients and is characterized by high intrahepatic levels of HBV DNA, hepatocyte ballooning with cholestasis, and a paucity of inflammatory cells.93 This represents the most severe presentation of recurrent disease and is rarely seen in the current era of prophylactic therapy.
FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
87 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 78 Liver biopsy is useful in monitoring disease severity and progression and in distinguishing recurrent hepatitis C virus (HCV) disease from other causes of liver enzyme elevations (grade 1, level C).
RATIONALE 78 Although recurrent HCV is more likely the longer the interval from LT, in practice, it is often difficult to distinguish between the histopathological appearances of a recurrent HCV infection and acute cellular rejection.
Posttransplant antiviral therapy is generally reserved for those showing evidence of progressive disease, which is manifested by the presence of moderate to severe necroinflammation or mild to moderate fibrosis, although this paradigm will change with more efficacious and less toxic antiviral therapy.100 The primary goal of post-LT antiviral therapy is the achievement of sustained viral clearance because this virological outcome is associated with fibrosis stabilization or regression and improved graft survival.101 The initiation of antiviral therapy is recommended when significant histological disease is present, although this paradigm would change with more efficacious and less toxic antiviral therapy.
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88 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 79 Prophylactic antiviral therapy has no current role in the management of HCV disease (grade 1, level A).
RATIONALE 79 Posttransplant antiviral therapy is generally reserved for those showing evidence of progressive disease, which is manifested by the presence of moderate to severe necroinflammation or mild to moderate fibrosis, although this paradigm will change with more efficacious and less toxic antiviral therapy.100 FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
89 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 80 Moderate acute rejection should be treated with increased maintenance immunosuppression and corticosteroid boluses, whereas lymphocyte-depleting drugs should be avoided (grade 1, level B).
RATIONALE 80 The impact of immunosuppressives on the progression of HCV is poorly understood, although some data suggest that anti-lymphocyte agents promote HCV-associated liver injury.
The pooled estimated rate of acute graft rejection occurring in patients receiving peginterferon and ribavirin is 5%, which is not significantly higher than the rate in untreated controls.102 However, alloimmune or plasma cell hepatitis, characterized histologically by an inflammatory infiltrate with abundant plasma cells in the setting of increased liver enzymes, has been described during antiviral therapy.103 This is most likely a variant of allograft rejection and responds to the discontinuation of interferon and the amplification of immunosuppression in most cases FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
90 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 81 Antiviral therapy is indicated for significant histological disease: grade 3 or higher inflammatory activity and/or stage 2 or higher fibrosis (on a scale of 4) or cholestatic hepatitis. Peginterferon and ribavirin are the current drugs of choice. The risks and benefits of triple therapy with protease inhibitors are to be determined. The goal of antiviral therapy is the achievement of a sustained virological response, and this confers a survival benefit (grade 1, class B).
RATIONALE 81 Posttransplant antiviral therapy is generally reserved for those showing evidence of progressive disease, which is manifested by the presence of moderate to severe necroinflammation or mild to moderate fibrosis, although this paradigm will change with more efficacious and less toxic antiviral therapy.100 The primary goal of post-LT antiviral therapy is the achievement of sustained viral clearance because this virological outcome is associated with fibrosis stabilization or regression and improved graft survival.101 The initiation of antiviral therapy is recommended when significant histological disease is present, although this paradigm would change with more efficacious and less toxic antiviral therapy.
With the recent approval of the first-generation protease inhibitors, telaprevir and boceprevir, it is anticipated that triple therapy (peginterferon, ribavirin, and either telaprevir or boceprevir) will evolve into the new standard of care over the next few years for LT recipients infected with genotype 1 virus.
FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
91 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 82 Retransplantation for recurrent HCV disease should be considered selectively (grade 2, level B).
RATIONALE 82 Recurrent HCV infection is invariable among patients who are viremic at LT, the majority of whom will have histological evidence of recurrent hepatitis within the first year after LT.94 Although the progression of fibrosis in HCV-infected LT recipients is highly variable, in the absence of antiviral therapy, the median time to the development of cirrhosis is 8 to 10 years, whereas an estimated 30% will develop cirrhosis within 5 years of LT.95 The risk of decompensation is 15% to 30% within the first year of the onset of cirrhosis, and the mortality risk is 40% to 55% within 6 to 12 months of the decompensating event. Recurrent HCV cirrhosis is the most frequent cause of graft loss in this population.96 Patient survival and graft survival are reduced in HCV-infected patients versus HCV-negative patients, with a 5-year patient survival rate of approximately 70%.97, 98 FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
92 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 83 Primary biliary cirrhosis LT recipients should be routinely monitored for associated autoimmune diseases (eg, thyroid disease) and bone density (grade 2, level B).
RATIONALE 83 Immunological abnormalities (eg, elevated immunoglobulins and autoantibodies) persist after transplantation. Recipients remain at risk for associated conditions, such as sicca syndrome, osteoporosis, and thyroid disease, so screening should be included in the follow-up.
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93 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 84 For those with histological evidence of recurrent disease, treatment with ursodeoxycholic acid at 10 to 15 mg/kg/day (grade 2, level B) may be considered, and although its use is associated with the improvement of liver tests, no impact on graft survival has been documented (grade 2, level B). There is no indication for offering prophylaxis with ursodeoxycholic acid to patients with normal liver histology (grade 2, level B).
RATIONALE 84 Recurrent PBC is diagnosed by liver histology: recurrent disease may occur in the presence of normal liver tests, and neither the presence nor the titer of anti-mitochondrial antibodies correlates with the presence or degree of recurrence.105, 106 The reported incidence of recurrent PBC varies from 4% to 33% (the average is 18%).104 Although the use of cyclosporine is associated with less severe recurrence and corticosteroids may be associated with less recurrence, there are insufficient data to recommend a preferred immunosuppressive regimen.106 The impact of the recurrence of PBC on graft function and survival is minimal for the first decade after transplantation, with end-stage disease affecting less than 5%. There is no evidence that routine protocol biopsy in PBC LT recipients will improve outcomes. Ursodeoxycholic acid at a dose of 10 to 15 mg/kg/day is associated with an improvement in liver tests, but there are no data to show benefits in patient or graft survival.104 FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
94 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 85 Although there are few data on prevention, it is recommended that those patients grafted for primary sclerosing cholangitis (PSC) in the presence of chronic ulcerative colitis (CUC) have an annual colonoscopy with mucosal biopsy (grade 2, level B).
RATIONALE 85 In patients with chronic ulcerative colitis (CUC), colitis may improve or deteriorate after transplantation.107 PSC LT recipients with CUC are at greater risk of developing colonic polyps and cancer and should have an annual colonoscopy. There is no evidence for the optimal screening approach in PSC LT recipients without CUC, but many advocate an annual colonoscopy in this group also.
FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
95 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 86 Although there is no evidence for recommending a particular immunosuppressive regimen in patients undergoing transplantation for autoimmune hepatitis (AIH), it is prudent to maintain patients on long-term, low-dose corticosteroids in addition to routine immunosuppression (with attention to maintaining bone health; grade 2, level B).
RATIONALE 86 Outcomes after transplantation for AIH are good. Patients should be closely monitored for evidence of recurrence via liver tests every 6 months.109 Protocol liver biopsy should be considered at 5 yearly intervals. The reported outcome rates for recurrent AIH are highly variable. Although the majority of patients with putative recurrent AIH will respond clinically, serologically, and histologically to increased immunosuppression, some will progress to end-stage graft failure and may require retransplantation.
FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
96 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 87 All patients with a prior diagnosis of alcoholic liver disease (ALD) should be encouraged to remain abstinent from alcohol (grade 1, level B).
RATIONALE 87 The best prospective study showed that 80% of ALD LT recipients either did not drink or consumed only small amounts occasionally in the first 5 years.111 Conversely, in the remaining 20%, there were various patterns of harmful drinking. Anecdotal reports suggest that patients who relapse to harmful drinking are at risk for alcoholic hepatitis, delirium tremens, alcoholic pancreatitis, and reduced survival.110 Furthermore, the causes of death for the patients who returned to heavy consumption of alcohol tended to be liver-related, whereas abstinent ALD patients died of cardiovascular disease and malignant tumors.
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97 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 88 Patients should be encouraged to enter therapy or counseling if they relapse into alcohol use (grade 1, level C).
RATIONALE 88 Anecdotal reports suggest that patients who relapse to harmful drinking are at risk for alcoholic hepatitis, delirium tremens, alcoholic pancreatitis, and reduced survival.110 Furthermore, the causes of death for the patients who returned to heavy consumption of alcohol tended to be liver-related, whereas abstinent ALD patients died of cardiovascular disease and malignant tumors.
FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
98 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 89 All patients with a prior diagnosis of alcoholic liver disease (ALD) who are users of tobacco should be encouraged to undertake smoking cessation (grade 1, level B).
RATIONALE 89 Furthermore, the causes of death for the patients who returned to heavy consumption of alcohol tended to be liver-related, whereas abstinent ALD patients died of cardiovascular disease and malignant tumors. The stratification of cardiovascular deaths and new-onset cancers of the aerodigestive tract in patients undergoing LT for ALD suggests a causal linkage with cigarette smoking.
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99 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 90 Careful attention should be given to the risk of cardiovascular disease and/or new-onset cancers of the aerodigestive tract, especially in cigarette smokers (grade 1, level A).
RATIONALE 90 Furthermore, the causes of death for the patients who returned to heavy consumption of alcohol tended to be liver-related, whereas abstinent ALD patients died of cardiovascular disease and malignant tumors. The stratification of cardiovascular deaths and new-onset cancers of the aerodigestive tract in patients undergoing LT for ALD suggests a causal linkage with cigarette smoking.
FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
100 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 91 The confirmation of recurrent or de novo nonalcoholic fatty liver disease (NAFLD), the recognition of fibrosis, and the exclusion of alternate causes of elevated liver chemistry tests require liver biopsy (grade 1, level B).
RATIONALE 91 New-onset or recurrent NAFLD/NASH may present with elevated liver aminotransferases. Distinguishing NAFLD/NASH from other causes of elevated liver tests in the post-LT patient requires liver biopsy. NAFLD/NASH arising in the liver allograft, whether new-onset or recurrent, may lead to fibrosis.115 FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
101 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 92 No specific recommendations regarding the prevention or treatment of nonalcoholic fatty liver disease (NAFLD) or nonalcoholic steatohepatitis (NASH) in LT recipients can be made other than general recommendations to avoid excessive gains in bodyweight and control hypertension and diabetes (grade 1, level C).
RATIONALE 92 Risk factors for post-LT NASH/NAFLD are familiar as the hallmarks of metabolic syndrome: body mass index before and after LT, DM, systemic hypertension, hyperlipidemia, and steatosis on an allograft biopsy sample. Among patients who undergo LT on account of NASH-associated or cryptogenic cirrhosis, 50% to 70% will gain excessive weight in 1 year.115 FORWARD BACK BACK TO RECOMMENDATIONS LIST AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
102 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS RECOMMENDATIONS RECOMMENDATION 93 LT recipients with an incisional hernia should be instructed to recognize incarcerated hernias and advised to seek immediate medical assistance (grade 1, level B).
RATIONALE 93 Hepatic artery stenosis, biliary cast syndrome, and bilomas have already been discussed with respect to abnormal liver tests. Incisional hernia is a common late complication after LT. Postoperative weight gain exacerbates the risk.
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103 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS FULL TEXT The following is the complete content of this practice guideline. For an alternate printable version in the original publication layout, please use the “Web Site” link above.
Long-Term Management of the Successful Adult Liver Transplant: 2012 Practice Guideline by AASLD and the American Society of Transplantation Michael R. Lucey,1 Norah Terrault,2 Lolu Ojo,3 J. Eileen Hay,4 James Neuberger,5 Emily Blumberg,6 and Lewis W. Teperman7 Received: October 8, 2012; accepted October 20, 2012 DOI: 10.1002/lt.23566 1Division of Gastroenterology and Hepatology, Department of Medicine, University of Wisconsin School of Medicine and Public Health, Madison, WI; 2Gastroenterology Division, Department of Medicine, University of California San Francisco, San Francisco, CA; 3Division of Nephrology, Department of Medicine, University of Michigan, Ann Arbor, MI; 4Mayo Clinic, Rochester, MN; 5Liver Unit, Queen Elizabeth Hospital, Birmingham, United Kingdom; 6Division of Infectious Diseases, University of Pennsylvania School of Medicine, Philadelphia, PA; and 7Department of Surgery, NYU Transplant Associates, New York, NY The development of this practice guideline was funded by AASLD.
All AASLD Practice Guidelines are updated annually. If you are viewing a Practice Guideline that is more than 12 months old, please visit www.aasld.org for an update in the material. This practice guideline was approved by the American Association for the Study of Liver Diseases on August 4, 2012 and by the American Society of Transplantation on September 19, 2012.
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104 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS FULL TEXT Abbreviations: AIH: autoimmune hepatitis ALD: alcoholic liver disease BMD: bone mineral density CKD: chronic kidney disease CMV: cytomegalovirus CNI: calcineurin inhibitor CUC: chronic ulcerative colitis DM: diabetes mellitus EBV: Epstein-Barr virus ESRD: end-stage renal disease FDA: Food and Drug Administration GRADE: Grading of Recommendations Assessment, Development, and Evaluation HAART: highly active antiretroviral therapy HbA1c: hemoglobin A1c HBIG: hepatitis B immune globulin HBV: hepatitis B virus HCC: hepatocellular carcinoma HCV: hepatitis C virus HIV: human immunodeficiency virus HLA: human leukocyte antigen LT: liver transplantation mTOR: mammalian target of rapamycin NAFLD: nonalcoholic fatty liver disease NASH: nonalcoholic steatohepatitis NODM: new-onset diabetes mellitus PBC: primary biliary cirrhosis PSC: primary sclerosing cholangitis PTLD: posttransplant lymphoproliferative disorder TB: tuberculosis FORWARD BACK AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
105 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS FULL TEXT PREAMBLE This practice guideline has been approved by the American Association for the Study of Liver Diseases and the American Society of Transplantation. These recommendations provide a data-supported approach to management of adult patients who have successfully undergone liver transplantation. They are based on the following: (1) a formal review and analysis of recently published world literature on the topic (via a MEDLINE search); (2) A Manual for Assessing Health Practices and Designing Practice Guidelines (American College of Physicians)1; (3) guideline policies,2 including the American Association for the Study of Liver Diseases policy on the development and use of practice guidelines and the American Gastroenterological Association policy statement on guidelines3; and (4) the experience of the authors in the specified topic.
Intended for use by physicians and health care providers working with adult recipients of liver transplantation (LT), these recommendations suggest preferred approaches to the diagnostic, therapeutic, and preventive aspects of care. They are intended to be flexible, in contrast to standards of care, which are inflexible policies to be followed in every case. Specific recommendations are based on relevant published information.
TABLE 1. GRADE STRENGTH OF RECOMMENDATION CRITERIA 1. Strong Factors influencing the strength of the recommendation include the quality of the evidence, the presumed patient-important outcomes, and the cost.
2. Weak There is variability in the preferences and values or more uncertainty. The recommendation is made with less certainty, or the cost or resource consumption is higher.
QUALITY OF EVIDENCE CRITERIA A. High Further research is unlikely to change confidence in the estimate of the clinical effect.
B. Moderate Further research may change confidence in the estimate of the clinical effect.
C. Low Further research is very likely to affect confidence in the estimate of the clinical effect.
To more fully characterize the available evidence supporting the recommendations, the American Association for the Study of Liver Diseases Practice Guidelines Committee has adopted the classification used by the Grading of Recommendations Assessment, Development,and Evaluation (GRADE) workgroup with minor modifications (Table 1).4 In the GRADE system,the strength of a recommendation is classified as (1) strong or (2) weak. The quality of evidence supporting a strong or weak recommendation is designated by 1 of 3 levels: (A) high, (B) moderate, or (C) low. FORWARD BACK AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
106 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS FULL TEXT LT AS A TREATMENT FOR END-STAGE LIVER DISEASE LT is the treatment of choice for patients with decompensated cirrhosis, acute liver failure, small hepatocellular carcinomas (HCCs), or acute liver failure. The success of LT has meant that there is a growing cohort of LT recipients throughout the world. From 1985 through 2011, approximately 100,000 persons in the United States underwent LT. On December 30, 2011, there were 30,000 LT recipients who were alive and had survived at least 5 years, and there were more than 16,000 recipients with 10 or more years’ survival. These long-term survivors are at risk of early death and increased morbidity. The purpose of this guideline is to assist in the management of adult recipients of LT, identify the barriers to maintaining their health, and make recommendations on the ways to best prevent or ameliorate these barriers. This guideline focuses on management beyond the first 90 days after transplantation.
MORTALITY AFTER LT The greatest proportion of deaths or retransplants after LT occur soon after transplantation. The causes of death and graft loss vary according to the interval from transplantation, with infection and intraoperative and perioperative causes accounting for nearly 60% of deaths and graft losses in the first posttransplant year. After the first year, death due to acute infections declines, whereas malignancies and cardiovascular causes account for a greater proportion of deaths. The recurrence of the pretransplant condition, especially hepatitis C virus (HCV) or autoimmune liver disease, is an increasingly important cause of graft loss the longer the patient survives transplantation for these etiologies. Today, death (or a need for retransplantation) attributable to acute or chronic allograft rejection is uncommon throughout the first 10 years after transplantation.
MORBIDITY AFTER LT The transplanted liver becomes partially tolerant of immune-mediated injury, so the requirement for immunosuppression declines after the first 90 days. Although some LT recipients may eventually achieve operational tolerance (ie, maintenance without immunosuppressant medications), this is rare. Most patients receive immunosuppression throughout the life of the allograft.5 The continued use of immunosuppression carries inevitable consequences: an increased risk of bacterial, viral, and fungal infections, which can be recurrent or newly acquired; metabolic complications such as hypertension, diabetes mellitus (DM), hyperlipidemia, obesity, and gout; and hepatobiliary or extrahepatic de novo cancers [including posttransplant lymphoproliferative disorder (PTLD)]. The combination of the complications of immunosuppression and the recurrence of the underlying liver disease translates into a heavy burden of ill heath for many LT recipients. An analysis of a longitudinal US database of 36,847 LT recipients indicated that the prevalence of kidney failure [defined as a glomerular filtration rate of 29 mL/minute/1.73 m2 of body surface area or less or the development of endstage renal disease (ESRD)] was 18% at 5 years and 25% at 10 years.6 LT recipients have at-risk cardiovascular profiles with a high prevalence of hypertension requiring antihypertensive medications, recurrent DM and new-onset diabetes mellitus (NODM), and hyperlipidemia requiring lipid-lowering agents. Cardiovascular disease and renal failure are the leading nonhepatic causes of morbidity and mortality late after LT (Table 2). The recurrence of the original disease, such as a chronic HCV infection, primary biliary cirrhosis (PBC), primary sclerosing cholangitis (PSC), autoimmune hepatitis (AIH), or HCC, can cause ongoing morbidity and mortality. Many of the patients undergoing LT have a past or present history of addictions, especially to alcohol, cigarettes, or both, which may also persist with harmful effects on patients’ health, often by interacting with other risk factors already mentioned. FORWARD BACK AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
107 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS FULL TEXT An assessment of the quality of life after LT has shown that although quality measures improve in LT patients in most domains in comparison with their status before transplantation, LT recipients continue to have many deficits in comparison with age-matched control populations; these are manifested as worsening physical symptoms, fatigue, and a greater sense of being unwell.7 Through the reduction of cardiovascular risks, the suppression or eradication of specific infections, improved surveillance for cancer, and the prevention or treatment of recurrent liver diseases, both the quantity and the quality of post-LT life can be improved. In these guidelines, we show how a concentrated effort to moderate immunosuppression, manage recurrent disease, and ameliorate metabolic complications of immunosuppression is required to convert short-term success into sustained success for an extended healthy life.
TABLE 2. PREVALENCE OF CARDIOVASCULAR RISK FACTORS AND CKD IN LT RECIPIENTS BEYOND THE FIRST POSTTRANSPLANT YEAR PREVALENCE RATE Cardiovascular risk factor Metabolic syndrome Systemic hypertension DM Obesity Dyslipidemia Cigarette smoking 50%—60% 40%—85% 10%—64% 24%—64% 40%—66% 10%—40% CKD (stage 3-4)† 30%—80% End-stage kidney disease 5%—8% Any 3 of the following: hypertension, obesity, dyslipidemia, and DM. †Estimated glomerular filtration rate. 15 to <60 mL/minute/1.73 m2.
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108 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS FULL TEXT COMPLICATIONS OF PORTAL HYPERTENSION AFTER LT Typically, clinical features of liver failure and portal hypertension resolve rapidly after LT, and they are not usual after the first 3 months. The exception is splenomegaly, which may persist for years. Variceal hemorrhage is very unusual unless the patient has an occluded portal vein. The late emergence of hepatic encephalopathy in a patient with a functioning liver allograft suggests the development of clandestine cirrhosis or a persistent portosystemic shunt. Late-onset ascites or peripheral edema may indicate stenosis of the inferior vena cava or portal vein anastomosis. Persistent late ascites in a patient with a recurrent HCV infection is a poor prognostic sign. TABLE 3. CAUSES OF LIVER TEST ABNORMALITIES IN THE ASYMPTOMATIC RECIPIENT ALLOGRAFT PARENCHYMAL DAMAGE Immune-mediated disease (rejection and de novo AIH) Recurrent disease (HCV, HBV, PBC, PSC, AIH, and others) Drug toxicity (including immunosuppressive drugs) Alcohol and other toxins De novo infection (including de novo HBV and HCV) Space-occupying lesion (recurrent cancer) De novo or recurrent NAFLD BILIARY DAMAGE Biliary strictures (anastomotic strictures, hepatic artery thrombosis or stenosis, and others) Biliary stones/cast syndrome Recurrent PSC VASCULAR DISEASE Hepatic artery thrombosis Portal or hepatic vein thrombosis METABOLIC DISEASE IN THE ALLOGRAFT Gilbert’s syndrome NONHEPATIC DISEASE MIMICKING LIVER DISEASE Hemolysis causing raised indirect bilirubin levels Bone disease causing raised alkaline phosphatase levels NONHEPATIC DISEASE CAUSING LIVER ABNORMALITIES Celiac disease Diabetes FORWARD BACK AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
109 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS FULL TEXT LIVER TESTS Liver tests are routinely monitored after LT. When liver tests are elevated for a healthy recipient, the course of action will depend on the severity and type of abnormality (cholestatic, hepatitic, or other). Clinical challenges arise when liver tests are normal in the presence of graft damage or conversely abnormal in an asymptomatic LT recipient. The many causes of liver test abnormalities in the asymptomatic recipient are shown in Table 3. More than 1 cause may coexist in the same patient. When abnormal liver tests are recognized in a healthy, asymptomatic LT recipient, it is reasonable to repeat the tests in 1 to 2 weeks. A decision to investigate further should be based on the persistence and severity of the liver test abnormalities. Investigations should include a thorough history and examination, appropriate laboratory tests, and Doppler ultrasound of the liver. It should not be assumed without appropriate histological confirmation that abnormal liver tests represent immunemediated damage.
Elevated alkaline phosphatase, total bilirubin, and aminotransferase levels may arise from the late appearance of biliary anastomotic strictures due to thrombosis or stenosis of the hepatic artery or to recurrent PSC or PBC. Appropriate biliary imaging includes endoscopic retrograde cholangiopancreatography, magnetic resonance cholangiopancreatography, and/or ultrasound. Biliary cast syndrome refers to a severe form of intrahepatic bile duct ischemic injury unique to post-LT patients,8 and it is associated with hepatic artery thrombosis and the use of a split liver, including partial grafts derived from living donors and, more commonly, from donation after cardiac death donors. Biliary cast syndrome may resolve with repeated clearance of bile duct either percutaneously or endoscopically.
RECOMMENDATIONS: 1. The frequency of monitoring with liver tests should be individualized by the transplant center according to the time from LT, the complications from LT, the stability of serial test results, and the underlying cause (grade 1, level A). 2. Depending on the pattern of liver tests, magnetic resonance imaging, computed tomography, endoscopic retrograde cholangiopancreatography, and sonography may be appropriate (grade 1, level A). 3. Liver histology should be obtained when parenchymal injury is suspected as the cause of abnormal liver tests (grade 1, level A).
VASCULAR THROMBOSIS Hepatic artery thrombosis (HAT) or stenosis may present clinically after 3 months, as: • intrahepatic non-anastomotic strictures and/or sterile or infected fluid collections within the liver, sometimes referred to as bilomas, • ischemic cholangiopathy or • biliary cast syndrome.
The combination of hepatic artery thrombosis and biliary complications usually requires retransplantation.9 RECOMMENDATIONS: 4. Bilomas and biliary cast syndrome should be managed in a center with expertise in LT medicine, radiology, and biliary endoscopy (grade 1, level A). 5. Hepatic artery thrombosis or stenosis is most readily assessed initially by Doppler ultrasound, but angiography is usually required to confirm the diagnosis and plan therapy (grade 1, level B).
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110 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS FULL TEXT IMMUNOSUPPRESSION The choice of immunosuppression depends on the following: • Indication for transplantation: the choice of immunosuppression may affect disease recurrence (eg, HCV, malignancy, or autoimmune disease).
• Comorbidities.
• Drug side effects: calcineurin inhibitors (CNIs) may cause renal impairment. • Likelihood of pregnancy: mycophenolate and mammalian target of rapamycin (mTOR) inhibitors such as sirolimus are potential teratogens. • History of severe or recurrent rejection. • Prior experience with the various immunosuppressive agents. • History of or risk for cancers. • History of or risk for infections.
There is no reliable marker for determining the effective level of immunosuppression; therefore, the choice of the agent (or agents) and doses given will be determined by the clinical, laboratory, and histological response. The CNI dose is generally determined by the drug level; the target levels after 3 months are 5 to 10 ng/mL for tacrolimus and 100 to 150 ng/mL for cyclosporine (both are whole blood trough levels). The target whole blood trough level for sirolimus is 5 ng/mL. The need for therapeutic drug monitoring for mycophenolate is uncertain. Table 4 describes drug-drug interactions involving the commonly used immunosuppressant medications. Common side effects of immunosuppressants are presented in Table 5. The majority of LT recipients need lifelong immunosuppression to maintain graft function. A very small number of LT recipients develop operational tolerance to the allograft and do not require long-term immunosuppression.5 TABLE 4. MAJOR DRUG-DRUG INTERACTIONS INVOLVING IMMUNOSUPPRESSIVE AGENTS ANTIMICROBIALS CNIs mTOR INHIBITORS MYCOPHENOLATE Fluoroquinolones (primarily ofloxacin > ciprofloxacin) Increased levels Macrolides (erythromycin > clarithromycin > azithromycin) Markedly increased levels Markedly increased levels Rifamycins (rifampin > rifabutin) Markedly decreased levels Markedly decreased levels Linezolid Increased myelosuppression Increased myelosuppression and platelet decrease Triazoles (ketoconazole/ voriconazole/posaconazole > itraconazole/fluconazole) Increased levels Increased levels (voriconazole contraindicated) Ganciclovir/valganciclovir Increased myelosuppression Increased myelosuppression FORWARD BACK AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
111 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS FULL TEXT TABLE 5. UNWANTED SIDE EFFECTS OF IMMUNOSUPPRESSIVES SIDE EFFECT CORTICOSTEROIDS CNIs mTOR INHIBITORS MYCOPHENOLATE MOFETIL Kidney injury -+++ + (proteinuria) -Bone disease +++ ---Gastrointestinal +/---+ Bone marrow suppression ---+ Pulmonary fibrosis --+ -Hypercholesterolemia + + +++ -Diabetes ++ + (tacrolimus) --Hypertension + ++ + -LATE REJECTION Late rejection is defined as rejection that has its onset more than 90 days after transplantation. Traditionally, 2 forms have been recognized: cellular rejection (also known as acute cellular rejection and late-onset rejection) and ductopenic rejection (also known as vanishing bile duct syndrome). Both forms of rejection are, until the late stages, asymptomatic, and the diagnosis is made through the investigation of abnormal liver tests; the diagnosis can be confirmed only on the basis of histology. For both cellular rejection and ductopenic rejection, the Banff criteria have been adopted to define the nature and severity.10 Liver tests in patients with late-onset cellular rejection show nonspecific abnormalities with a rise in serum bilirubin and aminotransferases. Histologically, cellular rejection is characterized by the triad of inflammatory bile duct damage, subendothelial inflammation of the portal, central, or perivenular veins, and a predominantly lymphocytic portal inflammatory infiltrate with neutrophils and eosinophils in addition. The focus of inflammation may be portal, central, or both, but the central component is more prominent and frequently occurs as pure centrilobular necroinflammation (isolated central perivenulitis). Late acute rejection differs from early acute cellular rejection by having fewer classic histological features. Risk factors leading to late-onset cellular rejection include the following: • Reduction of immunosuppression (whether iatrogenic or due to noncompliance). • Pre-LT autoimmune liver disease. • Concurrent administration of interferon (for HCV treatment). The differential diagnosis includes infection, recurrent and de novo autoimmune disease, and drug toxicity; it may sometimes be difficult to distinguish cellular rejection from HCV infection, and indeed, the two often coexist. In mild cases of cellular rejection, an increase in maintenance levels of immunosuppression may be sufficient, whereas in histologically moderate or severe cases, the treatment should be a short course of increased immunosuppression (eg, methyl prednisone at 500 mg/day or prednisolone at 200 mg/day for 3 days) followed by an increase in the baseline immunosuppression. A full response (defined as a return to normal liver tests) is seen in only approximately half of patients, with approximately 25% developing a further episode of cellular rejection and 25% developing ductopenic rejection. FORWARD BACK AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
112 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS FULL TEXT Ductopenic rejection is seen most commonly in the first year but may occur at any time. Recent data suggest that humoral alloreactivity mediated by antibodies against donor human leukocyte antigen (HLA) molecules, acting in concert with cellular mechanisms, may play a role in the development of ductopenia (a process known as antibody-mediated rejection).11 The onset of ductopenia is usually insidious, with a progressive rise in liver tests with a cholestatic picture (a rise in alkaline phosphatase and gamma-glutamyl transpeptidase followed by a progressive rise in serum bilirubin). In late cases, the recipient may complain of pruritus and jaundice. A liver biopsy sample with at least 10 portal tracts is advisable in order to establish with confidence that injury to and loss of bile ducts have occurred. In the early stages of ductopenic rejection, there may be a cellular infiltrate, but more commonly, the characteristic features include the progressive loss of bile ducts from the portal tracts and cholestasis; in late stages, foamy macrophages may be seen. Risk factors for ductopenic rejection include the following: • Recurrent and unresponsive cellular rejection. • Transplantation for autoimmune disease. • Exposure to interferon. • Loss of a previous graft to ductopenic rejection. The differential diagnosis includes recurrent disease (PBC or PSC) and drug toxicity. The treatment of ductopenic rejection is increased immunosuppression, and an increase in or switch to tacrolimus may be effective in some early cases. Conversely, especially when fewer than 50% of the portal tracts contain bile ducts, the condition progresses to graft failure. RECOMMENDATIONS 6. Immunosuppressive drugs for LT recipients should be prescribed and monitored only by those with knowledge and expertise in that area. The choice of agents will depend on many factors, and no one regimen can be recommended for any patient (grade 2, level A). 7. Every patient’s immunosuppressive regimen should be reviewed at least every 6 months and modified as required with the goal of minimizing long-term toxicities (grade 1, level B). 8. Rejection can be reliably diagnosed only on the basis of liver histology; a biopsy sample should be taken before treatment initiation and classified according to the Banff criteria (grade 1, level A). 9. Although the long-term withdrawal of all immunosuppression can be achieved in a small number of patients, this should be undertaken only with select recipients and under close supervision (grade 2, level C). FORWARD BACK AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
113 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS FULL TEXT PROMOTING HEALTH AFTER LT RECOMMENDATIONS 10. Frequent handwashing reduces the risk of infection with pathogens acquired by direct contact, including Clostridium difficile, community-acquired viral infections, and pathogens found in soil (grade 1, level A). 11. Shoes, socks, long-sleeve shirts, and long pants should be worn for activities that will involve soil exposure and tick exposure and also to avoid unnecessary sun exposure (grade 1, level A). 12. During periods of maximal immunosuppression, LT recipients should avoid crowds to minimize exposures to respiratory illnesses (grade 1, level A). 13. Work in high-risk areas, such as construction, animal care settings, gardening, landscaping, and farming, should be reviewed with the transplant team to develop appropriate strategies for the prevention of high-risk exposures (grade 2, level A). 14. LT recipients should avoid the consumption of water from lakes and rivers (grade 1, level A). 15. LT recipients should avoid unpasteurized milk products and raw and undercooked eggs and meats (particularly uncooked pork, poultry, fish, and seafood; grade 1, level A). 16. LT recipients should avoid high-risk pets, which include rodents, reptiles, chicks, ducklings, and birds (grade 1, level A). 17. Travel by LT recipients, especially to developing countries, should be reviewed with the transplant team a minimum of 2 months before departure to determine optimal strategies for the reduction of travel-related risks (grade 1, level A). 18. LT recipients should take precautions to prevent vector (including mosquito) -borne diseases. These include avoiding going out during peak mosquito feeding times (dawn and dusk) and using N,N-diethyl-meta-toluamide–containing insect repellants (grade 1, level A). 19. LT recipients should undertake a thorough review of hobbies to assess potential infectious disease risks, particularly those associated with outdoor hobbies (grade 2, level A). 20. All LT recipients should be educated about the importance of sun avoidance and sun protection through the use of a sun block with a sun protection factor of at least 15 and protective clothing. They should be encouraged to examine their skin on a regular basis and report any suspicious or concerning lesions to their physicians (grade 1, level A). 21. Because of the strong association of lung, head, and neck cancers with smoking, the sustained cessation of smoking is the most important preventative intervention (grade 1, level A). 22. For female LT recipients of a child-bearing age, preconception counseling about contraception and the risks and outcomes of pregnancy should start in the pretransplant period and should be reinforced after transplantation (grade 1, level A).
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114 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS FULL TEXT BONE HEALTH Bone loss and fracturing are seen with 2 distinct phases after LT. In the first 4 postoperative months, there is accelerated bone loss in almost all liver recipients, regardless of the pretransplant bone mineral density (BMD), that is consistent with the effects of corticosteroids and possibly CNIs.12 After the first 4 postoperative months with normal allograft function, bone metabolism improves, and in the osteopenic patient, there will be a gain in bone mass over the next postoperative years with a gradual reduction in the incidence of fractures.12,13 In patients with preexisting osteopenia or pretransplant fracturing, this early, rapid bone loss results in a high susceptibility to fracturing, mainly at sites of trabecular bone (vertebrae and ribs), especially in the first year after LT, but there is a smaller but steady cumulative increase in fracturing. Table 6 outlines the evaluation of the metabolic bone status of LT recipients with osteopenia. In the early years after LT, BMD should be measured annually in osteopenic patients and every 2 to 3 years in patients with normal BMD. Later screening depends on risk factors. TABLE 6. EVALUATION OF THE METABOLIC BONE STATUS OF THE LT RECIPIENT WITH OSTEOPENIA Assessment of bone pain or fractures Dietary intake of protein and calcium Serum calcium, phosphorus, and parathormone levels 25-hydroxyvitamin D level 24-hour urinary calcium (200-300 mg/day) Gonadal status: free testosterone (males) or menopausal status (females) Thyroid function BMD: lumbar spine and hips Spinal radiographs (thoracolumbar) If it is necessary to confirm a positive calcium balance.
In order to diminish factors that promote bone loss, glucocorticoids should be reduced or discontinued as soon as possible after LT. Calcium supplements are recommended for all LT recipients with (or at risk of) osteopenia, and all patients should receive 1000 to 1200 mg of elemental calcium daily to optimize bone remodeling and mineralization. Vitamin D levels should be maintained at a serum level of at least 30 ng/mL, and most LT patients will require supplementation (generally 400-1000 IU/day). The serum levels of 25-hydroxyvitamin D must be checked to assess the adequacy of replacement, even with supplementation, and should be rechecked annually or more frequently if a deficiency is diagnosed. Many issues with regard to bisphosphonate therapy for LT recipients remain to be defined: the optimal duration of therapy, the optimal doses of bisphosphonates, whether oral or intravenous therapy is better, and the LT population most likely to benefit. Despite these caveats, we suggest that bisphosphonate therapy should be considered in the following circumstances: • T-score less than -2.5 or atraumatic fractures. • T-score between -1.5 and -2.5 and other risk factors. FORWARD BACK AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
115 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS FULL TEXT Oral alendronate at 70 mg weekly is an appropriate starting point, although other oral agents may, however, be equally efficacious. If oral therapy is not tolerated, intravenous zolendronic acid or ibandronate can be used. The optimal duration of therapy is unknown, although oral alendronate has been given with good effect for 10 years for postmenopausal osteoporosis. Hormone replacement therapy is an alternative in postmenopausal women.
RECOMMENDATIONS 23. In the first 5 years after transplantation, screening by BMD should be done yearly for osteopenic patients and every 2 to 3 years for patients with normal BMD; thereafter, screening depends on the progression of BMD and on risk factors (grade 2, level B). 24. If osteopenic bone disease is confirmed or if atraumatic fractures are present, then patients should be assessed for risk factors for bone loss; in particular, this should include an assessment of calcium intake and 25-hydroxyvitamin D levels, an evaluation of gonadal and thyroid function, a full medication history, and thoracolumbar radiography (grade 1, level A). 25. The osteopenic LT recipient should perform regular weight-bearing exercise and receive calcium and vitamin D supplements (grade 1, level A). 26. Bisphosphonate therapy should be considered in LT recipients with osteoporosis or recent fractures (grade 1, level A). SYSTEMIC DISEASE KIDNEY DISEASE The majority of LT recipients who survive the first 6 months develop chronic kidney disease (CKD).6 Predialysis CKD prevalence rates in this population range from 30% to 80%. The wide range of reported incidences is partly due to the different thresholds used to define CKD and the various durations of posttransplant observation. The cumulative risk of ESRD that requires maintenance dialysis therapy or kidney transplantation is 5% to 8% during the first 10 years after LT.6,14,15 Furthermore, 1.0% of all kidney transplants currently in the United States are undertaken for LT recipients who subsequently developed ESRD.16 The etiology of CKD in the LT population is multifactorial (Table 2) and includes chronic exposure to CNIs, hypertension, DM, obesity, atherosclerosis, hyperlipidemia, chronic HCV infection, pretransplant renal dysfunction, and perioperative acute kidney injury. CKD is associated with a 4.48 relative risk of death more than 1 year after LT in comparison with recipients without CKD.6,17 A serum creatinine elevation is a late and insensitive indicator of CKD in this population. An estimating equation that has been shown to have reasonable precision should be routinely used. Both the 4-variable Modification of Diet in Renal Disease equation and the Chronic Kidney Disease Epidemiology Collaboration formula are superior to serum creatinine alone and 24-hour urinary creatinine clearance in estimating renal function.18,19 Increased proteinuria (spot protein-to-creatinine ratio >0.3) may be absent even in the presence of advanced CKD because of the antiproteinuric effect of CNIs. Proteinuria is best assessed by the measurement of the concentration ratio of protein to creatinine in a spot urine specimen.20 Aggressive blood pressure control and the use of agents that block the renin-angiotensin-aldosterone system are key foundations of CKD treatment in the nontransplant population and would be expected to have beneficial effects in LT recipients. A reduction in the dosage or a complete withdrawal of CNIs several months to years after LT is a common practice aimed at ameliorating the progression of CKD. These renal-sparing maintenance protocols typically rely on sirolimus or everolimus, often in combination with mycophenolate, to prevent acute rejection; others FORWARD BACK AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
116 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS FULL TEXT use steroids and mycophenolate or azathioprine.21-24 Renal function is more likely to be preserved if CNI withdrawal is instituted when the estimated glomerular filtration rate is between 40 and 50 mL/minute/1.73 m2.25 LT recipients with ESRD who subsequently receive a living or deceased kidney transplant have a 44% to 60% reduction in long-term mortality in comparison with their dialysis-treated counterparts.26,27 TABLE 7. FACTORS ASSOCIATED WITH THE CLINICAL FEATURES OF METABOLIC SYNDROME FACTOR CORTICOSTEROIDS TACROLIMUS CYCLOSPORINE SIROLIMUS CHRONIC HCV Abdominal obesity + ----Dyslipidemia + + + +++ -Systemic hypertension + ++ ++ + -Insulin-resistant DM +++ ++ + -++ RECOMMENDATIONS 27. Monitoring of renal function in LT recipients for the detection and management of CKD should use an estimating equation to evaluate the glomerular filtration rate (grade 1, level B). 28. Urinary protein quantification using the concentration ratio of protein to creatinine in a spot urine specimen should be evaluated at least once yearly (grade 1, level B). 29. The reduction or withdrawal of CNI-associated immunosuppression is an appropriate response to the development of CKD in LT recipients (grade 1, level A). 30. Kidney transplantation from deceased or living donors is beneficial in improving survival and should be considered the optimal therapy for LT recipients who develop ESRD (grade 1, level A). METABOLIC SYNDROME The clinical features of metabolic syndrome, either alone or in combination, contribute to post-LT morbidity and mortality. The clinical factors related to LT that exacerbate metabolic syndrome are shown in Table 7. DM The spectrum of hyperglycemia after LT includes preexisting DM and NODM, some of which is transient in the perioperative period. Insulin-requiring DM that is present at the time of transplantation virtually always persists after LT, and many patients on oral hypoglycemic agents need a conversion to insulin early after LT. In LT recipients followed beyond 1 year, estimates of the prevalence of NODM vary from 5% to 26%. Diabetogenic factors after LT include corticosteroids, CNIs (tacrolimus more than cyclosporine), HCV infection, and metabolic syndrome.28-33 NODM tends to remit over time, especially as corticosteroids are withdrawn and the tacrolimus dosage is reduced, and patients may go from insulin therapy to oral hypoglycemic agents to diet control only over the years. Because stringent glycemic control significantly reduces morbidity and mortality in diabetic patients, it seems reasonable to assume that LT recipients would similarly benefit. The goals of the long-term management of diabetes after LT are not substantially different from the goals for nontransplant patients (Table 8). There is controversy regarding the appropriate target level of hemoglobin A1c (HbA1c), and consequently, our recommendation of a threshold of <7.0% rather than <6.0% reflects the view that the more demanding standard may confer no additional advantage. When insulin requirements are low, oral agents may be substituted if allograft function is normal. Metformin or FORWARD BACK AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
117 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS FULL TEXT a sulfonylurea may be used in LT recipients with normal renal function, whereas sulfonylureas such as glipizide and glimepiride are preferable if there is any deterioration in renal function. The safety of thiazolidinediones in LT recipients is unproven. Retrospective data sets and a small prospective study suggest that the conversion of immunosuppression from tacrolimus to cyclosporine improves glycemic control in patients with established DM and NODM.33 RECOMMENDATIONS 31. The treatment of DM after LT should aim for an HbA1c target goal of <7.0% with a combination of lifestyle modifications and pharmacological agents as appropriate (grade 1, level B). 32. When high-dose corticosteroids are administered, insulin therapy is the most effective and safest agent with which to control hyperglycemia; however, as the interval from LT extends, patients with NODM may experience a decline in insulin requirements, and oral hypoglycemic agents may be appropriate if allograft function is normal (grade 1, level C). 33. Metformin or sulfonylureas may be used in LT recipients with normal renal function, whereas sulfonylureas such as glipizide and glimepiride are preferable if there is any deterioration of renal function (grade 1, level C). 34. Consideration can be given to the conversion of immunosuppression from tacrolimus to cyclosporine in LT recipients TABLE 8. LONG-TERM MANAGEMENT OF DM (NEW-ONSET OR PREEXISTING) AFTER LT INTERVENTION FREQUENCY Diagnosis Fasting plasma glucose Every 3 months in the first year and then annually Monitoring Self-monitoring of blood glucose HbA1c Diabetic complications Microalbuminuria Tailoring of immunosuppression (especially if there is poor control): discontinuation of steroids and change from tacrolimus to cyclosporine Lipid levels Review every 3 months Every 3 months (intervention at ≥7.0%) Annual screening (retinopathy) Annual screening Annual evaluation Treatment For all patients Dietary and lifestyle modification: exercise and weight loss (if the patient is obese) Control of hypertension and dyslipidemia Depending on glycemic control Insulin† Oral agent or agents Insulin ± oral agent (if there is poor control) This should be interpreted with care for patients with anemia or renal impairment.
†Refer patients to an endocrinologist once insulin is started.
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118 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS FULL TEXT HYPERTENSION Hypertension in LT recipients increases the risk of fatal and nonfatal cardiovascular disease events and CKD.15 Although there are no clinical trials of antihypertensive therapy in LT recipients, it is prudent to target a blood pressure treatment goal of 130/80 mm Hg in LT recipients with systemic hypertension.34 For the management of hypertensive LT recipients, immunosuppression leading to hypertension, such as CNIs and corticosteroids, should be minimized under the direction of the transplant center.35 Lifestyle modifications, including weight loss in overweight recipients (see the discussion on obesity) and the restriction of dietary salt intake, are appropriate nonpharmacological interventions.34 Home measurement of blood pressure is encouraged. If lifestyle modification and a reduction of immunosuppression do not achieve the target blood pressure goal, antihypertensive medications should be introduced. Calcium channel blockers such as amlodipine and nifedipine may be more effective in LT recipients because they counteract the vasoconstrictive effect of CNIs.36 The non-dihydropyridine calcium channel blockers (verapamil and diltiazem) should be used with caution because they may increase the bioavailability of CNIs significantly. Beta-blockers are equally as effective as calcium channel blockers in the treatment of hypertension among LT recipients.36 Angiotensin-converting enzyme inhibitors, angiotensin receptor blockers, and direct renin inhibitors should be used as first-line antihypertensive therapy in LT recipients with DM, CKD, and/or significant proteinuria. Monitoring of potassium levels is necessary when these drugs are used in conjunction with CNIs (particularly tacrolimus). Because of the increased risk of electrolyte abnormalities, thiazide or loop diuretics should be used with caution. The combination of diuretics with other classes of antihypertensive medication may be particularly effective in some LT recipients because diuretics tend to mitigate the volume retention associated with CNIs and/or advanced CKD that commonly coexists in hypertensive LT recipients. RECOMMENDATIONS 35. The treatment of hypertension should aim for a target goal of 130/80 mm Hg with a combination of lifestyle modifications and pharmacological agents as appropriate (grade 1, level A). 36. Angiotensin-converting enzyme inhibitors, angiotensin receptor blockers, and direct renin inhibitors should be used as first-line antihypertensive therapy in LT recipients with DM, CKD, and/or significant proteinuria (grade 1, level A).
HYPERLIPIDEMIA Dyslipidemia occurs in up to 70% of LT recipients (a prevalence much higher than that before transplantation) and is a major risk factor for cardiovascular morbidity and mortality (Table 2).37,38 Although age, body weight, and genetics have some influence, medications— especially CNIs, mTOR inhibitors, and glucocorticoids— are the major influences on the high prevalence of dyslipidemia in LT recipients. Furthermore, 3-hydroxy-3-methyl-glutaryl-coenzyme A reductase inhibitors and CNIs share metabolic pathways and have significant drug-drug interactions (Table 4). The measurement of blood lipids after a 14-hour fast is recommended annually for healthy LT recipients. Table 9 shows a plan for the stepwise treatment of dyslipidemia after LT.
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119 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS FULL TEXT TABLE 9. GENERAL PLAN FOR THE STEPWISE MANAGEMENT OF DYSLIPIDEMIA Elevated low-density lipoprotein cholesterol level >100 mg/dL (with or without elevated triglycerides) 1. Therapeutic lifestyle and dietary changes 2. Statins 3. Addition of ezetimibe Hypertriglyceridemia with normal cholesterol 1. Fish oil at 1000 mg twice daily to 4 g daily if tolerated 2. Fibric acid derivatives Refractory hyperlipidemia: consider changes in immunosuppression 1. Conversion of cyclosporine to tacrolimus 2. CNI reduction (eg, add mycophenolate mofetil) 3. Discontinuation of sirolimus RECOMMENDATIONS 37. The measurement of blood lipids after a 14-hour fast is recommended annually for healthy LT recipients. An elevated low-density lipoprotein cholesterol level >100 mg/dL, with or without hypertriglyceridemia, requires therapy. If therapeutic lifestyle and dietary changes are not enough, statin therapy should be introduced. Suboptimal control with statins can be improved by the addition of ezetimibe (grade 2, level B). 38. Isolated hypertriglyceridemia is first treated with omega-3 fatty acids (up to 4 g daily if tolerated). If this is not sufficient for control, gemfibrozil or fenofibrate can be added, although patients must be followed carefully for side effects, especially with the concomitant use of statins and CNIs (grade 2, level C). NUTRITION AND OBESITY (BODY MASS INDEX >30 kg/m2) Weight accumulation is common after LT. In American and European cohorts, approximately 20% of lean patients become obese (body mass index >30 kg/m2) in the first 2 to 3 years after LT; this phenomenon is driven by the restoration of health and the stimulation of appetite by medicines such as corticosteroids.39,40 RECOMMENDATIONS 39. All LT patients require ongoing dietary counseling to avoid obesity (grade 1, level C). 40. Among LT recipients who become severely or morbidly obese and fail behavioral weight-loss programs, bariatric surgery may be considered, although the optimal procedure and its timing with respect to transplantation remain to be defined (grade 1, level C).
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120 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS FULL TEXT ONCOLOGY DE NOVO CANCER The incidence of de novo cancer is higher among LT recipients versus an age- and sex-matched nontransplant control population41 (Table 10). The cumulative incidence of de novo cancer after LT increases from 3% to 5% at 1 to 3 years to 11% to 20% at 10 years after LT.42, 43 Cutaneous malignancies are the most common form of malignancy in recipients of solid organ transplants, but cigarette smokers are at increased risk of developing lung cancer and oropharyngeal cancer, and the rate of colon cancer is increased in patients undergoing transplantation for PSC because of the comorbid risk from inflammatory bowel disease.42-45 The oncogenic risk due to viral infections [eg, Epstein-Barr virus (EBV) leading to PTLD] is discussed in the section on viral infections. The American Cancer Society guideline on screening for cervical cancer recommends that women who are immunosuppressed on account of solid organ transplantation “may need to be screened more often (than every 3 to 5 years). They should follow the recommendations of their healthcare team.”46 Careful prospective surveillance accompanied by lifestyle modifications to protect the skin and to quit smoking improves outcomes for LT recipients.43, 44 TABLE 10. RELATIVE RISKS OF DE NOVO MALIGNANCIES IN LT RECIPIENTS VERSUS A SEX- AND AGE-MATCHED POPULATION MALIGNANCY RELATIVE RISK Skin cancers Squamous and basal cell carcinoma Melanoma 20%-70% 2%-5% (estimate) Lymphoma 10%-30% Oropharyngeal cancer, including esophageal cancer 3%-14% (as high as 25% if the prior diagnosis was alcoholic cirrhosis) Lung cancer 1.7%-2.5% Colorectal cancer 25%-30% if ulcerative colitis is present Kidney cancer 5%-30% RECURRENT OR PERSISTENT CANCER The proportion of patients undergoing LT for HCC has increased significantly in the past decade. Rates of recurrence at 4 years are 10% for patients with tumors within the Milan criteria and 40% to 60% for patients with tumors outside the Milan criteria.47 Tumor recurrence reduces long-term survival after LT for HCC. Accumulating data suggest that once postoperative healing is complete, the substitution of sirolimus for a CNI reduces the risk of recurrence of HCC.48 Guidelines for surveillance after LT, including the choice of the surveillance method, the intervals between surveillance tests, and the duration of surveillance, have not been established for patients undergoing transplantation for known HCC or for patients with incidental HCC found in the explanted liver.47 A reasonable plan is for the patient to undergo abdominal and chest computed tomography every 6 months for 3 years after LT. The serial measurement of alpha-fetoprotein is a useful adjunct for patients who had an elevated alpha-fetoprotein level before transplantation or ablation therapy. Any suspicious lesion discovered on surveillance should be characterized fully, and biopsy should be included when the diagnosis is in doubt. Ablation with radiofrequency is the best treatment for small solitary recurrences. FORWARD BACK AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
121 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS FULL TEXT RECOMMENDATIONS 41. All LT recipients should see a dermatologist after transplantation to assess cutaneous lesions, with at least an annual evaluation by a dermatologist 5 years or more after transplantation (grade 1, level A).
42. Patients with PSC and inflammatory bowel disease or other established risk factors for colorectal cancer should undergo an annual screening colonoscopy with biopsies. Colectomy, including continence-preserving pouch operations, should be considered when colonic biopsy reveals dysplasia (grade 1, level B).
43. For patients without prior HCC who develop recurrent cirrhosis of the allograft, surveillance for de novo HCC should be undertaken with abdominal imaging every 6 to 12 months (grade 1, level A).
44. An immunosuppressant regimen that includes sirolimus (started several weeks after transplantation) should be considered for patients undergoing transplantation for HCC (grade 2, level B). 45. Resection or ablation is usually the treatment of choice for a solitary extrahepatic metastasis or an intrahepatic recurrence of HCC (grade 1, level B).
REPRODUCTIVE HEALTH Menstruation and probably fertility return by 10 months in 90% of premenopausal females after successful LT and in some patients as early as 1 to 2 months.49-51 Free testosterone levels increase in males after LT, but the recovery of male gonadal function is often incomplete. LT has limited efficacy for curing pretransplant sexual dysfunction in either men or women.52 Sildenafil is beneficial and well tolerated by male LT recipients with erectile dysfunction.53 Pregnancy in the LT recipient has risks to both the mother and the fetus.51,54 Although the numbers of pregnancies reported are relatively small, pregnancies completing the first trimester successfully generally proceed to a live birth, although there is a higher incidence of prematurity (29%-50%) and low birth weight (17%-57%).51,54 Neonatal deaths or birth defects are not more frequent in comparison with the general population (except when the mother is on mTOR inhibitors).55,56 The maternal risks include hypertension and pre-eclampsia, which occur more commonly in comparison with the general population.57 Maternal deaths following pregnancy in LT recipients are rare and occur at a rate similar to that in the general population. The National Transplant Pregnancy Registry guidelines51 recommend the female LT recipients postpone conception until • At least 1 year after LT. • Allograft function is stable. • Medical comorbidities such as diabetes and hypertension are well controlled. • Immunosuppression is at a low maintenance level.
The choice of immunosuppression should be made before conception. All immunosuppressive drugs cross the placenta and enter the fetal circulation with resulting concerns about teratogenicity and fetal loss. Table 11 shows the Food and Drug Administration (FDA) safety categories for drugs in pregnancy. Generally, CNIs (class C drugs), prednisone (class B), and azathioprine (class D) appear to be safe.54 The newer agents should be avoided if possible; in the National Transplant Pregnancy Registry,51 more structural abnormalities have been seen in babies born to mothers on mTOR inhibitors or mycophenolic acid, especially when they are used in early pregnancy.55,56 European guidelines for renal recipients advise discontinuing mTOR inhibitors at least 6 weeks before conception.58 An early diagnosis of pregnancy is desirable to maximize positive pregnancy outcomes. Immunosuppression should FORWARD BACK AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
122 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS FULL TEXT be maintained during pregnancy to avoid rejection, and drug levels of CNIs should be monitored with dose adjustments for the increasing blood volume during the second half of pregnancy.59 Allograft function and CNI serum levels should be monitored frequently until delivery. Screening the mother for urinary tract infections, the presence of cytomegalovirus (CMV) and toxoplasmosis, hypertension, gestational diabetes, and pre-eclamzpsia, along with serial assessments of fetal growth, is mandatory. Allograft dysfunction during pregnancy warrants appropriate investigation, including liver biopsy in selected patients, to assess for rejection. The pregnant patient with acute cellular rejection is treated in the same manner as the nonpregnant patient. There are no contraindications to vaginal delivery. Allograft function and drug levels should be checked weekly for at least 1 month after birth or until the patient is stable, especially if adjustments were made during the pregnancy or allograft dysfunction arose late in the pregnancy. Although the known benefits of breast feeding probably outweigh the theoretical risks, no definitive recommendations regarding breast feeding can be made. Low levels of immunosuppressive drugs may be found in breast milk. Contraception with whatever method is favored by the LT recipient should start before sexual activity is resumed. RECOMMENDATIONS 46. Pregnancy in an LT recipient should be managed by a high-risk obstetrician in coordination with the transplant hepatologist (grade 1, level C). 47. Pregnancy should be delayed for 1 year after LT and occur at a time with good, stable allograft function, with maintenance immunosuppression, and with good control of any medical complications such as hypertension and diabetes (grade 1, level B). 48. The ideal immunosuppression for pregnancy is tacrolimus monotherapy, which should be maintained at therapeutic levels throughout pregnancy; cyclosporine, azathioprine, and prednisone may also be used if they are necessary (grade 1, level B). 49. Allograft function and CNI serum levels are monitored every 4 weeks until 32 weeks, then every 2 weeks, and then weekly until delivery (grade 1, level B). 50. Contraception should begin before the resumption of sexual activity, although no particular form of contraception can be recommended over another (grade 2, level B). FORWARD BACK TABLE 11. FDA SAFETY CATEGORIES FOR DRUGS USED DURING PREGNANCY A. Controlled studies in women fail to demonstrate a risk to the fetus in the first trimester (and there is no evidence of risk in later trimesters), and the possibility of fetal harm appears remote. B. Animal reproduction studies have not demonstrated fetal risk, but there are no controlled studies in pregnant women, OR animal studies have shown an adverse effect that has not been confirmed in controlled studies in women in the first trimester.
C. Animal studies have revealed adverse effects on the fetus, and there are no controlled studies in women, OR studies in women and animals are not available. Give the drug only if the potential benefit justifies the risk. D. There is positive evidence of human fetal risk, but benefits from use in pregnant women may be acceptable despite the risk. X. A definitive fetal risk exists, and the drug is contraindicated in women who are or may become pregnant.
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123 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS FULL TEXT INFECTIOUS DISEASE GENERAL OVERVIEW The interval from the third to sixth month after LT is a high-risk period because of the occurrence of infections with opportunistic pathogens: herpes viruses (especially CMV, herpes zoster and simplex, and EBV), fungi (including Aspergillus and Cryptococcus), and unusual bacterial infections such as Nocardia, Listeria, and mycobacteria. The implementation of prophylactic antimicrobials, the avoidance of high-risk exposures, and the minimization of immunosuppression may reduce the occurrence of these pathogens.60 After the sixth posttransplant month, the risk of infection is lower, and this is related to the reduction of immunosuppression. From 3 to 24 months after LT, in the standard-risk LT recipient (ie, no augmented immunosuppression or specific environmental exposures), the most common infections are intra-abdominal or in the lower respiratory tract or infections by community-acquired pathogens such as enteric gram-negative infections, Streptococcus pneumonia, and respiratory viruses.60 Rare infections related to immunosuppression, such as the reactivation of John Cunningham polyomavirus resulting in progressive multifocal leukoencephalopathy, are not reviewed here. Table 12 shows an outline of prophylactic strategies for countering common organisms that affect LT recipients. Table 4 outlines the drug-drug interactions involving anti-infectives and immunosuppressive agents. TABLE 12. PROPHYLACTIC STRATEGIES FOR COMMON ORGANISMS THAT AFFECT LT RECIPIENTS ORGANISM AGENT/DOSAGE DURATION COMMENTS CMV Donor-positive/ recipient-negative Recipient-positive Valganciclovir (900 mg/day) or intravenous ganciclovir (5 mg/kg/day) Valganciclovir (900 mg/day), intravenous ganciclovir, or weekly CMV viral load monitoring and antiviral initiation when viremia is identified 3-6 months 3 months Valganciclovir is not FDA-approved for LT. Prolonged-duration regimens are effective in kidney transplantation.
Valganciclovir is not FDA-approved for LT.
Fungi Fluconazole (100-400 mg daily), itraconazole (200 mg twice daily), caspofungin (50 mg daily), or liposomal amphotericin (1 mg/kg/day) 4-6 weeks? (optimal duration unknown) Reserve for high-risk individuals (pretransplant fungal colonization, renal replacement therapy, massive transfusion, choledochojejunostomy, reoperation, retransplantation, or hepatic iron overload).
P. jirovecii (P. carinii) Trimethoprim sulfamethoxazole (single strength daily or double strength 3 times per week), dapsone (100 mg daily), or atovaquone (1500 mg daily) 6-12 months (optimal duration unknown) A longer duration of therapy should be considered for patients on augmented immunosuppression. Lifelong therapy should be considered for HIV-infected recipients.
TB (latent infection) Isoniazid (300 mg daily) 9 months Monitor for hepatotoxicity.
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124 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS FULL TEXT RECOMMENDATIONS 51. An assessment for infections following LT should take into account the intensity of immunosuppression, the timing of the presentation, the environmental and donor exposures, the recipient’s history of both symptomatic and latent infections, and the utilization of prophylactic antimicrobials and immunizations (grade 1, level A). 52. Attention should be paid to potential drug interactions when new antimicrobial therapies are initiated (grade 1, level A).
CMV CMV remains the most significant opportunistic pathogen affecting LT recipients and produces diverse clinical manifestations and significant morbidity and mortality.61,62 The most common clinical syndromes include viremia, bone marrow suppression, and involvement of the gastrointestinal tract and liver. Risk factors for CMV61,62 include the following: • CMV-seropositive donor organ (especially in the absence of prior immunity, ie, a CMV-seronegative recipient). • Augmented immunosuppression (especially with the use of anti-lymphocyte antibodies or high-dose mycophenolate). • Allograft rejection. • Coinfection with other immunomodulating viruses (eg, human herpesviruses 6 and 7), bacteria, or fungi.
The diagnosis of CMV includes the detection of the virus in conjunction with the recognition of an associated clinical syndrome.61,62 Patients who are not receiving prophylactic antivirals and are at increased risk for CMV (because of a CMV-seropositive donor and/or treatment for rejection) may be monitored for evidence of infection with nucleic acid testing (polymerase chain reaction). Typically, CMV occurs in the first 3 months in the absence of prophylaxis. However, because of current standard prophylactic strategies, it now presents later after the cessation of prophylaxis, frequently in the first year or after the augmentation of immunosuppression. Currently, routine screening for CMV is not recommended while patients are receiving prophylaxis. After the completion of prophylaxis, some centers have adopted a hybrid approach using nucleic acid testing to screen for infections in the highest risk patients. However, there is no clear evidence to support the screening of asymptomatic patients at this time. The detection of viremia by either nucleic acid testing (polymerase chain reaction) or the pp65 antigenemia assay is recommended for the diagnosis of an active CMV infection.61,62 Typically, the viral load correlates with the severity of disease and can be a marker of the response to therapy. Some individuals, especially those with hepatitis or gastrointestinal disease, may exhibit low-level or no viremia despite a symptomatic infection and require tissue biopsy for the diagnosis of CMV disease to be made. Finally, some LT recipients exhibit low-level viremia without symptomatic disease. The treatment of CMV should be started whenever recipients are symptomatic, have a tissue injury, or have persistent or increasing viremia.61,63,64 All LT recipients with a symptomatic CMV infection and/or end organ disease should receive antiviral therapy and have their immunosuppression reduced until viremia and all symptoms have resolved. Patients with low-level viremia (this is difficult to define because of laboratory variability) should be assessed for symptoms and, if they are asymptomatic, should have immunosuppression reduction as tolerated and viral load testing repeated. If the viral load rises and/or symptoms develop, treatment should be administered. Options for antiviral treatment include intravenous ganciclovir (5 mg/kg twice daily adjusted for renal impairment) and oral valganciclovir (900 mg twice daily adjusted for renal impairment) for mild to moderate disease if no significant gastrointestinal involvement is assumed (note: valganciclovir is not approved for use in LT). For those FORWARD BACK AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
125 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS FULL TEXT patients with more severe disease or gastrointestinal involvement, intravenous ganciclovir is preferable. A minimum of 2 weeks of treatment is recommended for those patients with rapid resolution of symptoms, but treatment should be continued until there is complete resolution of both symptoms and viremia. Whenever possible, a reduction in immunosuppression should be combined with antiviral therapy. Ganciclovir resistance is uncommon in solid organ transplant recipients. RECOMMENDATIONS 53. High-risk recipients (CMV-seronegative recipients of CMV-seropositive donor organs) should receive prophylaxis with ganciclovir or valganciclovir for a minimum of 3 months after transplantation (grade 1, level B). 54. The treatment of LT recipients with CMV should be maintained until viremia and all symptoms have resolved (grade 2, level B). 55. Prophylaxis against CMV should be resumed whenever LT recipients receive anti-lymphocyte therapy for the treatment of rejection and should be continued for 1 to 3 months after the treatment of rejection (grade 2, level B). 56. The treatment of a CMV infection should consist of the following: a. Consideration of immunosuppression reduction. b. High-dose intravenous ganciclovir or oral valganciclovir in individuals with mild to moderate disease without gastrointestinal involvement or a reduced capacity for absorption. c. A minimum of 2 weeks of treatment. Treatment should be continued to complete the resolution of all symptoms and viremia (grade 1, level A). 57. Resistant virus should be suspected in patients with a history of prolonged ganciclovir or valganciclovir exposure who have a persistent or progressive infection despite treatment with high-dose intravenous ganciclovir (grade 1, level A). In such instances, genotypic assays should be performed, and consideration should be given to the initiation of foscarnet with or in substitution for ganciclovir (grade 1, level B). EBV/PTLD EBV-associated PTLD is an uncommon but serious complication of LT with an incidence in adults of 0.9% to 2.9%.64,65 Risk factors include a primary EBV infection, CMV donor-recipient mismatch or CMV disease, and augmented immunosuppression, especially with anti-lymphocyte antibodies.66 It is uncertain whether the etiology of liver disease influences the development of PTLD.67,68 The association of PTLD with EBV infection is variable in adult LT recipients; later onset PTLD is less likely to be EBV-associated.64,67,68 Manifestations of PTLD include lymphadenopathy, cytopenias, unexplained fever, and disturbances of the gastrointestinal tract, lungs, spleen, and central nervous system.
The diagnosis of PTLD requires a high index of suspicion and should be considered in high-risk individuals who present with undiagnosed fever or unexplained lymphadenopathy or cytopenias.66–69 Radiographic studies can identify sites of involvement, especially when pulmonary or intra-abdominal sites are involved. The detection of EBV viremia with nucleic acid testing is not diagnostic of EBV-associated PTLD. The initial treatment of PTLD is a reduction of immunosuppression.65–69 If there is no clinical response within 2 to 4 weeks, additional therapies, including anti-CD20 humanized chimeric monoclonal antibodies (rituximab), surgical therapy, radiation therapy, and cytotoxic chemotherapy, may be required. The addition of antiviral therapy has not been proven to affect outcomes. FORWARD BACK AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
126 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS FULL TEXT RECOMMENDATIONS 58. PTLD should be considered in LT recipients (especially high-risk individuals) who present with unexplained fever, lymphadenopathy, or cytopenias (grade 1, level A). 59. Although EBV may be associated with the development of PTLD, the detection of EBV viremia is not diagnostic for PTLD; a histopathological diagnosis is required (grade 1, level A).
FUNGAL INFECTIONS Risk factors for fungal infections after LT include preoperative fungal colonization, massive transfusion requirements (>40 U of blood products), choledochojejunostomy, reoperation, retransplantation, hepatic iron overload, renal replacement therapy, and extended intervals of intensive care immediately before LT. The epidemiology of invasive fungal infections in LT recipients has shifted over the past 2 decades, with a decrease in Candida infections and an increasing incidence of Aspergillus infections.70 The recognition of an invasive fungal infection after 90 days is challenging. Blood cultures are relatively insensitive for the diagnosis of many fungal infections, including Candida species, for which the (1,3)-β-D-glucan test is an inconsistent measure.71,72 Aspergillus is especially difficult to diagnose with noninvasive testing.70 The sensitivity and specificity of galactomannan in either blood or bronchoalveolar lavage from LT recipients with presumed pulmonary aspergillosis are variable.72-74 Serum and cerebrospinal cryptococcal antigen testing is a sensitive tool for the rapid diagnosis of cryptococcal infections in organ transplant recipients.75 The isolation of Cryptococcus from a site other than cerebrospinal fluid should prompt lumbar puncture to rule out central nervous system involvement. Urinary histoplasmosis and Blastomyces antigens have been useful for the diagnosis of disseminated histoplasmosis and blastomycosis, respectively.76,77 The treatment of fungal infections includes antifungal drug therapy as well as a reduction of immunosuppressive therapy. The choice of antifungal agents varies with the pathogen and the site of involvement, as shown in Table 13.
RECOMMENDATIONS 60. The diagnosis of fungal infections may require diagnostic biopsy for pathological and microbiological confirmation (grade 1, level A). a. Blood cultures are most helpful for the diagnosis of Candida bloodstream infections (class 1, level B) and Blastomyces (grade 1, level B). b. Cryptococcal antigen testing of cerebrospinal fluid or blood is most helpful for the diagnosis of Cryptococcus (grade 1, level B). c. Urinary histoplasmosis and Blastomyces antigens are useful for the diagnosis of disseminated histoplasmosis and blastomycosis, respectively (grade 1, level B). 61. A cautious reduction of immunosuppression should be initiated to prevent immune reconstitution syndrome, especially for cryptococcal infections (grade 1, level B). PNEUMOCYSTIS JIROVECII (PNEUMOCYSTIS CARINII) P. jirovecii (formerly called P. carinii) is an uncommon pathogen in LT recipients, primarily because of the widespread use of antimicrobial prophylaxis after LT.78 Pneumocystis should be suspected in individuals presenting with respiratory symptoms, hypoxemia (often exacerbated by exercise), and fever.78 Classic radiographic findings include bilateral interstitial infiltrates. The diagnosis is confirmed by the identification of the organism by a cytological examination of induced sputum or bronchoalveolar lavage fluid. High-dose trimethoprim-sulfamethoxazole (administered orally or intravenously at 15-20 mg/kg/day in divided doses and adjusted for renal dysfunction) is the drug of choice.78 Corticosteroids (40-60 mg of prednisone or its equivalent) should be used in conjunction with antimicrobial therapy for patients with significant hypoxia (partial pressure of arterial oxygen <70 mm Hg on room air). The minimal duration of antimicrobial therapy is 14 days, but more severe infections may merit longer courses of treatment.
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127 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS FULL TEXT TABLE 13. PREFERRED ANTIFUNGAL AGENTS ORGANISM/ DISEASE AGENT COMMENTS Candida Triazoles (fluconazole, Itraconazole, voriconazole, and posaconazole), echinocandins (eg, caspofungin, micafungin, and anidulafungin), or amphotericin B and analogues Candida glabrata and Candida krusei may be resistant to triazoles (especially fluconazole). Differentiate colonization from infection. The duration of therapy varies with the site of infection. Aspergillus Triazoles (voriconazole is the drug of choice; itraconazole and posaconazole are also active), caspofungin, or amphotericin B and analogues The duration of therapy is dependent on the response to therapy.
Cryptococcus Amphotericin B and analogues in combination with 5-flucytosine for 2 weeks followed by fluconazole (400-800 mg/day) for 8 weeks and then fluconazole (200 mg/day) for 6-12 months Cautiously reduce immunosuppression. Patients with isolated pulmonary disease may not require amphotericin induction. The duration varies with the response. Blastomycosis Itraconazole (200 mg twice daily) for mild to moderate disease and amphotericin B and analogues for severe disease The standard duration is 6-12 months. Coccidiomycosis Fluconazole (400-800 mg daily), itraconazole (200 mg twice daily), or amphotericin B and analogues Amphotericin should be used for more severe disease and should be considered when there is central nervous system involvement. The standard duration is 6-12 months with chronic suppression thereafter. Histoplasmosis Itraconazole (200 mg twice daily) or amphotericin B and analogues for 2 weeks of induction followed by fluconazole The minimum duration is 12 months RECOMMENDATIONS 62. All LT recipients should receive prophylaxis against P. jirovecii with trimethoprim-sulphamethoxazole (single strength daily or double strength 3 times per week) for a minimum of 6 to 12 months after transplantation (grade 1, level A). Atovaquone and dapsone are the preferred alternatives for patients who are intolerant of trimethoprim sulfamethoxazole (grade 1, level B). 63. Trimethoprim-sulphamethoxazole is the drug of choice for the treatment of P. jirovecii pneumonia. Intravenous pentamidine is the preferred alternative for patients intolerant of trimethoprim- sulphamethoxazole with more severe infections (grade 1, level A). 64. Patients with clinical signs and symptoms or radiological features suggestive of P. jirovecii pneumonia should undergo sputum sampling or bronchoalveolar lavage with a cytological examination using a silver or Giemsa stain, polymerase chain reaction, or a specific antibody stain to identify the organism (grade 1, level A). FORWARD BACK AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
128 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS FULL TEXT TUBERCULOSIS (TB) Several risk factors for the development of symptomatic TB after LT have been identified: a prior infection with TB; intensified immunosuppression (especially anti–T lymphocyte therapies); DM; and coinfections with CMV, mycoses, P. jirovecii, and Nocardia.79 Donor-derived TB is rare. The diagnosis of TB may be confounded by an increased incidence of atypical presentations (especially extrapulmonary infections involving diverse organs and locations).80,81 Standard antituberculous regimens for drug-susceptible isolates include 2 months of 4-drug therapy with isoniazid, rifampin, ethambutol, and pyrazinamide followed by an additional 4 months of isoniazid and rifampin. Patients with central nervous system involvement, bone and joint disease, or disseminated infections may warrant longer courses of treatment.80- 82 The use of anti-TB agents in LT recipients is complicated by hepatotoxicity and by the significant drug-drug interactions with immunosuppressive agents, which lead to the potential for hepatotoxicity associated with antitubercular chemotherapy. Because of the risk of a marked reduction in CNI and mTOR inhibitor levels with rifampin coadministration, the doses of CNIs will need to be increased 2- to 5-fold at the initiation of treatment.79 Rifabutin may be substituted for rifampin to reduce the impact on drug levels, or non–rifampin-containing regimens can be considered, although the duration of treatment will need to be extended. RECOMMENDATIONS 65. The treatment of active TB should include the initiation of a 4-drug regimen using isoniazid, rifampin, pyrazinamide, and ethambutol (under the assumption of susceptible TB) with adjustments in accordance with subsequent culture results. This may be tapered to 2 drugs (isoniazid and rifampin) after 2 months (under the assumption of no resistance) and continued for a minimum of 4 additional months (grade 1, level B). 66. Close monitoring for rejection and hepatotoxicity is imperative while LT recipients receive anti-TB therapy (grade 1, level A). HUMAN IMMUNODEFICIENCY VIRUS (HIV) HIV-infected patients with well-controlled infections have undergone transplantation with success, although aggressive HCV recurrence has been problematic in LT recipients coinfected with HCV.83 HIV-infected patients maintained on highly active antiretroviral therapy (HAART) after transplantation do not experience an increase in opportunistic infections. The use of HAART in LT recipients is complicated by significant drug-drug interactions with immunosuppressive agents, which lead to a risk of cyclosporine or tacrolimus toxicity or inadequate immunosuppression.84 LT recipients with HCV-HIV coinfections have a higher frequency and severity of acute cellular rejection.85 The CNI doses and the frequency of their administration need to be reduced markedly in LT recipients receiving HAART containing protease inhibitors.83,84 In contrast, those receiving nonnucleoside reverse transcriptase inhibitors (especially efavirenz) will require higher doses of CNIs. RECOMMENDATIONS 67. HIV-infected LT recipients receiving HAART require frequent monitoring of CNI levels because of the significant interaction between antiretrovirals and CNIs (grade 1, level A). 68. HIV-infected LT recipients receiving HAART should be followed with scheduled HIV viral loads and T lymphocyte subset counts (grade 1, level A). 69. Standard prophylaxis for CMV is recommended for HIV-infected LT recipients receiving HAART, and lifelong Pneumocystis pneumonia prophylaxis is the norm (grade 1, level A). 70. Standard HIV-specific prophylaxis for low CD4 counts should be used (grade 1, level A).
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129 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS FULL TEXT IMMUNIZATIONS Appropriate advice regarding vaccination after LT has been reviewed by Danzinger-Isakov et al.86 and is also reviewed in Table 14. LT recipients should avoid live virus vaccines because of concerns about the dissemination of infections. Vaccine-related rejection has not been associated with immunization following LT. RECOMMENDATIONS 71. All LT recipients should receive an annual influenza vaccination (grade 1, level B). 72. All LT recipients should avoid live virus vaccines (grade 1, level A). 73. Re-immunization is indicated for some vaccines, notably the influenza vaccine (annually) and the pneumococcal vaccine (every 3-5 years; no class or level provided). (grade 1, level A).
TABLE 14. RECOMMENDED IMMUNIZATIONS FOR ADULT LT RECIPIENTS BEFORE TRANSPLANTATION AFTER TRANSPLANTATION Influenza Pneumococcus Hepatitis A virus† HBV† Tetanus/diphtheria/acellular pertussis‡ Human papilloma virus§ Varicella virus◊ Zoster Influenza Pneumococcus NOTE: Transplant recipients may also receive the following vaccines safely: the meningococcal vaccine, the inactivated Salmonella Typhi vaccine (Typhim Vi intramuscular vaccine), the Japanese encephalitis vaccine, and the Vibrio cholera vaccine. Live virus vaccines should be avoided after transplantation.
The pneumococcal vaccine should be repeated every 3 to 5 years after the initial administration. † Ideally, hepatitis A virus and HBV immunizations should be administered before transplantation. There are no guidelines regarding posttransplant immunization, although these vaccines are safe after transplantation. HBV antibody levels should be measured annually after transplantation, with boosters considered for waning immunity. ‡ The tetanus/diphtheria/acellular pertussis vaccine can also be safely administered after transplantation. § This vaccine is indicated for females up to the age of 26 years. It can be safely administered after transplantation.
◊This vaccine can be administered safely before transplantation to nonimmune individuals. It should not be administered after transplantation. This vaccine is indicated for individuals who are 60 years old or older. No studies have been conducted in patients with cirrhosis. It should not be administered after transplantation. FORWARD BACK AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
130 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS FULL TEXT VIRAL HEPATITIS HEPATITIS B VIRUS (HBV) Chronic HBV accounts for less than 10% of transplants performed in the United States and Western Europe, whereas in Asia, it is the most common indication for LT. Importantly, in the last decade, there has been a shift in the primary indication for LT among HBV-infected patients, with HCC more frequent than end-stage liver disease.87 This trend reflects the efficacy of antiviral therapy in preventing complications of cirrhosis as well as the increased prioritization of HCC for LT. The survival for patients undergoing transplantation for HBV is excellent, and HBV ranks among the best of all indications for LT. The improvements in patient and graft survival evident over the past 10 to 15 years reflect the advances in therapeutics to prevent and control HBV infections after LT. The combination of hepatitis B immune globulin (HBIG) and nucleos(t)ide analogues can prevent recurrent infections in almost all HBV-infected LT patients.88-91 The combination of HBIG and nucleos(t)ide analogues is superior to HBIG alone. The individualization of prophylactic combination therapy can be undertaken on the basis of pretransplant clinical and virological characteristics. For example, low-dose intramuscular HBIG is much less expensive and avoids painful side effects associated with intravenous HBIG. The discontinuation of HBIG is generally reserved for patients at low risk for HBV recurrence.92 A recurrent infection is manifest with persistently detectable HBV DNA and hepatitis B surface antigen in serum and is usually due to a failure of prophylactic therapy. Liver biopsy is useful for assessing the severity of HBV recurrence and the progression of fibrosis. Fibrosing cholestatic HBV is a unique histological variant observed in LT recipients and is characterized by high intrahepatic levels of HBV DNA, hepatocyte ballooning with cholestasis, and a paucity of inflammatory cells.93 This represents the most severe presentation of recurrent disease and is rarely seen in the current era of prophylactic therapy. RECOMMENDATIONS 74. Long-term prophylactic therapy using a combination of antiviral agents and low-dose HBIG on demand or at fixed intervals can effectively prevent HBV recurrence rates in ≥ 90% of transplant recipients (grade 1, level B). 75. In patients with low or undetectable HBV DNA levels before transplantation and an absence of high-risk factors for recurrence, HBIG can be discontinued, and long-term treatment with antivirals (single or in combination) can be used as an alternative prophylactic strategy (grade 2, level B). 76. Lifelong antiviral therapy should be used to treat patients with recurrent HBV infections. Combination antiviral therapy is superior to monotherapy when drugs with a low genetic barrier to resistance are used, whereas the discontinuation of HBIG is generally reserved for patients at low risk for HBV recurrence (grade 1, level B).
77. Retransplantation for recurrent HBV is appropriate when treatment strategies to prevent or treat recurrent HBV disease are available (grade 1, level C). HCV Recurrent HCV infection is invariable among patients who are viremic at LT, the majority of whom will have histological evidence of recurrent hepatitis within the first year after LT.94 Although the progression of fibrosis in HCV-infected LT recipients is highly variable, in the absence of antiviral therapy, the median time to the development of cirrhosis is 8 to 10 years, whereas an estimated 30% will develop cirrhosis within 5 years of LT.95 FORWARD BACK AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
131 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS CONTENTS The risk of decompensation is 15% to 30% within the first year of the onset of cirrhosis, and the mortality risk is 40% to 55% within 6 to 12 months of the decompensating event. Recurrent HCV cirrhosis is the most frequent cause of graft loss in this population. 96 Patient survival and graft survival are reduced in HCV-infected patients versus HCV-negative patients, with a 5-year patient survival rate of approximately 70%.97,98 HCV-infected recipients have higher rates of graft loss when the allograft is from an older donor.99 There is a higher risk of cirrhosis when HCV-infected LT recipients develop acute rejection that requires treatment or comorbid CMV hepatitis.100 Although recurrent HCV is more likely the longer the interval from LT, in practice, it is often difficult to distinguish between the histopathological appearances of a recurrent HCV infection and acute cellular rejection. The impact of immunosuppressives on the progression of HCV is poorly understood, although some data suggest that anti-lymphocyte agents promote HCV-associated liver injury. Post-LT diabetes, insulin resistance, and (more inconsistently) steatosis have been associated with a higher risk of rapid progression to advanced fibrosis. Posttransplant antiviral therapy is generally reserved for those showing evidence of progressive disease, which is manifested by the presence of moderate to severe necroinflammation or mild to moderate fibrosis, although this paradigm will change with more efficacious and less toxic antiviral therapy.100 The primary goal of post-LT antiviral therapy is the achievement of sustained viral clearance because this virological outcome is associated with fibrosis stabilization or regression and improved graft survival.101 The initiation of antiviral therapy is recommended when significant histological disease is present, although this paradigm would change with more efficacious and less toxic antiviral therapy. The pooled estimated rate of acute graft rejection occurring in patients receiving peginterferon and ribavirin is 5%, which is not significantly higher than the rate in untreated controls.102 However, alloimmune or plasma cell hepatitis, characterized histologically by an inflammatory infiltrate with abundant plasma cells in the setting of increased liver enzymes, has been described during antiviral therapy.103 This is most likely a variant of allograft rejection and responds to the discontinuation of interferon and the amplification of immunosuppression in most cases. With the recent approval of the first-generation protease inhibitors, telaprevir and boceprevir, it is anticipated that triple therapy (peginterferon, ribavirin, and either telaprevir or boceprevir) will evolve into the new standard of care over the next few years for LT recipients infected with genotype 1 virus. Currently, neither protease inhibitor is approved for use in transplant recipients. There are significant drug-drug interactions between HCV protease inhibitors and CNIs and probably mTOR inhibitors as well. The prospect of interferon-free protocols is also of great interest because of the possibility that interferon induces an alloimmune response in some LT recipients.
RECOMMENDATIONS 78. Liver biopsy is useful in monitoring disease severity and progression and in distinguishing recurrent HCV disease from other causes of liver enzyme elevations (grade 1, level C). 79. Prophylactic antiviral therapy has no current role in the management of HCV disease (grade 1, level A). 80. Moderate acute rejection should be treated with increased maintenance immunosuppression and corticosteroid boluses, whereas lymphocyte-depleting drugs should be avoided (grade 1, level B). 81. Antiviral therapy is indicated for significant histological disease: grade 3 or higher inflammatory activity and/or stage 2 or higher fibrosis (on a scale of 4) or cholestatic hepatitis. Peginterferon and ribavirin are the current drugs of choice. The risks and benefits of triple therapy with protease inhibitors are to be determined. The goal of antiviral therapy is the achievement of a sustained virological response, and this confers a survival benefit (grade 1, class B). 82. Retransplantation for recurrent HCV disease should be considered selectively (grade 2, level B).
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132 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS FULL TEXT PBC PBC is an excellent indication for liver replacement with the one of the highest rates of risk-adjusted outcome. 104 Immunological abnormalities (eg, elevated immunoglobulins and autoantibodies) persist after transplantation. Recipients remain at risk for associated conditions, such as sicca syndrome, osteoporosis, and thyroid disease, so screening should be included in the follow-up. Recurrent PBC is diagnosed by liver histology: recurrent disease may occur in the presence of normal liver tests, and neither the presence nor the titer of anti-mitochondrial antibodies correlates with the presence or degree of recurrence.105,106 The reported incidence of recurrent PBC varies from 4% to 33% (the average is 18%).104 Although the use of cyclosporine is associated with less severe recurrence and corticosteroids may be associated with less recurrence, there are insufficient data to recommend a preferred immunosuppressive regimen.106 The impact of the recurrence of PBC on graft function and survival is minimal for the first decade after transplantation, with end-stage disease affecting less than 5%. There is no evidence that routine protocol biopsy in PBC LT recipients will improve outcomes. Ursodeoxycholic acid at a dose of 10 to 15 mg/kg/day is associated with an improvement in liver tests, but there are no data to show benefits in patient or graft survival.104 RECOMMENDATIONS 83. PBC LT recipients should be routinely monitored for associated autoimmune diseases (eg, thyroid disease) and bone density (grade 2, level B). 84. For those with histological evidence of recurrent disease, treatment with ursodeoxycholic acid at 10 to 15 mg/kg/day (grade 2, level B) may be considered, and although its use is associated with the improvement of liver tests, no impact on graft survival has been documented (grade 2, level B). There is no indication for offering prophylaxis with ursodeoxycholic acid to patients with normal liver histology (grade 2, level B). PSC PSC is an excellent indication for LT with good long-term outcomes. Recipients with a Roux loop are at increased risk for recurrent cholangitis; those few who have a retained native bile duct are at risk for cholangiocarcinoma. In patients with chronic ulcerative colitis (CUC), colitis may improve or deteriorate after transplantation.107 PSC LT recipients with CUC are at greater risk of developing colonic polyps and cancer and should have an annual colonoscopy. There is no evidence for the optimal screening approach in PSC LT recipients without CUC, but many advocate an annual colonoscopy in this group also. Recurrent PSC is seen in up to 50% of patients at 5 years, with graft loss due to recurrent PSC occurring in as many as 25% of patients with recurrent PSC.108 The diagnosis of recurrent PSC is based on a combination of biochemical, radiological, and histological findings in particular, multiple nonanastomotic biliary strictures or characteristic liver histology, and the exclusion of other causes such as infections or ischemia secondary to thrombosis of the hepatic artery. Risk factors for recurrent PSC include male sex, an intact colon before or during transplantation, a history of steroid-resistant or recurrent rejection, active CUC after transplantation, the use of anti-lymphocyte therapy for the treatment of cellular rejection, sex mismatch between the donor and the recipient, CMV infection, and the presence of specific HLA haplotypes (eg, HLA-DRB108). In PSC recipients with CUC, prophylactic colectomy does not reduce the risk of recurrent PSC. There is insufficient evidence to support maintaining corticosteroids in patients undergoing transplantation for PSC. FORWARD BACK AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
133 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS FULL TEXT RECOMMENDATION 85. Although there are few data on prevention, it is recommended that those patients grafted for PSC in the presence of CUC have an annual colonoscopy with mucosal biopsy (grade 2, level B). AIH Outcomes after transplantation for AIH are good. Patients should be closely monitored for evidence of recurrence via liver tests every 6 months.109 Protocol liver biopsy should be considered at 5 yearly intervals. The reported outcome rates for recurrent AIH are highly variable. Although the majority of patients with putative recurrent AIH will respond clinically, serologically, and histologically to increased immunosuppression, some will progress to end-stage graft failure and may require retransplantation. RECOMMENDATION 86. Although there is no evidence for recommending a particular immunosuppressive regimen in patients undergoing transplantation for AIH, it is prudent to maintain patients on long-term, low-dose corticosteroids in addition to routine immunosuppression (with attention to maintaining bone health; grade 2, level B).
ALCOHOLIC LIVER DISEASE (ALD) Although ALD patients selected for LT have a survival rate similar to that of recipients without ALD,110 post-LT mortality is increased in recipients with comorbid ALD and HCV. There is a wide variation in the reported rates of alcohol relapse by ALD patients after LT (10%-90%). The best prospective study showed that 80% of ALD LT recipients either did not drink or consumed only small amounts occasionally in the first 5 years.111 Conversely, in the remaining 20%, there were various patterns of harmful drinking. Anecdotal reports suggest that patients who relapse to harmful drinking are at risk for alcoholic hepatitis, delirium tremens, alcoholic pancreatitis, and reduced survival.110 Furthermore, the causes of death for the patients who returned to heavy consumption of alcohol tended to be liver-related, whereas abstinent ALD patients died of cardiovascular disease and malignant tumors. The stratification of cardiovascular deaths and new-onset cancers of the aerodigestive tract in patients undergoing LT for ALD suggests a causal linkage with cigarette smoking.17 RECOMMENDATIONS 87. All patients with a prior diagnosis of ALD should be encouraged to remain abstinent from alcohol (grade 1, level B). 88. Patients should be encouraged to enter therapy or counseling if they relapse into alcohol use (grade 1, level C). 89. All patients with a prior diagnosis of ALD who are users of tobacco should be encouraged to undertake smoking cessation (grade 1, level B). 90. Careful attention should be given to the risk of cardiovascular disease and/or new-onset cancers of the aerodigestive tract, especially in cigarette smokers (grade 1, level A). FORWARD BACK AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
134 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS FULL TEXT NONALCOHOLIC STEATOHEPATITIS (NASH)/ NONALCOHOLIC FATTY LIVER DISEASE (NAFLD) It appears that NASH-associated cirrhosis is the fourth most common cause of liver failure leading to LT in the United States, and it is predicted that by 2020-2030, NASH-associated cirrhosis will become the most common indication for LT.112 NAFLD and NASH, both recurrent and de novo, are common after LT.37,113-115 Immunosuppressant agents may contribute to metabolic syndrome: corticosteroids and tacrolimus promote diabetes, sirolimus promotes hyperlipidemia, and cyclosporine and tacrolimus promote systemic hypertension. Risk factors for post-LT NASH/ NAFLD are familiar as the hallmarks of metabolic syndrome: body mass index before and after LT, DM, systemic hypertension, hyperlipidemia, and steatosis on an allograft biopsy sample. Among patients who undergo LT on account of NASH-associated or cryptogenic cirrhosis, 50% to 70% will gain excessive weight in 1 year.115 New-onset or recurrent NAFLD/NASH may present with elevated liver aminotransferases. Distinguishing NAFLD/NASH from other causes of elevated liver tests in the post-LT patient requires liver biopsy. NAFLD/ NASH arising in the liver allograft, whether new-onset or recurrent, may lead to fibrosis.115 Cirrhosis associated with fat accumulation in the allograft is uncommon in the first 5 years after LT. No effect on patient or graft survival has been observed among LT recipients with new-onset or re-emergent NAFLD/NASH, although most studies have been short in duration. Although there are no good data to support one immunosuppressive regimen over another in patients who undergo transplantation for NASH/cirrhosis or cryptogenic cirrhosis, minimizing corticosteroids appears prudent. Renal impairment is more common in those undergoing transplantation for NAFLD. RECOMMENDATIONS 91. The confirmation of recurrent or de novo NAFLD, the recognition of fibrosis, and the exclusion of alternate causes of elevated liver chemistry tests require liver biopsy (grade 1, level B). 92. No specific recommendations regarding the prevention or treatment of NAFLD or NASH in LT recipients can be made other than general recommendations to avoid excessive gains in body weight and control hypertension and diabetes (grade 1, level C). LATE SURGICAL COMPLICATIONS Hepatic artery stenosis, biliary cast syndrome, and bilomas have already been discussed with respect to abnormal liver tests. Incisional hernia is a common late complication after LT. Postoperative weight gain exacerbates the risk. RECOMMENDATION 93. LT recipients with an incisional hernia should be instructed to recognize incarcerated hernias and advised to seek immediate medical assistance (grade 1, level B).
ACKNOWLEDGMENT This practice guideline was produced in collaboration with the American Association for the Study of Liver Diseases Practice Guidelines Committee, which provided extensive peer review of the manuscript. The members of the committee include Jayant A. Talwalkar, M.D., M.P.H. (chair); Keith D. Lindor, M.D. (board liaison); Sumeet Asrani, M.D.; Hari S. Conjeevaram, M.D., M.S.; David A. Gerber, M.D.; Marlyn J. Mayo, M.D.; Raphael B. Merriman, M.D., M.R.C.P.; Gerald Y. Minuk, M.D.; Alexander Monto, M.D.; Michael K. Porayko, M.D.; Benjamin L. Shneider, M.D.; Tram T. Tran, M.D.; and Helen S. Yee, Pharm.D.
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135 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS REFERENCES 1. Eddy DM. A Manual for Assessing Health Practices and Designing Practice Guidelines. Philadelphia, PA: American College of Physicians; 1996. 2. Committee to Advise the Public Health Service on Clinical Practice Guidelines, Institute of Medicine; Field MJ, Lohr KN, eds. Clinical Practice Guidelines: Directions of a New Program. Washington, DC: National Academies Press; 1990. 3. American Gastroenterological Association policy statement on the use of medical practice guidelines by managed care organizations and insurance carriers. Gastroenterology 1995;108:925-926. 4. Guyatt GH, Oxman AD, Vist GE, Kunz R, Falck-Ytter Y, Alonso-Coello P, Schu¨nemann HJ; for GRADE Working Group. GRADE: an emerging consensus on rating quality of evidence and strength of recommendations. BMJ 2008;336:924-926. 5. Devlin J, Doherty D, Thomson L, Wong T, Donaldson P, Portmann B, Williams R. Defining the outcome of immunosuppression withdrawal after liver transplantation. Hepatology 1998;27:926-933. 6. Ojo AO, Held PJ, Port FK, Wolfe RA, Leichtman AB, Young EW, et al. Chronic renal failure after transplantation of a nonrenal organ. N Engl J Med 2003;349: 931-940. 7. Tome S, Wells JT, Said A, Lucey MR. Quality of life after liver transplantation. A systematic review. J Hepatol 2008;48:567-577. 8. Shah JN, Haigh WG, Lee SP, Lucey MR, Brensinger CM, Kochman ML, et al. Biliary casts after orthotopic liver transplantation: clinical factors, treatment, biochemical analysis. Am J Gastroenterol 2003;98:1861-1867. 9. Safdar N, Said A, Lucey MR, Knechtle SJ, D’Alessandro A, Musat A, et al. Infected bilomas in liver transplant recipients: clinical features, optimal management, and risk factors for mortality. Clin Infect Dis 2004;39: 517-525. 10. Banff Working Group, Demetris AJ, Adeyi O, Bellamy CO, Clouston A, Charlotte F, et al. Liver biopsy interpretation for causes of late liver allograft dysfunction. Hepatology 2006;44:489-501. 11. Musat AI, Agni RM, Wai PY, Pirsch JD, Lorentzen DF, Powell A, et al. The significance of donor-specific HLA antibodies in rejection and ductopenia development in ABO compatible liver transplantation. Am J Transplant 2011;11:500-510. 12. Guichelaar MM, Kendall R, Malinchoc M, Hay JE. Bone mineral density before and after OLT: long-term followup and predictive factors. Liver Transpl 2006;12: 1390-1402. 13. Guichelaar MM, Schmoll J, Malinchoc M, Hay JE. Fractures and avascular necrosis before and after orthotopic liver transplantation: long-term follow-up and predictive factors. Hepatology 2007;46:1198-1207. 14. Gonwa TA, Mai ML, Melton LB, Hays SR, Goldstein RM, Levy MF, Klintmalm GB. End-stage renal disease (ESRD) after orthotopic liver transplantation (OLTX) using calcineurin-based immunotherapy: risk of development and treatment. Transplantation 2001;72: 1934-1939. 15. Paramesh AS, Roayaie S, Doan Y, Schwartz ME, Emre S, Fishbein T, et al. Post-liver transplant acute renal failure: factors predicting development of end-stage renal disease. Clin Transplant 2004;18:94-99.
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136 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS REFERENCES 22. Yoshida EM, Marotta PJ, Greig PD, Kneteman NM, Marleau D, Cantarovich M, et al. Evaluation of renal function in liver transplant recipients receiving daclizumab (Zenapax), mycophenolate mofetil, and a delayed, lowdose tacrolimus regimen vs. a standard-dose tacrolimus and mycophenolate mofetil regimen: a multicenter randomized clinical trial. Liver Transpl 2005;11: 1064-1072. 23. Neuberger JM, Mamelok RD, Neuhaus P, Pirenne J, Samuel D, Isoniemi H, et al.; for ReSpECT Study Group. Delayed introduction of reduced-dose tacrolimus, and renal function in liver transplantation: the ’ReSpECT’ study. Am J Transplant 2009;9:327-336. 24. Nair S, Eason J, Loss G. Sirolimus monotherapy in nephrotoxicity due to calcineurin inhibitors in liver transplant recipients. Liver Transpl 2003;9:126-129. 25. Saner FH, Cicinnati VR, Sotiropoulos G, Beckebaum S. Strategies to prevent or reduce acute and chronic kidney injury in liver transplantation. Liver Int 2012;32: 179-188. 26. Demirci G, Becker T, Nyibata M, Lueck R, Bektas H, Lehner F, et al. Results of combined and sequential liver-kidney transplantation. Liver Transpl 2003;9: 1067-1078. 27. Fabrizi F, Dixit V, Martin P, Messa P. Chronic kidney disease after liver transplantation: recent evidence. Int J Artif Organs 2010;33:803-811. 28. Kuo HT, Sampaio MS, Ye X, Reddy P, Martin P, Bunnapradist S. Risk factors for new-onset diabetes mellitus in adult liver transplant recipients, an analysis of the Organ Procurement and Transplant Network/ United Network for Organ Sharing database. Transplantation 2010;89:1134-1140. 29. Chen T, Jia H, Li J, Chen X, Zhou H, Tian H. New onset diabetes mellitus after liver transplantation and hepatitis C virus infection: meta-analysis of clinical studies. Transpl Int 2009;22:408-415. 30 Laryea M, Watt KD, Molinari M, Walsh MJ, McAlister VC, Marotta PJ, et al. Metabolic syndrome in liver transplant recipients: prevalence and association with major vascular events. Liver Transpl 2007;13: 1109-1114. 31 Delgado-Borrego A, Liu YS, Jordan SH, Agrawal S, Zhang H, Christofi M, et al. Prospective study of liver transplant recipients with HCV infection: evidence for a causal relationship between HCV and insulin resistance. Liver Transpl 2008;14:193-201. 32 Demirci MS, Toz H, Yilmaz F, Ertilav M, Asci G, Ozkahya M, et al. Risk factors and consequences of posttransplant diabetes mellitus. Clin Transplant 2010;24: E170-E177. 33. Heisel O, Heisel R, Balshaw R, Keown P. New onset diabetes mellitus in patients receiving calcineurin inhibitors: a systematic review and meta-analysis. Am J Transplant 2004;4:583-595. 34. Chobanian AV, Bakris GL, Black HR, Cushman WC, Green LA, Izzo JL Jr, et al.; for National Heart, Lung, and Blood Institute Joint National Committee on Prevention, Detection, Evaluation and Treatment of High Blood Pressure and National High Blood Pressure Education Program Coordinating Committee. The seventh report of the Joint National Committee on Prevention, Detection, Evaluation, and Treatment of High Blood Pressure: the JNC 7 report. JAMA 2003;289:2560-2572. 35. Textor SC, Taler SJ, Canzanello VJ, Schwartz L, Augustine JE. Posttransplantation hypertension related to calcineurin inhibitors. Liver Transpl 2000;6:521-530. 36. Galioto A, Semplicini A, Zanus G, Fasolato S, Sticca A, Boccagni P, et al. Nifedipine versus carvedilol in the treatment of de novo arterial hypertension after liver transplantation: results of a controlled clinical trial. Liver Transpl 2008;14:1020-1028. 37. Watt KD, Charlton MR. Metabolic syndrome and liver transplantation: a review and guide to management. J Hepatol 2010;53:199-206. 38. Laish I, Braun M, Mor E, Sulkes J, Harif Y, Ben Ari Z. Metabolic syndrome in liver transplant recipients: prevalence, risk factors, and association with cardiovascular events. Liver Transpl 2011;17:15-22.
39 Richards J, Gunson B, Johnson J, Neuberger J. Weight gain and obesity after liver transplantation. Transpl Int 2005;18:461-466. 40. Everhart JE, Lombardero M, Lake JR, Wiesner RH, Zetterman RK, Hoofnagle JH. Weight change and obesity after liver transplantation: incidence and risk factors. Liver Transpl Surg 1998;4:285-296. 41. Engels EA, Pfeiffer RM, Fraumeni JF Jr, Kasiske BL, Israni AK, Snyder JJ, et al. Spectrum of cancer risk among US solid organ transplant recipients. JAMA 2011;306: 1891-1901. FORWARD BACK References (cont.) AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
137 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS REFERENCES 42. Watt KD, Pedersen RA, Kremers WK, Heimbach JK, Sanchez W, Gores GJ. Long-term probability of and mortality from de novo malignancy after liver transplantation. Gastroenterology 2009;137:2010-2017. 43. Finkenstedt A, Graziadei IW, Oberaigner W, Hilbe W, Nachbaur K, Mark W, et al. Extensive surveillance promotes early diagnosis and improved survival of de novo malignancies in liver transplant recipients. Am J Transplant 2009;9:2355-2361. 44. Herrero JI, Pardo F, D’Avola D, Alegre F, Rotellar F, Inarrairaegui M, et al. Risk factors of lung, head and neck, esophageal, and kidney and urinary tract carcinomas after liver transplantation: the effect of smoking withdrawal. Liver Transpl 2011;17:402-408. 45. Nishihori T, Strazzabosco M, Saif MW. Incidence and management of colorectal cancer in liver transplant recipients. Clin Colorectal Cancer 2008;7:260-266. 46. Simon S. New screening guidelines for cervical cancer. for-cervical-cancer. Published March 14, 2012. Accessed October 2012. 47. Roberts JP. Tumor surveillance—what can and should be done? Screening for recurrence of hepatocellular carcinoma after liver transplantation. Liver Transpl 2005;11 (suppl 2):S45-S46. 48. Liang W, Wang D, Ling X, Kao AA, Kong Y, Shang Y, et al. Sirolimus-based immunosuppression in liver transplantation for hepatocellular carcinoma: a metaanalysis. Liver Transpl 2012;18:62-69. 49. Mass K, Quint EH, Punch MR, Merion RM. Gynecological and reproductive function after liver transplantation. Transplantation 1996;62:476-479. 50. Cundy TF, O’Grady JG, Williams R. Recovery of menstruation and pregnancy after liver transplantation. Gut 1990;31:337-338. 51. Armenti VT, Radomski JS, Moritz MJ, Gaughan WJ, Gulati R, McGrory CH, Coscia LA. Report from the National Transplantation Pregnancy Registry (NTPR): outcomes of pregnancy after transplantation. Clin Transpl 2005:69-83. 52. Burra P, Germani G, Masier A, De Martin E, Gambato M, Salonia A, et al. Sexual dysfunction in chronic liver disease: is liver transplantation an effective cure? Transplantation 2010;89:1425-1429. 53. Ho JK, Ko HH, Schaeffer DF, Erb SR, Wong C, Buczkowski AK, et al. Sexual health after orthotopic liver transplantation. Liver Transpl 2006;12:1478-1484. 54. Coffin CS, Shaheen AA, Burak KW, Myers RP. Pregnancy outcomes among liver transplant recipients in the United States: a nationwide case-control analysis. Liver Transpl 2010;16:56-63. 55. Sifontis NM, Coscia LA, Constantinescu S, Lavelanet AF, Moritz MJ, Armenti VT. Pregnancy outcomes in solid organ transplant recipients with exposure to mycophenolate mofetil or sirolimus. Transplantation 2006;82:1698-1702. 56. Anderka MT, Lin AE, Abuelo DN, Mitchell AA, Rasmussen SA. Reviewing the evidence for mycophenolate mofetil as a new teratogen: case report and review of the literature. Am J Med Genet A 2009;149A: 1241-1248. 57. Heneghan MA, Selzner M, Yoshida EM, Mullhaupt B. Pregnancy and sexual function in liver transplantation. J Hepatol 2008;49:507-519. 58. EBPG Expert Group on Renal Transplantation. European best practice guidelines for renal transplantation. Section IV: Long-term management of the transplant recipient. IV.10. Pregnancy in renal transplant recipients. Nephrol Dial Transplant 2002;17(suppl 4):50-55. 59. Christopher V, Al-Chalabi T, Richardson PD, Muiesan P, Rela M, Heaton ND, et al. Pregnancy outcome after liver transplantation: a single-center experience of 71 preg-nancies in 45 recipients. Liver Transpl 2006;12: 1138-1143. 60. San Juan R, Aguado JM, Lumbreras C, Díaz-Pedroche C, López-Medrano F, Lizasoain M, et al.; for RESITRA Network (Spain). Incidence, clinical characteristics and risk factors of late infection in solid organ transplant recipients: data from the RESITRA study group. Am J Transplant 2007;7:964-971. 61. Humar A, Snydman D; for AST Infectious Diseases Community of Practice. Cytomegalovirus in solid organ transplant recipients. Am J Transplant 2009;9(suppl 4):S78-S86. 62. Razonable RR. Cytomegalovirus infection after liver transplantation. Liver Transpl 2010;16(suppl 2):S45-S53. 63. Kotton CN, Kumar D, Caliendo AM, Asberg A, Chou S, Snydman DR, et al.; for Transplantation Society International CMV Consensus Group. International consensus guidelines on the management of cytomegalovirus in solid organ transplantation. Transplantation 2010; 89:779-795. FORWARD BACK References (cont.) AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
138 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS REFERENCES 64. Asberg A, Humar A, Rollag H, Jardine AG, Mouas H, Pescovitz MD, et al.; for VICTOR Study Group. Oral valganciclovir is noninferior to intravenous ganciclovir for the treatment of cytomegalovirus disease in solid organ transplant recipients. Am J Transplant 2007;7:2106-2113. 65. Burra P, Buda A, Livi U, Rigotti P, Zanus G, Calabrese F, et al. Occurrence of post-transplant lymphoproliferative disorders among over thousand adult recipients: any role for hepatitis C infection? Eur J Gastroenterol Hepatol 2006;18:1065-1070. 66. Jain A, Nalesnik M, Reyes J, Pokharna R, Mazariegos G, Green M, et al. Posttransplant lymphoproliferative disorders in liver transplantation: a 20-year experience. Ann Surg 2002;236:429-436. 67. Allen U, Preiksaitis J; for AST Infectious Diseases Community of Practice. Epstein-Barr virus and posttransplant lymphoproliferative disorder in solid organ transplant recipients. Am J Transplant 2009;9 (suppl 4):S87-S96. 68. Kremers WK, Devarbhavi HC, Wiesner RH, Krom RA, Macon WR, Habermann TM. Post-transplant lymphoproliferative disorders following liver transplantation: incidence, risk factors and survival. Am J Transplant 2006;6(pt 1): 1017-1024. 69. Koch DG, Christiansen L, Lazarchick J, Stuart R, Willner IR, Reuben A. Posttransplantation lymphoproliferative disorder— the great mimic in liver transplantation: appraisal of the clinicopathologic spectrum and the role of Epstein-Barr virus. Liver Transpl 2007;13:904-912. 70. Singh N, Wagener MM, Marino IR, Gayowski T. Trends in invasive fungal infections in liver transplant recipients: correlation with evolution in transplantation practices. Transplantation 2002;73:63-67. 71. Berenguer J, Buck M, Witebsky F, Stock F, Pizzo PA, Walsh TJ. Lysis-centrifugation blood cultures in the detection of tissue-proven invasive candidiasis. Disseminated versus single-organ infection. Diagn Microbiol Infect Dis 1993;17:103-109 72. Alexander BD, Smith PB, Davis RD, Perfect JR, Reller LB. The (1,3)β-D-glucan test as an aid to early diagnosis of invasive fungal infections following lung transplantation. J Clin Microbiol 2010;48:4083-4088. 73. Hope WW, Walsh TJ, Denning DW. Laboratory diagnosis of invasive aspergillosis. Lancet Infect Dis 2005;5: 609-622. 74. Fortun J, Martín-Dávila P, Alvarez ME, Norman F, Sanchez-Sousa A, Gajate L, et al. False-positive results of Aspergillus galactomannan antigenemia in liver transplant recipients. Transplantation 2009;87:256-260. 75. Wu G, Vilchez RA, Eidelman B, Fung J, Kormos R, Kusne S. Cryptococcal meningitis: an analysis among 5,521 consecutive organ transplant recipients. Transpl Infect Dis 2002;4:183-188. 76. Durkin M, Witt J, Lemonte A, Wheat B, Connolly P. Antigen assay with the potential to aid in diagnosis of blastomycosis. J Clin Microbiol 2004;42:4873-4875. 77. Connolly PA, Durkin MM, Lemonte AM, Hackett EJ, Wheat LJ. Detection of Histoplasma antigen by a quantitative enzyme immunoassay. Clin Vaccine Immunol 2007;14: 1587-1591. 78. Martin SI, Fishman JA; for AST Infectious Diseases Community of Practice. Pneumocystis pneumonia in solid organ transplant recipients. Am J Transplant 2009;9 (suppl 4):S227-S233. 79. Aguado JM, Torre-Cisneros J, Fortun J, Benito N, Meije Y, Doblas A, Munoz P. Tuberculosis in solid-organ transplant recipients: consensus statement of the Group for the Study of Infection in Transplant Recipients (GESITRA) of the Spanish Society of Infectious Diseases and Clinical Microbiology. Clin Infect Dis 2009; 48:1276-1284. 80. Subramanian A, Dorman S; for AST Infectious Diseases Community of Practice. Mycobacterium tuberculosis in solid organ transplant recipients. Am J Transplant 2009;9 (suppl 4):S57-S62. 81. Yehia BR, Blumberg EA. Mycobacterium tuberculosis infection in liver transplantation. Liver Transpl 2010; 16:1129-1135. 82. Blumberg HM, Burman WJ, Chaisson RE, Daley CL, Etkind SC, Friedman LN, et al.; for American Thoracic Society, Centers for Disease Control and Prevention, and Infectious Diseases Society. American Thoracic Society/ Centers for Disease Control and Prevention/Infectious Diseases Society of America: treatment of tuberculosis. Am J Respir Crit Care Med 2003;167:603-662. 83. Blumberg EA, Stock P; for AST Infectious Diseases Community of Practice. Solid organ transplantation in the HIV-infected patient. Am J Transplant 2009;9 (suppl 4):S131-S135. FORWARD BACK References (cont.) AASLD PRACTICE GUIDELINE Long-Term Management of Adult Liver Transplant: 2012 Practice Guideline by AASLD and AST © 2012 The American Association for the Study of Liver Diseases, All rights reserved.
139 CONTENTS FULL TEXT REFERENCES WEB SITE RECOMMENDATIONS REFERENCES 84. Frassetto LA, Browne M, Cheng A, Wolfe AR, Roland ME, Stock PG, et al. Immunosuppressant pharmacokinetics and dosing modifications in HIV-1 infected liver and kidney transplant recipients. Am J Transplant 2007;7:2816-2820. 85. Castells L, Escartı´n A, Bilbao I, Len O, Allende H, Vargas V, et al. Liver transplantation in HIV-HCV coinfected patients: a case-control study. Transplantation 2007;83:354-358. 86. Danzinger-Isakov L, Kumar D; for AST Infectious Diseases Community of Practice. Guidelines for vaccination of solid organ transplant candidates and recipients. Am J Transplant 2009;9(suppl 4):S258-S262. 87. KimWR, Terrault NA, Pedersen RA, Therneau TM, Edwards E, Hindman AA, Brosgart CL. Trends in waiting list registration for liver transplantation for viral hepatitis in the United States. Gastroenterology 2009;137:1680-1686. 88. Loomba R, Rowley AK, Wesley R, Smith KG, Liang TJ, Pucino F, Csako G. Hepatitis B immunoglobulin and lamivudine improve hepatitis B-related outcomes after liver transplantation: meta-analysis. Clin Gastroenterol Hepatol 2008;6:696-700. 89. Rao W, Wu X, Xiu D. Lamivudine or lamivudine combined with hepatitis B immunoglobulin in prophylaxis of hepatitis B recurrence after liver transplantation: a meta-analysis. Transpl Int 2009;22:387-394. 90. Katz LH, Paul M, Guy DG, Tur-Kaspa R. Prevention of recurrent hepatitis B virus infection after liver transplantation: hepatitis B immunoglobulin, antiviral drugs, or both? Systematic review and meta-analysis. Transpl Infect Dis 2010;12:292-308. 91. Dan YY, Wai CT, Yeoh KG, Lim SG. Prophylactic strategies for hepatitis B patients undergoing liver transplant: a cost-effectiveness analysis. Liver Transpl 2006; 12:736-746. 92. Fox AN, Terrault NA. The option of HBIG-free prophylaxis against recurrent HBV. J Hepatol 2012;56: 1189-1197. 93. Lucey MR, Graham DM, Martin P, Di Bisceglie A, Rosenthal S, Waggoner JG, et al. Recurrence of hepatitis B and delta hepatitis after orthotopic liver transplantation. Gut 1992;33:1390-1396. 94. Terrault NA, Berenguer M. Treating hepatitis C infection in liver transplant recipients. Liver Transpl 2006; 12: 1192-1204. 95. Berenguer M, Prieto M, Rayón JM, Mora J, Pastor M, Ortiz V, et al. Natural history of clinically compensated hepatitis C virus-related graft cirrhosis after liver transplantation. Hepatology 2000;32(pt 1):852-858. 96. Kalambokis G, Manousou P, Samonakis D, Grillo F, Dhillon AP, Patch D, et al. Clinical outcome of HCV related graft cirrhosis and prognostic value of hepatic venous pressure gradient. Transpl Int 2009;22: 172-181. 97. Forman LM, Lewis JD, Berlin JA, Feldman HI, Lucey MR. The association between hepatitis C infection and survival after orthotopic liver transplantation. Gastroenterology 2002;122:889-896. 98. Neumann UP, Berg T, Bahra M, Puhl G, Guckelberger O, Langrehr JM, Neuhaus P. Long-term outcome of liver transplants for chronic hepatitis C: a 10-year follow-up. Transplantation 2004;77:226-231. 99. Lake JR, Shorr JS, Steffen BJ, Chu AH, Gordon RD, Wiesner RH. Differential effects of donor age in liver transplant recipients infected with hepatitis B, hepatitis C and without viral hepatitis. Am J Transplant 2005;5: 549-557. 100. Wiesner RH, Sorrell M, Villamil F; for International Liver Transplantation Society Expert Panel. Report of the first International Liver Transplantation Society expert panel consensus conference on liver transplantation and hepatitis C. Liver Transpl 2003;9:S1-S9. 101. Veldt BJ, Poterucha JJ, Watt KD, Wiesner RH, Hay JE, Kremers WK, et al. Impact of pegylated interferon and ribavirin treatment on graft survival in liver transplant patients with recurrent hepatitis C infection. Am J Transplant 2008;8:2426-2433. 102. Xirouchakis E, Triantos C, Manousou P, Sigalas A, Calvaruso V, Corbani A, et al. Pegylated-interferon and ribavirin in liver transplant candidates and recipients with HCV cirrhosis: systematic review and meta-analysis of prospective controlled studies. J Viral Hepat 2008;15: 699-709. 103. Fiel MI, Agarwal K, Stanca C, Elhajj N, Kontorinis N, Thung SN, Schiano TD. Posttransplant plasma cell hepatitis (de novo autoimmune hepatitis) is a variant of rejection and may lead to a negative outcome in patients with hepatitis C virus. Liver Transpl 2008;14: 861-871.
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References (cont.) BACK |
1135 | https://medievalbritain.com/type/medieval-life/occupations/medieval-knight/ | Medieval knights had to go through years of training in the use of weapons, horsemanship, and warfare. Frequently, members of the noble class, knights were responsible for defending their feudal lord’s territory from rivals and keeping the local serfdom in line with the lord’s rule.
They were an elite class of warriors that used weapons such as longswords and lances and specialized in armored combat from horseback. They conducted themselves according to the Code of Chivalry, which covered courtly etiquette and valor in battle and devotion to the Christian faith.
History of the Knight
The origins of knighthood can be traced back to the Greek hippeis (ἱππεῖς) and Roman eques. In Sparta, the hippeus was the royal guard of honor, 300 Spartan youths who had served in the Krypteia and/or were Olympic champions. They served as heavily armed infantry soldiers in the King’s bodyguard. During Roman times, legionary cavalry was originally recruited exclusively from the ranks of the patricians. It was provided with a sum of money by the state to purchase a horse for military service and its fodder.
The first knights appeared during the reign of Charlemagne in the 8th century, and the older Carolingian ceremony of presenting a young man with weapons influenced the emergence of knighthood ceremonies. In the Middle Ages in Europe, knighthood was conferred upon mounted warriors. In the course of the 12th-century, knighthood became a social rank, with the military role of fully armored cavalryman gaining a separate term, “man-at-arms” (not all men-at-arms were knights).
The first military orders of knighthood were founded shortly after the First Crusade of 1099. These were the Knights of the Holy Sepulchre and the Knights Hospitaller, which were followed by the Order of Saint Lazarus (1100), the Knights Templars (1118), and the Teutonic Knights (1190).
By the Late Middle Ages, the rank had become associated with chivalry ideals, a code of conduct for the perfect courtly Christian warrior. When new methods of warfare began to be used, these rendered classical knights in armor obsolete.
How Many Knights Were There in a Medieval Army?
The number of knights in a medieval army varied depending on its location, time period, and purpose. During the early Middle Ages, the number of knights in an army was relatively small, as these were usually wealthy and privileged individuals who could afford expensive armour and horses. They were also required to provide their own equipment and serve in the army for a limited time.
As time progressed, the number of knights in an army increased, as more individuals of lower social status were able to acquire the necessary equipment and training to become knights. By the 14th century, knights made up a significant portion of the European armies, but they still represented only a fraction of the total number of soldiers. A typical medieval army could range from a few hundred to tens of thousands of soldiers, depending on the size and purpose of the campaign.
Training to Become a Knight
Higher nobles would grant the vassals portions of land in return for their loyalty, protection, and service. They also provided their knights with lodging, food, armor, weapons, horses, and money. In exchange, knights provided military service that usually lasted 40 days a year.
A knight would start their life in a castle as a Page and then move up to a Squire’s role. They had to be born of nobility – typically sons of knights or lords, although there were cases of commoners knighted as a reward for extraordinary military service. The children were cared for by noble foster-mothers in castles until they reached age seven when they were given the title of pageand turned over to the castle’s lords’ care. Their early training included hunting and academic studies. They would also assist older knights in battle, carrying and cleaning armor, taking care of the horses, and packing the baggage.
When boys turned 15, he became a squire after swearing on a sword consecrated by a bishop or priest. Squires continued training in combat and were allowed to own armor and had to master the “seven points of agilities.” Riding, swimming and diving, shooting different types of weapons, climbing, participation in tournaments, wrestling, fencing, long jumping, and dancing.
Upon turning 21, the squire was eligible to be knighted. The knighting ceremony was usually held during one of the great feasts or holidays, like Christmas or Easter, and sometimes at the wedding of a noble or royal.
The Chivalric Code
Chivalry developed as an early standard of professional ethics for knights. During medieval times, this grew from simple military professionalism into a social code, including the values of gentility, nobility, and treating others reasonably.
Ramon Llull‘s Book of the Order of Chivalry (1275) shows that by the end of the 13th century, chivalry entailed peculiar duties, such as riding warhorses, jousting, attending tournaments, holding Round Tables and hunting, as well as aspiring to the more æthereal virtues of “faith, hope, charity, justice, strength, moderation, and loyalty.”
Chivalry and religion were mutually influenced during the period of the Crusades, with clergy instituting religious vows which required knights to use their weapons chiefly for the protection of the weak and defenseless, especially women and orphans, and of churches.
Books about Knights and Medieval Life
Medieval Warrior: Weapons, Technology, And Fighting Techniques
Amazon
Chivalry: The Everyday Life of the Medieval Knight
Amazon
The Time Traveler's Guide to Medieval England
Amazon
A Catholic Quest for the Holy Grail
Amazon
Life in a Medieval City
Amazon
Medieval Castles of England and Wales
Amazon
Swords: An Artist's Devotion
Amazon
Ancient and Medieval Siege Weapons: A Fully Illustrated Guide
Amazon
More Medieval Occupations
Medieval Minstrel
Medieval minstrels sang, played musical instruments, and told engaging stories. Here’s what life was like for a minstrel in the Middle Ages.
Medieval Miller
Millers were some of the most important tradesmen in the Middle Ages. Learn more about this medieval profession and how millers lived.
Medieval Butcher
Middle Ages butchers prepared meat, fish, and fowl for the people in a castle or a city. They sometimes had stalls in a marketplace.
Medieval Wheelwright
Medieval candlemakers made candles from materials such as fat, tallow and beeswax.
Medieval Shoemaker
Medieval candlemakers made candles from materials such as fat, tallow and beeswax.
Medieval Shipwrights and Shipmaking
Being a sailor in the middle ages meant living a lonely and difficult life, as they would often set sail for months or even a year at a time.
Posted inOccupations |
1136 | https://www.jamesspring.com/news/common-compression-spring-uses/ | Published Time: 2018-11-15T15:00:46+00:00
Compression Spring Uses | How Do Compression Springs Work?
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Home » Blog » Most Common Compression Spring Uses
Most Common Compression Spring Uses
November 15, 2018
Compression spring uses are many. It is the most common type of spring and is designed to function when a load is applied to it. Made of spring steel, or another non-ferrous metal, a compression spring is an elastic coil that very efficiently builds up energy when loaded. When you think compression spring, think mattress springs, upholstery springs or even a pogo stick—each carry a load.
Compression springs, as the name suggests, absorb force or provide resistance when the spring coil is being compressed. In an unloaded state, the spaces between the spring are visible through its coil. When loaded, the spaces between the coils are compressed, creating the necessary energy to carry the load. However, for a compression spring to operate effectively, the coils must never compress to the point that they are in contact with each other. If so, then the load is either too heavy or the spring is shot or both, and its service life has ended.
What are common compression spring uses?
A compression spring comes in many configurations, shapes and sizes. They are conical, hourglass and barrel of varying lengths—and can be used as a shock absorber, vibration damper, a pure energy accumulator, or force generator. Because of their versatility, how efficiently they store energy and the multiple shapes in which they can be manufactured, these types of mechanical springs have a wide selection of applications.
Consider your car. In the automotive industry, compression springs are used throughout the vehicle. As the springs can be manufactured to tight tolerances, they are used in the engine around shafts or placed in small holes while in a vertical position. Look under any vehicle at the suspension system. It is compression springs that absorb the shock of the road while driving to provide a comfortable ride. The springs also help to absorb various types of vibrations and resonances while driving. They also add to your personal comfort as part of most seating assemblies.
Medical applications for compression springs are used as parts in assembling a device. They also implanted straight into a patient’s body. For example, so specialized are the springs for arterial widening—not only the materials but also for the extremely small dimensions to .030 mm.
High energy rated industrial compression springs is used in the manufacturing of mining machinery, too, for drilling and other related equipment.
How indispensable are compression springs?
Original equipment manufacturers (OEMs) that produce products for various industries that use compression springs need compression springs in their machinery in order to produce those parts and products. Compression springs are used in such industrial sectors as manufacturing, transportation, construction, agriculture, petrochemical, and aeronautical.
Compression spring applications in the everyday products, machines, and devices we use in home, office, on the job, or in school, are so varied and common we don’t even realize how much a part of our lives compression springs are. They are found in pens and notebooks, mattresses, flashlights, tools, toys, armchairs, small electronic devices, countertop kitchen equipment, electrical light switches, precision instruments, lawn mowers, washing machines, and more. Industry-specific spring applications are easy to overlook as well – for example, in the medical field, medical compression springs are used for a variety of equipment, from syringes to inhalers.
Compression spring uses are seemingly endless. From automotive springs to medical compression springs,these springs play an important role in all facets of today’s industrial world.They allow for the efficient and safe operation of machines, devices, tools and components, and undoubtedly is the main reason that they have become the most commonly used spring on the market.
To learn more about our products and industries served, contact us at James Spring & Wire Company today.
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A graphic organizer used to compare and contrast what a student's life was like with the virus and without it.
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1138 | https://www.justquant.com/uncategorized/absolute-value/ | Absolute Value Equations and Inequalities
How to Calculate Remainders of large numbers
Divisors of a Number – Sum and Product of Divisors
How to Calculate Remainders of large numbers
Divisors of a Number – Sum and Product of Divisors
Whats is Absolute value?
Absolute value (or Modulus) represented by |x| of a real number x is the non-negative value of x without regard to its sign.
For example:
Absolute value of -3 => |-3| = 3
Absolute value of 7 => |7| = 7
Absolute value as distance?
We can also think of absolute value of a number as its distance from zero. For example, consider this number line,
Here, the integer 3 is at a distance of 3 units from number 0. Similarly number -3 is also at distance of 3 units from 0. Since distance is always positive, we can say that,
The Distance of 3 from 0 => |3| = 3 &
The Distance of -3 from 0 => |-3| = 3
The Distance of 3 from 0 => |3| = 3 &
The Distance of -3 from 0 => |-3| = 3
Now, if |x| = 3, then x is at a distance of 3 units from 0. From the number line, we get that the absolute of x = 3 or x = -3
|x| = 3
x = 3 or x = -3
In general, if |x| = a => x = a or x = -a. (Since x is at a distance of ‘a’ units from 0).
If |x| = a is the distance of x from 0, then what is |x-a|?
Distance of x from a number ‘a’ on the number line can be represented by |x-a|.
Solve the equation |x-2| = 3.
|x-2| = 3 implies x is at a distance of 3 units from 2. Representing this on the number line,
|x-2| = 3 implies x is at a distance of 3 units from 2. Representing this on the number line,
The numbers on the number line which are at a distance of 3 units from 2 are 5 and -1. Therefore, x = 5 or x = -1 are the values of the equation |x-2| = 3.
Solve the inequality |x-2| < 3
|x-2| = 3 implies x is at a distance less than 3 units from 2. Representing this on the number line,
|x-2| = 3 implies x is at a distance less than 3 units from 2. Representing this on the number line,
From the number line, we see that all the points between -1 and 5 are at a distance less than 3 units from 2. Hence, the values taken by x is given by -1 < x < 5.
If you know what is |x| = a, can you explain what is meant by |x-a| + |x-b| ?
From our understanding of absolute value so far, we know that
|x – a| represents the distance of x from a. Similarly |x – b| represents the distance of x from b. Implying that |x-a| + |x-b| is the sum of the distances of x from both a and b! :). This is evident from the number line,
Solve for x, |x+2| + |x-3| = 7
=> |x-(-2)| + |x-3| = 7
i.e we have to calculate the sum of the distances of x from -2 and 3.
=> |x-(-2)| + |x-3| = 7
i.e we have to calculate the sum of the distances of x from -2 and 3.
Representing this on the number line,
From the image, we see that 4 is at distance of 1 unit from 3 and 6 units from -2. Hence, the sum of the distances of 4 from 3 and -2 is 7.
Similarly the sum of the distances of -3 from 4 and 3 is 7.
Therefore, x = 4 or x = -3 are the values of the equation |x+2| + |x-3| = 7.
When is the distance of x from ‘a’ and ‘b’ minimum?
We know that the sum of the distances of x from a and b is represented in terms of absolute value by |x – a| + |x – b|.
Now, |x – a| + |x – b| is minimum when x lies between a and b. The minimum value is given by |b-a|.
What is the minimum value of |x+2| + |x-3|?
=> Minimum value = |3-(-2)| = 5, which occurs for -2 ≤ x ≤ 3
=> Minimum value = |3-(-2)| = 5, which occurs for -2 ≤ x ≤ 3
Solving Absolute Value Equations
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Triangle Inequality
The Triangle Inequality says that in a nondegeneratetriangle:
That is, the sum of the lengths of any two sides is larger than the length of the third side. In degenerate triangles, the strict inequality must be replaced by "greater than or equal to."
The Triangle Inequality can also be extended to other polygons. The lengths can only be the sides of a nondegenerate -gon if for . Expressing the inequality in this form leads to , where is the sum of the , or . Stated in another way, it says that in every polygon, each side must be smaller than the semiperimeter.
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Contents
1 Context and method
1.1 Applicability
1.2 Complex replacement
1.3 Linear time-invariant operator
1.4 Problem setting and ERF method
2 Example
3 Comparison with method of undetermined coefficients
4 Generalized exponential response formula
4.1 Example
5 Application examples
5.1 Motion of object hanging from a spring
5.2 Electrical circuits
5.3 Complex gain and phase lag
6 References
7 External links
Exponential response formula
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| Differential equations |
| Scope |
| Fields | Natural sciences Engineering | | Astronomy Physics Chemistry Biology Geology | | Applied mathematics | | Continuum mechanics Chaos theory Dynamical systems | | Social sciences | | Economics Population dynamics | List of named differential equations |
| Classification |
| Types | | | Ordinary Partial Differential-algebraic Integro-differential Fractional Linear Non-linear | | By variable type | | Dependent and independent variables Autonomous Coupled / Decoupled Exact Homogeneous / Nonhomogeneous | | Features | | Order Operator Notation | |
| Relation to processes Difference (discrete analogue) Stochastic + Stochastic partial Delay |
| Solution |
| Existence and uniqueness Picard–Lindelöf theorem Peano existence theorem Carathéodory's existence theorem Cauchy–Kowalevski theorem |
| General topics Initial conditions Boundary values + Dirichlet + Neumann + Robin + Cauchy problem Wronskian Phase portrait Lyapunov / Asymptotic / Exponential stability Rate of convergence Series / Integral solutions Numerical integration Dirac delta function |
| Solution methods Inspection Method of characteristics Euler Exponential response formula Finite difference (Crank–Nicolson) Finite element + Infinite element Finite volume Galerkin + Petrov–Galerkin Green's function Integrating factor Integral transforms Perturbation theory Runge–Kutta Separation of variables Undetermined coefficients Variation of parameters |
| People |
| List Isaac Newton Gottfried Leibniz Jacob Bernoulli Leonhard Euler Joseph-Louis Lagrange Józef Maria Hoene-Wroński Joseph Fourier Augustin-Louis Cauchy George Green Carl David Tolmé Runge Martin Kutta Rudolf Lipschitz Ernst Lindelöf Émile Picard Phyllis Nicolson John Crank |
| v t e |
In mathematics, the exponential response formula (ERF), also known as exponential response and complex replacement, is a method used to find a particular solution of a non-homogeneous linear ordinary differential equation of any order. The exponential response formula is applicable to non-homogeneous linear ordinary differential equations with constant coefficients if the function is polynomial, sinusoidal, exponential or the combination of the three. The general solution of a non-homogeneous linear ordinary differential equation is a superposition of the general solution of the associated homogeneous ODE and a particular solution to the non-homogeneous ODE. Alternative methods for solving ordinary differential equations of higher order are method of undetermined coefficients and method of variation of parameters.
Context and method
[edit]
Applicability
[edit]
The ERF method of finding a particular solution of a non-homogeneous differential equation is applicable if the non-homogeneous equation is or could be transformed to form ; where are real or complex numbers and is homogeneous linear differential equation of any order. Then, the exponential response formula can be applied to each term of the right side of such equation. Due to linearity, the exponential response formula can be applied as long as the right side has terms, which are added together by the superposition principle.
Complex replacement
[edit]
Complex replacement is a method of converting a non-homogeneous term of equation into a complex exponential function, which makes a given differential equation a complex exponential.
Consider differential equation .
To make complex replacement, Euler's formula can be used;
Therefore, given differential equation changes to . The solution of the complex differential equation can be found as , from which the real part is the solution of the original equation.
Complex replacement is used for solving differential equations when the non-homogeneous term is expressed in terms of a sinusoidal function or an exponential function, which can be converted into a complex exponential function differentiation and integration. Such complex exponential function is easier to manipulate than the original function.
When the non-homogeneous term is expressed as an exponential function, the ERF method or the undetermined coefficients method can be used to find a particular solution. If non-homogeneous terms can not be transformed to complex exponential function, then the Lagrange method of variation of parameters can be used to find solutions.
Linear time-invariant operator
[edit]
The differential equations are important in simulating natural phenomena. In particular, there are numerous phenomena described as high order linear differential equations, for example the spring vibration, LRC circuit, beam deflection, signal processing, control theory and LTI systems with feedback loops.
Mathematically, the system is time-invariant if whenever the input has response then for any constant "a", the input has response . Physically, time invariance means system’s response does not depend on what time the input begins. For example, if a spring-mass system is at equilibrium, it will respond to a given force in the same way, no matter when the force was applied.
When the time-invariant system is also linear, it is called a linear time-invariant system (LTI system). Most of these LTI systems are derived from linear differential equations, where the non-homogeneous term is called the input signal and solution of the non-homogeneous equations is called the response signal. If the input signal is given exponentially, the corresponding response signal also changes exponentially.
Considering the following th order linear differential equation
and denoting
where are the constant coefficients, produces differential operator , which is linear and time-invariant and known as the LTI operator. The operator, is obtained from its characteristic polynomial;
by formally replacing the indeterminate s here with the differentiation operator
Therefore, the equation (1) can be written as
Problem setting and ERF method
[edit]
Considering LTI differential equation above, with exponential input , where and are given numbers. Then, a particular solution is
provide only that .
Proof: Due to linearity of operator , the equation can be written as
On the other hand, since
substituting this into equation (3), produces
Therefore, is a particular solution to the non-homogeneous differential equation.
Thus, the above equation for a particular response is called the exponential response formula (ERF) for the given exponential input.
In particular, in case of , a solution to equation (2) is given by
and is called the resonant response formula.
Example
[edit]
Let's find the particular solution to 2nd order linear non-homogeneous ODE;
The characteristic polynomial is . Also, the non-homogeneous term, can be written as follows
Then, the particular solutions corresponding to and , are found, respectively.
First, considering non-homogeneous term, . In this case, since and .
from the ERF, a particular solution corresponding to can be found.
: .
Similarly, a particular solution can be found corresponding to .
Let's find a particular solution to DE corresponding to 3rd term;
In order to do this, equation must be replaced by complex-valued equation, of which it is the real part:
Applying the exponential response formula (ERF), produces
and the real part is
Therefore, the particular solution of given equation, is
Comparison with method of undetermined coefficients
[edit]
The undetermined coefficients method is a method of appropriately selecting a solution type according to the form of the non-homogeneous term and determining the undetermined constant, so that it satisfies the non-homogeneous equation. On the other hand, the ERF method obtains a special solution based on differential operator. Similarity for both methods is that special solutions of non-homogeneous linear differential equations with constant coefficients are obtained, while form of the equation in consideration is the same in both methods.
For example, finding a particular solution of with the method of undetermined coefficients requires solving the characteristic equation . The non-homogeneous term is then considered and since is not a characteristic root, it puts a particular solution in form of , where is undetermined constant. Substituting into the equation to determine the tentative constant yields
therefore
The particular solution can be found in form:
On the other hand, the exponential response formula method requires characteristic polynomial to be found, after which the non-homogeneous terms is complex replaced. The particular solution is then found using formula
Generalized exponential response formula
[edit]
The exponential response formula method was discussed in case of . In the case of , the resonant response formula is also considered.
In the case of , we will discuss how the ERF method will be described in this section.
Let be a polynomial operator with constant coefficients, and its -th derivative. Then ODE
: , where is real or complex.
has the particular solution as following.
. In this case, a particular solution will be given by .(exponent response formula)
but . In this case, a particular solution will be given by .(resonant response formula)
but . In this case, a particular solution will be given by
Above equation is called generalized exponential response formula.
Example
[edit]
To find a particular solution of the following ODE;
the characteristic polynomial is .
By the calculating, we get the following:
Original exponential response formula is not applicable to this case due to division by zero. Therefore, using the generalized exponential response formula and calculated constants, particular solution is
Application examples
[edit]
Motion of object hanging from a spring
[edit]
Object hanging from a spring with displacement . The force acting is gravity, spring force, air resistance, and any other external forces.
From Hooke’s law, the motion equation of object is expressed as follows;
where is external force.
Now, assuming drag is neglected and , where (the external force frequency coincides with the natural frequency). Therefore, the harmonic oscillator with sinusoidal forcing term is expressed as following:
Then, a particular solution is
Applying complex replacement and the ERF: if is a solution to the complex DE
then will be a solution to the given DE.
The characteristic polynomial is , and , so that . However, since , then . Thus, the resonant case of the ERF gives
Electrical circuits
[edit]
Considering the electric current flowing through an electric circuit, consisting of a resistance (), a capacitor (), a coil wires (), and a battery (), connected in series.
This system is described by an integral-differential equation found by Kirchhoff called Kirchhoff’s voltage law, relating the resistor , capacitor , inductor , battery , and the current in a circuit as follows,
Differentiating both sides of the above equation, produces the following ODE.
Now, assuming , where . ( is called resonance frequency in LRC circuit). Under above assumption, the output (particular solution) corresponding to input can be found. In order to do it, given input can be converted in complex form:
The characteristic polynomial is , where . Therefore, from the ERF, a particular solution can be obtained as follows;
Complex gain and phase lag
[edit]
Considering the general LTI system
where is the input and are given polynomial operators, while assuming that . In case that , a particular solution to given equation is
Considering the following concepts used in physics and signal processing mainly.
The amplitude of the input is . This has the same units as the input quantity.
The angular frequency of the input is . It has units of radians/time. Often it will be referred to it as frequency, even though technically frequency should have units of cycles/time.
The amplitude of the response is . This has the same units as the response quantity.
The gain is . The gain is the factor that the input amplitude is multiplied by to get the amplitude of the response. It has the units needed to convert input units to output units.
The phase lag is . The phase lag has units of radians, i.e. it’s dimensionless.
The time lag is . This has units of time. It is the time that peak of the output lags behind that of the input.
The complex gain is . This is the factor that the complex input is multiplied by to get the complex output.
References
[edit]
^ a b c Miller, Haynes; Mattuck, Arthur (June 2004), Differential Equations, vol. IMSCP-MD5-9ca77abee86dc4bbaef9e2d6b157eaa9, pp. 50–56, hdl:1721.1/34888
^ a b c Wirkus, Stephen A.; Swift, Randal J.; Szypowski, Ryan S. (2016), A Course in Differential Equations with Boundary Value Problems, Second Edition, Textbooks in Mathematics (2nd ed.), Chapman and Hall/CRC, pp. 230–238, ISBN 978-1498736053
^ a b Charles L, Phillips (2007), Signals, Systems, And Transforms, Prentice Hall, pp. 112–122, ISBN 978-0-13-198923-8
^ a b Coddington, Earl A.; Carlson, Robert (1997), Linear Ordinary Differential Equations (PDF), pp. 3–80, ISBN 0-89871-388-9
^ Ralph P. Grimaldi (2000). "Nonhomogeneous Recurrence Relations". Section 3.3.3 of Handbook of Discrete and Combinatorial Mathematics. Kenneth H. Rosen, ed. CRC Press. ISBN 0-8493-0149-1.
^ a b Edwards, C. Henry; Penney, David E. (2008), ELEMENTARY DIFFERENTIAL EQUATIONS, Pearson Prentice Hall, pp. 100–193, ISBN 978-0-13-239730-8
External links
[edit]
Operators and the exponential response formula
Generalized exponential response formula
| v t e Differential equations |
| Classification | | | | --- | | Operations | Differential operator Notation for differentiation Ordinary Partial Differential-algebraic Integro-differential Fractional Linear Non-linear Holonomic | | Attributes of variables | Dependent and independent variables Homogeneous Nonhomogeneous Coupled Decoupled Order Degree Autonomous Exact differential equation On jet bundles | | Relation to processes | Difference (discrete analogue) Stochastic + Stochastic partial Delay | |
| Solutions | | | | --- | | Existence/uniqueness | Picard–Lindelöf theorem Peano existence theorem Carathéodory's existence theorem Cauchy–Kowalevski theorem | | Solution topics | Wronskian Phase portrait Phase space Lyapunov stability Asymptotic stability Exponential stability Rate of convergence Series solutions Integral solutions Numerical integration Dirac delta function | | Solution methods | Inspection Substitution Separation of variables Method of undetermined coefficients Variation of parameters Integrating factor Integral transforms Euler method Finite difference method Crank–Nicolson method Runge–Kutta methods Finite element method Finite volume method Galerkin method Perturbation theory | |
| Examples | List of named differential equations List of linear ordinary differential equations List of nonlinear ordinary differential equations List of nonlinear partial differential equations |
| Mathematicians | Isaac Newton Gottfried Wilhelm Leibniz Leonhard Euler Jacob Bernoulli Émile Picard Józef Maria Hoene-Wroński Ernst Lindelöf Rudolf Lipschitz Joseph-Louis Lagrange Augustin-Louis Cauchy John Crank Phyllis Nicolson Carl David Tolmé Runge Martin Kutta Sofya Kovalevskaya |
Retrieved from "
Category:
Ordinary differential equations
Hidden category:
CS1: long volume value
Exponential response formula
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Molecular Shapes & Valence Bond Theory
MO Theory: Bond Order
Molecular Shapes & Valence Bond Theory
MO Theory: Bond Order: Videos & Practice Problems
Video Lessons
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Topic summary
Bond order is a key concept in understanding molecular stability and bond strength, indicating the number of chemical bonds between a pair of atoms. A bond order greater than zero suggests a stable molecule, while a bond order of zero implies instability. The bond order can be determined by setting up a molecular orbital diagram and applying the bond order formula:
The molecular orbital diagram helps distribute electrons into bonding and antibonding orbitals to calculate this value. Additionally, bond order correlates with bond strength and length; as bond order increases, the bond strengthens and shortens. Visually, bond order can be inferred from Lewis structures: a single bond has a bond order of one, a double bond has a bond order of two, and a triple bond has a bond order of three, reflecting the number of shared electron pairs.
Molecular Orbital Diagrams can be used to determine the bond order of a molecule.
Bond Order
1
concept
Bond Order Calculation
Video duration:
1m
Play a video:
Bond Order Calculation Video Summary
Molecular orbital diagrams are essential tools for understanding the bond order of a molecule, which quantifies the number of electrons participating in bonds between two elements. A bond order greater than 0 indicates that the compound is stable and exists, while a bond order of 0 signifies instability and non-existence of the compound. As the bond order increases, both the stability and strength of the bond enhance. This relationship is crucial: stronger bonds correspond to shorter bond lengths, illustrating an inverse relationship between bond strength and bond length.
To calculate bond order, one must first construct a molecular orbital diagram. This involves distributing electrons among the bonding molecular orbitals and the antibonding molecular orbitals. The bond order can then be determined using the formula:
Bond Order =
where represents the number of bonding electrons and denotes the number of antibonding electrons. By filling in the molecular orbital diagram and applying this formula, one can accurately ascertain the bond order for any given molecule, providing insights into its stability and bond characteristics.
2
example
Bond Order Example
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Bond Order Example Video Summary
To determine the bond order of the nitroxide ion (NO-), we begin by constructing a molecular orbital diagram based on the less electronegative element, which is nitrogen. The molecular orbital configuration for nitrogen includes the following levels:
σ2s
σ2s
π2p
σ2p
π2p
σ2p
Nitrogen, being in group 5A, has 5 valence electrons, while oxygen, in group 6A, has 6 valence electrons. The additional negative charge from the ion indicates that there is one extra electron, leading to a total of 12 valence electrons (5 from nitrogen + 6 from oxygen + 1 from the negative charge).
Next, we fill the molecular orbitals with these 12 electrons. The filling order is as follows:
σ2s: 2 electrons
σ2s: 2 electrons
π2p: 4 electrons (2 in each π orbital)
σ2p: 2 electrons
π2p: 2 electrons (1 in each π orbital)
Now, we can calculate the bond order using the formula:
Bond Order (BO) = ½ (Number of bonding electrons - Number of antibonding electrons)
In this case, the bonding electrons are found in the σ2s, σ2p, and π2p orbitals, totaling 8 bonding electrons (2 from σ2s, 2 from σ2p, and 4 from π2p). The antibonding electrons are in the σ2s and π2p orbitals, totaling 4 antibonding electrons (2 from σ2s and 2 from π2p).
Substituting these values into the bond order formula gives:
BO = ½ (8 - 4) = ½ (4) = 2
Thus, the bond order for the nitroxide ion (NO-) is 2, indicating a double bond character between nitrogen and oxygen.
3
concept
Bond Order and Type of Bond
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Bond Order and Type of Bond Video Summary
Bond order is a crucial concept in understanding the strength and stability of bonds between two elements in a compound. It quantifies the number of chemical bonds formed between a pair of atoms. A visual representation of a molecule, such as a Lewis dot structure, allows us to easily determine the bond order by examining the connections between the elements.
For instance, if a molecule exhibits a single bond between two elements, the bond order is 1. In contrast, a double bond indicates a bond order of 2. Taking a specific example, when carbon is triple bonded to nitrogen, the bond order is 3 due to the presence of three shared electron pairs.
To calculate bond order more systematically, one can utilize a molecular orbital diagram. This involves identifying the number of bonding and antibonding electrons. The bond order can be calculated using the formula:
Bond Order = (Number of Bonding Electrons - Number of Antibonding Electrons) / 2
By applying this formula or simply analyzing the Lewis structure, one can effectively determine the bond order, which is essential for predicting the properties and reactivity of the molecule.
4
example
Bond Order and Type of Bond Example
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Bond Order and Type of Bond Example Video Summary
To determine the number of bonds connecting the nitrogen atoms in the N2²⁻ ion, we can utilize molecular orbital (MO) theory and calculate the bond order. In this case, nitrogen has five valence electrons, and since there are two nitrogen atoms, we start with a total of 10 valence electrons. Additionally, the N2²⁻ ion has 2 extra electrons, bringing the total to 12 electrons.
In the MO diagram for nitrogen, the relevant molecular orbitals are arranged as follows: σ2s, σ2s, π2p, σ2p, π2p, and σ2p.
We fill the molecular orbitals with the 12 electrons as follows:
σ2s: 2 electrons
σ2s: 0 electrons
π2p: 4 electrons (2 in each of the two degenerate π orbitals)
σ2p: 2 electrons
π2p: 0 electrons
σ2p: 0 electrons
This results in a total of 8 bonding electrons (from σ2s, π2p, and σ2p) and 4 antibonding electrons (from σ2s and π2p). The bond order (B.O.) can be calculated using the formula:
B.O. = ½ (Bonding Electrons - Antibonding Electrons) = ½ (8 - 4) = ½ (4) = 2.
This bond order of 2 indicates that there is a double bond between the two nitrogen atoms in the N2²⁻ ion. Thus, we conclude that the nitrogen atoms are connected by a double bond, which is consistent with the visualization of the ion.
Apply Molecular Orbital Theory to determine the bond order of HHe+ion.
2
4
3
1
Problem
Apply molecular orbital theory to predict which species has the strongest bond. a) O2 b) O2– c) O2+ d) All the bonds are equivalent
A
O2
B
O2-
C
O2+
D
All the bonds are equivalent
7
Problem
Using Molecular Orbital Theory, answer the following questions dealing with carbon mononitride, CN.
A
CN is diagmagnetic
B
CN– is paramagnetic
C
If an electron is removed to give CN+, the bond length decreases.
D
The π2p orbital is the highest energy orbital containing an electron in CN.
E
If an electron is added to give CN–, the bond order increases.
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Bond order is a way to determine the stability and strength of a chemical bond. It is calculated by taking the difference between the number of bonding electrons and the number of antibonding electrons, then dividing that number by two. Here's a step-by-step guide:
Draw the Lewis structure for the molecule.
Determine the number of bonding electrons (these are typically the electrons shared between atoms in a bond).
Determine the number of antibonding electrons (these are electrons in antibonding orbitals, often denoted by an asterisk, like σ or Π).
Subtract the number of antibonding electrons from the number of bonding electrons.
Divide the result by two.
The formula for bond order (BO) is:
For example, in a diatomic nitrogen (N≡N) molecule, there are three bonds (six bonding electrons) and no antibonding electrons, so the bond order is 3. Higher bond orders indicate stronger, shorter bonds between atoms.
Bond order is a concept in molecular chemistry that represents the number of chemical bonds between a pair of atoms. In essence, it indicates the stability of a bond. For a simple molecule like diatomic nitrogen (), where two nitrogen atoms are connected by a triple bond, the bond order is three. This is calculated by taking the number of bonding electrons (in this case, six for the three bonds) and subtracting the number of anti-bonding electrons (which is zero for ), then dividing by two.
Bond order can also be fractional when considering resonance structures. For example, in benzene (), each carbon-carbon bond has a bond order of 1.5 due to the delocalization of electrons across the ring structure.
Higher bond orders generally mean stronger, shorter bonds. A bond order of zero implies that no bond exists between the atoms, and typically, a molecule with such a bond order is not stable. Understanding bond order is crucial for predicting molecular stability, reactivity, and the strength of bonds within different molecules.
Bond order refers to the number of chemical bonds between a pair of atoms in a molecule. It is a way of expressing the stability and strength of the bond. In simple terms, the higher the bond order, the stronger and more stable the bond between the atoms. For example, a single bond has a bond order of 1, a double bond has a bond order of 2, and a triple bond has a bond order of 3.
To calculate the bond order in molecular orbital theory, you subtract the number of electrons in antibonding orbitals from the number of electrons in bonding orbitals and then divide by two. The formula looks like this:
This calculation can help predict the stability of molecules and the lengths of the bonds between atoms. A bond order of zero implies that the bond is not stable and is unlikely to exist under normal conditions.
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Open AccessArticle
Geometrically Constructed Family of the Simple Fixed Point Iteration Method
by
Vinay Kanwar
Vinay Kanwar
SciProfiles Scilit Preprints.org Google Scholar
1,
Puneet Sharma
Puneet Sharma
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Ioannis K. Argyros
Ioannis K. Argyros
SciProfiles Scilit Preprints.org Google Scholar
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Ramandeep Behl
Ramandeep Behl
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Christopher Argyros
Christopher Argyros
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Ali Ahmadian
Ali Ahmadian
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Mehdi Salimi
Mehdi Salimi
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1
University Institute of Engineering and Technology, Panjab University, Chandigarh 160014, India
2
Department of Mathematics, Goswami Ganesh Dutta Sanatan Dharma College, Chandigarh 160030, India
3
Department of Mathematics Sciences, Cameron University, Lawton, OK 73505, USA
4
Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia
5
Department of Computer Science, University of Oklahoma, Norman, OK 73071, USA
6
Institute of IR 4.0, The National University of Malaysia, Bangi 43600, UKM, Malaysia
7
Department of Mathematics & Statistics, McMaster University, Hamilton, ON L8S 4K1, Canada
8
Center for Dynamics and Institute for Analysis, Faculty of Mathematics, Technische Universität Dresden, 01062 Dresden, Germany
Author to whom correspondence should be addressed.
Mathematics 2021, 9(6), 694;
Submission received: 8 February 2021 / Revised: 10 March 2021 / Accepted: 19 March 2021 / Published: 23 March 2021
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by a linear approximation.
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Abstract
This study presents a new one-parameter family of the well-known fixed point iteration method for solving nonlinear equations numerically. The proposed family is derived by implementing approximation through a straight line. The presence of an arbitrary parameter in the proposed family improves convergence characteristic of the simple fixed point iteration as it has a wider domain of convergence. Furthermore, we propose many two-step predictor–corrector iterative schemes for finding fixed points, which inherit the advantages of the proposed fixed point iterative schemes. Finally, several examples are given to further illustrate their efficiency.
Keywords:
fixed point method; nonlinear equation; banach space; order of convergence
MSC:
47H99; 49M15; 65G99; 65H10
1. Introduction
The fixed point iteration is probably the simplest and most important root-finding algorithm in numerical analysis [1,2]. The fixed point methods and fixed point theorems have many applications in mathematics and engineering. One way to study numerical ordinary differential solvers and Runge–Kutta methods is to convert them as fixed point iterations. The well-known Newton’s method [2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17] is also a special case of an iteration method. The fixed point theory has been voluminously used as a tool to find the solution of function-differential equations. Furthermore, the fixed point problems are equivalent to root-finding problems and sometimes easier to analyze while posing some strange and cute problems by themselves.
Suppose that we wish to find the approximate solution of the nonlinear equationwhere is a sufficiently differentiable function with simple zeros.
This can be rewritten to obtain an equation of the formin such a way that any solution of (2), which is a fixed point, is a root of the original Equation (1). Root-finding problems and fixed-point problems are equivalent classes in the following sense:
Geometrically, the fixed point occurs where the graph of intersects the graph of the straight line . Starting from a suitable approximation and consider that the recursive process:is called a fixed point iteration method. This method is locally linearly convergent if for all .
2. Geometric Derivation of the Family
Assume that Equation (2) has a fixed point at . Letrepresent the graph of the function .
Let be an initial guess to the required fixed point and be the corresponding point on the graph of the function . The idea is to approximate nonlinear function by a linear approximation. Therefore, we assume thatbe a linear approximation to the curve , where is the slope of the straight line. The expression (5) can be rewritten as , and this line is passing through the points and . More details can be found in the following Figure 1:
The point of intersection of (5) with the straight line will be a required fixed point and let be this point of intersection. Therefore, at the point of intersection, the expression (5) yields
Without loss of generality, the general form of the above expression (6) can be written as follows:
Next, we want to demonstrate the convergence order of the proposed iterative scheme (7). Therefore, we rewrite the expression (7) in the following way:or we can say that . If is a fixed point of , then
So, we conclude that is a root of .
Theorem 1.
Let and be continuous functions on the interval . In addition, we assume that and for all , then (i) has a unique solution ; (ii) for any initial guess , the iteration will converge to α.
Proof.
First of all, we will prove the first part. Since exists in the interval , therefore, this implies that exists in the interval . For any two points , we havewhich yields further
Let us suppose that there are two solutions and of in the interval . So, we have
From (8), we further have
If , then which contradicts the fact that . Therefore, we have
Hence, has a unique solution in [a, b].
Next, we move to the second part.
If , then . This further implies that
Therefore, and hence .
Repeating the above process inductively, one gets .
Furthermore, one can have
Continuing inductively
Hence, the sequence of converges to . □
Theorem 2.
Let be an analytic function in the region containing the fixed point . In addition, we assume that initial guess is sufficiently close to the required fixed point for guaranteed convergence. Then, the proposed scheme (7) has at least linear convergence
In the case , we obtainsince , then the scheme (7) reaches at least the second order of convergence.
Proof.
Suppose . We can writeby Taylor’s expansion in the neighborhood of fixed point “”.
Therefore, one gets
As , we get and .
Substituting these values in (10), one can have
Furthermore, if , then .
This implies that scheme (7) has at least second-order convergence. □
Special Cases
Here, we shall consider the role of the parameter and derive the various following formulas as follows:
1.
: For , Formula (7) corresponds to the classical fixed point method .
2.
: For , where , Formula (7) corresponds to the following well-known Schaefer’s iteration scheme
3.
: For , where is a real sequence in , Formula (7) corresponds to the following well-known Mann’s iteration
4.
: By inserting , in scheme (7), one achieves the following well-known Kranselski’s iteration denoted by for the computational results (see also more recent work on this iteration in the book ).
Similarly, we can derive several other formulas by taking different specific values of m. Furthermore, we proposed the following new schemes on the basis of some standard means of two quantities and of same signs:
5.
: Geometric mean-based fixed point formula is given by
6.
: Harmonic mean-based fixed point formula is defined by
7.
: Centroidal mean-based fixed point formula is mentioned as follows:
8.
: The following fixed point formula based on the Heronian mean is defined as
9.
: The fixed point formula based on Contra-harmonic is depicted as follows:
Remark 1.
Geometric mean-based fixed point formula and Heronian mean formula are applicable for finding positive fixed points only.
3. Two-Step Iterative Schemes
In this section, we present a new two-step predictor–corrector iterative schemes using the modified fixed point methods as predictor. There are several two-point [1,22,23] and multi-point [24,25] iterative schemes in the literature for finding the fixed points. Here, we mention some of them as follows:
1.
: Ishikawa has proposed the following iterative scheme:where and are sequences of positive numbers in as a generalization of the Mann iteration scheme. We denote this method as for the computational work and choose .
2.
: Agarwal et al. have proposed the following iteration scheme defined aswhere and are sequences of positive numbers in . We call this scheme by for the computational work and consider . For , it reduces to the well-known Mann iteration scheme.
3.
: Thianwan defined the following two-step iteration scheme aswhere and are sequences of positive numbers in . We denote this method as for the computational work and choose . This scheme is also known as modification of Mann’s method.
Modified Schemes
These elementary schemes allow us to propose the following iterative schemes with any of the proposed methods as the first predictor step and these existing methods as the second step. For the sake of simplicity, we consider some of the special cases as a predictor part. Therefore, we have the following modified schemes depicted in the Table 1.
4. Numerical Examples
The theoretical results developed in the previous sections are tested in this section. We choose our methods by substituting , , and in the proposed scheme (7), denoted by , and , respectively. In addition, we select methods and from special cases and , respectively.
In order to check the effectiveness of our results, we consider five different types of nonlinear problems which are illustrated in examples (1)–(5). In the Table 2, we compared them with classical fixed point method. In addition, we contrast our method to the existing Ishikawa and Agarwal methods, and the results are mentioned in the Table 3 and Table 4, respectively. Finally, we compared them with classical Mann’s and Thianwan methods and computational depicted in the Table 5. In all the tables, we mentioned the results after twelve iterations (i.e., ) with .
Additionally, we obtain the computational order of convergence by adopting the pursuing techniquesor the approximate computational order of convergence (ACOC)
Computations are performed with the package with multiple precision arithmetic. The stands for .
Example 1.
Let us consider the following standard test problem
The corresponding fixed point iterative method is given as follows:
The required zero of expression (16) and fixed point for (17) is with initial guess .
Example 2.
We choose the following expression for the comparison with other different fixed point methods
We can easily obtained the following fixed point iterative method based on expression (18)
The required zero of expression (18) and fixed point for (19) is with initial guess .
Example 3.
Here, we assume the following expression
Based on the expression (20), we have the following fixed point iterative method:
The required zero of expression (20) and fixed point for (21) is . We select as the initial guess for comparison.
Example 4.
Assume another test problem as follows
Corresponding to expression (22), we have
as the fixed point iterative method. The required zero of expression (22) and fixed point for (23) is . We assume the starting point for contrast.
Example 5.
Here, we assume another expression
We have the following expression for the fixed point method
The required zero of expression (24) and fixed point for (25) is . We consider as the initial guess for comparison.
5. Role of the Parameter ‘m’
The presence of the arbitrary slope in the proposed family has the following characteristics:
1.
: Since implies that . Therefore, the parameter ‘’ ensures that the fixed point divides the interval between and internally in the ratio or , otherwise, there will be an external division and hence, .
2.
: Since . As is the sufficient condition for the convergence of modified fixed point method, then we have
This further implies that
This is the interval of convergence of our proposed scheme (7). As , so (26) represents a wider domain of convergence in contrast to the classical fixed point method . In particular for (arithmetic mean), (26) gives the following interval of convergence as
Therefore, the arithmetic mean formula has a bigger interval of convergence as compared to simple fixed point method.
Remark 2.
For , we have different ways to choose ; however, we have to select in such a way that the fixed point iteration method converges to its fixed point. We shall illustrate it by taking the following two examples:
Example 6.
We choose the following expression for the comparison of simple fixed point method with the modified fixed point method, namely, arithmetic mean formula
Corresponding to expression (27), one has
One of the required zero of expression (27) and fixed point for (28) is with the initial guess . For , the fixed point method diverges for the interval . Here, and implies . This further implies , which violates the condition of for all . On the other hand, the interval of convergence for arithmetic mean formula is and clearly lies with in this interval. For , Formula (7) becomes . This further gives
The modified arithmetic mean fixed point method converges to for the initial guess .
Example 7.
Let us consider the general square root finding problem by fixed point methods. We wish to compute square roots of . This is equivalent to find the roots of . For example, let . Therefore, the corresponding function becomes
Consider the following two possible rearrangements of as
1.
2.
One of the required zero of expression (29) and fixed point for the above two sequences is with initial guess . The first considered sequence diverges as , and the second one converges for the simple fixed point method as , for all . Let us discuss the first sequence further. We have
The interval of convergence for arithmetic mean formula is and clearly lies within this interval. For , Formula (7) becomes .
This further implies
The modified arithmetic mean fixed point method converges to for the initial guess , since for all .
Remark 3.
We can do better sometimes with the selection of initial points or the convergence rate or order, if we consider iteration functions using similar information. As an example, consider Newton’s method defined for all by
Then, the celebrated Newton–Kantorovich semi-local convergence criterion [13,14] is given by
where
and l is the Lipschitz constant in the condition
for all for some D. Then, as an example in the case of Example (7), (ii) Newton’s method (30) coincides with the modified arithmetic mean fixed point method, but since , and , condition (31) is satisfied for , which includes S, and if then, , so the convergence is quadratic faster than for the given (only linear) in the case of modified arithmetic mean.
6. Conclusions
Motivated by geometrical considerations, we developed a one-parameter class of fixed point iteration methods for generating a sequence approximating fixed points of nonlinear equations. These methods are more specialist than a number of earlier popular methods. Sufficient convergence criteria have been provided as well as the convergence order. Numerical examples further demonstrate the efficiency as well as the superiority of the new methods over earlier ones using similar convergence information. The convergence order of Theorem 2 is confirmed in Table 2 by using COC or ACOC. These schemes can also be extended for finding the fixed points of nonlinear systems.
Author Contributions
V.K., P.S., I.K.A. and R.B.: conceptualization; methodology; validation; writing—original draft preparation; writing—review and editing. C.A., A.A. and M.S.: review and editing. All authors have read and agreed to the published version of the manuscript.
Funding
This research received no external funding.
Institutional Review Board Statement
Not applicable.
Informed Consent Statement
Not applicable.
Data Availability Statement
Not applicable.
Conflicts of Interest
The authors declare no conflict of interest.
References
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Figure 1. The graph of approximate nonlinear function by a linear approximation.
Figure 1. The graph of approximate nonlinear function by a linear approximation.
Table 1. Some modified schemes based on Ishikawa’s, Agarwal and Thianwan as corrector, respectively.
Table 1. Some modified schemes based on Ishikawa’s, Agarwal and Thianwan as corrector, respectively.
| Predictor | Ishikawa’s | Agarwal | Thianwan |
--- --- |
| Corrector | Corrector | Corrector |
| | | | |
| called by | | | |
| | | | |
| known by | | | |
| | | | |
| denoted by | | | |
| | | | |
| called by | | | |
| | | | |
| known by | | | |
Table 2. Comparison of different fixed point methods on Examples (1)–(5) with .
Table 2. Comparison of different fixed point methods on Examples (1)–(5) with .
| Examples | E.C. | FIM | KM | GM | HM | OM1 | OM2 | OM3 | OM4 |
--- --- --- --- --- |
| R.E. |
| |
| (1) | | | | | | | | | |
| | | | | | | | | |
| | | | | | | | | | |
| (2) | | | | | | | | | |
| | | | | | | | | |
| | | | | | | | | | |
| (3) | | | | | | | | | |
| | | | | | | | | |
| | | | | | | | | | |
| (4) | | | | | | | | | |
| | | | | | | | | |
| | | | | | | | | | |
| (5) | | | | | | | | | |
| | | | | | | | | |
| | | | | | | | | | |
FIM, E.C. and R.E. stand for classical fixed point method, errors between two consecutive iterations and residual errors in the corresponding function by using the obtained fixed point, respectively.
Table 3. Comparison of our methods with existing Ishikawa method based on number of iterations.
Table 3. Comparison of our methods with existing Ishikawa method based on number of iterations.
| Examples | E.C. | | IGM | IHM | IOM1 | IOM2 | IOM3 |
--- --- --- --- |
| R.E. |
| (1) | | | | | | | |
| | | | | | | |
| (2) | | | | | | | |
| | | | | | | |
| (1) | | | | | | | |
| | | | | | | |
| (4) | | | | | | | |
| | | | | | | |
| (5) | | | | | | | |
| | | | | | | |
IS stands for Ishikawa’s scheme. From the above numerical results, we concluded that our methods and , have smaller absolute residual errors and smaller errors difference between two iterations as compared to the original Ishikawa method in all the examples. On the other hand, our methods, and , have similar computational results to Ishikawa method.
Table 4. Comparison of our methods with standard Agarwal scheme after number of iterations.
Table 4. Comparison of our methods with standard Agarwal scheme after number of iterations.
| Examples | E.C. | AS | AGM | AHM | AOM1 | AOM2 | AOM3 |
--- --- --- --- |
| R.E. |
| (1) | | | | | | | |
| | | | | | | |
| (2) | | | | | | | |
| | | | | | | |
| (3) | | | | | | | |
| | | | | | | |
| (4) | | | | | | | |
| | | | | | | |
| (5) | | | | | | | |
| | | | | | | |
AS stands for Agarwal’s scheme. We deduced from the obtained numerical results that our methods, , and , have better numerical results as compared to the classical Agarwal scheme in examples 1, 2 and 5. In addition, our methods have similar numerical results to the Agarwal method in the case of examples 3 and 4.
Table 5. Comparison of our methods with classical Mann’s and Thianwan method after number of iterations.
Table 5. Comparison of our methods with classical Mann’s and Thianwan method after number of iterations.
| Examples | E.C. | MS | TS | TGM | THM | TOM1 | TOM2 | TOM3 |
--- --- --- ---
| R.E. |
| (1) | | | | | | | | |
| | | | | | | | |
| (2) | | | | | | | | |
| | | | | | | | |
| (3) | | | | | | | | |
| | | | | | | | |
| (4) | | | | | | | | |
| | | | | | | | |
| (5) | | | | | | | | |
| | | | | | | | |
MS and TS stand for Mann’s and Thianwan’s schemes, respectively. On the basis of computational results, we inferred that our methods, and , have better performance in the form of absolute residual error and smaller error difference between two iterations as compared to the classical Mann’s and Thianwan schemes.
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Kanwar, V.; Sharma, P.; Argyros, I.K.; Behl, R.; Argyros, C.; Ahmadian, A.; Salimi, M. Geometrically Constructed Family of the Simple Fixed Point Iteration Method. Mathematics 2021, 9, 694.
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Kanwar V, Sharma P, Argyros IK, Behl R, Argyros C, Ahmadian A, Salimi M. Geometrically Constructed Family of the Simple Fixed Point Iteration Method. Mathematics. 2021; 9(6):694.
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Kanwar, Vinay, Puneet Sharma, Ioannis K. Argyros, Ramandeep Behl, Christopher Argyros, Ali Ahmadian, and Mehdi Salimi. 2021. "Geometrically Constructed Family of the Simple Fixed Point Iteration Method" Mathematics 9, no. 6: 694.
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Kanwar, V., Sharma, P., Argyros, I. K., Behl, R., Argyros, C., Ahmadian, A., & Salimi, M. (2021). Geometrically Constructed Family of the Simple Fixed Point Iteration Method. Mathematics, 9(6), 694.
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Kanwar, V.; Sharma, P.; Argyros, I.K.; Behl, R.; Argyros, C.; Ahmadian, A.; Salimi, M. Geometrically Constructed Family of the Simple Fixed Point Iteration Method. Mathematics 2021, 9, 694.
AMA Style
Kanwar V, Sharma P, Argyros IK, Behl R, Argyros C, Ahmadian A, Salimi M. Geometrically Constructed Family of the Simple Fixed Point Iteration Method. Mathematics. 2021; 9(6):694.
Chicago/Turabian Style
Kanwar, Vinay, Puneet Sharma, Ioannis K. Argyros, Ramandeep Behl, Christopher Argyros, Ali Ahmadian, and Mehdi Salimi. 2021. "Geometrically Constructed Family of the Simple Fixed Point Iteration Method" Mathematics 9, no. 6: 694.
APA Style
Kanwar, V., Sharma, P., Argyros, I. K., Behl, R., Argyros, C., Ahmadian, A., & Salimi, M. (2021). Geometrically Constructed Family of the Simple Fixed Point Iteration Method. Mathematics, 9(6), 694.
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1143 | https://math.stackexchange.com/questions/2386605/how-to-negate-predicates | Stack Exchange Network
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How to negate predicates?
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0
$\begingroup$
How can I go about negating predicates? It's asking me to shift a negation in as far inside the predicate as possible.
$$\forall x ((x \ge 100) \lor (x < 100))$$
I am quite new to discrete mathematics so would greatly appreciate a walkthrough. Thanks!
discrete-mathematics
propositional-calculus
predicate-logic
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edited Aug 8, 2017 at 12:15
Rodrigo de Azevedo
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Oliver KOliver K
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$\endgroup$
1
2
$\begingroup$ $ \neg \forall x.P(x) \iff \exists x.\neg P(x)$ $\endgroup$
mrp
– mrp
2017-08-08 12:04:43 +00:00
Commented Aug 8, 2017 at 12:04
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2 Answers 2
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2
$\begingroup$
hint
$$\forall \to \;\;\exists $$ $$\ge \to \;\;<$$ $$\lor \to \;\;\land $$
so the negation is
$$\exists x \;: x <100 \;\; \land \;\; x\ge 100$$
remark
Your proposition is always true (tautology), thus its negation is always false (contradiction).
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edited Aug 8, 2017 at 12:21
answered Aug 8, 2017 at 12:11
hamam_Abdallahhamam_Abdallah
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$\endgroup$
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$\begingroup$
If you jut want the symbol which represents logical negation, you can visit Wikipedia's symbology page and see that you can use either
$$!$$
Or
$$¬$$
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answered Aug 8, 2017 at 12:14
user3141592user3141592
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compound Summary
Carbon Monoxide
PubChem CID
281
Structure
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Molecular Formula
CO
Synonyms
carbon monoxide
carbon monooxide
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Computed by PubChem 2.2 (PubChem release 2025.04.14)
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1010305
Associated Article
Vegard, L. Struktur und Leuchtfaehigkeit von festem Kohlenoxyd.. Zeitschrift fuer Physik 1930;61:185-190. DOI: 10.1007/BF01339658
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Hermann-Mauguin space group symbol
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Hall space group symbol
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Space group number
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a
5.63 Å
b
5.63 Å
c
5.63 Å
α
90 °
β
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0.333333333333333
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2.1.1 IUPAC Name
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Computed by Lexichem TK 2.7.0 (PubChem release 2025.04.14)
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2.1.2 InChI
InChI=1S/CO/c1-2
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2.1.3 InChIKey
UGFAIRIUMAVXCW-UHFFFAOYSA-N
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PubChem
2.2 Molecular Formula
CO
Computed by PubChem 2.2 (PubChem release 2025.04.14)
Australian Industrial Chemicals Introduction Scheme (AICIS); CAMEO Chemicals; ILO-WHO International Chemical Safety Cards (ICSCs); PubChem
2.3 Other Identifiers
2.3.1 CAS
630-08-0
Australian Industrial Chemicals Introduction Scheme (AICIS); CAMEO Chemicals; CAS Common Chemistry; ChemIDplus; DrugBank; EPA Acute Exposure Guideline Levels (AEGLs); EPA Chemical Data Reporting (CDR); EPA Chemicals under the TSCA; EPA DSSTox; European Chemicals Agency (ECHA); FDA Global Substance Registration System (GSRS); Hazardous Substances Data Bank (HSDB); Human Metabolome Database (HMDB); ILO-WHO International Chemical Safety Cards (ICSCs); New Zealand Environmental Protection Authority (EPA); NJDOH RTK Hazardous Substance List; Occupational Safety and Health Administration (OSHA); Risk Assessment Information System (RAIS); The National Institute for Occupational Safety and Health (NIOSH); USDA APHIS Chemical Effects Database
35907-63-2
European Chemicals Agency (ECHA)
2.3.2 Related CAS
147965-81-9
Compound: Carbon monoxide, dimer
CAS Common Chemistry
2.3.3 Deprecated CAS
153929-54-5, 155399-52-3, 162342-48-5, 167416-30-0, 18421-60-8, 192819-80-0, 1976049-23-6, 201230-82-2, 724693-69-0, 740776-49-2, 807306-74-7, 82063-46-5
ChemIDplus; EPA Chemicals under the TSCA
153929-54-5, 155399-52-3, 162342-48-5, 167416-30-0, 18421-60-8, 192819-80-0, 724693-69-0, 740776-49-2, 807306-74-7
EPA DSSTox
2.3.4 European Community (EC) Number
211-128-3
European Chemicals Agency (ECHA)
685-227-9
European Chemicals Agency (ECHA)
2.3.5 UNII
7U1EE4V452
FDA Global Substance Registration System (GSRS)
2.3.6 UN Number
1016 (CARBON MONOXIDE)
CAMEO Chemicals
9202 (CARBON MONOXIDE, REFRIGERATED LIQUID (CRYOGENIC LIQUID))
CAMEO Chemicals; Emergency Response Guidebook (ERG)
1016 (Carbon monoxide, compressed)
Emergency Response Guidebook (ERG)
1016
ILO-WHO International Chemical Safety Cards (ICSCs); The National Institute for Occupational Safety and Health (NIOSH)
9202
The National Institute for Occupational Safety and Health (NIOSH)
2.3.7 ChEBI ID
CHEBI:17245
ChEBI
2.3.8 ChEMBL ID
CHEMBL1231840
ChEMBL
2.3.9 DrugBank ID
DB11588
DrugBank
2.3.10 DSSTox Substance ID
DTXSID5027273
EPA DSSTox
2.3.11 HMDB ID
HMDB0001361
Human Metabolome Database (HMDB)
2.3.12 ICSC Number
0023
ILO-WHO International Chemical Safety Cards (ICSCs)
2.3.13 KEGG ID
C00237
KEGG
D09706
KEGG
2.3.14 Metabolomics Workbench ID
50833
Metabolomics Workbench
2.3.15 NCI Thesaurus Code
C76742
NCI Thesaurus (NCIt)
2.3.16 Nikkaji Number
J1.401.546I
Japan Chemical Substance Dictionary (Nikkaji)
2.3.17 RTECS Number
FG3500000
The National Institute for Occupational Safety and Health (NIOSH)
2.3.18 Wikidata
Q2025
Wikidata
2.3.19 Wikipedia
Carbon monoxide
Wikipedia
2.4 Synonyms
2.4.1 MeSH Entry Terms
Carbon Monoxide
Monoxide, Carbon
Medical Subject Headings (MeSH)
2.4.2 Depositor-Supplied Synonyms
carbon monoxide
carbon monooxide
Carbonic oxide
Carbon oxide (CO)
carbon(II) oxide
630-08-0
Exhaust gas
Flue gas
monoxide
Kohlenmonoxid
Koolmonoxyde
Kohlenoxyd
Wegla tlenek
Oxyde de carbone
Carbone (oxyde de)
Carboneum oxygenisatum
Carbonio (ossido di)
HSDB 903
EINECS 211-128-3
UNII-7U1EE4V452
CHEBI:17245
7U1EE4V452
Carbon monoxide, compressed
[CO]
DTXSID5027273
EC 211-128-3
(CO)
CARBON MONOXIDE (MART.)
CARBON MONOXIDE [MART.]
CARBON MONOXIDE (EP MONOGRAPH)
CARBON MONOXIDE [EP MONOGRAPH]
Monoxide, Carbon
Kohlenoxid
Carbon monoxide (DOT)
DTXCID407273
211-128-3
35907-63-2
CARBON MONOXIDE (13C; 18O)
carbon monoxide (acgih:osha)
carbonoxide
Carbon monoxide [USAN]
Carbon monoxide (USAN)
Kohlenoxyd [German]
Koolmonoxyde [Dutch]
Kohlenmonoxid [German]
Wegla tlenek [Polish]
Oxyde de carbone [French]
C#O
Carbone (oxyde de) [French]
Carbonio (ossido di) [Italian]
NA9202
UN1016
.carbon monoxide
Carbon monoxide 10% by volume or more
CARBON MONOXIDE [MI]
CARBON MONOXIDE [HSDB]
CARBON MONOXIDE [VANDF]
CHEMBL1231840
CARBON MONOXIDE [WHO-DD]
c0369
CARBONEUM OXYGENISATUM [HPUS]
DB11588
NS00075458
Q2025
C00237
D09706
Carbon monoxide, compressed [UN1016] [Poison gas]
Carbon monoxide, refrigerated liquid (cryogenic liquid)
Carbon monoxide, refrigerated liquid (cryogenic liquid) [NA9202] [Poison gas]
PubChem
3 Chemical and Physical Properties
3.1 Computed Properties
Property Name
Property Value
Reference
Property Name
Molecular Weight
Property Value
28.010 g/mol
Reference
Computed by PubChem 2.2 (PubChem release 2025.04.14)
Property Name
XLogP3-AA
Property Value
0.7
Reference
Computed by XLogP3 3.0 (PubChem release 2025.04.14)
Property Name
Hydrogen Bond Donor Count
Property Value
0
Reference
Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14)
Property Name
Hydrogen Bond Acceptor Count
Property Value
1
Reference
Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14)
Property Name
Rotatable Bond Count
Property Value
0
Reference
Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14)
Property Name
Exact Mass
Property Value
27.994914619 Da
Reference
Computed by PubChem 2.2 (PubChem release 2025.04.14)
Property Name
Monoisotopic Mass
Property Value
27.994914619 Da
Reference
Computed by PubChem 2.2 (PubChem release 2025.04.14)
Property Name
Topological Polar Surface Area
Property Value
1 Ų
Reference
Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14)
Property Name
Heavy Atom Count
Property Value
2
Reference
Computed by PubChem
Property Name
Formal Charge
Property Value
0
Reference
Computed by PubChem
Property Name
Complexity
Property Value
10
Reference
Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14)
Property Name
Isotope Atom Count
Property Value
0
Reference
Computed by PubChem
Property Name
Defined Atom Stereocenter Count
Property Value
0
Reference
Computed by PubChem
Property Name
Undefined Atom Stereocenter Count
Property Value
0
Reference
Computed by PubChem
Property Name
Defined Bond Stereocenter Count
Property Value
0
Reference
Computed by PubChem
Property Name
Undefined Bond Stereocenter Count
Property Value
0
Reference
Computed by PubChem
Property Name
Covalently-Bonded Unit Count
Property Value
1
Reference
Computed by PubChem
Property Name
Compound Is Canonicalized
Property Value
Yes
Reference
Computed by PubChem (release 2025.04.14)
PubChem
3.2 Experimental Properties
3.2.1 Physical Description
Carbon monoxide is a colorless, odorless gas. Prolonged exposure to carbon monoxide rich atmospheres may be fatal. It is easily ignited. It is just lighter than air and a flame can flash back to the source of leak very easily. Under prolonged exposure to fire or intense heat the containers may violently rupture and rocket.
CAMEO Chemicals
Carbon monoxide, refrigerated liquid (cryogenic liquid) appears as a colorless cryogenic liquid. Prolonged exposure to carbon monoxide rich atmospheres may be fatal. Contact with the liquid can cause severe frostbite. Less dense than air. Easily ignited and a flame can flash back to the source of a leak very easily. Burns with a violet flame. Under prolonged exposure to fire or intense heat the containers may rupture violently and rocket. It is used in organic synthesis, metallurgy, and a fuel.
CAMEO Chemicals
NKRA; Gas or Vapor
EPA Chemical Data Reporting (CDR)
Colorless, odorless gas. [Note: Shipped as a nonliquefied or liquefied compressed gas.] [NIOSH]
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
Liquid
Human Metabolome Database (HMDB)
ODOURLESS TASTELESS COLOURLESS COMPRESSED GAS.
ILO-WHO International Chemical Safety Cards (ICSCs)
Colorless, odorless gas.
Occupational Safety and Health Administration (OSHA)
Colorless, odorless gas. [Note: Shipped as a nonliquefied or liquefied compressed gas.]
The National Institute for Occupational Safety and Health (NIOSH)
3.2.2 Color / Form
Colorless gas [Note: Shipped as a nonliquefied or liquefied compressed gas].
NIOSH. NIOSH Pocket Guide to Chemical Hazards & Other Databases CD-ROM. Department of Health & Human Services, Centers for Disease Prevention & Control. National Institute for Occupational Safety & Health. DHHS (NIOSH) Publication No. 2005-151 (2005)
Hazardous Substances Data Bank (HSDB)
3.2.3 Odor
Odorless
NIOSH. NIOSH Pocket Guide to Chemical Hazards & Other Databases CD-ROM. Department of Health & Human Services, Centers for Disease Prevention & Control. National Institute for Occupational Safety & Health. DHHS (NIOSH) Publication No. 2005-151 (2005)
Hazardous Substances Data Bank (HSDB)
3.2.4 Taste
Tasteless
O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Whitehouse Station, NJ: Merck and Co., Inc., 2006., p. 294
Hazardous Substances Data Bank (HSDB)
3.2.5 Boiling Point
-312.7 °F at 760 mmHg (USCG, 1999)
U.S. Coast Guard. 1999. Chemical Hazard Response Information System (CHRIS) - Hazardous Chemical Data. Commandant Instruction 16465.12C. Washington, D.C.: U.S. Government Printing Office.
CAMEO Chemicals
-313 °F at 760 mmHg (NIOSH, 2024)
CAMEO Chemicals
-191.5
MSDS
DrugBank
-191.5 °C
Lide, D.R. CRC Handbook of Chemistry and Physics 88TH Edition 2007-2008. CRC Press, Taylor & Francis, Boca Raton, FL 2007, p. 3-88
Hazardous Substances Data Bank (HSDB)
-191 °C
ILO-WHO International Chemical Safety Cards (ICSCs)
-313 °F
Occupational Safety and Health Administration (OSHA); The National Institute for Occupational Safety and Health (NIOSH)
3.2.6 Melting Point
-326 °F (USCG, 1999)
U.S. Coast Guard. 1999. Chemical Hazard Response Information System (CHRIS) - Hazardous Chemical Data. Commandant Instruction 16465.12C. Washington, D.C.: U.S. Government Printing Office.
CAMEO Chemicals
-337 °F (NIOSH, 2024)
CAMEO Chemicals
-205.02 °C
Lide, D.R. CRC Handbook of Chemistry and Physics 88TH Edition 2007-2008. CRC Press, Taylor & Francis, Boca Raton, FL 2007, p. 3-88
Hazardous Substances Data Bank (HSDB)
-56.5 °C
Human Metabolome Database (HMDB)
-205 °C
ILO-WHO International Chemical Safety Cards (ICSCs)
-337 °F
Occupational Safety and Health Administration (OSHA); The National Institute for Occupational Safety and Health (NIOSH)
3.2.7 Flash Point
Flammable gas
ILO-WHO International Chemical Safety Cards (ICSCs)
NA (Gas)
The National Institute for Occupational Safety and Health (NIOSH)
3.2.8 Solubility
2 % (NIOSH, 2024)
CAMEO Chemicals
Soluble in benzene
O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Whitehouse Station, NJ: Merck and Co., Inc., 2006., p. 3-88
Hazardous Substances Data Bank (HSDB)
Freely absorbed by a concentrated solution of cuprous chloride in hydrochloric acid or ammonium hydroxide; Appreciably soluble in some organic solvents, such as ethyl acetate, chloroform, acetic acid; solubility in methanol and ethanol about 7 times as great as in water.
O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Whitehouse Station, NJ: Merck and Co., Inc., 2006., p. 249
Hazardous Substances Data Bank (HSDB)
Sparingly soluble in water: 3.3 ml/100 ml at 0 °C, 2.3 ml/100 ml at 20 °C
O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Whitehouse Station, NJ: Merck and Co., Inc., 2006., p. 249
Hazardous Substances Data Bank (HSDB)
1.48 mg/mL at 25 °C
Human Metabolome Database (HMDB)
Solubility in water, ml/100ml at 20 °C: 2.3 (sparingly soluble)
ILO-WHO International Chemical Safety Cards (ICSCs)
2%
The National Institute for Occupational Safety and Health (NIOSH)
3.2.9 Density
0.791 at -312.7 °F (USCG, 1999) - Less dense than water; will float
U.S. Coast Guard. 1999. Chemical Hazard Response Information System (CHRIS) - Hazardous Chemical Data. Commandant Instruction 16465.12C. Washington, D.C.: U.S. Government Printing Office.
CAMEO Chemicals
1.250 g/L at 0 °C/4 °C
O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Whitehouse Station, NJ: Merck and Co., Inc., 2006., p. 294
Hazardous Substances Data Bank (HSDB)
Density at critical point = 301.0 kg/cu m
Bierhals J; Ullmann's Encyclopedia of Industrial Chemistry. (2002). New York, NY: John Wiley and Sons; Carbon Monoxide. Online Posting Date: March 15, 2001.
Hazardous Substances Data Bank (HSDB)
Density of liquid = 788.6 kg/cu m at 81.63 K
Bierhals J; Ullmann's Encyclopedia of Industrial Chemistry. (2002). New York, NY: John Wiley and Sons; Carbon Monoxide. Online Posting Date: March 15, 2001.
Hazardous Substances Data Bank (HSDB)
Density of solid, hexagonal = 929 kg/cu m at 65 K
Bierhals J; Ullmann's Encyclopedia of Industrial Chemistry. (2002). New York, NY: John Wiley and Sons; Carbon Monoxide. Online Posting Date: March 15, 2001.
Hazardous Substances Data Bank (HSDB)
0.791 at -312.7 °F
Occupational Safety and Health Administration (OSHA)
0.97(relative gas density)
The National Institute for Occupational Safety and Health (NIOSH)
3.2.10 Vapor Density
0.97 (NIOSH, 2024) - Lighter than air; will rise (Relative to Air)
CAMEO Chemicals
0.968 (Air = 1)
O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Whitehouse Station, NJ: Merck and Co., Inc., 2006., p. 294
Hazardous Substances Data Bank (HSDB)
Relative vapor density (air = 1): 0.97
ILO-WHO International Chemical Safety Cards (ICSCs)
0.97
Occupational Safety and Health Administration (OSHA)
3.2.11 Vapor Pressure
greater than 35 atm (NIOSH, 2024)
CAMEO Chemicals
1.55X10+8 mm Hg at 25 °C
Yaws CL; Handbook of Vapor Pressure. Vol 1: C1-C4 Compounds. Houston, TX: Gulf Pub Co (1994)
Hazardous Substances Data Bank (HSDB)
35 atm
Occupational Safety and Health Administration (OSHA); The National Institute for Occupational Safety and Health (NIOSH)
3.2.12 LogP
0.83
HANSCH,C ET AL. (1995)
Human Metabolome Database (HMDB)
3.2.13 Henry's Law Constant
Henry's Law constant = 1.04 atm-cu m/mol at 25 °C (reported as 57978.5 atm/mol fraction)
Yaws CL et al; Chem Eng; 106: 102-105 (1999)
Hazardous Substances Data Bank (HSDB)
3.2.14 Autoignition Temperature
1128 °F (USCG, 1999)
U.S. Coast Guard. 1999. Chemical Hazard Response Information System (CHRIS) - Hazardous Chemical Data. Commandant Instruction 16465.12C. Washington, D.C.: U.S. Government Printing Office.
CAMEO Chemicals
1292 °F (700 °C)
Fire Protection Guide to Hazardous Materials. 13 ed. Quincy, MA: National Fire Protection Association, 2002., p. 49-39
Hazardous Substances Data Bank (HSDB)
605 °C
ILO-WHO International Chemical Safety Cards (ICSCs)
3.2.15 Viscosity
Viscosity gas at 273 K = 16.62 uN s/sq m
Bierhals J; Ullmann's Encyclopedia of Industrial Chemistry. (2002). New York, NY: John Wiley and Sons; Carbon Monoxide. Online Posting Date: March 15, 2001.
Hazardous Substances Data Bank (HSDB)
3.2.16 Heat of Combustion
-4.343 BTU/lb = -2,412 cal/g = -101X10+5 joules/kg
U.S. Coast Guard, Department of Transportation. CHRIS - Hazardous Chemical Data. (1999). Available from, as of Sept 22, 2009:
Hazardous Substances Data Bank (HSDB)
3.2.17 Heat of Vaporization
Latent: 92.8 BTU/lb = 51.6 cal/g = 2.16X10+5 joules/kg
U.S. Coast Guard, Department of Transportation. CHRIS - Hazardous Chemical Data. (1999). Available from, as of Sept 22, 2009:
Hazardous Substances Data Bank (HSDB)
3.2.18 Surface Tension
9.8 mN/m (of the liquid at 80 K)
Bierhals J; Ullmann's Encyclopedia of Industrial Chemistry. (2002). New York, NY: John Wiley and Sons; Carbon Monoxide. Online Posting Date: March 15, 2001.
Hazardous Substances Data Bank (HSDB)
3.2.19 Ionization Potential
14.01 eV
Occupational Safety and Health Administration (OSHA); The National Institute for Occupational Safety and Health (NIOSH)
3.2.20 Refractive Index
Refractive index of gas = 1.0003364 at 273 K and 546.1 nm
Bierhals J; Ullmann's Encyclopedia of Industrial Chemistry. (2002). New York, NY: John Wiley and Sons; Carbon Monoxide. Online Posting Date: March 15, 2001.
Hazardous Substances Data Bank (HSDB)
3.2.21 Other Experimental Properties
Decomposes into carbon and carbon dioxide at 400-700 °C, at lower temperatures when in contact with catalytic surfaces
O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Whitehouse Station, NJ: Merck and Co., Inc., 2006., p. 294
Hazardous Substances Data Bank (HSDB)
Burns in air with bright blue flame; top pressure: 1500 psi; heat capacity at 20 °C: 6.95 Cal/mole/deg C; heat value/cu m: 3033 kcal; heat of formation: -26.39 Kcal/mol; above 800 °C equilibrium reaction favors carbon monoxide formation; hopcalite (a mixture of the oxides of manganese and copper) catalyzes decomposition at room temp, as does palladium on silica gel
O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Whitehouse Station, NJ: Merck and Co., Inc., 2006., p. 294
Hazardous Substances Data Bank (HSDB)
Specific vol: 13.8 cu ft/lb at 70 °F
Lewis, R.J. Sr.; Hawley's Condensed Chemical Dictionary 15th Edition. John Wiley & Sons, Inc. New York, NY 2007., p. 235
Hazardous Substances Data Bank (HSDB)
Triple point temp = 68.15 K at 15.35 kPa
Bierhals J; Ullmann's Encyclopedia of Industrial Chemistry. (2002). New York, NY: John Wiley and Sons; Carbon Monoxide. Online Posting Date: March 15, 2001.
Hazardous Substances Data Bank (HSDB)
For more Other Experimental Properties (Complete) data for Carbon monoxide (19 total), please visit the HSDB record page.
Hazardous Substances Data Bank (HSDB)
3.3 SpringerMaterials Properties
Dunham energy parameter
Gibbs energy
Moessbauer quadrupole effect
Acentric factor
Adsorbate coverage
Band structure
Boiling point
Centrifugal distortion
Chemical diffusion
Chemical shift
Composition
Compressibility
Core level transition
Corrosion
Critical point
Crystal structure
Density
Diffusion
Diffusive flux
Electric dipole moment
Electronic state
Electronic structure
Enthalpy
Entropy
Excess enthalpy
Fine structure
Formation energy
Formation enthalpy
Formation entropy
Fusion temperature
Geothermal energy content
Heat capacity
Heat of combustion
Heat of solution
Heat of sublimation
Heat transfer coefficient
Hindering potential
Hyperfine structure
Internuclear distance
Lineshape
Magnetic permeability
Mean free path
Melting curve
Melting temperature
Mixing enthalpy
Molar mass
Molecular dipole moment
Molecular structure
Moment of inertia
Nuclear quadrupole coupling
Nuclear quadrupole moment
Nuclear quadrupole resonance spectroscopy
Parity
Phase diagram
Phase transition
Photoemission
Photoemission spectroscopy
Potential energy
Quadrupole coupling
Quantum number
Rotation-vibration spectrum
Rotational energy
Rotational excitation cross section
Self-diffusion
Sound absorption
Sound propagation
Sound velocity
Spin-rotation constant
Surface tension
Thermal conductivity
Thermal expansion coefficient
Transition enthalpy
Transition entropy
Transition frequency
Valence band
Vapor pressure
Vapor-liquid equilibrium
Vibrational ground state
Vibrational mode frequency
Virial coefficient
Viscosity
SpringerMaterials
3.4 Chemical Classes
Toxic Gases & Vapors -> Chemical Asphyxiants
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
3.4.1 Pesticides
Rodenticides
EU Pesticides Database
Active substance -> EU Pesticides database: Not approved
EU Pesticides Database
4 Spectral Information
4.1 Mass Spectrometry
4.1.1 GC-MS
Source of Spectrum
AA-0-3-1
Copyright
Copyright © 2020-2025 John Wiley & Sons, Inc. All Rights Reserved.
Thumbnail
SpectraBase
4.2 IR Spectra
4.2.1 Vapor Phase IR Spectra
Source of Spectrum
Sigma-Aldrich Co. LLC.
Source of Sample
Sigma-Aldrich Co. LLC.
Catalog Number
295116
Copyright
Copyright © 2021-2025 Sigma-Aldrich Co. LLC. - Database Compilation Copyright © 2021-2025 John Wiley & Sons, Inc. All Rights Reserved.
Thumbnail
SpectraBase
5 Related Records
5.1 Related Compounds with Annotation
Follow these links to do a live 2D search or do a live 3D search for this compound, sorted by annotation score. This section is deprecated (see here for details), but these live search links provide equivalent functionality to the table that was previously shown here.
PubChem
5.2 Related Compounds
Same Connectivity Count
11
Same Parent, Connectivity Count
15
Same Parent, Exact Count
5
Mixtures, Components, and Neutralized Forms Count
17587
Similar Compounds (2D)
View in PubChem Search
Similar Conformers (3D)
View in PubChem Search
PubChem
5.3 Substances
5.3.1 PubChem Reference Collection SID
481107745
PubChem
5.3.2 Related Substances
All Count
38775
Same Count
431
Mixture Count
38344
PubChem
5.3.3 Substances by Category
PubChem
5.4 Other Relationships
Tobacco Smoke (part of)
PubChem
5.5 Entrez Crosslinks
PubMed Count
107
Protein Structures Count
361
Taxonomy Count
5
OMIM Count
4
Gene Count
145
PubChem
5.6 NCBI LinkOut
NCBI
6 Chemical Vendors
PubChem
7 Drug and Medication Information
7.1 Drug Indication
Used as a marker of respiratory status in spirometry tests,. Food additive for pigment fixation in meat.
DrugBank
Open Targets
7.2 Clinical Trials
7.2.1 ClinicalTrials.gov
ClinicalTrials.gov
7.2.2 NIPH Clinical Trials Search of Japan
NIPH Clinical Trials Search of Japan
7.3 Therapeutic Uses
MEDICATION (VET): Euthanasia of dogs and cats can be carried out in a carbon monoxide chamber, but there are a number of precautions and guidelines for proper use of such chambers.
Booth, N.H., L.E. McDonald (eds.). Veterinary Pharmacology and Therapeutics. 5th ed. Ames, Iowa: Iowa State University Press, 1982., p. 1061
Hazardous Substances Data Bank (HSDB)
There are over 350 variants to normal human hemoglobin. In the hemoglobin S variant, sickling takes place when deoxyhemoglobin S in the red blood cell reaches a critical level and causes intracellular polymerization. Oxygenation of the hemoglobin S molecules in the polymer, therefore, should lead to a change in molecular shape, breakup of the polymer and unsickling of the cell. Carbon monoxide was considered at one time to be potentially beneficial, because it ultimately would reduce the concentration of deoxyhemoglobin S by converting part of the hemoglobin to carboxyhemoglobin. Exposure to carbon monoxide, however, was not considered to be an effective clinical treatment, because high carboxyhemoglobin levels (>20%) were required. /Former use/
Environmental Health Criteria 213: Carbon Monoxide pp. 1-12 (1999) by the International Programme on Chemical Safety (IPCS) under the joint sponsorship of the United Nations Environment Programme, the International Labour Organisation and the World Health Organization
Hazardous Substances Data Bank (HSDB)
/EXPER THER/ Cardiopulmonary bypass (CPB) is associated with pulmonary inflammation and dysfunction. This may lead to acute lung injury and acute respiratory distress syndrome with increased morbidity and mortality. The authors hypothesized that inhaled carbon monoxide before initiation of CPB would reduce inflammatory response in the lungs. In a porcine model, a beating-heart CPB was used. The animals were either randomized to a control group, to standard CPB, or to CPB plus carbon monoxide. In the latter group, lungs were ventilated with 250 ppm inhaled carbon monoxide in addition to standard ventilation before CPB. Lung tissue samples were obtained at various time points, and pulmonary cytokine levels were determined. Hemodynamic parameters were largely unaffected by CPB or carbon monoxide inhalation. There were no significant differences in cytokine expression in mononuclear cells between the groups throughout the experimental time course. Compared with standard CPB animals, carbon monoxide significantly suppresses tumor necrosis factor-alpha and interleukin-1beta levels (P<0.05) and induced the antiinflammatory cytokine interleukin 10 (P<0.001). Carbon monoxide inhalation modulates effector caspase activity in lung tissue during CPB. The results demonstrate that inhaled carbon monoxide significantly reduces CPB-induced inflammation via suppression of tumor necrosis factor alpha, and interleukin-1beta expression and elevation of interleukin 10. Apoptosis induced by CPB was associated with caspase-3 activation and was significantly attenuated by carbon monoxide treatment. Based on the observations of this study, inhaled carbon monoxide could represent a potential new therapeutic modality for counteracting CPB-induced lung injury.
PMID:18497603
Goebel U et al; Anesthesiology 108 (6): 1025-36 (2008)
Hazardous Substances Data Bank (HSDB)
7.4 Reported Fatal Dose
5000 ppm over 5 minutes for an adult human. (T26)
Toxin and Toxin Target Database (T3DB)
8 Food Additives and Ingredients
8.1 Associated Foods
FooDB
9 Agrochemical Information
9.1 Agrochemical Category
Pesticide active substances -> Rodenticides
EU Pesticides Database
9.2 EU Pesticides Data
Active Substance
carbon monoxide
Status
Not approved [Reg. (EC) No 1107/2009]
Legislation
2008/967
ADI
Not Applicable
ARfD
Not Applicable
AOEL
Not Applicable
EU Pesticides Database
10 Pharmacology and Biochemistry
10.1 Pharmacodynamics
Carbon monoxide is used to measure the diffusing capacity for carbon monoxide (DLCO), also known as the transfer factor for carbon monoxide. It is a measure of the gas transfer from inspired gas to the circulatory system (red blood cells in particular). It is used in a particular pulmonary function test called "the single-breath test". DLCO, measured for clinical and research purposes almost exclusively by the single-breath method is an important and very useful pulmonary function test. It is useful in the evaluation of patients with dyspnea, obstructive lung diseases, restrictive lung diseases, and in patients with diseases of the pulmonary vasculature. The measurement of DLCO using carbon monoxide is representative of the surface area available, the volume of blood present in the pulmonary capillaries, as well as the thickness of the alveolar-capillary membrane. Conditions that increase DLCO: Heart failure, erythrocythemia, alveolar hemorrhage, asthma Conditions that decrease DLCO: emphysema, pulmonary fibrosis, pulmonary hypertension, pulmonary embolism In addition to the above uses, carbon monoxide (CO) is increasingly being accepted in recent years as a protective molecule with important signaling capabilities in both physiological/homeostatic and pathophysiological situations. The endogenous production of CO occurs via the activity of constitutive (heme oxygenase 2) and inducible (heme oxygenase 1) heme oxygenase enzymes, which are both responsible for the breakdown of heme. Through the generation of its products, which in addition to carbon monoxide, includes the biliary pigments biliverdin, bilirubin and ferrousiron, the heme oxygenase 1 system also have an essential role in the regulation of the stress response and in cell adaptation to injury. Preclinical studies have suggested potential benefits of carbon monoxide in cardiovascular disease, inflammatory disorders, as well as organ transplantation.
DrugBank
10.2 MeSH Pharmacological Classification
Gasotransmitters
Endogenously produced lipid-soluble gaseous molecules which function as neurotransmitters and signal mediators targeting ION CHANNELS and transporters. (See all compounds classified as Gasotransmitters.)
Medical Subject Headings (MeSH)
Antimetabolites
Drugs that are chemically similar to naturally occurring metabolites, but differ enough to interfere with normal metabolic pathways. (From AMA Drug Evaluations Annual, 1994, p2033) (See all compounds classified as Antimetabolites.)
Medical Subject Headings (MeSH)
10.3 ATC Code
V - Various
V04 - Diagnostic agents
V04C - Other diagnostic agents
V04CX - Other diagnostic agents
V04CX08 - Carbon monoxide
WHO Anatomical Therapeutic Chemical (ATC) Classification
10.4 Absorption, Distribution and Excretion
Absorption
Although CO is not one of the respiratory gases, the similarity of physico-chemical properties of CO and oxygen (O2) permits an extension of the findings of studies on the kinetics of transport of O2 to those of CO. The rate of formation and elimination of COHb, its concentration in blood, and its catabolism is controlled by numerous physical factors and physiological mechanisms. The absorption of carbon monoxide from the consumption treated food products is not significant. Risk of CO toxicity from the packaging process or from consumption of CO-treated meats is negligible.
DrugBank
Carbon monoxide is eliminated through the lungs when air free of carbon monoxide is inhaled.
Sax, N.I. Dangerous Properties of Industrial Materials. 6th ed. New York, NY: Van Nostrand Reinhold, 1984., p. 643
Hazardous Substances Data Bank (HSDB)
... /Carbon monoxide/ readily crosses placenta.
Gilman, A.G., T.W. Rall, A.S. Nies and P. Taylor (eds.). Goodman and Gilman's The Pharmacological Basis of Therapeutics. 8th ed. New York, NY. Pergamon Press, 1990., p. 1620
Hazardous Substances Data Bank (HSDB)
Carbon monoxide is not a cumulative poison in the usual sense. Carboxyhemoglobin is fully dissociable, and once exposure has been terminated, the pigment will revert to oxyhemoglobin. Liberated carbon monoxide is eliminated via the lungs.
Amdur, M.O., J. Doull, C.D. Klaasen (eds). Casarett and Doull's Toxicology. 4th ed. New York, NY: Pergamon Press, 1991., p. 268
Hazardous Substances Data Bank (HSDB)
The absorption of carbon monoxide is said not to occur, but its absorption followed by oxidation within the epidermis has not been excluded.
Hayes, W.J., Jr., E.R. Laws Jr., (eds.). Handbook of Pesticide Toxicology Volume 1. General Principles. New York, NY: Academic Press, Inc., 1991., p. 139
Hazardous Substances Data Bank (HSDB)
For more Absorption, Distribution and Excretion (Complete) data for Carbon monoxide (13 total), please visit the HSDB record page.
Hazardous Substances Data Bank (HSDB)
10.5 Metabolism / Metabolites
Metabolism of the dihalomethanes leads to dehalogenation, and the end product is carbon monoxide ... The carbon monoxide appears to arise from a formyl halide intermediate resulting from the loss of one halide atom from the halocarbon. This intermediate as an alternative to losing carbon monoxide can covalently bind to cellular protein or lipid.
Amdur, M.O., J. Doull, C.D. Klaasen (eds). Casarett and Doull's Toxicology. 4th ed. New York, NY: Pergamon Press, 1991., p. 692
Hazardous Substances Data Bank (HSDB)
In addition to exogenous sources, humans are also exposed to small amounts of carbon monoxide produced endogenously. In the process of natural degradation of hemoglobin to bile pigments, in concert with the microsomal reduced nicotinamide adenine dinucleotide phosphate (NADPH) cytochrome P-450 reductase, two heme oxygenase isoenzymes, HO-1 and HO-2, catalyse the oxidative breakdown of the alpha-methene bridge of the tetrapyrrole ring of heme, leading to the formation of biliverdin and carbon monoxide. The major site of heme breakdown, and therefore the major organ for production of endogenous carbon monoxide, is the liver. The spleen and the erythropoietic system are other important catabolic generators of carbon monoxide ... Other hemoproteins, such as myoglobin, cytochromes, peroxidases and catalase, contribute approximately 20-25% to the total amount of carbon monoxide generated. Approximately 0.4 mL carbon monoxide/hr is formed by hemoglobin catabolism, and about 0.1 mL/hr originates from non-hemoglobin sources.
Environmental Health Criteria 213: Carbon Monoxide pp. 1-12 (1999) by the International Programme on Chemical Safety (IPCS) under the joint sponsorship of the United Nations Environment Programme, the International Labour Organisation and the World Health Organization.
Hazardous Substances Data Bank (HSDB)
Any disturbance leading to increased destruction of red blood cells and accelerated breakdown of other haemoproteins would lead to increased production of carbon monoxide. Hematomas, intravascular hemolysis of red blood cells, blood transfusion and ineffective erythropoiesis will all elevate the carbon monoxide concentration in the blood. Degradation of red blood cells under pathological conditions such as anemias (hemolytic, sideroblastic, sickle cell), thalassaemia, Gilbert's syndrome with hemolysis and other hematological diseases will also accelerate carbon monoxide production.
Environmental Health Criteria 213: Carbon Monoxide pp. 1-12 (1999) by the International Programme on Chemical Safety (IPCS) under the joint sponsorship of the United Nations Environment Programme, the International Labour Organisation and the World Health Organization.
Hazardous Substances Data Bank (HSDB)
The primary factors that determine the final level of carboxyhemoglobin are: the amount of inspired carbon monoxide; minute alveolar ventilation at rest and during exercise; endogenous carbon monoxide production; blood volume; barometric pressure; and the relative diffusion capability of the lungs. The rate of diffusion from the alveoli and the binding of carbon monoxide with the blood hemoglobin are the steps limiting the rate of uptake into the blood.
WHO; Environ Health Criteria 13: Carbon Monoxide p.35 (1979)
Hazardous Substances Data Bank (HSDB)
For more Metabolism/Metabolites (Complete) data for Carbon monoxide (6 total), please visit the HSDB record page.
Hazardous Substances Data Bank (HSDB)
10.6 Biological Half-Life
The half-life of carbon monoxide at room air temperature is 3-4 hours. 100% oxygen reduces the half-life to 30-90 minutes; hyperbaric oxygen at 2.5 atm (atmosphere units) with 100% oxygen reduces it to 15-23 minutes.
DrugBank
Elimination 1/2 life: 5-6 hours (shortened by administration of oxygen); [TDR, p. 283]
TDR - Ryan RP, Terry CE, Leffingwell SS (eds). Toxicology Desk Reference: The Toxic Exposure and Medical Monitoring Index, 5th Ed. Washington DC: Taylor & Francis, 1999., p. 283
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
The half-time of carbon monoxide disappearance from blood under normal recovery conditions while breathing air showed considerable between-individual variance. For carboxyhemoglobin concentrations of 2-10%, the half-time ranged from 3 to 5 hr; others reported the range to be 2-6.5 hr for slightly higher initial concentrations of carboxyhemoglobin.
Environmental Health Criteria 213: Carbon Monoxide pp. 1-12 (1999) by the International Programme on Chemical Safety (IPCS) under the joint sponsorship of the United Nations Environment Programme, the International Labour Organisation and the World Health Organization.
Hazardous Substances Data Bank (HSDB)
The biological half-life of carbon monoxide concentration in the blood of sedentary adults is about 2-5 hr. The elimination of carbon monoxide becomes slower with time, and the lower the initial level of carboxyhemoglobin, the slower the rate of excretion.
International Labour Office. Encyclopedia of Occupational Health and Safety. Vols. I&II. Geneva, Switzerland: International Labour Office, 1983., p. 396
Hazardous Substances Data Bank (HSDB)
10.7 Mechanism of Action
In respiratory testing, the diffusing capacity for carbon monoxide (DLCO) is a measure of the ability of gas to transfer from the alveoli across the alveolar epithelium and the capillary endothelium to the red blood cells. The DLCO depends not only on the area and thickness of the blood-gas barrier but additionally on the volume of blood in the pulmonary capillaries. The distribution of alveolar volume and ventilation also has an impact on the measurement. DLCO is measured by sampling end-expiratory gas for carbon monoxide (CO) after patients inspire a small and safe amount of exogenous CO, hold their breath, and exhale. Measured DLCO is adjusted for alveolar volume (which is estimated from dilution of helium) and the patient’s hematocrit level. DLCO is reported as mL/min/mm Hg and as a percentage of a predicted value. Carbon monoxide exerts effects on cell metabolism through both hypoxic and non-hypoxic modes of action. Both mechanisms of action are thought to be the result of the ability of carbon monoxide to bind strongly to heme and alter the function and/or metabolism of heme proteins. The binding affinity of carbon monoxide for hemoglobin is more than 200 times greater than that of oxygen for hemoglobin. The formation of carboxyhemoglobin (COHb) decreases the O2 carrying capacity of blood and disrupts the release of O2 from Hb for its use in tissues. Through similar mechanisms, carbon monoxide diminishes the O2 storage in muscle cells by binding to and displacing O2 from, myoglobin. Though all human tissues are vulnerable to carbon monoxide-induced hypoxic injury, those with the highest O2 demand are especially vulnerable, including the brain and heart. Most of the non-hypoxic mechanisms of action of carbon monoxide have been thought to be due to binding of carbon monoxide to heme in proteins other than Hb. The most notable targets of carbon monoxide include components of many important physiological regulatory systems, including brain and muscle oxygen storage and use(myoglobin, neuroglobin); nitric oxide cell signaling (e.g., nitric oxide synthase, guanylyl cyclase); prostaglandin cell signaling (cyclooxygenase, prostaglandin H synthase); energy metabolism and mitochondrial respiration (cytochrome c oxidase, cytochrome c, NADPH oxidase); steroid and drug metabolism (cytochrome P450); cellular redox balance and reactive oxygen species (ROS; catalase, peroxidases); and numerous transcription factors (e.g., neuronal PAS domain protein, NPAS2, implicated in regulation of circadian rhythm). In meat processing, carbon monoxide reacts with myoglobin, to form carboxymyoglobin, imparting a red appearance to the meat.
DrugBank
The binding of carbon monoxide to hemoglobin, producing carboxyhemoglobin and decreasing the oxygen carrying capacity of blood, appears to be the principal mechanism of action underlying the induction of toxic effects of low-level carbon monoxide exposures. The precise mechanisms by which toxic effects are induced via carboxyhemoglobin formation are not understood fully but likely include the induction of a hypoxic state in many tissues of diverse organ systems. Alternative or secondary mechanisms of carbon monoxide-induced toxicity (besides carboxyhaemoglobin) have been hypothesized, but none has been demonstrated to operate at relatively low (near-ambient) carbon monoxide exposure levels.
Environmental Health Criteria 213: Carbon Monoxide pp. 1-12 (1999) by the International Programme on Chemical Safety (IPCS) under the joint sponsorship of the United Nations Environment Programme, the International Labour Organisation and the World Health Organization.
Hazardous Substances Data Bank (HSDB)
Carbon monoxide is known to react with a variety of metal-containing proteins found in nature. Carbon monoxide-binding metalloproteins present in mammalian tissues include oxygen carrier proteins such as hemoglobin and myoglobin as well as metalloenzymes (oxidoreductases) such as cytochrome c oxidase, cytochromes of the P-450 type, tryptophan oxygenase and dopamine hydroxylase. These metalloproteins contain iron and/or copper centers at their active sites that form metal-ligand complexes with carbon monoxide in competition with molecular oxygen. Carbon monoxide and oxygen form complexes with metalloenzymes only when the iron and copper are in their reduced forms (Fe(II), Cu(I)). ... The competitive relationship between carbon monoxide and oxygen for the active site of intracellular hemoproteins is usually described by the Warburg partition coefficient (R), which is the carbon monoxide/oxygen ratio that produces 50% inhibition of the oxygen uptake of the enzyme, or, in the case of myoglobin, a 50% decrease in the number of available oxygen binding sites.
Environmental Health Criteria 213: Carbon Monoxide pp. 1-12 (1999) by the International Programme on Chemical Safety (IPCS) under the joint sponsorship of the United Nations Environment Programme, the International Labour Organisation and the World Health Organization
Hazardous Substances Data Bank (HSDB)
Carbon monoxide ... reacts /in the blood stream/ with hemoglobin to form carboxyhemoglobin, a form which is incapable of combining with oxygen. Exposure to air containing 0.4% of carbon monoxide for 20-30 min results in the conversion of 70% of the hemoglobin in the blood to carboxyhemoglobin.
Humphreys, D.J. Veterinary Toxicology. 3rd ed. London, England: Bailliere Tindell, 1988., p. 81
Hazardous Substances Data Bank (HSDB)
Carbon monoxide binds tightly to the reduced form of iron in hemoglobin, reducing the delivery of oxygen to tissues. Although this has for many years been thought to be the sole mechanism of toxicity of carbon monoxide, there is evidence to suggest that carbon monoxide also binds to cytochrome a + a3.
Amdur, M.O., J. Doull, C.D. Klaasen (eds). Casarett and Doull's Toxicology. 4th ed. New York, NY: Pergamon Press, 1991., p. 28
Hazardous Substances Data Bank (HSDB)
For more Mechanism of Action (Complete) data for Carbon monoxide (8 total), please visit the HSDB record page.
Hazardous Substances Data Bank (HSDB)
10.8 Human Metabolite Information
10.8.1 Cellular Locations
Cytoplasm
Extracellular
Human Metabolome Database (HMDB)
10.8.2 Metabolite Pathways
Acute Intermittent Porphyria
Congenital Erythropoietic Porphyria (CEP) or Gunther Disease
Hereditary Coproporphyria (HCP)
Porphyria Variegata (PV)
Porphyrin Metabolism
Human Metabolome Database (HMDB)
10.9 Biochemical Reactions
Rhea - Annotated Reactions Database
PubChem
10.10 Transformations
NORMAN Suspect List Exchange
11 Use and Manufacturing
11.1 Uses
Sources/Uses
An incomplete combustion product of carbon-containing materials and an emission of internal combustion engines; [ACGIH] Sewer construction workers can be poisoned by carbon monoxide migrating through the soil from nearby use of explosives. [Appl Occup Environ Hyg 2002 Mar;17(3):152-3]
ACGIH - Documentation of the TLVs and BEIs, 7th Ed. Cincinnati: ACGIH Worldwide, 2020.
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
Industrial Processes with risk of exposure
Aluminum Producing [Category: Industry]
Steel Producing [Category: Industry]
Forging [Category: Heat or Machine]
Molding and Core Making [Category: Foundry]
Metal Preparation and Pouring [Category: Foundry]
Shakeout, Cleaning, and Finishing [Category: Foundry]
Welding [Category: Weld]
Heat Treating [Category: Heat or Machine]
Smelting Copper or Lead [Category: Industry]
Gas Welding and Cutting [Category: Weld]
Mining [Category: Industry]
Sewer and Wastewater Treatment [Category: Industry]
Firefighting [Category: Other]
Cement Producing [Category: Industry]
Burning Natural Polymers [Category: Burn]
Welding Over Coatings [Category: Weld]
Burning Synthetic Polymers [Category: Burn]
Burning Celluloid [Category: Burn]
Farming (Respiratory Hazards) [Category: Industry]
Glass Manufacturing [Category: Industry]
Metal Extraction and Refining [Category: Industry]
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
Activities with risk of exposure
Ceramics making [Category: Hobbies]
Lost-wax casting [Category: Hobbies]
Sculpturing plastics [Category: Hobbies]
Glassblowing [Category: Hobbies]
Smoking cigarettes [Category: Food & Drugs]
Burning biomass fuel for cooking and heating [Category: Environments]
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
Reducing agent in metallurgical operations; in organic synthesis especially in the Fischer-Tropsch process for petroleum-type products and in the oxo reaction; in the manuf of metal carbonyls
O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Whitehouse Station, NJ: Merck and Co., Inc., 2006., p. 294
Hazardous Substances Data Bank (HSDB)
Unisolated component of gaseous fuels, eg, water gas; chem int for phosgene, methanol, acetic acid, acrylic acid, synthetic fuels (non-US use), dimethylformamide, oxo alcohols via aldehydes (eg, butyl alcohol), methyl formate, alkyl carbonates, and silicon carbide fibers; comonomer in ethylene-carbon monoxide copolymer; reducing agent in iron ore processing; purification agent for nickel via nickel carbonyl; chem int for other metal carbonyls, eg, tungsten carbonyl; chem int for ethylene glycol (former use)
SRI
Hazardous Substances Data Bank (HSDB)
Carbon monoxide is increasingly being used on a very large scale for the production of chemical intermediates. It is used in the production of hydrogen, heterogeneous catalysts, pure metals, phosgene, aluminum chloride, acetic acid, acetic anhydride, formic acid, methyl formate, N,N-dimethylformamide, acrylic acid, propanoic acid and as a reducing agent in blast furnaces; a large variety of chemicals, ranging from saturated hydrocarbons to oxygenated compounds are produced using syngas (CO and H2) as a feedstock.
Bierhals J; Ullmann's Encyclopedia of Industrial Chemistry. (2002). New York, NY: John Wiley and Sons; Carbon Monoxide. Online Posting Date: March 15, 2001.
Hazardous Substances Data Bank (HSDB)
Organic synthesis (methanol, ethylene, isocyanates, aldehydes, acrylates, phosgene), fuels (gaseous), metallurgy (special steels, reducing oxides, nickel refining), zinc white pigments.
Lewis, R.J. Sr.; Hawley's Condensed Chemical Dictionary 15th Edition. John Wiley & Sons, Inc. New York, NY 2007., p. 235
Hazardous Substances Data Bank (HSDB)
Carbon monoxide is a major atmospheric pollutant in urban areas, chiefly from exhaust of internal combustion engines, but also from improper burning of various other fuels. (L960)
Toxin and Toxin Target Database (T3DB)
11.1.1 Use Classification
Hazard Classes and Categories -> Teratogens, Flammable - 4th degree
NJDOH RTK Hazardous Substance List
11.1.2 Industry Uses
Processing aids, not otherwise listed
Plasticizer
Intermediates
Fuels and fuel additives
Intermediate
EPA Chemical Data Reporting (CDR)
11.1.3 Consumer Uses
Processing aids, not otherwise listed
Agricultural chemicals (non-pesticidal)
Intermediate
Fuels and fuel additives
Intermediates
EPA Chemical Data Reporting (CDR)
11.2 Methods of Manufacturing
Separation from synthesis gas, eg, water or coke oven gas, by either absorption by salt solutions, eg, cuprous ammonium salts, or by low temperature condensation or fractionation; reaction of carbon dioxide and coke
SRI
Hazardous Substances Data Bank (HSDB)
Produced on industrial scale by partial oxidation of hydrocarbon gases from natural gas or by gasification of coal and coke. Conveniently prepared in lab by heating calcium carbonate with zinc dust; by dehydration of formic acid with H2SO4.
O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Whitehouse Station, NJ: Merck and Co., Inc., 2006., p. 294
Hazardous Substances Data Bank (HSDB)
Carbon monoxide is formed by the incomplete combustion of carbonaceous materials or by the reduction of carbon dioxide. It can also be formed by the decomposition of organic compounds such as aldehydes. Carbon monoxide can also be recovered from the off-gas of several industrial processes such as blast furnace processes or calcium carbide synthesis.
Ullmann's Encyclopedia of Industrial Chemistry. 6th ed.Vol 1: Federal Republic of Germany: Wiley-VCH Verlag GmbH & Co. 2003 to Present, p. V6 460 (2003)
Hazardous Substances Data Bank (HSDB)
Incomplete combustion of carbonaceous materials is commonly employed for the manufacture of commercial quantities of carbon monoxide. ... In most production processes the initial product is a gas mixture containing carbon monoxide. The most important processes are: 1) gasification of coal, 2) steam reforming/CO2 reforming (for light hydrocarbons up to naphtha) and 3) partial oxidation of hydrocarbons (for hydrocarbons heavier than naphtha). To obtain the pure gases, the mixtures must be separated. The main gas that is produced together with carbon monoxide is hydrogen. Many applications need a synthesis gas that is a combination of carbon monoxide with hydrogen. Therefore the separation of carbon monoxide or the adjustment of the carbon monoxide hydrogen ratio are the following production steps. The carbon monoxide can be separated by various procedures: 1) reversible complexation (copper ammonium salt wash) at elevated pressure, followed by desorption by pressure release, 2) cryogenic separation: low-temperature partial condensation and fractionation; liquid methane scrubbing and separation, 3) pressure-swing adsorption and 4) permeable membranes.
Ullmann's Encyclopedia of Industrial Chemistry. 6th ed.Vol 1: Federal Republic of Germany: Wiley-VCH Verlag GmbH & Co. 2003 to Present, p. V6 460 (2003)
Hazardous Substances Data Bank (HSDB)
(1) Made almost pure by placing a mixture of oxygen and carbon dioxide in contact with incandescent graphite, coke, or anthracite. (2) Action of steam on hot coke or coal (water gas) or on natural gas (synthesis gas). In the latter case, carbon dioxide is removed by absorption in amine solution, and the hydrogen and carbon monoxide separated in a low-temperature unit. (3) By-product in chemical reactions. (4) Combustion of organic compound with limited amount of oxygen, as in automobile cylinders. (5) Dehydration of formic acid.
Lewis, R.J. Sr.; Hawley's Condensed Chemical Dictionary 15th Edition. John Wiley & Sons, Inc. New York, NY 2007., p. 235
Hazardous Substances Data Bank (HSDB)
11.3 Formulations / Preparations
Grade: commercial, 98.0-99.0%; C.P., 99.0-99.5%; ultra-high purity, 99.8%; research, 99.97-99.99%
Kirk-Othmer Encyclopedia of Chemical Technology. 3rd ed., Volumes 1-26. New York, NY: John Wiley and Sons, 1978-1984., p. V4 790
Hazardous Substances Data Bank (HSDB)
Available as two-component mixtures in air, argon, helium, hydrogen, nitrogen and carbon dioxide.
CHEMCYCLOPEDIA 1986 p.214
Hazardous Substances Data Bank (HSDB)
11.4 U.S. Production
Aggregated Product Volume
2019: 10,000,000,000 - <20,000,000,000 lb
2018: 10,000,000,000 - <20,000,000,000 lb
2017: 10,000,000,000 - <20,000,000,000 lb
2016: 5,000,000,000 - <10,000,000,000 lb
EPA Chemical Data Reporting (CDR)
(1977) At least 2.8x10+12 g (incl captive mfr)
SRI
Hazardous Substances Data Bank (HSDB)
(1982) Probably greater than 9.08x10+6 g
SRI
Hazardous Substances Data Bank (HSDB)
1 million tons annually (Europe)
European Commission, ESIS; IUCLID Dataset, Carbon monoxide (630-08-0) p.3 (2000 CD-ROM edition). Available from, as of October 20, 2009:
Hazardous Substances Data Bank (HSDB)
Production volumes for non-confidential chemicals reported under the Inventory Update Rule.
Year
Production Range (pounds)
Year
1986
Production Range (pounds)
1 billion
Year
1990
Production Range (pounds)
1 billion
Year
1994
Production Range (pounds)
1 billion
Year
1998
Production Range (pounds)
1 billion
Year
2002
Production Range (pounds)
1 billion
US EPA; Non-confidential Production Volume Information Submitted by Companies for Chemicals Under the 1986-2002 Inventory Update Rule (IUR). Carbon Monoxide (630-08-0). Available from, as of November 23, 2009:
Hazardous Substances Data Bank (HSDB)
For more U.S. Production (Complete) data for Carbon monoxide (6 total), please visit the HSDB record page.
Hazardous Substances Data Bank (HSDB)
11.5 U.S. Exports
(1984) 1.14X10+13 g /Carbon Dioxide, Nitrous Oxide, and Carbon Monoxide/
BUREAU OF THE CENSUS. U.S. EXPORTS, SCHEDULE E, 1984 p.2-94
Hazardous Substances Data Bank (HSDB)
11.6 General Manufacturing Information
Industry Processing Sectors
Industrial Gas Manufacturing
Petroleum Refineries
All Other Basic Organic Chemical Manufacturing
Petrochemical Manufacturing
Construction
EPA Chemical Data Reporting (CDR)
EPA TSCA Commercial Activity Status
Carbon monoxide: ACTIVE
EPA Chemicals under the TSCA
12 Identification
12.1 Analytic Laboratory Methods
Method: NIOSH 6604, Issue 1; Procedure: electrochemical sensor; Analyte: carbon monoxide; Matrix: air; Detection Limit: 1 ppm.
CDC; NIOSH Manual of Analytical Methods, 4th ed. Carbon Monoxide (630-08-0). Available from, as of November 23, 2009:
Hazardous Substances Data Bank (HSDB)
Method: OSHA ID-209; Procedure: direct-reading instrument - datalogger; Analyte: carbon monoxide; Matrix: air; Detection Limit: 1.2 ppm (qualitative), 4.1 ppm (quantitative).
U.S. Department of Labor/Occupational Safety and Health Administration's Index of Sampling and Analytical Methods. Carbon Monoxide (630-08-0). Available from, as of November 23, 2009:
Hazardous Substances Data Bank (HSDB)
Method: OSHA ID-210; Procedure: gas chromatograph and discharge ionization detector; Analyte: carbon monoxide; Matrix: air; Detection Limit: 0.12 ppm (qualitative), 0.40 ppm (quantitative).
U.S. Department of Labor/Occupational Safety and Health Administration's Index of Sampling and Analytical Methods. Carbon Monoxide (630-08-0). Available from, as of November 23, 2009:
Hazardous Substances Data Bank (HSDB)
ANALYTE: CARBON MONOXIDE; MATRIX: AIR; PROCEDURE: INFRARED ABSORPTION SPECTROPHOTOMETRY.
U.S. Department of Health, Education Welfare, Public Health Service. Center for Disease Control, National Institute for Occupational Safety Health. NIOSH Manual of Analytical Methods. 2nd ed. Volumes 1-7. Washington, DC: U.S. Government Printing Office, 1977-present., p. VI 112-1
Hazardous Substances Data Bank (HSDB)
For more Analytic Laboratory Methods (Complete) data for Carbon monoxide (12 total), please visit the HSDB record page.
Hazardous Substances Data Bank (HSDB)
12.2 Clinical Laboratory Methods
SPECTROPHOTOMETRIC METHOD COMMONLY USED IN CLINICAL LABORATORIES.
DUBOWSKY KM, LUKE JL; MEASUREMENT OF CARBOXYHEMOGLOBIN AND CARBON MONOXIDE IN BLOOD; ANN CLIN LAB SCI 3: 53 (1973)
Hazardous Substances Data Bank (HSDB)
ANALYTE: CARBON MONOXIDE; MATRIX: BLOOD; PROCEDURE: GAS CHROMATOGRAPHY.
U.S. Department of Health, Education Welfare, Public Health Service. Center for Disease Control, National Institute for Occupational Safety Health. NIOSH Manual of Analytical Methods. 2nd ed. Volumes 1-7. Washington, DC: U.S. Government Printing Office, 1977-present., p. V1 113-1
Hazardous Substances Data Bank (HSDB)
12.3 NIOSH Analytical Methods
CARBON MONOXIDE 6604
NIOSH Manual of Analytical Methods
13 Safety and Hazards
13.1 Hazards Identification
ERG Hazard Classes
Toxic/poison by inhalation (TIH/PIH)
Emergency Response Guidebook (ERG)
13.1.1 GHS Classification
1 of 6
View All
Pictogram(s)
Signal
Danger
GHS Hazard Statements
H220 (87.3%): Extremely flammable gas [Danger Flammable gases]
H221 (12.9%): Flammable gas [Danger Flammable gases]
H280 (77.8%): Contains gas under pressure; may explode if heated [Warning Gases under pressure]
H331 (99%): Toxic if inhaled [Danger Acute toxicity, inhalation]
H360 (63.1%): May damage fertility or the unborn child [Danger Reproductive toxicity]
H360D (37.2%): May damage the unborn child [Danger Reproductive toxicity]
H372 (100%): Causes damage to organs through prolonged or repeated exposure [Danger Specific target organ toxicity, repeated exposure]
Precautionary Statement Codes
P203, P210, P222, P260, P261, P264, P270, P271, P280, P304+P340, P316, P318, P319, P321, P377, P381, P403, P403+P233, P405, P410+P403, and P501
(The corresponding statement to each P-code can be found at the GHS Classification page.)
ECHA C&L Notifications Summary
Aggregated GHS information provided per 613 reports by companies from 30 notifications to the ECHA C&L Inventory. Each notification may be associated with multiple companies.
Information may vary between notifications depending on impurities, additives, and other factors. The percentage value in parenthesis indicates the notified classification ratio from companies that provide hazard codes. Only hazard codes with percentage values above 10% are shown. For more detailed information, please visit ECHA C&L website.
European Chemicals Agency (ECHA)
13.1.2 Hazard Classes and Categories
Flam. Gas 1 (87.3%)
Press. Gas (Comp.) (77.8%)
Acute Tox. 3 (99%)
Repr. 1A (63.1%)
Repr. 1A (37.2%)
STOT RE 1 (100%)
European Chemicals Agency (ECHA)
Flam. Gas 1 (100%)
Press. Gas (Comp.) (100%)
Acute Tox. 3 (100%)
Repr. 1A (100%)
STOT RE 1 (100%)
European Chemicals Agency (ECHA)
View More...
13.1.3 NFPA Hazard Classification
1 of 2
View All
NFPA 704 Diamond
NFPA Health Rating
3 - Materials that, under emergency conditions, can cause serious or permanent injury.
NFPA Fire Rating
4 - Materials that rapidly or completely vaporize at atmospheric pressure and normal ambient temperature or that are readily dispersed in air and burn readily.
NFPA Instability Rating
0 - Materials that in themselves are normally stable, even under fire conditions.
Hazardous Substances Data Bank (HSDB)
13.1.4 Health Hazards
Inhalation causes headache, dizziness, weakness of limbs, confusion, nausea, unconsciousness, and finally death. 0.04% conc., 2-3 hr. or 0.06% conc., 1 hr.- headache and discomfort; with moderate exercise, 0.1-0.2% will produce throbbing in the head in about 1/2 hr., a tendency to stagger in about 1 1/2 hr., and confusion of the mind, headache, and nausea in about 2 hrs. 0.20-25% usually produces unconsciousness in about 1/2 hr. Inhalation of a 0.4% conc. can prove fatal in less than 1 hr. Inhalation of high concentrations can cause sudden, unexpected collapse. Contact of liquid with skin will cause frostbite. (USCG, 1999)
U.S. Coast Guard. 1999. Chemical Hazard Response Information System (CHRIS) - Hazardous Chemical Data. Commandant Instruction 16465.12C. Washington, D.C.: U.S. Government Printing Office.
CAMEO Chemicals
Excerpt from ERG Guide 168 [Carbon Monoxide (Refrigerated Liquid)]:
TOXIC; Extremely Hazardous. Inhalation extremely dangerous; may be fatal. Contact with gas, liquefied gas or cryogenic liquids may cause burns, severe injury and/or frostbite. Odorless, will not be detected by sense of smell. (ERG, 2024)
2024 Emergency Response Guidebook,
CAMEO Chemicals
ERG 2024, Guide 168 (Carbon monoxide, refrigerated liquid (cryogenic liquid))
· TOXIC; Extremely Hazardous.
· Inhalation extremely dangerous; may be fatal.
· Contact with gas, liquefied gas or cryogenic liquids may cause burns, severe injury and/or frostbite.
· Odorless, will not be detected by sense of smell.
Emergency Response Guidebook (ERG)
ERG 2024, Guide 119 (Carbon monoxide, compressed)
· TOXIC; may be fatal if inhaled or absorbed through skin. Some may cause severe skin burns and eye damage.
· Contact with gas or liquefied gas may cause burns, severe injury and/or frostbite.
· Fire will produce irritating, corrosive and/or toxic gases.
· Runoff from fire control or dilution water may cause environmental contamination.
Emergency Response Guidebook (ERG)
13.1.5 Fire Hazards
Special Hazards of Combustion Products: Asphyxiation due to carbon dioxide production may result.
Behavior in Fire: Flame has very little color. Containers may explode in fire. (USCG, 1999)
U.S. Coast Guard. 1999. Chemical Hazard Response Information System (CHRIS) - Hazardous Chemical Data. Commandant Instruction 16465.12C. Washington, D.C.: U.S. Government Printing Office.
CAMEO Chemicals
Excerpt from ERG Guide 168 [Carbon Monoxide (Refrigerated Liquid)]:
EXTREMELY FLAMMABLE. CAUTION: Flame can be invisible. Use an alternate method of detection (thermal camera, broom handle, etc.) May be ignited by heat, sparks or flames. Containers may explode when heated. Vapor explosion and poison hazard indoors, outdoors or in sewers. Vapors from liquefied gas are initially heavier than air and spread along ground. Vapors may travel to source of ignition and flash back. Runoff may create fire or explosion hazard. (ERG, 2024)
2024 Emergency Response Guidebook,
CAMEO Chemicals
ERG 2024, Guide 168 (Carbon monoxide, refrigerated liquid (cryogenic liquid))
· EXTREMELY FLAMMABLE.
CAUTION: Flame can be invisible. Use an alternate method of detection (thermal camera, broom handle, etc.)
· May be ignited by heat, sparks or flames.
· Containers may explode when heated.
· Vapor explosion and poison hazard indoors, outdoors or in sewers.
· Vapors from liquefied gas are initially heavier than air and spread along ground.
· Vapors may travel to source of ignition and flash back.
· Runoff may create fire or explosion hazard.
Emergency Response Guidebook (ERG)
ERG 2024, Guide 119 (Carbon monoxide, compressed)
· Flammable; may be ignited by heat, sparks or flames.
· May form explosive mixtures with air. Ethylene oxide (UN1040) may react explosively even in the absence of air.
· Those substances designated with a (P) may polymerize explosively when heated or involved in a fire.
· Vapors from liquefied gas are initially heavier than air and spread along ground.
· Vapors may travel to source of ignition and flash back.
· Some of these materials may react violently with water.
· Cylinders exposed to fire may vent and release toxic and flammable gas through pressure relief devices.
· Containers may explode when heated.
· Ruptured cylinders may rocket.
· Runoff may create fire or explosion hazard.
Emergency Response Guidebook (ERG)
Extremely flammable. Heating will cause rise in pressure with risk of bursting. Gas/air mixtures are explosive.
ILO-WHO International Chemical Safety Cards (ICSCs)
13.1.6 Hazards Summary
Carbon monoxide is a colorless, nonirritating, odorless, and tasteless gas. It is found in both outdoor and indoor air.
Agency for Toxic Substances and Disease Registry (ATSDR)
Carboxyhemoglobin 1/2 life in the body = 4 hours in room air and 60-90 minutes in 100% oxygen; Nonsmoker normal = 2%; Smoker = 5-7%; Neurological symptoms = 30%; Death = 40-50%; Carbon monoxide is the most common cause of chemical asphyxia. There is limited positive data that carbon monoxide causes low birth weight and fetal death in humans. In animals, there is strong positive data that it causes birth defects and neonatal mortality and limited positive data that it causes testicular damage. [ATSDR Case Studies, #29] Decreased exercise time to onset of angina or ischemia was observed at COHb levels as low as 3% and increased ventricular arrhythmias at COHb levels of 6%. [ACGIH] Early on, symptoms include headache, dizziness, and disorientation. More prolonged or severe hypoxia is accompanied by a varying combination of tremor, chorea, spasticity, dystonia, rigidity, and bradykinesia. Recovery from the hypoxia may be incomplete. Residual dementia, spasticity, cortical blindness, and parkinsonian features are relatively common. [Ladou, p. 430]
ACGIH - Documentation of the TLVs and BEIs, 7th Ed. Cincinnati: ACGIH Worldwide, 2020.
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
13.1.7 Fire Potential
Flammable gas.
Fire Protection Guide to Hazardous Materials. 13 ed. Quincy, MA: National Fire Protection Association, 2002., p. 49-39
Hazardous Substances Data Bank (HSDB)
13.2 Safety and Hazard Properties
13.2.1 Acute Exposure Guideline Levels (AEGLs)
13.2.1.1 AEGLs Table
AEGLs
10 min
30 min
60 min
4 hr
8 hr
AEGLs
AEGL 1: Notable discomfort, irritation, or certain asymptomatic non-sensory effects. However, the effects are not disabling and are transient and reversible upon cessation of exposure (Unit: ppm)
10 min
NR
30 min
NR
60 min
NR
4 hr
NR
8 hr
NR
AEGLs
AEGL 2: Irreversible or other serious, long-lasting adverse health effects or an impaired ability to escape (Unit: ppm)
10 min
420
30 min
150
60 min
83
4 hr
33
8 hr
27
AEGLs
AEGL 3: Life-threatening health effects or death (Unit: ppm)
10 min
1,700
30 min
600
60 min
330
4 hr
150
8 hr
130
EPA Acute Exposure Guideline Levels (AEGLs)
13.2.1.2 AEGLs Notes
NR = Not recommended due to insufficient data
AEGLs Status: Final
EPA Acute Exposure Guideline Levels (AEGLs)
13.2.2 Flammable Limits
Lower: 12% by volume; Upper: 75% by volume
Fire Protection Guide to Hazardous Materials. 13 ed. Quincy, MA: National Fire Protection Association, 2002., p. 49-39
Hazardous Substances Data Bank (HSDB)
Flammability
Flammable Gas
The National Institute for Occupational Safety and Health (NIOSH)
13.2.3 Lower Explosive Limit (LEL)
12 % (USCG, 1999)
U.S. Coast Guard. 1999. Chemical Hazard Response Information System (CHRIS) - Hazardous Chemical Data. Commandant Instruction 16465.12C. Washington, D.C.: U.S. Government Printing Office.
CAMEO Chemicals
12.5 % (NIOSH, 2024)
CAMEO Chemicals
12.5%
Occupational Safety and Health Administration (OSHA); The National Institute for Occupational Safety and Health (NIOSH)
13.2.4 Upper Explosive Limit (UEL)
75 % (USCG, 1999)
U.S. Coast Guard. 1999. Chemical Hazard Response Information System (CHRIS) - Hazardous Chemical Data. Commandant Instruction 16465.12C. Washington, D.C.: U.S. Government Printing Office.
CAMEO Chemicals
74 % (NIOSH, 2024)
CAMEO Chemicals
74%
Occupational Safety and Health Administration (OSHA); The National Institute for Occupational Safety and Health (NIOSH)
13.2.5 Critical Temperature & Pressure
Critical Pressure: 35 Atmospheres; Critical Temperature: -139 °C
O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Whitehouse Station, NJ: Merck and Co., Inc., 2006., p. 294
Hazardous Substances Data Bank (HSDB)
13.2.6 Physical Dangers
The gas mixes well with air, explosive mixtures are easily formed. The gas penetrates easily through walls and ceilings.
ILO-WHO International Chemical Safety Cards (ICSCs)
13.2.7 Explosive Limits and Potential
Lower 12.5% and upper 74.2% by volume.
International Program on Chemical Safety/Commission of the European Communities; International Chemical Safety Card on Carbon monoxide (April 2007).
Hazardous Substances Data Bank (HSDB)
A dangerous fire hazard when exposed to flame.
Lewis, R.J. Sr. (ed) Sax's Dangerous Properties of Industrial Materials. 11th Edition. Wiley-Interscience, Wiley & Sons, Inc. Hoboken, NJ. 2004., p. 708
Hazardous Substances Data Bank (HSDB)
Gas/air mixtures are explosive.
International Program on Chemical Safety/Commission of the European Communities; International Chemical Safety Card on Carbon monoxide (April 2007). Available from, as of November 17, 2009:
Hazardous Substances Data Bank (HSDB)
Explosive limits , vol% in air: 12.5-74.2
ILO-WHO International Chemical Safety Cards (ICSCs)
13.2.8 OSHA Standards
Permissible Exposure Limit: Table Z-1 8-hr Time Weighted Avg: 50 ppm (55 mg/cu m).
29 CFR 1910.1000 (USDOL); U.S. National Archives and Records Administration's Electronic Code of Federal Regulations. Available from, as of November 2, 2009:
Hazardous Substances Data Bank (HSDB)
Vacated 1989 OSHA PEL TWA 35 ppm (40 mg/cu m); Ceiling limit 200 ppm (229 mg/cu m) is still enforced in some states.
NIOSH. NIOSH Pocket Guide to Chemical Hazards. DHHS (NIOSH) Publication No. 97-140. Washington, D.C. U.S. Government Printing Office, 1997., p. 361
Hazardous Substances Data Bank (HSDB)
13.2.9 NIOSH Recommendations
Recommended Exposure Limit: 10 Hr Time-Weighted Avg: 35 ppm (40 mg/cu m).
NIOSH. NIOSH Pocket Guide to Chemical Hazards & Other Databases CD-ROM. Department of Health & Human Services, Centers for Disease Prevention & Control. National Institute for Occupational Safety & Health. DHHS (NIOSH) Publication No. 2005-151 (2005)
Hazardous Substances Data Bank (HSDB)
Recommended Exposure Limit: Ceiling Value: 200 ppm (229 mg/cu m).
NIOSH. NIOSH Pocket Guide to Chemical Hazards & Other Databases CD-ROM. Department of Health & Human Services, Centers for Disease Prevention & Control. National Institute for Occupational Safety & Health. DHHS (NIOSH) Publication No. 2005-151 (2005)
Hazardous Substances Data Bank (HSDB)
13.3 First Aid Measures
Inhalation First Aid
Fresh air, rest. Administration of oxygen may be needed. Artificial respiration may be needed. Refer immediately for medical attention.
ILO-WHO International Chemical Safety Cards (ICSCs)
13.3.1 First Aid
Excerpt from NIOSH Pocket Guide for Carbon monoxide:
Eye: FROSTBITE - If eye tissue is frozen, seek medical attention immediately; if tissue is not frozen, immediately and thoroughly flush the eyes with large amounts of water for at least 15 minutes, occasionally lifting the lower and upper eyelids. If irritation, pain, swelling, lacrimation, or photophobia persist, get medical attention as soon as possible.
Skin: FROSTBITE - If frostbite has occurred, seek medical attention immediately; do NOT rub the affected areas or flush them with water. In order to prevent further tissue damage, do NOT attempt to remove frozen clothing from frostbitten areas. If frostbite has NOT occurred, immediately and thoroughly wash contaminated skin with soap and water.
Breathing: RESPIRATORY SUPPORT - If a person breathes large amounts of this chemical, move the exposed person to fresh air at once. If breathing has stopped, perform artificial respiration. Keep the affected person warm and at rest. Get medical attention as soon as possible. (NIOSH, 2024)
CAMEO Chemicals
ERG 2024, Guide 168 (Carbon monoxide, refrigerated liquid (cryogenic liquid))
General First Aid:
· Call 911 or emergency medical service.
· Ensure that medical personnel are aware of the material(s) involved, take precautions to protect themselves and avoid contamination.
· Move victim to fresh air if it can be done safely.
· Administer oxygen if breathing is difficult.
· If victim is not breathing:
-- DO NOT perform mouth-to-mouth resuscitation; the victim may have ingestedor inhaled the substance.
-- If equipped and pulse detected, wash face and mouth, then give artificial respiration using a proper respiratory medical device (bag-valve mask, pocket mask equipped with a one-way valve or other device).
-- If no pulse detected or no respiratory medical device available, provide continuouscompressions. Conduct a pulse check every two minutes or monitor for any signs of spontaneous respirations.
· Remove and isolate contaminated clothing and shoes.
· For minor skin contact, avoid spreading material on unaffected skin.
· In case of contact with substance, remove immediately by flushing skin or eyes with running water for at least 20 minutes.
· For severe burns, immediate medical attention is required.
· Effects of exposure (inhalation, ingestion, or skin contact) to substance may be delayed.
· Keep victim calm and warm.
· Keep victim under observation.
· For further assistance, contact your local Poison Control Center.
· Note: Basic Life Support (BLS) and Advanced Life Support (ALS) should be done by trained professionals.
Specific First Aid:
· In case of contact with liquefied gas, only medical personnel should attempt thawing frosted parts.
Emergency Response Guidebook (ERG)
ERG 2024, Guide 119 (Carbon monoxide, compressed)
General First Aid:
· Call 911 or emergency medical service.
· Ensure that medical personnel are aware of the material(s) involved, take precautions to protect themselves and avoid contamination.
· Move victim to fresh air if it can be done safely.
· Administer oxygen if breathing is difficult.
· If victim is not breathing:
-- DO NOT perform mouth-to-mouth resuscitation; the victim may have ingestedor inhaled the substance.
-- If equipped and pulse detected, wash face and mouth, then give artificial respiration using a proper respiratory medical device (bag-valve mask, pocket mask equipped with a one-way valve or other device).
-- If no pulse detected or no respiratory medical device available, provide continuouscompressions. Conduct a pulse check every two minutes or monitor for any signs of spontaneous respirations.
· Remove and isolate contaminated clothing and shoes.
· For minor skin contact, avoid spreading material on unaffected skin.
· In case of contact with substance, remove immediately by flushing skin or eyes with running water for at least 20 minutes.
· For severe burns, immediate medical attention is required.
· Effects of exposure (inhalation, ingestion, or skin contact) to substance may be delayed.
· Keep victim calm and warm.
· Keep victim under observation.
· For further assistance, contact your local Poison Control Center.
· Note: Basic Life Support (BLS) and Advanced Life Support (ALS) should be done by trained professionals.
Specific First Aid:
· In case of contact with liquefied gas, only medical personnel should attempt thawing frosted parts.
· In case of burns, immediately cool affected skin for as long as possible with cold water. Do not remove clothing if adhering to skin.
In Canada, an Emergency Response Assistance Plan (ERAP) may be required for this product. Please consult the shipping paper and/or the "ERAP" section.
Emergency Response Guidebook (ERG)
(See general first aid procedures)
Eye: Frostbite - If eye tissue is frozen, seek medical attention immediately; if tissue is not frozen, immediately and thoroughly flush the eyes with large amounts of water for at least 15 minutes, occasionally lifting the lower and upper eyelids. If irritation, pain, swelling, lacrimation, or photophobia persist, get medical attention as soon as possible.
Skin: Frostbite - Compressed gases may create low temperatures when they expand rapidly. Leaks and uses that allow rapid expansion may cause a frostbite hazard. Wear appropriate personal protective clothing to prevent the skin from becoming frozen.
Breathing: Respiratory support
The National Institute for Occupational Safety and Health (NIOSH)
13.4 Fire Fighting
Excerpt from ERG Guide 119 [Gases - Toxic - Flammable]:
DO NOT EXTINGUISH A LEAKING GAS FIRE UNLESS LEAK CAN BE STOPPED.
SMALL FIRE: Dry chemical, CO2, water spray or alcohol-resistant foam.
LARGE FIRE: Water spray, fog or alcohol-resistant foam. FOR CHLOROSILANES, DO NOT USE WATER; use alcohol-resistant foam. If it can be done safely, move undamaged containers away from the area around the fire. Damaged cylinders should be handled only by specialists.
FIRE INVOLVING TANKS: Fight fire from maximum distance or use unmanned master stream devices or monitor nozzles. Cool containers with flooding quantities of water until well after fire is out. Do not direct water at source of leak or safety devices; icing may occur. Withdraw immediately in case of rising sound from venting safety devices or discoloration of tank. ALWAYS stay away from tanks in direct contact with flames. (ERG, 2024)
2024 Emergency Response Guidebook,
CAMEO Chemicals
Excerpt from ERG Guide 168 [Carbon Monoxide (Refrigerated Liquid)]:
CAUTION: Flame can be invisible. Use an alternate method of detection (thermal camera, broom handle, etc.) DO NOT EXTINGUISH A LEAKING GAS FIRE UNLESS LEAK CAN BE STOPPED.
SMALL FIRE: Dry chemical, CO2 or water spray.
LARGE FIRE: Water spray, fog or regular foam. If it can be done safely, move undamaged containers away from the area around the fire.
FIRE INVOLVING TANKS: Fight fire from maximum distance or use unmanned master stream devices or monitor nozzles. Cool containers with flooding quantities of water until well after fire is out. Do not direct water at source of leak or safety devices; icing may occur. Withdraw immediately in case of rising sound from venting safety devices or discoloration of tank. ALWAYS stay away from tanks in direct contact with flames. (ERG, 2024)
2024 Emergency Response Guidebook,
CAMEO Chemicals
Shut off supply; if not possible and no risk to surroundings, let the fire burn itself out. In other cases extinguish with carbon dioxide, water spray, powder. In case of fire: keep cylinder cool by spraying with water. Combat fire from a sheltered position.
ILO-WHO International Chemical Safety Cards (ICSCs)
13.4.1 Fire Fighting Procedures
Stop flow of gas before extinguishing fire. Use water spray to keep fire-exposed containers cool. Fire situation may require evacuation.
Fire Protection Guide to Hazardous Materials. 13 ed. Quincy, MA: National Fire Protection Association, 2002., p. 49-39
Hazardous Substances Data Bank (HSDB)
Use powder or carbon dioxide.
ITII. Toxic and Hazardous Industrial Chemicals Safety Manual. Tokyo, Japan: The International Technical Information Institute, 1988., p. 108
Hazardous Substances Data Bank (HSDB)
If material on fire or involved in fire: Do not extinguish fire unless flow can be stopped. Use water in flooding quantities as fog. Cool all affected containers with flooding quantities of water. Apply water from as far a distance as possible.
Association of American Railroads; Bureau of Explosives. Emergency Handling of Hazardous Materials in Surface Transportation. Association of American Railroads, Pueblo, CO. 2005, p. 178
Hazardous Substances Data Bank (HSDB)
Let fire burn; shut off flow of gas and cool adjacent exposures with water. Extinguish (only if wearing self-contained breathing apparatus) with dry chemicals or carbon dioxide.
U.S. Coast Guard, Department of Transportation. CHRIS - Hazardous Chemical Data. Volume II. Washington, D.C.: U.S. Government Printing Office, 1984-5.
Hazardous Substances Data Bank (HSDB)
13.4.2 Firefighting Hazards
Flame has very little color. Containers may explode in fire.
U.S. Coast Guard, Department of Transportation. CHRIS - Hazardous Chemical Data. Volume II. Washington, D.C.: U.S. Government Printing Office, 1984-5.
Hazardous Substances Data Bank (HSDB)
Carbon monoxide is the most frequent cause of immediate fire deaths, and carbon monoxide poisoning should be suspected in every fire victim. Carbon monoxide levels at fires may reach 10%, which can raise carboxyhemoglobin levels in active firefighters without respiratory protection to 75% within 1 minute.
Ellenhorn, M.J. and D.G. Barceloux. Medical Toxicology - Diagnosis and Treatment of Human Poisoning. New York, NY: Elsevier Science Publishing Co., Inc. 1988., p. 820
Hazardous Substances Data Bank (HSDB)
Asphyxiation due to carbon dioxide production may result /from combustion/.
U.S. Coast Guard, Department of Transportation. CHRIS - Hazardous Chemical Data. Volume II. Washington, D.C.: U.S. Government Printing Office, 1984-5.
Hazardous Substances Data Bank (HSDB)
13.5 Accidental Release Measures
Public Safety: ERG 2024, Guide 168 (Carbon monoxide, refrigerated liquid (cryogenic liquid))
· CALL 911. Then call emergency response telephone number on shipping paper. If shipping paper not available or no answer, refer to appropriate telephone number listed on the inside back cover.
· Keep unauthorized personnel away.
· Stay upwind, uphill and/or upstream.
· Many gases are heavier than air and will spread along the ground and collect in low or confined areas (sewers, basements, tanks, etc.).
· Ventilate closed spaces before entering, but only if properly trained and equipped.
Emergency Response Guidebook (ERG)
Spill or Leak: ERG 2024, Guide 168 (Carbon monoxide, refrigerated liquid (cryogenic liquid))
· ELIMINATE all ignition sources (no smoking, flares, sparks or flames) from immediate area.
· All equipment used when handling the product must be grounded.
· Do not touch or walk through spilled material.
· Stop leak if you can do it without risk.
· Use water spray to reduce vapors or divert vapor cloud drift. Avoid allowing water runoff to contact spilled material.
· Do not direct water at spill or source of leak.
· If possible, turn leaking containers so that gas escapes rather than liquid.
· Prevent entry into waterways, sewers, basements or confined areas.
· Isolate area until gas has dispersed.
Emergency Response Guidebook (ERG)
Spill or Leak: ERG 2024, Guide 119 (Carbon monoxide, compressed)
· ELIMINATE all ignition sources (no smoking, flares, sparks or flames) from immediate area.
· All equipment used when handling the product must be grounded.
· Do not touch or walk through spilled material.
· Stop leak if you can do it without risk.
· Do not direct water at spill or source of leak.
· Use water spray to reduce vapors or divert vapor cloud drift. Avoid allowing water runoff to contact spilled material.
· FOR CHLOROSILANES, use alcohol-resistant foam to reduce vapors.
· If possible, turn leaking containers so that gas escapes rather than liquid.
· Prevent entry into waterways, sewers, basements or confined areas.
· Isolate area until gas has dispersed.
Emergency Response Guidebook (ERG)
13.5.1 Isolation and Evacuation
Excerpt from ERG Guide 119 [Gases - Toxic - Flammable]:
IMMEDIATE PRECAUTIONARY MEASURE: Isolate spill or leak area for at least 100 meters (330 feet) in all directions.
SPILL: See ERG Table 1 - Initial Isolation and Protective Action Distances on the UN/NA 1016 datasheet.
FIRE: If tank, rail tank car or highway tank is involved in a fire, ISOLATE for 1600 meters (1 mile) in all directions; also, consider initial evacuation for 1600 meters (1 mile) in all directions. (ERG, 2024)
2024 Emergency Response Guidebook,
CAMEO Chemicals
Excerpt from ERG Guide 168 [Carbon Monoxide (Refrigerated Liquid)]:
IMMEDIATE PRECAUTIONARY MEASURE: Isolate spill or leak area for at least 100 meters (330 feet) in all directions.
SPILL: See ERG Table 1 - Initial Isolation and Protective Action Distances on the UN/NA 9202 datasheet.
FIRE: If tank, rail tank car or highway tank is involved in a fire, ISOLATE for 800 meters (1/2 mile) in all directions; also, consider initial evacuation for 800 meters (1/2 mile) in all directions. (ERG, 2024)
2024 Emergency Response Guidebook,
CAMEO Chemicals
Evacuation: ERG 2024, Guide 168 (Carbon monoxide, refrigerated liquid (cryogenic liquid))
Immediate precautionary measure
· Isolate spill or leak area for at least 100 meters (330 feet) in all directions.
Spill
· See Table 1 - Initial Isolation and Protective Action Distances.
Fire
· If tank, rail tank car or highway tank is involved in a fire, ISOLATE for 800 meters (1/2 mile) in all directions; also, consider initial evacuation for 800 meters (1/2 mile) in all directions.
Emergency Response Guidebook (ERG)
Isolation
Small spill:
ISOLATE in all directions: 30 m (100 ft)
Large spill:
ISOLATE in all directions: 200 m (600 ft)
Emergency Response Guidebook (ERG)
Protection
Small spill:
PROTECT people from downwind during DAY time: 0.1 km (0.1 mi)
PROTECT people from downwind during NIGHT time: 0.2 km (0.1 mi)
Large spill:
PROTECT people from downwind during DAY time: 1.2 km (0.7 mi)
PROTECT people from downwind during NIGHT time: 3.9 km (2.4 mi)
Emergency Response Guidebook (ERG)
Evacuation: ERG 2024, Guide 119 (Carbon monoxide, compressed)
Immediate precautionary measure
· Isolate spill or leak area for at least 100 meters (330 feet) in all directions.
Spill
· For highlighted materials: see Table 1 - Initial Isolation and Protective Action Distances.
· For non-highlighted materials: increase the immediate precautionary measure distance, in the downwind direction, as necessary.
Fire
· If tank, rail tank car or highway tank is involved in a fire, ISOLATE for 1600 meters (1 mile) in all directions; also, consider initial evacuation for 1600 meters (1 mile) in all directions.
Emergency Response Guidebook (ERG)
13.5.2 Spillage Disposal
Evacuate danger area! Consult an expert! Personal protection: self-contained breathing apparatus. Remove all ignition sources. Ventilation.
ILO-WHO International Chemical Safety Cards (ICSCs)
13.5.3 Cleanup Methods
Ventilate area of leak or release to disperse gas. 2. Stop flow of gas. If source of leak is a cylinder and the leak cannot be stopped in place, remove the leaking cylinder to a safe place in the open air and repair the leak or allow the cylinder to empty.
Mackison, F. W., R. S. Stricoff, and L. J. Partridge, Jr. (eds.). NIOSH/OSHA - Occupational Health Guidelines for Chemical Hazards. DHHS(NIOSH) Publication No. 81-123 (3 VOLS). Washington, DC: U.S. Government Printing Office, Jan. 1981., p. 3
Hazardous Substances Data Bank (HSDB)
Use water spray to cool and disperse vapors and protect personnel. With cryogenic liquids, releases may require isolation or evacuation.
Fire Protection Guide to Hazardous Materials. 13 ed. Quincy, MA: National Fire Protection Association, 2002., p. 49-39
Hazardous Substances Data Bank (HSDB)
Gas leakage: By forced ventilation, maintain concentration of gas below the range of explosive mixture. Remove the tank or cylinder to an open area. Leave to bleed off in the atmosphere.
ITII. Toxic and Hazardous Industrial Chemicals Safety Manual. Tokyo, Japan: The International Technical Information Institute, 1988., p. 108
Hazardous Substances Data Bank (HSDB)
13.5.4 Disposal Methods
SRP: The most favorable course of action is to use an alternative chemical product with less inherent propensity for occupational harm/injury/toxicity or environmental contamination. Recycle any unused portion of the material for its approved use or return it to the manufacturer or supplier. Ultimate disposal of the chemical must consider: the material's impact on air quality; potential migration in soil or water; effects on animal and plant life; and conformance with environmental and public health regulations.
Hazardous Substances Data Bank (HSDB)
Incineration: Remove leaky cylinders to remote area to empty; then return to supplier with label indicating that repairs are needed. The waste carbon monoxide can be piped to an approved incinerator or the cylinder can be placed in a pit to burn carbon monoxide to carbon dioxide under controlled conditions.
United Nations. Treatment and Disposal Methods for Waste Chemicals (IRPTC File). Data Profile Series No. 5. Geneva, Switzerland: United Nations Environmental Programme, Dec. 1985., p. 132
Hazardous Substances Data Bank (HSDB)
13.5.5 Preventive Measures
If material not on fire and not involved in fire: Keep sparks, flames, and other sources of ignition away. Keep material out of water sources and sewers. Attempt to stop leak if without undue personnel hazard. Use water spray to knock-down vapors.
Association of American Railroads; Bureau of Explosives. Emergency Handling of Hazardous Materials in Surface Transportation. Association of American Railroads, Pueblo, CO. 2005, p. 178
Hazardous Substances Data Bank (HSDB)
Personnel protection: Avoid breathing vapors. Keep upwind ... Do not handle broken packages unless wearing appropriate personal protective equipment. Approach fire with caution.
Association of American Railroads; Bureau of Explosives. Emergency Handling of Hazardous Materials in Surface Transportation. Association of American Railroads, Pueblo, CO. 2005, p. 178
Hazardous Substances Data Bank (HSDB)
Evacuation: If fire becomes uncontrollable or container is exposed to direct flame consider evacuation of one-third (1/3) mile radius. If material leaking (not on fire) consider evacuation from downwind area based on amount of material spilled, location and weather conditions.
Association of American Railroads; Bureau of Explosives. Emergency Handling of Hazardous Materials in Surface Transportation. Association of American Railroads, Pueblo, CO. 2005, p. 178
Hazardous Substances Data Bank (HSDB)
SRP: Operations involving entry into tanks or closed vessels, and emergency situations, require consideration of potentially oxygen deficient, or "immediately dangerous to life and health" IDLH environments. This may necessitate use of a self-contained breathing apparatus (SCUBA), or a positive pressure supplied air respirator.
Hazardous Substances Data Bank (HSDB)
For more Preventive Measures (Complete) data for Carbon monoxide (20 total), please visit the HSDB record page.
Hazardous Substances Data Bank (HSDB)
13.6 Handling and Storage
13.6.1 Nonfire Spill Response
Excerpt from ERG Guide 119 [Gases - Toxic - Flammable]:
ELIMINATE all ignition sources (no smoking, flares, sparks or flames) from immediate area. All equipment used when handling the product must be grounded. Do not touch or walk through spilled material. Stop leak if you can do it without risk. Do not direct water at spill or source of leak. Use water spray to reduce vapors or divert vapor cloud drift. Avoid allowing water runoff to contact spilled material. FOR CHLOROSILANES, use alcohol-resistant foam to reduce vapors. If possible, turn leaking containers so that gas escapes rather than liquid. Prevent entry into waterways, sewers, basements or confined areas. Isolate area until gas has dispersed. (ERG, 2024)
2024 Emergency Response Guidebook,
CAMEO Chemicals
Excerpt from ERG Guide 168 [Carbon Monoxide (Refrigerated Liquid)]:
ELIMINATE all ignition sources (no smoking, flares, sparks or flames) from immediate area. All equipment used when handling the product must be grounded. Do not touch or walk through spilled material. Stop leak if you can do it without risk. Use water spray to reduce vapors or divert vapor cloud drift. Avoid allowing water runoff to contact spilled material. Do not direct water at spill or source of leak. If possible, turn leaking containers so that gas escapes rather than liquid. Prevent entry into waterways, sewers, basements or confined areas. Isolate area until gas has dispersed. (ERG, 2024)
2024 Emergency Response Guidebook,
CAMEO Chemicals
13.6.2 Safe Storage
Fireproof. Cool. Keep in a well-ventilated room.
ILO-WHO International Chemical Safety Cards (ICSCs)
13.6.3 Storage Conditions
Store in a cool, dry, well-ventilated location.
Fire Protection Guide to Hazardous Materials. 13 ed. Quincy, MA: National Fire Protection Association, 2002., p. 49-39
Hazardous Substances Data Bank (HSDB)
Remove the sources of ignition. Electric installation should be explosion-proof construction. Protect container against sunlight, and store in well-ventilated, safe areas.
ITII. Toxic and Hazardous Industrial Chemicals Safety Manual. Tokyo, Japan: The International Technical Information Institute, 1988., p. 108
Hazardous Substances Data Bank (HSDB)
13.7 Exposure Control and Personal Protection
Protective Clothing: ERG 2024, Guide 168 (Carbon monoxide, refrigerated liquid (cryogenic liquid))
· Wear positive pressure self-contained breathing apparatus (SCBA).
· Wear chemical protective clothing that is specifically recommended by the manufacturer when there is NO RISK OF FIRE.
· Structural firefighters' protective clothing provides thermal protection but only limited chemical protection.
· Always wear thermal protective clothing when handling refrigerated/cryogenic liquids.
Emergency Response Guidebook (ERG)
Protective Clothing: ERG 2024, Guide 119 (Carbon monoxide, compressed)
· Wear positive pressure self-contained breathing apparatus (SCBA).
· Wear chemical protective clothing that is specifically recommended by the manufacturer when there is NO RISK OF FIRE.
· Structural firefighters' protective clothing provides thermal protection but only limited chemical protection.
Emergency Response Guidebook (ERG)
Exposure Summary
Biological Exposure Indices (BEI) [ACGIH] - Carboxyhemoglobin in blood = 3.5% of hemoglobin at end of shift; carbon monoxide in end-exhaled air = 20 ppm at end of shift;
TIH (Toxic Inhalation Hazard) - Term used to describe gases and volatile liquids that are toxic when inhaled. Some are TIH materials themselves, e.g., chlorine, and some release TIH gases when spilled in water, e.g., chlorosilanes. [ERG 2016].
ACGIH - Documentation of the TLVs and BEIs, 7th Ed. Cincinnati: ACGIH Worldwide, 2020.
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
Maximum Allowable Concentration (MAK)
30.0 [ppm]
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
13.7.1 Recommended Exposure Limit (REL)
REL-TWA (Time Weighted Average)
35 ppm (40 mg/m³)
Occupational Safety and Health Administration (OSHA)
REL-C (Ceiling)
200 ppm (229 mg/m³)
Occupational Safety and Health Administration (OSHA)
TWA 35 ppm (40 mg/m 3) C 200 ppm (229 mg/m 3)
The National Institute for Occupational Safety and Health (NIOSH)
13.7.2 Permissible Exposure Limit (PEL)
50.0 [ppm]
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
PEL-TWA (8-Hour Time Weighted Average)
50 ppm (55 mg/m³)
Occupational Safety and Health Administration (OSHA)
TWA 50 ppm (55 mg/m 3) See Appendix G
The National Institute for Occupational Safety and Health (NIOSH)
13.7.3 Immediately Dangerous to Life or Health (IDLH)
1200 ppm (NIOSH, 2024)
CAMEO Chemicals
1200.0 [ppm]
Excerpts from Documentation for IDLHs: Other human data: It has been stated that a 1hour exposure to 1,000 to 1,200 ppm would cause unpleasant but no dangerous symptoms, but that 1,500 to 2,000 ppm might be a dangerous concentration after 1 hour [Henderson et al. 1921a, 1921b]. In general, a carboxyhemoglobin (COHb) level of 1020% will only cause slight headaches [NIOSH 1972] and a COHb of 1113% will have no effect on hand and foot reaction time, hand steadiness, or coordination [Stewart and Peterson 1970]. At a COHb of 35%, manual dexterity is impaired [Stewart 1975]. At 40% COHb, mental confusion, added to increasing incoordination, precludes driving an automobile [Stewart 1975]. A 30minute exposure to 1,200 ppm will produce a COHb of 1013% [NIOSH 1972].
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
1200 ppm
NIOSH. NIOSH Pocket Guide to Chemical Hazards & Other Databases CD-ROM. Department of Health & Human Services, Centers for Disease Prevention & Control. National Institute for Occupational Safety & Health. DHHS (NIOSH) Publication No. 2005-151 (2005)
Hazardous Substances Data Bank (HSDB); Occupational Safety and Health Administration (OSHA)
1200 ppm
See: 630080
The National Institute for Occupational Safety and Health (NIOSH)
13.7.4 Threshold Limit Values (TLV)
25.0 [ppm]
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
8 hr Time Weighted Avg (TWA): 25 ppm.
American Conference of Governmental Industrial Hygienists. Threshold Limit Values of Chemical Substances and Biological Exposure Indices, ACGIH, Cincinnati, OH 2009, p. 18
Hazardous Substances Data Bank (HSDB)
Excursion Limit Recommendation: Excursions in worker exposure levels may exceed 3 times the TLV-TWA for no more than a total of 30 minutes during a work day, and under no circumstances should they exceed 5 times the TLV-TWA, provided that the TLV-TWA is not exceeded.
American Conference of Governmental Industrial Hygienists. Threshold Limit Values of Chemical Substances and Biological Exposure Indices, ACGIH, Cincinnati, OH 2009, p. 5
Hazardous Substances Data Bank (HSDB)
Biological Exposure Index (BEI): Determinant: carboxyhemoglobin in blood; Sampling Time: end of shift; BEI: 3.5% of hemoglobin. Determinant: carbon monoxide in end-exhaled air; Sampling Time: end of shift; BEI: 20 ppm. The determinant may be present in biological specimens collected from subjects who have not been occupationally exposed, at a concentration which could affect interpretation of the result. Such background concentrations are incorporated in the BEI value. The determinant is nonspecific, since it is also observed after exposure to other chemicals.
American Conference of Governmental Industrial Hygienists. Threshold Limit Values of Chemical Substances and Biological Exposure Indices, ACGIH, Cincinnati, OH 2009, p. 101
Hazardous Substances Data Bank (HSDB)
25 ppm as TWA; BEI issued.
ILO-WHO International Chemical Safety Cards (ICSCs)
TLV-TWA (Time Weighted Average)
25 ppm
Occupational Safety and Health Administration (OSHA)
13.7.5 Occupational Exposure Limits (OEL)
EU-OEL
23 mg/m
ILO-WHO International Chemical Safety Cards (ICSCs)
13.7.6 Emergency Response Planning Guidelines
Emergency Response: ERG 2024, Guide 168 (Carbon monoxide, refrigerated liquid (cryogenic liquid))
CAUTION: Flame can be invisible. Use an alternate method of detection (thermal camera, broom handle, etc.)
· DO NOT EXTINGUISH A LEAKING GAS FIRE UNLESS LEAK CAN BE STOPPED.
Small Fire
· Dry chemical, CO2 or water spray.
Large Fire
· Water spray, fog or regular foam.
· If it can be done safely, move undamaged containers away from the area around the fire.
Fire Involving Tanks
· Fight fire from maximum distance or use unmanned master stream devices or monitor nozzles.
· Cool containers with flooding quantities of water until well after fire is out.
· Do not direct water at source of leak or safety devices; icing may occur.
· Withdraw immediately in case of rising sound from venting safety devices or discoloration of tank.
· ALWAYS stay away from tanks in direct contact with flames.
Emergency Response Guidebook (ERG)
Emergency Response: ERG 2024, Guide 119 (Carbon monoxide, compressed)
· DO NOT EXTINGUISH A LEAKING GAS FIRE UNLESS LEAK CAN BE STOPPED.
Small Fire
· Dry chemical, CO2, water spray or alcohol-resistant foam.
Large Fire
· Water spray, fog or alcohol-resistant foam.
· FOR CHLOROSILANES, DO NOT USE WATER; use alcohol-resistant foam.
· If it can be done safely, move undamaged containers away from the area around the fire.
· Damaged cylinders should be handled only by specialists.
Fire Involving Tanks
· Fight fire from maximum distance or use unmanned master stream devices or monitor nozzles.
· Cool containers with flooding quantities of water until well after fire is out.
· Do not direct water at source of leak or safety devices; icing may occur.
· Withdraw immediately in case of rising sound from venting safety devices or discoloration of tank.
· ALWAYS stay away from tanks in direct contact with flames.
Emergency Response Guidebook (ERG)
ERPG-1: 200 ppm - one hour exposure limit: 1 = mild transient health effects or objectionable odor [AIHA]
ERPG-2: 350 ppm - one hour exposure limit: 2 = impaired ability to take protective action [AIHA]
ERPG-3: 500 ppm - one hour exposure limit: 3 = life threatening health effects [AIHA]
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
13.7.7 Other Standards Regulations and Guidelines
Emergency Response Planning Guidlines (ERPGs) for carbon monoxide:
ERPG
Maximum Airborne Concentration
ERPG
The ERPG-1: The maximum airborne concentration below which it is believed nearly all individuals could be exposed for up to 1 hour without experiencing more than mild, transient adverse health effects or without perceiving a clearly defined objectionable odor.
Maximum Airborne Concentration
200 ppm
ERPG
The ERPG-2: The maximum airborne concentration below which it is believed nearly all individuals could be exposed for up to 1 hour without experiencing or developing irreversible or other serious health effects or symptoms that could impair an individual's ability to take protective action.
Maximum Airborne Concentration
350 ppm
ERPG
The ERPG-3: The maximum airborne concentration below which it is believed nearly all individuals could be exposed for up to 1 hour without experiencing or developing life-threatening health effects.
Maximum Airborne Concentration
500 ppm
American Industrial Hygiene Association; Emergency Response Planning Guidelines & Workplace Enviromental Exposure Levels. Fairfax, VA 2009, p. 24
Hazardous Substances Data Bank (HSDB)
National Primary Ambient Air Quality Standards (NAAQS): Carbon monoxide: short term concentration averaging 35 ppm (40 mg/cu m) over 1 hr and 9 ppm (10 mg/cu m) over 8 hrs (not to be exceeded more than once a year).
American Society of Heating, Refrigerating and Air-Conditioning Engineers, Inc.; ASHRAE Standard: Ventilation for Acceptable Indoor Air Quality. ANSI/ASHRAE Standard 62.1-2007 and ASHRAE Addenda a,b,e,f, and h to ANSI/ASHRAE Standard 62.1-2007, 2008 Supplement. ASHRAE Customer Service. 1791 Tullie Circle NE, Atlanta, GA.
Hazardous Substances Data Bank (HSDB)
In the USA, a new voluntary standard for carbon monoxide detectors was published in 1992 by the Underwriters Laboratories (UL Standard 2034) and revised in 1995. This standard provides alarm requirements for detectors that are based on both the carbon monoxide concentration and the exposure time. It is designed so that an alarm is activated within 90 min of exposure to 110 mg/cu m (100 ppm), within 35 min of exposure to 230 mg/cu m (200 ppm) or within 15 min of exposure to 460 mg/cu m (400 ppm) (i.e., when exposures are equivalent to 10% carboxyhaemoglobin.
Environmental Health Criteria 213: Carbon Monoxide pp. 1-12 (1999) by the International Programme on Chemical Safety (IPCS) under the joint sponsorship of the United Nations Environment Programme, the International Labour Organisation and the World Health Organization.
Hazardous Substances Data Bank (HSDB)
The following WHO guideline values (ppm values rounded) and periods of time-weighted average exposures have been determined in such a way that the carboxyhemoglobin level of 2.5% is not exceeded, even when a normal subject engages in light or moderate exercise: 100 mg/cu m (87 ppm) for 15 min; 60 mg/cu m (52 ppm) for 30 min; 30 mg/cu m (26 ppm) for 1 hr; 10 mg/cu m (9 ppm) for 8 hr.
Environmental Health Criteria 213: Carbon Monoxide pp. 1-12 (1999) by the International Programme on Chemical Safety (IPCS) under the joint sponsorship of the United Nations Environment Programme, the International Labour Organisation and the World Health Organization.
Hazardous Substances Data Bank (HSDB)
13.7.8 Inhalation Risk
A harmful concentration of this gas in the air will be reached very quickly on loss of containment.
ILO-WHO International Chemical Safety Cards (ICSCs)
13.7.9 Effects of Short Term Exposure
The substance may cause effects on the blood. This may result in carboxyhaemoglobinemia and cardiac disorders. Medical observation is indicated. Exposure at high levels could cause death.
ILO-WHO International Chemical Safety Cards (ICSCs)
13.7.10 Effects of Long Term Exposure
The substance may have effects on the cardiovascular system and central nervous system. This may result in cardiovascular disorders and nervous system impairment. May cause toxicity to human reproduction or development.
ILO-WHO International Chemical Safety Cards (ICSCs)
13.7.11 Personal Protective Equipment (PPE)
Excerpt from NIOSH Pocket Guide for Carbon monoxide:
Skin: FROSTBITE - Compressed gases may create low temperatures when they expand rapidly. Leaks and uses that allow rapid expansion may cause a frostbite hazard. Wear appropriate personal protective clothing to prevent the skin from becoming frozen.
Eyes: FROSTBITE - Wear appropriate eye protection to prevent eye contact with the liquid that could result in burns or tissue damage from frostbite.
Wash skin: No recommendation is made specifying the need for washing the substance from the skin (either immediately or at the end of the work shift).
Remove: WHEN WET (FLAMMABLE) - Work clothing that becomes wet should be immediately removed due to its flammability hazard (i.e., for liquids with a flash point <100 °F).
Change: No recommendation is made specifying the need for the worker to change clothing after the workshift.
Provide: FROSTBITE WASH - Quick drench facilities and/or eyewash fountains should be provided within the immediate work area for emergency use where there is any possibility of exposure to liquids that are extremely cold or rapidly evaporating. (NIOSH, 2024)
CAMEO Chemicals
Compressed gases may create low temperatures when they expand rapidly. Leaks and uses that allow rapid expansion may cause a frostbite hazard. Wear appropriate personal protective clothing to prevent the skin from becoming frozen. Wear appropriate personal protective clothing to prevent the skin from becoming frozen.
NIOSH. NIOSH Pocket Guide to Chemical Hazards & Other Databases CD-ROM. Department of Health & Human Services, Centers for Disease Prevention & Control. National Institute for Occupational Safety & Health. DHHS (NIOSH) Publication No. 2005-151 (2005)
Hazardous Substances Data Bank (HSDB)
Wear appropriate eye protection to prevent eye contact with the liquid that could result in burns or tissue damage from frostbite.
NIOSH. NIOSH Pocket Guide to Chemical Hazards & Other Databases CD-ROM. Department of Health & Human Services, Centers for Disease Prevention & Control. National Institute for Occupational Safety & Health. DHHS (NIOSH) Publication No. 2005-151 (2005)
Hazardous Substances Data Bank (HSDB)
Quick drench facilities and/or eyewash fountains should be provided within the immediate work area for emergency use where there is any possibility of exposure to liquids that are extremely cold or rapidly evaporating.
NIOSH. NIOSH Pocket Guide to Chemical Hazards & Other Databases CD-ROM. Department of Health & Human Services, Centers for Disease Prevention & Control. National Institute for Occupational Safety & Health. DHHS (NIOSH) Publication No. 2005-151 (2005)
Hazardous Substances Data Bank (HSDB)
Respirator Recommendations: Up to 350 ppm
Assigned Protection Factor (APF)
Respirator Recommendation
Assigned Protection Factor (APF)
APF = 10
Respirator Recommendation
Any supplied-air respirator.
NIOSH. NIOSH Pocket Guide to Chemical Hazards & Other Databases CD-ROM. Department of Health & Human Services, Centers for Disease Prevention & Control. National Institute for Occupational Safety & Health. DHHS (NIOSH) Publication No. 2005-151 (2005)
Hazardous Substances Data Bank (HSDB)
For more Personal Protective Equipment (PPE) (Complete) data for Carbon monoxide (8 total), please visit the HSDB record page.
Hazardous Substances Data Bank (HSDB)
(See personal protection and sanitation codes)
Skin: Frostbite - Compressed gases may create low temperatures when they expand rapidly. Leaks and uses that allow rapid expansion may cause a frostbite hazard. Wear appropriate personal protective clothing to prevent the skin from becoming frozen.
Eyes: Frostbite - Wear appropriate eye protection to prevent eye contact with the liquid that could result in burns or tissue damage from frostbite.
Wash skin: No recommendation
Remove: When wet (flammable)
Change: No recommendation
Provide: Frostbite wash - Quick drench facilities and/or eyewash fountains should be provided within the immediate work area for emergency use where there is any possibility of exposure to liquids that are extremely cold or rapidly evaporating.
The National Institute for Occupational Safety and Health (NIOSH)
13.7.12 Respirator Recommendations
NIOSH
Up to 350 ppm:
(APF = 10) Any supplied-air respirator
Up to 875 ppm:
(APF = 25) Any supplied-air respirator operated in a continuous-flow mode
Up to 1200 ppm:
(APF = 50) Any air-purifying, full-facepiece respirator (gas mask) with a chin-style, front- or back-mounted canister providing protection against the compound of concern†
(APF = 50) Any self-contained breathing apparatus with a full facepiece
(APF = 50) Any supplied-air respirator with a full facepiece
Emergency or planned entry into unknown concentrations or IDLH conditions:
(APF = 10,000) Any self-contained breathing apparatus that has a full facepiece and is operated in a pressure-demand or other positive-pressure mode
(APF = 10,000) Any supplied-air respirator that has a full facepiece and is operated in a pressure-demand or other positive-pressure mode in combination with an auxiliary self-contained positive-pressure breathing apparatus
Escape:
(APF = 50) Any air-purifying, full-facepiece respirator (gas mask) with a chin-style, front- or back-mounted canister providing protection against the compound of concern†
Any appropriate escape-type, self-contained breathing apparatus
Important additional information about respirator selection
The National Institute for Occupational Safety and Health (NIOSH)
13.7.13 Preventions
Fire Prevention
NO open flames, NO sparks and NO smoking. Closed system, ventilation, explosion-proof electrical equipment and lighting. Use non-sparking handtools.
ILO-WHO International Chemical Safety Cards (ICSCs)
Exposure Prevention
AVOID ALL CONTACT!
ILO-WHO International Chemical Safety Cards (ICSCs)
Inhalation Prevention
Use ventilation, local exhaust or breathing protection.
ILO-WHO International Chemical Safety Cards (ICSCs)
13.8 Stability and Reactivity
13.8.1 Air and Water Reactions
Highly flammable.
CAMEO Chemicals
13.8.2 Reactive Group
Reducing Agents, Weak
CAMEO Chemicals
13.8.3 Reactivity Alerts
Highly Flammable
CAMEO Chemicals
13.8.4 Reactivity Profile
Bromine trifluoride and carbon monoxide react explosively at high temperatures or concentrations [Mellor 2 Supp. 1:166 1956]. The same is true for various oxidizers such as: chlorine dioxide, oxygen (liquid), peroxodisulfuryl difluoride. The product of the reaction between lithium and carbon monoxide, lithium carbonyl, detonates violently with water, igniting the gaseous products [Mellor 2, Supp. 2:84 1961]. Potassium and sodium metals behave similarly. Cesium oxide, iron(III) oxide, and silver oxide all react, in the presence of moisture, at ambient temperatures with carbon monoxide causing ignition, [Mellor, 1941, vol. 2, 487]. Contact of very cold liquefied gas with water may result in vigorous or violent boiling of the product and extremely rapid vaporization due to the large temperature differences involved. If the water is hot, there is the possibility that a liquid "superheat" explosion may occur. Pressures may build to dangerous levels if liquid gas contacts water in a closed container [Handling Chemicals Safely 1980].
CAMEO Chemicals
Contact of very cold liquefied gas with water may result in vigorous or violent boiling and extremely rapid vaporization. If the water is hot, a liquid "superheat" explosion may occur. Pressures may build to dangerous levels if the liquid contacts water in a closed container [Handling Chemicals Safely 1980]. Reacts explosively with bromine trifluoride at high temperatures or concentrations [Mellor 2, Supp. 1:166 1956]. The same is true for various oxidizers such as: chlorine dioxide, oxygen (liquid), peroxodisulfuryl difluoride. Reacts with lithium to give lithium carbonyl, which detonates violently with water, igniting the gaseous products [Mellor 2, Supp 2:84 1961]. Potassium and sodium metals behave similarly. Cesium oxide, iron(III) oxide, and silver oxide all react, in the presence of moisture, at ambient temperatures with carbon monoxide causing ignition, [Mellor, 1941, vol. 2, 487].
CAMEO Chemicals
13.8.5 Hazardous Reactivities and Incompatibilities
May react vigorously with oxygen, acetylene, chlorine, fluorine, nitrous oxide.
International Program on Chemical Safety/Commission of the European Communities; International Chemical Safety Card on Carbon monoxide (April 2007). Available from, as of November 17, 2009:
Hazardous Substances Data Bank (HSDB)
Strong oxidizers, bromine trifluoride, chlorine trifluoride, lithium.
NIOSH. NIOSH Pocket Guide to Chemical Hazards & Other Databases CD-ROM. Department of Health & Human Services, Centers for Disease Prevention & Control. National Institute for Occupational Safety & Health. DHHS (NIOSH) Publication No. 2005-151 (2005)
Hazardous Substances Data Bank (HSDB)
... Explosion /occurred/ during reduction of iron oxide with carbon monoxide /due to/ formation of pentacarbonyliron at temperatures between 0 and 150 °C.
Bretherick, L. Handbook of Reactive Chemical Hazards. 4th ed. Boston, MA: Butterworth-Heinemann Ltd., 1990, p. 1136
Hazardous Substances Data Bank (HSDB)
Carbon monoxide is exothermically oxidized over silver oxide, and the temperature may attain 300 °C.
Bretherick, L. Handbook of Reactive Chemical Hazards. 4th ed. Boston, MA: Butterworth-Heinemann Ltd., 1990, p. 16
Hazardous Substances Data Bank (HSDB)
For more Hazardous Reactivities and Incompatibilities (Complete) data for Carbon monoxide (8 total), please visit the HSDB record page.
Hazardous Substances Data Bank (HSDB)
13.9 Transport Information
13.9.1 DOT Emergency Guidelines
If ... THERE IS NO FIRE, go directly to the Table of Initial Isolation and Protective Action Distances /(see table below)/ ... to obtain initial isolation and protective action distances. IF THERE IS A FIRE, or IF A FIRE IS INVOLVED, go directly to the appropriate guide /(see guide(s) below)/ and use the evacuation information shown under PUBLIC SAFETY. /Carbon monoxide; Carbon monoxide, compressed; Carbon monoxide, refrigerated liquid (cryogenic liquid)/
Table: Table of Initial Isolation and Protective Action Distances for Carbon monoxide; Carbon monoxide, compressed; Carbon monoxide, refrigerated liquid (cryogenic liquid)
Small Spills (from a small package or small leak from a large package)
Small Spills (from a small package or small leak from a large package)
First ISOLATE in all Directions 30 meters (100 feet)
Then PROTECT persons Downwind during DAY: 0.1 kilometers (0.1 miles)
Then PROTECT persons Downwind during NIGHT: 0.1 kilometers (0.1 miles)
Small Spills (from a small package or small leak from a large package)
Large Spills (from a large package or from many small packages)
Small Spills (from a small package or small leak from a large package)
First ISOLATE in all Directions 150 meters (500 feet)
Then PROTECT persons Downwind during DAY: 0.7 kilometers (0.5 miles)
Then PROTECT persons Downwind during NIGHT: 2.7 kilometers (1.7 miles)
U.S. Department of Transportation. 2008 Emergency Response Guidebook. Washington, D.C. 2008300, 341
Hazardous Substances Data Bank (HSDB)
/GUIDE 119: GASES - TOXIC - FLAMMABLE/ Health: TOXIC; may be fatal if inhaled or absorbed through skin. Contact with gas or liquefied gas may cause burns, severe injury and/or frostbite. Fire will produce irritating, corrosive and/or toxic gases. Runoff from fire control may cause pollution. /Carbon monoxide; Carbon monoxide, compressed/
U.S. Department of Transportation. 2008 Emergency Response Guidebook. Washington, D.C. 2008
Hazardous Substances Data Bank (HSDB)
/GUIDE 119: GASES - TOXIC - FLAMMABLE/ Fire or Explosion: Flammable; may be ignited by heat, sparks or flames. May form explosive mixtures with air. Those substances designated with a "P" may polymerize explosively when heated or involved in a fire. Vapors from liquefied gas are initially heavier than air and spread along ground. Vapors may travel to source of ignition and flash back. Some of these materials may react violently with water. Cylinders exposed to fire may vent and release toxic and flammable gas through pressure relief devices. Containers may explode when heated. Ruptured cylinders may rocket. Runoff may create fire or explosion hazard. /Carbon monoxide; Carbon monoxide, compressed/
U.S. Department of Transportation. 2008 Emergency Response Guidebook. Washington, D.C. 2008
Hazardous Substances Data Bank (HSDB)
/GUIDE 119: GASES - TOXIC - FLAMMABLE/ Public Safety: CALL Emergency Response Telephone Number ... As an immediate precautionary measure, isolate spill or leak area for at least 100 meters (330 feet) in all directions. Keep unauthorized personnel away. Stay upwind. Many gases are heavier than air and will spread along ground and collect in low or confined areas (sewers, basements, tanks). Keep out of low areas. Ventilate closed spaces before entering. /Carbon monoxide; Carbon monoxide, compressed/
U.S. Department of Transportation. 2008 Emergency Response Guidebook. Washington, D.C. 2008
Hazardous Substances Data Bank (HSDB)
For more DOT Emergency Guidelines (Complete) data for Carbon monoxide (17 total), please visit the HSDB record page.
Hazardous Substances Data Bank (HSDB)
13.9.2 DOT ID and Guide
1016 119 9202 168 (cryogenic liquid)
The National Institute for Occupational Safety and Health (NIOSH)
13.9.3 Shipping Name / Number DOT/UN/NA/IMO
IMO 2.3; Carbon monoxide, compressed; Carbon monoxide, refrigerated liquid (cryogenic liquid)
Hazardous Substances Data Bank (HSDB)
NA 9202; Carbon monoxide, refrigerated liquid (cryogenic liquid)
Hazardous Substances Data Bank (HSDB)
UN 1016; Carbon monoxide, compressed
Hazardous Substances Data Bank (HSDB)
13.9.4 Standard Transportation Number
49 201 90; Carbon monoxide
Hazardous Substances Data Bank (HSDB)
13.9.5 Shipment Methods and Regulations
No person may /transport,/ offer or accept a hazardous material for transportation in commerce unless that person is registered in conformance ... and the hazardous material is properly classed, described, packaged, marked, labeled, and in condition for shipment as required or authorized by ... /the hazardous materials regulations (49 CFR 171-177)./
49 CFR 171.2; U.S. National Archives and Records Administration's Electronic Code of Federal Regulations. Available from, as of February 15, 2006:
Hazardous Substances Data Bank (HSDB)
The International Air Transport Association (IATA) Dangerous Goods Regulations are published by the IATA Dangerous Goods Board pursuant to IATA Resolutions 618 and 619 and constitute a manual of industry carrier regulations to be followed by all IATA Member airlines when transporting hazardous materials.
International Air Transport Association. Dangerous Goods Regulations. 47th Edition. Montreal, Quebec Canada. 2006., p. 158
Hazardous Substances Data Bank (HSDB)
The International Maritime Dangerous Goods Code lays down basic principles for transporting hazardous chemicals. Detailed recommendations for individual substances and a number of recommendations for good practice are included in the classes dealing with such substances. A general index of technical names has also been compiled. This index should always be consulted when attempting to locate the appropriate procedures to be used when shipping any substance or article.
International Maritime Organization. International Maritime Dangerous Goods Code. London, UK. 2004., p. 42
Hazardous Substances Data Bank (HSDB)
13.9.6 DOT Label
Poison Gas Flammable Gas
CAMEO Chemicals
13.9.7 EC Classification
Symbol: F+, T; R: 12-23-48/23-61; S: 53-45; Note: E
ILO-WHO International Chemical Safety Cards (ICSCs)
13.9.8 UN Classification
UN Hazard Class: 2.3; UN Subsidiary Risks: 2.1
ILO-WHO International Chemical Safety Cards (ICSCs)
13.10 Regulatory Information
The Australian Inventory of Industrial Chemicals
Chemical: Carbon monoxide
Australian Industrial Chemicals Introduction Scheme (AICIS)
California Safe Cosmetics Program (CSCP) Reportable Ingredient
Hazard Traits - Cardiovascular Toxicity; Developmental Toxicity; Neurotoxicity; Reproductive Toxicity
Authoritative List - ATSDR Neurotoxicants; EC Annex VI CMRs - Cat. 1A; OEHHA RELs; Prop 65
Report - regardless of intended function of ingredient in the product
California Safe Cosmetics Program (CSCP) Product Database
Status Regulation (EC)
2008/967
EU Pesticides Database
REACH Registered Substance
Status: Active Update: 27-04-2023
European Chemicals Agency (ECHA)
New Zealand EPA Inventory of Chemical Status
Carbon monoxide: HSNO Approval: HSR001056 Approved with controls
New Zealand Environmental Protection Authority (EPA)
13.11 Other Safety Information
Chemical Assessment
IMAP assessments - Carbon monoxide: Human health tier II assessment
Australian Industrial Chemicals Introduction Scheme (AICIS)
13.11.1 History and Incidents
During power outages after hurricanes, survivors can be at risk for carbon monoxide (CO) poisoning if they use portable generators improperly. On September 13, 2008, Hurricane Ike struck the coast of Texas, leaving approximately 2.3 million households in the southeastern portion of the state without electricity. Six days later, 1.3 million homes were still without electrical power. To assess the impact of storm-related CO exposures and to enhance prevention efforts, CDC analyzed data from five disparate surveillance sources on CO exposures reported during September 13--26 in counties of southeast Texas that were declared disaster areas by the federal government. ... One data source, Texas poison centers, received reports of 54 persons with storm-related CO exposures during the surveillance period. Another data source, the Undersea and Hyperbaric Medical Society (UHMS) hyperbaric oxygen treatment database, reported that 15 persons received hyperbaric oxygen treatment for storm-related CO poisoning. Medical examiners, public health officials, and hospitals in Texas reported that seven persons died from storm-related CO poisoning. Among the data sources, the percentage of reported storm-related CO exposures caused by improper generator use ranged from 82% to 87%.
PMID:19680219
CDC; MMWR Morb Mortal Wkly Rep 58 (31): 845-9 (2009). Available from, as of November 29, 2009;
Hazardous Substances Data Bank (HSDB)
After Hurricane Ike's landfall in September 2008, major power outages were associated with an epidemic of CO poisoning from electrical generators, as expected. Staff at Memorial Hermann Hospital-Texas Medical Center treated or telephone-triaged cases from the Houston area. A review of the details of those cases forms the basis of this report. ... Memorial Hermann Hospital-Texas Medical Center staff treated or triaged 37 individuals exposed to CO from gasoline-powered electrical generators in 13 incidents in the first 36 hours after landfall of the hurricane. Notably, 54% (20 of 37) of the patients were under the age of 18 years. Symptoms ranged from mild to severe, with 1 child dying at the scene. Eleven patients were treated with hyperbaric oxygen. Among 9 incidents in which the reason for generator use was determined, 5 were due to generators powering video games or televisions to watch movies or programs.
PMID:19482736
Fife CE et al; Pediatrics 123 (6): e1035-8 (2009);
Hazardous Substances Data Bank (HSDB)
The incidence and mechanisms of carbon monoxide exposure during the first 5 days after Hurricane Rita /are reported/, as experienced by a Disaster Medical Assistance Team staffing the only open health care facility in the Beaumont, Texas region after the storm. Improper placement of portable generators in indoor locations or proximate to home air conditioning intake systems were completely responsible for the 21 exposures including 5 fatalities, 1 brain dead, 2 transfers for hospitalization, and 13 treated and released.
PMID:17976553
Cukor J, Restuccia M; J Emerg Med 33 (3): 261-4 (2007). Available from, as of November 29, 2009;
Hazardous Substances Data Bank (HSDB)
On September 21, 2003, a violent explosion destroyed an underground distillation tower at the Isotec chemical manufacturing plant in Miami Township, OH, injuring one worker. The explosion ruptured a carbon monoxide gas pipe and led to a precautionary evacuation of about 2000 residents. The Isotec facility manufactures rare forms of oxygen and nitrogen, known as stable isotopes, which are used in research and medicine.
U.S. Chemical Safety Board; Isotec/Sigma Aldrich Nitric Oxide Explosion. Available from, as of September 15, 2009:
Hazardous Substances Data Bank (HSDB)
...The New York State Department of Health (NYSDOH) documented 234 events during January 2000--June 2004 in which CO releases resulted from underground utility cable fires (also known as CO burnout events). ... The findings underscore the need for preventive actions, such as installation of CO detectors in central locations in homes and businesses. ... Kings County (Brooklyn). In December 2003, an underground cable burnout caused CO to seep into a block of two-family homes. No one was injured; however, at least 20 residents were evacuated overnight as utility workers turned off the electricity. After the electricity was turned off, approximately 65 firefighters from 12 fire companies extinguished the fire. The street was excavated and blowers were installed to disperse the CO. Queens County (Flushing). In March 2003, an underground cable burnout released CO into the basement of a nursing home. The utility company reported a reading of 300 parts per million (ppm) CO; the recommended indoor air level for CO is <10 ppm for any 8-hour period and <25 ppm for any 1-hour period (5). Immediate response and venting by the utility company eliminated any need to evacuate nursing home residents. Kings County (Brooklyn). In February 2003, a total of 25 children and staff were evacuated from a private school after an underground cable burnout caused elevated CO levels. The fire department measured 115 ppm CO. An emergency crew from the utility company vented the building and allowed evacuees to return after 1 hour. Bronx County (Bronx). In January 2001, an underground cable burnout caused CO to seep into a laundromat. Four customers were exposed to CO and experienced gastrointestinal symptoms (e.g., nausea and vomiting). All four were treated at a hospital; two were admitted. A hazardous materials crew responded, and the building was evacuated. Bronx County (Bronx). In May 2000, CO from an underground cable burnout entered the basement of a medical center through conduits and the ventilation intake. CO levels in the basement were 1,300 ppm. An unknown number of building occupants were evacuated. The utility company used blowers to vent the basement. The evacuation lasted 3 hours. Surveillance Data. During January 2000--June 2004, NYSDOH reported 234 CO burnout events. All occurred in the New York City (NYC) metropolitan area in the following counties: Queens County (73 [31%]), New York County (72 [31%]), Kings County (59 [25%]), Bronx County (29 [12%]), and Richmond County (one [<1%]). The majority of these events (214 [91%]) occurred in commercial or residential areas. More than half (130 [56%]) occurred during November--February. Twelve of the burnout events resulted in injury to 37 persons, of whom 28 (76%) were members of the general public, five (13%) were firefighters, and four (11%) were of unknown affiliation. The injuries most frequently sustained included dizziness and other central nervous system symptoms and gastrointestinal irritation. Twenty-eight persons (76%) were treated at a hospital; of these, two (7%) were admitted, and 26 (93%) were treated and released. In addition, seven (19%) of the 37 injured persons were treated on the scene; for two (5%) persons, disposition was unknown. No fatalities were reported. Of the 234 events, 220 (94%) were known to have involved ordered evacuations in which at least 3,855 persons were evacuated (range: one to 810 persons). The average length of these ordered evacuations was 1.7 hours (range: <1--12 hours).
CDC; Underground Utility Cable Fires --- New York, January 2000--June 2004. MMWR 53 (39): 920-2 (2004). Available from, as of September 15, 2009
Hazardous Substances Data Bank (HSDB)
13.11.2 Special Reports
Environmental Health Criteria 213: Carbon Monoxide pp. 1-12 (1999). EHC are designed for scientists and administrators responsible for the establishment of safety standards and regulations and provide basic scientific risk evaluations of a wide range of chemicals and groups of chemicals.[Available from, as of November 28, 2009:
Hazardous Substances Data Bank (HSDB)
European Commission, ESIS; IUCLID Dataset, Carbon monoxide (630-08-0) (2000 CD-ROM edition) contains information on use, toxicology, and environmental effects of this chemical as supplied to the European Union by industry.[Available from, as of October 20, 2009:
Hazardous Substances Data Bank (HSDB)
14 Toxicity
14.1 Toxicological Information
CDC-ATSDR Toxicological Profile
Agency for Toxic Substances and Disease Registry (ATSDR)
14.1.1 Toxicity Summary
It is a product of the incomplete combustion of carbon-containing fuels and is also produced by natural processes or by biotransformation of halomethanes within the human body. With external exposures to additional carbon monoxide, subtle effects can begin to occur, and exposure to higher levels can result in death. The health effects of carbon monoxide are largely the result of the formation of carboxyhemoglobin (COHb), which impairs the oxygen carrying capacity of the blood ... During typical daily activities, people encounter carbon monoxide in a variety of microenvironments - while travelling in motor vehicles, working at their jobs, visiting urban locations associated with combustion sources, or cooking or heating with domestic gas, charcoal or wood fires - as well as in tobacco smoke. ... Studies of human exposure have shown that motor vehicle exhaust is the most important source for regularly encountered elevated carbon monoxide levels ... The workplace is another important setting for carbon monoxide exposures ... Certain industrial processes can expose workers to carbon monoxide produced directly or as a byproduct ... Carbon monoxide is absorbed through the lungs, and the concentration of carboxyhemoglobin will depend ... mainly on the concentrations of inspired carbon monoxide and oxygen ... and will also depend on the duration of exposure, pulmonary ventilation, and the concentration of carboxyhemoglobin originally present ... In addition to its reaction with hemoglobin, carbon monoxide combines with myoglobin, cytochromes, and metalloenzymes such as cytochromoe c oxidase and cytochrome P-450 ... The binding of carbon monoxide to hemoglobin, producing carboxyhemoglobin and decreasing the oxygen carrying capacity of blood, appears to be the principal mechanism of action underlying the induction of toxic effects of low-level carbon monoxide exposures. The precise mechanisms by which toxic effects are induced ... are not understood fully but likely include the induction of a hypoxic state in many tissues of diverse organ systems ... A unique feature of carbon monoxide exposure, therefore, is that the blood carboxyhemoglobin level represents a useful biological marker of the dose that the individual has received ... The formation of carboxyhemoglobin is a reversible process; however, because of the tight binding of carbon monoxide to hemoglobin, the elimination half-time is quite long, ranging from 2 to 6.5 hr ... The level of carboxyhemoglobin in the blood may be determined directly by blood analysis or indirectly by measuring carbon monoxide in exhaled breath ... Decreased oxygen uptake and the resultant decreased work capacity under maximal exercise conditions have clearly been shown to occur ... However, of greater concern at more typical ambient carbon monoxide exposure levels are certain cardiovascular effects (i.e., aggravation of angina symptoms during exercise) likely to occur in a smaller, but sizeable, segment of the general population. This group, chronic angina patients, is currently viewed as the most sensitive risk group for carbon monoxide exposure effects ... The adverse health consequences of low level carbon monoxide exposure to patients with ischemic heart disease are very difficult to predict in the at-risk population of individuals with heart disease ... At high carbon monoxide concentrations, excessive increases in hemoglobin and hematocrit may impose an additional workload on the heart and compromise blood flow to the tissues ... It is unlikely that carbon monoxide has any direct effects on lung tissue except for extremely high concentrations associated with carbon monoxide poisoning ... Occupational or accidental exposure to the products of combustion and pyrolysis, particularly indoors, may lead to acute decrements in lung function if the carboxyhemoglobin levels are high. It is difficult, however, to separate the potential effects of carbon monoxide from those due to other respiratory irritants in the smoke and exhaust ... Of special note are those individuals who are taking drugs with primary or secondary depressant effects that would be expected to exacerbate carbon monoxide-related neurobehavorial decrements. Other groups at possible increased risk for carbon monoxide-induced neurobehavorial effects are the aged and ill ... Under normal circumstances, the brain can increase blood flow or tissue oxygen extraction to compensate for the hypoxia caused by exposure to carbon monoxide ...
Environmental Health Criteria 213: Carbon Monoxide pp. 1-12 (1999) by the International Programme on Chemical Safety (IPCS) under the joint sponsorship of the United Nations Environment Programme, the International Labour Organisation and the World Health Organization.
Hazardous Substances Data Bank (HSDB)
... Studies in several laboratory animal species provide strong evidence that maternal carbon monoxide exposures ... produce reductions of birth weight, cardiomegaly, delays in behavorial development and disruptions in cognitive function ... Laboratory animal studies suggest that enzyme metabolism of xenobiotic compounds may be affected by carbon monoxide exposure ... The decreases in xenobiotic metabolism shown with carbon monoxide exposure might be important to individuals receiving treatment with drugs ... Tissues of highly active oxygen metabolism, such as heart, brain, liver, kidney, and muscle, may be particularly sensitive to carbon monoxide poisoning. There are reports ... of effects on liver, kidney, bone and the immune capacity of the lung and spleen. It is generally agreed that the severe tissue damage occurring during acute carbon monoxide poisoning is due to one of more of the following: (1) ischemia resulting from the formation of carboxyhemoglogin, (2) inhibition of oxygen release from oxyhemoglobin, (3) inhibition of oxygen release from oxyhemoglobin, (3) inhibition of cellular cytochrome function (e.g., cytochrome oxidases) and (4) metabolic acidosis ... Whereas certain data also suggest that perinatal effects (e.g., reduced birth weight, slowed post-natal developments, sudden infant death syndrome) are associated with carbon monoxide exposure, insufficient evidence exists by which to either qualitatively confirm such an association in humans or establish any pertinent exposure-effect relationships ... There remains little direct information on the possible enhancement of carbon monoxide toxicity by concomitant drug use or abuse ... The greatest evidence for a potentially important interaction of carbon monoxide comes from studies with alcohol in both laboratory animals and humans, where at least additive effects have been obtained. The significance of this is augmented by the high probable incidence of combined alcohol use and carbon monoxide exposure ... Besides being a source of carbon monoxide for smokers as well as non-smokers, tobacco smoke is also a source of other chemicals with which environmental carbon monoxide could interact ... On the basis of known effects described, patients with reproducible exercise-induced ischemia appear to be the best established as a sensitive group within the general population that is at increased risk for experiencing health effects of concern (i.e., decreased exercise duration due to exacerbation of cardiovascular symptoms) at ambient or near-ambient carbon monoxide concentrations ... Decrements in exercise duration in the healthy population would therefore be of concern mainly to competing athletes, rather than to ordinary people carrying out the common activities of daily life. It can be hypothesized, however, from both clinical and theoretical work and from experimental research on laboratory animals, that certain other groups in the population may be at probable risk from exposure to carbon monoxide. Identifiable probable risk groups can be categorized by gender differences; by age ...; by genetic variations ...; by pre-existing diseases ...; or by the use of medications, recreational drugs or alterations in environment ... Unfortunately, little empirical evidence is currently available by which to specify health effects associated with ambient or near-ambient carbon monoxide exposure to these probable risk groups ...
Environmental Health Criteria 213: Carbon Monoxide pp. 12-18 (1999) by the International Programme on Chemical Safety (IPCS) under the joint sponsorship of the United Nations Environment Programme, the International Labour Organisation and the World Health Organization.
Hazardous Substances Data Bank (HSDB)
... Carbon monoxide is responsible for a large percentage of the accidental poisonings and deaths reported throughout the world each year ... Outdoors, concentrations of carbon monoxide are highest near street intersections, in congested traffic, near exhaust gases from internal combustion engines and from industrial sources, and in poorly ventilated areas such as parking garages and tunnels. Indoors, carbon monoxide concentrations are highest in workplaces or in homes that have faulty or poorly vented combustion appliances or downdrafts or backdrafts. The symptoms and signs of acute carbon monoxide poisoning correlate poorly with the level of carboxyhemoglobin measured at the time of arrival at the hospital ... Neurological symptoms of carbon monoxide poisoning can ocur, such as headache, dizziness, weakness, nausea, confusion, disorientation and visual disturbances. Exertional dyspnea, increases in pulse and respiratory rates and syncope are observed with continuous exposure ... When carboxyhemoglobin levels are higher than 50%, convulsions and cardiopulmonary arrest may occur. Complications occur frequently in carbon monoxide poisoning (immediate death, myocardial impairment, hypotension, arrhythmias, pulmonary edema). Perhaps the most insidious effect of carbon monoxide poisoning is the delayed development of neuropyschiatric impairment ... and the neurobehavioral consequences, especially in children. Carbon monoxide poisoning during pregnancy results in high risk for the mother, by increasing the short-term complications rate and for the fetus by causing fetal death, developmental disorders, and cerebral anoxic lesions. Furthermore, the severity of fetal intoxication cannot be assessed by the maternal rate. Carbon monoxide poisoning occurs frequently, has severe consequences, including immediate death, involves complications and late sequelae and is often overlooked ...
Environmental Health Criteria 213: Carbon Monoxide pp. 18-19 (1999) by the International Programme on Chemical Safety (IPCS) under the joint sponsorship of the United Nations Environment Programme, the International Labour Organisation and the World Health Organization.
Hazardous Substances Data Bank (HSDB)
Carbon monoxide possesses a higher affinity than oxygen for hemoglobin, leading to the formation of carboxyhemoglobin, this provoking anoxemia. Carbon monoxide also binds to myoglobin, impairing its ability to utilize oxygen. It can also bind to cytochrome c oxidase, though with a lesser affinity than oxygen. This interferes with aerobic metabolism and efficient ATP synthesis. As a result, cells switch to anaerobic metabolism, causing anoxia, lactic acidosis, and eventual cell death. Carbon monoxide also causes endothelial cell and platelet release of nitric oxide, and the formation of oxygen free radicals. This results in lipid peroxidation, leading to edema and necrosis within the brain. (L961)
Toxin and Toxin Target Database (T3DB)
14.1.2 RAIS Toxicity Values
Inhalation Acute Reference Concentration (RfCa) (mg/m^3)
23
Inhalation Acute Reference Concentration Reference
CALEPA
Risk Assessment Information System (RAIS)
14.1.3 NIOSH Toxicity Data
The National Institute for Occupational Safety and Health (NIOSH)
14.1.4 Carcinogen Classification
Carcinogen Classification
No indication of carcinogenicity to humans (not listed by IARC).
Toxin and Toxin Target Database (T3DB)
14.1.5 Health Effects
Chronic exposure to low levels of carbon monoxide may cause persistent headaches, lightheadedness, depression, confusion, memory loss, and nausea and vomiting. (L961)
Toxin and Toxin Target Database (T3DB)
14.1.6 Exposure Routes
The substance can be absorbed into the body by inhalation.
ILO-WHO International Chemical Safety Cards (ICSCs)
inhalation, skin and/or eye contact (liquid)
The National Institute for Occupational Safety and Health (NIOSH)
Inhalation (L960)
Toxin and Toxin Target Database (T3DB)
14.1.7 Symptoms
Inhalation Exposure
Shortness of breath. Headache. Weakness. Irregular heartbeat. Chest pain. Nausea. Dizziness. Unconsciousness.
ILO-WHO International Chemical Safety Cards (ICSCs)
headache, tachypnea, nausea, lassitude (weakness, exhaustion), dizziness, confusion, hallucinations; cyanosis; depressed S-T segment of electrocardiogram, angina, syncope
The National Institute for Occupational Safety and Health (NIOSH)
Early symptoms of acute carbon monoxide poisoning are nonspecific and include headaches, nausea, and fatigue. Symptoms may progress to tachycardia and hypertension. The central nervous system is one of the organ systems most sensitive to poisoning and symptoms displayed include dizziness, ataxia, confusion, convulsions, unconsciousness, respiratory arrest, and even death. (L961)
Toxin and Toxin Target Database (T3DB)
14.1.8 Target Organs
Cardiovascular (Heart and Blood Vessels), Death, Developmental (effects while organs are developing), Hematological (Blood Forming), Neurological (Nervous System), Respiratory (From the Nose to the Lungs)
Agency for Toxic Substances and Disease Registry (ATSDR)
cardiovascular system, lungs, blood, central nervous system
The National Institute for Occupational Safety and Health (NIOSH)
14.1.9 Adverse Effects
Neurotoxin - Parkinsonism
Other Poison - Chemical Asphyxiant
Reproductive Toxin - A chemical that is toxic to the reproductive system, including defects in the progeny and injury to male or female reproductive function. Reproductive toxicity includes developmental effects. See Guidelines for Reproductive Toxicity Risk Assessment.
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
14.1.10 Acute Effects
ChemIDplus
14.1.11 Toxicity Data
LC50 (rat) = 1,807 ppm/4H
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
14.1.12 Treatment
Carbon monoxide poisoning is first treated by immediate removal from the source of exposure. High-flow or 100% oxygen should then be administered by a nonrebreather reservoir oxygen mask. Oxygen hastens the dissociation of carbon monoxide from hemoglobin, improving tissue oxygenation by reducing carbon monoxides biological half-life. Hyperbaric oxygen may also be used, as it increases carboxyhemoglobin dissociation to a greater extent than normal oxygen. (L961)
Toxin and Toxin Target Database (T3DB)
14.1.13 Interactions
Combined exposure to carbon monoxide plus hydrogen cyanide had an additive effect in rats, as evidenced by increases in mortality rate. Results from this series of experiments showed that the exposed animals died at lower carbon monoxide concentrations as the levels of hydrogen cyanide increased. In the presence of hydrogen cyanide, carboxyhemoglobin at equilibrium was less than that measured in the absence of hydrogen cyanide; however, the initial rate of carboxyhemoglobin formation was the same. This apparent depressive effect of hydrogen cyanide on carboxyhemoglobin formation may explain the reason for the low carboxyhemoglobin levels (<50%) seen in some people who died in a fire.
Environmental Health Criteria 213: Carbon Monoxide pp. 1-12 (1999) by the International Programme on Chemical Safety (IPCS) under the joint sponsorship of the United Nations Environment Programme, the International Labour Organisation and the World Health Organization.
Hazardous Substances Data Bank (HSDB)
Combined exposure of rats to carbon monoxide plus nitric oxide for 3 hr caused a significant (P < 0.01) increase in mean methemoglobin levels when compared with methemoglobin levels in rats exposed to nitric oxide alone. No significant changes were observed in blood carboxyhemoglobin levels compared with exposure to carbon monoxide alone or to carbon monoxide plus nitric oxide. Combined exposure also caused significant behavioral changes..
Environmental Health Criteria 213: Carbon Monoxide pp. 1-12 (1999) by the International Programme on Chemical Safety (IPCS) under the joint sponsorship of the United Nations Environment Programme, the International Labour Organisation and the World Health Organization.
Hazardous Substances Data Bank (HSDB)
In the study of interactions of intraperitoneal carbon monoxide administration with psychoactive drugs on operant behavior of mice, d-amphetamine, chlorpromazine, nicotine, diazepam and morphine were studied in addition to alcohol and pentobarbital. As with alcohol, a suggestion of greater than additive effects was obtained from combinations of carbon monoxide with both d-amphetamine and chlorpromazine; however, in these cases, the differences from additivity did not reach statistical significance. Effects of carbon monoxide in combination with nicotine, caffeine and morphine were additive.
Environmental Health Criteria 213: Carbon Monoxide pp. 1-12 (1999) by the International Programme on Chemical Safety (IPCS) under the joint sponsorship of the United Nations Environment Programme, the International Labour Organisation and the World Health Organization.
Hazardous Substances Data Bank (HSDB)
A large interaction of carbon monoxide exposure and alcohol administration /was observed/ on operant behavior in animals. Mice, trained to lever press for water reinforcement, were tested with 1.1 g alcohol/kg body weight and various doses of carbon monoxide, alone and in combination. An unusual feature of this study was that both alcohol and carbon monoxide were administered by intraperitoneal injection. A dose of alcohol that had little effect on rates of lever pressing when given alone resulted in large rate-decreasing effects when given in combination with doses of carbon monoxide that also had no effects when given alone. Typically, behavioral effects of carbon monoxide alone were not seen under these test conditions until carboxyhemoglobin saturations greater than 40-50% were obtained. Thus, alcohol about doubled the acute toxicity of carbon monoxide in this study.
Environmental Health Criteria 213: Carbon Monoxide pp. 1-12 (1999) by the International Programme on Chemical Safety (IPCS) under the joint sponsorship of the United Nations Environment Programme, the International Labour Organisation and the World Health Organization.
Hazardous Substances Data Bank (HSDB)
For more Interactions (Complete) data for Carbon monoxide (12 total), please visit the HSDB record page.
Hazardous Substances Data Bank (HSDB)
14.1.14 Antidote and Emergency Treatment
Treatment includes 100% oxygen and, in severe cases, hyperbaric oxygen. The half-life of carboxyhemoglobin is 6 hours at room air, 1.5 hours with 100% oxygen, and 23 minutes at three atmospheres of pressure.
Haddad, L.M., Clinical Management of Poisoning and Drug Overdose. 2nd ed. Philadelphia, PA: W.B. Saunders Co., 1990., p. 303
Hazardous Substances Data Bank (HSDB)
The prompt administration of oxygen is critical to maternal and fetal survival. In gestationally appropriate pregnancies, it is reasonable to use indicators of adequate fetal oxygenation central nervous system responsiveness (heart rate and variability), in addition to responses of the mother and her laboratory findings, in adjusting or terminating oxygen therapy. To ensure adequate treatment of the fetus, it has been recommended that the mother receive oxygen therapy for five times as long as it is expected to require to return her carbon monoxide concentrations to normal; this is how long it may take for fetal levels to normalize. The maternal carboxyhemoglobin elimination rate can be increased from a half-life of 2 to 3 hours to 3/4 of an hour by breathing 100% oxygen; fetal carboxyhemoglobin half-lives are expected to decrease from 6 to 7 hours to 2 to 4 hours by the use of maternal oxygen therapy. The fetal rate of elimination remains slower than that of the mother.
Haddad, L.M., Clinical Management of Poisoning and Drug Overdose. 2nd ed. Philadelphia, PA: W.B. Saunders Co., 1990., p. 427-8
Hazardous Substances Data Bank (HSDB)
Immediate first aid: Remove patient from contact with the material. Ensure that adequate decontamination has been carried out. If patient is not breathing, start artificial respiration, preferably with a demand-valve resuscitator, bag-valve-mask device, or pocket mask, as trained. Perform CPR as necessary. Immediately flush contaminated eyes with gently flowing water. Do not induce vomiting. If vomiting occurs, lean patient forward or place on left side (head-down position, if possible) to maintain an open airway and prevent aspiration. Keep patient quiet and maintain normal body temperature. Obtain medical attention. /Carbon Monoxide and Related Compounds/
Currance, P.L. Clements, B., Bronstein, A.C. (Eds).; Emergency Care For Hazardous Materials Exposure. 3Rd edition, Elsevier Mosby, St. Louis, MO 2005, p. 427
Hazardous Substances Data Bank (HSDB)
Basic treatment: Establish a patent airway (oropharyngeal or nasopharyngeal airway, if needed). Suction if necessary. Watch for signs of respiratory insufficiency and assist ventilations if necessary. Administer 100% oxygen by nonrebreather mask at 10 to 15 L/min. Monitor for shock and treat if necessary ... . Monitor for signs of an acute myocardial infarction and treat if necessary. Monitor for pulmonary edema and treat if necessary ... . Anticipate seizures and treat if necessary ... . For eye contamination, flush eyes immediately with water. Irrigate each eye continuously with 0.9% saline (NS) during transport ... . /Carbon Monoxide and Related Compounds/
Currance, P.L. Clements, B., Bronstein, A.C. (Eds).; Emergency Care For Hazardous Materials Exposure. 3Rd edition, Elsevier Mosby, St. Louis, MO 2005, p. 428
Hazardous Substances Data Bank (HSDB)
For more Antidote and Emergency Treatment (Complete) data for Carbon monoxide (8 total), please visit the HSDB record page.
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14.1.15 Medical Surveillance
This clinical policy focuses on critical issues concerning the management of adult patients presenting to the emergency department (ED) with acute symptomatic carbon monoxide (CO) poisoning. The subcommittee reviewed the medical literature relevant to the questions posed. The critical questions are: Should hyperbaric oxygen (HBO(2)) therapy be used for the treatment of patients with acute CO poisoning; and Can clinical or laboratory criteria identify CO-poisoned patients who are most or least likely to benefit from this therapy? Recommendations are provided on the basis of the strength of evidence of the literature. Level A recommendations represent patient management principles that reflect a high degree of clinical certainty; Level B recommendations represent patient management principles that reflect moderate clinical certainty; and Level C recommendations represent other patient management strategies that are based on preliminary, inconclusive, or conflicting evidence, or based on committee consensus. This clinical policy is intended for physicians working in hospital-based EDs. .
Wolf SJ et al; J Emergency Nursing 34 (2): 19-32 (2008); Available from; as of November 29, 2009:
Hazardous Substances Data Bank (HSDB)
Patients with acute carbon monoxide poisoning commonly have cognitive sequelae. ... A double-blind, randomized trial /was conducted/ to evaluate the effect of hyperbaric-oxygen treatment on such cognitive sequelae. METHODS: ... Patients with symptomatic acute carbon monoxide poisoning /were randomly assigned/ in equal proportions to three chamber sessions within a 24-hour period, consisting of either three hyperbaric-oxygen treatments or one normobaric-oxygen treatment plus two sessions of exposure to normobaric room air. Oxygen treatments were administered from a high-flow reservoir through a face mask that prevented rebreathing or by endotracheal tube. Neuropsychological tests were administered immediately after chamber sessions 1 and 3, and 2 weeks, 6 weeks, 6 months, and 12 months after enrollment. The primary outcome was cognitive sequelae six weeks after carbon monoxide poisoning. RESULTS: The trial was stopped after the third of four scheduled interim analyses, at which point there were 76 patients in each group. Cognitive sequelae at six weeks were less frequent in the hyperbaric-oxygen group (19 of 76 [25.0 percent]) than in the normobaric-oxygen group (35 of 76 [46.1 percent], P=0.007), even after adjustment for cerebellar dysfunction and for stratification variables (adjusted odds ratio, 0.45 [95 percent confidence interval, 0.22 to 0.92]; P=0.03). The presence of cerebellar dysfunction before treatment was associated with the occurrence of cognitive sequelae (odds ratio, 5.71 [95 percent confidence interval, 1.69 to 19.31]; P=0.005) and was more frequent in the normobaric-oxygen group (15 percent vs. 4 percent, P=0.03). Cognitive sequelae were less frequent in the hyperbaric-oxygen group at 12 months, according to the intention-to-treat analysis (P=0.04). CONCLUSIONS: Three hyperbaric-oxygen treatments within a 24-hour period appeared to reduce the risk of cognitive sequelae 6 weeks and 12 months after acute carbon monoxide poisoning.
Weaver LK et al; New Eng J Med 347 (14): 1057-67 (2002). Available from, as of November 29, 2009:
Hazardous Substances Data Bank (HSDB)
The aim of this study was to make a retrospective descriptive analysis of the features of children with acute carbon monoxide poisoning (COP). We evaluated 74 children (43 girls, 31 boys; age range 1 to 17.8 years) who were consecutively admitted to our emergency unit and hospitalized with accidental acute COP between June 2003 and June 2005. All patients received normobaric oxygen therapy until their carboxyhemoglobin (COHb) levels were decreased below 2% and their symptoms resolved. Thirty-eight of 74 patients (51.4%) also received hyperbaric oxygen (HBO) therapy as indicated by signs and symptoms or COHb levels. COHb levels were significantly higher and hospitalization period was longer in the children who had abnormal neurological findings (p<0.05 for both). All patients showed complete recovery without neurological sequelae except one who had visual impairment at discharge, and antiepileptic therapy was started because of epilepsy after seven months...
Yarar C et al; Turk J Pediat 50 (3): 235-41 (2008). Available from, as of November 29, 2009:
Hazardous Substances Data Bank (HSDB)
Clinical studies suggest that all patients admitted to hospital with moderate to severe CO poisoning should routinely undergo ECG and serial evaluation of cardiac markers, and that those with positive signs of myocardial cytonecrosis or preexisting ischemic heart disease should also undergo echocardiography. A finding of myocardial damage in patients with CO poisoning seems to indicate an unfavorable long-term prognosis, although it needs further confirmation.
Rastelli G et al; Giornale italiano di cardiologia 10 (4): 227-33 (2009). Available from, as of November 29, 2009:
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For more Medical Surveillance (Complete) data for Carbon monoxide (7 total), please visit the HSDB record page.
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14.1.16 Human Toxicity Excerpts
/HUMAN EXPOSURE STUDIES/ The most extensive human studies on the cardiorespiratory effects of carbon monoxide are those involving the measurement of oxygen uptake during exercise. Healthy young individuals were used in most of the studies evaluating the effects of carbon monoxide on exercise performance; healthy older individuals were used in only two studies. In all of these studies, oxygen uptake during submaximal exercise for short durations (5-60 min) was not affected by carboxyhemoglobin levels as high as 15-20%. Under conditions of short-term maximal exercise, however, statistically significant decreases (3-23%) in maximal oxygen uptake ( Func {V dot} O2 max) were found at carboxyhemoglobin levels ranging from 5% to 20%. In another study, the critical level at which carboxyhemoglobin marginally influenced Func {V dot} O2 max (P < 0.10) was approximately 4.3%. ... There is a linear relationship between decline in Func {V dot} O2 max and increase in carboxyhemoglobin that can be expressed as % decrease in Func {V dot} O2 max = 0.91 (% COHb) + 2.2. Short-term maximal exercise duration has also been shown to be reduced (3-38%) at carboxyhemoglobin levels ranging from 2.3% to 7%.
Environmental Health Criteria 213: Carbon Monoxide pp. 1-12 (1999) by the International Programme on Chemical Safety (IPCS) under the joint sponsorship of the United Nations Environment Programme, the International Labour Organisation and the World Health Organization.
Hazardous Substances Data Bank (HSDB)
/HUMAN EXPOSURE STUDIES/ Numerous studies have demonstrated that an increase in carboxyhemoglobin is associated with a compensatory increase in brain blood flow. Two sets of studies /were conducted/ on 14 and 12 young healthy men, respectively, measuring brain blood flow by the method of impedance plethysmography. In the first study, subjects were transiently exposed to various concentrations of carbon monoxide. In the second study, the exposure lasted 4 hr. The exposures produced carboxyhemoglobin levels up to 18.4%. The variation of the brain blood flow response among subjects was large and statistically significant. The authors speculated that changes in carbon monoxide-induced brain blood flow might be related to behavioural effects.
Environmental Health Criteria 213: Carbon Monoxide pp. 1-12 (1999) by the International Programme on Chemical Safety (IPCS) under the joint sponsorship of the United Nations Environment Programme, the International Labour Organisation and the World Health Organization.
Hazardous Substances Data Bank (HSDB)
/HUMAN EXPOSURE STUDIES/ 15 men /were exposed/ to 7652 mg C18O/cu m (6683 ppm) for 3.1-6.7 min at rest. Both arterial and venous blood carboxyhemoglobin levels were determined frequently during and for 10 min following the exposures. Except for the Haldane constant (M), which was assumed to be 245, all other physiological parameters of the CFK (Coburn, Forster, Kane) equation were measured for each individual from the very beginning. Arterial carboxyhemoglobin was considerably higher than the venous carboxyhemoglobin. The rate of increase in blood carboxyhemoglobin and the arterial-venous carboxyhemoglobin differences varied widely among individuals. The peak arterial carboxyhemoglobin concentration at the end of exposure ranged from 13.9 to 20.9%. The peak venous carboxyhemoglobin concentration reached during the recovery period ranged from 12.4 to 18.1%. The arterial-venous carboxyhemoglobin difference ranged from 2.3 to 12.1% carboxyhemoglobin. These increases in venous and arterial carboxyhemoglobin were not predicted accurately by the CFK equation. Venous blood carboxyhemoglobin levels were overestimated, whereas arterial blood carboxyhemoglobin levels were significantly and consistently underestimated by the CFK equation. Thus, exposure of such organs as brain or heart to carboxyhemoglobin may be substantially higher than expected during transient carbon monoxide exposure.
Environmental Health Criteria 213: Carbon Monoxide pp. 1-12 (1999) by the International Programme on Chemical Safety (IPCS) under the joint sponsorship of the United Nations Environment Programme, the International Labour Organisation and the World Health Organization
Hazardous Substances Data Bank (HSDB)
/HUMAN EXPOSURE STUDIES/ The rate of formation of carboxyhemoglobin /was measured. in healthy young males at a low (approximately 45 W) and moderate (approximately 90 W) exercise load. Individuals were exposed to 3400 mg carbon monoxide/cu m (3000 ppm) for 3 min at rest followed by three intermittent exposures, ranging from 3400 mg carbon monoxide/cu m (3000 ppm) for 1 min at low exercise to 764 mg carbon monoxide/cu m (667 ppm) at moderate exercise. The CFK equation underpredicted the increase in carboxyhemoglobin for the exposures at rest and the first exposure at exercise, whereas it overpredicted the carboxyhaemoglobin increase for the latter two exposures at exercise. The carboxyhemoglobin concentration after all exposures reached approximately 10%. The measured and predicted carboxyhemoglobin values differed by <1 percentage point. The slight shift of the measured carboxyhemoglobin dissociation curve from the predicted curve was attributed to a delay in the delivery of carboxyhaemoglobin to the blood sampling point, a dorsal hand vein.
Environmental Health Criteria 213: Carbon Monoxide pp. 1-12 (1999) by the International Programme on Chemical Safety (IPCS) under the joint sponsorship of the United Nations Environment Programme, the International Labour Organisation and the World Health Organization
Hazardous Substances Data Bank (HSDB)
For more Human Toxicity Excerpts (Complete) data for Carbon monoxide (90 total), please visit the HSDB record page.
Hazardous Substances Data Bank (HSDB)
14.1.17 Non-Human Toxicity Excerpts
/LABORATORY ANIMALS: Acute Exposure/ Carbon monoxide (CO) poisoning is a major cause of brain injury and mortality; delayed neurological syndrome (DNS) is encountered in survivors of acute CO exposure. The toxic effects of CO have been attributed to oxidative stress induced by hypoxia. Heme oxygenase-1 (HO-1) is the inducible heme oxygenase isoform, and its induction acts as an important cellular defense mechanism against oxidative stress, cellular injury and disease. In this study, we examined the functional roles of HO-1 induction in a rat model of CO-exposured hippocampal injury. We report that acute CO exposure produces severe hippocampal injury in rats. However, hemin pretreatment reduced both the CO-induced rise in hippocampal water content and levels of neuronal damage in the hippocampus; survival rates at 24 h were significantly improved. Upregulation of HO-1 by hemin pretreatment resulted in a significant decrease in hippocampal levels of malondialdehyde (MDA), a marker of oxidative stress; levels of pro-apoptotic caspase-3 were also reduced. In contrast, inhibition of HO activity by administration of tin protoporphyrin IX (SnPP, a specific inhibitor of HO) abolished the neuroprotective effects of HO-1 induction. These data suggested that the upregulation of endogenous HO-1 expression therefore plays a pivotal protective role in CO neurotoxicity. Though the precise mechanisms underlying hemin-mediated HO-1 induction and neuroprotection are not known, these may involve the anti-oxidant and anti-apoptotic effects of HO-1 enzyme activity.
Guan L et al; Toxicology 262 (2): 146-52 (2009). Available from, as of November 29, 2009:
Hazardous Substances Data Bank (HSDB)
/LABORATORY ANIMALS: Acute Exposure/ Inhalation of high levels of carbon monoxide, leading to carboxyhemoglobin concentrations greater than 10-15%, has been reported to cause a number of systemic effects in laboratory animals as well as effects in humans suffering from acute carbon monoxide poisoning. Tissues of highly active oxygen metabolism, such as heart, brain, liver, kidney and muscle, may be particularly sensitive to carbon monoxide poisoning.
Environmental Health Criteria 213: Carbon Monoxide pp. 1-12 (1999) by the International Programme on Chemical Safety (IPCS) under the joint sponsorship of the United Nations Environment Programme, the International Labour Organisation and the World Health Organization.
Hazardous Substances Data Bank (HSDB)
/LABORATORY ANIMALS: Acute Exposure/ Rats were exposed to high levels of carbon monoxide (1700-2900 mg/cu m [1500-2500 ppm]) for 90 min. The common carotid artery and jugular vein serving one side of the brain of the rats were occluded by indwelling catheter. The usual changes noted were hypotension, bradycardia, hypothermia, altered blood glucose, unconsciousness, cerebral edema and behavioral evidence of central nervous system damage. The authors found that both greatly depressed and greatly elevated glucose during and/or after carbon monoxide exposure increased morbidity and mortality. Added hypothermia induced during carbon monoxide exposure was found to reduce morbidity/enhance survival, whereas rapid rewarming was found to be the best strategy to reduce morbidity/enhance survival post-carbon monoxide exposure.
Environmental Health Criteria 213: Carbon Monoxide pp. 1-12 (1999) by the International Programme on Chemical Safety (IPCS) under the joint sponsorship of the United Nations Environment Programme, the International Labour Organisation and the World Health Organization.
Hazardous Substances Data Bank (HSDB)
/LABORATORY ANIMALS: Acute Exposure/ Carbon monoxide hypoxia results in an increase in cerebral blood flow, and this has been demonstrated by a number of investigators. The cerebral blood flow responses to carbon monoxide hypoxia /were examined/ in anesthetized dogs ... . A carboxyhemoglobin level as low as 2.5% resulted in a small, but significant, increase in cerebral blood flow to 102% of control. With reductions in oxygen carrying capacity of 5, 10, 20 and 30% (5, 10, 20 and 30% carboxyhemoglobin), cerebral blood flow increased to approximately 105, 110, 120 and 130% of control, respectively. At each of these levels, cerebral oxygen consumption remained unchanged. At carboxyhemoglobin levels above 30%, cerebral blood flow increased out of proportion to the decrease in oxygen carrying capacity, but the brain could no longer maintain cerebral oxygen consumption constant. At a carboxyhemoglobin level of 50%, cerebral blood flow increased to about 200% of control. These findings are in general agreement with those /that/ ... demonstrated that as carboxyhemoglobin increased to 20, 50 and 65%, cerebral blood flow increased to 200, 300 and then 400%, respectively, in cats. These cerebral blood flow increases at 20% carboxyhemoglobin are higher than those reported in /the previous/ study, but the reason for this is not known ... A 50-150% of control increase in cerebral blood flow with carboxyhemoglobin levels of 30-70% /was demonstrated/, and /another study/ showed a 26% increase in cerebral blood flow with a carboxyhemoglobin level of 20%. These findings also indicate that cerebral blood flow increases progressively with increasing carboxyhemoglobin concentrations and that cerebral oxygen consumption is maintained constant even at a carboxyhemoglobin level of 30%. This has important implications regarding the behavioral and electrophysiological consequences of carbon monoxide exposure ... At levels above /~40%/, the brain cannot increase blood flow enough to compensate for decreased tissue oxygen delivery. At these high carboxyhemoglobin levels, then, behavioral and neurophysiological abnormalities should be quite evident. At lower carboxyhemoglobin levels, these abnormalities should not be seen, because the brain can increase its blood flow or oxygen extraction to maintain a constant cerebral oxygen consumption ...
Environmental Health Criteria 213: Carbon Monoxide pp. 1-12 (1999) by the International Programme on Chemical Safety (IPCS) under the joint sponsorship of the United Nations Environment Programme, the International Labour Organisation and the World Health Organization.
Hazardous Substances Data Bank (HSDB)
For more Non-Human Toxicity Excerpts (Complete) data for Carbon monoxide (79 total), please visit the HSDB record page.
Hazardous Substances Data Bank (HSDB)
14.1.18 Non-Human Toxicity Values
LC50 Rat inhalation 1807 ppm/4 hr
Lewis, R.J. Sr. (ed) Sax's Dangerous Properties of Industrial Materials. 11th Edition. Wiley-Interscience, Wiley & Sons, Inc. Hoboken, NJ. 2004., p. 708
Hazardous Substances Data Bank (HSDB)
LC50 Rat inhalation 4600-5000 ppm 30 min
European Commission, ESIS; IUCLID Dataset, Carbon monoxide (630-08-0) p.22 (2000 CD-ROM edition). Available from, as of October 20, 2009:
Hazardous Substances Data Bank (HSDB)
LC50 Mouse inhalation 2444 ppm/4 hr
Lewis, R.J. Sr. (ed) Sax's Dangerous Properties of Industrial Materials. 11th Edition. Wiley-Interscience, Wiley & Sons, Inc. Hoboken, NJ. 2004., p. 708
Hazardous Substances Data Bank (HSDB)
LC50 ICR Mouse inhalation ca. 8000 ppm for 30 min
European Commission, ESIS; IUCLID Dataset, Carbon monoxide (630-08-0) p.24 (2000 CD-ROM edition). Available from, as of October 20, 2009:
Hazardous Substances Data Bank (HSDB)
For more Non-Human Toxicity Values (Complete) data for Carbon monoxide (6 total), please visit the HSDB record page.
Hazardous Substances Data Bank (HSDB)
14.1.19 Populations at Special Risk
Chronic angina patients, is currently viewed as the most sensitive risk group for carbon monoxide exposure effects.
Environmental Health Criteria 213: Carbon Monoxide pp. 1-12 (1999) by the International Programme on Chemical Safety (IPCS) under the joint sponsorship of the United Nations Environment Programme, the International Labour Organisation and the World Health Organization.
Hazardous Substances Data Bank (HSDB)
Patients with reproducible exercise-induced ischaemia appear to be best established as a sensitive group within the general population that is at increased risk for experiencing health effects of concern (i.e., decreased exercise duration due to exacerbation of cardiovascular symptoms) at ambient or near-ambient carbon monoxide exposure concentrations that result in carboxyhaemoglobin levels down to 3%. A smaller sensitive group of healthy individuals experiences decreased exercise duration at similar levels of carbon monoxide exposure, but only during short-term maximal exercise. Decrements in exercise duration in the healthy population would therefore be of concern mainly to competing athletes, rather than to ordinary people carrying out the common activities of daily life.
Environmental Health Criteria 213: Carbon Monoxide pp. 1-12 (1999) by the International Programme on Chemical Safety (IPCS) under the joint sponsorship of the United Nations Environment Programme, the International Labour Organisation and the World Health Organization.
Hazardous Substances Data Bank (HSDB)
Men with chronic bronchitis or asthma resist effect of carbon monoxide very badly and course of carbon monoxide poisoning is unfavorably influenced by alcoholism, obesity, and chronic disease of heart. Chronic vascular disease increases the damage done to basal ganglia.
Hamilton, A., and H. L. Hardy. Industrial Toxicology. 3rd ed. Acton, Mass.: Publishing Sciences Group, Inc., 1974., p. 243
Hazardous Substances Data Bank (HSDB)
The fetus may be extremely susceptible to effects of carbon monoxide, and the gas readily crosses the placenta.
Gilman, A.G., T.W. Rall, A.S. Nies and P. Taylor (eds.). Goodman and Gilman's The Pharmacological Basis of Therapeutics. 8th ed. New York, NY. Pergamon Press, 1990., p. 1620
Hazardous Substances Data Bank (HSDB)
For more Populations at Special Risk (Complete) data for Carbon monoxide (9 total), please visit the HSDB record page.
Hazardous Substances Data Bank (HSDB)
14.1.20 Protein Binding
Binds with very high affinity to hemoglobin in humans.
DrugBank
14.2 Ecological Information
14.2.1 Ecotoxicity Values
USDA APHIS Chemical Effects
Schafer, E. W., Jr., and W. A. Bowles Jr. 2004. Toxicity, Repellency of Phytotoxicity of 979 Chemicals to Birds, Mammals and Plants. Research Report No. 04-01. National Wildlife Research Center, Fort Collins, Colorado. 118 pp.
Schafer, E. W., Jr., W. A. Bowles Jr., and J. Hurlbut. 1983. The acute oral toxicity and repellency and hazard potential of 998 chemicals to one or more species of wild and domestic birds. Archives of Environmental Contamination and Toxicology 12:355-382.
USDA APHIS Chemical Effects Database
14.2.2 Ecotoxicity Excerpts
/AQUATIC SPECIES/ At concentrations of 28-350 mg/L in the air and a light intensity of 18 klux, carbon monoxide had no effect on growth of Chlorella cultures. However, at 4.5 klux, the lowest concentration of CO inhibited the algal growth rate.
European Commission, ESIS; IUCLID Dataset, Carbon monoxide (630-08-0) p.18 (2000 CD-ROM edition). Available from, as of November 11, 2009:
Hazardous Substances Data Bank (HSDB)
/OTHER TERRESTRIAL SPECIES/ Ladybirds and stick insects exposed to high levels of carbon monoxide (80% CO, 20% O) for < 10 days, all survived. When exposed for > 10 days, they died.
European Commission, ESIS; IUCLID Dataset, Carbon monoxide (630-08-0) p.20 (2000 CD-ROM edition). Available from, as of October 20, 2009:
Hazardous Substances Data Bank (HSDB)
/OTHER TERRESTRIAL SPECIES/ At levels of 100 ppm (50%), CO had negligible effects on the behavior of Enchytraeus species, Arion fasciatus, Tracheoniscus rathkei, Diploiulus species, Liobunum calcar, and several other forest litter invertebrates. Also, it had no effect on the health and biological functions of the various organisms.
European Commission, ESIS; IUCLID Dataset, Carbon monoxide (630-08-0) p.20 (2000 CD-ROM edition). Available from, as of October 20, 2009:
Hazardous Substances Data Bank (HSDB)
/PLANTS/ Leaves of 35 species of temperate and tropical plants absorbed CO in light from air containing 6 ppm CO at an average rate of 190 nL/kg of fresh wt. CO uptake by 9 species, having widely different rates of absorption, was proportional to CO concn in the range 0 to 100 ppm. Absorbed CO was metabolized either by oxidation to carbon dioxide and fixation as such or by reduction and incorporation into serine. CO had various effects on the photosynthesis of leaves of different species, acting like an inhibitory at concn as low as 65 ppm, or exerting no influence, or even permitting an increase in net CO2 fixation at 99% CO because of the absence of oxygen.
European Commission, ESIS; IUCLID Dataset, Carbon monoxide (630-08-0) p.21 (2000 CD-ROM edition). Available from, as of October 20, 2009:
Hazardous Substances Data Bank (HSDB)
14.2.3 Natural Pollution Sources
NATURAL SOURCES SUCH AS ATMOSPHERIC OXIDN OF METHANE, FOREST FIRES, TERPENE OXIDN & OCEAN (WHERE MICROORGANISMS PRODUCE CARBON MONOXIDE) ARE RESPONSIBLE FOR ABOUT 90% OF ATMOSPHERIC CARBON MONOXIDE; HUMAN ACTIVITY PRODUCES ABOUT 10%.
Gilman, A. G., L. S. Goodman, and A. Gilman. (eds.). Goodman and Gilman's The Pharmacological Basis of Therapeutics. 6th ed. New York: Macmillan Publishing Co., Inc. 1980., p. 1641
Hazardous Substances Data Bank (HSDB)
A small amount of carbon monoxide is produced normally in the body. This endogenous carbon monoxide is sufficient in amount to maintain a carbon monoxide hemoglobin saturation of about 0.4 to 0.7 percent. In some persons with blood disease, such as hemolytic anemia, the carbon monoxide saturation may reach 6 percent
PATTY. INDUS HYG & TOX 3RD ED VOL2A, 2B, 2C, 1981-1982 p.4117
Hazardous Substances Data Bank (HSDB)
14.2.4 Artificial Pollution Sources
WATER HEATERS ARE A COMMON SOURCE OF CARBON MONOXIDE.
Hamilton, A., and H. L. Hardy. Industrial Toxicology. 3rd ed. Acton, Mass.: Publishing Sciences Group, Inc., 1974., p. 239
Hazardous Substances Data Bank (HSDB)
MOTOR VEHICLES ACCOUNT FOR ABOUT 55 TO 60% OF GLOBAL MAN-MADE EMISSIONS OF CARBON MONOXIDE.
International Labour Office. Encyclopedia of Occupational Health and Safety. Vols. I&II. Geneva, Switzerland: International Labour Office, 1983., p. 396
Hazardous Substances Data Bank (HSDB)
SINCE MOST...POLYMERIC MATERIALS CONTAIN CARBON, CARBON MONOXIDE IS ONE OF THE PRIMARY GASES GENERATED FROM THE HEATING AND BURNING OF THESE MATERIALS /PLASTICS/.
Doull, J., C.D. Klaassen, and M. D. Amdur (eds.). Casarett and Doull's Toxicology. 2nd ed. New York: Macmillan Publishing Co., 1980., p. 551
Hazardous Substances Data Bank (HSDB)
Concentrations as high as 30% have been measured in automobile exhaust gas, although 7% is more common. Pyrolysis of some vinyl plastics results in the production of appreciable concentrations of carbon monoxide. Natural gas associated with petroleum deposits has no carbon monoxide but in processing natural gas (e.g., cracking), carbon monoxide may be produced. As distributed, manufactured gas commonly has a carbon monoxide content between 2 and 15% (by volume)
GOSSELIN. CTCP 5TH ED 1984 p.III-94
Hazardous Substances Data Bank (HSDB)
For more Artificial Pollution Sources (Complete) data for Carbon monoxide (8 total), please visit the HSDB record page.
Hazardous Substances Data Bank (HSDB)
14.2.5 Environmental Fate
ATMOSPHERIC FATE: A photochemical model was used to quantify the sensitivity of the tropospheric oxidants ozone (O3) and OH to changes in methane (CH4), carbon monoxide (CO), and NO emissions and to perturbations in climate and stratospheric chemistry. In most cases, incr CH4 and CO emissions will suppress OH (neg coefficients) in incr O3 (pos coefficients) except in areas where NO and O3 influenced by pollution are sufficient to incr OH. In most regions, NO, CO, and CH4 emission incr will suppress OH and incr O3, but these trends may be opposed by stratospheric O3 depletion and climate change.
Thompson AM, Stewart RW; Atmos Environ 23 (3): 519-32 (1989)
Hazardous Substances Data Bank (HSDB)
14.2.6 Other Environmental Concentrations
Unblended non filter cigarettes were made of the leaf and cutter of 5 kinds of bright tobacco cultivars and smoked to a 30 mm butt length on a smoking machine. Large variations were observed in the rates of formation of CO among the different kinds of tobacco. Leaf cigarette CO values ranged from 15.7 to 22.9 mg/cigarette, while cutter CO values ranged from 13.9 to 19.4 mg/cigarette. The CO formation rate was a more influential factor determining the amount of CO in mainstream smoke than the wt loss of the cigarette during puffs. Correlation coefficients were calculated for rate of CO formation and ethanolbenzene extract, hexane extract, nicotine, or potassium. The highest was with potassium (-0.95). The rate of formation of CO was mainly dependent on the potassium content of the tobacco and could be estimated from the amounts of potassium, total carbon, and lignin. The rates of formation of CO increased with a rise in combustion temperature, which in turn rose as the potassium content of the tobacco decr.
Yamamoto T et al; Beitr Tabakforsch Int 14 (3): 163-9 (1989)
Hazardous Substances Data Bank (HSDB)
Environmental tobacco smoke was analyzed after smoking of research cigarettes by a machine in an experimental chamber 13.6 cu m in volume. The ventilation rate was 3.55 air changes per hour. Air removed for sampling added about 0.5 air changes per hour. One cigarette was lit every 30 min and was smoked with a 35 ml puff of 2 sec every minute until extinguished after about 12 min. Mainstream smoke was vented to the outside of the chamber. Additional tests were performed with one cigarette smoked every 15 min and with several commercial cigarette brands. Carbon monoxide concentrations averaged 2.48 + or - 0.2 mg/cu m in the first series of 9 tests and 1.79 + or - 0.81 mg/cu m in a similar series. With one cigarette every 15 min the carbon monoxide concentrations averaged 4.76 + or - 0.21 mg/cu m. The airborne yield per cigarette was 67 mg of carbon monoxide. Concentrations of carbon monoxide varied in a saw toothed form with the pattern of smoking one cigarette every 30 min. The ratio of the average maximum to the minimum concentration was about 3. The average concentration of carbon monoxide was about 65 to 70% of the maximum concentration. The ventilation time of carbon monoxide corresponded to the predetermined air exchange rate of about 4 per hour. Concentrations of carbon monoxide using commercial brands of cigarettes in the chamber and in a tavern setting were similar to those produced by the research cigarettes.
Lofroth G et al; Environ Sci Technol 23 (5): 610-4 (1989)
Hazardous Substances Data Bank (HSDB)
14.2.7 Probable Routes of Human Exposure
...LARGE QUANTITIES OF CARBON MONOXIDE GAS RELEASED BY BURNING CHARCOAL CAN RESULT IN SEVERE POISONING OR DEATH. HIBACHIS SHOULD NEVER BE USED AS A SOURCE OF HEAT IN SLEEPING QUARTERS.
Arena, J. M. Poisoning: Toxicology, Symptoms, Treatments. Fourth Edition. Springfield, Illinois: Charles C. Thomas, Publisher, 1979., p. 240
Hazardous Substances Data Bank (HSDB)
CAR EXHAUST CONTAINS 1 TO 7% CARBON MONOXIDE. THIS IS WELL INTO...TOXIC RANGE...
Arena, J. M. Poisoning: Toxicology, Symptoms, Treatments. Fourth Edition. Springfield, Illinois: Charles C. Thomas, Publisher, 1979., p. 240
Hazardous Substances Data Bank (HSDB)
Occupational exposure to increased ambient carbon monoxide has been a major /concern/ to firefighters, traffic police, coal miners, coke oven and smelter workers, caisson workers, toll both attendants, and transportation mechanics. As commuting distances increase, workers driving to and from work are exposed to more ambient carbon monoxide.
Sullivan, J.B. Jr., G.R. Krieger (eds.). Hazardous Materials Toxicology-Clinical Principles of Environmental Health. Baltimore, MD: Williams and Wilkins, 1992., p. 540
Hazardous Substances Data Bank (HSDB)
15 Associated Disorders and Diseases
Comparative Toxicogenomics Database (CTD); Open Targets; Therapeutic Target Database (TTD)
Associated Occupational Diseases with Exposure to the Compound
Asphyxiation, chemical [Category: Acute Poisoning]
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
Disease
References
Disease
Smoking
References
PubMed: 6890513, 12194923, 8294547, 15649624, 10636262, 9137998, 19222926, 15878312, 18666198, 18029489, 19036546
Geigy Scientific Tables, 8th Rev edition, pp. 165-177. Edited by C. Lentner, West Cadwell, N.J.: Medical education Div., Ciba-Geigy Corp., Basel, Switzerland c1981-1992.
Human Metabolome Database (HMDB)
16 Literature
16.1 Consolidated References
PubChem
16.2 NLM Curated PubMed Citations
PubChem
16.3 Springer Nature References
Springer Nature
Springer Nature
16.4 Thieme References
Thieme Chemistry
16.5 Wiley References
Wiley
16.6 Chemical Co-Occurrences in Literature
PubChem
16.7 Chemical-Gene Co-Occurrences in Literature
PubChem
16.8 Chemical-Disease Co-Occurrences in Literature
PubChem
16.9 Chemical-Organism Co-Occurrences in Literature
PubChem
17 Patents
17.1 Depositor-Supplied Patent Identifiers
PubChem
Link to all deposited patent identifiers
PubChem
17.2 WIPO PATENTSCOPE
Patents are available for this chemical structure:
PATENTSCOPE (WIPO)
17.3 Chemical Co-Occurrences in Patents
PubChem
17.4 Chemical-Disease Co-Occurrences in Patents
PubChem
17.5 Chemical-Gene Co-Occurrences in Patents
PubChem
17.6 Chemical-Organism Co-Occurrences in Patents
PubChem
18 Interactions and Pathways
18.1 Protein Bound 3D Structures
RCSB Protein Data Bank (RCSB PDB)
View 361 proteins in NCBI Structure
PubChem
18.1.1 Ligands from Protein Bound 3D Structures
PDBe Ligand Code
CMO
PDBe Structure Code
1BZR
PDBe Conformer
Protein Data Bank in Europe (PDBe)
18.2 Chemical-Target Interactions
Comparative Toxicogenomics Database (CTD); Toxin and Toxin Target Database (T3DB)
DrugBank
18.3 Pathways
PubChem
19 Biological Test Results
19.1 BioAssay Results
PubChem
20 Taxonomy
E. coli Metabolome Database (ECMDB); ECI Group, LCSB, University of Luxembourg; LOTUS - the natural products occurrence database; Natural Product Activity and Species Source (NPASS)
21 Classification
21.1 MeSH Tree
Medical Subject Headings (MeSH)
21.2 NCI Thesaurus Tree
NCI Thesaurus (NCIt)
21.3 ChEBI Ontology
ChEBI
21.4 KEGG: ATC
KEGG
21.5 WHO ATC Classification System
WHO Anatomical Therapeutic Chemical (ATC) Classification
21.6 ChemIDplus
ChemIDplus
21.7 CAMEO Chemicals
CAMEO Chemicals
21.8 UN GHS Classification
GHS Classification (UNECE)
21.9 NORMAN Suspect List Exchange Classification
NORMAN Suspect List Exchange
21.10 EPA DSSTox Classification
EPA DSSTox
21.11 EPA TSCA and CDR Classification
EPA Chemicals under the TSCA
21.12 LOTUS Tree
LOTUS - the natural products occurrence database
21.13 EPA Substance Registry Services Tree
EPA Substance Registry Services
21.14 MolGenie Organic Chemistry Ontology
MolGenie
21.15 Chemicals in PubChem from Regulatory Sources
PubChem
22 Information Sources
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1145 | https://boepublic.parkhill.k12.mo.us/attachments/df5d5ff6-abd1-42ab-9da5-00194accb46a.pdf | [Type here] High School Geometry Curriculum Course Description: This course involves the integration of logical reasoning and spatial visualization skills. It includes a study of deductive proofs and applications from Algebra, an intense study of polygons, and an introduction to Trigonometry. Students will be required to “think visually” while transferring information to real life problems. Scope and Sequence: Time Frame (blocks) Unit 11 Basics of Geometry 7 Parallel and Perpendicular Lines 6 Transformations 9 Congruent Triangles 6 Quadrilaterals and Other Polygons 5 Similarity 11 Right Triangles and Trigonometry 11 Circumference, Area, Surface Area, and Volume 8 Probability Remainder Circles Board Approved: 2 | P a g e Curriculum Revision Tracking Spring, 2022 ● Adopted DESE Priority Standards ● Included Learning Targets and Success Criteria ● Updated units from: o Unit 1: Basic Geometry o Unit 2: Properties of 2-Dimensional Figures o Unit 3: Similarity o Unit 4: Transformations of Symmetry o Unit 5: Trigonometric Ratios and Pythagorean Theorem o Unit 6: Measurements of 2-Dimensional and 3-Dimensional Figures o Unit 7: Properties of Circles o Unit 8: Probability to o Unit 1: Basics of Geometry o Unit 2: Parallel and Perpendicular Lines o Unit 3: Transformations o Unit 4: Congruent Triangles o Unit 5: Quadrilaterals and Other Polygons o Unit 6: Similarity o Unit 7: Right Triangles and Trigonometry o Unit 8: Circumference, Area, Surface Area, and Volume o Unit 9: Probability o Unit 10: Circles ● Updated scope, sequence and pacing to match newly updated units Spring, 2018 ● Unit 6: Changed title from “Properties of 3-Dimensional Figures” to “Measurements of 2-Dimensional and 3-Dimensional Figures” Board Approved: 3 | P a g e ● Changed order of units from o Unit 1: Basic Geometry o Unit 2: Properties of 2-Dimensional Figures o Unit 3: Properties of Circles o Unit 4: Similarity o Unit 5: Measurements of 2-Dimensional and 3-Dimensional Figures o Unit 6: Transformations of Symmetry o Unit 7: Trigonometric Ratios and Pythagorean Theorem o Unit 8: Probability to o Unit 1: Basic Geometry o Unit 2: Properties of 2-Dimensional Figures o Unit 3: Similarity o Unit 4: Transformations of Symmetry o Unit 5: Trigonometric Ratios and Pythagorean Theorem o Unit 6: Measurements of 2-Dimensional and 3-Dimensional Figures o Unit 7: Properties of Circles o Unit 8: Probability ● Unit 2 o Changed “Topic 3: Perimeter and Area of Polygons” to “Perimeter and Area of 2- D Figures” and moved it to Topic 1 of Unit 6: Measurements of 2-Dimensional and 3-Dimensional Figures ● Unit 4 o Added priority standard Geo.GPE.B Board Approved: 4 | P a g e Unit 1: Basics of Geometry Subject: Geometry Grade: 9, 10, 11, 12 Name of Unit: Basics of Geometry Length of Unit: 11 blocks Overview of Unit: Students will learn precise definitions of line segment and angle, which are based on the undefined notions of point and line. They will make formal geometric constructions and find perimeters and areas of polygons in the coordinate plane. Students will also use geometric shapes, their measures, and their properties to describe objects. Students will then use precise definitions and prove geometric theorems. Priority Standards for unit: ● G.CO.D.11 Construct geometric figures using various tools and methods. Supporting Standards for unit: ● G.CO.A.1 Define angle, circle, perpendicular line, parallel line, line segment and ray based on the undefined notions of point, line, distance along a line and distance around a circular arc. ● G.CO.C.8 Prove theorems about lines and angles. Unwrapped Concepts (Students need to know) Unwrapped Skills (Students need to be able to do) Bloom’s Taxonomy Levels Webb's DOK geometric figures using various tools and methods Construct Create 4 Essential Questions: 1. How can you represent a three-dimensional figure with a two-dimensional drawing? 2. What are the building blocks of geometry? 3. How can you describe the attributes of a segment or angle? 4. How can you make a conjecture and prove that it is true? Board Approved: 5 | P a g e Enduring Understanding/Big Ideas: 1. Three-dimensional objects can be represented with a two-dimensional figure using special drawing techniques. 2. Geometry is a mathematical system built on accepted facts, basic terms, and definitions. 3. Number operations can be used to find and compare the lengths of segments and the measures of angles. 4. Special angle pairs can be used to identify geometric relationships and to find angle measures. 5. Formulas can be used to find the midpoint and length of any segment in the coordinate plane. 6. Perimeter and area are two different ways of measuring the size of geometric figures. 7. Some mathematical relationships can be described using a variety of if-then statements. 8. Algebraic properties of equality are used in geometry to solve problems and justify reasoning. 9. Given information, definitions, properties, postulates, and previously proven theorems can be used as reasons in proof. Unit Vocabulary: Academic Cross-Curricular Words Content/Domain Specific Plane Distance Formula point line line segment angle acute angle right angle obtuse angle complementary angles supplementary angles conditional statement hypothesis conclusion equivalent statements biconditional statement conjecture proof theorem Board Approved: 6 | P a g e Resources for Vocabulary Development: Textbook Resources Unit Postulates and Theorems: 1.1 Ruler Postulate 1.2 Segment Addition Postulate 1.3 Protractor Postulate 1.4 Angle Addition Postulate 2.1 Two Point Postulate 2.2 Line-Point Postulate 2.3 Line Intersection Postulate 2.4 Three Point Postulate 2.5 Plane-Point Postulate 2.6 Plane-Line Postulate 2.7 Plane Intersection Postulate 2.8 Linear Pair Postulate 2.1 Properties of Segment Congruence 2.2 Properties of Angle Congruence 2.3 Right Angles Congruence Theorem 2.4 Congruent Supplements Theorem 2.5 Congruent Complements Theorem 2.6 Vertical Angles Congruence Theorem Board Approved: 7 | P a g e Big Ideas Chapters 1 & 2: Basics of Geometry & Reasoning and Proofs Standard Topic & Section Suggested # of Blocks Learning Target Success Criteria G.CO.A.1, G.CO.D.11 1.1 Points, Lines, and Planes & 1.2 Measuring and Constructing Segments 1 Use defined terms and undefined terms, and measure and construct line segments. · I can describe a point, a line, and a plane. · I can define and name segments and rays. · I can sketch intersections of lines and planes. · I can measure a line segment. · I can copy a line segment. · I can explain and use the Segment Addition Postulate. G.CO.D.11 1.3 Using Midpoint and Distance Formulas 1 Find midpoints and lengths of segments. · I can find lengths of segments. · I can construct a segment bisector. · I can find the midpoint of a segment. G.CO.A.1, G.CO.D.11 1.5 Measuring and Constructing Angles 1 Measure, construct, and describe angles. · I can measure and classify angles. · I can construct congruent angles. · I can find angle measures. · I can construct an angle bisector. 1.6 Describing Pairs of Angles 1 Identify and use pairs of angles. · I can identify complementary and supplementary angles. · I can identify linear pairs and vertical angles. · I can find angle measures in pairs of angles. Review & Assess 2 Review & Assess Board Approved: 8 | P a g e 2.4 Algebraic Reasoning (+ if-then & converse from 2.1) 1 Use properties of equality to solve problems. · I can write conditional statements · I can identify algebraic properties of equality. · I can use algebraic properties of equality to solve equations. · I can use properties of equality to solve for geometric measures. G.CO.C.8 2.5 Proving Statements about Segments and Angles 1 Prove statements about segments and angles. · I can explain the structure of a two‐ column proof. · I can write a two‐column proof. · I can identify properties of congruence. G.CO.C.8 2.6 Proving Geometric Relationships 1 Prove geometric relationships. · I can prove geometric relationships by writing flowchart proofs. · I can prove geometric relationships by writing paragraph proofs. Review & Assess 2 Review & Assess Engaging Scenarios Engaging Scenarios: There are several options for engaging scenarios for this unit. They may be appropriate as a means for introducing concepts in the unit, as a unit-long project for attaching learning to, or as a culminating experience at the end of the unit. ● Performance Task found in the Big Ideas Assessment Book (accessible in the physical Assessment Book, or the digital version in the Assess area of the Featured Chapter Resources after selecting the appropriate Chapter). ● Performance Task found at the end of the chapter in Big Ideas. ● Performance Task found in the Teach area of the Featured Chapter Resources after selecting the appropriate Chapter. Board Approved: 9 | P a g e Unit 2: Parallel and Perpendicular Lines Subject: Geometry Grade: 9, 10, 11, 12 Name of Unit: Parallel and Perpendicular Lines Length of Unit: 7 blocks Overview of Unit: Students will learn the precise definition of parallel line. They will prove theorems about lines and angles and make formal geometric constructions. Students will also partition directed line segments and use the slope criteria for parallel and perpendicular lines. Priority Standards for unit: ● G.CO.D.11 Construct geometric figures using various tools and methods. Supporting Standards for unit: ● G.CO.A.1 Define angle, circle, perpendicular line, parallel line, line segment and ray based on the undefined notions of point, line, distance along a line and distance around a circular arc. ● G.CO.C.8 Prove theorems about lines and angles. ● G.GPE.B.4 Prove the slope criteria for parallel and perpendicular lines and use them to solve problems. ● G.GPE.B.5 Find the point on a directed line segment between two given points that partitions the segment in a given ratio. Unwrapped Concepts (Students need to know) Unwrapped Skills (Students need to be able to do) Bloom’s Taxonomy Levels Webb's DOK geometric figures using various tools and methods Construct Create 4 Essential Questions: 1. How do you prove that two lines are parallel or perpendicular? 2. What is the sum of the measures of the angles of a triangle? 3. How do you write the equation of a line in the coordinate plane? Board Approved: 10 | P a g e Enduring Understanding/Big Ideas: 1. Definitions establish meanings and remove possible misunderstanding. 2. Some attributes of geometric figures, such as length, area, volume, and angle measure, are measurable. Units are used to describe these attributes. 3. A coordinate system on a line is a number line on which points are labeled Corresponding to the real numbers. 4. A coordinate system in a plane is formed by two perpendicular number lines called the x- and y-axes, and the quadrants they form. Unit Vocabulary: Academic Cross-Curricular Words Content/Domain Specific parallel lines parallel planes transversal corresponding angles alternate interior angles alternate exterior angles consecutive interior angles perpendicular bisector Resources for Vocabulary Development: Textbook Resources Board Approved: 11 | P a g e Unit Postulates and Theorems: 3.1 Parallel Postulate 3.2 Perpendicular Postulate 3.1 Corresponding Angles Theorem 3.2 Alternate Interior Angles Theorem 3.3 Alternate Exterior Angles Theorem 3.4 Consecutive Interior Angles Theorem 3.5 Corresponding Angles Converse 3.6 Alternate Interior Angles Converse 3.7 Alternate Exterior Angles Converse 3.8 Consecutive Interior Angles Converse 3.9 Transitive Property of Parallel Lines 3.10 Linear Pair Perpendicular Theorem 3.11 Perpendicular Transversal Theorem 3.12 Lines Perpendicular to a Transversal 3.13 Slopes of Parallel Lines 3.14 Slopes of Perpendicular Lines Board Approved: 12 | P a g e Big Ideas Chapter 3 Parallel and Perpendicular Lines Standard Topic & Section Suggested # of Blocks Learning Target Success Criteria G.CO.A.1 3.1 Pairs of Lines and Angles 1 Understand lines, planes, and pairs of angles. · I can identify lines and planes. · I can identify parallel and perpendicular lines. · I can identify pairs of angles formed by transversals. G.CO.A.1, G.CO.C.8 3.2 Parallel Lines and Transversals 1 Prove and use theorems about parallel lines. · I can use properties of parallel lines to find angle measures. · I can prove theorems about parallel lines. G.CO.C.8, G.CO.D.11 3.3 Proofs with Parallel Lines 1 Prove and use theorems about identifying parallel lines. · I can use theorems to identify parallel lines. · I can construct parallel lines. · I can prove theorems about identifying parallel lines. G.CO.A.1, G.CO.C.8 3.4 Proofs with Perpendicular Lines 1 Prove and use theorems about perpendicular lines. · I can find the distance from a point to a line. · I can construct perpendicular lines and perpendicular bisectors. · I can prove theorems about perpendicular lines. Board Approved: 13 | P a g e G.GPE.B.4, G.GPE.B.5 3.5 Equations of Parallel and Perpendicular Lines 1 Partition a directed line segment and understand slopes of parallel and perpendicular lines. · I can partition directed line segments using slope. · I can use slopes to identify parallel and perpendicular lines. · I can write equations of parallel and perpendicular lines. · I can find the distance from a point to a line. Review & Assess 2 Engaging Scenario Engaging Scenarios: There are several options for engaging scenarios for this unit. They may be appropriate as a means for introducing concepts in the unit, as a unit-long project for attaching learning to, or as a culminating experience at the end of the unit. ● Performance Task found in the Big Ideas Assessment Book (accessible in the physical Assessment Book, or the digital version in the Assess area of the Featured Chapter Resources after selecting the appropriate Chapter). ● Performance Task found at the end of the chapter in Big Ideas. ● Performance Task found in the Teach area of the Featured Chapter Resources after selecting the appropriate Chapter. ● Safe Crossings Project: Students will work on building a model bridge to a school over a busy road - the bridge must be perpendicular to the existing sidewalk and then create a parallel sidewalk on the other side of the road. (worksheet can be found in the district Geometry Shell Course) Board Approved: 14 | P a g e Unit 3: Transformations Subject: Geometry Grade: 9, 10, 11, 12 Name of Unit: Transformations Length of Unit: 6 blocks Overview of Unit: Students will understand congruence and similarity in terms of transformations. Students will learn that rigid motions preserve distance and angle measure, whereas nonrigid transformations may change the shape or size of a figure. This chapter also establishes the approach of using rigid motions to identify congruent figures. Priority Standards for unit: ● G.CO.A.5 Demonstrate the ability to rotate, reflect or translate a figure, and determine a possible sequence of transformations between two congruent figures. ● G.CO.B.6 Develop the definition of congruence in terms of rigid motions. ● G.CO.D.11 Construct geometric figures using various tools and methods. Supporting Standards for unit: ● G.CO.A.2 Represent transformations in the plane and describe them as functions that take points in the plane as inputs and give other points as outputs. ● G.CO.A.3 Describe the rotational symmetry and lines of symmetry of two- dimensional figures. ● G.CO.A.4 Develop definitions of rotations, reflections, and translations in terms of angles, circles, perpendicular lines, parallel lines, and line segments. ● G.SRT.A.1 Construct and analyze scale changes of geometric figures. ● G.SRT.A.2 Use the definition of similarity to decide if figures are similar and to solve problems involving similar figures. Board Approved: 15 | P a g e Unwrapped Concepts (Students need to know) Unwrapped Skills (Students need to be able to do) Bloom’s Taxonomy Levels Webb's DOK the ability to rotate, reflect or translate a figure, and determine a possible sequence of transformations between two congruent figures. Demonstrate Knowledge 1 the definition of congruence in terms of rigid motions. Develop Create 4 geometric figures using various tools and methods. Construct Create 4 Essential Questions: 1. How can you change a figure’s position without changing its size and shape? 2. How can you change a figure’s size without changing its shape? 3. How can you represent a transformation in the coordinate plane? 4. How do you recognize congruence and similarity in figures? Enduring Understanding/Big Ideas: 1. Transformations may be described geometrically or by coordinates. Symmetries of figures may be defined and classified by transformations. 2. A coordinate system in a plane is formed by two perpendicular number lines, called the x-and y-axes, and the quadrants they form. It is possible to verify some complex truths using deductive reasoning in combination with distance, midpoint, and slope formulas. 3. Visualization can help you see the relationships between two figures and help you connect properties of real objects with two-dimensional drawings of these objects. Board Approved: 16 | P a g e Unit Vocabulary: Academic Cross-Curricular Words Content/Domain Specific transformation translation rigid motion reflection glide reflection line symmetry rotation rotational symmetry congruent figures dilation scale factor similar figures Resources for Vocabulary Development: Textbook Resources Unit Postulates and Theorems: 4.1 Translation Postulate 4.2 Reflection Postulate 4.3 Rotation Postulate 4.1 Composition Theorem 4.2 Reflections in Parallel Lines Theorem 4.3 Reflections in Intersecting Lines Theorem Board Approved: 17 | P a g e Big Ideas Chapter 4: Transformations Standard Topic & Section Suggested # of Blocks Learning Target Success Criteria G.CO.A.2, G.CO.A.4, G.CO.A.5 4.1 Translations 1 Understand translations of figures. · I can translate figures. · I can write a translation rule for a given translation. · I can explain what a rigid motion is. · I can perform a composition of translations on a figure. G.CO.A.2, G.CO.A.4, G.CO.A.5 4.2 Reflections 1 Understand reflections of figures. · I can reflect figures. · I can perform compositions with reflections. · I can identify line symmetry in polygons. G.CO.A.2, G.CO.A.3, G.CO.A.4, G.CO.A.5, G.CO.D.11, G.CO.B.6 4.3 Rotations & 4.4 Congruence and Transformations 1 Understand rotations of figures and understand congruence transformations. · I can rotate figures. · I can perform compositions with rotations. · I can identify rotational symmetry in polygons. · I can identify congruent figures. · I can describe congruence transformations. · I can use congruence transformations to solve problems. Board Approved: 18 | P a g e G.CO.A.2, G.CO.D.11, G.SRT.A.1, G.SRT.A.2 4.5 Dilations & 4.6 Similarity and Transformations 1 Understand dilations of figures and similarity transformations. · I can identify dilations. · I can dilate figures. · I can solve real‐life problems involving scale factors and dilations. · I can perform similarity transformations. · I can describe similarity transformations. · I can prove that figures are similar. Review & Assess 2 Engaging Scenario Engaging Scenarios: There are several options for engaging scenarios for this unit. They may be appropriate as a means for introducing concepts in the unit, as a unit-long project for attaching learning to, or as a culminating experience at the end of the unit. ● Performance Task found in the Big Ideas Assessment Book (accessible in the physical Assessment Book, or the digital version in the Assess area of the Featured Chapter Resources after selecting the appropriate Chapter). ● Performance Task found at the end of the chapter in Big Ideas. ● Performance Task found in the Teach area of the Featured Chapter Resources after selecting the appropriate Chapter. ● Students will be given parameters of a shape and will create a picture by rotating and transforming the shape (using Geogebra). Board Approved: 19 | P a g e Unit 4: Congruent Triangles Subject: Geometry Grade: 9, 10, 11, 12 Name of Unit: Congruent Triangles Length of Unit: 9 blocks Overview of Unit: Students will prove theorems about triangles and use the definition of congruence in terms of rigid motions to show that two triangles are congruent. Students will explain how the criteria for triangle congruence follows from the definition of congruence, and they will use congruence and similarity criteria for triangles to solve problems and to prove relationships in geometric figures. Students will also use coordinates to prove theorems algebraically. Priority Standards for unit: ● G.CO.D.11 Construct geometric figures using various tools and methods. ● G.SRT.B.4 Use congruence and similarity criteria for triangles to solve problems and to prove relationships in geometric figures. ● G.GPE.B.3 Use coordinates to prove geometric theorems algebraically. Supporting Standards for unit: ● G.CO.B.7 Develop the criteria for triangle congruence from the definition of congruence in terms of rigid motions. ● G.CO.C.9 Prove theorems about triangles. Unwrapped Concepts (Students need to know) Unwrapped Skills (Students need to be able to do) Bloom’s Taxonomy Levels Webb's DOK geometric figures using various tools and methods. Construct Apply 2 congruence and similarity criteria for triangles to solve problems and to prove relationships in geometric figures. Use Understand 1 coordinates to prove geometric theorems algebraically. Use Understand 1 Board Approved: 20 | P a g e Essential Questions: 1. How do you identify corresponding parts of congruent triangles? 2. How do you show that two angles are congruent? 3. How can you tell whether a triangle is isosceles or equilateral? Enduring Understanding/Big Ideas: 1. Visualization can help you connect properties of real objects with two-dimensional drawings of these objects. 2. Definitions establish meanings and remove possible misunderstanding. Other truths are more complex and difficult to see. It is often possible to verify complex truths by reasoning from simpler ones by using deductive reasoning. Unit Vocabulary: Academic Cross-Curricular Words Content/Domain Specific interior angles exterior angles corresponding parts base base angles hypotenuse coordinate proof Resources for Vocabulary Development: Textbook Resources Board Approved: 21 | P a g e Unit Postulates and Theorems: 5.1 Triangle Sum Theorem 5.2 Exterior Angle Theorem Corollary 5.1 Corollary to the Triangle Sum 5.3 Properties of Triangle Congruence 5.4 Third Angles Theorem 5.5 Side-Angle-Side (SAS) Congruence Theorem 5.6 Base Angles Theorem 5.7 Converse of the Base Angles Theorem Corollary 5.2 Corollary to the Base Angles Theorem Corollary 5.3 Corollary to the Converse of the Base Angles Theorem 5.8 Side-Side-Side (SSS) Congruence Theorem 5.9 Hypotenuse-Leg (HL) Congruence Theorem 5.10 Angle-Side-Angle (ASA) Congruence Theorem 5.11 Angle-Angle-Side (AAS) Congruence Theorem Board Approved: 22 | P a g e Big Ideas Chapter 5: Congruent Triangles Standard Topic & Section Suggested # of Blocks Learning Target Success Criteria 5.1 Angles of Triangles 1 Prove and use theorems about angles of triangles. · I can classify triangles by sides and by angles. · I can prove theorems about angles of triangles. · I can find interior and exterior angle measures of triangles. G.CO.B.7, G.CO.C.9, G.CO.D.11 5.2 Congruent Polygons & 5.3 Proving Triangle Congruence by SAS 1 Understand congruence in terms of rigid motions, and prove and use the Side-Angle-Side Congruence Theorem · I can use rigid motions to show that two triangles are congruent. · I can identify corresponding parts of congruent polygons. · I can use congruent polygons to solve problems. · I can use rigid motions to prove the SAS Congruence Theorem. · I can use the SAS Congruence Theorem. G.CO.C.9, G.CO.D.11 5.4 Equilateral and Isosceles Triangles 1 Prove and use theorems about isosceles and equilateral triangles. · I can prove and use theorems about isosceles triangles. · I can prove and use theorems about equilateral triangles. G.CO.C.9, G.CO.D.11 5.5 Proving Triangle Congruence by SSS & HL 1 Prove and use the Side‐Side‐Side Congruence Theorem. · I can use rigid motions to prove the SSS Congruence Theorem. · I can use the SSS Congruence Theorem. · I can use the Hypotenuse‐Leg Congruence Theorem. Board Approved: 23 | P a g e G.CO.C.9, G.CO.D.11 5.6 Proving Triangle Congruence by ASA and AAS 1 Prove and use the Angle‐Side‐Angle Congruence Theorem and the Angle‐Angle‐ Side Congruence Theorem. · I can use rigid motions to prove the ASA Congruence Theorem. · I can prove the AAS Congruence Theorem. · I can use the ASA and AAS Congruence Theorems. G.CO.C.9, G.CO.D.11, G.SRT.B.4 5.7 Using Congruent Triangles 1 Use congruent triangles in proofs and to measure distances. · I can use congruent triangles to prove statements. · I can use congruent triangles to solve real‐life problems. · I can use congruent triangles to prove constructions. G.CO.C.9, G.GPE.B.3 5.8 Coordinate Proofs 1 Use coordinates to write proofs. · I can place figures in a coordinate plane. · I can write plans for coordinate proofs. · I can write coordinate proofs. Review & Assess 2 Engaging Scenario Engaging Scenarios: There are several options for engaging scenarios for this unit. They may be appropriate as a means for introducing concepts in the unit, as a unit-long project for attaching learning to, or as a culminating experience at the end of the unit. ● Performance Task found in the Big Ideas Assessment Book (accessible in the physical Assessment Book, or the digital version in the Assess area of the Featured Chapter Resources after selecting the appropriate Chapter). ● Performance Task found at the end of the chapter in Big Ideas. ● Performance Task found in the Teach area of the Featured Chapter Resources after selecting the appropriate Chapter. Board Approved: 24 | P a g e Unit 5: Quadrilaterals and Other Polygons Subject: Geometry Grade: 9, 10, 11, 12 Name of Unit: Quadrilaterals and Other Polygons Length of Unit: 6 blocks Overview of Unit: Students will prove theorems about parallelograms. They will also use coordinates and properties of trapezoids and kites to find measures. Priority Standards for unit: ● G.SRT.B.4 Use congruence and similarity criteria for triangles to solve problems and to prove relationships in geometric figures. ● G.CO.C.10 Prove theorems about polygons. Unwrapped Concepts (Students need to know) Unwrapped Skills (Students need to be able to do) Bloom’s Taxonomy Levels Webb's DOK congruence and similarity criteria for triangles to solve problems and to prove relationships in geometric figures. Use Understand 1 theorems about polygons. Prove Create 4 Essential Questions: 1. How can you find the sum of the measures of polygon angles? 2. How can you classify quadrilaterals? 3. How can you use coordinate geometry to prove general relationships? Enduring Understanding/Big Ideas: 1. Some attributes of geometric figures, such as length, area, volume, and angle measure, are measurable. Units are used to describe these attributes. 2. Definitions establish meanings and remove possible misunderstanding. Other truths are more complex and difficult to see. It is often possible to verify complex truths by reasoning from simpler ones using deductive reasoning. Board Approved: 25 | P a g e 3. A coordinate system in a plane is formed by two perpendicular lines, called the x- and y-axes. It is possible to verify some complex truths using deductive reasoning in combination with Distance, Midpoint, and Slope formulas. Unit Vocabulary: Academic Cross-Curricular Words Content/Domain Specific diagonal equilateral polygon equiangular polygon parallelogram rhombus rectangle square trapezoid base angles isosceles trapezoid midsegment of a trapezoid kite Resources for Vocabulary Development: Textbook Resources Unit Postulates and Theorems: 7.1 Polygon Interior Angles Theorem Corollary 7.1 Corollary to the Polygon Interior Angles Theorem 7.2 Polygon Exterior Angles Theorem 7.3 Parallelogram Opposite Sides Theorem 7.4 Parallelogram Opposite Angles Theorem 7.5 Parallelogram Consecutive Angles Theorem 7.6 Parallelogram Diagonals Theorem 7.7 Parallelogram Opposite Sides Converse 7.8 Parallelogram Opposite Angles Converse 7.9 Opposite Sides Parallel and Congruent Theorem 7.10 Parallelogram Diagonals Converse Corollary 7.2 Rhombus Corollary Corollary 7.3 Rectangle Corollary Corollary 7.4 Square Corollary Board Approved: 26 | P a g e 7.11 Rhombus Diagonals Theorem 7.12 Rhombus Opposite Angles Theorem 7.13 Rectangle Diagonals Theorem 7.14 Isosceles Trapezoid Base Angles Theorem 7.15 Isosceles Trapezoid Base Angles Converse 7.16 Isosceles Trapezoid Diagonals Theorem 7.17 Trapezoid Midsegment Theorem 7.18 Kite Diagonals Theorem 7.19 Kite Opposite Angles Theorem Board Approved: 27 | P a g e Big Ideas Chapter 7: Quadrilaterals and Other Polygons Standard Topic & Section Suggested # of Blocks Learning Target Success Criteria 7.1 Angles of Polygons 1 Find angle measures of polygons. · I can find the sum of the interior angle measures of a polygon. · I can find interior angle measures of polygons. · I can find exterior angle measures of polygons. G.CO.C.10, G.SRT.B.4 7.2 & 7.3 Properties of Parallelograms 1 Prove and use properties of parallelograms and prove that a quadrilateral is a parallelogram. · I can prove properties of parallelograms. · I can use properties of parallelograms. · I can solve problems involving parallelograms in the coordinate plane. · I can identify features of a parallelogram. · I can prove that a quadrilateral is a parallelogram. · I can find missing lengths that make a quadrilateral a parallelogram. · I can show that a quadrilateral in the coordinate plane is a parallelogram. Board Approved: 28 | P a g e G.CO.C.10, G.SRT.B.4 7.4 Properties of Special Parallelograms 1 Explain the properties of special parallelograms. · I can identify special quadrilaterals. · I can explain how special parallelograms are related. · I can find missing measures of special parallelograms. · I can identify special parallelograms in a coordinate plane. G.CO.C.10, G.SRT.B.4 7.5 Properties of Trapezoids and Kites (+ Midsegment Theorem from 6.5) 1 Use properties of trapezoids and kites to find measures. · I can identify trapezoids and kites. · I can use properties of trapezoids and kites to solve problems. · I can find the length of the midsegment of a trapezoid. · I can explain the hierarchy of quadrilaterals. Review & Assess 2 Engaging Scenario Engaging Scenarios: There are several options for engaging scenarios for this unit. They may be appropriate as a means for introducing concepts in the unit, as a unit-long project for attaching learning to, or as a culminating experience at the end of the unit. ● Performance Task found in the Big Ideas Assessment Book (accessible in the physical Assessment Book, or the digital version in the Assess area of the Featured Chapter Resources after selecting the appropriate Chapter). ● Performance Task found at the end of the chapter in Big Ideas. ● Performance Task found in the Teach area of the Featured Chapter Resources after selecting the appropriate Chapter. Board Approved: 29 | P a g e Unit 6: Similarity Subject: Geometry Grade: 9, 10, 11, 12 Name of Unit: Similarity Length of Unit: 5 blocks Overview of Unit: Students will understand properties of similar figures and prove theorems involving similarity. Then they will use similar triangles to prove the slope criteria for parallel and perpendicular lines. Students will also learn to construct a point along a directed line segment that partitions the segment in a given ratio. Priority Standards for unit: ● G.SRT.B.4 Use congruence and similarity criteria for triangles to solve problems and to prove relationships in geometric figures. ● G.CO.D.11 Construct geometric figures using various tools and methods. Supporting Standards for unit: Unwrapped Concepts (Students need to know) Unwrapped Skills (Students need to be able to do) Bloom’s Taxonomy Levels Webb's DOK congruence and similarity criteria for triangles to solve problems and to prove relationships in geometric figures. Use Understand 1 geometric figures using various tools and methods. Construct Apply 2 Essential Questions: 1. How do you use proportions to find side lengths in similar polygons? 2. How do you show two triangles are similar? 3. How do you identify corresponding parts of similar triangles? Board Approved: 30 | P a g e Enduring Understanding/Big Ideas: 1. Two geometric figures are similar when corresponding lengths are proportional and corresponding angles are congruent. 2. Definitions establish meanings and remove possible misunderstanding. Other truths are more complex and difficult to see. It is often possible to verify complex truths by reasoning from simpler ones by using deductive reasoning. 3. Visualization can help you see the relationships between two figures and help you connect the properties of real objects with two-dimensional drawings of these objects. Unit Vocabulary: Academic Cross-Curricular Words Content/Domain Specific similar figures similarity transformation corresponding parts proportion Resources for Vocabulary Development: Textbook Resources Unit Postulates and Theorems: 8.1 Perimeters of Similar Polygons 8.2 Areas of Similar Polygons 8.3 Angle-Angle (AA) Similarity Theorem 8.4 Side-Side-Side (SSS) Similarity Theorem 8.5 Side-Angle-Side (SAS) Similarity Theorem 8.6 Triangle Proportionality Theorem 8.7 Converse of the Triangle Proportionality Theorem 8.8 Three Parallel Lines Theorem 8.9 Triangle Angle Bisector Theorem Board Approved: 31 | P a g e Big Ideas Chapter 8: Similarity Standard Topic & Section Suggested # of Blocks Learning Target Success Criteria G.SRT.A.1, G.SRT.A.2 8.1 Similar Polygons 1 Understand the relationship between similar polygons · I can use similarity statements. · I can find corresponding lengths in similar polygons. · I can find perimeters and areas of similar polygons. · I can decide whether polygons are similar. G.SRT.A.4, G.SRT.B.4, G.GPE.B.4, G.GPE.B.5 8.2 Proving Triangle Similarity by AA & 8.3 Proving Triangle Similarity by SSS and SAS 1 Understand and use the Angle‐ Angle Similarity Theorem and understand and use additional triangle similarity theorems. · I can use similarity transformations to prove the Angle‐Angle Similarity Theorem. · I can use angle measures of triangles to determine whether triangles are similar. · I can prove triangle similarity using the Angle‐ Angle Similarity Theorem. · I can solve real‐life problems using similar triangles. · I can use the SSS and SAS Similarity Theorems to determine whether triangles are similar. · I can use similar triangles to prove theorems about slopes of parallel and perpendicular lines. Board Approved: 32 | P a g e G.CO.D.11, G.SRT.B.4 8.4 Proportionality Theorems 1 Understand and use proportionality theorems. · I can use proportionality theorems to find lengths in triangles. · I can find lengths when two transversals intersect three parallel lines. · I can find lengths when a ray bisects an angle of a triangle. Review & Assess 2 Engaging Scenario Engaging Scenarios: There are several options for engaging scenarios for this unit. They may be appropriate as a means for introducing concepts in the unit, as a unit-long project for attaching learning to, or as a culminating experience at the end of the unit. ● Performance Task found in the Big Ideas Assessment Book (accessible in the physical Assessment Book, or the digital version in the Assess area of the Featured Chapter Resources after selecting the appropriate Chapter). ● Performance Task found at the end of the chapter in Big Ideas. ● Performance Task found in the Teach area of the Featured Chapter Resources after selecting the appropriate Chapter. Board Approved: 33 | P a g e Unit 7: Right Triangles and Trigonometry Subject: Geometry Grade: 9, 10, 11, 12 Name of Unit: Right Triangles and Trigonometry Length of Unit: 11 blocks Overview of Unit: Students will prove and use theorems involving similarity. They will also define trigonometric ratios and solve problems involving right triangles. Priority Standards for unit: ● G.SRT.B.4 Use congruence and similarity criteria for triangles to solve problems and to prove relationships in geometric figures. ● G.SRT.C.7 Use trigonometric ratios and the Pythagorean Theorem to solve right triangles. ● G.CO.C.10 Prove theorems about polygons. Supporting Standards for unit: ● G.CO.C.9 Prove theorems about triangles. ● G.SRT.C.5 Understand that side ratios in right triangles define the trigonometric ratios for acute angles. ● G.SRT.C.6 Explain and use the relationship between the sine and cosine of complementary angles. ● G.SRT.C.8 Derive the formula A = 1/2 ab sin(C) for the area of a triangle. Unwrapped Concepts (Students need to know) Unwrapped Skills (Students need to be able to do) Bloom’s Taxonomy Levels Webb's DOK congruence and similarity criteria for triangles to solve problems and to prove relationships in geometric figures. Use Understand 1 trigonometric ratios and the Pythagorean Theorem to solve right triangles Use Understand 1 theorems about polygons. Prove Create 4 Board Approved: 34 | P a g e Essential Questions: 1. How do you find a side length or angle measure in a right triangle? 2. How do trigonometric ratios relate to similar right triangles? Enduring Understanding/Big Ideas: 1. Some attributes of geometric figures, such as length, area, volume, and angle measure, are measurable. Units are used to describe these attributes. 2. Two geometric figures are similar when corresponding lengths are proportional and corresponding angles are congruent. Unit Vocabulary: Academic Cross-Curricular Words Content/Domain Specific angle of depression angle of elevation cosine law of cosines law of sines Pythagorean triple sine tangent Resources for Vocabulary Development: Textbook Resources Unit Postulates and Theorems: 9.1 Pythagorean Theorem 9.2 Converse of the Pythagorean Theorem 9.3 Pythagorean Inequalities Theorem 9.4 45°-45°-90° Triangle Theorem 9.5 30°-60°-90° Triangle Theorem 9.6 Right Triangle Similarity Theorem 9.7 Geometric Mean (Altitude) Theorem 9.8 Geometric Mean (Leg) Theorem 9.9 Law of Sines 9.10 Law of Cosines Board Approved: 35 | P a g e Big Ideas Chapter 9: Right Triangles and Trigonometry Standard Topic & Section Suggested # of Blocks Learning Target Success Criteria G.CO.C.9, G.CO.C.10 9.1 The Pythagorean Theorem 1 Understand and apply the Pythagorean Theorem. · I can list common Pythagorean triples. · I can find missing side lengths of right triangles. · I can classify a triangle as acute, right, or obtuse given its side lengths. 9.2 Special Right Triangles 1 Understand and use special right triangles. · I can find side lengths in 45°‐ 45°‐ 90° triangles. · I can find side lengths in 30°‐ 60°‐ 90° triangles. · I can use special right triangles to solve real‐life problems. G.SRT.B.4 9.3 Similar Right Triangles 1 Use proportional relationships in right triangles. · I can explain the Right Triangle Similarity Theorem. · I can find the geometric mean of two numbers. · I can find missing dimensions in right triangles. Review & Assess 2 Board Approved: 36 | P a g e G.SRT.C.5, G.SRT.C.6, G.SRT.C.8 9.4&5 Trigonometric Ratios 2 Understand and use the tangent, sine, and cosine ratios. · I can explain the tangent ratio. · I can find tangent ratios. · I can use tangent ratios to solve real‐life problems. · I can explain the sine and cosine ratios. · I can find sine and cosine ratios. · I can use sine and cosine ratios to solve real‐life problems. G.SRT.C.7 9.6 Solving Right Triangles 1 Find unknown side lengths and angle measures of right triangles. · I can explain inverse trigonometric ratios. · I can use inverse trigonometric ratios to approximate angle measures. · I can solve right triangles. · I can solve real‐life problems by solving right triangles. Review & Assess 2 Engaging Scenario Engaging Scenarios: There are several options for engaging scenarios for this unit. They may be appropriate as a means for introducing concepts in the unit, as a unit-long project for attaching learning to, or as a culminating experience at the end of the unit. ● Performance Task found in the Big Ideas Assessment Book (accessible in the physical Assessment Book, or the digital version in the Assess area of the Featured Chapter Resources after selecting the appropriate Chapter). ● Performance Task found at the end of the chapter in Big Ideas. ● Performance Task found in the Teach area of the Featured Chapter Resources after selecting the appropriate Chapter. Board Approved: 37 | P a g e Unit 8: Circumference, Area, Surface Area, and Volume Subject: Geometry Grade: 9, 10, 11, 12 Name of Unit: Circumference, Area, Surface Area, and Volume Length of Unit: 11 blocks Overview of Unit: Students will explain and use the formulas for the circumference and area of a circle. They will derive the fact that the length of the arc intercepted by an angle is proportional to the radius and explain and use radian measure. Students will derive the formula for the area of a sector. They will also apply geometric concepts in modeling situations. Students will identify the shapes of cross sections of solids and solids generated by rotations of two-dimensional objects. They will explain and use volume formulas for cylinders, pyramids, cones, and spheres. Students will also apply geometric concepts in modeling situations, such as finding densities of various solids. Priority Standards for unit: ● G.GMD.A.2 Use volume formulas for cylinders, pyramids, cones, spheres, and composite figures to solve problems. Supporting Standards for unit: ● G.CO.A.1 Define angle, circle, perpendicular line, parallel line, line segment and ray based on the undefined notions of point, line, distance along a line and distance around a circular arc. ● G.C.B.4 Derive the formula for the length of an arc of a circle. ● G.GMD.A.1 Give an informal argument for the formulas for the circumference of a circle, area of a circle, volume of a cylinder, pyramid, and cone. ● G.C.B.5 Derive the formula for the area of a sector of a circle. Unwrapped Concepts (Students need to know) Unwrapped Skills (Students need to be able to do) Bloom’s Taxonomy Levels Webb's DOK volume formulas for cylinders, pyramids, cones, spheres, and composite figures to solve problems. Use Apply 2 Essential Questions: 1. How do you find the area and perimeter of polygons and circles? 2. How do area and perimeters of similar polygons compare? Board Approved: 38 | P a g e 3. How can you determine the intersection of a solid and a plane? 4. How do you find the surface area and volume of a solid? Enduring Understanding/Big Ideas: 1. Some attributes of geometric figures, such as length, area, volume, angle measure, are measurable. Units are used to describe these attributes. 2. Two geometric figures are similar when corresponding lengths are proportional and corresponding angles are congruent. Areas of similar figures are proportional to the squares of their corresponding lengths. 3. Visualization can help you connect properties of real objects with two-dimensional drawings of these objects. 4. Some attributes of geometric figures, such as length, area, volume, and angle measure, are measurable. Units are used to describe these attributes. Unit Vocabulary: Academic Cross-Curricular Words Content/Domain Specific Density Volume Cross section Prism Pyramid Cone Sphere Cylinder adjacent arcs apothem arc length central angle concentric circles congruent arcs composite figure diameter major arc minor arc radius sector of a circle segment of a circle face polyhedron surface area Resources for Vocabulary Development: Textbook Resources Board Approved: 39 | P a g e Big Ideas Chapters 11 & 12: Circumference and Area & Surface Area and Volume Standard Topic & Section Suggested # of Blocks Learning Target Success Criteria G.CO.A.1, G.C.B.4, G.GMD.A.1, G.C.B.5 11.1 Circumference and Arc Length & 11.2 Areas of Circles and Sectors 1 Understand circumference, arc length, and radian measure, and find areas of circles and areas of sectors of circles. · I can use the formula for the circumference of a circle to find measures. · I can find arc lengths and use arc lengths to find measures. · I can solve real‐life problems involving circumference. · I can explain radian measure and convert between degree and radian measure. · I can use the formula for area of a circle to find measures. · I can find areas of sectors of circles. · I can solve problems involving areas of sectors. G.GMD.A.1 11.3 Areas of Polygons 1 Find angle measures and areas of regular polygons, given the radius. · I can find areas of rhombuses and kites. · I can find angle measures in regular polygons. · I can find areas of regular polygons. · I can explain how the area of a triangle is related to the area formulas for rhombuses, kites, and regular polygons. Review & Assess 1 Board Approved: 40 | P a g e G.GMD.B.3, G.GMD.B.4 12.1 Cross Sections of Solids & 12.7 Solids of Revolution 1 Describe and draw cross sections, and sketch and use solids of revolution. · I can describe attributes of solids. · I can describe and draw cross sections. · I can solve real‐life problems involving cross sections. · I can sketch and describe solids of revolution. · I can find surface areas and volumes of solids of revolution. · I can form solids of revolution in the coordinate plane. G.GMD.A.1, G.GMD.A.2 12.2, 12.3, 12.4 Surface Area of Prisms, Cylinders, Cones, and Pyramids 2 Find and use surface areas of prisms, cylinders, and cones. · I can find surface areas of prisms. · I can find surface areas of cylinders. · I can find surface areas of cones. G.GMD.A.1, G.GMD.A.2 12.2, 12.3, 12.4 Volume of Prisms, Cylinders, Cones, and Pyramids 2 Find and use volumes of prisms, cylinders, cones, and pyramids · I can find volumes of prisms. · I can find volumes of cylinders. · I can find volumes of cones. · I can find volumes of pyramids. · I can find volumes of composite solids. G.GMD.A.2 12.5 Surface Areas and Volumes of Spheres 1 Find and use surface areas and volumes of spheres. · I can find surface areas of spheres. · I can find volumes of spheres. · I can find the volumes of composite solids. Review & Assess 2 Board Approved: 41 | P a g e Engaging Scenario Engaging Scenarios: There are several options for engaging scenarios for this unit. They may be appropriate as a means for introducing concepts in the unit, as a unit-long project for attaching learning to, or as a culminating experience at the end of the unit. ● Performance Task found in the Big Ideas Assessment Book (accessible in the physical Assessment Book, or the digital version in the Assess area of the Featured Chapter Resources after selecting the appropriate Chapter). ● Performance Task found at the end of the chapter in Big Ideas. ● Performance Task found in the Teach area of the Featured Chapter Resources after selecting the appropriate Chapter. Board Approved: 42 | P a g e Unit 9: Probability Subject: Geometry Grade: 9, 10, 11, 12 Name of Unit: Probability Length of Unit: 8 blocks Overview of Unit: Students will understand independent probability and conditional probability, use them to interpret data, and use probability rules to find probabilities of compound events. Priority Standards for unit: ● G.CP.A.2 Understand the definition of independent events and use it to solve problems. ● G.CP.A.5 Recognize and explain the concepts of conditional probability and independence in a context. Supporting Standards for unit: ● G.CP.A.1 Describe events as subsets of a sample space using characteristics of the outcomes, or as unions, intersections or complements of other events. ● G.CP.A.3 Calculate conditional probabilities of events. ● G.CP.A.4 Construct and interpret two-way frequency tables of data when two categories are associated with each object being classified. Use the two- way table as a sample space to decide if events are independent and to approximate conditional probabilities. ● G.CP.A.6 Apply and interpret the Addition Rule for calculating probabilities. ● G.CP.A.7 Apply and Interpret the general Multiplication Rule in a uniform probability model. ● G.CP.A.8 Use permutations and combinations to solve problems. Unwrapped Concepts (Students need to know) Unwrapped Skills (Students need to be able to do) Bloom’s Taxonomy Levels Webb's DOK the definition of independent events and use it to solve problems. Understand Understand 1 and explain the concepts of conditional probability and independence in a context. Recognize Understand 1 Board Approved: 43 | P a g e Essential Questions: 1. What is the difference between experimental probability and theoretical probability? 2. What is a frequency table? 3. What does it mean for an event to be random? Enduring Understanding/Big Ideas: 1. Probability describes the likelihood that an event will occur. The probability of an event can range from 0 (impossible) to 1 (certain). Experimental probability is based on observation or trials of an experiment, while theoretical probability is based on what should happen mathematically. Combinations and permutations can be used to count the number of possible outcomes. ● Probability is a measure of the likelihood that an event will occur ● Counting techniques can be used to find all of the possible ways to complete different tasks or choose items from a list. ● The probability of compound events can be found by using the probability of each part of the compound event. 2. A frequency table is a data display that shows how often an item appears in a particular category. Frequency tables can be used to calculate the relative frequencies of each item. A two-way frequency table, or contingency table, displays the frequencies of data in two different categories. Contingency tables can be used to find conditional probabilities. ● Tables can be used to organize data by frequency and find probabilities. ● Two-way frequency tables can be used to organize data and identify sample spaces to approximate probabilities. ● Tables, tree diagrams, and formulas can be used to find conditional probability. 3. A random event has no bias or inclination toward any particular outcome. Random number tables and electronic random number generators can be used to model random events. In order to reach a fair decision, each possible choice must have the same probability of being selected. Expected value uses theoretical probability to tell you what you can expect in the long run, which can help you make more informed decisions. 4. Probability can be used to make fair decisions based on prior experience. Board Approved: 44 | P a g e Unit Vocabulary: Academic Cross-Curricular Words Content/Domain Specific Combination Conditional probability Dependent events Experimental probability Theoretical probability Independent events Mutually exclusive events Permutation Sample Space Resources for Vocabulary Development: Textbook Resources Board Approved: 45 | P a g e Big Ideas Chapter 13: Probability Standard Topic & Section Suggested # of Blocks Learning Target Success Criteria G.CP.A.1 13.1 Sample Spaces and Probability 1 Find sample spaces and probabilities of events. · I can list the possible outcomes in a sample space. · I can find theoretical probabilities. · I can find experimental probabilities. G.CP.A.4 13.2 Two‐Way Tables and Probability 1 Use two‐way tables to represent data and find probabilities. · I can make two‐way tables. · I can find and interpret relative frequencies and conditional relative frequencies. · I can use conditional relative frequencies to find probabilities. G.CP.A.3, G.CP.A.5 13.3 Conditional Probability 1 Find and use conditional probabilities. · I can explain the meaning of conditional probability. · I can find conditional probabilities. · I can make decisions using probabilities. G.CP.A.2, G.CP.A.5 13.4 Independent and Dependent Events 1 Understand and find probabilities of independent and dependent events. · I can explain how independent events and dependent events are different. · I can determine whether events are independent. · I can find probabilities of independent and dependent events. Board Approved: 46 | P a g e G.CP.A.6, G.CP.A.7 13.5 Probability of Disjoint and Overlapping Events 1 Find probabilities of disjoint and overlapping events. · I can explain how disjoint events and overlapping events are different. · I can find probabilities of disjoint events. · I can find probabilities of overlapping events. · I can solve real‐life problems using more than one probability rule. G.CP.A.8 13.6 Permutations and Combinations 1 Count permutations and combinations. · I can explain the difference between permutations and combinations. · I can find numbers of permutations and combinations. · I can find probabilities using permutations and combinations. Review & Assess 2 Engaging Scenario Engaging Scenarios: There are several options for engaging scenarios for this unit. They may be appropriate as a means for introducing concepts in the unit, as a unit-long project for attaching learning to, or as a culminating experience at the end of the unit. ● Performance Task found in the Big Ideas Assessment Book (accessible in the physical Assessment Book, or the digital version in the Assess area of the Featured Chapter Resources after selecting the appropriate Chapter). ● Performance Task found at the end of the chapter in Big Ideas. ● Performance Task found in the Teach area of the Featured Chapter Resources after selecting the appropriate Chapter. Board Approved: 47 | P a g e Unit 10: Circles Subject: Geometry Grade: 9, 10, 11, 12 Name of Unit: Circles Length of Unit: Remainder of blocks in semester with 2 for review and assessment Overview of Unit: Students will understand and apply theorems about circles. They will translate between geometric descriptions and equations for circles and parabolas. Students will also write coordinate proofs involving circles. Priority Standards for unit: ● G.CO.D.11 Construct geometric figures using various tools and methods. ● G.GPE.B.3 Use coordinates to prove geometric theorems algebraically. Supporting Standards for unit: ● G.C.A.1 Prove that all circles are similar using similarity transformations. ● G.C.A.2 Identify and describe relationships among inscribed angles, radii, and chords of circles. ● G.CO.A.1 Define angle, circle, perpendicular line, parallel line, line segment and ray based on the undefined notions of point, line, distance along a line and distance around a circular arc. ● G.GPE.A.1 Derive the equation of a circle. Unwrapped Concepts (Students need to know) Unwrapped Skills (Students need to be able to do) Bloom’s Taxonomy Levels Webb's DOK geometric figures using various tools and methods. Construct Apply 2 coordinates to prove geometric theorems algebraically. Use Understand 1 Board Approved: 48 | P a g e Essential Questions: 1. How can you prove relationships between angles and arcs in a circle? 2. When lines intersect a circle, or within a circle, how do you find the measures Of resulting angles, arcs, and segments? 3. How do you find the equation of a circle in the coordinate plane? Enduring Understanding/Big Ideas: 1. Definitions establish meanings and remove possible misunderstanding. Other truths are more complex and difficult to see. It is often possible to verify complex truths by reasoning from simpler ones by using deductive reasoning. 2. Some attributes of geometric figures, such as length, area, volume, and angle measure, are measurable. Units are used to describe these attributes. 3. It is possible to verify some complex truths on the coordinate plane using deductive reasoning in combination with distance, midpoint, and slope formulas. Unit Vocabulary: Academic Cross-Curricular Words Content/Domain Specific chord inscribed angle intercepted arc point of tangency secant standard form of an equation of a circle tangent to a circle Resources for Vocabulary Development: Textbook Resources Board Approved: 49 | P a g e Unit Postulates and Theorems: 10.1 Arc Addition Postulate 10.1 Tangent Line to Circle Theorem 10.2 External Tangent Congruence Theorem 10.3 Congruent Circles Theorem 10.4 Congruent Central Angles Theorem 10.5 Similar Circles Theorem 10.6 Congruent Corresponding Chords Theorem 10.7 Perpendicular Chord Bisector Theorem 10.8 Perpendicular Chord Bisector Converse 10.9 Equidistant Chords Theorem 10.10 Measure of an Inscribed Angle Theorem 10.11 Inscribed Angles of a Circle Theorem 10.12 Inscribed Right Triangle Theorem 10.13 Inscribed Quadrilateral Theorem 10.14 Tangent and Intersected Chord Theorem 10.15 Angles Inside the Circle Theorem 10.16 Angles Outside the Circle Theorem 10.17 Circumscribed Angle Theorem 10.18 Segments of Chords Theorem 10.19 Segments of Secants Theorem 10.20 Segments of Secants and Tangents Theorem Board Approved: 50 | P a g e Big Ideas Chapter 10: Circles Standard Topic & Section Suggested # of Blocks Learning Target Success Criteria G.CO.A.1, G.CO.D.11 , G.C.A.2 10.1 Lines and Segments That Intersect Circles Remainder of semester Identify lines and segments that intersect circles and use them to solve problems. · I can identify special segments and lines that intersect circles. · I can draw and identify common tangents. · I can use properties of tangents to solve problems. G.CO.A.1, G.C.A.1, G.C.A.2 10.2 Finding Arc Measures Understand arc measures and similar circles. · I can find arc measures. · I can identify congruent arcs. · I can prove that all circles are similar. G.C.A.2 10.3 Using Chords Understand and apply theorems about chords. · I can use chords of circles to find arc measures. · I can use chords of circles to find lengths. · I can describe the relationship between a diameter and a chord perpendicular to a diameter. · I can find the center of a circle given three points on the circle. G.CO.D.11 , G.C.A.2 10.4 Inscribed Angles and Polygons Use properties of inscribed angles and inscribed polygons. · I can find measures of inscribed angles and intercepted arcs. · I can find angle measures of inscribed polygons. · I can construct a square inscribed in a circle. Board Approved: 51 | P a g e G.C.A.2 10.5 Angle Relationships in Circles Understand angles formed by chords, secants, and tangents. · I can identify angles and arcs determined by chords, secants, and tangents. · I can find angle measures and arc measures involving chords, secants, and tangents. · I can use circumscribed angles to solve problems. G.C.A.2 10.6 Segment Relationships in Circles Use theorems about segments of chords, secants, and tangents. · I can find lengths of segments of chords. · I can identify segments of secants and tangents. · I can find lengths of segments of secants and tangents. G.GPE.A.1, G.GPE.B.3 10.7 Circles in the Coordinate Plane Understand equations of circles. · I can write equations of circles. · I can find the center and radius of a circle. · I can graph equations of circles. · I can write coordinate proofs involving circles. Review & Assess 2 Engaging Scenario Engaging Scenarios: There are several options for engaging scenarios for this unit. They may be appropriate as a means for introducing concepts in the unit, as a unit-long project for attaching learning to, or as a culminating experience at the end of the unit. ● Performance Task found in the Big Ideas Assessment Book (accessible in the physical Assessment Book, or the digital version in the Assess area of the Featured Chapter Resources after selecting the appropriate Chapter). ● Performance Task found at the end of the chapter in Big Ideas. ● Performance Task found in the Teach area of the Featured Chapter Resources after selecting the appropriate Chapter. Board Approved: 52 | P a g e Unit of Study Terminology Appendices: All Appendices and supporting material can be found in this course’s shell course in the District’s Learning Management System. Assessment Leveling Guide: A tool to use when writing assessments in order to maintain the appropriate level of rigor that matches the standard. Big Ideas/Enduring Understandings: Foundational understandings teachers want students to be able to discover and state in their own words by the end of the unit of study. These are answers to the essential questions. Engaging Experience: Each topic is broken into a list of engaging experiences for students. These experiences are aligned to priority and supporting standards, thus stating what students should be able to do. An example of an engaging experience is provided in the description, but a teacher has the autonomy to substitute one of their own that aligns to the level of rigor stated in the standards. Engaging Scenario: This is a culminating activity in which students are given a role, situation, challenge, audience, and a product or performance is specified. Each unit contains an example of an engaging scenario, but a teacher has the ability to substitute with the same intent in mind. Essential Questions: Engaging, open-ended questions that teachers can use to engage students in the learning. Priority Standards: What every student should know and be able to do. These were chosen because of their necessity for success in the next course, the state assessment, and life. Supporting Standards: Additional standards that support the learning within the unit. Topic: These are the main teaching points for the unit. Units can have anywhere from one topic to many, depending on the depth of the unit. Unit of Study: Series of learning experiences/related assessments based on designated priority standards and related supporting standards. Unit Vocabulary: Words students will encounter within the unit that are essential to understanding. Academic Cross-Curricular words (also called Tier 2 words) are those that can be found in multiple content areas, not just this one. Content/Domain Specific vocabulary words are those found specifically within the content. |
1146 | https://puzzling.stackexchange.com/questions/106739/how-to-get-the-least-number-of-flips-to-a-plastic-chips-to-get-a-certain-figure | visual - How to get the least number of flips to a plastic chips to get a certain figure? - Puzzling Stack Exchange
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How to get the least number of flips to a plastic chips to get a certain figure?
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Asked 4 years, 8 months ago
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The puzzle is as follows:
(Transcription with grammar errors fixed)
A game consists of the following:
The challenge is to flip the chips in such a way so that they are as in state 2. You start in state 1. A movement is valid when you flip a chip, and the two that are next to it, at the same time.
What is the minimum number of movements to do so?
(A and B are the two states)
A B A B A B
B B --> A A
A B A B A B
The alternatives given are:
8 movements
9 movements
5 movements
6 movements
This puzzle doesn't have an official answer. But I'm getting 8. I don't know if my answer is right or reasonable.
Does a method exist to solve these kinds of puzzles? The thing here is that the figure makes a closed loop and because the only allowed movement is to flip three chips that means one plus the other two which are next to it.
Thus what I did was begin at the top left, to try to make the figure be entirely the cherry shade instead of the orange one.
Once I made this, then my next move was to cancel the necessary cherry chips by flipping them.
In the end this yielded me 8 movements. But is this the least?
Can someone help me with a drawing or sketch to better understand this?. Because the intended method for solution which it would help me best is the kind which doesn't rely too heavily on advanced mathematics, but something which could be done without too many manipulation of variables. Does it exist a way or a suggested approach on these lines?.
For reference, I got this from my Logical games book sheet from 2000's. It seems to be an adaptation from a reprinted copy of Martin Gardner's Puzzle carnival book of the 1970s.
visual
optimization
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edited Jan 20, 2021 at 14:19
RobPratt
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asked Jan 20, 2021 at 3:17
Chris Steinbeck BellChris Steinbeck Bell
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Suppose there are n≥3 n≥3 chips. It's easy to see that n n moves is possible, making each move exactly once (this flips each chip three times). Since making the same move twice cancels out, any optimal solution also involves making each move at most once.
Since each chip needs to be flipped an odd number of times, in any optimal solution each chip is flipped either once or three times. But the only way a chip can be flipped three times in an optimal solution is if all three moves involving that chip are used once each. If this happens, both the adjacent chips are flipped at least twice, so they must also be flipped three times.
It follows that in any optimal solution, either every chip is flipped three times or every chip is flipped exactly once. The former case takes n n moves, since each move flips three chips and there are 3 n 3 n total flips. The latter case takes exactly n/3 n/3 moves for the same reason; this is only possible if n n is a multiple of 3 3. (It is possible in this case, by dividing the chips into consecutive groups of three, and flipping each group.)
So the answer is n n if n n is not a multiple of 3 3, and n/3 n/3 if it is.
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answered Jan 20, 2021 at 12:45
Especially LimeEspecially Lime
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2 This is a very neat argument that I had not seen before.Jaap Scherphuis –Jaap Scherphuis 2021-01-20 13:10:01 +00:00 Commented Jan 20, 2021 at 13:10
An important implicit observation here: the order of moves in sequences doesn't matter.aschepler –aschepler 2021-01-20 21:42:59 +00:00 Commented Jan 20, 2021 at 21:42
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You can solve the problem via a system of eight linear equations, one per chip, with eight binary variables, one per possible movement. The order of operations doesn’t matter, and there is no reason to perform the same movement twice because the two steps would just cancel. So the number of movements required is at most 8. Because each movement changes the parity of the number of red chips, there must be an even number of movements, ruling out 9 (again) and 5. It turns out that
8 is the minimum, achieved by making each possible movement once.
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edited Jan 20, 2021 at 5:18
answered Jan 20, 2021 at 3:28
RobPrattRobPratt
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How does this argument rule out 6?Misha Lavrov –Misha Lavrov 2021-01-21 00:14:29 +00:00 Commented Jan 21, 2021 at 0:14
1 It doesn't. The "It turns out that" conclusion is a result of solving the corresponding integer linear programming problem.RobPratt –RobPratt 2021-01-21 00:18:33 +00:00 Commented Jan 21, 2021 at 0:18
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First, we can prove that every puzzle in this setup (with any starting state, and any ending state) has a solution.
To see this, suppose that we do five flips centered at the chips marked with an x or X in one of the diagrams below:
x . x x x . . x x X . x . X . x . X x x . . x x
x x . x X . . . x x x x . . . X x .
. X . X . x . x x x x . x . x . x x x x . x . X
This has the effect of toggling the chip marked with capital X, and leaving all other chips as they were. Applying several of these sequences lets us make any change we like (possibly inefficiently).
A solution by the strategy above can take up to 40 moves. But we can simplify any solution: the steps can be reordered without changing the result, and if we make the same flip twice, we can skip it instead. After skipping repeated flips, and sorting in some canonical order, there are only 2 8=256 possible simplified sequences of moves: for each possible flip, either we did it once or we didn't do it at all.
We know that from every starting state, we can reach all 256 ending states by some sequence of moves. But there's only 256 simplified sequences of moves. Therefore, there's only one simplified sequence of moves per ending state - every simplified sequence of moves has a different effect.
In particular, there's only one simplified sequence of moves that solves the puzzle in that question. It has 8 moves. The only other solutions must be unsimplified, which means they have even more moves: simplifying makes a solution shorter, because we cancel out repeated moves.
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edited Jan 20, 2021 at 21:17
bobble
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answered Jan 20, 2021 at 21:08
Misha LavrovMisha Lavrov
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This is a really nice argument.Especially Lime –Especially Lime 2021-01-22 17:53:17 +00:00 Commented Jan 22, 2021 at 17:53
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RobPratt gave the correct mathematical approach. To elaborate further, let's call the eight chips A,B,⋯,H A,B,⋯,H:
A H G B F C D E A B C H D G F E
If we assign 0 to yellow and 1 to pink, flipping a chip is equivalent to adding 1 modulo 2, because 0+1=1(mod 2)0+1=1(mod2) and 1+1=0(mod 2)1+1=0(mod2).
Now, let a,b,⋯,h a,b,⋯,h denote the number of flip operations, such that a a is the number of flips on A A and its neighbors H,A,B H,A,B, b b is for A,B,C A,B,C, and so on. Then A A gets flipped h+a+b h+a+b times, B B is flipped a+b+c a+b+c times, and so on. Based on the initial and final states of A,B,⋯,H A,B,⋯,H, we can set up a system of linear equations modulo 2:
0+h+a+b=1(mod 2)1+a+b+c=0(mod 2)0+b+c+d=1(mod 2)1+c+d+e=0(mod 2)0+d+e+f=1(mod 2)1+e+f+g=0(mod 2)0+f+g+h=1(mod 2)1+g+h+a=0(mod 2)0+h+a+b=1(mod2)1+a+b+c=0(mod2)0+b+c+d=1(mod2)1+c+d+e=0(mod2)0+d+e+f=1(mod2)1+e+f+g=0(mod2)0+f+g+h=1(mod2)1+g+h+a=0(mod2)
You can solve this just like a plain system of equations, keeping in mind that, in modulo 2 system,
0+0=1+1=0−0=1−1=0 0+1=1+0=0−1=1−0=1 0+0=1+1=0−0=1−1=0 0+1=1+0=0−1=1−0=1
So, for example, subtracting between the first two equations gives h+c=0(mod 2)h+c=0(mod2).
If you solve it without mistakes, you should get a=b=⋯=h=1 a=b=⋯=h=1 for this specific problem.
This is the general approach to standard "chip flipping" / "turn the lights on and off"-type of problems. One kind of complication is when the system is NOT uniquely solvable (many solutions or no solutions). Dealing with it is a mathematically deep subject, and Jaap's web page on this topic has a discussion on this problem. Also check out other questions tagged lights-out on Puzzling.
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answered Jan 20, 2021 at 7:12
BubblerBubbler
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Gee I didn't know that lights out problem had a deep mathematical insight. The sort of intended method which I was asking was some more like in layman terms or a more visual approach rather than mathematical. I arrived to the same answer as Rob, it just happens that I had a typo which I'm correcting it right away. I'm not very savvy with modular arithmetics so I can't understand your method well.Chris Steinbeck Bell –Chris Steinbeck Bell 2021-01-20 09:06:53 +00:00 Commented Jan 20, 2021 at 9:06
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1147 | https://www.nature.com/articles/s41565-024-01813-z | Skip to main content
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Article
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Published:
Superionic conducting vacancy-rich β-Li3N electrolyte for stable cycling of all-solid-state lithium metal batteries
Weihan Li
ORCID: orcid.org/0000-0002-5768-96411,2na1,
Minsi Li1,2na1,
Shuo Wang3na1,
Po-Hsiu Chien
ORCID: orcid.org/0000-0002-1607-12714na1,
Jing Luo1,
Jiamin Fu1,
Xiaoting Lin
ORCID: orcid.org/0000-0002-1218-18951,
Graham King
ORCID: orcid.org/0000-0003-1886-72545,
Renfei Feng
ORCID: orcid.org/0000-0001-8566-41615,
Jian Wang
ORCID: orcid.org/0000-0001-5184-70235,
Jigang Zhou5,
Ruying Li1,
Jue Liu
ORCID: orcid.org/0000-0002-4453-910X4,
Yifei Mo
ORCID: orcid.org/0000-0002-8162-46293,6,
Tsun-Kong Sham
ORCID: orcid.org/0000-0003-1928-66972 &
…
Xueliang Sun
ORCID: orcid.org/0000-0003-0374-12451,7
Nature Nanotechnology
volume 20, pages 265–275 (2025)Cite this article
27k Accesses
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Abstract
The advancement of all-solid-state lithium metal batteries requires breakthroughs in solid-state electrolytes (SSEs) for the suppression of lithium dendrite growth at high current densities and high capacities (>3 mAh cm−2) and innovation of SSEs in terms of crystal structure, ionic conductivity and rigidness. Here we report a superionic conducting, highly lithium-compatible and air-stable vacancy-rich β-Li3N SSE. This vacancy-rich β-Li3N SSE shows a high ionic conductivity of 2.14 × 10−3 S cm−1 at 25 °C and surpasses almost all the reported nitride-based SSEs. A Li- and N-vacancy-mediated fast lithium-ion migration mechanism is unravelled regarding vacancy-triggered reduced activation energy and increased mobile lithium-ion population. All-solid-state lithium symmetric cells using vacancy-rich β-Li3N achieve breakthroughs in high critical current densities up to 45 mA cm−2 and high capacities up to 7.5 mAh cm−2, and ultra-stable lithium stripping and plating processes over 2,000 cycles. The high lithium compatibility mechanism of vacancy-rich β-Li3N is unveiled as intrinsic stability to lithium metal. In addition, β-Li3N possesses excellent air stability through the formation of protection surfaces. All-solid-state lithium metal batteries using the vacancy-rich β-Li3N as SSE interlayers and lithium cobalt oxide (LCO) and Ni-rich LiNi0.83Co0.11Mn0.06O2 (NCM83) cathodes exhibit excellent battery performance. Extremely stable cycling performance is demonstrated with high capacity retentions of 82.05% with 95.2 mAh g−1 over 5,000 cycles at 1.0 C for LCO and 92.5% with 153.6 mAh g−1 over 3,500 cycles at 1.0 C for NCM83. Utilizing the vacancy-rich β-Li3N SSE and NCM83 cathodes, the all-solid-state lithium metal batteries successfully accomplished mild rapid charge and discharge rates up to 5.0 C, retaining 60.47% of the capacity. Notably, these batteries exhibited a high areal capacity, registering approximately 5.0 mAh cm−2 for the compact pellet-type cells and around 2.2 mAh cm−2 for the all-solid-state lithium metal pouch cells.
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Main
The emerging all-solid-state lithium metal batteries offer new opportunities to replace the flammable liquid electrolytes and meet the demanding energy and power densities as well as high safety compared with the traditional lithium-ion batteries1,2,3,4,5. However, the current research and development of all-solid-state lithium metal batteries encounter serious challenges with lithium metal anode for practical application, when using oxides, sulfides and halides as solid-state electrolytes (SSEs)6,7. Not only serious side reactions occur at the interface between lithium metal and the low-voltage-unstable sulfide- and halide-based SSEs, but also lithium dendrites easily propagate through the grain boundaries of the stable oxides, especially at high current densities8,9,10,11,12,13.
To overcome the important interfacial challenges, the application of an effective coating layer or interlayer design to stabilize the lithium metal anode represents a provisional remedy, such as protective layers14,15, lithiophilic layers16,17 and interface design18,19. It is crucial to acknowledge that the adoption of these coating strategies and interlayers frequently involves complex methodologies or escalates material costs and the intricacy of the fabrication process, which may limit their practical deployment. Moreover, the critical current density for lithium symmetric cells using these strategies is typically capped below 3 mA cm−2, failing to meet the demands of rapid electric vehicle charging (required current density, 1–10 mA cm−2, and required areal capacity, 3 mAh cm−2)20. Therefore, the imperative to formulate an SSE that is both highly conductive ionically and demonstrates robust stability over successive lithium stripping/plating cycles cannot be overstated. Zhu et al. discovered that nitride anion chemistry shows unique stability against lithium metal21. The promising Li3N SSE is proven thermodynamically stable against lithium metal for all-solid-state lithium metal batteries application22. Utilizing Li3N as a modification layer on garnet SSEs, J. B. Goodenough et al. achieved stable operation of all-solid-state lithium metal batteries at 40 °C for up to 300 cycles23. In a parallel advancement, Xu et al. utilized β-Li3N as an interfacial layer to mitigate the incompatibility issues between halide SSEs and lithium metal, further highlighting the stability of β-Li3N in contact with lithium24. Among two Li3N phases, α-Li3N and β-Li3N, α-Li3N has been extensively researched25,26,27, showing high ionic conductivities in the range of 10−4 S cm−1 to 10−3 S cm−1 at room temperature, following the initial structural elucidation by Zintl and Brauer28, and later confirmed by Rabenau and Schulz29,30. In addition, the β-Li3N phase, with its crystal structure detailed by Beister et al.31 in the context of phase transformations under pressure, was found by Chen et al.27 to exhibit superior ionic conductivity (~2 × 10−4 S cm−1 at room temperature) when prepared via ball milling. Nevertheless, α-Li3N’s high electronic conductivity (measured at 2.6 × 10−7 S cm−1) exceeds the desirable threshold (10−10 S cm−1 to 10−12 S cm−1) for preventing lithium dendrite formation at high currents32. By contrast, Li3N possesses much lower electronic conductivity (measured at 4.5 × 10−10 S cm−1). The low lithium and nitrogen vacancy concentrations in the inherent lattice structures of Li3N lead to low ionic conductivity and inadequate resistance against lithium dendrite formation. Consequently, this culminates in suboptimal electrochemical performance of all-solid-state lithium batteries, characterized by restricted cycling life, pronounced lithium dendrite proliferation and diminished capacities at elevated current densities33. As theoretical calculation predicts, the lithium-ion diffusion behaviour in Li3N relies on the lithium vacancies in crystal structure27. Therefore, it is a rational strategy to increase the vacancy defects in Li3N to achieve high ionic conductivity and stable lithium stripping/plating behaviour for high-performance all-solid-state lithium metal batteries.
We report a highly lithium-compatible, air-stable and superionic conducting vacancy-rich β-Li3N SSE. The vacancy-rich β-Li3N SSE presents a high room-temperature ionic conductivity of 2.14 × 10−3 S cm−1, surpassing almost all the reported nitride-based SSEs. The fast lithium-ion migration behaviour originates from a vacancy-mediated superionic diffusion mechanism, unrevealed by combined Rietveld refinement of synchrotron-based X-ray diffraction (SXRD) and time-of-flight (TOF) neutron diffraction data, density functional theory (DFT) calculation and ab initio molecular dynamics (AIMD) simulation. In addition, this vacancy-rich β-Li3N SSE achieves breakthroughs in resistance against lithium metal dendrite growth and enables high critical current densities of up to 45 mA cm−2 and excellent performance cycling of lithium symmetric cells for over 4,000 h at 7.5 mA cm−2 and 7.5 mAh cm−2. The stability mechanism of vacancy-rich β-Li3N SSE towards lithium metal is unveiled by DFT calculation and scanning transmission X-ray microscopy (STXM). Furthermore, this vacancy-rich β-Li3N SSE possesses high air stability in dry rooms, clarified by in situ and operando XRD studies. Owing to the chemical stability against lithium metal, we demonstrate a long-cycling and high-areal-capacity all-solid-state lithium metal batteries design, which uses a vacancy-rich β-Li3N SSE layer to enable the use of lithium metal anode and uses LiCoO2 (LCO) and Ni-rich LiNi0.83Co0.11Mn0.06O2 (NCM83) as cathodes. The all-solid-state lithium metal batteries demonstrate high capacity retentions of 82.05% with 95.2 mAh g−1 over 5,000 cycles at 1.0 C for LCO and 92.5% with 153.6 mAh g−1 over 3,500 cycles at 1.0 C for NCM83. The all-solid-state lithium metal batteries exhibit reasonable rapid charge and discharge capabilities up to 5.0 C, retaining 60.47% of the capacity. These batteries show a high areal capacity of approximately 5.0 mAh cm−2 for the pellet-type cell with an area of 0.785 cm2 and around 2.2 mAh cm−2 for the all-solid-state lithium metal pouch cells, measuring 2.5 × 2 cm2.
Results
Superionic conducting mechanism
The ball-milling method is chosen to create high pressure for obtaining pure β-Li3N from the commercial mixed phased Li3N and a ball-milling speed of 400 rpm is chosen to prepare pure β-Li3N (Fig. 1a and Supplementary Note 1). Pure β-Li3N was obtained by ball milling at 400 rpm for 8 h, denoted by β-Li3N-400rpm-8h. The Arrhenius plots of the commercial Li3N and the β-Li3N-400rpm-8h are shown in Fig. 1b. The β-Li3N-400rpm-8h SSE demonstrated a high room-temperature (25 °C) ionic conductivity of 1.92 × 10−3 S cm−1, which is around two orders of magnitude higher than the commercial Li3N. The activation energy (0.377 eV) for the β-Li3N-400rpm-8h SSE is lower than that of commercial Li3N (that is, 0.389 eV). When the ball-milling time further increases from 8 h to 16 h and 24 h with a constant speed of 400 rpm, the ionic conductivity further increases. The corresponding β-Li3N SSEs prepared by different ball-milling times are denoted by β-Li3N-400rpm-16h and β-Li3N-400rpm-24h (Fig. 1b). A ball-milling time of 16 h leads to an optimized ionic conductivity of 2.14 × 10−3 S cm−1 at 25 °C and the lowest activation energy of 0.371 eV (Fig. 1c). The achieved ionic conductivity is among the highest values reported for not only pure Li3N but also all nitride SSEs so far (Fig. 2a), showing great promise to achieve high energy and power densities for all-solid-state lithium metal batteries21,34. In addition to the phase transformation, the improvement of ionic conductivity of the ball-milled β-Li3N compared with the commercial Li3N was related to the vacancy-mediated lithium-ion diffusion mechanisms27.
The crystal structures of commercial Li3N and ball-milled β-Li3N studied by SXRD and TOF neutron diffraction with a focus on lithium and nitrogen vacancies were refined using the Rietveld method, as shown in Fig. 1d,e, Supplementary Figs. 1–4 and Supplementary Tables 1–8. The commercial Li3N was determined as a mixture of 63.1(9) wt% α-phase and 36.9(7) wt% β-phase, and the refined crystal structures of the α- and β-phases are shown in Supplementary Fig. 1. In the commercial Li3N, the Li(1) sites in the α-phase (1b) and the β-phase (2b) are fully occupied. Conversely, the Li(2) sites (2c in α-phase; 4f in β-phase) and N(3) sites (1a in α-phase; 2c in β-phase) are partially occupied. Subsequent calculations reveal that the lithium and nitrogen vacancy concentrations for both phases in the commercial Li3N are minimal: 0.7(4)% at the Li(2) 2c site (amounting to 0.5(4)% across all lithium sites) and 0.5(2)% at the N(3) 1a site in the α-phase, and 0.5(4)% at the Li(2) 4f site (equivalent to 0.3(4)% across all lithium sites) and 0.3(2)% at the N(3) 2c site in the β-phase. In the case of the β-Li3N-400rpm-8h sample, the purely constituted β-Li3N demonstrated an augmented lithium vacancy concentration at the 4f site, approximately 6.2(6)% (translating to 4.1(6)% across all lithium sites) as indicated in Supplementary Fig. 2 and Supplementary Tables 3 and 4. The lithium and nitrogen vacancy concentrations in the vacancy-rich β-Li3N generally escalated with prolonged ball-milling durations, yet plateaued post 16 h. Specifically, the β-Li3N-400rpm-16h sample, hereinafter referred to as vacancy-rich β-Li3N, showcased peak vacancy concentrations, with lithium vacancies approximated at 8.1(2)% at the Li(2) 4f sites (around 5.4(2)% for all lithium sites) and nitrogen vacancies at roughly 5.4(1)% at the N 2c sites (Fig. 1e). The presence of lithium vacancies at the 4f sites, rather than the 2b sites, can be attributed to the comparatively weaker bonding between N3− and Li+ at the 4f sties, as well as the low lithium vacancy formation energy at the 4f site compared with the 2b site (Fig. 1e and Supplementary Note 2).
Apparent correlations can be concluded that the higher lithium and nitrogen vacancy concentrations can lead to a higher concentration of mobile lithium ions, lower lithium-ion diffusion barriers (that is, lower activation energy) and thus higher ionic conductivity as shown in Figs. 1f and 2b, Supplementary Fig. 5 and Supplementary Note 3. The lithium diffusion mechanism in vacancy-rich β-Li3N with different concentrations of lithium and nitrogen vacancies was further studied by AIMD simulation. When vacancies were introduced, fast ionic conduction was observed in a three-dimensional channel as shown in Fig. 2d,e. The faster ionic conduction was achieved for β-Li3N with the increased vacancy concentration in AIMD simulations (Fig. 2c). When total vacancy concentrations increased from 2.7% to 5.6% in vacancy-rich β-Li3N, the activation energy decreased from 0.28 ± 0.04 eV to 0.25 ± 0.03 eV with extrapolated lithium-ion conductivity increased from 2.0 × 10−3 S cm−1 to 4.2 × 10−3 S cm−1 at 300 K. The result of accelerated lithium-ion diffusion in the crystal lattice by vacancies is consistent with our experimental results that the ionic conductivity increased from 2.05 × 10−5 S cm−1 to 2.14 × 10−3 S cm−1 when vacancy concentration increased from 0.5% to 8.1% at Li(2) 4f sites (from 0.3(4)% to 5.4(2)% for total lithium sites). The observed enhancement in lithium-ion diffusion within the crystal lattice, attributed to the presence of vacancies, agrees with the experimental results. Specifically, the ionic conductivity increases from 2.05 × 10−5 S cm−1 to 2.14 × 10−3 S cm−1 as the vacancy concentration at the Li(2) 4f sites increased from 0.5% to 8.1% (translating to an increase from 0.3(4)% to 5.4(2)% across all lithium sites). According to the DFT calculations, the formation energy of vacancy-rich β-Li3N across varying concentrations increases from 1.25 eV for Li2.92N0.97 (vacancy concentration 2.7%) to 2.67 eV for Li2.83N0.94 (vacancy concentration 5.6%) and 5.44 eV for Li2.67N0.89 (vacancy concentration 11.1%) (Supplementary Fig. 6). This high formation energy for the vacancy-rich β-Li3N may explain the limit for the achievable vacancy concentration observed in our experimental sample, which exhibited a vacancy concentration of approximately 5.4%. A deeper analysis with scanning electron microscopy (SEM) probing the interrelationship between lithium-ion diffusion, particle size and vacancy concentration pinpoints an elevated vacancy concentration as a crucial determinant of enhanced lithium-ion diffusion within the β-Li3N SSEs (Supplementary Fig. 7 and Supplementary Note 4).
Chemical stability towards lithium metal and air
According to the calculations (Fig. 3a and Supplementary Fig. 8), almost all well-known SSEs are unstable with lithium metal owing to the reduction of central cations. By contrast, only vacancy-rich β-Li3N is stable against lithium metal anode and shows a stable electrochemical window of 0–0.48 V. Meanwhile, as shown in Supplementary Fig. 9, the time-resolved electrochemical impedance spectroscopy (EIS) spectra of the Li/vacancy-rich β-Li3N/Li symmetric cell remained almost unchanged for 30 h, confirming the thermodynamic stability of vacancy-rich β-Li3N towards lithium metal. SEM images, depicted in Fig. 3b,c, show the morphology of pristine and lithium-interacted vacancy-rich β-Li3N, revealing minimal morphological changes and suggesting an absence of surface reactions with lithium metal. Ex situ X-ray absorption near-edge structure (XANES) analysis further corroborates the stability of vacancy-rich β-Li3N SSE against lithium, as illustrated in Fig. 3d. The N K-edge XANES spectrum characterizes electron transitions from nitrogen’s 1s orbital to the vacant electronic states in the conduction band, with a notable peak at approximately 398 eV indicative of 1s to π transitions and broader peaks around 400 eV and 403 eV corresponding to 1s to σ transitions35,36. The consistency of nitrogen K-edge XANES spectra before and after lithium contact underscores the SSE’s substantial chemical stability towards lithium metal. Enhanced spatial resolution XANES spectra through STXM, shown in Supplementary Fig. 10a,b, further validate these findings, showing the preserved particle morphology of vacancy-rich β-Li3N upon lithium interaction. These STXM images, captured at the pre-edge of the N K-edge absorption (395 eV), and the nitrogen K-edge XANES spectra derived from the surface of β-Li3N particles post-lithium contact (data were collected at photon energies spanning from 395 eV to 418 eV (refs. 37,38); Supplementary Fig. 10c), exhibit no notable chemical changes, affirming the material’s resistance to reactions with lithium. The consistency across various surface locations of a β-Li3N particle (Supplementary Fig. 10b) reinforces the absence of surface reactions, with only minor spectral fluctuations attributed to the STXM technique’s photon energy resolution limits.
Beyond its thermodynamic stability against lithium metal, the air stability of vacancy-rich β-Li3N stands as another essential trait for practical handling. To assess the air stability of the vacancy-rich β-Li3N sample upon atmospheric exposure, we used operando and in situ X-ray diffraction (XRD) combined with ex situ soft X-ray XANES techniques, targeting both the N K-edge and O K-edge. These methods were chosen given that XRD offers sensitivity to crystal structures, while soft X-ray XANES collected in the total electron yield (TEY) mode provides insights into surface chemistry. Figure 3e and Supplementary Fig. 11 depict operando XRD results for the vacancy-rich β-Li3N sample when exposed to ambient air with a relative humidity of 25%. Notably, the emergence of LiOH impurity was observed approximately 1–1.5 h post air exposure, attributable to its interaction with atmospheric moisture, as inferred from the operando XRD and ex situ XANES studies (Supplementary Figs. 11a and 12). After the 10 h exposure duration, the dominant β-Li3N phase remained largely intact, accompanied by a consistent proportion of the LiOH impurity. It is postulated that the formation of LiOH acted as a protective barrier, limiting the further exposure of β-Li3N to moisture, as corroborated by the operando XRD and ex situ XANES analyses (Supplementary Figs. 11a and 12). Analogous observations have been documented for α-Li3N single crystals39. The chemical resilience of the vacancy-rich β-Li3N sample under dry rooms (dew points −50 °C to −60 °C, equivalent to <0.3% relative humidity) is ascertained via in situ XRD and ex situ XANES. Figure 3f and Supplementary Fig. 11b show the in situ XRD data for samples stored for up to 150 h in a dry room environment. Noteworthy is the broad hump around 26 degrees 2θ, attributed to Kapton tape used during sample preparation, absent of distinct, sharp XRD peaks. Notably, these samples consistently exhibited the β-Li3N phase with uniform peak intensities and showed no additional peaks suggestive of crystalline LiOH or other secondary crystalline phases. Furthermore, the ex situ XANES data, depicted in Supplementary Fig. 12c,d (Supporting Information), verify the presence of the LiOH layer on the β-Li3N surface. This LiOH layer likely acts as a protective barrier, mitigating further exposure of the β-Li3N to moisture. Figure 3g compares the lithium-ion conductivity of vacancy-rich β-Li3N at 25 °C after different atmospheric/humidity exposures. High ionic conductivities above 10−3 S cm−1 were maintained after different exposure conditions, even for the sample exposed to 3–5% humidity for 150 h. The findings indicate that β-Li3N exhibits stability under conditions of low humidity, both in a dry room environment and during prolonged storage, rendering it suitable for integration into large-scale industrial manufacturing processes. However, exposure to high-humidity environments and direct contact with water present potential hazards for this vacancy-rich β-Li3N. To mitigate these safety concerns, it is advisable to implement coating strategies designed to minimize risks40. The criteria for selecting appropriate coating materials include compatibility with vacancy-rich β-Li3N, resistance to moisture intrusion, high ionic conductivity and minimal electronic conductivity.
All-solid-state lithium metal batteries
Determined by the direct current polarization method (Supplementary Fig. 13), this vacancy-rich Li3N shows a low electronic conductivity of ~4.5 × 10−10 S cm−1, which was promising to ensure stable lithium stripping and plating behaviour even at high current densities and capacities (Supplementary Note 5). All-solid-state lithium symmetric cells were fabricated to investigate the lithium stripping and plating behaviours, as depicted in Supplementary Fig. 14. To assess the performance of commercially available Li3N, an all-solid-state Li/commercial-Li3N/Li symmetric cell was evaluated at current densities and capacity of 0.1 mA cm−2 and 0.1 mAh cm−2. Figure 4a reveals an initial overpotential of approximately 0.5 V, attributable to the low ionic conductivity (specifically, 2.05 × 10−5 S cm−1). In subsequent stripping/plating cycles, this overpotential rapidly decreased to nearly 0 V within 80 h, suggesting serious lithium dendrite growth within the commercial Li3N SSE layer and a heightened propensity for short-circuiting. In stark contrast, the all-solid-state Li/vacancy-rich β-Li3N/Li symmetric cell showcased outstanding electrochemical performance under identical cycling conditions. Figure 4b and Supplementary Figs. 15, 16 and 17a elucidate that both the initial overpotential and the EIS spectra of this symmetric cell remained consistent even after 4,000 h of cycling. Figure 4c demonstrates the cell’s stable potential profiles, enduring a notable current density up to 7.5 mA cm−2 and a commendable capacity of 7.5 mAh cm−2. While the overpotential profiles manifested fluctuations between 7.5 and 15 mA cm−2 – which suggests potential lithium dendrite formation under these rigorous conditions – the consistent subsequent stripping/plating performance at 1 mA cm−2 and 1 mAh cm−2 indicated the non-existence of a hard short circuit. The Li/vacancy-rich β-Li3N/Li symmetric cell showcased a transient adaptability to heightened current densities and capacities, withstanding up to 15 mA cm−2 and 15 mAh cm−2, respectively41. After this rigorous assessment, continuous cycling at 1 mA cm−2 and 1 mAh cm−2 continued for 3,000 h, as presented in Supplementary Fig. 18. Analyses via EIS, XANES and SEM analyses collectively further confirm the exceptional stability under high current densities, as evident in Supplementary Fig. 9 and Supplementary Note 6. For preset critical capacities of 1 mAh cm−2 and 3 mAh cm−2, peak critical current densities of 45 mA cm−2 and 32.5 mA cm−2 were documented, as illustrated in Supplementary Figs. 18 and 19. Compared with existing all-solid-state lithium symmetric cells utilizing sulfide, nitride and oxide SSEs, and interlayer design, the Li/vacancy-rich β-Li3N/Li symmetric cell exhibited substantial enhancements in critical current densities and cycling capacities, as concisely illustrated in Fig. 4d. For the pursuit of high energy and power densities, extended lithium stripping/plating cycling was evaluated at various conditions. Figure 4e,f and Supplementary Fig. 15 demonstrate that the Li/vacancy-rich β-Li3N/Li symmetric cells maintained consistent cycling performance over 4,000 h at both 3 mA cm−2 (capacity 3 mAh cm−2) and 7.5 mA cm−2 (capacity 7.5 mAh cm−2). In addition, Supplementary Figs. 17 and 19 showcase thinner layers of vacancy-rich β-Li3N (15 mg, approximately 0.25 mm), highlighting the importance of minimizing overpotential in the Li/vacancy-rich β-Li3N/Li symmetric cells for practical application in full cells. The cells exhibited low overpotentials and sustained cycling performance across 750 h at current densities of 0.1 mA cm−2 (capacity 0.1 mAh cm−2) and 7.5 mA cm−2 (capacity 7.5 mAh cm−2), and stable cycling through 500 cycles at 45 mA cm−2 (capacity 1 mAh cm−2) and 32.5 mA cm−2 (capacity 3 mAh cm−2), along with stable cycling over 500 cycles at both 45 mA cm−2 (and 1 mAh cm−2) and 32.5 mA cm−2 (and 3 mAh cm−2). Furthermore, Supplementary Fig. 20 showcases the remarkable cycling stability of the Li/vacancy-rich β-Li3N/Cu asymmetric all-solid-state cells across 1,000 cycles at various conditions: 0.1 mA cm−2 and 0.1 mAh cm−2, 3 mA cm−2 and 3 mAh cm−2, and finally, at 7.5 mA cm−2 and 7.5 mAh cm−2.
All-solid-state lithium metal batteries were fabricated with LiCoO2 (LCO) and Ni-rich LiNi0.83Co0.11Mn0.06O2 (NCM83) serving as cathodes. Halide SSEs, Li3InCl6 and Li3YCl6, were paired with this vacancy-rich β-Li3N to form the SSE layer, while lithium metal constituted the anode, resulting in LCO or NCM83/Li3InCl6/Li3YCl6/vacancy-rich β-Li3N/Li full cells (Supplementary Figs. 14 and 21, and Supplementary Note 7). The LCO/Li3InCl6/Li3YCl6/vacancy-rich β-Li3N/Li full cells with an LCO loading of 8.92 mg cm−2 exhibit stable cycling performance (Fig. 5 and Supplementary Fig. 22). The full cell exhibited an impressive initial discharge capacity of 139.2 mAh g−1 coupled with a high initial Coulombic efficiency (ICE) of 96.9% at a rate of 0.05 C. At a rate of 0.1 C, the cell achieved a reversible capacity of 133.3 mAh g−1 and maintained an elevated Coulombic efficiency of 99%. Notably, the discernible phase-transition-induced voltage plateaus alluded to minimal polarization, ensuring efficient Li⁺ ion transport across the LCO and SSE interface. As depicted in Fig. 5c, at cycling rates of 0.5 C, 1.0 C, 2.0 C, 3.0 C and 4.0 C, the reversible capacities were sustained at 124 mAh g−1, 116 mAh g−1, 102 mAh g−1, 73 mAh g−1 and 46 mAh g−1, respectively. The compatibility of the lithium metal with the vacancy-rich β-Li3N layer is further underscored by the prolonged cycling life of the full cell, as illustrated in Fig. 5b,d–f. When subjected to a cycling rate of 0.1 C, the cell demonstrated consistent capacity retention, delivering 124 mAh g−1 across 250 cycles. At higher current densities of 0.5 C and 1.0 C, excellent long-term cycling stability with high reversible capacity (115.5 mAh g−1 over 1,000 cycles at 0.5 C and 95.21 mAh g−1 over 5,000 cycles at 1.0 C) and high capacity retention (93.6% over 1,000 cycles at 0.5 C and 82.05% over 5,000 cycles at 1.0 C) were demonstrated. Supplementary Fig. 22 (Supporting Information) illustrates the average cycling performance at rates of 0.1 C, 0.5 C and 1.0 C across ten cells for each condition, confirming the reproducibility and robustness of the all-solid-state lithium metal cells’ performance.
Figure 6a presents the voltage profiles of the NCM83/Li3InCl6/Li3YCl6/vacancy-rich β-Li3N/Li full cells at a rate of 0.1 C (where 1 C = 200 mA g−1). The ICE registers at 87.6%, and the Coulombic efficiency of subsequent cycles approaches approximately 100%, as depicted in Fig. 6b, Supplementary Fig. 23 and Supplementary Note 8. The reversible capacity of full cells was ~207 mAh g−1 and then maintained at ~195 mAh g−1 and ~190 mAh g−1 over 100 and 200 cycles, respectively. Another full cell with an NCM83 loading of 3.82 mg cm−2 achieved fast charging and discharging performance up to 5.0 C and can reach 83.77%, 73.90%, 68.63%, 64.31% and 60.47% of the reversible capacity at 1.0 C, 2.0 C, 3.0 C, 4.0 C and 5.0 C, respectively (Fig. 6c and Supplementary Fig. 24). We also demonstrated ultra-long cycling life of all-solid-state lithium metal batteries as shown in Fig. 6d. At 1.0 C, the full cell presented a high reversible capacity of ~142 mAh g−1 and an ultra-high capacity retention of 92.5% over 3,500 cycles, suggesting high chemical stability and high compatibility of vacancy-rich β-Li3N layers to lithium metal during long cycling life. The full cells with a substantial NCM83 loading of 30.31 mg cm−2 exhibited an impressive initial areal capacity of 5.42 mAh cm−2. They achieved a commendable ICE of 85.1% and sustained a robust areal capacity of approximately 4.88 mAh cm−2 over 100 cycles (Fig. 6e). Supplementary Fig. 25 details the average cycling performance of full cells with an NCM83 loading of 8.92 mg cm−2 at 0.1 C and 1.0 C, as well as those with a high areal loading of 30.31 mg cm−2, based on testing 10 cells for each cycling condition, demonstrating high reproducibility for the all-solid-state lithium metal cells. Given the pragmatic demands of electric vehicles (EVs), all-solid-state lithium metal pouch cells emerge as a potent approach, aiming for an elevated energy density (approaching 500 Wh kg−1) and ensuring an extended driving range (surpassing 300 miles) for EVs. All-solid-state lithium metal pouch cells are fabricated through the dry-film technique (Fig. 6f,g and Supplementary Fig. 26). Notably, this all-solid-state lithium metal pouch cell presented a remarkable ICE of 86.2% and sustained a capacity nearing 2.11 mAh cm−2 across 100 cycles. The average cycling performance of these all-solid-state pouch cells, demonstrating their high reproducibility, is documented in Supplementary Fig. 25.
The stable electrochemical performance observed in all-solid-state lithium metal batteries can be attributed to several key factors: the outstanding structural resilience and lithium metal compatibility of the vacancy-rich β-Li3N SSE throughout the electrochemical cycling, the robust interfacial stability within the layered SSE architecture and the effective compatibility of Li3InCl6 with the used cathode materials. The structural stability of this vacancy-rich β-Li3N SSE during electrochemical cycling is confirmed by ex situ SXRD, XANES, SEM and EDX mapping results, as depicted in Supplementary Figs. 27–38, Supplementary Tables 9–12 and Supplementary Note 9. To further extend the study’s scope to a broader range of applications, additional commercially available SSEs such as Li6PS5Cl are investigated. In all-solid-state lithium metal battery configurations comprising LCO or NCM83/Li6PS5Cl/LYC/β-Li3N/Li, remarkable electrochemical stability is observed at both 0.1 C and 1 C cycling rates as shown in Supplementary Fig. 39 and Supplementary Note 10. To address the critical requirement for air stability in SSEs for practical applications, the vacancy-rich β-Li3N SSEs, after exposure to ambient air for 10 h (hereafter referred to as β-Li3N-air-10h), still exhibit high lithium-ion diffusion characteristics, good electrochemical performance and robust interfacial stability, as depicted in Supplementary Figs. 40–43 and Supplementary Note 11. Considering cost as a critical factor for practical application, the expense associated with β-Li3N SSE has been evaluated against that of other conventional sulfide, halide and oxide SSEs, as depicted in Supplementary Fig. 44. This comparison reveals that the cost of β-Li3N SSE is on par with that of widely used commercial SSEs, such as Li6PS5Cl, Li3InCl6, La3Li7O12Zr2 (LLZO) and Li6.4La3Zr1.4Ta0.6O12 (LLZTO). Furthermore, it is anticipated that the cost of β-Li3N SSE could decrease substantially with the adoption of large-scale manufacturing processes, underscoring its potential viability for the battery industry at large. In summary, the demonstrated high ionic conductivity at ambient temperature, exceptional cycling performance, substantial interfacial stability and low cost together highlight the remarkable air stability and vast potential for practical applications of the vacancy-rich β-Li3N SSE.
Conclusion
In summary, we reported a vacancy-mediated superionic diffusion mechanism and a superionic conducting, highly lithium-compatible and air-stable vacancy-rich β-Li3N SSE. The vacancy concentration is optimized to around 8.1% lithium vacancies at Li(2) site and 5.4% nitrogen vacancies. The high concentration of vacancies results in a high ionic conductivity of up to 2.14 × 10−3 S cm−1, surpassing almost all nitride SSEs. This vacancy-mediated fast lithium-ion migration mechanism is unravelled by refined crystal structures of SXRD and TOF neutron diffraction, DFT calculations and AIMD simulations. Furthermore, the high lithium compatibility mechanism of vacancy-rich β-Li3N is unveiled by DFT calculation and STXM and promises excellent application of vacancy-rich β-Li3N in all-solid-state lithium symmetric cells and all-solid-state lithium metal batteries. The Li/vacancy-rich β-Li3N/Li lithium symmetric cells make breakthroughs in ultra-high critical current densities of 45 mA cm−2 and 32.5 mA cm−2 for the fixed capacities of 1 mAh cm−2 and 3 mAh cm−2. In the case of a fixed stripping/plating time of 1 h, the critical current density and capacity reached 7.5 mA cm−2 and 7.5 mAh cm−2, respectively. The cell also withstood a temporary demanding current density of 15 mA cm−2 and a capacity of 15 mAh cm−2. The lithium symmetric cells further delivered stable cycling performance for over 4,000 h at 0.1 mA cm−2, 3 mA cm−2 and 7.5 mA cm−2 with a fixed stripping/plating time of 1 h. Owing to the high stability of vacancy-rich β-Li3N towards lithium metal, all-solid-state lithium metal batteries coupled with LCO and NCM83 cathodes, in the configuration of LCO or NCM83/halide/vacancy-rich β-Li3N/Li, delivered excellent electrochemical performance. When coupled with an LCO cathode, the LCO/halide/vacancy-rich β-Li3N/Li all-solid-state lithium metal batteries showed excellent cycling stability at both low and high current densities (93.7% capacity retention over 250 cycles at 0.1 C, 93.6% capacity retention over 1,000 cycles at 0.5 C and 82.05% capacity retention over 5,000 cycles at 1.0 C). The NCM83/halide/vacancy-rich β-Li3N/Li full cells exhibit moderate rapid charge and discharge capabilities up to 5.0 C, maintaining 60.47% of the initial capacity. Furthermore, these cells achieve a robust capacity retention of 92.5% with a specific capacity of 153.6 mAh g−1 over 3,500 cycles at 1.0 C. The cells also deliver an impressive areal capacity of approximately 5.0 mAh cm−2 for pellet-type cells with an area of 0.785 cm2 and around 2.2 mAh cm−2 for the all-solid-state lithium metal pouch cells measuring 2.5 × 2 cm2. In addition, the air stability of this vacancy-rich β-Li3N SSE in dry room is clarified by in situ and operando XRD and promises commercialization of vacancy-rich β-Li3N for all-solid-state lithium metal batteries. On the basis of our results, this vacancy-rich β-Li3N SSE is a highly promising candidate for enabling the use of lithium metal anodes in all-solid-state lithium metal batteries, which is a key to achieve high energy density and meets the ever-increasing requirements of high specific energy for fast-developing EVs market.
Methods
Preparation of the β-Li3N SSEs through the ball-milling method
Commercial lithium nitride (Li3N, Alfa Aesar, 99.4%) was ball milled in a ZrO2 pot with ZrO2 balls (ϕ = 5 mm; the mass ratio of balls to Li3N was 40). The ball milling was performed using a planetary ball-milling apparatus (Retsch planetary ball mill PM 200) at a series of ball-milling speeds and a series of ball-milling times. All the preparation processes were carried out with an Ar atmosphere.
Notes: The energy transferred to the materials in the ball-milling process can be calculated based on the equation below46:
$$\Delta {E}_{\mathrm{b}}=-{m}_{\mathrm{b}}\left[\frac{{\omega }_{\mathrm{J}}^{3}\left({R}_{\mathrm{J}}-\frac{{d}_{\mathrm{b}}}{2}\right)}{{\omega }_{\mathrm{P}}}+{\omega }_{\mathrm{P}}{\omega }_{\mathrm{J}}{R}_{\mathrm{P}}\right]\left({R}_{\mathrm{J}}-\frac{{d}_{\mathrm{b}}}{2}\right)$$
where mb is the mass of balls, db is the diameter of balls, RJ is the internal radius of jars, RP is the distance between the rotating plate centre and the jar centre, ωP is the angular velocity of the main plate and ωJ is the angular velocity of jars.
The ball-milling parameters used in this study are as follows: mb, 0.4 g; db, 5 mm; RJ, 70.0 mm; RP, 23.2 mm; ({\omega }_{\mathrm{J}}/{\omega }_{\mathrm{P}}=-2). The speed shown in the study is the angular velocity of the main plate, ωP. The energy transferred to the materials in the ball-milling process with a speed of 400 rpm is calculated to be 0.292 J, which is large enough to create lithium vacancies and nitrogen vacancies in β-Li3N (lithium vacancy formation energy, 0.81 eV at 4f sites, and nitrogen vacancy formation energy, 3.28 eV).
Preparation of Li3InCl6 and Li3YCl6 halide SSEs
Li3InCl6
Stoichiometric amounts of lithium chloride (LiCl, Alfa Aesar, 99.9% purity) and indium chloride (InCl3, Alfa Aesar, 99.99% purity) were combined and dissolved in deionized water under ambient conditions (outside a glove box environment). The resultant solution was subsequently dried under vacuum to yield the precursor hydrate. This precursor was then subjected to thermal treatment at 200 °C for 4 h under vacuum conditions to synthesize Li3InCl6 (ref. 47).
Li3YCl6
Lithium chloride (LiCl, Alfa Aesar, 99.9% purity) and yttrium chloride (YCl3, Alfa Aesar, 99.99% purity) were measured in stoichiometric ratios and combined. The mixture was then homogenized in a zirconia (ZrO2) milling jar containing ZrO2 milling balls (diameter = 5 mm, with a ball-to-mixture mass ratio of 40:1) using a planetary ball mill. The milling operation was conducted at a speed of 500 rpm for a duration of 24 h under an Ar atmosphere to ensure inert conditions during the process48.
Morphology characterizations
The morphology was characterized with by a Hitachi S-4800 field-emission scanning electron microscope.
Ionic conductivity measurements
Ionic conductivities were measured by a.c. impedance spectroscopy. Typically, 100 mg of powder samples was placed between two stainless-steel rods with 10 mm diameter and pressed at 3 tons (~375 MPa). The thickness of the pellets was 1.25 mm. The procedures were performed inside an Ar-filled glove box. EIS was performed within the temperature range of −25–55 °C using one potentiostat (BioLogic SP-200) from 7 MHz to 1 Hz with an amplitude of 10 mV.
Small-area pellet-type all-solid-state lithium metal cells
Lithium symmetric cells
All cell preparation procedures were executed within an argon-filled glove box to maintain an inert atmosphere. For the fabrication of Li/Li3N/Li symmetric cells, approximately 100 mg of Li3N was positioned within a polytetrafluoroethylene (PTFE) mould of 10 mm diameter and subsequently compressed under a force of 3 tons, resulting in a pellet with an approximate thickness of 1.25 mm, as illustrated in Supplementary Fig. 14. Subsequent to the positioning of lithium metal foils on each side of the electrolyte layer, a force of 0.1 ton was applied to ensure intimate contact between the lithium foils and the Li3N layers. The assembly, comprising the PTFE mould with the pressed SSE layer and lithium metal foils, was then installed within custom-fabricated cells for electrochemical cycling. All-solid-state lithium symmetric cells were cycled without any external pressure.
Full cells
For the preparation of cathode composites, commercial LCO and Ni-rich LiNi0.83Co0.11Mn0.06O2 (NCM83), both sourced from China Automotive Battery Research Institute, were manually ground in conjunction with the synthesized Li3InCl6. A mortar and pestle were used for the grinding process, which lasted for 5 min, maintaining a weight ratio of 70:30 between the active material and Li3InCl6. For cell fabrication, 60 mg Li3InCl6 powder47 and 60 mg Li3YCl6 powder48 were placed into a PTFE die with a diameter of 10 mm and pressed at 3 tons (Li3InCl6 pellet thickness, ~0.36 mm; Li3YCl6 pellet thickness, ~0.36 mm) as shown in Supplementary Fig. 14. Subsequently, 10 mg of the cathode composite powder was dispersed on one side of the electrolyte and pressed again at 3 tons (LCO composite cathode thickness, 0.033 mm; NCM83 composite cathode thickness, 0.034 mm). Before attaching lithium metal anode (capacity ~18 mAh cm−2) on the other side of the electrolyte layer, a thin layer of Li3N (25 mg) was set to face lithium metal anode (Li3N pellet thickness ~0.31 mm). To ensure optimal contact between the lithium foils and the Li3N layers, a pressure of 0.1 ton was applied to the cells before cycling. The PTFE die with pressed SSE and cathode layers and lithium metal foil were then placed inside the in-house fabricated cells for cycling. All-solid-state lithium metal cells were cycled in a galvanostatic mode (that is, a constant current protocol) within the voltage range of 2.7–4.2 V (versus Li+/Li) for LCO and 2.7–4.3 V (versus Li+/Li) for NCM83 using a Neware battery test system and Landt cyclers at room temperature. The galvanostatic charge–discharge studies of LCO and NCM83 cathodes were conducted at different current densities. In the configuration involving a cathode loading of 8.92 mg cm−2, the cells were cycled without the application of any external pressure. However, for configurations with a higher cathode loading of 30.31 mg cm−2, an external pressure of 20 MPa was exerted during cycling to ensure optimal contact and performance.
All-solid-state lithium metal pouch cells
All cell preparation procedures were performed inside an Ar-filled glove box. The dry-film processes were used to fabricate halide SSEs, Li3N and cathode films. A mixture of 99.5 wt% SSEs (or cathode composites as described in the section on ‘Small-area pellet-type all-solid-state lithium metal cells’) and 0.5 wt% PTFE (MSE Supplies LLC, particle size ~480 µm) was weighed and combined in an agate mortar heated at 80 °C until a cohesive dough-like consistency was achieved. Subsequently, this dough was placed between two stainless-steel foils at 100 °C, calendared to produce a flexible film of desired thickness and then tailored to the necessary area. Specifications for the films are as follows: NCM83 film (mass ~90 mg, ~160 µm thickness, 2.5 cm × 2 cm area), Li3InCl6 film (mass ~80 mg, ~100 µm thickness, 3 cm × 2.5 cm area), Li3YCl6 film (mass ~80 mg, ~100 µm thickness, 3 cm × 2.5 cm area) and Li3N film (mass ~33 mg, ~100 µm thickness, 2.7 cm × 2.3 cm area).
For the assembly of the dry-film-based all-solid-state full cells, a multistep lamination process was adopted:
1.
The Li3InCl6 film was sandwiched between stainless-steel foils and pressed at 3 tons, producing a stand-alone, rigid film.
2. 2.
The Li3YCl6 film was overlaid on the Li3InCl6 film, pressed between stainless-steel foils at 3 tons, resulting in a bi-layered composite film.
3. 3.
The Li3N film was aligned atop the Li3YCl6 side of the bi-layered composite, pressed between stainless-steel foils at 3 tons, yielding a tri-layered composite film.
4. 4.
The NCM83 cathode film was positioned on the Li3InCl6 side of the tri-layered composite, pressed between stainless-steel foils at 3 tons, generating a quad-layered composite film.
5. 5.
Lastly, a lithium foil film (~90 µm thickness, 2.6 cm × 2.2 cm area) was layered on the Li3N side of the quad-layered composite. This assembly was further enveloped by stainless-steel foils (25.4 µm thickness) and then hermetically sealed within aluminium foil bags under vacuum to produce the final pouch cell.
6. 6.
To ensure optimal interfacial contact among the cathode, SSE and lithium foil layers, the pouch cells were subjected to a pressure of 20 MPa throughout the cycling process.
Synchrotron-based X-ray powder diffraction, scanning transmission X-ray microscopy and X-ray absorption near-edge structure
The synchrotron-based X-ray powder diffraction was carried out at the Very Sensitive Elemental and Structural Probe Employing Radiation from a Synchrotron (VESPERS) and Brockhouse X-ray Diffraction and Scattering (BXDS) beamlines at the Canadian Light Source. The STXM was conducted at the SM beamline of the Canadian Light Source, which is equipped with a 25 nm outermost-zone zone plate (CXRO, Berkeley Lab). The diffraction-limited spatial resolution for this zone plate is 30 nm. The samples were raster scanned with synchronized detection of transmitted X-rays to generate images. Chemical imaging and XANES spectra are obtained using image sequence (stack) scans over a range of photon energies at the N K-edge. STXM data were analysed using the aXis2000 software package ( which allows for detailed interactive processing of the images and fitting of the X-ray absorption spectra. XANES from interesting locations were extracted from image stacks using image mask, which only selects the regions of interest. For the assessment of the chemical stability of β-Li3N to lithium metal using STXM, all sample preparation and mounting procedures were executed in an Ar-filled glove box. Initially, pristine vacancy-rich β-Li3N particles were dispersed onto a TEM grid, which was subsequently utilized for STXM analysis. In a parallel set-up, an additional set of pristine vacancy-rich β-Li3N particles was sandwiched between two lithium foils (diameter 10 mm). This assembly was subjected to a pressure of 1 ton to ensure intimate contact between the lithium foil and the β-Li3N particles. Following a 24 h interval, the β-Li3N particles in direct contact with the lithium foil surface were collected and spread onto a separate TEM grid in preparation for subsequent STXM analysis. To mitigate the risk of contamination from ambient air, all TEM grid-mounted samples were transported to the STXM measurement chamber in an Ar atmosphere. The N K-, O K-, Ni L3,2-, Co L3,2- and Mn L3,2-edge XANES spectra were collected in the TEY mode at the Spherical Grating Monochromator (SGM) beamline at the Canadian Light Source. In addition, the In L3- and Y L3-edge spectra were collected in the TEY mode at the Soft X-ray Microcharacterization Beamline (SXRMB) beamline at the Canadian Light Source.
Neutron total scattering
The room-temperature neutron diffraction was performed at the NOMAD beamline (BL-1B) at the Spallation Neutron Source (SNS) at Oak Ridge National Laboratory (ORNL). Powdered (~0.15 g) samples were packed into 3 mm quartz capillaries and sealed with epoxy in a glove box filled with dry argon. The acquisition time was 1 h for each sample. The background was subtracted from the acquired data followed by normalization against the vanadium rod.
DFT calculation
We performed all DFT calculations using Vienna Ab initio Simulation Package (VASP)49 within projector augmented-wave50 approach. Generalized gradient approximation with the Perdew–Burke–Ernzerhof functional51 was used in our DFT calculation. The plane-wave energy cut-off, k-points density and other convergence parameters used in all static DFT calculations were consistent with the Materials Project52.
Calculation of Li and N vacancies
Lithium has two unique sites in the hexagonal structure β-Li3N, of which one is the tetrahedral 4f site and another is the trigonal planar 2b site (Fig. 1). To study the site preference of Li vacancy, we calculated the formation energy of single Li vacancy. A 3 × 3 × 2 supercell model of β-Li3N, which contains 36 formula units (f.u.) of Li3N and a total of 144 atoms, was used. A single neutral Li was removed from either 2b or 4f site in the supercell, and then the formation energy of Li vacancy was calculated as
$$\Delta {E}_{{\rm{Li}}\_{\rm{vac}}}=E[{{\rm{Li}}}_{107}{{\rm{N}}}_{36}]+E[{\rm{Li}}]-E[{{\rm{Li}}}_{108}{\rm{N}}_{36}]$$
where E[Li108N36] represents the total energies of the pristine supercell of β-Li3N, E[Li107N36] represents the supercell with a single Li vacancy and E[Li] is the bulk energy of Li metal normalized to a single Li atom. Similarly, we calculated the formation energy of single N vacancy as
$$\Delta {E}_{{\rm{N}}\_{\rm{vac}}}=E[{{\rm{Li}}}_{108}{{\rm{N}}}_{35}]+E[{\rm{N}}]-E[{{\rm{Li}}}_{108}{{\rm{N}}}_{36}]$$
where E[Li108N35] represents the supercell with a single N vacancy at 2c sites and E[N] is the energy of N2 gas normalized to a single N atom.
The formation energy of vacancy-rich Li3N is calculated as
$$\varDelta {E}_{x\cdot {\rm{Li}}3{\rm{N}}\_{\rm{vac}}}=E[{{\rm{Li}}}_{108-3x}{{\rm{N}}}_{36-x}]+(1-x/36)\cdot E[{{\rm{Li}}}_{108}{{\rm{N}}}_{36}]$$
where E[Li108−3xN36−x] represents the supercell with a total of x Li3N units of vacancies, specifically 3 Li vacancy at the 4f site and N vacancy at the 2b site.
Calculation of structural ordering
The atomistic configurations of Li2.92N0.97 and Li2.83N0.94 with the Li and N partial occupancy at 4f and 2c sites, respectively, were determined using the same ordering procedure as in our previous studies53,54. One and two formula units of Li3N were removed from the 3 × 3 × 2 supercell models, to model Li2.92N0.97 and Li2.83N0.94, respectively. For each composition, we generated 10,000 structures by randomly removing the Li (4f) and N (2c) from the supercell models. Among these structures, 50 symmetrically distinctive configurations with minimal electrostatic energies were selected for DFT calculations, and the lowest-energy structure was identified as the ground state for other further calculations.
AIMD simulation
We performed AIMD simulations in 3 × 3 × 2 supercell models of defected β-Li3N structures using NVT ensemble with the Nosé–Hoover thermostat55. Non-spin mode, a time step of 2 fs and a single Γ-centred k-point were used as in previous studies. In each simulation, the structures were first heated from 100 K to the target temperatures (600–1,600 K) at a constant rate during a period of 2 ps. Then the AIMD simulations were performed for a duration of 100–300 ps. The ionic diffusivity D was calculated as the mean square displacement (MSD) over a time interval Δt:
$$D=\frac{1}{2Nd\Delta t}\mathop{\sum }\limits_{i=1}^{N}{\langle {[{{\bf{r}}}_{i}(t+\Delta t)-{{\bf{r}}}_{i}(t)]}^{2}\rangle }_{t}$$
where d = 3 is the dimension of the diffusion, N is the total number of diffusion ions and ri(t) is the displacement of the ith ion at time t. The ionic conductivity was calculated according to the Nernst–Einstein relationship:
$$\sigma =\frac{{n}{{q}}^{2}}{{{k}}_{\mathrm{B}}{T}}{D}$$
where n is the mobile ion volume density, q is the ionic charge and kB is the Boltzmann constant. Arrhenius relation is used to get activation energy and to extrapolate ionic conductivity at desired temperature:
$$\sigma {{T}}={\sigma }_{0}\exp \begin{array}{l}(\frac{-{E}_{\mathrm{a}}}{{k}_{\mathrm{B}}T})\end{array}$$
where Ea is the activation energy and σ0 is the pre-exponential factor. Given that ion hopping is a stochastic process, the statistical deviations of the conductivity were evaluated according to the number of hopping events in our previous report56. At each temperature, the AIMD simulations were performed for 100–300 ps until the ionic diffusivity converged with a relative standard deviation between 20% and 30%.
Calculation of electrochemical window
The electrochemical window was evaluated using the method in our previous studies57,58,59. A grand potential phase diagram was used to identify the phase equilibria of a given phase in equilibrium with Li reservoir at chemical potential μLi referenced to Li metal. The stable electrochemical window of the phase was estimated as the range of μLi, where the phase is neither oxidized nor reduced.
Data availability
Data supporting the findings from this work are available within this article and the Supplementary Information. All other relevant data supporting findings are available from the corresponding authors on request. Source data are provided with this paper.
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Acknowledgements
This work was funded by the Natural Sciences and Engineering Research Council of Canada (NSERC), the Canada Research Chair Program, the Canada Foundation for Innovation (CFI), the Ontario Research Fund, the Canadian Light Source (CLS) at the University of Saskatchewan and the University of Western Ontario. CLS was supported by CFI, NSERC, NRC, CHIR and the University of Saskatchewan. W.L. and M.L. acknowledge the receipt of support from the CLSI Graduate and Post-Doctoral Student Travel Support Program. W.L. appreciates the funding support from Mitacs Accelerate Fellowships. We also appreciate the help of the beamline scientists of SGM and SXRMB beamlines at the Canadian Light Source: T. Regier, J. Dynes, Z. Arthur, M. Shakouri, Q. Xiao and A. Paterson. Y.M. acknowledges the funding support from National Science Foundation award number 1940166 and the computational facilities from the University of Maryland supercomputing resources and the Maryland Advanced Research Computing Center (MARCC). Part of this work was conducted at the NOMAD beamlines at ORNL’s Spallation Neutron Source, which was sponsored by the Scientific User Facilities Division, Office of Basic Sciences, US Department of Energy. J. Liu would like to thank the partial financial support from ORNL LDRD number 10761.
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Author notes
These authors contributed equally: Weihan Li, Minsi Li, Shuo Wang, Po-Hsiu Chien.
Authors and Affiliations
Department of Mechanical and Materials Engineering, Western University, London, Ontario, Canada
Weihan Li, Minsi Li, Jing Luo, Jiamin Fu, Xiaoting Lin, Ruying Li & Xueliang Sun
2. Department of Chemistry and Soochow-Western Centre for Synchrotron Radiation Research, Western University, London, Ontario, Canada
Weihan Li, Minsi Li & Tsun-Kong Sham
3. Department of Materials Science and Engineering, University of Maryland, College Park, MD, USA
Shuo Wang & Yifei Mo
4. Neutron Scattering Division, Oak Ridge National Laboratory, Oak Ridge, TN, USA
Po-Hsiu Chien & Jue Liu
5. Canadian Light Source, Saskatoon, Saskatchewan, Canada
Graham King, Renfei Feng, Jian Wang & Jigang Zhou
6. Maryland Energy Innovation, University of Maryland, College Park, MD, USA
Yifei Mo
7. Eastern Institute for Advanced Study, Eastern Institute of Technology, Ningbo, China
Xueliang Sun
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Contributions
Supervision: X.S., T.-K.S., Y.M. and J. Liu. Conceptualization: W.L., M.L., X.S. and T.-K.S. Methodology: W.L., M.L., S.W., P.-H.C., J. Liu, Y.M., T.-K.S. and X.S. Investigation: W.L., M.L., S.W., P.-H.C., J. Luo, J.F., X.L., G.K., R.F., J.W., J.Z. and R.L. Visualization: W.L., M.L., S.W. and P.-H.C. Writing (original draft): W.L., M.L., S.W. and P.-H.C. Writing (review and editing): W.L., M.L., S.W., P.-H.C., J. Luo, X.S., T.-K.S., Y.M. and J. Liu. All authors discussed the results and commented on the paper.
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Jue Liu, Yifei Mo, Tsun-Kong Sham or Xueliang Sun.
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Li, W., Li, M., Wang, S. et al. Superionic conducting vacancy-rich β-Li3N electrolyte for stable cycling of all-solid-state lithium metal batteries.
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1148 | https://www.youtube.com/watch?v=xEZIoJmyA4g | Solving a trigonometric equation using the zero product property
Brian McLogan
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👉 Learn how to solve trigonometric equations using the zero product property. The zero product property states that when the product of two quantities is equal to 0, then either of the quantities is zero. When solving factored trigonometric expressions which equals zero, each of the terms of the trigonometric expression is equated to zero and evaluated accordingly.
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Transcript:
so in this case ladies and gentlemen we are kind of set up pretty nicely because instead of having to factor something or anything else we already have a zero product property set up so now i can just say that i have cosine of 2x equals 0 and then 2 cosine of x plus 1 equals 0. now we're trying to find all the solutions so we are going to be using our plus 2 pi or plus pi or whatever else is going to be the case so here i can solve this so and then divide by two so you can say cosine of x equals negative one half and over here i have cosine of 2x well we already have it equal to zero so let's go ahead and determine our solutions so to determine my solutions i'll go back to my unit circle and i'll say that cosine of two x equals zero well that occurs at two points you could say at pi halves right that's at 0 1 and at 3 pi halves so we could say 2x equals pi over 2 and then you could say now if i go to pi over 2 to go from this solution you could also do 2x equals 3 pi over 2. well remember ladies and gentlemen we're trying to find all the solutions so to get from this solution to this solution what i have to add pi and if i add pi again it takes me to the next one so rather than writing both the solutions i can just say 2x equals pi halves plus pi times r all right now let's go into when is cosine equal to one half well we've done this already a couple times today that's at the angle pi over three and then also at the angle of um four pi five pi over three oh it is a negative you're right so therefore it's two pi over three and then also my other angle which is four pi over 3. so i can say x equals now the thing is about 4 pi over 3 4 pi over 3 if i add pi to it that's going to take me over here which is not a solution which is not another solution so the only way to go from my i'm sorry my 2 pi over 3 to the next solution is i have to add a full revolution all right the same thing for 4 pi over 3. i can't add any number constantly around there to give me the other solution so i'm going to have to add another 2 pi so in this case i'm going to have x equals 2 pi over 3 plus 2 pi r and x equals 4 pi over 3 plus 2 pi r now in this case since we have a double angle to find my solutions of x i'm going to have to divide by 2. so therefore x equals pi over 4 plus pi halves r okay it's not don't think of this as the 2x just think of this as cosine of x equals zero when does cosine of x equal zero it equals zero at pi halves and at three pi halves but if i go to pi halves plus pi that takes me to three pi halves okay so it doesn't matter what if it's a double angle a half angle and so forth you just solve for the cosine of your angle when it equals zero okay then you apply your double angle half angle whatever else we're going to be doing okay so don't worry don't let this affect your answer until you get to this point once you're able to figure out what x is right then you can do oh it's 2x divide by two now okay the same thing this could be like this could be x over two you still do the exact same answer until you get to this last point and then you multiply by two to get x by itself okay anybody have any last questions good okay |
1149 | https://proofwiki.org/wiki/Integer_as_Sum_of_Three_Squares | Integer as Sum of Three Squares
From ProofWiki
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Contents
1 Theorem
1.1 Sequence
2 Proof
2.1 Sufficient Condition
2.2 Necessary Condition
3 Sources
Theorem
Let r be a positive integer.
Then r can be expressed as the sum of three squares if and only if it is not of the form:
: 4n(8m+7)
for some m,n∈Z≥0.
Sequence
The sequence of positive integers that cannot be expressed as the sum of at most 3 squares begins:
: 7,15,23,28,31,39,47,55,60,…
Proof
Sufficient Condition
Suppose r is not of the form 4n(8m+7).
Then we need to show that it can always be expressed as the sum of three squares.
| | |
--- |
| | This theorem requires a proof. You can help Pr∞fWiki by crafting such a proof. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{ProofWanted}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
Necessary Condition
From Square Modulo 8, the squares modulo 8 are 0,1 and 4.
So the sum of three squares can be congruent modulo 8 to any of the values 0,1,2,3,4,5 or 6, but not 7.
So no number of the form 8m+7 can be the sum of three squares.
Now, suppose that ∃n≥1,m≥0 such that:
: 4n(8m+7)=x2+y2+z2
As the left hand side is congruent modulo 4 to 0, and as squares modulo 4 are either 0 or 1, it must be that x,y and z are all even.
Putting x=2x1,y=2y1,z=2z1, we get:
: 4n−1(8m+7)=x21+y21+z21
If n−1>1 then x1,y1 and z1 are all still even, and the argument can be repeated:
: 4n−2(8m+7)=x22+y22+z22
Thus we descend through all powers of 4 till 8m+7 itself is expressed as the sum of three squares.
But this is impossible, as we saw above.
So the assumption that 4n(8m+7) can be expressed as the sum of three squares is false.
■
Sources
1971: George E. Andrews: Number Theory ... (previous) ... (next): 3-5 The Use of Computers in Number Theory: Exercise 4
1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): 3
1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): 4
1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): 3
1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): 4
1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): Lagrange's theorem: 1. (J.L. Lagrange, 1772)
2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): Lagrange's theorem: 1. (J.L. Lagrange, 1772)
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1150 | https://www.cuemath.com/geometry/apothem/ | Apothem
Tim and Sam were finding the area of various regular polygons. Tim divided the polygons into triangles and was trying to calculate the area of the polygons, which took him too long. On the other hand, Sam found the area of the polygons easily, as he knew the length of the apothem.So let's learn about apothem today!
Before we get started, check out this interesting simulation to identify the apothem for various polygons. This apothem calculator will help you understand the lesson better.
Lesson Plan
| | |
--- |
| 1. | What Is Meant by Apothem? |
| 2. | Important Notes on Apothem |
| 3. | Solved Examples on Apothem |
| 4. | Challenging Questions on Apothem |
| 5. | Interactive Questions on Apothem |
What Is Meant by Apothem?
Apothem is a line drawn from the center of any polygon to the midpoint of one of the sides.
Formulas Used to Calculate the Apothem Length
The apothem formula, when the side length is given is:
| |
| (a) = (\begin{align}\frac{S}{2\,\, \text{tan}\left ( \frac{180}{n} \right )}\end{align}) |
Where,
(a) = apothem length
(s) = side length
(n) = number of sides of a polygon.
The apothem formula , when the radius is given is:
| |
| (a) = (r.cos\frac{180}{n}) |
(r) = radius.
(n) = number of sides
(Cos) = cosine function which is calculated in degrees.
We can use the apothem area formula of a polygon to calculate the length of the apothem.
| |
| (A) = (\dfrac{1}{2}aP) |
Where,
(A) = area of the polygon
(a) = apothem.
(P) = perimeter
How to Calculate Area of a Polygon Using Apothem?
To calculate the area of a polygon with the help of apothem, we use the formula:
(A) = (\dfrac{1}{2}aP)
Where,
(a) = apothem.
(P) = perimeter.
Example: Find the area of a regular hexagon, if the side length is (5) inches, and the apothem is (3) inches.
(A) = (\dfrac{1}{2}aP)
As we know perimeter:
[\begin{align}P &= [\text{side length}]\times [\text{no. of sides}]\&= 5\times6 = 30\end{align}]
After the perimeter is calculated, we use it in the formula of Area = (A) = (\dfrac{1}{2}aP)
[\begin{align}A &= \dfrac{1}{2}aP \
& = \dfrac{1}{2} \left ( 3\right )\left (30 \right )\& = 45 \text{ inches}^2\end{align}]
Solved Examples
| |
| Example 1 |
Help Bryan find the length of the apothem of a regular pentagon of side = (10) inches and area (150\sqrt{3}\,\,\text {inches}^{2}).
Solution
Given,
(L) = (10) inches.
(A) = (150\sqrt{3}\,\,\text {inches}^{2})
So, the perimeter will be (P) = (10\times 5) = (50) inches.
[\begin{align}A&=\dfrac{1}{2}aP\
150\sqrt{3}&= \dfrac{1}{2}a\times 50\
25a&=150\sqrt{3}\
a&=\dfrac{150\sqrt{3}}{25}\
a&=6\sqrt{3}\text{ inches}\end{align}]
Therefore, Apothem = (6\sqrt{3}) inches.
| |
| (\therefore) (a) = (6\sqrt{3}) inches |
| |
| Example 2 |
Can you help Dylan calculate the area of a regular hexagon of side 8 inches and apothem (4\sqrt{3}), using the area of polygon formula?
Solution
Given,
side length = 8 inches.
perimeter (P) = (L\times n)
= (8\times 6)
= (48) inches
Area of polygon = [\begin{align}A& = \dfrac{1}{2}aP\&= \dfrac{1}{2}\times4\sqrt{3}\times 48\&= 96\sqrt{3} \text{ inches}^2\end{align}]
| |
| (\therefore) The area of the polygon is (96\sqrt{3} \text{ inches}^2) |
| |
| Example 3 |
Emily's teacher asked her to calculate the area of a regular hexagon, whose apothem is 7 inches and perimeter 48 inches.
Solution
Given,
(a) = 7 inches.
(P) = 48 inches.
[\begin{align}&\frac{1}{2}aP\
&=\frac{1}{2}\times 7\times 48\
&=\frac{1}{2}\times 336\
&=\frac{336}{2}\
&=168 \text{ inches}^2\end{align}]
| |
| (\therefore) (A) = (168 \text{ inches}^2) |
| |
| Example 4 |
Can you help Jose calculate the length of the apothem of a square, which has a side length of 3 inches?
Solution
The formula for calculating apothem, when side length is given is:
(a) = (\begin{align}\frac{S}{2 \,\,\text{tan}\left ( \frac{180}{n} \right )}\end{align})
[\begin {align}a &=\dfrac{S}{2\,\,\text{tan}\left ( \dfrac{180}{n} \right)}\
&=\dfrac{3}{2\,\,\text{tan}\left ( \dfrac{180}{4} \right)}\
&=\dfrac{3}{2\,\,\text{tan} 45}\
&=\dfrac{3}{2\times 1}\
&=\dfrac{3}{2}\
&=1.5\text{ inches}\end{align}]
| |
| (\therefore) The length of apothem is (1.5\text{ inches}) |
Interactive Questions
Here are a few activities for you to practice.
Select/Type your answer and click the "Check Answer" button to see the result.
Let's Summarize
We hope you enjoyed learning about apothem with the simulations and practice questions. Now you will be able to easily solve problems related to the apothem.
About Cuemath
At Cuemath, our team of math experts is dedicated to making learning fun for our favorite readers, the students!
Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic.
Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we at Cuemath believe in.
Frequently Asked Questions (FAQs)
1. Is the apothem the same as the radius?
Apothem is also a radius, but when we talk about radius, we usually refer to a circle or a sphere.
However, when we talk about apothem, it can be any other polygon as well, such as the square, triangle, or hexagon.
2. Is the Apothem Equal to the Side Length?
No, an apothem's length is not always equal to its side length. However, if we know the side length of a polygon, the apothem can be calculated.
3. What is the Apothem of a Square?
The apothem of a square is equal to half of its side length. |
1151 | https://brainly.com/question/17033454 | [FREE] consider the perfect square trinomial identity:a^2+2ab+b^2=(a+b)^2 - brainly.com
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consider the perfect square trinomial identity:a^2+2ab+b^2=(a+b)^2
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Explanation
x² + 10x + 25
a² +2ab + b² = (a + b)²
b² = 25
b = √25
b = 5
2ab = 2 x 5 = 10x
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The perfect square trinomial identity states that a 2+2 ab+b 2=(a+b)2. This means a quadratic expression can be factored as the square of a binomial. An example with specific values shows the identity in action, demonstrating how it simplifies mathematical expressions.
Explanation
The perfect square trinomial identity states that:
a 2+2 ab+b 2=(a+b)2
This means that when you have a quadratic expression in the form of a 2+2 ab+b 2, it can be factored into the square of a binomial, specifically (a+b).
Explanation of Terms:
a and b represent any real numbers.
2ab is the term that represents twice the product of a and b.
Perfect Square Trinomial is a polynomial that can be expressed as the square of a binomial.
Example:
To see this in action, let’s consider an example with specific numbers:
Let a=x and b=5.
Substitute these values into the identity:
(x)2+2(x)(5)+(5)2
=x 2+10 x+25
According to our identity, this expression can also be factored as:
(x+5)2
Conclusion:
This identity helps in simplifying expressions and is useful in various areas of mathematics, including factoring and solving quadratic equations. Understanding how to apply and recognize this identity enhances algebraic skills.
Examples & Evidence
For instance, if you take a=2 and b=3, then according to the identity: 2 2+2⋅2×3+3 2=(2+3)2 simplifies to 4+12+9=25, which is indeed equal to 5 2.
The identity is established in algebra as a fundamental property, which can be proven through expansion and rearrangement of terms, affirming its reliability in mathematical practices.
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Community Answer 5.0 2 The polynomial below is a perfect square trinomial of the form A^2-2AB+B^2 4x^2-20x+25
Community Answer 4.7 23 the polynomial below is a perfect square trinomial of the form A^2-2AB+B^2 4x^2-16x+16
Community Answer 4.4 4 Perfect square trinomial: (a – b)2 = (a – b)(a – b) = a2 – 2ab + b2 Find the product of (k – 9)2 using the perfect square trinomial rule shown on the left.
Community Answer 5.0 2 The polynomial below is a perfect square trinomial of the form a^2-2ab+b^2 9x^2-30x+25?
Community Answer 5.0 50 The polynomial below is a perfect square trinomial of the form A^2-2AB+B^2. 9x^2-36x+16
Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer
Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)?
Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD
Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four?
New questions in Mathematics
Write each number in Roman numerals. (a) 3000 (b) 2170
Solve the equation. 2 z(z−4)+3 z 2=5 z 2+z+2
The elements of a set may be sets themselves, as in {1, {1, 3}, {2}, 4}. Explain why the set {} is not the same set as {{}}. A. The set {} is the set containing no elements, while the set {{}} is a set containing the empty set. B. The set {} is the set containing the empty set and the empty set is a set with no elements, while the set {{}} is a set containing a set 0. C. The set {} is the set containing no elements, while the set {{}} is a set containing a set 0. D. The set {} is the set containing the empty set and the empty set is a set with no elements, while the set {{}} is a set containing the element 0.
(a) Bill runs 11 miles in 91 minutes. How many minutes does he take per mile? (b) It takes 39 pounds of seed to completely plant a 6-acre field. How many acres can be planted per pound of seed?
Evaluate (f∘g)(x) and write the domain in interval notation. Write the answer in the intervals as an integer or simplified fraction. f(x)=x−1 xg(x)=x 2−16 9 The domain of (f∘g)(x) in interval notation is.
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1152 | https://ibmathsresources.com/2019/11/30/hollow-cubes-and-hypercubes-investigation/ | Hollow Cubes and Hypercubes investigation – IB Maths Resources from Intermathematics
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Hollow Cubes and Hypercubes investigation
Hollow Cubes investigation
Hollow cubes like the picture above [reference] are an extension of the hollow squares investigation done previously. This time we can imagine a 3 dimensional stack of soldiers, and so try to work out which numbers of soldiers can be arranged into hollow cubes.
Therefore what we need to find is what numbers can be formed from a 3-b 3
Python code
We can write some Python3 code to find this out (this can be run here):
for k in range(1,200):
for a in range(0, 100):
for b in range(0,100):
if a3-b3 == k :
print(k,a,b)
This gives the following: (the first number is the number of soldiers and the 2 subsequent numbers are the 2 cubes).
1 1 0
7 2 1
8 2 0
19 3 2
26 3 1
27 3 0
37 4 3
56 4 2
61 5 4
63 4 1
64 4 0
91 6 5
98 5 3
117 5 2
124 5 1
125 5 0
127 7 6
152 6 4
169 8 7
189 6 3
We could perhaps investigate any patterns in these numbers, or explore how we can predict when a hollow cube has more than one solution. I’ll investigate which numbers can be written as both a hollow square and also a hollow cube.
Hollow squares and hollow cubes
list1=[]
for a in range(2, 50):
for b in range(2,50):
if a2-b2 !=0:
if a2-b2 > 0:
list1.append(a2-b2)
list2=[]
for j in list1:
for c in range(2,50):
for d in range(2,50):
if c3-d3 == j:
list2.append(c3-d3)
print(list2)
This returns the following numbers which can all be written as both hollow squares and hollow cubes.
[56, 91, 19, 117, 189, 56, 208, 189, 217, 37, 279, 152, 117, 448, 513, 504, 448, 504, 387, 665, 504, 208, 875, 819, 936, 817, 61, 999, 988, 448, 728, 513, 189, 1216, 936, 784, 335, 469, 1323, 819, 1512, 1352, 1197, 992, 296, 152, 1519, 1512, 1197, 657, 1664, 1323, 1647, 1736, 1701, 1664, 936, 504, 2107, 1387, 1216, 1027, 91, 2015, 279, 2232]
Hollow squares, cubes and hypercubes
Taking this further, can we find any number which can be written as a hollow square, hollow cube and hollow hypercube (4 dimensional cube)? This would require our soldiers to be able to be stretch out into a 4th dimensional space – but let’s see if it’s theoretically possible.
Here’s the extra code to type:
list1=[]
for a in range(2, 200):
for b in range(2,200):
if a2-b2 !=0:
if a2-b2 > 0:
list1.append(a2-b2)
list2=[]
for j in list1:
for c in range(2,200):
for d in range(2,200):
if c3-d3 == j:
list2.append(c3-d3)
print(list2)
for k in list2:
for e in range(2,200):
for f in range(2,200):
if k == e4-f4:
print(k)
Very pleasingly this does indeed find some solutions:
9919: Which can be formed as either 100 2-9 2 or 22 3-9 3 or 10 4-3 4.
14625: Which can be formed as either 121 2-4 2 or 25 3-10 3 or 11 4-2 4.
Given that these took some time to find, I think it’ll require a lot of computer power (or a better designed code) to find any number which is a hollow square, hollow cube, hollow hypercube and hollow 5-dimensional cube, but I would expect that there is a number out there that satisfies all criteria. Maybe you can find it?
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1153 | https://www.tiger-algebra.com/drill/2x_3x=90/ | Copyright Ⓒ 2013-2025
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Solution - Linear equations with one unknown
Other Ways to Solve
Step by Step Solution
Rearrange:
Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :
2x+3x-(90)=0
Step by step solution :
Step 1 :
Pulling out like terms :
1.1 Pull out like factors :
5x - 90 = 5 • (x - 18)
Equation at the end of step 1 :
Step 2 :
Equations which are never true :
2.1 Solve : 5 = 0
This equation has no solution.
A a non-zero constant never equals zero.
Solving a Single Variable Equation :
2.2 Solve : x-18 = 0
Add 18 to both sides of the equation :
x = 18
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1154 | https://chrisj.math.gatech.edu/20s/m1553/materials/070-linear_independence-2_5-web.pdf | Section 2.5 Linear Independence Motivation Sometimes the span of a set of vectors is “smaller” than you expect from the number of vectors.
Span{v, w} v w Span{u, v, w} v w u This means that (at least) one of the vectors is redundant: you’re using “too many” vectors to describe the span.
Notice in each case that one vector in the set is already in the span of the others—so it doesn’t make the span bigger.
Today we will formalize this idea in the concept of linear (in)dependence.
Linear Independence Definition A set of vectors {v1, v2, . . . , vp} in Rn is linearly independent if the vector equation x1v1 + x2v2 + · · · + xpvp = 0 has only the trivial solution x1 = x2 = · · · = xp = 0. The set {v1, v2, . . . , vp} is linearly dependent otherwise.
In other words, {v1, v2, . . . , vp} is linearly dependent if there exist numbers x1, x2, . . . , xp, not all equal to zero, such that x1v1 + x2v2 + · · · + xpvp = 0.
This is called a linear dependence relation or an equation of linear dependence.
Like span, linear (in)dependence is another one of those big vocabulary words that you absolutely need to learn. Much of the rest of the course will be built on these concepts, and you need to know exactly what they mean in order to be able to answer questions on quizzes and exams (and solve real-world problems later on).
Linear Independence Definition A set of vectors {v1, v2, . . . , vp} in Rn is linearly independent if the vector equation x1v1 + x2v2 + · · · + xpvp = 0 has only the trivial solution x1 = x2 = · · · = xp = 0. The set {v1, v2, . . . , vp} is linearly dependent otherwise.
Note that linear (in)dependence is a notion that applies to a collection of vectors, not to a single vector, or to one vector in the presence of some others.
Checking Linear Independence Question: Is 1 1 1 , 1 −1 2 , 3 1 4 linearly independent?
Equivalently, does the (homogeneous) the vector equation x 1 1 1 + y 1 −1 2 + z 3 1 4 = 0 0 0 have a nontrivial solution? How do we solve this kind of vector equation?
1 1 3 1 −1 1 1 2 4 row reduce 1 0 2 0 1 1 0 0 0 So x = −2z and y = −z. So the vectors are linearly dependent, and an equation of linear dependence is (taking z = 1) −2 1 1 1 − 1 −1 2 + 3 1 4 = 0 0 0 .
[interactive] Checking Linear Independence Question: Is 1 1 −2 , 1 −1 2 , 3 1 4 linearly independent?
Equivalently, does the (homogeneous) the vector equation x 1 1 −2 + y 1 −1 2 + z 3 1 4 = 0 0 0 have a nontrivial solution?
1 1 3 1 −1 1 −2 2 4 row reduce 1 0 0 0 1 0 0 0 1 The trivial solution x y z = 0 0 0 is the unique solution. So the vectors are linearly independent.
[interactive] Linear Independence and Matrix Columns In general, {v1, v2, . . . , vp} is linearly independent if and only if the vector equation x1v1 + x2v2 + · · · + xpvp = 0 has only the trivial solution, if and only if the matrix equation Ax = 0 has only the trivial solution, where A is the matrix with columns v1, v2, . . . , vp: A = | | | v1 v2 · · · vp | | | .
This is true if and only if the matrix A has a pivot in each column.
▶The vectors v1, v2, . . . , vp are linearly independent if and only if the matrix with columns v1, v2, . . . , vp has a pivot in each column.
▶Solving the matrix equation Ax = 0 will either verify that the columns v1, v2, . . . , vp of A are linearly independent, or will produce a linear dependence relation.
Important Linear Independence Criterion Suppose that one of the vectors {v1, v2, . . . , vp} is a linear combination of the other ones (that is, it is in the span of the other ones): v3 = 2v1 −1 2v2 + 6v4 Then the vectors are linearly dependent: 2v1 −1 2v2 −v3 + 6v4 = 0.
Conversely, if the vectors are linearly dependent 2v1 −1 2v2 + 6v4 = 0.
then one vector is a linear combination of (in the span of) the other ones: v2 = 4v1 + 12v4.
Theorem A set of vectors {v1, v2, . . . , vp} is linearly dependent if and only if one of the vectors is in the span of the other ones.
Linear Independence Another criterion Theorem A set of vectors {v1, v2, . . . , vp} is linearly dependent if and only if one of the vectors is in the span of the other ones.
Equivalently: Theorem A set of vectors {v1, v2, . . . , vp} is linearly dependent if and only if you can remove one of the vectors without shrinking the span.
Indeed, if v2 = 4v1 + 12v3, then a linear combination of v1, v2, v3 is x1v1 + x2v2 + x3v3 = x1v1 + x2(4v1 + 12v3) + x3v3 = (x1 + 4x2)v1 + (12x2 + x3)v3, which is already in Span{v1, v3}.
Conclusion: v2 was redundant.
Linear Independence Increasing span criterion Theorem A set of vectors {v1, v2, . . . , vp} is linearly dependent if and only if one of the vectors is in the span of the other ones.
Better Theorem A set of vectors {v1, v2, . . . , vp} is linearly dependent if and only if there is some j such that vj is in Span{v1, v2, . . . , vj−1}.
Equivalently, {v1, v2, . . . , vp} is linearly independent if for every j, the vector vj is not in Span{v1, v2, . . . , vj−1}.
This means Span{v1, v2, . . . , vj} is bigger than Span{v1, v2, . . . , vj−1}.
A set of vectors is linearly independent if and only if, every time you add another vector to the set, the span gets bigger.
Translation Linear Independence Increasing span criterion: justification Better Theorem A set of vectors {v1, v2, . . . , vp} is linearly dependent if and only if there is some j such that vj is in Span{v1, v2, . . . , vj−1}.
Why?
Take the largest j such that vj is in the span of the others. Then vj is in the span of v1, v2, . . . , vj−1. Why? If not (j = 3): v3 = 2v1 −1 2v2 + 6v4 Rearrange: v4 = −1 6 2v1 −1 2v2 −v3 so v4 works as well, but v3 was supposed to be the last one that was in the span of the others.
Linear Independence Pictures in R2 [interactive 2D: 2 vectors] [interactive 2D: 3 vectors] Span{v} v One vector {v}: Linearly independent if v ̸= 0.
Linear Independence Pictures in R2 [interactive 2D: 2 vectors] [interactive 2D: 3 vectors] Span{v} Span{w} v w One vector {v}: Linearly independent if v ̸= 0.
Two vectors {v, w}: Linearly independent ▶Neither is in the span of the other.
▶Span got bigger.
Linear Independence Pictures in R2 [interactive 2D: 2 vectors] [interactive 2D: 3 vectors] Span{v} Span{w} Span{v, w} v w u One vector {v}: Linearly independent if v ̸= 0.
Two vectors {v, w}: Linearly independent ▶Neither is in the span of the other.
▶Span got bigger.
Three vectors {v, w, u}: Linearly dependent: ▶u is in Span{v, w}.
▶Span didn’t get bigger after adding u.
▶Can remove u without shrinking the span.
Also v is in Span{u, w} and w is in Span{u, v}.
Linear Independence Pictures in R2 [interactive 2D: 2 vectors] [interactive 2D: 3 vectors] Span{v} v w Two collinear vectors {v, w}: Linearly dependent: ▶w is in Span{v}.
▶Can remove w without shrinking the span.
▶Span didn’t get bigger when we added w.
Observe: Two vectors are linearly dependent if and only if they are collinear.
Linear Independence Pictures in R2 [interactive 2D: 2 vectors] [interactive 2D: 3 vectors] Span{v} v w u Three vectors {v, w, u}: Linearly dependent: ▶w is in Span{u, v}.
▶Can remove w without shrinking the span.
▶Span didn’t get bigger when we added w.
Observe: If a set of vectors is linearly dependent, then so is any larger set of vectors!
Linear Independence Pictures in R3 [interactive 3D: 2 vectors] [interactive 3D: 3 vectors] v w Span{v} Span{w} Two vectors {v, w}: Linearly independent: ▶Neither is in the span of the other.
▶Span got bigger when we added w.
Linear Independence Pictures in R3 [interactive 3D: 2 vectors] [interactive 3D: 3 vectors] v w u Span{v} Span{w} Span{v, w} Three vectors {v, w, u}: Linearly independent: span got bigger when we added u.
Linear Independence Pictures in R3 [interactive 3D: 2 vectors] [interactive 3D: 3 vectors] v w x Span{v} Span{w} Span{v, w} Three vectors {v, w, x}: Linearly dependent: ▶x is in Span{v, w}.
▶Can remove x without shrinking the span.
▶Span didn’t get bigger when we added x.
Poll Are there four vectors u, v, w, x in R3 which are linearly depen-dent, but such that u is not a linear combination of v, w, x? If so, draw a picture; if not, give an argument.
Poll Yes: actually the pictures on the previous slides provide such an example.
Linear dependence of {v1, . . . , vp} means some vi is a linear combination of the others, not any.
Linear Dependence and Free Variables Theorem Let v1, v2, . . . , vp be vectors in Rn, and consider the matrix A = | | | v1 v2 · · · vp | | | .
Then you can delete the columns of A without pivots (the columns corresponding to free variables), without changing Span{v1, v2, . . . , vp}. The pivot columns are linearly independent, so you can’t delete any more columns.
This means that each time you add a pivot column, then the span increases.
Let d be the number of pivot columns in the matrix A above.
▶If d = 1 then Span{v1, v2, . . . , vp} is a line.
▶If d = 2 then Span{v1, v2, . . . , vp} is a plane.
▶If d = 3 then Span{v1, v2, . . . , vp} is a 3-space.
▶Etc.
Upshot Linear Dependence and Free Variables Justification Why? If the matrix is in RREF: A = 1 0 2 0 0 1 3 0 0 0 0 1 then the column without a pivot is in the span of the pivot columns: 2 3 0 = 2 1 0 0 + 3 0 1 0 + 0 0 0 1 and the pivot columns are linearly independent: 0 0 0 = x1 1 0 0 + x2 0 1 0 + x4 0 0 1 = x1 x2 x4 = ⇒x1 = x2 = x4 = 0.
Linear Dependence and Free Variables Justification Why? If the matrix is not in RREF, then row reduce: A = 1 7 23 3 2 4 16 0 −1 −2 −8 4 RREF 1 0 2 0 0 1 3 0 0 0 0 1 The following vector equations have the same solution set: x1 1 2 −1 + x2 7 4 −2 + x3 23 16 −8 + x4 3 0 4 = 0 x1 1 0 0 + x2 0 1 0 + x3 2 3 0 + x4 0 0 1 = 0 We conclude that 23 16 −8 = 2 1 2 −1 + 3 7 4 −2 + 0 3 0 4 and that x1 1 2 −1 + x2 7 4 −2 + x4 3 0 4 = 0 has only the trivial solution.
Linear Independence Two more facts Fact 1: Say v1, v2, . . . , vn are in Rm. If n > m then {v1, v2, . . . , vn} is linearly dependent: the matrix A = | | | v1 v2 · · · vn | | | .
cannot have a pivot in each column (it is too wide).
This says you can’t have 4 linearly independent vectors in R3, for instance.
A wide matrix can’t have linearly independent columns.
Fact 2: If one of v1, v2, . . . , vn is zero, then {v1, v2, . . . , vn} is linearly dependent. For instance, if v1 = 0, then 1 · v1 + 0 · v2 + 0 · v3 + · · · + 0 · vn = 0 is a linear dependence relation.
A set containing the zero vector is linearly dependent.
Summary ▶A set of vectors is linearly independent if removing one of the vectors shrinks the span; otherwise it’s linearly dependent.
▶There are several other criteria for linear (in)dependence which lead to pretty pictures.
▶The columns of a matrix are linearly independent if and only if the RREF of the matrix has a pivot in every column.
▶The pivot columns of a matrix A are linearly independent, and you can delete the non-pivot columns (the “free” columns) without changing the span of the columns.
▶Wide matrices cannot have linearly independent columns.
These are not the official definitions!
Warning |
1155 | https://stats.stackexchange.com/questions/248678/mean-and-variance-after-subtracting-mean-and-dividing-by-standard-deviation | self study - Mean and variance after subtracting mean and dividing by standard deviation - Cross Validated
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Mean and variance after subtracting mean and dividing by standard deviation
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That is mean, variance and standard deviation.
My question is how to get from point 4 to 5 and also with the variance from point 6 to 7.
self-study
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edited Nov 30, 2016 at 8:22
Tim
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asked Nov 29, 2016 at 19:42
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2 Divide top and bottom by n n.Nick Cox –Nick Cox 2016-11-29 19:45:20 +00:00 Commented Nov 29, 2016 at 19:45
That is the last step, I want to know how to get rid of Xi. (one step earlier)Daniel –Daniel 2016-11-29 19:50:43 +00:00 Commented Nov 29, 2016 at 19:50
2 In each case, the left-most equation defines a statistic of z z (i.e. z¯z¯, s 2 z s z 2). Substitute "x x" for "z z" to get definitions for the the corresponding x x statistics. The steps you are asking about seem to reduce to these definitions.GeoMatt22 –GeoMatt22 2016-11-29 20:02:26 +00:00 Commented Nov 29, 2016 at 20:02
2 Please add the [self-study] tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck.gung - Reinstate Monica –gung - Reinstate Monica 2016-11-29 20:07:14 +00:00 Commented Nov 29, 2016 at 20:07
1 Well I understand how to calculate all three of these subjects with specific numbers. Only problem I have encountered are these steps. It is about proving that variance and standard deviation are equal when 1. But i don´t understand the transition where the sum and Xi disappears.Daniel –Daniel 2016-11-29 20:14:48 +00:00 Commented Nov 29, 2016 at 20:14
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These equations represent a particularly obscure way to make some important points that everybody ought to understand. I will therefore provide an indirect answer by highlighting the fundamentals (1-4 below), demonstrating them, and then applying them in what amounts to an equivalent proof.
When you add a constant a a to all data x i x i, the mean of the new values is a a plus the mean of the old values. This should be obvious, because adding a a to each of n n values adds n a n a to the sum. When the sum is divided by n n to get the mean, n a n a is divided by n n to show n a/n=a n a/n=a is added to the sum.
When you multiply each x i x i by a constant b b, the mean of the new values is b b times the original mean. This truly is obvious (it's a direct application of distributive and commutative laws of arithmetic).
When you add a constant a a to all data, the variance is unchanged. This is because the variance is the average of the squared residuals, (x i−x¯)2(x i−x¯)2. By (1), x¯x¯ increases by a a and that exactly cancels the addition of a a to each x i x i, whence the residuals are unchanged. Consequently the variance is unchanged.
When you multiply all data by a constant b b, the variance is multiplied by b 2 b 2. Since (3) tells us each x i x i as well as their mean x¯x¯ are multiplied by b b, the residuals x i−x¯x i−x¯ are also multiplied by b b. Consequently the squared residuals are multiplied by b 2 b 2 and so (exactly as in (2)) the mean squared residual is multiplied by b 2 b 2.
The equations in the question attempt to demonstrate that the mean and variance of z i z i are zero and one, respectively, when the z i z i are formed by standardizing the data: that is, −x¯−x¯ is first added to the data (giving the residuals) and those results are divided by the square root of the variance. Call the square root s s, so the variance is s 2 s 2.
Here, then, is an alternative to the equations in the question:
By (1), the mean after the first step is x¯−x¯=0 x¯−x¯=0.
By (2), the mean remains zero upon division by the square root of the variance. (This should remind you of step "5" in the question.)
By (3), the variance is unchanged after the first step.
By (4), the variance s 2 s 2 is divided by the square of s s in the second step: but that just divides the variance by itself (step "7" in the question), giving s 2/s 2=1 s 2/s 2=1, QED.
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answered Nov 29, 2016 at 20:37
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It made the issue much more clear but I still struggle with the 6 to 7 transition. What is b in step "6" how do I end up with sx^2/sx^2? Especially how do I get the upper part of the fraction from (Xi-X)^2 to sx^2 Daniel –Daniel 2016-11-29 22:01:33 +00:00 Commented Nov 29, 2016 at 22:01
2 Your image might be rendering differently on different machines, for I see no "b b" anywhere in the equations. Step 7, as a previous commenter has noted, merely substitutes the formula for s 2 x=∑i(x i−x¯)2 n s x 2=∑i(x i−x¯)2 n in ∑i(x i−x¯)2 n s 2 x=∑i(x i−x¯)2 n 1 s 2 x=s 2 x s 2 x.∑i(x i−x¯)2 n s x 2=∑i(x i−x¯)2 n 1 s x 2=s x 2 s x 2. whuber –whuber♦ 2016-11-29 22:15:29 +00:00 Commented Nov 29, 2016 at 22:15
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I'm not sure if this is what you wanted, but I'll give it a shot.
First, we declare these known relationships:
x¯=∑n i=1 x i n x¯=∑i=1 n x i n
∑n i=1 x¯=n x¯∑i=1 n x¯=n x¯
s 2 x=∑n i=1(x 1−x¯)2 n s x 2=∑i=1 n(x 1−x¯)2 n
So, starting with (4):
∑n i=1 x i n s x−∑n i=1 x¯n s x=(∑n i=1 x i n⋅1 s x)−n x¯n s x=(x¯⋅1 s x)−n x¯n s x=x¯s x−n x¯n s x∑i=1 n x i n s x−∑i=1 n x¯n s x=(∑i=1 n x i n⋅1 s x)−n x¯n s x=(x¯⋅1 s x)−n x¯n s x=x¯s x−n x¯n s x.
We get (5). QED.
Starting with (6):
∑n i=1(x 1−x¯)2 n s 2 x=(∑n i=1(x 1−x¯)2 n⋅1 s 2 x)=(s 2 x⋅1 s 2 x)=s 2 x s 2 x∑i=1 n(x 1−x¯)2 n s x 2=(∑i=1 n(x 1−x¯)2 n⋅1 s x 2)=(s x 2⋅1 s x 2)=s x 2 s x 2.
We get (7). QED.
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answered Nov 29, 2016 at 22:35
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Welcome to the site, @ElmerVillanueva. Please be cautious about answering [self-study] questions. Our policy is to provide hints to help the OP get unstuck, not to do their homework for them. You can find the full statement of our policies here.gung - Reinstate Monica –gung - Reinstate Monica 2016-11-29 22:42:37 +00:00 Commented Nov 29, 2016 at 22:42
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1156 | https://www.rainbow-nursery.info/wp-content/uploads/2014/12/Common-Opposites.pdf | Common Opposites -‐ Antonyms Vocabulary Word List A absent -‐ present abundant -‐ scarce accept -‐ decline, refuse accurate -‐ inaccurate admit -‐ deny advantage -‐ disadvantage against -‐ for agree -‐ disagree alive -‐ dead all -‐ none, nothing ally -‐ enemy always -‐ never ancient -‐ modern answer -‐ question antonym -‐ synonym apart -‐ together appear -‐ disappear, vanish approve -‐ disapprove arrive -‐ depart artificial -‐ natural ascend -‐ descend attic -‐ cellar attractive -‐ repulsive awake -‐ asleep B backward -‐ forward bad -‐ good beautiful -‐ ugly before -‐ after begin -‐ end below -‐ above bent -‐ straight best -‐ worst better -‐ worse, worst big -‐ little, small black -‐ white blame -‐ praise bless -‐ curse bitter -‐ sweet borrow -‐ lend bottom -‐ top boy -‐ girl brave -‐ cowardly build -‐ destroy bold -‐ meek, timid borrow -‐ lend bound -‐ unbound, free boundless -‐ limited bright -‐ dim, dull brighten -‐ fade broad -‐ narrow C calm -‐ windy, troubled can -‐ cannot, can't capable -‐ incapable captive -‐ free careful -‐ careless cheap -‐ expensive cheerful -‐ sad, discouraged, dreary clear -‐ cloudy, opaque clever -‐ stupid clockwise -‐ counterclockwise close -‐ far, distant closed -‐ ajar, open clumsy -‐ graceful cold -‐ hot combine -‐ separate come -‐ go comfort -‐ discomfort common -‐ rare conceal -‐ reveal contract -‐ expand cool -‐ warm correct -‐ incorrect, wrong courage -‐ cowardice create -‐ destroy crooked -‐ straight cruel -‐ kind compulsory -‐ voluntary courteous -‐ discourteous, rude D dangerous -‐ safe dark -‐ light day -‐ night daytime -‐ nighttime dead -‐ alive decline -‐ accept, increase deep -‐ shallow definite -‐ indefinite demand -‐ supply despair -‐ hope dim -‐ bright disappear -‐ appear diseased -‐ healthy down -‐ up downwards -‐ upwards dreary -‐ cheerful dry -‐ moist, wet dull -‐ bright, shiny decrease -‐ increase discourage -‐ encourage dusk -‐ dawn E early -‐ late east -‐ west easy -‐ hard, difficult empty -‐ full encourage -‐ discourage end -‐ begin, start enter -‐ exit even -‐ odd expand -‐ contract export -‐ import exterior -‐ interior external -‐ internal F fade -‐ brighten fail -‐ succeed false -‐ true famous -‐ unknown far -‐ near fast -‐ slow fat -‐ thin feeble -‐ strong, powerful few -‐ many find -‐ lose first -‐ last float -‐ sink foolish -‐ wise fore -‐ aft free -‐ bound, captive fold -‐ unfold forget -‐ remember found -‐ lost fresh -‐ stale frequent -‐ seldom friend -‐ enemy for -‐ against fortunate -‐ unfortunate full -‐ empty G generous -‐ stingy gentle -‐ rough get -‐ give giant -‐ tiny, small, dwarf girl -‐ boy give -‐ receive, take glad -‐ sad, sorry gloomy -‐ cheerful go -‐ stop good -‐ bad, evil grant -‐ refuse great -‐ tiny, small, unimportant grow -‐ shrink guest -‐ host guilty -‐ innocent H happy -‐ sad hard -‐ easy hard -‐ soft harmful -‐ harmless harsh -‐ mild hate -‐ love haves -‐ have-‐nots healthy -‐ diseased, ill, sick heaven -‐ hell heavy -‐ light help -‐ hinder here -‐ there hero -‐ coward high -‐ low hill -‐ valley hinder -‐ help honest -‐ dishonest horizontal -‐ vertical hot -‐ cold humble -‐ proud I ill -‐ healthy, well immense -‐ tiny, small important -‐ trivial in -‐ out include -‐ exclude increase -‐ decrease inferior -‐ superior inhale -‐ exhale inner -‐ outer inside -‐ outside intelligent -‐ stupid, unintelligent interesting -‐ boring interior -‐ exterior interesting -‐ dull, uninteresting internal -‐ external intentional -‐ accidental J join -‐ separate junior -‐ senior just -‐ unjust justice -‐ injustice K knowledge -‐ ignorance known -‐ unknown L landlord -‐ tenant large -‐ small last -‐ first laugh -‐ cry lawful -‐ unlawful, illegal lazy -‐ industrious leader -‐ follower left -‐ right lend -‐borrow lengthen -‐ shorten lenient -‐ strict left -‐ right less -‐ more light -‐ dark, heavy like -‐ dislike, hate likely -‐ unlikely limited -‐ boundless little -‐ big long -‐ short loose -‐ tight lose -‐ find loss -‐ win loud -‐ quiet love -‐ hate low -‐ high loyal -‐ disloyal M mad -‐ happy, sane major -‐ minor many -‐ few mature -‐ immature maximum -‐ minimum melt -‐ freeze merry -‐ sad messy -‐ neat minor -‐ major minority -‐ majority miser -‐ spendthrift misunderstand -‐ understand more -‐ less N nadir -‐ zenith narrow -‐ wide near -‐ far, distant neat -‐ messy, untidy never -‐ always new -‐ old night -‐ day nighttime -‐ daytime no -‐ yes noisy -‐ quiet none -‐ some north -‐ south O obedient -‐ disobedient odd -‐ even offer -‐ refuse old -‐ young old -‐ new on -‐ off open -‐ closed, shut opposite-‐ same, similar optimist -‐ pessimist out -‐ in outer -‐ inner over -‐ under P past -‐ present patient -‐ impatient peace -‐ war permanent -‐ temporary plentiful -‐ scarce plural -‐ singular poetry -‐ prose polite -‐ rude, impolite possible -‐ impossible poverty -‐ wealth, riches powerful -‐ weak pretty -‐ ugly private -‐ public prudent -‐ imprudent pure -‐ impure, contaminated push -‐ pull Q qualified -‐ unqualified question -‐ answer quiet -‐ loud, noisy R raise -‐ lower rapid -‐ slow rare -‐ common regular -‐ irregular real -‐ fake rich -‐ poor right -‐ left, wrong right-‐side-‐up -‐ upside-‐down rough -‐ smooth rude -‐ courteous S safe -‐ unsafe same -‐ opposite satisfactory -‐ unsatisfactory secure -‐ insecure scatter -‐ collect separate -‐ join, together serious -‐ trivial second-‐hand -‐ new shallow -‐ deep shrink -‐ grow sick -‐ healthy, ill simple -‐ complex, hard singular -‐ plural sink -‐ float slim -‐ fat, thick slow -‐ fast sober -‐ drunk soft -‐ hard some -‐ none sorrow -‐ joy sour -‐ sweet sow -‐reap straight -‐ crooked start -‐ finish stop -‐ go strict -‐ lenient strong -‐ weak success -‐ failure sunny -‐ cloudy synonym -‐ antonym sweet -‐ sour T take -‐ give tall -‐ short tame -‐ wild them -‐ us there -‐ here thick -‐ thin tight -‐ loose, slack tiny -‐ big, huge together -‐ apart top -‐ bottom tough -‐ easy, tender transparent -‐ opaque true -‐ false truth -‐ flasehood, lie, untruth U under -‐ over unfold -‐ fold unknown -‐ known unqualified -‐ qualified unsafe -‐ safe up -‐ down upside-‐down -‐ right-‐side-‐up upstairs -‐ downstairs us -‐ them useful -‐ useless V vacant -‐ occupied vanish -‐ appear vast -‐ tiny victory -‐ defeat virtue -‐ vice visible -‐ invisible voluntary -‐ compulsory W war -‐ peace wax -‐ wane weak -‐ strong white -‐ black wide -‐ narrow win -‐ lose wisdom -‐ folly, stupidity within -‐ outside wrong -‐ right wet -‐ dry Y yes -‐ no yin -‐ yang young -‐ old Z zip -‐ unzip zenith -‐ nadir |
1157 | https://www.youtube.com/watch?v=eL6pUU09UbU | Quadratic Functions: Introduction (What do the Coefficients tell us?)
FerranteMath
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Posted: 20 Mar 2017
This video defines the meaning of coefficients in a quadratic function. The quadratic coefficient (a) describes which direction the parabola opens; the linear coefficient (b) in combination with the quadratic coefficient can be used to determine the equation for the axis of symmetry; and the constant (c) describes the y-intercept of the graphed parabola.
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Transcript:
foreign we're going to go over quadratic functions introduction part three but we're going to over I'm going to go over something called the coefficients and talk about what exactly the coefficients themselves mean so remember I told you that the form the standard form for a quadratic function is f of x again which means the same thing as Y is equal to a times x squared we said that was the quadratic function the quadratic term remember plus BX and we call this the linear term quadratic term linear term plus the C the a the B and the C technically not C but we're going to include it in coefficients anyway the ADB and the C are what we call the coefficients for a quadratic function now what is a coefficient let me just write this in here really quickly code efficients the A to B and the C actually tell us something about the quadratic or the parabola self okay and that's what I want to go over in this particular video now the a the number that's in front of the variable X right actually coefficients are numbers that are in front of variables to begin with okay so that's one definition of coefficients but in this particular case the a for example in ax squared plus BX plus C will tell you the direction that the parabola opens now what does that mean the direction the parabola opens so for example we know that the parabola opens up like this one right or a parabola could open downward it can also open sideways but again we're not going to talk about that in in this particular class now if the a value is positive the parabola opens upward okay and which would make sense by saying then that if the parabola is or the coefficient is a negative coefficient the parabola will open downward okay so right away we'll be able to tell something about the parabola just by looking at the equation itself let me give you an example if I have the quadratic function f of x is equal to 2X squared plus 4X Plus 1. I know immediately that this Parabola is going to open upward in that direction because the a here is a positive 2. okay so again positive number opens upward so that's what the a can tell me all right now more information the a and the B together can tell me what's called the axis of symmetry axis of symmetry now what is the axis of symmetry remember one of the things that we talked about in terms of a parabola is that you can actually cut it exactly in half right and that one side of the parabola is a mirror image you just took it and you just pivoted on that axis flip it like that you'll get that side okay so this flips over to become this this area where you actually do the pivoting is called the axis of symmetry all right so that's an important term axis of symmetry and the axis of symmetry is some type of equation it's usually an x value like X is equal to 10 or X is equal to 5. and how do you find the axis of symmetry it's basically this particular formula right here negative B over 2A all right so X is equal to negative B over 2A and that will give you the axis of symmetry let's go back up to this example again so what would the axis of symmetry be on this parabola well let's go ahead and put in the numbers negative B in this case is positive four over two times again notice I'm always putting whatever value I'm replacing I'm putting it in parentheses please please do that that's always a helpful thing 2 times 2. that's going to give me a negative 4. over 4 or negative 1. now what I know about this particular equation or this Parabola is that the axis of symmetry is at negative one again let's choose this as our example here's my x y axis [Music] and now there's negative one negative two negative three positive one positive two positive three so according to this now I know that my Parabola opens upward and its axis of symmetry what cuts it right in half is at negative one now this is a little misleading I had to draw it somewhere the last thing I what I don't know about this is I don't know where this particular vertex is right [Music] what is a vertex you asked well the vertex is the top or the bottom the top or the bottom of the parabola all right the top is known as the maximum and the bottom is known as the minimum whoops there we go so in this particular equation I don't know whether it's doing this or this or this or even down here right that's what I don't know yet is there any way to figure that out there is I'm going to say that for another video but I will give you one more hint and that's the final coefficient which is the c what does the C tell you the C will tell you where the y-intercept of your Parabola is located okay now in this particular case we know that the y-intercept is one so now what I can do is go back to my parabola positive one positive 2 positive 3 positive 4. I know that this Parabola is going to cross the axis of symmetries at negative one and the y-intercept will be at this point here so I should probably make that a little wider all right but again it could be like a really skinny thing it just kind of gives me a little bit of an idea though what the parabola is looking like okay I hope that was helpful that was a little bit long-winded but just remember the a the B and the C can give you some important information about what your Parabola looks like hope that was helpful |
1158 | https://www.ck12.org/flexi/chemistry/enthalpy/what-is-the-relationship-between-h-and-u/ | Flexi answers - What is the relationship between ∆H and ∆U? | CK-12 Foundation
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What is the relationship between ∆H and ∆U?
Flexi Says:
The relationship between ∆H (change in enthalpy) and ∆U (change in internal energy) is given by the equation:
Δ H=Δ U+P Δ V
Where:
∆H is the change in enthalpy
∆U is the change in internal energy
P is the pressure of the system
∆V is the change in volume
This equation is derived from the first law of thermodynamics and is applicable for processes occurring at constant pressure.
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Is it true that Delta H accounts for work against the atmospheric pressure in an open system, Delta H equals qv, Delta E and Delta H are equal in value when there is no change in gas volume, and Delta E is the value calculated in a bomb calorimeter?Calculate ΔH for the following reaction: N2 + 3H2 → 2NH3. Average Bond Energies N≡N, 941 kJ/mol H–H, 432 kJ/mol N–H, 391 kJ/mol Now calculate ΔH for the reaction. Reactant Bonds Broken = 2237 kJ Product Bonds Formed = 2346 kJWhat is the change in enthalpy?
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1159 | https://www.khanacademy.org/math/cc-fifth-grade-math/divide-fractions/imp-dividing-fractions-and-whole-numbers-word-problems/e/division-with-fractions-and-whole-numbers | Use of cookies
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1160 | https://cims.nyu.edu/~tjl8195/CMUMC_POTD_Book_JustTheSolutions.pdf | incredibly sus draft lmfao sup how was your day incredibly sus draft lmfao sup how was your day The Carnegie Mellon University Math Club Problem of the Day Curated by Thomas Lam Courant Institute of Mathematical Sciences incredibly sus draft lmfao sup how was your day Copyright© Thomas Lam, 2025. Reproduction with proper credit is fine and encouraged.
All rights reserved.
Published by me.
First published August 2025.
Printed nowhere because this is intended to be an ebook but you can print it if you really want to.
Written with L AT EX and way too much Asymptote.
incredibly sus draft lmfao sup how was your day 4 This is a version of the book that removes the problems and hints, leaving only the solutions.
Index Solution to Problem 1 Solution to Problem 2 Solution to Problem 3 Solution to Problem 4 Solution to Problem 5 Solution to Problem 6 Solution to Problem 7 Solution to Problem 8 Solution to Problem 9 Solution to Problem 10 Solution to Problem 11 Solution to Problem 12 Solution to Problem 13 Solution to Problem 14 Solution to Problem 15 Solution to Problem 16 Solution to Problem 17 Solution to Problem 18 Solution to Problem 19 Solution to Problem 20 incredibly sus draft lmfao sup how was your day 5 Solution to Problem 21 Solution to Problem 22 Solution to Problem 23 Solution to Problem 24 Solution to Problem 25 Solution to Problem 26 Solution to Problem 27 Solution to Problem 28 Solution to Problem 29 Solution to Problem 30 Solution to Problem 31 Solution to Problem 32 Solution to Problem 33 Solution to Problem 34 Solution to Problem 35 Solution to Problem 36 Solution to Problem 37 Solution to Problem 38 Solution to Problem 39 Solution to Problem 40 Solution to Problem 41 Solution to Problem 42 Solution to Problem 43 Solution to Problem 44 incredibly sus draft lmfao sup how was your day 6 Solution to Problem 45 Solution to Problem 46 Solution to Problem 47 Solution to Problem 48 Solution to Problem 49 Solution to Problem 50 Solution to Problem 51 Solution to Problem 52 Solution to Problem 53 Solution to Problem 54 Solution to Problem 55 Solution to Problem 56 Solution to Problem 57 Solution to Problem 58 Solution to Problem 59 Solution to Problem 60 Solution to Problem 61 Solution to Problem 62 Solution to Problem 63 Solution to Problem 64 Solution to Problem 65 Solution to Problem 66 Solution to Problem 67 Solution to Problem 68 incredibly sus draft lmfao sup how was your day 7 Solution to Problem 69 Solution to Problem 70 Solution to Problem 71 Solution to Problem 72 Solution to Problem 73 Solution to Problem 74 Solution to Problem 75 Solution to Problem 76 Solution to Problem 77 Solution to Problem 78 Solution to Problem 79 Solution to Problem 80 Solution to Problem 81 Solution to Problem 82 Solution to Problem 83 Solution to Problem 84 Solution to Problem 85 Solution to Problem 86 Solution to Problem 87 Solution to Problem 88 Solution to Problem 89 Solution to Problem 90 Solution to Problem 91 Solution to Problem 92 incredibly sus draft lmfao sup how was your day 8 Solution to Problem 93 Solution to Problem 94 Solution to Problem 95 Solution to Problem 96 Solution to Problem 97 Solution to Problem 98 Solution to Problem 99 Solution to Problem 100 Solution to Problem 101 Solution to Problem 102 Solution to Problem 103 Solution to Problem 104 Solution to Problem 105 Solution to Problem 106 Solution to Problem 107 Solution to Problem 108 Solution to Problem 109 Solution to Problem 110 Solution to Problem 111 Solution to Problem 112 Solution to Problem 113 Solution to Problem 114 Solution to Problem 115 Solution to Problem 116 incredibly sus draft lmfao sup how was your day 9 Solution to Problem 117 Solution to Problem 118 Solution to Problem 119 Solution to Problem 120 Solution to Problem 121 Solution to Problem 122 Solution to Problem 123 Solution to Problem 124 Solution to Problem 125 Solution to Problem 126 Solution to Problem 127 Solution to Problem 128 Solution to Problem 129 Solution to Problem 130 Solution to Problem 131 Solution to Problem 132 Solution to Problem 133 Solution to Problem 134 Solution to Problem 135 Solution to Problem 136 Solution to Problem 137 Solution to Problem 138 Solution to Problem 139 Solution to Problem 140 incredibly sus draft lmfao sup how was your day 10 Solution to Problem 141 Solution to Problem 142 Solution to Problem 143 Solution to Problem 144 Solution to Problem 145 Solution to Problem 146 Solution to Problem 147 Solution to Problem 148 Solution to Problem 149 Solution to Problem 150 Solution to Problem 151 Solution to Problem 152 Solution to Problem 153 Solution to Problem 154 Solution to Problem 155 Solution to Problem 156 Solution to Problem 157 Solution to Problem 158 Solution to the Grand Finale incredibly sus draft lmfao sup how was your day CHAPTER 1 Solutions incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 1 12 Solution 1 The cups started with the same volume. Since the cups end with the same volume, the net amount of milk transferred to the tea must equal the net amount of tea transferred to the milk. That is, the contamination is equal.
■ Remark: We did not need to use the fact that the cups were stirred. Indeed, the problem still holds true if we do not stir, by the same logic.
Source: This is a classic. I first heard this in an old book of puzzles more than a decade ago.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 2 13 Solution 2 We claim that the answer is 2 3.
For each triangle we place, we associate it with some empty space near it. If we can show that the area of this empty space is at least half the area of the triangle, then we have the at-least-2/3 ratio.
P Suppose we place down a triangle such as the gray one in the above image. We claim that at least 1 2 of the red area is blank space. This would solve the problem because this red area cannot be associated to any other triangle that we place in this way. That is, we’re guaranteed that we’re not ”double-counting any blank space”.
First, observe that at most one triangle can intersect the red region. This is because any triangle that does so must contain the point P in its interior. Call the triangle that intersects the red region (if any) T.
Next, we observe that to maximize that fraction of the red region’s area covered by T, it must be the case that T’s bottom-left vertex lies on the bottom-left edge of the region. This is not difficult to argue: If this were not the case, then the area covered will increase when T is “pushed” down and/or to the left.
Thus the case of maximal area coverage must look like the diagram below.
T incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 2 14 We see that there are two uncovered “red triangles”. It remains to show that the sum of their areas is at most half the area of the original red region. Indeed, if their side lengths are x and y, then the sum x + y is fixed. Their areas are proportional to x2 + y2, so we must minimize the quantity x2 + y2. By the QM-AM inequality, the minimum exists and occurs when x = y. It is not hard to see that when this holds, exactly half the red region’s area is covered.
The ratio 2/3 is achieved below.
■ [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 3 15 Solution 3 We claim the answer is 4.
Construction Draw a tetrahedron around the sun. For each side of the tetrahedron, place a large planet that contains that entire side but does not intersect the sun.
Since the boundary of the tetrahedron is fully contained in the union of the planets, no ray of light will escape the sun.
The issue is that the planets may intersect.
To fix this, take a planet and apply a homothety or dilation on it, centered at the sun. This enlarges the planet, but in return we can sent it as far away from the other planets as we want. By applying homotheties to each planet, we can ensure that no two planets intersect. Moreover, any ray of light blocked by a planet will still be blocked upon homothety. This finishes the construction.
Minimality We want to show that 3 is impossible.
Associate each possible direction for a ray of light with a point on the surface of a sphere S centered at the sun. Observe that if we have a planet centered at some point P, and P ′ is the intersection of S with the light ray that goes though P, then the set of all light rays blocked by this planet may be viewed as a strict subset of the open hemisphere on S centered at P ′.
Thus, it is sufficient to show that 3 open hemispheres cannot cover the surface of S.
Suppose we could, and call the hemispheres H1, H2, and H3. The boundary of H1 is a great circle C1 that is not covered. Likewise, the boundary of H2 is another great circle C2 that must intersect C1 at two diametrically opposite points A and B. However, the third hemisphere H3 cannot cover both of the uncovered points A and B, contradiction.
■ Remarks: The problem is slightly more challenging if the sun were instead a sphere whose surface emitted rays of light. (Note that such rays need not be collinear with the sun’s center.) I leave this as an exercise.
In general, n + 1 is the least number of n-dimensional planets required to shield a source of light in n dimensions.
A natural question follow-up is as follows: Among all configurations of four planets that incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 3 16 satisfy the problem, what is the minimum possible value of Largest Planet Radius Smallest Planet Radius (or rather, the infimum of this quantity)? The following explicit construction gives an upper bound of (5 + √ 24)3 ≈970: • A planet of radius q 8 3 centered at (5 + √ 24) · (1, 1, 1) • A planet of radius q 8 3 · (5 + √ 24) centered at (5 + √ 24) · (−1, −1, 1) • A planet of radius q 8 3 · (5 + √ 24)2 centered at (5 + √ 24)2 · (1, −1, −1) • A planet of radius q 8 3 · (5 + √ 24)3 centered at (5 + √ 24)3 · (−1, 1, −1) You can find a picture of this construction on the front cover of this book.
Source: I have no idea, I first heard it at AMSP [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 4 17 Solution 4 Let us apply the principle given in the hint. Label some points as shown.
A B C D E First we slide the vertex at C towards D, which is a direction that is parallel to AB. This does not change the triangle’s area by the principle.
A B C D E Next, we apply the principle again by sliding the vertex at B to E. This direction is parallel to AD.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 4 18 A B C D E We conclude that the area of the original triangle, △ABC, is exactly the area of △AED, which is clearly 2 .
■ Source: Catriona Agg [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 5 19 Solution 5 Remark: This may or may not be a true story. Whether Kaz took my quarters or not is omitted and left as an exercise.
Source: I saw this in a puzzle book a long time ago.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 6 20 Solution 6 Alice can pass whereas Bob is doomed to fail.
Proof that Alice can pass First she constructs the reflection of A over B by drawing the following four circles.
A B A′ Calling the reflection A′, she then draws the circle centered at A′ that passes through A, and marks its intersections with the circle centered at A.
A B A′ We claim that these intersections both lie on the perpendicular bisector of M, where M is the midpoint of AB!
This can be shown easily with coordinate geometry, but there is also a clean approach.
Without loss of generality let us assume that AB = 1. Call one of the intersections P, and construct N on AA′ for which △PAN ∼△A′NA.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 6 21 A A′ P N B We know that AP A′P = 1 2, so by the similarity, AN AP = 1 2. But AP = 1, so AN = 1 2, which implies that N is the midpoint of segment AB, which has length 1.
With the claim proven, Alice may finish constructing M by drawing a circle at each intersection that passes through A.
A B A′ M The claim implies that the intersection of these two circles must be M.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 6 22 Proof that Bob will fail To show that Bob will fail, we will use the idea behind projective transformations. If you are familiar with projective geometry, then the proof is summarized in one line: “There exists a projective transformation that fixes A and B but moves the constructed midpoint M, and projective transformations send lines to lines”. Otherwise, fear not. There is a perfectly elementary way to visualize the reasoning.
Let us assume for contradiction that Bob can construct the midpoint using only a straight-edge. Then Bob can find a finite algorithm for constructing the midpoint that consists of a combination of the following two steps: • Pick an arbitrary point in the plane.
• Take two marked points or intersections and draw the line that connects them.
Bob wins if he can mark the midpoint as the point of intersection of two of the lines he draws.
But crucially, we moreover must have that this marked point will remain the midpoint even if we perturb the arbitrary points that Bob picked. Otherwise, there would be no guarantee that Bob’s scheme will construct the midpoint.
Of course, Bob must pick an arbitrary point in the plane at some point in his construction, else there is only one line he could ever draw: The one between A and B.
This is the vulnerability we will exploit. To be precise, we will show that we can move the arbitrary points that Bob chooses such that the supposed midpoint also moves.
Let us say, for instance, that Bob’s construction is as shown below, with the blue triangles representing points that were selected arbitrarily.
To help illustrate the point, I have also drawn this diagram on a real-life table.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 6 23 The intuitive idea is that an observer that perceives this construction from a different angle (in 3D space) will still see a construction that uses straight lines, but the relative lengths of the perceived segments may not be preserved. Alternatively, one can think of this routine as taking a picture of the construction from an angled camera. If the angle is chosen ap-propriately, then Bob’s purported “midpoint” could fail to be the midpoint in the picture captured by the angled camera. This is demonstrated below.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 6 24 Formally speaking, we are executing a central projection in which we select a point F in space (the “eye”, where we locate ourselves to view the construction), select an angled plane P (the “lens”, where the angled version of the construction is drawn), and then map every point X in the plane of the construction to the intersection of line FX with plane P.
When all points in lines in Bob’s construction are transformed to the new plane P in this way, we can obtain a picture in which the purported midpoint is not the midpoint of AB.
More precisely, the images of the arbitrarily-selected points have moved to a set of new loca-tions for which Bob’s construction will construct a point that is not the midpoint, showing that his construction could have failed.
■ Source: Famous [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 7 25 Solution 7 The trick is to write 9N = 10N −N. If N = d1d2 . . . dn with d1 < d2 < . . . < dn, then 9N can be though of as the result of the following stacked subtraction.
d1 d2 d3 . . .
dn−1 dn 0 − d1 d2 . . .
dn−2 dn−1 dn ?
?
?
. . .
?
?
?
The first n −1 digits of the answer are clear: They are d1, d2 −d1, d3 −d2, · · · , dn−1 −dn−2, which is valid because the digits are increasing. As for the last two columns, they are instead (dn −1) −dn−1 and 10 −dn, which is valid because the digits are strictly increasing.
From here, the digit sum is evidently d1 + (d2 −d1) + . . . + (dn−1 −dn−2) + (dn−1 −1 −dn−1) + (10 −dn) = −1 + 10 = 9 .
Source: Kvant Magazine?
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 8 26 Solution 8 We present four different solutions. If you must read only one solution, I highly recom-mend the fourth one.
First Solution Let y = √5 −x = 5 −x2. Then we have: x2 + y = 5 y2 + x = 5 Subtracting, we get that (x + y −1)(x −y) = 0, so either x = y or x + y = 1. Thus, either x2 + x −5 = 0 or x2 −x −4 = 0. This gives the four possibilities: x = −1 ± √ 21 2 , 1 ± √ 17 2 Two of these solutions happen to be extraneous, and so the solutions are x = −1 + √ 21 2 , 1 − √ 17 2 .
■ Second Solution First, after squaring both sides, we obtain 5 −x = x4 −10x2 + 25 which we can rearrange as P(x) := x4 −10x2 + x + 20 = 0.
We would like to factor the polynomial P(x0).
To find a suitable factor, observe that f(x) := √5 −x and g(x) := 5 −x2 are function inverses, and so the graph of g(x) is obtained by reflecting the graph for f(x) over the line y = x. In particular, the graph of f(x) intersects the line y = x exactly where g(x) does (and it is simple to argue that such intersections must exist by the Intermediate Value Theorem), so one class of solutions to f(x) = g(x) are those values of x for which f(x) = x. That is, x = √5 −x. Hence we should expect that x2 + x −5 is a factor of the polynomial P(x). Indeed it is, and we will find that P(x) = (x2 + x −5)(x2 −x −4).
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 8 27 Setting each factor to 0, we can then proceed as in the first solution.
■ Third Solution Let g(x) := 5 −x2. We may write the given equation as x = 5 −(5 −x2)2, or x = g(g(x)). Observe that if x satisfies x = g(x), then g(g(x)) = g(x) = x, so one class of solutions to the original equation is given by those x satisfying x = g(x) = 5 −x2. So we expect x2 + x −5 to be a factor of the polynomial P(x) as defined in the second solution.
We proceed as in the second solution.
■ Fourth Solution Eyeing the equation obtained after squaring both sides, 5 −x = x4 −10x2 + 25, we make the insane leap that this is a quadratic. That is, a quadratic in 5.
52 −(1 + 2x2)5 + (x4 + x) = 0 We now may apply the quadratic formula to solve for 5. This gives 5 = (1 + 2x2) ± p (1 + 2x2)2 −4(x4 + x) 2 = (1 + 2x2) ± p (1 + 4x2 + 4x4 −4x4 −4x 2 = (1 + 2x2) ± p (1 −4x + 4x2 2 = (1 + 2x2) ± (1 −2x) 2 Casing on the sign of (1 −2x), we get two possible quadratics which admit four possible solutions in total. We get the desired answer after testing them all.
■ Source: Probably Titu Andreescu.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 9 28 Solution 9 We first recall that ex = ∞ X k=0 xk k!
for all x. Taking x = −1 gives 1 e = ∞ X k=0 (−1)k k!
, and so n!
e = ∞ X k=0 (−1)k n!
k!.
The quantity A := Pn k=0(−1)k n!
k! is clearly an integer. Note that when 0 ≤k ≤n −2, we have that n!
k! is necessarily even, so the parity of A is determined by the last two terms in the sum, which add to n −1 (up to a sign). Hence, if n is even then A is odd, and when n is odd then A is even.
Let f := P∞ k=n+1(−1)k n!
k! be the rest of the sum. It is not hard to show that |f| < 1 (by, say, writing a geometric series as an upper bound). Now case on the parity of n.
• If n is even, then f < 0 because the first and largest term, (−1)n+1 1 n+1, is negative, and the remaining terms decrease too rapidly to change the sign of the partial sum.
(Rigorously, you can bound the rest of the sum via a geometric series.) It follows that n!
e = ⌊A + f⌋= A −1, which is even because A is odd.
• If n is odd, then f > 0 by a similar argument, and so n!
e = ⌊A + f⌋= A, which is even because A is even.
So n!
e is always even.
■ If you wish to see the details on some of the claimed bounds, this is for you. To see that |f| < 1, write |f| ≤ ∞ X k=n+1 n!
k!.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 9 29 When k ≥n + 1, we have that n!
k! = 1 (n + 1)(n + 2) · · · (k) ≤ 1 (n + 1)k−n, so |f| ≤ ∞ X k=n+1 1 (n + 1)k−n = 1 n+1 1 − 1 n+1 = 1 n < 1.
The other bound I used without proof was that ∞ X k=n+2 (−1)k n!
k! < 1 n + 1.
For this, we simply use the result we just proved — that P∞ k=n+1 n!
k! < 1 n — but replace n with n + 1. This gives P∞ k=n+2 (n+1)!
k!
< 1 n+1. Thus ∞ X k=n+2 (−1)k n!
k! ≤ ∞ X k=n+2 n!
k! = 1 n + 1 ∞ X k=n+2 (n + 1)!
k!
< 1 (n + 1)2 < 1 n + 1.
This shows the bound and that it is, in fact, quite loose.
Source: I saw this on Math Stack Exchange.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 10 30 Solution 10 Become God and delete the river by slamming the two landmasses together.
A B A B This decreases the length of the shortest path from A to B by 1 (why?). Since the shortest distance is now clearly 5, the original minimum distance was 6 .
To determine where to place the bridge, we draw the straight-line path from A to B in the transformed problem.
A B Then, for this path to correspond naturally to a valid path in the original problem with the river, we must place the vertical bridge where this path intersects the blue segment (where the river used to be).
A B To be precise, the bridge must be placed 4 3 miles east of Town A.
■ [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 11 31 Solution 11 The line from (0, 1 −t) to (t, 0) is given by y = t−1 t x + 1 −t = x + 1 − t + x t . For some fixed x ∈[0, 1], we are interested in finding the line (i.e. the value of t) that obtains the maximum possible y value at that x. By AM-GM x + 1 − t + x t ≤x + 1 −2 r t · x t = x + 1 −2√x = (1 −√x)2 with equality obtained at t = √x. So the equation of the curve is given by y = (1 −√x)2.
This magically rearranges to √x + √y = 1 .
■ Source: Back in middle school, I had graph paper and I was really bored.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 12 32 Solution 12 In this solution, the curly brackets {·} denote multisets instead of sets.
As per the hint, we must find a positive integer S such that there is exactly one ordered pair of positive integers (P, n) for which there exist two distinct ways to partition S into a sum of n positive integers whose product is P.
We claim that the only possible value for S is S = 12.
Claim 1: S = 12 is a valid solution to the problem.
Various mathematical acquaintances and I have tried various approaches for demonstrat-ing this without an extreme amount of casework. Alas, although there were a few successes, they were just not as elegant or short as getting our hands dirty and listing out (essentially) every possible partition of 12. So let’s just do it.
We can skip partitions of 12 into a sum of 1 or 2 integers. That is, it is impossible for n = 1 or n = 2. For n = 1, there is obviously only one way for an integer to equal 12, so we need not consider n = 1. As for n = 2, you will have a difficult time finding two distinct solutions to the system a + b = S and ab = P.
Thus we may start checking partitions of 12 from n = 3.
The chart begins on the next page.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 12 33 n Partition of S = 12 into n terms Product (P) 3 1,1,10 10 3 1,2,9 18 3 1,3,8 24 3 1,4,7 28 3 1,5,6 30 3 2,2,8 32 3 2,3,7 42 3 2,4,6 48 3 2,5,5 50 3 3,3,6 54 3 3,4,5 60 3 4,4,4 64 4 1,1,1,9 9 4 1,1,2,8 16 4 1,1,3,7 21 4 1,1,4,6 24 4 1,1,5,5 25 4 1,2,2,7 28 4 1,2,3,6 36 4 1,2,4,5 40 4 1,3,3,5 45 4 1,3,4,4 48 4 2,2,2,6 48 4 2,2,3,5 60 4 2,2,4,4 64 4 2,3,3,4 72 4 3,3,3,3 81 (Continued on next page) incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 12 34 n Partition of S = 12 into n terms Product (P) 5 1,1,1,1,8 8 5 1,1,1,2,7 14 5 1,1,1,3,6 18 5 1,1,1,4,5 20 5 1,1,2,2,6 24 5 1,1,2,3,5 30 5 1,1,2,4,4 32 5 1,2,2,2,5 40 5 1,2,2,3,4 48 5 2,2,2,2,4 64 5 2,2,2,3,3 72 6 1,1,1,1,1,7 7 6 1,1,1,1,2,6 12 6 1,1,1,1,3,5 15 6 1,1,1,1,4,4 16 6 1,1,1,2,2,5 20 6 1,1,1,2,3,4 24 6 1,1,2,2,2,4 32 6 1,1,2,2,3,3 36 6 1,2,2,2,2,3 48 6 2,2,2,2,2,2 64 7 1,1,1,1,1,1,6 6 7 1,1,1,1,1,2,5 10 7 1,1,1,1,1,3,4 12 7 1,1,1,1,2,2,4 16 7 1,1,1,1,2,3,3 18 7 1,1,1,2,2,2,3 24 7 1,1,2,2,2,2,2 32 8 1,1,1,1,1,1,1,5 5 8 1,1,1,1,1,1,2,4 8 8 1,1,1,1,1,1,3,3 9 8 1,1,1,1,1,2,2,3 12 8 1,1,1,1,2,2,2,2 16 9 1,1,1,1,1,1,1,1,4 4 9 1,1,1,1,1,1,1,2,3 6 9 1,1,1,1,1,1,2,2,2 8 10 1,1,1,1,1,1,1,1,1,3 3 10 1,1,1,1,1,1,1,1,2,2 4 11 1,1,1,1,1,1,1,1,1,1,2 2 12 1,1,1,1,1,1,1,1,1,1,1,1 1 We see that there is exactly one pair of rows that have the same value of n and product P.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 12 35 Thus S = 12 is a valid solution.
It is important to note the details of this solution: The value of n is 4 and the two possible multisets of ages are {1, 3, 4, 4} and {2, 2, 2, 6}.
Claim 2: No value of S less than 12 can be a valid solution.
Suppose some S < 12 is a valid solution.
Then there exists n, P, and two distinct multisets of ages {x1, x2, · · · , xn} and {y1, y2, · · · , yn} that both have sum S and product P.
(The two multisets need to be unique, but we won’t need this.) But now the two multisets {12 −S, x1, x2, · · · , xn} and {12 −S, y1, y2, · · · , yn} contain positive integers, have the same number of elements, have the same product, and each sum to 12. This pair of multisets is a different one from the one we found before — {1, 3, 4, 4} and {2, 2, 2, 6} — because they share a common element (12 −S), whereas the pair of multisets we found before does not. This contradicts the validity of S = 12 as a solution.
Claim 3: No value of S greater than 12 can be a valid solution.
To rule out S = 13, note that there are at least two pairs of multisets with the same number of elements, same product, and sum 13: • {1, 1, 3, 4, 4} and {1, 2, 2, 2, 6} • {2, 2, 9} and {1, 6, 6} To rule out any S ≥14, we have the following general construction of two such pairs: • {1, 3, 4, 4, 15 −S} and {2, 2, 2, 6, 15 −S} • {2, 2, 9, 15 −S} and {1, 6, 6, 15 −S} In conclusion, there is exactly one value of S that works: S = 12. That is: • Beth’s favorite number is 12.
• Alice’s favorite number is the product, 48 , and is the number that must fill the blank.
• The two possible multisets of stuffed animal turtle ages are {1, 3, 4, 4} and {2, 2, 2, 6}.
■ incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 12 36 Remarks: As is apparent, the proof is not pretty, and I do not know of a more elegant proof.
But what would a solution look like? That is, how would one go about finding the answer of S = 12?
For sure, the easiest way to do this is to tell the computer to do it. But if we are stuck on an island, then an intuitive line of reasoning could begin by observing that there are two forces at play. The first is one that we have identified in the proof above: If S is too large, then there are “too many partitions of S”, and we can expect that there will be multiple pairs of partitions with the same cardinality and product. In contrast, if S is too small, then there will be no such pairs.
To more carefully formalize this, we may begin by classifying the possible values of S into three categories.
• Let us say that S = k has multiplicity m if there are exactly m distinct pairs of partitions of k with the same number of elements and product.
• S = k is a non-solution if it has multiplicity 0.
• S = k is a strong solution if it has multiplicity 1.
• S = k is a weak solution if it has multiplicity at least 1.
Our goal is to seek a value of S that is a strong solution. We can show that small values of S are non-solutions and large values of S are weak solutions, so that a strong solution must be a carefully-selected in-between value which strikes some sort of balance.
Indeed, we can observe that the sequence of multiplicities for the values S = 1, 2, 3, · · · must be increasing. This is because if S = k has multiplicity m, then we may append the element 1 to each of the 2m partitions of k to obtain a collection of m pairs of partitions of k +1 satisfying the required conditions. Hence the multiplicity of S = k +1 must be at least m, the multiplicity of S = k.
It follows that if there exists a strong solution, then the smallest weak solution must be a strong solution. Thus, for simplicity, one can focus on finding the smallest weak solu-tion, i.e. the smallest value of S which admits two distinct partitions {x1, x2, · · · , xn} and {y1, y2, · · · , yn} with the same product. The advantage of this assumption is that for S to be minimal, it cannot be the case that the two partitions have an element in common. Oth-erwise, we could simply remove said common element from both partitions. The products of the new multisets will still be equal, as will the sum, which will have decreased.
From this assumption, we can eliminate quite a few possible values for the product P when S is minimal. For example, we now know that at least one of the partitions of S in question cannot have a 1, and this fact can be used to show that P cannot be a product of three primes. (Sketch: If it were, then, letting the primes be p, q, r, one of the two partitions incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 12 37 must be {p, q, r}, since this is the only one that does not use 1’s. The other partition is either {1, 1, pqr} or {1, p, qr} up to reoredering, so we either have p + q + r = 1 + 1 + pqr or p+q+r = 1+p+qr, neither of which are possible because you can show that p+q+r < 2+pqr and p + q + r < 1 + p + qr.) This fact is extremely helpful, as now some of the first few possible values for P are 16, 24, 32, 36, 48, 54, 56, 60, · · · . The “correct” value, P = 48, is not too far down this list.
So, while this methodology is far from concrete, a hopeful solver could reasonably come up with P = 48 provided that there is an efficient way to eliminate values of P that do not correspond to weak solutions for S.
One way to approach this is by finding a simple upper bound for S. It is plausible to stumble upon an upper bound of 13 by observing that S = 13 is a weak solution, witnessed by the pair of partitions {1, 6, 6} and {2, 2, 9}. Though, finding this requires a knack for intelligent guesswork that I do not possess. Is there another approach? This, I leave to the reader. I hope the ideas presented here were interesting nonetheless.
Source: John Conway [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 13 38 Solution 13 We divide the 3 × 3 square into the following three pieces.
We now rearrange the pieces by moving the top triangle to the bottom and the right triangle to the left side.
From this, it is clear that the shaded region occupies 1/10 of the area of this figure. Since the area of the original square was 9, the shaded area is 9/10 ■.
Source: Adapted from the 2002 Lomonosov Tournament incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 13 39 [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 14 40 Solution 14 Let y = 2x. Then x = log2 y = log y log 2, so we are solving: y + 3 log 2 log y = 5 Move the y over and take the log: 3 log 2 log y = 5 −y log 3 log 2 log y = log(5 −y) log 2 log y · log 3 = log(5 −y) log 2 · log 3 = log(5 −y) log(y) Now it is clear that y = 2, 3 are solutions. This corresponds to x = 1, log2 3, respectively. So the other solution is log2 3 .
■ [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 15 41 Solution 15 Draw a “finish line” on the hiking trail, so that we may view the game as occurring in “laps”. The fireman’s strategy is amusingly simple: On every lap, extinguish every tree until the pyromaniac sets a tree on fire (and do nothing for the rest of the lap). That’s it!
To show that this works, consider the binary number N formed by viewing each burning tree as a 1, each extinguished tree as a 0, and reading the digits from the end to the start.
So the first tree is worth 1, the second tree is worth 2, etc. until the last tree, which is worth 22021.
If the pyromaniac doesn’t set any trees on fire during a lap, then the fireman’s strategy extinguishes everything, meaning we’re already done. So we can assume that the pyromaniac tries to set something on fire on every lap.
In this case, we have on every lap that the last tree whose state is changed is some tree that the pyromaniac sets on fire (because the fireman does nothing after some tree is set on fire). Since this tree has the highest place value in the binary representation of N among all trees whose states were changed, we deduce that N must have increased during the lap.
N cannot increase forever because it is capped by 22022 −1. So eventually it must neces-sarily hit 22022 −1, meaning that all trees are on fire at the end of the lap. The fireman can then extinguish all the trees in one lap.
■ Remarks: There is also an inductive approach. Credits to “InductionEnjoyer” for spotting this. We strengthen the problem to showing that for any tree T, the firefighter can always reach the state where the firefighter and pyromaniac are leaving tree T, and all trees are extinguished.
Clearly this is true if there is only 1 tree. Assume that the firefighter can complete their goal at any particular tree if there are n trees. Now suppose there are n + 1 trees.
The firefighter applies the inductive hypothesis to the n trees other than T, so that all trees other than T are extinguished and the firefighter and pyromaniac are leaving the tree before T, so that the next tree they arrive at is T. There are now two cases.
• T is on fire at this time. Then the firefighter simply extinguishes it and they move on from T, completing the induction.
• T is not on fire at this time. Then the only way to prevent the firefighter’s goal from being completed is for the pyromaniac to set T on fire. Since the pyromaniac can’t extinguish T, the firefighter then repeats their strategy so that they once again reach the state where they reach T with all other trees extinguished. Now the firefighter incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 15 42 extinguishes T and moves on from T, completing the induction.
It turns out that the strategy generated by this induction is the same as the explicit strategy that we constructed from before.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 16 43 Solution 16 If you are like me and unable to pull a construction out of thin air, here is a cute approach.
Suppose there were indeed a decomposition f(x) = e(x) + o(x) for some even e(x) and an odd o(x). This equation holds for all x, so if I replace x with −x then it should still be true: f(−x) = e(−x) + o(−x) This simplifies to f(−x) = e(x) −o(x).
But now we have ( f(x) = e(x) + o(x) f(−x) = e(x) −o(x) , which is a system of equations in the two “variables” e(x) and o(x)! Solving, we get: e(x) = f(x) + f(−x) 2 o(x) = f(x) −f(−x) 2 This is what e(x) and o(x) would have to be if the described decomposition existed. It remains to verify that these are indeed even and odd respectively. Fortunately, they are!
■ Source: Folklore [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 17 44 Solution 17 Let the speeds of the Red October and USS Dallas be v1 and v2, respectively. So, v2 > v1 > 0. Without loss of generality, we may say that the Red October was detected at the origin (0, 0).
The locus of points that the Red October can be at some time t is a circle centered at (0, 0) whose radius is expanding at a rate of v1. Tracking down the Red October means that we must eliminate every possible angle that it could have taken by traversing this expanding circle a full 2π radians. Evidently we must first make contact with the expanding circle to begin traversing its circumference, so the first step of our strategy is to send our ship straight towards (0, 0) until we hit the imaginary expanding circle. We say that this occurs at time t = 0 and that, without loss of generality, we make contact with the circle at (R0, 0).
We now derive parametric equations in polar form that describe our path from here on out. We will follow the path of the expanding circle counter-clockwise, and this necessitates that our distance from (0, 0) is increasing at a rate of v1. Thus the equation for r(t) is: r(t) = v1t + R0 It remains to find θ(t), i.e. our angular velocity.
We obtain this by using the fact that we are traveling at speed v2.
We can quickly derive a formula for our speed in parametric form: We know that x = r cos θ hence x′ = r′ cos θ −rθ′ sin θ. Similarly y′ = r′ sin θ + rθ′ cos θ. Squaring these equations and adding, we obtain v2 2 = (r′)2 + r2(θ′)2. Clearly r′(t) = v1 and r(t)2 = (v1t + R0)2. Hence: v2 2 = v2 1 + (v1t + R0)2θ′(t)2 θ′(t) = p v2 2 −v2 1 v1t + R0 Here it is clear why we needed v2 > v1.
Integrating: θ(t) = θ(0) + Z t 0 p v2 2 −v2 1 v1s + R0 , ds = sv2 v1 2 −1 log v1 R0 t + 1 It remains to verify that we obtain θ(t) = 2π in finite time. Indeed we do, since log increases to ∞(albeit at a turtle’s pace)! Thus we indeed eventually traverse the entire circle, thereby checking every possible angle.
Ergo, our strategy eventually lets us crash into the Red October.
■ Source: I saw this on the Data Genetics blog.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 18 45 Solution 18 At t = 0, let P start at C and Q start at A. We then let P move toward D at speed |CD| and let move Q move toward B at speed |AB|. Thus, P and Q end up at D and B respectively at time t = 1.
It suffices to prove that the problem statement holds for all time t.
CLAIM: [ABP] and [CQD] change linearly with time.
Proof. View AB as the base of △PAB, so that the height is the altitude from P. Note then that the height is changing linearly with time because P is moving along a line. Thus so is the area. The logic for △QCD is the same.
CLAIM: We are done.
Proof. [APB] + [CQD] = [ABCD] holds at the beginning of time and at the end of time.
But by the previous claim, the quantity [APB] + [CQD] changes linearly with time. We deduce that in fact, this quantity must be constant with time, because it takes the same value [ABCD] at two distinct times.
Thus [APB] + [CQD] = [ABCD] for all time, which is what we wanted.
■ Source: I saw this in some random corner of AoPS.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 19 46 Solution 19 Yes we can.
• Begin by setting three fires at once: Two on both ends of the 1-minute rope, and a third on one end of the 2-minute rope.
• After 30 seconds have elapsed, the 1-minute rope will be burnt up, and there will be 90 seconds left on the 2-minute rope. At this point in time, set the other end of the 2-minute rope on fire.
• After 45 seconds have elapsed, the remaining 90 seconds of the 2-minute rope will have burnt up.
In sum, this procedure achieves 30 + 45 = 75 seconds of burning time, as needed.
■ Source: Classic [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 20 47 Solution 20 We claim that Sydney cannot reach (1, 0).
Call a point (a, b) even iff a and b have the same parity. We claim Sydney can only reach even points.
To see this, assume Sydney moves from an even to an odd point. Without loss of generality let us suppose that the even point was (0, 0), and that she moves to an odd point (a, b). Then the equation of the perpendicular bisector, which can be written as 2(ax + by) = a2 + b2, must pass through an integer point (x, y).
But then 2(ax + by) would be even, so a2 + b2 would have to be even, so a + b must be even, so a and b have the same parity. This contradicts the premise that (a, b) is odd.
Since (1, 0) is an odd point, Sydney cannot reach it.
■ Source: Thomas Lam, USAMTS [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 21 48 Solution 21 It is still impossible.
Credits to “axcaea” for this approach.
Let us begin by proving some well-known lemmas. The interested reader should note that their proofs may be simplified using the fact that Z[i] is a UFD.
Lemma 1 Suppose that x and y can each be written as the sum of two squares. Then the same is true of their product.
Proof.
Write x = a2 + b2 = |a + bi|2 and y = c2 + d2 = |c + di|2. Then xy = |(a + bi)(c + di)|2 = |ac −bd + (ad + bc)i|2 = (ac −bd)2 + (ad + bc)2, which completes the argument.
□ Lemma 2 Suppose that x is even and can be written as the sum of two squares. Then the same is true of x 2 Proof.
Write x = a2 + b2. If a and b are both even then we may simply divide each side by 4 to see that x 4 = a 2 2 + b 2 2 is a sum of two squares, thus (x/4)(12 + 12) is a sum of two squares by the previous lemma. If a and b are both odd, then write x = a2 + b2 = |a + bi|2 = 1 2|(1 + i)(a + bi)|2 = 1 2|a −b + (a + b)i|2 = (a −b)2 + (a + b)2 2 , so x/2 = a−b 2 2 + a+b 2 2.
□ Back to the problem. We claim that S := np d, q d : d is an odd sum of two coprime squares, and p ≡q (mod 2) o is the set of all points reachable by Sydney. Of course, it is not necessary for us to show that all points in S are indeed reachable. For the sake of the problem, it is sufficient to show that Sydney cannot escape S, since (1, 0) ̸∈S. So we shall just prove this direction and leave the reachability as an exercise to the curious reader.
Let (p/d, q/d) ∈S. Suppose ax + by + c = 0 is the equation of a line that connects two lattice point. It is sufficient to show that the reflection of (p/d, q/d) over this line will remain in S.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 21 49 Since ax + by + c = 0 has an integer solution for x and y, it must be the case that gcd(a, b) | c. By dividing out by the GCD, we may assume without loss of generality that gcd(a, b) = 1.
We now leave it to the reader to check that the coordinates of the reflection in question are: p(a2 −b2) −2b(aq + c) d(a2 + b2) , −q(a2 −b2) −2a(bp + c) d(a2 + b2) There are now two cases to check.
1. If 2 | ab, then since gcd(a, b) = 1, exactly one of a, b is even and the other is odd. So d(a2 + b2) is odd. But d is a sum of two squares, so by the first lemma we may deduce that d(a2 + b2) is also a sum of two squares.
As for the numerators, the fact that a2 −b2 is odd implies that the first numerator has the same parity as p, and the second numerator has the same parity as q. These have the same parity by the assumption that p ≡q (mod 2). We conclude that the reflection is in S in this case.
2. If 2 ∤ab, then d(a2 + b2) is even, but a quick mod 4 argument reveals that it has only one factor of two. The numerator also has a factor of 2, since a2 −b2 must be even. So we may reduce each fraction by dividing by 2 to write the coordinates as p(a2 −b2)/2 −b(aq + c) d(a2 + b2)/2 , −q(a2 −b2)/2 −a(bp + c) d(a2 + b2)/2 .
Now, d(a2 + b2)/2 must be odd, and since d(a2 + b2) is a sum of two squares, the same is true for d(a2 + b2)/2 by the second lemma.
As for the numerators, a2 −b2 is divisible by 4, so both p(a2 −b2)/2 and q(a2 −b2)/2 are even. And, since a, b ≡1 (mod 2), we have that b(aq + c) ≡aq + c ≡q + c ≡p + c ≡bp + c ≡a(bp + c) (mod 2).
So again, the numerators have the same parity, and thus the reflection is in S.
This completes the proof.
■ Remarks: This problem has a funny history.
My original solution to this, claiming the answer of “yes”, was incorrect as it had fallen for the trap mentioned by the first hint. After outsourcing the problem, we found at least two more proofs claiming an answer of “yes” and at least two more proofs claiming an answer of “no”, all of which were wrong.
Source: Me!
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 22 50 Solution 22 The shaded area is obtained by simply adding up the areas of the first and third circles and subtracting out the second circle! So the answer is 512π + 482π −422π = 3141π .
■ [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 23 51 Solution 23 The key observation is that ( √ x2 + 1−x)( √ x2 + 1+x) = 1. So if we multiply both sides by ( √ x2 + 1 −x), we get y + p y2 + 1 = √ x2 + 1 −x.
Likewise, if we instead multiplied both sides by ( p y2 + 1 −y), we’d get x + √ x2 + 1 = p y2 + 1 −y.
Adding these two equations together, we conclude that x + y = 0.
■ Source: The earliest source I could find was the 1985 Norway Math Olympiad [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 24 52 Solution 24 The idea is as follows: Tile R2 with a triangular packing of radius-10 circles. We show that we can translate this tiling so that each crewmate is in a circle. This is done by picking a random translation of the tiling, and proving that it covers more than 9 crewmates in expectation. We then select those circles containing crewmates to be the buoys, and we are guaranteed that we use no more than 10 buoys because there are only 10 crewmates.
To wit, if we pick a random translation of the tiling (by, e.g. choosing one circle’s center uniformly at random from some fundamental domain of the tiling, and extending this to a full tiling with a consistent orientation), then the probability that a particular crewmate is saved is given by the efficiency of the tiling, i.e. the “ratio” of the plane taken up by the circles’ interiors.
To find this “ratio”, we take a nice fundamental domain such as the shape of the colored regions above. We now determine what fraction of this domain is taken up by the circle.
Using underhanded tricks, this can be done very quickly. Specifically, the total area can be found by dividing the circle into 6 sectors and rearrarranging them so that the region becomes two equilateral triangles. From this argument, we may find the ratio to be π 2 √ 3 ≈0.9069.
To finish, we denote by An the event that crewmate n is saved. By linearity of expectation, we deduce that the expected number of crewmates saved upon picking a random arrangement of buoys is given by: E 10 X n=1 1An = 10 X n=1 E1An = 10 X n=1 P(An) ≈10(0.907) ≈9.07 > 9 So there must exist an arrangement that saves more than 9 crewmates. That’s equivalent to saving all the crewmates, hence we have proven that it is possible.
■ incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 24 53 Remark: It is possible to do better than 10! If n is the most number of crewmates that can be saved if they all fall into the water, then according to the paper papers/paper13.pdf, we have that 12 ≤n ≤44.
Source: Naoki Inaba [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 25 54 Solution 25 Minsung can escape within 100 seconds.
In the optimal strategy, we always make Minsung move in the direction that is “most angled toward the center/spawnpoint”. Hence, Minsung’s optimal strategy is to actually move at a “right angle” from the center. That is, he always faces in a direction v such that v is perpendicular to the line segment connecting the center and Minsung.
Inductively, by using the Pythagorean theorem, we deduce that after t seconds, Minsung will be √ t feet away from the center. Solving the equation √ t = 10 for t, we get that Minsung might murder us all after t = 100 seconds.
■ Source: Classic.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 26 55 Solution 26 Suppose the B coins I use to make the A cents are worth c1, c2, · · · , cB in cents. Then: c1 + c2 + . . . + cB = A The A coins we will use to make B dollars are as follows: • c1 coins worth 100 c1 • c2 coins worth 100 c2 • · · · • cB coins worth 100 cB Indeed, there are c1 + c2 + . . . + cB = A coins here. Moreover, their total worth is c1 100 c1 + c2 100 c2 + . . . + cB 100 cB = 100B which is B dollars.
■ [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 27 56 Solution 27 The answer is no. The identity function on R is the only ring endomorphism on R. Let f : R →R be a ring endomorphism. We will show that f(x) = x for all x ∈R.
Step 1: We show that f is the identity on rationals.
Clearly f(2) = f(1) + f(1) = 2. Inductively we see that f(n) = n for all naturals n.
Moreover f(0 + 0) = f(0) + f(0), so f(0) = 0, and from here we see that f(n + −n) = f(n) + f(−n) so that f(−n) = −f(n). Thus f is the identity on integers. Lastly, for any rational m/n, where m is an integer and n is natural, we may write f(m/n) + f(m/n) + . . . + f(m/n) = f(m/n + m/n + . . . + m/n) = f(m), where the “. . .” represents continuing on for n terms. Here we may deduce that f(m/n) = f(m)/n = m/n, so indeed f is the identity on rationals.
Step 2: We show that f sends positive reals to non-negative reals.
If x > 0, then x = y2 for some y. Hence f(x) = f(y2) = f(y)f(y) ≥0.
Step 3: Now we may finish.
Take any real x, and suppose f(x) ̸= x. Then, appealing to the symmetry f(−x) = −f(x), we may suppose without loss of generality that f(x) < x.
Now find a rational q with f(x) < q < x.
Since q < x, we have x −q > 0, so by Claim 2 we have that f(x −q) ≥0. So f(x) ≥ f(q) = q by Claim 1. But f(x) < q, contradiction.
■ [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 28 57 Solution 28 We claim that the numbers are actually equal.
For each lit triangle of perimeter 2019, add 1 to each of its sides. Then: 1. It is still a triangle, because if a + b > c, then (a + 1) + (b + 1) > (c + 1) 2. The perimeter is now 2022.
So it becomes a lit triangle of perimeter 2022.
We claim that every lit triangle of perimeter 2022 can be obtained in this way! This would show that there is a one-to-one correspondence, so that the number of lit triangles of perimeters 2019 and 2022 respectively are equal.
To see this, suppose otherwise. Then there is a lit triangle of perimeter 2022 that cannot be obtained using the above procedure. This entails that if we subtract 1 from each side, then we obtain an invalid triangle.
Take such a triangle with sides a, b, and c, so that a + b + c = 2022. Without loss of generality, let us assume that c is the longest side. Then a + b > c, and moreover c would be the longest side of the hypothetical triangle with sides (a −1), (b −1), and (c −1). This would be a valid triangle if and only if (a −1) + (b −1) > (c −1). We are assuming that it is not valid, hence (a −1) + (b −1) ≤(c −1), or a + b ≤c + 1. But a + b > c, so c + 1 ≤a + b ≤c + 1. We deduce that a + b = c + 1.
Adding c to both sides, we obtain 2022 = a+b+c = 2c+1, thus 2022 is odd, contradiction.
■ Source: 2022 ICMC, Constantinos Papachristoforou [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 29 58 Solution 29 It suffices to find the dimension of V0, the space of n×n magic squares with magic number exactly equal to 0, via the natural isomorphism V0 ⊕R ∼ = V where V is the space of magic squares.
We now define a linear transformation T : Rn×n →R2n+1 as follows: For A ∈Rn2, T(A) is the column vector whose components are the sums of the rows, columns, and diagonals of A, except the last row. That is, the components are the sums along the following lines: We note that the null space of T is precisely V0. By the rank-nullity theorem, it follows that: dim V0 = n2 −rk T So it suffices to find the rank of T. In fact, we claim that T has full rank.
To prove this, we only need to show that for every line above, we can find a matrix such that that line has sum 1 (... or any non-zero real) and all other lines have sum 0. What’s nice is that throwing out the last row makes this quite feasible!
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 29 59 As suggested in the above graphic, the scheme is as follows: • For any row, we can find a cell that lies on neither diagonal and then place a 1 in it.
Then, place a −1 in the bottommost cell below 1.
• For a column that is neither the first nor last, we may simply place down a 1 on its bottommost cell. Otherwise, we can do a strange thing, as shown, provided that n ≥4.
The case n = 3 is scary, and is also shown.
• For a diagonal, we may simply build a 2 × 2 square of 1’s and −1’s, provided that n ≥4. The case n = 3 is not hard.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 29 60 This proves that, indeed, T has full rank, so that dim V0 = n2 −(2n + 1) = n2 −2n −1.
Hence, dim V = 1 + dim V0 = n2 −2n for n ≥3, whereas for n = 1, 2 we see that the dimension is 1.
■ [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 30 61 Solution 30 If the square “rotates” n times, then the angle between two consecutive squares is π 2n. By geometrical and symmetrical reasoning, we see that if the side length of some square is x then the next side length y must satisfy y sin π 2n + y cos π 2n = x thus the ratio between the side lengths of two consecutive squares is just 1 sin( π 2n)+cos( π 2n). After n “rotations”, we see that the side length of the last square is given by 1 (sin( π 2n)+cos( π 2n)) n. It remains to compute limn→∞ sin π 2n + cos π 2n −2n.
Letting x = π 2n, this limit will be equal to limx→0 (sin x + cos x)−π x , if it exists. We may write this as: exp h lim x→0 −π x log (sin x + cos x) i Applying Taylor’s Theorem, we may now write sin x = x+o(x) and cos x = 1+o(x) to write the desired as: = exp h lim x→0 −π x log (1 + x + o(x)) i We may moreover write log(1 + y) = y + o(y), and we may take here y = x + o(x) to obtain: = exp h lim x→0 −π x (x + o(x) + o(x + o(x))) i = exp h lim x→0 −π x (x + o(x)) i = exp lim x→0 −π −πo(x) x Which is exp(−π) = e−π by definition of little-o.
■ Remarks: The knowledgeable reader may have found this problem familiar! Indeed, it bears a strong resemblance to the following problem: On each corner of a unit square lies an ant. Starting at the same time, each ant moves directly towards their counter-clockwise neighbor at a speed of 1. When they meet at the center, how far has each ant travelled?
The limiting curve that you have witnessed in this problem is precisely the path taken by these ants, albeit cut short after 90◦have been traversed about the center.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 30 62 We can derive the equation of the curve in a cute way. First, observe that it should not be important that the ants move at a constant speed — as long as they all share the same speed, it should be the case that they trace out the same curve. Now, overlay the complex plane unto the square, with the square’s center being 0. If an ant is at z ∈C, then it is moving towards iz. The observation lets us take the ant’s speed at such a location to be |iz −z|, so that its path z(t) will modeled by the differential equation z′(t) = iz(t) −z(t) = (i −1)z(t).
This naturally solves as z(t) = z(0)e(i−1)t.
With this parametrization of the curve, we have that t lives in [0, ∞). Writing it as z(t) = z(0)e−teit, we see that the distance to the center exponentially decays whereas the angular velocity is constant. This formula directly shows that the distance to the center upon a quarter rotation is given by |z(0)|e−π/2, which is consistent with our deductions in the original problem’s solution.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 31 63 Solution 31 Let t be the total amount of tea (in cups), and m be the total amount of milk (in cups).
Then t + m = n, where n is the number of cups. Moreover, we are given that t 4 + m 6 = 1.
We know that there’s some positive amount of tea. After all, it’s not a department milk!
So, t > 0, and in particular, t/4 > t/6. Thus 1 = m 6 + t 4 > m 6 + t 6 = n 6 so that 6 > n.
Similarly, we know there’s some positive amount of milk, because there is some “contam-ination” as stated in the problem. So m > 0 and in particular m/4 > m/6. This gives us 1 = m 6 + t 4 < m 4 + t 4 = n 4 so that 4 < n.
Since 4 < n < 6, we must have n = 5 .
I leave it as an exercise to demonstrate that there indeed exist 5 cups of milk-contaminated tea that satisfies the problem’s conditions.
■ Source: Folklore [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 32 64 Solution 32 You can argue that the longest such stick length is equal to the minimum possible length of the red segment shown below.
a b θ With the angle θ marked as shown, it is clear that this length is given by a cos θ + b sin θ. One can use calculus to minimize this quantity. Alternatively, by H¨ older’s inequality with exponents 3/2 and 3, we can more directly write a2/3 cos2/3 θ 3/2 + b2/3 sin2/3 θ 3/2!2/3 cos2/3 θ 3 + sin2/3 θ 31/3 ≥a2/3 + b2/3.
This rearranges to a cos θ + b sin θ ≥(a2/3 + b2/3)3/2 , and by examining the equality case one can see that this lower bound is indeed obtainable.
■ Remarks: If you have not heard of it, this is an easy variant of the Moving Sofa Problem, which asks for the area of the largest region that can be passed through this bend of the hallway. Mathematicians have had a lot of fun drawing interesting telephone-like shapes that can get through, but the problem is still officially open at the time of writing. Recently, Jineon Baek has claimed a very promising resolution to the problem ( abs/2411.19826).
Here’s a fun variant: What is the maximum area of a rectangle that can be carried through such a bend?
Source: Folklore [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 33 65 Solution 33 Claim: The set of all P ∈R[x] that are the sum of two squares is closed under multiplication.
Proof.
Take P = A2 + B2 and Q = C2 + D2. Then by magic: PQ = |A + iB|2|C + iD|2 = |(AC −BD) + i(BC + AD)|2 = (AC −BD)2 + (BC + AD)2 □ Without loss of generality suppose P is monic. Since P ≥0, its roots come in conjugate pairs, and by multiplying such corresponding terms in the factorization of P, we see that P is a product of quadratics x2 + bx + c, each of which is non-negative. From discriminant analysis it follows that b2 −4c ≤0, so particularly we may write x2 + bx + c = (x + b/2)2 + p c −(b/2)22 which is a sum of two squares. Now inductively apply the claim!
■ [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 34 66 Solution 34 We compute Clara’s probability of winning. If she goes second or third with probability 2 3, then the first shooter (either Alex or Blaire) must shoot the other expert markswoman or else she guarantees her own death. Then it will be Clara’s turn, and her probability of winning will be exactly the probability that she lands the shot on the other living duelist, i.e.
1 2.
If Clara goes first with probability 1 3, it actually doesn’t matter who she decides to fire at.
If she shoots at Alex, Blaire, or even herself (!), she will lose if she lands the hit. Conditioned on the event that she misses (with probability 1 2), we reduce to the case in which Clara goes third, and we know her survival probability here is 1 2.
Altogether, Clara’s odds of winning are 2 3 · 1 2 + 1 3 · 1 2 · 1 2 = 5 12.
By symmetry it follows that Alex and Blaire each have survival odds of 7 24. Despite having the worst aim, Clara is the most likely one to get out alive.
■ Source: I found this in a puzzle book.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 35 67 Solution 35 Solution 1 Let X be the set of duelists and define the map f : X →X via: f : x 7→duelist shot by x If we assume that nobody is alive, then f is a surjection. By finiteness it follows that f is actually a bijection.
Thus we may view f as a permutation that can be decomposed into cycles. Since 31415 is odd, there is at least one odd cycle (x1 x2 . . . xk). k ̸= 1 because nobody shoots themselves, so k ≥3.
Now, by virtue of the shooting cycle, we see that d(x1, x2) > d(x2, x3) > . . . > d(xk−1, xk) > d(xk, x1) > d(x1, x2), contradiction.
■ Solution 2 Assume for contradiction that everyone dies. First, observe that there are 31415 bullets and 31415 people. So it cannot be the case that someone is shot more than once, since then someone must be shot less than once, i.e. not at all.
By finiteness, we may find the two duelists that are of the shortest distance apart. By minimality, these two duelists shoot each other. By the observation, nobody else shoots these two duelists. Thus we may essentially ignore these two duelists. Inductively repeating this argument, we eventually end up with one person. They must die, but they can’t shoot themselves, contradiction.
■ Source: Folklore [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 36 68 Solution 36 The miracle point in question is the incenter.
The credits for this approach go to “tenth”. We will start by showing a strong variant of the converse.
Lemma 1 Let △ABC be a triangle with incenter I. Let X, Y be points on the boundary of △ABC. Then cutting △ABC along segment XI and then along segment IY will divide the area and perimeter of △ABC into the same ratio.
I A B C Proof.
The key idea is to subdivide the regions into triangles along the angle bisectors, as shown.
I A B C If we view each triangle as having height equal to the inradius r (and thus having a base along the perimeter of the triangle), then we can see that its area is r 2 multiplied by the length of the perimeter it occupies. Thus, for any triangle, the ratio Area Occupied Perimeter is constant, being equal to r/2. It follows that the ratio Red Area Red Occupied Perimeter is also r/2, and the blue region is no different.
□ incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 36 69 With this proven, we are ready to solve the problem swiftly and with style. Suppose that we cut the pizza along segment XY , and that this cut divides the area and perimeter of the pizza into the same ratio.
Consider also cutting the pizza along segments XI and IY . As we have just shown, this also divides the pizza’s area and perimeter into the same ratio.
But both the XY cut and the XI and IY cuts divide the pizza’s perimeter into the same ratio! Thus, by the previous two paragraphs, they must divide the pizza’s area into the same ratio.
This can only be possible if △XY I has zero area. That is, X, I, Y are collinear, so XY passes through the incenter.
■ Source: This is called Haider’s Theorem.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 37 70 Solution 37 The missing digit is 4.
The key property that we will use is that an integer n and the sum of the digits of n will always have the same remainder upon division by 9. Let us calculate the remainder of 229 upon division by 9.
One simple way to determine this is to seek a pattern in the remainders among the powers of 2. If you know modular arithmetic, this gives a swift way to evaluate the remainder: 229 = 4 · 89 ≡4 · (−1)9 = −4 ≡5 (mod 9) So the remainder is 5.
We know that if all ten possible digits were present, then the digit sum would be 45, which is divisible by 9. So, the missing digit must be a digit d such that 5 + d is divisible by 9. There is only one digit for which this holds: d = 4 .
■ Source: Classic [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 38 71 Solution 38 The situation encountered by the Hearthians is possible, and it turns out that Desmine must be missing.
The key ideas are: • If you do the exploration procedure with a distance that is not an integer multiple of the distance along a quarter turn on the planet, then your path is likely too chaotic to expect being able to reunite with many other Hearthians.
• Doing the exploration procedure with a number of quarter turns around the planet that is congruent to 0, 1, or 3 mod 4 will return you to where you started. If instead it is congruent to 2, you actually end up on the antipode of where you started.
This may be hard to believe, so I will sketch out why this is true. Imagine that a Hearthian starts at the north pole.
– If this Hearthian returns to the north pole after travelling x km forwards, then evidently they will still arrive at the north pole again after turning, travelling x km forwards, turning, and travelling x km forwards again.
– If this Hearthian does a single quarter-turn around the planet by walking forwards x km, then they will arrive at the equator after the first leg of their exploration procedure. After turning 90◦counter-clockwise, the direction they face will be aligned with the equator, and so they will remain on the equator after waking x km forwards. Finally, after turning 90◦counter-clockwise again, they will be facing north, and so walking x km forwards will get them back to the north pole.
– If this Hearthian does a half-turn around the planet by walking forwards x km, then they will find themselves at the south pole after the first leg of their explo-ration procedure. The direction they turn does not matter: After travelling x km again in any direction, they arrive at the north pole, thus after yet another x km in any direction, they end their journey at the south pole.
– If this Hearthian does a three quarters-turn around the planet by walking for-wards x km, then as in the one quarter-turn case, they will arrive at the equator.
Following the same logic reveals that they indeed will end up at the north pole at the end of their journey.
So doing the procedure with 1, 3, 4, 5, 7, 8, 9, 11, 12, ... quarter turns will take you to where you started. Inspired by the subsequence of numbers 3, 4, 5, 7, 8, 9, we can take the quarter-turn length to be 10 so that Desmine (the one with the 60) ends up at the antipode but everyone else ends up reunited back at the ship.
■ incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 38 72 Remarks: The radius of the planet is not unique. For example, the radius could also be such that the length of a quarter-turn is 10 3 .
It is also true that Desmine is the only Hearthian that could be missing as a result of this procedure. I will spare the details, but it turns out that if we assume that the ship is at the north pole, then the z-coordinate (which I take to be up and down) of one’s location at the end of the exploration procedure can be modeled by the function f(θ) := cos3 θ + sin2 θ, where θ is the “angle” traversed in in a single leg of the procedure. It follows that showing uniqueness reduces to proving that if θ is such that exactly 6 of the 7 values in the list [f(3θ), f(4θ), · · · , f(9θ)] are equal to each other, then the odd element out must be f(6θ).
Unfortunately, this seems difficult to prove without computational aids.
Source: Konstantin Knop [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 39 73 Solution 39 The statement is false! We present two different methodologies for constructing an un-countable totally-ordered family of subsets of N.
Solution 1 Using a bijection, it is sufficient to tackle the problem when “natural” is replaced with “rational”. For each x ∈R we take the Dedekind cut Dx := {q < x : q ∈Q}. We can then take our family to be F := {Dx : x ∈R}.
Clearly this is uncountable and is a totally-ordered family of subsets of the countable set Q.
■ Solution 2 Using a bijection, it is sufficient to answer the problem replacing the naturals with the nodes of an infinite binary tree.
The family of infinite paths starting from the root is uncountable, but not totally-ordered. To fix this, we simply add to each path all elements to the “left” of the path.
■ Source: I saw this on Math Stack Exchange [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 40 74 Solution 40 We may save 9. Clearly this is the maximum we can guarantee saving because there is no guarantee that the first prisoner can get their hat color right, by virtue of having no information about their hat. To save everyone else, the first prisoner says “black” if they see an odd number of black hats, and “white” otherwise.
Then the second prisoner can deduce their hat color — if they see an odd number of black hats, then they must be wearing white. Otherwise, they must be wearing black. Now consider any prisoner P thereafter. By listening to the correct guesses of the prisoners behind them and counting the hats in from of them, P will be able to compute the number B of black hats excluding P’s and the first prisoner’s hats. So the number of black hats excluding just the first prisoner is either B or B +1, depending on whether P has a white or black hat, and this can be disambiguated by the first prisoner’s information on whether this number is even or odd.
■ Source: Classic [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 41 75 Solution 41 Recall that a permutation can be either odd or even. A permutation is even if it is created via an even number of swaps (“transpositions”), and odd otherwise. It is a theorem that no permutation can be both odd and even, so this is a well-defined characterization.
A key consequence is that if you take an even permutation and perform one additional swap, then the result is an odd permutation, and vice versa.
Starting from the back of the line, number the prisoners from 1 to 100. In some order, number the hats from 1 to 101. Mark the warden as “prisoner 101” and give him the missing hat.
Using the idea of even and odd permutations, the plan is as follows: Viewing the 101 hats as a permutation of the integers from 1 to 101, prisoner 1 will guess that the permutation is even and guess their hat according to this assumption. That’s it.
Note that prisoner 1 sees all hats except their own and the warden’s, so from their per-spective, there are exactly two possible sequences for the hats, and they differ by exactly one swap. Thus, of the two possibilities that prisoner 1 sees, one represents an even permutation and the other is odd. Hence their assumption that the permutation is even corresponds to a well-defined guess for their hat.
For example, if there are 4 prisoners and prisoner 1 sees ?
2 3 4, then prisoner 1 guesses 1, because this corresponds to the identity permutation (in which prisoner 1 wears hat 1 and the warden wears hat 5), which is even.
Prisoner 1 may or may not get shot. If they are not shot, then everyone knows that the permutation is indeed even. Otherwise, it must be odd.
Now we get to prisoner 2. We claim that prisoner 2 knows either prisoner 1’s hat or the warden’s hat. To see this: • If prisoner 1 was not shot, then obviously by virtue of hearing their correct guess, prisoner 2 knows prisoner 1’s hat.
• If otherwise prisoner 1 was shot, then let the hat they guessed be A. Let the hat they’re actually wearing be B. The two possibilities were that either prisoner 1 was wearing A and the warden was wearing B, or the other way around. By virtue of hearing the shot, we know that the former was not the case, so it must be the latter, which implies that the warden is wearing A. That is, the warden is wearing the hat guessed by prisoner 1.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 41 76 Thus, there are only two hat colors that prisoner 2 does not know: That of their own, and that of one other person. So, prisoner 2 is also guessing between two possible sequences, and they differ by a single swap. Since they know the parity of the permutation, they may disambiguate between the two possibilities and guess correctly.
Inductively, prisoner k can guess correctly. This is because prisoners 2, 3, · · · , k −1 all guess correctly, and by the same logic, we can argue that prisoner k can either deduce prisoner 1’s hat or the warden’s. So again, prisoner k needs to decide between two possible sequences of hats that differ by a single swap, which can be disambiguated because prisoner k knows the parity of the permutation.
In all, we see that all prisoners after prisoner 1 will be guaranteed survival. This is clearly the best possible outcome because prisoner 1 has no such guarantee of survival no matter the strategy. Hence 99 prisoners may be saved.
■ [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 42 77 Solution 42 Let X be the set of all possible hat sequences. Define an equivalence relation ∼on X as follows: A ∼B if and only if A and B are eventually the same. That is, they only differ in finitely many places.
The relation ∼partitions X into equivalence classes.
The prisoners, in the planning phase, will apply the axiom of choice to agree on a representative of each class.
When the game starts, the hats form some sequence S ∈X. Every prisoner, by virtue of being able to see the tail of the sequence S, knows the equivalence class of S under ∼, and can therefore obtain the agreed-upon representative T of the class [S]∼. Every prisoner will then guess their hat in accordance to the sequence T.
Since S ∼T, we have that S and T will eventually be the same. That is, S and T will differ only in finitely many places, so only finitely many prisoners can die under this scheme.
■ [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 43 78 Solution 43 Solution 1 An incredible approach uses L’Hˆ opital’s rule. Suppose such a function f existed. Then 2 = lim x→∞f(x) = lim x→∞ f(x)ex ex = lim x→∞ d dxf(x)ex d dxex (L’Hˆ opital’s rule) = lim x→∞ f ′(x)ex + f(x)ex ex (Product rule) = lim x→∞f ′(x) + f(x) = 1 + 2 = 3, contradiction.
To rigorously justify the application of L’Hˆ opital’s rule, we need to check that • f(x)ex is differentiable and approaches ∞as x →∞, • ex is differentiable and approaches ∞as x →∞, and • limx→∞ f′(x)ex+f(x)ex ex exists and is finite.
Of course, all these are true.
■ Solution 2 The main purpose of this problem was to show off Solution 1, but it is more of an amusing parlor trick then an instructive methodology. A more typical approach is as follows. Suppose for contradiction that such a function f exists. Then there exists N > 0 so large that • |f(x) −2| < 1 for all x ≥N, and • |f ′(x) −1| < 1 2 for all x ≥N.
Now take a = N and b = N + 100. By the Mean Value Theorem, there exists c ∈(a, b) such that f ′(c) = f(b) −f(a) b −a = f(b) −f(a) 100 .
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 43 79 Since c ≥N, we have that |f ′(c)−1| < 1 2. Thus f(b)−f(a) 100 −1 < 1 2. In particular, f(b)−f(a) 100 > 1 2, and so f(b) −f(a) > 50. But now 50 < f(b) −f(a) ≤|f(b) −f(a)| ≤|f(b) −2| + |2 −f(a)| < 2 because a, b ≥N, contradiction.
■ Remarks: Let’s consider an alternate variant that can’t be treated using the approach in Solution 2: Suppose that f : R →R is differentiable such that there exists the limit lim x→∞f(x) + f ′(x) = L, with L finite. Prove that lim x→∞f(x) = L and lim x→∞f ′(x) = 0.
The intent behind this formulation is to force the methodology used in Solution 1. At first glance, this may seem successful. However, it is not so straightforward. To wit, here is an incorrect solution.
“If we apply L’Hˆ opital’s rule as in Solution 1, we can write lim x→∞f(x) = lim x→∞ f(x)ex ex ?
= lim x→∞ f(x)ex + f ′(x)ex ex = lim x→∞f(x) + f ′(x) = L, as needed.” The error is that the application of L’Hˆ opital’s rule was not justified. The issue is that we do not know that f(x)ex →∞as x →∞.
To apply L’Hˆ opital’s rule, we would be just as happy if it were the case that f(x)ex →−∞ as x →∞. So it suffices to prove that f(x) is either bounded from below or bounded from above over all x ≥0. This is because if, for example, f(x) ≥−M for all x ≥0, then we may take g(x) := f(x) + M + 1 so that g(x) ≥1. We have that g(x) + g′(x) →L + M + 1 and g(x)ex →∞as x →∞, so we may apply the trick with L’Hˆ opital’s rule to g!
Suppose for contradiction that f(x) is unbounded from above and unbounded from below over x ≥0. Then it is not hard to show that there exists an increasing sequence of local maxima xn of f, with xn →∞and f(xn) →∞. Intuitively, this is because for f to be unbounded from above and unbounded from below, it must obtain higher and higher peaks (as well as lower and lower valleys), and we select xn to be the x-coordinates of these peaks.
It then follows that ∞> L = lim x→∞f(x) + f ′(x) = lim n→∞f(xn) + f ′(xn) = lim n→∞f(xn) = ∞, contradiction.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 44 80 Solution 44 We claim that there is no winning strategy because the game must end in a draw assuming perfect play from both players. This is because the game is isomorphic to Tic-Tac-Toe.
The first observation is that there are exactly 8 ways to make 15. That is, there are exactly 8 three-element subsets of {1, 2, · · · , 9} that sum to 15. Here they are: • 1, 5, 9 • 2, 5, 8 • 3, 5, 7 • 4, 5, 6 • 3, 4, 8 • 2, 4, 9 • 2, 6, 7 • 1, 6, 8 It turns out that these eight triplets of digits are exactly the triplets that show up among the rows, columns, and two diagonals of a 3 × 3 magic square!
2 7 6 9 5 1 4 3 8 Thus, if we view the game between Ana and Beth as taking turns claiming digits from this magic square, then this reduces to Tic-Tac-Toe because the eight possible lines of victory in the board are exactly the eight possible obtainable sums by the claim.
■ Remarks: This is a remarkable connection. Though, some may argue that this may not entirely be a coincidence! See this blog post for an explanation.
Source: Probably John Conway [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 45 81 Solution 45 Let us first tackle the cube’s “height”, i.e. the length of its projection unto the z-axis.
Note that the entirety of this height is traversed via the following path along the cube’s edges, from the bottom to the top.
The desired “height” of the cube is given by the sum of the “heights” of these three edges.
For easier analysis, we may shift these edges downwards so that they protrude from the bottom-most vertex.
As in the diagram, we call the red, orange, and yellow edges x, y, and z, respectively. Let the “height” of the x, y, and z edges be hx, hy, and hz, respectively. Then the “height” of the cube is hx + hy + hz.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 45 82 Putting this aside, let us now tackle the cube’s “shadow”. If you view a cube from any perspective, you always see three faces (or less), each being a rhombus. The cube’s shadow is no different — it is a hexagon which can be partitioned into three rhombi, each being the projection of a different lower face.
As in the diagram, we label the area of the “shadow” under the face spanned by the y and z edges as Ax. Defining Ay and Az similarly, we see that the total area of the cube’s “shadow” is Ax + Ay + Az.
We claim that hx = Ax. To prove this, we orient our perspective to view the cube from the “side”, so that the y and z edges coincide to form a line segment orthogonal to x.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 45 83 Mark the angle θ as in the diagram. Then, from the lengths triangle formed on the left, we have that sin(90◦−θ) = hx 1 .
The triangle formed on the right, on the other hand, displays a proportional relationship between areas instead of lengths. To be precise, it relates the area of the face spanned by y and z (which is 1) to the area of the shadow of this face (which is Ax) via the cosine of the angle between this face and the ground. We get that cos θ = Ax 1 .
But sin(90◦−θ) = cos θ, so indeed hx = Ax.
By symmetrical reasoning, we have hy = Ay and hz = Az, and so we may conclude that hx + hy + hz = Ax + Ay + Az. That is, the “height” is numerically equal to the area of the shadow.
■ Remarks: The following generalization is true.
Theorem 1 Let m, n ∈N. Suppose that a unit cube C lies in Rm × Rn with some orientation.
Then the m-dimensional measure of the projection of C unto Rm (the subspace formed by the first m components) is equal to the n-dimensional measure of the projection of C unto Rn (the subspace formed by the last n components).
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 45 84 To build up to the proof of this, let us first present two lemmas, both of which are fascinating results in and of themselves.
Lemma 1 (Sylvester’s Determinant Identity) Let A be an m × n matrix and B be an n × m matrix. Then det(Im + AB) = det(In + BA), where Ik denotes the k × k identity matrix.
Proof.
Consider the block-form (m + n) × (m + n) matrix Im A −B In .
We evaluate the determinant of this matrix in two different ways. Using “row reduction”, we have on one hand that det Im A −B In = det Im A −B + BIm In + BA = det Im A 0 In + BA = det(In + BA).
On the other hand, we can use “column reduction” to get that det Im A −B In = det Im + AB A −B + InB In = det Im + AB A 0 In = det(Im + AB).
□ Lemma 2 (Complementary submatrices of a unitary matrix have same determinant) Let U := A B C D be an (m + n) × (m + n) unitary matrix, where A is an m × m square matrix and D is an n × n square matrix. Then | det A| = | det D|.
Proof.
We have U TU = ATA + CTC ATB + CTD BTA + DTC BTB + DTD and UU T = AAT + BBT ACT + BDT CAT + DBT CCT + DDT .
Since U is unitary, both of the above products must be the (m+n)×(m+n) identity matrix.
In particular, it follows that BTB + DTD = In and AAT + BBT = Im. Now, by Sylvester’s Determinant Identity, | det A|2 = det(AAT) = det(Im −BBT) = det(In −BTB) = det(DTD) = | det D|2.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 45 85 □ It turns out that this lemma implies the theorem.
Proof.
Without loss of generality, we may let 0 be a vertex of C. Then there are m + n edges protruding from 0, which we may view as a set of m + n orthonormal vectors v1, v2, · · · , vm+n ∈Rm+n.
Let U = [v1 v2 · · · vm+n]. That is, U is the matrix whose ith column is vi. Then U is a unitary matrix.
There are m+n m m-dimensional faces of C that include 0, since each such face corresponds to the “span” of a selection of m of the vectors v1, · · · , vm+n.
It happens to be the case that the projections unto Rm of all m+n m such m-dimensional faces will partition the projection of C unto Rm. We defer the work of reasoning this out to the reader (who hopefully can think in n+m dimensions...). What’s important is that due to this, the m-dimensional measure of the projection of C unto Rm is the sum of the measures of the projections of the m+n m m-dimensional faces. Analogously, the n-dimensional measure of the projection of C unto Rn is the sum of the measures of the projections of the m+n n n-dimensional faces.
To prove this, we use an approach that mirrors the solution to the original problem!
Instead of pairing a 2D face with a 1D edge of equal measure after projection, we can pair an m-dimensional face with an n-dimensional face of equal measure after projection.
We pair these faces in the obvious way: Let {vi : i ∈S} be a selection of m vectors that determine an m-dimensional face F of C, for a subset S ⊆{1, 2, · · · , m + n} of m indices.
Then we may pair this with the n-dimensional face F ′ spanned by the unused n vectors, i.e. {vi : i ̸∈S}. We claim that the projection of F unto Rm has the same measure as the projection of F ′ unto Rn.
The projection of F unto Rm is the m-dimensional parallelepiped formed by the m vectors in {vi : i ∈S}, but with their last n components removed. That is, the m vectors that determine this parallelepiped are given by {(vi,1, vi,2, · · · , vi,m) : i ∈S}. Naturally, these vectors form an m × m matrix A whose determinant | det A| is the m-dimensional measure of the parallelepiped. Moreover, A is the submatrix of U whose columns’ indices are given by S and whose rows are the first m rows.
Similarly, the projection of F ′ unto Rn is the n-dimensional parallelepiped formed by the n vectors {vi : i ̸∈S}, but with their first m components removed. With these com-ponents erased, these n vectors form an n × n matrix D whose determinant | det D| is the n-dimensional measuure of the parallelpiped.
Crucially, A and D are complementary square submatrices of U in the sense that the rows incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 45 86 and columns that they extract are all distinct. It follows by Lemma 2 that | det A| = | det D| (rearranging rows and columns as necessary so that it may be applied), and this concludes the proof.
□ [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 46 87 Solution 46 Suppose the center of the circle is (0, 0). Let the radius of the circle be r. Then the equation of the parabola is given by y = x2 +r. Moreover, the line y = √ 3x must be tangent to the parabola. It follows that the quadratic √ 3x = x2 + r must have exactly one root, and hence must be a perfect square trinomial. This occurs exactly when r = √ 3 2 2 = 3 4 .
■ [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 47 88 Solution 47 On each square of an m×n board, place a biased coin that comes up heads with probability x. Flip all the coins.
Let A be the event that in each of the n columns, there exists a tails. Let B be the event that in each of the m rows, there exists a heads. Note that P(A) = (1 −xm)n, and P(B) = (1 −(1 −x)n)m. Moreover, A and B encompass the entire probability space! That is, no matter how the coins flip, either A happens or B happens or both. Thus P(A) + P(B) ≥P(A ∪B) = P(Ω) = 1, which is the desired inequality.
■ Source: I saw this problem in the Czech Republic Math Olympiad, but it may be older.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 48 89 Solution 48 For simplicity, let us replace “100” with “1”, so that the red block’s temperature begins at 1 degree and the blue block’s temperature begins at 0 degrees. We’ll show that the blue block’s temperature can get arbitrarily close to 1.
Represent the blocks’ temperatures as an ordered pair (Red Temperature, Blue Temperature), so that at the start their temperatures are (1, 0). We first claim that we can go from (1, 0) to (3/8, 5/8). To see this, split each block into halves, so that the temperatures are 1 0 1 0, with the red blocks on the left. Hit the top blocks together to get 1/2 1/2 1 0 , and now hit each block with the block diagonally opposite them to get 1/4 3/4 3/4 1/4.
Lastly, hit the bottom two blocks together to get 1/4 3/4 1/2 1/2.
Merging back, we indeed end up with (3/8, 5/8).
In general, suppose we have found a way to go from (1, 0) to (1 −x, x). Then by scaling the temperatures up, we can go from (a −b, 0) → (1 −x)(a −b), x(a −b) , and by adding a temperature of b to all blocks, we see that we can go from (a, b) → (1 −x)a + xb, xa + (1 −x)b .
Call this the X-procedure. Let’s retry the previous procedure we came up with by using the X-procedure in place of hitting two blocks together normally.
1 0 1 0 incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 48 90 Apply the X-procedure on the top blocks.
1 −x x 1 0 Apply the X-procedure between the top-left and bottom-right.
(1 −x)2 x 1 x(1 −x) Apply the X-procedure between the bottom-left and top-right.
(1 −x)2 2x −x2 1 −x + x2 x(1 −x) Apply the X-procedure between the bottom blocks.
(1 −x)2 2x −x2 something awful 2x −3x2 + 2x3 Bringing the blocks together again, the temperature of the blue block is then 2x −2x2 + x3.
Let this expression be f(x).
We have shown that, if we can execute the heat transfer (1, 0) →(1 −x, x), then we can execute the heat transfer (1, 0) →(1 −f(x), f(x)). Inductively, it follows that we can execute the heat transfer (1, 0) →(1 −f (n)(x), f (n)(x)) for all n, where f (n) is f composed with itself n times. It remains to prove that lim n→∞f (n)(1/2) ?
= 1.
First we show that the limit exists. Indeed, by writing f(x) = x + x(1 −x)2, we see that f(x) ≥x for all x ∈[0, 1], which entails that the sequence {f (n)(1/2)}n must be increasing.
So the desired limit exists and is some real number L. Now, write f (n+1)(1/2) = f(f (n)(1/2)) = 2f (n)(1/2) −2f (n)(1/2)2 + f (n)(1/2)3.
Sending n →∞gives L = 2L −2L2 + L3, so either L = 0 or L = 1. The case L = 0 is quite obviously deranged, so the limit is L = 1, concluding the proof.
■ Source: Puzzling Stack Exchange [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 49 91 Solution 49 We are given that 1 12 + 1 22 + 1 32 + · · · = π2 6 .
(∗) Multiply each side by 1/4, we can get 1 22 + 1 42 + 1 62 + · · · = π2 24.
(∗∗) By subtracting (∗∗) from (∗), we conclude that 1 12 + 1 32 + 1 42 + · · · = π2 6 −π2 24 = π2 8 .
■ [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 50 92 Solution 50 As suggested by the hint, Ari first flips a fair coin to decide which card she chooses. The remainder of her strategy is as follows: If the real number on that card is t, then Ari will guess “Higher!” with a probability P(t) that we will define later.
Let us show that this works. Suppose that the two cards chosen by Beth are x and y, with x > y. Then the probability that Ari wins under this strategy is 1 2 · P(x) + 1 2 · (1 −P(y)).
If you work out the algebra, this probability is strictly greater than 1 2 exactly when P(x) > P(y). So Ari’s scheme reduces to the following problem: Find a function P : R →[0, 1] such that whenever x < y, we have P(x) < P(y), i.e. P is strictly increasing.
This is clearly possible! For example, Ari can choose P(x) := 1 2 + arctan(x) π .
■ Source: Classic?
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 51 93 Solution 51 Claim: The hydra can be moved only to rooms that are multiples of 6.
Proof. For each integer n, we assign a weight wn to room n as follows: wn := 1, n ≡0 (mod 6) √ 2, n ≡1 (mod 6) −1 + √ 2, n ≡2 (mod 6) −1, n ≡3 (mod 6) − √ 2, n ≡4 (mod 6) 1 − √ 2, n ≡5 (mod 6) Observe that wn−1 + wn+1 = wn for all integers n. Thus the sum of the weights of the rooms occupied by the hydra heads (counting multiplicity) does not change with every move you make. Since we start with one hydra head in room 0, the starting weight sum is 1, so at the end, if there are only h > 0 hydra heads left in some room n (and nowhere else), then h · wn = 1. By our choice of weights, this is only possible if n is a multiple of 6 and h = 1.
Claim: All multiples of 6 may be achieved.
Proof. The proof is by painful example. Here is one (half of a) procedure that works for moving the hydra from 0 to 6.
-1 0 1 2 3 4 5 6 7 0 1 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 1 0 1 0 0 0 0 0 1 1 1 0 1 0 0 0 0 0 2 0 0 1 0 0 0 0 0 2 0 1 0 1 0 0 0 0 2 0 1 1 0 1 0 0 0 2 1 0 2 0 1 0 0 0 2 1 1 1 1 1 0 0 0 2 0 2 0 1 1 0 0 0 1 1 1 0 1 1 0 0 0 0 2 0 0 1 1 0 0 0 0 2 0 1 0 2 0 0 The last row is symmetrical with respect to the midpoint between 0 and 6, so by sym-metrically reversing the steps we have done, we will end up with a single hydra head in room incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 51 94 6. Inductively, this implies that all multiples of 6 can be reached.
■ Remarks: Here are other interesting approaches that work: • One could have used the weights wn := ωn, where ω is a primitive 6th root of unity.
Then the sum of all weights is always equal to 1, due to the miraculous identity ωn−1 + ωn+1 = ωn, which can be easily verified “visually”. Of course, this entails that the hydra can only remain in room numbers that are multiples of 6.
• For any particular moment in time, let an be the number of hydras in room n. Then the distribution of hydras may be represented as a “polynomial” P(x) = X n∈Z anxn.
At the start, we have P(x) = 1. Then, every move adds or subtracts a multiple of x2 −x + 1.
Thus x2 −x + 1 divides P(x) −1 (or rather, x2 −x + 1 divides the numerator of P(x) −1 when expressed as an irreducible rational function). It follows that P(eiπ/3) = 1 at all times.
If all hydra heads end up in some room k, then P will take the form P(x) = akxk. So akekiπ/3 = 1. This is possible only when ak = 1 and k is a multiple of 6.
Source: Andreas Blass, Seven Trees in One [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 52 95 Solution 52 Cut T along the inradii. “Swapping” the two orange and purple pieces below will execute the desired reflection.
■ Remarks: Cutting T along the circumradii does not work if T is obtuse! Unfortunately, this approach cannot be salvaged as far as I’m aware.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 53 96 Solution 53 It turns out that such a pair of dice exists. In fact, there is only one such pair of dice!
A 6-sided die whose sides are a, b, c, d, e, f may be represented as the polynomial xa +xb + xc + xd + xe + xf, in the sense that the coefficient of xn is the number of ways that n may be achieved. This sort of property is preserved if we consider multiple dice by multiplying the corresponding polynomials!
For instance, the standard 6-sided die is represented as x + x2 + x3 + x4 + x5 + x6. Thus, the possible outcomes for rolling two standard 6-sided dice is represented by the polynomial (x + x2 + x3 + x4 + x5 + x6)2. Our goal is to write this polynomial as the product of two other polynomials P(x) and Q(x), each with positive coefficients summing to 6 (and with no constant term since we want positive integer sides for the dice). Manipulating: (x + x2 + x3 + x4 + x5 + x6)2 = x2(x6 −1)2 (x −1)2 = x2(x −1)2(x2 + x + 1)2(x + 1)2(x2 −x −1)2 (x −1)2 = x2(x2 + x + 1)2(x + 1)2(x2 −x + 1)2 To ensure that P and Q have no constant term, they should each get a factor of x. Next, if we plug in 1 into the above expression, we get something that looks like (12)(32)(22)(12).
This tells us that each of P and Q need to take a factor of x2 + x + 1 and a factor of x + 1 in order to have a coefficient sum of 6. It remains to donate the two (x2 −x + 1) factors. But if we divide those evenly then we just end up with the original dice. So the only reasonable distribution of the factors is as follows: P(x) = x(x + 1)(x2 + x + 1) Q(x) = x(x + 1)(x2 + x + 1)(x2 −x + 1)2 Expanding using Mathematica or something, this gives the generating functions: x4 + x3 + x3 + x2 + x2 + x1 x8 + x6 + x5 + x4 + x3 + x1 Magically, these satisfy the conditions we need! Hence there indeed (uniquely) exists another such pair of dice, and their sides are 1, 2, 2, 3, 3, 4 and 1, 3, 4, 5, 6, 8.
■ Source: This pair of dice is known as the Sichermann Dice.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 54 97 Solution 54 If we imagine walking along the sides of the tridecagon, then our movement along each side may be represented as a vector, and these 13 vectors sum to 0. We may enforce that the angles of the tridecagon are multiples of 20 degrees by taking the vectors to be 18th roots of unity.
To ensure convexity, the order in which we arrange the roots of unity to form the tridecagon must be by increasing or decreasing argument. Moreover, we cannot use a root of unity twice, otherwise this scheme would force the two repetitions of the root of unity to be adjacent, and hence forming a single side rather than two.
So, the problem reduces to finding a 13 distinct 18th roots of unity that sum to 0. This further reduces to just finding 5 such roots of unity that sum to 0.
The claim is that if we can find 5 such roots of unity, then two of them sum to 0 and the other three also sum to 0 (and thus form an equilateral triangle in the complex plane). By inspection, this would indeed imply that our desired tridecagon is unique up to similarity.
Let ζ be a primitive 18th root of unity, and suppose that ζe1 + ζe2 + ζe3 + ζe4 + ζe5 = 0 for distinct integer powers 0 ≤ei < 18. If we assume that e5 = 0, then it suffices to prove that either 9 ∈{e1, e2, e3, e4} or 6, 12 ∈{e1, e2, e3, e4} (why?).
Let P(x) = xe1 +xe2 +xe3 +xe4 +1. Then P ∈Q[x] and ζ is a root of P, so the cyclotomic polynomial Φ18 = x6 −x3 + 1 divides P. Let P/Φ18 = Q.
Let Q = Q0 + Q1 + Q2, where Qi is the polynomial formed by the terms of Q whose exponents are congruent to i mod 3. Similarly let P = P0 + P1 + P2. Then Φ18Q0 + Φ18Q1 + Φ18Q2 = P0 + P1 + P2.
The terms on the LHS whose exponents are multiples of 3 are precisely those terms in the polynomial Φ18Q0 because Φ18’s terms all have degrees that are multiples of 3. Thus Φ18Q0 = P0. Similarly, Φ18Q1 = P1 and Φ18Q2 = P2.
This tells us that P0(ζ) = P1(ζ) = P2(ζ) = 0. Translating back to English, what this means is that among the 5 roots of unity we’ve chosen to sum to 0, • the ones that are of the form ζ3k sum to 0, • the ones that are of the form ζ3k+1 sum to 0, and • the ones that are of the form ζ3k+2 sum to 0.
The five roots of unity are divided among these three “classes” of roots.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 54 98 To finish, note that none of these classes may have all five of the chosen roots, because then these roots are five vertices of a regular hexagon, and hence have no hope of summing to 0. And, trivially, none of these classes may have exactly one of the roots. It follows that the only possible distribution for the roots is “0, 2, 3”, in some order. If you think hard, this is exactly what we wanted to show.
■ A remark from “tenth”: If you want to show that x6 −x3 + 1 is irreducible for the sake of lowering the amount of “technology” used in this proof, then here you go!
The observation to make is that, over the field F3, we amusingly have that x6 −x3 + 1 = (x + 1)6. This motivates looking at the shifted polynomial (x −1)6 −(x −1)3 + 1. If this is irreducible, then x6 −x3 + 1 must be too.
By the observation, all coefficients of this shifted polynomial are divisible by 3 (sans the leading coefficient). Also it is not hard to see that the constant term is not divisible by 9.
Thus, this polynomial is irreducible by Eisenstein’s Criterion. Voil` a!
Source: Math Prize for Girls [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 55 99 Solution 55 Get a paper plate of the same radius as the original plate. When you fold the rectangular paper, fold the paper plate along the same crease!
The folded rectangular paper clearly still fits on the folded paper plate. Moreover, the folded paper plate must fit on the original plate. By “transitivity of fitting”, the folded rectangular paper still fits on the original plate.
■ Remarks: By the exact same reasoning, the problem still holds when the paper is non-rectangular!
Source: Art of Problem Solving Forums [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 56 100 Solution 56 Here is a clever approach that minimizes computation. The expected fraction is given by the integral R 1 0 x 1 x dx. To evaluate this slickly, we use the crazy identity x 1 x = ∞ X n=1 x · 1 0, 1 n i(x).
This is because of the motto “Baka takes a bite, and if x ≤1/2 then Baka takes another bite, and if x ≤1/3 then Baka takes another bite, ...” and so on.
Now integrate and apply monotone convergence: Z 1 0 x 1 x dx = Z 1 0 ∞ X n=1 x · 1 0, 1 n i(x) dx = ∞ X n=1 Z 1 0 x · 1 0, 1 n i(x) dx = ∞ X n=1 Z 1/n 0 x dx = ∞ X n=1 1 2n2 = π2 12 ■ Remarks: What a crazy answer from a numberless problem!
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 57 101 Solution 57 The plan is “simple”: If n ships have been bombed so far, then the Queen should board the ship with probability pn that we shall choose later.
We will construct the sequence {pn}∞ n=0 so as to ensure that the probability of survival is at least 1 −ε.
Seeing this plan, suppose that the Insurrection has b bombs and chooses to bomb ships x1 < x2 < . . . < xb ∈N. The probability that this kills the Queen is given by: (1 −p0)x1−1p0 + (1 −p0)x1(1 −p1)x2−x1−1p1 +(1 −p0)x1(1 −p1)x2−x1(1 −p2)x3−x2−1p2 + . . .
≤p0 + p1 + p2 + . . .
Taking p0 = ε2, p1 = ε3, p2 = ε4, · · · we get that P(Queen Dies) ≤ ε2 1 −ε < ε, so indeed the Queen gets out alive with probability at least 1 −ε.
■ [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 58 102 Solution 58 Suppose the mirror were a distance L away from Cherie. (We will see that L does not matter!) Let the mirror have infinite height. The minimum height that the mirror could be is given by the height spanned by Cherie’s perceived image in the mirror.
Now replace the mirror with glass and place a doppelganger of Cherie across the glass at distance L.
By similar triangles, we can compute the height spanned by Cherie’s image to be exactly half her real height.
■ Source: Me [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 59 103 Solution 59 Note that a parabola cannot cover all but a finite length of a given ray unless the ray is parallel to the axis of symmetry (This follows from a rate of growth argument; x2 has higher order of growth than any linear function ax + b.).
Find an infinite set of rays, none of which are parallel to each other. A finite covering of the plane must cover these rays.
For each ray there must exist a parabola whose axis of symmetry is parallel to it. Other-wise every parabola will only intersect at most a finite length of the ray, which is bad because rays are infinite and we only have a finite number of parabolas.
But no two rays are parallel, so no parabola can have axis of symmetry parallel to more than one ray. It follows that there are at least as many parabolas as rays. Contradiction, because there are infinitely many rays.
■ Source: VJIMC [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 60 104 Solution 60 This solution is a highly visual one. Let us start by discovering what precisely is going on behind the scenes.
Let’s start with one flip.
Now, let us keep flipping until the next piece we flip overlaps with the first piece that we flipped.
From this point forward, we need to be more careful. As we keep going, any black/gray line incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 60 105 represents a cut in the cake as it would in real life. Without further ado, let us do another flip, but slowly!
Left: I’ve outlined in purple the slice that we’re about to flip.
Center: I’ve executed the flip.
Right: The outline is removed.
Notice that we not only inverted the colors. We also had to reflect the piece! That’s what it means to flip a slice of cake. Consequently, the very first cut we’ve ever made has moved as a consequence of this flip.
Let’s keep doing this for a bit so you can see the pattern.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 60 106 We have arrived at yet another critical point. What happens now?
Here’s the big revelation: After this point, no additional cuts in the cake are ever made! We’re now just taking existing pieces (two at a time) and flipping them over. See for yourself!
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 60 107 At last, we can piece together the story of what’s really going on: There are always two “small pieces” together, and each flip “moves” one of those small pieces to be next to the next small piece, via flipping over that small piece with the adjacent “big piece”.
With this insight in hand, we may now proceed to form a proof.
Let’s start from the beginning, with all the cuts we’re ever going to make already filled in. I’ll also number the slices.
Instead of doing a flip, we are going to do a flip and a rotation, such that the two small slices are always at the top of the cake.
This sequence of moves executes a permutation on the slices! Since a permutation on a finite set must have a finite order, we see that by repeating this permutation over and over again, we eventually must have all the pieces end up where they started.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 60 108 But are the orientations (i.e. the colors) of the pieces correct? They might not be. If so, then the long sequence of moves thus far simply executes an algorithm that flips some of the slices. Hence, by executing repeating all those moves just once more, we will flip those same slices, thereby ending with a cake whose top has returned to its original color.
■ [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 61 109 Solution 61 Suppose that every initial configuration of lights may be solved. Then there exists a func-tion f from the set of initial configurations (which has size 216) to the set of possible solutions (which, as in Hint 1, has size 216) for which f(x) is a solution to the initial configuration x.
But f must be an injection, because the same solution cannot solve two different boards.
We deduce that actually f is a bijection. This implies that no initial board can be solved in two different ways.
We arrive at a contradiction by looking at the all-lights-off board, in which one solution is to do nothing, and another solution is to press the following lights: ■ Remarks: There is a much more remarkable fact about the 4×4 standard Lights Out puzzle.
Theorem 1 (Chasing Lights) Suppose that a given 4 × 4 Lights Out puzzle has a solution. Then the puzzle may be solved by the following naive process: 1. Start with the first row.
2. For each light that is on in this row, press the light below it.
3. Repeat Step 1 for the next row.
There is a natural proof that involves some linear algebra. Here is a more elementary one.
Proof.
Take such a solvable puzzle. It suffices to show that it has a solution that does not involve pressing any lights in the top row. This is because if no lights are to be pressed in the first row, then the lights that are pressed in the second row of the solution must precisely be those lights that are under a light that is on in the first row. After pressing those lights, incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 61 110 the solution no longer involves pressing any lights in the second row, and we repeat this reasoning for each subsequent row.
Take a solution. If the solution involves pressing the first light in the first row, then add the following light-presses to the solution: We can do this because pressing these lights does not ultimately change the state of any of the lights.
Likewise, if the solution involves pressing the second light in the first row, then add the following light presses to the solution: The cases in which the solution involves pressing the third or fourth light in the first row are handled symmetrically. Hence we can obtain a solution that does not press any light in the first row.
□ The theory shown here is merely the tip of the iceberg. For more, I highly recommend the article Two Reflected Analyses of Lights Out by ´ Oscar Mart´ ın and Crist´ obal Pareja-Flores.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 62 111 Solution 62 There are two cases to consider: When n is odd, and when n is even.
Case 1: n is odd We proceed via a sort of “induction”. Clearly, if all lights are off at the start, then the puzzle is solved in an easy run by virtue of being already solved. This is the “base case”.
Now for the “inductive step”, we show that if some solvable puzzle is solved in an easy run, then this property is preserved after pressing any light. Since every solvable puzzle can be obtained by starting from an empty board and pressing a finite sequence of lights, this would solve the problem.
We say that two distinct lights are neighboring or neighbors if they are in the same row or the same column.
Consider a puzzle that is solved in one easy run. Then all its lights are turned off, so every light has an even number of neighbors that are on (regardless of whether that light is on or off!). Now press a light L. We need only show that all lights still satisfy this property.
Take a light K distinct from L.
• If K is not a neighbor of L, then exactly two neighbors of K are toggled, so its number of neighbors that are on has changed by either −2, 0, or 2. Hence the number of such neighbors remains even.
• If K is a neighbor of L, then n −1 of its neighbors are toggled. Since n −1 is even, the parity of the number of neighbors that are on must remain even.
As for L itself, it has 2n −2 neighbors, all of which are toggled after pressing L, and 2n −2 is evidently even.
Case 2: n is even There are multiple approaches to this case. Here, I present the one that I believe to be the most interesting.
The first observation is that every initial configuration of lights constitutes a solvable puzzle. This is because we can toggle any individual light L by pressing L and all neighbors of L. By reasoning as in the previous problem, it follows that every puzzle has a unique solution.
Incidentally, the pattern of lights that need to be pressed to toggle a light L is the same pattern of lights that are toggled when L is pressed. This motivates an interesting notion incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 62 112 of duality: For a board of lights X, let X′ be the board of lights where a light is on iff it is one of the lights pressed in the solution to X. Then X and X′ have the same effect on each other: Toggling a light on X′ corresponds to pressing the corresponding light on X, and toggling a light on X corresponds to pressing the corresponding light on X′!
This symmetry between the board X and its dual board, X′, presents several miracles: • By definition, the on-lights in X′ constitute the solution to X.
But likewise, the on-lights in X constitute the solution to X′.
• Duality is an involution: We have that (X′)′ = X. This follows by the previous bullet and the uniqueness of solutions.
• Duality is linear: For boards X and Y , we have that X′ + Y ′ = (X + Y )′. Here, addition is done in the sense that “On” is 1 and “Off” is 0, over the field F2 (so that 1 + 1 = 0).
Now let us exploit this symmetry for an elegant proof. For a board X, let E(X) be the board obtained after an easy run. We need to show that E(E(X)) = 0, where 0 is the all-off board. But a board is solved in one easy run exactly when it is equal to its dual (i.e.
E(Y ) = 0 ⇐ ⇒Y = Y ′), so it suffices to show that E(X) = E(X)′.
Intuitively speaking, the argument is as follows: An easy run on X simply changes the solution X′ by adding X to it, and by duality we can argue that it also changes the original board by adding X′ to it. More precisely, we first observe that E(X)′ = X′ + X.
(∗) That is, the solution to the board after an easy run is changed by an addition of X, which is evident because an easy run consist of pressing those lights that are on in X, and pressing each such light corresponds to a simple toggle in the dual.
Now, we take the dual of each side of (∗) to obtain E(X) = X + X′, where we have applied the fact that duality is an involution and is linear. But now we are done since E(X)′ = X′ + X = X + X′ = E(X).
■ Remarks: This was the main result in a research project I did in 10th grade. It turns out that the result itself is more well-known than I thought it was back then, but I still quite like the ideas behind the proof.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 62 113 There’s more, though! Going back to the standard Lights Out rules, where pressing a light toggles only that light and its orthogonally-adjacent neighbors, we can still define the notion of an easy run. It turns out that in this case, easy runs still have the potential for solving puzzles. To be specific: In 2011, Bruce Torrence proved that if n is such that an n × n standard Lights Out puzzle can be solved for every possible initial configuration of lights, then repeatedly applying easy runs will eventually solve any puzzle.
Source: Me [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 63 114 Solution 63 We proceed by induction. The case n = 1 is kinda easy. Now suppose that we may solve the all-on Lights Out for n −1 lights.
Fix a light L. Then by the hypothesis, we may press some lights so that all lights other than L are toggled off. If this process toggles L off, then we already win. Thus we may assume that the lights we pressed ended up toggling only those n −1 lights besides L.
The same logic may be applied to all other lights, so we may assume that we have the power to toggle any n −1 lights of our choice.
Claim 1: We may assume that n is odd.
This is because if n is even, then we may toggle all n subsets of n −1 lights. This toggles every light n −1 times, which is odd, so this constitutes a solution.
Claim 2: We have the power to toggle any two lights of our choice at once.
Suppose the two lights we’d like to toggle are L and K. Toggle all lights except L, and then toggle all lights except K. Tada!
Claim 3: There is a vertex with even degree.
If all vertices have odd degree, then the sum of the degrees is odd (because n is odd by the assumption permitted by Claim 1). But the sum of the degrees is twice the number of edges by the handshake lemma, and hence must be even, contradiction.
Now we may use Claims 2 and 3 to finish. Start with all lights on.
By Claim 3, find a vertex/light L with even degree.
Press it.
This turns L and all neighbors of L off. L has an even number of neighboring lights, so by Claim 2, we are able to toggle all of L’s neighbors back to on. This means that L is off whereas all the other n−1 lights are on. As we established at the beginning, we may toggle all those n −1 lights off.
■ [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 64 115 Solution 64 Instead of ants turning around upon collision, it is equivalent to let them pass through each other.
Part (a) Under this new framing, it is clear that 22 ants fall off the left end, whereas 20 ants all off the right end.
Part (b) Under this new framing, every ant on the left passes through (i.e. “collides”) with every ant on the right (though, this isn’t true under the old framing). Thus there are 20×22 = 440 “collisions” in total.
■ Remarks: Can you figure out which ant(s) endure the most collisions?
Source: Classic [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 65 116 Solution 65 Draw a line L that intersects that graph at three points A, B, C. The x-coordinates a, b, c of these points must be the roots of the cubic polynomial x3 −(mx+n), where mx+n is the equation of the line L. By Vieta’s Formulae, it follows that a + b + c = 0. Thus the center of mass of the three points A, B, and C lies on the y-axis! Construct said center of mass.
Repeating this procedure, we now have two points on the y-axis, and so by connecting them we will have constructed the y-axis. Construction of the x-axis quickly follows.
To construct the center of mass, one way is as follows: Since you can construct paral-lelograms, you can definitely construct A + B + C, where the “0 vector” can be anywhere you want. By scaling, you can then construct A+B+C 3 , which must be the center of mass no matter where you place the “0 vector”.
■ Source: Heard this from an internet acquaintance, who in turn saw this on Reddit.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 66 117 Solution 66 Invent a second king that’s trying to get from the left side to the right side. The second king moves like the first king, but steps only on burning squares.
We claim that the first king can reach the top side if and only if the second king cannot reach the right side. If this is true, then we can conclude by symmetry that the probability is exactly 50%.
To prove this, we first modify the chessboard by shifting each row so that both king’s movements consist of simply moving to an adjacent square.
Red: Square that is on fire Viewing the board as such, the proof becomes quite simple.
Suppose that the first king can reach the top edge. Then his path clearly blocks the second king from reaching the right edge.
Conversely, suppose that the second kind cannot reach the right edge. Then the first king can find a path to the top edge by following the boundary of the set of all squares that can be reached by the second king.
■ Remarks: What we have done was reduce the problem to the game of Hex, which is played on a board of hexagons.
The Hex Theorem states that a draw is impossible in the game of Hex. A fascinating fact is that this theorem is equivalent to Brouwer’s Fixed Point theorem!
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 67 118 Solution 67 If none of the numbers are lying, then any three numbers that lie in three different rows and columns must sum to the perimeter of the outer rectangle. Note that in the following two cases, the sum of the indicated numbers is the same, and equal to 42: 14 16 12 18 14 10 16 18 14 14 16 12 18 14 10 15 18 14 If any of those green numbers were lying, then these sums wouldn’t agree. So they’re all telling the truth, and moreover the outer perimeter is 42.
The liar is either the 12 or one of the 18s. To narrow it down, sum these three numbers: 14 16 12 18 14 10 15 18 14 These three sum to 48. Treasonous! We know that the green 16 and green 14 are truthful, thus the 18 marked in red is the liar, and really should be 12.
■ Source: Georgia Southern Math Tournament [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 68 119 Solution 68 Booster can find Mario, guaranteed.
The core mechanism behind Booster’s strategy is the following claim: Claim: If Mario is behind curtain n, and Booster opens curtains m, m−1, m−2, ..., 1 for some m > n with the same parity as n, then Booster will find Mario.
Proof.
Act out the procedure with your fingers until you are convinced.
□ From this, we get the next claim.
Claim: For any positive integer N, if it is assumed that Mario is behind one of the first N curtains, then Booster has an algorithm to catch Mario.
Proof.
First, Booster guesses that Mario is on an even parity. Then Booster opens every curtain from 2N to 1 (the choice of 2N is extremely sub-optimal). This would find Mario if he’s indeed on an even parity. If Booster does not find Mario, then we now know that Mario was on an odd parity. In fact, he still is. More specifically, Mario must be behind an odd curtain between 1 and N + 2N = 3N. So if Booster opens every curtain from 4N + 1 to 1, then he will find Mario.
If Booster still doesn’t find Mario, then Booster has verified that Mario was not in the first N curtains.
□ Now we construct Booster’s algorithm.
0. Let N = 1. Let C be the number of curtains opened thus far.
1. Booster checks if Mario started within the first N curtains by assuming that Mario is currently within the first N + C curtains, and using the claim to test this hypothesis.
2. If Mario is not found, we increment N, update the value of C, and loop back to Step 1.
Note that Step 1 works because if Booster has opened C curtains so far, then Mario must be within the first N + C curtains, assuming that Mario started within the first N curtains.
Eventually, N will equal Mario’s starting curtain, and then on that loop of the algorithm, Booster will catch Mario. It just might take a very long time.
■ incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 68 120 Remarks: As per the hint, Booster’s strategy can be viewed as constructing increasingly-long diagonal “barriers” on an infinite grid. One of these diagonals must “catch” Mario, who is descending downwards diagonally.
Red: Mario’s position Black: Curtain checked by Booster [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 69 121 Solution 69 Let us instead suppose that A and B are only just out of reach — say, at most 1.5 inches apart.
A B 1 in Now, since A and B are not too far apart, we may identify two points C and D that are both within one inch of both A and B. (Also, we must ensure that C and D are within 1 inch.) A B C D 1 in Next, we extend ray AC slightly to a point E. Likewise we extend ray BD slightly to a point F. (We must ensure that E and F are within 1 inch.) A B C D E F 1 in incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 69 122 If the steps thus far were followed sufficiently delicately, then our straightedge will be long enough to connect C with F and E with D. We label their intersections with AD and CB respectively as X and Y , respectively.
A B C D E F X Y 1 in Lastly, we mark the intersection of XY and EF as P.
A B C D E F X Y P 1 in As suggested above, we claim that P lies on AB, so that we may connect A with P and P with B to construct the line segment between A and B.
This follows from Pappus’s Theorem from projective geometry. To be specific, let Z be the intersection of AB and EF. Then Pappus tells us that the points X = AD ∩CF, Y = CB ∩ED, and Z = AB ∩EF are collinear. This implies that AB, EF, and XY are concurrent at the point Z. Hence, in fact, Z = P, so P is collinear with A and B.
Our work thus far gives a scheme for connecting two points that are just out of reach — say, at most 1.5 inches apart. Now we may finish with absurdity: If we can connect any two points that are at most 1.5 inches apart, then this means that we can simulate a 1.5-inch straightedge by using a 1-inch straightedge. Scaling up the argument, it follows that we can use a 1.5-inch straightedge to simulate a (1.5)2-inch straightedge. So our 1-inch straightedge incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 69 123 can simulate a (1.5)2-inch straightedge. Inductively, we deduce that we can simulate a (1.5)n-inch straightedge for all positive integers n. That is, arbitrarily large straightedges can be simulated. In particular, we must be able to connect any two points in the plane, no matter their distance.
■ Remarks: If you enjoyed that, here is a variant. You are on a plane and have been tasked with drawing the ray − → AB until it hits a point C far into the distance using your straightedge (which, in this problem, is as long as you would like). Unfortunately, about halfway through, you have come to a standstill: A sleeping cat!
This is a serious problem. If you were to continue extending this ray, your pen would touch the cat and wake it up, which is unacceptable. Can you find a way to continue the ray past the cat? That is, using only your straightedge, can you construct the rest of the ray’s extension (sans the area around the cat)?
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 70 124 Solution 70 Emily wins (provided that the table can fit at least one quarter).
She starts by placing a quarter right in the center of the table. Then, if Sydney places a quarter centered at a point P, then Emily will place a quarter at the reflection of P about the center. It is clear that whenever Sydney can move, Emily must be able to execute her move as well, so Emily cannot lose. So she has to win because the game eventually ends.
■ [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 71 125 Solution 71 Let the integral be I, so that I = Z ∞ 0 e−x2−1/x2 dx.
Then, by substituting x with 1/x, we get I = Z ∞ 0 1 x2e−x2−1/x2 dx.
Now add the above two equalities to obtain 2I = Z ∞ 0 1 + 1 x2 e−x2−1/x2 dx.
At first this looks dumb. But if we rewrite the integrand as 2I = Z ∞ 0 1 + 1 x2 e−(x−1/x)2−2 dx, then miraculously we see that the u-substitution u = x−1/x is applicable because du dx = 1+ 1 x2!
Now we have that 2I = Z ∞ −∞ e−u2−2 du = √π e2 by the Gaussian integral (note the new limits on the integral). Thus I = √π 2e2 .
■ Remarks: This is also a textbook application of the miraculous Glasser’s Master Theorem.
Theorem 1 (Glasser’s Master Theorem) Suppose that f : R →R is integrable. (That is, R ∞ −∞|f(x)| dx < ∞.) Then Z ∞ −∞ f x −1 x dx = Z ∞ −∞ f(x) dx.
Essentially, if the integrand is a function of x −1 x, then x −1 x can be replaced with x without any fuss.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 72 126 Solution 72 Define a rational ball to be a ball (i.e. a circle) whose radius is rational and whose center has rational coordinates. Evidently, there exist countably many rational balls.
For each “Y-set”, draw three pairwise-disjoint rational balls containing its endpoints, such that each of the balls does not intersect either of the other two “branches” of the Y-set.
Define the map f which sends each “Y-set” to the 3-tuple of rational balls around its endpoints (in any order). Evidently, the number of such tuples is countable.
As suggested in the hints, we were motivated to try and construct an injection from the collection of Y-sets to a countable set, and the initial hope is that f is the desired injection. Unfortunately, f is not an injection. There could exist distinct Y-sets x, y for which f(x) = f(y), as shown below.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 72 127 However, I claim that we can’t do better! That is, there do not exist distinct Y-sets x, y, z for which f(x) = f(y) = f(z). That is, there do not exist three distinct Y-sets whose endpoints circles are identical. Proving this is sufficient for showing that the domain of f is at most countable, which is what we want to show.
To see this, suppose that we have indeed found three such distinct Y-sets.
Inside each ball, erase all parts of the Y-sets, and replace those parts with radii to the center.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 72 128 The condition that each ball does not intersect any of the other two branches ensures that after this operation, the three sets are still homeomorphic to the letter Y.
Now, place a house on the center of each circle and a utility at the “3-way crossing point” of each Y-set.
By the assumption that none of the Y-sets cross each other (i.e. are disjoint), we see that we have constructed a solution to the three utilities problem (!!!), contradiction.
■ [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 73 129 Solution 73 The shortest distance between two points is a straight line, but there is no straight line between a cube’s opposite vertices that does not exit the cube’s surface. Fortunately, this issue can be solved by unfolding the cube.
No matter how you unfold the cube, the ant will be at least a “knight’s move” away from the opposite vertex, and this distance is √ 5.
■ [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 74 130 Solution 74 We begin by computing the shortest distance to the opposite vertex.
Let us unfold the box as below, where some faces of the box are repeated.
The ant begins at the red point and seeks to reach the vertex opposite this point, which can be represented by any of six green points in the above unfolding, as depicted.
With some thought, it is not hard to see that the six paths above (drawn with dashed lines) are the only six sensible paths to the opposite vertex, and the shortest paths among these have length 2 √ 2. This is the shortest distance to the opposite vertex.
This is crucial for our arguments, as we can notice that all points on the four 1 × 2 faces are with 2 √ 2 of the ant’s starting position.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 74 131 We deduce that the point on the surface that is farthest from the ant must lie somewhere on the 1 × 1 face opposite the ant.
Let us now unfold the box in a different way, as below.
P A B C D E F incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 74 132 Here, it is the ant’s starting vertex which is being duplicated (with six different labels from A to F), and the purple point P is an arbitrary point on the opposite 1 × 1 face. As before, there are six sane paths that the ant could choose to travel to P, as shown. We seek to select P so that the shortest of these paths is maximized in length.
To that end, we make two observations: 1. The distances from A and F to P in the above net will always be strictly greater than the distances from B and E to P, respectively. So we may disregard the paths from A and F.
2. If the point P maximizes the shortest path, then it must lie on the blue segment.
Otherwise, the shortest path may be increased by moving P slightly towards the blue segment.
From this, we find that the point P that maximizes the distance is given by the intersection of the blue segment and the perpendicular bisector of segment BC.
P A B C D E F Working out the coordinate geometry or otherwise, we find that P must be a quarter of the incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 74 133 way up the blue segment, and its distance from any of B, C, D, or E will be √ 130 4 .
■ Remarks: • √ 130 4 is indeed larger than the distance to the opposite vertex, 2 √ 2, but just barely — the difference is about 0.022.
• For a 1 × 1 × 1 box, the farthest point is the opposite vertex. For a 1 × 1 × 2 box, this is not so. Is there a critical value 1 ≤A ≤2 for which the farthest point for a 1×1×A box switches from being the opposite vertex to a point strictly on the opposite face?
Yes, there is, and this value is A = 3+ √ 17 4 .
Source: This is called Kotani’s Ant.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 75 134 Solution 75 You cannot stop me from assuming that P is monic, so let us do so. Suppose that the roots of P are r1, · · · , rn, so that we may write P(x) = n Y i=1 (x −ri) by the Fundamental Theorem of Algebra. Then by the product rule, we have that P ′(x) = n X i=1 P(x) x −ri .
Let z be a root of P ′(x), so that n X i=1 P(z) z −ri = 0.
Either z is a root of P (in which there is nothing to prove), or we may divide each side by P(z) to obtain n X i=1 1 z −ri = 0.
Now take the conjugate of each side, n X i=1 1 z −ri = 0 and rationalize (...realize?) the fractions by multiplying the top and bottom of each by (z −ri) to obtain n X i=1 z −ri |z −ri|2 = 0.
This now rearranges to z = n X i=1 1/|z −ri|2 Pn j=1 1 |z−rj|2 !
ri, which implies that z is a convex combination of the {ri} because n X i=1 1/|z −ri|2 Pn j=1 1 |z−rj|2 = 1.
■ Source: This is the Gauss-Lucas Theorem.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 76 135 Solution 76 This will be a “proof by demonstration” because I cannot be bothered to formalize this.
Let us suppose that this is our polygon.
The key idea is to draw the segments obtained by “sliding down three sides”.
The dashed segments partition the top region into parallelograms. As for the bottom region, it is a smaller polygon with 180-degree rotational symmetry, so we may induct down on the number of sides to partition it into parallelograms.
■ [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 77 136 Solution 77 Part (a) Instead of Heads/Tails, use 0/1 for the sides of the fair coin.
Let a binary expansion of p be p = (0.d1d2d3 · · · )2. We keep flipping our fair coin until the binary sequence generated by our results is different from d1, d2, d3, · · · .
At any step, we will stop flipping with 50% odds, so we must eventually stop flipping the coin almost surely (i.e. with probability 1). We claim that our final flip is 0 (i.e. Heads) with probability p.
Indeed, our final flip is 0 iff we stopped at the kth flip with dk = 1. So the probability is given by X k:dk=1 1 2k since there is a probability of 1/2k that we stopped at the kth flip. By definition of binary expansion, this sum is exactly p.
Part (b) Flip the coin twice. If the results are the same, then flip the coin twice again. Keep doing this until the results are different, which will eventually happen with probability 1. We claim that when this happens, these two flips are equally likely to be HT or TH. Indeed, this is true because HT and TH are equally likely to occur among any two flips, so conditioned on the event that we got either HT or TH, the odds of getting either are 50 : 50.
■ Remarks: By combining both parts, we see that one can simulate any biased coin using any biased coin, as long as all probabilities are strictly between 0 and 1. Moreover, the simulation is possible even if the provided biased coin’s probability of flipping heads is unknown.
The problem also shows that you can simulate a fair coin by using a fair coin, but this is not as groundbreaking.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 78 137 Solution 78 Part (a) We claim that A > 4 is large enough.
First, some housekeeping: Suppose the side lengths of our collection of squares are given by a1, a2, · · · . We may assume countability, because the sum of uncountably many positive reals cannot be finite. We can also assume that ai < 1 for all i, otherwise this is dumb. Since P∞ i=1 a2 i = A > 4, we may find n so large that Pn i=1 a2 i > 4.
From the above, we see that we may replace “collection” with “finite collection” in the problem (at least, for proving that A > 4 works). We want this so that we may apply an inductive argument.
This also lets us strengthen the claim we are proving as follows: If the sum of the areas of a finite collection of squares is 4Ns2, and none of the squares has area greater than s2, then we may cover N squares of side length s with them. This would solve the problem.
It’s sufficient to prove the case for s = 1 by scaling.
Again, let the side lengths be a1, · · · , an, so that Pn i=1 a2 i > 4N. Suppose a1 is the largest sidelength. Find k ∈N for which 1 2k ≤a1 < 1 2k−1. Subdivide the N unit squares into N(2k)2 squares of side length 1/2k.
We’ll use the square with side length a1 to cover exactly one of these.
Now there are 4kN −1 more squares of side length 1/2k to cover. Moreover, the sum of the areas of the rest of the squares we may use for covering is n X i=2 a2 i > 4N −a2 i > 4N − 1 4k−1 = 4(4kN −1) 1 2k 2 .
Thus, by taking s = 1/2k in the claim we’re proving and using induction (the number of covering squares we may use has decreased), we can cover the rest.
Part (b) We can show that 0 < a < 1/4 is small enough by using the same argument as above, replacing > with < and replacing “cover” with “fit”.
■ [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 79 138 Solution 79 Magellan captains a ship A and orders some of his crew on ship B to follow him as he sails forward. Once they’re a quarter of the way along the equator, both ships A and B are halfway through supplies, so Magellan plunders the supplies of ship B so that ship A has full supplies.
Once Magellan is halfway across the equator, where he again is halfway through his supplies, he magically orders the third ship C to start sailing backwards, so that once Magellan is 3/4 done with his circumnavigation, he will rendezvous with ship C, which is halfway through their supplies. Since Magellan has no more supplies, he steals ship C’s supplies so that he finishes the circumnavigation.
■ [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 80 139 Solution 80 A nice candidate for such a function would be f(x) = ∞ X n=0 x3n (3n)!, but this is not in closed form.
The trick is that we may write f(x) as f(x) = ex + eωx + eω2x 3 where ω is a primitive third root of unity! To see why this works, expand ex, eωx, and eω2x into power series, add them up, and watch the magic.
To simplify, we may write f(x) = 1 3 ex + e −1 2 + √ 3 2 i x + e −1 2 − √ 3 2 i x = 1 3 ex + e−x/2 cis √ 3 2 x !
+ e−x/2 cis − √ 3 2 x !!
= 1 3 ex + 2e−x/2 cos √ 3 2 x !!
.
This is in a nice closed form! Also, one might notice that some of these terms here are useless, so for maximal simplicity we may take the function e−x/2 cos √ 3 2 x !
, and this works!
■ [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 81 140 Solution 81 Let pk be the kth prime. For all n ∈N, define Sn := {1/n} ∪ 1 n + 1 + 1 kpn : k ∈N, 1 n + 1 + 1 kpn < 1 n .
Let S = S∞ n=1 Sn. (Note that S is closed because it contains all its accumulation points, which all have the form 1/n.) Then, identifying the points on the boundary of a circle with the interval (0, 1], we define our convex set K to be the convex hull of S.
To construct a Venn-diagram with N copies of K, we take (for i = 1, 2, · · · , N) K(i) to be K “rotated back by 1/i”, so that the “1/i” point of K(i) lies on top of the “1” point of K. Analogously, let us define S(i) n := −1/i + Sn and S(i) := −1/i + S = S∞ n=1 S(i) n . We claim that K(2), K(3), · · · , K(N+1) forms the desired Venn diagram.
To see that our construction works, consider any subset A of {2, 3, · · · , N + 1}. We claim that there exists x ∈(0, 1] such that: • x ̸= 1/i for i = 2, 3, · · · , N + 1 (this ensures that x is not an accumulation point!) • x ∈S(i) for all i ∈A • x ̸∈S(i) for all i ̸∈A Let m = Q i∈A pi. Then we simply take x = 1 mk where k is large enough so that x < 1 N(N+1).
The bound x < 1 N(N+1) ensures that x ∈ 1 i , 1 i−1 −1 i for i = 2, · · · , N + 1.
Consider i ∈A. To see that x ∈S(i), we claim that x ∈S(i) i−1. Indeed, 1 i + 1 mk ∈Si−1 because mk is large enough to ensure 1 i + 1 mk < 1 i−1, and pi divides mk. Thus after rotating back we have 1 mk ∈S(i) i−1.
Consider i ̸∈A. Then x ̸∈S(i) because we may follow the reasoning above, and then note that pi does not divide mk.
We conclude that T i∈A K(i) ∩T i̸∈A(K(i))c is non-empty for all A. Hence we have formed a Venn diagram using N copies of K.
■ Remarks: The convex set we have constructed looks something like the figure on the next page.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 81 141 The red points are the points represented by {1/n : n ∈N}. As you may have noticed, not all of them are drawn due to a shortcoming of the algorithm used to generate the figure.
CMU Alumni Isaac Browne and Edward Hou have shown a stronger construction that solves an infinite version of the problem, that is, they found a sequence {Ki}i∈N of congruent open convex sets such that for any finite A ⊆N, the intersection T i∈A Ki ∩T i̸∈A Kc i is non-empty. This was proposed as Problem 12424 for the American Mathematical Monthly.
You can find a solution here (“A Universal Venn Diagram”).
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 82 142 Solution 82 ■ Remarks: What if Timmie is even pickier than usual, and doesn’t want any part of the crust, not even a point? It turns out that it is still possible to satisfy Timmie! It’s quite a bit harder though. See the paper for some ways to do it, and more!
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 83 143 Solution 83 The problem in question is one variant of the “Pizza Theorem”. We begin by proving a key lemma.
Lemma 1 Suppose that ABCD is an orthodiagonal, cyclic quadrilateral. Then AB2 + CD2 = BC2 + AD2 = 4R2, where R is the radius of the circumcircle of ABCD.
Proof.
A B C D A′ Swap chords AB and AD, so that A becomes A′. We claim that ∠A′BC is a right angle, so that A′C is a diameter. Indeed, observe that ∠A′BD = ∠ADB = ∠ACB, So ∠A′BC = ∠A′BD + ∠DBC = ∠ACB + ∠DBC = 90◦.
From this claim, we now have A′B2 + BC2 = A′C2 = 4R2 and A′D2 + CD2 = A′C2 = 4R2.
Substituting back A′B = AD and A′D = AB proves the desired equalities.
□ We now may resolve the original problem. Fix some arbitrary “right-angled cross”.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 83 144 A0 B0 C0 D0 X If we “rotate” the cross by some angle θ, then a certain amount of area is displaced.
A(θ) B(θ) C(θ) D(θ) A0 B0 C0 D0 X Note that if we can prove that the shaded area depends only on the radius R and the angle θ, then we are done!
We find the area by integration. Let A(t) be the point on the arc between A0 and A(θ) such that ∠A0XA(t) = t. So, A(0) = A0 and, well, A(θ) = A(θ)... Define B(t), C(t), and D(t) similarly.
By polar integration, the area of the “A”-shaded region is given by the integral Z θ 0 1 2XA(t)2 dt, where XA(t) denotes the length of the segment connecting X and A(t). Hence the total shaded area is 1 2 Z θ 0 XA(t)2 + XB(t)2 + XC(t)2 + XD(t)2 dt.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 83 145 But the segments XA(t), XB(t), XC(t), XD(t) are equally spaced, at angles of 90◦. Thus by the lemma, we have that XA(t)2 + XB(t)2 + XC(t)2 + XD(t)2 = 4R2.
Therefore the area is just 1 2 R θ 0 4R2 dt = 2R2θ. This indeed depends only on R and θ, proving the Pizza Theorem.
■ Remarks: A more “romantic” version of the Pizza Theorem (and, perhaps, the version that is more popular) goes as follows: Let N ≥2. If 2N cuts are made at equal angles through a point, dividing the pizza into 4N slices, then taking alternate slices will share the pizza among two people. More succinctly, the number of slices created in this manner must be one of 8, 12, 16, 20, · · · .
Unfortunately, this version of the Pizza Theorem is not implied by the version we have proven, as shown by the case depicted above with 12 slices (N = 3).
On the bright side, we can still apply the same calculus approach: It is sufficient to show that for any point X in a circle, if we draw 2N rays at equal angles emanating from X that hit the circumference at points A1, A2, · · · , A4N, then 2N X i=1 XA2 i is a constant in the sense that it does not depend on which 2N rays we draw.
The following beautiful approach is due to “tenth”.
Assume that the labeling of the points A1, A2, · · · , A2N is in counter-clockwise order.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 83 146 A1 A2 A3 A4 A5 A6 X M1 M2 M3 O We first pair up the terms in the sum as 2N X i=1 XA2 i = N X i=1 XA2 i + XA2 i+N.
The key idea is that XA2 i + XA2 i+N = (XAi −XAi+N)2 + 2(XAi)(XAi+N).
The product (XAi)(XAi+N) is the power of X with respect to the circle (to be precise, it is R2 −OX2), so it is a constant. As for (XAi −XAi+N)2, this is the square of the distance between X and the midpoint of AiAi+N, which we shall denote as Mi for i = 1, 2, · · · , N. It is now sufficient to prove that the sum N X i=1 XM 2 i is a constant.
The conclusion follows by two miracles.
Miracle 1: The midpoints M1, M2, · · · , MN form a regular N-gon! Proof: ∠OMiX = 90◦ for all i by virtue of Mi being the midpoint of a chord, so the points M1, M2, · · · , MN are concyclic with O and X. Then the regularity of polygon M1M2 · · · MN follows from the fact that the cuts were made at equal angles.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 83 147 Miracle 2: This lemma exists!
Lemma 2 Let X, M1, M2, · · · , MN be points. Then the sum N X i=1 XM 2 i depends only on the distance between X and the centroid of the points M1, M2, · · · , MN.
I leave it to the reader to verify this (...or perhaps this fact will come up in a different problem in this book, where it shall be proven?). Now, since M1M2 · · · MN is regular, its centroid coincides with its circumcenter, which is the midpoint of OX. So the sum N X i=1 XM 2 i depends only on the distance between X and this midpoint, which is a constant OX 2 .
This completes the proof. But what goes wrong when N = 1? The issue is that when N = 1, the centroid of the “regular 1-gon” M1 (which is M1 itself) no longer coincides with the circumcenter of △OXM1, due in part to the fact that a regular 1-gon does not have a well-defined circumcenter. A similar issue occurs for N = 2, but this case is handled just fine by the original problem.
Even More Remarks: The Pizza Theorem generalizes slightly to very poorly-made pizzas.
Consider, for example, the following pizza with D8 symmetry.
Amazingly, when we divide this pizza into 8 slices as shown, we once again obtain equal shares by alternating slices.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 83 148 To demonstrate this, we begin by drawing a circle centered at the center of this oddly-shaped pizza, large enough to contain the point through which the cuts intersect.
By the Pizza Theorem, the pizza contained inside the circle is equally shared. So it remains to show that the pizza outside the circle is equally shared. This is far easier to argue once we remove the inside of the circle.
The divisions of the remaining pizza consist of 8 segments from the circle to the crust. As we move a pair of opposite such segments towards the middle (such as the two blue segments above), we can see that neither color gains nor loses pizza. Doing this for every such pair, we may reach a configuration in which the cuts are perfectly symmetrical, so that we may conclude that, in fact, the amount of pizza of each color is equal.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 83 149 In general, if we include more cuts, then the shape of the pizza must exhibit an appropriate amount of symmetry in accordance to the version of Pizza Theorem that is applied.
Source: Pizza Theorem [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 84 150 Solution 84 The answer is yes. Here is a construction.
0 1 0 1 0 0 1 0 2 0 0 2 0 2 0 The numbers indicate the sizes of the slices. Realistically, Beth cannot make a slice of size 0, but she could make those slices have some negligible size such as 0.01 instead. This does not change the strategy.
Let us now describe Beth’s strategy. There are two cases, based on the first slice Allison chooses.
Case 1: Allison chooses a slice of size 0 If this is the case, then Beth two-colors the remaining slices in an alternating fashion.
0 1 0 1 0 1 0 2 0 0 2 0 2 0 She then determines which color contains more pizza than the other, and proceeds to take slices of only that color. This ensures that she gets all slices of said color, and Allison gets the slices of the other color.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 84 151 All slices are of integer size and the total size of the pizza is 9, so there necessarily will be one color that contains strictly more than half the pizza (to be precise, one color must have size at least 5).
Case 2: Allison chooses a slice of positive size 0 A 1 B 0 C 1 D 0 E 0 F 1 G 0 H 2 I 0 J 0 K 2 L 0 M 2 N 0 O Beth’s strategy here is more complicated. Refer to the above coloring of the pizza which partitions the pizza into three colored “wedges”. We also have labeled all 15 pieces with letters for convenience.
• Beth starts by taking the 0 slice that is adjacent to the edge of the colored wedge that Allison ate from. For example, if Allison ate slice I, then Beth takes slice J.
• From then on, there are three cases: – If they have finished eating one colored wedge, and the other two colored wedges are untouched, then Beth takes the slice from the wedge of smaller size.
For example, if the green wedge is all finished, and both the red and blue wedges are untouched, then Beth takes slice E.
– If Beth can avoid eating from an untouched wedge, then she will.
– If neither of the above points apply, Beth simply “copies” Allison by always taking the slice adjacent to the one that Allison picks.
The key idea is that this strategy ensures that Beth can claim two things: incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 84 152 • The largest wedge untouched by Allison’s first bite (either the blue or green wedge) • Either the other wedge untouched by Allison’s first bite, or the rest of the wedge that Allison first ate from.
This will always ensure that Beth gets at least 5 9 of the pizza!
Let us go through a quick example: If Allison starts with slice N, then Beth takes slice O. Beth can now guarantee that she will get slices G and I, and she will also either get slices B and D or the slice L. Indeed, if Allison takes M, then Beth can claim L, and if otherwise Allison wants to prevent Beth from claiming L, then Allison must take slices A and then C, conceding to Beth the B and D slices. In any case, Allison will eventually be forced to take F or J because it will be Allison’s turn once the red and blue wedges are finished, so Beth can guarantee herself all positive slices of the green wedge.
■ Remarks: 5 9 is the best Beth can ensure! This was conjectured by Peter Winkler and proven by Knauer, Micek, and Ueckerdt in 2011. See the following paper for the proof and more: [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 85 153 Solution 85 The map 1/x is a bijection on (0, ∞). Thus, the minimum of xx is the minimum of (1/x)1/x, which is equal to 1 x1/x. This minimum is equal to the reciprocal of the maximum value of x1/x. Thus the maximum value of x1/x is the reciprocal of the minimum of xx, which is 1/M.
■ Source: Shamelessly stolen from user “juliankuang” of AoPS , who allegedly came up with this in the shower.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 86 154 Solution 86 Suppose for contradiction that such a polygon existed. Imagine that I start on one of the vertices, and then traverse the perimeter until I come back to where I started.
Let L, R, U, and D be the number of units I move left, right, up, and down, respectively.
Since I came back to where I started, we have L = R and U = D. And, since the sides alternate between horizontal and vertical, we must have L + R = U + D. By some algebra, we may deduce from these facts that L = R = U = D.
Since L + R + U + D is the perimeter, we conclude that the perimeter is divisible by 4.
But 314 is not divisible by 4, contradiction.
■ Source: The Brilliant.org community. Rest in peace.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 87 155 Solution 87 My friends can indeed arrange themselves in such a way, and 15 friends is sufficient to do this.
The gray square is the original mirror room, the black dot is me, and the 15 dark blue squares are my friends. All other squares depict reflections of the mirror room, and the green rays are the possible lines of sight from me to a reflection of myself. Each such ray is blocked by a friend, as needed.
To be more precise: If we view the room as [0, 1]2 and I decide to stand at (a, b), then my friends will stand at the following locations: incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 87 156 • The four corners • (a, 0), (a, 1), (0, b), and (1, b) • (1 −a, 0), (1 −a, 1), (0, 1 −b), and (1, 1 −b) • (1 −a, b), (a, 1 −b), and (1 −a, 1 −b) Now, why does this work? First, note that the coordinates of any of my reflections will take the form (2m ± a, 2n ± b) for integers m and n. The key claim is that there will be a friend blocking my line of sight to (2m ± a, 2n ± b) at precisely the midpoint! (Or the midpoint is a reflection of me, in which case we induct downwards.) Indeed, the midpoint of the segment connecting (a, b) and (2m ± a, 2n ± b) is (m or m + a, n or n + b). This gives 16 cases since we also must consider the parities of m and n.
When all cases are reduced to an equivalent point inside the unit square, we obtain the 16 coordinates of me and and my 15 friends.
■ Source: The Leningrad Olympiad, but it is likely more famous.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 88 157 Solution 88 Me ■ Remarks: The least number of sides you can get for a counterexample is 8.
Probably.
Someone claimed a proof but didn’t give one. Maybe you can prove it!
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 89 158 Solution 89 We begin with an intuitive sketch for the process. Let A, B, C, D be the bottoms of the table’s legs, in counter-clockwise order.
1. Place the table somewhere so that the two opposite corners A and C touch the floor.
It’s alright if the B and D legs clip through the floor.
2. Pivot the table about AC until the B and D are the same vertical distance above/below the floor. Assume without loss of generality that they’re both above the floor after the pivoting, and let the points on the floor under B and D at this time be B′ and D′.
View these two points as fixed.
3. (The key step) “Rotate” the table continuously so that A and C are always touching the floor, and B and D are always the same distance vertically above the floor, until points A and C arrive at points B′ and D′ respectively.
4. Now B and D must be under the floor (Why?), so by the Intermediate Value Theorem there must have existed a time during the “rotation” during which both of them are on the floor.
This is not completely rigorous since it is not clear that we can “rotate” the table continuously in the manner described. To finish, we must sketch out this process precisely. That is: Given that A and C are on the floor, and B and D are the same vertical distance above the floor, we can move A, B, C, and D continuously so that ABCD remains a square of the same size, A and C always remain on the floor, B and D always remain the same vertical distance above/below the floor, and A and C switch positions at the end.
In general, if it is only given that the floor is continuous, then this is quite a difficult issue to tackle. We will see that the Lipschitz condition on the floor will make this much more feasible.
First, note that it suffices to find a continuous motion of the two points A and C so that they stay on the graph, switch positions, and their distance is fixed throughout. If so, then at all times there are unique positions for B and D so that ABCD is a square and B, D are the same distance above the graph. The composition of continuous functions entails that these positions move continuously.
The following key result allows us to “rotate” C around A (and vice versa).
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 89 159 Lemma 1 Let f : R2 →R be a 1-Lipschitz continuous function and fix a point x0 ∈R2. Fix R > 0 and let S be the set of points in R2 × R = R3 at a distance R from (x0, f(x0)).
For each θ ∈R, let V (θ) := {x0 + (t cos θ, t sin θ) : t > 0} be the ray emanating from x0 at the angle θ. Then, for all θ ∈R, there exists a unique point g(θ) ∈R3 on the surface S whose projection unto R2, (g1(θ), g2(θ)), lies in V (θ). Moreover, g : R →R3 is a continuous function.
The rigorous statement above is quite atrocious. Intuitively, the picture to have in mind is as follows: Draw a sphere around some point on the graph of f. Then the sphere should intersect the graph at some curvy “ring”.
Proof.
For existence and uniqueness of the point g(θ), it is sufficient to consider the case θ = 0 by symmetry, in which case we need only consider the cross section of f and S obtaining by intersecting these surfaces with the xz-plane.
To wit, we may rephrase the problem as follows: Let f : R →R be a 1-Lipschitz continuous function, let x0 ∈R, let R > 0, and let S be the circle of radius R centered at (x0, f(x0)). Then S intersects the graph of f at exactly two points: One to the “left” of x0, and one to the “right” of x0. Particularly, the case θ = 0 concerns itself with the existence and uniqueness of such an intersection to the “right” of x0.
Roughly speaking, the existence is quite simple and comes from applying the Intermediate Value Theorem or the Jordan Curve Theorem properly. For uniqueness, we suppose there are two distinct intersections at (y1, f(y1)) and (y2, f(y2)) with y1, y2 > x0. Observe that y1 ≥x0 + R √ 2, otherwise the two points x0 and y1 fail the Lipschitz condition on f. The same is true for y2. But now we may argue that the circle S is “too steep” between x0 + R √ 2 and x0 + R, so that the points y1, y2 fail the Lipschitz condition.
g(θ) must be continuous because f is.
□ To be precise, the sort of “rotation” that this Lemma allows us to execute is as follows: We may slide C continuously along the intersection between the graph of f and the surface S of points of distance √ 2 from A. And, viewed from above, this motion for C will appear to be “circling around A”.
We may now describe the procedure for swapping A and C.
1. “Rotate” C about A as described above until C is a distance of √ 2 from its starting position. An application of the Intermediate Value Theorem will show that this is possible.
2. “Rotate” A about C until A arrives at the starting position of C.
3. “Rotate” C about A until C arrives at the starting location of A.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 89 160 This completes the proof.
■ Remarks: The proof sketch works quite well in practice. If you have a well-shaped table that is wobbling on an uneven ground, it can be stabilized by “rotating” the table.
The Lemma is extremely false when f is not assumed to be 1-Lipschitz (Do you see why?).
See the paper for the general proof.
One of the authors of said paper happens to be the Youtuber “Mathologer”! Naturally, he made a video about the problem at
A very nice exercise from the video (which is certainly nicer to rigorously reason about than the problem!) is as follows: Given any bounded figure, prove that there exists a square all of whose sides are tangent to the figure.
Lastly, the solution I wrote completely ignored any issues that could occur with the legs or top of the table intersecting the graph of f. I will leave it to you to think about whether or not this could be a problem.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 90 161 Solution 90 We claim that 6 soldiers is the minimum number of soldiers required to ensure the capture.
Sufficiency of 6 soldiers For each positive integer n, let f(n) be the greatest positive integer for which n soldiers can guarantee the capture of any criminal hiding in a tree-shaped palace of f(n) rooms. We seek to show that f(6) ≥1000.
For a room R of a tree-shaped palace, the branches from R are the connected components of rooms that arise when R is deleted from the palace. For example, if S is a neighbor of R, then the set of all rooms that can be reached from S without passing through R is a branch of R. The number of branches of R is equal to the degree of R.
The key observation is as follows.
Claim: Suppose that within a tree-shaped palace, there exists a path of distinct rooms R1, R2, · · · , Rm that, when deleted from the palace, will result in the palace being split into smaller connected components of rooms, each with no more than f(n) rooms. Then n + 1 soldiers can guarantee the capture of a criminal in this palace.
This is because, given the existence of such a path of rooms R1, · · · , Rm, a strategy for the n + 1 soldiers is as follows: 1. Soldier 1 waits at room R1.
2. For each room R besides R2 that is adjacent to R1, the other n soldiers check if the criminal lies within the branch of R1 that lies past R. This branch has no more than f(n) rooms, so n soldiers are sufficient.
3. Soldier 1 advances to room R2, and the rest of the n soldiers checks each branch of R2 besides the ones that contain R1 and R3.
4. Soldier 1 advances to room R3, and we continue until all rooms are searched.
This observation lets us prove the essential result we require.
Claim: For each positive integer n, we have f(n + 1) ≥3f(n) + 3.
To show this, we take a tree-shaped palace of 3f(n) + 3 rooms and show that there exists a path of rooms that splits the palace into components of size no more than f(n).
We generate this path via the following algorithm.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 90 162 Main Algorithm 1. Start with any room. Call it S1. Set m = 1.
2. Consider the branches of Sm that do not include the tentative path thus far (i.e.
S1, · · · , Sm−1).
3. If each such branch has size at most f(n), then we terminate the algorithm, with the desired path being S1, · · · , Sm.
4. If there is exactly one branch of Sm of size greater than f(n), then we take Sm+1 to be the neighbor of Sm leading into this branch, increment m, and loop back to Step 2.
5. Otherwise, there are exactly two branches with size greater than f(n), and particularly their sizes are at least f(n)+1. It follows that the branch of Sm containing the tentative path thus far has size at most 3f(n)+3−(f(n)+1+f(n)+1+1) = f(n). Armed with this deduction, we cancel the current path and construct a new one as follows: Run the subroutine (described below) on each of the two branches of size at least f(n) + 1 to obtain two paths from R := Sm, each of which splits their respective branches into further branches of size at most f(n).
6. Concatenate these two paths with R to form the desired path and terminate.
Subroutine 1. Let R1 be the room leading into the branch of R on which we call this subroutine. The branch has at least f(n) + 1 rooms and at most 3f(n) + 3 −(f(n) + 1 + 1) = 2f(n) + 1 rooms. Set m = 1.
2. Consider the branches of Rm that do not include the path thus far (i.e. R1, · · · , Rm−1).
3. If each such branch has size at most f(n), then we terminate the algorithm, with the desired path being R1, · · · , Rm.
4. Otherwise, there is exactly one branch with more than f(n) rooms. Take Rm+1 to be the neighbor of Rm that leads into this branch, increment m, and loop back to Step 2.
The correctness of this algorithm is mostly self-evident, though I should justify the ex-haustion of cases. In the main algorithm, the number of branches of Sm (excluding the one with the tentative path) of size greater than f(n) can only be 0, 1, or 2. If there were 3, then there are at least 3(f(n) + 1) + 1 = 3f(n) + 4 rooms, where we have also included Sm in the count, which is bogus. Similarly, in the subroutine, the number of branches of Rm (excluding the one with the tentative path) of size greater than f(n) can only be 0 or 1, as if there were 2, then the branch on which we call the subroutine would have at least 2(f(n) + 1) + 1 = 2f(n) + 3 rooms, which is not the case.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 90 163 This algorithm proves the essential result, which we shall now use to compute a lower bound on f(6). It is not hard to see that f(1) = 3. It follows that • f(2) ≥3(3 + 1) = 12, • f(3) ≥3(12 + 1) = 39, • f(4) ≥3(39 + 1) = 120, • f(5) ≥3(120 + 1) = 363, and • f(6) ≥3(363 + 1) = 1092.
In particular, f(6) ≥1000, which is what we wanted to show.
Necessity of 6 soldiers We will construct a tree-shaped palace of size at most 1000, inside of which a lucky criminal could evade 5 soldiers.
First, if four rooms are arranged in a “Y” shape, then one soldier is clearly insufficient for capturing a criminal hiding in such an arrangement of rooms. Let us call this structure Y1, and call the room in the middle the “central room” of Y1.
Recursively, for each positive integer n, we build the structure Yn+1 as follows: Place a room R, which we take to be the central room of Yn+1. Then, we place three copies of Yn and attach their central rooms R1, R2, R3 to R.
Assume that n soldiers are insufficient to guarantee capture of a criminal inside a palace in the shape Yn. We claim that n + 1 soldiers are insufficient to guarantee capture of a criminal inside a palace of shape Yn+1.
Indeed, this is not too difficult to reason out. Let the branches of R that contain R1, R2, and R3 be T1, T2, and T3, respectively. The criminal, who knows how to dodge n soldiers in a Yn-shaped palace with positive probability, employs the following “strategy”: • Pick a branch of R to hide in (T1, T2, or T3) at random.
• With sufficient luck, dodge any search of the branch that uses at most n soldiers.
• If n + 1 soldiers are all present in a common branch, then with sufficient luck, this branch is not the same branch that the criminal is in. So the criminal can decide to switch to the third unoccupied branch with 50% odds.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 90 164 We claim that with sufficient luck, the criminal can dodge any search of Yn+1 that uses n+1 soldiers.
Since the criminal could be hiding in T1, any search that is guaranteed to work must still be guaranteed to work if the criminal promises to stay in T1. That is, the search of Yn+1 must include an exhaustive search of Tn, and for this, n soldiers is insufficient. Hence, the search pattern must at some point involve all n + 1 soldiers inside T1 simultaneously. The same is true for T2 and T3.
Take the first time that all soldiers are within some Ti for some i, and assume for ease that i = 1. Then, the criminal will not be in T1 with sufficient luck, and by the criminal’s strategy, they could now be hiding in either T2 or T3. It is impossible to verify which one the criminal is in without taking all n + 1 soldiers and putting them all in one of these branches at some point — say, T2. Then, with sufficient luck, the criminal will actually have been in T3, and at this point in time, they have the opportunity to switch to T1. The soldiers cannot ascertain whether the criminal is in T1 or T3, and this cycle may continue forever.
Inductively, we have shown that n soldiers is insufficient for searching a Yn-shaped palace.
In remains to compute the size of Y5. Since |Y1| = 4 and |Yn+1| = 3|Yn|+1, we may compute |Y2| = 13, |Y3| = 40, |Y4| = 121, and |Y5| = 364. Since 364 ≤1000, we have proven the necessity of 6 soldiers.
■ Source: Leningrad Math Olympiad [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 91 165 Solution 91 Part (a) It is unlikely that you had trouble with this part, but I will spell it out anyways. If the radius of Gloria’s house is R, then the fence is a circle of radius R + 1. Their perimeters are 2πR and 2π(R + 1), respectively, so the difference is 2π, no matter the value of R.
Part (b) This part is much more interesting. It turns out that the fence is obtained by pushing each side of the polygon one foot “outwards”, and then connecting the obtained segments via circular arcs.
The sum of the length of those “outwards” segments (marked in blue) is evidently the perimeter of the house. Hence the length of the fence is longer by exactly the sum of the lengths of the arcs, and it can be seen that they may combine to form a circle of unit radius.
Thus the difference must still be 2π.
Part (c) For appropriate sets E ⊆R2, we denote by P(E) its perimeter. For a line L and a set or point A, let us write projL(A) for the orthogonal projection of A unto L. We will make use of the following incredibly useful theorem.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 91 166 Theorem 1 Let K ⊆R2 be a convex set and let (U, V ) be a random unit vector whose angle is uniformly distributed. If L is the line through (0, 0) and (U, V ), then E length(projL(K)) = C · P(K) for some universal constant C.
Proof.
Let φ = (φ1, φ2) : [0, T] →R2 be a parametrization of the boundary ∂K. Then the perimeter of K may be expressed as P(K) = Z T 0 ∥φ′(t)∥dt.
Our goal is to massage the expected length into the right hand side.
Note that the point projLφ(t) stays on line L as t varies from 0 to T, and in doing so traces out an interval on L, visiting each point on that interval twice! (This is because it has to go back and forth.) This technically induces a parametrization of the projection of K, and so if we associate L with the real line then we can find the “length” of this curve to be: 2 length(projL(K)) = Z T 0 d dt(U, V ) · φ′(t) dt To reiterate, we need to include the factor of 2 on the left hand side since we’re double-counting the interval length when we view it as the length of the “curve traced out by the projection”.
Anyways this is pretty nice because we can now mess with the right side to obtain = Z T 0 |Uφ′ 1(t) + V φ′ 2(t)| dt.
Taking the expectation gives 2E length(projL(K)) = E Z T 0 |Uφ′ 1(t) + V φ′ 2(t)| dt.
And hey, expectations are just integrals, and the integrand is non-negative, so we can swap the expectation and integral by Tonelli’s Theorem to arrive at = Z T 0 E |Uφ′ 1(t) + V φ′ 2(t)| dt.
This expectation is actually quite nice to compute! In order to evaluate it, we switch gears to geometry: The expectation is the expected distance from the origin of the projection of φ′(t) unto the line L. We can instead view this as taking a random point on a circle centered incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 91 167 at (0, 0) and radius ∥φ′(t)∥and computing the expected absolute value of its x-coordinate. I don’t actually need to compute this to prove the theorem, but I’ll do it anyway. By four-fold symmetry this is just 2 π R π/2 0 ∥φ′(t)∥cos θ dθ = 2 π∥φ′(t)∥. So our nasty integral is really just: = Z T 0 2 π∥φ′(t)∥dt = 2 πP(K) Thus the theorem has been proven, and we have shown that C = 1/π.
□ This beautiful result has many applications, and demolishing this problem is just one of them. Here is the argument: The projection of the fence unto a line L will always be 2 feet greater in length than the projection of the house unto L. Thus, when L is selected at random, then the expected lengths will differ by exactly 2 as well. By the Theorem, we conclude that the perimeters differ by exactly 2 · 1 C, where 1 C = π.
■ Remarks: How can we be assured that the perimeter can be computed as R T 0 ∥φ′(t)∥dt for an appropriate parametrization ϕ, which may not even be differentiable? In the unlikely event that you’ve bothered to ponder such a question, I shall answer it because I feel obligated to put my masters degree in mathematics to good use. Let K be a convex bounded set. It is clear that its boundary, ∂K, can be parametrized (i.e. “traced in a continuous way”, by e.g. taking a ray emanating from inside K, marking its intersection with the boundary and “spinning” the ray), and roughly speaking, this means that ∂K is a curve.
Curves always have a notion of length, which is computed by using an increasingly large number of segments that approximate the curve.
There is a theorem which states that basically every curve can be traced out with a careful selection of a parametrization φ : [0, T] →R2 such that the length of the curve can be found by integrating the “speed” of the parametrization over time, i.e.
R T 0 ∥φ′(t)∥dt.
It’s possible for the “speed” to not exist at some points. For example, a square has sharp corners, and since a parametrization isn’t traveling in any certain “direction” at such corners, we cannot define its “speed”. But this is alright, since as long as there aren’t too many such “corners”, the integral can still be computed.
If you’re wondering, we can also define perimeters for (most) arbitrary sets! For any (measurable) E ⊆R2, we can define its perimeter as sup Z E div ϕ dx : ϕ ∈C∞ c (R2; R2) and ∥ϕ∥∞≤1 .
For sufficiently nice sets E, it can be shown that this definition is consistent with simpler definitions of perimeter via the divergence theorem.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 92 168 Solution 92 Part (a) This is a classic Pigeonhole Principle argument. Partition the rectangle into six 1 × 2 dominoes. Then two of the 7 points must lie in the same domino. Since such dominoes have a diameter of √ 5, these two points must be at most √ 5 apart.
Part (b): Solution 1 The problem is still true if there were only 6 points in the rectangle.
Subdivide the 4×3 rectangle into 1×1 cells, and color the cells like a checkerboard. Clearly the six points must lie in different cells, and the cells in which they lie cannot be orthogonally adjacent to one another. From some inspection, it follows that the six cells in which the six points lie must be all the same color — either all white or all black.
Without loss of generality we may assume that they lie in the black cells. Draw five red squares of side length 3 2 as above. These squares cover the black cells, thus they contain the six points. By the Pigeonhole Principle, two of the points lie in a common red square. Since each red square has diameter 3 2 √ 2 < √ 5, we are done.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 92 169 Part (b): Solution 2 (From “asbodke”) The Pigeonhole Principle can be applied directly by using the clever partition shown below.
■ [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 93 170 Solution 93 As suggested by the hints, we view the piles of stones as stacks of boxes, sorted from largest to smallest. For example, if n = 4 and the pile sizes are 3, 2, and 5, then we represent this configuration with the following diagram.
Each column represents a different pile, and the number of boxes in each column is the number of stones in the corresponding pile.
As the process described in the problem statement progresses, we will update the diagram in such a way that the columns will always be in decreasing order. If we do this, then a monovariant will appear, which is more easily seen if we rotate the diagram 45◦!
How does this diagram change when we take one stone from each pile to form a new one?
One way to represent this process is via two steps. For the first step, we take the red boxes shown below (which represents one stone from each pile) and “rotate them over” to the other side (to form the new pile), incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 93 171 − → and then we take all the other boxes (what remains of the other piles) and shift them “one slot” to the right.
− → In sum, the first step of the process can be viewed as a set of simultaneous “cycles”.
− → For the second step, we simply sort the piles’ sizes to be in decreasing order. In terms of the diagram, this would entail rearranging the columns to be in order of decreasing height.
However, it’s more revealing to view the sorting process as letting all boxes slide down!
− → This hence completes the visual representation of the process. Our goal is to show that upon repeating these two steps, the arrangement of boxes eventually forms the pattern shown above: a “perfect staircase”.
With this visual representation in hand, the proof is incredibly slick.
We make two observations: incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 93 172 1. Under the first step of the process, every box’s altitude remains the same. For fancy points, we can say that this implies that the total gravitational potential energy of the boxes does not change.
2. Under the second step, the altitudes of boxes can only decrease. In other words: Since we let gravity act on the boxes, the total gravitational potential energy energy can only decrease (though it could stay the same).
Hence the total gravitational potential energy is a decreasing monovariant.
The desired “perfect staircase” arrangement is clearly the arrangement with the minimum total gravita-tional potential energy. Hence, it remains to prove that if the arrangement is not a “perfect staircase”, then the total gravitational energy must eventually decrease.
If not, then boxes never slide down, so the second step does not move any boxes. Hence only the first step moves boxes around, in the “cycles” shown on the previous page. Find the first such “cycle” of boxes from the bottom that isn’t full, i.e. has a space not occupied by a box.
Since the diagram isn’t a “perfect staircase” by assumption, the cycle above this one must have a box, which we shall color red.
These two cycles have lengths r and r+1 for some integer r, so in particular their lengths are relatively prime. Thus, with enough iterations of the cycling, the red box must eventually hover over the empty space which we know to exist in the cycle below it. When this happens, gravity will pull it down to decrease the potential energy of the system, contradicting the assumption that this never occurs.
■ incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 93 173 Remarks: I stole the beautiful idea behind this proof from the expository paper https: //arxiv.org/pdf/1503.00885.pdf of V. Drensky, which contains more results.
Source: This is called Bulgarian Solitaire.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 94 174 Solution 94 The side lengths of the rectangle are irrelevant. Let the vertices of the rectangle be A, B, C and D. Let the center be P. Then the set of points inside the rectangle that are closer to P than A is given by cutting the rectangle along the perpendicular bisector of AP.
A B C D P Arguing in the same way for B, C, and D, we find that the set of points inside the rectangle that are closer to P than any of A, B, C, or D is given by the shaded region below.
A B C D P The desired probability is given by the fraction of the rectangle’s area that is taken up by the shaded region. To determine this fraction, divide the rectangle into quarter rectangles as shown.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 94 175 A B C D P Exactly half of each of these quarter rectangles are shaded! This is because each of the four dashed lines are perpendicular bisectors. We conclude that the whole rectangle is exactly half-shaded. Thus the probability is 1 2 .
■ [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 95 176 Solution 95 Part (a) Let E = {x ∈[0, 1] : x ≤f(x)} and x0 = sup E.
Claim 1: x0 ≤f(x0) This follows from writing x ≤f(x) ≤f(x0) for x ∈E, and then taking the sup on the LHS to get x0 ≤f(x0).
Claim 2: x0 ≥f(x0) From x0 ≤f(x0) we have by the increasing condition that f(x0) ≤f(f(x0)), so f(x0) ∈E by definition of E, hence f(x0) ≤x0 by definition of x0.
From the two claims, we have f(x0) = x0, so x0 is a fixed point.
■ Part (b) Let F = {E ∈P(X) : E ⊆f(E)} and E0 = S E∈F E.
Claim 1: E0 ⊆f(E0) This follows from writing E ⊆f(E) ⊆f(E0) for E ∈F, and then taking the union on the LHS E0 ⊆f(E0).
Claim 2: E0 ⊇f(E0) From E0 ⊆f(E0) we have by the increasing condition that f(E0) ⊆f(f(E0)), so f(E0) ∈ F by definition of F, hence f(E0) ⊆E0 by definition of E0.
From the two claims, we have f(E0) = E0, so E0 is a fixed point.
■ Remarks: These two solutions are basically the same.
Source: Probably a classic exercise [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 96 177 Solution 96 Let f(n) be the number of digits of n that are at least 5. By the hint, we have that the desired sum is rational exactly when P∞ n=0 f(2n) 2n is. Now note that we may express f(m) as the sum f(m) = ∞ X k=0 1the 10k place is ≥5(m), so the desired sum is ∞ X k=0 ∞ X n=0 1 2n · 1the 10k place is ≥5(2n).
But The 10k place of 2n is ≥5 ⇐ ⇒2n mod 10k −2n mod 5 · 10k−1 5 · 10k−1 = 1 ⇐ ⇒2n mod 10k −2n+1 mod 10k 2 5 · 10k−1 = 1.
Hence ∞ X n=0 1 2n · 1the 10k place is ≥5(2n) = 1 5 · 10k−1 ∞ X n=0 2n mod 10k 2n −2n+1 mod 10k 2n+1 = 1 5 · 10k−1, and this is clearly rational once we sum over k.
■ Remarks: I’m too lazy to perturb these computations to get the actual answer, but suppos-edly the sum comes out to 10 9 .
One person misread the problem as determining whether ∞ X n=0 o(n) 2n is rational, and appar-ently the solution to this is also nice.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 97 178 Solution 97 Imagine connecting the tops of everyone’s heads with line segments, as shown.
This creates a periodic “line graph”.
Then each occurrence of “taller” coincides with a “peak” along the line graph, whereas each occurrence of “shorter” coincides with a “valley”. The key insight from this visualization is that between any two consecutive valleys, there exists exactly one peak! This proves that the number of valleys and peaks are equal.
We deduce that exactly 5 people said “shorter”, because 5 people said “taller”. Hence, the number of people who say “in-between” is 25 −5 −5 = 15 .
■ Remarks: Did you spot the joke in the problem statement?
Source: Math Hour Olympiad [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 98 179 Solution 98 The answer is 10 inches.
The turtle cannot crawl more than 10 inches Model the turtle’s crawling time via the interval [0, 6]. Since the turtle is watched at all times, there exists an enthusiast watching during [0, 1] and another during [5, 6]. We claim that we may select 8 other enthusiasts whose watching intervals cover the rest of [0, 6]. To see this, select the latest enthusiast that starts watching at a time in (0, 1]. If their interval is [a, a + 1], then 0 < a ≤1 and a is maximal (in the sense that there is no enthusiast that watches [a′, a′ + 1] with 0 < a′ ≤1 and a′ > a).
Now select another enthusiast that watches the interval [b, b + 1] where 1 < b < a + 1 (which exists by maximality of a; indeed, if there were no such enthusiast, then a small moment of time after a would be left unwatched). These two watchers, together, cover the interval [1, 2]. By repeating this reasoning, we may find 2 × 3 = 6 more enthusiasts that cover the three intervals [2, 3], [3, 4], and [4, 5].
We have hence found 10 enthusiasts that watch the whole interval. During each of their watching intervals, the turtle can move at most one inch. Thus a seemingly rough bound on the most the turtle can move is 10 inches.
The turtle could crawl 10 inches In fact, 10 inches can be obtained. See the diagram below. The red and blue rectangles represent the watching-intervals of ten enthusiasts. During each of the 10 marked orange periods of time, let the turtle move 1 inch forward, and otherwise let the turtle stay still.
0 6/5 12/5 18/5 24/5 6 ■ Remarks: The solution does not change if we interpret time intervals as being open rather than closed.
A possible concern is that the solution breaks if there are infinitely many turtle enthusiasts observing the turtle (in which case we are not guaranteed a maximum).
In this case, we can modify the solution as follows. Fix ϵ > 0. Then, since the turtle’s incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 98 180 movement is continuous on a compact interval, it is in particular uniformly continuous, so there exists δ > 0 such that the turtle never moves more than ϵ within a time interval of length δ. Now extend all enthusiasts’ watching intervals by δ in each direction to form open intervals of length 1 + 2δ. By compactness of [0, 6] we may select a finite number subcollection of enthusiasts whose extended watching intervals cover [0, 6]. Rerunning the logic of the solution, we find 10 of these enthusiasts whose watching intervals cover [0, 6]. It follows that the turtle cannot crawl more than 10 + 20ϵ inches. Sending ϵ →0+ gives the expected conclusion.
Source: Mathematical Circles (Russian Experience) [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 99 181 Solution 99 The possible values for t are t = 0 and t = 1/n for n ∈N.
Let the continuous and increasing function f : [0, 1] →[0, 1] with f(0) = 0 and f(1) = 1 represent the turtle’s movement. Observe that a value of t works if and only if for every such function f, we can find x such that f(x + t) = f(x) + t.
The claimed values of t work Clearly t = 0 works. As for t = 1/n, let us suppose that for some f, there does not exist x ∈[0, 1 −1 n] for which f(x + 1/n) = f(x) + 1/n. Then the continuous functions f(x + 1/n) and f(x) + 1/n never intersect in [0, 1 −1 n], hence one of these functions is strictly greater than the other for all x in [0, 1 −1 n].
The first case is that f(x + 1/n) > f(x) + 1/n for all x ∈[0, 1 −1 n]. If so, then 1 = f(1) > f n −1 n + 1 n > f n −2 n + 2 n > f n −3 n + 3 n > . . . > f(0) + n n = 1, contradiction. If instead f(x + 1/n) < f(x) + 1/n for all x ∈[0, 1 −1 n], then we accordingly obtain 1 = f(1) < f n −1 n + 1 n < f n −2 n + 2 n < f n −3 n + 3 n < . . . < f(0) + n n = 1, again a contradiction.
All other possible values of t fail Suppose t ̸= 0 and t ̸= 1/n for any n ∈N.
We wish to construct an f for which f(x + t) ̸= f(x) + t for all x ∈[0, 1 −t]. Equivalently, we want f(x) ̸= f(x −t) + t for all x ∈[t, 1], so we want the graphs of f(x) and f(x −t) + t to never intersect. The graph of f(x −t) + t is that of f(x), except it is shifted horizontally to the right by t and vertically upwards by t. We can hence think of this problem as the following “game”: We start at (0, 0) and a doppelganger at (t, t) mimics our movements. The goal is the reach (1, 1) without ever crossing the doppelganger’s path.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 99 182 (0, 0) (1, 1) (t, t) To achieve this, begin by drawing gridlines at all multiples of t, subdividing the square into t × t cells and some residual rectangles along the upper and right edges.
(0, 0) (1, 1) (t, t) If we wish to dodge the doppelganger, we must always stay above it or always stay below it.
The choice does not matter by symmetry, so let us choose to stay above it. To “make room” above the doppelganger, the idea is to stay low for as long as possible before rising above (t, t). Repeating this idea for getting above (2t, 2t), (3t, 3t), etc. will successfully get us to the last gridline, at which point we’ll be able to make an unobstructed beeline for (1, 1), provided that we’ve made a sufficient amount of room for ourselves.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 99 183 (0, 0) (1, 1) (t, t) Note that the importance of t not “dividing 1 evenly” is so that the last segment can be drawn.
Intuitively, since the last “square” in the upper-right corner is smaller than the others, we can plan the path so that the doppelganger is forced to go under (1, 1).
A more explicit construction is given by Edward Hou: We can take f(x) = x + t 4 sin2 πx t −x sin2 π t .
See for an interactive plot.
■ Source: Stolen from AoPS [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 100 184 Solution 100 Let’s label the trains with the following letters.
Let’s make space for N to move all the way to the left via the following sequence: 1. A up 2 2. X right 1 3. C up 1 4. E up 3 5. B left 2 6. V right 1, up 2 Once we move N to the left, we can move O and P down so that X is free.
7. N left 3 8. O down 1 9. P down 1 incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 100 185 10. X out!
■ Source: Scott Kim, ThinkFun Railroad Rush Hour [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 101 186 Solution 101 Let the squares at “odd coordinates” be the vertices of a graph, and connect two adjacent such squares with an edge if and only if there is some train running between them.
This forms a graph with 10112 vertices and 10112 −1 edges. This is because there is a 1-to-1 correspondence formed by taking any of the 10112 −1 vertices whose square is occupied (i.e.
all except the one with the empty cell — hence why it is important that a corner is empty!) and corresponding it to the train running through it.
Suppose we can prove that the graph is a tree. Then there is a unique path from the empty cell to the top-right cell where the red block is. By construction of the graph we must be able to push each train along the edges of the path until we free up the square in front of the red train, solving the puzzle.
To show that the graph is a tree, note that since we’ve already shown that the number of vertices exceeds the number of edges by exactly 1, it remains to prove that the graph has no loops.
A loop would look something like this.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 101 187 The inside of the loop must be filled with 1 × 2 trains, and since the single empty cell lies in a corner cell, that empty cell won’t be in this interior. So the area enclosed within the loop must be even. We claim that this cannot be the case.
Consider the lattice grid formed by the centers of all cells. The lattice points that lie on the trains connect to form a polygon whose sides lengths are all even. By Pick’s Theorem applied to this polygon, we have A = I + B 2 −1, where A is the area enclosed by the polygon, I is the number of lattice points strictly inside the polygon (colored red), and B is the number of lattice points on the boundary of the polygon (colored blue). Our goal is to show that I is odd.
To that end, we simply must argue that A is even and B is divisible by 4. That A is even follows quickly from the fact that all side lengths of the polygon are even (which entails that it can be subdivided into 2 × 2 squares). To show that B is divisible by 4, start by scaling down the polygon by a factor of 2.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 101 188 We’re done if we can prove that the scaled-down polygon’s perimeter is divisible by 2. At last, we can show this without further reduction. As suggested by the arrows, we see that if we were to traverse the boundary of the polygon, then the number of times we move upwards is the same as the number of times we move downwards, so the total vertical length must be even. Similarly, the total horizontal length must be even. This concludes the proof.
■ incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 101 189 Remarks: A similar problem that uses the same key claim (that a loop of dominoes must enclose a region of odd area) appeared in the Math Hour Olympiad. Thanks to Dr. Jonah Ostroff for kindly providing a snippet of the official solution that proves this crucial result without Pick’s Theorem.
The interior is built from 2 × 2 squares and so its area is a multiple of 4.
The original interior area is smaller, because it does not include the “inner half” of the cycle dominoes. Assuming a clockwise cycle, if the cycle contains N dominoes, has T CCW turns and T + 4 CW turns, then each of the N line segments contributes area 1 to this “inner half” (half the area of a 2 × 1 domino), except when a CW turn double-counts area 1/4 or CCW turn under-counts area 1/4. Thus, the total “inner half” area is 2M −T + 4 4 + T 4 = 2M −1.
Therefore, the original interior area is 4K −(2M −1), which is odd.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 102 190 Solution 102 Label the trains as follows.
X A B C D E M N O P U V The X, A, B, and C trains form a “cycle”. The difference between this problem and Problem 100 is that the cycle is “oriented” the other way. The key insight is that the orientation of the cycle determines whether U or V can be moved out easily. In Problem 100, when the cycle was oriented the other way, V could easily move upwards. Thus, by symmetry, it must be the case that U can easily be moved downwards in the current problem.
With this observation, we are motivated to try moving U out of the way so that we have room to shuffle the X, A, B, and C trains around and reverse the cycle.
Let’s start by getting U deeper down.
1. U right 1 2. C up 2 3. X left 2 4. B left 1 5. A down 2 6. U right 1 down 3 incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 102 191 X A B C D E M N O P U V Now, notice that we can drive U into the bottom-right corner.
7. X right 2 8. C down 2 9. M left 3 10. O up 2 11. P up 2 12. U right 3 X A B C D E M N O P U V 13. A up 2 14. B right 1 15. E up 2 16. V left 1 17. N left 2 18. U down 2 19. O down 2 20. P down 2 21. M right 3 incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 102 192 X A B C D E M N O P U V We’ve successfully made room to reverse the cycle!
22. C up 2 23. X left 1 24. A up 3 25. X right 1 26. B right 1 27. C down 3 28. X left 1 29. A down 2 X A B C D E M N O P U V From the intuition outlined at the start, we now hope that V can be moved around more easily. This is indeed true — we can move V upwards enough to let it join U in the bottom-right corner.
30. M left 3 31. O up 2 32. P up 2 33. U up 2 34. N right 2 35. V right 2 incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 102 193 X A B C D E M N O P U V 36. E down 2 37. C down 2 38. B left 2 39. D down 2 40. M left 1 41. A up 2 42. V up 2 X A B C D E M N O P U V 43. N left 2 44. U down 2 45. V right 3 46. A down 2 47. B right 2 48. C up 2 49. O down 2 50. M right 3 51. D up 2 52. E up 2 53. N left 2 54. U left 2 55. V down 2 56. P down 2 57. M right 1 incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 102 194 X A B C D E M N O P U V We arrive at what seems to be a dead end. Amazingly, the key to making more progress is to reverse the cycle again!
58. A up 2 59. X right 1 60. C up 3 61. X left 1 62. B left 1 63. A down 3 64. X right 1 65. C down 2 X A B C D E M N O P U V This allows us to move V to new places.
66. M left 3 67. O up 2 67. P up 2 68. V up 2 69. U right 2 70. N right 2 71. E up 1 72. N left 2 73. A down 2 74. B left 1 75. V left 3 incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 102 195 X A B C D E M N O P U V 76. O down 4 77. P down 2 78. X right 2 79. M right 3 80. V up 3 81. B right 2 82. E down 1 83. C down 1 84. X left 4 85. V down 2 86. M left 3 87. V right 2 up 2 right 1 X A B C D E M N O P U V At this point, we can conclude that a solution exists as follows: If we were to move O up 2, X right 2, C up 1, B left 1, A up 2, N right 1, and E down 2, then we reach a symmetrical position except for the cycle. However, it’s easy to reverse the cycle without changing anything. So by an argument of symmetry, we can reduce to Problem 100. The solution this generates is quite long, but fortunately there is a short finish from the current position.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 102 196 88. O up 4 89. X right 2 90. E up 1 91. C up 1 92. B left 2 93. U left 1 up 2 left 2 94. O down 3 95. P down 1 96. X out!
X A B C D E M N O P U V ■ Remarks: I was lucky enough to have encountered this puzzle as a toddler. Today, I still think this is the best sliding puzzle to ever exist. Who could possibly expect that the subtle difference between Problem 100 and Problem 102 could make such a devilish disparity in difficulty?
Sadly, the product in which these puzzles appear is no longer in stores. Thus, even though this isn’t a math problem, I wanted to include this beautiful creation in the POTD collection to help give it the attention and renown that it deserves. I hope you’ll forgive me.
Video Solution: Source: Scott Kim, ThinkFun Railroad Rush Hour [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 103 197 Solution 103 The answer is 150. One can arrive at this easily by focusing on just two bottles of beer.
We claim that we can get exactly three drinks from these bottles. Indeed, we can drink these two bottles of beer, take an empty glass from our friend, trade the three empty bottles in for a full beer, drink it (that’s the 3rd drink!), and then return that empty bottle to our friend.
We’re left with nothing! Doing this for every pair, we end up with 3 2 × 100 = 150 drinks that have been drunk.
To see that we cannot do better, let’s begin by making a small simplification: View our friend as a mechanism for allowing us to have a negative number of empty bottles, as long as in the end the number of empty bottles is non-negative.
Suppose that over the course of the alcohol-fueled night, we drink A beers and execute the trading operation B times. Every time we drink a beer, the number of empty bottles increases by 1, and every time we do the trading operation we lose 3 empty bottles. So we have A −3B empty bottles. To pay back the friend, this must be a non-negative quantity, so A ≥3B.
On the other hand, we cannot drink more beers than 100 plus the number of beers obtained from trading, so A ≤100 + B. Hence 3B ≤A ≤100 + B, which gives B ≤50. Hence A ≤100 + B ≤150.
■ [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 104 198 Solution 104 The procedure can be done if and only if n is even.
It is possible when n is even It suffices to prove that we can take any two pancakes and have them flipped in-place without changing anything else. Consider two pancakes A and B. Then, ignoring all other pancakes, we can notate the configuration as A B , where denotes an empty pan. Specifically it is currently my friend’s empty pan. We first flip A into the 3rd pan (use overhead bars to denote the flipped state).
B A Then we flip B into the first pan.
B A Then we flip A back into the second pan. And, well, you kinda just keep going since there’s only one move at each step that makes any progress.
B A A B A B A B Done! Repeating this procedure, we can flip two pancakes in-place at a time.
n must be even for the procedure to be possible The key observation is that the number of flipped pancakes has the same parity as the executed permutation, where we view a pancake flip as a swapping of two “pancakes”: one actual pancake and one “phantom pancake” that always resides on the empty pan. Indeed, with every move, the number of (not-phantom) flipped pancakes changes by 1, and the parity of the permutation changes from even to odd and vice versa.
If we are able to reach the goal, then since the permutation of the pancakes in the goal situation is the identity, which is even, it must follow that the number of flipped pancakes needs to be even by the observation. All pancakes are flipped, so n needs to be even.
■ Remarks: The original problem’s setting consisted of a tape-recorder, n+1 reels and n tapes.
Unfortunately I’m not quite ancient enough to make sense of this. I hope you agree that incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 104 199 pancakes are a much more fun and tastier context for the mathematics at hand.
Source: Mathematical Circles (Russian Experience) [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 105 200 Solution 105 Part (a) The naive “square packing” works just fine.
Part (b) As ludicrous as it sounds, you can fit 401 coins in the box. The scheme involves alternating between “upright” and “upside-down” triangles of three coins, as shown.
· · · Computing the number of coins this scheme allows us to fit is an instructive exercise in elementary geometry. First, we determine the period of the packing, i.e. how long it takes for the pattern of coins to repeat. The following diagram shows half a repeating segment.
Connecting centers with points of tangency and labelling various lengths gives us the follow-ing diagram.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 105 201 C B A 1 2 x √ 3 2 1 2 1 2 1 2 1 2 We have that AB = x and BC = 1 2 + 1 2 = 1, so AC = √ 1 −x2. Using the fact that the height of the box is 2, we can write the equation 2 = 1 2 + √ 1 −x2 + √ 3 2 + 1 2.
Solving for the value of x gives x = 1 2 p 4 √ 3 −3. So the length of one period of the packing is 2 x + 1 2 , or p 4 √ 3 −3 + 1.
It turns out that 6 coins every p 4 √ 3 −3+1 is every so slightly denser than 2 coins every 1. Indeed, a quick calculator computation shows that 6 √ 4 √ 3−3+1 exceeds 2 1 by about 0.0121.
It remains to do some housekeeping to prove that exactly 401 coins can be fit in the box with this scheme, but I’ll spare you the details.
■ Remarks: See for a Desmos visualiza-tion.
As far as I am aware, nobody knows whether or not 401 is the most number of coins we can fit. If it is, a proof of this seems quite hard.
Source: I saw this on Puzzling Stack Exchange [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 106 202 Solution 106 Beth wins.
In general, for an m × n bar, Beth wins if and only if m and n are both odd. Beth’s winning strategy is to do whatever the she wants. This is because the game always ends in mn −1 moves. Indeed, this is due to the fact that the number of pieces goes up by 1 with every move.
■ Remarks: It’s very common to use induction, but this very clearly isn’t necessary. Another form of this problem is as follows: Prove that a jigsaw takes the same number of moves to complete no matter what you do, where a move consists of joining two pieces together.
Source: Folklore [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 107 203 Solution 107 Every positive integer either starts with 1 or will gain a digit when multiplied by 5.
Starting with 50 = 1, we will perform 2023 multiplications by 5. 1414 of these multiplications gain a digit, so 1414 of the obtained powers of 5 will not start with 1. Thus 2023−1414 = 609 of them do start with 1.
■ Source: I stole the idea from a Mildorf Mock AIME [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 108 204 Solution 108 We’ll instead prove that 9 X n=1 1 10f(n/10) !
+ 9 X n=1 1 10f −1(n/10) !
≤99 100.
This is proven pictorially with the following diagram.
The black curved line is the graph of f. The red rectangles each have width 1 10, so the sum of their areas is P9 n=1 1 10f(n/10). Likewise, the blue rectangles’ areas sum to P9 n=1 1 10f −1(n/10).
Their total area is bounded by the area of the square minus the uncovered gray square in the bottom-left corner. This gives the upper bound 1 − 1 100 = 99 100, as needed.
■ Remarks: The bound cannot be attained, but it is tight.
Source: Leningrad Math Olympiad [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 109 205 Solution 109 Clearly true for n = 1. Now let n be the first integer for which the tens digit of 3n is odd.
Since the tens digit of 3n−1 is even, it follows that the ones digit of 3n−1 must be a digit d for which 10 ≤3d < 19. Thus d = 4, 5, 6. However it is not hard to find that the ones digit of a power of 3 can only be 1, 3, 7, or 9, contradiction.
■ Source: I saw this on Brilliant back in the stone age [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 110 206 Solution 110 I present the proof by Burns and Hasselblatt ( BurnsHasselblattRevised-1.pdf). Denote by f (i) the i-fold composition of f and assume that n ≥2 (the case n = 1 is not so interesting). We begin with the following lemma.
Lemma 1 Suppose I0, I1, · · · , In−1 are subintervals of I such that f(Ii) covers Ii+1 for i = 0, 1, 2, · · · , n −2, and f(In−1) covers I0. Then we can find x ∈I0 such that f (i)(x) ∈Ii for all i and f (n)(x) = x.
Proof.
Essentially, you just start with I0 and pull it back n times via f to get a smaller interval.
We have that f(In−1) ⊇I0, so f −1(I0) ⊆In−1. f −1(I0) is a subset of In−1 that gets mapped onto I0, but before we pull back further, we want to make this subset an interval.
f −1(I0) is not necessarily an interval, but it’s certainly a union of intervals! Pick one such interval that gets mapped onto I0, and call it Jn−1. (Why does Jn−1 exist?) We now have a subinterval Jn−1 ⊆In1 with f(Jn−1) = I0. Pull back Jn−1 to get a subset f −1(Jn−1) ⊆In−2. Again, f −1(Jn−1) is a union of intervals, and we can pick one of them that gets mapped onto Jn−1 and call it Jn−2.
This gives a subinterval Jn−2 ⊆In−2 with f(Jn−2) = Jn−1, and we can repeat this to get a subinterval Jn−3 ⊆In−3 with f(Jn−3) = Jn−2, and so on! In the end, we find J0 ⊆I0, J1 ⊆J1, · · · , and Jn−1 ⊆In−1 such that the restrictions f : J0 →J1 f : J1 →J2 f : J2 →J3 · · · f : Jn−2 →Jn−1 f : Jn−1 →I0 are all surjective! Hence the n-fold composition f (n) : J0 →I0 is also surjective. Recalling that J0 ⊆I0 and that J0 and I0 are both intervals, a simple application of the Intermediate Value Theorem gives the existence of some x ∈J0 for which f (n)(x) = x. Due to how we chose the Ji intervals, this point x satisfies all the desired properties.
□ The way we use this lemma to solve the problem is extremely cool. Let’s suppose that the point a ∈I has period 3. Let b = f(a) and c = f(b). Without loss of generality, we can assume that a < b < c. Then: incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 110 207 • f((a, b)) covers (b, c), and • f((b, c)) covers (a, c).
In particular, it is both true that f((b, c)) covers (a, b) and f((b, c)) covers (b, c). Thus, if we make the following choices for I0, · · · , In−1: • I0 := (a, b) • I1 := (b, c) • I2 := (b, c) • I3 := (b, c) • . . .
• In−2 := (b, c) • In−1 := (b, c) then f(Ii) covers Ii+1 for all 0 ≤i ≤n −2 and f(In−1) covers I0.
So by the lemma, f (n) has a fixed point x in I0 with f(x) ∈I1. But I0 = (a, b) and I1 = (b, c) are disjoint, so clearly x and f(x) cannot be the same. In fact, f (i)(x) ∈Ii = (b, c) for all 1 ≤i ≤n −1, so x ̸= f (i)(x) for any such i! So n is the first time that x gets sent back to itself. That is, we found a point of period n.
■ Source: This is a special case of Sharkovsky’s Theorem [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 111 208 Solution 111 Thanks to Alan Abraham for the following approach. Begin with the following dissection.
A The dissection also works when n is even, and would look something like this.
A The utility of this dissection is that the blue triangles can be joined to form a copy of P.
Both the odd and even cases are shown below.
What remains is n −1 red isosceles triangles. We aim to show that the sum of their areas is twice that of P.
Scale the diagram so that the radius of P (the distance from the center of P to a vertex of P) is 1. Then the area of P is given by n · 1 2 sin 2π n .
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 111 209 As for the n −1 red isosceles triangles, they are all similar. The lengths of their legs are given by {|z −1| : zn = 1, z ̸= 1}, and the angle formed by the legs is 2π n . So the sum of their areas is X zn=1,z̸=1 1 2|z −1|2 sin 2π n = 1 2 sin 2π n X zn=1,z̸=1 |z −1|2 = 1 2 sin 2π n X zn=1 |z −1|2 = 1 2 sin 2π n X zn=1 (z −1)(z −1) = 1 2 sin 2π n X zn=1 2 −z −z = 1 2 sin 2π n X zn=1 2 (Roots of unity sum to zero) = 2 × n 2 sin 2π n , which is indeed twice the area of P.
■ Remarks: A purely dissective proof can be found on this page.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 112 210 Solution 112 As a consequence of the looping condition, we see that the amount of money gained or lost between any two rooms is independent of the path taken.
Suppose I start from the northwest room with no money. Note that any room that is not the southeast room can be reached in 7 steps, so I could not possibly have more than $7 in any of these rooms since my money can go up by at most $1 with each step. Thus I could only have observed having $8 in the southeast room.
Considering the 8-step path EESSSSEE, I must end up with $8 via this path, since any path that reaches the southeast room must do so. So on each step of this 8-step path, I must gain a dollar. It follows that I would lose a dollar if I exit the center room via its north door.
■ Remarks: The ideas in this problem bear a resemblance to the arguments used with conser-vative vector fields.
Source: Math Hour Olympiad [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 113 211 Solution 113 I pick a degree-(k −1) polynomial P(x) such that P(0) is my favorite number. For each integer x with 1 ≤x ≤n, I give friend x the ordered pair (x, P(x)). Any k of these points is enough to uniquely identify P, and hence let my friends deduce P(0). But knowing k −1 of the points is never enough. In fact, it won’t give any information regarding P(0).
■ [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 114 212 Solution 114 Part (a) As mentioned in the hint, the key insight is that if you start moving in the direction perpendicular to the segment connecting you and the center of the circle, then the lion can never catch you.
In fact, you could change directions as much as you want, as long as when you change direction, you begin moving in the direction perpendicular to the segment connecting you and the center.
This motivates the following strategy: • Orient yourself so that you’re perpendicular to the segment connecting you and the center.
• Run r1 units forwards.
• Orient yourself so that you’re perpendicular to the segment connecting you and the center.
• Run r2 units forwards.
• Orient yourself so that you’re perpendicular to the segment connecting you and the center.
• Run r3 units forwards.
• etc.
Provided that we can keep running in this way forever, this strategy will work. To ensure that this strategy works forever, the numbers r1, r2, · · · must be chosen so that we never exit the cage and that we run indefinitely. Assume for convenience that the radius of the cage is 10.
• To ensure that we never exit the cage, note that by iteratively applying the Pythagorean Theorem, our squared distance to the center of the cage is given by P∞ i=1 r2 i . So we require that P∞ i=1 r2 i < 102.
• To ensure that we run indefinitely, we must plan to run an infinite distance. That is, we require that P∞ i=1 ri = +∞.
From these conditions, we choose ri = 1 i , which works!
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 114 213 · · · Part (b) Impose coordinate axes centered at the center of cage. The first lion chases your projection unto the x axis, and the second lion chases your projection unto the y axis. Once each lion has caught up with these projections, they move towards you in such a way that the first lion’s x-coordinate always the matches yours, and the second lion’s y-coordinate always matches yours.
I’ll leave it to you to convince yourself that this works.
■ [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 115 214 Solution 115 If we change the frame of reference so that the rain is stationary, then our velocity has a component of 10 in a 60◦direction, and our goal position moves along a 60◦-sloped line at the same speed of 10. We see that, provided these restrictions, reaching the goal is equivalent to reaching the line. Minimizing the rain encountered is equivalent to finding the shortest path to this line. This is given by orthogonal projection, and some vector arithmetic shows that we should run at a speed of 20.
10 20 10 ■ Source: Someone posted this in a Discord server and I stole it.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 116 215 Solution 116 We begin by working under the assumption that an even number of servers are on.
1. Press all four buttons. (If not successful, we now know exactly two servers are on.) 2. Press two diagonally-opposite buttons.
3. Press all four buttons. (If not successful at this point, we now know that there are exactly two servers on, and that they are adjacent.) 4. Press two adjacent buttons.
5. Press all four buttons. (If not successful at this point, we now know that there are exactly two servers on, and that they are diagonally opposite.) 6. Press two diagonally-opposite buttons.
7. Press all four buttons.
If not successful at this point, then our assumption that there were an even number of servers was wrong. Thus we may win in 8 more steps as follows: 8. Press any button. (We now know that there are an even number of servers that are on.) 9. Repeat steps 1-7.
Hence we may guarantee success within 15 steps.
■ Remark: Apparently you can still win if there instead are 2n servers for positive integer n.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 117 216 Solution 117 We claim that everyone is equally likely to try the grape juice last (!).
Lemma 1 Imagine a frog at the integer 0. It repeatedly hops left and right until it either reaches −a, where it wins, or reaches b, where it loses. Then the probability that the frog wins is b a+b.
Proof.
There are a good number of ways to approach this. Here is a short one. Let p be the desired probability. Let Mn be the martingale representing the frog’s location after n steps (and it is indeed a martingale because the expectation of its change at each step is 0).
We endow it with the stopping time τ := inf{k ∈N : Mk ∈{−a, b}}. It is now easy to justify the application of Doob’s Optional Stopping Theorem, which entails that EM0 = EMτ. But EM0 = 0, and EMτ = p(−a) + (1 −p)b. Solving for p gives p = b a+b.
□ We can now solve the original problem. Let us index my friends and I via the integers from 0 to 2023, where I am labelled with 0, and we will consider the friend at some 1 ≤n ≤2023.
We will show that friend n tries the grape juice last with probability 1/2023.
There are exactly two ways in which friend n could try the grape juice last. Either • the grape juice reaches friend n −1, then goes around the other way to friend n + 1, without ever reaching friend n, or • the grape juice reaches friend n + 1, and then goes around the other way to friend n −1, without ever reaching friend 0.
We compute the probability of the first case. For convenience, allow negative indices, taking all indices mod 2024. The probability that the grape juice reaches friend n −1 before it reaches friend n + 1 ≡−(2023 −n) is given by 2023−n 2022 by the lemma. From here, the probability of the grape juice reaching friend n + 1 ≡−(2023 −n) before reaching friend n is 1 2023, by the lemma again. Thus the probability of the first case occurring is 2023−n (2022)(2023).
Analogously, the probability of the second case occurring is given by n−1 (2022)(2023). Summing the cases, we conclude that the probability that friend n tries the grape juice last is 2023 −n (2022)(2023) + n −1 (2022)(2023) = 1 2023 as claimed.
■ incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 117 217 Remarks: See for what appears to be a clean, computation-less solution.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 118 218 Solution 118 From classical right triangle geometry, the length of the segment connecting the corner of the wall and the ladder’s midpoint is always half the ladder’s length. So it’s constant.
Hence the shape traced is a circle. Specifically, it is a quarter circle centered at the corner of the wall.
■ [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 119 219 Solution 119 We claim that such a regular n-gon exists only for n = 4. Obviously squares exist, so n = 4 certainly works. Now let us rule out n ≥5.
Suppose that n ≥5 and that there is a regular n-gon with lattice vertices. Rotating each vertex 90◦inwards about the previous vertex, we form a smaller regular n-gon with lattice vertices, which is a contradiction since we may descend in this way infinitely.
Finally, we rule out n = 3. Suppose there were an equilateral triangle △ABC where A, B, C are lattice points. Multiply all the coordinates of A, B, and C by 3. Then, take the two trisection points on each of the three sides of △ABC. These are lattice points and they form a regular hexagon. But we ruled out n = 6 from the previous analysis, contradiction.
■ [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 120 220 Solution 120 This approach seems novel enough to justify crediting myself for it. We claim that 6 is the best we can do.
To see that it is obtainable, take the 6 lines that pass through two opposite vertices of a given icosahedron.
We now show that 7 is impossible. First we prove the following weird lemma that does not seem to have any relevance to the problem whatsoever.
Lemma 1 Color the edges of K7 graph red and blue. Consider a “Lights Out”-type game in which we may “press” any vertex to toggle the color of all edges emanating from that vertex (from red to blue and vice versa).
Then there exists a sequence of moves that will result in there existing a monochromatic subgraph.
Proof.
Pick any vertex v. Press some of the other vertices so that all edges from v are red.
Pick another vertex w. Since there 5 other vertices, there exist 3 of them, x, y and z, such that edges wx, wy, and wz are the same color.
If this common color is red, then w, x, y, z, and v form the desired monochromatic subgraph, which will be all red. If otherwise the common color is blue, then these vertices will form a blue such monochromatic subgraph after we press v.
□ With this totally irrelevant lemma proven, we may proceed to solve the original problem.
Suppose there exist 7 distinct lines through the origin that form the same angles with incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 120 221 each other. Pick unit vectors v1, v2, · · · , v7, one in the direction of each of these 7 lines (there will be two legal choices for each). By the equal angle condition, the quantity |vi · vj| is the same constant c for all distinct i and j.
Form a graph on these 7 vectors, coloring the edge between vi and vj red if vi · vj = c, and coloring it blue if otherwise vi · vj = −c.
Note that if we were to replace vi with −vi, then all the edges emanating from vi in this graph will toggle colors. By the irrelevant lemma, we may make a sequence of such replacements such that among the vectors {v1, v2, · · · , v7}, there will exist five distinct vectors v, x, y, z, and w for which v · x = v · y = v · z = w · x = w · y = w · z = ±c.
Note that from v · x = v · y = v · z, we have that v is perpendicular to the plane formed by x, y, and z. Indeed, this is because v · (x −y) = v · (x −z) = 0. Similarly, w is also perpendicular to the plane formed by x, y, and z. This can only happen if v and w lie on the same line through the origin (because the space of vectors orthogonal to both x −y and x −z has dimension 1), contradicting how we chose the vectors.
■ [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 121 222 Solution 121 Whenever the question mentions “the apple that’s X...”, this implies that there is exactly one apple that satisfies the condition X. From this, we can work backwards to get some information on the apples, starting from the green apple.
• “...the apple that’s cheaper than the apple that’s green” implies that there is only one apple that is cheaper than the green apple, so the green apple is $2 and “the” apple in question is $1.
• “...the apple that’s smaller than [the $1 apple]” implies that the $1 apple is the second-smallest, and “the” apple in question is the smallest.
• “the apple that that costs more than [the smallest apple]” implies that the smallest apple is $4 and “the” apple in question is $5.
• “the apple that’s bigger than [the $5 apple]” implies that the $5 apple is the second-largest (i.e. fourth-smallest) and “the” apple in question is the largest apple. Moreover, the “it” in “given that it is red” refers to this apple so the largest apple is red.
We may collate this information into the following “logic grid”.
$1 $2 $3 $4 $5 smallest X X X O X 2nd-smallest O X X X X 3rd-smallest X X X 4th-smallest X X X X O 5th-smallest X X X To resolve the ambiguity, we use the colors of the apples! The largest apple is given to be red whereas the green apple is $2, so the largest apple and the $2 apple are different apples. This lets us place one more X in the grid, and we conclude that the red apple (i.e.
the largest apple) is $3 .
■ Source: Jack Lance [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 122 223 Solution 122 Here is one possible argument: Suppose that the cube is 1 × 1 × 1. We can tilt the cube 45◦(any less is also perfectly fine) so that its shadow looks like a 1× √ 2 rectangle, as shown below.
Rotating the cube very slightly in the direction of either of the pictured arrows, the shadow becomes slightly vertically elongated. Any amount of vertical elongation will be enough to be able to fit a 1 × 1 square with room to spare.
Since there is room to spare, the 1 × 1 square can be enlarged to some (1 + ε) × (1 + ε) square (still with room to spare), and drilling a hole through this square gives the desired hole, through which we can pass through a cube of side length 1 + ε.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 122 224 For a more constructive/explicit approach, let’s tilt the cube so that one vertex lies directly above the opposite vertex. Then the cube’s shadow will be a regular hexagon ABCDEF.
But what is the side length of this hexagon? You can find that the segment AC (and ditto for BD, CE, etc.) lies directly under a face diagonal of the cube, which is actually parallel to the shadow’s plane. So AC = √ 2 and using 30-60-90 triangles will tell us that the side length is AB = q 2 3.
Now to geometrically represent the hole we plan to drill, let us inscribe a square WXY Z into ABCDEF. To maximize the area of this square, a good guess is to inscribe it so that WZ is parallel to a side of the hexagon. See the diagram.
A B C D E F W X Y Z If the side length of the square is x, then CW = x √ 3. So BW = q 2 3 − x √ 3. But now x = WZ = AB + BW cos 60◦+ AZ cos 60◦= AB + BW = 2 r 2 3 −x √ 3.
So (1+ √ 3)x = 2 √ 2 which solves as x = √ 6− √ 2 ≈1.035, which is very slightly larger than 1, so it is indeed barely possible to drill a square hole into a unit cube that will fit a larger cube.
■ Remark: We can actually do better than √ 6− √ 2. You can think about how to achieve this or look it up.
Remark 2: It is known that the problem still holds for any other platonic solid! Mathemati-cians have no idea if this extends to all convex polyhedra.
Source: Classic, known as Prince Rupert’s Cube [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 123 225 Solution 123 Let the positive integer be n, and suppose it has k digits. Assume for contradiction that all digits of n are at least 6, and all digits of n2 are at most 4.
A natural inequality to write is n ≥666 . . . 66, but it will be helpful for later if we can improve this bound. Indeed, we can do better by examining the units digit: If n ends in a 6, then n2 will end in a 6 as well, which is not less than 5, so this cannot be. Similarly, n could not end in 7 since otherwise n2 will end in 9. So n ≥666 . . . 68.
More mathematically, this entails that 10k ≥n ≥6 · 10k −1 9 + 2 = 2 3 · 10k + 4 3, so 102k ≥n2 ≥4 9 · 102k + 16 9 · 10k + 16 9 > 4 · 102k −1 9 .
The number 4 · 102k−1 9 is simply the 2k-digit number 444 . . . 4. So n is somehow strictly greater than this, and cannot exceed the (2k + 1)-digit number 100 . . . 0, while also having all of its digits between 0 and 4. This is evidently impossible.
■ [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 124 226 Solution 124 I present two solutions.
Solution 1 Let’s say the width of the bottle is 1. Then the bottle on the left shows that the amount of wine in the bottle is 1 × 1 3 = 1 3. Now, instead of finding the total area of the bottle, notice that the amount of empty space must be the same on the left and on the right, and in the bottle on the right, the empty space is a 1 × 1 2 rectangle whose area is 1 2.
Thus, the total area of the bottle is 1 3 + 1 2 = 5 6, and so the bottle is 1/3 5/6 = 2 5 full.
■ Solution 2 ■ Source: I saw this on the MindYourDecisions channel.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 125 227 Solution 125 Consider the permutation which maps each student to the student that holds their midterm exam. Since each permutation can be decomposed into disjoint cycles, it suffices to solve the problem under the assumption that this permutation is just a cycle. That is, we may assume that the students are S1, S2, . . . , Sn, and that Sk holds Sk+1’s exam (where we identify Sn+1 := S1).
The procedure for resolving this case is best described graphically. There are two cases depending on the parity of the number of students in the cycle, n. In either case (depicted as n = 10 and n = 11 below), we first swap along the solid red lines and then swap along the dashed blue lines.
1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 11 ■ Source: Leningrad Olympiad [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 126 228 Solution 126 Solution 1 This proof follows one of the official solutions from the contest in which this problem appeared. We will use the notation [XY Z] to denote the area of △XY Z. First we prove the following lemma.
Lemma 1 Let △X′Y ′Z′ be a congruent copy of △XY Z which is rotated by 180◦and then translated. Then Area(△XY Z ∩△X′Y ′Z′) ≤2 3Area(△XY Z).
More succinctly, any centrally symmetric subset of a triangle takes up at most 2 3 of the triangle’s area.
Proof.
Without loss of generality we may assume △XY Z is equilateral with side length 1. There are two cases.
First Case: △XY Z ∩△X′Y ′Z′ is a parallelogram, two of whose vertices are X and X′ (without loss of generality). In the interest of maximizing the area of intersection we may assume that X′ lies on side Y Z. Let a = |Y X′| and b = |X′Z|, so that a + b = 1. The area of the parallelogram is now [XY Z] −|Y X′|2[XY Z] −|X′Z|2[XY Z] = (1 −a2 −b2)[XY Z], and from a2 + b2 ≥1 2(a + b)2 = 1 2 we obtain an upper bound of 1 2[XY Z], which is certainly ≤2 3[XY Z].
Second Case: △XY Z ∩△X′Y ′Z′ is a hexagon. If this hexagon is removed from △XY Z, then we are left with three equilateral triangles of side lengths a, b, and c. Using the fact that the hexagon is centrally symmetric, it can be seen that a + b + c = 1. The area of the hexagon is thus [XY Z] −a2[XY Z] −b2[XY Z] −c2[XY Z] = (1 −a2 −b2 −c2)[XY Z].
From the QM-AM inequality, a2 + b2 + c2 is minimized subject to a + b + c = 1 exactly when a = b = c = 1 3. So the area of the hexagon is at most 1 −3 32 [XY Z] = 2 3[XY Z].
□ Now we return to the original problem. Let the triangle be T , and let the ellipse be E with center O. Rotate T 180◦about O to obtain a triangle T ′.
Consider the regions R := T ∩E and R′ := T ′ ∩E. These regions are shown below.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 126 229 O O Note that R and R′ are congruent, and each have area A. Moreover R ∪R′ is a subset of the ellipse, hence Area(R ∪R′) ≤E.
The area of the union may be expressed as Area(R ∪R′) = Area(R) + Area(R′) −Area(R ∩R′) = 2A −Area(R ∩R′).
But by the lemma, Area(R ∩R′) ≤2 3T. We conclude that 2A −2 3T ≤E, which rearranges to T 3 + E 2 ≥A, as needed.
■ We present an alternate solution starting on the next page.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 126 230 Solution 2 (Sketch) This solution is on the lengthy side, but it showcases a nice technique known as visual calculus.
To set up, let us first assume without loss of generality that the ellipse is a circle. We will show that the minimum value for T 3 + E 2 −A is 0. To avoid technicalities we will take the following facts for granted: • There exists a minimum value for T 3 + E 2 −A.
• In a configuration which achieves this minimum value, the circle intersects the triangle 6 times — twice per side.
Take a configuration which achieves the minimum value of T 3 + E 2 −A. Then it follows that any perturbation to the configuration cannot decrease the value of T 3 + E 2 −A. This principle is the basis for the following deductions.
Call the triangle △XY Z. One way to perturb the configuration is to expand the triangle slightly by a dilation centered at a vertex (say, X), as depicted below.
X Y Z Y ′ Z′ I J ∆h Let’s start with a slightly informal argument using Calculus. Label two of the intersections as I and J as shown.
If we perturb the triangle in this way continuously in time so that the perturbation in height ∆h increases with rate d∆h dt = 1, then by the Fundamental Theorem of Calculus (seen incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 126 231 more easily by rotating the above diagram 90◦), we have dT dt = Y Z and dA dt = IJ, and so d dt T 3 + E 2 −A = Y Z 3 + 0 −IJ.
But d dt T 3 + E 2 −A = 0 because T 3 + E 2 −A is minimized, so we conclude that Y Z 3 −IJ = 0.
If the reader finds this argument suspicious, we can obtain the same result with elementary arguments: Suppose it were the case that IJ > Y Z 3 . As shown in the diagram from before, we expand side Y Z slightly to Y ′Z′, increasing the triangle’s height by a small ∆h. ∆h should be chosen to be so small that it is negligible compared to the difference IJ −Y Z 3 .
Then: • The quantity T increases by the area of trapezoid Y ZZ′Y ′, which is essentially Y Z·∆h for small ∆h.
• E does not change.
• A increases by just the area of the striped blue region, which is essential IJ · ∆h.
So the quantity T 3 + E 2 −A changes by Y Z 3 −IJ · ∆h. But since T 3 + E 2 −A is minimized, this can be possible only if Y Z 3 −IJ ≥0, contradicting the assumption that IJ > Y Z 3 . We similarly can show that IJ < Y Z 3 is impossible, so IJ = Y Z 3 .
Running the same logic for the other sides, we conclude that exactly a third of each of the triangle’s sides must lie inside the circle.
A second way to perturb the configuration is to take a side and rotate it slightly around the center of the circle. We can think of this as moving the chord IJ around the circle, as shown below.
I J I′ J′ incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 126 232 It is clear that E doesn’t change. It turns out that A doesn’t change either (why?), thus only T changes. The change in T is depicted below.
X Y Z Y ′ Z′ I J I′ J′ K M O Denote the intersection of IJ and I′J′ as K, and let the midpoint of segment IJ be M. In the perturbation, the area is increased by the area of △Y Y ′K (in red, shaded) and decreased by the area of △ZZ′K (in blue, striped).
If the angle of rotation ∆θ for the perturbation is negligibly small, then M and K are basically the same point, and so the lengths Y K, Y ′K, Y M are morally indistinguishable.
Hence Area(△Y Y ′M) = 1 2(Y K)(Y ′K) sin(∆θ) ≈1 2|Y M|2 sin(∆θ) and, similarly, Area(△Y Y ′M) = 1 2(ZK)(Z′K) sin(∆θ) ≈1 2|ZM|2 sin(∆θ).
(The estimate sin(∆θ) ≈∆θ is applicable but unnecessary.) It follows that T changes by ≈1 2(|Y M|2 −|ZM|2) sin(∆θ), which must be non-negative by the hypothesis that T 3 + E 2 −A is minimized, thus Y M ≥ZM.
By a completely symmetrical argument, Y M ≤ZM. Hence M is the midpoint of side Y Z.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 126 233 By applying this argument to the other two sides, we conclude that the center of the circle, O, is the circumcenter of △XY Z (!). Moreover, in view of the deduction from the first perturbation, we see that the circle must divide each side of △XY Z into thirds.
To finish, consider the below quadrilateral QRST.
X Y Z Q R S T Since the circle divides each side into thirds, we have XQ = QR and XT = TS. Thus QRST is a trapezoid. But the only cyclic trapezoids are isosceles trapezoids, so QR = ST and thus XY = XZ. Analogously, Y X = Y Z, so △XY Z is equilateral.
The below diagram is hence the minimal configuration.
Computing T, E, and A, we find that T 3 + E 2 −A = 0, thus 0 is the minimum value, as desired.
■ Remark: In the official solution, the ellipse (scaled to be a circle) was perturbed instead, but making this work is trickier.
Source: HMIC [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 127 234 Solution 127 Move a pencil along the edges of the star in the manner described by the diagram on the next page.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 127 235 incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 127 236 Note that the pencil’s orientation captures the sum of the angles of the star that the pencil has rotated through. At the end of the procedure, we see that the pencil’s orientation has reversed, completing exactly half a rotation. Thus the sum of the angles is 180◦.
■ Remark: This result holds for any “thin” star, no matter how many vertices it has. For “thicker” stars, like the one shown below, the same methodology can be applied to quickly compute the sum of the angles.
There is also a similar (and possibly more well-known) procedure for showing that, in any convex polygon, the sum of the external angles is 360◦.
Source: Classic [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 128 237 Solution 128 Suppose we found such a partition into n sequences, n > 1. Let their initial terms be a1, · · · , an and their common differences be d1, · · · , dn. Since the progressions partition the positive integers, we have for all complex |x| < 1 that ∞ X k=1 xk = n X j=1 ∞ X k=1 xaj+kdj, or x 1 −x | {z } LHS = n X j=1 xaj 1 −xdj | {z } RHS .
Assume without loss of generality that dn is the largest common difference. Then, by the condition that all common differences are distinct, we have in particular that xan 1−xdn is the only term in RHS with a pole at x = e2πi/dn. (All other terms will be continuous at x = e2πi/dn.) So RHS has a pole at x = e2πi/dn. In other words, lim x→e2πi/dn |RHS| = +∞.
However, the only pole of LHS is at x = 1, so lim x→e2πi/dn |LHS| ̸= +∞.
The two limits are in contradiction.
■ Source: This is an exercise in Stein and Shakarchi’s complex analysis textbook.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 129 238 Solution 129 The key property of centroids that we require is as follows: For distinct points x1, x2, · · · , xn in the plane, the minimum value for the sum of squared distances n X i=1 |x −xi|2 occurs exactly when x is the centroid of {x1, x2, · · · , xn}. We will prove this afterwards.
Let r > 0 be Yohane’s visual radius.
Let the footprints be located at the points x1, x2, · · · , xn. Assume for contradiction that Yohane never stops moving. We claim that the quantity f(x) := n X i=1 min |x −xi|2, r2 is a strictly decreasing monovariant.
To see this, suppose that Yohane is currently at the point x = a, and that the footprints in Yohane’s visual radius at this point are x1, · · · , xk, relabeling the indices as necessary.
We’ll split the sum for f(x) into two parts, f(x) = k X i=1 min |x −xi|2, r2 + n X i=k+1 min |x −xi|2, r2 , and analyze the change in each part separately as Yohane moves from x = a to the centroid of {x1, · · · , xk}, which we will call g.
For the first part, we use the aforementioned key property of centroids to find that k X i=1 |a −xi|2 > k X i=1 |g −xi|2, (1) where the inequality is strict because g is the unique minimizer of the sum of squared distances to the points in {x1, x2 · · · , xk}. Now, on one hand, it is plain to see that k X i=1 |g −xi|2 ≥ k X i=1 min |g −xi|2, r2 .
(2) On the other hand, since all the points in {x1, x2, · · · , xk} are in Yohane’s visual radius at x = a, we have |a −xi|2 ≤r for all 1 ≤i ≤k. Hence k X i=1 min |a −xi|2, r2 = k X i=1 |a −xi|2.
(3) incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 129 239 Combining (1), (2), and (3), we deduce that k X i=1 min |a −xi|2, r2 > k X i=1 min |g −xi|2, r2 , (∗) which shows that the first part must (strictly) decrease.
As for the other part of the sum, we see that x = a maximizes X i = k + 1n min |x −xi|2, r2 since all terms are exactly r2, which is as large as possible, so Yohane’s movement cannot increase this part of the sum. That is, n X i=k+1 min |a −xi|2, r2 ≥ n X i=k+1 min |g −xi|2, r2 .
(∗∗) Summing (∗) and (∗∗) gives f(a) > f(g). This proves that f(x) is a strictly decreasing monovariant, as claimed.
To finish, note since f(x) is strictly decreasing, it follows that it takes on infinitely many values while Yohane moves, which in turn implies that Yohane visits infinitely many points (as opposed to revisiting certain points). But there are only finitely many centroids of subsets of {x1, x2, · · · , xn}, which are the only possible locations that Yohane can move to. This is a contradiction.
■ For completeness we now prove the key property of centroids that we’ve used. In fact, we will prove the following more general statement.
Lemma 1 Let x1, x2, · · · , xn be points in an inner product space. Let g be the centroid of these points, g = 1 n n X i=1 xi.
Then the value of n X i=1 |x −xi|2 depends only on |x −g|, and in particular it is a strictly increasing function of |x −g|.
Proof.
This can be proven by a direct computation, but we’ll avoid the mess that comes incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 129 240 with this via a more refined approach. Write n X i=1 |x −xi|2 = n X i=1 |x −g + g −xi|2 = n X i=1 ⟨x −g + g −xi, x −g + g −xi⟩ = n X i=1 |x −g|2 + 2⟨x −g, g −xi⟩+ |g −xi|2 = n|x −g|2 + 2 x −g, n X i=1 g −xi + + n X i=1 |g −xi|2, and note that Pn i=1 g −xi = 0. It follows that n X i=1 |x −xi|2 − n X i=1 |g −xi|2 = n|x −g|2, which proves the lemma.
□ As a remark, the same computation shows that Z E |x −y|2 dµ(y) − Z E |g −y|2 dµ(y) = µ(E)|x −g|2 for any measure µ on RN and any measurable set E ⊆RN, where the centroid g of µ is defined as g := 1 µ(E) Z E y dµ(y).
When µ is “mass”, this is the parallel axis theorem from physics.
Source: I stole this from someone, but I do not know of an original source.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 130 241 Solution 130 Let the triangle be △ABC and let the unique lattice point in its interior be P. By Pick’s Theorem, the triangles △ABP, △BCP, and △CAP all have an area of 1 2. In particular, their areas are all equal. This is a defining property of the centroid, so P is the centroid.
■ [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 131 242 Solution 131 Solution 1 ■ Solution 2 Thanks to “InductionEnjoyer” for the following hilarious construction.
■ incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 131 243 Remark: The original problem can be solved without making any triangles, and can also be solved without any rods intersecting. See
html for the constructions.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 132 244 Solution 132 The first series diverges for θ = 2π/3. For this value of θ, we have that sin n2θ cos nθ = ( 0, n ≡0 (mod 3) − √ 3 4 , otherwise , so that P∞ n=1 sin n2θ cos nθ n = −∞.
The second series converges. Consider the partial sum PN n=1 cos n2θ sin nθ n and apply the product-to-sum formula to see that N X n=1 cos n2θ sin nθ n = N X n=1 sin(n(n + 1)θ) −sin((n −1)nθ) n .
Now split into two sums and do an index shift: = N X n=1 sin(n(n + 1)θ) n − N−1 X n=0 sin(n(n + 1)θ) n + 1 = sin(N(N + 1)θ) N + N−1 X n=1 sin(n(n + 1)θ) n2 + n Sending N →+∞we find that ∞ X n=1 cos n2θ sin nθ n = ∞ X n=1 sin(n(n + 1)θ) n2 + n , which converges uniformly by using the bound | sin(n(n + 1)θ)| ≤1.
■ Source: AMM, C.E. Stanaitis [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 133 245 Solution 133 If the smaller square has area 16, then the shaded red triangle has area 8.
The red triangle, together with the gray triangle, takes up half the area of the larger square (!), thus the larger square has area 2(1 + 8) = 18 .
■ Source: Math Kangaroo [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 134 246 Solution 134 No (finite) number of pirates is sufficient!
Say there are n pirates. My strategy is as follows: 1. Move towards the shore until I am ε away from the edge of the lake. (Here, ε > 0 is a very small distance that will depend on n.) 2. Pick n + 1 disjoint arcs each satisfying the following property: If the interval has no pirate in it, then I am guaranteed to escape by moving straight to the interval’s midpoint.
3. Since n + 1 > n, one of these intervals must be empty. This is the only time I need to check where the pirates are.
4. Move towards that interval’s midpoint to escape!
All we need to do is pick ε > 0 so that Step 2 is possible. This entails selecting n disjoint arcs A1, A2, . . . , An+1 such that for all j, my distance to the midpoint of Aj is at most the length of Aj.
A1 A2 A3 P incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 134 247 A procedure for choosing ε and constructing these arcs is as follows. For ease, take the convention that arcs include their endpoints (and are thus closed).
Step 1 The idea here is to “take ε = 0”. Take a point P on the boundary of the circle, and we call an arc A safe if dist(P, midpoint(A)) ≤length(A).
We claim that safe arcs can be disjoint from P and contained in an arbitrarily small neigh-borhood of P. (To be more precise: Any arc with endpoint P, no matter how small, will contain a safe arc.) To see that this claim is true, choose a point M on the circle that is as close to P as you desire. Then, since any chord is shorter than the minor arc that it subtends, the arc with midpoint M and length MP will not contain P.
M P This arc is safe. Moreover it is evident that this arc “shrinks” to P as we move M towards P, which shows that the safe arcs can indeed be arbitrarily small and arbitrarily close to P.
This entails the claim.
Step 2 From Step 1, we can pick a safe arc A1 disjoint from P. But by the claim from Step 1, we can fit another safe arc A2 in the gap between P and A1. We can do this as much as we want, thus we can generate n + 1 safe arcs A1, A2, . . . , An+1, all of which are disjoint!
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 134 248 Step 3 Now we move P a very small distance ε > 0 away from the boundary. The danger in doing so is that the condition we desire on the arcs, dist(P, midpoint(Aj)) ≤length(Aj), may no longer be satisfied.
However, by taking ϵ to be sufficiently small, the amount that each of the distances dist(P, midpoint(Aj)) change by, over all j, can be made to be arbitrarily small! Thanks to this, the condition will be satisfied if we simply expand each arc Aj ever so slightly, to become a new arc A′ j centered at the same midpoint. If these arc expansions are small enough, then the new arcs A′ 1, A′ 2, . . . , A′ n+1 will still be disjoint. This completes the proof.
■ Remark: We have shown that no finite number of pirates is sufficient.
What about a countably infinite number of pirates? Shockingly, this is still not sufficient!
Showing this requires a more technical argument. Let the radius of the lake be 1. Let O be the center of the lake. Enumerate the pirates via the sequence {p1, p2, p3, . . . }. Then follow this procedure: 1. Move towards the edge of the lake until you are x1 away from the edge of the lake, where x1 > 0 is to be chosen later. Let your current location be the point A1. Set n = 1.
2. We will now “dodge pirate pn”. To do so, draw the chord through An which is per-pendicular to OAn, and move in the direction along this chord which faces away from pirate pn.
An pn Move in this direction until you are xn+1 > 0 away from the boundary, where xn+1 > 0 is to be chosen later. Let the point you arrive at be An+1.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 134 249 3. Increment n and loop back to the previous step, repeating infinitely.
This procedure constructs a continuous path. To verify that it has finite length, observe, using power of a point, that |AnAn+1|2 < (2 −xn)xn < 2xn.
(∗) Then we have the upper bound |OA1| + ∞ X n=1 |AnAn+1| ≤(1 −x1) + ∞ X n=1 √ 2xn on the length of the path. For this to converge, we can impose that xn ≤ 1 4n.
With this, the path must converge to a point, and since xn →0, the limit point lies on the boundary of the lake. In other words, the path will reach the edge of the lake at a point A∞. Fixing n, we claim that pirate pn cannot have reached A∞at this time.
To show this, note that when we reach An, the “worst case” position for pn will be the point Mn, defined as the point on the boundary such that O, An, and Mn are collinear in that order. Now: • Since the remainder of our path has length P∞ k=n |AkAk+1|, the set of points that pn can reach before we reach the shore is the arc with length 2 P∞ k=n |AkAk+1| and midpoint Mn.
• Once we reach An+1, the remainder of our path has length P∞ k=n+1 |AkAk+1|, so a superset of the set of points that the path can end at is given by the disk with center An+1 and radius P∞ k=n+1 |AkAk+1|.
Now rotate the diagram so that ray − − − − − → AnAn+1 points in the direction of the positive x-axis.
For the two sets described in the above two bullet points to be disjoint, it is sufficient to require that their sets of x-coordinates are disjoint.
• When we are at An, the x-coordinate of pn is exactly 0. Hence, the maximum pos-sible x-coordinate that can be reached by pn by the time we reach An+1 is given by sin (|AnAn+1|). After that, a loose upper bound on the maximum possible x-coordinate reached by pn when we reach A∞is sin (|AnAn+1|) + ∞ X k=n+1 |AkAk+1|.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 134 250 • When we are at An+1, our x-coordinate is exactly |AnAn+1|. Hence a loose lower bound on the minimum possible x-coordinate that we could end up at is given by |AnAn+1| − ∞ X k=n+1 |AkAk+1|.
Hence it suffices to achieve the inequality sin (|AnAn+1|) + ∞ X k=n+1 |AkAk+1| < |AnAn+1| − ∞ X k=n+1 |AkAk+1|, or 2 ∞ X k=n+1 |AkAk+1| < |AnAn+1| −sin (|AnAn+1|) , (∗∗) for all n. The next goal is to convert this to another sufficient inequality that is written in terms of the sequence {xn}n.
First we tame the left hand side of (∗∗). Thanks to the earlier observation (∗), we have that 2 ∞ X k=n+1 |AkAk+1| ≤2 √ 2 ∞ X k=n+1 √xk.
Now we tame the right hand side of (∗∗). From Taylor expansion it is not too hard to verify the bound x −sin x ≥x3 7 for all 0 ≤x ≤1. It follows that |AnAn+1| −sin (|AnAn+1|) ≥1 7|AnAn+1|3.
To continue, observe by n −1 applications of the Pythagorean Theorem that |OA1|2 + n−1 X k=1 |AkAk+1|2 = |OAn|2 = (1 −xn)2, and similarly |OA1|2 + n X k=1 |AkAk+1|2 = |OAn+1|2 = (1 −xn+1)2.
Subtracting these two equations and some factoring gives |AnAn+1|2 = xn(2 −xn) −xn+1(2 −xn+1).
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 134 251 Since xn, xn+1 ∈(0, 1), it follows that |AnAn+1|2 ≥xn −2xn+1.
Hence 1 7|AnAn+1|3 ≥1 7(xn −2xn+1)3/2.
Gathering the pieces, we finally arrive at the sufficient inequality 2 √ 2 ∞ X k=n+1 √xk ≤1 7(xn −2xn+1)3/2.
I leave it to the reader to verify that taking xn = 1 100n2n works.
So we have proven that for this choice of {xn}n, pn could not reach A∞by the time we do. But n was arbitrary, so the constructed path will successfully avoid the infinitely many pirates and escape.
This leads to a rather uncomfortable conclusion! On one hand, |N| pirates is insufficient.
On the other hand, |R| pirates is clearly enough. So the minimum number of pirates is either |R| or some cardinality strictly between |N| and |R|. I shall let you discover which is true.
Source: Puzzling Stack Exchange [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 135 252 Solution 135 We prove the contrapositive. Suppose that all subset sums of S = {a1, a2, · · · , an} are distinct. Then n Y i=1 (1 + xai) < ∞ X k=0 xk = 1 1 −x for all 0 < x < 1. Thus n X i=1 log(1 + xai) < −log(1 −x).
Dividing by x and integrating, we find that n X i=1 Z 1 0 log(1 + xai) x dx < Z 1 0 −log(1 −x) x dx.
The RHS evaluates to π2 6 (Sketch: Taylor expand!), so we focus on the LHS. Using the substitution u = xai we have du = aixai−1 dx = aiu1−1 ai dx, so Z 1 0 log(1 + xai) x dx = Z 1 0 log(1 + u) u 1 ai · du aiu1−1 ai = 1 ai Z 1 0 log(1 + u) u du.
This evaluates to 1 ai · π2 12 (Sketch: Taylor expand!). All in all we have n X i=1 1 ai · π2 12 < π2 6 , hence Pn i=1 1 ai < 2 as needed.
■ Remark: The 2 in the problem is tight, by taking S to be {1, 1 2, 1 4, 1 8, . . . , 1 2n} for increasingly large values of n.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 136 253 Solution 136 Let the disk have radius 1 and have center O. If the disk is partitioned into two congruent sets A and B, then O lies in one of these sets, say A. Let O′ be the point in the other set, B, that corresponds to O under the congruency.
O O′ Consider the diameter perpendicular to OO′.
The endpoints of this diameter cannot belong to B because they lie outside the unit circle centered at O′, so they instead belong to A. That is, there exists a segment of length 2 whose endpoints are in A and whose midpoint is O.
By the congruency, it follows that there must be a segment of length 2 whose endpoints are in B and whose midpoint is O′. But every segment of length 2 is a diameter, and the midpoint of every diameter is O, so certainly it could never be O′.
■ Source: Putnam [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 137 254 Solution 137 Let us call the required condition on the cake slices the fairness condition.
We start by cutting the cake into a 2:3 ratio. This ratio is far from tight — it just needs to be somewhere strictly in-between 1:2 and 1:1.
For every next cut, we always bisect the largest piece, but miss slightly so that (1) all piece sizes are distinct, and (2) the fairness condition holds. (Ensuring that all piece sizes are distinct is the key idea!) Our very first cut satisfied these two criteria, so it remains to show that this condition keeps holding inductively. Suppose that at a certain point, the sizes of the smallest and largest pieces are m and M, respectively. By inductive hypothesis, we know two crucial facts: (a) The largest piece is the only piece with size M (because all sizes are distinct!).
(b) M/2 < m If we were to bisect the piece of size M, then we get two new pieces of sizes M/2. By (b), these are the two new smallest pieces. It remains to transfer a small amount of mass from one of these pieces to the other to get sizes of M/2 −ϵ and M/2 + ϵ, while satisfying the required properties.
We must first ensure that double the size of the smallest piece, which is 2(M/2 −ϵ) or M −2ϵ, is greater than the size of the new largest piece. By (a), the new largest piece’s size cannot be M, so it is strictly smaller than M. Hence we can choose ϵ small enough so that M −2ϵ is larger. Thus we can satisfy the fairness condition for all ϵ small enough.
Now it remains to ensure all piece sizes are distinct. Since all piece sizes were distinct before the cut, I need only ensure that M/2 + ϵ is not equal to the size of any existing piece.
Indeed, since I have finitely many pieces, I can always ensure this by a careful choice of ϵ.
This completes the induction.
■ Remarks: I found this problem on Math Stack Exchange (
com/questions/2882265/optimal-strategy-for-cutting-a-sausage). The link contains some more interesting discussion on the problem.
For example, the above solution shows that the ratio r between the sizes of the largest and smallest pieces can be ensured to always satisfy r < 2. But can we do better than 2?
That is, can we find a different strategy so that the ratio r can be ensured to always satisfy incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 137 255 r < α where α is a constant number that’s even smaller than 2? The answer is no — 2 is the best you can do.
The accepted answer in the link also provides an interesting explicit construction for the cuts: If we view the cake as the interval [0, 1], then you can make the nth cut at {log2(2n+1)}, where the curly brackets denote the fractional part, {x} := x −⌊x⌋.
Source: Math Stack Exchange [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 138 256 Solution 138 Call my friends Emily and Sydney, number the gold pieces from 1 to 101, and for each piece i pick a distribution of pieces so that Emily and Sydney get the same weight of gold.
Now construct a 101 × 101 matrix M as follows: Mi,j = 0, i = j 1, Emily gets piece j if I take piece i −1, Sydney gets piece j if I take piece i Visually, row i depicts the pieces that would be in Emily’s share with 1’s if I choose piece i.
Let v be the 101 × 1 column vector that stores the weights of the gold pieces. Then by construction of M, Mv = 0. We would like to show that v is a scalar multiple of 1, the column vector consisting only of ones. Since M1 = 0, we have that 1 is in the null space of M, so it is sufficient to prove that the null space of M has dimension 1. This will force v, which is also in the null space, to be in the space spanned by 1, which is what we need.
By rank-nullity, we need only show that M has rank 100. To do this, take the upper 100 × 100 submatrix of M, and call it M ′. We are done if M ′ has full rank. We show this by proving that det M ′ ̸= 0. Since det M ′ is an integer, it is further sufficient to show that det M ′ is odd. In the pursuit of this, we may view the elements of M ′ as elements of F2, so that we hence need to show that det M ′ = 1. Over F2, M ′ is a matrix with 0’s on the diagonal and 1’s everywhere else. By a standard formula for determinant, we see that det M ′ is equivalent to the number of derangements in S100, which is odd because 100 is even.
■ Source: I forgot where I stole this from but it does seem to be relatively well-known.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 139 257 Solution 139 No matter the strategy, the odds will be 1/13. One way to convince yourself of this is to observe that if we stop at any time, then the probability that the top is an ace is always equal to the probability that the bottom card is an ace. So we may reformulate the game to an equivalent one as follows: Deal cards until you say stop, and then you win if the bottom card is an ace.
Well, the bottom card never changes, so the probability of victory will always be the same.
■ [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 140 258 Solution 140 It suffices to prove that AB ⊥CD, for if so, then a completely symmetrical argument shows that AC ⊥BD and AD ⊥BC, which will entail that any one of the four points will be the orthocenter of the triangle formed by the other three.
Let K and L be two perpendicular lines in the plane, and rotate them counter-clockwise by θ. Let f(θ) be the length of the orthogonal projection of AB unto the rotated K, and let g(θ) be the length of the orthogonal projection of CD unto the rotated L.
L K A B C D f(0) g(0) θ L K A B C D f(θ) g(θ) Notice that if there exists an angle θ for which f(θ) = g(θ), then there will exist a square whose sides will pass through the four points A, B, C, and D. (This is not quite an equivalent condition as it only accounts for one possible assignment of the four points to the square’s four sides.) Since the hypothesis asserts the contrary, we have f(θ) ̸= g(θ) for all θ.
If we let α be the angle at which AB intersects K, and β be the angle at which CD intersects L, then it is not hard to discover that f(θ) = |AB| · | cos(α + θ)| and g(θ) = |CD| · | cos(β + θ)|.
Since these two quantities are never equal, we may write |AB| |CD| ̸= cos(β + θ) cos(α + θ) for all θ, which in turn implies that the function θ 7→cos(β+θ) cos(α+θ) cannot have full range (image) over the domain on which it is defined. If the zeroes of θ 7→cos(β +θ) and θ 7→cos(α+θ) do incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 140 259 not coincide, then this function will have zeroes and attain the limits ±∞at singularities, which will give it full range by continuity. Thus cos(α + θ) and cos(β + θ) must have the same zeroes, which can occur only when α and β differ by a multiple of π.
This implies that the angle formed between AB and CD will be π 2 plus some multiple of π, thus they are perpendicular, as needed.
■ [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 141 260 Solution 141 Solution 1 For each 1 ≤i ≤n, let vi ∈Rn be the vector whose kth component is p φ(k) if k | i, and 0 otherwise. Let A be the matrix whose ith column is vi. For example, when n = 5, A = p φ(1) p φ(2) p φ(3) p φ(4) p φ(5) 0 p φ(2) 0 p φ(4) 0 0 0 p φ(3) 0 0 0 0 0 p φ(4) 0 0 0 0 0 p φ(5) .
Then vi · vj = n X k=1 1(k | i and k | j) · p φ(k) = n X k=1 1k|gcd(i,j) · φ(k) = X k|gcd(i,j) φ(k) = gcd(i, j), where in the last line we applied the identity m = P k|m φ(k) with gcd(i, j) in place of m.
It follows that the matrix in the original question, whose (i, j) entry was gcd(i, j), is given by ATA. Its determinant is thus the square of the determinant of A. But A is upper triangular with diagonal entries p φ(1), p φ(2), . . . , p φ(n), so det A = p φ(1)φ(2) . . . φ(n) and the desired determinant is the square of the right hand side, which was what was sought.
■ Solution 2 Here I present the proof given in the original paper by Smith which proved this determi-nant identity.
Let n = Qk i=1 pki i . Then φ(n) = n k Y i=1 1 −1 pi = X S⊆[k] (−1)|S|n Y j∈S 1 pj .
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 141 261 Let aS := n Q j∈S 1 pj , so φ(n) = X S⊆[k] (−1)|S|aS.
(∗) Now, for each set ∅⊂S ⊆[k] we add (−1)|S| times column aS to the last column. Focus on what happens to the last row. From this column operation, we see from (∗) that the (n, n) entry will become X S⊆[k] (−1)|S| gcd(n, aS) = X S⊆[k] (−1)|S|aS = φ(n).
It remains to prove that (m, n) entry is 0 for each 1 ≤m < n, so that we may conclude by argument of induction.
The (m, n) entry is given by X S⊆[k] (−1)|S| gcd(m, aS).
We must show that this sum is 0. Since m < n, there exists a prime pj such that vpj(m) < vpj(n). This prime is the catalyst for causing cancellations in the sum: We claim that (−1)|S| gcd(m, aS) + (−1)|S∪{j| gcd(m, aS∪{j}) = 0 for each S ⊆[k] with j ̸∈S. This will complete the proof.
We need only show that gcd(m, aS) = gcd(m, aS∪{j}), or gcd m, n Y i∈S 1 pi !
?
= gcd m, n · 1 pj Y i∈S 1 pi !
.
(?) If pj ∤m, evidently they are equal, since removing a factor of pj from n will not change the gcd. If pj | m, then we need only check that the exponent of pj is the same on both sides of (?). Indeed, note that vpj(m) < vpj(n) = vpj n Y i∈S 1 pi !
(∗∗) and thus vpj(m) ≤vpj n Y i∈S 1 pi !
−1 = vpj n · 1 pj Y i∈S 1 pi !
(∗∗∗) From both (∗∗) and (∗∗∗) we conclude that there are vpj(m) factors of pj on both sides of (?), as needed.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 142 262 Solution 142 Call the gloves Glove A and Glove B. To mix all the three bowls, I can use the following procedure. All gloves are to be put on my dominant hand, which shall do all the mixing.
1. Wear Glove A, and then wear Glove B on top of that. Then I’ll mix the the first bowl with this.
2. I’ll take off Glove B and then mix the second bowl using only Glove A.
3. Lastly, I turn Glove B inside-out and wear it on top of Glove A. I mix the third bowl with this.
Tada!
■ Remark: Those who recognize the problem’s premise have likely inferred that I’ve deemed its original context quite unsavory for a general audience. This is true, and rewriting the problem has proven to be quite bothersome. Indeed, it took me more than a year to come up with a more family-friendly setting!
Here are two highly mathematical problems relying on the same premise, for your perusal.
• m employees from company A are meeting with n companies from company B for a company merger. Every employee from A must shake hands with every employee from company B. Assuming that all employees are germaphobes, what is the least number of gloves needed to accomplish this?
• n germaphobic graduate students are getting to know each other. Every pair of grad-uate students must shake hands. What is the least number of gloves needed to accom-plish this?
Source: Classic [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 143 263 Solution 143 Let Leilani be the first guard to complete their loop around the museum (or, any such guard that completes their loop at the first time in which a guard’s loop is completed).
Consider a point T in time at which Leilani is in the final room in her tour of the museum before she returns to her assigned room.
We claim that at time T, no guard is watching their assigned room. Indeed, Leilani is not watching her assigned room. Now consider any other guard, and suppose for contradiction that this guard is watching their assigned room. Then there are two possibilities: Either (1) the guard has not yet started their tour, or (2) the guard has finished their tour.
Since Leilani was the first guard to complete their tour, (2) is impossible. So (1) must be the case. That is, the guard has stayed put up to time T. But Leilani visited this guard’s assigned room without encountering them, a contradiction.
■ Remark: You can also consider the last guard to start their tour, and then study a point in time when this guard has just exited their assigned room.
Source: Leningrad Mathematical Olympiad, abridged slightly by me [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 144 264 Solution 144 Part (a) The answer is yes. Take T to be the following triangle formed by three vertices of a regular heptagon H.
A B C D E F G To win with this triangle, Amber only needs to choose 7 points that form a regular heptagon congruent to H. This works because any 2-coloring of the vertices of H must contain a congruent copy of T.
One way to reason this is as follows: Since 7 is odd, two adjacent vertices must have the same color. Without loss of generality we may assume that these vertices are D and E (following the labels in the diagram) and that they are red. If B is red then △BDE is a red triangle congruent to T, so we may assume that B is blue. Similarly we may assume G is blue. Now if C is blue then △BGC is a blue triangle congruent to T, so we may assume C (and, symmetrically, F) are red. With this, △CFD is a red triangle congruent to T.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 144 265 Part (b) I thank “wen” for finding this construction. The answer is yes. Take T to be an equilateral triangle with height 1 (hence a side length of 2 √ 3). Then, to prevent Amber from winning, Beth employs the following simple strategy: If Amber chooses the point (x, y), then Beth colors (x, y) red if ⌊x⌋is even, and otherwise colors (x, y) blue if ⌊x⌋is odd.
In this way, Beth essentially colors the xy-plane in alternating “stripes” of red and blue.
-5 -4 -3 -2 -1 0 1 2 3 4 5 Beth’s strategy, with some sample copies of T Amber can never win because T is “too wide” to fit inside one stripe, but “too thin” for its vertices to land in two different same-color stripes.
More concretely, a copy of T can fit inside a red stripe if and only if there exists a line such that the orthogonal projection of T unto this line has length less than 1. It is not too hard to see that the minimum possible value for the length of this projection (the “minimum width” of T) is given by the length of the shortest height of T, which is 1. So this case is not possible.
On the other hand, for a copy of T to have vertices across two different red stripes, say, A in one stripe and B, C in the other, we have that the height from A to BC crosses the width of a blue stripe. So this height has length greater than 1, which is impossible since the heights of △ABC have lengths exactly 1.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 144 266 The same reasoning holds for the blue stripes, exhausting all reasonable possibilities for a monochromatic copy of T to exist, completing the proof. In fact, this argument shows that any triangle for which such a coloring will result in a win for Beth must have all of its heights to have length both ≥1 and ≤1, and it is easy to see that the equilateral triangle of height 1 is the only such triangle.
■ [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 145 267 Solution 145 Since each side is 2π-periodic and is invariant under a reflection about x = π, it is sufficient to consider 0 ≤x ≤π. In fact, since a reflection over x = π/2 negates sin(cos x) but does nothing to cos(sin x), it is sufficient to consider 0 ≤x ≤π/2, where sin(cos x) is non-negative.
For such x, we may write sin(cos x) ≤cos x ≤cos(sin x), (∗) where we have applied the inequality sin y ≤y for y ≥0, twice, and have used the fact that cos is decreasing over [0, π/2].
Since the equality case of sin y ≤y occurs only when y = 0, we could only have equality in (∗) if cos x = 0 and x = 0 hold simultaneously, which cannot be the case, so it is strict.
■ [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 146 268 Solution 146 There are six points of tangency, distributed symmetrically over the surface of the sphere, and so they form the vertices of a regular octahedron with edge length √ 2. The answer is the circumradius of one of its sides, which is q 2 3.
■ Source: Me, for CMIMC [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 147 269 Solution 147 This video by polylog ( does a very nice job at showcasing a construction for the dice. Give it a watch if you’d prefer a more visual explanation.
It suffices to construct a finite sequence of elements in {1, 2, ..., n} such that the subse-quence π(1), π(2), . . . , π(n) appears equally many times over all permutations π : {1, · · · , n} → {1, · · · , n}.
We go by induction. Obviously, by taking the sequence 1, 2, · · · , n, we can construct a sequence such that the one-element subsequence (i) appears equally many times over all i.
Now assume that for some 1 ≤k < n, we have constructed a sequence A for which the sequence a1, a2, · · · , ak appears as a subsequence of A equally many times over all possible selection of k distinct elements a1, a2, · · · , ak ∈{1, 2, · · · , n}.
Let π1, π2, · · · , πn! be the n! permutations on {1, 2, · · · , n}. Denote by πi(A) the sequence whose jth element is πi(Aj). That is, it is simply the sequence A but with its elements relabeled according to πi. Now we take the sequence B := π1(A)⌢π2(A)⌢· · ·⌢πn!(A), where ⌢denotes concatenation of sequences. We claim that the sequence b1, b2, · · · , bk+1 appears as a subsequence of B equally many times over all possible selection of k +1 distinct elements b1, b2, · · · , bk+1 ∈{1, 2, · · · , n}.
It is easier to see this by reverting back to the probabilistic view: Let us select a random subsequence b of B which consists of k + 1 distinct integers. Then we claim that every selection and ordering of these k + 1 integers is equally likely.
There are two ways to pick such a subsequence b.
• The first way is that all elements of b fall into the same πi(A) sequence for some i. If we must choose b in this way, then it is equally likely to fall into any πi(A) for any 1 ≤i ≤n!. Since π1, · · · , πn! runs through all permutations, this forces every selection and ordering of b to be equally likely.
• The second way is that the first way does not occur. That is, each πi(A) contains at most k elements of b. Suppose we restrict the probability space to a specific way to distribute the elements in this way (e.g. consider only subsequences b with 3 elements in π7(A), 2 elements in π8(A), and k −4 elements in π9(A)). By definition of A, every possible selection and ordering of the elements of b within some πi(A) is equally likely.
This holds for all i, so every selection and ordering of b is equally likely.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 147 270 This completes the induction.
■ [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 148 271 Solution 148 Consider a gridline segment. If it’s on the border, then it can be claimed and counted towards the perimeter by claiming the square that it borders. Otherwise, the segment will either not count towards either player’s perimeter (when both adjacent squares are claimed by the same player) or it will count equally towards both player’s perimeters (when the adjacent squares are claimed by different players).
Thus, no interior gridline segment will help either player win.
The only factor that contributes to the difference in the players’ perimeters is the number of segments claimed on the boundary of the grid. Thus the best strategy entails claiming as many such segments as possible.
The corner squares are worth the most since they each contain two boundary segments.
Ashley and Beth must first rush to claim as many of these as possible, and they will be tied in doing so because four is even. Then, since there are an even number of non-corner squares along the boundary (4 × 2021, to be exact), they will also be tied in claiming the number of such squares. So they will tie in the end. This fully decides the game since, as discussed, nothing that occurs in the interior of the board actually matters. So they will tie.
■ Source: Math Hour Olympiad [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 149 272 Solution 149 Part (a) We take the n 2 points that have exactly two components equal to 1, with all other components equal to 0.
For any two distinct such points, their “1-components” either “overlap” at one component, resulting in one possible distance ( √ 2), or they do not “overlap”, resulting in a second possible distance ( √ 4).
Part (b) We use the construction for Part (a) for the next dimension, Rn+1! The key observation is that the n+1 2 points in the constructed set are coplanar (i.e. lie in a common n-dimensional subspace), and this is because they all lie in the hyperplane x1 + x2 + · · · + xn+1 = 2. This hyperplane is a copy of Rn, thus this corresponds to a two-distance set of size n+1 2 in Rn.
■ [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 150 273 Solution 150 Part (a) The answer is neither!
We will biject each way that Ai could go broke first to a way that Beth can go broke that has the same probability of occurring. This will prove that Ai and Beth are equally likely to have gone broke first.
The bijection is simple: For a sequence of coin flips where Ai goes broke first, flip every heads to a tails and vice versa, then switch Ai’s and Beth’s sequences of coinflips.
It is easy to see that this results in a game where Beth goes bankrupt first instead of Ai, and that this is a bijection.
To see that they have the same probability of occurring, observe that when Ai and Beth go bankrupt, the difference between the tails flipped and heads flipped by Ai is equal to the difference between the heads flipped and tails flipped by Beth, because they start with the same amount of money. Thus the total number of heads flipped is equal to the total number of tails flipped. Ergo, switching all heads to tails and vice versa does not change the total number of heads and tails.
Part (b) We use the same bijection in which we flip every heads to a tails and vice versa. However, we now note that in the new sequence, the probability decreases.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 150 274 This is because in the original sequence, Beth goes broke last, meaning there were more heads than tails. The transformed sequence will then have more tails than heads. Since heads are more likely, the original sequence will always be more likely than the transformed sequence.
We deduce that Beth is more likely to go broke last. So Ai is more likely to go broke first.
■ Source: Peter Winkler [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 151 275 Solution 151 Beth needs to only spend two dollars.
Beth first obtains an upper bound on the largest coefficient by asking for P(1). Let k be the number of digits of P(1). Beth now knows that every coefficient has at most k digits.
So, Beth now asks Angela to hand over the value of P(10k), and this forces Angela to quite literally write down all the coefficients plainly.
For example, if P(x) = 31x2 + 41x + 59, then Angela will end up giving Beth the value of P(1000), which is 31041059. The selection of k ensures that no two coefficients “collide” upon computing P(10k).
■ Remarks: Beth could have used any base in place of 10. If Beth were allowed to request P(a) for a a real number, then in theory one dollar would be enough by simply asking or P(π). Though, this would require infinitely precise computation to ensure victory.
Source: Classic [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 152 276 Solution 152 A clean approach is as follows. To tie a rope around two nails consists of a sequence of the following four actions: • (L) Wrap it once counter-clockwise around the left nail.
• (L−1) Wrap it once clockwise around the left nail.
• (R) Wrap it once counter-clockwise around the right nail.
• (R−1) Wrap it once clockwise around the right nail.
A sequence of such actions can be simplified if inverse operations are adjacent. For example, LR−1RR simplifies to LR, and this represents the weight of the painting untying some of its loops. If a sequence can fully simplify into an empty sequence, such as LRR−1L−1, then this represents the knot failing and the painting falling.
We aim to generate such a sequence of these actions that cannot be simplified to an empty sequence, but would collapse to an empty sequence if either all “left” actions (L, L−1) are removed or all “right” actions (R, R−1) are removed, as this corresponds to removal of the left nail or right nail.
Indeed, the commutator LR−1L−1R works! This corresponds to the following picture.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 152 277 You can verify visually that this painting will indeed fall if either nail is removed.
■ Remarks: See the paper for more problems of this flavor. (It comes with nice pictures!) [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 153 278 Solution 153 We claim BEE is located here: Suppose for contradiction that BEE is located elsewhere. If the location of BEE is known, then for any sequence of cells of the form EE or EE, we may deduce that the blank square is an E, since we are guaranteed that BEE cannot appear more than once. Call this principle “tripling”.
Consider the two “chains” of squares: incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 153 279 (The chain of shaded squares leading out left of the diagonally-adjacent pair of E’s is colored red. The chain of shaded squares on the right is colored blue. The square above the bottom-most E is colored purple.) We claim that the purple square is E. Suppose not. The BEE must be somewhere, and the BEE’s B cannot be on both the red and blue chains at once. So one of the chains does not contain the BEE’s B. Without loss of generality, suppose it is the red chain. Then by tripling, starting from the two diagonally-adjacent E’s, we must keep placing E’s along the red chain until we reach the purple square, which is a B, thus forming a second BEE, contradiction.
Using the same idea, it is not hard to deduce also that one of the red or blue squares diagonally adjacent to the purple square is also an E. Without much harm to the generality of the argument, let us place the E on the blue square.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 153 280 Since we have assumed for contradiction that the BEE we’re looking for is not where we claimed it is, we know that the pink square is an E. From here we need to do a bit of gruntwork to expand to a more workable structure. First, observe by a simple inspection that none of the three purples squares shown below can be a B, otherwise two BEEs will be created among these squares.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 153 281 Next, the below purple square must be an E, otherwise by three triplings we see that the orange squares are E and then the yellow square must be an E, creating two BEEs.
(The colors of the shaded squares, in “book order”, are: orange, purple, orange, yellow) At this point, we see that if any of the four below purple squares are B, then the other three must be E, creating two BEEs. So in fact, all of them are E.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 153 282 What we have accomplished with this work is the creation of a ”square” of E’s. The major claim is that we can always expand any square of E’s in any direction. Note that in general it is safe to assume that the empty squares above exist.
Indeed, we have for free that the three below purple squares are E, incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 153 283 and that, as before, all of the four purple squares below must simultaneously be E.
We have thus proven the major claim. By iteratively applying the major claim, we may fill the following squares with E’s.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 153 284 From here, it is now not difficult to argue that all of the remaining blank squares must be the same letter, i.e. either all B’s or all E’s. It follows that BEE does not appear in the grid, contradiction.
■ Source: Me!
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 154 285 Solution 154 Let the quartic be P(x) = ax4 + bx3 + cx2 + dx + e.
• Adding a linear function to P(x) corresponds to an affine transformation of the plane, which preserves ratios along lines. Thus we may add −dx −e to assume WLOG that P(x) takes the form ax4 + bx3 + cx2.
• By vertical scaling, which also preserves ratios along lines, we may divide by a to assume that P(x) takes the form x4 + bx3 + cx2.
• −b is the sum of the roots of P, and by translation we can assume that this sum is zero. So we may assume that P(x) takes the form x4 + cx2.
• By a horizontal scaling by a factor of 1 √ |c|, followed by another vertical scaling, we may assume that P(x) is either x4 + x2 or x4 −x2.
The second derivatives of these two candidates are 12x2 + 2 and 12x2 −2. Since there are two inflection points, we may eliminate the first candidate. The inflection points then occur at the roots of x2 −1/6. Now write x4 −x2 + 5/36 = (x2 −1/6)(x2 −5/6).
This shows that the line connecting these inflection points is given by y = −5/36, and that the other intersections of this line with the graph of x4 −x2 occur at x = ± p 5/6. The desired ratio is then p 5/6 −(− p 1/6) p 1/6 −(− p 1/6) = √ 5 + 1 2 = φ as needed.
■ Source: This is called “Lin McMullin’s Theorem”.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 155 286 Solution 155 Part (a) We guess that f(x) takes the form axb. If f ′(x) = f −1(x), then abxb−1 = a−1/bx1/b.
So we wish to solve the system ( ab = a−1/b b −1 = 1/b .
The second equation immediately gives b = φ. So it remains to solve for a in φa = a−1/φ.
Dividing by a gives a−1/φ−1 = φ or a−φ = φ. So a = φ −1 φ . We conclude that f(x) = φ−1/φxφ is a solution.
■ Part (b) Step 0 Since x ∈(0, ∞), we must have f −1(x) > 0, and so f ′(x) > 0 for all x. Moreover we must have f(0+) = 0 and f(∞) = ∞in order for f −1 to be well-defined for all x > 0. In view of this we may treat f to be of type f : [0, ∞] →[0, ∞].
Step 1 The key tool is that if we know that f(x) > axb for x ∈[0, T], then we have the following deductions: f(x) > axb for 0 ≤x ≤T = ⇒f −1(x) < x a 1/b for 0 ≤x ≤f(T) = ⇒f ′(x) < x a 1/b for 0 ≤x ≤f(T) = ⇒f(x) < Z x 0 t a 1/b dt for 0 ≤x ≤f(T) = ⇒f(x) < b (b + 1)a1/bx b+1 b for 0 ≤x ≤f(T) We similarly have that f(x) < axb for 0 ≤x ≤T = ⇒f(x) > b (b + 1)a1/bx b+1 b for 0 ≤x ≤f(T).
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 155 287 For this to be of use, we need to select a T so that [0, T] = [0, f(T)]. This motivates locating a non-trivial fixed point of f.
Suppose for contradiction that f has no fixed point in (0, ∞). Then either f(x) > x or f(x) < x for all 0 < x < ∞. If f(x) > x = 1x1 then f(x) < 1 1+1 x 1 1+1 1 = 1 2x2 for all such x, and it is not too hard to see that this is contradictory with f(x) > x. Similarly the hypothesis f(x) < x leads to a contradiction. So a fixed point x0 ∈(0, ∞) exists.
Step 2 x0 is, in fact, the unique fixed point in (0, ∞). To see this, write f ′′(x) = d dxf −1(x) = 1 f ′(f −1(x)) > 0 to deduce that f is strictly convex for x > 0. Now, as f(0) = 0 and f(x0) = x0, it must follow that f(x) < x for 0 < x < x0, and f(x) > x for x0 < x < ∞.
Step 3 We now repeatedly apply Step 1 to the inequality f(x) < x over the interval [0, x0]. Let a0 = 1, b0 = 1, and, recursively, define an := bn−1 (bn−1 + 1)a 1 bn−1 n−1 and bn := bn−1 + 1 bn−1 for n ≥1. Then f(x) < a0xb0 for x ∈[0, x0], and so Step 1 tells us that f(x) > a1xb1 for x ∈[0, x0]. Proceeding inductively, we discover that a2k+1xb2k+1 < f(x) < a2kxb2k, x ∈[0, x0] for all k. In the next step we will see that as k →∞, the upper and lower bounds will squeeze f to the function φ −1 φ xφ.
Step 4 We will now show that the sequences {an}n and {bn}n converge. It is classical that {bn}n converges to φ and so I will omit the proof of this. As for {an}n, we will apply the following lemma which I will prove later.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 155 288 Lemma 1 Suppose that the sequences {rn}n and {sn}n converge to r and s respectively, and that |r| < 1. Then the sequence {un}n recursively defined by un := rn−1un−1 + sn−1 converges for any initial data u0.
To apply this lemma, apply the logarithm to the definition of an to find that log an = log bn−1 (bn−1 + 1) − 1 bn−1 log an−1.
We apply the lemma to the sequence {log an}n, noting that − 1 bn−1 →−1 φ and that −1 φ < 1.
This tells us that log an converges, hence so does an.
To find the value of the limit of an, we simply send n →∞in the definition of an. This, combined with some algebra, will give us that limn→∞an = φ−1 φ, which is what we expected.
From the convergence of these sequences, we may conclude by the squeeze rule that f(x) = φ−1 φxφ for all x ∈[0, x0].
Step 5 Since x0 is a fixed point, x0 = f(x0) = φ−1 φxφ 0 and so we may now solve for x0. Working out the algebra, we find that x0 = φ.
Step 6 We know from strict convexity that f(x) > x over (x0, ∞) = (φ, ∞). It follows that, for any large T ≫φ of our choice, we have that f(x) > 1 T x2 for all 0 < x < T. Indeed, for 0 < x ≤φ you can verify that f(x) = φ−1/φxφ > 1 T x2, and for φ < x < T we have f(x) > x > 1 T x2.
Hence, if we define the sequences {an}n and {bn} as in Step 3, with initial data a0 = 1 T and b0 = 2, then by Step 1, we must have f(x) < a1xb1 for 0 < x < f(T). But f(T) > 1 T T 2 = T, so f(x) < a1xb1 holds for 0 < x < T.
Inductively, as in Step 3, we thus obtain a2kxb2k < f(x) < a2k+1xb2k+1, x ∈[0, T) incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 155 289 for all k. By Step 4, the upper and lower bounds converge and squeeze f(x) so that we get f(x) = φ−1/φxφ for all 0 < x < T. But T was arbitrary, so this conclusion holds for all 0 < x < ∞. This completes the main proof.
Step 7 Finally, we prove the lemma. The proof I present is a bit weird, and it turns out to be slightly cleaner to re-index the sequences: Suppose rn →r and sn →s with |r| < 1, and recursively define un = rnun−1 + sn. Then we claim un converges for any u0.
We in fact claim that un → s 1−r.
Since sn 1−rn → s 1−r, it is sufficient to show that un − sn 1−rn →0.
Before we begin to do this, it will be important for later to demonstrate that {un}n is bounded. First, pick some R > 0 strictly between |r| and 1. Then there is some N large enough so that |rn| < R < 1 for all n ≥N. For all such n we have |un| ≤R|un−1| + S where S is an upper bound on |sn|. By induction, we have that |un| ≤hn−N(|uN|) where h(z) := Rz + S, and since |R| < 1 the Banach Fixed Point theorem applied to h shows that limn→∞hn−N(|uN|) exists, proving that un is bounded.
We return to the proof that un − sn 1−rn →0. Fix ε > 0 and find Nε such that for all n > Nε, the following hold: • sn 1−rn − s 1−r < ε 10 (In particular we seek sn 1−rn − sn−1 1−rn−1 < ε) • Nε ≥N, so that |rn| < R < 1.
Then, for n > Nε, we have the following bound. un − sn 1 −rn = rnun−1 + sn − sn 1 −rn = rnun−1 −rnsn 1 −rn = |rn| · un−1 − sn 1 −rn ≤|rn| · un−1 − sn 1 −rn + |rn| · sn 1 −rn − sn−1 1 −rn−1 ≤|rn| · un−1 − sn 1 −rn + |rn|ε.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 155 290 Inductively, un − sn 1 −rn ≤|rnrn−1 . . . rNε+1|· uNε − sNε 1 −rNε +|rn|ε+|rnrn−1|ε+· · ·+|rnrn−1 . . . rNε+1|ε.
Now, we will demonstrate that this is small. For the first term, both un and sn 1−rn are bounded sequences, hence so is un − sn 1−rn . Moreover |rk| ≤R < 1 for all n ≤k ≤Nε+1. So, if M is an upper bound on un − sn 1−rn , then |rnrn−1 . . . rNε+1| · uNε − sNε 1 −rNε ≤Rn−NεM.
For the other terms, we again apply the bound |rk| ≤R to find that |rn|ε + |rnrn−1|ε + · · · + |rnrn−1 . . . rNε+1|ε ≤ε n−Nε X k=1 Rk ≤ε ∞ X k=0 Rk ≤ ε 1 −R.
In all, we have for all n > Nε that un − sn 1 −rn ≤Rn−NεM + ε 1 −R.
Sending n →∞, lim sup n→∞ un − sn 1 −rn ≤ ε 1 −R.
But ε was arbitrary, so un − sn 1−rn →0.
■ Remark: I’ve taken the time to write this argument in detail because most popular sources which mention this problem do not seem to bother with proving the uniqueness of the solution. This is perfectly understandable since any proof of uniqueness will likely be quite technical.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 156 291 Solution 156 The answer is 1 φ where φ is the golden ratio. Let Pn k=0 akxk be our polynomial, so that ak ∈{0, 1} for all k and 0 = n X k=0 akzk.
Since z ̸= 0, we may divide by a power of z so that the constant term is 1. That is, we may assume WLOG that a0 = 1. Thus 0 = 1 + n X k=1 akzk.
There are now two cases.
Case 1: a1 = 0, i.e. there is no z1 on the RHS.
Then −1 = n X k=2 akzk, and so we may use the rough bound 1 ≤ n X k=2 ak|z|k ≤ ∞ X k=2 |z|k = |z|2 1 −|z|.
So |z|2 + |z| −1 ≥0 which directly implies |z| ≥1/φ.
Case 2: a1 = 1 Then we have 0 = 1 + z + n X k=2 akzk.
Multiplying by 1 −z on each side gives 0 = 1 −z2 + n X k=3 bkzk where bk ∈{−1, 0, 1}. Now there is no z term, and the coefficients of −1 are not an issue for the argument in Case 1, so we may repeat the argument in Case 1 to deduce again that |z| ≥1/φ.
This shows that 1/φ is a lower bound. To “obtain” 1/φ, note that z = −1/φ is a solution to 0 = 1 + z + z3 + z5 + z7 + . . . .
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 156 292 So we can expect that for large n, there is a root of 1 + z + z3 + z5 + z7 + . . . + z2n+1 that is quite close to −1/φ.
If you want the murky details, here you go. Fix ε > 0. Let fn(z) = 1 + Pn k=0 z2k+1 and gn(z) = P∞ k=n+1 z2k+1, so that z = −1/φ is a root of fn+gn. Since |−1/φ| < 1, we have that fn + gn is holomorphic around a neighborhood Dr(−1/φ) of −1/φ by studying the radius of convergence. For a choice of 0 < r < ε small enough we can guarantee that fn + gn does not vanish on ∂Dr(−1/φ), so |fn + gn| ≥δ > 0 over ∂Dr(−1/φ). Moreover we see that gn →0 uniformly on Dr(−1/φ) so we may pick an n so large that 2|gn| < δ on ∂Dr(−1/φ). For this n we have 2|gn| ≤δ < |fn + gn| ≤|fn| + |gn| or |gn| < |fn|, over ∂Dr(−1/φ). Of course, both fn and gn are holomorphic in Dr(−1/φ), so by Rouch´ e’s theorem fn and fn + gn have the same number of roots in Dr(−1/φ). We conclude that fn, which is a polynomial, has at least one root in Dr(−1/φ) ⊆Dε(−1/φ).
But ε was arbitrary.
■ Source: Putnam, modified slightly [Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 157 293 Solution 157 We first take u = 1 + xφ to convert to I = 1 φ Z ∞ 1 1 uφ(u −1) φ−1 φ du In order to make the integrand’s structure more symmetrical, we now take v = 1/u to write I as I = 1 φ Z 1 0 vφv φ−1 φ v2(1 −v) φ−1 φ dv.
Miraculously, using, φ2 −φ −1 = 0, this simplifies as = 1 φ Z 1 0 1 (1 −v) φ−1 φ dv.
This evaluates to 1 φ · 1 φ−1, which is just 1, as needed.
■ Remark: We are quite confident that p = φ is the unique value of p that solves Z ∞ 0 1 (1 + xp)p dx = 1.
This can be seen visually using Desmos by either graphing the function f(x) = R 50 0 1 (1+tx)x dt (which is computationally expensive) or using a few substitutions to discover that, for p > 1, Z ∞ 0 1 (1 + xp)p dx = Γ p −1 p Γ 1 p Γ(1 + p) , and then graphing Γ(x−1 x)Γ( 1 x) Γ(1+x) (by implementing the Gamma function as (x −1)!). This is equal to the original integral only on (1, ∞) due to convergence issues on (0, 1). I do not know of an elegant way to rigorously demonstrate the uniqueness of p = φ.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 158 294 Solution 158 Let n ≥1 be the degree of P, and let x1, . . . , xn be the roots of P.
Black Magic Claim: For all real x ∈[0, 3], we have |x(x −1)(x −2)(x −3)| ≤1, with equality iff x ∈{3± √ 5 2 }.
Proof.
Write x(x −1)(x −2)(x −3) = (x2 −3x)(x2 −3x + 2) = (x2 −3x + 1)2 −1.
So the inequality is equivalent to showing that |x2 −3x + 1| ∈[0, √ 2|. An analysis of the extreme values of the quadratic x2 −3x+1 over [0, 3] reveal that, in fact, 0 ≤|x2 −3x+1| < √ 2, and the equality case |x2 −3x + 1| = 0 occurs exactly at the roots of the quadratic, x = 3± √ 5 2 .
□ With the claim, the proof is amazingly short: note that by the claim applied to each xi, |P(0)P(1)P(2)P(3)| = n Y i=1 |xi(xi −1)(xi −2)(xi −3)| ≤1.
On the other hand, since P has no integer root, we know |P(0)P(1)P(2)P(3)| ≥1.
Thus equality holds everywhere. In particular, equality holds in the black magic claim for each xi, so xi ∈{3± √ 5 2 } for all i. The end is simple: in order for P to have integer coefficients, there must be an equal amount of 3+ √ 5 2 and 3− √ 5 2 among its roots, so in particular 3+ √ 5 2 is a root. Thus P 3+ √ 5 2 = 0.
■ Remarks: To say that this problem is shrouded in mystery would be an understatement.
Back in my high school years, I put this in my personal collection of interesting problems, but I didn’t mark its source. To this day, despite my best efforts to track down its origins, I haven’t the slightest clue where it came from, much less who invented it.
Unfortunately, I also didn’t know how to solve it. In several layers of outsourcing, some very dedicated solvers came up with a variety of convoluted but fascinating methodologies.
It was only after quite some time that someone on AoPS sent me the “”true” solution given above: a completely elementary proof of the statement that was not more than half a page.
Yet, almost comically, they did not know the source of the proof — a black magic solution from nowhere to a problem that came from nowhere.
Source: I have no idea.
[Back to Index] incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 158 295 Solution to the Grand Finale After reading all the problems, it is evident that there is something funky going on. The 12 problems appear to be “positioned” in some way, with their answers labeled using the letters A, B, C, D, E, F, G, H, I, J, K, and L. The biggest hint as to what is going on is revealed in Problem 160: • “Let B be the answer to the problem that is counter-clockwise adjacent to (and at the same altitude as) this one...” • “The ’counter-clockwise’ direction is from the perspective of an observer looking down from above...” This suggests that the problems are positioned as points in 3D space, and connected to each other in some way in order to determine which problems are “adjacent”.
The exact shape formed by these problems and their connections must be inferred from some other details. These include: • There are exactly 12 problems.
• One problem is labeled the “abyss”, and one problem is labeled the “peak”.
• Problems 162, 164, and 166 imply that there is only one problem below them.
• Problems 161, 163, 165, 167, 169 all mention the “two adjacent problems below” them.
• Problem 170, the “peak”, says there are exactly five problems adjacent to it.
The number five is a particularly damning piece of evidence: This is an icosahedron.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 158 296 (A is the bottom-most vertex. Above A is the regular pentagon BCDEF, whose vertices are written in counter-clockwise order from when looking from above. Above this is pentagon GHIJK, also in counter-clockwise order. L is the top-most vertex.) Using the relative positionings implied by the problem statements, we can match up letters with problem numbers. The assignment is confirmed when one notices that the sequences A, B, . . . , K, L and 159, 160, . . . , 169, 170 both correspond to paths connecting adjacent ver-tices starting from the bottom and ending at the top. Note that the letters G, H, I, J, and L are never mentioned in the problems, but their existence and locations can be inferred from the previous points.
The arrows in the above diagram represent the dependencies required to solve each prob-lem. However, none of the answers are given, and the only problem with no dependencies, Problem 159 (A), has been corrupted and hence cannot be solved. Thus, we will need to get creative.
Problem 159 (A) There is no information here, so the value of A must be inferred from the problems that reference it.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 158 297 Problem 160 (F) We claim that WZ = B 3 √ 2.
For the sake of elegance, scale the diagram down by a factor of B so that WX = 1.
W X Y Z P Q 1 1 1 1 1 x y It is clear that △WXY is equilateral and △XPY is isosceles. Chasing angles, we find that ∠WY Z = 90◦and ∠ZY Q = 30◦. If we let WZ = x and Y Q = y, then by the sine area formula for triangles, x = WZ ZQ = [WY Z] [ZY Q] = 1 2(Y W)(Y Z) sin 90◦ 1 2(Y Z)(Y Q) sin 30◦= 2 y, so xy = 2. By Law of Cosines on △Y QW, (x + 1)2 = 12 + y2 + y, or x2 + 2x = y2 + y. Substituting y = 2/x gives x4 + 2x3 −2x −4 = 0 which factors as (x3 −2)(x + 2) = 0. Thus x = 3 √ 2. After undoing the scaling by B from the beginning of this solution, we get WZ = B 3 √ 2, as claimed.
Now logB(WZ) = 1 + 1 3 logB 2. Since this is rational, logB 2 must be rational. So B is a rational power of 2. Since the answer to Problem 162 (B) is a positive integer, B must in fact be 2k for a positive integer k (note that k ̸= 0 because the base of a logarithm cannot be 1). Given k, we can then write 1 + 1 3 logB 2 = 1 + 1 k = k + 1 k = p q.
No matter the value of k, it must be the case that p = k and q = k + 1. So |p −q| = 1, giving F = 1 .
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 158 298 Problem 161 (G) This is actually quite subtle, and so we will come back to this later.
Problem 162 (B) We claim that A = n!.
Assign each square a rank based on how far north-east they are, with the long diagonal being rank 0.
-5 -4 -3 -2 -1 0 -4 -3 -2 -1 0 1 -3 -2 -1 0 1 2 -2 -1 0 1 2 3 -1 0 1 2 3 4 0 1 2 3 4 5 (Ranks for n = 6) Observe that no snake can occupy more than two rank-0 squares. Now let us focus on the squares of positive rank, from 1 to n −1.
• One of the n snakes must reach the top-right-most square, i.e. the sole square of rank n −1.
• That snake will occupy one of the squares of rank n −2, leaving just one more square of rank n −2 unoccupied by them. That square must be occupied by one of the other n −1 snakes.
• The two snakes from the previous two bullet points must occupy two squares of rank n −3, leaving just one more square of rank n −3 unoccupied by them. That square must be occupied by one of the other n −2 snakes.
• . . .
• The n −1 snakes from the previous n −1 bullet points must occupied n −1 squares of rank 1, leaving just one more square of rank 1 unoccupied by them. That square must be occupied by the last remaining snake.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 158 299 From this reasoning, we deduce that for each 0 ≤k ≤n −1, there is exactly one snake that occupies k squares of positive rank. There are n! ways to assign these snakes to the n squares of rank 0, and this uniquely determines how the snake covers the squares! This is because the identity of the snake corresponding to k = 0, i.e. the snake that does not reach rank 1, uniquely determines how the other n −1 snakes must occupy rank 1.
k = 4 k = 3 k = 0 k = 5 k = 1 k = 2 Then, one of these n −1 snakes corresponds to k = 1, meaning that this snake does not reach rank 2, and this uniquely determines how the other n −2 snakes must occupy rank 2.
k = 4 k = 3 k = 0 k = 5 k = 1 k = 2 Inductively we obtain the uniqueness of the covering for the positive rank squares, and by an entirely symmetrical argument we get the uniqueness of the covering for the negative rank squares as well. This completes the proof that there are exactly n! coverings.
Hence A = B!.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 158 300 Problem 163 (H) We immediately get the information that B and C are perfect squares greater than 1.
The lengths of the numbers {bn}n≥1 in base √ B are {⌊log√ B bn⌋+ 1}n≥1 = {⌊n log√ B b⌋+ 1}n≥1, and their lengths in base √ C are {⌊log√ C bn⌋+ 1}n≥1 = {⌊n log√ C b⌋+ 1}n≥1.
So the sequences {⌊n log√ B b⌋}n≥1 and {⌊n log√ C b⌋}n≥1 partition N. We can now rely on asymptotics: the former sequence has density 1 log√ B b in N, and the latter has density 1 log√ C b, thus 1 log√ B b + 1 log√ C b = 1.
(See also Rayleigh’s Theorem: Now, 1 = logb √ B + logb √ C = logb √ BC, which entails that b = √ BC. So H = √ BC.
Problem 164 (C) P X Y Z By Ptolemy’s theorem on quadrilateral PXZY , PX · ZY + PY · XZ = PZ · XY.
But XY = Y Z = XZ, so PX + PY = PZ. Hence C = A + K.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 158 301 Problem 165 (I) By the factorization a3 + b3 + c3 −3abc = (a + b + c)(a2 + b2 + c2 −ab −bc −ca), we have the implication a + b + c = 0 = ⇒a3 + b3 + c3 = 3abc. Applying this to x −y, y −z, and z −x, we find that µ = (x −y)(y −z)(z −x). So C + D −F = (x −y)(y −z)(z −x).
It is difficult to make progress with this without knowing C, D, and F, so we must move on.
Problem 166 (D) We claim that A must be a multiple of 6, and that the answer is (A/6)!. A few example tilings are shown below.
To argue that A must be a multiple of 6, first note that an angle of the A-gon can only be partitioned by the angles of at most two rhombi, because all angles are strictly less than incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 158 302 180◦and 180◦ 60◦= 3. It follows that, among the triangles and rhombi that share a side with the A-gon (which can be thought of as the “first layer”), the consecutive sequences of rhombi must be connected by a sequences of sides that are all mutually parallel. From this, it is not hard to deduce that there exists at least one equilateral triangle in this “first layer”. Then, via a computation or otherwise, it can be shown that the first layer must have exactly six equilateral triangles, positioned symmetrically, which forces A to be divisible by 6.
More generally, if we let A = 6n, it can be seen that any such tiling of the 6n-gon must consist of n layers of rhombi and triangles. Each layer, starting with the outermost one, consists of exactly 6 equilateral triangles placed symmetrically, with the gaps in between them filled rhombi. When this layer is removed, we are left with a (not necessarily regular) polygon with 6 fewer sides.
Intuitively, you should view the first layer as “removing” those 6 sides marked by the equilateral triangles, and all other sides are translated along the sides of these equilateral triangles to form the boundary of the smaller polygon. This process can then be repeated for the smaller polygon, again and again, until we are left with the “0-sided polygon” at the center.
For the first layer, there are n ways to choose the positions of the equilateral triangles.
For the next layer, since there are now 6(n −1) sides, there will be (n −1) ways to choose the positions of the equilateral triangles. If we keep going, the conclusion is not too hard to infer: There will be n! ways in total to choose the positions of the triangles, and therefore, n! ways to tile the 6n-gon. Since n = A/6, we get D = (A/6)!.
Problem 167 (J) This is a bit tricky without knowing D and E, so we will return later.
Problem 168 (E) It is clear that E is a positive integer and that E ≥2. Unfortunately, we claim that no more information can be deduced about E from this problem alone.
This is because if the sizes of the fish are X1, X2, . . . , XE−1, and Y is their minimum, then for t ∈(0, 1), P(Y ≥t) = P(X1 ≥t, . . . , XE−1 ≥t) = E−1 Y i=1 P(Xi ≥t) = P(X1 ≥t)E−1 = (1 −t)E−1, so EY = Z 1 0 P(Y ≥t) dt = Z 1 0 (1 −t)E−1 dt = 1 E .
So the answer, E, is given by 1 E −1, reducing to E = E, telling us nothing.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 158 303 Problem 169 (K) The key idea is that rearranging the sides of a cyclic polygon does not change its area!
We can therefore rearrange the sides into the following far more pleasing octagon.
E F E F E F E F Subdividing along the dashed lines, it becomes plain to see that the area is E2 + F 2 + 2 √ 2EF.
When this is expressed as m + n √ 2, we have m + E2 + F 2 and n = 2EF, so m + n = E2 + F 2 = 2EF = (E + F)2. That is, K = (E + F)2. Since E and F are integers, this tells us that K is a perfect square! This will be crucial for later.
Problem 161 Revisited (G) We know that F = 1, so the only unknown dependency is B, which we know to be a positive integer. From Problem 160, we know that B is a power of 2 that is greater than 1.
From Problem 163, we know that B is a perfect square greater than 1. Thus B is a power of 4. Solving this problem will, at last, allow us to deduce the value of B.
By a reflection argument, the answer is essentially the length of the shortest path from the X′ to Y ′, where X′ is the reflection of X over Y Z and Y ′ is the reflection of Y over XZ.
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 158 304 X Y Z X′ Y ′ 40◦ 40◦ 40◦ 1 1 B B At first glance, the answer appears to simply be the length of segment X′Y ′, which can be found by the Law of Cosines. However, if segment X′Y ′ exits the interior of pentagon ZX′Y XY ′, then after reflecting back, this segment will correspond to a path that exits the triangle and fails to visit one of the segments ZX or ZY . This could occur if one segment is too long compare to the other one.
(It’s a bit hard to see, but segment Y ′X′ lies slightly above segment Y X′.) If this is the case, then the shortest path will instead be the union of the segments Y ′Y and Y X′, which has length 2 sin 40◦+ √ 1 + B2 −2B cos 40◦. It is inconceivable that this could ever be written in the form √n for integer n (though I must confess that I have no rigorous proof of this), so we must prevent this case from occurring.
Since F = 1 and B is an integer, XZ is the longer segment. We can now compute the threshold that B would have to exceed for segment X′Y ′ to exit the pentagon. This threshold is exactly when Y ′, Y, and X are collinear. In which case, we may compute ∠ZY ′X′ = 50◦ and ∠ZX′Y ′ = 10◦, and now Law of Sines applied to △ZX′Y ′ gives B sin 50◦= 1 sin 10◦ incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 158 305 or B = sin 50◦ sin 10◦≈4.411. So B cannot exceed 4. But we know that B is a power of 4 that is greater than 1, therefore B = 4 .
We now obtain an explosion of information: • The answer to this problem, by the Law of Cosines, is given by √ G = p 12 + 42 −2(1)(4) cos 120◦= √ 21, so G = 21 .
• From Problem 162, A = B! = 4! and so A = 24 .
• From Problem 166, D = (A/6)! = 4! and so D = 24 .
• From Problem 164, C = A+K = 24+K. From Problem 169, K = (E+F)2 = (E+1)2.
From Problem 163, C is a perfect square. Thus we have the factorization 24 = C −K = √ C 2 −(E + 1)2 = ( √ C + E + 1)( √ C −E −1).
These factors sum to 2 √ C which is even, so they must have the same parity. This gives two possible cases: (√ C + E + 1 = 12 √ C −E −1 = 2 or (√ C + E + 1 = 6 √ C −E −1 = 4 These cases solve to ( √ C, E) = (7, 4) and ( √ C, E) = (5, 0) respectively. However, from Problem 168, we know that E ≥2, which eliminates the second case and gives us E = 4 and C = 49 .
• Since K = (E + 1)2 = (4 + 1)2, we get K = 25 .
• From Problem 163, we know that H = √ BC = √ 4 · 49, so H = 14 .
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 158 306 Problem 165 Revisited (I) Now that we know that C = 49, D = 24, and F = 1, we have (x −y)(y −z)(z −x) = 72.
Up to cyclic symmetry there are two possible orderings for x, y and z: Either x < y < z or x > y > z. (Note that no two can be equal.) The latter is impossible since this then (x −y)(y −z)(z −x) would be negative, so x < y < z. Now write (y −x)(z −y)(z −x) = 72 so that all factors are positive. We see that the first two factors sum to the third. Studying the divisors of 72 = 23·32, particularly the power of 2, a parity analysis gives two possibilities: Either each of the three factors is even, or one of the factors is divisible by 23. The former possibility can be ruled out easily, and so the factors are thus 1, 8, 9 in some order. We are left with two cases: y −x = 1 z −y = 8 z −x = 9 or y −x = 8 z −y = 1 z −x = 9 Solving in terms of x, the first case gives (x, y, z) = (x, x + 1, x + 9) and the second case gives (x, y, z) = (x, x + 8, x + 9). In either case, the maximum among {x, y, z} is x + 9 and the minimum is x, and the difference between these extremes will always be 9. We conclude that I = 9 .
Problem 167 Revisited (J) We now know that D + E2 = 24 + 42 = 40. So we wish to determine the factorial n! that should be removed from the product (1!)(2!) . . . (40!) so that what remains is a perfect square. We may rewrite this product into the form (1!)(1! · 2) · (3!)(3! · 4) · . . . · (39!)(39! · 40) = (1! · 3! · . . . · 39!)2(2 · 4 · . . . · 40) = (1! · 3! · . . . · 39!)2 · 220 · 20!
= (1! · 3! · . . . · 39!)2(210)2 · 20!, which makes it clear that removing 20! will result in a perfect square, (1!·3!·. . .·39!)2(210)2.
(The problem of investigating if this is the unique solution is left to the reader!) Thus J = 20 .
incredibly sus draft lmfao sup how was your day CHAPTER 1. SOLUTIONS Solution to Problem 158 307 Problem 170 We gather up the answers to the 5 adjacent problems: • G = 21 • H = 14 • I = 9 • J = 20 • K = 25 Using the A1Z26 cipher, the numbers 21, 14, 9, 20, and 25 correspond to the letters U, N, I, T, and Y. The answer to the final problem is UNITY .
■ Indeed, the hidden purpose of the CMUMC POTD was to strengthen the sense of commu-nity within the CMU Math Club and, therefore, unify its members. I’m overjoyed to know that the countless fascinating problems that I’ve accumulated over the years have found such a fulfilling purpose. Dear reader, whether you are from CMU or another place, whether you’ve experienced the POTD during its lifespan or are solving from another time, I thank you for your participation.
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1996 AIME Problems/Problem 15
Problem
In parallelogram, let be the intersection of diagonals and . Angles and are each twice as large as angle , and angle is times as large as angle . Find .
Contents
[hide]
1 Problem
2 Solution
2.1 Solution 1 (trigonometry)
2.2 Solution 2 (trigonometry)
2.3 Solution 3
3 See also
Solution
Solution 1 (trigonometry)
Let . Then , , and . Since is a parallelogram, it follows that . By the Law of Sines on ,
Dividing the two equalities yields
Pythagorean and product-to-sum identities yield
and the double and triple angle () formulas further simplify this to
The only value of that fits in this context comes from . The answer is .
Solution 2 (trigonometry)
Define as above. Since , it follows that , and so . The Law of Sines on yields that
Expanding using the sine double and triple angle formulas, we have
By the quadratic formula, we have , so (as the other roots are too large to make sense in context). The answer follows as above.
Solution 3
We will focus on . Let , so . Draw the perpendicular from intersecting at . Without loss of generality, let . Then , since is the circumcenter of . Then .
By the Exterior Angle Theorem, and . That implies that . That makes . Then since by AA ( and reflexive on ), .
Then by the Pythagorean Theorem, . That makes equilateral. Then . The answer follows as above.
See also
1996 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 14Followed by
Final Problem
1•2•3•4•5•6•7•8•9•10•11•12•13•14•15
All AIME Problems and Solutions
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.
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1162 | https://peppyhare.github.io/r/notes/griffiths/ch7-2/ | Electromagnetic Induction
7.2: Electromagnetic Induction #
7.2.1: Faraday’s Law #
In 1831 Michael Faraday reported on a series of experiments, including three that (with some violence to history) can be characterized as follows:
Experiment 1:. He pulled a loop of wire to the right through a magnetic field (Fig 7.21a). A current flowed in the loop.
Experiment 2: He moved the magnet to the left, holding the loop still (Fig 7.21b). Again, a current flowed in the loop.
Experiment 3: With both the loop and the magnet at rest (Fig 7.21c), he changed the strength of the field (he used an electromagnet, and varied the current in the coil). Once again, current flowed in the loop.
The first experiment, of course, is a straightforward case of motional emf; according to the flux rule:
[\mathcal{E} = - \dv{\Phi}{t}]
I don’t think it will surprise you to learn that exactly the same emf arises in Experiment 2 - all that really matters is the relative motion of the magnet and the loop. Indeed, in the light of special relativity it has to be so. But Faraday knew nothing of relativity, and in classical electrodynamics this simple reciprocity is a remarkable coincidence. For if the loop moves, it’s a magnetic force that sets up the emf, but if the loop is stationary, the force cannot be magnetic - stationary charges experience no magnetic forces. In that case, what is responsible? What sort of field exerts a force on charges at rest? Well, electric fields do, of course, but in this case there doesn’t seem to be any electric field in sight.
Faraday had an ingenious inspiration:
[\textbf{A changing magnetic field induces an electric field}]
It is this induced electric field that accounts for the emf in Experiment 2. Indeed, if (as Faraday found empirically) the emf is again equal to the rate of change of the flux,
[\mathcal{E} = \oint \vec{E} \cdot \dd \vec{l} = - \dv{\Phi}{t} \tagl{7.14}]
then ( \vec{E} ) is related to the change in ( \vec{B} ) by the equation
[\oint \vec{E} \cdot \dd \vec{l} = - \int \pdv{\vec{B}}{t} \cdot \dd \vec{a} \tagl{7.15}]
This is Faraday’s law, in integral form. We can convert it to differential form by applying Stokes’ theorem:
[\curl \vec{E} = - \pdv{\vec{B}}{t} \tagl{7.16}]
Note that Faraday’s law reduces to the old rule ( \oint \vec{E} \cdot \dd \vec{l} = 0 ) (or, in differential form, ( \curl \vec{E} = 0 )) in the static case (constant ( \vec{B} )), as, of course, it should.
In Experiment 3, the magnetic field changes for entirely different reasons, but according to Faraday’s law an electric field will again be induced, giving rise to an emf ( - d \Phi / dt ). Indeed, one can subsume all three cases (and for that matter any combination of them) into a kind of universal flux rule:
Whenever (and for whatever reason) the magnetic flux through a loop changes, an emf [\mathcal{E} = - \pdv{\Phi}{t} \tagl{7.17}] will appear in the loop.
Many people call this “Faraday’s law.” Maybe I’m overly fastidious, but I find this confusing. There are really two totally different mechanisms underlying Eq. 7.17, and to identify them both as “Faraday’s law” is a little like saying that because identical twins look alike we ought to call them by the same name. In Faraday’s first experiment, it’s the Lorentz force law at work; the emf is magnetic. But in the other two it’s an electric field (induced by the changing magnetic field) that does the job. Viewed in this light, it is quite astonishing that all three processes yield the same formula for the emf. In fact, it was precisely this “coincidence” that led Einstein to the special theory of relativity - he sought a deeper understanding of what is, in classical electrodynamics, a peculiar accident. But that’s a story for chapter 12. In the meantime, I shall reserve the term “Faraday’s law” for electric fields induced by changing magnetic fields, and I do not regard Experiment 1 as an instance of Faraday’s law.
Example 7.5 #
A long cylindrical magnet of length \( L \) and radius \( a \) carries a uniform magnetization \( \vec{M} \) parallel to its axis. It passes at constant velocity \( \vec{v} \) through a circular wire ring of slightly larger diameter (Fig. 7.22). Graph the emf induced in the ring, as a function of time.
The magnetic field is the same as that of a long solenoid with surface current \( \vec{K}_b = M \vu{\phi} \) . So the field inside is \( \vec{B} = \mu_0 \vec{M} \) , except near the ends, where it starts to spread out. The flux through the ring is zero when the magnet is far away; it builds up to a maximum of \( \mu_0 M \pi a^2 \) as the leading end passes through; and it drops back to zero as the trailing end emerges (Fig. 7.23a). The emf is (minus) the derivative of \( \Phi \) with respect to time, so it consists of two spikes, as shown in Fig. 7.23b.
Keeping track of the signs in Faraday’s law can be a real headache. For instance, in Ex. 7.5 we would like to know which way around the ring the induced current flows. In principle, the right-hand rule does the job (we called ( \Phi ) positive to the left, in Fig. 7.22, so the positive direction for current in the ring is counter-clockwise, as viewed from the left; since the first spike in Fig. 7.23b is negative, the first current pulse flows clockwise, and the second counterclockwise). But there’s a handy rule, called Lenz’s law, whose sole purpose is to help you get the directions right:
Nature abhors a change in flux
The induced current will flow in such a direction that the flux it produces tends to cancel the change. (As the front end of the magnet in Ex. 7.5 enters the ring, the flux increases, so the current in the ring must generate a field to the right - it therefore flows clockwise.) Notice that it is the change in flux, not the flux itself, that nature abhors (when the tail end of the magnet exits the ring, the flux drops, so the induced current flows counterclockwise, in an effort to restore it). Faraday induction is a kind of “inertial” phenomenon: A conducting loop “likes” to maintain a constant flux through it; if you try to change the flux, the loop responds by sending a current around in such a direction as to frustrate your efforts. (It doesn’t succeed completely; the flux produced by the induced current is typically only a tiny fraction of the original. All Lenz’s law tells you is the direction of the flow.)
Example 7.6 #
The 'jumping ring' demonstration. If you wind a solenoidal coil around an iron core (the iron is there to beef up the magnetic field), place a metal ring on top, and plug it in, the ring will jump several feet in the air (Fig 7.24). Why?
Before you turn on the current, the flux through the ring was zero. Afterward a flux appeared (upward in the diagram), and the emf generated in the ring led to a current (in the ring) which, according to Lenz's law, was in such a direction that its field tended to cancel this new flux. This means that the current in the loop is opposite to the current in the solenoid. As opposite currents repel (as we saw in our Biot-Savart calculations in the last chapter), the ring flies off.
7.2.2: The Induced Electric Field #
Faraday’s law generalizes the electrostatic rule ( \curl \vec{E} = 0 ) to the time-dependent regime. The divergence of ( \vec{E} ) is still given by Gauss’s law (( \div \vec{E} = \frac{1}{\epsilon_0} \rho ) ). If ( \vec{E} ) is a pure Faraday field (due exclusively to a changing ( \vec{B} ) , with ( \rho = 0 ) ), then
[\div \vec{E} = 0 \qquad \curl \vec{E} = - \pdv{\vec{B}}{t}]
This is mathematically identical to magnetostatics
[\div \vec{B} = 0 \qquad \curl \vec{B} = \mu_0 \vec{J}]
Conclusion: Faraday-induced electric fields are determined by ( -(\partial \vec{B} / \partial t) ) in exactly the same way as magnetostatic fields are determined by ( \mu_0 \vec{J} ). The analog to Biot-Savart is
[\vec{E} = - \frac{1}{4 \pi} \int \frac{(\partial \vec{B} / \partial t) \cross \vu{\gr}}{\gr ^2} \dd \tau = - \frac{1}{4 \pi} \pdv{}{t} \int \frac{\vec{B} \cross \vu{\gr}}{\gr ^2} \dd \tau \tagl{7.18}]
and if symmetry permits, we can use all the tricks associated with Ampere’s law in integral form (( \oint \vec{B} \cdot \dd \vec{l} = \mu_0 I_{enc} )), only now it’s Faraday’s law in integral form:
[\oint \vec{E} \cdot \dd \vec{l} = - \dv{\Phi}{t} \tagl{7.19}]
The rate of change of (magnetic) flux through the Amperian loop plays the role formerly assigned to ( \mu_0 I_{enc} ).
Example 7.7 #
A uniform magnetic field \( \vec{B}(t) \), pointing straight up, fills the shaded circular region of Fig. 7.25. If \( \vec{B} \) is changing with time, what is the induced electric field?
\( \vec{E} \) points in the circumferential direction, just like the magnetic field inside a long straight wire carrying a uniform current density. Draw an Amperian loop of radius \( s \), and apply Faraday's law: [\oint \vec{E} \cdot \dd \vec{l} = E ( 2 \pi s) = - \dv{\Phi}{t} = - \dv{}{t} \left( \pi s^2 B(t) \right) = - \pi s^2 \dv{B}{t}] Therefore [\vec{E} = - \frac{s}{2} \dv{B}{t} \vu{\phi}] If \( \vec{B} \) is increasing, \( \vec{E} \) runs clockwise, as viewed from above.
Example 7.8 #
A line charge \( \lambda \) is glued to the rim of a wheel of radius \( b \), which is then suspended horizontally, as shown in Fig 7.26, so that it is free to rotate (the spokes are made of some nonconducting material - wood, maybe). In the central region, out to radius \( a \) , there is a uniform magnetic field \( \vec{B}_0 \), pointing up. Now someone turns the field off. What happens?
The changing magnetic field will induce an electric field, curling around the axis of the wheel. This electric field exerts a force on the charges at the rim, and the wheel starts to turn. According to Lenz's law, it will rotate in such a direction that its field tends to restore the upward flux. The motion, then, is counterclockwise, as viewed frrom above. Faraday's law, applied to the loop at radius \( b \), says [\oint \vec{E} \cdot \dd \vec{l} = E ( 2 \pi b) = - \dv{\Phi}{t} = - \pi ^2 \dv{B}{t}] or [\vec{E} = - \frac{a^2}{2b} \dv{B}{t} \vu{\phi}] The torque on a segment of length \( \dd \vec{l} \) is \( (\vec{r} \cross \vec{F}) \) , or \( b \lambda E \dd l \) . The total torque on the wheel is therefore [N = b \lambda \left( - \dv{\Phi}{t} = - \pi ^2 \dv{B}{t} \right) \oint \dd l = - b \lambda \pi a^2 \dv{B}{t}] and the angular momentum imparted to the wheel is [\int N \dd t = - \lambda \pi a^2 b \int_{B_0} ^0 \dd B = \lambda \pi a^2 b B_0] It doesn't matter how quickly or slowly you tum off the field; the resulting angular velocity of the wheel is the same regardless. (If you find yourself wondering where the angular momentum came from, you're getting ahead of the story! Wait for the next chapter.) Note that it's the electric field that did the rotating. To convince you of this, I deliberately set things up so that the magnetic field is zero at the location of the charge. The experimenter may tell you she never put in any electric field - all she did was switch off the magnetic field. But when she did that, an electric field automatically appeared, and it's this electric field that turned the wheel.
I must warn you, now, of a small fraud that tarnishes many applications of Faraday’s law: Electromagnetic induction, of course, occurs only when the magnetic fields are changing, and yet we would like to use the apparatus of magnetostatics (Ampere’s law, the Biot-Savart law, and the rest) to calculate those magnetic fields. Technically, any result derived in this way is only approximately correct. But in practice the error is usually negligible, unless the field fluctuates extremely rapidly, or you are interested in points very far from the source. Even the case of a wire snipped by a pair of scissors (Prob. 7.18) is static enough for Ampere’s law to apply. This regime, in which magnetostatic rules can be used to calculate the magnetic field on the right hand side of Faraday’s law, is called quasistatic. Generally speaking, it is only when we come to electromagnetic waves and radiation that we must worry seriously about the breakdown of magnetostatics itself.
Example 7.9 #
An infinitely long straight wire carries a slowly varying current \( I(t) \). Determine the induced electric field, as a function of the distance \( s \) from the wire
In the quasistatic approximation, the magnetic field is \( (\mu_0 I / 2 \pi s) \) and it circles around the wire. Like the __B__-field of a solenoid, __E__ here runs parallel to the axis. For the rectangular "Amperian loop" in Fig 7.27, Faraday's law gives: [\begin{aligned} \oint \vec{E} \cdot \dd \vec{l} & = E (s_0) l - E(s) l \ & = - \dv{}{t} \int \vec{B} \cdot \dd \vec{a} \ & = - \frac{\mu_0 l }{2\pi} \dv{I}{t} \int _{s_0} ^s \frac{1}{s} \dd s' \ & = - \frac{\mu_0 l}{2 \pi} \dv{I}{t} ( \ln \, s - \ln \, s_0 ) \end{aligned}] Thus [\vec{E}(s) = \left( \frac{\mu_0}{2\pi} \dv{I}{t} \ln \, s + K \right) \vu{z} \tagl{7.20}] where \( K \) is a constant (that is to say, it is independent of \( s \) - it might still be a function of \( t \) ). The actual value of \( K \) depends on the whole history of the function \( I(t) \) - we'll see some examples in Chapter 10. Equation 7.20 has the particular implication that \( E \) blows up as \( s \) goes to infinity. That can't be true... What's gone wrong? Answer: we have overstepped the limits of the quasistatic approximation. As we shall see in Chapter 9, electromagnetic "news" travels at the speed of light, and at large distances __B__ depends not on the current _now_, but on the current _as it was at some earlier time_ (indeed, a whole range of earlier times, since different points on the wire are different distances away). If \( \tau \) is the time it takes \( I \) to change substantially, then the quasistatic approximation should hold only for [s \ll c \tau \tagl{7.21}] and hence Eq. 7.20 simply does not apply, at extremely large \( s \).
7.2.3: Inductance #
Suppose you have two loops of wire, at rest (Fig 7.30). If you run a steady current ( I_1 ) around loop 1, it produces a magnetic field ( \vec{B}_1 ) . Some of the field lines pass through loop 2; let ( \Phi_2 ) be the flux of ( \vec{B}_1 ) through 2. You might have a tough time actually calculating ( \vec{B_1} ), but a glance at the Biot-Savart law,
[\vec{B}_1 = \frac{\mu_0}{4 \pi} I_1 \oint \frac{\dd \vec{l}_1 \cross \vu{\gr}}{\gr^2} ]
reveals one significant fact about this field: It is proportional to the current ( I_1 ) . Therefore, so too is the flux through loop 2:
[\Phi_2 = \int \vec{B}_1 \cdot \dd \vec{a}_2]
Thus
[\Phi_2 = M_{21} I_1 \tagl{7.22}]
where ( M_{21} ) is the constant of proportionality; it is known as the mutual inductance of the two loops.
There is a cute formula for the mutual inductance, which you can derive by expressing the flux in terms of the vector potential, and invoking Stokes’ theorem:
[\Phi_2 = \int \vec{B}_1 \cdot \dd \vec{a}_2 = \int (\curl \vec{A}_1) \cdot \dd \vec{a}_2 = \oint \vec{A}_1 \cdot \dd \vec{l}_2]
Now, according to Eq. 5.66,
[\vec{A}_1 = \frac{\mu_0 I_1}{4 \pi} \oint \frac{\dd \vec{l}_1}{\gr} ]
and hence
[\Phi_2 = \frac{\mu_0 I_1}{4 \pi} \oint \left( \oint \frac{\dd \vec{l}_1}{\gr} \right) \cdot \dd \vec{l}_2]
Evidently
[M_{21} = \frac{\mu_0}{4 \pi} \oint \oint \frac{\dd \vec{l}_1 \cdot \dd \vec{l}_2}{\gr} \tagl{7.23}]
This is the Neumann formula; it involves a double line integral - one integration around loop 1, the other around loop 2 (Fig 7.31). It’s not very useful for practical calculations, but it does reveal two important things about the mutual inductance:
( M_{21} ) is a purely geometrical quantity, having to do with the sizes, shapes, and relative positions of the two loops.
The integral in Eq. 7.23 is unchanged if we switch the roles of loops 1 and 2; it follows that
[M_{21} = M_{12} \tagl{7.24}]
This is an astonishing conclusion: Whatever the shapes and positions of the loops, the flux through 2 when we run a current I around 1 is identical to the flux through 1 when we send the same current I around 2. We may as well drop the subscripts and call them both M.
Example 7.10 #
A short solenoid (length \( l \) and radius \( a \), with \( n_1 \) turns per unit length) lies on the axis of a very long solenoid (radius \( b \) , \( n_2 \) turns per unit length) as shown in Fig 7.32. Current \( I \) flows in the short solenoid. What is the flux through the long solenoid?
Since the inner solenoid is short, it has a very complicated field; moreover, it puts a different flux through each turn of the outer solenoid. It would be a miserable task to compute the total flux this way. However, if we exploit the equality of the mutual inductances, the problem becomes very easy. Just look at the reverse situation: run the current \( I \) through the outer solenoid, and calculate the flux through the inner one. The field inside the long solenoid is constant [B = \mu_0 n_2 I] so the flux through a single loop of the short solenoid is [B \pi a^2 = \mu_0 n_2 I \pi a^2] There are \( n_1 l \) turns in all, so the total flux through the inner solenoid is [\Phi = \mu_0 \pi a^2 n_1 n_2 I] This is also the flux a current \( I \) in the short solenoid would put through the long one, which is what we set out to find. Incidentally, the mutual inductance, in this case, is [M = \mu_0 \pi a^2 n_1 n_2 l]
Suppose now, that you vary the current in loop 1. The flux through loop 2 will vary accordingly, and Faraday’s law says this changing flux will induce an emf in loop 2:
[\mathcal{E}_2 = - \dv{\Phi_2}{t} = - M \dv{I_1}{t} \tagl{7.25}]
(In quoting Eq. 7.22 - which was based on the Biot-Savart law - I am tacitly assuming that the currents change slowly enough for the system to be considered quasistatic.) What a remarkable thing: Every time you change the current in loop 1, and induced current flows in loop 2 - even though there are no wires connecting them!
Come to think of it, a changing current not only induces an emf in any nearby loops, it also induces an emf in the source loop itself (Fig 7.33). Once again, the field (and therefore the flux) is proportional to the current
[\Phi = L I \tagl{7.26}]
The constant of proportionality ( L ) is called the self inductance (or simply the inductance) of the loop. As with ( M ), it depends on the geometry (side and shape ) of the loop. If the current changes, the emf induced in the loop is
[\mathcal{E} = - L \dv{I}{t} \tagl{7.27}]
Inductance is measured in henries (H); a henry is a volt-second per ampere.
Example 7.11 #
Find the self inductance of a toroidal coil with rectangular cross-section (inner radius \( a \), outer radius \( b \), height \( h \)), that carries a total of \( N \) turns.
The magnetic field inside of a toroid is [B = \frac{\mu_0 N I}{2 \pi s} ]The flux through a single turn (Fig 7.34) is [\int \vec B \cdot \dd \vec a = \frac{\mu_0 N I}{2 \pi} h \int_a ^b \frac{1}{s} = \frac{\mu_0 N I h}{2 \pi} \ln \frac{b}{a} ] The total flux is \( N \) times this, so the self-inductance (Eq. 7.26) is [L = \frac{\mu_0 N^2 h}{2 \pi} \ln \left( \frac{b}{a} \right) \tagl{7.28}]
Inductance (like capacitance) is an intrinsically positive quantity. Lenz’s law, which is enforced by the minus sign in Eq. 7.27, dictates that the emf is in such a direction as to oppose any change in current. For this reason, it is called a back emf. Whenever you try to alter the current in a wire, you must fight against this back emf. Inductance plays somewhat the same role in electric currents that mass plays in mechanical systems: The greater ( L ), the harder it is to change the current, just as the larger the mass, the harder it is to change an object’s velocity.
Example 7.12 #
Suppose a current \( I \) is flowing around a loop, when someone suddenly cuts the wire. The current drops "instantaneously" to zero. This generates a whopping back emf, for although \( I \) may be small, \( d I / d t \) is enormous. (That's why you sometimes draw a spark when you unplug an iron or toaster - electromagnetic induction is desperately trying to keep the current going, even if it has to jump the gap in the circuit.) Nothing so dramatic occurs when you plug in a toaster or iron. In this case induction opposes the sudden increase in current, prescribing instead a smooth and continuous buildup. Suppose, for instance, that a battery (which supplies a constant emf \( \mathcal{E}_0 \) ) is connected to a circuit of resistance \( R \) and inductance \( L \) (Fig. 7.35). What current flows?
The total emf in this circuit is \( \mathcal{E}_0 \) from the battery plus \( -L (\dv{I}{t}) \) from the inductance. Ohm's law says [\mathcal{E}_0 - L \dv{I}{t} = I R] This is a first-order differential equation for \( I \) as a function of time. The general solution is [I(t) = \frac{\mathcal{E}_0}{R} + k e^{-(R/L) t}] where \( k \) is a constant to be determined by the initial conditions. In particular, if you close the switch at time \( t = 0 \), so \( I(0) = 0 \), then \( k = - \mathcal{E}_0 / R \), and [I(t) = \frac{\mathcal{E}_0}{R} \left( 1 - e^{-(R/L) t} \right) \tagl{7.29}] This function is plotted in Fig 7.36. Had there been no inductance in the circuit, the current would have jumped immediately to \( \mathcal{E}_0 / R \). In practice, every circuit has some self-inductance, and the current approaches \( \mathcal{E}_0 / R \) asymptotically. The quantity \( \tau = L / R \) is the __time constant__ for an LR circuit; it tells you how long the current takes to reach a substantial fraction \( (1 - 1/e) \) of its final value.
7.2.4: Energy in Magnetic Fields #
It takes a certain amount of energy to start a current flowing in a circuit. I’m not talking about the energy delivered to the resistors and converted into heat - that is irretrievably lost, as far as the circuit is concerned, and can be large or small, depending on how long you let the current run. What I am concerned with, rather, is the work you must do against the back emf to get the current going. This is fixed amount, and it is recoverable: you get it back when the current is turned off. In the meantime, it represents energy latent in the circuit; as we’ll see in a moment, it can be regarded as energy stored in the magnetic field.
The work done on a unit charge, against the back emf, in one trip around the circuit is ( - \mathcal{E} ) (the minus sign records the fact that this is the work done by you against the emf, not the work done by the emf). The amount of charge per unit time passing down the wire is I. So the total work done per unit time is
[\dv{W}{t} = - \mathcal{E}I = L I \dv{I}{t}]
If we start with zero current and build it up to a final value I, the work done (integrating the last equation over time) is
[W = \frac{1}{2} L I^2 \tagl{7.30}]
So, this is the energy stored in an inductor, or in any loop that has an inductance ( L ). It does not depend on how long we take to crank up the current, only on the geometry of the loop (in the form of ( L ) ) and the final current ( I ).
This is only really sensible for a system of conducting loops, but we can be a bit more general. We can express ( W ) by recalling that the flux ( \Phi ) through a loop (which is ( LI ) ) is
[\Phi = \int \vec{B} \cdot \dd \vec{a} = \int (\curl \vec{A}) \cdot \dd \vec{a} = \oint \vec{A} \cdot \dd \vec{l}]
where the line integral is around the perimeter of the loop. So, we have
[LI = \oint \vec{A} \cdot \dd \vec{l}]
and therefore
[W = \frac{1}{2} I \oint \vec{A} \cdot \dd \vec{l} = \frac{1}{2} \oint (\vec{A} \cdot \vec{I}) \dd l \tagl{7.31}]
We can pretty obviously generalize this to volume currents
[W = \frac{1}{2} \int _V (\vec{A} \cdot \vec{J}) \dd \tau \tagl{7.32}]
But we can do one better, expressing ( W ) entirely in terms of the magnetic field: ( \curl \vec{B} = \mu_0 \vec{J} ) lets us eliminate the current density from the picture
[W = \frac{1}{2 \mu_0} \int \vec{A} \cdot (\curl \vec{B}) \dd \tau \tagl{7.33}]
Integration by parts gets us to slap the derivative from B to A
[\div (\vec{A} \cross \vec{B}) = \vec{B} \cdot (\curl \vec{A}) - \vec{A} \cdot (\curl \vec{B})]
[\vec{A} \cdot (\curl \vec{B}) = \vec{B} \cdot \vec{B} - \div (\vec{A} \cross \vec{B}]
[W = \frac{1}{2\mu_0} \left( \int B^2 \dd \tau - \int \div (\vec{A} \cross \vec{B} \dd \tau \right) \ = \frac{1}{2\mu_0} \left( \int _V B^2 \dd \tau - \oint_S (\vec{A} \cross \vec{B} ) \cdot \dd \vec{a} \right) \tagl{7.34}]
Now, the integration in Eq. 7.32 is to be taken over the entire volume occupied by the current. But any region larger than this will do just as well, for ( \vec{J} ) is zero out there anyway. In Eq. 7.34, the larger the region we pick the greater is the contribution from the volume integral, and therefore the smaller is that of the surface integral (this makes sense: as the surface gets farther from the current, both A and B decrease). In particular, if we agree to integrate over all space, then the surface integral goes to zero, and we are left with
[W = \frac{1}{2 \mu_0} \int _{\text{all space}} B^2 \dd \tau \tagl{7.35}]
In view of this result, we say the energy is “stored in the magnetic field,” in the amount ( (B^2 / 2 \mu_0) ) per unit volume. This is a nice way to think of it, though someone looking at Eq. 7.32 might prefer to say that the energy is stored in the current distribution, in the amount ( \frac{1}{2} (\vec{A} \cdot \vec{J}) ) per unit volume. The distinction is one of bookkeeping; the important quantity is the total energy ( W ) , and we need not worry about where (if anywhere) the energy is “located.”
You might find it strange that it takes energy to set up a magnetic field - after all, magnetic fields themselves do no work. The point is that producing a magnetic field, where previously there was none, requires changing the field, and a changing B-field, according to Faraday, induces an electric field. The latter, of course, can do work. In the beginning, there is no ( \vec{E} ) , and at the end there is no ( \vec{E} ) ; but in between, while ( \vec{B} ) is building up, there is an ( \vec{E} ) , and it is against this that the work is done. (You see why I could not calculate the energy stored in a magnetostatic field back in Chapter 5.) In the light of this, it is extraordinary how similar the magnetic energy formulas are to their electrostatic counterparts:
Example 7.13 #
A long coaxial cable carries current \( I \) (the current flows down the surface of the inner cylinder, radius \( a \) , and back along the outer cylinder, radius \( b \) ) as shown in Fig 7.40. Find the magnetic energy stored in a section of length \( l \)
Ampere's law will tell us that __B__ between the surfaces is [\vec{B} = \frac{\mu_0 I}{2 \pi s } \vu{\phi}] and outside the cable, the field is zero. Eq. 7.35 gives us the volume energy density [\frac{1}{2 \mu_0} \left( \frac{\mu_0 I}{2 \pi s} \right)^2 = \frac{\mu_0 I^2}{8 \pi^2 s^2} ] The energy in a shell of length \( l \), radius \( s \) and thickness \( \dd s \) is [\left( \frac{\mu_0 I^2}{8 \pi ^2 s^2} \right)2 \pi l s \, \dd s = \frac{\mu_0 I^2 l}{4 \pi} \left( \frac{\dd s}{s} \right)] Integrating from \( a \) to \( b \) , we have [W = \frac{\mu_0 I^2 l}{4 \pi} \ln \left( \frac{b}{a} \right)] Incidentally, this suggests a very simple way to calculate the self-inductance of the cable. According to Eq. 7.30, the energy can also be written as \( \frac{1}{2} L I^2 \). Comparing the two expressions, [L = \frac{\mu_0 l}{2 \pi} \ln \left( \frac{b}{a} \right)] This method of calculating the self-inductance is especially useful when the current is not confined to a single path, but spreads over some surface or volume, so that different parts of the current enclose different amounts of flux. In such cases, it can be very tricky to get the inductance directly from Eq. 7.26, and it is best to let 7.30 _define_ L |
1163 | https://pubchem.ncbi.nlm.nih.gov/compound/Cesium-137 | Cesium-137 | Cs | CID 5486527 - PubChem
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compound Summary
Cesium-137
PubChem CID
5486527
Structure
Molecular Formula
Cs
Synonyms
Cesium-137
Caesium-137
137Cs radioisotope
10045-97-3
(137Cs)caesium
View More...
Molecular Weight
136.907089 g/mol
Computed by PubChem 2.2 (PubChem release 2025.05.06)
Element Name
Cesium
Dates
Create: 2005-08-08
Modify: 2025-08-30
Description
Cesium Cs 137 is a radioactive isotope of cesium with an atomic mass of 139 and potential application in radiotherapy. Cesium Cs 137 is prevalent due to its spontaneous production, which occurs as a result of nuclear fission of other radioactive materials, such as uranium and plutonium. This radionuclide has a relatively long half-life, 30 years, and decays by emitting beta particles. Both Cs 137 and its metastable nuclear isomer, barium-137m, emit gamma radiation of moderate energy and so are used in sterilization procedures in the food industry or in hospital environments.
NCI Thesaurus (NCIt)
Cesium is the chemical element with the symbol Cs and atomic number 55. Cesium-137 is a radioactive isotope of cesium with a half-life of 30.07 years. It is produced from the detonation of nuclear weapons and is produced in nuclear power plants. Cesium-137 was released to the atmosphere most notably from the 1986 Chernobyl meltdown. It is commonly used as a gamma-emitter in industrial applications such as moisture and density gauges, leveling gauges, flow meters, and other sensor equipment. Cesium-137 is water-soluble and extremely toxic in minute amounts. (L1111, L1124)
Toxin and Toxin Target Database (T3DB)
1 Structures
1.1 2D Structure
Structure Search
Get Image
Download Coordinates
Chemical Structure Depiction
Full screen Zoom in Zoom out
PubChem
2 Names and Identifiers
2.1 Computed Descriptors
2.1.1 IUPAC Name
cesium-137
Computed by Lexichem TK 2.7.0 (PubChem release 2025.05.06)
PubChem
2.1.2 InChI
InChI=1S/Cs/i1+4
Computed by InChI 1.07.3 (PubChem release 2025.05.06)
PubChem
2.1.3 InChIKey
TVFDJXOCXUVLDH-RNFDNDRNSA-N
Computed by InChI 1.07.3 (PubChem release 2025.05.06)
PubChem
2.1.4 SMILES
[137Cs]
Computed by OEChem 2.3.0 (PubChem release 2025.05.06)
PubChem
2.2 Molecular Formula
Cs
Computed by PubChem 2.2 (PubChem release 2025.05.06)
PubChem
2.3 Other Identifiers
2.3.1 CAS
10045-97-3
CAS Common Chemistry; EPA DSSTox
2.3.2 DSSTox Substance ID
DTXSID8040985
EPA DSSTox
2.3.3 NCI Thesaurus Code
C68812
NCI Thesaurus (NCIt)
2.3.4 Wikidata
Q27260451
Wikidata
2.4 Synonyms
2.4.1 MeSH Entry Terms
Cesium-137
137Cs radioisotope
Cs-137 radioisotope
Caesium-137
Medical Subject Headings (MeSH)
2.4.2 Depositor-Supplied Synonyms
Cesium-137
Caesium-137
137Cs radioisotope
10045-97-3
(137Cs)caesium
Cesium Cs-137
DTXSID8040985
Cesium, isotope of mass 137
Cs-137 radioisotope
CESIUM 137
137Cs
Cs-137
Cesium Cs 137
RefChem:124649
4T2E65IAR7
DTXCID701323549
137Cs isotope
Isotope:137Cs
137Cs (isotope)
Cesium-137 isotope
Isotope:Cesium-137
SCHEMBL29351853
Q27260451
PubChem
3 Chemical and Physical Properties
3.1 Computed Properties
Property Name
Property Value
Reference
Property Name
Molecular Weight
Property Value
136.907089 g/mol
Reference
Computed by PubChem 2.2 (PubChem release 2025.05.06)
Property Name
Hydrogen Bond Donor Count
Property Value
0
Reference
Computed by Cactvs 3.4.8.18 (PubChem release 2025.05.06)
Property Name
Hydrogen Bond Acceptor Count
Property Value
0
Reference
Computed by Cactvs 3.4.8.18 (PubChem release 2025.05.06)
Property Name
Rotatable Bond Count
Property Value
0
Reference
Computed by Cactvs 3.4.8.18 (PubChem release 2025.05.06)
Property Name
Exact Mass
Property Value
136.907089 Da
Reference
Computed by PubChem 2.2 (PubChem release 2025.05.06)
Property Name
Monoisotopic Mass
Property Value
136.907089 Da
Reference
Computed by PubChem 2.2 (PubChem release 2025.05.06)
Property Name
Topological Polar Surface Area
Property Value
0 Ų
Reference
Computed by Cactvs 3.4.8.18 (PubChem release 2025.05.06)
Property Name
Heavy Atom Count
Property Value
1
Reference
Computed by PubChem
Property Name
Formal Charge
Property Value
0
Reference
Computed by PubChem
Property Name
Complexity
Property Value
0
Reference
Computed by Cactvs 3.4.8.18 (PubChem release 2025.05.06)
Property Name
Isotope Atom Count
Property Value
1
Reference
Computed by PubChem
Property Name
Defined Atom Stereocenter Count
Property Value
0
Reference
Computed by PubChem
Property Name
Undefined Atom Stereocenter Count
Property Value
0
Reference
Computed by PubChem
Property Name
Defined Bond Stereocenter Count
Property Value
0
Reference
Computed by PubChem
Property Name
Undefined Bond Stereocenter Count
Property Value
0
Reference
Computed by PubChem
Property Name
Covalently-Bonded Unit Count
Property Value
1
Reference
Computed by PubChem
Property Name
Compound Is Canonicalized
Property Value
Yes
Reference
Computed by PubChem (release 2025.05.06)
PubChem
4 Related Records
4.1 Related Compounds with Annotation
Follow these links to do a live 2D search or do a live 3D search for this compound, sorted by annotation score. This section is deprecated (see the neighbor discontinuation help page for details), but these live search links provide equivalent functionality to the table that was previously shown here.
PubChem
4.2 Related Compounds
Same Connectivity Count
38
Mixtures, Components, and Neutralized Forms Count
3
Similar Compounds (2D)
View in PubChem Search
Similar Conformers (3D)
View in PubChem Search
PubChem
4.3 Related Element
Element Name
Cesium
Element Symbol
Cs
Atomic Number
55
PubChem Elements
4.4 Substances
4.4.1 PubChem Reference Collection SID
500768067
PubChem
4.4.2 Related Substances
All Count
54
Same Count
24
Mixture Count
30
PubChem
4.4.3 Substances by Category
PubChem
4.5 Entrez Crosslinks
PubMed Count
5
Taxonomy Count
2
Gene Count
10
PubChem
5 Chemical Vendors
PubChem
6 Pharmacology and Biochemistry
6.1 Metabolism / Metabolites
Cesium can be absorbed following ingestion, inhalation, or dermal exposure. Cesium behaves in a manner similar to potassium and distributes uniformly throughout the body. Gastrointestinal absorption from food or water is the principal source of internally deposited cesium in the general population. Essentially all cesium that is ingested is absorbed into the bloodstream through the intestines. Cesium tends to concentrate in muscles because of their relatively large mass. Cesium has been shown to compete with potassium for transport through potassium channels and can also substitute for potassium in activation of the sodium pump and subsequent transport into the cell. Like potassium, cesium is excreted from the body fairly quickly, mainly in the urine. In an adult, 10% is excreted with a biological half-life of 2 days, and the rest leaves the body with a biological half-life of 110 days. This means that if someone is exposed to radioactive cesium and the source of exposure is removed, much of the cesium will readily clear the body along the normal pathways for potassium excretion within several months. (L1126' L1850)
Toxin and Toxin Target Database (T3DB)
7 Use and Manufacturing
7.1 Uses
Cesium-137 is produced from the detonation of nuclear weapons and is produced in nuclear power plants. Cesium-137 was released to the atmosphere most notably from the 1986 Chernobyl meltdown. It is commonly used as a gamma-emitter in industrial applications such as moisture and density gauges, leveling gauges, flow meters, and other sensor equipment. Cesium-137 is also used in brachytherapy to treat various types of cancer. (L1111, L1126)
Toxin and Toxin Target Database (T3DB)
8 Safety and Hazards
8.1 Hazards Identification
8.1.1 DOT Hazard Classification
Radionuclide
Cesium-137
Hazard Class
Radioactive material
Placard/Label(s)
Notes
Atomic number: 55
Reportable quantity: 1 (.037) [Curie(Terabecquerels)]
US Code of Federal Regulations, Hazardous Materials, 49 CFR Part 172
9 Toxicity
9.1 Toxicological Information
9.1.1 Toxicity Summary
Highly penetrating gamma rays are the major cause of damage to tissues and internal organs following external overexposure to radioactive cesium. Once radioactive cesium is taken internally, cells of nearby tissues are at highest risk for damage due to the emission of beta particles. The ionizing radiation produced by cesium-137 causes cellular damage that includes DNA breakage, accurate or inaccurate repair, apoptosis, gene mutations, chromosomal change, and genetic instability. This leads to loss of normal cell and tissue homeostasis, and development of malignancy. Ionizing radiation that does not directly damage DNA can produce reactive oxygen intermediates that directly affect the stability of p53, an important enzyme in cell-cycle regulation, and produce oxidative damage to individual bases in DNA and point mutations by mispairing during DNA replication. (L1837, L1850)
Toxin and Toxin Target Database (T3DB)
9.1.2 Carcinogen Classification
Carcinogen Classification
1, carcinogenic to humans. (L135)
Toxin and Toxin Target Database (T3DB)
9.1.3 Health Effects
Cesium-137 presents external as well as internal health hazard, both from beta and gamma radiation. Cesium-137 is water-soluble and extremely toxic in minute amounts. The radioactivity of Cesium-137 can damage cells and cause cancer 10, 20 or 30 years from the time of ingestion, inhalation or absorption, provided sufficient material enters the body. Radioactive cesium overexposure can result in adverse effects such as reduced fertility, abnormal neurological development, genotoxicity, and damage to blood-forming organs(L1111, L1126, L1850)
Toxin and Toxin Target Database (T3DB)
9.1.4 Exposure Routes
Oral (L1126) ; inhalation (L1126)
Toxin and Toxin Target Database (T3DB)
9.1.5 Signs and Symptoms
Large amounts of cesium can cause hyperirritability and spasms. Exposure to high doses of ionizing radiation results in acute radiation syndrome, which can cause skin burns, hair loss, nausea, vomiting, dizziness, disorientation, low blood pressure, headache, fatigue, weakness, fever, birth defects, illness, infection, and death. (L1837, L1852, L1124, L1850)
Toxin and Toxin Target Database (T3DB)
9.1.6 Minimum Risk Level
Acute Radiation: 4 mSv (L134) Chronic Radiation: 1 mSv/yr (L134)
Toxin and Toxin Target Database (T3DB)
9.1.7 Treatment
Treatment reversing the effects of irradiation is currently not possible. Anaesthetics and antiemetics are administered to counter the symptoms of exposure, as well as antibiotics for countering secondary infections due to the resulting immune system deficiency. (L1852)
Toxin and Toxin Target Database (T3DB)
10 Associated Disorders and Diseases
Comparative Toxicogenomics Database (CTD)
11 Literature
11.1 Consolidated References
PubChem
11.2 NLM Curated PubMed Citations
Medical Subject Headings (MeSH)
11.3 Chemical Co-Occurrences in Literature
PubChem
11.4 Chemical-Gene Co-Occurrences in Literature
PubChem
11.5 Chemical-Disease Co-Occurrences in Literature
PubChem
11.6 Chemical-Organism Co-Occurrences in Literature
PubChem
12 Patents
12.1 Depositor-Supplied Patent Identifiers
PubChem
Link to all deposited patent identifiers
PubChem
12.2 Chemical Co-Occurrences in Patents
PubChem
12.3 Chemical-Disease Co-Occurrences in Patents
PubChem
12.4 Chemical-Gene Co-Occurrences in Patents
PubChem
12.5 Chemical-Organism Co-Occurrences in Patents
PubChem
13 Interactions and Pathways
13.1 Chemical-Target Interactions
Comparative Toxicogenomics Database (CTD)
14 Classification
14.1 MeSH Tree
Medical Subject Headings (MeSH)
14.2 NCI Thesaurus Tree
NCI Thesaurus (NCIt)
14.3 EPA DSSTox Classification
EPA DSSTox
14.4 EPA Substance Registry Services Tree
EPA Substance Registry Services
14.5 MolGenie Organic Chemistry Ontology
MolGenie
15 Information Sources
Filter by Source
CAS Common ChemistryLICENSE The data from CAS Common Chemistry is provided under a CC-BY-NC 4.0 license, unless otherwise stated. Cesium-137
EPA DSSToxLICENSE Caesium-137 CompTox Chemicals Dashboard Chemical Lists
Comparative Toxicogenomics Database (CTD)LICENSE It is to be used only for research and educational purposes. Any reproduction or use for commercial purpose is prohibited without the prior express written permission of NC State University. Cesium-137
NCI Thesaurus (NCIt)LICENSE Unless otherwise indicated, all text within NCI products is free of copyright and may be reused without our permission. Credit the National Cancer Institute as the source. NCI Thesaurus
Toxin and Toxin Target Database (T3DB)LICENSE T3DB is offered to the public as a freely available resource. Use and re-distribution of the data, in whole or in part, for commercial purposes requires explicit permission of the authors and explicit acknowledgment of the source material (T3DB) and the original publication. Cesium-137
PubChem ElementsLICENSE Cesium
US Code of Federal Regulations, Hazardous Materials, 49 CFR Part 172LICENSE Cesium-137
WikidataLICENSE CCZero cesium cs-137
PubChem
Medical Subject Headings (MeSH)LICENSE Works produced by the U.S. government are not subject to copyright protection in the United States. Any such works found on National Library of Medicine (NLM) Web sites may be freely used or reproduced without permission in the U.S. Cesium-137 MeSH Tree
EPA Substance Registry ServicesLICENSE EPA SRS List Classification
MolGenieLICENSE CC-BY 4.0 MolGenie Organic Chemistry Ontology
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CONTENTS
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1 Structures Expand this menu
2 Names and Identifiers Expand this menu
3 Chemical and Physical Properties Expand this menu
4 Related Records Expand this menu
5 Chemical Vendors
6 Pharmacology and Biochemistry Expand this menu
7 Use and Manufacturing Expand this menu
8 Safety and Hazards Expand this menu
9 Toxicity Expand this menu
10 Associated Disorders and Diseases
11 Literature Expand this menu
12 Patents Expand this menu
13 Interactions and Pathways Expand this menu
14 Classification Expand this menu
15 Information Sources
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目录
序言
1 与焦距和轴长的关系
2 相关资料
离心率
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外观
维基百科，自由的百科全书
此条目的主题是几何学。关于天文学中的偏心率，请见“轨道离心率”。
离心率（eccentricity，）又称偏心率，是指“圆锥曲线上任一点 到平面内一特定点 的距离”与“ 到平面内一不通过 的特定直线 的距离”之比。该特定点 称为焦点（focus），特定直线 称为准线（directrix）。
设一圆锥曲线 由 定义，其中 为焦点而 为准线（详见主条目圆锥曲线），则此时 称为 的离心率。
与焦距和轴长的关系
[编辑]
圆锥曲线之离心率与轴长有下述关系：
其中
半焦距
半长轴（椭圆）或半实轴（双曲线）
或采用较融贯的表法：
其中对椭圆取，对抛物线取，对双曲线取。
圆锥曲线依离心率之分类如下
圆：
椭圆：
抛物线：
双曲线：
相关资料
[编辑]
标准椭圆方程：
此时半长轴，半短轴，焦距，而且
标准双曲线方程：
此时半实轴，半虚轴，焦距，而且
| 查 论 编 几何学术语 |
| 点 | 顶点 交点 中点 角 + 同界角 极值点 最值点 临界点 驻点 鞍点 |
| 直线和曲线 | 线段 射线 直线 切线 （主）法线 副法线 曲线 圆锥曲线 双曲线 抛物线 正弦曲线 蚌线 蜗线 螺线（阿基米德螺线、等角螺线……） 摆线（最速降线问题） 悬链线 曳物线 渐开线 渐屈线 渐近线 测地线 边 周界 弦 弧 矢 垂直平分线 代数曲线 椭圆曲线 超椭圆 星形线 三尖瓣线 方圆形 勒洛三角形 |
| 平面图形 | 圆（广义圆） 椭圆 扇形 弓形 环形 多边形 三角形 四边形 五边形 六边形 多边形 正多边形 梯形 平行四边形 菱形 矩形 正方形 鹞形 卵形线 梭形 星形 五角星 六角星 |
| 立体图形 | 多面体 正多面体 四面体 长方体 立方体 平行六面体 棱柱 反棱柱 棱锥 棱台 圆柱体 圆锥 圆台 椭球（长球体、扁球体） 球体 球缺 球冠 球台 准线 母线 |
| 曲面 | 二次曲面 旋转曲面 抛物面 双曲面 马鞍面 球面 椭球面 类球面 环面 莫比乌斯带 流形 黎曼曲面 |
| 高维空间 | 超平面 超面 超曲面 胞 多胞形 超球体 超方形 超立方体 克莱因瓶 四维柱体柱 |
| 图形关系 | 相似 全等 对称 平行 垂直 相交 相切 镜像 旋转 反演 截面 缩放 |
| 三角形关系 | 相似三角形 全等三角形 |
| 量 | 距离 长度 周长 弧长 高度 面积 表面积 体积 容积 角度 曲率 挠率 离心率 凹凸性 有向曲面 可展曲面 直纹曲面 |
| 作图 | 尺 + 直尺 + 三角尺 圆规 尺规作图 二刻尺作图 |
| 分支 | 平面几何 立体几何 三角学 解析几何 微分几何 拓扑学 图论 折纸数学 欧几里得几何 非欧几里得几何（双曲几何、球面几何……） 分形 |
| 理论 | 定理 公理 定义 数学证明 |
| 分类 主题 共享资源 专题 |
| 查 论 编 引力的轨道 |
| 类型 | | | | --- | | 一般 | 盒轨道 捕获轨道 圆轨道（英语：Circular orbit） 椭圆轨道 / 高椭圆轨道 逃逸轨道 马蹄形轨道 双曲线轨道（英语：Hyperbolic trajectory） 倾斜轨道（英语：Inclined orbit） / 无倾斜轨道（英语：Non-inclined orbit） 拉格朗日点 吻切轨道 抛物线轨道 停泊轨道（英语：Parking orbit） 顺行/逆行 同步轨道 + 半同步 + 亚同步（英语：Subsynchronous orbit） 转移轨道 | | 地心 | 地球同步轨道 + 地球静止轨道 + 地球同步转移轨道 墓地轨道 低地球轨道 中地球轨道 高地球轨道 闪电轨道 近赤道轨道（英语：Near-equatorial orbit） 月球轨道 极轨道 太阳同步轨道 冻原轨道 | | 其他 | 火星 + 火星轨道 + 火星同步轨道（英语：Areosynchronous orbit） + 火星静止轨道（英语：Areostationary orbit） 拉格朗日点 + 远距离逆行轨道 + 晕轮轨道 + 利萨如轨道 绕月轨道 太阳 + 日心轨道 - 地球轨道 + 太阳同步轨道 | |
| 参数 | | | | --- | | 形状 大小 | e 离心率 a 半长轴 b 半短轴 Q, q 拱点 | | 方向 | i 倾角 Ω 升交点经度 ω 近心点幅角 ϖ 近心点经度 | | 位置 | M 平近点角 ν, θ, f 真近点角 E 偏近点角 L 平黄经 l 真黄经 | | 变化 | T 轨道周期 n 平均运动 v 轨道速度 t0 历元 | |
| 机动（英语：Orbital maneuver） | 双椭圆转移 碰撞规避（英语：Collision avoidance (spacecraft)） ΔV ΔV预算（英语：Delta-v budget） 重力助推 重力转向（英语：Gravity turn） 霍曼转移轨道 轨道倾角更改（英语：Orbital inclination change） 低能量转移 奥伯特效应 定相（英语：Orbit phasing） 火箭方程 交会 转位、对接和提取（英语：Transposition, docking, and extraction） |
| 航天动力学 | 天球坐标系统 特征能量（英语：Characteristic energy） 宇宙速度 星历表 赤道坐标系统 星下点轨迹（英语：Ground track） 希尔球 行星际运输网络（英语：Interplanetary Transport Network） 开普勒定律 拉格朗日点 N体问题 轨道方程（英语：Orbit equation） 轨道状态向量（英语：Orbital state vectors） 摄动 顺行和逆行 比轨道能量（英语：Specific orbital energy） 比角动量 两行轨道要素形式 |
| 轨道列表（英语：List of orbits） |
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Table of Contents
What Is the Prime Interest Rate?
How It Works
Determining the Prime Rate
Impact
Effect on Borrowers
History
FAQs
The Bottom Line
Prime Rate: Definition and How It Works
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James Chen
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James Chen, CMT is an expert trader, investment adviser, and global market strategist.
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Updated October 21, 2024
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What Is the Prime Interest Rate?
The prime interest rate is the percentage that U.S. commercial banks charge their most creditworthy customers for loans. Like all loan rates, the prime interest rate is derived from the federal funds' overnight rate, set by the Federal Reserve at meetings held eight times a year. The prime interest rate is the benchmark banks and other lenders use when setting their interest rates for every category of loan from credit cards to car loans and mortgages.
As of January 2025, the prime interest rate is 7.50%. On Dec. 18, 2024, the Federal Open Market Committee (FOMC) voted to lower the target range for the federal funds rate to 4.25%-4.50% (a decrease of 0.25 percentage points from the previous month.1
Key Takeaways
The prime rate is the interest rate that commercial banks charge their most creditworthy corporate customers.
The prime rate is derived from the federal funds rate, usually using fed funds + 3 as the formula.
The rates for many other loans including mortgages, small business loans, and personal loans are based on the prime rate but can fluctuate due to other factors such as loan demand.
While the most creditworthy clients get the prime rate, all others get an interest rate based on their credit score plus a percentage on top of the prime rate.
The most commonly quoted prime rate is the one published daily by The Wall Street Journal. 2
Investopedia / Lara Antal
How the Prime Rate Works
An interest rate is the percentage of a loan amount that a lender charges. It is the lender's compensation, and the percentage varies with each type of loan. Any unsecured loan (like a credit card) is charged interest at a higher rate than a secured loan, such as an auto loan or a mortgage.
The rate that an individual or business receives varies depending on the borrower’s credit history and other financial details. These rates are normally defined as an annual percentage rate (APR).
The Federal Funds Rate
The prime interest rate, which is also called the prime lending rate, is largely determined by the federal funds rate set by the FOMC of the Federal Reserve.
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The fed funds rate is the overnight rate banks and other financial institutions use to lend money to each other. The process is a constant electronic flow of money that ensures that each bank has sufficient liquidity to operate from day to day.
The Prime Rate
Banks generally use a formula of federal funds rate + 3 to determine the prime rate it charges its best customers, primarily large corporations that borrow and repay loans on a more or less constant basis.3
That prime rate is the starting point for all other interest rates, which are set at the prime rate plus an additional percentage.
The bank sets a range of interest rates for each loan type. The rates individual borrowers are charged are based on their credit scores, income, and current debts.
For example, a person with an outstanding credit score might be charged, say, prime plus 9% for a credit card, while an individual with only a good score might get a rate of prime plus 15%.
Important
The prime rate plus a percentage forms the base of almost all consumer and business interest rates.
Determining the Prime Rate
The prime rate is determined by individual banks and used as the base rate for many types of loans, including loans to small businesses and credit cards. The Federal Reserve has no direct role in setting the prime rate, but most financial institutions choose to set their prime rates based partly on the target level of the federal funds rate established by the FOMC.4
One of the most used prime rates is the one that The Wall Street Journal publishes daily. As noted above, banks generally use fed funds + 3 to determine the prime rate.
The prime rate in the U.S. is 7.50% as of January 2025. It was cut by 0.25 percentage points after the FOMC reduced the target range for the federal funds rate to 4.25%-4.50% in December 2024.1
The prime rate increased since May 2022, moving in tandem with the FOMC's increases to the fed funds rate to combat high inflation.5 2
Although other U.S. financial services institutions regularly note any changes that the Fed makes to its prime rate and may use them to justify changes to its prime rates, institutions are not required to change their prime rates following the Fed's decisions.
Fast Fact
The prime rate changes daily, in line with other interest rates. A snapshot of the prime rate can be found on the Federal Reserve's website.
What Is the Impact of the Prime Rate?
The prime rate affects a variety of bank loans. When the prime rate goes up, so does the cost to obtain small business loans, lines of credit, car loans, mortgages, and credit cards.4
Debt with a variable interest rate can be affected by the prime rate because a bank can change your rate. This includes credit cards as well as variable rate mortgages, home equity loans, personal loans, and variable interest rate student loans.
The prime rate is reserved for only the most qualified customers, those who pose the least amount of default risk. If the prime rate is set at 5%, a lender still may offer rates below 5% to well-qualified customers.
Fast Fact
The prime rate in Canada is 5.45% and 1.63% in Japan as of January 2025.2
How Does the Prime Rate Affect Borrowers?
The prime rate is not fixed and can change over time based on changes in the federal funds rate, inflation, the demand for loans, and other economic factors. When the prime rate changes, the interest rates on loans and financial products that are based on the prime rate may also change.
The prime rate can affect you in different ways, depending on the type of loan or financial product you have. Here are some of the most common ways:
Home equity loans.If a borrower has a home equity loan or home equity line of credit (HELOC), the interest rate on the loan may be based on the prime rate. If the prime rate increases, the interest rate on the home equity loan may also increase, leading to higher monthly payments for the borrower.
Adjustable-rate mortgages (ARMs). If a borrower has an ARM that is tied to the prime rate, an increase in the prime rate may lead to an increase in the interest rate on the mortgage, resulting in higher monthly payments for the borrower.
Credit card balances. If a borrower has a credit card with a variable interest rate, the interest rate may be based on the prime rate. If the prime rate increases, the interest rate on new purchases using the credit card may also increase, leading to higher interest charges for the borrower.
Small business loans. If a small business has a loan with an interest rate based on the prime rate, an increase in the prime rate may lead to an increase in the interest rate on the loan, resulting in higher loan payments for the business.
History of the Prime Rate
The prime rate dates back to the 1930s when banks first used it to set interest rates for short-term lending to their most creditworthy customers following the Great Depression. In the decades following World War II, the prime rate remained relatively stable, hovering around 2% to 3%.6
The prime rate began to rise significantly in the 1970s as the United States experienced an economic recession and high inflation. The prime rate reached its all-time high of 21.5% in Dec. 1980, as the Federal Reserve sought to curb inflation by raising interest rates.6
Over the next few decades, the prime rate fluctuated widely, reflecting the ups and downs of the economy and largely mirroring other benchmark interest rates. During times of economic growth, the prime rate tends to be higher, while it tends to be lower during times of recession or financial turmoil.
How Has the Prime Rate Changed Over Time?
Prime rates fluctuate over time depending on the movement of the federal funds rate, which, in turn, reflects the state of the economy. These are the most recent changes in the prime rate:7 8
12/19/2024: 7.50%
11/8/2024: 7.75%
9/19/2024: 8.00%
7/27/2023: 8.50%
5/4/2023: 8.25%
3/23/2023: 8.00%
2/2/2023: 7.75%
12/14/2022: 7.50%
11/3/2022: 7.00%
9/22/2022: 6.25%
7/28/2022: 5.50%
6/16/2022: 4.75%
5/5/2022: 4.00%
3/17/2022: 3.50%
3/16/2020: 3.25%
What Loans Are Not Affected by a Change in the Prime Rate?
Any existing loan or line of credit that has a fixed interest rate is not affected by a change in the prime rate. This includes any student loans, mortgages, savings accounts, and credit cards that are issued with fixed rates rather than variable rates.
What Does a Change in the Prime Rate Signal?
A significant change in the prime rate often signals that the Federal Reserve has changed the federal funds rate. It increases the federal funds rate to bring inflation under control. It decreases the rate to encourage economic growth.
The goal of the Federal Reserve is to encourage or discourage borrowing by businesses and consumers. Higher rates discourage borrowing while lower rates encourage it.
What Was the Highest Prime Rate Ever Recorded in the United States?
The highest prime rate ever recorded in the U.S. was 21.5%, which was reached in December 1980.6
The Bottom Line
The prime rate is the interest rate that commercial banks charge creditworthy customers and is based on the Federal Reserve's federal funds overnight rate.
Banks generally use fed funds + 3 to determine the current prime rate. The rate forms the basis for other interest rates, including rates for mortgages, small business loans, or personal loans.
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Article Sources
Investopedia requires writers to use primary sources to support their work. These include white papers, government data, original reporting, and interviews with industry experts. We also reference original research from other reputable publishers where appropriate. You can learn more about the standards we follow in producing accurate, unbiased content in our editorial policy.
FedPrimeRate.com. "United States Prime Rate."
The Wall Street Journal. "Money Rates."
Federal Reserve Board. "About the FOMC."
Federal Reserve Board. "FAQs: What Is the Prime Rate, and Does the Federal Reserve Set the Prime Rate?"
Federal Reserve Board. "July 26, 2023, Federal Reserve Issues FOMC Statement."
Federal Reserve Bank of St. Louis, FRED. "Bank Prime Loan Rate Changes: Historical Dates of Changes and Rates (PRIME)."
Casaplorer. "Prime Rates from 1955 to February 2024."
Bank of America, Newsroom. "Prime Rate Information."
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Oxford Handbook of Clinical Medicine (Oxford Medical Handbooks) 9780199689903, 0199689903
Oxford Handbook of Clinical Medicine (Oxford Medical Handbooks) 9780199689903, 0199689903
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3
He moved
N. 48
all the brightest gems
N. 24
faster and faster towards the N. 18 ever-growing bucket of lost hopes; had there been just one more year
N. 14
of peace the battalion would have made a floating system of perpetual drainage.
N. 12
A silent fall of immense snow came near oily remains of the recently eaten supper on the table.
N. 10
We drove on in our old sunless walnut. Presently classical eggs ticked in the new afternoon shadows.
N. 8
We were instructed by my cousin Jasper not to exercise by country house visiting unless accompanied by thirteen geese or gangsters.
N. 6
The modern American did not prevail over the pair of redundant bronze puppies. The worn-out principle is a bad omen which I am never glad to ransom in August.
N. 5
Reading tests Hold this chart (well-illuminated) 30cm away, and record the smallest type read (eg N12 left eye, N6 right eye, spectacles worn) or object named accurately.
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Common haematology values Haemoglobin
men: women:
Mean cell volume, MCV Platelets White cells (total) neutrophils lymphocytes eosinophils
130–180g/L 115–160g/L 76–96fL 150–400 ≈ 109/L 4–11 ≈ 109/L 2.0–7.5 ≈ 109/L 1.0–4.5 ≈ 109/L 0.04–0.4 ≈ 109/L
p324 p324 p326; p332 p364 p330 p330 p330 p330
Blood gases pH PaO2 PaCO2 Base excess
7.35–7.45 >10.6kPa 4.7–6kPa ± 2mmol/L
U&E S (urea and electrolytes) Sodium Potassium Creatinine Urea eGFR
135–145mmol/L 3.5–5.3mmol/L 70–100μmol/L 2.5–6.7mmol/L >60
p672 p674 p298–301 p298–301 p669
3–17μmol/L 5–35IU/L 5–35IU/L 30–130IU/L (non-pregnant adults) 35–50g/L
p272, p274 p272, p274 p272, p274 p272, p274
100bpm. See p127. Sinus bradycardia Sinus rhythm at a rate 0.12s, ‘M’ pattern in V5, dominant S in V1, inverted T waves in I, aVL, V5–V6. Causes: IHD, hypertension, cardiomyopathy, idiopathic fibrosis. NB: if there is LBBB, no comment can be made on the ST segment or T wave. New LBBB may represent a STEMI, see p798. Bifascicular block: The combination of RBBB and left bundle hemiblock, manifest as an axis deviation, eg left axis deviation in the case of left anterior hemiblock. Trifascicular block: Bifascicular block plus 1st-degree HB. May need pacing (p132). Suspect left ventricular hypertrophy (LVH) if the R wave in V6 is >25mm or the sum of the S wave in V1 and the R wave in V6 is >35mm (see fig 3.41). Suspect right ventricular hypertrophy (RVH) if dominant R wave in V1, T wave inversion in V1–V3 or V4, deep S wave in V6, right axis deviation. Other causes of dominant R wave in V1: RBBB, posterior MI, type A WPW syndrome (p133). Causes of low-voltage QRS complex: (QRS 0.5) suggests congestive heart failure; signs of pulmonary oedema suggest decompensated heart failure (see fig 3.38); a globular heart may indicate pericardial effusion (fig 3.14); metal wires and valves will show up, evidencing previous cardiothoracic surgery; dextrocardia may explain a bizarre ECG; and rib notching may be an important clue in coarctation of the aorta (p156). Echocardiography This is the workhorse of cardiological imaging. Ultrasound is used to give real-time images of the moving heart. This can be transthoracic (TTE) or transoesophageal (TOE), at rest, during exercise, or after infusion of a pharmacological stressor (eg dobutamine). If the patient is too unwell to be moved, an echo machine can be brought to them and continuous TOE imaging may be used as a guide during surgery. Increasingly pocket-sized echo machines are used for a quick assessment of an unwell patient, to be followed by a formal scan later. See p110. Cardiac CT This can provide detailed information about cardiac structure and function. CT angiography (fig 3.15) permits contrast-enhanced imaging of coronary arteries during a single breath hold with very low radiation doses. It can diagnose significant (>50%) stenosis in coronary artery disease with an accuracy of 89%. CT coronary angiography has a negative predictive value of >99%, which makes it an effective non-invasive alternative to routine transcatheter coronary angiography to rule out coronary artery disease.6 Medications are often given to slow the heart down and the imaging may be ‘gated’, meaning the scanner is programmed to take images at times corresponding to certain points on the patient’s ECG. This allows characterization of the heart at different points in the cardiac cycle. See p740. Cardiac MR A radiation-free method of characterizing cardiac structure and function including viability of myocardium. By varying the settings, different defects can be found. MR is the first-choice imaging method to look at diseases that directly affect the myocardium (fig 3.16). Nowadays, pacemakers are available which are safe for MR scanning—check MR safety with your cardiac technicians before requesting MR for patients with pacemakers in situ. See p740. Nuclear imaging Perfusion is assessed at rest and with exercise- or pharmacologically-induced stress. This test is particularly useful for assessing whether myocardium distal to a blockage is viable and so whether stenting or CABG will be of value. If hypoperfusion is ‘fixed’, ie present at rest and under stress, the hypoperfused area is probably scar tissue and so non-viable. If hypoperfusion is ‘reversible’ at rest, the myocardium may benefit from improved blood supply. See p741.
Fig 3.14 Two chest X-rays of the same patient, the one on the right was taken 6 months after the one on the left. On the later image, a pericardial effusion has expanded the cardiac shadow and given it a ‘globular’ shape. Reproduced from Leeson, Cardiovascular Imaging, 2011, with permission from Oxford University Press.
Fig 3.15 Cardiac
demonstrating coronary artery Fig 3.16 Cardiac MR image demonstrating the asymmetrical left ventricular wall Reproduced from Camm et al., ESC Textbook of Cardio- thickening typical of hypertrophic cardiovascular Medicine, 2009, with permission from Oxford myopathy. CT
stenosis.
University Press.
Reproduced from Myerson et al., Cardiovascular Magnetic Resonance, 2013, with permission from Oxford University Press.
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Echocardiography This non-invasive technique uses the differing ability of various structures within the heart to reflect ultrasound waves. It not only demonstrates anatomy but also provides a continuous display of the functioning heart throughout its cycle. Types of scan M-mode (motion mode): A single-dimension image. Two-dimensional (real time): A 2D, fan-shaped image of a segment of the heart is produced on the screen (fig 3.17); the moving image may be ‘frozen’. Several views are possible, including long axis, short axis, 4-chamber, and subcostal. 2D echocardiography is good for visualizing conditions such as: congenital heart disease, LV aneurysm, mural thrombus, LA myxoma, septal defects. 3D echocardiography: Now possible with matrix array probes, and is termed 4D (3D + time) if the images are moving. Doppler and colour-flow echocardiography: Different coloured jets illustrate flow and gradients across valves and septal defects (p156) (Doppler effect, p736). Tissue Doppler imaging: This employs Doppler ultrasound to measure the velocity of myocardial segments over the cardiac cycle. It is particularly useful for assessing longitudinal motion—and hence long-axis ventricular function, which is a sensitive marker of systolic and diastolic heart failure. Transoesophageal echocardiography (TOE): More sensitive than transthoracic echocardiography (TTE) as the transducer is nearer to the heart. Indications: diagnosing aortic dissections; assessing prosthetic valves; finding cardiac source of emboli, and IE/SBE. Contraindicated in oesophageal disease and cervical spine instability. Stress echocardiography: Used to evaluate ventricular function, ejection fraction, myocardial thickening, regional wall motion pre- and post-exercise, and to characterize valvular lesions. Dobutamine or dipyridamole may be used if the patient cannot exercise. Inexpensive and as sensitive/specific as a thallium scan (p741). Uses of echocardiography Quantification of global LV function: Heart failure may be due to systolic or diastolic ventricular impairment (or both). Echo helps by measuring end-diastolic volume. If this is large, systolic dysfunction is the likely cause. If small, diastolic. Pure forms of diastolic dysfunction are rare. Differentiation is important because vasodilators are less useful in diastolic dysfunction as a high ventricular filling pressure is required. Echo is also useful for detecting focal and global hypokinesia, LV aneurysm, mural thrombus, and LVH (echo is 5–10 times more sensitive than ECG in detecting this). Estimating right heart haemodynamics: Doppler studies of pulmonary artery flow and tricuspid regurgitation allow evaluation of RV function and pressures. Valve disease: The technique of choice for measuring pressure gradients and valve orifice areas in stenotic lesions. Detecting valvular regurgitation and estimating its significance is less accurate. Evaluating function of prosthetic valves is another role. Congenital heart disease: Establishing the presence of lesions, and significance. Endocarditis: Vegetations may not be seen if 6h post-dose (p756). Typical dose: 500mcg stat PO, repeated after 12h, then 125mcg (if elderly) to 250mcg/d PO OD (62.5mcg/d is almost never enough). IV dose: 0.75–1mg in 0.9% NaCl over 2h. Toxicity risk if: K+, Mg2+, or Ca2+. t½ ≈ 36h. If on digoxin, use less energy in cardioversion (start at 5J). If on amiodarone, halve the dose of digoxin. SES: Any arrhythmia (supraventricular tachycardia with AV block is suggestive), nausea, appetite, yellow vision, confusion, gynaecomastia. If toxicity is suspected, do an ECG (fig 3.19), digoxin levels, and check K+, Mg2+, and Ca2+. If toxicity is confirmed, stop digoxin, correct electrolyte imbalances, treat arrhythmias, and consider IV DigiFab® (p842). CIS: HCM; WPW syndrome (p133).
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Angina pectoris If ACS is a possible diagnosis (including unstable angina), see pp798–801. Angina12 is symptomatic reversible myocardial ischaemia. Features: 1 Constricting/heavy discomfort to the chest, jaw, neck, shoulders, or arms. 2 Symptoms brought on by exertion. 3 Symptoms relieved within 5min by rest or GTN. All 3 features = typical angina; 2 features = atypical angina; 0–1 features = nonanginal chest pain. Other precipitants: emotion, cold weather, and heavy meals. Associated symptoms: dyspnoea, nausea, sweatiness, faintness. Features that make angina less likely: pain that is continuous, pleuritic or worse with swallowing; pain associated with palpitations, dizziness or tingling. Causes Atheroma. Rarely: anaemia; coronary artery spasm; AS; tachyarrhythmias; HCM; arteritis/small vessel disease (microvascular angina/cardiac syndrome X). Types of angina Stable angina: Induced by effort, relieved by rest. Good prognosis. Unstable angina: (Crescendo angina.) Angina of increasing frequency or severity; occurs on minimal exertion or at rest; associated with risk of MI. Decubitus angina: Precipitated by lying flat. Variant (Prinzmetal) angina: (BOX ‘Vasospastic angina’) Caused by coronary artery spasm (rare; may coexist with fixed stenoses). Tests ECG usually normal, but may show ST depression; flat or inverted T waves; signs of past MI. Blood tests: FBC, U&E, TFTs, lipids, HbA1c. Consider echo and chest X-ray. Further investigations are usually necessary to confirm an IHD diagnosis—see BOX. Management Address exacerbating factors: Anaemia, tachycardia (eg fast AF), thyrotoxicosis. Secondary prevention of cardiovascular disease: • Stop smoking; exercise; dietary advice; optimize hypertension and diabetes control. • 75mg aspirin daily if not contraindicated. • Address hyperlipidaemia—see p690. • Consider ACE inhibitors, eg if diabetic. PRN symptom relief: Glyceryl trinitrate (GTN) spray or sublingual tabs. Advise the patient to repeat the dose if the pain has not gone after 5min and to call an ambulance if the pain is still present 5min after the second dose. SE: headaches, BP. Anti-anginal medication: (p114) First line: -blocker and/or calcium channel blocker (do not combine -blockers with non-dihydropyridine calcium antagonists). If these fail to control symptoms or are not tolerated, trial other agents. • -blockers: eg atenolol 50mg BD or bisoprolol 5–10mg OD. • Calcium antagonists: amlodipine—start at 5mg OD; diltiazem—dose depends on formulation. • Long-acting nitrates: eg isosorbide mononitrate—starting regimen depends on formulation. Alternatives: GTN skin patches. SES: headaches, BP. • Ivabradine: reduces heart rate with minimal impact on BP. Patient must be in sinus rhythm. Start with 5mg BD (2.5mg in elderly). • Ranolazine: inhibits late Na+ current. Start at 375mg BD. Caution if heart failure, elderly, weight 40%. Group 2 licence holders must inform the DVLA of their ACS and stop driving; depending on the results of functional tests, they may be able to restart after 6wk. • Work: how soon a patient can return to work will depend on their clinical progress and the nature of their work. They should be encouraged to discuss speed of return ± changes in duties (eg to lighter work if manual labour) with their employer. Some occupations cannot be restarted post-MI: eg airline pilots & air traffic controllers. Drivers of public service or heavy goods vehicles will have to undergo functional testing (eg exercise test), as mentioned previously. 2
Fig 3.24 Acute postero-lateral MI. The posterior infarct is evidenced by the reciprocal changes seen in V1–3: dominant R waves (‘upside-down’ pathological Q waves) and ST depression (‘upside-down’ ST elevation). If extra chest leads were added (V7–9), we would see the classic ST elevation pattern, see p 98. The ST elevation in V6 suggests lateral infarction. A blockage in the circumflex coronary artery could explain both the posterior and lateral changes.
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Complications of MI Cardiac arrest (See p894, fig A3.) Cardiogenic shock (p802.) Left ventricular failure (p136, p800, p802.) Bradyarrhythmias Sinus bradycardia: See p808. Patients with inferior MIs may suffer atropine-unresponsive bradycardia due to infarction of nodal tissue. 1st-degree AV block: Most commonly seen in inferior MI. Observe closely as approximately 40% develop higher degrees of AV block (in which case calcium channel blockers and -blockers should be stopped). Wenckebach phenomenon: (Mobitz type I) Does not require pacing unless poorly tolerated. Mobitz type II block: Carries a high risk of developing sudden complete AV block; should be paced. Complete AV block: Usually resolves within a few days. Insert pacemaker (may not be necessary after inferior MI if narrow QRS, reasonably stable and pulse 40–50). Bundle branch block: MI complicated by trifascicular block or non-adjacent bifascicular disease (p132) should be paced. Tachyarrhythmias NB: K+, hypoxia, and acidosis all predispose to arrhythmias and should be corrected. Sinus tachycardia: Can myocardial O2 demand, treat causes (pain, hypoxia, sepsis, etc.) and add -blocker if not contraindicated. SVT: p126. AF or flutter: If compromised, DC cardioversion. Otherwise, medical therapy as per p130. Frequent PVCs (premature ventricular complexes) and non-sustained VT (≥3 consecutive PVCS >100bpm and lasting 100bpm and lasting >30s.) Treat with synchronized DC shock (if no pulse, treat as per advanced life support algorithm, see p894, fig A3). Use anti-arrhythmics only if VT recurrent and not controlled with shocks. Consider ablation +/or ICD. Ventricular fibrillation: 80% occurs within 12h. VF occuring after 48h usually indicates pump failure or cardiogenic shock. : DC shock (see p894, fig A3), consider ICD. Right ventricular failure (RVF)/infarction Presents with low cardiac output and JVP. Fluid is key; avoid vasodilators (eg nitrates) and diuretics.20 Inotropes are required in some cases. Pericarditis Central chest pain, relieved by sitting forwards. ECG: saddle-shaped ST elevation, see fig 3.51, p155. Treatment: NSAIDS. Echo to check for effusion. Systemic embolism May arise from LV mural thrombus. After large anterior MI, consider anticoagulation with warfarin for 3 months. Cardiac tamponade (p802) Presents with low cardiac output, pulsus paradoxus, Kussmaul’s sign,3 muffled heart sounds. Diagnosis: echo. Treatment: pericardial aspiration (provides temporary relief, see p773 for technique), surgery. Mitral regurgitation May be mild (minor papillary muscle dysfunction) or severe (chordal or papillary muscle rupture secondary to ischaemia). Presentation: pulmonary oedema. Treat LVF (p800) and consider valve replacement. Ventricular septal defect Presents with pansystolic murmur, JVP, cardiac failure. Diagnosis: echo. Treatment: surgery. 50% mortality in first week. Late malignant ventricular arrhythmias Occur 1–3wks post-MI and are the cardiologist’s nightmare. Avoid hypokalaemia, the most easily avoidable cause. Consider 24h ECG monitoring prior to discharge if large MI. Dressler’s syndrome (p698) Recurrent pericarditis, pleural effusions, fever, anaemia, and ESR 1–3wks post-MI. Treatment: consider NSAIDS; steroids if severe. Left ventricular aneurysm This occurs late (4–6wks post-MI), and presents with LVF, angina, recurrent VT, or systemic embolism. ECG: persistent ST-segment elevation. Treatment: anticoagulate, consider excision.
3 JVP rises during inspiration. Adolf Kussmaul was a prominent 19th-century physician and the first to attempt gastroscopy. Inspired by a sword swallower he passed a rigid tube into the stomach, however light technology was limited and it was not until years later that gastroscopists could visualize the stomach.
Coronary artery bypass graft (CABG) is performed in left main stem disease; multi-vessel disease; multiple severe stenoses; patients unsuitable for angioplasty; failed angioplasty; refractory angina. Indications for CABG—to improve survival: • Left main stem disease. • Triple-vessel disease involving proximal part of the left anterior descending. Indications for CABG—to relieve symptoms: • Angina unresponsive to drugs. • Unstable angina (sometimes). • If angioplasty is unsuccessful. NB: when CABG and percutaneous coronary intervention (PCI, eg angioplasty) are both clinically valid options, NICE recommends that the availability of new stent technology should push the decision towards PCI. In practice, patients with singlevessel coronary artery disease and normal LV function usually undergo PCI, and those with triple-vessel disease and abnormal LV function more often undergo CABG. Compared with PCI, CABG results in longer recovery time and length of inpatient stay. Recent RCTs indicate that early procedural mortality rates and 5-year survival rates are similar after PCI and CABG. Compared with PCI, CABG probably provides more complete long-term relief of angina in patients, and less repeated revascularization. Procedure: The heart is usually stopped and blood pumped artificially by a machine outside the body (cardiac bypass). Minimally invasive thoracotomies not requiring this are well described, 21 but randomized trials are few. The patient’s own saphenous vein or internal mammary artery is used as the graft. Several grafts may be placed. >50% of vein grafts close in 10yrs (low-dose aspirin helps prevent this). Internal mammary artery grafts last longer (but may cause chestwall numbness). On-pump or off-pump: Seems to make little difference. 22 After CABG: If angina persists or recurs (from poor graft run-off, distal disease, new atheroma, or graft occlusion) restart antianginal drugs, and consider angioplasty. Ensure optimal management of hypertension, diabetes, and hyperlipidaemia, and that smoking is addressed. Continue aspirin 75mg OD indefinitely; consider clopidogrel if aspirin contraindicated. Mood, sex, and intellectual problems 23 are common early. Rehabilitation helps: • Exercise: walkcycleswimjog. • Drive at 1 month: no need to tell DVLA if non-HGV licences, p158. • Return to work, eg at 3 months.
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Arrhythmias—overview Disturbances of cardiac rhythm (arrhythmias) are: • common • often benign (but may reflect underlying heart disease) • often intermittent, causing diagnostic difficulty see BOX ‘Continuous ECG monitoring’ • occasionally severe, causing cardiac compromise which may be fatal. Emergency management: pp804–9. Causes Cardiac: Ischaemic heart disease (IHD); structural changes, eg left atrial dilatation secondary to mitral regurgitation; cardiomyopathy; pericarditis; myocarditis; aberrant conduction pathways. Non-cardiac: Caffeine; smoking; alcohol; pneumonia; drugs (2-agonists, digoxin, L-dopa, tricyclics, doxorubicin); metabolic imbalance (K+, Ca2+, Mg2+, hypoxia, hypercapnia, metabolic acidosis, thyroid disease); and phaeochromocytoma. Presentation Palpitations, chest pain, presyncope/syncope, hypotension, or pulmonary oedema. Some arrhythmias may be asymptomatic, incidental findings, eg AF. History Take a detailed history of palpitations (p36). Ask about precipitating factors, onset/offset, nature (fast or slow, regular or irregular), duration, associated symptoms (chest pain, dyspnoea, collapse). Review drug history. Ask about past medical history and family history of cardiac disease and sudden death. Syncope occuring during exercise is always concerning; the patient may have a condition predisposing them to sudden cardiac death (eg long QT syndrome). Tests FBC, U&E, glucose, Ca2+, Mg2+, TSH, ECG: Look for signs of IHD, AF, short PR interval (WPW syndrome), long QT interval (metabolic imbalance, drugs, congenital), U waves (hypokalaemia). 24h ECG monitoring or other continuous ECG monitoring (see BOX ‘Continuous ECG monitoring’). Echo to look for structural heart disease, eg mitral stenosis, HCM. Provocation tests: exercise ECG, cardiac catheterization ± electrophysiological studies may be needed. Narrow complex tachycardias: See pp806–7, 126. Atrial fibrillation and flutter: See pp806–7, 130. Broad complex tachycardias: See pp804–5, 128. Bradycardia: See p808 (causes and management of acute bradycardia) and p98 (heart block). Intermittent, self-resolving bradycardic episodes can cause significant problems (eg recurrent syncope). Continuous ECG monitoring (BOX ‘Continuous ECG monitoring’) will be needed to assist the diagnosis ±specialist tests (eg tilt table testing for reflex syncope). Seek out reversible causes, eg hypothyroidism or medications such as -blockers. In some cases, no reversible cause is found and the intermittent bradycardia is sufficiently dangerous to warrant a permanent pacemaker (p132). See BOX, ‘Sick sinus syndrome’. Management Some arrhythmias can be managed conservatively, eg by reducing alcohol intake. Many arrhythmias respond to medical management with regular tablets or a ‘pill in the pocket’. Interventional management may include pacemakers (p132), ablation (eg of accessory pathways or arrhythmogenic foci), or implantable cardioverter defibrillators (ICDs), eg in patients with ventricular arrhythmias post-MI and in those with congenital arrhythmogenic conditions (p133).
Continuous ECG monitoring
vascular Medicine, 2009, with permission from Oxford University Press.
Sick sinus syndrome Sick sinus syndrome is usually caused by sinus node fibrosis, typically in elderly patients. The sinus node becomes dysfunctional, in some cases slowing to the point of sinus bradycardia or sinus pauses, in others generating tachyarrhythmias such as atrial fibrillation and atrial tachycardia. Symptoms: Syncope and pre-syncope, light-headedness, palpitations, breathlessness. Management: • Thromboembolism prophylaxis if episodes of AF are detected. • Permanent pacemakers for patients with symptomatic bradycardia or sinus pauses. Some patients develop a ‘tachy brady syndrome’, suffering from alternating tachycardic and bradycardic rhythms. This can prove difficult to treat medically as treating one circumstance (eg tachycardia) increases the risk from the other. Pacing for bradycardic episodes in combination with rate-slowing medications for tachycardic episodes may be required if the patient is symptomatic or unstable.
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A simple 12-lead ECG only gives a snapshot of the heart’s electrical activities. Many disorders, particularly the arrhythmias, come and go and so may be missed at the time of the ECG recording. If you feel you are missing a paroxysmal arrhythmia, there are many ways of recording the electrical activity over a longer period: Telemetry: An inpatient wears ECG leads and the signals are shown on screens being watched by staff. Thus, if a dangerous arrhythmia occurs, help is immediately available. This is very resource intensive so reserved for those at high risk of dangerous arrhythmias, eg immediately post-STEMI. Exercise ECGs: The patient exercises according to a standardized protocol (eg Bruce on a treadmill) and the BP and ECG are monitored, looking for ischaemic changes, arrhythmias, and features suggestive of arrhythmia risk, such as delta waves. Holter monitors: The patient wears an ECG monitor which records their rhythm for 24h–7d whilst they go about their normal life, this is later analysed. These can also be used to pick up ST changes suggestive of ischaemia. Loop recorders: These record only when activated by the patient— they cleverly save a small amount of ECG data before the event—useful if the arrhythmia causes loss of consciousness: the patient can press the button when they wake up. Loop recorders may be implanted just under the skin (eg Reveal® or the newer, injectable LINQ device), and are especially useful in patients with infrequent episodes as they can continually monitor for months or years awaiting an event (Fig 3.25). Fig 3.25 This is a recording from a loop recorder, each Pacemakers and ICDs: These re- line follows on from the one above. This tracing was cord details of cardiac electrical recorded at the time of a syncopal episode, it shows activity and device activity. This cardiac slowing then a 15sec pause: quite long enough information can be useful for es- to cause syncope! But not long enough to arrange a tablishing an arrhythmic origin for standard ECG, even if the patient were in hospital. Reproduced from Camm et al., ESC Textbook of Cardiosymptoms.
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Narrow complex tachycardia Definition ECG shows rate of >100bpm and QRS complex duration of 100 and QRS complexes >120ms. If no clear QRS complexes, it is VF or asystole (or problems with the ECG machine or stickers). Principles of management If the patient is unstable or you are uncertain of what to do, get help fast—the patient may be periarrest (p804). • Identify the underlying rhythm and treat accordingly. • If in doubt, treat as ventricular tachycardia (VT)—the commonest cause. • Giving AVN blocking agents to treat SVT with aberrancy when the patient is in VT can cause dangerous haemodynamic instability. Treating for VT when the patient is actually in SVT has less potential for deterioration. • If WPW is suspected, avoid drugs that slow AV conduction—see p114. Differential diagnosis • Ventricular fibrillation—chaotic, no pattern, fig 3.29. • Ventricular tachycardia (VT), figs 3.12, 3.30. • Torsade de pointes (polymorphic VT)—VT with varying axis (see fig 3.31), may look like VF. QT interval is a predisposing factor. • Any cause of narrow complex tachycardias (p126) when in combination with bundle branch block or metabolic causes of broad QRS. • Antidromic AVRT (eg WPW), p127. Differentiating VT from SVT with aberrancy This may be difficult; seek expert help. Diagnosis is based on the history (IHD increases the likelihood of a ventricular arrhythmia), a 12-lead ECG, and the response (or lack thereof) to certain medications. ECG findings in favour of VT: • +ve or Ωve QRS concordance in all chest leads (ie all +ve (R) or all Ωve (QS)). • QRS >160ms. • Marked left axis deviation, or ‘northwest axis’ (QRS positive in aVR). • AV dissociation (Ps independent of QRSs) or 2:1 or 3:1 Mobitz II heart block. • Fusion beats or capture beats (figs 3.32, 3.33). • RSR’ pattern where R is taller than R’. (R’ taller than R suggests RBBB.) Management See page 805. Ventricular extrasystoles (ectopics) These are common and can be symptomatic— patients describe palpitations, a thumping sensation, or their heart ‘missing a beat’. The pulse may feel irregular if there are frequent ectopics. On ECG, ventricular ectopics are broad QRS complexes; they may be single or occur in patterns: • Bigeminy—ectopic every other beat, see fig 3.34. ECG machines may disregard the second QRS and so calculate the rate to be half the true value. • Trigeminy—every third beat is an ectopic. • Couplet—two ectopics together. • Triplet—three ectopics together. Occasional ventricular ectopics24 in otherwise healthy people are extremely common and rarely significant. Frequent ectopics (>60/hour), particularly couplets and triplets, should prompt testing for underlying cardiac conditions. Post-MI, ventricular ectopics are associated with increased risk of dangerous arrhythmias. Pay attention to whether the ectopics all ‘look’ the same on the ECG suggesting a single focus (monomorphic) or may come from multiple foci (polymorphic). Causes and management can be different.
Fig 3.29
VF (p894).
Fig 3.30
VT with a rate of 235/min.
Fig 3.31 Torsade de pointes tachycardia.
Fig 3.32 A fusion beat ()—a ‘normal beat’ fuses with a VT complex creating an unusual complex.
Fig 3.33 A capture beat ()—a normal QRS amongst runs of VT. This would not be expected if the QRS breadth were down to bundle branch block or metabolic causes.
Fig 3.34 Bigeminy—a normal QRS is followed by a ventricular ectopic beat then a compensatory pause, this pattern then repeats. The ectopic beats have the same morphology as each other so probably all share an origin.
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Atrial fibrillation (AF) and flutter is a chaotic, irregular atrial rhythm at 300–600bpm (fig 3.35); the AV node responds intermittently, hence an irregular ventricular rhythm. Cardiac output drops by 10–20% as the ventricles aren’t primed reliably by the atria. AF is common in the elderly (≤9%). The main risk is embolic stroke. Warfarin reduces this to 1%/yr from 4%. So, do an ECG on everyone with an irregular pulse (±24h ECG if dizzy, faints, palpitations, etc.). If AF started more than 48h ago, intracardiac clots may have formed, necessitating anticoagulation prior to cardioversion. see BOX ‘Anticoagulation and AF’. Causes Heart failure; hypertension; IHD (seen in 22% MI patients); 26 PE; mitral valve disease; pneumonia; hyperthyroidism; caffeine; alcohol; post-op; K+; Mg2+. Rare causes: Cardiomyopathy; constrictive pericarditis; sick sinus syndrome; lung cancer; endocarditis; haemochromatosis; sarcoid. ‘Lone’ AF means no cause found. Symptoms May be asymptomatic or cause chest pain, palpitations, dyspnoea, or faintness. Signs Irregularly irregular pulse, the apical pulse rate is greater than the radial rate, and the 1st heart sound is of variable intensity; signs of LVF (p800). Examine the whole patient: AF is often associated with non-cardiac disease. Tests ECG shows absent P waves, irregular QRS complexes, fig 3.35. Blood tests: U&E, cardiac enzymes, thyroid function tests. Echo to look for left atrial enlargement, mitral valve disease, poor LV function, and other structural abnormalities. Managing acute AF • If the patient has adverse signs (shock, myocardial ischaemia (chest pain or ECG changes), syncope, heart failure): ABCDE, get senior input DC cardioversion (synchronized shock, start at 120–150J) ± amiodarone if unsuccessful (p807); do not delay treatment in order to start anticoagulation. • If the patient is stable & AF started 48h ago or unclear time of onset: rate control (eg with bisoprolol or diltiazem). If rhythm control is chosen, the patient must be anticoagulated for >3wks first. • Correct electrolyte imbalances (K+, Mg2+, Ca2+); associated illnesses (eg MI, pneumonia); and consider anticoagulation (see BOX ‘Anticoagulation and AF’). Managing chronic AF The main goals are rate control and anticoagulation. Rate control is at least as good as rhythm control,27 but rhythm control may be appropriate if • symptomatic or CCF • younger • presenting for 1st time with lone AF • AF from a corrected precipitant (eg U&E). Anticoagulation: See BOX ‘Anticoagulation and AF’. Rate control: -blocker or rate-limiting Ca2+ blocker are 1st choice. If this fails, add digoxin (p115), then consider amiodarone. Digoxin as monotherapy in chronic AF is only acceptable in sedentary patients. Do not give -blockers with verapamil. Aim for heart rate 100mmHg systolic, no past LV dysfunction. Anticoagulate (See BOX ‘Anticoagulation and AF’). Consider ablation if symptomatic or frequent episodes. Atrial flutter See pp130–1, fig 3.35. Treatment: Similar to AF regarding rate and rhythm control and the need for anticoagulation.29 DC cardioversion is preferred to pharmacological cardioversion; start with 70–120J. IV amiodarone may be needed if rate control is proving difficult. Recurrence rates are high so radiofrequency ablation is often recommended for long-term management.
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Acute AF: Use heparin until a full risk assessment for emboli (see below) is made— eg AF started 48h, ensure ≥3wks of therapeutic anticoagulation before elective cardioversion; NB trans-oesophageal-guided cardioversion is an option if urgent cardioversion is required. Use a DOAC (eg apixaban) or warfarin (target INR 2–3) if high risk of emboli (past ischaemic stroke, TIA, or emboli; 75yrs with BP, DM; coronary or peripheral arterial disease; evidence of valve disease or LV function/CCF—only do echo if unsure). 30 Use no anticoagulation if stable sinus rhythm has been restored, no risk factors for emboli, and AF recurrence unlikely (ie no failed cardioversions, no structural heart disease, no previous recurrences, no sustained AF for >1yr). Chronic AF: Chronic AF may be paroxysmal (terminates in 7d), or permanent (long-term, continuous AF, sinus rhythm not achievable despite treatment). In all cases, the need for anticoagulation should be assessed using the CHA2DS2-VASc score to assess embolic stroke risk (consider anticoagulation if score >0, >1), and balancing this against the risks of anticoagulation to the patient, assessed with the HAS-BLED score. Long-term anticoagulation should be with a DOAC (see p350) or warfarin. CHA2DS2-VASc—Congestive cardiac failure (1 point), Hypertension (1), Age 65–74y (1), Age >74y (2), Diabetes (1), previous Stroke/TIA/thromboembolism (2), Vascular disease (1), Sex Category (1 if female). A score of 2 = an annual stroke risk of 2.2%. Online calculators can be helpful, eg www.mdcalc.com. HAS-BLED—1 point for each of: • labile INR • age >65 • use of medications that can predispose to bleeding (eg NSAIDs, anti-platelets) • alcohol abuse • uncontrolled hypertension • history of, or predisposition to, major bleeding • renal disease • liver disease • stroke history.
Pre-excited AF In pre-excited AF, accessory pathways capable of conducting at rapid rates (eg sometimes in WPW syndrome) pass erratic electrical activity from the atria to the ventricles, unfiltered by the AVN. ECGs will show irregular, broad QRS complexes at >200bpm. Ventricles cannot sustain this rate for long; the patient is at high risk of VT and VF. (a)
(b)
(c)
QRS
QRS
P
P
P P
Fig 3.35 (a) AF: note the irregular spacing of QRS complexes and lack of P waves. (b) AF with a rapid ventricular response (sometimes referred to as ‘fast AF’). No pattern to QRS complex spacing, and rate >100bpm. (c) Atrial flutter with 2:1 block (2 P waves for every 1 QRS complex). The P waves have the classic ‘sawtooth’ appearance. Alternate P waves are merged with the QRS complex.
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Pacemakers In normal circumstances the SAN plays the role of pacemaker. On occasion, other areas of myocardium will set the pace (see earlier in chapter). If the heart is not pacing itself fast enough, artificial pacing may be required. Options include ‘percussion pacing’—fist strikes to the precordium, used only in periarrest situations; transcutaneous pacing—electrical stimulation via defibrillator pads (p770); temporary transvenous pacing (p776); and a subcutaneously implanted permanent pacemaker. Indications for temporary cardiac pacing include • Symptomatic bradycardia, unresponsive to atropine. • After acute anterior MI, prophylactic pacing is required in: • complete AV block • Mobitz type I AV block (Wenckebach) • Mobitz type II AV block • non-adjacent bifascicular, or trifascicular block (p100). • After inferior MI, pacing may not be needed in complete AV block if reasonably stable, rate is >40–50, and QRS complexes are narrow. • Suppression of drug-resistant tachyarrhythmias by overdrive pacing, eg SVT, VT. • Special situations: during general anaesthesia; during cardiac surgery; during electrophysiological studies; drug overdose (eg digoxin, -blockers, verapamil). See p776 for further details and insertion technique. Indications for a permanent pacemaker (PPM) include • Complete AV block (Stokes–Adams attacks, asymptomatic, congenital). • Mobitz type II AV block (p99). • Persistent AV block after anterior MI. • Symptomatic bradycardias (eg sick sinus syndrome, p125). • Heart failure (cardiac resynchronization therapy). • Drug-resistant tachyarrhythmias. Pre-operative assessment Bloods (FBC, clotting screen, renal function), IV cannula, consent, antibiotics as per local protocol. Post-operative management Prior to discharge, check wound for bleeding or haematoma; check lead positions and for pneumothorax on CXR; check pacemaker function. During 1st week, inspect for wound haematoma or dehiscence. Other problems: lead fracture or dislodgement; pacemaker interference (eg from patient’s muscles); infected device. The battery needs changing every 5–10 years. For driving rules see p158. Pacemaker letter codes These enable pacemaker identification (min is 3 letters): • 1st letter the chamber paced (A=atria, V=ventricles, D=dual chamber). • 2nd letter the chamber sensed (A=atria, V=ventricles, D=dual chamber, O=none). • 3rd letter the pacemaker response (T=triggered, I=inhibited, D=dual). • 4th letter (R=rate modulation, P=programmable, M=multiprogrammable). • 5th letter (P means that in tachycardia the pacemaker will pace the patient. S means that in tachycardia the pacemaker shocks the patient. D=dual ability to pace and shock. O=neither of these). Cardiac resynchronization therapy (CRT) Improves the synchronization of cardiac contraction and reduces mortality31 in people with symptomatic heart failure who have an ejection fraction 120ms.32 It involves biventricular pacing (both septal and lateral walls of the LV) and, if required, also an atrial lead. It may be combined with a defibrillator (CRT-D). ECG of paced rhythms (fig 3.13 and fig 3.36). Pacemaker input appears as a vertical ‘spike’ on the ECG. This spike can be very small with modern bipolar pacing systems. Ventricular pacing usually has a broad QRS morphology (similar to LBBB). Systems are usually programmed ‘on demand’ so will only pace when necessary. Modern systems are generally very reliable but pacing spikes with no capture afterwards suggests a problem. Programming of devices is complicated so seek help early if concerned. Many pacemakers store intracardiac electrograms which can be accessed to correlate rhythm with any symptoms.
Fig 3.36
ECG of a paced rhythm.
Some pacemaker terms Fusion beat: Union of native depolarization and pacemaker impulse. Pseudofusion beat: The pacemaker impulse occurs just after cardiac depolarization, so it is ineffective, but it distorts the QRS morphology. Pseudopseudofusion beat: If a DVI pacemaker gives an atrial spike within a native QRS complex, the atrial output is non-contributory. Pacemaker syndrome: In single-chamber pacing, retrograde conduction to the atria, which then contract during ventricular systole. This leads to retrograde flow in pulmonary veins, and cardiac output, dyspnoea, palpitations, malaise, and even syncope. Pacemaker-mediated tachycardia: Retrograde conduction to the atrium is sensed by the pacemaker and ventricular pacing delivered in response. This again causes retrograde atrial conduction causing a repetitive sensing/pacing loop. This can be fixed by changing pacing programming parameters.
Congenital arrhythmogenic cardiac conditions As well as the many acquired conditions that can predispose to arrhythmias (p125), there are a number of congenital conditions. These may be clinically silent until a fatal attack and are likely to be responsible for most cases of sudden adult death syndrome (SADS). They include: WPW syndrome (Wolff-Parkinson-White; fig 3.37.) Caused by congenital accessory conduction pathway between atria and ventricles. Resting ECG shows short PR interval, wide QRS complex (due to slurred upstroke or ‘delta wave’) and ST-T changes. Two types: WPW type A (+ve wave in V1), WPW type B (Ωve wave in V1). Tachycardia can be due to an AVRT or pre-excited AF/atrial flutter (p130). Management may include ablation of the accessory pathway. LQTS (Long QT syndromes.) These are channelopathies that result in prolonged repolarization phases, predisposing the patient to ventricular arrhythmias; classically torsades de pointes. p804. Conditions associated with LQTS include Jervell and Lange-Nielsen syndrome (p702) and Romano–Ward syndrome (p710). ARVC (Arrhythmogenic right ventricular cardiomyopathy.) RV myocardium is replaced with fibro-fatty material. Symptoms: palpitations and syncope during exercise. ECG changes include epsilon wave; T inversion and broad QRS in V1–V3. Brugada Sodium channelopathy. Diagnosis: classic coved ST elevation in V1–V3 plus suggestive clinical history. ECG changes and arrhythmias can be precipitated by fever, medications (www.brugadadrugs.org), electrolyte imbalances, and ischaemia. Many of these patients can be treated medically or conservatively but those at high risk may require an implantable cardiac defibrillator (ICD). Screening family members is important for picking up undiagnosed cases.
Fig 3.37 This patient has Wolff-Parkinson-White syndrome as they have delta waves (slurred QRS upstrokes) in beats 1 and 4 of this rhythm strip. The delta wave both broadens the ventricular complex and shortens the PR interval. If a patient with WPW has AF, avoid AV node blockers such as diltiazem, verapamil, and digoxin—but flecainide may be used.
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Heart failure—basic concepts Definition Cardiac output is inadequate for the body’s requirements.33 Prevalence 1–3% of the general population; ~10% among elderly patients. 34 Key classifications Systolic failure: Inability of the ventricle to contract normally, resulting in cardiac output. Ejection fraction (EF) is 50%, this is termed HFpEF (heart failure with preserved EF). Causes: ventricular hypertrophy, constrictive pericarditis, tamponade, restrictive cardiomyopathy, obesity. NB: systolic and diastolic failure pathophysiology often coexists. Left ventricular failure (LVF): Symptoms: dyspnoea, poor exercise tolerance, fatigue, orthopnoea, paroxysmal nocturnal dyspnoea (PND), nocturnal cough (± pink frothy sputum), wheeze (cardiac ‘asthma’), nocturia, cold peripheries, weight loss. Right ventricular failure (RVF): Causes: LVF, pulmonary stenosis, lung disease (cor pulmonale, see p194). Symptoms: peripheral oedema (up to thighs, sacrum, abdominal wall), ascites, nausea, anorexia, facial engorgement, epistaxis. LVF and RVF may occur independently, or together as congestive cardiac failure (CCF). Acute heart failure: Often used exclusively to mean new-onset acute or decompensation of chronic heart failure characterized by pulmonary and/or peripheral oedema with or without signs of peripheral hypoperfusion. Chronic heart failure: Develops or progresses slowly. Venous congestion is common but arterial pressure is well maintained until very late. Low-output heart failure: Cardiac output is and fails to normally with exertion. Causes: • Excessive preload: eg mitral regurgitation or fluid overload (eg renal failure or too rapid IV infusions, particularly in the elderly and those with established HF). • Pump failure: systolic and/or diastolic HF (see above), heart rate (eg -blockers, heart block, post MI), negatively inotropic drugs (eg most antiarrhythmic agents). • Chronic excessive afterload: eg aortic stenosis, hypertension. Excessive preload can cause ventricular dilatation, this exacerbates pump failure. Excessive afterload prompts ventricular muscle thickening (ventricular hypertrophy), resulting in stiff walls and diastolic dysfunction. High-output heart failure: This is rare. Here, output is normal or increased in the face of needs. Failure occurs when cardiac output fails to meet these needs. It will occur with a normal heart, but even earlier if there is heart disease. Causes: anaemia, pregnancy, hyperthyroidism, Paget’s disease, arteriovenous malformation, beriberi. Consequences: initially features of RVF; later LVF becomes evident. Diagnosis Requires symptoms of failure (see above) and objective evidence of cardiac dysfunction at rest. For CCF, use the Framingham criteria. 35 Signs As described previously plus cyanosis, BP, narrow pulse pressure, pulsus alternans, displaced apex (LV dilatation), RV heave (pulmonary hypertension), signs of valve diseases. Severity can be graded using the New York classification (see BOX). Investigations According to NICE, 33 if ECG and B-type natriuretic peptide (BNP; p137) are normal, heart failure is unlikely, and an alternative diagnosis should be considered; if either is abnormal, then echocardiography (p110) is required. Tests FBC; U&E BNP; CXR (see fig 3.38); ECG; echo. ECG may indicate cause (look for evidence of ischaemia, MI, or ventricular hypertrophy). It is rare to get a completely normal ECG in chronic heart failure. Echocardiography is the key investigation. 36 It may indicate the cause (MI, valvular heart disease) and can confirm the presence or absence of LV dysfunction. Endomyocardial biopsy is rarely needed. Prognosis Poor with ~25–50% of patients dying within 5yrs of diagnosis. If admission is needed, 5yr mortality ≈75%. Be realistic: in one study, 54% of those dying in the next 72h had been expected to live for >6months.37
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(a)
(b) Fig 3.38 (a) The CXR in left ventricular failure. These features can be remembered as A B C D E. Alveolar oedema, classically this is perihilar ‘bat’s wing’ shadowing. Kerley B lines—now known as septal lines. These are variously attributed to interstitial oedema and engorged peripheral lymphatics. Cardiomegaly—cardiothoracic ratio >50% on a PA film. Dilated prominent upper lobe veins (upper lobe diversion). Pleural Effusions. Other features include peribronchial cuffing (thickened bronchial walls) and fluid in the fissures. (b) ‘Bat’s wing’, peri-hilar pulmonary oedema indicating heart failure and fluid overload.
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I Heart disease present, but no undue dyspnoea from ordinary activity. II Comfortable at rest; dyspnoea during ordinary activities. III Less than ordinary activity causes dyspnoea, which is limiting. IV Dyspnoea present at rest; all activity causes discomfort.
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Heart failure—management Acute heart failure This is a medical emergency (p800). Chronic heart failure Stop smoking. Stop drinking alcohol. Eat less salt. Optimize weight & nutrition.33 • Treat the cause (eg if dysrhythmias; valve disease). • Treat exacerbating factors (anaemia, thyroid disease, infection, BP). • Avoid exacerbating factors, eg NSAIDS (fluid retention) and verapamil (Ωve inotrope). • Annual ’flu vaccine, one-off pneumococcal vaccine. • Drugs: 1 Diuretics: Give loop diuretics to relieve symptoms, eg furosemide 40mg/24h PO or bumetanide 1–2mg/24h PO. Increase dose as necessary. SE: K+, renal impairment. Monitor U&E and add K+-sparing diuretic (eg spironolactone) if K+ 90%; specificity: 80–90%). The rises are greater with left than right heart failure and with systolic than diastolic dysfunction. What BNP threshold for diagnosing heart failure: If BNP >100ng/L, this ‘diagnoses’ heart failure better than other clinical variables or clinical judgement (history, examination, and CXR). BNP can be used to ‘rule out’ heart failure if 50ng/L does not exclude other coexisting diseases; conditions that can cause BNP rises include tachycardia, cardiac ischaemia, COPD, PE, renal disease, sepsis, hepatic cirrhosis, diabetes, and old age. Also, assays vary, so liaise with your lab. Prognosis in heart failure: The higher the BNP, the higher the cardiovascular and all-cause mortality (independent of age, NYHA class, previous MI, and LV ejection fraction) and the greater the risk of sudden death. So, a patient whose symptoms are currently well controlled may benefit from more aggressive treatment if their BNP if persistently raised.
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Left ventricular assist devices (LVADs) are increasingly used as bridging therapies for patients awaiting heart transplantation (fig 3.39). An internalized pump forces blood through tubing from the left ventricle to the aorta. To power the pump, the patient attaches the device to the mains electricity, and uses batteries when out and about. Patients with continuous (rather than pulsatile) flow LVADs have no pulse (fig 3.39c) and ascultation will reveal a loud, continuous, mechanical hum. If the patient collapses and there is no hum, resuscitation should include checking the LVAD power supply! (b)
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Hypertension Hypertension 44 is the most important risk factor for premature death and CVD; causing ~50% of all vascular deaths (8≈106/yr). Usually asymptomatic, so regular screening (eg 3-yrly) is a vital task—most preventable deaths are in areas without universal screening. 45 Defining hypertension BP has a skewed normal distribution (p751) within the population, and risk is continuously related to BP, so it is impossible to define ‘hypertension’. 46 We choose to select a value above which risk is significantly increased and the benefit of treatment is clear cut, see below. Don’t rely on a single reading—assess over a period of time (how long depends on the BP and the presence of other risk factors or end-organ damage). Confirm with 24hr ambulatory BP monitoring (ABPM); or a week of home readings. NB: the diagnostic threshold is lower ~135/85mmHg. Whom to treat All with BP ≥160/100mmHg (or ABPM ≥150/95mmHg). For those ≥140/90, the decision depends on the risk of coronary events, presence of diabetes, or end-organ damage; see fig 3.40. 44 The HYVET study showed that there is even substantial benefit in treating the over-80s. 47 Lower thresholds may be appropriate for young people—BP is on average lower in young people (eg 100–110/60–70 in 18-yearolds) and they have a ‘lifetime’ of risk ahead of them; but evidence to treat is lacking. White-coat hypertension Refers to an elevated clinic pressure, but normal ABPM (day average 200, diastolic>130mmHg) + bilateral retinal haemorrhages and exudates; papilloedema may or may not be present. Symptoms are common, eg headache ± visual disturbance. It requires urgent treatment, and may also precipitate acute kidney injury, heart failure, or encephalopathy, which are hypertensive emergencies. Untreated, 90% die in 1yr; treated, 70% survive 5yrs. It is more common in younger and in black subjects. Look hard for any underlying cause. Primary or ‘essential’ hypertension: (Cause unknown.) ~95% of cases. Secondary hypertension: ~5% of cases. Causes include: • Renal disease: the most common secondary cause. 75% are from intrinsic renal disease: glomerulonephritis, polyarteritis nodosa (PAN), systemic sclerosis, chronic pyelonephritis, or polycystic kidneys. 25% are due to renovascular disease, most frequently atheromatous (elderly cigarette smokers, eg with peripheral vascular disease) or rarely fibromuscular dysplasia (young ). • Endocrine disease: Cushing’s (p224) and Conn’s syndromes (p228), phaeochromocytoma (p228), acromegaly, hyperparathyroidism. • Others: coarctation (p156), pregnancy (OHCS p48), liquorice, drugs: steroids, MAOI, oral contraceptive pill, cocaine, amphetamines. Signs and symptoms Usually asymptomatic (except malignant hypertension, see earlier in topic). Headache is no more common than in the general population. Always examine the CVS fully and check for retinopathy. Are there features of an underlying cause (phaeochromocytoma, p228, etc.), signs of renal disease, radiofemoral delay, or weak femoral pulses (coarctation), renal bruits, palpable kidneys, or Cushing’s syndrome? Look for end-organ damage: LVH, retinopathy and proteinuria—indicates severity and duration of hypertension and associated with a poorer prognosis. Tests To confirm diagnosis: ABPM or home BP monitoring. To help quantify overall risk: Fasting glucose; cholesterol. To look for end-organ damage: ECG or echo (any LV hypertrophy? past MI?); urine analysis (protein, blood). To ‘exclude’ secondary causes: U&E (eg K+ in Conn’s); Ca2+ ( in hyperparathyroidism). Special tests: Renal ultrasound/arteriography (renal artery stenosis); 24h urinary meta-adrenaline (p228); urinary free cortisol (p225); renin; aldosterone; MR aorta (coarctation).
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Measuring BP with a sphygmomanometer • Use the correct size cuff. The cuff width should be >40% of the arm circumference. The bladder should be centred over the brachial artery, and the cuff applied snugly. Support the arm in a horizontal position at mid-sternal level. • Inflate the cuff while palpating the brachial artery, until the pulse disappears. This provides an estimate of systolic pressure. • Inflate the cuff until 30mmHg above systolic pressure, then place stethoscope over the brachial artery. Deflate the cuff at 2mmHg/s. • Systolic pressure: appearance of sustained repetitive tapping sounds (Korotkoff I). • Diastolic pressure: usually the disappearance of sounds (Korotkoff V). However, in some individuals (eg pregnant women) sounds are present until the zero point. In this case, the muffling of sounds, Korotkoff IV, should be used. State which is used for a given reading. For children, see OHCS p157. • For advice on using automated sphygmomanometers and a list of validated devices see
Managing suspected hypertension Clinic blood pressure 38°C. • Vascular phenomena (emboli, Janeway’s lesions, etc.). • Immunological phenomena (glomerulonephritis, Osler’s nodes, etc.). • Positive blood culture that does not meet major criteria. How to diagnose: Definite infective endocarditis: 2 major or 1 major and 3 minor or all 5 minor criteria.
Antibiotic therapy for infective endocarditis Prescribe antibiotics for infective endocarditis as follows.66 For more information on individual antibiotics, see tables 9.4–9.9, pp386–7. • Blind therapy—native valve or prosthetic valve implanted >1y ago: ampicillin, flucloxacillin and gentamicin. Vancomycin + gentamicin if penicillin-allergic. If thought to be Gram Ωve: meropenem + vancomycin. • Blind therapy—prosthetic valve: vancomycin + gentamicin + rifampicin. • Staphs—native valve: flucloxacillin for >4wks. If allergic or MRSA: vancomycin . • Staphs—prosthetic valves: flucloxacillin + rifampicin + gentamicin for 6wks (review need for gentamicin after 2wks). If penicillin-allergic or MRSA: vancomycin + rifampicin + gentamicin. • Streps—fully sensitive to penicillin: benzylpenicillin 1.2g/4h IV for 4–6wks.8 • Streps—less sensitive: benzylpenicillin + gentamicin; if penicillin allergic or highly penicillin resistant: vancomycin + gentamicin. • Enterococci: amoxicillin + gentamicin. If pen-allergic: vancomycin + gentamicin— for 4wks (6wks if prosthetic valve); review need for gentamicin after 2wks. • HACEK organisms (Haemophilus, Actinobacillus, Cardiobacterium, Eikenella, Kingella): ceftriaxone for 4wks with native valve or 6wks with prosthetic. • Fungal: Candida—amphotericin. Aspergillus—voriconazole.
Fig 3.46 Clubbing with endo- Fig 3.47 Splinter haemorrhag- Fig 3.48 Janeway’s lesions are carditis.
es are normally seen under the non-tender erythematous, haefingernails or toenails. They are morrhagic, or pustular spots, eg on the palms or soles. usually red-brown in colour.
8 If Strep bovis is cultured, do colonoscopy, as a colon neoplasm is the likely portal of entry (table 6.3, p249).
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Modified Duke criteria for infective endocarditis
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Diseases of heart muscle Acute myocarditis This is inflammation of myocardium, often associated with pericardial inflammation (myopericarditis).67 Causes: See table 3.3. Symptoms and signs: ACS-like symptoms, heart failure symptoms, palpitations, tachycardia, soft S1, S4 gallop (p44).Tests: ECG: ST changes and T-wave inversion, atrial arrhythmias, transient AV block, QT prolongation. Bloods: CRP, ESR, & troponin may be raised; viral serology and tests for other likely causes. Echo: diastolic dysfunction, regional wall abnormalities. Cardiac MR if clinically stable. Endomyocardial biopsy is gold standard. : Supportive. Treat the underlying cause. Treat arrhythmias and heart failure (p136). NSAID use is controversial. Avoid exercise as this can precipitate arrhythmias. Prognosis: 50% will recover within 4wks. 12–25% will develop DCM and severe heart failure. DCM can occur years after apparent recovery. Dilated cardiomyopathy (DCM) A dilated, flabby heart of unknown cause. Associations: alcohol, BP, chemotherapeutics, haemochromatosis, viral infection, autoimmune, peri- or postpartum, thyrotoxicosis, congenital (X-linked). Prevalence: 0.2%. Presentation: Fatigue, dyspnoea, pulmonary oedema, RVF, emboli, AF, VT. Signs: Pulse, BP, JVP, displaced and diffuse apex, S3 gallop, mitral or tricuspid regurgitation (MR/TR), pleural effusion, oedema, jaundice, hepatomegaly, ascites. Tests: Blood: BNP (p137), Na+ indicates a poor prognosis. CXR: cardiomegaly, pulmonary oedema. ECG: tachycardia, non-specific T-wave changes, poor R-wave progression. Echo: globally dilated hypokinetic heart and low ejection fraction. Look for MR, TR, LV mural thrombus. : Bed rest, diuretics, -blockers, ACE-i, anticoagulation, biventricular pacing, ICDs, LVADs, transplantation. Mortality: Variable, eg 40% in 2yrs. Hypertrophic cardiomyopathy (HCM) LV outflow tract (LVOT) obstruction from asymmetric septal hypertrophy. HCM is the leading cause of sudden cardiac death in the young. Prevalence: 0.2%. Autosomal dominant inheritance, but 50% are sporadic. 70% have mutations in genes encoding -myosin, -tropomyosin, and troponin T. May present at any age. Ask about family history of sudden death. Symptoms and signs: Sudden death may be the first manifestation of HCM in many patients (VF is amenable to implantable defibrillators), angina, dyspnoea, palpitation, syncope, CCF. Jerky pulse; a wave in JVP; double-apex beat; systolic thrill at lower left sternal edge; harsh ejection systolic murmur. Tests: • ECG: LVH; progressive T-wave inversion; deep Q waves (inferior + lateral leads); AF; WPW syndrome (p133); ventricular ectopics; VT. • Echo: asymmetrical septal hypertrophy; small LV cavity with hypercontractile posterior wall; midsystolic closure of aortic valve; systolic anterior movement of mitral valve. • MRI: see fig 3.16. • Cardiac catheterization helps assess: severity of gradient; coronary artery disease or mitral regurgitation, but may provoke VT. • Electrophysiological studies may be needed (eg if WPW, p133). • Exercise test ± Holter monitor (p125) to risk stratify. : -blockers or verapamil for symptoms (the aim is reducing ventricular contractility). Amiodarone (p130) for arrhythmias (AF, VT). Anticoagulate for paroxysmal AF or systemic emboli. Septal myomectomy (surgical or chemical (with alcohol) to LV outflow tract gradient) is reserved for those with severe symptoms. Consider implantable defibrillator—use to assess risk of sudden cardiac death. Mortality: 5.9%/yr if 14yrs. Poor prognostic factors: age 300mL; fig 3.14. ECG shows low-voltage QRS complexes and may have alternating QRS morphologies (electrical alternans). Echocardiography shows an echo-free zone surrounding the heart. Management: Treat the cause. Pericardiocentesis may be diagnostic (suspected bacterial pericarditis) or therapeutic (cardiac tamponade). See p773. Send pericardial fluid for culture, ZN stain/TB culture, and cytology. Constrictive pericarditis The heart is encased in a rigid pericardium.68 Causes: Often unknown (UK); elsewhere TB, or after any pericarditis. Clinical features: These are mainly of right heart failure with JVP (with prominent x and y descents, p43); Kussmaul’s sign (JVP rising paradoxically with inspiration); soft, diffuse apex beat; quiet heart sounds; S3; diastolic pericardial knock, hepatosplenomegaly, ascites, and oedema. Tests: CXR: small heart ± pericardial calcification. CT/MRI—helps distinguish from restrictive cardiomyopathy. Echo. Cardiac catheterization. Management: Surgical excision. Medical to address the cause and symptoms. Cardiac tamponade A pericardial effusion that raises intrapericardial pressure, reducing ventricular filling and thus dropping cardiac output.68 Can lead rapidly to cardiac arrest. Signs: Pulse, BP, pulsus paradoxus, JVP, Kussmaul’s sign, muffled S1 and S2. Diagnosis: Beck’s triad: falling BP; rising JVP; muffled heart sounds. ECG: low-voltage QRS ± electrical alternans. Echo is diagnostic: echo-free zone (>2cm, or >1cm if acute) around the heart ± diastolic collapse of right atrium and right ventricle. Management: Seek expert help. The pericardial effusion needs urgent drainage (p773). Send fluid for culture, ZN stain/TB culture, and cytology.
Fig 3.51 Pericarditis. Note the widespread ‘saddle-shaped’ ST elevation—particularly clear in V5 and V6.
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Adult congenital heart disease (ACHD) This is a growing area of cardiology as increasing numbers of children with congenital heart defects survive to adulthood, sometimes as a result of complex restructuring procedures which have their own physiological implications (see BOX ‘Patients with one ventricle’). ACHD69 patients are at increased risk of many conditions described elsewhere, for which many of the ‘standard’ investigations and therapies will apply: including arrhythmias (p124), heart failure (p134), and infective endocarditis (p150). Investigations Echocardiography (± bubble contrast) is first line. Increasingly, cardiac CT and MR are used to provide precise anatomical and functional information. Cardiac catheterization generates data on oxygen saturation and pressure in different vessels and chambers. Exercise testing assesses functional capacity. A few of the more common ACHDs are discussed below: Bicuspid aortic valve These work well at birth and go undetected. Many eventually develop aortic stenosis (needing valve replacement) ± aortic regurgitation predisposing to IE/SBE ± aortic dilatation/dissection. Intense exercise may accelerate complications, so do yearly echocardiograms on affected athletes. 70 Atrial septal defect (ASD) A hole connects the atria. • Ostium secundum defects: 80% cases; hole high in the septum; often asymptomatic until adulthood when a LR shunt develops. Shunting depends on the compliance of the ventricles. LV compliance decreases with age (esp. if BP), so augmenting LR shunting; hence dyspnoea/heart failure, typically aged 40–60yrs. • Ostium primum defects: associated with AV valve anomalies, eg in Down’s syndrome; present in childhood. Signs and symptoms: Chest pain, palpitations, dyspnoea. Arrhythmias incl. AF; JVP; wide, fixed split S2; pulmonary systolic flow murmur. Pulmonary hypertension may cause pulmonary or tricuspid regurgitation, dyspnoea and haemoptysis. Frequency of migraine. Simple tests: ECG: RBBB with LAD (primum defect) or RAD (secundum defect). CXR: small aortic knuckle, pulmonary plethora, atrial enlargement. Complications: • Reversal of left-to-right shunt, ie Eisenmenger’s complex: initial LR shunt leads to pulmonary hypertension which increases right heart pressures until they exceed left heart pressures, hence shunt reversal. This causes cyanosis as deoxygenated blood enters systemic circulation. • Paradoxical emboli eg causing CVAs (veinartery via ASD; rare). Treatment: May close spontaenously. If not, primum defects are usually closed in childhood. Secundum defects should be closed if symptomatic or signs of RV overload. Transcatheter closure is more common than surgical. Ventricular septal defect (VSD) A hole connects the ventricles. Causes: Congenital (prevalence 2:1000 births); acquired (post-MI). Symptoms: May present with severe heart failure in infancy, or remain asymptomatic and be detected incidentally in later life. Signs: Classically, a harsh pansystolic murmur is heard at the left sternal edge, with a systolic thrill, ± left parasternal heave. Smaller holes, which are haemodynamically less significant, give louder murmurs. Signs of pulmonary hypertension. Complications: AR, IE/SBE, pulmonary hypertension, Eisenmenger’s complex (above), heart failure from volume overload. Tests: ECG: normal, LAD, LVH, RVH. CXR: normal heart size ± mild pulmonary plethora (small VSD) or cardiomegaly, large pulmonary arteries and marked pulmonary plethora (large VSD). Cardiac catheter: step up in O2 saturation in right ventricle. Treatment: Initially medical as many close spontaneously. Indications for surgical closure: failed medical therapy, symptomatic VSD, shunt >3 : 1, SBE/IE. Endovascular closure may be possible. 71 Coarctation of the aorta Congenital narrowing of the descending aorta; usually occurs just distal to the origin of the left subclavian artery. More common in boys. Associations: Bicuspid aortic valve; Turner’s syndrome. Signs: Radiofemoral delay; weak femoral pulse; BP; scapular bruit; systolic murmur (best heard over the left scapula); cold feet. Complications: Heart failure from high afterload; IE; intracerebral haemorrhage. Tests: CT or MRI-aortogram; CXR may show rib notching as blood diverts down intercostal arteries to reach the lower body, causing these vessels to dilate and erode local rib bone. Treatment: Surgery, or balloon dilatation ± stenting. Tetralogy of Fallot See p157.
Tetralogy of Fallot (TOF) is the most common cyanotic congenital heart disorder (prevalence: 3–6 per 10 000). It is also the most common cyanotic heart defect that survives to adulthood, accounting for 10% of all ACHD. 72 It is believed to be due to abnormalities in separation of the truncus arteriosus into the aorta and pulmonary arteries early in gestation (fig 3.52). The ‘tetralogy’ of features are: 1 Ventricular septal defect (VSD). 2 Pulmonary stenosis. 3 Right ventricular hypertrophy. 4 The aorta overrides the VSD, accepting right heart blood. A few patients also have an ASD, which makes up the pentad of Fallot. Presentation: Severity of illness depends greatly on the degree of pulmonary stenosis. Infants may be acyanotic at birth, with a pulmonary stenosis murmur as the only initial finding. Gradually (especially after closure of the ductus arteriosus) they become cyanotic due to decreasing flow of blood to the lungs Fig 3.52 Tetralogy of Fallot. Reproduced from Thorne et al., and increasing right-to-left flow across the VSD. During a hypoxic spell, the child becomes restless and Adult Congenital Heart Disease, 2009, with permission from agitated. Toddlers may squat, which is typical of TOF, Oxford University Press. as it increases peripheral vascular resistance, thereby decreasing the degree of right to left shunt. Adult patients are often asymptomatic. In the unoperated adult patient, cyanosis is common, although extreme cyanosis or squatting is uncommon. In repaired patients, late symptoms include exertional dyspnoea, palpitations, clubbing, RV failure, syncope, and even sudden death. Investigations: ECG shows RV hypertrophy with a right bundle-branch block. CXR may be normal, or show the hallmark of TOF, which is the classic bootshaped heart (fig 3.53). Echocardiography can show the anatomy as well as the degree of stenosis. Cardiac CT and cardiac MRI can give valuable information for planning the surgery. 73 Management: Surgery is usually done before 1yr of age, with closure of the VSD and correction of pulmonary stenosis. Prognosis: Without surgery, mortality rate is ~95% by age 20. After repair, 85% of patients survive to 35yrs. Common problems in adulthood include pulmonary regurgitation, causing RV dilatation and failure; RV outflow Fig 3.53 Boot-shaped heart. tract obstruction; AR; LV dysfunction; and arCourtesy of Dr Edward Singleton. rhythmias.
Patients with one ventricle Many patients born with single-ventricle hearts (eg hypoplastic left heart syndrome) will undergo a Fontan procedure. This results in systemic venous blood flowing directly into the pulmonary arteries and the single ventricle being used to pump oxygenated blood into the aorta. The lack of a right heart results in many of the signs and symptoms of right heart failure and puts the patient at risk of rapid cardiac decompensation. When looking after these patients, seek advice from specialist ACHD centres.
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Fallot’s tetralogy: what the non-specialist needs to know
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Driving and the heart
Cardiovascular medicine
UK licences are inscribed ‘You are required by law to inform Drivers Medical Branch, DVLA, Swansea SA99 1AT at once if you have any disability (physical or medical), which
is, or may become likely to affect your fitness as a driver, unless you do not expect it to last more than 3 months’. It is the responsibility of drivers to inform the DVLA (the UK Driving and Vehicle Licensing Authority), and that of their doctors to advise patients that medical conditions74 (and drugs) may affect their ability to drive and for which conditions patients should inform the DVLA. Drivers should also inform their insurance company of any condition disclosed to the DVLA. If in doubt, ask your defence union. The following are examples of the guidance for holders of standard licences; different rules apply for group 2 vehicle licence-holders (eg lorries, buses). More can be found at Angina Driving must cease when symptoms occur at rest or with emotion. Driving may recommence when satisfactory symptom control is achieved. DVLA need not be notified. Angioplasty Driving must cease for 1wk, and may recommence thereafter provided no other disqualifying condition. DVLA need not be notified. MI If successfully treated with angioplasty, cease driving for 1 week provided urgent intervention not planned and LVEF (left ventricular ejection fraction) >40%, and no other disqualifying condition. Otherwise, driving must cease for 1 month. DVLA need not be notified. Dysrhythmias Including sinoatrial disease, AF/flutter, atrioventricular conduction defects, and narrow or broad complex tachycardias. Driving must cease if the dysrhythmia has caused or is likely to cause incapacity. Driving may recommence 4wks after successful control provided there is no other disqualifying condition. Pacemaker implant Stop driving for 1wk, the patient must notify the DVLA. Implanted cardioverter/defibrillator The licence is subject to annual review. Driving may occur when these criteria can be met: • 6 months have passed since ICD implanted for secondary prevention. • 1 month has passed since ICD implanted for primary prophylaxis. • The device has not administered therapy (shock and/or symptomatic antitachycardia pacing) within the last 6 months (except during testing). • No therapy (shock) in the last 2 years has been accompanied by incapacity (whether caused by the device or arrhythmia)—unless this was a result of device malfunction which has been corrected for at least 1 month or steps have been taken to avoid recurrence (eg ablation) which have been successful for at least 6 months. • A period of 1 month off driving must occur following any revision of the device (generator and/or electrode) or alteration of antiarrhythmics. • The device is subject to regular review with interrogation. • There is no other disqualifying condition. Syncope Simple faint: No restriction. Unexplained syncope: With probable cardiac aetiology—4wks off driving if cause identified and treated; otherwise 6 months off. Loss of consciousness or altered awareness associated with signs of seizure requires 6 months off driving. If the patient is known to be epileptic or has had another such episode in the preceeding 5yrs, they must abstain from driving for 1yr. See driving and epilepsy (BOX). Patients who have had a single episode of loss of consciousness with no cause found despite neurological and cardiac investigations, must abstain from driving for 6 months. Hypertension Driving may continue unless treatment causes unacceptable sideeffects. DVLA need not be notified.
• Epilepsy (the patient must have had at least two seizures in the last 5yrs). An epileptic patient who has suffered an epileptic attack while awake must not drive for 1yr from the date of the attack. Patients who have seizures that do not affect their consciousness (eg simple partial seizures) or seizures only during sleep may be allowed to drive. Being allowed to drive is conditional on the patient following medical advice and there not being reason to believe they are at high risk of further seizures. • TIA or stroke. These patients should not drive for at least 1 month. There is no need to inform the DVLA unless there is residual neurological defect after 1 month, eg visual field defect. If TIAS have been recurrent and frequent, a 3-month period free of attacks may be required. • Sudden attacks or disabling giddiness, fainting, or blackouts. • Chronic neurological conditions including multiple sclerosis, Parkinson’s (any ‘freezing’ or on–off effects), and motor neuron diseases. • Severe mental disorders; including serious memory problems and severe psychiatric illness. Those with dementia should only drive if the condition is mild (do not rely on armchair judgements: on-the-road trials are better). Encourage relatives to contact DVLA if a dementing relative should not be driving. GPs may desire to breach confidentiality (the GMC approves) and inform DVLA of demented or psychotic patients (tel. 01792 783686). Many elderly drivers (~1 in 3) who die in accidents are found to have Alzheimer’s. • A pacemaker, defibrillator, or antiventricular tachycardia device fitted. • Diabetes controlled by insulin or tablets. The main issues which may result in driving bans are impaired awareness of hypoglycaemia and impaired vision. • Angina while driving. • Any type of brain surgery, brain tumour. Severe head injury involving inpatient treatment at hospital. • Continuing/permanent difficulty in the use of arms or legs which affects ability to control a vehicle. • Dependence on or misuse of alcohol, illicit drugs, or chemical substances in the past 3yrs (do not include drink/driving offences). • Any visual disability which affects both eyes (do not declare short/long sight or colour blindness). Vision (new drivers) should be 6/9 on Snellen’s scale in the better eye and 6/12 on the Snellen scale in the other eye, wearing glasses or contact lenses if needed, and 3/60 in each eye without glasses or contact lenses. The above-listed rules apply to standard licences only, for group 2 entitlement (eg HGV drivers) see www.dvla.gov.uk/medical/ataglance.aspx.
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Other conditions: UK DVLA states it must be informed if a driver suffers from medical conditions including:
4 Chest medicine Contents Respiratory health 161 Investigations Bedside tests in chest medicine 162 Further investigations in chest medicine 164 Pulmonary conditions: Pneumonia 166 Specific pneumonias 168 Complications of pneumonia 170 Bronchiectasis 172 Cystic fibrosis (CF) 173 Lung tumours 174 Lung tumours: staging and treatment 176 Fungi and the lung 177 Asthma 178 Management of chronic asthma 182 Chronic obstructive pulmonary disease (COPD) 184 Acute respiratory distress syndrome (ARDS) 186 Respiratory failure 188 Pulmonary embolism (PE) 190 Pneumothorax 190 Pleural effusion 192 Obstructive sleep apnoea syndrome 194 Cor pulmonale 194 Sarcoidosis 196 Interstitial lung disease (ILD) 198 Extrinsic allergic alveolitis (EAA) 198 Idiopathic pulmonary fibrosis (IPF) 200 Industrial dust diseases 201
Fig 4.1 In 1948, the Medical Research Council published a landmark paper in the BMJ about streptomycin as a treatment for pulmonary TB. The paper was regarded as a milestone in the history of clinical trials and set a precedent for the use of randomization in controlled trials. Before this, bed rest alone had been standard treatment for patients with pulmonary TB. After the successes of penicillin, there was excitement in the discovery that streptomycin proved effective against the tubercle bacilli. Patients aged 15 to 30 with ‘acute progressive bilateral pulmonary tuberculosis of presumably recent origin, bacteriologically proved and unsuitable for collapse therapy’ were entered into the trial. The streptomycin and bed rest group did better initially but the development of resistance was soon recognized. This was a new phenomenon which had not then been seen with penicillin. This led to the notion that combination therapies were needed to overcome TB drug resistance. The ‘Edinburgh Method’, described in 1957, advocated the use of triple therapy. Reproduced from the BMJ, volume 2, Jan 1, © 1948, with permission from BMJ Publishing Group
We thank Dr Phillippa Lawson, our Specialist Reader, and William Flowers, our Junior Reader, for their contribution to this chapter.
Respiratory health
Larynx Superior lobe of left lung
Trachea Superior lobe of right lung Right primary bronchus
Left primary bronchus
Right secondary bronchus Right tertiary bronchus
Inferior lobe of right lung
Left secondary bronchus Left tertiary bronchus
Middle lobe Smaller bronchi
Primary bronchi Secondary bronchi Tertiary bronchi Smaller bronchi
Inferior lobe of left lung Smaller bronchi
Fig 4.2 Segmental anatomy of the lungs and main bronchi. The left lung has two lobes and the right has three.
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The lungs provide a vital physiological function in allowing gas exchange, but are also at the vanguard of a constant battle between host, pathogens, and pollutants. Respiratory medicine exemplifies how careful epidemiology, science, and randomized controlled trials have revolutionized our understanding of common diseases, leading to preventative measures and effective treatments. However, the importance of poverty and general improvements in public health cannot be underestimated. Rates of TB in the UK declined well before the introduction of BCG vaccination and streptomycin, largely due to improvements in sanitation and less dense living conditions. Public health campaigns and taxation have helped lower smoking rates, although reductions in lung cancer will lag behind for many years.
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Bedside tests in chest medicine There is no substitute for careful history taking and examination in making the ‘correct’ diagnosis. Tests should help clarify and assess severity. When examining the chest think about the anatomy, and the location of pathology (fig 4.2). Sputum examination Collect a good sample; if necessary ask a physiotherapist to help. Note the appearance: clear and colourless (chronic bronchitis), yellow-green or brown (pulmonary infection), red (haemoptysis), black (smoke, coal dust), or frothy white-pink (pulmonary oedema). Send the sample to the laboratory for microscopy, culture/sensitivity. If indicated, ask for ZN stain, and PCR. Peak expiratory flow (PEF) Measured by a maximal forced expiration through a peak flow meter. It correlates well with the forced expiratory volume in 1 second (FEV1) & is used as an estimate of airway calibre in asthma, but is effort-dependent. Pulse oximetry Allows non-invasive assessment of peripheral O2 saturation (SpO2). Useful for monitoring those who are acutely ill or at risk of deterioration. Target oxygen saturations are usually 94–98% in a well patient or 88–92% in those with certain pre-exisiting lung pathology (eg COPD). Oxygen saturation of 80% predicted 20mg/d, cochlear implant, occupation risk (eg welders), CSF fluid leaks. Vaccinate every 5yrs. CI: Pregnancy, lactation, T°, previous anaphylaxis to vaccine or one of its components.
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Table 4.2 Empirical treatment of pneumonia (check local policy) Clinical Antibiotic (further dosage Organisms setting details: pp386–7) Community-acquired Mild not Streptococcus pneumoniae Oral amoxicillin 500mg–1g/8h or clarithropreviously Haemophilus influenzae mycin 500mg/12h or doxycycline 200mg loading then 100mg/day (initially 5-day CURB 0–1 course) Streptococcus pneumoniae Oral amoxicillin 500mg–1g/8h + Moderate clarithromycin 500mg/12h or doxycycline Haemophilus influenzae CURB 2 200mg loading then 100mg/12h Mycoplasma pneumoniae If IV required: amoxicillin 500mg/8h + clarithromycin 500mg/12h (7-day course) Severe As above Co-amoxiclav 1 . 2g/8h IV or cephalosporin IV (eg cefuroxime 1 . 5g/8h IV) AND clarithromyCURB >3 cin 500mg/12h IV (7 days) Add flucloxacillin ± rifampicin if Staph suspected; vancomycin (or teicoplanin) if MRSA suspected. Treat for 10d (14–21d if Staph, Legionella, or Gram Ωve enteric bacteria suspected) Panton-Valentine Seek urgent help. Consider adding IV Leukocidin-producing linezolid, clindamycin, and rifampicin Staph. aureus (PVL-SA) Atypical Legionella pneumophilia Fluoroquinolone combined with clarithromycin, or rifampicin, if severe. See p168 Chlamydophila species Tetracycline Pneumocystis jirovecii High-dose co-trimoxazole (pp400–1) Hospital-acquired Aminoglycoside IV + antipseudomonal Gram-negative bacilli penicillin IV or 3rd-generation cephalosporin Pseudomonas IV (p387) Anaerobes Aspiration Streptococcus pneumoniae Cephalosporin IV + metronidazole IV Anaerobes Neutropenic patients Aminoglycoside IV + antipseudomonal peniGram-positive cocci cillin IV or 3rd-generation cephalosporin IV Gram-negative bacilli Fungi (p177) Consider antifungals after 48h
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Specific pneumonias Pneumococcal pneumonia The commonest bacterial pneumonia. Affects all ages, but is commoner in the elderly, alcoholics, post-splenectomy, immunosuppressed, and patients with chronic heart failure or pre-existing lung disease. Clinical features: Fever, pleurisy, herpes labialis. CXR shows lobar consolidation. If mod/severe check for urinary antigen. Treatment: amoxicillin, benzylpenicillin, or cephalosporin. Staphylococcal pneumonia May complicate influenza infection or occur in the young, elderly, intravenous drug users, or patients with underlying disease, eg leukaemia, lymphoma, cystic fibrosis (CF). It causes a bilateral cavitating bronchopneumonia. Treatment: flucloxacillin ± rifampicin, MRSA: contact lab; consider vancomycin. Klebsiella pneumonia Rare. Occurs in elderly, diabetics, and alcoholics. Causes a cavitating pneumonia, particularly of the upper lobes, often drug resistant. Treatment: cefotaxime or imipenem. Pseudomonas A common pathogen in bronchiectasis and CF. It also causes hospital-acquired infections, particularly on ITU or after surgery. Treatment: antipseudomonal penicillin, ceftazidime, meropenem, or ciprofloxacin + aminoglycoside. Consider dual therapy to minimize resistance. Mycoplasma pneumoniae Occurs in epidemics about every 4yrs. It presents insidiously with flu-like symptoms (headache, myalgia, arthralgia) followed by a dry cough. CXR: reticular-nodular shadowing or patchy consolidation often of one lower lobe, and worse than signs suggest. Diagnosis: PCR sputum or serology. Cold agglutinins may cause an autoimmune haemolytic anaemia. Complications: Skin rash (erythema multiforme, fig 12.22, p563), Stevens–Johnson syndrome, meningoencephalitis or myelitis; Guillain–Barré syndrome. Treatment: Clarithromycin (500mg/12h) or doxycycline (200mg loading then 100mg OD) or a fluroquinolone (eg ciprofloxacin or norfloxacin). Legionella pneumophila Colonizes water tanks kept at 38°C), chills, rigors, myalgia, dry cough, headache, diarrhoea, and dyspnoea—with an abnormal CXR and WCC. Respiratory failure is a complicating feature: ~20% progress to acute respiratory distress syndrome requiring invasive ventilation. 16 Mortality is 1–50%, depending on age, but no cases since 2004. Close contacts, or travel to an area with known cases should raise suspicion. The mechanism of transmission of SARS-CoV is human–human. Management: seek expert help. Largely supportive with good infection control measures. Middle East respiratory syndrome (MERS) is a viral respiratory disease caused by novel coronavirus (MERS CoV) and was first identified in 2012 in Saudi Arabia. Symptoms include fever, cough, shortness of breath, and gastrointestinal upset. Incubation period 14 days. Human-to-human transmission has been reported in most cases, but camels play a pivotal host role in animal-to-human transmission. Large outbreaks linked to healthcare facilities have been reported in the Middle East and South Korea. The World Health Organization has reported mortality as high as 36% in known cases.13
2 Therapeutic or prophylactic antivirals are said to be the most effective single intervention followed by vaccine and basic public health measures. 17 But oseltamivir resistance and unavailability of a suitable vaccine during the early stages of a pandemic make non-drug interventions all the more important.
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Avian influenza
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Complications of pneumonia Respiratory failure (See p188.) Type I respiratory failure (PaO2 6kPa. Be careful with O2 in COPD patients; check ABGS frequently, and consider elective ventilation if rising PaCO2 or worsening acidosis. Aim to keep SaO2 at 94–98%, PaO2 ≥8kPa. Hypotension May be due to a combination of dehydration and vasodilation due to sepsis. If systolic BP is 90mmHg. If systolic BP remains 1500 mutations have been identified). This is a ClΩ channel, and the defect leads to a combination of defective chloride secretion and increased sodium absorption across airway epithelium. The changes in the composition of airway surface liquid predispose the lung to chronic pulmonary infections and bronchiectasis. See OHCS (‘Paediatrics’, p162) for more detail. Clinical features Neonate: Failure to thrive; meconium ileus; rectal prolapse. Children and young adults: Respiratory: cough; wheeze; recurrent infections; bronchiectasis; pneumothorax; haemoptysis; respiratory failure; cor pulmonale. Gastrointestinal: pancreatic insufficiency (diabetes mellitus, steatorrhoea); distal intestinal obstruction syndrome (meconium ileus equivalent); gallstones; cirrhosis. Other: male infertility; osteoporosis; arthritis; vasculitis (p556); nasal polyps; sinusitis; and hypertrophic pulmonary osteoarthropathy (HPOA). Signs: cyanosis; finger clubbing; bilateral coarse crackles. Diagnosis Sweat test: Sweat sodium and chloride >60mmol/L; chloride usually > sodium. Genetics: Screening for known common CF mutations should be considered. Faecal elastase is a simple and useful screening test for exocrine pancreatic dysfunction. Tests Blood: FBC, U&E, LFT; clotting; vitamin A, D, E levels; annual glucose tolerance test (p206). Bacteriology: Cough swab, sputum culture. Radiology: CXR; hyperinflation; bronchiectasis. Abdominal ultrasound: Fatty liver; cirrhosis; chronic pancreatitis; Spirometry: Obstructive defect. Aspergillus serology/skin test (20% develop ABPA, p177). Biochemistry: Faecal fat analysis. Management Management should be multidisciplinary, eg physician, GP, physiotherapist, specialist nurse, and dietician, with attention to psychosocial as well as physical wellbeing. Chest: Physiotherapy (postural drainage, airway clearance techniques). Antibiotics are given for acute infective exacerbations and prophylactically. Chronic Pseudomonas infection is an important predictor of survival. Mucolytics may be useful (eg DNase, ie Dornase alfa, 2.5mg daily nebulized, or nebulized hypertonic saline). Bronchodilators. Annual CXR surveillance is recommended. Gastrointestinal: Malabsorption, GORD, distal obstruction syndrome. Pancreatic enzyme replacement; fat-soluble vitamin supplements (A, D, E, K); ursodeoxycholic acid for impaired liver function; cirrhosis may require liver transplantation. Other: Treatment of CF-related diabetes (screen annually with OGTT from 12yrs); screening/treatment of osteoporosis (DEXA bone scanning); arthritis, sinusitis, and vasculitis; fertility and genetic counselling. Advanced lung disease: Oxygen, diuretics (cor pulmonale); noninvasive ventilation; lung or heart/lung transplantation (post-transplant survival 5 years). Prognosis: Median survival is now ~ 41yrs in the UK, although a baby born today would expect to live longer.
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Lung tumours Carcinoma of the bronchus Second most common cancer in the UK, accounting for 13% of all new cancer cases and 27% of cancer deaths (40 000 cases/yr in UK).22 Incidence is increasing in women. Only 5% ‘cured’. Risk factors: Cigarette smoking (causes 90% of lung ca). Others: passive smoking, asbestos, chromium, arsenic, iron oxides, and radiation (radon gas). Histology: Clinically the most important division is between small cell (SCLC) and non-small cell (NSCLC). NSCLC: Squamous (35%); adenocarcinoma (27%), large cell (10%); adenocarcinoma in situ (rare, 3cm and >2cm distal to carina or any size if pleural involvement or obstructive pneumonitis extending to hilum, but not all the lung T3 Involves the chest wall, diaphragm, mediastinal pleura, pericardium, or 7cm diameter and nodules in same lobe T4 Involves mediastinum, heart, great vessels, trachea, oesophagus, vertebral body, carina, malignant effusion, or nodules in another lobe Regional nodes (N) N0 None involved (after mediastinoscopy) N1 Peribronchial and/or ipsilateral hilum N2 Ipsilateral mediastinum or subcarinal N3 Contralateral mediastinum or hilum, scalene, or supraclavicular Distant metastasis (M) M0 None M1 a) Nodule in other lung, pleural lesions, or malignant effusion; b) distant metastases present Stages I II IIIa IIIb IV Occult TX N0 M0 TIS/T1/T2 N0 M0 T1/T2 N1 M0 T3 N1 M0 T1–4 N3 M0 T1–4 N0–3 M1 or T3 N0 M0 or T1–3 N2 M0 or T4 N0–2 M0 Reproduced with permission from Edge, SB et al. (Eds.), AJCC Cancer Staging Manual, 7th Edition. New York: Springer; 2010.
Treatment NSCLC: Lobectomy (open or thoracoscopic) is the treatment of choice if medically fit and aim is curative intent or parenchymal sparing operation for patients with borderline fitness and smaller tumours ((T1a–b, N0, M0). Radical radiotherapy for patient with stage I, II, III NSCLC. Chemotherapy ± radiotherapy for more advanced disease. Regimens may be platinum based, eg with monoclonal antibodies targeting the epidermal growth factor receptor (cetuximab). SCLC: consider surgery with limited stage disease. Chemotherapy ± radiotherapy if well enough. Palliation: Radiotherapy is used for bronchial obstruction, SVC obstruction, haemoptysis, bone pain, and cerebral metastases. SVC stent + radiotherapy and dexamethasone for SVC obstruction. Endobronchial therapy: tracheal stenting, cryotherapy, laser, brachytherapy (radioactive source is placed close to the tumour). Pleural drainage/ pleurodesis for symptomatic pleural effusions. Drugs: analgesia; steroids; antiemetics; cough linctus; bronchodilators; antidepressants. Prognosis Non-small cell: 50% 2yr survival without spread; 10% with spread. Small cell: median survival is 3 months if untreated; 1–1½yrs if treated. Prevention Stop smoking (p93). Prevent occupational exposure to carcinogens.
Fig 4.9 Aspergillosis.
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Fungi and the lung Aspergillus This group of fungi affects the lung in five ways: 1 Asthma: Type I hypersensitivity reaction to fungal spores (p178). 2 Allergic bronchopulmonary aspergillosis (ABPA): Results from type I and III hypersensitivity reactions to Aspergillus fumigatus. Affects 1–5% of asthmatics, 2–25% of CF patients.26 Initially bronchoconstriction, then permanent damage occurs causing bronchiectasis (fig 4.9). Symptoms: wheeze, cough, sputum (plugs of mucus containing fungal hyphae, see p408), dyspnoea, and ‘recurrent pneumonia’. Investigations: CXR (transient segmental collapse or consolidation, bronchiectasis); Aspergillus in sputum; positive Aspergillus skin test and/ or Aspergillus-specific IgE RAST (radioallergosorbent test); positive serum precipitins; eosinophilia; raised serum IgE. Treatment: prednisolone 30–40mg/24h PO for acute attacks; maintenance dose 5–10mg/d. Itraconazole can be used in combination with corticosteroids. Bronchodilators for asthma. Sometimes bronchoscopic aspiration of mucus plugs is needed. 3 Aspergilloma (mycetoma): A fungus ball within a pre-existing cavity (often caused by TB or sarcoidosis). It is usually asymptomatic but may cause cough, haemoptysis (may be torrential), lethargy ± weight loss. Investigations: CXR (round opacity within a cavity, usually apical); sputum culture; strongly positive serum precipitins; Aspergillus skin test (30% +ve). Treatment (only if symptomatic): consider surgical excision for solitary symptomatic lesions or severe haemoptysis. Oral itraconazole and other antifungals have been tried with limited success. Local instillation of amphotericin paste under CT guidance yields partial success in carefully selected patients, eg in massive haemoptysis. 4 Invasive aspergillosis: Risk factors: 27 immunocompromise, eg HIV, leukaemia, burns, Wegener’s (p714), and SLE, or after broad-spectrum antibiotic therapy. Investigations: sputum culture; BAL; biopsy; serum precipitins; CXR (consolidation, abscess). Early chest CT and serial serum measurements of galactomannan (an Aspergillus antigen) may be helpful. Diagnosis may only be made at lung biopsy or autopsy. Treatment: voriconazole is superior to IV amphotericin. 28 Alternatives: IV miconazole or ketoconazole (less effective). Prognosis: 30% mortality. 5 Extrinsic allergic alveolitis (EAA): See p198. Other fungal infections Candida and Cryptococcus may cause pneumonia in the immunosuppressed (see p408).
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Asthma Asthma affects 5–8% of the population. It is characterized by recurrent episodes of dyspnoea, cough, and wheeze caused by reversible airways obstruction. Three factors contribute to airway narrowing: bronchial muscle contraction, triggered by a variety of stimuli; mucosal swelling/inflammation, caused by mast cell and basophil degranulation resulting in the release of inflammatory mediators; and increased mucus production. Symptoms Intermittent dyspnoea, wheeze, cough (often nocturnal), and sputum (see table 4.5). Precipitants: Cold air, exercise, emotion, allergens (house dust mite, pollen, fur), infection, smoking and passive smoking,29 pollution, NSAIDS, -blockers. Diurnal variation Symptoms or peak flow may vary over the day. Marked morning dipping of peak flow is common and can tip the balance into a serious attack, despite having normal peak flow (fig 4.12) at other times. Exercise: Quantify the exercise tolerance. Disturbed sleep: Quantify as nights per week (a sign of severe asthma). Acid reflux: 40–60% of those with asthma have reflux; treating it improves spirometry, but not necessarily symptoms.30 Other atopic disease: Eczema, hay fever, allergy, or family history? The home (especially the bedroom): Pets? Carpet? Feather pillows or duvet? Floor cushions and other ‘soft furnishings’? Job: If symptoms remit at weekends or holidays, work may provide the trigger (15% of cases are work-related—more for paint sprayers, food processors, welders, and animal handlers). 31 Ask the patient to measure their peak flow at intervals at work and at home (at the same time of day) to confirm this (see fig 4.13). Days per week off work or school. Signs Tachypnoea; audible wheeze; hyperinflated chest; hyper-resonant percussion note; air entry; widespread, polyphonic wheeze. Severe attack: Inability to complete sentences; pulse >110bpm; respiratory rate >25/min; PEF 33–50% predicted. Life-threatening attack: Silent chest; confusion; exhaustion; cyanosis (PaO2 20 pack year)
Widespread wheeze heard on auscultation Cardiac disease Unexplained low FEV1 or PEF
Normal PEF when symptomatic
Unexplained peripheral blood eosinophilia (Data from
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Table 4.5 Clinical features which increase or decrease probability of asthma in adults.
Suspected asthma in children
180
Intermediate probability
High probability
Chest medicine
Trial of asthma treatment
+
Low probability
Ω
Consider lung function tests/atopy
If successful, continue minimum effective dose. If unsuccessful, assess inhaler technique/ compliance
Consider referral Investigate/treat other cause
If no response to treatment, consider further investigation or onward referral
If no further improvement, consider onward referral Fig 4.10
BTS/SIGN British guideline on the management of asthma in children.
Data from Fig 1, p21:
Suspected asthma in adults Clinical investigation (spirometry or peak expiratory flow if spirometry not available)
High probability
Trial of asthma treatment
If successful, continue minimum effective dose. If unsuccessful, assess inhaler technique/ compliance
If no further improvement, consider onward referral Fig 4.11
Intermediate probability
Low probability
FEV1 /FVC
FEV1 /FVC
0.7
Investigate or treat other cause
Consider referral/ treat other cause
If no response to treatment, consider further investigation or onward referral
BTS/SIGN British guideline on the management of asthma in adults. Data from Fig 2, p25:
asthma-guideline-2014/
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Fig 4.12 Normal peak expiratory flow (PEF). Data from Nunn, AJ, Gregg, I. New regression equations for predicting peak expiratory flow in adults. BMJ 1989;298:1068–70.
Fig 4.13 Examples of serial peak flow charts.
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Management of chronic asthma Lifestyle Help to quit smoking (p93). Avoid precipitants. Weight loss if overweight. Check inhaler technique. Teach use of a peak flow meter to monitor PEF twice a day. Educate to enable self-management by altering their medication in the light of symptoms or PEF. Give specific advice about what to do in an emergency; provide a written action plan. Consider teaching relaxed breathing to avoid dysfunctional breathing 32 (Papworth method).3 British Thoracic Society guidelines (BTS33) Start at the step most appropriate to severity; moving up if needed, or down if control is good for >3 months. Rescue courses of prednisolone may be used at any time. For drug examples see table 4.6. • Step 1: Occasional short-acting inhaled 2-agonist as required for symptom relief. If used more than once daily, or night-time symptoms, go to Step 2. • Step 2: Add standard-dose inhaled steroid, eg beclometasone 200–800mcg/day, or start at the dose appropriate for disease severity, and titrate as required. • Step 3: Add long-acting 2-agonist (eg salmeterol 50mcg/12h by inhaler). If benefit—but still inadequate control—continue and dose of beclometasone to 800mcg/ day. If no effect then stop LABA and dose of beclometasone to 800mcg/day. Leukotriene receptor antagonist or oral theophylline may be tried. • Step 4: Consider trials of: beclometasone up to 2000mcg/day; modified-release oral theophylline; modified-release oral 2-agonist tablets; oral leukotriene receptor antagonist, in conjunction with previous therapy. • Step 5: Add regular oral prednisolone (1 dose daily, at the lowest possible dose). Continue with high-dose inhaled steroids. Refer for specialist input. Drugs 2-adrenoceptor agonists: Relax bronchial smooth muscle (CAMP), acting within minutes. Salbutamol is best given by inhalation (aerosol, powder, nebulizer), but may also be given PO or IV. SE: tachyarrhythmias, K+, tremor, anxiety. Long-acting inhaled 2-agonist (eg salmeterol, formoterol) can help nocturnal symptoms and reduce morning dips. They may be an alternative to steroid dose when symptoms are uncontrolled; doubts remain over whether they are associated with an increase in adverse events. 34 SE: as salbutamol, paradoxical bronchospasm. 35 Corticosteroids: Best inhaled to minimize systemic effects, eg beclometasone via spacer (or powder), but may be given PO or IV. They act over days to bronchial mucosal inflammation. Rinse mouth after inhaled steroids to prevent oral candidiasis. Oral steroids are used acutely (high-dose, short courses, eg prednisolone 40mg/24h PO for 7d) and longer term in lower dose (eg 5–10mg/24h) if control is not optimal on inhalers. Warn about SEs: p377. Aminophylline: (Metabolized to theophylline) acts by inhibiting phosphodiesterase, thus bronchoconstriction by CAMP levels. Try as prophylaxis, at night, PO, to prevent morning dipping. Stick with one brand name (bioavailability variable). Also useful as an adjunct if inhaled therapy is inadequate. In acute severe asthma, it may be given IVI. It has a narrow therapeutic ratio, causing arrhythmias, GI upset, and fits in the toxic range. Check theophylline levels (p756), and do ECG monitoring and check plasma levels after 24h if IV therapy is used. Anticholinergics: (Eg ipratropium, tiotropium.) May muscle spasm synergistically with 2-agonists but are not recommended in current guidelines for chronic asthma. They may be of more benefit in COPD. Cromoglicate (Mast cell stabilizer.) May be used as prophylaxis in mild and exerciseinduced asthma (always inhaled), especially in children. It may precipitate asthma. Leukotriene receptor antagonists: (Eg oral montelukast, zafirlukast.) Block the effects of cysteinyl leukotrienes in the airways by antagonizing the CystLT1 receptor. Anti-IgE monoclonal antibody: Omalizumab 36 may be of use in highly selected patients with persistent allergic asthma. Given as a subcutaneous injection every 2–4 wks depending on dose. Specialists prescribe only.
3 Integrated breathing and relaxation training (Papworth method) is psychological and physical: patients learn to drop their shoulders, relax their abdomen, and breathe calmly and appropriately.
Inhaled aerosol
Inhaled powder
Nebulized (supervised)
Salbutamol 100–200mcg/6h 200–400mcg/6h 2.5–5mg/6h Dose example: Airomir® is a CFC-free example of a breath-actuated inhaler Terbutaline 500mcg 2.5mg/mL Single dose Recommended regimen 500mcg/6h 5–10mg/6–12h Salmeterol 25mcg 50mcg — Dose/puff Recommended regimen 50–100mcg/12h 50–100mcg/12h — Tiotropium bromide (COPD) 2.5mcg 9mcg — Dose/puff Recommended regimen 25mcg daily 18mcg daily — Steroids (Clenil Modulite®=beclometasone; Pulmicort®=budesonide; Flixotide®=fluticasone) Fluticasone (Flixotide®) As for aerosol 50, 100, 250, & 250mcg/mL Doses available/puff 500mcg Recommended regimen Clenil Modulite® Doses available/puff Recommended regimen
100–250mcg/12h
50 & 100mcg 250mcg 200mcg/12h then 400mcg/12h then 1000mcg/12h
100–250mcg/12h 0.5–2mg/12h max 1mg/12h —
—
Available as a Turbohaler®; Autohalers® are an alternative (breath-actuated) and don’t need breathing coordination, eg Airomir® (salbutamol) and Qvar® (beclometasone). Accuhalers® deliver dry powders (eg Flixotide®, Serevent®). Systemic absorption (via the throat) is less if inhalation is through a large-volume device, eg Volumatic® or AeroChamber Plus® devices. The latter is more compact. Static charge on some devices reduces dose delivery, so wash in water before dose; leave to dry (don’t rub). It’s pointless to squirt many puffs into a device: it is best to repeat single doses, and be sure to inhale as soon as the drug is in the spacer. SE: local (oral) candidiasis (p377); rate of cataract if lifetime dose ≥2g beclometasone. 37
Prescribe beclometasone by brand name, and state that a CFC-free inhaler should be dispensed. This is because, dose for dose, Qvar® is twice as potent as the other available CFC-free brand (Clenil Modulite®). Any dose ≥250mcg ≈ significant steroid absorption: carry a steroid card; this recommendation is being widened, and lower doses (beclometasone) are now said to merit a steroid card (manufacturer’s information).
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Table 4.6 Adult doses of common inhaled drugs used in bronchoconstriction
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Chronic obstructive pulmonary disease (COPD) Definitions COPD is a common progressive disorder characterized by airway obstruction (FEV1 55yrs have TSH. Risk of progression to frank hypothyroidism is ~2%, and increases as TSH; risk doubles if thyroid peroxidase antibodies are present, and is also increased in men. Management: • Confirm that raised TSH is persistent (recheck in 2–4 months). • Recheck the history: if any non-specific features (eg depression), discuss benefits of treating (p220) with the patient—maybe they will function better. • Have a low threshold for carefully supervised treatment as your patient may not be so asymptomatic after all, and cardiac deaths may be prevented. Treat if: 1 TSH ≥10mu/L. 2 +ve thyroid autoantibodies. 3 Past (treated) Graves’. 4 Other organ-specific autoimmunity (type 1 DM, myasthenia, pernicious anaemia, vitiligo), as they are more likely to progress to clinical hypothyroidism. If TSH 4–10, and vague symptoms, treat for 6 months—only continue if symptoms improve (or the patient is trying to conceive). If the patient does not fall into any of these categories, monitor TSH yearly. • Risks from well-monitored treatment of subclinical hypothyroidism are small (but there is an risk of atrial fibrillation and osteoporosis if over-treated). Subclinical hyperthyroidism occurs when TSH, with normal T4 and T3. There is a 41% increase in relative mortality from all causes versus euthyroid control subjects—eg from AF and osteoporosis. Management: • Confirm that suppressed TSH is persistent (recheck in 2–4 months). • Check for a non-thyroidal cause: illness, pregnancy, pituitary or hypothalamic insufficiency (suspect if T4 or T3 are at the lower end of the reference range), use of TSH-suppressing medication, eg thyroxine, steroids. • If TSH 4yrs. If acromegaly occurs before bony epiphyses fuse (rare), gigantism occurs. Complications (May present with CCF or ketoacidosis.) • Impaired glucose tolerance (~40%), DM (~15%). • Vascular: BP, left ventricular hypertrophy (±dilatation/CCF), cardiomyopathy, arrhythmias. There is risk of ischaemic heart disease and stroke (?due to BP ± insulin resistance and GH-induced increase in fibrinogen and decrease in protein S). • Neoplasia: colon cancer risk; colonoscopy may be needed.21 Acromegaly in pregnancy (Subfertility is common.) Pregnancy may be normal; signs and chemistry may remit. Monitor glucose. Tests Glucose, Ca2+, and PO43Ω. GH: Don’t rely on random GH as secretion is pulsatile and during peaks acromegalic and normal levels overlap. GH also in: stress, sleep, puberty, and pregnancy. Normally GH secretion is inhibited by high glucose, and GH hardly detectable. In acromegaly GH release fails to suppress. • If basal serum GH is >0.4mcg/L (1.2mIU/L) and/or if IGF-I (p232), an oral glucose tolerance test (OGTT) is needed. If the lowest GH value during OGTT is above 1mcg/L (3mIU/L), acromegaly is confirmed. With general use of very sensitive assays, it has been said that this cut-off be decreased to 0.3mcg/L (0.9mIU/L).28 Method: Collect samples for GH glucose at: 0, 30, 60, 90, 120, 150min. Possible false +ves: puberty, pregnancy, hepatic and renal disease, anorexia nervosa, and DM. • MRI scan of pituitary fossa. • Look for hypopituitarism (p232). • Visual fields and acuity. • ECG, echo. Old photos if possible. Treatment Aim to correct (or prevent) tumour compression by excising the lesion, and to reduce GH and IGF-I levels to at least a ‘safe’ GH level of 600mOsmol/kg in Stage 1 (DI is excluded). • Free fluids until 07.30. Light breakfast at 06.30, no tea, no coffee, no smoking. Stage 1 Fluid deprivation (0–8h): for diagnosis of DI. Start at 08.00. • Empty bladder, then no drinks and only dry food. • Weigh hourly. If >3% weight lost during test, order urgent serum osmolality. If >300mOsmol/kg, proceed to Stage 2. If 600mOsmol/kg (ie normal). Stage 2 Differentiate cranial from nephrogenic DI. • Proceed if urine still dilute—ie urine osmolality 600mOsmol/kg in Stage 1 U:P ratio >2 (normal concentrating ability) Primary polydipsia Urine concentrates, but less than normal, eg >400–600mOsmol/kg Cranial DI Urine osmolality increases to >600mOsmol/kg after desmopressin (if equivocal an extended water deprivation test may be tried (no drinking from 18:00 the night before)) Nephrogenic DI No increase in urine osmolality after desmopressin
Syndrome of inappropriate ADH secretion (SIADH) In SIADH, ADH continues to be secreted in spite of low plasma osmolality or large plasma volume. Diagnosis requires concentrated urine (Na+ >20mmol/L and osmolality >100mOsmol/kg) in the presence of hyponatraemia and low plasma osmolality. Causes are numerous. See p673.
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The 8-hour water deprivation test
6 Gastroenterology Contents Healthy, enjoyable eating 244 The mouth 246 Procedures: Endoscopy and biopsy 248 Some presenting symptoms: Dysphagia 250 Nausea and vomiting 250 Dyspepsia and peptic ulcer disease 252 Gastro-oesophageal reflux disease (GORD) 254 Upper gastrointestinal bleeding 256 Diarrhoea 258 Constipation 260 Diseases and conditions: Ulcerative colitis (UC) 262 Crohn’s disease 264 Gastrointestinal malabsorption 266 Coeliac disease 266 Irritable bowel syndrome (IBS) 266 Nutritional disorders 268 Chronic pancreatitis 270 Carcinoma of the pancreas 270 Carcinoid tumours 271 Jaundice 272 Liver failure 274 Cirrhosis 276 Viral hepatitis 278 Alcoholism 280 Primary biliary cholangitis (PBC) 282 Primary sclerosing cholangitis (PSC) 282 Autoimmune hepatitis (AIH) 284 Non-alcoholic fatty liver disease (NAFLD) 285 Wilson’s disease/hepatolenticular degeneration 285 Liver tumours 286 Hereditary haemochromatosis (HH) 288 1-antitrypsin (A1AT) deficiency 290
Fig 6.1 Families are rarely what they seem: Otto, Aurelia, and Sylvia seem to be having a nice cup of tea, but Warren (the son and brother) is absent, Otto’s leg is missing, Aurelia is beside herself with anxiety, and neither is fully aware of the turmoil spiralling out of control in their unstable daughter, Sylvia. How the gut weaves in and out of our patients’ stories is one of gastroenterology’s perpetually fascinating and significant riddles. So whenever you are presented with an image in gastroenterology, ask what is missing, and try to work out the forces which are perpetuating or relieving symptoms. This is all very helpful, but it can never be relied on to tame or predict what happens next. So what did happen next? See BOX to find out.
We thank Dr Simon Campbell, our Specialist Reader for this chapter.
Lumen
243
We learn about gastroenterological diseases as if they were separate entities, independent species collected by naturalists, each kept in its own dark matchbox—collectors’ items collecting dust in a desiccated world on a library shelf. But this is not how illness works. Otto had diabetes, but refused to see a doctor until it was far advanced, and an amputation was needed. He needed looking after by his wife Aurelia. But she had her children Warren and Sylvia to look after too. And when Otto was no longer the bread-winner, she forced herself to work as a teacher, an accountant, and at any other job she could get. Otto’s illness manifested in Aurelia’s duodenum—as an ulcer. The gut often bears the brunt of other people’s worries. Inside every piece of a gut is a lumen1—the world is in the gut, and the gut is in the world. But the light does not always shine. So when the lumen filled with Aurelia’s blood, we can expect the illness to impact on the whole family. Her daughter knows where blood comes from (‘straight from the heart … pink fizz’). After Otto died, Sylvia needed long-term psychiatric care, and Aurelia moved to be near her daughter. The bleeding duodenal ulcer got worse when Sylvia needed electroconvulsive therapy. The therapy worked and now, briefly, Sylvia, before her own premature death, is able to look after Aurelia, as she prepares for a gastrectomy. The story of each illness told separately misses something; but even taken in its social context, this story is missing something vital—the poetry, in most of our patients lived rather than written—tragic, comic, human, and usually obscure—but in the case of this family not so obscure. Welling up, as unstoppable as the bleeding from her mother’s ulcer came the poetry of Sylvia Plath.
Gastroenterology
1 Lumen is Latin for light (hence its medical meaning of a tubular cavity open to the world at both ends), as well as being the SI unit of light flux falling on an object—ie the power to illuminate. All doctors have this power, whether by insightfully interpreting patients’ lives and illnesses to them, or by acts of kindness— even something so simple as bringing a cup of tea.
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Healthy, enjoyable eating ‘There’s a lot of people in this world who spend so much time watching their health that they haven’t the time to enjoy it.’ Josh Billings (1818–85). Updates to guidelines on healthy eating perhaps provide fodder for journalists who have been served a diet both rich and varied in apparently contradictory advice. Nonetheless, for many of our increasingly overweight population, simply eating less (eg 2500 calories/d for men and 2000 for women) and balancing intake across food groups seems a sensible start. Diet is of course not independent of lifestyle, and we should continue to promote a balanced diet in the context of the full range of public health messages. Unravelling these confounding threads in population-level data will always pose a challenge—while some studies show vegetarians may be less likely to die from ischaemic heart disease, is this effect because vegetarians in the UK are more likely to be non-smokers? Overly proscriptive application of such population-level data in advice given to individuals (as journalists may be prone to do) will always be flawed and risks drowning important fundamental concepts in a sea of cynicism. Current recommendations must take into account three facts • Obesity costs health services as much as smoking—1 in 4 UK adults is obese. • Diabetes mellitus is burgeoning: in some places prevalence is >7% (p206). • Past advice has not changed eating habits in large sections of the population. Advice is likely to focus on the following Body mass index: (BMI; table 6.1.) Aim for 18.5–25. Controlling quantity may be more important than quality. In hypertension, eating the ‘right’ things lowers BP only marginally, but controlling weight causes a more significant reduction. Base meals on starch: (Bread, rice, potatoes, pasta.) These provide a slower release form of carbohydrate compared to diets containing refined sugar, and beware of the high sugar contents of, eg soft drinks. Eat enough fruit and vegetables: Aiming for 5 portions a day. Eat foods high in fat, salt, or sugar infrequently. Eat some meat, fish, eggs, and beans: Aim for 2 portions of fish a week, including oily fish (those rich in omega-3 fatty acid, such as mackerel, herring, pilchards, salmon). Non-dairy sources of protein include beans and nuts. Aim to reduce intake of red or processed meat to 94cm in men and >80cm in women reflects omental fat and correlates better with risk than does BMI. BMI is still a valid way of comparing populations: average BMI in the USA is 28.8; in Japan, 22. A nation can be lean without being poor. As nations continue to adopt the lifestyle trends of the USA, this impacts sustainability.
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The risks of too much sugar Excess sugar causes caries, diabetes, obesity— which itself contributes to osteoarthritis, cancer, hypertension, and increased oxidative stress—so raising cardiovascular mortality and much more. Losing weight Motivational therapy. Consider referral to a dietician—a needsspecific diet may be best. In conjunction with exercise and diet strategies, targeted weight-loss can also be achieved successfully with psychotherapy. Drugs or surgery for obesity? The most desirable treatment for obesity is still primary prevention, but pharmacotherapy does work. Orlistat lowers fat absorption (hence SE of oily faecal incontinence)—see OHCS p514. Surgery: Carries potential for significant weight loss in appropriately selected patients but also significant morbidity (see p626).
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The mouth The diagnosis will often come out of your patient’s mouth, so open it! So many GI investigations are indirect...now is your chance for direct observation. Leucoplakia (fig 6.2) Is an oral mucosal white patch that will not rub off and is not attributable to any other known disease. It is a premalignant lesion, with a transformation rate, which ranges from 0.6% to 18%. Oral hairy leucoplakia is a shaggy white patch on the side of the tongue seen in HIV, caused by EBV. When in doubt, refer all intra-oral white lesions (see BOX). Aphthous ulcers (fig 6.3) 20% of us get these shallow, painful ulcers on the tongue or oral mucosa that heal without scarring. Causes of severe ulcers: Crohn’s and coeliac disease; Behçet’s (p694); trauma; erythema multiforme; lichen planus; pemphigus; pemphigoid; infections (herpes simplex, syphilis, Vincent’s angina, p712). : Minor ulcers: avoid oral trauma (eg hard toothbrushes or foods such as toast) and acidic foods or drinks. Tetracycline or antimicrobial mouthwashes (eg chlorhexidine) with topical steroids (eg triamcinolone gel) and topical analgesia. Severe ulcers: possible therapies include systemic corticosteroids (eg oral prednisolone 30–60mg/d PO for a week) or thalidomide (absolutely contraindicated in pregnancy). Biopsy any ulcer not healing after 3 weeks to exclude malignancy; refer to an oral surgeon if uncertain. Candidiasis (thrush) (fig 6.4) Causes white patches or erythema of the buccal mucosa. Patches may be hard to remove and bleed if scraped. Risk factors: Extremes of age; DM; antibiotics; immunosuppression (long-term corticosteroids, including inhalers; cytotoxics; malignancy; HIV). : Nystatin suspension 400 000U (4mL swill and swallow/6h). Fluconazole for oropharyngeal thrush. Cheilitis (angular stomatitis) Fissuring of the mouth’s corners is caused by denture problems, candidiasis, or deficiency of iron or riboflavin (vitamin B2). (fig 8.5, p327.) Gingivitis Gum inflammation ± hypertrophy occurs with poor oral hygiene, drugs (phenytoin, ciclosporin, nifedipine), pregnancy, vitamin C deficiency (scurvy, p268), acute myeloid leukaemia (p356), or Vincent’s angina (p712). Microstomia (fig 6.5) The mouth is too small, eg from thickening and tightening of the perioral skin after burns or in epidermolysis bullosa (destructive skin and mucous membrane blisters ± ankyloglossia) or systemic sclerosis (p552). Oral pigmentation Perioral brown spots characterize Peutz–Jeghers’ (p708). Pigmentation anywhere in the mouth suggests Addison’s disease (p226) or drugs (eg antimalarials). Consider malignant melanoma. Telangiectasia: Systemic sclerosis; Osler–Weber–Rendu syndrome (p708). Fordyce glands: (Creamy yellow spots at the border of the oral mucosa and the lip vermilion.) Sebaceous cysts, common and benign. Aspergillus niger colonization may cause a black tongue. Teeth (fig 6.6) A blue line at the gum–tooth margin suggests lead poisoning. Prenatal or childhood tetracycline exposure causes a yellow–brown discolouration. Tongue This may be furred or dry (xerostomia) in dehydration, drug therapy,3 after radiotherapy, in Crohn’s disease, Sjögren’s (p710), and Mikulicz’s syndrome (p706). Glossitis: Means a smooth, red, sore tongue, eg caused by iron, folate, or B12 deficiency (fig 8.27, p335). If local loss of papillae leads to ulcer-like lesions that change in colour and size, use the term geographic tongue (harmless migratory glossitis). Macroglossia: The tongue is too big. Causes: myxoedema; acromegaly; amyloid (p370). A ranula is a bluish salivary retention cyst to one side of the frenulum, named after the bulging vocal pouch of frogs’ throats (genus Rana). Tongue cancer: Appears as a raised ulcer with firm edges. Risk factors: smoking, alcohol.4 Spread: anterior ⅓ of tongue drains to submental nodes; middle ⅓ to submandibular nodes; posterior ⅓ to deep cervical nodes (see BOX, p599). Treatment: Radiotherapy or surgery. 5yr survival (early disease): 80%. When in doubt, refer. 3 Drugs causing xerostomia: ACE-i; antidepressants; antihistamines; antipsychotics; antimuscarinics/ anticholinergics; bromocriptine; diuretics; loperamide; nifedipine; opiates; prazosin; prochlorperazine, etc. 4 Betel nut (Areca catechu) chewing, common in South Asia, may be an independent risk factor.
White intra-oral lesions
247 • Carcinoma • Hairy oral leucoplakia • Lupus erythematosus • Smoking • Aphthous stomatitis • Secondary syphilis.
Gastroenterology
• Idiopathic keratosis • Leucoplakia • Lichen planus • Poor dental hygiene • Candidiasis • Squamous papilloma
Fig 6.2 Leucoplakia on the underside of the Fig 6.3 An aphthous ulcer inside the cheek. The tongue. It is important to refer leucoplakia be- name is tautological: aphtha in Greek means cause it is premalignant. ulceration.
Fig 6.4 White fur on an erythematous tongue caused by oral candidiasis. Oropharyngeal candidiasis in an apparently fit patient may suggest underlying HIV infection.
Fig 6.6 White bands on the teeth can be caused by excessive fluoride intake.
Fig 6.5 Microstomia (small, narrow mouth), eg from hardening of the skin in scleroderma which narrows the mouth. It is cosmetically and functionally disabling.1
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Endoscopy and biopsy Consent is needed for all these procedures; see p568. Upper GI endoscopy Indications: See table 6.2. Pre-procedure: Stop PPIS 2wks preop if possible ( pathology-masking). Nil by mouth for 6h before. Don’t drive for 24h if sedation is used. Procedure: Sedation optional, eg midazolam 1–5mg slowly IV (to remain conscious; if deeper sedation is needed, propofol via an anaesthetist (narrow therapeutic range)); nasal prong O2 (eg 2L/min; monitor respirations & oximetry). The pharynx may be sprayed with local anaesthetic before the endoscope is passed. Continuous suction must be available to prevent aspiration. Complications: Sore throat; amnesia from sedation; perforation (2 wks off ppi Sensitivity Specificity Invasive tests CLO test 95% 95% 95% 100% 95% 94% 83%
Gastroenterology
Non-invasive
95% 90% 95% 95% 92%
Histology Culture 13 C breath test Stool antigen Serology
The 13C breath test is the most accurate non-invasive Helicobacter test.
Differential diagnosis of dyspepsia • Non-ulcer dyspepsia • Oesophagitis/GORD
• Duodenal/gastric ulcer • Gastric malignancy
• Duodenitis • Gastritis (p257)
Dysphagia or >55yrs and persistent symptoms or ALARM Signs (see p252) Yes Upper GI endoscopy No further action
No • Stop drugs causing dyspepsia, eg NSAIDS • Lifestyle changes (p254) • Over-the-counter antacids, eg magnesImprovement ium trisilicate 10mL/8h PO • Review after 4wks No improvement
PPIS or H2 blockers for 4wks Ωve (eg omeprazole 20mg/24h PO or ranitidine 150mg/12h PO). Improvement No improvement
No further action
No further action
Test for H. pylori
Longer-term, lowdose treatment. Consider upper GI endoscopy Improvement
+ve
to eradicate H. pylori;† review after 4 wks
No improvement
No†
Urea breath test. H. pylori eradicated? Yes Proceed as for H. pylori -ve dyspepsia above Fig 6.14 See NICE dyspepsia guidelines.4
Nothing magical happens on the 55th birthday—this is simply an inflection point in population risk data. We should not be overly rigid in applying these rules to the patient in front of us—though those who hold the purse strings may at time seek to reduce costs by strict enforcement of such guidelines. †Don’t treat +ve cases of H. pylori more than twice. If still +ve refer for specialist opinion.
Fig 6.15 Endoscopic image of a duodenal ulcer ©Dr Jon Simmons.
253
254
Gastro-oesophageal reflux disease (GORD) is common, and caused by reflux of stomach contents (acid ± bile)8 causing troublesome symptoms and/or complications. If reflux is prolonged, it may cause oesophagitis (fig 6.16), benign oesophageal stricture, or Barrett’s oesophagus (fig 6.17 and p695; it is pre-malignant). Causes Lower oesophageal sphincter hypotension, hiatus hernia (see BOX), oesophageal dysmotility (eg systemic sclerosis), obesity, gastric acid hypersecretion, delayed gastric emptying, smoking, alcohol, pregnancy, drugs (tricyclics, anticholinergics, nitrates), Helicobacter pylori?9 Symptoms Oesophageal: Heartburn (burning, retrosternal discomfort after meals, lying, stooping, or straining, relieved by antacids); belching; acid brash (acid or bile regurgitation); waterbrash ( salivation: ‘My mouth fills with saliva’); odynophagia (painful swallowing, eg from oesophagitis or ulceration). Extra-oesophageal: Nocturnal asthma, chronic cough, laryngitis (hoarseness, throat clearing), sinusitis. Complications Oesophagitis, ulcers, benign stricture, iron-deficiency. Metaplasia dysplasianeoplasia: GORD may lead to Barrett’s oesophagus (p695; distal oesophageal epithelium undergoes metaplasia from squamous to columnar, fig 6.17). 0.1–0.4%/yr of those with Barrett’s progress to oesophageal cancer (higher if dysplasia is present). Oesophagitis from corrosives, NSAIDS, herpes, Candida; duodenal or gastric ulcers or cancers; non-ulcer dyspepsia; oesophageal spasm; cardiac disease. Tests Endoscopy if dysphagia, or if ≥55yrs old with alarm symptoms (p252) or with treatment-refractory dyspepsia. 24h oesophageal pH monitoring ± manometry help diagnose GORD when endoscopy is normal. Treatment Lifestyle: Weight loss; smoking cessation; small, regular meals; reduce hot drinks, alcohol, citrus fruits, tomatoes, onions, fizzy drinks, spicy foods, caffeine, chocolate; avoid eating < 3h before bed. Raise the bed head. Drugs: Antacids, eg magnesium trisilicate mixture (10mL/8h), or alginates, eg Gaviscon® (10–20mL/8h PO) relieve symptoms. Add a PPI, eg lansoprazole 30mg/24h PO. For refractory symptoms, add an H2 blocker and/or try twice-daily PPI. Avoid drugs affecting oesophageal motility (nitrates, anticholinergics, Ca2+ channel blockers—relax the lower oesophageal sphincter) or that damage mucosa (NSAIDS, K+ salts, bisphosphonates). Surgery: (Eg laparoscopic Nissen fundoplication, or novel options including laparoscopic insertion of a magnetic bead band or radiofrequency-induced hypertrophy.) These all aim to resting lower oesophageal sphincter pressure. Consider in severe GORD (confirm by pH-monitoring/manometry) if drugs are not working. Atypical symptoms (cough, laryngitis) are less likely to improve with surgery compared to patients with typical symptoms.
Gastroenterology
GORD
Fig 6.16 Upper GI endoscopy showing longi- Fig 6.17 Barrett’s oesophagus. tudinal mucosal breaks in severe oesophagitis.
©Dr A Mee.
©Dr A Mee. 8 The reflux of duodenal fluid, pancreatic secretions and bile may be as important as acid; it may respond to similar lifestyle measures, sucralfate (2g/12h PO), domperidone, or metoclopramide. 9 H. pylori association with GORD controversial, but eradication may help symptoms.
Hiatus hernia
Fig 6.19 CT chest (IV contrast) showing the rolling components of a hiatus hernia anterior to the oesophagus. Between the oesophagus and the vertebral column on the left-hand side is the aorta. ©Dr S Golding.
Fig 6.18 Hiatus hernia—sliding and rolling.
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Sliding hiatus hernia (80%) The gastro-oesophageal junction slides up into the chest—see fig 6.18. Acid reflux often happens as the lower oesophageal sphincter becomes less competent in many cases. Paraoesophageal hernia (rolling hiatus hernia) (20%) The gastro-oesophageal junction remains in the abdomen but a bulge of stomach herniates up into the chest alongside the oesophagus—see figs 6.18, 6.19. As the gastro-oesophageal junction remains intact, GORD is less common. Clinical features Common: 30% of patients >50yrs, especially obese women. Although most small hernias are asymptomatic, patients with large hernias may develop GORD. Imaging Upper GI endoscopy visualizes the mucosa (?oesophagitis) but cannot reliably exclude a hiatus hernia. Treatment Lose weight. Treat GORD. Surgery indications: intractable symptoms despite aggressive medical therapy, complications (see p254). Although paraoesophageal hernias may strangulate the risk of this drops dramatically after 65 yrs. Prophylactic repair is only undertaken in those considered at high risk, due to operative mortality (≈1–2%).
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Upper gastrointestinal bleeding Haematemesis is vomiting of blood. It may be bright red or look like coffee grounds. Melaena (Greek melas = black) means black motions, often like tar, and has a characteristic smell of altered blood. Both indicate upper GI bleeding. Take a brief history and Common causes Rare causes examine to assess severity. • Peptic ulcers • Bleeding disorders Ask about past GI bleeds; • Mallory–Weiss tear • Portal hypertensive dyspepsia/known ulcers; • Oesophageal varices gastropathy known liver disease or oes- • Gastritis/gastric erosions • Aorto-enteric fistula10 ophageal varices (p257); • Drugs (NSAIDS, aspirin, • Angiodysplasia dysphagia; vomiting; steroids, thrombolytics, • Haemobilia weight loss. Check drugs anticoagulants) • Dieulafoy lesion11 (see BOX on common and • Oesophagitis • Meckel’s diverticulum rare causes) and alcohol • Duodenitis • Peutz–Jeghers’ syndrome • Osler–Weber–Rendu use. Is there serious co- • Malignancy syndrome. morbidity (bad prognosis), • No obvious cause eg cardiovascular disease, respiratory disease, hepatic or renal impairment, or malignancy? Look for signs of chronic liver disease (p276) and do a PR to check for melaena. Is the patient shocked? Also: • Peripherally cool/clammy; capillary refill time >2s; urine output 100bpm). • Systolic BP 20mmHg. • Calculate the Rockall score (tables 6.6, 6.7). Acute management (p820.) Skill in resuscitation determines survival, so get good at this! Summary:5 start by protecting the airway and giving high-flow O2, then: Insert 2 large-bore (14–16G) IV cannulae and take blood for FBC (early Hb may be normal because haemodilution has not yet taken place), U&E (urea out of proportion to creatinine indicative of massive blood meal), LFT, clotting, and crossmatch. Give IV fluids (p821) to restore intravascular volume while waiting for crossmatched blood. If haemodynamically deteriorating despite fluid resuscitation, give group O RhΩve blood. Avoid saline if cirrhotic/varices. Insert a urinary catheter and monitor hourly urine output. Organize a CXR, ECG, and check ABG. Consider a CVP line to monitor and guide fluid replacement. Transfuse (with crossmatched blood if needed) if significant Hb drop (5cmH2O) at least hourly until stable. Arrange an urgent endoscopy (p248). If endoscopic control fails, surgery or emergency mesenteric angiography/ embolization may be needed. For uncontrolled oesophageal variceal bleeding, a Sengstaken–Blakemore tube may compress the varices, but should only be placed by someone with experience. Further management Anatomy is important in assessing risk of rebleeding. Posterior DUs are highest risk as they are nearest to the gastroduodenal artery. • Re-examine after 4h and consider the need for FFP if >4 units transfused. • Hourly pulse, BP, CVP, urine output (4hrly if haemodynamically stable may be OK). • Transfuse to keep Hb >70g/L; ensure a current valid group & save sample. • Check FBC, U&E, LFT, and clotting daily. • Keep nil by mouth if at high rebleed risk (see BOX ‘Management of peptic ulcer bleeds’ and p256)—ask the endoscopist. 10 A patient with an aortic graft repair and upper GI bleeding is considered to have an aorto-enteric fistula until proven otherwise: CT abdomen is usually required as well as endoscopy. 11 A Dieulafoy lesion is the rupture of an unusually big arteriole, eg in the fundus of the stomach.
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1 pt 60–79yrs BP >100mmHg Pulse >100/min Heart failure; ischaemic heart disease Post-endoscopy Mallory–Weiss tear; All other Diagnosis no lesion; no sign of diagnoses recent bleeding Signs of recent None, or dark red haemorrhage spot
on endoscopy
2 pts ≥80yrs BP 37.8°C Resting pulse 90 beats/min Haemoglobin >110g/L 105–110g/L 45mg/L) Data from Truelove et al., ‘Cortisone in ulcerative colitis’, BMJ; 2(4947): 1041–8.
Complications Acute: Toxic dilatation of colon (mucosal islands, colonic diameter >6cm) with risk of perforation; venous thromboembolism: give prophylaxis to all inpatients regardless of rectal bleeding (p350); K+ Chronic: Colonic cancer: risk related to disease extent and activity ≈5–10% with pancolitis for 20yrs. Neoplasms may occur in flat, normal-looking mucosa. To spot precursor areas of dysplasia, surveillance colonoscopy eg 1–5yrs (depending on risk), with multiple random biopsies or biopsies guided by differential uptake by abnormal mucosa of dye sprayed endoscopically. Treatment Goals are to induce, then maintain disease remission.7 Mild UC: • 5-ASA,17 eg mesalazine (=mesalamine) is the mainstay for remission-induction/maintenance. Given PR (suppositories or enemas) for distal disease (eg Pentasa® 1g daily); or PO for more extensive disease (eg Pentasa® 2g daily; once-daily dosing as effective as split dose; combine PR+PO if flare). • Topical steroid foams PR (eg hydrocortisone as Colifoam®), or prednisolone 20mg retention enemas (Predsol®) less effective than PR 5-ASA but may be added in addition. Moderate UC: If 4–6 motions/day, but otherwise well, induce remission with oral prednisolone 40mg/d for 1wk, then taper by 5mg/week over following 7wks. Then maintain on 5-ASA (SES: rash, haemolysis, hepatitis, pancreatitis, paradoxical worsening of colitis monitor FBC and U&E at start, then at 3 months, then annually). 17 5-aminosalicylic acid (5-ASA or mesalazine) must be stabilized in oral preparations to survive gastric pH. Alternatively, olsalazine is a dimer of 5-ASA or balsalazide is a prodrug, both of which are cleaved in the colon. Rare hypersensitivity reactions: worsening colitis, pancreatitis, pericarditis, nephritis.
Diagnosing IBD-unclassified (IBD-U) After full investigation, IBD may not obviously be Crohn’s or UC. IBD-U refers to isolated colonic IBD where the diagnosis remains unknown (small bowel involvement=Crohn’s). This situation is rare in adults but commoner in children. Over time the phenotype tends to become clearer (generally UC>Crohn’s). Colectomy ± pouch formation may be needed, though pouch failure rate is higher than in UC.
18 Day 3 stool frequency >8≈/day or frequency 3–8≈/day & CRP >45 =85% chance of colectomy this admission.
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Severe UC: If unwell and 6 motions/d, admit for: IV hydration/electrolyte replacement; IV steroids, eg hydrocortisone 100mg/6h or methylprednisolone 40mg/12h; rectal steroids, eg hydrocortisone 100mg in 100mL 0.9% saline/12h PR; thromboembolism prophylaxis (p350); ensure multiple stool MC&S/CDT to exclude infection. • Monitor T°, pulse, and BP—and record stool frequency/character on a stool chart. • Twice-daily exam: document distension, bowel sounds, and tenderness. • Daily FBC, ESR, CRP, U&E ± AXR. Consider blood transfusion (eg if Hb 45 or >6 stools/d, action is needed.18 Rescue therapy with ciclosporin or infliximab, can avoid colectomy, but involve surgeons early in shared care. • If improving, transfer to prednisolone PO (40mg/24h). Schedule maintenance infliximab if used for rescue, or azathioprine if ciclosporin rescue. • If fails to improve then urgent colectomy by d7–10—the challenge is not to delay surgery so long as to accumulate significant steroid exposure and debilitation that will delay post-surgical recovery. It’s time for immunomodulation if... Patients flare on steroid tapering or require ≥2 courses of steroids/year eg azathioprine (2–2.5mg/kg/d PO). 30% of patients will develop SES requiring treatment cessation including abdominal pain, nausea, pancreatitis, leucopenia, abnormal LFTS. Monitor FBC, U+E, LFT weekly for 4 wks, then every 4 wks for 3 months, then at least 3-monthly. Biologic therapy For patients intolerant of immunomodulation, or developing symptoms despite an immunomodulator, monoclonal antibodies to TNF (infliximab, adalimumab, golimumab) or to adhesion molecules involved in gut lymphocyte trafficking (vedolizumab) play an important role (see BOX ‘Therapies in Crohn’s disease’ p265). Surgery This is needed at some stage in ~20%, eg subtotal colectomy + terminal ileostomy for failure of medical therapy or fulminant colitis with toxic dilatation/ perforation. Subsequently completion proctectomy (permanent stoma) vs ileo–anal pouch. Pouches mean stoma reversal and the possibility of long-term continence but pouch opening frequency may still be around 6≈/day and recurrent pouchitis can be troublesome (give antibiotics, eg metronidazole + ciprofloxacin for 2wks).
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Crohn’s disease A chronic inflammatory disease characterized by transmural granulomatous inflammation affecting any part of the gut from mouth to anus (esp. terminal ileum in ~70%). Unlike UC, there is unaffected bowel between areas of active disease (skip lesions). Cause As with UC an inappropriate immune response against the (?abnormal) gut flora in a genetically susceptible individual.19 Prevalence 100–200/100 000. Incidence 10–20/100 000/yr; typically presents ~20–40yrs. Associations Smoking risk ≈3–4; NSAIDS may exacerbate disease. Symptoms Diarrhoea, abdominal pain, weight loss/failure to thrive. Systemic symptoms: fatigue, fever, malaise, anorexia. Signs Bowel ulceration (fig 6.22); abdominal tenderness/mass; perianal abscess/ fistulae/skin tags; anal strictures. Beyond the gut: (fig 6.21) Clubbing, skin, joint, & eye problems. Complications Small bowel obstruction; toxic dilatation (colonic diameter >6cm, toxic dilatation is rarer than in UC); abscess formation (abdominal, pelvic, or perianal); fistulae (present in ~10%), eg entero-enteric, colovesical (bladder), colovaginal, perianal, enterocutaneous; perforation; colon cancer; PSC (p282), malnutrition. Tests Blood: FBC, ESR, CRP, U&E, LFT, INR, ferritin, TIBC, B12, folate. Stool: MC&S and CDT (p258) to exclude eg C. difficile, Campylobacter, E. coli; faecal calprotectin is a simple, non-invasive test for GI inflammation with high sensitivity. Colonoscopy + biopsy: Even if mucosa looks normal. Small bowel: To detect isolated proximal disease by eg capsule endoscopy (p248, use dummy patency capsule 1st that disintegrates if it gets stuck); MRI increasingly used to assess pelvic disease and fistulae, small bowel disease activity and strictures; US in skilled hands can provide small bowel imaging. Treatment (See BOX.8) Find out how your patient deals with what may be a brutal disease (no intimacy...no sex...no hope...‘I live with this alone and will die alone’). With a collaborative approach, courage, attention to detail, and a dose of humour, this can change. Help quit smoking. Optimize nutrition. Assess severity: T°, pulse, ESR, WCC, CRP, + albumin may merit admission for IV steroids. Mild–moderate: Symptomatic but systemically well. Prednisolone 40mg/d PO for 1wk, then taper by 5mg every wk for next 7wks. An alternative dietary approach based upon ‘elemental’ or ‘polymeric’ diets is effective in children but less used for adults. Plan maintenance therapy (see BOX). Severe: Admit for IV hydration/electrolyte replacement; IV steroids, eg hydrocortisone 100mg/6h or methylprednisolone 40mg/12h; thromboembolism prophylaxis (p350); ensure multiple stool MC&S/CDT to exclude infection. • Monitor T°, pulse, BP, and record stool frequency/character on a stool chart. • Physical examination daily. Daily FBC, ESR, CRP, U&E, and plain AXR. • Consider need for blood transfusion (if Hb 3-fold above top end of normal. SE: rash. Avoid in people with known underlying malignancy. TB may reactivate when on infliximab, so screen patients before starting the treatment (CXR, PPD, interferon gamma release assay (IGRA)). Combined AZA and infliximab can efficacy of at 12 months, but there are long-term safety issues (eg increased lymphoma risk). Anti-integrin: Monoclonal antibodies targeting adhesion molecules involved in gut lymphocyte trafficking, eg vedolizumab, reduce disease activity and have a more gut-specific mechanism of activiy. Anti-IL12/23: Represents a new cytokine target with an emerging role in treatment, eg ustekinumab. Nutrition Enteral is preferred (eg polymeric diet); consider TPN as a last resort. Elemental diets: (Eg E028®.) Contain amino acids and can give remission. Low residue diets: Help symptoms in those with active disease or strictures. Surgery 50–80% need ≥1 operation in their life. It never cures. Indications: drug failure (most common); GI obstruction from stricture; perforation; fistulae; abscess. Surgical aims are: 1 resection of affected areas—but beware short bowel syndrome (p580) 2 to control perianal or fistulizing disease 3 defunction (rest) distal disease eg with a temporary ileostomy. Pouch surgery is avoided in Crohn’s ( risk of recurrence). Poor prognosis Age 1:1. Relative risk in 1st-degree relatives is 6≈. Presentation Stinking stools/steatorrhoea; diarrhoea; abdominal pain; bloating; nausea + vomiting; aphthous ulcers; angular stomatitis (p327, fig 8.5); weight; fatigue; weakness; osteomalacia; failure to thrive (children). ~30% less severe: may mimic IBS. Diagnosis Hb; RCDW (p325); B12, ferritin. Antibodies: anti-transglutaminase is single preferred test (but is an IgA antibody—check IgA levels to exclude subclass deficiency). Where serology positive or high index of suspicion proceed to duodenal biopsy while on a gluten-containing diet: expect subtotal villous atrophy, intra-epithelial WBCS + crypt hyperplasia. Where doubt persists, HLA DQ2 and DQ8 genotyping may help. Treatment Lifelong gluten-free diet—patients become experts. Rice, maize, soya, potatoes, and sugar are OK. Limited consumption of oats (≤50g/d) may be tolerated in patients with mild disease. Gluten-free biscuits, flour, bread, and pasta are prescribable. Monitor response by symptoms and repeat serology.9 Complications Anaemia; dermatitis herpetiformis (OHCS p588); osteopenia/osteoporosis; hyposplenism (offer ‘flu and pneumococcal vaccinations); GI T-cell lymphoma (rare; suspect if refractory symptoms or weight); risk of malignancy (lymphoma, gastric, oesophageal, colorectal); neuropathies.
Irritable bowel syndrome (IBS) denotes a mixed group of abdominal symptoms for which no organic cause can be found. Most are probably due to disorders of intestinal motility, enhanced visceral perception (the ‘brain–gut’ axis: see BOX ‘Managing IBS’), or microbial dysbiosis. Several diagnostic criteria exist (see BOX ‘Defining gastrointestinal dysfunction’ p261). Prevalence 10–20%; age at onset: 40yrs; : ≥2:1. Diagnosis Only diagnose IBS if recurrent abdominal pain (or discomfort) associated with at least 2 of: • relief by defecation • altered stool form • altered bowel frequency (constipation and diarrhoea may alternate). Other features: urgency; incomplete evacuation; abdominal bloating/distension; mucus PR; worsening of symptoms after food. Symptoms are chronic (>6 months), and often exacerbated by stress, menstruation, or gastroenteritis (post-infectious IBS). Signs: Examination may be normal, but general abdominal tenderness is common. Insufflation of air during lower GI endoscopy (not usually needed) may reproduce the pain. Think of other diagnoses if: Age >60yrs; history 70yrs old. Risk factors Smoking, alcohol, carcinogens, DM, chronic pancreatitis, waist circumference (ie adiposity), and possibly a high-fat and red or processed meat diet. Pathology Mostly ductal adenocarcinoma (metastasize early; present late). 60% arise in the pancreas head, 25% in the body, 15% tail. A few arise in the ampulla of Vater (ampullary tumour) or pancreatic islet cells (insulinoma, gastrinoma, glucagonomas, somatostatinomas (p223), VIPomas); both have a better prognosis. Genetics ~95% have mutations in the KRAS2 gene. The patient Tumours in the head of the pancreas present with painless obstructive jaundice. 75% of tumours in the body and tail present with epigastric pain (radiates to back and relieved by sitting forward). Either may cause anorexia, weight loss, diabetes, or acute pancreatitis. Rarer features: Thrombophlebitis migrans (eg an arm vein becomes swollen and red, then a leg vein); Ca2+; marantic endocarditis; portal hypertension (splenic vein thrombosis); nephrosis (renal vein metastases). Signs: Jaundice + palpable gallbladder (Courvoisier’s ‘law’, p272); epigastric mass; hepatomegaly; splenomegaly; lymphadenopathy; ascites. Tests Blood: Cholestatic jaundice. CA 19–9 (p531) is non-specific, but helps assess prognosis. Imaging: US or CT can show a pancreatic mass ± dilated biliary tree ± hepatic metastases. They can guide biopsy and help staging prior to surgery/stent insertion. ERCP/MRCP (p742) show biliary tree anatomy and may localize the site of obstruction. EUS (endoscopic sonography) is an emerging adjunct for diagnosis and staging. : Most ductal cancers present with metastatic disease; 100s. Or 3 out of 5 of the following: 1 Drug-induced liver failure 2 Age 40yrs old 3 >1wk from 1st jaundice to encephalopathy 4 PT >50s 5 Bilirubin ≥300μmol/L. Fulfilling these criteria predicts poor outcome in acute liver failure and should prompt consideration for transplantation (p277). Reproduced from O'Grady J et al. ‘Early indicators of prognosis in fulminant hepatic failure.’ Gastroenterology, 97(2):439–45, 1989 with permission from Elsevier.
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Hepatic encephalopathy: letting loose some false neurotransmitters
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Cirrhosis Cirrhosis (Greek kirrhos = yellow) implies irreversible liver damage. Histologically, there is loss of normal hepatic architecture with bridging fibrosis and nodular regeneration. Causes Most often chronic alcohol abuse, HBV, or HCV infection. Others: see BOX ‘Causes of cirrhosis’. Signs Leuconychia: white nails with lunulae undemarcated, from hypoalbuminaemia; Terry’s nails—white proximally but distal ⅓ reddened by telangiectasias; clubbing; palmar erythema; hyperdynamic circulation; Dupuytren’s contracture; spider naevi (fig 6.27); xanthelasma; gynaecomastia; atrophic testes; loss of body hair; parotid enlargement (alcohol); hepatomegaly, or small liver in late disease; ascites; splenomegaly. Complications Hepatic failure: Coagulopathy (failure of hepatic synthesis of clotting factors); encephalopathy (p259); hypoalbuminaemia (oedema); sepsis (pneumonia; septicaemia); spontaneous bacterial peritonitis (SBP); hypoglycaemia. Portal hypertension: Ascites (fig 6.28); splenomegaly; portosystemic shunt including oesophageal varices (± life-threatening upper GI bleed) and caput medusae (enlarged superficial periumbilical veins). HCC: risk. Tests Blood: LFT: or bilirubin, AST, ALT, ALP, GT. Later, with loss of synthetic function, look for albumin ± PT/INR. WCC & platelets indicate hypersplenism. Find the cause: ferritin, iron/total iron-binding capacity (p288); hepatitis serology (p278); immunoglobulins (p290); autoantibodies (ANA, AMA, SMA, p553); -feto protein (p286); caeruloplasmin in patients 250/mm3 indicates spontaneous bacterial peritonitis (see later in topic for treatment). Liver biopsy: (See p248.) Confirms the clinical diagnosis. Management General: Good nutrition is vital. Alcohol abstinence (p280). Avoid NSAIDS, sedatives, and opiates. Colestyramine helps pruritus (4g/12h PO, 1h after other drugs). Consider ultrasound ± -fetoprotein every 6 months to screen for HCC (p286) in those where this information will change management. Specific: For hepatitis-induced cirrhosis see p278. High-dose ursodeoxycholic acid in PBC (p282) may improve LFT and improve transplant-free survival. Penicillamine for Wilson’s disease (p285). Ascites: Fluid restriction (60kg and 0; 82% sensitive, 79% specific). This can be followed up with the easily remembered CAGE questions: Ever felt you ought to cut down on your drinking? Have people annoyed you by criticizing your drinking? Ever felt guilty about your drinking? Ever had an eye-opener in the morning? Those answering ‘yes’ to ≥2 may be exhibiting dependency (sensitivity 43–94%; specificity 70–97%), but accuracy does change according to background population. Those who refuse, or give unconvincing answers may have more to tell in their biochemistry: look for GT, ALT, MCV, AST:ALT>2, urea, platelets.
Managing alcoholic hepatitis The patient: Malaise; TPR; anorexia; D&V tender hepatomegaly ± jaundice; bleeding; ascites. Blood: WCC; platelets (toxic effect or hypersplenism); INR; AST; MCV; urea. Jaundice, encephalopathy or coagulopathy ≈ severe hepatitis. • Most need hospitalizing; urinary catheter and CVP monitoring may be needed. • Screen for infections ± ascitic fluid tap and treat for SBP (p276). • Stop alcohol consumption: for withdrawal symptoms, if chlordiazepoxide by the oral route is impossible, try lorazepam IM. • Vitamins: vit K: 10mg/d IV for 3d. Thiamine 100mg/d PO (high-dose B vitamins can also be given IV as Pabrinex®—1 pair of ampoules in 50mL 0.9% saline IVI over ½h). • Optimize nutrition (35–40kcal/kg/d non-protein energy). Use ideal body weight for calculations, eg if malnourished. • Don’t use low-protein diets even if severe encephalopathy is present. Give >1.2g/ kg/d of protein; this prevents encephalopathy, sepsis, and some deaths. • Daily weight; LFT; U&E INR. If creatinine , get help with this—HRS (p275). Na+ is common, but water restriction may make matters worse. • Steroids may confer benefit in those with severe disease. The Maddrey Discriminant Factor (DF) = (4.6 ≈ patient’s prothrombin time in sec – control time) + bilirubin (μmol/L) roughly reflects mortality. If Maddrey score >31 and encephalopathy then consider prednisolone 40mg/d for 5d tapered over 3wks. CI: sepsis; variceal bleeding. The largest study to date (STOPAH) showed only a non-significant trend towards benefit with this regimen. Prognosis: Mild episodes hardly affect mortality; if severe, mortality ≈ 50% at 30d. 1yr after admission for alcoholic hepatitis, 40% are dead...a sobering thought.
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Screening for unhealthy alcohol use
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Primary biliary cholangitis (PBC) Interlobular bile ducts are damaged by chronic autoimmune granulomatous31 inflammation causing cholestasis which may lead to fibrosis, cirrhosis, and portal hypertension. Cause Unknown environmental triggers (?pollutants, xenobiotics, non-pathogenic bacteria) + genetic predisposition (eg IL12A locus) leading to loss of immune tolerance to self-mitochondrial proteins. Antimitochondrial antibodies (AMA) are the hallmark of PBC. Prevalence 4/100 000. : ≈ 9:1. Risk if +ve family history (seen in 1–6%); many UTIS; smoking; past pregnancy; other autoimmune diseases; use of nail polish/hair dye. Typical age at presentation ~50yrs. The patient Often asymptomatic and diagnosed after incidental finding ALP. Lethargy, sleepiness, and pruritus may precede jaundice by years. Signs: Jaundice; skin pigmentation; xanthelasma (p691); xanthomata; hepatosplenomegaly. Complications: Those of cirrhosis (p276); osteoporosis is common. Malabsorption of fat-soluble vitamins (A, D, E, K) due to cholestasis and bilirubin in the gut lumen results in osteomalacia and coagulopathy; HCC (p286). Tests Blood: ALP, GT, and mildly AST & ALT; late disease: bilirubin, albumin, prothrombin time. 98% are AMA M2 subtype +ve, eg in a titre of 1:40 (see earlier in topic). Other autoantibodies (p553) may occur in low titres. Immunoglobulins are (esp. IgM). TSH & cholesterol or . Ultrasound: Excludes extrahepatic cholestasis. Biopsy: Not usually needed (unless drug-induced cholestasis or hepatic sarcoidosis need excluding); look for granulomas around bile ducts ± cirrhosis.31 Treatment Symptomatic: Pruritus: try colestyramine 4–8g/24h PO; naltrexone and rifampicin may also help. Diarrhoea: codeine phosphate, eg 30mg/8h PO. Osteoporosis prevention: p682. Specific: Fat-soluble vitamin prophylaxis: vitamin A, D, and K. Consider high-dose ursodeoxycholic acid (UDCA)Ωit may improve survival and delay transplantation. SE: weight. Monitoring: Regular LFT; ultrasound ± AFP twice-yearly if cirrhotic. Liver transplantation: (See p277.) For end-stage disease or intractable pruritus. Histological recurrence in the graft: ~17% after 5yrs; although graft failure can occur as a result of recurrence, it is rare and unpredictable. Prognosis Highly variable. The Mayo survival model is a validated predictor of survival that combines age, bilirubin, albumin, PT time, oedema, and need for diuretics.
Primary sclerosing cholangitis (PSC) Progressive cholestasis with bile duct inflammation and strictures (figs 6.30, 6.31). Symptoms/signs Pruritus ± fatigue; if advanced: ascending cholangitis, cirrhosis, and hepatic failure. Associations: • sex. • HLA-A1; B8; DR3. • AIH (p284); >80% of Northern European patients also have IBD, usually UC; this combination is associated with risk of colorectal malignancy. Cancers Bile duct, gallbladder, liver, and colon cancers are more common, so do yearly colonoscopy + ultrasound; consider cholecystectomy for gallbladder polyps.32 Tests ALP, then bilirubin; hypergammaglobulinaemia and/or IgM; AMA Ωve, but ANA, SMA, and ANCA may be +ve; see BOX and p553. ERCP (fig 6.30) or MRCP (fig 6.31) reveal duct anatomy and damage. Liver biopsy shows a fibrous, obliterative cholangitis. Treatment Liver transplant is the mainstay for end-stage disease; recurrence occurs in up to 30%; 5yr graft survival is >60%. Prognosis is worse for those with IBD, as 5–10% develop colorectal cancer post-transplant. Ursodeoxycholic acid may improve LFT but has not shown evidence of survival benefit. High doses, eg 25–30mg/ kg/d, may be harmful. Colestyramine 4–8g/24h PO for pruritus (naltrexone and rifampicin may also help). Antibiotics for bacterial cholangitis.
31 Other causes of liver granulomas: TB, sarcoid, infections with HIV (eg toxoplasmosis, CMV, mycobacteria), PAN, SLE, granulomatosis with polyangiitis, lymphoma, syphilis, isoniazid, quinidine, carbamazepine, allopurinol. Signs: PUO; LFT. 32 Usually gallbladder polyps are an incidental finding on ultrasound, and they can often be left if 40yrs old). Up to 40% present with acute hepatitis and signs of autoimmune disease, eg fever, malaise, urticarial rash, polyarthritis, pleurisy, pulmonary infiltration, or glomerulonephritis. The remainder present with gradual jaundice or are asymptomatic and diagnosed incidentally with signs of chronic liver disease. Amenorrhoea is common and disease tends to attenuate during pregnancy. Complications Those associated with cirrhosis (p276) and drug therapy. Tests Serum bilirubin, AST, ALT, and ALP all usually , hypergammaglobulinaemia (esp. IgG), +ve autoantibodies (see table 6.12). Anaemia, WCC, and platelets indicate hypersplenism. Liver biopsy: (See p248.) Mononuclear infiltrate of portal and periportal areas and piecemeal necrosis ± fibrosis; cirrhosis ≈ worse prognosis. MRCP: (See p742.) Helps exclude PSC if ALP disproportionately. Diagnosis Depends on excluding other diseases (no lab test is pathognomonic). Diagnostic criteria based on IgG levels, autoantibodies, and histology in the absence of viral disease are helpful. Sometimes diagnosis is a challenge—there is overlap with other chronic liver disease: eg PBC (p282), PSC (p282) and chronic viral hepatitis. Table 6.12 Classifying autoimmune hepatitis: types I–II I Seen in 80%. Typical patient: 80%. SES are a big problem (p376)—partly ameliorated by a switch to budesonide, eg in non-cirrhotic AIH. Liver transplantation: (See p277.) Indicated for decompensated cirrhosis or if there is failure to respond to medical therapy, but recurrence may occur. It is effective (actuarial 10yr survival is 75%). Prognosis Appears not to matter whether symptomatic or asymptomatic at presentation (10yr survival ~80% for both). The presence of cirrhosis at presentation reduces 10yr survival from 94% to 62%. Overlap syndromes: AIH-PBC (primary biliary cholangitis) overlap is worse than AIH-AIC (autoimmune cholangitis).
Associations of autoimmune hepatitis •Pernicious anaemia •Ulcerative colitis •Glomerulonephritis •Autoimmune thyroiditis
•Autoimmune haemolysis •Diabetes mellitus •PSC (p282) •HLA A1, B8, and DR3 haplotype.
33 Hepatotropic viruses (eg measles, herpes viruses) and some drugs appear to trigger AIH in genetically predisposed individuals exposed to a hepatotoxic milieu intérieur. Viral interferon can inactivate cytochrome P450 enzymes ( metabolism of ex- or endogenous hepatotoxins). Resulting modifications to proteins may generate autoantigens driving CD4 T-helper cell activation.
Wilson’s disease/hepatolenticular degeneration Wilson’s disease is a rare (3/100 000) inherited disorder of copper excretion with excess deposition in liver and CNS (eg basal ganglia). It is treatable, so screen all with cirrhosis. Genetics An autosomal recessive disorder of a copper transporting ATPase, ATP7B. Physiology Total body copper content is ~125mg. Intake ≈ 3mg/day (absorbed in proximal small intestine). In the liver, copper is incorporated into caeruloplasmin. In Wilson’s disease, copper incorporation into caeruloplasmin in hepatocytes and excretion into bile are impaired. Copper accumulates in liver, and later in other organs. Signs Children present with liver disease (hepatitis, cirrhosis, fulminant liver failure); young adults often start with CNS signs: tremor; dysarthria, dysphagia; dyskinesias; dystonias; dementia; Parkinsonism; ataxia/clumsiness. Mood: Depression/mania; labile emotions; libido; personality change. Ignoring these may cause years of needless misery: often the doctor who is good at combining the analytical and integrative aspects will be the first to make the diagnosis. Cognition: Memory; slow to solve problems; IQ; delusions; mutism. Kayser–Fleischer (KF) rings: Copper in iris (see 6 in following list); they are not invariable. Also: Haemolysis; blue lunulae (nails); arthritis; hypermobile joints; grey skin. Tests Equivocal copper studies need expert interpretation. 1 Urine: 24h copper excretion is high, eg >100mcg/24h (normal 1500 is not part of the picture). 3 Serum copper: typically 30). • If first-line empirical treatment fails, culture urine and treat according to antibiotic sensitivity. • In upper UTI, take a urine culture and treat initially with a broad-spectrum antibiotic according to local guidelines/sensitivities, eg co-amoxiclav. Hospitalization should be considered due to risk of antibiotic resistance. Avoid nitrofurantoin as it does not achieve effective concentrations in the blood. Pregnant women: Get expert help: UTI in pregnancy is associated with preterm delivery and intrauterine growth restriction. Asymptomatic bacteriuria should be confirmed on a second sample. Treat with an antibiotic. Refer to local guidance advice for antibiotic choice (avoid ciprofloxacin, trimethoprim in 1st trimester, nitrofurantoin in 3rd trimester). Confirm eradication. Men: • Treat lower UTI with a 7-day course of trimethoprim or nitrofurantoin (if eGFR >30). • If symptoms suggest prostatitis (pain in pelvis, genitals, lower back, buttocks) consider a longer (4-week) course of a fluoroquinolone (eg ciprofloxacin) due to ability to penetrate prostatic fluid. • If upper or recurrent UTI, refer for urological investigation. Catheterized patients: • All catheterized patients are bacteriuric. Send MSU only if symptomatic. Symptoms of UTI may be non-specific/atypical. Possible symptoms include fever, flank/ suprapubic pain, change in voiding pattern, vomiting, confusion, sepsis. • Change long-term catheter before starting an antibiotic. • Refer to local guidelines for initial antibiotic choice. Where possible use a narrowspectrum antibiotic according to culture sensitivity.
Urinary tract tuberculosis • A cause of sterile pyuria: dysuria, frequency, suprapubic pain but negative standard culture. Ask about malaise, fever, night sweats, weight loss, back/flank pain, visible haematuria (p393). • Can also cause an interstitial nephritis (p318) and renal amyloidosis (p315). Glomerulonephritis is rare. • Diagnose by microscopy with acid-fast techniques and mycobacterial culture of an early morning MSU and/or urinary tract tissue. • Treat with rifampicin and isoniazid for 6 months in conjunction with pyrazinamide and ethambutol for 2 months (see p394).
The ‘Piss Prophets’ Beware the fallacies, deceit and juggling of the piss-pot science used by all those who pretend knowledge of diseases by the urine. Thomas Brian, 1655. Medieval texts1 give the following maxims regarding urinary change and disease: • White or straw coloured urine = weak and cold liver and stomach • Foamy urine = eructation (belching) • Light coloured, turbid urine = mucus • Lead circle on thin urine = pathological melancholy • Bubbles on the surface = disease of the head • Watery urine = love sickness • Swampy, black, stinking urine = fatal • Lead coloured urine = a disintegrating uterus • Reddish, cloudy urine with bubbles = asthma or an irregular heart beat. 1 Uroscopy in Early Modern Europe by Michael Stolberg, Routledge, 2016, p53–6.
297
Renal medicine
Managing UTI
Renal medicine
298
Acute kidney injury (AKI): a clinical approach Definition Acute kidney injury (AKI) is a syndrome of decreased renal function, measured by serum creatinine or urine output, occurring over hours–days. It includes different aetiologies and may be multifactorial. Different definitions of AKI exist. In 2012, there was an attempt to amalgamate different diagnostic criteria into a single definition and staging system. The Kidney Diseases: Improving Global Outcomes (KDIGO) guidelines2 define AKI as: • rise in creatinine >26μmol/L within 48h. • rise in creatinine >1.5 ≈ baseline (ie before the AKI) within 7 days. • urine output 6 consecutive hours. The severity of AKI is then staged according to the highest creatinine rise or longest period/severity of oliguria (table 7.3). Table 7.3 Stage 1
KDIGO staging system for AKI
2 3
Serum creatinine >26.5μmol/L (0.3mg/dL) or 1.5–1.9 ≈ baseline 2.0–2.9 ≈ baseline >353.6μmol/L (4.0mg/dL) or >3.0 ≈ baseline or renal replacement therapy
Urine output 12h
Although there are limitations to the use of serum creatinine, including the effects of muscle mass and dilution, no other biomarker has been able to supersede it (yet). The clinical approach to AKI is shown in fig 7.3.3 Epidemiology AKI is common, occurring in up to 18% of hospital patients and ~50% of ICU patients. Risk factors for AKI include pre-existing CKD, age, male sex, and comorbidity (DM, cardiovascular disease, malignancy, chronic liver disease, complex surgery). Causes Commonest causes: 1 Sepsis. 2 Major surgery. 3 Cardiogenic shock. 4 Other hypovolaemia. 5 Drugs. 6 Hepatorenal syndrome. 7 Obstruction. Aetiology can be divided according to site (table 7.4) as: • pre-renal: perfusion to the kidney. • renal: intrinsic renal disease. • post-renal: obstruction to urine. Table 7.4 Aetiology of AKI Where? Pathology Pre-renal Vascular volume Cardiac output Systemic vasodilation Renal vasoconstriction Renal Glomerular (pp310–13)
Post-renal
Interstitial (p318) Vessels (pp314–5) Within renal tract Extrinsic compression
Example Haemorrhage, D&V, burns, pancreatitis Cardiogenic shock, MI Sepsis, drugs NSAIDs, ACE-i, ARB, hepatorenal syndrome Glomerulonephritis, ATN (prolonged renal hypoperfusion causing intrinsic renal damage) Drug reaction, infection, infiltration (eg sarcoid) Vasculitis, HUS, TTP, DIC Stone, renal tract malignancy, stricture, clot Pelvic malignancy, prostatic hypertrophy, retroperitoneal fibrosis
AKI
299
NEWS (see p892, fig A1) Consider critical care referral Is there a life-threatening complication?
Pulmonary oedema? Early referral to renal as may need dialysis (see below)
Examine Heart rate, BP, JVP, capillary refill, palpate for bladder
Treat hypovolaemia Bolus fluid 250–500mL (p300) until volume replete If 2L given without response, seek expert help (renal/HDU/ITU)
Monitor • Fluid balance—consider urinary catheter and hourly urine output • K+—check response to treatment and at least daily until creatinine falls • Observations—minimum every 4 hours • Lactate if signs of sepsis • Daily creatinine until (lags ~24 hours behind clinical response) Investigate • Urine dipstick (pre-catheter) and quantification of any proteinuria. Haematuria/ proteinuria may suggest intrinsic renal disease • USS within 24 hours (unless cause obvious or AKI improving). Small kidneys (6.5mmol/L or any ECG changes (p301)
300
Acute kidney injury (AKI): management Management of AKI requires diagnosis and treatment of the underlying aetiology: • Pre-renal: correct volume depletion and/or renal perfusion via circulatory/cardiac support, treat any underlying sepsis. • Renal: refer for likely biopsy and specialist treatment of intrinsic renal disease. • Post-renal: catheter, nephrostomy, or urological intervention. Common to all aetiologies of AKI is the need to manage fluid balance, acidosis, hyperkalaemia, and the timely recognition of those who may require renal replacement.
Renal medicine
Fluid balance Volume status: • Hypovolaemia: BP, urine volume, non-visible JVP, poor tissue turgor, pulse, daily weight loss. • Fluid overload: BP, JVP, lung crepitations, peripheral oedema, gallop rhythm. Hypotension may be relative in old age, vascular stiffness, untreated BP. JVP does not reflect intravascular volume if right-sided heart disease/failure. BP, skin turgor, capillary refill changes may be late—do not wait for them. Hypovolaemia, fluid resuscitation: • If hypovolaemic, renal perfusion will improve with volume replacement.4 • Care in cardiac disease (renal perfusion despite adequate circulating volume) and sepsis/third-spacing (extravascular volume). • Dynamic assessment is essential: examine before and after all fluid given to ensure an adequate response and to reduce the risk of fluid overload. 1 Give 500mL crystalloid over 15 min. 2 Reassess fluid state. Get expert help if unsure or if patient remains shocked. 3 Further boluses of 250–500mL crystalloid with clinical review after each. 4 Stop when euvolaemic or seek expert help when 2L given. Which crystalloid? Any crystalloid can be used (follow local guidelines). 0.9% (‘normal’) saline is non-buffered, contains chloride, and may cause hyperchloraemic acidosis. ‘Balanced’ or buffered crystalloids include Hartmann’s, Ringer’s lactate, and Plasma-Lyte®. Because they are ‘balanced’ they are often used preferentially. However, they contain 4–5mmol/L of K+ so caution if K+and oligo/anuria. What about colloid? Blood components should be used in resuscitation due to haemorrhage. Human albumin solutions may be given only under specialist advice in hepatorenal syndrome and as second line to crystalloids in septic shock. Hypervolaemia, fluid overload: Occurs due to aggressive fluid resuscitation, oliguria, and in sepsis due to capillary permeability. Monitor weight daily in patients receiving IV fluids. Treat with: • Oxygen supplementation if required. • Fluid restriction. Consider oral and IV volumes. Give antibiotics in minimal fluid and consider concentrated nutritional support preparations. • Diuretics. Only in symptomatic fluid overload. They are ineffective and potentially harmful if used to treat oliguria without fluid overload. • Renal replacement therapy (p306). AKI with fluid overload and oligo/anuria needs urgent referral to renal/critical care.
Acidosis • Mild = pH 7.30–7.36 (~bicarbonate >20mmol/L). • Moderate = pH 7.20–7.29 (~bicarbonate 10–19mmol/L). • Severe = pH 6.0mmol/L): 1 10mL of 10% calcium chloride2 (or 30mL of 10% calcium gluconate) IV via a big vein over 5–10min, repeated if necessary and if ECG changes persist. This is cardioprotective (for 30–60min) but does not treat K+ level. 2 Intravenous insulin (10u soluble insulin) in 25g glucose (50mL of 50% or 125mL of 20% glucose). Insulin stimulates intracellular uptake of K+, lowering serum K+ by 0.65–1.0mmol/L over 30–60min. Monitor hourly for hypoglycaemia (in 11–75% of treated patients) which may be delayed in renal impairment (up to 6 hours after infusion). 3 Salbutamol also causes an intracellular K+ shift but high doses are required (10–20mg via nebulizer) and tachycardia can limit use (10mg dose in IHD, avoid in tachyarrhythmias). 4 Definitive treatment requires K+ removal. If the underlying pathology cannot be corrected renal replacement may be indicated. Safe transfer to an offsite renal unit requires K+ 3 months, with implications for health.6 Classification Based on GFR category (table 7.5), the presence of albuminuria as a marker of kidney damage (table 7.6), and the cause of kidney disease (table 7.7). (Problems using formula to grade renal disease by eGFR p669). Table 7.5 Classification of CKD by GFR (mL/min/1.73m2) Category GFR Notes G1 >90 Only CKD if other evidence of kidney damage: protein/haematuria, pathology on biopsy/imaging, tubule disorder, transplant G2 60–89 G3a 45–59 Mild–moderate GFR G3b 30–44 Moderate–severe GFR G4 15–29 Severe GFR G5 105 High and optimal 90–104
GFR
75–89 stages, G2 Mild description 60–74 and range (ml/min G3a Mild– 45–59 moderate per G3b Moderate– 1.73 m2) severe 30–44 G4 G5
Severe Kidney failure
15–29
Low risk Moderate risk High risk Very high risk Extrapolated data
3 months—is there a previous creatinine on record? • Possible cause: Ask about previous UTI, lower urinary tract symptoms, PMH of BP, DM, IHD, systemic disorder, renal colic. Check drug history including when medications started. Family history including renal disease and subarachnoid haemorrhage. Systems review: look out for more than is immediately obvious, consider rare causes, ask about eyes, skin, joints, ask about symptoms suggestive of systemic disorder (‘When did you last feel well?’) and malignancy. • Current state: Patients may have symptomatic CKD if GFR 6mmol/L, eGFR >25%, or creatinine >30%: exclude other possible causes and consider a lower dose. Glycaemic control: Target HbA1C of ~53mmol/mol (7.0%) unless risk of hypoglycaemia, comorbidity or limited life expectancy. Lifestyle: Offer advice about exercise, healthy weight, and smoking cessation. Salt intake should be reduced to 6 months. Treating the original infection may make little difference to the arthritis.
Rheumatology
552
Autoimmune connective tissue diseases Included under this heading are SLE (p554), systemic sclerosis, Sjögren’s syndrome (p710), idiopathic inflammatory myopathies (myositis—see following topic), mixed connective tissue disease, relapsing polychondritis, and undifferentiated connective tissue disease and overlap syndromes. They overlap with each other, affect many organ systems, and often require immunosuppressive therapies (p376). Consider as a differential in unwell patients with multi-organ involvement, especially if no infection. Systemic sclerosis Features scleroderma (skin fibrosis), internal organ fibrosis, and microvascular abnormalities. Severe cases have a 40–50% mortality at 5 years. 90% are ANA positive and 30–40% have anticentromere antibodies (see BOX). Skin disease is limited or diffuse. Limited involves the face, hands, and feet (formally CREST syndrome). It is associated with anticentromere antibodies in 70–80%. Pulmonary hypertension is often present subclinically, and can become rapidly life-threatening, so should be looked for (: sildenafil, bosentan). Diffuse can involve the whole body. Antitopoisomerase-1 (SCL-70) antibodies in 40% and anti-RNA polymerase in 20%. Prognosis is often poor. Control BP meticulously. Perform annual echocardiogram and spirometry. Both limited and diffuse have the potential for organ fibrosis: lung, cardiac, GI, and renal (p314) but this occurs later in limited sub-set. Management: Currently no cure. Immunosuppressive regimens, including IV cyclophosphamide, are used for organ involvement or progressive skin disease. Trials of antifibrotic tyrosine kinase inhibitors are ongoing.41 Monitor BP and renal function. Regular ACE-i or A2RBS risk of renal crisis (p314). Raynaud’s phenomenon: (see p708). Mixed connective tissue disease Combines features of systemic sclerosis, SLE, and polymyositis and the presence of high titres of anti-U1-RNP antibodies. Relapsing polychondritis Rare condition with recurrent episodes of cartilage inflammation and destruction. Affects pinna (floppy ears), nasal septum, larynx (stridor), tracheobronchial tree (infections), and joints. Associations: Aortic valve disease, polyarthritis, and vasculitis. 30% have underlying rheumatic or autoimmune disease. Diagnosis is clinical. : Steroids, DMARDS or CPAP/tracheostomy for airway involvement.
Polymyositis and dermatomyositis Rare conditions characterized by insidious onset of progressive symmetrical proximal muscle weakness and autoimmune-mediated striated muscle inflammation (myositis), associated with myalgia ± arthralgia. Muscle weakness may also cause dysphagia, dysphonia (ie poor phonation, not dysphasia), or respiratory weakness. The myositis (esp. in dermatomyositis) may be a paraneoplastic phenomenon, commonly from lung, pancreatic, ovarian, or bowel malignancy. Screen for cancers. Dermatomyositis Myositis plus skin signs: •Macular rash (shawl sign is +ve if over back and shoulders). •Lilac-purple (heliotrope) rash on eyelids often with oedema (fig 12.26, p563). •Nailfold erythema (dilated capillary loops). •Gottron’s papules: roughened red papules over the knuckles, also seen on elbows and knees (pathognomonic if CK + muscle weakness). Malignancy in 30% cases. Extra-muscular signs In both conditions include fever, arthralgia, Raynaud’s, interstitial lung fibrosis and myocardial involvement (myocarditis, arrhythmias). Tests Muscle enzymes (ALT, AST, LDH, CK, & aldolase) in plasma; EMG shows characteristic fibrillation potentials; muscle biopsy confirms diagnosis (and excludes mimicking conditions). MRI shows muscle oedema in acute myositis. Autoantibody associations: (see BOX) anti-Mi2, anti-Jo1—associated with acute onset and interstitial lung fibrosis that should be treated aggressively. Differential diagnoses Carcinomatous myopathy, inclusion-body myositis, muscular dystrophy, PMR, endocrine/metabolic myopathy (eg steroids), rhabdomyolysis, infection (eg HIV), drugs (penicillamine, colchicine, statins, or chloroquine). Management Start prednisolone. Immunosuppressives (p376) and cytotoxics are used early in resistant cases. Hydroxychloroquine/topical tacrolimus for skin disease.
Plasma autoantibodies (Abs): disease associations
553
Always interpret in the context of clinical findings: Different antibodies have different disease asssociations. Rheumatological: Rheumatoid factor (RhF) positive in: RA
Infection (SBE/IE; hepatitis)
≤100% ≤100% 70% ≤50%
Mixed connective tissue disease SLE
Systemic sclerosis Normal
50% ≤40% 30% 2–10%
Anticyclic citrullinated peptide Ab (anti-CCP): 5 rheumatoid arthritis (~96% specificity) Antinuclear antibody (ANA) positive by immunofluorescence in: SLE >95% Systemic sclerosis 96% Autoimmune hepatitis 75% RA 30% Sjögren’s syndrome 68% Normal 0–2% ANA titres are expressed according to dilutions at which antibodies can be detected, ie 1:160 means antibodies can still be detected after the serum has been diluted 160 times. Titres of 1:40 or 1:80 may not be significant. The pattern of staining may indicate the disease (although these are not specific): • Homogeneous SLE • Nucleolar Systemic sclerosis • Speckled Mixed CT disease • Centromere Limited systemic sclerosis
Anti-double-stranded DNA (dsDNA): SLE (60% sensitivity, but highly specific). Antihistone Ab: drug-induced SLE (~100%). Antiphospholipid Ab (eg anti-cardiolipin Ab): antiphospholipid syndrome, SLE. Anticentromere Ab: limited systemic sclerosis. Anti-extractable nuclear antigen (ENA) antibodies (usually with +ve ANA): • Anti-Ro (SSA)
SLE, Sjögren’s syndrome, systemic sclerosis.
• Anti-La (SSB) • Anti-Sm • Anti-RNP • Anti Jo-1; Anti-Mi-2 • Anti-Scl70
Associated with congenital heart block. Sjögren’s syndrome, SLE (15%). SLE (20–30%). SLE, mixed connective tissue disease. Polymyositis, dermatomyositis. Diffuse systemic sclerosis.
Gastrointestinal: (For liver autoantibodies, see p284.) Antimitochondrial Ab (AMA): primary biliary cholangitis (>95%), autoimmune hepatitis (30%), idiopathic cirrhosis (25–30%). Anti-smooth muscle Ab (SMA): autoimmune hepatitis (70%), primary biliary cholangitis (50%), idiopathic cirrhosis (25–30%). Gastric parietal cell Ab: pernicious anaemia (>90%), atrophic gastritis (40%), ‘normal’ (10%). Intrinsic factor Ab: pernicious anaemia (50%). -gliadin Ab, antitissue transglutaminase, anti-endomysial Ab: coeliac disease. Endocrine: Thyroid peroxidase Ab: Hashimoto’s thyroiditis (~87%), Graves’ (>50%). Islet cell Ab (ICA), glutamic acid decarboxylase (GAD) Ab: type 1 diabetes mellitus (75%). Renal: Glomerular basement membrane Ab (anti-GBM): Goodpasture’s disease (100%). Antineutrophil cytoplasmic Ab (ANCA): • Cytoplasmic (cANCA), specific for serine proteinase-3 (PR3 +ve). Granulomatosis with polyangiitis (Wegener’s) (90%); also microscopic polyangiitis (30%), polyarteritis nodosa (11%). • Perinuclear (pANCA), specific for myeloperoxidase (MPO +ve). Microscopic polyangiitis (45%), Churg–Strauss, pulmonary-renal vasculitides (Goodpasture’s). Unlike immune-complex vasculitis, in ANCA-associated vasculitis no complement consumption or immune complex deposition occurs (ie pauci-immune vasculitis). 42 ANCA may also be +ve in UC/Crohn’s, sclerosing cholangitis, autoimmune hepatitis, Felty’s, RA, SLE, or drugs (eg antithyroid, allopurinol, ciprofloxacin). Neurological: Acetylcholine receptor Ab: myasthenia gravis (90%)(see p512). Anti-voltage-gated K +-channel Ab: limbic encephalitis. Anti-voltage-gated Ca 2+-channel Ab: Lambert–Eaton syndrome (see p512). Anti-aquaporin 4: neuromyelitis optica (Devic’s disease, p698). 5 Most centres now use anti-CCP antibodies for the initial workup of suspected RA.
Rheumatology
Sjögren’s syndrome Felty’s syndrome
554
Systemic lupus erythematosus (SLE)
Rheumatology
SLE is a multisystemic autoimmune disease. Autoantibodies are made against a variety of autoantigens (eg ANA) which form immune complexes . Inadequate clearance
of immune complexes results in a host of immune responses which cause tissue inflammation and damage. Environmental triggers play a part (eg EBV p405).43 Prevalence ~0.2%. : ≈ 9:1, typically women of child-bearing age. Commoner in African-Caribbeans, Asians, and if HLA B8, DR2, or DR3 +ve. ~10% of patients have a 1st- or 2nd-degree relative with SLE. Clinical features See BOX. Remitting and relapsing illness of variable presentation and course. Features often non-specific (malaise, fatigue, myalgia, and fever) or organ-specific and caused by active inflammation or damage. Other features include lymphadenopathy, weight loss, alopecia, nail-fold infarcts, non-infective endocarditis (Libman–Sacks syndrome), Raynaud’s (30%; see p708), stroke, and retinal exudates. Immunology >95% are ANA +ve. A high anti-double-stranded DNA (dsDNA) antibody titre is highly specific, but only +ve in ~60% of cases. ENA (p553) may be +ve in 20–30% (anti-Ro, anti-La, anti-Sm, anti-RNP); 40% are RhF +ve; antiphospholipid antibodies (anticardiolipin or lupus anticoagulant) may also be +ve. SLE may be associated with other autoimmune conditions: Sjögren’s (15–20%), autoimmune thyroid disease (5–10%). Diagnosis See BOX. Monitoring activity Three best tests: 1 Anti-dsDNA antibody titres. 2 Complement: C3, C4 (denotes consumption of complement, hence C3 and C4, and C3d and C4d, their degradation products). 3 ESR. Also: BP, urine for casts or protein (lupus nephritis, below), FBC, U&E, LFTs, CRP (usually normal) think of SLE whenever someone has a multisystem disorder and ESR but CRP normal. If CRP, think instead of infection, serositis, or arthritis. Skin or renal biopsies may be diagnostic. Drug-induced lupus Causes (>80 drugs) include isoniazid, hydralazine (if >50mg/24h in slow acetylators), procainamide, quinidine, chlorpromazine, minocycline, phenytoin, anti-TNF agents. It is associated with antihistone antibodies in >95% of cases. Skin and lung signs prevail (renal and CNS are rarely affected). The disease remits if the drug is stopped. Sulfonamides or the oral contraceptive pill may worsen idiopathic SLE. Management Refer: complex cases should involve specialist SLE/nephritis clinics. • General measures: High-factor sunblock. Hydroxychloroquine, unless contraindicated, reduces disease activity and improves survival. Screen for co-morbidities and medication toxicity. For skin flares, first trial topical steroids. • Maintenance: NSAIDs (unless renal disease) and hydroxychloroquine for joint and skin symptoms. Azathioprine, methotrexate, and mycophenolate as steroid-sparing agents. Belimumab (monoclonal antibody) used as an add-on therapy for autoantibody positive disease where disease activity is high.44 (See BOX.) • Mild flares: (No serious organ damage.) Hydroxychloroquine or low-dose steroids. • Moderate flares: (Organ involvement.) May require DMARDs or mycophenolate. Severe flares: If life- or organ-threatening, eg haemolytic anaemia, nephritis, severe pericarditis or CNS disease; urgent high-dose steroids, mycophenolate, rituximab, cyclophosphamide. MDT working vital for neuropsychiatric lupus (psychometric testing, lumbar puncture may be indicated). Lupus nephritis: (p314.) May require more intensive immunosuppression with steroids and cyclophosphamide or mycophenolate. BP control vital (e.g. ACE-i). Renal replacement therapy (p306) may be needed if disease progresses; nephritis recurs in ~50% post-transplant, but is a rare cause of graft failure.45 Prognosis: ~80% survival at 15 years.43 There is an increased long-term risk of CVD and osteoporosis. Antiphospholipid syndrome Can be associated with SLE (20–30%). Often occurs as a primary disease. Antiphospholipid antibodies (anticardiolipin & lupus anticoagulant, anti- 2 glycoprotein 1) cause CLOTS: Coagulation defect (arterial/venous), Livedo reticularis (p557), Obstetric (recurrent miscarriage), Thrombocytopenia. Thrombotic tendency affects cerebral, renal, and other vessels. Dx: Persistent antiphosphlolipid antibodies with clinical features. : Anticoagulation; seek advice in pregnancy.46
A favourite differential diagnosis, SLE mimics other illnesses, with wide variation in symptoms that may come and go unpredictably. Diagnose SLE47 in an appropriate clinical setting if ≥4 criteria (at least 1 clinical and 1 laboratory) or biopsy-proven lupus nephritis with positive ANA or anti-DNA. Clinical criteria 1 Acute cutaneous lupus: Malar rash/butterfly. Fixed erythema, flat or raised, over the malar eminences, tending to spare the nasolabial folds (fig 12.14). Occurs in up to 50%. Bullous lupus, toxic epidermal necrolysis variant of SLE, maculopapular lupus rash, photosensitive lupus rash, or subacute cutaneous lupus (non-indurated psoriasiform and/or annular polycyclic lesions that resolve without scarring). 2 Chronic cutaneous lupus: Discoid rash, erythematous raised patches with adherent keratotic scales and follicular plugging ± atrophic scarring (fig 12.15). Think of it as a three-stage rash affecting ears, cheeks, scalp, forehead, and chest: erythemapigmented hyperkeratotic oedematous papulesatrophic depressed lesions. 3 Non scarring alopecia: (In the absence of other causes.) 4 Oral/nasal ulcers: (In the absence of other causes.) 5 Synovitis: (Involving two or more joints or two or more tender joints with >30 minutes of morning stiffness.) 6 Serositis: a) Lung (pleurisy for >1 day, or pleural effusions, or pleural rub; b) pericardial pain for >1 day, or pericardial effusion, or pericardial rub, or pericarditis on ECG. 7 Urinanalysis: Presence of proteinuria (>0.5g/d ) or red cell casts. 8 Neurological features: Seizures; psychosis; mononeuritis multiplex; myelitis; peripheral or cranial neuropathy; cerebritis/acute confusional state in absence of other causes. 9 Haemolytic anaemia. 10 Leucopenia: (WCC 6 months, affecting mental and physical function, present >50% of the time, plus ≥4 of: myalgia (~80%), polyarthralgia, memory, unrefreshing sleep, fatigue after exertion >24h, persistent sore throat, tender cervical/axillary lymph nodes. Management principles are similar to fibromyalgia and include graded exercise and CBT. No pharmacological agents have yet been proved effective for chronic fatigue syndrome (see also OHCS p502).
Psychosocial risk factors for developing persisting chronic pain and long-term disability have been termed ‘yellow flags’: 59 Belief that pain and activity are harmful. Sickness behaviours such as extended rest. Social withdrawal. Emotional problems such as low mood, anxiety, or stress. Problems or dissatisfaction at work. Problems with claims for compensation or time off work. Overprotective family or lack of support. Inappropriate expectations of treatment, eg low active participation in treatment.
An existential approach to difficult symptoms The manner in which management is discussed is almost as important as the management itself, which should focus on education of the patient and their family and on developing coping strategies. Such a diagnosis may be a relief or a disappointment to the patient. Explain that fibromyalgia is a relapsing and remitting condition, with no easy cures, and that they will continue to have good and bad days. Reassure them that there is no serious underlying pathology, that their joints are not being damaged, and that no further tests are necessary, but be sympathetic to the fact that they may have been seeking a physical cause for their symptoms. We all at some stage come across a patient with difficult symptoms and an exasperating lack of pathology to explain them. Investigations are all normal, and medications do not seem to work. It is tempting to dismiss such patients as malingerers, but often this conclusion comes from the clinician approaching the problem from the wrong angle. The patient has symptoms that are real and disabling to them, and that will not improve without help. Perhaps a more pragmatic approach is to take advice from the Danish philosopher Kierkegaard who wrote to a friend in 1835, ‘What I really lack is to be clear in my mind what I am to do, not what I am to know ... The thing is to understand myself ... to find a truth which is true for me.’ Listen to the patient and accept their story. Then help them to focus on what they can do to improve their situation, and to move away from dwelling on finding a physical answer to their symptoms.
Pathogenesis of fibromyalgia The current hypothesis is that fibromyalgia is caused by aberrant peripheral and central pain processing. Two key features of the condition are allodynia (pain in response to a non-painful stimulus) and hyperaesthesia (exaggerated perception of pain in response to a mildly painful stimulus), examined for by palpation of tender points. Research is beginning to suggest that certain antidepressants can relieve pain and other symptoms, and especially those that have both serotonergic and noradrenergic activity (tricyclics and venlafaxine). Those acting on serotonergic receptors only are less effective. There is also some evidence to support the use of alternative therapies such as acupuncture and spa therapies, which have been postulated to act through similar spinal pain-modulatory pathways.60 Thus far, trials have involved relatively small numbers of patients or short time periods, and lack the power to draw strong conclusions. However, it is interesting to note that the CSF of patients with fibromyalgia appears to have increased levels of substance P, while levels of noradrenaline and serotonin metabolites are decreased. All three are neurotransmitters involved in descending pain-modulatory pathways in the spinal cord.61, 62 Evidence from PET imaging suggests that patients with fibromyalgia may have an abnormal central dopamine response to pain.63 The critical question is: is this cause or effect?
559
Rheumatology
Risk factors: yellow flags
Rheumatology
560
Systemic conditions causing eye signs The eye is host to many diseases: the more you look, the more you’ll see, and the more you’ll enjoy, not least because the eye is as beautiful as its signs are legion. Behçet’s (p694.) Systemic inflammatory disorder, HLA B27 association. Causes a uveitis amongst other systemic manifestations. Cause unknown. Granulomatous disorders Syphilis, TB, sarcoidosis, leprosy, brucellosis, and toxoplasmosis may inflame either the front chamber (anterior uveitis/iritis) or back chamber (posterior uveitis/choroiditis). Refer to an ophthalmologist. Systemic inflammatory diseases May manifest as iritis in ankylosing spondylitis and reactive arthritis; uveitis in Behçet’s; conjunctivitis in reactive arthritis; scleritis or episcleritis in RA, vasculitis, and SLE. Scleritis in RA and granulomatosis with polyangiitis (Wegener’s) may damage the eye. Refer urgently if eye pain. GCA causes optic nerve ischaemia presenting as sudden blindness. Keratoconjunctivitis sicca A reduction in tear formation, tested by the Schirmer filter paper test (3d or majory surgery in last 12wks Local tenderness along distribution of deep venous system Entire leg swollen Calf swelling >3cm compared with asymptomatic leg (measured 10cm below tibial tuberosity) Pitting oedema (greater in the symptomatic leg) Collateral superficial veins (non-varicose) Previously documented DVT Alternative diagnosis at least as likely as DVT
Score 1 point 1 point 1 point 1 point 1 point 1 point 1 point 1 point 1 point Ω2 points
Reprinted from the Lancet, 350, Wells PS et al., 'Value of assessment of pretest probability of deep-vein thrombosis in clinical management', 1795–8, Copyright 1997, with permission from Elsevier.
Travel and DVT Long-distance travel appears to be a risk factor for the development of venous thromboembolism (VTE). Data suggests this is not confined to air travel, increases with the duration of travel, and results in clinical thrombosis more often in travellers with pre-existing risk factors. Dehydration, immobilization, decreased oxygen tension, and prolonged sitting have all been suggested as contributory factors. The risk of developing a DVT from a long-distance flight has been estimated at 1 in 10 000 to 1 in 40 000 for the general population. • The incidence of DVT in high-risk groups has been shown to be 4–6% for flights >10h. Travellers with ≥1 risk factor should consider compression stockings. For high-risk individuals consider a single dose of prophylactic LMWH for flights >6h. • There is risk of PE associated with long-distance air travel. • Compression stockings may risk of DVT. • There is no evidence to support the use of prophylactic aspirin. • Risk reduction measures: leg exercises, increased water intake, and refraining from alcohol or caffeine during the journey.
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Surgery
Swollen legs Bilateral oedema implies systemic disease with venous pressure (eg right heart failure) or intravascular oncotic pressure (any cause of albumin, so test the urine for protein). It is dependent (distributed by gravity), which is why legs are affected early, but severe oedema extends above the legs. In the bed-bound, fluid moves to the new dependent area, causing a sacral pad. The exception is the local increase in venous pressure occurring in IVC obstruction: the swelling neither extends above the legs nor redistributes. Causes: •Right heart failure (p134). •Albumin (p686, eg renal or liver failure). •Venous insufficiency: acute, eg prolonged sitting, or chronic, with haemosiderin-pigmented, itchy, eczematous skin ± ulcers. •Vasodilators, eg nifedipine, amlodipine. •Pelvic mass (p57, p604). •Pregnancy—if BP + proteinuria, diagnose pre-eclampsia (OHCS p48): find an obstetrician urgently. In all the above, both legs need not be affected to the same extent. Unilateral oedema Pain ± redness implies DVT or inflammation, eg cellulitis or insect bites (any blisters?). Bone or muscle may be to blame, eg tumours; necrotizing fasciitis (p660); trauma (check for sensation, pulses, and severe pain esp. on passive movement: a compartment syndrome with ischaemic necrosis needs prompt fasciotomy). Impaired mobility suggests trauma, arthritis, or a Baker’s cyst (p694). Non-pitting oedema is oedema you cannot indent: see p35. Nine questions to ask 1 Is it both legs? 2 Is she pregnant? 3 Is she mobile? 4 Any trauma? 5 Any pitting (p35)? 6 Past diseases/on drugs? 7 Any pain? 8 Any skin changes? 9 Any oedema elsewhere? Tests Look for proteinuria (+hypoalbuminaemia ≈nephrotic syndrome). CCF? Treatment of leg oedema Treat the cause. Diuretics for all is not an answer. Elevating legs for dependent oedema (ankles higher than hips—do not just use footstools); raise the foot of the bed. Graduated support stockings may help (CI: ischaemia).
Surgery
580
Specific post-operative complications Laparotomy Wound may break down from a few days to a few weeks post-op (incidence ≈ 3.5%). Particular risk in the elderly, malnourished (eg cancer, IBD), if infection, uraemia, or haematoma is present, or in repeat laparotomies. Warning sign is a pink serous discharge. Always assume the defect involves the whole of the wound. Wound dehiscence may lead to a ‘burst abdomen’ with evisceration of bowel (mortality 15–30%). If you are on the ward when this happens, call your senior, put the viscera back into the abdomen, place a sterile dressing over the wound, and give IV antibiotics (eg piperacillin/tazobactam; see local guidelines). Allay anxiety, give parenteral pain control, set up an IVI, and return patient to theatre. Incisional hernia is a common late problem (20%), repairable by mesh insertion (if necessary). Biliary surgery Early: Iatrogenic bile duct injury, cholangitis, bile leakage, bleeding into the biliary tree (haemobilia—may lead to melaena or haematemesis); pancreatitis. Retained stones may be removed by ERCP (p742); if this is not available a ‘T-tube’ left in the bile duct at the time of closure allows free drainage to the exterior; unrelieved distal obstruction of the bile duct may result in fistula formation and chronic leakage of bile. If jaundiced, maintain a good urine output, monitor coagulation, and consider antibiotics. Late: Bile duct stricture; post-cholecystectomy syndrome (symptoms arising from alterations in bile flow due to loss of the reservoir function of the gallbladder). Thyroid surgery Early: Recurrent (± superior) laryngeal nerve palsy (hoarseness) can occur permanently in 0 . 5% and transiently in 1 . 5%—warn the patient that their voice will be different for a few days post-op because of intubation and local oedema (NB: pre-operative fibreoptic laryngoscopy should be performed to exclude pre-existing vocal cord dysfunction); thyroid storm (symptoms of severe hyperthyroidism—see p834); tracheal obstruction due to haematoma in the wound: relieve by immediate removal of stitches or clips using the cutter/remover that should remain at the beside; may require urgent surgery; hypoparathyroidism (p222); check plasma Ca2+ daily; transient drops in serum concentration are common, permanent in 2 . 5%. Late: Hypothyroidism; recurrent hyperthyroidism. Mastectomy Arm lymphoedema in up to 20% of those undergoing axillary node sampling or dissection. The risk of lymphoedema increases with the level of axillary dissection: risk is lower with level 1 dissection (remains inferior to pec. minor) compared to level 3 dissection (goes superior to pec. minor, rarely done); skin necrosis. Arterial surgery Bleeding; thrombosis; embolism; graft infection; MI; AV fistula formation. Complications of aortic surgery: Gut ischaemia; renal failure; respiratory distress; trauma to ureters or anterior spinal artery (leading to paraplegia); ischaemic events from distal emboli from dislodged thrombus; aorto-enteric fistula. Colonic surgery Early: Sepsis; ileus; fistulae; anastomotic leak (11% for radical rectal surgery); haemorrhage; trauma to ureters or spleen. Late: Adhesional obstruction (BOX). Small bowel surgery Short gut syndrome (best defined functionally, as malabsorption due to insufficient residual small bowel; adults with 150cm at risk). Diarrhoea and malabsorption (particularly of fats) lead to a number of metabolic abnormalities including deficiency in vitamins A, D, E, K, and B12, hyperoxaluria (causing renal stones), and bile salt depletion (causing gallstones). The management of short bowel syndrome is complex, aiming to correct metabolic abnormalities, optimize residual bowel function, and support nutrition (using parenteral route if necessary). Tracheostomy Mediastinitis; surgical emphysema. Later: stenosis. Splenectomy (p373.) Acute gastric dilatation (a serious consequence of not using a NGT, or to check that the one in place is working); thrombocytosis; sepsis. Lifetime sepsis risk is partly preventable with pre-op vaccines—ie Haemophilus type B, meningococcal, and pneumococcal (p407 & p167) and prophylactic penicillin. Genitourinary surgery Septicaemia (from instrumentation in the presence of infected urine)—consider a stat dose of gentamicin; urinoma—rupture of a ureter or renal pelvis leading to a mass of extravasated urine. Gastrectomy See p622. Prostatectomy p642. Haemorrhoidectomy p632.
When re-operating on the abdomen, the struggle against adhesions tests the farthest and darkest boundaries of patience of the abdominal surgeon and the assistant. The skill and persistence required to gently and atraumatically tease apart these fibrous bands that restrict access and vision makes any progression, no matter how slight, cause for subdued celebration. Perseverance is the name of this game. Surgical division of adhesions is known as adhesiolysis. Any surgical procedure that breaches the abdominal or pelvic cavities can predispose to the formation of adhesions, which are found in up to 90% of those with previous abdominal surgery; this is why we do not rush to operate on small bowel obstruction: the operation predisposes to yet more adhesions. Handling of the serosal surface of the bowel causes inflammation, which over weeks to years can lead to the formation of fibrous bands that tether the bowel to itself or adjacent structures—though adhesions can also form secondary to infection, radiation injury, and inflammatory processes such as Crohn’s disease. Their main sequelae are intestinal obstruction (the cause in ~60% of cases—see p610) and chronic abdominal or pelvic pain. Studies have shown that adhesiolysis may help relieve chronic pain, though for a small proportion of patients the pain never improves or even worsens after directed intervention. As far as prevention is concerned, the best approach is to avoid operating; laparoscopy compared with laparotomy reduces the rate of local adhesions. Insertion of synthetic films (eg hyaluronic acid/carboxymethyl membrane) to prevent adhesions to the anterior abdominal wall reduces incidence, extent, and severity of adhesions, but not incidence of obstruction or operative re-intervention.
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Surgery
Adhesions—legacy of the laparotomy, bane of the surgeon
Surgery
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Stoma care A stoma (Greek=mouth) is an artificial union between a conduit and the outside world—eg a colostomy, in which faeces are made to pass through an opening in the abdominal wall when a loop of colon is brought out onto the skin. NB: a stoma can also be made between two internal conduits (eg a choledochojejunostomy). Colostomies (Usually left illiac fossa and flush with the skin—fig 13.8.) May be temporary or permanent. Are they suitable for a laparoscopic operation? • Loop colostomy: A loop of colon is exteriorized and partially divided, forming two stomas that are joined together (the proximal end passes stool, the distal end passes mucus, see fig 13.6). A rod under the loop prevents retraction and may be removed after 7d. A loop colostomy is often temporary and performed to protect a distal anastomosis, eg after anterior resection. • End colostomy: The bowel is divided and the proximal end brought out as a stoma; the distal end may be: 1 resected, eg abdominoperineal (AP) resection (inspect the perineum for absent anus when examining a stoma) 2 closed and left in the abdomen (Hartmann’s procedure) 3 exteriorized, forming a ‘mucous fistula’. • Paul–Mikulicz colostomy: A double-barrelled colostomy in which the colon is divided completely (eg to excise a section of bowel). Each end is exteriorized as two separate stomas. Output: Colostomies ideally pass 1–2 formed motions/day into an adherent plastic pouch. Some may be managed with irrigation, thus avoiding a pouch. Incidence: 21 000 stomas/yr in UK (>50% are permanent). Most manage their stomas well. The cost for appliances is ~£1500/yr. If there is a skin reaction to the adhesive or pouch, a change of device may be all that is needed. Contact the stoma nurse. Ileostomies (Usually right illiac fossa.) Protrude from the skin and emit frequent fluid motions which contain active enzymes (so the skin needs protecting—see fig 13.7). Loop ileostomies can be formed to defunction the colon as a temporary measure eg during control of difficult perianal Crohn’s disease. End ileostomy follows total or subtotal colectomy, eg for UC; subsequent formation of ileal pouch-anal anastomosis (pouch of ileum is joined to the upper anal canal) can allow for stoma reversal. Alternative (non-stoma forming) surgery Low/ultralow anterior resection: All or part of the rectum is excised and the proximal colon anastomosed to the top of the anal canal (the lower the level of anastomosis, the higher the risk of complication). Transanal endoscopic microsurgery: Allows excision of small tumours within the rectum with preservation of sphincter function. Urostomies are fashioned after total cystectomy, bringing urine from the ureters to the abdominal wall via an ileal conduit that is usually incontinent. Formation of a catheterizable valvular mechanism may retain continence. Advances in urological surgery have seen an increase in continence-saving procedures such as orthotopic neobladder reconstruction, with good long-term continence rates. When choosing a stoma site, avoid: • Bony prominences (eg anterior superior iliac spine, costal margins). • The umbilicus. • Old wounds/scars—there may be adhesions beneath. • Skin folds and creases. • The waistline. • The site should be assessed pre-operatively by the stoma nurse, with the patient both lying and standing.
Complications of stomas
Psychological aspects of stoma care The physical and psychological aspects of stoma care must not be undervalued. Be alert to any vicious cycle in which a skin reaction leads to leakage and precipitates a fear of going out, or a fear of eating. This in turn may lead to poor nutrition and further skin reactions, resulting in further leakage and depression. These cycles can be circumvented by the stoma nurse, who is the expert in fitting secure, odourless devices and providing patients with a wealth of physical and psychological support, both pre and post operative (explaining what is going to happen, what the stoma will be like, and troubleshooting post-op problems). Early referral prevents problems. Without input from the stoma nurse, a patient may reject their colostomy, never attend to it, and develop deep-seated psychological and psychiatric problems.
Fig 13.7 An ileostomy sits proud, has prominent mucosal folds, and is often right-sided.
Fig 13.6 A loop colostomy with double-barrelled stoma and supporting ostomy rod.
Fig 13.8 A colostomy sits flush with the skin and is typically sited in the left iliac fossa.
Surgery
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Liaise early with the stoma nurse, starting pre-operatively. Early: • Haemorrhage at stoma site. • Stoma ischaemia—colour progresses from dusky grey to black. • High output (can lead to K+)—consider loperamide ± codeine to thicken output. • Obstruction secondary to adhesions (see p581). • Stoma retraction. Delayed: • Obstruction (failure at operation to close lateral space around stoma). • Dermatitis around stoma site (worse with ileostomy). • Stoma prolapse. • Stomal intussusception. • Stenosis. • Parastomal hernia (risk increases with time). NB: prophylactic mesh insertion at the time of stoma formation reduces this risk. • Fistulae. • Psychological problems.
Surgery
584
Nutritional support in hospital Over 25% of hospital inpatients may be malnourished. Hospitals can become so focused on curing disease that they ignore the foundations of good health—malnourished patients recover more slowly and experience more complications. See table 13.6. Why are so many hospital patients malnourished? 1 Increased nutritional requirements (eg sepsis, burns, surgery). 2 Increased nutritional losses (eg malabsorption, output from stoma). 3 Decreased intake (eg dysphagia, nausea, sedation, coma). 4 Effect of treatment (eg nausea, diarrhoea). 5 Enforced starvation (eg prolonged periods nil by mouth). 6 Missing meals (eg due to investigations—minimize meal time disruption). 7 Difficulty with feeding (eg lost dentures; no one available to assist). 8 Unappetizing food. Identifying at-risk patients Assess nutrition state (using eg Malnutrition Universal Screening Tool3) and weight on admission; reassess weekly thereafter. Involve dieticians early in those at risk. • History: Recent weight (>20%, accounting for fluid balance); recent reduced intake; diet change (eg recent change in consistency of food); nausea, vomiting, pain, diarrhoea which might have led to reduced intake. • Examination: State of hydration (p666): dehydration can go hand-in-hand with malnutrition, and overhydration can mask malnutrition. Evidence of malnutrition: skin hanging off muscles (eg over biceps); no fat between fold of skin; hair rough and wiry; pressure sores; sores at corner of mouth. Calculate body mass index (p244); BMI 2500kCal/d. Multiply kCal by a factor of 4.2. 6.25g of enteral protein gives 1g of nitrogen. Considering nitrogen balance is important because although catabolism is inevitable, replenishment is vital. Contains 5kCal/g. Contains 10kCal/g. Contains 4kCal/g. +500mL/d for each °C of pyrexia.
Electrolytes need to be considered, even if not on IVI.
Surgery
Table 13.6 Daily energy and nutritional requirements
Surgery
586
Parenteral (intravenous) nutrition Do not undertake parenteral feeding lightly: it has risks. Specialist advice is vital. It should only be considered if the patient is likely to become malnourished without it—this normally means that the gastrointestinal tract is not functioning (eg bowel obstruction), and is unlikely to function for at least 7d. Parenteral feeding may supplement other forms of nutrition (eg in short bowel syndrome or active Crohn’s disease, when nutrition cannot be sufficiently absorbed in the gut) or it can be used alone (total parenteral nutrition—TPN). Even if there is GI disease, studies show that enteral nutrition is safer, cheaper, and at least as efficacious as parenteral nutrition in the perioperative period.5 Administration Nutrition must be given via a dedicated central venous line (or peripherally inserted central catheter—PICC line) or via a dedicated lumen of a multilumen catheter (see figs 13.9 and 13.10). Requirements There are many different regimens for parenteral feeding. Most provide 2000kCal and 10–14g nitrogen in 2–3L; this usually meets a patient’s daily requirements (see table 13.6, p585). ~50% of calories are provided by fat and ~50% by carbohydrate. Regimens comprise vitamins, minerals, trace elements, and electrolytes; these will normally be included by the pharmacist. Complications • Sepsis: (eg Staphylococcus epidermidis and Staphylococcus aureus; Candida; Pseudomonas; infective endocarditis.) Look for spiking pyrexia and examine wound at tube insertion point. Stop PN, take line and peripheral cultures and give antibiotics via the line. If central venous line-related sepsis is suspected, the safest course of action is always to remove the line. Do not attempt to salvage a line when Staph. aureus or Candida infection has been identified. • Thrombosis: Central vein thrombosis may occur, resulting in pulmonary embolus or superior vena caval obstruction (p528). • Metabolic imbalance: Electrolyte abnormalities—see BOX ‘Refeeding syndrome’; deranged plasma glucose; hyperlipidaemia; deficiency syndromes (table 6.9, p268); acid-base disturbance (eg hypercapnia from excessive CO2 production). • Mechanical: Pneumothorax; embolism of IV line tip. Guidelines for success Liaise closely with line insertion team, nutrition team, and pharmacist. • Meticulous sterility. Do not use central venous lines for uses other than nutrition. Remove the line if you suspect infection. Culture its tip. • Review fluid balance at least twice daily, and requirements for energy and electrolytes daily. • Check weight, fluid balance, and urine glucose daily during establishment of parenteral nutrition. Check plasma glucose, creatinine and electrolytes (including calcium and phosphate), and FBC daily until stable. Check LFT and lipid clearance three times a week until stable. Check zinc and magnesium weekly. • Do not rush. Achieve the maintenance regimen in small steps. • Treat underlying conditions vigorously—eg sepsis may impede +ve nitrogen balance.
5 Enteral feeding promotes integrity of the gut mucosal barrier, thus preventing bacterial and endotoxin translocation across the gut wall, which can lead to multiple organ dysfunction and perpetuation of a systemic inflammatory response—even when the gut is not the primary source of pathology.
Refeeding syndrome
The venous system at the thoracic outlet When trying to judge the position of a central venous line tip on CXR (see fig 13.10) it helps to know the anatomical landmarks of the venous system (fig 13.9). The subclavian veins join the internal jugular veins behind the sternoclavicular joints to form the brachiocephalic veins. These come together behind the right 1st sternocostal joint to form the superior vena cava (SVC), which runs from this point to the right 3rd sternocostal joint. The right atrium starts here. Fig 13.9 Neck veins. Fig 13.10 Right arm
PICC (peripherally inserted central catheter) still with a wire in the lumen. This is a radiograph at the time of insertion to determine if placement is correct. The tip lies in the SVC—ie good positioning for TPN or long-term antibiotic therapy. The tip of a Hickman line, for cytotoxic administration, is better in the right atrium, to avoid possible irritation of the SVC and consequent thrombosis or stenosis. NB: this image was acquired in the angiography room, where radio-opaque material appears black (it is easier to see contrast media against a white background). A similar effect may be achieved by digitally inverting a standard X-ray.
Image courtesy of Prof. Peter Scally.
Surgery
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This is a life-threatening metabolic complication of refeeding via any route after a prolonged period of starvation. At-risk patients include those initiating artificial feeding (enteral or parenteral) after prolonged starvation, or with malignancy, anorexia nervosa, or alcoholism. As the body turns to fat and protein metabolism in the starved state, there is a drop in the level of circulating insulin (because of the paucity of dietary carbohydrates). The catabolic state also depletes intracellular stores of phosphate, although serum levels may remain normal (0.85–1.45mmol/L). When refeeding begins, the level of insulin rises in response to the carbohydrate load, and one of the consequences is to increase cellular uptake of phosphate. A hypophosphataemic state (60mL/min. To minimize the risk of nephrotoxicity, if serum creatinine is raised or eGFR 15mmol/L when starting the VRIII use 0.9% saline until 32 (p244). •ASA category ≥ III (p567) thus potentially unstable comorbidities—discuss with the anaesthetist. •Infection at the site of the operation.
Exposing patients to our learning curves? The jury is still out… All surgeons get better over time (for a while), as they perform new techniques with increasing ease and confidence. When Wertheim did his first hysterectomies, his first dozen patients died—but then one survived. He assumed it was a good operation, and pressed ahead. He was a brave man, and thousands of women owe their lives to him. But had he tried to do this today, he would have been stopped. The UK’s General Medical Council (GMC) and other august bodies tell us that we must protect the public by reporting doctors whose patients have low survival rates. The reason for this is partly ethical, and partly to preserve self-regulation. We have the toughest codes of practice and disciplinary procedures of any group of workers. It is assumed that doctors are loyal to each other out of self-interest, and that this loyalty is bad. This has never been tested formally, and is not evidence-based. We can imagine two clinical worlds: one of constant ‘reportings’ and recriminatory audits, and another of trust and team-work. Both are imperfect, but we should not assume that the first world would be better for our patients. When patients are sick with fear, they do not, perhaps, want to know everything. We may tell to protect ourselves. We may not tell to protect ourselves. Perhaps what we should do is, in our hearts, appeal to those 12 dead women-ofWertheim—a jury as infallible as sacrificial—and try to hear their reply. And to those who complain that in doing so we are playing God, it is possible to reply with some humility that, whatever it is, it does not seem like play. ‘It is amazing what little harm doctors do when one considers all the opportunities they have.’ M. Twain.
Surgery
593
Surgery
594
Lumps Examine the regional lymph nodes as well as the lump. If the lump is a node, examine its area of drainage. Always examine the circulation and nerve supply distal to any lump. History How long has it been there? Does it hurt? Any other symptoms, eg itch? Any other lumps? Is it getting bigger? Ever been abroad? Otherwise well? Physical exam Remember the 6 S’s: site, size, shape, smoothness (consistency), surface (contour/edge/colour), and surroundings. Other questions: Does it transilluminate (see next paragraph)? Is it fixed/tethered to skin or underlying structures (see BOX)? Is it fluctuant/compressible? Temperature? Tender? Pulsatile (US duplex may help)? Transilluminable lumps After eliminating as much external light as possible, place a bright, thin ‘pen’ torch on the lump, from behind (or at least to the side), so the light is shining through the lump towards your eye. If the lump glows red it is said to transilluminate—a fluid-filled lump such as a hydrocele is a good example. Lipomas These benign fatty lumps, occurring wherever fat can expand (ie not scalp or palms), have smooth, imprecise margins, a hint of fluctuance, and are not fixed to skin or deeper structures. Symptoms are only caused via pressure. Malignant change very rare (suspect if rapid growth/hardening/vascularization). Multiple scattered lipomas, which may be painful, occur in Dercum’s disease, typically in postmenopausal women. Sebaceous cysts Refer to either epidermal (fig 13.11) or pilar cysts (they are not of sebaceous origin and contain keratin, not sebum). They appear as firm, round, mobile subcutaneous nodules of varying size. Look for the characteristic central punctum. Infection is quite common, and foul pus exits through the punctum. They are common on the scalp, face, neck, and trunk. Treatment: Excision of cyst and contents. Lymph nodes Causes of enlargement: Infection: Glandular fever; brucellosis; TB; HIV; toxoplasmosis; actinomycosis; syphilis. Infiltration: Malignancy (carcinoma, lymphoma); sarcoidosis. Cutaneous abscesses Staphylococci are the most common organisms. Haemolytic streptococci only common in hand infections. Proteus is a common cause of nonstaphylococcal axillary abscesses. Below the waist, faecal organisms are common (aerobes and anaerobes). Treatment: Incise and drain. Boils (furuncles) are abscesses involving a hair follicle and associated glands. A carbuncle is an area of subcutaneous necrosis which discharges itself on to the surface through multiple sinuses. Think of hidradenitis suppurativa if recurrent inguinal or axillary abscesses. Rheumatoid nodules (fig 13.12) Collagenous granulomas which appear in established rheumatoid arthritis on the extensor aspects of joints—especially the elbows (fig 13.12). Ganglia Degenerative cysts from an adjacent joint or synovial sheath commonly seen on the dorsum of the wrist or hand and dorsum of the foot. May transilluminate. 50% disappear spontaneously. Aspiration may be effective, especially when combined with instillation of steroid and hyaluronidase. For the rest, treatment of choice is excision rather than the traditional blow from your bible (the Oxford Textbook of Surgery)! See fig 13.13. Fibromas These may occur anywhere in the body, but most commonly under the skin. These whitish, benign tumours contain collagen, fibroblasts, and fibrocytes. Dermoid cysts Contain dermal structures and are found at the junction of embryonic cutaneous boundaries, eg in the midline or lateral to the eye. See fig 13.14. Malignant tumours of connective tissue Fibrosarcomas, liposarcomas, leiomyosarcomas (smooth muscle), and rhabdomyosarcomas (striated muscle). These are staged using modified TNM system including tumour grade. Needle-core (Trucut®) biopsies of large tumours precede excision. Any lesion suspected of being a sarcoma should not be simply enucleated. Refer to a specialist. Neurofibromas See p514. Keloids Caused by irregular hypertrophy of vascularized collagen forming raised edges at sites of previous scars that extend outside the scar (fig 13.15). Common in dark skin. Treatment can be difficult. Intralesional steroid injections are a mainstay.
In or under the skin?
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Surgery
Intradermal Subcutaneous • Sebaceous cyst • Lipoma • Abscess • Ganglion • Dermoid cyst • Neuroma • Granuloma. • Lymph node. If a lump is intradermal, you cannot draw the skin over it, while if the lump is subcutaneous, you should be able to manipulate it independently from the skin.
Fig 13.11 Epidermal cyst.
Fig 13.12 Rheumatoid nodule.
Copyright www.dermnetnz.org, reproduced with permission.
Copyright www.dermnetnz.org, reproduced with permission.
Fig 13.14 Dermoid cyst.
Fig 13.13 Ganglion. Courtesy of John M Erikson, MD, Raleigh Hand Centre.
Reproduced from Lewis–Jones, Paediatric Dermatology, 2010, with permission from Oxford University Press.
Fig 13.15 Keloid scar. Courtesy of East Sussex Hospitals Trust.
596
Skin diagnoses not to be missed Malignant tumours 1 Malignant melanoma: (See fig 13.16.) : ≈ 1.3:1. UK incidence: ≥10:100 000/yr (up ≥200% in last 20yrs). Commonly affects younger patients early diagnosis is vital. Short periods of intense UV exposure is a major cause, particularly in the early years. May occur in pre-existing moles. If smooth, well-demarcated, and regular, it is unlikely to be a melanoma but diagnosis can be tricky. Most melanomas have features described by Glasgow 7-point checklist (table 13.8) and ABCDE critera (BOX), but not all. If in doubt, refer.
Surgery
Table 13.8 Glasgow 7-point checklist (refer if ≥3 points, or with 1 point if suspicious) Major (2 pts each) • Change in size • Change in shape • Change in colour
Minor (1 pt each) • Crusting or bleeding • Inflammation • Sensory change • Diameter >7mm (unless growth is in the vertical plane)
Superficial spreading melanomas (70%) grow slowly, metastasize later, and have better prognosis than nodular melanomas (10–15%) which invade deeply and metastasize early. Nodular lesions may be amelanotic in ~5%. Others: acral melanomas occur on palms, soles, and subungual areas; lentigo maligna melanoma evolves from pre-exisiting lentigo maligna. Breslow thickness (depth in mm), tumour stage, and presence of ulceration are important prognostic factors. : urgent excision can be curative. Chemotherapy gives a response in 10–30% with metastatic disease (OHCS p592). Ipilimumab, a human monoclonal antibody that blocks CTLA-4, an inhibitory T-cell receptor, has been shown to improve survival in patients with metastatic melanoma.5,6 2 Squamous cell cancer: Usually presents as an ulcerated lesion, with hard, raised edges, in sun-exposed sites. May begin in solar keratoses (see later in topic), or be found on the lips of smokers or in long-standing ulcers (=Marjolin’s ulcer). Metastasis to lymph nodes is rare, local destruction may be extensive. : excision + radiotherapy to treat recurrence/affected nodes. See fig 13.17. NB: the condition may be confused with a keratoacanthoma—a fast-growing, benign, self-limiting papule plugged with keratin. 3 Basal cell carcinoma: (AKA rodent ulcer) Nodular: typically a pearly nodule with rolled telangiectatic edge, on the face or a sun-exposed site. May have a central ulcer. See fig 13.18. Metastases are very rare. It slowly causes local destruction if left untreated. Superficial: lesions appear as red scaly plaques with a raised smooth edge, often on the trunk or shoulders. Cause: (most frequently) UV exposure. : excision; cryotherapy; for superficial BCCS topical flurouracil or imiquimod (see as for ‘Solar keratoses’). Pre-malignant tumours 1 Solar (actinic) keratoses appear on sun-exposed skin as crumbly, yellow-white crusts. Malignant change to squamous cell carcinoma may occur after several years. Treatment: cryotherapy; 5% fluorouracil cream or 5% imiquimod—work by causing: erythema vesiculation erosion ulceration necrosis healing epithelialization, leaving healthy skin unharmed. Warn patients of expected inflammatory reaction. See BNF for dosing. Alternatively: diclofenac gel (3%, use thinly twice-daily for ≤90d). 2 Bowen’s disease: Slow-growing red/brown scaly plaque, eg on lower legs. Histology: full-thickness dysplasia (carcinoma in situ). It infrequently progresses to squamous cell cancer. Penile Bowen’s disease is called Queyrat’s erythroplasia. Treatment: cryotherapy, topical fluorouracil (see as for ‘Solar keratoses’) or photodynamic therapy. 3 See also Kaposi’s sarcoma (p702); Paget’s disease of the breast (p708). Others •Secondary carcinoma: Most common metastases to skin are from breast, kidney, or lung. Usually a firm nodule, most often on the scalp. See also acanthosis nigricans (p562). •Mycosis fungoides: Cutaneous T-cell lymphoma usually confined to skin. Causes itchy, red plaques (Sézary syndrome-variant also associated with erythroderma). •Leucoplakia: This appears as white patches (which may fissure) on oral or genital mucosa (where it may itch). Frank carcinomatous change may occur. •Leprosy: Suspect in any anaesthetic hypopigmented lesion (p441). •Syphilis: Any genital ulcer is syphilis until proved otherwise. Secondary syphilis: papular rash—including, unusually, on the palms (p412).
ABCDE criteria for diagnosis of melanoma
597
Fig 13.16 Melanoma.
Fig 13.17 Squamous cell can- Fig 13.18 Basal cell carcinoma cer.
(BCC).
Surgery
Asymmetry Border—irregular Colour—non-uniform Diameter >7mm Elevation
Surgery
598
Lumps in the neck Don’t biopsy lumps until tumours within the head and neck have been excluded by an ENT surgeon. Culture all biopsied lymph nodes for TB. Diagnosis (See fig 13.19.) First, ask how long the lump has been present. If 20yrs, consider lymphoma (hepatosplenomegaly?) or metastases (eg from GI or bronchial or head and neck neoplasia), 8% are goitres (p600), and other diagnoses account for 7%. Tests Do virology and TB tests (p394). US shows lump consistency: cystic, solid, complex, vascular. CT defines masses in relation to their anatomical neighbours. CXR may show malignancy or, in sarcoid, reveal bilateral hilar lymphadenopathy. Consider fine-needle aspiration (FNA). Midline lumps •If patient is 20yrs old, it is probably a thyroid isthmus mass. •If it is bony hard, the diagnosis may be a chondroma (benign cartilaginous tumour). Submandibular triangle (Bordered by the mental process, mandible, and the line between the two angles of the mandible.) •If 20yrs, exclude malignant lymphadenopathy (eg firm and non-tender). Is TB likely? •If it is not a node, think of submandibular salivary stone, sialadenitis, or tumour (see BOX for Salivary gland pathology). Anterior triangle (Between midline, anterior border of sternocleidomastoid, and the line between the two angles of the mandible.) •Branchial cysts emerge under the anterior border of sternocleidomastoid where the upper third meets the middle third (age 40yrs)? •Laryngoceles are an uncommon cause of anterior triangle lumps. They are painless and may be made worse by blowing. These cysts are classified as external, internal, or mixed, and may be associated with laryngeal cancer. If pulsatile may be: •Carotid artery aneurysm, •Tortuous carotid artery, or •Carotid body tumours (chemodectoma). These are very rare, move from side to side but not up and down, and splay out the carotid bifurcation. They are usually firm and occasionally soft and pulsatile. They do not usually cause bruits. They may be bilateral, familial, and malignant (5%). Suspect in any mass just anterior to the upper third of sternomastoid. Diagnose by duplex USS (splaying at the carotid bifurcation) or digital computer angiography. : Extirpation by vascular surgeon. Posterior triangle (Behind sternocleidomastoid, in front of trapezius, above clavicle.) •Cervical ribs may intrude into this area. These are enlarged costal elements from C7 vertebra. The majority are asymptomatic but can cause Raynaud’s syndrome by compressing subclavian artery and neurological symptoms (eg wasting of 1st dorsal interosseous) from pressure on lower trunk of the brachial plexus. •Pharyngeal pouches can protrude into the posterior triangle on swallowing (usually left-sided). •Cystic hygromas (usually infants) arise from jugular lymph sac. These macrocystic lymphatic malformations transilluminate brightly. Treat by surgery or hypertonic saline sclerosant injection. Recurrence can be troublesome. •Pancoast’s tumour (see p708). •Subclavian artery aneurysm will be pulsatile.
External carotid artery
Parotid gland Parotid nodes Sternomastoid
Mylohyoid
Hypoglossal nerve Tonsillar node
Submandibular nodes
Occipital nodes Scalenus medius
Ant. and Post. digastric Submandibular gland
Internal jugular vein
Scalenus anterior Vagus nerve
Sternothyroid Subclavian artery
Common carotid artery Sternomastoid
Subclavian vein
Fig 13.19 Important structures in the head and neck.
Salivary gland pathology There are three pairs of major salivary glands: parotid, submandibular, and sublingual (there are many minor glands). History: Lumps; swelling related to food; pain. Examination: Note external swelling; look for secretions; bimanual palpation for stones. Examine VIIth nerve and regional lymph nodes. Cytology: Do FNA. Acute swelling Think of mumps and HIV. Recurrent unilateral pain and swelling is likely to be from a stone. 80% are submandibular. The classical story is of pain and swelling on eating—with a red, tender, swollen, but uninfected gland. The stone may be seen on plain X-ray or by sialography (fig 13.20). Distal stones are removed via the mouth but deeper stones may require excision of the gland. Chronic bilateral symptoms may coexist with dry eyes and mouth and autoimmune disease, eg hypothyroidism, Mikulicz’s or Sjögren’s syndrome (p706 & p710)—also bulimia or alcohol excess. Fixed swelling may be from a tumour/ALL (fig 8.49, Fig 13.20 Normal sialogram of the subp355), sarcoid, amyloid, granulomatosis with mandibular gland. Wharton’s (submandibular) duct opens into the mouth near polyangiitis , or be idiopathic. Salivary gland tumours (table 13.9) ‘80% are the frenulum of the tongue. in the parotid, 80% of these are pleomorphic adenomas, 80% of these are in the superficial lobe.’ Deflection of the ear outwards is a classic sign. Remove any salivary gland swelling for assessment if present for >1 month. VIIth nerve palsy means malignancy. Table 13.9 Types of salivary gland tumours Malignant Mucoepidermoid Acinic cell
Malignant Squamous or adeno ca Adenoid cystic ca
Pleomorphic adenomas often present in middle age and grow slowly. Remove by superficial parotidectomy. Adenolymphomas (Warthin’s tumour): usually older men; soft; treat by enucleation. Carcinomas: rapid growth; hard fixed mass; pain; facial palsy. Treatment: surgery + radiotherapy.
Surgery
Omohyoid
Hyoid bone Sternohyoid
Benign or malignant Cystadenolymphoma Pleomorphic adenoma
599
Surgery
600
Lumps in the thyroid If the thyroid (fig 13.21) is enlarged (=goitre), ask yourself: 1 Is the thyroid diffusely enlarged or nodular? 2 Is the patient euthyroid, thyrotoxic (p218), or hypothyroid (p220)? Diffuse goitre: Causes: Endemic (iodine deficiency); congenital; secondary to goitrogens (substances that iodine uptake); acute thyroiditis (de Quervain’s); physiological (pregnancy/puberty); autoimmune (Graves’ disease; Hashimoto’s thyroiditis). Nodular goitre: •Multinodular goitre (MNG): The most common goitre in the UK. 50% who present with a single nodule actually have MNG. Patients are usually euthyroid, but may become hyperthyroid (‘toxic’). MNG may be retro- or substernal. Hypothyroidism and malignancy within MNG are rare. Plummer’s disease is hyperthyroidism with a single toxic nodule (uncommon). •Fibrotic goitre: Eg Reidel’s thyroiditis. •Solitary thyroid nodule: typically cyst, adenoma, discrete nodule in MNG or malignant (~10%). Investigations Check TSH and USS (solid, cystic, complex or part of a group of lumps). If abnormal consider: •T4, autoantibodies (p216, eg if Hashimoto’s /Graves’, suspected). •CXR with thoracic inlet view (tracheal goitres and metastases?). •Radionuclide scans (fig 13.22) may show malignant lesions as hypofunctioning or ‘cold’, whereas a hyperfunctioning ‘hot’ lesion suggests adenoma. •FNA (fine-needle aspiration) and cytologyΩwill characterize lesion. A FNA finding of a follicular neoplasm can be challenging (15–30% malignant)Ωdiscuss with cytopathologist and perform molecular diagnostics where available; if any doubt, refer for surgery. What should you do if high-resolution ultrasound shows impalpable nodules? Such thyroid nodules can usually be observed provided they are: • 30yrs old, early menarche; late menopause; HRT; obesity; BRCA genes (p521); not breastfeeding; past breast cancer (metachronous rate ≈2%, synchronous rate ≈1%). Pathology Non-invasive ductal carcinoma in situ (DCIS) is premalignant and seen as microcalcification on mammography (unifocal or widespread). Non-invasive lobular CIS is rarer and tends to be multifocal. Invasive ductal carcinoma is most common (~70%) whereas invasive lobular carcinoma accounts for 10–15% of breast cancers. Medullary cancers (~5%) tend to affect younger patients while colloid/mucoid (~2%) tend to affect the elderly. Others: papillary, tubular, adenoid-cystic and Paget’s (p708). 60–70% of breast cancers are oestrogen receptor +ve, conveying better prognosis. ~30% over-express HER2 (growth factor receptor gene) associated with aggressive disease and poorer prognosis. Investigations (See p82 for history and examination.) All lumps should undergo ‘triple’ assessment: Clinical examination + histology/cytology + mammography/ultrasound; see fig 13.24. Staging: Stage 1: Confined to breast, mobile. Stage 2: Growth confined to breast, mobile, lymph nodes in ipsilateral axilla. Stage 3: Tumour fixed to muscle (but not chest wall), ipsilateral lymph nodes matted and may be fixed, skin involvement larger than tumour. Stage 4: Complete fixation of tumour to chest wall, distant metastases. Also TNM staging: (p523) T15cm, T4, fixity to chest wall or peau d’orange; N1, mobile ipsilateral nodes; N2, fixed nodes; M1, distant metastases. Treating local disease (Stage 1–2.)7 •Surgery: Removal of tumour by wide local excision (WLE) or mastectomy ± breast reconstruction + axillary node sampling/ surgical clearance or sentinel node biopsy (BOX ‘Sentinel node biopsy’). •Radiotherapy: Recommended for all patients with invasive cancer after WLE. Risk of recurrence decreases from 30% to 2yrs. Staging investigations should include CXR, bone scan, liver USS, CT/ 2+ MRI or PET-CT (p739), + LFTS and Ca . Radiotherapy (p526) to painful bony lesions (bisphosphonates, p677, may pain and fracture risk). Tamoxifen is often used in ER+ve; if relapse after initial success, consider chemotherapy. Trastuzumab should be given for HER2 +ve tumours, in combination with chemotherapy. CNS surgery for solitary (or easily accessible) metastases may be possible; if not—radiotherapy. Get specialist help for arm lymphoedema (try decongestive methods first). Preventing deaths •Promote awareness. •Screening: 2-view mammography every 3yrs for women aged 47–73 in UK has breast cancer deaths by 30% in women >50yrs.
Breast lump
603
‘Triple assessment’ 1 Clinical examination 2 Radiology: ultrasound for 35yrs old 3 Histology/cytology (FNA or core biopsy: US-guided core biopsy is best for new lumps)
Residual mass core biopsy
Solid lump core biopsy
Clear fluid discard fluid and reassure
Bloody fluid cytology
Malignant plan
Clear fluid discard fluid and reassure
Reassurance can be more emphatic if there is no family history and biopsy shows a non-proliferative lesion.
Fig 13.24 Triple assessment and investigation of a breast lump. US is more accurate at detecting invasive breast cancer, though mammography remains most accurate at detecting ductal carcinoma in situ (DCIS). MRI is used in the assessment of multifocal/bilateral disease and patients with cosmetic implants who are identified as high risk.
Sentinel node biopsy Decreases needless axillary clearances in lymph node Ωve patients. • Patent blue dye and/or radiocolloid injected into periareolar area or tumour. • A gamma probe/visual inspection is used to identify the sentinel node. • The sentinel node is biopsied and sent for histology ± immunohistochemistry; further clearance only if sentinel node +ve. Sentinel node identified in 90%. False Ωve rates 3 nodes +ve for breast cancer. Histological grade is also scored 1–3.
Surgery
Cystic lump aspirate
604
Abdominal masses As with any mass (see p594), determine size, site, shape, and surface. Find out if it is pulsatile and if it is mobile. Examine supraclavicular and inguinal nodes. Is the lump ballotable (like bobbing an apple up and down in water)?
Surgery
Right iliac fossa masses • Appendix mass/abscess • Caecal carcinoma • Crohn’s disease • Pelvic mass (see later in topic)
• Intussusception • TB mass • Amoebic abscess • Actinomycosis (p389)
• Transplanted kidney • Kidney malformation • Tumour in an undescended testis.
Abdominal distension Flatus, fat, fluid, faeces, or fetus (p57)? Fluid may be outside the gut (ascites) or sequestered in bowel (obstruction; ileus). To demonstrate ascites elicit signs of a fluid thrill and/or shifting dullness (p61). Causes of ascites • Malignancy • Infections—esp. TB • Albumin (eg nephrosis)
Ascites with portal hypertension • Portal nodes • CCF; pericarditis • Cirrhosis • Pancreatitis • Budd–Chiari syndrome (p696) • Myxoedema. • IVC or portal vein thrombosis.
Tests: Aspirate ascitic fluid (p764) for cytology, culture and albumin;7 US. Left upper quadrant mass Is it spleen, stomach, kidney, colon, pancreas, or a rare cause (eg neurofibroma)? Pancreatic cysts may be true (congenital; cystadenomas; retention cysts of chronic pancreatitis; cystic fibrosis) or pseudocysts (fluid in lesser sac from acute pancreatitis). Splenomegaly Causes are often said to be infective, haematological, neoplastic, etc, but grouping by associated feature is more useful clinically: Splenomegaly with fever With lymphadenopathy • InfectionHS (malaria, SBE/IE, • Glandular feverHS hepatitis,HS EBV,HS TB, CMV, HIV) • Leukaemias; lymphoma • Sjögren’s syndrome. • Sarcoid; malignancy.HS With arthritis With ascites • Sjögren’s syndrome • Carcinoma • Rheumatoid arthritis; SLE • Portal hypertension.HS • Infection, eg Lyme (p422) • Vasculitis/Behçet’s (p556). With anaemia With weight + CNS signs • Sickle-cell;HS thalassaemiaHS • Cancer; lymphoma • Leishmaniasis;HS leukaemiaHS • TB; arsenic poisoning • Pernicious anaemia (p334) • Paraproteinaemia.HS • POEMS syn. (p220). HS
With purpura • Septicaemia; typhus • DIC; amyloidHS • Meningococcaemia. With a murmur • SBE/IE • Rheumatic fever • Hypereosinophilia • AmyloidHS (p370). Massive splenomegaly • Malaria (hyper-reactivity after chronic exposure) • Myelofibrosis; CMLHS • Gaucher’s syndromeHS • Leishmaniasis.
=causes of hepatosplenomegaly.
Smooth hepatomegaly Hepatitis, CCF, sarcoidosis, early alcoholic cirrhosis (a small liver is typical later); tricuspid incompetence ( pulsatile liver). Craggy hepatomegaly Secondaries or 1° hepatoma. (Nodular cirrhosis typically causes a small, shrunken liver, not an enlarged craggy one.) Pelvic masses Fibroids, fetus, bladder, ovarian cysts or malignancies. Is it truly pelvic?—Yes, if by palpation you cannot get ‘below it’. Investigating lumps Check FBC (with film); CRP; U&E LFT; Ca2+; tumour markers only as appropriate. Imaging by CT or US (transvaginal approach may be useful); MRI also has a role, eg in assessment of liver masses (p286). Others: TB tests (p394). Biopsy to give a tissue diagnosis may be obtained using a fine needle guided by CT, US, or endoscopy. 7 Subtract fluid albumin from serum albumin to obtain serum-ascites albumin gradient (SAAG). Gradient 40% >120bpm >140bpm 30–40/min >35/min 5–15mL/h Negligible Confused Lethargic Crystalloid + blood
Assumes a body mass of 70kg. An adaptation of ‘Estimated blood loss based on initial presentation’ table from the 9th edition of the Advanced Trauma Life Support Manual. Adapted with permission from the American College of Surgeons.
Fig 13.26 Erect CXR showing air beneath the right hemidiaphragm, indicating presence of a pneumoperitoneum. Causes: • Bowel perforation (visible only in 75%) (fig 13.25). • Gas-forming infection, eg C. perfringens. • Iatrogenic, eg laparoscopic surgery (detectable on CXR up to 10d post-op). • Per vaginam (eg sexual activity). • Interposition of bowel between liver and diaphragm (Chilaiditi sign—not true free air). Image courtesy of Mr P. Paraskeva.
Surgery
608
Acute appendicitis Incidence Most common surgical emergency (lifetime incidence = 6%). Can occur at any age, though highest incidence is between 10–20yrs.8 It is rare before age 2 because the appendix is cone shaped with a larger lumen. Pathogenesis Gut organisms invade the appendix wall after lumen obstruction by lymphoid hyperplasia, faecolith, or filarial worms. This leads to oedema, ischaemic necrosis, and perforation. Presentation Classically periumbilical pain that moves to the RIF. Associated signs may include tachycardia, fever, peritonism with guarding and rebound or percussion tenderness in RIF. Pain on right during PR examination suggests an inflammed, lowlying pelvic appendix. Anorexia is an important feature; vomiting is rarely prominent—pain normally precedes vomiting in the surgical abdomen. Constipation is usual, though diarrhoea may occur. Additional signs: Rovsing’s sign (pain > in RIF than LIF when the LIF is pressed). Psoas sign (pain on extending hip if retrocaecal appendix). Cope sign (pain on flexion and internal rotation of right hip if appendix in close relation to obturator internus). Investigations Blood tests may reveal neutrophil leucocytosis and elevated CRP. US may help, but the appendix is not always visualized. CT has high diagnostic accuracy and is useful if diagnosis is unclear: it reduces Ωve appendicectomy rate. Variations in the clinical picture • Inflammation in a retrocecal/retroperitoneal appendix (2.5%) may cause flank or RUQ pain; its only sign may be tenderness on the right on PR. • The child with vague abdominal pain who will not eat their favourite food. • The shocked, confused octogenarian who is not in pain. • Appendicitis occurs in ~1/1000 pregnancies. Mortality is higher, especially from 20wks’ gestation. Perforation is more common, and increases fetal mortality. Pain is often less well localized (may be RUQ) and signs of peritonism less obvious. Hints • If a child is anxious, use their hand to press their tummy. • Check for recent viral illnesses and lymphadenopathy—mesenteric adenitis? • Don’t start palpating in the RIF (makes it difficult to elicit pain elsewhere). • Expect diagnosis to be wrong half the time. If diagnosis is uncertain, re-examine often. A normal appendix is removed in up to 20% of patients. Treatment Prompt appendicectomy (fig 13.27). Antibiotics: piperacillin/tazobactam 4.5g/8h, 1 to 3 doses IV starting 1h pre-op, reduces wound infections. Give a longer course if perforated. Laparoscopy: Has diagnostic and therapeutic advantages (if surgeon experienced), especially in women and the obese. It is not recommended in cases of suspected gangrenous perforation as the rate of abscess formation may be higher. Complications • Perforation is commoner if a faecolith is present and in young children, as the diagnosis is more often delayed. • Appendix mass may result when an inflamed appendix becomes covered with omentum. US/CT may help with diagnosis. Some advocate early surgery. Alternatively, initial conservative management—NBM and antibiotics. If the mass resolves, some perform an interval (ie delayed) appendicectomy. Exclude a colonic tumour (laparotomy or colonoscopy), which can present as early as the 4th decade. • Appendix abscess May result if an appendix mass fails to resolve but enlarges and the patient gets more unwell. Treatment usually involves drainage (surgical or percutaneous under US/CT-guidance). Antibiotics alone may bring resolution.
8 There is a second peak between 60–70yrs; older adults may present later with atypical symptoms.
Explaining the patterns of abdominal pain
609
Internal organs and the visceral peritoneum have no somatic innervation, so the brain attributes the visceral (splanchnic) signals to a physical location whose dermatome corresponds to the same entry level in the spinal cord. Importantly, there is no laterality to the visceral unmyelinated C-fibre pain signals, which enter the cord bilaterally and at multiple levels. Division of the gut according to embryological origin is the important determinant here: see table 13.11. Table 13.11 Somatic referral of abdominal pain Division points Proximal to 2nd part of duodenum Above to ⅔ along transverse colon Distal to above
Somatic referral Epigastrium Periumbilical Suprapubic
Arterial supply Coeliac axis Superior mesenteric Inferior mesenteric
Early inflammation irritates the structure and walls of the appendix, so a colicky pain is referred to the mid-abdomen—classically periumbilical. As the inflammation progresses and irritates the parietal peritoneum (especially on examination), the somatic, lateralized pain settles at McBurney’s point, ⅔ of the way along from the umbilicus to the right anterior superior iliac spine. These principles also help us understand patterns of referred pain. In pneumonia, the T9 dermatome is shared by the lung and the abdomen. Also, irritation of the underside of the diaphragm (sensory innervation is from above through the phrenic nerve, C3–5) by an inflamed gallbladder or a subphrenic abscess refers pain to the right shoulder: dermatomes C3–5.
• Ectopic (do a pregnancy test!) • UTI (test urine!) • Mesenteric adenitis • Cystitis.
• Cholecystitis • Diverticulitis • Salpingitis/PID • Dysmenorrhoea
• Crohn’s disease • Perforated ulcer • Food poisoning • Meckel’s diverticulum
Fig 13.27 Appendicectomy. Reproduced from McLatchie et al., Operative Surgery, 2006, with permission from Oxford University Press.
Surgery
Gut Fore Mid Hind
Surgery
610
Obstruction of the bowel Cardinal features of intestinal obstruction •Vomiting,9 nausea and anorexia.•Colic occurs early ( in long-standing obstruction). •Constipation may be absolute (ie no faeces or flatus passed) in distal obstruction; less pronounced if obstruction is high. •Abdominal distension as the obstruction progresses with active, ‘tinkling’ bowel sounds. The key decisions 1 Is it obstruction of the small or large bowel? In small bowel obstruction, vomiting occurs early, distension is less, and pain is higher in the abdomen; in large bowel obstruction, pain is more constant. The AXR plays a key role (fig 13.28 & p728). 2 Is there an ileus or mechanical obstruction? Ileus is functional obstruction from bowel motility (see BOX ‘Paralytic ileus or pseudo-obstruction?’ & p728). Bowel sounds are absent; pain tends to be less. 3 Is the obstructed bowel simple/closed loop/strangulated? Simple: one obstructing point and no vascular compromise. Closed loop: obstruction at two points (eg sigmoid volvulus) forming a loop of grossly distended bowel at risk of perforation. Strangulated: blood supply is compromised and the patient is iller than you would expect. There is sharper, more constant, and localized pain. Peritonism is the cardinal sign. There may be fever + WCC with other signs of mesenteric ischaemia (p620). Causes See table 13.12. Table 13.12 Causes of bowel obstruction Causes: small bowel Causes: large bowel • Adhesions (p581) • Colon ca (p616) • Hernias (p612) • Constipation (p260)
Rarer causes • Crohn’s stricture • Gallstone ileus (p634) • Intussusception • Diverticular stricture • TB (developing world) • Volvulus • Sigmoid (see BOX ‘Sigmoid • Foreign body volvulus’) • Caecal
Management • General principles: Cause, site, speed of onset, and completeness of obstruction determine definitive therapy: strangulation and large bowel obstruction require surgery; ileus and incomplete small bowel obstruction can be managed conservatively, at least initially. • Immediate action: ‘Drip and suck’—NGT and IV fluids to rehydrate and correct electrolyte imbalance (p668). Being NBM does not give adequate rest for the bowel because it can produce up to 9L of fluid/d. Also: analgesia, blood tests (inc. amylase, FBC, U&E), AXR, erect CXR, catheterize to monitor fluid status. • Further imaging: CT to establish the cause of obstruction (may show dilated, fluidfilled bowel and a transition zone at the site of obstruction—figs 13.29, 13.30). Oral Gastrografin® prior to CT can help identify level of obstruction and may have mild therapeutic action against mechanical obstruction. Consider investigating the cause of large bowel obstruction by colonoscopy but beware risk of perforation. • Surgery: Strangulation needs emergency surgery. Closed loop obstruction may be managed with surgery or endoscopic decompression attempted. Endoscopic stenting may be used for obstructing large bowel malignancies either in palliation or as a bridge to surgery in acute obstruction (p616). Small bowel obstruction secondary to adhesions should rarely lead to surgery—see BOX, p581.
9 Fermentation of the intestinal contents in established obstruction causes ‘faeculent’ vomiting. True ‘faecal’ vomiting is found when there is a colonic fistula with the proximal gut.
Paralytic ileus or pseudo-obstruction?
611
Paralytic ileus is adynamic bowel due to the absence of normal peristaltic contractions. Contributing factors include abdominal surgery, pancreatitis (or any localized peritonitis), spinal injury, hypokalaemia, hyponatraemia, uraemia, peritoneal sepsis and drugs (eg tricyclic antidepressants). Pseudo-obstruction resembles mechanical GI obstruction but with no obstructing lesion. Acute colonic pseudo-obstruction is called Ogilvie’s syndrome (p706), and clinical features are similar to that of mechanical obstruction. Predisposing factors: puerperium; pelvic surgery; trauma; cardiorespiratory and neurological disorders. Treatment: Neostigmine or colonoscopic decompression are sometimes useful. In chronic pseudo-obstruction weight loss from malabsorption is a problem.
Sigmoid volvulus occurs when the bowel twists on its mesentery, which can produce severe, rapid, strangulated obstruction (fig 13.28c). It tends to occur in the elderly, constipated, and comorbid patient, and is managed by insertion of a flatus tube or sigmoidoscopy. Sigmoid colectomy is sometimes required. If not treated successfully, it can progress to perforation and fatal peritonitis. (a)
(b)
(c)
Fig 13.28 (a) Small bowel obstruction: AXR shows central gas shadows with valvulae conniventes that completely cross the lumen and no gas in the large bowel. (b) Large bowel obstruction: AXR shows peripheral gas shadows proximal to the blockage (eg in caecum) but not in the rectum. (c) Sigmoid volvulus: there is a characteristic AXR with an ‘inverted U’ loop of bowel that looks a bit like a coffee bean. Images (a), (b), and (c) reproduced from Darby et al., Oxford Handbook of Medical Imaging, 2011, with permission from Oxford University Press.
Fig 13.29 Unenhanced axial CT of the abdomen showing multiple loops of dilated, fluid-filled small bowel in a patient with small bowel obstruction.
Fig 13.30 Axial CT of the abdomen post-oral con-
Image courtesy of Norwich Radiology Dept.
Image courtesy of Norwich Radiology Dept.
trast showing dilated loops of fluid and air-filled large bowel (contrast medium is in the small bowel).
Surgery
Sigmoid volvulus
Surgery
612
Abdominal hernias Definition The protrusion of a viscus or part of a viscus through a defect of the walls of its containing cavity into an abnormal position. See fig 13.31. Terminology: • Irreducible: contents cannot be pushed back into place (see p614 for technique). • Obstructed: bowel contents cannot pass—features of intestinal obstruction (p610). • Strangulated: ischaemia occurs—the patient requires urgent surgery. • Incarceration: contents of the hernial sac are stuck inside by adhesions. Care must be taken with reduction as it is possible to push an incarcerated hernia back into the abdominal cavity, giving the initial appearance of successful reduction. Inguinal hernia The commonest type in both & (but >>), p614. Femoral hernia Bowel enters the femoral canal, presenting as a mass in the upper medial thigh or above the inguinal ligament where it points down the leg, unlike an inguinal hernia which points to the groin. They occur more often in especially in middle age and the elderly. They are likely to be irreducible and to strangulate due to the rigidity of the canal’s borders. Anatomy:See fig 13..32 Differential diagnosis: (See p651.) 1 Inguinal hernia. 2 Saphena varix. 3 An enlarged Cloquet’s node (p615). 4 Lipoma. 5 Femoral aneurysm. 6 Psoas abscess. Treatment: Surgical repair is recommended. Herniotomy is ligation and excision of the sac, herniorrhaphy is repair of the hernial defect. Paraumbilical hernias occur just above or below the umbilicus. Risk factors are obesity and ascites. Omentum or bowel herniates through the defect. Surgery involves repair of the rectus sheath (Mayo repair). Epigastric hernias pass through linea alba above the umbilicus. Incisional hernias follow breakdown of muscle closure after surgery (11–20%). If obese, repair is not easy. Mesh repair has recurrence but infection over sutures. Spigelian hernias occur through the linea semilunaris at the lateral edge of the rectus sheath, below and lateral to the umbilicus. Lumbar hernias occur through the inferior or superior lumbar triangles in the posterior abdominal wall. Richter’s hernias involve bowel wall only—not the whole lumen. Maydl’s hernias involve a herniating ‘double loop’ of bowel. The strangulated portion may reside as a single loop inside the abdominal cavity. Littré’s hernias are hernial sacs containing strangulated Meckel’s diverticulum. Obturator hernias occur through the obturator canal. Typically there is pain along the medial side of the thigh in a thin woman. Sciatic hernias pass through the lesser sciatic foramen (a way through various pelvic ligaments). GI obstruction + a gluteal mass suggests this rare possibility. Sliding hernias contain a partially extraperitoneal structure (eg caecum on the right, sigmoid colon on the left). The sac does not completely surround the contents. Paediatric hernias include Umbilical hernias: (3% of live births). Are a result of a persistent defect in the transversalis fascia. Surgical repair rarely needed as most resolve by the age of 3. Indirect inguinal hernias (~ 4% of all infants due to patent processus vaginalis—prematurity is a risk factor; uncommon in infants—consider testicular feminization.) Surgical repair is required. Gastroschisis: Protrusion of the abdominal contents through a defect in the anterior abdominal wall to the right of the umbilicus. Prompt surgical repair required. Exomphalos: Abdominal contents are found outside the abdomen, covered in a three-layer membrane consisting of peritoneum, Wharton’s jelly, and amnion. Surgical repair less urgent because the bowel is protected by these membranes.
Surgery
613
Fig 13.31 Some examples of hernias.
Fig 13.32 The boundaries of the femoral canal are anteriorly the inguinal ligament; medially the lacunar ligament (and pubic bone); laterally the femoral vein (and iliopsoas); and posteriorly the pectineal ligament and pectineus. The canal contains fat and Cloquet’s node. The neck of the hernia is felt inferior and lateral to the pubic tubercle (inguinal hernias are superior and medial to this point).
Surgery
614
Inguinal hernias Indirect hernias pass through the internal inguinal ring and, if large, out through the external ring (fig. 13.33). Direct hernias push their way directly forward through the posterior wall of the inguinal canal, into a defect in the abdominal wall (Hesselbach’s triangle; medial to the inferior epigastric vessels and lateral to the rectus abdominus). Predisposing conditions: males (: ≈ 8:1), chronic cough, constipation, urinary obstruction, heavy lifting, ascites, past abdominal surgery (eg damage to the iliohypogastric nerve during appendicectomy). There are two landmarks to identify: the deep (internal) ring may be defined as being the mid-point of the inguinal ligament, ~1½ cm above the femoral pulse (which crosses the mid-inguinal point); the superficial (external) ring is a split in the external oblique aponeurosis just superior and medial to the pubic tubercle (the bony prominence forming the medial attachment of the inguinal ligament). Examination Look for previous scars; feel the other side (more common on the right); examine the external genitalia. Then ask: •Is the lump visible? If so, ask the patient to reduce it—if he cannot, make sure that it is not a scrotal lump. Ask him to cough. Appears above and medial to the pubic tubercle. •If no lump is visible, feel for a cough impulse. •Repeat the examination with the patient standing. Distinguishing direct from indirect hernias: This is loved by examiners but is of little clinical use—not least because repair is the same for both (see ‘Repairs’ later in topic). The best way is to reduce the hernia and occlude the deep (internal) ring with two fingers. Ask the patient to cough or stand—if the hernia is restrained, it is indirect; if not, it is direct. The ‘gold standard’ for determining the type of inguinal hernia is at surgery: direct hernias arise medial to the inferior epigastric vessels; indirect hernias are lateral. Indirect hernias: • Common (80%) • Can strangulate.
Direct hernias: • Less common (20%) • Reduce easily • Rarely strangulate.
Femoral hernias: • More frequent in females • Frequently irreducible • Frequently strangulate.
Irreducible hernias You may be called because a long-standing hernia is now irreducible and painful. It is always worth trying to reduce these yourself to prevent strangulation and necrosis (demanding prompt laparotomy). Learn how to do this from an expert, ie one of your patients who has been reducing his hernia for years. Then you will know how to act correctly when the emergency presents. Notice that such patients use the flat of the hand, directing the hernia from below, up towards the contralateral shoulder. Sometimes, as the hernia obstructs, reduction requires perseverance, which may be rewarded by a gurgle from the retreating bowel and a kiss from the attending spouse who had thought that surgery was inevitable. Repairs Weight loss (if over-weight) and stop smoking pre-op. Warn that hernias may recur and patients should be counselled about possibility of chronic pain postoperatively. Mesh techniques (eg Lichtenstein repair) have replaced older methods. In mesh repairs, a polypropylene mesh reinforces the posterior wall. Recurrence rate is less than with other methods (eg 100 000 per year in the UK). Laparoscopic repair gives similar recurrence rates. Methods include transabdominal pre-peritoneal (TAPP) in which the peritoneum is entered and the hernia repaired, and totally extraperitoneal (TEP), which decreases the risk of visceral injury. For benefits of laparoscopic surgery see p592. Return to work: Will depend upon surgical approach and patient—discuss this preoperatively. Rest for 4wks and convalescence over 8wks with open approaches, but laparoscopic repairs may allow return to manual work (and driving) after ≤2wks if all is well.
Anterior superior iliac spine Deep inguinal ring Indirect inguinal hernia Deep inguinal node
Inferior epigastric vessels
615
Direct inguinal hernia Course of spermatic cord Superficial ring Pubic tubercle Femoral hernia Cloquet’s node
Fig 13.33 Anatomy of the inguinal canal. Floor: Inguinal ligament and lacunar ligament medially; Roof: Fibres of transversalis, internal oblique; Anterior: External oblique aponeurosis + internal oblique for the lateral ⅓ ; Posterior: Laterally, transversalis fascia; medially, conjoint tendon.
The contents of the inguinal canal in the male • The external spermatic fascia (from external oblique), cremasteric fascia (from internal oblique and transverses abdominus), and internal spermatic fascia (from transversalis fascia) covering the cord. • The spermatic cord: • Vas deferens, obliterated processus vaginalis, and lymphatics. • Arteries to the vas, cremaster, and testis. • The pampiniform plexus and the venous equivalent of the above. • The genital branch of the genitofemoral nerve and sympathetic nerves. • The ilioinguinal nerve, which enters the inguinal canal via the anterior wall and runs anteriorly to the cord. NB: in the female the round ligament of the uterus is in place of the male structures. A hydrocele of the canal of Nuck is the female equivalent of a hydrocele of the cord.
Surgery
Nerve Femoral Artery Vein
Surgery
616
Colorectal carcinoma This is the 3rd most common cancer and 2nd most common cause of UK cancer deaths (16 000 deaths/yr). Usually adenocarcinoma. 86% of presentations are in those >60yrs old. Lifetime UK incidence: = 1 : 15; = 1 : 19. Predisposing factors Neoplastic polyps (see BOX & p520); IBD (UC and Crohn’s); genetic predisposition (15yrs old. Spread Local, lymphatic, by blood (liver, lung, bone) or transcoelomic. The TNM system (Tumour, Node, Metastases see table 13.13 and p523) is used to stage disease and is preferred to the older Dukes’ classification (Dukes A: limited to muscularis mucosae; Dukes B: extension through muscularis mucosae; Dukes C: involvement of regional lymph nodes). Surgery aims to cure and may survival times by up to 50%. In elective surgery, anastomosis is typically achieved at the 1st operation. Laparoscopic surgery has revolutionized surgery for colon cancer. It is as safe as open surgery and there is no difference in overall survival or disease recurrence. •Right hemicolectomy for caecal, ascending, or proximal transverse colon tumours. •Left hemicolectomy for tumours in distal transverse or descending colon. •Sigmoid colectomy for sigmoid tumours. •Anterior resection for low sigmoid or high rectal tumours. •Abdomino-perineal (AP) resection for tumours low in the rectum (8cm from anus): permanent colostomy and removal of rectum and anus. •Hartmann’s procedure in emergency bowel obstruction, perforation, or palliation (p582). •Transanal endoscopic microsurgery allows local excision through a wide proctoscope for localized rectal disease. Endoscopic stenting should be considered for palliation in malignant obstruction and as a bridge to surgery in acute obstruction. Stenting need for colostomy, has less complications than emergency surgery, shortens intensive care and total hospital stays, and prevents unnecessary operations. Surgery with liver resection may be curative if single-lobe hepatic metastases and no extrahepatic spread. Radiotherapy is mostly used in palliation for colonic cancer. It is occasionally used pre-op in rectal cancer to allow resection. Post-op radiotherapy is only used in patients with rectal tumours at high risk of local recurrence. Chemotherapy Adjuvant chemotherapy for stage 3 disease has been shown to reduce disease recurrence by 30% and mortality by 25%. Benefits for stage 2 disease are more marginal and warrant an individualized approach. The FOLFOX regimen has become standard (fluorouracil, folinic acid and oxaliplatin). Chemotherapy is also used in palliation of metastatic disease. Biological therapies: Bevacizumab (antiVEGF antibody) improves survival when added to combination therapy in advanced disease. Cetuximab and panitumumab (anti-EGFR agents) improve response rate and survival in KRAS wild-type metastatic colorectal cancer. Prognosis Survival is dependent on age and stage; for stage 1 disease, 5yr survival is ~75% but this drops to just 5% with diagnosis at stage 4, hence the imperative for effective screening (BOX).
Right side
Left side Splenic flexure 2%
617
Hepatic flexure 3%
Ascending colon 7%
Transverse colon 5%
Descending colon 3%
Caecum 14% Sigmoid colon 20%
Appendix 1%
Rectum 27% Anus 2%
more proximal neoplasms. White men tend to have more distal neoplasms. TNM staging in colorectal cancer
Table 13.13 Colorectal cancer: TNM staging Tx Primary tumour cannot be assessed Tis Carcinoma in situ T1 Invading submucosa T2 Invading muscularis propria T3 Invading subserosa and beyond (not other organs) T4 Invasion of adjacent structures
M0
Nodes cannot be assessed No node spread Metastases in 1–3 regional nodes Metastases in >3 regional nodes No distant spread
M1
Distant metastasis
Nx N0 N1 N2
Reproduced with permission from Edge, SB et al. (Eds.), AJCC Cancer Staging Manual, 7th Edition. New York: Springer; 2010.
TNM status used to define overall stage. This is complex with several important subtypes, but in essence, stage 1 disease is T1 or T2/N0/M0; stage 2 is T3 or T4/N0/M0; stage 3 is characterized by N1 or N2 but still M0; stage 4 is M1.
Polyps, the challenges of screening, and the NHS Polyps are growths that appear above the mucosa and can be inflammatory, hamartomatous, or neoplastic. Left in situ, polyps carry a risk of malignant transformation that will relate to size and histology (tubular or villous adenomas, esp. if >2cm). Patients with polyps may have no symptoms and thus a colonoscopy is required to detect and remove. Colonscopy allows the opportunity to detect colorectal cancer at an earlier stage when treatment may be more effective. However, population-based colonoscopic screening is costly and some studies have suggested that the test does not impact on deaths from right-sided cancers which are rarer and harder to detect (fig 13.34). Therefore, the NHS has introduced a one-off screening flexible sigmoidoscopy offered to all people in their 55th year. Trial results have shown the incidence of colorectal cancer in the intervention (screening) group is reduced by 33% and mortality from colorectal cancer is reduced by 43%. Number needed to screen to prevent one diagnosis (=191); or death (=489). In parallel, the NHS Bowel Cancer Screening Programme (introduced in 2006) offers colonoscopy to all men and women aged 60–75 who test positive for faecal occult blood (FOB) using a home testing kit performed every 2 years. This FOBstratification targets screening to those in the highest risk groups, permitting detection of more advanced adenomas and early stage cancers. The relative risk of death from colorectal cancer in patients undergoing screening is reduced by 16%. A 11% increase in incidence rates since 2006 for people aged 60–69 is almost certainly due to earlier detection through the screening programme.
Surgery
Other and unspecified 9%
Fig 13.34 Distribution of colorectal carcinomas. These are averages: black females tend to have
Surgery
618
Carcinoma of the oesophagus Incidence Australia 1000U/mL or around 3-fold upper limit of normal). The degree of elevation is not related to severity of disease. Amylase may be normal even in severe pancreatitis (levels starts to fall within 24–48h). It is excreted renally so renal failure will levels. Cholecystitis, mesenteric infarction, and GI perforation can cause lesser rises. Serum lipase is more sensitive and specific for pancreatitis (especially when related to alcohol), and rises earlier and falls later. ABG to monitor oxygenation and acid–base status. AXR: No psoas shadow (retroperitoneal fluid), ‘sentinel loop’ of proximal jejunum from ileus (solitary air-filled dilatation). Erect CXR helps exclude other causes (eg perforation). CT is the standard choice of imaging to assess severity and for complications. US (if gallstones + AST). ERCP if LFTS worsen. CRP >150mg/L at 36h after admission is a predictor of severe pancreatitis. Management Severity assessment is essential (see BOX and table 13.17). • Nil by mouth, consider NJ feeding (decrease pancreatic stimulation). Set up IVI and give lots of crystalloid, to counter third-space sequestration, until vital signs are satisfactory and urine flow stays at >30mL/h. Insert a urinary catheter and consider CVP monitoring. • Analgesia: pethidine 75–100mg/4h IM, or morphine (may cause Oddi’s sphincter to contract more, but it is a better analgesic and not contraindicated). • Hourly pulse, BP, and urine output; daily FBC, U&E, Ca2+, glucose, amylase, ABG. • If worsening: ITU, O2 if PaO2. In suspected abscess formation or pancreatic necrosis (on CT), consider parenteral nutrition ± laparotomy & debridement (‘necrosectomy’). Antibiotics may help in severe disease. • ERCP + gallstone removal may be needed if there is progressive jaundice. • Repeat imaging (usually CT) is performed in order to monitor progress. Any acute abdomen (p606), myocardial infarct. Early complications Shock, ARDS (p186), renal failure (give lots of fluid!), DIC, sepsis, Ca2+, glucose (transient; 5% need insulin). Late complications (>1wk.) Pancreatic necrosis and pseudocyst (fluid in lesser sac, fig 13.45), with fever, a mass ± persistent amylase/LFT; may resolve or need drainage. Abscesses need draining. Bleeding from elastase eroding a major vessel (eg splenic artery); embolization may be life-saving. Thrombosis may occur in the splenic/gastroduodenal arteries, or colic branches of the SMA, causing bowel necrosis. Fistulae normally close spontaneously. If purely pancreatic they do not irritate the skin. Some patients suffer recurrent oedematous pancreatitis so often that near-total pancreatectomy is contemplated. It can all be a miserable course.
Modified Glasgow criteria for predicting severity of pancreatitis
637
Three or more positive factors detected within 48h of onset suggest severe pancreatitis, and should prompt transfer to ITU/HDU. Mnemonic: PANCREAS. PaO2 Age Neutrophilia Calcium Renal function Enzymes Albumin Sugar
55yrs WBC >15 x 109/L 16mmol/L LDH >600iu/L; AST >200iu/L 10mmol/L
Republished with permission of Royal College of Surgeons of England, from Annals of the Royal College of Surgeons of England, Moore E M, 82, 16–17, 2002. Permission conveyed through Copyright Clearance Center, Inc.
These criteria have been validated for pancreatitis caused by gallstones and alcohol; Ranson’s criteria are valid for alcohol-induced pancreatitis, and can only be fully applied after 48h, which does have its disadvantages. Other criteria for assessing severity include the Acute Physiology and Chronic Health Examination (APACHE)-II, and the Bedside Index for Severity in Acute Pancreatitis (BISAP).
Fig 13.45 Axial CT of the abdomen (with IV and PO contrast media) showing a pancreatic pseudocyst occupying the lesser sac of the abdomen posterior to the stomach. It is called a ‘pseudocyst’ because it is not a true cyst, rather a collection of fluid in the lesser sac (ie not lined by epi/endothelium). It develops at ≥6wks. The cyst fluid is of low attenuation compared with the stomach contents because it has not been enhanced by the contrast media. Image courtesy of Dr Stephen Golding.
Surgery
Table 13.17
Surgery
638
Urinary tract calculi (nephrolithiasis) Renal stones (calculi) consist of crystal aggregates. Stones form in collecting ducts and may be deposited anywhere from the renal pelvis to the urethra, though classically at: 1 Pelviureteric junction 2 Pelvic brim 3 Vesicoureteric junction. Prevalence Common: lifetime incidence up to 15%. Peak age: 20–40yr. : ≈ 3:1. Types •Calcium oxalate (75%). •Magnesium ammonium phosphate (struvite/triple phosphate; 15%). •Also: urate (5%), hydroxyapatite (5%), brushite, cystine (1%), mixed. Presentation Asymptomatic or: 1 Pain: Excruciating spasms of renal colic ‘loin to groin’ (or genitals/inner thigh), with nausea/vomiting. Often cannot lie still (differentiates from peritonitis). Obstruction of kidney: felt in the loin, between rib 12 and lateral edge of lumbar muscles (like intercostal nerve irritation pain; the latter is not colicky, and is worsened by specific movements/pressure on a trigger spot). Obstruction of mid-ureter: may mimic appendicitis/diverticulitis. Obstruction of lower ureter: may lead to symptoms of bladder irritability and pain in scrotum, penile tip, or labia majora. Obstruction in bladder or urethra: causes pelvic pain, dysuria, strangury (desire but inability to void) ± interrupted flow. 2 Infection: Can coexist (risk if voiding impaired), eg UTI; pyelonephritis (fever, rigors, loin pain, nausea, vomiting); pyonephrosis (infected hydronephrosis) 3 Haematuria. 4 Proteinuria. 5 Sterile pyuria. 6 Anuria. Examination Usually no tenderness on palpation. May be renal angle tenderness especially to percussion if there is retroperitoneal inflammation. Tests FBC, U&E, Ca2+, PO43Ω, glucose, bicarbonate, urate. Urine dipstick: Usually +ve for blood (90%). MSU: MC&S. Further tests for cause: Urine pH; 24h urine for: calcium, oxalate, urate, citrate, sodium, creatinine; stone biochemistry (sieve urine & send stone). Imaging: Non-contrast CT is investigation of choice for imaging stones (99% visible) & helps exclude differential causes of an acute abdomen. A ruptured abdominal aortic aneurysm may present similarly. 80% of stones are visible on KUB XR (kidneys + ureters + bladder). Look along ureters for calcification over the transverse processes of the vertebral bodies. US an alternative for hydronephrosis or hydroureter. Initially: Analgesia, eg diclofenac 75mg IV/IM, or 100mg PR. (If CI: opioids) + IV fluids if unable to tolerate PO; antibiotics (eg piperacillin/tazobactam 4.5g/8h IV, or gentamicin) if infection. Stones 5mm/pain not resolving: Medical expulsive therapy: start at presentation; nifedipine 10mg/8h PO or -blockers (tamsulosin 0.4mg/d) promote expulsion and reduce analgesia requirements. Most pass within 48h (>80% after ~30d). If not, try extracorporeal shockwave lithotripsy (ESWL) (if 6 (eg with potassium citrate or sodium bicarbonate). •Cystine: vigorous hydration to keep urine output >3L/d and urinary alkalinization (as above-mentioned). Penicillamine is used to chelate cystine, given with pyridoxine to prevent vitamin B6 deficiency.
Questions to address when confronted by a stone
639
What is its composition? (See table 13.18.) Table 13.18 Types, causes, and X-ray appearance of renal stones
Magnesium ammonium phosphate (fig 13.47)
Urate (p680) Cystine (fig 13.48)
Appearance on X-ray Spiky, radio-opaque Smooth, may be large, radioopaque UTI (proteus causes alLarge, horny, ‘staghorn’, radiokaline urine and calcium opaque precipitation and ammonium salt formation) Hyperuricaemia Smooth, brown, radiolucent Renal tubular defect Yellow, crystalline, semi-opaque
Why has he or she got this stone now? • Diet: chocolate, tea, rhubarb, strawberries, nuts, and spinach all oxalate levels. • Season: variations in calcium and oxalate levels are thought to be mediated by vitamin D synthesis via sunlight on skin. • Work: can he/she drink freely at work? Is there dehydration? • Medications: precipitating drugs include: diuretics, antacids, acetazolamide, corticosteroids, theophylline, aspirin, allopurinol, vitamin C and D, indinavir. Are there any predisposing factors? For example: • Recurrent UTIS (in magnesium ammonium phosphate calculi). • Metabolic abnormalities: • Hypercalciuria/hypercalcaemia (p676): hyperparathyroidism, neoplasia, sarcoidosis, hyperthyroidism, Addison’s, Cushing’s, lithium, vitamin D excess. • Hyperuricosuria/plasma urate: on its own, or with gout. • Hyperoxaluria. • Cystinuria (p321). • Renal tubular acidosis (pp316–7). • Urinary tract abnormalities: eg pelviureteric junction obstruction, hydronephrosis (renal pelvis or calyces), calyceal diverticulum, horseshoe kidney, ureterocele, vesicoureteric reflux, ureteral stricture, medullary sponge kidney.11 • Foreign bodies: eg stents, catheters. Is there a family history? Risk of stones 3-fold. Specific diseases include X-linked nephrolithiasis and Dent’s disease (proteinuria, hypercalciuria, and nephrocalcinosis). Is there infection above the stone? Eg fever, loin tender, pyuria? This needs urgent intervention.
Fig 13.46 Calcium oxalate
Fig 13.47 Struvite stone.
monohydrate.
Image courtesy of Dr Glen Austin.
Fig 13.48 Cystine stone. Image courtesy of Dr Glen Austin.
Image courtesy of Dr Glen Austin.
11 Medullary sponge kidney is a typically asymptomatic developmental anomaly of the kidney mostly seen in adult females, where there is dilatation of the collecting ducts, which if severe leads to a sponge-like appearance of the renal medulla. Complications/associations: UTIs, nephrolithiasis, haematuria and hypercalciuria, hyperparathyroidism (if present, look for genetic markers of MEN type 2A, see p223).
Surgery
Type Causative factors Calcium oxalate (fig 13.46) Metabolic or idiopathic Calcium phosphate Metabolic or idiopathic
Surgery
640
Urinary tract obstruction Urinary tract obstruction is common and should be considered in any patient with impaired renal function. Damage can be permanent if the obstruction is not treated promptly. Obstruction may occur anywhere from the renal calyces to the urethral meatus, and may be partial or complete, unilateral or bilateral. Obstructing lesions are luminal (stones, blood clot, sloughed papilla, tumour: renal, ureteric, or bladder), mural (eg congenital or acquired stricture, neuromuscular dysfunction, schistosomiasis), or extra-mural (abdominal or pelvic mass/tumour, retroperitoneal fibrosis, or iatrogenic—eg post surgery). Unilateral obstruction may be clinically silent (normal urine output and U&E) if the other kidney is functioning. Bilateral obstruction or obstruction with infection requires urgent treatment. See p641. Clinical features • Acute upper tract obstruction: Loin pain radiating to the groin. There may be superimposed infection ± loin tenderness, or an enlarged kidney. • Chronic upper tract obstruction: Flank pain, renal failure, superimposed infection. Polyuria may occur due to impaired urinary concentration. • Acute lower tract obstruction: Acute urinary retention typically presents with severe suprapubic pain ± acute confusion (elderly); often acute on chronic (hence preceded by chronic symptoms , see next bullet point). Clinically: distended, palpable bladder containing ~600mL, dull to percussion. Causes include prostatic obstruction (usual cause in older ), urethral strictures, anticholinergics, blood clots eg from bladder lesion (‘clot retention’), alcohol, constipation, post-op (pain/inflammation/ anaesthetics), infection (p296), neurological (cauda equina syndrome, see p466). • Chronic lower tract obstruction: Symptoms: urinary frequency, hesitancy, poor stream, terminal dribbling, overflow incontinence. Signs: distended, palpable bladder (capacity may be >1.5L) ± large prostate on PR. Complications: UTI, urinary retention, renal failure (eg bilateral obstructive uropathy—see BOX ‘Obstructive uropathy’). Causes include prostatic enlargement (common); pelvic malignancy; rectal surgery; DM; CNS disease, eg transverse myelitis/MS; zoster (S2–S4). Tests Blood: U&E, creatinine, FBC, and prostate-specific antigen (PSA, p530).12 Urine: Dipstick and MC&S. Ultrasound (p744) is the imaging modality of choice for investigating upper tract obstruction: If there is hydronephrosis or hydroureter (distension of the renal pelvis and calyces or ureter), arrange a CT scan. This will determine the level of obstuction. NB: in ~5% of cases of obstruction, no distension is seen on US. Radionuclide imaging enables functional assessment of the kidneys. Treatment Upper tract obstruction: Nephrostomy or ureteric stent. NB: stents may cause significant discomfort and patients should be warned of this and other risks (see BOX ‘Problems of ureteric stenting’). -blockers help reduce stent-related pain (ureteric spasm). Pyeloplasty, to widen the PUJ, may be performed for idiopathic PUJ obstruction. Lower tract obstruction: Insert a urethral or suprapubic catheter (p762) to relieve acute retention. In chronic obstruction only catheterize patient if there is pain, urinary infection, or renal impairment; intermittent self-catheterization is sometimes required (p763). If in clot retention the patient will require a 3-way catheter and bladder washout. If >1L residual check U&E and monitor for post-obstructive diuresis (see BOX ‘Obstructive uropathy’). Monitor weight, fluid balance, and U&E closely. Treat the underlying cause if possible, eg if prostatic obstruction, start an -blocker (see p642). After 2–3 days, trial without catheter (TWOC, p763) may work (especially if 10yrs. Screening: DRE of prostate; transrectal US; PSA (see BOX ‘Advice to asymptomatic men’). Penile cancer Epidemiology: Rare in UK, more common in Far East and Africa, very rare in circumcised. Related to chronic irritation, viruses, smegma. Presentation: Chronic fungating ulcer, bloody/purulent discharge, 50% spread to lymph at presentation : Radiotherapy & irridium wires if early; amputation & lymph node dissection if late.
• Many men over 50 consider a PSA test to detect prostatic cancer. Is this wise? • The test is not very accurate, and we cannot say that those having the test will live longer—even if they turn out to have prostate cancer. Most men with prostate cancer die from an unrelated cause. • If the test is falsely positive, you may needlessly have more tests, eg prostate sampling via the back passage (causes bleeding and infection in 1–5% of men). • Only one in three of those with a high PSA level will have cancer. • You may be worried needlessly if later tests put you in the clear. • If a cancer is found, there’s no way to tell for sure if it will impinge on health. You might end up having a bad effect from treatment that wasn’t needed. • There is much uncertainty on treating those who do turn out to have prostate cancer: options are radical surgery to remove the prostate (risks erectile dysfunction and incontinence), radiotherapy, or hormones. • Screening via PSA has shown conflicting results. Some RCTS have shown no difference in the rate of death from prostate cancer, others have found reduced mortality, eg 1 death prevented per 1055 men invited for screening (if 37 cancers detected). Ultimately, you must decide for yourself what you want.
Prognostic factors in prostate cancer A number of prognostic factors help determine if ‘watchful waiting’ or aggressive therapy should be advised: •Pre-treatment PSA level. •Tumour stage (as measured by the TNM system; p523). •Tumour grade—Gleason score. Gleason grading is from 1 to 5, with 5 being the highest grade, and carrying the poorest prognosis. Gleason grades are decided by analysing histology from two separate areas of tumour specimen, and adding them to get the total Gleason score for the tumour, from 2 to 10. Scores 8–10 suggest an aggressive tumour; 5–7: intermediate; 2–4: indolent.
Benign diseases of the penis Balanitis Acute inflammation of the foreskin and glans. Associated with strep and staph infections. More common in diabetics. Often seen in young children with tight foreskins : Antibiotics, circumcision, hygiene advice. Phimosis The foreskin occludes the meatus. In young boys this causes recurrent balanitis and ballooning, but time (+ trials of gentle retraction) may obviate the need for circumcision. In adulthood presents with painful intercourse, infection, ulceration, and is associated with balanitis xerotica obliterans. Paraphimosis Occurs when a tight foreskin is retracted and becomes irreplaceable, preventing venous return leading to oedema and even ischaemia of the glans. Can occur if the foreskin is not replaced after catheterization. : Ask patient to squeeze glans. Try applying a 50% glucose-soaked swab (oedema may follow osmotic gradient). Ice packs and lidocaine gel may also help. May require aspiration/dorsal slit/circumcision.
Prostatitis May be acute or chronic. Usually those >35yrs. Acute prostatitis is caused mostly by S. faecalis and E. coli, also Chlamydia (and previously TB). Features: UTIs, retention, pain, haematospermia, swollen/boggy prostate on DRE. : Analgesia; levofloxacin 500mg/24h PO for 28d. Chronic prostatitis may be bacterial or non-bacterial. Symptoms as for acute prostatitis, but present for >3 months. Non-bacterial chronic prostatitis does not respond to antibiotics. Anti-inflammatory drugs, –blockers, and prostatic massage all have a place.
645
Surgery
Advice to asymptomatic men asking for a PSA blood test
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Bladder tumours >90% are transitional cell carcinomas (TCCS) in the UK. Adenocarcinomas and squamous cell carcinomas are rare in the West (the latter may follow schistosomiasis). UK incidence ≈ 1:6000/yr. : ≈ 5:2. Histology is important for prognosis: Grade 1— differentiated; Grade 2—intermediate; Grade 3—poorly differentiated. 80% are confined to bladder mucosa, and only ~20% penetrate muscle (increasing mortality to 50% at 5yrs). Presentation Painless haematuria; recurrent UTIs; voiding irritability. Associations Smoking; aromatic amines (rubber industry); chronic cystitis; schistosomiasis (risk of squamous cell carcinoma); pelvic irradiation. Tests • Cystoscopy with biopsy is diagnostic. • Urine: microscopy/cytology (cancers may cause sterile pyuria). • CT urogram is both diagnostic and provides staging. • Bimanual EUA helps assess spread. • MRI or lymphangiography may show involved pelvic nodes. Staging See table 13.19. Treating TCC of the bladder • Tis/Ta/T1: (80% of all patients) Diathermy via transurethral cystoscopy/transurethral resection of bladder tumour (TURBT). Consider a regimen of intravesical BCG (which stimulates a non-specific immune response) for multiple small tumours or high-grade tumours. Alternative chemotherapeutic agents include mitomycin, epirubicin and gemcitabine. 5yr survival ≈ 95%. • T2–3: Radical cystectomy is the ‘gold standard’. Radiotherapy gives worse 5yr survival rates than surgery, but preserves the bladder. ‘Salvage’ cystectomy can be performed if radiotherapy fails, but yields worse results than primary surgery. Post-op chemotherapy (eg M-VAC: methotrexate, vinblastine, doxorubicin, and cisplatin) is toxic but effective. Neoadjuvant chemotherapy with M-VAC or GC (gemcitabine and cisplatin) has improved survival compared to cystectomy or radiotherapy alone. Methods to preserve the bladder with transurethral resection or partial cystectomy + systemic chemotherapy have been tried, but long-term results are disappointing. If the bladder neck is not involved, orthotopic reconstruction rather than forming a urostoma is an option (both using ~40cm of the patient’s ileum), but adequate tumour clearance must not be compromised. The patient should have all these options explained by a urologist and an oncologist. • T4: Usually palliative chemo/radiotherapy. Chronic catheterization and urinary diversions may help to relieve pain. Follow-up History, examination, and regular cystoscopy: •High-risk tumours: Every 3 months for 2yrs, then every 6 months. •Low-risk tumours: First follow-up cystoscopy after 9 months, then yearly. Tumour spread Local to pelvic structures; lymphatic to iliac and para-aortic nodes; haematogenous to liver and lungs. Survival This depends on age at surgery. For example, the 3yr survival after cystectomy for T2 and T3 tumours is 60% if 65–75yrs old, falling to 40% if 75–82yrs old (operative mortality is 4%). With unilateral pelvic node involvement, only 6% of patients survive 5yrs. The 3yr survival with bilateral or para-aortic node involvement is nil. Complications Cystectomy can result in sexual and urinary malfunction. Massive bladder haemorrhage may complicate treatment or be a feature of disease treated palliatively. Determining the cause of bleeding is key. Consider alum solution bladder irrigation (if no renal failure) as 1st-line treatment for intractable haematuria in advanced malignancy: it is an inpatient procedure.
Reproduced with permission from Edge, SB et al. (Eds.), AJCC Cancer Staging Manual, 7th Edition. New York: Springer; 2010.
Is asymptomatic non-visible haematuria significant? Dipstick tests are often done routinely for patients on admission. If non-visible (previously microscopic) haematuria is found, but the patient has no related symptoms, what does this mean? Before rushing into a barrage of investigations, consider: • One study found that incidence of urogenital disease (eg bladder cancer) was no higher in those with asymptomatic microhaematuria than in those without. • Asymptomatic non-visible haematuria is the sole presenting feature in only 4% of bladder cancers, and there is no evidence that these are less advanced than malignancies presenting with macroscopic haematuria. • When monitoring those with treated bladder cancer for recurrence, non-visiblehaematuria tests have a sensitivity of only 31% in those with superficial bladder malignancy, in whom detection would be most useful. • Although 80% of those with flank pain due to a renal stone have microscopic haematuria, so do 50% of those with flank pain but no stone. The conclusion is not that urine dipstick testing is useless, but that results should not be interpreted in isolation. Unexplained non-visible haematuria in those >50yrs should be referred under the 2-week rule. Smokers and those with +ve family history for urothelial cancer may also be investigated differently from those with no risk factors. It is worth considering, that in a young, fit athlete, the diagnosis is more likely to be exercise-induced haematuria. Wise doctors work collaboratively with their patients. ‘Shall we let sleeping dogs lie?’ is a reasonable question for some patients.
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Table 13.19 TNM staging of bladder cancer (See also p523) Tis Carcinoma in situ Ta Tumour confined to epithelium Tumour in submucosa or lamina propria T1 Invades muscle T2 Extends into perivesical fat T3 Invades adjacent organs T4 No LN involved N0 N1–N3 Progressive LN involvement No metastases M0 Distant metastasis M1
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Urinary incontinence Think twice before inserting a urinary catheter. Carry out rectal examination to exclude faecal impaction. Is the bladder palpable after voiding (retention with overflow)? Is there neurological comorbidity: eg MS; Parkinson’s disease; stroke; spinal trauma? Incontinence in men Enlargement of the prostate is the major cause of incontinence: urge incontinence (see later in topic) or dribbling may result from partial retention of urine. TURP (p642) & other pelvic surgery may weaken the bladder sphincter and cause incontinence. Troublesome incontinence needs specialist assessment. Incontinence in women Often under-reported with delays before seeking help. • Functional incontinence: Ie when physiological factors are relatively unimportant. The patient is ‘caught short’ and too slow in finding the toilet because of (for example) immobility, or unfamiliar surroundings. • Stress incontinence: Leakage from an incompetent sphincter, eg when intraabdominal pressure rises (eg coughing, laughing). Increasing age and obesity are risk factors. The key to diagnosis is the loss of small (but often frequent) amounts of urine when coughing etc. Examine for pelvic floor weakness/prolapse/pelvic masses. Look for cough leak on standing and with full bladder. Stress incontinence is common in pregnancy and following birth. It occurs to some degree in ~50% of post-menopausal women. In elderly women, pelvic floor weakness, eg with uterine prolapse or urethrocele (OHCS p290), is a very common association. • Urge incontinence/overactive bladder syndrome: The urge to urinate is quickly followed by uncontrollable and sometimes complete emptying of the bladder as the detrusor muscle contracts. Urgency/leaking is precipitated by: arriving home (latchkey incontinence, a conditioned reflex); cold; the sound of running water; caffeine; and obesity. : urodynamic studies. Cause: detrusor overactivity (see table 13.20), eg from central inhibitory pathway malfunction or sensitization of peripheral afferent terminals in the bladder; or a bladder muscle problem. Check for organic brain damage (eg stroke; Parkinson’s; dementia). Other causes: urinary infection; diabetes; diuretics; atrophic vaginitis; urethritis. In both sexes incontinence may result from confusion or sedation. Occasionally it may be purposeful (eg preventing admission to an old people’s home) or due to anger. Management Effective treatment can have a huge impact on quality of life. Check for: UTI; DM; diuretic use; faecal impaction; palpable bladder; GFR. • Stress incontinence: Pelvic floor exercises are 1st line (8 contractions ≈3/d for 3 months). Intravaginal electrical stimulation may also be effective, but is not acceptable to many women. A ring pessary may help uterine prolapse, eg while awaiting surgical repair. Surgical options (eg tension-free vaginal tape) aim to stabilize the mid-urethra. Urethral bulking also available. Medical options: duloxetine 40mg/12h PO (50% have ≥50% in incontinence episodes). SE = nausea. • Urge incontinence: The patient (or carer) should complete an ‘incontinence’ chart for 3d to define the pattern of incontinence. Examine for spinal cord and CNS signs (including cognitive test, p64); and for vaginitis (if postmenopausal). Vaginitis can be treated with topical oestrogen therapy for a limited period. Bladder training (may include pelvic floor exercises) and weight loss are important. Drugs may help reduce night-time incontinence (see BOX) but can be disappointing. Consider aids, eg absorbent pad. If consider a condom catheter. Do urodynamic assessment (cystometry & urine flow rate measurement) before any surgical intervention to exclude detrusor overactivity or sphincter dyssynergia.
NB: desmopressin nasal spray 20mcg nocte reduces urine production and nocturia in overactive bladder. Unsuitable if elderly (SE: fluid retention, heart failure, Na+). Mind over bladder: •Void when you DON’T have urge; DON’T go to the bathroom when you do have urge. •Gradually extend the time between voiding. •Schedule your trips to toilet. •Stretch your bladder to normal capacity. • When urge comes, calm down and make it go using mind over bladder tricks.
Not all male urinary symptoms are prostate-related! Detrusor overactivity Men get this as well as women. Pressure-flow studies help diagnose this (as does detrusor thickness ≥2.9mm on US). Primary bladder neck obstruction A condition in which the bladder neck does not open properly during voiding. Studies in men and women with voiding dysfunction show that it is common. The cause may be muscular or neurological dysfunction or fibrosis. Diagnosis: Video-urodynamics, with simultaneous pressure-flow measurement, and visualization of the bladder neck during voiding. Treatment: Watchful waiting; -blockers (p642); surgery. Urethral stricture This may follow trauma or infection (eg gonorrhoea)—and frequently leads to voiding symptoms, UTI, or retention. Malignancy is a rare cause. Imaging: Retrograde urethrogram or antegrade cystourethrogram if the patient has an existing suprapubic catheter. Internal urethrotomy involves incising the stricture transurethrally using endoscopic equipment—to release scar tissue. Stents incorporate themselves into the wall of the urethra and keep the lumen open. They work best for short strictures in the bulbar urethra (anterior urethral anatomy, from proximal to distal: prostatic urethraposterior or membranous urethrabulbar urethrapenile or pendulous urethrafossa navicularismeatus).
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Table 13.20 Managing detrusor overactivity in urge incontinence Agents for detrusor overactivity Notes Improves frequency & urgency. Alternatives: Antimuscarinics: eg tolterodine SR 4mg/24h; SE: dry mouth, eyes/skin, solifenacin 5mg/24h (max 10mg); oxybutynin, drowsiness, constipation, tachycardia, but more SE unless transdermal route or modified release used; trospium or fesoterodine abdominal pain, urinary retention, (prefers M3 receptors). Avoid in myasthenia, sinusitis, oedema, weight, glaucoma and if glaucoma or UC are uncontrolled. precipitation. Up to 4mg/12h may be needed (unlicensed). Topical oestrogens Post-menopausal urgency, frequency + nocturia may occasionally be improved by raising the bladder’s sensory threshold. Systemic therapy worsens incontinence. 3 adrenergic agonist: mirabegron Consider if antimuscarinics are contrain50mg/24h; SE tachycardia; CI: severe HTN; dicated or clinically ineffective, or if SE Caution if renal/hepatic impairment. unacceptable. Intravesical botulinum toxin (Botox®) Consider if above medications ineffective. Percutaneous posterior tibial nerve Consider if drug treatment ineffective and stimulation (PTNS). (A typical treatment Botox® not wanted. PTNS delivers neuromoduconsists of ≈12 weekly 30 min sessions.) lation to the S2–S4 junction of the sacral nerve plexus. Neuromodulation via transcutaneous Sacral nerve stimulation inhibits the reflex electrical stimulation behaviour of involuntary detrusor contractions. Modulation of afferent input from Gabapentin (unlicensed). bladder Hypnosis, psychotherapy, bladder (These all require good motivation.) training Surgery (eg clam ileocystoplasty) Reserved for troublesome or intractable symptoms. The bladder is bisected, opened like a clam, and 25cm of ileum is sewn in.
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Lumps in the groin and scrotum Testicular lump = cancer until proved otherwise. Acute, tender enlargement of testis = torsion (p652) until proved otherwise. Diagnosing scrotal masses (fig 13.51) 1 Can you get above it? 2 Is it separate from the testis? 3 Cystic or solid? • Cannot get above: inguinoscrotal hernia (p614) or hydrocele extending proximally. • Separate and cystic: epididymal cyst. • Separate and solid: epididymitis/varicocele. • Testicular and cystic: hydrocele. Testicular and solid—tumour, haematocele, granuloma (p196), orchitis, gumma (p412). US may help. Epididymal cysts Usually develop in adulthood and contain clear or milky (spermatocele) fluid. They lie above and behind the testis. Remove if symptomatic. Hydroceles (Fluid within the tunica vaginalis.) Primary (associated with a patent processus vaginalis, which typically resolves during the 1st year of life) or secondary to testis tumour/trauma/infection. Primary hydroceles are more common, larger, and usually in younger men. Can resolve spontaneously. : Aspiration (may need repeating) or surgery: plicating the tunica vaginalis (Lord’s repair)/inverting the sac (Jaboulay’s repair). Is the testis normal after aspiration? If any doubt, do US. Epididymo-orchitis Causes: Chlamydia (eg if 3IU/mL)13 and -hCG are useful tumour markers and help monitor treatment (p531); check before & during . : Radical orchidectomy (inguinal incision; occlude the spermatic cord before mobilization to risk of intra-operative spread). Options are constantly updated (surgery, radiotherapy, chemotherapy). Seminomas are exquisitely radiosensitive. Stage 1 seminomas: orchidectomy + radiotherapy cures ~95%. Do close followup to detect relapse. Cure of NSGCT, even if metastases are present, is achieved by 3 cycles of bleomycin + etoposide + cisplatin. 5yr survival >90% in all groups. Encourage regular self-examination (prevents late presentation).
13 FP is not in pure seminoma; may also be in: hepatitis, cirrhosis, liver cancer, open neural tube defect.
Diagnosing groin lumps: lateral to medial thinking
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• Psoas abscess—may present with back pain, limp, and swinging pyrexia. • Neuroma of the femoral nerve. • Femoral artery aneurysm. • Saphena varix—like a hernia, it has a cough impulse. • Lymph node. • Femoral hernia. • Inguinal hernia. • Hydrocele or varicocele. • Also consider an undescended testis (cryptorchidism).
Fig 13.51 Diagnosis of scrotal masses. (=transilluminates: position of pen torch shown in image).
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Torsion of the testis The aim is to recognize this condition before the cardinal signs and symptoms are fully manifest, as prompt surgery saves testes. If surgery is performed in 24h it is 0–10%. If in any doubt, surgery is required. If suspected refer immediately to urology. Symptoms: Sudden onset of pain in one testis, which makes walking uncomfortable. Pain in the abdomen, nausea, and vomiting are common. Signs: Inflammation of one testis—it is very tender, hot, and swollen. The testis may lie high and transversely. Torsion may occur at any age but is most common at 11–30yrs. With intermittent torsion the pain may have passed on presentation, but if it was severe, and the lie is horizontal, prophylactic fixing may be wise. : The main one is epididymo-orchitis (p650) but with this the patient tends to be older, there may be symptoms of urinary infection, and more gradual onset of pain. Also consider tumour, trauma, and an acute hydrocele. NB: torsion of testicular or epididymal appendage (the hydatid of Morgagni—a remnant of the Müllerian duct)—usually occurs between 7–12yrs, and causes less pain. Its tiny blue nodule may be discernible under the scrotum. It is thought to be due to the surge in gonadotrophins which signal the onset of puberty. Idiopathic scrotal oedema is a benign condition usually between ages 2 and 10yrs, and is differentiated from torsion by the absence of pain and tenderness. Tests: Doppler US may demonstrate lack of blood flow to testis. Only perform if diagnosis equivocal—do not delay surgical exploration. Treatment: Ask consent for possible orchidectomy + bilateral fixation (orchidopexy)—see p568. At surgery expose and untwist the testis. If its colour looks good, return it to the scrotum and fix both testes to the scrotum.
Undescended testes Incidence About 3% of boys are born with at least one undescended testis (30% of premature boys) but this drops to 1% after the first year of life. Unilateral is four times more common than bilateral. (If bilateral then should have genetic testing.) • Cryptorchidism: Complete absence of the testis from the scrotum (anorchism is absence of both testes). • Retractile testis: The genitalia are normally developed but there is an excessive cremasteric reflex. The testis is often found at the external inguinal ring. : reassurance (examining while in a warm bath, for example, may help to distinguish from maldescended/ectopic testes). • Maldescended testis: May be found anywhere along the normal path of descent from abdomen to groin. • Ectopic testis: Most commonly found in the superior inguinal pouch (anterior to the external oblique aponeurosis) but may also be abdominal, perineal, penile, and in the femoral triangle. Complications of maldescended and ectopic testis Infertility; ≈40 increased risk of testicular cancer (risk remains after surgery but in cryptorchidism may be if orchidopexy performed before aged 10), increased risk of testicular trauma, increased risk of testicular torsion. Also associated with hernias (due to patent processus vaginalis in >90%, p613) and other urinary tract anomalies. Treatment of maldescended and ectopic testis restores (potential for) spermatogenesis; the increased risk of malignancy remains but becomes easier to diagnose. Surgery: Orchidopexy, usually dartos pouch procedure, is performed in infancy: testis and cord are mobilized following a groin incision, any processus vaginalis or hernial sac is removed and the testis is brought through a hole made in the dartos muscle into the resultant subcutaneous pouch where the muscle prevents retraction. Hormonal: Hormonal therapy, most commonly human chorionic gonadotrophin (hCG), is sometimes attempted if an undescended testis is in the inguinal canal.
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Aneurysms of arteries An artery with a dilatation >50% of its original diameter has an aneurysm; remember this is an ongoing process. True aneurysms are abnormal dilatations that involve all layers of the arterial wall. False aneurysms (pseudoaneurysms) involve a collection of blood in the outer layer only (adventitia) which communicates with the lumen (eg after trauma). Aneurysms may be fusiform (eg most AAAS) or sac-like (eg Berry aneurysms; fig 10.17 p479). Causes Atheroma, trauma, infection (eg mycotic aneurysm in endocarditis; tertiary syphilis—especially thoracic aneurysms), connective tissue disorders (eg Marfan’s, Ehlers–Danlos), inflammatory (eg Takayasu’s aortitis, p712). Common sites Aorta (infrarenal most common), iliac, femoral, and popliteal arteries. Complications Rupture; thrombosis; embolism; fistulae; pressure on other structures. Screening All at age 65yr are invited for screening in UK, decreases mortality from ruptured AAA. Ruptured abdominal aortic aneurysm (AAA) Death rates/year from ruptured AAAS rise with age: 125 per million in those aged 55–59; 2728 per million if over 85yrs. Symptoms & signs: Intermittent or continuous abdominal pain (radiates to back, iliac fossae, or groins; don’t dismiss this as renal colic), collapse, an expansile abdominal mass (it expands and contracts, unlike swellings that are purely pulsatile, eg nodes overlying arteries), and shock. If in doubt, assume a ruptured aneurysm. Unruptured AAA Definition: >3cm across. Prevalence: 3% of those >50yrs. : >3 : 1. Less common in diabetics. Cause: Degeneration of elastic lamellae and smooth muscle loss. There is a genetic component. Symptoms: Often none, they may cause abdominal/back pain, often discovered incidentally on abdominal examination (see BOX). Monitoring: RCTS have failed to demonstrate benefit from early endovascular repair (EVAR, see later in paragraph) of aneurysms 1cm/yr, or symptomatic aneurysms. Operative mortality: ~5%; complications include spinal or visceral ischaemia and distal emboli from dislodged thrombus debris. Studies show that age >80yrs should not, in itself, preclude surgery. Stenting (EVAR): Major surgery can be avoided by inserting an endovascular stent via the femoral artery. EVAR has less early mortality but higher graft complications, eg failure of stent-graft to totally exclude blood flow to the aneurysm—‘endoleak’. See fig 13.52.
Emergency management of a ruptured abdominal aneurysm Mortality—treated: 41% and improving; untreated: ~100%. Summon a vascular surgeon and an experienced anaesthetist; warn theatre. Do an ECG, and take blood for amylase, Hb, crossmatch (10–40U may eventually be needed). Catheterize the bladder. Gain IV access with 2 large-bore cannulae. Treat shock with O RhΩve blood (if not cross matched), but keep systolic BP ≤100mmHg to avoid rupturing a contained leak (NB: raised BP is common early on). Take the patient straight to theatre. Don’t waste time on X-rays: fatal delay may result, though CT can help in a stable patient with an uncertain diagnosis. Give prophylactic antibiotics, eg co-amoxiclav 625mg IV. Surgery involves clamping the aorta above the leak, and inserting a Dacron® graft (eg ‘tube graft’ or, if significant iliac aneurysm also, a ‘trouser graft’ with each ‘leg’ attached to an iliac artery).
Thoracic aortic dissection
Fig 13.52 Stenting: not an open or closed case … this is a digital subtraction angiogram showing correct positioning of an endovascular stent at the end of the procedure. Although less invasive than open repair, some are unsuited to this method, owing to the anatomy of their aneurysm. Lifelong monitoring is needed: stents may leak and the aneurysm progress (the risk can be reduced by coiling the internal iliac arteries, as shown). Image courtesy of Norwich Radiology Dept.
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Blood splits the aortic media with sudden tearing chest pain (± radiation to back). As the dissection extends, branches of the aorta occlude sequentially leading to hemiplegia (carotid artery), unequal arm pulses and BP, or acute limb ischaemia, paraplegia (anterior spinal artery), and anuria (renal arteries). Aortic valve incompetence, inferior MI, and cardiac arrest may develop if dissection moves proximally. Type A (70%) dissections involve the ascending aorta, irrespective of site of the tear, while if the ascending aorta is not involved it is called type B (30%). All patients with type A thoracic dissection should be considered for surgery: get urgent cardiothoracic advice. Definitive treatment for type B is less clear and may be managed medically, with surgery reserved for distal dissections that are leaking, ruptured, or compromising vital organs. Management: •Crossmatch 10U blood. •ECG & CXR (expanded mediastinum is rare). •CT or transoesophageal echocardiography (TOE). Take to ITU; hypotensives: keep systolic at ~100–110mmHg: labetalol (p140) or esmolol (t½ is ultra-short) by IVI is helpful here (calcium-channel blockers may be used if -blockers contraindicated). Acute operative mortality: 1 mL/kg/h; the minimum is >0. 5mL/kg/h. Give a fluid challenge, eg 500mL 0.9% saline over 1h (or half this volume in heart failure or the elderly), and recheck the urine output. If not catheterized, exclude retention; if catheterized, ensure the catheter is not blocked! Post-operative: Check the operation notes for intraoperative losses, and ensure you chart and replace added losses from drains, etc. Shock: Resuscitate with colloid or 0.9% saline via large-bore cannulae. Identify the type of shock (p790). Transpiration losses: (Fever, burns.) Beware the large amounts of fluid that can be lost unseen through transpiration. Severe burns in particular may require aggressive fluid resuscitation (p846).
Potassium in IV fluids • Potassium ions can be given with 5% glucose or 0.9% saline, usually 20mmol/L or 40mmol/L. • K+ may be retained in renal failure, so beware giving too much IV. • Gastrointestinal fluids are rich in K+, so increased fluid loss from the gut (eg diarrhoea, vomiting, high-output stoma, intestinal fistula) will need increased K+ replacement. The maximum concentration of K+ that is safe to infuse via a peripheral line is 40mmol/L, at a maximum rate of 20mmol/h in a cardiac monitored patient. Fluidrestricted patients may require higher concentrations or rates in life-threatening hypokalaemia. Faster rates risk cardiac dysrhythmias and asystole, and higher concentrations thrombophlebitis, depending on the size of the vein, so give concentrated solutions >40mmol/L via a central venous catheter, and use ECG monitoring for rates >10mmol/h. For symptoms and signs of hyper- and hypokalaemia see p674.
Clinical chemistry
Underfilled:
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Electrolyte physiology and the kidney The kidney Controls the homeostasis of a number of serum electrolytes (including Na+, K+, Ca2+, and PO43Ω), helps to maintain acid–base balance, and is responsible for the excretion of many substances. It also makes erythropoietin and renin, and hydroxylates 25-hydroxyvitamin D to 1,25-dihydroxyvitamin D (see p676 for Ca2+ and PO43Ω physiology). All of these functions can be affected in chronic kidney disease (p302), but it is the biochemical effects of kidney failure that are used to monitor disease progression. The renin–angiotensin–aldosterone system Plasma is filtered by the glomeruli, and Na+, K+, H+, and water are reabsorbed from this filtrate under the control of the renin–angiotensin–aldosterone system. Renin is released from the juxtaglomerular apparatus (fig 7.13, p316) in response to low renal flow and raised sympathetic tone, and catalyses the conversion of angiotensinogen (a peptide made by the liver) to angiotensin I. This is then converted by angiotensin-converting enzyme (ACE), which is located throughout the vascular tree, to angiotensin II. The latter has several important actions including efferent renal arteriolar constriction (thus perfusion pressure), peripheral vasoconstriction, and stimulation of the adrenal cortex to produce aldosterone, which activates the Na+/K+ pump in the distal renal tubule leading to reabsorption of Na+ and water from the urine, in exchange for K+ and H+. Glucose spills over into the urine when the plasma concentration > renal threshold for reabsorption (≈10mmol/L, but this varies between people, and is in pregnancy). Control of sodium Control is through the action of aldosterone on the distal convoluted tubule (DCT) and collecting duct to increase Na+ reabsorption from the urine. The natriuretic peptides ANP, BNP, and CNP (p137) contribute to Na+ homeostasis by reducing Na+ reabsorption from the DCT and inhibiting renin. A high GFR (see later in this topic) results in increased Na+ loss, and high renal tubular flow and haemodilution decrease Na+ reabsorption in the proximal tubule. Control of potassium Most K+ is intracellular, and thus serum K+ levels are a poor reflection of total body potassium. The concentrations of K+ and H+ in extracellular fluid tend to vary together. This is because these ions compete with each other in the exchange with Na+ that occurs across most cell membranes and in the distal convoluted tubule of the kidney, where Na+ is reabsorbed from the urine. Thus, if the H+ concentration is high, less K+ will be excreted into the urine. Similarly, K+ will compete with H+ for exchange across cell membranes and extracellular K+ will accumulate. Insulin and catecholamines both stimulate K+ uptake into cells by stimulating the Na+/K+ pump. Serum osmolality A laboratory measurement of the number of osmoles per kilogram of solvent. It is approximated by serum osmolarity (the number of osmoles per litre of solution) using the equation 2(Na+ + K+) + Urea + Glucose, since these are the predominant serum electrolytes. Normal serum osmolarity is 280–300mmol/L, which will always be a little less than the laboratory-measured osmolality—the osmolar gap. However, if the osmolar gap is greater than 10mmol/L, this indicates the presence of additional solutes: consider diabetes mellitus or high blood ethanol, methanol, mannitol, or ethylene glycol. Control of water Control is mainly via serum Na+ concentration, since water intake and loss are regulated to hold the extracellular concentration of Na+ constant. Raised plasma osmolality (eg dehydration or glucose in diabetes mellitus) causes thirst through the hypothalamic thirst centre and the release of antidiuretic hormone (ADH) from the posterior pituitary. ADH increases the passive water reabsorption from the renal collecting duct by opening water channels to allow water to flow from the hypotonic luminal fluid into the hypertonic renal interstitium. Low plasma osmolality inhibits ADH secretion, thus reducing renal water reabsorption. Glomerular filtration rate (GFR) Defined as the volume of fluid filtered by the glomeruli per minute (units mL/min), and is one of the primary measures of disease progression in chronic kidney disease. It can be estimated in a number of different ways (see BOX).
Calculating GFR is useful because it is a more sensitive indication of the degree of renal impairment than serum creatinine. Subjects with low muscle mass (eg the elderly, women) can have a ‘normal’ serum creatinine, despite a significant reduction in GFR. This can be important when prescribing nephrotoxic drugs, or drugs that are renally excreted, which may therefore accumulate to toxic levels in the serum. A number of methods for estimating GFR exist, all relying on a calculation of the clearance of a substance that is renally filtered and then not reabsorbed in the renal tubule. For example, the rate of clearance of creatinine can be used as a marker for the rate of filtration of fluid and solutes in the glomerulus because it is only slightly reabsorbed from the renal tubule. The more of the filtered substance that is reabsorbed, however, the less accurate the estimate of GFR. MDRD (Modification of Diet in Renal Disease Study Group): This provides an estimate of GFR from four simple parameters: serum creatinine, age, gender, and race (black/non-black). It is one of the best validated for monitoring patients with established moderately severe renal impairment,2 and most labs now routinely report estimated GFR (eGFR) using the MDRD equation on all U&E reports: Ω1 . 154 eGFR = 32 788 ≈ serum creatinine ≈ ageΩ0.203 ≈ [1 . 212 if black] ≈ [0 . 742 if female] However, a number of caveats exist, so that it is best used in monitoring declining renal function rather than labelling elderly patients with mild renal impairment: • It is not validated for mild renal impairment, and therefore its use for screening general populations is questionable. • Inter-individual variations (and thus confidence intervals) are wide, although for each individual variations are small so that a decline in eGFR over a number of serum samples is always significant. • Single results may be affected by variations in serum creatinine, such as after a protein-rich meal. Cockcroft–Gault equation: This provides an estimate of creatinine clearance. It is an improvement on the MDRD equation because it also takes into account the patient’s weight. However, 10% of creatinine is actively excreted in the tubules, and therefore creatinine clearance overestimates true GFR and underestimates renal impairment. Moreover, the equation assumes ideal body weight and is thus unreliable in the obese or oedematous. Also unreliable in unstable renal function. (140 Ω age) ≈ weight (kg) ≈ [0.85 if female] ≈ [1.212 if black] Creatinine clearance = 0 . 813 ≈ serum creatinine (μmol/L) Creatinine clearance can also be calculated by measuring the excreted creatinine in a 24h urine collection and comparing it with the serum creatinine concentration. However, the accuracy of collection is vital but often poor, making this an unreliable and inconvenient method. GFR can also be measured by injection of a radioisotope followed by sequential blood sampling (51Cr-EDTA) or by an isotope scan (eg DTPA 99Tc, p190). These methods allow a more accurate estimate of GFR than creatinine clearance, since smaller proportions of these substances are reabsorbed in the tubules. They also have the advantage of being able to provide split renal function. Inulin clearance: The gold standard for calculating GFR, because 100% of filtered inulin (not insulin) is retained in the luminal fluid and therefore reflects exactly the rate of filtration of water and solutes in the glomerulus. However, measuring inulin clearance again requires urine collection over several hours, and also a constant IV infusion of inulin, and is therefore inconvenient to perform.
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Estimating GFR
Clinical chemistry
670
Acid–base balance Arterial blood pH is closely regulated in health to 7.40 ± 0.05 by various mechanisms including bicarbonate, other plasma buffers such as deoxygenated haemoglobin, and the kidney. Acid–base disorders needlessly confuse many people, but if a few simple rules are applied, then interpretation and diagnosis are easy. The key principle is that primary changes in HCO3Ω are metabolic and in CO2 respiratory. See fig 14.2. A simple method 1 Look at the pH, is there an acidosis or alkalosis? • pH 7.45 is an alkalosis. 2 Is the CO2 abnormal? (Normal range 4.7–6.0kPa.) If so, is the change in keeping with the pH? • CO2 is an acidic gas—is CO2 raised with an acidosis, lowered with an alkalosis? If so, it is in keeping with the pH and thus caused by a respiratory problem. If there is no change, or an opposite one, then the change is compensatory. 3 Is the HCO3Ω abnormal? (Normal concentration 22–28mmol/L.) If so, is the change in keeping with the pH? Ω Ω • HCO3 is alkaline—is HCO3 raised with an alkalosis, lowered with an acidosis? If so, the problem is a metabolic one. 4 Is the PaO2 abnormal? Interpret in the context of the FiO2. An example Your patient’s blood gas shows: pH 7.05, CO2 2.0kPa, HCO3Ω 8.0mmol/L. There is an acidosis. The CO2 is low, and thus it is a compensatory change. The HCO3Ω is low and is thus the primary change, ie a metabolic acidosis. The anion gap Estimates unmeasured plasma anions (‘fixed’ or organic acids such as phosphate, ketones, and lactate—hard to measure directly). It is calculated as the difference between plasma cations (Na+ and K+) and anions (ClΩ and HCO3Ω). Normal range: 10–18mmol/L. It is helpful in determining the cause of a metabolic acidosis. Metabolic acidosis pH, HCO3Ω Causes of metabolic acidosis and an increased anion gap: Due to increased production, or reduced excretion, of fixed/organic acids. HCO3Ω falls and unmeasured anions associated with the acids accumulate. • Lactic acid (shock, infection, tissue ischaemia). • Urate (renal failure). • Ketones (diabetes mellitus, alcohol). • Drugs/toxins (salicylates, biguanides, ethylene glycol, methanol). Causes of metabolic acidosis and a normal anion gap: Due to loss of bicarbonate or ingestion of H+ ions (ClΩ is retained). • Renal tubular acidosis. • Diarrhoea. • Drugs (acetazolamide). • Addison’s disease. • Pancreatic fistula. • Ammonium chloride ingestion. Metabolic alkalosis pH, HCO3Ω • Vomiting. • K+ depletion (diuretics). • Burns. • Ingestion of base. Respiratory acidosis pH, CO2 • Type 2 respiratory failure due to any lung, neuromuscular, or physical cause (p188). • Most commonly chronic obstructive pulmonary disease (COPD). Look at the PaO2. It will probably be low. Is oxygen therapy required? Use controlled O2 (Venturi connector) if COPD is the underlying cause, as too much oxygen may make matters worse (p189). Beware exhaustion in asthma, pneumonia, and pulmonary oedema, which can present with this picture when close to respiratory arrest. A normal or high PaCO2 is worrying. These patients require urgent ITU review for ventilatory support.
671
Clinical chemistry
Respiratory alkalosis pH, CO2 A result of hyperventilation of any cause. CNS causes: Stroke; subarachnoid bleed; meningitis. Others: Mild/moderate asthma; anxiety; altitude; T°; pregnancy; pulmonary emboli (reflex hyperventilation); drugs, eg salicylates. Terminology To aid understanding, we have used the terms acidosis and alkalosis, where a purist would sometimes have used acidaemia and alkalaemia. Technically acidaemia is the state of having a low blood pH, whereas acidosis refers to the processes which generate H+, leading to the acidaemia.
Fig 14.2 The shaded area represents normality. This method is very powerful. The result represented by point ≈, for example, indicates that the acidosis is in part respiratory and in part metabolic. Seek a cause for each.
Clinical chemistry
672
Hypernatraemia Signs and symptoms Lethargy, thirst, weakness, irritability, confusion, coma, and fits, along with signs of dehydration (p666). Laboratory features: Na+, PCV, alb, urea. Causes Usually due to water loss in excess of Na+ loss: • Fluid loss without water replacement (eg diarrhoea, vomit, burns). • Diabetes insipidus (p240). Suspect if large urine volume. This may follow head injury, or CNS surgery, especially pituitary. • Osmotic diuresis (for diabetic coma, see p832). • Primary aldosteronism: rarely severe, suspect if BP, K+, alkalosis (HCO3Ω). • Iatrogenic: incorrect IV fluid replacement (excessive saline). Management Give water orally if possible. If not, give glucose 5% IV slowly (1L/6h) guided by urine output and plasma Na+. Use 0.9% saline IV if hypovolaemic, since this causes less marked fluid shifts and is hypotonic in a hypertonic patient. Avoid hypertonic solutions.
Hyponatraemia Plasma Na+ concentration depends on the amount of both Na+ and water in the plasma. Hyponatraemia therefore does not necessarily imply Na+ depletion. Assessing fluid status is the key to diagnosis (see fig 14.3). Signs and symptoms Look for anorexia, nausea, and malaise initially, followed by headache, irritability, confusion, weakness, GCS, and seizures, depending on the severity and rate of change in serum Na+. Cardiac failure or oedema may help to indicate the cause. Hyponatraemia also increases the risk of falls in the elderly. 3 Causes See fig 14.3. Artefactual causes include: •blood sample was from a drip arm •high serum lipid/protein content causing serum volume, with Na+ concentration but normal plasma osmolality •if hyperglycaemic (20mmol/L) add ~4.3mmol/L to plasma Na+ for every 10mmol/L rise in glucose above normal. Iatrogenic hyponatraemia If 5% glucose is infused continuously without adding 0.9% saline, the glucose is quickly used, rendering the fluid hypotonic and causing hyponatraemia, esp. in those on thiazide diuretics, women (esp. pre-menopausal), and those undergoing physiological stress (eg post-operative, septic). In some patients, only marginally low plasma Na+ levels cause serious effects (eg ~128mmol/L)—don’t attribute odd CNS signs to non-existent strokes/TIAS if Na+. Management • Correct the underlying cause; never base treatment on Na+ concentration alone. The presence of symptoms, the chronicity of the hyponatraemia, and state of hydration are all important. Replace Na+ and water at the same rate they were lost. • Asymptomatic chronic hyponatraemia, fluid restriction is often sufficient if asymptomatic, although demeclocycline (ADH antagonist) may be required. If hypervolaemic (cirrhosis, CCF), treat the underlying disorder first. • Acute or symptomatic hyponatraemia, or if dehydrated, cautious rehydration with 0.9% saline may be given, but do not correct changes rapidly as central pontine myelinolysis2 may result. Maximum rise in serum Na+ 15mmol/L per day if chronic, or 1mmol/L per hour if acute. Consider using furosemide when not hypovolaemic to avoid fluid overload. • Vasopressor receptor antagonists (‘vaptans’, eg tolvaptan) promote water excretion without loss of electrolytes, and appear to be effective in treating hypervolaemic and euvolaemic hyponatraemia but are expensive. 4 In emergency: (Seizures, coma) seek expert help. Consider hypertonic saline (eg 1.8% saline) at 70mmol Na+/h ± furosemide. Aim for a gradual increase in plasma Na+ to ≈125mmol/L. Beware heart failure and central pontine myelinolysis.2 2 Central pontine myelinolysis: irreversible and often fatal pontine demyelination seen in malnourished alcoholics or rapid correction of Na+. There is subacute onset of lethargy, confusion, pseudobulbar palsy, para- or quadriparesis, ‘locked-in’ syndrome, or coma.
Hyponatraemia
673
Is the patient dehydrated?
Yes Na+ and H2O are lost via kidneys: • Addison’s dis. • Renal failure, eg diuretic phase of renal failure; nephrocalcinosis or medullary cystic disease • Diuretic excess • Osmolar diuresis (glucose; urea)
No Na+ and H2O are lost other than via the kidneys: • Diarrhoea • Vomiting • Fistulae • Burns • Rectal villous adenoma • Small bowel obstruction • Trauma • Cystic fibrosis • Heat exposure
No Is the patient oedematous? Yes
No Is the urine osmolality >100mmol/kg?
• Nephrotic syndrome • Cardiac failure • Liver cirrhosis (hyponatraemia may precede oedema) • Renal failure
Yes • SIADH (see BOX)
No • Water overload • Severe hypothyroidism • Glucocorticoid insufficiency
Fig 14.3 Hyponatraemia.
Syndrome of inappropriate ADH secretion (SIADH) An important, but over-diagnosed, cause of hyponatraemia. The diagnosis requires concentrated urine (Na+ > 20mmol/L and osmolality > 100mosmol/kg) in the presence of hyponatraemia (plasma Na+ < 125mmol/L) and low plasma osmolality (< 260mosmol/kg), in the absence of hypovolaemia, oedema, or diuretics. Causes: • Malignancy: lung small-cell, pancreas, prostate, thymus, or lymphoma. • CNS disorders: meningoencephalitis, abscess, stroke, subarachnoid or subdural haemorrhage, head injury, neurosurgery, Guillain–Barré, vasculitis, or SLE. • Chest disease: TB, pneumonia, abscess, aspergillosis, small-cell lung cancer. • Endocrine disease: hypothyroidism (not true SIADH, but perhaps due to excess ADH release from carotid sinus baroreceptors triggered by cardiac output). • Drugs: opiates, psychotropics, SSRIs, cytotoxics. • Other: acute intermittent porphyria, trauma, major abdominal or thoracic surgery, symptomatic HIV. Treatment: Treat the cause and restrict fluid. Consider salt ± loop diuretic if severe. Demeclocycline is used rarely. Vasopressin receptor antagonists (‘vaptans’, p672) are an emerging class of drug used in SIADH and other types of hyponatraemia.
Clinical chemistry
Yes Is urinary Na+ >20mmol/L?
Clinical chemistry
674
Hyperkalaemia A plasma potassium >6.5mmol/L is a potential emergency and needs urgent assessment (see p301). The worry is of myocardial hyperexcitability leading to ventricular fibrillation and cardiac arrest. First assess the patient—do they look unwell, is there an obvious cause? If not, could it be an artefactual result? Concerning signs and symptoms Include a fast irregular pulse, chest pain, weakness, palpitations, and light-headedness. ECG: (see fig 14.4) tall tented T waves, small P waves, a wide QRS complex (eventually becoming sinusoidal), and ventricular fibrillation. Artefactual results: If the patient is well, and has none of the above-mentioned findings, repeat the test urgently as it may be artefactual, caused by: •haemolysis (difficult venepuncture; patient clenched fist) •contamination with potassium EDTA anticoagulant in FBC bottles (do FBCS after U&ES) •thrombocythaemia (K+ leaks out of platelets during clotting) •delayed analysis (K+ leaks out of RBCS; a particular problem in a primary care setting due to long transit times to the lab).5 Causes • Oliguric renal failure. • Addison’s disease (see p226). • K+-sparing diuretics. • Massive blood transfusion. • Rhabdomyolysis (p319). • Burns. • Metabolic acidosis (DM). • Drugs, eg ACE-i, suxamethonium. • Excess K+ therapy. • Artefactual result (see earlier ‘Artefactual results’). Treatment in non-urgent cases Treat the underlying cause; review medications. • Polystyrene sulfonate resin (eg Calcium Resonium® 15g/8h PO) binds K+ in the gut, preventing absorption and bringing K+ levels down over a few days. If vomiting prevents PO administration, give a 30g enema, followed at 9h by colonic irrigation. Emergency treatment If there is evidence of myocardial hyperexcitability, or K+ is >6.5mmol/L, get senior assistance, and treat as an emergency (see p301).
Hypokalaemia If K+ 2.5mmol/L, no symptoms.) Give oral K+ supplement (≥80mmol/24h, eg Sando-K® 2 tabs/8h). Review K+ after 3 days. If taking a thiazide diuretic, and K+ >3.0 consider repeating and/or K+-sparing diuretic. If severe: (3.5mmol/L and symptomatic: 1 Correct dehydration: If dehydrated give IV 0.9% saline. 2 Bisphosphonates: These prevent bone resorption by inhibiting osteoclast activity. A single dose of pamidronate lowers Ca2+ over 2–3d; maximum effect is at 1wk. Infuse slowly, eg 30mg in 300mL 0.9% saline over 3h via a largish vein. Max dose 90mg (see table 14.4). Zoledronic acid is significantly more effective in reducing serum Ca2+ than previously used bisphosphonates.8 Usually, a single dose of 4mg IV (diluted to 100mL, over 15min) will normalize plasma Ca2+ within a week. SE, flu symptoms, PO43Ω, bone pain, myalgia, nausea, vomiting, headache, lymphocytopenia, Mg2+, Ca2+, seizures. 3 Further management: Chemotherapy may help in malignancy. Steroids are used in sarcoidosis, eg prednisolone 40–60mg/d. Salmon calcitonin acts similarly to bisphosphonates, and has a quicker onset of action, but is now rarely used. NB: the use of furosemide is contentious, as supporting RCT evidence is scant.9,10 It helps to promote renal excretion of Ca2+, but can exacerbate hypercalcaemia by worsening dehydration. Thus it should only be used once fully rehydrated, and with concomitant IV fluids (eg 0.9% saline 1L/4–6h). Avoid thiazides. Table 14.4 Disodium pamidronate doses Calcium (mmol/L; corrected) Single-dose pamidronate (mg)
4
15–30 30–60 60–90 90
Clinical chemistry
Albumin raised
677
Clinical chemistry
678
Hypocalcaemia Apparent hypocalcaemia may be an artefact of hypoalbuminaemia (p676). Signs and symptoms See BOX.11 Mild: cramps, perioral numbness/paraesthesiae. Severe: carpopedal spasm (especially if brachial artery compressed, Trousseau’s sign; see fig 14.6), laryngospasm, seizures. Neuromuscular excitability may also be demonstrated by tapping over parotid (facial nerve) causing facial muscles to twitch (Chvostek’s sign; see fig 14.7). Cataract if chronic hypocalcaemia. ECG: Long QT interval. Causes With PO43Ω With or PO43Ω • Chronic kidney disease (p302). • Vitamin D deficiency. • Hypoparathyroidism (incl thyroid • Osteomalacia (ALP). • Acute pancreatitis. or parathyroid surgery, p222). • Pseudohypoparathyroidism (p222). • Over-hydration. • Acute rhabdomyolysis. • Respiratory alkalosis (total Ca2+ is normal, but ionized Ca2+ due • Hypomagnesaemia. to pH symptomatic). Treatment • Mild symptoms: Give calcium 5mmol/6h PO, with daily plasma Ca2+ levels. • In chronic kidney disease: See p302. May require alfacalcidol, eg 0.5–1mcg/24h PO. • Severe symptoms: Give 10mL of 10% calcium gluconate (2.25mmol) IV over 30min, and repeat as necessary. If due to respiratory alkalosis, correct the alkalosis.
Features of hypocalcaemia ‘SPASMODIC ’ Spasms (carpopedal spasms = Trousseau’s sign) Perioral paraesthesiae Anxious, irritable, irrational Seizures Muscle tone in smooth muscle—hence colic, wheeze, and dysphagia Orientation impaired (time, place, and person) and confusion Dermatitis (eg atopic/exfoliative) Impetigo herpetiformis (Ca2+ and pustules in pregnancy—rare and serious) Chvostek’s sign; choreoathetosis; cataract; cardiomyopathy (long QT interval on ECG).
Fig 14.6 Trousseau’s sign: on inflating the cuff, Fig 14.7 Chvostek’s sign: the corner of the the wrist and fingers flex and draw together mouth twitches when the facial nerve is tapped (carpopedal spasm). over the parotid.
Magnesium Magnesium is distributed 65% in bone and 35% in cells; plasma concentration tends to follow that of Ca2+ and K+. Hypomagnesaemia Causes paraesthesiae, ataxia, seizures, tetany, arrhythmias. Digitalis toxicity may be exacerbated. Causes: Diuretics, severe diarrhoea, ketoacidosis, alcohol abuse, total parenteral nutrition (monitor weekly), Ca2+, K+, and PO43Ω. Treatment: If needed, give magnesium salts, PO or IV (eg 8mmol MgSO4 IV over 3min to 2h, depending on severity, with frequent Mg2+ levels). Hypermagnesaemia Rarely requires treatment unless severe (>7.5mmol/L). Causes: Renal failure or iatrogenic (eg excessive antacids). Signs: If severe: neuromuscular depression, BP, pulse, hyporeflexia, CNS & respiratory depression, coma.
Zinc Zinc deficiency This may occur in parenteral nutrition or, rarely, from a poor diet (too few cereals and dairy products; anorexia nervosa; alcoholism). Rarely it is due to a genetic defect. Symptoms: Alopecia, dermatitis (look for red, crusted skin lesions especially around nostrils and corners of mouth), night blindness, diarrhoea. Diagnosis: Therapeutic trial of zinc (plasma levels are unreliable as they may be low, eg in infection or trauma, without deficiency).
Selenium An essential element present in cereals, nuts, and meat. Low soil levels in some parts of Europe and China cause deficiency states. Required for the antioxidant glutathione peroxidase, which harmful free radicals. Selenium is also antithrombogenic, and is required for sperm motility proteins. Deficiency may increase risk of neoplasia and atheroma, and may lead to a cardiomyopathy or arthritis. Serum levels are a poor guide. Toxic symptoms may also be found with over-energetic replacement.
679
Clinical chemistry
Phosphate Hypophosphataemia Common and of little significance unless severe (2L/d). Unlike most other renal calculi, existing uric acid stones can often be dissolved with either systemic Fig 14.8 Urate stone. ©Dr G. Austin. or topical alkalinizing agents. Potassium citrate or potassium bicarbonate at a dose titrated to alkalinize the urine to a pH of 6–7 dissolves some urate stones. If hyperuricosuria, consider dietary management ± allopurinol (xanthine oxidase inhibitor).
Clinical chemistry
681
Clinical chemistry
682
Metabolic bone diseases: osteoporosis Osteoporosis implies reduced bone mass. It may be 1° (age-related) or 2° to another condition or drugs. If trabecular bone is affected, crush fractures of vertebrae are common (hence the ‘littleness’ of little old ladies and their dowager’s hump); if cortical bone is affected, long bone fractures are more likely, eg femoral neck: the big cause of death and orthopaedic expense (80% hip fractures in the UK occur in women >50yrs). Prevalence (In those >50yrs): 6%, 18%. Women lose trabeculae with age, but in men, although there is reduced bone formation, numbers of trabeculae are stable and their lifetime risk of fracture is less. Risk factors Age-independent risk factors for 1° osteoporosis: parental history, alcohol >4 units daily, rheumatoid arthritis, BMI 75yrs. Lifestyle measures should apply to all (including those at risk but not yet osteoporotic). Lifestyle measures: • Quit smoking and reduce alcohol consumption. • Weight-bearing exercise may increase bone mineral density.15 • Balance exercises such as tai chi reduce risk of falls. • Calcium and vitamin D-rich diet (use supplements if diet is insufficient—see ‘Pharmacological measures’ later in this topic). • Home-based fall-prevention programme, with visual assessment and a home visit. NB: hip-protectors are unreliable for preventing fractures.16 Pharmacological measures: • Bisphosphonates: alendronic acid is 1st line (10mg/d or 70mg/wk; not if eGFR 30min and wait 30min before eating or other drugs. (SE: photosensitivity; GI upset; oesophageal ulcers—stop if dysphagia or abdo pain; rarely, jaw osteonecrosis). • Calcium and vitamin D: rarely used alone for prophylaxis, as questionable efficacy and some evidence of a small CV risk. Offer if evidence of deficiency, eg calcium 1g/d + vit D 800U/d. Target serum 25-hydroxy-vitamin D level ≥75nmol/L. • Strontium ranelate: due to an increased risk of cardiac problems it should only be used in those with severe intolerance of other agents and without cardiovascular disease. • Hormone replacement therapy (HRT) can prevent (not treat) osteoporosis in postmenopausal women. Relative risk of breast cancer is 1 . 4 if used >10yrs; CV risk. • Raloxifene is a selective oestrogen receptor modulator (SERM) that acts similarly to HRT, but with breast cancer risk. • Teriparatide (recombinant PTH) is useful in those who suffer further fractures despite treatment with other agents. There is a potential risk of renal malignancy. • Calcitonin may reduce pain after a vertebral fracture. • Testosterone may help in hypogonadal men by promoting trabecular connectivity. • Denosumab, a monoclonal Ab to RANK ligand, given SC twice yearly reabsorption.
DEXA bone densitometry: WHO osteoporosis criteria
683
It is better to scan the hip than the lumbar spine. Bone mineral density (g/cm2) is compared with that of a young healthy adult. The ‘T-score’ is the number of standard deviations (SD, p751) the bone mineral density (BMD) is from the youthful average. Each decrease of 1 SD in BMD ≈ 2.6-fold in risk of hip fracture. T-score >0
BMD is better than the reference.
0 to Ω1 Ω1 to Ω2.5 Ω2.5 or worse
BMD is in the top 84%: no evidence of osteoporosis.
Osteopenia. Risk of later osteoporotic fracture. Offer lifestyle advice. Osteoporosis. Offer lifestyle advice and treatment (p682). Repeat DEXA in 2yrs.
Some indications for DEXA: • NICE suggests DEXA if previous low-trauma fracture, or for women ≥ 65yrs with one or more risk factors for osteoporosis, or younger if two or more. The benefits of universal screening for osteoporosis remain unproven, but some authorities recommend this for men and women over 70—and earlier if risk factors are present.17 • DEXA is not needed pre-treatment for women over 75yrs if previous low-trauma fracture, or ≥ 2 present of rheumatoid arthritis, alcohol excess, or positive family history. • Prior to giving long-term prednisolone (eg 3 months at >5mg/d). Steroids cause osteoporosis by promoting osteoclast bone resorption, muscle mass, and Ca2+ absorption from the gut. • Men or women with osteopenia if low-trauma, non-vertebral fracture. • Bone and bone-remodelling disorders (eg parathyroid disorders, myeloma, HIV, esp. if on protease inhibitors).
Osteoporosis risk factors: ‘SHATTERED’ Steroid use of >5mg/d of prednisolone. Hyperthyroidism, hyperparathyroidism, hypercalciuria. Alcohol and tobacco use . Thin (BMI 40% of skeleton involved), and osteosarcoma (10yrs—suspect if sudden onset or worsening of bone pain).19 Radiology X-ray Localized enlargement of bone. Patchy cortical thickening with sclerosis, osteolysis, and deformity (eg osteoporosis circumscripta of the skull). Affinity for axial skeleton, long bones, and skull. Bone scan may reveal ‘hot spots’. Blood chemistry Ca2+ and PO43Ω normal; ALP markedly raised. Treatment If analgesia fails, alendronic acid may be tried to reduce pain and/or deformity. It is more effective than etidronate or calcitonin, and as effective as IV pamidronate. Follow expert advice.
Clinical chemistry
686
Plasma proteins The plasma contains a number of proteins including albumin, immunoglobulins, 1antitrypsin, 2-macroglobulin, caeruloplasmin, transferrin, low-density lipoprotein (LDL), fibrinogen, complement, and factor VIII. The most abundant is albumin (see fig 14.12). Albumin Synthesized in the liver; t½ ≈ 20d. It binds bilirubin, free fatty acids, Ca2+, and some drugs. Low albumin: Results in oedema, and is caused by: •synthesis: liver disease, acute phase response (due to vascular permeability—eg sepsis, trauma, surgery), malabsorption, malnutrition, malignancy •loss: nephrotic syndrome, protein-losing enteropathy, burns •haemodilution: late pregnancy, artefact (eg from ‘drip’ arm). Also posture (5g/L if upright) and genetic variations. High albumin: Causes are dehydration; artefact (eg stasis). Immunoglobulins (Antibodies) are synthesized by B cells. Five isoforms Ig A,D,E,G,M exist in humans, and IgG is the most abundant circulating form. Specific monoclonal band in paraproteinaemia (see p370). Diffusely raised in chronic infections, TB, bronchiectasis, liver cirrhosis, sarcoidosis, SLE, RA, Crohn’s disease, 1° biliary cirrhosis, hepatitis, and parasitaemia. Low in nephrotic syndrome, malabsorption, malnutrition, and immune deficiency states (eg severe illness, renal failure, diabetes mellitus, malignancy, or congenital). Acute phase response The body responds to a variety of insults with, among other things, the synthesis, by the liver, of a number of proteins (normally present in serum in small quantities)—eg 1-antitrypsin, fibrinogen, complement, haptoglobin, and CRP. A concomitant reduction in albumin level, is characteristic of conditions such as infection, malignancy (especially 2-fraction), trauma, surgery, and inflammatory disease. CRP So called because it binds to a polysaccharide (fraction C) in the cell wall of pneumococci. Levels help monitor inflammation/infection (normal 200mg/L; see table 14.6.
Urinary proteins Urinary protein loss >150mg/d is pathological (p294). Albuminuria Usually caused by renal disease (p294). Microalbuminuria: Urinary protein loss between 30 and 300mg/d (so not visible on normal dipstick) and may be seen with diabetes mellitus, BP, SLE, and glomerulonephritis (see p314 for role in DM). Can also be quantified by measuring the urinary albumin:creatinine ratio (A:CR), usually a first-in-the-morning spot urine sample. A level >30mg/mmol indicates albuminuria, and microalbuminuria is defined as >2.5mg/mmol in men and >3.5 in women. This is a useful screening test in diabetics, and subjects with reduced eGFR. Note some labs measure total urinary protein not albumin—a P:CR of 50, is equivalent to an A:CR of 30.20 Bence Jones protein Consists of light chains excreted in excess by some patients with myeloma (p368). They are not detected by dipsticks and may occur with normal serum electrophoresis. Haemoglobinuria Caused by intravascular haemolysis (p336). Myoglobinuria Caused by rhabdomyolysis (p319).
Fig 14.12 A normal electrophoretic scan.
Normal-to-slight elevation Viral infection Steroids/oestrogens Ulcerative colitis
687
SLE
Morbid obesity Atherosclerosis
Clinical chemistry
Table 14.6 C-reactive protein (CRP) Marked elevation Bacterial infection Abscess Crohn’s disease Connective tissue diseases (except SLE) Neoplasia Trauma Necrosis (eg MI)
Clinical chemistry
688
Plasma enzymes Reference intervals vary between laboratories. See p752 for a guide to normal values. Raised levels of specific enzymes can be a useful indicator of a disease. However, remember that most can be raised for other reasons too. Levels may be raised due to cellular damage, cell turnover, cellular proliferation (malignancy), enzyme induction, and clearance. The major causes of raised enzymes: Alkaline phosphatase (Several distinguishable isoforms exist, eg liver and bone.) • Liver disease (suggests cholestasis; also cirrhosis, abscess, hepatitis, or malignancy). • Bone disease (isoenzyme distinguishable, reflects osteoblast activity) especially Paget’s, growing children, healing fractures, bone metastases, osteomalacia, osteomyelitis, chronic kidney disease, and hyperparathyroidism. • Congestive cardiac failure (moderately raised). • Pregnancy (placenta makes its own isoenzyme). Alanine and aspartate aminotransferase (ALT and AST) • Liver disease (suggests hepatocyte damage). • AST also in MI, skeletal muscle damage (especially crush injuries), and haemolysis. -Amylase • Acute pancreatitis (smaller rise in chronic pancreatitis as less tissue remaining). • Also: severe uraemia, diabetic ketoacidosis, severe gastroenteritis, and peptic ulcer. Creatine kinase (CK) A raised CK does not necessarily mean an MI. • Myocardial infarction (p118; isoenzyme ‘CK-MB’. Diagnostic if CK-MB >6% of total CK, or CK-MB mass >99 percentile of normal). CK returns to baseline within 48h (unlike troponin, which remains raised for ~10 days), useful for detecting re-infarction. • Muscle damage (rhabdomyolysis, p319; prolonged running; haematoma; seizures; IM injection; defibrillation; bowel ischaemia; myxoedema; dermatomyositis, p552)— and drugs (eg statins). Gamma-glutamyl transferase (GGT, GT) • Liver disease (particularly alcohol-induced damage, cholestasis, drugs). Lactate dehydrogenase (LDH) • Myocardial infarction (p118). • Liver disease (suggests hepatocyte damage). • Haemolysis (esp. sickle cell crisis), pulmonary embolism, and tumour necrosis. Troponin • Subtypes troponin T and troponin I are used clinically. • Cardiac damage or strain (MI—p118, pericarditis, myocarditis, PE, sepsis, CPR). • Chronic kidney disease (troponin T only; elevation less marked; aetiology unknown).
Hepatic drug metabolism is mainly by conjugation or oxidation. The oxidative pathways are catalysed by the family of cytochrome P450 isoenzymes, the most important of which is the CYP 3A4 isoenzyme. The cytochrome P450 pathway may be either induced or inhibited by a range of commonly used drugs and foods (table 14.7). This can lead to important interactions or side-effects. For example, phenytoin reduces the effectiveness of the contraceptive pill due to more rapid oestrogen metabolism, and ciprofloxacin retards the metabolism of methylxanthines (aminophylline) which leads to higher plasma levels and potentially more side-effects. The BNF contains a list of the major interactions between drugs. Table 14.7 Common inhibitors and inducers of cytochrome P450 isoenzymes Enzyme inducers Phenytoin Rifampicin Carbamazepine Alcohol St John’s wort Barbiturates
Enzyme inhibitors SSRIS Amiodarone Ciprofloxacin Diltiazem Isoniazid Verapamil Macrolides Omeprazole HIV protease inhibitors Grapefruit juice Imidazole and triazole antifungal agents
689
Clinical chemistry
Enzyme inducers and inhibitors
Clinical chemistry
690
Hyperlipidaemia Lipids travel in blood packaged with proteins as lipoproteins. There are four classes: chylomicrons and VLDL (mainly triglyceride), LDL (mainly cholesterol), and HDL (mainly phospholipid) (for abbreviations see footnote3). The evidence that cholesterol is a major risk factor for cardiovascular disease (CVD) is undisputed (‘4S’ STUDY,21 WOSCOPS,22 CARE STUDY, 23 HEART PROTECTION STUDY 24) and indeed it may even be the ‘green light’ that allows other risk factors to act.25 Half the UK population have a serum cholesterol putting them at significant risk of CVD. HDL appears to correlate inversely with CVD. Who to screen for hyperlipidaemia NB: full screening requires a fasting lipid profile. Those at risk of hyperlipidaemia: •Family history of hyperlipidaemia. •Corneal arcus 2. • Hypoalphalipoproteinaemia (Tangier disease): HDL, chol 20mins without a reversible cause. Ask for the opinion of others in the resuscitation team. Resuscitation decisions Consider, discuss, and record CPR decisions: • at the request of a patient with capacity • as part of end-of-life care (p12, p536) • in deteriorating, severe illness. Your patient should be involved in decisions about CPR (unless it would cause physical or psychological harm). Explain your clinical decision to them, including futility. Do not make judgements about the quality of life of others based on your own perception. 1 Meta-analysis fails to show that adrenaline increases survival to hospital discharge ( gov/pubmed/24193240). RCT results are awaited (Paramedic 2: The Adrenaline Trial ISRCTN 73485024).
Useful doses for the new doctor These pages outline the typical adult doses of drugs that a foundation doctor will be called upon to prescribe. Refer to local guidelines first. If in any doubt, consult a drug formulary (eg British National Formulary www.bnf.org) especially if eGFR or weight
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1167 | https://www.subscriptionflow.com/2025/05/guide-to-multi-unit-pricing/ | Strategic Bundling: A Guide to Multi-Unit Pricing
In the modern business landscape, markets are getting highly competitive, and navigating innovative ideas to stand out by building customer loyalty. One notable strategy that has garnered the attention of business owners especially in the SaaS niche is the multi-unit pricing strategy. This pricing strategy revolves around incentivizing customers by offering discounts on bulk buying or availing bundled offers. From ‘buy one get one free’ to ‘buy 6 at the price of 3’, these tactics are helpful to create a perceived value that aligns with customer preferences. But the question is: what drives this perceived value? How do businesses sort out optimal prices and balance them with profitability and customer appeal?
In this blog, we will discuss the concept of multi-unit pricing and how it helps to attract potential customers in the SaaS industry. Moreover, we will talk about the benefits, challenges, best practices and future prospects for multi-unit pricing implementation.
Understanding Multi-unit Pricing
Multiple-unit pricing is a business practice where a business offers consumers a discounted price for a specific number of units of a good/service. For SaaS businesses, this means offering discounts on subscriptions for larger teams and enterprises or long-term commitments. This strategy incentivizes customers to purchase more, driving revenue and growth for the business.
Multi-unit pricing is a strategic business tactic to manipulate the psyche of buyers into buying several products, which they assume they are getting at lower than actual rates. This strategy can be effective for service-based businesses that offer tiered pricing plans where customers can upgrade to premium features or additional services at discounted rates. Interestingly, multi-unit strategy is designed by thinking from a consumer’s viewpoint, thereby making them attracted to bundled offers.
A consumer believes they are saving up by making bulk purchases. They think they are making a profitable deal, as buying one item will be more expensive than buying the same category of items in bulk.
In the SaaS landscape, this could mean customers opting for annual subscription plans rather than monthly ones to save on costs. In general, products are sold at a predetermined price point, often at or below the maximum retail price. In order to boost sales and gain more customers, retailers employ promotional strategies such as discounts, freebies, or bundled deals. SaaS businesses can offer bundled services, such as premium support features or free calls to agents, to upsell customers.
For instance, a retailer might offer a free washing powder with the purchase of a washing machine or a “buy 2 get 1 free” deal on packs of washing powder. Similarly, a software company might offer a free trial or discounted rate for the first month of subscription to engage new customers. This strategy aims to clear inventory, drive sales, and create a perceived value for customers, thus increasing loyalty and retention.
By bundling products or offering freebies, retailers can differentiate themselves and stay competitive in the market. Software businesses can use multi-unit pricing to create a smooth bond with customers, thereby minimizing churn and increasing customer lifetime value.
Exploring the Benefits of a Multi-Unit Pricing Strategy
Multiple-unit pricing strategy has a variety of benefits to offer, such as cost-effectiveness for consumers, reduced costs for businesses, improved revenue and sales, strategic market footprint, and competitive edge. For SaaS businesses, these benefits can be crucial, as they can drive subscription renewals and upselling opportunities. Here we offer an elaborated version of each of the benefits:
Cost-Effectiveness
Multiple unit pricing satisfies customer needs as they make multiple purchases at a lower per-unit price, thereby getting more value for less price. SaaS customers may appreciate discounted rates for annual subscription plans or bulk licenses. This encourages them to buy more as they make a profitable deal, thus leading to loyalty and satisfaction.
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Reduced Costs
The costs of bearing inventory are 20%-30% of the total cost of inventory. By clearing out stock with consumer-friendly deals, businesses can sell high quantities of products in one go, thereby reducing the time that they had to wait in order to sell individual items.
Lower stock carrying costs can offset volume discounts, driving increased profit margins through faster inventory turnover and increased sales. For SaaS businesses, this could mean that they may have to incur lower costs associated with larger customers or handling fewer larger transactions.
Increased Revenue
Bulk purchases can be game changers for any business, as customers are incentivized to buy in large quantities, thereby boosting average sales values and increasing revenues. Software businesses can benefit from increased revenues by offering discounts on large-scale signups for services, leading to increasing average revenue per user. If the profits compress, higher sales can compensate for the difference. Moreover, these prices can stimulate impulse buying, opening doors to additional revenue streams for a business.
Market Position
Multiple unit pricing engages old customers and attracts new buyers to buy products at discounted prices. It convinces them to buy large volumes of products. This approach increases sales, increases customer engagement with your products, and enhances your brand reputation in the highly competitive market.
Data-Driven Decisions
Analysis of bulk purchase data offers insights about customer behaviors, trends, and preferences. This data can help improve future decisions and strategies, thus keeping the brand up to date with the latest consumer buying patterns, and also improves marketing strategies to a more targeted consumer base. For software businesses, it could mean user insights that can help in informed product development and marketing tactics.
Reasons Why SaaS Businesses Use Multi-Unit Pricing
Several reasons contribute to why businesses employ a multi-unit pricing strategy. SaaS businesses can use multi-unit pricing to increase memberships, increase ARPU, and enhance customer lifetime value. This includes increased sales, upselling and cross-selling opportunities, market penetration via the launch of new products, and offering customized deals to attract buyers.
Improved Sales
Multiple unit pricing is employed to encourage sales of products and services. Goods and services are often promoted via this strategy as it improves brand presence in the market and enhances business-customer relationships. Multi-pricing bundles are introduced on holidays, such as Christmas Eve and New Year’s, encouraging bulk purchasing and discounted prices for consumers. By rendering tiered pricing plans, SaaS companies cater to varying customer segments and increase sales.
Market Penetration
In a highly dynamic market, new products take time to compete with existing products. For SaaS companies launching new software or features, multi-unit pricing can be an effective way to attract early adopters and gain attention. In order to enter the competition, sellers employ a penetration strategy, thereby easing customer base expansion. It helps the business to gain market share and boost sales for that particular product.
Win-Win for Businesses
Multi-unit pricing enables businesses to craft personalized deals that drive sales and improve business customer terms. When handling large orders, companies often add additional user licenses or premium support at no charge to reduce per-unit cost, making the deal more compelling to the buyers. In B2B negotiations, multi-unit pricing helps businesses to respond easily to customer needs. For instance, a seller may agree to offer 100 units free with the purchase of 500 units, or a software company may offer discounts on annual subscriptions. In this way, mutually beneficial agreements are reached upon and help build profitable outcomes for businesses.
Analyzing the Challenges of Multi-Pricing Implementation
For businesses in the current landscape, implementation of a multi-pricing strategy can be a challenging task and can complicate business operations. SaaS companies in particular must navigate the intricacies of tiered plans and feature bundling. Here are some top challenges for this type of strategy implementation:
Sense of Confusion
When too many items are bundled as one deal, it can complicate the purchase decision and confuse the buyer. For example, offering bundled products with variations can lead to indecision among customers. To avoid doing so, it is best to simplify offers. Thus, offering few similar types of products is important to avoid buyer frustration.
Flexibility is risked
Another challenge is lack of flexibility in the offers. It means that a one-size-fits-all offer can affect purchasing power on the customer’s part. Those who seek relevant products or software features may feel uncomfortable purchasing services or products that are dissimilar and non-complementary; thus, it will pose a challenge for the business. Businesses must offer customized deals to meet the needs of those who require complementary products or those of a similar type or category.
Communication Gap
Improper communication with regard to offers and price bundling strategies can create misunderstandings. If a business does not educate and inform customers of the savings a customer attains from making a bundled purchase, it can reduce the customer’s interest in your products. Similarly, software companies must clearly communicate the value proposition of product plans to avoid confusion. Businesses must ensure they mitigate any gaps in communicating the offers and seek expert guidance on effective marketing tactics.
Pricing perception
Pricing research and testing must be conducted to ensure products/services are priced strategically in bundled offers. Pricing perception has a crucial role to play here, as products or services sold at a higher price as bundles than when sold separately can create a fear of overspending among customers. Also, if products are cheaply priced, it may create a sense of quality compromise the customer is making. Thus, addressing this challenge requires thorough product research, customer preferences, and expert guidance to avoid wrong perceptions.
Future Prospects in Multi-Unit Pricing Strategy
As more businesses utilize multi-unit pricing to mold their pricing decisions and clear stocks, there is a range of emerging trends that are taking over the business ecosystem. Read on to explore various emerging trends:
Personalization of Prices
One notable trend is the rise of personalized prices as per customer needs. Businesses are now carefully examining data analytics to tailor bundles to customer choices. For example, Netflix can offer a mix of free dramas and music videos as a bundled offer based on user history and preferences.
Subscription models
Subscription models are gaining popularity as more businesses in e-commerce, media, and software niches are adopting membership-based pricing reliant on tiers. With the help of varying tier categorization, customers can opt for the best-suited plan as per affordability. This in turn can lock in revenue streams from various customer segments based on tier memberships.
Demand-Based Prices
Businesses are increasingly adopting real-time price adjustments based on significant factors such as inventory, demand, competitors’ prices, and more. This approach allows customer loyalty and engagement. For instance, a ride-sharing company may charge high prices during peak hours and reduced prices during off-peak hours.
Pricing automation
Use of artificial intelligence-driven platforms has allowed easier price optimization. Automated tools can help figure out the right prices based on data and AI algorithms. This in turn is helping businesses to rule out pricing and revenue maximization schemes that can be beneficial for businesses.
Top Tips for Smooth Multi-Pricing Implementation
Businesses can use the given tips to stay market competitive and fine-tune prices based on real-time analytics and business needs. Here are a few tips, such as
Regular review and adjustment of pricing tiers to meet the market demand and align with customer expectations.
Trying and testing various pricing tiers, such as freemium models, usage-based pricing, and more, to reach an optimal solution for business needs and pricing SaaS products.
Monitor the key metrics such as churn rate, customer lifetime value, and revenue growth to encourage consistent revenue and avoid customer base erosion.
Conduct thorough market research and gauge customer segmentation to maximize revenues via a multi-pricing strategy. Identification of upsell/cross-sell opportunities must be considered.
How can SubscriptionFlow Help
SubscriptionFlow can help to opt for smooth implementation of a multi-unit pricing strategy. Our team, with its years of technical know-how and expertise, can maintain focus on how you price your products. By using pricing automation and gauging data-focused insights, your business can benefit and grow exponentially.
Bottom Line
Optimizing pricing plans can favor software-based businesses operating in the contemporary landscape. By using a multi-pricing strategy, SaaS businesses can drive growth and multiply revenue. Moreover, bundling up premium features with core offerings can contribute to business success. Utilizing well-curated pricing strategies powered by real-time data analytics can help you stay ahead of the curve and maximize profits. Employing automation can allow business growth and seamless financial operations. With our help, you can implement the right business pricing strategy that meets the needs of your business model. Not sure how to begin?
Contact us and learn more.
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Takeaways from This Article
Understanding Multi-unit Pricing
Exploring the Benefits of a Multi-Unit Pricing Strategy
Reasons Why SaaS Businesses Use Multi-Unit Pricing
Analyzing the Challenges of Multi-Pricing Implementation
Future Prospects in Multi-Unit Pricing Strategy
Top Tips for Smooth Multi-Pricing Implementation
How can SubscriptionFlow Help
Bottom Line |
1168 | https://www.geeksforgeeks.org/python/integer-to-binary-string-in-python/ | Integer to Binary String in Python
Last Updated :
23 Jul, 2025
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We have an Integer and we need to convert the integer to binary string and print as a result. In this article, we will see how we can convert the integer into binary string using some generally used methods.
: 77 : 0b1001101: Here, we have integer 77 which we converted into binary string
Using bin()
bin() function is the easiest way to convert an integer into a binary string. It returns the binary with a "0b" prefix to indicate it's a binary number.
Python
````
n = 77
b = bin(n)
print(type(b),b)
````
n = 77 n = 77
b = bin(n) b = bin n
print(type(b),b) print type b b
Output
```
0b1001101
```
Using format()
format() function lets you convert a number to binary without the "0b" prefix. It gives you more control over formatting.
Python
````
n = 81
b = format(n, 'b')
print(type(b),b)
````
n = 81 n = 81
b = format(n, 'b') b = format n 'b'
print(type(b),b) print type b b
Output
```
1010001
```
Using Bitwise operations
This method builds the binary string manually by dividing the number by 2 and collecting remainders. It's slower but helps you understand how binary conversion works.
Python
````
n = 42
orig = n
b = ''
while n > 0:
b = str(n % 2) + b
n //= 2
b = b if b else '0'
print(type(b),b)
````
n = 42 n = 42
orig = n orig = n
b = '' b = ''
while n > 0: while n> 0
b = str(n % 2) + b b = str n% 2 + b
n //= 2 n//= 2
b = b if b else '0' b = b if b else '0'
print(type(b),b) print type b b
Output
```
101010
```
Explanation: We store the original number in orig, then use a loop to divide it by 2 and prepend each remainder (n % 2) to the string b. This builds the binary representation from right to left. If the number is 0, we set b to "0".
Related articles:
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Python Bitwise Operators
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Inward Rectifier K+ Currents and Kir2.1 Expression in Renal Afferent and Efferent Arterioles
Lisa Chilton
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1 Smooth Muscle Research Group, University of Calgary, Calgary, Alberta, Canada
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1 Smooth Muscle Research Group, University of Calgary, Calgary, Alberta, Canada
Correspondence: Dr. Rodger D. Loutzenhiser, Department of Pharmacology and Therapeutics, University of Calgary Faculty of Medicine, 3330 Hospital Drive N.W., Calgary, Alberta T2N 4N1, Canada. Phone: 403-220-8860; Fax: 403-270-2211; E-mail: rloutzen@ucalgary.ca
Received 2007 Jan 11; Accepted 2007 Jul 24.
Copyright © 2008 by the American Society of Nephrology
PMC Copyright notice
PMCID: PMC2391029 PMID: 18178799
Abstract
The afferent and efferent arterioles regulate the inflow and outflow resistance of the glomerulus, acting in concert to control the glomerular capillary pressure and glomerular filtration rate. The myocytes of these two vessels are remarkably different, especially regarding electromechanical coupling. This study investigated the expression and function of inward rectifier K+ channels in these two vessels using perfused hydronephrotic rat kidneys and arterioles and myocytes isolated from normal rat kidneys. In afferent arterioles pre-constricted with angiotensin II, elevating [K+]0 from 5 to 15 mmol/L induced hyperpolarization (−27 ± 2 to −41 ± 3 mV) and vasodilation (6.6 ± 0.9 to 13.1 ± 0.6 μm). This manipulation also attenuated angiotensin II-induced Ca 2+ signaling, an effect blocked by 100 μmol/L Ba 2+. By contrast, elevating [K+]0 did not alter angiotensin II-induced Ca 2+ signaling or vasoconstriction in efferent arterioles, even though a significant hyperpolarization was observed (from −30 ± 1 to −37 ± 3 mV, P = 0.003). Both vessels expressed mRNA for Kir2.1 and exhibited anti-Kir2.1 antibody labeling. Patch-clamp measurements revealed prominent inwardly rectifying and Ba 2+-sensitive currents in afferent and efferent arteriolar myocytes. Our findings indicate that both arterioles express an inward rectifier K+ current, but that modulation of this current alters responsiveness of only the afferent arteriole. The expression of Kir in the efferent arteriole, a resistance vessel whose tone is not affected by membrane potential, is intriguing and may suggest a novel function of this channel in the renal microcirculation.
Vascular smooth muscle exhibits considerable regional heterogeneity in vasomotor mechanisms and ion channel expression patterns. Understanding circulatory regulation at the vascular level requires knowledge of this smooth muscle diversity and its impact on regional control of vascular resistance. The renal microcirculation represents an extreme example of such heterogeneity. The myocytes of the preglomerular afferent arteriole and postglomerular efferent arteriole differ remarkably in Ca 2+ entry mechanisms, myosin expression, and the mechanisms modulating vascular reactivity.1–6 Further regional heterogeneity has been noted for ion channel expression in the juxtamedullary versus cortical efferent arterioles7 and in the activating mechanisms of the contractile pericytes of the descending vasa recta.8 For example, angiotensin II (AngII) responses of the afferent arteriole and descending vasa recta involve membrane depolarization and voltage-gated Ca 2+ entry and are attenuated by L-type Ca 2+ channel antagonists. In contrast, AngII responses of efferent arterioles are not related to membrane potential and involve a voltage-independent Ca 2+ entry mechanism unaffected by L-type Ca 2+ channel antagonists.1,2,4,9
Potassium channels are prominent regulators of smooth muscle function. The inward rectifier K+ channel (Kir) exhibits marked regional differences in its expression and involvement in vascular regulation.10,11 In general, Kir expression and the influence of Kir on membrane potential are reported to be greater in smaller resistance vessels than in larger conduit arteries.12,13 Such a trend may also occur in the kidney. Our laboratory found indirect evidence that Kir plays a prominent role in the afferent arteriole,14 whereas Prior et al.15 found no evidence that Kir influences tone or membrane potential in the upstream arcuate artery, a conduit vessel that does not contribute to renal vascular resistance. Recently, Cao et al.16 demonstrated that the pericytes of the postglomerular descending vasa recta express a Kir current and exhibit K+-induced hyperpolarization. The expression of Kir and its role in the efferent arteriole are unknown.
In this study, we investigated these issues using the in vitro perfused hydronephrotic rat kidney model for in situ contractile and membrane potential measurements, as well as afferent and efferent arterioles isolated from normal rat kidneys for studies of Ca 2+ signaling and to examine the expression of Kir2.1 by reverse transcriptase–PCR and in situ antibody labeling. We used freshly dispersed myocytes obtained from individually isolated afferent and efferent arterioles to study barium-sensitive Kir currents. Our findings indicate that Kir2.1 is expressed in and contributes to the regulation of membrane potential in the myocytes of both the afferent and efferent arterioles; however, the impact of Kir modulation on Ca 2+ signaling, smooth muscle activation, and contractile tone differs in these two vessels.
RESULTS
Figure 1 compares the effects of elevating [K+]o from 5 to 15 mmol/L on the responses of cortical afferent and efferent arterioles to AngII. K+-induced vasodilation is a functional marker for the presence of Kir.10 As shown, this manipulation elicited different responses. In the afferent arteriole, 0.1 nmol/L AngII reduced diameters from 15.1 ± 0.3 to 6.6 ± 0.9 μm (n = 7; P = 0.0001) and elevating [K+]o reversed the vasoconstriction, increasing diameters to 13.1 ± 0.6 μm (P = 0.0003 versus AngII alone). Returning [K+]o to 5 mmol/L restored diameters to 5.8 ± 0.6 μm. The efferent arteriole did not exhibit K+-induced vasodilation. Thus, 0.1 nmol/L AngII reduced efferent arteriole diameter from 14.3 ± 2.2 to 8.2 ± 1.8 μm (n = 7; P = 0.003; Figure 1B), and 15 mmol/L KCl had no effect (8.1 ± 1.5 μm; P = 0.86).
Figure 1.
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Effects of 15 mmol/L KCl on contractile response of afferent (A) and efferent (B) arterioles to AngII (0.1 nmol/L), as observed in the in vitro perfused hydronephrotic rat kidney model. Original tracings (left) and summary data (n = 7; right) illustrate vasodilator response of afferent arteriole to 15 mmol/L KCl and lack of response of efferent arteriole.
Figure 2 illustrates the effects of 15 mmol/L K+ on AngII-induced Ca 2+ signaling in cortical afferent and efferent arterioles isolated from normal rat kidneys. A higher concentration of AngII (10 nmol/L) was used in these in vitro studies to produce a robust Ca signal and facilitate analysis. As shown (Figure 2A), 10 nmol/L AngII increased the fura-2 ratio in afferent arterioles to 142 ± 2% of basal (1.11 ± 0.02 to 1.58 ± 0.05; n = 6; P< 0.0001). The application of 15 mmol/L K+ produced a rapid decrease in [Ca 2+], reducing the fura-2 ratio to 108 ± 2% of basal (1.20 ± 0.04; P = 0.0001 versus AngII alone). Returning [K+]o to 5 mmol/L restored the AngII-induced increase in [Ca 2+] (1.41 ± 0.04; P = 0.004 versus 15 mmol/L KCl). Figure 2B depicts the effects of KCl on AngII signaling in the efferent arteriole. AngII (10 nmol/L) increased the fura-2 ratio from 1.17 ± 0.03 to 1.42 ± 0.03 (n = 6; P = 0.0002); however, increasing [K+]o from 5 to 15 mmol/L had no effect (1.40 ± 0.03; P = 0.65 versus AngII alone).
Figure 2.
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Effects of 15 mmol/L KCl on AngII-induced calcium signaling in afferent (A) and efferent (B) arterioles isolated from normal rat kidneys. As shown in original tracings (left) and mean data (right), 15 mmol/L KCl reduced AngII-evoked increase in calcium in the afferent (A) but not in the efferent (B) arteriole.
We previously showed that barium (10 to 100 μmol/L) elicited membrane depolarization and vasoconstriction and prevented K+-induced vasodilation in the afferent arteriole of the hydronephrotic kidney.14 As illustrated in Figure 3, barium had comparable effects on [Ca 2+] in afferent arterioles isolated from the normal kidney. The fura-2 ratio was 1.08 ± 0.05 (n = 5) in controls and 1.38 ± 0.14 (P = 0.08) after the addition of 30 μmol/L barium and increased to 1.70 ± 0.07 (P< 0.0001) upon the addition of 100 μmol/L barium. Barium also prevented the effects of 15 mmol/L KCl on [Ca 2+] (fura-2 ratio 1.65 ± 0.06; P = 0.60 versus barium alone). In contrast to the afferent arteriole, barium had no effect on [Ca 2+] in the efferent arteriole (fura-2 ratio 1.21 ± 0.05 versus 1.22 ± 0.05; P = 0.89; n = 5; Figure 3, right). These two vessels differ in the role of membrane depolarization and voltage-gated Ca 2+ channels in AngII-induced vasoconstriction and Ca 2+ signaling.1,4 Thus, the lack of effect of 15 mmol/L KCl on contractile responses and of barium on Ca 2+ signaling in the efferent arteriole might reflect this unique characteristic, rather than a lack of expression of Kir.
Figure 3.
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Barium evokes an increase in calcium in the afferent arteriole and blocks the response of this vessel to 15 mmol/L KCl (compare with Figure 2A). Barium has no effect on intracellular calcium in the efferent arteriole (far right).
To address this issue, we determined the effects of 15 mmol/L KCl on the membrane potential of in situ afferent and efferent arterioles. In these experiments, kidneys were first treated with 0.1 nmol/L AngII, the arterioles were impaled, and membrane potential and diameter were monitored during the application of 15 mmol/L KCl. AngII reduced afferent arteriolar diameters from 15.4 ± 2.0 to 8.3 ± 0.4 μm (n = 7). After AngII treatment, afferent arteriolar membrane potentials averaged −27 ± 2 mV (n = 7). The application of 15 mmol/L KCl was associated with vasodilation to 12.2 ± 0.7 μm (P = 0.003; Figure 4A, left) and hyperpolarization to −41 ± 3 mV (P = 0.001; Figure 4A, right). These data are presented in Figure 4B as individual diameter and membrane potentials in AngII before and after treatment with 15 mmol/L KCl. Note the relation between diameter and membrane potential (R = −0.72, P = 0.004).
Figure 4.
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(A) Simultaneous measurements of diameter (left) and membrane potential (right) in afferent arterioles of the in vitro perfused hydronephrotic rat kidney during 0.1 nmol/L AngII-induced constriction and after 15 mmol/L KCl-induced dilation. (B) Individual diameter and membrane potential responses. Note that the vasodilatory response to KCl was associated with hyperpolarization.
The results of similar experiments on efferent arterioles are depicted in Figure 5. AngII (0.1 nmol/L) reduced diameters from 12.4 ± 1.1 to 6.7 ± 0.6 μm (n = 10). Membrane potentials in the AngII-treated efferent arterioles were −30 ± 1 mV. KCl (15 mmol/L) did not alter diameter (7.2 ± 0.9 μm; P = 0.62; Figure 5A, left) but produced a significant hyperpolarization (−37 ± 3 mV; P = 0.003; Figure 5A, right). Individual efferent arteriolar diameter and membrane potential responses are presented in Figure 5B. Note that the membrane potential responses were dissociated from changes in diameter (R = −0.29, P = 0.2). Thus, K+ elicited hyperpolarization in both vessels but altered vasoconstriction and Ca 2+ signaling in only the afferent arteriole.
Figure 5.
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(A) Simultaneous measurements of diameter (left) and membrane potential (right) in efferent arterioles of the in vitro perfused hydronephrotic rat kidney during 0.1 nmol/L AngII-induced constriction and after the application of 15 mmol/L KCl. (B) Individual diameter and membrane potential responses. Note that KCl elicited a significant hyperpolarization in the efferent arteriole but that this response was not associated with vasodilation.
Reverse transcriptase–PCR studies demonstrated message for Kir2.1 in both vessels (Figure 6). Amplification with the primers for Kir2.1 yielded a single 329-bp product using cDNA from both vessels. Sequencing confirmed that this product corresponded to the 862 to 1190 region of the rat Kir2.1 mRNA. No product was seen in the absence of reverse transcriptase. To evaluate the expression of Kir2.1 protein, we used an antibody directed against Kir2.1. Figure 7 depicts afferent and efferent arterioles treated with anti–α-smooth muscle actin (α-SMA; Figure 7, A and C) and anti-Kir2.1 (Figure 7, B and D). Note the anti-Kir2.1 labeling in both α-SMA–positive and -negative (arrows) cells, the latter possibly reflecting Kir2.1 in endothelial cells. Together, these data suggest that both the afferent and efferent arterioles express message for Kir2.1 and Kir2.1 protein.
Figure 6.
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Reverse transcriptase–PCR demonstrates the expression of mRNA encoding Kir2.1 in both afferent and efferent arterioles. Standard ladder is shown on far left. Distilled water control and kidney homogenate (Kid) are shown in the two lanes on the far right. Note that no signal is seen in arteriolar preparations in the absence of reverse transcriptase (-RT).
Figure 7.
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(A and B) anti–α-SMA and anti-Kir2.1 antibody labeling of afferent arteriole. (C and D) Labeling of efferent arteriole. Note anti-Kir2.1 antibody labeling of both α-SMA–positive (myocytes) and α-SMA–negative (arrows) cells in each vessel. (E) α-SMA labeling. (F) Lack of nonspecific labeling using anti-Kir2.1 antibody preabsorbed with Kir2.1 peptide antigen (afferent arteriolar segments).
Finally, we measured whole-cell currents in freshly dispersed myocytes from isolated afferent and efferent arterioles. Figure 8 illustrates the protocol and presents examples of current tracings. Figure 9 presents mean data obtained from 11 afferent and 11 efferent arteriolar myocytes. As shown, myocytes from both afferent (left) and efferent (right) arterioles exhibited Ba 2+-sensitive inwardly rectifying currents in response to a voltage ramp (−120 to 60 mV; Figure 8, middle). The bottom tracings in Figure 8 and summary of mean data in Figure 9C illustrate the difference currents, obtained by subtracting the current measured in 50 μmol/L Ba 2+ from the control currents. The mean currents in Figure 9 are corrected for cell capacitance. As is often the case, the small outward component of the barium-sensitive current is not readily apparent from these data. The small size of the renal arteriolar myocytes and difficulties in obtaining stable patch recordings contribute to this difficulty. Afferent and efferent arteriolar myocytes had capacitances of 6.5 ± 0.5 and 4.9 ± 0.4 pF (P = 0.021), respectively. The barium-sensitive currents clearly display the typical inwardly rectifying characteristics that identify this current as Kir. As shown in Figure 9C, the barium-sensitive current densities were significantly larger in the afferent myocytes (12.6 ± 2.4 versus 6.1 ± 1.5 pA/pF for afferent and efferent myocytes at −100 mV; P = 0.022).
Figure 8.
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Examples of original tracings of whole-cell currents obtained in freshly dispersed myocytes isolated from single afferent (left) and efferent (right) arterioles in 20 mmol/L extracellular K+. (Middle) Four-second ramp protocol. Cells were held in voltage-clamp mode at −40 mV and ramped from −120 mV to 60 mV. Note that the inward current is sensitive to 50 μmol/L barium (top). Lower tracings depict examples of difference currents, obtained by subtracting currents obtained in presence of 50 μmol/L barium from the control currents.
Figure 9.
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(A and B) Current-voltage plots of mean currents in afferent (A) and efferent (B) myocytes (n = 11). Control currents are shown in solid symbols; currents seen in presence of 50 μmol/L barium are indicated by open symbols. (C) Difference currents (control − barium) for afferent (circles) and efferent (squares) arteriolar myocytes.
DISCUSSION
Using diverse and complimentary approaches, this study demonstrated that afferent and efferent arterioles of cortical nephrons express Kir2.1. A previous study,14 using the hydronephrotic kidney preparation, suggested that Kir is a dominant current in the afferent arteriole. The present study extended these findings using direct approaches in afferent arterioles isolated from normal kidneys. The observation that Kir2.1 is also expressed in the cortical efferent arteriole and that elevated extracellular K+ alters membrane potential but not vascular reactivity in this vessel is a novel finding. It is intriguing that this vessel, which does not exhibit voltage-dependent Ca 2+ signaling and does not seem to be regulated by membrane potential, exhibits a Kir current. This finding may suggest a novel role of this channel in the renal microcirculation.
We previously reported that Kir is the dominant current setting membrane potential of the afferent arteriole under basal conditions.14 The present study extended those observations, providing direct evidence that Kir2.1 is expressed in the myocytes of the afferent arteriole and that these cells exhibit a prominent barium-sensitive inwardly rectifying current. This study is the first to demonstrate this current in the contractile vascular myocytes of the afferent arteriole. Consistent with this observation, studies by Kurtz and Penner17 and Leichtle et al.18 reported similar currents in renin-producing juxtaglomerular (JG) cells. Friis et al.,19 however, did not find an inward rectifying current in JG cells. The reasons for these differing observations are unclear; however, JG cells may be recruited from the vascular myocytes,20 which, as this study demonstrated, do express Kir.
The presence of K+-induced hyperpolarization and vasodilation are hallmarks of Kir current in intact vessels.10 Accordingly, the observations of this response in in situ afferent arterioles and the corresponding effects of K+ on Ca 2+ signaling in isolated afferent arterioles provide independent confirmation of functioning Kir currents in intact vessels. Previous studies showed that gene deletion of Kir2.1 eliminates K+-induced vasodilation, implicating this specific isoform in this characteristic response.21 Our observation that Kir 2.1 is expressed in renal arterioles is thus consistent with this view. The underlying mechanism is thought to involve an interaction between external K+ ions and polyamine binding sites within the channel pore, such that an elevation in [K+]o at the outer vestibule of the pore shifts the position of the polyamine and reduces inward rectification.22 Reducing [K+]o has the opposite effect of increasing inward rectification, and we showed that this manipulation evokes afferent arteriolar vasoconstriction,14 further implicating Kir current as the dominant basal K+ current in this vessel. Elevation of [K+]o can also elicit hyperpolarization by stimulating the electrogenic Na+/K+ ATPase15,23; however, we found that the K+-induced vasodilation of the afferent arteriole is insensitive to ouabain but blocked by barium,14 suggesting a predominant role of Kir in this response. The ability of barium to block the effects of 15 mmol/L [K+]o on the Ca 2+ responses of the afferent arteriole (Figure 3) is consistent with this interpretation.
Efferent arteriolar myocytes also exhibited a prominent barium-sensitive, inwardly rectifying current and expressed message for Kir2.1 and Kir2.1 protein. Moreover, elevated extracellular [K+] elicited a significant hyperpolarization of in situ efferent arterioles. This K-induced hyperpolarization mirrors the response of the afferent arteriole and would be consistent with an augmentation of the outward Kir current by external K+; however, we have not eliminated a possible contribution of the electrogenic Na+/K+-ATPase to membrane potential response of the efferent arteriole to elevated [K+]. In contrast to the afferent arteriole, the K+-induced efferent arteriolar hyperpolarization was not associated with vasodilation or altered Ca 2+ signaling. The differing responses of these two arterioles to hyperpolarization may be explained by their differing activation mechanisms. Whereas AngII evokes depolarization4 and activates L-type Ca 2+ channels in the afferent arteriole,1 this agonist activates nifedipine-insensitive Ca 2+ entry in the efferent arteriole.1 Indeed, efferent arterioles of cortical nephrons do not express L-type Ca 2+ channels,7 and the contractile response of these vessels to AngII is dissociated from membrane depolarization.4 Accordingly, we suggest that the divergent responses of these two vessels to K+-induced hyperpolarization simply reflect their differing dependence on voltage-dependent Ca 2+ entry. Similarly, the lack of effect of barium on Ca 2+ signaling in the efferent arteriole (Figure 3, right) may reflect the lack of depolarization-induced Ca 2+ entry. We previously demonstrated that 30 μmol/L barium depolarizes the afferent arteriole.14 Although the membrane potential response of the efferent arteriole to barium was not measured in this study, it is known that depolarization induced by an elevation in extracellular [K+] (30 to 80 mmol/L) does not stimulate Ca 2+ entry or vasoconstriction in this vessel.1,5,6
Our finding that Kir is expressed in afferent and efferent arterioles is consistent with the suggestion that this channel is preferentially expressed in resistance vessels. The finding that the postglomerular efferent arterioles express Kir also agrees with a recent report by Cao et al.16 that Kir currents are observed in the contractile pericytes of the postglomerular descending vasa recta. In contrast, Prior et al.15 demonstrated a lack of Kir in the renal arcuate artery, a conduit vessel. Others have shown Kir expression to exhibit regional heterogeneity within vascular beds and suggested that expression is inversely related to vessel diameter or order. Thus, Quayle et al.12 found Kir channel density was greater in myocytes from smaller versus larger coronary arteries, and Edwards et al.13 found that K+-induced hyperpolarization was greater in smaller versus larger segments of rat cerebral vessels. This study, in concert with the observations of Prior et al.15 and Cao et al.,16 suggest a similar distribution within the renal vascular bed.
The preferential expression of Kir channels in resistance vessels suggests an importance in the regulation of blood flow and BP; however, the physiologic roles of Kir in the renal microcirculation are not understood. In the cerebral and skeletal muscle vascular beds, elevated interstitial [K+] is an important signal associated with increased activity and is thought to trigger increased blood flow in part via Kir.24 It is not known whether such a mechanism is important in the kidney. It has been suggested that endothelium-derived K+ is a hyperpolarizing factor (EDHF) and that K+ efflux through endothelial charybdotoxin- and apamin-sensitive K+ channels elevates [K+]o to evoke hyperpolarization, in part, by Kir.25 Responses attributed to EDHF are commonly observed in resistance vessels, but not all vessels exhibiting such responses express Kir channels.26 Moreover, we found that the charybdotoxin- and apamin-sensitive EDHF response of the afferent arteriole is insensitive to barium, ruling out a contribution of Kir.27 A recent study28 suggested that Kir may be involved in pressure-induced depolarization and myogenic signaling, in that hypotonic cell swelling reduced Kir currents, implicating regulation by mechanosensitive mechanisms. Our previous attempts to ascertain the involvement of Kir in afferent arteriolar myogenic signaling were compromised by the marked vasoconstrictor response to barium,14 and a possible role of Kir in this response remains an intriguing possibility.
When considering the role of K+ channels in the vasculature, one generally focuses on their potential involvement in the membrane-dependent regulation of tone. This is reasonable when considering the role of Kir currents in the afferent arteriole; however, what could be the function of Kir in the efferent arteriole, a vessel that is not modulated by membrane potential? An interesting possibility relates to recent suggestions that Kir plays an essential role in transmitting electrical signals along the microvasculature, an important characteristic of small vessels expressing Kir channels. The transmission of such signals is blocked by barium, suggesting an essential role of Kir channels.29–32 Whether Kir plays a physiologic role in the efferent arteriole or is simply a vestigial current in this vessel is not known; however, one might speculate that its presence in this vessel could facilitate the transmission of electrical signals to the upstream preglomerular afferent arteriole, where electrical signals might modulate preglomerular tone and GFR.
Although the physiologic role of Kir in the renal microcirculation is not resolved, this study demonstrates that this current is expressed in the myocytes of both afferent and efferent arterioles. Modulation of the outward Kir current alters tone and Ca 2+ signaling in the afferent arterioles but not in the efferent arteriole, a vessel that lacks electromechanical coupling. The finding that Kir is expressed in the afferent arteriole and that its modulation alters vessel reactivity is consistent with a traditional view that, by affecting membrane potential, Kir plays an important role in the regulation of vasomotor tone and blood flow in the resistance circulation. Our finding that this current is also prominent in the myocytes of the efferent arteriole, a vessel whose tone is not modulated by membrane potential, may suggest a role in the microcirculation that extends beyond this traditional view.
CONCISE METHODS
All animal protocols were approved by the University of Calgary Animal Care Committee in accordance with the Canadian Council on Animal Care. The in vitro perfused hydronephrotic rat kidney was used to assess contractile and membrane potential responses. Unilateral hydronephrosis was induced in male Sprague-Dawley rats by ligation of the left ureter. After 6 to 8 wk, the rats were anesthetized, the renal artery was cannulated in vivo, and the kidney was excised with continuous perfusion. Kidneys were perfused in vitro with modified DMEM (Life Technologies, Gaithersburg, MD) containing (in mmol/L) 1.6 Ca 2+, 26 bicarbonate, 5 glucose, 1 pyruvate, and 5 HEPES. Ibuprofen (10 μmol/L) was added to eliminate the effects of renal prostanoids.33 Perfusion pressure within the renal artery was held at 80 mmHg. Vessel diameters were measured by on-line image processing, and membrane potentials were measured using a dual-pipette system as described previously.4,14,34 Elevated KCl solutions were prepared by isotonic substitution for NaCl.
Afferent and efferent arterioles were isolated from the renal cortex (excluding the juxtamedullary region) of normal rat kidneys using the gel perfusion technique previously described.1 For Ca 2+ signaling, fura-2–loaded vessels were superfused (2 ml/min) in a custom chamber held at 37°C and equilibrated with 5% CO 2. The 340/380 emission fluorescence ratio was used as a qualitative index of cellular Ca 2+ signaling as described previously.1 The expression of Kir2.1 was examined using a nested PCR approach. Total RNA was extracted (RNEasy micro kit; Qiagen, Mississauga, Ontario), and an outer reaction (20 cycles) was performed using primers corresponding to nucleotides 758 to 777 (gttcgatagcggaatcgac) and 1214 to 1234 (cctggttgtggagatctatgc). The nested reaction (35 cycles) was performed using primers corresponding to nucleotides 862 to 883 (gacaatgcagacttgaaatcg) and 1168 to 1190 (ctctctggaactccgttctcac) to yield a 329-bp amplicon. The products were sequenced at the University Sequencing Facility (www.sequencing.ucalgary.ca). For determination of the presence of immunoreactive Kir2.1 protein, vessels were fixed (1% formalin) and treated with 2% Triton X-100. Primary antibodies against α-SMA (mouse monoclonal, A2547; Sigma, St. Louis, MO) and Kir2.1 (rabbit polyclonal, AB5374; Chemicon, Temecula, CA) and secondary antibodies (Cy3–anti-rabbit [Jackson Laboratories, Bar Harbor, ME] and Alexa 488–anti-mouse [Molecular Probes, Eugene, OR]) were used. For ruling out nonspecific interactions, vessels were treated with Kir2.1 antibody pre-equilibrated with the peptide antigen.
Patch-clamp studies were performed on myocytes isolated from single afferent and efferent arterioles, obtained as described previously. Only cells exhibiting the characteristic spindle-shape morphology for the afferent arteriole and elongated smooth muscle morphology with characteristic bifurcated ends for the efferent arteriole (see Loutzenhiser and Loutzenhiser1) were studied. Borosilicate glass pipettes were pulled to a 2- to 3-μm tip diameter, fire polished to a resistance of 4 to 6 MΩ, and coated with Sylgard polymer. Pipettes were filled with an internal solution containing (in mmol/L) 120 K-gluconate, 20 KCl, 1.5 ATP-Mg, 0.1 GTP, 10 HEPES, 5 EGTA, 2 MgCl 2, and 0.1 CaCl 2 (pH 7.2). The external solution contained 105 mmol/L NaCl, 20 mmol/L KCl, 2 mmol/L MgSO 4, 1.5 mmol/L CaCl 2, 0.2 μmol/L nifedipine, 5 μmol/L glibenclamide, and 10 mmol/L HEPES (pH 7.4). A 4-s voltage ramp protocol (0.045 mV/ms, −120 to 60 mV) was applied in the whole-cell configuration using pClamp (v8) and an Axoclamp 200B amplifier (Axon Instruments, Union City, CA). Data were acquired at 10 KHz with low-pass filter at 5 KHz. Voltages were not corrected for the liquid junction potential (15 mV, calculated using pClamp).
Data are expressed as means ± SEM. Differences between means for single comparisons were evaluated by paired or unpaired t test. P< 0.05 was considered significant. For multiple measurements, ANOVA followed by Bonferroni t test were applied to assess significance.
Disclosures
None.
Acknowledgments
This study was supported by grants from the Canadian Institutes for Health Research, the Heart and Stroke Foundation of Alberta and Nunavet, the Natural Sciences and Engineering Research Council, and the Alberta Heritage Foundation for Medical Research (AHFMR). R.L. is an AHFMR Scientist, and L.C. was supported by studentships from AHFMR and the Medical Research Council and is an Honorary Killam Memorial Scholar.
This work was previously presented, in part, at the annual meeting of the American Society of Nephrology (November 12 through 17, 2003; San Diego, CA) and published in abstract form.35
We gratefully acknowledge the technical assistance of Bob Winkfein in designing the PCR studies. We also thank Dr. Robert Clark and Dr. Philip Aaronson for advice in the preparation of this article.
Published online ahead of print. Publication date available at www.jasn.org.
L.C. and K.L. contributed equally to this work.
References
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1170 | https://uspas.fnal.gov/materials/12MSU/xverse_dynamics.pdf | USPAS notes Barletta,Spentzouris,Harms 1 Overview on Magnetic fields When accelerating particles in a linear accelerator, the machine must get longer to reach higher final particle energies. At high enough energies, the linac becomes awkwardly long.
One solution is to take particles exiting an acceleration section and bend them back around to the entrance of the acceleration section so that the accelerator can be used repeatedly.
The magnetic fields of dipole magnets are used to bend particles. Dipole magnets can be used to keep particles moving in a circular trajectory, either to store them (storage ring), or to make them repeatedly pass through an acceleration section inserted into the circular path (circular accelerator). Circular accelerators or storage rings do not always have a perfectly circular shape, other shapes such as a rounded triangle can be used since the beam trajectory still closes on itself. However, a perfect circle has more uniform bending and so may be the easiest example with which to start.
In order for a particle to move in a circle, there must be a centripetal force. To move a negatively charged particle in counterclockwise circular motion as shown below, the magnetic field must be directed out of the plane of the figure. The magnetic force on the particle is given by ⃗ F = q⃗ v × ⃗ B, where q is the charge of the particle (including the sign of the charge), ⃗ v is the particle velocity, and ⃗ B is the magnetic field through which the particle travels. The direction of the particle velocity at any point in the circular trajectory is tangent to the circle. The force, velocity and magnetic field vectors are mutually orthogonal. Given that the force is directed radially inward, and that the velocity is tangent to the trajectory, the magnetic field must be directed upward, perpendicular to the plane containing the circular particle trajectory.
F v 1 As an example, consider an antiproton (same charge as an electron) with a total energy of 9 GeV. Suppose we want to bend it around a circle (storage ring) of circumference 500 m.
Calculating how much bend per length is needed; deflection meter = 2π radian 500 m = 0.013 rad/m 2 Field required to Bend particles A calculation of basic importance in the manipulation of charged particle beams is the determination of the deflection of a particle by a magnetic or electric field. The force on a particle must have a component perpendicular to the direction of motion in order to bend its trajectory.
+ + + + + + + + + + + + −−−−−−−−−−−−−− θ θ v B E Figure 1: The left figure shows a negatively charged particle getting bent in an electrostatic field. The right figure shows a negatively charged particle getting bent in a magnetic field.
The magnetic field points out of the page.
Particles in a beam typically have large momenta in the direction motion, p∥. As a high energy particle travels through a field, the transverse momentum, p⊥, imparted by the field to turn the particle is much smaller than the momentum in the direction of motion.
p p ∆θ p Figure 2: An sketch showing parallel and perpendicular components of particle motion (scale of perpendicular momentum exaggerated for clarity).
In other words, in traveling through the bending field, the particle is turned through a 2 small angle. So, the small angle approximation may be used to simplify calculations.
tan (∆θ) ≃sin (∆θ) ≃∆θ ≃p⊥ p∥ ≃p⊥ p Writing the change in the transverse momentum as ∆p, the longitudinal momentum as p, and the angle change imparted by the field as ∆θ: ∆θ = ∆p p = 1 p ∆p ∆t ∆t ∆z∆z = 1 p ∆z ∆t ∆p ∆t ∆z where ∆z is the short distance traveled through the field in a time ∆t. Going to differential notation: ∆θ = 1 p dz dt dp dt dz = 1 pv F dz The total angle change through a field region of length L: θ = 1 pv Z L 0 F dz Suppose we want to bend a negatively charged particle with total energy 9 GeV and charge of magnitude e through an angle of 0.013 rad as it travels one meter in the direction of motion. This particle has a beta of β = .995, and its energy over charge is 9GeV e = 9 × 109 volts. Let’s find the strength needed from an electric field to bend the particle. Let the direction of beam motion be ˆ z, and the desired deflection be in the ˆ x 3 direction. If a particle is to be bent by an electric field, the field must be directed in the −ˆ x direction, or, FEˆ x = −eEx(−ˆ x) = eExˆ x.
θtotal = 1 pv Z L 0 eExdz Solving for the integrated electric field, Z L 0 Exdz = θpv e = θ(γmc2)β2 e ExL = (13 × 10−3rad)(9 × 109eV)(.99) e where γmv = γmβc is the momentum of the particle, γmc2 is the energy of the particle, β = v c, and in this case, β2 = .99. For a one meter long electrostatic device, the following field is needed: Ex = 116 MV m If a particle is to be bent by a magnetic field instead, the field must be directed in the ˆ y direction, or, FBˆ x = −evˆ z × Byˆ y = evByˆ x.
θtotal = 1 pv Z L 0 evBydz = e p Z L 0 Bydz Note that the angular deflection a particle will experience is proportional to its charge and inversely proportional to its momentum. A single bending dipole may be used as a spectrometer, spreading out the particles in a beam according to their momenta. Solving the previous equation for the integrated magnetic field, Z L 0 Bydz = θp e = θ(γmc2)β ec 4 ByL = (.95)(13 × 10−3rad)(9 × 109)V 3 × 108m/s For a one meter dipole magnet, the following field is needed: By = 0.37 T The extra factor of v = βc in the magnetic force makes a large difference in the scale of the needed magnetic field compared to electric field.
3 Magnetic rigidity The ’magnetic rigidity’ of a beam is a convenient parameter defined as the following: p e = Bρ where p is the magnitude of the particle momentum, e is the charge of the particle, B is magnetic field, and ρ is the bending radius of a particle immersed in a magnetic field B.
The ratio of p to e describes the ’stiffness’ of a beam, it can be considered as a measure of how much angular deflection results when a particle travels through a given magnetic field. For a specific magnetic field, the greater the momentum of a particle, the less it will be bent as it travels through that field. The greater the charge of a particle, the more it will be bent as it travels through a given magnetic field.
This ratio can be put in a convenient form if both p and e are multiplied by c, the speed of light. Then, Bρ = p e = pc ec. In SI units this is: (Bρ)[Tesla m] = cp ce [joule] [C m/s] = q E2 total −E2 rest (3 × 108)(1.6 × 10−19) [joule] [C m/s] since cp = q E2 total −E2 rest. It is convenient to specify cp in units of MeV. To keep the expression in SI units, it must be multiplied by the joule to MeV conversion factor. Then we have (still in SI units): (Bρ)[Tesla m] = cp[MeV ] (3 × 108)m/s (1.6 × 10−19[joule]) 10−6[MeV ] 5 Upon simplification this is: (Bρ)[Tesla m] = cp[MeV] 300[MeV] [joule s] [m coul] = cp[MeV] 300[MeV][Tesla m] Checking the units: [joule s] [m coul] = [nt m s] [m coul] = [nt] [coulm s ] = [Tesla m] Example 1: An electron beam with total energy 6 GeV. If the rest energy of the particles is small compared to the total energy, the total energy may be used for cp to good approximation. Then it is easy to calculate Bρ; in this case, Bρ = 6000/300 Tesla m= 20 Tesla m. A 2 Tesla magnet would then have a bending radius of 0.1 meter.
Alternatively, if a ring were made up completely of dipoles, for a ring radius of 1000 m, the field in the dipoles would have to be Bρ ρ = 20/1000 = .02 Tesla.
Example 2: Fermilab Booster at injection (protons). The total energy cannot be used, the particles are not relativistic enough. The kinetic energy of the particles at injection is 400 MeV. So, the numerator is given by pc = q (1338)2 −9382 MeV. The denominator is just c, the e cancels with the e in MeV to put the ratio in SI units.
pc ec = 9.54 × 108 3 × 108 = 3.18 T-m 4 Finding an equation for transverse motion The design of an accelerator or beamline defines a desired trajectory for the particle motion. (From now on, I’ll use the term ’machine’ to mean accelerator, storage ring, or beamline.) This desired trajectory is called the ’reference trajectory’ or ’ideal trajectory’.
In the case of a circular path it is often called the ’reference orbit’ of the particle. Since each particle is supposed to follow this path, the parameters of interest for the motion become the deviation of a particle from this path, rather than its absolute coordinates with respect to a fixed reference. Generally, convenient particle coordinates describe errors in the motion; deviations of position and direction from the ideal trajectory. A particle also potentially has a deviation of its momentum from the ideal momentum, and an error in longitudinal location. However, for the purpose of understanding transverse 6 motion, for now it will be assumed that all particles have the design momentum, and ideal longitudinal position.
Finding an equation of motion is often done by writing a force equation. The harmonic oscillator is a familiar example. For the case of a spring with spring constant k this is F = ma = −kx, or d2x dt2 + k mx = 0.
Let’s find the equation of transverse motion in a simple case for a particle undergoing motion in a magnetic field at a constant energy. We make a bunch of assumptions, which usually reflect the situation of a high energy particle in a ring or transport line. First, the energy is constant (or varying slowly, and so we don’t worry about it). Secondly, the magnetic field only has components transverse to the direction of motion, ⃗ B = Bxˆ x + Byˆ y. Thirdly, the component of the particle velocity along the direction of motion, vs, is much greater than the transverse components of particle velocity, vs >> vx, vy. Since the only force is magnetic, the force equation is, F = dp dt = γm⃗ ¨ R = e(v × B) where v is the velocity of the particle, m is the particle mass, ⃗ R is the transverse position vector which can also be written as ⃗ R = (ρ + x)ˆ x + yˆ y with ρ the radius of curvature. We are assuming the charge of a proton here, +e. Letting v ∼vs, the cross-product v × B is: v × B = ˆ x ˆ y ˆ z 0 0 v Bx By 0 = −vByˆ x + vBxˆ y If ρ is constant, then we have: γmd⃗ R2 d2t = γm dx2 d2t ˆ x + dy2 d2t ˆ y !
= −evByˆ x + evBxˆ y Now let’s convert from a time derivative to a derivative with respect to the beam direction, s. Use s = vt, then d dt = ds dt d ds = v d ds. Then: v2dx2 d2s = −1 γmevBy v2dy2 d2s = 1 γmevBx 7 These differential equations can be written: dx2 d2s + e pBy = 0 dy2 d2s −e pBx = 0 The case of motion through an ideal quadrupole magnet will be considered. The ideal particle goes through the centers of the dipole and quadrupole magnets of the beamline.
The dipole field strength is required to bend particles to follow the reference trajectory, and affects all particles the same way. Since we are considering only deviations from the reference trajectory, bending magnets and sections with no magnets can be neglected in this simplest description of particle motion. On the other hand, the quadrupole magnets have strength that increases linearly with position from the center of the magnet, providing a linear restoring force to particles with position and direction errors. The magnetic quadrupole field components have the form: Bx = ∂Bx ∂y y + ∂Bx ∂x x By = ∂By ∂x x + ∂By ∂y y In the absence of coupling, Bx = ∂Bx ∂y y By = ∂By ∂x x Using the curl equation, ∇× B = 0, we find that ∂Bx ∂y = ∂By ∂x , and we can abbreviate the magnetic field gradient as ∂By ∂x = B ′.
The equations describing motion through a quadrupole may be written: dx2 d2s + e pB ′ !
x = 0 8 dy2 d2s − e pB ′ !
y = 0 In a high energy particle accelerator, a particle passes from one magnetic element to another, and so the magnetic force on the particle changes with longitudinal location, s, around the ring (or through the beamline). So, a simple model of linear motion (no magnetic elements of higher order than a quadrupole) through a beamline or accelerator is Hill’s equation; x ′′ + K(s)x = 0 where the derivative is with respect to longitudinal position, not time. There is a similar equation for motion in the y direction. The restoring force is a function of longitudinal position, and varies according to the quadrupole magnets through which the particle traverses. The general solution to Hill’s equation is the following: x(s) = A q β(s) cos (ψ(s)) + B q β(s) sin (ψ(s)) where A and B are arbitrary coefficients that depend on initial conditions. The function β(s) describes how the amplitude of motion varies with longitudinal position, and it depends on the distribution and strength of the quadrupoles. The function ψ(s) describes how the phase of the motion evolves along the particle trajectory, and also depends on the distribution and strength of the quadrupoles. Note that since there is transverse motion in both the x and y directions, there are two sets of beta functions, βx and βy, and separate phase evolutions ψx and ψy for each direction.
Hill’s equation looks like an oscillator equation, although the amplitude and phase evolution of the ’oscillation’ vary with longitudinal position. The transverse motion of a particle through a series of quadrupoles and other machine components is called a betatron oscillation. It is the quadrupole magnets that determine the character of the betatron oscillations.
5 Machine tune and the Twiss parameters The beta function β(s) and the phase advance of the betatron oscillation ψ(s) are not independent. Their relationship can be found by stuffing the solution to Hill’s equation back into Hill’s equation. The result is terms that go as either cos (ψ(s)) or sin (ψ(s)).
Once the terms are grouped, the coefficients of cos (ψ(s)) and sin (ψ(s)) must 9 independently vanish since they are orthogonal functions. Applying this requirement to the coefficient of the sine term leads to a relationship between the phase advance, ψ(s), and the beta function, β(s). Application to the cosine term leads to an equation governing the evolution of the beta function (or beam envelope). For simplicity, write the solution to Hill’s equation as x(s) = A q β(s) cos (ψ(s)) Take the derivative of the position x(s) twice: x ′′(s) = − A 4(β(s)) 3 2 (β ′)2 cos (ψ(s)) + A 2 q β(s) (β ′′) cos (ψ(s)) − A q β(s) β ′ψ ′ sin (ψ(s)) −A q β(s))(ψ ′)2 cos (ψ(s)) − A q β(s))(ψ ′′) sin (ψ(s)) Substitute x(s) and x ′(s) into Hill’s equation, x ′′ + K(s)x = 0, and group the cosine and sine terms, A − 1 4(β(s)) 3 2 (β ′)2 + 1 2√ β(s)β ′′ − q β(s)(ψ ′)2 + Kx q β(s) cos (ψ(s)) + A − 1 √ ββ ′ψ ′ −√βψ ′′ sin (ψ(s)) = 0 Set the coefficient of the sine term to zero, β ′ψ ′ −βψ ′′ = 0 d(βψ ′) ds = 0 βψ ′ = constant Let the constant be one (it is unconstrained, so we can choose it), then, ψ(s) = Z ds β 10 If the machine is a ring, the total phase advance for one revolution is called the ’tune’, ν: ν = 1 2π I ds β The tune, ν, is the total number of betatron oscillations per turn around the machine.
There is a betatron tune for each transverse plane, νx and νy. The horizontal and vertical betatron oscillations in a (now decommissioned) accelerator called the ’Main Ring’ are shown in Fig. 3. The horizontal (vertical) beam position is plotted as a function of location in the ring. For viewing purposes, the ring is ’cut’ at a chosen location and unfolded into a linear picture, ending once again at the initial location.
Figure 3: Horizontal and vertical closed orbits in the (now extinct) Main Ring at Fermilab.
The alpha function, α(s), is defined to be proportional to the slope of the beta function; α(s) = −1 2 dβ(s) ds The gamma function, γ(s), is a combination of α and β that simplifies the notation in some common expressions; γ(s) = 1 + α(s)2 β(s) The three machine functions β(s), α(s), γ(s) are referred to as the ’Twiss parameters’.
11 6 Transfer maps Suppose that a particle has an initial position x0 and angle x ′ 0 at a machine location with Twiss parameters β0, α0, and γ0. When the particle goes through an orbit of a storage ring (or other repeated group of magnets), its new position x and angle x ′ may be found using the following transfer map; M = cos (∆ψ) + α0 sin (∆ψ) β0 sin (∆ψ) −γ0 sin (∆ψ) cos (∆ψ) −α0 sin (∆ψ) !
where ∆ψ is the change in phase from the initial location to the final location at the end of the repeated period. If the repeated period is an orbit, then ∆ψ is 2πν, where ν is the betatron tune.
M = cos (2πν) + α0 sin (2πν) β0 sin (2πν) −γ0 sin (2πν) cos (2πν) −α0 sin (2πν) !
When a particle propagates from an arbitrary initial location with Twiss parameters β1, α1, and γ1 to another arbitrary location with Twiss parameters β2, α2, and γ2, then the evolution of the position and angle coordinates of a particle can be found using the following transfer map; M12 = β2 β1[cos (ψ12) + α1 sin (ψ12)] (β1β2) 1 2 sin (ψ12) −1+α1α2 (β1β2) 1 2 sin (ψ12) + α1−α2 (β1β2) 1 2 cos (ψ12) β1 β2 1 2 cos (ψ12) −α2 sin (ψ12) where ψ12 is the phase advance in going from location one to location two.
6.1 FODO example The Twiss parameters, β, α, and γ may be related to magnet element parameters such as focal length, f, and drift length, L by equating equivalent transfer maps. Each magnetic element has a transfer map that allows the calculation of the position and angle of an exiting particle given its incoming position and angle. When magnetic elements are connected together to make a string of different elements (transport line), the transfer map of the entire string may be found by matrix multiplication of the individual transfer maps. Finally, this may be equated to the corresponding transfer map written using the Twiss parameters. A classic calculation for a FODO cell will be done as an example. The pattern of magnets is denoted by the abbreviation ’FODO’. This stands for Focusing 12 quadrupole (F), drift space (or equivalently a bending magnet, O), Defocusing quadrupole (D), drift or bend (O). A ’lattice’ is the pattern of magnets in an entire machine. A simple lattice would be to repeat the FODO sequence again and again. For calculation purposes, the repeated cell would start and end in the middle of an element to insure matching of the beta and alpha functions at the cell boundary. This way, the cell can fairly be considered a repeated section. The FODO cell will start in the middle of the focusing quadrupole for this example, as shown in Fig. 4. Half a focusing quadrupole has half the strength of the original, and so twice the focal length.
L L 2f f 2f Figure 4: FODO cell, starting and ending in the center of the focusing quadrupole.
The maximum size of the beta function, βmax, will be in the middle of the focusing quadrupole, while the minimum size of the beta function, βmin will be in the middle of the defocusing quadrupole. The alpha function, α, will be zero at the centers of the quadrupoles. Since α is given by the slope of the beta function, it will be zero at the locations where the beta function has a minimum or maximum. The matrix for a repeated section written in terms of the Twiss parameters at the start (end) of a repeated section is cos (∆ψ) + α sin (∆ψ) β sin (∆ψ) −γ sin (∆ψ) cos (∆ψ) −α sin (∆ψ) !
= cos (∆ψ) βmax sin (∆ψ) − 1 βmax sin (∆ψ) cos (∆ψ) !
(1) The matrix for the repeated section may also be calculated based on the thin lens elements in that section, multiplying the matrices from the upstream to downstream ends. 1 0 −1 2f 1 ! 1 L 0 1 ! 1 0 1 f 1 ! 1 L 0 1 ! 1 0 −1 2f 1 !
= 1 −L2 2f2 2L + L2 f −L 2f2 + L2 4f3 1 −L2 2f2 (2) To find the phase advance, ∆ψ, equate the traces of the matrices from Eq. 1 and Eq. 2.
2 cos (∆ψ) = 2 −L2 f 2 13 = 2 1 − √ 2L 2f ! 1 + √ 2L 2f !
Use the identity, cos (∆ψ) = 1 −2 sin2 ∆ψ 2 !
= 1 + √ 2 sin ∆ψ 2 !! 1 − √ 2 sin ∆ψ 2 !!
So that, sin ∆ψ 2 !
= L 2f Once the phase advance is known, βmax can be found by equating matix elements M12: βmax sin (∆ψ) = 2L 1 + L 2f !
= 2L 1 + sin ∆ψ 2 !!
and with the help of some algebra and trigonometric identities, βmax = 2L "1 + sin ( ∆ψ 2 ) sin (∆ψ) # = 2f "1 + sin ( ∆ψ 2 ) 1 −sin ( ∆ψ 2 ) # 1 2 One way to find βmin would be to follow this procedure again, but start the cell in the middle of the defocusing quadrupole, rather than the focusing quadrupole. Then, βmin = 2L "1 −sin ( ∆ψ 2 ) sin (∆ψ) # 14 References H. Wiedemann, ’Particle Accelerator Physics I’, Springer-Verlag, 2003. ISBN 3-540-00672-9 D.A. Edwards, M.J. Syphers, ’An introduction to the physics of high energy accelerators’, Wiley, 1993. ISBN 0-471-55163-5 Karl L. Brown, Roger V. Servranckx, ’Optics modules for circular accelerator design’, SLAC-PUB-3957, (1986) 15 |
1171 | https://www.frontiersin.org/journals/oral-health/articles/10.3389/froh.2025.1520067/full | Your new experience awaits. Try the new design now and help us make it even better
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Front. Oral Health, 27 January 2025
Sec. Oral Health and Nutrition
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This article is part of the Research TopicNutrition and Oral Health: At the Micro-LevelView all 5 articles
Hypovitaminosis and its association with recurrent aphthous stomatitis: a comprehensive review of clinical correlations and diagnostic considerations
Alessio Rosa1Giovanni Cianconi1Riccardo De Angelis1Alberto Maria Pujia2Claudio Arcuri3
1Department of Chemical Science and Technologies, Dentistry, University of Rome Tor Vergata, Rome, Italy
2Department of Biomedicine and Prevention, University of Rome Tor Vergata, Rome, Italy
3Department of Clinical Sciences and Translational Medicine, University of Rome Tor Vergata, Rome, Italy
Background: Hypovitaminosis, or vitamin deficiency, has been increasingly recognized as a potential contributing factor in the development of recurrent aphthous stomatitis (RAS), a condition characterized by the periodic formation of painful ulcers in the oral mucosa.
Materials and methods: This mini review includes a literature search on PubMed, Web of Science, and Scopus databases using keywords “hypovitaminosis AND aphthous ulcers.”
Results: There is a growing body of evidence supporting the link between various vitamin deficiencies—particularly vitamins B12, C, and folate—and the prevalence of RAS, with implications for both diagnosis and management.
Conclusion: This review aims to outline the clinical and biochemical findings associated with hypovitaminosis in individuals presenting with RAS, emphasizing the diagnostic importance of recognizing vitamin deficiencies in these patients and exploring possible therapeutic approaches.
1 Introduction
Recurrent aphthous stomatitis (RAS) is a common oral mucosal disorder presenting as painful, recurrent ulcerations of the oral cavity. RAS lesions typically appear as multiple, small, shallow, irregular ulcers with well-defined borders surrounded by erythematous haloes. While the etiology remains multifactorial, an emerging area of study highlights the role of hypovitaminosis in RAS, particularly deficiencies in vitamins B12, C, and folate. Etiological factors such as gastrointestinal malabsorption, systemic diseases, and dietary insufficiencies can impair vitamin absorption, contributing to hypovitaminosis. Vitamin deficiencies can compromise mucosal integrity, impacting cellular repair and immune response (1–3). Recognition of hypovitaminosis as a possible underlying factor in RAS offers a valuable diagnostic and therapeutic opportunity for clinicians involved in patient care.
To definitively diagnose hypovitaminosis and differentiate overlapping clinical features of vitamin deficiencies, laboratory investigations and biochemical analyses are essential. For vitamin B12 deficiency, serum vitamin B12 levels (<200 pg/ml diagnostic, 200–300 pg/ml requiring further testing), elevated methylmalonic acid (MMA), homocysteine levels, and complete blood count (CBC) findings such as macrocytosis and hypersegmented neutrophils are critical (4). Testing for intrinsic factor and parietal cell antibodies aids in identifying pernicious anemia as a cause. For vitamin C deficiency, plasma ascorbic acid levels (<0.2 mg/dl diagnostic) and erythrocyte antioxidant capacity provide reliable markers. Folate deficiency diagnosis involves serum folate (<3 ng/ml), red blood cell (RBC) folate levels (<140 ng/ml), and elevated homocysteine levels, with normal MMA helping differentiate from B12 deficiency (5–9). A comprehensive approach, including simultaneous measurement of B12, folate, and vitamin C, along with advanced tools like genetic testing for metabolic polymorphisms (e.g., MTHFR for folate metabolism), ensures accurate diagnosis and effective management. This multifaceted strategy prevents reliance solely on clinical findings and supports tailored treatment plans for conditions like recurrent aphthous stomatitis (RAS) (10–12).
The role of vitamin B12, vitamin C, and folate in the etiopathogenesis of recurrent aphthous stomatitis (RAS) can be detailed by exploring their biological functions, contributions to oral mucosal health, and involvement in metabolic pathways. Vitamin B12 is crucial for DNA synthesis, red blood cell formation, and maintaining neuronal health. Its deficiency impairs cellular repair and mucosal integrity, leading to increased susceptibility to RAS. B12 is metabolized through the cobalamin pathway, requiring gastric intrinsic factor for absorption, and is primarily excreted through bile (13–15). Disruptions in these processes, such as in malabsorption syndromes or pernicious anemia, can exacerbate deficiencies. Vitamin C plays a vital role in collagen synthesis and tissue repair by acting as a cofactor for prolyl and lysyl hydroxylase enzymes, which stabilize collagen structure. Deficiency impairs wound healing, increases mucosal fragility, and contributes to the development of painful ulcers associated with RAS (16, 17). Vitamin C is absorbed in the small intestine via active transport and is excreted through urine, with renal reabsorption maintaining physiological levels. Folate (Vitamin B9) is essential for nucleotide synthesis, DNA repair, and methylation processes, supporting rapid cell turnover in the oral mucosa. Folate deficiency disrupts epithelial cell renewal and compromises mucosal integrity, predisposing individuals to RAS (18). Folate undergoes hepatic metabolism into its active form, tetrahydrofolate, and is primarily excreted through urine. Understanding these vitamins' metabolic pathways and their excretion dynamics helps clarify their roles in RAS pathogenesis and guides targeted therapeutic interventions to restore mucosal health and prevent recurrence (19).
2 Clinical signs and symptoms
Vitamin B12 Deficiency:
○ Glossitis with mucosal atrophy
○ Increased mucosal sensitivity and erythema
○ Painful ulcerative lesions of the oral mucosa
Vitamin C Deficiency:
○ Delayed wound healing
○ Gingival bleeding and friability
○ Higher prevalence of small, painful ulcers
Folate Deficiency:
○ Mucosal pallor
○ Erosions and ulcerations along with a burning sensation
○ Increased risk of developing RAS in immunocompromised individuals
Therapeutic approaches for addressing hypovitaminosis should be discussed in detail, emphasizing adequate dosage, frequency, duration, and alternative delivery methods, tailored to the specific vitamin deficiency and the patient's stage of life. For vitamin B12 deficiency, oral supplementation (500–1,000 mcg daily) or intramuscular injections (1,000 mcg weekly for 4–6 weeks, followed by monthly doses) are recommended, with higher doses for severe deficiencies or absorption disorders such as pernicious anemia. Maintenance dosing typically includes 1,000 mcg monthly via injection or 1,000 mcg orally daily. For vitamin C deficiency, 100–500 mg daily is effective for mild cases, increasing to 1,000–2,000 mg daily for severe deficiencies, with gradual tapering to 100–200 mg daily for maintenance. For folate deficiency, 400–800 mcg daily is recommended, increasing to 1–5 mg daily for severe deficiencies or during pregnancy, with maintenance doses adjusted to 400 mcg daily. Adjustments based on Recommended Dietary Allowances (RDAs) should be provided for specific life stages: infancy, childhood, adulthood, pregnancy, and old age. For example, pregnant women require higher folate doses (600–800 mcg daily), while older adults may need increased B12 due to reduced absorption. Alternative delivery options, such as fortified foods or intramuscular/intravenous administration for severe cases, should be explored. This comprehensive therapeutic strategy ensures effective treatment and maintenance of adequate vitamin levels across varying needs and life stages (20).
Biochemical analysis findings, such as serum vitamin concentrations, provide valuable insights into the severity of hypovitaminosis and its association with RAS. Diagnostic workups should incorporate these findings to ensure accurate assessments, particularly in patients unresponsive to standard therapies (21–25).
3 Materials and methods
The PICO question was defined as follows:
• Population (P): Patients with hypovitaminosis
• Intervention (I): Measurement of vitamin levels and RAS prevalence
• Comparison (C): Patients without vitamin deficiencies
• Outcomes (O): Incidence and severity of RAS
A review of literature search was conducted on PubMed, Web of Science, and Scopus using the Boolean AND connector: “hypovitaminosis AND aphthous ulcers” (“hypovitaminosis”[MeSH Terms] OR [“hypovitaminosis”(All Fields)] OR “vitamin deficiency”[All Fields] AND [“aphthous”[All Fields] AND “ulcers”[All Fields]]. Abstracts were screened for relevance by three independent reviewers, with relevant articles selected for data extraction. Results from each database were presented separately in a tabulated format for clarity.
4 Results
The literature search yielded four studies meeting the inclusion criteria Table 1:
Tabel et al.: Case-control study on 40 individuals with documented RAS revealed 75% of participants exhibited deficiencies in vitamin B12 and folate levels. Clinical findings included glossitis, pale mucosa, and recurrent ulcerative lesions, significantly more frequent among patients with lower serum vitamin levels (26).
Padayatty et al.: Cohort of 25 patients with RAS reported vitamin C deficiency significantly associated with delayed ulcer healing and increased gingival bleeding. This study supports the role of vitamin C in mucosal repair and integrity, suggesting supplementation may aid ulcer resolution (26).
Hugar et al.: Observational study highlighted the association between folate deficiency and RAS severity. Patients with low serum folate levels showed more frequent and severe RAS lesions compared to those with normal folate levels, underscoring the importance of folate in cellular repair processes (27).
Freitas et al.: Evaluated the effects of vitamin supplementation on RAS resolution and observed a marked reduction in ulcer frequency among patients receiving vitamins B12, C, and folate. This finding suggests that vitamin supplementation may play a therapeutic role for RAS patients (24).
Table 1
Table 1. Table of included studies.
5 Discussion
Evidence increasingly supports hypovitaminosis as a critical factor in RAS pathogenesis. Key points for consideration include:
Vitamin Pathways and Genetic Influences:
○ Research into vitamin metabolic pathways and genetic predispositions may clarify the mechanisms predisposing individuals to mucosal breakdown.
Innovative Diagnostic Tools:
○ Non-invasive serum vitamin assessments and genetic screenings could aid early detection and improve treatment outcomes.
Longitudinal Data:
○ Long-term studies assessing vitamin supplementation's impact on RAS recurrence are needed. These studies should explore optimal dosages, duration, and frequency.
Therapeutic Approaches:
○ Supplementation strategies must specify adequate dosage, frequency, and duration. Alternatives such as fortified foods or intramuscular injections for severe deficiencies warrant discussion.
Multidisciplinary Collaboration:
○ Collaboration among nutritionists, dentists, and oral medicine specialists ensures comprehensive care, from diagnosis to management (28).
6 Stanley's classification of RAS
RAS is classified into three main types according to Stanley's classification:
Minor RAS Figure 1:
○ The most common type, accounting for approximately 80% of cases.
○ Lesions are small (<10 mm), round or oval, and shallow with erythematous borders.
○ Typically heal within 7–14 days without scarring.
○ Common sites include the non-keratinized mucosa such as the inner cheeks and lips (29).
Major RAS Figure 2:
○ Less frequent, accounting for around 10%–15% of cases.
○ Lesions are larger (>10 mm), deeper, and more painful.
○ Healing may take weeks to months and often results in scarring.
○ Frequently observed on the soft palate, tonsillar fauces, and lips (30, 31).
Herpetiform RAS Figure 3:
○ The least common type, comprising 5%–10% of cases (5, 20).
○ Characterized by numerous small (1–3 mm) ulcers that can coalesce to form larger irregular lesions.
○ Lesions may occur on both keratinized and non-keratinized mucosa.
○ Heals within 7–10 days without scarring.
Figure 1
Figure 1. Minor RAS.
Figure 2
Figure 2. Major RAS.
Figure 3
Figure 3. Herpetiform RAS.
7 Conclusion
Hypovitaminosis represents a critical factor in the etiology and management of RAS. Understanding the multifactorial etiology of RAS, including gastrointestinal and systemic conditions, can guide diagnostic efforts. Identification of vitamin deficiencies, particularly involving vitamins B12, C, and folate, is essential for formulating effective diagnostic and therapeutic strategies. Further research into advanced diagnostic tools, genetic influences, and long-term supplementation protocols will enhance clinical approaches and outcomes. Multidisciplinary management remains key to optimizing care for individuals affected by this challenging oral condition.
Author contributions
AR: Conceptualization, Data curation, Investigation, Software, Writing – original draft, Writing – review & editing. GC: Methodology, Writing – original draft, Writing – review & editing. RD: Formal Analysis, Methodology, Software, Writing – original draft, Writing – review & editing. AP: Resources, Validation, Visualization, Writing – original draft, Writing – review & editing. CA: Project administration, Supervision, Validation, Writing – original draft, Writing – review & editing.
Funding
The author(s) declare that no financial support was received for the research, authorship, and/or publication of this article.
Conflict of interest
The authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest.
Generative AI statement
The author(s) declare that no Generative AI was used in the creation of this manuscript.
Publisher's note
All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article, or claim that may be made by its manufacturer, is not guaranteed or endorsed by the publisher.
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Keywords: hypovitaminosis, aphthous ulcers, vitamin deficiency, recurrent aphthous stomatitis, oral health
Citation: Rosa A, Cianconi G, De Angelis R, Pujia AM and Arcuri C (2025) Hypovitaminosis and its association with recurrent aphthous stomatitis: a comprehensive review of clinical correlations and diagnostic considerations. Front. Oral. Health 6:1520067. doi: 10.3389/froh.2025.1520067
Received: 30 October 2024; Accepted: 9 January 2025;
Published: 28 January 2025.
Edited by:
Shamimul Hasan, Jamia Millia Islamia, India
Reviewed by:
Saba Khan, Darshan Dental College and Hospital, India
Sajad Buch, International Medical University, Malaysia
Copyright: © 2025 Rosa, Cianconi, De Angelis, Pujia and Arcuri. This is an open-access article distributed under the terms of the Creative Commons Attribution License (CC BY). The use, distribution or reproduction in other forums is permitted, provided the original author(s) and the copyright owner(s) are credited and that the original publication in this journal is cited, in accordance with accepted academic practice. No use, distribution or reproduction is permitted which does not comply with these terms.
Correspondence: Alessio Rosa, alessio.rosa.21@alumni.uniroma2.eu
Disclaimer: All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article or claim that may be made by its manufacturer is not guaranteed or endorsed by the publisher. |
1172 | https://web.itu.edu.tr/~guneshasa/viscous/blt.pdf | HIGH RENOLDS NUMBER FLOW BOUNDARY LAYERS (Re ∞) BOUNDARY LAYER Thin region adjacent to surface of a body where viscous forces dominate over inertia forces Re = Re >> 1 inertia forces viscous forces ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ Boundary layer separation Wake: viscous effects not important vorticity not zero Flow field around an arbitrary shape Inner flow Strong viscous effects Outer flow Viscouseffect s negligible Vorticity zero (Inviscid potential flow) BOUNDARY LAYER THEORY Steady ,incompressible 2-D flow with no body forces. Valid for laminar flow O.D.E for To solve eq. we first ”assume” an approximate velocity profile inside the B.L Relate the wall shear stress to the velocity field Typically the velocity profile is taken to be a polynomial in y, and the degree of fluid this polynominal determines the number of boundary conditions which may be satisfied EXAMPLE: LAMINAR FLOW OVER A FLAT PLATE: 0 2 1 ( 2 ) d dU dx dx U τ θ δ θ θ ρ + + = 0 ( ) n u dy τ ∂ ∼ ( ) x θ 2 ( ) u a b c f U η η η = + + = U∞ U ≈ 0,99U∞ Re UL ν = High Reynolds Number Flow • Laminar boundary layer predictable • Turbulent boundary layer poor predictability • Controlling parameter • To get two boundary layer flows identical match Re (dynamic similarity) • Although boundary layer’s and prediction are complicated,simplify the N-S equations to make job easier 2-D , planar flow u = , x, y= , u v v U∞ = , x y L Dimensionless gov. eqs.
X ; Y; “Naïve” way of solving problem for If you drop the viscous term Euler’s eqs. (inviscid fluid) 2 2 2 2 1 ( ) Re P u t x y y x y ν ν ν ν ν ν ∂ ∂ ∂ ∂ ∂ ∂ + + = − + + ∂ ∂ ∂ ∂ ∂ ∂ .
0 V → ∇ = 2 2 2 2 viscous terms 1 ( ) Re u u u P u u u t x y x x y ν ∂ ∂ ∂ ∂ ∂ ∂ + + = − + + ∂ ∂ ∂ ∂ ∂ ∂ 2 P P U ρ ∞ = 1 0 Re → R e →∞ • We can not satisfy all the boundary B.C.s because order of eqs. Reduces by 1 Inside B-L can not get rid of viscous terms Derivation of B-L eqs. From the N-S eqs • Physically based argument :determine the order of terms in N-S • Limiting procedure as Re ∞ eqs. and throw out small terms U∞ U (x,y) y L U∞ δ 1 100 L δ δ = 〈 Assumption 1 Term Order v 1 L δ δ = 〈〈 u x ∂ ∂ y ν ∂ ∂ x ν ∂ ∂ 2 2 u y ∂ ∂ du dt (1) 1 (1) = 1 δ δ = 1 δ δ = 2 1 δ 1 u u x ∂ = ∂ δ (1) (1) Neglect since of order >>>1 2 2 2 2 1 ( ) Re P u t x y y x y ν ν ν ν ν ν ∂ ∂ ∂ ∂ ∂ ∂ + + = − + + ∂ ∂ ∂ ∂ ∂ ∂ 2 2 2 2 1 ( ) Re u u u P u u u t x y x x y ν ∂ ∂ ∂ ∂ ∂ ∂ + + = − + + ∂ ∂ ∂ ∂ ∂ ∂ 1 1 δδ = (1) 1 (1) = 2 δ 2 (1) (1) 2 (1) ( ) δ 2 (1) ( ) δ Also for y –direction ( ) (1) (1) ( ) δ δ ( ) ( ) ( ) ( ) δ δ δ δ 2 2 2 ( ){ } (1) ( ) ( ) δ δ δ δ δ + ( ) δ small relative to To good approximation pressure at the edge of B-L. is equal to pressure on boundary layer.
• Time – dependant known from the other flow • Pressure at all points is the same • Only need to consider x-direction B-L. eqs.
P y ∂ ∂ ( ) δ P x ∂ ∂ (1) ( ) P P x ≅ ( , ) P P x t ≅ Prandtl (1904) 0 u v x y ∂ ∂ + = ∂ ∂ Outer flow (inviscid) y x 2-D planar 1) 2) 2 2 1 u u u P u u v t x y x y ν ρ ∂ ∂ ∂ ∂ ∂ + + = − + ∂ ∂ ∂ ∂ ∂ Governing eqs.for B.L B-L.eqs.
still non-linear but parabolic type unknows u,v (x,y,t) known from the potential flow ( , ) P P x t ≅ Need B.C.s & I.C.(time dependant) • 2-D, steady BCs • u= =0 at y=0 • u=u(y) at x=0 • u= (x) y (y ) marching condition • B-L. eqs. can be solved exactly for several cases • Can approximate solution for other cases Limitation of B.L egs.: where they fail?
(1) Abrupt chances ν U∞ ∞ δ (2) Eqs. are not applicable near the leading edge 1 L δ δ = 〈〈 L is small invalid (3) Where the flow separates not valid beyond the separation point Separation point Bernouilli eqs. =constant ρ 1 1 2 0 2 dP dV U dx dx ρ + = 2 constant 2 P V ρ + = Valid along the streamlines substitute the B.L eqs u,v can be found known 0 dp dx = 1 dP dU U dx dx ρ − = SIMILARITY SOLUTION TO B.L. EQS Example 1 Flow over a semi-infinite flat plate Zero pressure gradient p = constant Steady ,laminar & U=constant ( ) 0 dp dx = • Bernouilli eqs. outsideB.L U=constant , Governing (B.L. eqs.) become 2 2 u u u u v x y y ν ∂ ∂ ∂ + = ∂ ∂ ∂ U y x 2 1 .
2 p U cons ρ + = 0 dp dx = 0 u x y ν ∂ ∂ + = ∂ ∂ (1) (2) B.C.
• y=0 u= v =0 (no-slip) & y ∞ , u U • x=0 u=U Blasuis(1908) : 1.Introduce the stream function (x,y) • Recall ; ψ u y ψ ∂ = ∂ x ψ ν ∂ = −∂ note that satisfies cont. eqs. substitute intoB.L. mom. Eqs ψ 2 2 3 2 3 .
.
y x y x y y ψ ψ ψ ψ ψ ν ∂ ∂ ∂ ∂ ∂ − = ∂ ∂∂ ∂ ∂ ∂ (2’) • Now, assume that we have a similarity “stretching” variable, which has all velocity profiles on plate scaling on .
i.e ( , , ) g U x δ ν ∞ = δ ( ) u y f U δ ∞ = y δ x dimensional analysis ( ) (Re) U x g g x δ ν ∞ = = 2 1 ( ) Re δ ∼ δ ν ∼ x U ν δ ∞ ∼ [ ] 2 .
.
m m s m s m = 1 Rex x δ ∼ both ( ) δ Viscous dif. Depth y η δ = Re U x ν ∞ = 5 x U ν δ ∞ ≈ Let [-] similarity variable U y x η ν ∞ = ( ) u f U η = Use similarity profile assumption to turn 2 P.D.E 1 O.D.E 0 0 0 ( ) ( ) y y x udy Uf dy Uf d U η ν ψ η η η = = = ∫ ∫ ∫ x fix e d u y ψ = ∂ = ∂ ψ 0 ( ) ( ) U x f d U xF η ψ ν η η ν η = = ∫ ( ) F η ( ) U xF ψ ν η = ( ) U xF ψ ν η = U y x η ν ∞ = 0 0 y udy ψ ψ − = ∫ d dy dx y x ψ ψ ψ ∂ ∂ = + ∂ ∂ •Now, substitute into P.D.E for (x,y) to get O.D.E for F( ) ψ η ' 1 2 U F U xF x x x ν ψ η ν ∞ ∞ ∂ ∂ = + ∂ ∂ ' dF F dη = 2 '' 2 d F F dη = 2 '' 2 U U F y x ψ ν ∞ ∞ ∂ = ∂ 1 1 1 2 2 U y x x x x η η ν ∞ ∂ = − = − ∂ ' 1 ( ) 2 U F F x x ν ψ η ∞ ∂ = − ∂ ' ' U U xF U F y x ψ ν ν ∞ ∞ ∞ ∂ = = ∂ 2 '' 2 U F x y x ψ η ∞ ∂ = − ∂∂ 2 3 ''' 3 U F y x ψ ν ∞ ∂ = ∂ Substituting into eq. (2’) 2 1 1 2 2 1 '( ''') ( ) ( ') ( ) '' ''' 2 2 U U U U U F F F F U F F x x x x ν η η ν ν ν ∞ ∞ ∞ ∞ ∞ ∞ ⎡ ⎤⎡ ⎤ − − − = ⎢ ⎥⎢ ⎥ ⎣ ⎦⎣ ⎦ or 2 '' 2 U F x η ∞ − 2 1 ' 2 U F x ∞ − 2 1 '' 2 U F F x ∞ + 2 '' ' U F F x η ∞ = ''' F 1 ''' '' 0 2 F FF + = blasius eq. 3rd order , non linear ODE 0 0 0 y y u y ψ = = ∂ = = ∂ Note: for BVP ''' '' 0 F FF + = 2 U y x η ν ∞ = BC’s are At y=0 u=v=0 0 η = BC 1) 0 ' 0 U F η ∞ = = F’(0)=0 BC 2) 0 0 y ν = = 1 ( ') 0 2 U F F x ν η ∞ − − = F(0)=0 BC 3) (x,y ∞) U∞ y U y ψ ∞ →∞ ∂ → ∂ ' U F U η ∞ ∞ →∞= '( ) F η →∞ 1 '( ) 1 F ∞= Or At x=0 F’( )=1 same with BC 3) Matching B.C • Solution to blasius eg a)power series b)runge-kutta • results tabulated form for F,F’,F’’,etc p.g 121 F( η ) dimensionless function u U∞ = 0 ' x U F U η = ∞ ∞ →∞= ∞ # # 0 0 0 0.33206 F’’= 0.33206 From the solution 5.0 3.28329 0.99155 0.01591 U y x η ν ∞ = F ' u F U∞ = '' F # # • Velocity profile 5 [ ] 12 1 ( ' ) 2 1 Re ' 2 x U F F x x F F U ν ψ ν η ν η ∞ − ∞ ∂ = − = − ∂ = − 1 (5 1 3.28) 2 1 0.86 Rex U x x U ν η ν ν ∞ ∞ ∞ ∞ →∞ = − = Rex U ν ∞ 0.8 η U y x γ ∞ = 5 F’= uU∞ Shear stress distribution along the flat plate ( ) ( , ) u x y y x u u y x y ν τ µ τ ν τ µ ∂ ∂ = + ∂ ∂ ∂ ∂ ∂ ≅ ∂ ∂ ∂ 4 6 1 Re 10 0.00865 100 1 For Re 10 0.000865 1000 x x For U U ν ν ∞ ∞ ∞ ∞ = ⇒ = ≈ = ⇒ = ≈ At the wall (y=0) 0 0 ( ) y u x y τ µ = ∂ = ∂ ( ) w x τ 2 0 2 0 0 ( ) '' y U x U F y x η ψ τ µ µ ν ∞ ∞ = = ∂ = = ∂ 3 0( ) ''(0) U x F x τ µ ν ∞ = Distribution along the wall 0.332 Non dimensionalize : 0 2 2 ''(0) 0.664 . Re 1 Re Re 2 f x x x F U x C U τ ν ρ ∞ = = = = 0.664 f C Ux ν = Friction coef.
0 Note : 0 x τ ν → ⇒ →∞ →∞ y x B.L eqs.are not valid near the leading edge x Up to the point we are considering Drag force acting on the flat plate We have to integrate shear stress 0 0 unit width ( ) x per d D F τ ζ ζ ↓ = ∫ 3 2 1.328( ) D F b U x µρ ∞ = x τ x D dimensionless drag coef.( C ) we have 2 wetted sides 2 5 6 2 A=2bx 1 2 1.328 valid for laminar flow i.e for Re 5.10 to 10 D D D x x F C U A C Re ρ ∞ = = < Width normal to the blackboard 6 x for Re >10 turbulent drag becomes considerably greater → Boundary Layer Thickness : δ at 5 0.99 (Table) 5 5 Re Re x x U u y y x U U U x x x η η δ ν δ δ ν ν ∞ ∞ ∞ = = ⇒ = → = ≅ ≅ = :defined as the distance from the wall for which u=0.99U δ ∞ Boundary Layer Parameter (thicknesses) Most widely used is but is rather arbitrary y= when u=0.99 U δ δ ∞ hard to establish more physical parameters are needed Displacement thickness: δ U∞ U∞ δ δ δ an imaginary displacement of fluid from the surface to account for “lost” mass flow in boundary layer .
0 0 0 0 0 or ( ) (1 ) tot y U m udy U dy U dy U dy u U U u dy dy U δ δ ρ δ ρ ρ ρ ρ ρ δ ρ ρ δ ∞ ∞ ∞ ∞ ∞ ∞ ∞ = − ∞ ∞ ∞ ∞ ∞ = = = − = − = − ∫ ∫ ∫ ∫ ∫ ∫ if . always by definition cons ρ δ δ = > Momentum thickness: θ U∞ θ an imaginary displacement of fluid of velocity to account for “lost” momentum due to the formation of a boundary layer velocity profile U∞ 2 0 0 Mass flow in B.L Possible momentum actual momentum ( ) ( ) U udy U udy u ρ θ ρ ρ ∞ ∞ ∞ ∞ = − ∫ ∫ "lost" momentum 0 (1 ) will occur in B.L eqs.
u u dy U U θ ∞ ∞ ∞ = − ∫ V arious thinknesses defined above are,to som e extend,an indication of the distance no over w hich viscous effects extend.
, ( ) only > ( ) D efinition is sam e tes(rem aks) f x always δ θ δ δ θ ∗ ∗ > ∗ ∗ or ZPG ,A PG ,FPG ,turbulance 5 From flat plate analysis Rex x δ ≅ [ ] [ ] 0 5 5 ' 0 0 5 0 (1 ) (1 ) (1 ) 1.72 5 3.283 Re Re x y x u and dy u u u remember y d d x x u x x d F d u u u xx x x F u x x δ η η η η η δ η ν ν ν ν δ ν η ∞ ∞ ∞ = = ∞ ∞ ∞ ∞ = − = ⇒ = = − = − − = − = ∫ ∫ ∫ 0 (5) 3.283 1.72 Re 0.664 , (1 ) Re y F x x u u x Similarly d u u x δ δ θ ∞ ∞ = = = − = ∫ δ δ θ δ δ θ FALKNER-SKAN SIMILARITY SOLUTIONS Stagnation-point flow (Hiemenz flow) Similarity methods Flow over a flat plate (Blasius flow) ( , ) x y η ⇒ Falkner & Skan (1931) Ægeneral similarity solution of the B-L eqs.
Family of similarity solutions to the 2-D,steady B-L egs. Look for general similarity solutions of the form ( , ) ( ) '( ) where (1) ( ) - unspecified function of x which will be determined later ( ) u x y U x f x y x η ζ η ζ = = (2) ( , ) ( ) ( ) ( ) check : u= ( ) ( ) x y U x x f U x x y ψ ψ ζ η ζ ∂ = = ∂ 1 '( ) ( ) f x η ζ 2 2 2 2 3 2 3 B.L eqs. (3) or in terms of ( , ) (3') B.C.s no-slip, smooth matching u u dU u u v U x y dx y dU x y U y x y x y dx y ν ψ ψ ψ ψ ψ ψ ν ∂ ∂ ∂ + = + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ − = + ∂ ∂∂ ∂ ∂ ∂ ( ) y x ζ η = Substitute eq .(2) into (3’) 2 2 2 ( ) ( ) ( ) ( , ) ' , " ( 1 ' ' ) df d f U x x f x y f f d d Uf u y dU d df d f U f dU d d f U f U f x dx dx dx U x dx dx d dx d y d d dx dx dx ψ ζ η ψ η η ψ ψ ζ η ζ ζ η η ψ ζ ζ ζ η ζ ζ η ζ ζ = = = = ∂ = = ∂ ∂ = + + ∂ = − ∂ = + − ∂ = − [ ] 2 2 ' ' ' ' 1 = ' ' ' " " dU f Uf f U x y x y x dx x dU df dU d f U dU U d f f Uf dx d f x y d x dx dx x dx ψ ψ η ζ η η ζ ψ ζ η ζ ⎛ ⎞ ∂ ∂ ∂ = ∂ ∂ ∂ = = = + ⎜ ⎟ ∂∂ ∂ ∂ ∂ ∂ ⎝ ⎠ ⎡ ⎤ ∂ + = + − − ∂∂ ⎢ ⎥ ∂ ⎣ ⎦ [ ] 2 2 3 3 2 2 2 2 2 2 " "' ' " Substitute above results into (3') ' ' " "' 1 ( ') " " "' ( ' ) " ' Uf Uf y y dU dU U f y U f y U d d f U f dx d U dU U Uf f f U f f U f dx dx dc dx dU dU d dU U U f U ff U ff U f dx dx dx dx dU u f dx dx η ζ ζ ν ψ ζ ψ ζ ζ ζ η ζ ζ ζ ν ζ ζ η ζ ∂ ∂ = = ∂ ∂ ⎡ ⎤ ⎡ ⎤ − − + − = + ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ ∂ = = + ∂ = − − ∂ ∂ ( ) ( ) 2 2 2 2 ( ) " "' To put the eq. into standard form, multiply by "' " 1 ' 0 (4) d U d dU U U ff U f dx dx U dU f U ff f dx dx α β ζ ν ζ ζ ζ ν ζ ζ ζ ν ν ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ + + − = ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ ⎦ − = ⎣ + Transformed gov. Eq.
If a similarity solution exists, eq.(4) must be an ODE for the function f in terms of η.
So, coefficieuts α & β must be constant for a similarity solution ( ) 2 Falker-Skan "' " 1 ' eq. ( 0 5) f ff f α β ⎡ ⎤ + + − = ⎣ ⎦ B.C same as for flat plate (0) '(0) 0 '( ) 1 : BCs don't depend on , Exact solutions to the B-L. Eqs. May be obtained by pursuing the following PRO remark Step CEDURE : Select & . (a p ar 1 f f f η α β α β = = →∞→ 2 ticular flow configuration is considered this will not be known a priori but will be exident when step 2 is completed).
: Determine U ( ) , (x) , (x) (6a-b) Step 2 d dU U dx dx ζ ζ α ζ ν ζ β ν = = '' ' 2 ' ' : Determine the function f ( )which is the solution of the following problem ''' 1 ( ) 0 with BCs (0) (0) 0 , ( ) 1 as Calculate the stream fun Step ctionin 3 Step h 4 p : y f ff f f f f η α β η η ⎡ ⎤ + + − = ⎣ ⎦ = = → →∞ N ( ) ( ) 2 2 sical coord.
( , ) ( ) ( ) ( ) in step #2 , instead of working with eqs. 6 a-b) (6a)' Remar E 2 (6b) k ' Flate P xample #1 step la te (ZPG) #1 y x y U x x f x dU d U dx dx η ψ ζ ζ ζ β ζ ν α β ν ⎛ ⎞ ⎜ ⎟ = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ = = − ( ) 2 ' means that flat plate at ZPG 1 = , =0 2 (6a)' 0 (6b)' ( ) 0 (6 ) leads to step #2 U=con st 0 .
this d U dx dU dx dU x b dx α β ζ ν ζ ν ζ = = ≠ ⇒ = ⇒ 2 2 (6 )' 1 : ''' '' 0 (0) '(0) 0 2 ; ' 1 compare with Blasius s Step #3 Step # olution ( , ) ( ) , 4 ( ) d x a dx U U f ff f f y f x U y x x x U y x y U f U f U x U x y U x f y ζ ν ν ζ η η ν ν ψ ζ ζ ν ψ ν ν ζ = → = + = = = = →∞ → ⎛ ⎞ ⎜ ⎟ ⎛ ⎞ ⎜ ⎟ = = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎜ ⎟ ⎝ ⎠ = = same as Blasius solution U x ν ⎛ ⎞← ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ( ) ( ) ( ) 2 2 FLOW OVER WEDGE 1, arbitrary constant (6a' ) Exampl 2 2 e #2 Step (7) #1 d U U x dx α β ζ ν β ζ ν β = = = − ⇒ = − 2 2 (6 ') Divide eq. (6b') by (7) ln ln ln outer flow is that over a wedge of angle (Fig.) 2 1 1 2 ( ) dU U dx x U x dU b dx U cx x c β β ζ νβ β πβ β β β − = − = + ⇒ = = − 2(1 ) 2 2 2 2 2 1 2 (9) Solve the BVP ''' '' 1 ( ') 0 (0) '(0) 0 ' 1 Solve numerically to get ( ), '( ), ' (2 ) ( ) Step #3 '( ) dU c x dx f ff f f f as f f c f x x f β β β β β ζ νβ ζ νβ β β η η η η ν β ζ − − − − − = = − ⎡ ⎤ + + − = ⎣ ⎦ = = →∞ → − = ( ) ( ) ( ) ( )( ) 1 2 1 2 : Go back to the physical c Step oordinate ( , ) ( ) ( ) = 2 ( ) 2 / 4 y y x y U x x f c x f x x c β β β ψ ζ β ν ζ β ν − −− − ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ = − ⎜ ⎟ ⎜ ⎟ − ⎝ ⎠ ⎝ ⎠ STAGNATION-POINT ; FLO = W 1 1 β α = Flow over a wedge Let 1 β → = 2 Eq. (8) gives, ( ) (9) ( ) "' " 1 ( ') 0 (0) '(0) 0 as '( ) 1 ( , ) x / Not U x cx x f ff f c f f f y x y c f c ν ζ η η ψ ν ν = → = + + − = = = →∞ → ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ e: See Hiemenz flow Exact solution to the full Navier-Stokes equations obtained by Hiemenz for a stagnation point.
FLOW IN A CONVERGENT CHANNEL 0 , 1 α β = = Boundary layer flow on the wall of a convergent channel.
Exercise: pg. 132.
Solve the BVP (F-S. eq.) More on similarity solutions to the B.L.
Evans (1968) “Laminar Boundary Layers” Numerical Solutions Finite differences H.B. Keller (1978) Ann. Rev. of Fluid Mech. Vol.10.pp. 417-433 Finite Element Methods, Finite Volume Methods Spectral (Element) Methods APPROXIMATE SOLUTIONS: Solve exact eq. approximately Von Karman Momentum Integral Eqn (General Momentum Integral Equation for Boundary Layer) Develop an eqn. which can accept "approximate" vel. profiles as input & yield accurate (close, but approximate) shear stress , , as output. Integrate the differen I tial B-L. eqs. Approac ea: h: d a δ δ θ 2 2 cross the B-L. 0 y Start with B-L. eqs. 0 . 0 , 0 First note u u dU u u v U x y dx y u v x y B C y u v y u U δ ν δ ≤ ≤ ∂ ∂ ∂ + = + ∂ ∂ ∂ ∂ ∂ + = ∂ ∂ = = = = ( ) ( ) 2 2 0 0 0 ( ) ( ) Substitute into B.L eq. & integrate from y=0 to y= . 2 u uv v u v u uv u continuity y y y y v u y x uv u dU u u dy dy U dy x y dx x y δ δ δ δ ν ∂ ∂ = − ∂ ∂ ∂ ∂ ∂ ∂ = − = + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ + = + ∂ ∂ ∂ ∂ ∫ ∫ ∫ ( ) 0 0 0 ? (1) (2) (3) (4) Consider term (2) 0 ( , ) dy uv dy uv U v x y δ δ δ δ ∂ = = − ∂ ∫ ∫ 0 0 0 0 0 2 2 0 0 0 0 0 0 Integrate cont. eq. dy 0 ( , ) 0 0 Integrate term (4 ( , ) ) y y y u v u u dy dy dy v x U U dy x y x x u u u u u dy dy y y y y y y d x y v u d δ δ δ δ δ δ δ δ δ δ τ µ δ = = = ∂ ∂ ∂ ∂ + = ⇒ + − = = − ∂ ∂ ∂ ∂ ⎛ ⎞ ∂ ∂ ∂ ∂ ∂ ∂ = = = − ⎜ ⎟ ∂ ∂ ∂ ∂ ∂ ∂ ⎝ ⎠ = ⇒ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ( ) 2 0 0 0 0 0 2 0 0 0 0 0 ( ) (4) Term(1) 2 ( ) B-L. eq. becomes u u u dy dy x x u dU dy U U dy x dx uU u dU U u dy x u dy x U dy u x x dx δ δ δ δ δ δ δ τ τ ν µ ρ τ ρ ∂ ∂ ⇒− = − ⇒ = ∂ ∂ ∂ − − ∂ ⎛ ⎞ ∂ ∂ = = − ⎜ ⎟ ∂ ∂ ∂ ∂ ⎠ = ∂ ⎝ ∂ ∫ ∫ ∫ ∫ ∫ ∫ ∫ 0 dy δ ∫ =0 ( ) ( ) ( ) 2 0 0 0 0 0 2 0 0 0 2 ( ) Thus, get uU u dU dU dy dy u dy U dy x x dx dx dU u uU dy u U dy x dx u x δ δ δ δ δ δ τ ρ τ ρ ∂ ∂ − + − = − ∂ ∂ ∂ − + − = − ∂ ∂ − ∂ ∫ ∫ ∫ ∫ ∫ ∫ ( ) 0 uU dy δ ∫ Using Leibnitz’s rule permits the order of integ. & dif. to be interchanged 2 2 0 2 0 0 1 u u u dU U dy U dy x U U U dx δ δ τ ρ ⎛ ⎞ ∂ ⎛ ⎞ − + − = − ⎜ ⎟ ⎜ ⎟ ∂ ⎝ ⎠ ⎝ ⎠ ∫ ∫ Multiply by -1 & factor U terms out of integrals, 2 0 0 0 ( ) ( ) 1 1 x x u u dU u U dy U dy x U U dx U δ δ θ δ τ ρ ⎡ ⎤ ⎢ ⎥ ∂ ⎛ ⎞ ⎛ ⎞ − + − = ⎢ ⎥ ⎜ ⎟ ⎜ ⎟ ∂ ⎝ ⎠ ⎝ ⎠ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ∫ ∫ 2 0 2 2 0 2 ( ) 2 , ( ) only Divide eq. by U & get ( 2) ( 2 1 2 ) f f d dU d dU U U x dx dU d U U x x dx x dx C d dU or H H C dx U dx x U dx U τ θ δ θ τ θ δ ρ θ θ θ θ θ τ θ θ δ θ ρ + + ∂ + = ∂ ↓ ∂ ∂ + → ∂ + = ∂ + = = = 0 2 H= shape factor 1 2 U ρ Ordinary Differential eq. for θ(x) & is called von Karman Momentum Integral eqn. or Generalized momentum integral equation To solve the integral eq. we first “assume” an approximate velocity profile, i.e. one that “fits” & has proper “shape” and satisfies the proper B.C we do thin by using similarity concept again & writing potential similarity velocity profiles in terms of the variable , & apply B.C & get particular form.
evaluate θ(x), δ(x) and from their definitions.
integral equation can be solved for the B.L. thickness, δ(x) ( ) y x η δ = 0 τ η 0 ( )n u dy τ ∂ ∼ U y δ = An approximate velocity profile, for example 2 u a b c U η η = + + 0 2 1 ( 2 ) d dU dx dx U τ θ δ θ θ ρ + + = ( ) x θ 2 ( ) u a b c f U η η η = + + = 2 u a by cy = + + Steady ,incompressible 2-D flow with no body forces. Valid for laminar and turbulent flow O.D.E for To solve eq. we first ”assume” an approximate velocity profile inside the B.L Relate the wall shear stress to the velocity field Typically the velocity profile is taken to be a polynomial in y, and the degree of this polynomial determines the number of boundary conditions which may be satisfied EXAMPLE: LAMINAR FLOW OVER A FLAT PLATE: laminar profile later as an example or B.C 1-)u=0 at y=0 ( =0) a=0 b=2 2-)u=U at y= ( =1) 1=b+c c=-1 3-) at y= ( =1) 0=b+2c Now use the approximate velocity profile to obtain terms in the momentum integral eq.
NOTE: Using the approximate velocity profile across the B.L will reduce the momentum integral to an O.D.E for the B.L thickness, δ (x).
2 2 2 2( ) ( ) u y y U η η δ δ = − = − 0 (1 ) u dy U δ δ = − ∫ ( ) y x η δ = dy dη δ = 1 1 2 0 0 (1 ) (1 2 ) u d d U η δ δ η δ η η η = = − = − + ∫ ∫ 1 2 3 0 1 ( ) 3 3 δ δ δ η η η = − + = 0 u y ∂ = ∂ η δ δ η η or 1 0 0 (1 ) (1 ) u u u u dy d U U U U η δ θ δ η = = − = − ∫ ∫ 1 2 2 0 (2 )(1 2 )d θ δ η η η η η = − − + ∫ 2 15 θ δ = 0 0 0 0 1 2 y du du d du U dy d dy d η η η τ µ µ µ µ η δ η δ = = = = = = = (2 2 U η − 0 ) η= 2 0 0 2 2( ) ( ) 2 y y y U U y τ η µ δ δ δ = ∂⎡ ⎤ = − = ⎢ ⎥ ∂⎣ ⎦ 0 2 1 ( 2 ) d dU dx U dx U τ θ δ θ ρ + + = Momentum Integral eq. becomes For a flow over a flat plate U=const.
ODE for (x).solve (x)first then Solving for , , , , 2 2 4 1 2 2 ( ) ( ) 15 3 15 d dU U dx U dx U U δ δ δ µ ν δρ δ + + = = 0 dU dx = δ δ 2 2 15 d dx U δ ν δ = 0 , , δ θ τ δ 2 0 0 15 15 2 x x d dx U U δ ν δ ν δ δ = ⇒ = ∫ ∫ 5.477 30 5.477 Rex x x x U U ν ν δ = = = Rex Ux ν = 1.826 3 x U δ ν δ = = 2 0.73 15 x U δ ν θ = = 0 2 0.73 1 Re 2 f x C U τ ρ = = 0 2 U τ µ ν = Comparing to (exact) blasius solution note:2nd order profile but it should be zero Additional BCs need to be imposed (=0 for flat plate) 5,477 1.095 5 blasius δ δ = = 1,826 1.061 1,72 blasius δ δ = = 10% ∼ 2 2 2 2 ( ,0) 0 u U x y δ ∂ = − ≠ ∂ 0.73 1.099 0.664 B θ θ = = 2 3 4 u A B C D E U η η η η = + + + + u 0 y u x ν = ∂ + ∂ 2 2 0 0 1 y y u P u y x y ν ρ = = ∂ ∂ ∂ = − + ∂ ∂ ∂ 2 2 0 1 y u P dU U y x dx ν ρ = ∂ ∂ = = − ∂ ∂ BC#5 at y= all higher derivates should also be zero at y= for a smooth transition from the B-L. to the outer flow Note: 2nd order profile BC#4 by employing 3rd order profile , i.e the above condt. may be imposed More accurate results are obtained Flat Plate at zero incidence Vel. Dist.
δ 2 2 0 u y ∂ = ∂ δ 2 2( ) ( ) u y y U δ δ = − 2 2 2 2 ( ,0) 0 u U x y δ ∂ = − ≠ ∂ 2 3 u a b c d U η η η = + + + ( ) ( ) u y f f U η δ = = ( ) f η η = , Note 1: Once the variation of is known, viscous drag on the surface can be evaluated by integration over the area of the flat plate.
Note 2: B-L thickness at transition air ( ) less than 1% of development length,x.
viscous effects are confined to a very thin layer near surface of body 2 ( ) 2 f η η η = − 3 3 1 ( ) 2 2 f η η η = − 3 4 ( ) 2 2 f η η η η = − + ( ) sin( ) 2 f π η η = 4.64 Rex δ = 5.84 Rex δ = 4.80 δ = 0.647 Re f x x C = 0.685 Re f x x C = 0.65 f C = 0 τ 5 Re 5.10 x = 30m U s = 0.24 x m = ν 5.48 0.00775 Rex x δ = = 0.00775 1.86 x mm δ = = ← Boundary layer seperation Separation wake formation increase in drag total force exerted on body in direction of fluid motion Boundary layers have a tendency to separate and form wake Wake leads to large streamwise pressure differentials across the body results in substantial pressure drag (form drag) For large Re ( or higher) bluff bodies (e.g circular cylinder) pressure drag constitutes almost all the total drag Total drag = pressure drag + viscous drag due to shear stress along the surface due to pressure differences caused by separation of flow Wake 4 10 2) AIRFOILS-LIFT drops sharply “STALL” due to separation force normal to flow direction Shape of streamlines near point of separation 3) FLAT PLATE ~ No separation REMARK: After separation point ,external (decelerating) stream ceases to flow nearly parallel to the boundary surface S chord t No separation Condition for separation Pressure gradient , >0 adverse pressure gradient (decelerating external stream) increasing pressure in the flow direction <0 favourable P.G and =0 (zero pressure gradient) NOTE: pressure gradient along a B.L is determined by the outer flow (Bern. Eq.) Separation occurs only for APG condition o Momentum contained in the fluid layers adjacent to surface will be insufficient to overcome the force exerted by the pressure gradient , so that a region of reverse flow occurs.
dP dx dP dx dP dx dP dx 1 dU dP U dx dx ρ = − i.e at some point downstream, the APG will cause the fluid layers adjacent to the surface to flow in a direction opposite to that of the outer flow B.L separation velocity profiles in a B.L near separation Note :shear stress changes its sign after separation Definition of separation point = point at which the shear (or velocity gradient) vanishes ( ,0) 0, for separation u x y ∂ = ∂ δ 0 0 u y ∂ > ∂ 0 0 u y ∂ = ∂ 0 0 u y ∂ < ∂ . point sep • Question show that separation can occur only in region of adverse pressure gradient !
Steady state B.L eqs.
If <0 the same u 0 y u v x = ∂ + ∂ 2 2 0 0 1 y y u P u y x y ν ρ = = ∂ ∂ ∂ = − + ∂ ∂ ∂ 2 2 0 y u dP y dx µ = ∂ = ∂ 3 3 0 0 y u y = ∂ = ∂ dP dx 2 2 u y ∂ ∂ 2 2 u P y x ∂ ∂ = ∂ ∂ 0 y = y δ = 2 2 0 u y ∂ = ∂ 0 u y ∂ > ∂ 0 u y ∂ = ∂ 0 u = u U = u y ∂ ∂ y u U ?
case constant slope Case APG PI= point of inflection where 0 P x ∂ = ∂ y y y 2 2 0 u y ∂ = ∂ u y ∂ ∂ u 0 P x ∂ > ∂ u y ∂ ∂ 2 2 0 u y ∂ = ∂ y 2 2 0 u y ∂ = ∂ 2 2 0 wall u dP y dx µ ∂ = > ∂ PI y y u U PI Control of separation by suction Control of separation by variable geometry and by blowing How to calculate the separation point ?
Goldstein Stewartson The Karman – Pohlhausen Approximate Method Fourth order polynomial for u (y). Pohlhausen (1921) Step #1 :coefs. a,b,c,d,e, in general, will be functions of x, so that solutions which are not similar may be obtained.
2 3 4 u a b c d e U η η η η = + + + + y η δ = δ 2 2 ( ) u U x dU y dx ν ∂ = − ∂ y=0 y= u=0 u=U 0 y u y δ = ∂ = ∂ 1 dp dx µ = 2 2 0 y u y δ = ∂ = ∂ 2 2 0 2 2 2 0 1 1 ( ) ( ) 1 = y u u u y y y u U dU dx η δ η δ η δ η ν = = ∂ ∂ ∂ ∂ ∂ = = ∂ ∂ ∂ ∂ ∂ ∂ = − ∂ 2 2 2 impose B.C.s =0 0=a :dimensionless variable; a measure of pressure gradient in outer flow ( ) =0 2 u dU U dx η δ η η ν Λ ∂ = −Λ = − = ∂ 1 1=a+b+c+d+e =1 0=b+2c+3d+4e =1 0=2c+6d+12e solution a=0 b=2+ c=- d=-2+ e=1-6 2 2 6 c η η η = Λ Λ Λ Λ → 2 3 3 where F( )=1-(1+ )(1- ) Pohlhausen parameter G( )= (1-(x)= -12 1 ( ) ( ) (1 ) ) 2 6 u F G dU dx U η η η η η η ν η δ η = ≤ Λ Λ Λ ≤ + Note : for velocity profile corresponds to a flat plate Plot function F( ) & G( ) 0 Λ = η η 1 η F( ) η G( ) η 0.25 0.016 1 1 1 12 Λ < − 0 Λ < 0 Λ = 0 Λ > 12 Λ > u U η th u 0: ( ) Flat surface in which the represantation is a 4 order polynominal U u >12 1 vel. in B.L. is not expected to exceed that of the outer flow locally.
U F η Λ = = Λ > So must be less than 12 <-12 negative velocity reverse flow.B.L. theory is not applicable after separation Λ Λ ⇒ ∴ δ 1 0 0 1 3 3 0 1 2 0 ( ) (1 ) (1 ) 3 = (1+ )(1- ) (1 ) ( ) (2) 6 10 120 momentum thickness 37 (x)= (1 ) ( ) (3) 315 945 9072 wall u u x dy d U U d u u d U U δ δ δ η δ η η η η η δ θ δ η δ = − = − Λ Λ ⎡ ⎤ − − = − ⎢ ⎥ ⎣ ⎦ Λ Λ − = − − ∫ ∫ ∫ ∫ N 0 0 0 0 shear stress : ( ) (2 ) 6 b U u U U η τ τ µ τ µ δ η δ = ∂ Λ = = + ∂ Step#2 Displacement thickness Uθ ν Step #3 Plug into the general momentum eq. Multiply the mom. Eq. by 0 2 2 0 2 2 2 2 2 2 2 (2 ) or 1 ( ) (2 ) (5) 2 (x)= evaluate each term in terms of (x) 37 ( ) ( ) 315 945 9072 3 (10 120 ( ) U d dU dx dx U d dU U dx dx U dU dx dU dU K K x d dx x x τ θ θ θ θ θ δ ν ν µ τ θ θ δ θ ν θ ν µ δ ν θ θ ν θ θ δ ν δ + + = + + = Λ Λ Λ Λ = Λ = − − Λ = Λ − = = 2 2 0 ) ( ) (6) 37 ( ) 315 945 9072 f( ) f(x) but K=K(x) f(K) 37 ( ) , g(K)=(2+ )( ) 6 315 945 9072 f K g K U τ θ µ = Λ Λ − − Λ → ⇒ Λ Λ Λ = − − [ ] [ ] { } 0 2 2 2 (2 ) 6 1 ( ) 2 ( ) ( ) (7) 2 where K= ( ) , let us take Z= as the new dependent variable so that K=Z and the mom,int. becomes U 2 ( ) 2 ( ) ( U ) or U d U f K K g K dx dU K x dx d dZ U Now dx dZ g K f K K H K dx dx τ µ δ θ ν θ ν θ ν Λ = + + + = = = − + = st (8) H(K) is known (1 order nonlinear , ODE for Z , solve numericallay , start x=0 stop =-12 ) ( H K = → Λ [ ] separation ) but complex H( ) ODE for Z(x) - mom. int. reduces to above form IVP for OD for any (x) K & H(K) may be evalu ted E a Λ Λ → 0.0783 0.47 H(K) 0.0783 0.47 H(K) K H(K)=0.47-6K (9) approximation Linear in K over the range of interest 5 6 0 6 5 Mom. Int. eq. becomes U 0.47 6 ( ) 0.47 6 or 1 ( ) 0.47 U 0.47 ( ) ( ) ( ) Mom. int. may be expressed interms of this quadra x dZ dU K H K Z dx dx d ZU Z x U d U dx x ς ς = − = = − = = ∫ 2 2 5 6 0 ture then , since Z= , the value of will be 0.47 ( ) ( ) (10) ( ) x x U d U x θ θ ν ν θ ς ς = ∫ Procedure: Potential flow problem should be solved to yield the outer velocity U(x) (for a given boundary shape) Use eq. (10) to evaluate the momentum thickness ( ) x θ 2 2 2 Pr parameter (x) may be evaluated from the relation (11) difficult to find (x) found (x), (x) is evaluated from eq. (3) 3 37 (x) 7 ( ) ( ) 315 945 = (315 072 945 9 essure havi dU K x x n d g δ θ δ θ ν Λ Λ = = − − Λ Λ Λ Λ − − 2 0 ) and eq. (2) 9072 3 ( ) 10 120 u ( ) ( ) vel. distribution eq (1) U shear stress at the surface is given by eq. (4) (2 ) 6 F G U u δ δ δ η η τ µ Λ Λ = − = + Λ ← Λ = + In practice it is difficult to evaluate the quality from eq (11) unless is a constant Instead : choose specific functions and use foregoing eqs. to determine the outer-flow vel. & hence the nature of the boundary shape EXAMPLE Karman-Pohlhausen approx. applied to the case of flow over a flat plate ( ) x Λ Λ ( ) x Λ 2 x 2 x x tan eq. (10) 0.47 =0.686 =0.686 U Re 0 eq. (11) 0 ( ) x U cons t U dU dx dU dx ν ν θ θ θ δ ν = → = → = ⇒ ⇓ Λ = 0 0 2 x 37 x 5.84 eq. (3) (x)= =5.84 315 U Re 3 1.75 .(2) 10 Re 0.686 .(4) 2 1 Re 2 x x x From x eq U eq U ν θ δ δ δ δ δ τ τ µ δ ρ → = ⇒ = → = ⇒ = ⇒ = Θ Exact 0.664 4th order vel. pr 0.686 2nd order vel. pr 0.73 3.5% error STABILITY OF STEADY FLOWS Boundary – Layers Instabilities Usually laminar flow becomes turbulent flow EXAMPLE: Flow over a circular cylinder ~ 82D ~108D Re D C Laminar B.L Turbulent B.L • Due to vel. profile difference between lam. & turb. flow Significant drop in the drag coefficent D C • x y flow V(y) is known undisturbed flow "base flow" v=0 parallel ⎫ ⎬ ⎭ Linear Stability Analysis: The Method of Small Perturbations Introduce arbitrary small (infinitesimal) disturbance into the flow eqs. & determine whether this disturbance grows or decays with time if the disturbance grows with time, the flow (the B.L) will be classified as unstable if the disturbance decays with time, the flow (the B.L) will be classified as stable marginal stability (neutral): the disturbance neither grows nor decays Non linear stability analysis: no restriction on disturbance size A1 Introduce small disturbance to the velocity profile u(x,y,t) = V(y) + u’(x,y,t) 0 ( , , ) = ( ) + '( , , ) ( , , ) 0 '( , , ) ( , , ) ( ) '( , , ) u x y t V y u x y t v x y t v x y t p x y t p x p x y t = + = + 0 u' ' p' 1 ; 1 ; 1 V V p v where << << << A2 Substitute A1 into the N-S eqs. & continuity 2 0 2 2 2 2 2 2 2 2 2 ' ' 0 ' ' ' 1 ' ' ' ; ( ') '( ) ( ) ( ) ' ' ' 1 ' ' ' ; ( ') ' ( ) u v x y u u dV u dp u u x V u v t x dy y dx x y v v v dp v v y dp V dx y V u v t x x dy x y ν ρ ν ρ ∂ ∂ + = ∂ ∂ ∂ ∂ ∂ ∂ ∂ + + + + = − + + + + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ + + + = − + + ∂ ∂ ∂ ∂ ∂ ∂ ∂ A3 When the perturbation is zero, the above eqs. reduce to 2 0 2 1 0 dp d V dx dy ν ρ = − + Undisturbed flow (parallel) A4 Drop term A3 in x-mom. Eq.
Since the perturbation is assumed to be small, products of all primed quantities may be neglected as being small Thus , Linearized eqs. governing the motion of the disturbances are 2 2 2 2 2 2 2 2 ' ' 0 ' ' 1 ' ' ' ; ' ( ) ' ' 1 ' ' ' ; ( ) u v x y u u dV dp u u X V v t x dy dx x y v v dp v v Y V t x dy x y ν ρ ν ρ ∂ ∂ + = ∂ ∂ ∂ ∂ ∂ ∂ + + = − + + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ + = − + + ∂ ∂ ∂ ∂ Introduce a perturbation stream - function (to reduce number or eqs. by one) ψ A5 ' , '= x u v y ψ ψ ∂ ∂ = − ∂ ∂ In terms of this stream function the governing eqs. become 2 2 3 3 2 3 2 2 3 3 2 3 2 1 ' ( ) 1 ' ( ) dV dp V y t x y x dy dx x y y dp V x t x dy x x y ψ ψ ψ ψ ψ ν ρ ψ ψ ψ ψ ν ρ ∂ ∂ ∂ ∂ ∂ + − = − + + ∂∂ ∂∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ − − = − − + ∂∂ ∂ ∂ ∂∂ A6 Eliminate the pressure term by forming mixed derivative, above two eqs. above two eqs. may be reduced to one , 2 ' p x y ∂ ∂∂ 2 2 2 4 4 4 2 2 2 4 2 2 4 ( )( ) ( 2 ) d V V t x y x dy x y y x x ψ ψ ψ ψ ψ ψ ν ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ + + − = + + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ Stream function for the disturbance must satisfy this linear , 4th order , PDE A7 Since the disturbance under consideration is arbitrary in form, Perturbation stream function may be represented by the following Fourier – lntegral: ( ) 0 ( , , ) ( ) c:time coefficient i x ct x y t y e d α ψ φ α ∞ − = ∫ [ ] : & positive (inverse wavelength) 2 = m wave length of the disturbances real α π λ α 6 -i ct -i ct i : time variation e if c > 0 e as t disturbance will grow unstable in general co r i note c c c i α α = + → → →∞ →∞ → -i ct i i mplex number: if c < 0 e 0 as t disturbance will decay stable c = 0 ne α → → →∞ → → utrally stable (c=0) Plug in A6 yields the integro – differential equation: ( ) 2 i (x-ct) 0 2 2 4 2 4 2 4 4 i (x-ct) 0 ( " ) " e ( ''' 2 " , i 1 i 1 " , ""= ,..
) e .
i c i V i d d d d d y V d y α α α α φ α φ αφ α ν φ α φ φ α φ φ φ φ α ∞ ∞ ⎡ ⎤ − + − − ⎣ ⎦ ⎡ ⎤ = − + − = ⎣ ⎦ = = ∫ ∫ Above equation should be valid for arbitrary . , the integrand should vanish (because eq. should be valid for arbitrary disturbance) Thus α 2 2 4 (V-c)( "-)-V = ( '''' 2 " ) (A) i Orr-Sommerfield equation ν φ α φ φ φ α φ α φ α − + B.C disturbance should vanish at the surface y=0 and at the edge of the Boundary Layer ' ( , 0, ) 0 , '( , 0, ) 0 ' ( , , ) '( , , ) 0 as y u x y t v x y t u x y t v x y t = = = = = = →∞ 0 0 interms of the stream function (y) ' 0 ' 0 y y u y v x ψ ψ ψ = = ∂ = = → ∂ ∂ = − = → ∂ '(0) 0 (B) (0) 0 '( ) ( ) 0 as y y y φ φ φ φ = = = → →∞ Solution of the Orr – Sommerfeld Equation Undisturbed vel. profile V(y) and disturbance wavelength is specified α ( ) & known V y α . (A) with BC.(B)represent an eigenvalue problem for the time coefficient , c , 0 flow stable 0 flow unstable ( ) , 0 r i i i i Eq c c i c c c i x ct c α = + < ⇒ > ⇒ − = neutral stablity ⇒ i (x-ct) (y) e α ψ φ = unstable stable , 0 i stable c < Re Uα ν = Recritical α δ Steady laminar flow can become another steady lam. flow H T L T . Pr Ra Gr = Diagram: Stability Orszag (1971): 0 0 0 i i i c c c < > < V α typical stability – calculation result for fixed , is varied. Then, by considering all possible values of the undisturbed B.L vel. (which less than the outer –flow vel.) a stability diagram is constructed V α possible values of ( ) in the range 0 ( ) ( ) All V y V y U x ≤ ≤ Flow over a flat surface Re 420 cr cr Uα ν = = unstable boundary stability 420 α δ Re Uα ν = 0.34 575 Recr Schlichting = Re >420 arbitrary disturbance will be unstable.
manifest themselves in the form of turbulence FREE – SHEAR FLOWS (LAYERS) Unaffected by walls Develop and spread in an open ambient fluid Possess vel. gradient created upstream mechanism viscous diffusion convective deceleration ⇔ EXAMPLE: 1) The free-shear layer between parallel moving streams: 2 U 1 U 1 U interface y 1 1 0 2 2 u ρ µ ρ µ 2 U u - shaped free-shear layer is due to viscous diffusion S At x=0 , upper free stream lower free stream 1 2 U meets as x=0 U ⎫ ⎬ ⎭ 1 2 & U uniform U For each stream , can define a Blasius – type similarity variable Lock(1951) – two different fluids with physical parameters 1 1 2 2 ( , ) & ( , ) ρ µ ρ µ 1 1 j 1 , , j=1,2 2 2 ( ) j j j j j j j u U y f x U U x f η ν ψ ν η ′ = = = Following the same procedure as in derivation of Blasius equation, one can obtain Blasius-type eq. for each layer ''' '' 0 j=1,2 j j j f f f + = 1 '( ) 1 asymptotic approach to the two stream veloci i s t e f +∞= . . 1) B C s 2 2 2 2 1 1 1 U y (- ) U ' U as + u f u U η η → ∞ → →−∞ ⇒ → → = → →∞ . . 2) B C s 1 2 1 2 Kinematics equality , and at the interface u u v v = = j 1 2 0 1 2 1 2 1 2 1 2 0 '(0) '(0) 0 u u (0) (0) 0 x x f f u f f v v η ψ ψ = → = ≠ = = ∂ ∂ = = = ⇒ = ∂ ∂ . . 3) B C s 1 1 2 1 2 j 1 1 1 1 1 1 1 1 1 1 1 1 0 0 (0) (0) or 2 ' '' (1) 2 i y y U u u y y x U u f U U f y y x µ µ η ν η µ µ µ η ν = ∂ ∂ = = ∂ ∂ ∂ ∂ ∂ = = ∂ ∂ ∂ Equality of shear stress at the interface 1 2 2 2 1 2 2 0 2 2 1 1 2 2 1 2 1 1 1 2 '' (2) 2 1 1 (1)=(2) ''(0) ''(0) ''(0) ''(0) y U u U f y x f f f f µ µ ν ρ µ µ µ ρ µ ν ν = ∂ = ∂ ⇒ = → = 2 2 1 2 1 1 ''(0) ''(0) k= f k f ρ µ ρ µ = 1 2 1 2 : k=1 (identical fluids) ; : a gas flowing over a liquid k>>1 ex Case 1 Case 2 . air-water Most p interf ractic ace k 60000 k 2 al cas s 45 e ρ ρ µ µ = = ≈ ⇒ ≈ TURBULENCE INTRODUCTION LAMINAR FLOW : Smooth , orderly flow limited to finite values of critical parameters: Re, Gr, Ta, Ri Beyond the critical parameter, Laminar flow is unstable a new flow regime turbulent flow Transition Laminar Turbulent x Characteristics 1) Disorder : not merely white noise but has spatial structure (Random variations) 2) Eddies : (or fluid packets of many sizes) Large & small varies continuously from shear – layer thickness down to the Kolmogorov length scale , 3) Enhanced mixing in laminar flow molecular action mixing in turbulent flow turbulent eddies actively about in 3-D and cause rapid diffusion of mass, momentum & energy Heat transfer & friction are greatly enhanced compared to Lam. Flow 4) Fluctuations : (in pressure, vel. & temp. ) Velocity fluctuates in all three directions 5) Self-sustaining motion: Once trigged turbulent flow can maintain. Itself by producing new eddies to replace those lost by viscous dissipation δ 3 14 3 ( ) L U ν δ = Experimental measurement : Hot-wire anemometer measure fluctuations in velocity via heat transfer Examine change in resistance assoc. with temp. (use wire ~ 0.0001” dia.) u t u t u t Laminar B.L Shedding cylinder Turbulent B.L Mathematical Description N-S eqs. do apply to turbulent flow Direct Numerical Simulation :Solve the N-S eqs. directly using computers Problem: wide range of flow scales involved solutions requires supercomputers and even then are limited to very low Reynolds numbers Mesh points : beyond the capacity of present computers (trillions) Eq. Turbulent flow in a pipe 7 22 d At Re 10 requires 10 numerical operatious computation would take thousand years to complete (for the fine details of the turbulent flow) = → ⇒ Direct numerical simulation DNS Because of complexity of the fluctuations, a purely numerical computation of turbulent flow has only been possible in a few special cases.
Therefore, consider time average of turbulent motion Difficulties in setting up eqs. of motion for mean motion Turbulent fluctuations coupled with mean motion Time averaging N-S additional terms (determined by turbulent fluctuations) Additional unknowns in computation of mean motion We have more unknowns than eqs.
To close system of eqs. of motion need additional eqs ⇒ These eqs. can no longer be set up purely from the balances of mass momentum & energy But, they are model eqs. which model relation between the fluctuations & mean motion called turbulence modelling central problem in computing the mean motion of turbulent flows Mean Motion & Fluctuations , time average value u ' u Decompose the motion into a mean motion & a fluctuating motion ' ' ' ' u u u v v v w w w p p p = + = + = + = + compressible turbulent flows = ' ; ' In T T T ρ ρ ρ + = + Average is formed as the time average at a fixed point in space 0 0 1 integral is to be taken over a sufficently large time interval T so that ( ) t T t u u dt u f t T + = ← ≠ ∫ 12 2 0 Characterization of fluctuation RMS 1 ( ) T u u u d t T ⇒ ⎫ ⎧ ⎪ ⎪ = − ⎨ ⎬ ⎪ ⎪ ⎩ ⎭ ∫ ' ( ) ' ( ) u g t u u u f t = = + = definition time average of fluctuating quautities are zero i.e. ' 0 , ' 0 , ' 0 , ' 0 assume that mean motion indep. of time steady turbulent flow By u v w p First = = = = ⇒ u t steady unsteady Turb. flow u u t steady unsteady Lam . flow ' , ' , ' influence the progrees of mean motion , , , so that mean motion exhibit an apparent increase in resistance aganist deformation.Increased apparent viscosity all Fluctuations u v w u v w is cenral of theoretical considerations on turbulent flow , + , . . u , ; ' ' ; ' 0 x of computation u u u v u v u v u v u udx udx uv Rul u v u v u v x es = + = = ∂ ∂ = = = + = ∂ ∂ ∫ ∫ x ' ' xy xy xy lam tur u u v y τ τ τ µ ρ ∂ = + = − ∂ Additional shear stress (Reynolds stress) N N ( ') ( ')= ' ' ' ' ' ' ' : ' 0 u v uv u u v v uv uv vu u v uv Ex uv u v u v = + + + + + = + ≠ Physical Interpretation of ' ' as a stress a)Consider fluid particle moving up from 1 to 2 ' 0 ' 0 (since u v v u ρ > < 1 2 turb particle has velocity deficit i.e ) ' ' 0 0 cel. of flow at 2 b)if particle moves down from u u u v de τ < < ⇒ > ⇒ turb 2 to 1 ' 0 ' 0 (particle has excess vel.) ' ' 0 0 accel. of flow at 1 v u u v τ < > ∴ < ⇒ > ⇒ 2 1 ' v ' u 2 1 Momentum exchange Turbulent shear stress is higher Basic Eqs. for Mean Motion of Turbulent Flows Consider flows with constant properties Continuity equation (1) ' ' of (1) u v w u u u x y z u u u Time averaging x x x ∂ ∂ ∂ + + = + ∂ ∂ ∂ ∂ ∂ ∂ − = + ∂ ∂ ∂ (2) =0 ' ' ' (3) Also , using (1) 0 time average values &fluctuations satisfy laminar flow continuity eq Momentum Eqs.(Re y u v w x y z u v w x y z Both ∂ ∂ ∂ + + ∂ ∂ ∂ ∂ ∂ ∂ + + = ∂ ∂ ∂ 2 Incomp. N-S eqs. ( ( . ) ) nolds eqs.) - (4) V V V p V t ρ µ ∂ + ∇ = ∇+ ∇ ∂ J G J G J G Substitute ' ' ' ' into N-S s 1) eg u u u v v v w w w p p p = + = + = + = + 2) Time average the equations 3) Drop-out terms which average to zero . Use “Rules of Computation” 2 2 2 ' ' 0 0 terms which are linear in fluctuating quantities 0 ' 0 ' ' 0 terms which are quadratic in fluctuating quantities 0 u u t x u u v ∂ ∂ = = ← ⇒ ∂ ∂ ≠ ≠ ← ⇒ 2 2 2 2 Resultant eqs. (called Reynolds eqs.) ( ) ( ) ' ' ' ' ' ( ) ' ' ' ' ' ( ) u u u p u v w u x y z u u v u w x y z u v v v w x y x v v v p u v w v x y z z y ρ ρ ρ µ ρ µ ∂ ∂ ∂ + + ∂ ∂ ∂ ∂ ∂ + + = − + ∂ ∂ ∂ ∂ ∂ + ∇ − ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ + + = − + ∇ − ∂ ∂ + ∂ ∂ ∂ ∂ ∂ 2 2 ( ) treat unsteady "fluctuations" add ' ' ' ' ' ( itional terms due to turbulent as added stresses call ) ed w w w p u v w w x u w v w w x y y z z z µ ρ ρ ∂ ∂ ∂ + + ∂ ∂ ∂ ∂ ∂ ∂ ∂ + + = − + ∇ − ∂ ∂ ∂ ∂ ∴ ⇒ fluctuating motion momentum Reynolds stresses(turbulent stresses) exchange due to fluctuations "stresses" ⇒ ⇒ N 2 xx xy Re stress apparent turbulent viscous stresses laminar stresses Complete stresses consist of 2 ' fluctuatios ( ) ' ' ,.......
ynolds u p u x u v u v y x σ µ ρ τ µ ρ ∂ = − + − → ∂ ∂ ∂ = + − ∂ ∂ In general , Reynolds stresses dominate over viscous stresses, except for regions directly at the wall Closure problem too few eqs : 4 too many unknowns : 10 Figure some way to approximate Reynolds stresses Objective : Establish relationship between Reynolds stresses & mean motions, i.e , , u v w model eqs. must be developed turbulence models or turbulence modeling.
model equations contain empirical elements ⇒ ∴ lam turb . cos Attempt to approximate a "turbulent" viscosity idea : Since ' ' viscosity A Eddy vis ity u u y y u Let u v y Eddy τ µ ν ρ τ ρ ρ ν − ∂ ∂ = = ∂ ∂ ∂ = = − ∂ ⇒ ∈ ∈>> 6 : how to model ?
For some situations const.
In general . ( , , , .) In general, many wild Pr guesses are made, not many work oblem u const f u y etc y ∈ ⇒∈≈ ∂ ∈≠ ⇒∈= ∂ Energy Equation Consider the energy equation for incompressible flow with constant properties 2 p DT c k T Dt ρ = ∇ + Φ Taking the time-average of the energy eq. , we obtain following eq. for the average temp.
field T ( , , ) ( ) convection p x y z T T T c u v w x y z ρ = ⎫ ∂ ∂ ∂ + + ⎬ ∂ ∂ ∂ ⎭ G G J G 2 2 2 2 2 2 p 2 2 2 2 2 =k( + + ) molecular heat transport ' ' ' ' ' ' - c ( ) turbulent heat transport("apparent" heat conduction) + 2( ) 2( ) 2( ) ( + ) ( + ) ( T T T x y z u T v T w T x y z u v w u v u w v x y z y x z x z ρ µ ⎫ ∂ ∂ ∂ ⎬ ∂ ∂ ∂ ⎭ ⎫ ∂ ∂ ∂ + + ⎬ ∂ ∂ ∂ ⎭ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ + + + + + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ 2 + ) direct dissipation w y ⎫ ⎡ ⎤ ∂ ⎪ ⎬ ⎢ ⎥ ∂ ⎪ ⎣ ⎦⎭ The same eq.holds for the average temp. fields as for laminar temp. fields, apart from two additional terms "apparent" heat conduction div( ' ') "turbulent" dissipation , V T ρ ⇒ ∈ JJ G 2 2 2 2 2 2 2( ) 2( ) 2( ) ( + ) ( + ) ( + ) u v w u v u w v w x y z y x z x z y ρ µ ⎫ ⎡ ⎤ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ⎪ ∈= + + + + + ⎢ ⎥⎬ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ⎢ ⎥⎪ ⎣ ⎦⎭ In turbulent flows mechanical energy is transformed into internal energy in two different ways: a) Direct dissipation : transfer is due to the viscosity (as in laminar flow) b) Turbulent dissipation : transfer is due to the turbulent fluctuations The Turbulence Kinetic Energy Equation (K-equation) Many attemps have been made to add “turbulence conservation” relations to the time-averaged continuity, momentum and energy equations derived.
A relation for the turbulence kinetic energy K of fluctuations.
( ) 1 2 3 1 1 2 2 Einstein summation notation, ( , , ) ( , , ) i i i K u u v v w w u u u u u u u v w ′ ′ ′ ′ ′ ′ ′ ′ ≡ + + = = = A conservation relation for K can be derived by forming the mechanical energy equation i.e., dot product of ui ve ith momentum equation subtract instantaneous mechanical energy equation from its time averaged value.
Result: Turbulence kinetic energy relation for an incompressible fluid. N I III II V IV 1 2 j i j j i j i i j j j i i j i j i i j i u DK p u u u u u Dt x x u u u u u u x x x x x x ρ ν ν ′ ⎡ ⎤ ∂ ′ ⎛ ⎞ ∂ ′ ′ ′ ′ ′ = − + − + ⎢ ⎥ ⎜ ⎟ ∂ ∂ ⎝ ⎠ ⎢ ⎥ ⎣ ⎦ ⎡ ⎤ ⎛ ⎞ ⎛ ⎞ ′ ′ ′ ′ ′ ∂ ∂ ∂ ∂ ∂ ∂⎢ ⎥ ′ ⎜ ⎟ ⎜ ⎟ + − + ⎢ ⎥ ⎜ ⎟ ⎜ ⎟ ∂ ′ ′ ′ ′ ′ ∂ ∂ ∂ ∂ ∂ ⎝ ⎠ ⎝ ⎠ ⎢ ⎥ ⎣ ⎦ I. Rate of change of turbulent (kinetic) energy II. Convective diffusion of turbulence energy III. Production of turbulent energy IV. Viscous diffusion (work done by turbulence viscous stresses) V. Turbulent viscous dissipation Reynolds stress equation: conservation equations for Reynolds stresses see F. White pg. 406 2-D Turbulent Boundary Layer Equations Just as laminar flows, turbulent flows at high Re also have boundary layer character, i.e. large lateral changes and small longitudinal changes in flow properties.
Ex.: Pipe flow, channel flow, wakes and jets. δ(x)<<x y x Same approximations as in laminar boundary layer analysis, v u x y ∂ ∂ << << ∂ ∂ Assume that mean flow structure is 2D 2 0 0 but 0 w w z ∂ ′ = = ≠ ∂ Basic turbulent equations (Reynolds equations) reduce to Continuity: 0 (1) 1 x-momentum: (2) : free stream velocity Thermal energy: (3) where e e e p u v x y dU u u u v U x y dx y U T T q u c u v x y y y τ ρ ρ τ τ ∂ ∂ + = ∂ ∂ ∂ ∂ ∂ + ≈ + ∂ ∂ ∂ ⎛ ⎞ ∂ ∂ ∂ ∂ + ≈ + ⎜ ⎟ ∂ ∂ ∂ ∂ ⎝ ⎠ = N turbulent flux molecular flux (4) p u u v y T q k c v T y µ ρ ρ ∂ ′ ′ − ∂ ∂ ′ ′ = − ∂ Above equations closely resemble the laminar flow equations except that τ and q contain turbulent shear stress and turbulent heat flux (Reynolds Stress) must be modelled. y-momentum equation reduces to 2 (5) p v y y ρ ′ ∂ ∂ ≈− ∂ ∂ Integrating over the boundary layer yields: 2 ( ) e p p x v ρ ′ ≈ − Unlike laminar flow, p varies slightly across the boundary layer due to velocity fluctuations normal to the the wall 2 .
p v const ρ ′ + ≈ Note: : wall pressure no-slip v 0 ( ) w e w p p p x ′ ⇒ ≡ ⇒ = e e e dp U dU ρ ≈− Bernoulli equation in the (inviscid) free stream Boundary Conditions: Free stream conditions Ue(x) and Te(x) are known.
No-slip, no jump: ( ,0) ( ,0) 0 , ( ,0) ( ) Free stream matching: ( , ) , ( , ) ( ) w e T e u x v x T x T x u x U T x T x δ δ = = = = = The velocity and thermal boundary layer thicknesses (δ, δT) are not necessarily equal u v if a suitable correlation for total shear τ is known. but depend upon the Pr, as in laminar flow. Eqs. 1 and 2 can be solved for Turbulent Boundary Layer Integral Relations: The integral momentum equation has the identical form as laminar flow ( ) 2 0 2 2 1 , H= (momentum shape factor) 1 f e w e e e e e c dU d H dx U dx U u u dy U U u dy U τ θ θ ρ δ θ θ δ ∞ + + = = ⎛ ⎞ = − ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ = − ⎜ ⎟ ⎝ ⎠ ∫ Turbulent velocity profile is more complicated in shape and many different correlations have been proposed.
Example: Turbulent pipe flow Often used correlation is the empirical power-law velocity profile R x r 1/ 1 n c u r V R ⎛ ⎞ = − ⎜ ⎟ ⎝ ⎠ n=f(Re) for many practical flows n = 7 104 5 6 8 7 105 106 n Re=ρVD/µ 0 0 1.0 1.0 r/R laminar n=6 n=8 n=10 Turbulent profile c u V Turbulent profiles are much “flatter” than laminar profile Flatness increases with Reynolds number (i.e., with n) Turbulent velocity profile(s): The inner, outer, and overlap layers.
Key profile shape consist of 3 layers Inner layer: very narrow region near the wall (viscous sublayer) viscous (molecular) shear dominates laminar shear stress is dominant, random eddying nature of flow is absent Outer layer: turbulent (eddy) shear (stress) dominates Overlap layer: both types of shear important; profile smoothly connects inner and outer regions. Example: Structure of turbulent flow in a pipe R r 0 τlam τtur τ pipe wall τw τ(r) Shear stress 0 R r Vc Viscous sublayer overlap layer outer layer Average velocity Inner law: ( , , , ) (1) w u f y τ ρ µ = Velocity profile would not depend on free stream parameters.
Outer law: ( , , , , ) (2) e e w dp U u g y dx τ ρ δ − = Wall acts as a source of retardation, independent of µ.
Overlap law: (3) inner outer u u = We specify inner and outer functions merge together smoothly.
Dimensionless Profiles: The functional forms in Eqs.(1)-(3) are determined from experiment after use of dimensional analysis. Primary Dimensions: (mass, length, time) : 3 Eq.(1) : 5 variables Π groups : 5-3 = 2 (dimensionless parameters) Proper dimensionless inner law: 1/2 w ; = u yv f v v τ ν ρ ⎛ ⎞ ⎛ ⎞ = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ Variable v [m/s] called wall friction velocity. v is used a lot in turbulent flow analyses.
Outer law using Π - theorem: w , ; = e e U u dp y g v dx δ ξ ξ δ τ − ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ Often called velocity defect law, with e U u − being “defect” or retardation of flow due to wall effects. At any given position x, defect g(y/δ) will depend on local pressure gradient ξ. Let ξ have some particular value. Then overlap function requires Overlap law: -e U u v y y f g v v δ ν δ δ ⎛ ⎞ ⎛ ⎞ = = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ From functional analysis: both f and g must be logarithmic functions.
Thus, in overlap layer: 1 Inner variables: ln 1 Outer variables: ln e u yv B v k U u y A v k ν δ = + − = − + Where K and B are near-universal constants for turbulent flow past smooth, impermeable walls. K≈0.41 , B≈5.0 pipe flow measurements, data correlations A varies with pressure gradient ξ (perhaps with other parameters also).
Let , and y u yv u v ν + + = = Inner layer details, Law of the wall.
At very small y, velocity profile is linear.
5: w u y or u y y τ µ + + + ≤ = = Example: Thickness of viscous sublayer 5 : viscous length scale of a turbulent boundary layer sub v v ν ν δ = Flat plate airfoil data: v=1.24 m/s , νair≈1.51x10-5 m2/s Between 5 ≤y+≤30 buffer layer.
Velocity profile is neither linear nor logarithmic but is a smooth merge between two. Spalding (1961) single composite formula.
( ) ( ) 2 3 1 2 6 KB Ku Ku Ku y u e e Ku + + + + + − + ⎡ ⎤ ⎢ ⎥ = + −− − − ⎢ ⎥ ⎣ ⎦ Notes: 1 1 1 1 1 1 0 0 n c n c u r V R V du r dr n R R du r R dr du r dr − ⎛ ⎞ = − ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ = − − ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ = = ∞ = ≠ Power law profile cannot be valid near the wall.
Power law profile cannot be precisely valid near the centreline. However, it does provide a reasonable approximation to measured velocity profiles across most of the pipe.
Example: Water at 20 °C (ρ=998 kg/m3), ν=1.004x10-6 m2/s Q=0.04 m3/s D=0.1m 2.59 / dp kPa m dx = δs = ? thickness of viscous sublayer?
centreline velocity, Vc = ?
ratio of turbulent to laminar shear stress, τturb/τlam = ? at a point midway between the centreline and pipe wall i.e., at r = 0.025 m.
Law of the wall valid 5 y± ≤ viscous sublayer 5 5 5 5 s s s w yv y v y y v v ν δ ν δ δ ν τ ρ ± ± = ≤ = = ⇒ = = = Pressure drop and wall shear stres in a fully developed pipe flow is related by 4 w l p D τ ∆= (Valid for both laminar & turbulent flow) (Exercise: Obtain the above equation considering the force balance of a fluid element) 3 2 2 3 6 5 (0,1)(2,59.10 ) 64,8 / 4 4(1 ) 64,8 / So, v 0,255 / 998 / 5.1,004.10 1,97.10 0,02 0,255 w s D p Pa N m l m N m m s kg m m mm τ δ − − ∆ = = = = = = = ≅ Imperfections on pipe wall will protrude into this sublayer and affect some of the characteristics of flow(i.e.,wall shear stres & pressure drop) 3 2 2 5 6 5 0,04 / 5,09 / (0,1) / 4 5,09.(0,1) Re 5,07.10 1,004.10 Re 5,07.10 8,4 Q m s V m s A m VD n π ν − = = = = = = = ⇒ = Power-law profile 1/8,4 1/ 0 2 2 2 2 c (1 ) .
(1 ) (2 ) 2 ( 1)(2 1) V 2 V ( 1)(2 1) 8,4: V 1,186 1,186(5,09) 6,04 / c R n c c c u r V R r Q AV udA V r dr R n Q R V n n n Q R V n n n V m s π π π ≅ − = = = − = + + = ∴ = + + = = = = ∫ ∫ Recall that Vc=2V for laminar pipe flow: 0,025 ?
turb lam r m τ τ = = Shear stres distribution throughout the pipe 2 wr D τ τ = (Valid for laminar or turbulent flow) r R=D/2 2 1/ (1 )/ (1 8,4)/8,4 0,025 2(64,8).0,025 ( 0,025) 32,4 / 0,1 32,4 ; (1 ) (1 ) 6,04 0,025 (1 ) 26,5 8,4(0,05) 0,05 lam turb n n n c lam c r r N m V du r du r u V dr R dr nR R du dr τ τ τ τ τ µ − − = = = = = + = = − = − ⇒ = − − = − − = − 6 2 ( ) (1,004.10 ).(998).( 26,5) 0,0266 / 32,4 0,0266 1220 0,0266 lam turb lam du du dr dr N m τ µ νρ τ τ − = − = − = − − = − = = As expected turb lam τ τ >> Thus Turbulent Boundary Layer on a Flat Plate Problem of flow past a sharp flat plate at high Re has been studied extensively, numerous formulas have been proposed for friction factor.
-curve fits of data -use of Momentum Integral Equation and/or law of the wall -numerical computation using models of turbulent shear Momentum Integral Analysis 2 0 ( .) 2 f w C dp d U const dx dx U τ θ ρ = = = = Momentum Interal Equation valid for either laminar or turbulent flow.
For turbulent flow a reasonable approximation to the velocity profile ( / ) u f y U δ = Functional relationship describing the wall shear stress Need to use some empirical relationship For laminar flow 0 w y u y τ µ = ∂ = ∂ Example: Turbulent flow of an incompressible fluid past a flat plate Boundary layer velocity profile is assumed to be 1/7 ( ) u y U δ = ←power law profile suggested by Prandtl (taken From pipe data!) Reasonable approximation of experimentally observed profiles, except very near the plate, 0 !
y u y = ∂ = ∞ ∂ Laminar Turbulent 1 0 0 1 y η δ = 1/7 ( ) u y U δ = Assume shear stress aggrees with experimentally determined formula 1/4 2 1/ 4 0,045Re or 0,0225 ( ) Re f w C U U U δ ν τ ρ δ δ ν − ⎧ ⎫ = = ⎨ ⎬ ⎩ ⎭ = Determine; , , and as a function of x.
w δ δ θ τ What is the friction drag coefficient CD,f=?
Momentum Integral Equation (with U=constant) 1/7 1/7 2 1 1 1/7 1/7 0 0 1/ 4 1/ 4 1/ 4 1/ 4 0 0 1/5 4/5 ; ( ) 2 7 (1 ) (1 ) (1 ) 72 7 0,0225Re 0,0225( ) 72 0,231( ) 0,370( ) f w o x C d y u y dx U U u u u u dy d d U U U U d dx U d dx U x U δ δ τ θ η η ρ δ δ δ θ δ η δ η η η δ ν δ ν δ δ ν δ ∞ − = = = = = = − = − = − = = = = = ∫ ∫ ∫ ∫ ∫ or in dimensionless form 1/5 0,370 Rex x δ = Boundary layer at leading edge of plate is laminar but in practice,laminar boundary layer often exists over a relatively short portion of plate. error associated with starting turbulent boundary layer with =0 at x=0 can be negligible.
δ ∴ 1 1 1/7 0 0 0 1/5 (1 ) (1 ) (1 ) 8 0,0463 Rex u u dy d d U U x δ δ δ η δ η η δ ∞ = − = − = − = = ∫ ∫ ∫ 1/5 4/5 1/5 1/ 4 2 2 1/5 4/5 1/5 1/5 7 0,0360( ) 72 0,036 Re 0,0288 0,0225 (0,37)( / ) Re 0,058 Re x w x f x x U x U U U U x C ν θ δ θ θ δ δ ν ρ τ ρ ν = = = < < ⎡ ⎤ = = ⎢ ⎥ ⎣ ⎦ = Friction drag on one side of plate,Df 2 1/5 0 2 1/5 1/5 2 4/5 1/5 1/ 2 1/ 2 (0,0288 ) ( ) 0,0360 where A=b.l area of plate Re 0,0720 1 Re 2 Turbulent flow: ( ) ~ ; ( ) ~ Laminar flow: ( ) ~ ; ( ) ~ l l f w o f l f Df l w w D b dx b U dx Ux A D U D C U A x x x x x x x x ν τ ρ ρ ρ δ τ δ τ − − = = = = = ∫ ∫ Note:Results presented in this example are valid only in the range of validity of original data, assumed velocity profile & shear stres.
The range covers smooth flat plates with 5x105<Rel<107 See Fig 6-20 (White, page 432) Example 1 : Momentum Integral Equation-Approximate vel. profile .
( 0) dU U const dx = ⇒ = 2 1 1 1 1 ( ) For 0 1/ 2 2 1 at & 0 at 0 3 2 0, 4/3 4 : 0 1/ 2 3 Similarly, w d dx U u y f U f a b f f a b u U τ θ ρ η η δ η η η η η η = = = ≤ ≤ = + = = = = ∴ = = = ≤ ≤ 1 1/ 2 1 0 0 1/ 2 0 0 1 2 1 for 1 3 3 2 4 4 1 2 1 2 (1 ) (1 ) ( )(1 ) 3 3 3 3 3 3 0,1574 u 4 = 3 4 0,1574 3 4 0,1574 3 w y u U u u d d d U U u U y d dx U d dx U η η η θ δ η δ η η η δ η η δ τ µ µ µ η δ δ ν δ ν δ δ = = = + ≤ < = − = − + + − − = ∂ ∂ = = ∂ ∂ = = ⇒ ∫ ∫ ∫ 0 2 ( ) 4,12 0,648 1 Re 2 w f x x x U C U δ ν δ τ ρ = = = ∫ u U 2U/3 / 2 δ δ U l 4l a) U 4l l b) 4 2 , , 2 , D,b , , , , 1 2 1,328 1,328 & 4 Re , , is the same 1,328 1,328 C Re 2 4 l l D a D a D a D a D a D b D b F U AC C A l U U A U F C l l F C ρ ν ρ ν = = = = = = = = a) The shear stres decreases with distance from the leading edge of the plate. Thus, even though the plate area is the same for case (a) or (b), the average shear stress (and the drag) is greater for case (a).
Example 2 : Viscous drag in thin plate Example 3: Thin flat plate in water tunnel Parabolic velocity profile: 2 2 5 5 6 0 w 0 0 0 2( ) ( ) 2 1,6.(0,3) Re 4,8.10 5.10 10 Flow is laminar Viscos drag 2 (2 sides of plate) .
(2 2 ) l L D w y u y y U Ul F bdx u u U y y η η η η δ δ ν τ η µ τ µ µ η η δ − = = = = − = − = = = < ∴ = = ∂ ∂ ∂ = = = − ∂ ∂ ∂ ∫ 0 0 2 5,48 Re 2 4 8 2 5,48 5,48 1,62 x L L D D U x U U dx b U UL F bdx b U x F N µ δ δ µ µ µ δ ν ν = = = = = = ∫ ∫ b=1 m L=0, 3m U x δ(x) 2 3 0 2 2 2 0 . , (0.30.3)0.7 0.063 / ( ) ( 2 ) :effective area of the duct (allowing for the decreased flowrate in the b.l.) Thus, ( 2 ) 0.09 inlet inlet Continuity eq for incompressible flow Q d U m s Q Q x UA U d A d d δ δ = = = = = = − = − = [ ] 0 2 0.3 2 d d m δ δ ⇒ = + = + [ ] [ ] [ ] 5 1.510 1.72 1.72 0.00796 0.3 0.0159 0. m 0.7 ( 3 ) 328 d x x x x U m d x m m ν δ − = = = = = + ≅ |
1173 | https://askfilo.com/user-question-answers-smart-solutions/phosphorus-pentachloride-pcl5-reacts-with-alcohols-to-give-3335363632313932 | Phosphorus pentachloride (PCl5) reacts with alcohols to give alkyl chlori..
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Phosphorus pentachloride (PCl5) reacts with alcohols to give alkyl chlorides. The general reactions are:
```markdown
3 R-OH + PX3 \rightarrow 3 R-X + H3PO3
R-OH + PCl5 \rightarrow R-Cl + HCl + POCl3
```
Explain or describe this reaction process.
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Explanation of the reaction between phosphorus pentachloride and alcohols
Phosphorus pentachloride (PCl5) is a chlorinating agent that converts alcohols (R-OH) into alkyl chlorides (R-Cl). The reaction proceeds as follows:
When an alcohol reacts with PCl5, the hydroxyl group (-OH) is replaced by a chlorine atom (-Cl).
Byproducts of the reaction are hydrogen chloride (HCl) and phosphoryl chloride (POCl3).
The balanced chemical equation for a typical monohydric alcohol is:
R−O H+PC l 5→R−Cl+H Cl+POC l 3
For a reaction with phosphorus trihalides (PX3), the generalized equation is:
3 R−O H+P X 3→3 R−X+H 3P O 3
Reaction steps overview:
The lone pair on the oxygen of the alcohol attacks the phosphorus atom in PCl5, forming an intermediate complex.
This intermediate subsequently loses a proton (H+), which results in the replacement of the -OH group by a chlorine atom.
The side products HCl and POCl3 are formed during the process.
This reaction is useful in organic synthesis for preparing alkyl chlorides from their corresponding alcohols.
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Phosphorus pentachloride (PCl5) reacts with alcohols to give alkyl chlorides. The general reactions are: markdown 3 R-OH + PX3 \rightarrow 3 R-X + H3PO3 R-OH + PCl5 \rightarrow R-Cl + HCl + POCl3 Explain or describe this reaction process.
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Question Text Phosphorus pentachloride (PCl5) reacts with alcohols to give alkyl chlorides. The general reactions are: markdown 3 R-OH + PX3 \rightarrow 3 R-X + H3PO3 R-OH + PCl5 \rightarrow R-Cl + HCl + POCl3 Explain or describe this reaction process.
Updated On Jun 24, 2025
Topic All topics
Subject Smart Solutions
Class Class 9
Answer Type Text solution:1
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1174 | https://www.damtp.cam.ac.uk/user/tong/kintheory/two.pdf | Kinetic Theory The purpose of this section is to lay down the foundations of kinetic theory, starting from the Hamiltonian description of 1023 particles, and ending with the Navier-Stokes equation of fluid dynamics. Our main tool in this task will be the Boltzmann equation.
This will allow us to provide derivations of the transport properties that we sketched in the previous section, but without the more egregious inconsistencies that crept into our previous attempt. But, perhaps more importantly, the Boltzmann equation will also shed light on the deep issue of how irreversibility arises from time-reversible classical mechanics.
2.1 From Liouville to BBGKY Our starting point is simply the Hamiltonian dynamics for N identical point particles.
Of course, as usual in statistical mechanics, here is N ridiculously large: N ⇠O(1023) or something similar. We will take the Hamiltonian to be of the form H = 1 2m N X i=1 ~ p 2 i + N X i=1 V (~ ri) + X i<j U(~ ri −~ rj) (2.1) The Hamiltonian contains an external force ~ F = −rV that acts equally on all parti-cles. There are also two-body interactions between particles, captured by the potential energy U(~ ri −~ rj). At some point in our analysis (around Section 2.2.3) we will need to assume that this potential is short-ranged, meaning that U(r) ⇡0 for r ≫d where, as in the last Section, d is the atomic distance scale.
Hamilton’s equations are @~ pi @t = −@H @~ ri , @~ ri @t = @H @~ pi (2.2) Our interest in this section will be in the evolution of a probability distribution, f(~ ri, ~ pi; t) over the 6N dimensional phase space.
This function tells us the proba-bility that the system will be found in the vicinity of the point (~ ri, ~ pi). As with all probabilities, the function is normalized as Z dV f(~ ri, ~ pi; t) = 1 with dV = N Y i=1 d3rid3pi Furthermore, because probability is locally conserved, it must obey a continuity equa-tion: any change of probability in one part of phase space must be compensated by – 14 – a flow into neighbouring regions. But now we’re thinking in terms of phase space, the “r” term in the continuity equation includes both @/@~ ri and @/@~ pi and, corre-spondingly, the velocity vector in phase space is (˙ ~ ri, ˙ ~ pi). The continuity equation of the probability distribution is then @f @t + @ @~ ri · ⇣ ˙ ~ rif ⌘ + @ @~ pi · ⇣ ˙ ~ pif ⌘ = 0 where we’re using the convention that we sum over the repeated index i = 1, . . . , N.
But, using Hamilton’s equations (2.2), this becomes @f @t + @ @~ ri · ✓@H @~ pi f ◆ −@ @~ pi · ✓@H @~ ri f ◆ = 0 ) @f @t + @f @~ ri · @H @~ pi −@f @~ pi · @H @~ ri = 0 This final equation is the Liouville’s equation.
It is the statement that probability doesn’t change as you follow it along any trajectory in phase space, as is seen by writing the Liouville equation as a total derivative, d f dt = @f @t + @f @~ ri · ˙ ~ ri + @f @~ pi · ˙ ~ pi = 0 To get a feel for how probability distributions evolve, one often evokes the closely related Liouville’s theorem2. This is the statement that if you follow some region of phase space under Hamiltonian evolution, then its shape can change but its volume remains the same. This means that the probability distribution on phase space acts like an incompressible fluid. Suppose, for example, that it’s a constant, f, over some region of phase space and zero everywhere else. Then the distribution can’t spread out over a larger volume, lowering its value. Instead, it must always be f over some region of phase space. The shape and position of this region can change, but not its volume.
The Liouville equation is often written using the Poisson bracket, {A, B} ⌘@A @~ ri · @B @~ pi −@A @~ pi · @B @~ ri With this notation, Liouville’s equation becomes simply @f @t = {H, f} 2A fuller discussion of Hamiltonian mechanics and Liouville’s theorem can be found in the lectures on Classical Dynamics.
– 15 – It’s worth making a few simple comments about these probability distributions. Firstly, an equilibrium distribution is one which has no explicit time dependence: @f @t = 0 which holds if {H, f} = 0. One way to satisfy this is if f is a function of H and the most famous example is the Boltzmann distribution, f ⇠e−βH. However, notice that there is nothing (so-far!) within the Hamiltonian framework that requires the equilibrium distribution to be Boltzmann: any function that Poisson commutes with H will do the job. We’ll come back to this point in Section 2.2.2.
Suppose that we have some function, A(~ ri, ~ pi), on phase space. The expectation value of this function is given by hAi = Z dV A(~ ri, ~ pi)f(~ ri, ~ pi; t) (2.3) This expectation value changes with time only if there is explicit time dependence in the distribution. (For example, this means that in equilibrium hAi is constant). We have dhAi dt = Z dV A@f @t = Z dV A ✓@f @~ pi @H @~ ri −@f @~ ri @H @~ pi ◆ = Z dV ✓ −@A @~ pi @H @~ ri + @A @~ ri @H @~ pi ◆ f (2.4) where we have integrated by parts to get to the last line, throwing away boundary terms which is justified in this context because f is normalized which ensures that we must have f ! 0 in asymptotic parts of phase space. Finally, we learn that dhAi dt = Z dV {A, H} f = h{A, H}i (2.5) This should be ringing some bells. The Poisson bracket notation makes these expres-sions for classical expectation values look very similar to quantum expectation values.
2.1.1 The BBGKY Hierarchy Although we’re admitting some ignorance in our description of the system by consider-ing a probability distribution over N-particle phase space, this hasn’t really made our life any easier: we still have a function of ⇠1023 variables. To proceed, the plan is – 16 – to limit our ambition. We’ll focus not on the probability distribution for all N parti-cles but instead on the one-particle distribution function. This captures the expected number of partcles lying at some point (~ r, ~ p). It is defined by f1(~ r, ~ p; t) = N Z N Y i=2 d3rid3pi f(~ r,~ r2, . . . ,~ rN, ~ p, ~ p2, . . . ~ pN; t) Although we seem to have singled out the first particle for special treatment in the above expression, this isn’t really the case since all N of our particles are identical.
This is also reflected in the factor N which sits out front which ensures that f1 is normalized as Z d3rd3p f1(~ r, ~ p; t) = N (2.6) For many purposes, the function f1 is all we really need to know about a system. In particular, it captures many of the properties that we met in the previous chapter. For example, the average density of particles in real space is simply n(~ r; t) = Z d3p f1(~ r, ~ p; t) (2.7) The average velocity of particles is ~ u(~ r; t) = Z d3p ~ p mf1(~ r, ~ p; t) (2.8) and the energy flux is ~ E(~ r; t) = Z d3p ~ p mE(~ p)f1(~ r, ~ p; t) (2.9) where we usually take E(~ p) = p2/2m. All of these quantities (or at least close relations) will be discussed in some detail in Section 2.4.
Ideally we’d like to derive an equation governing f1. To see how it changes with time, we can simply calculate: @f1 @t = N Z N Y i=2 d3rid3pi @f @t = N Z N Y i=2 d3rid3pi {H, f} Using the Hamiltonian given in (2.1), this becomes @f1 @t = N Z N Y i=2 d3rid3pi " − N X j=1 ~ pj m · @f @~ rj + N X j=1 @V @~ rj · @f @~ pj + N X j=1 X k<l @U(~ rk −~ rl) @~ rj · @f @~ pj # – 17 – Now, whenever j = 2, . . . N, we can always integrate by parts to move the derivatives away from f and onto the other terms. And, in each case, the result is simply zero because when the derivative is with respect to ~ rj, the other terms depend only on ~ pi and vice-versa. We’re left only with the terms that involve derivatives with respect to ~ r1 and ~ p1 because we can’t integrate these by parts. Let’s revert to our previous notation and call ~ r1 ⌘~ r and ~ p1 ⌘~ p. We have @f1 @t = N Z N Y i=2 d3rid3pi " −~ p m · @f @~ r + @V (~ r) @~ r · @f @~ p + N X k=2 @U(~ r −~ rk) @~ r · @f @~ p # = {H1, f1} + N Z N Y i=2 d3rid3pi N X k=2 @U(~ r −~ rk) @~ r · @f @~ p (2.10) where we have defined the one-particle Hamiltonian H1 = p2 2m + V (~ r) (2.11) Notice that H1 includes the external force V acting on the particle, but it knows nothing about the interaction with the other particles. All of that information is included in the last term with U(~ r −~ rk). We see that the evolution of the one-particle distribution function is described by a Liouville-like equation, together with an extra term. We write @f1 @t = {H1, f1} + ✓@f1 @t ◆ coll (2.12) The first term is sometimes referred to as the streaming term. It tells you how the particles move in the absence of collisions. The second term, known as the collision integral, is given by the second term in (2.10). In fact, because all particles are the same, each of the (N −1) terms in PN k=2 in (2.10) are identical and we can write ✓@f1 @t ◆ coll = N(N −1) Z d3r2d3p2 @U(~ r −~ r2) @~ r · @ @~ p Z N Y i=3 d3rid3pi f(~ r,~ r2, . . . , ~ p, ~ p2, . . . ; t) But now we’ve got something of a problem. The collision integral can’t be expressed in terms of the one-particle distribution function. And that’s not really surprising. As the name suggests, the collision integral captures the interactions – or collisions – of one particle with another. Yet f1 contains no information about where any of the other particles are in relation to the first. However some of that information is contained in the two-particle distribution function, f2(~ r1,~ r2, ~ p1, ~ p2; t) ⌘N(N −1) Z N Y i=3 d3rid3pi f(~ r1,~ r2, . . . , ~ p1, ~ p2, . . . ; t) – 18 – With this definition, the collision integral is written simply as ✓@f1 @t ◆ coll = Z d3r2d3p2 @U(~ r −~ r2) @~ r · @f2 @~ p (2.13) The collision term doesn’t change the distribution of particles in space. This is captured by the particle density (2.7) which we get by simply integrating n = R d3pf1. But, after integrating over R d3p, we can perform an integrating by parts in the collision integral to see that it vanishes. In contrast, if we’re interested in the distribution of velocities – such as the current (2.8) or energy flux (2.9) – then the collision integral is important.
The upshot of all of this is that if we want to know how the one-particle distribution function evolves, we also need to know something about the two-particle distribution function. But we can always figure out how f2 evolves by repeating the same calculation that we did above for f1. It’s not hard to show that f2 evolves by a Liouville-like equation, but with a corrected term that depends on the three-particle distribution function f3. And f3 evolves in a Liouville manner, but with a correction term that depends on f4, and so on. In general, the n-particle distribution function fn(~ r1, . . .~ rn, ~ p1, . . . ~ pn; t) = N!
(N −n)!
Z N Y i=n+1 d3rid3pi f(~ r1, . . .~ rN, ~ p1, . . . ~ pN; t) obeys the equation @fn @t = {Hn, fn} + n X i=1 Z d3rn+1d3pn+1 @U(~ ri −~ rn+1) @~ ri · @fn+1 @~ pi (2.14) where the e↵ective n-body Hamiltonian includes the external force and any interactions between the n particles but neglects interactions with any particles outside of this set, Hn = n X i=1 ✓~ p 2 i 2m + V (~ ri) ◆ + X i<jn U(~ ri −~ rj) The equations (2.14) are known as the BBGKY hierarchy.
(The initials stand for Bogoliubov, Born, Green, Kirkwood and Yvon). They are telling us that any group of n particles evolves in a Hamiltonian fashion, corrected by interactions with one of the particles outside that group. At first glance, it means that there’s no free lunch; if we want to understand everything in detail, then we’re going to have to calculate everything. We started with the Liouville equation governing a complicated function f of N ⇠O(1023) variables and it looks like all we’ve done is replace it with O(1023) coupled equations.
– 19 – However, there is an advantage is working with the hierarchy of equations (2.14) because they isolate the interesting, simple variables, namely f1 and other lower fn.
This means that the equations are in a form that is ripe to start implementing various approximations. Given a particular problem, we can decide which terms are important and, ideally, which terms are so small that they can be ignored, truncating the hierarchy to something manageable. Exactly how you do this depends on the problem at hand.
Here we explain the simplest, and most useful, of these truncations: the Boltzmann equation.
2.2 The Boltzmann Equation “Elegance is for tailors” Ludwig Boltzmann In this section, we explain how to write down a closed equation for f1 alone. This will be the famous Boltzmann equation. The main idea that we will use is that there are two time scales in the problem. One is the time between collisions, ⌧, known as the scattering time or relaxation time. The second is the collision time, ⌧coll, which is roughly the time it takes for the process of collision between particles to occur. In situations where ⌧≫⌧coll (2.15) we should expect that, for much of the time, f1 simply follows its Hamiltonian evolution with occasional perturbations by the collisions. This, for example, is what happens for the dilute gas. And this is the regime we will work in from now on.
At this stage, there is a right way and a less-right way to proceed. The right way is to derive the Boltzmann equation starting from the BBGKY hierarchy. And we will do this in Section 2.2.3. However, as we shall see, it’s a little fiddly. So instead we’ll start by taking the less-right option which has the advantage of getting the same answer but in a much easier fashion. This option is to simply guess what form the Boltzmann equation has to take.
2.2.1 Motivating the Boltzmann Equation We’ve already caught our first glimpse of the Boltzmann equation in (2.12), @f1 @t = {H1, f1} + ✓@f1 @t ◆ coll (2.16) But, of course, we don’t yet have an expression for the collision integral in terms of f1. It’s clear from the definition (2.13) that the second term represents the change in – 20 – momenta due to two-particle scattering. When ⌧≫⌧coll, the collisions occur occasion-ally, but abruptly. The collision integral should reflect the rate at which these collisions occur.
Suppose that our particle sits at (~ r, ~ p) in phase space and collides with another particle at (~ r, ~ p2). Note that we’re assuming here that collisions are local in space so that the two particles sit at the same point. These particles can collide and emerge with momenta ~ p 0 1 and ~ p 0 2. We’ll define the rate for this process to occur to be Rate = !(~ p, ~ p2|~ p 0 1, ~ p 0 2) f2(~ r,~ r, ~ p, ~ p2) d3p2d3p0 1d3p0 2 (2.17) (Here we’ve dropped the explicit t dependence of f2 only to keep the notation down).
The scattering function ! contains the information about the dynamics of the process.
It looks as if this is a new quantity which we’ve introduced into the game. But, using standard classical mechanics techniques, one can compute ! for a given inter-atomic potential U(~ r). (It is related to the di↵erential cross-section; we will explain how to do this when we do things better in Section 2.2.3). For now, note that the rate is proportional to the two-body distribution function f2 since this tells us the chance that two particles originally sit in (~ r, ~ p) and (~ r, ~ p2).
We’d like to focus on the distribution of particles with some specified momentum ~ p. Two particles with momenta ~ p and ~ p2 can be transformed in two particles with momenta ~ p 0 1 and ~ p 0 2. Since both momenta and energy are conserved in the collision, we have ~ p + ~ p2 = ~ p 0 1 + ~ p 0 2 (2.18) p2 + p2 2 = p0 2 1 + p0 2 2 (2.19) There is actually an assumption that is hiding in these equations. In general, we’re considering particles in an external potential V . This provides a force on the particles which, in principle, could mean that the momentum and kinetic energy of the particles is not the same before and after the collision. To eliminate this possibility, we will assume that the potential only varies appreciably over macroscopic distance scales, so that it can be neglected on the scale of atomic collisions. This, of course, is entirely reasonable for most external potentials such as gravity or electric fields. Then (2.18) and (2.19) continue to hold.
While collisions can deflect particles out of a state with momentum ~ p and into a di↵erent momentum, they can also deflect particles into a state with momentum ~ p.
– 21 – This suggests that the collision integral should contain two terms, ✓@f1 @t ◆ coll = Z d3p2d3p0 1d3p0 2 h !(~ p 0 1, ~ p 0 2|~ p, ~ p2) f2(~ r,~ r, ~ p 0 1, ~ p 0 2) −!(~ p, ~ p2|~ p 0 1, ~ p 0 2)f2(~ r,~ r, ~ p, ~ p2) i The first term captures scattering into the state ~ p, the second scattering out of the state ~ p.
The scattering function obeys a few simple requirements. Firstly, it is only non-vanishing for scattering events that obey the conservation of momentum (2.18) and energy (2.19). Moreover, the discrete symmetries of spacetime also give us some im-portant information. Under time reversal, ~ p ! −~ p and, of course, what was coming in is now going out. This means that any scattering which is invariant under time reversal (which is more or less anything of interest) must obey !(~ p, ~ p2|~ p 0 1, ~ p 0 2) = !(−~ p 0 1, −~ p 0 2| −~ p, −~ p2) Furthermore, under parity (~ r, ~ p) ! (−~ r, −~ p). So any scattering process which is parity invariant further obeys !(~ p, ~ p2|~ p 0 1, ~ p 0 2) = !(−~ p, −~ p2| −~ p 0 1, −~ p 0 2) The combination of these two means that the scattering rate is invariant under exchange of ingoing and outgoing momenta, !(~ p, ~ p2|~ p 0 1, ~ p 0 2) = !(~ p 0 1, ~ p 0 2|~ p, ~ p2) (2.20) (There is actually a further assumption of translational invariance here, since the scat-tering rate at position −~ r should be equivalent to the scattering rate at position +~ r).
The symmetry property (2.20) allows us to simplify the collision integral to ✓@f1 @t ◆ coll = Z d3p2d3p0 1d3p0 2 !(~ p 0 1, ~ p 0 2|~ p, ~ p2) h f2(~ r,~ r, ~ p 0 1, ~ p 0 2) −f2(~ r,~ r, ~ p, ~ p2) i (2.21) To finish the derivation, we need to face up to our main goal of expressing the collision integral in terms of f1 rather than f2. We make the assumption that the velocities of two particles are uncorrelated, so that we can write f2(~ r,~ r, ~ p, ~ p2) = f1(~ r, ~ p)f1(~ r, ~ p2) (2.22) This assumption, which sometimes goes by the name of molecular chaos, seems innocu-ous enough. But actually it is far from innocent! To see why, let’s look more closely – 22 – at what we’ve actually assumed. Looking at (2.21), we can see that we have taken the rate of collisions to be proportional to f2(~ r,~ r, ~ p1, ~ p2) where p1 and p2 are the momenta of the particles before the collision. That means that if we substitute (2.22) into (2.21), we are really assuming that the velocities are uncorrelated before the collision. And that sounds quite reasonable: you could imagine that during the collision process, the velocities between two particles become correlated. But there is then a long time, ⌧, before one of these particles undergoes another collision. Moreover, this next collision is typically with a completely di↵erent particle and it seems entirely plausible that the velocity of this new particle has nothing to do with the velocity of the first. Nonethe-less, the fact that we’ve assumed that velocities are uncorrelated before the collision rather than after has, rather slyly, introduced an arrow of time into the game. And this has dramatic implications which we will see in Section 2.3 where we derive the H-theorem.
Finally, we may write down a closed expression for the evolution of the one-particle distribution function given by @f1 @t = {H1, f1} + ✓@f1 @t ◆ coll (2.23) with the collision integral ✓@f1 @t ◆ coll = Z d3p2d3p0 1d3p0 2 !(~ p 0 1, ~ p 0 2|~ p, ~ p2) h f1(~ r, ~ p 0 1)f1(~ r, ~ p 0 2) −f1(~ r, ~ p)f1(~ r, ~ p2) i (2.24) This is the Boltzmann equation. It’s not an easy equation to solve! It’s a di↵erential equation on the left, an integral on the right, and non-linear. You may not be surprised to hear that exact solutions are not that easy to come by. We’ll see what we can do.
2.2.2 Equilibrium and Detailed Balance Let’s start our exploration of the Boltzmann equation by revisiting the question of the equilibrium distribution obeying @f eq/@t = 0. We already know that {f, H1} = 0 if f is given by any function of the energy or, indeed any function that Poisson commutes with H. For clarity, let’s restrict to the case with vanishing external force, so V (r) = 0.
Then, if we look at the Liouville equation alone, any function of momentum is an equilibrium distribution. But what about the contribution from the collision integral?
One obvious way to make the collision integral vanish is to find a distribution which obeys the detailed balance condition, f eq 1 (~ r, ~ p 0 1)f eq 1 (~ r, ~ p 0 2) = f eq 1 (~ r, ~ p)f eq 1 (~ r, ~ p2) (2.25) – 23 – In fact, it’s more useful to write this as log(f eq 1 (~ r, ~ p 0 1)) + log(f eq 1 (~ r, ~ p 0 2)) = log(f eq 1 (~ r, ~ p)) + log(f eq 1 (~ r, ~ p2)) (2.26) How can we ensure that this is true for all momenta? The momenta on the right are those before the collision; on the left they are those after the collision. From the form of (2.26), it’s clear that the sum of log f eq 1 must be the same before and after the collision: in other words, this sum must be conserved during the collision. But we know what things are conserved during collisions: momentum and energy as shown in (2.18) and (2.19) respectively. This means that we should take log(f eq 1 (~ r, ~ p)) = β (µ −E(~ p) + ~ u · ~ p ) (2.27) where E(p) = p2/2m for non-relativistic particles and µ, β and ~ u are all constants.
We’ll adjust the constant µ to ensure that the overall normalization of f1 obeys (2.6).
Then, writing ~ p = m~ v, we have f eq 1 (~ r, ~ p) = N V ✓β 2⇡m ◆3/2 e−βm(~ v−~ u)2/2 (2.28) which reproduces the Maxwell-Boltzmann distribution if we identify β with the inverse temperature. Here ~ u allows for the possibility of an overall drift velocity. We learn that the addition of the collision term to the Liouville equation forces us to sit in the Boltzmann distribution at equilibrium.
There is a comment to make here that will play an important role in Section 2.4.
If we forget about the streaming term {H1, f1} then there is a much larger class of solutions to the requirement of detailed balance (2.25). These solutions are again of the form (2.27), but now with the constants µ, β and ~ u promoted to functions of space and time. In other words, we can have f local 1 (~ r, ~ p; t) = n(~ r, t) ✓β(~ r, t) 2⇡m ◆3/2 exp ⇣ −β(~ r, t)m 2 [(~ v −~ u(~ r, t)]2⌘ (2.29) Such a distribution is not quite an equilibrium distribution, for while the collision integral in (2.23) vanishes, the streaming term does not. Nonetheless, distributions of this kind will prove to be important in what follows. They are said to be in local equilibrium, with the particle density, temperature and drift velocity varying over space.
The Quantum Boltzmann Equation Our discussion above was entirely for classical particles and this will continue to be our focus for the remainder of this section. However, as a small aside let’s look at how – 24 – things change for quantum particles. We’ll keep the assumption of molecular chaos, so f2 ⇠f1f1 as in (2.22). The main di↵erence occurs in the scattering rate (2.17) for scattering ~ p1 + ~ p2 ! ~ p 0 1 + ~ p 0 2 which now becomes Rate = !(~ p, ~ p2|~ p 0 1, ~ p 0 2) f1(~ p1)f1(~ p2){ 1 ± f1(~ p 0 1)} {1 ± f(~ p 0 2)} d3p2d3p0 1d3p0 2 The extra terms are in curly brackets. We pick the + sign for bosons and the −sign for fermions. The interpretation is particularly clear for fermions, where the number of particles in a given state can’t exceed one. Now it’s not enough to know the probability that initial state is filled. We also need to know that probability that the final state is free for the particle to scatter into: and that’s what the {1 −f1} factors are telling us.
The remaining arguments go forward as before, resulting in the quantum Boltzmann equation ✓@f1 @t ◆ coll = Z d3p2d3p0 1d3p0 2 !(~ p 0 1, ~ p 0 2|~ p, ~ p2) h f1(~ p 0 1)f1(~ p 0 2){1 ± f1(~ p)} {1 ± f1(~ p2} −f1(~ p)f1(~ p2) {1 ± f1(~ p 0 1)} {1 ± f1(~ r, ~ p 0 2)} i To make contact with what we know, we can look again at the requirement for equi-librium. The condition of detailed balance now becomes log ✓ f eq 1 (~ p 0 1) 1 ± f eq 1 (~ p 0 1) ◆ + log ✓ f eq 1 (~ p 0 2) 1 ± f eq 1 (~ p 0 2) ◆ = log ✓ f eq 1 (~ p) 1 ± f eq 1 (~ p) ◆ + log ✓ f eq 1 (~ p2) 1 ± f eq 1 (~ p2) ◆ Which is again solved by relating each log to a linear combination of the energy and momentum. We find f eq 1 (~ p) = 1 e−β(µ−E(~ p)+~ u·~ p) ⌥1 which reproduces the Bose-Einstein and Fermi-Dirac distributions.
2.2.3 A Better Derivation In Section 2.2.1, we derived an expression for the collision integral (2.24) using intuition for the scattering processes at play. But, of course, we have a mathematical expression for the collision integral in (2.13) involving the two-particle distribution function f2. In this section we will sketch how one can derive (2.24) from (2.13). This will help clarify some of the approximations that we need to use. At the same time, we will also review some basic classical mechanics that connects the scattering rate ! to the inter-particle potential U(r).
– 25 – We start by returning to the BBGKY hierarchy of equations. For simplicity, we’ll turn o↵the external potential V (~ r) = 0. We don’t lose very much in doing this because most of the interesting physics is concerned with the scattering of atoms o↵each other.
The first two equations in the hierarchy are ✓@ @t + ~ p1 m · @ @~ r1 ◆ f1 = Z d3r2d3p2 @U(~ r1 −~ r2) @~ r1 · @f2 @~ p1 (2.30) and ✓@ @t + ~ p1 m · @ @~ r1 + ~ p2 m · @ @~ r2 −1 2 @U(~ r1 −~ r2) @~ r1 · @ @~ p1 −@ @~ p2 /◆ f2 = (2.31) Z d3r3d3p3 ✓@U(~ r1 −~ r3) @~ r1 · @ @~ p1 + @U(~ r2 −~ r3) @~ r2 · @ @~ p2 ◆ f3 In both of these equations, we’ve gathered the streaming terms on the left, leaving only the higher distribution function on the right. To keep things clean, we’ve sup-pressed the arguments of the distribution functions: they are f1 = f1(~ r1, ~ p1; t) and f2 = f2(~ r1,~ r2, ~ p1, ~ p2; t) and you can guess the arguments for f3.
Our goal is to better understand the collision integral on the right-hand-side of (2.30).
It seems reasonable to assume that when particles are far-separated, their distribution functions are uncorrelated. Here, “far separated” means that the distance between them is much farther than the atomic distance scale d over which the potential U(r) extends. We expect f2(~ r1,~ r2, ~ p1, ~ p2; t) ! f1(~ r1, ~ p1; t)f1(~ r2, ~ p2; t) when |~ r1 −~ r1| ≫d But, a glance at the right-hand-side of (2.30) tells us that this isn’t the regime of interest. Instead, f2 is integrated @U(r)/@r which varies significantly only over a region r d. This means that we need to understand f2 when two particles get close to each other.
We’ll start by getting a feel for the order of magnitude of various terms in the hierarchy of equations. Dimensionally, each term in brackets in (2.30) and (2.31) is an inverse time scale. The terms involving the inter-atomic potential U(r) are associated to the collision time ⌧coll.
1 ⌧coll ⇠@U @~ r · @ @~ p This is the time taken for a particle to cross the distance over which the potential U(r) varies which, for short range potentials, is comparable to the atomic distance scale, d, – 26 – itself and ⌧coll ⇠d ¯ vrel where ¯ vrel is the average relative speed between atoms. Our first approximation will be that this is the shortest time scale in the problem. This means that the terms involving @U/@r are typically the largest terms in the equations above and determine how fast the distribution functions change.
With this in mind, we note that the equation for f1 is special because it is the only one which does not include any collision terms on the left of the equation (i.e.
in the Hamiltonian Hn). This means that the collision integral on the right-hand side of (2.30) will usually dominate the rate of change of f1. (Note, however, we’ll meet some important exceptions to this statement in Section 2.4). In contrast, the equation that governs f2 has collision terms on both the left and the right-hand sides. But, importantly, for dilute gases, the term on the right is much smaller than the term on the left. To see why this is, we need to compare the f3 term to the f2 term. If we were to integrate f3 over all space, we get Z d3r2d3p3 f3 = Nf2 (where we’ve replaced (N −2) ⇡N in the above expression). However, the right-hand side of (2.31) is not integrated over all of space. Instead, it picks up a non-zero contribution over an atomic scale ⇠d3. This means that the collision term on the right-hand-side of (2.31) is suppressed compared to the one on the left by a factor of Nd3/V where V is the volume of space. For gases that we live and breath every day, Nd3/V ⇠10−3 −10−4. We make use of this small number to truncate the hierarchy of equations and replace (2.31) with ✓@ @t + ~ p1 m · @ @~ r1 + ~ p2 m · @ @~ r2 −1 2 @U(~ r1 −~ r2) @~ r1 · @ @~ p1 −@ @~ p2 /◆ f2 ⇡0 (2.32) This tells us that f2 typically varies on a time scale of ⌧coll and a length scale of d.
Meanwhile, the variations of f1 is governed by the right-hand-side of (2.30) which, by the same arguments that we just made, are smaller than the variations of f2 by a factor of Nd3/V . In other words, f1 varies on the larger time scale ⌧.
In fact, we can be a little more careful when we say that f2 varies on a time scale ⌧coll. We see that – as we would expect – only the relative position is a↵ected by the – 27 – collision term. For this reason, it’s useful to change coordinate to the centre of mass and the relative positions of the two particles. We write ~ R = 1 2(~ r1 + ~ r2) , ~ r = ~ r1 −~ r2 and similar for the momentum ~ P = ~ p1 + ~ p2 , ~ p = 1 2(~ p1 −~ p2) And we can think of f2 = f2(~ R,~ r, ~ P, ~ p; t). The distribution function will depend on the centre of mass variables ~ R and ~ P in some slow fashion, much as f1 depends on position and momentum. In contrast, the dependence of f2 on the relative coordinates ~ r and ~ p is much faster – these vary over the short distance scale and can change on a time scale of order ⌧coll.
Since the relative distributions in f2 vary much more quickly that f1, we’ll assume that f2 reaches equilibrium and then feeds into the dynamics of f1. This means that, ignoring the slow variations in ~ R and ~ P, we will assume that @f2/@t = 0 and replace (2.32) with the equilibrium condition ✓~ p m · @ @~ r −@U(~ r) @~ r · @ @~ p ◆ f2 ⇡0 (2.33) This is now in a form that allows us to start manipulating the collision integral on the right-hand-side of (2.30). We have ✓@f1 @t ◆ coll = Z d3r2d3p2 @U(~ r1 −~ r2) @~ r1 · @f2 @~ p1 = Z d3r2d3p2 @U(~ r) @~ r · @ @~ p1 −@ @~ p2 / f2 = 1 m Z |~ r1−~ r2|d d3r2d3p2 (~ p1 −~ p2) · @f2 @~ r (2.34) where in the second line the extra term @/@~ p2 vanishes if we integrate by parts and, in the third line, we’ve used our equilibrium condition (2.33), with the limits on the integral in place to remind us that only the region r d contributes to the collision integral.
A Review of Scattering Cross Sections To complete the story, we still need to turn (2.34) into the collision integral (2.24).
But most of the work simply involves clarifying how the scattering rate !(~ p, ~ p2|~ p 0 1, ~ p 0 2) is defined for a given inter-atomic potential U(~ r1 −~ r2). And, for this, we need to review the concept of the di↵erential cross section.
– 28 – b b δ dσ dΩ θ φ Figure 4: The di↵erential cross section.
Let’s think about the collision between two particles.
They start with momenta ~ pi = m~ vi and end with momenta ~ p 0 i = m~ v 0 i with i = 1, 2. Now let’s pick a favourite, say particle 1. We’ll sit in its rest frame and consider an onslaught of bombarding particles, each with velocity ~ v2 −~ v1. This beam of incoming particles do not all hit our favourite boy at the same point. Instead, they come in randomly distributed over the plane perpendicular to ~ v2 −~ v1. The flux, I, of these incoming particles is the number hitting this plane per area per second, I = N V |~ v2 −~ v1| Now spend some time staring at Figure 4. There are a number of quantities defined in this picture. First, the impact parameter, b, is the distance from the asymptotic trajectory to the dotted, centre line.
We will use b and φ as polar coordinates to parameterize the plane perpendicular to the incoming particle. Next, the scattering angle, ✓, is the angle by which the incoming particle is deflected. Finally, there are two solid angles, dσ and d⌦, depicted in the figure. Geometrically, we see that they are given by dσ = bdbdφ and d⌦= sin ✓d✓dφ The number of particles scattered into d⌦in unit time is Idσ. We usually write this as I dσ d⌦d⌦= Ib db dφ (2.35) – 29 – Figure 5: On the left: a point particle scattering o↵a hard sphere. On the right: a hard sphere scattering o↵a hard sphere.
where the di↵erential cross section is defined as 0 0 0 0 dσ d⌦ 0 0 0 0 = b sin ✓ 0 0 0 0 db d✓ 0 0 0 0 = 1 2 0 0 0 0 d(b2) d cos ✓ 0 0 0 0 (2.36) You should think of this in the following way: for a fixed (~ v2 −~ v1), there is a unique relationship between the impact parameter b and the scattering angle ✓and, for a given potential U(r), you need to figure this out to get |dσ/d⌦| as a function of ✓.
Now we can compare this to the notation that we used earlier in (2.17). There we talked about the rate of scattering into a small area d3p0 1d3p0 2 in momentum space. But this is the same thing as the di↵erential cross-section.
!(~ p, ~ p2; ~ p 0 1, ~ p 0 2) d3p0 1d3p0 2 = |~ v −~ v2| 0 0 0 0 dσ d⌦ 0 0 0 0 d⌦ (2.37) (Note, if you’re worried about the fact that d3p0 1d3p0 2 is a six-dimensional area while d⌦is a two dimensional area, recall that conservation of energy and momenta provide four restrictions on the ability of particles to scatter. These are implicit on the left, but explicit on the right).
An Example: Hard Spheres In Section 1.2, we modelled atoms as hard spheres of diameter d. It’s instructive to figure out the cross-section for such a hard sphere.
In fact, there are two di↵erent calculations that we can do. First, suppose that we throw point-like particles at a sphere of diameter d with an impact parameter b d/2 From the left-hand diagram in Figure 5, we see that the scattering angle is ✓= ⇡−2↵, where b = d 2 sin ↵= d 2 sin ✓⇡ 2 −✓ 2 ◆ = d 2 cos ✓ 2 – 30 – or b2 = d2 4 cos2 ✓ 2 = d2 8 (1 + cos ✓) From (2.36), we then find the di↵erential cross-section 0 0 0 0 dσ d⌦ 0 0 0 0 = d2 16 The total cross-section is defined as σT = 2⇡ Z ⇡ 0 d✓sin ✓dσ d⌦= ⇡ ✓d 2 ◆2 This provides a nice justification for the name because this is indeed the cross-sectional area of a sphere of radius d/2.
Alternatively, we could consider two identical hard spheres, each of diameter d, one scattering o↵the other. Now the geometry changes a little, as shown in the right-hand diagram in Figure 5. The impact parameter is now the distance between the centres of the spheres, and given by b = 2 ⇥d 2 sin ↵ Clearly we now need b d. The same calculation as above now gives σT = ⇡d2 This is the same e↵ective cross-sectional area that we previously used back in Section 1.2 when discussing basic aspects of collisions.
Almost Done With this refresher course on classical scattering, we can return to the collision integral (2.34) in the Boltzmann equation.
✓@f1 @t ◆ coll = Z |~ r1−~ r2|d d3r2d3p2 (~ v1 −~ v2) · @f2 @~ r We’ll work in cylindrical polar coordinates shown in Figure 6. The direction parallel to ~ v2 −~ v1 is parameterized by x; the plane perpendicular is parameterised by φ and – 31 – v −v 1 2 v −v 1 2 ’ ’ b b d x x 1 2 φ Figure 6: Two particle scattering the impact parameter b. We’ve also shown the collision zone in this figure. Using the definitions (2.35) and (2.37), we have ✓@f1 @t ◆ coll = Z d3p2 |~ v1 −~ v2| Z dφ db b Z x2 x1 @f2 @x = Z d3p2d3p0 1d3p0 2 !(~ p 0 1, ~ p 0 2|~ p, ~ p2) [f2(x2) −f2(x1)] It remains only to decide what form the two-particle distribution function f2 takes just before the collision at x = x1 and just after the collision at x = x2. At this point we invoke the assumption of molecular chaos. Just before we enter the collision, we assume that the two particles are uncorrelated. Moreover, we assume that the two particles are once again uncorrelated by the time they leave the collision, albeit now with their new momenta f2(x1) = f1(~ r, ~ p1; t)f1(~ r, ~ p2; t) and f2(x2) = f1(~ r, ~ p 0 1; t)f1(~ r, ~ p 0 2; t) Notice that all functions f1 are evaluated at the same point ~ r in space since we’ve assumed that the single particle distribution function is suitably coarse grained that it doesn’t vary on scales of order d. With this final assumption, we get what we wanted: the collision integral is given by ✓@f1 @t ◆ coll = Z d3p2d3p0 1d3p0 2 !(~ p 0 1, ~ p 0 2|~ p, ~ p2) h f1(~ r, ~ p 0 1)f1(~ r, ~ p 0 2) −f1(~ r, ~ p)f1(~ r, ~ p2) i in agreement with (2.24).
– 32 – 2.3 The H-Theorem The topics of thermodynamics and statistical mechanics are all to do with the equi-librium properties of systems.
One of the key intuitive ideas that underpins their importance is that if you wait long enough, any system will eventually settle down to equilibrium. But how do we know this? Moreover, it seems that it would be rather tricky to prove: settling down to equilibrium clearly involves an arrow of time that dis-tinguishes the future from the past. Yet the underlying classical mechanics is invariant under time reversal.
The purpose of this section is to demonstrate that, within the framework of the Boltzmann equation, systems do indeed settle down to equilibrium. As we described above, we have introduced an arrow of time into the Boltzmann equation. We didn’t do this in any crude way like adding friction to the system. Instead, we merely assumed that particle velocities were uncorrelated before collisions. That would seem to be a rather minor input but, as we will now show, it’s enough to demonstrate the approach to equilibrium.
Specifically, we will prove the “H-theorem”, named after a quantity H introduced by Boltzmann. (H is not to be confused with the Hamiltonian. Boltzmann originally called this quantity something like a German E, but the letter was somehow lost in translation and the name H stuck). This quantity is H(t) = Z d3rd3p f1(~ r, ~ p; t) log(f1(~ r, ~ p; t)) This kind of expression is familiar from our first Statistical Mechanics course where we saw that the entropy S for a probability distribution p is S = −kBp log p. In other words, this quantity H is simply S = −kBH The H-theorem, first proven by Boltzmann in 1872, is the statement that H always decreases with time. The entropy always increases. We will now prove this.
As in the derivation (2.4), when you’re looking at the variation of expectation values you only care about the explicit time dependence, meaning dH dt = Z d3rd3p (log f1 + 1)@f1 @t = Z d3rd3p log f1 @f1 @t – 33 – where we can drop the +1 because R f1 = N is unchanging, ensuring that R @f1/@t = 0.
Using the Boltzmann equation (2.23), we have dH dt = Z d3rd3p log f1 ✓@V @~ r · @f1 @~ p −~ p m · @f1 @~ r + ✓@f1 @t ◆ coll ◆ But the first two terms in this expression both vanish. You can see this by integrating by parts twice, first moving the derivative away from f1 and onto log f1, and then moving it back. We learn that the change in H is governed entirely by the collision terms dH dt = Z d3rd3p log f1 ✓@f1 @t ◆ coll = Z d3rd3p1d3p2d3p0 1d3p0 2 !(~ p 0 1, ~ p 0 2|~ p1, ~ p2) log f1(~ p1) ⇥ h f1(~ p 0 1)f1(~ p 0 2) −f1(~ p1)f1(~ p2) i (2.38) where I’ve suppressed ~ r and t arguments of f1 to keep things looking vaguely reasonable I’ve also relabelled the integration variable ~ p ! ~ p1. At this stage, all momenta are integrated over so they are really nothing but dummy variables. Let’s relabel 1 $ 2 on the momenta. All the terms remain unchanged except the log. So we can also write dH dt = Z d3rd3p1d3p2d3p0 1d3p0 2 !(~ p 0 1, ~ p 0 2|~ p1, ~ p2) log f1(~ p2) ⇥ h f1(~ p 0 1)f1(~ p 0 2) −f1(~ p1)f1(~ p2) i (2.39) Adding (2.38) and (2.39), we have the more symmetric looking expression dH dt = 1 2 Z d3rd3p1d3p2d3p0 1d3p0 2 !(~ p 0 1, ~ p 0 2|~ p1, ~ p2) log [f1(~ p1) f1(~ p2)] ⇥ h f1(~ p 0 1)f1(~ p 0 2) −f1(~ p1)f1(~ p2) i (2.40) Since all momenta are integrated over, we’re allowed to just flip the dummy indices again. This time we swap ~ p $ ~ p 0 in the above expression. But, using the symmetry property (2.20), the scattering function remains unchanged3. We get dH dt = −1 2 Z d3rd3p1d3p2d3p0 1d3p0 2 !(~ p 0 1, ~ p 0 2|~ p1, ~ p2) log [f1(~ p 0 1) f1(~ p 0 2)] 3An aside: it’s not actually necessary to assume (2.20) to make this step. We can get away with the weaker result Z d3p0 1d3p0 2 !(~ p 0 1, ~ p 0 2|~ p1, ~ p2) = Z d3p0 1d3p0 2 !(~ p1, ~ p2|~ p 0 1, ~ p 0 2) which follows from unitarity of the scattering matrix.
– 34 – ⇥ h f1(~ p 0 1)f1(~ p 0 2) −f1(~ p1)f1(~ p2) i (2.41) Finally, we add (2.40) and (2.41) to get dH dt = −1 4 Z d3rd3p1d3p2d3p0 1d3p0 2 !(~ p 0 1, ~ p 0 2|~ p1, ~ p2) ⇥ h log [f1(~ p 0 1) f1(~ p 0 2)] −log [f1(~ p1) f1(~ p2)] ih f1(~ p 0 1)f1(~ p 0 2) −f1(~ p1)f1(~ p2) i (2.42) The bottom line of this expression is a function (log x −log y)(x −y). It is positive for all values of x and y. Since the scattering rate is also positive, we have the proof of the H-theorem.
dH dt 0 , dS dt ≥0 And there we see the arrow of time seemingly emerging from time-invariant Hamiltonian mechanics! Clearly, this should be impossible, a point first made by Loschmidt soon after Boltzmann’s original derivation. But, as we saw earlier, everything hinges on the assumption of molecular chaos (2.22). This was where we broke time-reversal symmetry, ultimately ensuring that entropy increases only in the future. Had we instead decided in (2.21) that the rate of scattering was proportional to f2 after the collision, again assuming f2 ⇠f1f1 then we would find that entropy always decreases as we move into the future.
There is much discussion in the literature about the importance of the H-theorem and its relationship to the second law of thermodynamics. Notably, it is not particularly hard to construct states which violate the H-theorem by virtue of their failure to obey the assumption of molecular chaos. Nonetheless, these states still obey a suitable second law of thermodynamics4.
The H-theorem is not a strict inequality. For some distributions, the entropy remains unchanged. From (2.42), we see that these obey f1(~ p 0 1)f1(~ p 0 2) −f1(~ p1)f1(~ p2) But this is simply the requirement of detailed balance (2.25). And, as we have seen al-ready, this is obeyed by any distribution satisfying the requirement of local equilibrium (2.29).
4This was first pointed out by E. T. Jaynes in the paper “Violation of Boltzmann’s H Theorem in Real Gases”, published in Physical Review A, volume 4, number 2 (1971).
– 35 – 2.4 A First Look at Hydrodynamics Hydrodynamics is what you get if you take thermodynamics and splash it. You know from your first course on Statistical Mechanics that, at the most coarse grained level, the equilibrium properties of any system are governed by the thermodynamics. In the same manner, low energy, long wavelength, excitations of any system are described by hydrodynamics.
More precisely, hydrodynamics describes the dynamics of systems that are in local equilibrium, with parameters that vary slowly in space in time. As we will see, this means that the relevant dynamical variables are, in the simplest cases, • Density ⇢(~ r, t) = m n(~ r, t) • Temperature T(~ r, t) • Velocity ~ u(~ r, t) Our goal in this section is to understand why these are the relevant variables to describe the system and to derive the equations that govern their dynamics.
2.4.1 Conserved Quantities We’ll start by answering the first question: why are these the variables of interest? The answer is that these are quantities which don’t relax back down to their equilibrium value in an atomic blink of an eye, but instead change on a much slower, domestic time scale. At heart, the reason for they have this property is that they are all associated to conserved quantities. Let’s see why.
Consider a general function A(~ r, ~ p) over the single particle phase space. Because we live in real space instead of momentum space, the question of how things vary with ~ r is more immediately interesting. For this reason, we integrate over momentum and define the average of a quantity A(~ r, ~ p) to be hA(~ r, t)i = R d3p A(~ r, ~ p)f1(~ r, ~ p; t) R d3p f1(~ r, ~ p; t) However, we’ve already got a name for the denominator in this expression: it is the number density of particles n(~ r, t) = Z d3p f1(~ r, ~ p; t) (2.43) – 36 – (As a check of the consistency of our notation, if you plug the local equilibrium dis-tribution (2.29) into this expression, then the n(~ r, t) on the left-hand-side equals the n(~ r, t) defined in (2.29)). So the average is hA(~ r, t)i = 1 n(~ r, t) Z d3p A(~ r, ~ p)f1(~ r, ~ p; t) (2.44) It’s worth making a couple of simple remarks. Firstly, this is di↵erent from the average that we defined earlier in (2.3) when discussing Liouville evolution. Here we’re inte-grating only over momenta and the resulting average is a function of space. A related point is that we’re at liberty to take functions which depend only on ~ r (and not on ~ p) in and out of the h·i brackets. So, for example, hnAi = nhAi.
We’re interested in how the average of A changes with time. We looked at this kind of question for Liouville evolution earlier in this section and found the answer (2.5).
Now we want to ask the same question for the Boltzmann equation. Before we actually write down the answer, you can guess what it will look like: there will be a streaming term and a term due to the collision integral. Moreover, we know from our previous discussion that the term involving the collision integral will vary much faster than the streaming term.
Since we’re ultimately interested in quantities which vary slowly, this motivates look-ing at functions A which vanish when integrated against the collision integral. We will see shortly that the relevant criterion is Z d3p A(~ r, ~ p) ✓@f1 @t ◆ coll = 0 We’d like to find quantities A which have this property for any distribution f1. Using our expression for the collision integral (2.23), we want Z d3p1d3p2d3p0 1d3p0 2 !(~ p 0 1, ~ p 0 2|~ p, ~ p2) A(~ r, ~ p1) h f1(~ r, ~ p 0 1)f1(~ r, ~ p 0 2) −f1(~ r, ~ p)f1(~ r, ~ p2) i = 0 This now looks rather similar to equation (2.38), just with the log f replaced by A. In-deed, we can follow the steps between (2.38) and (2.41), using the symmetry properties of !, to massage this into the form Z d3p1d3p2d3p0 1d3p0 2 !(~ p 0 1, ~ p 0 2|~ p1, ~ p2) h f1(~ p 0 1)f1(~ p 0 2) −f1(~ p1)f1(~ p2) i ⇥ h A(~ r, ~ p1) + A(~ r, ~ p2) −A(~ r, ~ p 0 1) −A(~ r, ~ p 0 2) i = 0 – 37 – Now it’s clear that if we want this to vanish for all distributions, then A itself must have the property that it remains unchanged before and after the collision, A(~ r, ~ p1) + A(~ r, ~ p2) = A(~ r, ~ p 0 1) + A(~ r, ~ p 0 2) (2.45) Quantities which obey this are sometimes called collisional invariants. Of course, in the simplest situation we already know what they are: momentum (2.18) and energy (2.19) and, not forgetting, the trivial solution A = 1. We’ll turn to each of these in turn shortly. But first let’s derive an expression for the time evolution of any quantity obeying (2.45).
Take the Boltzmann equation (2.23), multiply by a collisional invariant A(~ r, ~ p) and integrate over R d3p. Because the collision term vanishes, we have Z d3p A(~ r, ~ p) ✓@ @t + ~ p m · @ @~ r + ~ F · @ @~ p ◆ f1(~ r, ~ p, t) = 0 where the external force is ~ F = −rV . We’ll integrate the last term by parts (remem-bering that the force ~ F can depend on position but not on momentum). We can’t integrate the middle term by parts since we’re not integrating over space, but nonethe-less, we’ll also rewrite it. Finally, since A has no explicit time dependence, we can take it inside the time derivative. We have @ @t Z d3p Af + @ @~ r · Z d3p ~ p mAf − Z d3p ~ p m · @A @~ r f − Z d3p ~ F · @A @~ p f = 0 Although this doesn’t really look like an improvement, the advantage of writing it in this way is apparent when we remember our expression for the average (2.44). Using this notation, we can write the evolution of A as @ @thnAi + @ @~ r · hn~ vAi −nh~ v · @A @~ r i −nh~ F · @A @~ p i = 0 (2.46) where ~ v = ~ p/m. This is our master equation that tells us how any collisional invariant changes. The next step is to look at specific quantities. There are three and we’ll take each in turn Density Our first collisional invariant is the trivial one: A = 1. If we plug this into (2.46) we get the equation for the particle density n(~ r, t), @n @t + @ @~ r · (n~ u) (2.47) – 38 – where the average velocity ~ u of the particles is defined by ~ u(~ r, t) = h~ vi Notice that, once again, our notation is consistent with earlier definitions: if we pick the local equilibrium distribution (2.29), the ~ u(~ r, t) in (2.29) agrees with that defined above.
The result (2.47) is the continuity equation, expressing the conservation of particle number. Notice, however, that this is not a closed expression for the particle density n: we need to know the velocity ~ u as well.
It’s useful to give a couple of extra, trivial, definitions at this stage. First, although we won’t use this notation, the continuity equation is sometimes written in terms of the current, ~ J(~ r, t) = n(~ r, t) ~ u(~ r, t). In what follows, we will often replace the particle density with the mass density, ⇢(~ r, t) = mn(~ r, t) Momentum Our next collisional invariant is the momentum. We substitute A = m~ v into (2.46) to find @ @t(mnui) + @ @rj hmnvjvii −hnFii = 0 (2.48) We can play around with the middle term a little. We write hvjvii = h(vj −uj)(vi −ui)i + uihvji + ujhvii −iiuj = h(vj −uj)(vi −ui)i + uiuj We define a new object known as the pressure tensor, Pij = Pji = ⇢h(vj −uj)(vi −ui)i This tensor is computing the flux of i-momentum in the j-direction. It’s worth pausing to see why this is related to pressure. Clearly, the exact form of Pij depends on the distribution of particles. But, we can evaluate the pressure tensor on the equilibrium, Maxwell-Boltzmann distribution (2.28). The calculation boils down to the same one we did in our first Statistical Physics course to compute equipartition: you find Pij = nkBTδij (2.49) – 39 – which, by the ideal gas law, is proportional to the pressure of the gas.
Using this definition – together with the continuity equation (2.47) – we can write (2.48) as ⇢ ✓@ @t + uj @ @rj ◆ ui = ⇢ mFi −@ @rj Pij (2.50) This is the equation which captures momentum conservation in our system. Indeed, it has a simple interpretation in terms of Newton’s second law. The left-hand-side is the acceleration of an element of fluid. The combination of derivatives is sometimes called the material derivative, Dt ⌘@ @t + uj @ @rj (2.51) It captures the rate of change of a quantity as seen by an observer swept along the streamline of the fluid. The right-hand side of (2.50) includes both the external force ~ F and an additional term involving the internal pressure of the fluid. As we will see later, ultimately viscous terms will also come from here.
Note that, once again, the equation (2.50) does not provide a closed equation for the velocity ~ u. You now need to know the pressure tensor Pij which depends on the particular distribution.
Kinetic Energy Our final collisional invariant is the kinetic energy of the particles. However, rather than take the absolute kinetic energy, it is slightly easier if we work with the relative kinetic energy, A = 1 2m (~ v −~ u)2 If we substitute this into the master equation5 (2.46), the term involving the force vanishes (because hvi −uii = 0). However, the term that involves @E/@ri is not zero because the average velocity ~ u depends on ~ r. We have 1 2 @ @th⇢(~ v −~ u)2i + 1 2 @ @ri h⇢vi(~ v −~ u)2i −⇢hvi @uj @ri (vj −uj)i = 0 (2.52) 5There is actually a subtlety here. In deriving the master equation (2.46), we assumed that A has no explicit time dependence, but the A defined above does have explicit time dependence through ~ u(~ r, t). Nonetheless, you can check that (2.46) still holds, essentially because the extra term that you get is ⇠h(~ v −~ u) · @~ u/@ti = h~ v −~ ui · @~ u/@t = 0.
– 40 – At this point, we define the temperature, T(~ r, t) of our non-equilibrium system. To do so, we fall back on the idea of equipartition and write 3 2kBT(~ r, t) = 1 2mh(~ v −~ u(~ r, t))2i (2.53) This coincides with our familiar definition of temperature for a system in local equilib-rium (2.29), but now extends this to a system that is out of equilibrium. Note that the temperature is a close relative of the pressure tensor, TrP = 3⇢kBT/m.
We also define a new quantity, the heat flux, qi = 1 2m⇢h(vi −ui) (~ v −~ u)2i (2.54) (This actually di↵ers by an overall factor of m from the definition of ~ q that we made in Section 1. This has the advantage of making the formulae we’re about to derive a little cleaner). The utility of both of these definitions becomes apparent if we play around with the middle term in (2.52). We can write 1 2m⇢hvi(~ v −~ u)2i = 1 2m⇢h(vi −ui) (~ v −~ u)2i + 1 2m⇢uih(~ v −~ u)2i = qi + 3 2⇢uikBT Invoking the definition of the pressure tensor (2.49), we can now rewrite (2.52) as 3 2 @ @t(⇢kBT) + @ @ri ✓ qi + 3 2⇢uikBT ◆ + mPij @uj @xi = 0 Because Pij = Pji, we can replace @uj/@ri in the last term with the symmetric tensor known as the rate of strain (and I promise this is the last new definition for a while!) Uij = 1 2 ✓@ui @rj + @uj @ri ◆ (2.55) Finally, with a little help from the continuity equation (2.47), our expression for the conservation of energy becomes ⇢ ✓@ @t + ui @ @ri ◆ kBT + 2 3 @qi @ri + 2m 3 UijPij = 0 (2.56) It’s been a bit of a slog, but finally we have three equations describing how the particle density n (2.47), the velocity ~ u (2.50) and the temperature T (2.56) change with time.
It’s worth stressing that these equations hold for any distribution f1. However, the – 41 – set of equations are not closed. The equation for n depends on ~ u; the equation for ~ u depends on Pij and the equation for T (which is related to the trace of Pij) depends on a new quantity ~ q. And to determine any of these, we need to solve the Boltzmann equation and compute the distribution f1. But the Boltzmann equation is hard! How to do this?
2.4.2 Ideal Fluids We start by simply guessing a form of the distribution function f1(~ r, ~ p; t). We know that the collision term in the Boltzmann equation induces a fast relaxation to equilibrium, so if we’re looking for a slowly varying solution a good guess is to take a distribution for which (@f1/@t)coll = 0. But we’ve already met distribution functions that obey this condition in (2.29): they are those describing local equilibrium. Therefore, our first guess for the distribution, which we write as f (0) 1 , is local equilibrium f (0) 1 (~ r, ~ p; t) = n(~ r, t) ✓ 1 2⇡mkBT(~ r, t) ◆3/2 exp ✓ − m 2kBT(~ r, t) [(~ v −~ u(~ r, t)]2 ◆ (2.57) where ~ p = m~ v. In general, this distribution is not a solution to the Boltzmann equation since it does not vanish on the streaming terms. Nonetheless, we will take it as our first approximation to the true solution and later see what we’re missing.
The distribution is normalized so that the number density and temperature defined in (2.43) and (2.53) respectively coincide with n(~ r, t) and T(~ r, t) in (2.29). But we can also use the distribution to compute Pij and ~ q. We have Pij = kBn(~ r, t)T(~ r, t) δij ⌘P(~ r, t) δij (2.58) and ~ q = 0. We can substitute these expressions into our three conservation laws. The continuity equation (2.47) remains unchanged. Written in terms for ⇢= mn, it reads ✓@ @t + uj @ @rj ◆ ⇢+ ⇢@ui @ri = 0 (2.59) Meanwhile, the equation (2.50) governing the velocity flow becomes the Euler equation describing fluid motion ✓@ @t + uj @ @rj ◆ ui + 1 ⇢ @P @ri = Fi m (2.60) and the final equation (2.56) describing the flow of heat reduces to ✓@ @t + uj @ @rj ◆ T + 2T 3 @ui @ri = 0 (2.61) – 42 – These set of equations describe the motion of an ideal fluid. While they are a good starting point for describing many properties of fluid mechanics, there is one thing that they are missing: dissipation. There is no irreversibility sown into these equations, no mechanism for the fluid to return to equilibrium.
We may have anticipated that these equations lack dissipation. Their starting point was the local equilibrium distribution (2.57) and we saw earlier that for such distribu-tions Boltzmann’s H-function does not decrease; the entropy does not increase. In fact, we can also show this statement directly from the equations above. We can combine (2.59) and (2.60) to find ✓@ @t + uj @ @rj ◆ (⇢T −3/2) = 0 which tells us that the quantity ⇢T −3/2 is constant along streamlines. But this is the requirement that motion along streamlines is adiabatic, not increasing the entropy. To see that this is the case, you need to go back to your earlier statistical mechanics or thermodynamics course6. The usual statement is that for an ideal gas, an adiabatic transformation leaves V T 3/2 constant. Here we’re working with the density ⇢= mN/V and this becomes ⇢T −3/2 is constant. Note, however, that in the present context ⇢and T are not numbers, but functions of space and time: we are now talking about a local adiabatic change.
Sound Waves It is also simple to show explicitly that one can set up motion in the ideal fluid that doesn’t relax back down to equilibrium. We start with a fluid at rest, setting ~ u = 0 and ⇢= ¯ ⇢and T = ¯ T, with both ¯ ⇢and ¯ T constant. We now splash it (gently). That means that we perturb the system and linearise the resulting equations. We’ll analyse these perturbations in Fourier modes and write ⇢(~ r, t) = ¯ ⇢+ δ⇢e−i(!t−~ k·~ r) and T(~ r, t) = ¯ T + δT e−i(!t−~ k·~ r) (2.62) Furthermore, we’ll look for a particular kind of perturbation in which the fluid motion is parallel to the perturbation. In other words, we’re looking for a longitudinal wave ~ u(~ r, t) = ˆ ~ k δu e−i(!t−~ k·~ r) (2.63) The linearised versions of (2.59), (2.60) and (2.61) then read !
|~ k| δ⇢= ¯ ⇢δu 6See, for example, the discussion of the Carnot cycle in the lectures on Statistical Physics.
– 43 – !
|~ k| δu = kB ¯ T m¯ ⇢δ⇢+ kB m δT !
|~ k| δT = 2 3 ¯ Tδu There is one solution to these equations with zero frequency, ! = 0. These have δu = 0 while δ⇢= −¯ ⇢and δT = ¯ T. (Note that this notation hides a small ✏. It really means that δ⇢= −✏¯ ⇢and δT = ✏¯ T. Because the equations are linear and homogeneous, you can take any ✏you like but, since we’re looking at small perturbations, it should be small). This solution has the property that P = mnkBT is constant. But since, in the absence of an external force, pressure is the only driving term in (2.60), the fluid remains at rest, which is why δu = 0 for this solution.
Two further solutions to these equations both have δ⇢= ¯ ⇢, δT = 2 3 ¯ T and δu = !/|~ k| with the dispersion relation ! = ±vs|~ k| with vs = r 5kB ¯ T 3m (2.64) These are sound waves, the propagating version of the adiabatic change that we saw above: the combination ⇢T −3/2 is left unchanged by the compression and expansion of the fluid. The quantity vs is the speed of sound.
2.5 Transport with Collisions While it’s nice to have derived some simple equations describing fluid mechanics, as we’ve seen they’re missing dissipation. And, since the purported goal of these lectures is to understand how systems relax back to equilibrium, we should try to see what we’ve missed.
In fact, it’s clear what we’ve missed. Our first guess for the distribution function was local equilibrium f (0) 1 (~ r, ~ p; t) = n(~ r, t) ✓ 1 2⇡mkBT(~ r, t) ◆3/2 exp ✓ − m 2kBT(~ r, t) [(~ v −~ u(~ r, t)]2 ◆ (2.65) We chose this on the grounds that it gives a vanishing contribution to the collision integral. But we never checked whether it actually solves the streaming terms in the Boltzmann equation. And, as we will now show, it doesn’t.
– 44 – Using the definition of the Poisson bracket and the one-particle Hamiltonian H1 (2.11), we have @f (0) 1 @t −{H1, f (0) 1 } = @f (0) 1 @t + ~ F · @f (0) 1 @~ p + ~ v · @f (0) 1 @~ r Now the dependence on ~ p = m~ v in local equilibrium is easy: it is simply @f (0) 1 @~ p = −1 kBT (~ v −~ u)f (0) 1 Meanwhile all ~ r dependence and t dependence of f (0) 1 lies in the functions n(~ r, t), T(~ r, t) and ~ u(~ r, t). From (2.65) we have @f (0) 1 @n = f (0) 1 n @f (0) 1 @T = −3 2 f (0) 1 T + m 2kBT 2(~ v −~ u)2f (0) 1 @f (0) 1 @~ u = m kBT (~ v −~ u)f (0) 1 Using all these relations, we have @f (0) 1 @t −{H1, f (0) 1 } = 1 n ˜ Dtn + ✓m(~ v −~ u)2 2kBT 2 −3 2T ◆ ˜ DtT + m kBT (~ v −~ u) · ˜ Dt~ u − 1 kBT ~ F · (~ v −~ u) / f (0) 1 (2.66) where we’ve introduced the notation ˜ Dt which di↵ers from the material derivative Dt in that it depends on the velocity ~ v rather than the average velocity ~ u, ˜ Dt ⌘@ @t + ~ v · @ @~ r = Dt + (~ v −~ u) · @ @~ r Now our first attempt at deriving hydrodynamics gave us three equations describing how n (2.59), ~ u (2.60) and T (2.61) change with time. We substitute these into (2.66).
You’ll need a couple of lines of algebra, cancelling some terms, using the relationship P = nkBT and the definition of Uij in (2.55), but it’s not hard to show that we ultimately get @f (0) 1 @t −{H1, f (0) 1 } = 1 T ✓ m 2kBT (~ v −~ u)2 −5 2 ◆ (~ v −~ u) · rT (2.67) + m kBT ✓ (vi −ui)(vj −uj) −1 3(~ v −~ u)2δij ◆ Uij / f (0) 1 – 45 – And there’s no reason that the right-hand-side is zero. So, unsurprisingly, f (0) 1 does not solve the Boltzmann equation. However, the remaining term depends on rT and @~ u/@~ r which means that we if we stick to long wavelength variations in the temperature and velocity then we almost have a solution. We need only add a little extra something to the distribution f1 = f (0) 1 + δf1 (2.68) Let’s see how this changes things.
2.5.1 Relaxation Time Approximation The correction term, δf1, will contribute to the collision integral (2.24). Dropping the ~ r argument for clarity, we have ✓@f1 @t ◆ coll = Z d3p2d3p0 1d3p0 2 !(~ p 0 1, ~ p 0 2|~ p1, ~ p2) [f1(~ p 0 1)f1(~ p 0 2) −f1(~ p1)f1(~ p2)] = Z d3p2d3p0 1d3p0 2 !(~ p 0 1, ~ p 0 2|~ p1, ~ p2) h f (0) 1 (~ p 0 1)δf1(~ p 0 2) + δf(~ p 0 1)f (0) 1 (~ p 0 2) −f (0) 1 (~ p1)δf1(~ p2) −δf(~ p1)f (0) 1 (~ p2) i where, in the second line, we have used the fact that f (0) 1 vanishes in the collision integral and ignored quadratic terms ⇠δf 2 1. The resulting collision integral is a linear function of δf1. But it’s still kind of a mess and not easy to play with.
At this point, there is a proper way to proceed. This involves first taking more care in the expansion of δf1 (using what is known as the Chapman-Enskog expansion) and then treating the linear operator above correctly. However, there is a much easier way to make progress: we just replace the collision integral with another, much simpler function, that captures much of the relevant physics. We take ✓@f1 @t ◆ coll = −δf1 ⌧ (2.69) where ⌧is the relaxation time which, as we’ve already seen, governs the rate of change of f1. In general, ⌧could be momentum dependent. Here we’ll simply take it to be a constant.
The choice of operator (2.69) is called the relaxation time approximation. (Sometimes it is referred to as the Bhatnagar-Gross-Krook operator). It’s most certainly not exact.
– 46 – In fact, it’s a rather cheap approximation. But it will give us a good intuition for what’s going on. With this replacement, the Boltzmann equation becomes @(f (0) 1 + δf1) @t −{H1, f (0) 1 + δf1} = −δf1 ⌧ But, since δf1 ⌧f (0) 1 , we can ignore δf1 on the left-hand-side. Then, using (2.67), we have a simple expression for the extra contribution to the distribution function δf1 = −⌧ 1 T ✓ m 2kBT (~ v −~ u)2 −5 2 ◆ (~ v −~ u) · @T @~ r + m kBT ✓ (vi −ui)(vj −uj) −1 3(~ v −~ u)2δij ◆ Uij / f (0) 1 (2.70) We can now use this small correction to the distribution to revisit some of the transport properties that we saw in Section 1.
2.5.2 Thermal Conductivity Revisited Let’s start by computing the heat flux qi = 1 2m⇢h(vi −ui) (~ v −~ u)2i (2.71) using the corrected distribution (2.68). We’ve already seen that the local equilibrium distribution f (0) 1 gave ~ q = 0, so the only contribution comes from δf1. Moreover, only the first term in (2.70) contributes to (2.71). (The other is an odd function and vanishes when we do the integral). We have ~ q = −rT This is the same phenomenological law that we met in (1.12). The coefficient is the thermal conductivity and is given by = m⌧⇢ 2T Z d3p (~ vi −~ ui)2(~ v −~ u)2 m 2kBT (~ v −~ u)2 −5 2 / f (0) 1 = m⌧⇢ 6T m 2kBT hv6i0 −5 2hv4i0 / In the second line, we’ve replaced all (v −u) factors with v by performing a (~ r-dependent) shift of the integration variable. The subscript h·i0 means that these aver-ages are to be taken in the local Maxwell-Boltzmann distribution f (0) 1 with u = 0. These – 47 – integrals are simple to perform. We have hv4i0 = 15k2 BT 2/m2 and hv6i0 = 105k3 BT 3/m3, giving = 5 2⌧nk2 BT The factor of 5/2 here has followed us throughout the calculation. The reason for its presence is that its the specific heat at constant pressure, cp = 5 2kB.
This result is parameterically the same that we found earlier in (1.13). (Although you have to be a little careful to check this because, as we mentioned after (2.54), the definition of heat flux di↵ers and, correspondingly, , di↵ers by a factor of m.
Moreover, the current formula is written in terms of slightly di↵erent variables. To make the comparison, you should rewrite the scattering time as ⌧⇠1/mσn p hv2i, where σ is the total cross-section and hv2i ⇠T/m by equipartition). The coefficient di↵ers from our earlier derivation, but it’s not really to be trusted here, not least because the only definition of ⌧that we have is in the implementation of the relaxation time approximation.
We can also see how the equation (2.56) governing the flow of temperature is related to the more simplistic heat flow equation that we introduced in (1.14). For this we need to assume both a static fluid ~ u = 0 and also that we can neglect changes in the thermal conductivity, @/@~ r ⇡0. Then equation (2.56) reduces to the heat equation ⇢kB @T @t = −2 3r2T 2.5.3 Viscosity Revisited Let’s now look at the shear viscosity. From our discussion in Section 1, we know that the relevant experimental set-up is a fluid with a velocity gradient, @ux/@z 6= 0. The shear viscosity is associated to the flux of x-momentum in the z-direction. But this is precisely what is computed by the o↵-diagonal component of the pressure tensor, Pxz = ⇢h(vx −ux)(vz −uz)i We’ve already seen that the local equilibrium distribution gives a diagonal pressure tensor (2.58), corresponding to vanishing viscosity. What happens if we use the cor-rected distribution (2.68)? Now only the second term in (2.70) contributes (since the first term is an odd function of (v −u)). We write Pij = P δij + ⇧ij (2.72) – 48 – where the extra term ⇧ij is called the stress tensor and is given by ⇧ij = m⌧⇢ kBT Ukl Z d3p (vj −uj)(vi −ui) ✓ (vk −ul)(vk −ul) −1 3(~ v −~ u)2δkl ◆ f (0) 1 = m⌧⇢ kBT Ukl hvivjvkvli0 −1 3δklhvivjv2i0 / Before we compute ⇧ij, note that it is a traceless tensor. This is because the first term above becomes hv2vkvli0 = δjkhv2vxvxi0 which is easily calculated to be hv2v2 xi0 = 5k2 BT 2/m2 = 1 3hv4i0. Moreover, ⇧ij depends linearly on the tensor Uij. These two facts mean that ⇧ij must be of the form ⇧ij = −2⌘ ✓ Uij −1 3δijr · ~ u ◆ (2.73) In particular, if we set up a fluid gradient with @ux/@z 6= 0, we have ⇧xz = −⌘@ux @z which tells us that we should identify ⌘with the shear viscosity. To compute it, we return to a general velocity profile which, from (2.73), gives ⇧xz = m⌧⇢ kBT Ukl hvxvzvkvli0 −1 3δklhvxvzv2i0 / = m⌧⇢ kBT (Uxz + Uzx)hvxvzvxvzi0 = 2m⌧⇢ 15kBT Uxzhv4i0 Comparing to (2.73), we get an expression for the coefficient ⌘, ⌘= nkBT⌧ Once again, this di↵ers from our earlier more naive analysis (1.11) only in the overall numerical coefficient. And, once again, this coefficient is not really trustworthy due to our reliance on the relaxation time approximation.
The scattering time ⌧occurs in both the thermal conductivity and the viscosity. Tak-ing the ratio of the two, we can construct a dimensionless number which characterises our system. This is called the Prandtl number, Pr = cp⌘ – 49 – With cp the specific heat at constant pressure which takes the value cp = 5kB/2 for a monatomic gas. Our calculations above give a Prandtl number Pr = 1. Experimental data for monatomic gases shows a range of Prandtl numbers, hovering around Pr ⇡2/3.
The reason for the discrepancy lies in the use of the relaxation time approximation. A more direct treatment of the collision integral, thought of as a linear operator acting on δf1, gives the result Pr = 2/3, in much better agreement with the data7.
2.6 A Second Look: The Navier-Stokes Equation To end our discussion of kinetic theory, we put together our set of equations governing the conservation of density, momentum and energy with the corrected distribution function. The equation of motion for density fluctuations doe not change: it remains, @⇢ @t + r · (⇢~ u) = 0 (2.74) Meanwhile the equation for momentum (2.50) now has an extra contribution from the stress tensor contribution (2.72). Moreover, we typically assume that, to leading order, variations in the viscosity can be neglected: r⌘⇡0. Written in vector notation rather than index notation, the resulting equation is ✓@ @t + ~ u · r ◆ ~ u = ~ F m −1 ⇢rP + ⌘ ⇢r2~ u + ⌘ 3⇢r(r · ~ u) (2.75) This is the Navier-Stokes equation. Finally, we have the heat conduction equation. We again drop some terms on the grounds that they are small. This time, we set r⇡0 and Uij⇧ij ⇡0; both are small at the order we are working to. We’re left with ⇢ ✓@ @t + ~ u · r ◆ T −2 3r2T + 2m 3 P r · ~ u = 0 We can again look at fluctuations of these equations about a static fluid with ⇢= ¯ ⇢, T = ¯ T and ~ u = 0. Longitudinal fluctuations (2.62) and (2.63) now give rise to the linearised equations of motion, !δ⇢= ¯ ⇢|~ k|δu !δu = kB ¯ T m¯ ⇢|~ k|δ⇢+ kB m |~ k|δT −i4⌘|~ k|2 3¯ ⇢ δu !δT = 2 3 ¯ T|~ k|δu −i2|~ k|2 3kB ¯ ⇢δT 7You can read about this improved calculations in the lectures by Daniel Arovas.
– 50 – Notice that terms involving transport coefficients ⌘and each come with a factor of i; this is a sign that they will give rise to dissipation. To compute the frequencies of the di↵erent modes, it’s best to think of this as an eigenvalue problem for !/|~ k|; the coefficients of the various terms on the right-hand-side define a 3 ⇥3 matrix M, with det M = 2i 3 |~ k|4 ¯ T m¯ ⇢ and Tr M = −i ✓4 3⌘+ 2 3 kB ◆|~ k|2 ¯ ⇢ The product of the three eigenvalues is equal to det M. We know that for the ideal fluid, the eigenvalues are zero and ! = ±vs|~ k| where vs is the sound speed computed in (2.64). Let’s first look at the eigenvalue that was zero, corresponding to fluctuations of constant pressure. Working to leading order in and ⌘, we must have −v2 s|~ k|2! = det M ) ! = −2i 5 kB ¯ ⇢|~ k|2 The purely imaginary frequency is telling us that these modes are damped. The ! ⇠ i|~ k|2 is characteristic of di↵usive behaviour.
The remaining two modes are related to the sound waves. These too will gain a dispersive contribution, now with ! = ±vs|~ k| −iγ (2.76) Using the fact that the sum of the eigenvalues is equal to the trace, we find γ = ✓2 3⌘+ 2 15 kB ◆|~ k|2 ¯ ⇢ (2.77) The fluctuations above are all longitudinal. There are also two shear modes, whose fluctuations are in a direction perpendicular to the velocity. It is simple to check that the linearised equations are solved by δ⇢= δT = 0 and δ~ u ·~ k, with the frequency given by ! = −i ⌘|~ k|2 ¯ ⇢ Once again, we see that these modes behave di↵usively.
Navier Stokes Equation and Liquids Our derivation of the Navier-Stokes equation relied on the dilute gas approximation.
However, the equation is more general than that. Indeed, it can be thought of as the – 51 – most general expression in a derivative expansion for momentum transport (subject to various requirements). In fact, there is one extra parameter that we could include: ⇢ ✓@ @t + ~ u · r ◆ ~ u = ⇢~ F m −rP + ⌘r2~ u + ⇣⌘ 3 + ⇣ ⌘ r(r · ~ u) where ⇣is the bulk viscosity which vanished in our derivation above. Although the equation above governs transport in liquids, we should stress that first-principles com-putations of the viscosity (and also thermal conductivity) that we saw previously only hold in the dilute gas approximation.
– 52 – |
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Rules for Divisibility of 7, 11, and 12
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Divisibility Rules for 7, 11, and 12
In our previous lesson, we discussed the divisibility rules for 2, 3, 4, 5, 6, 9, and 10. In this lesson, we are going to talk about the divisibility tests for numbers 7, 11, and 12. The reason why I separated them is that the divisibility rules for 7, 11, and 12 are a little bit more advanced. However, I promise you that after learning their respective rules and applying them to some practice problems, you will realize that they are not that difficult. In fact, they are actually fun!
Divisibility Rule for 7
Rule: Take the last digit and cross it out from the original number. Then double it. Subtract it from the “new” number which is the original number excluding the last digit. If the difference is divisible by 7, then the original number must also be divisible by 7. If at first application, the result is not obviously divisible by 7, you can repeat the process as needed until you reach a two-digit number that can easily be determined if it is divisible by 7 or not.
Example 1: True or False. The number 6,895 is divisible by 7.
Solution: Let’s take the last digit of 6,895 which 5 then double it, thus 2(5)=10. Now, subtract the “new” number (old number excluding the last digit) by twice the last digit, we have 689−10=679. Is 679 divisible by 7? We can perform long division. But the good thing is that we can perform the process again and again until we reach a two-digit number because it is much easier to know if it’s divisible or not by 7.
Let’s repeat the process one more time and see what we will get. Remember we ended up at 679 from the last step. Moving on, the last digit of 679 is 9. If we double it, we get 2(9)=18. The remaining number formed when we get rid of the last digit is 67. If we subtract 67 by 18, we obtain 67−18=49.
Since 49 is divisible by 7, therefore the original number 6,895 must also be divisible by 7. So, the answer is True. ✔︎
Example 2: Multiple Choice. Which number is divisible by 7?
Note: There is only one correct answer.
A) 18,046
B) 11,749
C) 20,704
D) 21,011
I understand that the procedure can be tricky at first but the more you use it, the more it becomes much easier. Below are easy-to-follow steps that I hope can help to cement in your memory.
Steps to Check for the Divisibility of 7
Drop the last digit of the number then double the digit that we dropped.
Subtract it from the new number formed by removing the last digit of the original number.
Repeat the process until the number is reduced to two digits.
If the two-digit number is divisible by 7, then the original number is divisible by 7. Otherwise, it is not.
Solution: In an actual multiple questions test, you may want to randomly select an option (a letter) to solve because it is possible that you can stumble upon the correct answer right away, therefore, saving you a lot of time. But in this lesson, we will go from A to D for the sake of practice.
◉ Testing Option A: 18,046
Drop the last digit of 18,046 which becomes 1,804 then double the digit that we dropped, so we have 2(6)=12.
Subtract the new number by the double of the last digit: 1,804–12=1,792. We have reduced the original five-digit number to a four-digit number. Remember, we want it to be reduced to a two-digit number. Let’s repeat the process.
Drop the last digit of 1,792 which becomes 179 then double the digit the we dropped, so we have 2(2)=4.
Subtract the new number by the double of the last digit: 179–4=175. We have reduced it now to a three-digit number. Let’s do it one more time!
Drop the last digit of 175 which becomes 17 then double the digit that we removed, thus 2(5)=10.
Subtract the new number by twice the last digit: 17−10=7.
Since 7 is divisible by 7, then the original number which is 18,046 is also divisible by 7. So, option A is the correct answer. ✔︎
The final answer is option A.
I will leave it to you as an exercise as to why options B, C, and D are NOT divisible by 7. However, I will still provide you with a shortened solution below. I highly encourage you to perform the exercise not only for more practice but also because it is as satisfying to show that a number is not divisible by 7.
You Try!
◉ Testing Option B: 11,749
Answer
Original number: 11,749
1,174−2(9)=1,174−18=1,156
115−2(6)=115−12=103
10−2(3)=10−6=4
Since 4 is not divisible by 7 then 11,749 is also not divisible by 7. ✘
◉ Testing Option C: 20,704
Answer
Original number: 20,704
2,070−2(4)=2,070−8=2,062
206−2(2)=206−4=202
20−2(2)=20−4=16
Since 16 is not divisible by 7 then 20,704 is also not divisible by 7. ✘
◉ Testing Option D: 21,011
Answer
Original number: 21,011
2,101−2(1)=2,101−2=2,099
209−2(9)=209−18=191
19−2(1)=19−2=17
Since 17 is not divisible by 7 then the original number which is 21,011 is not divisible by 7 as well. ✘
Example 3: Select all that apply. Which numbers are divisible by 7?
Note: There can be more than one answer.
A) 5,544
B) 3,110
C) 54,810
D) 34,125
Solution: I am sure that at this point you have already mastered the steps on how to check if a number is divisible by 7 or not. With that said, I will be using a shortened solution.
◉ Testing Option A: 5,544
We are testing if 5,544 is divisible by 7.
554−2(4)=554−8=546
54−2(6)=54−12=42
Because 42 can be divided by 7 then the original number 5,544 is also divisible by 7. ✔︎
◉ Testing Option B: 3,110
We are checking if 3,110 is divisible by 7.
311−2(0)=311−0=311
31−2(1)=31−2=29
Since 29 cannot be divided by 7 then the original number 3,110 is not divisible by 7 either. ✘
◉ Testing Option C: 54,810
Let’s examine if 54,810 is divisible by 7.
5,481−2(0)=5,481−0=5,481
548−2(1)=548−2=546
54−2(6)=54−12=42
The algorithm has reduced the original number into a two-digit number which is 42 that is divisible by 7. It means that the original number 54,810 must also be divisible by 7. ✔︎
◉ Testing Option D: 34,125
Let’s determine if 34,125 is divisible by 7.
3,412−2(5)=3,412−10=3,402
340−2(2)=340−4=336
33−2(6)=33−12=21
We have reduced the original five-digit number into a two-digit number 21 that is divisible by 7. It implies that the original number 34,125 should be divisible by 7 as well. ✔︎
So in summary, options A, C, and D are divisible by 7.
Divisibility Rule for 11
Rule: From the left to right of a number, take the first digit, and attach an addition symbol to its left. Then subtract it by the next digit, then add the result by the third digit, and subtract again the result by the fourth digit, and so on and so forth. If the answer is divisible by 11, then the original number is divisible by 11.
Condensed Rule: Alternately add and subtract the digits of a number from left to right. If the answer is divisible by 11, then the original number is divisible by 11.
Standard Rule: Take the alternating sum of the digits of a number. If the result is a multiple of 11, the number is divisible by 11.
NOTE: All the rules above mean the same thing. The first two rules are more instructive in nature while the last one is the rule that you may encounter in your textbook or being taught by your teacher.
Example 1: True or False. The number 9,581 is divisible by 11.
The rule is actually quite simple. We will add and subtract, then repeat the pattern until all the digits of the number are assigned with plus and minus symbols from left to right. After setting it up, we simplify it. If the result is a multiple of 11, then the original number is also divisible by 11.
Here is the set-up:
+9−5+8−1
Step 1: +9−5=4
4+8−1
Step 2: 4+8=12
12−1
Step 3: 12−1=11
11
Since the final result is 11 and a multiple of 11, then the original number which is 9,581 is divisible of 11. Thus, our final answer is True. ✔︎
Example 2: Multiple Choice. Which number is divisible by 11?
Note: There is only one correct answer.
A) 98,517
B) 79,829
C) 82,709
D) 50,453
We will check the divisibility of each number from option A to option D.
◉ Checking Option A: 98,517
Let’s set it up by taking the alternating sum of the digits of the number.
9−8+5−1+7
Then, we simplify.
(9−8)+5−1+7
1+5−1+7
(1+5)−1+7
6−1+7
(6−1)+7
5+7
12
The final result is 12 which is not a multiple of 11. Therefore, the original number 98,517 is not divisible by 11. ✘
◉ Checking Option B: 79,829
Set it up by writing the alternating sum of the digits.
7+9−8+2−9
Simplify.
(7+9)−8+2−9
16−8+2−9
(16−8)+2−9
8+2−9
(8+2)−9
10−9
1
Since the final answer (1) is not divisible by 11, therefore the original number 79,829 is also not divisible by 11. ✘
◉ Checking Option C: 82,709
We first construct the alternating sum of the digits of the number.
8−2+7−0+9
Then simplify from left to right. No need to worry about the Order of Operations since we are only dealing with addition and subtraction.
(8−2)+7−0+9
6+7−0+9
(6+7)−0+9
13−0+9
(13−0)+9
13+9
22
Since the final result is 22 which is a multiple of 11, it implies that the original number 82,709 is divisible by 11. Therefore, the final answer is C. ✔︎
☞ There is no need to check for Option D because we have already found the correct answer.
The final answer is option C.
Example 3: Which numbers are divisible by 11? Select all that apply.
Note: There can be more than one answer.
A) 69,245
B) 73,186
C) 843,210
D) 918,071
Solution:
◉ Testing Option A: 69,245 if it is divisible by 11
6−9+2−4+5
6−9+2−4+5
−3+2−4+5
−3+2−4+5
−1−4+5
−1−4+5
−5+5
0
Since 0 is a multiple of 11, therefore 69,245 is divisible by 11. ✔︎
◉ Testing Option B: 73,186 if it is divisible by 11
7−3+1−8+6
7−3+1−8+6
4+1−8+6
4+1−8+6
5−8+6
5−8+6
−3+6
3
Since 3 is not a multiple of 11, thus 73,186 is not divisible by 11. ✘
◉ Testing Option C: 843,210 if it is divisible by 11
8−4+3−2+1−0
8−4+3−2+1−0
4+3−2+1−0
4+3−2+1−0
7−2+1−0
7−2+1−0
5+1−0
5+1−0
6−0
6
Since 6 is not a multiple of 11, hence 843,210 is not divisible by 11. ✘
◉ Testing Option D: 918,071 if it is divisible by 11
9−1+8−0+7−1
9−1+8−0+7−1
8+8−0+7−1
8+8−0+7−1
16−0+7−1
16−0+7−1
16+7−1
16+7−1
23−1
22
Since 22 is a multiple of 11, it implies that 918,071 is divisible by 11. ✔︎
In summary, options A and D are divisible by 11.
Divisibility Rule for 12
Rule: A number is divisible by 12 if it is both divisible by 3 and 4.
A number is divisible by 3 if the sum of its digits is divisible by 3.
A number is divisible by 4 if the last two digits of the number are divisible by 4.
Example 1: True or False. The number 7,512 is divisible by 12.
Solution:
The first step is to check if it is divisible by 3. We will first add all the digits of the number of 7,512.
7,512
7+5+1+2=15
Since 15 is divisible by 3, therefore 7,512 is also divisible by 3.
The last step is to test if the number formed by the last two digits of the original number is divisible by 4, then it is divisible by 4.
7,512
Since 12 is divisible by 4, then 7,512 is divisible by 4.
Therefore, because the original number 7,512 is both divisible by 3 and 4, then it must be divisible by 12. ✔︎
Example 2: Multiple Choice. Which number is divisible by 12?
Note: There is only one correct answer.
A) 527,037
B) 981,128
C) 746,936
D) 49,9920
Solution:
There is a faster way to test for the divisibility of 12. Remember, a number is divisible by 12 if 3 and 4 can both divide it. Since it is much quicker to test for the divisibility of 4 than 3 because for the former you just have to look at the last two digits of the number and check if it is a multiple of 4, and the latter will take slightly more time because you will have to add all the digits of the number and check if the sum is divisible by 3. Therefore, we will check first for the divisibility of 4 followed by the divisibility of 3. The other way around is a little bit more time-consuming.
◉ Testing Option A: 527,037 for divisibility of 12
The last two digits of 527,037 is 37 is not a multiple of 4. Therefore, it is not divisible by 4. There is no need to check for the divisibility of 3 since it fails on one of the two requirements. Thus, 527,037 is not divisible by 12. ✘
◉ Testing Option B: 981,128 for divisibility of 12
The last two digits of 981,128 is 28 which is a multiple of 4 that makes it divisible of 4. Now let’s check if it is divisible by 3 by adding all its digits, thus 9+8+1+1+2+8=29. Since, the sum 29 is not divisible by 3, then the number itself is also not divisible by 3. Because 981,128 cannot be divided by both 3 and 4, that means the two requirements are not met, hence the original number is not divisible by 12. ✘
◉ Testing Option C: 746,936 for divisibility of 12
The number 36 is the last two digits of 746,936. And it is a multiple of 4 which makes the original number divisible by 4. Now for divisibility of 3. Add all the digits of 746,936, we get 7+4+6+9+3+6=35. The sum of the digits is not divisible by 3. It follows that the number is also not divisible by 3. Because one of the two required conditions are not met (both are not true), then 746,936 is not divisible by 12. ✘
◉ Testing Option D: 49,9920 for divisibility of 12
The number 20 is the last two digits of 49,9920 which is clearly a multiple of 4, thus makes 49,9920 divisible by 4. Adding up all the digits of the number: 4+9+9+9+2+0=33. The sum 33 can be divided by 3 and so 49,9920 is divisible by 3. Since, the original number is both divisible by 3 and 4, it must also be divisible by 12. ✔︎
The final answer is option D.
Example 3: Which numbers are divisible by 12? Select all that apply.
Note: There can be more than one answer.
A) 344,888
The number 88 is the last two digits of 344,888 which is clearly a multiple of 4, thus divisible by 4.
The sum of the digits of 344,888 is calculated as 3+4+4+8+8+8=35. But 35 is obviously not divisible by 3.
Since 344,888 is only found to be divisible by 4 but not by 3, failing one of the two requirements implies that the original number is not divisible by 12. ✘
B) 521,340
The last two digits of 521,340 formed the number 40 which is a multiple of 4, thus can be divided by 4.
Adding up its digits we get 5+2+1+3+4+0=15. The sum 15 is divisible by 3.
Since 521,340 is both divisible by 3 and 4, then it must be divisible by 12. ✔︎
C) 842,652
The number 52 is the last two digits of the number which is clearly divisible by 4.
The sum of the digits is 8+4+2+6+5+2=27. The number 27 is divisible by 3.
Since 842,652 are both divisible by 3 and 4, then it should also be divisible by 12. ✔︎
D) 676,968
The last two digits 68 are divisible by 4.
The sum of the digits 6+7+6+9+6+8=42 is divisible by 3.
Because the original number can both be divided by 3 and 4, then it must also be divisible by 12.
You might also like these tutorials:
Divisibility Rules for 2, 3, 4, 5, 6, 9, and 10
Tags:
Introductory Algebra,
Lessons
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1176 | https://www.saspublishers.com/article/12178/download/ | Available online at 4354
Scholars Journal of Applied Medical Sciences (SJAMS) ISSN 2320-6691 (Online)
Sch. J. App. Med. Sci., 2016; 4(12C):4354-4357 ISSN 2347-954X (Print)
©Scholars Academic and Scientific Publisher
(An International Publisher for Academic and Scientific Resources)
www.saspublishers.com DOI: 10.36347/sjams.2016.v04i12.033
Correlation of vaginal pH, cytology and vaginal maturation value in diagnosis of atrophic vaginitis in post menopausal women
Uruj Jahan 1, Yashodhara Pradeep 2, Nuzhat Hussain 3
1
Department of Obstetrics and Gynecology, GSVM Medical College, Kanpur, Uttar Pradesh, India
2
Department of Obstetrics and Gynecology, King George Medical University, Lucknow, Uttar Pradesh, India
3
Department of Pathology, King George Medical University, Lucknow, Uttar Pradesh, India
Corresponding author
Dr. Uruj Jahan
Email: druruj@gmail.com
Abstract: The objective of this study is to detect the correlation among vaginal pH, cytology and vaginal maturation value (VMV) in diagnosis of atrophic vaginitis in postmenopausal women. 100 women were enrolled out of which 49 were symptomatic having symptoms of atrophic vaginitis and 51 postmenopausal otherwise healthy women who had no symptoms of atrophic vaginitis. All women underwent vaginal pH, cytology and vaginal maturation value assessment in addition to routine history and examination. In women with atrophic symptoms, age and duration of menopause was significantly higher than women without this symptoms. There was highly significant correlation was found between vaginal pH and parabasa cells. Similarly highly significant inverse correlation was detected between vaginal pH and VMV. This study confirms that high vaginal pH is associated with elevated parabasal cells and low vaginal maturation value in postmenopausal women of atrophic vaginitis.
Keywords: atrophic vaginitis, menopause, pH, vaginal maturation value, parabasal cell.
INTRODUCTION
Menopause is a universal phenomenon and principal health concern of menopausal women includes vasomotor symptoms, atrophic vaginitis, osteoporosis, cardiovascular disease, cognitive decline and sexual problems . Atrophic vaginitis is inflammation of vagina due to thinning and shrinking of vaginal tissue as well as decrease lubrication, which is caused by decreased level of estrogen. Declining level of estrogen; increases tissue fragility , urogenital infection, vaginal dryness and vaginal tissue trauma . 10-40% of patients experience urogenital atrophy after menopause although only 25% of these reports with symptom to the gynaecologist. Generally the prevalence of vaginal atrophy has been reported to range from 7% to 57% in healthy peri and post menopausal women. Approximately 40% of women with vaginal atrophy reports dyspareunia [3-5].
Estrogen deficiency may cause reduction in superficial epithelial cells and increment of parabasal cells in the vagina. It also results in less exfoliation of cells which leads to less release of glycogen and reduced conversion to lactic acid by the vaginal flora3. Lactic acid maintains the vaginal pH at about 3.5to 4.5 . This acidic pH is an important component of a women´s non specific defense against pathogens . Vaginal pH is increased in postmenopausal women. The weighted average of vaginal pH is 6in post menopausal women not receiving estrogen therapy . Vaginal maturation value (VMV) is calculated from the ratio of superficial, intermediate and parabasal cells count in vaginal smear which is used to detect vaginal atrophy and estrogen deficiency in post menopausal women .
So, in present study, we are presenting the correlation between vaginal pH, cytology and VMV in post menopausal women.
METHODS
Hundred post menopausal women attending gynecology OPD, KGMU, Lucknow were included in the study. Detailed history and examination of all patients was done. Demographic characteristics including age, religion, socio-economic status, education, parity werenoted. Symptoms of atrophic vaginitis such as dryness, itching, burning, soreness, discharge per vaginum, dyspareunia, burning micturition, painful micturition and incontinence were enquired. Women declining consent to enroll in study,
Original Research Article Uruj Jahan et al., Sch. J. App. Med. Sci., Dec 2016; 4(12C):4354-4357
Available online at 4355
having infection of vagina, malignancies of genital tract, women with systemic diseases, Previous vaginal surgery involving more than 1/3rd of vagina, positive amine test and women with history of current or past exposure to estrogen progestrone replacement or vaginal estrogen therapy were excluded. In addition all women underwent vaginal pH measurement and vaginal smear was taken for cytology and maturation value assessment.
PH estimation : Vaginal pH level was measured by pH indicator paper with colour scale which range from pH 2 to 10.5 developed by Merck specialities Pvt. Ltd. This device is composed of foils containing Nitrizine yellow for pH testing. Paper is contacted to vagina for 5 seconds and colour is compared with colour scale and thus pH value is determined.
Cytology Evaluation : Cytological evaluation was performed by vaginal smear collected from lateral wall of mid third of vagina and mounted on a slide. Smear is immediately fixed in an alcohol ether dip for 1 hour and then stained with Papanicoloau stain. Each slide is evaluated in department of Pathology, KGMU Lucknow. In a total of 100 exfoliated vaginal cells, parabasal cells, intermediate cells, superficial cells were counted and results were expressed as the maturation value . Parabasal cell are small rounded cell with large nuclei comprising50 – 70% of the total cell size. Intermediate cell have small nuclei with round cell. Superficial cells have smallernuclei, rectangular cell membrane with abundant cytoplasm with nucleous comprising 10 – 20% of the cell.
Superfecial cells, Intermediate cellsand parabasal cells were assigned a point value of 1, 0.5 and 0 respectively. The number of cells in each category will be multiplied by point valuesand all three results will be added to arrive at a maturation value. A value of 0-49 indicates low estrogen effect. Value of 50 – 64 indicate moderate estrogen effect and value of 65 –100 indicate high estrogen effect . All examination will be interpreted by same pathologist without prior knowledge of subject data.
Statistical Analysis :
The statistical analysis was done using SPSS version 15.0 statistical analysis software. The value was represented in number (%), mean + SD and p value.
RESULTS
Out of 100 women enrolled, 49 were symptomatic and 51 were having no symptoms of atrophic vaginitis. The most common symptoms were dryness (98%), itching (89.8%), burning (61.2%) and dyspareunia (32.7%). Other less common symptoms were burning and increased frequency of micturation, discharge per vaginum and prolapse. Mean age in symptomatic and asymptomatic women was 55.35 + 7.03 and 50.99 + 4.83 years respectively which was statistically significant (p=0.007). Mean duration of menopause in symptomatic women was 7.33 + 6.03 while in asymptomatic women it was 4.33 + 3.28 which was also statistically significant(p<0.001). There was no significant difference between groups regarding parity and body mass index (BMI) (table 1).
Majority of symptomatic women (30.6%) had vaginal pH range from 6- 7 as compared to asymptomatic women in which most women (54.9%) had vaginal pH range 5.1 –6. Mean pH in symptomatic and asymptomatic women was 6.67+0.95 and 5.49+0.75 respectively, which was statistically significant (p < 0.001) (table 2).
Mean parabasal cell count was 2.31 + 3.27 in symptomatic women but asymptomatic women had no parabasal cells which was also statistically significant (p < 0.001). Intermediate cell count is more in symptomatic women while superficial cell count was maximum in asymptomatic women and difference of cell counts was found statistically significant in both groups ( p <0.001). Mean VMV in symptomatic and asymptomatic women was 56.65 + 6.14 and 84.12 + 11.08 respectively and difference was found to be statistically significant (p<0.001) (table 3).
No subjects having VMV <49 had pH <5 and no subjects with VMV >65 had pH >8. Mean pH in VMV category <49, 50 – 64 and >65 was 7.82+0.85, 6.28 + 0.79 and 6.02 + 0.82 respectively which was found to be statistically significant. In this study 28 women (60.8%) with VMV >65 had pH < 6 and 10 out of 13 (76.9%) patients with severe atrophic vaginitis (VMV < 49) had pH >7 while 10 out of 41 women (23.8%) with mild atrophic vaginitis (VMV 50-64) had pH above 7, suggesting women with high vaginal pH had low VMV score (table 4). Significant inverse correlation was found between vaginal pH and VMV (p<0.001).
Table 1: Demographic Profile of Menopausal women
Symptomatic
(n=49)
Asymptomatic
(n=51)
P value
Mean Age 55.35 + 7.03 50.99 +_ 4.83 0.007
Mean Duration of menopause 7.33 +6.03 4.33+ 3.28 0.001
Mean Parity 3.4 +2.1 3.6 +2.2 0.13
Mean BMI 21.67 + 1.73 21.27 +1.59 0.603 Uruj Jahan et al., Sch. J. App. Med. Sci., Dec 2016; 4(12C):4354-4357
Available online at 4356
Table2: Distribution of pH in menopausal women
Ph
Symptomatic
(n=49)
Asymptomatic
(n=51)
P value
4.5 – 5 6(12.2%) 14(27.5%)
P<0.001
5.1 – 6 13(26.5%) 28(54.9%)
6.1 – 7 15(30.6%) 08(15.7%
7.1 – 8.0 10(20.4%) 1(2%)
8 5(10.2%)
Mean pH of all women 6.08+10.41 6.67 + 0.95 5.49+0.75
Table 3: Comparison of cytology in menopausal women
Findings
Symptomatic
(n=49)
Asymptomatic
(n=51)
P value
Mean SD Mean SD
<0.001
Parabasal cells 2.31 3.27 0.0 0.0
Intermediate cells 79.06 16.33 31.65 21.98
Superficial cells 18.53 17.92 68.38 21.96
VMV 56.65 6.14 84.12 11.08
Table 4: Correlation of pH and VMV in menopausal women
pH VMV Category
VMV<49
(n=13)
VMV 50 -64
(n=41)
VMV >65
(n=46)
4.5 – 5.0 0 04(9.8) 03(6.5)
5.1 – 6.0 2(23.1%) 12(29.3%) 25 (54.3%)
6.1 – 7.0 1(7.7%) 15(36.6%) 17(37%)
7.1 – 8.0 8(61.5%) 8(19.5%) 1(2.71%)
8.0 2(15.4%) 02(4.33%) 0
Mean pH 7.82±0.35 6.28+ 0.79 6.0 + 0.82
DISCUSSION :
In our study there was significant correlation among age, duration of menopause and atrophic vaginitis but no significant difference was found regarding parity and BMI which is comparable to study done by Pinar Yoruket al in which women with atrophic symptoms, the age was higher than those without symptoms and no significant difference was found regarding parity and BMI . Similarly in study done by Sebestian et al.; no significant difference of BMI was found . While from previous studies it is known that BMI can influence serum estrogen (E2) value, vaginal pH and consecutively vaginal mucosal health .
In present study mean pH in symptomatic patients was significantly higher than asymptomatic patients.i.e. 6.67 + 0.95 and 5.49 + 0.75 respectively ( p<0.001) which is comparable to study done by Pinar Yoruk et al.; in which women with urogenitalatrophy had mean pH of 6.5±0.48 and was significantly higher than women without these symptoms . Davila et al.;
also demonstrated that vaginal pH is solid predictor of maturation value and most reliable indicator of urogenital atrophy . Meta-analysis of 16 reports by Roy et al.; confirmed that vaginal pH reflects circulating estradiol level .Therefore with the use of vaginal pH value; it is possible to detect atrophic vaginitis.
In current study parabasal cells are present only in symptomatic women and mean was 2.31 + 3.27. A study done by Fantl et al.; showed that mean parabasal count in 70post menopausal women with urinary incontinence was 18 ± 27% which decrease to 0 + 1% on estrogen supplementation . In this study, mean VMV in symptomatic women was significantly higher than women without atrophic vaginitis. Similarly, study done by Pinar Yoruk et al.; also represented that difference in mean VMV in symptomatic and asymptomatic women was 34.7±16.2 and 83.8±9.4 which was statistically significant .Vander Linden et al.; Notelovitz et al.; and Sartori
et al.; also described on estrogen supplementation, number of parabasal cells decrease in postmenopausal women with atrophic vaginitis [14-16]. In symptomatic women 61.2% had vaginal pH above 6 while in asymptomatic only 17.6% had vaginal pH above 6. In women with elevated parabasal cell count, mean pH was 6.8 which is comparable to study done by Shawna Brizzolara et al.; which conclude vaginal pH above 6 significantly correlates with high level of parabasal cells (>20%) . Hustin et al.; also noticed that increasing Uruj Jahan et al., Sch. J. App. Med. Sci., Dec 2016; 4(12C):4354-4357
Available online at 4357
age was associated with progressive decline in maturation value and higher parabasalcell .
In study done by Pinar Yoruk et al.; also, there was highly significant inverse correlation between vaginal pH and VMV (10). Thus vaginal MV is similar to vaginal pH in identification of patients with atrophic vaginitis even in presence of vaginal inflammation.
CONCLUSIONS:
So in the present study we concluded that high vaginal pH is associated with elevated parabasal cell count and low vaginal maturation value (VMV) in post menopausal women of atrophic vaginitis.
DECLARATIONS
Funding: no funding sources
Conflict of interest: none declared
REFERENCES
Berek and Novak’s Gynecology 14 th edition, Menopause 1325.
Bachmann G, Eberg GA, Burd ID. Vulvovaginal complaints. In: Lobo RA, editor. Treatment of the postmenopausal Woman: Basic and Clinical Aspects. Philadelphia PA: Lippincott Williams and Wilkins.1999: 195-201.
Mac Bride MB, Rhodes DJ, Shuster LT. Vulvovaginal atrophy. Mayo ClinProc 2010; 85(1): 87-94.
Levine KB, Williams RE, Hartmann KE. Vulvovaginal atrophy is strongly associated with female sexual dysfunction among sexually active postmenopausal women. Menopause. 2008 Jul 1; 15(4):661-6.
Dennerstein L, Dudley EC, Hopper JL, Guthrie JR, Burger HG. A prospective population-based study of menopausal symptoms. Obstetrics & Gynecology. 2000 Aug 23; 96(3):351-8.
Goswami PK, Samant M, Srivastava R, Khale A. Atrophic vaginitis. International Research Journal of Pharmacy 2013, 4(11): 17 – 19.
Milsom I, Arvidsson L, Ekelund P, Molander U, Eriksson O. Factors influencing vaginal cytology, pH and bacterial flora in elderly women. Acta obstetricia et gynecologica Scandinavica. 1993 Jan 1; 72(4):286-91.
Roy S, Caillouette JC, Roy T, Faden JS. Vaginal pH is similar to follicle-stimulating hormone for menopause diagnosis. American journal of obstetrics and gynecology. 2004 May 31; 190(5):1272-7.
Meisels A. The maturation value. Acta cytologica. 1966 Dec; 11(4):249-.
Yoruk P., Uygur M, Erenus M., Eren F: The Role of vaginal maturation value assessment in prediction of vaginal pH, serum FSH and E 2
level: Marmara Medical Journal 2006; 19(2): 52 – 57.
Carranza-Lira S, MacGregor-Gooch AL. Differences in vaginal dryness according to hormone therapy schedule using pH test strip in two groups of postmenopausal women. International journal of fertility and women's medicine. 2003 Dec; 49(2):88-90.
Davila GW, Karapanagiotou I, Woodhouse S, Singh A, Huber K, Zimberg S, Seiler J. Are women with urogenital atrophy symptomatic? Obstetrics & Gynecology. 2001 Apr 30; 97(4):S48.
Fantl JA, Bump RC, Robinson D, McCLISH DK, Wyman JF, Continence Program for Women Research Group. Efficacy of estrogen supplementation in the treatment of urinary incontinence. Obstetrics & Gynecology. 1996 Nov 30; 88(5):745-9.
Van der Linden MC, Gerretsen G, Brandhorst MS, Ooms EC, Kremer CM, Doesburg WH. The effect of estriol on the cytology of urethra and vagina in postmenopausal women with genito-urinary symptoms. European Journal of Obstetrics & Gynecology and Reproductive Biology. 1993 Sep 30; 51(1):29-33.
Notelovitz M. Estrogen therapy in the management of problems associated with urogenital ageing: a simple diagnostic test and the effect of the route of hormone administration. Maturitas. 1995 Dec 31; 22:S31-3.
Sartori MG, Baracat EC, Girao MJ, Goncalves WJ, Sartori JP, de Lima GR. Menopausal genuine stress urinary incontinence treated with conjugated estrogens plus progestogens. International Journal of Gynecology &Obstetrics. 1995 May 31; 49(2):165-9.
Brizzolara S, Killeen J, Severino R. Vaginal pH and parabasal cells in postmenopausal women. Obstetrics & Gynecology. 1999 Nov 1; 94(5, Part 1):700-3.
Hustin J, Van den Eynde JP. Cytologic evaluation of the effect of various estrogens given in postmenopause. Acta cytologica. 1977; 21(2):225. |
1177 | https://en.wikipedia.org/wiki/Alexander_Fleming | Jump to content
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Contents
1 Early life and education
2 Scientific contributions
2.1 Antiseptics
2.2 Discovery of lysozyme
2.3 Discovery of penicillin
2.3.1 Experiment
2.3.2 Reception and publication
2.3.3 Purification and stabilisation
2.3.4 Medical use and mass production
2.3.5 Antibiotic resistance
3 Personal life
4 Death
5 Awards and legacy
6 Myths
6.1 The Fleming myth
6.2 The Churchills
7 See also
8 References
9 Further reading
10 External links
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From Wikipedia, the free encyclopedia
Scottish physician and microbiologist (1881–1955)
For other people named Alexander Fleming, see Alexander Fleming (disambiguation).
| Sir Alexander Fleming FRS FRSE FRCS |
| Fleming in his laboratory, c. 1943 |
| Born | (1881-08-06)6 August 1881 Darvel, Ayrshire, Scotland |
| Died | 11 March 1955(1955-03-11) (aged 73) London, England |
| Resting place | St Paul's Cathedral |
| Alma mater | Royal Polytechnic Institution St Mary's Hospital Medical School |
| Known for | Discovery of penicillin and lysozyme |
| Spouses | Sarah Marion McElroy (m. 1915; died 1949) Amalia Koutsouri-Vourekas (m. 1953) |
| Awards | Nobel Prize in Physiology or Medicine (1945) Cameron Prize for Therapeutics of the University of Edinburgh (1945) Actonian Prize (1949) |
| Scientific career |
| Fields | Bacteriology immunology |
| Institutions | St Mary's Hospital, London |
| |
| Signature |
Sir Alexander Fleming FRS FRSE FRCS (6 August 1881 – 11 March 1955) was a Scottish physician and microbiologist, best known for discovering the world's first broadly effective antibiotic substance, which he named penicillin. His discovery in 1928 of what was later named benzylpenicillin (or penicillin G) from the mould Penicillium rubens has been described as the "single greatest victory ever achieved over disease". For this discovery, he shared the Nobel Prize in Physiology or Medicine in 1945 with Howard Florey and Ernst Chain.
He also discovered the enzyme lysozyme from his nasal discharge in 1922, and along with it a bacterium he named Micrococcus lysodeikticus, later renamed Micrococcus luteus.
Fleming was knighted for his scientific achievements in 1944. In 1999, he was named in Time magazine's list of the 100 Most Important People of the 20th century. In 2002, he was chosen in the BBC's television poll for determining the 100 Greatest Britons, and in 2009, he was also voted third "greatest Scot" in an opinion poll conducted by STV, behind only Robert Burns and William Wallace.
Early life and education
Born on 6 August 1881 at Lochfield farm near Darvel, in Ayrshire, Scotland, Alexander Fleming was the third of four children of farmer Hugh Fleming and Grace Stirling Morton, the daughter of a neighbouring farmer. Hugh Fleming had four surviving children from his first marriage. He was 59 at the time of his second marriage to Grace, and died when Alexander was seven.
Fleming went to Loudoun Moor School and Darvel School, and earned a two-year scholarship to Kilmarnock Academy before moving to London, where he attended the Royal Polytechnic Institution. After working in a shipping office for four years, the twenty-year-old Alexander Fleming inherited some money from an uncle, John Fleming. His elder brother, Tom, was already a physician and suggested to him that he should follow the same career, and so in 1903, the younger Alexander enrolled at St Mary's Hospital Medical School in Paddington (now part of Imperial College London); he qualified with an MBBS degree from the school with distinction in 1906.
Fleming, who was a private in the London Scottish Regiment of the Volunteer Force from 1900 to 1914, had been a member of the rifle club at the medical school. The captain of the club, wishing to retain Fleming in the team, suggested that he join the research department at St Mary's, where he became assistant bacteriologist to Sir Almroth Wright, a pioneer in vaccine therapy and immunology. In 1908, he gained a BSc degree with gold medal in bacteriology, and became a lecturer at St Mary's until 1914.
Commissioned lieutenant in 1914 and promoted captain in 1917, Fleming served throughout World War I in the Royal Army Medical Corps, and was Mentioned in Dispatches. He and many of his colleagues worked in battlefield hospitals at the Western Front in France.
In 1918 he returned to St Mary's Hospital, where he was elected Professor of Bacteriology of the University of London in 1928. In 1951 he was elected the Rector of the University of Edinburgh for a term of three years.
Scientific contributions
Antiseptics
During World War I, Fleming with Leonard Colebrook and Sir Almroth Wright joined the war efforts and practically moved the entire Inoculation Department of St Mary's to the British military hospital at Boulogne-sur-Mer. Serving as a temporary lieutenant of the Royal Army Medical Corps, he witnessed the death of many soldiers from sepsis resulting from infected wounds. Antiseptics, which were used at the time to treat infected wounds, he observed, often worsened the injuries. In an article published in the medical journal The Lancet in 1917, he described an ingenious experiment, which he was able to conduct as a result of his own glassblowing skills, in which he explained why antiseptics were killing more soldiers than infection itself during the war. Antiseptics worked well on the surface, but deep wounds tended to shelter anaerobic bacteria from the antiseptic agent, and antiseptics seemed to remove beneficial agents produced that protected the patients in these cases at least as well as they removed bacteria, and did nothing to remove the bacteria that were out of reach. Wright strongly supported Fleming's findings, but despite this, most army physicians over the course of the war continued to use antiseptics even in cases where this worsened the condition of the patients.
Discovery of lysozyme
At St Mary's Hospital, Fleming continued his investigations into bacteria culture and antibacterial substances. As his research scholar at the time V. D. Allison recalled, Fleming was not a tidy researcher and usually expected unusual bacterial growths in his culture plates. Fleming had teased Allison of his "excessive tidiness in the laboratory", and Allison rightly attributed such untidiness as the success of Fleming's experiments, and said, "[If] he had been as tidy as he thought I was, he would not have made his two great discoveries."
In late 1921, while Fleming was maintaining agar plates for bacteria, he found that one of the plates was contaminated with bacteria from the air. When he added nasal mucus, he found that the mucus inhibited the bacterial growth. Surrounding the mucus area was a clear transparent circle (1 cm from the mucus), indicating the killing zone of bacteria, followed by a glassy and translucent ring beyond which was an opaque area indicating normal bacterial growth. In the next test, he used bacteria maintained in saline that formed a yellow suspension. Within two minutes of adding fresh mucus, the yellow saline turned completely clear. He extended his tests using tears, which were contributed by his co-workers. As Allison reminisced, saying, "For the next five or six weeks, our tears were the source of supply for this extraordinary phenomenon. Many were the lemons we used (after the failure of onions) to produce a flow of tears... The demand by us for tears was so great, that laboratory attendants were pressed into service, receiving threepence for each contribution."
His further tests with sputum, cartilage, blood, semen, ovarian cyst fluid, pus, and egg white showed that the bactericidal agent was present in all of these. He reported his discovery before the Medical Research Club in December and before the Royal Society the next year but failed to stir any interest, as Allison recollected:
I was present at this [Medical Research Club] meeting as Fleming's guest. His paper describing his discovery was received with no questions asked and no discussion, which was most unusual and an indication that it was considered to be of no importance. The following year he read a paper on the subject before the Royal Society, Burlington House, Piccadilly and he and I gave a demonstration of our work. Again with one exception little comment or attention was paid to it.
Reporting in the 1 May 1922 issue of the Proceedings of the Royal Society B: Biological Sciences under the title "On a remarkable bacteriolytic element found in tissues and secretions", Fleming wrote:
In this communication I wish to draw attention to a substance present in the tissues and secretions of the body, which is capable of rapidly dissolving certain bacteria. As this substance has properties akin to those of ferments I have called it a "Lysozyme", and shall refer to it by this name throughout the communication. The lysozyme was first noticed during some investigations made on a patient suffering from acute coryza.
This was the first recorded discovery of lysozyme. With Allison, he published further studies on lysozyme in October issue of the British Journal of Experimental Pathology the same year. Although he was able to obtain larger amounts of lysozyme from egg whites, the enzyme was only effective against small counts of harmless bacteria, and therefore had little therapeutic potential. This indicates one of the major differences between pathogenic and harmless bacteria. Described in the original publication, "a patient suffering from acute coryza" was later identified as Fleming himself. His research notebook dated 21 November 1921 showed a sketch of the culture plate with a small note: "Staphyloid coccus from A.F.'s nose." He also identified the bacterium present in the nasal mucus as Micrococcus Lysodeikticus, giving the species name (meaning "lysis indicator" for its susceptibility to lysozymal activity). The species was reassigned as Micrococcus luteus in 1972. The "Fleming strain" (NCTC2665) of this bacterium has become a model in different biological studies. The importance of lysozyme was not recognised, and Fleming was well aware of this, in his presidential address at the Royal Society of Medicine meeting on 18 October 1932, he said:
I choose lysozyme as the subject for this address for two reasons, firstly because I have a fatherly interest in the name, and, secondly, because its importance in connection with natural immunity does not seem to be generally appreciated.
In his Nobel lecture on 11 December 1945, he briefly mentioned lysozyme, saying, "Penicillin was not the first antibiotic I happened to discover." It was only towards the end of the 20th century that the true importance of Fleming's discovery in immunology was realised as lysozyme became the first antimicrobial protein discovered that constitute part of our innate immunity.
Discovery of penicillin
Main article: History of penicillin
One sometimes finds what one is not looking for. When I woke up just after dawn on September 28, 1928, I certainly didn't plan to revolutionize all medicine by discovering the world's first antibiotic, or bacteria killer. But I suppose that was exactly what I did.
— Alexander Fleming
Experiment
By 1927, Fleming had been investigating the properties of staphylococci. He was already well known from his earlier work, and had developed a reputation as a brilliant researcher. In 1928, he studied the variation of Staphylococcus aureus grown under natural condition, after the work of Joseph Warwick Bigger, who discovered that the bacterium could grow into a variety of types (strains). On 3 September 1928, Fleming returned to his laboratory having spent a holiday with his family at Suffolk. Before leaving for his holiday, he inoculated staphylococci on culture plates and left them on a bench in a corner of his laboratory. On his return, Fleming noticed that one culture was contaminated with a fungus, and that the colonies of staphylococci immediately surrounding the fungus had been destroyed, whereas other staphylococci colonies farther away were normal, famously remarking "That's funny". Fleming showed the contaminated culture to his former assistant Merlin Pryce, who reminded him, "That's how you discovered lysozyme."[page needed] He identified the mould as being from the genus Penicillium. He suspected it to be P. chrysogenum, but a colleague Charles J. La Touche identified it as P. rubrum. (It was later corrected as P. notatum and then officially accepted as P. chrysogenum; in 2011, it was resolved as P. rubens.)
The laboratory in which Fleming discovered and tested penicillin is preserved as the Alexander Fleming Laboratory Museum in St. Mary's Hospital, Paddington. The source of the fungal contaminant was established in 1966 as coming from La Touche's room, which was directly below Fleming's.
Fleming grew the mould in a pure culture and found that the culture broth contained an antibacterial substance. He investigated its anti-bacterial effect on many organisms, and noticed that it affected bacteria such as staphylococci and many other Gram-positive pathogens that cause scarlet fever, pneumonia, meningitis and diphtheria, but not typhoid fever or paratyphoid fever, which are caused by Gram-negative bacteria, for which he was seeking a cure at the time. It also affected Neisseria gonorrhoeae, which causes gonorrhoea, although this bacterium is Gram-negative. After some months of calling it "mould juice" or "the inhibitor", he gave the name penicillin on 7 March 1929 for the antibacterial substance present in the mould.
Reception and publication
Fleming presented his discovery on 13 February 1929 before the Medical Research Club. His talk on "A medium for the isolation of Pfeiffer's bacillus" did not receive any particular attention or comment. Henry Dale, the then Director of National Institute for Medical Research and chair of the meeting, much later reminisced that he did not even sense any striking point of importance in Fleming's speech. Fleming published his discovery in 1929 in the British Journal of Experimental Pathology, but little attention was paid to the article. His problem was the difficulty of producing penicillin in large amounts, and moreover, isolation of the main compound. Even with the help of Harold Raistrick and his team of biochemists at the London School of Hygiene & Tropical Medicine, chemical purification was futile. "As a result, penicillin languished largely forgotten in the 1930s", as Milton Wainwright described.
As late as in 1936, there was no appreciation for penicillin. When Fleming talked of its medical importance at the Second International Congress of Microbiology held in London, no one believed him. As Allison, his companion in both the Medical Research Club and international congress meeting, remarked the two occasions:
[Fleming at the Medical Research Club meeting] suggested the possible value of penicillin for the treatment of infection in man. Again there was a total lack of interest and no discussion. Fleming was keenly disappointed, but worse was to follow. He read a paper on his work on penicillin at a meeting of the International Congress of Microbiology, attended by the foremost bacteriologists from all over the world. There was no support for his views on its possible future value for the prevention and treatment of human infections and discussion was minimal. Fleming bore these disappointments stoically, but they did not alter his views or deter him from continuing his investigation of penicillin.
In 1941, the British Medical Journal reported that "[Penicillin] does not appear to have been considered as possibly useful from any other point of view."
Purification and stabilisation
In Oxford, Ernst Chain and Edward Abraham were studying the molecular structure of the antibiotic. Abraham was the first to propose the correct structure of penicillin. Shortly after the team published its first results in 1940, Fleming telephoned Howard Florey, Chain's head of department, to say that he would be visiting within the next few days. When Chain heard that Fleming was coming, he remarked "Good God! I thought he was dead."
Norman Heatley suggested transferring the active ingredient of penicillin back into water by changing its acidity. This produced enough of the drug to begin testing on animals. There were many more people involved in the Oxford team, and at one point the entire Sir William Dunn School of Pathology was involved in its production. After the team had developed a method of purifying penicillin to an effective first stable form in 1940, several clinical trials ensued, and their amazing success inspired the team to develop methods for mass production and mass distribution in 1945.
Fleming was modest about his part in the development of penicillin, describing his fame as the "Fleming Myth" and he praised Florey and Chain for transforming the laboratory curiosity into a practical drug. Fleming was the first to discover the properties of the active substance, giving him the privilege of naming it: penicillin. He also kept, grew, and distributed the original mould for twelve years, and continued until 1940 to try to get help from any chemist who had enough skill to make penicillin. Sir Henry Harris summed up the process in 1998 as: "Without Fleming, no Chain; without Chain, no Florey; without Florey, no Heatley; without Heatley, no penicillin." The discovery of penicillin and its subsequent development as a prescription drug mark the start of modern antibiotics.
Medical use and mass production
In his first clinical trial, Fleming treated his research scholar Stuart Craddock who had developed severe infection of the nasal antrum (sinusitis). The treatment started on 9 January 1929 but without any effect. It probably was due to the fact that the infection was with influenza bacillus (Haemophilus influenzae), the bacterium which he had found unsusceptible to penicillin. Fleming gave some of his original penicillin samples to his colleague-surgeon Arthur Dickson Wright for clinical test in 1928. Although Wright reportedly said that it "seemed to work satisfactorily", there are no records of its specific use. Cecil George Paine, a pathologist at the Royal Infirmary in Sheffield and former student of Fleming, was the first to use penicillin successfully for medical treatment. He cured eye infections (conjunctivitis) of one adult and three infants (neonatal conjunctivitis) on 25 November 1930.
Fleming also successfully treated severe conjunctivitis in 1932. Keith Bernard Rogers, who had joined St Mary's as medical student in 1929, was captain of the London University rifle team and was about to participate in an inter-hospital rifle shooting competition when he developed conjunctivitis. Fleming applied his penicillin and cured Rogers before the competition. It is said that the "penicillin worked and the match was won." However, the report that "Keith was probably the first patient to be treated clinically with penicillin ointment" is no longer true as Paine's medical records showed up.
There is a popular assertion both in popular and scientific literature that Fleming largely abandoned penicillin work in the early 1930s. In his review of André Maurois's The Life of Sir Alexander Fleming, Discoverer of Penicillin, William L. Kissick went so far as to say that "Fleming had abandoned penicillin in 1932... Although the recipient of many honors and the author of much scientific work, Sir Alexander Fleming does not appear to be an ideal subject for a biography." This is false, as Fleming continued to pursue penicillin research. As late as in 1939, Fleming's notebook shows attempts to make better penicillin production using different media. In 1941, he published a method for assessment of penicillin effectiveness. As to the chemical isolation and purification, Howard Florey and Ernst Chain at the Radcliffe Infirmary in Oxford took up the research to mass-produce it, which they achieved with support from World War II military projects under the British and US governments.
By mid-1942, the Oxford team produced the pure penicillin compound as yellow powder. In August 1942, Harry Lambert (an associate of Fleming's brother Robert) was admitted to St Mary's Hospital due to a life-threatening infection of the nervous system (streptococcal meningitis). Fleming treated him with sulphonamides, but Lambert's condition deteriorated. He tested the antibiotic susceptibility and found that his penicillin could kill the bacteria. He requested Florey for the isolated sample. Florey sent the incompletely purified sample, which Fleming immediately administered into Lambert's spinal canal. Lambert showed signs of improvement the very next day, and completely recovered within a week. Fleming published the clinical case in The Lancet in 1943.
Upon this medical breakthrough, Allison informed the British Ministry of Health of the importance of penicillin and the need for mass production. The War Cabinet was convinced of the usefulness upon which Sir Cecil Weir, Director General of Equipment, called for a meeting on the mode of action on 28 September 1942. The Penicillin Committee was created on 5 April 1943. The committee consisted of Weir as chairman, Fleming, Florey, Sir Percival Hartley, Allison and representatives from pharmaceutical companies as members. The main goals were to produce penicillin rapidly in large quantities with collaboration of American companies, and to supply the drug exclusively for Allied armed forces. By D-Day in 1944, enough penicillin had been produced to treat all the wounded of the Allied troops.
Antibiotic resistance
Fleming also discovered very early that bacteria developed antibiotic resistance whenever too little penicillin was used or when it was used for too short a period. Almroth Wright had predicted antibiotic resistance even before it was noticed during experiments. Fleming cautioned about the use of penicillin in his many speeches around the world. On 26 June 1945, he made the following cautionary statements: "the microbes are educated to resist penicillin and a host of penicillin-fast organisms is bred out ... In such cases the thoughtless person playing with penicillin is morally responsible for the death of the man who finally succumbs to infection with the penicillin-resistant organism. I hope this evil can be averted." He cautioned not to use penicillin unless there was a properly diagnosed reason for it to be used, and that if it were used, never to use too little, or for too short a period, since these are the circumstances under which bacterial resistance to antibiotics develops.
It had been experimentally shown in 1942 that S. aureus could develop penicillin resistance under prolonged exposure. Elaborating the possibility of penicillin resistance in clinical conditions in his Nobel Lecture, Fleming said:
The time may come when penicillin can be bought by anyone in the shops. Then there is the danger that the ignorant man may easily underdose himself and by exposing his microbes to non-lethal quantities of the drug make them resistant.
It was around that time that the first clinical case of penicillin resistance was reported.
Personal life
On 24 December 1915, Fleming married a trained nurse, Sarah Marion McElroy of Killala, County Mayo, Ireland. Their only child, Robert Fleming (1924–2015), became a general medical practitioner. After his first wife's death in 1949, Fleming married Amalia Koutsouri-Vourekas, a Greek colleague at St. Mary's, on 9 April 1953; she died in 1986.
Fleming came from a Presbyterian background, while his first wife Sarah was a (lapsed) Roman Catholic. It is said that he was not particularly religious, and their son Robert was later received into the Anglican church, while still reportedly inheriting his two parents' fairly irreligious disposition.
When Fleming learned of Robert D. Coghill and Andrew J. Moyer patenting the method of penicillin production in the United States in 1944, he was furious, and commented:
I found penicillin and have given it free for the benefit of humanity. Why should it become a profit-making monopoly of manufacturers in another country?
From 1921 until his death in 1955, Fleming owned a country home named "The Dhoon" in Barton Mills, Suffolk.
Death
On 11 March 1955, Fleming died at his home in London of a heart attack. His ashes are buried in St Paul's Cathedral.
Awards and legacy
Fleming's discovery of penicillin changed the world of modern medicine by introducing the age of useful antibiotics; penicillin has saved, and is still saving, millions of people around the world.
The laboratory at St Mary's Hospital where Fleming discovered penicillin is home to the Fleming Museum, a popular London attraction. His alma mater, St Mary's Hospital Medical School, merged with Imperial College London in 1988. The Sir Alexander Fleming Building on the South Kensington campus was opened in 1998, where his son Robert and his great-granddaughter Claire were presented to the Queen; it is now one of the main preclinical teaching sites of the Imperial College School of Medicine.
His other alma mater, the Royal Polytechnic Institution (now the University of Westminster) has named one of its student halls of residence Alexander Fleming House, which is near to Old Street.
Fleming, Florey and Chain jointly received the Nobel Prize in Medicine in 1945. According to the rules of the Nobel committee, a maximum of three people may share the prize. Fleming's Nobel Prize medal was acquired by the National Museums of Scotland in 1989 and is on display after the museum re-opened in 2011.
He was a member of the Pontifical Academy of Sciences.
He was elected a Fellow of the Royal Society (FRS) in 1943.
He was awarded the Hunterian Professorship by the Royal College of Surgeons of England.
He was knighted as a Knight Bachelor by King George VI in 1944.
He was awarded the Medal for Merit by the President of the United States.
He was made a Grand Cross of the Legion of Honour by the French Republic.
He was made a Grand Cross of the Order of the Phoenix of Greece.
He was made a Knight Grand Cross of the Order of Alfonso X the Wise (Spain) in 1948.
In 1999, Time magazine named Fleming one of the 100 Most Important People of the 20th century, stating:
It was a discovery that would change the course of history. The active ingredient in that mould, which Fleming named penicillin, turned out to be an infection-fighting agent of enormous potency. When it was finally recognized for what it was, the most efficacious life-saving drug in the world, penicillin would alter forever the treatment of bacterial infections. By the middle of the century, Fleming's discovery had spawned a huge pharmaceutical industry, churning out synthetic penicillins that would conquer some of mankind's most ancient scourges, including syphilis, gangrene and tuberculosis.
The importance of his work was recognized by the placement of an International Historic Chemical Landmark plaque at the Alexander Fleming Laboratory Museum in London on 19 November 1999.
When 2000 was approaching, at least three large Swedish magazines ranked penicillin as the most important discovery of the millennium.
In 2002, Fleming was named in the BBC's list of the 100 Greatest Britons following a nationwide vote.
A statue of him stands outside the main bullring in Madrid, Plaza de Toros de Las Ventas. It was erected by subscription from grateful matadors, as penicillin greatly reduced the number of deaths in the bullring.
Flemingovo náměstí is a square named after Fleming in the university area of the Dejvice community in Prague.
A secondary school is named after him in Sofia, Bulgaria.
In Athens, a small square in the downtown district of Votanikos is named after Fleming and bears his bust. There are also a number of streets in greater Athens and other towns in Greece named after either Fleming or his Greek second wife Amalia.
In mid-2009, he was commemorated on a new series of banknotes issued by the Clydesdale Bank; his image appears on the new issue of £5 notes.
In 2009, Fleming was voted third greatest Scot in an opinion poll conducted by STV, behind only Scotland's national poet Robert Burns and national hero William Wallace.
91006 Fleming, an asteroid in the Asteroid Belt, is named after him.
Fleming metro station, on the Thessaloniki Metro system, takes its name from Fleming Street on which it is located.
Sir Alexander Fleming College, a British school in Trujillo, northern Peru
He and Howard Florey were jointly awarded the Cameron Prize for Therapeutics of the University of Edinburgh in 1945.
Rue Alexander Fleming in the borough of Saint-Laurent in Montreal is named in his honour.
The Fleming crater on the moon is named after him and the Scottish astronomer Williamina Fleming.
Mount Fleming in New Zealand's Paparoa Range was named after him in 1970 by the Department of Scientific and Industrial Research.
Biomedical Sciences Research Center "Alexander Fleming", a research organization in Greece established in the vision of his wife Amalia Fleming.
In March 2025, a gable-end mural featuring a portrait of Fleming was unveiled in the centre of Darvel. It is the work of Glasgow-based artist Rogue One.
Myths
The Fleming myth
By 1942, penicillin, produced as pure compound, was still in short supply and not available for clinical use. When Fleming used the first few samples prepared by the Oxford team to treat Harry Lambert who had streptococcal meningitis, the successful treatment was major news, particularly popularised in The Times. Wright was surprised to discover that Fleming and the Oxford team had not been mentioned, though Oxford was attributed as the source of the drug. Wright wrote to the editor of The Times, which eagerly interviewed Fleming, but Florey prohibited the Oxford team from seeking media coverage. As a consequence, only Fleming was widely publicised in the media, which led to the misconception that he was entirely responsible for the discovery and development of the drug. Fleming himself referred to this incident as "the Fleming myth."
The Churchills
The popular story of Winston Churchill's father paying for Fleming's education after Fleming's father saved young Winston from death is false. According to the biography, Penicillin Man: Alexander Fleming and the Antibiotic Revolution by Kevin Brown, Alexander Fleming, in a letter to his friend and colleague Andre Gratia, described this as "A wondrous fable." Nor did he save Winston Churchill himself during World War II. Churchill was saved by Lord Moran, using sulphonamides, since he had no experience with penicillin, when Churchill fell ill in Carthage in Tunisia in 1943. The Daily Telegraph and The Morning Post on 21 December 1943 wrote that he had been saved by penicillin. He was saved by the new sulphonamide drug sulphapyridine, known at the time under the research code M&B 693, discovered and produced by May & Baker Ltd, Dagenham, Essex – a subsidiary of the French group Rhône-Poulenc. In a subsequent radio broadcast, Churchill referred to the new drug as "This admirable M&B".
See also
Fleming Prize Lecture
People on Scottish banknotes
References
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^ in October 1943 Abraham proposed a molecular structure which included a cyclic formation containing three carbon atoms and one nitrogen atom, the β-lactam ring, not then known in natural products. This structure was not immediately published due to the restrictions of wartime secrecy, and was initially strongly disputed, by Sir Robert Robinson among others, but it was finally confirmed in 1945 by Dorothy Crowfoot Hodgkin using X-ray analysis." Oxford Dictionary of National Biography; "Abraham, Sir Edward Penley"
^ Lowe, Gordon (13 May 1999). "Obituary: Sir Edward Abraham". The Independent. London. Archived from the original on 4 October 2013.
^ Yanes, Javier (6 August 2018). "Fleming and the Difficult Beginnings of Penicillin: Myth and Reality". OpenMind. Retrieved 7 June 2020.
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^ "Norman Heatley". The Independent. London. 23 January 2004. Archived from the original on 16 November 2020. Retrieved 7 June 2020.
^ Henry Harris, Howard Florey and the development of penicillin, a lecture given on 29 September 1998, at the Florey Centenary, 1898–1998, Sir William Dunn School of Pathology, Oxford University (sound recording)
^ Conly, J. M.; Johnston, B. L. (2005). "Where are all the new antibiotics? The new antibiotic paradox". Canadian Journal of Infectious Diseases and Medical Microbiology. 16 (3): 159–160. doi:10.1155/2005/892058. PMC 2095020. PMID 18159536.
^ Wainwright, M.; Swan, H. T. (1987). "The Sheffield penicillin story". Mycologist. 1 (1): 28–30. doi:10.1016/S0269-915X(87)80022-8.
^ a b Wainwright, Milton (1990). "Besredka's "antivirus" in relation to Fleming's initial views on the nature of penicillin". Medical History. 34 (1): 79–85. doi:10.1017/S0025727300050286. PMC 1036002. PMID 2405221.
^ Wainwright, M. (1987). "The history of the therapeutic use of crude penicillin". Medical History. 31 (1): 41–50. doi:10.1017/s0025727300046305. PMC 1139683. PMID 3543562.
^ Wainwright, M.; Swan, H. T. (January 1986). "C.G. Paine and the earliest surviving clinical records of penicillin therapy". Medical History. 30 (1): 42–56. doi:10.1017/S0025727300045026. PMC 1139580. PMID 3511336.
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^ Heaman, Elsbeth A. (2003). St Mary's: The History of a London Teaching Hospital. Montreal, Que.: McGill-Queen's University Press. p. 212. ISBN 978-0-7735-7086-3. OCLC 144085272.
^ Marko, Vladimir (2020). "Penicillin". From Aspirin to Viagra: Stories of the Drugs that Changed the World. Springer. pp. 105–106. doi:10.1007/978-3-030-44286-6_5. ISBN 978-3-030-44286-6. OCLC 1164582807. S2CID 241636139.
^ a b Rossiter, Peter (2005). "Keith Bernard Rogers". The BMJ. 331 (7516): 579. doi:10.1136/bmj.331.7516.579-c. PMC 1200632.
^ Maurois, André (1963). The Life of Sir Alexander Fleming: Discoverer of Penicillin. Penguin Books. p. 156.ISBN 1-199-30814-5
^ Aronson, J. K. (1992). "Penicillin". European Journal of Clinical Pharmacology. 42 (1): 1–9. doi:10.1007/BF00314911. PMID 1541305. S2CID 62877498.
^ Kyle, Robert A.; Steensma, David P.; Shampo, Marc A. (2015). "Howard Walter Florey – Production of Penicillin". Mayo Clinic Proceedings. 90 (6): e63–64. doi:10.1016/j.mayocp.2014.12.028. PMID 26046419.
^ Shama, Gilbert (2017). "Miracle near 34th street: Wartime Penicillin Research at St John's University, NY". Endeavour. 41 (4): 217–220. doi:10.1016/j.endeavour.2017.09.003. PMID 29055651.
^ Morin, Robert B.; Gorman, Marvin (2014). Penicillins and Cephalosporins. Academic Press. pp. xxii. ISBN 978-1-4832-7719-6.
^ Ward, John W.; Warren, Christian (2006). Silent Victories: The History and Practice of Public Health in Twentieth-Century America. Oxford University Press. p. 50. ISBN 978-0-19-974798-6.
^ Kissick, William L. (1959). "The Life of Sir Alexander Fleming, Discoverer of Penicillin". The Yale Journal of Biology and Medicine. 32 (2): 140. PMC 2604061.
^ Wainwright, Milton (2002). "Fleming's unfinished". Perspectives in Biology and Medicine. 45 (4): 529–538. doi:10.1353/pbm.2002.0065. PMID 12388885. S2CID 32684352.
^ Fleming, A. (1942). "In-vitro Tests of Penicillin Potency". The Lancet. 239 (6199): 732–733. doi:10.1016/S0140-6736(00)70368-0.
^ Bickel, L. Florey: The Man Who Made Penicillin, Sun Books, Melbourne, 1972.
^ Abraham, E. P.; Chain, E.; Holiday, E. R. (1942). "Purification and Some Physical and Chemical Properties of Penicillin". British Journal of Experimental Pathology. 23 (3): 103–119. PMC 2065494.
^ Cairns, H.; Lewin, W. S.; Duthie, E. S.; Smith, Honor V. (1944). "Pneumococcal Meningitis Treated with Penicillin". The Lancet. 243 (6299): 655–659. doi:10.1016/S0140-6736(00)77085-1.
^ Fleming, Alexander (1943). "Streptococcal Meningitis treated With Penicillin". The Lancet. 242 (6267): 434–438. doi:10.1016/S0140-6736(00)87452-8.
^ Mathews, John A. (2008). "The Birth of the Biotechnology Era: Penicillin in Australia, 1943–80". Prometheus. 26 (4): 317–333. doi:10.1080/08109020802459306. S2CID 143123783.
^ Baldry, Peter (1976). The Battle Against Bacteria: A Fresh Look. CUP Archive. p. 115. ISBN 978-0-521-21268-7.
^ Richards, A. N. (1964). "Production of penicillin in the United States (1941–1946)". Nature. 201 (4918): 441–445. Bibcode:1964Natur.201..441R. doi:10.1038/201441a0. PMID 14164615. S2CID 4296757.
^ Fishman, Neil; Infectious Diseases Society of America; Pediatric Infectious Diseases Society (April 2012). "Policy statement on antimicrobial stewardship by the Society for Healthcare Epidemiology of America (SHEA), the Infectious Diseases Society of America (IDSA), & the Pediatric Infectious Diseases Society (PIDS)" (PDF). Infection Control & Hospital Epidemiology. 33 (4): 322–327. doi:10.1086/665010. PMID 22418625. S2CID 24828623. Archived (PDF) from the original on 9 October 2022.
^ Rosenblatt-Farrell, Noah (2009). "The Landscape of Antibiotic Resistance". Environmental Health Perspectives. 117 (6): 244–150. doi:10.1289/ehp.117-a244. PMC 2702430. PMID 19590668.
^ Rammelkamp, Charles H.; Maxon, Thelma (1942). "Resistance of Staphylococcus aureus to the Action of Penicillin". Proceedings of the Society for Experimental Biology and Medicine. 51 (3): 386–389. doi:10.3181/00379727-51-13986. S2CID 87530495.
^ Plough, Harold H. (1945). "Penicillin Resistance of Staphylococcus Aureus and its Clinical Implications". American Journal of Clinical Pathology. 15 (10): 446–451. doi:10.1093/ajcp/15.10.446. PMID 21005048.
^ Biographical Index of Former Fellows of the Royal Society of Edinburgh 1783–2002 (PDF). The Royal Society of Edinburgh. July 2006. ISBN 0-902198-84-X. Archived from the original (PDF) on 24 January 2013. Retrieved 9 May 2016.
^ Penicillin Man: Alexander Fleming and the Antibiotic Revolution. The History Press. September 2005. ISBN 0-7509-3153-1.
^ US 2423873, Coghill, Robert D. & Moyer, Andrew J., "Method for production of increased yields of penicillin", published 15 July 1947
^ BartonMills.net local history Archived 28 September 2022 at the Wayback Machine. Retrieved 17 October 2016.
^ Roberts, Michael; Ingram, Neil (2001). Biology (2nd, illustrated ed.). Nelson Thornes. p. 105. ISBN 0-7487-6238-8. Retrieved 4 March 2012. Penicillin is just one of a very large number of drugs which today are used by doctors to treat people with diseases.
^ "100,000 visitors in 6 days". National Museums Scotland. 3 August 2011. Archived from the original on 23 February 2012. Retrieved 4 March 2012.
^ "No. 36544". The London Gazette (Supplement). 2 June 1944. p. 2566.
^ "People of the century". P. 78. CBS News. Simon & Schuster, 1999
^ Santesmases, María Jesús (18 December 2017). The Circulation of Penicillin in Spain: Health, Wealth and Authority. Springer. p. 39. ISBN 978-3-319-69718-5. Retrieved 7 July 2020.
^ "Alexander Fleming – Time 100 People of the Century". Time. 29 March 1999. Archived from the original on 16 October 2007.
^ "Discovery and Development of Penicillin". International Historic Chemical Landmarks. American Chemical Society. Retrieved 21 August 2018.
^ "Great Britons – Top 100". BBC. Archived from the original on 4 December 2002. Retrieved 19 July 2017.
^ a b Lewine, Edward (2007). Death and the Sun: A Matador's Season in the Heart of Spain. Houghton Mifflin Harcourt. p. 123.
^ "Banknote designs mark Homecoming". BBC News. 14 January 2008. Archived from the original on 25 January 2009. Retrieved 20 January 2009.
^ "Robert Burns voted Greatest Scot". STV Group. 30 November 2009. Archived from the original on 6 November 2018. Retrieved 7 February 2016.
^ "Place name detail: Mount Fleming". New Zealand Gazetteer. Land Information New Zealand. Retrieved 21 August 2022.
^ Toom, Sarah (7 April 2025). "Mural renews town's love of 'hero' Alexander Fleming". BBC News. Retrieved 7 April 2025.
^ Gaynes, Robert (2017). "The Discovery of Penicillin – New Insights After More Than 75 Years of Clinical Use". Emerging Infectious Diseases. 23 (5): 849–853. doi:10.3201/eid2305.161556. PMC 5403050.
^ a b Dufour, Héloïse D.; Carroll, Sean B. (2013). "History: Great myths die hard". Nature. 502 (7469): 32–33. doi:10.1038/502032a. PMID 24137644.
^ Ho, David (29 March 1999). "Bacteriologist Alexander Fleming". Time. ISSN 0040-781X. Retrieved 17 October 2020.
^ Selwyn, Sydney (1980). "Howard Florey: the making of a great scientist". Journal of Medical Microbiology. 13 (3): 483. doi:10.1099/00222615-13-3-483.
^ e.g., The Philadelphia Inquirer, 17 July 1945: Brown, Penicillin Man, note 43 to Chapter 2
^ 14 November 1945; British Library Additional Manuscripts 56115: Brown, Penicillin Man, note 44 to Chapter 2
^ see Wikipedia Discovery of penicillin article entry for 1920
^ a b Lesch JE (2007). "Chapter 7: M&B 693". The first miracle drugs: how the sulfa drugs transformed medicine (PDF). Oxford: Oxford University Press. pp. 158–159. ISBN 978-0-19-518775-5. Archived from the original (PDF) on 21 July 2019.
^ A History of May & Baker 1834–1984, Alden Press 1984.[ISBN missing]
Further reading
The Life Of Sir Alexander Fleming, Jonathan Cape, 1959. Maurois, André.
Nobel Lectures, the Physiology or Medicine 1942–1962, Elsevier Publishing Company, Amsterdam, 1964
An Outline History of Medicine. London: Butterworths, 1985. Rhodes, Philip.
The Cambridge Illustrated History of Medicine. Cambridge, England: Cambridge University Press, 1996. Porter, Roy, ed.
Penicillin Man: Alexander Fleming and the Antibiotic Revolution, Stroud, Sutton, 2004. Brown, Kevin.
Alexander Fleming: The Man and the Myth, Oxford University Press, Oxford, 1984. Macfarlane, Gwyn
Fleming, Discoverer of Penicillin, Ludovici, Laurence J., 1952
The Penicillin Man: the Story of Sir Alexander Fleming, Lutterworth Press, 1957, Rowland, John.
External links
Wikiquote has quotations related to Alexander Fleming.
Wikimedia Commons has media related to Alexander Fleming.
Alexander Fleming Obituary
Alexander Fleming on Nobelprize.org including the Nobel Lecture, 11 December 1945 Penicillin
Some places and memories related to Alexander Fleming
Newspaper clippings about Alexander Fleming in the 20th Century Press Archives of the ZBW
| Academic offices |
| Preceded by Alastair Sim | Rector of the University of Edinburgh 1951–1954 | Succeeded by Sydney Smith |
| v t e Laureates of the Nobel Prize in Physiology or Medicine |
| 1901–1925 | 1901: Emil Behring 1902: Ronald Ross 1903: Niels Finsen 1904: Ivan Pavlov 1905: Robert Koch 1906: Camillo Golgi / Santiago Ramón y Cajal 1907: Alphonse Laveran 1908: Élie Metchnikoff / Paul Ehrlich 1909: Emil Kocher 1910: Albrecht Kossel 1911: Allvar Gullstrand 1912: Alexis Carrel 1913: Charles Richet 1914: Róbert Bárány 1915 1916 1917 1918 1919: Jules Bordet 1920: August Krogh 1921 1922: Archibald Hill / Otto Meyerhof 1923: Frederick Banting / John Macleod 1924: Willem Einthoven 1925 |
| 1926–1950 | 1926: Johannes Fibiger 1927: Julius Wagner-Jauregg 1928: Charles Nicolle 1929: Christiaan Eijkman / Frederick Gowland Hopkins 1930: Karl Landsteiner 1931: Otto Warburg 1932: Charles Scott Sherrington / Edgar Adrian 1933: Thomas Morgan 1934: George Whipple / George Minot / William Murphy 1935: Hans Spemann 1936: Henry Dale / Otto Loewi 1937: Albert Szent-Györgyi 1938: Corneille Heymans 1939: Gerhard Domagk 1940 1941 1942 1943: Henrik Dam / Edward Doisy 1944: Joseph Erlanger / Herbert Gasser 1945: Alexander Fleming / Ernst Chain / Howard Florey 1946: Hermann Muller 1947: Carl Cori / Gerty Cori / Bernardo Houssay 1948: Paul Müller 1949: Walter Hess / António Egas Moniz 1950: Edward Kendall / Tadeusz Reichstein / Philip Hench |
| 1951–1975 | 1951: Max Theiler 1952: Selman Waksman 1953: Hans Krebs / Fritz Lipmann 1954: John Enders / Thomas Weller / Frederick Robbins 1955: Hugo Theorell 1956: André Cournand / Werner Forssmann / Dickinson W. Richards 1957: Daniel Bovet 1958: George Beadle / Edward Tatum / Joshua Lederberg 1959: Severo Ochoa / Arthur Kornberg 1960: Frank Burnet / Peter Medawar 1961: Georg von Békésy 1962: Francis Crick / James Watson / Maurice Wilkins 1963: John Eccles / Alan Hodgkin / Andrew Huxley 1964: Konrad Bloch / Feodor Lynen 1965: François Jacob / André Lwoff / Jacques Monod 1966: Francis Rous / Charles B. Huggins 1967: Ragnar Granit / Haldan Hartline / George Wald 1968: Robert W. Holley / Har Khorana / Marshall Nirenberg 1969: Max Delbrück / Alfred Hershey / Salvador Luria 1970: Bernard Katz / Ulf von Euler / Julius Axelrod 1971: Earl Sutherland Jr. 1972: Gerald Edelman / Rodney Porter 1973: Karl von Frisch / Konrad Lorenz / Nikolaas Tinbergen 1974: Albert Claude / Christian de Duve / George Palade 1975: David Baltimore / Renato Dulbecco / Howard Temin |
| 1976–2000 | 1976: Baruch Blumberg / Daniel Gajdusek 1977: Roger Guillemin / Andrew Schally / Rosalyn Yalow 1978: Werner Arber / Daniel Nathans / Hamilton O. Smith 1979: Allan Cormack / Godfrey Hounsfield 1980: Baruj Benacerraf / Jean Dausset / George Snell 1981: Roger Sperry / David H. Hubel / Torsten Wiesel 1982: Sune Bergström / Bengt I. Samuelsson / John Vane 1983: Barbara McClintock 1984: Niels Jerne / Georges Köhler / César Milstein 1985: Michael Brown / Joseph L. Goldstein 1986: Stanley Cohen / Rita Levi-Montalcini 1987: Susumu Tonegawa 1988: James W. Black / Gertrude B. Elion / George H. Hitchings 1989: J. Michael Bishop / Harold E. Varmus 1990: Joseph Murray / E. Donnall Thomas 1991: Erwin Neher / Bert Sakmann 1992: Edmond Fischer / Edwin G. Krebs 1993: Richard J. Roberts / Phillip Sharp 1994: Alfred G. Gilman / Martin Rodbell 1995: Edward B. Lewis / Christiane Nüsslein-Volhard / Eric F. Wieschaus 1996: Peter C. Doherty / Rolf M. Zinkernagel 1997: Stanley B. Prusiner 1998: Robert F. Furchgott / Louis Ignarro / Ferid Murad 1999: Günter Blobel 2000: Arvid Carlsson / Paul Greengard / Eric Kandel |
| 2001–present | 2001: Leland H. Hartwell / Tim Hunt / Paul Nurse 2002: Sydney Brenner / H. Robert Horvitz / John E. Sulston 2003: Paul Lauterbur / Peter Mansfield 2004: Richard Axel / Linda B. Buck 2005: Barry Marshall / Robin Warren 2006: Andrew Fire / Craig Mello 2007: Mario Capecchi / Martin Evans / Oliver Smithies 2008: Harald zur Hausen / Luc Montagnier / Françoise Barré-Sinoussi 2009: Elizabeth Blackburn / Carol W. Greider / Jack W. Szostak 2010: Robert G. Edwards 2011: Bruce Beutler / Jules A. Hoffmann / Ralph M. Steinman (posthumously) 2012: John Gurdon / Shinya Yamanaka 2013: James Rothman / Randy Schekman / Thomas C. Südhof 2014: John O'Keefe / May-Britt Moser / Edvard Moser 2015: William C. Campbell / Satoshi Ōmura / Tu Youyou 2016: Yoshinori Ohsumi 2017: Jeffrey C. Hall / Michael Rosbash / Michael W. Young 2018: James P. Allison / Tasuku Honjo 2019: Gregg L. Semenza / Peter J. Ratcliffe / William Kaelin Jr. 2020: Harvey J. Alter / Michael Houghton / Charles M. Rice 2021: David Julius / Ardem Patapoutian 2022: Svante Pääbo 2023: Katalin Karikó / Drew Weissman 2024: Victor Ambros / Gary Ruvkun |
| v t e 1945 Nobel Prize laureates |
| Chemistry | Artturi Ilmari Virtanen (Finland) |
| Literature (1945) | Gabriela Mistral (Chile) |
| Peace | Cordell Hull (United States) |
| Physics | Wolfgang Pauli (Austria) |
| Physiology or Medicine | Alexander Fleming (Great Britain) Ernst Boris Chain (Great Britain) Howard Florey (Australia) |
| Nobel Prize recipients 1940 1941 1942 1943 1944 1945 1946 1947 1948 1949 1950 |
| v t e Fellows of the Royal Society elected in 1943 |
| Fellows | Shanti Swaroop Bhatnagar Ivan De Burgh Daly Patrick Alfred Buxton John Augustine Edgell Arthur James Ewins Arthur Felix Alexander Fleming John Jacob Fox William Michael Herbert Greaves Sydney Harland George Armand Robert Kon Andrew McCance Wilder Penfield Guy Ellcock Pilgrim Reginald Stradling Charles Sykes John Lighton Synge George Frederick James Temple Alexander du Toit Solly Zuckerman |
| Foreign | Victor Goldschmidt Bernardo Houssay |
| v t e History of infectious diseases |
| Individual diseases | Cholera HIV/AIDS Leprosy Malaria Plague Polio Smallpox Syphilis Tuberculosis Typhoid fever Yellow fever |
| People | Alexander Fleming Maurice Hilleman Edward Jenner Robert Koch Louis Pasteur Jonas Salk Ignaz Semmelweis John Snow |
| Major epidemics | Black Death (1346–1353) Spanish flu (1918–1920) HIV/AIDS pandemic (1981–present) COVID-19 pandemic (2019–present) |
| Other | Discovery of disease-causing pathogens Smallpox eradication Social history of viruses |
| Authority control databases |
| International | ISNI VIAF GND FAST WorldCat |
| National | United States France BnF data Japan Italy Australia Czech Republic Spain Portugal Netherlands Norway Latvia Croatia Chile Greece Korea Sweden Poland Israel Catalonia Belgium |
| Academics | |
| Artists | |
| People | Netherlands Trove Deutsche Biographie DDB |
| Other | IdRef SNAC + 2 Yale LUX |
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1178 | https://www.liveism.com/live-concept.php?q=%E5%88%86%E6%95%B8%E6%9C%80%E7%B0%A1%E6%A0%B9%E5%BC%8F%E8%88%87%E5%88%86%E6%AF%8D%E6%9C%89%E7%90%86%E5%8C%96 | 分數最簡根式與分母有理化 - Live 多媒體數學觀念典 Online
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第二章 平方根與畢氏定理
§2-2 根式的運算
(5) 分數最簡根式與分母有理化
分數最簡根式與分母有理化
分數最簡根式：若根號內為分數或分母含有根號的根式，如： √2 3 2 3 、 4√5 4 5 ，此時都不是最簡根式。將分母化成不帶有根號的根式之過程，就稱為分母有理化或稱根式有理化。
例 將 √2 3 2 3 、 4√5 4 5 化為最簡根式。
解√2 3=√2 3×2 3=√2×3 3×3=√6 3 2=√6√3 2=√6 3 2 3=2 3×2 3=2×3 3×3=6 3 2=6 3 2=6 3
4√5=4√5×√5√5=4×√5√5×√5=4√5(√5)2=4√5 5 4 5=4 5×5 5=4×5 5×5=4 5(5)2=4 5 5
觀念影片
5
#### (5)分數最簡根式與分母有理化 7:02
前往其他章節
（一） 乘法公式與多項式
1-1 乘法公式
1-2 多項式與其加減運算
1-3 多項式的乘除運算
（二） 平方根與畢氏定理
2-1 平方根與近似值
2-2 根式的運算
2-3 畢氏定理(勾股定理)
（三） 因式分解
3-1 因式、倍式與因式分解
3-2 提出公因式與分組分解
3-3 利用乘法公式做因式分解
3-4 利用十字交乘法做因式分解
（四） 一元二次方程式
4-1 因式分解解一元二次方程式
4-2 配方法與公式解
4-3 一元二次方程式應用問題
國中數學第三冊
（二） 平方根與畢氏定理
2-2 根式的運算
(1) 根式的介紹與簡記
(2) 根式的運算規則─交換律、結合律、分配律
(3) 根式的乘除運算與代數式之推導
(4) 根式的化簡與最簡根式
(5) 分數最簡根式與分母有理化
(6) 同類方根與根式的加減運算
(7) 利用平方差公式有理化分母
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1179 | https://www.betterworks.com/magazine/compensation-strategy-examples/ | Performance Management
6 Employee Compensation Strategy Examples to Inspire Yours
By Michelle Gouldsberry
April 22, 2024
9 minute read
Table of Contents
What is a compensation strategy?
Cash
Benefits
Non-monetary compensation
Compensation strategy examples: 6 different ways to approach pay
Market-based compensation
Pay for performance
Skill-based compensation
Profit-sharing
Equity compensation
Cost-of-living adjustments and merit pay
8 elements of a tailored compensation strategy
Alignment with business strategy and values
Assessment and management of compensation ranges
Alignment with talent strategy
Fairness and equity
Transparency
Budgetary constraints
Legal and regulatory considerations
Evolution based on company growth
How to design and implement your compensation strategy
Align pay strategy to business strategy
Conduct market research
Define compensation structures
Design for legal compliance
Set performance evaluation metrics
Develop a benefits package
Communicate and implement your strategy
Monitor, evaluate, and adjust
Elevate your compensation strategy
Share
Table of Contents
What is a compensation strategy?
Cash
Benefits
Non-monetary compensation
Compensation strategy examples: 6 different ways to approach pay
Market-based compensation
Pay for performance
Skill-based compensation
Profit-sharing
Equity compensation
Cost-of-living adjustments and merit pay
8 elements of a tailored compensation strategy
Alignment with business strategy and values
Assessment and management of compensation ranges
Alignment with talent strategy
Fairness and equity
Transparency
Budgetary constraints
Legal and regulatory considerations
Evolution based on company growth
How to design and implement your compensation strategy
Align pay strategy to business strategy
Conduct market research
Define compensation structures
Design for legal compliance
Set performance evaluation metrics
Develop a benefits package
Communicate and implement your strategy
Monitor, evaluate, and adjust
Elevate your compensation strategy
Your compensation strategy is a critical component of your overall HR strategy, strongly influencing your ability to attract, motivate, and retain employees.
More than half of employers (64%) are taking proactive steps to guarantee fair pay practices, according to Payscale’s 2024 Compensation Best Practices report. And in 2024, more employers are rewarding employees for high performance and acquiring competitive skills than in 2023, signaling a shift toward more strategic and intentional compensation practices.
Organizations approach compensation in a wide variety of ways, and learning about different compensation strategy examples can help you choose the right model to support your strategic goals, drive employee satisfaction, and improve overall performance.
What is a compensation strategy?
Your compensation strategy spells out the principles you plan to follow when paying your employees in alignment with the company’s goals, competitive positioning for talent, and overall cultural values. Your strategy will drive a broad range of pay-related decisions, including cash payments, benefits, and nonmonetary compensation.
“Oftentimes, compensation decisions are made ad hoc and over time as a company grows,’ says Caitlin Collins, organizational psychologist and program strategy director at Betterworks. “ There is not a solid framework to guide decisions, but it is important to have rigor around defining compensation bands and the reasoning behind all forms of compensation. Flexibility is also important so that companies can respond to market needs.”
Cash
Cash compensation, including salaries, wages, bonuses, and commissions, is the core of your company’s pay strategy. For most businesses, this represents their single greatest expense, and it’s also the major driver in conversations about pay with individual employees.
Benefits
Employee benefits include health and wellness plans, retirement savings options, and initiatives aimed at promoting a better health and work-life balance, such as free gym memberships. These components add to total compensation and support employee well-being and job satisfaction.
Non-monetary compensation
Compensation extends beyond cash and equivalents. Non-monetary compensation, such as career development opportunities, recognition programs, and a supportive work environment, play a crucial role in employee satisfaction and retention. These aspects can significantly impact an employee’s decision to stay with an organization long-term.
Compensation strategy examples: 6 different ways to approach pay
Businesses adopt diverse strategies to align pay structures with their top-level goals, organizational culture, and the competitive market environment. These strategies not only aim to attract and retain talent but also motivate employees towards higher performance.
Let’s explore six compensation strategy examples, each offering a different approach to structuring pay and benefits to meet the specific needs of both the organization and its workforce.
Market-based compensation
A market-based compensation strategy sets salaries based on what other businesses are paying for similar roles in the same industry and geographic area. This approach ensures competitiveness and fairness, attracting skilled professionals by offering compensation that reflects the rates available elsewhere.
Some employers take it a step further. Paying employees above their market value is a deliberate choice many companies make to attract high-quality talent. For instance, a tech company might regularly analyze salary surveys and industry reports to adjust its pay scales. Company leaders decide to pay employees at the 75th percentile compared to other firms in the region, making it an attractive employer for skilled professionals.
“Some companies pay above market rate as a retention and attraction strategy,” says Caitlin Collins, program strategy director at Betterworks. “Pay is one part of total compensation and companies should consider other perks that would incentivize and meet the diverse needs of employees, such as company stock, generous or unlimited paid time off, generous insurance benefits and 401k matching, and flexible working.”
Pay for performance
A pay-for-performance strategy ties employee compensation directly to performance levels, often through bonuses, commissions, or other financial incentives. This model motivates employees to excel in their roles, as their financial rewards are directly linked to their contributions to the company’s success.
In many cases, pay for performance is role-based, and requires employees in those roles to achieve a high level of performance to offset a low base salary. For example, a sales organization may offer commissions or bonuses to salespeople who exceed their quarterly sales targets, directly linking compensation to individual output and success.
Organizations need to be clear on what they base their decision on,” Caitlin explains. ” What key performance indicators (KPIs) are being measured? Do the top 10% receive these? The system has to be seen as fair and just by employees.”
Skill-based compensation
Skill-based pay rewards employees for the range and depth of their skills and competencies, rather than their job title or position. A skill-based performance strategy encourages ongoing learning and development, as employees are incentivized to acquire new skills that can lead to higher pay.
For instance, a software development firm might increase the salary of developers as they achieve certifications in new programming languages or technologies, incentivizing continuous learning and adaptation to emerging trends.
“With skills as the new currency of work, it’s critical that organizations move beyond the identification and evaluation of skills to embedding skills development into performance plans and empowering managers to coach their employees, “ Caitlin says. “This is where modern performance management systems offer a distinct advantage. Managers and employees work together to identify the skills most useful to the business in the employee’s role, make the development of those skills an integral part of a performance plan, and then provide a framework for ensuring that the skills are properly developed and leveraged by the employee.”
Profit-sharing
Profit-sharing involves distributing a portion of the company’s profits to employees, linking their compensation to the organization’s success. This approach fosters a sense of ownership among employees, aligning their interests with the company’s goals and encouraging teamwork and collaboration.
For example, a manufacturing company might distribute an annual bonus to all employees, calculated as a percentage of the net profits for the year, thereby fostering a sense of ownership and aligning individual efforts with the company’s profitability.
“Part of the goal of profit-sharing — and this is true for merit increases, too — is to enable companies to reinforce their values by rewarding collaborative actions that reflect those values,” Caitlin says.
Equity compensation
Equity compensation offers employees a share of the company’s equity, such as stock options or restricted stock units (RSUs). This long-term incentive plan is often used by startups and growth-stage companies, which often lack the cash flow necessary for standard compensation practices. Additionally, equity compensation aligns employees’ interests with the success of the business and provides the potential for significant financial rewards as the company grows.
Cost-of-living adjustments and merit pay
Combining cost-of-living adjustments (COLA) with merit pay allows businesses to address inflation and individual performance. COLA ensures that employees’ salaries keep pace with inflation, while merit pay rewards exceptional performance, providing a balanced approach to compensation that addresses both external economic factors and internal achievements.
For example, a corporation may automatically adjust salaries annually based on the local cost of living increase, and also provide merit increases to employees who exceed their performance targets, ensuring fair compensation that rewards both loyalty and excellence.
8 elements of a tailored compensation strategy
A comprehensive compensation strategy includes elements that align with the organization’s goals and the expectations of its workforce.
Alignment with business strategy and values
A successful compensation strategy must be in sync with the company’s broader business strategy and core values. This alignment ensures that compensation practices reinforce the company’s mission and objectives, encouraging behaviors that drive the organization forward.
Assessment and management of compensation ranges
Managing pay ranges meticulously ensures that salaries remain competitive and equitable. This vigilance prevents pay compression and ensures that the organization keeps pace with market trends, maintaining its appeal to both current and prospective employees.
Alignment with talent strategy
Compensation is a key driver of strategic talent outcomes. “Identify how your compensation strategy serves your business strategy,” Caitlin says. “Is your business looking to rapidly grow, remain stable, or penetrate a new market and how does your total compensation package support that?”
Identify skill and talent gaps required to achieve tomorrow’s business objectives and incentivize filling those gaps through employee development and/or talent attraction. Consistently and regularly review your compensation and talent strategies to ensure they align with each other and with the changing needs and objectives of the business. This approach affects an organization’s ability to retain top talent, making compensation a key player in the battle for skilled employees.
Fairness and equity
The drive for fairness and equity in compensation is fundamental. “Not only does a compensation strategy need to adhere to local and government laws around equal pay for equal work but it should also be transparent and well understood by employees. When employees view compensation as a ‘black box,’ it creates barriers because they may focus on whether they’re being treated fairly instead of focusing on their potential growth and development.
“Consider whether you are paying people fairly within what the market demands,” Caitlin says. Additionally, this practice supports diversity, equity, and inclusion efforts and builds trust within the workforce, as employees feel confident that their compensation reflects their value to the organization without bias.
Transparency
Transparency in compensation practices fosters an environment of trust and openness. By demystifying compensation, organizations can empower employees with the knowledge of how pay decisions are made, enhancing employee satisfaction and engagement.
Budgetary constraints
When designing your compensation strategy, consider your specific business circumstances. Acknowledging and navigating budgetary constraints is crucial, especially for smaller or growing companies. Take a pragmatic approach to compensation that considers financial limitations to scaling your program and promoting sustainable growth.
Legal and regulatory considerations
Legal and regulatory frameworks can significantly influence compensation strategies, from minimum wage requirements to evolving pay equity and transparency laws. In the state of California, you have to turn in an audit where you show pay equity. Compliance with these legal requirements not only avoids penalties but also reinforces the organization’s commitment to fairness and equity.
Evolution based on company growth
Every business will grow and change over time, and so should your compensation strategy. By adapting compensation practices as your company progresses, you can ensure that the strategy remains relevant and effective in attracting and retaining the talent you need to grow.
How to design and implement your compensation strategy
By following these steps, you can create a compensation strategy that not only attracts the talent you need but also motivates employees to contribute their best towards the company’s success.
Align pay strategy to business strategy
Make sure your pay plan helps your company achieve its big-picture goals. Think about what you want to accomplish as a business, and design your pay system to encourage the behaviors that will get you there. For example, if innovation is key to your business, consider bonuses for creative ideas.
Conduct market research
Look into what other companies in your industry are paying their employees. Use surveys and reports to get an idea of the going rates for different jobs. This will help you set competitive salaries that attract the right talent without overspending.
Define compensation structures
Create clear rules for how much you pay for different jobs within your company. This includes deciding on base salaries, and when and how employees can earn more through bonuses or stock options. Make sure these pay ranges are fair and match the job’s importance and the employee’s skills.
Design for legal compliance
Stay up-to-date on laws about pay and working hours. Follow rules about minimum wage, overtime, and equal pay for equal work. Following the law avoids fines and helps your employees feel they’re treated fairly.
Set performance evaluation metrics
Decide how you’ll measure success in your company, such as sales targets for a salesperson or project deadlines for a developer. Clear goals help employees understand what’s expected of them and how they can grow their paychecks by helping the company succeed.
Develop a benefits package
Think beyond just salaries. Offer benefits like health insurance, retirement plans, college loan repayment assistance, or flexible working hours. These extras can make a big difference in attracting and keeping employees happy and healthy, which is good for your business in the long run.
Communicate and implement your strategy
Talk openly with your employees about how the compensation system works. Explain why you’ve chosen this strategy and how it benefits both the company and them. When rolling out changes, consider doing it in stages to make the transition smoother and to get feedback at each step.
Monitor, evaluate, and adjust
Keep an eye on how well your compensation strategy is working. Are you attracting and keeping the talent you need? Are your employees motivated and hitting their targets? Be ready to tweak your strategy based on what you find and any changes in your business or the market.
Elevate your compensation strategy
As businesses continue to navigate the complexities of the modern work environment, the lessons drawn from these diverse compensation strategy examples offer a roadmap for creating more cohesive, motivated, and high-performing teams. By placing employee well-being and organizational values at the heart of compensation planning, companies can cultivate a work environment that not only meets the immediate needs of their employees but also supports long-term sustainability and growth.
In this dynamic interplay between strategy and compensation, the ultimate goal remains clear: to build a thriving, resilient organization where both the company and its employees can achieve their fullest potential.
Want to learn more? Discover 5 ways compensation practices harm your employees and company.
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1180 | https://physics.stackexchange.com/questions/327934/what-is-so-special-about-spontaneous-symmetry-breaking-time-reversal-example | Stack Exchange Network
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What is so special about spontaneous symmetry breaking? (time reversal example)
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I have serious trouble understanding the concept of spontaneous symmetry breaking (in condensed matter specifically). Let's take time reversal in magnetic systems as an example.
Ferromagnetism is said to spontaneously breaks time reversal symmetry. As I understand it, time reversal symmetry can be understood with a time reversal operator $\mathcal{T}$ that reverses the sign of all momentum and spin, so that $\mathcal{T} S_{zi} =- S_{zi} \mathcal{T}$.
We take an hamiltonian that can generate ferromagnetism like Ising $$ H_0 = -J \sum_{} S_{zi} \ S_{zj} $$ and notice that $[\mathcal{T}, H_0] = 0$ because there are two spin operators. So no symmetry breaking here.
Apparently, a spontaneous symmetry breaking is manifested in the asymmetry of the ground state rather than that of the hamiltonian. There are two ground states for the hamiltonian one with all spins up ${\left|\left. \uparrow \right>\right.}^{\otimes n}$ and one with all spins down ${\left|\left. \downarrow \right>\right.}^{\otimes n}$. Since $\mathcal{T} {\left|\left. \uparrow \right>\right.}^{\otimes n} = {\left|\left. \downarrow \right>\right.}^{\otimes n}$, time reversing keeps you in the ground state, so no symmetry breaking here.
A point that is often made is that passing below the critical temperature breaks the time reversal symmetry because you will find the system exhibits non vanishing magnetization and thus is in a particular ground state not in a superposition of both. But this is only the case because some noise in the environment (it could also be an irregularity in the system) caused the system to choose a particular direction. We could simply add that noise into the model by saying
$$H= H_0 + \delta H$$ with $[\mathcal{T}, \delta H] \neq 0$. Then the total hamiltonian is not symmetric.
Is that the essence of the so called "spontaneous symmetry breaking"? If it is, what is so special about it?
Couldn't we just say that below the critical temperature the system is greatly susceptible (literally since susceptibilities are discontinuous) to small perturbations?
Is there a rigorous definition of what a spontaneous symmetry breaking is?
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asked Apr 20, 2017 at 20:57
user140255user140255
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Indeed, one of the definitions of spontaneous symmetry breaking is in terms of its susceptibility:
Suppose we add a symmetry breaking perturbation $h \; \delta H$ to our Hamiltonian (as you do), if $$ \lim_{h \to 0} \lim_{N \to \infty} \langle m \rangle \neq 0 $$ then we say our system has spontaneous symmetry breaking.
(Note: $N$ is the number of spins in our system. Indeed, on a mathematical level, non-analyticities can only arise in the thermodynamic limit.)
What is special is that any arbitrarily small perturbation will do. Imagine you have a million spins. If the state is originally in a symmetric state (i.e. not symmetry broken yet), then even if I just apply an arbitrarily small magnetic field on a single spin, the whole system will choose that orientation.
You suggest that the fact one in principle needs the environment to 'make the choice' that this is not really spontaneous. It is true that in that philosophical sense of the word, the direction of magnetization is not 'spontaneous'. But what can be called spontaneous in the universe? If I perfectly balance an egg, then the direction it will eventually roll when it loses its balance is spontaneous (or not spontaneous) in exactly the same sense. And note that once the egg has rolled down (and stopped), the tiny perturbations in the air which influenced its original direction are now no longer sufficient to change its position. I.e.: after the `spontaneous' process, the system is now stable.
The same thing happens in the above magnet: once it has chosen a direction of magnetization, then changing the applied magnetic field on that single spin I mentioned before will not change the total magnetization. So in that sense it is not true that it is so susceptible! One needs to apply an extensive magnetic field (i.e. a field that acts on most of the spins) to change the direction of the magnetization.
That is what is so funny about these systems:
An arbitrarily small perturbation can create a magnetization, but it cannot change it!
On a more quantum-mechanical note, if one has a Hamiltonian whose ground state should display spontaneous symmetry breaking, then if one takes the ground state to be in a symmetric superposition (which one can always do), then this state has ridiculously long entanglement. These are called cat states (in reference to Schrodinger's cat). This is a natural consequence of the above: an interaction with a single spin has to influence all spins at once, which is only possible if every single spin is entangled with every other spin. An example is the state $|\uparrow \uparrow \uparrow \cdots \rangle + |\downarrow \downarrow \downarrow \cdots \rangle$. (Indeed: an interaction with a single spin will collapse this 'cat state' to a product state, and then it is clear that any subsequent single-spin interaction cannot flip the state to the other product state.) Indeed, the way symmetry breaking phases are classified in one spatial dimension is in terms of these entanglement properties [Schuch et al., 2010].
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edited Apr 20, 2017 at 22:09
answered Apr 20, 2017 at 22:02
Ruben VerresenRuben Verresen
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$\begingroup$ Great answer thanks! So, is this "one-way" sensitivity to perturbations equivalent to the the phenomenon called hy hysteresis? $\endgroup$
user140255
– user140255
2017-04-21 00:08:32 +00:00
Commented Apr 21, 2017 at 0:08
$\begingroup$ Not really. A simple difference is for example that hysteresis is even relevant when you are applying a global (i.e. extensive) magnetic field. $\endgroup$
Ruben Verresen
– Ruben Verresen
2017-04-21 00:14:57 +00:00
Commented Apr 21, 2017 at 0:14
$\begingroup$ Another question, do you know if there's a way to prove that the definition you give is equivalent to the one given by GaragePhys below? $\endgroup$
user140255
– user140255
2017-04-21 14:51:33 +00:00
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You've already mentioned the exact definition of spontaneous symmmetry breaking:
Spontaneous symmetry breaking for a system that is described by a hamiltonian $H$ with ground state $\left| g \right\rangle$ happens where there is a symmetry transformation of $H$ that doesn't leave the ground state invariant $$[T,H] = 0 \text{ but } T \left| g \right\rangle \neq 0. $$ Much like a stick that's standing on its tip can be rotated around itself but will eventually fall back to a "ground state" that doesn't have this rotational symmetry anymore.
The example you are quoting is a statistical theory, so the fluctuations that drive the system into a state of non-vanishing magnetisation below the critical temperature are already built in from the start.
Spontaneous symmetry breaking is a key ingredient in the Standard Model of particle physics, it is used to explain why the particles that mediate the weak force are massive (W and Z bosons, this is remarkable because you can't achieve that by just putting a mass term in the Lagrangian!).
But it also works the other way around, in that it explains why sometimes there are massless modes in a system (see Goldstone bosons).
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answered Apr 20, 2017 at 22:19
GaragePhysGaragePhys
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$\begingroup$ Thanks for your answer! I am not sure if it's a typo, do you mean $ T \left g \right \rangle \neq \left g \right \rangle$ instead? Because I can't see any reason why it could give 0. Take the case of an hamiltonian symmetric under parity like an harmonic oscillator. Then applying the parity operator on the ground state gives back the ground state not 0. $\endgroup$
user140255
– user140255
2017-04-21 00:20:33 +00:00
Commented Apr 21, 2017 at 0:20
$\begingroup$ No, that's not a typo, what you mean by the symbol $T$ is the generator of the symmetry, which is an infinitesimal version of it. For example if you have rotations in 3d space acting on the vector: $\left| x \right\rangle = \begin{pmatrix} 0\ 0\x\end{pmatrix}$ Let's consider the transformation: $\endgroup$
GaragePhys
– GaragePhys
2017-04-21 05:36:16 +00:00
Commented Apr 21, 2017 at 5:36
$\begingroup$ $$\left| x \right\rangle\rightarrow R \left| x \right\rangle \text{ with } R = \begin{pmatrix} cos(\alpha) & sin(\alpha) & 0 \ -sin(\alpha) & cos(\alpha) & 0 \ 0& 0& 1 \end{pmatrix} \approx \alpha \underbrace{\begin{pmatrix}0 & 1 & 0 \ -1 &0 & 0 \ 0& 0& 1 \end{pmatrix}}_{ = T} + \begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0& 0& 1 \end{pmatrix}$$ Here $\left| x \right\rangle$ is clerly invariant under $R$, which is expressed by $T \left| x \right\rangle= 0$ Here you'd find more about it: \url{en.wikipedia.org/wiki/Symmetry_in_quantum_mechanics}. $\endgroup$
GaragePhys
– GaragePhys
2017-04-21 05:36:21 +00:00
Commented Apr 21, 2017 at 5:36
$\begingroup$ I understand what you mean! Also, do you know where I can find a comprehensive explanation of what Goldstone bosons are (considering I'm from condensed matter and not familiar with field theory formalism) $\endgroup$
user140255
– user140255
2017-04-21 18:06:19 +00:00
Commented Apr 21, 2017 at 18:06
1
$\begingroup$ The complete title is 'An introduction to quantum field theory', it's the standard reference for QFT. $\endgroup$
GaragePhys
– GaragePhys
2017-04-21 19:58:21 +00:00
Commented Apr 21, 2017 at 19:58
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1181 | https://www.youtube.com/watch?v=hwNyMO8SMVM | L1 vs "L"1, Null sets & functions, Almost Everywhere vs Norm Convergence | Intro to Analysis
Dr Mihai Nica
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Transcript:
last time we ended on the Cliffhanger of L1 is not even a norm space because okay by L1 I mean ordinary L1 because the norm of something being zero does not mean it's the zero Vector right there are some things who are not the zero Vector also known as not the zero function because the the functions are fancy vectors um whose Norm is zero right and we talked about this definition of a null function so a function whose integral of the absolute value is zero and some examples are like the indicator functions of uh the rationals things like that and the problem is that ruins our day because now we have things whose Norm is zero but are not zero and so technically Capital L1 is not a normed space because the norm does not satisfy the the only thing whose Norm is zero is the zero vector and we're going to do a little sneaky sneak and replace Capital L1 with script curly l L one and we're going to fix the problem and fixing the problem involves these things called equivalence classes and as a little warmup I'm hoping you've seen some equivalence classes before which your hand if if you've seen an equivalence class before okay fantastic so this is like a little warmup on uh an equivalence class to see if uh you can cut through all the jargon and understand what it means it's a pretty sometimes there's a lot of notation but it's a nice idea of dividing your space into equivalence classes so uh what I did here is I listed the three properties that a relationship has to have for it to be called an equivalence class or an equivalence relation and then the equivalence classes are the things that are all equivalent to one element and you write them with this little square so for example in my example it's an equivalence on z uh so I made this new equivalence equivalence 2 which is on Z and so like the integers and open square bracket one close Square bracket means all the things that are equivalent to one those are the equivalence classes um when you have such a situation you can do nice things with it but to see if you guys are have seen this kind of thing before I made a little Dopey problem well you tell me this equivalence class has a name that you've heard before uh and I want you to type it in and then we can we can go from there so I'll give you a minute to think about this uh let me know if you need more time to read if if all this if you're Rusty on all this stuff um another thing you can do if you already know the answer you can check how would you prove these three things about equivalent 2 um on your homework which has just been posted as of this morning um you have to prove this equivalence thing for something we're doing with integrals you'll see what it is in a second uh but it's a good check how do you prove these three things for some rule like this okay okay let me let me bump this [Music] up in equivalent to it every element has to be equivalent to itself yeah oh it says it's SL yes in every ele every element oh in I guess the set or is equivalent to itself sure I thought I'm today and then I remember that for 20 seconds left put in what you think this set is called to regular people who don't like notation so every time I [Music] have okay let's take a look what is this thing called the set of things that are equivalent to the number one is called odd numbers odd underscore numbers odd numbers ah this person was off they said even it's odd okay uh the odd integers the odd numbers very nice odd integers okay so this is the odd numbers that's that's correct uh odd numbers um some other people got it even though you weren't graded right or I'm so so this is the idea we started with all the infinite integers and we can divide it into two sets the even numbers and the odd numbers how do you do it you make equivalence thing you say two numbers are equivalent if they differ by an even number and all the numbers that are equivalent to one are the odd numbers and you can check that this satisfies all three of these things every number is equivalent to itself right because z a minus a would be Z eight Zer is divisible by two I saw a Twitter post the other day where someone's math teacher told them that zero is not an even number and then they had to like appeal to like higher and higher levels of math professors until they agreed that zero is an even number anyway zero is an even number um so uh uh everything is equivalent to itself uh a if a is equivalent to B then B is equalent to a because a minus B and B minus a it that's one is divisible by two the other one is negative of it it's also divisible by two and finally if you add up two numbers that are divisible by two then you'll still get something divisible by two which will prove number three so you have to do like a little mini proof but it works and then you have these equivalent classes and the nice thing about equivalent classes is as long as it obeys all these things you can do things with the equivalence classes so you can work instead of working with all the possible integers you can work with the equivalence classes of the integers and you can do addition and stuff with them um so for in our Dopey example you can do the addition of 0 + 1 this is adding zero represents the even numbers even numbers and this is odd numbers and you can add them together so this is all in this uh till 2 business what happens if you add the even numbers and the odd numbers well what we're trying to say is even an even number plus an odd number is another odd number so you're going to get the 0 + 1 which is one which is the odd numbers again so you can take the operation of addition which was on all the integers and now it's on these equivalent classes um sometimes people write the equivalence classes is this set so we get this new set where we can do stuff and people write it like this they do Z and then they do a big divided by the equivalence class and this means modulo the equivalence relationship and this is the set of equivalent spes if you're doing it with this one it turns out there's only two equivalent spes there's only even an odd so this is the even numbers and the odd numbers and they're equipped with all the regular things like you can add them you can multiply them by scalers anything you can do to an integer you can now do to these things by just upgrading so for example you can if you want to do Lambda 1 where Lambda is a number then you get Lambda now Lambda has to be an integer because if you multiply by a non- integer you'll get a non- integer that's not allowed on Z right what what are the things you can do for numbers in z uh you can multiply by Lambda okay um so for example if you multiply by two you're going to get zero the even numbers okay so this I I hope you've seen something like this before this should be sort of a recap um but this is like vaguely how it works we took all Infinity things we took some relationship and then we created equivalence classes and that was like a simpler space and it kind of distills down the things we care about what do we care about if we only care about the remainder when you divide by two then two things that have an even remainder are the same and so the only things we care about are even and odd numbers all right that was like a little bit of a warm up uh any questions or comments about this is this jogging people's memory for how this stuff works did you guys take like an algebra class on like rings and stuff no okay all right um yeah I think proof sets of numbers you do it in proof sets of numbers no you don't do it in well do you do this in proof sets of numbers but not I did that when I classes we did yeah classes in relation yes I see yeah well okay I'll a ring is is things that are like numbers ring so a ring is anything you can you can add the stuff together you can multiply and things have inverses and and this is the ring called Z2 which is this you can make a ring called zp for any P you use Z mod two things are equivalent if they are a multiple of p and p is a prime number then you get the numbers uh zero one all the way up to P minus one and uh this this is a useful ring in cryptography this is how they do everything in cryptography when you send messages you you make them mod P and then you do a bunch of theorems in in P to like laugh at people when they can't Factor your Prime and like your coding system works or something um okay uh so let's call the ring I don't know maybe you did this without calling it a ring we didn't really we saw it like mostly to Dos I see I see okay well anyway what do you need to know for this course you need to have heard of this uh module thing cuz this is how we are going to rescue L1 right L1 has this problem that two uh functions they might not be the same function but their integral is uh you know the integral of something is zero but it's not zero so how do we rescue it we're going to rescue here's our attempt to rescue it we're going to Define uh equivalent up to Le big integral so this is how we are going to rescue rescue L1 we're going to Define equivalent let's call it Le I don't know I'm probably only to write this one time CU it's annoying but just so you don't mix it up with the other one um what does it mean for two functions to be equivalent well well when it was Zero the things that are equivalent to zero are the null functions right the ones who's integrate to Z so what are two two functions equivalent well we'll take the difference of the functions G and we'll integrate it and we want to get zero so if you have two functions and you subtract them and take the absolute value and that integrates to zero then we say they're equivalent okay and as an example as an example f is equivalent to zero the zero function if and only if f is a null function that was the definition of a null function so the null functions are the ones that are equivalent to zero so now we have an equivalence relationship and we're kind of saying we only care about things up to like if if the function f is equivalent to Z then it might as well be zero for everything we want to do like we're not going to distinguish and be like well that function isn't zero no it's it's good enough it counts it's close enough to zero so what we're going to do is we're going to take L1 right that's the set of all functions this is all possible functions that are leig integrable possible integrable functions and we are going to do the quotient by the equivalent right and this means modulo equivalent okay and so what is left are you have equivalence classes of functions this is the set of all equivalence classes so an equival class looks like square brackets F and it's the set of functions G such that uh the integral of Fus g equals z and a good I think a really good example is if you look at the equivalence class of zero this is is the set of null functions right so instead of saying zero is literally the number the function that's always zero we're talking about the equivalence class which is all the functions that are null functions and we treat that whole group of functions all together as one thing and that space this space is called this space is called Script L1 okay and this rescues the problem before we had the problem that the only thing whose Norm uh sorry the problem was that some things they're normal zero but they weren't the zero function but now all the things that were zero those are the null functions those are all one thing that we call like the zero set right the the equivalence class of zero and so this thing is okay now the norm makes sense and so you know to be Crystal Clear we have to move all our operations so L1 of R is a normed space right and you have the operations if you want to add up two equivalence classes so somebody tells you I want to know what is it the plus of this equivalence class plus that equivalence class well you just take the equivalence class of f plus G that's exactly like the even numbers and the odd numbers right if someone says add the even numbers and the odd numbers then you take the you take some representative from the even numbers some representative of the odd numbers add them together and then take that class uh similarly if you multiply by a scalar same deal so that's Lambda f huh and if you want to do the norm this is I guess the last thing to do so you take the equivalence class you take the norm in L1 what you do is that's just the norm of f in ordinary L1 and this last one I guess technically is the only one that you would have to check something the other two as long as it's an equivalence relationship will work but this one maybe it's not defined like maybe if you take if I take an f and then my cousin from Romania takes a G from the same equivalence class and we both go to compute the norm we get two different answers like that would be a problem right that would mean this thing is not defined but fortunately uh you always get the same answer uh so always same answer same answer and let's try spelling answer a little better okay and um so I guess the norm of f in L1 equals the norm of G in L1 for all FG uh in in some equivalence class so if you take two different functions from the same equivalence class they have the same L1 Norm that's like a fun little exercise that you have to check and all of this stuff proving that it's an equivalence relationship and this kind of thing I left it for you as a homework Pro it's not too trick okay so that's how we're going to ask yeah can you just scroll back up really quickly you wrote The N functions as yes zero the so the equivalent class of zero so it it's equivalent Okay so if if you're equivalent to zero then you're a null function that's like the definition of a n function yeah right okay so yeah what's the like what's the slash ah the slash okay right okay so this line you would read out loud as modulo and then you would modulo by the equivalence relation so that's why I did this analogy over here with the uh even numbers so if equivalent means the difference is divisible by two and then you do Z this modulo line so that means look at the equivalence classes mod the equivalence then so for even and odd there were two of them there's the equivalent class of zero there's the equivalent class of one it happens those the only two things that's because every other number is in one of those two equivalence classes in our situation okay there's still infinitely many equivalence classes right so there's the equivalence class uh like you could say it's the set of all this thing let's make this an equal sign this thing so it's the set of all possible equivalence classes so it's a set of sets right so it's the set of all possible equivalence classes of f right there's still infinitely many of them it's just that we've taken all infinitely many functions and we've grouped together all the functions that are uh whose difference integrates to zero and we said that's one equal class this is another equal class which is just like taking all the numbers and saying all these even numbers those are all one equivalent St all these odd numbers those are another equivalent St um it just so happens that there's still infinitely many left over yeah okay great question any any other questions so it's it's a very weird idea if you've never seen it before if you've seen it before a little bit the more you see it the less weird it feels and you're like ah yeah of course you know 20 years after you see it you'll be like ah of course Mak when you teach the course on an advanced analysis you be like oh yes of course okay so here's the trippy thing so now it's it saves it right so now L1 really is an actual uh Norm Vector space and everything is fine um here's the trippy thing here's the trippy thing if if I say okay so now we have some f in the real the actual space L1 right and I say what is the value the value of it at let's say xal 0.5 right this is like a thing right if you have a function one of the most basic things you want to do is like plug in xal 0.5 all right unfortunately there's a problem with this okay what is the problem yeah a whole class of functions to choose from to Value yeah exactly right and okay well so this is it could that could be a problem right this is not just one function it's a whole class of functions and that was a thing before right but then before it was okay cuz like everything in in the class agreed so if you have to do something to the class and you pick one guy you better hope that it's going to be the same answer no matter you get this is like the cousin in Romania problem okay but is picking out xals the value at x equal 0.5 is that the same for every element of the class and the answer is definitely not um and that's because you can you can do like f plus the function uh which is uh let's say 1 at x = 0.5 and 0 at X not equal to 0.5 this is the classic nest of null functions right the indicator of the set 0.5 this is a null function and now you have added a null function to F so this thing is still in the equivalence class of f right their difference is this stupid null function and so this guy is definitely still in F okay let's do it in Black so this is still an F but it has a different value right whatever the value of f is this the value of this thing is going to be f + one so uh I don't know let's call this function G so G of 0.5 = F of 0.5 + 1 but both still in F right and so you can no longer if you want to work in the real space L1 you can no longer ask for the value at a single point excuse me in fact you can no longer ask for the value like at the rationals like that could all change um if you want to ask for the values at some points it has to be like a a chunky part of the real line that has some non zero width sort of so you can ask for like all the values between 0 and 0.5 uh that sort of you can't change all of those without changing the set so this is like a real a real issue yeah so every number between 0 and 0.5 yeah wouldn't this be a problem if a function deviates from that of find Point yeah right so again you can mess up all the all the numbers between 0 and 0.5 you can mess up countably many of them and in fact you can mess up uncountably many of them as long as you do it in like a smart way but you you can't like you can't mess with a whole interval all at once because then the integral will change and you'll be out of the so it's kind of like it's it's actually a little bit trippy if you think about it right it's like what what the hell is this thing anyway right it's there's so many functions in there there's so many null functions that you could add on you can change the value at like many many points um but just not a lot of points and I think the the the the way your brain should try to resolve this is that sets of measure zero like countable sets are just so wimpy compared to like actual chunky sets like 0 to one like 0 to one is so much bigger than the rational numbers that somebody coming along and being like aha I messed up your function all the rationals you're like well you're you're an idiot like there's so many more like you you didn't do a lot and it really is because the the like Infinity between all the numbers between zero and one is just way way bigger than any countable Infinity right um so yeah okay it's kind of like sad but that's that's the the universe okay yeah when you define the Le equivalence yes it was with when their um the integral of the difference is zero that's right is that integral on a specific domain as it Z to no so okaywhere yeah so this is the book does everything leig measure on R and um everything still makes sense I think if you if you want to like imagine L1 of 01 you have the same all the same arguments apply and then you'll like have like one less Infinity in your mind to deal with so I think I think if you if that helps you to think about that yeah it it does in the original definition it is saying along all the real numbers that that's true yes right MH MH yeah uh right yeah so so L1 I mean the whole book book just does everything with L1 of R and it's really not a problem because all the integral functions are made of limits of uh step functions so and step functions can are only nonzero on a finite interval so the way you get to all of our is by doing limits while you go like further and further out in fact you have one homework problem like that as well um but yeah so if you understand L1 of 01 going to L1 of R you should just think about it okay we just make the 0 to one we make 1 to two we make two three and then we like add them together in a smart way then we'll have L1 of R so L1 of R should not scare you more than countable sums of L1 of intervals um and everything in L1 of an interval is already a countable sum so it's it's really not any extra work yeah okay yeah great great question okay so now hopefully you guys appreciate the weirdo quote I'll read it one more time now it will make sense uh from the book right let's let's bring this thing back up okay right uh we can only wait that's not the one that's this one okay so the whole construction might seem artificial that's what we just did this weird modulea thing in practice we do not often distinguish between script L and L1 and formulate everything in L1 so what this is saying is okay actually I just showed you how this stuff works in practice what we're going to do is we're going to always work in capital L1 we're not going to worry about the script L1 we know it's there and we're just going to keep in the back of our mind that when something is Norm zero it's not actually literally zero it's just a null function okay that's really the biggest thing and you're going to see as we go on today we're going to learn about almost everywhere convergence and we're going to see that being a null function is the same as being zero almost everywhere okay and so we're just going to keep that in our mind so we're always going to work in capital L1 not in script L1 um so this does not cause any real difficulties as long as you're aware of the problem that's why I'm telling you now you're aware okay actually when dealing with script L1 new difficulties arise for instance note that for a function f in L1 it's not actually one function it's an equivalence speci functions you cannot specify the value at a point that is what I just showed you okay all right so we're going to we're going to move on because now we we have like built up all the tools now of the big integral and now we can start to talk about convergence in this space in L1 and we talk about convergence and norm and we're also going to see this almost everywh thing and we're going to see um this is me putting on my probability hat let me let me write down what this says on the thing we're going to see uh that actually you can replace all of this uh stuff about null functions we're going to see that f is a null function a null function that's going to be the same as f ofx equals z quote almost everywhere almost everywhere what does almost everywhere mean uh it means that if you picked a random number then if you did F of U and you ask what is the probability that F of U equals z so U is some random number between 0 and one now I'm assuming that the function is on really on 01 um but uh that's fine okay well let's say F on 01 to cover myself okay so F of U is some random number between 0 and 1 U Is Random between 0 and 1 you apply F to it you get some other number what's the probability equal to zero well it will be 100% 100% so the null functions they are zero sort of almost everywhere that's what the almost everywhere means in the sense that if you ask for the probability that it's not zero that is a 0% chance the probability that F of U is not equal to zero is 0% it's not literally everywhere but it's almost everywhere and in probability sometimes people instead the say almost everywhere they say almost surely almost surely right and almost surely is this idea it's 100% chance but that doesn't mean literally everywhere because there are some sets that are not empty but have 0% chance and that's exactly what we're trying to say don't worry about those okay um by the way almost everywhere you'll see it very often as AE because people are lazy in particular your homework okay question called me almost everywhere is that the same thing ah this is great okay so this is uh Le almost everywhere okay leig almost everywhere technically everything we're doing is leig and why is it leig almost everywhere it's because it's U is the uniform distribution you can have other measures mu which would be a fun project for someone to do mu and if you have mu almost everywhere that is telling you the probability of f f of x equals 0 is 100% where X is drawn according to me okay something went crazy with my uh pen here okay let's try this everywhere and okay for a lot of random variables like a gaan or something it's exactly the same as the big almost everywhere but if you have a random variable that uh has different probability structure then be just having different chances everywhere then it can get weird so like giving you is like a Cantor set or something that it's like some other thing um but yeah same idea great great question just a measure me means any measure that I'm thinking of okay yeah and and in this class we're going to think about the L B one M okay yeah okay other questions okay so let's let's start moving we're going to we're going to move into this we're going to do that the two sections we're going to work on today are convergence in norm and convergence almost everywhere so section 2.6 is Convergence in Norm in Norm which in just to warn you in the textbook they do i n which to me is like too lazy like come on it's like six letters uh and we're going to do 2.7 which is Convergence AE and that one I'm on board with because almost everywhere is a lot more okay um so we're going to start with these things and we're going to see that all this stuff about n functions it really comes out in this like almost everywhere and uh in Norm thing okay so the first thing we're going to show theorem 2.6.5 uh if FN converges to F in Norm so again this is like the usual thing we did in chapter one right that means the norm of FN minus F goes to zero or the integral of FN minus f um goes to zero then what then the integrals are equal then the integral of FN converges to the integral of f which is a nice property so if they're converging their integrals as numbers also converging and the proof is very simple it's like one liner just by triangle inequality which we've done many versions of so far if you look at the difference of FN minus F that is less than or equal to the integral of FN minus F this is exactly one of the inequalities we proved along our journey to defining all the stuff about the liting integral so sort of all our hard work all these theorems we Prov that were like really annoying and we have to like worry about oh f is some sum of uh step functions and blah blah blah it's just like now we use them and they do exactly what we need uh right and this integral goes to zero that's exactly what it means so you're less than something going to zero then you also go to zero yeah is any function or St function no here FN is any integral function yeah right yeah so you should think of a the step functions were like the Lego that built everything and now we have everything built and we can like play with like a full Lego house so we don't need to worry about the B okay uh all right okay here's another good one next one in the textbook theum 2.6.6 which says if uh f is Sim equal to fub1 plus FS2 plus dot dot dot and they don't have to be step functions so now they can be any integral functions right and that's because again last class we proved that if you make a sequence of integraal functions and add them up like this that's what I called super integrable but that was just the integrable functions again so now you can this is one way you can build a new integral function is by adding together integral functions that you want according to bullet points A and B of that um and so last time we were had the Sim equal we had part A and Part B and now I'm telling you that it implies that uh then the sum of Nal 1 to Infinity of FN converges to F in Norm right so kind of this like special equal sign that we've been like floating around for so long we had to use it to define the norm but now that we've defined the norm it really means it implies convergence in Norm so great and the proof again is very simple um because uh given any Epsilon greater than zero by definition of this funny sin equals uh find uh n n and not so that the sum of the integrals of FN are uh less than Epsilon right so I think one of the two conditions is that this this infinite sum is summable so make it less than Epsilon and that will be for n bigger than n uh and then you have then have Again by all the triangle in equalities we proved that F minus the sum of the first n knot well that will be less than the sum of N greater than n of absolute value of FN Which is less than Epsilon so the norm of this Fus the sum is becoming less than Epsilon and we can do that for any Epson greater than zero therefore they can converge down to zero Norm uh I believe I'm the theorem we invoked here is theorem 2.3.1 way back when uh which was you know several classes ago now uh okay so that's convergence in norm and like really we're just checking like the things that we want convergence Norm to do we we already know about convergence and Norm from chapter one and now we're just like seeing it in this Le space and it does all the things that we think it should do okay questions or comments yeah this might be obvious how did we get the inequality is it just in the theorem it uh yeah the theorem says I'll tell you what the theorem says the theorem let me let me let me remind you what the theor says the theorem says I'm going to use G's uh so we don't get mix up it says if G is G1 plus G2 plus dot dot dot then the integral of G is uh less than the integral of the sum of the GN or something like that I think that and this was when we proved last class and here what I did was it just moved the first n and not terms over and then it worked thank yeah so okay no that's a good a good clarifying question okay yeah any any other all right and so now we're going to get to a little more of an exciting thing which is 2.7 which is Convergence almost everywhere and this starts to get a little more slippery so convergence almost everywhere is kind of like pointwise convergence pointwise convergence is Convergence everywhere and this is Convergence almost everywhere and so we're going to like dig into what this actually means and you're going to see how the norm plays with these almost a rare things okay we're going to start with this definition that I already gave you but in the book it comes later on it's the definition of a null set 2.7.1 a null set a null set and uh that means X is a null set that means the integral of the indicator function of x equals zero so in the book they say its characteristic function is a null function so it's the sets where you look at the function that's one on the set and zero otherwise that is uh integrates to zero yeah [Music] just sorry what meure me yes right these are sometimes called measure zero null set means X is measure zero the measure of a set I can tell you what the measure of a set is the measure of some set X is the integral of x so you want to measure how big a set is you make the function that is one the set and you integrate it so for example the measure of the interval 0 to a half you take a function that is one between 0 and a half and zero everywhere else you integrate that up it's going to be a half actually because it's a step function in this case so it's okay um and so the measure will be half just defition right so yeah this is I think I mentioned this once but let me mention it again now U because now that we're kind of through the hard part of the big integral a lot of courses start with measures so they say like how do we Define the measure what's the measure of an interval blah blah blah and then they build up using the measure they build the Le integral this book goes the other way so they start with the Le integral and they have this tricky Sim equals infinite something and now let's you skip over a bunch of stupid technicalities with the measures uh and like you can take a whole course just on measures so when I was a undergrad we took two courses we took one whole course on measure it was like a third year course and we had another course in fourth year analysis you guys only get one so that's why we kind of skip some of this stuff but it is a great project if someone wants to do something like that and for example you can make a non-measurable set that's really fun or you can talk about these weirdo measures like the canor measure that is like super weird okay I can't resist we have to we have to take a second where's my mouse okay I'm going to search for the right thing devil staircase what is the devil staircase all right so if you search on Google for the devil staircase it shows you some stupid thing in Arizona that's not that's not what you want uh you want to click on Wikipedia and you want to find the math one the gter function look other uses a singular function mathematics the caner function this is the coolest one okay you click on this one okay this some weirdo function what does this function do this is a function whose derivative wherever it's derivative exists is zero everywhere it derivative is always zero and yet the function is increasing the function and it's continuous it goes from 0 to one it's increasing uh and it's derivative wherever it exists by the way and its derivative exists almost everywhere now you know what almost everywhere means so the only places where its derivative does not exist is a null set the probability that its derivative is zero you pick a random point the probability that it's differentiable is 100% And Its derivative is zero everywhere and yet this function goes from 0 to one somehow the answer is it has this tricky set called the canor set and it's increasing only on the canor set in fact this is the cumulative distribution function of some kind of random canor set so you pick a random point in the caner set that's the cumulative distribution function so this is a great project if someone wants to do this and in fact um in the textbook they have as an exercise to prove some stuff about the canor set so that's a great place to start you do those exercises and like look out how to do them and then that'll be a good place so this is a really fun a fun function yeah this is everyone's coolest function look here's how they make it how do you make a function it's a limit of other functions right it's a good thing we're doing the little big integral or nothing would make sense um but you make you start up with a a function that's kind of boring and then you add more and more pieces to it and in the limit it converges to this uh weiro function what else can I say about properties I don't know lack of absolute continuity more definitions fract oh it has a fractal volume okay great amazing okay uh I'll stop there nobody asked me any questions about the canor function I didn't I didn't prepare okay all right uh other questions or comments yeah not function no you can actually ask me okay you you've said a few times that the integral functions are the limits of a sequence of functions yeah are those functions that um in this sequence are those the the first nend terms of the sum yeah what I mean is I mean literally like if you go back to the definition the definition was F is this kind of thing FS1 plus FS2 plus dot dot dot so I mean these functions and it's a limit in a very particular sense it has to satisfy A and B where A and B were like the definition of this thing and and then the point is that these F1 f2s in the definition they were uh step functions so it's it really is like we have the simplest possible things the step function fun and we built all the integ functions by doing this specific type of limit of those things now that we've uh defined everything you're going to see that it really is just saying that it converges in norm and it converges almost everywhere we're going to prove both of those things soon um but it's really this specific type of limit and like that constructs all the integral functions out of limits of these like baby functions yeah okay yeah you're talking about the indicator function right defition in this definition I'm talking about the inic function yes that's what one means yeah oh I'll warn you guys some books they do a kai kai X because that's not confusing uh but yeah just watch out and some people call this the characteristic function U you also have to watch out cu characteristic function also means some other thing so like you know I think the moral of the story is mathematicians are bad at naming things so watch out uh be ready okay all right uh okay where are we any any other questions okay so we have these null sets what does convergence almost everywhere means it means convergence except for a null set that's what the almost in almost everywhere means uh so yeah the definition uh f equals g almost everywhere a e means if you look at the set of X where f ofx is not equal to G of X the set where they're not equal that is a null set okay and again putting on my probability hat you would say f equals g almost everywhere it means the probability that F of U is not equal to G of u g of U equals zero right so if if you pick a random point and you ask are they equal there's a 100% chance the answer is yes they're equal there are some points where they're not equal but those points have a 0% chance of getting chosen all right that's almost everywhere uh and that those are n sets here is a cool fact about n set that has uh very interesting proof uh in the book um which is theorem 2.7.2 any subset of a null set is still a null set a very reasonable theorem any subset of a null set is still a null set okay and actually uh the only hard part about this is if you take a subset of a null set maybe it's a set who's not integ at all right so the real the real meat of this theorem is question let's say a is subset of B and uh the integral of the indicator function of b equals z so B is a null set is a uh even integrable is the integral of a does that exist cuz if it exists then that function is less than that function and then it's fine but it might not even exist we saw last time there are some non integrable functions you can make them with the a of Choice by some crazy method so you got to be careful and so the textbook has a really funny proof um which is goes like this uh okay I guess I have to let me translate it uh okay in the textbook it's written with functions so in the textbook they do it like this they do 2 point uh okay there's some 2.6.3 and they say if F or if G is less than F and the integral of f equals 0 then G is integrable and the integral of G is zero okay let's write this one out is integrable and the integral of g equals zero uh in absolute value okay and I think I think it actually needs to be an absolute value let's do it if G is less than f there we go okay um if you know this one then that one follows by putting f is the indicator function of B and G is the indicator function of a uh but let's do this one this has a really funny proof here's the claim that looks so bizarre is that g is equal s equal to the following infinite series f plus f plus f plus f plus dot dot dot you add up FS forever and this seems like totally crazy right how can you if you add up something forever this is like what you learn in uh like you know infinite numbers 101 if you add up the same number over and over again it is not going to converge you can't just do 1 plus 1 plus 1 plus 1 plus one and hope it converges the numbers have to get smaller or something as you go down or else it's not going to converge so what is going on here well actually the integral of f is zero so f is mostly all zeros there is one special you can add up over and over again and that's zero and F is mostly zero so this thing actually will work out because f is mostly zero so it looks crazy but it works because f is mostly zero um we have to check A and B right A and B properties A and B which are the heart of this construction with the same equals and property a says the sum of the integral of f so it's F1 plus F2 plus F3 here we have FFF repeated so the sum from Nal 1 to infin converges so the sum of the integrals converges well the integral is zero here so it does converge right 0 plus 0 plus 0 you add up the integral it converges yes it's all zeros it's all zeros okay property B says that uh G of X has to be equal to the sum of uh F uh of x uh whenever whenever whenever sum converges absolutely okay but when can the sum converge absolutely you're adding up infinitely many copies of a number it converges absolutely exactly when f ofx equals z so we only have to check that this thing equals that thing when the absolute value of f ofx equals z and when the absolute value of f ofx equals z uh it is true that g of x equals this stuff why is it true well G is less than F what numbers are less than zero the only possible thing is that g was zero so uh and this is true since G ofx equal Z whenever absolute value of f ofx equals Zer so this is like a really funny proof that if you have a null function you can add up the null function together infinitely many times and it will be Sim equal to some other functions um right and in fact what's happening is that these two functions they just both happen to be zero on the same set and there's some nasty dumb stuff on not the set where there's zero but that nasty dumb stuff is an is a null set the set where f is not zero is some tiny tiny set it's mostly it's almost everywhere zero okay and this proves that the subset of a n set is still integrable uh the integral is zero because the integral of G by the definition is the sum from Nal 1 to Infinity of the integrals of FN which is zero you add up zero infinitely many times still zero okay so that's nice A little proof okay okay and um there's another proof that uses the same technique called theorem 2.7.4 which says that f equals g almost everywhere if and only if the integral of f minus g equals z so two functions are equal almost whatever you are if and only if their difference is zero uh this is a bit like saying the things that are in those equivalence classes that we started the class with those are all functions that are equal almost everywhere this is kind of the intuition I was trying to give anyway uh so those equivalence classes you can mess up a small number of points but you can't mess up more than a null set number of points okay and here's the proof the proof is uh exactly this this this Dopey trick the proof so here I'm going to do two proofs uh let's suppose f equals g almost everywhere if f equals g almost everywhere then uh what I'm going to do is let the B be the set where they are not equal uh okay and that is a null set because if they are equal almost everywhere it means that that's a null set and then I'm going to claim that the indicator function of the set B is Sim equal absolute value of Fus G Plus absolute value of Fus G Plus absolute value of Fus G forever plus do again this is one of those silly things how could it be uh that uh this holds oh I did I did an oopsy it's the other way around okay hold on that's the next part that's coming up it's this one uh Fus G is uh Sim equal the indicator of B plus the indicator of B plus the indicator of B forever okay how could this be true again it's one of those things you're adding up infinitely many copies of the same number over and over again so of course this doesn't make any sense except for when you're not in the set B when you're not in the set b b stands for bad set by the way so when when you're not in the bad set then f equals g and when f equals g the left hand side is zero and then you're adding up 0 0 0 0 it all makes sense it's exactly like the proof we just did so uh so uh check conditions check conditions the sum of the integral of the indicator of B that infinite sum is zero since it's a null set that's a so that is fine and B uh F minus G of x equals Zer for X not in B by definition of v of V uh which is the only place where the right hand side converges so in other words we have to check that the two things are equal when the right hand side converges and it only converges off of the bad set on the Bad set this is zero when you get zeros it's all fine so it's the same trick as before okay and so this thing is that sum then the integral is that sum the integral of Fus G is the sum of the integrals 1B forever which is zero okay so that's that's if they're equal almost everywhere then they are the norm is zero the integral the norm of f is z and the reverse is the same thing using Del let me let me make these like more high levels we don't get these mixed up with the other ones so I'm proving both ways of this proof like that and all you do is the opposite now in this case we're going to prove the indicator of B is less than Fus G Plus Fus G forever like I originally wrote down by accident and in this case if the integral of Fus G is zero then that will show that this stuff converges where could it possibly not converge the only place it converges is when Fus G is exactly zero that's exactly off the set B um and so it all works uh show I want to say by similar argument to save us a little bit of writing okay so there you go I think this is a really good intuition there almost everywhere business uh the norm is zero if and only if they're equal almost everywhere okay uh okay here's one more actually let me stop if there's any questions here's one more property I'll leave this one as an exercise it's very similar to all these other ones uh which is or you can go read it from the book theorem 2.7.5 if uh FN goes to F in norm and FN goes to G in Norm right so if L1 was like true truly a normed space and like really obeyed this thing with like the norm is zero if it only if it's a zero function then you can only have one type of limit right in Norm but because L1 is not a norm space like it could be that f and g are just two completely different functions that can happen but the thing is they are two completely different functions but they are equal almost everywhere then f equals g almost everywhere okay and the proof again is not too hard you can just do some tryangle quality and like use the things before it's like a one a one L but let's uh let's skip it because we're about to come to a fun probability example uh questions okay so we have equal almost everywhere we have this Norm stuff there's I told you what equal almost everywhere means let me tell you now what converence almost everywhere means definition 2.7.6 convergence almost everywhere and what does it mean FN goes to F converges to F almost everywhere means the set of X where FN of X converges to F ofx uh I guess the set where it doesn't converge let's say let's write in big letters not not this is a null set and in probability language you would say the probability that FN of U equal FN let's say the limit is 100% uh 1 so you have some sequence of functions and the probability that it's converging to the f is 100% uh this should just say F of U okay I realized somehow we got zoomed out let's let's go back so there's a 100% chance that FN converges to F just there's like maybe some couple bad points where it doesn't yeah sorry I don't really understand what you mean by the not the not I mean the limit of FN is not F ofx let me write it with a not equal we have okay we have a symbol for this yes the limit as n goes to Infinity of FN of U is not equal to F of U yeah um I guess if the limit doesn't exist that counts as not equal just that's the only thing you got to be careful of okay yeah that that is more clear okay that's convergence almost everywhere so now we have two types of convergence we have convergence almost everywhere we have convergence and norm and we've got a couple of little toolbox theorems that have said things like look like this norm and almost everywhere are kind of similar right like something converging in norm and something being equal almost everywhere is like pretty much the same thing so you might think convergence almost everywhere is kind of the same as convergence in Norm but unfortunately you are wrong okay let me and I'm going to give you an example so this is an example example where uh FN goes to F almost everywhere but but FN does not go to F in nor so here's the example and okay you could do this the textbook way draw the picture here are the functions and then like you know it's a it's a fact like we're done but I'm going to tell you the fun probability story that goes with this okay because it's not just uh some functions it really is about what can you do with random variables when do random variables converge and what do they mean and again almost everywhere is kind of like converging with a 100% probability and converging in Norm means in expected value the expected values converge and to do this I'm going to set up a probability of having an infinite sequence of coin flows so this is the the only little setup we got to do set up I'm going to make a a an infinite sequence of coin flips of coin flips so an infinite sequence like heads Tails heads Heads dot dot dot and it goes on forever and I'm going to do this for every possible X in 01 so so far up until now all of our probability all the little probability stories have been happening on the interval 01 and this story is just like that story it's something that happens on 01 and I'm going to tell you how you make an infinite sequence of Point FS from a point x and 01 here's how you do it write how write X in binary okay so you give me a point x in binary let's say you give me the point uh 13 and I think then x equals 0.11 011 0 it's some sequence in binary right and what I'm going to do is if the first one is a one the first sequence is a heads if the second one is a one the second one is a heads and just match up one with heads and two with Tails match one means heads and zero means tails and it turns out that this works perfectly well all of the coin flips will be equally likely to the heads or tails all of the if you pick a random number in 01 all of the binary digits are equally likely to be zero and one not very surprising okay so this is how you can assign to a sequence of heads and taals it's the same thing as a point in zero one and now here are my functions FN I guess before I do the functions is anything any questions about the setup okay so here's the function FN FN is going to be the function so X is the input like X is the infinite sequence and FN is we are going to gamble on the sequence and FN is how much money we're going to get after we gamble yeah like so x x is a number in zero and one but I want you to interpret X as an infinite list of heads and tails because I'm going to Define it in terms of heads and tails even though it's a function of X so I'm thinking of X both as a number in 0 and one and as an infinite list and I'm trying to tell you that those are the same thing if you just do the infinite binary decomposition yeah you said the binary decomposition successfully encapsulates the one half probability that's right yeah is kind of confused by why that I mean I intuitively yes okay let me tell you what it's great okay let's do let's do the first coin flip let's look at the set of binary sequences that have a zero in the first one and the set of things that have a one in the first one and you'll see that on the real number line so 01 is the real number line if you look at the set of X that start with a zero it's everything over here is to start with zero start with zero and if you look at the things that start with one it's everything over here start with one um so the first coin flip is equally likely to be heads and tails because they're a half a half if you look at the first two coin flips you'll get a division into four and it'll be 0 0 that'll be 01 and then it'll be like one Zer and one one and it'll be divided up into four and if for any number of PES you look at you divide it into 2 to the N yeah no it's it's a very nice property uh and it's quite nice I'll tell you uh because we're in I told you something about the Cantor set and I can't resist the Cantor set has something to do instead of coins that are heads or tails you look at three sided things that are uh heads Tails or sideways oh actually that's perfect yeah it's heads Tails or sideways and then you do something to it to make it the can set so the canor set you write out things as not as binary but intern the 0.0122 one01 and then you do something okay I won't spoil what the something is okay okay uh great question yeah any anybody else okay so what is FN of X it's going to be we're going to gamble this is our gamble our gamble payoffs ha G A gamble I was grading maybe I shouldn't say this I was grading somebody's assignment today uh in my other programming class and there was a gambler and they made a typo in their code and they called it a gamlet which I think is adorable that's like a baby Gambler right okay so our gamblers pay off out of the that was ready okay so our our gamblers payoff and here's what it's going to be our Gambler is going to bet it's FN we're going to bet on N heads coming up in a row okay we're feeling lucky we're like 10 heads in a row let's go go okay so how much money do we get and it's a fair at a fair casino at a fair Casino I'll tell you so if we if we lose then we lose $1 we're only going to bet $1 uh we're going to bet $1 so if we lose if it doesn't come up first n heads in a row then we're going to lose $1 minus one and this is if it's not if not uh heads heads heads heads heads at the beginning so we don't care even though it's an infinite sequence we only care about what happens in the first first end everything that happens after the first end we just don't care about like we only get to see the first end heads and if they're all heads then we're going to win and if we don't uh we're going to lose and we're going to lose a dollar if we don't uh if it is heads heads heads Heads then we're going to have a big payday here because getting n heads in a row is quite unlucky uh or you have to be quite lucky to get n heads in a row so the casino is going to give us a big payout to make it a fair casino this is a fair Casino they offer even odds bets does anyone know how much money the casino would pay you if you if it's a fair casino and you got to get heads heads heads heads heads in a row end times how big is uh n is n you're you can handle letters yeah yeah is it two to the N it's two to the N yes 2 to the n and actually more so 2 to the N is the probability of it 1 over 2 the N is the probability of it happening so the casino has to pay you out 2 to the N to make it fair uh there is one other small thing is actually it's 2 to the end minus one cuz you have to pay like an entry fee of a dollar you always lose your entry fee and then you get two to the end if you win you can also think about this if the casino if you go to a casino and you say I want to bet on heads to come up 10 times in a row or if you go to the roulette table because they don't actually have heads of Tails they have roulette you say I want to bet I get 10 blacks in a row the casino will say look man you can bet on red or black and then you can bet again but you can't just bet on the next 10 you have to bet on the next one but fortunately what you do is you just bet on black and then you double your bet every time okay so you bet black is going to come up if it wins you double if you win again you double if you win again you double and how many times did you double n times so that's why it's to you double your bet n times uh if you win then you you walk away and you win if at any point you lose then you lose your whole bet and you you Le in yourself okay so there's a very nice function FN and actually I will I will draw a picture for you of FN of X now FN of X you can think about as a function on 01 right X is in 01 it's a very simple function it looks like this it's 2 nus1 on the first interval of length 1 over 2 the n and it's zero or it's negative 1 everywhere else it's netive 1ga 1 so it's it's actually the function which is1 if X is bigger than 1 2 the n and it's 2 the N -1 if x is between 0 and 1 2 the n and you can see that because in our binary decomposition what numbers start with heads heads heads heads Heads it's uh only the numbers that are less than one over two end I think I might have flipped heads and tails maybe it has to be Tails Tails Tails Tails Tails I it probably okay okay look very simple fix you could just I'm doing it this way wait no okay I had it right okay wait look see I said it technically correct okay so I guess we're betting on heads heads heads heads Heads is the sequence 00000000 Z not this equals one1 okay good that is fine this is what I want does have to be around zero around one it could be around yeah if you flip heads and tails it'll be on the other side but then it's annoying to write down and you have to do a one minus I don't have room for that I hit the margin okay okay um so that's what the function FM look this is actually a relative of the sliding Spike functions we've seen a few times in the course so far right you have a very narrow Spike and the spike is getting bigger and bigger as the width gets smaller and smaller I'll tell you what the expected value of FN of U that means flip a fair sequence of points use an honest fair sequence of points expected value of FN of U is the expected value of our bet and in integral language this is the integral of FN of X DX or in as this textbook would say it's just the integral of FN and does anyone know what the expected value would be or the integral I'll let you think for a second you go to the casino you make a bunch of fair bets what is your expected value at the end of the day Zero yeah it's zero it's zero and in fact you know if you make Fair bets at a casino how could your expected value be anything other than zero every single bet you made was Zero I made n bets put them together still zero um so it's not surprising if you make a finite number of bets at a casino that is fair your expected value will be zero and you can also check right this height of this Spike exactly cancels out the negative that is quite wide um on the other hand let's look at the limit as n goes to Infinity of FN of X and I would say as n is getting bigger and bigger what is happening so for from for some fixed sequence of heads Tails heads tails what is the limit of FN what is happening and I claim this is -1 except for one point there's only one point where it's not minus one except for one single value of x right and this is saying as you bet on more and more heads to come up eventually you will lose you will not be heads forever right it will be minus one unless there's one exception which is what if it is heads forever huh you could get heads forever uh so if x equals heads heads heads heads Heads forever there is exactly one point which is heads heads heads forever that's the point x equals z so this this limit is equal to1 except for uh at zero and so if you ask for what is the probability that the limit as n goes to Infinity of FN of U equals minus1 it's a 100% chance there's a 100% chance this betting strategy will leave you with a profit of negative 1 so as you make as you bet on more and more heads to come up you're going to have a profit of negative one with a 100% chance even though your profit on average is zero all right so this is an example the functions FN they go to1 almost everywhere but uh the integral of FN uh does not go they don't go to negative 1 let's say FN FN does not go to uh1 in Norm uh and you can you can see that because the integral of FN equals z always okay so this is kind of like a lottery right like the lottery becomes so so unlikely and the payoffs become so large that you just always lose your you pay for the lottery ticket you pay a dollar you just get nothing so far not exciting I'm going to do one simple thing and I'm going to blow your mind we're going to flip it with a negative sign let's make the function GN which is minus FN and this is a function which is you gain a dollar unless it's heads heads Heads unless heads heads heads and when heads heads Heads comes up then you lose uh2 to the end doll okay if heads heads Heads uh this strategy before it was a very simple betting strategy you just double your bet all the time you look like uh like you're you're just being crazy this one very civilized betting strategy this is called the Martin Gale batting strategy Martin Gale why is it called the Martin Gale batting strategy there's a fun story there's a city in France called Mar a Martin if you're angle siiz like I am okay and and people from Marte would come to the casino in Paris and do this stupid betting strategy and they thought they were Invincible okay what is the betting strategy you bet on on black and if you lose you're like oh I'm down a dollar now I'm going to bet $2 and then if you lose now you bet $4 and if you lose again now you bet 8 you see you keep doubling your bet so that when you eventually win it cancels out all your previous losses and you end up with plus $1 and you will actually you will get a plus $1 very high probability chance right unless it comes up heads heads heads heads Heads what are the odds of that uh you get plus one if it does come up heads heads heads heads Heads then you uh uh lose a lot of money and here's the beautiful thing about this betting strategy it has a 100% chance if you bet for long enough of gaining you a dollar right the limit as n goes to Infinity of GN of a u it's equal to plus one with probability 100% okay 100% so this is a betting strategy that gains you a dollar with 100% strategy that's a lot more exciting than losing a dollar with 100% strategy that's why I had to bring it up okay so so here's a betting strategy you can gain a dollar with a 100% chance the only problem is the expected value is zero it's still zero you cannot escape and there's a whole theory of Martin SC which are these expected value zero processes and like when you can take limits of them in things so a whole area of probability again this could make for a nice project is's a very nice book called probability with Martin Gales you could do something from that book um okay the thing that I had to show you was a function so GN converges to plus one uh almost everywhere but uh GN does not converge to plus one in Norm because the integral of GN is is always zero okay so there you go almost everywhere and in Norm they're not exactly the same thing you can converge in almost everywhere but not in Norm questions or comments about that one let me let me leave you with a question now it's your turn to do a question what about the opposite so you've seen an example where you converge almost everywhere but you don't converge in Norm can you converge the other way is it possible that you converge in Norm but not almost everywhere is it possible um is is it possible to do the reverse uh yes it's possible or no it's not possible and again this is one just based on Vibes what do you think can you try to think of an example where it's converging in Norm but it is not converging almost everywhere that's what we going to do can you think of such an example I'll give you I guess like two minutes to brainstorm an example before I tell you the trick here is that we need need to that's so conver Norm means the absolute the integral goes to zero the integral of the absolute value goes to zero yeah Z I think it is if I have actually I'm thinking if we have a function that is like one with this the measure isn't zero but the is zero there you look Z no wait never mind no it's not okay so there's like 10 seconds left put in your whatever your Vibes are telling you and uh well see you know I think I'm going to be the single wrong in this class all right let's see what you guys said eight out of 11 people said no it's not possible the answer in analysis class it's always possible there's always some nasty one I should have Tred your and I will I will even tell you what the sequence is that does it and we're going to go over it next next class next class yeah because we have two minutes left here's the function here's the function that does it you can for every sequence so for every every sequence uh of heads and tails every every finite sequence let's say finite sequence of heads Tails can invent this function it's the function f heads Tails whatever you put in your favorite sequence right so you your sequence goes here okay you tell me that sequence and I'm going to make the function which is one if it starts if x starts with if x starts with your sequence and it's zero otherwise and as an example the function we had before FN what I called FN before was something like the sequence F of heads heads heads heads Heads n times in a row multipli by 2 to the N you got a payout in in this function you only got a payout of size one before you got a payout of size 2 the N uh minus one so old example was so you can bet on basically you go to the casino and you bet on any PR you pick your favorite sequence and you're betting that it comes up but you only get $1 if it comes up you get Zer if it it uh doesn't come up which is actually rather generous of the casino because you're not paying anything right there's no negatives um this is a function if you look if you enumerate these right so there's two for every n for sequences of length n there's two to the N such guys if you enumerate all the guys so you let GN be an enumeration of f of all sequences so line up all the sequences and count them up the first sequence the second sequence the third sequence uh the first two sequences are probably heads and tails then it's heads heads Heads Tails Tails heads Tails tails and so on do the L three ones this sequence GN it will go to zero in Norm GN goes to zero in Norm which is not hard to see the integral of this thing will go to zero um but but uh but GN of X the limit does not exist n goes to Infinity does not exist for all X so for every sequence this limit does not exist it goes one zero infinitely often it will sometimes be one and sometimes be zero no matter what point you're at and we're going to see exactly why next class question here if we're mapping the sequences to the interval 01 have similar cality wouldn't that immediately exclude the possibility an enumeration of the sequences by H yeah that's a great question but I'm only looking at finite sequences oh that's right I'm not this is a great point I don't want I don't want you can't do F of an infinite sequence it's F of a finite sequence and I want you to enumerate all finite sequences yeah that's great I'll add the word finite be an enumeration of all finite sequences there you go that's a great okay we'll see this one next class have a great weekend the homework is posted make sure you go take a look and don't forget to write up a nice solution for homework one to get the Boost uh okay |
1182 | http://www.phys.ufl.edu/courses/phy3221/spring11/lecture1.pdf | 1 PHY3221 Detweiler Lecture #1 Relativity Time Dilation Some Taylor expansions: I had intended to go over these Taylor expansions again in class. They are useful for the homework.
1 1 −x = 1 + x + O(x2), √ 1 −x = 1 −1 2x + O(x2), 1 √1 −x = 1 + 1 2x + O(x2) The mathematical notation O(x2) represents something that is smaller than some number times x2, when x is very small. Then, x2 will usually be comparable to the error in these approximations. In this class, when x is very small, you are allowed to just use 1/(1 −x) ≈1 + x √ 1 −x ≈1 −1 2x 1/ √ 1 −x ≈1 + 1 2x.
Try them out on your calculator for x = 10−1, 10−2, 10−3, 10−4 and 10−5 to see how the approximation improves while x decreases. See if the error is comparable to x2.
Details of Time Dilation Definition A thought experiment: is an experiment that we can imagine doing, but which would not be possible under reasonable circumstances.
Thought experiments allow us to test our understanding and to sharpen our physical intuition. Formulating a thought experiment is particularly useful when the laws of physics, as understood, appear superficially to be paradoxical, self-contradictory or inconsistent.
Definition An Event: is a particular place at a particular time. Something might happen at that place and at that time, such as the snap of my fingers, or a firecracker going off.
Different frames of reference might give different coordinates for the location and time of an event.
A thought experiment: A horizontal mirror is a height h above a source-detector of light.
2 h ∗∗ #1 Alice starts a pulse of light from the emitter at event #1.
The light then bounces offthe mirror and is reflected back to the detector where it is received at event #2. Alice measures the time interval to be ∆t0 = t2 −t1 = dist/speed = 2h/c (1) 2 between event #1(emission) and event #2 (detection).
Definition Proper time interval: Notice that in Alice’s frame of reference the events #1 and #2 occurred at the same location in space but at different times. This is important: If two events occur at the same location, then the time interval measured between these two events in that frame of reference is special and called the proper time interval. The smallest time interval between these two events is always the proper time interval. In any other frame of reference, the events would occur at different locations and the time interval would be longer than the proper time interval.
Alice’s experimental apparatus is actually on a train going through a station with a speed v. In the station Bob sees the source-detector move a distance L before the light is detected.
2 d d L h ∗ ∗ #1 Let Bob be in the primed frame of reference.
Notice that in Bob’s frame of refer-ence event #1 and event #2 do not occur at the same location in space—and this is not at all surprising because the train is moving through Bob’s station.
Assume that Bob measures a change in time ∆t′ between the events #1 and #2. Then the distance between the events, in Bob’s frame, is just L = v∆t′.
Bob measures the length of the light’s path and obtains distance = 2 p h2 + L2/4 = 2 p h2 + v2∆t′2/4, which is longer than the distance 2h that Alice measured. From the principle of relativity Bob knows that the speed of light is c, so he measures ∆t′ = dist/speed = 2 p h2 + v2∆t′2/4 c .
Now, Bob sits back and does some algebra: squaring both side gives ∆t′2 = 4(h2 + v2∆t′2/4) c2 , and solving for ∆t′2 yields ∆t′2 1 −v2 c2 = 4h2 c2 or ∆t′2 = 4h2/c2 (1 −v2/c2) = ∆t2 (1 −v2/c2) (Use eqn. 1 on the previous page.) or ∆t′ = ∆t p 1 −v2/c2 3 So in Bob’s frame of reference the light traveled farther than it did in Alice’s frame of reference. But the speed of light must be the same in the two frames of reference. The unmistakeable conclusion is that the time interval ∆t′ that Bob measures between events #1 and #2 is longer than the time interval that Alice measures.
The time interval between two event, as measured in the frame of reference where the events occur at the same place, is called the proper time ∆t0 between the events. Any other observer who is moving with respect to this particular frame will measure a longer time interval ∆t′ = ∆t0 p 1 −v2/c2 .
Because Alice’s time interval is shorter than Bob’s time interval, Bob concludes that Alice’s watch ticked offfewer seconds and therefore runs slower than Bob’s own watch.
This consequence of special relativity is called time dilation.
This situation is actually symmetrical: Bob could do a similar experiment in the station.
And Alice could go by in the train with a velocity v. But, from Alice’s frame of reference, Bob would appear to have a velocity −v. The analysis of the time intervals is just that same as above, except that the two events would occur at the same location only in Bob’s frame of reference. and he would measure the proper time interval ∆t0. Alice would measure a longer time interval ∆t = ∆t0/ p 1 −v2/c2, and she would conclude that Bob’s clock was more running slower than her own watch.
Time dilation might seem paradoxical, but here is a statement that is true: When Bob carefully observes Alice’s watch, it will appear (to him) to run slower than his own watch.
Here is another statement that is true: When Alice carefully observes Bob’s watch, it will appear (to her) to run slower than her own watch.
This conclusion is surprising. It appears to be a logical paradox and to contradict our everyday experiences involving time, trains and watches. But, it has been tested experimen-tally and describes the way Nature behaves.
To summarize: The proper time ∆t0 is the smallest time interval between two events that any inertial (non-accelerating) observer can measure. Any inertial observer for which these same two events do not occur at the same location, will measure a longer time interval ∆t = ∆t0/ p 1 −v2/c2, where v is the magnitude of the relative speed between the frames of reference. |
1183 | https://www.mdpi.com/1422-0067/23/9/4874 | Epigenetic Mechanisms of Epidermal Differentiation
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Table of Contents
Abstract
Introduction
Search Strategy
The Role of Epigenetics in Health
The Role of Epigenetic Mechanisms Involved in Epidermal Differentiation
The Role of Epigenetics in Inflammatory Skin Diseases
Conclusions
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Open Access Review
Epigenetic Mechanisms of Epidermal Differentiation
by
Chiara Moltrasio
Chiara Moltrasio
SciProfilesScilitPreprints.orgGoogle Scholar
1,2,,†,
Maurizio Romagnuolo
Maurizio Romagnuolo
SciProfilesScilitPreprints.orgGoogle Scholar
1,3,† and
Angelo Valerio Marzano
Angelo Valerio Marzano
SciProfilesScilitPreprints.orgGoogle Scholar
1,3
1
Dermatology Unit, Fondazione IRCCS Ca’ Granda Ospedale Maggiore Policlinico of Milan, 20122 Milan, Italy
2
Department of Medical Surgical and Health Sciences, University of Trieste, 34137 Trieste, Italy
3
Department of Pathophysiology and Transplantation, Università degli Studi di Milano, 20122 Milan, Italy
Author to whom correspondence should be addressed.
†
These authors contributed equally to this work.
Int. J. Mol. Sci.2022, 23(9), 4874;
Submission received: 28 March 2022 / Revised: 24 April 2022 / Accepted: 27 April 2022 / Published: 28 April 2022
(This article belongs to the Special Issue Functional Defects of Keratinocytes in Inflammatory Skin Diseases)
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Abstract
Keratinocyte differentiation is an essential process for epidermal stratification and stratum corneum formation. Keratinocytes proliferate in the basal layer of the epidermis and start their differentiation by changing their functional or phenotypical type; this process is regulated via induction or repression of epidermal differentiation complex (EDC) genes that play a pivotal role in epidermal development. Epidermal development and the keratinocyte differentiation program are orchestrated by several transcription factors, signaling pathways, and epigenetic regulators. The latter exhibits both activating and repressive effects on chromatin in keratinocytes via the ATP-dependent chromatin remodelers, histone demethylases, and genome organizers that promote terminal keratinocyte differentiation, and the DNA methyltransferases, histone deacetylases, and Polycomb components that stimulate proliferation of progenitor cells and inhibit premature activation of terminal differentiation-associated genes. In addition, microRNAs are involved in different processes between proliferation and differentiation during the program of epidermal development. Here, we bring together current knowledge of the mechanisms controlling gene expression during keratinocyte differentiation. An awareness of epigenetic mechanisms and their alterations in health and disease will help to bridge the gap between our current knowledge and potential applications for epigenetic regulators in clinical practice to pave the way for promising target therapies.
Keywords:
keratinocytes; epidermal differentiation complex; epigenetic regulators; ATP-dependent chromatin remodeler; DNA methyltransferases; histone modifications; polycomb proteins; microRNAs; psoriasis; atopic dermatitis
1. Introduction
The epidermis constitutes the skin surface and is composed of specialized epithelial cells called keratinocytes, which originate from two pools of quiescent epidermal stem cells (qESC), one living in the basal layer of the interfollicular epidermis (IFE) and the other, bulge stem cells, located in the hair follicle of the sebaceous gland that gives rise to keratinocytes of the hair follicle (HF) lineage. From qESC, transient amplifying (TA) cells generate “mature” keratinocytes residing in the basal layer of the epidermis or in the hair bulge (region of the outer root sheath) . Progenitor cells of the basal layer undergo a programed process of proliferation and differentiation, culminating in the formation of a cornified envelope consisting of corneocytes, and terminally differentiated, enucleated, keratinocytes . Epidermal growth and differentiation are driven by the accurate expression of multiple genes regulated by several transcriptional and post-transcriptional mechanisms . This strict control is often challenged by external biological or environmental factors, such as UV light , and/or modulated by physiological factors (e.g., aging) . Collectively, genomic alterations result in a broad deregulation of gene expression and the dysfunction of signal transduction pathways that control proliferation and several cellular functions. In addition to changes in DNA sequence and structure, it has become increasingly evident that all cell processes can be deeply orchestrated by epigenetic mechanisms . Epigenetic modifications refer to reversible changes in gene expression without alteration of the DNA sequence and occur throughout all stages of development or in response to environmental factors . On the other hand, recent research has raised the notion that epigenetic mechanisms could mediate “stable” changes in tissue function. These studies converged on a set of common enzymatic modifications to chromatin structure that can up- or down-regulate gene expression. Thus, epigenetic mechanisms play a crucial role in the control of cellular functions and can help to explain the relationships between a genetic background and effects of the environment on the susceptibility to different diseases , including skin disorders . During last years, it has been shown that epigenetic mechanisms are involved in the growth and differentiation of keratinocytes by regulating a gene-rich region at 1q21 known as the epidermal differentiation complex (EDC) . These genes encode involucrin, loricrin and small proline-rich proteins, as well as several calcium binding proteins and S100A proteins that play a pivotal role in epithelial tissue development and repair by regulating the terminal differentiation program of keratinocytes through signal transduction events . During terminal differentiation, a pool of epigenetic regulators influences the programing of epidermal keratinocytes via induction or repression of EDC genes, thus playing a crucial role in epidermal development. In particular, the program of epidermal development and keratinocyte differentiation is governed by several transcription factors (the most important of which is p63), signaling pathways (Notch, Wnt, Bmp, Hedgehog, and so on), and epigenetic regulators. The latter exhibits both activating and repressive effects on chromatin in keratinocytes and includes the following: ATP-dependent chromatin remodeler Brg1 and Mi-2β; histone demethylase Jmjd3 and Setd8 and genome organizer Satb1 that promote terminal keratinocyte differentiation; DNA methyltransferase DNMT1; histone deacetylases HDAC1/2; Polycomb proteins (in particular, Bmi1 and Ezh1/2) that stimulate the proliferation of progenitor cells and inhibit the premature activation of terminal differentiation-associated genes [11,12,13,14,15]. Despite the advancement of knowledge, many aspects of epigenetic control of the gene expression programs in keratinocytes remain to be elucidated.
In the present review, we aim to summarize the available evidence on activating and repressive effects on chromatin that promote epidermal differentiation. In the first section, we review the normal functions of epigenetic mechanisms and their role in epidermal differentiation. The second section focuses on our current understanding of the different types of epigenetic changes in psoriasis and atopic dermatitis—as emblematic examples of inflammatory skin diseases—providing some specific examples of each.
In summary, we provide an overview of the epigenetic mechanisms underlying epidermal differentiation in health and disease and we conclude with a discussion of the potential implications for clinical practice.
2. Search Strategy
Criteria for Paper Selection
To provide an overview of the current state of knowledge regarding the topic of epidermal differentiation, papers were selected from those included in the electronic databases PubMed/MEDLINE, Google Scholar, Scopus, and Web of Science over the past 20 years. The following search terms were used: keratinocytes, epigenetics, epigenetic regulators, methylation, histones, epidermal differentiation, epidermal differentiation complex, microRNAs, psoriasis, atopic dermatitis, inflammatory skin diseases, environmental factors. Each selected paper was analyzed, the data extracted, and presented in the main text to provide an overview of the main aspects of epigenetic regulation that influences the programming of epidermal keratinocytes.
3. The Role of Epigenetics in Health
Epigenetics refers to changes in gene expression without alteration of the DNA sequence. These changes, which make up the so called “epigenome”, include modifications to DNA and the histone components of nucleosomes as well as expression of noncoding RNAs (ncRNAs). Since epigenetic signatures help determine if genes are turned on or off, it influences the production of encoded proteins, thus ensuring cell-type-specific expression . Epigenetic mechanisms of the control of gene expression include several levels of regulation that are based, among others, on: (i) methylation and hydroxymethylation; (ii) histone covalent modifications, such as the addition or removal of methyl- or acetyl- groups, at specific genome regions; (iii) changes in nucleosome positioning and ATP-dependent chromatin remodeling; (iv) RNA regulation . The three main epigenetic mechanisms are DNA methylation, histone modification, and miRNA. These mechanisms are responsible for the initiation and maintenance of a gene expression profile within a series of cellular processes, including cell differentiation, embryogenesis, and genomic imprinting . DNA methylation involves the transfer of a methyl group into the C5 position of cytosine to form 5-methylcytosine and promote gene expression regulation by either recruiting proteins involved in gene repression or inhibiting the binding of transcription factor(s) to DNA . During development stages, the pattern of DNA methylation changes, resulting in a plastic and dynamic landscape involving both de novo DNA methylation and demethylation. DNA methylation is regulated by a family of DNA methyltransferase (DNMTs): DNMT1, DNMT2, DNMT3A, DNMT3B, and DNMT3L . DNMT1 usually methylates CpG islands on hemimethylated DNA and is localized in replication foci during the S phase of the cell cycle. In line with this function, DNMT1 is responsible for maintaining DNA methylation but can catalyze de novo DNA methylation in specific genomic contexts (Figure 1a).
Figure 1. (a) The concurrence of DNA methylation and demethylation. Exported by Biorender.com (accessed on 1 April 2022). (b) Histone modifications including acetylation and methylation of lysine. Created in Biorender.com. (c) MicroRNA formation and their role in protein degradation. Created in Biorender.com.
Histone modification is another key epigenetic mechanism. Histone complexes are composed of two H2A-H2B dimers and a H3-H4 tetramer to form the nucleosome, and they may be chemically modified through the action of enzymes to regulate gene transcription. The most common modifications are the methylation of arginine or lysine residues, or the acetylation of lysine (Figure 1b).
Methylation can modulate the interaction between transcription factors and other proteins with nucleosomes , while lysine acetylation eliminates a positive charge on lysine, thereby weakening the electrostatic attraction between histone and DNA, thus resulting in partial unwinding of the DNA to make it more accessible for gene expression . Mature microRNA (miRNA) is another key player in the epigenetic picture (Figure 1c).
MiRNAs are a group of short (~22 nucleotides), single stranded, non-coding RNAs that can regulate the expression of other protein-coding gene at the post-transcriptional level. MiRNAs function via base-pairing with complementary sequences within messenger RNA (mRNA); after this binding, mRNA molecules are silenced by one or more of the following processes: (i) mRNA cleavage, (ii) destabilization of the mRNA through de-adenylation of its poly(A) tail, and (iii) translational inactivation . In rare cases, miRNAs can mediate the activation of gene expression profiles under specific conditions .
Below, we review the mechanisms that control chromatin remodeling, with special emphasis on keratinocyte-specific gene expression programs in skin development (data illustrating the role of chromatin remodeling factors in epidermal differentiation are shown in Table 1) and in inflammatory skin diseases, such as psoriasis and atopic dermatitis.
Table 1. Effect of knockout/down and overexpression of chromatin-modifying protein on keratinocyte/epidermal growth and differentiation.
4. The Role of Epigenetic Mechanisms Involved in Epidermal Differentiation
4.1. Epigenetic Regulation by ATP-Dependent Chromatin Remodeling
ATP-dependent chromatin remodeling complexes use the energy of ATP hydrolysis to alter chromatin architecture by repositioning, assembling, and restructuring nucleosomes. These complexes fall into one of four families: switch/sucrose non-fermentable (SWI/SNF), Mi-2/nucleosome remodeling and deacetylase (NuRD) complex, imitation switch (ISWI)/SNF2L, and INO80 . The SWI/SNF family, also called the BAF complex (Brg/Brm associated factor), is thought to regulate gene expression by altering nucleosome positioning and structure. The human SWI/SNF complex contains either Brg1 or hBrm (Brahma) as ATPase subunits that can play different roles in various cellular processes including proliferation and differentiation; moreover, these molecules also contain bromodomains that allow binding to acetylated-lysine residues within histone H3 and H4 tails . Genetics studies have demonstrated a crucial role of Brg1 in the development of skin; indeed, Brg1 deficiency gravely impairs the final stage of keratinocyte terminal differentiation, leading to skin barrier defects . Moreover, Brg1 promotes the relocation of the EDC locus from the nuclear periphery toward the nuclear interior and into the compartment enriched by nuclear speckles, which is associated with overexpression of the EDC genes .
Additionally, regulatory subunits of the SWI/SNF complex such as the actin-like 6A protein/Brahma-associated factor (ACTL6a/BAF53A), along with the catalytic subunits (Brg1, Brm and BAF250a), are involved in suppressing differentiation and maintaining the progenitor state in epidermal cells .
ATP-dependent chromatin remodelers of the CHD group are characterized by a chromodomain that specifically binds to methylated lysine residues; the ATPase subunits within this family include Chd1-9, but Chd3 and Chd4 (also known as Mi-2β) have a prominent role in NuRD complex. The chromatin remodeler Mi-2β is crucial for the self-renewal of epidermal precursors during the earlier phases of embryogenesis, and its ablation leads to defective basal layer formation, whereas its loss during the later stages of embryogenesis leads to impaired induction and development of the hair follicles .
4.2. Epigenetic Regulation by Histone Methylation
Histone methylation is the modification of certain amino acids in a histone protein by the addition of one, two, or three methyl groups. The site-specific methylation and demethylation of histone residues are catalyzed by methyltransferases and demethylases, respectively. Histone methylation is generally associated with transcriptional repression; however, methylation of some lysine and arginine residues of histones results in transcriptional activation . Trimethylation of histone H3 at the lysine 4 position (H3K4me3) leads to actively transcribed genes as well as H3K4Me2/1, H3K79me2/3 and H3K36me3 .
In contrast, the chromatin of an inactive gene is enriched in H3K9me2/3, H3K27me2/3 and H4K20me3 modifications . The Jumonji domain-containing protein-3 (JMJD3), also known as lysine-specific demethylase 6B (KDM6B), is a histone demethylase that regulates the trimethylation of histone H3 on lysine 27 (H3K27me3), and it has been studied extensively in immune diseases, cancer, and tumor development . Recent studies illustrated that JMJD3 plays a pivotal role in the cell fate determination of pluripotent and multipotent stem cells (MSCs), enhancing their self-renewal ability and reducing the differentiation capacity of embryonic stem cells (ESCs) and MSCs into specialized cells . The role of H3K27me3 in human epidermal progenitor cells has been revealed by inactivation of JMJD3, which led to a blockade of progenitor cell differentiation . The genes normally induced during epidermal differentiation were repressed in epidermal progenitor cells lacking JMJD3 protein, and their induction was associated with H3K27me3 de-methylation in their promoter regions, indicating that epigenetic transcriptional upregulation by JMJD3 controls mammalian epidermal differentiation .
Histone H4 mono-methylation at lysine 20 is regulated by Setd8 histone methyltransferase. Setd8 is a transcriptional target of c-Myc, a protein involved in cellular epidermal differentiation, and it is involved in many physiological processes, including cell cycle, chromatin condensation, apoptosis, tumorigenesis, and epithelial to mesenchymal transition . The loss of Setd8 methyltransferase led to inhibition of progenitor cell proliferation in the basal layer of the epidermis with simultaneous impairment of epidermal cell proliferation and differentiation . Moreover, Setd8 acts as regulator of p63 expression, a master regulator of epidermal development , and when Setd8 is lost, epidermal cells fail to express p63 and exhibit impaired terminal differentiation, resulting in skin apoptosis .
In HaCaT cells, spontaneously immortalized human keratinocytes, knockout of Suv39h1, a gene encoding a H3K9 histone methyltransferase, provoked a marked reduction in the level of tri-methylation at the ninth lysine residue of the histone H3 protein and promoted changes in the expression of several keratinocyte-specific genes . Suv39h1 knockout also led to induction of genes encoding differentiation markers such as keratin 10, desmoglein 1, S100A8, and late cornified envelope (LCE1) proteins, whereas expression of genes associated with undifferentiated keratinocytes, such as keratin 14 and S100A6, remained unaltered, or at most, slightly decreased . Overall, the overexpression concerned mainly genes of the middle/late differentiation stage reinforcing the keratinocyte differentiation program.
Lysine demethylase Jarid1 family members (Jarid1a, Jarid1b, Jarid1c, and Jarid1d) demethylase the fourth lysine residue of histone H3 and participates in multiple repressive transcriptional complexes, although the specific regulatory effects on chromatin remain undefined . The role of Jarid1b in cell differentiation has recently received increased attention; it positively controls epidermal cell differentiation both in vitro and in vivo, increasing expression of mesenchymal-epithelial transition (MET)-related genes such as Ovol1 (Ovo Like Transcriptional Repressor 1), which is regulated in turn by the PI3K-AKT signaling pathway . Jarid1b activates PI3K/AKT/Ovol1 by controlling suppressors of this pathway such as Ship1 (SH2-containing inositol-5′-phosphatase 1); knockdown of Ship1 activates the downstream PI3K-AKT pathway and enhanced Ovol1 expression in HaCaT cells, whereas overexpression of Ship1 led to decreased Ovol1 expression . Thus, Sun et al., demonstrated that Jarid1b negatively regulates Ship1 expression by directly marking its promoter to modulate H3K4me3 enrichment, and consequently, to drive the regulation of keratinocyte differentiation .
4.3. Epigenetic Regulation by the Genome Organizer Satb1
Special AT-rich sequence-binding protein-1 (Satb1) is a global chromatin organizer and transcription factor and is critical for integrating higher-order chromatin architecture with gene regulation .
Satb1 regulates gene expression by acting as a “docking site” for several chromatin remodelers and by recruiting corepressors or coactivators directly to promoter regions , thus establishing specific three-dimensional conformations in tissue-specific gene loci . Satb1 is expressed in basal epidermal keratinocytes and promotes cell differentiation through higher-order chromatin folding and transcriptional regulation of the EDC locus , and its depletion in a mouse skin model caused thinning of the epidermis accompanied by strong alteration of the expression of terminal differentiation-associated genes in keratinocytes .
4.4. Epigenetic Regulation by DNA Methylation and Hidroxymethylation
DNMT1is both a maintaining and de novo DNA methylation enzyme that is enriched in undifferentiated epidermal progenitor cells, where it is required to retain proliferative potential and suppress differentiation; after the embryonic stages, it remains confined to the basal epidermal layer containing proliferating keratinocytes . Khavari et al. used human 3D culture and DNMT1 knockdown in primary human keratinocytes to demonstrate a prominent role for DNMT1 in the maintenance of epidermal progenitor cells and epidermal tissue renewal .
Epidermal depletion of DNMT1 coexists with overexpression of genes associated with cell cycle arrest such as Cdk inhibitors (e.g., cycling-dependent kinase inhibitors p16INK4a and p15INK4B); however, cyclin-dependent kinase 4 and 6 (Cdk4/6) partially rescues the effect of DNMT1 impairment on cell proliferation .
These findings support the crucial role of DNMT1 in preserving epidermal progenitor cell identity and remodeling of this pattern during terminal differentiation .
Some gene promoters remain active when methylated through the binding of CCAAT-enhancer-binding proteins (C/EBPα) that contribute to the opening of the chromatin structure with consequent gene activation, whereas 5-aza-cytidine promotes the inhibition of DNA methylation and prevents the expression of a subset of genes during calcium-induced differentiation in culture . 5-azacytidine also differentially affects gene expression within the EDC locus in normal human keratinocytes; indeed, Elder and Zhao demonstrated small proline-rich protein 1/2 (PRR1/2) and involucrin were overexpressed and S100A2 was under-expressed when compared to controls . These data suggests that DNA methylation can regulate both gene activation and repression with a lack of correlation between methylation status and gene expression level, probably in line with the required balance between proliferation/quiescence of progenitor cells.
Finally, 5-hydroxymethylcytosine (5hmC) is the first oxidative product in the active demethylation of 5-methylcytosine (5mC) (Figure 1a). Three ten-eleven translocation (TET) enzymes (TET1/2/3) catalyze the hydroxylation of DNA (5mC) to 5-hydroxymethylcytosine (5hmC) and can further catalyze oxidation of 5hmC to 5-formylcytosine (5fC) and then to 5-carboxycytosine (5caC) . It has been proposed that 5-hydroxymethylcytosine, the first product in the active demethylation of 5mC, is prevalent in embryonic stems cells and reduced levels of TET1 and subsequently 5hmC causes the impaired self-renewal of stem cells . It has also been hypothesized that conversion of 5mC to 5hmC by TETs blocks the repressive methyl-CpG-binding domain (MBD) and DNMT proteins , promoting the activation of gene expression; however, this role of this process in the control of gene expression in keratinocyte progenitor cells remains unclear.
4.5. Epigenetic Regulation by Histone Deacetylation
Histone deacetylases (HDACs) are a class of enzymes that remove acetyl groups from an ε-N-acetyl lysine amino acid on a histone, resulting in a more closed chromatin structure and repression of gene expression . HDACs play a crucial role in coordinating the crosstalk between signaling pathways with chromatin remodeling and transcription factors to orchestrate gene expression . HDAC1 and HDAC2 are involved in hair follicle formation as well as epidermal development and stratification. Their role in the control of the gene expression program in epidermal progenitor cells and epidermal differentiation has been demonstrated by LeBouef et al. who showed that HDAC1/2 null epidermis failed to differentiate, remaining single-layered . The same authors also demonstrated that HDAC1/2 directly mediate the repressive activity of p63 in keratinocytes and independently suppressed p53 activity via deacetylation of the p53 protein .
Further evidence for the role of HDACs was obtained after depletion of HDAC1/2 from the basal layer of mouse epidermis resulted in increased histone acetylation and enhanced keratinocyte proliferation, leading to hyperplasia in addition to apoptosis and hair follicle dystrophy .
Zhu et al. demonstrated for adult mouse epidermis that homozygous epidermal HDAC1/2 codeletion—in association with a null p53 expression—partially restored epidermal proliferation, indicating that the effects of HDAC1/2 deletion on keratinocyte proliferation are mediated in part via increased p53 activity. Moreover, p16 deletion did not significantly rescue epidermal thickness, proliferation, and/or apoptosis of HDAC1/2 null epidermis, but its loss enabled survival of HDAC1/2-deficient keratinocytes by preventing their senescence. Finally, the same authors demonstrated that double HDAC1/2 homozygous deletion caused hyperacetylation of c-Myc. Taken together, these findings indicate that HDAC1/2 play three distinct roles in adult epidermis: (i) deacetylation of p53 to promote progenitor cell proliferation, (ii) deacetylation of c-Myc to prevent premature differentiation, and (iii) restraint of p16 to repress senescence and allow long-term maintenance of progenitor cells .
4.6. Epigenetic Regulation by Polycomb Group (PcG) Proteins
Polycomb group (PcG) proteins are transcriptional repressors with a key role in stem cell identity and differentiation. PcG proteins are evolutionarily conserved and act within complexes, called Polycomb repressive complexes (PRCs).
PRC1 mono-ubiquitinates histone H2A on lysine 119 (H2AK119Ub1) and PRC2 catalyzes trimethylation of lysine 27 on histone H3 (H3K27me2/3) . Chromobox 4 (Cbx4) is a component of the PRC1 complex and plays an important role in the maintenance of quiescence in human epidermal progenitor cells .
Bmi-1 is another PRC1 component. It is expressed in the basal layers of human epidermis and is implicated in the control of cell survival by altering cell cycle regulatory protein expression, such as p16 and p19, and inhibiting apoptosis [12,61].
PRC2 is the most widely conserved PcG complex in multi-cellular eukaryotes and is composed by four core subunits with an intrinsic histone methyltransferase activity; the catalytic subunit, known as Ezh2 (enhancer of zeste homolog 2), contains the signature SET domain commonly found in lysine methyltransferases . It has been shown that Ezh2 as well as Ezh1 are involved in the regulation of EDC genes in epidermal cell progenitors, providing strong evidence for the involvement of this epigenetic actor in the temporal and spatial keratinocyte terminal differentiation program . On the other hand, Ezh1/2 knockout mouse models do not show any epidermal defects, suggesting that the loss of their activities in epidermal keratinocytes might be compensated by other PRC2 components . The transcriptional repressive function of Ezh1/2 are potentiated by Jarid2, which also acts as a component of the PRC2 complex ; depletion of Jarid2 in mouse epidermis inhibits proliferation and promotes differentiation of progenitor cells postnatally. Jarid2 deficiency in keratinocytes reduces H3K27m3, resulting in delayed hair follicle cycling because of decreased proliferation of hair follicle stem cells and their progeny . Thus, similar to Ezh1/2, Jarid2 is required for the maintenance of cell proliferation and inhibition of differentiation in epidermal stem and progenitor cells .
4.7. Epigenetic Regulation by microRNAs
MicroRNAs (miRNAs) are a group of short (~22 nucleotides), single stranded, non-coding RNAs, which can regulate the expression of other protein-coding gene at post-transcriptional level. In most cases, miRNAs interact with the 3′ UTR of target mRNAs to suppress expression , but they can also activate gene expression under specific conditions . In human skin, different miRNAs have been reported to play a pivotal role in epidermal development by regulating differentiation , and several studies have provided evidence of differential patterns of miRNAs expression in epidermis and hair follicle, which may be indicative of differences in the epigenetic control of these two populations .
There are several reports documenting the role of miRNAs in epidermal development and differentiation; for example, miR203, is induced in terminally differentiated keratinocytes, both in vitro and in vivo. Through the regulation of p63 expression, miR203 maintained the proliferative and differentiation potential of basal keratinocytes as well as precursor cells . Besides miR-203, miR-23b was found to be a second miRNA highly expressed in differentiated keratinocytes, identifying it as a differentiation marker for human skin .
A total of nine miRNAs—miR-203, miR-95, miR-210, miR-224, miR-26a, miR-200a, miR-27b, miR-328, and miR-376a—were found not only to be induced in terminally differentiated keratinocytes in vitro and in vivo but were also functionally involved in the differentiation process .
Another miRNA, miR-214, when overexpressed, led to a reduction in epidermal thickness, lower keratinocyte proliferation rate, and hair follicle loss .
These examples suggest that several miRNAs are involved in different processes between proliferation and differentiation in the program of epidermal development.
5. The Role of Epigenetics in Inflammatory Skin Diseases
In the field of dermatology, skin disorders notably affect skin integrity and impair epidermal differentiation. Epigenetic mechanisms have been extensively studied in cutaneous T-cell lymphoma (CTCL) , but in recent years, the causative role of epigenetics is also emerging in inflammatory skin diseases, such as psoriasis and atopic dermatitis . The involvement of epigenetic determinants in these conditions has been arduously studied, and the great amount of published data reflects their importance in the etiology of these diseases. A dysregulation of multiple epigenetic mechanisms, including aberrant DNA methylation, alterations in histone modifications, and miRNA expression have been found to play a crucial role in the pathogenetic scenarios of these skin conditions. Notably, differences in epigenetic signatures could be observed not only between the skin of healthy control and lesional skin of psoriatic patients but also between a patient’s lesional, perilesional, and unaffected skin [73,74,75,76], indicating different phases of the disease may have unique epigenetic signatures.
5.1. The Role of Epigenetics in Psoriasis
Psoriasis is a chronic, immune-mediated inflammatory skin disease with a multifactorial etiology involving accelerated proliferation and abnormal differentiation of keratinocytes . It is now considered an autoinflammatory keratinization disease (AIKD), a term that encompasses disorders with mixed pathomechanisms of autoinflammation and autoimmunity .
Epigenetic changes, including DNA methylation, histone modification, and noncoding RNA regulation have been reported to be involved in the complex pathogenesis of psoriasis (Figure 2).
Figure 2. Epigenetic modifications in psoriasis: DNA methylation, histone modifications and microRNAs, including the microRNAs that can be considered as potential therapeutic targets and biomarkers for severity or prognosis. Created in Biorender.com.
5.1.1. DNA Methylation
Roberson et al. first identified global DNA methylation in psoriatic skin lesions when compared to controls, in which more than 1000 differentially methylated CpG islands (CGIs) were detected and, among them, twelve mapped to the epidermal differentiation complex or nearby genes upregulated in psoriasis. Moreover, the authors highlighted the reversible nature of DNA methylation, suggesting that this epigenetic mechanism is a dynamic actor in this disease .
Chandra et al., in an epigenome-wide DNA methylation study, found several differentially methylated loci, overlapped with different PSORS regions, involved in the regulation of pathogenetic gene expression, such as S100A9 (S100 calcium binding protein A9), SELENBP1 (selenium binding protein 1), CARD14 (caspase recruitment domain family member 14), KAZN (kazrin, periplakin interacting protein), and PTPN22 (protein tyrosine phosphatase non-receptor type 22), indicating the potential role of DNA methylation in regulating specific histopathological features commonly seen in psoriasis .
A candidate gene approach has been employed to define the promoter methylation profile of several psoriasis risk loci and, according to gene ontology, CpG hyper/hypomethylation coincided with genes involved in many processes impaired in the pathogenesis of psoriasis, such as cell cycle, apoptosis, immune system regulation, cell communication, and signal transduction . Loss of methylation has been unraveled at the promoter region of SHP-1 (Src homology region 2 domain-containing phosphatase-1), a protein tyrosine phosphatase (PTPs) that regulates several cellular processes, including growth and cell differentiation , and at the promoter level of p16 , p15 and p21, which encode negative regulators of the cell cycle in hematopoietic stem cells of psoriatic patients .
In addition, hypermethylation has been detected in the promoter sequences of PDCD5 (programed cell death 5) and TIMP2 (TIMP metallopeptidase inhibitor 2) genes, which are involved in apoptosis and the maintenance of tissue homeostasis by suppressing the proliferation of quiescent tissues, respectively, thus regulating the proliferation of keratinocytes . Moreover, the promoter of SFRP4 (secreted frizzled related protein) gene, a negative regulator of the Wnt signaling pathway, has been found heavily hypermethylated in psoriasis skin lesions and its consequent downregulation contributes to epidermal hyperplasia .
Finally, genome-wide association studies have identified hundreds of hypermethylated genes in psoriasis, and among them are immune-associated genes, such as TLR-7 (Toll-like receptor 7) and IRAK1 (interleukin 1 receptor associated kinase 1) .
5.1.2. Histone Modification
Evidence of unbalanced histone modification in psoriatic skin is also beginning to increase. H3K9 dimethylation is decreased in keratinocytes from psoriasis patients, correlating with IL-23 overexpression and supporting its relevance for this disease. Indeed, the actin polymerizing molecule N-WASP controls IL-23 expression in keratinocytes in response to TNF-α by regulating the degradation of the histone methyltransferases G9a and GLP and H3K9 dimethylation of the IL-23 promoter, leading to a chronic skin inflammation with an IL-23 inflammatory profile . Zhang et al. demonstrated that mRNA expression levels for histone deacetylase HDAC1, histone methyltransferase SUV39H1, and EZH2 were all increased in psoriatic peripheral blood mononuclear cells (PBMCs), correlating with higher keratinocyte proliferation, and contributing to psoriatic hyperplasia . It has also been reported that H3K27 demethylation, via Jmjd3, regulates Th17 cell differentiation and expression of several inflammatory cytokines, establishing a pathogenetic role for this epigenetic mechanism in psoriasis; indeed, Jmjd3 directly bound to and reduced H3K27 trimethylation levels in the genomic region of Rorc (RAR related orphan receptor C), which encodes the master Th17 transcription factor Rorγt and Th17 cytokine genes such as IL-17, IL-17f, and IL-22 .
5.1.3. MicroRNAs
Changes in miRNA expression have also been found in psoriatic skin lesions and more than 250 miRNAs have been implicated in the pathogenesis of this skin disorder . MiR-203 is exclusively expressed by keratinocytes and its upregulation in psoriatic skin lesions is associated with the downregulation of its target, SOCS-3 (suppressor of cytokine signaling 3), which is involved in inflammatory responses and keratinocyte functions . Subsequently, Zibert et al. discovered one downregulated and nine upregulated miRNAs in non-lesional psoriatic skin, including miR-21, miR-205, miR-221, and miR-222, suggesting their role in early phases of psoriasis . Moreover, inhibition of miR-21 reduces disease severity in patient-derived psoriatic skin xenotransplants in mice as well as in a psoriasis-like mouse model, suggesting that miR-21 could represent a potential therapeutic target for the treatment of psoriasis .
MiR-146a is another microRNA that appears upregulated in psoriasis. It can promote TNF expression and is positively correlated with IL-17-driven inflammation in keratinocytes, revealing a key role with therapeutic potential in the pathogenesis of psoriasis [92,93]. In addition, miR-155 has been identified as a potential therapeutic target for psoriasis ; indeed, it is markedly increased in lesional skin and PBMCs of psoriasis patients and plays several crucial roles in keratinocyte proliferation and apoptosis inhibition—through the phosphatase and tension homolog deleted on chromosome 10 (PTEN) signaling pathway—as well as in inflammatory pathways [94,95,96]. MiR-31 and miR-210 have also been shown to contribute to inflammation in psoriatic skin plaques by regulating the production of inflammatory cytokines/chemokines, enhancing keratinocyte proliferation , and inducing Th17 and Th1 cell differentiation , respectively.
Some miRNA clusters have also been also implicated in the pathogenesis of psoriasis; for example, cytokine-induced overexpression of the miR-17-92 cluster can promote keratinocyte proliferation, contributing to the development of psoriasis-like inflammation .
Finally, serum levels of miR-33, miR-126, and miR-143 have been proposed as potential biomarkers for disease severity or the prognosis of psoriasis .
5.2. The Role of Epigenetics in Atopic Dermatitis
Atopic dermatitis is one of the most common inflammatory skin diseases worldwide with a recurrent, chronic-relapsing clinical course characterized by an impairment of the skin barrier function supported by the (epi-)dermal immune system and the microbiome of the skin .
Epigenetic modifications have also been recognized in this disease and are mainly mediated by DNA methylation and non-coding RNAs [101,102] (Figure 3).
Figure 3. Epigenetic modifications in atopic dermatitis: DNA methylation and microRNAs, including the microRNAs that can be considered as potential therapeutic targets. Created in Biorender.com.
5.2.1. DNA Methylation
Regarding DNA methylation changes in atopic dermatitis, Rodriguez et al. used an epigenome-wide approach to identify 127 differentially methylated CpG sites (DMSs) between atopic dermatitis lesional skin and controls. Methylation in several of these regions correlated with genes encoding keratins located within the keratin cluster and altered S100A gene expression within the EDC . The same authors confirmed S100A2, A7, A8, A9, and A15 overexpression in atopic dermatitis lesional skin; these expression changes correlated with DNA hypermethylation of one CpG within S100A5, indicating a coregulatory mechanism through methylation. Similarly, KRT6A and KRT6B mRNA overexpression depended on decreased methylation of a single CpG in KRT6A . Both these keratins belong to the group of “alternative pathway keratins”, and it is supposed that this alternative pathway provides a physiological response for epidermal wound healing after injury . A single differentially methylated CpG site within OAS2 (2′-5′-oligoadenylate synthetase 2) coupled with altered co-expression was also observed for OAS1 (2′-5′-oligoadenylate synthetase 1), OAS2, and OAS3 (2′-5′-oligoadenylate synthetase 3) , which belong to a family of proteins—induced by interferons—that synthesizes 2′,5′-oligoadenylates and are involved in the innate immune response to viral infection . Further, CD36 has been found upregulated upon skin barrier disruption as well as in other inflammatory skin conditions [107,108] that contribute to terminal keratinocyte differentiation .
Collectively, these findings have confirmed—in lesional skin of atopic dermatitis patients—an altered methylation pattern affecting key genes for keratinocyte differentiation, proliferation, and the immune response .
The TSLP (thymic stromal lymphopoietin) gene encodes a hemopoietic cytokine that plays a crucial role in promoting a T helper type 2 (TH2) cell response. It appears to be a central player in the development of asthma and atopic dermatitis and is being considered as a potential therapeutic target for the treatment of such diseases . In a study conducted by Luo et al. , the authors investigated TSLP methylation status, and the results showed—in lesional skin from patients with atopic dermatitis when compared to control—hypomethylation of the TSLP gene at the promoter level, confirming that DNA hypomethylation contributes to TSLP overexpression.
In another study conducted by Olisova et al., DMSs were observed within genes involved in several atopic dermatitis-related processes, including immune response regulation, lymphocyte activation, cell proliferation, apoptosis, and epidermis differentiation, highlighting marked epigenetic involvement in the development of this disease .
Finally, a study investigating the relationship between genetic and epigenetic changes in the filaggrin (FLG) gene—essential for the regulation of epidermal homeostasis—in PBMCs of patients with atopic dermatitis, indicated that the association between loss-of-function mutations in the FLG gene and eczema was orchestrated by DNA methylation; in particular, at an 86% methylation level, filaggrin haploinsufficient subjects had about a 6-fold increased risk of eczema when compared to those with wild type FLG .
5.2.2. MicroRNAs
Several microRNAs have an active part in the pathogenetic scenario of atopic dermatitis. Upregulation of miR-10a-5p has been demonstrated to impair keratinocyte proliferation and migration through HAS3 (hyaluronan synthase 3), a damage-associated positive regulator of keratinocyte proliferation and migration that is the direct target of miR-10a-5p . Gu et al. demonstrated miR-29b upregulation in lesional skin and sera from atopic dermatitis patients. This microRNA promotes keratinocyte apoptosis by inhibiting Bcl2L2 (Bcl-2-like protein 2) protein, contributing to epithelial barrier dysfunction typically impaired in atopic dermatitis .
Other microRNAs that play a critical role in atopic dermatitis are the following: (i) miR-124 that regulates inflammatory responses in keratinocytes and chronic skin inflammation through the NF-κB pathway ; (ii) miR-143 that decreases IL-13 activity and the inflammatory cascade through targeting IL-13Rα1 in epidermal keratinocytes ; (iii) miR-146, a mediator of inflammation, which is regulated by inflammatory mediators such as interleukin 1 and TNF-α and operates in a “negative regulatory loop” to modulate the inflammatory response ; (iv) miR-155, which regulates both cytokine responses and epithelial barrier function by targeting PKIα . Among these, miR-124 and miR-143 can be regarded as potential novel therapeutic targets in AD patients [115,116].
Interestingly, miR-17-5p, miR-20a, miR-21, and miR-106b are upregulated in both atopic dermatitis and psoriasis skin lesions, whereas miR-122a, miR-133a-133b, miR-133b, miR-215, and miR-326 are downregulated in both diseases . The similar expression of miRNAs in both atopic dermatitis and psoriasis lesions is consistent with the common clinical features of these diseases (Figure 4).
Figure 4. MicroRNAs involved in both psoriasis and atopic dermatitis.
6. Conclusions
The central dogma of biology holds that cell information flows from DNA to RNA to proteins ; this concept has now been completely confuted due to the role of epigenetics in regulating gene expression. The Greek prefix epi- (ἐπι- “over, outside of, around”) in epigenetics implies features that are “on top of” or “in addition to” the traditional genetic basis for inheritance , and the term “epigenetic” has become one of the newest emerging and interesting fields in the scientific world. Epigenetics refers to enzymatic modifications—DNA methylation, histone acetylation/deacetylation, and RNA regulation—to the chromatin structure that can up- or down-regulate gene expression in the absence of DNA sequence changes . More and more data demonstrate that epigenetic mechanisms are involved in the regulation of multiple aspects of epidermal growth and differentiation. Moreover, epigenetic modifications can be activated by environmental factors, upon which their action appears to help epidermal cells adapt to long-term physiological changes . To note, the skin neuroendocrine system acts by defending and maintaining the structural and functional integrity of the skin as well as systemic homeostasis . Epigenetic regulators modulate both local and higher-order chromatin structure in epidermal keratinocytes, affecting the function of the genes, mainly those in EDC, that are associated with proliferation and differentiation processes in keratinocytes as well as their progenitor cells.
DNA methyltransferase DNMT1, histone deacetylases HDAC1/2, and Polycomb components (Cbx4, Bmi1, Ezh1/2) act as repressive chromatin regulators stimulating proliferation of progenitor cells, and some of these also inhibit premature activation of terminal differentiation-associated genes. In contrast, a group of chromatin remodelers such as histone demethylases, ATP-dependent chromatin remodeler Brg1, and genome organizer Satb1 promote terminal keratinocyte differentiation and also exhibit effects on cell proliferation. Finally, microRNAs are also involved in different processes between proliferation and differentiation in the program of epidermal development.
As demonstrated by several studies concerning skin diseases, changes in the epigenetic signature in the epidermis can contribute to the pathogenesis of these diseases. More importantly, in contrast to genetic changes—which are difficult to reverse—epigenetic modifications are pharmaceutically reversible; thus, the emerging tools of epigenetics can be used as preventive, diagnostic, and therapeutic markers .
In the past few decades, development of epigenetic drugs—so-called “epidrugs”—has achieved significant progress; thus, epigenetic therapy is a promising new therapeutical solution that targets the main causative epigenetic mechanisms involved in inflammatory/autoinflammatory diseases. Future efforts in this direction will help to bridge the gap between our current knowledge and potential applications for epigenetic regulators in clinical practice to pave the way for promising tailored treatments.
Author Contributions
Conceptualization, C.M. and M.R.; formal Analysis, C.M. and M.R.; data curation, C.M. and M.R.; writing—original draft preparation, C.M. and M.R.; writing—review and editing, C.M. and M.R.; supervision—A.V.M. All authors have read and agreed to the published version of the manuscript.
Funding
This research received no external funding.
Institutional Review Board Statement
Not applicable.
Informed Consent Statement
Not applicable.
Data Availability Statement
Not applicable.
Conflicts of Interest
The authors declare no conflict of interest.
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Figure 1. (a) The concurrence of DNA methylation and demethylation. Exported by Biorender.com (accessed on 1 April 2022). (b) Histone modifications including acetylation and methylation of lysine. Created in Biorender.com. (c) MicroRNA formation and their role in protein degradation. Created in Biorender.com.
Figure 2. Epigenetic modifications in psoriasis: DNA methylation, histone modifications and microRNAs, including the microRNAs that can be considered as potential therapeutic targets and biomarkers for severity or prognosis. Created in Biorender.com.
Figure 3. Epigenetic modifications in atopic dermatitis: DNA methylation and microRNAs, including the microRNAs that can be considered as potential therapeutic targets. Created in Biorender.com.
Figure 4. MicroRNAs involved in both psoriasis and atopic dermatitis.
Table 1. Effect of knockout/down and overexpression of chromatin-modifying protein on keratinocyte/epidermal growth and differentiation.
| Gene/Modified Gene | Effect on Chromatin | Effect on Keratinocyte/Epidermal Growth and Differentiation |
:---:
| Brg1 | Alteration of the chromatin architecture by repositioning, assembling, and restructuring nucleosomes | Overexpression of the EDC genes |
| Brg1 knockout | Alteration of the chromatin architecture by repositioning, assembling, and restructuring nucleosomes | Impairment of the final stage of keratinocyte terminal differentiation |
| SWI/SNF complex | Alteration of the chromatin architecture by repositioning, assembling, and restructuring nucleosomes | Suppress differentiation and promote the maintenance of the progenitor state in epidermal cells |
| Mi-2β | Alteration of the chromatin architecture by repositioning, assembling, and restructuring nucleosomes | Self-renewal of epidermal precursors during the earlier phases of embryogenesis; defective basal layer formation |
| Mi-2β knockout | Alteration of the chromatin architecture by repositioning, assembling, and restructuring nucleosomes | Impaired induction and development of the hair follicles |
| JMJD3 knockdown | H3K27me3 ↑ | Blocked progenitor cell differentiation |
| JMJD3 overexpression | H3K27me3 ↓ | Enhanced expression of epidermal differentiation markers |
| Setd8 knockout | H3K20me1 ↓ | Inhibition of progenitor cell proliferation; impaired differentiation [31,32] |
| Suv39H1 knockout | H3K9me3 ↓ | Induction of genes encoding differentiation markers |
| Jarid1b knockdown | H3K4me3 ↑ | Delayed differentiation |
| Jarid1b overexpression | H3K4me3 ↓ | Reduced proliferation; enhanced differentiation |
| Satb1 | Integration of higher-order chromatin architecture with gene regulation | Induction of cell differentiation |
| Satb1 knockout | Integration of higher-order chromatin architecture with gene regulation | Altered expression of terminal differentiation-associated genes |
| DNMT1 knockdown | DNA methylation | Maintenance of epidermal progenitor cells and epidermal tissue renewal |
| 5-hydroxymethylcytosine | Demethylation 5mC | Impairment of self-renewal of stem cells |
| HDAC1/2 knockout | acH3 ↑ | Enhanced proliferation; epidermal hyperplasia; disturbed hair follicle differentiation |
| Jarid2 knockout | H3K27me3 ↓ | Inhibition of proliferation; premature differentiation |
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Moltrasio, C.; Romagnuolo, M.; Marzano, A.V. Epigenetic Mechanisms of Epidermal Differentiation. Int. J. Mol. Sci.2022, 23, 4874.
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Moltrasio, C., Romagnuolo, M., & Marzano, A. V. (2022). Epigenetic Mechanisms of Epidermal Differentiation. International Journal of Molecular Sciences, 23(9), 4874.
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1184 | https://math.stackexchange.com/questions/3264693/why-does-multiplication-by-purely-imaginary-numbers-regardless-of-magnitude-no | vector spaces - Why does multiplication by purely imaginary numbers, regardless of magnitude, not cause purely rotation? - Mathematics Stack Exchange
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Why does multiplication by purely imaginary numbers, regardless of magnitude, not cause purely rotation?
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For context, I have been relearning a lot of math through the lovely website Brilliant.org. One of their sections covers complex numbers and tries to intuitively introduce Euler's Formula and complex exponentiation by pulling features from polar coordinates, trigonometry, real number exponentiation, and vector space transformations.
While I am now decently familiar with how complex exponentiation behaves (i.e. inducing rotation), I am slightly confused by the following.
2 3 z 2 3 z can be viewed as stretching the complex number z z by 2 3 2 3. This could be rewritten as 8 z 8 z. Therefore, Brilliant.org suggests that exponentiation of real numbers can be thought of as stretching a vector just like real number multiplication would. (check - understood)
Brilliant.org then demonstrates that multiplying z 1 z 1 by another complex number z 2 z 2 is equivalent to first stretching z 1 z 1 by the magnitude of z 2 z 2 and then rotating z 1 z 1 by the angle that z 2 z 2 creates with the real axis counterclockwise. (check - understood)
However, this is where I get confused. Why does, for example, 2 2 i∗z 2 2 i∗z cause purely rotation of z but 2 i∗z 2 i∗z does not (i.e. it causes stretching, too, in addition to rotation)?
To me, the fact that 2(2 i+3)2(2 i+3) causes both rotation and stretching makes perfect sense because we can rewrite this as (2 3)∗(2(2 i))(2 3)∗(2(2 i)). As previously noted by Brilliant.org, exponentiation by real numbers can thought of as stretching.
Here is the crux of my issue:
I understand that the magnitude of the imaginary number in the exponent (for example, the ′2′′2′ in e 2 i e 2 i ) can be thought of as a rate of speed...but why does this interpretation 'drop' when we are doing something like 2 i∗z 2 i∗z. i.e. Why is the 2 2 in 2 i∗z 2 i∗z not also treated like a rate of rotation but instead treated like a magnitude of stretching ?
My math skill is not particularly high level so if anyone can offer as much of an intuitive answer as possible, it would be greatly appreciated!
Edit 1: I guess another way of expressing this question is as follows:
Why does a duality exist between real number exponentiation and real number multiplication but a duality does not exist between imaginary number exponentiation and imaginary number multiplication (i.e. imaginary number multiplication can cause stretching in addition to rotation)?
Edit 2: While I accept that Euler's formula is a way of proving that exponentiation of purely imaginary numbers has a magnitude of 1 and therefore does not invoke stretching, that is not the sort of answer I am looking for. My question is aimed at identifying what was specified in Edit 1.
Edit 3: Here is a picture that helps clarify my point of confusion.
Edit 4: The question that was asked in this post Which general physical transformation to the number space does exponentiation represent? is sort of the theme that I am going for. The answer that was given to this post, however, omits a reference to the complex numbers.
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edited Jun 19, 2019 at 2:35
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asked Jun 16, 2019 at 20:47
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2 Do you mean e 2 i z e 2 i z where yo wrote 2^(2i)z?coffeemath –coffeemath 2019-06-16 20:57:41 +00:00 Commented Jun 16, 2019 at 20:57
1 From what I understand about complex exponentiation, it could be any base. Conceptually, the question should still be the same. (Yes, I know that the rate of rotation changes depending on which base I use)S.C. –S.C. 2019-06-16 20:59:09 +00:00 Commented Jun 16, 2019 at 20:59
1 Right, one could define 2 2 i z 2 2 i z as e 2 ln 2 i z e 2 ln2 i z which just has a different angle.coffeemath –coffeemath 2019-06-16 21:02:27 +00:00 Commented Jun 16, 2019 at 21:02
1 The magnitude of e i θ=cos θ+i sin θ e i θ=cosθ+i sinθ is 1 1, so you're stretching by a factor of 1 1, which is the same as not stretching.saulspatz –saulspatz 2019-06-16 21:05:08 +00:00 Commented Jun 16, 2019 at 21:05
hmmm, while I accept that is a quick way of proving this to be the case, I feel like this answer masks the truth that I am looking for. It's the failure in duality (which I mentioned in my edit at the bottom of the page) that is specifically confusing to me.S.C. –S.C. 2019-06-16 21:09:06 +00:00 Commented Jun 16, 2019 at 21:09
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To briefly address your specific question about "duality" first, there is no "duality" (not sure what is precisely meant by that term here) between complex exponentiation and complex multiplication, unless the arguments of both are real. This is because complex multiplication is commutative; complex exponentiation is not - neither is real exponentiation (e i x(e i x is not the same as i x e i x e). Take the second line:
c a+b i⋅z↔(a+b i)⋅z c a+b i⋅z↔(a+b i)⋅z
stretching and rotating↔stretching and rotating stretching and rotating↔stretching and rotating
Yes, (a+b i)⋅z(a+b i)⋅z corresponds to a transformation consisting of some stretching and some rotation, but the stretching is due to both multiplying by a a and multiplying by b i b i.
Now, let's break down intuitively what different operations on the complex plane are (I know you said you already went over some of this in your question, but I think it will lead into the explanation for complex exponentiation nicely).
Complex addition is the same as vector addition: we add the components. Think of this intuitively by imagining each complex number as an arrow: adding two complex numbers is like sticking one of their arrows on the end of the other. Another way: think of adding a complex number not as a static operation, but as a transformation. Adding the number (a+b i)(a+b i) is the same as shifting the origin of the complex plane onto the point (−a−b i)(−a−b i). Take a second to imagine why that's true. Thus (a+b i)(a+b i) is a function with respect to addition, which maps every point in the complex plane to another point in the complex plane, (a+bi) away.
Complex multiplication corresponds to both a stretch and a rotation (usually).
(a+b i)⋅(c+d i)=(a c−b d)+(a d+b c)(a+b i)⋅(c+d i)=(a c−b d)+(a d+b c)
A better way to think about this is in terms of Euler's formula: represent your two complex numbers as polar coordinates, and multiplication becomes much clearer:
r 1 e i θ 1⋅r 2 e i θ 2=r 1 r 2 e i(θ 1+θ 2)r 1 e i θ 1⋅r 2 e i θ 2=r 1 r 2 e i(θ 1+θ 2)
So we can imagine complex multiplication of two numbers is taking the angles they make with the x axis, adding those two angles to get the angle of your new number, then multiplying the magnitudes of the two original numbers to get the magnitude of your new number. Think of complex multiplication by (a+b i)(a+b i) as two, very dynamic transformations composted together: first stretching the entire complex plane by a factor of a 2+b 2−−−−−−√a 2+b 2, then rotating the entire complex plane by a factor of tan−1(b a)tan−1(b a). Thus (a+b i)(a+b i) can also be thought of as a function with respect to multiplication: it maps every point in the complex plane to another point in the complex plane by combination of a rotation and a stretch.
What kind of function is "complex exponentiation"? We define it as follows: a b+c⋅i a b+c⋅i=a b⋅a c⋅i a b⋅a c⋅i where a c⋅i=e c⋅i log a a c⋅i=e c⋅i loga, etc. based on Euler's formula. e c⋅i log a e c⋅i loga is e e raised to a complex number and is thus a rotation. Note that we can imagine complex exponentiation again as a sort of dynamic transformation, squishing and mapping the complex plane to a new location.
Let's break this down into two cases. The first is the only one your question asks about.
1) The base of the exponent is real. As can be seen from above, every "complex exponentiation" is a transformation of the complex plane consisting of two transformations: first, a stretch by some factor, and then a rotation. This is very similar to complex multiplication, which begs the question, when do these two things behave in the same way? You called it "duality," I'm not going to call it that because that word means something specific in linear algebra, I'll just call this similarity "similarity."
Multiplying x x by (a+b i)→(a+b i)→ Stretch by a 2+b 2−−−−−−√a 2+b 2, rotate by tan−1(b a)tan−1(b a) radians.
Raising x x to the (a+b i)(a+b i) power →→ Stretch by x a x a, rotate by b⋅i log x b⋅i logx radians.
Note that the two are very dissimilar - exponentiation is dependent on the base whereas multiplication is not. This is a result of the effect that multiplication by a complex number is something called a linear transformation ( raising something to a complex power is most certainly not.
2) The base of the exponent is complex. This gets a little more complicated, because raising a complex number to a real power corresponds in part to a rotation, so separating which parts are the rotation and which parts are the stretching is a little annoying and won't give much insight here. Complex exponentiation of complex numbers is really funky, giving rise to all sorts of weird fractal shapes when we consider which complex numbers get really large when we raise them to a complex power, and which ones don't - this is related to how the Mandelbrot set is formed ( The point is that, again, the magnitude of the stretch and rotation is dependent on x, meaning that this is not a linear transformation.
If you want to gain an intuition for how complex exponentiation of complex numbers works, I recommend you play around with some functions with this grapher:
So, to return to what I mentioned at the beginning, the reason why you're not seeing a "duality" in case 3 is that there's no "duality" in case 2 to begin with - yes, both complex exponentiation and complex multiplication correspond to both a rotation and a stretch, but the rotation and stretch for each behave in very different ways. Complex multiplication is a linear transform, complex exponentiation is not.
I also enjoy brilliant.org's courses; if you're interested, I would recommend you check out their course on linear algebra next ( This is the first answer I've actually posted. I would love feedback from anyone if they have it.
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answered Jun 19, 2019 at 4:37
Laurel TurnerLaurel Turner
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Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer.dantopa –dantopa 2019-06-19 04:49:56 +00:00 Commented Jun 19, 2019 at 4:49
is another important point the fact that the "Stretching Event" caused by multiplication is different than the "Stretching Event" caused by real number exponentiation? For example, the "0.5" in 2.5 times x is a different sort of half than the "0.5" in x raised the 2.5. i.e. that 0.5 stretches the space differently depending on whether or not we are talking about exponentiation or multiplication.S.C. –S.C. 2019-06-19 14:59:08 +00:00 Commented Jun 19, 2019 at 14:59
Yes, the arguments mean different things. Complex multiplication corresponds to a general stretching of your basis - imagine the plane as a lattice; stretch it out. Complex exponentiation corresponds to a stretching that is dependent on location. It's very rare that exponentiation and multiplication by the same argument correspond to the same transformation. My intuition is that this should only be true when the argument is 1. Go to davidbau.com/conformal/#z and compare the functions (2+i)z vs z^(2+i).Laurel Turner –Laurel Turner 2019-06-19 20:14:55 +00:00 Commented Jun 19, 2019 at 20:14
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2 2 i z 2 2 i zdoes cause a stretching of z z. It causes a stretching by the magnitude of 2 2 i 2 2 i. And the magnitude of 2 2 i 2 2 i is one. So, it stretches by a factor of 1 1. If you choose to deny that stretching by a factor of 1 1 is stretching, well, then that's where your problem is.
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answered Jun 19, 2019 at 2:48
Gerry MyersonGerry Myerson
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is another important point the fact that the "Stretching Event" caused by multiplication is different than the "Stretching Event" caused by real number exponentiation? For example, the "0.5" in 2.5 times x is a different sort of half than the "0.5" in x raised the 2.5. i.e. that 0.5 stretches the space differently depending on whether or not we are talking about exponentiation or multiplication.S.C. –S.C. 2019-06-19 15:01:01 +00:00 Commented Jun 19, 2019 at 15:01
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The failure of duality is that there was never really duality there in the first place.
It's true that in most cases, the vector from the origin to c a+b i z c a+b i z is rotated and stretched (or shrunk) relative to the vector from the origin to z.z.
But sometimes it is not stretched or shrunk: specifically, when a=0.a=0.
It is also true that in most cases, the vector from the origin to (a+b i)z(a+b i)z is rotated and stretched (or shrunk) relative to the vector from the origin to z.z.
But sometimes that vector is not stretched or shrunk either: specifically, when a 2+b 2=1.a 2+b 2=1.
The key observation to me is that when you write something like c a+b i,c a+b i, you identify a point in the complex plane using the parameters a a and b b somewhat like polar coordinates. Parameter a a dictates the radius r r, parameter b b dictates the angle θ.θ.
When you write a+b i,a+b i, however, the parameters a a and b b act like Cartesian coordinates x x and y y of a point in the complex plane.
Polar coordinates don't work like Cartesian coordinates, and vice versa. They both identify points in a plane, that's mostly what they have in common. So when you wrote that both c a+b i z c a+b i z and (a+b i)z(a+b i)z stretch a vector as well as rotating it, that was (mostly!) true, but the way that a a and b b contributed to making it be a rotation or a stretching was completely different in each case.
What really makes multiplication not stretch or shrink a vector, is you have to make sure the thing you multiply z z by is a complex number on the "unit circle." You can get this either by choosing r=1 r=1 in polar coordinates (which corresponds to a=0 a=0 in the formula c a+b i c a+b i) or by choosing x 2+y 2=1.x 2+y 2=1.
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answered Jun 19, 2019 at 2:54
David KDavid K
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is another important point the fact that the "Stretching Event" caused by multiplication is different than the "Stretching Event" caused by real number exponentiation? For example, the "0.5" in 2.5 times x is a different sort of half than the "0.5" in x raised the 2.5. i.e. that 0.5 stretches the space differently depending on whether or not we are talking about exponentiation or multiplication.S.C. –S.C. 2019-06-19 15:01:19 +00:00 Commented Jun 19, 2019 at 15:01
Yes, that is a true point. The numbers get converted to “stretching factors” in very different ways. For example, if you put −2−2 in the exponent of your multiplier you shrink the vector, but if you just multiply by −2−2 the vector flips 180 180 degrees around and gets longer. David K –David K 2019-06-19 20:21:29 +00:00 Commented Jun 19, 2019 at 20:21
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Let z o z o and z d z d be two complex number, lets call the first one operator and second one operand, and let z′d z d′ be the transformed version( state after it being operated by z o z o ) of z d z d i.e.
z′d=z o×z d=|z o|e i a r g(z o)×|z d|e i a r g(z d)=|z o z d|e i(a r g(z o)+a r g(z d))z d′=z o×z d=|z o|e i a r g(z o)×|z d|e i a r g(z d)=|z o z d|e i(a r g(z o)+a r g(z d))
From here all your results get derived:
If z o z o is purely real, then a r g(z o)=0,π a r g(z o)=0,π when z o z o are positive and negative respectively. In the case when z o z o is positive
z′d=|z o z d|e i(a r g(0+a r g(z d))=|z o z d|e i(a r g(z d))=|z o|z d z d′=|z o z d|e i(a r g(0+a r g(z d))=|z o z d|e i(a r g(z d))=|z o|z d
the number z d z d is scaled by factor of |z d||z d|
For the case when z o z o is negative real number,
z′d=|z o z d|e i(π+a r g(z d))=−|z o|z d z d′=|z o z d|e i(π+a r g(z d))=−|z o|z d
In this case, the number is rotatedπ π radians, you can its flipped and scaled by the factor of |z o||z o|, but no one is wrong here, because only positive real numbers can be expressed in the terms of real exponents, in that case, what I've said conform with the website.
If z o z o is complex number, z′d=|z o z d|e i(a r g(z o)+a r g(z d))z d′=|z o z d|e i(a r g(z o)+a r g(z d)) as you can see here, the magnitude of z d z d is scaled by |z o||z o| and in addition to that argument of complex number has been changed by the addition of argument of operator, so here the number is rotated and scaled.
Now, coming to your question, for multiplication of 2 2 i 2 2 i with z z , if I denote z′z′ as the transformed version, then
z′=2 2 i z=|2 2 i|e a r g(2 2 i)z=(1)e a r g(2 2 i)z z′=2 2 i z=|2 2 i|e a r g(2 2 i)z=(1)e a r g(2 2 i)z
As you can see here, its magnitude is one, while in the case when z o=2 i z o=2 i, |2 i|=2,a r g(2 i)=π 2|2 i|=2,a r g(2 i)=π 2, so here you will rotation and scaling.
Coming to your crux-points, If z o z o is e 2 i e 2 i, its implication is |z o|=1|z o|=1 which implies it can't scale the complex number, and it can only rotated z d z d by 2 2 radians. When z o=2 i,|z o|=2,a r g(2 i)=π 2 z o=2 i,|z o|=2,a r g(2 i)=π 2, so it is scaled two times and rotated half pi radians.
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edited Jun 19, 2019 at 2:56
answered Jun 19, 2019 at 2:51
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Think of ot this way. 2 i z=2(i z)2 i z=2(i z), so you have two steps here: z↦i z z↦i z, which is a pure rotation and (i z)↦2(i z)(i z)↦2(i z), which is a pure stretching. Of course, you can also write 2 i z=i(2 z)2 i z=i(2 z) in which case you'll do the stretching first, but the result will be the same.
Now, when you multiply by 2 2 in the exponent, something different is going on. Instead of x y z=(x y)z x y z=(x y)z, you have x y z=(x y)z x y z=(x y)z, so e 2 i z=(e i)2 z=e i(e i z)e 2 i z=(e i)2 z=e i(e i z), i.e., you apply the rotation given by e i e i twice, not follow or precede it by stretching two times. That's why the rotation remains a rotation but the angle doubles.
Of course, you can write it also as (e 2)i z(e 2)i z, but this way you will have to say that you apply the stretching (z↦e 2 z z↦e 2 z) imaginary (i i) number of times. There is no intuitive physical meaning of applying some transformation i i times but, since you want to maintain the usual laws for powers as much as you can, you are forced to say that stretching done i i times becomes a rotation. Strange, but since there is no contradiction with common sense (common sense just says nothing about applying a transformation imaginary number of times), possible.
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answered Jun 19, 2019 at 3:21
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Write z=r 1 e i θ 1 z=r 1 e i θ 1. When multiplying w=r 2 e i θ 2 w=r 2 e i θ 2 by z z, we stretch by a factor of r 1 r 1, and rotate by an angle of θ 1 θ 1. This is easy to see, since z w=r 1 r 2 e i(θ 1+θ 2)z w=r 1 r 2 e i(θ 1+θ 2).
If z z is purely imaginary, so that we can take θ 1=π 2 θ 1=π 2, then we get rotation by π 2 π 2 and stretching by r 1 r 1.
Now for multiplication by z r i z r i, we have z r i=e r i ln z z r i=e r i lnz. Depending on the value of ln z lnz, we have various possibilities. But we do know there will be no stretching. For, the complex number e r i ln z e r i lnz has magnitude 1 1.
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edited Jun 19, 2019 at 18:55
answered Jun 19, 2019 at 3:27
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7Which general physical transformation to the number space does exponentiation represent?
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2Does a purely imaginary number have a corresponding "angle" in polar coordinate system?
2Why are "imaginary numbers" imaginary?
7Which general physical transformation to the number space does exponentiation represent?
5If ∣∣z 1+z 2 z 1−z 2∣∣=1|z 1+z 2 z 1−z 2|=1, then what is z 1 z 2 z 1 z 2
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0Complex numbers, set of values for which z will be purely real or imaginary.
3Why does the imaginary part disappear when you take the magnitude square of a complex number?
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1186 | https://www.weareteachers.com/area-and-perimeter/ | Every product is independently selected by our team of teacher-reviewers and editors. Things you buy through our links may earn us a commission.
28 Creative Area and Perimeter Activities
Teach kids these practical everyday skills.
As young kids learn about shapes and measurement, they soon enough come to measure the perimeters and areas of those shapes. There are shapes all around us. In fact, most of us live in homes that are made of a lot of rectangles. These rectangles, like rooms, are often measured, and perimeters and areas are calculated for things like buying rugs or flooring. The same is true for painting those rectangular walls. So, calculating area and perimeter is a pretty practical skill. Here are some interesting area and perimeter activities and explorations to help students learn about perimeter and area in hands-on, motivating ways.
FREE PRINTABLES
Math Vocabulary Word List and Cards
Our free printable math word list is a great classroom reference, and the accompanying word cards can be used to create a math word wall or for vocabulary activities.
Area and Perimeter Activities
1. Tiling Area and Perimeter Chart
As a first exploration with both perimeter and area, put out a set of square counting tiles or even real tiles. Circle students up, give them six tiles each, and have them build along with you. First, put out four tiles and explain how this is a model of a floor. Explain how we count the outside of the tile’s edges to determine the perimeter. Then show how we count the tiles inside to determine the area. Draw a picture of this on chart paper and label “area” and “perimeter.”
Let students experiment with their tiles, making different kinds of rectangles and figuring out their areas and perimeters. Have them share back what they find, and add these ideas to the chart. This can become a good class reference as you continue working with these concepts.
2. Graph Paper Person
Graph paper is great for exploring area and perimeter because the squares are easily counted. Give students colored pencils and/or crayons and a piece of graph paper. Have them create a “drawing” of a person using only colored-in squares on the graph paper. Then ask them to determine the perimeter and area of their person and write both on the paper along with the calculations.
3. Sticky Note Measurement
Provide students with pads of colorful sticky notes and drawing paper. Ask them to create interesting designs or shapes using this colorful material. After some free exploration, direct them to clear the paper and create a design according to the perimeter and area you give them. You might ask for a rectangle that has a perimeter of 12 or a square that has an area of 16. After a while, let students make their own and then calculate their area and perimeter. Then take turns sharing the area and perimeter and see if the rest of the class can make the same thing.
4. Math Letter Mosaic
Break out the sticky notes again. Explain to students that they can create a mosaic using this material on a piece of large drawing paper. You will give them a capital letter to create, like an L or R. Once they do this, they measure and record the perimeter and area on a separate sheet of paper. It’s great if you can assign letters that will spell out a sentence. These make a great bulletin board or class sign.
5. LEGO Land
Everyone loves LEGO! Provide a set of LEGO to students and have them freely explore building flat designs. Then, in a box or bag, put a set of cards with areas and perimeters written on them. Students take turns pulling a card, then using their LEGO bricks, they build a flat design with that exact perimeter and area.
6. Floor Tile–Style Field Trip
Take a mini field trip to another part of the school. Bring along yardsticks, rulers, measuring tapes, and painter’s tape. Many school floors in classrooms, cafeterias, and hallways are made of linoleum tiles. Partner students up and give them a roll of painter’s tape. Take the class to a large area to work together. Partners should create an interesting shape with the tiles and tape. Have them then calculate the area and perimeter, with each floor tile being a unit of measure. You might not need the measuring tools, but you have them just in case.
7. Dream House Room Design
Provide each student with graph paper, a pencil, scissors, a glue stick, and colored paper. Have them each draw a rectangle that represents a room in their dream house. Then cut out colored paper to represent furniture like tables, beds, sofas, and chairs. Arrange these in the room and glue them down. On each piece, write the perimeter and area of that piece as well as the perimeter and area of the room. You can add more rooms each session to build a total dream house floor plan.
8. Perimeter and Area Museum
Using tiles, cubes, sticky notes, attribute blocks, and any other manipulatives, explain to students that they will be building an area-and-shape museum on their desks. They should have at least five exhibits at the museum, each one showing a different perimeter or area. Use small pieces of construction paper and write the measurements on them. Fold them in half and stand them up as exhibit signs. Finally, have students take a museum walk and visit all the exhibits.
9. Perimeter and Area Hunt
Kids love scavenger hunts. Here’s one that helps them practice perimeter and area measures. Have each student make a flat model using multi-link cubes according to a perimeter and area you give them individually. Then take these models and place them around the room, labeling each one with a letter. Students get a piece of lined paper, pencil, and clipboard and go around the room determining the perimeter and area of each model. They write it next to the corresponding letter on their paper. When finished, review them together while holding the model up as a reference.
10. Vegetable Garden Planner
Give each student a piece of graph paper, colored pencils or crayons, and a pencil. Discuss what kinds of vegetables are grown in gardens. Ask students to use their graph paper to plan out a garden with at least five different things growing there. Determine how much of the garden they want to have for each vegetable, and draw and color in the areas for each. At the bottom of the plan, have a key that tells what each color means and how big that part of the garden is in terms of area and perimeter.
11. Geo Board Explore
Geo boards are a great way to explore shapes and measurements. Give each student a geo board and a pack of rubber bands. Begin by allowing students free exploration for five minutes. Then ask them to create a shape and calculate the perimeter and area. Finally, students take turns giving measurements to the class of the shape they made, like, “Make a triangle with a perimeter of 16,” or “Make a rectangle with an area of 8.” Everyone tries it on their geo boards and holds them up to show their answers.
12. How Many Rectangles Can You Make?
Here’s one of the area and perimeter activities that can stretch some thinking. Pair up students and give them 16 tiles. Explain, “You have 16 tiles. I want you to find how many different rectangles you can make using no more or no less than 16 tiles each time.” Have partners record these on a piece of graph paper, numbering them and writing the perimeter and area of each. Share together at the end of the session.
13. Draw a Card, Draw a Rectangle
On a set of blank index cards, write a series of task cards. Each card should list a perimeter, an area, or an area and a perimeter. Give each student a piece of graph paper and a pencil. Ask them to choose a card from the deck. They should draw whatever the card specifies on the paper. When finished, have it checked, and then students can trade cards to try some others.
14. How Big Are the Books?
Classrooms are full of books. Most are rectangles. Give each student lined paper, a pencil, and a ruler. For these area and perimeter activities, explain that they are to go around the room and select three books. They then write the titles down and measure the perimeter and area of each, recording it next to the title. Share back to see which book had the biggest and smallest perimeter and area. To extend this, send students two at a time to the library to find even bigger or smaller books to measure.
15. Tangram Time
Tangrams are always fun to play with. For these area and perimeter activities, print them out and let students explore and create for five minutes. Then focus back and have students create a tangram design on a piece of graph paper. Trace it, color it, and then find its perimeter and area. Suggest that students create at least two designs that are very different from each other.
16. The Area of My Pizza
Draw a couple of different-size circles on the board. Explain how you can find the area of a circle using the formula Area = 3.14 (pi) x radius (r)squared. Demonstrate how to measure the radius, and then use a calculator to do the operations in the formula. Using lids from cans or a compass, have students make at least two circular pizzas on a piece of drawing paper and with crayons “top” their pizzas with things like sausage, peppers, and so on. After decorating the pizzas, they should use a ruler and calculator to determine the area of each one and write it underneath the pizzas.
17. Graph Paper Names
Kids are always interested in their own names. Give students graph paper, colored pencils, and/or crayons. Explain that they should write their names in block letters using graph paper and color them in. Once each student completes their area and perimeter activities, ask them to calculate the area and perimeter of each letter and write that information under each letter.
18. Mystery Box
Put a collection of flat objects like books, notebooks, file folders, brochures, maps, box lids, stamps, stickers, and so on in a cardboard box. Students close their eyes and reach into the box and pull out an object. They take the object to their desks, write down the name of the object on a piece of lined paper, and then measure its perimeter and area, recording it next to its name. When finished, they return the object to the box and pick a new one. At the end, have students share their measurements for each object as you name them one by one.
19. Same/Same
Give each student a piece of graph paper and a pencil. The simple challenge is this: Can you draw two shapes that have the same area and perimeter but are different? Is it possible? How many can you make? Students draw these and share them on a bulletin board at the end. This is a great exploration no matter the results. Everyone will be thinking hard and measuring away!
20. Number Card Design
Have students pick two number cards from a deck of cards and then multiply them to determine the area. Now they take a piece of graph paper and try to draw a shape with that exact area. You can do the same with perimeter by having them pick two cards, one of which will show width and the other length.
21. Run Out of Room
Pair students up and give them crayons, graph paper to share, a pencil, and a pair of dice or number cubes. Each player chooses a color crayon to use. Players take turns rolling the dice. The two numbers that come up must be multiplied to create an area. The player then takes the color they chose and draws and colors in a shape of that area on the graph paper. The next player does the same thing. Players keep repeating this procedure until they can no longer fit any shapes on the paper. They add up the areas of their shapes, and the player with the most area covered is the winner.
22. Land Rush
Stake your claim! There’s a land rush out in Area Acres. It’s a small place—in fact, only ants live there. You can stake your claim, but you can’t have a parcel of land with an area of more than 24 square centimeters or less than 12 square centimeters. Those are the rules. Using centimeter graph paper, have students draw two parcels of land that meet the requirements, and record their areas and perimeters.
23. Cheez-It Designs
Cheez-Its are not only tasty, they are also perfect little squares, which means you can tile with them and create some rectangles or other shapes. Give each student a couple of handfuls of crackers and a paper towel to build on. How motivating, especially if after you determine the perimeter and area, you get to eat them!
24. BIG Rectangles
Pair students up and give them some colored chalk. Have them draw a fairly big rectangle on the playground. Provide measuring tapes and ask them to work together to measure the perimeter and write it under the rectangle. Strategize on how they might determine the area. They could draw a grid on top and try to count squares and partial squares, adding them up to get a final answer, or just multiply the length by the width.
25. Magic Carpets
Here’s an opportunity to measure something a little larger. Bring in some rugs or carpet samples from a local carpet store. Pair up students and give them rulers, yardsticks, and tape measures. Tape a piece of paper with a letter on each rug. Spread these out around the room. Assign a letter for each pair to start at. They should use their measuring tools to discover the perimeter and area of each rug and record it on a piece of lined paper. When finished with one, they rotate to the next until they’ve measured all the rugs and shared their findings.
26. Self-Portrait Picture Frame Perimeter
Bring in a framed picture or poster. Show students how you would measure the perimeter using a ruler or measuring tape. Give students a piece of drawing paper and ask them to draw a perimeter frame with a pattern design on it. Inside the frame, they should draw a self-portrait or a portrait of someone they admire. They then measure the perimeter of the frame and record it on the back of the paper.
27. Centimeters and Inches at the Pool
Give each student a ruler that has centimeters and inches on it or two rulers, one with centimeters and the other with inches. Ask students to draw a bird’s-eye view of a swimming pool on a piece of drawing paper and color it in. They can add deck chairs, umbrellas, tables, and so on around the pool. When finished, have them measure the perimeter and area in both inches and centimeters and record it on the same paper. Have students compare and discuss the differences in the numbers.
28. Design Estimation
We rarely carry measurement tools around with us—we typically estimate instead. Here’s one of the easy area and perimeter activities to get some practice with that.
Draw a fairly large, closed shape with straight lines on the board. Ask the class to take out whiteboards and markers, estimate the perimeter and area of your drawing, and record it. They can come up to the board to get a closer look, but they can’t use a measuring tool or touch the board. Once they have an estimate, they should share it back and explain their thinking.
Next, take a ruler and measure a couple of the lines of your drawing and tell the class the measurements. Based on the new information, they can revise their estimates or leave them as is. Finally, ask students to take turns coming up to help measure the rest of the design and get a final answer.
Get your free Math Word List and cards printables!
So much of math teaching is actually language teaching as well. There are quite a few math vocabulary words kids must understand in order to learn math concepts. That’s where these free printable math vocabulary lists and cards come in handy. There are two lists: one for grades K–3 and one for grades 4–6.
If you liked these area and perimeter activities, check out Active Math Games and Activities for Kids.
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1187 | http://hyperphysics.phy-astr.gsu.edu/hbase/faia2.html | Mass on incline with angled force
Force on Inclined Mass ====================== Though describable by the motion equations and the standard model of friction, this example requires care because under different conditions the mass can move up or down the incline, or it can be just held in place by the applied force. In the diagrams above, it is assumed that the mass is moving up the incline, so that the frictional resistance acts downward to oppose it. If the mass is moving downward, then the frictional resistance force must be reversed in direction.Application of Newton's second law to mass on incline. A force of Newtons is applied at an angle of degrees above the horizontal to a mass of kg which sits on an incline of angle degrees. There is a coefficient of friction μ = between the mass and the incline. For a sufficient applied force so that the mass moves upward, the net force is given by the expression and the resultant upward acceleration is given by ### Acceleration =m/s 2 The expressions above are valid only for the case where the applied force is sufficient to give an upward acceleration. If the applied force is too small to accelerate the mass upward, then there is a range of forces where the mass will not move, and if the incline is steep enough there is a range where the mass will accelerate downward. In the latter case, the friction force must be reversed in direction in the expression above since it will then act upward to oppose the downward motion. If the acceleration above is negative, that reversal was done in the calculation of the number. If the acceleration is zero,then it is in the range of conditions where the force will hold the mass on the incline, neither moving up nor down. ### Expressions Remove frictionIndex Newton's laws Standard mechanics problems
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1189 | https://mabouali.com/2024/02/04/piecewise-linear-functions-part-i/ | Piecewise-Linear Functions: Part I – Moe’s Homepage
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Piecewise-Linear Functions: Part I
Posted on February 4, 2024 February 10, 2024 by mabouali
We are going to divide this document in two parts. The first part, this document, we are going to discuss piecewise linear functions (PLFs) and how to implement one in Python. The second part, Part II, we will discuss how to fit the coefficients, detect the break-points locations, and even optimize the number of the knots using python.
Table of Contents
Toggle
Piecewise-Linear Functions (PLF)
What are they?
How do you implement them in Python?
Can I drop the last condition?
What if multiple conditions evaluate as True?
One last note on numpy’s piecewise?
Are Piecewise-Linear Function continuous?
Rewriting PLF using a [pseudo]-single equation
How are the equations in two forms related?
What’s Next?
References
Piecewise-Linear Functions (PLF)
What are they?
As the name suggests, piecewise-Linear function consists of couple of lines. Essentially you will have a line, i.e. , fitting your data at different intervals. For example, consider the or absolute value of . If you plot this function it looks like:
Instead of writing this function using absolute value, i.e. , you can show it as a piecewise-linear function:
In the above example, we are representing with two line segments: One for the interval where and another for the interval where .
You might ask: But why should we do that? Using looks much easier. Well, in this case you are absolutely right. Using Absolute value is a much more concise way of showing this specific function. But you are not always this lucky. There are functions that you are left with no option but to only show them as multiple different segments.
Let’s have a look at this function, which looks like a letter “W”:
This one could be represented as a piecewise-linear function:
This time you have 4 segments and 4 line equations. Although you can say that using absolute value, you can simplify the representation a bit as follows:
But we already know that the absolute value function is representing two straight line.
Notice that although We are saying Piecewise-Linear Function, notice that “piecewise” and “linear” are mixed into one word. We are emphasizing that each piece is a line function or better to say an affine function. This should not be misunderstood with Linear Functions at all. These are two different concept. For one, a linear function must satisfy . For example, none of the piecewise-linear function mentioned above are Linear Function. So, have this in mind that these two concepts are different
Piecewise-Linear Function and Linear Function are different concepts;
How do you implement them in Python?
Now as of the question of how do we implement them in Python? Well, it depends. For example, the first function could be implemented as:
```python
import numpy as np
def my_piecewise_function(x):
return np.abs(x)
```
But we already see that things are not always that easy. So, another way of implementing this function is:
```python
import numpy as np
def my_piecewise_function(x):
if x < 0:
return -x
else:
return x
```
But there is even a better way to do it, and that’s using numpy.piecewise as follows:
```python
import numpy as np
def my_piecewise_function(x):
return np.piecewise(
x, # The value at which the piecewise function should be evaluated
[ # A list of conditions
x < 0,
x >=0
],
[ # One function corresponding to ech of the condition above.
lambda x: -x,
lambda x: x
]
)
x = np.linspace(-1, 1, 51)
np.all(np.abs(x) == my_piecewise_function_1(x)) # will print True
```
This is the minimum that you need to implement a piecewise function:
values for which the piecewise function needs to be evaluated,
a list of conditions that defines how many segments you have, and
A list of functions for each segment.
If the first condition is true, the first function from the list is returned; If the second condition is true, the second function is returned.
That’s it.
Let’s implement the second function:
```python
import numpy as np
def my_piecewise_function_2(x):
return np.piecewise(
x, # The value at which the piecewise function should be evaluated
[ # A list of conditions
x < -0.5,
(-0.5 <= x) & (x < 0),
(0 <= x) & (x < 0.5),
(0.5 <= x)
],
[ # One function corresponding to ech of the condition above.
lambda x: -x - 0.5,
lambda x: x + 0.5,
lambda x: -x + 0.5,
lambda x: x - 0.5
]
)
```
Can I drop the last condition?
Yes, you can. If you have one condition less than the number of the function, the last function in the list will be used when all the conditions evaluate as “False”. Try this:
```python
import numpy as np
def my_piecewise_function_2(x):
return np.piecewise(
x, # The value at which the piecewise function should be evaluated
[ # A list of conditions
x < -0.5,
(-0.5 <= x) & (x < 0),
(0 <= x) & (x < 0.5),
# (0.5 <= x)
],
[ # One function corresponding to ech of the condition above.
lambda x: -x - 0.5,
lambda x: x + 0.5,
lambda x: -x + 0.5,
lambda x: x - 0.5
]
)
x1 = np.linspace(0, 1.5, 151)
x2 = np.linspace(-1.5, 0, 151)
np.all(np.abs(x1-0.5) == my_piecewise_function_2(x1)) # will be True
np.all(np.abs(-x2-0.5) == my_piecewise_function_2(x2)) # will be True
```
You will get the same results as before.
What if multiple conditions evaluate as True?
It is important to recognize that the list of conditions are not evaluated like a switch-case or if-elif-else. In another word, the following:
python
np.piecewise(
x,
[
cond1,
cond2,
cond3,
],
[
func1,
func2,
func3,
func4
]
)
IS NOT EQUIVALENT TO
python
if cond1:
func1
elif cond2:
func2:
elif cond3:
func3
else:
func4
If multiple condition in the numpy piecewise function evaluates as true, the value corresponding to the function that matches that last condition that evaluates to true is returned.
Confusing? Look at the following:
```python
def my_wrong_piecewise_function(x):
return np.piecewise(
x, # The value at which the piecewise function should be evaluated
[ # A list of conditions
0.5 <= x,
0 <= x,
-0.5 <= x,
x < -0.5,
],
[ # One function corresponding to ech of the condition above.
lambda x: 1,
lambda x: 2,
lambda x: 3,
lambda x: 4,
]
)
my_piecewise_function_3(2) # will return 3
```
You might thing that the above function should return 1 for as input, because is the first condition that evaluates to “True”; however, 3 is returned because the last condition that evaluates to “True” is and the function associated to that condition returns 3.
So, those conditions, are like switch-case implementation in C/C++ when you forget “break” statement.
So, be careful how you implement the conditions.
One last note on numpy’s piecewise?
Although this document is about piecewise-linear, so we are assuming each piece is an affine function, as you have noticed the numpy’s piecewise, does not impose any restriction on what the function for each piece should be. You can use that to implement any function; they can be none-linear. Or they can be even not a function, let’s say you do need to perform a different action depending on which piece or segment you are.
Are Piecewise-Linear Function continuous?
They don’t have to. They can be non-continuous. But in this writing we will focus on the continuous one.
Rewriting PLF using a [pseudo]-single equation
this new form is helpful when fitting a PLF. If the PLF is non-continuous, all you have to do is to treat it as bunch of separate line fitting. No big deal. The issue is when the PLF is continuous. You need to guarantee the continuousness of the function. So, let’s see how we can do that. We are going to first rewrite the equation in a slight different format. Let’s say we have the following PLF:
This PLF has segments or pieces: counting from zero: segment 0 to segment . The intercept for i-th segment ( ) is defined by and the slope for it is defined by . We are going to rewrite the above equation as:
where:
This might look like we were able to rewrite the PLF as a single equation! well, it depends how you interpret it. The definition for is still using multiple condition.
The more important question is: How is this guaranteeing the pieces to be continuous? Notice that are called break-points, tie-points, knot-points, or simply knots. To check for continuity, just try the functions at the knots, i.e. evaluate that the limit of that function when approach a knot, from left and right is the same, i.e.:
Ok! let me make it easy. Just evaluate at for both equations, i.e. the equation specified for condition and the one that is for . Ignore that for the first one it is said .
Let me do it for . in this case all the where are zero. is 1. When approaching from left, and you have:
When you are approaching from the right side you have: and you get:
You can see that both equation evaluate to the same value; hence, you have your continuity guaranteed.
How are the equations in two forms related?
So, how are and related to ? Essentially, if you want to know the slopes and intercepts of each segment, how can we get those?
Remember that there were segments and we were numbering them from 0 to , so when we say i-th segment, remember that . That’s is we are counting from 0. Don’t get confused like that episode from Emily in Paris with the elevator and floor numbering, at least not in this document. Now that we have that cleared, the slope for i-th segment is:
In another word, the slope for the first segment the slope is: , for the second segment the slope is: , and so on.
And how about the intercepts? You can recompute the intercepts using the following equation:
so, we have:
,
,
, and
…
What’s Next?
As mentioned in the beginning, we are dividing this document into two parts. This document, Part I, focused on what Piecewise-Linear Functions (PLFs) are and we tried to ease you in with the mathematics.
In the next document, Part II, we will focus on how to estimate the coefficients, i.e. when you have a data set to fit. We are going to divide the problem into three different types. In the first type, we assume we already know the break points (how many there are and where they are), i.e. all the are known. In the second type, we are going to assume we just know how many break points there are; but we don’t know where they are. In the third type, pretty much we just know that we want to fit a PLF. We don’t know how many segments there are and where they are.
References
Piecewise Linear Function – Wikipedia – (here) – last accessed: Feb. 1st, 2024
Numpy’s API Reference (here) – last accessed: Feb. 1st, 2024
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1190 | https://www.grammar.com/anyone_vs._any_one | Anyone vs. Any one
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Anyone vs. Any one
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Have you ever wondered what the difference is between anyone and any one?
Consider the sentences below;
If any one of your friends knows, please tell them to keep quiet.
Has anyone seen my wallet?
Both of these sentences have used the words anyone and any one correctly, question is can you use them correctly? If you can’t figure out when to use which of these, keep reading as this article will tell you everything about the two words.
Anyone as pronoun:
Anyone is a pronoun and is used to call upon a noun. It means any person at all; anybody:
Did anyone see the accident?
Anyone as a pronoun meaning “anybody” or “any person at all” is written as one word.It is used when there are no qualifications to the grouping. Something could belong to anyone if there are no distinguishing marks or unique factors.
Does anyone have the correct time?
Any one as adjective phrase:
Any one is a combination of two words which is generally not listed in dictionaries except perhaps to distinguish the differences with anyone. Any one is a term that means any single object or person.
Any one of your buddies, if he's careless enough, could turn out to be your enemy.
Adjective phrase that refers to any single member of a group (of either people or things). Any one is commonly followed by the preposition of.
She never admitted that any one of her pupils, even the ones who were unmistakably tone deaf, were deficient in musical talent.
Examples:
Anyone willing to part with between £20,000 and £200,000 for a week’s holiday can live in the homes of the rich and famous while they are away, order their staff around, and make use of their cars, tennis courts and swimming pools. [The Guardian]
Impressively, everything is available to tweak and tune if you so desire from the very start but the stock settings on each car are good enough to get you around any one of the current 110 track variations. [Scottish Daily Record]
In the United States, spiritual mobilization has long been anyone’s game. [The New York Times]
Anyone or any one:
Anyone or any one, both are grammatically singular, regardless of meaning. But there is a difference in meaning between the one- and two-word versions: when you type anyone, you're referring to people; when you type any one you may be referring to people, but not necessarily--it depends on what follows or what is understood.For example, perhaps you mean 'any one of the customers' (in which case you are referring to people); or maybe you mean 'any one of the petunias' or (in which case you are not referring to people). In sum, any one means one of a group (of people or things), rather than one person (anyone) or a bunch of people (everyone).
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Gleidison Lima I just didn't know it was possible the use of "any one". Thanks for the tip! Like Reply 1 5 years ago
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AWe had eaten when she arrived.
BThey have been waiting for the train.
CI will be finished my work soon.
DShe was reading a book when the phone rang.
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1191 | http://www.hanlonmath.com/pdfFiles/643EquationsofLines-Derive.pdf | Hanlonmath 800.218.5482 bill@hanlonmath.com 1 Equations of Lines – Derivations If you know how slope is defined mathematically, then deriving equations of lines is relatively simple. We will start off with the equation for slope, normally designated by the letter m, and derive the Point Slope form of a line, Slope Intercept, and General Form of an Equation of a Line. Slope y2 −y1 x2 −x1 = m Using the slope formula, I will substitute x for x2 and y for y2, that results in y −y1 x −x1 = m Multiply both sides by CD; x – x1; y – y1 = m(x – x1) That simple multiplication by the denominator in the slope formula yields y – y1 = m(x – x1) which is the Point Slope Form of a Line The Point Slope Form of a Line is generally used for writing/finding an equation of a line. Example 1 Find an equation of a line that passes through (2, –5) with slope 3. Using y – y1 = m(x – x1), substitute the values of x and y from the ordered pair and the value of the slope into the equation. y – (–5) = 3(x – 2) this is an answer, but I could simplify further y + 5 = 3(x + 2) y + 5 = 3x + 6 y = 3x + 1 Example 2 Find an equation of a line that passes through (2, 1) and (4, – 9). Using y – y1 = m(x – x1), I can substitute the values of x and y from either ordered pair. The slope was not given, but I can find it by the slope formula. m = −9 −1 4 −2 = −10 2 = −5 Using the ordered pair (2, 1) and substituting Hanlonmath 800.218.5482 bill@hanlonmath.com 2 y – 1 = – 5(x – 2) y – 1 = – 5x + 10 y = – 5x + 11 If I did enough problems like examples 1 & 2 and simplified them by solving for y, we would see that always results in an equivalent equation the form of y = mx + b. The b is the result of using the Distributive Property and combining like terms. In both cases you can see the slope is the coefficient of the x. Interestingly, when solving the Point Slope Form of an Equation of a Line for y, it always results in an equation of the form y = mx + b. The y-intercept of a graph, where a graph crosses the y-axis, occurs when the value of x is 0. In the equation y = mx + b, when x = 0, then y = b. So b is called the y-intercept and it is where the graph crosses the y=axis. y = mx + b is called the Slope Intercept Form of a Line The Slope Intercept Form of a Line is normally used for graphing. The graph of a linear equation is a line - a line is defined by two points. If one point is given to us, the y-intercept, we can find another point by using the slope. Example 3 Graph y = 3x + 2 Example 4 Graph y = – 2 3 x + 1 The y-intercept is 2, located at (0, 2) The slope is 3 or 3/1. That means from (0,2), I go up 3 and over 1 to find another point. The y-intercept is 1, located at (0, 1) The slope is –2/3. That means from (0, 1), I go down 2 and over 3. OR I could have gone up 2 and over –3 from (0, 1) Hanlonmath 800.218.5482 bill@hanlonmath.com 3 By knowing the Slope Intercept Form of an Equation of a Line, I can graph it by inspection rather than making an x-y chart and plugging in points. Again and again we see, the more math you know, the easier it gets. Example 5 Find the slope and y-intercept of y = 4x – 2. By inspection, the slope is 4 or 4/1 and the y-intercept occurs at (0, – 2) If I were to graph that equation, I would place a point on (0, – 2), then from there count up 4 and go over 1 and place another point. Drawing a line through those two points is the graph of the equation. Now if I took an equation, such as in Examle 5, y = 4x – 2 and placed all the variables on one side and the constant on the other side of the equation, y = 4x – 2 would look like 4x – y = 2 An equation in the form, Ax + By = C is said to be the General Form of an Equation of a Line. Equations are often written in this form. The question is, how much information can I gleam from an equation written in General Form? Well, we already now that the intercepts occur when the other variable is zero. That is, the y-intercept occurs when x = 0 and the x-intercept occurs when y = 0. So by looking at an equation in General Form, we can readily identify where the graph crosses the x and y axes. Example 6 Find the x and y intercepts of 2x + 5y = 10 The y-intercept occurs when x = 0, if x = 0, then 5y = 10 or yint = 2. The x-intercept occurs when y = 0, if y = 0, then 2x = 10 or xint = 5 The xint occurs at (5, 0) and the yint occurs at (0, 2) Example 7 Find the x and y intercept of 3x – 2y = 6 If x = 0, then – 2y = 6 or yint = – 3 (0, – 3) is the yint If y = 0, then 3x = 6, or xint = 2 (2, 0) is the xint In examples 6 and 7, those two intercepts could be plotted and a line draw through those two points for the graph of the line. Hanlonmath 800.218.5482 bill@hanlonmath.com 4 If we looked further at those two equations, we might be able to find more information by inspection. Lets look at the last two equations written in General Form and solve the for y, placing them in Slope Intercept Form. 2x + 5y = 10 3x – 2y = 6 5y = – 2x + 10 – 2y = – 3x + 6 y = – 2 5 x + 2 y = 3 2 x – 3 In these two problems, the y-intercepts are the same as we found using the General Form and since the equations are now in Slope Intercept Form, we can see the slopes are – 2 5 and 3 2 respectively. Looking at the two equations in General Form; 2x + 5y = 10 and 3x – 2y = 6, is there a pattern that might suggest I could find the slopes of those two lines without solving for y? Let’s see 2x + 5y = 10, slope = – 2 5 and 3x – 2y = 6, slope = 3 2 It appears if I put the coefficient of the x over the coefficient of the y and take the opposite sign, that’s the slope. That observation suggests that if we have an equation in General Form, in Ax + By = C, the slope = – A B Let’s show that more formally Ax + By = C Solving for y, we have By = –Ax + C y = – A B x + C B We can see that when we solve the General Form of an equation of a line for y, the coefficient of x will always be – A B , the slope. Example 8 Find the x and y intercepts and the slope of 4x + 3y = 12 by inspection. xint occurs when y = 0, 4x = 12, xint = 3 yint occurs when x = 0, 3y = 12, yint = 4 Hanlonmath 800.218.5482 bill@hanlonmath.com 5 The slope is – A B , – 4 3 Example 9 Find the xint, yint, slope and graph 5x – 3y = 15 By inspection, the xint = 3, the yint = – 5, and m = – 5 −3 = 5 3 Summary We can see all these equations of lines are related and came directly from the definition of slope. The Point Slope form of a line came directly from the slope formula. The Slope Intercept form of a line came directly from the Point Slope (solving for y), and the General Form of an equation of a line came from placing the x and y terms on the same side of the equation. Knowing those formulas allows us to find the xint, yint,, slope and graph by inspection. |
1192 | https://www.youtube.com/watch?v=k5Tlg27JDtc | Mesh Current Problems - Electronics & Circuit Analysis
The Organic Chemistry Tutor
9880000 subscribers
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Description
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Posted: 28 May 2019
This electronics video tutorial explains how to analyze circuits using mesh current analysis. it explains how to use kirchoff's voltage laws combined with ohm's law to calculate the electric current flowing through each resistor as well as the electric potential of each node.
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in this video we're going to solve this circuit using mesh current analysis so what we have here on the left is a 90 volt battery r1 is 9 ohms r2 is 6 ohms r3 is 8 ohms and we have a 5 amp current source so using mesh current analysis calculate the current flowing through each resistor as well as the electric potentials at these points let's say this is the ground with an electric potential of zero volts what is the potential let's say at point a b and c so go ahead feel free to pause the video use mesh current analysis to solve this circuit so what we need to do is identify the currents in each loop let's call this loop one and the current that flows through it we're gonna say is i1 the current that flows through loop two is i2 now the direction of these currents is not really important because we know in the second loop there's a current flowing in this direction so you can just make the directions of each current in each loop you could just put them in a clockwise direction and as long as you follow the pattern it's going to work out now you need to be familiar with kirchhoff's voltage law which basically states that the sum of the voltages around a closed loop circuit adds up to zero now you also need to be familiar with the polarity sides for instance as you travel around the circuit if you go from a negative value to a positive value the electric potential is increasing so this is a voltage rise so you're going to use a positive v value if you travel let's say through an element going from the positive sign to the negative side the electric potential is decreasing so you need to use a negative v sign so feel free to write that down in your notes somewhere because we're gonna use that often so let's say we're starting from this point and we're going to go around the first loop as we travel through the battery notice that we're going from the negative terminal to the positive terminal so that's a voltage rise so we're going to use positive v let's say vb4 the voltage across the battery now going through this resistor this is the positive side of the resistor this is going to be the negative side current always flows from a high potential to a low potential so that's going to be a voltage drop resistors tend to be associated with a voltage drop because they consume energy from a circuit whereas battery tend to provide energy to the circuit so typically they're associated with a voltage rise now this is going to be v1 the voltage across r1 and then as we travel through r2 that's going to be another voltage drop current is going to flow from vb from a high potential to a low potential since vb is higher than zero volts this side should be positive so we're going to use minus v2 going through r2 it's another voltage drop vb is higher than the ground value so this is the equation that we have using kirchhoff's voltage law now keep in mind that voltage is equal to current times resistance according to ohm's law now we already know what vb is vb is equal to 90. v1 is going to be i1 times r1 v2 is the current that flows through r2 times r2 now the current that flows through r2 it's really the difference between i1 and i2 we have i1 going down in this direction i2 going up in that direction so the actual current that is flowing through resistor 2 is the difference between i1 and i2 so now let's simplify what we have we don't know the value of i1 but we know that r1 is 9 ohms so this is going to be 9 times i1 r2 is 8 ohms actually i take that back r2 is not 8 ohms but it's 6 ohms so let's correct that mistake now let's distribute the six so we have 90 minus nine times i one minus six times i1 plus six times i2 our next step is to combine like terms so negative 9 minus 6 that's going to be negative 15. i'm going to take the 90 and move it to the right side it's positive on the left side but it's going to be negative on the right side now i'm going to simplify this equation i can divide everything by negative three so i'm gonna have five i one minus two i two equal positive thirty now what is the value of i2 i2 is going in the clockwise direction but we have a current of 5 amps going in the counterclockwise direction so therefore i2 is not 5 rather it's a negative 5 amps because it's opposite to the direction to the actual direction of the current so we can replace i2 with negative 5 amps negative 2 times negative 5 that's positive 10 and now let's subtract both sides by 10. so we have 5 times i1 is equal to 30 minus 10 which is 20. dividing both sides by 5 we can see that i1 is going to be 20 divided by 5 which is 4 amps so now that we have the current flowing in the first loop and we also have the current flowing in the second loop we have everything that we need to solve for this circuit so let me just clear out a few things so we have a current of 4 amps flowing through resistor 1 and we have a current of 5 amps flowing through resistor 3. so 4 and five is going to add up to nine so that's nine amps of current is flowing through resistor two so now that we have the currents flowing in each resistor we can now calculate the electric potential at points a b and c to calculate it at point a we really don't have to do much because the voltage of the battery is 90. so the positive terminal is going to be 90 volts higher than the negative terminal and the potential anywhere along this line is zero so this is the potential at a now what about the potential at b what's the answer v2 is equal to i2 times r2 that is the voltage across resistor 2 is equal to the current flowing through it times resistor 2. now voltage is the electric potential difference between two points in this case the ground and vb so v2 is going to be vb minus the ground which is zero i2 is nine amps r2 is six ohms so the voltage or the electric potential rather at point b is 54 volts now what about at point c how can we find the answer now the first thing we need to realize is that the current is going to flow from a high electric potential to a low electric potential so the potential at point c is going to be higher than point b so using this formula v equals ir we'll say v3 is equal to i3 times r3 the voltage across resistor 3 is going to be the difference between the potential at point c and the potential at point b which is 54 volts i3 is 5 amps r3 is 8 ohms 5 times 8 is 40 and if you add 54 to both sides the potential at point c is going to be 94 volts so that's how we can calculate the electric potential at every point in a circuit as well as all of the currents flowing through each resistor in a circuit here's another example with two batteries and five resistors go ahead and calculate the current through each resistor as well as the electric potential at every point so let's say the ground is given a value of zero we already know that this point will have a potential of 130 and this point has to have a potential of 50 volts now in the first loop we're going to call this current i1 and in the second loop that's going to be i2 and then i3 will be in the third loop this is going to be the positive sign of resistor 1 this will be the negative sign and this will be the positive sign of resistor two this will be the negative sign since this is the ground which has the lowest potential so now let's analyze the first loop started from this point so as we go through the battery that's a voltage rise through resistor one that's a voltage drop and through resistor two that's a voltage drop so this is going to be the voltage of the battery that's a voltage rise so we're going to give it a positive value we have a voltage drop across r1 so that's going to be a negative value and the voltage drop across r2 which is also negative so that's the first equation that we have using kirchhoff's voltage law the voltage of the battery is 130. v1 is going to be i1 times r1 r1 is 10 so v1 is 10 times i1 v2 is going to be r2 which is 8 times the difference between i1 and i2 now let's go ahead and distribute the eight so this is going to be negative 10 i1 minus 8i2 i mean minus 8i1 plus 8i2 and now let's combine these two terms so negative 10 minus 8 that's negative 18 and then plus 8i2 and let's take the 130 move it to the other side where it's going to be negative 130. now our next step is to divide everything by negative 2 just to simplify the equation so negative 18 divided by negative 2 that's going to be positive 9 8 divided by negative 2 that's negative 4 and negative 130 divided by negative 2 is positive 65. so we're going to save this equation for now we'll use it later now let's focus on our second loop so this is going to be positive that's going to be negative across resistor this is the positive terminal of resistor 4 and this is going to be the negative terminal of it so starting from this point as we go up through r2 that's going to be a voltage rise so i'm going to have positive v2 and then as we go through resistor 3 that's a voltage drop so that's going to be negative v3 and then going through resistor 4 that's also a voltage drop because we're going from positive to negative so that's minus v4 so v2 we know it's going to be r2 which is 8 times the difference between i1 and i2 v3 that's i3 times r3 i mean i2 times r3 i2 flows through r3 based on what we have here so it's going to be 5 times i2 v4 the voltage across r4 we have a 2 ohm resistor and it's going to be the difference between these two currents so it's i2 minus i3 and this is going to equal 0. now let's simplify so we're going to have 8 i1 minus 8i2 minus 5i2 and then distributing the negative 2 this is going to be negative 2 i2 plus 2 i3 and that's going to equal 0. so we can combine these three those three values so negative eight minus five that's negative thirteen minus two that's negative fifteen so we have this formula eight i one minus fifteen i two plus 2 i3 all of that is going to equal 0. so we'll save that for later now let's focus on the third loop we're going to make this side positive and this side negative so going up through resistive four that is a voltage rise going from negative to positive so that's going to be positive v4 going through a resistor five that's a voltage drop so negative v5 now going through this battery we're going from the positive side to the negative side a positive value is higher than a negative value so the potential is decreasing so we're going to say this is negative 50. so in this case this battery is being charged by the 130 volt battery since current is going this way if that is the correct direction of the current if the current is going this way then this battery is being discharged right now we don't know but assuming if the current is going this way then it's being charged up now if i three is negative that means that the current is going in this direction which it could be now v4 we're going to replace it with the 2 ohm resistor times the difference between i2 and i3 v5 is going to be r5 times the current flowing through it which is i3 so that's 5 i3 now let's distribute the two so this is going to be 2i2 minus two i three minus five i three and let's move the negative fifty to the other side so now let's combine like terms this is gonna give us two i two minus seven i3 and that's going to equal 50. so now we have three equations which is enough to find the values of the three unknown variables so how can we solve this system of three equations what do you think we need to do what i would recommend doing is canceling i1 in the first two equations that's going to give us an equation with i2 and i3 and then we'll combine that with equation 3. so what we're going to do is we're going to multiply this equation by 8. so 9 times 8 that's 72 4 times 8 is 32 and then 65 times eight that's going to be 520 now we're going to multiply the equation in red by negative nine eight times negative nine that's going to give us negative 72 negative 15 times negative nine that's going to give us positive 135 and then zero times negative 9 is 0. so adding these two equations those values will cancel and so we're going to have negative 32 plus 135 you know what i forgot something i forgot to multiply i3 that is 2i3 by negative 8. so that's going to be negative 18i3 and that's equal to 0. so now adding these two negative 32 plus 135 that's going to be 102 i2 i mean 103 rather i'm making too many mistakes today and then we're going to have minus 18 i3 and that's going to equal 520 now let's combine this equation with this one here so let's cancel i2 i'm going to multiply the equation in green by 103 and the equation of white by negative 2. so i will get negative 206 i2 and positive 206 i2 so i can cancel those variables so let's start with this one 2 times 103 that's going to be 206 i2 and then negative 7 times 103 that's going to be negative 721 times i3 and then 50 times 103 that's 5150. now multiplying this equation by 2 i'm going to have negative 206 i2 negative 18 times negative 2 that's going to be positive 36 i3 and then 520 times negative 2 that's going to be negative 10 40. now let's go ahead and add those two equations together so we're going to have negative 721 i3 plus 36 i3 which is negative 6 85 i3 5150 minus 1040 that's going to be 4110 dividing both sides by negative 685 this will give us i3 so i3 is equal to negative 6 amps so because i three is negative that means that we have a current flowing in this direction so the current shouldn't be flowing from a negative value to a positive value it always flows from a high potential to a low potential so we need to reverse the signs of r5 so this should be the positive side and this should be the negative side let's call this point point a and point b so the potential at point b should be less than 50 volts since current is flowing in that direction now let's go ahead and calculate the other currents so let's use this equation before we multiply it by 103 to calculate i2 so it's 2 times i2 minus 7 times i3 and i3 is negative 6 and this is going to equal 50. negative 7 times negative 6 that's positive 42 subtracting both sides by 42 we'll have 50 minus 42 which is 8. dividing both sides by 2 we can see that i2 is equal to 4 amps so the current flowing through a resistor three that's going to be four amps so let me just put this here i'm gonna have to erase a few things just to have some extra space so that's a four amp current right there now notice that we have six amps flowing threes this to five four amps flowing through resistor three so they will combine and form a 10 amp current flowing through resistor four because four plus six will add up to ten what just happened sometimes this computer acts up now let's calculate i one using this equation so this is going to be 9 times i1 minus 4 times i2 which is 4 and that's equal to 65. 4 times 4 is 16. so adding 16 to both sides we'll have 65 plus 16 which is 81 and then dividing both sides by 9 81 divided by 9 is 9. so that's the value of i1 it's 9 amps so the current flowing through r1 that's going to be 9 amps so what is the current flowing through r2 so let's draw a picture we have nine amps of current entering point a four amps is leaving point a that means that we have five amps that must be leaving as well so there's a 5 amp current that is flowing through resistor 2. now let's calculate the electric potential at point a and b to find it at point a we need to find the voltage across resistor 2 which is the current that flows through it times r2 so the voltage across resistor 2 is the difference between the potential at a and the potential of the ground which is zero the current that flows to it is five and the resistance is eight so five times eight is forty so that is the electric potential at point a so i'm going to highlight this in green now just for the sake of discussion notice that across resistor 1 we see that the positive side is at a higher potential than the negative side the negative side is at a potential of 40. the positive side is at a potential of which is how it should be now let's focus on calculating the potential at point b the best way to do that is to find the voltage across resistor 4. so v4 is going to equal the current through it times r4 so v4 is going to be the potential at point b minus the ground potential the current is 10 amps and the resistance is 2. 10 times 2 is 20. so this is twenty volts notice that the difference between these two potentials is twenty which is equal to i times r five times four the difference between these two potentials is 30 which is equal to i times r 6 times 5 which is 30. so everything should make sense if you did everything correctly the voltage across each resistor should be the difference in the electric potential values and that should equal the current times the resistance so you can check that to make sure everything is correct so that's basically it for this video hopefully this really helped you to be able to solve complex circuits like this one thanks again for watching |
1193 | https://askfilo.com/user-question-answers-smart-solutions/as-a-power-of-2-3132353237303431 | 16 ^ { - 2 } as a power of 2 . | Filo
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16 ^ { - 2 } as a power of 2 .
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1 6−2 as a power of 2 .
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Updated on: Sep 28, 2024
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Text solutionVerified
Concepts:
Exponents, Powers of numbers
Explanation:
To express 1 6−2 as a power of 2, we first need to express 16 as a power of 2. Then we can apply the exponent rule (a m)n=a m⋅n.
Step by Step Solution:
Step 1
Express 16 as a power of 2: 16=2 4.
Step 2
Substitute 2 4 for 16 in the original expression: 1 6−2=(2 4)−2.
Step 3
Apply the exponent rule (a m)n=a m⋅n: (2 4)−2=2 4⋅(−2)=2−8.
Final Answer:
1 6−2=2−8
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Question Text 1 6−2 as a power of 2 .
Updated On Sep 28, 2024
Topic All topics
Subject Smart Solutions
Class Class 8
Answer Type Text solution:1
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1194 | https://rexresearch1.com/PuzzlesRiddlesLibrary/300MathematicalPatternPuzzles.pdf | 300+ Mathematical Pattern Puzzles Number Pattern Recognition & Reasoning Chris McMullen, Ph.D.
Improve Your Math Fluency Series ISBN-13: 978-1512044287 ISBN-10: 1512044288 www.chrismcmullen.com www.improveyourmathfluency.com 300+ Mathematical Pattern Puzzles Number Pattern Recognition & Reasoning Improve Your Math Fluency Series Chris McMullen, Ph.D. Copyright © 2015 Mathematics > Popular > Patterns Entertainment > Puzzles > Math Games Contents Introduction 5 1 Ordered 9 2 Patterned 13 3 Alternating 17 4 Cyclic 23 5 Additive & Multiplicative 29 6 Multiple Operations 33 7 Digits 39 8 Prime Numbers 45 9 Fibonacci Inspired 49 10 Roman Numerals 53 11 Powers 59 12 Factorials 63 13 Negative Numbers 71 14 Fractions 75 15 Algebraic 85 16 Visual 95 17 Arrays 101 18 Analogies 109 Answers 117 About the Author 169 Introduction Enjoy solving a variety of mathematical pattern puzzles, including prime num-bers, Fibonacci-like series, visual pat-terns, analogies, arrays, and more. This book starts out easy with basic patterns and simple puzzles, and the level of difficulty steadily grows so that people of all ages and abilities can enjoy many of the patterns and puzzles in this book. Each chapter begins with a brief intro-duction (and, where necessary, a quick review) of the relevant concepts, followed by 2-3 examples of pattern puzzles with explanations. The answer, along with an explana-tion, to every puzzle can be found at the end of the book. Some chapters incorporate ideas from earlier chapters. For example, once prime numbers are introduced in Chapter 8 and the Fibonacci sequence is introduced in Chapter 9, these concepts may appear in a subsequent chapter. Examples Following are a few examples of the kinds of patterns that you can find in this book. Fill in the blanks. Example 1. 9, 25, 49, 81, _, , , _ Example 2. 15, 15, 10, 30, 30, 20, 60, 60, 40, , , _ Example 3. Check your answers on the next page. Answer to Example 1. Square odd numbers: 3 × 3 = 9, 5 × 5 = 25, 7 × 7 = 49, and 9 × 9 = 81. The next four numbers are 11 × 11 = 121, 13 × 13 = 169, 15 × 15 = 225, and 17 × 17 = 289. Answer to Example 2. The first 3 numbers are 15, 15, and 10. Double these to get 30, 30, and 20. Double those to get 60, 60, and 40. The next three numbers are 120, 120, and 80. Answer to Example 3. The black square advances 3 spaces clockwise along the large square: 1 Ordered The patterns in Chapter 1 follow a common recognizable order, like whole numbers (1, 2, 3, 4, 5...), the alphabet (a, b, c, d, e...), or geometric shapes (triangle, square, pentagon, hexagon, heptagon...). Example 1. These even numbers progress in order, beginning with 6. 6, 8, 10, 12, 14, 16, 18... The next two numbers are 20 and 22. Example 2. These letters of the alphabet appear in order, skipping every other letter, beginning with m. m, o, q, s, u... It’s really m, skip n, o, skip p, q, skip r, s, skip t, u, skip v. The next two letters are w and y (since x is skipped). Example 3. This visual pattern consists of polygons. Count the number of sides. The triangle has 3 sides, the square has 4 sides, the pentagon has 5 sides, etc. , , , , ... The next two shapes are an octagon (8 sides) and nonagon (9 sides): , #1 7, , 11, _, , 17, 19, _ #2 5, _, 15, , _, 30, , 40 #3 _, 600, 700, , 900, , _, 1200 #4 25, 50, , , 125, _, 175, #5 , 65, 77, _, , 113, , 137 #6 b, d, f, , j, , , p #7 c, g, k, , s, ___ #8 _, , _, , #9 , , _, , , _ #10 15, _, 13, , 11, _, , 8 #11 _, 27, 24, 21, , 15, _, #12 125, _, 115, , , 100, _, 90 #13 , 1075, 1050, , _, , 950, 925 #14 , w, v, , t, s, r, ___ #15 q, , m, , i, , e, c #16 y, v, , p, , , g, d #17 , _, , , , __ #18 _, , , , , _ #19 A, , I, O, U #20 d, f, g, h, , , l, m, n, , q 2 Patterned The sequences of Chapter 2 follow a repeating pattern. For example, 1, 7, 5, 1, 7, 5, 1, 7, 5 repeats the digits 1, 7, and 5. Example 1. These digits alternate: five, then three, five again, then three again. 5, 3, 5, 3, 5, 3, 5... The next two numbers are 3 and 5. Example 2. These letters follow a pattern: 1 Q, then 2 E’s, 1 Q, 2 E’s, 1 Q, 2E’s, etc. Q, E, E, Q, E, E, Q, E, E, Q... The next two letters are E and E. Example 3. This pattern grows. Each num-ber gains one digit in order. The number 9 turns into 98, then 987, followed by 9876, and so on. 9, 98, 987, 9876, 98765, 987654... The next two numbers are 9876543 and 98765432. #1 0, 1, 0, 1, 0, , , ___ #2 t, , , , t, y, t, y #3 A, a, , a, A, a, A, a, , ___ #4 2, 4, 8, 2, 4, 8, 2, , , ___ #5 z, x, q, z, x, q, z, x, , , ___ #6 , , , , , , , _, , _ #7 9, 9, 4, 9, 9, 4, 9, , , ___ #8 3, 5, 3, 3, 5, 3, 3, , , ___ #9 G, g, g, G, g, g, , , , G #10 F, F, e, F, , e, F, F, , ___ #11 , , , , , , , _, , _ #12 2, 5, 3, 9, 2, 5, 3, 9, 2, 5, , , ___ #13 1, 7, 6, 4, 8, 1, 7, 6, 4, 8, 1, 7, 6, , , ___ #14 D, b, P, q, D, b, P, q, D, b, P, , , ___ #15 C, e, H, j, c, E, h, J, C, e, H, , , ___ #16 8, 2, 8, 8, 2, 8, 8, 8, 2, 8, 8, 8, 8, 2, 8, 8, 8, , , ___ #17 6, , 888, __, 1010101010, 111111111111, 12121212121212, _____ #18 2, 23, 234, 2345, _, _, _ #19 1, 31, ___, 7531, 97531, 1197531, __, __ #20 4, 4, 3, 12, 12, 9, 36, 36, 27, _, , _, 3 Alternating An alternating sequence has two (or more) patterns intertwined. For example, consider the two patterns 1, 2, 3, 4... and A, B, C, D... They can be combined together in a single alternating pattern like 1, A, 2, B, 3, C, 4, D... Example 1. This pattern alternately merg-es one sequence of odd numbers with another sequence consisting of every oth-er letter of the alphabet. One pattern is 1, 3, 5, 7, 9, 11, 13... The second pattern is b, d, f, h, j, l, n... See how these merge together below. 1, b, 3, d, 5, f, 7, h, 9, j, 11, l... The next two elements are 13 (following 11) and n (following the letter l). Example 2. The first pattern of this alternating sequence counts by four: 4, 8, 12, 16, 20... The second pattern subtracts three: 36, 33, 30, 27, 24... 4, 36, 8, 33, 12, 30, 16, 27, 20, 24... The next two numbers are 24 (adding 4 to 20) and 21 (subtracting 3 from the previous 24). Example 3. Both of these patterns consist of letters. The first has letters progressing forward: A, B, C, D, E... The second has letters going backward: M, L, K, J, I... A, M, B, L, C, K, D, J, E, I... The next two letters are F (coming after E) and H (coming before I). #1 11, 99, 22, 88, , , 44, 66, , , 66, 44 #2 3, 18, 6, 16, 9, 14, 12, 12, , , , ___ #3 6, 24, 12, , , 24, 24, 24, 30, , ___ #4 5, 6, 7, 9, 9, 12, 11, 15, , , , ___ #5 10, 50, 15, 40, 20, 30, 25, , , , ___ #6 p, q, o, r, n, s, m, , , , ___ #7 , , , , V, P, T, Q, R, R, P #8 c, f, d, , , f, f, f, g, f, h, , ___ #9 J, 29, L, 26, N, 23, P, 20, , , , ___ #10 7, z, , , 15, t, 19, , , n #11 , , , , , , , , _, _ #12 , , , , , , , , , , #13 , , , , , , , , , #14 →, ↑, ↓, ↓, ←, ↑, ↑, ↓, →, ↑, , , , ___ #15 ↑, ←, ←, ↖, ↓, ↑, →, ↗, ↑, →, , , , ___ #16 Q, z, X, w, , z, , w, , , X, w, Q #17 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, , , , ___ #18 D, e, f, F, H, g, j, H, L, i, n, , , , ___ #19 30, 70, 25, 35, 60, 35, 40, 50, 45, 45, 40, 55, 50, 30, 65, 55, , , , ___ #20 K, h, Q, L, i, P, M, j, O, N, k, N, O, l, M, P, , , , ___ 4 Cyclic A pattern that repeats itself in some way is cyclic. For example, time is cyclic. The hours vary from 1 to 12 and then start over again at 1. That is, after 12 o’clock comes 1 o’clock. Example 1. These 3 digits cycle in the same order repeatedly: 418, then 184, and then 841. 418, 184, 841, 418, 184, 841... The next two numbers are 418 and 184. Example 2. The single-digit odd numbers are cyclic, where after 9 we continue again with 1. 1, 3, 5, 7, 9, 1, 3, 5... The next two numbers are 7 and 9. Example 3. This pattern continues in half hour increments of time. 10:00, 10:30, 11:00, 11:30, 12:00, 12:30, 1:00, 1:30... The next two times are 2:00 and 2:30. #1 927, 279, 792, 927, 279, , _, , _ #2 3841, 8413, 4138, 1384, 3841, _, _, _, _ #3 4, 6, 8, 0, 2, 4, 6, , , , ___ #4 , 88, __, 0000, 11111, 222222, ___, __ #5 123, 132, 231, , _, 321, 123, 132, , _ #6 act, cta, tac, act, cta, , _, , _ #7 spot, __, _, tspo, spot, pots, _, _ #8 u, w, , , c, , g, , k #9 ccccccccc, bbbbbbbb, __, _, yyyyy, __, www, _ #10 Roygbiv, rOygbiv, __, royGbiv, __, __, roygbiV, Roygbiv, __ #11 0, 11, _, 3333, 222, , 0, 11, , 3333, _ #12 CIRCLE, IRCLEC, _, CLECIR, _, ECIRCL, _, IRCLEC, _ #13 10:45, 11:15, _, _, 12:45, _, _, 2:15 #14 9:30, _, 11:00, 11:45, _, _, 2:00, _ #15 Friday, __, _, Monday, _, Wednesday, __, Friday #16 October, November, _, _, February, __, April, _ #17 north, _, south, west, north, __, _, __ #18 east, northeast, north, _, west, southwest, _, southeast, _, _ #19 north, southeast, west, northeast, south, northwest, _, __, _, __ #20 90°, 180°, 270°, 360°, 90°, , , , 5 Additive & Multiplicative The patterns of Chapter 5 involve one of the four basic arithmetic operations: addition, subtraction, multiplication, or division. (However, the operations are not merged together until Chapter 6.) Example 1. This sequence is made by adding 4 repeatedly, beginning with 5. 5, 9, 13, 17, 21, 25, 29... The next two numbers are 33 and 37. Example 2. This pattern involves multi-plying the previous answer by 2. 3, 6, 12, 24, 48... The next two numbers are 96 and 192. Example 3. This sequence is made by subtracting 8 each time. 70, 62, 54, 46, 38, 30, 22... The next two numbers are 14 and 6. #1 6, 12, 18, 24, , _, , _ #2 72, 64, 56, 48, , _, , _ #3 , _, 4, 8, 16, 32, , _ #4 7, 13, , _, 31, 37, , _ #5 640, 320, 160, 80, , _, , _ #6 , 37, 41, _, , 53, 57, _ #7 , _, 38, 31, , 17, 10, _ #8 25, , _, , , 800, 1600, 3200 #9 6, _, 16, , 26, _, 36, #10 _, 20, , 58, , 96, _, 134 #11 9, 18, , _, 45, , 63, _ #12 , 40, 34, 28, _, , 10, _ #13 8, 16, , 64, _, , 512, #14 _, 21, 33, , 57, 69, _, #15 768, 384, 192, 96, _, , _, #16 212, _, 186, 173, , 147, , _ #17 , , 18, __, 162, 486, 1458, _ #18 192, , _, 264, 288, 312, , #19 _, , 50, 250, 1250, 6250, _, _ #20 2187, 729, 243, 81, _, , _, 6 Multiple Operations Multiple arithmetic operations (addition, subtraction, multiplication, and division) are merged together in these sequences. For example, the same sequence may in-volve both addition and multiplication. Example 1. This sequence involves multi-plying the previous term by 2 and then adding 1. For example, 2 × 3 + 1 = 7 and 2 × 7 + 1 = 15. 3, 7, 15, 31, 63, 127... The next two numbers are 255 (because 2 × 127 + 1 = 255) and 511 (since 2 × 255 + 1 = 511). Note that you can also make this pattern by adding 4, then adding 8, then adding 16, then adding 32, each time adding twice as much as previously: 3 + 4 = 7, 7 + 8 = 15, 15 + 16 = 31, 31 + 32 = 63... Example 2. This pattern is made by adding 2, then adding 3, then adding 2, then adding 3, continuing to alternate. For example, 4 + 2 = 6, 6 + 3 = 9, 9 + 2 = 11, and 11 + 3 = 14. 4, 6, 9, 11, 14, 16, 19... The next two numbers are 21 (since 19 + 2 = 21) and 24 (since 21 + 3 = 24). Example 3. This sequence is made by alternately multiplying by 2 and dividing by 3. Beginning with 324, multiplying by 2 makes 648, dividing by 3 makes 216, multiplying by 2 makes 432, dividing by 3 makes 144, and so on. 324, 648, 216, 432, 144... The next two numbers are 288 (since 144 × 2 = 288) and 96 (since 288 / 3 = 96). (Note that the slash can be used for division: 288 / 3 = 288 ÷ 3 = 96.) #1 2, 7, 22, 67, 202, __, _, , #2 3, 9, 8, 14, 13, 19, 18, , _, , _ #3 1022, 510, 254, 126, 62, , _, , _ #4 1, 3, 6, 18, 36, 108, 216, _, _, _, _ #5 101, 99, 95, 93, 89, , _, , _ #6 256, 128, 144, 72, 88, 44, 60, , _, , _ #7 3, 5, 9, , 33, _, , 257, #8 5, 7, 15, 17, 35, 37, 75, 77, 155, _, , , _ #9 1, 3, 6, 10, 15, 21, , _, , _ #10 4, 6, 9, 13, 18, 24, , _, , _ #11 100, 51, 98, 53, 95, 56, 91, 60, , _, , _ #12 8, 4, 12, 6, 24, 12, 60, 30, 180, 90, 630, , , , #13 50, 52, 47, 55, 43, 59, 38, 64, 32, , _, , _ #14 10, 50, 20, 100, 70, 350, 320, 1600, _, __, __, __ #15 1, 3, 6, 9, 18, 22, 44, 49, 98, _, , , #16 2, 2, 1, 2, 1, 3, 2, 8, 7, 35, 34, , , , #17 a, d, c, f, e, h, g, j, i, l, k, , , , ___ #18 1, 2, 6, 9, 10, 14, 17, 18, 22, 25, 26, 30, 33, __, _, , #19 2, 5, 10, 9, 12, 24, 23, 26, 52, 51, 54, 108, 107, , , , #20 8, 6, 24, 12, 10, 40, 20, 18, 72, 36, 34, 136, 68, , , , 7 Digits Each sequence in Chapter 7 involves changing one or more digits of a multi-digit number. For example, the number 384 has three digits: The hundreds digit is 3, the tens digit is 8, and the units digit is 4. A possible pattern is to raise the hundreds digit by 1 and reduce the tens digit by 2. If so, the next number would be 464 (since 3 + 1 = 4 and 8 – 2 = 6), followed by 544 (since 4 + 1 = 5 and 6 – 2 = 4). Example 1. In this sequence of three-digit numbers, the first digit decreases, the second digit remains unchanged, and the third digit increases. For example, 951 becomes 852 as the 9 decreases to 8 and the 1 increases to 2. Watch how the first digit changes from 8, 7, 6..., the middle digit is always 5, and the last digit changes from 1, 2, 3... 951, 852, 753, 654, 555... The next two numbers are 456 and 357. Example 2. The digits in this sequence cycle through a pattern. They rearrange themselves in order. 123, 132, 213, 231, 312... The next two numbers are 321 and 123. Example 3. In this sequence, the digits add up to 9. As examples, look at 63, where 6 + 3 = 9, and 108, where 1 + 0 + 8 = 9. 63, 72, 81, 90, 108, 117... The next two numbers are 126 and 135. (Note that the digits of 99 add up to 18. Although the 1 and 8 of 18 add up to 9, this example is only counting the first sum of the digits, which is why 99 is skipped.) #1 807, 716, 625, 534, , _, , #2 29, 47, 65, 83, 21, 49, _, , _, #3 122, 223, 324, 425, _, , , _ #4 369, 396, _, 693, 936, , 369, , #5 5678, 5687, 5768, 5786, _, 5876, 6578, _, 6758, _, _, 6875 #6 3456, 3465, _, 5463, 6435, 6453, 3456, _, _, _ #7 d8, f7, h6, j5, , _, _, #8 y3C, v5d, s7E, p9f, m1G, __, _, , #9 T8e2, r7Q4, P6E6, n5q8, L4e0, j3Q2, _, _, _, _ #10 24680, 24681, 24691, 24791, 25791, 35791, 35792, 35702, __, _, _, _ #11 89, 98, 179, 188, 197, 269, 278, _, , , _ #12 10, 11, 20, 12, 21, 30, 13, , _, , _ #13 39, 57, 75, 93, 1119, _, 1155, 1173, 1191, 1317, __, __, __ #14 117, 144, 171, 225, 252, 333, 414, _, _, _, 1116, _ #15 13, 35, 57, 79, 135, 357, 579, _, , #16 892, 128, 783, 237, 674, 346, 565, 455, , , , #17 , Atc, _, Cta, Tac, Tca, , Atc, , Cta #18 _, TOSP, TPOS, TPSO, TSOP, _, SOPT, SOTP, _, SPTO, _, STPO #19 bit, diq, fin, hik, jih, __, _, piy, , , vip #20 Ae, Bd, Cc, Db, Ea, Aad, Abc, , _, , #21 123, 342, 534, 456, 675, 867, 789, 908, 190, 012, 231, __, _, , 8 Prime Numbers A prime number is a positive integer that is evenly divisible only by itself and the number one. For example, 7 is prime because it can only be factored as 7 × 1, while 6 is not prime because it can be factored as 2 × 3 in addition to 6 × 1. The first several prime numbers are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31... Example 1. This sequence is formed by skipping every other prime number starting with 5. The pattern is 5, skip 7, 11, skip 13, 17, skip 19, 23, skip 29, 31, etc. 5, 11, 17, 23, 31... The next two numbers are 41 and 47 (since 37 and 43 are skipped). Example 2. This pattern doubles each prime number. For example, 2 × 2 = 4, 2 × 3 = 6, and 2 × 5 = 10. 4, 6, 10, 14, 22... The next two numbers are 26 (since 2 × 13 = 26) and 34 (since 2 × 17 = 34). #1 29, 31, 37, 41, 43, , _, , _ #2 4, 6, 8, 9, 10, 12, 14, 15, , _, , _ #3 , 89, _, 79, , 71, _, 61 #4 2, 3, 7, 11, 17, 19, 29, 31, 41, , _, , _ #5 3, 4, 6, 8, 12, 14, , _, , _ #6 11, 31, 41, 61, 71, , _, , _ #7 3, 5, 7, 23, 29, 41, 43, 47, , _, , _, 113 #8 9, 21, 39, 57, 87, 111, , _, , #9 21, 25, 33, 37, 45, 57, 61, _, , _, #10 4, 10, 22, 34, 46, 62, _, , _, #11 101, 103, 107, 109, 113, _, , , _ #12 102, 103, 105, 107, 111, 113, , , _, #13 3, 5, 5, 7, 11, 13, 17, 19, 29, 31, 41, 43, , _, , _ #14 2, 3, 5, 7, 11, 23, 29, 41, 43, 47, 61, 67, 83, , _, , _ 9 Fibonacci Inspired The Fibonacci sequence adds consecutive terms together. The Fibonacci sequence begins with 0 and 1, adds these together to make 1, then 1 + 1 makes 2, followed by 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8, 5 + 8 = 13, and so on. The patterns in Chapter 9 are not neces-sarily the Fibonacci sequence itself, but they do involve performing arithmetic operations (addition, subtraction, multi-plication, or division) on the previous terms. In this way, these patterns are Fibonacci-inspired. Example 1. The Fibonacci sequence be-gins with 0, 1, and then adds the two previous elements together to make the next element. As examples, 0 + 1 = 1, 1 + 1 = 2, 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8, and 5 + 8 = 13. 0, 1, 1, 2, 3, 5, 8, 13, 21, 34... The next two numbers are 55 (since 21 + 34 = 55) and 89 (since 34 + 55 = 89). Example 2. This sequence multiplies the previous two numbers instead of adding them. For example, 2 × 3 = 6 and 3 × 6 = 18. 2, 3, 6, 18, 108... The next two numbers are 1944 (since 18 × 108 = 1944) and 209952 (since 108 × 1944 = 209952). Example 3. This sequence is similar to the Fibonacci sequence except that it adds the last 3 numbers rather than the last 2. For example, 0 + 1 + 2 = 3, 1 + 2 + 3 = 6, and 2 + 3 + 6 = 11. 0, 1, 2, 3, 6, 11, 20, 37... The next two numbers are 68 (since 11 + 20 + 37 = 68) and 125 (since 20 + 37 + 68 = 125). #1 2, 2, 4, 6, 10, 16, , _, , _ #2 , 3, _, 7, 11, 18, 29, 47, , _ #3 6, , 15, _, 39, 63, , 165, , 432 #4 309, 191, 118, 73, 45, 28, _, , _, #5 1, 2, 2, 4, 8, 32, _, _, _ #6 1, 2, 5, 13, 34, 89, 233, _, , , _ #7 1, 1, 2, 4, 7, 13, 24, 44, 81, 149, _, , , #8 2, 3, 3, 4, 5, 7, 10, 15, 23, 36, , , , #9 1, 3, 5, 9, 15, 25, 41, 67, 109, 177, , , , #10 0, 1, 2, 2, 3, 5, 7, 10, 15, 22, 32, 47, 69, 101, , , , 10 Roman Numerals This chapter involves Roman numerals. The table below provides a quick re-fresher of Roman numerals. I = 1 XI = 11 L = 50 II = 2 XIV = 14 LX = 60 III = 3 XV = 15 XC = 90 IV = 4 XVI = 16 C = 100 V = 5 XIX = 19 CD = 400 VI = 6 XX = 20 D = 500 VII = 7 XXXIX = 39 CM = 900 VIII = 8 XL = 40 M = 1000 IX = 9 XLIX = 49 MM = 2000 X = 10 MMCDXLVIII = 2448 The same symbol can appear as many as three times in a row. For example, XXX equals 30. If a smaller symbol appears left of a larger symbol, it subtracts. For example, XL is 40 and XCV is 95 (but not VC, as X rather than V is used before C). Otherwise, smaller numbers appear to the right, in which case they add. For example, XXXIII equals 33. As a last example of writing Roman numerals, XLV is 45, as the X subtracts 10 from 50, while the V adds 5. Example 1. This sequence begins with Roman numeral X (ten) and counts upward by ones. X, XI, XII, XIII, XIV, XV, XVI, XVII, XVIII... The next two numbers are IX (nineteen) and XX (twenty). Example 2. This pattern counts Roman numerals by fives. V, X, XV, XX, XXV, XXX, XXXV, XL, XLV, L, LV... The next two numbers are LX (60) and LXV (65). Example 3. This pattern is made by adding 2, then adding 3, then adding 2, then adding 3, continuing to alternate. For example, IV + II = VI (or 4 + 2 = 6), VI + III = IX (or 6 + 3 = 9), and IX + II = XI (or 9 + 2 = 11), and XI + III = XIV (or 11 + 3 = 14). IV, VI, IX, XI, XIV, XVI, XIX... The next two numbers are XXI (since 19 + 2 = 21) and XXIV (since 21 + 3 = 24). #1 II, IV, , VIII, X, _, , XVI, , XX #2 III, VI, XII, XXIV, XLVIII, _, _, _, __ #3 CVI, CV, CIV, CIII, CII, CI, _, _, _, _ #4 XXV, L, LXXV, C, CXXV, CL, _, _, _, _ #5 C, CC, CCC, CD, D, DC, , _, , #6 MCMLX, MCMLXV, MCMLXX, MCMLXXV, MCMLXXX, MCMLXXXV, _, _, _, _ #7 DLXX, DLX, DL, DXL, DXXX, DXX, _, _, _, _ #8 IX, VII, XIV, XII, XIX, XVII, XXIV, XXII, _, _, _, _ #9 II, VI, III, IX, VI, XVIII, XV, XLV, XLII, _, _, _, _ #10 VII, IX, XII, XVI, XXI, XXVII, _, _, _, _ #11 II, IV, VI, IX, XI, XV, XX, XL, LI, _, _, _, _ #12 CC, CD, CI, CL, CM, CV, CX, DC, DI, _, _, _, _ #13 III, V, IX, XVII, XXXIII, LXV, CXXIX, _, _, _, _ #14 I, I, II, III, V, VIII, XIII, XXI, _, _, _, _ #15 II, III, V, VII, XI, XIII, XVII, _, _, _, _ 11 Powers A number raised to a power has an exponent. For example, in 34, the number 3 is raised to the power of 4, where 4 is called the exponent. When a number is raised to a power, it means that the number is multiplied by itself that many times. For example, 34 equals 3 × 3 × 3 × 3 (four threes are multiplied together). An exponent of 2 is called a square. For example, 52 is read as “five squared,” and means 5 × 5. Thus, 52 = 25. An exponent of 3 is called a cube. For example, 23 is read as “two cubed,” and means 2 × 2 × 2. Thus, 23 = 8. Any nonzero number raised to the power of 0 equals 1. For example, 80 = 1. If you would like to know why, this is explained in Chapter 15. Example 1. This sequence involves cubing numbers (raising them to the power of three). For example, 13 = 1 × 1 × 1 = 1, 23 = 2 × 2 × 2 = 8, 33 = 3 × 3 × 3 = 27, and 43 = 4 × 4 × 4 = 64. 1, 8, 27, 64, 125, 216... The next two numbers are 343 (since 73 = 7 × 7 × 7 = 343) and 512 (since 83 = 8 × 8 × 8 = 512). Example 2. This pattern consists of powers of three. For example, 30 = 1, 31 = 3, 32 = 3 × 3 = 9, 33 = 3 × 3 × 3 = 27, and 34 = 3 × 3 × 3 × 3 = 81. 1, 3, 9, 27, 81... The next two numbers are 243 (since 35 = 3 × 3 × 3 × 3 × 3 = 243) and 729 (since 36 = 3 × 3 × 3 × 3 × 3 × 3 = 729). Note that you could also make this sequence by multiplying the previous term by 3. #1 __, 4, 9, 16, _, 36, , #2 1, , , 8, 16, 32, 64, _, _ #3 1, 100, 10000, 1000000, __, ___, ___, ___ #4 1, 16, 81, 256, 625, 1296, _, _, _, _ #5 4, 16, 36, 64, 100, 144, _, _, _, _ #6 1, 4, 27, 256, 3125, _, __ #7 1, 16, 256, 4096, _, _ #8 0, 3, 8, 15, 24, 35, _, , _, #9 4, 9, 25, 49, 121, 169, _, , , _ #10 2, 8, 18, 32, 50, 72, , , _, 12 Factorials An integer followed by an exclamation mark (!) is called a factorial. For example, 4! is read as “four factorial.” The factorial notation means to multiply the given number by successively smaller numbers until reaching the number one (except for 0! which will be explained shortly). For example, 4! means 4 times 3 times 2 times 1, which equals 4 × 3 × 2 × 1 = 24. As another example, 3! = 3 × 2 × 1 = 6. Observe that 4! = 4 × 3! (since 24 = 4 × 6). Note that 0! is defined to equal 1. The reason that 0! = 1 is so that 1! can follow the rule N! = N (N – 1)! for all positive integers (N > 0). This way, 1! = 1 (1 – 1)! = 0! since 1! and 0! both equal 1. Example 1. This sequence is made from factorials. For example, 0! = 1, 1! = 1, 2! = 2 × 1 =2, 3! = 3 × 2 × 1 = 6, 4! = 4 × 3 × 2 × 1 = 24. 1, 1, 2, 6, 24, 120... The next two numbers are 720 (since 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720) and 5040 (since 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040). Note that you could also make this pattern by multiplying the first number by 1 (1 × 1 = 1), the second number by 2 (1 × 2 = 2), the third number by 3 (2 × 3 = 6), the fourth number by 4 (6 × 4 = 24), the fifth number by 5 (24 × 5 = 120), and so on. Example 2. The double factorial involves just odd or even numbers. For example, 0!! = 1, 2!! = 2, 4!! = 4 × 2 = 8, 6!! = 6 × 4 × 2 = 48, and 8!! = 8 × 6 × 4 × 2 = 384. 1, 2, 8, 48, 384, 3840... The next two numbers are 46,080 (since 12!! = 12 × 10 × 8 × 6 × 4 × 2 = 46,080) and 645,120 (since 14!! = 14 × 12 × 10 × 8 × 6 × 4 × 2 = 645,120). Example 3. This sequence features a formula for combinations in probability. This formula is: 𝑁𝑁!
(𝑁𝑁 – 𝑀𝑀)! 𝑀𝑀! In this sequence, N = 8 and M grows from 0 to 8. For example, 8!
(8 −0)! 0! = 8!
8! 0! = 1 8!
(8 −1)! 1! = 8!
7! 1! = 8 8!
(8 −2)! 2! = 8!
6! 2! = 28 8!
(8 −3)! 3! = 8!
5! 3! = 56 8!
(8 −4)! 4! = 8!
4! 4! = 70 8!
(8 −5)! 5! = 8!
3! 5! = 56 1, 8, 28, 56, 70, 56, 28... The next two numbers are 8 and 1. Note that there is a simple geometric way to generate this same sequence. It's called Pascal's triangle. Study the triangle illustrated on the following page. Each number on the inside of this triangle comes from adding the two numbers above it. If you happen to know about algebra, yet another way to make this triangle is to foil out (x + y)N. #1 1, 6, 120, 5040, 362880, ____, ___ #2 1, 3, 15, 105, 945, _, _, _, _ #3 1, 1, 2, 3, 8, 15, 48, 105, 384, _, _, _, _ #4 2, 2, 4, 12, 48, 240, _, _, _, _ #5 1, 1, 4, 36, 576, 14400, ___, ___ #6 1, 7, 21, 35, 35, __, _, #7 1, 9, 36, , , , , 36, 9, 1 #8 1, 4, 10, 20, 35, 56, _, _, _, _ #9 2, 3, 8, 30, 144, 840, 5760, _, _, _, _ #10 3628800, 1814400, 604800, 151200, 30240, 5040, _, _, _, _ 13 Negative Numbers This chapter involves negative numbers. Recall that subtracting a negative number equates to addition. For example, 3 – (–2) = 3 + 2 = 5. However, adding a negative number is like subtraction. For example, 7 + (–5) = 7 – 5 = 2. Two more examples of adding or subtracting with negative numbers include –6 + 3 = –3 and –1 – 4 = –5. When multiplying two numbers together, the answer is negative if the two numbers have opposite sign and positive if they have the same sign. For example, 3 × (–4) = –12, –3 × 4 = –12, 3 × 4 = 12, and –3 × (–4) = 12. Example 1. This sequence repeatedly subtracts 2, starting with 6. 6, 4, 2, 0, –2, –4, –6... The next two numbers are –8 (since –6 – 2 = –8) and –10 (since –8 – 2 = –10). Example 2. In this sequence, each number is multiplied by –3. 1, –3, 9, –27, 81, –243... The next two numbers are 729 (since –3 times –243 = 729) and –2187 (since –3 times 729 = –2187). Example 3. This sequence is formed by taking the prime numbers and making every third number negative. 2, 3, –5, 7, 11, –13, 17... The next two numbers are 19 and –23. #1 –28, –24, –20, –16, –12, –8, , _, , #2 –2, 4, –8, 16, –32, 64, _, , , _ #3 3, –5, 7, –7, 11, –9, 15, –11, , , _, #4 –57, 43, –42, 35, –27, 27, –12, 19, , , , #5 1, –3, –15, 45, 33, –99, –111, 333, 321, _, _, _, _ #6 0, 1, –1, 2, –3, 5, –8, 13, –21, , _, , #7 –3, 4, –4, 3, –4, 4, –3, 4, –4, 3, –4, _, , , _ #8 20, 19, 17, 14, 10, 5, , , _, #9 3, 9, –9, 27, –27, –27, 81, –81, –81, –81, 243, –243, –243, , , , #10 –6, –9, –15, –21, –33, –39, –51, –57, –69, , , , 14 Fractions The sequences of Chapter 14 involve fractions. Following is a brief refresher of some basic properties of fractions. A fraction consists of a numerator and a denominator. For example, in 2 5, the 2 is referred to as the numerator and the 5 is called the denominator. When adding or subtracting fractions, first find a common denominator. The lowest common denominator can be found by examining the factors of each denominator. For example, in 5 6 and 7 8, the lowest com-mon denominator is 24. To see this, factor 6 as 2 × 3 and 8 as 2 × 2 × 2. Both 6 and 8 share a common factor of 2. The lowest common denominator is 2 × 2 × 2 × 3 = 24, as both 6 and 8 can be made from this set of factors. If you have trouble finding the lowest common denominator, an easier way to find a common denominator (but not necessarily the lowest) is to multiply the two denominators together. For example, with 5 6 and 7 8, if you multiply 6 × 8 = 48, you find that 48 is a possible common denominator (whereas the lowest com-mon denominator in this case is 24). Once you have a common denominator, multiply both the numerator and denomi-nator of each fraction by the number needed to make that common denomi-nator. For example, for 5 6 and 7 8, multiply 5 6 by 4 4 and 7 8 by 3 3 to make 20 24 (since 5 × 4 = 20 and 6 × 4 = 24) and 21 24 (since 7 × 3 = 21 and 8 × 3 = 24). After you have a common denominator, you can add or subtract (as necessary) the numerators. For example, 5 6 + 7 8 = 20 24 + 21 24 = 41 24 (since 20 + 21 = 41). Note that the com-mon denominator remains unchanged. Fractions can sometimes be reduced. This is the case when both the numerator and denominator share a common factor. It is considered good form to reduce fractions. For example, 42 24 reduces to 7 4, since both the numerator and denominator are divisible by 6 (that is, 42 ÷ 6 = 7 and 24 ÷ 6 = 4). As another example, consider the fraction 18 21. This can be reduced because both 18 and 21 are divisible by 3: 18 ÷ 3 = 6 and 21 ÷ 3 = 7. Therefore, 18 21 reduces to 6 7. A fraction is said to be improper if the numerator exceeds the denominator. For example, 7 4 is an improper fraction since 7 is greater than 4. This can alternatively be expressed as the mixed number 1 3 4. This is because 7 4 can be written as 4 4 + 3 4, and 4 4 equals 1. So 7 4 is 1 plus 3 4, or 1 3 4. When multiplying two fractions, simply multiply the numerators together and multiply the denominators together. For example, 2 5 × 3 4 = 6 20, which reduces to 3 10. This is because 2 × 3 = 6 and 5 × 4 = 20, and because both 6 and 20 are divisible by 2 (since 6 ÷ 2 = 3 and 20 ÷ 2 = 10). The reciprocal of a fraction is obtained by reversing the roles of the numerator and denominator. For example, the reciprocal of 3 4 equals 4 3. The reciprocal of an integer is simply 1 divided by the integer. For example, the reciprocal of 2 is 1 2. In order to divide two fractions, multiply the first number (the dividend) by the reciprocal of the second number (the divisor). For example, in 5 6 ÷ 2 3, the first number ቀ 5 6ቁ is called the dividend and the second number ቀ 2 3ቁ is called the divisor. The reciprocal of 2 3 is 3 2. Therefore, 5 6 ÷ 2 3 is equivalent to 5 6 × 3 2, which equals 15 12 (since 5 × 3 = 15 and 6 × 2 = 12), which reduces to 5 4 (since 15 ÷ 3 = 5 and 12 ÷ 3 = 4). Example 1. In this sequence, the denomi-nator grows by one. 1, 1 2 , 1 3 , 1 4 , 1 5 , 1 6... The next two numbers are 1 7 and 1 8. Example 2. Here the numerators and denominators both grow by one. For example, 2 5 becomes 3 6 (since 2 + 1 = 3 and 5 + 1 = 6), but 3 6 reduces to 1 2. Next is 4 7 (since 4 + 1 = 5 and 6+1 = 7), followed by 5 8 and 6 9. However, 6 9 reduces to 2 3. 2 5 , 1 2 , 4 7 , 5 8 , 2 3 , 7 10 , 8 11... The next two numbers are 3 4 (since 9 12 reduces to 3 4) and 10 13. Note that the sequence is really 2 5 , 3 6 , 4 7 , 5 8 , 6 9 , 7 10 , 8 11, but 3 6 reduces to 1 2 and 6 9 reduces to 2 3. Example 3. This sequence consists of 1 16 , 2 16 , 3 16 , 4 16 , 5 16, and so on, but expressed as decimals. For example, 1 16 is 0.0625 and 2 16 (equivalent to 1 8) is 0.125. 0.0625, 0.125, 0.1875, 0.25, 0.3125, 0.375... The next two numbers are 0.4375 and 0.5. Example 4. This sequence adds 1 3 to the previous number. 1 3 , 2 3 , 1,1 1 3 , 1 2 3 , 2, 2 1 3 ... The next two numbers are 2 2 3 and 3. #1 2 3 , 4 5 , 6 7 , 8 9 , 10 11, , , , #2 1 3 , 5 6 , 1 12 , 23 24 , 1 48 , 95 96, , , , #3 1 4 , 6 7 , 3 8 , 10 11 , 5 12 , 14 15 , 7 16 , 18 19, , , , #4 1 16 , 1 8 , 3 16 , 1 4 , 5 16 , 3 8 , 7 16, , , , #5 0.5, 0.25, 0.125, 0.0625, 0.03125, 0.015625, __, ___, ___, ___ #6 1 2 , 1 1 2 , 4 1 2 , 13 1 2 , 40 1 2 , 121 1 2, _, __, _, __ #7 1 5 , 13 15 , 1 8 15 , 2 1 5 , 2 13 15 , 3 8 15 , 4 1 5, __, _, __, _ #8 1 12 , 1 6 , 1 4 , 1 3 , 5 12 , 1 2, , , , #9 1 2 , 1 3 , 5 6 , 1 1 6 , 2, 3 1 6 , 5 1 6 , 8 1 3 , 13 1 2, __, _, __, _ #10 4 3 7 , 5 5 9 , 6 7 11 , 9 7 13 , 8 11 15 , 13 9 17 , 10 15 19 , 17 11 21, __, _, __, _ #11 0.125, 0.5, 0.875, 1.25, 1.625, __, _, __, _ #12 2 5 , 3 7 , 5 11 , 7 15 , 11 23 , 13 27 , 17 35, , , , #13 1 2 , 1 3 , 2 9 , 4 27 , 8 81 , 16 243, , , , #14 1 4 , 9 16 , 25 36 , 49 64 , 81 100, , , , #15 1, 2, 1 1 2 , 1 1 3 , 1, 3 4 , 7 13 , 8 21 , 9 34, , , , ___ #16 1 2 , 1 6 , 5 12 , 1 12 , 1 3 , 0, 1 4 , – 1 12 , 1 6 , – 1 6, 1 12, , , , #17 0. 5 ത, 1, 1. 4 ത, 1. 8 ത, 2. 3 ത, 2. 7 ത, 3. 2 ത, 3. 6 ത, _, _, _, _ #18 1 3 , 1 2 , 3 4 , 1 1 8 , 1 11 16 , 2 17 32, _, __, _, __ #19 1 4 , 0.5, 75%, 1, 1.25, 150%, 1 3 4 , 2, 225%, _, __, _, __ #20 1 2 , 1 3 , 1 1 2 , 2 9 , 6 3 4 , 8 243 , 205 1 32, _ 15 Algebraic Chapter 15 involves basic algebra. Fol-lowing is a brief review of relevant basic algebra terminology and rules. In the expression 3𝑥𝑥+ 2, the symbol 𝑥𝑥 is called the variable, while the numbers 3 and 2 are constants. The number 3 is called the coefficient of the variable 𝑥𝑥. The plus sign (+) separates two terms: One term is 3𝑥𝑥, the other is 2. In the expression 3𝑥𝑥5, the exponent 5 is called a power. Here, 𝑥𝑥 is raised to the fifth power. When two different powers of the same variable are multiplied together, add the powers. For example, 𝑥𝑥3𝑥𝑥4 = 𝑥𝑥7 because 3 + 4 = 7. When two different powers of the same variable are divided, subtract the powers. For example, 𝑥𝑥8 𝑥𝑥5 = 𝑥𝑥3 because 8 – 5 = 3. This is why 𝑥𝑥0 = 1. For example, consider 𝑥𝑥3 𝑥𝑥3 = 1 (since any nonzero number di-vided by itself equals 1). According to the rule, 𝑥𝑥3 𝑥𝑥3 = 𝑥𝑥3−3 = 𝑥𝑥0. The last two equa-tions can only both be true if 𝑥𝑥0 = 1. When distributing across parentheses, multiply every term in parentheses by the outside expression. For example, 3𝑥𝑥(𝑥𝑥2 −2) = 3𝑥𝑥3 −6𝑥𝑥 since 3𝑥𝑥(𝑥𝑥2) = 3𝑥𝑥3 and 3𝑥𝑥(−2) = −6x. As another example, −2(𝑥𝑥−4) = −2𝑥𝑥+ 8 since −2(𝑥𝑥) = −2𝑥𝑥 and −2(−4) = 8. The opposite of distributing is factoring. For example, 6𝑥𝑥4 + 9𝑥𝑥2 can be factored by pulling out the common term of 3𝑥𝑥2. The result equals 3𝑥𝑥2(2𝑥𝑥2 + 3), since 3𝑥𝑥2(2𝑥𝑥2) = 6𝑥𝑥4 and 3𝑥𝑥2(3) = 9𝑥𝑥2. Example 1. This sequence consists of powers of 𝑥𝑥, beginning with 𝑥𝑥0, which equals one. Next is 𝑥𝑥1, which equals 𝑥𝑥, followed by 𝑥𝑥2, 𝑥𝑥3, 𝑥𝑥4, and so on. 1, 𝑥𝑥, 𝑥𝑥2, 𝑥𝑥3, 𝑥𝑥4, 𝑥𝑥5... The next two elements are 𝑥𝑥6 and 𝑥𝑥7. Example 2. This pattern grows one term at a time, beginning with 𝑥𝑥12, where each new terms adds 1 to the coefficient and reduces the power by 1. Add 2𝑥𝑥11 to get 𝑥𝑥12 + 2𝑥𝑥11. Then add 3𝑥𝑥10 to get 𝑥𝑥12 + 2𝑥𝑥11 + 3𝑥𝑥10. The next term to add is 4𝑥𝑥9, which gives 𝑥𝑥12 + 2𝑥𝑥11 + 3𝑥𝑥10 +4𝑥𝑥9. 𝑥𝑥12, 𝑥𝑥12 + 2𝑥𝑥11, 𝑥𝑥12 + 2𝑥𝑥11 + 3𝑥𝑥10, 𝑥𝑥12 + 2𝑥𝑥11 + 3𝑥𝑥10 + 4𝑥𝑥9, 𝑥𝑥12 + 2𝑥𝑥11 + 3𝑥𝑥10 + 4𝑥𝑥9 + 5𝑥𝑥8... The next two elements are 𝑥𝑥12 + 2𝑥𝑥11 + 3𝑥𝑥10 + 4𝑥𝑥9 + 5𝑥𝑥8 + 6𝑥𝑥7 and 𝑥𝑥12 + 2𝑥𝑥11 + 3𝑥𝑥10 + 4𝑥𝑥9 + 5𝑥𝑥8 + 6𝑥𝑥7 + 7𝑥𝑥6. Example 3. This sequence begins with 2𝑥𝑥 and 3𝑥𝑥, and multiplies the previous two terms together. The next element is 6𝑥𝑥2. Now multiply 3𝑥𝑥 by 6𝑥𝑥2 to make 18𝑥𝑥3. Next, 6𝑥𝑥2 times 18𝑥𝑥3 makes 108𝑥𝑥5. 2𝑥𝑥, 3𝑥𝑥, 6𝑥𝑥2, 18𝑥𝑥3, 108𝑥𝑥5... The next two elements are 1944𝑥𝑥8 (since 18 × 108 = 1944 and 𝑥𝑥3𝑥𝑥5 = 𝑥𝑥8) and 209952𝑥𝑥13 (since 108 × 1944 = 209952 and 𝑥𝑥5𝑥𝑥8 = 𝑥𝑥13). #1 𝑥𝑥10, 10𝑥𝑥9, 90𝑥𝑥8, 720𝑥𝑥7, __, _, __, _ #2 120, 120𝑥𝑥, 60𝑥𝑥2, 20𝑥𝑥3, 5𝑥𝑥4, __, _, __, _ #3 4𝑥𝑥, 9𝑥𝑥2, 16𝑥𝑥4, 25𝑥𝑥8, 36𝑥𝑥16, __, _, __, _ #4 2, 3𝑥𝑥2, 5𝑥𝑥4, 7𝑥𝑥6, 11𝑥𝑥8, 13𝑥𝑥10, 17𝑥𝑥12, __, _, __, _ #5 𝑥𝑥, 2 𝑥𝑥2, 𝑥𝑥3 6 , 24 𝑥𝑥5, 𝑥𝑥8 120, 720 𝑥𝑥13, , , , #6 2𝑥𝑥4, 7𝑥𝑥11, 16𝑥𝑥22, 29𝑥𝑥37, 46𝑥𝑥56, __, _, __, _ #7 (𝑎𝑎+ 2)3, (𝑏𝑏+ 4)6, (𝑐𝑐+ 8)12, (𝑑𝑑+ 16)24, __, _, __, _ #8 𝑥𝑥4 24, 𝑥𝑥3 12, 𝑥𝑥2 4 , 𝑥𝑥, , , , , 15120 𝑥𝑥4 #9 2𝑎𝑎, 𝑐𝑐4, 8𝑒𝑒, 𝑔𝑔16, 32𝑖𝑖, _, , , _ #10 7, 23, 32, 2·5, 11, 22·3, 13, 2·7, 3·5, 24, _, __, _, __ #11 5𝑥𝑥2, 7𝑥𝑥3, 11𝑥𝑥5, 19𝑥𝑥9, 35𝑥𝑥17, 67𝑥𝑥33, _, __, _, __ #12 3𝑥𝑥2, 𝑥𝑥5, 4𝑥𝑥3, 2𝑥𝑥6, 5𝑥𝑥4, 3𝑥𝑥7, 6𝑥𝑥5, _, __, _, __ #13 𝑥𝑥, 2𝑥𝑥 1 2, 𝑥𝑥2 2 , 2𝑥𝑥 3 2 3 , 𝑥𝑥5 5 , 2𝑥𝑥 5 2 5 , 𝑥𝑥10 10 , 2𝑥𝑥 7 2 7 , 𝑥𝑥17 17 , 2𝑥𝑥 9 2 9 , 𝑥𝑥26 26 , 2𝑥𝑥 11 2 11 , 𝑥𝑥37 37 , , , , #14 8𝑥𝑥 1 8 3 , 2𝑥𝑥 1 4, 8𝑥𝑥 3 8 5 , 4𝑥𝑥 1 2 3 , 8𝑥𝑥 5 8 7 , 𝑥𝑥 3 4, 8𝑥𝑥 7 8 9 , _, __, _, __ #15 2𝑥𝑥+ 1, 𝑥𝑥−3, 3𝑥𝑥−2, 4𝑥𝑥−5, 7𝑥𝑥−7, 11𝑥𝑥−12, 18𝑥𝑥−19, 29𝑥𝑥−31, _, __, _, __ #16 𝑥𝑥, 𝑦𝑦, 𝑥𝑥2, 𝑥𝑥𝑥𝑥, 𝑦𝑦2, 𝑥𝑥3, 𝑥𝑥2𝑦𝑦, 𝑥𝑥𝑦𝑦2, 𝑦𝑦3, _, __, _, __ #17 2𝑥𝑥5, 7𝑥𝑥3, 5𝑥𝑥11, 15𝑥𝑥7, 11𝑥𝑥23, 27𝑥𝑥13, 17𝑥𝑥35, _, __, _, __ #18 3𝑥𝑥+ 4 = 𝑥𝑥2, 3𝑥𝑥= 𝑥𝑥2 −4, 6𝑥𝑥= 2𝑥𝑥2 −8, 6𝑥𝑥−4 = 2𝑥𝑥2 −12, 12𝑥𝑥−8 = 4𝑥𝑥2 −24, 12𝑥𝑥−12 = 4𝑥𝑥2 −28, 24𝑥𝑥−24 = 8𝑥𝑥2 −56, 24𝑥𝑥−28 = 8𝑥𝑥2 −60, ______, ______, ______, ______ #19 𝑥𝑥3, 𝑥𝑥4 + 3𝑥𝑥3, 𝑥𝑥5 + 4𝑥𝑥4 + 5𝑥𝑥3, 𝑥𝑥6 + 5𝑥𝑥5 + 6𝑥𝑥4 + 11𝑥𝑥3, 𝑥𝑥7 + 6𝑥𝑥6 + 7𝑥𝑥5 + 13𝑥𝑥4 + 20𝑥𝑥3, 𝑥𝑥8 + 7𝑥𝑥7 + 8𝑥𝑥6 + 15𝑥𝑥5 + 23𝑥𝑥4 + 38𝑥𝑥3, ______, ______ #20 1, 𝑥𝑥+ 𝑦𝑦, 𝑥𝑥2 + 2𝑥𝑥𝑥𝑥+ 𝑦𝑦2, 𝑥𝑥3 + 3𝑥𝑥2𝑦𝑦+ 3𝑥𝑥𝑦𝑦2 + 𝑦𝑦3, 𝑥𝑥4 + 4𝑥𝑥3𝑦𝑦+ 6𝑥𝑥2𝑦𝑦2 + 4𝑥𝑥𝑦𝑦3 + 𝑦𝑦4, 𝑥𝑥5 + 5𝑥𝑥4𝑦𝑦+ 10𝑥𝑥3𝑦𝑦2 + 10𝑥𝑥2𝑦𝑦3 + 5𝑥𝑥𝑦𝑦4 + 𝑦𝑦5, 𝑥𝑥6 + 6𝑥𝑥5𝑦𝑦+ 15𝑥𝑥4𝑦𝑦2 + 20𝑥𝑥3𝑦𝑦3 + 15𝑥𝑥2𝑦𝑦4 + 6𝑥𝑥𝑦𝑦5 + 𝑦𝑦6, ______, ______ 16 Visual The patterns in this chapter involve pictures. Solving these puzzles involves visual pattern recognition. Example 1. The black square moves one space clockwise along the large square. Example 2. This shape rotates 45° clockwise. Example 3. Add the top and left numbers to make the right number. 3 7 4 4 1 0 6 0 2 2 8 2 0 1 2 #1 Shade the correct triangles in the last figure. #2 Draw the shape that comes next. #3 Draw the shape that comes next. #4 Shade the correct square in the last figure. #5 Fill in the missing number. 5 2 3 6 5 1 1 1 7 4 1 3 5 #6 Draw the shape that comes next. #7 Draw the shape that comes next. #8 Fill in the missing number. #9 Shade the correct square in the last figure. #10 Draw the shape that comes next. 2 8 4 3 5 1 5 2 8 4 7 4 9 #11 Draw the shape that comes next. #12 Draw the shape that comes next. #13 Draw the shape that comes next. #14 Color the correct squares in the last figure. #15 Shade the correct squares in the last figure. #16 Fill in the missing number. 3 4 2 8 1 9 5 6 7 9 5 7 7 9 3 6 10 20 13 #17 Color the correct squares in the last figure. #18 Draw the shape that comes next. #19 Draw the shape that comes next. #20 Color the correct squares in the last figure. 17 Arrays In this chapter, the patterns come in the form of arrays. Each array has rows and columns. Some of the arrays are numeri-cal, while others have visual elements. Example 1. In the array that follows, odd numbers are arranged in order from left to right, top to bottom. 3 5 7 9 11 13 15 17 19 Example 2. In the following array, the top row is in order (2, 3, and 4), the middle row doubles the top row (2 × 2 = 4, 3 × 2 = 6, and 4 × 2 = 8), and the bottom row squares the top row (22 = 4, 32 = 9, and 42 = 16). 2 3 4 4 6 8 4 9 16 Example 3. The pattern , , and is used to make the following array. Instructions: Fill in the missing elements. #1# 8 16 24 32 40 56 72 #2# 95 84 73 62 40 29 18 #3# 3 6 12 4 8 5 20 #4# 2 7 17 3 19 5 13 #5# 4 3 8 9 5 1 7 #6# 4 6 9 11 16 18 25 27 36 38 49 64 #7# #8# 14 17 20 10 13 16 18 7 9 12 5 6 #9# 1 1 0 2 7 14 3 27 26 52 64 126 #10# II IV VIII XVI XXXII CXXVIII DXII #11# 1 2 5 6 4 3 8 7 9 10 12 11 #12# 2 3 5 9 2049 8193 17 32769 16385 33 513 129 #13# 5 8 14 17 23 26 32 44 #14# 1 2 3 4 2 3 5 7 3 5 8 4 8 20 #15# 2 1 1 8 5 3 21 18 Analogies These puzzles involve analogies. They are solved with logical inference. In particu-lar, they involve finding the similarities between relationships and using that to predict further similarity between them. In this book, analogies will be expressed with the following notation: AA : aa :: BB : bb This reads as, “AA is to aa as BB is to bb.” The first colon (:) equates to the phrase “is to.” The pair of colons (::) in the middle equates to the word “as.” The last colon (:) equates to the phrase “is to.” AA and aa are related to one another. BB and bb are related to one another in a similar way. The two relationships are analogous to one another; i.e. they form an analogy. The concept of the analogy as well as the notation are illustrated with the following examples. Example 1. finger : hand :: toe : ___ A finger is to a hand as a toe is to a foot. A hand has 5 fingers and a foot has 5 toes. Therefore, the word “foot” completes this analogy. Example 2. 5 : 25 :: 7 : ___ 5 is to 25 as 7 is to 49. The expression 5 : 25 is related to 7 : 49 by squaring: 52 = 25 and 72 = 49. Example 3. → : ↔ :: ↑ : ___ → is to ↔ as ↑ is to ↕. The horizontal arrow turns into a horizontal double arrow. Therefore, the vertical arrow turns into a vertical double arrow. Instructions: Select the answer to each multiple choice question that best com-pletes the analogy. #1 three : 27 :: four : (A) 8 (B) 12 (C) 16 (D) 64 #2 #3 hand : wrist :: foot : _ (A) ankle (B) leg (C) sole (D) toe : : : : ( A ) ( B ) ( C ) ( D ) #4 81 : 9 :: 36 : _ (A) 2 (B) 3 (C) 6 (D) 12 #5 century : year :: meter : ___ (A) centimeter (B) inch (C) millimeter (D) yard #6 ADE : 145 :: CFH : __ (A) 2 (B) 367 (C) 368 (D) 389 #7 : : : : ( A ) ( B ) ( C ) ( D ) #8 moon : earth :: Venus : __ (A) Mercury (B) Mars (C) sun (D) Uranus #9 circle : circumference :: square : _ (A) area (B) diameter (C) diagonal (D) perimeter #10 (3, 4, 5) : 35 :: (6, 7, 8) : __ (A) 42 (B) 48 (C) 86 (D) 104 #11 uncle : nephew :: aunt : __ (A) cousin (B) daughter (C) mother (D) niece #12 (10, 11) : 101 :: (100, 101) : (A) 110 (B) 111 (C) 1001 (D) 1101 #13 horse : foal :: deer : __ (A) calf (B) colt (C) fawn (D) filly #14 rectangle : square :: parallelogram : _ (A) octahedron (B) quadrilateral (C) rhombus (D) trapezoid #15 5 8 : 1 1 4 :: 3 4 : __ (A) 1 1 4 (B) 1 1 2 (C) 2 1 4 (D) 2 1 2 #16 #17 ቀ8, 2 3ቁ : 4 :: ቀ81, 3 4ቁ : (A) 6 (B) 9 (C) 18 (D) 27 #18 COCCOON : C3O3N : : COCOA : __ (A) A3C3O (B) C2O2A (C) C2OA2 (D) C6H12O6 : : : : ( C ) ( B ) ( A ) ( D ) #19 circle : sphere : : square : __ (A) cube (B) cylinder (C) diamond (D) rectangle #20 PUZZLES : OTYYKDR :: PATTERNS : __ (A) OZSSDQMR (B) OBYYGXFJ (C) QZXXHPJK (D) QBUUFSO Answers 1 Ordered (1) 7, 9, 11, 13, 15, 17, 19, 21 Count odd numbers, starting with 7. (2) 5, 10, 15, 20, 25, 30, 35, 40 Count by 5. (3) 500, 600, 700, 800, 900, 1000, 1100, 1200 Count by 100. (4) 25, 50, 75, 100, 125, 150, 175, 200 Count by 25. (5) 53, 65, 77, 89, 101, 113, 125, 137 Count by 12. (6) b, d, f, h, j, l, n, p Skip every other letter (b, not c, d, not e, f, not g, h, not i, etc.). (7) c, g, k, o, s, w Skip three letters (c, skip d-e-f, g, skip h-i-j, k, skip l-m-n, etc.). (8) , , , , Draw polygons which have an even number of sides. (9) , , , , , Add one star point. (10) 15, 14, 13, 12, 11, 10, 9, 8 Count backwards. (11) 30, 27, 24, 21, 18, 15, 12, 9 Subtract 3. (12) 125, 120, 115, 110, 105, 100, 95, 90 Subtract 5. (13) 1100, 1075, 1050, 1025, 1000, 975, 950, 925 Subtract 25. (14) x, w, v, u, t, s, r, q Write the alphabet backwards, starting with x. (15) q, o, m, k, i, g, e, c Skip every other letter, going backwards. (16) y, v, s, p, m, j, g, d Skip two letters, going backwards (y, skip x-w, v, skip u-t, s, skip r-q, etc.). (17) , , , , , Draw a polygon with one less side. (18) , , , , , Draw one less star point. (19) A, E, I, O, U Write the vowels in order. (20) d, f, g, h, j, k, l, m, n, p, q Write the consonants in order, beginning with d (thus e, i, and o are skipped). 2 Patterned (1) 0, 1, 0, 1, 0, 1, 0, 1 Alternate 0 and 1. (2) t, y, t, y, t, y, t, y Alternate t and y. (3) A, a, A, a, A, a, A, a, A, a Alternate A and a. (4) 2, 4, 8, 2, 4, 8, 2, 4, 8, 2 Repeat 2, 4, and 8. (5) z, x, q, z, x, q, z, x, q, z, x Repeat z, x, and q. (6) , , , , , , , , , Repeat heart, star, and moon. (7) 9, 9, 4, 9, 9, 4, 9, 9, 4, 9 Repeat 9, 9, and 4. (8) 3, 5, 3, 3, 5, 3, 3, 5, 3, 3 Repeat 3, 5, and 3. (9) G, g, g, G, g, g, G, g, g, G Repeat G, g, and g. (10) F, F, e, F, F, e, F, F, e, F Repeat F, F, and e. (11) , , , , , , , , , Repeat smile, frown, and smile. (12) 2, 5, 3, 9, 2, 5, 3, 9, 2, 5, 3, 9, 2 Repeat 2, 5, 3, and 9. (13) 1, 7, 6, 4, 8, 1, 7, 6, 4, 8, 1, 7, 6, 4, 8, 1 Repeat 1, 7, 6, 4, and 8. (14) D, b, P, q, D, b, P, q, D, b, P, q, D, b Repeat D, b, P, and q. (15) C, e, H, j, c, E, h, J, C, e, H, j, c, E Repeat C, e, H, j, c, E, h, and J. (16) 8, 2, 8, 8, 2, 8, 8, 8, 2, 8, 8, 8, 8, 2, 8, 8, 8, 8, 8, 2 Steadily increase the number of 8’s between the 2’s (one 8, 2, two 8’s, 2, three 8’s, 2, etc.). (17) 6, 77, 888, 9999, 1010101010, 111111111111, 12121212121212, 1313131313131313 Write one 6, two 7’s, three 8’s, four 9’s, five 10’s, etc. (18) 2, 23, 234, 2345, 23456, 234567, 2345678 Add one more digit to the list, in order. (19) 1, 31, 531, 7531, 97531, 1197531, 131197531, 15131197531 Insert the next odd number on the left (insert 11 to the left of 97531 to get 1197531, for example, then insert 13 to the left of this to get 131197531). (20) 4, 4, 3, 12, 12, 9, 36, 36, 27, 108, 108, 81, 324 Triple 4, 4, and 3 to get 12, 12, and 9 (4 × 3 = 12 and 3 × 3 = 9). Triple 12, 12, and 9 to get 36, 36, and 27 (12 × 3 = 36 and 9 × 3 = 27). Repeat this pattern. 3 Alternating (1) 11, 99, 22, 88, 33, 77, 44, 66, 55, 55, 66, 44 The sequence 11, 22, 33, etc. is mixed with 99, 88, 77, etc. (2) 3, 18, 6, 16, 9, 14, 12, 12, 15, 10, 18, 8 The sequence 3, 6, 9, etc. adds 3, while the sequence 18, 16, 14, etc. subtracts 2. (3) 6, 24, 12, 24, 18, 24, 24, 24, 30, 24, 36 The sequence 6, 12, 18, etc. adds 6, while the sequence 24, 24, 24, etc. repeats the same number over and over. (4) 5, 6, 7, 9, 9, 12, 11, 15, 13, 18, 15, 21 The sequence 5, 7, 9, etc. consists of odd numbers, while the sequence 6, 9, 12, etc. adds 3. (5) 10, 50, 15, 40, 20, 30, 25, 20, 30, 10, 35 The sequence 10, 15, 20, etc. adds 5, while the sequence 50, 40, 30, etc. subtracts 10. (6) p, q, o, r, n, s, m, t, l, u, k The sequence p, o, n, etc. is the reverse alphabet, while the sequence q, r, s, etc. is forward. (7) Z, N, X, O, V, P, T, Q, R, R, P The sequence Z, X, V, T, etc. skips every other letter going backwards, while the sequence N, O, P, etc. goes forward without skipping. (8) c, f, d, f, e, f, f, f, g, f, h, f, i The sequence c, d, e, etc. is the alphabet beginning with c, while the sequence f, f, f, etc. repeats f over and over. (9) J, 29, L, 26, N, 23, P, 20, R, 17, T, 14 The sequence J, L, N, etc. skips every other letter, while the sequence 29, 26, 23, etc. subtracts 3. (10) 7, z, 11, w, 15, t, 19, q, 23, n The sequence 7, 11, 15, etc. adds 4, while the sequence z, w, t, etc. skips two letters going backward (z, skip y-x, w, skip v-u, t, skip s-r, q, etc.). (11) , , , , , , , , , The sequence square, hexagon, octagon, etc. adds 2 sides, while the sequence octagon, heptagon, hexagon, etc. subtracts 1 side. (12) , , , , , , , , , , The sequence triangle, square, pentagon, etc. adds 1 side, while the sequence hexagon, hexagon, hexagon, etc. repeats the same shape over and over. (13) , , , , , , , , , The sequence star 4, star 6, star 8, etc. adds 2 star points, while the sequence octagon, heptagon, hexagon, etc. subtracts 1 side. (14) →, ↑, ↓, ↓, ←, ↑, ↑, ↓, →, ↑, ↓, ↓, ←, ↑ The sequence right (→), down (↓), left (←), up (↑), etc. rotates 90° clockwise, while the sequence up (↑), down (↓), up (↑), down (↓), etc. flips repeatedly. (15) ↑, ←, ←, ↖, ↓, ↑, →, ↗, ↑, →, ←, ↘, ↓, ↓ The sequence 90° (↑), 180° (←), 270° (↓), etc. rotates 90° counterclockwise, while the sequence 180° (←), 135° (↖), 90° (↑), etc. rotates 45° clockwise. (16) Q, z, X, w, Q, z, X, w, Q, z, X, w, Q This is just Q, z, X, w repeating itself. (17) 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1 The sequence 1, 0, 1, 0, 1, etc. simply alternates 1’s and 0’s, and this sequence is merged together with the sequence 0, 0, 1, 0, 0, 1 etc. (which repeats 0, 0, and 1). (18) D, e, f, F, H, g, j, H, L, i, n, J, P, k, r The sequence D, f, H, j, L, etc. skips every other letter and alternates case, while the sequence e, F, g, H, I, etc. is the alphabet changing case. (19) 30, 70, 25, 35, 60, 35, 40, 50, 45, 45, 40, 55, 50, 30, 65, 55, 20, 75, 60, 10 Three patterns are merged together: The sequence 30, 35, 40, etc. adds 5, the sequence 70, 60, 50, etc. subtracts 10, and the sequence 25, 35, 45, etc. adds 10. (20) K, h, Q, L, i, P, M, j, O, N, k, N, O, l, M, P, m, L, Q, n Three patterns are merged together: The sequence K, L, M, etc. is the uppercase alphabet, the sequence h, i, j, etc. is the lowercase alphabet, and the sequence Q, P, O, etc. is the uppercase alphabet backwards. 4 Cyclic (1) 927, 279, 792, 927, 279, 792, 927, 279, 792 These digits cycle in the order 927, 279, and 792 re-peatedly. (2) 3841, 8413, 4138, 1384, 3841, 8413, 4138, 1384, 3841 These digits cycle in the order 3841, 8413, 4138, and 1384 repeatedly. (Move the first digit to the end each time. For example, 3841 turns into 8413.) (3) 4, 6, 8, 0, 2, 4, 6, 8, 0, 2, 4 These single-digit even numbers cycle in order. (4) 7, 88, 999, 0000, 11111, 222222, 3333333, 44444444 The digits 7, 8, 9, 0, 1, 2, etc. cycle in order, increasing the number of digits by one each time. (5) 123, 132, 231, 213, 312, 321, 123, 132, 231, 213 The digits 1, 2, and 3 cycle through all possible 3-digit permutations: 123, 132, 231, 213, 312, and 321. (Every other one, 123, 231, and 312 has the digits 1, 2, 3, 1, and 2 in order.) (6) act, cta, tac, act, cta, tac, act, cta, tac These letters cycle in the order act, cta, and tac repeatedly. (7) spot, pots, otsp, tspo, spot, pots, otsp, tspo These letters cycle in the order spot, pots, otsp, and tspo repeatedly. (Move the first letter to the end each time. For example, spot turns into pots.) (8) u, w, y, a, c, e, g, i, k The alphabet is cyclic, if after z comes a. This pattern skips every other letter: u, skip v, w, skip x, y, skip z, a, skip b, c, etc. (9) ccccccccc, bbbbbbbb, aaaaaaa, zzzzzz, yyyyy, xxxx, www, vv The alphabet appears in reverse, with z following a, removing one letter each time. (10) Roygbiv, rOygbiv, roYgbiv, royGbiv, roygBiv, roygbIv, roygbiV, Roygbiv, rOygbiv The letters roygbiv are repeated, with one uppercase letter. The placement of the uppercase letter advances each time, returning to the beginning after reaching the end. (Note that roygbiv is an acronym for red, orange, yellow, green, blue, indigo, and violet, which are the colors of the primary rainbow from top to bottom. However, it’s not necessary to know this acronym to solve the puzzle.) (11) 0, 11, 222, 3333, 222, 11, 0, 11, 222, 3333, 222 This sequence increases from 0, 11, 222, to 333, then decreases 222, 11, to 0, and continues to increase and decrease this way. (12) CIRCLE, IRCLEC, RCLECI, CLECIR, LECIRC, ECIRCL, CIRCLE, IRCLEC, RCLECI The first letter of one word becomes the last letter of the next word. For example, move the I of IRCLEC to the end, forming RCLECI. (13) 10:45, 11:15, 11:45, 12:15, 12:45, 1:15, 1:45, 2:15 Add 30 minutes. (14) 9:30, 10:15, 11:00, 11:45, 12:30, 1:15, 2:00, 2:45 Add 45 minutes. (15) Friday, Saturday, Sunday, Monday, Tuesday, Wednesday, Thursday, Friday The days of the week appear in order. (16) October, November, December, January, February, March, April, May The months appear in order. (17) north, east, south, west, north, east, south, west The compass directions cycle in the order north, east, south, and west. (18) east, northeast, north, northwest, west, southwest, south, southeast, east, northeast Rotate 45° counterclockwise. (19) north, southeast, west, northeast, south, northwest, east, southwest, north, southeast Rotate 135° clockwise. (20) 90° (↑), 180° (←), 270° (↓), 360° (→), 90° (↑), 180° (←), 270° (↓), 360° (→), 90° (↑) Rotate 90° counterclockwise. 5 Additive & Multiplicative (1) 6, 12, 18, 24, 30, 36, 42, 48 Add 6. (2) 72, 64, 56, 48, 40, 32, 24, 16 Subtract 8. (3) 1, 2, 4, 8, 16, 32, 64, 128 Double each number. (4) 7, 13, 19, 25, 31, 37, 43, 49 Add 6. (5) 640, 320, 160, 80, 40, 20, 10, 5 Cut in half. (6) 33, 37, 41, 45, 49, 53, 57, 61 Add 4. (7) 52, 45, 38, 31, 24, 17, 10, 3 Subtract 7. (8) 25, 50, 100, 200, 400, 800, 1600, 3200 Double each number. (9) 6, 11, 16, 21, 26, 31, 36, 41 Add 5. (10) 1, 20, 39, 58, 77, 96, 115, 134 Add 19. (11) 9, 18, 27, 36, 45, 54, 63, 72 Add 9. (12) 46, 40, 34, 28, 22, 16, 10, 4 Subtract 6. (13) 8, 16, 32, 64, 128, 256, 512, 1024 Double each number. (14) 9, 21, 33, 45, 57, 69, 81, 93 Add 12. (15) 768, 384, 192, 96, 48, 24, 12, 6 Cut in half. (16) 212, 199, 186, 173, 160, 147, 134, 121 Subtract 13. (17) 2, 6, 18, 54, 162, 486, 1458, 4374 Multiply by 3. (18) 192, 216, 240, 264, 288, 312, 336, 360 Add 24. (19) 2, 10, 50, 250, 1250, 6250, 31250, 156250 Multiply by 5. (20) 2187, 729, 243, 81, 27, 9, 3, 1 Divide by 3. 6 Multiple Operations (1) 2, 7, 22, 67, 202, 607, 1822, 5467, 16402 Multiply by 3 and add 1. For example, 3 × 2 + 1 = 6 + 1 = 7 and 3 × 7 + 1 = 21 + 1 = 22. (2) 3, 9, 8, 14, 13, 19, 18, 24, 23, 29, 28 Add 6, subtract 1, add 6, subtract 1, repeating this pattern. For example, 3 + 6 = 9, 9 – 1 = 8, 8 + 6 = 14, and 14 – 1 = 13. (3) 1022, 510, 254, 126, 62, 30, 14, 6, 2 Divide by 2 and subtract 1. For example, 1022 ÷ 2 – 1 = 511 – 1 = 510 and 510 ÷ 2 – 1 = 255 – 1 = 254. (4) 1, 3, 6, 18, 36, 108, 216, 648, 1296, 3888, 7776 Triple, double, triple, double, repeating this pattern. For example, 1 × 3 = 3, 3 × 2 = 6, 6 × 3 = 18, and 18 × 2 = 36. (5) 101, 99, 95, 93, 89, 87, 83, 81, 77 Subtract 2, subtract 4, subtract 2, subtract 4, repeating this pattern. For example, 101 – 2 = 99, 99 – 4 = 95, 95 – 2 = 93, and 93 – 4 = 89. (6) 256, 128, 144, 72, 88, 44, 60, 30, 46, 23, 39 Divide by 2, add 16, divide by 2, add 16, repeating this pattern. For example, 256 ÷ 2 = 128, 128 + 16 = 144, 144 ÷ 2 = 72, and 72 + 16 = 88. (7) 3, 5, 9, 17, 33, 65, 129, 257, 513 Multiply by 2 and subtract 1. For example, 3 × 2 – 1 = 6 – 1 = 5 and 5 × 2 – 1 = 10 – 1 = 9. (8) 5, 7, 15, 17, 35, 37, 75, 77, 155, 157, 315, 317, 635 Add 2, next double and add 1, add 2, next double and add 1, repeating this pattern. For example, 5 + 2 = 7, 7 × 2 + 1 = 15, 15 + 2 = 17, 17 × 2 + 1 = 35, 35 + 2 = 37, and 37 × 2 + 1 = 75. (9) 1, 3, 6, 10, 15, 21, 28, 36, 45, 55 Add 2, add 3, add 4, add 5, add 6, etc., always adding one more. For example, 1 + 2 = 3, 3 + 3 = 6, 6 + 4 = 10, and 10 + 5 = 15. (10) 4, 6, 9, 13, 18, 24, 31, 39, 48, 58 Add 2, add 3, add 4, add 5, add 6, etc., always adding one more. For example, 4 + 2 = 6, 6 + 3 = 9, 9 + 4 = 13, 13 + 5 = 18, etc. (11) 100, 51, 98, 53, 95, 56, 91, 60, 86, 65, 80, 71 This alternating sequence has two patterns merged together. One subtracts 2, subtracts 3, subtracts 4, etc. (100, 98, 95, 91, etc.), while the other adds 2, adds 3, adds 4, etc. (51, 53, 56, 60, etc.). (12) 8, 4, 12, 6, 24, 12, 60, 30, 180, 90, 630, 315, 2520, 1260, 11340 Cut in half, multiply by 3, cut in half, multiply by 4, cut in half, multiply by 5, etc. For example, 8 ÷ 2 = 4, 4 × 3 = 12, 12 ÷ 2 = 6, 6 × 4 = 24, 24 ÷ 2 = 12, 12 × 5 = 60, 60 ÷ 2 = 30, and 30 × 6 = 180. (13) 50, 52, 47, 55, 43, 59, 38, 64, 32, 70, 25, 77, 17 This alternating sequence has two patterns merged together. One subtracts 3, subtracts 4, subtracts 5, etc. (50, 47, 43, 38, etc.), while the other adds 3, adds 4, adds 5, etc. (52, 55, 59, 64, etc.). (14) 10, 50, 20, 100, 70, 350, 320, 1600, 1570, 7850, 7820, 39100 Multiply by 5, subtract 30, multiply by 5, subtract 30, repeating this pattern. For example, 10 × 5 = 50, 50 – 30 = 20, 20 × 5 = 100, and 100 – 30 = 70. (15) 1, 3, 6, 9, 18, 22, 44, 49, 98, 104, 208, 215, 430 Add 2, double, add 3, double, add 4, double, add 5, double, add 6, repeating this pattern. For example, 1 + 2 = 3, 3 × 2 = 6, 6 + 3 = 9, 9 × 2 = 18, 18 + 4 = 22, 22 × 2 = 44, 44 + 5 = 49, and 49 × 2 = 98. (16) 2, 2, 1, 2, 1, 3, 2, 8, 7, 35, 34, 204, 203, 1421, 1420 Multiply by 1, subtract 1, multiply by 2, subtract 1, multiply by 3, subtract 1, multiply by 4, subtract 1, multiply by 5, subtract 1, repeating this pattern. For example, 2 × 1 = 2, 2 – 1 = 1, 1 × 2 = 2, 2 – 1 = 1, 1 × 3 = 3, 3 – 1 = 2, 2 × 4 = 8, 8 – 1 = 7, 7 × 5 = 35, and 35 – 1 = 34. (17) a, d, c, f, e, h, g, j, i, l, k, n, m, p, o Skip 2 letters, go back 1, skip 2 letters, go back 1, repeating this pattern. For example, a, skip b-c, d, c, skip d-e, f, e, skip f-g, h, g, skip h-i, j, i, etc. (18) 1, 2, 6, 9, 10, 14, 17, 18, 22, 25, 26, 30, 33, 34, 38, 41, 42 Add 1, add 4, add 3, add 1, add 4, add 3, repeating this pattern. For example, 1 + 1 = 2, 2 + 4 = 6, 6 + 3 = 9, 9 + 1 = 10, 10 + 4 = 14, 14 + 3 = 17, 17 + 1 = 18, 18 + 4 = 22, and 22 + 3 = 25. (19) 2, 5, 10, 9, 12, 24, 23, 26, 52, 51, 54, 108, 107, 110, 220, 219, 222 Add 3, double, subtract 1, add 3, double, subtract 1, repeating this pattern. For example, 2 + 3 = 5, 5 × 2 = 10, 10 – 1 = 9, 9 + 3 = 12, 12 × 2 = 24, and 24 – 1 = 23. (20) 8, 6, 24, 12, 10, 40, 20, 18, 72, 36, 34, 136, 68, 66, 264, 132, 130 Subtract 2, multiply by 4, cut in half, subtract 2, multiply by 4, cut in half, repeating this pattern. For example, 8 – 2 = 6, 6 × 4 = 24, 24 ÷ 2 = 12, 12 – 2 = 10, 10 × 4 = 40, and 40 ÷ 2 = 20. 7 Digits (1) 807, 716, 625, 534, 443, 352, 261, 170 The first digit decreases (8, 7, 6, etc.), the middle digit increases (0, 1, 2, etc.), and the last digit decreases (7, 6, 5, etc.). (2) 29, 47, 65, 83, 21, 49, 67, 85, 23, 41 The first digit cycles through 2, 4, 6, 8, while the last digit cycles through 9, 7, 5, 3, 1. (3) 122, 223, 324, 425, 526, 627, 728, 829 The first digit increases (1, 2, 3, etc.), the middle digit is always 2, and the last digit increases (2, 3, 4, etc.). (4) 369, 396, 639, 693, 936, 963, 369, 396, 639 This sequence cycles through all 6 permutations of the digits 3, 6, and 9: 369, 396, 639, 693, 936, and 963. (5) 5678, 5687, 5768, 5786, 5867, 5876, 6578, 6587, 6758, 6785, 6857, 6875 This sequence cycles through all 12 permutations of the digits 5, 6, 7, and 8. (6) 3456, 3465, 5436, 5463, 6435, 6453, 3456, 3465, 5436, 5463 The second digit is always 4. The remaining digits cycle through all 6 permutations of the digits 3, 5, and 6. (7) d8, f7, h6, j5, l4, n3, p2, r1 The letter skips every other letter of the alphabet (d, skip e, f, skip g, h, skip i, etc.), while the number decreases by 1 (8, 7, 6, etc.) (8) y3C, v5d, s7E, p9f, m1G, j3h, g5I, d7j, a9K The first letter skips two letters going backward (y, skip x-w, v, skip u-t, s, skip r-q, p, etc.), the number cycles through single-digit odd numbers (1, 3, 5, 7, 9), and the last letter advances one while changing case (C, d, E, f, G, h, etc.). (9) T8e2, r7Q4, P6E6, n5q8, L4e0, j3Q2, H2E4, f1q6, D0e8, b9Q0 The first letter skips one letter going backward while alternating case (T, skip S, r, skip q, P, skip O, n, skip m, L, etc.), the first number decreases by 1 (8, 7, 6, etc.) with 9 following 0, the second letter cycles through e, Q, E, and q, and the last digit cycles through single-digit even numbers (0, 2, 4, 6, 8). (10) 24680, 24681, 24691, 24791, 25791, 35791, 35792, 35702, 35802, 36802, 46802, 46803 Increase the 5th digit by 1, increase the 4th digit by 1, increase the 3rd digit by 1, increase the 2nd digit by 1, increase the 1st digit by 1, increase the 5th digit by 1, and so on (where increasing 9 by 1 changes it to zero). For example, 35791 changes to 35792 as the 5th digit increase from 1 to 2, 35792 changes to 35702 as the 4th digit changes from 9 to 0, and 35702 changes to 35802 as the 3rd digit increases from 7 to 8. (11) 89, 98, 179, 188, 197, 269, 278, 287, 296, 359, 368 These are the numbers, in order, where the digits add up to 17. For example, 8 + 9 = 17, 9 + 8 = 17, 1 + 7 + 9 = 17, and 1 + 8 + 8 = 17. (12) 10, 11, 20, 12, 21, 30, 13, 22, 31, 40, 14 First, write down all the two-digit numbers where the digits add up to 1 (the only answer is 10). Next, write down all the two-digit numbers in order where the digits add up to 2 (11 and 20). Next, write down all the two-digit numbers in order where the digits add up to 3 (12, 21, and 30). Next, write down all the two-digit numbers in order where the digits add up to 4 (13, 22, 31, and 40). Next, write down all the two-digit numbers in order where the digits add up to 5 (14, 23, 32, 41, and 50). (13) 39, 57, 75, 93, 1119, 1137, 1155, 1173, 1191, 1317, 1335, 1353, 1371 These are the numbers, in order, composed of odd digits where the digits add up to 12. For example, 3 + 9 = 12, 5 + 7 = 12, 7 + 5 = 12, 9 + 3 = 12, 1 + 1 + 1 + 9 = 12, and 1 + 1 + 3 + 7 = 12. (14) 117, 144, 171, 225, 252, 333, 414, 441, 522, 711, 1116, 1125 These are numbers with at least two repeating digits where the digits add up to 9. For example, 1 + 1 + 7 = 9, 1 + 4 + 4 = 9, 1 + 7 + 1 = 9, and 2 + 2 + 5 = 9. (At least two of the digits must be the same.) (15) 13, 35, 57, 79, 135, 357, 579, 1357, 3579, 13579 These numbers have odd digits increasing in order. For example, 357 has odd digits increasing in order. (16) 892, 128, 783, 237, 674, 346, 565, 455, 456, 564, 347, 673 Two sequences are merged together. One is 892, 783, 674, etc., where the first digit decreases by 1 (8, 7, 6, etc.), the middle digit decreases by 1 (9, 8, 7, etc.), and the last digit increases by 1 (2, 3, 4, etc.). The other is 128, 237, 346, etc., where the first digit increases by 1 (1, 2, 3, etc.), the middle digit increases by 1 (2, 3, 4, etc.), and the last digit decreases by 1 (8, 7, 6, etc.). (17) Act, Atc, Cat, Cta, Tac, Tca, Act, Atc, Cat, Cta This sequence cycles through the 6 permutations of the letters a, c, and t, where the first letter is capitalized. (18) TOPS, TOSP, TPOS, TPSO, TSOP, TSPO, SOPT, SOTP, SPOT, SPTO, STOP, STPO This sequence cycles through 12 of the 24 permutations of the letters T, O, P, and S. (19) bit, diq, fin, hik, jih, lie, nib, piy, riv, tis, vip The first letter skips every other letter (b, skip c, d, skip e, f, skip g, etc.), the middle letter is always i, and the last letter skips 2 letters going backward (t, skip s-r, q, skip p-o, n, skip m-l, k, skip j-i, etc.), where a follows z (going y, skip z-a, b). (20) Ae, Bd, Cc, Db, Ea, Aad, Abc, Acb, Ada,Bac, Bbb Let a = 1, b = 2, c = 3, d = 4, and e = 5. Using these numerical values, each group of letters adds up to 6. For example, A + e = 1 + 5 = 6, B + d = 2 + 4 = 6, C + c = 3 + 3 = 6, Aad = 1 + 1 + 4 = 6, Bac = 2 + 1 + 3 = 6, and Bbb = 2 + 2 + 2 = 6. (21) 123, 342, 534, 456, 675, 867, 789, 908, 190, 012, 231, 423, 345, 564, 756 Each number consists of 3 consecutive digits. The first number has the digits in order, the second number has the smallest digit last and the largest digit in the middle, and the third number has the smallest digit in the middle and the largest digit first. This pattern repeats over and over. 8 Prime Numbers (1) 29, 31, 37, 41, 43, 47, 53, 59, 61 These prime numbers appear in order. Note that 33 = 3 × 11, 35 = 5 × 7, 39 = 3 × 13, 45 = 3 × 3 × 5, 49 = 7 × 7, 51 = 3 × 17, 55 = 5 × 11, and 57 = 3 × 19. (2) 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21 These are the non-prime numbers. That is, the prime numbers 2, 3, 5, 7, 11, 13, 17, and 19 have been skipped. (3) 97, 89, 83, 79, 73, 71, 67, 61 These prime numbers are in reverse order. Note that 95 = 5 × 19, 93 = 3 × 31, 91 = 7 × 13, 87 = 3 × 29, 85 = 5 × 17, 81 = 3 × 3 × 3 × 3, 77 = 7 × 11, 75 = 3 × 5 × 5, 69 = 3 × 23, 65 = 5 × 13, and 63 = 3 × 3 × 7. (4) 2, 3, 7, 11, 17, 19, 29, 31, 41, 43, 53, 59, 67 Skip every third prime number: 2, 3, skip 5, 7, 11, skip 13, 17, 19, skip 23, 29, 31, skip 37, 41, etc. (5) 3, 4, 6, 8, 12, 14, 18, 20, 24, 30 Add 1 to each prime number. For example, 2 + 1 = 3, 3 + 1 = 4, 5 + 1 = 6, 7 + 1 = 8, 11 + 1 = 12, and 13 + 1 = 14. (6) 11, 31, 41, 61, 71, 101, 131, 151, 181 These prime numbers end with 1. Note that 21 = 3 × 7, 51 = 3 × 17, 81 = 3 × 3 × 3 × 3, 91 = 7 × 13, 111 = 3 × 37, 121 = 11 × 11, 141 = 3 × 47, 161 = 7 × 23, and 171 = 3 × 3 × 19. (7) 3, 5, 7, 23, 29, 41, 43, 47, 61, 67, 83, 89, 113 These are the prime numbers where the digits add up to an odd number. For example, 23’s digits add up to 5, 47’s digits add up to 11, and 113’s digits add up to 5. (8) 9, 21, 39, 57, 87, 111, 129, 159, 183, 213 Every other prime number is tripled: Skip 2, 3 × 3 = 9, skip 5, 7 × 3 = 21, skip 11, 13 × 3 = 39, skip 17, 19 × 3 = 57. (9) 21, 25, 33, 37, 45, 57, 61, 73, 81, 85, 93 Multiple the prime number by 2 and then subtract 1. For example, 11 × 2 – 1 = 22 – 1 = 21, 13 × 2 – 1 = 26 – 1 = 25, 17 × 2 – 1 = 34 – 1 = 33, and 19 × 2 – 1 = 38 – 1 = 37. (10) 4, 10, 22, 34, 46, 62, 82, 94, 118, 134 Double every other prime number: 2 × 2 = 4, skip 3, 5 × 2 = 10, skip 7, 11 × 2 = 22, skip 13, 17 × 2 = 34, etc. (11) 101, 103, 107, 109, 113, 127, 131, 137, 139 These prime numbers appear in order. (12) 102, 103, 105, 107, 111, 113, 117, 119, 123, 129 Add 100 to each prime number: 2 + 100 = 102, 3 + 100 = 103, 5 + 100 = 105, 7 + 100 = 107, 11 + 100 = 111, etc. (13) 3, 5, 5, 7, 11, 13, 17, 19, 29, 31, 41, 43, 59, 61, 71, 73 These are pairs of consecutive odd-numbered primes. For example, 3 and 5 are consecutive odd-numbered primes. Another example is 5 and 7. Another pair is 11 and 13. (14) 2, 3, 5, 7, 11, 23, 29, 41, 43, 47, 61, 67, 83, 89, 101, 113, 131 These are the prime numbers where the digits add up to a prime number. For example, the 1 and 1 of 11 add up to 2, the 2 and 3 of 23 add up to 5, the 2 and 9 of 29 add up to 11, the 4 and 1 of 41 add up to 5, and the 4 and 3 of 43 add up to 7. 9 Fibonacci Inspired (1) 2, 2, 4, 6, 10, 16, 26, 42, 68, 110 Add consecutive numbers: 2 + 2 = 4, 2 + 4 = 6, 4 + 6 = 10, 6 + 10 = 16, 10 + 16 = 26, etc. (2) 1, 3, 4, 7, 11, 18, 29, 47, 76, 123 Add consecutive numbers: 1 + 3 = 4, 3 + 4 = 7, 4 + 7 = 11, 7 + 11 = 18, etc. (3) 6, 9, 15, 24, 39, 63, 102, 165, 267, 432 Add consecutive numbers: 6 + 9 = 15, 9 + 15 = 24, 15 + 24 = 39, 24 + 39 = 63, etc. (4) 309, 191, 118, 73, 45, 28, 17, 11, 6, 5 Subtract consecutive numbers: 309 – 191 = 118, 191 – 118 = 73, 118 – 73 = 45, etc. (5) 1, 2, 2, 4, 8, 32, 256, 8192, 2097152 Multiply consecutive numbers: 1 × 2 = 2, 2 × 2 = 4, 2 × 4 = 8, 4 × 8 = 32, etc. (6) 1, 2, 5, 13, 34, 89, 233, 610, 1597, 4181, 10946 Skip every other number of the Fibonacci sequence: 1, skip 1, 2, skip 3, 5, skip 8, 13, skip 21, etc. (7) 1, 1, 2, 4, 7, 13, 24, 44, 81, 149, 274, 504, 927, 1705 Add the 3 previous numbers: 1 + 1 + 2 = 4, 1 + 2 + 4 = 7, 2 + 4 + 7 = 13, 4 + 7 + 13 = 24, etc. (8) 2, 3, 3, 4, 5, 7, 10, 15, 23, 36, 57, 91, 146, 235 Add consecutive numbers and subtract 2: 2 + 3 – 2 = 3, 3 + 3 – 2 = 4, 3 + 4 – 2 = 5, 4 + 5 – 2 = 7, 5 + 7 – 2 = 10, etc. (9) 1, 3, 5, 9, 15, 25, 41, 67, 109, 177, 287, 465, 753, 1219 Add consecutive numbers and add 1: 1 + 3 + 1 = 5, 3 + 5 + 1 = 9, 5 + 9 + 1 = 15, 9 + 15 + 1 = 25, 15 + 25 + 1 = 41, etc. (10) 0, 1, 2, 2, 3, 5, 7, 10, 15, 22, 32, 47, 69, 101, 148, 217, 318, 466 Add the 3 previous numbers and subtract the middle of those three numbers. Put another way, add the previous number to the number two places before it. For example, 0 + 1 + 2 – 1 = 2 (or simply 0 + 2 = 2), 1 + 2 + 2 – 2 = 3 (or 1 + 2 = 3), 2 + 2 + 3 – 2 = 5 (or 2 + 3 = 5), 2 + 3 + 5 – 3 = 7 (or 2 + 5 = 7), 3 + 5 + 7 – 5 = 10 (or 3 + 7 = 10), 5 + 7 + 10 – 7 = 15 (or 5 + 10 = 15), and 7 + 10 + 15 – 10 = 22 (or 7 + 15 = 22). 10 Roman Numerals (1) II, IV, VI, VIII, X, XII, XIV, XVI, XVIII, XX These are even numbers (2, 4, 6, etc.). (2) III, VI, XII, XXIV, XLVIII, XCVI, CXCII, CCCLXXXIV, DCCLXVIII Double each number (3 × 2 = 6, 6 × 2 = 12, 12 × 2 = 24, 24 × 2 = 48, etc.). (3) CVI, CV, CIV, CIII, CII, CI, C, XCIX, XCVIII, XCVII Count backwards (106, 105, 104, 103, 102, etc.). Note that 99 is correctly written as XCIX and not IC. (4) XXV, L, LXXV, C, CXXV, CL, CLXXV, CC, CCXXV, CCL Count by 25 (25, 50, 75, 100, 125, etc.). (5) C, CC, CCC, CD, D, DC, DCC, DCCC, CM, M Count by 100 (100, 200, 300, 400, 500, etc.). (6) MCMLX, MCMLXV, MCMLXX, MCMLXXV, MCMLXXX, MCMLXXXV, MCMXC, MCMXCV, MM, MMV Count by 5 (1960, 1965, 1970, 1975, 1980, etc.). (7) DLXX, DLX, DL, DXL, DXXX, DXX, DX, D, CDXC, CDLXXX Subtract 10 (570, 560, 550, 540, 530, etc.). (8) IX, VII, XIV, XII, XIX, XVII, XXIV, XXII, XXIX, XXVII, XXXIV, XXXII Subtract 2, add 7, subtract 2, add 7, repeating this pattern (9 – 2 = 7, 7 + 7 = 14, 14 – 2 = 12, 12 + 7 = 19, 19 – 2 = 17, 17 + 7 = 24, etc.). (9) II, VI, III, IX, VI, XVIII, XV, XLV, XLII, CXXVI, CXXIII, CCCLXIX, CCCLXVI Multiply by 3, subtract 3, multiply by 3, subtract 3, repeating this pattern (2 × 3 = 6, 6 – 3 = 3, 3 × 3 = 9, 9 – 3 = 6, 6 × 3 = 18, 18 – 3= 15, etc.). (10) VII, IX, XII, XVI, XXI, XXVII, XXXIV, XLII, LI, LXI Add 2, add 3, add 4, add 5, etc. (7 + 2 = 9, 9 + 3 = 12, 12 + 4 = 16, 16 + 5 = 21, 21 + 6 = 27, etc.). (11) II, IV, VI, IX, XI, XV, XX, XL, LI, LV, LX, XC, CI These are the Roman numerals that can be made with exactly two letters, in numerical order. (12) CC, CD, CI, CL, CM, CV, CX, DC, DI, DL, DV, DX, II These are the Roman numerals that can be made with exactly two letters, in alphabetical order. (13) III, V, IX, XVII, XXXIII, LXV, CXXIX, CCLVII, DXIII, MXXV, MMXLIX Multiply the previous number by 2 and subtract 1: (2 × 3 – 1 = 5, 2 × 5 – 1 = 9, 2 × 9 – 1 = 17, 2 × 17 – 1 = 33, 2 × 33 – 1 = 65, etc.). (14) I, I, II, III, V, VIII, XIII, XXI, XXXIV, LV, LXXXIX, CXLIV This is the Fibonacci sequence (Chapter 9): 1, 1, 2, 3, 5, 8, 13, 21, etc. (15) II, III, V, VII, XI, XIII, XVII, XIX, XXIII, XXIX, XXXI These are prime numbers (Chapter 8): 2, 3, 5, 7, 11, 13, 17, etc. 11 Powers (1) 1, 4, 9, 16, 25, 36, 49, 64 Square consecutive numbers: 12 = 1, 22 = 4, 32 = 9, 42 = 16, 52 = 25, etc. (2) 1, 2, 4, 8, 16, 32, 64, 128, 256 Raise 2 to the next power: 20 = 1, 21 = 2, 22 = 4, 23 = 8, 24 = 16, 25 = 32, etc. (3) 1, 100, 10000, 1000000, 100000000, 10000000000, 1000000000000, 100000000000000 Raise 10 to the next even power: 100 = 1, 102 = 100, 104 = 10000, 106 = 100000, etc. (4) 1, 16, 81, 256, 625, 1296, 2401, 4096, 6561, 10000 These are powers of 4: 14 = 1, 24 = 16, 34 = 81, 44 = 256, 54 = 625, etc. (5) 4, 16, 36, 64, 100, 144, 196, 256, 324, 400 Square even numbers: 22 = 4, 42 = 16, 62 = 36, 82 = 64, 102 = 100, etc. (6) 1, 4, 27, 256, 3125, 46656, 823543 Raise each number to the power of itself: 11 = 1, 22 = 4, 33 = 27, 44 = 256, 55 = 3125, etc. (7) 1, 16, 256, 4096, 65536, 1048576 Raise 4 to the next even power: 40 = 1, 42 = 16, 44 = 256, 46 = 4096, 48 = 65536, etc. (8) 0, 3, 8, 15, 24, 35, 48, 63, 80, 99 These are squares minus one: 12 – 1 = 0, 22 – 1 = 3, 32 – 1 = 8, 42 – 1 = 15, 52 – 1 = 24, etc. (9) 4, 9, 25, 49, 121, 169, 289, 361, 529, 841 Square prime numbers (see Chapter 8): 22 = 4, 32 = 9, 52 = 25, 72 = 49, 112 = 121, etc. (10) 2, 8, 18, 32, 50, 72, 98, 128, 162, 200 Double each square: 2 × 12 = 2, 2 × 22 = 8, 2 × 32 = 18, 2 × 42 = 32, 2 × 52 = 50, etc. 12 Factorials (1) 1, 6, 120, 5040, 362880, 39916800, 6227020800 Skip every other factorial: 1! = 1, skip 2!, 3! = 6, skip 4!, 5! = 120, skip 6, 7! = 5040, etc. (2) 1, 3, 15, 105, 945, 10395, 135135, 2027025, 34459425 These are double factorials with odd numbers: 1!! = 1, 3!! = 3 × 1 = 3, 5!! = 5 × 3 × 1 = 15, 7!! = 7 × 5 × 3 × 1 = 105, 9!! = 9 × 7 × 5 × 3 × 1 = 945, etc. (3) 1, 1, 2, 3, 8, 15, 48, 105, 384, 945, 3840, 10395, 46080 Each number starts with a factorial and divides by the previous number: 1, 1! ÷ 1 = 1, 2! ÷ 1 = 2, 3! ÷ 2 = 3, 4! ÷ 3 = 8, 5! ÷ 8 = 15, 6! ÷ 15 = 48, 7! ÷ 48 = 105, 8! ÷ 105 = 384, etc. (4) 2, 2, 4, 12, 48, 240, 1440, 10080, 80640, 725760 Multiply each factorial by two: (0!) × 2 = 1 × 2 = 2, (1!) × 2 = 1 × 2 = 2, (2!) × 2 = 2 × 2 = 4, (3!) × 2 = 6 × 2 = 12, (4!) × 2 = 24 × 2 = 48, etc. (5) 1, 1, 4, 36, 576, 14400, 518400, 25401600 Square each factorial: (0!)2 = 12 = 1, (1!) 2 = 12 = 1, (2!) 2 = 22 = 4, (3!) 2 = 62 = 36, (4!) 2 = 242 = 576, etc. (6) 1, 7, 21, 35, 35, 21, 7, 1 This sequence is made from the combination formula for N = 7: 7!
7! 0! = 1 7!
3! 4! = 35 7!
6! 1! = 7 7!
2! 5! = 21 7!
5! 2! = 21 7!
1! 6! = 7 7!
4! 3! = 35 7!
0! 7! = 1 (7) 1, 9, 36, 84, 126, 126, 84, 36, 9, 1 This sequence is made from the combination formula for N = 9: 9!
9! 0! = 1 9!
4! 5! = 126 9!
8! 1! = 9 9!
3! 6! = 84 9!
7! 2! = 36 9!
2! 7! = 36 9!
6! 3! = 84 9!
1! 8! = 9 9!
5! 4! = 126 9!
0! 9! = 1 (8) 1, 4, 10, 20, 35, 56, 84, 120, 165, 220 One way to make this sequence is to read along a diagonal of Pascal's triangle, as illustrated below: 𝑥𝑥+ 𝑦𝑦 𝑥𝑥2 + 2𝑥𝑥𝑥𝑥+ 𝑦𝑦2 𝑥𝑥3 + 3𝑥𝑥2𝑦𝑦+ 3𝑥𝑥𝑦𝑦2 + 𝑦𝑦3 𝑥𝑥4 + 4𝑥𝑥3𝑦𝑦+ 6𝑥𝑥2𝑦𝑦2 + 4𝑥𝑥𝑦𝑦3 + 𝑦𝑦4 𝑥𝑥5 + 5𝑥𝑥4𝑦𝑦+ 10𝑥𝑥3𝑦𝑦2 + 10𝑥𝑥2𝑦𝑦3 + 5𝑥𝑥𝑦𝑦4 + 𝑦𝑦5 𝑥𝑥6 + 6𝑥𝑥5𝑦𝑦+ 15𝑥𝑥4𝑦𝑦2 + 20𝑥𝑥3𝑦𝑦3 + 15𝑥𝑥2𝑦𝑦4 + 6𝑥𝑥𝑦𝑦5 + 𝑦𝑦6 𝑥𝑥7 + 7𝑥𝑥6𝑦𝑦+ 21𝑥𝑥5𝑦𝑦2 + 35𝑥𝑥4𝑦𝑦3 + 35𝑥𝑥3𝑦𝑦4 + 21𝑥𝑥2𝑦𝑦5 + 7𝑥𝑥𝑦𝑦6 + 𝑦𝑦7 𝑥𝑥8 + 8𝑥𝑥7𝑦𝑦+ 28𝑥𝑥6𝑦𝑦2 + 56𝑥𝑥5𝑦𝑦3 + 70𝑥𝑥4𝑦𝑦4 + 56𝑥𝑥3𝑦𝑦5 + 28𝑥𝑥2𝑦𝑦6 + 8𝑥𝑥𝑦𝑦7 + 𝑦𝑦8 Another way to make this sequence is to add 3, add 6, add 10, add 15, add 21, etc. (The pattern 3, 6, 10, 15, 21, etc. is made by adding 3, 4, 5, 6, etc.) For example, 1 + 3 = 4, 4 + 6 = 10, 10 + 10 = 20, 20 + 15 = 35, and 35 + 21 = 56. (9) 2, 3, 8, 30, 144, 840, 5760, 45360, 403200, 3991680, 43545600 Add consecutive factorials: 2! + 1! = 3, 3! + 2! = 8, 4! + 3! = 30, 5! + 4! = 144, 6! + 5! = 840, etc. (10) 3628800, 1814400, 604800, 151200, 30240, 5040, 720, 90, 10, 1 This sequence is made by dividing 10! by a smaller factorial: 10! / 1! = 3628800, 10! / 2! = 1814400, 10! / 3! = 604800, 10! / 4! = 151200, 10! / 5! = 30240, etc. 13 Negative Numbers (1) –28, –24, –20, –16, –12, –8, –4, 0, 4, 8 Add 4: –28 + 4 = –24, –24 + 4 = –20, –20 + 4 = –16, etc. (2) –2, 4, –8, 16, –32, 64, –128, 256, –512, 1024 Multiply by –2: –2 × (–2) = 4, 4 × (–2) = –8, –8 × (–2) = 16, 16 × (–2) = –32, etc. (3) 3, –5, 7, –7, 11, –9, 15, –11, 19, –13, 23, –15 Two sequences are merged together. One sequence adds 4 (3 + 4 = 7, 7 + 4 = 11, 11 + 4 = 15, etc.), while the other sequence subtracts 2 (–5 – 2 = –7, –7 – 2 = –9, –9 – 2 = –11, etc.): The sequence 7, 11, 15, 19, 23, etc. is mixed with the sequence –5, –7, –9, –11, –13, etc. (4) –57, 43, –42, 35, –27, 27, –12, 19, 3, 11, 18, 3 Two sequences are merged together. One sequence adds 15 (–57 + 15 = –42, –42 + 15 = –27, –27 + 15 = –12, etc.), while the other sequence subtracts 8 (43 – 8 = 35, 35 – 8 = 27, 27 – 8 = 19, etc.): The sequence –57, –42, –27, –12, etc. is mixed with the sequence 43, 35, 27, 19, etc. (5) 1, –3, –15, 45, 33, –99, –111, 333, 321, –963, –975, 2925, 2913 Multiply by –3, subtract 12, multiply by –3, subtract 12, repeating this pattern: 1 × (–3) = –3, –3 – 12 = –15, –15 × (–3) = 45, 45 – 12 = 33, 33 × (–3) = –99, –99 – 12 = –111, etc. (6) 0, 1, –1, 2, –3, 5, –8, 13, –21, 34, –55, 89, –144 Make every other element of the Fibonacci sequence negative (Chapter 9). (7) –3, 4, –4, 3, –4, 4, –3, 4, –4, 3, –4, 4, –3, 4, –4 Repeat the pattern 3, 4, 4, 3, 4, 4 over and over, making every other number negative. (8) 20, 19, 17, 14, 10, 5, –1, –8, –16, –25 Subtract 1, subtract 2, subtract 3, subtract 4, subtract 5, etc.: 20 – 1 = 19, 19 – 2 = 17, 17 – 3 = 14, 14 – 4 = 10, 10 – 5 = 5, 5 – 6 = –1, –1 – 7 = –8, etc. (9) 3, 9, –9, 27, –27, –27, 81, –81, –81, –81, 243, – 243, –243, –243, –243, 729, –729 This pattern involves powers of 3. 31, 32, –(32), 33, –(33), –(33), 34, –(34), –(34), –(34), 35, –(35), –(35), –(35), –(35), etc. There is one more negative number each time the power increases. (10) –6, –9, –15, –21, –33, –39, –51, –57, –69, –87, –93, –111, –123 These are prime numbers (Chapter 8) multiplied by –3: 2 × (–3) = –6, 3 × (–3) = –9, 5 × (–3) = –15, 7 × (–3) = –21, 11 × (–3) = –33, 13 × (–3) = –39 etc. 14 Fractions (1) 2 3 , 4 5 , 6 7 , 8 9 , 10 11 , 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏, 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏, 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏, 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 Increase each numerator and denominator by 2. For example, the 2 of 2 3 becomes 2 + 2 = 4 and the 3 of 2 3 becomes 3 + 2 = 5, making 4 5. Similarly, 6 7 turns into 8 9, since 6 + 2 = 8 and 7 + 2 = 9. (2) 1 3 , 5 6 , 1 12 , 23 24 , 1 48 , 95 96 , 𝟏𝟏 𝟏𝟏𝟏𝟏𝟏𝟏, 𝟑𝟑𝟑𝟑𝟑𝟑 𝟑𝟑𝟑𝟑𝟑𝟑, 𝟏𝟏 𝟕𝟕𝟕𝟕𝟕𝟕, 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 Double the denominator (3, 6, 12, 24, etc.). The numera-tor alternates between 1 and the denominator minus 1. (3) 1 4 , 6 7 , 3 8 , 10 11 , 5 12 , 14 15 , 7 16 , 18 19 , 𝟗𝟗 𝟐𝟐𝟐𝟐, 𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐, 𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐, 𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐 For the numerator, add 5, divide by 2, add 7, divide by 2, add 9, divide by 2, add 11, divide by 2, repeating this pattern (1 + 5 = 6, 6 ÷ 2 = 3, 3 + 7 = 10, 10 ÷ 2 = 5, 5 + 9 = 14, 14 ÷ 2 = 7, 7 + 11 = 18, etc.). For the denominator, add 3, add 1, add 3, add 1, repeating this pattern (4 + 3 = 7, 7 + 1 = 8, 8 + 3 = 11, 11 + 1 = 12, etc.). (4) 1 16 , 1 8 , 3 16 , 1 4 , 5 16 , 3 8 , 7 16 , 𝟏𝟏 𝟐𝟐, 𝟗𝟗 𝟏𝟏𝟏𝟏, 𝟓𝟓 𝟖𝟖, 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 Add 1 16: 1 16 + 1 16 = 1 8, 1 8 + 1 16 = 3 16, 3 16 + 1 16 = 1 4, etc. (5) 0.5, 0.25, 0.125, 0.0625, 0.03125, 0.015625, 0.0078125, 0.00390625, 0.001953125, 0.0009765625 Divide by 2: 0.5 ÷ 2 = 0.25, 0.25 ÷ 2 = 0.125, 0.125 ÷ 2 = 0.0625, etc. (6) 1 2 , 1 1 2 , 4 1 2 , 13 1 2 , 40 1 2 , 121 1 2, 𝟑𝟑𝟑𝟑𝟑𝟑 𝟏𝟏 𝟐𝟐, 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏 𝟐𝟐, 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑 𝟏𝟏 𝟐𝟐, 𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗 𝟏𝟏 𝟐𝟐 Add 30, add 31, add 32, add 33, add 34, etc.: 1 2 + 1 = 1 1 2 13 1 2 + 27 = 40 1 2 1 1 2 + 3 = 4 1 2 40 1 2 + 81 = 121 1 2 4 1 2 + 9 = 13 1 2 121 1 2 + 243 = 364 1 2 (7) 1 5 , 13 15 , 1 8 15 , 2 1 5 , 2 13 15 , 3 8 15 , 4 1 5, 𝟒𝟒 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏, 𝟓𝟓 𝟖𝟖 𝟏𝟏𝟏𝟏, 𝟔𝟔 𝟏𝟏 𝟓𝟓, 𝟔𝟔 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 Add 2 3: 1 5 + 2 3 = 13 15 , 13 15 + 2 3 = 23 15 (or 1 and 8 15), 23 15 + 2 3 = 11 5 (or 2 and 1 5), 11 5 + 2 3 = 43 15 (or 2 and 13 15), 43 15 + 2 3 = 53 15 (or 3 and 8 15), etc. (8) 1 12 , 1 6 , 1 4 , 1 3 , 5 12 , 1 2 , 𝟕𝟕 𝟏𝟏𝟏𝟏, 𝟐𝟐 𝟑𝟑, 𝟑𝟑 𝟒𝟒, 𝟓𝟓 𝟔𝟔 Add 1 12: 1 12 + 1 12 = 1 6 , 1 6 + 1 12 = 1 4 , 1 4 + 1 12 = 1 3 , 1 3 + 1 12 = 5 12, etc. (9) 1 2 , 1 3 , 5 6 , 1 1 6 , 2, 3 1 6 , 5 1 6 , 8 1 3 , 13 1 2, 𝟐𝟐𝟐𝟐 𝟓𝟓 𝟔𝟔, 𝟑𝟑𝟑𝟑 𝟏𝟏 𝟑𝟑, 𝟓𝟓𝟓𝟓 𝟏𝟏 𝟔𝟔, 𝟗𝟗𝟗𝟗 𝟏𝟏 𝟐𝟐 Add consecutive fractions (like the Fibonacci sequence described in Chapter 9): 1 2 + 1 3 = 5 6, 1 3 + 5 6 = 7 6 (or 1 and 1 6), 5 6 + 7 6 = 2, 7 6 + 2 = 19 6 (or 3 and 1 6), 2 + 19 6 = 31 6 (or 5 and 1 6), etc. (10) 4 3 7 , 5 5 9 , 6 7 11 , 9 7 13 , 8 11 15 , 13 9 17 , 10 15 19 , 17 11 21, 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐, 𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐, 𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐, 𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐 The denominators are consecutive odd numbers (7, 9, 11, 13, etc.). Two more patterns alternate between the numerators and whole numbers. One pattern is 4, 5, 6, 7, 8, etc. with the first whole number, second numera-tor, third whole number, fourth numerator, etc. The other is 3, 5, 7, 9, 11, etc. with the first numerator, second whole number, third numerator, fourth whole number, etc. (11) 0.125, 0.5, 0.875, 1.25, 1.625, 2, 2.375, 2.75, 3.125 Add 0.375: 0.125 + 0.375 = 0.5, 0.5 + 0.375 = 0.875, 0.875 + 0.375 = 1.25, etc. (12) 2 5 , 3 7 , 5 11 , 7 15 , 11 23 , 13 27 , 17 35 , 𝟏𝟏𝟏𝟏 𝟑𝟑𝟑𝟑, 𝟐𝟐𝟐𝟐 𝟒𝟒𝟒𝟒, 𝟐𝟐𝟐𝟐 𝟓𝟓𝟓𝟓, 𝟑𝟑𝟑𝟑 𝟔𝟔𝟔𝟔 The numerators are prime numbers (2, 3, 5, 7, 11, 13, 17, etc.). The denominator is twice the numerator plus 1: 2 × 2 + 1 = 5, 3 × 2 + 1 = 7, 5 × 2 + 1 = 11, 7 × 2 + 1 = 15, 11 × 2 + 1 = 23, etc. (13) 1 2 , 1 3 , 2 9 , 4 27 , 8 81 , 16 243 , 𝟑𝟑𝟑𝟑 𝟕𝟕𝟕𝟕𝟕𝟕, 𝟔𝟔𝟔𝟔 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐, 𝟏𝟏𝟏𝟏𝟏𝟏 𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔, 𝟐𝟐𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 Multiply by 2 3: 1 2 × 2 3 = 1 3 , 1 3 × 2 3 = 2 9 , 2 9 × 2 3 = 4 27, etc. (14) 1 4 , 9 16 , 25 36 , 49 64 , 81 100 , 𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏𝟏𝟏, 𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏𝟏𝟏, 𝟐𝟐𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐𝟔𝟔, 𝟐𝟐𝟐𝟐𝟐𝟐 𝟑𝟑𝟑𝟑𝟑𝟑 The numerators are odd numbers squared (12 = 1, 32 = 9, 52 = 25, 72 = 49, etc.). The denominators are even numbers squared (22 = 4, 42 = 16, 62 = 36, 82 = 64, etc.). (15) 1, 2, 1 1 2 , 1 1 3 , 1, 3 4 , 7 13 , 8 21 , 9 34, 𝟐𝟐 𝟏𝟏𝟏𝟏, 𝟏𝟏𝟏𝟏 𝟖𝟖𝟖𝟖, 𝟏𝟏 𝟏𝟏𝟏𝟏, 𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐𝟐𝟐 The numerators are integers (1, 2, 3, etc.). The denominators are elements of the Fibonacci sequence described in Chapter 9 (1, 1, 2, 3, 5, 8, 13, etc.). The sequence is: 1 1 = 1 5 5 = 1 2 1 = 2 6 8 = 3 4 3 2 = 1 1 2 7 13 4 3 = 1 1 3 8 21 (16) 1 2 , 1 6 , 5 12 , 1 12 , 1 3 , 0, 1 4 , – 1 12 , 1 6 , – 1 6 , 1 12 , – 𝟏𝟏 𝟒𝟒, 𝟎𝟎, – 𝟏𝟏 𝟑𝟑, – 𝟏𝟏 𝟏𝟏𝟏𝟏 Two sequences are merged together. Each sequence involves subtracting 1 12. The first sequence is: 1 2 – 1 12 = 5 12, 5 12 – 1 12 = 1 3 , 1 3 – 1 12 = 1 4 , 1 4 – 1 12 = 1 6, etc. The second sequence is 1 6 – 1 12 = 1 12 , 1 12 – 1 12 = 0, 0 – 1 12 = – 1 12 , – 1 12 – 1 12 = – 1 6, etc. (17) 0. 5 ത, 1, 1. 4 ത, 1. 8 ത, 2. 3 ത, 2. 7 ത, 3. 2 ത, 3. 6 ത, 𝟒𝟒. 𝟏𝟏 ഥ, 𝟒𝟒. 𝟓𝟓 ഥ, 𝟓𝟓, 𝟓𝟓. 𝟒𝟒 ഥ Add 0. 4 ത. For example: 0. 5 ത+ 0. 4 ത= 1, 1 + 0. 4 ത= 1. 4 ത, 1. 4 ത+ 0. 4 ത= 1. 8 ത, and 1. 8 ത+ 0. 4 ത= 2. 3 ത. Note: The bar over a digit indicates a repeating decimal. For example, 0. 5 ത= 0.5555555555555555555555555555555555555... (18) 1 3 , 1 2 , 3 4 , 1 1 8 , 1 11 16 , 2 17 32, 𝟑𝟑 𝟓𝟓𝟓𝟓 𝟔𝟔𝟔𝟔, 𝟓𝟓 𝟖𝟖𝟖𝟖 𝟏𝟏𝟏𝟏𝟏𝟏, 𝟖𝟖 𝟏𝟏𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐𝟐𝟐, 𝟏𝟏𝟏𝟏 𝟒𝟒𝟒𝟒𝟒𝟒 𝟓𝟓𝟓𝟓𝟓𝟓 Multiply by 3 2: 1 3 × 3 2 = 1 2 , 1 2 × 3 2 = 3 4 , 3 4 × 3 2 = 9 8 (or 1 and 1 8), 9 8 × 3 2 = 27 16 (or 1 and 11 16), etc. (19) 1 4 , 0.5, 75%, 1, 1.25, 150%, 1 3 4 , 2, 225%, 𝟐𝟐 𝟏𝟏 𝟐𝟐, 𝟐𝟐. 𝟕𝟕𝟕𝟕, 𝟑𝟑𝟑𝟑𝟑𝟑%, 𝟑𝟑 𝟏𝟏 𝟒𝟒 Add 0.25, alternately expressing the answer as a fraction, decimal, or percentage: 1 4 + 0.25 = 0.5, 0.5 + 0.25 = 75%, 75% + 0.25 = 1, 1 + 0.25 = 1.25, 1.25 + 0.25 = 150%, 150% + 0.25 = 7 4 (or 1 and 3 4), 7 4 + 0.25 = 2, etc. (20) 1 2 , 1 3 , 1 1 2 , 2 9 , 6 3 4 , 8 243 , 205 1 32, 𝟐𝟐𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 Divide consecutive fractions: 1 2 ÷ ቀ 1 3ቁ = 3 2 (or 1 and 1 2), 1 3 ÷ ቀ 3 2ቁ= 2 9 , 3 2 ÷ ቀ 2 9ቁ= 27 4 (or 6 and 3 4), 2 9 ÷ ቀ 27 4 ቁ = 8 243, etc. 15 Algebraic (1) 𝑥𝑥10, 10𝑥𝑥9, 90𝑥𝑥8, 720𝑥𝑥7, 𝟓𝟓𝟓𝟓𝟓𝟓𝟓𝟓𝒙𝒙𝟔𝟔, 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝒙𝒙𝟓𝟓, 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝒙𝒙𝟒𝟒, 𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝒙𝒙𝟑𝟑 Multiply by 10, multiply by 9, multiply by 8, multiply by 7, etc., while also dividing by 𝑥𝑥. For example: 𝑥𝑥10 ∙10 𝑥𝑥= 10𝑥𝑥9 90𝑥𝑥8 ∙8 𝑥𝑥= 720𝑥𝑥7 10𝑥𝑥9 ∙9 𝑥𝑥= 90𝑥𝑥8 720𝑥𝑥7 ∙7 𝑥𝑥= 5040𝑥𝑥6 (2) 120, 120𝑥𝑥, 60𝑥𝑥2, 20𝑥𝑥3, 5𝑥𝑥4, 𝒙𝒙𝟓𝟓, 𝒙𝒙𝟔𝟔 𝟔𝟔, 𝒙𝒙𝟕𝟕 𝟒𝟒𝟒𝟒, 𝒙𝒙𝟖𝟖 𝟑𝟑𝟑𝟑𝟑𝟑 Divide by 1, divide by 2, divide by 3, divide by 4, etc.: 120 ÷ 1 = 120, 120 ÷ 2 = 60, 60 ÷ 3 = 20, 20 ÷ 4 = 5, 5 ÷ 5 = 1, 1 ÷ 6 = 1/6, etc. Also increase the power each time: 1, 𝑥𝑥, 𝑥𝑥2, 𝑥𝑥3, 𝑥𝑥4, etc. (3) 4𝑥𝑥, 9𝑥𝑥2, 16𝑥𝑥4, 25𝑥𝑥8, 36𝑥𝑥16, 𝟒𝟒𝟒𝟒𝒙𝒙𝟑𝟑𝟑𝟑, 𝟔𝟔𝟔𝟔𝒙𝒙𝟔𝟔𝟔𝟔, 𝟖𝟖𝟖𝟖𝒙𝒙𝟏𝟏𝟏𝟏𝟏𝟏, 𝟏𝟏𝟏𝟏𝟏𝟏𝒙𝒙𝟐𝟐𝟐𝟐𝟐𝟐 The coefficients are squares (22 = 4, 32 = 9, 42 = 16, 52 = 25, etc.) and the exponents are powers of 2 (20 = 1, 21 = 2, 22 = 4, 23 = 8, 24 = 16, etc.). (4) 2, 3𝑥𝑥2, 5𝑥𝑥4, 7𝑥𝑥6, 11𝑥𝑥8, 13𝑥𝑥10, 17𝑥𝑥12, 𝟏𝟏𝟏𝟏𝒙𝒙𝟏𝟏𝟏𝟏, 𝟐𝟐𝟐𝟐𝒙𝒙𝟏𝟏𝟏𝟏, 𝟐𝟐𝟐𝟐𝒙𝒙𝟏𝟏𝟏𝟏, 𝟑𝟑𝟑𝟑𝒙𝒙𝟐𝟐𝟐𝟐 The coefficients (2, 3, 5, 7, 11, 13, etc.) are prime numbers (Chapter 8). The powers are even numbers (0, 2, 4, etc.), where 𝑥𝑥0 = 1. (5) 𝑥𝑥, 2 𝑥𝑥2, 𝑥𝑥3 6 , 24 𝑥𝑥5, 𝑥𝑥8 120, 720 𝑥𝑥13, 𝒙𝒙𝟐𝟐𝟐𝟐 𝟓𝟓𝟓𝟓𝟓𝟓𝟓𝟓, 𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒 𝒙𝒙𝟑𝟑𝟑𝟑, 𝒙𝒙𝟓𝟓𝟓𝟓 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑, 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑 𝒙𝒙𝟖𝟖𝟖𝟖 The Fibonacci sequence (1, 2, 3, 5, 8, 13, etc.) described in Chapter 9 forms the exponents. The constant is a factorial (1! = 1, 2! = 2, 3! = 6, 4! = 24, 5! = 120, 6! = 720, etc.), which is defined in Chapter 12. The variable and constant alternate position between the numerator and denominator. (6) 2𝑥𝑥4, 7𝑥𝑥11, 16𝑥𝑥22, 29𝑥𝑥37, 46𝑥𝑥56, 𝟔𝟔𝟔𝟔𝒙𝒙𝟕𝟕𝟕𝟕, 𝟗𝟗𝟗𝟗𝒙𝒙𝟏𝟏𝟏𝟏𝟏𝟏, 𝟏𝟏𝟏𝟏𝟏𝟏𝒙𝒙𝟏𝟏𝟏𝟏𝟏𝟏, 𝟏𝟏𝟏𝟏𝟏𝟏𝒙𝒙𝟏𝟏𝟏𝟏𝟏𝟏 For the coefficients, add 5, add 9, add 13, add 17, etc.: 2 + 5 = 7, 7 + 9 = 16, 16 + 13 = 29, 29 + 17 = 46, 46 + 21 = 67, etc. For the exponents, add 7, add 11, add 15, add 19, etc.: 4 + 7 = 11, 11 + 11 = 22, 22 + 15 = 37, 37 + 19 = 56, 56 + 23 = 79, etc. (7) (𝑎𝑎+ 2)3, (𝑏𝑏+ 4)6, (𝑐𝑐+ 8)12, (𝑑𝑑+ 16)24, (𝒆𝒆+ 𝟑𝟑𝟑𝟑)𝟒𝟒𝟒𝟒, (𝒇𝒇+ 𝟔𝟔𝟔𝟔)𝟗𝟗𝟗𝟗, (𝒈𝒈+ 𝟏𝟏𝟏𝟏𝟏𝟏)𝟏𝟏𝟏𝟏𝟏𝟏, (𝒉𝒉+ 𝟐𝟐𝟐𝟐𝟐𝟐)𝟑𝟑𝟑𝟑𝟑𝟑 Advance the letter (𝑎𝑎, 𝑏𝑏, 𝑐𝑐, etc.), double the constant (2, 4, 8, 16, etc.), and double the power (3, 6, 12, 24, etc.). (8) 𝑥𝑥4 24, 𝑥𝑥3 12, 𝑥𝑥2 4 , 𝑥𝑥, 𝟓𝟓, 𝟑𝟑𝟑𝟑 𝒙𝒙, 𝟐𝟐𝟐𝟐𝟐𝟐 𝒙𝒙𝟐𝟐, 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝒙𝒙𝟑𝟑, 15120 𝑥𝑥4 Divide by 𝑥𝑥 2, divide by 𝑥𝑥 3, divide by 𝑥𝑥 4, divide by 𝑥𝑥 5, etc.: 𝑥𝑥4 24 ÷ 𝑥𝑥 2 = 𝑥𝑥4 24 ∙2 𝑥𝑥= 𝑥𝑥3 12 𝑥𝑥3 12 ÷ 𝑥𝑥 3 = 𝑥𝑥3 12 ∙3 𝑥𝑥= 𝑥𝑥2 4 𝑥𝑥2 4 ÷ 𝑥𝑥 4 = 𝑥𝑥2 4 ∙4 𝑥𝑥= 𝑥𝑥 𝑥𝑥÷ 𝑥𝑥 5 = 𝑥𝑥∙5 𝑥𝑥= 5 5 ÷ 𝑥𝑥 6 = 5 ∙6 𝑥𝑥= 30 𝑥𝑥 30 𝑥𝑥÷ 𝑥𝑥 7 = 30 𝑥𝑥∙7 𝑥𝑥= 210 𝑥𝑥2 (9) 2𝑎𝑎, 𝑐𝑐4, 8𝑒𝑒, 𝑔𝑔16, 32𝑖𝑖, 𝒌𝒌𝟔𝟔𝟔𝟔, 𝟏𝟏𝟏𝟏𝟖𝟖𝒎𝒎, 𝒐𝒐𝟐𝟐𝟐𝟐𝟐𝟐, 𝟓𝟓𝟓𝟓𝟐𝟐𝒒𝒒 The constant and variable alternate position between the base and exponent. The variable skips every other letter of the alphabet (𝑎𝑎, skip 𝑏𝑏, 𝑐𝑐, skip 𝑑𝑑, 𝑒𝑒, skip 𝑓𝑓, 𝑔𝑔, etc.). The constant is a power of 2: 21 = 2, 22 = 4, 23 = 8, 24 = 16, etc. (10) 7, 23, 32, 2·5, 11, 22·3, 13, 2·7, 3·5, 24, 17, 2·32, 19, 22·5 These are the prime factors of every number: 7 is prime, 8 factors as 2 × 2 × 2 (or 23), 9 factors as 3 × 3 (or 32), 10 factors as 2 × 5, 11 is prime, 12 factors as 2 × 2 × 3 (or 22 × 3), 13 is prime, etc. (11) 5𝑥𝑥2, 7𝑥𝑥3, 11𝑥𝑥5, 19𝑥𝑥9, 35𝑥𝑥17, 67𝑥𝑥33, 𝟏𝟏𝟏𝟏𝟏𝟏𝒙𝒙𝟔𝟔𝟔𝟔, 𝟐𝟐𝟐𝟐𝟐𝟐𝒙𝒙𝟏𝟏𝟏𝟏𝟏𝟏, 𝟓𝟓𝟓𝟓𝟓𝟓𝒙𝒙𝟐𝟐𝟐𝟐𝟐𝟐, 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝒙𝒙𝟓𝟓𝟓𝟓𝟓𝟓 Double the previous exponent and subtract 1 to get the new exponent: 2 × 2 – 1 = 3, 3 × 2 – 1 = 5, 5 × 2 – 1 = 9, 9 × 2 – 1 = 17, etc. The coefficient is twice the exponent plus 1: 2 × 2 + 1 = 5, 3 × 2 + 1 = 7, 5 × 2 + 1 = 11, 9 × 2 + 1 = 19, etc. (12) 3𝑥𝑥2, 𝑥𝑥5, 4𝑥𝑥3, 2𝑥𝑥6, 5𝑥𝑥4, 3𝑥𝑥7, 6𝑥𝑥5, 𝟒𝟒𝒙𝒙𝟖𝟖, 𝟕𝟕𝒙𝒙𝟔𝟔, 𝟓𝟓𝒙𝒙𝟗𝟗, 𝟖𝟖𝒙𝒙𝟕𝟕 The coefficient equals the previous exponent minus 1: 2 – 1 = 1, 5 – 1 = 4, 3 – 1 = 2, 6 – 1 = 5, 4 – 1 = 3, etc. The exponent equals the previous coefficient plus 2: 3 + 2 = 5, 1 + 2 = 3, 4 + 2 = 6, 2 + 2 = 4, 5 + 2 = 7, etc. (13) 𝑥𝑥, 2𝑥𝑥1/2, 𝑥𝑥2 2 , 2𝑥𝑥 3 2 3 , 𝑥𝑥5 5 , 2𝑥𝑥 5 2 5 , 𝑥𝑥10 10 , 2𝑥𝑥 7 2 7 , 𝑥𝑥17 17 , 2𝑥𝑥 9 2 9 , 𝑥𝑥26 26 , 2𝑥𝑥 11 2 11 , 𝑥𝑥37 37 , 𝟐𝟐𝒙𝒙 𝟏𝟏𝟏𝟏 𝟐𝟐 𝟏𝟏𝟏𝟏, 𝒙𝒙𝟓𝟓𝟓𝟓 𝟓𝟓𝟓𝟓, 𝟐𝟐𝒙𝒙 𝟏𝟏𝟏𝟏 𝟐𝟐 𝟏𝟏𝟏𝟏, 𝒙𝒙𝟔𝟔𝟔𝟔 𝟔𝟔𝟔𝟔 Two sequences are merged together. Get the powers of one sequence by adding 1, adding 3, adding 5, adding 7, etc.: 1 + 1 = 2, 2 + 3 = 5, 5 + 5 = 10, 10 + 7 = 17, 17 + 9 = 26, 26 + 11 = 37, 37 + 13 = 50, etc. Get the powers of the other sequence by adding 1: 1 2 + 1 = 3 2 , 3 2 + 1 = 5 2 , 5 2 + 1 = 7 2 , 7 2 + 1 = 9 2, etc. In either case, the coef-ficient is the reciprocal of the exponent: (1)−1 = 1, ቀ 1 2ቁ −1 = 2, (2)−1 = 1 2, ቀ 3 2ቁ −1 = 2 3, (5)−1 = 1 5, ቀ 5 2ቁ −1 = 2 5, (10)−1 = 1 10, etc. (14) 8𝑥𝑥 1 8 3 , 2𝑥𝑥1/4, 8𝑥𝑥 3 8 5 , 4𝑥𝑥 1 2 3 , 8𝑥𝑥 5 8 7 , 𝑥𝑥3/4, 8𝑥𝑥 7 8 9 , 𝟒𝟒𝒙𝒙 𝟓𝟓, 𝟖𝟖𝒙𝒙𝟗𝟗/𝟖𝟖 𝟏𝟏𝟏𝟏, 2𝒙𝒙 𝟓𝟓 𝟒𝟒 𝟑𝟑, 𝟖𝟖𝒙𝒙 𝟏𝟏𝟏𝟏 𝟖𝟖 𝟏𝟏𝟏𝟏 Add 1 8 to each power ( 1 8 + 1 8 = 1 4 , 1 4 + 1 8 = 3 8 , 3 8 + 1 8 = 1 2 , 1 2 + 1 8 = 5 8, etc.). To get the coefficient, add 1 4 to the power and find its reciprocal. For example, 1 8 + 1 4 = 3 8 and ቀ 3 8ቁ −1 = 8 3. Similarly, 1 4 + 1 4 = 1 2 and ቀ 1 2ቁ −1 = 2. (15) 2𝑥𝑥+ 1, 𝑥𝑥−3, 3𝑥𝑥−2, 4𝑥𝑥−5, 7𝑥𝑥−7, 11𝑥𝑥− 12, 18𝑥𝑥−19, 29𝑥𝑥−31, 𝟒𝟒𝟒𝟒𝒙𝒙−𝟓𝟓𝟓𝟓, 𝟕𝟕𝟕𝟕𝒙𝒙−𝟖𝟖𝟖𝟖, 𝟏𝟏𝟏𝟏𝟏𝟏𝒙𝒙−𝟏𝟏𝟏𝟏𝟏𝟏, 𝟏𝟏𝟏𝟏𝟏𝟏𝒙𝒙−𝟐𝟐𝟐𝟐𝟐𝟐 Add consecutive expressions: 2𝑥𝑥+ 1 + 𝑥𝑥−3 = 3𝑥𝑥−2, 𝑥𝑥− 3 + 3𝑥𝑥−2 = 4𝑥𝑥−5, 3𝑥𝑥−2 + 4𝑥𝑥−5 = 7𝑥𝑥−7, etc. (16) 𝑥𝑥, 𝑦𝑦, 𝑥𝑥2, 𝑥𝑥𝑥𝑥, 𝑦𝑦2, 𝑥𝑥3, 𝑥𝑥2𝑦𝑦, 𝑥𝑥𝑦𝑦2, 𝑦𝑦3, 𝒙𝒙𝟒𝟒, 𝒙𝒙𝟑𝟑𝒚𝒚, 𝒙𝒙𝟐𝟐𝒚𝒚𝟐𝟐, 𝒙𝒙𝒚𝒚𝟑𝟑 These are the powers of the variable terms when you foil (𝑥𝑥+ 𝑦𝑦) raised to an integer power. For example, (𝑥𝑥+ 𝑦𝑦)1 = 𝑥𝑥+ 𝑦𝑦, (𝑥𝑥+ 𝑦𝑦)2 = 𝑥𝑥2 + 2𝑥𝑥𝑥𝑥+ 𝑦𝑦2, and (𝑥𝑥+ 𝑦𝑦)3 = 𝑥𝑥3 + 3𝑥𝑥2𝑦𝑦+ 3𝑥𝑥𝑦𝑦2 + 𝑦𝑦3. (17) 2𝑥𝑥5, 7𝑥𝑥3, 5𝑥𝑥11, 15𝑥𝑥7, 11𝑥𝑥23, 27𝑥𝑥13, 17𝑥𝑥35, 𝟑𝟑𝟑𝟑𝒙𝒙𝟏𝟏𝟏𝟏, 𝟐𝟐𝟐𝟐𝒙𝒙𝟒𝟒𝟒𝟒, 𝟓𝟓𝟓𝟓𝒙𝒙𝟐𝟐𝟐𝟐, 𝟑𝟑𝟑𝟑𝒙𝒙𝟔𝟔𝟔𝟔 The coefficients and exponents are alternately prime numbers (see Chapter 8): 2, 3, 5, 7, 11, 13, etc. The other number (coefficient or exponent) is twice the prime number plus one (2 × 2 + 1 = 5, 3 × 2 + 1 = 7, 5 × 2 + 1 = 11, 7 × 2 + 1 = 15, 11 × 2 + 1 = 23, etc.). (18) 3𝑥𝑥+ 4 = 𝑥𝑥2, 3𝑥𝑥= 𝑥𝑥2 −4, 6𝑥𝑥= 2𝑥𝑥2 −8, 6𝑥𝑥−4 = 2𝑥𝑥2 −12, 12𝑥𝑥−8 = 4𝑥𝑥2 −24, 12𝑥𝑥−12 = 4𝑥𝑥2 −28, 24𝑥𝑥−24 = 8𝑥𝑥2 −56, 24𝑥𝑥−28 = 8𝑥𝑥2 −60, 𝟒𝟒𝟒𝟒𝒙𝒙−𝟓𝟓𝟓𝟓= 𝟏𝟏𝟏𝟏𝒙𝒙𝟐𝟐−𝟏𝟏𝟏𝟏𝟏𝟏, 𝟒𝟒𝟒𝟒𝒙𝒙−𝟔𝟔𝟔𝟔= 𝟏𝟏𝟏𝟏𝒙𝒙𝟐𝟐−𝟏𝟏𝟏𝟏𝟏𝟏, 𝟗𝟗𝟗𝟗𝒙𝒙−𝟏𝟏𝟏𝟏𝟏𝟏= 𝟑𝟑𝟑𝟑𝒙𝒙𝟐𝟐−𝟐𝟐𝟐𝟐𝟐𝟐, 𝟗𝟗𝟗𝟗𝒙𝒙−𝟏𝟏𝟏𝟏𝟏𝟏= 𝟑𝟑𝟑𝟑𝒙𝒙𝟐𝟐−𝟐𝟐𝟐𝟐𝟐𝟐 Subtract 4 from both sides, multiply both sides by 2, subtract 4 from both sides, multiply by 2, repeating this pattern. For example: 3𝑥𝑥+ 4 = 𝑥𝑥2 Subtract 4: 3𝑥𝑥+ 4 −4 = 𝑥𝑥2 −4 3𝑥𝑥= 𝑥𝑥2 −4 Multiply by 2: 2(3𝑥𝑥) = 2(𝑥𝑥2 −4) 6𝑥𝑥= 2𝑥𝑥2 −8 Subtract 4: 6𝑥𝑥−4 = 2𝑥𝑥2 −8 −4 6𝑥𝑥−4 = 2𝑥𝑥2 −12 Multiply by 2: 2(6𝑥𝑥−4) = 2(2𝑥𝑥2 −12) 12𝑥𝑥−8 = 4𝑥𝑥2 −24 (19) 𝑥𝑥3, 𝑥𝑥4 + 3𝑥𝑥3, 𝑥𝑥5 + 4𝑥𝑥4 + 5𝑥𝑥3, 𝑥𝑥6 + 5𝑥𝑥5 + 6𝑥𝑥4 + 11𝑥𝑥3, 𝑥𝑥7 + 6𝑥𝑥6 + 7𝑥𝑥5 + 13𝑥𝑥4 + 20𝑥𝑥3, 𝑥𝑥8 + 7𝑥𝑥7 + 8𝑥𝑥6 + 15𝑥𝑥5 + 23𝑥𝑥4 + 38𝑥𝑥3, 𝒙𝒙𝟗𝟗+ 𝟖𝟖𝒙𝒙𝟖𝟖+ 𝟗𝟗𝒙𝒙𝟕𝟕+ 𝟏𝟏𝟏𝟏𝒙𝒙𝟔𝟔+ 𝟐𝟐𝟐𝟐𝒙𝒙𝟓𝟓+ 𝟒𝟒𝟒𝟒𝒙𝒙𝟒𝟒+ 𝟔𝟔𝟔𝟔𝒙𝒙𝟑𝟑, 𝒙𝒙𝟏𝟏𝟏𝟏+ 𝟗𝟗𝒙𝒙𝟗𝟗+ 𝟏𝟏𝟏𝟏𝒙𝒙𝟖𝟖+ 𝟏𝟏𝟏𝟏𝒙𝒙𝟕𝟕+ 𝟐𝟐𝟐𝟐𝒙𝒙𝟔𝟔+ 𝟒𝟒𝟒𝟒𝒙𝒙𝟓𝟓+ 𝟕𝟕𝟕𝟕𝒙𝒙𝟒𝟒+ 𝟏𝟏𝟏𝟏𝟏𝟏𝒙𝒙𝟑𝟑 Raise the first power by 1 (𝑥𝑥3, 𝑥𝑥4, 𝑥𝑥5, etc.). The coefficient of the second term is 1 less than the power of the first term (4 – 1 = 3, 5 – 1 = 4, 6 – 1 = 5, etc.). The power of each term decreases by 1. For the third coefficients and onward, add the last 2 coefficients together (for example, with 𝑥𝑥7 + 6𝑥𝑥6 + 7𝑥𝑥5 + 13𝑥𝑥4 + 20𝑥𝑥3, 1 + 6 = 7, 6 + 7 = 13, and 7 + 13 = 20). (20) 1, 𝑥𝑥+ 𝑦𝑦, 𝑥𝑥2 + 2𝑥𝑥𝑥𝑥+ 𝑦𝑦2, 𝑥𝑥3 + 3𝑥𝑥2𝑦𝑦+ 3𝑥𝑥𝑦𝑦2 + 𝑦𝑦3, 𝑥𝑥4 + 4𝑥𝑥3𝑦𝑦+ 6𝑥𝑥2𝑦𝑦2 + 4𝑥𝑥𝑦𝑦3 + 𝑦𝑦4, 𝑥𝑥5 + 5𝑥𝑥4𝑦𝑦+ 10𝑥𝑥3𝑦𝑦2 + 10𝑥𝑥2𝑦𝑦3 + 5𝑥𝑥𝑦𝑦4 + 𝑦𝑦5, 𝑥𝑥6 + 6𝑥𝑥5𝑦𝑦+ 15𝑥𝑥4𝑦𝑦2 + 20𝑥𝑥3𝑦𝑦3 + 15𝑥𝑥2𝑦𝑦4 + 6𝑥𝑥𝑦𝑦5 + 𝑦𝑦6, 𝒙𝒙𝟕𝟕+ 𝟕𝟕𝒙𝒙𝟔𝟔𝒚𝒚+ 𝟐𝟐𝟐𝟐𝒙𝒙𝟓𝟓𝒚𝒚𝟐𝟐+ 𝟑𝟑𝟑𝟑𝒙𝒙𝟒𝟒𝒚𝒚𝟑𝟑+ 𝟑𝟑𝟑𝟑𝒙𝒙𝟑𝟑𝒚𝒚𝟒𝟒+ 𝟐𝟐𝟐𝟐𝒙𝒙𝟐𝟐𝒚𝒚𝟓𝟓+ 𝟕𝟕𝒙𝒙𝒚𝒚𝟔𝟔+ 𝒚𝒚𝟕𝟕, 𝒙𝒙𝟖𝟖+ 𝟖𝟖𝒙𝒙𝟕𝟕𝒚𝒚+ 𝟐𝟐𝟐𝟐𝒙𝒙𝟔𝟔𝒚𝒚𝟐𝟐+ 𝟓𝟓𝟓𝟓𝒙𝒙𝟓𝟓𝒚𝒚𝟑𝟑+ 𝟕𝟕𝟕𝟕𝒙𝒙𝟒𝟒𝒚𝒚𝟒𝟒+ 𝟓𝟓𝟓𝟓𝒙𝒙𝟑𝟑𝒚𝒚𝟓𝟓+ 𝟐𝟐𝟐𝟐𝒙𝒙𝟐𝟐𝒚𝒚𝟔𝟔+ 𝟖𝟖𝒙𝒙𝒚𝒚𝟕𝟕+ 𝒚𝒚𝟖𝟖 Multiply (𝑥𝑥+ 𝑦𝑦) raised to an integer power. For exam-ple, (𝑥𝑥+ 𝑦𝑦)1 = 𝑥𝑥+ 𝑦𝑦, (𝑥𝑥+ 𝑦𝑦)2 = 𝑥𝑥2 + 2𝑥𝑥𝑥𝑥+ 𝑦𝑦2, and (𝑥𝑥+ 𝑦𝑦)3 = 𝑥𝑥3 + 3𝑥𝑥2𝑦𝑦+ 3𝑥𝑥𝑦𝑦2 + 𝑦𝑦3. 16 Visual (1) The center triangle is always black. A second triangle appears black on the left, right, top, left, right, top, etc. (2) Rotate counterclockwise 90°. (3) Two patterns are merged together: One is triangle, square, pentagon, hexagon, heptagon, etc., while the other is always a pentagon. Also, every third shape is gray. (4) Rotate counterclockwise 90°. Subtract the right number from the left number to get the top number: 5 – 2 = 3, 6 – 5 = 1, 11 – 7 = 4, and 13 – 5 = 8. (6) Copy the triangle and flip it upside down, joining this flipped triangle to the original to make the diamond. Do the same thing to the square to get the rectangle, and to the pentagon to find the answer. (7) Draw the top of a T, add the bottom of the T, add the top of a new smaller T to the right of the first one, add the bottom of this T, add the top of a third yet smaller T to the right of the previous T, etc. 5 2 3 6 5 1 11 7 4 13 5 8 2 8 4 3 5 15 28 4 7 4 36 9 (5) (8) Multiply two numbers to get the third number. The third number appears at the right, then the top, then the left, then the right, etc. (9) Rotate 90° counterclockwise. (10) Rotate 90° about a vertical axis. (11) Take one step right. Take one step down. Take a second step down. Take one step right. Take one step down. Take a second step down. Repeat this pattern. (12) Cut the width in half, cut the height in half, cut the width in half, cut the height in half, etc. Also, shade white, gray, black, white, gray, black, etc. (13) Rotate 90° about a horizontal axis. (14) The black square advances 2 spaces along a clockwise direction. Ignoring the black square, the pattern flips horizontally, vertically, horizontally, vertically, and so on. If the black square coincides with a gray square (as in the fifth and sixth diagrams), the square appears black. (15) Black out the top row, then the middle row, then the bottom row, then the top row again, etc. The left column is blacked out for the first three, the middle column is blacked out for the second three, and the right column is blacked out for the next three (only 2 of these are shown). (16) Multiply the top numbers and subtract the bottom left number to get the bottom right number: 3 × 4 – 5 = 7, 1 × 9 – 3 = 6, 5 × 6 – 10 = 20, and 7 × 9 – 13 = 50. 3 4 5 7 7 9 13 50 5 6 10 20 20 20 1 9 3 6 2 8 7 9 (17) The black square advances 2 spaces along a clockwise path. The gray squares advances 1 space along a smaller clockwise path. (18) One arrow rotates 135° clockwise (↖,→,↙,↑,↘, and ←), while the other arrow rotates 90° counterclockwise (→,↑,←,↓,→, and ↑). (19) The vertical line of 3 squares remains constant. The fourth square moves along a counterclockwise path (down, down, over to the bottom, over to the bottom right, up, etc.). (20) Rotate the right half with the top coming forward, rotate the top half with the front going to the right, and repeat this pair of rotations over and over. (In the original cube, the left and right faces are black, the front and back faces are white, and the top and bottom faces are gray.) 17 Arrays (1) 8 16 24 32 40 48 56 64 72 Multiples of 8 appear in order from left to right, top to bottom: 8 × 1 = 8, 8 × 2 = 16, 8 × 3 = 24, 8 × 4 = 32, 8 × 5 = 40, etc. (2) 95 84 73 62 51 40 29 18 7 Subtract 11, working left to right, top to bottom: 95 – 11 = 84, 84 – 11 = 73, 73 – 11 = 62, 62 – 11 = 51, etc. (3) 3 6 12 4 8 16 5 10 20 Double the left column to make the middle column (3 × 2 = 6, 4 × 2 = 8, and 5 × 2 = 10). Double the middle column to make the right column (6 × 2 = 12, 8 × 2 = 16, and 10 × 2 = 20). (4) 2 7 17 3 11 19 5 13 23 Prime numbers (Chapter 8) are arranged top to bottom, left to right: 2, 3, 5, 7, 11, 13, 17, 19, and 23. (5) 4 3 8 9 5 1 2 7 6 This is a magic square: The numbers of every row (4 + 3 + 8 = 15, 9 + 5 + 1 = 15, and 2 + 7 + 6 = 15), every column (4 + 9 + 2 = 15, 3 + 5 + 7 = 15, and 8 + 1 + 6 = 15), and every diagonal (4 + 5 + 6 = 15 and 2 + 5 + 8 = 15) add up to 15. (6) 4 6 9 11 16 18 25 27 36 38 49 51 64 66 81 83 The first and third columns are perfect squares: 22 = 4, 32 = 9, 42 = 16, 52 = 25, 62 = 36, etc. The second and fourth columns are 2 more than the previous columns. For example, 4 + 2 = 6, 9 + 2 = 11, 16 + 2 = 18, and 25 + 2 = 27. (7) The pattern repeats over and over from left to right, top to bottom. (8) 14 17 19 20 10 13 16 18 7 9 12 15 5 6 8 11 The numbers 5 thru 20 appear in order (5, 6, 7, 8, 9, etc.). After 5, the next diagonal is 6 and 7, followed by 8, 9, and 10, etc. (9) 1 1 0 0 2 8 7 14 3 27 26 52 4 64 63 126 The second column equals the first column cubed (13 = 1, 23 = 8, 33 = 27, and 43 = 64). The third column is one less than the second column (1 – 1 = 0, 8 – 1 = 7, 27 – 1 = 26, and 64 – 1 = 63). The last column is twice the third column (0 × 2 = 0, 7 × 2 = 14, 26 × 2 = 52, and 63 × 2 = 126). (10) II IV VIII XVI XXXII LXIV CXXVIII CCLVI DXII Powers of 2 appear in Roman numerals (Chapter 10): 21 = 2 (II), 22 = 4 (IV), 23 = 8 (VIII), 24 = 16 (XVI), 25 = 32 (XXXII), etc. (11) 1 2 5 6 4 3 8 7 9 10 13 14 12 11 16 15 The numbers 1-16 appear in order. Divide the large 4×4 square into 4 smaller 2×2 squares. Write the numbers 1-4 in the top left 2×2 square in a clockwise fashion. Write the numbers 5-8 in the top right 2×2 square in a clockwise fashion. Write the numbers 9-12 in the bottom left 2×2 square in a clockwise fashion. Similarly, write the numbers 13-16 in the bottom right 2×2 square in a clockwise fashion. (12) 2 3 5 9 2049 4097 8193 17 1025 32769 16385 33 513 257 129 65 Beginning with 2, multiply the previous number by 2 and subtract 1: 2 × 2 – 1 = 3, 3 × 2 – 1 = 5, 5 × 2 – 1 = 9, 9 × 2 – 1 = 17, 17 × 2 – 1 = 33, 33 × 2 – 1 = 65, etc. Write these numbers in a clockwise spiral from outside to inside. (13) 5 8 14 17 23 26 32 35 41 44 50 Add 3 to each number, working left to right, top to bottom: 5 + 3 = 8, 8 + 3 = 11, 11 + 3 = 14, 14 + 3 = 17, 17 + 3 = 20, etc. When a number coincides with a black square, simply don’t write it (11, 20, 29, 38, and 47 are hidden by black squares). (14) 1 2 3 4 2 3 5 7 3 5 8 12 4 8 13 20 Add the number directly above a cell to the number diagonally up and to the left of the cell. For example, 1 + 2 = 3, 2 + 3 = 5, 3 + 4 = 7, 3 + 5 = 8, and 5 + 7 = 12. This last example is illustrated below: 5 7 8 12 (15) 2 1 1 8 5 3 34 21 13 The Fibonacci sequence (Chapter 9) appears right to left, top to bottom: 1 + 1 = 2, 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8, 5 + 8 = 13, 13 + 8 = 21, and 13 + 21 = 34.
18 Analogies (1) three : 27 :: four : 64 (D) 64 Raise the first number to the third power: 33 = 27 and 43 = 64. (2) Rotate the first shape 90° clockwise. (3) hand : wrist :: foot : ankle (A) ankle The wrist joins the hand to the arm, while the ankle joins the foot to the leg. (4) 81 : 9 :: 36 : 6 (C) 6 Squareroot the first number: √81 = 9 and √36 = 6. (Take only the positive root.) (5) century : year :: meter : centimeter (A) centimeter A century lasts 100 times longer than a year, while a meter is 100 times longer than a centimeter. (There are 100 years in one century, and 100 centimeters in one meter.) (6) ADE : 145 :: CFH : 368 : : :: (C) (C) 368 A is the 1st letter of the alphabet, D is the 4th letter, and E is the 5th letter. Put these together to make 145. Similarly, C is the 3rd letter, F is the 6th letter, and H is the 8th letter, which makes 368. (7) Flip the first image about a vertical axis: The second image is the reflection of the first. (Tip: hold this page up to a mirror.) (8) moon : earth :: Venus : sun (C) sun The moon orbits the earth, while Venus orbits the sun. (9) circle : circumference :: square : perimeter (D) perimeter The distance around a circle is its circumference, while the distance around a square is its perimeter. (10) (3, 4, 5) : 35 :: (6, 7, 8) : 104 (D) 104 Add the first two numbers and multiply by the third: (3 + 4) × 5 = 7 × 5 = 35 and (6 + 7) × 8 = 13 × 8 = 104. (11) uncle : nephew :: aunt : niece (D) niece Here is an example: John’s brother is Kyle. Jane’s sister is Kate. John and Jane get married, and have two kids, Luke and Lisa. : : :: (C) Kyle is Luke’s uncle; Luke is Kyle’s nephew. Kate is Lisa’s aunt; Lisa is Kate’s niece. (Luke’s uncle is his father’s brother; Kyle’s nephew is his brother’s son. Lisa’s aunt is her mother’s sister; Kate’s niece is her sister’s daughter.) (12) (10, 11) : 101 :: (100, 101) : 1001 (C) 1001 The binary numbers 10 and 11 are the numbers 2 and 3 in the decimal system. Add 2 and 3 to make 5, which is 101 in binary. The binary numbers 100 and 101 equate to 4 and 5 in decimal form. Add 4 and 5 to make 9, which is 1001 in binary. Note: Binary numbers are made of 0’s and 1’s. The following table shows the relationship between decimal numbers and binary numbers: Decimal Binary 1 1 2 10 3 11 4 100 5 101 6 110 7 111 8 1000 9 1001 10 1010 (13) horse : foal :: deer : fawn (C) fawn A baby horse is called a foal, while a baby deer is called a fawn. (14) rectangle : square :: parallelogram : rhombus (C) rhombus If you make the sides of a rectangle have equal length, you get a square. Similarly, if you make the sides of a parallelogram have equal length, you get a rhombus. (15) 5 8 : 1 1 4 :: 3 4 : 1 1 2 (B) 1 1 2 Double the first number: 5/8 × 2 = 10/8 = 5/4 = 1 + 1/4 and 3/4 × 2 = 6/4 = 3/2 = 1 + 1/2.
(16) Rotate the right half with the top coming to the front, then rotate the bottom half with the front going to the right. (The simpler solution is to change white to black, gray to white, and black to gray.) (17) ቀ8, 2 3ቁ : 4 :: ቀ81, 3 4ቁ : 27 (D) 27 Raise the first number to the power of the fraction: (8)2/3 = 4 and (81)3/4 = 27. (Take the positive root in the latter case.) (18) COCCOON : C3O3N : : COCOA : C2O2A (B) C2O2A The word COCCOON has 3 C’s, 3 O’s, and 1 N, forming C3O3N. The word COCOA has 2 C’s, 2 O’s, and 1 A, : : :: (A) forming C2O2A. (These are not chemical formulas, even though they might resemble such.) (19) circle : sphere : : square : cube (A) cube Here is one way to look at this: If you slice a sphere down the middle, you obtain a circular cross section. If you slice a cube down the middle parallel to one face, you obtain a square cross section. (20) PUZZLES : OTYYKDR :: PATTERNS : OZSSDQMR (A) OZSSDQMR Replace each letter with the letter that comes before it in the alphabet. For example, replace B with A, replace C with B, and replace D with C. Also, replace A with Z. In the word PATTERNS, P becomes O, A becomes Z, the T’s become S’s, the E becomes D, the R becomes Q, the N becomes M, and the S becomes R, forming OZSSDQMR. (This is a cryptogram.) About the Author Chris McMullen is a physics instructor at Northwestern State University of Louisiana and also an author of academic books. Whether in the classroom or as a writer, Dr. McMullen loves sharing knowledge and the art of motivating and engaging students. He earned his Ph.D. in phenomenological high-energy physics (particle physics) from Oklahoma State University in 2002. Originally from California, Dr. McMullen earned his Master's degree from California State University, Northridge, where his thesis was in the field of electron spin resonance. As a physics teacher, Dr. McMullen observed that many students lack fluency in fundamental math skills. In an effort to help students of all ages and levels master basic math skills, he published a series of math workbooks on arithmetic, fractions, and algebra called the Improve Your Math Fluency Series. Dr. McMullen has also published a variety of science books, including introductions to basic astronomy and chemistry concepts in addition to physics textbooks. Dr. McMullen is very passionate about teaching. Many students and observers have been impressed with the transformation that occurs when he walks into the classroom, and the interactive engaged discussions that he leads during class time. Dr. McMullen is well-known for drawing monkeys and using them in his physics examples and problems, using his creativity to inspire students. A stressed out student is likely to be told to throw some bananas at monkeys, smile, and think happy physics thoughts. Improve Your Math Fluency This series of math workbooks is geared toward practicing essential math skills: • Algebra and trigonometry • Fractions, decimals, and percentages • Long division • Multiplication and division • Addition and subtraction Science Books Dr. McMullen has published a variety of science books, including: • Basic astronomy concepts • Basic chemistry concepts • Creative physics problems • Calculus-based physics Puzzle Books Chris McMullen enjoys solving puzzles. His favorite puzzle is Kakuro (kind of like a cross between cross-word puzzles and Sudoku). He once taught a three-week summer course on puzzles. In addition to the pattern puzzle book that you are reading right now, Chris McMullen has coauthored several word scramble books. This includes a cool idea called VErBAl ReAcTiONS. A VErBAl ReAcTiON expresses word scrambles so that they look like chemical reactions. Here is an example: 2 C + U + 2 S + Es → S U C C Es S The left side of the reaction indicates that the answer has 2 C’s, 1 U, 2 S’s, and 1 Es. Rearrange CCUSSEs to form SUCCEsS. Each answer is not merely a word, it’s a chemical word. A chemical word is made up not of letters, but of elements of the periodic table. In this case, SUCCEsS is made up of sulfur (S), uranium (U), carbon (C), and Einsteinium (Es). Another example of a chemical word is GeNiUS. It’s made up of germanium (Ge), nickel (Ni), uranium (U), and sulfur (S). If you enjoy anagrams and like science or math, these puzzles are tailor-made for you. |
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NH PBS - NatureWorks - Tracheophytes
The George Washington University - Trachelophytes - Vascular Plants
Paleontological Research Institution - Digital Atlas of Ancient Life - Introduction to Vascular Plant Structure
Biology LibreTexts - Seedless Vascular Plants
PNAS - Evolution of vascular plants through redeployment of ancient developmental regulators
Milne Library - Inanimate Life - Vascular plant anatomy: primary growth
The Royal Society - The relationships of vascular plants (PDF)
National Center for Biotechnology Information - PubMed Central - Plant vascular development: mechanisms and environmental regulation
National Park Service - Glen Canyon National Recreation Area - Vascular Plant
Also known as: Tracheophyta, tracheophyte
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Article History
Also called:
: tracheophyte
Key People:
: Edward Charles Jeffrey
: George Bentham
: Douglas Houghton Campbell
Related Topics:
: lower vascular plant
: seed plant
: lycopsid
: sphenopsid
: psilopsid
See all related content
vascular plant, any of some 260,000 species of plants with vascular systems, including all of the conspicuous flora of Earth today. Plant vascular systems consist of xylem, concerned mainly with the conduction of water and dissolved minerals, and phloem, which functions mainly in the conduction of foods, such as sugar. Tracheophyte, meaning “tracheid plant,” refers to the water-conducting cells (called tracheids, or tracheary elements) that show spiral bands like those in the walls of the tracheae, or air tubes, of insects.
Formerly a taxonomic division or phylum, the group comprises a tremendous diversity of plants, including the majority of seedless plants and all seed plants. The lower vascular plants, all of which are herbaceous, can be divided into two groups of seedless plants: the lycophytes (club mosses, spike mosses, and quillworts) and the ferns (including horsetails and whisk ferns). The two groups of seed-bearing plants are the gymnosperms (cycads, ginkgo, pines and other conifers, and gnetophytes) and the angiosperms (flowering plants). See bryophyte for plants that lack vascular systems.
Tracheophytes are believed to have originated from the green algae (Chlorophyta). The earliest fossils are from Silurian rocks more than 400,000,000 years old.
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Plants: From Cute to Carnivorous
The Editors of Encyclopaedia BritannicaThis article was most recently revised and updated by Melissa Petruzzello. |
1196 | https://www.youtube.com/watch?v=l_khO0MLGKQ | Statics 7.61 - Draw the shear and moment diagrams for the beam.
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Posted: 8 Jul 2021
Question: Draw the shear and moment diagrams for the beam.
Problem 7-61 from:
Engineering Mechanics: Statics, 14th edition
Russell C. Hibbeler
Thank you guys for watching.
Please let me know if you have any comments or suggestions and also if you have any other problem you would like to go over with.
48 comments
Transcript:
Intro welcome back everyone to learning edition today we're in aesthetics and we're going to solve this problem 761 okay it says draw the sheer moment diagrams for them okay so we're Free body diagram going to apply the same method that i used for the last video so basically we're not following the method that bulk uses so in order to do this what we're going to do is that we're going to draw our free body diagram and we're going to try to simplify our beam as much as we can with our reaction forces okay so let's start by drawing the free body diagram so we will have our beaming here we will have our 20 kips going down another 20 kips going down at the other end then we will have our reaction forces that will have a roller at a we'll have a1 then we'll have a p net b okay so we'll have b y and we also will have bx so we'll have it like this a fourth in here but as you guys can see none of the other forces are going into the x directions meaning that this bx is going to be equal to zero so we can totally ignore it in this case then we'll have the distributed load that goes from a to b okay it's high it's 4 kips per feet okay so that's our distributed load and our distances are from my end to my point a it's 15 feet from a to b it's equal to 30 feet and then from b to my other end will be equal to 15 feet okay so this is basically our first free body diagram and what we can do with it is that we're going to apply some material forces like any type of static problem right so we got uh summative forces in the y we're going to assume that going up is positive and what do we have for we have negative 20 kips then we'll have positive a y then we'll have positive by i'm gonna leave the distributed load for the n then i'll have negative 20 kips and then i'll have negative the distributed level how much is this distributed level we need to know the area right so we know that the height is four and the distance that is recurring is 30 feet okay so we'll have four times three that'll give me 12 we add the zero so it'll be 120. okay and all this should be equal to zero if we solve for a y plus b y we will end up having 120 minus so negative 120 minus 40 that's minus 160 if we move it to the other side will be positive 160 caps okay so i said minus 40 because i got negative 20 and negative 20 that's 40 okay now the next thing that i can do is the summary of moments around either point a or point b sum in the counter clockwise is positive and i'm just going to pick b just because okay so we're going to choose this point over here my point b for my moment so let's just start with this 20 kips holding here and this for like that will want to rotate me counterclockwise therefore is positive 20 multiplied by the distance so the distance will be 30 plus 15 that will be equal to 45 then i will have a y since it's in the opposite direction then i'll have negative a y multiplied by the distance which is equal to 30 feet that one we can clearly see it then we'll have our distributed load for our distributed load i'm going to simplify it just for this step but we're not going to simplify for our further steps so if we simplify this it will be a float in the middle which will be equal to these 120 that we already calculated okay so we can basically say that if it's since it's going down the same as here so going counterclockwise positive so i'll have positive 120 and the distance with respect to my point b should be equal to half of our distance which is steady so half of that which is equal to 15 okay now my force b y doesn't have um distance respect to my point b and then i'll have this 20 kilopascal that keeps kilopounds to going here so you want to rotate me clockwise so i have negative 20 multiplied by the distance which is another 15 feet all this should be equal to zero okay so let's try to simplify these numbers and what we'll have is that we'll have 20 times 45 that will be equal to 900 then i'll have negative 30 a y then i'll have positive 120 times 15 that will give me 1800 and then all that negative 20 times 15 that will give me 300 and all this should be equal to zero let's solve for a y now what do we have well we'll have 900 so we got 900 plus 1800 minus 1300 divided by 30 okay so if we do this we'll end up having the a y is equal to 80 hips as well okay so [Music] from here we can find b y so basically b y how can we find b y using this equation in here we'll end up finding that this is 80 kips 2 okay because 80 plus 80 is equal to 160 cubes and what we're going to do is that we're going to take this free body diagram without simplifying without simplifying our distributed load and we're gonna copy it in here we're going to replace the values for a y and b y for the values that we know which are 80 and 80 keeps each one okay and by knowing this now we're going to use this values that we know and we're going to have Shear diagram key parts in order to draw our free body diagram and you know and what these key parts i mean is like every single time i see something interesting on this beam i will draw a line and you will see guys why so i'll draw a line in here because i see these negative 20 pounds going in there same goes at the end where i got this 20 cabs at the end i have 80 keeps going up in this point and i have 80 keeps going up in this point okay and in between these two guys i have this really load so i'm going to leave it like this for now and let's start by drawing the shear diver okay so the shear diagram should look something like what so we're going to start in here at zero but then right away we end up with negative 20 cubs so i'm going to draw from 0 to negative 20 gibbs so i'm going to place negative 20 then nothing happens so i keep on running running running running and nothing happens until here so i'm going to keep those negative 20 all the way until that point okay then after that right away i go up 80 kips so i'm going to go up 80. so if we were at negative 20 we go up 80 we should have positive 60. so we're here at positive 60 okay so this is 60. now if we pay attention we have a distributed load that is starting to go down so and we know that the total of this is 120. so we slowly will go down until this point and we're going to go down 120. so if i was at positive 60 and then i go 120 down i should get up in here to negative 60. so let's do that so we're going to do this and we're going to have negative 16 here now after i end up with my distributor load right away i have these 80 kilo pounds going up so if i'm negative 60 and i go up 80 i should end up a positive 20. okay then i keep on running running running running into this beam until i find these negative 20 keeps so i'm gonna go running until i go down 20 games now how do we know if we did this diagram correct we should always start at zero end up at zero okay so now that we know that this is our shear diagram and the last thing i like to do is that okay and here i got a negative um area in here i have a positive area also in here i have a positive area and then i got a negative area in here the next thing we're going to do is our moment diagram okay so this is my moment diagram so we will start with this negative area therefore my moment should start going down remember these is from to go from cheer to go to moment the moment is the integral of this cheer so if we have a negative and we the constant like in here negative 20 and we apply an integral we'll end up with a negative slope so we'll go from zero all the way to here how much is this well in order to know this we have to do we have to know how much is this area so that area i'm just gonna call it area one how much is that well we know the height is 20. so how much is the distance well the distance is giving here by 15 so we got 15 and if we plug this into the calculator which you get 300 kips times fifth okay so we'll end up that this over here is equal to 300 okay we can put the units later on then we'll have a positive area this area over here and how much is that well that area i'm going to call it area 2 will be equal to my the area of a triangle so have one half my height which is 60 times my base so how much is this base so let's go up so it's exactly in the middle of that 30 feet so it should be 15 feet multiply by 15 and if we plug this into our calculator that's 60 times 15 divided by 2 we get 450 okay so we got a positive area of 450 kips but remember we were already in a negative area so we have to go up into a positive so let's draw this line in here as well because we will need it to know where to stop and there is another part we already have a slope in here so this is loop is something like m x plus b right so it's just a linear slope the next slope will be um quadratic slope how do we know what type of quadratic is it like this or is it like this well it depends on how this triangle is positioned so in here we start having a lot of area a lot of area until we have almost little adding of area so since we have that what we would do is that we would do a lot of area that we're adding until we're almost not adding anything okay so that's how we draw these three parts these diagrams uh this um moment diagram in this case what we have is that we're going to start adding a little little little or in this case subtract a little little and then we will add a lot and then how much is this area well this area is the same so if we went up this amount of area and then we will go down this amount of area we should end up in the same place so we should end up in here now we'll remember we start little and then a lot at the end so have a little little little little going down and then a lot at the end okay we should end up at 300 and you may ask what is this peak pointing here well this peak point should be area 2 minus area 1 so we got 450 minus 300 that should be equal to 115 here so we should end up in 150 positive now guess what in here we have the positive but the same areas here same area but in positive direction so we'll have a similar shape by going positive therefore this is how my moment diagram should look first negative then we'll have a lot of positive and then we'll end up with negative and remember how do we know if this is correct we'll start at zero and double zero okay so this is all for this video guys i hope you guys like it if you guys have any questions or any comments please just post it down below and i'll see you guys in the next one you |
1197 | https://en.wikipedia.org/wiki/Wolf,_goat_and_cabbage_problem | Published Time: 2004-11-05T22:29:37Z
Wolf, goat and cabbage problem - Wikipedia
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1 The story
2 Solution
3 Occurrence and variations
4 See also
5 References
6 External links
Wolf, goat and cabbage problem
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From Wikipedia, the free encyclopedia
River crossing puzzle
Illuminated illustration depicting the wolf, goat and cabbage problem in the Ormesby Psalter, dating to 1250–1330
The wolf, goat, and cabbage problem is a river crossing puzzle. It dates back to at least the 9th century, and has entered the folklore of several cultures.
The story
[edit]
A farmer with a wolf, a goat, and a cabbage must cross a river by boat. The boat can carry only the farmer and a single item. If left unattended together, the wolf would eat the goat, or the goat would eat the cabbage. How can they cross the river without anything being eaten?
Solution
[edit]
The two solutions with the vertical axis denoting time, and brown, grey, green and beige paths denoting the wolf, goat, cabbage and boat, respectively
The first step that must be taken is to let the goat go across the river, as any other actions will result in the goat or the cabbage being eaten. When the farmer returns to the original side, he has the choice of taking either the wolf or the cabbage across next. If he takes the wolf across, he would have to return to get the cabbage, resulting in the wolf eating the goat. If he takes the cabbage across second, he will need to return to get the wolf, resulting in the cabbage being eaten by the goat. The dilemma is solved by taking the wolf (or the cabbage) over and bringing the goat back. Now he can take the cabbage (or the wolf) over, and finally return to fetch the goat.
An animation of the solution
His actions in the solution are summarized in the following steps:
Take the goat over
Return empty-handed
Take the wolf or cabbage over
Return with the goat
Take whichever wasn't taken in step 3 over
Return empty-handed
Take the goat over
There are seven crossings: four forward and three back.
The key to the solution is realizing that one can bring things back (emphasized above). This is often unclear from the wording of the story, but never forbidden. Knowing this will make the problem easy to solve even by small children. The focus of the puzzle is not just task scheduling, but creative thinking, similarly to the Nine dots puzzle.
Visualisation of the moves possible in the puzzle. Uppercase letters denote the Fox, Goose and Beans at the destination, and lowercase ones denote them at the origin. Movement of each object is represented by a coordinate axis. All the 8 valid and invalid placements are shown as vertices of a cube, and all 12 movements as its edges. Invalid moves are crossed out, leaving the 2 solutions shown in blue and purple.
Occurrence and variations
[edit]
The puzzle is one of a number of river crossing puzzles, where the object is to move a set of items across a river subject to various restrictions.
In the earliest known occurrence of this problem, in the medieval manuscript Propositiones ad Acuendos Juvenes, the three objects are a wolf, a goat, and a cabbage, but other cosmetic variations of the puzzle also exist, such as: wolf, sheep, and cabbage;, p. 26 fox, chicken, and grain; fox, goose and corn; and panther, pig, and porridge. The logic of the puzzle, in which there are three objects, A, B, and C, such that neither A and B nor B and C can be left together, remains the same.
Another version of the puzzle stemming from a Chinese legend is recorded in an 18th-century painted panel by Japanese artist Maruyama Ōkyo, in the collection of the British museum. According to the legend, when a tiger has three cubs, one of them will be a leopard rather than a tiger, and more fierce than the others. Following this legend, the subject of a tiger with her cubs became a traditional subject for art in east Asia. The depiction by Ōkyo shows the tiger family crossing a river, with the mother carrying one cub across the river at a time. This depicts a puzzle equivalent to the puzzle of the wolf, goat, and cabbage, asking how the mother can do this without leaving the leopard cub alone with any of the other tiger cubs. The same variation of the puzzle has also been recorded as a koan of Ryōan-ji, a Zen temple in Kyoto.
The puzzle has been found in the folklore of African-Americans, Cameroon, the Cape Verde Islands, Denmark, Ethiopia, Ghana, Italy, Romania, Russia, Scotland, the Sudan, Uganda, Zambia, and Zimbabwe., pp. 26–27; It has been given the index number H506.3 in Stith Thompson's motif index of folk literature, and is ATU 1579 in the Aarne–Thompson classification system.
The puzzle was a favorite of Lewis Carroll, and has been reprinted in various collections of recreational mathematics., p. 26.
In his 'Arabian Nights' memoir, Meetings with Remarkable Men, the metaphysical Magus, G. I. Gurdjieff cites this riddle as "The Wolf, the goat and the cabbage". He notes, "This popular riddle clearly shows that...not solely by means of the ingenuity which every normal man should have, but that in addition he must not be lazy nor spare his strength, but must cross the river extra times for the attainment of his aim."
Variations of the puzzle also appear in the adventure game Broken Sword: The Sleeping Dragon, the Nintendo DS puzzle game Professor Layton and the Curious Village, and in The Simpsons episode "Gone Maggie Gone", where Homer has to get across a river with Maggie, Santa's Little Helper, and a jar of rat poison that looks like candy. In the Class of 3000 episode "Westley Side Story", Sunny and his students perform a similar exercise involving a chicken, a coyote and a sack of corn. The Between the Lions episode "Farmer Ken's Puzzle" portrays it being made into a computer game with a cat, a hen, and a sack of seeds.
Interactive chicken, fox and grain problem.
In the Bull episode "Justice for Cable", Benny begins a riddle with "a man has a fox, a duck, and a bag of beans". Bull inexplicably declares "There is no answer", and everyone believes him.
In some parts of Africa, variations on the puzzle have been found in which the boat can carry two objects instead of only one. When the puzzle is weakened in this way it is possible to introduce the extra constraint that no two items, including A and C, can be left together., p. 27.
In the Star Trek: Prodigy episode "Time Amok", a holographic version of Kathryn Janeway employs the tale (here told as the chicken, fox and grain problem) to teach the crew of the USS Protostar how to work together.
See also
[edit]
Missionaries and cannibals problem
References
[edit]
^ Pressman, Ian; David Singmaster (June 1989). ""The Jealous Husbands" and "The Missionaries and Cannibals"". The Mathematical Gazette. 73 (464). The Mathematical Association: 73–81. doi:10.2307/3619658. JSTOR 3619658. S2CID 116924808.
^ Jump up to: a b c d e Ascher, Marcia (February 1990). "A River-Crossing Problem in Cross-Cultural Perspective". Mathematics Magazine. 63 (1). Mathematical Association of America: 26–29. doi:10.2307/2691506. JSTOR 2691506.
^ Gurdjieff, G. I. (1963). Meetings with Remarkable Men (1st English ed.). London: Routledge & Kegan Paul. pp. 4–5.
^ Alcuin's Transportation Problems and Integer Programming Archived 2011-07-19 at the Wayback Machine, Ralf Borndörfer, Martin Grötschel, and Andreas Löbel, preprint SC-95-27 (November 1995), Konrad-Zuse-Zentrum für Informationstechnik Berlin.
^ The Classic River Crossing Puzzle Archived 2008-06-17 at the Wayback Machine
^ Mary Jane Sterling, Math Word Problems for Dummies, p. 313
^ Stewart, Ian (1998). The Magical Maze. Phoenix. ISBN 0-7538-0514-6.
^ Sung, Hou-Mei (2004). "Tiger with cubs: A rediscovered Ming court painting". Artibus Asiae. 64 (2): 281–293. JSTOR 3250187.
^ "A tiger's tale: British Museum buys rare Japanese screen". Art Fund. 20 October 2006. Retrieved 2021-01-08.
^ Goto, Seiko; Naka, Takahiro (2015). Japanese Gardens: Symbolism and Design. Routledge. p. 20. ISBN 9781317411642.
^ Evans-Pritchard, E. E. (1962). "235. Three Zande Texts". Man. 62: 149–152. doi:10.2307/2796709. JSTOR 2796709.
^ "Carrying a Wolf, a Goat, and a Cabbage across the Stream. Metamorphoses of ATU 1579", Piret Voolaid, Folklore: Electronic Journal of Folklore 35 (2007), pp. 111–130. Tartu: Eesti Kirjandusmuuseum.
^ p. 17, Rediscovered Lewis Carroll Puzzles, Lewis Carroll, compiled by Edward Wakeling, Courier Dover Publications, 1996, ISBN 0-486-28861-7.
^ "Springfield! Springfield!".
External links
[edit]
Goat, Cabbage and Wolf A Javascript simulation
Fox, Chook and Corn A simulation without Javascript needed
Retrieved from "
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Wolf, goat and cabbage problem
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1198 | https://arxiv.org/pdf/1011.1868 | arXiv:1011.1868v2 [cs.DS] 17 Nov 2010
Asymptotically Optimal Randomized Rumor Spreading
Benjamin Doerr ∗ Mahmoud Fouz †
August 11, 2018
Abstract
We propose a new protocol solving the fundamental problem of disseminating a piece of information to all members of a group of n players. It builds upon the classical randomized rumor spreading protocol and several extensions. The main achievements are the following: Our protocol spreads the rumor to all other nodes in the asymptotically optimal time of (1 + o(1)) log 2 n. The whole process can be implemented in a way such that only O(nf (n)) calls are made, where f (n) = ω(1) can be arbitrary. In contrast to other protocols suggested in the literature, our algorithm only uses push operations, i.e., only informed nodes take active actions in the network. To the best of our knowledge, this is the first randomized push algorithm that achieves an asymptotically optimal running time.
∗
Department 1: Algorithms and Complexity, Max-Planck-Institut f¨ ur Informatik, Saarbr¨ ucken, Germany
†
Faculty of Computer Science, Universit¨ at des Saarlandes, Saarbr¨ ucken, Germany 1 Introduction
Transmitting a piece of information to all nodes of a network is a classical problem in computer science. A protocol surprisingly powerful is called randomized rumor spreading , see, e.g., Feige, Peleg, Raghavan, and Upfal , Frieze and Grimmett , Karp, Schindelhauer, Shenker, and V¨ ocking . It proceeds in rounds as follows: in each round, each node that already knows the piece of information (“rumor”) chooses a communication partner uniformly at random and sends her a copy of this rumor. In spite of being that simple, this protocol succeeds in spreading the rumor to all nodes of a complete graph in (1 + o(1))(log 2 n + ln n) rounds with high probability, that is, with probability 1 − o(1). In addition, due to its randomized nature, it is highly robust against different types of transmission or node failures. This makes it an interesting alternative to deterministic protocols, which can reduce the broadcast time to log 2(n), but at the price of suffering greatly from failures. A clear disadvantage of this most simple version of randomized rumor spreading is the enormous number of Θ( n log n) calls that are necessary. This problem was overcome in the seminal work of Karp et al. . They present two variations of the randomized rumor spreading protocol which spread the rumor with a total number of O(n log log n) messages only while still using O(log n)rounds only. A central ingredient are so-called pull operations, which allow nodes not yet informed to call random nodes and ask for news. Pull operations, however, have the disadvantage that they create network traffic even if there is no news to be spread. Therefore the assumption underlying the analysis of Karp et al. is that there is constantly new information injected in the network. In this work, we present an alternative solution to the problem. It completely avoids the problematic pull operations. It achieves a broadcast time of (1 + o(1)) log 2 n and it uses a total number O(nf (n)) calls, where f = ω(1) can be any function tending to infinity arbitrarily slow. This is very close to the theoretically optimal values of ⌈log 2 n⌉ rounds and n − 1 calls. Due to its randomized nature, we still have reasonable robustness. For example, if a constant fraction of the nodes chosen uniformly at random crashes at arbitrary times, the time needed to inform all properly working nodes increases by at most a constant factor (depending on the failure rate). The only point in which we assume the protocol to be more powerful compared to previous works is that we discard the address-obliviousness. That is, we assume that each node has a unique label chosen arbitrarily from some ordered set (e.g., the integers). This seems to be a reasonable assumption in many settings.
1.1 The protocol of Karp et al.
As described above, Karp et al. showed how to modify the simple randomized rumor spreading protocol such that instead of Θ( n log n) messages only O(n log log n) are sufficient to spread a rumor. Roughly speaking, their protocol proceeds as follows. The rumor is equipped with a time stamp (or age counter) in such a way that all nodes that receive the rumor also know for how many rounds it has been in the network. In each round, each node chooses a random other node as a communication partner. The communication then proceeds in both directions, that is, any partner who knows the rumor forwards it to the other partner. It is shown that after log 3 n + Θ(log log n)rounds of this protocol, all nodes know the rumor with high probability. In addition, a rumor is transmitted in this time interval at most O(n log log n) times. Note that this way of counting tacitly ignores all communication effort which does not result in a rumor to be sent. In particular, all calls between two uninformed nodes that arise due to pull operations are ignored. The way this is usually justified is by assuming that there is sufficient traffic in the network due to regular insertions of new rumors. Still, we feel that this is slightly 1dissatisfying. Note that when using pull operations, there is no way to avoid such communication overhead—a node that did not receive a rumor recently has no way of finding out whether there are rumors around that justify starting pull operations or not. Moreover, even nodes that did receive a rumor recently cannot be sure that there is no new rumor that would justify starting pull operations again. Karp et al. also prove lower bounds, which, roughly speaking, show that if in each round all communication is restricted to random matchings of communication partners (i) any address-oblivious algorithm has to make Ω( n log log n) calls and (ii) that any algorithm informing all but a
o(1) fraction of the vertices in logarithmic time has to make ω(n) calls.
1.2 Our results
The first lower bound stated in the previous paragraph suggests that asking for an address-oblivious protocol may result in only a limited performance being achievable. In addition, one might also wonder if really many broadcasting problems ask for address-oblivious protocols, or if not rather in the majority of settings each participant naturally has a unique addresses, simply to organize the transport of a message to an addressee. In this work, we shall skip the requirement of address-obliviousness. However, we shall keep the concept of contacting random neighbors without any preference, as this seems to be the key to obtaining good broadcasting times, robustness and small number of calls in all previous works. Contrary to the model of Karp et al. , we do not perform pull operations. That is, all transmissions are initiated by nodes that know the rumor. In consequence, the only direction of informing is from the initiator of the transmission to its addressee, which is chosen uniformly at random, though not always independently. We do allow, however, two-way communication, in that the addressee acknowledges his readiness to receive the rumor or the fact that he already knows the rumor. Such a mechanism makes sense anyway, because it allows to reduce the amount of data sent through the network (if the addressee cannot receive the rumor or already knows it, we do not need to send it). In practice, most communication protocols (e.g., the standard network protocol FTP) allow some kind of two-way communication to ensure an error-free transmission. In this, as we think, natural setting, we propose a protocol that needs only (1 + o(1)) log 2 n
rounds and nf (n) calls, where f = ω(1) can be chosen arbitrarily. Note that no protocol that only uses push operations can work in less than ⌈log 2 n⌉ rounds or using less than n − 1 calls. More precisely, we have the following trade-off between rounds and messages. For all f :
N → N, we give a protocol that needs only log 2(n) + f (n) + O(f (n)−1 log n) rounds with high probability and O(nf (n)) calls. In terms of run-time, this is optimal for f (n) = Θ( √log n), leading to log 2 n + O(√log n) rounds and O(n√log n) calls. The protocol is very simple. For the presentation, let us assume that the nodes are numbered from 1 to n, even though what we really need is only that nodes are able (i) to compute the label of a node chosen uniformly at random and (ii) given a label of a node, to compute a uniquely defined successor along a cyclic order of the labels (here the label plus one, modulo n). Let f : N → N be given (to formulate the tradeoff scenario). Then the protocol works as follows. Each newly informed node sends its first message to another node chosen uniformly at random. From then on, it does the following. If the previous message was sent to a node that was not informed yet, then the next message is sent to the successor of that node in the cyclic order. Otherwise, the next message is sent again to a node chosen uniformly at random. After having encountered f (n) nodes that were already informed, the node stops and does not transmit the rumor anymore. This protocol can be interpreted as a variant of the quasirandom rumor spreading 2protocol investigated in [6, 7], where in addition all nodes have the same cyclic permutation and they re-start at a random position whenever they call a node that is already informed (up to f (n)−1times). The main technical difficulty in the analysis of the proposed protocols stems from the fact that the transmission of messages at each node is not independent, and thus, many classical tools cannot be employed. The key to the solution here is to exploit the existing independence stemming from communications started with random partners. In summary, our result shows that considerable improvements over the fully independent rumor spreading protocol are possible if we skip the requirement that the protocol is address-oblivious. It thus seems worthwhile questioning whether the address-obliviousness assumption is really needed in previous applications of the protocol. From the methodological side, our result again shows that spicing up randomized algorithms with well-chosen dependencies can yield additional gains. It may make the theoretical analysis more complicated, but not so much the algorithm itself.
1.3 Disclaimers
Applications: For reasons of space, we have not given extensive details on randomized rumor spreading and its applications. The seminal papers Feige et al. and Karp et al. contain great discussions of this, better than we could possibly do here. For reports on the actual use of such protocols, see Demers, Greene, Hauser, Irish, Larson, Shenker, Sturgis, Swinehart, and Terry , Hedetniemi, Hedetniemi, and Liestman and Kempe, Dobra, and Gehrke .
Other network topologies: Randomized rumor spreading can be used on all types of net-work topologies. Nodes then choose their communication partners at random from the set of their neighbors. For many network topologies, broadcast times logarithmic in the number of nodes have been shown. Besides the complete graph, they include hypercubes , random graphs G(n, p )with p ≥ (1 + ε) ln( n)/n and certain expander graphs . Recently, rumor spreading was also shown to be doable in poly-logarithmic time for social networks modelled by preferential attach-ment graphs and for graphs of bounded conductance . For Cayley graphs and random geometric graphs , the (in another sense) near-optimal bounds of O(diam( G) + log n) are known. In spite of these results, we concentrated ourselves on the setting where each node has a direct way to communicate with each other node. The reason is that we feel that this is a sufficiently interesting and useful case on its own. Also, of course, it is the setting in which it is easiest to experiment with new ideas. Recall that the concept of reducing the number of messages was also first demonstrated on complete graphs by Karp et al. . Only much later, similar results were obtained for other network topologies, e.g., by Berenbrink, Els¨ asser, and Friedetzky , Els¨ asser and Els¨ asser and Sauerwald .
2 The Hybrid Protocol
Let G = ( V, E ) be the complete, undirected graph on n nodes. We assume that the nodes of the complete graph are ordered and denote by i the i-th node according to that order. Our goal is to spread a ‘rumor’ known initially to one node to all nodes in V . We call the node initiating the rumor the starting node . A rumor can be transmitted along each edge of the graph in both directions. Every transmission along an edge is always initiated by a node that knows the rumor. We count every contact of a node to another node as a call . We assume that two nodes never call a node exactly simultaneously even if they both call the same node in the same round. Hence, a node is always only informed by a single node. 3We introduce a simple algorithm that for a certain instantiation achieves, up to lower order terms, an optimal running time. The algorithm is related to the quasi-random protocol by Doerr et al. . In this quasi-random protocol, every node v is equipped with a cyclic permutation
πv : V → V of all nodes in V . Once a node v becomes informed, it chooses one position on its list uniformly at random. This is the node v contacts first in the next round. In each following round,
v contacts πv(u) where u is the node it contacted in the previous round. Note that different nodes can have different permutations. Our hybrid protocol differs from this quasi-random protocol in two aspects. First, we assume that the permutations of all nodes are identical. Second, we introduce the notion of a restart : if a node calls an already informed node, it chooses a random communication partner in the next round instead of informing the next node according to the permutation. Furthermore, each node stops informing if it calls an informed node after the R-th random call, where R can be a function of n. By this rule we can bound the total number of calls made. This aspect of keeping the number of calls small was not discussed in Doerr et al. . A detailed description of the hybrid protocol is given in Algorithm 1. The only exception is the starting node. This node does not select a random communication partner at the beginning, but starts informing its own successor node (according to the given permutation) immediately. Only after it encountered the first informed node does it then proceed according to Algorithm 1.
Algorithm 1: Procedure started by newly informed node in the hybrid protocol
let R ∈ Z+ be number of random calls per node
for i = 1 to R do
select node j uniformly at random;
while j not informed // iteration counts as call even if j informed
do
inform j;
j ← j + 1;
We will analyze how long it takes for a rumor initiated by any node to reach all nodes under the proposed protocol. We give almost matching upper and lower bounds. Apart from the running time, we will also analyze the number of calls made.
2.1 Running Time And Number of Calls
We give an upper bound and an almost matching lower bound on the number of rounds and calls needed by the protocol to spread a rumor from an arbitrary starting node to all nodes of the complete graph.
2.1.1 Upper Bound Theorem 1. Let ε > 0 be an arbitrarily small constant. With probability 1 − o(1) , the hybrid model with R random calls per node informs all nodes in
log 2 n + (1 + ε) ln( n)/R + R + h(n), if R ≤ √ln n
log 2 n + (2 + ε)√ln( n), if R ≥ √ln n
rounds, where h(n) is a function of arbitrarily slow growth. This uses n(R + 1) calls.
4Note that by adjusting the stopping parameter R, we get a tradeoff between the number of rounds needed to inform all nodes and the number of calls. Before analyzing the protocol for general R, we describe two special cases that achieve an (almost) optimal number of rounds and communications, respectively. For R = √ln n, we achieve, up to a lower order term, an optimal running time while using only O(n√ln n) calls.
Corollary 1. Let ε > 0 be an arbitrarily small constant. With probability 1 − o(1) , the hybrid model with R = √ln n informs all nodes in log 2 n + (2 + ε)√ln n rounds. This uses 2(1 + 2 ε)n√ln n
communications.
For R = 1, we get a very simple broadcasting protocol that, up to constant factors, is both optimal in terms of rounds needed as well as the number of calls.
Corollary 2. Let ε > 0 be an arbitrarily small constant. With probability 1 − o(1) , the hybrid model with R = 1 informs all nodes in log 2 n + (1 + ε) ln n rounds. This uses 2n calls.
Before we prove Theorem 1, we make two observations that will prove useful for the analysis.
Fact 1. The hybrid model is always at least as fast as the quasi-random model implemented with identical lists.
This simple, but useful observation follows from the fact that in the hybrid model every node acts as in the quasi-random model until it encounters an informed node. In this case, since we assumed all lists to be the same, the node becomes useless in the quasi-random model as all successive nodes on its list will have also been informed once it tries to call them. In the hybrid model, however, the node can still potentially inform uninformed nodes.
Fact 2. If a node is delayed , i.e., halted for a number of rounds, then the protocol can only become slower.
It follows that, in our upper bound analysis, we can assume that a node is delayed any number of rounds.
Proof of Theorem 1. We first analyze the time until all nodes are informed. We distinguish three phases of the rumor spreading process. The first phase lasts for log 2 n + h(n) rounds where h(n) is an arbitrarily slowly growing function. By Fact 2, we can assume that every node is delayed to the second phase once it contacts an informed node. Since this delayed protocol remains at least as fast as the hybrid model with R = 1, it follows from Fact 1 that it is still faster than the quasi-random model implemented with identical lists. Fountoulakis and Huber showed that the quasi-random model informs (1 − ε)n nodes for an arbitrarily small constant ε > 0 with probability 1 − o(1) in this phase. Thus, we get the same result for our delayed protocol. The second phase lasts for R rounds. By our delaying assumption, every node that is informed in the first phase will remain active for at least R − 1 rounds before the second phase ends. The crucial observation is that every informed node that is still active either informs an uninformed node in a single round or calls a random node in the next round. The former happens at most εn
times in total. We conclude that at the end of the second phase the nodes will have contributed at least (1 − ε)nR − εn ≥ (1 − 2ε)nR random calls (including the random calls made in the first phase). We show that then the largest interval of uninformed nodes is at most (1 + 3 ε) ln( n)/R .Let I be an interval of length (1 + 3 ε) ln( n)/R . Then, the probability that no node in I becomes 5informed in the second phase by these random calls is at most
(
1 − (1 + 3 ε) ln n
nR
)(1 −2ε)nR
≤ exp ( −(1 − 2ε)(1 + 3 ε) ln n)= n−1−ε+6 ε2
= n−1−ε′
,
for some constant ε′ > 0 (when ε is sufficiently small). Hence, by a union bound argument, it follows that there is no completely uninformed interval of length (1 + 3 ε) ln( n)/R after the second phase with probability at least 1 − n−ε′
for some constant ε′ > 0. In the last phase, all the remaining uninformed intervals are filled up. This takes at most the length of the largest uninformed interval which is (1 + 3 ε) ln( n)/R by the previous argument. Using a simple union bound, we can bound the total failure probability by o(1). It remains to bound the number of calls. Note that each node calls at most R informed nodes in total. Hence, we use at most n calls to inform all nodes and, in addition, at most nR calls until all nodes stop informing.
2.1.2 Lower Bound
In this section, we show that the upper bound from the previous section is essentially sharp.
Theorem 2. Let ε > 0. If the hybrid model with R random calls per node is run for less than
log 2(n) + (1 − ε) ln( n)/R + 1
2
R, if R ≤ √2(1 − ε) ln n,
log 2(n) + √2(1 − ε) ln n, if R ≥ √2(1 − ε) ln n
rounds, then with probability 1 − exp( −nΘ( ε)) not all nodes are informed. Proof. Let ∆ = min {(1 − ε) ln( n)/R + 1
2
R, √2(1 − ε) ln n} and T = log 2 n + ∆. We first analyze the probability that one (specific) vertex u with a distance of more than T from the starting node (in the cyclic order) becomes informed in the first T rounds. We call the event that a node chooses another node to inform uniformly at random a random call . Hence, random calls occur as first calls of a node after the node encountered an already informed node. Clearly, u remains uninformed if for all i ≤ T all random calls happening at time T − i avoid u and the i vertices to the left of it. We say that u is unaffected by such a random call. We show that for ε′ > 0(i) with probability at least 1 − 3n−ε′
log 2 n, u is unaffected by random calls happening in rounds 1 to (1 − ε′) log 2 n,(ii) with probability at least (1 − ε′ log 2(n)+∆
n
)n, u is unaffected by random calls happening in rounds (1 − ε′) log 2 n + 1 to log 2 n,(iii) with probability at least ∏min {R, ∆}
j=1
(
1 − ∆−j
n
)n
, u is unaffected by random calls happening in the remaining rounds log 2(n) + 1 to log 2(n) + ∆. We start with analyzing the effect of random calls happening in a first phase lasting for (1 −
ε′) log 2 n rounds. In this phase, at most n1−ε′
nodes can become informed since the number of informed nodes can at most double in each round. Thus, we have at most n1−ε′
random calls in this phase. The probability that a particular of these calls affects u is at most T /n . Using a 6simple union bound, we conclude that the probability that any random call affects u is at most
n1−ε′
T /n ≤ 3n−ε′
log 2 n.For random calls happening in the second phase consisting of rounds (1 − ε′) log 2 n + 1 to log 2 n,we argue as follows. We bound from above the number of random calls by n and the probability of each one affecting u by ( ε′ log 2(n) + ∆) /n . Since these are too many random decisions for a simple union bound, we use the fact that each random call addresses a node chosen independently from previous random decisions. This yields Pr( u is unaffected in second phase) ≥
(
1 − ε′ log 2(n)+∆
n
)n
.
Note that in the third phase we have at most min {R, ∆} random calls per node. The probability that the i-th random call of a node in round log 2(n)+1 to log 2(n)+∆ affects u is at most (∆ −i)/n .Hence, the probability that u is unaffected in the third phase is at least
(∏min {R, ∆}
j=1
(1 − ∆−j
n
)
)n
.We distinguish two cases: If min {R, ∆} = R, then Pr( u is unaffected in third phase) ≥
R
∏
j=1
(
1 − ∆ − j
n
)n
≥ exp(
R
∑
j=1
−∆ + j − (∆ − j)2/n ) (1)
≥ (1 − o(1)) exp(
R
∑
j=1
−∆ + j) (2)
≥ (1 − o(1)) exp( −R∆ + R(R + 1) /2) ≥ (1 − o(1)) n−1+ ε,
where (1) follows from the fact that for x ≤ 1
2
, we have 1 − x ≥ e−x−x2
, (2) follows from our assumption R ≤ ∆ ≤ O(√ln n), and the last inequality follows from the definition of ∆. Similarly, if min {R, ∆} = ∆, then Pr( u is unaffected in third phase) ≥
∆
∏
j=1
(
1 − ∆ − j
n
)n
≥ (1 − o(1)) exp( −∆2/2) ≥ (1 − o(1)) n−(1 −ε).
We set ε′ := ε/ 4. Since all random calls choose their addressees independently, we have Pr( u remains unaffected in all three phases)
≥ (1 − 3n−ε′
log 2(n))
(
1 − ε′ log 2 n+∆
n
)n min {R, ∆}
∏
j=1
(
1 − ∆ − j
n
)n
≥ (1 − o(1)) exp( −2ε′ log 2 n − (1 − ε) ln n) ≥ n−1+Θ( ε).
Let k = n
T+1
− 1. Let u1, . . . , u k be nodes each having distance more than T from each other and from the starting node (in the cyclic order). We argue that, with sufficiently high probability, one such node will remain uninformed after the first T rounds. Let Ui denote the event that node
ui is informed. Note that since these nodes have a distance of T from each other, a random call that informs one such node ui can not lead to the informing of any other node uj during the first T
rounds. Hence, these events are negatively correlated : if some nodes are informed, the probability that another one is also informed decreases, or formally, Pr( U | U1, . . . , U j ) ≤ Pr( U ). We compute Pr(no node remains uninformed) ≤ Pr( U1 ∧ · · · ∧ Uk) ≤ ∏
1≤j≤k
Pr( Uj ) ≤ (1 − n−1+Θ( ε))k ≤ exp( −nΘ( ε)).
7References
P. Berenbrink, R. Els¨ asser, and T. Friedetzky. Efficient randomized broadcasting in random regular networks with applications in peer-to-peer systems. In 27th ACM Symposium on Principles of Distributed Computing (PODC) , pp. 155–164, 2008. M. Bradonjic, R. Els¨ asser, T. Friedrich, T. Sauerwald, and A. Stauffer. Efficient broadcast on random geometric graphs. In Proceedings of the 21st Annual ACM-SIAM Symposium on Discrete Algorithms (SODA) , pp. 1412–1421, 2010. F. Chierichetti, S. Lattanzi, and A. Panconesi. Rumor spreading in social networks. In 36 th International Colloquium on Automata, Languages and Programming (ICALP) , pp. 375–386, 2009. F. Chierichetti, S. Lattanzi, and A. Panconesi. Almost tight bounds for rumour spreading with conductance. In 42 th Proceedings of the ACM Symposium on Theory of Computing (STOC) ,pp. 399–408, 2010. A. Demers, D. Greene, C. Hauser, W. Irish, J. Larson, S. Shenker, H. Sturgis, D. Swinehart, and D. Terry. Epidemic algorithms for replicated database maintenance. In Proc. of the 6th ACM Symposium on Principles of Distributed Computing (PODC) , pp. 1–12, 1987. B. Doerr, T. Friedrich, and T. Sauerwald. Quasirandom rumor spreading. In Proceedings of the 19th Annual ACM-SIAM Symposium on Discrete Algorithms (SODA) , pp. 773–781, 2008. B. Doerr, T. Friedrich, and T. Sauerwald. Quasirandom rumor spreading: Expanders, push vs. pull, and robustness. In Proc. of the 36th International Colloquium on Automata, Languages and Programming (ICALP) , pp. 366–377, 2009. R. Els¨ asser. On the Communication Complexity of Randomized Broadcasting in Random-like Graphs. In 18th ACM Symposium on Parallelism in Algorithms and Architectures (SPAA) ,pp. 148–157, 2006. R. Els¨ asser and T. Sauerwald. Broadcasting vs. mixing and information dissemination on cayley graphs. In Proceedings of the 24th International Symposium on Theoretical Aspects of Computer Science (STACS) , pp. 163–174, 2007. R. Els¨ asser and T. Sauerwald. On the Power of Memory in Randomized Broadcasting. In 19th ACM-SIAM Symp. on Disc. Alg. (SODA) , pp. 218–227, 2008. U. Feige, D. Peleg, P. Raghavan, and E. Upfal. Randomized broadcast in networks. Random Structures and Algorithms , 1:447–460, 1990. N. Fountoulakis and A. Huber. Quasirandom rumor spreading on the complete graph is as fast as randomized rumor spreading. SIAM Journal on Discrete Mathematics , 23:1964–1991, 2009. A. Frieze and G. Grimmett. The shortest-path problem for graphs with random arc-lengths.
Discrete Applied Mathematics , 10:57–77, 1985. 8 S. M. Hedetniemi, S. T. Hedetniemi, and A. L. Liestman. A survey of gossiping and broad-casting in communication networks. Networks , 18:319–349, 1988. R. Karp, C. Schindelhauer, S. Shenker, and B. V¨ ocking. Randomized Rumor Spreading. In
Proc. of the 41st IEEE Symposium on Foundations of Computer Science (FOCS) , pp. 565–574, 2000. D. Kempe, A. Dobra, and J. Gehrke. Gossip-based computation of aggregate information. In
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1199 | https://bmcnutr.biomedcentral.com/articles/10.1186/s40795-018-0231-1 | A systematic review of methods to assess intake of saturated fat (SF) among healthy European adults and children: a DEDIPAC (Determinants of Diet and Physical Activity) study | BMC Nutrition | Full Text
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A systematic review of methods to assess intake of saturated fat (SF) among healthy European adults and children: a DEDIPAC (Determinants of Diet and Physical Activity) study
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Published: 08 May 2018
A systematic review of methods to assess intake of saturated fat (SF) among healthy European adults and children: a DEDIPAC (Determinants of Diet and Physical Activity) study
Fiona RiordanORCID: orcid.org/0000-0003-2572-47291,
Roisin McGann2,
Ciara Kingston3,
Ivan J. Perry1,
Matthias B. Schulze4,
Lene Frost Andersen5,
Anouk Geelen6,
Pieter van’t Veer6,
Simone J. P. M. Eussen7,
Martien C. J. M. Van Dongen7,
Nicole E. G. Wijckmans-Duysens7&
…
Janas M. Harrington1
Show authors
BMC Nutritionvolume 4, Article number:21 (2018) Cite this article
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Abstract
Background
Dietary fat is an essential macronutrient. However, saturated fact has been associated with negative health outcomes including cardiovascular disease. Shifting consumption from saturated fat to unsaturated fats and limiting the level of saturated fat in the diet has been recommended. Currently, there is no standard method to measure saturated fat intake in etiologic studies. Therefore, it is difficult to obtain a reliable picture of saturated fat intake in Europe. To inform the development of the DEDIPAC (DEterminants of DIet and Physical Activity) toolbox of methods, we aimed to identify the assessment methods and specific instruments which have been used to assess saturated fat intake among children or adults in pan-European studies.
Methods
Three electronic databases were searched for English language studies of any design which assessed intake of saturated fat. Reference lists were hand-searched. Studies were included if they were conducted in two or more European countries, and involved healthy, free-living children and adults.
Results
The review identified 20 pan-European studies which assessed saturated fat intake. Food Frequency Questionnaires (n = 8) and diet records (n = 7) were most common, followed by 24-h recalls (n = 5). Methods differed in portion size estimation and the composition data which was used to calculate nutrient intake. Of the instruments used in more than two European countries, five Food Frequency Questionnaires had been specifically tested for validity to assess saturated fat intake; four among adults (Food4me, PURE, IMMIDIET, Health, Alcohol and Psychosocial factors in Eastern Europe (HAPIEE)) and one among children (used by Piqueras et al.).
Conclusions
A standardised approach to portion size estimation and a common source of food composition data are required to measure saturated fat intake across Europe effectively. Only five instruments had been used in more than two European countries and specifically tested for validity to assess saturated fat intake. These instruments may be most appropriate to evaluate intake of saturated fat in future pan-European studies. However, only two instruments had been tested for validity in more than one European country. Future work is needed to assess the validity of the identified instruments across European countries.
Peer Review reports
Background
Dietary fat is an essential macronutrient, providing a source of energy and facilitating the absorption of fat-soluble dietary components such as vitamins . Saturated fatty acids (SFA) have been associated with the development of non-communicable diseases, including cardiovascular disease (CVD) [2,3,4]. The World Health Organisation (WHO) Global Strategy on Diet and Physical Activity recommends shifting consumption from saturated fat (SF) to unsaturated fats, and limiting the level of SF in the diet . The Food and Agriculture Organisation expert consultation on fat and fatty acids in human nutrition has proposed that SFA be replaced by Monounsaturated Fatty Acids (MUFA) and Polyunsaturated Fatty Acids (PUFA) in the diet to reduce the risk of Coronary Heart Disease .
The role of SF in the diet has recently been the subject of debate. Some studies suggest SF increases levels of beneficial high-density lipoprotein (HDL). However, whether this offsets the effect of detrimental low-density lipoprotein (LDL), and consequently the risk of CVD, is unclear [6, 7]. To better understand the role of SF in the development of chronic disease there is a need for dietary assessment methods which can measure SF and its contribution to daily energy intake in a reliable and consistent way. However, a number of factors have made cross-country comparisons of macronutrient intake difficult: differences in the methods used to assess dietary intake, different approaches to portion size estimation, and the type of food composition databases (FCD) used to calculate SF intake.
In recent years there has been growing emphasis on the standardisation of food classification systems, including Food Composition Databases (FCD), between European countries. This has been the focus of a number of European projects [8,9. Report of the third annual meeting of the FLAIR Eurofoods-Enfant project, 10- 12 November 1993. Vilamoura, Portugal. Wageningen. FLAIR Eurofoods-Enfant Project (February) 1994."),10,11,12,13:S37–50."),14:S6–14."),15:S102-7.
.")], including The Innovative Dietary Assessment Methods in Epidemiological Studies and Public Health (IDAMES) project, which aims to develop new methods to assess dietary intake in Europe [16. Dietary Assessment Methods: State of the Art Report. German Institute of Human Nutrition (DIfE); 2010.")]. The European Food Safety Authority (EFSA) has recommended the standardized 24-HDR recall method, EPIC-Soft (now known as GloboDiet) [17:1435."), 18 Guidance on the EU Menu methodology. EFSA Journal. 2014;12(12):3944.")]. However, there are no agreed standards with respect to the assessment of macronutrients, including SF, for monitoring purposes or aetiological studies.
Partly in recognition of the lack of agreed standards and methodologies, the DEDIPAC: “DEterminants of DIet and Physical Activity” project , aimed to create a toolbox of dietary assessment methods which may be most appropriate to use in pan-European studies [19, 20:1307–14.")]. The purpose of the current systematic literature review is to identify the assessment methods and specific instruments which have been used to measure intake of SF in European children or adults in more than one European country.
Methods
Data sources and study selection
This review adheres to the guidelines of the Preferred Reporting Items for Systematic Reviews and Meta-Analyses (PRISMA) Statement. The protocol for the review can be accessed from the PROSPERO (CRD42014014175) . A systematic literature search was conducted for pan-European studies which assessed the intake of SF. SF are fatty acids where the fatty acid chain have predominantly single bonds. They can be classified as short, medium, long and very long chain, and are mainly provided in the diet by animal dairy fats, along with some oils, palm oil and coconut oil . Three databases, PubMed, EMBASE and Web of Science, were searched by FR and RM. Search terms included terms for fats (e.g. dietary fat/s, saturated fat/s, dietary fatty acid/s, saturated fatty acid/s, volatile fatty acid/s, non-essential fatty acid/s, trans fatty acid/s, short chain fatty acid/s, trans fat/s, animal fat/s, lipid/s), along with keywords for dietary and caloric intake, and terms for European countries. A full copy of the EMBASE search strategy is included in Additional file1: Figure S1. All searches were limited to literatures in English published from 1990 through to 15th March 2017.
Titles and abstract screening of the articles was conducted by FR and RM. In the event of any uncertainty regarding inclusion, the full text of an article was sourced and reviewed. If FR and RM disagreed on article inclusion during full text review then they consulted a third author, JMH. To be included, studies had to be published in a peer-reviewed scientific journal, conducted in two or more European countries, as defined by the Council of Europe , and report on the intake of macronutrient SF. Therefore, studies were excluded if they only reported on fat as a food product (e.g. fat-based spreads, fats and oils). SF intake had to be measured at the individual level. Therefore studies which assessed SF intake at the household level or through analysis of biological samples were excluded. Studies had to be conducted among free-living, healthy populations. If study participants were hospital-based or belonged to a disease or societal sub-group they were excluded. The review was not limited by study design; studies with baseline intervention data, and case-control studies where intake was measured in population-based controls, were included (Fig.1).
Fig. 1
Flow diagram showing study selection process for the review
Full size image
Reference lists of all included papers were hand-searched for additional publications. The names of European projects listed in the DEDIPAC Inventory of Relevant European Studies, were also used to search the databases. If necessary, study authors were contacted to request a copy of full paper, or the instrument or questionnaire.
Data extraction and quality assessment
A data extraction form was created and piloted. This form recorded the following information from included studies: design, number and names of European countries involved, sample size (total and number for each country), age range of the included population, the method used and its description (including frequency categories for Food Frequency Questionnaires (FFQs), details of nutrient intake assessment, details of portion estimation), mode of administration, and details on the validation or reproducibility. Double extraction on each article was carried out by FR and RM. If necessary, further information on the methods was obtained from reference list of the originally included articles.
In line with a previous review of methods to assess fruit and vegetable (F&V) intake , the aim of the current review was to identify instruments. Therefore the quality of each included article was not appraised as part of the current review. Instead, information from the appropriate validation study was extracted by MvD, SE and NW. For an instrument to be considered suitable to assess intake in a pan-European study it had to meet two criteria: 1. Tested for validity; 2.Used in more than two countries as part of the same study. These two countries had to represent at least one country from at least three of the Southern, Northern, Eastern, Western European regions as defined by the United Nations [24/
. Accessed 6 May 2017.")]. Table1 shows the results of this assessment.
Table 1 Summary of all studies identified to assess saturated fat: design, population studied, and dietary assessment instruments used and details of validation and/or reproducibility. Studies were selected to be included in this review based on the following two criteria: (1) the instrument was tested for validity and (2) the instrument was used in more than two countries simultaneously which represent a range of European regions
Full size table
Results
Description of the included studies
In total 10,076 papers were identified. After removing duplicates 7519 remained. Following title and abstract screening and full text review, 82 primary research articles were retained. These articles were organised by the European project to which they belonged. If they did not belong to a project they were grouped as ‘Other’ (n = 6) (see Fig. 1 for a breakdown). ‘Study’ refers to the larger project, rather than individual articles based on the same project and methodology. Of the 82 articles retained, 26 provided a detailed description of the project or the method in question. These 26 articles were selected for the current review, which equated to 1–3 articles per project. A further 11 articles were sourced from reference lists [14, 25,26,27,28,29,30,31,32,33,34].
In total, 37 articles [14, 27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62] from 20 studies were included in the review. This number included articles from the original search (n = 26), and from reference lists (n = 11). Articles in which the instrument was tested for validity were also recorded (n = 11) [58, 63,64,65,66,67,68,69,70,71,72]. The characteristics of the included studies are described in Table 1. They comprised of large pan-European studies (n = 11) and smaller studies conducted in 2–4 countries (n = 6). Four studies assessed intake of SF in children [60,61,62], or adolescents , and 13 assessed intake among adults [35, 40, 41, 46,47,48,49,50, 54,55,56,57, 73].
Dietary assessment methods
Types of methods
Four approaches to measure SF intake were identified: FFQs, 24 h recalls (24-HDRs), dietary record/diet diaries, or dietary history methods. Most studies used FFQs or 24-HDRs. Several studies used instruments which had been tested for validity: IMMIDIET, European Prospective Investigation into Cancer and Nutrition (EPIC), Prospective Urban Rural Epidemiology (PURE), Healthy Lifestyle in Europe by Nutrition in Adolescence (HELENA), and Identification and prevention of Dietary- and lifestyle-induced health EFfects In Children and infantS (IDEFICS). One study instrument was based on a FFQ which had been used as part of a different study and previously tested for validity . Countries in the EPIC study did not use a common FFQ instrument, therefore only the EPIC-Soft instrument is discussed in this review.
According to the two criteria (Table 1) six study instruments were appropriate to assess intake of SF in future pan-European studies. Two, the EPIC 24-HDR instrument EPIC-Soft, and the cross-check dietary history method used by the Seven Countries Study, had been used to measure intake among adult populations. The HELENA-DIAT instrument had been used among adolescents. The IDEFICS FFQ and 24-HDR, and the FFQ used by Piqueras et al. had been used among children. Table 1 shows the instruments which met the two criteria. Some of these instruments were also used to assess intake of other macronutrients, and/or also met the criteria to assess intake of F&V or Sugar Sweetened Beverages (SSBs) as determined from two previous reviews. This is also indicated in Table 1.
Validation
Of the studies which assessed instrument validity and thus fulfilled inclusion criterion 1 (Table 1), only two instruments, EPIC-Soft and the IDEFICS 24-HDR , had been tested for validity in more than one country: Belgium, the Czech Republic, France, the Netherlands and Norway in the case of EPIC-Soft ; Belgium and Spain in the case of the IDEFICS 24-HDR .
Five instruments, all FFQs [31, 42, 43, 49, 72, 75], had been tested specifically for validity to assess SF intake, in Sweden (Piqueras et al. ), UK (Food4Me [42, 43]; Health, Alcohol and Psychosocial factors in Eastern Europe (HAPIEE) ), and Poland (PURE ), and Belgium (IMMIDIET ). FFQs , repeated 24-HDRs [31, 49, 72] or diet/food records [42, 75] were used as reference methods. Validity was assessed by crude correlations [31, 42, 49, 72], de-attenuated correlation coefficients [49, 72], mean or median differences in SF consumption [49, 72, 75], exact level of agreement of SF consumption [31, 49, 75], or Bland and Altman’s plots [31, 42, 43, 49].
In four studies, the instrument reproducibility was also tested. Reproducibility was assessed by correlations [43, 49, 72], mean/median differences, or intraclass correlation coefficient (ICC) between subsequent assessments of FFQs. References for validation studies are provided in Table 1. Where available, results of the statistical assessments are provided in Table2.
Table 2 Information on the validity and reliability of SF consumption
Full size table
Instruments tested for validity
Details of the five instruments, HAPIEE, Food4Me, PURE, IMMIDIET FFQs, and the FFQ used by Piqueras et al., are summarised in Table3. The HAPIEE FFQ had moderate agreement (0.4–0.6) for SF intake with a 7-day diet diary (7DD) (Spearman’s rank correlation: men, r = 0.43; women, r = 0.56) , as did the Food4Me FFQ which was tested for validity using a 4-day weighed food record (Pearson’s crude correlation: r = 0.48) . The Food4Me FFQ also had good agreement (> 0.6) with the EPIC Norfolk FFQ (r = 0.71) . The FFQ used as part of the PURE study had low to moderate agreement with 4 repeated 24-HDR (urban: r = 0.42; rural: r = 0.39) whereas the IMMIDIET FFQ had low agreement (< 0.4) with five repeated 24-HDR (men: r = 0.21; women = 0.34) . Of the instruments used among children, the FFQ used by Piqueras et al. had good to moderate agreement with 10 repeated 24-HDRs (men: r = 0.58; women: r = 0.54) .
Table 3 Summary of the instruments which were validated (n = 5) for assessment of saturated fat
Full size table
Macronutrient assessment
The identified instruments measured the intake of SFA or SF in grams [42, 50, 55, 56, 62], grams/day [34, 40, 41, 45, 49, 57] or % contribution to daily energy intake [34, 35, 42, 46, 50, 53,54,55, 60, 62, 76], % of total fat [48, 58], and ratio of unsaturated to saturated fats . Of the five instruments which were tested specifically for validity to assess SF, two reported SF as grams/day (HAPIEE , PURE [32, 49]), two as grams (Piqueras et al. , Food4Me [42,43,44]), and three as % energy/day (IMMIDIET [31, 46], Piqueras et al. , Food4Me [42,43,44]).
EPIC used the EPIC Nutrient DataBase (country-specific food composition data standardised across countries) to calculate intake of SF from dietary data . Mulder et al. (Seven Countries Study), determined SF intake by buying food products which represented the average daily intake in a cohort and analysing these products for composition of SF . The remaining studies used local food composition tables (FCT) from participating countries to calculate intake. In some cases, one FCT was used as the main source of composition data. For example, the German Food Code and Nutrient Data Base (Bundeslebensmittelschlussel) was used by the HELENA study and supplemented with information from the Belgian FCT [59, 77].
Food frequency questionnaires (FFQs)
Characteristics of the FFQ instruments are summarised in Table4. Where this information was available, the number of FFQs items ranged from to 43 to 322. Most FFQs recorded habitual consumption over the previous year, with the exception of the IDEFICS FFQ [30, 61]. This FFQ assessed intake over a typical week during the previous month. Almost all FFQs were paper-based and self-administered. All were semi-quantitative, and assessed portion size either through specifying a standard portion size on the FFQ for the food item in question [41, 49], or asking participants to consult photos , or use household measures . The number of pre-coded frequency categories on the FFQs ranged from 3 to 11.
Table 4 Summary of FFQs and their characteristics
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Diet records
Six studies used diet records or diaries (Table5) to assess SF intake. Portions were estimated using a photo book [33, 34, 50, 57], during the interview (portion description provided by a dietician) , using food models [33, 50], or by weighing foods [34, 48, 76]. Most of the identified records were three or seven day records, with one exception, the Zinc Effects on Nutrient/nutrient Interactions and Trends in Health and Aging (ZENITH) study. This study used a four recall day method, over two weekday and two weekend days. All records were self-administered.
Table 5 Summary of diet records and their characteristics
Full size table
Dietary history
A dietary history approach is an interview-based approach used to record usual intake, asking an individual to recall a typical intake patterns, typically over a longer period (e.g. 6 months) . The Seven Countries Study used a cross-check dietary history method conducted by face-to-face interview. The dietary history recorded diet intake in the month preceding the interview. This method had been tested for reproducibility, albeit not specifically to assess SF intake . Usual food consumption pattern was recorded (i.e. foods consumed at breakfast, lunch, dinner and between meals) on a daily basis during week and weekend days. A list of all foods was compiled from this record. Interviewers then recorded what was eaten on a daily, weekly, or monthly basis. A checklist with an extensive number of foods was also used to record the frequencies and amount of foods consumed. Portion size was estimated using different approaches: Finland: photos; The Netherlands: portable scale; Italy: artificial models of different foods in Italy), and also weighed .
24 hour dietary recalls (24-HDRs)
In total, six 24-HDRs were identified, three of which were computerised [61, 73, 80]. Their characteristics are summarised in Table6. Portion size was estimated for all of identified 24-HDRs, using household measures [80, 81] or photographs [60, 61, 80, 81]. EPIC-Soft, estimated portion sizes using six quantification methods. Two 24-HDRs were tested for validity: HELENA-DIAT , which was compared with 1-day food records and tested for reproducibility across administrative modes (self-administration and by interview), and IDEFICS Self-Administered Children and Infants Nutrition Assessment (SACINA) which was tested for validity using the doubly labelled water technique .
Table 6 Summary of 24-HDR and their characteristics
Full size table
Discussion
All four main assessment methods; FFQs, 24-HDRs, diet records/diaries and diet history methods have been used in pan-European studies to measure intake of SF. Of the 20 studies identified, most assessed intake of SF among adults (n = 16), and few measured intake among adolescents or children (n = 4). While FFQs were most common (n = 8), they differed in terms of the approach used to determine portion size and calculate macronutrient intake. Only one identified study, EPIC, used a standardised database as a source of food composition data.
If intake of fat sub-types such as SF and their relationship with disease are to be studied in a standardised way across European countries, it is essential to identify valid instruments. Six study instruments met two criteria (1.the instrument was tested for validity, and; 2. used in more than two European countries) to assess SF among adults in pan-European studies: the EPIC-Soft 24-HDR, HAPIEE, Food4Me, IMMIDIET, and PURE FFQs, and the SENECA 3-day record. However, only four of these, all FFQs (HAPIEE, IMMIDIET, PURE and Food4Me) had been specifically tested for validity to assess SF intake. Two of these (HAPIEE and Food4Me FFQs) were found to have moderate agreement with diet records. The PURE FFQ had low to moderate agreement with repeated 24-HDRs, and the IMMIDIET FFQ had low agreement with repeated 24-HDRs. Only one identified instrument had been used among adolescents, HELENA-DIAT, but this had not been tested for validity to assess SF intake. Finally, the 24-HDR and FFQ used by the IDEFICS study, and the FFQ used by Piqueras et al. had been used to measure SF intake among children. Of the two, only the FFQ used by Piqueras et al. had been tested for validity to assess SF intake. This instrument had good to moderate agreement with repeated 24-HDRs.
All instruments which had been tested for validity to assess SF intake, had done so using food records (4 and 7 day) [42, 75] or 24-HDRs [31, 49, 72] as the reference method. However, using these methods as a reference assumes they are superior in terms of assessing true SF intake. No specific biomarkers for SF exist, therefore, the validity of a 24-HDR to assess true intake cannot be determined. This raises an important question: whether the identified instruments are valid to specifically assess SF intake. Another important consideration is the fact that the level of macronutrient intake may be affected by the source of FCD used for calculations . Ideally pan-European studies would use a common data collection instrument tested for validity, a common approach to portion size estimation, and a standardised source of composition data to calculate intake of SF.
As with previous reviews the results will contribute to the DEDIPAC toolbox of dietary intake assessment methods. The two criteria used in this review, are only an initial approach to identifying suitable instruments. Other factors, including the existing evidence with respect to instrument validity together with instrument feasibility, should be taken into consideration when deciding the appropriateness of an instrument to assess intake of SF in a pan-European population. Only two instruments had been tested for validity in more than one European country. To determine which instruments may be most appropriate, will require further work to test validity across countries. Most identified instruments were also included in two previous reviews on methods to assess intake of F&V and SSBs . Exceptions were the ZENITH 4-day recall method, PURE FFQ, and the FFQs used by Van Oostrom et al., and Piqueras et al. Overall these two reviews identified a greater number, and variety, of instruments. While this review was limited to pan-European studies, this is not to suggest that other instruments used as part of non-European studies, could not be used to assess intake across Europe.
The review has a number of strengths and limitations. A comprehensive search strategy identified all pan-European studies measuring intake of SF among children or adults, and the instruments used by these studies. In addition to searching databases, reference lists were hand-searched and study authors were contacted to identify further instruments. A copy of the instrument was sourced in order to accurately describe each instrument. Although the search was comprehensive, it is possible that all relevant articles were not identified. Furthermore, the search was limited to English-language papers. Where a copy of the original instrument or article could not be sourced, the description may be limited, although the results can still be used as a reference. The quality of the identified instruments was not assessed as part of this review. It is important to emphasize that the current review only provides an initial selection of instruments that may be most appropriate to assess SF across European countries. A decision on appropriateness will depend upon instrument validity, which requires further research. Not all SF may be detrimental to health [83, 84]. In light of this, a final limitation of the review may be the focus on total SF intake. Assessing different SFs or subgroups and their relation to health, and reviewing instruments which examine and report on these differences may be an important next step. The majority of the identified instruments evaluated SF as one class. Only one FFQ, used in the IMMIDIET study [31, 46], assessed the intake of SF by sub-types. Lastly, it is important to consider the fact that the identified instruments rely on available food composition data for analysis; the assessment of SF may lag behind changes in food production and composition. FFQs may need to be updated in line with such changes e.g. adding new foods, changing numbers on answer options.
Conclusion
This review has identified a range of methods to assess intake of SF, FFQs being the most common method used. Key differences exist between the instruments which are currently available to assess SF intake. In order to standardise and harmonise assessment methods between European countries, and increase the accuracy with which intake of SF is measured, it is essential that (1) an agreed method and approach to portion size estimation is used and (2) this is used in conjunction with a standardised source of composition data. This review has indicated five instruments, all FFQs (Food4me, PURE, IMMIDIET, HAPIEE, and FFQ used by Piqueras et al.) which meet both criteria, and were tested for validity to assess SF intake. These instruments may be most suitable to assess intake of SF among healthy populations across Europe. These methods have been used in pan-European populations which encompass a range of European regions, and should be considered by future studies which focus on evaluating SF intake. However, these instruments have only been tested for validity in one country. Future work is needed to test the validity of these instruments across European countries.
Abbreviations
24-HDR:
24 Hour Dietary Recall
CVD:
Cardiovascular disease
DD:
7-day diet diary
DD:
Diet Diary
DEDIPAC:
DEterminants of DIet and Physical Activity
DR:
Dietary Record
EFSA:
European Food Safety Authority
EPIC:
European Prospective Investigation into Cancer and Nutrition
EYHS:
European Youth Heart Study
F&V:
Fruit and vegetable
FCD:
Food composition database
FCT:
Food composition table
FFQ:
Food Frequency Questionnaire
HAPIEE:
Health, Alcohol and Psychosocial factors in Eastern Europe
HDL:
High-density lipoprotein
HELENA:
Healthy Lifestyle in Europe by Nutrition in Adolescence
ICC:
Intraclass correlation coefficient
IDAMES:
Innovative Dietary Assessment Methods in Epidemiological Studies and Public Health
IDEFICS:
Dietary- and lifestyle-induced health EFfects In Children and infantS
LDL:
Low-density lipoprotein
MUFA:
Monounsaturated Fatty Acids
PRISMA:
Preferred Reporting Items for Systematic Reviews and Meta-Analyses
PUFA:
Polyunsaturated Fatty Acids
PURE:
Prospective Urban Rural Epidemiology
SACINA:
Self-Administered Children and Infants Nutrition Assessment
SENECA:
Survey in Europe on Nutrition and the Elderly; a Concerted Action
SF:
Saturated Fat
SFA:
Saturated Fatty Acids
SSBs:
Sugar-sweetened beverages
WFR:
Weighed Food Record
WHO:
World Health Organisation
WHO-MONICA:
The World Health Organization Multinational MONItoring of trends and determinants in CArdiovascular disease
ZENITH:
Zinc Effects on Nutrient/nutrient Interactions and Trends in Health and Aging
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Funding
The preparation of this paper was supported by the DEterminants of DIet and Physical ACtivity (DEDIPAC) knowledge hub. This work is supported by the Joint Programming Initiative ‘Healthy Diet for a Healthy Life’. The funding agencies supporting this work are: The Health Research Board (HRB), Ireland (DEDIPAC/2013/1). The funders had no role in the design of the study and collection, analysis, and interpretation of data and in writing the manuscript.
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Authors and Affiliations
School of Public Health, University College Cork, Western Road, Cork, Ireland
Fiona Riordan,Ivan J. Perry&Janas M. Harrington
School of Public Health, Physiotherapy and Population Science, University College Dublin, Dublin, Ireland
Roisin McGann
School of Food and Nutritional Sciences, University College Cork, Cork, Ireland
Ciara Kingston
Department of Molecular Epidemiology, German Institute of Human Nutrition Potsdam-Rehbruecke, Nuthetal, Germany
Matthias B. Schulze
Department of Nutrition, Institute of Basic Medical Sciences, University of Oslo, Oslo, Norway
Lene Frost Andersen
Division of Human Nutrition, Wageningen University and Research, Wageningen, Netherlands
Anouk Geelen&Pieter van’t Veer
Department of Epidemiology of the Faculty of Health, Medicine and Life Sciences, Maastricht University, Maastricht, The Netherlands
Simone J. P. M. Eussen,Martien C. J. M. Van Dongen&Nicole E. G. Wijckmans-Duysens
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2. Roisin McGannView author publications Search author on:PubMedGoogle Scholar
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4. Ivan J. PerryView author publications Search author on:PubMedGoogle Scholar
5. Matthias B. SchulzeView author publications Search author on:PubMedGoogle Scholar
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8. Pieter van’t VeerView author publications Search author on:PubMedGoogle Scholar
9. Simone J. P. M. EussenView author publications Search author on:PubMedGoogle Scholar
10. Martien C. J. M. Van DongenView author publications Search author on:PubMedGoogle Scholar
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FR planned and conducted the review, drafted and revised the paper. RM planned and conducted the review, and drafted the paper. CK, IJP, AG, PV, LFA, MS and JMH contributed to the planning, drafted and revised the paper. SE, MVD and NWD conducted the review of validation data, drafted and revised the paper. All authors read and approved the final manuscript.
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Riordan, F., McGann, R., Kingston, C. et al. A systematic review of methods to assess intake of saturated fat (SF) among healthy European adults and children: a DEDIPAC (Determinants of Diet and Physical Activity) study. BMC Nutr4, 21 (2018).
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Keywords
Saturated fat
Dietary assessment
Europe
DEDIPAC
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