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16000 | https://www.nhvweb.net/nhhs/science/rquinn/files/2014/04/Ch26-Homework-Solutions.pdf | Chapter 26 The Magnetic Field Conceptual Problems 1 • [SSM] When the axis of a cathode-ray tube is horizontal in a region in which there is a magnetic field that is directed vertically upward, the electrons emitted from the cathode follow one of the dashed paths to the face of the tube in Figure 26-30. The correct path is (a) 1, (b) 2, (c) 3, (d) 4, (e) 5. Determine the Concept Because the electrons are initially moving at 90° to the magnetic field, they will be deflected in the direction of the magnetic force acting on them. Use the right-hand rule based on the expression for the magnetic force acting on a moving charge B v F r r r × = q , remembering that, for a negative charge, the force is in the direction opposite that indicated by the right-hand rule, to convince yourself that the particle will follow the path whose terminal point on the screen is 2. ) (b is correct. 2 •• We define the direction of the electric field to be the same as the direction of the force on a positive test charge. Why then do we not define the direction of the magnetic field to be the same as the direction of the magnetic force on a moving positive test charge? Determine the Concept The direction of the force depends on the direction of the velocity. We do not define the direction of the magnetic field to be in the direction of the force because if we did, the magnetic field would be in a different direction each time the velocity was in a different direction. If this were the case, the magnetic field would not be a useful construct. 3 • [SSM] A flicker bulb is a light bulb that has a long, thin flexible filament. It is meant to be plugged into an ac outlet that delivers current at a frequency of 60 Hz. There is a small permanent magnet inside the bulb. When the bulb is plugged in the filament oscillates back and forth. At what frequency does it oscillate? Explain. Determine the Concept Because the alternating current running through the filament is changing direction every 1/60 s, the filament experiences a force that changes direction at the frequency of the current. 4 • In a cyclotron, the potential difference between the dees oscillates with a period given by T = 2πm qB ( ). Show that the expression to the right of the equal sign has units of seconds if q, B and m have units of coulombs, teslas and kilograms, respectively. 2493 Chapter 26 2494 Determine the Concept Substituting the SI units for q, B, and m yields: s 1 m kg s m kg kg m s s m kg kg m C s N C kg m A N C kg T C 2 = ⋅ ⋅ = ⋅ ⋅ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ ⋅ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ = ⋅ 5 • A 7Li nucleus has a charge equal to +3e and a mass that is equal to the mass of seven protons. A 7Li nucleus and a proton are both moving perpendicular to a uniform magnetic field . The magnitude of the momentum of the proton is equal to the magnitude of the momentum of the nucleus. The path of the proton has a radius of curvature equal to R B r p and the path of the 7Li nucleus has a radius of curvature equal to RLi. The ratio Rp/RLi is closest to (a) 3/1, (b) 1/3, (c) 1/7, (d) 7/1, (e) 3/7, (f) 7/3. Determine the Concept We can use Newton’s 2nd law for circular motion to express the radius of curvature R of each particle in terms of its charge, momentum, and the magnetic field. We can then divide the proton’s radius of curvature by that of the 7Li nucleus to decide which of these alternatives is correct. Apply Newton’s 2nd law to the lithium nucleus to obtain: R v m qvB 2 = ⇒ qB mv R = For the 7Li nucleus this becomes: eB p R 3 Li Li = (1) For the proton we have: eB p R p p = (2) Divide equation (2) by equation (1) and simplify to obtain: Li p Li p Li p 3 3 p p eB p eB p R R = = Because the momenta are equal: 3 Li p = R R ⇒ ) (a is correct. 6 • An electron moving in the +x direction enters a region that has a uniform magnetic field in the +y direction. When the electron enters this region, it will (a) be deflected toward the +y direction, (b) be deflected toward the –y direction, (c) be deflected toward the +z direction, (d) be deflected toward the –z direction, (e) continue undeflected in the +x direction. The Magnetic Field 2495 Determine the Concept Application of the right-hand rule indicates that a positively charged body would experience a downward force and, in the absence of other forces, be deflected downward. Because the direction of the magnetic force on an electron is opposite that of the force on a positively charged object, an electron will be deflected upward. ) (c is correct. 7 • [SSM] In a velocity selector, the speed of the undeflected charged particle is given by the ratio of the magnitude of the electric field to the magnitude of the magnetic field. Show that E B in fact does have the units of m/s if E and B are in units of volts per meter and teslas, respectively. Determine the Concept Substituting the SI units for E and B yields: s m C m s C C m A m A N C N = ⋅ == ⋅ = ⋅ 8 • How are the properties of magnetic field lines similar to the properties of electric field lines? How are they different? Similarities Differences 1. Their density on a surface perpendicular to the lines is a measure of the strength of the field 2. The lines point in the direction of the field 3. The lines do not cross. 1. Magnetic field lines neither begin nor end. Electric field lines begin on positive charges and end on negative charges. 2. Electric forces are parallel or anti-parallel to the field lines. Magnetic forces are perpendicular to the field lines. 9 • True or false: (a) The magnetic moment of a bar magnet points from its north pole to its south pole. (b) Inside the material of a bar magnet, the magnetic field due to the bar magnet points from the magnet’s south pole toward its north pole. (c) If a current loop simultaneously has its current doubled and its area cut in half, then the magnitude of its magnetic moment remains the same. (d) The maximum torque on a current loop placed in a magnetic field occurs when the plane of the loop is perpendicular to the direction of the magnetic field. Chapter 26 2496 (a) False. By definition, the magnetic moment of a small bar magnet points from its south pole to its north pole. (b) True. The external magnetic field of a bar magnet points from the north pole of the magnet to south pole. Because magnetic field lines are continuous, the magnet’s internal field lines point from the south pole to the north pole. (c) True. Because the magnetic dipole moment of a current loop is given by n NIAr r = μ , simultaneously doubling the current and halving its area leaves the magnetic dipole moment unchanged. (d) False. The magnitude of the torque acting on a magnetic dipole moment is given θ μ τ sin B = where θ is the angle between the axis of the current loop and the direction of the magnetic field. When the plane of the loop is perpendicular to the field direction θ = 0 and the torque is zero. 10 •• Show that the von Klitzing constant, h e2 , gives the SI unit for resistance (the ohm) h and e are in units of joule seconds and coulombs, respectively. Determine the Concept The von Klitzing resistance is related to the Hall resistance according to where H K nR R = 2 K e h R = . Substituting the SI units of h and e yields: Ω A V s C C J C s J 2 = = = ⋅ 11 ••• The theory of relativity states that no law of physics can be described using the absolute velocity of an object, which is in fact impossible to define due to a lack of an absolute reference frame. Instead, the behavior of interacting objects can only be described by the relative velocities between the objects. New physical insights result from this idea. For example, in Figure 26-31 a magnet moving at high speed relative to some observer passes by an electron that is at rest relative to the same observer. Explain why you are sure that a force must be acting on the electron. In what direction will the force point at the instant the north pole of the magnet passes directly underneath the electron? Explain your answer. Determine the Concept From relativity; this is equivalent to the electron moving from right to left at speed v with the magnet stationary. When the electron is directly over the magnet, the field points directly up, so there is a force directed out of the page on the electron. The Magnetic Field 2497 Estimation and Approximation 12 • Estimate the maximum magnetic force per meter that Earth’s magnetic field could exert on a current-carrying wire in a 20-A circuit in your house. Picture the Problem Because the magnetic force on a current-carrying wire is given by B L F r r r × = I , the maximum force occurs when θ = 90°. Under this condition, . ILB F = max The maximum force per unit length that the Earth’s magnetic field could exert on a current-carrying wire in your home is given by: IB L F = ⎥ ⎦ ⎤ max For a 20-A circuit and B = 0.5 × 10−4 T: ( )( ) mN/m 1 T 10 5 .
0 A 20 4 max = × = ⎥ ⎦ ⎤ − L F 13 •• Your friend wants to be magician and intends to use Earth’s magnetic field to suspend a current-carrying wire above the stage. He asks you to estimate the minimum current needed to suspend the wire just above Earth’s surface at the equator (where Earth’s magnetic field is horizontal). Assume the wire has a linear mass density of 10 g/m. Would you advise him to proceed with his plans for this act? Picture the Problem Because the magnetic force on a current-carrying wire is given by B L F r r r × = I , the maximum force occurs when θ = 90°. Under this condition, . In order to suspend the wire, this magnetic force would have to be equal in magnitude to the gravitational force exerted by Earth on the wire: ILB F = max 0 g m = −F F or, 0 = −mg ILB Letting the upward direction be the +y direction, apply to the wire to obtain: 0 = ∑ y F Solving for I yields: B g L m LB mg I ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = = where m/L is the linear density of the wire. Chapter 26 2498 Substitute numerical values and evaluate I: ( ) kA 2 T 10 5 .
0 m/s 81 .
9 g/m 10 4 2 ≈ × = − I You should advise him to develop some other act. A current of 2000 A would overheat the wire (which is a gross understatement). The Force Exerted by a Magnetic Field 14 • Find the magnetic force on a proton moving in the +x direction at a speed of 0.446 Mm/s in a uniform magnetic field of 1.75 T in the +z direction. Picture the Problem The magnetic force acting on a charge is given by B v F r r r × = q . We can express v r and B r , form their vector (″cross″) product, and multiply by the scalar q to find F r . The magnetic force acting on the proton is given by: B v F r r r × = q Express v r : ( )i v ˆ Mm/s 446 .
0 = r Express : B r ( )k B ˆ T 75 .
1 = r Substitute numerical values and evaluate F r : ( ) ( ) ( ) [ ] ( )j k i F ˆ pN 125 .
0 ˆ T 75 .
1 ˆ Mm/s 446 .
0 C 10 602 .
1 19 − = × × = − r 15 • A point particle has a charge equal to –3.64 nC and a velocity equal to . Find the force on the charge if the magnetic field is i m/s 10 75 .
2 3 × (a) 0.38 T ˆ j , (b) 0.75 Tˆ i + 0.75 T ˆ j , (c) 0.65 T , and (d) 0.75 T . ˆ i ˆ i + 0.75 T ˆ k Picture the Problem The magnetic force acting on the charge is given by . We can express v B v F r r r × = q r and B r , form their vector (also known as the ″cross″) product, and multiply by the scalar q to find F r . Express the force acting on the charge: B v F r r r × = q Substitute numerical values to obtain: ( ) ( ) [ ] B i F r r × × − = ˆ m/s 10 75 .
2 nC 64 .
3 3 The Magnetic Field 2499 (a) Evaluate F r for = 0.38 T : B r j ˆ ( ) ( ) [ ( ) ] ( )k j i F ˆ N 8 .
3 ˆ T 38 .
0 ˆ m/s 10 75 .
2 nC 64 .
3 3 μ − = × × − = r (b) Evaluate F r for = 0.75 T + 0.75 T : B r i ˆ j ˆ ( ) ( ) [ ( ) ( ) { }] ( )k j i i F ˆ N 5 .
7 ˆ T 75 .
0 ˆ T 75 .
0 ˆ m/s 10 75 .
2 nC 64 .
3 3 μ − = + × × − = r (c) Evaluate F r for = 0.65 T : B r i ˆ ( ) ( ) ( ) [ ] 0 ˆ T 65 .
0 ˆ m/s 10 75 .
2 nC 64 .
3 3 = × × − = i i F r (d) Evaluate F r for B r = 0.75 T + 0.75 T : i ˆ k ˆ ( ) ( ) [ ( ) ( ) ] ( )j k i i F ˆ N 5 .
7 ˆ T 75 .
0 ˆ T 75 .
0 ˆ m/s 10 75 .
2 nC 64 .
3 3 μ = + × × − = r 16 • A uniform magnetic field equal to 1.48 T is in the +z direction. Find the force exerted by the field on a proton if the velocity of the proton is (a) , (b) , (c) , and (d) . i ˆ km/s 7 .
2 j ˆ km/s 7 .
3 k ˆ km/s 8 .
6 j i ˆ km/s 3.0 ˆ km/s 0 .
4 + Picture the Problem The magnetic force acting on the proton is given by . We can express v B v F r r r × = q r and B r , form their vector product, and multiply by the scalar q to find F r . The magnetic force acting on the proton is given by: B v F r r r × = q (a) Evaluate F r for v = 2.7 km/s : r i ˆ ( ) ( ) ( ) [ ] ( )j k i F ˆ N 10 4 .
6 ˆ T 48 .
1 ˆ km/s 7 .
2 C 10 602 .
1 16 19 − − × − = × × = r (b) Evaluate F r for v = 3.7 km/s : r j ˆ ( ) ( ) ( ) [ ] ( )i k j F ˆ N 10 8 .
8 ˆ T 48 .
1 ˆ km/s 7 .
3 C 10 602 .
1 16 19 − − × = × × = r Chapter 26 2500 (c) Evaluate F r for = 6.8 km/s : v r k ˆ ( ) ( ) ( ) [ ] 0 ˆ T 48 .
1 ˆ km/s 8 .
6 C 10 602 .
1 19 = × × = − k k F r (d) Evaluate F r for : j i ˆ km/s 0 .
3 ˆ km/s 0 .
4 + = v r ( ) ( ) ( ) { } ( ) [ ] ( ) ( ) j i k j i F ˆ N 10 5 .
9 ˆ N 10 1 .
7 ˆ T 48 .
1 ˆ km/s 0 .
3 ˆ km/s 0 .
4 C 10 602 .
1 16 16 19 − − − × − × = × + × = r 17 • A straight wire segment that is 2.0 m long makes an angle of 30º with a uniform 0.37-T magnetic field. Find the magnitude of the force on the wire if the wire carries a current of 2.6 A. Picture the Problem The magnitude of the magnetic force acting on a segment of wire carrying a current I is given by θ sin B I F l = where is the length of the segment of wire, B is the magnetic field, and θ is the angle between direction of the current in the segment of wire and the direction of the magnetic field. l Express the magnitude of the magnetic force acting on the segment of wire: θ sin B I F l = Substitute numerical values and evaluate F: ( )( )( ) N 96 .
0 30 sin T 37 .
0 m 0 .
2 A 6 .
2 = ° = F 18 • A straight segment of a current-carrying wire has a current element L I r , where I = 2.7 A and . The segment is in a region with a uniform magnetic field given by . Find the force on the segment of wire. j i L ˆ cm 0 .
4 ˆ cm 0 .
3 + = r i ˆ T 3 .
1 Picture the Problem We can use B L F r r r × = I to find the force acting on the wire segment. Express the force acting on the wire segment: B L F r r r × = I Substitute numerical values and evaluate F r : ( ) ( ) ( ) [ ] ( ) ( )k i j i F ˆ N 14 .
0 ˆ T 3 .
1 ˆ cm 0 .
4 ˆ cm 3.0 A 7 .
2 − = × + = r The Magnetic Field 2501 19 • What is the force on an electron that has a velocity equal to when it is in a region with a magnetic field given by ? j i ˆ 10 3.0 ˆ m/s 10 0 .
2 6 6 × − × k i ˆ T 0.60 ˆ T 80 .
0 + Picture the Problem The magnetic force acting on the electron is given by . B v F r r r × = q The magnetic force acting on the proton is given by: B v F r r r × = q Substitute numerical values and evaluate F r : ( ) ( ) { [ ( ) } ( ) ] ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )k j i k j i i k j k k j i j i F ˆ pN 58 .
0 ˆ pN 13 .
0 ˆ pN 19 .
0 ˆ pN 577 .
0 ˆ pN 128 .
0 ˆ pN 192 .
0 ˆ pN 192 .
0 ˆ pN 384 .
0 ˆ pN 128 .
0 ˆ pN 192 .
0 T ˆ 4 .
0 ˆ 6 .
0 ˆ 8 .
0 ˆ Mm/s 3 ˆ Mm/s 2 C 10 602 .
1 19 − − − = − − − = − + − + − + − = − + × − × − = − r 20 •• The section of wire shown in Figure 26-32 carries a current equal to 1.8 A from a to b. The segment is in a region that has a magnetic field whose value is . Find the total force on the wire and show that the total force is the same as if the wire were in the form of a straight wire directly from a to b and carrying the same current. k ˆ T 2 .
1 Picture the Problem We can use B F r l r r × = I to find the force acting on the segments of the wire as well as the magnetic force acting on the wire if it were a straight segment from a to b. Express the magnetic force acting on the wire: cm 4 cm 3 F F F r r r + = Evaluate : cm 3 F r ( ) ( ) ( ) [ ] ( )( ) ( )j j k i F ˆ N 0648 .
0 ˆ N 0648 .
0 ˆ T 2 .
1 ˆ cm 0 .
3 A 8 .
1 cm 3 − = − = × = r Evaluate : cm 4 F r ( ) ( ) ( ) [ ] ( )i k j F ˆ N 0864 .
0 ˆ T 2 .
1 ˆ cm 0 .
4 A 8 .
1 cm 4 = × = r Chapter 26 2502 Substitute to obtain: ( ) ( ) ( ) ( )j i i j F ˆ mN 65 ˆ mN 86 ˆ N 0864 .
0 ˆ N 0648 .
0 − = + − = r If the wire were straight from a to b: ( ) ( )j i ˆ cm 0 .
4 ˆ cm 0 .
3 + = l r The magnetic force acting on the wire is: ( ) ( ) ( ) [ ] ( ) ( ) ( ) ( ) ( )j i i j k j i F ˆ mN 65 ˆ mN 86 ˆ N 0864 .
0 ˆ N 0648 .
0 ˆ T 2 .
1 ˆ cm 0 .
4 ˆ cm 0 .
3 A 8 .
1 − = + − = × + = r in agreement with the result obtained above when we treated the two straight segments of the wire separately. 21 •• A straight, stiff, horizontal 25-cm-long wire that has a mass equal to 50 g is connected to a source of emf by light, flexible leads. A magnetic field of 1.33 T is horizontal and perpendicular to the wire. Find the current necessary to ″float″ the wire, that is, when it is released from rest it remains at rest. Picture the Problem Because the magnetic field is horizontal and perpendicular to the wire, the force it exerts on the current-carrying wire will be vertical. Under equilibrium conditions, this upward magnetic force will be equal to the downward gravitational force acting on the wire. Apply to the wire: 0 vertical = ∑F 0 mag = −w F Express : mag F B I F l = mag because θ = 90°. Substitute for to obtain: mag F 0 = −mg B Il ⇒ B mg I l = Substitute numerical values and evaluate I: ( )( ) ( )( ) A 5 .
1 T 33 .
1 cm 25 m/s 81 .
9 g 50 2 = = I 22 •• In your physics laboratory class, you have constructed a simple gaussmeter for measuring the horizontal component of magnetic fields. The setup consists of a stiff 50-cm wire that hangs vertically from a conducting pivot so that its free end makes contact with a pool of mercury in a dish below (Figure 26-33). The mercury provides an electrical contact without constraining the movement of the wire. The wire has a mass of 5.0 g and conducts a current downward. (a) What is the equilibrium angular displacement of the wire from vertical if the horizontal component of the magnetic field is 0.040 T and the current is 0.20 A? The Magnetic Field 2503 (b) What is the sensitivity of this gaussmeter? That is, what is the ratio of the output to the input (in radians per tesla). Picture the Problem The magnetic field is out of the page. The diagram shows the gaussmeter displaced from equilibrium under the influence of the gravitational force , the magnetic force g r m m F r , and the force exerted by the conducting pivot F r . We can apply the condition for translational equilibrium in the x direction to find the equilibrium angular displacement of the wire from the vertical. In Part (b) we can solve the equation derived in Part (a) for B and evaluate this expression for the given data to find the horizontal magnetic field sensitivity of this gaussmeter. θ θ g mr y m F r F r x θ θ (a) Apply to the wire to obtain: 0 = ∑ x F 0 sin m = + − = ∑ θ mg F Fx φ sin m B I F l = or, because φ = 90°, B I F l = m The magnitude of the magnetic force acting on the wire is given by: Substitute for Fm to obtain: 0 sin = + − θ mg B Il (1) Solving for θ yields: ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ = − mg B Il 1 sin θ Substitute numerical values and evaluate θ: ( )( )( ) ( )( ) mrad 82 7 .
4 679 .
4 m/s 81 .
9 kg 0050 .
0 T 040 .
0 m 50 .
0 A 20 .
0 sin 2 1 = ° = ° = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ = − θ (b) The sensitivity of this gaussmeter is the ratio of the output to the input: B θ = y sensitivit Chapter 26 2504 rad/T 0 .
2 T 040 .
0 mrad 82 y sensitivit = = Substitute numerical values and evaluate the sensitivity of the gaussmeter: 23 •• [SSM] A 10-cm long straight wire is parallel with the x axis and carries a current of 2.0 A in the +x direction. The force on this wire due to the presence of a magnetic field r B is 3.
. If this wire is rotated so that it is parallel with the y axis with the current in the +y direction, the force on the wire becomes − . Determine the magnetic field 0 N ˆ j + 2.0 N ˆ k 3.0 N ˆ i −2.0 N ˆ k r B . Picture the Problem We can use the information given in the 1st and 2nd sentences to obtain an expression containing the components of the magnetic field . We can then use the information in the 1 B r st and 3rd sentences to obtain a second equation in these components that we can solve simultaneously for the components of B r . Express the magnetic field in terms of its components: B r k j i B ˆ ˆ ˆ z y x B B B + + = r (1) Express F r in terms of : B r ( )( ) ] [ ( ) ( ) ( ) ( ) ( ) k B j B k B j B i B i k B j B i B i B I F y z z y x z y x ˆ m A 20 .
0 ˆ m A 20 .
0 ˆ ˆ ˆ ˆ m A 20 .
0 ˆ ˆ ˆ ˆ m 10 .
0 A 0 .
2 ⋅ + ⋅ − = + + × ⋅ = + + × = × = r l r r ( ) N 0 .
3 m A 20 .
0 = ⋅ − z B and ( ) N 0 .
2 m A 20 .
0 = ⋅ y B Equate the components of this expression for F r with those given in the second sentence of the statement of the problem to obtain: Noting that BBx is undetermined, solve for Bz B and BBy: T 15 − = z B and T 10 = y B When the wire is rotated so that the current flows in the positive y direction: ( )( ) ] [ ( ) ( ) ( ) ( ) ( ) k B i B k B j B i B j k B j B i B j B I F x z z y x z y x ˆ m A 20 .
0 ˆ m A 20 .
0 ˆ ˆ ˆ ˆ m A 20 .
0 ˆ ˆ ˆ ˆ m 10 .
0 A 0 .
2 ⋅ − ⋅ = + + × ⋅ = + + × = × = r l r r The Magnetic Field 2505 ( ) N 0 .
2 m A 20 .
0 − = ⋅ x B and ( ) N 0 .
3 m A 20 .
0 − = ⋅ − z B Equate the components of this expression for F r with those given in the third sentence of the problem statement to obtain: Solve for BBx and Bz B to obtain: T 10 = x B and, in agreement with our results above, T 15 − = z B Substitute in equation (1) to obtain: ( ) ( ) ( )k j i B ˆ T 15 ˆ T 10 ˆ T 10 − + = r 24 •• A 10-cm long straight wire is parallel with the z axis and carries a current of 4.0 A in the + z direction. The force on this wire due to a uniform magnetic field r B is −0.20 N ˆ i + 0.20 N ˆ j . If this wire is rotated so that it is parallel with the x axis with the current is in the +x direction, the force on the wire becomes . Find r 0.20 ˆ k N B . Picture the Problem We can use the information given in the 1st and 2nd sentences to obtain an expression containing the components of the magnetic field . We can then use the information in the 1 B r st and 3rd sentences to obtain a second equation in these components that we can solve simultaneously for the components of B r . Express the magnetic field in terms of its components: B r k j i B ˆ ˆ ˆ z y x B B B + + = r (1) Express F r in terms of : B r ( )( ) ] [ ( ) ( ) ( ) ( ) ( ) i B j B k B j B i B k k B j B i B k B I F y y z y x z y x ˆ m A 40 .
0 ˆ m A 40 .
0 ˆ ˆ ˆ ˆ m A 40 .
0 ˆ ˆ ˆ ˆ m 1 .
0 A 0 .
4 ⋅ − ⋅ = + + × ⋅ = + + × = × = r l r r ( ) N 20 .
0 m A 40 .
0 = ⋅ y B Equate the components of this expression for F r with those given in the second sentence of the statement of the problem to obtain: and ( ) N 20 .
0 m A 40 .
0 = ⋅ x B Noting that BBz is undetermined, solve for Bx B and BBy: T 50 .
0 = x B and T 50 .
0 = y B Chapter 26 2506 When the wire is rotated so that the current flows in the positive x direction: ( )( ) ] [ ( ) ( ) ( ) ( ) ( ) k B j B k B j B i B i k B j B i B i B I F y z z y x z y x ˆ m A 40 .
0 ˆ m A 40 .
0 ˆ ˆ ˆ ˆ m A 40 .
0 ˆ ˆ ˆ ˆ m 10 .
0 A 0 .
4 ⋅ + ⋅ − = + + × ⋅ = + + × = × = r l r r ( ) 0 m A 40 .
0 = ⋅ − z B and ( ) N 2 .
0 m A 40 .
0 = ⋅ y B Equate the components of this expression for F r with those given in the third sentence of the problem statement to obtain: Solve for BBz and By B to obtain: 0 = z B and, in agreement with our results above, T 50 .
0 = y B Substitute in equation (1) to obtain: ( ) ( )j i B ˆ T 50 .
0 ˆ T 50 .
0 + = r 25 •• [SSM] A current-carrying wire is bent into a closed semicircular loop of radius R that lies in the xy plane (Figure 26-34). The wire is in a uniform magnetic field that is in the +z direction, as shown. Verify that the force acting on the loop is zero. Picture the Problem With the current in the direction indicated and the magnetic field in the z direction, pointing out of the plane of the page, the force is in the radial direction and we can integrate the element of force dF acting on an element of length dℓ between θ = 0 and π to find the force acting on the semicircular portion of the loop and use the expression for the force on a current-carrying wire in a uniform magnetic field to find the force on the straight segment of the loop. Express the net force acting on the semicircular loop of wire: segment straight loop ar semicircul F F F r r r + = (1) The Magnetic Field 2507 Express the force acting on the straight segment of the loop: j RIB k B i RI B I F ˆ 2 ˆ ˆ 2 segment straight − = × = × = r l r r Express the force dF acting on the element of the wire of length dℓ: θ IRBd B Id dF = = l Express the x and y components of dF: θ cos dF dFx = and θ sin dF dFy = θ θ d IRB dFy sin = Because, by symmetry, the x component of the force is zero, we can integrate the y component to find the force on the wire: and j RIB j d RIB j F F y ˆ 2 ˆ sin ˆ 0 loop ar semicircul = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = = ∫ π θ θ r Substitute in equation (1) to obtain: 0 ˆ 2 ˆ 2 = − = j RIB j RIB F r 26 ••• A wire bent in some arbitrary shape carries a current I. The wire is in a region with a uniform magnetic field r B . Show that the total force on the part of the wire from some arbitrary point on the wire (designated as a) to some other arbitrary point on the wire (designated as b) is r F = I r L × r B , where r L is the vector from point a to point b. In other words, show that the force on an arbitrary section of the bent wire is the same as the force would be on a straight section wire carrying the same current and connecting the two endpoints of the arbitrary section. Picture the Problem We can integrate the expression for the force acting on an element of the wire of length F r d L r d from a to b to show that .
B L F r r r × = I Express the force acting on the element of the wire of length F r d : L r d B L F r r r × = Id d Integrate this expression to obtain: ∫ × = b a Id B L F r r r Because and I are constant: B r B L B L F r r r r r × = × ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = ∫ I d I b a where L r is the vector from a to b. Chapter 26 2508 Motion of a Point Charge in a Magnetic Field 27 • [SSM] A proton moves in a 65-cm-radius circular orbit that is perpendicular to a uniform magnetic field of magnitude 0.75 T. (a) What is the orbital period for the motion? (b) What is the speed of the proton? (c) What is the kinetic energy of the proton? Picture the Problem We can apply Newton’s 2nd law to the orbiting proton to relate its speed to its radius. We can then use T = 2πr/v to find its period. In Part (b) we can use the relationship between T and v to determine v. In Part (c) we can use its definition to find the kinetic energy of the proton. (a) Relate the period T of the motion of the proton to its orbital speed v: v r T π 2 = (1) Apply Newton’s 2nd law to the proton to obtain: r v m qvB 2 = ⇒ qB mv r = Substitute for r in equation (1) and simplify to obtain: qB m T π 2 = Substitute numerical values and evaluate T: ( ) ( )( ) ns 87 ns 4 .
87 T 75 .
0 C 10 602 .
1 kg 10 673 .
1 2 19 27 = = × × = − − π T (b) From equation (1) we have: T r v π 2 = Substitute numerical values and evaluate v: ( ) m/s 10 7 .
4 m/s 10 67 .
4 ns 4 .
87 m 65 .
0 2 7 7 × = × = = π v (c) Using its definition, express and evaluate the kinetic energy of the proton: ( )( ) MeV 11 J 10 1.602 eV 1 J 10 82 .
1 m/s 10 67 .
4 kg 10 673 .
1 19 12 2 7 27 2 1 2 2 1 = × × × = × × = = − − − mv K The Magnetic Field 2509 28 • A 4.5-keV electron (an electron that has a kinetic energy equal to 4.5 keV) moves in a circular orbit that is perpendicular to a magnetic field of 0.325 T. (a) Find the radius of the orbit. (b) Find the frequency and period of the orbital motion. Picture the Problem (a) We can apply Newton’s 2nd law to the orbiting electron to obtain an expression for the radius of its orbit as a function of its mass m, charge q, speed v, and the magnitude of the magnetic field B. Using the definition of kinetic energy will allow us to express r in terms of m, q, B, and the electron’s kinetic energy K. (b) The period of the orbital motion is given by T = 2πr/v. Substituting for r (or r/v) from Part (a) will eliminate the orbital speed of the electron and leave us with an expression for T that depends only on m, q, and B. The frequency of the orbital motion is the reciprocal of the period of the orbital motion. (a) Apply Newton’s 2nd law to the orbiting electron to obtain: r v m qvB 2 = ⇒ qB mv r = Express the kinetic energy of the electron: 2 2 1 mv K = ⇒ m K v 2 = Substituting for v in the expression for r and simplifying yields: qB Km m K qB m r 2 2 = = Substitute numerical values and evaluate r: ( ) ( ) ( )( ) mm 70 .
0 mm 696 .
0 T 325 .
0 C 10 .602 1 eV J 10 1.602 kg 10 109 .
9 keV 5 .
4 2 19 19 31 = = × ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ × × = − − − r (b) Relate the period of the electron’s motion to the radius of its orbit and its orbital speed: v r T π 2 = Because qB mv r = : qB m v qB mv T π π 2 2 = = Chapter 26 2510 Substitute numerical values and evaluate T: ( ) ( )( ) ns 11 .
0 s 10 099 .
1 T 325 .
0 C 10 1.602 kg 10 109 .
9 2 10 19 31 = × = × × = − − − π T The frequency of the motion is known as the cyclotron frequency and is the reciprocal of the period of the electron’s motion: GHz 1 .
9 ns 110 .
0 1 1 = = = T f 29 •• A proton, a deuteron and an alpha particle in a region with a uniform magnetic field each follow circular paths that have the same radius. The deuteron has a charge that is equal to the charge a proton has, and an alpha particle has a charge that is equal to twice the charge a proton has. Assume that mα = 2md = 4mp. Compare (a) their speeds, (b) their kinetic energies, and (c) the magnitudes of their angular momenta about the centers of the orbits. Picture the Problem We can apply Newton’s 2nd law to the orbiting particles to derive an expression for their orbital speeds as a function of their charge, their mass, the magnetic field in which they are moving, and the radii of their orbits. We can then compare their speeds by expressing their ratios. In Parts (b) and (c) we can proceed similarly starting with the definitions of kinetic energy and angular momentum. (a) Apply Newton’s 2nd law to an orbiting particle to obtain: r v m qvB 2 = ⇒ m qBr v = p p p m Br q v = , (1) α α α m Br q v = , and (2) d d d m Br q v = (3) The speeds of the orbiting particles are given by: Divide equation (2) by equation (1) and simplify to obtain: ( ) 2 1 p p p p p p p 4 2 = = = = m e em m q m q m Br q m Br q v v α α α α α or p 2 v v = α The Magnetic Field 2511 Divide equation (3) by equation (1) and simplify to obtain: ( ) 2 1 p p d p p d p p d d p d 2 = = = = m e em m q m q m Br q m Br q v v or p d 2 v v = Combining these results yields: p d 2 2 v v v = = α (b) Using the expression for its orbital speed derived in (a), express the kinetic energy of an orbiting particle: m r B q m qBr m mv K 2 2 2 2 2 2 1 2 2 1 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = = The kinetic energies of the three particles are given by: p 2 2 2 p p 2m r B q K = , (4) α α α m r B q K 2 2 2 2 = , and (5) d 2 2 2 d d 2m r B q K = (6) Dividing equation (7) by equation (6) and simplifying yields: ( ) ( ) p p 2 p 2 2 p p 2 p 2 2 2 p 2 1 2 2 2 2 1 p 1 4 2 K K m e m e m q m q m r B q m r B q K K = ⇒ = = = = α α α α α α Divide equation (8) by equation (6) and simplify to obtain: ( ) d p 2 1 p 2 p 2 d 2 p p 2 d p 2 2 2 p d 2 2 2 d p d 2 2 2 2 K K m e m e m q m q m r B q m r B q K K = ⇒ = = = = Combining these results yields: p d 2 K K K = = α Chapter 26 2512 r v m L p p p = , r v m L α α α = , and r v m L d d d = (c) The angular momenta of the orbiting particles are given by: ( )( ) 2 4 p p p 2 1 p p p p = = = v m v m r v m r v m L L α α α Express the ratio of Lα to Lp: or p 2L L = α ( )( ) 1 2 p p p 2 1 p p p d d p d = = = v m v m r v m r v m L L Express the ratio of Ld to Lp: or p d L L = Combining these results yields: p d 2 2 L L L = = α 30 •• A particle has a charge q, a mass m, a linear momentum of magnitude p and a kinetic energy K. The particle moves in a circular orbit of radius R perpendicular to a uniform magnetic field B r . Show that (a) p = BqR and (b) K = 1 2 B2q2R 2 / m . Picture the Problem We can use the definition of momentum to express p in terms of v and apply Newton’s 2nd law to the orbiting particle to express v in terms of q, B, R, and m. In Part (b) we can express the particle’s kinetic energy in terms of its momentum and use our result from Part (a) to show that .
2 2 2 2 1 m R q B K = (a) Express the momentum of the particle: mv p = (1) Apply to the orbiting particle to obtain: c radial ma F = ∑ R v m qvB 2 = ⇒ m qBR v = Substitute for v in equation (1) to obtain: qBR m qBR m p = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = The Magnetic Field 2513 (b) Express the kinetic energy of the orbiting particle as a function of its momentum: m p K 2 2 = Substitute our result for p from Part (a) to obtain: ( ) m R B q m qBR K 2 2 2 2 2 2 = = 31 •• [SSM] A beam of particles with velocity r v enters a region that has a uniform magnetic field r B in the +x direction. Show that when the x component of the displacement of one of the particles is 2π(m/qB)v cos θ, where θ is the angle between and r v r B , the velocity of the particle is in the same direction as it was when the particle entered the field. Picture the Problem The particle’s velocity has a component v1 parallel to B r and a component v2 normal to B r . v1 = v cosθ and is constant, whereas v2 = v sinθ , being normal to , will result in a magnetic force acting on the beam of particles and circular motion perpendicular to B r B r . We can use the relationship between distance, rate, and time and Newton’s 2nd law to express the distance the particle moves in the direction of the field during one period of the motion. Express the distance moved in the direction of by the particle during one period: B r T v x 1 = (1) Express the period of the circular motion of the particles in the beam: 2 2 v r T π = (2) Apply Newton’s 2nd law to a particle in the beam to obtain: r v m B qv 2 2 2 = ⇒ m qBr v = 2 Substituting for v2 in equation (2) and simplifying yields: qB m m qBr r T π π 2 2 = = Because v1 = v cosθ, equation (1) becomes: ( ) θ π π θ cos 2 2 cos v qB m qB m v x ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = 32 •• A proton that has a speed equal to 1.00 × 106 m/s enters a region that has a uniform magnetic field that has a magnitude of 0.800 T and points into the Chapter 26 2514 page, as shown in Figure 26-35. The proton enters the region at an angle θ = 60º. Find the exit angle φ and the distance d. Picture the Problem The trajectory of the proton is shown to the right. We know that, because the proton enters the uniform field perpendicularly to the field, its trajectory while in the field will be circular. We can use symmetry considerations to determine φ. The application of Newton’s 2nd law to the proton while it is in the magnetic field and of trigonometry will allow us to conclude that r = d and to determine the value of d. From symmetry, it is evident that the angle θ in Figure 26-35 equals the angle φ: ° = 60 φ Apply to the proton while it is in the magnetic field to obtain: c radial ma F = ∑ r v m qvB 2 = ⇒ qB mv r = Use trigonometry to obtain: ( ) r d 2 2 1 30 sin 90 sin = = ° = − ° θ Solving for d yields: r = d Substitute for r to obtain: qB mv d = Substitute numerical values and evaluate d: ( )( ) ( )( ) mm 1 .
13 T 800 .
0 C 10 602 .
1 m/s 10 00 .
1 kg 10 673 .
1 19 6 27 = × × × = = − − r d 33 •• [SSM] Suppose that in Figure 26-35, the magnetic field has a magnitude of 60 mT, the distance d is 40 cm, and θ is 24º. Find the speed v at which a particle enters the region and the exit angle φ if the particle is a (a) proton and (b) deuteron. Assume that md = 2mp. The Magnetic Field 2515 Picture the Problem The trajectory of the proton is shown to the right. We know that, because the proton enters the uniform field perpendicularly to the field, its trajectory while in the field will be circular. We can use symmetry considerations to determine φ. The application of Newton’s 2nd law to the proton and deuteron while they are in the uniform magnetic field will allow us to determine the values of vp and vd. (a) From symmetry, it is evident that the angle θ in Figure 26-35 equals the angle φ: ° = 24 φ Apply to the proton while it is in the magnetic field to obtain: c radial ma F = ∑ p 2 p p p p r v m B v q = ⇒ p p p p m B r q v = (1) Use trigonometry to obtain: ( ) r d 2 66 sin 90 sin = ° = − ° θ Solving for r yields: ° = 66 sin 2 d r Substituting for r in equation (1) and simplifying yields: ° = 66 sin 2 p p p m Bd q v (2) Substitute numerical values and evaluate vp: ( )( )( ) ( ) m/s 10 3 .
1 66 sin kg 10 1.673 2 m 40 .
0 mT 60 C 10 602 .
1 6 27 19 p × = ° × × = − − v (b) From symmetry, it is evident that the angle θ in Figure 26-35 equals the angle φ: ° = 24 φ independently of whether the particles are protons or deuterons. For deuterons equation (2) becomes: ° = 66 sin 2 d d d m Bd q v Chapter 26 2516 Because and : p d 2m m = p d q q = ( ) ° = ° ≈ 66 sin 4 66 sin 2 2 p p p p d m Bd q m Bd q v Substitute numerical values and evaluate vd: ( )( )( ) ( ) m/s 10 3 .
6 66 sin kg 10 1.673 4 m 40 .
0 mT 60 C 10 602 .
1 5 27 19 d × = ° × × = − − v 34 •• The galactic magnetic field in some region of interstellar space has a magnitude of 1.00 × 10–9 T. A particle of interstellar dust has a mass of 10.0 μg and a total charge of 0.300 nC. How many years does it take for the particle to complete revolution of the circular orbit caused by its interaction with the magnetic field? Picture the Problem We can apply Newton’s 2nd law of motion to express the orbital speed of the particle and then find the period of the dust particle from this orbital speed. Assume that the particle moves in a direction perpendicular to B r . The period of the dust particle’s motion is given by: v r T π 2 = Apply to the particle: c ma F = ∑ r v m qvB 2 = ⇒ m qBr v = Substitute for v in the expression for T and simplify: qB m qBr rm T π π 2 2 = = Substitute numerical values and evaluate T: ( ) ( )( ) y 10 64 .
6 Ms 31.56 y 1 s 10 094 .
2 T 10 00 .
1 nC 300 .
0 kg/g 10 g 10 0 .
10 2 3 11 9 3 6 × = × × = × × × = − − − π T Applications of the Magnetic Force Acting on Charged Particles 35 • [SSM] A velocity selector has a magnetic field that has a magnitude equal to 0.28 T and is perpendicular to an electric field that has a magnitude equal to 0.46 MV/m. (a) What must the speed of a particle be for that particle to pass through the velocity selector undeflected? What kinetic energy must (b) protons and (c) electrons have in order to pass through the velocity selector undeflected? The Magnetic Field 2517 Picture the Problem Suppose that, for positively charged particles, their motion is from left to right through the velocity selector and the electric field is upward. Then the magnetic force must be downward and the magnetic field out of the page. We can apply the condition for translational equilibrium to relate v to E and B. In (b) and (c) we can use the definition of kinetic energy to find the energies of protons and electrons that pass through the velocity selector undeflected. (a) Apply to the particle to obtain: 0 = ∑ y F 0 mag elec = −F F or 0 = −qvB qE ⇒ B E v = Substitute numerical values and evaluate v: m/s 10 6 .
1 m/s 10 64 .
1 T 28 .
0 MV/m 46 .
0 6 6 × = × = = v (b) The kinetic energy of protons passing through the velocity selector undeflected is: ( )( ) keV 14 J 10 1.602 eV 1 J 10 26 .
2 m/s 10 64 .
1 10 673 .
1 19 15 2 6 kg 27 2 1 2 p 2 1 p = × × × = × × = = − − − v m K (c) The kinetic energy of electrons passing through the velocity selector undeflected is: ( )( ) eV 7 .
7 J 10 1.602 eV 1 J 10 23 .
1 m/s 10 64 .
1 10 109 .
9 19 18 2 6 kg 31 2 1 2 e 2 1 e = × × × = × × = = − − − v m K 36 •• A beam of protons is moving in the +x direction with a speed of 12.4 km/s through a region in which the electric field is perpendicular to the magnetic field. The beam is not deflected in this region. (a) If the magnetic field has a magnitude of 0.85 T and points in the + y direction, find the magnitude and direction of the electric field. (b) Would electrons that have the same velocity as the protons be deflected by these fields? If so, in what direction would they be deflected? If not, why not? Chapter 26 2518 Picture the Problem Because the beam of protons is not deflected; we can conclude that the electric force acting on them is balanced by the magnetic force. Hence, we can find the magnetic force from the given data and use its definition to express the electric field. (a) Use the definition of electric field to relate it to the electric force acting on the beam of protons: q elec elec F E r r = Express the magnetic force acting on the beam of protons: k j i F ˆ ˆ ˆ mag qvB B qv = × = r Because the electric force must be equal in magnitude but opposite in direction: ( )( )( ) ( )k k k qvB F ˆ N 10 689 .
1 ˆ T 85 .
0 km/s 4 .
12 C 10 .602 1 ˆ 15 19 elec − − × − == × − = − = r Substitute in the equation for the electric field to obtain: ( ) ( )k k E ˆ kV/m 11 C 10 .602 1 ˆ N 10 689 .
1 19 15 elec − = × × − = − − r (b) Because both and would be reversed, electrons are not deflected either. mag F r elec F r 37 •• The plates of a Thomson q/m apparatus are 6.00 cm long and are separated by 1.20 cm. The end of the plates is 30.0 cm from the tube screen. The kinetic energy of the electrons is 2.80 keV. If a potential difference of 25.0 V is applied across the deflection plates, by how much will the point where the beam strikes the screen displaced? Picture the Problem Figure 26-18 is reproduced below. We can express the total deflection of the electron beam as the sum of the deflections while the beam is in the field between the plates and its deflection while it is in the field-free space. We can, in turn, use constant-acceleration equations to express each of these deflections. The resulting equation is in terms of v0 and E. We can find v0 from the kinetic energy of the beam and E from the potential difference across the plates and their separation. In Part (b) we can equate the electric and magnetic forces acting on an electron to express B in terms of E and v0. The Magnetic Field 2519 Express the total deflection Δy of the electrons: 2 1 y y y Δ + Δ = Δ (1) where Δy1 is the deflection of the beam while it is in the electric field and Δy2 is the deflection of the beam while it travels along a straight-line path outside the electric field. Use a constant-acceleration equation to express Δy1: ( ) 2 2 1 1 t a y y Δ = Δ (2) where Δt = x1/v0 is the time an electron is in the electric field between the plates. Apply Newton’s 2nd law to an electron between the plates to obtain: y ma qE = ⇒ m qE ay = Substitute for ay in equation (2) to obtain: 2 0 1 2 1 1 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = Δ v x m qE y (3) Express the vertical deflection Δy2 of the electrons once they are out of the electric field: 2 2 t v y yΔ = Δ (4) Use a constant-acceleration equation to find the vertical speed of an electron as it leaves the electric field: ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + = Δ + = 0 1 1 0 0 v x m qE t a v v y y y Substitute in equation (4) to obtain: 2 0 2 1 0 2 0 1 2 mv x qEx v x v x m qE y = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = Δ (5) Chapter 26 2520 2 0 2 1 2 0 1 2 1 mv x qEx v x m qE y + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = Δ Substitute equations (3) and (5) in equation (1) to obtain: or ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = Δ 2 1 2 0 1 2 x x mv qEx y (6) Use the definition of kinetic energy to express the square of the speed of the electrons: 2 0 2 1 mv K = ⇒ m K v 2 2 0 = Express the electric field between the plates in terms of their potential difference: d V E = Substituting for E and in equation (6) and simplifying yields: 2 0 v ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = 2 1 1 2 1 1 2 2 2 2 Δ x x dK qVx x x m K m x d V q y Substitute numerical values and evaluate Δy: ( )( )( ) ( )( ) mm 37 .
7 cm 0 .
30 2 cm 00 .
6 keV 80 .
2 cm 20 .
1 2 cm 00 .
6 V 5.0 2 C 10 602 .
1 Δ 19 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + × = − y 38 •• Chlorine has two stable isotopes, 35Cl and 37Cl. Chlorine gas which consists of singly-ionized ions is to be separated into its isotopic components using a mass spectrometer. The magnetic field strength in the spectrometer is 1.2 T. What is the minimum value of the potential difference through which these ions must be accelerated so that the separation between them, after they complete their semicircular path, is 1.4 cm? Picture the Problem The diagram below represents the paths of the two ions entering the magnetic field at the left. The magnetic force acting on each causes them to travel in circular paths of differing radii due to their different masses. We can apply Newton’s 2nd law to an ion in the magnetic field to obtain an expression for its radius and then express their final separation in terms of these radii that, in turn, depend on the energy with which the ions enter the field. We can connect their energy to the potential through which they are accelerated using the work-kinetic energy theorem and relate their separation Δs to the accelerating potential difference ΔV. The Magnetic Field 2521 Express the separation Δs of the chlorine ions: ( ) 35 37 2 r r s − = Δ (1) Apply Newton’s 2nd law to an ion in the magnetic field of the mass spectrometer: r v m qvB 2 = ⇒ qB mv r = (2) Relate the speed of an ion as it enters the magnetic field to the potential difference through which it has been accelerated: 2 2 1 mv V q = Δ ⇒ m V q v Δ = 2 Substitute for v in equation (2) to obtain: 2 2 2 qB V m m V q qB m r Δ = Δ = Use this equation to express the radii of the paths of the two chlorine isotopes to obtain: 2 35 35 2 qB V m r Δ = and 2 37 37 2 qB V m r Δ = Substitute for r35 and r37 in equation (1) to obtain: ( )⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − Δ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ Δ − Δ = Δ 35 37 2 35 2 35 2 1 2 2 2 2 m m q V B qB V m qB V m s Solving for ΔV yields: ( ) 2 35 37 2 2 2 2 m m s qB V − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛Δ = Δ Chapter 26 2522 Substitute numerical values and evaluate ΔV: ( )( ) ( ) ( ) ( ) MV 12 .
0 kg 10 1.66 35 37 m T C 10 65 .
5 u 35 u 37 2 2 cm 4 .
1 T 2 .
1 C 10 1.602 Δ 27 2 2 2 24 2 2 2 19 = × − ⋅ ⋅ × = − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × = − − − V 39 •• [SSM] In a mass spectrometer, a singly ionized 24Mg ion has a mass equal to 3.983 × 10–26 kg and is accelerated through a 2.50-kV potential difference. It then enters a region where it is deflected by a magnetic field of 557 G. (a) Find the radius of curvature of the ion’s orbit. (b) What is the difference in the orbital radii of the 26Mg and 24Mg ions? Assume that their mass ratio is 26:24. Picture the Problem We can apply Newton’s 2nd law to an ion in the magnetic field to obtain an expression for r as a function of m, v, q, and B and use the work-kinetic energy theorem to express the kinetic energy in terms of the potential difference through which the ion has been accelerated. Eliminating v between these equations will allow us to express r in terms of m, q, B, and ΔV. Apply Newton’s 2nd law to an ion in the magnetic field of the mass spectrometer: r v m qvB 2 = ⇒ qB mv r = (1) Apply the work-kinetic energy theorem to relate the speed of an ion as it enters the magnetic field to the potential difference through which it has been accelerated: 2 2 1 mv V q = Δ ⇒ m V q v Δ = 2 Substitute for v in equation (1) and simplify to obtain: 2 2 2 qB V m m V q qB m r Δ = Δ = (2) (a) Substitute numerical values and evaluate equation (2) for 24Mg : ( )( ) ( )( ) cm 3 .
63 T 10 557 C 10 1.602 kV 50 .
2 kg 10 983 .
3 2 2 4 19 26 24 = × × × = − − − r The Magnetic Field 2523 (b) Express the difference in the radii for 24Mg and 26Mg: 24 26 r r r − = Δ Substituting for r26 and r24 and simplifying yields: ( ) ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = − = − = 1 24 26 Δ 2 1 24 26 Δ 2 1 Δ 2 1 Δ 2 Δ 2 Δ 24 24 24 24 26 2 24 2 26 q Vm B m m q V B m m q V B qB V m qB V m r Substitute numerical values and evaluate Δr: ( )( ) cm 58 .
2 1 24 26 C 10 1.602 kg 10 983 .
3 kV 50 .
2 2 T 10 557 1 Δ 19 26 4 = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − × × × = − − − r 40 •• A beam of singly ionized 6Li and 7Li ions passes through a velocity selector and enters a region of uniform magnetic field with a velocity that is perpendicular to the direction of the field. If the diameter of the orbit of the 6Li ions is 15 cm, what is the diameter of the orbit for 7Li ions? Assume their mass ratio is 7:6. Picture the Problem We can apply Newton’s 2nd law to an ion in the magnetic field of the spectrometer to relate the diameter of its orbit to its charge, mass, velocity, and the magnetic field. If we assume that the velocity is the same for the two ions, we can then express the ratio of the two diameters as the ratio of the masses of the ions and solve for the diameter of the orbit of 7Li. Apply Newton’s 2nd law to an ion in the field of the spectrometer: r v m qvB 2 = ⇒ qB mv r = Express the diameter of the orbit: qB mv d 2 = The diameters of the orbits for 6Li and 7Li are: qB v m d 6 6 2 = and qB v m d 7 7 2 = Assume that the velocities of the two ions are the same and divide the 2nd of these diameters by the first to obtain: 6 7 6 7 6 7 2 2 m m qB v m qB v m d d = = Chapter 26 2524 Solve for and evaluate d7: ( ) cm 18 cm 15 6 7 6 6 7 7 = = = d m m d 41 •• Using Example 26- 6, determine the time required for a 58Ni ion and a 60Ni ion to complete the semicircular path. Picture the Problem The time required for each of the ions to complete its semicircular paths is half its period. We can apply Newton’s 2nd law to an ion in the magnetic field of the spectrometer to obtain an expression for r as a function of the charge and mass of the ion, its velocity, and the magnetic field. Express the time for each ion to complete its semicircular path: v r T t π = = Δ 2 1 Apply Newton’s 2nd law to an ion in the field of the spectrometer: r v m qvB 2 = ⇒ qB mv r = Substitute for r to obtain: qB m t π = Δ ( ) ( )( ) s 7 .
15 T 120 .
0 C 10 1.602 kg 10 6606 .
1 58 Δ 19 27 58 μ π = × × = − − t Substitute numerical values and evaluate Δt58 and Δt60: and ( ) ( )( ) s 3 .
16 T 120 .
0 C 10 1.602 kg 10 6606 .
1 60 Δ 19 27 60 μ π = × × = − − t 42 •• Before entering a mass spectrometer, ions pass through a velocity selector consisting of parallel plates that are separated by 2.0 mm and have a potential difference of 160 V. The magnetic field strength is 0.42 T in the region between the plates. The magnetic field strength in the mass spectrometer is 1.2 T. Find (a) the speed of the ions entering the mass spectrometer and (b) the difference in the diameters of the orbits of singly ionized 238U and 235U. The mass of a 235U ion is 3.903 × 10–25 kg. Picture the Problem We can apply a condition for equilibrium to ions passing through the velocity selector to obtain an expression relating E, B, and v that we can solve for v. We can, in turn, express E in terms of the potential difference V between the plates of the selector and their separation d. In (b) we can apply The Magnetic Field 2525 Newton’s 2nd law to an ion in the bending field of the spectrometer to relate its diameter to its mass, charge, velocity, and the magnetic field. (a) Apply to the ions in the crossed fields of the velocity selector to obtain: 0 = ∑ y F 0 mag elec = −F F or 0 = −qvB qE ⇒ B E v = Express the electric field between the plates of the velocity selector in terms of their separation and the potential difference across them: d V E = Substituting for E yields: dB V v = Substitute numerical values and evaluate v: ( )( ) m/s 10 9 .
1 m/s 10 905 .
1 T 42 .
0 mm 2.0 V 160 5 5 × = × = = v (b) Express the difference in the diameters of the orbits of singly ionized 238U and 235U: 235 238 d d d − = Δ (1) Apply to an ion in the spectrometer’s magnetic field: c radial ma F = ∑ r v m qvB 2 = ⇒ qB mv r = Express the diameter of the orbit: qB mv d 2 = The diameters of the orbits for 238U and 235U are: qB v m d 238 238 2 = and qB v m d 235 235 2 = Substitute in equation (1) to obtain: ( ) 235 238 235 238 2 2 2 m m qB v qB v m qB v m d − = − = Δ Chapter 26 2526 Substitute numerical values and evaluate Δd: ( )( ) ( )( ) cm 1 T 2 .
1 C 10 1.602 u kg 10 6606 .
1 u 235 u 238 m/s 10 905 .
1 2 Δ 19 27 5 = × ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ × − × = − − d 43 •• [SSM] A cyclotron for accelerating protons has a magnetic field strength of 1.4 T and a radius of 0.70 m. (a) What is the cyclotron’s frequency? (b) Find the kinetic energy of the protons when they emerge. (c) How will your answers change if deuterons are used instead of protons? Picture the Problem We can express the cyclotron frequency in terms of the maximum orbital radius and speed of the protons/deuterons. By applying Newton’s 2nd law, we can relate the radius of the particle’s orbit to its speed and, hence, express the cyclotron frequency as a function of the particle’s mass and charge and the cyclotron’s magnetic field. In Part (b) we can use the definition of kinetic energy and their maximum speed to find the maximum energy of the emerging protons. (a) Express the cyclotron frequency in terms of the proton’s orbital speed and radius: r v v r T f π π 2 2 1 1 = = = (1) Apply Newton’s 2nd law to a proton in the magnetic field of the cyclotron: r v m qvB 2 = ⇒ qB mv r = (2) Substitute for r in equation (1) and simplify to obtain: m qB mv qBv f π π 2 2 = = (3) Substitute numerical values and evaluate f: ( )( ) ( ) MHz 21 MHz 3 .
21 kg 10 673 .
1 2 T 4 .
1 C 10 602 .
1 27 19 = = × × = − − π f (b) Express the maximum kinetic energy of a proton: 2 max 2 1 max mv K = From equation (2), is given by: max v m qBr v max max = The Magnetic Field 2527 Substitute for and simplify to obtain: max v 2 max 2 2 2 1 2 max 2 1 max r m B q m qBr m K ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = Substitute numerical values and evaluate : max K ( ) ( ) ( ) MeV 46 MeV 46.0 J 10 1.602 eV 1 J 10 37 .
7 m 7 .
0 kg 10 673 .
1 T 4 .
1 C 10 602 .
1 19 -12 2 27 2 2 19 2 1 max = = × × × = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ × × = − − − K (c) From equation (3) we see that doubling m halves f: MHz 11 protons 2 1 deuterons = = f f From our expression for Kmax we see that doubling m halves K: MeV 23 protons 2 1 deuterons = = K K 44 •• A certain cyclotron that has a magnetic field whose magnitude is 1.8 T is designed to accelerate protons to a kinetic energy of 25 MeV. (a) What is the cyclotron frequency for this cyclotron? (b) What must the minimum radius of the magnet be to achieve this energy? (c) If the alternating potential difference applied to the dees has a maximum value of 50 kV, how many revolutions must the protons make before emerging with kinetic energies of 25 MeV? Picture the Problem We can express the cyclotron frequency in terms of the maximum orbital radius and speed of the protons be accelerated in the cyclotron. By applying Newton’s 2nd law, we can relate the radius of the proton’s orbit to its speed and, hence, express the cyclotron frequency as a function of the its mass and charge and the cyclotron’s magnetic field. In Part (b) we can use the definition of kinetic energy express the minimum radius required to achieve the desired emergence energy. In Part (c) we can find the number of revolutions required to achieve this emergence energy from the energy acquired during each revolution. (a) Express the cyclotron frequency in terms of the proton’s orbital speed and radius: r v v r T f π π 2 2 1 1 = = = Apply Newton’s 2nd law to a proton in the magnetic field of the cyclotron: r v m qvB 2 = ⇒ qB mv r = (1) Chapter 26 2528 Substitute for v and simplify to obtain: m qB mv qBv f π π 2 2 = = Substitute numerical values and evaluate f: ( )( ) ( ) MHz 27 kg 10 673 .
1 2 T 8 .
1 C 10 602 .
1 27 19 = × × = − − π f (b) Using the definition of kinetic energy, relate emergence energy of the protons to their velocity: 2 2 1 mv K = ⇒ m K v 2 = Substitute for v in equation (1) and simplify to obtain: qB Km m K qB m r 2 2 = = Substitute numerical values and evaluate rmin: ( )( ) ( )( ) cm 40 T 8 .
1 C 10 602 .
1 kg 10 673 .
1 MeV 25 2 19 27 = × × = − − r (c) Express the required number of revolutions N in terms of the energy gained per revolution: rev MeV 25 E N = Because the beam is accelerated through a potential difference of 50 kV twice during each revolution: keV 100 2 rev = Δ = V q E Substitute the numerical value of and evaluate N: rev E rev 10 5 .
2 keV/rev 100 MeV 25 2 × = = N 45 •• Show that for a given cyclotron the cyclotron frequency for accelerating deuterons is the same as the frequency for accelerating alpha particles is half the frequency for accelerating protons in the same magnetic field. The deuteron has a charge that is equal to the charge a proton has, and an alpha particle has a charge that is equal to twice the charge a proton has. Assume that mα = 2md = 4mp. Picture the Problem We can express the cyclotron frequency in terms of the maximum orbital radius and speed of a particle being accelerated in the cyclotron. By applying Newton’s 2nd law, we can relate the radius of the particle’s orbit to its The Magnetic Field 2529 speed and, hence, express the cyclotron frequency as a function of its charge-to-mass ratio and the cyclotron’s magnetic field. We can then use data for the relative charges and masses of deuterons, alpha particles, and protons to establish the ratios of their cyclotron frequencies. Express the cyclotron frequency in terms of a particle’s orbital speed and radius: r v v r T f π π 2 2 1 1 = = = Apply Newton’s 2nd law to a particle in the magnetic field of the cyclotron: r v m qvB 2 = ⇒ qB mv r = Substitute for r to obtain: m q B mv qBv f π π 2 2 = = (1) Evaluate equation (1) for deuterons: d d d d 2 2 m e B m q B f π π = = d d 2 2 2 2 2 m e B m e B m q B f π π π α α α = = = and α f f = d Evaluate equation (1) for alpha particles: d d d 2 1 p p p 2 2 2 2 2 f m e B m e B m q B f = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = = = π π π and α f f f = = d p 2 1 Evaluate equation (1) for protons: 46 ••• Show that the radius of the orbit of a charged particle in a cyclotron is proportional to the square root of the number of orbits completed. Picture the Problem We can apply Newton’s 2nd law to the orbiting charged particle to obtain an expression for its radius as a function of its particle’s kinetic energy. Because the energy gain per revolution is constant, we can express this kinetic energy as the product of the number of orbits completed and the energy gained per revolution and, hence, show that the radius is proportional to the square root of the number of orbits completed. Chapter 26 2530 Apply Newton’s 2nd law to a particle in the magnetic field of the cyclotron: r v m qvB 2 = ⇒ qB mv r = (1) Express the kinetic energy of the particle in terms of its speed and solve for v: 2 2 1 mv K = ⇒ m K v 2 = (2) Noting that the energy gain per revolution is constant, express the kinetic energy in terms of the number of orbits N completed by the particle and energy Erev gained by the particle each revolution: rev NE K = (3) Substitute equations (2) and (3) in equation (1) to obtain: 2 1 rev rev 2 2 1 2 1 2 N qB mE mNE qB mK qB m K qB m r = = = = or 2 1 N r ∝ Torques on Current Loops, Magnets, and Magnetic Moments 47 • [SSM] A small circular coil consisting of 20 turns of wire lies in a region with a uniform magnetic field whose magnitude is 0.50 T. The arrangement is such that the normal to the plane of the coil makes an angle of 60º with the direction of the magnetic field. The radius of the coil is 4.0 cm, and the wire carries a current of 3.0 A. (a) What is the magnitude of the magnetic moment of the coil? (b) What is the magnitude of the torque exerted on the coil? Picture the Problem We can use the definition of the magnetic moment of a coil to evaluate μ and the expression for the torque exerted on the coil B μ τ r r r × = to find the magnitude of τ. (a) Using its definition, express the magnetic moment of the coil: 2 r NI NIA π μ = = Substitute numerical values and evaluate μ: ( )( ) ( ) 2 2 2 m A 30 .
0 m A 302 .
0 m 040 .
0 A 0 .
3 20 ⋅ = ⋅ = = π μ The Magnetic Field 2531 (b) Express the magnitude of the torque exerted on the coil: θ μ τ sin B = Substitute numerical values and evaluate τ : ( )( ) m N 13 .
0 60 sin T 50 .
0 m A 302 .
0 2 ⋅ = ° ⋅ = τ 48 • What is the maximum torque on a 400-turn circular coil of radius 0.75 cm that carries a current of 1.6 mA and is in a region with a uniform magnetic field of 0.25 T? Picture the Problem The coil will experience the maximum torque when the plane of the coil makes an angle of 90° with the direction of B r . The magnitude of the maximum torque is then given by B μ τ = max . The maximum torque acting on the coil is: B μ τ = max Use its definition to express the magnetic moment of the coil: 2 r NI NIA π μ = = Substitute to obtain: B r NI 2 max π τ = Substitute numerical values and evaluate τ : ( )( ) ( ) ( ) m N 28 T 25 .
0 cm 75 .
0 mA 6 .
1 400 2 max ⋅ = = μ π τ 49 • [SSM] A current-carrying wire is in the shape of a square of edge-length 6.0 cm. The square lies in the z = 0 plane. The wire carries a current of 2.5 A. What is the magnitude of the torque on the wire if it is in a region with a uniform magnetic field of magnitude 0.30 T that points in the (a) +z direction and (b) +x direction? Picture the Problem We can use B μ τ r r r × = to find the torque on the coil in the two orientations of the magnetic field. Express the torque acting on the coil: B μ τ r r r × = Express the magnetic moment of the coil: k k μ ˆ ˆ 2 IL IA ± = ± = r Chapter 26 2532 (a) Evaluate τ r for in the +z direction: B r ( ) 0 ˆ ˆ ˆ ˆ 2 2 = × ± = × ± = k k B IL k B k IL τ r ( ) ( )( ) ( ) ( )j j i k B IL i B k IL ˆ m mN 7 .
2 ˆ T 30 .
0 m 060 .
0 A 5 .
2 ˆ ˆ ˆ ˆ 2 2 2 ⋅ ± = ± = × ± = × ± = τ r (b) Evaluate τ r for in the +x direction: B r and m N 10 7 .
2 3 ⋅ × = − τ r 50 • A current-carrying wire is in the shape of an equilateral triangle of edge-length 8.0 cm. The triangle lies in the z = 0 plane. The wire carries a current of 2.5 A. What is the magnitude of the torque on the wire if it is in a region with a uniform magnetic field of magnitude 0.30 T that points in the (a) +z direction and (b) +x direction? Picture the Problem We can use B μ τ r r r × = to find the torque on the equilateral triangle in the two orientations of the magnetic field. Express the torque acting on the coil: B μ τ r r r × = Express the magnetic moment of the coil: k μ ˆ IA ± = r Relate the area of the equilateral triangle to the length of its side: ( ) 2 2 1 4 3 2 3 2 1 altitude base L L L A = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = × = Substitute to obtain: k μ ˆ 4 3 2I L ± = r (a) Evaluate τ r for in the +z direction: B r ( ) 0 ˆ ˆ 4 3 ˆ ˆ 4 3 2 2 = × ± = × ± = k k k k τ IB L B I L r The Magnetic Field 2533 (b) Evaluate τ r for in the +x direction: B r ( ) ( ) ( )( ) ( )j j i k IB L i B k I L τ ˆ m N 10 1 .
2 ˆ 4 T 30 .
0 A 5 .
2 m 080 .
0 3 ˆ ˆ 4 3 ˆ ˆ 4 3 3 2 2 2 ⋅ × ± = ± = × ± = × ± = − r and m N 10 1 .
2 3 ⋅ × = − τ r 51 •• A rigid wire is in the shape of a square of edge-length L. The square has mass m and the wire carries current I. The square lies on flat horizontal surface in a region where there is a magnetic field of magnitude B that is parallel to two edges of the square. What is the minimum value of B so that one edge of the square will lift off the surface? Picture the Problem One edge of the square will lift off the surface when the magnitude of the magnetic torque acting on it equals the magnitude of the gravitational torque acting on it. The condition for liftoff is that the magnitudes of the torques must be equal: grav mag τ τ = (1) Express the magnetic torque acting on the square: B IL B 2 mag = = μ τ Express the gravitational torque acting on one edge of the square: mgL = grav τ Substituting in equation (1) yields: mgL B IL = min 2 ⇒ IL mg B = min 52 •• A rectangular current-carrying 50-turn coil, as shown in Figure 26-36, is pivoted about the z axis. (a) If the wires in the z = 0 plane make an angle θ = 37º with the y axis, what angle does the magnetic moment of the coil make with the unit vector ? (b) Write an expression for in terms of the unit vectors and , where is a unit vector in the direction of the magnetic moment. ˆ i ˆ n ˆ i ˆ j ˆ n (c) What is the magnetic moment of the coil? (d) Find the torque on the coil when there is a uniform magnetic field r B = 1.5 T in the region occupied by the coil. (e) Find the potential energy of the coil in this field. (The potential energy is zero when θ = 0.) ˆ j Chapter 26 2534 Picture the Problem The diagram shows the coil as it would appear from along the positive z axis. The right-hand rule for determining the direction of n has been used to establish n as shown. We can use the geometry of this figure to determine θ and to express the unit normal vectorn. The magnetic moment of the coil is given by and the torque exerted on the coil by ˆ ˆ ˆ n μ ˆ NIA = r B μ τ r r r × = . Finally, we can find the potential energy of the coil in this field from B μ r r ⋅ − = U . x y I n ˆ θ ° 37 (a) Noting that θ and the angle whose measure is 37° have their right and left sides mutually perpendicular, we can conclude that: ° = 37 θ (b) Use the components of to express nin terms of and : n ˆ ˆ i ˆ j ˆ j i j i j i j n i n n y x ˆ 60 .
0 ˆ 80 .
0 ˆ 602 .
0 ˆ 799 .
0 ˆ 37 sin ˆ 37 cos ˆ ˆ ˆ − = − = ° − ° = + = (c) Express the magnetic moment of the coil: n μ ˆ NIA = r Substitute numerical values and evaluate μ r : ( )( )( )( ) ( ) ( ) ( ) ( )j i j i j i μ ˆ m A 25 .
0 ˆ m A 34 .
0 ˆ m A 253 .
0 ˆ m A 335 .
0 ˆ 602 .
0 ˆ 799 .
0 cm 0 .
48 A 75 .
1 50 2 2 2 2 2 ⋅ − ⋅ = ⋅ − ⋅ = − = r (d) Express the torque exerted on the coil: B μ τ r r r × = The Magnetic Field 2535 Substitute for and to obtain: μ r B r ( ) ( ) { } ( ) ( )( ) ( )( ) ( )k j j j i j j i τ ˆ m N 50 .
0 ˆ ˆ m N 379 .
0 ˆ ˆ m N 503 .
0 ˆ T 5 .
1 ˆ m A 253 .
0 ˆ m A 335 .
0 2 2 ⋅ = × ⋅ − × ⋅ = × ⋅ − ⋅ = r (e) Express the potential energy of the coil in terms of its magnetic moment and the magnetic field: B μ r r ⋅ − = U Substitute for and and evaluate U: μ r B r ( ) ( ) { } ( ) ( )( ) ( )( ) J 38 .
0 ˆ ˆ m N 379 .
0 ˆ ˆ m N 503 .
0 ˆ T 5 .
1 ˆ m A 253 .
0 ˆ m A 335 .
0 2 2 = ⋅ ⋅ + ⋅ ⋅ − = ⋅ ⋅ − ⋅ − = j j j i j j i U 53 •• [SSM] For the coil in Problem 52 the magnetic field is now r B = 2.0 T . Find the torque exerted on the coil when is equal to (a) , (b) ˆ j n ˆ ˆ i ˆ j , (c) − ˆ j , and (d) ˆ i 2 + ˆ j 2 . Picture the Problem We can use the right-hand rule for determining the direction of to establish the orientation of the coil for value of and n ˆ n ˆ B μ τ r r r × = to find the torque exerted on the coil in each orientation. (a) The orientation of the coil is shown to the right: x y n ˆ Evaluate τ r for = 2.0 T and = : B r j ˆ n ˆ i ˆ ( )( )( ) ( ) ( )( ) ( ) ( )k k j i j i B n NIA B μ τ ˆ m N 84 .
0 ˆ m N 840 .
0 ˆ ˆ m N 840 .
0 ˆ T 0 .
2 ˆ cm 0 .
48 A 75 .
1 50 ˆ 2 ⋅ = ⋅ = × ⋅ = × = × = × = r r r r Chapter 26 2536 (b) The orientation of the coil is shown to the right: x y n ˆ Evaluate τ r for = 2.0 T and = : B r j ˆ n ˆ j ˆ ( )( )( ) ( ) ( )( ) 0 ˆ ˆ m N 840 .
0 ˆ T 0 .
2 ˆ cm 0 .
48 A 75 .
1 50 ˆ 2 = × ⋅ = × = × = × = j j j j B n NIA B μ τ r r r r (c) The orientation of the coil is shown to the right: x y n ˆ Evaluate τ for = 2.0 T and = − : r B r j ˆ n ˆ j ˆ ( )( )( ) ( ) ( )( ) 0 ˆ ˆ m N 840 .
0 ˆ T 0 .
2 ˆ cm 0 .
48 A 75 .
1 50 ˆ 2 = × ⋅ − = × − = × = × = j j j j B n NIA B μ τ r r r r (d) The orientation of the coil is shown to the right: x y n ˆ Evaluate τ r for = 2.0 T and = ( + )/ B r j ˆ n ˆ i ˆ j ˆ 2 : ( )( )( )( ) ( ) ( )( ) ( )( ) ( )k j j j i j j i B n NIA B μ τ ˆ m N 59 .
0 ˆ ˆ m N 594 .
0 ˆ ˆ m N 594 .
0 ˆ T 0 .
2 ˆ ˆ 2 cm 0 .
48 A 75 .
1 50 ˆ 2 ⋅ = × ⋅ + × ⋅ = × + = × = × = r r r r 54 •• A small bar magnet has a length equal to 6.8 cm and its magnetic moment is aligned with a uniform magnetic field of magnitude 0.040 T. The bar magnet is then rotated through an angle of 60 about an axis perpendicular to its length The observed torque on the bar magnet has a magnitude of 0.10 N⋅m. The Magnetic Field 2537 (a) Find the magnetic moment of the magnet. (b) Find the potential energy of the magnet. Picture the Problem Because the small magnet can be modeled as a magnetic dipole; we can use the equation for the torque on a current loop to find its magnetic moment. (a) Express the magnitude of the torque acting on the magnet: θ μ τ sin B = Solve for μ to obtain: θ τ μ sin B = Substitute numerical values and evaluate μ: ( ) 2 m A 9 .
2 60 sin T 040 .
0 m N 10 .
0 ⋅ = ° ⋅ = μ (b) The potential energy of the magnet is given by: θ μ μ cos B B U − = ⋅ − = r r Substitute numerical values and evaluate U: ( )( ) mJ 58 60 cos T 040 .
0 m A 887 .
2 2 − = ° ⋅ − = U 55 •• A wire loop consists of two semicircles connected by straight segments (Figure 26-37). The inner and outer radii are 0.30 m and 0.50 m, respectively. A current of 1.5 A is in this wire and the current in the outer semicircle is in the clockwise direction. What is the magnetic moment of this current loop? Picture the Problem We can use the definition of the magnetic moment to find the magnetic moment of the given current loop and a right-hand rule to find its direction. Using its definition, express the magnetic moment of the current loop: IA = μ Express the area bounded by the loop: ( ) ( ) 2 inner 2 outer 2 inner 2 outer 2 1 2 R R R R A − = − = π π π Substitute for A to obtain: ( ) 2 inner 2 outer 2 R R I − = π μ Chapter 26 2538 Substitute numerical values and evaluate μ: ( ) ( ) ( ) [ ] 2 2 2 m A 38 .
0 m 30 .
0 m 50 .
0 2 A 5 .
1 ⋅ = − = π μ Apply the right-hand rule for determining the direction of the unit normal vector (the direction of μ) to conclude that μ r points into the page. 56 •• A wire of length L is wound into a circular coil that has N turns. Show that when the wire carries a current I, the magnetic moment of the coil has a magnitude given by IL2/(4πN). Picture the Problem We can use the definition of the magnetic moment of a coil to find the magnetic moment of a wire of length L that is wound into a circular coil of N loops. We can find the area of the coil from its radius R and we can find R by dividing the length of the wire by the number of turns. Use its definition to express the magnetic moment of the coil: NIA = μ (1) Express the circumference of each loop: R N L π 2 = ⇒ N L R π 2 = where R is the radius of a loop. The area of the coil is given by: 2 R A π = Substituting for A and simplifying yields: 2 2 2 4 2 N L N L A π π π = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = Substitute for A in equation (1) and simplify to obtain: N IL N L NI π π μ 4 4 2 2 2 = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = 57 •• [SSM] A particle that has a charge q and a mass m moves with angular velocity ω in a circular path of radius r. (a) Show that the average current created by this moving particle is ωq/(2π) and that the magnetic moment of its orbit has a magnitude of 1 2 qωr2 . (b) Show that the angular momentum of this particle has the magnitude of mr2ω and that the magnetic moment and angular momentum vectors are related by r μ = q 2m ⎛ ⎝ ⎞ ⎠ r L , where L r is the angular momentum about the center of the circle. The Magnetic Field 2539 Picture the Problem We can use the definition of current and the relationship between the frequency of the motion and its period to show that I = qω/2π . We can use the definition of angular momentum and the moment of inertia of a point particle to show that the magnetic moment has the magnitude .
2 2 1 r qω μ = Finally, we can express the ratio of μ to L and the fact that μ r and L r are both parallel to ω r to conclude that = (q/2m) μ r L r . (a) Using its definition, relate the average current to the charge passing a point on the circumference of the circle in a given period of time: qf T q t q I = = Δ Δ = Relate the frequency of the motion to the angular frequency of the particle: π ω 2 = f Substitute for f to obtain: π ω 2 q I = From the definition of the magnetic moment we have: ( ) 2 2 1 2 2 r q r q IA ω π π ω μ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = = (b) Express the angular momentum of the particle: ω I L = The moment of inertia of the particle is: 2 mr I = Substituting for I yields: ( ) ω ω 2 2 mr mr L = = Express the ratio of μ to L and simplify to obtain: m q mr r q L 2 2 2 2 1 = = ω ω μ ⇒ L m q 2 = μ Because and μ r L r are both parallel to : ω r L μ r r m q 2 = 58 ••• A hollow non-conducting cylinder has length L and inner and outer radii Ri A uniformly charged non-conducting cylindrical shell (Figure 26-38) has length L, inner and outer radii Ri and Ro, respectively, a charge density ρ and an Chapter 26 2540 angular velocity ω about its axis. Derive an expression for the magnetic moment of the cylinder. Picture the Problem We can express the magnetic moment of an element of charge dq in a cylinder of length L, radius r, and thickness dr, relate this charge to the length, radius, and thickness of the cylinder, express the current due to this rotating charge, substitute for A and dI in our expression for μ and then integrate to complete our derivation for the magnetic moment of the rotating cylinder as a function of its angular velocity. Express the magnetic moment of an element of charge dq in a cylinder of length L, radius r, and thickness dr: dI r AdI d 2 π μ = = (1) Relate the charge dq in the cylinder to the length of the cylinder, its radius, and thickness: rdr L dq ρ π 2 = The current due to this rotating charge is given by: ( ) dr r L rdr L dq dI ω ρ ρ π π ω π ω = = = 2 2 2 Substitute for dI in equation (1) and simplify to obtain: ( ) dr r L rdr L r d 3 2 ρπω ρω π μ = = Integrate r from Ri to R0 to obtain: ( ) 4 i 4 0 4 1 3 0 i R R L dr r L R R − = = ∫ ρπω ρπω μ Because and μ r ω r are parallel: ( )ω μ r r 4 i 4 0 4 1 R R L − = ρπ 59 ••• [SSM] A uniform non-conducting thin rod of mass m and length L has a uniform charge per unit length λ and rotates with angular speed ω about an axis through one end and perpendicular to the rod. (a) Consider a small segment of the rod of length dx and charge dq = λdr at a distance r from the pivot (Figure 26-40). Show that the average current created by this moving segment is ωdq/(2π) and show that the magnetic moment of this segment is 1 2 λωr2dx . (b) Use this to show that the magnitude of the magnetic moment of the rod is 1 6 λωL 3. (c) Show that the magnetic moment r μ and angular momentum r L are related by r μ = Q 2m ⎛ ⎝ ⎞ ⎠ r L , where Q is the total charge on the rod. The Magnetic Field 2541 Picture the Problem We can follow the step-by-step outline provided in the problem statement to establish the given results. (a) Express the magnetic moment of the rotating element of charge: AdI d = μ (1) The area enclosed by the rotating element of charge is: 2 x A π = Express dI in terms of dq and Δt: t dx t dq dI Δ = Δ = λ where Δt is the time required for one revolution. The time Δt required for one revolution is: ω π 2 1 = = Δ f t Substitute for Δt and simplify to obtain: dx dI π λω 2 = Substituting for dI in equation (1) and simplifying yields: ( ) dx x dx x d 2 2 1 2 2 λω π λω π μ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = (b) Integrate dμ from x = 0 to x = L to obtain: 3 6 1 0 2 2 1 L dx x L λω λω μ = = ∫ (c) Express the angular momentum of the rod: ω I L = where L is the angular momentum of the rod and I is the moment of inertia of the rod with respect to the point about which it is rotating. Express the moment of inertia of the rod with respect to an axis through its end: 2 3 1 mL I = where L is now the length of the rod. Substitute to obtain: ω 2 3 1 mL L = Chapter 26 2542 Divide the expression for μ by L to obtain: m L mL L L 2 2 3 1 3 6 1 λ ω λω μ = = or, because Q = λL, L m Q 2 = μ Because ω r and are parallel: ω L r r I = L μ r r M Q 2 = 60 ••• A non-uniform, non-conducting thin disk of mass m, radius R, and total charge Q has a charge per unit area σ that varies as σ0r/R and a mass per unit area σm that is given by (m/Q) σ. The disk rotates with angular speed ω about its central axis. (a) Show that the magnetic moment of the disk has a magnitude 1 5πωσ 0R 4 which can be alternatively rewritten as 3 10ωQR 2 . (b) Show that the magnetic moment r μ and angular momentum r L are related by L μ r r M Q 2 = . Picture the Problem We can express the magnetic moment of an element of current dI due to a ring of radius r, and thickness dr with charge dq. Integrating this expression from r = 0 to r = R will give us the magnetic moment of the disk. We can integrate the charge on the ring between these same limits to find the total charge on the disk and divide μ by Q to establish the relationship between them. In Part (b) we can find the angular momentum of the disk by first finding the moment of inertia of the disk by integrating r2dm between the same limits used above. (a) Express the magnetic moment of an element of the disk: AdI d = μ The area enclosed by the rotating element of charge is: 2 x A π = The Magnetic Field 2543 Express the element of current dI: ( ) dr r R rdr R r dA f t dA t dq dI 2 0 0 2 2 ω σ π σ π ω σ σ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = = Δ = Δ = Substitute for A and dI and simplify to obtain: dr r R dr r R r d 4 0 2 0 2 πω σ ω σ π μ = = Integrate dμ from r = 0 to r = R to obtain: 4 0 5 1 0 4 0 R dr r R R πω σ πω σ μ = = ∫ (1) The charge dq within a distance r of the center of the disk is given by: dr r R dr R r r dr r dq 2 0 0 2 2 2 πσ σ π σ π = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = = Integrate dq from r = 0 to r = R to obtain: 2 0 3 2 0 2 0 2 R dr r R Q R πσ πσ = = ∫ (2) Divide equation (1) by Q to obtain: 10 3 2 2 0 3 2 4 0 5 1 R R R Q ω πσ πω σ μ = = and 2 10 3 R Qω μ = (3) (b) Express the moment of inertia of an element of mass dm of the disk: ( ) dr r QR m dr r Q R r m rdr Q m r dA r dm r dI 4 0 3 0 2 m 2 2 2 2 2 σ π σ π π σ σ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = = = Integrate dI from r = 0 to r = R to obtain: 4 0 0 4 0 5 2 2 R Q m dr r QR m I R σ π σ π = = ∫ Chapter 26 2544 Divide I by equation (2) and simplify to obtain: 2 2 0 3 2 4 0 5 3 5 2 R Q m R R Q m Q I = = πσ σ π and 2 5 3 R m I = Express the angular momentum of the disk: ω ω 2 5 3 mR I L = = Divide equation (3) by L and simplify to obtain: m Q mR R Q L 2 2 5 3 2 10 3 = = ω ω μ ⇒ L m Q 2 = μ Because is in the same direction as ω μ r r : L μ r r m Q 2 = 61 ••• [SSM] A spherical shell of radius R carries a constant surface charge density σ. The shell rotates about its diameter with angular speed ω. Find the magnitude of the magnetic moment of the rotating shell. Picture the Problem We can use the result of Problem 57 to express μ as a function of Q, M, and L. We can then use the definitions of surface charge density and angular momentum to substitute for Q and L to obtain the magnetic moment of the rotating shell. Express the magnetic moment of the spherical shell in terms of its mass, charge, and angular momentum: L M Q 2 = μ Use the definition of surface charge density to express the charge on the spherical shell: 2 4 R A Q πσ σ = = Express the angular momentum of the spherical shell: ω ω 2 3 2 MR I L = = Substitute for L and simplify to obtain: ω πσ ω πσ μ 4 3 4 2 2 3 2 2 4 R MR M R = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = The Magnetic Field 2545 62 ••• A uniform solid uniformly charged sphere of radius R has a volume charge density ρ. The sphere rotates about an axis through its center with angular speed ω. Find the magnitude of the magnetic moment of this rotating sphere. Picture the Problem We can use the result of Problem 57 to express μ as a function of Q, M, and L. We can then use the definitions of volume charge density and angular momentum to substitute for Q and L to obtain the magnetic moment of the rotating sphere. Express the magnetic moment of the solid sphere in terms of its mass, charge, and angular momentum: L M Q 2 = μ Use the definition of volume charge density to express the charge of the sphere: 3 3 4 R V Q πρ ρ = = Express the angular momentum of the solid sphere: ω ω 2 5 2 MR I L = = Substitute for Q and L and simplify to obtain: ω πρ ω πρ μ 5 15 4 2 3 3 4 5 2 2 R MR M R = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = 63 ••• A uniform thin uniformly charged disk of mass m, radius R, and uniform surface charge density σ rotates with angular speed ω about an axis through its center and perpendicular to the disk (Figure 26-40). The disk in in a region with a uniform magnetic field r B that makes an angle θ with the rotation axis. Calculate (a) the magnitude of the torque exerted on the disk by the magnetic field and (b) the precession frequency of the disk in the magnetic field. Picture the Problem We can use its definition to express the torque acting on the disk, Example 26-11 to express the magnetic moment of the disk, and the definition of the precession frequency to find the precession frequency of the disk. (a) The magnitude of the net torque acting on the disk is: τ = μBsinθ where μ is the magnetic moment of the disk. From Example 26-11: ω πσ μ 4 4 1 r = Substitute for μ in the expression for τ to obtain: θ ω πσ τ sin 4 4 1 B r = Chapter 26 2546 (b) The precession frequency Ω is equal to the ratio of the torque divided by the spin angular momentum: ω τ I = Ω For a solid disk, the moment of inertia is given by: 2 2 1 mr I = Substitute for τ and I to obtain: θ πσ ω θ ω πσ sin 2 sin Ω 2 2 2 1 4 4 1 m B r mr B r = = Remarks: Note that the precession frequency is independent of ω. The Hall Effect 64 • A metal strip that is 2.00-cm wide and 0.100-cm thick carries a current of 20.0 A in region with a uniform magnetic field of 2.00 T, as shown in Figure 26-41. The Hall voltage is measured to be 4.27 μV. (a) Calculate the drift speed of the free electrons in the strip. (b) Find the number density of the free electrons in the strip. (c) Is point a or point b at the higher potential? Explain your answer. Picture the Problem We can use the Hall effect equation to find the drift speed of the electrons and the relationship between the current and the number density of charge carriers to find n. In (c) we can use a right-hand rule to decide whether a or b is at the higher potential. (a) Express the Hall voltage as a function of the drift speed of the electrons in the strip: Bw v V d H = ⇒ Bw V v H d = Substitute numerical values and evaluate vd: ( )( ) mm/s 107 .
0 mm/s 1068 .
0 cm 00 .
2 T 2.00 V 27 .
4 d = = = μ v (b) Express the current as a function of the number density of charge carriers: d nAqv I = ⇒ d Aqv I n = Substitute numerical values and evaluate n: ( )( )( )( ) 3 28 19 m 10 85 .
5 mm/s 1068 .
0 C 10 .602 1 cm 100 .
0 cm 2.00 A 0 .
20 − − × = × = n The Magnetic Field 2547 (c) Apply a right-hand rule to l r I and B r to conclude that positive charge will accumulate at a and negative charge at b and therefore b a V V > . The Hall effect electric field is directed from a toward b. 65 •• [SSM] The number density of free electrons in copper is 8.47 × 1022 electrons per cubic centimeter. If the metal strip in Figure 26-41 is copper and the current is 10.0 A, find (a) the drift speed vd and (b) the potential difference Va – Vb. Assume that the magnetic field strength is 2.00 T. Picture the Problem We can use A nqv I d = to find the drift speed and to find the potential difference V Bw v V d H = a – Vb . (a) Express the current in the metal strip in terms of the drift speed of the electrons: A nqv I d = ⇒ nqA I v = d Substitute numerical values and evaluate vd: ( )( )( )( ) m/s 10 68 .
3 m/s 10 685 .
3 cm 100 .
0 cm 00 .
2 C 10 602 .
1 cm 10 8.47 A 0 .
10 5 5 19 3 22 d − − − − × = × = × × = v (b) The potential difference b a V V − is the Hall voltage and is given by: Bw v V V V b a d H = = − Substitute numerical values and evaluate b a V V − : ( )( )( ) V 47 .
1 cm 00 .
2 T 00 .
2 m/s 10 685 .
3 5 μ = × = − − b a V V 66 •• A copper strip has 8.47 × 1022 electrons per cubic centimeter is 2.00-cm wide, is 0.100-cm thick, and is used to measure the magnitudes of unknown magnetic fields that are perpendicular to it. Find the magnitude of B when the current is 20.0 A and the Hall voltage is (a) 2.00 μV, (b) 5.25 μV, and (c) 8.00 μV. Picture the Problem We can use Bw v V d H = to express B in terms of VH and to eliminate the drift velocity v A nqv I d = d and derive an expression for B in terms of VH, n, and t. Chapter 26 2548 Relate the Hall voltage to the drift velocity and the magnetic field: Bw v V d H = ⇒ w v V B d H = Express the current in the metal strip in terms of the drift velocity of the electrons: A nqv I d = ⇒ nqA I v = d Substitute for vd and simplify to obtain: H H H H V I nqt Iw nqwtV Iw nqAV w nqA I V B = = = = Substitute numerical values and simplify to obtain: ( )( )( ) ( ) H 2 5 H 19 3 22 s/m 10 7845 .
6 A 0 .
20 cm 100 .
0 C 10 602 .
1 cm 10 47 .
8 V V B × = × × = − − (a) Evaluate B for VH = 2.00 μV: ( )( ) T 36 .
1 V 00 .
2 s/m 10 7845 .
6 2 5 = × = μ B (b) Evaluate B for VH = 5.25 μV: ( )( ) T 56 .
3 V 25 .
5 s/m 10 7845 .
6 2 5 = × = μ B (c) Evaluate B for VH = 8.00 μV: ( )( ) T 43 .
5 V 00 .
8 s/m 10 7845 .
6 2 5 = × = μ B 67 •• Because blood contains ions, moving blood develops a Hall voltage across the diameter of an artery. A large artery that has a diameter of 0.85 cm can have blood flowing through it with a maximum speed of 0.60 m/s. If a section of this artery is in a magnetic field of 0.20 T, what is the maximum potential difference across the diameter of the artery? Picture the Problem We can use Bw v V d H = to find the Hall voltage developed across the diameter of the artery. Relate the Hall voltage to the flow speed of the blood vd, the diameter of the artery w, and the magnetic field B: Bw v V d H = The Magnetic Field 2549 Substitute numerical values and evaluate VH: ( )( )( ) mV 0 .
1 cm 85 .
0 T 20 .
0 m/s 60 .
0 H = = V 68 •• The Hall coefficient RH is a property of conducting material (just as resistivity is). It is defined as RH = Ey/(JxBBz), where Jx is x component of the current density in the material, Bz B is the z component of the magnetic field, and Ey is the y component resulting Hall electric field. Show that the Hall coefficient is equal to 1/(nq), where q is the charge of the charge carriers (–e if they are electrons). (The Hall coefficients of monovalent metals, such as copper, silver, and sodium are therefore negative.) Picture the Problem Let the width of the slab be w and its thickness t. We can use the definition of the Hall electric field in the slab, the expression for the Hall voltage across it, and the definition of current density to show that the Hall coefficient is also given by 1/(nq). The Hall coefficient is: z x y B J E R = Using its definition, express the Hall electric field in the slab: w V Ey H = The current density in the slab is: d nqv wt I J x = = Substitute for Ey and Jx and simplify to obtain: z z wB nqv V B nqv w V R d H d H = = Express the Hall voltage in terms of vd, B, and w: w B v V z d H = Substitute for VH and simplify to obtain: nq wB nqv w B v R z z 1 d d = = 69 •• [SSM] Aluminum has a density of 2.7 × 103 kg/m3 and a molar mass of 27 g/mol. The Hall coefficient of aluminum is R = –0.30 × 10–10 m3/C. (See Problem 68 for the definition of R.) What is the number of conduction electrons per aluminum atom? Chapter 26 2550 Picture the Problem We can determine the number of conduction electrons per atom from the quotient of the number density of charge carriers and the number of charge carriers per unit volume. Let the width of a slab of aluminum be w and its thickness t. We can use the definition of the Hall electric field in the slab, the expression for the Hall voltage across it, and the definition of current density to find n in terms of R and q and M N n A a ρ = , to express na. Express the number of electrons per atom N: a n n N = (1) where n is the number density of charge carriers and na is the number of atoms per unit volume. From the definition of the Hall coefficient we have: z x y B J E R = Express the Hall electric field in the slab: w V Ey H = The current density in the slab is: d nqv wt I J x = = Substitute for Ey and Jx in the expression for R to obtain: z z wB nqv V B nqv w V R d H d H = = Express the Hall voltage in terms of vd, B, and w: w B v V z d H = Substitute for VH and simplify to obtain: nq wB nqv w B v R z z 1 d d = = ⇒ Rq n 1 = (2) Express the number of atoms na per unit volume: M N n A a ρ = (3) Substitute equations (2) and (3) in equation (1) to obtain: A N qR M N ρ = The Magnetic Field 2551 Substitute numerical values and evaluate N: ( ) 4 mol atoms 10 6.022 m kg 10 7 .
2 C m 10 0.30 C 10 1.602 mol g 27 23 3 3 3 10 19 ≈ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ × − × − = − − N General Problems 70 • A long wire parallel to the x axis carries a current of 6.50 A in the +x direction. The wire occupies a region that has a uniform magnetic field r B = 1.35 T ˆ j . Find the magnetic force per unit length on the wire. Picture the Problem We can use the expression for the magnetic force acting on a wire ( ) to find the force per unit length on the wire. B F r l r r × = I Express the magnetic force on the wire: B F r l r r × = I Substitute for and to obtain: l r I B r ( ) ( )j i F ˆ T 35 .
1 ˆ A 50 .
6 × = l r and ( ) ( j i F ˆ T 35 .
1 ˆ A 50 .
6 × = l ) r Simplify to obtain: ( )( ) ( )k j i F ˆ N/m 78 .
8 ˆ ˆ N/m 78 .
8 = × = l r 71 • An alpha particle (charge +2e) travels in a circular path of radius 0.50 m in a region with a magnetic field whose magnitude is 0.10 T. Find (a) the period, (b) the speed, and (c) the kinetic energy (in electron volts) of the alpha particle. (The mass of an alpha particle is 6.65 × 10–27 kg.) Picture the Problem We can express the period of the alpha particle’s motion in terms of its orbital speed and use Newton’s 2nd law to express its orbital speed in terms of known quantities. Knowing the particle’s period and the radius of its motion we can find its speed and kinetic energy. Chapter 26 2552 (a) Relate the period of the alpha particle’s motion to its orbital speed: v r T π 2 = (1) Apply Newton’s 2nd law to the alpha particle to obtain: r v m qvB 2 = ⇒ m qBr v = Substitute for v in equation (1) and simplify to obtain: qB m m qBr r T π π 2 2 = = Substitute numerical values and evaluate T: ( ) ( )( ) s 3 .
1 s 30 .
1 T 10 .
0 C 10 602 .
1 2 kg 10 65 .
6 2 19 27 μ μ π = = × × = − − T (b) Solve equation (1) for v: T r v π 2 = Substitute numerical values and evaluate v: ( ) m/s 10 4 .
2 m/s 10 409 .
2 s 30 .
1 m 50 .
0 2 6 6 × = × = = μ π v (c) The kinetic energy of the alpha particle is: ( )( ) MeV 12 .
0 J 10 1.602 eV 1 J 10 930 .
1 m/s 10 409 .
2 kg 10 65 .
6 19 14 2 6 27 2 1 2 2 1 = × × × = × × = = − − − mv K 72 •• The pole strength qm of a bar magnet is defined by l r r m q = μ , where r μ is the magnetic moment of the magnet and r l is the position of the north-pole end of the magnet relative to the south-pole end. Show that the torque exerted on a bar magnet in a uniform magnetic field r B is the same as if a force B q r m + is exerted on the north-pole of the magnetic and a force B q r m − is exerted on the south-pole. The Magnetic Field 2553 Picture the Problem The configuration of the magnet and field are shown in the figure. We’ll assume that a force B r m q + is exerted on the north-pole end and a force B r m q − is exerted on the south-pole end and show that this assumption leads to the familiar expression for the torque acting on a magnetic dipole. B θ μ Assuming that a force B r m q + is exerted on the north-pole end and a force B r m q − is exerted on the south- pole end, express the net torque acting on the bar magnet: θ θ θ τ sin sin 2 sin 2 m m m l l l Bq Bq Bq = − − = θ μ θ μ τ sin sin B B = = l l r Substitute for qm to obtain: or B μ τ r r r × = 73 •• [SSM] A particle of mass m and charge q enters a region where there is a uniform magnetic field r parallel with the x axis. The initial velocity of the particle is , so the particle moves in a helix. (a) Show that the radius of the helix is r = mv B r v = v0xˆ i + v0y ˆ j 0y/qB. (b) Show that the particle takes a time Δt = 2πm/qB to complete each turn of the helix. (c) What is the x component of the displacement of the particle during time given in Part (b)? Picture the Problem We can use B v F r r r × = q to show that motion of the particle in the x direction is not affected by the magnetic field. The application of Newton’s 2nd law to motion of the particle in yz plane will lead us to the result that r = mv0y /qB. By expressing the period of the motion in terms of v0y we can show that the time for one complete orbit around the helix is t = 2πm/qB. (a) Express the magnetic force acting on the particle: B v F r r r × = q Chapter 26 2554 Substitute for v r and and simplify to obtain: B r ( ) ( ) ( k k i j i i i j i F ˆ ˆ 0 ˆ ˆ ˆ ˆ ˆ ˆ ˆ 0 0 0 0 0 0 B qv B qv B qv B qv B v v q y y y x y x − = − = × + × = × + = ) r i.e., the motion in the direction of the magnetic field (the x direction) is not affected by the field. Apply Newton’s 2nd law to the particle in the plane perpendicular to (i.e., the yz plane): i ˆ r v m B qv y y 2 0 0 = (1) Solving for r yields: qB mv r y 0 = (b) Relate the time for one orbit around the helix to the particle’s orbital speed: y v r t 0 2 Δ π = Solve equation (1) for v0y: m qBr v y = 0 Substitute for v0y and simplify to obtain: qB m m qBr r t π π 2 2 Δ = = (c) Because, as was shown in Part (a), the motion in the direction of the magnetic field (the x direction) is not affected by the field, the x component of the displacement of the particle as a function of t is: ( ) t v t x x o = For t = : t Δ ( ) qB mv qB m v t x x x o o 2 2 π π = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = Δ 74 •• A metal crossbar of mass m rides on a parallel pair of long horizontal conducting rails separated by a distance L and connected to a device that supplies constant current I to the circuit, as shown in Figure 26-42. The circuit is in a region with a uniform magnetic field B r whose direction is vertically downward. The Magnetic Field 2555 There is no friction and the bar starts from rest at t = 0. (a) In which direction will the bar start to move? (b) Show that at time t the bar has a speed of (BIL/m)t. Picture the Problem We can use a constant-acceleration equation to relate the velocity of the crossbar to its acceleration and Newton’s 2nd law to express the acceleration of the crossbar in terms of the magnetic force acting on it. We can determine the direction of motion of the crossbar using a right-hand rule or, equivalently, by applying B F r l r r × = I . at v v + = 0 or, because v0 = 0, at v = (a) Using a constant-acceleration equation, express the velocity of the bar as a function of its acceleration and the time it has been in motion: Use Newton’s 2nd law to express the acceleration of the rail: m F a = where F is the magnitude of the magnetic force acting in the direction of the crossbar’s motion. Substitute for a to obtain: t m F v = Express the magnetic force acting on the current-carrying crossbar: ILB F = Substitute to obtain: t m ILB v = (b) Because the magnetic force is to the right and the crossbar starts from rest, the motion of the crossbar will also be toward the right. 75 •• [SSM] Assume that the rails Problem 74 are frictionless but tilted upward so that they make an angle θ with the horizontal, and with the current source attached to the low end of the rails. The magnetic field is still directed vertically downward. (a) What minimum value of B is needed to keep the bar from sliding down the rails? (b) What is the acceleration of the bar if B is twice the value found in Part (a)? Picture the Problem Note that with the rails tilted, F r still points horizontally to the right (I, and hence , is out of the page). Choose a coordinate system in which down the incline is the positive x direction. Then we can apply a condition l r Chapter 26 2556 for translational equilibrium to find the vertical magnetic field B r needed to keep the bar from sliding down the rails. In Part (b) we can apply Newton’s 2nd law to find the acceleration of the crossbar when B is twice its value found in (a). F B Mg Fn θ θ (a) Apply to the crossbar to obtain: 0 = ∑ x F 0 cos sin = − θ θ B I mg l Solving for B yields: θ tan l I mg B = and v ˆ tan u B θ l r I mg − = where is a unit vector in the vertical direction. v ˆ u (b) Apply Newton’s 2nd law to the crossbar to obtain: ma mg B I = − θ θ sin cos ' l Solving for a yields: θ θ sin cos g m B' I a − = l Substitute B′ = 2B and simplify to obtain: θ θ θ θ θ θ sin sin sin 2 sin cos tan 2 g g g g m I mg I a = − = − = l l Note that the direction of the acceleration is up the incline. 76 •• A long, narrow bar magnet that has magnetic moment r μ parallel to its long axis is suspended at its center as a frictionless compass needle. When placed in region with a horizontal magnetic field r B , the needle lines up with the field. If it is displaced by a small angle θ, show that the needle will oscillate about The Magnetic Field 2557 its equilibrium position with frequency f = 1 2π μB I , where I is the moment of inertia of the needle about the point of suspension. Picture the Problem We’re being asked to show that, for small displacements from equilibrium, the bar magnet executes simple harmonic motion. To show its motion is SHM we need to show that the bar magnet experiences a linear restoring torque when displaced from equilibrium. We can accomplish this by applying Newton’s 2nd law in rotational form and using a small angle approximation to obtain the differential equation for simple harmonic motion. Once we have the differential equation of motion we can identify ω and express f. Apply Newton’s 2nd law to the bar magnet: 2 2 sin dt d I B θ θ μ = − where the minus sign indicates that the torque acts in such a manner as to align the magnet with the magnetic field and I is the moment of inertia of the magnet. For small displacements from equilibrium, θ << 1 and: θ θ ≈ sin Hence our differential equation of motion becomes: θ μ θ B dt d I − = 2 2 Thus for small displacements from equilibrium we see that the differential equation describing the motion of the bar magnet is the differential equation of simple harmonic motion. Solve this equation for d2θ/dt2 to obtain: θ ω θ μ θ 2 2 2 − = − = I B dt d where I B μ ω = Relate f to ω to obtain: I B f μ π π ω 2 1 2 = = 77 •• A straight conducting wire whose length is 20 m is parallel to the y axis and is moving in the +x direction with a speed of 20 m/s in a region with a magnetic field given by . (a) Because of this magnetic force, electrons move to one end of the wire leaving the other end positively charged, until the k ˆ T 50 .
0 Chapter 26 2558 electric field due to this charge separation exerts a force on the conduction electrons that balances the magnetic force. Find the magnitude and direction of this electric field in the steady state situation. (b) Which end of the wire is positively charged and which end is negatively charged? (c) Suppose the moving wire is 2.0-m long. What is the potential difference between its two ends due to this electric field? Picture the Problem (a) We can use a condition for translational equilibrium to relate E r to F r . In Part (c) we can apply the definition of electric field in terms of potential difference to evaluate the difference in potential between the ends of the moving wire. (a) Sum the forces acting on an electron under steady-state conditions to obtain: 0 = + F E r r q ⇒ q F E r r − = The magnetic force on an electron in the conductor is given by: ( ) j k i k i B v F ˆ ˆ ˆ ˆ ˆ qvB qvB B qv q − = × = × = × = r r r Substituting for F r and simplifying yields: j j E ˆ ˆ vB q qvB = − − = r Substitute numerical values and evaluate E r : ( )( ) ( )j j E ˆ V/m 10 ˆ T 50 .
0 m/s 20 = = r (b) Because the electric force acting on the conduction electrons is in the +y direction, the end of the wire that is in the +y direction becomes negatively charged and the end of the wire that is in the −y direction becomes positively charged. The positive end has the lesser y coordinate. (c) The potential difference between the ends of the wire is: ( )( ) V 20 m 0 .
2 V/m 0 .
10 Δ Δ = = = y E V 78 ••• A circular loop of wire that has a mass m and carries a constant current I is in a region with a uniform magnetic field. It is initially in equilibrium and its magnetic moment is aligned with the magnetic field. The loop is given a small angular displacement about an axis through it center and perpendicular to the magnetic field and then released. What is the period of the subsequent motion? (Assume that the only torque exerted on the loop is due to the magnetic field and that there are no other forces acting on the loop.) The Magnetic Field 2559 Picture the Problem We’re being asked to show that, for small displacements from equilibrium, the circular loop executes simple harmonic motion. To show its motion is SHM we must show that the loop experiences a linear restoring torque when displaced from equilibrium. We can accomplish this by applying Newton’s 2nd law in rotational form and using a small angle approximation to obtain the differential equation for simple harmonic motion. Once we have the differential equation we can identify ω and express the period T of the motion. Apply Newton’s 2nd law to the loop: 2 2 inertia sin dt d I IAB θ θ = − where the minus sign indicates that the torque acts in such a manner as to align the loop with the magnetic field and Iinertia is the moment of inertia of the loop. For small displacements from equilibrium, θ << 1 and: θ θ ≈ sin Hence, our differential equation of motion becomes: θ θ IAB dt d I − = 2 2 inertia Thus for small displacements from equilibrium we see that the differential equation describing the motion of the current loop is the differential equation of simple harmonic motion. Solve this equation for d2θ/dt2 to obtain: θ θ inertia 2 2 I IAB dt d − = . Noting that the moment of inertia of a hoop about its diameter is 2 2 1 mR , substitute for Iinertia and simplify to obtain: θ ω θ π θ π θ 2 2 2 1 2 2 2 2 − = − = − = m B I mR B R I dt d where m IB π ω 2 = The period T of the motion is related to the angular frequency ω: ω π 2 = T Substituting for ω and simplifying yields: IB m T π 2 = Chapter 26 2560 79 ••• A small bar magnet has a magnetic moment r μ that makes an angle θ with the x axis. The magnet is in a region that has a non-uniform magnetic field given by . Using ( ) ( ) j i B y B x B y x + = ˆ r x U Fx ∂ ∂ − = , y U Fy ∂ ∂ − = and z U Fz ∂ ∂ − = , show that there is a net magnetic force on the magnet that is given by j i F ˆ ˆ y B x B y y x x ∂ ∂ + ∂ ∂ = μ μ r . Picture the Problem We can express μ r in terms of its components and calculate U from and μ r B r using B μ r r ⋅ − = U . Knowing U we can calculate the components of F r using Fx = −dU/dx and Fy = −dU/dy. Express the net force acting on the magnet in terms of its components: j i F ˆ ˆ y x F F + = r (1) Express in terms of its components: μ r k j i μ ˆ ˆ ˆ z y x μ μ μ + + = r Express the potential energy of the bar magnetic in the nonuniform magnetic field: ( ) ( ) ( ) ( ) ( ) ( ) y B x B y B x B U y y x x y x z y x μ μ μ μ μ − − = + ⋅ + + − = ⋅ − = j i k j i B μ ˆ ˆ ˆ ˆ ˆ r r Because is constant but depends on x and y: μ r B r ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∂ ∂ = − = x B dx dU F x x x μ and ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ∂ ∂ = − = y B dy dU F y y y μ Substitute in equation (1) to obtain: j i F ˆ ˆ y B x B y y x x ∂ ∂ + ∂ ∂ = μ μ r 80 •• A proton, a deuteron and an alpha particle all have the same kinetic energy. They are moving in a region with a uniform magnetic field that is perpendicular to each of their velocities. Let Rp, Rd, and Rα be the radii of their circular orbits, respectively. The deuteron has a charge that is equal to the charge a proton has, and an alpha particle has a charge that is equal to twice the charge a proton has. Find the ratios Rd/Rp and Rα/Rp. Assume that mα = 2md = 4mp. The Magnetic Field 2561 Picture the Problem We can apply Newton’s 2nd law to an orbiting particle to obtain an expression for the radius of its orbit R as a function of its mass m, charge q, speed v, and the magnitude of the magnetic field B. Apply Newton’s 2nd law to an orbiting particle to obtain: r v m qvB 2 = ⇒ qB mv r = Express the kinetic energy of the particle: 2 2 1 mv K = ⇒ m K v 2 = Substitute for v in the expression for r and simplify to obtain: Km qB m K qB m r 2 1 2 = = (1) Using equation (1), express the ratio Rd/Rp: 2 2 2 1 2 1 p p p d d p p p d d p d = = = = m m e e m m q q Km B q Km B q R R Using equation (1), express the ratio Rα /Rp: 1 4 2 2 1 2 1 p p p p p p p = = = = m m e e m m q q Km B q Km B q R R α α α α α 81 ••• Your forensic chemistry group, working closely with the local law enforcement agencies, has acquired a mass spectrometer similar to that discussed in the text. It employs a uniform magnetic field that has a magnitude of 0.75 T. To calibrate the mass spectrometer, you decide to measure the masses of various carbon isotopes by measuring the position of impact of the various singly ionized carbon ions that have entered the spectrometer with a kinetic energy of 25 keV. A wire chamber with position sensitivity of 0.50 mm is part of the apparatus. What will be the limit on its mass resolution (in kg) for ions in this mass range, that is those whose mass is on the order of that of a carbon atom? Picture the Problem We can apply Newton’s 2nd law, with the force on a moving charged particle in a magnetic field as the net force, to an ion in the spectrometer to obtain an expression for the radius of its trajectory as a function of its Chapter 26 2562 momentum. We can then use the definition of kinetic energy to eliminate the speed of the ion from the expression for the radius of its trajectory. Differentiating the expression for the range (twice the radius of curvature) of the ions with respect to their mass will yield the mass resolution for ions whose masses are roughly 19.9 × 10−27 kg. We’ll assume that the carbon atoms are singly ionized. Apply Newton’s 2nd law to an ion in the spectrometer to obtain: r v m qvB 2 = ⇒ qB mv r = (1) where q is the charge of the ion, m is its mass, and r is the radius of curvature of its path. From the definition of kinetic energy we have: 2 2 1 mv E = ⇒ m E v 2 = Substituting for v in equation (1) and simplifying yields: qB mE qB m E m r 2 2 = = (2) The range R of the ions is twice their radius of curvature: qB mE qB mE R 8 2 2 = = (3) Differentiate R with respect to m to obtain: ( ) 2 2 2 2 2 1 8 8 8 B mq E qB m E m qB E m dm d qB E qB mE dm d dm dR = = = = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = Solving for dm yields: E B mq dR B mq E dR dm 2 2 2 2 2 2 = = Substitute numerical values and evaluate dm: ( ) ( )( ) ( ) kg 10 0 .
1 eV C 10 602 .
1 keV 25 2 T 80 .
0 C 10 602 .
1 kg 10 9 .
19 mm 50 .
0 28 19 2 2 19 27 − − − − × = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ × × × × = dm |
16001 | https://www.youtube.com/watch?v=pIa9N-n8Vss | Percent Yield Problems - AP Chemistry Complete Course - Lesson 6.4
Jeremy Krug (krugslist)
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Posted: 24 Sep 2020
In this video, Mr. Krug shows students how to solve stoichiometry problems involving percent yield.
Transcript:
welcome back to ap chemistry i'm jeremy krug and in this video we are continuing our discussion of moles and stoichiometry and talking about percent yield problems in this particular video now when we say percent yield percent yield is determined by taking the actual yield and dividing it by the theoretical yield and then multiplying that by a hundred now when we say actual yield that's the amount of product that we actually produce in a chemical reaction in the laboratory it's determined experimentally so that means we have to go into the lab and actually carry out the reaction and see how much we really or actually make now the theoretical yield this is the amount of product that we calculate that should be produced in a reaction and it's determined using a stoichiometry problem all of the problems that you saw me work in the last couple of videos we were calculating theoretical yields okay all of those were theoretical yields that we calculated you know on paper and pencil you can calculate how much you should make you know that's theoretical pretty much always or almost always the actual yield is going to be a little bit less and sometimes a lot less than the theoretical yield and that's because there are sometimes factors that keep us kept from producing as much as we can like for example there might be side reactions there was a problem that i showed you earlier where we had the potassium reacting with water okay now that is a reaction that's quite uh spectacular rather explosive however potassium doesn't just react with water it also reacts with oxygen in the air so sometimes there are competing side reactions that prevent us from having the full effect or the full 100 percent yield there may be other reactions going on and that happens a lot in chemistry there are other side reactions that we have to think about so let's try an example here in an experiment a chemist calculates that 16.37 grams of calcium carbonate should be produced however after the experiment is complete only 14.08 grams are collected so what's the percent yield in this process once again it's it's a fairly simple calculation you know percent yield is just actual divided by a theoretical times a hundred so the actual yield it says is the 14.08 grams and we know that because it says that's how much are collected that's how much we really collect how much we really actually produce the theoretical yield well that's the 16.37 grams because that's the calculation that should be produced so that goes in the denominator now we just multiply that by a hundred and on the calculator we get an answer of about 86.01 percent so usually percent yield can be a very simple calculation now sometimes there's a little bit more involved for example let's say that we have an experimental process that we've done many many times and we know that over the long term it tends to have about a 72.0 percent yield but we want to make we really want to make 10 grams of the product well that means that the chemist should strive to produce what theoretical yield during the process so do you understand that the problem here we know it's a 72 percent yield we really want 10 grams so we should be shooting for more than that shouldn't we because we know that only 72 percent will actually be made we're going to use the same equation here this time we know what the percent yield is it's the 72.0 so i can i can plug that in right there to that part of the equation we also know what the actual yield is going to be it tells us we're going to be making 10.0 grams in reality so that goes on top here and we don't know what the theoretical yield is that's what the question is asking so that's going to be our unknown our x and of course we have a times 100 in here so this is just an algebra problem at this point i'm going to divide both sides by a hundred and we're going to get 0.72 equals 10 divided by x and let's cross multiply to make this problem a little bit easier to solve so we have 0.72 x equals 10 and now i can divide both sides by .72 i get that x equals 13.9 grams so i should be shooting for 13.9 grams so i can really make the 10 grams now here's another possibility let's say we have some calcium hydroxide produced here we're going to take some calcium drop it in water and make some calcium hydroxide along with hydrogen and let's say a student drops a five gram chunk of solid calcium into some water after filtering and drying the product the student finds that 7.76 grams of calcium hydroxide has been recovered what was the percent yield well notice that we're given the actual yield aren't we we're told that the student actually gets from the laboratory 7.76 grams of calcium hydroxide but does the problem tell us what the theoretical yield is of the calcium hydroxide it doesn't does it it just tells us that we start with five grams of calcium down here at the beginning so we're actually going to have to calculate the theoretical yield using a stoichiometry process so i'm going to set this up here i'm going to balance the equation of course with that 2 there in front of water and we're going to start with 5 grams of calcium and let's figure out what the theoretical yield of calcium hydroxide is so grams of calcium hydroxide at the end and we're going to go through my three-step process that i told you in the last couple of videos so in the first conversion factor got to have grams on the bottom one mole on top and for grams from the periodic table that's about 40.08 so grams are out in the second process or the second step rather we're going to have the mole ratio so calcium on the bottom and calcium hydroxide on the top and this looks like it's a one to one ratio this time both of the coefficients are one so it's a one to one calcium goes out top and bottom we're in moles of calcium hydroxide we want to be in grams of calcium hydroxide so let's convert to grams our last conversion factor one mole goes on the bottom and grams goes on the top and when you add up the individual atomic masses you know one calcium two oxygens and two hydrogens we get about 74.10 grams in one mole of that so we cancel moles now and we can do the arithmetic 5.00 divided by 40.08 times 74.10 and we get that the theoretical yield is 9.24 grams okay now the actual yield was 7.76 so now we can figure out the percent yield we just take the yeah there's a theoretical yield so we take the 7.76 divided by the 9.24 and of course multiply by 100 and when you key that into your calculator it looks like the percent yield is 84.0 percent and so at this point you should be able to work several types of percent yield problems hope you have smashed the like button if you haven't done so already hope you learned something from the video at the very least and enjoyed it if you hit the like button then the world the word will get out about my youtube videos here about ap chemistry and this complete online ap chemistry course so that other folks can take advantage of it as well hope you subscribe as well so you can keep up and and follow me here on youtube join me again in our next lesson where we can learn some more chemistry together |
16002 | https://www.ck12.org/flexi/cbse-math/derivatives-of-trigonometric-functions/what-is-the-derivative-of-cosine/ | Flexi answers - What is the derivative of cosine? | CK-12 Foundation
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16004 | https://www.vedantu.com/jee-main/physics-average-speed-formula | JEE Main
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How to Calculate Average Speed When Two Speeds Are Given
Average Speed Formula is a core topic in JEE Main Physics, especially in kinematics chapters where you must distinguish between speed and velocity. Every time an object covers different segments of distance at various speeds or spends unequal periods travelling at different rates, the concept of average speed becomes essential to solve numericals quickly and accurately.
In JEE kinematics, average speed measures the total distance travelled by a particle divided by total time taken, irrespective of direction. For example, if a car travels 120 km in 2 hours, its average speed is 60 km/h regardless of route or motion reversals. You'll frequently see this principle in both one-dimensional and two-dimensional motion scenarios.
What is the Average Speed Formula?
The direct formula for average speed in Physics is:
Average Speed = Total Distance Travelled / Total Time Taken
Let stot denote total distance and ttot denote total time. Then:
vavg =
Units: Always use SI units – distance in metres (m), time in seconds (s), so average speed in m/s. For practical applications like vehicles, km/h is common (1 m/s = 3.6 km/h).
Difference Between Average Speed and Average Velocity
A common JEE trap is to confuse these two quantities, especially in questions involving return trips or zig-zag motion. Here’s a focused contrast:
| Aspect | Average Speed | Average Velocity |
---
| Definition | Total distance / total time | Total displacement / total time |
| Quantity Type | Scalar | Vector |
| Value | Always positive | Can be zero/positive/negative |
| Depends On | Path followed | Initial & final positions only |
For more distinctions, see Difference Between Speed and Velocity and Average Velocity Formula from Vedantu.
Average Speed Formula in Special Cases (Two or More Speeds)
In JEE Main, the most frequent applications involve an object moving at two different speeds for equal distances. This scenario is a classic trap—do NOT simply average the two speeds. Instead:
If an object travels distance d at speed v1 and again distance d at v2:
Average speed =
This formula comes from total distance = 2d, and total time = d/v1 + d/v2.
For three equal distances at speeds v1, v2, v3:
Average speed =
When distances are unequal or time intervals differ, return to the basic formula: sum the total distance, sum the total time, then compute their ratio. Check the details in fundamental kinematics at Motion in One Dimension.
Exam-Focused Numericals Using Average Speed Formula
Let’s address a classic JEE-type question using the above approach:
An object covers 60 km at 30 km/h and another 60 km at 90 km/h. What’s the average speed for the entire trip?
Given equal distances (d = 60 km), use:
Average speed = km/h
Final result: 45 km/h
What if problem data gives different times at different speeds, for example: 30 min at 80 km/h, then 1 hour at 60 km/h? Compute each distance, sum them, and divide by total time.
Distance1 = 0.5 h × 80 km/h = 40 km
Distance2 = 1.0 h × 60 km/h = 60 km
Total distance = 100 km, total time = 1.5 h
Average speed = 100 km / 1.5 h = 66.67 km/h
Practise such mixed examples in JEE Kinematics Mock Test and Motion in 2D Dimensions.
Common Pitfalls and Quick Tips
Never sum speeds and divide by number of segments unless time intervals are equal.
Total distance is always path length, never displacement.
SI units are mandatory; convert km/h to m/s if needed (1 km/h = 5/18 m/s).
Check if segments are for equal distances or equal times—formula differs.
In round-trip travel, average speed is not zero unless net distance is zero.
For tricky scenarios, revisit Distance and Displacement and Uniform and Non-Uniform Motion for clarity.
Applications of Average Speed Formula in JEE Physics
Calculating speeds for vehicles, trains, or projectiles in single or multi-stage trips.
Solving race and chase problems where velocities change mid-route.
Understanding Earth's rotation effects, as in Rotational Motion scenarios.
Interpreting graphs in Displacement and Velocity-Time Graphs and extracting mean values from graphical data.
Relating average speed to other kinematic quantities for series problems, e.g., Average Acceleration Formula.
Vedantu provides a range of conceptual and problem-solving resources covering the average speed formula and related motion concepts to elevate your JEE Main preparation. Regular practice ensures you internalise what formula applies in each scenario, reducing the risk of mistakes under exam pressure.
To consolidate, always recall: average speed depends on entire path and total time, not just initial and final position. For full mastery, review mock tests and formula-based lists in Vedantu’s kinematics section.
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FAQs on Average Speed Formula Explained with Examples
What is the formula for average speed in physics?
Average speed in physics is calculated by dividing the total distance travelled by the total time taken.
Formula for Average Speed:
- Average Speed = Total Distance / Total Time
- Units: metres per second (m/s) or kilometres per hour (km/h)
This formula helps in solving numerical problems on speed, especially for JEE, NEET, and board exams.
How do you calculate average speed if two speeds are given for equal distances?
When two different speeds are involved for covering equal distances, the average speed is calculated using the harmonic mean.
Step-by-step method:
1. Add the two speeds (v1 and v2)
2. Multiply the two speeds
3. Use the formula:
Average Speed = (2 × v1 × v2) / (v1 + v2)
This ensures you are not simply taking their arithmetic mean, which is a common mistake.
How to find average speed with 2 distance and time?
To find average speed when two different distances and times are given, follow these steps:
Method:
- Add the total distance covered: D = d1 + d2
- Add the total time taken: T = t1 + t2
- Apply formula: Average Speed = Total Distance / Total Time
This process works for any number of trips or different speed values.
What is the difference between average speed and average velocity?
Average speed is a measure of total distance travelled per unit time, while average velocity is the total displacement (straight line from start to finish) per unit time.
Key differences:
- Average speed is always positive; velocity can be zero or negative.
- Average speed is a scalar; velocity is a vector.
- Average velocity considers direction; speed does not.
Understanding this distinction is important for exams and physics concepts.
How do I convert average speed to km/h?
To convert average speed from metres per second (m/s) to kilometres per hour (km/h), simply multiply by 3.6.
Conversion formula:
- km/h = m/s × 3.6
For example, if average speed = 5 m/s, then km/h = 5 × 3.6 = 18 km/h.
This conversion is commonly needed in physics numericals and real-life problems.
Why can't you just add two speeds and divide by 2 for average speed?
You cannot simply add two speeds and divide by 2 when distances or times are not the same.
Reason:
- The average speed formula depends on total distance and total time.
- If time spent at each speed is different, or distances aren't equal, the true average must consider both.
- Use the harmonic mean for equal distances:
Average Speed = (2 × v1 × v2) / (v1 + v2)
Simply dividing by 2 leads to incorrect answers, which is a common exam error.
What is the average speed if 8 km is covered in 30 minutes?
If you travel 8 km in 30 minutes, your average speed is calculated by dividing distance by time.
Calculation:
- Distance = 8 km
- Time = 30 min = 0.5 hours
- Average Speed = 8 km / 0.5 h = 16 km/h
This method uses the standard average speed formula applied to real-world situations.
Can average speed be greater than instantaneous speed?
In general, average speed can be equal to or less than the maximum instantaneous speed, but not greater.
Explanation:
- Instantaneous speed refers to speed at a specific moment.
- Average speed is overall, over the whole journey.
- At times, instantaneous speed may be higher or lower, but average speed is the total distance over total time.
This distinction is tested frequently in exams for conceptual understanding.
When does average velocity equal average speed?
Average velocity equals average speed only when the total displacement equals total distance—typically when motion is in a straight line and there is no change in direction.
Key cases:
- Straight line motion in one direction
- No reversal or deviation in path
This condition helps avoid confusion in numerical problems.
What are some common mistakes students make when calculating average speed?
Common mistakes in average speed problems include:
Averaging two speeds by simple addition instead of using the correct formula
Mixing up units (e.g., hours with minutes)
Not using total distance and total time
Confusing average speed with average velocity
Forgetting to convert minutes to hours (or vice versa) when units must match
Avoiding these errors is vital for JEE, NEET, and board exams.
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16005 | https://ijretina.com/index.php/ijretina/article/download/91/45/ | Published by : INAVRS | International Journal of Retina 20 19 ; 2; 2; 80
International Journal of Retina ( IJRETINA ) 20 19 , Volume 2, Number 2.
P-ISSN. 2614 -8684, E -ISSN.2614 -8536
Challenges in Cytomegalovirus (Cmv) Retinitis Management
Denisa Rosati 1, Sauli Ari Widjaja 1,2 , Ima Yustiarini 1,2 , Randi Montana 1,3 , W imbo Sasono 1,2 , Muhammad
Firmansjah 1,2 , Ady Dwi Prakosa 1,2 , Moestidjab 1,2 , Gatut Suhendro 1,2
1Department of Ophthalmology, Faculty of Medicine Universitas Airlangga, Dr. Soetomo Gene ral Academic Hospital Surabaya, Indonesia
2Vitreoretinal Division, Department of Ophthalmology, Faculty of Medicine Universitas Airlangga,
Dr. Soetomo General Academic Hospital Surabaya, Indonesia
3Infection Immunology Division, Department of Ophthalmology, Faculty of Medicine Universitas Airlangga,
Dr. Soetomo General Academic Hospital Surabaya, Indonesia
ABSTRACT
Introduction : HIV infection can manifest in a variety of ways in and around the eyes and it is most commonly
due to retinal microvasculopathy, neoplasm and also opportunistic infection. Those usually occur associated
with a significantly reduced CD4 T -cell counts. In this era of Highly Active Anti Retroviral Therapy (HAART)
has caused a major decreasing of the ocular involvement prevalence itself.
Case report : A 31 year -old -male came with blurred vision on the right eye, which has started 3 years ago
and slowly worsened. Central scotoma also presented previously. Patient wa s an HIV -AIDS, that placed him
on HAART. CD4+ T -lymphocyte count was 3 cells/mm 3. The initial visual acuity was light perception and
fundus examination showed Roth spots, massive exudates and hemorrhages covering the optic disc and
decreased foveal reflex. Laboratory examination revealed positive Rubella and anti -CMV immunoglobulin -
G (IgG). He also suffered from lung tuberculosis and took tuberculosis medication regularly. Patient was
diagnosed with Cytomegalovirus (CMV) retinitis based on history of illnes s, fundus examination as well as
laboratory testing and given oral induction valganciclovir 900 mg once daily for 3 weeks followed by
maintenance dosage.
Result : After valganciclovir induction, there was significant changes with decreased peripapillary ex udates,
hemorrhages and vasculitis, but the optic disc appeared pale. The patient also had bicytopenia due to
valganciclovir therapy that complicate his condition and passed away after 3 months follow up.
Conclusion : CMV retinitis is reported to occur in patient with extreme CD4 count usually less than 50
cells/mm 3. The sooner of proper treatment would likely following better outcome. Making diagnosis of
immunosuppresed patient with ocular manifestations was challenging so that comprehensive eye
examination in HIV -infected individuals should be conducted. Oral valganciclovir could give satisfactory
response to decrease the pro gression of retinitis but risk of blindness may still occur.
Keywords: cytomegalovirus, CMV retinitis, valganciclovir, HIV -AIDS, CD4+ T -lymphocyte
Cite This Article : ROSATI, denisa. Challenges in Cytomegalovirus (CMV) Retinitis Management. International Journal of
Retina, [S.l.], v. 2, n. 2, sep. 2019. ISSN 2614 -8536. Available at:
ijretina.com/index.php/ijretina/article/view/91
INTRODUCTION
Cor respondence to:
Denisa Rosati
Department of Ophthalmology,
Faculty of Medicine Universitas
Airlangga, Dr. Soetomo General
Academic Hospital Surabaya,
Indonesia
denisarosati88@gmail.com
Human immunodeficiency virus (HIV) remains
a major public health problem with 56,000 new
HIV infections per year in the United States .1 In
Indonesia, HIV/AIDS reported case was 3,679 in
2016 .2 It is estimated that about 70 to 80% of
adult HIV/AIDS patients will experience ocular
complication during their illness .3 The majori ty
of ocular involvement in HIV is HIV retinopathy
and Cytomegalo virus (CMV) retinitis . HIV
retinopathy is a non -infectious microvascular
disorder characterized by cotton wool spots,
microaneurysms, retinal hemorrhages, Roth
spots and telangiectatic vascular changes .481 Published by: INAVRS | International Journal of Retina 2019 ; 2; 2;
The most common manifestation of this disorder is cotton
wool spots. Unlike infective lesion, cotton wool spot in HIV
retinopathy are transient, not visually threatening and tend
to disappear within 6 -12 weeks .5 Retinal hemorrhages are
seen less frequently in appro ximately 30% of patients with
advanced HIV/AIDS .6 It may appear as flame -shaped areas
when they affect the nerve fiber layer and as dot -and -blot
patterns when they affect the deeper layers of the retina .7 It
can be differentiated from CMV retinitis by the presence of
fewer hemorrhages and the absence of subtle iritis or
vitritis.
CMV retinitis is the most common ocular opportunistic
infection that potentially blinding of patients with HIV -
AIDS. This disease occurred in up to one -third of HIV -
infected patie nts before the invention of highly active anti
retroviral therapy (HAART) and significantly associated with
CD4+ T-lymphocyte cell count <50 cells/mm 3.8,9 The
incidence of CMV retinitis in the post -HAART era is
estimated to be at most 5.6 cases /100 persons/year .10 CMV
retinitis is characterized by typical white, crumbly areas of
retinal necrosis and hemorrhage which is sight threatening
if originate in posterior pole. Cotton wool spot is an early
manifestation of CMV retinitis .9 The lesions tend to e nlarge
and coalesce over time, forming large, wedge -shaped areas
of involvement .11
The clinical forms of CMV retinitis divided as typical form
which appear as white spots with many hemorrhages,
atypical form which appear as a zone of thinned retina and
small dot infiltrate without hemorrhages, perivascular form
or “frosted branch angiitis” and optic neuropathy which has
the worst prognosis .12 Patient with vision disturbance may
have irreversible vision loss because of direct damage to the
macula and optic nerve, retinal detachment even after CMV
retinitis has resolved and immune recovery uveitis .13 CMV
retinitis can be treated with ganciclovir or foscarnet,
administered systemically or intravitreally .14 Another dr ug
of choice is valganciclovir, an orall y administered monovalyl
ester prodrug of ganciclovir . Induction therapy typically 900
mg once daily for 2 -3 weeks followed by maintenance
therapy 450 mg once daily .10
CASE REPORT
A 31 -year -old man referred to the outpatient clinic with
painless visual loss on the right eye since three years ago,
started with black dot in the center of his vision, that slowly
worsening by time. Patient had been diagnosed with HIV
since 4 years and treated with highly active an ti retroviral
therapy (HAART) such as Emtrici tabine/Tenofovir and
Lopinavir/Ritonavir . This patient also suffered from lung
tuberculosis and had t aken anti -tuberculosis fixed -dose
combinations (FDCs) for five months. Discrete painless
papulo -nodular lesions found over face and neck, assessed
as pruritic pa pular eruption a nd molluscum contagiosum
(Figure 1 ).
Figure 1. Discrete papulonodular lesions (pruritic papular
eruption and moluscum contagiosum)
The visual acuity was light pe rception on the right eye
with mid -dilated pupil, and relative afferent pupillary defect
(RAPD) was present at presentation. There was no
inflammatory sign on anterior segment . Funduscopy
examination revealed massive soft exudates and
hemorrhages in poster ior pole covering the optic disc,
Roth’s spot, necrotic lesion due to vasculitis and also
reduced foveal reflex (Figure 2). The left eye wa s within
normal limit. Laboratory test showed low CD4 count of 3
cells/mm 3, anemia 8.5 g/dl, positive IgG rubella and anti -
CMV. Chest x -ray revealed infiltrates with suspicious of lung
tuberculosis.
Figure 2. Fundus photograph both eyes before oral
valganciclovir therapy. Right eye showed massive soft
exudates and hemorrhages in posterior pole
The patient was diagnosed with cytomegalovirus (CMV)
retinitis and treated with oral valganciclovir, 900 mg once
daily for 3 weeks induction therapy and followed by 45 0 mg
once daily for maintenance therapy. Fu nduscopy
examination was done at the end of induction therapy,
showed significant changes with decreased peripapillary
exudates, hemorrhages and vasculitis, bu t the optic disc
appeared pale (Figure 3). Published by: INAVRS | International Journal of Retina 2019 ; 2; 2; 82
Optical Coherence Tomography (OCT) examination showed
macular thickening, intraretinal and subretinal fluid (Figure
4).
Shortly after the end of induction therapy, patient was
admitted at Department of Internal Medicine ward due to
chronic diarrhea and bicytopenia . Hemoglobin decreased
to 6.5 g/dl and white blood cells (WBC) count was 1.430. He
had to get PRC transfusion and at that time, considering his
weak condition, Internal Medicine Department suggested
to postpone valganciclovir maintenance therapy since it
appeared to b e the underlying cause of his bi cytopenia.
During hospitalized, visual acuity decreased to no light
perception.
Figure 3. Fundus photograph both eyes after oral
valganciclovir inductio n therapy. Right eye showed
significant decreasing of exudates and hemorrhages
Figure 4. Macular OCT both eyes before valganciclovir
induction therapy (4A) and after valganciclovir induction
therapy (4B)
One week after , valganciclovir maintenance therapy was
continued. During one week follow up, there was no
changes in visual acuity, fundusco py showed a pale optic
disc, necrotic lesion and decreased exudates with minimal
hemorrhages. Patient passed away after 3 months
evaluation with remained ocular condition.
DISCUSSION
HIV/AIDS is undoubtedly a multi systemic disease and
ocular involvement occurs in up to 70% of cases during t he
natural history of infection. Ocular manifestations of HIV -
associated spectrum are very broad and extend from a
simple blepharitis to blindness .12 The two most common
posterior segment ocular manifestations of HIV/AI DS are
HIV retinopathy and cytomegalovirus (CMV) retinitis. In this
case, the patient complained of gradually painless visual
loss with central scotoma and also massive exudates and
hemorrhages on funduscopy. HIV retinopathy itself is an
occlusive microangiopathy ,15 which presents as cotton wool
spots, microane urysms and retinal hemorrhages.
However, cotton wool spot associated with HIV
retinopathy are usually superficial, smaller lesions that
resolve within few months .16 Patients with HIV retinopa thy
rarely have immediate vision loss but there may be damage
to the retinal nerve fiber layer, decrease colour vision and
also visual field defect .17 On the other hand, CMV lesions
tend to enlarge and coalesce over time to form larger areas
of involvement . Some individuals may complain of blurred
vision, scotomas, flashlights or floaters. However,
approximately 15% of infected patients are often
asymptomatic despite the presence of extensive or vision
threatening CMV retinitis .18
CMV retinitis is the most common cause of blindness in
patient with HIV -AIDS. The location of infected retina,
determine the risk for vision loss. Posterior retinitis
threatens the macula and optic nerve and anterior retina
increases the risk of retinal detachment .15 In this case, the
infected retina was posteriorly, which include macula and
optic nerve. Hence, the risk of vision threatening was
greater. The patient also developed permanent and
irreversible visual impairment. Laboratory tests showed a
severe decline of CD4+ cell co unt to only 3 cells/μL. CD4+
cell count of less than 50 cells/μL of patient with HIV/AIDS
is a major risk factor for having active CMV retinitis
infection .(19,20) Data from the Longitudinal Study of the
Ocular Complications of AIDS (LSOCA) showed that 90%
patients with CMV retinitis had a recent CD4+ cell count of
<50 cells/μL and 85% was using Anti retroviral Therapy
(ART) prior to CMV diagnosis .21 Conducting routine
ophthalmology examinations especially funduscopy on
patients presenting for ART initiation with advanced disease
is very important .21,22
Before HAART, treatment of CMV retinitis was a lifelong
treatment with specific anti CMV therapy that likely to
relapsing of retinitis after two until three weeks of
discontinuation .23
4A 4B 83 Published by: INAVRS | International Journal of Retina 2019 ; 2; 2;
In this HAART era, immune function that is improved by
HAART make the cessation of all anti -CMV therapy without
reactivation of CMV retinitis possible .24 Despite such
reports, anti -CMV chemotherapies are still involved
especially ganciclovir, foscarnet and cidofovir .25,30 In this
case, patient was given oral valganciclovir as induction and
maintenance therapy. It was the only dru g of choice that
was available in our center. Historically, the most utilized
antiviral agent has been intravenous ganciclovir. In an
attempt to make an oral preparation with convenient
dosing that has the safety profile, efficacy and
bioavailability compa rable to ganciclovir , the prodrug was
developed .26 A randomized control trial study in 2002
showed that valganciclovir , the valine ester of ganciclovir,
was found to be as effective as intravenous ganciclovir in
more convenient way . In this study, the part icipants were
given oral valganciclovir 900 mg twice daily as induction
therapy and the remaining received intravenous ganciclovir
5 mg/kg for three weeks .27 The main adverse effe cts of both
drugs were diarrhea (30%), neutropenia and anemia
(20%) .28 This p atient also had been admitted to internal
medicine ward due to chronic diarrhea and bicy topenia. So
that, patient on oral valganciclovir therapy must undergo
complete both ophthalmic and systemic evaluations
periodically. Depending the medication used, com plete
blood count, chemistries and intraocular pressure must be
checked. Also dilated eye examinations should be
performed daily to weekly initially, then 2 weeks after
induction therapy. Patients should be undergone CD4
counts and viral load studies. As w ell. The other reliable
treatment for CMV retinitis is int raocular sustained release
ganciclovir implants, which have been very effective in
treating CMV retinitis but it is not readily available. 29
In CMV retinitis management, it is often hard to choose
the best therapy for the patient considering the side effects.
Oral valgancyclovir has been a favourite because it doesn’t
need IV administration and hospitalization. On the other
hand, the cost of the drug is very high, longterm therapy
needs strictly go od compliance and it causes some
toxicities that worsen the condition of immunosuppresed
patient. In a point that the patient get remarkable decrease
of myelosuppression, sometimes we need to postpone
therapy. Consequences, the infection might relapse or
remain. It should be avoided if hemoglobin is <8 g%,
absolute neutrophil count is less than 500 cells/ L, and the
platelet count is less than 25,000/ L. 30 In this case, we
needed to postpone valganciclovir maintenance therapy
due to bicytopenia. Delay of therapy administration
bothered the evaluation of medication result. Patient’s
visual acuity did not improve after 3 weeks maintenance
therapy.
Besides, patient got sight -threatening lesions that close to
the macula and optic nerve head. The cho ice of therapy of
this condition is injection of 2 mg gancyclovir or 2.4 mg
foscarnet. Those medications were not available in our
center. On the foll ow up examination after valganciclovir
induction therapy, fundus evaluati on showed a remarkable
changes. Exudates and hemorrhages were significantly
decreased and the optic disc was easier to evaluate. A pale
disc with sclerotic vascularities appeared and his visual
acuity was no light perception. As optic atrophy had already
set in the vision did not improve even though the
chorioretinitis patches had resolved .31
Visual loss adds to the overwhelming social and economic
burden not only for the patient and family itself but also
society. We support routine funduscopic examination that
has to be included in the standard WHO care package for
HIV -infected patients with advanced disease .32
CONCLUSION
Ocular involvement in HIV/AIDS infected patients is very
common with broad spectrum of manifestations including
non infectious, infectious and neoplasm. CMV retinitis with
involvement of posterior pole or owing retinal detachment
is the major factor that causes blindness. As CMV retinitis
still exists in the HAART era, we need to conduct
ophtalmological examinations as part of routine HIV care.
Furthermore, st andard treatment guidelines for HIV/AIDS
patients with CD4+ cell counts < 100 cells/μL should
include ophtalmological screening.
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Lansky A, Brooks JT, DiN enno E, Heffelfinger J, et. al.
Epidemiology of HIV in the United States. JAIDS. J
Acquir Immune Defic Syndr 2010. Vol 53: 55 –61: S64 -
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Epidemiology review Indonesia. 2016 -2017. P 26
Abu EK, Abokyi S, Yeboah DO, et. al. Retinal
Microvasculo pathy is common HIV/AIDS patients : A
cross sectional study at the cape coast teaching
hospital, Ghana. Hindawi Journal of Ophtalmology
Pp 1 -5
Govender P, Hansraj R, Naidoo KS. Ocular
manifestations of HIV/AIDS : A literature review (Part
2). S Afr Optom 2011. Vol 70(2) : 81 -88
Verma N, Kearney J. Ocular manifestations of AIDS.
Pap New Guin Med J 1996. Vol 39 196 -199
Faia LJ, Bakri S, DooHo BK, Nader M. HIV. Emedicine
website. [Online] Available from:
com/OPH/topic417.htm.
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Bhatia RS. Ophthalmic manifestations of AIDS. J Ind
Acad Clin Med 200 2. Vol 3 85 -88
Sun HY, Peng XY, Li D, Mao FF, You QS, Jonas JB.
Cytomegalovirus retinitis in patients with AIDS before
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Ophthalmol 2014. Vol 24(2):209 -215
Chen C, Guo CG, Meng L, Yu J, et.al. Comparative
analysis of cytomegalovirus retinitis and microvascular
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10 :1296 -1400
Stewart, Michael. Optimal management of
cytomegal ovirus retinitis in patients with AIDS. Clin
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Cheung TW, Teich SA. Cytomegalovirus infection in
patients with HIV infection. Mount Sin J Med 1999.
Vol 66 113 - 124
Chiotan C, Radu L, Serban R, et.al . Posterior segment
ocular manifestations of HIV/AIDS patients. J Med Life
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Heiden D, Ford N, Wilson D, et. al. Cytome - galovirus
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McCannel C. Cytomegalovirus Retinitis. Retina and
Vitreous Section 12 . San Fransisco : American
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Jabs DA. Ocular manifestations of HIV infection. Trans
Am Ophthalmol Soc 1995. Vol. 93:623
Holland GN. AIDS and opht halmology: The first quarter
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Agarwal A, Singh R, Sharma A, et.al . Ocular
manifestations in patients with human
immunodeficiency virus infection in the pre -HAART
versus the HAART era in the North Indian pop ulation.
Ocular Immunol Inflamm 2016 :396 -404
Cheung TW, Teich SA. Cytomegalovirus infection in
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Ausayakhun S, Keenan JD, Ausayakhun S. Clinical
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153(5):923 -931.
Leenasiri makul P, Liu Y, Jirawison C, et. al. Risk factors
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Tun N, Smithuis FM, London N, et.al . Mortality in
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16006 | https://openstax.org/books/contemporary-mathematics/pages/5-key-concepts | Skip to ContentGo to accessibility pageKeyboard shortcuts menu
Contemporary Mathematics
Key Concepts
Contemporary MathematicsKey Concepts
Search for key terms or text.
Key Concepts
## 5.1 Algebraic Expressions
Algebra is useful because it allows us to understand many situations in real life by modeling them with expressions.
Algebraic expressions are the building blocks of algebra. From algebraic expressions we can create algebraic equations.
Algebraic expressions are often simplified and evaluated using the four arithmetic operations.
## 5.2 Linear Equations in One Variable with Applications
Solving linear equations means discovering what the value of the variable in a linear equation represents in the given conditions.
When solving a linear equation, most often you will have one solution; however, a linear equation may have no solutions or infinitely many solutions.
## 5.3 Linear Inequalities in One Variable with Applications
Inequalities can be used when the possible values (answers) in a certain situation are numerous, or when the exact value (answer) is not known, but it is known to be within a range of possible values.
Linear inequalities can be represented using a number line or using interval notation.
## 5.4 Ratios and Proportions
A ratio is a comparison of two numbers. The ratio of two numbers and can be written as: to OR : OR the fraction /.
All fractions are ratios, but not all ratios are fractions. Ratios make part to part, part to whole, and whole to part comparisons. Fractions make part to whole comparisons only.
When two ratios are equal, we say they are in proportion or are proportional.
Setting up proportions allows us to solve many various situations where three of the four values of the proportion are known.
## 5.5 Graphing Linear Equations and Inequalities
Linear equations can be represented graphically on a rectangular coordinate system.
Solving linear equations in two variables means finding the point where two lines intersect. There are three possibilities: The lines intersect at exactly one point; the lines do not intersect (they are parallel); or the lines intersect everywhere (they are the same line).
Solving linear inequalities in two variables means finding a region of possible answers. Every point in this region will make both inequalities true statements.
Plotting points is a standard way to help graph linear equations and linear inequalities.
## 5.6 Quadratic Equations In One Variable with Applications
A quadratic equation is an algebraic equation where the highest power (degree) of the equation is two.
To solve a quadratic equation is to find the value(s) that when substituted in for the variables, will make the equation equal to zero.
There can be two, one, or no solutions to any quadratic equation.
There are several methods to solve a quadratic equation. These methods include factoring quadratic equations, graphic quadratic equations, using the square root method, and using the quadratic formula.
## 5.7 Functions
A relation is any set of ordered pairs . All of the -values of the set are the domain, and all of the -values of the set are the range.
A relation is a function if each -value in the domain is assigned to exactly one element in the range. A -value in the range can have more than one -value assigned to it; but each -value can only be assigned to one -value.
For the function is the name of the function, is the domain value variable, and is the range value variable.
The vertical line test is a test that can be done on the graph of a relation to determine if it is a function.
## 5.8 Graphing Functions
Every linear function can be graphically represented by a unique line that shows all the solutions of the equation.
The points where the graph of a line intersects the -axis and -axis are called the intercepts of the line.
Most lines will have one -intercept and one -intercept. Only if the line is straight vertical (no -intercept) or straight horizontal (no -intercept) will it not have both intercepts. Note that a line that is straight vertical is not a function, but a line that is straight horizontal is a function.
Since any two points determine a straight line, any linear function can be graphed if both intercepts are known.
The slope of a linear function is the ratio of the vertical change divided by the horizontal change. It is often referred to as .
A formula for finding the slope of linear functions is for any two points of the linear function and .
## 5.9 Systems of Linear Equations in Two Variables
To solve a system of linear equations means finding the point or points where the two linear equations intersect.
Two lines can intersect at one point, no points if they are parallel, or every point if they are the same equation.
Systems of linear equations can be solved by graphing, by using substitution, or by using the elimination method.
## 5.10 Systems of Linear Inequalities in Two Variables
To solve a system of linear inequalities means to find the area(s) where the points in that area make all the linear inequalities true.
Systems of linear inequalities can be solved by graphing the linear equations associated with the inequalities, then 'testing' points to see whether the values of the point make the equation true or not.
## 5.11 Linear Programming
Linear programming is a mathematical technique to solve problems involving finding maximums or minimums where a linear function is limited by various constraints.
An objective function is a linear function in two or more variables that describes the quantity that needs to be maximized or minimized.
In linear programming, a constraint is a restriction that affects the maximum or minimum values of an objective function.
Through the creation of objective functions and restraints, a linear system can be developed and solved through linear programming.
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Book title: Contemporary Mathematics
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Probability a pair of pairs of rows have the same vector sum
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Let X X be a random square n n by n n matrix with X i,j∈{0,1}X i,j∈{0,1}. What is the probability that there is a distinct pair of pairs of rows which have the same vector sum?
If you add two rows elementwise then each entry has 0 0 with probability 1/4 1/4, 2 2 with probability 1/4 1/4 and 1 1 with probability 1/2 1/2 and they are independent. So if the two pairs are disjoint the probability they are the same is (1/4 2+1/2 2+1/4 2)n=(3/8)n(1/4 2+1/2 2+1/4 2)n=(3/8)n. If the two pairs have one row in common the probability is 1/2 n 1/2 n.
How can you use these facts to give the final probability?
Clarification:
If we let r i r i be row i i then I want i,j,k,ℓ i,j,k,ℓ such that r i+r j=r k+r ℓ r i+r j=r k+r ℓ with rule 1) that we can't have both i∈{k,ℓ}i∈{k,ℓ} and j∈{k,ℓ}j∈{k,ℓ} and rule 2) that both i≠j i≠j and k≠ℓ k≠ℓ hold.
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edited Jan 19, 2014 at 14:12
user115998user115998
asked Jan 15, 2014 at 10:34
user115998user115998
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this question, is, as it stands badly defined, what is a vector sum?Lost1 –Lost1 2014-01-17 10:47:33 +00:00 Commented Jan 17, 2014 at 10:47
1 @Lost1: I assume that the vector sum of (x 1,x 2,…,x n)(x 1,x 2,…,x n) and (y 1,y 2,…,y n)(y 1,y 2,…,y n) is (x 1+y 1,x 2+y 2,…,x n+y n)(x 1+y 1,x 2+y 2,…,x n+y n).ShreevatsaR –ShreevatsaR 2014-01-17 10:48:32 +00:00 Commented Jan 17, 2014 at 10:48
@ShreevatsaR Thank you. Yes.user115998 –user115998 2014-01-17 10:48:54 +00:00 Commented Jan 17, 2014 at 10:48
you want 2 rows a a and b b and 2 rows c c and d d to have the same vector sum? i see.Lost1 –Lost1 2014-01-17 10:49:27 +00:00 Commented Jan 17, 2014 at 10:49
@Lost1 Yes that is right.user115998 –user115998 2014-01-17 11:04:24 +00:00 Commented Jan 17, 2014 at 11:04
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We'll answer (approximately) the question for m×n m×n matrices for clarity, and just set m=n m=n at the end.
As you say, for the sum of any two rows i 1 i 1 and i 2 i 2, for a particular column j j, the j j th entry in the sum of the two rows is 0 0 w.p. 1 4 1 4, 1 1 w.p. 2 4 2 4, and 2 2 w.p. 1 4 1 4, and different columns are independent.
So the probability that two disjoint pairs of rows have the same sum is the probability that in each column position j j, each of the two pairs has the same sum, which is p=(1 4 2+1 2 2+1 4 2)n=(3/8)n p=(1 4 2+1 2 2+1 4 2)n=(3/8)n. The probability that two pairs of rows that share a row in common have the same sum is the probability that in each of the j j positions, both of the "other" rows have the same element, which is q=(1/2)n q=(1/2)n.
Now, to find the probability that there exist some two pairs {i,j}≠{k,l}{i,j}≠{k,l} with the same sum, we can use the inclusion-exclusion principle. To the first approximation, the probability is just the sum over all such pairs of pairs:
∑{i,j}≠{k,l}Pr(r i+r j=r k+r l)∑{i,j}≠{k,l}Pr(r i+r j=r k+r l)
To avoid overcounting pairs-of-pairs, let us say we only count those tuples (i,j,k,l)(i,j,k,l) that are in some canonical order, say i<j i<j, k<l k<l, and the two pairs (i,j)(i,j) and (k,l)(k,l) are themselves distinct and in lexicographic order. Then
with {i,j}{i,j} and {k,l}{k,l} disjoint, there are m(m−1)(m−2)(m−3)8 m(m−1)(m−2)(m−3)8 such pairs of pairs — you can arrive at this number as 1 2(m 2)(m−2 2)1 2(m 2)(m−2 2) or as 3(m 3)3(m 3), or whatever, and
with one row in common (either i=k i=k so that k=i<j<l k=i<j<l) or j=k j=k so that i<j=k<l i<j=k<l) or j=l j=l so that i<k<l=j i<k<l=j), there are m(m−1)(m−2)2 m(m−1)(m−2)2 such pairs of pairs.
So the above sum is
m(m−1)(m−2)(m−3)8 p+m(m−1)(m−2)2 q.m(m−1)(m−2)(m−3)8 p+m(m−1)(m−2)2 q.
This is strictly an upper bound on the probability, as it overcounts cases where multiple pairs of pairs each have the same sum. To refine the sum, we need to consider those cases (the second-order term in the inclusion-exclusion sum).
What is the probability of situations where multiple terms in the above sum are positive (i.e. happen simultaneously)?
There are many ways in which this can happen, but note that terms like p 2 p 2, p q p q, q p q p or q 2 q 2 are all exponentially smaller than the first-order sum. So for large n n, it is not worth it to care about such terms. The only ways in which multiple events can happen, whose probability is not exponentially smaller, are the ways in which two rows being equal lead to multiple pairs-of-pairs having equal sum. So let's look at these for the second- (and also third- and higher-) order sums.
If two rows a a and b b are equal, then for every pair-of-pairs like {a,i}{a,i} and {b,i}{b,i}, we'll have r a+r i=r b+r i r a+r i=r b+r i. And there are (m−2)(m−2) such pairs-of-pairs. And (m 2)(m 2) possible pairs (a,b)(a,b). So the entire inclusion-exclusion sum, if we care about just the (1/2)n(1/2)n term, would be:
(m 2)q((m−2)−(m−2 2)+(m−3 3)−…)=(m 2)q(1−(1−1)m−2)=(m 2)q(m 2)q((m−2)−(m−2 2)+(m−3 3)−…)=(m 2)q(1−(1−1)m−2)=(m 2)q
This matches the direct calculation as: the probability of some two rows being equal is roughly (m 2)(1/2)n(m 2)(1/2)n.
So the answer (to a very high degree of approximation) can be taken to be the above. In fact, for most practical purposes we can approximate further: the above sum is asymptotically (for large m m and n n)
∼m 2 2 1 2 n∼n 2 2 n+1∼m 2 2 1 2 n∼n 2 2 n+1
when m=n m=n.
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edited Jan 26, 2014 at 9:21
answered Jan 17, 2014 at 11:44
ShreevatsaRShreevatsaR
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Is this also an upper bound?user115998 –user115998 2014-01-17 11:45:46 +00:00 Commented Jan 17, 2014 at 11:45
@user115998: Yes, the first term of the inclusion-exclusion is guaranteed to be an upper bound (Bonferroni inequalities. I've added that to the answer.ShreevatsaR –ShreevatsaR 2014-01-17 11:48:00 +00:00 Commented Jan 17, 2014 at 11:48
This must be a stupid question but.. there is a pair of pairs with one a row in common that have the same sum if and only if there are two rows that are identical right? So why isn't the probability (n 2)2−n(n 2)2−n for this case?user115998 –user115998 2014-01-17 12:02:55 +00:00 Commented Jan 17, 2014 at 12:02
@user115998: I considered the probability for a particular pair of pairs, say {i,j}{i,j} and {j,k}{j,k}. For these two, the probability is 2−n 2−n, as we must have r i=r k r i=r k. Now, the number of such (i,j,k)(i,j,k) I counted separately: we can pick j j in m m ways, and then {i,k}{i,k} in (m−1 2)(m−1 2) ways; that's how I got m(m−1)(m−2)/2 m(m−1)(m−2)/2. (Or, pick {i,j,k}{i,j,k} in (m 3)(m 3) ways, and then choose which is the "common member" in 3 3 ways -- again, 3(m 3)=m(m−1)(m−2)/2 3(m 3)=m(m−1)(m−2)/2.) Oh but I understand your question... let me think.ShreevatsaR –ShreevatsaR 2014-01-17 12:07:17 +00:00 Commented Jan 17, 2014 at 12:07
1 @Lost1: Yes, that's intuitively clear: with 2 n 2 n possible vectors, the chances that some two pairs of rows are in a "special" configuration is going to be tiny.ShreevatsaR –ShreevatsaR 2014-01-17 13:06:54 +00:00 Commented Jan 17, 2014 at 13:06
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16008 | https://www.geeksforgeeks.org/maths/absolute-frequency/ | Absolute Frequency - GeeksforGeeks
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Absolute Frequency
Last Updated : 23 Jul, 2025
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Absolute frequency is key idea in statistics that is essential to data analysis. It describes how frequently a specific value or category occurs within a dataset.
Students must comprehend absolute frequency since it establishes the foundation for more complex statistical analysis, including probability distributions, inferential statistics, and data visualization. It is part of a wider branch of Mathematics the "Number Theory in Discrete Mathematics".
Table of Content
What is Absolute Frequency?
Comprehending Absolute Frequency
Calculating Absolute Frequency
Representing Absolute Frequency
Absolute Frequency vs Relative Frequency
Absolute Frequency vs Cumulative Frequency
Applications of Absolute Frequency
Examples on Absolute frequency
What is Absolute Frequency?
The statistical term ‘Absolute frequency’ means how often a certain piece of data occurs in a dataset. In other words, it is the simple counting of a certain number. Normally represented by an integer, absolute frequency becomes the first-level measurement within statistics.
Key Features
Counting how often something occurs is referred to as absolute frequency.
Even if it is a simple statistical measure, absolute frequency can be used to do other calculations like finding out the relative frequency.
One way that people often use absolute frequency is through visualizations represented in graphs to show how often one thing happens relative to another.
The term "absolute frequency" is also known as "raw count" or simply "frequency."
Comprehending Absolute Frequency
Absolute frequency is often a component of basic data collection. For example, If 15 friends were to be surveyed about their favourite sport, 5 of them would prefer soccer and another 6 would prefer basketball while the remaining 4 would go for tennis therefore one would be able to determine the absolute frequency of each sport. Regarding sport’s absolute frequencies “Soccer” has a count of 5, Basketball dominates at 6 cases while Tennis only amounts to 4 instances.
However, it's important to note that as the number of respondents increases, the distribution of their responses is likely to change. Despite this, the way we count how many times each response occurs remains simple.
Displaying Survey Results.For example, when 50 people were asked as a sample group one question concerning the means of transport that they would prefer, it is found that such services as bus, bike, car or even walking have different rates in terms of absolute figures which can be displayed on a bar chart. This visual representation helps to quickly identify the most and least common preferences among the respondents.
Calculating Absolute Frequency
Imagine a classroom where a teacher wants to analyze the distribution of students' exam scores to identify patterns in performance.
The teacher collects the scores of 11 students, resulting in the following dataset: 85, 90, 85, 70, 90, 85, 75, 90, 80, 70, 75.
After collecting the data, the teacher organizes it into a table displaying the absolute frequencies.
| Score | Frequency |
--- |
| 70 | 2 |
| 75 | 2 |
| 80 | 1 |
| 85 | 3 |
| 90 | 3 |
There are several observations we can make from the table displaying absolute frequency. For instance, a significant portion of the class (over 50%) scored either 85 or 90, indicating a concentration of high performers. While, only a few students scored 70 or 80, suggesting that most students are achieving higher scores.
Representing Absolute Frequency
Bar Graph: Represents frequencies using bars.
Pie Chart: Shows frequencies as proportions of a whole.
Below is the visual representation of the above data as Bar graph and Pie chart:
1 / 2
Concepts Related to Absolute Frequency
Cumulative Frequency: This represents the sum of the absolute frequencies of all values up to a certain point in the dataset.
Relative Frequency: This measures the proportion of times a value occurs relative to the total number of observations.
Absolute Frequency vs Relative Frequency
Here’s a comparison of Absolute Frequency and Relative Frequency:
| Aspect | Absolute Frequency | Relative Frequency |
---
| Definition | Number of times a particular value occurs in a dataset | Ratio of the absolute frequency to the total number of observations |
| Formula | Count of each value | A b s o l u t e F r e q u e n c y T o t a l N u m b e r o f O b s e r v a t i o n s\frac{Absolute\ Frequency}{Total\ Number\ of\ Observations}T o t a l N u mb er o f O b ser v a t i o n s A b so l u t e F re q u e n cy |
| Expression | Integer count | Fraction or percentage |
| Advantages | Simplicity - Direct insight into raw data counts | Proportional insight - Standardization for comparison |
| Disadvantages | Limited for comparative insight - No proportional context | Doesn’t show exact counts, just how common something is |
| Best Used When | You need exact counts - Analyzing specific data | Comparing proportions - Analyzing relative data sizes |
Example: Considering the old dataset as discussed above - Dataset: [85, 90, 78, 85, 90, 92, 85, 78, 85, 90].
| Score | Absolute Frequency | Relative Frequency |
---
| 70 | 2 | 2/11≈0.18(18%)2/11\ \approx\ 0.18\ (18 \%)2/11≈0.18(18%) |
| 75 | 2 | 2/11≈0.18(18%)2/11\ \approx\ 0.18\ (18 \%)2/11≈0.18(18%) |
| 80 | 1 | 1/11≈0.09(09%)1/11\ \approx\ 0.09\ (09 \%)1/11≈0.09(09%) |
| 85 | 3 | 3/11≈0.27(27%)3/11\ \approx\ 0.27\ (27 \%)3/11≈0.27(27%) |
| 90 | 3 | 3/11≈0.27(27%)3/11\ \approx\ 0.27\ (27 \%)3/11≈0.27(27%) |
By inspecting relative frequencies, students who scored 85 were almost 27%, while those who scored 90 were about the same. In simple language, this implies that more than half of the class pooled together either 85 or 90. Such revelations aid teachers to understand better the distribution of performance within their classroom hence guiding them in choosing what teachings methods or resource allocation approaches to use.
Learn More aboutRelative Frequency
Absolute Frequency vs Cumulative Frequency
Here’s a comparison of Absolute Frequency and Relative Frequency:
| Aspect | Absolute Frequency | Cumulative Frequency |
---
| Definition | Count of occurrences of each distinct value | Running total of frequencies up to a certain point |
| Purpose | Understand the distribution of individual values | Determine the proportion of data below a certain value |
| Calculation | Count the occurrences of each value | Sum the absolute frequencies sequentially |
| Example Calculation | For values [1, 2, 2, 3]: 1: 1, 2: 2, 3: 1 | For values [1, 2, 2, 3]: 1: 1, 2: 3, 3: 4 |
| Visualization | Bar charts, histograms | Cumulative frequency curves, ogives |
| Data Representation | Individual frequencies | Accumulated frequencies |
| Use in Analysis | Identifying mode, creating histograms | Calculating percentiles, quartiles |
| Change Over Dataset | Does not accumulate, each value is standalone | Always increases or stays the same |
| Example Application | Number of books read by each student | Total number of books read up to a certain number of students |
Example: For the dataset [50, 60, 70, 60, 80, 70, 90, 100, 60, 50]
| Value | Absolute Frequency | Cumulative Frequency |
---
| 50 | 2 | 2 |
| 60 | 3 | 5 |
| 70 | 2 | 7 |
| 80 | 1 | 8 |
| 90 | 1 | 9 |
| 100 | 1 | 10 |
Learn More aboutCumulative Frequency
Applications of Absolute Frequency
Various application of Absolute Frequency are:
Survey Analysis: Understanding preferences and trends.
Quality Control: Monitoring defects or issues in manufacturing.
Market Research: Analyzing consumer behavior and product popularity.
Education: Assessing student performance and learning outcomes.
Examples on Absolute frequency
Example 1: In a study of 80 households, data was collected on the number of pets each household owns.
| Number of Pets | Frequency |
--- |
| 0 | 20 |
| 1 | 35 |
| 2 | 15 |
| 3 | 7 |
| 4 | 3 |
Calculate the total number of households surveyed.
Solution:
Total number of households surveyed is the sum of all the frequencies:
Total number of households = 20 + 35 + 15 + 7 + 3 = 80
Example 2: What is the modal number of pets, and what is its frequency?
Solution:
Modal number of pets is the number that occurs most frequently in the dataset. To find the mode, we identify the highest frequency in the data:
Number of Pets 0: Frequency 20
Number of Pets 1: Frequency 35
Number of Pets 2: Frequency 15
Number of Pets 3: Frequency 7
Number of Pets 4: Frequency 3
Highest frequency is 35, which corresponds to the number of pets being 1.
Therefore, the modal number of pets is 1, with a frequency of 35.
Learn More aboutMode of Dataset
Example 3: A market research survey asked 100 customers about their preferred type of smartphone. The results were,
| Smartphone Type | Frequency |
--- |
| Android | 45 |
| iPhone | 30 |
| Windows Phone | 15 |
| Other | 10 |
What is the relative frequency of iPhone users?
Solution:
Total number of customers surveyed is 100.
Relative frequency of iPhone users = 30 / 100 = 0.30 or 30%
Example 4: What is the proportion of customers who prefer iPhones compared to those who prefer Android phones?
Solution:
To find the proportion of customers who prefer iPhones compared to those who prefer Android phones, use the following steps:
(i)Determine the absolute frequencies for iPhone and Android:
iPhone: 30
Android: 45
(ii)Calculate the proportion of iPhone users compared to Android users:
Proportion = F r e q u e n c y o f i P h o n e F r e q u e n c y o f A n d r o i d\frac{Frequency\ of\ iPhone}{Frequency\ of\ Android}F re q u e n cy o f A n d ro i d F re q u e n cy o f i P h o n e = 30 45\frac{30}{45}45 30=2 3≈0.67=\frac{2}{3} \approx\ 0.67=3 2≈0.67
Therefore, the proportion of customers who prefer iPhones compared to those who prefer Android phones is approximately 0.67, or 67%.
Example 5: In a study of 50 people, data was collected on the number of favorite fruit of people. Here’s the data we collected:
| Favorite Fruit Name | Frequency |
--- |
| Apple | 15 |
| Banana | 20 |
| Orange | 10 |
| Grapes | 5 |
What is the absolute frequency of each fruits?
Solution:
The absolute frequency of each fruit is the number of times it was chosen by the respondents. Thus:
Absolute frequency of Apples is 15.
Absolute frequency of Bananas is 20.
Absolute frequency of Oranges is 10.
Absolute frequency of Grapes is 5.
Example 6: What is the proportion of people who prefer Apple compared to those who prefer Grapes?
Solution:
To find the proportion of people who prefer apple compared to those who prefer grapes, use the following steps:
(i)Determine the absolute frequencies for apple and grapes:
Apple: 15
Grapes: 5
(ii)Calculate the proportion of favorite fruit Apple compared to favorite fruit Grapes:
Proportion = F r e q u e n c y o f A p p l e F r e q u e n c y o f G r a p e s\frac{Frequency\ of\ Apple}{Frequency\ of\ Grapes}F re q u e n cy o f G r a p es F re q u e n cy o f A ppl e = 15 5\frac{15}{5}5 15=3=3=3
Therefore, the proportion of people who prefer apple compared to those who prefer grapes is 3.
Practice Question on Absolute Frequency
Ques 1:Given the dataset: [3, 7, 3, 2, 3, 5, 7, 2, 3, 5, 7, 7, 3, 5]. Calculate the absolute frequency for each distinct value.
Ques 2:Given the dataset: [6, 8, 6, 7, 8, 9, 6, 8, 9, 8, 6, 9, 9, 9]. Determine the mode using absolute frequency.
Ques 3:Given a dataset of student scores: [56, 67, 78, 45, 67, 89, 78, 90, 67, 45, 78, 90, 90, 67, 45]. Calculate the absolute frequency for each score range: 40-49, 50-59, 60-69, 70-79, 80-89, 90-99.
Ques 4:Consider the following survey results on the number of books read in a month by a group of people: [1, 0, 2, 1, 3, 0, 1, 4, 2, 3, 0, 1, 2, 3, 4]
Construct a frequency distribution table.
Identify the most common number of books read.
Ques 5:Compare the absolute frequencies of the following two datasets:
Dataset A: [2, 3, 2, 4, 5, 3, 2, 4, 2, 5, 4, 2, 3, 4]
Dataset B: [1, 2, 1, 3, 4, 1, 2, 3, 1, 4, 3, 1, 2, 3]
Ques 6:Given the dataset: [5, 6, 7, 5, 6, 8, 7, 5, 6, 8, 7, 6, 5]
Calculate the cumulative absolute frequency.
Conclusion
In statistics, the absolute frequency is an important tool despite being so simple. It is essential during early stages of data analysis and also enables more complicated measures such as relative as well as cumulative frequencies to be obtained subsequently. The concept of absolute frequency is fundamental when talking about distribution of data because it underpins every other statistical measure that follows afterwards while also being key in understanding other disciplines ranging from market marketing to research.
Absolute frequency comprehension is inevitable if you want to make decisions based on data, whether you are analyzing consumer views or appraising performance levels.
Related Article:
Frequency Distribution
Types of Frequency distribution
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16009 | https://www.drivingline.com/articles/on-the-trail-black-bear-pass/ | On the Trail: Black Bear Pass | DrivingLine
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On the Trail: Black Bear Pass
December 1, 2020
/ Story By Mark Masker
If you're looking for an easy off-road cruise that's kick back and mellow, it ain't Black Bear Pass. However, if you want lots of treacherous driving to really test your driver's chops and a real sense of accomplishment, you've come to the right place. It's infamous in the off-road world for its dicey obstacles. The trail tests your nerves as much as your ability. On the Trail tackled it and as you'll see in the video, it was no joke.
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Men in Black
Collin Coates from Built2Wander hosted Evo Manufacturing's Mel Wade and 3D Offroad's Matt Thompson for a hard day's off-roading through Black Bear Pass. Their weapons of choice: Triton JL on 40-inch Nitto Trail Grapplers with Bilsteincoilovers and one-ton Dana 60 axles, 2020 Gladiator aka (“Blaze JT”) also on 40-inch Trail Grapplers, and Mel Wade’s daughter’s Jeep JL on 37-inch Trail Grapplers. Note: a low center of gravity really helps on Black Bear's switchbacks. Mel tells us that stock vehicles actually work better here because of that.
Blacklist
Look at your kids' breakfast cereal. See all those little marshmallows in it? Imagine the cereal is a mountain and all of those marshmallows are switchbacks. That's Black Bear Pass. Just don't expect a toy at the end. As Mel points out in the video, any minor mishap can turn into a very big deal here. Here more than other places, preparedness is even more vital. We're up at the 13,000 foot level so food, water, and sunscreen are musts for this run. Not only do you need all the basics you usually take on a trip, but recovery gear like straps, shackles, and winch hooks are essential too. That way, the person behind you can help if you run into trouble on these switchbacks. Cold weather gear is also a good idea.
Black Out
Geared up and ready to go, our team headed out. The notorious trail starts at the 11,000 foot level at Red Mountain Pass on the Ouray side and travel is one way, except for the annual Jeep Jamboree where the route is reversed. It's also only open from mid-July to early Fall – weather permitting. Black Bear Pass used to be marked with a sign that read, "Telluride --> City of Gold; 12 Miles, 2 Hours; You Don't Have to Be Crazy to Drive this Road--but it Helps; Jeeps Only." At least, until local authorities got tired of replacing it due to constant thievery. Even the beginning has spectacular views, as you'll see. From the meadows, the road rises to the summit of Black Bear Pass at 12,840 feet. That portion is less about driving skill, more about taking in the view.
Men in Black II
And at the summit of the pass, the view does not disappoint. Mel noted at this point that stock vehicles do well here because they're not overbuilt. Meaning, they've not been widened out, which makes navigating the switchbacks ahead more difficult. Big tires and lifted suspension are great and all, but the wider turning radius they bring into the mix isn't helpful here. And neither is the mods' top heavy nature. Plus, a longer wheelbase makes getting off-camber very easy, which is not a good thing. After lunch at the summit, the crew got back to business.
The Empire Strikes Black
In the words of Han Solo, "Here's where the fun begins." Remember all of those switchbacks we kept bringing up earlier? This is where they start. If there's a time to turn back, it's now. There's a nice, friendly sign that even warns you about the 1000-foot drop offs, off-camber switchbacks, and loose shale awaiting you up ahead. From the summit, the guys drove down the backside of the pass, noting a patch where someone went off-trail to do some donuts. It's that sort of behavior which gets trails closed, so staying on the designated trail is important not just for you but other drivers.
As they hit the shelf roads, Mel reiterated the importance of paying attention and respecting the obstacles; if you don't they can really bite you in the backside. Hitting the wrong line could make you tipsy in a hurry in terrain like this as well. Walking through rough patches and planning your descents through them is an excellent idea here, as you'll see on the most treacherous parts of the trail.
The Black Keys
One of which is The Steps. At this point our trio faced that particularly challenging part of the trip and Mel walked Collin's and Matt's vehicles through it. If you want an idea of how treacherous it can be, check out the wonky angles the axles take as the Jeeps wind their way through it. It's a gnarly combo plate of shale, drop offs, blind trail, and snaky path. The Steps are not to be tread lightly. And neither is the next part.
The Steps were preschool compared to the education awaiting the team next. This last stretch is the toughest part of the whole trail. It comes down into Telluride proper, passing the electric plant on the way; it also takes you to Bridal Veil Falls, the highest waterfall in all of Colorado.
Heart of Darkness
How difficult, you ask? Swearing scary, that's how. Up until this point, we didn't have to bleep anything. As Collin hit the big kid stretch of switches and drop offs, he hit the throttle on the potty talk, he was so moved by the experience. The narrowness is why this trail is one-way during the few months it's open each year.
The team inched its way down the sliver of a trail, hugging the shale wall as tightly as they could. Concentration is key here. Collin saw just how key when he rolled a little too aggressively into the first ultra tight switchback. Black Bear Pass is a very treacherous and unforgiving trail. Take it slow and easy. If you have to make a two- or three-point turn to get through it safely, you should.
Lesson learned, Collin applied the knowledge to the remaining seven hardcore hairpins ahead. The reward for all of this patience and focus came when the guys got to the power plant ahead. Built in 1907 for the Smuggler-Union Mining Company, it sits atop the spectacular Bridal Veil Falls. Getting here is hard work through tough terrain but the reward is worth it.
Fade to Black
Making the descent into the Telluride area, all three of the guys were stoked. Anxiety faded but the happiness didn't. Collin and Matt crossed Black Bear Pass off of their respective bucket lists. We think you should too.
Click here to watch the entire video.
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16010 | https://www.wyzant.com/resources/answers/806331/discrete-mathematics | Discrete Mathematics | Wyzant Ask An Expert
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Discrete MathComputer ScienceComputer EngineeringMathematicsDiscrete Mathematics
Moe T.
asked • 12/20/20
Discrete Mathematics
Prove that if n is a perfect square, then n + 2 is not a perfect square.
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All perfect squares are 0 or 1 mod 4. This is easy to show. If an integer z is even, it may be written as 2n, n an integer. Then, (2n)^2 = 4n^2. 4 |4n^2, so the perfect square mod 4 = 0
If an integer z is odd, it may be written as 2n+1. Then, (2n+1)^2 = 4n^2 + 4n + 1 = 4(n^2 _+ n) + 1, so the perfect square mod n = 1.
If x mod 4 = 0 or 1, then x+2 mod 4 = 2 or 3. Thus, x+2 mod 4 is not equal to 0 or 1, so it can't be a perfect square.
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16011 | https://www.quora.com/What-are-the-possible-solutions-for-sin-2-x-cos-2-x-1 | What are the possible solutions for [math]\sin^2 x+ \cos ^{2} x = 1 [/math]? - Quora
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What are the possible solutions for [math]\sin^2 x+ \cos ^{2} x = 1 [/math]?
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Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views
·8y
Originally Answered: How can I prove \sin^2x +\cos^2x =1? ·
This identity often turns up in mathematics problems.
Whenever it does, I say to my students:
“Why does it equal 1?”
I expect my students to reply with the following explanation.
Then I KNOW they UNDERSTAND!
Using the following idea, this almost proves itself!
Most students just “know” this result but it is really pleasing when they really “understand” it by showing me the above proof.
Continue Reading
This identity often turns up in mathematics problems.
Whenever it does, I say to my students:
“Why does it equal 1?”
I expect my students to reply with the following explanation.
Then I KNOW they UNDERSTAND!
Using the following idea, this almost proves itself!
Most students just “know” this result but it is really pleasing when they really “understand” it by showing me the above proof.
Upvote ·
99 34
9 3
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What are the solutions for [math]\sqrt{x^2}=1[/math]?
Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views
·1y
The following is just an ordinary algebraic equation which we can solve…
2x – 8 = 4
2x = 12
x = 6
…and this is the only value for x for which the equation is true.
However, the following is a special sort of equation called an IDENTITY…
2x – 8 = 2(x – 4)
It is always true for all values of x
Continue Reading
The following is just an ordinary algebraic equation which we can solve…
2x – 8 = 4
2x = 12
x = 6
…and this is the only value for x for which the equation is true.
However, the following is a special sort of equation called an IDENTITY…
2x – 8 = 2(x – 4)
It is always true for all values of x
Upvote ·
9 7
Buddha Buck
Studied at University at Buffalo · Upvoted by
Pepito Moropo
, B.S. / M.S. Mathematics · Author has 5.8K answers and 16.9M answer views
·9y
Originally Answered: How can I prove \sin^2x +\cos^2x =1? ·
How you prove that identity depends greatly on how you think of sine and cosine.
If you think of sine and cosine as ratios of sides of a right triangle (as in high school, where they teach sine as opposite over hypotenuse), then you get a right triangle with sides [math]a, b, c; a^2+b^2=c^2[/math] (the latter by Pythagorean triangle), and [math]\sin \theta = \frac{a}{c}, \cos\theta = \frac{b}{c}, \sin^2\theta + \cos^2\theta = (\frac{a}{c})^2 + (\frac{b}{c})^2 = \frac{a^2}{c^2} + \frac{b^2}{c^2} = \frac{a^2+b^2}{c^2} = \frac{c^2}{c^2} = 1[/math].
If you think of sine and cosine as the coordinates of a point on the unit ci
Continue Reading
How you prove that identity depends greatly on how you think of sine and cosine.
If you think of sine and cosine as ratios of sides of a right triangle (as in high school, where they teach sine as opposite over hypotenuse), then you get a right triangle with sides [math]a, b, c; a^2+b^2=c^2[/math] (the latter by Pythagorean triangle), and [math]\sin \theta = \frac{a}{c}, \cos\theta = \frac{b}{c}, \sin^2\theta + \cos^2\theta = (\frac{a}{c})^2 + (\frac{b}{c})^2 = \frac{a^2}{c^2} + \frac{b^2}{c^2} = \frac{a^2+b^2}{c^2} = \frac{c^2}{c^2} = 1[/math].
If you think of sine and cosine as the coordinates of a point on the unit circle (parameterized by the arclength of the circle), then by the definition of the unit circle, every point satisfies [math]x^2+y^2=1[/math], so the point math [/math] does too, so [math]\sin^2\theta+\cos^2\theta = 1[/math].
Sine and cosine can also be defined as independent solutions to the differential equation [math]f’’ = -f[/math], with [math]\sin 0 = 0, \sin’ 0 = 1, \cos 0 = 1, \cos’ 0 = 0[/math]. Since there are only two independent solutions to the equation, and it’s easy to see that [math]f^{(n)}[/math] is a solution, it must be the case that [math]\sin x, \sin’ x, \sin’’ x[/math] cannot be independent solutions. In fact, [math]\sin’’ x = -\sin x[/math], so [math]\sin’ 0 = 1, \sin’’ 0 = 0[/math], so [math]\sin’x = \cos x, \cos’ x = -\sin’ x[/math]. From this we can implicitly differentiate [math]\sin^2 x + \cos^2 x[/math] to get [math]2\sin x\sin’ x + 2\cos x\cos’ x = 2\sin x\cos x + 2\cos x(-\sin x) = 0[/math]. So the value of [math]\sin^2x + \cos^2x[/math] is a constant, and evaluated at 0 we get [math]\sin^2 0 + \cos^2 0 = 0^2 + 1^2 = 0+1 = 1[/math], so [math]\sin^2 x + \cos^2 x = 1[/math].
Sine and cosine can also be defined by the power series [math]\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots = \sum_{i=0}{\infty} (-1)^n\frac{x^{2n+1}}{(2n+1)!}[/math], [math]\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots = \sum_{i=0}{\infty}(-1)^n\frac{x^{2n}}{(2n)!}[/math]. A careful expansion of those power series in the expression [math]\sin^2 x + \cos ^2 x[/math] will show all of the terms involving [math]x^n[/math] cancel, leaving just the constant term [math]1[/math] as the value.
Upvote ·
99 32
9 1
Jack Fraser-Govil
Doctor of Physics, Writer of Code, Player of Games · Upvoted by
Benjamin Hardisty
, U. of Utah, PhD in Mathematical Modeling · Author has 2.6K answers and 51.5M answer views
·Updated 8y
Originally Answered: How can I prove \sin^2x +\cos^2x =1? ·
Consider the triangle:
We have:
[math]\sin{\theta} = \frac{b}{c}[/math]
[math]\cos{\theta} = \frac{a}{c}[/math]
Therefore we can square and add:
[math]\sin^2{\theta} + \cos^2{\theta}=\frac{b^2}{c^2} + \frac{a^2}{c^2} = \frac{a^2 + b^2}{c^2}[/math]
However, from Pythagoras, we have:
[math]a^2 + b^2 = c^2[/math]
So:
[math]\sin^2{\theta} + \cos^2{\theta} = \frac{c^2}{c^2} = 1[/math]
This is a fairly incomplete proof (it assumes the geometric interpretation of [math]\cos{\theta}[/math], and hence that we can use Pythagoras, for example), but it works to get a feeling for it.
Continue Reading
Consider the triangle:
We have:
[math]\sin{\theta} = \frac{b}{c}[/math]
[math]\cos{\theta} = \frac{a}{c}[/math]
Therefore we can square and add:
[math]\sin^2{\theta} + \cos^2{\theta}=\frac{b^2}{c^2} + \frac{a^2}{c^2} = \frac{a^2 + b^2}{c^2}[/math]
However, from Pythagoras, we have:
[math]a^2 + b^2 = c^2[/math]
So:
[math]\sin^2{\theta} + \cos^2{\theta} = \frac{c^2}{c^2} = 1[/math]
This is a fairly incomplete proof (it assumes the geometric interpretation of [math]\cos{\theta}[/math], and hence that we can use Pythagoras, for example), but it works to get a feeling for it.
Upvote ·
99 79
9 2
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Theoretical Hydrodynamics ========================= L. M. Milne-Thomson Courier Corporation, Apr 22, 2013 - Science - 768 pages This classic text offers a thorough, clear and methodical introductory exposition of the mathematical theory of fluid motion, useful in applications to both hydrodynamics and aerodynamics. Departing radically from traditional approaches, the author bases the treatment on vector methods and notation with their natural consequence in two dimensions — the complex variable. New features in this edition include: a chapter bringing together various exact treatments of two-dimensional motion with a free surface in a gravitational field, followed by one dealing with approximations (mostly linearized) relevant to this but with emphasis on waves; a chapter on tensor methods applied to the flow of viscous fluids; a chapter on flow with small Reynolds' number, including an account of a novel application of the complex variable to Stokes' flow; and an outline of the theory of two-dimensional laminar flow in a boundary layer. Prerequisites are restricted to a knowledge of elementary calculus since any additional mathematics is introduced as required, making this a self-contained treatment. Nearly 400 diagrams help illustrate the text and over 600 exercises are collected into sets of examples at the end of each chapter. More » Preview this book »
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Theoretical Hydrodynamics
Louis Melville Milne-Thomson
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Theoretical Hydrodynamics
Louis Melville Milne-Thomson
No preview available - 1996
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aerofoilangleangular velocityBernoulli's theoremboundary layercentrecirclecircular cylindercomplex potentialconsiderconstantcontourcoordinatescorrespondingcoshcross-sectioncurvedensitydirectiondistancedoubletellipseellipsoidequation of continuityequation of motionfixedfluidforcefree streamlinefree surfaceGauss's theoremgivengivesHenceholomorphicinfiniteinfinityintegralirrotationalirrotational motionkinetic energyLaplace's equationliquidmappingmovingnormalobtainedP₁parallelparticleperpendicularplanepositionpressureProveradiusreal axisregionrotationsatisfiedscalarsinhsolidsolutionspeedspherestagnation pointsteady motionstream functionstrengthtangenttheoremtransformationuniform streamunitvanishesvectorvelocity potentialviscousvortex linesvorticitywallwavex-axisz-planezeroαζдидрдудфдх
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Applied mathematician Louis Melville Milne-Thomson (1891-1974) taught for several decades at the Royal British Naval Academy in Greenwich. In his retirement he served as a visiting professor at research facilities around the world.
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Title Theoretical Hydrodynamics
Dover Books on Physics
AuthorL. M. Milne-Thomson
Edition reprint, revised
Publisher Courier Corporation, 2013
ISBN 0486318699, 9780486318691
Length 768 pages
SubjectsScience
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Physics
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General
Science / Physics / General
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16013 | https://www.ablebits.com/office-addins-blog/excel-cumulative-sum-running-total/ | Ablebits blog
Excel
Calculations
Excel Cumulative Sum - easy way to calculate running total
by Svetlana Cheusheva, updated on
This short tutorial shows how a usual Excel Sum formula with a clever use of absolute and relative cell references can quickly calculate a running total in your worksheet.
A running total, or cumulative sum, is a sequence of partial sums of a given data set. It is used to show the summation of data as it grows with time (updated every time a new number is added to the sequence).
This technique is very common in everyday use, for example to calculate the current score in games, show year-to-date or month-to-date sales, or compute your bank balance after each withdrawal and deposit. The following examples show the fastest way to calculate running total in Excel and plot a cumulative graph.
How to calculate running total (cumulative sum) in Excel
To calculate a running total in Excel, you can use the SUM function combined with a clever use of absolute and relative cells references.
For example, to calculate the cumulative sum for numbers in column B beginning in cell B2, enter the following formula in C2 and then copy it down to other cells:
In your running total formula, the first reference should always be an absolute reference with the $ sign ($B$2). Because an absolute reference never changes no matter where the formula moves, it will always refer back to B2. The second reference without the $ sign (B2) is relative and it adjusts based on the relative position of the cell where the formula is copied. For more information about Excel cell references, please see Why use dollar sign ($) in Excel formulas.
So, when our Sum formula is copied to B3, it becomes SUM($B$2:B3), and returns the total of values in cells B2 to B3. In cell B4, the formula turns into SUM($B$2:B4), and totals numbers in cells B2 to B4, and so on:
In a similar manner, you can use the Excel SUM function to find the cumulative sum for your bank balance. For this, enter deposits as positive numbers, and withdrawals as negative numbers in some column (column C in this example). And then, to show the running total, enter the following formula in column D:
Strictly speaking, the above screenshot shows not exactly a cumulative sum, which implies summation, but some sort of "running total and running difference" Anyway, who cares about the right word if you've got the desired result, right? :)
At first sight, our Excel Cumulative Sum formula looks perfect, but it does have one significant drawback. When you copy the formula down a column, you will notice that the cumulative totals in the rows below the last cell with a value in column C all show the same number:
To fix this, we can improve our running total formula a bit further by embedding it in the IF function:
The formula instructs Excel to do the following: if cell C2 is blank, then return an empty string (blank cell), otherwise apply the cumulative total formula.
Now, you can copy the formula to as many cells as you want, and the formula cells will look empty until you enter a number in the corresponding row in column C. As soon as you do this, the calculated cumulative sum will appear next to each amount:
How to make a cumulative graph in Excel
As soon as you've calculated the running total using the Sum formula, making a cumulative chart in Excel is a matter of minutes.
Select your data, including the Cumulative Sum column, and create a 2-D clustered column chart by clicking the corresponding button on the Insert tab, in the Charts group:
In the newly created chart, click the Cumulative Sum data series (orange bars in this example), and right click to select Change Series Chart Type... from the context menu.
If you are using a recent version of Excel 2013 or Excel 2016, select the Combo chart type, and click on the first icon (Clustered Column - Line) at the top of Change Chart Type dialog:
Or, you can highlight the Custom Combination icon, and choose the line type you want for the Cumulative Sum data series (Line with Markers in this example):
In Excel 2010 and earlier, simply select the desired line type for the Cumulative Sum series, which you've selected on the previous step:
4. Click OK, and evaluate your Excel cumulative chart:
5. Optionally, you can right-click the Cumulative Sum line in the chart, and select Add Data Labels from the context menu:
As the result, your Excel cumulative graph will look similar to this:
To embellish your Excel cumulative chart further, you can customize the chart and axes titles, modify the chart legend, choose other chart style and colors, etc. For the detailed instructions, please see our Excel charts tutorial.
This is how you do a running total in Excel. If you are curious to learn a few more useful formulas, check out the below examples. I thank you for reading and hope to see you again soon!
You may also be interested in
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49 comments
Carolyn Pollack says:
Hi there...I know this is an old post but putting this out there.Is there a way to have a running total in one column, let's say column D, with "debits" in column B and "credits" in column C?
Is there a formula I can plug into Excel to add from Column C and subtract items from Column B for a running total in Column D?
Thanks
Alexander Trifuntov (Ablebits Team) says:
Hello Carolyn!If I understand your task correctly, the following formula should work for you:
=SUM($C$2:C2) - SUM($B$2:B2)
2. Qasim says:
Hey,Anyone know to find overall total from different sheets?
Alexander Trifuntov (Ablebits Team) says:
Hi! If I got you right, this guide will help you with your task: 3-D reference in Excel: reference the same cell or range in multiple worksheets.
3. Kris says:
The formula =SUM($L$7:L7) is working for me in an area formatted as a table; however, whenever I add a new row, the new row will have the correct formula, but what had previously been the final cell in that column will change to include the new row. Each subsequent adding of a row will screw up what had previously been the final row.The formulas in three cells in a column will go from:=SUM($L$7:L7)=SUM($L$7:L8)=SUM($L$7:L9)
To (when adding a forth row)=SUM($L$7:L7)=SUM($L$7:L8)=SUM($L$7:L10)=SUM($L$7:L10)
Adding a fifth row will turn to=SUM($L$7:L7)=SUM($L$7:L8)=SUM($L$7:L11)=SUM($L$7:L11)=SUM($L$7:L11)
Is there a fix?
Alexander Trifuntov (Ablebits Team) says:
Hello Kris!If you are working with an ever-changing dataset, you may want to make the range dynamic. This means that it automatically expands to accommodate newly added records. You can find the examples and detailed instructions here: How to create and use dynamic named range in Excel.
4. Bill S says:
Editing a spreadsheet that has a cumulative sum column
I have a spreadsheet into which I have entered a series of transactions, and alongside each transaction amount I have a formula which gives the running total.
Lets say, at some point in the future, I notice that the order of the transactions is incorrect and I need to move one of them up to its correct position. So I insert a blank row at the appropriate point and I can then copy the transaction that is in the wrong place into this blank row, and then delete the original row. This works fine.
However, I had expected to be able to MOVE the transaction into the blank row by dragging, rather than COPY it and delete the original. When I MOVE the transaction, the relative reference does not update.
For example if row 4 has a cumulative total formula of =SUM($B$2:B4), and row 7 has a cumulative total formula of =SUM($B$2:B7), and I want to move the current row 7 transaction into the position currently occupied by the row 4 transaction, if I use "copy and delete", the formula copied from what was row 7 has its formula correctly changed to =SUM($B$2:B4) to reflect its new position.
If, instead I MOVE the row, then I end up with the formula =SUM($B$2:B8) in row 4. The relative reference has not been updated. (The reason it is now 8 not 7 is because it was updated when I added the new row at 4.)
Obviously it is no great hardship having to use "copy and delete", but I have to do this row rearrangement quite often, and I find myself using the MOVE method by mistake and then wondering why my spreadsheet is wrong.
Does anyone have any suggestions?
Alexander Trifuntov (Ablebits Team) says:
Hi! Maybe this article will be helpful: How to copy formula in Excel with or without changing references.
5. Robert says:
Great information here.This is probably a simple one for the illuminati:Suppose column C tracks profit gain and column D tracks profit losses. What would be the formula for column E tracking the ongoing balance?It seems like it should be =SUM($C$2:C2)-SUM($D$2:D2) but I've apparently missed something.
Alexander Trifuntov (Ablebits Team) says:
Hi! If you copy this formula down the column, you will get the running profit total. To get just the profit total, change the cell references in the SUM formula. For example,
=SUM($C$2:C20)-SUM($D$2:D20)
6. phoebe radford says:
How do I calculate the average of the total month at day 1, day 2, day 3 etc?
Alexander Trifuntov (Ablebits Team) says:
Hi!If I understand your task correctly, you can use the AVERAGEIFS function and specify the desired days as the criteria.
7. summer says:
Hi I'm trying to get a running accm. total for revenue over various sheets on an excel file. For each week I get the revenue and the following week I get another total and I want to keep adding them together as I pull new sheets. How would I do that?for example;sheet one -- total revenuesheet two -- total revenue +total revenue of sheet onesheet three -- total revenue + (sheet two total revenue + sheet one total revenue)sheet four -- total revenue + (sheet three total revenue + sheet two total revenue + sheet one total revenue)
Alexander Trifuntov (Ablebits Team) says:
Hi!Here is the article that may be helpful to you: 3-D reference in Excel: reference the same cell or range in multiple worksheets.
8. Laura says:
I want to set up a spreadsheet so that it totals values per week, i.e. on 12/2/22, I can see the sum of the values from 11/26/22-12/2/22. Then next week, I want it to total the sum of the values from 12/3/22-12/9/22 and so on and so forther.
Alexander Trifuntov (Ablebits Team) says:
Hi!Write in cell B1 the formula SUM(A1:A7). Then select the range B1:B7 and drag the fill handle down the column. See for more information: How to copy formula in Excel.
9. Dillip Kumar says:
HiIs there any function that will do running total values of same date and repeat again when new date starts
Alexander Trifuntov (Ablebits Team) says:
Hi!You can calculate the amount for a certain date using the SUMIF function. When calculating the running total, use it instead of the SUM function. I hope it’ll be helpful.
10. Ag says:
A B C D E1 10 102 22 323 0 224 32 325 16 486 0 167 0 0COUNTIF(B1:B7;">0")RESULT = 3 from Column B
Require formula whereResult = 1Which is determined as follows:In Column C - Helper columnIF(B1+B2;"=0";0;B1+B3)RESULT is 1
Formula required to getRESULT =1Without the Helper Column C
Thank you.
Alexander Trifuntov (Ablebits Team) says:
Hi!I don’t know how you can get result 1. But the formula should be written like this
IF(B1+B2=0;0;B1+B3)
Read about using IF function in this tutorial.
11. Kevin says:
How would you calculate the cumulative totals if the dates in column A were not in order? Thanks!
Ashish kumar agrawal says:
1 oct 1998 to today how much will be the total cumulative interest is 18per cent per month cumulativly if amount is 10000
2. john says:
have you ever found an answer for this
12. J C says:
Thank you! My question:I have a complex table. Therefore, I am not aiming at creating additional column with cumulative sum.Can I then, ask Excel to plot a cumulative sum line chart? i.e. the dataseries entry will specify cummulative figures from a (yet to be cumulatively summed) column?thanks.
Alexander Trifuntov (Ablebits Team) says:
Hi!A graph in Excel is built according to the data that is written in certain cells.
13. David says:
When I follow your example exactly, it doesn't work for me!When I have got the formulas in each of the cells, eg in C4 it is =SUM($B$2:B4) it only adds B2and B4, missing out B3.For cell B5, the formula is =SUM($B$2:B5) but it only adds B2 and B5, missing out B3 and B4.Am I doing something wrong?
Alexander Trifuntov (Ablebits Team) says:
Hello!The formula =SUM($B$2:B5) sums cells B2 through B5. You can check this using a calculator.You can learn more about SUM function in Excel in this article on our blog.
14. James Walsh says:
Is there a way to pull a specific date based on the cumulative total. As in, when did the total pass $1700? Is there a function to return that date?
I am working on a data set where I am projecting X number of real estate developments per year. Lets say at 6 per year. I want to show in what year that the total passes development 14 is built? (This would be year 3). I cannot figure out the function.
Thanks!
Alexander Trifuntov (Ablebits Team) says:
Hello!Column A contains dates, column B - amounts, column C - cumulative total. The date when the cumulative total exceeded a certain amount can be determined by the formula
=INDEX(A1:A100,MATCH(1700,C1:C100,1)+1)
Hope this is what you need.
15. Robert Coates says:
Hi Everyone,
I am looking to calculate a maximum cumulative average of at least 60 cells in length from within a larger range of cells. So, if in a range of data (A1:A101), with the cumulative average running from cell A1 to cell A101, how would I find out the section within this range that has the highest cumulative average? To qualify, it has to be at least 60 cells long.
Thank you
Ian says:
Based on 100 rows of data -In column B, work out the average for your range,soB1 = =Average(B1:B60)B2 - =Average(B2:B61)
Continue this down for 41 rows so all accumulations are covered.
In column C, use the Rank option to see which set is the highest/lowest group of values.= B1 =RANK(B1,$B$1:$B$41,0) all the way down to C40This will give you the highest ranked, =RANK(B13,$B$1:$B$41,1) Changing the 0 to a 1 will give you the lowest ranked.
16. Kristin Simpson says:
I'm creating a spread sheet to track provisional credit issued and then funds recouped. What I want to be able to do is enter a total into column A say 200.00 then in column B enter the return credit of 50.00 and have that amount auto subtract so that column A now reads 150.00.
To further complicate it in rare situations I have a third column C where I will on occassion need to subtract from column B without impacting the total in column A. So if I put 25.00 in column C. Column B becomes 25.00 but column A stays 150.00.
Alexander Trifuntov (Ablebits Team) says:
Hello Kristin!If a value is written in cell A1, then the formula can no longer be written to it. This has already been discussed on the blog many times. So your question about the formula in cell A1 does not make sense. Only VBA Macro Can Solve the Problem
17. alan coombs says:
Hi,I have 2 columns.First column is text of various items , with varying amounts of the same items (1-100).Second column is just numbers relating to data in first column.I am trying to get a running total in the second column for each item in the first column.Any help appreciated.
Alexander Trifuntov (Ablebits Team) says:
Hello Alan!If I understand your task correctly, if the name of the item is recorded in column A, the quantity is recorded in column B, then running total can be calculated by the formula
=SUMIF($A$3:A30,A3,$B$3:B30)
I hope my advice will help you solve your task.
18. rpgosnell says:
if $ sign is used, what happens if a new column is inserted to the left.
Svetlana Cheusheva says:
Absolute cell references (with the $ sign) do change when you add or remove rows and/or columns to reflect a new position of the referenced cell, so your reference will change accordingly.
19. Tangiean Leon says:
I have columns with different items and then prices and then a total what I want is for it to automatically sum up the total of the previous square with the new amount put in. For exampleB C D EITEM, DESCRIPTION, AMOUNT, TOTALFood, Bagel , $3.00, =$3.00Drink, Tea ,$ 2.00, =$5.00 etc...
But I want it to sum it up automatically, right now I am having to do it individually =SUM(E1,D2) can anyone help?
Gennady Terekhov (Ablebits Team) says:
If I understand your task correctly, please try to do the following:
Enter the following formula in cell E2;
Just select the cell where you've entered the formula and drag the fill handle (a small square at the lower right-hand corner of the selected cell) down.
Hope this will help.
20. MASHOK says:
my sheet column head is description (A:A), debit(B:B), credit(C:C), Balance (D:D). In description if i write "Deposit" then the value will put on debit column, and if write "Credited", the value will put on credit column. But how to ensure if i write "credit" in A3, THE VALUE 500 will only can be post in Credit column C3, B3 should not accept any input if the description is 'CREDITED'. OR if the description is "deposit" the value cannot be post in Credit clumn C3. How to apply, pls.
Gennady Terekhov (Ablebits Team) says:
Hello,
If I understand your task correctly, please create a custom Data Validation rule for columns B and C using these formulas:
column B=A1="Deposit"
column C=A1="Credited"
We have an article on our blog that describes how to use Data Validation in Excel. Please have a look at it.
Hope you’ll find it helpful.
Post a comment
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Contents
1 Definition and classification
2 Electronic configuration
3 Characteristic properties
3.1 Coloured compounds
3.2 Oxidation states
3.3 Magnetism
3.4 Catalytic properties
3.5 Physical properties
4 See also
5 References
Transition metal
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From Wikipedia, the free encyclopedia
Series of chemical elements
Transition metals in the periodic table
| | | |
---
| Hydrogen | | Helium |
| Lithium | Beryllium | | Boron | Carbon | Nitrogen | Oxygen | Fluorine | Neon |
| Sodium | Magnesium | | Aluminium | Silicon | Phosphorus | Sulfur | Chlorine | Argon |
| Potassium | Calcium | | Scandium | Titanium | Vanadium | Chromium | Manganese | Iron | Cobalt | Nickel | Copper | Zinc | Gallium | Germanium | Arsenic | Selenium | Bromine | Krypton |
| Rubidium | Strontium | | | Yttrium | Zirconium | Niobium | Molybdenum | Technetium | Ruthenium | Rhodium | Palladium | Silver | Cadmium | Indium | Tin | Antimony | Tellurium | Iodine | Xenon |
| Caesium | Barium | Lanthanum | Cerium | Praseodymium | Neodymium | Promethium | Samarium | Europium | Gadolinium | Terbium | Dysprosium | Holmium | Erbium | Thulium | Ytterbium | Lutetium | Hafnium | Tantalum | Tungsten | Rhenium | Osmium | Iridium | Platinum | Gold | Mercury (element) | Thallium | Lead | Bismuth | Polonium | Astatine | Radon |
| Francium | Radium | Actinium | Thorium | Protactinium | Uranium | Neptunium | Plutonium | Americium | Curium | Berkelium | Californium | Einsteinium | Fermium | Mendelevium | Nobelium | Lawrencium | Rutherfordium | Dubnium | Seaborgium | Bohrium | Hassium | Meitnerium | Darmstadtium | Roentgenium | Copernicium | Nihonium | Flerovium | Moscovium | Livermorium | Tennessine | Oganesson |
| |
| Part of a series on the |
| Periodic table |
| Periodic table forms 18-column 32-column Alternative and extended forms |
| Periodic table history D. Mendeleev + 1871 table + 1869 predictions Discovery of elements Naming and etymology + for people + for places + controversies (in East Asia) Systematic element names |
| Sets of elements |
| By periodic table structure Groups (1–18) 1 (alkali metals) 2 (alkaline earth metals) 3 4 5 6 7 8 9 10 11 12 13 14 15 (pnictogens) 16 (chalcogens) 17 (halogens) 18 (noble gases) Periods (1–7, ...) 1 2 3 4 5 6 7 8+ Blocks (s, p, d, f, ...) + Atomic orbitals + Aufbau principle |
| By metallic classification Metals alkali alkaline earth transition post-transition lanthanide actinide Metalloids + dividing metals and nonmetals Nonmetals nonmetal halogen noble gas |
| By other characteristics Coinage metals Platinum-group metals Precious metals Refractory metals Heavy metals Light metals Native metals Noble metals Main-group elements Rare-earth elements Transuranium elements Major, minor and trans- actinides |
| Elements |
| List of chemical elements by abundance (in human body) by atomic properties by isotope stability by symbol |
| Properties of elements Relative atomic mass Crystal structure Electron affinity configuration Electronegativity (Allen, Pauling) Goldschmidt classification Nutrition Valence |
| Data pages for elements Abundance Atomic radius Boiling point Critical point Density Elasticity Electrical resistivity Electron affinity / configuration Electronegativity Hardness Heat capacity / of fusion / of vaporization Ionization energy Melting point Oxidation state Speed of sound Thermal conductivity / expansion coefficient Vapor pressure |
| Category Chemistry Portal |
| v t e |
In chemistry, a transition metal (or transition element) is a chemical element in the d-block of the periodic table (groups 3 to 12), though the elements of group 12 (and less often group 3) are sometimes excluded. The lanthanide and actinide elements (the f-block) are called inner transition metals and are sometimes considered to be transition metals as well.
They are lustrous metals with good electrical and thermal conductivity. Most (with the exception of group 11 and group 12) are hard and strong, and have high melting and boiling temperatures. They form compounds in any of two or more different oxidation states and bind to a variety of ligands to form coordination complexes that are often coloured. They form many useful alloys and are often employed as catalysts in elemental form or in compounds such as coordination complexes and oxides. Most are strongly paramagnetic because of their unpaired d electrons, as are many of their compounds. All of the elements that are ferromagnetic near room temperature are transition metals (iron, cobalt and nickel) or inner transition metals (gadolinium).
English chemist Charles Rugeley Bury (1890–1968) first used the word transition in this context in 1921, when he referred to a transition series of elements during the change of an inner layer of electrons (for example n = 3 in the 4th row of the periodic table) from a stable group of 8 to one of 18, or from 18 to 32. These elements are now known as the d-block.
Definition and classification
The 2011 IUPAC Principles of Chemical Nomenclature describe a "transition metal" as any element in groups 3 to 12 on the periodic table. This corresponds exactly to the d-block elements, and many scientists use this definition. In actual practice, the f-block lanthanide and actinide series are called "inner transition metals". The 2005 Red Book allows for the group 12 elements to be excluded, but not the 2011 Principles.
The IUPAC Gold Book defines a transition metal as "an element whose atom has a partially filled d sub-shell, or which can give rise to cations with an incomplete d sub-shell", but this definition is taken from an old edition of the Red Book and is no longer present in the current edition.
In the d-block, the atoms of the elements have between zero and ten d electrons.
Transition metals in the d-block
| Group | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| Period 4 | 21Sc | 22Ti | 23V | 24Cr | 25Mn | 26Fe | 27Co | 28Ni | 29Cu | 30Zn |
| 5 | 39Y | 40Zr | 41Nb | 42Mo | 43Tc | 44Ru | 45Rh | 46Pd | 47Ag | 48Cd |
| 6 | 71Lu | 72Hf | 73Ta | 74W | 75Re | 76Os | 77Ir | 78Pt | 79Au | 80Hg |
| 7 | 103Lr | 104Rf | 105Db | 106Sg | 107Bh | 108Hs | 109Mt | 110Ds | 111Rg | 112Cn |
Published texts and periodic tables show variation regarding the heavier members of group 3. The common placement of lanthanum and actinium in these positions is not supported by physical, chemical, and electronic evidence, which overwhelmingly favour putting lutetium and lawrencium in those places. Some authors prefer to leave the spaces below yttrium blank as a third option, but there is confusion on whether this format implies that group 3 contains only scandium and yttrium, or if it also contains all the lanthanides and actinides; additionally, it creates a 15-element-wide f-block, when quantum mechanics dictates that the f-block should only be 14 elements wide. The form with lutetium and lawrencium in group 3 is supported by a 1988 IUPAC report on physical, chemical, and electronic grounds, and again by a 2021 IUPAC preliminary report as it is the only form that allows simultaneous (1) preservation of the sequence of increasing atomic numbers, (2) a 14-element-wide f-block, and (3) avoidance of the split in the d-block. Argumentation can still be found in the contemporary literature purporting to defend the form with lanthanum and actinium in group 3, but many authors consider it to be logically inconsistent (a particular point of contention being the differing treatment of actinium and thorium, which both can use 5f as a valence orbital but have no 5f occupancy as single atoms); the majority of investigators considering the problem agree with the updated form with lutetium and lawrencium.
The group 12 elements zinc, cadmium, and mercury are sometimes excluded from the transition metals. This is because they have the electronic configuration [ ]d10s2, where the d shell is complete, and they still have a complete d shell in all their known oxidation states. The group 12 elements Zn, Cd and Hg may therefore, under certain criteria, be classed as post-transition metals in this case. However, it is often convenient to include these elements in a discussion of the transition elements. For example, when discussing the crystal field stabilization energy of first-row transition elements, it is convenient to also include the elements calcium and zinc, as both Ca2+ and Zn2+ have a value of zero, against which the value for other transition metal ions may be compared. Another example occurs in the Irving–Williams series of stability constants of complexes. Moreover, Zn, Cd, and Hg can use their d orbitals for bonding even though they are not known in oxidation states that would formally require breaking open the d-subshell, which sets them apart from the p-block elements.
The 2007 (though disputed and so far not reproduced independently) synthesis of mercury(IV) fluoride (HgF4) has been taken by some to reinforce the view that the group 12 elements should be considered transition metals, but some authors still consider this compound to be exceptional. Copernicium is expected to be able to use its d electrons for chemistry as its 6d subshell is destabilised by strong relativistic effects due to its very high atomic number, and as such is expected to have transition-metal-like behaviour and show higher oxidation states than +2 (which are not definitely known for the lighter group 12 elements). Even in bare dications, Cn2+ is predicted to be 6d87s2, unlike Hg2+ which is 5d106s0.
Although meitnerium, darmstadtium, and roentgenium are within the d-block and are expected to behave as transition metals analogous to their lighter congeners iridium, platinum, and gold, this has not yet been experimentally confirmed. Whether copernicium behaves more like mercury or has properties more similar to those of the noble gas radon is not clear. Relative inertness of Cn would come from the relativistically expanded 7s–7p1/2 energy gap, which is already adumbrated in the 6s–6p1/2 gap for Hg, weakening metallic bonding and causing its well-known low melting and boiling points.
Transition metals with lower or higher group numbers are described as 'earlier' or 'later', respectively. When described in a two-way classification scheme, early transition metals are on the left side of the d-block from group 3 to group 7. Late transition metals are on the right side of the d-block, from group 8 to 11 (or 12, if they are counted as transition metals). In an alternative three-way scheme, groups 3, 4, and 5 are classified as early transition metals, 6, 7, and 8 are classified as middle transition metals, and 9, 10, and 11 (and sometimes group 12) are classified as late transition metals.
The heavy group 2 elements calcium, strontium, and barium do not have filled d-orbitals as single atoms, but are known to have d-orbital bonding participation in some compounds, and for that reason have been called "honorary" transition metals. The same is likely true of radium.
The f-block elements La–Yb and Ac–No have chemical activity of the (n−1)d shell, but importantly also have chemical activity of the (n−2)f shell that is absent in d-block elements. Hence they are often treated separately as inner transition elements.
Electronic configuration
Main article: Electron configuration
The general electronic configuration of the d-block atoms is noble gasd0–10ns0–2np0–1. Here "[noble gas]" is the electronic configuration of the last noble gas preceding the atom in question, and n is the highest principal quantum number of an occupied orbital in that atom. For example, Ti (Z = 22) is in period 4 so that n = 4, the first 18 electrons have the same configuration of Ar at the end of period 3, and the overall configuration is [Ar]3d24s2. The period 6 and 7 transition metals also add core (n − 2)f14 electrons, which are omitted from the tables below. The p orbitals are almost never filled in free atoms (the one exception being lawrencium due to relativistic effects that become important at such high Z), but they can contribute to the chemical bonding in transition metal compounds.
The Madelung rule predicts that the inner d orbital is filled after the valence-shell s orbital. The typical electronic structure of transition metal atoms is then written as [noble gas]ns2(n − 1)dm. This rule is approximate, but holds for most of the transition metals. Even when it fails for the neutral ground state, it accurately describes a low-lying excited state.
The d subshell is the next-to-last subshell and is denoted as (n − 1)d subshell. The number of s electrons in the outermost s subshell is generally one or two except palladium (Pd), with no electron in that s sub shell in its ground state. The s subshell in the valence shell is represented as the ns subshell, e.g. 4s. In the periodic table, the transition metals are present in ten groups (3 to 12).
The elements in group 3 have an ns2(n − 1)d1 configuration, except for lawrencium (Lr): its 7s27p1 configuration exceptionally does not fill the 6d orbitals at all. The first transition series is present in the 4th period, and starts after Ca (Z = 20) of group 2 with the configuration [Ar]4s2, or scandium (Sc), the first element of group 3 with atomic number Z = 21 and configuration [Ar]4s23d1, depending on the definition used. As we move from left to right, electrons are added to the same d subshell till it is complete. Since the electrons added fill the (n − 1)d orbitals, the properties of the d-block elements are quite different from those of s and p block elements in which the filling occurs either in s or in p orbitals of the valence shell. The electronic configuration of the individual elements present in all the d-block series are given below:
First (3d) d-block Series (Sc–Zn)
| Group | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| Atomic number | 21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 | 29 | 30 |
| Element | Sc | Ti | V | Cr | Mn | Fe | Co | Ni | Cu | Zn |
| Electronconfiguration | 3d14s2 | 3d24s2 | 3d34s2 | 3d54s1 | 3d54s2 | 3d64s2 | 3d74s2 | 3d84s2 | 3d104s1 | 3d104s2 |
Second (4d) d-block Series (Y–Cd)
| Atomic number | 39 | 40 | 41 | 42 | 43 | 44 | 45 | 46 | 47 | 48 |
| Element | Y | Zr | Nb | Mo | Tc | Ru | Rh | Pd | Ag | Cd |
| Electronconfiguration | 4d15s2 | 4d25s2 | 4d45s1 | 4d55s1 | 4d55s2 | 4d75s1 | 4d85s1 | 4d105s0 | 4d105s1 | 4d105s2 |
Third (5d) d-block Series (Lu–Hg)
| Atomic number | 71 | 72 | 73 | 74 | 75 | 76 | 77 | 78 | 79 | 80 |
| Element | Lu | Hf | Ta | W | Re | Os | Ir | Pt | Au | Hg |
| Electronconfiguration | 5d16s2 | 5d26s2 | 5d36s2 | 5d46s2 | 5d56s2 | 5d66s2 | 5d76s2 | 5d96s1 | 5d106s1 | 5d106s2 |
Fourth (6d) d-block Series (Lr–Cn)(Configurations predicted for Mt–Cn)
| Atomic number | 103 | 104 | 105 | 106 | 107 | 108 | 109 | 110 | 111 | 112 |
| Element | Lr | Rf | Db | Sg | Bh | Hs | Mt | Ds | Rg | Cn |
| Electronconfiguration | 7s27p1 | 6d27s2 | 6d37s2 | 6d47s2 | 6d57s2 | 6d67s2 | 6d77s2 | 6d87s2 | 6d97s2 | 6d107s2 |
A careful look at the electronic configuration of the elements reveals that there are certain exceptions to the Madelung rule. For Cr as an example the rule predicts the configuration 3d44s2, but the observed atomic spectra show that the real ground state is 3d54s1. To explain such exceptions, it is necessary to consider the effects of increasing nuclear charge on the orbital energies, as well as the electron–electron interactions including both Coulomb repulsion and exchange energy. The exceptions are in any case not very relevant for chemistry because the energy difference between them and the expected configuration is always quite low.
The (n − 1)d orbitals that are involved in the transition metals are very significant because they influence such properties as magnetic character, variable oxidation states, formation of coloured compounds etc. The valence s and p orbitals (ns and np) have very little contribution in this regard since they hardly change in the moving from left to the right in a transition series. In transition metals, there are greater horizontal similarities in the properties of the elements in a period in comparison to the periods in which the d orbitals are not involved. This is because in a transition series, the valence shell electronic configuration of the elements do not change. However, there are some group similarities as well.
Characteristic properties
There are a number of properties shared by the transition elements that are not found in other elements, which results from the partially filled d shell. These include
the formation of compounds whose colour is due to d–d electronic transitions
the formation of compounds in many oxidation states, due to the relatively low energy gap between different possible oxidation states
the formation of many paramagnetic compounds due to the presence of unpaired d electrons. A few compounds of main-group elements are also paramagnetic (e.g. nitric oxide, oxygen)
Most transition metals can be bound to a variety of ligands, allowing for a wide variety of transition metal complexes.
Coloured compounds
Colour in transition-series metal compounds is generally due to electronic transitions of two principal types.
charge transfer transitions. An electron may jump from a predominantly ligand orbital to a predominantly metal orbital, giving rise to a ligand-to-metal charge-transfer (LMCT) transition. These can most easily occur when the metal is in a high oxidation state. For example, the colour of chromate, dichromate and permanganate ions is due to LMCT transitions. Another example is that mercuric iodide, HgI2, is red because of a LMCT transition.
A metal-to-ligand charge transfer (MLCT) transition will be most likely when the metal is in a low oxidation state and the ligand is easily reduced.
In general charge transfer transitions result in more intense colours than d–d transitions.
d–d transitions. An electron jumps from one d orbital to another. In complexes of the transition metals the d orbitals do not all have the same energy. The pattern of splitting of the d orbitals can be calculated using crystal field theory. The extent of the splitting depends on the particular metal, its oxidation state and the nature of the ligands. The actual energy levels are shown on Tanabe–Sugano diagrams.
In centrosymmetric complexes, such as octahedral complexes, d–d transitions are forbidden by the Laporte rule and only occur because of vibronic coupling in which a molecular vibration occurs together with a d–d transition. Tetrahedral complexes have somewhat more intense colour because mixing d and p orbitals is possible when there is no centre of symmetry, so transitions are not pure d–d transitions. The molar absorptivity (ε) of bands caused by d–d transitions are relatively low, roughly in the range 5–500 M−1cm−1 (where M = mol dm−3). Some d–d transitions are spin forbidden. An example occurs in octahedral, high-spin complexes of manganese(II), which has a d5 configuration in which all five electrons have parallel spins; the colour of such complexes is much weaker than in complexes with spin-allowed transitions. Many compounds of manganese(II) appear almost colourless. The spectrum of [Mn(H2O)6]2+ shows a maximum molar absorptivity of about 0.04 M−1cm−1 in the visible spectrum.
Oxidation states
A characteristic of transition metals is that they exhibit two or more oxidation states, usually differing by one. For example, compounds of vanadium are known in all oxidation states between −1, such as [V(CO)6]−, and +5, such as VO3−4.
Main-group elements in groups 13 to 18 also exhibit multiple oxidation states. The "common" oxidation states of these elements typically differ by two instead of one. For example, compounds of gallium in oxidation states +1 and +3 exist in which there is a single gallium atom. Compounds of Ga(II) would have an unpaired electron and would behave as a free radical and generally be destroyed rapidly, but some stable radicals of Ga(II) are known. Gallium also has a formal oxidation state of +2 in dimeric compounds, such as [Ga2Cl6]2−, which contain a Ga-Ga bond formed from the unpaired electron on each Ga atom. Thus the main difference in oxidation states, between transition elements and other elements is that oxidation states are known in which there is a single atom of the element and one or more unpaired electrons.
The maximum oxidation state in the first row transition metals is equal to the number of valence electrons from titanium (+4) up to manganese (+7), but decreases in the later elements. In the second row, the maximum occurs with ruthenium (+8), and in the third row, the maximum occurs with iridium (+9). In compounds such as [MnO4]− and OsO4, the elements achieve a stable configuration by covalent bonding.
The lowest oxidation states are exhibited in metal carbonyl complexes such as Cr(CO)6 (oxidation state zero) and [Fe(CO)4]2− (oxidation state −2) in which the 18-electron rule is obeyed. These complexes are also covalent.
Ionic compounds are mostly formed with oxidation states +2 and +3. In aqueous solution, the ions are hydrated by (usually) six water molecules arranged octahedrally.
Magnetism
Main article: Magnetochemistry
Transition metal compounds are paramagnetic when they have one or more unpaired d electrons. In octahedral complexes with between four and seven d electrons both high spin and low spin states are possible. Tetrahedral transition metal complexes such as [FeCl4]2− are high spin because the crystal field splitting is small so that the energy to be gained by virtue of the electrons being in lower energy orbitals is always less than the energy needed to pair up the spins. Some compounds are diamagnetic. These include octahedral, low-spin, d6 and square-planar d8 complexes. In these cases, crystal field splitting is such that all the electrons are paired up.
Ferromagnetism occurs when individual atoms are paramagnetic and the spin vectors are aligned parallel to each other in a crystalline material. Metallic iron and the alloy alnico are examples of ferromagnetic materials involving transition metals. Antiferromagnetism is another example of a magnetic property arising from a particular alignment of individual spins in the solid state.
Catalytic properties
The transition metals and their compounds are known for their homogeneous and heterogeneous catalytic activity. This activity is ascribed to their ability to adopt multiple oxidation states and to form complexes. Vanadium(V) oxide (in the contact process), finely divided iron (in the Haber process), and nickel (in catalytic hydrogenation) are some of the examples. Catalysts at a solid surface (nanomaterial-based catalysts) involve the formation of bonds between reactant molecules and atoms of the surface of the catalyst (first row transition metals utilize 3d and 4s electrons for bonding). This has the effect of increasing the concentration of the reactants at the catalyst surface and also weakening of the bonds in the reacting molecules (the activation energy is lowered). Also because the transition metal ions can change their oxidation states, they become more effective as catalysts.
An interesting type of catalysis occurs when the products of a reaction catalyse the reaction producing more catalyst (autocatalysis). One example is the reaction of oxalic acid with acidified potassium permanganate (or manganate (VII)). Once a little Mn2+ has been produced, it can react with MnO4− forming Mn3+. This then reacts with C2O4− ions forming Mn2+ again.
Physical properties
As implied by the name, all transition metals are metals and thus conductors of electricity.
In general, transition metals possess a high density and high melting points and boiling points. These properties are due to metallic bonding by delocalized d electrons, leading to cohesion which increases with the number of shared electrons. However the group 12 metals have much lower melting and boiling points since their full d subshells prevent d–d bonding, which again tends to differentiate them from the accepted transition metals. Mercury has a melting point of −38.83 °C (−37.89 °F) and is a liquid at room temperature.
See also
Scholia has a profile for transition metal (Q19588).
Inner transition element, a name given to any member of the f-block
Main-group element, an element other than a transition metal
Ligand field theory a development of crystal field theory taking covalency into account
Crystal field theory a model that describes the breaking of degeneracies of electronic orbital states
Post-transition metal, a metallic element to the right of the transition metals in the periodic table
References
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^ Bury, C. R. (1921). "Langmuir's theory of the arrangement of electrons in atoms and molecules". J. Am. Chem. Soc. 43 (7): 1602–1609. doi:10.1021/ja01440a023.
^ Bury, Charles Rugeley. Encyclopedia.com Complete dictionary of scientific biography (2008).
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^ IUPAC, Compendium of Chemical Terminology, 5th ed. (the "Gold Book") (2025). Online version: (2006–) "transition element". doi:10.1351/goldbook.T06456
^ Scerri, Eric R. (2020). The Periodic Table: Its Story and Its Significance. New York, NY. ISBN 978-0-19-091436-3. OCLC 1096234740.{{cite book}}: CS1 maint: location missing publisher (link)
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^ Wittig, Jörg (1973). "The pressure variable in solid state physics: What about 4f-band superconductors?". In H. J. Queisser (ed.). Festkörper Probleme: Plenary Lectures of the Divisions Semiconductor Physics, Surface Physics, Low Temperature Physics, High Polymers, Thermodynamics and Statistical Mechanics, of the German Physical Society, Münster, March 19–24, 1973. Advances in Solid State Physics. Vol. 13. Berlin, Heidelberg: Springer. pp. 375–396. doi:10.1007/BFb0108579. ISBN 978-3-528-08019-8.
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^ a b c Jensen, William B. (2015). "The positions of lanthanum (actinium) and lutetium (lawrencium) in the periodic table: an update". Foundations of Chemistry. 17: 23–31. doi:10.1007/s10698-015-9216-1. S2CID 98624395. Archived from the original on 30 January 2021. Retrieved 28 January 2021.
^ a b c Scerri, Eric (18 January 2021). "Provisional Report on Discussions on Group 3 of the Periodic Table" (PDF). Chemistry International. 43 (1): 31–34. doi:10.1515/ci-2021-0115. S2CID 231694898. Archived (PDF) from the original on 13 April 2021. Retrieved 9 April 2021.
^ Thyssen, P.; Binnemans, K. (2011). "Accommodation of the Rare Earths in the Periodic Table: A Historical Analysis". In Gschneidner, K. A. Jr.; Bünzli, J-C.G; Vecharsky, Bünzli (eds.). Handbook on the Physics and Chemistry of Rare Earths. Vol. 41. Amsterdam: Elsevier. pp. 1–94. doi:10.1016/B978-0-444-53590-0.00001-7. ISBN 978-0-444-53590-0.
^ Barber, Robert C.; Karol, Paul J; Nakahara, Hiromichi; Vardaci, Emanuele; Vogt, Erich W. (2011). "Discovery of the elements with atomic numbers greater than or equal to 113 (IUPAC Technical Report)". Pure Appl. Chem. 83 (7): 1485. doi:10.1351/PAC-REP-10-05-01.
^ Karol, Paul J.; Barber, Robert C.; Sherrill, Bradley M.; Vardaci, Emanuele; Yamazaki, Toshimitsu (22 December 2015). "Discovery of the elements with atomic numbers Z = 113, 115 and 117 (IUPAC Technical Report)". Pure Appl. Chem. 88 (1–2): 139–153. doi:10.1515/pac-2015-0502.
^ Pyykkö, Pekka (2019). "An essay on periodic tables" (PDF). Pure and Applied Chemistry. 91 (12): 1959–1967. doi:10.1515/pac-2019-0801. S2CID 203944816. Retrieved 27 November 2022.
^ Fluck, E. (1988). "New Notations in the Periodic Table" (PDF). Pure Appl. Chem. 60 (3): 431–436. doi:10.1351/pac198860030431. S2CID 96704008. Archived (PDF) from the original on 25 March 2012. Retrieved 24 March 2012.
^ Scerri, Eric (2009). "Which Elements Belong in Group 3?". Journal of Chemical Education. 86 (10): 1188. Bibcode:2009JChEd..86.1188S. doi:10.1021/ed086p1188. Retrieved 1 January 2023.
^ Chemey, Alexander T.; Albrecht-Schmitt, Thomas E. (2019). "Evolution of the periodic table through the synthesis of new elements". Radiochimica Acta. 107 (9–11): 771–801. doi:10.1515/ract-2018-3082. S2CID 104470619.
^ Cotton, F. Albert; Wilkinson, G.; Murillo, C. A. (1999). Advanced Inorganic Chemistry (6th ed.). New York: Wiley, ISBN 978-0-471-19957-1.
^ Tossell, J.A. (1 November 1977). "Theoretical studies of valence orbital binding energies in solid zinc sulfide, zinc oxide, and zinc fluoride". Inorganic Chemistry. 16 (11): 2944–2949. doi:10.1021/ic50177a056.
^ Farberovich, O. V.; Kurganskii, S. I.; Domashevskaya, E. P. (1980). "Problems of the OPW Method. II. Calculation of the Band Structure of ZnS and CdS". Physica Status Solidi B. 97 (2): 631–640. Bibcode:1980PSSBR..97..631F. doi:10.1002/pssb.2220970230.
^ Singh, Prabhakar P. (1994). "Relativistic effects in mercury: Atom, clusters, and bulk". Physical Review B. 49 (7): 4954–4958. Bibcode:1994PhRvB..49.4954S. doi:10.1103/PhysRevB.49.4954. PMID 10011429.
^ Wang, Xuefang; Andrews, Lester; Riedel, Sebastian; Kaupp, Martin (2007). "Mercury Is a Transition Metal: The First Experimental Evidence for HgF4". Angew. Chem. Int. Ed. 46 (44): 8371–8375. doi:10.1002/anie.200703710. PMID 17899620.
^ Jensen, William B. (2008). "Is Mercury Now a Transition Element?". J. Chem. Educ. 85 (9): 1182–1183. Bibcode:2008JChEd..85.1182J. doi:10.1021/ed085p1182.
^ Fernández, Israel; Holzmann, Nicole; Frenking, Gernot (2020). "The Valence Orbitals of the Alkaline-Earth Atoms". Chemistry: A European Journal. 26 (62): 14194–14210. doi:10.1002/chem.202002986. PMC 7702052. PMID 32666598. S2CID 220529532.
^ Pyykkö, Pekka; Desclaux, Jean-Paul (1979). "Relativity and the Periodic System of Elements". Accounts of Chemical Research. 12 (8): 276–281. doi:10.1021/ar50140a002.
^ a b Miessler, G. L. and Tarr, D. A. (1999) Inorganic Chemistry, 2nd edn, Prentice-Hall, p. 38-39 ISBN 978-0-13-841891-5
^ Jørgensen, Christian (1973). "The Loose Connection between Electron Configuration and the Chemical Behavior of the Heavy Elements (Transuranics)". Angewandte Chemie International Edition. 12 (1): 12–19. doi:10.1002/anie.197300121.
^ Matsumoto, Paul S (2005). "Trends in Ionization Energy of Transition-Metal Elements". Journal of Chemical Education. 82 (11): 1660. Bibcode:2005JChEd..82.1660M. doi:10.1021/ed082p1660.
^ Hogan, C. Michael (2010). "Heavy metal" in Encyclopedia of Earth. National Council for Science and the Environment. E. Monosson and C. Cleveland (eds.) Washington DC.
^ Orgel, L.E. (1966). An Introduction to Transition-Metal Chemistry, Ligand field theory (2nd. ed.). London: Methuen.
^ Protchenko, Andrey V.; Dange, Deepak; Harmer, Jeffrey R.; Tang, Christina Y.; Schwarz, Andrew D.; Kelly, Michael J.; Phillips, Nicholas; Tirfoin, Remi; Birjkumar, Krishna Hassomal; Jones, Cameron; Kaltsoyannis, Nikolas; Mountford, Philip; Aldridge, Simon (16 February 2014). "Stable GaX2, InX2 and TlX2 radicals". Nature Chemistry. 6 (4): 315–319. Bibcode:2014NatCh...6..315P. doi:10.1038/nchem.1870. PMID 24651198.
^ Greenwood, Norman N.; Earnshaw, Alan (1997). Chemistry of the Elements (2nd ed.). Butterworth-Heinemann. doi:10.1016/C2009-0-30414-6. ISBN 978-0-08-037941-8. p. 240
^ Figgis, B.N.; Lewis, J. (1960). Lewis, J.; Wilkins, R.G. (eds.). The Magnetochemistry of Complex Compounds. Modern Coordination Chemistry. New York: Wiley Interscience. pp. 400–454.
^ Kovacs KA, Grof P, Burai L, Riedel M (2004). "Revising the Mechanism of the Permanganate/Oxalate Reaction". J. Phys. Chem. A. 108 (50): 11026–11031. Bibcode:2004JPCA..10811026K. doi:10.1021/jp047061u.
| v t e Periodic table |
| Periodic table forms | Alternatives Extended periodic table |
| Sets of elements | | | | | | | | | | --- --- --- --- | | By periodic table structure | | | | --- | | Groups | 1 (Hydrogen and alkali metals) 2 (Alkaline earth metals) 3 4 5 6 7 8 9 10 11 12 13 (Triels) 14 (Tetrels) 15 (Pnictogens) 16 (Chalcogens) 17 (Halogens) 18 (Noble gases) | | Periods | 1 2 3 4 5 6 7 8+ + Aufbau + Fricke + Pyykkö | | Blocks | Aufbau principle | | | By metallicity | | | | --- | | Metals | Lanthanides Actinides Transition metals Post-transition metals | | Metalloids | Lists of metalloids by source Dividing line | | Nonmetals | Noble gases | | | Other sets | Platinum-group metals (PGM) Rare-earth elements Refractory metals Precious metals Coinage metals Noble metals Heavy metals Native metals Transuranium elements Superheavy elements Major actinides Minor actinides | |
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| See also | IUPAC + nomenclature + systematic element name Trivial name Dmitri Mendeleev |
| Category WikiProject |
| v t e Periodic table |
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Solving Absolute Value Inequalities Interactive Activity - Print and Digital
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"This resource was extremely helpful in a station rotation where my students were reviewing different concepts for their end of trimester exam. It was basic enough that it worked for review, I'm not sure how it would've worked if the concept was brand new to them."
Meghan E.
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Description
This drag-and-drop activity is designed for Google Slides™ and Google Classroom™. The printable scaffolded worksheet version of this activity allows for flexible implementation.
Students will solve absolute value linear inequalities then match the inequalities to their solutions expressed in inequality notation and as number lines.
INCLUDES:
3 slides with 4 problems each
Slide 1: < and >
Slide 2: ≤ and ≥
Slide 3: Special Solutions
1 optional concept check slide with 2 free-response questions
A printable version in PDF form is also included.
The activity is in Google Slides format. (The printable version is in PDFformat.)
You need a free Google account to access this resource, add this activity to your Google Drive™, or assign this activity using Google Classroom.
Primary Standards Addressed
2A.6F from Texas Essential Knowledge and Skills (TEKS)
⭐️ Bundle & save for a discount ⭐️
Purchase Absolute Value Functions - Unit 2 Bundleto get this activity plus so much more...
5 activities for Google Slides™ (with printable versions)
4 paper-based activitiesin PDF format
1 classroom-based activityin PowerPoint (.pptx) format
7 lessonswith guided notes, practice, exit tickets, and warmups
a mid-unit reviewand editable mid-unit assessment
a unit reviewand editable unit test
3 concept organizers
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Visit the FAQs section, submit a help ticket, or email allison@mathbeach.combefore leaving feedback.
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Solving Absolute Value Inequalities Interactive Activity - Print and Digital
Rated 4.9 out of 5, based on 10 reviews
4.9(10 ratings)
Math Beach Solutions
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Absolute Value Functions - Unit 2 Bundle - Algebra 2 Curriculum
Get a complete, ready-to-print unit covering topics including absolute value function transformations, solving absolute value equations, and solving absolute value inequalities.⭐⭐⭐ Now with lesson videos! ⭐⭐⭐UNIT OVERVIEW:The study of absolute value function transformations (v-shaped graphs) provide
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This complete Algebra 2 curriculum includes standards-aligned planning guides for both TEKS and CCSS. (Texas teachers, I also have a specialized TEKS curriculum map crafted specifically for you!)What You Get: 2,000+ Pages of Ready-to-Print Resources You'll have 89 lessons with guided notes, warm-ups
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This activity bundle contains 72 interactive practice activities designed to use with Google Slides™ and Google Classroom™ or as printed card sorts, puzzles, and scaffolded worksheets. INCLUDES:72 activities in Google Slides format and Print (PDF)Download any Google Slides activity in PowerPoint (.p
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Rated 5 out of 5
March 6, 2025
This resource was extremely helpful in a station rotation where my students were reviewing different concepts for their end of trimester exam. It was basic enough that it worked for review, I'm not sure how it would've worked if the concept was brand new to them.
Meghan E.
21 reviews
Grades taught:9th
Rated 5 out of 5
September 11, 2024
Great activity! The concept check at the end allowed them to think deeper about it.
Adrienne D.
402 reviews
Grades taught:9th
Rated 4 out of 5
January 2, 2024
It was a good activity and helped my kids practice with graphing their solutions on the number line.
Jessica P.
28 reviews
Grades taught:9th, 10th
Rated 5 out of 5
September 18, 2023
This was a great activity. The last slide with the special solutions really had my students thinking about problem, not just following a process.
Kara H.
623 reviews
Grades taught:11th
Rated 5 out of 5
May 9, 2023
My students loved using these resources in class for engagement. Thanks
Tracy Barnes
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Grades taught:8th
Rated 5 out of 5
September 10, 2021
Thank you for sharing. My students loved it!
Amy Ballou
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Rated 5 out of 5
January 11, 2021
This was great virtual practice for my students after learning the topic. Super easy to put on Google Classroom!
Christine E.
237 reviews
Grades taught:10th
Rated 5 out of 5
January 4, 2021
This is a great activity for distance learning!
Elizabeth C.
206 reviews
Grades taught:10th
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Questions & Answers
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Question
|October 4, 2022 from
Bryan D.
Will I be able to edit the file? I would love to add interval notation either at the beginning of the slides or add an additional slide. If so how would I go about this?
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|October 4, 2022 from Math Beach Solutions
(TPT Seller)
Hi Bryan, thanks for reaching out. Additional slides can be added to the Slides file, but the text on the existing slides is not editable. New text boxes can be added; however, students will be able to interact with any new text boxes. Please feel free to reach out again in the Q&A or email allison@mathbeach.com if you have any other questions.
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### Math Beach Solutions
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Use the ⭐ Follow ⭐ button to connect with Math Beach on TPT. Thanks for visiting! 👀 What you'll find in the shop... ✔️ Algebra 1, Algebra 2, and Geometry Curriculum ✔️ NEW Print and Paperless Algebra 1 Modules ✔️ Skills Spiral Workbooks ✔️ Print + Digital Interactive Activities ✔️ Lessons Notes & Practice 👋 Welcome! I'm Allison, the founder of Math Beach Solutions. When I first started teaching, I didn't have a textbook - just the standards and ideas for a curriculum framework. I was swamped with creating notes, practice worksheets, and tests on top of the never-ending grading and paperwork teachers are all too familiar with. With time, I realized that implementing textbook-based resources is just as frustrating. They were supposedly standards-aligned, but contained many extraneous topics and lacked instruction of any kind for some topics! I was still spending my evenings creating new resources or desperately searching for a quick lesson for the next day. Over the years, I invested in studying standards, sequencing curriculum, and developing standards-aligned resources for use in my own classroom. I dreamed of developing a curriculum with time-saving standards-alignment guides to ease the burden of delivering standards-aligned instruction. The dream behind Math Beach Solutions was brought to fruition in 2018 when I began my journey as a full-time teacher-author. I am committed to creating engaging resources that enrich student learning while saving precious time when lesson planning.
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16016 | https://coolconversion.com/area/square%20kilometer-to-square%20mile | Square Kilometers to Square Miles (km²→mi²) Converter
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Home › Area › square kilometers to square miles
Convert Square Kilometers To Square Miles
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1 square kilometer = 0.386102 square miles Formula: multiply the value in square kilometers by the conversion factor '0.386102158542'. So, 1 square kilometer = 1 × 0.386102158542 = 0.386102158542 square miles. #### Conversion of 1 square kilometer to other area units 1 square kilometer = 247 acres 1 square kilometer = 10000 ares 1 square kilometer = 100 hectares 1 square kilometer = 1 × 10 10 square centimeters 1 square kilometer = 1.076 × 10 7 square feet 1 square kilometer = 1.55 × 10 9 square inches 1 square kilometer = 1000000 square meters 1 square kilometer = 1200000 square yards 1 square kilometer = 1.076 × 10 7 square feet By coolconversion.com
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Definition of Square Kilometer
Square Kilometer: A Measure of Vast Areas
The square kilometer (km²) is a unit of area measurement used to quantify large expanses of land or territory. It represents an area equal to a square with sides of 1 kilometer each. Square kilometers are commonly abbreviated as "km²" and are extensively employed in geography, urban planning, environmental studies, and land management.
Conversions to Other Units of Measurement:
Square Meter (m²): 1 square kilometer is equal to 1,000,000 square meters. This conversion is frequently used when dealing with smaller subdivisions of large areas. For example, a national park covering 2 square kilometers has an area of 2,000,000 square meters.
Hectares: 1 square kilometer is equal to 100 hectares. This conversion is often used in agriculture and land development projects. For instance, a farm spanning 5 square kilometers occupies 500 hectares of land.
Square Miles: 1 square kilometer ≈ 0.386102 square miles. This conversion is widely utilized in countries that primarily use the imperial system of measurement. For example, a desert covering 100 square kilometers extends over approximately 38.6 square miles.
Acres: 1 square kilometer ≈ 247.105381 acres. This conversion is relevant when dealing with land area measurements, especially in rural and agricultural contexts. For instance, a forest reserve covering 10 square kilometers encompasses approximately 2471.05 acres.
Square Yards (yd²): 1 square kilometer is equal to ≈ 1,195,990.046 square yards. This conversion is useful for understanding area measurements in smaller units, particularly in construction and real estate. For example, a city covering 100 square kilometers occupies ≈ 119,599,004.6 square yards.
The square kilometer plays a crucial role in facilitating accurate measurements and calculations, particularly when dealing with vast territories or large-scale projects.
Definition Area of a square with sides 1 kilometre (1,000 meters).Exact factor 1 km² = 1,000,000 m² (exact)Common equivalents
1 km² = 100 ha (exact)
1 km² ≈ 0.386102159 mi²
Sources
BIPM — SI Brochure
NIST SP 811 — Appendix B.9
Definition of Square Mile
Square Mile: A Unit of Vast Area
The square mile (mi²) is a unit of area measurement commonly used to quantify large territories, land areas, and regions. It represents an area equal to a square with sides of one mile each. Square miles are abbreviated as "mi²" and find significant applications in geography, urban planning, real estate, and land management.
Conversions to Other Units of Measurement:
Square Kilometer (km²): 1 square mile is approximately equal to 2.59 square kilometers. This conversion is often used in international contexts or when comparing areas of different countries. For example, a national park covering 100 square miles has an area of approximately 259 square kilometers.
Hectares: 1 square mile is equal to approximately 259 hectares. This conversion is commonly used in agriculture and land development projects. For instance, a farm spanning 10 square miles occupies approximately 2590 hectares of land.
Acres: 1 square mile is equivalent to 640 acres. This conversion is frequently employed in rural and agricultural contexts. For example, a ranch covering 5 square miles encompasses approximately 3200 acres of land.
Square Yards (yd²): 1 square mile ≈ 3,097,600.00 square yards. This conversion is useful for understanding area measurements in smaller units, particularly in construction and real estate. For example, a city covering 50 square miles ≈ 154,880,000.0 square yards.
Square Feet (ft²): 1 square mile is equal to approximately 27,878,400 square feet. This conversion is relevant for detailed measurements of area, especially in urban planning and construction projects. For example, a planned residential development covering 2 square miles encompasses approximately 55,756,800 square feet of land.
The square mile plays a crucial role in facilitating accurate measurements and calculations, particularly when dealing with large-scale geographic areas or urban developments.
Definition Area of a square with sides 1 statute mile (1,609.344 meters).Exact factor 1 mi² = 2,589,988.110336 m² (exact)Common equivalents
1 mi² = 640 acres (exact)
1 mi² = 3,097,600 yd² (exact)
Sources
NIST SP 811 — Appendix B.9
Conversion Formula
To calculate a square kilometer value to the corresponding value in square mile, just multiply the quantity in square kilometer by 0.38610215854245 (the conversion factor). Here is the formula:
Value in square miles = value in square kilometer × 0.38610215854245
Suppose you want to convert 1 square kilometer into square miles. Using the conversion formula above, you will get:
Value in square mile = 1 × 0.38610215854245 = 0.386102 square miles
Step-by-Step Guide: Convert Square Kilometers to Square Miles
Learn how to convert square kilometers into square miles using a simple multiplication with the conversion factor. You can also explore our unit conversion calculators for other measurement types.
How to Convert Square Kilometers (km²) to Square Miles (mi²)
Conversion Factor: 1 square kilometer = 0.38610215854245 square miles
Example Calculation: 1 square kilometer × 0.38610215854245 = 0.386102 square miles
How to Convert Square Miles back to Square Kilometers
Reverse Conversion Factor: 1 square mile = 2.58999 square kilometers
Example Calculation: 0.386102 square mile × 2.58999 = 1 square kilometers
Frequently Asked Questions About 1 Square Kilometer Conversion
How Much Is 1 Square Kilometer In square miles?
1 square kilometer ≈ 0.386102 square miles.
What Is The Formula to Convert Square Kilometer To square miles?
Multiply the area value by 0.38610215854245. Formula: square kilometer × 0.38610215854245 = square miles.
How Do I Convert From Square Kilometer To Square Miles?
To convert square kilometer to square miles, multiply the square kilometer value by 0.38610215854245. For example: 1 square kilometer × 0.38610215854245 = 0.386102 square miles.
What Is The Conversion Factor Between Square Kilometer And square miles?
The conversion factor is 0.38610215854245. This means 1 square kilometer equals 0.38610215854245 square miles. For other area conversions, visit our Area converter homepage.
Is There A Quick Way To Mentally Estimate Square Kilometer To square miles?
Yes! Since the conversion factor is 0.386102, you can multiply by approximately 0.4 for quick mental calculations.
Square kilometers to square miles conversion table
The table below contains pairs of values from square kilometers to square miles with high precision calculations.
Square kilometers to square miles
0 to 10
| Square kilometers | = | square miles |
---
| 0.005 km² | = | 0.00193051 mi² |
| 0.01 km² | = | 0.00386102 mi² |
| 0.02 km² | = | 0.00772204 mi² |
| 0.03 km² | = | 0.0115831 mi² |
| 0.04 km² | = | 0.0154441 mi² |
| 0.05 km² | = | 0.0193051 mi² |
| 0.06 km² | = | 0.0231661 mi² |
| 0.07 km² | = | 0.0270272 mi² |
| 0.08 km² | = | 0.0308882 mi² |
| 0.09 km² | = | 0.0347492 mi² |
| 0.1 km² | = | 0.0386102 mi² |
| 1 5 km² | = | 0.0772204 mi² |
| 0.3 km² | = | 0.115831 mi² |
| 0.4 km² | = | 0.154441 mi² |
| 1 2 km² | = | 0.193051 mi² |
| 0.6 km² | = | 0.231661 mi² |
| 0.7 km² | = | 0.270272 mi² |
| 0.8 km² | = | 0.308882 mi² |
| 0.9 km² | = | 0.347492 mi² |
| 1 km² | = | 0.386102 mi² |
| 2 km² | = | 0.772204 mi² |
| 3 km² | = | 1.15831 mi² |
| 4 km² | = | 1.54441 mi² |
| 5 km² | = | 1.93051 mi² |
| 6 km² | = | 2.31661 mi² |
| 7 km² | = | 2.70272 mi² |
| 8 km² | = | 3.08882 mi² |
| 9 km² | = | 3.47492 mi² |
| 10 km² | = | 3.86102 mi² |
⬇ CSV
Square kilometers to square miles
10 to 100000
| Square kilometers | = | square miles |
---
| 10 km² | = | 3.86102 mi² |
| 20 km² | = | 7.72204 mi² |
| 30 km² | = | 11.5831 mi² |
| 40 km² | = | 15.4441 mi² |
| 50 km² | = | 19.3051 mi² |
| 60 km² | = | 23.1661 mi² |
| 70 km² | = | 27.0272 mi² |
| 80 km² | = | 30.8882 mi² |
| 90 km² | = | 34.7492 mi² |
| 100 km² | = | 38.6102 mi² |
| 200 km² | = | 77.2204 mi² |
| 300 km² | = | 115.831 mi² |
| 400 km² | = | 154.441 mi² |
| 500 km² | = | 193.051 mi² |
| 600 km² | = | 231.661 mi² |
| 700 km² | = | 270.272 mi² |
| 800 km² | = | 308.882 mi² |
| 900 km² | = | 347.492 mi² |
| 1000 km² | = | 386.102 mi² |
| 2000 km² | = | 772.204 mi² |
| 3000 km² | = | 1158.31 mi² |
| 4000 km² | = | 1544.41 mi² |
| 5000 km² | = | 1930.51 mi² |
| 6000 km² | = | 2316.61 mi² |
| 7000 km² | = | 2702.72 mi² |
| 8000 km² | = | 3088.82 mi² |
| 9000 km² | = | 3474.92 mi² |
| 10000 km² | = | 3861.02 mi² |
| 100000 km² | = | 38610.2 mi² |
⬇ CSV
Square miles to square kilometers
0.0005 to 1
| Square miles | = | square kilometers |
---
| 0.005 mi² | = | 0.0129499 km² |
| 0.01 mi² | = | 0.0258999 km² |
| 0.02 mi² | = | 0.0517998 km² |
| 0.03 mi² | = | 0.0776996 km² |
| 0.04 mi² | = | 0.1036 km² |
| 0.05 mi² | = | 0.129499 km² |
| 0.06 mi² | = | 0.155399 km² |
| 0.07 mi² | = | 0.181299 km² |
| 0.08 mi² | = | 0.207199 km² |
| 0.09 mi² | = | 0.233099 km² |
| 0.1 mi² | = | 0.258999 km² |
| 1 5 mi² | = | 0.517998 km² |
| 0.3 mi² | = | 0.776996 km² |
| 0.4 mi² | = | 1.036 km² |
| 1 2 mi² | = | 1.29499 km² |
| 0.6 mi² | = | 1.55399 km² |
| 0.7 mi² | = | 1.81299 km² |
| 0.8 mi² | = | 2.07199 km² |
| 0.9 mi² | = | 2.33099 km² |
| 1 mi² | = | 2.58999 km² |
| 2 mi² | = | 5.17998 km² |
| 3 mi² | = | 7.76996 km² |
| 4 mi² | = | 10.36 km² |
| 5 mi² | = | 12.9499 km² |
| 6 mi² | = | 15.5399 km² |
| 7 mi² | = | 18.1299 km² |
| 8 mi² | = | 20.7199 km² |
| 9 mi² | = | 23.3099 km² |
| 10 mi² | = | 25.8999 km² |
⬇ CSV
Square miles to square kilometers
1 to 10000
| Square miles | = | square kilometers |
---
| 10 mi² | = | 25.8999 km² |
| 20 mi² | = | 51.7998 km² |
| 30 mi² | = | 77.6996 km² |
| 40 mi² | = | 103.6 km² |
| 50 mi² | = | 129.499 km² |
| 60 mi² | = | 155.399 km² |
| 70 mi² | = | 181.299 km² |
| 80 mi² | = | 207.199 km² |
| 90 mi² | = | 233.099 km² |
| 100 mi² | = | 258.999 km² |
| 200 mi² | = | 517.998 km² |
| 300 mi² | = | 776.996 km² |
| 400 mi² | = | 1036 km² |
| 500 mi² | = | 1294.99 km² |
| 600 mi² | = | 1553.99 km² |
| 700 mi² | = | 1812.99 km² |
| 800 mi² | = | 2071.99 km² |
| 900 mi² | = | 2330.99 km² |
| 1000 mi² | = | 2589.99 km² |
| 2000 mi² | = | 5179.98 km² |
| 3000 mi² | = | 7769.96 km² |
| 4000 mi² | = | 10360 km² |
| 5000 mi² | = | 12949.9 km² |
| 6000 mi² | = | 15539.9 km² |
| 7000 mi² | = | 18129.9 km² |
| 8000 mi² | = | 20719.9 km² |
| 9000 mi² | = | 23309.9 km² |
| 10000 mi² | = | 25899.9 km² |
| 100000 mi² | = | 258999 km² |
⬇ CSV
Square Kilometers to Square Miles (km²→mi²) Converter
97000 acre to square foot
46000 square meter to are
3500 are to square inch
950 are to square foot
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16018 | https://math.stackexchange.com/questions/414925/gauss-lemma-for-polynomials-and-divisibility-in-mathbb-z-and-mathbb-q | abstract algebra - Gauss Lemma for Polynomials and Divisibility in $\mathbb Z$ and $\mathbb Q$. - Mathematics Stack Exchange
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Gauss Lemma for Polynomials and Divisibility in Z Z and Q Q.
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I am working through Gauss Lemma and various corollaries of it. In the book Algebra of Michael Artin, I have a question to the proof of the following Theorem:
Theorem.
(a) Let f,g f,g be polynomials in Q[x]Q[x], and let f 0,g 0 f 0,g 0 be the associated primitive polynomials in Z[x]Z[x]. If f f divides g g in Q[x]Q[x], then f 0 f 0 divides g 0 g 0 in Z[x]Z[x].
(b) Let f f be a primitive polynomial in Z[x]Z[x], and let g g be any polynomial with integer coefficients. Suppose that f f divides g g in Q[x]Q[x], say g=f q g=f q, with q∈Q[x]q∈Q[x]. Then q∈Z[x]q∈Z[x], and hence f f divides g g in Z[x]Z[x].
(c) Let f,g f,g be polynomials in Z[x]Z[x]. If they have a common nonconstant factor in Q[x]Q[x], then they have a common nonconstant factor in Z[x]Z[x] too.
Proof: To prove (a), we may clear denominators so that f f and g g become primitive. Then (a) is a consequence of (b). To prove (b), we apply (3.1) in order to write the quotient in the form q=c q 0 q=c q 0, where q 0 q 0 is primitive and c∈Q c∈Q. By Gauss's Lemma, f q 0 f q 0 is primitive, and the equation g=c f q 0 g=c f q 0 shows that it is the primitive polynomial g 0 g 0 associated to g g. Therefore g=c g 0 g=c g 0 is the expression for g g referred to in Lemma (3.1), and c c is the content of g g. Since g∈Z[x]g∈Z[x], it follows that c∈Z c∈Z, hence that q∈Z[x]q∈Z[x]. Finally, to prove (c), suppose that f,g f,g have a common factor h h in Q[x]Q[x]. We may assume that h h is primitive, and then by (b) h h divides both f f and g g in Z[x]Z[x].
The following theorem are proofed in the chapter before.
(3.1) Lemma. Every nonzero polynomial f(x)∈Q[x]f(x)∈Q[x] can be written as a product
f(x)=c f 0(x),f(x)=c f 0(x),
where c c is a rational number and f 0(x)f 0(x) is a primitive polynomial in Z[x]Z[x].
(3.3) Theorem. Gauss's Lemma: A product of primitive polynomials in Z[x]Z[x] is primitive.
I have a question to the proof of part (c) of the Theorem. Why we can assume that the common factor h h is primitive? The notion of primitive polynomial is just definied for polynomials in Z[x]Z[x], and h h is in Q[x]Q[x]? So to me this steps makes no sense...
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edited Jun 8, 2013 at 19:43
Alex Wertheim
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asked Jun 8, 2013 at 19:32
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If we multiply h∈Q[x]h∈Q[x] by an integer constant with just enough prime factors to clear denominators, then h h becomes a primitive polynomial in Z[x]Z[x]. This new h h is still a common factor of f f and g g in Q[x]Q[x] (because integer constants are units in Q[x]Q[x]). So we can assume we started with such a primitive h h as a common factor of f f and g g.
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answered Jun 8, 2013 at 19:47
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4Divisibility in Z[X]Z[X]
1Number of ring endomorphisms of Z[ζ n]Z[ζ n]
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10Is Gauss's lemma valid for polynomials with coefficients in a GCD domain?
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1Using Gauss Lemma to reason about closure properties of Z[X]Z[X].
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1Primitive g∣a f⇒g∣f,g∣a f⇒g∣f, for 0≠a∈0≠a∈ UFD A, f,g∈A[x],f,g∈A[x], by Gauss's Lemma
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Int J Mol Sci
. 2021 Apr 28;22(9):4642. doi: 10.3390/ijms22094642
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The Chemistry of Reactive Oxygen Species (ROS) Revisited: Outlining Their Role in Biological Macromolecules (DNA, Lipids and Proteins) and Induced Pathologies
Celia Andrés Juan
Celia Andrés Juan
1 Cinquima Institute and Department of Organic Chemistry, Faculty of Sciences, Valladolid University, Paseo de Belén, 7, 47011 Valladolid, Spain; celia.andres.juan@uva.es
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1, José Manuel Pérez de la Lastra
José Manuel Pérez de la Lastra
2 Institute of Natural Products and Agrobiology, CSIC-Spanish Research Council, Avda. Astrofísico Fco. Sánchez, 38206 La Laguna, Spain
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2,, Francisco J Plou
Francisco J Plou
3 Institute of Catalysis and Petrochemistry, CSIC-Spanish Research Council, 28049 Madrid, Spain; fplou@icp.csic.es
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3, Eduardo Pérez-Lebeña
Eduardo Pérez-Lebeña
4 Sistemas de Biotecnología y Recursos Naturales, 47625 Valladolid, Spain; info@glize.eu
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4
Editor: Steffen Reinbothe
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1 Cinquima Institute and Department of Organic Chemistry, Faculty of Sciences, Valladolid University, Paseo de Belén, 7, 47011 Valladolid, Spain; celia.andres.juan@uva.es
2 Institute of Natural Products and Agrobiology, CSIC-Spanish Research Council, Avda. Astrofísico Fco. Sánchez, 38206 La Laguna, Spain
3 Institute of Catalysis and Petrochemistry, CSIC-Spanish Research Council, 28049 Madrid, Spain; fplou@icp.csic.es
4 Sistemas de Biotecnología y Recursos Naturales, 47625 Valladolid, Spain; info@glize.eu
Correspondence: jm.perezdelalastra@csic.es
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Steffen Reinbothe: Academic Editor
Received 2021 Apr 10; Accepted 2021 Apr 26; Collection date 2021 May.
© 2021 by the authors.
Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (
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PMCID: PMC8125527 PMID: 33924958
Abstract
Living species are continuously subjected to all extrinsic forms of reactive oxidants and others that are produced endogenously. There is extensive literature on the generation and effects of reactive oxygen species (ROS) in biological processes, both in terms of alteration and their role in cellular signaling and regulatory pathways. Cells produce ROS as a controlled physiological process, but increasing ROS becomes pathological and leads to oxidative stress and disease. The induction of oxidative stress is an imbalance between the production of radical species and the antioxidant defense systems, which can cause damage to cellular biomolecules, including lipids, proteins and DNA. Cellular and biochemical experiments have been complemented in various ways to explain the biological chemistry of ROS oxidants. However, it is often unclear how this translates into chemical reactions involving redox changes. This review addresses this question and includes a robust mechanistic explanation of the chemical reactions of ROS and oxidative stress.
Keywords: ROS, oxidative stress, macromolecules
1. Introduction
In chemistry, a free radical (FR) is a relatively stable species that contains one or more unpaired electrons and can react with other molecules, either by donating its unpaired electron to another molecule or by taking it away from another molecule to increase stability. In this way it converts the molecule with which it reacts into another FR, so a common feature of FR reactions is the chain process: one radical gives rise to another radical. This process only ceases when two FRs react with each other.
In a biological context, reactive oxygen species (ROS) are formed as a natural by-product of cellular aerobic metabolism. Mitochondrial respiration is a significant cause of reactive oxygen species (ROS) . In addition to mitochondria, ROS are produced by a variety of enzymes such as NADPH oxidases (NOXs), xanthine oxidase, nitric oxide synthase, and peroxisomal constituents . They are also produced by ionizing and UV radiation, as well as by the metabolism of a wide range of drugs and xenobiotics. In the endoplasmic reticulum, oxidants are released during the folding of proteins and the formation of disulphide bonds. They are highly reactive chemical molecules derived from the ability of the O 2 molecule to accept electrons , generating subsequent unstable molecules such as superoxide anion (•O 2−), hydrogen peroxide (H 2 O 2), hydroxyl radical (OH−), and singlet oxygen (1 O 2−), produced by all kinds of cells. Stationary (physiological) levels of ROS are intrinsic to the normal functioning of cells, fulfilling the functions of cell signalling and homeostasis . On the other hand, when they are produced in excess or when cellular defences are not able to metabolise them, oxidative stress (OS) damage occurs .
2. Genesis and Effects of ROS
All aerobic organisms need oxygen O 2 for efficient energy production. Molecular O 2 contains two unpaired electrons, but it is weakly reactive because they are located in different molecular orbitals and have a parallel spin. Consequently, oxygen preferentially accepts electrons one at a time. The ultimate goal of the electron transport chain ETC in the inner mitochondrial membrane (complexes I to V) is the reduction of the oxygen molecule to produce water. The first step of the O 2 reduction reaction occurs spontaneously (Figure 1A).
Figure 1.
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Gibbs free energy of O 2 reduction yielding water (A) and the anion superoxide (B). Gibbs free energy calculated with Gaussian 09 software, revision D.01, method B3LYP/6-31G(d). Authors cited Note 1, after References.
Univalent reduction of O 2 gives rise to the anion superoxide (Figure 1B), which results from the addition of an electron filling one of the two uncompleted molecular orbitals, leaving a charged ionic species with a single unpaired electron and a net negative charge of −1 (Figure 2).
Figure 2.
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Lewis structure of the anion superoxide.
Superoxide •O 2− is detrimental and is mainly produced as a by-product of mitochondrial respiration (especially in Complexes I and III, in the electron transport chain ETC) , where a small percentage of the electrons in the ETC chain escape from it, as well as by several other enzymes, which catalyse the electron transfer directly to molecular oxygen under strongly reducing conditions, as occurs in the mitochondrial matrix. It is also generated in the immune system to eliminate invading micro-organisms. In phagocytes, the enzyme NADPH oxidase produces •O 2− in large quantities for use in the oxygen-dependent destruction mechanisms of invading pathogens .
The hydroxyl radical •OH (Figure 3A–C), the most powerful ROS oxidant, is formed during the Haber–Weiss reaction , by the Fenton reaction or by decomposition of peroxynitrite , and has a very short half-life (10−9 s) and high reactivity.
Figure 3.
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Reactive species ROS and RNS formed in the mitochondrial matrix by the Haber–Weiss reaction (A), the Fenton reaction (B) or by decomposition of peroxynitrite (C).
O 2 and ROS reduction are reactions in which the final product is water and have a negative Gibbs free energy, ∆G o ≤ 0, so they occur spontaneously . A small percentage of the electrons in the ETC transport chain, 0.1 to 2% of cases, pass through the chain and reach the mitochondrial matrix, where they prematurely reduce molecular oxygen O 2 to superoxide ion •O 2−, the precursor of most reactive oxygen species.
3. Dual Role of ROS: Physiological and Pathological
ROS can be induced by exogenous sources such as tobacco, pollution, smoke, drugs, xenobiotics or ionising radiation, leading to irreversible effects on tissue development in animals and plants. In these, abiotic factors such as lack of water or high temperature can influence their emission. ROS play an important role in physio- and pathological processes. Mitochondria are particularly susceptible to oxidative damage, as electrons escaping from the ETC electron transport chain in the inner membrane react with oxygen to produce a superoxide anion. This anion is unstable and cannot cross membranes, but it is rapidly converted to hydrogen peroxide, which is permeable to the membrane. It can then undergo the Fenton reaction to produce the hydroxyl radical, which is highly reactive in the mitochondrial matrix. Elevated levels of ROS lead to increased mitochondrial DNA (mtDNA) damage.
In eukaryotic cells, ROS are mainly produced by biochemical reactions in mitochondrial cellular respiration processes (complex I and III, located in inner membrane). Transmembrane NADPH oxidases (NOXs) and the mitochondrial electron transport chain (ETC) are the major endogenous enzymatic sources of •O 2− and H 2 O 2. However, they also come from sources such as peroxisomes, xanthine oxidase (XO), lipo- and cyclo-oxygenase and cytochrome P450 in endoplasmic reticulum . In mitochondria, physiological levels of •O 2− and H 2 O 2 participate in redox signalling, but their production is significantly enhanced during oxidative stress conditions, producing an imbalance between ROS production and the action of the antioxidant defence system. They include SOD enzymes that reduce O 2 into H 2 O 2 (Zn-Cu SOD or SOD1 in cytosol and intermembrane space in mitochondria and Mn-SOD or SOD2 in matrix mitochondria), catalase, glutathione peroxidases and thioredoxin reductase that convert levels of H 2 O 2 into H 2 O and O 2.
During episodes of environmental stress, oxidative damage (OS) in cells can increase dramatically, causing damage to their structures . Generally, the harmful effects of ROS in the cell include (i) damage on DNA or RNA ; (ii) lipid peroxidation of polyunsaturated fatty acids (such as membrane phospholipids) ; and (iii) oxidation of proteins , Figure 4. They cause irreversible damage to DNA, lipids and enzymes present in the cell cytosol, as they oxidise and modify cellular components and prevent them from carrying out their original functions.
Figure 4.
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ROS action on DNA, lipids and proteins lead to DNA base oxidation, lipid peroxidation and protein carbonylation, respectively. Unpaired electron.
4. ROS Damage on DNA
Nuclear and mtDNA have a separate evolutionary origin, as it is derived from the genomes of bacteria that were engulfed by the early ancestors of today’s eukaryotic cells. In extant organisms, most proteins present in mitochondria (in mammals, some 1500 different types) are encoded by nuclear DNA. mtDNA is particularly susceptible to reactive oxygen species generated by the respiratory chain , due to its proximity, despite being packaged with proteins as protective as those of nuclear chromatin. Its mutations can lead to a variety of diseases, as well as to the ageing process and pathologies associated with advanced age. Evidence suggests a link between ageing and mitochondrial genome dysfunction .
DNA damage refers to physico-chemical changes in DNA , which can affect the interpretation and transmission of genetic information. It is damaged by a series of exogenous and endogenous stresses, which cause different forms of molecular modification (Figure 5). The DNA damage caused by ROS, from endogenous or exogenous sources, has been a significant breakthrough in carcinogenesis research over the last 20 years . In DNA, ROS reacts with nitrogenous bases and deoxyribose, causing significant oxidative reactions. This can lead to mutations, carcinogenesis, apoptosis, necrosis and hereditary diseases. DNA fragmentation occurs, forced by the rupture of nucleosomes (fundamental structures for the organisation of DNA within chromosomes), thus causing problems in the compaction and coiling of DNA within chromatin. Chromatin plays an important role in the regulation of gene transcription , and thus alterations in its functional properties may result in errors leading to mutagenesis.
Figure 5.
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DNA damage caused by ROS.
Oxidation of DNA can lead to alterations in DNA bases or double helix breaks, among other mutagenic alterations . Hydroxyl radical stress causes direct damage to DNA, mainly by strand excision, and causes oxidative damage to the pyrimidine and purine bases. This process starts by radical-induced abstraction of a proton from any position of the deoxyribose and can result in many products.
In deoxyribose, •OH radical easily leads to hydrogen abstraction (also named hydrogen atom transfer HAT), forming different products and ribose fragments (Figure 6).
Figure 6.
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Abstraction of hydrogen at different carbons of deoxyribose by the •OH radical. 5′H > 4′H > 3’H = 2´H = 1´H.
Several steps of the oxidation of deoxyribose at the C-4 position leads to a radical carbon stabilised by resonance with the oxygen in the ring. There is little experimental information available on which of the hydrogen atoms of DNA deoxyribose reacts with the hydroxyl radical. Experiments by Balasubramanian et al. provided a clear structural basis for rationalising the abstraction site preference. Indeed, the authors have clearly demonstrated the existence of a good correlation between the reactivity of the different sites and their solvent. This order of reactivity parallels the solvent exposure of the deoxyribose hydrogens .
The addition of an O 2 gives the peroxyl radical of the sugar, which is transformed into hydroperoxide and undergoes a transposition reaction with expansion of the ring, which subsequently degrades to different products such as enamine propenal derivatives (Figure 7A–G).
Figure 7.
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Mechanisms of oxidative damage to DNA-deoxyribose. The reaction with oxygen leads to several transposition reactions with expansion of the ring, which subsequently degrades to different products (A–G). (A) The reaction is initiated by abstraction of the hydrogen on the C-4 of deoxyribose by the hydroxyl radical or any other radical present in the medium to form the radical on the carbon; (B) The carbon radical reacts with O 2 present in the reaction medium and transforms into the peroxyl radical which evolves into the hydroperoxide derivative; (C) The alkyl hydroperoxide formed undergoes a rearrangement, i.e., a migration from one atom or group of atoms to another within the same molecule, in this case with ring expansion to a six-linked ring and formation of the carbocation; (D) The generated carbocation is stabilised by delocalisation of the positive charge with the two adjacent oxygens and formation of the oxonium cation; (E) Dehydration with ring opening to form the enamine derivative which evolves to the unsaturated imine by loss of the phosphate residue; (F) Addition of water on the carbon and formation of the hydroxy acetal derivative that fragments to generate the acrylaldehyde-derived base; (G) In low oxygen environments the radical evolves to the oxonium cation and nucleophilic attack by a water molecule, then decomposes into the free base and the various fragments .
The mechanisms of oxidative damage to DNA bases involves abstraction and addition reactions by free radicals, with the formation of carbon-centred radicals. The hydroxyl radical causes direct DNA damage, by oxidation of pyrimidine and purine bases.
In thymine the abstraction of a methyl hydrogen from the 5-position by the hydroxyl radical generates a resonance-stabilised carbon radical (Figure 8), which provides the hydroxymethylene derivative, after treatment with oxygen and followed by reduction.
Figure 8.
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Abstraction of a hydrogen from the methyl group at position 5 by OH– .
The main site for the reaction of the hydroxyl radical with pyrimidines is the double bond at the C5–C6 position (Figure 9). Thymine after addition of the hydroxyl radical and reaction with oxygen is transformed into the hydroxy hydroperoxide derivative, which can be reduced to the diol or can undergo an opening process, as shown in the figure and subsequent cyclisation to the hydroxyhydantoin derivative.
Figure 9.
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Reaction of the hydroxyl radical with pyrimidines is the double bond at the C5–C6 position.
Among other reactions, hydroxyl radicals can be added to the C-8 position of guanine and generate a radical on the nitrogen at position 7 (Figure 10), which can be reduced by the addition of an electron and a proton to an unstable intermediate, ultimately giving the ring-opening fragmentation product. Alternatively, it can undergo oxidation to give the derivative 8-hydroxyguanine. Hydroxylation of the C-8 position of the guanine derivative of DNA is the most well-studied DNA lesion induced by hydroxyl radical attack.
Figure 10.
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Hydroxylation of the C-8 position of the guanine derivative of DNA, generating degradation products .
Mutations or deletions in the mitochondrial genome accumulate as cells mature, and are assumed to be triggered by long-term oxidative stress . It has been shown that these mechanisms induce mtDNA disruption in neuronal degenerative disorders such as Alzheimer’s disease, Parkinson’s disease, and amyotrophic lateral sclerosis (ALS). Furthermore, these neuronal diseases facilitate the progression of additional DNA degradation during ageing, resulting in disease symptoms earlier than would be anticipated. This deterioration mechanism may result in more mtDNA mutation or depletion.
In addition to oxidizing DNA bases, ROS may cause DNA strand breakage due to free radical attacks on the DNA sugar-phosphate backbone . DNA strand breaks may be identified as single-strand or double-strand breaks, and they can be mutagenic or clastogenic. A network of events known as the DNA damage response (DDR) is triggered in response to DNA damage . DNA harm detection, checkpoint activation, cell cycle arrest, and finally repair, apoptosis, and immune clearance are all part of this reaction. Relevant pathways are triggered depending on the type of the DNA lesion to enable in the recognition of damaged regions and their repair.
Mitochondrial-based diseases are a group of disorders caused by their dysfunction and have unique characteristics because their function is critical to physiological cellular metabolism. In essence, mutations in mtDNA alter a careful balance between the production of reactive oxygen species (ROS) and their neutralisation by enzymes such as superoxide dismutase, catalase, glutathione peroxidase and others. Some mutations that increase ROS production ultimately induce oxidative damage to the mitochondrial matrix, inhibiting proteins such as Sirtuin3 (antioxidant and anti-tumour protein) and reducing other redundant antioxidant defences, such as SOD2 .
Aging mitochondria are the critical factor in the origin of neurodegenerative diseases. In the brains of individuals with Alzheimer’s disease, there is high oxidative damage to nDNA and mtDNA, approximately 10-fold greater than in people without the disease. In Huntington’s disease, the mutant huntingtin protein causes mitochondrial dysfunction, with higher levels of ROS and increased oxidative stress .
During ageing, changes in mtDNA are well documented in the literature. Such changes include an accumulation of oxidised base pairs, base mismatches, strand breaks and deletions. Specific mutations and haplogroups of mtDNA emerge as predictors of both lifespan and risk of various age-associated diseases: cancer, diabetes, heart failure, sarcopenia and Alzheimer’s and Parkinson’s disease . In neurodegenerative diseases, including amyotrophic lateral sclerosis (ALS), lipid peroxidation products and abnormal protein aggregation (the amyloid beta peptide) are biomarkers of oxidative stress .
In mitochondria, H+ delivered to the intermembrane space is returned to the matrix during the oxidative phosphorylation reaction, so there is a reducing environment within the latter, facilitating the conversion of ADP into ATP (Figure 11).
Figure 11.
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The ADP to ATP reaction in the mitochondria has a negative Gibbs free energy, ∆G ≤ 0, due to the reducing environment, and occurs spontaneously.
Electrons from the ETC transport chain, in a small percentage of cases (0.1 to 2%), pass through the chain and reduce O2 in the mitochondrial matrix prematurely and incompletely to superoxide anion •O 2−, precursor of the reactive oxygen species .
Mitochondrial genome mutations accumulate in somatic tissues during normal ageing, as well as increased ROS production and biomarkers of ROS damage . Demonstrating causality between the two has been difficult, as have direct links to ageing and longevity. The landscape of mitochondrial DNA mutations is open to spatiotemporal fluctuations throughout life, and it is likely that all individuals harbour mitochondrial DNA mutations, albeit at undetectable levels. Therefore, mitochondrial ROS production is highly dynamic and variable between cells .
5. Lipid Peroxidation Caused by ROS
Cell membranes are sensitive to radical damage due to the presence of polyunsaturated fatty acids. Another main effect of ROS is lipid peroxidation, which occurs when membrane phospholipids are brought into contact with an ROS oxidising agent. In this reaction, the free radical oxidises an unsaturated lipid chain, leading to the formation of a hydroperoxidised lipid and an alkyl radical. This lipoperoxidation results in alterations of the membrane structure, affecting its fluidity and damaging its integrity . This process is initiated by the attack of a hydroxyl radical at one of the above-mentioned bis-allelic positions in the fatty acid side chains, leading to the generation of an alkyl radical.
The association of oxygen-derived free radicals with polyunsaturated fatty acids results in a number of extremely reactive electrophilic aldehydes during the process of lipid peroxidation. This effect happens as a result of continuing free radical chain reactions before they are terminated. Studies establishing a correlation between this type of oxidative disruption, neurodegeneration, and disease offered a wealth of information .
Peroxidation, which leads to atherosclerosis, is believed to occur inside blood vessel walls rather than (or to a lesser extent) in LDL circulating in the blood. Just as LDL can enter vessel walls, minimally modified LDL (LDL with some oxidation but not enough to be recognized by scavenger receptors) can escape back into the circulation. As a result, the sensitivity of circulating LDL to peroxidation can be a potentially valuable biomarker, suggestive of peroxidation in blood vessels .
The initiation phase of lipid peroxidation (Figure 12). If a free radical attacks a carbon of the aliphatic chain of a fatty acid, hydrogen abstraction of the methylene group (-CH2-) attached to a carbon flanked by double bonds of a polyunsaturated fatty acid occurs, with the formation of a radical species. The radicals formed are stabilised by resonance with the double bond. In the propagation phase, a chain reaction occurs with the extension of the damage and the formation of further radical species. The radical formed in the first phase reacts with oxygen and forms a peroxyl radical (LOO), which can react with other adjacent polyunsaturated fatty acids to form a hydroperoxide and an alkyl radical, thus causing a chain reaction and damage to an increasing number of fatty acids.
Figure 12.
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Mechanism of lipid peroxidation. The radical on the carbon reacts with an oxygen molecule to generate a peroxyl radical (R-O-O), which can abstract a new hydrogen atom from a double allylic C-H bond in the adjacent fatty acid side chain.
Lipid peroxidation can give rise to toxic breakdown products such as hydroxynonenal. Once formed, the peroxy radical cycles to a four-linked cyclic peroxide. The process increases with the number of unsaturations of the polyunsaturated fatty acid. Thus, when free radicals attack arachidonic acid (with four unsaturations) the initial abstraction of a hydrogen atom can occur at three points in the carbon chain of the fatty acid, because it has three methylene groups (-CH2-) attached to a carbon atom flanked by double bonds, thus increasing the complexity of the peroxidation reaction. The fatty acid undergoes a further reaction with oxygen followed by a break in the cycle to give hydroperoxynonenal, which is reduced to hydroxynonenal (Figure 13). Patients of Alzheimer’s disease have higher amounts of lipid-peroxidation compounds like 4-hydroxy-2-nonenal or acrolein in their brains , and elevated lipid peroxidation can also be seen in their cerebrospinal fluid and plasma. Increased oxidative stress and cell harm in this disease can be triggered by the association of transition metals, amyloid-peptide, and lipid peroxidation, according to recent studies.
Figure 13.
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Toxic breakdown products from lipid peroxidation.
Peroxidation of arachidonic acid leads to cyclic peroxides such as isoprostanoids in addition to hydroperoxides . Alternatively, they can cycle together with the addition of a second oxygen molecule. These intermediates generate malondialdehyde (MDA) via a retro-Diels–Alder reaction. This MDA does react with DNA bases and causes mutagenic lesions (Figure 14).
Figure 14.
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Lipid peroxidation of an arachidonic acid ester molecule.
In lipid peroxidation, toxic aldehydes such as malonaldehyde (Figure 15A–C) and hydroxynonenal are formed, which react with the -NH2 of proteins and DNA bases to form adducts that can cause mutations. Malondialdehyde MDA can covalently bind to amino groups of two different proteins, especially on lysine residues, or with two amino groups of the same protein and after water elimination forms imine derivatives or Schiff bases. It can also react with DNA bases and cause mutagenic lesions. MDA reacts with the amino group of guanine and after dehydration forms an enamine derivative, which eventually cycles to the six-linked ring after loss of a water molecule. With hydroxynonenal, the reaction is initiated by a 1,4 addition of the primary amino group of the guanine to the aldehyde-unsaturated position and subsequent formation of the imine and the cycle (Figure 15).
Figure 15.
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Mechanism of malonaldehyde reaction with proteins (A,B) and DNA (C).
6. Protein and Enzyme Damage Caused by ROS
Proteins and enzymes are large, complex molecules that perform critical functions in the body. They are encoded by nuclear and mitochondrial DNA and do most of the work in cells. They are necessary for the structure, function and regulation of the body’s tissues and organs. Another critical aspect of OS is the damage triggered to the structural integrity of proteins, causing loss of catalytic activity of several enzymes and paralysis in the regulation of metabolic pathways. In the last 20 years, research linking lipid peroxidation and neurodegeneration has increased dramatically, particularly with the advent of proteomics, as each disease has become better understood. Not only in neurodegenerative diseases, but also in tumours, this recent area of research has given critical knowledge about protein modifications.
Unlike nucleic acids, oxidised proteins must be hydrolysed or processed by the proteasome to prevent their diffusion in the metabolic network or their interaction with other proteins. The effects of ROS on proteins are several: (i) oxidation of amino acid residues, (ii) cleavage of peptide bonds and (iii) aggregation between proteins. A wide range of diseases have been linked to the presence of oxidised proteins, such as Alzheimer’s disease, rheumatoid arthritis and others .
The nature of free radical-mediated protein damage depends on the amino acid composition (Figure 16).
Figure 16.
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Mechanism of protein oxidation. The abstraction of hydrogen from the protein by the hydroxyl radical generates the alkyl radical, stabilised by resonance with the carboxyl function (A). The alkyl radical reacts with oxygen to form the peroxide radical (B). The peroxide radical abstracts another hydrogen from an adjacent protein and a hydroperoxide and an alkyl radical are formed (C). The hydroperoxide is reduced to an alkoxy radical in the presence of ferrous iron (D). Hydrogen abstraction from an adjacent protein by the alkoxyl radical forms hydroxy amino acid derivatives (E). The alkoxy radical upon cleavage generates different protein carboxy radicals and alkyl radicals (F). In the absence or at low oxygen levels the alkyl radicals form protein aggregates (G).
7. Neutrophil Oxidative DNA Damage: Consequences for Human Health
ROS can pass through bacterial membranes and damage their nucleic acids, proteins and cell membranes. Consequently, bacterial pathogens employ various strategies to avoid the negative consequences of ROS production, such as directly preventing their secretion by using secreted effector proteins or toxins that interfere with the translocation of the NADPH oxidase complex, or even interfering with the signalling pathways required for their activation . Several bacterial enzymes, such as superoxide dismutases (SOD), catalases and peroxiredoxins, are used to transform reactive species into products with lower toxicity. Catalases and peroxiredoxins act as H 2 O 2 scavengers . Free iron is necessary for the Fenton reaction to occur, so bacteria use a number of mechanisms to sequester it or to control its uptake in response to ROS in the environment . DNA damage is a key consequence of ROS action and was thought to be the main mechanism of bacterial killing by ROS, particularly at the concentrations found in mammalian tissues. Oxidation of DNA bases by OH− can produce several harmful by-products, and oxidation of ribose can induce strand breaks in bacterial DNA .
Neutrophils are normally found in the bloodstream (where they live for about a day), and in humans they constitute 40–70% of all white blood cells. They are formed from stem cells in the bone marrow and differentiate into neutrophil-killer and neutrophil-agent subpopulations, forming an essential part of the innate immune system. Neutrophils are short-lived and highly mobile, as they can be found in tissues where other cells are not. During the initial phase of a bacterial infection or in some cancers, neutrophils are one of the first cells to migrate to the site of inflammation, through blood vessels and then interstitial tissue, following chemical signals such as interleukin-8 (IL-8) and peroxide H 2 O 2 in a process called chemotaxis .
Neutrophils are highly mobile and rapidly congregate at the focus of infection, attracted by cytokines expressed by activated endothelium, mast cells and macrophages. They recruit and activate other cells of the immune system, playing a key role in the first line of defence against invading pathogens, with three methods of attacking micro-organisms: phagocytosis (ingestion), degranulation (release of antimicrobials) and generation of neutrophil extracellular traps (NETs).
NADPH oxidase is an enzyme complex found in the cell membrane, in the outer space. It is found in the membranes of phagosomes used by neutrophilic white blood cells to engulf micro-organisms. This enzyme catalyses the production of a superoxide anion by transferring an electron to oxygen from NADPH (Figure 17).
Figure 17.
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Mechanism of the reaction of NADPH with molecular oxygen and 2 electrons, producing NAPD+, the radical superoxide anion and one proton.
ROS generated by NADPH oxidase play an important role in host antimicrobial defence and inflammation, activating animal immune responses and plant signalling. Superoxide anion directly kills bacteria and fungi, as the virulence of many pathogens is attenuated when their superoxide dismutase (SOD) genes are knocked out. In essence, insufficient activity can lead to increased susceptibility to microorganisms, but excessive action can lead to oxidative stress and cell damage. Careful regulation of NADPH oxidase activity is crucial for maintaining a healthy level of ROS in the body .
Vascular NADPH oxidases are regulated by a number of hormones and factors involved in vascular remodelling and disease, such as thrombin, platelet-derived growth factor (PDGF), tumour necrosis factor (TNF-α), lactosylceramide, IL-1 and oxidised LDL, by agonists and by arachidonic acid . Mutations that alter the function of NADPH oxidase led to chronic granulomatous disease, characterised by severe infections and inflammatory disorders .
The formation of lipid hydroperoxides by neutrophil-derived ROS can also cause DNA damage. This process will produce a number of highly reactive side products, such as epoxides and aldehydes. Malondialdehyde and 4-hydroxynonenal, for example, have been widely researched and shown to be extremely DNA reactive and mutagenic .
Moreover, neutrophils play crucial roles in liver repair by promoting the phenotypic conversion of proinflammatory Ly6ChiCX3CR1lo monocytes/macrophages to proresolving Ly6CloCX3CR1hi macrophages. ROS expressed by neutrophils are important mediators that trigger this phenotypic conversion to promote liver repair. This conversion is prevented by neutrophil depletion through anti-Ly6G antibody, genetic deficiency of granulocyte colony-stimulating factor or genetic deficiency of NADPH oxidase 2 (Nox2) . Adoptive transfer of WT neutrophils in place of Nox2-/rescues the altered phenotypic conversion of macrophages in neutrophil-depleted mice .
8. ROS-Mediated Activity of Antimicrobial Peptides
Antimicrobial peptides (AMPs) are mainly gene-encoded peptides, generally positively charged, with molecular weights less than 10 kDa. The majority of AMPs have small amino acid residue spans, ranging from 5 to 40, displaying diverse amino acid sequences and secondary structures forming amphipathic helixes. AMPs can bind to the microbial cell wall or membrane through electrostatic and hydrophobic interactions, disrupting the membrane . AMPs are made with a marginal amount of energy and biomass, due to their limited molecular size, and are synthesized in a fast and versatile manner [56,57]. Unlike standard antibiotics, AMPs typically work on a large variety of pathogens, including parasites, enveloped viruses, fungi, gram negative and positive bacteria, and cancer cells .
Furthermore, unlike traditional antibiotics, which target a certain metabolic enzyme and can cause microbe resistance, AMPs destroy microorganisms primarily through a process requiring membrane destruction, which is normally difficult for microbes to overcome resistance . The amino acid sequence as well as the molecular conformation determine the antimicrobial spectra of individual peptides. Several attempts have been made to establish AMPs as therapeutic agents, mainly for the treatment of external infections such as oral mucositis-related infections, chronic lung infection associated with cystic fibrosis, diabetic ulcers, ocular infections, and buco-dental infections; however, none have entered clinical application yet .
Recently, the antimicrobial activity of antimicrobial peptides has been related to the production of ROS. For example, PMAP-23, a member of the cathelicidin family in pigs, induces a Ca 2+-dependent NADH oxidation that dramatically increased oxidized NADH levels, suggesting that elevated oxidation was induced by mitochondrial Ca 2+ . Reduced mitochondrial Ca 2+ levels inhibited NADH oxidation, which is consistent with the assumption that NADH oxidation necessitates the entrance of Ca 2+ ions into the mitochondria. These findings were consistent with PMAP-23 inducing a Ca 2+-dependent NADH oxidation. Increased levels of ROS disrupt intracellular redox homeostasis decreasing GSH levels and changes the intracellular environment to a more oxidative state. PMAP-23 induced apoptosis in C. albicans cells via mitochondrial Ca 2+-induced ROS. The overproduction of mitochondrial ROS compromised C. albicans’ intracellular redox homeostasis and decreased glutathione levels in fungal cells, resulting in oxidative stress .
Antimicrobial strategies such as ROS may be a feasible choice for coping with the increasing antimicrobial resistance situation. ROS have the potential to be a powerful weapon in the fight against microbial infections since they are toxic to a wide variety of pathogens by inducing oxidative stress . ROS-induced oxidative stress may destroy cellular macromolecules like DNA, resulting in breakdown and, finally, microbial cell death. The potential application of ROS-based techniques as therapeutic therapies for microbial infections remains difficult, but at least some of them could become well known and adequately developed to enter clinical practice [59,60].
9. Oxidative Phosphorylation and Mitochondrial Uncoupling
Inside cell and mitochondria, the harmful effects of ROS include structural changes in DNA and RNA, peroxidation of membrane phospholipids (composed of polyunsaturated fatty acids) and oxidation of cellular proteins that perform numerous functions inside the cell. It is therefore important to keep ROS levels at physiological values, preventing them from increasing and triggering the expression of the mitochondrial protein UCP2 on its inner membrane . The uncoupling protein (UCP) family acts as a proton channel or transporter, housed in the mitochondrial inner membrane. Therefore, it is able to dissipate the gradient from the mitochondrial matrix to the mitochondrial intermembrane space. The energy lost in dissipating the proton gradient generates heat, thus linking UCPs to thermogenesis.
UCPs are located in the same mitochondrial inner membrane as ATP synthase, which is also a proton channel. Both proteins work by introducing protons from the intermembrane space into the matrix, UCP2 generating heat and the other synthesizing adenosine triphosphate ATP from adenosine diphosphate ADP and inorganic phosphate, the last step of oxidative phosphorylation OSPHOS . OXPHOS is based on the transport of H+ protons from the mitochondrial intermembrane space to its matrix, thus contributing to the reduction of molecular O 2 to H 2 O. The disruption of this process is known as the Warburg effect .
Around 1920, Otto Heinrich Warburg and his group deduced that glucose and oxygen deprivation in tumour cells leads to lack of energy and ultimate apoptosis of the cells. Biochemist Herbert Grace Crabtree extended Warburg’s research by discovering environmental or genetic influences. Crabtree observed that Saccharomyces cerevisiae prefers glucose fermentation to ethanol over aerobic respiration in the presence of a high concentration of the former . Warburg observed something similar in tumours: cancer cells tend to use glucose for energy even under aerobic conditions, coining the term “aerobic glycolysis”. It was hypothesized that dysfunctional mitochondria may be the cause of the higher rate of glycolysis observed in tumour cells, as well as a predominant cause of their development and proliferation . This Warburg reprogramming occurs in many different types of cancer and is related to the Gordian knot in cancer treatment . Cancer cells cause glycolysis to predominate over mitochondrial oxidative phosphorylation as a source of energy storage. Aerobic glycolysis with lactate generation is an adaptive strategy to the new metabolic needs of neoplastic cells and causes an increasingly aggressive cancer phenotype . Thus, the origin of most cancer is now considered to be directly linked to mitochondrial dysfunction, which in turn is caused by ROS-induced oxidative mechanisms. There is therefore a direct link between ROS and cancer, which has also been observed in other diseases, such as Alzheimer’s and Type 2 diabetes .
Mitochondrial respiration is coupled to ATP synthesis via this ADP phosphorylation step, i.e., protons introduced by ATP synthase are used to reduce molecular O 2 to H 2 O. Activation of the UCP2 protein occurs by an increase in ROS emitted , thus providing an important mechanism for limiting the production of ROS, such as superoxide anion •O 2−, Figure 18.
Figure 18.
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Schematic representation of oxidative phosphorylation and mitochondrial uncoupling. Electrons derived from glucose and fatty acid metabolism flow through complexes I-IV of the ETC electron transport chain in the mitochondrial inner membrane. The energy gradient of this process is used to pump protons (H+) from complexes I-IV to the intermembrane space. The resulting H+ gradient sustains the membrane potential ΔΨm, which drives ATP synthase (and subsequent oxidative phosphorylation). ATP and ADP are exchanged between the matrix and the cytoplasm via the adenine nucleotide translocase ANT. UCP2-induced proton uptake reduces ΔΨm values and a decrease in ATP production. This limitation of ΔΨm accelerates electron transport and mitochondrial respiration, limits the likelihood of electron leakage and the production of superoxide anion.
In neoplastic cells, mitochondrial metabolism is modified to meet increased energy needs and the requirements associated with rapid and uncontrolled proliferation. One of the most prominent changes occurs in cellular energy metabolism, and is known as the.
An additional benefit of the Warburg effect is the detour of ETC substrates to decrease mitochondrial ROS production . Cancer cells exhibit increased levels of intracellular ROS with complex biological effects. ROS induce genomic instability and stimulate cancer cell growth and survival by inhibiting their apoptosis. UCP2 expression is associated with cancer and modulates energy metabolism in response to elevated ROS levels . Overexpression of UCP2 is found in leukaemia, ovarian, bladder, oesophageal, testicular, colorectal, kidney, pancreatic, lung and prostate tumours . The effect of UCP2 on ATP concentrations varies according to cell type. The β-pancreatic cells undergo impaired ATP production with increased UCP2 activity, associating this process with cell degeneration, decreased insulin secretion and the onset of type II diabetes mellitus . The increased number of mitochondria increases the combined concentration of ADP and ATP, resulting in an increase in ATP when the proton leakage mechanism is inhibited.
Zhang et al., 2006, demonstrate that, consistent with in vitro studies, overexpression of UCP2 leads to tumour development in vivo in an orthotopic model of breast cancer. Genipin, a phytochemical molecule, suppresses the tumorigenic properties of UCP2, mediated by downregulation of reactive oxygen species and downregulation of UCP2. This study demonstrates that: (i) the Warburg effect is mediated by UCP2; (ii) this protein is overexpressed in breast cancer and many other cancers; (iii) it promotes tumorigenic properties in vitro and in vivo; and (iv) genipin suppresses the tumour-promoting function of UCP2 .
10. ROS-Mediated Damage in Endoplasmic Reticulum (ER) and in Haemolytic Diseases
The endoplasmic reticulum (ER) is an organelle present in most eukaryotic cells, specialised in protein folding and transport. Alterations in the ER environment leads to the accumulation of misfolded proteins and this seriously affects various cell signalling processes, such as reduction-oxidation (redox) homeostasis, energy production, inflammation, differentiation and apoptosis. Reactive oxygen species (ROS) have been linked to cellular stress, and play a key role in many cellular processes occurring in the cytosol and in various organelles, such as the ER and mitochondria. Alteration of ROS homeostasis in the ER is sufficient to cause ER stress. It is also unclear how changes in the protein folding environment in the ER causes oxidative stress. Furthermore, it is unknown how ROS production and protein misfolding induce cell apoptosis and contribute to several degenerative diseases .
The unfolded protein response (UPR) is a group of adaptive signalling pathways whose function is to resolve protein misfolding and restore an efficient environment for protein folding. UPR signalling arises in response to protein misfolding in the ER. Prolonged ER stress leads to activation of the pro-apoptotic UPR, which plays a critical role in physiological and pathological conditions .
Haemolytic anaemias are caused by haemolysis, i.e., the abnormal breakdown of red blood cells (RBCs), either in the blood vessels (intravascular) or elsewhere in the human body (extravascular). During its course, red blood cells are destroyed faster than they can be made. Oxidative stress aggravates the symptoms of many diseases, including haemolytic anaemias, as it has been found in sickle cell anaemia and thalassaemia, glucose-6-phosphate dehydrogenase deficiency, hereditary spherocytosis, congenital dyserythropoietic anaemias and paroxysmal nocturnal haemoglobinuria. Although oxidative stress is not the origin of these diseases, OS of erythroid cells plays a crucial role in haemolysis due to ineffective erythropoiesis in the bone marrow and poor survival of red blood cells in the circulation. In addition, platelets and polymorphonuclear white blood cells (PMN) are also exposed to OS .
Correction of anaemia by red blood cell transfusions or iron supplementation may increase the oxidative stress burden. This situation suggests that both iron and redox status should be monitored during treatment, using red blood cells as biomarkers .
11. Conclusions
Oxidative stress is an early event in the aetiology of several diseases, as biomarkers of oxidative stress appear early in their development. The high chemical reactivity of ROS makes them very effective weapons against most biomolecules. Oxidants are implicated in an increasing number of varied and dynamic processes in molecular and cell biology research, where the underlying chemical pathways are often unknown. We here aimed to link chemistry and biology, and to put what could be considered a solid mechanistic foundation. We hope this review can help students and researchers to understand the mechanism of these highly reactive molecules and their role as mediators of oxidative modifications of cellular components.
Author Contributions
Conceptualization, C.A.J. and E.P.-L.; investigation, C.A.J. and E.P.-L.; writing—review and editing, C.A.J.; J.M.P.d.l.L.; E.P.-L. and F.J.P.; supervision, C.A.J. and J.M.P.d.l.L. All authors have read and agreed to the published version of the manuscript.
Funding
This research was funded by “Junta de Castilla y León”, projects FEDER-VA115P17, and VA149G18); by project APOGEO (Cooperation Program INTERREG-MAC 2014–2020, with European Funds for Regional Development-FEDER); by “Agencia Canaria de Investigación, Innovación y Sociedad de la Información (ACIISI) del Gobierno de Canarias”, project ProID2020010134; by CajaCanarias, project 2019SP43 and by Spanish Ministry of Economy and Competitiveness (Grant PID2019-105838RB-C31).
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. The funders had no role in the design of the study; in the collection, analyses, or interpretation of data; in the writing of the manuscript, or in the decision to publish the results.
Footnotes
Publisher’s Note: MDPI stays neutral with regard to jurisdictional claims in published maps and institutional affiliations.
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Back to Top |
16020 | https://www.youtube.com/watch?v=doDQVdBPQtI | Oxidation and reduction | Chapter 5 - Shriver & Atkins’ Inorganic Chemistry (5th Edition)
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Posted: 4 Sep 2025
Chapter 5 of Shriver & Atkins’ Inorganic Chemistry (Fifth Edition) provides a thorough exploration of oxidation–reduction chemistry, beginning with the concepts of electron transfer, oxidation numbers, and redox couples. The chapter introduces reduction half-reactions as a systematic way to balance and analyze redox processes, showing how spontaneity can be predicted through Gibbs free energy changes, standard potentials, and the construction of galvanic cells. Trends in standard potentials are rationalized by thermodynamic cycles, hydration enthalpies, and periodic properties, leading to the electrochemical series that ranks species as oxidizing or reducing agents. The Nernst equation is presented as a tool for calculating cell potentials under nonstandard conditions and for connecting redox equilibria to equilibrium constants. Redox stability is then examined in aqueous systems, highlighting the influence of pH, water stability, atmospheric oxygen, and the processes of disproportionation and comproportionation, with examples ranging from copper(I) chemistry to hypochlorite and manganese oxoanions. Complexation and solubility equilibria are shown to shift redox potentials, linking redox chemistry to coordination compounds and solubility products. The chapter introduces diagrammatic methods—Latimer diagrams for standard potentials, Frost diagrams for oxidation state stability, and Pourbaix (E–pH) diagrams for environmental and corrosion chemistry—providing powerful visual tools for predicting reactivity and stability across conditions. Environmental implications are discussed through the chemistry of natural waters, iron cycling, and atmospheric oxidation, while industrial applications are addressed through the extraction of metals. The Ellingham diagram is introduced to rationalize high-temperature reductions by carbon and carbon monoxide, explaining processes like smelting and blast furnace operation, while modern methods such as electrochemical extraction (Hall–Héroult process for aluminum) and oxidative methods for halogens and sulfur are also detailed. By integrating thermodynamics, electrochemistry, environmental chemistry, and industrial metallurgy, this chapter builds a complete framework for understanding redox processes from fundamental concepts to real-world applications.
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Shriver and Atkins Inorganic Chemistry Chapter 5 summary, oxidation and reduction explained, redox reactions electron transfer chemistry, oxidation states and redox couples, half reactions and balancing redox equations, standard reduction potentials and spontaneity, galvanic cells Gibbs free energy connection, trends in redox potentials periodic table, electrochemical series oxidizing and reducing agents, Nernst equation applications cell potentials, pH dependence of redox reactions Pourbaix diagram, water stability hydrogen oxygen potentials, atmospheric oxidation iron copper rusting, disproportionation and comproportionation chemistry, redox stability and complex formation, solubility product and redox equilibria, Latimer diagrams for redox data, Frost diagrams oxidation state stability, Pourbaix diagrams environmental chemistry, natural waters redox stability, Ellingham diagrams metallurgy reduction of oxides, smelting and blast furnace iron extraction, Hall Héroult process aluminum electrolysis, Claus process sulfur oxidation, chlorine and halogen electrochemical extraction, inorganic chemistry textbook summaries
Transcript:
Have you ever wondered, you know, why batteries work or why iron rusts or even how something like gold gets pulled out of solid rock? Turns out it all comes down to a really fundamental idea, redux chemistry. Welcome to the deep dive, your shortcut to being wellinformed. Today we're diving deep into the world of oxidation and reduction reactions, and we're drawing our insights straight from a classic text, Shrivever and Actton's inorganic chemistry. Right. And our mission today is pretty clear. We want to break down these uh sometimes complex ideas step by step. We want to make them really accessible. Whether you're say a college student getting ready for an exam or maybe just someone curious about the chemical forces shaping our world, we'll try to demystify the jargon, help you visualize these processes, no screen needed, and show you how it all connects, you know, from environmental stability right through to big industrial processes. Okay, so here's the plan. We'll start with the basics. electron transfer, how we measure the sort of push and pull of these reactions. Then we'll dig into how things like pH or other molecules floating around can change chemical stability. And finally, we'll look at some really powerful visual tools diagrams that act like maps for redux behavior, plus the amazing ways we actually use this chemistry to get hold of elements vital to modern life. Ready to unpack this? Let's do it. Okay, so at its heart, redux chemistry is all about electrons moving around. Can you give us the uh the core definition? Absolutely. Think of it simply. Oxidation means a species loses electrons. Reduction means it gains electrons. But the key thing, the really crucial insight is they never happen alone, ever. They're always paired up, always happening at the same time in what we call a redux reaction. Ah, okay. So if one thing is losing electrons, something else must be gaining them. No electrons left to hang around. Exactly. Right. We often talk about a redux couple that's just an oxidized species and its reduced partner like um H+ and H2 gas or the zooian 2 plus ions and solid zinc metal. And we can think about a full redux reaction by splitting it conceptually at least into two half reactions. One shows the electron loss, that's oxidation. The other shows the electron gain reduction. Helps keep track. And balancing these reactions, making sure everything adds up in the equation. Sounds like that can get a bit involved like a systematic process. Yeah, you got to make sure every atom and crucially every charge is accounted for. Often involves uh balancing things like oxygen first by adding water molecules, then hydrogen using H+ if it's acidic or O and water if it's basic. And only then the very end you balance the overall charge by adding electrons like a precise recipe you follow. Okay, so we've got the what's electrons moving. But the big question for me is always why do they move? What makes some reactions just go spontaneously while others need I don't know a push? That's a fantastic question. It gets right to the heart of it. We're talking about a reaction's natural tendency to proceed. Chemists call it spontaneity. From a thermodynamics angle, a reaction is favorable or spontaneous. If it's Gibbs energy change, that's E or G, is negative. Under standard conditions, if that standard gigs energy RG degrees, is negative, it means the equilibrium constant K. That tells you the product reactant ratio at the end. Exactly. It means K is greater than one. Products are favored. the reaction wants to go that way. So what's the electrochemical way to measure this this spontaneity? That would be the standard potential E degree. Think of E degree as the electrochemical measure of that spontaneity for a half reaction. It's directly linked to Gibbs energy by a fundamental equation. Your G degrees EA degrees here RG is just the number of electrons transferred in the reaction and F is Faraday's constant. Basically the charge of a whole mole of electrons. So a more positive E degree means a more negative energy degree which means it's more spontaneous. It's like the voltage in a battery. More volts more driving force. And isn't there a benchmark a zero point for all these E degree values? Yes, absolutely. By convention the standard potential for the hydrogen couple H plus H2 is defined as exactly 0 volts at all temperatures. That gives us a fixed reference point to measure everything else against. So how do you actually measure these potentials in the lab? What does the setup look like? You use something called a galvanic cell. Essentially, it's a setup where a chemical reaction generates an electric current like a battery. Basically, this lets us measure that potential difference, that voltage. In any cell like this, reduction in the gain of electrons always happens at the electrode we call the cathode. Mhm. And oxidation, the loss of electrons happens at the anode. And the really crucial rule here, an overall reaction is thermodynamically favorable. It wants to happen if it's standard cell potential. E degree is positive and E degrees is just the difference between the reduction potentials of the two half reactions involved. Okay. And here's where it gets really interesting, right? You take the zinc couple ZN2 plus ZN, its E degree is.76 volt. But if you combine that with our hydrogen reference electrode, which is 0 volt, right? The overall E degree cell becomes positive 0.76 volts. And that positive value tells us zinc metal has a thermodynamic tendency to reduce H+ ions. Or putting it plainly, stick zinc metal in acid and poof, it dissolves, making hydrogen gas. Exactly. That positive E degree cell predicts the reaction. That's a great illustration, but it makes you wonder what actually makes one metal's E degree so different from anothers. Why is zinc nagged 0.76V and something else different? Excellent point. These E degree values aren't just random numbers. They're the result of a delicate balance really of atomic properties. We have to consider the energy needed to uh turn the solid metal into gas atoms. That's called the atomization enalpy. Then the energy needed to rip an electron off that gas atom. The ionization energy. Okay. And finally the energy that gets released when that positive gas ion gets surrounded and stabilized by water molecules. That's the hydration enthalpy. Wow. Okay. So it's like a thermodynamic tugof-war between those different energy steps. Precisely. Take lithium for example. The limplus line couple has a very very negative E degrees mitigate 3.04 04 vol super reactive. Now compare that to silver egg plus egg that has a positive E degrees plus80 volt much less reactive. Lithium actually has a higher ionization energy than silver which might make you think it'd be harder to oxidize. But the lithium ion light plus a is tiny just 90 pm because it's so small it interacts incredibly strongly with water. It releases a huge amount of energy when it hydrates. It's this massive negative hydration enalpy that overwhelms the other factors and makes lithium such a powerful reducing agent so willing to give up its electron. Silver on the other hand has a high ionization energy partly because its inner d electrons don't shield the outer electron very well. This makes it harder to remove that electron contributing to its positive E degrees and explaining why silver doesn't dissolve and dilute acids like zinc or lithium do. So these subtle atomic properties really combine to dictate how reactive a metal is. Building on that, chemists have actually compiled all these standard potentials into what they call the electrochemical series. Yes. And this series is incredibly handy. It's basically a ranked list of redux couples, usually ordered from the most positive E degree values down to the most negative. A really big positive E degree means the oxidized form in that couple is a strong oxidizing agent. It really wants electrons. Think florine gas F2 at the top with plus 2.87V. Conversely, a really big negative E degrees means the reduced form is a strong reducing agent. It wants to give away electrons like our friend lithium metal lie way down at NASA 3.04 VP. And here's the key rule of thumb. The reduced member of any couple tends to be able to reduce the oxidized member of any couple that sits above it in the series. Okay, but and this seems really important. This series tells us what can happen thermodynamically speaking. It doesn't tell us how fast it will happen. Right? That is a absolutely crucial distinction. Kinetics versus thermodynamics. A reaction might be hugely favorable on paper, have a really positive E degrees, but it could be incredibly slow in practice. Maybe it needs a catalyst, or maybe there's just a high energy barrier to get it started. Got it? Take permaganate MO4 as an example. In acidic solution, it's a strong oxidizing agent. Essing the series, you can see it can thermodynamically oxidize iron. Fe2 plus E degree for F3 plus A2 plus is plus.77V or chloride ions CL degrees for CL2ZL is plus 1.36V but it cannot oxidize serium th C3+ because the E degrees for C4 plus E3+ is even higher plus 1.76V. This also has practical implications. It tells you why you can't use hydrochloric acid HDL to acidify a peraganate titration. The peranganate would just chew up the chloride. You have to use something like sulfuric acid instead because sulfate is much harder to oxidize. Right? Ral chemistry. Okay. So that's all under standard conditions, but you said it yourself. Standard conditions, one bar pressure, exactly one molar concentrations. They're pretty rare in the real world. So what does this all mean when things aren't perfectly standard? That's where the Nerst equation comes into play. It's a really powerful tool. It lets us calculate the actual cell potential EC cell under any conditions, any concentration, any pressure. It does this by incorporating the reaction quotient Q. or Q. It's like the equilibrium constant K, but it reflects the current ratio of products to reactants, not necessarily the equilibrium one. The Nerst equation basically adjusts the standard potential based on how far the current conditions are from equilibrium. And what's truly striking is how sensitive the equilibrium position is to potential. Even a small change in E degree can cause enormous shifts in the equilibrium constant K. We're talking orders of magnitude. A potential of just plus 2 volts can mean K is like 10 ^ of 34 almost complete reaction. While 90 volts means K is 10 the minus 34 basically no reaction. Tiny voltage changes have huge consequences for where the equilibrium lies. Wow. Okay. Now thinking about real world conditions especially in biology or environmental science water is everywhere and many reactions involve hydrogen ions. So I'm guessing the acidity the pH of the water must have a massive impact. You guessed right. a huge impact. Many many redux reactions, especially in aquous solution, involve the transfer of H+ ions. This means their electro potentials are inherently pH dependent. The nerds equation actually shows this directly. Generally, as the pH increases, meaning it gets less acidic, more basic, the potential tends to decrease. It becomes more negative or less positive. Take the perchloric keklloride couple again. At pH0, strongly acidic, E° is plus 1.201 V. But at neutral pH7, it drops quite a bit down to plus 0.788V. It clearly shows perchlorate as a much stronger oxidizing agent in acid than in neutral water. And this pH dependence is so important, especially in biological systems which operate near pH7 that biochemists often use a different standard state, the biological standard state reference specifically to pH7. So it sounds like water isn't just a passive background solvent then. It can actually get involved in the redux chemistry itself, like as a reactant. Oh, absolutely. Water is definitely not just sitting there. It can act as both an oxidizing agent and a reducing agent. It could be reduced to hydrogen gas H2. This happens when it reacts with metals that have very negative standard potentials. Think sodium potassium, the S block metals. They get oxidized, water gets reduced, and you produce hydrogen gas. Interestingly though, some metals that should react with water thermodynamically like aluminum or iron or even copper don't always corrode away instantly in moist air. That's because of passivation. They form a very thin, tough, invisible layer of oxide on their surface that protects the metal underneath from further reactions, like a natural shield. Oh, that's clever. Like rust, but sometimes protective. What about water acting the other way as a reducing agent? It can do that, too. Yes, water can be oxidized to oxygen gas, O2. But this requires a pretty strong oxidizing agent to force it to happen. For example, cobalt thryons CO3 plus scion have a very high potential plus 1.92V and will readily oxidize water to O2 while getting reduced themselves to CO2 plus ions. Now here that's the kinetics thing again. Many strong oxidizers like peranganate or dromate should thermodynamically oxidize water. Their potentials are above the plus 1.23V needed at standard conditions. But often that reaction is surprisingly slow. There's a significant kinetic barrier and over potential especially for forming that O bond in oxygen. This slow kinetics is actually why solutions of these strong oxidizers can be relatively stable in water even though they should react. This leads us to the idea of the stability field of water, doesn't it? Can you explain that? Yes. The stability field is basically a region on a potential versus pH graph. Within this region, water is thermodynamically stable. It won't spontaneously oxidize to O2 or reduce to H2. Any chemical species whose own potential and pH conditions fall outside this water stability field is in principle unstable in water. It will tend to either oxidize water or be oxidized by water or reduce water or be reduced by water. It's a fundamental constraint in aquous chemistry. Okay, so water itself is a factor. What about just air? If you have a solution open to the air, dissolved oxygen must be a player, right? A massive player. Atmospheric oxygen O2 dissolved in water is actually a pretty potent oxidizing agent. Think about iron 2 ions F2 plus Sarah. In water with no oxygen, they're stable. But expose that solution to air. Especially at low pH, the dissolved O2 readily oxidizes the F2 plus to F3 plus FEST. The overall cell potential is quite positive plus B.46V. This is fundamentally why iron 3, often as rust, like F23 or FOH, is the dominant form of iron you find in the Earth's crust, which is exposed to air and water. It's also why copper roofs slowly turn green over time. That green patina is a layer of basic copper carbonate or sulfate formed by slow oxidation by air, CO2, and acid rain. Again, it's often a passive layer protecting the copper underneath. Fascinating. Now, let's talk about some slightly stranger cases where an element reacts with itself. Ah, yes. Disproportionation and comproportionation. They sound complicated, but the idea is neat. Disproportionation is when a species containing an element in an intermediate oxidation state reacts such that some of it gets oxidized, its oxidation number goes up and some gets reduced, its oxidation number goes down simultaneously. A classic example is copper I ion plus I in water. It's unstable. It spontaneously disproportionates into copper ions Cu2 plus I and solid copper metal CS the E degree cell for this is positive so it really wants to happen manganese MNZ is another example in acid it rapidly disproportionates into MMZ and MM that's comproportionation it's where you have the same element in two different oxidation states reacting to form a product where the element has an intermediate oxidation state for instance silver ions Ag2 plus which are quite unstable will react with solids silver metal AGS to form the much more stable silver ions egg plus I both starting materials end up as egg plus I's okay so far we've mostly talked about the metal ion itself but you hinted earlier that what it's bonded to the lians can make a huge difference to its redux behavior a massive difference this is the lian effect complexation forming bonds with lians can dramatically shift standard potentials the reason is that lians often bind with different strengths to the oxidize versus the reduced form of the metal ion. If a lian binds much much more strongly to say the oxidized form, it stabilizes that oxidized form. This makes it harder to reduce the metal. Effectively, it shifts the reduction potential to a more negative value compared to the simple metal ion in water. Can you give us an example of how that plays out? Sure. Let's look at iron again. The standard potential for the simple aqua complex couple FeOH263 plus FOH262 plus is plus.77V. Now replace the water leggings with cyanide ions CN. The potential for the SEN63 ACN64 couple drops significantly down to plus.36V. That's a drop of 41 volts. It tells us that cyanide binds much more strongly to FA and to FIA 2 about 10 million times more strongly actually. This stabilizes the fi state relative to fi and making the cyanide complex much harder to reduce than the simple aqua ion. That's a huge difference just from changing the surrounding molecule. So it seems like everything's connected. Even something like solubility, how much a compound dissolves can be linked back to these potentials. That seems like a hidden link. It absolutely is linked. You can actually use standard cell potentials to figure out the solubility product Ksp for sparingly soluble compounds, things that barely dissolve. The approach involves cleverly combining two different half reactions. One involving the dissolved metal ion and the other involving the solid insoluble compound containing that ion. The difference in their standard potentials can be directly related mathematically to the Ksp value. This is really important for things like environmental chemistry. Knowing the Ksp of say plutonium hydroxide P4 which can be calculated from potentials tells you how its solubility changes with pH which is critical for understanding its potential migration in the environment. Okay, this is a lot of information about potentials and conditions. H how do chemists actually like summarize all this in a way that's easy to grasp? Are there shortcuts? There are. Diagrams are incredibly useful here. One of the first chemists often turn to is the Latimer diagram. It's a really concise linear way to show the standard potentials connecting various oxidation states of a single element under specific conditions, usually acidic or basic. You write the species out in a line, typically from the highest oxidation state on the left to the lowest on the right. Then you write the standard potential value in volts above the line connecting each adjacent pair of species. For example, for chlorine and acid, you might see Cl4, then a potential, then CL3, then another potential, and so on all the way down to Cl. So it's like a chemical road map showing the voltage steps between oxidation states. Exactly. But there's a crucial point. If you want the potential for a step that skips over an intermediate species, say going directly from CO3 to HLO, you cannot just add the potentials for the intervening steps. That doesn't work. You have to go back to Gibbs energies, convert each individual E degrees to a or fee degree, add the relevant air degree values together, and then convert that total A degree back into an overall E degrees for the non-adjacent step. It's an extra calculation, but essential for accuracy. Latimer diagrams also give you a quick visual check for disproportionation. Look at any species in the middle. If the potential on its right for its reduction is higher or more positive than the potential on its left for the reduction of the species to it, then that middle species is thermodynamically unstable and will tend to disproportionate. Right? Left means unstable. Got it? That seems handy. What about other diagrams? Another really useful one, perhaps even more visually intuitive for stability, is the Frost diagram, sometimes called a Frost Eworth diagram. Here you plot a quantity related to Gibbs energy, specifically negation number on the y-axis against the oxidation number n itself on the x-axis. The beauty of a frost diagram lies in its visual interpretation. The species whose point lies lowest on the diagram is the most thermodynamically stable oxidation state for that element under those conditions like pH0 or pH14. Lowest point and most stable. Okay. And the slope of the line connecting any two points on the diagram is equal to the standard potential E degrees for the couple formed by those two species. A steeper upward slope means a higher more positive potential. So you can visually compare oxidizing and reducing strengths just by looking at slopes. So you can literally see stability and potential just by looking at the shape of the plot pretty much. And you can spot disproportionation very easily too. Any species whose point lies above the straight line connecting its two neighboring oxidation states forming a sort of peak or convex shape is unstable with respect to disproportionation. Conversely, if an intermediate species lies below the line connecting its neighbors forming a valley or a concave shape, then the two neighboring species will tend to comproportionate to form that stable intermediate. Can you give an example? Sure. If you look at the frost diagram for manganese and acid, you'll see that the point for MN3+ sits clearly above the line connecting MN2 plus and MO2. That tells you instantly MN3+ wants to disproportionate into MN2 plus and MO2. For nitrogen, you might see that N2O lies below the line connecting NH4 plus N3 and HNO3 N plus5. This suggests that ammonium nitrate could potentially comproportionate to form N2O, which is indeed something that can happen sometimes explosively. Okay, Latimer and Frost give us great views of individual elements, but what about mapping stability across different conditions, especially for environmental or geological settings? That's where the Porbay diagram or EP diagram really shines. This is the go-to map for geocchemists and corrosion scientists. A bore diagram plots potential E on the Y-axis versus PH on the X-axis. The lines on the diagram divide it into regions and each region represents the conditions where a particular species like a metal ion, an oxide or the pure metal is the most thermodynamically stable form in contact with water. The lines themselves represent equilibria between species. You have different types. Horizontal lines separate species that are related only by electron transfer. The equilibrium doesn't depend on pH. Think F3 plus Fe2 plus S. Vertical lines separate species related only by proton transfer with no change in oxidation state. like F3+ aq becoming solid FOH3 as pH increases. And sloped lines represent equilibria involving both electron transfer and proton transfer. For example, the line separating solid FOH3 from dissolved Fe2 plus Bay. And our old friend, the stability field of water shows up on these diagrams too, right? Absolutely. The poor bay diagram always includes the two lines representing the limits of water stability. The lower line where water could be reduced to H2 and the upper line where it could be oxidized to O2. Any species whose stability region lies outside these lines is thermodynamically capable of reacting with water. Porve diagrams let us understand for instance why iron corrods under certain conditions but might be passivated under others. Or consider iron in a lake near the surface with plenty of oxygen high potential and maybe neutral pH. The porbay diagram might show insoluble iron oxides or hydroxides like Fe are stable. So iron precipitates out. But down in the deep oxygen pore sediments low potential, the diagram might show that soluble Fe2+ is the stable form. So the iron oxide sediment can get reduced and dissolve, allowing Fe2+ to diffuse upwards, potentially completing a cycle. It really helps map out these environmental processes. That's incredibly powerful for understanding how our world actually works. Okay, let's shift gears slightly. We know all this theory. How do we actually use redux chemistry in the real world particularly for getting elements metals especially out of their natural ores? Well, controlling redux is fundamental to extractive metallergy. Most metals we use don't occur naturally as pure elements. They're found combined with oxygen as oxides or sulfur as sulfides or other elements. They're in an oxidized state. So to get the pure metal, we usually need to reduce them. This goes way back. Think the Bronze Age, the Iron Age. Smelting iron ore with carbon in a blast furnace is a classic massive scale redux reaction. Carbon acts as the reducing agent, pulling oxygen away from the iron oxide at high temperatures. Carbon, often in the form of coke or the carbon monoxide it produces at high temperature, is still the dominant reducing agent for many important metals like iron, zinc, and tin. There's even a process, the pigeon process, that uses carbon to reduce magnesium oxide. How do engineers figure out if carbon reduction will even work for a specific metal oxide and at what temperature? They use another type of diagram, the Elium diagram. This is a crucial tool in metal energy. LEM diagrams plot the standard Gibbs energy of formation, the FG degrees for various oxides against temperature. You'll see lines for metal oxides like FeO, ZNO, Al23, and also lines for the oxidation of carbon like C to CO or C to CO2. Here's how you use it to see if carbon can reduce a specific metal oxide say ZN O you find the line for Z to no formation and the line for carbon oxidation usually C to CO at high temps. If the carbon oxidation line lies below the metal oxide line at a given temperature it means the gives energy change for the reduction reaction ZNO plus CN plus CO will be negative at that temperature. So the reduction is thermodynamically favorable. The point where the lines cross tells you the minimum temperature at which carbon reduction becomes spontaneous. Brazen know this happens around 1200° C. So these diagrams provide real guidance for industrial processes. Are there limitations? Oh definitely. Ellingham diagrams tell you if it's possible thermodynamically. They don't guarantee it works well in practice. For example, reducing aluminum oxide Al23 with carbon requires incredibly high temperatures where aluminum metal itself would vaporize. Not practical. And for titanium oxide, carbon tends to form titanium carbide tai instead of pure titanium metal. So this carbon reduction method called pyro metalergy works great for some metals like iron and zinc but not for everything. Okay. So reduction is key for many metals. Are there cases where we use oxidation to extract or purify elements? Yes, absolutely. Sometimes the element we want is already in a reduced state and we need to oxidize it. A major example is sulfur. Huge amounts of hydrogen sulfide H2S are found in natural gas. To get elemental sulfur we use the claws process. It carefully oxidizes the H2S using controlled amounts of air in two stages to produce pure liquid sulfur and water. It's a much cleaner way than older methods. And think about gold. It often occurs in tiny amounts in rock. To extract it, the ore is treated with a solution containing cyanide ions and air oxygen. The oxygen oxidizes the gold metal AU to O plus ions. Normally, this wouldn't happen easily, but the cyanide ions immediately grab the O plus ions, forming a very stable complex ion, O CN2. This complex formation pulls the equilibrium over, making the oxidation of gold favorable. Later, this gold complex is reduced back to pure gold metal, usually using zinc powder. Cyanide sounds a bit hazardous. It is, and handling it requires extreme care due to its toxicity. But it's incredibly effective at selectively leeching gold. Another example of oxidation is getting halogens like bromine and iodine. They exist naturally as broomemide, Br and iodinins in seawater or brine deposits. To get the pure elements Br2 and I2, these solutions are treated with chlorine gas CL2 which is a stronger oxidizing agent and oxidizes the BrNI ions. Okay, so we have chemical reduction, chemical oxidation. What if those aren't suitable or efficient enough? Can we just use electricity? Exactly. That's electrochemical extraction or electrolysis. We use electrical energy to force a non-spontaneous redux reaction to occur. This is essential for highly reactive metals that are very difficult to reduce chemically. Aluminum is the prime example. The Hall Herald process involves dissolving aluminum oxide Al23 in molten cryolyte, a sodium aluminum fluoride mineral at around 950° se. Then a powerful electric current has passed through. At the negative electrode cathode, the aluminum ions are reduced to molten aluminum metal. At the positive carbon electrodes anodess, oxide ions are oxidized reacting with the carbon to form COO and CO2. It consumes enormous amounts of electricity which is why aluminum smelters are usually located where power is cheap. And this works for nonmetals too like the hallogens. Yes, especially for the most reactive ones. Chlorine gas CL2 is produced industrially by the electrolysis of concentrated sodium chloride solution. Brian, now here's an interesting challenge. When you electrolyze water solutions, water itself could be oxidized to oxygen gas. Thermodynamically, oxidizing water is actually slightly easier than oxidizing chloride ions. But remember that over potential concept, the oxidation of water to O2 often has a high over potential on typical electrode materials. It requires a significantly higher voltage than the theory predicts to make it happen at a reasonable rate. This kinetic barrier works in our favor. It allows us to selectively oxidize the chlorine ions to produce Cl2 gas instead of O2 along with hydrogen gas and sodium hydroxide as valuable co-products and for the most reactive element of all florine F2. Such a powerful oxidizing agent that it would instantly react with water. You simply cannot make it by electrolying aquous solutions. It has to be produced by electrolying a molten mixture of potassium fluoride KF dissolved in anhydra hydrogen fluoride at Chichow. Wow. Okay. We've really covered a lot of ground here. We journeyed through the uh the core ideas of redux chemistry that fundamental electron transfer how we measure with potentials how things like pH and lians dramatically change stability. We explored those powerful diagrams latimer frost porb that help us visualize and predict chemical behavior like maps and we saw how all this knowledge underpins crucial industrial processes for actually extracting the elements that build our modern world. That's right. And hopefully you now have a much deeper appreciation for this sort of hidden electron economy that's running constantly behind the scenes. It drives everything from, you know, rust forming on a bridge to the energy stored in your phone's battery to the incredibly sophisticated chemical strategies chemists and engineers use to isolate the elements we rely on. This deep dive should equip you to not just observe these things, but to really start understanding the underlying mechanisms and the broader implications. So, what does this all mean for you listening right now? Maybe think about the intricate balance of redux reactions happening constantly inside your own body, keeping you alive. Or consider the amazing ways scientists are designing brand new materials, catalysts, and energy storage solutions, all by learning how to precisely control the flow of these tiny electrons. The world around us truly is just one vast, constantly unfolding redux reaction. And hopefully we've given you some tools to explore its secrets further. Thank you for joining us on this deep dive and a warm thank you from the last minute lecture team for making knowledge your superpower. |
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5
How can I simply prove that sin(π/2-x) =cosx?
Pankaj Kumar
B Tech from National Institute of Technology, Patna (Graduated 2022)
·
6y
To prove this, use sine Subtraction formula.
i.e, sin(a-b)= sin(a)cos(b)-cos(a)sin(b)
Here a=π/2 and b=x
sin(π/2-x) = sin(π/2)cos(x)-cos(π/2)sin(x)
= 1×{cos(x)}-{0×sin(x)}
= cos(x)
Hence proved
Related questions
How can I simply prove that cos(π/2-x)=sinx?
What is the proof of sin x = cos (x+π/2)?
How can I prove (simply) that cos (π/2+x) =-sin x?
How can I prove (simply) that sin (π-x) =sin x?
How can I simply prove that cos(π-x)=-cosx?
Vikas Gupta
M.Sc. in Mathematics, Rani Durgavati University (Graduated 2016)
·
6y
The simplest one would be to draw a right triangle (Let’s say ΔABC right angled at B) and then name any of the two acute angle ‘x’(Say A). Now find cos x for this triangle -> cos x = AB/AC (Base/Hypotenuse).
Now π/2 -x will give you the value of other angle C, since both acute angles are complementary (A and C give you the sum π/2).
Finding sin π/2 -x would mean finding sin C which is equal to AB/AC (Perpendicular/Hypotenuse).
So you can see that AB/AC is equal to both sin C=sin π/2 -x and cos A = cos x.
There are many more methods. This should be the easiest one.
Manjunath Subramanya Iyer
I am a retired bank officer teaching maths
·
Author has 7.2K answers and 10.4M answer views
·
6y
In the right angled triangle ABC right angled at say B, let A = x°. For A = x°, BC is the opposite side, AB is the adjacent side and AC is the hypotenuse. Hence cos A = cos x = adjacent side/ hypotenuse = AB/AC.
So, cos x = AB/AC…(1)
As A = x° and B = 90° = π/2, C = π/2 - x.
For C, AB is the opposite side, AC remaining the hypotenuse.
So, sin C = sin(π/2-x) = AB/AC …….(2)
From (1) and (2) we can see that sin (π/2-x) and cos x are both AB/AC.
Hence sin(π/2 - x) = cos x.
Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist.
·
Author has 6.8K answers and 52.8M answer views
·
6y
Related
How can I simply prove that cos(π/2-x)=sinx?
I don’t think you would be satisfied just with values substituted into some obscure formula that you probably have not met in your studies yet, so I am going to give you an explanation that you will fully understand very easily!
This is ANY right-angled triangl...
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Related questions
How can I prove that
cos
x
sin
x
sin
x
cos
2
x
+
sin
3
x
=
cos
x
?
How do we prove that
s
i
n
−
1
X
+
c
o
s
−
1
X
=
π
2
?
[How do I prove that
[
1
−
cos
(
x
)
+
i
×
sin
(
x
)
]
n
=
2
n
×
sin
(
x
2
)
n
×
[
cos
(
n
(
π
−
x
)
2
)
+
i
×
sin
(
n
(
π
−
x
)
2
)
]
?](
How do I prove
sin
2
x
=
2
sin
x
cos
x
?
For all
x
in [0,
π
2
, How do I prove that
cos
(
sin
x
)
sin
(
cos
x
)
?
Assistant
Bot
·
1y
To prove that
sin
(
π
2
−
x
)
=
cos
x
, we can use the co-function identity from trigonometry. Here's a simple proof:
Understanding the Unit Circle
: The sine and cosine functions can be understood using the unit circle. The angle
π
2
−
x
represents a rotation in the counterclockwise direction from the positive x-axis.
2. Coordinates on the Unit Circle
:
- The coordinates of a point on the unit circle corresponding to an angle
x
are
(
cos
x
,
sin
x
)
.
- The angle
π
2
−
x
corresponds to a point that is
π
2
radians (or 90 degrees) counterclockwise
To prove that
sin
(
π
2
−
x
)
=
cos
x
, we can use the co-function identity from trigonometry. Here's a simple proof:
Understanding the Unit Circle
: The sine and cosine functions can be understood using the unit circle. The angle
π
2
−
x
represents a rotation in the counterclockwise direction from the positive x-axis.
2. Coordinates on the Unit Circle
:
- The coordinates of a point on the unit circle corresponding to an angle
x
are
(
cos
x
,
sin
x
)
.
- The angle
π
2
−
x
corresponds to a point that is
π
2
radians (or 90 degrees) counterclockwise from the angle
x
.
3. Finding the Coordinates
:
- When you rotate by
π
2
radians from the angle
x
, the new coordinates become
(
sin
x
,
cos
x
)
. This is because:
The x-coordinate becomes the y-coordinate (which is
sin
x
).
+ The y-coordinate becomes the negative x-coordinate (which is
cos
x
).
Conclusion
: Thus, the sine of the angle
π
2
−
x
(which corresponds to the new coordinates) is equal to the cosine of
x
:
sin
(
π
2
−
x
)
=
cos
x
This relationship is a fundamental identity in trigonometry and can be verified using the unit circle or by looking at the definitions of sine and cosine in terms of right triangles.
Girija Warrier
Studied at Sufficiently Educated
·
Author has 5.9K answers and 13.9M answer views
·
Updated 3y
Related
How can I simply prove that cos(π/2-x)=sinx?
I’m proving it geometrically..
In the above image, the initial ray is rotated 90° anti clockwise direction, it takes position OC.
Then, x° is rotated in clockwise direction, which is = -x°
⇒ ( 90° - x) = y° . . . . . .(1)
AC is drawn // BO
Hence, <C becomes 90°
⇒ ABOC is a rectangle.
⇒ AC = BO . . . . . . . .(2)
In triangle ABO
Cos y = OB/OA
But cos y = cos(90° - x) ( by (1) )
⇒ cos( 90° ...
I’m proving it geometrically..
In the above image, the initial ray is rotated 90° anti clockwise direction, it takes position OC.
Then, x° is rotated in clockwise direction, which is = -x°
⇒ ( 90° - x) = y° . . . . . .(1)
AC is drawn // BO
Hence, <C becomes 90°
⇒ ABOC is a rectangle.
⇒ AC = BO . . . . . . . .(2)
In triangle ABO
Cos y = OB/OA
But cos y = cos(90° - x) ( by (1) )
⇒ cos( 90° ...
Doctor Sachidanand Das
PhD,Former Scientific Officer-G, Government of India
·
Author has 11.3K answers and 16M answer views
·
Updated 6y
Related
How can I simply prove that cos(π/2-x)=sinx?
Proof 1:
The simplest way to prove
cos(π/2 - x) = sin x
is to put A = π/2 , B = x in the trigonometric formula
cos(A-B) = cos A . cos B + sin A . sin B……………………………….(1)
and obtain
cos(π/2 - x) = cos π/2 . cos x + sin π/2 . sin x……………………….(2)
Substituting cos π/2 = 0 and sin π/2 = 1 in (2),
cos (π/2 - x) = 0 . cos x + 1 . sin x=0+sin x
∴cos (π/2 - x) = sin x (Proved)
Proof 2:
Let ABC be a triangle right-angled at B. Let AB be the base and AC the hypotenuse. If we denote the angle C by x, the base angle A = (π/2 - x) so that A + B + C = π/2 - x + π/2 + x =π or 180° .
Now for the base angle A, BC is the perpe
Proof 1:
The simplest way to prove
cos(π/2 - x) = sin x
is to put A = π/2 , B = x in the trigonometric formula
cos(A-B) = cos A . cos B + sin A . sin B……………………………….(1)
and obtain
cos(π/2 - x) = cos π/2 . cos x + sin π/2 . sin x……………………….(2)
Substituting cos π/2 = 0 and sin π/2 = 1 in (2),
cos (π/2 - x) = 0 . cos x + 1 . sin x=0+sin x
∴cos (π/2 - x) = sin x (Proved)
Proof 2:
Let ABC be a triangle right-angled at B. Let AB be the base and AC the hypotenuse. If we denote the angle C by x, the base angle A = (π/2 - x) so that A + B + C = π/2 - x + π/2 + x =π or 180° .
Now for the base angle A, BC is the perpendicular.
∴ cos A = cos (π/2 - x) = base/hypotenuse = AB/AC …………..(3)
For the angle C, AB is the perpendicular and therefore
sin C = sin x = perpendicular/hypotenuse = AB/AC…………….(4)
Equating (3) and (4),
cos (π/2 - x) = sin x (Proved)
Proof 3:
Use the Euler’s formula
eⁱᶿ = cos θ + i sin θ
which defines the symbol eⁱᶿ for any real value of θ . Here i = √-1 .
∴ We can put θ = (π/2 - x) in the formula and write
e^i(π/2 - x) = cos (π/2 - x) + i sin (π/2 - x)
Or, e^iπ/2 . e^(-ix) = cos (π/2 - x) + i sin (π/2 - x)
Now e^iπ/2 = cos π/2 + i sin π/2 = 0 + i.1 = i and e^(-ix) = cos x - i sinx
∴i .( cos x - i sin x) = cos (π/2 - x) + i sin (π/2 - x)
Or, i cos x + sin x = cos (π/2 - x) + i sin (π/2 - x) [Since i² =-1]
Equating the real and imaginary parts,
cos (π/2 - x) = sin x (Proved)
and cos x = sin (π/2 - x)
Concluding Remarks:
Of the three methods presented here for proving the given assertion , the preferred method should be the Proof 1. This is because it is simple, straight forward and fast. Can be done mentally by an average student in about 30 seconds. In Proof 2, there is room for confusion as to which is the base, which is the right perpendicular to be taken. Besides one needs to spend extra time to draw a triangle, mark the sides, the angles, etc. Proof 3 is fine; but not many are comfortable or good at working with complex functions. The method entails more algebra than the other methods; but it gives a bonus, namely: it proves the formula cos x = sin (π/2 - x).
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Bill Crean
Solved the Holey Cube problem without calculus.
·
Author has 5.1K answers and 6.4M answer views
·
Updated 1y
Originally Answered: How do we prove that sin(x) = cos (π/2 - x)?
·
A right triangle, a, b, c (c, hypotenuse), x a non-90 degree angle and the other will be 90-x (or pi/2-x), then, LHS = Sin(x) = a/c = Cos(pi/2 - x) = RHS. QED.
Raymond Beck
Former Infantry Sergeant
·
Author has 42.4K answers and 10.3M answer views
·
1y
Originally Answered: How do we prove that sin(x) = cos (π/2 - x)?
·
use the trig identity
cos(A-B) = cosAcosB + sinAsinB with A=90, B=x
cos(90-x) = cos90cosx + sin90sinx
= 0+(1)sinx
=sinx
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Alireza Shariati
Math major college student
·
Author has 1.4K answers and 972.5K answer views
·
Updated 3y
Related
How do you verify sin (pi -x) =sinx?
Consider the angle 0∘<α<90∘ on the unit circle; this angle can form a right-angled triangle with the base cosα and the height sinα.
Now consider the angle θ=180∘, rotate α degrees clockwise, and you'll have the angle 180∘−α on the unit circle, which would form another right-angled triangle same as the aforementioned triangle, reflected over the sine axis; hence the height of it would also be sinα.
Image Source
Consider the angle 0∘<α<90∘ on the unit circle; this angle can form a right-angled triangle with the base cosα and the height sinα.
Now consider the angle θ=180∘, rotate α degrees clockwise, and you'll have the angle 180∘−α on the unit circle, which would form another right-angled triangle same as the aforementioned triangle, reflected over the sine axis; hence the height of it would also be sinα.
Image Source
Works at Quora (product)
·
Upvoted by
Shubhankar Datta
, Master of Science Mathematics, Jadavpur University (2022) · Author has 145 answers and 516.5K answer views
·
8y
Related
Why is
2
sin
(
x
)
cos
(
x
)
=
sin
(
2
x
)
true?
You can prove it using Euler’s formula (eiθ=cosθ+isinθ) and the identity (a+b)(a−b)=a2−b2:
2sinθcosθ
=2eiθ−e−iθ2ieiθ+e−iθ2
=ei2θ−e−i2θ2i
=sin2θ
You can also prove it in an elegant way using plane geometry (with a unit circle):
we have use:
base angles of an isosceles triangle
exterior angle of a triangle equals the sum of two opposite angles
definitions of sine and cosine
You can prove it using Euler’s formula (eiθ=cosθ+isinθ) and the identity (a+b)(a−b)=a2−b2:
2sinθcosθ
=2eiθ−e−iθ2ieiθ+e−iθ2
=ei2θ−e−i2θ2i
=sin2θ
You can also prove it in an elegant way using plane geometry (with a unit circle):
we have use:
base angles of an isosceles triangle
exterior angle of a triangle equals the sum of two opposite angles
definitions of sine and cosine
Dean Rubine
Been doing high school math since high school, circa 1975
·
Author has 10.6K answers and 23.7M answer views
·
Updated 6y
Related
How can I simply prove that cos(π/2-x)=sinx?
cos(π2−x)+isin(π2−x)=ei(π/2−x)=eiπ/2e−ix=i(cosx−isinx)
=sinx+icosx
Equating respective real parts,
sinx=cos(π2−x)
Former Retired Teacher.
·
Author has 3.2K answers and 4M answer views
·
6y
sin(A-B)= sinAcosB-cosAsinB and sin π/2 = 1, and cos π/2 =0.
So sin(π/2 - x)= sin π/2cosx -cosπ/2sinx =cosx
Related questions
How can I simply prove that cos(π/2-x)=sinx?
What is the proof of sin x = cos (x+π/2)?
How can I prove (simply) that cos (π/2+x) =-sin x?
How can I prove (simply) that sin (π-x) =sin x?
How can I simply prove that cos(π-x)=-cosx?
How can I prove that
cos
x
sin
x
sin
x
cos
2
x
+
sin
3
x
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16022 | https://your-online.ru/en/molar-molecular-weight/CdCl2 | Molar and molecular weight of CdCl2
| Element | Total atoms | Atomic mass | Total mass |
--- --- |
| Cd (Cadmium) | 1 | 112.411 | 112.411 |
| Cl (Chlorine) | 2 | 35.453 | 70.906 |
| | | | 183.317 |
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16023 | http://astronomy.nmsu.edu/mchizek/105/LABS/EarthDensity.pdf | Name: Date: 9 Estimating the Earth’s Density 9.1 Introduction We know, based upon a variety of measurement methods, that the density of the Earth is 5.52 grams per cubic centimeter. [This value is equal to 5520 kilograms per cubic meter.
Your initial density estimate in Table 9.3 should be a value similar to this.] This density value clearly indicates that Earth is composed of a combination of rocky materials and metallic materials.
With this lab exercise, we will obtain some measurements, and use them to calculate our own estimate of the Earth’s density. Our observations will be relatively easy to obtain, but they will involve contacting someone in the Boulder, Colorado area (where the University of Colorado is located) to assist with our observations. We will then do some calculations to convert our measurements into a density estimate.
As we have discussed in class, and in previous labs this semester, we can calculate the density of an object (say, for instance, a planet, or more specifically, the Earth) by knowing that object’s mass and volume. It is a challenge, using equipment readily available to us, to determine the Earth’s mass and its volume directly. [There is no mass balance large enough upon which we can place the Earth, and if we could what would we have available to “balance” the Earth?] But we have through the course of this semester discussed physical processes which relate to mass. One such process is the gravitational attraction (force) one object exerts upon another.
The magnitude of the gravitational force between two objects depends upon both the masses of the two objects in question, as well as the distance separating the centers of the two objects. Thus, we can use some measure of the Earth’s gravitational attraction for an object upon its surface to ultimately determine the Earth’s mass. However, there is another piece of information that we require, and that is the distance from the Earth’s surface to its center: the Earth’s radius.
We will need to determine both the MASS of the Earth and the RADIUS of the Earth.
Since we will use the magnitude of Earth’s gravitational attraction to determine Earth’s mass, and since this magnitude depends upon the Earth’s radius, well first determine Earth’s circumference (which will lead us to the Earth’s radius and then to the Earth’s volume) and then determine the Earth’s mass.
9.2 Determining Earth’s Radius Earlier this semester you read (or should have read!) in your textbook the description of Eratosthenes’ method, implemented two-thousand plus years ago, to determine Earth’s circumference. Since the Earth’s circumference is related to its radius as: Circumference = 2 × π × RADIUS (with π = “pi” = 3.141592) 101 and the Earth’s volume is a function of its radius: VOLUME = (4/3) × π × RADIUS3 We will implement Eratosthenes’ circumference measurement method and end up with an estimate of the Earth’s radius.
Now, what measurements did Eratosthenes use to estimate Earth’s circumference? Er-atosthenes, knowing that Earth is spherical in shape, realized that the length of an object’s shadow would depend upon how far in latitude (north-or-south) the object was from being directly beneath the Sun. He measured the length of a shadow cast by a vertical post in Egypt at local noon on the day of the northern hemisphere summer solstice (June 20 or so).
He made a measurement at the point directly beneath the Sun (23.5 degrees North, at the Egyptian city Syene), and at a second location further north (Alexandria, Egypt). The two shadow lengths were not identical, and it is that difference in shadow length plus the knowl-edge of how far apart the the two posts were from each other (a few hundred kilometers), that permitted Eratosthenes to calculate his estimate of Earth’s circumference.
As we conduct this lab exercise we are not in Egypt, nor is today the seasonal date of the northern hemisphere summer solstice (which occurs in June), nor is it locally Noon (since our lab times do not overlap with Noon). But, nonetheless, we will forge ahead and estimate the Earth’s circumference, and from this we will estimate the Earth’s radius.
TASKS: • Take a post outside, into the sunlight, and measure the length of the post with the tape measure.
• Place one end of the post on the ground, and hold the post as vertical as possible.
• Using the tape measure provided, measure to the nearest 1/2 centimeter the length of the shadow cast by the post; this shadow length should be measured three times, by three separate individuals; record these shadow lengths in Table 9.1.
• You will be provided with the length of a post and its shadow measured simultaneously today in Boulder, Colorado.
• Proceed through the calculations described after Table 9.1, and write your answers in the appropriate locations in Table 9.1. (15 points) 9.3 ANGLE DETERMINATION: With a bit of trigonometry we can transform the height and shadow length you measured into an angle. As shown in Figure 9.1 there is a relationship between the length (of your shadow in this situation) and the height (of the shadow-casting pole in this situation), where: 102 Table 9.1: Angle Data Location Post Height Shadow Length Angle (cm) (cm) (Degrees) Las Cruces Shadow #1 Las Cruces Shadow #2 Las Cruces Shadow #3 Average Las Cruces Angle: Boulder, Colorado TANGENT of the ANGLE = far-side length/ near-side length Since you know the length of the post (the near-side length, which you have measured) and the length of the shadow (the far-side length, which you have also measured, three separate times), you can determine the shadow angle from your measurements, using the ATAN, or TAN−1 capability on your calculator (these functions will give you an angle if you provide the ratio of the height to length): ANGLE = ATAN (shadow length / post length) or ANGLE = TAN−1(shadow length / post length) Figure 9.1: The geometry of a vertical post sitting in sunlight.
Calculate the shadow angle for each of your three shadow-length measure-ments, and also for the Boulder, Colorado shadow-length measurement. Write these angle values in the appropriate locations in Table 9.1. Then calculate the average of the three Las Cruces shadow angles, and write the value on the “Average Las Cruces Angle” line.
The angles you have determined are: 1) an estimate of the angle (latitude) difference between Las Cruces and the latitude at which the Sun appears to be directly overhead (which is currently ∼12 degrees south of the equator since we are experiencing early northern autumn), and 2) the angle (latitude) difference between Boulder, Colorado and the latitude 103 at which the Sun appears to be directly overhead. The difference (Boulder angle minus Las Cruces angle) between these two angles is the angular (latitude) separation between Las Cruces and Boulder, Colorado.
We will now use this information and our knowledge of the actual distance (in kilometers) between Las Cruces’ latitude and Boulder’s latitude. This distance is: 857 kilometers north-south distance between Las Cruces and Boulder, Colorado In the same way that Eratosthenes used his measurements (just like those you have made today), we can now determine an estimate of the Earth’s circumference from: EARTH CIRCUMFERENCE (kilometers) = 857 kilometers × (360o)/(Boulder angle —Avg LC Angle) = 857 × [360o/( − )] = km (5 points) Using your calculated Boulder Shadow Angle and your Average Las Cruces Shadow Angle values, calculate the corresponding EARTH CIRCUMFERENCE value, and write it below: AVERAGE EARTH CIRCUMFERENCE = kilometers (5 points) The CIRCUMFERENCE value you have just calculated is related to the RADIUS via the equation: EARTH CIRCUMFERENCE = 2 × π × EARTH RADIUS which can be converted to RADIUS using: EARTH RADIUS = RE = EARTH CIRCUMFERENCE / (2 × π) For your calculated CIRCUMFERENCE, calculate that value of the Radius (in units of kilometers) in the appropriate location below: AVERAGE EARTH RADIUS VALUE = RE = kilometers (5 points) Convert this radius (RE) from kilometers to meters, and enter that value in Table 9.3. (Note we will use the radius in meters the rest of the lab.) You have now obtained one important piece of information (the radius of the Earth) needed for determining the density of Earth. We will, in a bit, use this radius value to calculate the Earth’s volume. Next, we will determine Earth’s mass, since we need to know both the Earth’s volume and its mass in order to be able to calculate the Earth’s density.
104 9.4 Determining the Earth’s Mass The gravitational acceleration (increase of speed with increase of time) that a dropped object experiences here at the Earth’s surface has a magnitude defined by the Equation (thanks to Sir Isaac Newton for working out this relationship!) shown below: Acceleration (meters per second per second) = G × ME/RE2 Where ME is the mass of the Earth in kilograms, RE is the radius of the Earth in units of meters, and the Gravitational Constant, G = 6.67 x 10−11 meters3/(kg-seconds2). You have obtained several estimates, and calculated an average value of RE, above. However, you currently have no estimate for ME. You can estimate the Earth’s mass from the measured acceleration of an object dropped here at the surface of Earth; you will now conduct such an exercise.
A falling object, as shown in Figure 9.2, increases its downward speed at the constant rate “X” (in units of meters per second per second). Thus, as you hold an object in your hand, its downward speed is zero meters per second. One second after you release the object, its downward speed has increased to X meters per second. After two seconds of falling, the dropped object has a speed of 2X meter per second, after 3 seconds its downward speed is 3X meters per second, and so on. So, if we could measure the speed of a falling object at some point in time after it is dropped, we could determine the object’s acceleration rate, and from this determine the Earth’s mass (since we know the Earth’s radius). However, it is difficult to measure the instantaneous speed of a dropped object.
Figure 9.2: The distance a dropped object will fall during a time interval t is proportional to t2. A dropped object speeds up as it falls, so it travels faster and faster and falls a greater distance as t increases.
We can, however, make a different measurement from which we can derive the dropped object’s acceleration, which will then permit us to calculate the Earth’s mass. As was pointed 105 out above, before being dropped the object’s downward speed is zero meters per second. One second after being dropped, the object’s downward speed is X meters per second. During this one-second interval, what was the object’s AVERAGE downward speed? Well, if it was zero to begin with, and X meters per second after falling for one second, its average fall speed during the one-second interval is: Average Fall speed during first second = (Zero + X) / 2 = X/2 meters per second, which is just the average of the initial (zero) and final (X) speeds.
At an average speed of X/2 meters per second during the first second, the distance traveled during that one second will be: (X/2) (meters per second) × 1 second = (X/2) meters, since: DISTANCE = AVERAGE SPEED × TIME = 1/2 × ACCELERATION x TIME2 So, if we measure the length of time required for a dropped object to fall a certain distance, we can calculate the object’s acceleration.
Tasks: • Using a stopwatch, measure the amount of time required for a dropped object (from the top of the Astronomy Building) to fall 9.0 meters (28.66 feet). Different members of your group should take turns making the fall-time measurements; write these fall time values for two “drops” in the appropriate location in Table 9.2. (10 points for a completed table) • Use the equation: Acceleration = [2.0 x Fall Distance] / [(Time to fall)2] and your measured Time to Fall values and the measured distance (9.0 meters) of Fall to determine the gravitational acceleration due to the Earth; write these acceleration values (in units of meters per second per second) in the proper locations in Table 9.2.
• Now, knowing the magnitude of the average acceleration that Earth’s gravity imposes upon a dropped object, we will now use the “Gravity” equation to get ME: Gravitational acceleration = G × ME/RE2 (where RE must be in meters!) By rearranging the Gravity equation to solve for ME, we can now make an estimate of the Earth’s mass: ME = Average Acceleration × [1000 × RE]2 / G = (5 points) [The factor of 1000 here converts your Average Radius determined in the previous section in units of kilometers to a radius in units of meters.] Write the value of ME (in kilograms) in Table 9.3 below.
106 Table 9.2: Time of Fall Data Time to Fall Fall Distance Acceleration Object Drop #1 9 meters Object Drop #2 9 meters Average = 9.5 Determining the Earth’s Density Now that we have estimates for the mass (ME) and radius (RE) of the Earth, we can easily calculate the density: Density = Mass/Volume. You will do this below.
Tasks: • Calculate the volume (VE) of the Earth given your determination of its radius (in meters!): VE = (4/3) × π × RE3 and write this value in the appropriate location in Table 9.3 below.
• Divide your value of ME (that you entered in Table 9.3) by your estimate of VE that you just calculated (also written in Table 9.3): the result will be your estimate of the Average Earth Density in units of kilograms per cubic meter. Write this value in the appropriate location in Table 9.3.
• Divide your AVERAGE ESTIMATE OF EARTH’S DENSITY value that you just calculated by the number 1000.0; the result will be your estimated Earth density value in units of grams per cubic centimeter (the unit in which most densities are tabulated).
Write this value in the appropriate location in Table 9.3.
Table 9.3: Data for the Earth Estimate of Earth’s Radius: m (7 points) Estimate of Earth’s Mass: kg (7 points) Estimate of Earth’s Volume: m3 (7 points) Estimate of Earth’s Density: kg/m3 (7 points) Density of the Earth: gm/cm3 (7 points) 107 9.6 IN-LAB QUESTIONS: 1. Is your calculated value of the Earth’s density GREATER THAN, or LESS THAN, or EQUAL TO the actual value (see the Introduction) of the Earth’s density? If your calculated density value is not identical to the known Earth density value, calculate the “percent error” of your calculated density value compared to the actual density value (5 points): PERCENT ERROR = 100% × (CALCULATED DENSITY −ACTUAL DENSITY) ACTUAL DENSITY = (9) 2. You used the AVERAGE Las Cruces shadow angle in calculating your estimate of the Earth’s density (which you wrote down in Table 9.3). If you had used the LARGEST of the three measured Las Cruces shadow angles shown in Table 9.1, would the Earth density value that you would calculate with the LARGEST Las Cruces shadow angle be larger than or smaller than the Earth density value you wrote in Table 9.3? Think before writing your answer!
Explain your answer. (7.5 points) 3. If the Las Cruces to Boulder, Colorado distance was actually 2000 km in length, but your measured fall times did not change from what you measured, would you have calculated a larger or smaller Earth density value? Explain the reasoning for your answer. (7.5 points) 4. If we had conducted this experiment on the Moon rather than here on the Earth, would your measured values (fall time, angles and angle difference between two locations separated 108 north-south by 857 kilometers) be the same as here on Earth, or different? Clearly explain your reasoning. [It might help if you draw a circle representing Earth and then draw a circle with 1/4th of the radius of the Earth’s circle to represent the Moon.] 109 110 Name: Earth Density Lab Ast 105 Take-Home Questions (35 points) 1. Using the post-height (length) and post and shadow lengths your Lab TA will provide to you at the end of the lab meeting, fill in the Table 9.4 below.
Table 9.4: Angle Data Location Post Height Shadow Length Angle (cm) (cm) (Degrees) Las Cruces Angle: Boulder, Colorado Angle Now re-calculate the following numbers: Earth Circumferance = kilometers Earth Radius = kilometers Answer Questions #2, #3, and #4 below in typed form on a separate sheet of paper, and then type your response to item #5. These will be handed in with the remainder of your lab materials.
2. List in an appropriate order the steps involved in determining the Earth’s density in this lab (this includes both measurements and calculations).
3. Using the “Earth Radius” value you calculated using the provided-by-your-TA Las Cruces and Boulder post and shadow lengths (see Table 9.4 above): Calculate a new estimate of Earth’s density using these values as your starting point.
How does your answer compare to your AVERAGE DENSITY OF EARTH value in Table 9.3 of the lab?
Do you believe this new density value is a worse, or better estimate than you calculated using the values measured during your lab? Why?
4. TYPE a 1.5-2 page LAB REPORT in which you will address the following topics: a) the estimated density value you arrived at was likely different from the actual Earth density value of 5.52 grams per cubic centimeter; describe 2 or 3 potential errors in your measurements that could possibly play a role in generating your incorrect estimated density value b) describe 2-3 ways in which you could improve the measurement techniques used in lab; keep in mind that NMSU is a state-supported school and thus we do not have infinite resources to purchase expensive sophisticated equipment, so your suggestions should not be too expensive 111 c) describe what you have learned from this lab, what aspects of the lab surprised you, what aspects of the lab worked just as you thought they would, etc.
112 |
16024 | https://ocw.mit.edu/courses/18-01sc-single-variable-calculus-fall-2010/resources/clip-2-chain-rule-revisited/ | Clip 2: Chain Rule, Revisited | Single Variable Calculus | Mathematics | MIT OpenCourseWare
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1. Differentiation
Part A: Definition and Basic Rules
Part B: Implicit Differentiation and Inverse Functions
Exam 1
2. Applications of Differentiation
Part A: Approximation and Curve Sketching
Part B: Optimization, Related Rates and Newton's Method
Part C: Mean Value Theorem, Antiderivatives and Differential Equa
Exam 2
3. The Definite Integral and its Applications
Part A: Definition of the Definite Integral and First Fundamental
Part B: Second Fundamental Theorem, Areas, Volumes
Part C: Average Value, Probability and Numerical Integration
Exam 3
4. Techniques of Integration
Part A: Trigonometric Powers, Trigonometric Substitution and Com
Part B: Partial Fractions, Integration by Parts, Arc Length, and
Part C: Parametric Equations and Polar Coordinates
Exam 4
5. Exploring the Infinite
Part A: L'Hospital's Rule and Improper Integrals
Part B: Taylor Series
Final Exam
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18.01SC | Fall 2010 | Undergraduate
Single Variable Calculus
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Syllabus
1. Differentiation
Part A: Definition and Basic Rules
Part B: Implicit Differentiation and Inverse Functions
Exam 1
2. Applications of Differentiation
Part A: Approximation and Curve Sketching
Part B: Optimization, Related Rates and Newton's Method
Part C: Mean Value Theorem, Antiderivatives and Differential Equa
Exam 2
3. The Definite Integral and its Applications
Part A: Definition of the Definite Integral and First Fundamental
Part B: Second Fundamental Theorem, Areas, Volumes
Part C: Average Value, Probability and Numerical Integration
Exam 3
4. Techniques of Integration
Part A: Trigonometric Powers, Trigonometric Substitution and Com
Part B: Partial Fractions, Integration by Parts, Arc Length, and
Part C: Parametric Equations and Polar Coordinates
Exam 4
5. Exploring the Infinite
Part A: L'Hospital's Rule and Improper Integrals
Part B: Taylor Series
Final Exam
Session 21: Review for Exam 1
Clip 2: Chain Rule, Revisited
» Accompanying Notes (PDF)
From Lecture 7 of 18.01 Single Variable Calculus, Fall 2006
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0:21 PROFESSOR: Right now, we're finishing up
0:25 with the first unit, and I'd like
0:27 to continue in this lecture, lecture seven,
0:31 with some final remarks about exponents.
0:45 So what I'd like to do is just review something
0:49 that I did quickly last time, and make
0:51 a few philosophical remarks about it.
0:53 I think that the steps involved were maybe a little tricky,
0:57 and so I'd like to go through it one more time.
1:00 Remember, that we were talking about this number a_k,
1:03 which is (1 + 1/k)^k.
1:08 And what we showed was that the limit as k
1:11 goes to infinity of a_k was e.
1:16 So the first thing that I'd like to do
1:19 is just explain the proof a little bit more clearly
1:22 than I did last time with fewer symbols,
1:28 or at least with this abbreviation of the symbol
1:31 here, to show you what we actually did.
1:35 So I'll just remind you of what we did last time,
1:41 and the first observation was to check,
1:45 rather than the limit of this function,
1:47 but to take the log first.
1:49 And this is typically what's done
1:51 when you have an exponential, when you have an exponent.
1:54 And what we found was that the limit here
1:57 was 1 as k goes to infinity.
2:03 So last time, this is what we did.
2:05 And I just wanted to be careful and show you
2:07 exactly what the next step is.
2:09 If you exponentiate this fact; you take e to this power,
2:14 that's going to tend to e^1, which is just e.
2:21 And then, we just observe that this is the same as a_k.
2:26 So the basic ingredient here is that e^ln a = a.
2:32 That's because the log function is the inverse
2:36 of the exponential function.
2:38 Yes, question?
2:38 STUDENT: [INAUDIBLE]
2:54 PROFESSOR: So the question was, wouldn't the log of this
2:58 be 0 because a_k is tending to 1.
3:01 But a_k isn't tending to 1.
3:03 Who said it was?
3:06 If you take the logarithm, which is what we did last time,
3:10 logarithm of a_k is indeed k ln(1 + 1/k).
3:17 That does not tend to 0.
3:18 This part of it tends to 0, and this part tends to infinity.
3:22 And they balance each other, 0 times infinity.
3:26 We don't really know yet from this expression,
3:28 in fact we did some cleverness with limits and derivatives,
3:32 to figure out this limit.
3:33 It was a very subtle thing.
3:34 It turned out to be 1.
3:37 All right?
3:38 Now, the thing that I'd like to say
3:40- I'm sorry I'm going to erase this aside here
3:43- but you need to go back to your notes
3:46 and remember that this is what we did last time.
3:48 Because I want to have room for the next
3:50 comment that I want to make on this little blackboard here.
3:54 What we just derived was this property here,
3:57 but I made a claim yesterday, and I just
4:01 want to emphasize it again so that we realized
4:04 what it is that we're doing.
4:07 I looked at this backwards.
4:08 One way you can think of this is we're evaluating this limit
4:11 and getting an answer.
4:13 But all equalities can be read both directions.
4:16 And we can write it the other way:
4:17 e equals the limit, as k goes to infinity, of this expression
4:25 here.
4:26 So that's just the same thing.
4:28 And if we read it backwards, what we're saying
4:30 is that this limit is a formula for e.
4:35 So this is very typical of mathematics.
4:38 You want to always reverse your perspective all the time.
4:40 Equations work both ways, and in this case,
4:43 we have two different things here.
4:46 This e was what we defined as the base,
4:49 which when you graph e^x, you get slope 1 at 0.
4:54 And then it turns out to be equal to this limit, which
4:56 we can calculate numerically.
4:59 If you do this on your calculators, you, of course,
5:02 will have a way of programming in this number
5:05 and evaluating it for each k.
5:07 And you'll have another button available to evaluate this one.
5:10 So another way of saying it is it
5:12 that there's a relationship between these two things.
5:14 And all of calculus is a matter of getting these relationships.
5:19 So we can look at these things in several different ways.
5:21 And indeed, that's what we're going
5:23 to be doing at least at the end of today
5:25 in talking about derivatives.
5:27 A lot of times when we talk about derivatives,
5:29 we're trying to look at them from several perspectives
5:32 at once.
5:34 Okay, so I have to keep on going with exponents,
5:37 because I have one loose end.
5:40 One loose end that I did not cover yet.
5:44 There's one very important formula that's left,
5:48 and it's the derivative of the powers.
5:51 We actually didn't do this - well we
5:54 did it for rational numbers r.
5:57 So this is the formula here.
6:00 But now we're going to check this for all real numbers, r.
6:06 So including all the irrational ones as well.
6:09 This is also good practice for using base e
6:13 and using logarithmic differentiation.
6:16 So let me do this by our two methods
6:20 that we can use to handle exponential type problems.
6:26 So method one was base e.
6:32 So if I just rewrite this base e again,
6:34 that's just this formula over here.
6:36 x^r = (e^ln x)^r, which is e^r ln x.
6:50 Okay, so now I can differentiate this.
6:55 So I get that d/dx (x^r), now I'm going to use prime
7:04 notation, because I don't want to keep on writing that d/dx
7:07 here; (e^(r ln x))'.
7:13 And now, what I can do is I can use the chain rule.
7:18 The chain rule says that it's the derivative
7:20 of this times the derivative of the function.
7:24 So the derivative of the exponential is just itself.
7:29 And the derivative of this guy here,
7:31 well I'll write it out once, is (r ln x)'.
7:39 So what's that equal to?
7:42 Well, e^(r ln x) is is just x^r.
7:45 And this derivative here is-- Well the derivative of r is 0.
7:53 This is a constant.
7:54 It just factors out.
7:56 And ln x now has derivative-- What's the derivative of ln x?
8:02 1/x, so this is going to be times r/x.
8:06 And now, we rewrite it in the customary form, which is r,
8:10 we put the r in front, x^(r-1).
8:13 Okay?
8:13 So I just derived the formula for you.
8:19 And it didn't matter whether r was rational or irrational,
8:23 it's the same proof.
8:25 Okay so now I have to show you how method two works as well.
8:29 So let's do method two, which we call
8:34 logarithmic differentiation.
8:39 And so here I'll use a symbol, say u, for x^r,
8:43 and I'll take its logarithm.
8:44 That's r ln x.
8:50 And now I differentiate it.
8:51 I'll leave that in the middle, because I
8:54 want to remember the key property
8:55 of logarithmic differentiation.
8:57 But first I'll differentiate it.
8:58 Later on, what I'm going to use is that this is the same
9:01 as u'/u.
9:02 This is one way of evaluating a logarithmic derivative.
9:06 And then the other is to differentiate
9:08 the explicit function that we have over here.
9:10 And that is just, as we said, r/x.
9:16 So now, I multiply through, and I get u' = ur/x which is just
9:25 x^r r/x, which is just what we did before.
9:29 It's r x^(r-1).
9:33 Again, you can now see by comparing
9:36 these two pieces of arithmetic that they're basically
9:40 the same.
9:41 Pretty much every time you convert to base e
9:43 or you do logarithmic differentiation,
9:45 it'll amount to the same thing, provided
9:46 you don't get mixed up.
9:48 You generally have to introduce a new symbol here.
9:51 On the other hand, you're dealing with exponents there.
9:55 It's worth it to know both points of view.
10:00 All right, so now I want to make one last remark before we
10:07 finish with exponents.
10:09 And, I'll try to sell this to you in a lot of ways
10:16 as the course goes on, but one thing that I
10:19 want to try to emphasize is that the natural logarithm really
10:23 is natural.
10:27 So, I claim that the natural log is natural.
10:39 And the example that we're going to use for this illustration
10:45 is economics.
10:53 Okay?
10:54 So let me explain to why the natural log is the one that's
10:58 natural for economics.
11:00 If you are imagining the price of a stock that you own
11:06 goes down by a dollar, that's a totally meaningless statement.
11:11 It depends on a lot of things.
11:13 In particular, it depends on whether the original price
11:15 was a dollar or 100 dollars.
11:18 So there's not much meaning to these absolute numbers.
11:22 It's always the ratios that matter.
11:25 So, for example, I just looked up an hour ago,
11:29 the London Exchange closed, and it was down 27.9,
11:42 which as I said, is pretty meaningless
11:44 unless you know what the actual total of this index is.
11:50 It turns out it was 6,432.
11:54 So the change in the price, divided
11:57 by the price, which in this case is 27.9 / 6,432,
12:03 is what matters.
12:07 And, in this case, it happens to be .43%.
12:11 All right?
12:12 That's what happened today.
12:14 And similarly, if you take the infinitesimal of this,
12:18 people think of days as being relatively small increments
12:21 when you're investing in a stock,
12:23 you would be interested in the infinitesimal sense,
12:27 you would be interested in p'/p.
12:28 The derivative of p divided by p.
12:33 That's just (ln p)'.
12:38 So this is the - let me put a little box around it
12:42- the formula of logarithmic differentiation.
12:45 But let me just emphasize that it has an actual significance,
12:49 and it's the one that's used by economists and people who
12:52 are modeling prices of things all the time.
12:54 They never use absolute prices when there are large swings.
12:58 They always use the log of the price.
13:01 And there's no point in using log base 10, or log base 2.
13:07 Those give you junk.
13:08 They give you an extra factor of log 2.
13:11 It's the natural log that's the obvious one to use.
13:14 It's completely straightforward that this
13:18 is a simpler expression than using log base 10
13:21 and having a factor of natural log of 10 there,
13:24 which would just mess everything up.
13:26 All right, so this is just one illustration.
13:29 Anything that has to do with ratios
13:31 is going to encounter logarithms.
13:36 All right, so that's pretty much it.
13:41 That's all I want to say for now anyway.
13:45 There's lots more to say, but we'll
13:47 be saying it when we do applications of derivatives
13:50 in the second unit.
13:51 So now, what I'd like to do is to start a review.
13:54 I'm just going to run through what we did in this unit.
13:57 I'll tell you approximately what I
13:59 expect from you on the test that's coming up tomorrow.
14:06 And, well, so let's get started with that.
14:14 So this is a review of Unit One.
14:27 And I'm just going to put on the board all of the things
14:32 that you need to think about, anyway, keep in your head.
14:35 And there are what are called general formulas
14:41 for derivatives.
14:45 And then there are the specific ones.
14:51 And let me just remind you what the general formulas are.
14:55 There's what you do to differentiate
14:58 a sum, a multiple of a function, the product
15:04 rule, the quotient rule.
15:08 Those are several general formulas.
15:11 And then there's one more, which is
15:13 the chain rule, which I'm going to say just
15:15 a little bit more about.
15:17 So the derivative of a function of a function
15:21 is the derivative of the function times the derivative
15:26 of the other function.
15:27 So here, I've abbreviated u is u(x).
15:33 Right, so this is one of two ways of writing it.
15:36 The other way is also one that you can keep in mind
15:39 and you might find easier to remember.
15:42 It's probably a good idea to remember both formulas.
15:46 And then the last type of general formula
15:49 that we did was implicit differentiation.
15:56 Okay?
15:59 So when you do implicit differentiation,
16:03 you have an equation and you don't
16:06 try to solve for the unknown function.
16:09 You just put it in its simplest form and you differentiate.
16:13 So, we actually did this, in particular, for inverses.
16:20 That was a very, very key method for calculating
16:23 the inverses of functions.
16:25 And it's also true that logarithmic differentiation
16:28 is of this type.
16:31 This is a transformation.
16:33 We're differentiating something else.
16:34 We're transforming the equation by taking its logarithm
16:37 and then differentiating.
16:40 Okay, so there are a number of different ways this is applied.
16:45 It can also be applied, anyway, these are two of them.
16:48 So maybe in parenthesis.
16:50 These are just examples.
16:53 All right.
16:54 I'll try to give examples of at least a few of these rules
16:59 later.
16:59 So now, the specific functions that you know how
17:05 to differentiate: well you know how to differentiate now x^r
17:08 thanks to what I just did.
17:11 We have the sine and the cosine functions,
17:15 which you're responsible for knowing
17:17 what their derivatives are.
17:19 And then other trig functions like tan and secant.
17:26 We generally don't bother with cosecants and cotangents,
17:29 because everything can be expressed in terms of these
17:32 anyway.
17:33 Actually, you can really express everything
17:35 in terms of sines and cosines.
17:36 But what you'll find is that it's
17:38 much more convenient to remember the derivatives of these
17:41 as well.
17:42 So memorize all of these.
17:45 All right, and then we had e^x and ln x.
17:49 And we had the inverses of the trig functions.
17:53 These were the two that we did: the arctangent and the arcsine.
18:00 So those are the ones you're responsible for.
18:02 You should have enough time, anyway, to work out anything
18:06 else, if you know these.
18:09 All right, so basically the idea is
18:11 you have a bunch of special formulas.
18:13 You have a bunch of general formulas.
18:14 You put them together, and you can
18:16 generate pretty much anything.
18:20 Okay, so let's do a few examples before going on
18:24 with the review.
18:41 Okay, so I do want to do a few examples in sort of increasing
18:48 level of difficulty in how you would
18:50 combine these things together.
18:51 So first of all, you should remember
18:55 that if you differentiate the secant function, that's just
19:00- oh I just realized that I wanted to say something
19:03 else before - so forget that.
19:06 We'll do that in a second.
19:08 I wanted to make some general remarks.
19:10 So there's one rule that you discussed in my absence, which
19:17 is the chain rule.
19:19 And I do want to make just a couple of remarks
19:21 about the chain rule now to remind you of what it is,
19:26 and also to present some consequences.
19:30 So, a little bit of extra on the chain rule.
19:39 The first thing that I want say is that we didn't really
19:43 fully explain why it's true.
19:46 And I do want to just explain it by example, okay?
19:54 So imagine that you have a function which is, say,
19:59 10x + b.
20:02 All right?
20:02 So y = 10x + b.
20:04 Then obviously, y is changing 10 times as fast as b, right?
20:09 The issue is this number here, dy/dx, is 10.
20:18 And now if x is a function of something,
20:20 say t, shifted by some other constant here, then dx/dt = 5.
20:34 Now all the chain rule is saying is that if y is going 10 times
20:38 as fast as t, I'm sorry as x, and x is going 5 times
20:44 as fast as t, then y is going 50 times as fast as t.
20:50 And algebraically, all this means is if I plug
20:53 in and substitute, which is what the composition of the two
20:57 functions amounts to, 10(5t + a) + b and I multiply it out,
21:04 I get 50t + 10a + b.
21:09 Now these terms don't matter.
21:11 The constant terms don't matter.
21:13 The rate is 50.
21:14 And so the consequence, if we put them together,
21:17 is that dy/dt = 105, which is 50.
21:30 All right, so this is in a nutshell
21:31 why the chain rule works.
21:33 And why these rates multiply.
21:39 The second thing that I wanted to say about the chain rule
21:42 is that it has a few consequences that
21:45 make some of the other rules a little easier
21:47 to remember or possibly to avoid.
21:50 The messiest rule in my humble opinion
21:54 is the quotient rule, which is kind of a nuisance to remember.
21:59 So let me just remind you, if you
22:01 take just the reciprocal of a function,
22:03 and you differentiate it, there's
22:05 another way of looking at this.
22:08 And it's actually the way that I use,
22:09 so I want to encourage you to think about it this way too.
22:12 This is the same as (v^(-1))'.
22:15.
22:16 And now instead of using the quotient rule, which
22:18 we could've used, we can use the chain rule here
22:23 with the power -1, which works by the power law.
22:29 So what is this equal to?
22:30 This is equal to -v^(-2) v'.
22:38 So here, I've applied the chain rule rather than
22:42 the quotient rule.
22:47 And similarly, suppose I wanted to derive the full quotient
22:54 rule.
22:54 Well, now this may or may not be easier.
22:57 But this is one way of remembering what's going on.
22:59 If you convert it to uv^(-1) and you differentiate that,
23:05 now I can use the product rule on this.
23:09 Of course, I have to use the chain rule and this rule
23:11 as well.
23:13 So what do I get?
23:15 I get u', the inverse, plus u, and then I have
23:21 to differentiate the v inverse.
23:22 That's the formula right up here.
23:24 That's -v^(-2) v'.
23:30 So that's one way of doing it.
23:33 This actually explains the funny minus sign
23:35 when you differentiate v in the formula.
23:38 The other formula, the other way that we did it,
23:41 was by putting this over a common denominator.
23:44 The common denominator was v^2.
23:49 This comes from this v v^(-2).
23:51 And then the second term is -uv'.
23:57 And the first term, we have to multiply by an extra factor
24:00 of v, because we have a v^2 in the denominator.
24:02 So it's u'v. All right, so this is the quotient rule as we
24:07 wrote it down in lecture, and this is just another way
24:11 of remembering it or deriving it without remembering it,
24:13 if you just remember the chain rule and the product rule.
24:16 Okay, so you'll find that in many contexts,
24:19 it's easier to do one or the other.
24:25 Okay, so now I'm ready to differentiate the secant
24:29 and a few such functions.
24:30 So we'll do some examples here here.
24:36 So here's the secant function, and I
24:39 want to use that formula up there for the reciprocal.
24:44 This is the way I think of it.
24:48 This is the cosine function to the power -1.
24:53 And so, the formula here is just what?
24:58 It's just -(cos x)^(-2) times -sin x.
25:20 So now this is usually written in a different fashion,
25:22 so that's why I'm doing this for a reason actually.
25:25 Which is although there are several formulas for things,
25:27 with trig functions, there are usually
25:29 five ways of writing something.
25:31 So I'm writing this one down so that you
25:33 know what the standard way of presenting it is.
25:36 So what happens here is that we have two minus signs
25:39 cancelling.
25:40 And we get sin x / cos^2 x.
25:44 That's a perfectly acceptable answer,
25:46 but there's a customary way in which is written.
25:49 It's written (1 / cos x) (sin x / cos x).
25:55 And then we get rid of the denominators
25:57 by rewriting it in terms of secant and tangent,
26:00 so sec x tan x.
26:04 So this is the form that's generally
26:07 used when you see these formulas written in textbooks.
26:11 And so you know, you need to watch out,
26:14 because if you ever want to use this kind of calculus,
26:16 you'll have not be put off by all the secants and tangents.
26:22 All right, so getting slightly more complicated,
26:26 how about if we differentiate ln(sec x)?
26:37 If you differentiate the natural log,
26:39 that's just going to be (sec x)' / sec x.
26:49 And plugging in the formula that we
26:51 had before, that's sec x tan x / sec x, which is tan x.
27:00 So this one also has a very nice form.
27:03 And you might say that this is kind of an ugly function,
27:07 but the strange thing is that the natural log was invented
27:14 before the exponential by a guy named Napier, exactly
27:19 in order to evaluate functions like this.
27:21 These are the functions that people cared about a lot,
27:25 because they were used in navigation.
27:28 You wanted to multiply sines and cosines together
27:32 to do navigation.
27:34 And the multiplication he encoded using a logarithm.
27:38 So these were invented long before people even
27:40 knew about exponents.
27:42 And it was a surprise, actually, that it
27:44 was connected to exponents.
27:46 So the natural log was invented before the log base 10
27:48 and everything else, exactly for this kind of purpose.
27:52 Anyway, so this is a nice function,
27:54 which was very important, so that your ships wouldn't
27:58 crash into the reef.
28:03 Okay, let's continue here.
28:05 So there's another kind of function
28:08 that I want to discuss with you.
28:10 And these are the kinds in which there's
28:12 a choice as to which of these rules to apply.
28:19 And I'll just give a couple of examples of that.
28:25 There usually is a better and a worse way,
28:27 so let me illustrate that.
28:38 Okay, yet another example.
28:41 I hope you've seen some of these before.
28:43 Say (x^10 + 8x)^6.
28:51 So it's a little bit more complicated than what
28:52 we had before, because there were several more symbols here.
29:00 So what should we do at this point?
29:03 There's one choice which I claim is a bad idea,
29:06 and that is to expand this out to the 6th power.
29:10 That's a bad idea, because it's very long.
29:13 And then your answer will also be very long.
29:15 It will fill the entire exam paper, for instance.
29:19 Yeah?
29:20 STUDENT: Can you use the chain rule?
29:21 PROFESSOR: Chain rule.
29:21 That's it.
29:22 We use the chain rule.
29:23 So fortunately, this is relatively easy
29:26 using the chain rule.
29:27 We just think of this box as being the function.
29:30 And we take 6 times this guy to the 5th,
29:34 times the derivative of this guy, which is 10x^9 + 8.
29:41 And this is, filling this in, it's x^10 + 8x.
29:43 And that's it.
29:46 That's all you need to do differentiate things like this.
29:50 The chain rule is very effective.
29:55 STUDENT: [INAUDIBLE]
29:59 PROFESSOR: That's a good question.
30:01 So I'm not really willing to answer too many questions
30:04 about what's going to be on the exam.
30:07 But the question that was just asked
30:09 is exactly the kind of question I'm very happy to answer.
30:13 Okay, the question was, in what form is-- what
30:18 form is an acceptable answer?
30:20 Now in real life, that is a really serious question.
30:23 When you ask a computer a question
30:25 and it gives you 500 million sheets of printout,
30:28 it's useless.
30:31 And you really care what form answers are in,
30:33 and indeed, somebody might really
30:35 care what this thing to the 6th power is,
30:39 and then you would be forced to discuss things in terms
30:42 of that other functional form.
30:46 For the purposes of this exam, this is okay form.
30:50 And, in fact, any correct form is an okay form.
30:54 I recommend strongly that you not try to simplify things
30:57 unless we tell you to.
30:59 Sometimes it will be to your advantage to simplify things.
31:04 Sometimes we'll say simplify.
31:08 It takes a good deal of experience
31:10 to know when it's really worth it to simplify expressions.
31:13 Yes?
31:13 STUDENT: [INAUDIBLE]
31:19 PROFESSOR: Right, so turning to this example.
31:23 The question is what is this derivative?
31:25 And here's an answer.
31:27 That's the end of the problem.
31:29 This is a more customary form.
31:31 But this is answer is okay.
31:37 Same issue.
31:38 That's exactly the point.
31:40 Yes?
31:41 STUDENT: [INAUDIBLE]
31:51 PROFESSOR: The question is, do you have to show the work?
31:59 Do you have to show the work?
32:00 Well if I ask you what is d/dx of sec x,
32:04 then if you wrote down this answer
32:06 or you wrote down this answer showing no work,
32:09 that would be acceptable.
32:11 If the question was derive the formula for this
32:15 from the formula for the derivative of the cosine
32:18 or something like that, then it would not be acceptable.
32:21 You'd have to carry out this arithmetic.
32:24 So, in other words, typically this
32:28 will come up, for instance, in various contexts.
32:32 You just basically have to follow directions.
32:34 Yes?
32:35 STUDENT: [INAUDIBLE]
32:41 PROFESSOR: The next question is, are you
32:43 expected to be able to prove what
32:44 the derivative of the sine function is?
32:46 The short answer to that is yes.
32:49 But I will be getting to that when I discuss
32:51 the rest of the material here.
32:54 We're almost there.
32:58 Okay, so let me just finish these examples
33:02 with one last one.
33:04 And then we'll talk about this question
33:06 of things like the derivative of the sine function,
33:10 and deriving it.
33:12 So the last example that I'd like to write down is the one
33:15 that I promised you in the first lecture,
33:18 namely to differentiate e^(x tan^(-1) x).
33:26 Basically you're supposed to be able to differentiate
33:28 any function.
33:29 So this is the one that we mentioned at the beginning.
33:32 So here it is.
33:34 Let's do it.
33:37 So what is it?
33:38 Well, it's just equal to - I have to differentiate.
33:45 I have to use the chain rule - it's
33:47 equal to the exponential times the derivative
33:52 of this expression here.
33:58 That's the chain rule.
33:59 That's the first step.
34:01 And now I have to apply the product rule here.
34:06 So I have e^(x tan^(-1) x).
34:10 And I differentiate the first factor, so I get tan^(-1) x.
34:15 Add to it what happens when I differentiate
34:17 the second factor, leaving alone the x.
34:19 So that's x / (1+x^2).
34:24 And that's it.
34:26 That's the end of the problem.
34:28 It wasn't that hard.
34:30 Of course, it requires you to remember all of the rules,
34:35 and a lot of formulas underlying them.
34:37 So that's consistent with what I just told you.
34:39 I told you that you wanted to know this.
34:42 I told you that you needed to know this product rule,
34:44 and that you needed to know the chain rule.
34:50 And I guess there was one more thing,
34:51 the derivative of e^x came into play there.
34:55 So of these formulas, we used four of them
34:59 in this one calculation.
35:03 Okay, so now what other things did we talk about in Unit One?
35:15 So the main other thing that we talked about
35:23 was the definition of a derivative.
35:33 And also there was sort of a goal
35:40 which was to get to the meaning of the derivative.
35:51 So these are things - so we had a couple of ways of looking
35:56 at it, or at least a couple that I'm
35:59 going to emphasize right now.
36:01 But first, let me remind you what the formula is.
36:06 The derivative is the limit as delta x goes to 0 of (f(x +
36:13 delta x) - f(x)) / delta x.
36:19 So that's it, and this is certainly
36:22 going to be a central focus here.
36:25 And you want to be able to recognize this formula
36:29 in a number of ways.
36:42 So, how do we use this?
36:44 Well one thing we did was we calculated a bunch
36:48 of these rates of change.
36:51 In fact, more or less, they're the ones which
36:53 are written right over here.
36:55 This list of functions here.
36:57 Now, which ones did we start out with just straight
37:01 from the definition here?
37:03 Which of these things?
37:04 There were a whole bunch of them.
37:06 So we started out with a function 1/x.
37:09 We did x^n.
37:11 We did sine x.
37:14 We did cosine x.
37:16 Now there was a little bit of subtlety with sine
37:19 x and cosine x.
37:21 We got them using something else.
37:25 We didn't quite get them all the way.
37:26 We got them using the case x = 0.
37:31 We got them from the derivative at x = 0,
37:34 we got the formulas for the derivatives of sine and cosine.
37:37 But that was an argument which involved plugging in sin
37:40(x + delta x), and running through.
37:44 So that's one example.
37:45 We also did a^x.
37:50 And that may be it.
37:53 Oh yeah, I think that's about it.
37:58 That may be about it.
38:00 No.
38:00 It isn't.
38:01 Okay, so let me make a connection here which you
38:03 probably haven't yet made, which is that we did it for (u v)'.
38:10 And we also did it for (u / v)'.
38:15 So sorry, I shouldn't write primes,
38:17 because that's not consistent with the claim there.
38:20 I differentiated the product; I differentiated the quotient
38:24 using the same delta x notation.
38:28 I guess I forgot that because I wasn't there when it happened.
38:32 So look, these are the ones that you do by this.
38:36 And, of course, you might have to reduce them to other things.
38:39 These involve using something else.
38:42 This one involves using the slope of this function at 0,
38:46 just the way the sine and the cosine did.
38:48 This one involves the slopes of the individual functions, u
38:52 and v. And this one also involves the individual--
38:54 So, in other words, it doesn't get you
38:56 all the way through to the end, but it's
38:58 expressed in terms of something simpler in each of these cases.
39:03 And I could go on.
39:05 We didn't do these in class, but you're certainly--
39:09 e^x is a perfectly okay one on one of the exams.
39:12 We ask you for 1/x^2.
39:14 In other words, I'm not claiming that it's
39:16 going to be one on this list, but it certainly
39:18 can be any one of these.
39:19 But we're not going to ask you to go
39:21 all the way through to the beginning in these formulas.
39:26 There are also some fundamental limits that I certainly
39:28 want you to know about.
39:31 And these you can derive in reverse.
39:34 So I will describe that now.
39:58 So let me also emphasize the following thing: I want
40:06 to read this backwards now.
40:18 This is the theme from the very beginning of this lecture.
40:21 Namely, if you're given the function f,
40:25 you can figure out its derivative by this formula
40:27 here.
40:28 That is the formula for this in terms of what's
40:29 on the right hand side.
40:30 On the other hand, you can also use
40:34 the formula in that direction, and if you
40:45 know the slope of something, you can figure out
40:48 what the limit is.
40:49 For example, I'll use the letter x here,
40:54 even though it's cheating.
40:56 Maybe I'll call it delta x so it's clearer to you.
40:59 Maybe I'll call it u.
41:06 Suppose you look at this limit here.
41:10 Well, I claim that you should recognize that is
41:14 the derivative with respect to u of the function e^u at u = 0,
41:19 which of course we know to be 1.
41:22 So this is reading this formula in reverse.
41:25 It's recognizing that one of these limits -
41:27 let me rewrite this again here - one of these so-called
41:35 difference quotient limits is a derivative.
41:39 And since we know a formula for that derivative,
41:42 we can evaluate it.
41:49 And lastly, there's one other type of thing
41:54 which I think you should know.
41:57 These are the ones you do with difference quotients.
41:59 There are also other formulas that you
42:01 want to be able to derive.
42:03 You want to be able to derive formulas
42:19 by implicit differentiation.
42:27 In other words, the basic idea is
42:30 to take whatever equation you've got
42:32 and simplify it as much as possible,
42:36 without insisting that you solve for y.
42:41 That's not necessarily the most appropriate way
42:44 to get the rate of change.
42:45 The much simpler formula is sin y = x.
42:51 And that one is easier to differentiate implicitly.
42:59 So I should say, do this kind of thing.
43:02 So that's, if you like, a typical derivation
43:05 that you might see.
43:08 And then there's one last type of problem that you'll face,
43:13 and it's the other thing that I claim we discussed.
43:21 And it goes all the way back to the first lecture.
43:26 So the last thing that we'll be talking about is tangent lines.
43:33 All right?
43:34 The geometric point of view of a derivative.
43:38 And we'll be doing more of this in next the unit.
43:41 So first of all, you'll be expected
43:44 to be able to compute the tangent line.
43:52 That's often fairly straightforward.
43:56 And the second thing is to graph y' ,
44:03 the derivative of a function.
44:07 And the third thing, which I'm going
44:09 to throw in here, because I regard it
44:11 in a sort of geometric vein, although it's got
44:14 an analytical aspect to it.
44:16 So this is a picture.
44:18 This is a computation.
44:20 And if you combine the two together,
44:23 you get something else.
44:24 And so this is to recognize differentiable functions.
44:37 Alright, so how do you do this?
44:40 Well, we really only have one way of doing it.
44:43 We're going to check the left and right tangents.
44:54 They must be equal.
44:59 So again, this is a property that you
45:04 should be familiar with from some of your exercises.
45:06 And the idea is simply, that if the tangent line exists,
45:09 it's the same from the right and from the left.
45:14 Okay, now I'm going to just do one example here
45:20 from this sort of qualitative sketching skill
45:25 to give you an example here.
45:27 And what I'm going to do is I'm going
45:28 to draw a graph of a function like this.
45:34 And what I want to do underneath is draw
45:38 the graph of the derivative.
45:41 So this is the function y = f(x),
45:45 and here I'm going to draw the graph of the function y =
45:48 f'(x) right underneath it.
45:56 So now, let's think about what it's supposed to look like.
46:00 And the one step that you need to make in order to do this,
46:05 is to draw a few tangent lines.
46:08 I'm just going to draw one down here.
46:13 And I'm going to draw one up here.
46:18 Now, the tangent lines here - notice
46:22 that the slope of these tangent lines are all positive.
46:27 So everything I draw down here is
46:28 going to be above the x-axis.
46:33 Furthermore, as I go further to the left,
46:36 they get steeper and steeper.
46:37 So they're getting higher and higher.
46:39 So the function is coming down like this.
46:44 It starts up there.
46:45 Maybe I'll draw it in green to illustrate the graph here.
46:50 So that's this function here.
46:56 As we get farther out, it's getting flatter and flatter.
46:59 So it's leveling off, but above the axis like that.
47:06 So one of the things to emphasize
47:08 is, you should not expect the derivative
47:10 to look like the function.
47:12 It's totally different.
47:13 It's keeping track at each point of its tangent line.
47:17 On the other hand, you should get some kind of physical feel
47:19 for it, and we'll be practicing this more in the next unit.
47:23 So let me give you an example of a function which
47:25 does exactly this.
47:27 And it's the function y = ln x.
47:33 If you differentiate it, you get y' = 1/x.
47:38 And this plot above is, roughly speaking, the logarithm.
47:44 And this plot underneath is the function 1/x.
47:50 We still have time for one question.
47:53 And so, fire away.
47:58 Yes?
48:03 STUDENT: [INAUDIBLE]
48:04 PROFESSOR: The question is, can you
48:05 show how you derive the inverse tangent of x.
48:09 So that's in a lecture.
48:13 I'm happy to do it right now, but it's going
48:17 to take me a whole two minutes.
48:20 So, here's how you do it. y = tan^(-1) x.
48:27 And now this is hopeless to differentiate,
48:30 so I rewrite it as tan y = x.
48:34 And now I have to differentiate that.
48:38 So when I differentiate it, I get
48:42 the derivative of tan y with respect
48:44 to x-- with respect to y.
48:46 That's 1 / (1 + y^2) times y'.
48:51 So this is a hard step.
48:52 That's the chain rule.
48:53 And on the left side I get 1.
48:55 So I'm doing this super fast because we
48:57 have thirty seconds left.
49:00 But this is the hard step right here.
49:02 And it needs for you to know that d/dy tan
49:04 y is equal to one over-- Oh, bad bad bad, secant squared.
49:14 I was ahead of myself so fast.
49:22 So here's the identity.
49:24 So you need have known this in advance.
49:28 And that's the input into this equation.
49:30 So now, what we have is that y' = 1 / sec^2 y y,
49:44 which is the same thing as cos^2 y.
49:51 Now, the last bit of the problem is
49:54 to rewrite this in terms of x.
49:57 And that you have to do with a right triangle.
50:02 If this is x and this is 1, then the angle
50:05 is y, because the tangent of y is x.
50:09 So this expresses the fact that the tangent of y is x.
50:14 And then the hypotenuse is the square root of 1 + x^2.
50:21 And so the cosine is 1 divided by that.
50:27 So this thing is 1 divided by the square root of 1 + x^2,
50:30 the quantity squared.
50:36 So, and then the last little bit here, since I'm racing along,
50:40 is that it's 1 / (1 + x^2), which I incorrectly wrote over
50:45 here.
50:46 Okay, so good luck on the test.
50:48 See you tomorrow.
Course Info
Instructor
Prof. David Jerison
Departments
Mathematics
As Taught In
Fall 2010
Level
Undergraduate
Topics
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Calculus
Differential Equations
Learning Resource Types
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notes Lecture Notes
theaters Lecture Videos
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theaters Problem-solving Videos
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16025 | https://www.youtube.com/watch?v=roVX_44a7OU | Transversal Two Column Proofs
Mr. Janes Math
673 subscribers
145 likes
Description
14779 views
Posted: 15 Oct 2014
Transversal Two Column Proofs
9 comments
Transcript:
Hey, this is Mr. James. In this video, we're going to be learning how to write two column proofs with transversals and the angles that are adjacent to a transversal. In the past, we've done some two column proofs, but those have largely centered around algebra and substitution and a little bit of geometry, but for these proofs, we're going to use 100% geometry and, you know, maybe a little bit of algebra, but mostly geometry here. So, let's just jump right in and try a few. So just like the other proofs, we have a given and and what we're trying to prove. So for this first one, given M is parallel to N. Let me mark that my diagram. All right. And we're trying to prove that angle one is congruent to angle five. So let me me just kind of mark those right there. I'm trying to prove that those two are congruent. Well, wait a second. They're they're corresponding angles, and I know that corresponding angles are always going to be congruent. So those are those two are kind of connected because they're they're corresponding angles. And that's all I need to do the proof. So let's go ahead and write that that two column proof. So remember in the beginning of any proof, we're going to write down what we're given. M is parallel to N. That's given. And then I can skip straight to saying that angle one is congruent to angle five. Why is that? Well, if you flip to your your note sheet, uh your theorem sheet, the the second side, and scroll down, here are all the different theorems that we can use in our proof. And well, I said they were corresponding. And so this says if the lines are parallel, and they are, then those two angles are congruent. And so since the lines are parallel then I can say that those two angles are congruent by the corresponding angle theorem. Pretty straightforward, right? So the first one's pretty straightforward. Uh let's take a look at the second one. All right, start the same way. Given M is parallel to N. Let's mark that in the diagram. All right. And prove angle one is congruent to angle five. Well, that's the same thing as before. So, let's let's mark that. Angle one is congruent to angle five. But there's a difference here. This says that you can't use corresponding angles. So, I can't do the same thing as before because I can't use corresponding angles. What are we going to do? Well, we can do something a little different here. We're going to use more than just one relationship between angles uh to do this. We're going to use two or maybe three. So, start by looking at this diagram up here and circling all the angles that are congruent to one and five. Did you do it? Three and seven are also congruent to one and five. They all have the same angle measure. And so, what I can do is instead of going straight from one to five, I can say, well, one is congruent to three because they're vertical angles. and three is congruent to five because they're alternate interior angles. And then since one's congruent to three, which is congruent to five, then one is congruent to five. Okay, and that's all we're going to do. Let's write our proof. Okay, remember we always start out with the parallel lines that are given. Remember, that's very important that those lines are parallel. Otherwise, we couldn't we couldn't do any of this. Um then angle one is congrent to angle three just like I said before because those are vertical angles remember right up here vertical angles. Uh and then I can say that angle three is congrent to angle five. Okay again it's because they're alterior. And now my last step is going to be to show that one is congrent to five. That this chain all fits together. Now what do we call that? Well, if you go back to your your theorem sheet right here, it says transitive property. If A is congruent to B and B is congruent to C, then A is congruent to C. That's kind of like what we have here. We have we have uh angle one is congrent to angle three, which is congruent to angle five. So, we can shorten that and just say angle one is congrent to angle five. And that's called using the transitive property. So the last step on our proof will be the transitive property. There's that end of proof. Now I have a feeling that you can do the third one by yourself. So try and pause the video here and do it by yourself. All right, let's see what you've done. So hopefully you should have started out with M is parallel to N. That's given. That's how we start all of our proofs. But next we have to prove that angle 3 plus angle 6 equals 180°. So, let's take a look at those up here. Here's angle three. There's angle six. Well, those look like same side interior angles. So, if we look back at our sheet here, they're same side interior angles. If the lines are parallel, then they add together to equal 180. That's exactly what we're going to do here. Okay. And so, our last step is going to be that they add up to equal 180 because they're same side interior angles. Perfect. This next one and the last one on this page is kind of like the last one we just did because we need to prove that angle 3 plus angle 6 equals 180°. However, it's kind of like the second one we did because we have a restriction. We can't use same side interior just as like in the second one. We couldn't use corresponding angles. So, we're going to have to find another way around that. And maybe we'll have to use more than one step here. But let's mark our diagram and come back to the proof in a second. So let's see here. We've got that m is parallel to n and that angle 3 plus angle six as up to 180°. And please don't mark these the same way, three and six, because they're not congruent. Maybe put one mark on three, another mark on six. Maybe use two different colors, but don't mark them the same way. They're not congruent. H. So I can't use same side interior angles. What am I going to do? Well, let's start the same way we started the last one. Let's start by writing down um let's start by writing down the angles that are congruent to three. So, three, one is congrent to three, five is congruent to three, and seven is congrent to three. Those are all congruent the same angle measure. So, what if I said, okay, well, three is congruent to seven because they're corresponding angles. And what if I said that seven and six were supplementary because well, that's a linear pair. Okay, and that would work. I'm not going straight from three to six. I'm using seven as kind of a bridge in between uh three and six. I think we could do that. Let's go write the proof now. So, of course, you start your proof off every single time with your givens. Next, we're going to write that first step that three is congruent to seven, right? And why is that? That's because they're they're corresponding. Okay? And next, we're going to write that seven and six added together equal 180°. And why is that? It's because it's because they're a linear pair. Okay, linear pair. And then since three is congruent to seven, what I can do is just take this three and substitute it in for seven. And now I've got that angle 3 plus angle 6 is 180°. Now, I wouldn't call this transitive because I'm not doing the the thing up here where I'm chaining one to three to to to five. Um, but I am substituting that three in for the seven to get my final answer here. So, we call that substitution property. And that ends this proof. Now that you've kind of got the fundamentals down, let's kick it up a notch. Look at the back page. We've got this crazy quadruple transversal thing going on here that's kind of hard to work with, but we're going to get through it and I'll show you some tips. Let's hop right in and try the first proof. Given M is parallel to N and P is parallel to Q. All right, so let's mark that M is parallel to N and P is parallel to Q. Remember to use different markings to show that P and Q go together and M and N go together. And I've got to prove got to prove that angle 13 is congruent to angle two. So angle 13 is congruent to angle two. All right. Well, this is kind of hard to see. Uh let's let's talk about some tips for how to work with this big picture here. Right. So here's a bigger picture of our quadruple transversal mess. Uh and I've got 13 and two um marked. And so what's the relationship between 13 and two? Well, if it's hard to see, start by maybe rotating your picture around like this. And then take another piece of paper or your hand and cover up any intersections that you're not using. So, don't worry about any of this right now. Don't worry about that. And now we can when we're looking at just the this picture right here, we can see that hey, 13 and two are on the exterior and that they're on alternate sides. So 13 and two are alternate exterior angles. You can use this technique any time throughout this uh this video to help you see things a little more clearly. So now that I know that 13 and two are linked because they are alternate exterior angles, I can go ahead and write my proof. Again, I start with the givens and then I end with angle 13 is congruent to angle two. Why? because they're alternate exterior angles. Perfect. Let's take a look at the second one on the back page. Given M is parallel to N is P is parallel to Q. Okay, that's like all of them. We need to prove that 13 is congruent to angle five. So, let's go mark our diagram. So, parallel parallel and I want to show that angle 13 is congruent to angle five. Oh, those are kind of far apart. How am I going to do that? Well, there's no one reason that will connect them. So, we're going to have to use uh another angle kind of as a bridge in between them. So, just like on the first page, let's start circling things that are congruent to 13. Um well, 11 is congruent to to 13 and 11's congruent to five. Okay. Well, why is why is 13 and 11 congruent? Well, those are those are alternate interior angles. That's what those are. Alternate interior. And why is 11 congruent to five? Those are actually corresponding angles. And if you're having trouble seeing it, I'll show you uh using the trick I showed you before. So again, if you're having a little trouble, kind of cover up the the lines using look at 1311. See how they are on the interior of those two lines? and then alternate sides of that transversal. Alter interior to look at 11 and five. It might be useful to rotate your your picture a little bit and cover up the intersections you're not using. Okay. So, just look at those two. And now we can kind of see that hey just on oops sorry we can see that just on that transversal that 11 and five are corresponding they're both in the kind of uh bottom right corner of both of those intersections and it makes it a little easier if you just cover up the intersections you're not looking at. All right let's go ahead and write the proof. We always start with the the givens. Okay. And then we said that angle 13 was congrent to angle 11. Why was that? Oh, that's because of alter interior. That's right. Those are alterior angles. Uh and then we said that angle 11 was congrent to angle five. And that's cuz they are corresponding angles. All right. And then we've got to put it all together and say that angle 13 since it's congrent to 11 and 11 is congruent to five. That 13 is congruent to five. And what's that called when we kind of chain them all together like that? That's the the transitive property. Now, you might be thinking, Mr. James, is that the only way to do the proof? And it's not. You can find any way you can uh to link together 13 and five to show they're congruent. You could have said that 13 is congruent to two because they are um alternate exterior. And then you could have said that two is congruent to five because they're corresponding. And your proof doesn't have to have just two steps to it. Two or three steps. You could have said, uh, 13's congruent to 15 because they're vertical angles. Then 15's congruent to three because they're alternate interior angles. Then you could have said that three is congruent to two because they're vertical angles again. And then two is congrent to five because they're corresponding angles. So any number of steps is fine as long as there's a log logical progression from 13 to five. Now for our last proof. If you think you can do it yourself, pause the video and try it. Otherwise, here we go. Let's start by marking the diagram. I need to show that 14 and seven are supplementary. Again, don't mark them using the same color or the same number of uh of arcs there to show that they're not congruent. They're supplementary. How am I going to do that, though? Well, I need somewhere in between. Hm. Well, I know that 14 and 9 are congruent because they're both alternate exterior angles. And then I know that 9 and 7 are supplementary because they are uh same side interior angles. So, I bet I can write a proof out of that. And again, if you don't see the alternate exterior and the same side interior, make sure you're turning your paper and covering things up until you are able to see those relationships. Right? So, here are my statements and reasons from above. Remember, to get what we're trying to prove at the very end, we had to substitute 14 in for 9 and use the substitution property. Q D |
16026 | https://medium.com/@nayla.khaleel202/constrained-optimization-a-basic-introduction-c9cec7075d1d | Sitemap
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Constrained Optimization – A basic Introduction
Naila Ansari
7 min readApr 21, 2023
Welcome to my blog! In this blog we gonna discuss constrained optimization, it’s usage, it’s need and fields in which it is required plus how to tackle with constrained optimization problem. Probably, somehow and somewhere, every field in this world uses constrained optimization in direct or indirect way. And at the end of this blog, you will gain all the basic knowledge about ‘Constrained Optimization’.
As we all know that in business as well as in real world problems, sometimes there are a lot of feasible solutions present or sometimes there are limited resources and we have to choose in between them. So, in these cases the real challenge is to find the best feasible or optimal solution that is beneficial regarding every aspect. So this is what Constrained Optimization is used for.
What is Constrained Optimization?
Constrained optimization refers to the process of optimizing an objective function subject to one or more constraints. In other words, it involves finding the maximum or minimum value of a function while satisfying a set of constraints on the values that the function can take.
Constraints can take many forms, such as limits on the values that variables can take, relationships between variables, or requirements that the solution meet certain criteria.
Solving a constrained optimization problem can be more challenging than solving an unconstrained optimization problem because the solution must satisfy the constraints in addition to optimizing the objective function.
Now that we have gained the idea of Constrained Optimization, let us discuss the fields in which it is used often.
Fields in which Constrained Optimization is used:
Constrained optimization is used in a wide range of fields, including:
Engineering: In engineering, constrained optimization is used to optimize designs while satisfying constraints such as cost, weight, and size.
Economics: In economics, constrained optimization is used to optimize resource allocation and to model the behavior of individuals and firms.
Finance: In finance, constrained optimization is used to optimize investment portfolios while taking into account risk and return.
Operations research: In operations research, constrained optimization is used to optimize processes and to solve complex problems in logistics, transportation, and supply chain management.
Environmental science: In environmental science, constrained optimization is used to optimize resource management and to minimize the impact of human activities on the environment.
Machine learning: In machine learning, constrained optimization is used to train models while satisfying constraints such as regularization and sparsity.
Physics: In physics, constrained optimization is used to optimize the performance of devices such as lasers and sensors while satisfying constraints such as power consumption and noise.
These are just a few examples of the many fields where constrained optimization is used. It is a powerful tool for solving complex problems where there are constraints on the solution.
The question arises here that:
Why do we do Constrained Optimization?
Constrained optimization is done because in many real-world problems, there are constraints that need to be taken into account when making decisions. These constraints can arise due to physical, economic, social, or other reasons. By incorporating these constraints into the optimization problem, we can find solutions that are feasible and satisfy the constraints while still optimizing the objective function.
For Example:
• In engineering design, there are often constraints on the size, weight, cost, and performance of a system. By using constrained optimization, we can design systems that meet these constraints while still optimizing the performance of the system.
In finance, there are constraints on the amount of capital available, the risk tolerance of the investor, and the regulations that govern the investment. By using constrained optimization, we can construct investment portfolios that meet these constraints while still optimizing the return on investment.
In operations research, there are constraints on the availability of resources, the capacity of facilities, and the demands of customers. By using constrained optimization, we can plan and schedule operations that meet these constraints while still optimizing the efficiency of the operations.
In summary, this approach can be used in many fields to solve complex problems where there are limits on resources or other factors that must be taken into account when making decisions.
Suppose that in case, you have to solve a constrained optimization problem. So what necessary information you must have? Let’s discuss that too.
How to tackle Constrained Optimization problems?
There are several approaches to tackling constrained optimization problems:
1. Analytical methods:
Some constrained optimization problems can be solved analytically by using calculus and linear algebra. For example, Lagrange multipliers can be used to find the optimum of a function subject to one or more constraints.
2. Linear programming:
Linear programming is a widely used method for solving constrained optimization problems. It involves optimizing a linear objective function subject to linear constraints. Linear programming is often used in operations research, finance, and other fields.
3. Nonlinear programming:
Nonlinear programming is a method for solving constrained optimization problems where the objective function or constraints are nonlinear. Nonlinear programming is more complex than linear programming but can handle more complex problems.
4. Genetic algorithms:
Genetic algorithms are a type of optimization algorithm inspired by the process of natural selection. They can be used to solve complex optimization problems, including constrained optimization problems.
5. Particle swarm optimization:
Particle swarm optimization is a population-based optimization algorithm inspired by the behavior of bird flocks and fish schools. It can be used to solve constrained optimization problems and has been applied in many fields.
6. Simulated annealing:
Simulated annealing is a stochastic optimization algorithm inspired by the annealing process in metallurgy. It can be used to solve complex optimization problems, including constrained optimization problems.
The choice of method depends on the specific problem at hand, the available computational resources, and the desired level of accuracy. In practice, a combination of methods may be used to solve a constrained optimization problem.
Sometimes the problem is complex or there is lack of time and we can’t solve it manually, so in these cases, we optimize and solve constrained optimization problems on online apps, which are discussed below
Online Platforms to perform Constrained Optimization:
There are many online platforms that can be used to solve constrained optimization problems. Here are a few popular ones:
1. MATLAB Optimization Toolbox:
MATLAB is a powerful programming language that is widely used for scientific computing. The Optimization Toolbox in MATLAB provides a set of tools for solving optimization problems, including constrained optimization problems.
2. WolframAlpha:
WolframAlpha is a computational knowledge engine that can solve a wide range of mathematical problems, including constrained optimization problems. You can enter your problem directly into the search box and get a solution.
3. Excel Solver:
Microsoft Excel has a built-in optimization tool called Solver that can solve constrained optimization problems. You can use Solver by selecting the "Solver" option from the "Data" tab in Excel.
4. Gurobi Optimization:
Gurobi Optimization is a commercial optimization software package that can solve a wide range of optimization problems, including constrained optimization problems. They offer a free academic license for students and researchers.
5. NEOS Server:
The NEOS Server is a free online service for solving optimization problems. It provides a wide range of solvers for different types of optimization problems, including constrained optimization problems. You can submit your problem to the NEOS Server and get a solution.
These are just a few examples of online platforms that can be used for constrained optimization. There are many other options available, depending on your specific needs and constraints.
Summary:
So now you are aware of what the ‘Constrained Optimization’ actually is, it’s the process of optimizing an objective function subject to one or more constraints. We have discussed that constraints can take many forms and that constrained optimization is used in a wide range of fields. You are now well aware of it’s need, how it is tackled manually and numerically and analytically. But surely, there is always much more to know! So in my next blog, I will relate ‘Constrained Optimization’ with Linear Algebra.
So, stay tuned for more!
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## Written by Naila Ansari
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I love to write! Let's discover together that how actually mathematics governs the business and computer world🤝🏻
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16027 | https://emedicine.medscape.com/article/262063-medication | No Results
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Placenta Previa Medication
Medication Summary
No medication is of specific benefit to a patient with placenta previa. Tocolysis may be cautiously considered in some circumstances in order to administer antenatal corticosteroids. A review article concluded that there is no improvement in perinatal outcome with prolonged tocolytics, and tocolysis beyond 48 hours is not clinically indicated.
Encourage patients with known placenta previa to maintain intake of iron and folate as a safety margin in the event of bleeding.
There have been studies that report using prothrombin complex and recombinant factor VII to control hemorrhage associated with obstetric complications and placenta previa.
Tocolytics
Class Summary
Tocolytic agents prevent preterm labor or contractions. Some specialists advocate tocolytics to promote the time for expectant management of symptomatic placenta previa. They should only be used after consultation with an obstetrician.
Magnesium sulfate
Magnesium sulfate
Magnesium sulfate is a nutritional supplement in hyperalimentation. It is a cofactor in enzyme systems involved in neurochemical transmission and muscular excitability. In adults, 60-180 mEq of potassium, 10-30 mEq of magnesium, and 10-40 mEq of phosphate per day may be necessary for optimum metabolic response.
Administer magnesium sulfate intravenously (IV) or intramuscularly (IM) for seizure prophylaxis in preeclampsia. Use the IV route for a quicker onset of action in true eclampsia. Discontinue treatment as soon as the desired effect is obtained. Repeat the doses, depending on the continuing presence of a patellar reflex and adequate respiratory function.
Corticosteroids
Class Summary
Steroids may be administered after consultation with a gynecologist, if vaginal bleeding is mild and intermittent, if the patient is not in labor, and if gestation is less than 37 weeks.
Dexamethasone acetate (Baycadron)
Dexamethasone acetate (Baycadron)
Dexamethasone may be given to promote the development of the lungs in the fetus.
Betamethasone (Celestone)
Betamethasone (Celestone)
Betamethasone helps to promote fetal lung maturity.
Adrenergic Agonists
Class Summary
These agents have sympathomimetic vasopressor activity.
Terbutaline (Brethine)
Terbutaline (Brethine)
Terbutaline acts directly on beta2-receptors to relax uterine contractions.
References
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Milosevic J, Lilic V, Tasic M, Radovic-Janosevic D, Stefanovic M, Antic V. [Placental complications after a previous cesarean section]. Med Pregl. 2009 May-Jun. 62(5-6):212-6. [QxMD MEDLINE Link].
Ananth CV, Smulian JC, Vintzileos AM. The effect of placenta previa on neonatal mortality: a population-based study in the United States, 1989 through 1997. Am J Obstet Gynecol. 2003 May. 188(5):1299-304. [QxMD MEDLINE Link].
Iyasu S, Saftlas AK, Rowley DL, Koonin LM, Lawson HW, Atrash HK. The epidemiology of placenta previa in the United States, 1979 through 1987. Am J Obstet Gynecol. 1993 May. 168(5):1424-9. [QxMD MEDLINE Link].
Kim LH, Caughey AB, Laguardia JC, Escobar GJ. Racial and ethnic differences in the prevalence of placenta previa. J Perinatol. 2012 Apr. 32 (4):260-4. [QxMD MEDLINE Link].
Williams MA, Mittendorf R. Increasing maternal age as a determinant of placenta previa. More important than increasing parity?. J Reprod Med. 1993 Jun. 38(6):425-8. [QxMD MEDLINE Link].
Ananth CV, Wilcox AJ, Savitz DA, Bowes WA Jr, Luther ER. Effect of maternal age and parity on the risk of uteroplacental bleeding disorders in pregnancy. Obstet Gynecol. 1996 Oct. 88(4 Pt 1):511-6. [QxMD MEDLINE Link].
Becker RH, Vonk R, Mende BC, Ragosch V, Entezami M. The relevance of placental location at 20-23 gestational weeks for prediction of placenta previa at delivery: evaluation of 8650 cases. Ultrasound Obstet Gynecol. 2001 Jun. 17(6):496-501. [QxMD MEDLINE Link].
Hill LM, DiNofrio DM, Chenevey P. Transvaginal sonographic evaluation of first-trimester placenta previa. Ultrasound Obstet Gynecol. 1995 May. 5(5):301-3. [QxMD MEDLINE Link].
Wexler P, Gottesfeld KR. Early diagnosis of placenta previa. Obstet Gynecol. 1979 Aug. 54(2):231-4. [QxMD MEDLINE Link].
Jansen CHJR, van Dijk CE, Kleinrouweler CE, et al. Risk of preterm birth for placenta previa or low-lying placenta and possible preventive interventions: A systematic review and meta-analysis. Front Endocrinol (Lausanne). 2022. 13:921220. [QxMD MEDLINE Link]. [Full Text].
Zaki ZM, Bahar AM, Ali ME, Albar HA, Gerais MA. Risk factors and morbidity in patients with placenta previa accreta compared to placenta previa non-accreta. Acta Obstet Gynecol Scand. 1998 Apr. 77(4):391-4. [QxMD MEDLINE Link].
Frederiksen MC, Glassenberg R, Stika CS. Placenta previa: a 22-year analysis. Am J Obstet Gynecol. 1999 Jun. 180(6 pt 1):1432-7. [QxMD MEDLINE Link].
Zlatnik MG, Cheng YW, Norton ME, Thiet MP, Caughey AB. Placenta previa and the risk of preterm delivery. J Matern Fetal Neonatal Med. 2007 Oct. 20(10):719-23. [QxMD MEDLINE Link].
Creasy RK, Resnik R, Iams J , Lockwood C, Moore T, Greene M. Placenta previa, placenta accreta, abruptio placentae, and vasa previa. Creasy and Resnik's Maternal-Fetal Medicine: Principles and Practice. 7th ed. Saunders: Philadelphia, PA; 2014. 732-742.
Bose DA, Assel BG, Hill JB, Chauhan SP. Maintenance tocolytics for preterm symptomatic placenta previa: a review. Am J Perinatol. 2011 Jan. 28(1):45-50. [QxMD MEDLINE Link].
Dola CP, Garite TJ, Dowling DD, Friend D, Ahdoot D, Asrat T. Placenta previa: does its type affect pregnancy outcome?. Am J Perinatol. 2003 Oct. 20(7):353-60. [QxMD MEDLINE Link].
Leerentveld RA, Gilberts EC, Arnold MJ, Wladimiroff JW. Accuracy and safety of transvaginal sonographic placental localization. Obstet Gynecol. 1990 Nov. 76(5 Pt 1):759-62. [QxMD MEDLINE Link].
Sherman SJ, Carlson DE, Platt LD, Medearis AL. Transvaginal ultrasound: does it help in the diagnosis of placenta previa?. Ultrasound Obstet Gynecol. 1992 Jul 1. 2(4):256-60. [QxMD MEDLINE Link].
Ghi T, Contro E, Martina T, Piva M, Morandi R, Orsini LF, et al. Cervical length and risk of antepartum bleeding in women with complete placenta previa. Ultrasound. Obstet Gynecol. Feb 2009. 33(2):209-12.
Smith RS, Lauria MR, Comstock CH, et al. Transvaginal ultrasonography for all placentas that appear to be low-lying or over the internal cervical os. Ultrasound Obstet Gynecol. 1997 Jan. 9(1):22-4. [QxMD MEDLINE Link].
Morlando M, Sarno L, Napolitano R, et al. Placenta accreta: incidence and risk factors in an area with a particularly high rate of cesarean section. Acta Obstet Gynecol Scand. 2013 Apr. 92(4):457-60. [QxMD MEDLINE Link].
Warshak CR, Eskander R, Hull AD, et al. Accuracy of ultrasonography and magnetic resonance imaging in the diagnosis of placenta accreta. Obstet Gynecol. 2006 Sep. 108(3 Pt 1):573-81. [QxMD MEDLINE Link].
Ueno Y, Kitajima K, Kawakami F, Maeda T, Suenaga Y, Takahashi S, et al. Novel MRI finding for diagnosis of invasive placenta praevia: evaluation of findings for 65 patients using clinical and histopathological correlations. Eur Radiol. 2014 Apr. 24(4):881-8. [QxMD MEDLINE Link].
Allen BC, Leyendecker JR. Placental evaluation with magnetic resonance. Radiol Clin North Am. 2013 Nov. 51(6):955-66. [QxMD MEDLINE Link].
Silver, R. Abnormal placentation: Placenta previa, vasa previa, and placenta accreta. Obstet Gynecolol. 2015. 126:654-68.
Soyama H, Miyamoto M, Sasa H, Ishibashi H, Yoshida M, Nakatsuka M, et al. Effect of routine rapid insertion of Bakri balloon tamponade on reducing hemorrhage from placenta previa during and after cesarean section. Arch Gynecol Obstet. 2017 Jun 24. [QxMD MEDLINE Link].
Besinger RE, Moniak CW, Paskiewicz LS, Fisher SG, Tomich PG. The effect of tocolytic use in the management of symptomatic placenta previa. Am J Obstet Gynecol. 1995 Jun. 172(6):1770-5; discussion 1775-8. [QxMD MEDLINE Link].
Bhide A, Prefumo F, Moore J, Hollis B, Thilaganathan B. Placental edge to internal os distance in the late third trimester and mode of delivery in placenta praevia. BJOG. 2003 Sep. 110(9):860-4. [QxMD MEDLINE Link].
Vergani P, Ornaghi S, Pozzi I, Beretta P, Russo FM, Follesa I, et al. Placenta previa: distance to internal os and mode of delivery. Am J Obstet Gynecol. 2009 Sep. 201(3):266.e1-5. [QxMD MEDLINE Link].
Blackwell, SC. Timing of delivery for women with stable placenta previa. Semin Perinatol. 2011. 35:249-51.
Machado LS. Emergency peripartum hysterectomy: Incidence, indications, risk factors and outcome. N Am J Med Sci. 2011 Aug. 3(8):358-61. [QxMD MEDLINE Link]. [Full Text].
Choi SJ, Song SE, Jung KL, Oh SY, Kim JH, Roh CR. Antepartum risk factors associated with peripartum cesarean hysterectomy in women with placenta previa. Am J Perinatol. Jan 2008. 25(1):37-41.
Masamoto H, Uehara H, Gibo M, Okubo E, Sakumoto K, Aoki Y. Elective use of aortic balloon occlusion in cesarean hysterectomy for placenta previa percreta. Gynecol Obstet Invest. 2009. 67(2):92-5.
Jain V, Bos H, Bujold E. Guideline No. 402: Diagnosis and Management of Placenta Previa. J Obstet Gynaecol Can. 2020 Jul. 42 (7):906-17.e1. [QxMD MEDLINE Link].
Gagnon R, Morin L, Bly S, et al. Guidelines for the management of vasa previa. J Obstet Gynaecol Can. 2009 Aug. 31(8):748-60. [QxMD MEDLINE Link].
Tables
| | |
--- |
| Morbidities | Relative Risk |
| Antepartum bleeding | 10 |
| Need for hysterectomy | 33 |
| Blood transfusion | 10 |
| Septicemia | 5.5 |
| Thrombophlebitis | 5 |
| Endometritis | 6.6 |
Morbidities
Relative Risk
Antepartum bleeding
10
Need for hysterectomy
33
Blood transfusion
10
Septicemia
5.5
Thrombophlebitis
5
Endometritis
6.6
Contributor Information and Disclosures
Ronan Bakker, MD Maternal-Fetal Medicine Specialist, The Perinatal Center
Ronan Bakker, MD is a member of the following medical societies: American College of Obstetricians and Gynecologists, American Medical Association
Disclosure: Nothing to disclose.
Ronald M Ramus, MD Professor of Obstetrics and Gynecology, Director, Division of Maternal-Fetal Medicine, Virginia Commonwealth University School of Medicine
Ronald M Ramus, MD is a member of the following medical societies: American College of Obstetricians and Gynecologists, American Institute of Ultrasound in Medicine, Medical Society of Virginia, Society for Maternal-Fetal Medicine
Disclosure: Nothing to disclose.
John G Pierce, Jr, MD Chairman of Women’s Health and Medical Specialties, Liberty University College of Osteopathic Medicine; Obstetrician/Gynecologist, Women’s Health of Central Virginia
John G Pierce, Jr, MD is a member of the following medical societies: American College of Obstetricians and Gynecologists, Association of Professors of Gynecology and Obstetrics, Christian Medical and Dental Associations, Medical Society of Virginia, Society of Laparoscopic and Robotic Surgeons
Disclosure: Nothing to disclose.
Carl V Smith, MD The Distinguished Chris J and Marie A Olson Chair of Obstetrics and Gynecology, Professor, Department of Obstetrics and Gynecology, Senior Associate Dean for Clinical Affairs, University of Nebraska Medical Center
Carl V Smith, MD is a member of the following medical societies: American College of Obstetricians and Gynecologists, American Institute of Ultrasound in Medicine, Association of Professors of Gynecology and Obstetrics, Central Association of Obstetricians and Gynecologists, Society for Maternal-Fetal Medicine, Council of University Chairs of Obstetrics and Gynecology, Nebraska Medical Association
Disclosure: Nothing to disclose.
Saju Joy, MD, MS Associate Director, Division Chief of Maternal-Fetal Medicine, Department of Obstetrics and Gynecology, Carolinas Medical Center
Saju Joy, MD, MS is a member of the following medical societies: American College of Obstetricians and Gynecologists, American Institute of Ultrasound in Medicine, Society for Maternal-Fetal Medicine, American Medical Association
Disclosure: Nothing to disclose.
Matthew M Finneran, MD Resident Physician, Department of Obstetrics and Gynecology, Carolinas Healthcare System
Disclosure: Nothing to disclose.
Pamela L Dyne, MD Professor of Clinical Medicine/Emergency Medicine, University of California, Los Angeles, David Geffen School of Medicine; Attending Physician, Department of Emergency Medicine, Olive View-UCLA Medical Center
Pamela L Dyne, MD is a member of the following medical societies: American Academy of Emergency Medicine, American College of Emergency Physicians, and Society for Academic Emergency Medicine
Disclosure: Nothing to disclose.
Patrick Ko, MD Clinical Assistant Professor, Department of Emergency Medicine, New York University Medical School; Assistant Program Director, Department of Emergency Medicine, North Shore University Hospital
Patrick Ko, MD is a member of the following medical societies: American College of Emergency Physicians and Society for Academic Emergency Medicine
Disclosure: Nothing to disclose.
Deborah Lyon, MD Director, Division of Gynecology, Associate Professor, Department of Obstetrics and Gynecology, University of Florida Health Science Center at Jacksonville
Deborah Lyon, MD is a member of the following medical societies: American College of Obstetricians and Gynecologists, Association of American Medical Colleges, Association of Professors of Gynecology and Obstetrics, and Florida Medical Association
Disclosure: Nothing to disclose.
John G Pierce Jr, MD Associate Professor, Departments of Obstetrics/Gynecology and Internal Medicine, Medical College of Virginia at Virginia Commonwealth University
John G Pierce Jr, MD is a member of the following medical societies: American College of Obstetricians and Gynecologists, Association of Professors of Gynecology and Obstetrics, Christian Medical & Dental Society, Medical Society of Virginia, and Society of Laparoendoscopic Surgeons
Disclosure: Nothing to disclose.
Joseph J Sachter, MD, FACEP Consulting Staff, Department of Emergency Medicine, Muhlenberg Regional Medical Center
Joseph J Sachter, MD, FACEP is a member of the following medical societies: American Academy of Emergency Medicine, American College of Emergency Physicians, American College of Physician Executives, American Medical Association, and Society for Academic Emergency Medicine
Disclosure: Nothing to disclose.
Ryan A Stone, MD Fellow, Department of Obstetrics and Gynecology, Maternal-Fetal Medicine Section, Wake Forest University Health Sciences
Ryan A Stone, MD is a member of the following medical societies: Academic Pediatric Association, American College of Obstetricians and Gynecologists, American Medical Association, and Society for Maternal-Fetal Medicine
Disclosure: Nothing to disclose.
Francisco Talavera, PharmD, PhD Adjunct Assistant Professor, University of Nebraska Medical Center College of Pharmacy; Editor-in-Chief, Medscape Drug Reference
Disclosure: Medscape Salary Employment
Lorene Temming, MD Resident Physician, Department of Obstetrics and Gynecology, Carolinas Medical Center
Lorene Temming, MD is a member of the following medical societies: American College of Obstetricians and Gynecologists and North Carolina Medical Society
Disclosure: Nothing to disclose.
Young Yoon, MD Associate Director, Assistant Professor, Department of Emergency Medicine, Mount Sinai Medical Center
Young Yoon, MD is a member of the following medical societies: Society for Academic Emergency Medicine
Disclosure: Nothing to disclose.
Mark Zwanger, MD, MBA Assistant Professor, Department of Emergency Medicine, Jefferson Medical College of Thomas Jefferson University
Mark Zwanger, MD, MBA is a member of the following medical societies: American Academy of Emergency Medicine, American College of Emergency Physicians, and American Medical Association
Disclosure: Nothing to disclose.
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16028 | https://www.gauthmath.com/solution/Bbz8qL7zx_R/Area-bounded-by-curve-y-k-s-i-n-x-between-x-and-x-2-is2-k-s-q-u-n-i-t-0-k-2-2-s- | Solved: Area bounded by curve y = k s i n x between x = π and x = 2 π [Math]
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Image 1: Area bounded by curve y = k s i n x between x = π and x = 2 π , is2 k s q . u n i t
0
k 2 2 s q . u n i t
k s q . u n i t
A
0
B
2 k s q . u n i t
C
k 2 2 s q . u n i t
D
k s q . u n i t
Area bounded by curve y = k s i n x between x = π and x = 2 π , is2 k s q . u n i t 0 k 2 2 s q . u n i t k s q . u n i t A 0 B 2 k s q . u n i t C k 2 2 s q . u n i t D k s q . u n i t
Show transcript
Expert Verified Solution
Answer
The correct option is B, which states that the required area is $$2k$$2 k square units.
Explanation
To determine the area bounded by the curve $$y = k \sin x$$y=k sin x between $$x = \pi$$x=π and $$x = 2\pi$$x=2 π, we follow these steps:
Set up the integral for the area. The area $$A$$A under the curve from $$x = \pi$$x=π to $$x = 2\pi$$x=2 π can be expressed as:
$$A = k \int_{\pi}^{2\pi} \sin x \, dx$$A=k∫π 2 πsin x d x
Evaluate the integral. The integral of $$\sin x$$sin x is $$-\cos x$$−cos x. Therefore, we compute:
$$A = k \left[ -\cos x \right]_{\pi}^{2\pi}$$A=k[−cos x]π 2 π
Substitute the limits into the integral. We calculate $$-\cos(2\pi)$$−cos(2 π) and $$-\cos(\pi)$$−cos(π):
$$A = k \left[ -\cos(2\pi) - (-\cos(\pi)) \right]$$A=k[−cos(2 π)−(−cos(π))]
Since $$\cos(2\pi) = 1$$cos(2 π)=1 and $$\cos(\pi) = -1$$cos(π)=−1, we have:
$$A = k \left[ -1 - (1) \right] = k \left[ -1 + 1 \right] = k \left[ -1 + 1 \right] = k \left[ 2 \right]$$A=k[−1−(1)]=k[−1+1]=k[−1+1]=k
Simplify the expression. Thus, the area becomes:
$$A = 2k$$A=2 k
Therefore, the area bounded by the curve $$y = k \sin x$$y=k sin x between $$x = \pi$$x=π and $$x = 2\pi$$x=2 π is $$2k$$2 k square units.
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16029 | https://hinative.com/questions/5441185 | Quality Point(s): 0
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IN mathematics, there often appears the phrase "not drawn to scale". what does it mean?
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It means the diagram shown is not exact to the measurements given. For example, if it's something like a square or rectangle that shows that one side is about 3 cm but it says 'not drawn to scale', it means that that side is not actually 3cm/drawn accurately.
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It's not just that it isn't the same size. It's that the proportions are not necessarily the same. So just because one side in the drawing is twice as long as the other side you can't assume that that is the actual ratio
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16030 | https://math.stackexchange.com/questions/654460/equilateral-triangle-and-120-angle | Skip to main content
Equilateral triangle and 120° angle
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P is the angle's vertex. With A,B we form an equilateral triangle on the same direction the angle's rays open up.
How can I prove the triangle's third vertex C is always on the bisector's line?
geometry
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edited Jan 28, 2014 at 13:17
DonAntonio
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asked Jan 28, 2014 at 11:44
user124432user124432
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I agree with David plus: how are the lines AP,BP constructed? Or better: what is point P ?
– DonAntonio
Commented
Jan 28, 2014 at 11:50
Yes, but A and B can be anywhere on those lines.
– user124432
Commented
Jan 28, 2014 at 11:50
120° angle is given, A and B points of the triangle is on the legs of the angle, the question is: C is always on the angle bisector?
– user124432
Commented
Jan 28, 2014 at 11:54
Ooooh, that's waay another way to correctly write the question, @user124432. You should probably add that P is the angle's vertex and that with A,B we form an equilateral triangle on the same direction the angle's rays open up, Now, prove the triangle's third vertex C is always on the bisector's line
– DonAntonio
Commented
Jan 28, 2014 at 11:55
1
Thanks @DonAntonio, I've corrected my question.
– user124432
Commented
Jan 28, 2014 at 12:02
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As pointed out by some comments, the problem has been originally misstated. It should be: we are given a 120 degrees angle with vertex P and an equilateral triangle ABC inscribed in it, such that the vertices A and B lie on the sides of the angle and C is in its interior, as in the drawing. Show that the line PC bisects the angle APB (i.e., that the angles APC and CPB are both 60 degrees).
Proof. Consider the quadrangle ACBP. It is cocyclic (its vertices lie on the same circle) since the sum of the angles at its opposite vertices P and C is 180 degrees. Thus the angles APC and CPB are equal, since they are inscribed angles in the same circle that are facing equal segments. QED
From the proof you can see that it is enough to assume that ABC is isosceles, i.e. AC=BC, and that the sum of the angles and P and C is 180 degrees.
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edited Jan 29, 2014 at 23:27
answered Jan 28, 2014 at 12:14
Gil BorGil Bor
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You seem to understand my question. Can you give me more details?
– user124432
Commented
Jan 28, 2014 at 12:51
1
I don't get it: of course ∠APC=∠BPC It is given that they both equal 60∘ ! Did I miss something here?
– DonAntonio
Commented
Jan 28, 2014 at 13:10
And anyway, how does that help to solve the problem? Oh, well: the answer's been accepted so who cares...
– DonAntonio
Commented
Jan 28, 2014 at 13:17
There seem to have been a confusion about the statement of this nice problem. I have now added to my answer a restatement. See if it's clearer now.
– Gil Bor
Commented
Jan 28, 2014 at 17:50
@DonAntonio: The diagram shows the bisector of angle APB, not (or only in hindsight) the line PC. The problem is to prove that C lies on the said bisector.
– TonyK
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Jan 28, 2014 at 18:09
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Draw two perpendicular lines from C to PA and PB, and the intersecting points are D and E, as shown below:
What we need to prove is that △CDA≌△CEB, therefore CD=CE, and since any point on an angle bisector has equal distance to the two side of the angle, we finish proving that C is on the angle bisector of ∠APB.
It is easy to get that CA=CB, ∠CDA=∠CEB=90∘. We need to get one more condition, which is ∠CAD=∠CBE.
Apparently, ∠CAB=∠CPB=60∘, therefore C、A、P、B are concyclic. Thus, ∠ACP=∠ABP. We can see that ∠CAD=∠CPA+∠ACP=60∘+∠ABP=∠CBA+∠ABP=∠CBE.
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answered Aug 18, 2014 at 13:29
qsmyqsmy
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Call M to the intersection point of PC,AB , and observe that
ΔAPM∼ΔBPMby s.a.s., since applying the angle bisector theorem we get
AMMB=APPB,on the triangleΔAPB
This means PC⊥AB (why?) , so both ΔCBP,ΔCAP are right triangles (and, in fact, golden ones 30−60−90).
Now just remember that an angle's bisector is the locus of all points on the plane that are equidistintant from both angle's rays.
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answered Jan 28, 2014 at 12:04
DonAntonioDonAntonio
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PC⊥AB??? What if A≈P? Perhaps I'm missing something...
– David Mitra
Commented
Jan 28, 2014 at 12:32
@DavidMitra, I don't know what A≅P means in this geometric setup.
– DonAntonio
Commented
Jan 28, 2014 at 12:33
Anyway, PC⊥AB follows from ΔAPM∼ΔBPM⟹∠AMP=∠BMP=90∘, since ∠AMB=180∘; by construction.
– DonAntonio
Commented
Jan 28, 2014 at 12:35
What if the point A is close to (or even equal to) the point P? The angle between AB and PC will then be approximately (equal to) 60∘.
– David Mitra
Commented
Jan 28, 2014 at 12:35
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Apparently, △APM need not be similar to △BPM. What are the two congruent sides referred to in your answer? It seems you're assuming AP is congruent to BP; but I don't think that was given.
– David Mitra
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condition for pair of straight line equation
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While determining the condition for the pair of straight line equation
a x 2+2 h x y+b y 2+2 g x+2 f y+c=0 a x 2+2 h x y+b y 2+2 g x+2 f y+c=0
i.e a x 2+2(h y+g)x+(b y 2+2 f y+c)=0 a x 2+2(h y+g)x+(b y 2+2 f y+c)=0
x=−2(h y+g)2 a±(h y+g)2−a(b y 2+2 f y+c)−−−−−−−−−−−−−−−−−−−−−−−√a x=−2(h y+g)2 a±(h y+g)2−a(b y 2+2 f y+c)a
x=−2(h y+g)2 a±(h 2−a b)y 2+2(h g−a f)y+(g 2−a c)−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√a x=−2(h y+g)2 a±(h 2−a b)y 2+2(h g−a f)y+(g 2−a c)a
The terms inside square root need to be a perfect square. I understand this. What I do not understand is when the inside square root terms ,quadratic in y is taken to be zero. Because of which its determinant 4(h g−a f)2−4(h 2−a b)(g 2−a c)=0 4(h g−a f)2−4(h 2−a b)(g 2−a c)=0 becomes the condition for the pair of straight line equation.
I am stuck here. Can someone please help. Thanks.
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asked Nov 14, 2019 at 17:12
Rajesh MarndiRajesh Marndi
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If the condition
a x 2+2 h x y+b y 2+2 g x+2 f y+c=0 a x 2+2 h x y+b y 2+2 g x+2 f y+c=0
represents the product of two lines then the set of solutions for this conditions should be one solution point, void or infinite solution points associated to the cases in which we have the two lines intersection, two lines parallel and two lines coincident. With this idea we follow obtaining
x=g h−a f±2(g+h y)2−a(b y 2+2 f y+c)−−−−−−−−−−−−−−−−−−−−−−−√2 a x=g h−a f±2(g+h y)2−a(b y 2+2 f y+c)2 a
If the intersection point is unique then the condition is
(g+h y)2−a(c+2 f y+b y 2)=0(g+h y)2−a(c+2 f y+b y 2)=0
which gives
y=g h−a f±a 2 f 2+a b g 2−2 a f g h+a c h 2−a 2 b c−−−−−−−−−−−−−−−−−−−−−−−−−−−−√a b−h 2 y=g h−a f±a 2 f 2+a b g 2−2 a f g h+a c h 2−a 2 b c a b−h 2
but again if the intersection point is unique we should have
a 2 f 2+a b g 2−2 a f g h+a c h 2−a 2 b c=0 a 2 f 2+a b g 2−2 a f g h+a c h 2−a 2 b c=0
or
c=a f 2+b g 2−2 f g h a b−h 2 c=a f 2+b g 2−2 f g h a b−h 2
The condition for distinct parallel lines follow as
a b−h 2=0 a b−h 2=0
Another approach:
Assuming a≠0 a≠0 and dividing a x 2+2 h x y+b y 2+2 g x+2 f y+c=0 a x 2+2 h x y+b y 2+2 g x+2 f y+c=0 by a a we have
x 2+b′y 2+c′+2 f′y+2 g′x+2 h′x y=(x+c 1 y+c 2)(x+d 1 y+d 2)x 2+b′y 2+c′+2 f′y+2 g′x+2 h′x y=(x+c 1 y+c 2)(x+d 1 y+d 2)
after equating coefficients and solving for c 1,c 2,d 1,d 2 c 1,c 2,d 1,d 2 we have the conditions
{h′2−b′>0 c′=−f′2−b′g′2−2 f′g′h′h′2−b′{h′2−b′>0 c′=−f′2−b′g′2−2 f′g′h′h′2−b′
such that the two lines equivalence is feasible.
or equivalently
{h 2−a b>0 c=a f 2+b g 2−2 f g h a b−h 2{h 2−a b>0 c=a f 2+b g 2−2 f g h a b−h 2
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edited Nov 15, 2019 at 13:11
answered Nov 14, 2019 at 19:10
CesareoCesareo
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Could you specify if it's necesssary to use c=a f 2+b g 2−2 f g h a b−h 2 c=a f 2+b g 2−2 f g h a b−h 2 to conclude that for there to not be a meeting point, c has to be undefined and thus h 2−a b=0 h 2−a b=0? That can be deducted straight from y=g h−a f±a 2 f 2+a b g 2−2 a f g h+a c h 2−a 2 b c√a b−h 2 y=g h−a f±a 2 f 2+a b g 2−2 a f g h+a c h 2−a 2 b c a b−h 2, (as in, there wouldn't be a defined y-coordinate for a meeting point if h 2=a b h 2=a b), and I was wondering why it was derived from looking at the expression for c, since this seems more direct and intuitive. Great answer, anyway!harry –harry 2021-08-10 01:50:22 +00:00 Commented Aug 10, 2021 at 1:50
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For A≥0,A≥0,
A x 2+B x+C=A(x+B/2 A)2+C−B 2/4 A A x 2+B x+C=A(x+B/2 A)2+C−B 2/4 A
will be perfect square for all real values of x x iff
C−B 2/4 A=0 C−B 2/4 A=0
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answered Nov 14, 2019 at 17:29
lab bhattacharjeelab bhattacharjee
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If I am not wrong , A x 2+B x+C=(A−−√x+B A√2)2−(B 2−4 A C 4 A)A x 2+B x+C=(A x+B A 2)2−(B 2−4 A C 4 A). Hence L.H.S is always a perfect square whenever B 2−4 A C=0 B 2−4 A C=0.Rajesh Marndi –Rajesh Marndi 2019-11-15 03:29:57 +00:00 Commented Nov 15, 2019 at 3:29
@Rajesh, I have consciously kept A A outside the square.lab bhattacharjee –lab bhattacharjee 2019-11-15 04:07:20 +00:00 Commented Nov 15, 2019 at 4:07
Thanks for the reply. R.H.S=A(x+B 2 A)2−(B 2−4 A C 4 A)=0 A(x+B 2 A)2−(B 2−4 A C 4 A)=0 , why is that only if the 2nd term is removed, it will be a perfect square. That is, it is not necessary B 2−4 A C=0 B 2−4 A C=0, will make it a perfect square, for e.g 4 2−7=3 2,6 2−11=5 2,23 2−45=22 2 4 2−7=3 2,6 2−11=5 2,23 2−45=22 2 and so on....Rajesh Marndi –Rajesh Marndi 2019-11-15 13:38:58 +00:00 Commented Nov 15, 2019 at 13:38
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The equation (ax+by+cz)(dx+ey+fz)=0 when multiplied out is a homogeneous quadratic in x, y, z, and therefore it is the equation of a conic.
May suppose then that the quadratic is the conic whose equation is a x 2+b y 2+c z 2+2 f y z+2 g z x+2 h x y=o a x 2+b y 2+c z 2+2 f y z+2 g z x+2 h x y=o.
The condition for the conic to be two straight lines (assumed distinct) is that the determinant of coefficients vanishes viz ∣∣∣∣a h g h b f g f c∣∣∣∣=0|a h g h b f g f c|=0
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edited Nov 27, 2019 at 4:39
answered Nov 26, 2019 at 10:05
Trevor FrancisTrevor Francis
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Hi, welcome to MSE! In the future, using MathJax to format your posts can make them easier to read, and thus appreciate. :)jgon –jgon 2019-11-26 23:27:04 +00:00 Commented Nov 26, 2019 at 23:27
Thank you; have amended as suggested Trevor Francis –Trevor Francis 2019-11-27 05:55:25 +00:00 Commented Nov 27, 2019 at 5:55
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1Prove that x 2+y 2−1 x 2+y 2−1 can't be a product of two linear equations
0When does 2 h x y+2 g x+2 f y+c=0 2 h x y+2 g x+2 f y+c=0, with h≠0 h≠0, represent a pair of straight lines?
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. 1986 Sep;77(3):625-31.
doi: 10.1093/jnci/77.3.625.
Differential expression of neural isozymes by human medulloblastomas and gliomas and neuroectodermal cell lines
P M Zeltzer,S L Schneider,P J Marangos,M H Zweig
PMID: 3462406
DOI: 10.1093/jnci/77.3.625
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Differential expression of neural isozymes by human medulloblastomas and gliomas and neuroectodermal cell lines
P M Zeltzer et al. J Natl Cancer Inst.1986 Sep.
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. 1986 Sep;77(3):625-31.
doi: 10.1093/jnci/77.3.625.
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P M Zeltzer,S L Schneider,P J Marangos,M H Zweig
PMID: 3462406
DOI: 10.1093/jnci/77.3.625
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For the determination of their possible utility as tumors markers, 2 neural-associated isozymes, neuron-specific enolase [(NSE) EC 4.2.1.11] and creatine kinase BB [(CK-BB) EC 2.7.3.2], were quantitated by radioimmunoassay in human neuroectodermal-derived cell lines, primary brain tumors, and sera and cerebrospinal fluid (CSF) from brain tumor patients. The NSE content of neuroblastoma cell lines was more than sixfold that of the glioma and medulloblastoma lines; the CK-BB content of neuroblastoma and medulloblastoma lines was fourfold to nineteen-fold that of the glioma and other lines. Expression of NSE in neuroblastoma cell lines was not related to time in culture and was cell line specific. NSE in ex vivo medulloblastomas was raised six to ten times that in astrocytomas and gliomas, although no significant differences were noted for the CK-BB content. Serum and CSF NSE levels were markedly raised above control values in 10 of 29 and 6 of 10 cases of astrocytoma, respectively. Raised CK-BB levels in serum (greater than 10 ng/ml) and CSF (greater than 12 ng/ml) were found in 9 of 18 and 2 of 10 patients, respectively. These data suggest that NSE is preferentially expressed by neuroblastoma lines and medulloblastomas and that NSE and CK-BB may have clinical utility as markers for prognosis, diagnosis, and monitoring of response to therapy.
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The diagnostic and prognostic value of pretreatment serum creatine kinase BB levels in patients with neuroblastoma.Ishiguro Y, Kato K, Akatsuka H, Ito T.Ishiguro Y, et al.Cancer. 1990 May 1;65(9):2014-9. doi: 10.1002/1097-0142(19900501)65:9<2014::aid-cncr2820650922>3.0.co;2-s.Cancer. 1990.PMID: 2372768
[The value of neuron specific enolase (NSE) in patients with brain tumors].Taomoto K, Kokunai T, Okuda T, Tamaki N, Matsumoto S, Hashimoto N.Taomoto K, et al.No To Shinkei. 1987 Feb;39(2):169-73.No To Shinkei. 1987.PMID: 3828153 Japanese.
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[Neuron-specific enolase].Kato K.Kato K.Tanpakushitsu Kakusan Koso. 1984 Nov;29(12 Suppl):1101-16.Tanpakushitsu Kakusan Koso. 1984.PMID: 6396721 Review.Japanese.No abstract available.
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16034 | https://byjus.com/physics/value-of-g-on-moon/ | The acceleration experienced by a freely falling object due to the gravitational force of the massive body is called acceleration due to gravity and is represented by g measured using SI unit m/s2. The value of g depends on the mass of the massive body and its radius and its value varies from one body to another. The value of g on the moon is constant.
| |
| Table of Contents Acceleration Due to Gravity on Moon Calculate Acceleration due to Gravity on Moon Frequently Asked Questions – FAQs |
Acceleration due to gravity on moon
The acceleration due to gravity on moon or the value of g on moon is 1.625 m/s2.
Calculate acceleration due to gravity on moon
The acceleration due to gravity formula is given by
G is the universal gravitational constant, G = 6.674×10-11m3kg-1s-2.
M is the mass of the massive body measured using kg.
R is the radius of the massive body measured using m.
g is the acceleration due to gravity measured using m/s2.
The mass of the moon is 7.35×1022kg.
The radius of the moon is 1.74×106m
Substituting the values in the formula we get-
Thus, the value of g on moon is g=1.625 m/s2.
The acceleration due to gravity also follows the unit of acceleration
Hope you got to know the value of g on the moon along with acceleration due to the gravity formula, definition, calculation, and SI units.
Physics Related Topics:
| |
| Relation Between G And g |
| Relation Between Escape Velocity And Orbital Velocity |
| Gravity Waves |
| Mass Energy Equivalence |
Watch the full summary of the chapter Gravitation Class 9
6,121
Frequently Asked Questions – FAQs
Q1
What does the value 9.8 m/s2 for acceleration due to gravity imply?
The value 9.8 m/s2 for acceleration due to gravity implies that for a freely falling body the velocity changes by 9.8 m/s every second.
Q2
Does mass have any effect on acceleration due to gravity?
The acceleration due to gravity is independent of the mass of the body.
Q3
What is the formula to calculate the force of attraction between two objects?
If m1 and m2 are the two masses separated by a distance d. According to the universal law of gravitation, the force of attraction between them is
F = Gm1m2/d2
Q4
What is free fall?
Freefall is defined as a situation when a body is moving only under the influence of the earth’s gravity. Since external force is acting on the ball, the motion will be accelerated.
Q5
What is the value of g on the moon?
The value of acceleration due to gravity on the moon is one-sixth of its value on the earth. It is equal to 1.625 m/s2.
Watch the video and solve important questions in the chapter Gravitation Class 9 Science
6,153
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16035 | https://math.stackexchange.com/questions/4636046/show-mathbbex-frac1-pp-if-mathbbpx-k-p1-pk | probability - Show $\mathbb{E}(X)=\frac{1-p}{p}$ if $\mathbb{P}(X=k)=p(1-p)^k$ - Mathematics Stack Exchange
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Show E(X)=1−p p if P(X=k)=p(1−p)k
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This is a discrete math problem, E(X) is the expected value of X: Let X be a random variable with P(X=k)=(1−p)k p, k=0,1,2,3,…, 0≤p≤1. Show that E(X)=1−p p.
I think we're supposed to use generating functions. So far, I have E(X)=∑∞k=0(k(1−p)k p)=0+(1−p)p+2(1−p)2 p+3(1−p)3 p+…. To try and construct that sequence: 1 1−(1−p)=1 p=1+(1−p)+(1−p)2+(1−p)3+(1−p)4+….
The derivative of 1 p is −1 p 2=(1−p)+2(1−p)2+3(1−p)3+…, so multiply by p to get p in every term, −p p 2=−1 p=(1−p)p+2(1−p)2 p+3(1−p)3 p+…, but that would mean E(X)=−1 p, so I must have made a mistake somewhere. I think maybe the last step where I multiplied by p to multiply every term by p, am I allowed to do that if p isn't a constant?
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edited Feb 9, 2023 at 20:04
ericeric
asked Feb 9, 2023 at 19:55
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1 Welcome to MSE! Please write your question in MathJax.bananapeel22 –bananapeel22 2023-02-09 20:00:16 +00:00 Commented Feb 9, 2023 at 20:00
1 Hint: Factor out (1−p)p from that sum 0+(1−p)p+2(1−p)2 p+3(1−p)3 p+… for E[X] to get E[X]=(1−p)p[1+2(1−p)+3(1−p)2+⋯]. Can you recognize the sum in square brackets? If not, can you figure out the value of the quantity in square brackets?Dilip Sarwate –Dilip Sarwate 2023-02-09 20:09:05 +00:00 Commented Feb 9, 2023 at 20:09
1 check the "derivative step" in your answer carefully.Math-fun –Math-fun 2023-02-09 20:10:55 +00:00 Commented Feb 9, 2023 at 20:10
$P(X=k)=p(1-p)^k is PMF of geometric random variable of second kindkludg –kludg 2023-02-10 00:44:14 +00:00 Commented Feb 10, 2023 at 0:44
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I would recommend you to notice that:
∞∑k=0 k(1−p)k p=∞∑k=1 k(1−p)k p=p(1−p)∞∑k=1 k(1−p)k−1 where the last series resembles the derivative of a geometric series.
Indeed, if |x|<1, the next series converges: ∞∑x=0 x n=1+x+x 2+…=1 1−x
Differentiating both sides, we get that: ∞∑n=1 n x n−1=1+2 x+3 x 2+…=1(1−x)2 which is exactly what we are looking for. Gathering the previous results, it yields that: ∞∑k=0 k(1−p)k p=p(1−p)1(1−(1−p))2=1−p p just as desired.
Hopefully this helps!
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answered Feb 9, 2023 at 20:11
Átila CorreiaÁtila Correia
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Thanks so much for the help eric –eric 2023-02-09 20:23:21 +00:00 Commented Feb 9, 2023 at 20:23
@eric you are welcome! I am glad I could help.Átila Correia –Átila Correia 2023-02-09 20:24:09 +00:00 Commented Feb 9, 2023 at 20:24
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Let g(a)=∞∑k=0 a k,
where a∈(0,1). According to this (convergent geometric series), we have that:
g(a)=1 1−a.
Notice that:
g′(a)=d d a∞∑k=0 a k−1=∞∑k=0 d d a a k−1=∞∑k=1 k a k−1,
and that
g′(a)=0⋅(1−a)−1⋅(−1)(1−a)2=1(1−a)2.
Hence:
∞∑k=1 k a k−1=1(1−a)2.
Let's pose a=1−p. Upon the aforementioned premises, we have that: E[X]=∞∑k=1 k P(X=k)=∞∑k=1 k p(1−p)k==p(1−p)∞∑k=1 k(1−p)k−1==p(1−p)∞∑k=1 k a k−1==p(1−p)⋅1(1−a)2==p(1−p)⋅1 p 2=1−p p.
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edited Feb 9, 2023 at 20:29
answered Feb 9, 2023 at 20:18
the_candymanthe_candyman
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appreciate the help thank you eric –eric 2023-02-09 20:28:57 +00:00 Commented Feb 9, 2023 at 20:28
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Taking a look at your work, you are correct to write: 1 p=1+(1−p)+(1−p)2+(1−p)3+(1−p)4+⋯ However, when you took the derivative wrt p, you differentiated the RHS incorrectly. You should have: −1 p 2=0+[−1]+[−2(1−p)]+[−3(1−p)2]+[−4(1−p)3]+⋯ since the derivative of (1−p) is −1, the derivative of (1−p)2 is 2(1−p)(−1), and in general the derivative of (1−p)k is k(1−p)k−1(−1). Clearing the negative sign from both sides gets you: 1 p 2=1+2(1−p)+3(1−p)2+4(1−p)3+⋯ To get the RHS to look like E(X)=(1−p)p+2(1−p)2 p+3(1−p)3 p+⋯, you multiply both sides by (1−p)p. The LHS then becomes 1−p p, as desired.
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answered Feb 10, 2023 at 3:20
grand_chatgrand_chat
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The moment generating function is G(z)=∑k≥0 P(k)z k=∑k≥0 p(1−p)k z k=p 1−(1−p)z, by the formula for the sum of a geometric series. Then E X=∑k≥0 k P(k)=G′(1), and since G′(z)=p(1−p)(1−(1−p)z)2, we have that E X=(1−p)/p.
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answered Feb 10, 2023 at 5:55
Per MattsonPer Mattson
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16036 | https://medium.com/operations-research-bit/linear-algebra-optimization-5a62760341c9 | Linear Algebra, Optimization. Linear algebra is used to solve… | by ORB, Operations Research Bit | Operations Research Bit | Medium
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Linear algebra is used to solve optimization problems. For example, linear algebra can be used to find the shortest path between two points, or to allocate resources efficiently.
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Linear algebra is a branch of mathematics that deals with vectors, matrices, and linear transformations. It is a fundamental tool in many fields, including operations research, computer science, and engineering.
Computer Science: Linear algebra is used in computer science to develop algorithms. For example, linear algebra can be used to sort data, or to find the eigenvalues of a matrix.
Engineering: Linear algebra is used in engineering to design systems. For example, linear algebra can be used to design bridges, or to optimize the performance of a machine.
What is linear algebra? Linear algebra is the study of linear equations and systems of linear equations. A linear equation is an equation of the form Ax = b, where A is a matrix, x is a vector, and b is a scalar. A system of linear equations is a set of linear equations that are all solved for the same unknowns.
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Why is linear algebra important? Linear algebra is important because it is a fundamental tool in many fields. It is used in operations research to solve optimization problems, in computer science to develop algorithms, and in engineering to design systems. Linear algebra can be used to:
Solve optimization problems
Develop algorithms
Design systems
Some of the most important concepts in linear algebra include:
Vectors: A vector is a list of numbers. Vectors can be added and multiplied by scalars.
Matrices: A matrix is a rectangular array of numbers. Matrices can be added, subtracted, multiplied, and inverted.
Linear Transformations: A linear transformation is a function that takes a vector and maps it to another vector. Linear transformations can be represented by matrices.
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Linear algebra is a powerful tool that can be used to solve problems in a variety of fields. By understanding the concepts of linear algebra, you can become a more effective problem-solver.
Applications of Linear Algebra. Linear algebra has many applications in operations research. For example, linear algebra can be used to:
Solving Optimization Problems: Linear algebra can be used to solve optimization problems. For example, linear algebra can be used to find the shortest path between two points, or to allocate resources efficiently.
Developing Algorithms: Linear algebra can be used to develop algorithms. For example, linear algebra can be used to sort data, or to find the eigenvalues of a matrix.
Designing Systems:Linear algebra can be used to design systems. For example, linear algebra can be used to design bridges, or to optimize the performance of a machine.
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Structures and Operations
Vectors, which are lists of numbers. Matrices, which are rectangular arrays of numbers. Finally, linear transformations, which are functions that take vectors and map them to other vectors.
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Vectors. A vector is a list of numbers. Vectors can be added and multiplied by scalars. They can be used to represent points, lines, and other geometric objects. By understanding the concepts of vectors, you can become a more effective problem-solver.
Vector addition: The sum of two vectors is the vector that is obtained by adding the corresponding components of each vector.
Scalar multiplication: The product of a vector and a scalar is the vector that is obtained by multiplying each component of the vector by the scalar.
Linear Algebra operations:
Addition: The sum of two vectors is a vector that has the same magnitude and direction as the sum of the vectors’ components.
Subtraction: The difference of two vectors is a vector that has the same magnitude and direction as the difference of the vectors’ components.
Scalar multiplication: The scalar multiplication of a vector by a scalar is a vector that has the same direction as the original vector but has a magnitude that is multiplied by the scalar.
Matrix multiplication: The product of two matrices is a matrix that has the same dimensions as the product of the matrices’ dimensions.
Matrices. A matrix is a rectangular array of numbers. Matrices can be added, subtracted, multiplied, and inverted. Some of the most common types include:
Square matrices: A square matrix is a matrix that has the same number of rows and columns.
Diagonal matrices: A diagonal matrix is a square matrix that has all zeros except for the diagonal elements.
Identity matrices: An identity matrix is a square matrix that has all ones on the diagonal and zeros elsewhere.
Triangular matrices: A triangular matrix is a matrix that has all zeros below (or above) the diagonal.
Matrices are a powerful tool that can be used to represent and manipulate data. By understanding the concepts of matrices, you can become a more effective problem-solver.
Matrix addition: The sum of two matrices is the matrix that is obtained by adding the corresponding components of each matrix.
Matrix subtraction: The difference of two matrices is the matrix that is obtained by subtracting the corresponding components of each matrix.
Matrix multiplication: The product of two matrices is the matrix that is obtained by multiplying the corresponding components of each matrix.
Matrix inversion: The inverse of a matrix is the matrix that, when multiplied by the original matrix, gives the identity matrix.
Linear Transformations. Linear transformations are a powerful tool that can be used to represent and manipulate geometric objects. A linear transformation is a function that takes a vector and maps it to another vector. Linear transformations can be represented by matrices.
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Linear Programming Problems
LP problems are used in a wide variety of applications, including production planning, transportation scheduling, and financial portfolio management.
Problems can be solved using a variety of methods, including the graphical method, the simplex method, and the interior point method. The simplex method is a common way to solve LP problems, and it is implemented in many software packages.
Problem. A company produces two products, A and B. Each product requires a certain amount of time and money to produce. The company has a limited amount of time and money available, so it needs to determine how many units of each product to produce in order to maximize its profits.
Decision variables: The decision variables in this problem are the number of units of product A and the number of units of product B that the company should produce.
Objective function: The objective function in this problem is to maximize the company’s profits. The profits for each unit of product A are $10 and the profits for each unit of product B are $15. So, the objective function is:
maximize 10x + 15y
Constraints. The company has a limited amount of time and money available, so there are two constraints in this problem. The first constraint is that the total time spent producing product A and product B cannot exceed 100 hours. The second constraint is that the total money spent producing product A and product B cannot exceed $5000.
x + 2y <= 100
10x + 5y <= 5000
Solution. The optimal solution to this problem is to produce 25 units of product A and 50 units of product B. This solution maximizes the company’s profits, which is equal to $1250.
Example 2: We have two machines, A and B, that can be used to produce widgets. Machine A can produce 10 widgets per hour, and machine B can produce 15 widgets per hour. We have a total of 120 hours of machine time available. We want to maximize the number of widgets that we produce, subject to the constraints that we have a limited amount of machine time and that we cannot produce more widgets than the demand.
from scipy.optimize import linprog
c Set the objective function
objective = [-10, -15]
A Set the constraints
constraints =
b =
b x,y Set the starting point
bounds = [(0, 100), (0, 150)]
Solve the problem
solution = linprog(objective, constraints,b, bounds=bounds)
Print the solution
print('Optimal number of widgets:', round(solution.fun-1, ndigits=2),
'\nx machine a:', int(solution.x),
'\ny machine b:', int(solution.x)x0 machine a: 70
x1 machine b: 150
Topics in Operations Research
Experimentation, ‘np.rand and Other Ways to Make Data’
Financial Analysis
Computational Thinking and Reasoning
Topics in Operations Research
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16037 | https://pmc.ncbi.nlm.nih.gov/articles/PMC2658837/ | Post-pyloric feeding - PMC
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World J Gastroenterol
. 2009 Mar 21;15(11):1281–1288. doi: 10.3748/wjg.15.1281
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Post-pyloric feeding
Eva Niv
Eva Niv
1 Eva Niv, Zvi Fireman, Department of Gastroenterology, Hillel Yaffe Medical Center, Hadera 38100, Israel
2 Eva Niv, Nachum Vaisman, The Unit of Clinical Nutrition, Tel Aviv Sourasky Medical Center, Tel Aviv, 6974444, Israel
Find articles by Eva Niv
1,2, Zvi Fireman
Zvi Fireman
1 Eva Niv, Zvi Fireman, Department of Gastroenterology, Hillel Yaffe Medical Center, Hadera 38100, Israel
2 Eva Niv, Nachum Vaisman, The Unit of Clinical Nutrition, Tel Aviv Sourasky Medical Center, Tel Aviv, 6974444, Israel
Find articles by Zvi Fireman
1,2, Nachum Vaisman
Nachum Vaisman
1 Eva Niv, Zvi Fireman, Department of Gastroenterology, Hillel Yaffe Medical Center, Hadera 38100, Israel
2 Eva Niv, Nachum Vaisman, The Unit of Clinical Nutrition, Tel Aviv Sourasky Medical Center, Tel Aviv, 6974444, Israel
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1,2
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1 Eva Niv, Zvi Fireman, Department of Gastroenterology, Hillel Yaffe Medical Center, Hadera 38100, Israel
2 Eva Niv, Nachum Vaisman, The Unit of Clinical Nutrition, Tel Aviv Sourasky Medical Center, Tel Aviv, 6974444, Israel
Author contributions: Niv E, Fireman Z and Vaisman N performed the literature review and wrote the paper.
Correspondence to: Eva Niv, MD, Department of Gastro-enterology, Hillel Yaffe Medical Center, PO Box 169, Hadera 38100, Israel. niv_em@netvision.net.il
Telephone: +972-4-6304480
Fax: +972-4-6304408
Received 2008 Dec 26; Revised 2009 Feb 15; Accepted 2009 Feb 22; Issue date 2009 Mar 21.
©2009 The WJG Press and Baishideng. All rights reserved.
PMC Copyright notice
PMCID: PMC2658837 PMID: 19294757
Abstract
Postpyloric feeding is an important and promising alternative to parenteral nutrition. The indications for this kind of feeding are increasing and include a variety of clinical conditions, such as gastroparesis, acute pancreatitis, gastric outlet stenosis, hyperemesis (including gravida), recurrent aspiration, tracheoesophageal fistula and stenosis in gastroenterostomy. This review discusses the differences between pre- and postpyloric feeding, indications and contraindications, advantages and disadvantages, and provides an overview of the techniques of placement of various postpyloric devices.
Keywords: Postpyloric feeding, Nasojejunal feeding, Nasojejunal tube, Jejunostomy, Nasoenteric tube, Percutaneous endoscopic gastrostomy-jejunostomy tube, Percutaneous endoscopic jejunostomy
INTRODUCTION
According to both European and American guidelines for enteral and parenteral nutrition, enteral feeding is the preferred method of nutritional support in patients who have a functioning gastrointestinal (GI) tract but cannot maintain an adequate oral intake[1,2]. Enteral nutrition prevents GI mucosal atrophy, keeps intestinal integrity and prevents bacterial translocation from the GI lumen to the rest of the body, by maintaining normal permeability of the GI mucosal barrier[3–6]. In addition, it is less expensive and has significantly fewer complications than parenteral nutrition[1,2].
The enteral route traditionally delivered nutrition directly into the stomach via a nasogastric tube or gastrostomy (prepyloric feeding). The concept of postpyloric feeding has been developed over the past few decades and has become a part of the routine practice of nutritional teams in many countries. A wide variety of postpyloric nutrition devices are currently available, including different types of nasoduodenal and nasojejunal tubes and jejunostomies.
What are the differences between pre- and postpyloric feeding approaches? Why is the location of the tip of a nutritional device before or after the pylorus so important? The current review will discuss the major differences between these two methods of enteral nutrition in order to provide essential information for every nutritionist and gastroenterologist in making the right choice for every specific case. This review will provide a comprehensive overview of accepted indications and contraindications of postpyloric feeding, based on existing studies and guidelines. In addition, various devices for postpyloric feeding, as well as different techniques for their insertion, their advantages and disadvantages will be discussed.
PHYSIOLOGICAL DIFFERENCES BETWEEN PRE- AND POSTPYLORIC FEEDING
There are several physiological differences between pre- and postpyloric feeding. The first major difference is a mechanical one. Postpyloric delivery of food significantly reduces the likelihood of aspiration/vomiting caused by gastroesophageal reflux, especially in the case of intrajejunal and not intraduodenal feeding. The second major difference is the neurohormonal effect of food that is supplied directly to the small intestine or the duodenum, compared to intragastric supply. It has different effects on pancreatico-biliary secretions and on small bowel and gallbladder motility. Ledeboer et al have demonstrated that intraduodenal feeding causes a stronger GI response than intragastric feeding. It stimulates gallbladder contractions, accelerates small bowel transit time, and increases cholecystokinin and pancreatic polypeptide release. Intrajejunal feeding has a completely different effect[8,9]. The classic work in a canine model by Ragins et al has demonstrated that jejunal feeding does not stimulate pancreatic secretion, as is seen in intragastric or intraduodenal delivery of food, which increases the volume and changes the content of pancreatic secretions. These results have been supported by animal and human studies on models of acute pancreatitis[11–14].
Unfortunately, almost all these studies have focused on the influence of intrajejunal feeding on the pancreas and failed to address the intriguing issue of its impact on small bowel function. Data on the changes in the levels of relevant hormones and changes in the motor pattern of the small and large intestine are scarce.
Table 1 demonstrates differences between gastric and jejunal feeding.
Table 1.
Differences between gastric and jejunal feeding
Nasogastric tubeNasojejunal tube
Indications Anorexia, dysphagia, odynophagia Gastroparesis, gastric outlet obstruction, recurrent aspirations, severe pancreatitis, hyperemesis gravida, proximal enteric fistula, postoperative anastomotic gastroenteric stenosis
Insertion technique Easy access, no need for endoscopic or radiological study or medication Needs endoscope or prokinetic agents
Costs Much cheaper because: 1. low cost of the tube; 2. may be inserted by a nurse More expensive because: 1. costly equipment; 2. requires insertion by physician
Physiology More physiological, keep normal motility and hormonal profile Less controlled motility and hormonal control. Less pancreatic stimulation if inserted after the Trietz ligament
Feeding mode Bolus or continuous. Pump is not mandatory Continuous only. Pump is mandatory in most cases
Risk of aspiration High in patients with GER and swallowing impairments Less frequent but not absolutely prevented
Clogging rate Rare thanks to larger diameter of tube Frequent
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ACCESS ROUTES FOR POSTPYLORIC FEEDING
The access routes for postpyloric feeding include nasoduodenal and nasojejunal tubes and jejunostomy. Nasoenteric tubes may be placed manually or with endoscopic, radiological or surgical guidance. Nasoenteric tubes are a good choice for short-term feeding but have many drawbacks for long-term management. They tend to recoil into the stomach, become clogged, cause nasal pressure sores, and can be pushed out of place accidentally. As such, jejunostomy is the preferred option for long-term postpyloric feeding. Feeding by jejunostomy generally requires surgical placement, although endoscopic or fluoroscopic placement for jejunostomy has been successful in some medical centers with adequate experience[15,16]. In cases for which a surgical jejunostomy is considered, the benefits and tolerability of surgical jejunostomy and postpyloric feeding can be assessed by temporarily placing a nasojejunal tube or by inserting a jejunal extension of a percutaneous endoscopic gastrostomy-jejunostomy (PEG-J).
There are several kinds of nasoenteric tubes made from various materials (e.g. polyurethane and polyvinylchloride), that have different diameters (8-12 French), with and without guide wires, and with and without weight at their tips. Unweighted tubes of smaller diameter (8 French) are used for endoscopic insertion to ensure a proper passage through a working channel of the endoscope. The length of a nasoenteric tube ranges from 140 cm to 220 cm.
Nasoduodenal tubes and duodenostomies have been in common use in the past, but cumulative experience has shown that the duodenal route is very problematic because of the tendency of a nasoduodenal tube to recoil back into the stomach, as well as the strong stimulatory effect of intraduodenal feeding on pancreatic secretions. In addition, a feeding formula tends to flow to the stomach because of duodenogastric reflux[17–19]. Thus, intraduodenal feeding is contraindicated in the setting of recurrent aspiration and severe pancreatitis, which are the most common indications for postpyloric feeding. As a result, intraduodenal feeding is no longer routinely used in most medical centers and will not be discussed further.
INDICATIONS FOR POSTPYLORIC FEEDING
There are several clinical situations in which postpyloric feeding is preferable to the intragastric route. One of most common indications is gastroparesis that is not responsive to prokinetics[20,21]. This situation is most frequently encountered in the early postoperative setting or in critical care patients. Theoretically, postpyloric feeding would appear to be an attractive option in critically ill patients because of the frequently present problems of gastroparesis and aspirations[22–25]. It is, however, associated with significant cost and risks of nasojejunal tube insertion. It has been accepted widely in the past that every critically ill patient should be fed postpylorically, and this approach has been investigated in many studies and meta-analyses[26–29]. One of the most interesting of these was by Montejo et al, who conducted a multicenter, prospective, randomized, single-blind study on 110 patients with similar characteristics who were randomized to be fed pre- or postpylorically. The authors concluded that the nutritional results were similar in both groups, and therefore, the routine use of a nasojejunal tube in critically ill patients is not justified. It may, however, have a role in selected patients with high gastric residuals on nasogastric feeding, or with various conditions of the GI tract, such as severe pancreatitis. Boulton-Jones et al have investigated postpyloric feeding in selected groups of 138 critically ill patients who suffered from burn injury, severe pancreatitis, sepsis, postoperative gastroparesis, and vomiting induced by bone marrow transplantation and chemotherapy. The results of that study demonstrated good nutritional results in all the patients. On the basis of these and additional studies that have been published since then, the current European and American guidelines for enteral and parenteral nutrition[1,2] support nasojejunal feeding only in selected groups of critically ill patients with one of indications mentioned in this section (i.e. gastroparesis, recurrent aspirations, severe hyperemesis, and severe acute pancreatitis).
Another common indication for postpyloric feeding is recurrent aspiration caused by severe gastroesophageal reflux disease (GERD) in bedridden patients[22,23]. One of the classic studies on this subject was by Montecalvo et al. Thirty-eight patients were randomly assigned to feeding by nasogastric or nasojejunal tube. There were no documented aspirations in the postpyloric feeding group compared to two aspirations in the nasojejunal tube group. It is important to define whether episodes of aspiration are truly caused by GERD or are the result of disorders in swallowing.
Nasojejunal tube insertion has become routine practice in many hospitals in cases of severe pancreatitis. This kind of feeding enables the provision of enteral nutrition with less stimulation of pancreatic secretions and less exacerbation of inflammation in the pancreas. There are four phases of pancreatic secretions: (1) basal - very little pancreatic secretion during fasting; (2) cephalic - mildly increased secretion when the individual looks at food; (3) gastric - increased pancreatic secretion initiated by gastric distention with food and mediated by gastrin and acid; and (4) duodenal - extensive stimulation of pancreatic secretions initiated by the entry of chyme and acid into the duodenum, and mediated by secretin and cholecystokinin. The classic work of Ragins et al in a canine model has demonstrated that intragastric or intraduodenal delivery of food increases the volume and changes the content of pancreatic secretions. In contrast, jejunal feeding does not stimulate pancreatic secretions[11,32]. Since it is very important to provide pancreatic rest during acute pancreatitis, the idea of intrajejunal feeding has become very attractive and it has been investigated in animal models of acute pancreatitis and in several prospective randomized controlled human studies that have compared nasojejunal feeding with total parenteral nutrition (TPN)[12,13,34]. The consensus is that nasojejunal feeding has a good and even better clinical outcome and time of recovery from severe pancreatitis than those with TPN. Moreover, TPN is more expensive and associated with more complications than intrajejunal feeding, giving further advantage to the latter. There have been only a few studies on the formula of choice for nasojejunal feeding in severe pancreatitis[35–37]. Most of these studies support polymeric formulas, but some show advantages for elemental or semi-elemental formulas. Polymeric formulas are preferred in most centers because of their lower cost.
A rare but important indication is a proximal enteric fistula. For example, if a fistula is located in the esophagus/stomach/duodenum (usually tracheo-esophageal fistula), a nasojejunal tube will supply food more distally and make it possible to provide food enterally as an alternative to parenteral nutrition.
A relatively newly defined indication is hyperemesis gravida. Parenteral nutrition had previously been indicated in some cases of severe hyperemesis gravida with significant weight loss. Two small studies have described the possibility of nasoenteric tube feeding in these women[38,39]. A pioneer study by Vaisman et al has examined the feasibility and efficacy of nasojejunal feeding in 11 pregnant women with severe hyperemesis gravida that persisted despite in-hospital anti-emetic treatment. The nasojejunal feeding approach proved to be effective, reducing vomiting within the first 48 h, with complete resolution after 5 d in most of the women. More prospective studies are needed to validate this promising method.
Postpyloric feeding is the only route for enteral feeding in pyloric or duodenal outlet stenosis. This condition is common in malnourished oncological patients with gastric or pancreatic cancers who are waiting for definitive or palliative surgery, and who are required to improve their nutritional status prior to undergoing surgery.
Another common situation is the postoperative setting after Bilroth II or Whipple procedures. Postoperative transient edema in a gastroenteric anastomosis might create a significant problem in gastric emptying. Temporary insertion of a nasojejunal tube below the anastomosis will provide an enteral feeding route for these patients until the edema resolves. In some cases of difficult GI anastomosis, the preventive intraoperative insertion of a nasojejunal tube is recommended to enable early postoperative enteral feeding.
CONTRAINDICATIONS FOR POSTPYLORIC FEEDING
The major contraindication for postpyloric feeding is an obstruction in different parts of the GI tract (esophagus, gastric outlet or intestine). An endoscopic nasojejunal tube or an endoscopic jejunostomy are contraindicated in some clinical scenarios because of the inability of inserting the gastroscope postpylorically, but surgical jejunostomy may still be indicated, as in the case of complete obstruction of the esophagus/stomach/duodenum. Endoscopic nasojejunal tube insertion may nevertheless be an option in some cases of partial obstruction of the upper GI tract because it is possible to push the tip of the tube far beyond the location of the endoscope, and the procedure might even be done blindly beyond a visible stricture. The feasibility of inserting an endoscopic nasojejunal tube depends on the degree of stenosis. Even a pinpoint passage that is sufficient for passage of a guide wire permits the insertion of a nasojejunal tube. Of course, surgical jejunostomy does not require any passage of an endoscope through the GI tract, which provides more possibilities for applying this kind of technique.
The most important absolute contraindication for all kinds of postpyloric feeding is bowel obstruction or perforation/leakage. Therefore, exact information about the GI tract’s mechanical problems, previous GI tract surgery, imaging of the GI tract and verification of GI tract patency must be obtained before postpyloric feeding can be considered.
Contraindications for jejunostomy, but not for a nasojejunal tube, are significant ascites, coagulopathy, peritoneal dialysis, and peritoneal metastasis. For endoscopic insertion of jejunostomy, there are additional contraindications, such as morbid obesity and the inability to transilluminate through the abdominal wall or to see a digital imprint.
TECHNIQUES OF INSERTION
Nasoenteric tube placement
Nasoenteric tubes may be placed by using manual (blind) techniques or with the aid of fluoroscopy or endoscopy. Nasojejunal tubes for surgical patients may be placed during laparotomy. There are several manual techniques for nasojejunal tube placement. Usually, a nasoenteric tube (8-9 French) is inserted with a guide wire and a weighted tip is inserted into the stomach using the usual technique for nasogastric tube insertion. The patient is then asked to change his/her position to right lateral decubitus and the tube is pushed through the pylorus. The guide wire should be removed at the end of the procedure. Several techniques have been developed to facilitate the passage of the tube through the pylorus, among them air insufflation of the stomach[44,45], pH-sensor feeding tube guidance[46,47], and prokinetic agents, such as intravenous erythromycin (250-500 mg)[48–51] or 10 mg metoclopramide[52–54]. For example, a very interesting randomized, double-blind, placebo-controlled study has been published by Griffith et al. Thirty-six critical ill patients were randomized to receive a single bolus of intravenous erythromycin (500 mg) or saline before placement of 10-French feeding tubes, using a standardized active bedside protocol. The conclusion of the study was very impressive, with a 93% success rate in the erythromycin group versus 55% in the placebo group. In contrast, a study by Gharpure et al, with a similar design, on a group of critically ill children demonstrated no clinical advantage with intravenous erythromycin (10 mg/kg) versus saline in facilitation of transpyloric passage of nasojejunal tubes.
There is no consensus on the best technique of manual insertion of a nasojejunal tube because of the great variety of success rates (30%-95%) reported in many studies carried out in different centers[44,46,48,52]. The advantage of weighted over unweighted tubes is uncertain, although it is a widely accepted belief.
The nasojejunal feeding tube is commonly placed endoscopically, which allows placement under direct vision[56–59]. Its major disadvantage is the requirement of a complete gastroscopy, which increases the cost and duration of the procedure, the risks related to intravenous sedation, and the number of possible complications associated with gastroscopy, such as perforation and dental injury. The high success rate of this procedure (93%-98%), however, makes it very attractive[56–59]. The technique is simple: after a gastroscope is placed deeply in the duodenum, a flexible unweighted 8-French nasojejunal tube with a guide wire is advanced through a working channel of the endoscope and pushed deep into the jejunum, beyond the tip of the endoscope during simultaneous withdrawal of the endoscope. When the procedure has been completed, the guide wire is removed and a feeding tube is passed from the mouth to the nose by means of a plastic device. Some centers also use a drag technique in which a suture is tied to the end of a feeding tube, which is then passed into the stomach via the nasopharynx. This suture is dragged with the endoscope snare or forceps from the stomach to the duodenum. Once the tube is in position, the suture is released and the endoscope is withdrawn. This procedure is less successful because the feeding tube frequently moves back into the stomach when the endoscope is removed.
A new technique of nasoenteric tube insertion has become very popular. It involves a transnasal thin endoscope that is inserted transnasally into the stomach and then into the duodenum[60–62]. A thin guide wire is inserted though a working channel while the endoscope is removed, after which a feeding tube is placed over the guide wire, which is then removed.
Fluoroscopic techniques of nasoenteric tube placement require skilled radiological support and exposure to radiation. In addition, they necessitate changes in the patient’s position that may not be feasible for the critically ill. The success rate of radiological placement varies from 40% to 94%, depending on the local expertise of the staff in different medical centers[57,63–65].
Whatever the technique that is used for nasojejunal tube placement, proper position of the nasoenteric feeding tubes must be verified radiographically before the feeding is initiated. Clinicians should not rely on the accepted ways of checking nasogastric tube position, because it is impossible to adequately hear the entrance of air injected through the tube into the jejunum, and to distinguish its erroneous placement in the stomach/esophagus/lungs. In addition, air insufflation of the jejunum is unsafe.
Jejunostomy placement
Most jejunostomies are placed at least 20 cm beyond the ligament of Treitz (a point of transition of the duodenum to the jejunum) because of the increased rate of complications of duodenostomy compared with jejunostomy. A jejunostomy may be inserted with endoscopic assistance (percutaneous endoscopic jejunostomy; PEJ) or surgically (surgical jejunostomy). A PEJ may be inserted indirectly via a previously placed gastrostomy (PEG-J)[16,67,68] or directly[15,69–71].
For the placement of a PEG-J, a feeding tube long enough to pass beyond the pylorus is inserted through an existing PEG tube. The tip of the feeding tube is then grasped with the biopsy forceps of the endoscope and the tube is pushed as far as possible into the duodenum. Extra tubing length is left within the stomach to allow peristalsis to pull the tip of the feeding tube past the ligament of Treitz. Although this procedure is simple, its major disadvantage is the tendency of the feeding tube to return back into the stomach during the withdrawal of the gastroscope. In addition, the feeding tube tends to dislodge from the outer gastrostomy.
An enteroscope or colonoscope should be placed into the proximal jejunum for direct PEJ placement[15,69–71]. One of the most common techniques includes the insertion of a 19-gauge needle into the jejunal lumen at the site of the transillumination or a finger indentation marking the jejunal loop that is closest to the abdominal wall. The needle should be snared tightly, fixing the small bowel against the abdominal wall. The plastic sheath with stylet should then be inserted adjacent to the 19-gauge needle and snared by a wire loop that has been removed from the needle. An insertion wire should then be passed through the plastic sheath and grasped with a snare. The rest of the procedure is similar to the PEG’s pull technique: the gastroscope together with a wire is pulled out through the duodenum, stomach, esophagus and mouth. The insertion wire is then secured to the loop at the end of the feeding tube with an internal jejunal bolster and the assembly is pulled through the mouth all the way to the duodenum. The tube is pulled through an incision in the abdominal wall, sufficiently tight to compress the jejunal wall against the anterior abdominal wall. Intrajejunal tube placement is then verified by a second gastroscopy. Finally, a skin disk is secured to the outside portion of the feeding tube to ensure the creation of a tract between the skin and jejunal lumen. It is important to avoid excess tension when approximating the jejunum to the abdominal wall, so as to prevent pressure sores of the skin or jejunal mucosa.
For patients in whom endoscopy is contraindicated, jejunal feeding tubes can be placed with radiological guidance. Access is obtained at a previous gastrostomy site or by direct jejunal punctures. With this method, the stomach and the jejunum are insufflated with air via a nasogastric or nasojejunal tube, and the location of internal organs is identified by means of ultrasound or fluoroscopy to ensure that no organs lie between the jejunum and the abdominal wall. A needle is inserted through the abdominal wall into the jejunal lumen and a guide wire is inserted through the needle. The needle is removed, the tract is dilated, and a feeding tube is placed over the guide wire and secured.
Surgical placement of a jejunostomy can be performed by a needle catheter or by Witzel techniques. A needle catheter jejunostomy is placed during laparotomy for surgical patients who need short-term enteral support[75,76]. A purse-string suture is placed in the bowel wall, through which a large-bore needle is tunneled subserosally for several centimeters before entering the bowel lumen. A 5-, 7-, or 9-French feeding catheter with a flexible stylet is inserted through this needle and advanced distally into the bowel. The needle is removed and the purse string is tied. Next, a 3-5 cm Witzel tunnel is created in the abdominal wall proximal to the catheter insertion. A second large-bore needle is inserted through the abdominal wall and the feeding catheter and stylet are passed through the needle to the skin. The needle and stylet are then removed and the intestine is fastened to the anterior abdominal wall to prevent leakage.
The Witzel jejunostomy is another open-surgery method. A tube is placed through an incision in the anterior abdominal wall and a tunneled incision is made in the jejunal wall. The adherence of jejunum to the abdominal wall is ensured by sutures.
Some centers perform laparoscopic jejunostomy. Duh et al have used this technique in 36 patients who could not undergo gastrostomy, with a good rate of success.
COMPLICATIONS AND DISADVANTAGES OF POSTPYLORIC FEEDING
There are various complications of postpyloric feeding. Some of them are specific to a specific device (nasojejunal tube versus jejunostomy) and others are universal for all kinds of postpyloric feeding techniques. Tables 2 and 3 specify common and uncommon complications of nasojejunal and jejunostomy feedings. The common complications of nasojejunal tubes are as follows: failure of nasojejunal tube placement (the rate depends on the technique of insertion), displacement of the tube, clogging of the tube, mild transient epistaxis, nasal mucosal irritation, feeding-related diarrhea, abdominal cramping, and hyperglycemia. The common complications of jejunostomy include pain and infection at the jejunostomy site, displacement of the jejunostomy, clogging, feeding-related diarrhea, abdominal cramping, hyperglycemia, transient pneumoperitoneum immediately after the insertion (in most cases, without any clinical significance), and leakage around the jejunostomy site[80,81]. It is essential to take into account any existing risks of intravenous sedation and gastroscopy as well as the risks of anesthesia and surgery. There is a possibility that the patient will experience abdominal cramping, hyperperistalsis and diarrhea whatever device is used for this kind of feeding. The considerable costs of postpyloric devices compared to prepyloric ones need to be taken into account as well.
Table 2.
Potential complications of nasojejunal tube feeding
Common (> 10%)
Failure of placement1
Displacement
Clogging of the tube
Mild transient epistaxis
Irritation of nasal, pharyngeal or esophageal areas
Feeding-related diarrhea
Abdominal cramping
Metabolic complication, such as hyperglycemia
Uncommon (< 10%)
Otitis media
Nasal mucosal pressure sores
Esophageal ulcers
Risks of intravenous sedation and gastroscopy
Sinusitis
Misplacement (pulmonary or intracranial intubation)
Dumping-like symptoms
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1
Depends on the technique of insertion.
Table 3.
Potential complications of jejunostomy feeding
Common (> 10%)
Pain at the jejunostomy site
Skin infection of the jejunostomy site
Feeding-related diarrhea
Abdominal cramping
Clogging of tube
Transient pneumoperitoneum immediately after the insertion (but it has no clinical significance in most cases)
Metabolic complication, such as hyperglycemia
Displacement of jejunostomy
Leakage around the jejunostomy
Uncommon (< 10%)
Failure of placement
Misplacement
Gastric hemmorhage
Perforation of internal organs during the placement and peritonitis
Colocutaneous fistula
Persistent jejunocutaneous fistula after the removal of jejunostomy
Risks of intravenous sedation and gastroscopy or risks of anesthesia and surgery
Hemorrhage at jejunostomy site
Pressure sore due to skin disk of jejunostomy
Dumping-like symptoms
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Although the list of possible complications is a long one, most of them might be successfully avoided by using proper techniques of placement and management of the post-pyloric devices. For example, a misplacement of a nasojejunal tube and subsequent aspiration may be detected and avoided by radiological verification of the tube’s location before feeding is started. The displacement of a nasojejunal tube may be prevented by proper fixation. Nasoenteric tubes tend to be blocked because they are usually longer and of finer bore. They are especially susceptible to being obstructed by crushed medications, viscous feeds and inadequate flushing. Therefore, these tubes should be flushed every 4-6 h, always before and after usage, and dense feeds and medications should be avoided. In the event of clogging, a tube can usually be unblocked by flushing it with hot water, coca-cola or pancreatic enzymes. The sudden influx of a hyperosmotic formula is likely to lead to abdominal cramping, hyperperistalsis and diarrhea since the jejunum relies on controlled delivery of isotonic substrates. An intrajejunal feeding is less physiological compared with an intragastric one. The ability of the stomach to distend and contain a large amount of food all at once is a great advantage compared to the limited distension capability of the jejunum. Some patients who are fed postpylorically may develop symptoms similar to dumping syndrome, i.e. faintness, palpitations, sweating, tachycardia, rebound hypoglycemia, and diarrhea. Therefore, intrajejunal feeding should always be carried out continuously by pump and not by boluses. The recommended actions for cases of diarrhea are to exclude other possible causes, to decrease the rate of feeding, and to consider a change in formula to a less osmotic one and one that contains fibers.
FORMULAS FOR POSTPYLORIC FEEDING
The mode of administration, the appropriate formula and the rate of administration are important features for successful postpyloric feeding. The preferable kind of formula has yet to be determined, and there are few studies that have addressed this issue[35,36,83,84]. Some of them advocate elemental and semi-elemental feeds and others support polymeric solutions. Lacking sufficient data, each medical center develops its own protocol.
Postpyloric feeds for children have traditionally been elemental or hydrolyzed and less viscous because of the narrow lumen of the tubes needed to pass the pylorus, although polymeric feeds have also been tolerated. For adults, polymeric formulas are usually chosen except for patients with malabsorptive disorders or lymph duct problems.
As mentioned earlier, the sudden influx of a hyperosmotic feed is likely to lead to abdominal cramping, hyperperistalsis, diarrhea and symptoms similar to dumping syndrome, since the jejunum relies on a controlled delivery of isotonic substrates. It is worth repeating that postpyloric feeds should be administered continuously by pump. The initial rate of administration should be slow and increased gradually. Parenteral support is sometimes used as caloric intake is gradually increased until the target caloric intake has been reached.
CONCLUSION
The postpyloric route is a promising method of enteral feeding. In some cases, it is the only feasible way of maintaining enteral input and avoiding parenteral nutrition. Knowledge on the indications, contraindications, advantages and disadvantages and experience with the placement and replacement of different kinds of postpyloric devices should be an essential part of training of gastroenterologists and nutritionists. Further research on the physiological differences between intragastric and intra-jejunal food supply, including hormonal and enzymatic changes, is warranted.
Peer reviewer: Nick P Thompson, MD, Department of Medicine, Freeman Hospital, Newcastle Upon Tyne, NE7 7DN, United Kingdom
S- Editor Li LF L- Editor Kerr C E- Editor Lin YP
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Abstract
INTRODUCTION
PHYSIOLOGICAL DIFFERENCES BETWEEN PRE- AND POSTPYLORIC FEEDING
ACCESS ROUTES FOR POSTPYLORIC FEEDING
INDICATIONS FOR POSTPYLORIC FEEDING
CONTRAINDICATIONS FOR POSTPYLORIC FEEDING
TECHNIQUES OF INSERTION
COMPLICATIONS AND DISADVANTAGES OF POSTPYLORIC FEEDING
FORMULAS FOR POSTPYLORIC FEEDING
CONCLUSION
References
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16038 | https://www.droracle.ai/articles/163344/what-is-the-transverse-pericardial-sinus-and-what-is-its-function | Select Language
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What is the function of the transverse pericardial sinus?
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From the Research
The transverse pericardial sinus functions as a crucial anatomical passage within the pericardium, allowing for the isolation of great vessels during cardiac surgery, as evident from the study published in the Journal of Clinical Medicine in 2023 1.
The transverse pericardial sinus is a tunnel-like space located posterior to the ascending aorta and pulmonary trunk but anterior to the superior vena cava and right pulmonary veins.
Its primary clinical significance lies in cardiac surgery, where surgeons can pass a finger or surgical instrument through this space to separate the aorta and pulmonary artery from the rest of the heart structures, facilitating the placement of vascular clamps during procedures like valve replacements or coronary artery bypass grafting.
This anatomical feature exists because of how the heart develops embryologically, with the great vessels becoming enveloped by the pericardial sac while maintaining this potential space.
Understanding the transverse pericardial sinus is essential for cardiac surgeons as it provides a natural surgical plane that helps minimize trauma to surrounding structures during complex cardiac operations.
Key points to consider include:
The transverse pericardial sinus can be classified into four types: Concave, Wine-type, Straight, and Convex, according to a study published in the Journal of Clinical Medicine in 2023 1.
The most common type of transverse sinus is Concave, with a median volume of 14.8 mL and a median length of 52.8 mm, as reported in the same study 1.
The clinical anatomy of the pericardial sinuses has not been thoroughly studied, but recent research aims to provide a classification of the oblique and transverse sinuses, as seen in the study published in the Journal of Clinical Medicine in 2023 1.
The transverse pericardial sinus is also discussed in other studies, such as the one published in Clinical Anatomy in 2017 2, which explores the anatomy of the transverse pericardial sinus and associated recesses, but the most recent and highest quality study is the one published in the Journal of Clinical Medicine in 2023 1.
References
1
Research
Morphology and Anatomical Classification of Pericardial Cavities: Oblique and Transverse Sinuses.
Journal of clinical medicine, 2023
2
Research
Computerized tomography of the transverse pericardial sinus: Normal or pathologic?
Clinical anatomy (New York, N.Y.), 2017
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16039 | https://www.merriam-webster.com/thesaurus/WEARY?pronunciation&lang=en_us&dir=a&file=ambien03 | WEARY Synonyms: 345 Similar and Opposite Words | Merriam-Webster Thesaurus
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Est. 1828
Thesaurus
adjective
as in tired
as in bored
as in tiring
verb
as in tobore
as in towear
as in tired
as in bored
as in tiring
as in to bore
as in to wear
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Synonyms of weary
weary 1 of 2
adjective
ˈwir-ē
Definition of weary
1
as in tired
depleted in strength, energy, or freshness I am just too weary to do any more work tonight
Synonyms & Similar Words
Relevance
tired
exhausted
wearied
drained
worn
fatigued
jaded
dead
beaten
aweary
beat
spent
limp
bleary
burnt-out
done
tapped out
bushed
loggy
worn to a frazzle
prostrate
wiped out
knackered
pooped
worn-out
logy
played out
all in
burned-out
sleepy
overworked
washed-out
weakened
overtaxed
tuckered (out)
enfeebled
debilitated
enervated
broken-down
drowsy
sapped
sluggish
lethargic
overfatigued
heavy
run-down
enervate
Antonyms & Near Antonyms
unwearied
rested
fresh
rejuvenated
refreshed
relaxed
active
tireless
strong
energetic
strengthened
invigorated
revitalized
peppy
vitalized
weariless
See More
2
as in bored
having one's patience, interest, or pleasure exhausted I am totally weary of this constant bickering
Synonyms & Similar Words
tired
bored
wearied
sick
jaded
frustrated
fed up
sick and tired
exhausted
annoyed
uninterested
disinterested
dispirited
irritated
disgusted
apathetic
dejected
demoralized
fatigued
exasperated
surfeited
disheartened
glutted
sated
satiated
drained
discouraged
beat
limp
blasé
burnt-out
bushed
enervated
done in
nauseated
worn-out
world-weary
burned-out
blase
repulsed
played out
tuckered (out)
Antonyms & Near Antonyms
interested
engaged
absorbed
engrossed
stimulated
intrigued
rapt
excited
galvanized
energized
invigorated
amused
animated
vitalized
charmed
enlivened
enthralled
enchanted
fascinated
delighted
hypnotized
pleased
entertained
beguiled
bewitched
captivated
mesmerized
thrilled
See More
3
as in tiring
causing weariness, restlessness, or lack of interest a weary march through a lot of boring facts and figures
Synonyms & Similar Words
tiring
boring
wearying
slow
old
stupid
dull
dusty
heavy
dry
wearisome
annoying
tiresome
irritating
tedious
monotonous
tame
drab
pedestrian
leaden
stale
dreary
uninteresting
arid
stodgy
ponderous
humdrum
numbing
pallid
stuffy
exhausting
colorless
drudging
jejune
monochromatic
ho-hum
flat
jading
mind-numbing
sterile
barren
grey
blank
spiritless
blah
gray
soggy
unimaginative
prosy
draining
irksome
dullish
prosaic
earthbound
bothersome
uninspiring
uneventful
fatiguing
unrewarding
unexciting
inanimate
pedantic
wearing
discouraging
aseptic
ordinary
debilitating
unspectacular
unsensational
plodding
suspenseless
common
cumbersome
undramatic
dispiriting
tepid
demoralizing
vapid
pleasureless
enervating
commonplace
pokey
disheartening
lumbering
unnewsworthy
palling
longsome
unsurprising
poky
unexceptional
enfeebling
Antonyms & Near Antonyms
interesting
involving
engaging
wonderful
intriguing
absorbing
surprising
gripping
wondrous
engrossing
amazing
marvellous
riveting
inspiring
marvelous
astonishing
exciting
awesome
stimulating
spectacular
stirring
amusing
astounding
sensational
rousing
entertaining
exhilarating
fabulous
moving
energizing
thrilling
breathtaking
electrifying
attractive
charming
animating
attracting
poignant
alluring
beguiling
fascinating
touching
enchanting
invigorating
enlivening
enthralling
entrancing
bewitching
galvanizing
diverting
rip-roaring
eye-opening
hair-raising
tantalizing
provocative
captivating
suspenseful
arresting
mesmerizing
spellbinding
See More
weary
2 of 2
verb
1
as in to bore
to make weary and restless by being dull or monotonous these constant complaints are really wearying me
Synonyms & Similar Words
bore
tire
wear
drain
put to sleep
exhaust
jade
discourage
fatigue
pall
wear out
burn out
enervate
debilitate
enfeeble
dishearten
wash out
disable
tucker (out)
deject
dispirit
fag
demoralize
do in
Antonyms & Near Antonyms
busy
engage
interest
grip
absorb
stimulate
strengthen
energize
intrigue
excite
invigorate
entertain
enliven
engross
fascinate
activate
amuse
galvanize
busy
animate
attract
vitalize
charm
beguile
enthrall
enthral
captivate
occupy
allure
preoccupy
involve
hypnotize
bewitch
enchant
monopolize
stir
rouse
immerse
mesmerize
rally
pump up
See More
2
as in to wear
to use up all the physical energy of a whole day of hard physical labor had thoroughly wearied her
Synonyms & Similar Words
wear
tire
kill
exhaust
drain
fatigue
bust
wear out
break
knock out
harass
burn out
wash out
waste
frazzle
fag
outwear
weaken
tucker (out)
wear to a frazzle
enervate
sap
do in
do up
debilitate
enfeeble
Antonyms & Near Antonyms
strengthen
energize
activate
rest
relax
rejuvenate
invigorate
unwind
vitalize
Synonym Chooser
How is the word weary distinct from other similar verbs?
Some common synonyms of weary are exhaust, fatigue, jade, and tire. While all these words mean "to make or become unable or unwilling to continue," weary stresses tiring until one is unable to endure more of the same thing.
wearied of the constant arguing
When is it sensible to use exhaust instead of weary?
The meanings of exhaust and weary largely overlap; however, exhaust implies complete draining of strength by hard exertion.
shoveling snow exhausted him
In what contexts can fatigue take the place of weary?
The synonyms fatigue and weary are sometimes interchangeable, but fatigue suggests great lassitude from excessive strain or undue effort.
fatigued by the day's chores
Where would jade be a reasonable alternative to weary?
While the synonyms jade and weary are close in meaning, jade suggests the loss of all freshness and eagerness.
appetites jaded by overindulgence
When is tire a more appropriate choice than weary?
Although the words tire and weary have much in common, tire implies a draining of one's strength or patience.
the long ride tired us out
Example Sentences
Examples are automatically compiled from online sources to show current usage.Read More Opinions expressed in the examples do not represent those of Merriam-Webster or its editors. Send us feedback.
Recent Examples of weary
Adjective
Messengers, all of us, speaking the word God left us in this weary land.—Ashley M. Jones, The Atlantic, 14 Sep. 2025 Underneath the stark floodlights streamed a procession of weary travelers in T-shirts and jeans, reaching into the bottom of a white coach bus for their oversize duffel bags.—ProPublica, 13 Sep. 2025
Verb
Johnny works 40 minutes past one and goes to bed wearied by the Sisyphean task.—Rafaela Bassili, Vulture, 1 Apr. 2025 Amid all this morphing, what has stayed constant is that COVID has been, in one way or another, wearying in a bone-deep way.—Meghan Bartels, Scientific American, 7 Mar. 2025 See All Example Sentences for weary
Recent Examples of Synonyms for weary
1. tired
2. bored
3. tiring
4. bore
5. wear
6. exhausted
7. boring
8. tire
Adjective
If tired, float or tread water until out of the rip current. — NC Weather Bot, Charlotte Observer, 21 Sep. 2025
Or, the formula comes in a radiant finish, which both makeup artists love for dull or tired skin in need of a pick-me-up. — Lily Wohlner, Allure, 19 Sep. 2025
Definition of tired
Adjective
They are easily trained, thanks to how smart the breed is, but require plenty of mental stimulation to prevent them from getting bored. — Rachael O'Connor, MSNBC Newsweek, 23 Sep. 2025
This adaptation of an Ibsen classic reimagines a bored Scandinavian newlywed into a lesbian love triangle and a whole lot of chaos. — Brittany Allen, Literary Hub, 23 Sep. 2025
Definition of bored
Adjective
Without a strict hierarchy or single decision-maker, our process can sometimes be slow, messy, or even tiring. — Caterina De Biasio, Vogue, 11 Sep. 2025
The film is a mess, opaque in its argument and tiring in its effortful weirdness, and yet in its best moments has a hypnotic pull. — Richard Lawson, HollywoodReporter, 7 Sep. 2025
Definition of tiring
Verb
Her mix of funky statement pieces and essentials strikes the ideal balance—proof that wearable doesn’t have to mean boring. — Minty Mellon, Vogue, 27 Sep. 2025
The monochrome noir look is a default of fashion house designers and global creatives—basic is far from boring in Milan. — Abbey Hudetz, Travel + Leisure, 27 Sep. 2025
Definition of bore
Verb
Johnson styled it simply with some silver and black cocktail rings, wearing her signature bangs and waist-length hair long and loose. — Anna Cafolla, Vogue, 26 Sep. 2025
The contestants come out wearing their final looks, and Michael Kors says that everyone’s collections were thoughtful and well-constructed, and then the judging begins. — Roxana Hadadi, Vulture, 26 Sep. 2025
Definition of wear
Adjective
Founders may feel emotionally exhausted or addicted to the thrill of building. — Anuradha Gupta, Forbes.com, 19 Sep. 2025
From severe morning sickness to complications that required hospitalization, the experience left me exhausted and anxious about the possibility of going through it again. — Harriette Cole, Mercury News, 18 Sep. 2025
Definition of exhausted
Adjective
And producers are looking again for the serious, boring money with no strings attached. — Ed Meza, Variety, 25 Sep. 2025
The resulting larvae both look and act like screws, boring and twisting into the animal while feasting on its living flesh. — Beth Mole, ArsTechnica, 24 Sep. 2025
Definition of boring
Verb
Ali covered his face and tried to absorb Frazier’s metronomic body blows, aiming to tire him. — Vann R. Newkirk II, The Atlantic, 16 Sep. 2025
Incoming waves in rapid succession can tire even an experienced swimmer quickly. — Anna Skinner, MSNBC Newsweek, 15 Sep. 2025
Definition of tire
Browse Nearby Words
wear to a frazzle
weary
wearying
See all Nearby Words
Articles Related to weary
### 'Weary' and 'Wary': Use With Caution Beware of the difference.### The Tired History of 'Fatigue' A relatively new word for an ancient feeling
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Nglish: Translation of weary for Spanish Speakers
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16040 | https://www.thesassymathteacher.com/order-of-operations/ | Published Time: 2023-09-24T16:25:09+00:00
4 Fun Order of Operations Activities - The Sassy Math Teacher
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4 Fun Order of Operations Activities
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Are you on the hunt for fun order of operations activities that will help your 7th and 8th grade students not only understand but also master the order of operations? Well, your quest ends here! In this blog post, I’m giving you the TEA on my four most popular order of operations activities that will not only make learning order of operations a breeze but also keep your students actively engaged in the process.
You are going to want to incorporate these ideas into your lesson plans. I’m sharing 4 different waysyou can practice this important concept with your students. Students need lots of practice!
Building a Strong Foundation
Before we delve into the activities themselves, let’s remind ourselves why understanding the order of operations is crucial. The order of operations is the set of rules that dictate the sequence in which different operations should be performed. Without a solid grasp of these rules, students may find themselves lost when dealing with complex numerical expressions.
Activity 1:Order of Operations Activity– Surgery Edition
As a devoted fan of the TV show Grey’s Anatomy, I couldn’t resist the temptation to incorporate surgery into my order of operations lesson last year. The Surgery Edition activity adds an element of excitement and drama to the learning process.
How It Works
Students move around the room, tackling a variety of expressions that need to be simplified using the order of operations. Each expression represents a patient in need of surgery.
When they arrive at an answer, they turn their attention to a board where they’ll find their answer and determine which “patient” is in each “operating room.” This adds an element of suspense and intrigue.
To intensify the experience, consider playing some Grey’s Anatomy background music to set the mood and immerse students in the surgical atmosphere.
This activity not only reinforces the order of operations but also injects an element of fun and suspense that students will undoubtedly appreciate. It’s a great way to make abstract math concepts come to life with an engaging resource.
Activity 2:Order of Operations Relay Race
Collaboration is a valuable skill, and the Order of Operations Relay Race is the perfect way to encourage teamwork and critical thinking. Even those it’s called a race, it is not necessarily a fast-paced game. Here’s how you can organize this exciting relay race in your classroom:
How It Works
Divide students into teams and have them sit in a vertical row.
Each team works together to simplify one expression at a time. The twist? Each student can only complete one operation before passing the expression to the student behind them.
The game continues until the expression is completely simplified. The student who finishes the last step brings it to you for verification.
The first team to arrive at the correct answer earns 5 points, while the other teams get 1 point each.
If a team’s answer is incorrect, provide constructive feedback by highlighting where the mistake occurred. Encourage them to go back to their group and try again.
While this may seem like a race, students will soon realize that accuracy takes precedence over speed. This activity fosters not only mathematical understanding but also teamwork and communication skills, which are essential for success both in and out of the classroom.
Fun Activity 3:Order of Operations Game
Who said learning couldn’t be combined with a dash of Thanksgiving-themed fun? The Order of Operations Game is a crowd-pleaser because it combines learning with excitement. It’s the perfect review activity.
How It Works
Students work in small groups to tackle challenging order of operations math problems. The game board features a turkey-themed oven where each oven is worth a mysterious point value.
After each challenge, one team member brings their solution for verification. If the answer is correct, the team gets to choose an oven for their virtual turkey.
The catch is that the point value of each oven is unknown, adding an element of chance and strategy to the game.
This game not only reinforces the order of operations but also adds a layer of competition and suspense that students will find irresistibly entertaining. It’s a great way to keep the learning atmosphere festive and engaging.
Activity 4:Order of Operations Coloring Worksheet
Who doesn’t love a good fall-themed activity? This Order of Operations Coloring Worksheet is best suited for higher level classes because it is both challenging and aesthetically pleasing, making it the perfect extension activity.
How It Works
Students are tasked with simplifying a series of expressions, with each answer corresponding to a specific color on a coloring page.
As they progress through the problems, they fill in the coloring page accordingly.
In the end, they’ll have a beautifully colored creation that doubles as classroom decoration. This not only reinforces the order of operations but also appeals to students’ creative sides.
This activity is an excellent way to combine math with art, allowing students to express their understanding in a creative and visually appealing manner. It’s also a great option for those who prefer working individually and at their own pace.
The Power of Immediate Feedback
One of the remarkable features of these order of operation activities is their ability to break down each step, allowing you to pinpoint and address mistakes early on. Immediate feedback is the key to effective learning, as it helps students understand and correct their errors.
Additional Resources and Tips
While these activities are incredibly effective in teaching the order of operations, there are many other resources and strategies you can employ to reinforce this essential math concept. Here are some additional tips and ideas:
– Interactive Online Tools: Explore online resources and interactive tools that allow students to practice the order of operations in a digital format. Many websites and apps offer engaging activities and games.
– Real-World Applications: Connect the order of operations to real-world scenarios to make it more relevant for your students. For example, discuss how engineers use these principles when designing structures or how scientists apply them in experiments.
– Peer Teaching: Encourage students to explain the order of operations to their peers. Teaching a concept to someone else reinforces their own understanding and can lead to meaningful discussions.
– Hands-On Manipulatives: Use physical objects, such as manipulative materials or even everyday items like fruits or candies, to demonstrate the order of operations. This hands-on approach can make the concept more tangible.
– Practice, Practice, Practice: Like any math skill, mastering the order of operations requires practice. Provide plenty of opportunities for students to have additional practice with a variety of expressions and problems.
Fun Order of Operations Activities Conclusion
Incorporating these engaging order of operation activities into your curriculum can make a significant difference in your students’ understanding and retention of this essential math concept. The combination of hands-on experiences, teamwork, creativity, and immediate feedback creates a dynamic learning environment that fosters deeper comprehension.
If you try any of thesefun order of operations activities, I hope they provide meaningful practice and I’d love to hear all about it!
Stay tuned for my next blog post, which will be packed with even more valuable tips and resources for teaching the order of operations.
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16041 | https://www.rapidtables.com/convert/length/mile-to-feet.html | Home›Conversion›Length conversion›Miles to feet
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How to convert miles to feet
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The distance d in feet (ft) is equal to the distance d in miles (mi) times 5280:
d(ft) = d(mi) × 5280
Example
Convert 20 mi to feet:
d(ft) = 20mi × 5280 = 105600ft
How many feet in a mile
One mile is equal to 5280 feet:
1mi = 1mi × 5280 = 5280ft
How many miles in a foot
One foot is equal to 1/5280 miles:
1ft = 1ft/5280 = 1.893939e-4mi
How to convert 10mi to feet
Multiply 10 miles by 5280 to get feet:
10mi = 10mi × 5280 = 52800ft
Miles to feet conversion table
| Miles (mi) | Feet (ft) |
--- |
| 0.01 mi | 52.8 ft |
| 0.1 mi | 528 ft |
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| 2 mi | 10560 ft |
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| 10 mi | 52800 ft |
| 20 mi | 105600 ft |
| 30 mi | 158400 ft |
| 40 mi | 211200 ft |
| 50 mi | 264000 ft |
| 60 mi | 316800 ft |
| 70 mi | 369600 ft |
| 80 mi | 422400 ft |
| 90 mi | 475200 ft |
| 100 mi | 528000 ft |
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16042 | https://it.wikipedia.org/wiki/Velocit%C3%A0_al_suolo | Velocità al suolo - Wikipedia
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La velocità al suolo (in lingua inglese Ground Speed, abbreviata in GS) è la velocità orizzontale di un aeromobile rispetto al terreno. Un aereo con assetto verticale avrebbe una velocità al suolo pari a zero. Le informazioni visualizzate ai passeggeri attraverso il sistema di intrattenimento spesso danno la velocità al suolo piuttosto che la velocità rispetto all'aria.
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La velocità al suolo può essere determinata dalla somma vettoriale della proiezione al suolo della velocità vera del velivolo rispetto all'aria, più la velocità del vento e la sua direzione; un vento contrario sottrae la velocità di avanzamento, mentre un vento in coda la aggiunge ad esso. Venti in altri angoli rispetto alla prua avranno componenti sia di vento contrario o di vento in coda, così come una componente di vento laterale. Con le velocità degli aerei commerciali correnti, la velocità del vento pesa solitamente per una piccola percentuale.
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16043 | https://www.youtube.com/watch?v=q7jHR9ar1Fo | The sum and product of finite sequences
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16044 | https://ems.press/content/serial-article-files/36987 | EMS Surv. Math. Sci. 1 (2014), 249–282 DOI 10.4171/EMSS/6
EMS Surveys in Mathematical Sciences
c© European Mathematical Society
The Green–Tao theorem: an exposition
David Conlon ∗
, Jacob Fox ∗∗
, and Yufei Zhao ‡
Abstract. The celebrated Green-Tao theorem states that the prime numbers contain arbitrarily long arithmetic progressions. We give an exposition of the proof, incorporating several simplifications that have been discovered since the original paper.
Mathematics Subject Classification (2010 ). 11–02, 05–02, 11B30, 11B25, 11A41.
Keywords. Green-Tao theorem, primes in arithmetic progression, transference principles.
1. Introduction
In 2004, Ben Green and Terence Tao proved the following celebrated theorem, re-solving a folklore conjecture about prime numbers.
Theorem 1.1 (Green-Tao). The prime numbers contain arbitrarily long arithmetic pro-gressions.
Our intention is to give a complete proof of this theorem. Although there have been numerous other expositions [21, 22, 28, 29, 43, 45, 47], we were prompted to write this note because of our recent work [7, 51] simplifying one of the key technical ingredients in the proof. Together with work of Gowers , Reingold, Trevisan, Tulisiani, and Vadhan , and Tao , there have now been substantial simplifications to almost every aspect
∗The first author was supported by a Royal Society University Research Fellowship.
∗∗ The second author was supported by a Packard Fellowship, NSF Career Award DMS-1352121, an Alfred P. Sloan Fellowship, and an MIT NEC Corporation Award.
‡The third author was supported by a Microsoft Research PhD Fellowship. D. Conlon, Mathematical Institute, Oxford OX2 6GG, United Kingdom E-mail: david.conlon@maths.ox.ac.uk J. Fox, Department of Mathematics, MIT, Cambridge, MA 02139-4307 E-mail: fox@math.mit.edu Y. Zhao, Department of Mathematics, MIT, Cambridge, MA 02139-4307 E-mail: yufeiz@math.mit.edu
250 D. Conlon, J. Fox, and Y. Zhao
of the proof. We have chosen to collect these simplifications and present an up-to-date exposition in order to make the proof more accessible. A key element in the proof of Theorem 1.1 is Szemerédi’s theorem on arithmetic progressions in dense subsets of the integers. To state this theorem, we define the upper density of a set A ⊆ N to be
lim sup
N→∞
|A ∩ [N ]|
N , where [N ] := {1, 2, . . . , N }.
Theorem 1.2 (Szemerédi). Every subset of N with positive upper density contains arbi-trarily long arithmetic progressions.
Szemerédi’s theorem is a deep and important result and the original proof is long and complex. It has had a huge impact on the subsequent development of combinatorics and, in particular, was responsible for the introduction of the regularity lemma , now a cornerstone of modern combinatorics. Numerous different proofs of Szemerédi’s theo-rem have since been discovered and all of them have introduced important new ideas that grew into active areas of research. The three main modern approaches to Szemerédi’s theorem are by ergodic theory [13, 15], higher order Fourier analysis [17, 18], and hyper-graph regularity [20, 31, 34, 35, 46]. However, none of these approaches are easy. We shall therefore assume Szemerédi’s theorem as a black box and explain how to derive the Green-Tao theorem using it. As the set of primes has density zero, Szemerédi’s theorem does not immediately im-ply the Green-Tao theorem. Nevertheless, Erd˝ os famously conjectured that the density of the primes alone should guarantee the existence of long APs. 1 Specifically, he conjec-tured that any subset A of N with divergent harmonic sum, i.e., ∑
a∈A
1/a = ∞, must contain arbitrarily long APs. This conjecture is widely believed to be true, but it has yet to be proved even in the case of 3-term APs. 2
If not by density considerations, how do Green and Tao prove their theorem? The answer is that they treat Szemerédi’s theorem as a black box and show, through a trans-ference principle , that a Szemerédi-type statement holds relative to sparse pseudorandom subsets of the integers, where a set is said to be pseudorandom if it resembles a random set of similar density in terms of certain statistics or properties. We refer to such a state-ment as a relative Szemerédi theorem . Given two sets A and S with A ⊆ S, we define the relative upper density of A in S to be lim sup N →∞ |A ∩ [N ]| / |S ∩ [N ]|.
Relative Szemerédi theorem. (Informally) If S is a (sparse) set of integers satisfying certain pseudorandomness conditions and A is a subset of S with positive relative density, then A contains arbitrarily long APs.
1For brevity, we will usually write AP for arithmetic progression and k-AP for a k-term AP.
2A recent result of Sanders is within a hair’s breadth of verifying Erd˝ os’ conjecture for 3-APs. Sanders proved that every 3-AP-free subset of [N]has size at most O(N(log log N)6/log N), which is just slightly shy of the logarithmic density barrier that one wishes to cross (see Bloom for a recent improvement). In the other direction, Behrend constructed a 3-AP-free subset of [N]of size N e −O(√log N). There is some evidence to suggest that Behrend’s lower bound is closer to the truth (see ). For longer APs, the gap is much larger. The best upper bound, due to Gowers , is that every k-AP-free subset of [N]has size at most
N/ (log log N)ckfor some ck>0(though for k= 4 there have been some improvements ). The Green–Tao theorem: an exposition
251 To prove the Green-Tao theorem, it then suffices to show that there is a set of “almost primes” containing, but not much larger than, the primes which satisfies the required pseu-dorandomness conditions. In the work of Green and Tao, there are two such conditions, known as the linear forms condition and the correlation condition. The proof of the Green-Tao theorem therefore falls into two parts, the first part being the proof of the relative Szemerédi theorem and the second part being the construction of an appropriately pseudorandom superset of the primes. Green and Tao credit the contem-porary work of Goldston and Yıldırım for the construction and estimates used in the second half of the proof. Here we will follow a simpler approach discovered by Tao . The proof of the relative Szemerédi theorem also splits into two parts, the dense model theorem and the counting lemma. Roughly speaking, the dense model theorem allows us to say that if S is a sufficiently pseudorandom set then any relatively dense subset
A of S may be “approximated” by a dense subset ˜A of N, while the counting lemma shows that the number of arithmetic progressions in A is close, up to a normalization factor, to the number of arithmetic progressions in ˜A. Since ˜A is a dense subset of N,Szemerédi’s theorem implies that ˜A contains arbitrarily long APs and this in turn implies that A contains arbitrarily long APs. This is also the outline we will follow in this paper, though for each part we will follow a different approach to the original paper. For the counting lemma, we will follow the recent approach taken by the authors in . This approach has significant advantages over the original method of Green and Tao, not least of which is that a weakening of the linear forms condition is sufficient for the relative Szemerédi theorem to hold. This means that the estimates involved in verifying the correlation condition may now be omitted from the proof. In , the dense model theorem was replaced with a certain sparse regularity lemma. However, as subsequently observed by Zhao , the original dense model theorem may also be used. To prove the dense model theorem, we will follow an elegant method de-veloped independently by Gowers and by Reingold, Trevisan, Tulsiani, and Vadhan . The 3-AP case of Szemerédi’s theorem was first proved by Roth in the 1950s. While Roth’s theorem, as this case is usually known, is already a very interesting and nontrivial result, the 3-AP case is substantially easier than the general result. In contrast, when proving a relative Szemerédi theorem by transferring Szemerédi’s theorem down to the sparse setting, the general case is not mathematically more difficult than the 3-AP case. However, as one might expect, the notation for the general case can be rather cumbersome. For this reason, we explain various aspects of the proof first for 3-APs and only afterwards discuss how it can be adapted to the general case. We begin, in Section 2, by presenting the Ruzsa-Szemerédi graph-theoretic approach to Roth’s theorem. In particular, we present a graph-theoretic construction that will mo-tivate the definition of the linear forms conditions, which we state in Sections 3 and 4, first for Roth’s theorem, then for Szemerédi’s theorem. The dense model theorem and the counting lemma are explained in Sections 5 and 6, respectively. We conclude the proof of the relative Szemerédi theorem in Section 7. In Sections 8 and 9, we will construct the rel-evant set of almost primes (or rather a majorizing measure for the primes) and show that 252 D. Conlon, J. Fox, and Y. Zhao
it satisfies the linear forms condition. We conclude with some remarks about extensions of the Green-Tao theorem.
2. Roth’s theorem via graph theory
One way to state Szemerédi’s theorem is that for every fixed k every k-AP-free subset of
[N ] has o(N ) elements. It is not hard to prove that this “finitary” version of Szemerédi’s theorem is equivalent to the “infinitary” version stated as Theorem 1.2. In fact, it will be more convenient to work in the setting of the abelian group
ZN := Z/N Z as opposed to [N ]. These two settings are roughly equivalent for studying
k-APs, with the only difference being that ZN allows APs to wrap around 0. For example,
N − 1, 0, 1 is a 3-AP in ZN , but not in [N ]. To deal with this issue, one simply embeds
[N ] into a slightly larger cyclic group so that no k-APs wrap around zero. Working in
ZN , we will now show how Roth’s theorem follows from a result in graph theory.
Theorem 2.1 (Roth). If A ⊆ ZN is 3-AP-free, then |A| = o(N ).
GA
X = ZN
Y = ZN Z = ZN
x
yz
x∼yiff
2x+y∈A
x∼ziff
x−z∈A
y∼ziff
−y−2z∈A
Figure 1. The construction in the proof of Roth’s theorem.
Consider the following graph construction (see Figure 1). Given A ⊆ ZN , we con-struct a tripartite graph GA whose vertex sets are X, Y , and Z, each with N vertices labeled by elements of ZN . The edges are constructed as follows (one may think of this as a variant of the Cayley graph for ZN generated by A):
•
(x, y ) ∈ X × Y is an edge if and only if 2x + y ∈ A;
•
(x, z ) ∈ X × Z is an edge if and only if x − z ∈ A;
•
(y, z ) ∈ Y × Z is an edge if and only if −y − 2z ∈ A.Observe that (x, y, z ) ∈ X × Y × Z forms a triangle if and only if all three of
2x + y, x − z, −y − 2zThe Green–Tao theorem: an exposition 253 are in A. These numbers form a 3-AP with common difference −x − y − z, so we see that triangles in GA correspond to 3-APs in A.However, we assumed that A is 3-AP-free. Does this mean that GA has no triangles? Not quite. There are still some triangles in GA, namely those that correspond to trivial
3-APs in A, i.e., 3-APs with common difference zero. So the triangles in GA are precisely those with x+y +z = 0 . This easily implies that every edge in GA is contained in exactly one triangle, namely the one that completes the equation x + y + z = 0 .What can we say about a graph where every edge is contained in exactly one triangle? The following result of Ruzsa and Szemerédi shows that it cannot have many edges.
Theorem 2.2 (Ruzsa-Szemerédi). If G is a graph on n vertices with every edge in exactly one triangle, then G has o(n2) edges.
Our graph GA has 3N vertices and 3N |A| edges (for every x ∈ X, there are exactly
|A| vertices y ∈ Y with 2x + y ∈ A and similarly for Y × Z and X × Z). So it follows by Theorem 2.2 that 3N |A| = o((3 N )2). Hence |A| = o(N ), proving Roth’s theorem. Theorem 2.2 easily follows from a result known as the triangle removal lemma , which says that if a graph on n vertices has o(n3) triangles, then it can be made triangle-free by removing o(n2) edges. Though both results look rather innocent, it is only recently [6, 10] that a proof was found which avoids the use of Szemerédi’s regularity lemma. We will not include a proof of Theorem 2.2 here, since this would lead us too far down the route of proving Szemerédi’s theorem. However, if our purpose was not to prove Roth’s theorem, then why translate it into graph-theoretic language in the first place? The reason is that the counting lemma and pseudorandomness conditions used for transferring Roth’s theorem to the sparse setting are most naturally phrased in terms of graph theory. 3
We will begin to make this explicit in the next section.
3. Relative Roth theorem
In this section, we describe the relative Roth theorem. We first give an informal statement.
Relative Roth Theorem. (Informally) If S ⊆ ZN satisfies certain pseudorandomness conditions and A ⊆ S is 3-AP-free, then |A| = o(|S|).
To state the pseudorandomness conditions, let p = |S| /N (which may decrease as a function of N ) and consider the graph GS . This is similar to GA, except that (x, y ) ∈
X × Y is now made an edge if and only if 2x + y ∈ S, etc. The pseudorandomness hypothesis now asks that the number of embeddings of K2,2,2 in GS (i.e., the number of tuples (x1, x 2, y 1, y 2, z 1, z 2) ∈ X × X × Y × Y × Z × Z where xiyj , x izj , y izj are all edges for all i, j ∈ { 1, 2}) be equal to (1 + o(1)) p12 N 6, where o(1) indicates a quantity that tends to zero as N tends to infinity. This is asymptotically the same as the expected number of embeddings of K2,2,2 in a random tripartite graph of density p or in the graph
3However, it is worth stressing that the bounds in the relative Roth theorem do not reflect the poor bounds given by the graph theoretic approach to Roth’s theorem. While graph theory is a convenient language for phrasing the transference principle, Roth’s theorem itself only appears as a black box and any bounds we have for this theorem transfer directly to the sparse version.
254 D. Conlon, J. Fox, and Y. Zhao
GS
X = ZN
Y = ZN Z = ZN
x
yz
x∼yiff
2x+y∈S
x∼ziff
x−z∈S
y∼ziff
−y−2z∈S
K2,2,2& subgraphs, e.g.,
Pseudorandomness hypothesis for S ⊆ ZN :
GShas asymptotically the expected number of embeddings of
Figure 2. Pseuodrandomness conditions for the relative Roth theorem.
GS formed from a random set S of density p. Assuming that p does not decrease too rapidly with N , it is possible to show that with high probability the true K2,2,2-count in these random graphs is asymptotic to this expectation. It is therefore appropriate to think of our condition as a type of pseudorandomness. For technical reasons, it is necessary to assume that this property of having the “cor-rect” count holds not only for K2,2,2 but also for every subgraph H of K2,2,2. That is, we ask that the number of embeddings of H into GS (with vertices of H mapped into their as-signed parts) be equal to (1 + o(1)) pe(H)N v(H). The full description is now summarized in Figure 2, although we will restate it in more formal terms later on. To see why this is a natural pseudorandomness hypothesis, we recall a famous result of Chung, Graham, and Wilson . This result says that several seemingly different notions of pseudorandomness for dense graphs (i.e., graphs with constant edge density) are equivalent. These notions are based, for example, on measuring eigenvalues, edge discrepancy, subgraph counts, or codegree distributions. One rather striking fact is that having the expected 4-cycle count turns out to be equivalent to all of the other definitions. For sparse graphs, these equivalences do not hold. While having the correct count for 4-cycles, which may be seen as the 2-blow-up of an edge (see Figure 3), still gives some control over the distribution of edges in the graph, this property is no longer strong enough to control the distribution of other small graphs such as triangles. This is where the pseudorandomness condition described above becomes useful, because knowing that The Green–Tao theorem: an exposition 255 we have approximately the correct count for K2,2,2, the 2-blow-up of a triangle, does allow one to control the distribution of triangles.
2-blow-up
−−−−−→ 2-blow-up
−−−−−→
Figure 3. The 2-blow-up of a graph is constructed by duplicating each vertex.
We now sketch the idea behind the proof of the relative Roth theorem. We begin by noting that Roth’s theorem can be rephrased as follows.
Theorem 3.1 (Roth). For every δ > 0, every A ⊆ ZN with |A| ≥ δN contains a 3-AP, provided N is sufficiently large.
By a simple averaging argument (attributed to Varnavides ), this version of Roth’s theorem is equivalent to the claim that A contains not just one, but many 3-APs.
Theorem 3.2 (Roth’s theorem, counting version). For every δ > 0, there exists
c = c(δ) > 0 such that every A ⊆ ZN with |A| ≥ δN contains at least cN 2 3-APs, provided N is sufficiently large.
To prove the relative Roth theorem from Roth’s theorem, assume that A ⊆ S ⊆ ZN
is such that |A| ≥ δ |S|. The first step is to show that there is a dense model ˜A for A.This is a dense subset of ZN such that | ˜A|/N ≈ | A|/|S| ≥ δ and ˜A approximates A in the sense of a certain cut norm. The second step is to use this cut norm condition to prove a counting lemma, which says that ˜A and A contain roughly the same number of 3-APs (after an appropriate normalization), i.e.,
(N/ |S|)3 |{ 3-APs in A}| ≈ |{ 3-APs in ˜A}| .
Since the counting version of Roth’s theorem implies that |{ 3-APs in ˜A}| ≥ cN 2, the relative Roth theorem is proved. This discussion is fairly accurate, except for one white lie, which is that it is more correct to think of ˜A as a weighted function from ZN to [0 , 1] than as a subset of ZN . It will therefore be more convenient to work with the following weighted version of Roth’s theorem. At this point, it is worth fixing some notation. We will write Ex1∈X1,...,x k ∈Xk
as a shorthand for |X1|−1 · · · | Xk|−1 ∑
x1∈X1
· · · ∑
xk∈Xk
. If the variables x1, . . . , x k or the sets X1, . . . , X k are understood, we will sometimes choose to omit them. We will also write o(1) to indicate a function that tends to zero as N tends to infinity, indicating further dependencies by subscripts when they are not understood.
Theorem 3.3 (Roth’s theorem, weighted version). For every δ > 0, there exists
c = c(δ) > 0 such that every f : ZN → [0 , 1] with Ef ≥ δ satisfies
Ex,d ∈ZN [f (x)f (x + d)f (x + 2 d)] ≥ c − oδ (1) . (1) Note that when f is {0, 1}-valued, i.e., f = 1 A is the indicator function of some set
A, this reduces to the counting version of Roth’s theorem. Up to a change of parameters, 256 D. Conlon, J. Fox, and Y. Zhao
Sets Functions Dense setting
A ⊆ ZN
|A| ≥ δN f : ZN → [0 , 1]
Ef ≥ δ
Sparse setting
A ⊆ S ⊆ ZN
|A| ≥ δ |S|
f ≤ ν : ZN → [0 , ∞)
Ef ≥ δ, Eν = 1 + o(1)
Table 1. Comparing the set version with the weighted version.
the counting version also implies the weighted version. Indeed, to deduce the weighted version from the counting version, let A = {x ∈ ZN | f (x) ≥ δ/ 2}. If Ef ≥ δ and
0 ≤ f ≤ 1, then |A| ≥ δN/ 2, so
Ex,d ∈ZN [f (x)f (x + d)f (x + 2 d)] ≥ (δ/ 2) 3Ex,d ∈ZN [1 A(x)1 A(x + d)1 A(x + 2 d)] .
By the counting version of Roth’s theorem, this is bounded below by a positive constant when N is sufficiently large. When working in the functional setting, we also replace the set S by a function
ν : ZN → [0 , ∞). This function ν, which we call a majorizing measure , will be nor-malized to satisfy 4
Eν = 1 + o(1) .
The subset A ⊆ S will be replaced by some function f : ZN → [0 , ∞) majorized by ν,that is, such that 0 ≤ f (x) ≤ ν(x) for all x ∈ ZN (we write this as 0 ≤ f ≤ ν). The hypothesis |A| ≥ δ|S| will be replaced by Ef ≥ δ. Note that ν and f can be unbounded, which is a major source of difficulty. The main motivating example to bear in mind is that when A ⊆ S ⊆ ZN , we take ν(x) = N
|S|
1S (x) and f (x) = ν(x)1 A(x), noting that if |A| ≥ δ|S| then Ef ≥ δ. We refer the reader to Table 1 for a summary of this correspondence. We can now state the pseudorandomness condition in a more formal way. We modify the graph GS to a weighted graph Gν , which, for brevity, we usually denote by ν. This is a weighted tripartite graph with vertex sets X = Y = Z = ZN and edge weights given by:
•
νXY (x, y ) = ν(2 x + y) for all (x, y ) ∈ X × Y ;
•
νXZ (x, z ) = ν(x − z) for all (x, z ) ∈ X × Z;
•
νY Z (y, z ) = ν(−y − 2z) for all (y, z ) ∈ Y × Z.We will omit the subscripts if there is no risk of confusion. The pseudorandomness con-dition then says that the weighted graph ν has asymptotically the expected H-density for any subgraph H of K2,2,2. For example, triangle density in ν is given by the expression
E[ν(x, y )ν(x, z )ν(y, z )] , where x, y, z vary independently and uniformly over X, Y , Z,
4We think of νas a sequence of functions ν(N), though we usually suppress the implicit dependence of ν
on N.
The Green–Tao theorem: an exposition 257 respectively. The pseudorandomness assumption requires, amongst other things, that this triangle density be 1+ o(1) , the normalization having accounted for the other factors. The full hypothesis, involving K2,2,2 and its subgraphs, is stated below.
Definition 3.4 (3-linear forms condition). A weighted tripartite graph ν with vertex sets
X, Y , and Z satisfies the 3-linear forms condition if
Ex,x ′∈X, y,y ′∈Y, z,z ′∈Z [ν(y, z )ν(y′, z )ν(y, z ′)ν(y′, z ′)ν(x, z )ν(x′, z )ν(x, z ′)ν(x′, z ′)
· ν(x, y )ν(x′, y )ν(x, y ′)ν(x′, y ′)] = 1 + o(1) (2) and also (2) holds when one or more of the twelve ν factors in the expectation are erased. Similarly, a function ν : ZN → [0 , ∞) satisfies the 3-linear forms condition 5 if
Ex,x ′,y,y ′,z,z ′∈ZN [ν(−y − 2z)ν(−y′ − 2z)ν(−y − 2z′)ν(−y′ − 2z′)ν(x − z)ν(x′ − z)
· ν(x − z′)ν(x′ − z′)ν(2 x + y)ν(2 x′ + y)ν(2 x + y′)ν(2 x′ + y′)] = 1 + o(1) (3) and also (3) holds when one or more of the twelve ν factors in the expectation are erased.
Remark. The 3-linear forms condition (3) for a function ν : ZN → [0 , ∞) is precisely the same as (2) for the weighted graph Gν .We can now state the relative Roth theorem formally.
Theorem 3.5 (Relative Roth). Suppose ν : ZN → [0 , ∞) satisfies the 3-linear forms condition. For every δ > 0, there exists c = c(δ) > 0 such that every f : ZN → [0 , ∞)
with 0 ≤ f ≤ ν and Ef ≥ δ satisfies
Ex,d ∈ZN [f (x)f (x + d)f (x + 2 d)] ≥ c − oδ (1) .
Moreover, c(δ) may be taken to be the same constant which appears in (1) .Remark. The rate at which the o(1) term in (3.5) goes to zero depends not only on δ but also on the rate of convergence in the 3-linear forms condition.
4. Relative Szemerédi theorem
As in the case of Roth’s theorem, we first state an equivalent version of Szemerédi’s theorem allowing weights.
Theorem 4.1 (Szemerédi’s theorem, weighted version). For every k ≥ 3 and δ > 0,there exists c = c(k, δ ) > 0 such that every f : ZN → [0 , 1] with Ef ≥ δ satisfies
Ex,d ∈ZN [f (x)f (x + d)f (x + 2 d) · · · f (x + ( k − 1) d)] ≥ c − ok,δ (1) . (4)
5We will assume that Nis odd, which simplifies the proof of Theorem 3.5 without too much loss in gener-ality. Theorem 3.5 holds more generally without this additional assumption on N.
258 D. Conlon, J. Fox, and Y. Zhao
The setup for the relative Szemerédi theorem is a natural extension of the previous section. Just as our pseudorandomness condition for 3-APs was related to the graph-theoretic approach to Roth’s theorem, the pseudorandomness condition in the general case is informed by the hypergraph removal approach to Szemerédi’s theorem [20, 31, 34, 35, 46]. Instead of constructing a weighted graph as we did for 3-APs, we now construct a weighted (k − 1) -uniform hypergraph corresponding to k-APs. For example, for 4-APs, the 3-uniform hypergraph corresponding to the majorizing measure ν : ZN → [0 , ∞) is
4-partite, with vertex sets W, X, Y, Z , each with N vertices labeled by elements of ZN .The weighted edges are given by:
•
νW XY (w, x, y ) = ν(3 w + 2 x + y) on W × X × Y ;
•
νW XZ (w, x, z ) = ν(2 w + x − z) on W × X × Z;
•
νW Y Z (w, y, z ) = ν(w − y − 2z) on W × Y × Z;
•
νXY Z (x, y, z ) = ν(−x − 2y − 3z) on X × Y × Z.The linear forms 3w + 2 x + y, 2w + x − z, w − y − 2z, −x − 2y − 3z are chosen because they form a 4-AP with common difference −w − x − y − z and each linear form depends on exactly three of the four variables. The pseudorandomness condition then says that the weighted hypergraph ν contains asymptotically the expected count of
H whenever H is a subgraph of the 2-blow-up of the simplex K(3) 4 . Here K(3) 4 is the complete 3-uniform hypergraph on 4 vertices, that is, with vertices {w, x, y, z } and edges
{wxy, wxz, wyz, xyz }, while the 2-blow-up of K(3) 4 is the 3-uniform hypergraph con-structed by duplicating each vertex in K(3) 4 and joining all those triples which correspond to edges in K(3) 4 . Explicitly, this 2-blow-up has vertex set {w1, w 2, x 1, x 2, y 1, y 2, z 1, z 2}
and edges wixj yk, w ixj zk, w iyj zk, x iyj zk for all i, j, k ∈ { 1, 2}.For general k, we are concerned with K(k−1)
k
, the complete (k − 1) -uniform hy-pergraph on k vertices, while the pseudorandomness condition again asks that a certain weighted k-partite (k−1) -uniform hypergraph contains asymptotically the expected count for every subgraph of the 2-blow-up of K(k−1)
k
. This 2-blow-up is constructed analo-gously to the 2-blow-up of K(3) 4 above and has 2k vertices and k2k−1 edges. For k-APs, the corresponding linear forms are given by the expressions ∑ki=1 (j−i)xi,for each j = k, k − 1, . . . , 1. The condition (5) below is now the natural extension of the
3-linear forms condition (3). When viewed as a hypergraph condition, it asks that the count for any subgraph of the 2-blow-up of K(k−1)
k
be close to the expected count.
Definition 4.2 (Linear forms condition). A function ν : ZN → [0 , ∞) satisfies the k-linear forms condition 6 if
Ex(0) 1 ,x (1) 1 ,...,x (0)
k,x (1)
k∈ZN
[ k∏
j=1
∏
ω∈{ 0,1}[k]{ j}
ν
( k∑
i=1
(j − i)x(ωi)
i
)nj,ω ]
= 1 + o(1) (5) for any choice of exponents nj,ω ∈ { 0, 1}.
6As in the footnote to Definition 3.4, in our proof of Theorem 4.3 we will make the simplifying assumption that Nis coprime to (k−1)! . In the proof of the Green-Tao theorem, one can always make this assumption.
The Green–Tao theorem: an exposition 259 Now we are ready to state the main result in the proof of the Green-Tao theorem.
Theorem 4.3 (Relative Szemerédi). Suppose k ≥ 3 and ν : ZN → [0 , ∞) satisfies the
k-linear forms condition. For every δ > 0, there exists c = c(k, δ ) > 0 such that every
f : ZN → [0 , ∞) with 0 ≤ f ≤ ν and Ef ≥ δ satisfies
Ex,d ∈ZN [f (x)f (x + d)f (x + 2 d) · · · f (x + ( k − 1) d)] ≥ c − ok,δ (1) . (6)
Moreover, c(k, δ ) may be taken to be the same constant which appears in (4) .Remark. The rate at which the o(1) term in (6) goes to zero depends not only on k and δ
but also on the rate of convergence in the k-linear forms condition for ν.Now we outline the proof of the relative Szemerédi theorem. This is simply a rephras-ing of the outline given after Theorem 3.2 for the unweighted version of the relative Roth theorem. We start with 0 ≤ f ≤ ν and Ef ≥ δ. In Section 5, we prove a dense model the-orem which shows that there exists another function ˜f : ZN → [0 , 1] which approximates
f with respect to a certain cut norm. 7 Note that ˜f is bounded (hence “dense” model) and
E ˜f ≥ δ − o(1) . In Section 6, we establish a counting lemma which says that the weighted
k-AP counts in f and ˜f are similar, that is,
Ex,d [f (x)f (x + d) · · · f (x + ( k − 1) d)] = Ex,d [ ˜f (x) ˜f (x + d) · · · ˜f (x + ( k − 1) d)] − o(1) .
The right-hand side is at least c(k, δ ) − ok,δ (1) by Szemerédi’s theorem (Theorem 4.1). Thus the relative Szemerédi theorem follows. We now begin the proof proper.
5. Dense model theorem
Given g : X ×Y → R, viewed as an edge-weighted bipartite graph with vertex set X ∪Y ,the cut norm of g, introduced by Frieze and Kannan (also see [30, Chapter 8]), is defined as
‖g‖ := sup
A⊆X,B ⊆Y
|Ex∈X,y ∈Y [g(x, y )1 A(x)1 B (y)] | . (7) For a weighted 3-uniform hypergraph g : X × Y × Z → R, we define
‖g‖ := sup
A⊆Y×Z, B ⊆X×Z, C ⊆X×Y
|Ex∈X,y ∈Y,z ∈Z [g(x, y, z )1 A(y, z )1 B (x, z )1 C (x, y )] | .
(The more obvious alternative, where we range A, B, C over subsets of X, Y, Z , respec-tively, gives a weaker norm that is not sufficient to guarantee a counting lemma.) More generally, given a weighted r-uniform hypergraph g : X1 × · · · × Xr → R, define
‖g‖ := sup |Ex1∈X1,...,x r ∈Xr [g(x1, . . . , x r )1 A1 (x−1)1 A2 (x−2) · · · 1Ar (x−r )] | , (8)
7In the original Green-Tao approach, they required ˜fand fto be close in a stronger sense related to the Gowers uniformity norm. The cut norm approach we present here requires less stringent pseudorandomness hypotheses for applying the dense model theorem but a stronger counting lemma.
260 D. Conlon, J. Fox, and Y. Zhao
where the supremum is taken over all choices of subsets Ai ⊆ X−i := ∏
j∈[r]{ i}
Xj ,
i ∈ [r], and we write
x−i := ( x1, x 2, . . . , x i−1, x i+1 , . . . , x r ) ∈ X−i
for each i. We extend this definition of cut norm to ZN : for any function f : ZN → R,define
‖f ‖,r := sup |Ex1,...,x r ∈ZN [f (x1 + · · · + xr )1 A1 (x−1)1 A2 (x−2) · · · 1Ar (x−r )] | ,
(9) where the supremum is taken over all A1, . . . , A r ⊆ Zr−1
N
. It is easy to see that this is a norm. Equivalently, it is the hypergraph cut norm applied to the weighted r-uniform hypergraph g : ZrN → R with g(x1, . . . , x r ) = f (x1 + · · · + xr ). For example,
‖f ‖,2 := sup
A,B ⊆ZN
|Ex,y ∈ZN [f (x + y)1 A(x)1 B (y)] | .
The main result of this section is the following dense model theorem (in this particular form due to the third author ). It gives a condition under which it is possible to approximate an unbounded (or sparse) function f by a bounded (or dense) function ˜f .
Theorem 5.1 (Dense model). For every > 0, there exists an ′ > 0 such that the following holds. Suppose ν : ZN → [0 , ∞) satisfies ‖ν − 1‖,r ≤ ′. Then, for ev-ery f : ZN → [0 , ∞) with f ≤ ν, there exists a function ˜f : ZN → [0 , 1] such that
‖f − ˜f ‖,r ≤ .Remark. One may take ′ = exp( −−C ) where C is some absolute constant (independent of r and, more importantly, N ). A more involved dense model theorem (using a norm based on the Gowers uniformity norm rather than the cut norm) was used by Green and Tao in . Its proof was sub-sequently simplified by Gowers and, independently, Reingold, Trevisan, Tulsiani, and Vadhan . Here we follow Gowers’ approach, but specialized to ‖·‖ ,r , which simplifies the exposition. It will be useful to rewrite Ex,y [f (x+y)1 A(x)1 B (y)] in the form 〈f, ϕ 〉 = Ex[f (x)ϕ(x)]
for some ϕ : ZN → R. We have, by a change of variable,
Ex,y [f (x + y)1 A(x)1 B (y)] = Ex,z [f (z)1 A(x)1 B (z − x)] = 〈f, 1A ∗ 1B 〉 ,
where the convolution is defined by h1 ∗ h2(z) := Ex[h1(x)h2(z − x)] . Let Φ2 denote the set of all functions that can be written as a convex combination of convolutions 1A ∗ 1B
with A, B ⊆ ZN . We then have, by convexity,
‖f ‖,2 = sup
A,B ⊆ZN
|〈 f, 1A ∗ 1B 〉| = sup
ϕ∈Φ2
|〈 f, ϕ 〉| .
More generally, given r functions h1, . . . , h r : Zr−1
N
→ R, define their generalized con-volution (h1, . . . , h r )∗ : ZN → R by
(h1, . . . , h r )∗(x) = Ey1,...,y r ∈ZN
y1+··· +yr=x
[h1(y2, · · · , y r )h2(y1, y 3, . . . , y r ) · · · hr (y1, · · · , y r−1)] .The Green–Tao theorem: an exposition 261 For example, when r = 2 , we recover the usual convolution (h1, h 2)∗ = h1 ∗ h2. We similarly have
‖f ‖,r = sup
A1,...,A r⊆Zr−1
N
|〈 f, (1 A1 , . . . , 1Ar )∗〉| = sup
ϕ∈Φr
|〈 f, ϕ 〉| ,
where Φr is the set of all functions ϕ : ZN → R that can be written as a convex combi-nation of generalized convolutions (1 A1 , 1A2 , . . . , 1Ar )∗ with A1, . . . , A r ⊆ Zr−1
N
. The next lemma establishes a key property of Φr .
Lemma 5.2. The set Φr is closed under multiplication, i.e., if ϕ, ϕ ′ ∈ Φr , then ϕϕ ′ ∈ Φr .Proof. It suffices to show that if ϕ = (1 A1 , · · · , 1Ar )∗ and ϕ′ = (1 B1 , . . . , 1Br )∗, where
A1, . . . , A r , B1, . . . , B r ⊆ Zr−1
N
, then ϕϕ ′ ∈ Φr . For any y = ( y1, . . . , y r ) ∈ ZrN , we write Σy = y1 + · · · + yr and y−i = ( y1, . . . , y i−1, y i+1 , . . . , y r ) ∈ Zr−1
N
. Then, for any
x ∈ ZN , we have
ϕ(x)ϕ′(x) = E y,y ′∈ZrN
Σy=Σ y′=x
[1 A1 (y−1)1 B1 (y′−1) · · · 1Ar (y−r )1 Br (y′−r )] = E y,z ∈ZrN
Σy=x, Σz=0
[1 A1 (y−1)1 B1 (y−1 + z−1) · · · 1Ar (y−r )1 Br (y−r + z−r )] = E y,z ∈ZrN
Σy=x, Σz=0
[1 A1∩(B1−z−1)(y−1) · · · 1Ar ∩(Br −z−r )(y−r )] = Ez∈ZrN
Σz=0
[(1 A1∩(B1−z−1), . . . , 1Ar ∩(Br −z−r ))∗(x)] .
This expresses ϕϕ ′ as a convex combination of generalized convolutions. Thus ϕϕ ′ ∈
Φr .For the rest of this section, we fix the value of r and simply write ‖·‖ for ‖·‖ ,r and
Φ for Φr . We have ‖f ‖ = sup ϕ∈Φ|〈 f, ϕ 〉| . An important role in the proof is played by the dual norm, which is defined by ‖ψ‖∗ = sup ‖f ‖≤ 1 〈f, ψ 〉. It follows easily from the definition that |〈 f, ψ 〉| ≤ ‖ f ‖ ‖ ψ‖∗.It is also easy to show that the unit ball for this dual norm is the convex hull of the union of Φ and −Φ. To see that the convex hull is contained in the unit ball, we note that each element of Φ ∪ (−Φ) is in the unit ball and apply the triangle inequality to deduce that the same holds for convex combinations. For the reverse implication, suppose that ψ
is in the unit ball of ‖·‖ ∗ but not in the convex hull of Φ ∪ (−Φ) . Then, by the separating hyperplane theorem, there exists f such that |〈 f, ϕ 〉| ≤ 1 for all ϕ ∈ Φ ∪ (−Φ) and
〈f, ψ 〉 > 1. But the first inequality implies that ‖f ‖ ≤ 1 and so, by the second inequality,
‖ψ‖∗ > 1, contradicting our assumption. By Lemma 5.2, this now implies that the unit ball for the dual norm is closed under multiplication. Thus, for every ϕ, ψ : ZN → R, we have ∥∥(ϕ/ ‖ϕ‖∗)( ψ/ ‖ψ‖∗)∥∥∗ ≤ 1, i.e.,
‖ϕψ ‖∗ ≤ ‖ ϕ‖∗ ‖ψ‖∗ . (10) Finally, we note that ‖·‖ ≤ ‖·‖ 1 and ‖·‖ ∞ ≤ ‖·‖ ∗. The first inequality follows since
‖f ‖ = sup
ϕ∈Φ
|〈 f, ϕ 〉| = sup
ϕ∈Φ
|Exf (x)ϕ(x)| ≤ Ex|f (x)| = ‖f ‖1 .262 D. Conlon, J. Fox, and Y. Zhao
The second inequality follows from duality or by letting x′ be a value for which ψ achieves its maximum and taking f (x) = N for x = x′ and 0 otherwise. It is then straightforward to verify that this function satisfies ‖f ‖ ≤ 1 and
‖ψ‖∗ ≥ |〈 f, ψ 〉| = |ψ(x′)| = ‖ψ‖∞ .
Proof of Theorem 5.1. We may assume without loss of generality that ≤ 110 . It suffices to show that there exists a function ˜f : ZN → [0 , 1 + / 2] with ‖f − ˜f ‖ ≤ / 2. Suppose, for contradiction, that no such ˜f exists. Let
K1 := { ˜f : ZN → [0 , 1 + / 2] } and K2 := {h : ZN → R | ‖ h‖ ≤ / 2}.
We can view K1 and K2 as closed convex sets in RN . By assumption, f / ∈ K1 + K2 :=
{ ˜f +h : ˜f ∈ K1, h ∈ K2}. Therefore, since K1+K2 is convex, the separating hyperplane theorem implies that there exists some ψ : ZN → R such that (a) 〈f, ψ 〉 > 1, and (b) 〈g, ψ 〉 ≤ 1 for all g ∈ K1 + K2.Note that since 0 ∈ K1, K 2, we have K1, K 2 ⊂ K1 + K2. Therefore, in (b), we may take
g = (1 + / 2)1 ψ> 0 ∈ K1, obtaining 〈1, ψ +〉 ≤ (1 + / 2) −1. Here x+ := max {0, x } and
ψ+(x) := ψ(x)+. On the other hand, ranging g over K2, we obtain ‖ψ‖∞ ≤ ‖ ψ‖∗ ≤
2/ , since if 〈g, ψ 〉 ≤ 1 for all g with ‖g‖ ≤ / 2, then 〈g, ψ 〉 ≤ 2/ for all g with ‖g‖ ≤ 1.By the Weierstrass polynomial approximation theorem, there exists some polynomial
P such that |P (x) − x+| ≤ / 8 for all x ∈ [−2/, 2/ ]. Let P (x) = pdxd + · · · +
p1x + p0 and R = |pd| (2 / )d + · · · + |p1| (2 / ) + |p0| (it is possible to take P so that
R = exp( −O(1) )). We write P ψ to mean the function on ZN defined by P ψ (x) = P (ψ(x)) . Using the triangle inequality, (10), and ‖ψ‖∗ ≤ 2/ , we have
‖P ψ ‖∗ ≤
d
∑
i=0
|pi| ‖ ψi‖∗ ≤
d
∑
i=0
|pi| (‖ψ‖∗)i ≤
d
∑
i=0
|pi| (2 / )i = R.
Therefore, since we are assuming that ‖ν − 1‖ ≤ ′,
|〈 ν − 1, P ψ 〉| ≤ ‖ ν − 1‖ ‖ P ψ ‖∗ ≤ ′R.
Since ‖ψ‖∞ ≤ 2/ , we have ‖P ψ − ψ+‖∞ ≤ / 8. Hence,
〈ν, P ψ 〉 ≤ 〈 1, P ψ 〉 + ′R ≤ 〈 1, ψ +〉 + / 8 + ′R ≤ (1 + / 2) −1 + / 8 + ′R.
Also, we have ‖ν‖1 = 〈ν, 1〉 ≤ ‖ ν − 1‖ + 1 ≤ 1 + ′, where we used 〈ν − 1, 1〉 ≤ ‖ν − 1‖ ‖ 1‖∗ and ‖1‖∗ = 1 . Thus,
〈f, ψ 〉 ≤ 〈 f, ψ +〉 ≤ 〈 ν, ψ +〉 ≤ 〈 ν, P ψ 〉 + ‖ν‖1 ‖P ψ − ψ+‖∞
≤ (1 + / 2) −1 + / 8 + ′R + (1 + ′)/ 8.
Since ≤ 110 , the right-hand side is at most 1 when ′ is made sufficiently small (e.g.,
′ = / (8 R)), but this contradicts (a) from earlier. The dense model theorem follows. The Green–Tao theorem: an exposition 263
6. Counting lemma
In this section, we prove the counting lemma. We will focus principally on the graph case, Theorem 6.2 below, since this case contains all the important ideas and is notationally simpler. The hypergraph generalization is then discussed towards the end of the section. For graphs, the counting lemma says that if two weighted graphs are close in cut norm, then they have similar triangle densities. To be more specific, we consider weighted tripartite graphs on the vertex set X ∪ Y ∪ Z, where X, Y , and Z are finite sets. Such a weighted graph g is given by three functions gXY : X × Y → R, gXZ : X × Z → R, and
gY Z : Y × Z → R, although we often drop the subscripts if they are clear from context. We write ‖g‖ = max {‖ gXY ‖ , ‖gXZ ‖ , ‖gY Z ‖}.We first consider the easier case of counting in dense (i.e., bounded weight) graphs (see, for example, ).
Proposition 6.1 (Triangle counting lemma, dense setting). Let g and ˜g be weighted tri-partite graphs on X ∪ Y ∪ Z with weights in [0 , 1] . If ‖g − ˜g‖ ≤ , then
|Ex∈X,y ∈Y,z ∈Z [g(x, y )g(x, z )g(y, z ) − ˜g(x, y )˜ g(x, z )˜ g(y, z )] | ≤ 3.
Proof. Unless indicated otherwise, all expectations are taken over x ∈ X, y ∈ Y , z ∈ Z
uniformly and independently. From the definition (7) of the cut norm, we have that
|Ex∈X,y ∈Y [( g(x, y ) − ˜g(x, y )) a(x)b(y)] | ≤ (11) for every function a : X → [0 , 1] and b : Y → [0 , 1] (since the expectation is bilinear in a
and b, the extrema occur when a and b are {0, 1}-valued, so (11) is equivalent to (7)). It follows that
|E[g(x, y )g(x, z )g(y, z ) − ˜g(x, y )g(x, z )g(y, z )] | ≤ ,
since the expectation has the form (11) if we fix any value of z. Similarly, we have
|E[˜ g(x, y )g(x, z )g(y, z ) − ˜g(x, y )˜ g(x, z )g(y, z )] | ≤
and
|E[˜ g(x, y )˜ g(x, z )g(y, z ) − ˜g(x, y )˜ g(x, z )˜ g(y, z )] | ≤ .
The result then follows from telescoping and the triangle inequality. This proof does not work in the sparse setting, when g is unbounded, since (11) re-quires a and b to be bounded. The main result of this section, stated next for graphs (the hypergraph version is stated towards the end of the section), gives a counting lemma as-suming 0 ≤ g ≤ ν for some ν satisfying the linear forms condition. This is one of the main results in our paper .
Theorem 6.2 (Relative triangle counting lemma). Let ν, g, ˜g be weighted tripartite graphs on X ∪ Y ∪ Z. Assume that ν satisfies the 3-linear forms condition (Definition 3.4),
0 ≤ g ≤ ν, and 0 ≤ ˜g ≤ 1. If ‖g − ˜g‖ = o(1) , then
|Ex∈X,y ∈Y,z ∈Z [g(x, y )g(x, z )g(y, z ) − ˜g(x, y )˜ g(x, z )˜ g(y, z )] | = o(1) .264 D. Conlon, J. Fox, and Y. Zhao
The proof uses repeated application of the Cauchy-Schwarz inequality, a standard technique in this area, popularized by Gowers [17, 18, 19, 20]. The key additional idea, introduced in [7, 8], is densification (see Figure 4). After several applications of the Cauchy-Schwarz inequality, it becomes necessary to analyze the 4-cycle density:
Ex,y,z,z ′ [g(x, z )g(x, z ′)g(y, z )g(y, z ′)] . To do this, one introduces an auxiliary weighted graph g′ : X × Y → [0 , ∞) defined by g′(x, y ) := Ez′ [g(x, z ′)g(y, z ′)] (this is basically the codegree function). Note that we benefit here from working with weighted graphs. The expression for the 4-cycle density now becomes Ex,y,z [g′(x, y )g(x, z )g(y, z )] .
x
yz
z′
g
g
E[g(x, z )g(x, z ′)g(y, z )g(y, z ′)] = E[g′(x, y )g(x, z )g(y, z )]
x
yz
g′
g′(x, y ) =
Ez′∈Z[g(x, z ′)g(y, z ′)]
g
g
Figure 4. The densification step in the proof of the relative triangle counting lemma.
At first glance, it seems that our reasoning is circular. Our aim was to estimate a certain triangle density expression but we have now returned to another triangle density expression. However, g′ behaves much more like a dense weighted graph with bounded edge weights, so what we have accomplished is to replace one of the “sparse” gXY by a “dense” g′
XY
. If we do this two more times, replacing gY Z and gXZ with dense counter-parts, the problem reduces to the dense case, which we already know how to handle. We begin with a warm-up showing how to apply the Cauchy-Schwarz inequality (there will be many more applications later on). The following lemma shows that the 3-linear forms condition on ν implies ‖ν − 1‖ = o(1) , which we need to apply the dense model theorem, Theorem 5.1.
Lemma 6.3. For any ν : X × Y → R,
‖ν − 1‖ ≤ (Ex,x ′∈X,y,y ′∈Y [( ν(x, y )−1)( ν(x′, y )−1)( ν(x, y ′)−1)( ν(x′, y ′)−1)]) 1/4.
(12)
Remark. The right-hand side of (12) is the Gowers uniformity norm of ν − 1. The lemma shows that the cut norm is weaker than the Gowers uniformity norm. To see ‖ν − 1‖ =
o(1) , we expand the right-hand side of (12) into an alternating sum of linear forms in ν,each being 1 + o(1) by the linear forms condition, so that the alternating sum cancels to
o(1) .The Green–Tao theorem: an exposition 265
Proof. By repeated applications of the Cauchy-Schwarz inequality, we have, for A ⊆ X
and B ⊆ Y ,
|Ex,y [( ν(x, y ) − 1)1 A(x)1 B (y)] |4
≤ ∣∣Ex[( Ey [( ν(x, y ) − 1)1 B (y)]) 21A(x)] ∣∣2
≤ ∣∣Ex[( Ey [( ν(x, y ) − 1)1 B (y)]) 2]∣∣2
= |Ex,y,y ′ [( ν(x, y ) − 1)( ν(x, y ′) − 1)1 B (y)1 B (y′)] |2
≤ Ey,y ′ [( Ex[( ν(x, y ) − 1)( ν(x, y ′) − 1)]) 21B (y)1 B (y′)]
≤ Ey,y ′ [( Ex[( ν(x, y ) − 1)( ν(x, y ′) − 1)]) 2]= Ex,x ′,y,y ′ [( ν(x, y ) − 1)( ν(x′, y ) − 1)( ν(x, y ′) − 1)( ν(x′, y ′) − 1)] .
The lemma then follows. The next lemma is crucial to what follows. It shows that in certain expressions a factor
ν can be deleted from an expectation while incurring only a o(1) loss.
Lemma 6.4 (Strong linear forms). Let ν, g, ˜g be weighted tripartite graphs on X ∪Y ∪Z.Assume that ν satisfies the 3-linear forms condition, 0 ≤ g ≤ ν, and 0 ≤ ˜g ≤ 1. Then
Ex∈X,y ∈Y,z,z ′∈Z [( ν(x, y ) − 1) g(x, z )g(x, z ′)g(y, z )g(y, z ′)] = o(1)
and the same statement holds if any subset of the four g factors are replaced by ˜g.Proof. We give the proof when none of the g factors are replaced. The other cases require only a simple modification. By the Cauchy-Schwarz inequality, we have
|Ex,y,z,z ′ [( ν(x, y ) − 1) g(x, z )g(x, z ′)g(y, z )g(y, z ′)] |2
≤ Ey,z,z ′ [( Ex[( ν(x, y ) − 1) g(x, z )g(x, z ′)]) 2g(y, z )g(y, z ′)] Ey,z,z ′ [g(y, z )g(y, z ′)]
≤ Ey,z,z ′ [( Ex[( ν(x, y ) − 1) g(x, z )g(x, z ′)]) 2ν(y, z )ν(y, z ′)] Ey,z,z ′ [ν(y, z )ν(y, z ′)] .
The second factor is at most 1 + o(1) by the linear forms condition. So it remains to ana-lyze the first factor. We have, by another application of the Cauchy-Schwarz inequality,
∣∣Ey,z,z ′ [( Ex[( ν(x, y ) − 1) g(x, z )g(x, z ′)]) 2ν(y, z )ν(y, z ′)] ∣∣2
= |Ex,x ′,y,z,z ′ [( ν(x, y )−1)( ν(x′, y )−1) g(x, z )g(x, z ′)g(x′, z )g(x′, z ′)ν(y, z )ν(y, z ′)] |2
= |Ex,x ′,z,z ′ [Ey [( ν(x, y )−1)( ν(x′, y )−1) ν(y, z )ν(y, z ′)] g(x, z )g(x, z ′)g(x′, z )g(x′, z ′)] |2
≤ Ex,x ′,z,z ′ [( Ey [( ν(x, y )−1)( ν(x′, y )−1) ν(y, z )ν(y, z ′)]) 2g(x, z )g(x, z ′)g(x′, z )g(x′, z ′)]
· Ex,x ′,z,z ′ [g(x, z )g(x, z ′)g(x′, z )g(x′, z ′)]
≤ Ex,x ′,z,z ′ [( Ey [( ν(x, y )−1)( ν(x′, y )−1) ν(y, z )ν(y, z ′)]) 2ν(x, z )ν(x, z ′)ν(x′, z )ν(x′, z ′)]
· Ex,x ′,z,z ′ [ν(x, z )ν(x, z ′)ν(x′, z )ν(x′, z ′)] .
Using the 3-linear forms condition, the second factor is 1 + o(1) and the first factor is o(1)
(expand everything and observe that all the terms are 1 + o(1) and the signs make all the
1’s cancel). 266 D. Conlon, J. Fox, and Y. Zhao
Proof of Theorem 6.2. If ν is identically 1, we are in the dense setting, in which case the theorem follows from Proposition 6.1. Now we apply induction on the number of
νXY , ν XZ , ν Y Z which are identically 1. By relabeling if necessary, we may assume with-out loss of generality that νXY is not identically 1. We define auxiliary weighted graphs
ν′, g ′, ˜g′ : X × Y → [0 , ∞) by
ν′(x, y ) := Ez [ν(x, z )ν(y, z )] ,g′(x, y ) := Ez [g(x, z )g(y, z )] ,
˜g′(x, y ) := Ez [˜ g(x, z )˜ g(y, z )] .
We refer to this step as densification . The idea is that even though ν and g are possibly unbounded, the new weighted graphs ν′ and g′ behave like dense graphs. The weights on ν′ and g′ are not necessarily bounded by 1, but they almost are. We cap the weights by setting g′∧1 := max {g′, 1} and ν′∧1 := max {ν′, 1} and show that the capping has negligible effect. We have
E[g(x, y )g(x, z )g(y, z )−˜g(x, y )˜ g(x, z )˜ g(y, z )] = E[gg ′−˜g˜g′] = E[g(g′−˜g′)]+ E[( g−˜g)˜ g′],
(13) where the first expectation is taken over x ∈ X, y ∈ Y, z ∈ Z and the other expectations are taken over X × Y (we will use these conventions unless otherwise specified). The second term on the right-hand side of (13) equals E[( g(x, y ) − ˜g(x, y ))˜ g(x, z )˜ g(y, z )] and its absolute value is at most ‖g − ˜g‖ = o(1) (here we use 0 ≤ ˜g ≤ 1 as in the proof of Proposition 6.1). So it remains to bound the first term on the right-hand side of (13). By the Cauchy-Schwarz inequality, we have
(E[g(g′ − ˜g′)]) 2 ≤ E[g(g′ − ˜g′)2] E[g] ≤ E[ν(g′ − ˜g′)2] E[ν]= Ex,y [ν(x, y )( Ez [g(x, z )g(y, z ) − ˜g(x, z )˜ g(y, z )]) 2] Ex,y [ν(x, y )] .
The second factor is 1+ o(1) by the linear forms condition. By Lemma 6.4, the first factor differs from
Ex,y [( Ez [g(x, z )g(y, z ) − ˜g(x, z )˜ g(y, z )]) 2] = E[( g′ − ˜g′)2] (14) by o(1) (take the difference, expand the square, and then apply Lemma 6.4 term-by-term). The 3-linear forms condition implies that E[ν′] = 1 + o(1) and E[ν′2] = 1 + o(1) .Therefore, by the Cauchy-Schwarz inequality, we have
(E[|ν′ − 1|]) 2 ≤ E[( ν′ − 1) 2] = o(1) . (15) We want to show that (14) is o(1) . We have
E[( g′ − ˜g′)2] = E[( g′ − ˜g′)( g′ − g′∧1)] + E[( g′ − ˜g′)( g′∧1 − ˜g′)] . (16) Since 0 ≤ g′ ≤ ν′, we have
0 ≤ g′ − g′∧1 = max {g′ − 1, 0} ≤ max {ν′ − 1, 0} ≤ | ν′ − 1|. (17) The Green–Tao theorem: an exposition 267 Using (15) and (17), the absolute value of the first term on the right-hand side of (16) is at most
E[( ν′ + 1) |ν′ − 1|] = E[( ν′ − 1) |ν′ − 1|] + 2 E[|ν′ − 1|] = o(1) .
Next, we claim that
‖g′∧1 − ˜g′‖ = o(1) . (18) Indeed, for any A ⊆ X and B ⊆ Y , we have
Ex,y [( g′∧1 − ˜g′)( x, y )1 A(x)1 B (y)] = E[( g′∧1 − ˜g′)1 A×B ]= E[( g′∧1 − g′)1 A×B ] + E[( g′ − ˜g′)1 A×B ].
By (17) and (15), the absolute value of the first term is at most E[|ν′ − 1|] = o(1) . The second term can be rewritten as
Ex,y,z [1 A×B (x, y )g(x, z )g(y, z ) − 1A×B (x, y )˜ g(x, z )˜ g(y, z )] ,
which is o(1) by the induction hypothesis (replace νXY , g XY , ˜gXY by 1, 1A×B , 1A×B ,respectively, and note that this increases the number of {νXY , ν XZ , ν Y Z } which are iden-tically 1). This proves (18). We now expand the second term on the right-hand side of (16) as
E[( g′ − ˜g′)( g′∧1 − ˜g′)] = E[g′g′∧1] − E[g′ ˜g′] − E[˜ g′g′∧1] + E[˜ g′2]. (19) We claim that each of the expectations on the right-hand side is E[(˜ g′)2] + o(1) . Indeed, we have
E[g′g′∧1] − E[(˜ g′)2] = Ex,y,z [g′∧1(x, y )g(x, z )g(y, z ) − ˜g′(x, y )˜ g(x, z )˜ g(y, z )] ,
which is o(1) by the induction hypothesis (replace νXY , g XY , ˜gXY by 1, g ′∧1, ˜g′, respec-tively, which by (18) satisfies ‖g′∧1 − ˜g′‖ = o(1) , and note that this increases the num-ber of {νXY , ν XZ , ν Y Z } which are identically 1). One can similarly show that the other expectations on the right-hand side of (19) are also E[(˜ g′)2] + o(1) . Thus (19) is o(1) and the theorem follows. The main difficulty in extending Theorem 6.2 to hypergraphs is notational. As dis-cussed in Section 4, to study k-APs, we consider (k − 1) -uniform k-partite weighted hy-pergraphs. The vertex sets will be denoted X1, . . . , X k (in application Xi = ZN for all i). We write X−i := X1×· · ·× Xi−1×Xi+1 ×· · ·× Xk and x−i := ( x1, . . . , x i−1, x i+1 , . . . , x k)
for any x = ( x1, . . . , x k) ∈ X1 × · · · × Xk. Then a weighted hypergraph g consists of functions g−i : X−i → R for each i = 1 , . . . , k . As before, we drop the subscripts if they are clear from context. We write ‖g‖ = max {‖ g−1‖ , . . . , ‖g−k‖}, where ‖g−i‖ is the cut norm of g−i defined in (8). The appropriate generalization of the 3-linear forms condition involves counts for the
2-blow-up of the simplex K(k−1)
k
. We say that a weighted hypergraph ν satisfies the
k-linear forms condition (the hypergraph version of Definition 4.2) if
Ex(0) 1 ,x (1) 1 ∈X1,...,x (0)
k,x (1)
k∈Xk
[ k∏
j=1
∏
ω∈{ 0,1}[k]{ j}
ν(x(ω)
−j
)
]
= 1 + o(1) 268 D. Conlon, J. Fox, and Y. Zhao
and also the same statement holds if any subset of the ν factors (there are k2k−1 such factors) are deleted. Here x(ω)
−j
:= ( x(ω1)1 , . . . , x (ωj−1)
j−1
, x (ωj+1 )
j+1
, . . . , x (ωk )
k
) ∈ X−j .The following theorem generalizes Theorem 6.2.
Theorem 6.5 (Relative simplex counting lemma). Let ν, g, ˜g be weighted (k −1) -uniform
k-partite weighted hypergraphs on X1 ∪ · · · ∪ Xk. Assume that ν satisfies the k-linear forms condition, 0 ≤ g ≤ ν and 0 ≤ ˜g ≤ 1. If ‖g − ˜g‖ = o(1) , then
|Ex1∈X1,...,x k ∈Xk [g(x−1)g(x−2) · · · g(x−k) − ˜g(x−1)˜ g(x−2) · · · ˜g(x−k)] | = o(1) .
The proof of Theorem 6.5 is a straightforward generalization of the proof of Theo-rem 6.2. We simply point out the necessary modifications and leave the reader to figure out the details (a full proof can be found in our paper , but it is perhaps easier to reread the graph case and think about the small changes that need to be made). The proof proceeds by induction on the number of ν−1, . . . , ν −k which are not iden-tically 1. When ν = 1 , we are in the dense setting and the proof of Proposition 6.1 easily extends. Now assume that ν−1 is not identically 1. We have extensions of Lemmas 6.3 and 6.4, where in the proof we have to apply the Cauchy-Schwarz inequality k − 1 times in succession. For the densification step, we define ν′, g ′, ˜g′ : X−1 → [0 , ∞) by
ν′(x−1) = Ex1∈X1 [ν(x−2) · · · ν(x−k)] ,g′(x−1) = Ex1∈X1 [g(x−2) · · · g(x−k)] ,
˜g′(x−1) = Ex1∈X1 [˜ g(x−2) · · · ˜g(x−k)] .
The rest of the proof works with minimal changes.
7. Proof of the relative Szemerédi theorem
We are now ready to prove the relative Szemerédi theorem using the dense model theorem and the counting lemma following the outline given in Section 4.
Proof of Theorem 4.3. The k-linear forms condition implies that ‖ν − 1‖,k −1 = o(1)
(by a sequence of k − 1 applications of the Cauchy-Schwarz inequality, following Lemma 6.3). By the dense model theorem, Theorem 5.1, we can find ˜f : ZN → [0 , 1]
so that ‖f − ˜f ‖,k −1 = o(1) .Let X1 = X2 = · · · = Xk = ZN . For each j = 1 , . . . , k , define the linear form
ψj : X−j → ZN by
ψj (x1, . . . , x j−1, x j+1 , . . . , x k) := ∑
i∈[k]{ j}
(j − i)xi.
Construct (k − 1) -uniform k-partite weighted hypergraphs ν, g, ˜g on X1 ∪ · · · ∪ Xk by setting
ν−j (x−j ) := ν(ψj (x−j )) , g−j (x−j ) := f (ψj (x−j )) , ˜g−j (x−j ) := ˜f (ψj (x−j )) The Green–Tao theorem: an exposition 269 (in the first definition, the left ν−j refers to the weighted hypergraph and the second ν
refers to the given function on ZN ). We claim that
‖ν−j − 1‖ = ‖ν − 1‖,k −1 (20) and
‖g−j − ˜g−j ‖ = ‖f − ˜f ‖,k −1 (21) for every j (in both (20) and (21) the left-hand side refers to the hypergraph cut norm (8) while the right-hand side refers to the cut norm (9) for functions on ZN ). We illustrate (21) in the case when k = j = 4 (the full proof is straightforward). The left-hand side of (21) equals
sup
A1,A 2,A 3⊆Z2
N
∣∣∣Ex1,x 2,x 3∈ZN [( f − ˜f )(3 x1 + 2 x2 + x3)1 A1 (x2, x 3)1 A2 (x1, x 3)1 A3 (x1, x 2)]
∣∣∣ ,
(22) while the right-hand side of (21) equals
sup
B1,B 2,B 3⊆Z2
N
∣∣∣Ex1,x 2,x 3∈ZN [( f − ˜f )( x1 + x2 + x3)1 B1 (x2, x 3)1 B2 (x1, x 3)1 B3 (x1, x 2)]
∣∣∣ .
(23) These two expressions are equal 8 up to a change of variables 3x1 ↔ x1 and 2x2 ↔ x2.It follows from (21) that ‖g − ˜g‖ = ‖f − ˜f ‖,k −1 = o(1) . Moreover, the k-linear forms condition for ν : ZN → [0 , ∞) translates to the k-linear forms condition for the weighted hypergraph ν. It follows from the counting lemma, Theorem 6.5, that
Ex1,...,x k ∈ZkN [g−1(x−1) · · · g−k(x−k)] = Ex1,...,x k ∈ZkN [˜ g−1(x−1) · · · ˜g−k(x−k)] + o(1) .
(24) The left-hand side is equal to
Ex1,...,x k ∈ZkN [f (ψ1(x−1)) · · · f (ψk(x−k))] = Ex,d ∈ZN [f (x)f (x+d) · · · f (x+( k−1) d)] ,
which can be seen by setting x = ψ1(x−1) and d = x1 + · · · + xk so that ψj (x−j ) =
x+( j−1) d. A similar statement holds for the right-hand side of (24). So (24) is equivalent to
Ex,d ∈ZN [f (x)f (x+d) · · · f (x+( k−1) d)] = Ex,d ∈ZN [ ˜f (x) ˜f (x+d) · · · ˜f (x+( k−1) d)]+ o(1) ,
which is at least c(k, δ ) − ok,δ (1) by Theorem 4.1, as desired.
8. Constructing the majorant
In this section, we use the relative Szemerédi theorem to prove the Green-Tao theorem. To do this, we must construct a majorizing measure for the primes that satisfies the linear forms condition.
8Here we use the assumption in the footnote to Definition 4.2 that Nis coprime to (k−1)! . Without this assumption, it can be shown that the two norms (22) and (23) differ by at most a constant factor depending on
k, which would also suffice for what follows.
270 D. Conlon, J. Fox, and Y. Zhao
Rather than considering the set of primes itself, we put weights on the primes, a common technique in analytic number theory. The weights we use will be related to the well-known von Mangoldt function Λ. This is defined by Λ( n) = log p if n = pk
for some prime p and positive integer k and Λ( n) = 0 if n is not a power of a prime (actually, the higher powers p2, p3, . . . play no role here and we will soon discard them from Λ). That these are natural weights to consider follows from the observation that the Prime Number Theorem is equivalent to ∑
n≤N
Λ( n) = (1 + o(1)) N .A difficulty with using Λ is that it is biased on certain residue classes. For example, every prime other than 2 is odd. This prevents us from making any pseudorandomness claims unless we can somehow remove these biases. This is achieved using the W-trick .Let w = w(N ) be any function that tends to infinity slowly with N . Let W = ∏
p≤w
p
be the product of primes up to w. The trick for avoiding biases mod p for any p ≤ w is to consider only those primes which are congruent to 1 (mod W ). In keeping with this idea, we define the modified von Mangoldt function by
˜Λ( n) :=
{ φ(W )
W
log( W n + 1) when W n + 1 is prime ,
0 otherwise .
The factor φ(W )/W is present since exactly φ(W ) of the W residue classes mod W have infinitely many primes and a strong form of Dirichlet’s theorem 9 tells us that the primes are equidistributed among these φ(W ) residue classes, i.e., ∑
n≤N
˜Λ( n) = (1 + o(1)) N
as long as w grows slowly enough with N . From now on, we will work with ˜Λ rather than
Λ. Our main goal is to prove the following result, which says that there is a majorizing measure for ˜Λ which satisfies the linear forms condition.
Proposition 8.1. For every k ≥ 3, there exists δk > 0 such that for every sufficiently large N there exists a function ν : ZN → [0 , ∞) satisfying the k-linear forms condition and ν(n) ≥ δk ˜Λ( n) for all N/ 2 ≤ n < N .
Using this majorant with the relative Szemerédi theorem, we obtain the Green-Tao theorem.
Proof of Theorem 1.1 assuming Proposition 8.1. Define f : ZN → [0 , ∞) by
f (n) = δk ˜Λ( n) if N/ 2 ≤ n < N and f (n) = 0 otherwise. By Dirichlet’s theorem, ∑
N/ 2≤n<N
f (n) = (1 /2 + o(1)) δkN , so Ef ≥ δk/3 for large N . Since 0 ≤ f ≤ ν and
ν satisfies the k-linear forms condition, it follows from the relative Szemerédi theorem, Theorem 4.3, that E[f (x)f (x + d) · · · f (x + ( k − 1) d)] ≥ c(k, δ k/3) − ok,δ (1) . There-fore, for sufficiently large N , we have f (x)f (x + d) · · · f (x + ( k − 1) d) > 0 for some
9In fact, Dirichlet’s theorem, or even the Prime Number Theorem, are not necessary to prove the Green-Tao theorem, though we assume them to simplify the exposition. Indeed, a weaker form of the Prime Number Theorem asserting that there are at least cN/ log Nprimes up to Nfor some c > 0suffices for our needs (this bound was first proved by Chebyshev and a famous short proof was subsequently found by Erd˝ os; see [1, Ch. 2]). Furthermore, in place of Dirichlet’s theorem, a simple pigeonhole argument shows that for each W,some residue class b(mod W)contains many primes (whereas we use Dirichlet’s theorem to take b= 1 ). The proof presented here can easily be modified to deal with general b, though the notation gets a bit more cumbersome as bcould vary with W. An analysis of this sort is necessary to prove a Szemerédi-type statement for the primes (see Section 10), since we do not then know how our subset of the primes is distributed on congruence classes. The Green–Tao theorem: an exposition
271
N/ 2 ≤ x < N and d 6 = 0 (the d = 0 terms contribute negligibly to the expectation). Since f is supported on [N/ 2, N ), we see that x, x + d, . . . , x + ( k − 1) d is not only an AP in ZN but also in Z, i.e., has no wraparound issues. Thus (x + jd )W + 1 for
j = 0 , . . . , k − 1 is a k-AP of primes. How do we construct the majorant ν for ˜Λ( n)? Recall that the Möbius function μ is defined by μ(n) = ( −1) ω(n) when n is square-free, where ω(n) is the number of prime factors of n, and μ(n) = 0 when n is not square-free. The functions Λ and μ are related by the Möbius inversion formula
Λ( n) = ∑
d|n
μ(d) log( n/d ).
In Green and Tao’s original proof, the following truncated version of Λ (motivated by ) was used to construct the majorant. For any R > 0, define
ΛR(n) := ∑
d|nd≤R
μ(d) log( R/d ).
Observe that if n has no prime divisors less than or equal to R, then ΛR(n) = log R.Tao later simplified the proof by using the following variant of ΛR, where the restric-tion d ≤ R is replaced by a smoother cutoff.
Definition 8.2. Let χ : R → [0 , 1] be any smooth, compactly supported function. Define
Λχ,R (n) := log R ∑
d|n
μ(d)χ
( log d
log R
)
.
In our application, χ will be supported on [−1, 1] , so only divisors d which are at most
R are considered in the sum. Note that ΛR above corresponds to χ(x) = max {1−| x|, 0},which is not smooth. The following proposition, which we will prove in the next section, gives a linear forms estimate for Λχ,R .
Proposition 8.3 (Linear forms estimate). Fix any smooth function χ : R → [0 , 1] sup-ported on [−1, 1] . Let m and t be positive integers. Let ψ1, . . . , ψ m : Zt → Z be fixed lin-ear maps, with no two being multiples of each other. Assume that R = o(N 1/(10 m)) grows with N and w grows sufficiently slowly with N . Let W := ∏
p≤w
p. Write θi := W ψ i +1 .Let B be a product ∏ti=1 Ii, where each Ii is a set of at least R10 m consecutive integers. Then
Ex∈B [Λ χ,R (θ1(x)) 2 · · · Λχ,R (θm(x)) 2] = (1 + o(1))
( W c χ log Rφ(W )
)m
, (25)
where o(1) denotes a quantity tending to zero as N → ∞ (at a rate that may depend on
χ, m, t, ψ1, . . . , ψ m, R, and w), and cχ is the normalizing factor
cχ :=
∫ ∞
0
|χ′(x)|2 dx. 272 D. Conlon, J. Fox, and Y. Zhao
Now we construct the majorizing measure ν and show that it satisfies the linear forms condition.
Proposition 8.4. Fix any smooth function χ : R → [0 , 1] supported on [−1, 1] with
χ(0) = 1 . Let k ≥ 3 and R := N k−12−k−3
. Assume that w grows sufficiently slowly with N and let W := ∏
p≤w
p. Define ν : ZN → [0 , ∞) by
ν(n) :=
{ φ(W )
W
Λχ,R (W n +1) 2
cχlog R
when N/ 2 ≤ n < N,
1 otherwise. (26)
Then ν satisfies the k-linear forms condition.
Note that while Λχ,R is not necessarily nonnegative, ν constructed in (26) is always nonnegative due to the square on Λχ,R .
Proof of Proposition 8.1 assuming Proposition 8.4. Take δk = k−12−k−4c−1
χ
. It suffices to verify that for N sufficiently large we have δk ˜Λ( n) ≤ ν(n) for all N/ 2 ≤ n < N . We only need to check the inequality when W n + 1 is prime, since ˜Λ( n) is zero otherwise. We have
log R = k−12−k−3 log N ≥ k−12−k−4 log( W N + 1) = cχδk log( W N + 1) ,
where the inequality holds for sufficiently large N provided w grows slowly enough. When W n + 1 is prime, we have Λχ,R (W n + 1) = log R, so
δk ˜Λ( n) = δk
φ(W )
W log( W n + 1) ≤ δk
φ(W )
W log( W N + 1) ≤ φ(W )
W
log Rcχ
= ν(n),
as claimed.
Proof of Proposition 8.4 assuming Proposition 8.3. We need to check that
Ex∈ZtN [ν(ψ1(x)) · · · ν(ψm(x))] = 1 + o(1) (27) whenever ψ1, . . . , ψ m, m ≤ k2k−1, are the linear forms that appear in (5) or any subset thereof. Note that no two ψi are multiples of each other. To use the two-piece definition of ν, we divide the domain ZN into intervals. Let
Q = Q(N ) be a slowly increasing function of N . Divide ZN into Q roughly equal intervals and form a partition of ZtN into Qt boxes, as follows:
Bu1,...,u t =
t
∏
j=1
([ uj N/Q, (uj + 1) N/Q ) ∩ ZN ) ⊆ ZtN , u1, . . . , u t ∈ ZQ.
Then, up to a o(1) error (due to the fact that the boxes do not all have exactly equal sizes), the left-hand side of (27) equals
Eu1,...,u t∈ZQ [Ex∈Bu1,...,u t [ν(ψ1(x)) · · · ν(ψm(x))]] .The Green–Tao theorem: an exposition 273 We say that a box Bu1,...,u t is good if, for each j ∈ [m], the set {ψj (x) : x ∈ Bu1,...,u t }
either lies completely in the subset [N/ 2, N ) of ZN or completely outside this subset. Otherwise, we say that the box is bad . We may assume Q grows slowly enough that
N/Q ≥ R10 m. From Proposition 8.3 and the definition of ν, we know that for good boxes,
Ex∈Bu1,...,u t [ν(ψ1(x)) · · · ν(ψm(x))] = 1 + o(1) .
For bad boxes, we use the bound ν(n) ≤ 1 + φ(W )
W
Λχ,R (W n +1) 2
cχlog R
. By expanding and applying (25) to each term, we find that
Ex∈Bu1,...,u t [ν(ψ1(x)) · · · ν(ψm(x))] = O(1)
(it is bounded in absolute value by 2m + o(1) ). It remains to show that the proportion of boxes that are bad is o(1) .Suppose Bu1,...,u t is bad. Then there exists some i such that the image of the box under ψi intersects both [N/ 2, N ) and its complement. This implies that there exists some (real-valued) x ∈ ∏tj=1 [uj N/Q, (uj + 1) N/Q ) ⊆ (R/N Z)t with ψi(x) = 0 or
N/ 2 (mod N ). Letting y = Qx/N , we see that y ∈ ∏tj=1 [uj , u j + 1) ⊆ (R/Q Z)t
satisfies ψi(y) = 0 or Q/ 2 (mod Q). This implies that ψi(u1, . . . , u t) is either O(1) or
Q/ 2 + O(1) (mod Q). Since ψi is a nonzero linear form, at most a O(1 /Q ) fraction of the tuples (u1, . . . , u t) ∈ ZtQ have this property. This can be seen by noting that if we fix all but one of the coordinates, there will be O(1) choices for the final coordinate for which ψi is in the required range. Taking the union over all i, we see that the proportion of bad boxes is O(1 /Q ) = o(1) .
9. Verifying the linear forms condition
In this section, we prove Proposition 8.3. There are numerous estimates along the way. To avoid getting bogged down with the rather technical error bounds, we first go through the proof while skipping some of these details (i.e., by only considering the “main term”). The approximations are then justified at the end, where we collect the error bound arguments. We note that all constants will depend implicitly on χ, m, t, ψ 1, . . . , ψ m.Expanding the definition of Λχ,R , we rewrite the left-hand side of (25) as
(log R)2m ∑
d1,d ′
1,...,d m,d ′
m∈N
m
∏
j=1
μ(dj )χ
( log dj
log R
)
μ(d′
j
)χ
( log d′
j
log R
) Ex∈B [1 dj ,d ′
j|θj(x)∀j
].
(28) Since μ(d) = 0 unless d is square-free, we only need to consider square-free
d1, d ′
1
, . . . , d m, d ′
m
. Also, since χ is supported on [−1, 1] , we may assume that
d1, d ′
1
, . . . , d m, d ′
m
≤ R. Let D denote the lcm of d1, d ′
1
, . . . , d m, d ′
m
. The width of the box B is at least R10 m in each dimension, so, by considering a slightly smaller box
B′ ⊆ B such that each dimension of B′ is divisible by D ≤ R2m, we obtain
Ex∈B [1 dj ,d ′
j|θj(x)∀j
] = Ex∈ZtD [1 dj ,d ′
j|θj(x)∀j
] + O(R−8m).274 D. Conlon, J. Fox, and Y. Zhao
Therefore, as there are at most R2m choices for d1, d ′
1
, . . . , d m, d ′
m
, we see that, up to an additive error of O(R−6m log 2m R), we may approximate (28) by
(log R)2m ∑
d1,d ′
1,...,d m,d ′
m∈N
m
∏
j=1
μ(dj )χ
( log dj
log R
)
μ(d′
j
)χ
( log d′
j
log R
) Ex∈ZtD [1 dj ,d ′
j|θj(x)∀j
].
(29) Let ϕ be the Fourier transform of exχ(x). That is,
exχ(x) =
∫
R
ϕ(ξ)e−ixξ dξ.
Substituting and simplifying, we have
χ
( log d
log R
)
=
∫
R
d− 1+ iξ
log R
ϕ(ξ) dξ.
We wish to plug this integral into (29). It helps to first restrict the integral to a compact interval I = [ − log 1/2 R, log 1/2 R]. By basic results in Fourier analysis (see, for exam-ple, [40, Chapter 5, Theorem 1.3]), since χ is smooth and compactly supported, ϕ decays rapidly, that is, ϕ(ξ) = OA((1 + |ξ|)−A) for any A > 0. It follows that for any A > 0,
χ
( log d
log R
)
=
∫
I
d− 1+ iξ
log R
ϕ(ξ) dξ + OA(d−1/ log R(log R)−A). (30) We write
zj := 1 + iξ j
log R and z′
j
:= 1 + iξ ′
j
log R .
We have χ(log d/ log R) = O(d−1/ log R) (we only need to check this for d ≤ R since χ
is supported on [−1, 1] ). Using (30), we have
m
∏
j=1
χ
( log dj
log R
)
χ
( log d′
j
log R
)
=
∫
I
· · ·
∫
Im
∏
j=1
d−zj
j
d′
j
−z′
j
ϕ(ξj )ϕ(ξ′
j
) dξ j dξ ′
j
OA
(log R)−Am∏
j=1
(dj d′
j
)−1/ log R
. (31) Using (31), we estimate (29) (error bounds are deferred to the end) by
(log R)2m
∫
I
· · ·
∫
I
∑
d1,d ′
1,...,d m,d ′
m∈N
Ex∈ZtD [1 dj ,d ′
j|θj(x)∀j
]
m
∏
j=1
μ(dj )d−zj
j
μ(d′
j
)d′
j
−z′
j
ϕ(ξj )ϕ(ξ′
j
) dξ j dξ ′
j
.
(32) We are allowed to swap the summation and the integrals because I is compact and the sum can be shown to be absolutely convergent (the argument for absolute convergence The Green–Tao theorem: an exposition 275 is similar to the error bound for (32) included towards the end of the section). Splitting
d1, d ′
1
, . . . , d m, d ′
m
in (32) into prime factors, we obtain (32) = (log R)2m
∫
I
· · ·
∫
I
∏
p
Ep(ξ) ·
m
∏
j=1
ϕ(ξj )ϕ(ξ′
j
) dξ j dξ ′
j
, (33) where ξ = ( ξ1, ξ ′
1
, . . . , ξ m, ξ ′
m
) ∈ I2m and Ep(ξ) is the Euler factor
Ep(ξ) := ∑
d1,d ′
1,...,d m,d ′
m∈{ 1,p }
Ex∈Ztp [1 dj ,d ′
j|θj(x)∀j
]
m
∏
j=1
μ(dj )d−zj
j
μ(d′
j
)d′
j
−z′
j
.
We have Ep(ξ) = 1 when p ≤ w (recall the W -trick, so p - θj (x) = W ψ j (x) + 1 for all
j when p ≤ w). When p > w , the expectation in the summand equals 1 if all dj , d ′
j
are
1, 1/p if dj d′
j
= 1 for all except exactly one j, and is at most 1/p 2 otherwise (here we assume that w is sufficiently large so that no two ψi are multiples of each other mod p). It follows that for p > w ,
Ep(ξ) = 1 − p−1
m
∑
j=1
(p−zj + p−z′
j
− p−zj −z′
j
) + O(p−2) = (1 + O(p−2)) E′
p
(ξ),
where, for any prime p,
E′
p
(ξ) :=
m
∏
j=1
(1 − p−1−zj )(1 − p−1−z′
j
)1 − p−1−zj −z′
j
.
It then follows that
∏
p
Ep(ξ) = ∏
p>w
(1 + O(p−2)) E′
p
(ξ) = (1 + O(w−1))
∏
p≤w
E′
p
(ξ)
−1
∏
p
E′
p
(ξ).
(34) Recall that the Riemann zeta function
ζ(s) := ∑
n≥1
n−s = ∏
p
(1 − p−s)−1
has a simple pole at s = 1 with residue 1 (a proof is included towards the end). This implies that
∏
p
E′
p
(ξ) =
m
∏
j=1
ζ(1 + zj + z′
j
)
ζ(1 + zj )ζ(1 + z′
j
) ≈
m
∏
j=1
zj z′
j
zj + z′
j
, (35) where ≈ denotes asymptotic equality. Here we use |z1| , |z′
1
| , . . . , |zm|, |z′
m
| =
O((log R)−1/2) as ξ1, ξ ′
1
, . . . ξ m, ξ ′
m
∈ I. For p ≤ w, we make the approximation
E′
p
(ξ) ≈ (1 − p−1)m. Hence,
∏
p≤w
E′
p
(ξ) ≈ ∏
p≤w
(1 − p−1)m =
( φ(W )
W
)m
. (36) 276 D. Conlon, J. Fox, and Y. Zhao
Substituting (34), (35), and (36) into (33), we find that (33) ≈ (log R)2m
( Wφ(W )
)m ∫
I
· · ·
∫
Im
∏
j=1
zj z′
j
zj + z′
j
ϕ(ξj )ϕ(ξ′
j
) dξ j dξ ′
j
. (37) It remains to estimate the integral
∫
I
∫
I
zj z′
j
zj + z′
j
ϕ(ξj )ϕ(ξ′
j
) dξ j dξ ′
j
= 1log R
∫
I
∫
I
(1 + iξ j )(1 + iξ ′
j
)2 + i(ξj + ξ′
j
) ϕ(ξj )ϕ(ξ′
j
) dξ j dξ ′
j
.
We can replace the domain of integration I = [ − log 1/2 R, log 1/2 R] by R with a loss of OA(log −A R) for any A > 0 due to the rapid decay of ϕ given by ϕ(ξ) = OA((1 +
|ξ|)−A). We claim
∫
R
∫
R
(1 + iξ )(1 + iξ ′)2 + i(ξ + ξ′) ϕ(ξ)ϕ(ξ′) dξdξ ′ =
∫ ∞
0
|χ′(x)|2 dx = cχ. (38) Using
12 + i(ξ + ξ′) =
∫ ∞
0
e−(1+ iξ )xe−(1+ iξ ′)x dx,
we can rewrite the left-hand side of (38) as
∫ ∞
0
(∫
R
ϕ(ξ)(1 + iξ )e−(1+ iξ )x dξ
)2
dx.
The expression in parentheses is −χ′(x), so (38) follows. Substituting (38) into (37) we arrive at the desired conclusion, Proposition 8.3.
Error estimates. Now we bound the error terms in the above analysis.
Simple pole of Riemann zeta function. Here is the argument showing that ζ(s) = (s−1) −1 +O(1) whenever Re s > 1 and s−1 = O(1) . We have (s−1) −1 = ∫ ∞
1
x−s dx .So
ζ(s) − 1
s − 1 =
∞
∑
n=1
n−s −
∫ ∞
1
x−s dx =
∞
∑
n=1
∫ n+1
n
(n−s − x−s) dx.
The n-th term on the right is bounded in magnitude by O(n−2). So the sum is O(1) .
Estimate (32) . We want to bound the difference between (32) and (29). This means bounding the contribution to (29) from the error term in (31). Taking absolute values everywhere, we bound these contributions by
A (log R)O(1) −A ∑
d1,d ′
1,...,d m,d ′
m
sq-free integers
Ex∈ZtD 1 dj ,d ′
j|θj(x)∀j
−1/ log R
= (log R)O(1) −A ∏
p
∑
d1,d ′
1,...,d m,d ′
m∈{ 1,p }
Ex∈Ztp 1 dj ,d ′
j|θj(x)∀j
−1/ log R.The Green–Tao theorem: an exposition 277 The expectation Ex∈Ztp [1 dj ,d ′
j|θj(x)∀j
] is 1 if all di and d′
i
are 1 and at most 1/p
otherwise. We continue to bound the above by
≤ (log R)O(1) −A ∏
p
1 + p−1 ∑
d1,d ′
1,...,d m,d ′
m∈{ 1,p }
not all 1’s
(d1d′
1
· · · dmd′
m
)−1/ log R
= (log R)O(1) −A ∏
p
(
1 + p−1(( p−1/ log R + 1) 2m − 1)
)
≤ (log R)O(1) −A ∏
p
(
1 − p−1−1/ log R)−O(1)
= (log R)O(1) −Aζ(1 + 1 / log R)O(1) .
So the difference between (32) and (29) is OA((log R)O(1) −A), which is small as long as we take A to be sufficiently large.
Estimate in (35) . We have |zj |, |z′
j
| = O(log −1/2 R) since |ξj |, |ξ′
j
| ≤ log 1/2 R. So
m
∏
j=1
ζ(1 + zj + z′
j
)
ζ(1 + zj )ζ(1 + z′
j
) =
m
∏
j=1
(( zj + z′
j
)−1 + O(1)) (z−1
j
O(1))( z′
j
−1
O(1)) = (1 + O(log −1/2 R))
m
∏
j=1
zj z′
j
zj + z′
j
. (39)
Estimate in (36) . If |z| log p = O(1) (which is the case for p ≤ w), then
1−p−1−z = 1 −p−1e−z log p = 1 −p−1(1+ O(|z| log p)) = (1 −p−1)(1+ O(|z|p−1 log p)) .
It follows that for all p ≤ w and ξ1, ξ ′
1
, . . . , ξ m, ξ ′
m
∈ I, we have
E′
p
(ξ) =
(
1 + O
( log pp log 1/2 R
))
(1 − p−1)m
and, hence, ∏
p≤w
E′
p
(ξ) =
(
1 + O
( w
log 1/2 R
)) ∏
p≤w
(1 − p−1)m. (40)
Estimate in (37) . Using (34), (39), and (40), we find that the ratio between the two sides in (37) is 1 + O(1 /w + w/ log 1/2 R) = 1 + o(1) , as long as w grows sufficiently slowly. 278 D. Conlon, J. Fox, and Y. Zhao
10. Extensions of the Green-Tao theorem
We conclude by discussing a few extensions of the Green-Tao theorem.
Szemerédi’s theorem in the primes. As noted already by Green and Tao , their method also implies a Szemerédi-type theorem for the primes. That is, every subset of the primes with positive relative upper density contains arbitrarily long arithmetic progres-sions. One elegant corollary of this result is that there are arbitrarily long APs where every term is a sum of two squares. This result follows from a combination of the well-known fact that every prime of the form 4n + 1 is a sum of two squares with Dirichlet’s theorem on primes in arithmetic progressions, which tells us that roughly half the primes are con-gruent to 1 (mod 4). Even this innocent-sounding corollary was open before Green and Tao’s paper.
Gaussian primes contain arbitrarily shaped constellations. The Gaussian integers is the set of all numbers of the form a + bi , where a, b ∈ Z. This set is a ring under the usual definitions of addition and multiplication for complex numbers. It is also a unique factorization domain, so it is legitimate to talk about the set of Gaussian primes. Tao proved that an analogue of the Green-Tao theorem holds for the Gaussian primes. We say that A ⊆ Zd contains arbitrary constellations if, for every finite set
F ⊆ Zd, there exist x ∈ Zd and t ∈ Z>0 such that x + tf ∈ A for every f ∈ F . Tao’s theorem then states that the Gaussian primes, viewed as a subset of Z2, contain arbitrary constellations. Just as the Green-Tao theorem uses Szemerédi’s theorem as a black box, this theorem uses the multidimensional analogue of Szemerédi’s theorem, first proved by Furstenberg and Katznelson . This states that every subset of Zd with positive upper density 10 contains arbitrary constellations. The Furstenberg-Katznelson theorem also fol-lows from the hypergraph removal lemma and the approach taken by Tao is to transfer this hypergraph removal proof to the sparse context. It may therefore be seen as a precursor to the approach taken here.
Multidimensional Szemerédi theorem in the primes. Let P denote the set of primes in
Z. It was shown recently by Tao and Ziegler and, independently, by Cook, Magyar, and Titichetrakun , that every subset of P d of positive relative upper density contains arbitrary constellations. A short proof was subsequently given in (though, like , it assumes some difficult results of Green, Tao, and Ziegler that we will discuss later in this section). Although both this result and Tao’s result on the Gaussian primes are multidimen-sional analogues of the Green-Tao theorem, they are quite different in nature. Informally speaking, a key difficulty in the second result is that there is a strong correlation between coordinates in P d (namely, that all coordinates are simultaneously prime), whereas there
10 A set A⊆Zdhas positive upper density if lim sup N→∞
∣∣A∩[−N, N ]d∣∣/(2 N+ 1) d>0. We say that
A⊆S⊆Zdhas positive relative upper density if lim sup N→∞
∣∣A∩S∩[−N, N ]d∣∣/∣∣S∩[−N, N ]d∣∣>
0.The Green–Tao theorem: an exposition
279 is no significant correlation between the real and imaginary parts of a typical Gaussian prime (after applying an extension of the W -trick).
The primes contain arbitrary polynomial progressions. We say that A ⊆ Z contains arbitrary polynomial progressions if, whenever P1, . . . , P k ∈ Z[X] are polynomials in one variable with integer coefficients satisfying P1(0) = · · · = Pk(0) = 0 , there is some x ∈ Z and t ∈ Z>0 such that x + Pj (t) ∈ A for each j = 1 , . . . , k . A striking generalization of Szemerédi’s theorem due to Bergelson and Leibman states that any subset of Z of positive upper density contains arbitrary polynomial progressions. To date, the only known proofs of this result use ergodic theory. For primes, an analogue of the Bergelson-Leibman theorem was proved by Tao and Ziegler . This result states that any subset of the primes with positive relative upper density contains arbitrary polynomial progressions. In particular, the primes themselves contain arbitrary polynomial progressions. It seems plausible that the simplifications out-lined here could also be used to simplify the proof of this theorem.
The number of k-APs in the primes. The original approach of Green and Tao (and the approach outlined in this paper) implies that for any k the number of k-APs of primes with each term at most N is on the order of N 2
log kN
. In subsequent work, Green, Tao, and Ziegler [25, 26, 27] showed how to determine the exact asymptotic. That is, they determine a constant ck such that the number of k-APs of primes with each term at most
N is (ck + o(1)) N 2
log kN
. More generally, they determine an asymptotic for the number of prime solutions to a broad range of linear systems of equations. The proof of these results also draws on the transference technique discussed in this paper but a number of additional ingredients are needed, most notably an inverse theo-rem describing the structure of those sets which do not contain the expected number of solutions to certain linear systems of equations. It is this result which is transferred to the sparse setting when one wishes to determine the exact asymptotic.
Acknowledgments. We thank Yuval Filmus, Mohammad Bavarian, and the anonymous referee for helpful comments on the manuscript.
References
M. Aigner and G. M. Ziegler, Proofs from The Book , fourth ed., Springer-Verlag, Berlin, 2010. F. A. Behrend, On sets of integers which contain no three terms in arithmetical progression ,Proc. Nat. Acad. Sci. U.S.A. 32 (1946), 331–332. V. Bergelson and A. Leibman, Polynomial extensions of van der Waerden’s and Szemerédi’s theorems , J. Amer. Math. Soc. 9(1996), 725–753. T. F. Bloom, A quantitative improvement for Roth’s theorem on arithmetic progressions ,arXiv:1405.5800. F. R. K. Chung, R. L. Graham, and R. M. Wilson, Quasi-random graphs , Combinatorica 9
(1989), 345–362.
280 D. Conlon, J. Fox, and Y. Zhao D. Conlon and J. Fox, Graph removal lemmas , Surveys in Combinatorics 2013, London Math. Soc. Lecture Note Ser., Cambridge University Press, 2013, pp. 1–50. D. Conlon, J. Fox, and Y. Zhao, A relative Szemerédi theorem , arXiv:1305.5440. Due to appear in Geom. Funct. Anal. D. Conlon, J. Fox, and Y. Zhao, Extremal results in sparse pseudorandom graphs , Adv. Math.
256 (2014), 206–290. B. Cook, A. Magyar, and T. Titichetrakun, A multidimensional Szemerédi theorem in the primes , arXiv:1306.3025. J. Fox, A new proof of the graph removal lemma , Ann. of Math. 174 (2011), 561–579. J. Fox and Y. Zhao, A short proof of the multidimensional Szemerédi theorem in the primes ,Amer. J. Math., to appear. A. Frieze and R. Kannan, Quick approximation to matrices and applications , Combinatorica
19 (1999), 175–220. H. Furstenberg, Ergodic behavior of diagonal measures and a theorem of Szemerédi on arith-metic progressions , J. Analyse Math. 31 (1977), 204–256. H. Furstenberg and Y. Katznelson, An ergodic Szemerédi theorem for commuting transforma-tions , J. Analyse Math. 34 (1978), 275–291. H. Furstenberg, Y. Katznelson, and D. Ornstein, The ergodic theoretical proof of Szemerédi’s theorem , Bull. Amer. Math. Soc. 7 (1982), 527–552. D. A. Goldston and C. Y. Yıldırım, Higher correlations of divisor sums related to primes. I. Triple correlations , Integers 3 (2003), A5, 66pp. W. T. Gowers, A new proof of Szemerédi’s theorem for arithmetic progressions of length four ,Geom. Funct. Anal. 8 (1998), 529–551. W. T. Gowers, A new proof of Szemerédi’s theorem , Geom. Funct. Anal. 11 (2001), 465–588. W. T. Gowers, Quasirandomness, counting and regularity for 3-uniform hypergraphs , Com-bin. Probab. Comput. 15 (2006), 143–184. W. T. Gowers, Hypergraph regularity and the multidimensional Szemerédi theorem , Ann. of Math. 166 (2007), 897–946. W. T. Gowers, Decompositions, approximate structure, transference, and the Hahn-Banach theorem , Bull. Lond. Math. Soc. 42 (2010), 573–606. B. Green, Long arithmetic progressions of primes , Analytic number theory, Clay Math. Proc., vol. 7, Amer. Math. Soc., Providence, RI, 2007, pp. 149–167. B. Green and T. Tao, The primes contain arbitrarily long arithmetic progressions , Ann. of Math. 167 (2008), 481–547. B. Green and T. Tao, New bounds for Szemerédi’s theorem. II. A new bound for r4(N ), Ana-lytic number theory, Cambridge Univ. Press, Cambridge, 2009, pp. 180–204. B. Green and T. Tao, Linear equations in primes , Ann. of Math. 171 (2010), 1753–1850. B. Green and T. Tao, The Möbius function is strongly orthogonal to nilsequences , Ann. of Math. 175 (2012), 541–566. B. Green, T. Tao, and T. Ziegler, An inverse theorem for the Gowers U s+1 [N ]-norm , Ann. of Math. 176 (2012), 1231–1372. The Green–Tao theorem: an exposition 281
B. Host, Progressions arithmétiques dans les nombres premiers (d’après B. Green et T. Tao) ,Astérisque (2006), Exp. No. 944, viii, 229–246, Séminaire Bourbaki. Vol. 2004/2005. B. Kra, The Green-Tao theorem on arithmetic progressions in the primes: an ergodic point of view , Bull. Amer. Math. Soc. 43 (2006), 3–23. L. Lovász, Large networks and graph limits , Amer. Math. Soc. Colloq. Publ., vol. 60, Amer. Math. Soc., Providence, RI, 2012. B. Nagle, V. Rödl, and M. Schacht, The counting lemma for regular k-uniform hypergraphs ,Random Structures Algorithms 28 (2006), 113–179. O. Reingold, L. Trevisan, M. Tulsiani, and S. Vadhan, New proofs of the Green-Tao-Ziegler dense model theorem: an exposition , arXiv:0806.0381. O. Reingold, L. Trevisan, M. Tulsiani, and S. Vadhan, Dense subsets of pseudorandom sets ,49th Annual IEEE Symposium on Foundations of Computer Science, IEEE Computer Society, 2008, pp. 76–85. V. Rödl and J. Skokan, Regularity lemma for k-uniform hypergraphs , Random Structures Algorithms 25 (2004), 1–42. V. Rödl and J. Skokan, Applications of the regularity lemma for uniform hypergraphs , Random Structures Algorithms 28 (2006), 180–194. K. F. Roth, On certain sets of integers , J. London Math. Soc. 28 (1953), 104–109. I. Z. Ruzsa and E. Szemerédi, Triple systems with no six points carrying three triangles ,Combinatorics (Proc. Fifth Hungarian Colloq., Keszthely, 1976), Vol. II, Colloq. Math. Soc. János Bolyai, vol. 18, North-Holland, Amsterdam, 1978, pp. 939–945. T. Sanders, On Roth’s theorem on progressions , Ann. of Math. 174 (2011), 619–636. T. Schoen and I. D. Shkredov, Roth’s theorem in many variables , Israel J. Math. 199 (2014), 287–308. E. M. Stein and R. Shakarchi, Fourier analysis: an introduction, Princeton Lectures in Anal-ysis, 1, Princeton University Press, Princeton, NJ, 2003. E. Szemerédi, On sets of integers containing no k elements in arithmetic progression , Acta Arith. 27 (1975), 199–245. T. Tao, A remark on Goldston-Yıldırım correlation estimates , available at:
dvi . T. Tao, Arithmetic progressions and the primes , Collect. Math. (2006), 37–88. T. Tao, The Gaussian primes contain arbitrarily shaped constellations , J. Anal. Math. 99
(2006), 109–176. T. Tao, Obstructions to uniformity and arithmetic patterns in the primes , Pure Appl. Math. Q.
2 (2006), 395–433. T. Tao, A variant of the hypergraph removal lemma , J. Combin. Theory Ser. A 113 (2006), 1257–1280. T. Tao, The dichotomy between structure and randomness, arithmetic progressions, and the primes , International Congress of Mathematicians. Vol. I, Eur. Math. Soc., Zürich, 2007, pp. 581–608. 282 D. Conlon, J. Fox, and Y. Zhao T. Tao and T. Ziegler, A multi-dimensional Szemerédi theorem for the primes via a correspon-dence principle , Israel J. Math., to appear. T. Tao and T. Ziegler, The primes contain arbitrarily long polynomial progressions , Acta Math. 201 (2008), 213–305. P. Varnavides, On certain sets of positive density , J. London Math. Soc. 34 (1959), 358–360. Y. Zhao, An arithmetic transference proof of a relative Szemerédi theorem , Math. Proc. Cam-bridge Philos. Soc. 156 (2014), 255–261. Received 17 April 2014; revised 28 July 2014 |
16045 | https://flexbooks.ck12.org/cbook/ck-12-cbse-math-class-9/section/8.1/related/lecture/quadrilateral-properties/ | Quadrilateral Properties | CK-12 Foundation
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16046 | https://www.engineeringtoolbox.com/bulk-modulus-elasticity-d_585.html | Engineering ToolBox - Resources, Tools and Basic Information for Engineering and Design of Technical Applications!
Bulk Modulus and Fluid Elasticities
Introduction to - and definition of - Bulk Modulus Elasticity commonly used to characterize the compressibility of fluids.
The Bulk Modulus Elasticity - or Volume Modulus - is a material property characterizing the compressibility of a fluid - how easy a unit volume of a fluid can be changed when changing the pressure working upon it. It is the ratio of thechange in unit pressure to the corresponding volume change per unit of volume.
The Bulk Modulus Elasticity can be calculated as
K = - dp / (dV / V0)
= - ( p1 - p0) / ((V1 - V0) / V0) (1)
K = Bulk Modulus of Elasticity (Pa, N/m2)
dp = differential change in pressure on the object (Pa, N/m2)
dV = differential change in volume of the object (m3)
V0 = initial volume of the object (m3)
p0 = initial pressure ( Pa, N/m2)
p1 = final pressure ( Pa, N/m2)
V1 = final volume ( m3)
The Bulk Modulus Elasticity can alternatively be expressed as
K = dp / (dρ / ρ0)
= (p1 - p0) / ((ρ1 - ρ0) / ρ0) (2)
where
dρ = differential change in density of the object (kg/m3)
ρ0 = initial density of the object (kg/m3)
ρ1 = final density of the object ( kg/m3)
An increase in the pressure will decrease the volume (1). A decrease in the volume will increase the density (2) .
The SI unit of the bulk modulus elasticity is N/m2 (Pa)
The imperial (BG) unit is lbf/in2 (psi)
1 lbf/in2(psi) = 6.894×103 N/m2 (Pa)
A large Bulk Modulus indicates a relative incompressible fluid.
Bulk Modulus Common Fluids
Bulk Modulus vs. Materials
| Fluid | Bulk Modulus - K - |
| Imperial Units - BG ( 105 psi, lbf/in2) | SI Units ( 109 Pa, N/m2) |
| Acetone | 1.34 | 0.92 |
| Benzene | 1.5 | 1.05 |
| Carbon Tetrachloride | 1.91 | 1.32 |
| Ethyl Alcohol | 1.54 | 1.06 |
| Gasoline | 1.9 | 1.3 |
| Glycerin | 6.31 | 4.35 |
| ISO 32 mineral oil | 2.6 | 1.8 |
| Kerosene | 1.9 | 1.3 |
| Mercury | 41.4 | 28.5 |
| Paraffin Oil | 2.41 | 1.66 |
| Petrol | 1.55 - 2.16 | 1.07 - 1.49 |
| Phosphate ester | 4.4 | 3 |
| SAE 30 Oil | 2.2 | 1.5 |
| Seawater | 3.39 | 2.34 |
| Sulfuric Acid | 4.3 | 3.0 |
| Water (10 oC) | 3.12 | 2.09 |
| Water - glycol | 5 | 3.4 |
| Water in oil emulsion | 3.3 | 2.3 |
1 GPa = 109 Pa (N/m2)
Stainless steel with Bulk Modulus 163×109 Pa is aprox. 80 times harder to compress than water with Bulk Modulus 2.15×109 Pa .
Example - Density of Seawater in the Mariana Trench
the deepest known point in the Earth's oceans - 10994 m .
The hydrostatic pressure in the Mariana Trench can be calculated as
p1 = (1022 kg/m3) (9.81 m/s2) (10994 m)
= 110 106 Pa (110 MPa)
The initial pressure at sea-level is 105 Pa and the density of seawater at sea level is 1022 kg/m3 .
The density of seawater in the deep can be calculated by modifying (2) to
ρ1 = ((p1 - p0) ρ0 + K ρ0) / K
= (((110×106 Pa) - (1×105 Pa)) (1022 kg/m3) + (2.34×109 Pa) (1022 kg/m3)) / ( 2.34×109 Pa)
= 1070 kg/m3
Note! - since the density of the seawater varies with dept the pressure calculation could be done more accurate by calculating in dept intervals.
Bulk Modulus of Water vs. Temperature
Bulk Modulus of Water vs. Temperatures
| Temperature (oC) | Bulk Modulus (109 Pa) |
| 0.01 | 1.96 |
| 10 | 2.09 |
| 20 | 2.18 |
| 30 | 2.23 |
| 40 | 2.26 |
| 50 | 2.26 |
| 60 | 2.25 |
| 70 | 2.21 |
| 80 | 2.17 |
| 90 | 2.11 |
| 100 | 2.04 |
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16047 | https://pathology.ubc.ca/files/2012/06/GYNEBOOK20111.pdf | Essentials of Gynecologic Cytology Gia-Khanh Nguyen Brenda Smith 2011 2 Essentials of Gynecologic Cytology Gia-Khanh Nguyen, MD, FRCPC University Of Alberta Edmonton, Alberta, Canada And Brenda Smith, BSc, RT, CT (ASCP) British Columbia Cancer Agency University Of British Columbia Vancouver, British Columbia, Canada Copyright © Gia-Khanh Nguyen. All rights reserved. This book was legally deposited at Archives and Library Canada and was given an ISBN: 978-0-9780929-4-8 3 Preface “Essentials of Gynecologic Cytology” is written for practicing pathologists in private laboratories and community hospitals, residents in Pathology and cytotechnologists who want a quick review of the cytopathology of the uterus and its annexae. It consists of nine chapters and eight of them are devoted to the evaluation of the Pap smear; a test that has been regarded as the most complicated to interpret in medicine. The text in this monograph is concise and contains relevant information for cytologic interpretation of cell samples from the above-mentioned anatomic sites. Lesions commonly encountered in day-to-day practice and lesions with distinct cytologic manifestations are briefly discussed and illustrated. The cytologic materials used for the illustrations were from two Canadian institutions: University of Alberta Hospital, Edmonton, Alberta and BC Cancer Agency, Vancouver, British Columbia. The Pap smears from the former were alcohol-fixed with commercial fixatives and those from the latter were air-dried, re-hydrated and then fixed in ethanol prior to staining. Therefore, some minor differences will be observed in some cell images. Most illustrations are taken from conventional Pap smears, with a few images taken from liquid-based preparations added for comparison. Some important differences in the cytologic findings between conventional and liquid-based preparations are mentioned elsewhere in the text. A small number of histologic images are also included for cytohistologic correlations. For improvement of the future editions of this monograph, constructive comments from the reader will be highly appreciated. Gia-Khanh Nguyen Brenda Smith Vancouver, BC, Canada Spring 2011 4 Contributors Katherine M. Ceballos, MD, FRCPC Clinical Assistant Professor Department of Pathology and Laboratory Medicine University of British Columbia, and Pathologist British Columbia Cancer Agency Vancouver, BC, Canada Gia-Khanh Nguyen, MD, FRCPC Professor Emeritus Department of Laboratory Medicine and Pathology University of Alberta Edmonton, AB, Canada Brenda Smith, BSc, RT, CT (ASCP) Clinical Instructor Department of Pathology and Laboratory Medicine University of British Columbia, and Cytotechnologist British Columbia Cancer Agency Vancouver, BC, Canada 5 Acknowledgments We wish to thank Dr. Jason Ford, Mrs. Helen Dyck and other members of The David Hardwick Pathology Learning Centre, Department of Pathology and Laboratory Medicine, The University of British Columbia, Vancouver, Canada, for their interest, enthusiasm and expertise in publishing our work online. Gia-Khanh Nguyen, MD, FRCPC Brenda Smith, BSc, RT, CT(ASCP) 6 Related material by the same author Essentials of needle aspiration cytology. Igaku-Shoin, New York, 1991 Essentials of exfoliative cytology. Igaku-Shoin, New York, 1992 Essentials of cytology, An atlas. Igaku-Shoin, New York, 1993 Critical issues in cytopathology. Igaku-Shoin, New York, 1996 Essentials of lung tumor cytology. UBC Pathology, Vancouver, 2008 Essentials of abdominal fine needle aspiration cytology. UBC Pathology, 2008 Essentials of head and neck cytology. UBC Pathology, 2009 Essentials of fluid cytology. UBC Pathology, 2010 Cytodiagnosis of breast lesions: A text and atlas. UBC Pathology, 2011 7 Important remarks In this monograph: • Cytologic materials (conventional Papanicolaou smears, Liquid-based preparations and Fine-needle aspirates) were stained by the Papanicolaou method or by another staining method as otherwise indicated. • Most cytologic images were taken under medium or high magnification. • Histologic sections were stained with hematoxylin and eosin or with another staining method as otherwise indicated. • Conventional Pap smear is abbreviated as CP smear. • Liquid-based preparation is abbreviated as LBP • The 2 terms “Cervical intraepithelial neoplasia” (CIN) and “Squamous intraepithelial lesion” (SIL) are used interchangeably in many sections: - Low-grade SIL (LSIL) as Low-grade CIN or CIN 1 - High-grade SIL (HSIL) as High-grade CIN or CIN 2 and CIN 3 8 Contents Chapter 1. Gynecologic cytology: historical development, current status, technical considerations and reporting 9 Chapter 2. Normal uterus and vagina 26 Chapter 3. Infections and nonneoplastic cellular changes 43 Chapter 4. Cervical squamous cell lesions 73 Chapter 5. Cervical glandular lesions 108 Chapter 6. Other cervical cancers and extrauterine cancers 141 Chapter 7. Endometrial lesions 150 Chapter 8. Vaginal lesions 172 Chapter 9. Uterine annexal mass lesions 181 9 Chapter 1 Gynecologic Cytology: Historical Development, Current status, Technical Considerations and Reporting Katherine M. Ceballos, Brenda Smith and Gia-Khanh Nguyen A. THE PAP TEST HISTORICAL DEVELOPMENT George Nicholas Papanicolaou has been regarded as the father of modern cytopathology. He was born in Greece in 1883, studied medicine in Athens and immigrated to the United States in 1913. In New York he first used exfoliated cells to study the estrous cycle in guinea pigs, and then he subsequently worked on human cells. In 1943 Papanicolaou and Traut published their book Diagnosis of Uterine Cancer by the Vaginal Smear that has had a strong impact on clinical practice worldwide. Also emerging in the field was Ruth Graham, one of the earliest and most renowned cytologists. Her paper on radiation changes of cancer cells of the uterine cervix has been regarded as a classic study, and her 1950 monograph entitled The Cytologic Diagnosis of Cancer has greatly contributed to the field of clinical cytology. In the 1950s, the cytology literature began to proliferate and Acta Cytologica commenced publication under George L. Wied with an editorial board consisting of Graham, Papanicolaou, Pundell and Reagan. Early volumes consisted of symposia devoted to definitions and criteria for cytodiagnosis. Four other international cytology journals have subsequently appeared, attesting to a need for advancement of cytopathology: Diagnostic Cytopathology in 1985, Cytopathology in 1990, Cancer Cytopathology in 1997 and Cytojournal in 2005. The practice of gynecologic cytology in the United States in the past 10 years has undergone important changes that have happened as a result of the publication of the article entitled “Lax laboratories” by Bogdanich in the Wall Street Journal on November 13, 1987, and the subsequent introduction of Bill HR 5471, the Clinical Laboratory Improvement Amendments of 1988 (CLIA ’88). As a result, numerous cytology laboratories in the United States lost their licenses, and several professional organizations have concentrated their efforts to evaluate the value of the Pap smear in cervical cancer screening. 10 CURRENT STATUS 1. THE BETHESDA SYSTEM The first Bethesda conference met on December 12-13, 1988 at the National Institute of Health in Maryland to consider methods of reporting gynecologic cytology in meaningful diagnostic terminologies. The second Bethesda conference was held on April 29-30, 1991 to discuss the problem of implementation, clarification of terminologies, the “less-than-optimal” category, adequacy criteria, and reporting. The third Bethesda conference held from April 30 to May 2, 2001 further revised the terminology. The Bethesda System has enjoyed widespread popularity, and it has been adopted almost worldwide to replace the outdated Papanicolaou numerical class system for reporting Pap smears. 2. LIQUID-BASED PREPARATIONS These preparations have been developed to improve preservation of cytologic detail and thus the diagnostic sensitivity of cervico-vaginal cytology. The ThinPrep TM Pap Test (formerly Cytyc Corp., now Hologic™ Inc) and the SurePath™ Pap test (BD Diagnostics) were approved by the Food and Drug Administration (FDA) of the United States. The FDA has approved ThinPrep as an alternative for the conventional Pap smear and approved claims that it is more effective than the conventional Pap smear in detecting cervical low-grade epithelial lesions. At the present time the ThinPrep Pap Tests has replaced nearly 100% of the cervical screening market in the United States. The morphology of squamous and glandular cells is preserved but the cellular patterns or arrangements associated with some lesions are altered, requiring slight modifications of cytodiagnostic criteria. Attaining proficiency in the interpretation of liquid based cytology will require a lot of time and effort from the cytology community worldwide. 3. CYTOLOGY AUTOMATION Automation has been in development since the 1950s. The Neopath AutoPap® 300 QC and the Neuromedical System PapNet were the first two devices that were approved by the FDA for quality control rescreening of negative Pap smears. The AutoPap System was the only device that was approved for primary screening. Recently, the Neuromedical Systems, Inc. has declared bankruptcy, and its intellectual property was sold to AutoCyte, Inc., now part of BD Diagnostics. Current FDA-approved devices for automated screening are the BD FocalPointTM Slide Profiler, and the ThinPrep Imaging System. These new technologies have increased the cost of medical care, and at this moment they are not considered to be cost-effective. 11 4. CERVICAL CANCER SCREENING Cervical cytology screening has been credited with the 70% decrease in cervical cancer mortality in the United States and Canada in the past 50 years. Prior to 2002 various organizations generally recommended that screening should start at the beginning of sexual activity. It should continue annually throughout life, and women with several prior negative Pap smears may have the test repeated at longer intervals. The revised cervical cancer screening guidelines of the American Cancer Society and those of the American College of Obstetricians and Gynecologists were published in 2002 and 2003. • Both organizations recommended that screening should start 3 years after sexual debut or by the age of 21, whichever comes first. • If a woman has 3 consecutive negative tests in the preceeding 10 years the screening may stop at age 70. • For women aged 30 and older dual screening by Pap smear and HPV testing to triage equivocal cases is an option. If both tests are negative, screening should not be repeated for 3 years. • Both organizations do not recommend cytology screening for hysterectomized women without a cervix, unless the surgery was performed for cervical premalignant or malignant lesions. Most European countries have established cervical cancer screening programs. Their recommended screening starts between the ages of 20 and 25 years and continues every 3 to 5 years until age 60 to 65. Many developing countries do not currently have any cervical cancer screening programs. For those countries, it has been estimated that the lifetime risk of cervical cancer could be reduced by up to 30% if a screening program uses a combination of Pap smear and HPV testing in women aged 30 to 59 at least once per lifetime. The prophylactic HPV vaccines for unexposed girls and women will likely to have an impact on the future cervical cancer screening. Currently, Gardasil, a vaccine targeting HPV types 6, 11, 16 and 18, and Cervarix, a vaccine targeting HPV types 16 and 18, are the two FDA-approved vaccines for females aged 9 to 26. The two HPV types 16 and 18 are responsible for about 70% of cervical cancers, but other HPV types that are not included in Gardasil and Cervarix are still responsible for the remaining 30% of cervical cancers. Therefore, women should continue to have cervical cancer screening by Pap smear regardless of their vaccination status. 12 5. QUALITY ASSURANCE QUALITY Quality assurance has always been an integral part of gynecologic cytology practice. It encompasses laboratory procedures, diagnostic accuracy and reliability with the goal to reduce false-positive and false-negative results. False-negative results may be due to: • Sampling error in: - collection device that does not sample lesional cells - failure of transfer of lesional cells to the glass slide • Laboratory error in: - detection - interpretation To minimize the laboratory component of these errors, a workload limit for each cytotechnologist screener and a re-examination of Pap smears is required: • Workload limit: 90 and 100 slides per 24 hours in Canada and The United States, respectively (minimum amount of time required to screen those slides is 8 hour). • Re-examination of cytologic materials: - prospective re-screening of 10% of all negative cases. - retrospective review of all negative Pap smears during 5 years before the diagnosis (5-year lookback) from patients with a newly diagnosed high-grade squamous intraepithelial lesion or cancer. - review of all cases having cytologic-histologic discrepancies. Results of re-examination of cytologic materials are statistically documented annually for evaluation of the laboratory performance. B. DIRECT ENDOMETRIAL CELL SAMPLES In the past fifty years, efforts have been made to design a simple endometrial sampling device for use in the physician’s office to obtain cells directly from the endometrium for cytologic examination. In 1943, Cary was the first investigator to report on the use of a metal canula and a syringe to aspirate endometrial cells. Morton et al. modified the Cary technique by including in the procedure an endometrial lavage with normal saline under positive pressure. The saline was then re-aspirated and processed for cytologic study. This sampling technique was never adopted because of the fear of seeding endometrial cancer cells into the peritoneal cavity. In 1955, Ayre promoted endometrial brushing to obtain endometrial cells. Butler and associates subsequently modified the Ayre brush and the technique for removing endometrial cells from the bristles of the brush. Despite a high diagnostic accuracy for endometrial carcinoma, the brushing technique was not widely adopted. In 1964, Dowling and Gravlee reported on the combination of 13 endometrial aspiration with uterine lavage under negative pressure to obtain endometrial cells. This technique did not produce consistently acceptable results because of the poor preservation of cell morphology, and it was subsequently discontinued. In 1974, Isaacs and co–workers promoted a relatively simple endometrial aspiration device to sample endometrial cells. This sampling technique was widely used and achieved a high diagnostic accuracy rate for endometrial cancer. In 1973, Milan and Markley reported on the use of a plastic helix to sample the endometrium by scraping. The Mi–Mark helix was widely adopted for many years, but it was subsequently found to be less efficient in detecting endometrial lesions. In recent years, a few endometrial scraping devices have been introduced to clinical practice. Of these, Endocyte and Endopap samplers have been the most popular ones. They are simple to use and cause little or no discomfort to patients and the cell samples obtained have a low rate of cellular inadequacy. C. FINE NEEDLE ASPIRATION Fine needle aspiration (FNA) during laparoscopy for cytologic evaluation of ovarian cystic lesions was originally performed by Mintz and associates in 1967. In 1971, Kjellgren and coworkers utilized this technique to classify ovarian cancers and published the results in their 1972, 1974 and 1979 publications. In 1979, Ramzy and Delaney reported on the cytologic features of epithelial ovarian tumors, and Sevin and his group reported on their FNA experience with gynecologic malignancies. In the early years of 1990s Trimor-Tritsch developed a transvaginal sonographic scorering system to distinguish a benign from malignant ovarian and adnexal cysts, and a transvaginal FNA has been used to diagnose cystic lesions with low and intermediate scores. FNA of small benign-appearing uterine adnexal masses detected during laparoscopy has now become a more routine procedure for patient care, and it proved to be a safe and economical method for diagnosis of those lesions. 14 TECHNICAL CONSIDERATIONS Specimen Procurement A. CONVENTIONAL CERVICAL/VAGINAL SMEARS The most commonly used cell collecting devices for conventional Pap smears are the modified Ayre spatula and the cytobrush. Cell samples from the uterine cervix and posterior vaginal fornix are usually collected, respectively, by the pointed end and the blunt end of an Ayre spatula. They are either deposited onto 2 glass slides or mixed together on one glass slide. The use of a cytobrush in conjunction with a spatula can help ensure an adequate representation of the squamocolumnar junction. Once the samples have been evenly spread on the glass slide, they are immediately fixed with a commercial spray fixative. To prevent the formation of thick cellular ridges, the slide should be held about 25 cm from the spray nozzle or floated within the fixative for 15 to 30 min. The slides are then air-dried and placed in a rigid container for mailing to a referral cytology laboratory. If the slide is broken during transportation, the cytologic material may be transferred to a new glass slide for examination by using a special technique. B. LIQUID-BASED PREPARATIONS Collecting devices for liquid-based preparations may be either a broom-like device or a plastic spatula in combination with a cytobrush. The ThinPrep Pap Test (HologicTM Inc) consists of rinsing the collection device into a vial of a methanol-based fixative media (PreservCyt®). The TP 2000 or 3000 processor uses a semi-automated technique for preparing monolayer cell samples. After gentle dispersion to homogenize the specimen, the solution is aspirated through a membranous filter which is subsequently pressed against a glass slide. Negative pressure and surface tension allow for transfer of the cells to the slide, which is then automatically placed in 95% ethanol for fixing before staining. After cell transfer, the filter assembly is discarded. The BD SurePath™ Pap test requires collection devices (broom, or plastic spatula and cytobrush) with detachable heads which are dropped into the ethanol-based collection vial. The automated processor homogenizes the specimen and dispenses it onto a density gradient reagent where cells are separated from interfering cell debris and inflammatory cells. Automated pipetting transfers the cell concentrate to small plastic chambers for gravity sedimentation onto glass slides and then subsequent Pap staining within each chamber. 15 C. DIRECT ENDOMETRIAL CELL SAMPLES Materials procured by Endopap or Endocyte samplers are spread on a glass slide and fixed in 95% ethanol or with a commercial spray fixative. The smears are then stained by the Papanicolaou method. Any minute tissue fragments obtained are fixed in 10% neutral buffered formalin for supplementary histological examination. D. FINE NEEDLE ASPIRATES Cytologic material obtained by fine needle aspiration is used to prepare direct smears and/or cytospin smears. The cell films are either wet-fixed with 95% ethanol for staining with the Papanicolaou technique or hematoxylin and eosin, or air-dried for staining by the May-Grϋnwald-Giemsa technique or by one of its modified methods (Giemsa and Diff-Quik methods). PAPANICOLAOU STAIN STAINING TECHNIQUE The Papanicolaou stain is the standard method for staining cervicovaginal cell samples worldwide. For optimal cytologic interpretation, an adequate well-prepared, well-stained and well-preserved cell sample is mandatory. The preparation of cell samples for cytodiagnosis consists of 3 basic steps: fixation, staining and mounting of slides. For Papanicolaou staining, fixation of cell samples with alcohol is mandatory. If the smear is air-dried, it must be rehydrated with either an isotonic saline solution or with a solution consisting of 50% glycerol and water prior to fixation in 95% ethanol and staining with the Papanicolaou method. This procedure satisfactorily preserves the morphology of squamous cells in cervicovaginal smears but the morphology of endocervical glandular cells is less satisfactorily preserved. The usual fixative is 95% ethanol, however substitutes such as 100% methanol, 80% isopropanol or denatured alcohol are suitable. The cell samples must be fixed for at least 15 to 30 minutes. If the slides are to be sent to a distant laboratory for cytologic evaluation, they should be air-dried after the fixation and carefully packed in a rigid container for mailing. At the laboratory, the slides will be re-immersed in 95% ethanol prior to staining. Papanicolaou stained smears may subsequently be stained with antibodies by routine immunocytochemical techniques without prior destaining of the cells with an acid-ethanol solution. 16 Papanicolaou staining consists of two main consecutive steps: nuclear staining with hematoxylin and cytoplasmic staining using Orange G and EA 36 or 50 polychrome. For nuclear staining two techniques are used depending on the laboratory preference: progressive and regressive methods. In the progressive method the cell nuclei are stained to the desired intensity with hematoxylin. In the regressive method the nuclei are over-stained with hematoxylin and then excess hematoxylin is removed with diluted HCl. The cytoplasm is stained with Orange G and EA 36 or 50 polychrome. To obtain good cellular detail the Papanicolaou staining solutions should be changed after staining 2000 slides or every 6 to 8 weeks, which ever comes first. They should be filtered daily to eliminate precipitates and "floaters". TROUBLESHOOTING Troubleshooting may be encountered in any aforementioned steps of cell sample preparation but they are most frequently encountered during the preparation and staining of the specimen. Common problems in Papanicolaou staining are summarized in Tables 1.1 and 1.2. Table 1.1. Nuclear staining problems Problem Possible Reason(s) Nuclei too dark 1. Overstaining with hematoxylin 2. Excess stain not adequately removed by rinsing with tap water 3. Inadequate rinsing in HCl solution 4. HCl concentration too weak 5. Ammonium chloride solution (or other bluing agent) too strong Nuclei too pale 1. Diluted hematoxylin solution 2. Inadequate time in hematoxylin 3. Polyethylene glycol coating not adequately removed prior to staining with hematoxylin 4. HCl not completely removed by tap water 5. HCl solution too concentrated 6. Slide dipped too long in HCl solution 7. Ammonium hydroxide solution too weak 8. Excessive time in chlorinated tap water 9. pH of tap water after hematoxylin not alkaline enough 17 Table 1.2. Cytoplasmic staining problems Problem Possible Reason(s) Inconsistent 1. Air-drying prior to fixation cytoplasmic staining 2. Polyethylene coating not adequately removed prior to staining 3. Slides left too long in ethanol rinses or clearing solution following OG/EA staining 4. Time in hematoxylin too long 5. Excess hematoxylin not removed prior to OG/EA staining 6. Inadequate rinsing between solutions 7. Inadequate rinsing following staining with dyes 8. Inadequate draining of slides between rinses 9. Inappropriate pH of tap water or EA solution 10. Change of cell pH by bacterial infection 11. Variable thickness of smear Cytoplasm too green Green dye too strong in EA solution Lack of contrasting Exhausted hematoxylin and EA dye cytoplasmic stain Hazy grey appearance of cells 1. Dehydrating & clearing solutions contaminated with water 2. Incomplete removal of polyethylene glycol coating prior staining Opaque white color Inadequate rinse after Scott tap water substitute on the back of slide Pink, orange or Oven temperature too high yellow slides "Cornflake" artifact Air bubbles entrapped on cell surfaces These 2 Tables are adapted with modifications from: Holmquist MD, Keebler CM. Cytopreparative techniques. In: Manual of Cytotechnology, Keebler CM, Somrak TM, eds. 7th ed, 1993. Chicago, ASCP Press, p 410-448. 18 REPORTING CERVICAL CYTOLOGY Since the introduction of cervical cancer screening programs in the 1940s, different classification systems have been used to report Pap smears. Prior to The Bethesda System (TBS), which was developed to standardize nomenclature and establish more consistent reports, the Papanicolaou numerical classification, the Reagan classification of “mild, moderate, severe dysplasia” and “carcinoma in situ”, and Richart’s concept of “cervical intraepithelial neoplasia” (CIN) have been used extensively to report cervicovaginal cytology worldwide. The original Papanicolaou system consisted of 5 classes: I (no atypia), II (atypia but no malignancy), III (suggestive of, but not conclusive for malignancy), IV (strongly suggestive of malignancy) and V (conclusive for malignancy) has proven to be difficult to reproduce. As a result, the dysplasia and CIN classification systems were more widely used. With the elimination of the Papanicolaou classification in 1988 by the first Bethesda Conference, the other two above-mentioned classifications are often still used in conjunction with TBS, as they provide histologic equivalents in cervical pathology. The 1988 TBS has been revised twice. The first review was in April 1991 and the second was in April/May 2001. When compared to the 1991 TBS, the latest TBS (2001) has several modifications, and the most important of which are listed below: • The type of the specimen, the utilization of an automated screening device and HPV testing. • Any preparation containing abnormal cells (Atypical squamous cells of undetermined significance (ASC-US), Atypical glandular cells (AGC) or worse is classified as satisfactory. • Cellularity criteria are given: if >75% of epithelial cells are obscured by blood or inflammatory exudates, the specimen is considered as “unsatisfactory”. • The ASC category is divided into 2 subtypes with more precise morphologic definitions and criteria for ASC-US and ASC, cannot rule out HSIL (ASC-H). • Individualization of Endocervical adenocarcinoma in situ and elimination of Atypical glandular cells of undetermined significance (AGUS), favor neoplastic and AGUS, favor reactive; and addition of atypical glandular cells, NOS, and AGC, favor neoplastic. Shortly after its introduction in 2001, the majority of cytology laboratories in the United States had adopted TBS 2001 for reporting Pap smears (85.5% by 2003). It is now used extensively around the world to report cervical cytology. 19 THE BETHESDA SYSTEM - 2001 The Bethesda System-2001 consists of several components, as outlined below, and it is recommended for reporting cervical cytology. SPECIMEN TYPE Indicate conventional (Pap smear) vs. liquid-based preparation versus other SPECIMEN ADEQUACY • Satisfactory for evaluation (describe presence or absence of endocervical or transformation zone component and other quality indicators, e.g., partially obscuring blood, inflammation, etc.) • Unsatisfactory for evaluation… (specify reason). • Specimen rejected/not processed (specify reason). • Specimen processed and examined, but unsatisfactory for evaluation of epithelial abnormality because of (specify reason). GENERAL CATEGORIZATION (Optional) • Negative for Intraepithelial Lesion or Malignancy. • Epithelial Cell Abnormality: See Interpretation/Result • Other: see Interpretation/Result (e.g. endometrial cells in a woman ≥40 yr of age). INTERPRETATION/RESULT A. Negative for Intraepithelial Lesion or Malignancy When there is no cellular evidence of neoplasia, state this in the General Categorization above and/or in the Interpretation/Result section of the report - whether or not there are organisms or other non-neoplastic findings. 1. Organisms: • Trichomonas vaginalis • Fungal organisms morphologically consistent with Candida spp. • Shift in flora suggestive of bacterial vaginosis. • Bacteria morphologically consistent with Actinomyces spp. • Cellular changes consistent with herpes simplex virus. 2. Other Non-neoplastic Findings (Optional to report; list not inclusive): • Reactive cellular changes associated with - inflammation (includes typical repair) - radiation - intrauterine device (IUD). • Glandular cells status posthysterectomy • Atrophy. 3. Other • Endometrial cells (in a woman ≥40 years of age) (specify if “negative for squamous intraepithelial lesion”) 20 B. Epithelial Cell Abnormalities 1. Squamous cell: • Atypical squamous cells - of undetermined significance (ASC-US). - cannot exclude HSIL (ASC-H). • Low-grade squamous intraepithelial lesion (LSIL) (encompassing: HPV/mild dysplasia/CIN1). • High-grade squamous intraepithelial lesion (HSIL) (encompassing: moderate and severe dysplasia, CIS, CIN 2 and CIN 3). • with features suspicious for invasion (if invasion is suspected). • Squamous cell carcinoma 2. Glandular Cell: • Atypical - endocervical cells (NOS or specify in comments) - endometrial cells (NOS or specify in comments) - glandular cells (NOS or specify in comments). • Atypical - endocervical cells, favor neoplastic. - glandular cells, favor neoplastic. • Endocervical adenocarcinoma in situ. • Adenocarcinoma - endocervical - endometrial - extrauterine - not otherwise specified (NOS) C. Other Malignant Neoplasm: (specify) ANCILLARY TESTING Provide a brief description of the test method(s) and report the result so that it is easily understood by the clinician. AUTOMATE REVIEW If specimen was examined by automated device, specify the device and the result. EDUCATIONAL NOTES AND SUGGESTIONS (optional) Suggestions should be concise and consistent with clinical follow-up guidelines published by professional organizations (references to relevant publications may be included). 21 SPECIMEN ADEQUACY Evaluation of specimen adequacy is important. There are 2 categories of specimen adequacy in The Bethesda System 2001: • Satisfactory for evaluation • Unsatisfactory for evaluation (lack of patient identification or unacceptable specimen due to slide broken beyond repair…) Depending on the specimen type, the estimated minimum numbers of well-preserved squamous cells required for a specimen to be regarded as adequate or satisfactory for cytologic evaluation are different: • 8,000 to 12,000 for a CP smear, and • 5,000 cells for a LBP. The numbers of squamous cells constitute an additional criterion besides the presence of at least 10 well-preserved endocervical or metaplastic squamous cells. Any specimen with abnormal cells is, by definition, satisfactory for evaluation. For obscuring factors, if a specimen has more than 75% of squamous cell nuclei obscured by white blood cells, blood, drying artifact, other, it should be termed unsatisfactory, assuming no abnormal cells are identified. (Fig.1.1). When 50% to 75% of the epithelial cells are obscured, a statement describing the specimen as partially obscured should follow the satisfactory term. Abundant cytolysis does not qualify the specimen as “unsatisfactory” unless nearly all nuclei are devoid of cytoplasm. (Fig.1.2). Fig. 1.1. An “unsatisfactory” CP smear showing abundant polymorphonuclear leukocytes obscuring over 75% of the squamous cell nuclei. 22 Fig.1.2 A CP smear with extensive cytolysis by Döderlein bacilli but still showing fairly well-preserved nuclei, some of which are surrounded by a small amount of cytoplasm. 23 REFERENCES American College of Obstetricians and Gynecologists. ACOG practice bulletin. Cervical Cytology Screening. Int J Gynaecol Obstet. 2003; 83: 237. Anderson GH. Cytologic screening programs. In Comprehensive Cytopathology, Bibbo M, ed, Philadelphia, Saunders, 1992, pp 48-58. Angstrom T, et al. The cytologic diagnosis of ovarian tumors by means of aspiration biopsy. Acta Cytol.1972;16:336. Babes A. Diagnostique du cancer du col uterin par les frottis. Press Med 1928; 36:451 Bales CE, Durfee GR. Cytologic techniques and principles of operation of a laboratory of cytology. In Diagnostic Cytology and its Histopathologic Bases, Koss LG, ed, 4th ed, 1992, Philadelphia, JB Lippincott, pp 1451. Bogdanich W. Lax laboratories: Hurried screening of Pap smears elevates error rate of the test for cervical cancer. Wall Street Journal, November 2, 1987, p.1. Bonfiglio TA. Gynecologic cytopathology. Historical perspective, current status, and future outlook. Pathology Case Reviews. 2005;10:98. Brown GG, Tao LC. Restoration of broken cytologic slide and creation of multiple slides from a single smear preparation. Acta Cytol 1992; 36:259-263. Chan JKC, Kung ITM. Hydration of air-dried smears with normal saline: application in fine needle aspiration cytologic examination. Am J Clin Pathol 1988; 89:30. Cibas ES. Cervical and vaginal cytology. In Cytology. Diagnostic principles and clinical correlates. 3nd edition, 2009, Philadelphia, Edinburgh, Saunders Elsevier, p.1. Cibas ES. Laboratory management. In Cytology. Diagnostic priciples and clinical correlates. 3nd edition, 2009. Philadelphia, Saunders Elsevier. p. 495. Davey DD, et al. Bethesda 2001 implementation and reporting rates 2003 practices of participants in the College of American Pathologists Interlaboratory Group Comparison Program in Cervicovaginal Cytology. Arch Pathol Lab Med. 2004; 128:1224. DeMay RM. The Pap Test. Chicago, ASCP Press, 2005. Graham RM. Effects of radiation on vaginal cells in cervical carcinoma. I. Description of cellular changes. II. Prognostic significance. Surg Gynecol Obstet 1947; 84:153. 24 Graham RM and Vicent Memorial Hospital Laboratory Staff. The cytologic diagnosis of cancer. Philadelphia, WB Saunders, 1950. Holmquist MD, Keebler CM. Cytopreparatory techniques. In The Manual of Cytotechnology, Keebler CM, Somrak TM, editors, 7th ed, 1993, Chicago, ASCP Press, p. 410. Kjellgren O, Angstrom T. Transvaginal and transrectal aspiration biopsy in the diagnosis and classification of ovarian tumors. In Aspiration Biopsy Cytology, Part 2, Cytology of infradiaphragmatic organs. Zajicek J, ed, Basel, Karger, 1979. Kline TS, Nguyen GK. The Bethesda System-with commentary. In Critical Issues in Cytopathology. New York, Igaku-shoin, 1996, p.11. Mintz M. Ponction de 94 kystes para-uterins sous coelioscopie et etude cytologique des liquides. Gynecologia. 1957;163 :61. Naylor B. The century for cytopathology. Acta Cytol 2000; 44:709. Ng WF, et al. Rehydration of air-dried smears with normal saline application in fluid cytology. Acta Cytol 1994; 38:56 Nguyen GK, Redburn J. Endometrial cytology by direct sampling. Its value and limitations in the diagnosis of endometrial lesions. Pathol Ann. 1995; 30(2):179. Nguyen GK, et al. Cervical squamous cell carcinoma and its precursor lesions: cytodiagnostic criteria and pitfalls. Anat Pathol. 1996;1:139. Papanicolaou GN. The diagnosis of early human pregnancy by the vaginal smear method. Proc Soc Exp Biol Med 1925; 22:436. Papanicolaou GN. New cancer diagnosis. Proc Third Race Betterment Conference, Battle Creek, Michigan 1928: 528. Papanicolaou GN, Traut HF. Diagnosis of uterine cancer by the vaginal smear. New York, Commonwealth Fund, 1943. Papanicolaou GN. Atlas of Exfoliative Cytology. Cambridge, Mass: The Commonwealth Fund, Harvard University Press; 1954. Ramzy I, Delaney M. Fine needle aspiration of ovarian masses. I. Correlative cytologic and histologic study of celomic epithelial neoplasms. Acta Cytol. 1979; 23:97. 25 Regan JW, et al. Cellular morphology of carcinoma in situ and dysplasia or atypical hyperplasia of the uterine cervix. Cancer. 1953;6:224. Richart RM. Cervical intraepithelial neoplasia: a review. Pathol Annu.1973; 8:301. Saslow D, et al. American Cancer Society guidelines for the early detection of cervical neoplasia and cancer. CA Cancer J Clin. 2002; 52:342. Sherlaw-Johnson C, et al. Evaluating cervical cancer screening programmmes for developing countries. Int J Cancer. 1997; 72:210. Sevin BU, et al. Fine needle aspiration cytology in gynecologic oncology. I Clinical aspects. Acta Cytol.1979; 23:227. Solomon D, Nayar R. The Bethesda System (2001) for reporting cervical cytology. 2nd ed, 2004, New York, Springer-Verlag Stockard CR, Papanicolaou GN. The existence of atypical oestrous cycle in the guinea pig-with a study of its histologic and physiological changes. Am J Anat 1917; 22: 2: 225. Stuart G, et al. Report of the 2003 Pan-Canadian forum on cervical cancer prevention and control. J Obstet Gynaecol Can. 2004;26:1004. The 1988 Bethesda System for reporting cervical/vaginal cytologic diagnoses. Diagn Cytopathol 1989; 5:331. The 1991 Bethesda System for reporting cervical/vaginal cytologic diagnoses. Diagn Cytopathol 1993; 9:235. Timor-Tritsch IE, et al. Puncture procedures utilizing transvaginal ultrasonic guidance. Ultrasound Obstet Gynecol. 1991;1: 144. Van Ballegooijen M, et al. Overview of important cervical cancer screening process values in European Union countries and tentative predictions of the corresponding effectiveness and cost-effectiveness. Eur J Cancer. 2000;36:2177. Wilbur DC, Henry MR, eds. College of American Pathologists Practical Guide to Gynecologic Cytopathology: Morphology, Management and Molecular Methods. Northfield, Illinois, College of American Pathologists, 2008. Yee H, et al. Transvaginal sonographic characterization combined with cytologic evaluation in the diagnosis of ovarian and adnexal cysts. Diagn Cytopathol. 1994;10: 107. 26 Chapter 2 Normal Uterus and Vagina Katherine M. Ceballos, Brenda Smith and Gia-Khanh Nguyen A. THE PAP TEST Cell samples from the uterine cervix are usually obtained with an Ayre–type spatula with a longer cervical tip, or with a cervical brush. Many heathcare providers use a combination of a spatula and an endocervical brush to sample the exocervix, cervical canal and T zone. To avoid air-drying artifactual changes that interfere with the cytologic evaluation, the cervical cell sample is immediately spread onto a glass slide and fixed with a commercial spray fixative. Cervical smears may also be wet-fixed with alcohol and are stained by the standard Papanicolaou method. In the Canadian province of British Columbia, cervical cancer screening is performed in one central cytology laboratory. All cervicovaginal smears are air-dried and submitted to the laboratory where they are rehydrated with 50% glycerol and water before staining with the Papanicolaou method. This technique satisfactorily preserves the morphology of cervical squamous cells but has more limited value with endocervical glandular cells. The normal uterine cervix consists of an exocervix and an endocervical canal. The exocervix is covered by a nonkeratinizing, stratified squamous epithelium and the endocervix is lined by a single layer of columnar epithelium with complex folding and a layer of reserve cells. Two types of glandular cells are identified: non-ciliated, mucous-secreting cells and ciliated cells that are very few in number. With advancing age, the distal part of the cervical canal is replaced by metaplastic squamous cells. The squamocolumnar junction is on the exocervix and the transformation zone (T-zone), also known as ectropion, is located between the original squamocolumnar junction and the inner border of metaplastic squamous epithelium. (Fig. 2.1). Endocrine and melanotic cells that can be visualized by immunohistochemistry are rarely observed in the cervix. For cervical cancer screening the T-zone should be sampled, as it is the site of origin of > 90% of cervical cancer and its precursor lesions. A representative cell sample from the T-zone commonly contains abundant squamous cells, metaplastic squamous cells and many glandular cells from the endocervical canal. The vagina mucosa is also covered by a layer of nonkeratinizing squamous epithelium that is similar to that of the exocervix and yields abundant squamous cells similar to those of the exocervix. 27 A B Fig.2.1. Histology of normal uterine cervix in women in reproductive ages. A. Squamocolumnar junction showing an abrupt change between the nonkeratinizing squamous epithelium of the exocervix and the endocervical glandular epithelium. B. Endocervical mucus-secreting glandular epithelium with basally located nuclei. REPRODUCTIVE AGE A cervical smear from a non-pregnant woman of reproductive age is cellular and shows numerous epithelial cells. Normal squamous cells exfoliate predominantly singly and endocervical glandular cells usually exfoliate singly or in sheets of different sizes. Vaginal smears show only squamous cells that are similar to those of the exocervix. Squamous cells of the cervix and vagina are classified as superficial, intermediate and parabasal, according to their characteristic features described below. A maturation index is expressed as a ratio of different types of squamous cells (parabasal: intermediate: superficial). During the proliferative phase of the menstrual cycle, the number of superficial cells increases gradually under the influence of increased levels of serum estrogen. Prior to ovulation the serum estrogen reaches its peak and superficial cells 28 predominate in the smear with the maturation index shifting to the right (for example 0:20:80). After ovulation and under the influence of an increasing level of serum progesterone, the intermediate squamous cells predominate in the smear, shifting the maturation index to the middle (for example 0:80:20). With the availability of sophisticated biochemical methods used for measurement of serum hormonal levels, a cytologic evaluation of maturation index by vaginal smear is currently no longer requested by endocrinologists or gynecologists. Normal endometrial cells may also be seen in normal Pap smears, depending on the phase of the menstrual cycle. Normal endometrial cells are described below. Main cytologic features of normal cervical cells are summarized as follows: • Superficial squamous cells are polygonal in shape with translucent, eosinophilic thin cytoplasm that may contain brownish keratohyalin granules. Their nuclei are pyknotic, centrally located and measure 16 to 20 µm2 in area. These cells are seen singly and in loose clusters. (Fig. 2.2). • Intermediate squamous cells are oval or polygonal in shape with translucent, eosinophilic or basophilic cytoplasm that commonly shows folding. Their nuclei are vesicular, have a fine chromatin and measure about 35 µm2 in area. Occasionally, superficial and intermediate squamous cells have a spindle-shape or display a long cytoplasmic extension or “tail” that is a rare collection of different types of intracytoplasmic filaments called Herxheimer’s spirals. • Parabasal squamous cells are rarely encountered in a smear from a premenopausal woman unless she is in the post partum period. They are commonly seen in post menopausal atrophy. They are seen singly and are oval in shape with opaque, basophilic cytoplasm. They have centrally located vesicular nuclei with fine chromatin which measure about 50 µm2 in area. • Endocervical glandular cells are columnar in shape with pale, abundant, mucinous cytoplasm and basally located vesicular nuclei displaying a granular chromatin and micronucleoli. A few are ciliated. In conventional Pap smears, endocervical cells are present singly or in monolayered sheets with characteristic honeycomb and picket-fence arrangements. Naked nuclei within mucus are a common finding, but they are not regarded as an evidence of Pap smear adequacy. Endocervical glandular cells commonly present singly in liquid-based preparations. (Fig. 2.3). • Metaplastic squamous cells arise from the reserve cells of endocervical columnar epithelium. They exfoliate singly or in pavement-like sheets and are polygonal or oval in shape. Their cytoplasm varies with the level of cell maturation and in immature cells it may be thin and vacuolated and with cytoplasmic extensions. It is waxy, basophilic or eosinophilic in mature cells. 29 Their vesicular nuclei have granular chromatin and have an area of about 50 µm2. (Fig. 2.4). A B Fig. 2.2. Normal cervical and vaginal squamous cells in a CP smear: A. Superficial, intermediate and metaplastic cells seen singly and in aggregates.A few smaller, oval parabasal cells are also present. B. One superficial cell with pyknotic nucleus and intracytoplasmic keratohyaline granules and 2 intermediate cells with vesicular nuclei. 30 A B Fig.2.3. Normal endocervical cells in a CP smear: A. Mucus secreting endocervical columnar cells in small epithelial fragments. B. Minute epithelial fragment consisting of ciliated glandular cells. A 31 B Fig. 2.4. Metaplastic squamous cells in a CP smear: A. A few smaller metaplastic cells admixed with superficial and intermediate cells. B. Mature metaplastic cells and immature metaplastic cells with intracytoplasmic mucous vacuoles. The uterine corpus consists of a thick smooth muscle wall and a triangular cavity lined by endometrium that is comprised of a columnar glandular epithelium supported by endometrial stroma, a specialized lamina propria. The endometrium consists of 2 layers: the basalis and the functionalis. The basalis layer is a thin deeper layer abutting the myometrium, and from this layer the endometrium regenerates after menstruation. The functionalis layer is located above the basalis layer and responds to ovarian hormones. It has a rapid growth during the first half of the menstrual cycle (proliferative phase) and it is characterized by straight and narrow glands with low columnar cells surrounded by spindle-shaped stromal cells with scant cytoplasm. During the second half of the cycle (secretory phase) it undergoes a maturation characterized by glandular secretion. The glands become more tortuous and the stroma is edematous. The superficial stromal cells, under the effect of progesterone, undergo a predecidual change that is characterized by an increased amount of cytoplasm and their nucleoli become visible. The endometrium in the lower uterine segment does not show the above-described cyclic changes. The length of the menstrual cycle varies considerably among female individuals, but it is 28 days long in most women. The proliferative phase is variable in length but the secretory phase is almost always 14 days long. Exfoliated endometrial cells are more abundant during the first 10 to12 days of the cycle, and they may be detectable in cervical vaginal smears. The normal endometrial cells are small, cuboidal cells with scant cytoplasm, round nuclei, chromatin clumping and small nucleoli, and they are commonly seen in cohesive clusters of different size. Superficial stromal cells resemble histiocytes and are present singly or in loosely cohesive sheets. Deep stromal cells appear as small loose clusters of spindle cells. Histiocytes and large masses of endometrial cells or 32 wreaths (exodus) are more commonly observed on the above-mentioned days of the cycle. (Fig. 2.5). A B C Fig. 2.5. Normal, spontaneously exfoliated endometrial cells in CP smears: A. A cluster of round endometrial epithelial cells showing scant cytoplasm. B. Superficial stromal cells, resembling histiocytes, present singly and in loose clusters. C. An endometrial wreath consisting of a large cluster of stromal cells (at the center) surrounded by a layer of endometrial epithelial cells. 33 An Ayre-type spatula with a longer tip or a cytobrush may inadvertently sample fragments of endometrium from the lower uterine segment (LUS). These LUS endometrial tissue fragments are more commonly seen in women with prior cervical cone biopsy and appear as large, thick cell sheets with folding. The epithelial cells at the periphery display nuclei in picket-fence arrangement. (Fig.2.6). Smaller endometrial cell clusters and aggregates of stromal cells are commonly present. A B Fig. 2.6. CP smear showing thick fragments and cell clusters of lower uterine segment endometrium scrapped by a cervical cell sampler with a longer tip. During pregnancy the placenta secretes large amounts of estrogen and progesterone, and as a result, intermediate squamous cells predominate in the smear, accounting for at least 80% of the total cell population. Intermediate cells are rich in glycogen and display an elongated, boat-shaped configuration (navicular 34 cells). Usually the smear consists entirely of intermediate squamous cells by the 4th or 5th month of pregnancy. (Fig. 2.7). In the postpartum period the Pap smear is predominated by parabasal cells. (Fig. 2.8). This is due to placental parturition and suppressed ovarian functions. Fig. 2.7. CP smear showing a few intermediate squamous cells and a navicular celll containing a large amount of intracytoplasmic glycogen that is yellowish in color. Fig. 2.8. LBP from a woman in postpartum period showing parabasal cells singly and in a sheet. During pregnancy the Pap smear may rarely show a few decidual cells. Decidual cells are of the same size as parabasal cells but occur in clusters and have thick, granular cytoplasm and larger oval or round nuclei without prominent nucleoli. (Fig. 2.9). It is important to note that decidual nodules may occur in the cervix during pregnancy and in women taking progesterone-rich oral contraceptives. An Arias-Stella reaction affects endometrial cells in early pregnancy but it may also occur in 35 endocervical canal. Arias-Stella cells are large cells with vacuolated cytoplasm, hyperchomatic multiple nuclei and prominent nucleoli. They may mimic cells derived from a clear cell adenocarcinoma. Cytotrophoblasts and multinucleated syncytiotrophoblasts are rarely observed, except in patients with threatened abortion. Fig. 2.9. A group of decidual cells in a CP smear. Cockleburrs are hematoidin crystal arrays that are often surrounded by histiocytes. They measure up to 100 µm in greatest dimension. They are more commonly found in pregnant women and rarely in nonpregnant women. (Fig 2.10). Fig. 2.10. Cockleburrs surrounded by histiocytes in a CP smear. 36 MENOPAUSE In early menopause the Pap smear is predominated by superficial squamous cells. This is caused by the development of nonovulated graafian follicles. As menopause progresses the smear is predominated by either intermediate cells or parabasal cells. (Fig. 2.11). A B C Fig. 2.11. Atrophic cervix: 37 A, B. A CP smear showing a parabasal cell predominant pattern. Some cells display eosinophilic cytoplasm. Abundant necrotic debris is present in the smear background. C. LBP showing parabasal cells in a clean background. Parabasal cells from atrophic vaginitis may exhibit nuclear enlargement, mimicking dyskaryotic squamous cells. A repeat Pap smear taken immediately after a course of topical treatment with estrogen cream (to induce cell maturation) will be helpful to solve this diagnostic dilemma, as any dyskaryotic squamous cells will remain unchanged, and normal parabasal cells will have matured. (Fig. 2.12) Fig. 2.12. Atrophic vaginitis showing in a CP smear degenerated polymorphonuclear leukocytes and rare parabasal cells with slightly enlarged nuclei. B. DIRECT ENDOMETRIAL CELL SAMPLES Endocyte sampler is more commonly used to obtain endometrial cells. The cytologic material obtained is spread on glass slides and fixed in 95% ethanol or with a commercial spray fixative. The smears are then stained by the Papanicolaou method. Excess cytologic material is fixed in 10% normal buffered formalin for supplementary cellblock preparation. Properly collected material usually contains numerous glandular fragments and clusters of endometrial stromal cells. Criteria for cellular adequacy of endometrial samples vary among investigators. With Endocyte samplers Byrne required the presence of at least 15 endometrial fragments to consider a sample satisfactory for cytologic evaluation. About 10% of all endometrial samples have been reported as inadequate for cytologic evaluation. Criteria for cellular adequacy vary with the types of endometrial lesions. For a nonmalignant lesion, the presence of at least 10 large fragments of endometrial epithelium, or clusters of endometrial cells on all available smears, is required for a confident cytodiagnosis. However, for an endometrial malignancy, the presence of 5 or 6 groups of well–preserved cancer cells with 5 to 10 cells in each group is adequate for a correct diagnosis. 38 The cytologic manifestations of a histologically normal endometrium in samples obtained by endometrial scraping with Endocyte and Endopap samplers are similar. About 4% to 10% of endometrial samples procured by these 2 devices show inadequate endometrial cells for evaluation. Reproductive endometrium yields large sheets of surface endometrial epithelium with folded, endometrial gland openings and cells in a honeycomb pattern, as well as several endometrial glandular elements and loosely clustered stromal cells with oval nuclei and ill–defined cytoplasm. The endometrial glands from a proliferative endometrium are straight and tubular in shape and consist of polygonal cells with oval nuclei and scant cytoplasm. Epithelial and stromal cells with mitotic figures are commonly found. The glandular cells from a secretory endometrium are composed of irregularly open glandular segments showing larger epithelial cells in honeycomb pattern, with more abundant cytoplasm and perinuclear halos. Epithelial and stromal cells with mitotic figures are practically not observed. (Figs. 2.13 to 2.17). Small endocervical epithelial sheets with mucus-secreting cells in honeycomb pattern are generally present A B Fig. 2.13. Histology of a proliferative endometrium showing tubularendometrial glands. 39 A B Fig. 2.14. Proliferative endometrium yields in direct endometrial sample. A. A monolayer sheet of endometrial surface epithelium with honeycomb pattern. B. Proliferative endometrial glands with tubular configuration and stromal cells. A 40 B Fig. 2.15. Histology of a secretory endometrium showing irregular, tortuous endometrial glands with secretion. Fig.2.16. Secretory endometrium showing in direct sampling secretory endometrial glands with irregularly open glands and scanty stromal cells. 41 Fig. 2.17. Clustered spindle-shaped stromal cells from a proliferative endometrium showing oval or elongated nuclei and ill-defined, slightly basophilic cytoplasm. Atrophic endometrium yields scant cellular material containing a few small surface epithelial sheets, some straight and narrow tubular endometrial glands and scattered stromal cells. Epithelial and stromal cells with mitotic figures are not identifiable. (Fig. 2.18). Fig. 2.18. Atrophic endometrium showing in CP smear 2 short endometrial tubular glands and an aggregate of stromal cells. 42 REFERENCES Byrne AJ. Endocyte endometrial smears in the cytodiagnosis of endometrial carcinoma. Acta Cytol 1990; 34: 373. Cibas ES. Cervical vaginal cytology. In Cytology. Diagnostic principles and clinical correlates. 3rd edition, 2009, Cibas ES, Ducatman BS, eds. Edinburgh, Sauders, p.1. DeMay RM. The Pap Test. Chicago, ASCP Press, 2005. Koss LG. Diagnostic Cytology and Its Histopathologic Bases, 3rd ed; 1979, Philadelphia, JB Lippincott, p.270, 510. Mckenzie P, et al. Cytology of body of uterus. In Diagnostic Cytopathology. 2nd ed, 2003, Gray W and McKee GT, eds. Philadelphia, Churchill Livingstone, p. 821. Nguyen GK, Kline TS. Essentials of exfoliative cytology. New York, Igaku-Shoin, 1992. Nguyen GK, Redburn J. Endometrial cytology by direct sampling. Its value and limitations in the diagnosis of endometrial lesions. Pathol Annu. 1992;30(2): 179. 43 Chapter 3 Infections and Nonneoplastic Cellular Changes Infections of the cervix and vagina are numerous and can be divided into those that are sexually transmitted and those that are not. Classic sexually transmitted diseases (STD) include syphilis, gonorrhea, chanchroid, lymphogranuloma venereum and granuloma inguinale; and material from these lesions are usually submitted to microbiology laboratories for culture and identification and not to cytology services. Other STDs are caused by Herpes simplex virus, Human papillomavirus, Trichomonas vaginalis and Chlamydia trachomatitis. These genital infections may be detected and/or suggested by routine cytologic examination of Pap smears. Nonvenereal infections are usually caused by an imbalance of the normal bacterial flora of the vagina. Only common infections with rather specific cytologic manifestations are discussed in this chapter. BACTERIAL INFECTIONS 1. Gonorrhea is caused by Neisseria gonorrhoeae, a Gram-negative diplococci. It can be asymptomatic or manifested by purulent vaginal discharges associated with a burning sensation. The bacteria may be seen within the cytoplasm of neutrophilic polymorphonuclear leukocytes in Papanicolaou-stained smears, but the infection is confirmed by bacterial culture. 2. Bacterial vaginosis is also called a “shift in flora”. It is a common, nonspecific cervicovaginitis and is often asymptomatic. On Pap smears characteristic “clue cells” are present in a filmy background of small coccobacilli. These are superficial and intermediate squamous cells covered by a layer of bacteria (coccobacilli) that obscures the cell membrane. These cells should be differentiated from “false-clue cells” that are also squamous cells covered with bacillary organisms. An absence of lactobacilli is evident. (Fig. 3.1). The Pap smear has 80% sensitivity and 87% specificity in the diagnosis of bacterial vaginosis. 44 Fig. 3.1. A “clue cell” covered with numerous coccobacilli in a CP smear. 3. Actinomycosis. Actinomyces normally reside in the female genital tract, so its presence is not an indicator of disease. Actinomycosis is characterized by a foul-smelling vaginal discharge containing sulfur granules. It is commonly caused by Actinomyces israelii in patients with IUDs or pessaries for contraception with a colonization rate of about 11% but this rate increases with the duration of use of the above-mentioned devices. These microorganisms are gram-positive and present as irregular, thick bundles or clusters of filaments (Gupta bodies). The smear background shows numerous polymorphonuclear leukocytes. (Fig.3.2). Fig. 3.2. Actinomyces in a CP smear showing a thick cluster of filamentous elements. 4. Granuloma inguinale is an uncommon disease. Material scrapped from the ulcerated lesion reveal inflammatory exudates with vacuolated macrophages containing Donovan bodies (safety pin-shaped, gram-negative microorganisms). They are best demonstrated by Giemsa stain. 45 5. Chlamydia trachomatitis is the 2nd most common STD in the Western world after HPV infection. There are about 4 million new cases diagnosed annually in the United States. The infection is usually asymptomatic in females and affects the cervix, uterus and its annexae, but not the vulva or vagina. The microorganism is an obligate intracellular parasite with 2 forms: the metabolically inactive form called the elementary body, and the metabolically active form called the reticulate body. The parasite mainly involves the endocervical columnar epithelium but may spread to the endometrium and fallopian tubes. On Pap smears the infection may be suspected by the presence of intracytoplasmic vacuoles containing aggregates of small coccoid bodies within columnar or metaplastic squamous cells. (Fig. 3.3). The diagnosis now is made by molecular testing. Metaplastic squamous cells with mucous globules may be mistaken for Chlamydial infected cells. Fig. 3.2. Chlamydial infection showing in a CP smear metaplastic squamous cells with vacuoles containing small coccoid bodies. 6. Follicular cervicitis is seen in about 50% of patients with Chlamydial infection, but the converse is not true. Numerous lymphoid cells at different stages of maturation and macrophages with tingible bodies are seen. (Fig.3.3). 46 A B Fig. 3.3. Benign lymphoid cells at different stages of maturation and a large histiocyte with tingible bodies in a CP smear from a follicular cervicitis. VIRAL INFECTIONS 1. HPV infection Genital HPV infection is common, almost exclusively sexually transmitted and selflimiting in young women. HPV is a member of papovirus family. It consists of an icosahedral viral particle (virion) containing an 8000 base pairs of double-stranded circular DNA molecule surrounded by an icosahedral protein capsid. The HPV DNA strand contains 6 E genes and a controlling regulatory region that code for viral replication process, and 2 L genes that code for capsule proteins. These E and L genes are named as “E” or “early” and “L” or “late” depending on the time of their expression in the course of the infection. The pathogenesis of HPV infection 47 was summarized by Cibas as follows. HPV requires a human host cell to replicate. In the female genital tract, the virus affects the squamous cells of vulva, vagina and cervix and the glandular cells of the endocervix. However, it targets mainly the metabolically active metaplastic squamous cells in the transformation zone of the cervix. HPV infection begins in the basal layers of the epithelium in which the HPV genome is established, with the expression of E genes. As the cells mature and move toward the surface, L1 and L2 genes are expressed. And with the progress of the infection, the viral DNA becomes established throughout the entire thickness of the epithelium, but intact virions are found only in the upper layers of the epithelium. HPVs are divided into low- and high-risk types: Low-risk HPVs are virtually never associated with cervical cancer and the most common types include 6, 11, 42 and 44. In these low-grade lesions the viral DNA does not integrate into the host genome and remains in the free episomal form. High-risk HPVs are the virus types that have been identified in CIN 2, CIN 3 and cancer, and the most common types are: 16, 18, 31, 33, 35, 45, 52 and 58. HPV types 16 and 18 usually integrate into the host genome and express large amounts of E6 and E7 proteins that block or inactivate tumor suppressor genes p53 and Rb (retinoblastoma), respectively: binding of E6 to p53 gene blocks the apoptosis of the cells, and binding of E7 to the Rb tumor suppressor protein pRb interferes with the cell cycle arrest. These bindings will lead to an uncontrolled cell proliferation. The transformed cells are capable of autonomous growth and susceptible to the acquisition of further mutations. The overall prevalence of high-risk HPV varies from continent to continent: 22.9% in Africa, 15.5% in the Americas, 6.6% in Europe and 8.3% in Asia. However, within the same continent, it varies from country to country. Its overall average prevalence in cytologically normal women is 10.4% worldwide. Young women have a high prevalence rate that is about 30% in the late teens and early twenties. It decreases through the reproductive years to as low as 5% and may rise again up to 10% to 20% in the late forties and fifties. The reasons for the second rise are not known with certainty. An increased number of new sexual partners in this age group or alteration in immune surveillance status has been suggested to explain this second rise. As a high percentage of women harbor these viruses but only a small number of them develop cancer, there is a suggestion of additional roles of risk factors. The most important risk factors include early age of first intercourse, multiple sexual partners, a male partner with multiple previous sexual partners and persistent infection by high-risk HPVs. Other risk factors include cigarette smoking, immunodeficiency status and genetic vulnerability. The peak incidence of CIN is about 30 years and that of invasive carcinoma is about 45 years. The precancerous changes referred to as CIN may begin as CIN 1 (flat condyloma) and progresses to higher grade CINs 2 and 3 or they may begin at the outset as 48 CIN 2 or 3, depending on the type of HPV infection (low- or high-risk virus) and other risk factors. According to some studies, CIN 1 regresses in 50% to 60%, persists in 30% and progresses to CIN 3 in 20% of cases. With progression, only 1% to 5% will eventually develop into an invasive cancer. With CIN 3, the rate of regression is about 33% and that of progression to invasive cancer is about 60% to 74%. Low-grade lesions are associated with replicative infections of HPV where the viral DNA is present as an episome outside of the host cell DNA. These reproductive infections allow completion of the viral life cycle and the production of whole infectious virions capable of infecting other cells. The koilocyte (characteristic for a CIN 1) is filled with complete virions ready to be discharged. Low-grade CINs typically run a benign course, resolve and clear to normal with nondetectable HPV in over 90% of cases. High-grade CINs and cancer may contain only the oncogenic portions of the HPV genome. These genes are most commonly integrated into the host’s DNA that then allows for uncontrolled expression and cellular replication. About 85% to 90% of CIN 1 lesions are caused by high-risk HPVs but they can also be caused by low-risk viral types. CIN 2 and CIN 3 are almost always caused by high-risk HPVs. CIN lesions have cytologic abnormalities that often reflect their severities. In Pap smears of patients showing only atypical squamous cells of undetermined significance, high-risk HPVs are found in about 50% of cases. In women with normal cervical cytology, 10% to 15% harbor high-risk HPVs, and of these, only about 10% will develop a high-grade CIN. HPV type 16 is more commonly associated with squamous cancers and HPV type 18 is more commonly associated with adenocarcinoma and neuroendocrine carcinoma of the cervix. The incidence of these high-risk types detected in cervical cancer in the United States range from 0.05% for HPV 42 to 54.5% for HPV 16. HPV types 6 and 11 are the commonest low-risk types that are most commonly associated with condylomas or genital warts. (Figs. 3.4 and 3.5). HPV types 16 and 18 are responsible for 70% of all cervical cancers; and together with the other high-risk types (45,31,33,35,52 and 58) are responsible for 87% of all cervical cancers. The reader is referred to Chapter 4 for illustrations of CINs and squamous cell carcinoma in Pap smears and biopsied tissues. 49 A B Fig. 3.4. Histology of a cervical condyloma. Fig.3.5. Koilocytes present in a CP smear of the lesion illustrated in Fig. 3.4. The usefulness of HPV testing as a screening test for cervical cancer is limited. As most sexually active women will contract a cervical HPV infection at some point in their lifetime, and cervical cytology will remain as the main test for this purpose. In 50 future years, with the introduction of HPV vaccines [Gardasil (targeting HPV types 6,11,16 and 18) and Cervarix (targeting HPV types 16 and 18)] administered to females aged 9 to 26, the number of high-grade CIN and carcinoma of the cervix, vulva and vagina is expected to decrease by 70%. Genital warts (caused by HPV types 6 and 11) are also expected to decrease by 90% subsequent to Gardasil vaccination. This vaccine also provides protection against genital warts and anal cancer in males. HPV Testing A few methods can be used to detect HPV infection in cytologic samples. 1. In situ hybridization -. This test is performed directly on cellular specimens and the infected cells can be visualized microscopically and correlated with morphologic abnormality. This assay is commercially available. 2. Polymerase chain reaction is the standard reference method for detecting and typing HPVs. It may also used to assess the viral load. The assay is commercially available. 3. Hybrid Capture 2 (hc2), “Digene® test” (Qiagen, Gaithersberg, Maryland) is the most commonly used method for detecting HPV DNA. It is an in vitro nucleic acid hybridization assay with signal amplification using microplate chemiluminescence for the qualitative detection of 18 types of HPV DNA in cervical specimens. The hc2 HPV DNA Test contains 18 RNA probes to various HPV types. The test can differentiate between 2 HPV DNA groups: low-risk HPV types 6,11,42,43,44 and high-risk HPV types 16,18,31,33,35,39,45, 51,52,56,58,59 and 68, but it cannot determine the specific HPV type present in the specimen. It can be performed on the residual cell samples collected for the ThinPrep® Pap Test or with the Standard Transport Medium™ (Qiagen). It is the only test approved by the US Food and Drug Administration for detecting HPV for patient care. The test has a high sensitivity and low specificity, as not all patients with positive results have CIN or invasive cancer. For high-volume sample-throughput testing, the hc2 HPV DNA test can be performed using the Rapid Capture® system Instrument Application, but only the oncogenic high-risk HPV Probe was approved for high-volume testing. As HPV infection is selflimiting in young women, this test is not useful in women under 30 years of age. Goal of HPV DNA testing The test is not intended for use as a screening tool in the general population. It is designed to augment existing methods for the detection of cervical disease and should be used in conjunction with clinical information. The test results should not be used as the sole basis for clinical assessment and treatment of patients. Its main utility is to screen patients with ASC-US Pap smear results to 51 determine the need for referral to colposcopy. However, results of the test are not intended to prevent women from proceeding to colposcopy. Interpretation HPV DNA test results should be interpreted in conjunction with clinical findings and data derived from other diagnostic procedures. • If the high-risk HPV probe is negative: there is a high probability that a high-grade CIN lesion will be not found at colposcopy. • If the high-risk HPV probe is positive: there is a low but increased probability that a high- grade CIN lesion or a more severe lesion will be detected at colposcopy. 2. Herpes simplex virus. HSV commonly infects cervix and vagina. The reported rate of genital herpes virus infection in North America range from 87 to 217 per 100,000, and it is caused by HSV-1 and HSV-2. The infection causes inflammatory epithelial ulcers. Multinucleated giant squamous cells with nuclear molding and intranuclear inclusions or chromatinic liquefaction with “ground-glass” appearance are seen at the ulcer borders and are characteristic for the infection. When the cytomorphologic changes are equivocal, an immunocytochemical staining of the infected cells with a commercial Herpes simplex antibody will be helpful for confirmation. (Fig. 3.6). This antibody is specific for HSV-2 but also cross-reacts with HSV-1. This immunocytochemical confirmatory test has a 91% sensitivity and 95% specificity. A 52 B C Fig. 3.6. Herpes simplex infection showing in a CP smear: A. Infected epithelial cells with intranuclear inclusions. B. Infected cells displaying multiple nuclei with “groundglass” change. C. Infected cell showing positives cytoplasmic reaction to Herpes simplex antibody. (Immunoperoxidase). 4. Cytomegalovirus infection. This infection is rarely detected by Pap smear. The infection affects endocervical and endometrial cells with production of characteristic large eosinophilic or amphophilic intranuclear inclusions. 53 FUNGAL INFECTION Candidiasis. Elements of Candida species are normally found in the cervix and vagina and are present in 3% of all Pap smears. Its presence is not indicative of a fungal infection requiring treatment. Candida albicans is the most common microorganism causing cervical vaginal candidiasis. Both yeasts and nonseptated pseudohyphae are seen. The budding yeasts are 3 to 7 um in greatest dimension and the pseudohyphae are eosinophilic to gray-brown. The pseudohyphae are formed by elongated budding that display constrictions along their length. They should be differentiated from elements of Trichophyton that may contaminate the Pap smears. (Fig.3.7). Fig.3.7. Vaginal candidiasis showing in a CP smear inflammatory exudate containing yeasts and nonseptated pseudohyphae of Candida species. PARASITIC INFECTION Trichomonas vaginalis is the most common STD of the lower female genital tract. Trichomonas vaginalis is a facultative anaerobic protozoan parasite without mitochondria or peroxisomes. The infection is characterized by a dense inflammatory exudate containing “pus balls” or large aggregates of polymorphonuclear leukocytes and Trichomonas vaginalis organisms. The organisms can be identified in routinely stained Pap smears. They are pear-shaped, oval or round cyanophilic organisms ranging in size from 15 to 30 um. The nucleus is pale, vesicular and eccentrically located. Intracytoplasmic eosinophilic granules are often present and flagella are usually not observed. Identification of Trichomonas vaginalis organisms in conventional Pap smears can be difficult, but its identification in liquid-based preparations is highly accurate and does not require a confirmatory test. The squamous cells present in the smear commonly show basophilic and eosinophilic cytoplasm and slightly enlarged hyperchromatic nuclei 54 with perinuclear haloes, mimicking ASC-US or LSIL cells. Leptothrix infection is a commonly associated infection with Trichomonas vaginalis. Elements of Leptothrix have a “spaghetti and meat balls” configuration. (Fig.3.8). Trichomonas vaginalis must be treated even if it is asymptomatic. A B Fig. 3.8. A CP smear showing Trichomonas vaginalis organisms with intracellular eosinophilic granules (A) and elements of Leptothrix (B). 55 INFLAMMATION-ASSOCIATED CELLULAR CHANGES Acute inflammation is often associated with changes in squamous cells. The affected cells are seen singly or in loose aggregates. They may show perinuclear halos or cytoplasmic vacuolization. Membrane rupture may be seen. The nuclei are enlarged, hyperchromatic with regular contours. (Fig. 3.9). The chromatin is clumped, fuzzy and may show karyorrhexis, karyolysis and karyopyknosis. The smear background contains numerous polymorphonuclear leukocytes. Marked cytolysis may be seen in cervicovaginal smears containing abundant Döderlein bacilli that produce enzymes causing destruction of intermediate squamous cells. Cellular debris and free lying rod-shaped bacilli are characteristic of the condition. This type of cytolysis is seen in conditions in which intermediate squamous cells are the predominant cell present such as with pregnancy, the secretory phase of the menstrual cycle, steroid therapy and postmenopause. A B Fig. 3.9. Perinuclear halos and slight nuclear enlargement in intermediate squamous cells caused by an acute inflammation seen in a CP smear. 56 NONNEOPLASTIC EPITHELIAL PROLIFERATIONS The cervical epithelium is under the effects of hormonal stimulation, inflammation or physical irritation. It may undergo hyperplasia, metaplasia and keratinization. A. Reserve cell hyperplasia Reserve cells form a discontinuous layer between the endocervical columnar cells and the basement membrane. These cells are capable of differentiating into either squamous cells or endocervical glandular cells and proliferate as a response to physical or chemical irritation. They are usually seen in clusters and show oval, bland nuclei with scant, ill-defined, vacuolated cytoplasm. Rarely, they present singly. They may also be seen in tissue fragments. (Fig. 3.10). A B Fig. 3.10. Reserve cell hyperplasia. A. Histology of endocervical mucosa showing reserve cell hyperplasia. B. A CP smear showing a cluster of hyperplastic reserve cells with oval, bland nuclei and scant cytoplasm. 57 B. Squamous metaplasia Hyperplastic reserve cells gradually transform into immature squamous cells that exfoliate in sheets or singly. The immature squamous cells have pale, vacuolated cytoplasm and may show cytoplasmic extensions or tails (spider cells). Their oval nuclei have fine chromatin and are slightly hyperchromatic. With time, immature metaplastic squamous cells change into mature squamous cells with more waxy, basophilic or eosinophilic cytoplasm. (Fig. 3.11). A B 58 C D Fig. 3.11. Endocervical epithelium with squamous metaplasia. A. Histology of the lesion. B. Metaplastic cells showing “spiderleg” cytoplasmic extensions, C. Metaplastic cells with some displaying intracytoplasmic mucinous vacuoles and D. Mature metaplastic cells with mild nuclear enlargement. (B - D: CP smears) C. Hyperkeratosis and Parakeratosis Hyperkeratosis is a protective process by which nonkeratinized squamous epithelium protects itself from injuries. The squamous epithelium becomes acanthotic and is covered by a thick layer of keratinous material characterized clinically as leukoplakia. It commonly occurs on the cervices of prolapsed uteri. Thick orangeophilic layers of anucleated squamous cells with keratohyaline granules are observed. (Fig.3.12). 59 Parakeratosis is a form of hyperkeratosis in which small round or spindle-shaped keratinized squamous cells retain their pyknotic nuclei. (Fig.3.13). These cells exfoliate singly or in loose clusters. It is important to differentiate these cells from pseudokeratotic cells that are also small squamous cells with eosinophilic or basophilic cytoplasm and nonpyknotic oval nuclei with fine chromatin. Both parakeratotic and pseudoparakeratotic cells are commonly seen in condylomatous lesions of the cervix and vagina. A B Fig. 3. 12. Hyperkeratotic cervical squamous epithelium. A. Histology of the lesion. B. An irregular aggregate of anucleated, orangeophilic squames in a CP smear. 60 A B Fig. 3.13. CP smear showing small parakeratotic cells with keratinized, orangeophilic or eosinophilic cytoplasm and pyknotic nuclei. D. Tubuloendometrial metaplasia is common, affecting about 30% of women. The lesion is located in the upper portion of endocervical canal, often in deep clefts. It may represent a response to injury. Cytologic material from this lesion may reveal ciliated cells with clear cytoplasm and abundant apical ciliae, secretory cells and intercalated cells with scant cytoplasm and thin, long nuclei (peg cells). E. Urothelial metaplasia may be seen on exocervical atrophic squamous epithelium in elderly patients. The exfoliated metaplastic cells display oval, bland nuclei with longitudinal grooves. (Fig.3.14) 61 Fig.3.14. CP smear showing a cluster of metaplastic urothelial cells showing thin cytoplasm and oval nuclei with some dispaying longitudinal grooves. F. Reactive Cellular Changes due to Inflammation and Repair Cells. Reactive cellular changes are benign in nature. They are often associated with inflammation, radiation, an IUD and other nonspecific causes. Criteria for reactive changes are not always well defined, and as the result, the interpretation may lack reproducibility. Repair cells are seen in Pap smears of patients with inflammatory epithelial ulcers, and with previous biopsy, cautery and cryosurgery of their uterine cervices. During the repair process the ulcer base is replaced by granulation tissue that is then covered by epithelial cells that proliferate from adjacent squamous or glandular epithelium. The repair cells present singly or in sheets, show cytoplasmic extensions and have single or double, regularly contoured nuclei with prominent nucleoli. They are spindle or columnar in configuration and may show intracytoplasmic vacuoles. Determination of the origin of repair cells (squamous versus glandular) is extremely difficult or impossible in the majority of cases. Repair cells commonly show nuclear enlargement (1, 1.5 or 2 times the area of the nucleus of a normal intermediate squamous cell). Endocervical cells may show a greater nuclear enlargement. Nuclei may be double or multiple with smooth contours, and mildly hyperchromatic with fine chromatin. Prominent single or multiple nucleoli are observed. The cell cytoplasm may display polychromasia, vacuolization and perinuclear halos without thick cytoplasmic rims. Squamous metaplastic cells display similar nuclear and cytoplasmic changes; and cytoplasmic processes (spider cells) may be observed. (Figs.3.16 and 3.17). 62 A B C Fig. 3.16. Repair epithelial cells in CP smears: A. Repair cells with cytoplasmic processes. 63 B. Repair cells with granular or vacuolated cytoplasm in a monolayered sheet. C. Repair cells with prominent nucleoli and cytoplasmic processes. Fig. 3.17. A monolayered sheet of repair squamous cells with conspicuous nucleoli in a LBP. G. Vitamin B12 and Folic acid Deficiency-induced Cellular Changes These cellular changes are characterized by an enlargement of squamous cells and their nuclei. The nuclei are single or double and slightly hyperchromatic with fine chromatin. The presence of hypersegmented nuclei within polymorphonuclear leukocytes on the smear is evidence supporting pernicious anemia. H. Radiation and Chemotherapy Effects Radiation and chemotherapy are routinely used in the treatment of patients with advanced solid cancers, lymphomas and leukemias. Their mechanisms of action on protein metabolism and mitosis on human cells differ, however they produce similar cellular changes. 1. Radiation effects Injury is caused by radiation-induced ionization of intracytoplasmic molecules from inhibition of DNA synthesis and destruction of cellular proteins and enzymes. The extent of cellular injury varies with the type of radiation, the duration of exposure and the radiosensitivity of the cells. Hematopoietic cells, germ cells, gastrointestinal cells and anaplastic tumors are highly susceptible to radiation injury because they have a high mitotic rate. Radiation changes may be classified as acute and chronic. 64 • Acute post radiation changes may appear a few days after the conclusion of treatment, persist for 6 to 8 weeks and then gradually subside. In the uterine cervix the changes affect mainly the squamous cells, however the endocervical glandular cells are also affected albeit to a lesser degree. The smear shows inflammatory exudates with cellular debris, histiocytes and polymorphonuclear leukocytes forming “pus balls”. The reactive epithelial cells are markedly increased in size without a substantial increase in the N/C ratio. Bizarre cell shapes may be seen and enlarged nuclei may show degenerative changes including nuclear vacuolization, pallor, wrinkling and smudging of chromatin. The nuclei vary in size, with some cell groups having both enlarged and normal-sized nuclei. Bi- or multinucleation and mild hyperchromasia are common and prominent single or multiple nucleoli are not uncommon. Cytoplasmic vacuolization, hyalinization or polychromasia may be present, as well as intracytoplasmic aggregates of leukocytes. Repair cells may also show radiation changes. (Figs.3.18 to 3.20). • Chronic radiation changes appear about 6 months after cessation of the initial radiotherapy and may persist for years. The Pap smear displays an atrophic pattern with parabasal cells and intermediate squamous cells with pleomorphic giant cells. Cellular changes as seen in acute postradiation injuries may still be seen but they are less prominent. A 65 B Fig. 3.18. CP smear showing squamous cells with radiation changes. A. An aggregate of squamous cells with slightly pleomorphic, hyperchromatic nuclei and polychromatic cytoplasm. B. A fragment of squamous epithelium shows large epithelial cells with mild nuclear enlargement, conspicuous nucleoli. Intracytoplasmic vacuoles are noted in some cells. Fig. 3.19. A cluster of large epithelial cells with enlarged nuclei, prominent nucleoli showing thick, polychromatic cytoplasm with cytoplasmic extensions suggesting repair cells with radiation changes in a CP smear. 66 Fig. 3.20. Squamous cells with radiation effects in a LBP. 2. Chemotherapeutic effects Many drugs used to treat malignant diseases are alkylating agents, which are derivatives of nitrogen mustard. They alter the cellular DNA, RNA and proteins by different mechanisms. The cellular changes caused by alkylating agents are similar to those caused by radiation but they are systemic. (Fig.3.21). In a Pap smear, the number of abnormal cells is much smaller than in the case of radiation therapy. The epithelial cells from a patient receiving immunosuppressive agents, such as azathioprine and corticosteroids to prevent rejection of transplanted organs, may show cellular atypia simulating dyskaryotic or malignant cells. Therefore, an awareness of the patient’s therapy is important to avoid a false-positive cytodiagnosis. A 67 B Fig.3.21. CP smear showing in A and B squamous cells with chemotherapeutic effects (Methotrexate) displaying enlarged, hyperchromatic nuclei. It is important to note that if during radiotherapy and after its conclusion malignant cells are still present, a persistence of the original tumor should be suspected and a tissue biopsy should be obtained for histologic confirmation. In the event of interpretative uncertainty (suspicious report) the patient should be kept under observation with repeating cell samples every 3 to 6 months. In the majority of the cases, the cellular abnormalities regress. If tumor cells reappear after a tumor-free interval of several months or years a tumor recurrence is suspected and it should be confirmed by tissue biopsy. It should be born in mind that patients with chemotherapy or immunosuppressive therapy have an increased risk of developing a second malignancy. Common cellular features of abnormal and nonneoplastic squamous cells of the cervix and vagina are summarized in Table 3.1. 68 Table 3.1: Common Cellular Features of Abnormal and Nonneoplastic Squamous Cells CELLULAR FEATURES INFLAMMATION ASSOCIATED CELLULAR CHANGES VITAMIN B12 OR FOLIC ACID DEFICIENCY RADIATION OR CHEMOTHERAPY REPAIR CELLS Architecture Singly Clusters, loose Singly Clusters,cohesive Singly Sheets, monolayered Singly Cytoplasm - Border - Perinuclear halo with thin cyto- plasmic rim - Vacuolization - Enlargement - Extensions Distinct +, small +/- + - Distinct +, small +/- + - Distinct +/- + ++ +/- Distinct - - ++ ++ Nucleus - Irregular contours - Enlargement - Multinucleation - Chromatin - +/- - Clumped, fuzzy - + - Granular,fine - ++ + Granular,fine - + - Fine Increased N/C ratio - + - - Inflammatory background +++ - ++ + Other -Intracytoplasmic vacuoles -Macronucleolus - - - - + +/- +/- + Adapted with modifications from Nguyen GK, Kline TS: Essentials of Cytology. An Atlas. New York, Igaku-Shoin, 1993, p. 4 and 6. 69 I. Other findings • “Cornflakes” or brown artifact “cornflaking” is due to evaporation of xylene before coverslipping with deposition of air on superficial squamous cells. Cornflaking is more commonly seen in conventional Pap smears than in liquid-based smear. (Fig.3.22). Fig. 3.22. CP smear showing superficial squamous cells with “cornflakes”. • Blue blobs represent condensed mucus, degenerated bare nuclei and precipitating hematoxylin. In postmenopausal women, they may represent parabasal/intermediate squamous cells with various degree of degeneration. Blue blobs appear as dark blue, round, oval, amorphous masses in Papanicolaou stained CP smear. (Fig.3.23). Fig. 3.23. A few “blue blobs”, 2 intermediate squamous cells and several polymorphonuclear leukocytes in a CP smear. 70 • Psammoma bodies are laminated calcified round bodies. (Fig. 3.24). They are rarely seen in Pap smears as they occur only about one in every 100,000 samples. In about 50% of patients a benign condition or no lesion is found. They have been identified in Pap smears of patients taking oral contraceptive pills, using an IUD, with salpingiosis or pelvic inflammatory disease. In other cases, in particular postmenopausal women, an ovarian papillary serous carcinoma is present. In those cases, the psammoma bodies are usually seen admixed with malignant epithelial cells that may wrap around some bodies. Therefore, further investigation is recommended in all patients showing psammoma bodies in their Pap smears to rule out the possibility of a clinically occult ovarian serous cancer. Fig. 3.24. A laminated round psammoma body in a CP smear. • Carpet beetle part is a contaminant from cotton applicator or tampon. It has a distinctive morphology permitting the correct identification when present. (Fig.3.25). Fig. 3.25. Carpet larva part with distinctive morphology in a CP smear. 71 • Curschmann spirals are rarely found in Pap smears. They are inspissated mucous threads within cervical glands or clefts and have no diagnostic significance. (Fig. 3.26). Fig 3.26. Curschmann spiral in a CP smear. REFERENCES Anderson GH, et al. Confirmation of genital Herpes Simplex Viral infection by an immunoperoxidase technique. Acta Cytol. 1985. 29: 695. Aslan DL, et al. The diagnosis of Trichomonas vaginalis in liquid-based Pap test: correlation with PCR. Diagn Cytopathol.2005; 32:341. Colgan TJ, et al. Reparative changes and the false-positive/false-negative Papanicolaou test: a study for the College of American Pathologists Interlaboratory Comparison in cervicovaginal cytology. Arch Pathol Lab Med. 2001; 125:134. DeMay RM. The Pap Test. Chicago, ASCP Press, 2005. de Sanjose S, et al. Worldwide prevalence and genotype distribution of cervical human papillomavirus DNA in women with normal cytology: a meta-analysis. Lancet Ifect Dis.2007; 7:453. Fu K. Biological basis for the interaction of chemotherapeutic agents and radiotherapy. Cancer. 1985; 55:2123. Frost JK. The Cells in Health and Disease. 2nd ed. New York, S karger, 1986. 72 Gierson G, et al. Epithelial repair and regeneration in uterine cervix. Acta Cytol. 1977; 21: 371. Iwa N, Noguchi H. Detection of herpes simplex virus DNA by in situ hybridization technique and polymerase chain reaction in Papanicolaou-stained cervicovaginal smears. Diagn Cytopathol. 2003; 29:246. Jones S, et al.The tip of the iceberg opportunistic screening for Chlamydia trachomatitis in asymptomatic patients attending a young people’s health clinic reveals a high prevalence-a pilot study. Sex Health. 2004; 1:115. Levine PH, et al. Atypical repair on Pap smears: clinicopathologic correlates in 647 cases. Diagn Cytopathol. 2005; 33:214. McMillan A. The detection of genital tract infection by Papanicolaou-stained tests. Cytopathology. 2006; 17:317. Melnikow J, et al. Natural history of cervical squamous intraepithelial lesions: a meta-analysis. Obstet Gynecol. 1998:92:727. Moscicki AB, et al. Regression of low-grade squamous intraepithelial lesions in young women. Lancet. 2004;364:1678. Murad TM. August C. Radiation induced atypia. A review. Diagn Cytopathol. 1895; 1:137. Shield PW. Chronic radiation effects: a correlative study of smears and biopsies from the cervix and vagina. Diagn Cytopathol. 1995;13:107. Schiffman M, et al. Human papillomavirus and cervical cancer. Lancet. 2007;370:890. Sodhani P, et al. Prevalence of bacterial vaginosis in community setting and role of the Pap smear in its detection. Acta Cytol. 2005; 49:634. Stoler MH. Testing for human papillomavirus: data driven implication for cervical neoplasia management. Clin Lab Med. 2003;23:596. 73 Chapter 4 Cervical Squamous Cell Lesions Since the 1950s with the introduction of nationwide cytologic screening programs for cervical cancer in the United States, the incidence of squamous cell carcinoma (SCC) of the uterine cervix has decreased from 32.6 in the 1940s to 8.3 per 100,000 women in 1983 and 1984. However, the disease is far from being eradicated, as about 11,070 new cases of cervical cancer are expected to be diagnosed in 2008. Studies over the past 5 decades have demonstrated that cervical SCCs develop through a multistep process involving preinvasive lesions. In the majority of cases, the tumor occurs as the end result of a series of epithelial changes, ranging from mild to severe atypia, at the T-zone of the cervix. These lesions have been given various names: dysplasia and carcinoma in situ, cervical intraepithelial neoplasia (CIN) and squamous intraepithelial lesions (SIL). In recent years, studies of cervical SILs and SCCs using molecular biology techniques have documented that the sexually transmitted human papillomaviruses (HPV) play an important role in the pathogenesis of these lesions. Of over 80 different types of HPV which have been identified, only about 40 are found in female anogenital tract lesions. The HPV types 6 and 11 are mainly associated with cervical condylomas and CIN 1 but only rarely with HSILs (CIN 2 and 3) and almost never with cervical SCCs. Among the high-risk viruses HPV type 16 is most commonly detected in cervical SCCs. Of the SILs, LSILs are extremely heterogeneous with regard to their association with HPV types. In contrast to HSILs, LSILs may be caused by any single or combined HPV types of low- or high-risk HPV types, while HSILs, in about 88% of cases, are associated with HPV types 16, 18 or 31; and only 7% of them contain more than one HPV–DNA type. SQUAMOUS INTRAEPITHELIAL LESIONS Cervical SILs have been generally regarded as precursors of cervical SCCs. SIL is predominantly a disease of women in their reproductive years. CIN lesions consist of three grades with grade 1 equivalent to mild dysplasia, grade 2 to moderate dysplasia, and grade 3 to severe dysplasia and carcinoma in situ (CIS). In TBS only two grades are recognized: LSIL that is equivalent to flat condyloma and CIN 1, and HSIL that is composed of CIN 2 and CIN 3 or CIS. 74 LSIL is defined as an intraepithelial lesion showing a preservation of differentiation, maturation and organization of squamous epithelium with mitoses confined to basal and parabasal epithelial layers, koilocytosis, dyskeratosis, multinucleation, and enlarged hyperchromatic nuclei. HSIL, on the other hand, is characterized by a lack of squamous differentiation, epithelial disorganization, and severe cellular dyskaryosis with the presence of mitoses throughout the entire or lower 2/3 of the epithelium. (Figs. 4.1 and 4.2). Two biomarkers p16 and Ki-67 or MIB1 are useful for identification of an SIL, but they are not reliable for grading it (HSIL versus LSIL). Ki-67 antibody shows positive nuclear staining in over 30% of nuclei in the upper epithelial layers of a SIL, and p16 is strongly expressed by nuclei and cytoplasm of dyskaryotic cells in both LSIL and HSIL associated with high-risk HPV types. (Fig. 4.3). Normal, atrophic, reactive and metaplastic squamous epithelia, in contrast, are negative for p16; and Ki-67 is expressed only by parabasal cells. LSILs caused by low-risk HPV tend to regress and those caused by high-risk HPV tend to persist and progress to HSILs or SCCs. In a critical literature review of CIN lesions, Östor has found that the rates of regression, persistence and progression to CIN 3 and to SCC for CIN 1 lesions were about 60%, 30%, 10% and 1%, respectively. For CIN 2 lesions the corresponding approximations were 40%, 40%, 20% and 5%, respectively. The likelihood of CIN 3 regression, persistence and progression to SCC were about 33%, 56% and over 12%, respectively. He has also noted that the probability of an atypical squamous epithelium progressing to a SCC increased with the severity of atypia but this did not occur in every case. In a small number of cases foci of CIN 3 arose de novo, higher up in the cervical canal or adjacent to those of CIN 1, and did not develop from the progression of CIN 1 lesions. This observation challenges the Richart's popular concept of CIN. In a more recent meta-analysis by Melnikow et al. fairly similar rates of regression and progression of CIN/SIL lesions were found: about 47% of LSILs regress, 21% progress to HSIL and 1.5% progress to invasive cancer. For HSILs, 35% regress and 14% progress to invasive neoplasms. A 75 B Fig. 4.1. Histology of LSIL: A. Flat condyloma with koilocytes. B. CIN 1/mild dysplasia. A B 76 C Fig. 4.2. Histology of HSIL. A. HSIL/CIN 2/Moderate squamous dysplasia showing focal squamous differentiation. B. HSIL/CIN 3/Severe squamous dysplasia/CIS consisting of small cells. C. HSIL/CIN 3/CIS consisting of large, pleomorphic cells. A B 77 C Fig. 4.3. Immunohistochemical features of normal squamous epithelium and HSIL: A. Normal cervical squamous epithelium is negative for p16, and only its parabasal cell nuclei express Ki-67. B, C. Dysplastic epithelium in this HSIL/CIN 2 showing positive nuclear and cytoplasmic reactions to p16 antibody (B), and over 30% of dyskaryotic nuclei express Ki-67 (C). (Avidin-biotin-complex). A. LSIL LSIL includes flat condyloma, mild dysplasia and CIN 1. LSIL is caused by a large number of different HPVs of low-risk and high-risk types. LSIL cells are found in about 2% of all Pap smears. The majority of women with LSIL Pap results have LSIL (CIN 1), but 18% of them are found to have HSIL (CIN 2 and 3) on cervical biopsy. 1. LSIL/Flat condyloma is characterized by the presence of koilocytes with dyskaryotic nuclei presenting singly and in sheets. These koilocytes are superficial and intermediate squamous cells displaying enlarged, hyperchromatic, single or multiple nuclei with granular or smudged chromatin pattern and irregular nuclear contours or membranes. The nuclei are surrounded by a perinuclear clear halo with a well–defined, thick cytoplasmic rim. The perinuclear halo is caused by degeneration of perinuclear cytoplasmic microorganelles caused by the HPV. Small or miniature parakeratotic and pseudoparakeratotic squamous cells are commonly seen, as well as dyskaryotic keratinizing squamous cells with atypical nuclei and thick, eosinophilic or orangeophilic cytoplasm. (Fig. 4.4). 78 A B C Fig. 4.4. LSIL/Flat condyloma: A, B. Classic koilocytes with dyskaryotic nuclei in a CP smear. C. A koilocyte with 2 nuclei in a LBP. 79 2. LSIL/Mild dysplasia/CIN 1 exfoliates superficial and intermediate squamous cells with enlarged, hyperchromatic and irregularly contoured nuclei. Nucleoli are absent and koilocytic changes are common. (Fig. 4.5). LSIL cells should be distinguished from squamous cells with inflammatory changes, repair cells, squamous cells with radiation and chemotherapy effects and atypical squamous cells of undetermined significance. The reader is referred to Chapter 3 for discussion and illustrations of the three first cell types. A B 80 C Fig. 4.5. LSIL: A, B. CP smears showing mildly dyskaryotic superficial and intermediate squamous cells. One of the cells in (B) shows koilocytic change with a poorly formed perinuclear halo. C. Mildly dyskaryotic squamous cells with some having perinuclear halos in a LBP. Management of LSIL The ASCUS/LSIL triage study has found that high-risk HPV types were detected in 85% of LSIL cases and that HPV DNA testing was not a useful triage strategy. Colposcopy is generally recommended for initial management of LSIL patients. For pregnant women, a colposcopically directed biopsy may be performed, but an endocervical tissue sampling is contraindicated. Otherwise, the colposcopy evaluation may be deferred to 6 weeks postpartum. B. HSIL HSILs include moderate and severe dysplasias, CIS, CIN 2 and CIN 3. HSIL accounts in about 0.5% of all Pap smears, and 97% of women with HSIL Pap result are positive for high-risk HPV. If left untreated about 14% of them will develop cervical invasive squamous cell carcinoma. 1. CIN 2 lesions exfoliate parabasal-type cells singly or in sheets with thick, well– defined cytoplasm and enlarged hyperchromatic nuclei showing smooth or irregular nuclear contours. The chromatin is evenly distributed and may be finely or coarsely granular. Nucleoli are absent and koilocytic change may be present. (Fig. 4.6). 81 A B C Fig. 4.6. HSIL/CIN 2 showing in CP smears: 82 A, B. Moderately dyskeratotic squamous cells of low intermediate/parabasal cell type with enlarged, hyperchromatic nuclei with irregular nuclear contours and no nucleoli. C. Similar moderately dyskaryotic squamous cells and normal intermediate cells in a LBP. 2. CIN 3 lesions consist of three main histologic patterns: large cell non-keratinizing, keratinizing and small cell types. A mixed cellular pattern is common. Cells exfoliated from a nonkeratinizing CIN 3 are large and pleomorphic. They show abundant, well– or ill–defined cytoplasm and enlarged, hyperchromatic nuclei displaying irregular nuclear contour. The nuclear chromatin is evenly distributed, and may be coarsely or finely granular. Cells in syncytial clusters are commonly encountered as well as epithelial fragments. (Figs. 4.7 and 4.8). Occasional cells with keratinizing cytoplasm may be observed. Nucleoli are absent, and the smear background is free of necrotic debris. Rarely, a nonkeratinizing CIN 3 is composed of spindle-shaped cells. (Fig. 4.9). A B Fig. 4.7. Cytology of HSIL/CIN 3, nonkeratinizing type in CP smears: 83 A. Markedly dyskaryotic medium-sized cells with hyperchromatic nuclei and irregular contours. B. A cohesive cluster of markedly dyskaryotic small cells showing nuclei with similar changes with those of the cells in A. A B Fig. 4.8. A, B. Large tridimensional clusters of markedly kyskaryotic squamous cells removed by cytobrush from a HSIL/nonkeratinizing CIN 3 in CP smears. 84 Fig. 4. 9. Nonkeratinizing HSIL/CIN 3 showing in a CP smear spindle-shaped cells with hyperchromatic, spindle nuclei. Keratinizing CIN 3 yields spindled-shaped cells with orangeophilic, well-defined, thick cytoplasm and enlarged hyperchromatic nuclei. They are chiefly seen as single cells and rarely in clusters. A small-cell CIN 3 exfoliates cells with hyperchromatic nuclei and scant, ill-defined cytoplasm, singly or in loose aggregates, with or without nuclear molding. (Fig. 4.10). A 85 B Fig. 4.10. Keratinizing markedly dyskaryotic pleomorphic squamous cells in a CP smear from a HSIL/keratinizing CIN 3. Management of HSIL For patients with a HSIL Pap result, colposcopic evaluation is mandatory, as most patients will have confirmed biopsies of CIN 2 or 3. However, for pregnant women, the colposcopy may be deferred to 6 weeks postpartum. In pregnant women a colposcopically directed biopsy may be performed but an endocervical tissue sampling is contraindicated. If a histologically confirmed CIN is not identified at colposcopy, all cytologic and histologic materials of the patients should be reviewed. If the cytologic diagnosis of HSIL is correct, a diagnostic excisional procedure should be performed. ATYPICAL SQUAMOUS CELLS Atypical squamous cells (ASC) are seen in less than 5% of all Pap smears and the ASC/SIL ratio is about 3:1. Patients with ASC diagnosis are found to have a CIN lesion on colposcopically directed cervical biopsy in 10% to 20% of cases. In The Bethesda System-2001, ASCs are divided in 2 categories: ASC of undetermined significance (ASC-US) and ASC, cannot exclude a high-grade squamous intraepithelial lesion (ASC-H). An ASC diagnosis is made when an SIL is suspected cytologically. Cytologic criteria for identification of ASC-US and ASC-H cells are somewhat subjective, and the diagnoses suffer high inter-observer and intra-observer variation rates. ASC-US represents about 90% of all ASC cases. 86 1. ASC-US ASC-US cells show cellular features that are more severe than those of squamous cells with reactive changes but less than those of a SIL. (Figs. 4.11 and 4.12). Thus, the diagnosis of ASC-US is made by exclusion of cells with known cytologic features. Cytologic criteria of ASC-US cells include: • ASC-US cells are of superficial or intermediate type with • Enlarged mononucleus or binuclei that are 2.5 to 3 times larger than the nucleus of a normal intermediate squamous cell (~35 µm2) • Slightly increased N/C ratio • Slightly hyperchromatic nuclei with irregular chromatin distribution. • Regular nuclear contours, but it may show focal irregularity. • Dense and eosinophilic or orangeophilic (keratinized) cytoplasm • Perinclear halo may be present • Nucleoli with repair features may be seen in ASC-US repair cells A B 87 C Fig. 4.11. ASC-US cells: A, B. Single and loosely clustered ASC-US cells in a CP smear showing smooth nuclear contours with minimally hyperchromatic nuclei and no nucleoli are seen. C. Similar ASC-US cells are seen in a LBP. A B Fig. 4.12. A, B. ASC-US cells with keratinized cytoplasm in CP smears. 88 Atrophic vaginitis, atypical parakeratosis and atypical repair may yield cells with features suggesting ASC-US changes. (Fig. 4.13). In atrophic vaginitis a short course of intravaginal treatment with estrogen cream for 4 to 7 days will be helpful to solve this diagnostic dilemma. This treatment will induce a maturation of squamous cells, but dyskaryotic cells will remain unchanged. A B 89 C Fig. 4.13. A-C: Atrophic vaginitis showing in CP smears ASC-US cells. Management Oncogenic (high-risk) HPV DNA testing is the preferred management for patients with ASC- US Pap results, especially when it can be performed concurrently. The test should not be performed in women younger than 30 years of age because the HPV infection in these patients is often caused by a mixture of low- and high-risk virus types, making the interpretation of the test results difficult, if not impossible. Follow-up with repeat Pap tests at 6-month intervals or immediate colposcopy is also acceptable. In pregnant women, the colposcopy may be deferred to 6 weeks postpartum. If the HPV DNA testing is positive for high-risk viruses, the patient should be referred to colposcopy. If the test is negative for high-risk viruses, she should be followed by a repeat Pap smear every 6 months for 2 years. If the cellular atypia is cleared within 2 years, she can return to routine annual screening. If HPV DNA testing is unavailable and if the cellular atypia persists over 2 years, she should be referred to colposcopy for further evaluation. 2. ASC-H ASC-H represents 5% to 10% of all ASC cases. ASC-H cells are metaplastic squamous cells with nuclear atypia that fall short of a definitive diagnosis of HSIL. (Fig. 4.14). Cytologic criteria of ASC-H cells include: • ASC-H cells are usually small in number. • The cells occur singly, in small groups or in epithelial fragment containing less than 10 cells. 90 • ASC-H cells are polygonal in shape and have the size of a squamous metaplastic cell and dense cytoplasm. • Their hyperchromatic nuclei are 1.5 to 2.5 times larger than that of a normal metaplastic squamous cell, irregular chromatin and mildly irregular contours. • No nucleoli. • The N/C ratio is increased and is about that of an HSIL cell. A B 91 C Fig. 4.14. ASC-H cells: A, B. ASC-H cells present singly and in loose aggregates in CP smears. C. Loosely clustered ASC-H cells in a LBP. ASC-H cells in a crowded sheet or epithelial fragments may show a loss of nuclear polarity. The cell cytoplasm has squamoid features, however. On smears these thick tissue fragments may display peripheral cells in vague palisades mimicking those of a cervical adenocarcinoma in situ. Nucleoli are virtually absent. However, it should be borne in mind that an adenocarcinoma, either in situ or invasive, may coexist with a squamous lesion of the cervix. Cytobrush bristles may remove large fragments of CIN 3 epithelium lining cervical glandular crypts, and these tissue fragments are difficult to distinguish from a large epithelial fragment containing ASC-H cells. Management Patient with ASC-H diagnosis should be referred to colposcopy as ASC-H has a positive predictive value for histologic CIN 2 or 3 much higher than that of ASC-US (50% versus 17%). If a CIN 2 or 3 is not found she should have a repeat Pap test in 6 months or a HPV DNA test. If her repeat Pap test result is ASC-US or worse, or if her HPV DNA is positive for high-risk viruses, she should be referred to a second colposcopy. If the patient is pregnant, the colposcopy may be deferred until 6 weeks postpartum. INVASIVE SQUAMOUS CELL CARCINOMA Invasive SCCs are the most common type of cervical cancers, accounting for 60 to 80% of all malignant tumors of the cervix. They occur mainly in adults with a peak incidence in the 5th and 6th decades of life. Their common clinical manifestation is abnormal vaginal bleeding that may occur spontaneously or following a sexual intercourse. Cervical SCCs are histologically classified as well- and poorly 92 differentiated (keratinizing and non-keratinizing SCCs). Cervical epithelium adjacent to SCC commonly shows foci of LSIL or HSIL. CYTOLOGIC MANIFESTATIONS OF CERVICAL SCC Cervical SCCs have distinctive cytologic manifestations. (Figs. 4.15 to 4.18). Common cytologic features of cervical SCC include: • Cancer cells with keratinized cytoplasm are seen predominantly singly. • Nonkeratinized cancer cells predominantly seen in small aggregates and in clusters. • Prominent nucleoli are present mainly in nonkeratinized tumor cells. • Necrotic debris or tumor diathesis is almost always observed in conventional Pap smears, but it is subtle or minimal in liquid-based preparations in which the necrotic debris is collected at the periphery of tumor cell groups (“clinging diathesis”). • Cells characteristic of SILs may be present, as SILs may coexist with SCC • Cervical adenocarcinoma in situ (AIS) cells may be seen if the AIS coexists with SCC. A 93 B C Fig. 4.15. Keratinizing SCC showing in a CP smear keratinizing, pleomorphic malignant squamous cells in a necrotic background (tumor diathesis). A 94 B Fig. 4.16. Keratinizing SCC showing: A. In CP smear pleomorphic malignant cells with keratinized cytoplasm and necrotic debris. A few tumor cells with tadpole configuration are present. B. Similar cancer cells in a LBP. A 95 B Fig. 4.17. Keratinizing SCC showing in CP smear spindle-shaped, “fiber” tumor cells, singly and in bundles. A B Fig. 4.18. Two poorly differentiated SCC showing in CP smears syncytial clusters of tumor cells with ill-defined, nonkeratinized cytoplasm, pleomorphic nuclei and tumor diathesis. 96 MICROINVASIVE SQUAMOUS CELL CARCINOMA Microinvasive squamous cell carcinoma (MICA) of the cervix is defined as an early invasive cancer up to 3 mm below the overlying basement membrane in which no vascular or lymphatic tumor invasion is identified. Cervical MICA, so-defined, has an incidence of pelvic lymph node metastasis lower than 1%. The cytologic manifestations of cervical MICA in CP smears include: • Pleomorphic, single HSIL cells with or without keratinization. • HSIL cells or malignant nonkeratinizing squamous cells in loose clusters, syncytia and hyperchromatic cell groups or tissue fragments. (Fig. 4.19). • Nuclei with irregular chromatin clumping and distribution. • Micro– and macronucleoli. • Small amount/focal tumor diathesis. Of these criteria, the first three are the most important ones, however they are not present in every single case. In the case of keratinizing MICA, keratinized malignant squamous cells with thick, orangeophilic cytoplasm are seen and tumor diathesis may be present. MICA may be suspected cytologically, but its definitive diagnosis must be made by careful histologic examination of the excisional cervical cone biopsy. A 97 B Fig. 4.19. Cervical microinvasive squamous cell carcinoma showing in CP smears: A. Slightly pleomorphic HSIL cells and tumor diathesis. B. Clustered malignant nonkeratinizing squamous cells with conspicuous nucleoli. 98 Table 4.1. Cytology Features of Low- and High-grade SILs and Invasive Squamous Cell Carcinoma (SCC). FEATURES LSIL/CIN 1 HSIL/CIN 2 HSIL/CIN 3 INVASIVE SCC Abnormal cell number + ++ +++ ++/variable Architecture Singly Clusters, loose Singly Clusters, loose Singly Clusters, loose Syncytia Singly Clusters, loose Syncytia Defined cytoplasm Perinuclear halo ++ ++ ++ +/- +/- - +/- - Increased N/C ratio Irregular nuclear contours + + ++ + +++ + + to ++ ++ Chromatin - Granular - Regular Nucleoli Fine + - Fine +/- - Fine - - Coarse/Dense - + Tumor diathesis - - - +++ Others - Cervical AIS cells - - +/- +/- Adopted with modifications from Nguyen GK, Kline TS. Essentials of Cytology. An Atlas. New York, Igaku-Shoin, 1993, p 3 and 4. Adenocarcinoma in situ (AIS) cells. VARIANTS OF SQUAMOUS CELL CARCINOMA 1. Verrucous carcinoma is a very rare and distinctive type of well-differentiated squamous cell carcinoma of the cervix and vagina. The tumor tends to grow slowly and recur locally but it does not metastasize. HPV type 6 has been identified in verrucous carcinoma by molecular techniques. Macroscopically, the tumor has a warty and fungating appearance. It is characterized histologically by an undulating hyperkeratinized surface consisting of pointed papillary projections. The tumor cells usually show normal-appearing nuclei or minimal nuclear atypia. In Pap smears, abundant benign-appearing squamouscells with keratinized cytoplasm admixed with abundant anucleated, keratinous squames are seen. 2. Papillary squamous (transitional) cell carcinoma is a rare variant of invasive squamous cell carcinoma of the cervix and vagina, with about 40 cases 99 reported in the literature. Microscopically, it can be graded as high- or low-grade tumor. In conventional Pap smears a high-grade neoplasm is characterized by single and clustered small cancer cells with scant cytoplasm and hyperchromatic nuclei, similar to those of a HSIL (CIN 3), admixed with necrotic debris and inflammatory cells (tumor diathesis). (Fig.4.20). A case of cervical low-grade papillary transitional cell carcinoma showing in Pap smear a few single and benign-appearing transitional cells with oval nuclei and well-defined, granular cytoplasm and a few syncytial, 3-dimensional, well-defined, cohesive clusters of medium-sized epithelial cells with granular, eosinophilic cytoplasm and oval nuclei with a loss of nuclear polarity has been reported. A B C Fig. 4.20. Papillary high-grade transitional cell carcinoma: A, B. Histology of the tumor. C: CP smear showing single and clustered tumor cells similar to those of a HSIL. 100 3. Lymphoepithelioma-like carcinoma of the cervix is also a rare tumor with lower rate of lymph node metastasis, potentially radiosensitive and better prognosis. Histologically, it is characterized by solid cords of poorly differentiated polygonal malignant cells with squamoid differentiation and a lymphocyte–rich stroma. In cervical smears, it shows numerous benign lymphoid cells and rare single and clustered malignant non–keratinizing, poorly differentiated or anaplastic epithelial cells with irregular nuclear membrane, hyperchromatic nuclei, coarse chromatin and prominent nucleoli. These cells are obscured by heavy inflammation and blood, a background resembles that of a menstrual smear. 4. Sarcomatous squamous carcinoma of the cervix uteri is rarely encountered. The tumor yields single and loosely clustered pleomorphic malignant cells that are difficult to differentiate from those of a soft tissue sarcoma. Immunohistochemical and electron microscopic studies of the tumor tissue obtained by biopsy are necessary for a correct diagnosis. SENSITIVITY, ACCURACY AND ERRORS In the screening of cervical cancer, about 85% to 90% of women in the general population have a normal Pap result, while about 10% of them show squamous cell atypia, and the remainder are diagnosed as having a SIL. Of these cervical SILs, 75% to 90% are low–grade lesions and the remainder are high–grade lesions. The cytodiagnosis of SIL is subjective and suffers remarkable interobserver variations. On rare occasions, a SIL cannot be graded as low– or high–grade, and a diagnosis of ungraded SIL has been made. Squamous cells with perinuclear haloes and normal nuclei are not specific for LSILs. (Fig. 4.21). These changes are non-specific, and HPV–DNA is often not detected within the cell cytoplasm by in situ hybridization. 101 Fig. 4.21. CP smear showing intermediate squamous cells with perinuclear haloes and normal, nondyskaryotic nuclei. The diagnostic accuracy rates of LSILs varied tremendously in different reported series. The sensitivity rate of the Pap test in detecting cervical SILs varied widely, ranging from 30% to 80%, with a mean of 47%. Its specificity rate ranged from 86% to 100%, with a mean of 95%. For LSILs, a correct cyto–histologic correlation was obtained in about 38% to 56% of cases, and in one series, the cervical biopsy showed HSIL in 12% and was unremarkable in 50% of the cases. In another series, a poor correlation between cytologic and histologic diagnoses of various grades of CIN was observed: 50% of patients with a cytodiagnosis of CIN 1 showed a higher grade CIN in biopsied cervical tissues, and the overall false-negative rate of cervical smears for CIN 2 and 3 was 19%. In Koss' experience, about 20% of cases with a cytodiagnosis of LSIL show HSIL in biopsied tissues. For HSILs, a cytodiagnostic accuracy rate of 85% to 100% has been reported. For invasive cervical cancers, the Pap test had a more variable sensitivity rate, ranging from 16% to 82%; and many patients had one or more negative test result. SCCs with keratinized surfaces are often underdiagnosed as scraping these tumors may yield only benign appearing keratinized squamous cells. It should be borne in mind that a false–negative cytodiagnosis is potentially dangerous as the cancer may be left untreated. A false–positive diagnosis is undesirable but it is less dangerous as the patient will be subsequently evaluated by colposcopy and cervical biopsy. False-positive diagnoses of cervical cancer occurred in 10% to 15% of cases, and the 3 most common errors were: • atrophic smear with benign atypia in a granular pseudonecrotic background, followed by • reparative changes, and • keratinizing HSIL. 102 REFERENCES Al–Nafussi AI, et al. 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The Bethesda System for Reporting Cervical Cytology. 2nd edition, 2004, New York, Springer-Verlag. Stuart G, et al. Report of the 2003 pan-Canadian forum on cervical cancer prevention. J Obstet Gynaecol Can. 2004;26:1004. 106 Sugimori H, et al. Cytology of microinvasive squamous cell carcinoma of the uterine cervix. Acta Cytol.1987;31:412. Tabbara S, et al. The Bethesda classification of squamous intraepithelial lesions: histologic, cytologic and viral correlation. Obstet Gynecol. 1992;79:338. The ALST Group. Human papillomavirus testing for triage of women with cytologic evidence of low-grade squamous intraepithelial lesions: baseline data from a randomized trial. J Natl Cancer Institute. 2000; 92:397. Walloch JL, et al. Effects of therapy on cytologic specimens. In Comprehensive Cytopathology, Bibbo M (ed.), Philadelphia, Saunders, 1992, p. 860. Wells M, et al. Epithelial tumours. WHO Classification of Tumours. Pathology and Genetics of Tumours of the Breast and Female Genital Organs, Tavassoli FA, Devilee P, eds, Lyon, IARC Press, 2003, p.262. Willet GD, et al. Correlation of the histological appearance of intraepithelial neoplasia of the cervix and human papillomavirus types. Int J Gynecol Pathol. 1989; 8:18. Wright TC, et al. 2001 Consensus guidelines for the management of women with cervical cytological abnormalities. JAMA. 2002; 287:2120. 107 Chapter 5 Cervical Glandular Lesions Cervical adenocarcinomas are believed to originate from the multipotential subcolumnar reserve cells of the endocervical canal. These tumors display complex growth patterns consisting of different cell types. Of these, adenocarcinoma of endocervical cell type is the most common. It may occur in a pure form or coexist with a squamous cell carcinoma. The etiology of cervical adenocarcinoma has not been fully elucidated. Recently, HPV types 16 and/or 18 have been identified in cervical adenocarcinoma tissues suggesting a common etiology with squamous cell cancer. In contrast to cervical squamous cell carcinoma, the sequential changes of cervical glandular epithelium leading to the development of adenocarcinoma have not been well documented. Currently, adenocarcinoma in situ (AIS) is widely accepted as the immediate precursor to cervical adenocarcinoma. ADENOCARCINOMA IN SITU Cervical AIS tends to occur in women in their 3th and 4th decades of life. Histologically, it involves the transformation zone of the cervix in the majority of cases and consists of three main cell types: endocervical, endometrioid and intestinal; with endocervical-type lesions being the most common. Lesions containing more than one cell type are not uncommon. Cervical AIS has been found in association with squamous dysplasia and squamous carcinoma in about 50% of cases. On the other hand, AIS is found only in about 5% of cervical HSILs. AIS cells are ER, PR, CEA negative and p16 and MIB1 positive. Depending on the degree of cellular differentiation, AIS may be classified as well and poorly differentiated. A well-differentiated AIS displays fairly distinctive cellular manifestations permitting its identification in a high percentage of cases, while a poorly differentiated tumor does not have any specific cytological pattern and may be readily mistaken for an invasive adenocarcinoma. (Fig. 5.1). 108 A B Fig. 5.1. Histology of cervical adenocarcinoma in situ. Well-differentiated AIS, endocervical cell type displays the following cytologic features in a CP smear: • Large sheets of malignant glandular epithelium with crowded columnar tumor cells showing nuclear stratification that is well visualized at the edges of the sheets. • Short strips of tumor cells with cytoplasm extending off the edges of the tumor cell sheets (feathering). (Fig. 5.2). • Short strips of tumor cells with palisading nuclei. • Rosettes of tumor cells. • Isolated tumor cells and cells in papillary clusters may be seen. • The smear background is free of necrotic debris or tumor diathesis. • The individual tumor cells are two to three times larger than normal endocervical glandular cells and have enlarged hyperchromatic nuclei with finely or coarsely granular chromatin pattern. • Nucleoli are absent in about 50% of cases 109 Atypical or malignant squamous cells may be seen if there is a coexisting squamous cell lesion that is usually present in about 50% of cases. Each of these features provides a key to an accurate cytodiagnosis which can be made in about 90% of cases. A B C 110 D E F 111 G Fig. 5.2. Cytology of cervical AIS in CP smear: A-C: Irregular monolayered sheet of tumor cells with cytoplasmic extensions or feathering. D-E: Strips of tumor cells with pseudostratified nuclei. F,G: Tumor cells forming rosettes. AIS, intestinal variant almost always coexists with AIS, endocervical type. Cytologically, the tumor cells are large and occur singly, in clusters and in large epithelial fragments. They demonstrate single intracytoplasmic mucous vacuoles, resembling colonic epithelial sheets. (Fig. 5.3). A 112 B C Fig. 5.3. Cervical AIS, intestinal variant. A. Histology of the tumor B, C: The tumor showing in CP smear: B. Epithelial fragment with elongated nuclei in vague palisade. C. Epithelial fragment with multiple round, clear spaces or vacuoles. DIFFERENTIAL DIAGNOSIS A few conditions may exfoliate cells mimicking those of cervical AIS. 1. Cervical endometriosis exfoliates cells that may be mistaken for those of cervical AIS. Efforts should be made to look for endometrial glandular fragments and clusters of endometrial stromal cells to avoid a false-positive diagnosis. 2. Tubal metaplasia, a fairly common lesion of the cervix, may yield cell clusters that are readily mistaken for those of AIS. Ciliated cells can be recognized in well-preserved cellular strips. (Fig.5.4). However, loss of cilia due to degenerative 113 changes is not uncommon. An awareness of tubal metaplasia and the potential for cytodiagnostic error is necessary to avoid an unnecessary cone biopsy. A B Fig. 5.4. Tubal metaplasia showing in CP smear ciliated columnar cell singly and in row. 3. Postcone biopsy smears may contain fragments of endometrium from lower uterine segment with crowded, hyperchromatic, and pseudostratified nuclei simulating those of cervical AIS. These endometrial epithelial fragments are usually mixed or surrounded by endometrial stromal cells. Clinical data will be helpful in avoiding a false-positive diagnosis in this setting. The reader is referred to Chapter 2 for illustrations of lower uterine segment endometrium. 4. Reactive endocervical cells. Cytologic criteria of reactive endocervical glandular cells include: 114 • Reactive endocervical cells tend to occur in flat sheets with minimal nuclear crowding. (Fig. 5.5). • Fairly abundant cytoplasm with well-defined cytoplasmic borders. • Slightly enlarged nuclei with fine chromatin and prominent nucleoli. • N/C ratio is within normal limits or only slightly increased. A B C 115 Fig. 5.5. Reactive endocervical glandular cells: A, B. In CP smear these cells have well-defined cytoplasm and multiple, enlarged nuclei with prominent nucleoli. C. Similar cells seen in a LBP. 5. Atypical cervical glandular cells. The reader is referred to the section on Atypical glandular cells below for discussion and illustrations. ATYPICAL GLANDULAR CELLS In The Bethesda System-2001 Atypical glandular cells (AGC) are defined as cells showing cellular changes that fall between those of definite benign reactive process and those of an unequivocal AIS or adenocarcinoma. AGCs are divided into 2 subtypes: AGC, NOS and AGC, favor neoplastic. AGCs are further divided into endocervical and endometrial types. AGC accounts for about 0.2% of all Pap tests, with about 30% of patients having a significant cervical lesion: 5% being AIS and adenocarcinoma, and 20% being CINs. 1. Atypical endocervical cells, NOS. The cytologic criteria of AGCs, NOS. include: • AGCs occur in sheets and strips with some cellular crowding and overlap. (Fig. 5.6). • Nuclei are enlarged, up to 3 to 5 times the area of normal endocervical nucleus. • Some variation in nuclear size and shape. • Fairly abundant, distinct cytoplasm. • Increased N/C ratio. • Mild nuclear hyperchromasia. • Nucleoli may be present. A 116 B C Fig. 5.6. A, B. CP smear showing a sheet of atypical endocervical glandular cells, NOS, displaying enlarged, slightly hyperchromatic nuclei and conspicuous nucleoli. C. Cervical biopsy in this case revealed atypical endocervical glandular epithelium with no definitive histologic features of an AIS. 2. Atypical endocervical cells, favor neoplastic, by definition, are AGCs with morphologic changes that qualitatively fall short of the cells derived from a cervical invasive or in situ adenocarcinoma. (Figs. 5.7 and 5.8). Their cytologic criteria include: • Cells exfoliate in sheets, clusters and strips with nuclear crowding and overlap. • Rare cell groups with rosette or acinar formation with feathering. • Hyperchromatic, enlarged nuclei. • Increased N/C ratios. • Relatively scant cytoplasm with ill-defined cell borders. • Nucleoli are rarely observed. 117 A B C Fig. 5.7. Clusters of atypical cervical glandular cells, favor neoplastic displaying nuclear crowding and overlapping. Nuclei in palisade are seen in A and B. 118 A B Fig. 5.8. A. CP smear showing atypical endocervical glandular cells, favor neoplastic, in a dyshesive cluster. The atypical cells show enlarged, hyperchromatic nuclei and ill-defined cytoplasm. The patient was subsequently found to have AIS on cervical biopsy. 3. Atypical endometrial glandular cells Diagnostic criteria of endometrial AGCs include: • Small glandular cells present singly or in rounded clusters. • Scant or moderately abundant or vacuolated cytoplasm. (Fig. 5.9). • Enlarged, hyperchromatic nuclei with 1 of the 2 additional nuclear changes below • Irregular nuclear contours, or • Prominent nucleoli These cellular changes may also be seen in association with endometrial polyp, chronic endometritis, endometrial hyperplasia and IUDs. The endometrial cell atypias can be difficult to identify because of cellular degeneration. On the other hand, normal endometrial cells exfoliated in menses may display reactive changes 119 with slight nuclear enlargement and pleomorphism or degeneration that could be misinterpreted as abnormal. As about 50% of postmenopausal women shedding atypical endometrial cells have a significant endometrial pathology including carcinoma. Therefore, it is more practical to lump all degrees of endometrial cell atypia into one category of “endometrial cell atypia” and rely on histologic examination of endometrial tissue samples for grading of endometrial cell atypias. The presence of a high maturation index of squamous cells or a tumor diathesis also represents a risk for malignancy. A B 120 C Fig. 5.9. CP smear showing 3-dimensional clusters of atypical endometrial cells. DIAGNOSTIC DIFFICULTY Cervical AGCs, favor neoplastic show changes that are qualitatively slightly more severe than those of AGCs, NOS. On several occasions AGCs are difficult to distinguish from atypical squamous cells, and in many patients with AGCs on Pap smears, the cervical biopsy revealed a squamous cell lesion or no significant epithelial abnormality. However, according to DeMay, AGCs constitute a high-risk finding that predicts adenocarcinoma in 5% to 10% of cases. Therefore, patients with Pap smears showing persistent AGCs, NOS or AGCs, favor neoplastic should undergo colposcopic examination with biopsy and fractional uterine curettages to rule out a squamous cell lesion of the cervix and glandular neoplasm of the uterus. Testing for oncogenic HPV-DNA may be of diagnostic help, as cervical adenocarcinoma is strongly associated with HPV types 16 and 18. INVASIVE ADENOCARCINOMA Invasive adenocarcinoma of the cervix occurs more frequently in the 5th decade of life and accounts for up to 25% of all cervical cancers. AIS is found at the edge of invasive cervical adenocarcinomas in 43% to 100% of patients. The tumor may be associated with an ovarian mucinous or endometrioid adenocarcinoma. It is p16, HPV16, HPV18, ER and PR positive in almost all cases. It may be well to poorly differentiated, and a mixed pattern consisting of areas of well and poor differentiation is not uncommon. Endocervical mucinous carcinomas are the most common type, accounting for 70% to 90% of cases followed by carcinoma of endometrioid type. Intestinal and signet-ring adenocarcinomas are rarely encountered. The neoplasm spreads first to pelvic structures then pelvic lymph nodes, ovaries, upper abdomen and distant organs. The 5-year survival rates of 121 patients with cervical adenocarcinoma vary with the tumor stage: 79% for stage I, 37% for stage II and less than 9% for stage III and IV tumors. Poor prognostic factors include: high stage, depth of invasion >5 mm, angiolymphatic invasion, over-expression of HER2 and elevated level of serum CA125. The tumor is usually graded as follows: • Well-differentiated or grade I: if it shows a glandular and papillary pattern and consists of columnar cells with uniform, oval nuclei; and <10% or less of the tumor has a solid growth pattern. (Fig.5.10). • Moderately differentiated or grade II: if 11% to 15% of the tumor displays a solid, cribriform, complex glandular growth pattern and is composed of tumor cells with round or irregular nuclei and prominent nucleoli. • Poorly differentiated or grade III: if the tumor shows pleomorphic cells. Fig. 5.10. Histology of a moderately differentiated cervical adenocarcinoma. A. Well-differentiated cervical adenocarcinoma has cytologic manifestations similar to those of AIS, endocervical type, previously described. B. Moderately differentiated cervical invasive adenocarcinoma usually has cytologic manifestations different from those of a well-differentiated tumor. Cytologic criteria include: • Abundant malignant glandular cells commonly forming acini, balls, sheets, papillary clusters, strips, rosettes or syncytia. (Fig. 5.11). • Dyshesive tumor cells are more commonly found (than AIS). • Oval or pleomorphic nuclei with normo- or hyperchromasia. • Finely or coarsely granular chromatin. • Single or multiple micro- or macronucleoli. • Cytoplasm is variable, ill defined, and rarely vacuolated. • Tumor diathesis is present in about 30% of cases. • Dyskaryotic or malignant squamous cells are found in about 20% of cases. 122 A B C Fig. 5.11. Clusters of fairly polygonal malignant glandular cells with ill-defined cytoplasm, hyperchromatic nuclei and small or inconspicuous nucleoli in CP smears from 3 cases of moderately differentiated endocervical adenocarcinoma. C. Poorly differentiated cervical adenocarcinoma exfoliates pleomorphic malignant glandular cells singly and in clusters. The tumor cells display pleomorphic nuclei, ill defined cytoplasm and prominent nucleoli. (Fig. 5.12). 123 Fig. 5.12. Clustered pleomorphic malignant glandular cells in a CP smear from a poorly differentiated cervical adenocarcinoma. DIFFERENTIAL DIAGNOSIS Cells from a cervical adenocarcinoma should be differentiated from cells derived from a reparative epithelium, endometrial adenocarcinoma and extrauterine cancer. The reader is referred to Chapter 2 in this book for discussion and illustration of repair epithelial cells. • Endometrial adenocarcinoma cells can be difficult to distinguish from those of a cervical glandular malignant tumor. In general, cells from an endometrial tumor are smaller than those of an endocervical cancer and they are usually associated with a large amount of necrotic debris. In difficult cases, a fractional uterine curettage should be done to distinguish these two lesions histologically. A well-differentiated endometrial adenocarcinoma invading the cervix may exfoliate tumor cells in sheets with nuclei in picket-fence at periphery, in strips with pseudostratified nuclei and in rosettes, as seen in cell samples from a cervical AIS. The reader is referred to Chapter 7 for discussion and illustrations of malignant endometrial cells. • Extrauterine adenocarcinoma cells may reach the cervix by traveling through the fallopian tubes and uterine cavity. These malignant cells are, in most cases, found in a smear background that is free of necrotic debris and inflammatory exudate. In doubtful cases, clinical data and fractional curettages of the uterus would be of diagnostic help. The reader is referred to Chapter 6 for a more detailed discussion on extrauterine cancers. 124 DIAGNOSTIC ACCURACY Cytodiagnosis of cervical AIS is challenging. In one series consisting of 94 patients with cervical AIS, 65 (69%) cases showed a glandular lesion and 29 (31%) displayed a squamous or unspecified lesion. For pure cervical AISs a diagnostic sensitivity of 40% to 69% has been reported. When a combined lesion consisting of AIS and HSIL the reported rate of detection of glandular cell abnormality was 16% to 23% only, depending on the types of cell preparation (CP smear or LBP). However, if the cell samples were diagnosed as having either an AIS plus HSIL or a HSIL, the sensitivity rates were about 63% and 74%, respectively. The diagnostic accuracy rate of invasive cervical adenocarcinomas in different reported series ranged from 86% to 97.4%. The cytologic detection of cervical adenosquamous carcinomas appears to be more difficult leading to a false-negative rate up to 55%. It is important to note that cervical biopsy is adequate for confirming an invasive adenocarcinoma but it is not adequate for diagnosing a cervical AIS which requires a deep cone biopsy for histologic confirmation. VARIANTS OF CERVICAL ADENOCARCINOMA Relatively common variants of cervical adenocarcinoma are mixed adenosquamous carcinoma, serous papillary carcinoma, and clear cell carcinoma. Adenoma malignum and adenoid cystic carcinoma are rare occurrences in this location. A. Glassy cell carcinoma is a poorly differentiated adenosquamous cell carcinoma and accounts for 1% to 2% of all cervical cancers. It more commonly occurs in relatively young patients, with a mean age of 41 and is HPV types 16 and 18 positive. It most often appears as a bulky exophytic mass consisting of nests and masses of polygonal cells with granular “glassy” cytoplasm and oval nuclei with prominent nucleoli. Patients with this type of tumor may show blood eosinophilia. In typical cases, the tumor presents in Pap smears as single and clustered large malignant epithelial cells with oval nuclei and prominent nucleoli, similar to those of a nonkeratinizing squamous cell carcinoma. (Fig. 5.13). Tumor cells with ground-glass cytoplasm may be found. In other cases the exfoliated tumor cells do not display a groundglass cytoplasm and a firm diagnosis of glassy cell carcinoma can be only made by tissue biopsy. 125 A B Fig. 5.13. Glassy cell carcinoma: A. Tumor histology showing cells with abundant, homogenous, eosinophilic, “glassy” cytoplasm, oval nuclei and prominent nucleoli. B. CP smear showing clusters of malignant epithelial cells with abundant, “glassy” or granular cytoplasm and large and oval nuclei. Small nucleoli are noted in some cells. B. Adenoid cystic carcinoma accounts for about 1% of all cervical adenocarcinomas. The tumor occurs mainly in elderly patients, but it may occur in patients under 50 years of age. The neoplasm commonly forms a polypoid friable mass and consists of small cancer cells forming clusters, cords and trabeculae with lumens containing hyaline eosinophilic material. The neoplasm may be associated with a cervical squamous cell carcinoma. It can be seen in Pap smears as clusters of small cells with scant cytoplasm and small, oval, hyperchromatic nuclei. Globules of basophilic material may be observed. (Fig.5.14). The tumor is HPV16 positive and has a poor prognosis. 126 A B Fig. 5.14. Adenoid cystic carcinoma: A. Tumor histology showing small tumor cells forming round spaces containing dense, hyaline eosinophilic material. B. A round hyaline body and smaller clusters or sheets of tumor cells with scant, ill-defined cytoplasm and round nuclei with conspicuous nucleoli seen in a CP smear. 127 C. Minimal deviation adenocarcinoma is a rare tumor accounting for about 1% of all primary cervical adenocarcinomas. Histologically, it may be divided into 3 types: cervical, endometrioid and non-specific. • Cervical minimal deviation adenocarcinoma (MDA), mucinous type is the most common type, occurring in young women between 32 to 42 years of age. It is usually sporadic but it may rarely occur synchronously or precede an ovarian tumor that is commonly mucinous in nature. The tumor is usually HPV negative and often missed by small cervical biopsy. Due to diagnostic delay, this neoplasm may be diagnosed at a high stage and therefore will have a poor prognosis. In about 50% of cases, a foci of moderately or poorly differentiated adenocarcinoma is present. In a Pap smear, a cervical MDA, mucinous type, yields sheets and clusters of glandular cells with monomorphic nuclei, small nucleoli and clear cytoplasm that may show wispy cytoplasmic extensions or tails. Higher grade tumor cells are not uncommonly found. (Figs. 5.15 to 5.17). Fig. 5.15. Histology of cervical MDA, mucinous type showing mucous glands with small, bland nuclei invading the cervical fibromuscular wall. 128 A B Fig. 5.16. Cervical MDA, mucinous type, showing in a CP smear: Irregular sheets and clusters of benign-appearing mucus secreting cells in honeycomb pattern. A 129 B C Fig. 5.17. In another case of cervical MDA, mucinous type, abundant tumor cells with relatively bland nuclei are seen (A and B). But a few more pleomorphic tumor cells with large, irregular nuclei and conspicuous nucleoli are present (C). The poor preservation of glandular cells is evident in this air-dried CP smear. • MDA, endometrioid type. The cytologic manifestations of this neoplasm haverecently been reported. This type of tumor exfoliates sheets of columnar glandular cells with low-grade oval nuclei in palisade at free borders. Similar tumor cells forming cell stripes with vague pseudostratified nuclei and rosettes, as seen in cervical AIS, are present. Tumor cells with higher nuclear grade may also be found. (Fig.5.18). Cytologic manifestations of a cervical MDA, non-specific type have not been reported so far. 130 A B C 131 D E Fig. 5.18. Minimal deviation adenocarcinoma, endometrioid type: A. Histology of the tumor. B-E. CP smear showing tumor cells in a large sheet with nuclei feathering and in palisade formation at the periphery (B), tumor cells with low-grade nuclei and cytoplasmic “tails” in a rosette and cell strip (C,D), and more pleomorphic malignant glandular cells with conspicuous nucleoli forming a rosette (E). D. Clear cell carcinoma accounts for about 4% of all cervical adenocarcinomas. About 2/3 of cases occur in young women who had an in utero exposure to diethylstilbestrol (DES). It may occur in older women without DES exposure. In about 50% of cervical clear cell carcinoma related to DES exposure a vaginal adenosis is present. Grossly, the tumor appears as a nodular or an ulcerated lesion. The 5- and 10-year survival rates of patients with clear cell carcinoma are 55% and 40%, respectively. Histologically, it is characterized by a papillary, microcystic, tubular or solid pattern. The tumor cells show a clear or eosinophilic, granular cytoplasm and often have a “hobnail” configuration. The nuclei are oval and show 132 prominent nucleoli. Cervical clear cell carcinoma shows in Pap smears irregular sheets and clusters of epithelial cells with clear, granular or vacuolated cytoplasm and oval nuclei with prominent nucleoli. (Fig. 5.19 and Fig. 5.20). A B C Fig. 5.19. Clear cell carcinoma: A. Histology of the tumor showing cells in “hobnail” pattern. 133 B, C. CP smear showing sheets of tumor cells with round nuclei, granular or opaque, well defined cytoplasm. Focal nuclear crowding is noted. A B C Fig. 5.20. Clear cell carcinoma: A. Histology of the tumor. 134 B, C. Large sheets of tumor cells with clear, vacuolated cytoplasm, oval nuclei and conspicuous nucleoli seen in a CP smear. E. Villoglandular carcinoma is a rare cervical cancer with low-grade nuclei and an excellent prognosis. It is composed of epithelial papillae with thick fibrovascular cores. It presents in Pap smears as monolayered sheets of malignant epithelial cells with folding and nuclear crowding. The nuclei are oval, hyperchromatic and show inconspicuous nucleoli. (Fig. 5.21). A B 135 C Fig. 5.21. Villoglandular carcinoma: A. Histology of the tumor. B-D. CP smear showing a large monolayered sheet of tumor epithelium with folding (B) and thick tumor cell clusters showing round, hyperchromatic monomorphic nuclei and inconspicuous nucleoli (C,D). F. Papillary serous carcinoma accounts for about 1% of all cervical carcinomas. The tumor is aggressive and has early pelvic and periaortic lymph node metastases. In some studies, tumors in young patients are HPV positive and those in older patients are HPV negative. Histologically the tumor is characterized by the presence of thin fibrovascular cores and covered by pleomorphic malignant glandular cells with prominent nucleoli. In a Pap smear, it displays tri-dimensional clusters of malignant glandular cells or irregular, monolayered sheets of similar cells. (Fig. 5.22). A 136 B C Fig. 5.22. Papillary serous carcinoma: A. Histology of the tumor. B, C. Tridimensional papillary clusters of tumor cells with prominent nucleoli seen in a CP smear from the lesion. 137 REFERENCES Akin MR, Nguyen GK. Cytologic manifestations of advanced endometrial adenocarcinomas in cervical-vaginal smears. Diagn Cytopathol. 1999; 20:108. Apgar BS, et al. Update on ASCCP consensus guidelines for abnormal cervical screening tests and cervical histology. Am Fam Physician. 2009;80:147. Ayer B, et al. Cytologic diagnosis of adenocarcinoma in situ of the cervix uteri and related lesions. I. Adenocarcinoma in situ. Acta Cytol.1987; 31:397. Ayer B, et al. The cytodiagnosis of adenocarcinoma in situ of the cervix. II Microinvasive adenocarcinoma. Acta Cytol.1988; 32:318. Bousfield L, et al. Expanded criteria for the diagnosis of adenocarcinoma in situ of the cervix and related lesions. Acta Cytol.1980; 24:283. Cibas ES. Cervical and vaginal cytology. In Cytology: diagnostic principles and clinical correlates. Cibas ES, Ducatman BS, eds. 3rd edition, 2009, Philadelphia, Saunders Elsevier, p 1. Chang WC, et al. Cytologic features of villoglandular adenocarcinoma of the uterine cervix. Comparison with typical endometrial adenocarcinoma with villoglandular component and papillary serous carcinoma. Cancer. 1999; 87:5. Costa MJ, et al. Cervicovaginal cytology in uterine adenocarcinoma and adenosquamous carcinoma. Comparison of cytologic and histologic findings. Acta Cytol.1991; 35: 137. Covell JL. Atypical glandular cells and endocervical adenocarcinoma in situ of the uterine cervix. Pathology Case Reviews. 2005;10:155. DeMay RM. The Pap Test. Chicago, ASCP Press. 2005. Gilks CB, et al. Adenoma malignum (minimal deviation adenocarcinoma) of the uterine cervix. A clinicopathological and immunohistochemical analysis of 26 cases. Am J Surg Pathol 1989; 13: 717. Gray JA, Nguyen GK. Can glassy cell carcinoma of the cervix be diagnosed by Pap smear?. Acta Cytol 2002; 46: 1168. Granter SR, Lee KR. Cytologic findings in minimal deviation adenocarcinoma (adenoma malignum) of the cervix. Am J Clin Pathol 1996;105:327. 138 Hanselaar AG, et al. Cytologic examination to detect clear cell adenocarcinoma of the vagina and cervix. Gynecol Oncol. 1999; 75: 338. Ishii K, et al. Cytologic and cytochemical features of adenoma malignum of the uterine cervix. Cancer 1999; 87: 245. Nguyen GK, Jeannot AB. Exfoliative cytology of in situ and microinvasive adenocarcinoma of the uterine cervix. Acta Cytol.1984; 28:461. Nguyen GK, Daya D. Cervical adenocarcinoma and related lesions. Cytodiagnostic criteria and pitfalls. Pathol Annu. 1993, 28(2):53. Nguyen GK, Daya D. Exfoliative cytology of papillary serous adenocarcinoma of the uterine cervix. Diagn Cytopathol. 1997;17:177. Novotny DB, et al. Tubal metaplasia. A frequent potential pitfall in the cytologic diagnosis of endocervical dysplasia on cervical smears. Acta Cytol. 1992; 236:1. Nunez C, et al. Glassy cell carcinoma of the uterine cervix. Cytopathologic and histopathologic studies of 5 cases. Acta Cytol.1985; 29:303. Odashiro AN, et al. Minimal deviation endometrioid adenocarcinoma of the uterine cervix: a report of 3 cases with exfoliative cytology. Diagn Cytopathol. 2006; 34:119. Oster AG, et al. Adenocarcinoma in situ of the cervix. Int J Gynecol Pathol.1984; 3:179. Powers CN, et al. Adenoid cystic basal cell carcinoma of the cervix: a potential pitfall in cervicovaginal cytology. Diagn Cytopathol 1996; 14: 172. Qizilbash AH. In situ and microinvasive adenocarcinoma of the endocervix. A clinical, cytologic and histologic study of 14 cases. Am J Clin Pathol.1975; 64:155. Ravinsky E, et al. Cytologic features of primary adenoid cystic carcinoma of the uterine cervix. A case report. Acta Cytol 1996; 40: 1304. Roberts JM, Thurloe JK. Comparative sensitivities of ThinPrep and Papanicolaou smear for adenocarcinoma in situ (AIS) and combined AIS/High-grade squamous intraepithelial lesion (HSIL): comparison with HSIL. Cancer (Cancer Cytopathol). 2007; 111:482. Ruba S, et al. Adenocarcinoma in situ of the uterine cervix. Screening and diagnostic errors in Papanicolaou smears. Cancer (Cancer Cytopathol). 2004;102:280. 139 Shin CH, et al. Cytologic and biopsy findings leading to conization in adenocarcinoma in situ of the cervix. Obstet Gynecol. 2002;100:271. Schnatz PF, et al. Clinical significance of atypical glandular cells on cervical cytology. Obstet Gynecol. 2006;107:701. Stuart G, et al. Report of the 2003 pan-Canadian forum on cervical cancer prevention. J Obstet Gynaecol Can. 2004; 287; 2120. Taft PD, et al. Cytology of clear cell adenocarcinoma of the genital tract in young females. A review of 95 cases from the registry. Acta Cytol.1974; 18:279. Valente PT, Schantz HD. The diagnosis of glandular abnormalities in cervical smears. Cytopathology. 1996; 1:39. Vogelsang PJ, et al. Exfoliative cytology of adenoma malignum (minimum deviation adenocarcinoma) of the uterine cervix. Diagn Cytopathol 1995; 13: 146. Vuong NP, et al. Adenoid cystic carcinoma associated with squamous carcinoma of the cervix uteri. Acta Cytol 1996; 40: 289. Wright TC, et al. 2002 consensus guidelines for the management of women with cervical cytological abnormalities. JAMA. 2002; 287; 2120. Yahr JJ, Lee KR. Cytologic finding of microglandular hyperplasia of the cervix. Diagn Cytopathol.1991; 7:248. Young RH, Scully RE. Invasive adenocarcinoma and related tumors of the uterine cervix. Semin Diagn Pathol.1990; 7:205. 140 Chapter 6 Other Cervical Cancers and Extrauterine Cancers PRIMARY CERVICAL CANCERS A. Neuroendocrine carcinoma of the cervix is a rare tumor that may occur alone or in association with a cervical adenocarcinoma of usual type. Histologically, these tumors are classified into 4 subtypes: typical and atypical carcinoid tumors, small cell carcinoma and large cell carcinoma with neuroendocrine differentiation. Carcinoid tumors are rare and highly aggressive with a 3-year survival rate of 12% to 33%. Large neuroendocrine carcinoma is also a very rare tumor. 1. Typical carcinoid tumor yields, in a CP smear, single and loosely clustered oval cells with plasmacytoid configuration and oval nuclei with chromatin clumping. (Fig. 6.1). A 141 B Fig. 6.1. Typical carcinoid tumor: A. Histology of the tumor. B. CP smear showing single and loosely clustered oval or polygonal tumor cells with eccentrically located nuclei. 2. Small-cell carcinoma (oat cell carcinoma) is an aggressive neoplasm, accounts for 2% to 5% of all cervical cancers and is strongly associated with HPV type 18. It is occasionally associated with Cushing syndrome or symptoms of other peptide hormones. It rarely coexists with SIL and its 5-year survival rate is 30% to 40%. It is histologically similar to small cell lung cancer and it yields in Pap smears small cancer cells with hyperchromatic nuclei with “salt and pepper” chromatin, nuclear molding and linear basophilic nuclear debris. (Fig. 6.2). A 142 B Fig. 6.2. Small cell carcinoma: A. Histology of the tumor. B. CP smear showing single and clustered tumor cells showing scant cytoplasm, round hyperchomatic nuclei with molding. B. Malignant mixed müllerian tumor (MMMT) is a very rare cervical cancer occurring in adult or elderly women (mean age: 50 to 65 years), often with a history of radiation therapy for cervical squamous cell carcinoma. Most MMMTs found in uterine cervix are actual endometrial MMMTs extending to the cervix. The tumors are classified as homologous and heterologous depending on the nature of its stromal neoplastic cells. The reader is referred to Chapter 7 for a more detailed discussion on MMMTs. These neoplasms commonly exfoliate malignant glandular cells in Pap smears, and stromal tumor cells are usually not observed. (Fig. 6.3). Fig. 6.3. Homologous MMMT showing in Pap smears a cohesive cluster of fairly monomorphic malignant glandular cells. 143 C. Stromal cell sarcoma of the cervix is a very rare tumor that tends to occur in postmenopausal women (mean age: 54 years). Histologically it is characterized by malignant, uniform cells or slightly pleomorphic cells with scant cytoplasm. It shows in Pap smears single and loosely clustered round or slightly pleomorphic malignant cells with scant cytoplasm, similar to the malignant stromal cells of a MMMT. The reader is referred to Chapter 7 for illustrations of this neoplasm. D. Other cervical nonepithelial cancers such as rhabdomyosarcoma, leiomyosarcoma, non-Hodgkin lymphoma, adenosarcoma, alveolar soft part sarcoma, granulocytic sarcoma are exceedingly rare tumors, however some of them have been cytologically reported in Pap smears. METASTATIC CANCERS Almost all malignant glandular neoplasms arising from extragenital organs can metastasize to the uterine cervix. Of these tumors, carcinomas of the breast, stomach and colon are the most common primaries while those arising from other anatomic sites such as the lung, pancreas, bladder, liver, kidney and gallbladder are sporadic. In one report consisting of 208 consecutive cases of primary and metastatic cervical adenocarcinomas, metastatic adenocarcinoma to the cervix (MAC) accounted for about 2 % of all cervical adenocarcinomas. Regardless of the primary site, nearly 90% of women with metastasis to the cervix have evidence of a disseminated cancer, and the most common symptom is vaginal bleeding, occurring in 75% of patients. The cytologic manifestations of MAC in Pap smears are rather distinctive and different from those of a usual primary endocervical adenocarcinoma. A metastatic moderately differentiated colonic adenocarcinoma yields abundant necrotic debris and irregular sheets of glandular cells with elongated nuclei in palisade at free borders and syncytial tumor cell clusters. (Fig. 6.4). Colonic adenocarcinoma cells are CK7 negative; and CK20, villin and CDX2 positive. Metastatic mammary duct carcinoma to the cervix yields malignant glandular cells in clusters and in Indian file arrangement. (Fig. 6.5). Breast cancer cells are ER, PR and gross cystic disease fluid protein fraction 15 positive. Endometrial carcinoma cells are ER and vimentin positive. 144 A B C Fig. 6.4. Metastatic well-differentiated colonic adenocarcinoma to the cervix: A. Histology of the tumor. B, C. CP smear showing irregular large sheets of tumor cells with cells at the periphery arranged in picket-fence. A large amount of necrotic debris is present. 145 Fig. 6.5. Metastatic breast carcinoma showing in CP smear an irregular cluster of malignant glandular cells with conspicuous nucleoli. 146 EXTRAUTERINE CANCERS Extrauterine cancers involving the abdominal peritoneum may traverse through the fallopian tubes and the uterine cavity to accumulate in the posterior vaginal fornix. A vaginal smear in this case may reveal tumor cells in irregular, tight tridimensional clusters. The smear background is more commonly free of tumor diathesis. The presence of psammoma bodies should alert the observer to the possibility of a papillary serous ovarian carcinoma, and effort should be made to identify psammoma bodies surrounded by malignant epithelial cells to confirm the diagnosis. However it should be born in mind that psammoma bodies may be seen in cervico-vaginal cell sample in patients without intra-abdominal cancer and in women with IUD. (Figs. 6.6 and 6.7). A B Fig. 6.6. Papillary serous ovarian carcinoma: A. Histology of the tumor. B. CP smear showing thick tridimensional clusters of malignant glandular cells in a clean smear background. 147 A B Fig. 6.7. Ovarian carcinoma: A. Borderline ovarian carcinoma showing in CP smear 3 psammoma bodies surrounded by low-grade epithelial tumor cells in a clean background. B. High-grade ovarian serous carcinoma showing in CP smear fairly pleomorphic malignant cells partially surrounding a psammoma body. 148 REFERENCES Ann-Foraker SH, Kawada CY. Cytodiagnosis of malignant mixed tumors of the uterus. Acta Cytol 1985; 29:137. Costa MJ, et al. Cervicovaginal cytology in carcinosarcoma, malignant mixed mullerian tumor of the uterus. Diagn Cytopathol 1992; 8: 33-40. Dabbs DJ. Immunohistology of metastatic carcinoma of unknown primary, p 180. In Diagnostic immunohistochemistry, D Dabbs, ed. 2nd edition, 2006, Philadelphia, Churchill Livingstone Elsevier. Deshpande AH, Munshi MM. Primary malignant melanoma of the cervix. Report of a case diagnosed by cervical scraping. Diagn Cytopathol 2001; 25: 108. Ferenzy A, Winkler B. Carcinoma and metastatic tumors of the cervix. In: Kurman RJ, ed. Blaustein’s Pathology of the Female Genital Tract 3rd ed. New York, Springer-Verlag, 1987, p. 218. Korhonen M, Stenback F. Adenocarcinoma metastatic to the uterine cervix. Gynecol Obstet Invest. 1984; 17: 57. Kurman RJ, et al. Tumors of the cervix, vagina, and vulva. In Atlas of tumor Pathology. Washington D.C., Armed Forces Institute of Pathology, 1992, p. 37. Raspollini MR, et al. Primary cervical adenocarcinoma with intestinal differentiation and colonic carcinoma metastatic to cervix. An investigation using CDX2 and limited immunohistochemical panel. Arch Pathol Lab Med. 2003; 127:1586. Reich O, et al. Exfoliative cytology of invasive neuroendocrine small cell carcinoma in a cervical cytology smear-A case report. Acta Cytol 1996; 40: 980. Wang X, et al. Cervical and peritoneal fluid cytology of uterine sarcoma. Acta Cytol 2002; 46: 465. Zhou C, et al. Small cell carcinoma of the uterine cervix: cytologic findings in 13 cases. Cancer 1998; 84:281. 149 Chapter 7 Endometrial Lesions The endometrium is an important target for hormonal stimulation and may be affected by a wide variety of disease processes. The resulting lesions are often manifested clinically by abnormal uterine bleeding, and exfoliate endometrial cells which, after a variable period of retention within the uterine cavity, pass through the cervical canal and tend to accumulate in the posterior vaginal fornix. They often show a variable degree of degenerative change rendering their morphologic evaluation difficult. However, when these cells are well–preserved a correct cytodiagnosis may be made. In the past fifty years efforts have been made to design a simple–to–use device for use in the physician's office to obtain cells directly from the endometrium for cytologic examination. Those devices included cannula for washing of uterine cavity, Mi-Mak plastic helix, Endopap sampler and Endocyte sampler. Of these, Endocyte samplers have been the most popular ones. They are simple to use and cause little or no discomfort to patients, and the cell samples obtained have a low rate of cellular inadequacy. A recent study with direct samplers has demonstrated that if the endometrial cytology is properly used, it would be cost–effective by substantially reducing the number of unnecessary endometrial curettages. However, to date, there is no known cytology screening program for endometrial cancer using direct endometrial samplers. BENIGN-APPEARING ENDOMETRIAL CELLS IN WOMEN OVER 40 YEARS OF AGE The presence of spontaneously exfoliated benign-appearing endometrial cells in Pap smears of women over 40 years of age is not regarded as an epithelial abnormality in The Bethesda System-2001. These cells are seen in small tridimensional, round clusters and they are present in less than 1% of all Pap smears. (Fig.7.1).In the majority of cases, these benign endometrial cells are from a normal and cycling endometrium, and in other cases their exfoliation is secondary to a benign endometrial polyp, an IUD and a hormonal replacement therapy. Only in about 1% of these women an endometrial hyperplasia or carcinoma is found. Therefore, the presence of spontaneously exfoliated benign-appearing endometrial cells in an asymptomatic woman does not constitute an indication for endometrial biopsy for histologic evaluation. LUS endometrial fragments scraped by a cervical cytology sampler do not belong to endometrial cells in this category. Clusters of atrophic endocervical cells, naked squamous cell nuclei or clustered histiocytes may be mistaken for benign endometrial cells. (Fig.7.2). 150 A B Fig. 7.1. Benign endometrial cells in LBP from a woman over 40 years of age with cycling endometrium. A. A tridimensional, ball-like cluster of endometrial stromal cells wrapped by endometrial glandular cells. B. A tridimensional cluster of endometrial glandular cells with scant cytoplasm. 151 Fig. 7.2. LBP showing clustered histiocytes mimicking endometrial cells. Note the bean-shaped nuclei and more defined and more abundant cytoplasm. ABNORMAL SHEDDING OF NORMAL-APPEARING ENDOMETRIAL CELLS Depending on a woman’s menstrual status, abnormal shedding of normal-appearing endometrial cells may have different endometrial pathologies, according to several studies. Normal appearing endometrial epithelial cells have scant cytoplasm and a small bland, round or oval nucleus that is of the same size as the nucleus of a normal intermediate squamous cell. These cells usually occur in small groups or clusters and have no conspicuous nucleoli. (Fig. 7.3). In a premenopausal woman, shedding of normal-appearing endometrial cells, epithelial and/or stromal types, beyond day 10 to 12 of the menstrual cycle is an abnormal finding that should be interpreted with caution in the light of clinical information. It can be secondary to an IUD, endometritis, anovulatory cycle, prior endometrial curettage or uterine endoscopy, endometrial polyp, hormonal therapy, submucosal myometrial leiomyoma, endometrial hyperplasia and rarely endometrial cancer. In a postmenopausal woman, hormonal therapy, endometrial polyp, endometrial hyperplasia and endometrial cancer are the main causes of abnormal shedding of normal-appearing endometrial cells. Of these etiologies, hormonal replacement therapy is the most common one; a benign endometrial polyp is found in 23% of patients, endometrial hyperplasia and endometrial carcinoma are found in 5% and 5% of cases, respectively. In another series, about 6% of postmenopausal women with endometrial carcinoma shed only normal-appearing endometrial cells. Thus, the presence of unexplained normal-appearing endometrial cells in Pap smears in an asymptomatic woman needs further investigation to rule out a significant 152 endometrial pathology, especially when clinical risk factors for endometrial carcinoma are present (hypertension, obesity, nulliparity and hormonal replacement therapy). It should be born in mind that a large percentage of patients with abnormal shedding of normal-appearing endometrial cells show no pathology in biopsied endometrial tissues. Fig. 7.3. A group of normal-appearing endometrial cells in a CP smear. The endometrial cell nuclei are of the same size with those of intermediate squamous cells. IUD-INDUCED CELLULAR CHANGES Women bearing IUDs usually show cellular atypias affecting endometrial glandular cells and cervical metaplastic squamous cells. These individuals may shed endometrial cells at any time of the menstrual cycle. By mechanical effects, reactive and regenerative endometrial cells are formed. These cells occur in groups or clusters and show cytoplasmic enlargement with intracytoplasmic vacuoles, conspicuous or prominent nucleoli thus mimicking malignant glandular cells. The cervical metaplastic squamous cells may show prominent nucleoli. Single endometrial cells with high N/C ratios and hyperchromatic nuclei with irregular nuclear membrane or contours mimicking HSIL/CIN 3 cells may be observed. (Fig. 7.4). It is important to note that these CIN 3-like cells are few in number and do not form hyperchromatic crowed groups as seen in cervical with HSIL/CIN 3, and there is an absence of other squamous cells with less dyskaryotic change, as usually observed in SIL cases. The differential diagnosis is more challenging when these CIN 3-like cells are present in abundance. (Fig. 7.5). 153 A B Fig. 7.4. CP smear showing IUD-induced cellular changes. A. A cluster of glandular cells with vacuolated cytoplasm. B. Four small HSIL-like cells. 154 A B Fig. 7.5. CP smear from a 25-year-old woman with IUD showing a large number of epithelial cells with IUD-induced changes, mimicking HSIL cells. Small cytoplasmic vacuoles are seen in a few cells. Colposcopy, endocervical curettage and long-term follow-up revealed no cervical SIL or any significant endometrial lesion. ENDOMETRIAL HYPERPLASIA Endometrial hyperplasia is the result of an increased or uninterrupted estrogenic stimulation and it is an important factor in endometrial carcinogenesis. However, not all cases of endometrial cancer are estrogen-related, as the tumor can also arise from an atrophic endometrium. Endometrial hyperplasias are classified into 3 main types: simple hyperplasia (cystic glandular hyperplasia), complex hyperplasia without atypia (adenomatous hyperplasia) and complex hyperplasia with atypia (atypical hyperplasia showing 155 remarkable nuclear abnormality). About 40% of patients with atypical endometrial hyperplasia will eventually develop endometrial adenocarcinoma. A. Endometrial hyperplasia without atypia and with atypia. 1. Endometrial hyperplasia without atypia yields in CP smears clustered normal-appearing endometrial cells in a well-estrogenized smear background. Simple and complex endometrial hyperplasias without atypia display in direct endometrial samples similar cellular manifestations. Large and irregularly dilated branching endometrial glandular fragments and cell clusters or clumps with no remarkable nuclear atypia are seen. Nuclei in palisade, nuclear crowding and overlapping and inconspicuous nucleoli may be observed. (Fig. 7.6). A B Fig. 7.6. Endometrial adenomatous/complex hyperplasia without atypia showing in direct endometrial sample an endometrial fragment displaying monomorphic oval or elongated, pseudostratified nuclei. 156 2. Endometrial hyperplasia with atypia is characterized by clustered atypical endometrial cells with enlarged nuclei, irregular nuclear contours, chromatin clumping, parachromatin clearing and nucleoli. However, all of the above-mentioned cellular changes are not present in a given case. The reader is referred to Chapter 5 for illustration of atypical endometrial glandular cells. Endometrial hyperplasia with atypia yields in direct samples irregularly dilated or branched endometrial glands and clusters or clumps of endometrial cells. The cell clumps display a loss of nuclear polarity, nuclear pleomorphism, hyperchromasia, chromatin clumping and clearing and prominent nucleoli. (Fig. 7.7). These cell clusters or clumps are indistinguishable from those of a well-differentiated or grade 1 endometrial adenocarcinoma. Large clusters of benign and slightly pleomorphic stromal cells are also commonly present. A B Fig. 7.7. Endometrial hyperplasia with atypia showing in direct sampling: A. A large cluster of epithelial cells with loss of nuclear polarity, slight nuclear pleomorphism, focal nuclear crowding and conspicuous nucleoli. B. Small clusters of atypical glandular cells with pleomorphic nuclei and conspicuous nucleoli. 157 ENDOMETRIAL ADENOCARCINOMA Endometrial adenocarcinoma is the most common malignancy of the female genital tract in North America. It occurs mainly in postmenopausal women in their 6th and 7th decades of life, with an average age of 60 years at the time of diagnosis. About 75% of the cases are seen in patients over 50 years in age. Patients younger than 40 years constitute about 5% of all cases. The tumor may arise from a hyperplastic endometrium (Type I tumor) or from a normal or atrophic endometrium showing carcinoma in situ (Type II tumor). Type I tumors are common (80%), and low-grade tumors have a favorable prognosis and tend to occur in obese, younger and perimenopausal women. Type II tumors tend to arise in thin, older, multiparous and postmenopausal women. They show a high histologic grade and have an aggressive behavior and poor prognosis. Epidemiological studies have identified a number of risk factors in patients with type I endometrial adenocarcinoma. The most important ones are atypical endometrial hyperplasia, obesity, unopposed estrogen effects, nulliparity and diabetes mellitus. Type II tumors usually lack those classic risk factors. In recent years efforts have been made to screen asymptomatic endometrial carcinoma cytologically by direct endometrial sampling. In one large study consisting of 2,586 asymptomatic perimenopausal and postmenopausal patients with high–risk factors revealed the prevalence of endometrial carcinoma of 6.9/1000. In another study of 747 asymptomatic women over 45 years of age, a prevalence of endometrial cancer of 4/1000 was found. However, there are no controlled data to justify a mass screening of endometrial carcinoma in low-risk patients. The most common clinical manifestation of endometrial carcinoma is an abnormal uterine bleeding. However, about 10% of the patients with early disease present with leukorrhea only. From the cancer screening point of view, routine Pap smears are not efficient in detecting endometrial carcinomas as the tests fail to detect cancer cells in about 50% of cases. Depending on the degree of differentiation of ordinary endometrial adenocarcinomas, their cytologic manifestations in Pap smears and in direct endometrial samples are similar. A well–differentiated tumor shows irregular clusters and groups of malignant epithelial cells displaying nuclear hyperchromasia, nuclear crowding and overlap, small or conspicuous nucleoli, as seen in an atypical endometrial hyperplasia. (Fig. 7.8). 158 A poorly differentiated neoplasm yields pleomorphic malignant glandular cells singly and in clusters that are readily identifiable. (Figs. 7.9 and 7.10). The smear background always contains a large amount of necrotic debris. Fragments of malignant glandular epithelium are evident in endometrial aspiration. (Fig.7.11). Cytologic features of endometrial adenocarcinomas in Pap smears include: • Tumor cells present singly and in small, tight clusters. • Scant, basophilic and often vacuolated cytoplasm. • Variation in nuclear size and loss of nuclear polarity. • Nuclei with moderate hyperchromasia and irregular chromatin distribution. • Prominent nucleoli with parachromatin clearing. • Increased nuclear and nucleolar sizes are observed with higher tumor grade. • Tumor diathesis variably present. A B Fig. 7. 8. Low-grade endometrial adenocarcinoma: A. Histology of the tumor. B. CP smear showing clustered monomorphic tumor cells with enlarged, hyperchromatic nuclei and small nucleoli. 159 A B Fig. 7.9. Poorly differentiated endometrial adenocarcinoma. A. Histology of the tumor. B. CP smear showing a large, cohesive cluster of tumor cells with pleomorphic nuclei, irregular chromatin clumping, parachromatin clearing and prominent nucleoli. 160 Fig. 7.10. Pleomorphic tumor cells with vacuolated cytoplasm in the CP smear of a patient with poorly differentiated endometrial adenocarcinoma. Fig. 7.11. Poorly differentiated endometrial adenocarcinoma showing in direct endometrial sample pleomorphic malignant epithelial cells with marked nuclear pleomorphism and prominent nucleoli. 161 OTHER ENDOMETRIAL CANCERS 1. Endometrial papillary serous carcinoma is a rare neoplasm. It yields in direct endometrial sampling large monolayered sheets of tumor cells with focal nuclear crowding and overlapping, and conspicuous nucleoli are present. (Figs. 7.12 and 7.13). A B 162 C Fig. 7.12. Papillary serous carcinoma: A. Histology of the tumor. B. Direct endometrial sample showing irregular large, thick sheets of tumor cells and C. A sheet of tumor cell displaying focal nuclear crowding and overlapping and inconspicuous nucleoli. A 163 B Fig. 7.13. A, B. In another case of papillary serous endometrial carcinoma the CP smear shows abundant 3-dimensional papillary clusters of tumor cells. 2. Squamous cell carcinoma of the endometrium is rare and consists of nonkeratinizing and keratinizing types. These two tumor types yield in direct samples single and clustered malignant squamous cells with keratinizing and nonkeratinizing malignant squamous cells, respectively. (Figs. 7.14 and 7.15). A 164 B Fig. 7.14. Keratinizing squamous cell carcinoma showing: A and B. Clustered and single malignant squamous cells with eosinophilic and keratinizing cytoplasm in a direct endometrial sample. Fig. 7.15. Nonkeratinizing squamous cell carcinoma showing in a direct endometrial sample a cohesive cluster of cancer cells with scant, slightly eosinophilic and defined cytoplasm and hyperchomatic, vesicular nuclei. 3. Small cell carcinoma is a very rare endometrial cancer. It shows single and clustered pleomorphic small cancer cells with scant cytoplasm, hyperchromatic nuclei without nucleoli and nuclear molding. (Fig. 7.16). 165 Fig. 7.16. Small cell carcinoma showing in direct endometrial sample tumor cells with nuclear molding. 4. Endometrial stromal sarcoma is a rare neoplasm. It occurs mainly in postmenopausal women. It may be classified as low– or high–grade, depending on the degree of cellular atypia and the number of mitotic figures. The cytologic manifestations of the tumor in vaginal pool smears and direct endometrial samples are similar and consist of numerous single or loosely clustered malignant round cells with oval or pleomorphic, hyperchromatic nuclei and scant, ill–defined cytoplasm. (Fig. 7.17). A 166 B Fig. 7.17. Low-grade endometrial stromal sarcoma: A. Tumor histology. B. Single and loosely clustered round tumor cells in a direct endometrial sample. 5. Malignant mixed müllerian tumor (MMMT). This rare tumor occurs predominantly in postmenopausal women in the 6th or 7th decade of life, with up to 30% having a history of exposure to radiation. Histologically, the tumor is classified as homologous or heterologous. A homologous tumor may yield in direct endometrial samples malignant epithelial and stromal cells and other malignant mesenchymal cells derived from the cells that are normally present in the uterus. (Fig. 7.18). A heterologous tumor may show in direct endometrial samples malignant epithelial and stromal cells and other malignant mesenchymal cells derived from cells that are not normally found in the uterus such as bone, cartilaginous, fat and striated muscle cells. It is important to note that in Pap smears only malignant glandular cells are identified in the majority of cases of endometrial MMMT. A 167 B C D Fig. 7.18. Homologous MMMT: A: Histology of the tumor. B-D. Malignant glandular/epithelial cells with prominent nuclei (B), malignant stromal cells present singly(C) and a syncytial cluster of malignant stromal cells with ill-defined cytoplasm and micronucleoli (D). 168 6. Leiomyosarcoma is a rare uterine tumor occurring mainly in adult women. The neoplasm peaks at ages 40 to 69, mean 54 years. Lung, bone and brain are the most common sites of metastatic deposits. When the tumor has invaded through the endometrium, malignant smooth muscle cells may be detected in cell samples collected from posterior vaginal fornix or in direct endometrial samples. 7. Metastatic cancers to the uterus are rare, with breast, gastrointestinal and kidney being the most common primary carcinomas followed by cutaneous melanoma. It may involve the myometrium but it may first appear in endometrial curettings, particularly in the case of lobular carcinoma of the breast. These neoplasms may be seen in cell samples collected from the posterior vaginal fornix or in direct endometrial samples. DIAGNOSTIC ACCURACY The Pap smear is not efficient in screening endometrial cancer as it will miss about 50% of cases. Criteria for assessing the cellular adequacy of direct endometrial samples vary among investigators. With Endocyte samplers, 10 to 15 endometrial fragments are required to consider a sample as adequate or representative for cytologic evaluation of a nonneoplastic lesion. However, for endometrial cancer, the presence of 5 or 6 groups of well-preserved cancer cells with 5 to 10 cells in each group is adequate for a confident diagnosis. Usually, about 10% of endometrial samples procured by Endocyte or Endopap samplers show inadequate cells for cytologic evaluation. The specificity of this technique is high, ranging from 81% to 100%, according to literature reviewed by Nguyen and Redburn. In the current practice of medicine, endometrial dating by direct endometrial cytology is not requested by gynecologists. The cytodiagnostic accuracy of endometrial carcinoma by direct endometrial sampling is high, ranging from 75% to 100%, according to several reports. Endometrial hyperplasia is difficult to identify cytologically and a broad diagnostic accuracy rate ranging from 31% to 97% have been reported. Finally, it should be borne in mind that patients with abnormal uterine bleeding, especially postmenopausal women, should have an endometrial biopsy or curettage to rule out an important endometrial pathology. 169 REFERENCES Al-Brahim N, Elavathil LJ. Metastatic lobular carcinoma to tamoxifen-associated endometrial polyp: case report and literature review. Ann Diagn Pathol. 2005;9:166. Becker SN, Wong JY. Detection of endometrial stromal sarcoma in cervicovaginal smears. Report of 3 cases. Acta Cytol 1981; 25:272. Byrne AJ. Endocyte endometrial smears in the cytodiagnosis of endometrial carcinoma. Acta Cytol 1990; 34: 373. Cibas ES. Cervical vaginal cytology. In Cytology. Diagnostic principles and clinical correlates. 3nd ed, 2009, Cibas ES, Ducatman BS, eds. Edinburgh, Sauders, p.1. DeMay RM. The Pap Test. Chicago, ASCP Press. 2005. DuBeshter B. Endometrial cancer: predictive value of cervical cytology. Gynecologic oncology. 1999;72:271. Fadare O, et al. The significance of benign endometrial cells in cervicovaginal smears. Adv Anat Pathol. 2005;12:274. Frable WJ. Screening for endometrial cancer?. Cancer (Cancer Cytopathol). 2008;114:219. Kapali M, et al. Routine endometrial sampling of asymptomatic premenopausal women shedding normal endometrial cells in Papanicolaou test is not cost effective. Cancer (Cancer Cytopathol). 2007;111:26. Kipp BR, et al. Direct uterine sampling with the Tao brush sampler using a liquid-based preparation method for the detection of endometrial cancer and atypical hyperplasia: a feasibility study. Cancer (Cancer Cytopathol). 2008:114:228. Mckenzie P, et al. Cytology of body of uterus. In Diagnostic Cytopathology. 2nd ed, 2003, Gray W and McKee GT, eds. Philadelphia, Churchill Livingstone, p. 821 Meisels A, Jolicoeur C. Criteria for the cytologic assessment of hyperplasias in endometrial samples obtained by the Endopap endometrial sampler. Acta Cytol 1985; 29: 297. Meisels A, et al. Endometrial hyperplasia and neoplasia. Cytologic screening with the Endopap endometrial sampler. J Reprod Med 1983; 28: 309. Nguyen GK, Redburn J. Endometrial cytology by direct sampling. Its value and limitations in the diagnosis of endometrial lesions. Pathol Annu. 1992; 30(2): 179. 170 Norimatsu Y, et al. Cellular features of endometrial hyperplasia and well differentiated adenocarcinoma using the Endocyte sampler. Diagnostic criteria based on cytoarchitecture of tissue fragments. Cancer(Cancer Cytopathol). 2006; 108:77. Pusiol T, et al. Prevalence and significance of psammoma bodies in cervicovaginal smears in cervical cancer screening program with emphasis on a case of primary bilateral ovarian psammocarcinoma. Cytojournal.2008;5:7. Ramzy I, Mody DR. Gynecologic cytology: practical considerations and limitations. Clin Lab Med 1991; 11: 271 Solomon D, Nayar R. The Bethesda System for Reporting Cervical Cytology. 2nd ed, 2004. New York, Springler. Zaman SS, et al. Efficacy of Endo–pap sampler in the detection of endometrial lesions: a review of 1983 cases. Acta Cytol 1993; 37:770 171 Chapter 8 Vaginal Lesions PROCUREMENT OF VAGINAL CELL SAMPLES Vaginal cell samples should be collected prior to all digital pelvic examinations as lubricant may obscure vaginal cell morphology. If an excessive mucous secretion is present, it should be removed by a cotton ball moistened with normal saline solution. Material is usually obtained from the posterior vaginal fornix of the patient in supine position. If a lesion is present, direct scraping of the lesion should be made. For detecting vaginal adenosis, circumvaginal scraping of the upper vagina or a four-quadrant downward scraping of the vaginal mucosa should be performed. The smears obtained are fixed with a commercial cytospray fixative for staining by the routine Papanicolaou method. CYTOLOGIC FINDINGS A. INFLAMMATORY LESIONS 1. Infectious diseases. Infections of the vagina are caused by the same microorganisms affecting the cervix. The reader is referred to Chapter 3 for discussion and illustrations of these disorders. 2. Noninfectious inflammatory lesions of the vagina include: atrophic vaginitis, malakoplakia and acute inflammation with granulation tissue formation of the vaginal vault following a total hysterectomy. Cytology of atrophic vaginitis is described in Chapter 2. Vaginal malakoplakia may be uni- or multifocal. Scraping of the lesion may reveal histiocytes containing round, basophilic, concentrically laminated, calcified Michaelis-Gutmann bodies that are diagnostic of the lesion. Granulation tissue in the vaginal vault following a total hysterectomy may show foreign body granuloma (suture material). A cell sample for the lesion may reveal inflammatory cells, repair squamous cells and suture material laden macrophages. (Fig. 8.1). 172 Fig. 8.1. A minute fragment of catgut surround by macrophages seen in a CP smear. B. NONNEOPLASTIC LESIONS 1. Vaginal endometriosis may be formed by implantation of viable endometrial cells discharged during the menstrual period. The lesion may appear as a bluish submucosal cyst with chocolate colored liquid contents. In typical cases of a scraping smear or fine needle aspiration, it yields fragments of endometrial epithelium, clusters of endometrial stromal cells, degenerated erythrocytes and a few hemosiderin laden macrophages. (Fig. 8.2). Fig. 8.2. Endometriosis in a vaginal scraping smear showing a fragment of benign endometrial epithelium. 173 2. Rectovaginal fistula may develop following a perforation of the rectal wall during a complicated total hysterectomy. Cell sample prepared from the vaginal secretion may reveal thick mucus containing columnar colonic epithelial cells, as well as digested food particles. (Fig. 8.3). A B Fig. 8.3. CP smear showing thick mucus containing columnar epithelial cells. 3. Vaginal adenosis is defined by the presence of either endocervical glandular epithelium or tuboendometrial epithelium in the vagina. It most commonly affects the upper third of the anterior vaginal wall and it is diagnosed histologically by tissue biopsy. Approximately 35% to 90% of female offspring with in utero exposure to diethylstilbestrol (DES), a synthetic estrogen that was commonly administered to women in early pregnancy to prevent spontaneous abortion in the late 1940s to the late 1950s, develop this lesion. However, vaginal adenosis is also found in women without in utero exposure to DES, and its incidence up to 41% has been reported. Vaginal adenosis containing endocervical-type epithelium is most 174 commonly encountered and it may also display squamous metaplasia. In a scraping smear, the lesion yields endocervical glandular cells singly, in loose clusters and in monolayered sheets. Metaplastic squamous cells may also be observed. (Fig. 8.4). Fig. 8.4. CP smear showing single and clustered columnar glandular cells exfoliated from a vaginal adenosis. 4. Glandular cells in posthysterectomy Pap smears Benign glandular cells similar to normal endocervical cells may be occasionally seen in Pap smears of patients with total hysterectomy. They are formed as the result of mucinous or glandular metaplasia of vaginal squamous epithelium, and they are more commonly seen in women with postoperative radiotherapy. (Fig. 8.5). Fig. 8.5. A cluster of benign glandular cells with mucus-filled clear cytoplasm in a CP smear of a woman with prior total radical hysterectomy and radiotherapy for endometrial carcinoma. C. PREMALIGNANT AND MALIGNANT LESIONS 175 1. Squamous intraepithelial lesions of the vagina or Vaginal intraepithelial neoplasia (VIN) are less common that those of the cervix. It is caused by HPV infection and is commonly associated with SIL or squamous cell carcinoma of the cervix or vulva. VIN lesions, or vaginal SILs, are histologically similar to those of the cervix and are also graded as VIN grade 1, 2 and 3 according to the criteria used for CIN or for LSIL and HSIL of the cervix. Vaginal SILs have cytomorphology similar to those of cervical SILs previously described in Chapter 3. 2. Postradiation dysplasia of vaginal mucosa more commonly develops following radiotherapy to the lower genital tract. The vaginal cells show radiation changes as described in Chapter 2. Postradiation dysplasia may develop after a latency period ranging from months to years. It exfoliates dyskaryotic squamous cells as seen in a cervical SIL. Diagnosis should be confirmed by tissue biopsy. (Fig. 8.6). Fig. 8.6. CP smear showing mildly dyskaryotic squamous cells from a postradiation dysplastic lesion of the vagina. 3. Primary carcinomas of the vagina are rare and account for 1% to 2% of female genital tract cancers. Of these, squamous cell carcinoma is the most common neoplasm. However, most vaginal squamous cancers represent an invasion of either a cervical or vulvar squamous carcinoma. i. Vaginal squamous cell carcinoma is most commonly of nonkeratinizing type, and its exfoliated cells are indistinguishable from those of the cervix of the same histologic type. ii. Vaginal adenocarcinoma of nonspecific type is extremely rare and in a scraping smear yields nonspecific malignant glandular cells. 176 iii. Clear cell adenocarcinoma (CCA) of the vagina accounts for 1% of all invasive carcinomas of the female genital tract. CCA developed as the result of intrauterine exposure to DES affects about 0.1% of patients up to 34 years of age. The mean age is 19.5 years with a range of 7 to 29 years. Vaginal CCA is often seen in association with adenosis but progression of adenosis to CCA has not been documented. CCA has 3 main histologic patterns: cystic, solid and papillary. Bulging of tumor cells in glandular lumens is referred to as a “hobnail pattern”, one of the characteristic histologic features of the tumor. In cytologic material, exfoliated tumor cells are present singly, in sheets and in aggregates with “bulging” cells. (Fig. 8.7). The cell cytoplasm is fragile, poorly stained, vacuolated and rich in glycogen. (Fig. 8.8). The reader is also referred to Chapter 5 for illustrations of more cases of CCA. A B 177 C Fig. 8.7. Vaginal clear cell carcinoma: A. Histology of the tumor showing a mixed glandular and solid pattern. B, C. CP smear showing sheets of tumor cells with clear, vacuolated or granular cytoplasm. Fig. 8.8. Vaginal CCA showing in CP smear an aggregate of tumor cells with prominent, large, hyperchromatic nuclei and ill-defined, granular cytoplasm displaying a vague ”bulging” pattern. 4. Primary nonepithelial malignant tumors of the vagina are very rare. Of these melanoma is the most common one in adult patients. In Pap smears it presents as single and loosely clustered pleomorphic malignant cells with variable cytoplasm that may contain intracytoplasmic melanin pigment granules (Fig. 8.9). When the tumor is amelanotic, a positive cytoplasmic reaction to S-100 protein, HMB-45 and MART-1 antibodies will confirm the diagnosis of melanoma. 178 Fig. 8.9. Vaginal melanotic melanoma showing in CP smear loosely clustered malignant cells with intracytoplasmic brownish melanin pigment granules. 5. Metastatic tumors are the most common tumor of the vagina. They may occur by direct extension or via hematogenous or lymphatic spread. The most common primary cancer sites are cervix, endometrium, ovary, colon and urinary bladder. These neoplasms are usually submucosal and may be diagnosed cytologically by transvaginal fine needle aspiration. 6. Benign tumors and Tumorlike lesions of the vagina include benign inclusion cyst (traumatic etiology), fibroepithelial polyp, fibroma, leiomyoma and hemangioma. These lesions are usually diagnosed by tissue biopsy and not by cytologic methods. 179 REFERENCES Benedet JL, et al. Primary invasive carcinoma of the vagina. Obstet Gynecol. 1983; 62:715. Davila RM, Miranda MC. Vaginal intraepithelial neoplasia and the Pap smear. Acta Cytol. 200: 44: 137. Ganesan R, et al. Vaginal adenosis in a patient on Tamoxifen therapy: a case report. Cytopathol. 1999; 10:127. Herbst AL, et al. Adenocarcinoma of the vagina. Association of maternal stilbestrol therapy with tumor appearance in young women. N Engl J Med. 1971; 284:878. Sodhani P, et al. Columnar and metaplastic cells in vault smears: cytologic and colposcopic study. Cytopathol. 1999;10:122. Tambouret R, et al. Benign glandular cells in post hysterectomy vaginal smear. Acta Cytol. 1998;42:1403. Tavassoli FA, Devilee P. WHO Classification of Tumours. Pathology and Genetics of Tumours of the Breast and Female genital organs (3rd ed). IARC Press: Lyon 2003. Vooijs GP, et al. The detection of vaginal adenosis and clear cell carcinoma. Acta Cytol. 1973; 17:59. 180 Chapter 9 Uterine Annexal Mass Lesions Uterine annexae include fallopian tubes and ovaries. Mass lesions of the uterine annexae are usually evaluated cytologically by fine needle aspiration. Depending on the clinical setting, a transabdominal, transvaginal or laparoscopic FNA is used. Cancer arising from the fallopian tube is rarely encountered in medical practice. It may exfoliate its cells into the uterine cavity and can be diagnosed by cytologic examination of material collected from the posterior vaginal fornix. INDICATIONS AND GOALS OF FNA OF UTERINE ANNEXAL MASS LESIONS 1. Confirming the benign nature of an ovarian cyst incidentally found during pregnancy and infertility evaluation. 2. Confirming an inoperable malignant ovarian tumor. 3. Documenting recurrence of a previously treated ovarian cancer. 4. Diagnosis of hydrosalpinx and fallopian tumors. 5. Diagnosis of a tubo-ovarian abscess CYTOLOGIC FINDINGS Fine needle aspirates are commonly obtained during laparoscopy. They may also be obtained by transvaginal, transrectal or transabdominal FNA under manual and/or ultrasound guidance. The cytologic samples obtained are prepared by direct smearing or by cytospin technique and stained by the Papanicolaou and/or Diff-Quik methods. Excessive material is used for cell block (CB) preparation for histologic evaluation. A. NONNEOPLASTIC LESIONS These lesions include follicular cyst, corpus luteum cyst, endometrioid cyst and simple cysts. 1. Follicular cysts can be solitary or multiple and are benign macroscopically and ultrasonographically. They are <3 to 8 cm in greatest dimension and lined by an inner layer of stratified granulosa cells and an outer layer of theca cells. In FNAs, it usually yields abundant isolated or clustered small, cuboidal benign cells with round 181 nuclei, coarse chromatin and scant or vacuolated cytoplasm. Mitosis and cells with pyknotic nuclei may be seen. (Fig. 9.1). Similar cells are also seen in an FNA of a granulosa cell tumor. Follicular cysts constitute a potential diagnostic pitfall as it may yield a high number of mitotic figures and bizarre appearing follicular cells. A B Fig. 9.1. Follicular cyst: A. Histology of a follicular cyst wall. B. FNA showing isolated, monomorphic, granulosa cells with bland, round nuclei and scant cytoplasm. 2. Corpus luteum cyst is unilocular and consists of large luteinized granulosa and theca interna cells. These cells are seen singly and in clusters in FNA. (Fig. 9.2). 182 Fig. 9.2: Cohesive clusters of luteinized granulosa cells showing bland, round nuclei and a moderate amount of granular cytoplasm in an FNA. 3. Endometriotic cyst is commonly called endometrioma which is a misnomer. This lesion is commonly discovered during laparoscopy for infertility work-up. Ovarian endometriotic cysts are bilateral in about 50% of cases and contain hemolysed blood (chocolate cyst). In FNAs, numerous hemosiderin-laden macrophages and hemolysed blood are found. (Fig. 9.3). Endometrial glandular and stromal cells are rarely seen in needle aspirates but minute fragments of endometrial tissue may be found in CB sections. A 183 B C Fig. 9.3. Endometrioid cyst: A. Histology of the endometrioid cyst. B, C. FNA showing fragments of benign epithelium admixed with hemolysed erythrocytes. 4. Simple cysts develop from invagination of mesothelium or surface epithelium of the ovary. They are usually small and most commonly seen in postmenopausal women. It can be seen in paraovarian or paratubal tissues. FNA of the lesion reveals clusters and large monolayered sheets of cuboidal epithelial cells or mesothelial cells with a honeycomb pattern. (Fig. 9.4). 184 Fig. 9.4. FNA showing monolayered benign epithelial sheets from a simple cyst. 5. Hydrosalpinx is caused by salpingitis and appears as a large cystic adnexal lesion. In an FNA it yields scanty cellular fluid containing a few benign ciliated cuboidal or columnar epithelial cells. (Fig. 9.5). Fig. 9.5. Ciliated epithelial cells in an FNA from a hydrosalpinx. (oil). 6. Tubo-ovarian abscess is the most severe result of pelvic inflammatory disease that is commonly caused by a Neisseria gonorrhoeae or Chlamydia trachomatitis ascending infection of the lower female genital tract. It yields in FNA purulent material consisting of numerous polymorphonuclear leukocytes and necrotic debris. B. EPITHELIAL OVARIAN TUMORS 185 These tumors can be benign or malignant, arise from the ovarian surface epithelium and account for 60% of all ovarian neoplasms. Ovarian carcinomas are the commonest and account for 80% to 90% of all primary ovarian cancers. Benign epithelial tumors are of the types serous cystadenoma, cystadenofibroma and mucinous cystadenoma. 1. Serous cystadenoma and cystadenofibroma account for about 20% of benign ovarian tumors. They are usually cystic and unilocular with clear fluid contents and can be bilateral in up to 20% of cases. The lesions are lined by a single layer of benign cuboidal epithelial cells with basally located nuclei. Columnar, clear or ciliated cells may be seen in FNAs of the lesions. Rarely, a few psammoma bodies are observed. (Fig. 9.6). A B Fig. 9.6. Serous cystadenoma: A. Histology of the tumor. B. FNA showing fragments of benign neoplastic epithelium. 2. Mucinous cystadenoma also accounts for about 20% of benign ovarian tumors. It is usually large and multiloculated and is rarely bilateral (2% to 3% of 186 cases). Its epithelial lining is smooth and of either endocervical or intestinal type. It shows in FNA a large amount of mucin and benign vacuolated mucus-secreting glandular cells that are seen singly, in sheets or ribbons. (Fig. 9.7). A B C Fig. 9.7. Mucinous cystadenoma: A. Histology of the lesion. B, C. FNA showing epithelial fragments, single and clustered mucus secreting benign epithelial cells. (Diff-Quik). 187 3. Brenner tumor accounts for about 2% of all ovarian tumors. The tumor is bilateral in 10% of cases and consists of nests of transitional or urothelial epithelium surrounded by fibrous tissue. Most Brenner tumors are benign and less than 2% of them are borderline or malignant. In FNA a benign Brenner tumor is characterized by single and clustered benign urothelial cells with defined, granular cytoplasm and oval nuclei. (Fig. 9.8). Nuclear grooves may be observed. A B Fig. 9.8. Benign Brenner tumor. A. Histology of the tumor. B. Tumor FNA showing single and clustered benign, spindle-shaped epithelial cells with bland, oval nuclei and ill-defined, granular cytoplasm. 4. Ovarian carcinomas consist of several types: • Serous cystadenocarcinomas are papillary and can be divided into borderline, 188 low-grade and high-grade tumors. These tumors are the most common ovarian cancers and usually occur in patients 40 to 60 years of ages. They are often bilateral, solid and cystic. Histologically, serous cystadenocarcinomas are composed of malignant non-mucus secreting cells forming papillary projections. Depending on the cellular atypia the tumor is graded as low- or high-grade. Stromal invasion is present and psammoma bodies are present in about one third of the cases. Low-grade serous cystadenocarcinoma and borderline serous tumors show morphologically similar neoplastic epithelium, but stromal invasion is absent in the latter. These tumors are rarely aspirated for initial diagnosis but they are sampled by FNA in recurrent and in inoperable tumors. Cytologically, a borderline serous tumor and low-grade serous cystadenocarcinoma are similar and consist of crowded sheets and irregular tridimensional clusters of epithelial cells with mild or moderate nuclear atypia, granular cytoplasm and psammoma bodies. (Fig. 9.9). A high-grade serous cystadenocarcinoma yields hypercellular material containing single and clustered large pleomorphic malignant epithelial cells with round nuclei and prominent nucleoli. Psammoma bodies may be observed. (Fig. 9.10). A 189 B Fig. 9.9. Low-grade serous carcinoma: A. Tumor histology. B. Tumor FNA showing cohesive groups and clusters of monomorphic tumor cells with small nucleoli. Fig. 9.10. High-grade serous carcinoma showing in FNA pleomorphic malignant cells with scant cytoplasm and prominent nucleoli admixed with a few macrophages. • Mucinous ovarian cystadenocarcinoma is less common than serous carcinoma and accounts for 5% to 10% of ovarian cancers. The tumor can occur at any age but is most commonly in patients 40 to 60 years of age. About 75% of the tumors are bilateral. It is large and mutiloculated and consists of solid areas of mucus-secreting malignant cells and areas with papillary formation. As in serous carcinomas, mucinous carcinomas are graded as low-and high-graded tumors. The borderline variant is cytologically similar to a low-grade tumor but it shows no stromal invasion histologically. 190 In FNA a borderline or low-grade mucinous cystadenocarcinoma shows clustered tumor cells with nuclear crowding and mild nuclear atypia. A high-grade tumor yields single and clustered pleomorphic malignant epithelial cells. Mucin production is not evident in routinely stained smears but can be demonstrated with mucicarmine and periodic acid-Shiff stain with prior diastase digestion. • Endometrioid carcinoma accounts for about 20% of ovarian cancers and is bilateral in about 30% of cases. About 30% of patients with this neoplasm have endometriosis and 25% of them have a coexisting endometrioid uterine carcinoma. In FNAs it yields numerous malignant cells singly and in crowded clusters, similar to those seen in an FNA of an ovarian serous cystadenocarcinoma. • Clear cell carcinoma is a rare neoplasm and may arise from a focus of endometriosis. It yields in FNA sheets of malignant epithelial cells with prominent nucleoli and clear cytoplasm. (Fig. 9.11). A B Fig. 9.11. Clear cell carcinoma: 191 A. Histology of the tumor. B. Tumor FNA showing a sheet of malignant epithelial cells with clear or vacuolated, ill-defined cytoplasm, large, oval nuclei and prominent nucleoli. (Diff-Quik). 5. Other Primary Ovarian Cancers • Germ cell tumors account for 20% of all ovarian tumors. They consist of several histologic types: teratoma (mature and immature), dysgerminoma, embryonal carcinoma, endodermal sinus tumor and choriocarcinoma. • Mature teratoma is the most common germ cell neoplasm. Histologically,as well as in FNA smears, it shows abundant anucleated squamous cells admixed with benign glandular cells. Immature teratoma is rare, malignant and usually solid. It is characterized by immature or embryonal tissue (usually neuroectodermal derivatives) admixed with benign elements as seen in its mature counterpart. • Dysgerminoma accounts for 1% to 5% of all ovarian cancers and about 40% of all malignant ovarian germ cell tumors. It occurs most frequently in young women under the age of 30 years. It yields in FNA single and loose aggregates of large tumor cells with ill-defined, variable cytoplasm and prominent nucleoli admixed with benign lymphoid cells. (Fig. 9.12). A “tiger-strip” background is only observed in air-dried smears with Romanowsky-type staining. A 192 B Fig. 9:12.Dysgerminoma: A. Histology of the tumor. B. Pleomorphic malignant cells with ill-defined, granular cytoplasm and prominent nucleoli in FNA of the above-illustrated tumor. (HE). • Embryonal carcinoma is a rare ovarian germ cell tumor. It is characterized cytologically by cohesive clusters of pleomorphic malignant cells with ill-defined cytoplasm and large nuclei with prominent nucleoli. (Fig. 9.13). Fig. 9.13. Cohesive cluster of pleomorphic malignant cells with prominent nucleoli in FNA of an ovarian embryonal carcinoma. (HE). • Endodermal sinus or Yolk-sac tumor is an uncommon ovarian germ cell tumor. Itshows in FNA cells resembling those of a poorly differentiated adenocarcinoma. Dense eosinophilic intracytoplasmic inclusions (alpha-fetoprotein) may be seen in some cells. A rare ovarian choriocarcinoma shows in FNA bizarre malignant cells that stain positively with human chorionic gonadotropin antibody. 193 • Sex Cord-Stromal tumors account for about 8% of all ovarian tumors and include neoplasms arising from granulosa cells, Sertoli-Leydig cells and fibroblastic cells. - Granulosa cell tumor of adult type accounts for almost all granulosa cell tumors in adults. It occurs more commonly in postmenopausal women and is a slow growing tumor with metastatic potential. It secretes estrogen causing endometrial hyperplasia. Some tumors have an associated ascitis that usually shows no tumor cells. In FNA several clusters and sheets of small or medium tumor cells with ill-defined cytoplasm, round nuclei and conspicuous nucleoli are seen. Tumor cells arranged in acini may be observed. (Fig. 9.14). Nuclear grooves may be seen but tumor cells forming Call-Exner bodies are rarely seen. These tumor cells are morphologically similar to those of a follicular ovarian cyst. - Sertoli-Leydig cell tumor and annular sex cord tumor yield cells that are cytologically similar to those of a granulosa cell tumor. -Thecoma and Fibroma are uncommon neoplasms that are usually non-functional and their needle aspirates are usually acellular. A 194 B Fig. 9.14. Granulosa cell tumor: A. Histology of the tumor. B. Single and clustered tumor cells with scant cytoplasm and oval nuclei showing inconspicuous nucleoli and occasional nuclear grooves. 6. Metastatic Tumors Tumors that most commonly metastasize to the ovary arise from the urogenital tract, colon, stomach and breast. About 15% to 20% of bilateral ovarian tumors are metastatic cancers. The well-known Krukenberg tumor is characterized by mucus-secreting signet-ring cells metastatic to the ovary. The majority of these tumors have a gastric primary. (Fig. 9.15). A 195 B Fig. 9.15. Metastatic gastric signet-ring cell carcinoma: A. Histology of the tumor. B. FNA showing thick mucus containing single and clustered malignant tumor cells with large intracytoplasmic mucinous vacuoles pushing the tumor cell nuclei to the cell periphery. A metastatic colonic adenocarcinoma yields in FNA sheets of malignant glandular cells in a “dirty” necrotic background. Tumor cell nuclei arranged in vague palisade may be visualized at the periphery of the aspirated tumor epithelial fragments. (Fig. 9.16). A 196 B Fig.9.16. Metastatic colonic adenocarcinoma to the ovary. A. Histology of the tumor. B. A cohesive malignant epithelial fragment in the tumor FNA. DIAGNOSTIC ACCURACY OF OVARIAN CANCERS An inadequate cell sample varying from 13% to 73% had been reported by some studies. According an extensive review by Cibas, the diagnostic sensitivity of malignancy in different reported series varied widely. It is low, at 26% to 40% if borderline ovarian tumors with a suspicious diagnosis were included with malignant tumors, and it ranges from 84% to 93% if those tumors are excluded from the malignant category. Both false-negative and false-positive diagnoses have been reported. The former occurs when a cellular sample from a follicular cyst shows abundant mitotic figures, and the latter happens mainly in borderline ovarian neoplasms. FALLOPIAN TUBE CARCINOMA Primary fallopian tube carcinoma is a very rare neoplasm. Most patients are postmenopausal and nulliparous. Abnormal vaginal discharge or bleeding is a common clinical manifestation. Histologically, the cancer is of papillary serous type. It may shed cancer cells into the uterine cavity and these cells may be accumulated in the posterior vaginal fornix. The tumor cells commonly display features of a serous carcinoma and often have a vacuolated cytoplasm. (Fig. 9.17). Psammoma bodies may be observed. A fallopian tube cancer should be suspected in an adult woman who shows malignant glandular cells in her Pap test and who also has a negative cone biopsy and endometrial curettage and who has no known primary cancer. Fallopian tube carcinoma may also be diagnosed cytologically by laparoscopic FNA. 197 Fig. 9.17. A fallopian adenocarcinoma showing in posterior vaginal fornix CP a loose cluster of malignant epithelial cells with a moderate amount of granular cytoplasm and prominent nucleoli. A tumor cell with vacuolated cytoplasm is present. 198 REFERENCES Angstrom T, et al. The cytologic diagnosis of ovarian tumors by means of aspiration biopsy. Acta Cytol.1972; 16:336. Benson PA. Cytologic diagnosis in primary carcinoma of fallopian tube. Case report and review. Acta Cytol. 1974; 18:429. Brenda JA, Zaleski S. Fine needle aspiration cytologic features of hepatic metastasis of granulose cell tumor of the ovary: differential diagnosis. Acta Cytol. 1988; 32:527. Cibas ES. Ovary. In Cytology. Diagnostic Principles and Clinical Correlates. 3nd ed, 2009, Cibas ES and Ducatman BS, eds, Philadelphia ,Saunders Elsevier, p.433. DeMay RM. The Pap Test, 2005, Chicago, ASCP Press. Ehya M, Lang WR. Cytology of granulosa cell tumor of the ovary. Am J Clin Pathol. 1986;85:402. Harai Y, et al. Clinical and cytologic aspects of primary fallopian tube carcinoma. A report of 10 cases. Acta Cytol. 1987; 31:834. King A, et al. Fallopian tube carcinoma: a clinicopathological study of 17 cases. Gynecol Oncol. 1989; 33: 351. Kjellgren O, et al. Fine needle aspiration in diagnosis and classification of ovarian carcinoma. Cancer.1971; 28: 967. Kjellgren O, Angstrom T. Transvaginal and transrectal aspiration biopsy in diagnosis and classification of ovarian tumors. In Aspiration Biopsy Cytology, Part 2, Cytology of Infradiaphragmatic Organs, Zajicek J, ed, Basel, S Karger, 1979. Kovacic J, et al. Aspiration cytology of normal structures and non-neoplastic cysts of the ovary. In Pathology of the Female Genital Tract, 2nd ed, 1982, Blaustein A, ed, p.716. Mintz M. Ponctions de 94 kystes para-uterins sous coelioscopie et etude cytologique des liquids. Gynaecologia.1967; 163:61. Mulvany NJ. Aspiration cytology of ovarian cyst and cystic neoplasms. A study of 235 aspirates. Acta Cytol.1996; 40:911. Nadji M, et al. Fine needle aspiration cytology in gynecologic oncology. II. Morpholgic aspects. Acta Cytol.1979; 23:380. 199 Nadji M. Aspiration cytology in diagnosis and assessment of ovarian neoplasms. In Tumors and Tumorlike Conditions of the Ovary. Roth LM and Czernobilsky B, eds. New York, Churchill Livingstone, 1985, p.153. Nguyen GK, Redburn J. Aspiration biopsy cytology of granulosa cell tumor of the ovary. Diagn Cytopathol. 1992; 8:253. Nunez C, Diaz JI. Ovarian follicular cysts: a potential source of false positive diagnoses in ovarian cytology. Diagn Cytopathol.1992; 8:532. Ramzy I, Delaney M. Fine needle aspiration of ovarian masses. I. Correlative cytologic and histologic study of celomic epithelial neoplasms. Acta Cytol. 1979; 23:97. Selvaggi SM. Fine needle aspiration cytology of ovarian follicle cysts with cellular atypia from reproductive-age patients. Diagn Cytopathol.1991; 7:189. Sevin BU, et al. Fine needle aspiration cytology in gynecologic oncology. I. Clinical aspects. Acta Cytol.1979; 23:277. Watson G, et al. Fine needle aspiration of benign and malignant gynecological lesions. In Diagnostic Cytopatholgy, 2nd ed, 2003, Gray W and Mckee GT, eds, Philadelphia, Churchill Livingstone, p.859. Yee H, et al. Transvaginal sonographic characterization combined with cytologic evaluation in the diagnosis of ovarian and adnexal cysts. Diagn Cytopathol. 1994;10:107. The End |
16048 | https://mocktheorytest.com/resources/how-do-you-calculate-speed-vs-time/ | How do you calculate speed vs distance vs time?
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Home›Advice›How do you calculate speed vs distance vs time?
How do you calculate speed vs distance vs time?
Sometimes you’ll need to estimate how long it will take to get somewhere. Maybe you’re planning a journey, maybe you’ve got a maths test. There are quick ways to calculate speed and time.
First, you’ll need to know how to change decimal time into hours, minutes and seconds and vice versa.
To convert standard time into decimal time, let’s take 3 hours 14 minutes 18 seconds. To convert to decimal is straightforward:
18 seconds /60 = 0.3 (i.e. 18 seconds is 0.3 of a minute). Add that to the 14 minutes
14.3 minutes / 60 = 0.238 (i.e. 14 minutes and 18 seconds is 0.238 of an hour)
Add this to the original 3 hours = 3.238 hours.
If you have your time in decimal format already, you do the calculations in reverse. Let’s say we have 2.74 hours. Remove the 2 and you’re left with .74. Multiply this by 60 to convert the decimal time to minutes = 44.4 minutes. Unfortunately, we have 0.4 minutes which doesn’t tell us the exact number of seconds. Take 0.4 and multiply it by 60 to convert to seconds = 24 seconds. Therefore we have 2 hours 44 minutes and 24 seconds.
How far will a car travel in a certain time, given a certain speed?
If you are calculating in km/h, simply replace mph with km/h.
If a car travels at 60mph, how far away will it be in one hour? The clue is in the question – 60 miles per hour means 60 miles in one hour. However, clarify with the person whether the distance they mean is the distance of the road or the distance as the crow flies. Roads are rarely straight for 60 miles, therefore the person could have driven 60 miles but only be 50 miles away from the initial starting point.
Distance = Speed Time
For example, a car is travelling at 45 miles per hour for 1 hour 40 minutes.
First, convert 1 hour 40 minutes into decimal time:
40 = 100 minutes. 0.67 hours. Add the 0.67 onto the original hour and you have 1.67 hours.
Therefore distance = 45 1.67 = 75 miles.
How fast is a car travelling if it covers a certain distance in a certain time?
Speed = Distance / Time
For example, a car travels 32 miles in 26 minutes. How fast is it going?
First, convert 26 minutes to hours: 26/60 = 0.43
Therefore 32/0.43 = 73.8 miles per hour.
How much time will it take for a car to travel a certain distance at a certain speed?
Time = Distance / Speed
For example, a car is travelling at an average of 57mph for 88 miles. How long does it take?
Therefore 88 / 57 = 1.544 hours.
We know we’ve got at least one hour, but we need to convert the 0.544 to minutes. Therefore, multiply 0.544 by 60 = 32.63.
Now we have our number of minutes, but it’s still in decimals and we want to know the number of seconds. Take the .63 and multiply by 60 = 37.9
Therefore, the total time taken is 1 hour 32 minutes, 37.9 seconds.
Other useful conversions
Miles = km / 1.60934
Km = miles 1.60934
For example, 32 miles in km is 32 1.60934 = 51.5
There are 1000 metres in 1 km and 1609.34 metres in 1 mile.
By Darren Cottingham
Darren has owned several companies in the automotive, advertising and education industries. He has run driving theory educational websites since 2010.
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16049 | https://www.youtube.com/watch?v=6KYp4zjqT6s | Combinations Formula: Counting the number of ways to choose r items from n items.
Dr. Trefor Bazett
506000 subscribers
469 likes
Description
26090 views
Posted: 22 Jul 2017
We have previously looked at the permutation formula to PICK r items from n items, when we care about the order they come out. Now we look at how to CHOOSE r items from n items, where we don't care about the order they come out. We derive the combination example formula and do an example
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Introduction in our various explorations of counting and probability so far we've typically cared about the order in which we pick different choices so for example we might talk about passwords where we care what is the first digit what is the second digit what is the third digit and so on but in a whole bunch of different applications where we're wanting to do counting or we'll want to do probability as well we don't care about the order in which things come up so let me take as Example an example I'm trying to assemble a team of people I'm trying to assemble a group of three people who are gonna work on some team and I've got this larger possible five people that I could choose the three people from I don't care who I choose first I don't care what you second the order in which I choose them doesn't matter and so the the keyword that's in this particular expression here is going to be this word choose and whenever you see the word choose you should be thinking I don't care which way they come up this is in contrast to the word pick that we've seen before if I say how many ways can I pick three people from five people where I'm caring who comes first and I care who comes second then that's one where order matters why don't you say I want to choose them doesn't matter who my team's going to be now if this problem was one where we didn't care about order then we could have done it by the method so we've seen before we would say something like this that for the first person that there's five possibilities that for the second person one's already used up so that there's four possibilities and that for the third one where you've used up - there's only three possibilities and then we've seen that this particular expression five times four times three was going to be the same thing as if I picked and we use the Peter to note it from five items a total of three items and and that this has the broader formula the larger number factorial divided by five minus R so in this case five minus three factorial and that those were all going to be the same so that's how I would have solved the old problem if I did care about order so my strategy now for for this problem if I if I return back to the original one is I'm going to take the number of ways that I could count this when I do care about the order and I need to divide it by the number of ways I can order it in other words I'm going to try and figure out what p53 is we've looked at that but I'm gonna then divide it out by whatever the number of ways that I can order this all right well if I've got three people in a team then how many ways are there to reorder them that's the problem we've actually looked at in the past there's there's three possibilities whoo-hoo-hoo I could put first and then there's two possibilities for who I could put second and there's only one possibility for who I can put third some of the words this p53 which we had computed before five times four times three on the top five times four times three on the top it divides out by the number of ways that I could reorder those three things three for the first option 2 for the second option and one for the third in other words what we have here is pick 5 comma 3 the number of ways I can pick three things from five things divided out by three factorial so for example if I had chosen Chris Christine and Cory then those were going to be the three people on my team then what I'm doing when I divide about by the three factorial is saying the choice where I go Chris Christine Cory in the choice Rhaego Chris Cory Christine where go pristine kritis Cory and all the different ways that I can add those up and and permute that order that all of those are the same it's the same three people on my team eventually and so I'm just dividing out by the different number of ways that I could reorder them so we think of this is picking is something where order matters so I can do that computation first that we've seen before and then I I don't care about order so I divide out by the number of ways to reorder it more Notation generally I can take the the lesson of this example and I can give a notation that is typically done like this it is usually written and on the top and then R on the bottom and you put brackets around this oh so this notation is is sometimes going to also be re-written as choose n comma R to be sort of compatible with our pick and comma R notation we've seen before but but the most common way to say how to choose R things from N Things is this notation with these brackets you put them with each other and effectively what it is going to be defined as is the way that I pick from n things are things divided out by the number of ways that I can rearrange and so R factorial on the bottom and then since I have a formula for this picking formula we know that that was going to be I'll leave the 1 over R factorial on the bottom of the front and I'll just deal with the pink portion this was an N factorial on the top divided by n minus R factorial or perhaps if I want to clean it up a little bit n factorial on the top divided by R factorial on the bottom multiplied by n minus R factorial also on the bottom and so this formula that I've got in here this is how I choose our objects are as the smaller one our objects from a total of n without caring about my order |
16050 | https://www.youtube.com/watch?v=bxOA6qukOoo | Nut Sort Level 71 Walkthrough | Nuts — Color Sort 71 solution
SolutionGuruji
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Posted: 24 May 2024
Nut Sort Level 71 Walkthrough
Nuts — Color Sort Puzzle Games
Play Brainteaser Nuts Color Sort Puzzle – a sorting game designed to engage and relax you with different levels.
In this addictive sort colors puzzle game, where every level offers an exciting challenge.
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16051 | https://www.teachstarter.com/us/teaching-resource/equivalent-ratios-mystery-picture-worksheet/ | Equivalent Ratios – Mystery Picture Worksheet | Teach Starter
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teaching resource
Equivalent Ratios – Mystery Picture Worksheet
Updated:31 May 2023
Practice how to find equivalent ratios with this color-by-number worksheet.
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Curriculum:CCSS, TEKS
Grade:6
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CCSS.MATH.CONTENT.6.RP.A.1
Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, "The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak." "For every ...
1. Grade 6
2. Standards for Mathematical Practice
3. Ratios & Proportional Relationships
4. Understand ratio concepts and use ratio reasoning to solve problems
CCSS.MATH.CONTENT.6.RP.A.2
Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0, and use rate language in the context of a ratio relationship. For example, "This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flou...
1. Grade 6
2. Standards for Mathematical Practice
3. Ratios & Proportional Relationships
4. Understand ratio concepts and use ratio reasoning to solve problems
CCSS.MATH.CONTENT.6.RP.A.3
Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
1. Grade 6
2. Standards for Mathematical Practice
3. Ratios & Proportional Relationships
4. Understand ratio concepts and use ratio reasoning to solve problems
MATH 6.4(C)
Give examples of ratios as multiplicative comparisons of two quantities describing the same attribute;
1. TEKS Math
2. TEKS Math 6
3. Math 6.4
Available on the Plus Plan
teaching resource
Equivalent Ratios – Mystery Picture Worksheet
Updated:31 May 2023
Practice how to find equivalent ratios with this color-by-number worksheet.
Editable:Google Slides
Non-Editable:PDF
Pages:1 Page
Curriculum:CCSS, TEKS
Grade:6
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Practice how to find equivalent ratios with this color-by-number worksheet.
Equivalent Ratios Worksheet
Equivalent ratios are ratios with the same value, even though the numbers in the ratio may differ. For example, the ratios 2:4 and 4:8 are equivalent because both have a value of 0.5 (or 1/2). To find equivalent ratios, you can multiply or divide both numbers in a ratio by the same number. If you are looking for an activity that your students can complete to practice this 6th-grade math skill. You have come to the right place!
With this worksheet, students will complete 36 problems where they must determine the missing number to complete the equivalent ratio. Students will then use the color key to color in the grid to reveal a mystery image.
An answer key is included with your download to make grading fast and easy!
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To save paper, we suggest printing the question pages double-sided.
Check out some of our other resources that cover ratios!
This resource was created by Cassandra Friesen, a teacher in Colorado and a Teach Starter Collaborator.
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Equivalent Ratios – Math Mazes
Practice how to find equivalent ratios with this set of math mazes.
1 page Grade: 6
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Equivalent Ratios – Worksheet
Use this worksheet to demonstrate an understanding of proportional relationships by calculating unknown values with equivalent ratios.
2 pages Grade: 6
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Writing Ratios – Task Cards
Practice how to write a ratio in different ways with this set of 24 task cards.
1 page Grade: 6
Curriculum
CCSS TEKS
CCSS.MATH.CONTENT.6.RP.A.1
Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, "The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak." "For every ...
1. Grade 6
2. Standards for Mathematical Practice
3. Ratios & Proportional Relationships
4. Understand ratio concepts and use ratio reasoning to solve problems
CCSS.MATH.CONTENT.6.RP.A.2
Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0, and use rate language in the context of a ratio relationship. For example, "This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flou...
1. Grade 6
2. Standards for Mathematical Practice
3. Ratios & Proportional Relationships
4. Understand ratio concepts and use ratio reasoning to solve problems
CCSS.MATH.CONTENT.6.RP.A.3
Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
1. Grade 6
2. Standards for Mathematical Practice
3. Ratios & Proportional Relationships
4. Understand ratio concepts and use ratio reasoning to solve problems
MATH 6.4(C)
Give examples of ratios as multiplicative comparisons of two quantities describing the same attribute;
1. TEKS Math
2. TEKS Math 6
3. Math 6.4
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16052 | https://pressbooks.bccampus.ca/introductorygeneralphysics2phys1207/chapter/14-2-temperature-change-and-heat-capacity/ | Skip to content
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Chapter 5 Temperature, Kinetic Theory, and the Gas Laws
5.4 Temperature Change and Heat Capacity
Summary
Observe heat transfer and change in temperature and mass.
Calculate final temperature after heat transfer between two objects.
One of the major effects of heat transfer is temperature change: heating increases the temperature while cooling decreases it. We assume that there is no phase change and that no work is done on or by the system. Experiments show that the transferred heat depends on three factors—the change in temperature, the mass of the system, and the substance and phase of the substance.
The dependence on temperature change and mass are easily understood. Owing to the fact that the (average) kinetic energy of an atom or molecule is proportional to the absolute temperature, the internal energy of a system is proportional to the absolute temperature and the number of atoms or molecules. Owing to the fact that the transferred heat is equal to the change in the internal energy, the heat is proportional to the mass of the substance and the temperature change. The transferred heat also depends on the substance so that, for example, the heat necessary to raise the temperature is less for alcohol than for water. For the same substance, the transferred heat also depends on the phase (gas, liquid, or solid).
HEAT TRANSFER AND ENERGY CHANGE
The quantitative relationship between heat transfer and temperature change contains all three factors:
where Q or ΔQ is the symbol for heat transfer, m is the mass of the substance, and is the change in temperature. The symbol cstands for specific heat and depends on the material and phase. The specific heat is the amount of heat necessary to change the temperature of 1.00 kg of mass by 1.00 oC. The specific heat cc is a property of the substance; its SI unit is or Recall that the temperature change ΔT is the same in units of kelvin and degrees Celsius. If heat transfer is measured in kilocalories, then the unit of specific heat is
Values of specific heat must generally be looked up in tables, because there is no simple way to calculate them. In general, the specific heat also depends on the temperature. Table 1 lists representative values of specific heat for various substances. Except for gases, the temperature and volume dependence of the specific heat of most substances is weak. We see from this table that the specific heat of water is five times that of glass and ten times that of iron, which means that it takes five times as much heat to raise the temperature of water the same amount as for glass and ten times as much heat to raise the temperature of water as for iron. In fact, water has one of the largest specific heats of any material, which is important for sustaining life on Earth.
Example 1: Calculating the Required Heat: Heating Water in an Aluminum Pan
A 0.500 kg aluminum pan on a stove is used to heat 0.250 litres of water from 20.0 oCto 80.0 oC(a) How much heat is required? What percentage of the heat is used to raise the temperature of (b) the pan and (c) the water?
Strategy
The pan and the water are always at the same temperature. When you put the pan on the stove, the temperature of the water and the pan is increased by the same amount. We use the equation for the heat transfer for the given temperature change and mass of water and aluminum. The specific heat values for water and aluminum are given in Table 1.
Solution
Because water is in thermal contact with the aluminum, the pan and the water are at the same temperature.
Calculate the temperature difference:
Calculate the mass of water. Because the density of water is one liter of water has a mass of 1 kg, and the mass of 0.250 liters of water is m w = 0.250 kg.
Calculate the heat transferred to the water. Use the specific heat of water in Table 1:
Calculate the heat transferred to the aluminum. Use the specific heat for aluminum in Table 1:
Compare the percentage of heat going into the pan versus that going into the water. First, find the total transferred heat:
Thus, the amount of heat going into heating the pan is
and the amount going into heating the water is
Discussion
In this example, the heat transferred to the container is a significant fraction of the total transferred heat. Although the mass of the pan is twice that of the water, the specific heat of water is over four times greater than that of aluminum. Therefore, it takes a bit more than twice the heat to achieve the given temperature change for the water as compared to the aluminum pan.
Figure 2. The smoking brakes on this truck are a visible evidence of the mechanical equivalent of heat.
Example 2: Calculating the Temperature Increase from the Work Done on a Substance: Truck Brakes Overheat on Downhill Runs
Truck brakes used to control speed on a downhill run do work, converting gravitational potential energy into increased internal energy (higher temperature) of the brake material. This conversion prevents the gravitational potential energy from being converted into kinetic energy of the truck. The problem is that the mass of the truck is large compared with that of the brake material absorbing the energy, and the temperature increase may occur too fast for sufficient heat to transfer from the brakes to the environment.
Calculate the temperature increase of 100 kg of brake material with an average specific heat of 800 J / kg • oC if the material retains 10% of the energy from a 10,000-kg truck descending 75.0 m (in vertical displacement) at a constant speed.
Strategy
If the brakes are not applied, gravitational potential energy is converted into kinetic energy. When brakes are applied, gravitational potential energy is converted into internal energy of the brake material. We first calculate the gravitational potential energy Mghthat the entire truck loses in its descent and then find the temperature increase produced in the brake material alone.
Solution
Calculate the change in gravitational potential energy as the truck goes downhill
Calculate the temperature from the heat transferred using Q = Mghand
where mis the mass of the brake material. Insert the values m = 100 kg and c = 800 J / kg • oC to find
= 92 oC.
Discussion
This temperature is close to the boiling point of water. If the truck had been traveling for some time, then just before the descent, the brake temperature would likely be higher than the ambient temperature. The temperature increase in the descent would likely raise the temperature of the brake material above the boiling point of water, so this technique is not practical. However, the same idea underlies the recent hybrid technology of cars, where mechanical energy (gravitational potential energy) is converted by the brakes into electrical energy (battery).
| Substances | Specific heat (c) |
--- |
| Solids | J/kg ºC | kcal/kg ºC2 |
| Aluminum | 900 | 0.215 |
| Asbestos | 800 | 0.19 |
| Concrete, granite (average) | 840 | 0.20 |
| Copper | 387 | 0.0924 |
| Glass | 840 | 0.20 |
| Gold | 129 | 0.0308 |
| Human body (average at 37 °C) | 3500 | 0.83 |
| Ice (average, -50°C to 0°C) | 2090 | 0.50 |
| Iron, steel | 452 | 0.108 |
| Lead | 128 | 0.0305 |
| Silver | 235 | 0.0562 |
| Wood | 1700 | 0.4 |
| Liquids |
| Benzene | 1740 | 0.415 |
| Ethanol | 2450 | 0.586 |
| Glycerin | 2410 | 0.576 |
| Mercury | 139 | 0.0333 |
| Water (15.0 °C) | 4186 | 1.000 |
| Gases 3 |
| Air (dry) | 721 (1015) | 0.172 (0.242) |
| Ammonia | 1670 (2190) | 0.399 (0.523) |
| Carbon dioxide | 638 (833) | 0.152 (0.199) |
| Nitrogen | 739 (1040) | 0.177 (0.248) |
| Oxygen | 651 (913) | 0.156 (0.218) |
| Steam (100°C) | 1520 (2020) | 0.363 (0.482) |
| Table 1. Specific Heats1 of Various Substances |
Note that Example 2 is an illustration of the mechanical equivalent of heat. Alternatively, the temperature increase could be produced by a blow torch instead of mechanically.
Example 3: Calculating the Final Temperature When Heat Is Transferred Between Two Bodies: Pouring Cold Water in a Hot Pan
Suppose you pour 0.250 kg of 20.0 oCwater (about a cup) into a 0.500-kg aluminum pan off the stove with a temperature of 150 oC.Assume that the pan is placed on an insulated pad and that a negligible amount of water boils off. What is the temperature when the water and pan reach thermal equilibrium a short time later?
Strategy
The pan is placed on an insulated pad so that little heat transfer occurs with the surroundings. Originally the pan and water are not in thermal equilibrium: the pan is at a higher temperature than the water. Heat transfer then restores thermal equilibrium once the water and pan are in contact. Because heat transfer between the pan and water takes place rapidly, the mass of evaporated water is negligible and the magnitude of the heat lost by the pan is equal to the heat gained by the water. The exchange of heat stops once a thermal equilibrium between the pan and the water is achieved. The heat exchange can be written as
Solution
Use the equation for heat transferto express the heat lost by the aluminum pan in terms of the mass of the pan, the specific heat of aluminum, the initial temperature of the pan, and the final temperature:
Express the heat gained by the water in terms of the mass of the water, the specific heat of water, the initial temperature of the water and the final temperature:
Note thatandand that they must sum to zero because the heat lost by the hot pan must be the same as the heat gained by the cold water:
This an equation for the unknown final temperature, Tf .
Bring all terms involving Tfon the left hand side and all other terms on the right hand side. Solve for Tf,
and insert the numerical values:
Discussion
This is a typical calorimetry problem—two bodies at different temperatures are brought in contact with each other and exchange heat until a common temperature is reached. Why is the final temperature so much closer to than The reason is that water has a greater specific heat than most common substances and thus undergoes a small temperature change for a given heat transfer. A large body of water, such as a lake, requires a large amount of heat to increase its temperature appreciably. This explains why the temperature of a lake stays relatively constant during a day even when the temperature change of the air is large. However, the water temperature does change over longer times (e.g., summer to winter).
TAKE-HOME EXPERIMENT: TEMPERATURE CHANGE OF LAND AND WATER
What heats faster, land or water?
To study differences in heat capacity:
Place equal masses of dry sand (or soil) and water at the same temperature into two small jars. (The average density of soil or sand is about 1.6 times that of water, so you can achieve approximately equal masses by using more water by volume.)
Heat both (using an oven or a heat lamp) for the same amount of time.
Record the final temperature of the two masses.
Now bring both jars to the same temperature by heating for a longer period of time.
Remove the jars from the heat source and measure their temperature every 5 minutes for about 30 minutes.
Which sample cools off the fastest? This activity replicates the phenomena responsible for land breezes and sea breezes.
Check Your Understanding
1: If 25 kJ is necessary to raise the temperature of a block from 25 oC to 30oC,how much heat is necessary to heat the block from 45 oC to 50 oC?
Summary
The transfer of heat Q that leads to a change ΔT in the temperature of a body with mass m is Q = m c ΔT, where c is the specific heat of the material. This relationship can also be considered as the definition of specific heat.
Conceptual Questions
1: What three factors affect the heat transfer that is necessary to change an object’s temperature?
2: The brakes in a car increase in temperature by ΔT when bringing the car to rest from a speed v. How much greater would ΔT be if the car initially had twice the speed? You may assume the car to stop sufficiently fast so that no heat transfers out of the brakes.
Problems & Exercises
1: On a hot day, the temperature of an 80,000-L swimming pool increases by 1.50 oC. What is the net heat transfer during this heating? Ignore any complications, such as loss of water by evaporation.
2: Show that
3: To sterilize a 50.0-g glass baby bottle, we must raise its temperature from 22.0 oC to 95.0 oC. How much heat transfer is required?
4: The same heat transfer into identical masses of different substances produces different temperature changes. Calculate the final temperature when 1.00 kcal of heat transfers into 1.00 kg of the following, originally at 20.0o C (a) water; (b) concrete; (c) steel; and (d) mercury.
5: Rubbing your hands together warms them by converting work into thermal energy. If a woman rubs her hands back and forth for a total of 20 rubs, at a distance of 7.50 cm per rub, and with an average frictional force of 40.0 N, what is the temperature increase? The mass of tissues warmed is only 0.100 kg, mostly in the palms and fingers. Remember the the work done by friction turns into heat energy. You need to find the work done by friction first. Remember your mechanics from the earlier chapters.
6: A 0.250-kg block of a pure material is heated from20.0oC to 65.0oC by the addition of 4.35 kJ of energy. Calculate its specific heat and identify the substance of which it is most likely composed.
7: Suppose identical amounts of heat transfer into different masses of copper and water, causing identical changes in temperature. What is the ratio of the mass of copper to water?
8: (a) The number of kilocalories in food is determined by calorimetry techniques in which the food is burned and the amount of heat transfer is measured. How many kilocalories per gram are there in a 5.00-g peanut if the energy from burning it is transferred to 0.500 kg of water held in a 0.100-kg aluminum cup, causing a 54.9 oC temperature increase? (b) Compare your answer to labeling information found on a package of peanuts and comment on whether the values are consistent.
9: Following vigorous exercise, the body temperature of an 80.0-kg person is 40.0 oC. At what rate in watts must the person transfer thermal energy to reduce the the body temperature to 37.0 oC in 30.0 min, assuming the body continues to produce energy at the rate of 150 W? 1 watt = 1 joule/second or 1 W = J/s. Remember that you will have to look up the specific heat capacity of a human. There is a table earlier on in this section. It is 3500 J/kg oC or 0.83 cal/ g oC.
10: Redo Question 1 but assume it is being heated by an electric heater that operates at 120.0 V. If the electric heater needs 15.0 amperes of current (that is often the maximum of a household circuit) how much time in seconds does it take to heat the water up that much? The temperature of an 80,000-L swimming pool increases by 1.50 oC. First find the net heat transfer during this heating? Ignore any complications, such as loss of water by evaporation.
11: Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150 MW by the radioactive decay of fission products. This heat transfer causes a rapid increase in temperature if the cooling system fails. 1 watt = 1 joule/second and 1 MW = 1 megawatt. (a) Calculate the rate of temperature increase in degrees Celsius per second oC/s if the mass of the reactor core is 1.60 x105 kg and it has an average specific heat of (b) How long would it take to obtain a temperature increase of 2000 oC which could cause some metals holding the radioactive materials to melt? (The initial rate of temperature increase would be greater than that calculated here because the heat transfer is concentrated in a smaller mass. Later, however, the temperature increase would slow down because the steel containment vessel would also begin to heat up.)
Figure 3. Radioactive spent-fuel pool at a nuclear power plant. Spent fuel stays hot for a long time. (credit: U.S. Department of Energy)
Footnotes
1 The values for solids and liquids are at constant volume and at except as noted.
2 These values are identical in units of
3 at constant volume and at except as noted, and at 1.00 atm average pressure. Values in parentheses are at a constant pressure of 1.00 atm.
Glossary
specific heat
: the amount of heat necessary to change the temperature of 1.00 kg of a substance by 1.00 ºC
Solutions
Check Your Understanding
1: The heat transfer depends only on the temperature difference. Since the temperature differences are the same in both cases, the same 25 kJ is necessary in the second case.
Problems & Exercises
1: 5.02 x 108 J
3: 3.07 x 103 J
5: 0.171 o C
7: 10.8
617 W as heat energy radiated because m c (3 degrees) = 840000 J in 30 minutes = 466 watts. But the human body continues to produce 150 W so the total power radiated has to be 466 W + 150 W = 617 W.
10: heat energy = electrical energy m c ΔT = I V t = 5.023×108 J so t=2.79×105 sec = 77.5 hours = 3.23 days. This is why most electrical eaters, such as your oven and clothes dryers use 220 V.
License
Douglas College Physics 1207 Copyright © August 22, 2016 by OpenStax is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted. |
16053 | https://en.wikipedia.org/wiki/Liaison_Committee_(House_of_Lords) | Liaison Committee (House of Lords) - Wikipedia
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The Liaison Committee is a select committee of the House of Lords. The Committee advises the House on financial and other resources required by the body's select committees and allocates those resources among them. It is also responsible for reviewing the work of the select committees, coordinating their work with those of the House of Commons, and considering requests to appoint Special Inqury committees.
Membership
[edit]
As of January 2025, the members of the committee are as follows:
| Member | Party |
--- |
| Lord Gardiner of Kimble0(Chair) | | Non-affiliated |
| Lord Bichard | | Crossbench |
| Lord Collins of Highbury | | Labour |
| Baroness Fookes | | Conservative |
| Baroness Garden of Frognal | | Liberal Democrat |
| Lord Haskel | | Labour |
| Earl Howe | | Conservative |
| Earl of Kinnoull | | Crossbench |
| Baroness Lawrence of Clarendon | | Labour |
| Lord Purvis of Tweed | | Liberal Democrat |
| Lord Smith of Hindhead | | Conservative |
External links
[edit]
Liaison Committee (UK Parliament page)
| hide v t e Current parliamentary committees of the United Kingdom |
| Commons | | Departmental | Business and Trade Culture, Media and Sport Defence Education Energy Security and Net Zero Environment, Food and Rural Affairs Foreign Affairs Health and Social Care Home Affairs Housing, Communities and Local Government International Development Justice Northern Ireland Affairs Science, Innovation and Technology Scottish Affairs Transport Treasury Welsh Affairs Work and Pensions | | Topical | Environmental Audit Petitions Public Accounts Public Administration and Constitutional Affairs Statutory Instruments Women and Equalities | | General | Panel of Chairs Public bill committees Delegated legislation committees European committees Northern Ireland Grand Committee Regional Affairs Committee Scottish Grand Committee Welsh Grand Committee | | Internal | Administration Finance Backbench Business Liaison Modernisation Procedure Selection Standards Privileges Works of Art | | Statutory | House of Commons Commission Administration Estimate Audit Members Estimate Members Estimate Audit Public Accounts Commission Speaker's Committee on the Electoral Commission Speaker's Committee for the Independent Parliamentary Standards Authority | |
| Lords | | Topical | Communications and Digital Constitution Delegated Powers and Regulatory Reform Economic Affairs European Affairs International Agreements International Relations and Defence Secondary Legislation Scrutiny Science and Technology | | Private | Hybrid Instruments Standing Orders | | Internal | Lords Commission Finance Liaison Conduct Procedure and Privileges Selection | | Domestic | Services Works of Art | |
| Joint | | Select | Consolidation Bills Human Rights National Security Strategy Statutory Instruments Tax Law Rewrite Bills | | Statutory | Ecclesiastical Committee Intelligence and Security | |
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16054 | https://www.youtube.com/watch?v=IxdoNRWSaFU | JEE 2019 If both the roots of the quadratic equation 𝑥^2−𝑚𝑥+4=0 are real and distinct
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If both the roots of the quadratic equation 𝑥^2−𝑚𝑥+4=0 are real and distinct and they lie in the interval [1,5] then 𝑚 lies in the interval
a. (4,5)
b. (−5,−4)
c. (5,6)
d. (3,4)
.
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Transcript:
hi all see this this is a problem regarding a quadratic equation okay but only it says that we are having a equation of the form x square minus mx plus 4 and also states that the root alpha and beta lies inside some given interval so here first we have to notice that uh coefficient of x square that means the value of a a x square b x plus c form so here you will see that that coefficient of x square a is positive so therefore if i plot it using a using our concept of coordinate geometry and if we draw the sketching you will find that in that curve it lies between one to five and obviously it comes from the top and its take a turn and became a parabola in this shape right so whenever e is greater than 0 the parabola or in the quadratic equation will be taking in this shape in the otherwise a less than 0 it will be in reverse direction okay so that's it so this is the given info so here we have to apply all the condition we can impose on him and find out that relation okay so first consider that given equation so here the equation says that it's x square minus mx plus 4 and that's equal to 0 that's the given equation now it says that all the roots are real here right so therefore that real and distinct root we just apply that condition real and distinct rule now for the case of real and distinct root we know that that discriminant must be greater than zero so discriminant is greater than 0 that's imply that b square that is m square here minus 4 c a 16 is greater than 0 that's imply that m plus 4 and m minus 4 is greater than 0 that implies that m is either greater than minus 4 greater than plus 4 and obviously a must be less than -4 that's the interval so we can consider a interval a in this case that represent minus infinity to minus 4 with a union of 4 to infinity so that gives a set of values of m we sat if satisfied then our roots became real and distinct so first condition is given but see that here we don't consider any condition stating that it lies between 1 and 5. now for example consider alpha and beta are the roots of this equation so alpha obviously greater than one and beta obviously greater than one and less than five so if i consider alpha maximum one minimum one and beta consider five so what is the midpoint midpoint definitely lies between one two 5 right so that's will be the second condition so let's apply that so let alpha and beta be the roots now as 1 less than equal to alpha comma beta less than equal to 5 so we can easily say that 1 less than equal to alpha plus beta divided by 2 less than 5 okay that's it now notice carefully that in this case we have to always think about the condition and stating that it's lies inside the interval one two five right so it's given close interval now as it is a close interval given so that's why we consider that uh equal to sign so let's go ahead so that's imply that 2 less than equal to alpha plus beta less than equal to 10. now observe carefully if this is the equation of my case right what is the sum of the roots minus b by a so therefore minus m divided by ah minus 1 or that step it is m so therefore it's implied that m lies between 2 less than equal to m less than equal to 10. so therefore we'll get my second interval b that states that aim actually lies between in that interval close interval 2 to 10 so that will be the second condition so let us go ahead and apply the third condition here let's go ahead okay now see at 1 and 5 right what is the function value function value is greater than zero that's the fourth and fifth third and fourth condition right so let's go ahead so here we have to consider that the function f one will be greater than zero now what is f x here we can consider the left hand side it is f x right for easy calculation we can consider that so let's apply and find this value of uh f 1 that gives you actually 1 minus m plus 4 is greater than 0 that's imply that 5 minus m is greater than 0 which imply that m is less than 5 so that's obviously my third condition and i consider the third condition is c the interval c then the next one a 5 is also greater than 0. now let us calculate a 5 so that gives you 25 minus 5 m plus 4 is greater than 0 that's imply that 29 is greater than 5m and in otherwise we imply that m is less than 29 by 5 and that's equal to 5.8 so that's obviously my last interval deep let's consider that okay now all this condition need to be satisfied so that our function are having root between alpha beta ah are lying between one and five so therefore that m must lie on the intersection of all those intervals so intersection of a b c and d so that's the desired and the final out result now obviously here all are given in a interval notation so somehow we have to find this intersection so to find out the intersection let us draw the number line so here i just mark it somehow at the midpoint and to represent the interval a i need to mark minus 4 and i need to mark plus 4 okay and that interval a is represented by that arrow mark so this is a and that's a so it's done let's concentrate on the second interval that is 2 to 10 so that's why i move somehow the midpoint as 2 and somewhere here it's 10 so therefore any values lies between 2 to 10 are represented by b so that's it now let's go ahead and consider that c interval am less than 5 so i just put a 0.5 and anything less than that will represent c and obviously that last one d that is i have to mark somewhere 5.8 and any values less than that will represent the d now see that by this arrow marks right what is the common area the common part is obviously between 4 to 5. so here that 4 to 5 we have to consider now see that exactly 4 to 5 if we consider that all cases aim is actually less than so we have open interval there and also 4 will be intersection when there will be open interval so therefore we'll say that this answer will gives you 4 to 5 and that's why 4 to 5 is the desired interval so that the given equation always have roots between our 1 and 5 and all the condition is satisfied now let's have a look in the detailed solution which obviously gives you a clear idea on that okay so that's it hope you understood the problem thank you |
16055 | https://www.wordreference.com/conj/esverbs.aspx?v=querer | Conjugación de querer - WordReference.com
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verbos -ar: modelo amar verbos -er: modelo temer verbos -ir: modelo partir querer - verbo modelo Verbos que siguen este modelo: 1. bienquerer 2. malquererFirefox and Chrome users: install a shortcut (Firefox or Chrome) then type "conj querer" in your address bar for the fastest conjugations. ### querer 'querer' es el modelo de su conjugación. infinitivo: gerundio: participio: pronominal:querer queriendo querido quererse ⇒definiciónen inglés en francés en portugués Open All Desktop View #### Indicativo | presente | | yo | quiero | | tú | quieres | | él, ella, usted | quiere | | nosotros, nosotras | queremos | | vosotros, vosotras | queréis | | ellos, ellas, ustedes | quieren | | vos | querés | | imperfecto ⓘTambién llamado: pretérito imperfecto o copretérito | | yo | quería | | tú | querías | | él, ella, usted | quería | | nosotros, nosotras | queríamos | | vosotros, vosotras | queríais | | ellos, ellas, ustedes | querían | | vos | querías | | pretérito ⓘTambién llamado: pretérito perfecto simple o pretérito indefinido | | yo | quise | | tú | quisiste | | él, ella, usted | quiso | | nosotros, nosotras | quisimos | | vosotros, vosotras | quisisteis | | ellos, ellas, ustedes | quisieron | | vos | quisiste | | futuro ⓘTambién llamado: futuro simple o futuro imperfecto | | yo | querré | | tú | querrás | | él, ella, usted | querrá | | nosotros, nosotras | querremos | | vosotros, vosotras | querréis | | ellos, ellas, ustedes | querrán | | vos | querrás | | condicional ⓘTambién llamado: condicional simple o pospretérito | | yo | querría | | tú | querrías | | él, ella, usted | querría | | nosotros, nosotras | querríamos | | vosotros, vosotras | querríais | | ellos, ellas, ustedes | querrían | | vos | querrías | #### Formas compuestas comunes | pretérito perfecto | | yo | he querido | | tú | has querido | | él, ella, usted | ha querido | | nosotros, nosotras | hemos querido | | vosotros, vosotras | habéis querido | | ellos, ellas, ustedes | han querido | | vos | has querido | | pluscuamperfecto ⓘTambién llamado: pretérito pluscuamperfecto | | yo | había querido | | tú | habías querido | | él, ella, usted | había querido | | nosotros, nosotras | habíamos querido | | vosotros, vosotras | habíais querido | | ellos, ellas, ustedes | habían querido | | vos | habías querido | | futuro perfecto | | yo | habré querido | | tú | habrás querido | | él, ella, usted | habrá querido | | nosotros, nosotras | habremos querido | | vosotros, vosotras | habréis querido | | ellos, ellas, ustedes | habrán querido | | vos | habrás querido | | condicional perfecto | | yo | habría querido | | tú | habrías querido | | él, ella, usted | habría querido | | nosotros, nosotras | habríamos querido | | vosotros, vosotras | habríais querido | | ellos, ellas, ustedes | habrían querido | | vos | habrías querido | #### Subjuntivo | presente | | yo | quiera | | tú | quieras | | él, ella, usted | quiera | | nosotros, nosotras | queramos | | vosotros, vosotras | queráis | | ellos, ellas, ustedes | quieran | | vos | quieras | | imperfecto ⓘTambién llamado: pretérito o pretérito imperfecto | | yo | quisiera oquisiese | | tú | quisieras oquisieses | | él, ella, usted | quisiera oquisiese | | nosotros, nosotras | quisiéramos oquisiésemos | | vosotros, vosotras | quisierais oquisieseis | | ellos, ellas, ustedes | quisieran oquisiesen | | vos | quisieras oquisieses | | futuro ⓘTambién llamado: futuro simple o futuro imperfecto | | yo | quisiere | | tú | quisieres | | él, ella, usted | quisiere | | nosotros, nosotras | quisiéremos | | vosotros, vosotras | quisiereis | | ellos, ellas, ustedes | quisieren | | vos | quisieres | #### Tiempos compuestos del subjuntivo | pretérito perfecto | | yo | haya querido | | tú | hayas querido | | él, ella, usted | haya querido | | nosotros, nosotras | hayamos querido | | vosotros, vosotras | hayáis querido | | ellos, ellas, ustedes | hayan querido | | vos | hayas querido | | pluscuamperfecto | | yo | hubiera o hubiese querido | | tú | hubieras o hubieses querido | | él, ella, usted | hubiera o hubiese querido | | nosotros, nosotras | hubiéramos o hubiésemos querido | | vosotros, vosotras | hubierais o hubieseis querido | | ellos, ellas, ustedes | hubieran o hubiesen querido | | vos | hubieras o hubieses querido | | futuro perfecto | | yo | hubiere querido | | tú | hubieres querido | | él, ella, usted | hubiere querido | | nosotros, nosotras | hubiéremos querido | | vosotros, vosotras | hubiereis querido | | ellos, ellas, ustedes | hubieren querido | | vos | hubieres querido | #### Imperativo | afirmativo | | | – | | (tú) | ¡quiere! | | (usted) | ¡quiera! | | (nosotros, nosotras) | ¡queramos! | | (vosotros, vosotras) | ¡quered! | | (ustedes) | ¡quieran! | | (vos) | ¡queré! | | negativo | | | – | | (tú) | ¡no quieras! | | (usted) | ¡no quiera! | | (nosotros, nosotras) | ¡no queramos! | | (vosotros, vosotras) | ¡no queráis! | | (ustedes) | ¡no quieran! | | (vos) | ¡no quieras! | #### Indicativo | pretérito anterior | | yo | hube querido | | tú | hubiste querido | | él, ella, usted | hubo querido | | nosotros, nosotras | hubimos querido | | vosotros, vosotras | hubisteis querido | | ellos, ellas, ustedes | hubieron querido | | vos | hubiste querido | Blue letters in conjugations are irregular forms. (example) Red letters in conjugations are exceptions to the model. (example) Grayed conjugations are not commonly used today. The pretérito perfecto indicativo or subjuntivo is often used in instead of the futuro perfecto, while the pretérito anterior is usually replaced by the pluscuamperfecto indicativo. An asterisk () next to vos conjugations indicates Central American spelling. Otherwise, it is the Argentine spelling. Report a problem / Infórmanos sobre cualquier problema. |
16056 | https://www.sciencedirect.com/science/article/abs/pii/0021951772902412 | Adsorption species of oxygen on the surfaces of transition metal oxides - ScienceDirect
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Journal of Catalysis
Volume 25, Issue 3, June 1972, Pages 398-406
Adsorption species of oxygen on the surfaces of transition metal oxides
Author links open overlay panel A.Bielański, M.Najbar
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Abstract
A method for the determination of the mean number of elementary electric charges acquired by one oxygen atom adsorbed on the surface of such oxides as NiO, CoO, MnO is proposed. It consists of the determination of the volume of oxygen gas adsorbed on a high surface area sample of strictly stoichiometric oxide and the subsequent analytical determination of the number of metal ions promoted to a higher oxidation state as the result of electron transfer from the adsorbent to the adsorbate. Using this method it has been shown that, in the case of high surface area nickel oxide, oxygen is chemisorbed at room temperature predominantly in the form of O− ions (irreversible adsorption). The percentage of reversibly adsorbed oxygen decreases with the time of contact. At 150 °C only the O− form was observed after a short period of adsorption. Slow formation of O 2− could be subsequently observed. The velocity of this latter process increases with an increase in temperature. In the case of high surface area cobalt oxide, the O 2− form predominates even at room temperature.
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References (17)
E.R.S. Winter
D.G. Tuck### J. Inorg. Nucl. Chem. (1964)
K. Kuchynka et al.### Collect. Czech. Chem. Commun. (1963)
R.P. Marcellini et al.
E.R.S. Winter### J. Catal. (1966)
A.W. Smith
P.C. Gravelle et al.### J. Chim. Phys. (1964)
Gravelle, P. C., Marty, G., and Teichner, S. J., private...
There are more references available in the full text version of this article.
Cited by (85)
Structural features of p-type semiconducting NiO as a co-catalyst for photocatalytic water splitting
2010, Journal of Catalysis Show abstract Perovskite-like NaTaO 3 photocatalysts, synthesized by sol–gel and solid-state methods, were loaded with NiO co-catalyst to enhance water splitting activity under UV illumination. Activity increased significantly with NiO loading and reached a maximum at 3 and 0.7 wt%, respectively, for the sol–gel and solid-state synthesized NaTaO 3. Beyond this point, photocatalytic activity decreased with further loading. Analysis using X-ray diffraction, high-resolution transmission electron microscopy, and diffuse reflectance spectroscopy shows that the interdiffusion of Na+ and Ni 2+ cations created a solid–solution transition zone on the outer sphere of NaTaO 3. For NiO contents less than 3 wt%, no NiO clusters appeared on the NaTaO 3 surface, and the reduction/oxidation pretreatment did not enhance photocatalytic activity. The high activity resulting from a low NiO loading suggests that the interdiffusion of cations heavily doped the p-type NiO and n-type NaTaO 3, reducing the depletion widths and facilitating charge transfers through the interface barrier.
### Characterization and Reactivity of Molecular Oxygen Species on Oxide Surfaces
1983, Advances in Catalysis Show abstract The surface oxygen species can conveniently be divided into two broad classes, i.e., mononuclear and molecular. Molecular oxygen species are also formed on the surface. The characterization of the adsorbed species has improved markedly as isotopic labeling with 17 O has become more widely used. Some novel forms of molecular oxygen species have been reported and, in particular, the reactivities of species have been studied. Molecular oxygen species have been also identified as intermediates in some biological reactions and are important as oxygen adducts in natural and artificial oxygen carriers. The chapter focuses on adsorbed molecular oxygen species, and shows how recent developments point the way toward an understanding of the role that they and the mononuclear forms of oxygen may play in oxidation reactions. There is a need for a holistic approach to the reaction mechanism which combines both a study of the intermediates by a variety of techniques coupled with an overall analysis of the reaction pathway. It is difficult to combine a general semiempirical approach with specialized characterization but such a synthesis is most likely to lead one to a better understanding of the complex surface phenomena.
### Photoemission studies of adsorbed oxygen and oxide layers
1982, Surface Science Reports Show abstract A comprehensive review is given about the enormous versatility of photoelectron spectroscopy to study the especially complex interaction of oxygen with metal surfaces and the nature of the reaction products. The great variety of well definable parameters of a photoemission experiment, e.g. energy, direction of incidence and polarization of the primary photon beam as well as the detection direction of the photocurrent, yields - through the distributions of energy, momentum and spin polarization of the photoelectrons - detailed insight in the kinetic, thermodynamic, electronic and structural aspects of oxygen adsorption on metal surfaces and incipient oxidation. Characteristic electron binding energies, multiplet and satellite structures of both the oxygen and substrate emission allow a distinction between possible states of adsorbed oxygen, i.e. condensed, molecularly and atomically adsorbed, and incorporated oxygen. Even a distinction between octahedral and tetrahedral oxygen coordination of oxide cations may be possible. Analysis of peak intensities (as a measure of coverages and concentrations) as a function of time and temperature provides information about the kinetics and thermodynamics of adsorbed layer and oxide formation. Angular resolved photoemission studies have led to the determination of absolute adsorption site geometries, individual ad-orbital symmetries and two-dimensional band structure formation within the oxygen overlayer. Measurement of the photoelectron spin-polarization offers a method to study surface magnetism, e.g. of ferromagnetic oxides. The determination of local work functions through the photoemission behavior of co-adsorbed rare gas atoms establishes a uniquely important tool to characterize heterogenous surfaces, e.g. oxygenated surfaces with coexisting oxygen states. Numerous different oxygen/metal systems are chosen to demonstrate the state of the art. Results from other surface spectroscopies and theoretical model calculations are, of course, considered and still open problems are named, e.g. the ionicity of the oxygen chemisorption bond. Problems inherent in sputter profiling through surface oxides as observed with photoemission are briefly addressed. This work is rounded by a list of about 600 references in alphabetic order of the reacting metals.
### Characterization and reactivity of mononuclear oxygen species on oxide Surfaces
1982, Advances in Catalysis Show abstract This chapter discusses the characterization and reactivity of mononuclear oxygen species on oxide surfaces. Measurement of the electron paramagnetic resonance (EPR) parameters gives information which at the simplest level indicates the presence of a paramagnetic species but which can be sufficiently detailed to identify and characterize the adsorbed species. Measurements of the variation of the intensity of the signal with time can give kinetic data on reactions occurring at the surface. Although most of the EPR measurements have been carried out at low temperatures, there is evidence from the examples for some of the species that are stable at much higher temperatures and may play an important role in catalytic reactions.
### Electrically-transduced chemical sensors based on two-dimensional nanomaterials
2019, Chemical Reviews
### Oxygen in Catalysis on Transition Metal Oxides
1979, Catalysis Reviews
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Copyright © 1972 Published by Elsevier Inc.
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16057 | https://chem.libretexts.org/Courses/Bellarmine_University/BU%3A_Chem_103_(Christianson)/Phase_3%3A_Atoms_and_Molecules_-_the_Underlying_Reality/10%3A_Molecular_Structure_and_Geometry/10.3%3A_VSEPR_Geometry | Skip to main content
10.3: VSEPR Geometry
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: Jun 5, 2019
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10.2: Formal Charge and Resonance
10.4: Geometry and Molecular Polarity
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( \newcommand{\kernel}{\mathrm{null}\,})
Skills to Develop
To use the VSEPR model to predict molecular geometries.
To predict whether a molecule has a dipole moment.
The Lewis electron-pair approach can be used to predict the number and types of bonds between the atoms in a substance, and it indicates which atoms have lone pairs of electrons. This approach gives no information about the actual arrangement of atoms in space, however. We continue our discussion of structure and bonding by introducing the valence-shell electron-pair repulsion (VSEPR) model (pronounced “vesper”), which can be used to predict the shapes of many molecules and polyatomic ions. Keep in mind, however, that the VSEPR model, like any model, is a limited representation of reality; the model provides no information about bond lengths or the presence of multiple bonds.
Molecular Geometry
The specific three dimensional arrangement of atoms in molecules is referred to as molecular geometry. We can describe molecular geometry in terms of the bond distances, angles, and relative arrangements in space (Figure 10.3.1). A bond angle is the angle between any two bonds that include a common atom, usually measured in degrees. A bond distance (or bond length) is the distance between the nuclei of two bonded atoms along the straight line joining the nuclei. Bond distances are measured in Ångstroms (1 Å = 10–10 m) or picometers (1 pm = 10–12 m, 100 pm = 1 Å).
Figure 10.3.1: Bond distances (lengths) and angles are shown for the formaldehyde molecule, H2CO.
There are various instrumental techniques such as X-Ray crystallography and other experimental techniques which can be used to tell us where the atoms are located in a molecule. Using advanced techniques, very complicated structures for proteins, enzymes, DNA, and RNA have been determined. Molecular geometry is critical to the chemistry of vision, smell, taste, drug reactions, and enzyme controlled reactions, to name a few.
Example 10.3.1: Carbon Tetrachloride
The Lewis structure of carbon tetrachloride provides information about connectivities, provides information about valence orbitals, and provides information about bond character.
However, the Lewis structure provides no information about the shape of the molecule, which is defined by the bond angles and the bond lengths. For carbon tetrachloride, each C-Cl bond length is 1.78Å and each Cl-C-Cl bond angle is 109.5°. Hence, carbon tetrachloride is tetrahedral in structure:
The VSEPR Model
The VSEPR model can predict the structure of nearly any molecule or polyatomic ion in which the central atom is a nonmetal, as well as the structures of many molecules and polyatomic ions with a central metal atom. The premise of the VSEPR theory is that electron pairs located in bonds and lone pairs repel each other and will therefore adopt the geometry that places electron pairs as far apart from each other as possible. This theory is very simplistic and does not account for the subtleties of orbital interactions that influence molecular shapes; however, the simple VSEPR counting procedure accurately predicts the three-dimensional structures of a large number of compounds, which cannot be predicted using the Lewis electron-pair approach.
We can use the VSEPR model to predict the geometry of most polyatomic molecules and ions by focusing only on the number of electron pairs around the central atom, ignoring all other valence electrons present. According to this model, valence electrons in the Lewis structure form groups, which may consist of a single bond, a double bond, a triple bond, a lone pair of electrons, or even a single unpaired electron, which in the VSEPR model is counted as a lone pair. Because electrons repel each other electrostatically, the most stable arrangement of electron groups (i.e., the one with the lowest energy) is the one that minimizes repulsions. Groups are positioned around the central atom in a way that produces the molecular structure with the lowest energy, as illustrated in Figure 10.3.2.
Figure 10.3.2: Electron Geometries for Species with Two to Six Electron Groups. Groups are placed around the central atom in a way that produces a molecular structure with the lowest energy, that is, the one that minimizes repulsions.
In the VSEPR model, the molecule or polyatomic ion is often given an AXmEn designation, where A is the central atom, X is a bonded atom, E is a nonbonding valence electron group (usually a lone pair of electrons), and m and n are integers. Each group around the central atom is designated as a bonding pair (BP) or lone (nonbonding) pair (LP). From the BP and LP interactions we can predict both the relative positions of the atoms and the bond angles. Using this information, we can describe the molecular geometry, the arrangement of the bonded atoms in a molecule or polyatomic ion.
Predicting Electron Pair Geometry and Molecular Structure
The following procedure uses VSEPR theory to determine the electron pair geometries and the molecular structures:
Draw the Lewis structure of the molecule or polyatomic ion.
Count the number of regions of electron density (lone pairs and bonds) around the central atom. A single, double, or triple bond counts as one region of electron density.
Identify the electron-pair geometry based on the number of regions of electron density: linear, trigonal planar, tetrahedral, trigonal bipyramidal, or octahedral (Figure 10.3.2).
Use the number of lone pairs to assign an AXmEn designation and determine the molecular geometry. Identify the LP–LP, LP–BP, or BP–BP interactions and predict deviations from ideal bond angles based on the principle that lone pairs occupy more space than multiple bonds, which occupy more space than single bonds.
We will illustrate the use of this procedure with several examples, beginning with atoms with two electron groups. In our discussion we will refer to Figure 10.3.2 and Figure 10.3.3, which summarize the common molecular geometries and idealized bond angles of molecules and ions with two to six electron groups.
Figure 10.3.3: Common Molecular Geometries for Species with Two to Six Electron Groups. Lone pairs are shown using a dashed line.
Two Electron Groups
Our first example is a molecule with two bonded atoms and no lone pairs of electrons, BeH2.
AX2: BeH2
The central atom, beryllium, contributes two valence electrons, and each hydrogen atom contributes one. The Lewis electron structure is
There are two electron groups around the central atom. We see from Figure 10.3.2 that the arrangement that minimizes repulsions places the groups 180° apart.
Both groups around the central atom are bonding pairs (BP). Thus BeH2 is designated as AX2.
From Figure 10.3.3 we see that with two bonding pairs, the molecular geometry that minimizes repulsions in BeH2 is linear.
AX2: CO2
The central atom, carbon, contributes four valence electrons, and each oxygen atom contributes six. The Lewis electron structure is
The carbon atom forms two double bonds. Each double bond is a group, so there are two electron groups around the central atom. Like BeH2, the arrangement that minimizes repulsions places the groups 180° apart.
Once again, both groups around the central atom are bonding pairs (BP), so CO2 is designated as AX2.
VSEPR only recognizes groups around the central atom. Thus the lone pairs on the oxygen atoms do not influence the molecular geometry. With two bonding pairs on the central atom and no lone pairs, the molecular geometry of CO2 is linear (Figure 10.3.3). The structure of CO2 is shown in Figure 10.3.2.1.
Three Electron Groups
AX3: BCl3
The central atom, boron, contributes three valence electrons, and each chlorine atom contributes seven valence electrons. The Lewis electron structure is
There are three electron groups around the central atom. To minimize repulsions, the groups are placed 120° apart (Figure 10.3.2).
All electron groups are bonding pairs (BP), so the structure is designated as AX3.
From Figure 10.3.3 we see that with three bonding pairs around the central atom, the molecular geometry of BCl3 is trigonal planar, as shown in Figure 10.3.2.
AX3: CO32−
The central atom, carbon, has four valence electrons, and each oxygen atom has six valence electrons. As you learned previously, the Lewis electron structure of one of three resonance forms is represented as
The structure of CO32− is a resonance hybrid. It has three identical bonds, each with a bond order of 113. We minimize repulsions by placing the three groups 120° apart (Figure 10.3.2).
All electron groups are bonding pairs (BP). With three bonding groups around the central atom, the structure is designated as AX3.
We see from Figure 10.3.3 that the molecular geometry of CO32− is trigonal planar with bond angles of 120°.
In our next example we encounter the effects of lone pairs and multiple bonds on molecular geometry for the first time.
AX2E: SO2
The central atom, sulfur, has 6 valence electrons, as does each oxygen atom. With 18 valence electrons, the Lewis electron structure is shown below.
There are three electron groups around the central atom, two double bonds and one lone pair. We initially place the groups in a trigonal planar arrangement to minimize repulsions (Figure 10.3.2).
There are two bonding pairs and one lone pair, so the structure is designated as AX2E. This designation has a total of three electron pairs, two X and one E. Because a lone pair is not shared by two nuclei, it occupies more space near the central atom than a bonding pair (Figure 10.3.4). Thus bonding pairs and lone pairs repel each other electrostatically in the order BP–BP < LP–BP < LP–LP. In SO2, we have one BP–BP interaction and two LP–BP interactions.
The molecular geometry is described only by the positions of the nuclei, not by the positions of the lone pairs. Thus with two nuclei and one lone pair the shape is bent, or V shaped, which can be viewed as a trigonal planar arrangement with a missing vertex (Figures 9.2.2.1 and 9.2.3). The O-S-O bond angle is expected to be less than 120° because of the extra space taken up by the lone pair.
Figure 10.3.4: The Difference in the Space Occupied by a Lone Pair of Electrons and by a Bonding Pair
As with SO2, this composite model of electron distribution and negative electrostatic potential in ammonia shows that a lone pair of electrons occupies a larger region of space around the nitrogen atom than does a bonding pair of electrons that is shared with a hydrogen atom.
Like lone pairs of electrons, multiple bonds occupy more space around the central atom than a single bond, which can cause other bond angles to be somewhat smaller than expected. This is because a multiple bond has a higher electron density than a single bond, so its electrons occupy more space than those of a single bond. For example, in a molecule such as CH2O (AX3), whose structure is shown below, the double bond repels the single bonds more strongly than the single bonds repel each other. This causes a deviation from ideal geometry (an H–C–H bond angle of 116.5° rather than 120°).
Four Electron Groups
One of the limitations of Lewis structures is that they depict molecules and ions in only two dimensions. With four electron groups, we must learn to show molecules and ions in three dimensions.
AX4: CH4
The central atom, carbon, contributes four valence electrons, and each hydrogen atom has one valence electron, so the full Lewis electron structure is
There are four electron groups around the central atom. As shown in Figure 10.3.2, repulsions are minimized by placing the groups in the corners of a tetrahedron with bond angles of 109.5°.
All electron groups are bonding pairs, so the structure is designated as AX4.
With four bonding pairs, the molecular geometry of methane is tetrahedral (Figure 10.3.3).
AX3E: NH3
In ammonia, the central atom, nitrogen, has five valence electrons and each hydrogen donates one valence electron, producing the Lewis electron structure
There are four electron groups around nitrogen, three bonding pairs and one lone pair. Repulsions are minimized by directing each hydrogen atom and the lone pair to the corners of a tetrahedron.
With three bonding pairs and one lone pair, the structure is designated as AX3E. This designation has a total of four electron pairs, three X and one E. We expect the LP–BP interactions to cause the bonding pair angles to deviate significantly from the angles of a perfect tetrahedron.
There are three nuclei and one lone pair, so the molecular geometry is trigonal pyramidal. In essence, this is a tetrahedron with a vertex missing (Figure 10.3.3). However, the H–N–H bond angles are less than the ideal angle of 109.5° because of LP–BP repulsions (Figure 10.3.3 and Figure 10.3.4).
AX2E2: H2O
Oxygen has six valence electrons and each hydrogen has one valence electron, producing the Lewis electron structure
There are four groups around the central oxygen atom, two bonding pairs and two lone pairs. Repulsions are minimized by directing the bonding pairs and the lone pairs to the corners of a tetrahedron Figure 10.3.2.
With two bonding pairs and two lone pairs, the structure is designated as AX2E2 with a total of four electron pairs. Due to LP–LP, LP–BP, and BP–BP interactions, we expect a significant deviation from idealized tetrahedral angles.
With two hydrogen atoms and two lone pairs of electrons, the structure has significant lone pair interactions. There are two nuclei about the central atom, so the molecular shape is bent, or V shaped, with an H–O–H angle that is even less than the H–N–H angles in NH3, as we would expect because of the presence of two lone pairs of electrons on the central atom rather than one. This molecular shape is essentially a tetrahedron with two missing vertices.
Five Electron Groups
In previous examples it did not matter where we placed the electron groups because all positions were equivalent. In some cases, however, the positions are not equivalent. We encounter this situation for the first time with five electron groups.
AX5: PCl5
Phosphorus has five valence electrons and each chlorine has seven valence electrons, so the Lewis electron structure of PCl5 is
There are five bonding groups around phosphorus, the central atom. The structure that minimizes repulsions is a trigonal bipyramid, which consists of two trigonal pyramids that share a base (Figure 10.3.2):
All electron groups are bonding pairs, so the structure is designated as AX5. There are no lone pair interactions.
The molecular geometry of PCl5 is trigonal bipyramidal, as shown in Figure 10.3.3. The molecule has three atoms in a plane in equatorial positions and two atoms above and below the plane in axial positions. The three equatorial positions are separated by 120° from one another, and the two axial positions are at 90° to the equatorial plane. The axial and equatorial positions are not chemically equivalent, as we will see in our next example.
AX4E: SF4
The sulfur atom has six valence electrons and each fluorine has seven valence electrons, so the Lewis electron structure is
With an expanded valence, this species is an exception to the octet rule.
There are five groups around sulfur, four bonding pairs and one lone pair. With five electron groups, the lowest energy arrangement is a trigonal bipyramid, as shown in Figure 10.3.2.
We designate SF4 as AX4E; it has a total of five electron pairs. However, because the axial and equatorial positions are not chemically equivalent, where do we place the lone pair? If we place the lone pair in the equatorial position, we have three LP–BP repulsions at 90°. If we place it in the axial position, we have two 90° LP–BP repulsions at 90°. With fewer 90° LP–BP repulsions, we can predict that the structure with the lone pair of electrons in the equatorial position is more stable than the one with the lone pair in the axial position. We also expect a deviation from ideal geometry because a lone pair of electrons occupies more space than a bonding pair.
Figure 10.3.5: Illustration of the Area Shared by Two Electron Pairs versus the Angle between Them
At 90°, the two electron pairs share a relatively large region of space, which leads to strong repulsive electron–electron interactions.
With four nuclei and one lone pair of electrons, the molecular structure is based on a trigonal bipyramid with a missing equatorial vertex; it is described as a seesaw. The Faxial–S–Faxial angle is 173° rather than 180° because of the lone pair of electrons in the equatorial plane.
AX3E2: BrF3
The bromine atom has seven valence electrons, and each fluorine has seven valence electrons, so the Lewis electron structure is
Once again, we have a compound that is an exception to the octet rule.
There are five groups around the central atom, three bonding pairs and two lone pairs. We again direct the groups toward the vertices of a trigonal bipyramid.
With three bonding pairs and two lone pairs, the structural designation is AX3E2 with a total of five electron pairs. Because the axial and equatorial positions are not equivalent, we must decide how to arrange the groups to minimize repulsions. If we place both lone pairs in the axial positions, we have six LP–BP repulsions at 90°. If both are in the equatorial positions, we have four LP–BP repulsions at 90°. If one lone pair is axial and the other equatorial, we have one LP–LP repulsion at 90° and three LP–BP repulsions at 90°:
Structure (c) can be eliminated because it has a LP–LP interaction at 90°. Structure (b), with fewer LP–BP repulsions at 90° than (a), is lower in energy. However, we predict a deviation in bond angles because of the presence of the two lone pairs of electrons.
The three nuclei in BrF3 determine its molecular structure, which is described as T shaped. This is essentially a trigonal bipyramid that is missing two equatorial vertices. The Faxial–Br–Faxial angle is 172°, less than 180° because of LP–BP repulsions (Figure 10.3.2.1).
Because lone pairs occupy more space around the central atom than bonding pairs, electrostatic repulsions are more important for lone pairs than for bonding pairs.
AX2E3: I3−
Each iodine atom contributes seven electrons and the negative charge one, so the Lewis electron structure is
There are five electron groups about the central atom in I3−, two bonding pairs and three lone pairs. To minimize repulsions, the groups are directed to the corners of a trigonal bipyramid.
With two bonding pairs and three lone pairs, I3− has a total of five electron pairs and is designated as AX2E3. We must now decide how to arrange the lone pairs of electrons in a trigonal bipyramid in a way that minimizes repulsions. Placing them in the axial positions eliminates 90° LP–LP repulsions and minimizes the number of 90° LP–BP repulsions.
The three lone pairs of electrons have equivalent interactions with the three iodine atoms, so we do not expect any deviations in bonding angles.
With three nuclei and three lone pairs of electrons, the molecular geometry of I3− is linear. This can be described as a trigonal bipyramid with three equatorial vertices missing. The ion has an I–I–I angle of 180°, as expected.
Six Electron Groups
Six electron groups form an octahedron, a polyhedron made of identical equilateral triangles and six identical vertices (Figure 10.3.2.)
AX6: SF6
The central atom, sulfur, contributes six valence electrons, and each fluorine atom has seven valence electrons, so the Lewis electron structure is
With an expanded valence, this species is an exception to the octet rule.
There are six electron groups around the central atom, each a bonding pair. We see from Figure 10.3.2 that the geometry that minimizes repulsions is octahedral.
With only bonding pairs, SF6 is designated as AX6. All positions are chemically equivalent, so all electronic interactions are equivalent.
There are six nuclei, so the molecular geometry of SF6 is octahedral.
AX5E: BrF5
The central atom, bromine, has seven valence electrons, as does each fluorine, so the Lewis electron structure is
With its expanded valence, this species is an exception to the octet rule.
There are six electron groups around the Br, five bonding pairs and one lone pair. Placing five F atoms around Br while minimizing BP–BP and LP–BP repulsions gives the following structure:
With five bonding pairs and one lone pair, BrF5 is designated as AX5E; it has a total of six electron pairs. The BrF5 structure has four fluorine atoms in a plane in an equatorial position and one fluorine atom and the lone pair of electrons in the axial positions. We expect all Faxial–Br–Fequatorial angles to be less than 90° because of the lone pair of electrons, which occupies more space than the bonding electron pairs.
With five nuclei surrounding the central atom, the molecular structure is based on an octahedron with a vertex missing. This molecular structure is square pyramidal. The Faxial–B–Fequatorial angles are 85.1°, less than 90° because of LP–BP repulsions.
AX4E2: ICl4−
The central atom, iodine, contributes seven electrons. Each chlorine contributes seven, and there is a single negative charge. The Lewis electron structure is
There are six electron groups around the central atom, four bonding pairs and two lone pairs. The structure that minimizes LP–LP, LP–BP, and BP–BP repulsions is
ICl4− is designated as AX4E2 and has a total of six electron pairs. Although there are lone pairs of electrons, with four bonding electron pairs in the equatorial plane and the lone pairs of electrons in the axial positions, all LP–BP repulsions are the same. Therefore, we do not expect any deviation in the Cl–I–Cl bond angles.
With five nuclei, the ICl4− ion forms a molecular structure that is square planar, an octahedron with two opposite vertices missing.
The relationship between the number of electron groups around a central atom, the number of lone pairs of electrons, and the molecular geometry is summarized in Figure 10.3.6.
Figure 10.3.6: The molecular structures are identical to the electron-pair geometries when there are no lone pairs present (first column). For a particular number of electron pairs (row), the molecular structures for one or more lone pairs are determined based on modifications of the corresponding electron-pair geometry.
Example 10.3.1
Using the VSEPR model, predict the molecular geometry of each molecule or ion.
PF5 (phosphorus pentafluoride, a catalyst used in certain organic reactions)
H3O+ (hydronium ion)
Given: two chemical species
Asked for: molecular geometry
Strategy:
Draw the Lewis electron structure of the molecule or polyatomic ion.
Determine the electron group arrangement around the central atom that minimizes repulsions.
Assign an AXmEn designation; then identify the LP–LP, LP–BP, or BP–BP interactions and predict deviations in bond angles.
Describe the molecular geometry.
Solution:
A The central atom, P, has five valence electrons and each fluorine has seven valence electrons, so the Lewis structure of PF5 is
B There are five bonding groups about phosphorus. The structure that minimizes repulsions is a trigonal bipyramid (Figure 9.2.6).
C All electron groups are bonding pairs, so PF5 is designated as AX5. Notice that this gives a total of five electron pairs. With no lone pair repulsions, we do not expect any bond angles to deviate from the ideal.
D The PF5 molecule has five nuclei and no lone pairs of electrons, so its molecular geometry is trigonal bipyramidal.
2. A The central atom, O, has six valence electrons, and each H atom contributes one valence electron. Subtracting one electron for the positive charge gives a total of eight valence electrons, so the Lewis electron structure is
B There are four electron groups around oxygen, three bonding pairs and one lone pair. Like NH3, repulsions are minimized by directing each hydrogen atom and the lone pair to the corners of a tetrahedron.
C With three bonding pairs and one lone pair, the structure is designated as AX3E and has a total of four electron pairs (three X and one E). We expect the LP–BP interactions to cause the bonding pair angles to deviate significantly from the angles of a perfect tetrahedron.
D There are three nuclei and one lone pair, so the molecular geometry is trigonal pyramidal, in essence a tetrahedron missing a vertex. However, the H–O–H bond angles are less than the ideal angle of 109.5° because of LP–BP repulsions:
Exercise 10.3.1
Using the VSEPR model, predict the molecular geometry of each molecule or ion.
XeO3
PF6−
NO2+
Answer a
: trigonal pyramidal
Answer b
: octahedral
Answer c
: linear
Example 10.3.2
Predict the molecular geometry of each molecule.
XeF2
SnCl2
Given: two chemical compounds
Asked for: molecular geometry
Strategy:
Use the strategy given in Example 9.2.1.
Solution:
A Xenon contributes eight electrons and each fluorine seven valence electrons, so the Lewis electron structure is
B There are five electron groups around the central atom, two bonding pairs and three lone pairs. Repulsions are minimized by placing the groups in the corners of a trigonal bipyramid.
C From B, XeF2 is designated as AX2E3 and has a total of five electron pairs (two X and three E). With three lone pairs about the central atom, we can arrange the two F atoms in three possible ways: both F atoms can be axial, one can be axial and one equatorial, or both can be equatorial:
The structure with the lowest energy is the one that minimizes LP–LP repulsions. Both (b) and (c) have two 90° LP–LP interactions, whereas structure (a) has none. Thus both F atoms are in the axial positions, like the two iodine atoms around the central iodine in I3−. All LP–BP interactions are equivalent, so we do not expect a deviation from an ideal 180° in the F–Xe–F bond angle.
D With two nuclei about the central atom, the molecular geometry of XeF2 is linear. It is a trigonal bipyramid with three missing equatorial vertices.
2. A The tin atom donates 4 valence electrons and each chlorine atom donates 7 valence electrons. With 18 valence electrons, the Lewis electron structure is
B There are three electron groups around the central atom, two bonding groups and one lone pair of electrons. To minimize repulsions the three groups are initially placed at 120° angles from each other.
C From B we designate SnCl2 as AX2E. It has a total of three electron pairs, two X and one E. Because the lone pair of electrons occupies more space than the bonding pairs, we expect a decrease in the Cl–Sn–Cl bond angle due to increased LP–BP repulsions.
D With two nuclei around the central atom and one lone pair of electrons, the molecular geometry of SnCl2 is bent, like SO2, but with a Cl–Sn–Cl bond angle of 95°. The molecular geometry can be described as a trigonal planar arrangement with one vertex missing.
Exercise 10.3.2
Predict the molecular geometry of each molecule.
SO3
XeF4
Answer a
: trigonal planar
Answer b
: square planar
Molecules with No Single Central Atom
The VSEPR model can be used to predict the structure of somewhat more complex molecules with no single central atom by treating them as linked AXmEn fragments. We will demonstrate with methyl isocyanate (CH3–N=C=O), a volatile and highly toxic molecule that is used to produce the pesticide Sevin. In 1984, large quantities of Sevin were accidentally released in Bhopal, India, when water leaked into storage tanks. The resulting highly exothermic reaction caused a rapid increase in pressure that ruptured the tanks, releasing large amounts of methyl isocyanate that killed approximately 3800 people and wholly or partially disabled about 50,000 others. In addition, there was significant damage to livestock and crops.
We can treat methyl isocyanate as linked AXmEn fragments beginning with the carbon atom at the left, which is connected to three H atoms and one N atom by single bonds. The four bonds around carbon mean that it must be surrounded by four bonding electron pairs in a configuration similar to AX4. We can therefore predict the CH3–N portion of the molecule to be roughly tetrahedral, similar to methane:
The nitrogen atom is connected to one carbon by a single bond and to the other carbon by a double bond, producing a total of three bonds, C–N=C. For nitrogen to have an octet of electrons, it must also have a lone pair:
Because multiple bonds are not shown in the VSEPR model, the nitrogen is effectively surrounded by three electron pairs. Thus according to the VSEPR model, the C–N=C fragment should be bent with an angle less than 120°.
The carbon in the –N=C=O fragment is doubly bonded to both nitrogen and oxygen, which in the VSEPR model gives carbon a total of two electron pairs. The N=C=O angle should therefore be 180°, or linear. The three fragments combine to give the following structure:
We predict that all four nonhydrogen atoms lie in a single plane, with a C–N–C angle of approximately 120°. The experimentally determined structure of methyl isocyanate confirms our prediction (Figure 10.3.8).
Figure 10.3.8: The Experimentally Determined Structure of Methyl Isocyanate
Certain patterns are seen in the structures of moderately complex molecules. For example, carbon atoms with four bonds (such as the carbon on the left in methyl isocyanate) are generally tetrahedral. Similarly, the carbon atom on the right has two double bonds that are similar to those in CO2, so its geometry, like that of CO2, is linear. Recognizing similarities to simpler molecules will help you predict the molecular geometries of more complex molecules.
Example 10.3.3
Use the VSEPR model to predict the molecular geometry of propyne (H3C–C≡CH), a gas with some anesthetic properties.
Given: chemical compound
Asked for: molecular geometry
Strategy:
Count the number of electron groups around each carbon, recognizing that in the VSEPR model, a multiple bond counts as a single group. Use Figure 10.3.3 to determine the molecular geometry around each carbon atom and then deduce the structure of the molecule as a whole.
Solution:
Because the carbon atom on the left is bonded to four other atoms, we know that it is approximately tetrahedral. The next two carbon atoms share a triple bond, and each has an additional single bond. Because a multiple bond is counted as a single bond in the VSEPR model, each carbon atom behaves as if it had two electron groups. This means that both of these carbons are linear, with C–C≡C and C≡C–H angles of 180°.
Exercise 10.3.3
Predict the geometry of allene (H2C=C=CH2), a compound with narcotic properties that is used to make more complex organic molecules.
Answer
: The terminal carbon atoms are trigonal planar, the central carbon is linear, and the C–C–C angle is 180°.
3D Geometries of ABn molecules
Understanding molecular geometry in three-dimensional space is an essential skill for chemists because geometry is so critical to molecular properties and function. With the VSEPR process, you should be able to name the correct geometry for a molecule, but you should also be able to visualize what that geometry looks like in real space. Figure 10.3.9 shows some three-dimensional depictions of geometries where a central atom A is bonded to two or more X atoms. Try to keep these images in mind when considering the geometries of real molecules!
| 6 | 5 | 4 | 3 | 2 |
| AX6 octahedral | AX5 trigonal bipyramidal | AX4 tetrahedral | AX3 trigonal planar | AX2 linear |
| 1 lone pair of electrons | | | | |
| AX5E square pyramidal | AX4E distorted tetrahedron | AX3E pyramidal | AX2E nonlinear | AXE linear |
| 2 lone pairs of electrons | | | |
| AX4E2 square planar | AX3E2 T-shaped | AX2E2 bent | AXE2 linear | |
Figure 10.3.9: Three-dimensional animations of common VSEPR molecular geometries
Summary
Lewis electron structures give no information about molecular geometry, the arrangement of bonded atoms in a molecule or polyatomic ion, which is crucial to understanding the chemistry of a molecule. The valence-shell electron-pair repulsion (VSEPR) model allows us to predict which of the possible structures is actually observed in most cases. It is based on the assumption that pairs of electrons occupy space, and the lowest-energy structure is the one that minimizes electron pair–electron pair repulsions. In the VSEPR model, the molecule or polyatomic ion is given an AXmEn designation, where A is the central atom, X is a bonded atom, E is a nonbonding valence electron group (usually a lone pair of electrons), and m and n are integers. Each group around the central atom is designated as a bonding pair (BP) or lone (nonbonding) pair (LP). From the BP and LP interactions we can predict both the relative positions of the atoms and the angles between the bonds, called the bond angles. From this we can describe the molecular geometry. The VSEPR model can be used to predict the shapes of many molecules and polyatomic ions, but it gives no information about bond lengths and the presence of multiple bonds. A combination of VSEPR and a bonding model, such as Lewis electron structures, is necessary to understand the presence of multiple bonds.
Contributors
: Mike Blaber (Florida State University)
: Robyn Rindge (Class of '98) who now works for PDI Dreamworks (look for his name in the credits of Shrek2.). Robyn drew these rotating molecules using Infini-D (MetaCreations).
: Paul Groves, chemistry teacher at South Pasadena High School and Chemmy Bear
: Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at
10.2: Formal Charge and Resonance
10.4: Geometry and Molecular Polarity |
16058 | https://opencw.aprende.org/high-school/mathematics/combinatorics-the-fine-art-of-counting/ | Combinatorics: The Fine Art of Counting | High School Mathematics | MIT OpenCourseWare
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Do you love math but get bored in math class? Then this is the course for you! Combinatorics is a fascinating branch of mathematics that applies to problems ranging from card games to quantum physics to the Internet. The only pre-requisite is basic algebra; however we will be covering a lot of material. A mathematically agile mind will be helpful.
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16059 | https://par.nsf.gov/servlets/purl/10403607 | Prediction of Core Electron Reactivity and High Oxidation States in Radium under High Pressure
Yihong Bai, a,§ Zhen Liu, a,§ Ling-Jun He, b Fernando Ortega, c Rafeed Khleif, c Yuanzheng Chen, d,b Yuanhui Sun, c, Dadong Yan, a, Maosheng Miao c,
a
Department of Physics, Beijing Normal University, Beijing 100875, China bBeijing Computational Science Research Center, Beijing 100193, China cDepartment of Chemistry and Biochemistry, California State University Northridge, California 91330, United States dSchool of Physical Science and Technology, Southwest Jiaotong University, Chengdu 610031, China § Y.B. and Z.L. contributed equally to this work.
ABSTRACT: Recent first-principles calculations and crystal structure predictions revealed a striking phenomenon that the 5 p
core electrons of Cs can be activated and form covalent bonds with F under high pressure, causing the formation of unconventional CsF n (n>1) compounds in which Cs can be oxidized up to +5 state. Despite many efforts, CsF n remains the only example that the core electrons of a main group element become reactive. Here, we conduct an extensive crystal structure and stability prediction on Ra −F compounds under high pressure, which successfully identifies a series of stable compounds with high F stoichiometry (RaF n, n = 3 ‒8). By various electronic structure and bonding feature analyses, we demonstrate that the 6p core electrons of Ra are activated in most of these compounds and involved in forming covalent bonds with neighboring F atoms. Many novel phenomena emerge in RaF n compounds whence the core electrons are activated, including oxidation states as high as +8, covalent Ra-F bonds, high-pressure molecular crystals, mixed-valence in RaF 3 and RaF 4 compounds, high coordination numbers, all of which point to a striking fact that Ra become a very different element under high pressure and exhibit very rich chemistry.
KEYWORDS : Radium polyfluoride, core electron chemistry, high oxidation states, first-principles calculation, crystal structure prediction.
Ta b le of Co nte n ts (TOC)
1§ INTRODUCTION
The oxidation state of elements in compounds is a key concept in chemistry and condensed-matter physics. 1–3 Ageneral doctrine of chemistry is that the oxidation state of an element is determined by the number of its valence electrons. Many elements such as alkali metals assume one typical oxidation state while many others may have several different states under ambient pressure. Finding a new oxidation state of an element greatly extends the scope of chemistry and often opens a route to many new materials. 3– 5Furthermore, achieving high oxidation state is an attractive goal in chemistry because the corresponding compounds could intrinsically be used as strong oxidants that can compel new types of chemical reactions. High oxidation states have been found in many species such as AuF 6 , IF 7 ,[XeF 8]2- and [WF 8] 2-.2,6 Very recently, an unprecedented +8 oxidation state of Ir and I atoms is predicted in pressure-driven stable IrF 8 and IF 8 compounds. 7,8 Besides Ir, another record of the highest oxidation state of metals in stable compounds under ambient condition is +7 as found in Re(VII)F 7. 9 Among all the species, high oxidation states in noble gases are particularly interesting because these elements have closed shell configurations and are very chemically inert. But once their closed shell electrons are activated by strong oxidants, as experimentally accomplished for the first time by Bartlett et al. who used PtF 6 to oxidize Xe and form Xe +[PtF 6 ]‒, 10 there are more electrons available to achieve higher oxidation states. To go one step further, it is an attractive conception to achieve higher oxidation state by activating the core electrons of main group elements other than noble gases. This turns out to be exceedingly difficult, and no attempt under ambient condition has led to a successful result. 11,12 Recently, with the use of first-principles calculations, we predicted that the inner-shell 5 p electrons of Cs were activated under high pressure and form thermodynamically stable CsF n (n=2-6) compounds, in which Cs atoms are oxidized beyond +1 valence state. 13 –15 This breaks the doctrine that the oxidation states originate only from the outermost valance electrons based on the shell model. 16 Moreover, high pressure might also cause the elements with semi-core 5 d electrons to participate in forming bonds. For example, Hg atom could assume +4 oxidation state by forming HgF 4 compound, 17 In contrast, the highest oxidation state of s-block elements currently reported is +5 in CsF 5 compound under high pressure, 13 although there are 9 electrons in 5s, 5 p and 6 s orbital of Cs atom in total. It is generally true that higher F composition in a compound correspond to high oxidation state. The previous studies show that Ba core electrons are much harder to be activated comparing with the core electrons of Cs. 18 In order to activate more electrons to participate in forming bonds at high-pressure, we need to look for an alkali or alkaline metal whose first and second ionization potentials are comparable or even lower than that of Cs. Radium (Ra) was discovered in 1898 by Pierre and Marie Curie, 19 and later was found almost ubiquitous in soils, water, geologic materials, plants, and foods at low concentrations. 20 Owing to its widely application in radiation therapy, 21 –23 Ra chemistry has been thoroughly explored since its discovery. 24 –28 However, in all of its compounds, Ra is always in +2 oxidation state, in accordance to the fact that it has two valence electrons in its 7s orbital. Ra is constantly being produced by the radioactive decay of Uranium (U) and Thorium (Th) in Earth ’s crust and mantle. 29 Therefore, the study of its high-pressure chemistry can also help to understand the decay of U and Th in Earth ’s interior that is believed the most important thermos-source of the Earth. In this work, aiming at exploring the potential high oxidation states of Ra, we conduct an extensive structure search based on first-principles calculations for F-rich RaF n
compounds with the compositions of n = 2 ‒8 under high pressure (0 ‒300 GPa). Our calculations show that Ra and F can form many stable compounds under pressure, including not only RaF 2 in which Ra is in the typical +2 oxidation state, but also higher F-composition compounds RaF n (n=3 –
8), in which Ra assume higher oxidation states due to the activation of inner-shell electrons. Such unusual chemical behavior leads to substantial changes of the chemical properties of Ra and the structures of the corresponding Ra ‒ Fcompounds. Remarkably, many Ra-F compounds contain molecular species under pressure, for example RaF 6consists of RaF 6 molecules in which Ra atom exhibits +6 oxidation state, exceeding the highest +5 state reported in s-block elements. In addition, RaF 8 is stable above 240 GPa, which is the most F-rich stoichiometry compound of s-block metal fluorides. Correspondingly, the coordination number of Ra reaches an unprecedented value of 12.
§ COMPUTATIONAL DETAILS
The structure searches of Ra ‒F compounds were carried out based on a global minimum search of the free energy landscape with respect to structural variations by combining particle swarm optimization (PSO) algorithm with first-principles calculations. 30,31 We searched the structures for RaF n (n=2 ‒8) compounds at 0, 10, 100, 200 and 300 GPa, respectively. By evaluating the total energy of these structures, 60% of them with lowest enthalpies, together with 40% newly generated structures, are used to produce the structures of next generation by the structure operators of PSO. Applications of the method in numerous previous works show high accuracy and efficiency. 13,17,32 –34 The geometry relaxations in the structure search and the formation enthalpies and electronic properties for the identified stable structures were calculated using density functional theory (DFT) as implemented in Vienna Ab initio Simulation Package (VASP) code. 35 The calculations were performed using a generalized gradient approximation (GGA) 36 within the framework of Perdew ‒Burke ‒
Ernzerhof exchange-correlation functional. 37 The electron-ion interaction is described by the projector augmented wave 38 method with 6s26p6 7s2 and 2s22p5 as valence electrons for Ra and F atoms, respectively. We adopted a
2"
" +
+
kinetic energy cutoff of 1400 eV for wave-function expansion and a k-point mesh of 2 π × 0.03 Å ‒1 or less for Brillouin zone integration. The structures (including lattice parameters and atomic positions) were fully optimized until the residual forces were converged within 0.02 eV/Å. The dynamical stability of predicted structures was determined by phonon calculations using the finite displacement approach 39 as implemented in the Phonopy code. 40 Bader ’sQuantum Theory of Atoms in Molecules (QTAIM) analysis is employed for charge transfer analysis. 41 The Integrated Crystal Hamilton Population (ICOHP) was calculated using the Stuttgart version of the tight-binding LMTOs, atomic sphere approximation (TB ‒LMTO ‒ASA) programme. 42
§ RESULTS AND DISCUSSION
The s tability of Ra ‒F compo un d s . The s tability of RaF n compound s i s examined by calculating the f o rmation enthal p ie s defined a s following:
ℎ! (RaF " ) = # ( % & ’ ! ) ) # ( % & ’ " ) ) ( ! $ " ) & ( ’ " )
(1) in w h ich ℎ! is the formation enthalpy per atom. We cho os e RaF 2 ( 䎐 䖠 3 䖠 , s ame cry stal form a s calcium f luo ride 43 ) a s a reactant in stead of Ra, becau s e of the sub stantial stability of Ra F2 throughout the s tudied p res sure range. The relative thermodynamic stabilitie s of Ra ‒F compound s can be a ss es sed by con vex hull s (Fig. 1a). The compound s locate on the s e line s are s table again s t decompo sitio n .44 The p res sure range s of the stable compo s ition are p res ented in F ig. 1b. The re s ult s show that be s ide sRaF 2 , Ra F 4 can become s table at ambient pre ss ure. Ho w ever, the oxi dation state of Ra in thi s compound i s still +2, becau s e it i s a mixture of chain-li k e Ra F2and F2 molecule. Under high pre ss ure, other Ra ‒Fcom pound s with different stoichiometry (Ra Fn, n
=3, 5‒8) might become stable. Ra F 3 become s stable at p re ss ure s of 48 G Pa and remain s s table up to 255 GPa. A s the pre s sure further increa se s , the more F-r ich stoichiometry structure s Ra F 5 , Ra F 6 , Ra F7 and RaF 8 become stable at pre ss ure s of 78 GPa, 111 GPa, 180 G P a, and 240 G Pa, re spectively. It should be noted that some stable Ra ‒F compound s undergo a s e r ie s of s tructural pha s e tran s ition s with varying p res sure ( Fig. 1b). All the predicted stable com pound s are found to be dynamically s table in vie w of the ab sen ce of imaginary phonon mode s in the whole Brillouin zone ( Figure S 1).
Figure 1. Convex hull and stability of RaF n compounds under pressure. (a) The enthalpies of formation (eV/atom) from 0 to 300 GPa with steps of 30 GPa. The solid lines show the convex hull and the dash lines connect data points. (b) Predicted stable pressure range for RaF n compounds.
Crystal structures. Selected structures for various compositions at their stable pressures are shown in Fig. 2, and the explicit structural information can be found in Table S1. RaF 3 adopts a structure with 䍠 2/ 䔀 symmetry in the pressure range from 48 to 219 GPa (Fig. 2a). The Ra ‒Fdistances in RaF 3 at 90 GPa are in the range of 2.389 ‒2.402 Å, shorter than that of RaF 2 (2.759 Å) at ambient pressure. 45 At higher pressure, RaF 3 undergoes a first-order transition to a structure with 䍠 2/ 䖠 symmetry, which exhibits a distinctive crystal structure that consists of two types of Ra atoms and F atoms. Ra 1 forms square-planar RaF 4 molecular units that is similar to the square-planar structures observed in XeF 4 , HgF 4 and AuF 4 .6,17,46 Ra 2 and neighboring F atoms present in the structure as non-bonded ions. The charge calculations and bonding analysis also reveal that Ra 1 is in +4 oxidation state whereas Ra 2 behaves as Ra 2+ ions. Therefore, the C2/m RaF 3 could be more accurately formulated as [Ra 4+ F 4]Ra 2+ (F ‒) 2. In the same fashion, the C2/c RaF 3 at lower pressure could be formulated as Ra 3+ F3 .In addition, the nearest F ‒F distance of 1.971 Å in 䍠 2/ 䖠
3
’ ’
’ ’
’
’
’
’
’
’
’
’
’
’
RaF 3 (240 GPa) is significantly larger than the F ‒F bond length of 1.404 Å in F 2 molecule. 47 The disappearance of F 2molecular feature in F rich compounds strongly indicates the involvement of inner-shell electrons in forming Ra ‒Fbonds.
Figure 2. Crystal structures of stable RaF n compounds. (a)
C2/ c RaF 3 at 90 GPa. (b) C2/ m RaF 3 at 240 GPa. (c) P2/ m
RaF 4 at 0 GPa. (d) C2/ m(I) RaF 4 at 30 GPa. (e) C2/m (II) RaF 4 at 120 GPa. (f) P1 RaF 4 at 300 GPa. (g) P1 RaF 5 at 90 GPa. (h) P1 RaF 6 at 120 GPa. (i) P1 RaF 7 at 180 GPa. (j)
P 1 RaF 8 at 240 GPa. the green and purple spheres demote Ra and F atoms, respectively. RaF 4 undergoes phase transitions from 䐰 2/ 䖠 to 䍠 2/ 䖠
(I) at 2 GPa (Fig. 2d), to 䍠 2/ 䖠 (II) at 88 GPa (Fig. 2e), then to 䐰 1 at 142 GPa (Fig. 2f), respectively. Under lower pressure, 䐰 2/ 䖠 RaF 4 contains edge sharing square-planar RaF 4 ribbons and F 2 molecules. Moreover, 䍠 2/ 䖠
(I) RaF 4 contains F 2 and linear RaF 2 molecules that is isostructural to XeF 248 . The Ra ‒F distance (2.518 Å at 30 GPa) in RaF 2 molecules is larger than the Cs ‒F bond length (2.358 Å at 20 GPa) in CsF 2 . Both P2/m and C2/m (I) RaF 4could be more accurately formulated as [Ra 2+ (F -)2 ]F 2 . At 120 GPa RaF 4 transforms into a 䍠 2/ 䖠 (II) symmetry structure and is composed of multiple equivalent RaF 4planar molecules, which are similar to those in 䍠 2/ 䖠
RaF 3. The major difference between them is that the molecules in RaF 4 are almost square-planar, whereas those in RaF 3 are largely distorted from perfect squares. While counting the oxidation state, the chemical formula of
C2/m (II) RaF 4 becomes Ra 4+ F4 . Under higher pressure, RaF 4 with 䐰 1 symmetry consists of two different units of RaF 8 and RaF 2 molecule. The RaF 8 unit is an 8-fold polyhedron, containing three types of Ra ‒F bond length (Ra ‒F1 , 1.880 Å; Ra ‒F2, 1.885 Å; Ra ‒F3, 2.212 Å). The nearest F ‒F distance is 1.958 Å, much larger than the covalent F ‒F bond length of 1.440 Å, indicating that RaF 4with 䐰 1 symmetry is a typical molecular crystal. The charge calculations (see below) also reveal that Ra is mixed-valent in 䐰 1 RaF 4 which should be formulated as [Ra 6+ F6 ]Ra 2+ (F - )2 .For more F-rich compounds RaF n (n=5 ‒8), all the identified structures with lowest enthalpy adopt 䐰 1
symmetry. RaF 5 contains two RaF 4 tetrahedrons and a F 2molecule (Fig. 2g) and it can be denoted as [Ra 4+ F 4] 2F2 . It has a very narrow stable pressure range from 78 to 113 GPa. As shown in Fig. 2h, RaF 6 is stable in a structure consisting of deformed octahedron molecules. At 120 GPa, six Ra ‒Fbonds show three different lengths of 1.964, 2.030, and 2.040 Å. In addition, the nearest F ‒F distance between two RaF 6 units is 1.984 Å, much larger than the covalent F ‒Fbond length of 1.440 Å, showing that RaF 6 is a molecular crystal. Therefore, its chemical formula is Ra 6+ F 6. RaF 7stabilizes into a structure consisting of F edge- and vertex-sharing RaF 12 polyhedra (Fig. 2i). Analogously, RaF 8 is also constructed as F edge-sharing RaF 12 polyhedra (Fig. 2j). The Ra ‒F distances in these polyhedra are in the range of 1.919 ‒2.140 Å. More accurate analysis of the bonding characteristics will be discussed later. Ra atoms in both compounds show only one valence.
Electronic structures and the nature of Ra ‒F bonds. To investigate the electronic structures of RaF n (n = 2 ‒8), we calculated the projected density of states (PDOS) by projecting electron wavefunctions onto the atomic orbitals. The PDOS of RaF 4 with 䐰 2/ 䖠 symmetry (0 GPa, Fig. 3a) and 䍠 2/ 䖠 (I) symmetry (30 GPa, Fig. S2b) shows small overlap between Ra 6 p and F 2 p states. This is because these structures are composed of RaF 2 ribbon/molecule and F 2molecule. Under high pressure, overlaps between Ra 6 p and F 2 p states near the Fermi level become more significant, suggesting charge transfers from Ra 6 p to F 2 p orbital in these RaF n compounds (Fig. 3 and S2a-e). As the result of charge transfer, large Ra 6 p components present above the Fermi level and its amount increases with increasing F composition. Furthermore, by analyzing the COHP 49 , we will show that once being activated, Ra 6 p electrons participate in forming chemical bonds with neighboring F atoms. Different to PDOS, COHP can be negative/positive which reveal the bonding/antibonding nature of the states. Among the selected Ra ‒F, Ra ‒Ra, and F ‒F pairs, COHP of Ra ‒F bond is the most negative one. The calculated COHP (Fig. 3d, f, h) shows that the overlaps between 6 p states of Ra and 2 p
states of its nearest neighboring F atom are around ‒10.5 eV, ‒13.0 eV, and ‒10.0 eV for 䍠 2/ 䖠 (II) RaF 4, 䐰 1
RaF 4, and 䐰 1 RaF 6, respectively. Moreover, the integrated COHPs (ICOHPs) up to the Fermi level can be used to measure the bong strength. The calculated ICOHPs are ‒
2.640 and ‒0.524 eV/pair for Ra ‒F1 and Ra ‒F2 bonds in
䍠 2/ 䖠 (II) RaF 4 (120 GPa) and are ‒2.697, ‒2.584, and ‒
0.818 eV/pair for three non-equivalent bonds in 䐰 1 RaF 4(300 GPa), respectively. The ICOHPs of Ra ‒F pair are found to be ‒2.695, ‒2.125 and ‒1.995 eV/pair for 䐰 1
RaF 6 (120 GPa). Comparing to the ‒0.335 eV/pair in RaF 2 (a typical ionic crystal) under 30 GPa, the bonds in these
4
’
Ra ‒F compounds are highly covalent. We also notice that the ICOHP of F ‒F pair in 䐰 1 RaF 6 is as low as ‒0.039 eV/pair, which confirms F atoms do not form bonds and the compound is formed by crystalizing RaF 6 molecules. The Ra-F ICOHP in RaF 5, RaF 7 and RaF 8 show an overall trend of enhancement with the increase of coordination number (Table S2). This is caused by the increasing number of activated inner-shell Ra 6 p electrons, which enhance the strength of interactions in Ra ‒F bonds.
Figure 3. Electronic structures and the nature of Ra ‒F bonds. (a) PDOS (state/eV/atom) for RaF 4 at 0 GPa. (b) PDOS for RaF 3 at 90 GPa. (c) PDOS and (d) COHP for RaF 4 at 120 GPa. (e) PDOS and (f) COHP for RaF 4 at 300 GPa. (g) PDOS and (h) COHP for RaF 6 at 120 GPa. ELFs of (i) RaF 4 at 0 GPa [(0 0 1) plane]. (j) RaF 4 at 120 GPa [(0 1 0) plane]. (k) RaF 4 at 300 GPa [(0 1 0) plane]. (l) RaF 6 at 120 GPa [Ra ‒F1‒F2 plane]. These bonding features of Ra ‒F compounds are also supported by the electron localization function (ELF) analysis. 50 –52 Larger ELF values usually correspond to inner shell or lone pair electrons and covalent bonds, whereas the ionic bonds correspond to small ELF values. As shown in Fig. 3i, the 䐰 2/ 䖠 RaF 4 is quite ionic in nature. However, large ELF values occur between nearest neighboring Ra and F atoms, indicating a covalent bonding character. More information about the bonding analysis can be found in Table S2 and Fig. S3. Ra oxidation states are beyond +2 state based on above analysis because Ra exhibits multiple covalent bonds with neighboring F atoms in RaF n (n = 3 ‒8). To further examine this view, we calculate the charges using Bader ’s quantum theory of atoms in molecules (QTAIM) 41 analysis. As shown in Fig. 4a, the charges of Ra steadily increase with higher F composition (selected at 120 GPa and 250 GPa), and are beyond +2 state in RaF n for compounds with n>2. We note that the Bader charges are usually notably smaller than the formal oxidation numbers, even for typical ionic crystals such as RaF 2 (1.68 at 30 GPa), and are more so for covalent or molecular compounds. As an independent validation, the wavefunction-based Mulliken and Löwdin charges of Ra –F (Fig. S4) are comparable to the evolutions of the density-derived Bader charge results in Fig. 4a. Fig. 4a reveals very different charge transfer trends at the two calculated pressures. At 120 GPa, the Bader charge increases with increasing F composition while n<6, except that RaF 6 shows a similar Bader charge as RaF 4 , which is consistent with the fact that Ra in RaF 5 ([Ra 4+ F4 ]2 F2 ) is actually in +4 state. The Bader charges in RaF 7 and RaF 8under 120 GPa are lower than that of RaF6. However, both two compounds are not stable at this pressure. Bader charges at 250 GPa show a striking feature, namely there are two very different values in RaF 3 and RaF 4 ,corroborating with the aforementioned mixed valence of Ra in these two compounds. If average Bader charges are used to represent RaF 3 and RaF 4, the Bader charges increase quite linearly with the increasing F composition (Fig. 4a), strongly suggesting the same increasement of the nominal oxidation state of Ra in these compounds. Thus, despite the actual charge values calculated by Bader ’s QTAIM method, the nominal oxidation state of Ra can be assigned. For example, Ra in RaF 6 is in +6 state, although the calculated Bader charges are 3.1 and 3.3 under 120 and 250 GPa. Furthermore, the Bader charge increases monotonically for n>5, indicating a continuously growing oxidation number. Therefore, it is reasonable to assign nominal oxidation states of +7 and +8 to Ra atoms in RaF 7 and RaF 8.In addition, the coordination numbers of Ra also increase with higher F composition, which also corroborate the high oxidation states, because the activation of more core electrons from Ra cause the formation of more Ra-F bonds and prevent the formation of F-F bonds. It is worth mentioning that the coordination number of Ra in RaF 8 is 12, which is one of the highest among all s-metal compounds as Ba 2+ in Ba(SbF 6 )2·5XeF 2. 48 The reactivity of the core electrons is driven by the change and reordering of the orbital energies of s/p-block electron of Ra and the p-block electrons of F. To reveal this point, we calculated the atomic orbital energy levels of F 2 p,Cs 5 p, Ba 5 p, and Ra 6 p, under high pressure using a He matrix model (Figure 4b). Although the energies of all
5
atomic orbitals increase with pressure, Cs 5 p and Ba 5 p
orbitals increase more significantly than F 2 p, surpassing the later at ~16 GPa and ~224 GPa, respectively, which agrees well with the previous atomic orbital calculations 13,18 and with the stable pressure regions of Cs-F compounds as predicted in the previous studies. 13 Similarly, Ra 6p
becomes higher in energy than F 2 p at ~92 GPa, which is consistent with the structure and stability predictions in this work. In addition, the Ra 6 p bands significantly broaden as the pressure increases, which further elevates the energy level of 6 p orbitals, causing more Ra 6 p electrons being transferred to F atoms. Both points are important for the gradual stabilization of F-rich compounds (RaF 6 à RaF 7 à
RaF 8) with increasing pressure. reactive and are prone to form strong covalent bonds with F atoms. Strikingly, RaF 6 exhibits a molecular crystal feature with +6 oxidation state of Ra. The coordination number of Ra in RaF 8 reaches 12, which is among the highest in all s-block metal fluorides. This coordination number has been observed before in a complex hexafluorantimoates Ba(SbF 6 )2·5XeF 2. 48
§ ASS OC I ATED CONTENT Supp orti ng In formatio n
The Supporting Information is available free of charge at . Phonon spectra, structural parameters, projected density of states, ICOHP of selected RaF n
compounds.
§ AUTHOR INFORMAT I ON
Corresponding Authors
E-mail: yuanhui.sun@csun.edu
E-mail: yandd@bnu.edu.cn
E-mail: mmiao@csun.edu
Figure 4. (a) Calculated Bader charge for Ra in RaF n
compounds at 120 GPa (blue line) and 250 GPa (red line). The number to the right of the symbol indicates the coordination number of Ra in each Ra ‒F compound. (b) Energies of F 2 p, Ra 6 p, Cs 5 p, and Ba 5 p orbitals as a function of the external pressure. Pressure effect is modeled by putting elements in a face-centered cubic supercell (3×3×3) He matrix, in which one He atom is replaced by the atom being examined. 53 Recent studies have illustrated that Cs can react with F and form unusual F-rich compounds under high pressure, exhibiting the chemical characteristics like p-block elements. 13,54 Ba ‒F compounds also show large portion of 5p components close to the Fermi level, although they do not exhibit p-block compound properties such as forming molecular crystals. Our results clearly show a new example that an s-metal can behave like a p-block element under pressure, due to the activation of its core electrons. Remarkably, the highest nominal oxidation number can be as high as +8, surpass the previously reported +5 in CsF 5 ,which sets a new record of oxidation number of alkali and alkaline metals.
ORCID
Yuanhui Sun: 0000-0003-2981-1133 Maosheng Miao: 0000-0001-9486-1204
Notes
The authors declare no competing financial interest.
§ ACKNOWLEDGMENT
Y.S. and M.M. acknowledge the support of NSF under the grant No. DMR-1848141, and computational resources provided by XSEDE (TG ‒DMR130005). M.M. also acknowledges the support of ACS PRF 50249 ‒UNI6 and the support of California State University Research, Scholarship and Creative Activity (RSCA) awards. D.Y. acknowledge the support of National Natural Science Foundation of China for the grant under Nos 21873015. Z.L. also acknowledge National Natural Science Foundation of China support for the funding under Nos 12004045. L.H. is supported by the China Postdoctoral Science Foundation under grants No. 2021M690326.
§ REFERENCES
(1)
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16060 | https://www.math.toronto.edu/mathnet/questionCorner/infinity.html | Question Corner -- Do Parallel Lines Meet At Infinity?
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Do Parallel Lines Meet At Infinity?
Asked by a student at St-Joseph Secondary School on October 5, 1997:
Could you help me prove that parallel lines meet at infinity or that infinity begins where parallel lines meet.
I am curious. Could this ever happen?
The answer to the question depends on exactly what kind of geometry you are dealing with and what "points" and "lines" mean.
If you are talking about ordinary lines and ordinary geometry, then parallel lines do not meet. For example, the line x=1 and the line x=2 do not meet at any point, since the x coordinate of a point cannot be both 1 and 2 at the same time.
In this context, there is no such thing as "infinity" and parallel lines do not meet.
However, you can construct other forms of geometry, so-called non-Euclidean geometries. For example, you can take the usual points of the plane and attach to them an additional point called "infinity" and consider all lines to also include this additional point. In this context, there is a single "infinity" location where all lines meet. In a geometry like this, all lines intersect at infinity, in addition to any finite point where they might happen to meet.
Or, you could attach not just one additional point, but a whole collection of additional points, one for each direction. Then you can consider two parallel lines to meet at the extra point corresponding to their common direction, whereas two non-parellel lines do not intersect at infinity but intersect only at the usual finite intersection point. This is called projective geometry, and is described in more detail in the answer to another question.
In summary, then: in usual geometry, parallel lines do not meet. There is no such thing as infinity, and it is wrong to say that parallel lines meet at infinity.
However, you can construct other geometric systems, whose "points" include not only the points of familiar geometry (describable as coordinate pairs (x,y)), but also other objects. These other objects can be constructed in various ways, as described in the discussion of projective geometry. In these other geometric systems, parallel lines may meet at a "point at infinity". Whether this is one single point or different points for different classes of parallel lines, depends on the particular geometric system you are considering.
You may also be interested in our answers and explanations page, which contains a discussion of the question does infinity exist?
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5
Why isn't
|
a
+
b
i
|
equal to
√
a
2
+
(
b
i
)
2
but instead is equal to
√
a
2
+
b
2
?
Max Sklar
I have either attended or taught at colleges for most of my adult life.
·
Upvoted by
David Joyce
, Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 1.9K answers and 3.8M answer views
·
8y
One definition of the modulus of a complex number is |z|=√z⋅¯z Here ¯z is the complex conjugate of z. If z=a+bi then ¯z=a−bi.
So z⋅¯z=(a+bi)(a−bi)=a2−(bi)2=a2+b2. Thus |z|=√a2−(bi)2.
This differs from your definition √a2+(bi)2=√a2−b2. The problem is that the modulus is meant to be an extension of the absolute value of a real number. It is meant to measure a distance from 0 and should have non-negative real values. Under your definition it is possible to have an imaginary value to the modulus. For example you would have |1+2i|=\sqrt{
One definition of the modulus of a complex number is |z|=√z⋅¯z Here ¯z is the complex conjugate of z. If z=a+bi then ¯z=a−bi.
So z⋅¯z=(a+bi)(a−bi)=a2−(bi)2=a2+b2. Thus |z|=√a2−(bi)2.
This differs from your definition √a2+(bi)2=√a2−b2. The problem is that the modulus is meant to be an extension of the absolute value of a real number. It is meant to measure a distance from 0 and should have non-negative real values. Under your definition it is possible to have an imaginary value to the modulus. For example you would have |1+2i|=√1−4=√3i. Moreover, the number √3 has no meaning with regards to the distance of z=1+2i from 0.
Under the standard definition |1+2i|=√1+4=√5 which is precisely the distance from (0,0) to (1,2) in the Cartesian plane.
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Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist.
·
Author has 6.8K answers and 52.8M answer views
·
Updated 8y
This is a question that puzzles many students, unfortunately especially during tests when they have never thought of doing it before!
The complex number z = a + ib drawn on an Argand diagram is like this
where a and b are just the same units of measurement (like centimetres).
Basically, starting from the origin, go a distance “a”centimetres along the real axis then go up a distance “b” centimetres, n
This is a question that puzzles many students, unfortunately especially during tests when they have never thought of doing it before!
The complex number z = a + ib drawn on an Argand diagram is like this
where a and b are just the same units of measurement (like centimetres).
Basically, starting from the origin, go a distance “a”centimetres along the real axis then go up a distance “b” centimetres, not a distance of “ib” centimetres!
The problem comes from labelling the axes like this below:
The distances on the imaginary axis are not measured i, 2i, 3i, 4i…
They ...
Assistant
Bot
·
1y
The expression |a+bi| represents the modulus (or absolute value) of the complex number a+bi, where a is the real part and b is the imaginary part.
To understand why |a+bi|=√a2+b2 and not √a2+(bi)2, let's break it down:
Definition of Modulus: The modulus of a complex number z=a+bi is defined as the distance from the origin in the complex plane, which is calculated using the Pythagorean theorem. This gives us:
|a+bi|=√a2+b2
Calculating (bi)2: If you were to compute (bi)2, you would get:
(bi)2=b2i2=b2(−1)=−b2
Therefore, (bi)2 is
The expression |a+bi| represents the modulus (or absolute value) of the complex number a+bi, where a is the real part and b is the imaginary part.
To understand why |a+bi|=√a2+b2 and not √a2+(bi)2, let's break it down:
Definition of Modulus: The modulus of a complex number z=a+bi is defined as the distance from the origin in the complex plane, which is calculated using the Pythagorean theorem. This gives us:
|a+bi|=√a2+b2
Calculating (bi)2: If you were to compute (bi)2, you would get:
(bi)2=b2i2=b2(−1)=−b2
Therefore, (bi)2 is not positive, and it changes the structure of the expression.
Why |a+bi|≠√a2+(bi)2: If you tried to use √a2+(bi)2, you would be calculating:
√a2+(bi)2=√a2−b2
This expression does not accurately represent the distance in the complex plane since it does not account for the imaginary unit correctly.
In summary, the correct formula for the modulus of a complex number is:
|a+bi|=√a2+b2
This formula reflects the geometric interpretation of the modulus as the distance from the point (a,b) to the origin (0,0) in the complex plane.
Pierre Reoxe
Lives in Massachusetts (2009–present)
·
Author has 282 answers and 533.2K answer views
·
8y
Hello,
To dumb things down a bit: The absolute value of any number is its distance from zero. Therefore, the absolute value of a2+i∗b2 would be…
The blue circle is all the arrangements of points that have the absolute value from the original expression if we include the imaginary number i. The distance of the red line is the real distance or absolute value. We can find this out using the Pythagorean theorem.
When we find the absolute value of a complex number, we have to consider the fact that the grid works in normal units. If we do choose to use the blue circle, then our calculations wo
Hello,
To dumb things down a bit: The absolute value of any number is its distance from zero. Therefore, the absolute value of a2+i∗b2 would be…
The blue circle is all the arrangements of points that have the absolute value from the original expression if we include the imaginary number i. The distance of the red line is the real distance or absolute value. We can find this out using the Pythagorean theorem.
When we find the absolute value of a complex number, we have to consider the fact that the grid works in normal units. If we do choose to use the blue circle, then our calculations would be terribly off and wouldn't make sense. This derives from the Pythagorean Theorem. Please see my other answers for more information on that topic.
Thank You
-Pierre
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Gaurav Patil
Studied Mathematics at University of Toronto
·
Updated 5y
We define the |z| function to emulate the comparison we can do in real numbers(distance between x and y as |x-y|). This comparison(in reals) leads to many interesting properties that allow us to intuitively apply the concept of real numbers towards understanding the world around us. Having similar comparison in complex numbers makes sense. Since we can compare real numbers, it makes sense to attach a real number to a given complex number with direct relation to the complex number. So we define |a+ib| as the distance from the origin. This allows us to say distance between complex numbers z and
We define the |z| function to emulate the comparison we can do in real numbers(distance between x and y as |x-y|). This comparison(in reals) leads to many interesting properties that allow us to intuitively apply the concept of real numbers towards understanding the world around us. Having similar comparison in complex numbers makes sense. Since we can compare real numbers, it makes sense to attach a real number to a given complex number with direct relation to the complex number. So we define |a+ib| as the distance from the origin. This allows us to say distance between complex numbers z and s is |z-s|.
The property defined by f(a+ib)=√a2+(bi)2=√a2−b2 is also well studied in math (In algebraic geometry), but the distance function has a lot more immediately visible applications in other subjects.
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Jasper Baltz
B.S. in Physics & Mathematics, Indiana University (Graduated 2022)
·
Author has 110 answers and 324.5K answer views
·
8y
The absolute value does not mean to make something positive, but instead, to find how far something is from zero.
Complex numbers are usually graphed similarly to an ordered pair. The x axis represents the real portion and the y axis represents the imaginary portion. Thus, a+bi is represented by the ordered pair (a, b).
Using the Pythagorean Theorem, you can find the distance of (a, b). The reason you don’t use a and bi is, for one, because you could easily receive an imaginary number as the result, which isn’t a representative distance.
César Leonardo Patiño Burgos
support teacher in science and English
·
Author has 140 answers and 310K answer views
·
8y
Because i^2 is -1
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Lukas Schmidinger
I have graduate CS and my studies included math courses.
·
Author has 27.7K answers and 14.9M answer views
·
8y
Originally Answered: Why isn't |a+bi|=\sqrt (a^(2) +(bi) ^(2)) but instead equal to \sqrt (a^(2) +b^(2))?
·
Because per definition |z| must be a non-negative real number. Now if 1<|a|<|b| than a2+(bi)2=a2−b2<0
therefor √a2+(bi)2∉R and this would go against the the definition.
Eric Dietrich
Studied Mathematics at Khan Academy
·
Author has 1.3K answers and 2.2M answer views
·
8y
Thanks for the A2A!
|x| is sometimes used to describe the distance of x to 0 on the number line, so it makes sense for |x+iy| is the distance from the origin, so from the distance formula:
|x+iy|=√x2+y2
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David Joyce
Ph.D. in Mathematics, University of Pennsylvania (Graduated 1979)
·
Upvoted by
Yair Livne
, Master's Mathematics, Hebrew University of Jerusalem (2007) and
Erik Bergland
, PhD Mathematics, Brown University (2024) · Author has 9.9K answers and 68.4M answer views
·
6y
Related
Why does
√
a
b
not equal
√
a
times
√
b
when
a
and
b
are negative?
As long as a and b are positive or zero, it is true that √ab=√a√b, but that’s no longer true when they’re both negative. You’re asking why.
For one thing, you can simply show it’s not true. Suppose that they’re both negative. That means they’re negations of positive numbers, a=−c and b=−d where c and d are positive. To begin with, we know √cd=√c√d. Then √a=√−c=i√c where i is our notation for √−1. Likewise, √b=√−d=i√d. Also, √ab=√(−c)(−d)=√cd. But \sqrt a\,\sqrt b=i\sqrt c\,i \sqrt d=i^2\sqrt a\,\sqrt
As long as a and b are positive or zero, it is true that √ab=√a√b, but that’s no longer true when they’re both negative. You’re asking why.
For one thing, you can simply show it’s not true. Suppose that they’re both negative. That means they’re negations of positive numbers, a=−c and b=−d where c and d are positive. To begin with, we know √cd=√c√d. Then √a=√−c=i√c where i is our notation for √−1. Likewise, √b=√−d=i√d. Also, √ab=√(−c)(−d)=√cd. But √a√b=i√ci√d=i2√a√b=−√a√b. Thus, √ab does not equal √a√b; instead, they’re negations of each other.
Although that answers your question, it doesn’t answer this question: What’s really going on here? In order to answer that, you need to consider more complex numbers than just negative numbers and i. Consider all the complex numbers z=x+iy one unit away from 0. Those numbers satisfy x2+y2=1 and form what is called the unit circle.
Every complex number z=x+iy on this unit circle can be written in the form z=cosθ+isinθ where θ is the angle from the positive real axis (the x-axis) to the point z measured counterclockwise. That angle is variously called the angle, argument, or direction of z. (If you’re familiar with complex exponentiation, you can also write z=eiθ.)
Multiplication of complex numbers on the unit circle is particularly simple. The product of two numbers cosθ+isinθ and cosφ+isinφ is equal to cos(θ+φ)+isin(θ+φ). (That comes from the addition rules for sine and cosine.) In other words, just add their angles. Of course, if you go past 360° (2π radians), you’ll need to subtract 360° (2π radians).
That implies that squaring a complex number on the unit circle corresponds to doubling the angle. For example, the angle of i is 90°, so the angle of i2 is 180°, in other words i2=−1. For another example, the angle of −i is 270°. so the angle of (−i)2 is 540°, which reduces to 180°, in other words (−i)2=−1.
Well, if squaring is doubling the angle, should taking the square root be halving the angle? For example, the angle of −1 is 180°, so the angle of √−1 is 90°, in other words √−1=i, which is where we started.
The problem comes from trying to define the square root all the way around the circle continuously. The square root of a complex number whose angle is just less than 360° ought to be a complex number whose angle is just less than 180°, that is, almost −1. But if you increase it to 360°, the square root jumps from 180° to 0°. (That jump corresponds to multiplying by −1.) The square root function cannot be defined on the entire complex plane as a continuous function. This non-continuity implies that the square root of the product of two negative numbers is the negation of their square roots. The jump in the definition of complex square roots introduces that negation.
Alan Bustany
Trinity Wrangler, 1977 IMO
·
Upvoted by
David Joyce
, Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 9.8K answers and 58.5M answer views
·
9y
Related
If
3
+
2
=
2
+
3
and
3
⋅
2
=
2
⋅
3
, why is
2
3
not equal to
3
2
?
If 3 + 2 = 2 + 3 and 3 2 = 2 3, why is 2^3 not equal to 3^2?
For the same reason that:
Washing socks then polishing shoes = polishing shoes then washing socks; and
Pairing socks then pairing shoes = pairing shoes then pairing socks; but
Putting on socks then putting on shoes is not equal to putting on shoes then putting on socks.
Some binary functions, including addition and multiplication, are commutative. Others, including exponentiation, are not.
Listing examples of commutative operators does not imply anything about the next operator you mention even if you do keep all the arguments of the o
If 3 + 2 = 2 + 3 and 3 2 = 2 3, why is 2^3 not equal to 3^2?
For the same reason that:
Washing socks then polishing shoes = polishing shoes then washing socks; and
Pairing socks then pairing shoes = pairing shoes then pairing socks; but
Putting on socks then putting on shoes is not equal to putting on shoes then putting on socks.
Some binary functions, including addition and multiplication, are commutative. Others, including exponentiation, are not.
Listing examples of commutative operators does not imply anything about the next operator you mention even if you do keep all the arguments of the operators the same.
Lav Kumar
tries to solve using real life examples
·
Updated 8y
Related
Can
(
a
+
b
)
2
be equal to
a
2
+
b
2
?
Consider a square of side length a+b.
You can vary the values of a and b by varying the lines inside the bigger square.
You have to get rid of the two ‘ba’ blocks
You can clearly see that for achieving that you will have to move the lines to their extreme limits i.e extreme left and up( implying a=0) or extreme right and bottom( implying b=0).
Or you can simply say that the square side is 0 making every inner blocks 0. This implies a=b=0.
Consider a square of side length a+b.
You can vary the values of a and b by varying the lines inside the bigger square.
You have to get rid of the two ‘ba’ blocks
You can clearly see that for achieving that you will have to move the lines to their extreme limits i.e extreme left and up( implying a=0) or extreme right and bottom( implying b=0).
Or you can simply say that the square side is 0 making every inner blocks 0. This implies a=b=0.
PhD in mathematics.
·
Author has 141 answers and 279.9K answer views
·
9y
Related
Why does
√
(
a
2
+
b
2
)
not equal to
a
+
b
?
The answer is simple. Equivalent statement is: Why
√
a
+
b
≠
√
a
+
√
b
. Whatever you expect to happen is the property of a linear function. A function
f
is linear then
f
(
a
+
b
)
=
f
(
a
)
+
f
(
b
)
. Square root is not a linear function. Therefore, you cannot have such relationship.
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16062 | https://en.wikipedia.org/wiki/Series_and_parallel_springs | Contents
Series and parallel springs
In mechanics, two or more springs are said to be in series when they are connected end-to-end or point to point, and they are said to be in parallel when they are connected side-by-side; in both cases, so as to act as a single spring:
Series | and | Parallel
| and |
More generally, two or more springs are in series when any external stress applied to the ensemble gets applied to each spring without change of magnitude, and the amount of strain (deformation) of the ensemble is the sum of the strains of the individual springs. Conversely, they are said to be in parallel if the strain of the ensemble is their common strain, and the stress of the ensemble is the sum of their stresses.
Any combination of Hookean (linear-response) springs in series or parallel behaves like a single Hookean spring. The formulas for combining their physical attributes are analogous to those that apply to capacitors connected in series or parallel in an electrical circuit.
Formulas
Equivalent spring
The following table gives formulas for the spring that is equivalent to an ensemble (or system) of two springs, in series or in parallel, whose spring constants are
k
1
{\displaystyle k_{1}}
and
k
2
{\displaystyle k_{2}}
. (The compliance
c
{\displaystyle c}
of a spring is the reciprocal
1
/
k
{\displaystyle 1/k}
of its spring constant.)
Quantity | In Series | | In Parallel
Equivalent spring constant | 1keq=1k1+1k2{\displaystyle {\frac {1}{k_{\mathrm {eq} }}}={\frac {1}{k_{1}}}+{\frac {1}{k_{2}}}} | | keq=k1+k2{\displaystyle k_{\mathrm {eq} }=k_{1}+k_{2}}
Equivalent compliance | ceq=c1+c2{\displaystyle c_{\mathrm {eq} }=c_{1}+c_{2}} | | 1ceq=1c1+1c2{\displaystyle {\frac {1}{c_{\mathrm {eq} }}}={\frac {1}{c_{1}}}+{\frac {1}{c_{2}}}}
Deflection (elongation) | xeq=x1+x2{\displaystyle x_{\mathrm {eq} }=x_{1}+x_{2}} | | xeq=x1=x2{\displaystyle x_{\mathrm {eq} }=x_{1}=x_{2}}
Force | Feq=F1=F2{\displaystyle F_{\mathrm {eq} }=F_{1}=F_{2}} | | Feq=F1+F2{\displaystyle F_{\mathrm {eq} }=F_{1}+F_{2}}
Stored energy | Eeq=E1+E2{\displaystyle E_{\mathrm {eq} }=E_{1}+E_{2}} | | Eeq=E1+E2{\displaystyle E_{\mathrm {eq} }=E_{1}+E_{2}}
Partition formulas
Quantity | In Series | In Parallel
Deflection (elongation) | x1x2=k2k1=c1c2{\displaystyle {\frac {x_{1}}{x_{2}}}={\frac {k_{2}}{k_{1}}}={\frac {c_{1}}{c_{2}}}} | x1=x2{\displaystyle x_{1}=x_{2}\,}
Force | F1=F2{\displaystyle F_{1}=F_{2}\,} | F1F2=k1k2=c2c1{\displaystyle {\frac {F_{1}}{F_{2}}}={\frac {k_{1}}{k_{2}}}={\frac {c_{2}}{c_{1}}}}
Stored energy | E1E2=k2k1=c1c2{\displaystyle {\frac {E_{1}}{E_{2}}}={\frac {k_{2}}{k_{1}}}={\frac {c_{1}}{c_{2}}}} | E1E2=k1k2=c2c1{\displaystyle {\frac {E_{1}}{E_{2}}}={\frac {k_{1}}{k_{2}}}={\frac {c_{2}}{c_{1}}}}
Derivations of spring formulas (equivalent spring constant)
Equivalent Spring Constant (Series)
When putting two springs in their equilibrium positions in series attached at the end to a block and then displacing it from that equilibrium, each of the springs will experience corresponding displacementsx1andx2for a total displacement ofx1+ x2. We will be looking for an equation for the force on the block that looks like:Fb=−keq(x1+x2).{\displaystyle F_{b}=-k_{\mathrm {eq} }(x_{1}+x_{2}).\,}The force that each spring experiences will have to be same since otherwise, the springs would buckle. Moreover, this force will be the same asFb. This means thatF1=−k1x1=F2=−k2x2=Fb.{\displaystyle F_{1}=-k_{1}x_{1}=F_{2}=-k_{2}x_{2}=F_{b}.\,}Working in terms of the absolute values, we can solve forx1{\displaystyle x_{1}\,}andx2{\displaystyle x_{2}\,}:x1=F1k1,x2=F2k2{\displaystyle x_{1}~=~{\frac {F_{1}}{k_{1}}}\,,\qquad x_{2}~=~{\frac {F_{2}}{k_{2}}}},and similarly,x1+x2=Fbkeq{\displaystyle x_{1}~+~x_{2}~=~{\frac {F_{b}}{k_{\mathrm {eq} }}}}.Substitutingx1{\displaystyle x_{1}\,}andx2{\displaystyle x_{2}\,}into the latter equation, we findF1k1+F2k2=Fbkeq{\displaystyle {\frac {F_{1}}{k_{1}}}~+~{\frac {F_{2}}{k_{2}}}~=~{\frac {F_{b}}{k_{\mathrm {eq} }}}}.Now remembering thatF1=F2=Fb{\displaystyle F_{1}~=~F_{2}~=~F_{b}}, we arrive at1keq=1k1+1k2.{\displaystyle {\frac {1}{k_{\mathrm {eq} }}}={\frac {1}{k_{1}}}+{\frac {1}{k_{2}}}.\,}
The force that each spring experiences will have to be same since otherwise, the springs would buckle. Moreover, this force will be the same as Fb. This means that
Working in terms of the absolute values, we can solve for
x
1
{\displaystyle x_{1}\,}
and
x
2
{\displaystyle x_{2}\,}
:
and similarly,
Substituting
x
1
{\displaystyle x_{1}\,}
and
x
2
{\displaystyle x_{2}\,}
into the latter equation, we find
Now remembering that
F
1
F
2
F
b
{\displaystyle F_{1}~=~F_{2}~=~F_{b}}
, we arrive at
Equivalent Spring Constant (Parallel)
Both springs are touching the block in this case, and whatever distance spring 1 is compressed has to be the same amount spring 2 is compressed.The force on the block is then:Fb{\displaystyle F_{b}\,}=F1+F2{\displaystyle =F_{1}+F_{2}\,}=−k1x−k2x{\displaystyle =-k_{1}x-k_{2}x\,}So the force on the block isFb=−(k1+k2)x.{\displaystyle F_{b}=-(k_{1}+k_{2})x.\,}Which is how we can define the equivalent spring constant askeq=k1+k2.{\displaystyle k_{\mathrm {eq} }=k_{1}+k_{2}.\,} | Fb{\displaystyle F_{b}\,} | =F1+F2{\displaystyle =F_{1}+F_{2}\,} | | =−k1x−k2x{\displaystyle =-k_{1}x-k_{2}x\,}
Fb{\displaystyle F_{b}\,} | =F1+F2{\displaystyle =F_{1}+F_{2}\,}
| =−k1x−k2x{\displaystyle =-k_{1}x-k_{2}x\,}
The force on the block is then:
Fb{\displaystyle F_{b}\,} | =F1+F2{\displaystyle =F_{1}+F_{2}\,}
| =−k1x−k2x{\displaystyle =-k_{1}x-k_{2}x\,}
So the force on the block is
Which is how we can define the equivalent spring constant as
Compressed Distance
In the case where two springs are in parallel it is immediate that:x1=x2{\displaystyle x_{1}=x_{2}\,}F1=−k1x1{\displaystyle F_{1}=-k_{1}x_{1}}andF2=−k2x2.{\displaystyle F_{2}=-k_{2}x_{2}.\,} | x1=x2{\displaystyle x_{1}=x_{2}\,} | F1=−k1x1{\displaystyle F_{1}=-k_{1}x_{1}}andF2=−k2x2.{\displaystyle F_{2}=-k_{2}x_{2}.\,} | In the case where two springs are in series, the force of the springs on each other are equal:F1=F2{\displaystyle F_{1}=F_{2}\,}−k1x1=−k2x2.{\displaystyle -k_{1}x_{1}=-k_{2}x_{2}.\,}From this we get a relationship between the compressed distances for thein seriescase:x1x2=k2k1.{\displaystyle {\frac {x_{1}}{x_{2}}}={\frac {k_{2}}{k_{1}}}.\,} | F1=F2{\displaystyle F_{1}=F_{2}\,} | −k1x1=−k2x2.{\displaystyle -k_{1}x_{1}=-k_{2}x_{2}.\,}
x1=x2{\displaystyle x_{1}=x_{2}\,}
F1=−k1x1{\displaystyle F_{1}=-k_{1}x_{1}}andF2=−k2x2.{\displaystyle F_{2}=-k_{2}x_{2}.\,}
F1=F2{\displaystyle F_{1}=F_{2}\,}
−k1x1=−k2x2.{\displaystyle -k_{1}x_{1}=-k_{2}x_{2}.\,}
x1=x2{\displaystyle x_{1}=x_{2}\,}
F1=−k1x1{\displaystyle F_{1}=-k_{1}x_{1}}andF2=−k2x2.{\displaystyle F_{2}=-k_{2}x_{2}.\,}
F1=F2{\displaystyle F_{1}=F_{2}\,}
−k1x1=−k2x2.{\displaystyle -k_{1}x_{1}=-k_{2}x_{2}.\,}
From this we get a relationship between the compressed distances for the in series case:
Energy Stored
For theseriescase, the ratio of energy stored in springs is:E1E2=12k1x1212k2x22,{\displaystyle {\frac {E_{1}}{E_{2}}}={\frac {{\frac {1}{2}}k_{1}x_{1}^{2}}{{\frac {1}{2}}k_{2}x_{2}^{2}}},\,}but there is a relationship between x1and x2derived earlier, so we can plug that in:E1E2=k1k2(k2k1)2=k2k1.{\displaystyle {\frac {E_{1}}{E_{2}}}={\frac {k_{1}}{k_{2}}}\left({\frac {k_{2}}{k_{1}}}\right)^{2}={\frac {k_{2}}{k_{1}}}.\,}For theparallelcase,E1E2=12k1x212k2x2{\displaystyle {\frac {E_{1}}{E_{2}}}={\frac {{\frac {1}{2}}k_{1}x^{2}}{{\frac {1}{2}}k_{2}x^{2}}}\,}because the compressed distance of the springs is the same, this simplifies toE1E2=k1k2.{\displaystyle {\frac {E_{1}}{E_{2}}}={\frac {k_{1}}{k_{2}}}.\,}GATO Ignace notes about spring associations
but there is a relationship between x1 and x2 derived earlier, so we can plug that in:
For the parallel case,
because the compressed distance of the springs is the same, this simplifies to
GATO Ignace notes about spring associations
See also
References |
16063 | http://www.thphys.nuim.ie/staff/jvala/MP462_notes.pdf | Lecture notes MP462 Quantum Mechanics I Autumn Semester 2008 Jiri Vala Department of Mathematical Physics National University of Ireland, Maynooth 1 Introduction 1.1 Origins of quantum In this chapter, we will be interested in essential logical steps that led to quantum mechanics. We understand that quantum mechanics is a physical theory (the most complete we have), meaning it is a mathematically consis-tent theoretical construct which is able to explain and predict certain exper-imental observations which can not be explained within the framework of classical mechanics and classical (i.e. Maxwell) theory of electromagnetism.
The specific focus of this section will be on development of the concept of wave-particle duality for both electromagnetic field and matter in the early times of quantum mechanics. We do not intend to present detailed or com-plete history of quantum theory which is beyond the scope of this course and can be found elsewhere and neither we present complete overview of unique features of quantum mechanics in this lecture as these will be subjects of the other lectures of this course.
Further reading: The early stages of the development of quantum me-chanics with detailed discussion of relevant experiments is presented in D.
Bohm, Quantum Theory, 1951. Also an excelent (and more historical) ac-count of the same period (and particularly an important role played by Einstein) can be found in A. Pais, Subtle is the Lord: The Science and the Life of Albert Einstein, 1983.
1 1.1.1 Black-body radiation problem A black-body is a body that absorbes all the electromagnetic radiation which falls on its surface. Rather than an ideally absorbing body, it can be physically realized as a hollow cavity in which electromagnetic radiation is trapped. The radiation is in thermal equilibrium with the surounding ma-terial, that is, with material oscillators of the cavity wall which continuously emit and absorb waves of any frequency. The higher is the temperature, the higher is the mean frequency of radiation inside the cavity.
The electromagnetic energy distribution inside the cavity as a function of frequency and temperature U(ν, T) is proportional to the intensity of the radiation per unit solid angle I(ν, T) = c 4πU(ν, T). This can be measured through a small opening in the cavity (negligible compared to the area of the cavity surface) which allows the radiation to escape from the cavity with-out causing significant alteration of the radiation distribution inside. These measurents, which have been carried out since late 19th century (e.g. par-ticularly noticeable experiments are those by Rubens and Kurlbaum (1900) due to their great precision and wide range of frequencies), revealed a single peak in the frequency dependence of the intensity I(ν, T). The experimen-tal data however could not be entirely explained within the framework of classical physics. This discrepancy between experimental data and classical theoretical concepts constitutes what we know as the black-body radiation problem. Let us see first classical attempts by Wien (1896), Rayleigh and Jeans (1900) before going in more detail to quantum hypothesis by Planck (1900) which is historically considered the beginning of quantum theory.
In 1896, Wien used empirical data and thermodynamical arguments to explain the black-body radiation problem. He proposed the following dis-tribution which is now known as the Wien law: I(ν, T) = aν3e−bν/kBT (1) where a and b are empirical constants, and kB is the Boltzmann constant (1.38 × 10−23JK−1).
Wien’s law provided agreed very well with the experimental data for high frequencies, but clearly deviated from the experimental data at short frequencies where it showed lower intensity I(ν, T) than observed. Moreover, introduction of empirical constants (without really providing their physical interpretation) made this theory semiempirical.
In an atempt to explain new observations by Rubens and Kurlbaum in 1900, Rayleight used classical electromagnetism and theormodynamics to derive the radiation distribution from the first principles. His work comple-mented by additional clarifications by Jeans resulted in the Rayleigh-Jeans law: I(ν, T) = 2ν2 c2 kBT (2) The formula well reproduces experimental observations at low frequencies but rapidly diverges at high frequencies. This behaviour, i.e.monotonic in-crease of radiation intensity with frequency, is known as ultraviolet catas-trophe. Clearly this can not describe any realistic black-body radiation law.
The same year, Planck attacked the black-body problem by proposing quantum hypothesis, i.e. that the energy of electromagnetic radiation of a given frequency is emitted and absorbed in elementary packets called quanta, so the energy of the radiation of a given frequency ν is given as: En = nhν (3) h is a new fundamental constant, known as the Planck constant (6.626 × 10−34), and n is an integer. Planck then obtained the black-body radiation distribution now know as the Planck law in the following form: I(ν, T) = 2hν3 c2 e−hν/kBT 1 −e−hν/kBT (4) which perfectly reproduces experimental observations.
Outline of the Planck law derivation.
The probability W(n) that an classical oscillator has energy corresponding to its n-th harmonic (i.e. n-th value) is given as W(n) ∝e−E/kBT . Assuming that the same form holds also in the case of a quantum mechanical oscillator, whose energy is En = nhν, the probability is W(n) ∝e−En/kBT . The normalization of this probability, which ensures that P∞ n=0 W(n) = 1, then lead to the equation W(n) = e−nhν/kBT P∞ n=0 e−nhν/kBT = e−nhν/kBT (1 −e−hν/kBT ) (5) To get the left hand side (i.e. l.h.s.) of the equation, we used the formula P∞ n=0(e−x)n = (1 −e−x)−1.
Now we can calculate the mean energy < E >= ∞ X n=0 EnW(n) = hν(1 −e−hν/kBT ) ∞ X n=0 ne−nhν/kBT (6) which can be rewritten using the relation P∞ n=0 ne−nx = −d dx P∞ n=0 e−nx = −d dx 1 1−e−x = e−x (1−e−x)2 as < E >= hνe−hν/kBT 1 −e−hν/kBT (7) and multiplying this by the density of oscillators with the frequency ν , which is given as 8π c3 ν2, we get the energy distribution U(ν) and from that the desired intensity as given by Eq. (4).(Remark: derivation of the U(ν) is presented in D. Bohm, Quantum Theory, 1951; we may add this later to these notes, however it is not essential in the present context).
The classical results by Wien and by Rayleigh and Jeans agree with experimental data respectively at high and low frequencies. Hence, it must be possible to reconstruct both classical laws as appropriate limits of the Planck law. For small frequencies, i.e.
hν kBT << 1, we can use the Taylor series to approximate e−hν/kBT as 1 −−hν kBT . Straightforward substitution of this relation into Eq.(4) gives the Rayleight-Jeans law I(ν, T) = 2ν2 c2 kBT.
In the opposite limit, i.e.
hν kBT >> 1, the denominator in Eq.(4) can be approximated 1 −e−hν/kBT ∼1 and thus neglected. Eq.(4) then acquires the form of the Wien law I(ν, T) = 2hν3 c2 e−hν/kBT . Note that the empirical constant a and b now got meaning.
1.2 Heat capacities problem For a few years after its conception, the Planck quantization hypothesis lived only within the narrow field of black-body elelctromagnetic radiation.
In 1906, Einstein extended it also to material systems and used it to resolved the problem of the specific heat capacities CV = ∂ ∂T of solids. The rele-vant experimental observations showed that at high temperature the specific heat capacity CV of solid materials converged to a plateau where CV = kB while at low temperatures it converges to zero. To explain the temperature dependence of CV , Einstein proposed that material oscilators (i.e. atoms bind in crystal structure) are quantized and, in contrast to electromagnetic waves which can have various frequencies, it permits only one frequency for all oscillators. The mean energy is given by Eq.(7) and the specific heat capacity is then expressed as CV = ∂< E > ∂T = h2ν2 kBT e−hν/kBT (1 −e−hν/kBT )2 (8) The high temperature limit when hν kBT →0 then properly reproduces ex-perimental observations giving CV = kB. The low temperature behaviour is given as CV = h2ν2 kBT 2 e hν kBT .
Precise measurements reveal that the ex-perimental data obtained for realistic materials differ from theoretical low temperature behaviour; this is explained by mutual coupling of the material oscillators neglected in Einstein’s approach.
1.3 Photoelectric effect Photoelectric effect was first observed by Lennard (1902) and theoretically explained by Einstein (1905) be proposing that light consists of particles called photons. The photoelectric effect is the process where ultraviolet (UV) light which falls on a metalic surface ejects electrons with the kinetic energy 1 2mev2 from the metal. Here me and v are the electron mass and velocity respectively. Important is that the kinetic energy of outgoing electrons only depend on frequency of the incoming radiation but not on its intensity.
Obviously, the electrons absorb the energy only in quanta, so that their kinetic energy is given as 1 2mev2 = hν −W where W is a work function of the metal (it is a material constant which characterizes the binding energy of electron in the metal) and hν is the energy of the UV light. In 1905, Einstein proposed that this experimental observations can be explain if we accept that the reason why the light energy is absorbed in quanta is that light consists of particles, photons, each carrying energy quantum hν.
1.4 Compton scattering The Compton scattering, discovered experimentally by Compton (1922), is a process when an electromagnetic radiation with the initial momentum ⃗ pin scatters through a stationary electron. The momentum of the outgoing elec-tron is ⃗ Pout. The electromagnetic radiation leaves the scattering event with the momentum ⃗ pout along the trajectory which is deflected from the original trajectory by an angle θ. The wavelength of the outgoing radiation is shifted compared to the incoming radiation as ∆λ = λout −λin. This experiment is direct demonstration of the particle-like character of electromagnetic field.
1.5 de Broglie hypothesis In 1923, de Broglie postulated particle-wave duality to material particles.
In analogy to light which can be regarded as both electromagnetic waves and photons, every massive particle of the momentum ⃗ p can be regarded as a wave whose wavelength is given as λ = h |⃗ p| (9) where ⃗ p = m⃗ v is a momentum. It is worth to note that this hypothesis was not based on any existing experimental observations but it rather predicted that the wave-like phenomena of material particles to be seen in future.
This happened in experiments by Davisson and Germes in 1927, who ob-served difraction of electrons from a crystal. With the de Broglie hypothesis, the wave-particle duality becomes an essential property of every quantum mechanical system.
1.6 Other developments Rutherford experiments (1911-1913) with scattering of α particles through atoms revealed that the most of the atomic mass is concentrated in the nu-clei which is very small. This led to the idea of planetary model of atom in which electrons are orbiting around the positively charged nucleus along Ke-pler orbits. Clearly this idea was not sustainable on the ground of classical physics as orbiting electron would be loosing energy by irradiating electro-magnetic waves and would eventually collaps to the nucleus, making atoms and thus matter quite unstable.
The experiments by Franck and Hertz (1913) showed that spectral terms En/h are stable and definitive for each atom. Consequently Bohr postulated that a new model of atom in which electrons are orbiting only on fixed discrete trajectories. This represents an important step towards the quantum theory of atoms and molecules which we will encounter towards the end of the course.
Other important developments include formulation of matrix mechan-ics (1925), and also development of Schroedinger wave mechanics (1926).
It has been shown later that both approaches are equivalent. We will be discussing these within the other lectures of this course along with other strange features of quantum mechanics including e.g. uncertainty relations, entanglement, measurement problem etc.
2 Mathematical foundations A state of a physical system is represented by a vector in the Hilbert space.
Composition of Hilbert spaces Let H1 be a subspace of the Hilbert space H. For ∀| ψ⟩orthogonal to all the vectors in H1 form a vector space H⊥ 1 (an orthogonal complement H1) .
Let H1, H2, ..., Hn be subspaces of H (n can be infinite). An element of H1 can be written as a linear combination of vectors from H1, H2, ..., Hn for all i ̸= k = 1, 2, ..., n and for all | ψi⟩∈Hi and | ψk⟩∈Hk such that ⟨ψi|ψk⟩= 0, we say that H is a direct sum of (mutually orthogonal) sub-spaces H1, H2, ..., Hn H = H1 ⊕H2 ⊕... ⊕Hn = k=n M j=1 Hk (10) Important role in quantum mechanics is played by tensor product of Hilbert spaces. Consider H1 and H2 of the dimension as N1 and N2 re-spectively. Let | ψ(j) n ⟩, n = 1, 2, ..., N and j = 1, 2, form a basis of Hj. The vectors | ψ(1) n ⟩| ψ(2) m ⟩can be identified with elements of the orthonormal basis of the Hilbert space H, with the dimension dim(H) = N1 × N2 | φ⟩= | ψ(1) n ⟩| ψ(2) m ⟩ (11) where j = (n −1)N2 + m. The scalar product is defined as ⟨φj|φj′⟩= ⟨ψ(1) n |ψ(1) n′ ⟩⟨ψ(2) n |ψ(2) n′ ⟩= δnn′δmm′ = δjj′ (12) We say that H is a tensor product of H1 and H2, H = H1 ⊗H2.
For ∀φ ∈H = H1 ⊗H2) can be written as | φ⟩= N X j=1 aj| φj⟩= N1 X n=1 N2 X m=1 = anm| ψ(1) n ⟩| ψ(2) m ⟩ (13) and if anm = αnβm, then | φ⟩= | A⟩| B⟩where | A⟩= PN1 n=1 αn| ψ(1) n ⟩∈H and | B⟩= PN2 m=1 βn| ψ(2) n ⟩∈H. we say that ketφ is a direct product of the vectors | A⟩and | B⟩.
We can define concepts analogous for real functions also for vector func-tions. Let ξ ∈C, a complex vector function is a map ξ →| ψ(ξ)⟩where | ψ(ξ)⟩is a vector. For example we can define a derivative through the con-cept of limit as done in mathematical analysis. We can use this to define derivative of an inner product d dξ ⟨φ(ξ)|ψ(ξ)⟩= ( d dξ ⟨φ(ξ) |)| ψ(ξ)⟩+ ⟨φ(ξ) |( d dξ | ψ(ξ)⟩) (14) where the first term of r.h.s.
corresponds to the bra vector of the ket d dξ| φ(ξ)⟩.
Examples of Hilbert spaces 1. Un space is n-dimensional Hilbert space formed by all (1×n)-matrices whose entries are complex numbers | ψ⟩= a1 a2 ...
an (15) where multiplication by a complex scalar and sum are carried out in a usual way. The inner product is defined as ˆ A = a∗ 1a∗ 2...a∗ n a1 a2 ...
an = n X j=1 a∗ jbj (16) 2. Space l2 is an infinite dimensional vector space of (1 × ∞)-matrices such that P∞ j=1 |aj|2 < ∞.
3. Space L2(a, b) is formed by all complex functions A(ξ) of a real variable ξ that are square integrable (in the sense of the Lebesgue integral), i.e.
Z b a |A(ξ)|2dξ < ∞ (17) The scalar product is defined as follows ⟨A|B⟩= Z b a A(ξ)∗B(ξ)dξ (18) 2.1 Operator theory Definition.
An operator ˆ A between the Hilbert spaces U and V is a func-tion which to every vector | ψ⟩∈U assigns a vector | ψ⟩′ ∈V. We write | ψ⟩′ = ˆ A| ψ⟩ (19) ˆ U is called the domain of the operator ˆ A, D( ˆ A). The set R( ˆ A) = {| ψ⟩′ = ˆ A| ψ⟩|∀| ψ⟩∈U} is called the range of the operator ˆ A.
Examples.
We list several simple examples of operators: 1. The operator ˆ A1 = λ ∈C acts on a vector | ψ⟩as ˆ A1| ψ⟩= λ| ψ⟩.
Note that if λ = 1, this operator acts as the identity operator ˆ 1: ˆ 1| ψ⟩= | ψ⟩.
2. ˆ A2| ψ⟩= 1 √ ⟨ψ|ψ⟩| ψ⟩normalizes a vector | ψ⟩.
3. ˆ A3| ψ⟩= ⟨ψ|ψ⟩| ψ⟩multiplies a vector | ψ⟩by the square of its norm.
4. ˆ A4 = | φ⟩⟨φ | acts as ˆ A4| ψ⟩= ⟨φ|ψ⟩| φ⟩.
The operators ˆ A4 and ˆ A3 look similar but there is actually a funda-mental difference between them. We can see this if we apply them on a quantum state | ψ⟩= c1| ψ1⟩+ c2| ψ2⟩: ˆ A4| ψ⟩= ˆ A4(c1| ψ1⟩+ c2| ψ2⟩) = c1 ˆ A4| ψ1⟩+ c2 ˆ A4| ψ2⟩ (20) here we say that the operator ˆ A4 is linear. However no such relation holds for ˆ A3 as we can see from the following ˆ A3| ψ⟩= ⟨ψ|ψ⟩| ψ⟩= (|c1|2⟨ψ1|ψ1⟩+ c∗ 1c2⟨ψ1|ψ2⟩+ c1c∗ 2⟨ψ2|ψ1⟩+ |c2|2⟨ψ2|ψ2⟩)| ψ⟩̸= c1 ˆ A3| ψ1⟩+ c2 ˆ A3| ψ2⟩= c1⟨ψ1|ψ1⟩| ψ1⟩+ c2⟨ψ2|ψ2⟩| ψ2⟩ (21) Linearity and basic algebraic properties of operators.
Linearity of operators is a very important property as operators relevant to quantum mechanics are linear operators. We say that an operator ˆ A is linear iffit satisfies ˆ A X i ci| ψ⟩= X i ci ˆ A| ψ⟩ (22) In this sense, quantum mechanics is a linear theory.
Let ˆ A, ˆ B and ˆ C be linear operators. We say that 1. ˆ A and ˆ B are equal, ˆ A = ˆ B, if ˆ A| ψ⟩ = ˆ B| ψ⟩ D( ˆ A) = D( ˆ B).
(23) for ∀| ψ⟩.
2. ˆ C is a sum of ˆ A and ˆ B, ˆ C = ˆ A + ˆ B, if ˆ C| ψ⟩= ˆ A| ψ⟩+ ˆ B| ψ⟩.
(24) for ∀| ψ⟩.
3. ˆ C is a product of ˆ A and ˆ B, ˆ C = ˆ A ˆ B, ˆ C| ψ⟩= ˆ A ˆ B| ψ⟩= ˆ A( ˆ B| ψ⟩) = ˆ A| ˆ Bψ⟩.
(25) for ∀| ψ⟩.
Specifically, note that ˆ A2 = ˆ A ˆ A and analogously ˆ An = ˆ A ˆ An−1. If a function f can be expanded in a power series f(ξ) = X n anξn (26) where ξ is a variable, then by the function of an operator f( ˆ A) we mean f( ˆ A) = X n an ˆ An (27) A very useful example is an exponential function e ˆ A = ∞ X n=0 1 n!
ˆ An (28) Commutator and anti-commutator.
In contrast to numbers, real or complex, a product of operators is generally not commutative which means that ˆ A ˆ B ̸= ˆ B ˆ A.
For example, let us have three vectors | x⟩, | y⟩, and | z⟩, and the operators ˆ Rx and ˆ Ry such that ˆ Rx| x⟩ = | x⟩ ˆ Rx| y⟩ = | z⟩ ˆ Rx| z⟩ = −| y⟩ ˆ Ry| x⟩ = −| z⟩ ˆ Ry| y⟩ = | y⟩ ˆ Ry| z⟩ = | x⟩ (29) then ˆ Rx ˆ Ry| z⟩= ˆ Rx| x⟩= | x⟩̸= ˆ Ry ˆ Rx| z⟩= −ˆ Ry| y⟩= −| y⟩ (30) An operator [ ˆ A, ˆ B] = ˆ A ˆ B −ˆ B ˆ A is called commutator of the operators ˆ A and ˆ B. If [ ˆ A, ˆ B] = 0, we say that ˆ A and ˆ B commute. If [ ˆ A, ˆ B] = 0 then also [f( ˆ A), f( ˆ B)] = 0.
An operator { ˆ A, ˆ B} = ˆ A ˆ B + ˆ B ˆ A is called anticommutator of the opera-tors ˆ A and ˆ B. If { ˆ A, ˆ B} = 0, we say that ˆ A and ˆ B anticommute.
Basic properties of commutators and anticommutators are summarized below: [ ˆ A, ˆ B] = −[ ˆ B, ˆ A] (31) [ ˆ A, ˆ B + ˆ C] = [ ˆ A, ˆ B] + [ ˆ A, ˆ C] (32) [ ˆ A, ˆ B ˆ C] = [ ˆ A, ˆ B] ˆ C + ˆ B[ ˆ A, ˆ C] = { ˆ A, ˆ B} ˆ C −ˆ B{ ˆ A, ˆ C} (33) [ ˆ A ˆ B, ˆ C] = ˆ A[ ˆ B, ˆ C] + [ ˆ A, ˆ C] ˆ B = ˆ A{ ˆ B, ˆ C} −{ ˆ A, ˆ C} ˆ B (34) { ˆ A, ˆ B} = { ˆ B, ˆ A} (35) { ˆ A, ˆ B ˆ C} = { ˆ A, ˆ B} ˆ C −ˆ B[ ˆ A, ˆ C] = ˆ B{ ˆ C, ˆ A} −[ ˆ B, ˆ A] ˆ C (36) { ˆ A ˆ B, ˆ C} = ˆ A{ ˆ B, ˆ C} −[ ˆ A, ˆ C] ˆ B = { ˆ C, ˆ A} ˆ B −ˆ A[ ˆ B, ˆ C] (37) Another important property is given by the Jacobi identity: [ ˆ A[ ˆ B, ˆ C]] + [ ˆ B[ ˆ C, ˆ A]] + [ ˆ C[ ˆ A, ˆ B]] = 0 (38) Classes of operators An operator ˆ A is called bounded iff∃β > 0 such that: ∥ˆ A| ψ⟩∥≤β∥| ψ⟩∥ (39) for ∀| ψ⟩∈D( ˆ A).
The symbol ∥.∥means the norm.
The infimum of β is called the norm of an operator. In finite dimensional Hilbert spaces, all operators are bounded. Sum and product of bounded operators is a bounded operator because ∥ˆ A + ˆ B∥≤∥ˆ A∥+ ∥ˆ B∥ ∥ˆ A ˆ B∥≤∥ˆ A∥∥ˆ B∥ (40) Specially ∥λ ˆ A∥= |λ|∥ˆ A∥ (41) An operator ˆ A is called symmetric if ⟨ψ1| ˆ Aψ2⟩= ⟨ˆ Aψ1|ψ2⟩ (42) for ∀| ψ1⟩, | ψ2⟩∈D( ˆ A) (dense in the Hilbert space H).
An operator ˆ A is called hermitian if it is bounded and symmetric. It is called antihermitian if the operator i ˆ A is hermitian.
Let ˆ A be a bounded operator with the domain dense in the Hilbert space H, then there is an adjoint operator ˆ A† such that ⟨ψ1| ˆ A†ψ2⟩= ⟨ˆ Aψ1|ψ2⟩ (43) that is ⟨ψ1| ˆ A†ψ2⟩= ⟨ψ2| ˆ Aψ1⟩∗ (44) for ∀| ψ1⟩, | ψ2⟩∈D( ˆ A).
The following identities hold ∥ˆ A†∥ = ∥ˆ A∥ (45) ( ˆ A†)† = ˆ A (46) ( ˆ A + ˆ B)† = ˆ A† + ˆ B† (47) ( ˆ A ˆ B)† = ˆ B† ˆ A† (48) (λ ˆ A)† = λ∗ˆ A† (49) and, for some fixed vectors | φ1⟩and | φ2⟩we can define an operator ˆ Aj,k = | φj⟩⟨φk | (where j, k ∈{1, 2}) then ˆ A† j,k = ˆ Ak,j (50) An operator ˆ A is self-adjoint if ˆ A† = ˆ A (51) These operators are particularly important in quantum mechanics as they represent observable physical quantities; hence we call these operators ob-servables. As we will see later, they possess real spectrum, i.e. their eigen-values are real numbers R.
An operator ˆ A is positive if ⟨ψ | ˆ A| ψ⟩≥0for∀| ψ⟩.
An operator ˆ A is normal if [ ˆ A, ˆ A†] = 0.
Let ˆ A be an operator. If there exist an operator ˆ A−1 such that ˆ A ˆ A−1 = ˆ A−1 ˆ A = ˆ 1, then it is called an inverse operator to ˆ A. The following proper-ties hold ( ˆ A ˆ B)−1 = ˆ B−1 ˆ A−1 (52) ( ˆ A†)−1 = ( ˆ A−1)†.
(53) Remark: in finite dimensional space ˆ A ˆ B = 1 implies ˆ B ˆ A = 1 but this is generally not true in the infinite dimensional spaces and that is why we need both ˆ A ˆ A−1 and ˆ A−1 ˆ A above.
An operator ˆ U is called unitary if ˆ U† = ˆ U−1 and D( ˆ U) = H.
This is a very important class of operators as they are formal solution of the Sch¨ odinger equation and thus represent quantum dynamics. Also important groups of symmetries are unitary.
The unitary operators are isometric ⟨ˆ Uψ1| ˆ Uψ2⟩= ⟨ψ1|ψ2⟩.Also, let the set {| ψi⟩} is a basis of the Hilbert space H and ˆ U is a unitary operator, then { ˆ U| ψi⟩} is also a basis of H.
The fundamental theorem by Wigner significantly restricts what oper-ators can represent symmetry operation on the Hilbert space of physical states. Consider maps of H onto itself | ψ⟩→| ψ′⟩such that |⟨ψ1|ψ2⟩= ⟨ψ′ 1|ψ′ 2⟩, ∀| ψ1,2⟩∈H, then | ψ′⟩= eiω(ψ) ˆ U| ψ⟩where ˆ U is a unitary operator which is either linear or antilinear and ω(ψ) ∈R is a phase that may depend on | ψ⟩.
A bounded operator ˆ P satisfying ˆ P = ˆ P † = ˆ P 2 (54) is called a projection operator or simply projector.
This class of opera-tors is very important in quantum mechanics, particularly in the context of quantum measurement.
Let ˆ P1 be a projector, then ˆ P2 = ˆ 1 −ˆ P1 is also a projector, with the following properties: ˆ P1 + ˆ P2 = ˆ 1 (55) ˆ P1 ˆ P2 = 0 (56) where the earlier is called the completeness relation (se below) and the latter expresses orthogonality.
For every vector | ψ⟩∈H the following holds | ψ⟩= | ψ1⟩+ | ψ2⟩ ⟨ψ1|ψ2⟩= 0 | ψ1⟩= ˆ P1| ψ⟩ (57) | ψ2⟩= ˆ P2| ψ⟩ (58) All vectors | ψj⟩, (j = 1, 2) for a subspace Hj ⊂H, that is the projectors ˆ P1 and ˆ P2 assign to each vector from H its projection onto the subspace H1 and H2 respectively. These projections are mutually orthogonal as can be see from the relations above.
If the projector operators ˆ P1 and ˆ P2 satisfy ˆ P1 ˆ P2 = ˆ P1 then the corre-sponding subspaces satisfy H1 ⊂H2; we then say ˆ P1 ≤ˆ P2.
Let | ψk⟩be a normalized vector, then the operator ˆ Pk = | ψk⟩⟨ψk | (59) is projection onto one-dimensional space spanned by all vectors linearly de-pendent on | ψk⟩.
If vectors {| ψk⟩} form a basis of H1 then X k ˆ Pk = X k | ψk⟩⟨ψk | (60) is projection operator onto H1. Furthermore if H1 = H then this projection is an identity operator X k | ψk⟩⟨ψk | = ˆ 1 (61) This relation known as the completeness relation has a prominent role in quantum mechanics.
Composition of operators Let H1 ⊂H. We say that H1 is invariant with respect to an operator ˆ A if ∀| ψ1⟩∈H1, ˆ A| ψ1⟩∈H1.
We say that H1 reduces ˆ A if the subspace H1 and its orthogonal com-plement H⊥ 1 are invariant with respect to ˆ A.
Let Hi, i = 1, 2, ..., k, reduce the operator ˆ A. Let us define operators ˆ Ai with domain D( ˆ Ai) = Hi such that ∀| ψi⟩∈Hi, ˆ Ai| ψi⟩= ˆ A| ψi⟩. Then ˆ A| ψi⟩= k X i=1 ˆ Ai| ψi⟩ (62) We say that ˆ A is a direct sum of the operators ˆ Ai ˆ A = ˆ A1 ⊕ˆ A2 ⊕... ⊕ˆ Ak = k M i=1 ˆ Ai (63) Example: An operator ˆ A in the Hilbert space U5 ˆ A = a11 a12 0 0 0 a21 a22 0 0 0 0 0 a33 a34 a35 0 0 a43 a44 a45 0 0 a53 a54 a55 (64) is the direct sum ˆ A = ˆ A1 ⊕ˆ A2 of the operators ˆ A1 and ˆ A2 ˆ A1 = a11 a12 a21 a22 (65) ˆ A2 = a33 a34 a35 a43 a44 a45 a53 a54 a55 (66) where ˆ A1 and ˆ A2 act in the Hilbert spaces U2 and U3 respectively.
The following properties of the direct sum are important Tr( ˆ A1 ⊕ˆ A2) = Tr ˆ A1 + Tr ˆ A2 (67) det( ˆ A1 ⊕ˆ A2) = det ˆ A1.det ˆ A2 (68) If ˆ A = ˆ A1 ⊕ˆ A2 and ˆ B = ˆ B1 ⊕ˆ B2, and ˆ A1 and ˆ B1 act in the same Hilbert space, and so does ˆ A2 and ˆ B2 , then ˆ A ˆ B = ˆ A1 ˆ B1 ⊕ˆ A2 ˆ B2 .
A particularly important in the context of quantum mechanics is the direct product of operators which describes the composition of operators acting on distinct subsystems of a quantum mechanical system. Let H = H1 ⊗H2, and ˆ Aj, j = 1, 2 is an operator in Hj with the domain D( ˆ Aj). An operator ˆ A = ˆ A1 ⊗ˆ A2 in H is a direct product of ˆ A1 and ˆ A2 and is defined as ˆ A| ψ1⟩| ψ2⟩= | ˆ A1ψ1⟩| ˆ A2ψ2⟩ (69) for ∀| ψj⟩∈D( ˆ Aj), and its domain is D( ˆ A) = D( ˆ A1)D( ˆ A2).
A useful example of a direct product of operators can be again given within the finite dimensional case. Let us have the operators ˆ B = b11 b12 b13 b21 b22 b23 b31 b32 b33 (70) ˆ C = c11 c12 c21 c22 (71) the direct product is then constructed as ˆ A = ˆ B ⊗ˆ C = b11c11 b11c12 b12c11 b12c12 b13c11 b13c12 b11c21 b11c22 b12c21 b12c22 b13c21 b13c22 b21c11 b21c12 b22c11 b22c12 b23c11 b23c12 b21c21 b21c22 b22c21 b22c22 b23c21 b23c22 b31c11 b31c12 b32c11 b32c12 b33c11 b33c12 b31c21 b31c22 b32c21 b32c22 b33c21 b33c22 (72) We of course notice that ˆ A = ˆ B ⊗ˆ C = b11 ˆ C b12 ˆ C b13 ˆ C b21 ˆ C b22 ˆ C b23 ˆ C b31 ˆ C b32 ˆ C b33 ˆ C (73) Eigenvalues and eigenvectors.
Let | ψα⟩be a nonzero vector such that ˆ A| ψα⟩= α| ψα⟩ (74) then α is called an eigenvalue of ˆ A and | ψα⟩is called an eigenvector of ˆ A corresponding to the eigenvalue α. If there is n > 1 vectors satisfying this equation for α we then say that eigenvalue α is n-fold degenerate.
The eigenvalues of a self-adjoint operator ˆ A, which represent physical quantities in quantum mechanics, are real numbers: α⟨ψα|ψα⟩= ⟨ψα| ˆ Aψα⟩= ⟨ˆ Aψα|ψα⟩∗= α∗⟨ψα|ψα⟩ (75) and thus α = α∗which is indeed possible only for real numbers.
Eigenvectors of self-adjoint operators corresponding to distinct eigenval-ues are orthogonal. If β ̸= α is also an eigenvalue of ˆ A then ⟨ψα| ˆ Aψβ⟩= β⟨ψα|ψβ⟩and also ⟨ψα| ˆ Aψβ⟩= ⟨ψβ| ˆ Aψα⟩∗= α⟨ψβ|ψα⟩∗= α⟨ψα|ψβ⟩ (76) which implies that ⟨ψα|ψβ⟩= 0.
Operator is uniquely defined from its action on the basis vectors of the Hilbert space. Let B = {| ψj⟩} is a basis of H(= D( ˆ A)), ˆ A| ψj⟩= X k | ψk⟩⟨ψk | ˆ A| ψj⟩= X k Akj| ψk⟩ (77) where Akj = ⟨ψk | ˆ A| ψj⟩are the matrix elements of the operator ˆ A in the matrix representation given by the basis B.
For practical calculations we write the operator ˆ A as ˆ A = X kj | ψk⟩⟨ψk | ˆ A| ψj⟩⟨ψj | = X kj Akj| ψk⟩⟨ψj | (78) We can now define an operator by its eigenrepresentation. Assume that the eigenvectors of ˆ A define a basis of the Hilbert space, B = {| ψj⟩} , then Akj = ⟨ψk | ˆ A| ψj⟩= αjδkj. Operator in its eigenrepresentation is a diagonal matrix with eigenvalues on the diagonal ˆ A = X kj | ψk⟩⟨ψk | ˆ A| ψj⟩⟨ψj | = X kj Akj| ψk⟩⟨ψj | (79) = X j αj| ψj⟩⟨ψj | = X j αj ˆ Ej (80) where ˆ Ej is a projector to a one-dimensional space defined by | ψj⟩. We call the last expression also a spectral decomposition of the operator ˆ A.
Process of obtaining eigenvalues and eigenvectors of operators is the most important part of solving quantum mechanical systems. Let A be a matrix representation of the operator ˆ A. To find the eigenvalues (and eigenvectors), we are looking for unitary matrix U that would transform the matrix A into its diagonal form D, D = U†AU. Two operators commute iffthere exist a unitary transformation which diagonalizes them simultaneously.
In practice, to get the eigenvalues, we have to solve the characteristic (or secular) equation defined as det(A −λ1) = 0 (81) where 1 is the identity matrix.
Given a spectral decomposition of an operator ˆ A = P j αj ˆ Ej. A function of the operator ˆ A can be defined as f( ˆ A) = P j f(αj) ˆ Ej. Specially if f = 1 we get the completeness relation 1 = P j ˆ Ej.
Complete set of observables.
Let ˆ A(1), ˆ A(2), ..., ˆ A(M) be a set of mutu-ally commuting operators. We say that these operators form a complete set of commuting operators if each of the nonzero operators ˆ I(12...M) j1j2...jM = ˆ I(1) j1 ˆ I(2) j2 ...ˆ I(M) jM (82) project to ne-dimensional subspace, i.e.
for each M-tuple of eigenvalues α(1) j1 α(2) j2 ...α(M) jM , there is maximally one linearly independent vector | α(1) j1 , α(2) j2 , ..., α(M) jM ⟩ (83) satisfying ˆ A(k)| α(1) j1 , α(2) j2 , ..., α(M) jM ⟩= α(k) jk | α(1) j1 , α(2) j2 , ..., α(M) jM ⟩ (84) where k = 1, ..., M and ∥| α(1) j1 , α(2) j2 , ..., α(M) jM ⟩∥= 1. In this case the projec-tors are explicitly ˆ I(12...M) j1j2...jM = | α(1) j1 , α(2) j2 , ..., α(M) jM ⟩⟨α(1) j1 , α(2) j2 , ..., α(M) jM | (85) Mutually commuting operators ˆ A(1), ˆ A(2), ..., ˆ A(M) form a complete set of commuting operators iffan arbitrary operator commuting with all ˆ A(k), k = 1, ..., M is a function of the operators ˆ A(1), ˆ A(2), ..., ˆ A(M).
Generalization to the continuous spectrum Let ˆ A is an arbitrary operator with a continuous spectrum ˆ A| α⟩= α| α⟩. The normalization is given as ⟨α|α′⟩= δ(α −α′) where δ(α −α′) which we can understand as a generalization of the Kronecker delat for the continuous case (more details about the delta function as presented at lectures can be found in C. Cohen-Tannoudji, Quantum Mechanics II, the Appendix II).
The spectral decomposition in the continuous case is defined as ˆ A = Z αmax αmin α| α⟩⟨α |dα (86) and the completeness relation is Z αmax αmin | α⟩⟨α |dα = 1 (87) We can use the completeness relation to define or change the representation of a state vector.
We can define a complex function of a real variable α, a wavefunction, ψ(α) = ⟨α|ψ⟩ (88) and the inner product of wavefunctions ⟨ψ1|ψ2⟩= Z αmax αmin ψ∗ 1(α)ψ2(α) (89) Remark: The mathematical foundations of quantum mechanics are well (though slightly differently) presented in C. Cohen-Tannoudji et al., Quan-tum Mechanics I, Chapter II.
3 Formalism of quantum mechanics Classically, the result of any measurement of a system of N particles can be predicted if the values of 3N coordinates and 3N momenta are known. A state of a system is known if results of all independent measurements are known.
Quantum mechanically, it is impossible to carry out simultaneous mea-surement of coordinate and momenta with arbitrary accuracy. A state of a quantum mechanical system is determined by the most complete measure-ment, i.e. simultaneous measurement of all independent physical quantities which are compatible. These quantities form the complete set of observables.
The quantities A and B are compatible if the measurement of A in no way disturbs the measurement of B. If A is compatible with B, then B is compatible with A. A and B can be measured simultaneously.
For measurements of some observables, the results form a discrete spec-trum which is characteristic for a given observable of the system.
Values {a1}, {a2}, ... , {an}, ... which can be obtained for measurement of the complete set of observables {A} are called eigenvalues of {A}.
Example: For a spinless particle the complete set of observables is given by the energy ˆ E, angular momentum ˆ l and a component of angular momen-tum ˆ lz, { ˆ E, ˆ l, ˆ lz}. The corresponding eigenvalues form the set {Ej, lj, mj}.
3.1 The concept of filter The devices defined below will allow us to present formalism of quantum theory in a unified fashion. A good example of these devices is Stern-Gerlach apparatus which is well described in C. Cohen-Tannoudji et al., Quantum Mechanics I, Chapter 4.
Separator S{A} . A separator is a device such that the measurement of the complete set of observables on a system which passed through its j-th channel always gives the eigenstate {aj}.
Recorder R{A} .
This devices records the channel through which the system has passed.
Measurement apparatus M{A} . This apparatus consists of the sepa-rator S{A} and the recorder R{A}.
Filter FP{a}j . The filter is a separator with shutters which close some of the channels.
3.2 Filter with ideal resolution F{a}j A filter with ideal resolution is the filter with only one channel open.
Measurement of the complete set of observables {A} on a system which has passed through the filter F{a}j is the eigenvalue {a}j. A studied system is associated with the Hilbert space H.
We assign a projector ˆ P{a}j to the filter F{a}j. The projector projects onto the one-dimensional subspace represented by a normalized vector | {a}j⟩: ˆ P{a}j = | {a}j⟩⟨{a}j | (90) We assign a vector | {a}j⟩to a system which has passed through the filter F{a}j . We say that a given system is in the eigenstate {A} corresponding to the eigenvalue {a}j, or briefly to the state | {a}j⟩= | {A} = {a}j⟩.
Eigenstates {A} corresponding to different eigenvalues are related to different vectors and we will require these vectors to be orthogonal ⟨{a}j|{a}i⟩= δji (91) A system prepared in a state | {a}i⟩will pass through the filter F{a}j iff j = i: ˆ P{a}j| {a}i⟩= | {a}j⟩⟨{a}j|{a}i⟩= δji| {a}j⟩ (92) If the system passes through the filter, the state on l.h.s. is nonzero and describes the state of the system after passage through the filter. The coef-ficient δji in front of the normalized vector | {a}j⟩on the r.h.s. is the proba-bility that the measurement of {A} on the state | {a}i⟩gives the eigenvalue {a}j.
Consider now a complete set of observables {B} such that {B} ̸= {A}, i.e. some observables in {B} are incompatible with some in {A}, then for all {a}i, the result of the measurement {B} on the state | {a}i⟩can not be predicted with certainty.
Now our goal will be to predict probability that a system in the state | {a}i⟩will pass through the filter F{b}j. We first apply the relevant projector onto the initial state ˆ P{b}j| {a}i⟩= | {b}j⟩⟨{b}j|{a}i⟩= A{a}i→{b}j| {b}j⟩ (93) Interpreting A{a}i→{b}j = ⟨{b}j|{a}i⟩as the desired probability (i.e. a real number) would not allow interference phenomena. These troubles disappear if we define the desired probability as W{a}i→{b}j = |A{a}i→{b}j|2 (94) The quantity A{a}i→{b}j = ⟨{b}j|{a}i⟩is called the probability amplitude for the transition from the state | {a}i⟩to the state | {b}j⟩.
Important remarks.
• A{a}i→{b}j = ⟨{b}j|{a}i⟩implies that W{a}i→{b}j = W{b}j→{a}i, i.e.
the probability to find {b}j when the measurement {B} is carried out on the state | {a}i⟩is teh same as the probability to find {a}i when the measurement {A} is carried out on the state | {b}j⟩.
• The fact that some value {b}j will be found when measurement of {B} is carried out is expressed as X j W{a}i→{b}j = 1 (95) that is X j ⟨{a}i|{b}j⟩⟨{b}j|{a}i⟩= 1 (96) and as this is valid for any ket | {a}i⟩then X j | {b}j⟩⟨{b}j | = 1 (97) This important relation is known as the completeness relation. It is essential fo mathematical structure of quantum theory and important for practical calculations. For example, it allows to write the vector | {a}i⟩as the expansion in terms of the set {| {b}j⟩} | {a}i⟩ = X j | {b}j⟩⟨{b}j|{a}i⟩= X j ⟨{b}j|{a}i⟩| {b}j⟩ = X j A{a}i→{b}j| {b}j⟩ (98) • At this point, it seems obvious to associate the probability W{a}i→{b}j with the square of the norm of the vector ˆ P{b}j| {a}i⟩: W{a}i→{b}j = || ˆ P{b}j| {a}i⟩||2 = ⟨{a}i | ˆ P{b}j| {a}i⟩ (99) Note that here we used the idempotence of the projection operator, ˆ P 2 = ˆ P.
• Eigenstate {A} with eigenvalue {a}i can be described not only by a normalized vector | {a}i⟩but an arbitrary vector | {a}′ i⟩= λ| {a}i⟩ where λ ̸= 0. We then get for the probability W{a}i→{b}j = ⟨{a}′ i | ˆ P{b}j| {a}′ i⟩ ⟨{a}′ i|{a}′ i⟩ (100) • Vectors linearly independent on | {a}i⟩can not be describe the eigen-state {A} with the eigenvalue {a}i. Let | ψ⟩̸= 0 describes such a state, then using Eq.(100) for {A} = {B}, i = j, we get W{a}i→{a}i = |⟨ψ|{a}i⟩|2 ⟨ψ|ψ⟩ ≤⟨ψ|ψ⟩⟨{a}i|{a}i⟩ ⟨ψ|ψ⟩ = 1 (101) We used that ⟨{a}i|{a}j⟩= δij, and the Schwartz inequality |⟨φ|ψ⟩| ≤ ||φ||.||ψ|| where the equality in the last expression is valid only if | φ⟩ and | ψ⟩are linearly dependent.
Note that when the measurement of {A} on the system which passed through the filter F{a}i is carried out , it always gives the eigenvalue {a}i with certainty: W{a}i→{a}i = 1.
In conclusion, no two linearly independent vectors in H can describe the same state.
3.3 Filter with finite resolution We assign to a filter FP{b}j a projector ˆ PP{b}j with the following properties • If the system enters the filter in a state | φ⟩then it will pass through with the following probability W = ⟨φ | ˆ PP{b}j| φ⟩ ⟨φ|φ⟩ (102) and will be in the final state ˆ PP{b}j| φ⟩ (103) where ˆ P 2 P{b}j = ˆ PP{b}j (idempotence).
Consider first a filter with infinitely low resolution, i.e. all channels are open and it is impossible to distinguish through which the system passed through.
This filter is described by an operator of identity ˆ PP{b}j = P j | {b}j⟩⟨{b}j | = P j ˆ P{b}j.
• Let the initial state be | {a}i⟩, then the state after the filter is | ψ⟩= ˆ PP{b}j| {a}i⟩ (104) and the probability is W{a}i→P{b}j = ⟨{a}i | ˆ PP{b}j| {a}i⟩ (105) where W{a}i→P{b}j = X j W{a}i→{b}j (106) 3.3.1 Quantum effects Up to this point, our discussion of the formalism of quantum theory has been understandable in classical terms.Now we proceed to quantum effects and particularly the interference effect.
What is the probability that a measurement apparatus M{C} after the filter FP{b}j records the value {c}k if the initial state was | {a}i⟩?
The state of the system after FP{b}j is described by an unnormal-ized state | ψ⟩= ˆ PP{b}j| {a}i⟩.
The probability that the result of mea-surement {C} on the system in the state | ψ⟩will be {c}k is given by || ˆ P{c}k ˆ PP{b}j| {a}i⟩||2: W{a}i→P{b}j→{c}k = ⟨ψ | ˆ P{c}k| ψ⟩ (107) that is W{a}i→P{b}j→{c}k = | X j ⟨{c}k|{b}j⟩⟨{b}j|{a}i⟩|2 = | X j A{a}i →{b}jA{b}j →{c}k|2 (108) The last equation documents the origin of quantum interference effect.
To provide a concrete example, we focus on the case when only two channels {b}1 and {b}2 are open in the filter FP{b}j. The probability that a mea-surement apparatus M{C} after the filter F{b}1+{b}2 records the value {c}k if the initial state was | {a}i⟩is given as W{a}i→{b}1+{b}2→{c}k = = |A{a}i→{b}1A{b}1→{c}k + A{a}i→{b}2A{b}2→{c}k|2 = |α1β1 + α2β2|2 = = |α1β1|2 + |α2β2|2 + 2Re(α1β1α∗ 2β∗ 2) (109) where α1 = A{a}i→{b}1, β1 = A{b}1→{c}k, α2 = A{a}i→{b}2, β2 = A{b}2→{c}k.
If we shut of of the channels in the filter FP{b}j, we obtain the following probabilities W{a}i→{b}1→{c}k = |α1β1|2 W{a}i→{b}2→{c}k = |α2β2|2 These terms in Eq.(109), correspond to probabilities that a classical particle will pass either through the channel {b}1 or {b}2. Quantum particle how-ever passes through both channels simultaneously leading to the quantum interference effect represented by the last term in Eq(109), 2Re(α1β1α∗ 2β∗ 2).
We can observe the interference effect for example in the Young double slit experiment (detailed discussion of this experiment can for example be found in Feynmann’s Lectures in Physics, Vol. III, Chapter 1).
Principle of superposition and reduction of state.
Let vectors | ψ1⟩ and | ψ2⟩describe two positive states of a system, then the system can also exist in a state | ψ⟩= α1| ψ1⟩+ α2| ψ2⟩.
If we found {b}k by the measurement, then the system has collapsed to the state | {B} = {b}k⟩. This is also called reduction of a state.
3.4 Uncertainty relation Uncertainty of a result of measuring an observable ˆ A in a given state is characterized by fluctuation ∆a = p < ˆ A2 > −< ˆ A >2, where ∆a = 0 if the state is an eigenstate of ˆ A.
Only compatible, i.e. commuting, observables can be measured simulta-neously with arbitrary precision. Let us have two incompatible observables, ˆ A and ˆ B, with the commutation relation [ ˆ A, ˆ B] = i ˆ C. We will now derive the uncertainty relation.
We first introduce the following operators ∆ˆ A = ˆ A −¯ a (110) ∆ˆ A = ˆ A −¯ a (111) where ¯ a = ⟨ψ | ˆ A| ψ⟩and ¯ b = ⟨ψ | ˆ B| ψ⟩and define vectors | φ⟩= ∆ˆ A| ψ⟩ (112) | χ⟩= ∆ˆ B| ψ⟩ (113) The fluctuations of the observables ˆ A and ˆ B in the state | ψ⟩satisfy (∆a)2(∆b)2 = ⟨ψ |(∆ˆ A)2| ψ⟩⟨ψ |(∆ˆ B)2| ψ⟩=∥φ ∥2∥χ ∥2 (114) and according to the Schwartz inequality ∥φ ∥2∥χ ∥2≥|⟨φ|χ⟩|2 ⟨φ|χ⟩ = ⟨ψ |∆ˆ A∆ˆ B| ψ⟩= 1 2(⟨ψ |{∆ˆ A, ∆ˆ B}| ψ⟩+ ⟨ψ |[∆ˆ A, ∆ˆ B]| ψ⟩) = = 1 2(⟨ψ |{∆ˆ A, ∆ˆ B}| ψ⟩+ i⟨ψ | ˆ C| ψ⟩) (115) As the operators ˆ A and ˆ B are self-adjoint operators with real eigenvalues, we can rewrite |⟨φ|χ⟩|2 = 1 4[(⟨ψ |{∆ˆ A, ∆ˆ B}| ψ⟩)2 + (⟨ψ | ˆ C| ψ⟩)2] (116) The inequality ∥φ ∥2∥χ ∥2≥|⟨φ|χ⟩|2 remains satisfied if we neglect the term 1 4(⟨ψ |{∆ˆ A, ∆ˆ B}| ψ⟩)2 in the equation above. We obtain then ∆a∆b ≥1 2|⟨ψ | ˆ C| ψ⟩| (117) Example: Let ˆ A = ˆ X and ˆ B = ˆ P, the one-dimensional coordinate and momentum operators respectively. The canonical commutation relation is [ ˆ X, ˆ P] = iℏ, so ˆ C = ℏ.
We define ∆ˆ X = ˆ X −x0 and ∆ˆ P = ˆ P −p0, and following the derivation above, we can easily formulate the Heisenberg uncertainty relation ∆x∆p ≥ℏ 2 (118) In mutidimentional systems, the following holds ∆xi∆pj ≥ℏ 2δij.
3.5 Energy, time and Schr¨ odinger equation Time evolution of quantum mechanical systems described by a state vector | ψ(t)⟩is governed by the Schr¨ odinger equation iℏd| ψ(t)⟩ dt = ˆ H| ψ(t)⟩ (119) where the operator ˆ H is called the Hamiltonian and represents the total energy of the system, i.e. ˆ H = ˆ T + ˆ V where ˆ T s the kinetic energy and ˆ V is the potential energy.
The general solution of the Schr¨ odinger equation can be obtained by separation of variables and integration from the initial time t0 to the final time tf. We consider the most general case of the Hamiltonian which is explicitly time dependent, ˆ H(t), Z tf t0 d| ψ(t)⟩ | ψ(t)⟩= −i ℏ Z tf t0 ˆ H(t)dt [ln(| ψ(t)⟩)]tf t0 = −i ℏ Z tf t0 ˆ H(t)dt ln| ψ(tf)⟩−ln| ψ(t0)⟩= −i ℏ Z tf t0 ˆ H(t)dt | ψ(tf)⟩= e−i ℏ R tf t0 ˆ H(t)dt| ψ(t0)⟩= ˆ U(t0, tf) (120) The operator ˆ U(t0, tf) = e−i ℏ R tf t0 ˆ H(t)dt is the quantum evolution operator which propagates quantum mechanical system from the initial state at time t0 to the final state at time tf. If the Hamiltonian is not explicitely time dependent, i.e. ˆ H ̸= ˆ H(t), the expression for the evolution operator acquires a simpler form ˆ U(t0, tf) = e−i ℏˆ Ht (121) Remark: The formalism of quantum mechanics discussed in the Sec. 3 is well (though slightly differently) treated in C. Cohen-Tannoudji et al., Quantum Mechanics I, Chapter III.
4 Applications to simple quantum mechanical sys-tems 4.1 One-particle systems Hilbert space.
Classically, to characterize a one-particle system we need three coordinates {x1, x2, x3} and three momenta {p1, p2, p3}.
Quantum mechanically, these are represented by self-adjoint operators ˆ Xj and ˆ Pj, j = 1, 2, 3, such that [ ˆ Xj, ˆ Xk] = 0, [ ˆ Pj, ˆ Pk] = 0 and [ ˆ Xj, ˆ Pk] = iℏδjk. The Hilbert space will have tensor product structure H = H1⊗H2⊗H3 = h⊗h⊗h and the corresponding operators will become ˆ X1 = ˆ X ⊗ˆ 1 ⊗ˆ 1, ˆ P1 = ˆ P ⊗ˆ 1 ⊗ˆ 1, ˆ X2 = ˆ 1 ⊗ˆ X ⊗ˆ 1, ˆ P1 = ˆ 1 ⊗ˆ P ⊗ˆ 1, ˆ X3 = ˆ 1 ⊗ˆ 1 ⊗ˆ X, ˆ P1 = ˆ 1 ⊗ˆ 1 ⊗ˆ P where ˆ 1 is an identity operator and the self-adjoint operators ˆ X and ˆ P are defined so that they satisfy the canonical commutation relation [ ˆ X, ˆ P] = iℏ.
The values of ˆ X are continuous and from (−∞, ∞).
The most suitable Hilbert space on which these conditions for ˆ X are satisfied is the one on which the operator ˆ X forms a complete set of com-muting operators h = L2(R). This Hilbert space is formed by all complex functions A(ξ) of a real variable ξ for which there exists a (Lebesgue) integral R ∞ −∞|A(ξ)|2dξ < ∞.
Let us have the vectors | A⟩, | B⟩∈L2(R), where | A⟩= A(ξ) and | B⟩= B(ξ) are square integrable functions R ∞ −∞|A(ξ)|2dξ < ∞and R ∞ −∞|B(ξ)|2dξ < ∞(meaning their norm is finite). The linear combination of these vectors is defined as α| A⟩+ β| B⟩= αA(ξ) + βB(ξ), and the inner product is defined as ⟨A|B⟩= Z ∞ −∞ A∗(ξ)B(ξ)dξ (122) Coordinate and momentum representations.
The spectral represen-tation of ˆ X is given as ˆ X = Z ∞ −∞ x| x⟩⟨x |dx (123) and the completeness relation is ˆ X = Z ∞ −∞ | x⟩⟨x |dx = 1 (124) Every vector | φ⟩∈h is in the coordinate representation (or X-representation) described by a wavefunction φ(x) = ⟨x|φ⟩ (125) which are coefficients in the expansion of the vector | φ⟩in terms of the eigenstates of the operator ˆ X: | φ⟩= R ∞ −∞| x⟩⟨x|φ⟩dx = R ∞ −∞φ(x)| x⟩dx.
For | φ⟩1, | φ⟩2 ∈h, the inner product is defined as ⟨φ1|φ2⟩= Z ∞ −∞ φ∗ 1(x)φ2(x)dx (126) The operator ˆ P has to satisfy the canonical commutation relation [ ˆ X, ˆ P] = iℏ, i.e. ˆ X ˆ P| φ⟩−ˆ P ˆ X| φ⟩= iℏ| φ⟩. In the coordinate representation, this is x ˆ P (X)φ(x) −ˆ P (X)xφ(x) = iℏφ(x) (127) which is satisfied by ˆ P (X) = −iℏd dx (128) This operator is self-adjoint.
For ∀p ∈R, there is a solution of the equation −iℏd dxψp(x) = pψp(x) (129) and every solution linearly depends on function ψp(x) = 1 √ 2πℏ e i ℏpx (130) which satisfies the normalization condition Z ∞ −∞ ψ∗ p′(x)ψp(x)dx = δ(p −p′) (131) Similarly Z ∞ −∞ ψ∗ p(x′)ψp(x)dx = δ(x −x′) (132) ψp(x) is an eigenfunction of ˆ P in the coordinate representation corre-sponding to the eigenvalue p. The completeness relation is Z ∞ −∞ | p⟩⟨p |dp = 1 (133) and the spectral representation of ˆ P is ˆ P = Z ∞ −∞ p| p⟩⟨p |dp (134) Every vector | φ⟩∈h in the momentum representation (or P-representation) corresponds to the wavefunction φ(P)(p) = ⟨p|φ⟩ (135) where φ(P)(p) = Z ∞ −∞ ⟨p|x⟩⟨x|φ⟩dx = 1 √ 2πℏ Z ∞ −∞ e−i ℏpxφ(x)dx (136) This relation shows that the wavefunction φ(P)(p) which describes the vector | φ⟩in the momentum representation is related to φ(x) which describes the same vector in the coordinate representation by the Fourier transform.
Let us have specially | φ⟩= ˆ X| ψ⟩, ˆ X(P)ψP (x) = ⟨p | ˆ X| x⟩= 1 √ 2πℏ Z ∞ −∞ e−i ℏpx⟨x | ˆ X| ψ⟩dx = = 1 √ 2πℏ Z ∞ −∞ e−i ℏpxxψ(x)dx = iℏ √ 2πℏ d dp Z ∞ −∞ e−i ℏpxψ(x)dx (137) that is ˆ X(P)ψP (x) = iℏd dpψP (x) (138) In the momentum representation, the operator ˆ X is therefore defined as ˆ X(P) = iℏd dp.
Schr¨ odinger equation in coordinate representation.
So far we have dealt with the Hamiltonian (in the context of time evolution) on an ab-stract operator level (end of the Sect. 3). Now we would like to arrive to a concrete form and representation of the Hamiltonian. We start from a classical Hamiltonian, which we will quantize by using the correspondence principle, i.e. we assign relevant quantum operators to classical variables.
The classical Hamiltonian has the following form H(⃗ x, ⃗ p) = ⃗ p2 2m + V (⃗ x) (139) where the first term represents the kinetic energy and the second term is the potential energy. By assigning the operator ˆ ⃗ X to the classical coordinate ⃗ x and the operator ˆ ⃗ P to the classical momentum ⃗ p, we obtain (note that we deal in general with three-dimensional case) ˆ H = ˆ ⃗ P 2m + V ( ˆ ⃗ X) (140) In order to proceed to the coordinate representation, we apply the canonical quantization, that is, we perform the following substitutions ˆ ⃗ X →⃗ x (141) ˆ ⃗ P →−iℏ 3 X j=1 ∂2 ∂x2 j = −iℏ∆ (142) The quantum Hamiltonian in the coordinate representation then becomes ˆ H = −ℏ2 2m∆+ V ( ˆ ⃗ X) (143) In the special case that the potential energy V ( ˆ ⃗ X) = 0, we say that the Hamiltonian describes a free particle.
4.2 Fourier transform Our presentation of the Fourier transform followed C. Cohen-Tannoudji et al., Quantum Mechanics II, Appendix I.
4.3 δ function C. Cohen-Tannoudji et al., Quantum Mechanics II, Appendix II.
4.4 Free particle C. Cohen-Tannoudji et al., Quantum Mechanics I, Chapter I, Sections B and C.
4.5 Particle in time-independent scalar potential C. Cohen-Tannoudji et al., Quantum Mechanics I, Chapter I, Section D and Appendix HI.
4.6 One-dimensional harmonic oscillator C. Cohen-Tannoudji et al., Quantum Mechanics I, Chapter V.
Remark: Please note the the lecture notes may contain typos. It would be appreciated if you let me know in the case you find any. Also, if there is anything you consider unclear or incomplete, or if you have any questions regarding the notes please do not hesitate to contact me. |
16064 | https://ocw.mit.edu/courses/18-02-multivariable-calculus-fall-2007/0cf928b8ae3520051bc44e3e2828d90c_18_022007L29.pdf | MIT OpenCourseWare 18.02 Multivariable Calculus, Fall 2007 Please use the following citation format: Denis Auroux. 18.02 Multivariable Calculus, Fall 2007. (Massachusetts Institute of Technology: MIT OpenCourseWare). (accessed MM DD, YYYY). License: Creative Commons Attribution-Noncommercial-Share Alike. Note: Please use the actual date you accessed this material in your citation. For more information about citing these materials or our Terms of Use, visit: MIT OpenCourseWare 18.02 Multivariable Calculus, Fall 2007 Transcript – Lecture 29 So, first of all, there is a handout here with some notes on what I'm going to explain today, giving you a bit more details. So, the goals for today is we're going to see two things. One is how to prove the divergence theorem, and the second one is actually what it says physically. So, we're going to see how it relates to partial differential equations that come up naturally in physics such as the diffusion equation, the heat equation, convection equation, and so on. So that's what these notes are about. If you don't have them yet, then you can just pick them up at the end of class. And the other thing is, well, my office hour tomorrow afternoon is canceled because I won't be around. But, I'm available this afternoon after this class if you want. Otherwise I'll be reachable by e-mail, and all that. OK, so remember we left things with this statement of the divergence theorem. So, the divergence theorem gives us a way to compute the flux of a vector field for a closed surface. OK, it says if I have a closed surface, s, bounding some region, D, and I have a vector field defined in space, so that I can try to compute the flux of my vector field through my surface. Double integral of F.dS or F.ndS if you want, and to set this up, of course, I need to use the geometry of the surface depending on what the surface is. We've seen various formulas for how to set up the double integral. But, we've also seen that if it's a closed surface, and if a vector field is defined everywhere inside, then we can actually reduce that to a calculation of the triple integral of the divergence of F inside, OK? So, concretely, if I use coordinates, let's say that the coordinates of my vector field are, sorry, the components are P, Q, and R dot ndS, then that will become the triple integral of, well, so, divergence is P sub x plus Q sub y plus R sub z. OK, so by the way, how to remember this formula for divergence, and other formulas for other things as well. Let me just tell you quickly about the del notation. So, this guy usually pronounced as del, rather than as pointy triangle going downwards or something like that, it's a symbolic notation for an operator. So, you're probably going to complain about putting these guys into a vector. But, let's think of partial with respect to x, with respect to y, and with respect to z as the components of some formal vector. Of course, it's not a real vector. These are not like anything. These are just symbols. But, so see for example, the gradient of function, well, if you multiply this vector by scalar, which is a function, then you will get partial, partial x of f, partial, partial y of f, partial, partial z, f, well, that's the gradient. That seems to work. So now, the interesting thing about divergence is I can think of divergence as del dot a vector field. See, if I do the dot product between this guy and my vector field P, Q, R, well, it looks like I will indeed get partial, partial x of P plus partial Q partial y plus partial R partial z. That's the divergence. OK, and of course, similarly, when we have two variables only, x and y, we could have thought of the same notation, just with a two component vector, partial, partial x, partial, partial y. So, now, this is like of slightly limited usefulness so far. It's going to become very handy pretty soon because we are going to see curl. And, the formula for curl in the plane was kind of complicated. But, if you thought about it in terms of this, it was actually the determinant of del and f. And now, in space, we are actually going to do del cross f. But, I'm getting ahead of things. So, let's not do anything with that. Curl will be for next week. Just getting you used to the notation, especially since you might be using it in physics already. So, it might be worth doing. OK, so the other thing I wanted to say is, what does this theorem say physically? How should I think of this statement? So, I think I said that very quickly at the end of last time, but not very carefully. So, what's the physical interpretation of a divergence field? So, I want to claim that the divergence of a vector field corresponds to what I'm going to call the source rate, which is somehow the amount of flux generated per unit volume. So, to understand what that means, let's think of what's called an incompressible fluid. OK, so an incompressible fluid is something like water, for example, where a fixed mass of it always occupies the same amount of volume. So, guesses are compressible. Liquids are incompressible, basically. So, if you have an incompressible fluid flow --- well, so, again, what that means is really, given mass occupies always a fixed volume. Then, well, let's say that we have such a fluid with velocity given by our vector field. OK, so we're thinking of F as the velocity and maybe something containing water, a pipe, or something. So, what does the divergence theorem say? It says that if I take a region in space, let's call it D, sorry, D is the inside, and S is the surface around it, well, so if I sum the divergence in D, well, I'm going to get the flux going out through this surface, S. I should have mentioned it earlier. The convention in the divergence theorem is that we orient the surface with a normal vector pointing always outwards. OK, so now, we know what flux means. Remember, we've been describing, flux means how much fluid is passing through this surface. So, that's the amount of fluid that's leaving the region, D, per unit time. And, of course, when I'm saying that, it means I'm counting everything that's going out of D minus everything that's coming into D. That's what the flux measures. So, now, if there is stuff coming into D or going out of D, well, it must come from somewhere. So, one possibility would be that your fluid is actually being compressed or expanded. But, I've said, no, I'm looking at something like water that you cannot squish into smaller volume. So, in that case, the only explanation is that there is something it here that actually is sucking up water or producing more water. And so, integrating the divergence gives you the total amount of sources minus the amount of syncs that are inside this region. So, the divergence itself measures basically the amount of sources or syncs per unit volume in a given place. And now, if you think about it that way, well, it's basically the divergence theorem is just stating something completely obvious about all the matter that is leaving this region must come from somewhere. So, that's basically how we think about it. Now, of course, if you're doing 8.02, then you might actually have seen the divergence theorem already being used for things that are more like force fields, say, electric fields and so on. Well, I'll try to say a few things about that during the last week of classes. But, then this kind of interpretation doesn't quite work. OK, any questions, generally speaking, before we move on to the proof and other applications? Yes? Oh, not the gradient. So, yeah, not the gradient. So, yeah, the divergence of F measures the amount of sources or syncs in there. Well, what makes it happen? If you want, in a way, it's this theorem. Or, in another way, if you think about it, try to look at your favorite vector fields and compute their divergence. And, if you take a vector field where maybe everything is rotating, a flow that's just rotating about some axis, then you'll find that its divergence is zero. If you, sorry? No, divergence is not equal to the gradient. Sorry, there's a dot here that maybe is not very big, but it's very important. OK, so you take the divergence of a vector field. Well, you take the gradient of a function. So, if the gradient of a function is a vector, the divergence of a vector field is a function. So, somehow these guys go back and forth between. So, I should have said, with new notations comes new responsibility. I mean, now that we have this nice, nifty notation that will let us do gradient divergence and later curl in a unified way, if you choose this notation you have to be really, really careful what you put after it because otherwise it's easy to get completely confused. OK, so divergence and gradients are completely different things. The only thing they have in common is that both are what's called a first order differential operator. That means it involves the first partial derivatives of whatever you put into it. But, one of them goes from functions to vectors. That's gradient. The other one goes from vectors to functions. That's divergence. And, curl later will go from vectors to vectors. But, that will be later. Let's see, more questions? No? OK, so let's see, so how are we going to actually prove this theorem? Well, if you remember how we prove Green's theorem a while ago, the answer is we're going to do it exactly the same way. So, if you don't remember, then I'm going to explain. OK, so the first thing we need to do is actually a simplification. So, instead of proving the divergence theorem, namely, the equality up there, I'm going to actually prove something easier. I'm going to prove that the flux of a vector field that has only a z component is actually equal to the triple integral of, well, the divergence of this is just R sub z dV. OK, now, how do I go back to the general case? Well, I will just prove the same thing for a vector field that has only an x component or only a y component. And then, I will add these things together. So, if you think carefully about what happens when you evaluate this, you will have some formula for ndS, and when you do the dot product, you'll end up with the sum, P times something plus Q times something plus R times something. And basically, we are just dealing with the last term, R times something, and showing that it's equal to what it should be. And then, we the three such terms together. We'll get the general case. OK, so then we get the general case by summing one such identity for each component. I should say three such identities, one for each component, whatever. Now, let's make a second simplification because I'm still not feeling confident I can prove this right away for any surface. I'm going to do it first or what's called a vertically simple region. OK, so vertically simple means it will be something which I can setup an integral over the z variable first easily. So, it's something that has a bottom face, and a top face, and then some vertical sides. OK, so let's say first what happens if the given region, D, is vertically simple. So, vertically simple means it looks like this. It has top. It has a bottom. And, it has some vertical sides. So, if you want, if I look at it from above, it projects to some region in the xy plane. Let's call that R. And, it lives between the top face and the bottom face. Let's say the top face is z equals z2 of (x, y). Let's say the bottom face is z equals z1(x, y). OK, and I don't need to know actual formulas. I'm just going to work with these and prove things independently of what the formulas will be for these functions. OK, so anyway, a vertically simple region is something that lives above a part of the xy plane, and is between two graphs of two functions. So, let's see what we can do in that case. So, the right-hand side of this equality, so that's the triple integral, let's start computing it. OK, so of course we will not be able to get a number out of it because we don't know, actually, formulas for anything. But at least we can start simplifying because the way this region looks like, I should say this is D, tells me that I can start setting up the triple integral at least in the order where I integrate first over z. OK, so I can actually do it as a triple integral with Rz dz dxdy or dydx, doesn't matter. So, what are the bounds on z? See, this is actually good practice to remember how we set up triple integrals. So, remember, when we did it first over z, we start by fixing a point, x and y, and for that value of x and y, we look at a small vertical slice and see from where to where we have to go. Well, we start at z equals whatever the value is at the bottom, so, z1 of x and y. And, we go up to the top face, z2 of x and y. Now, for x and y, I'm not going to actually set up bounds because I've already called R the quantity that I'm integrating. So let me change this to, let's say, U or something like that. If you already have an R, I mean, there's not much risk for confusion, but still. OK, so we're going to call U the shadow of my region instead. So, now I want to integrate over all values of x and y that are in the shadow of my region. That means it's a double integral over this region, U, which I haven't described to you. So, I can't actually set up bounds for x and y. But, I'm going to just leave it like this. OK, now you see, if you look at how you would start evaluating this, well, the inner integral certainly is not scary because you're integrating the derivative of R with respect to z, integrating that with respect to z. So, you should get R back. OK, so triple integral over D of Rz dV becomes, well, we'll have a double integral over U of, so, the inner integral becomes R at the point on the top. So, that means, remember, R is a function of x, y, and z. And, in fact, I will plug into it the value of z at the top, so, z of xy minus the value of R at the point on the bottom, x, y, z1 of x, y. OK, any questions about this? No? Is it looking vaguely believable? Yeah? OK. So, now, let's compute the other side because here we are stuck. We won't be able to do anything else. So, let's look at the flux integral. OK, we have to look at the flux of this vector field through the entire surface, S, which is the whole boundary of D. So, that consists of a lot of pieces, namely the top, bottom, and the sides. OK, so the other side -- So, let me just remind you, S is bottom plus top plus side of this vector field, dot ndS equals, OK, so what do we have? So first, we have to look at the bottom. No, let's start with the top actually. Sorry. OK, so let's start with the top. So, just remind you, let's do all of them. So, let's look at the top first. So, we need to set up the flux integral for a vector field dot ndS. We need to know what ndS is. Well, fortunately for us, we know that the top face is going to be the graph of some function of x and y. So, we've seen a formula for ndS in this kind of situation, OK? We have seen that ndS, sorry, so, just to remind you this is the graph of a function z equals z2 of x, y. So, we've seen ndS for that is negative partial derivative of this function with respect to x, negative partial z2 with respect to y, one, dxdy. OK, and, well, we can't compute these guys, but it's not a big deal because if we do the dot product with <0, 0, R> dot ndS, that will give us, well, if you dot this with zero, zero, R, these terms go away. You just have R dxdy. So, that means that the double integral for flux through the top of R vector field dot ndS becomes double integral of the top of R dxdy. Now, how do we evaluate that, actually? Well, so R is a function of x, y, z. But we said, we have only two variables that we're going to use. We're going to use x and y. We're going to get rid of z. How do we get rid of z? Well, if we are on the top surface, z is given by this formula, z2 of x, y. So, I plug z equals z2 of x, y into the formula for R, whatever it may be. Then, I integrate dxdy. And, what's the range for x and y? Well, my surface sits exactly above this region U in the xy plane. So, it's double integral over U, OK? Any questions about how I set up this flux integral? No? OK, let me close the door, actually. OK, so we've got one of the two terms that we had over there. Let's try to get the others. [LAUGHTER] No comment. OK, so, we need to look, also, at the other parts of our surface for the flux integral. So, the bottom, well, it will work pretty much the same way, right, because it's the graph of a function, z equals z1 of x, y. So, we should be able to get ndS using the same method, negative partial with respect to x, negative partial with respect to y, one dxdy. Now, there's a small catch. OK, we have to think of it carefully about orientations. So, remember, when we set up the divergence theorem, we need the normal vectors to point out of our region, which means that on the top surface, we want n pointing up. But, on the bottom face, we want n pointing down. So, in fact, we shouldn't use this formula here because that one corresponds to the other orientation. Well, we could use it and then subtract, but I was just going to say that ndS is actually the opposite of this. So, I'm going to switch all my signs. OK, that's the other side of the formula when you orient your graph with n pointing downwards. Now, if I do things the same way as before, I will get that <0, 0, R> dot ndS will be negative R dxdy. And so, when I do the double integral over the bottom, I'm going to get the double integral over the bottom of negative R dxdy, which, if I try to evaluate that, well, I actually will have to integrate. Sorry, first I'll have to substitute the value of z. The value of z that I will want to plug into R will be given by, now, z1 of x, y. And, the bounds of integration will be given, again, by the shadow of our surface, which is, again, this guy, U. OK, so we seem to be all set, except we are still missing one part of our surface, S, because we also need to look at the sides. Well, what about the sides? Well, our vector field, <0, 0, R>, is actually vertical. It's parallel to the z axis. OK, so my vector field does something like this everywhere. And, why that makes it very interesting on the top and bottom, that means that on the sides, really not much is going on. No matter is passing through the vertical sides. So, the side -- The sides are vertical. So, <0, 0, R> is tangent to the side, and therefore, the flux through the sides is just going to be zero. OK, no calculation needed. That's because, of course, that's the reason why a simplified first things so that my vector field would only have a z component, well, not just that but that's the main place where it becomes very useful. So, now, if I compare my double integral and, sorry, my triple integral and my flux integral, I get that they are, indeed, the same. Well, that's the statement of the theorem we are trying to prove. I shouldn't erase it, OK? [LAUGHTER] So, just to recap, we've got a formula for the triple integral of R sub z dV. It's up there at the very top. And, we got formulas for the flux through the top and the bottom, and the sides. And, when you add them together, you get indeed the same formula, top plus bottom -- -- plus sides of, OK, and so we have, actually, completed the proof for this part. Now, well, that's only for a vertically simple region, OK? So, if D is not vertically simple, what do we do? Well, we cut it into vertically simple pieces. OK so, concretely, I wanted to integrate over a solid doughnut. Then, that's not vertically simple because here in the middle, I have not only does top in this bottom, but I have this middle face. So, the way I would cut my doughnut would be probably I would slice it not in the way that you'd usually slice the doughnut or a bagel, but at it's probably more spectacular if you think that it's a bagel. Then, a mathematician's way of slicing it is like this into five pieces, OK? And, that's not very convenient for eating, but that's convenient for integrating over it because now each of these pieces has a well-defined top and bottom face, and of course you've introduced some vertical sides for two reasons. One is that we've said the flux through them is zero anyway. So, who cares? Why bother? But, also, if you sum the flux through the surface of each little piece, well, you will see that you will be integrating twice over each of these vertical cuts. Once, when you do this piece, you will be taking the flux through this red guy with normal vector pointing to the right, and once, when you take this middle little piece, you will be taking the flux through that cut surface again, but now with normal vector pointing the other way around. So, in fact, you'll be summing the flux through these guys twice with opposite orientations. They cancel out. Well, and again, because of what you are doing actually, the integral was just zero anyway. So, it didn't matter. But, even if it hadn't simplified, that would do it for us. OK, so that's how we do it with the general region. And then, as I said at the beginning, when we can do it for a vector field that has only a z component, well, we can also do it for a vector field that has only an x or only a y component. And then, we sum together and we get the general case. So, that's the end of the proof. OK, so you see, the idea is really the same as for Green's theorem. Yes? Oh, there's only four pieces, thank you. Yes, there's three kinds of mathematicians: those who know how to count, and those who don't. Well, OK. So, OK, now I hope that you can see already the interest of this theorem for the divergence theorem for computing flux integrals just for the sake of computing flux integrals like might happen on the problem set or on the next test. But let me tell you also why it's important physically to understand equations that had been observed empirically well before they were actually understood in terms of how things go. So, let's look at something called the diffusion equation. So, let me explain to you what it does. So, the diffusion equation is something that governs, well, what's called diffusion. Diffusion is when you have a fluid in which you are introducing some substance, and you want to figure out how that thing is going to spread out, the technical term is diffuse, into the ambient fluid. So, for example, that governs the motion of, say, smoke in the air, or if you put dye in the solution or things like that. That will tell you, say that you drop some ink into a glass of water. Well, you can imagine that obviously it will get diluted into there. And, that equation will tell you how exactly over time this thing is going to spread out and start filling the entire glass. So, what's the equation? Well, we need, first, to know what the unknown will be. So, it's a partial differential equation, OK? So the unknown is a function, and the equation will relate the partial derivatives of that function to each other. So, u, the unknown, will be the concentration at a given point. And, of course, that depends on the point where you are. So, that depends on x, y, z, the location where you are. But, since the goal is also to understand how things spread over time, it should also depend on time. Otherwise, we're not going to get very far. And, in fact, what the equation will give us is the derivative of u with respect to t. It will tell us how the concentration at a given point varies over time in terms of how the concentration varied in space. So, it will relate partial u partial t to partial derivatives with respect to x, y, and z. [APPLAUSE] OK, [LAUGHTER] so what's the equation? The equation is partial u partial t equals some constant. Let me call it constant k times something I will call del squared u, which is also called the Laplacian of u, and what is that? Well, that means, OK, so just to scare you, del squared is this, which means it's the divergence of gradient u that we've used this notation for gradient. OK, so if you actually expand it in terms of variables, that becomes partial u over partial x squared plus partial squared u over partial y squared plus partial squared u over partial z squared. OK, so the equation is this equals that. OK, so that's a weird looking equation. And, I'm going to have to explain to you, where does it come from? OK, but before I do that, well, let me point out actually that the equation is not just the diffusion equation. It's also known as the heat equation. And, that's because exactly the same equation governs how temperature changes over time when you have, again, so, sorry I should have been actually more careful. I should have said this is in air that's not moving, OK? OK, and same thing with the solution. If you drop some ink into your glass of water, well, if you start stirring, obviously it will mix much faster than if you don't do anything. OK, so that's the case where we don't actually, the fluid is not moving. And, the heat equation now does the same, but for temperature in a fluid that's at rest, that's not moving. It tells you how the heat goes from the warmest parts to the coldest parts, and eventually temperatures should somehow even out. So, in the heat equation, that would just be, this u would just measure the temperature for temperature of your fluid at a given point. Actually, it doesn't have to be a fluid. It could be a solid for that heat equation. So, for example, say that you have a big box made of metal or something, and you do some heating at one side. You want to know how quickly is the other side going to get hot? Well, you can use the equation to figure out, you know, say that you have a metal bar, and you keep one side at 400° because it's in your oven. How quickly will the other side get warm? OK, so it's the same equation for both phenomena even though they are, of course, different phenomena. Well, the physical reason why they're the same is actually that phenomena are different, but not all that much. They involve, actually, how the atoms and molecules are actually moving, and hitting each other inside this medium. OK, so anyway, what's the explanation for this? So, to understand the explanation, and given what we've been doing, we should have a vector field in there. So, I'm going to think of the flow of, well, let's imagine that we are doing motion of smoke in air. So, that's the flow of the smoke: that means at every point, it's a vector whose direction tells me in which direction the smoke is actually moving. And, its magnitude tells me how fast it's moving, and also what amount of smoke is moving. So, there's two things to understand. One is how the disparities in the concentration between different points causes the flow to be there, how smoke will flow from the regions where there's more smoke to the regions where there's less smoke. And, that's actually physics. But, in a way, it's also common sense. So, physics and common sense tell us that the smoke will flow from high concentration towards low concentration regions. OK, so if we think of this function, U, that measures concentration, that means we are always going to go in the direction where the concentration decreases the fastest. Well, what's that? That's negative the gradient. So, F is directed along minus gradient u. Now, how big is F going to be? Well, they are, you have to come up with some intuition for how the two are related. And, the easiest relation I can think of is that they might be just proportional. So, the steeper the differences in concentration, the faster the flow will be, or the more there will be flow. And, if you try to think about it as molecules moving in random directions, you will see it makes actually complete sense. Anyway, it should be part of your physics class, not part of what I'm telling you. So, I'm just going to accept that the flow is just proportional to the gradient of u. So, if you want, the differences between concentrations of different points are very small, then the flow will be very gentle. And, if on the other hand you have huge disparities, then things will try to even out faster. OK, so that's the first part. Now, we need to understand the second part, which is once we know how flow is going, how does that affect the concentration? If smoke is going that way, then it means the concentration here should be decreasing. And there, it should be increasing. So, that's the relation between F and partial u partial t. At that part is actually math, namely, the divergence theorem. So, let's try to understand that part more carefully. So, let's take a small piece of a small region in space, D, bounded by a surface, S. So, I want to figure out what's going on in here. So, let's look at the flux out of D through S. Well, we said that this flux would be given by double integral on S of F dot n dS. So, this flux measures how much smoke is passing through S per unit time. That's the amount of smoke through S per unit time. But now, how can I compute that differently? Well, I can compute it just by looking at the total amount of smoke in this region, and seeing how much it changes. If I'm gaining or losing smoke, it means it must be going up there. Well, unless I have a smoker in here, but that's not part of the data. So, that should be, sorry, that's the same thing as the derivative with respect to t of the total amount of smoke in the region, which is given by the triple integral of u. If I integrate the concentration of smoke, which means the amount of smoke per unit volume over d, I will get the total amount of smoke in d, except, well, let's see. This flux is counted positively if we go out of d. So, actually, it's minus the derivative. This is the amount of smoke that we are losing per unit time. OK, so now we are almost there. Well, let me actually -- Because we know yet another way to compute this guy using the divergence theorem. Right, so this part here is just common sense and thinking about what it means. The divergence theorem tells me this is also equal to the triple integral, d, of div f dV. So, what I got is that the triple integral over d of div F dV equals this derivative. Well, let's think a bit about this derivative so, see, you are integrating function over x, y, and z. And then, you are differentiating with respect to t. I claim that you can actually switch the order in which you do things. So, when we think about it, is, here, you are taking the total amount of smoke and then see how that changes over time. So, you're taking the derivative of the sum of all the small amounts of smoke everywhere. Well, that will be the sum of the derivatives of the amounts of smoke inside each little box. So, we can actually move the derivatives into the integral. So, let's see, I said minus d dt of triple integral over d udV. And, now I'm saying this is the same as the triple integral in d of du dt dv. And the reason why this is going to work is you should think of it as d dt of a sum of u of some values. You plug in the values of your points at that given time times the small volume. You sum them, and then you take the derivative. And now, you see, the derivative of this sum is the sum of the derivatives. yi, zi, t, so, if you think about what the integral measures, that's actually pretty easy. And it's because this variable here is not the same as the variables on which we are integrating. That's why we can do it. OK, so now, if we have this for any region, d. So, we have a function of x, y, z, t, and we have another function here. And whenever we integrate them anywhere, we get the same answer. Well, that must mean they're the same. Just think of what happens if you just take d to be a tiny little box. You will get roughly the value of div f at that point times the volume of the box. Or, you will get roughly the value of du dt at that point times the value of a little box. So, the values must be the same. Well, let me actually explain that a tiny bit better. So, what I get is that one over, let me divide by the volume of D, sorry. I promise, I'm done in a minute. Is the same thing as one over volume D of negative du dt, dV. So, that means the average value, OK, maybe that's the best way of telling it, the average of div f in D is equal to the average of minus partial u partial t in D. And, that's true for any region, D, not just for some regions, but for, really, any region I can think of. So, the outcome is that actually the divergence of f is equal to minus du dt. And, that's another way to think about what divergence means. The divergence, we said, is how much stuff is actually expanding, flowing out. That's how much we're losing. And so, now, if you combine this with that, you will get that du dt is minus divergence f, which is plus K del squared u. OK, so you combine this guy with that guy, and you get the diffusion equation. OK, that's the end. Happy Thanksgiving break. Try to go to recitation tomorrow if you can so that your TA's are not too lonely. And, enjoy the break. See you next week. |
16065 | https://artofproblemsolving.com/wiki/index.php/Rearrangement_Inequality?srsltid=AfmBOopTibKRaTAAOCLsPfulADl1xSJ-1JsadIHrh3B7LhUcaSL8Fnt2 | Art of Problem Solving
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Rearrangement Inequality
The Rearrangement Inequality states that, if is a permutation of a finiteset (in fact, multiset) of real numbers and is a permutation of another finite set of real numbers, the quantity is maximized when and are similarly sorted (that is, if is greater than or equal to exactly of the other members of , then is also greater than or equal to exactly of the other members of ). Conversely, is minimized when and are oppositely sorted (that is, if is less than or equal to exactly of the other members of , then is also greater than or equal to exactly of the other members ).
Contents
[hide]
1 Introductory
2 Intermediate
2.1 Proof of the Rearrangement Inequality
3 Uses
4 See Also
5 External Links
Introductory
Consider the following simple application: suppose you are involved in the hold-up of a convenience store. You note, as you are emptying the register, that there are different numbers of each denomination (penny, nickel, dime, quarter, dollar bill, five dollar bill, ten dollar bill and twenty dollar bill) in the register. When would your take be maximized? Certainly, you would hope that there would be the largest number of twenty dollar bills, then the next largest number of tens, etc. Meanwhile, you would find yourself very disappointed if there were more pennies than nickels, more nickels than dimes, and so on. This is a simple application of the rearrangement inequality. It is also an application of the greedy algorithm, so one possible interpretation of the rearrangement inequality is that sometimes, the greedy algorithm works.
Intermediate
Proof of the Rearrangement Inequality
The proof of the Rearrangment Inequality can be handled with proof by contradiction. Only the maximization form is proved here; the minimization proof is virtually identical.
Before we begin the proof properly, it is useful to consider the case where . Without loss of generality, sort and so that and . By hypothesis, . Expanding and taking some terms to the other side of the inequality, we get , as desired.
Now for the general case. Again, without loss of generality, set and ; and suppose the grouping that maximizes the desired sum of products is not the one that pairs with , with , and so on. This means that there is at least one instance where is paired with while is paired with , where and . However, using the technique seen above to prove the inequality for , we can see that the sum of products can only increase if we instead pair with and with (unless both a's or both b's are equal, in which case either we can choose another pair of products or note that the current arrangement is actually identical to the desired one), which contradicts our assumption that the arrangement we had was already the largest one.
Note: The minimization equality can be very easily proved by noting that if we have the set , ordered in decreasing order and the set , ordered in increasing order, then the maximum sum is just . Thus, by negating all values the inequality follows.
Uses
The Rearrangement Inequality has a wide range of uses, from MathCounts level optimization problems to Olympiad level inequality problems. A relatively simple example of its use in solving higher-level problems is found in the proof of Chebyshev's Inequality. It is particularly useful in that it does not require any terms of either sequence to be positive or negative, unlike the power-mean family of inequalities.
See Also
Chebyshev's Inequality
Power Mean Inequality
External Links
The Rearrangement Inequality by Dragos Hrimiuc
Retrieved from "
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Algebra
Inequalities
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16066 | https://askfilo.com/user-question-answers-smart-solutions/the-centre-of-circle-r-2cos-theta-3134353935383833 | Question asked by Filo student
The centre of circle r=2cos theta
Views: 5,125 students
Updated on: Nov 27, 2024
Text SolutionText solutionverified iconVerified
Concepts:
Polar coordinates, Circle, Geometry
Explanation:
To find the center of the circle given in polar coordinates as r=2cos(θ), we can convert this equation into Cartesian coordinates. The relationship between polar and Cartesian coordinates is given by x=rcos(θ) and y=rsin(θ). First, we multiply both sides of the polar equation by r: r2=2rcos(θ). In Cartesian coordinates, r2=x2+y2 and rcos(θ)=x. Thus, we can rewrite the equation as: x2+y2=2x. Rearranging gives us: x2−2x+y2=0. Completing the square for the x terms, we have: (x−1)2+y2=1. This represents a circle with center at (1,0) and radius 1.
Step by Step Solution:
Step 1
Start with the polar equation: r=2cos(θ).
Step 2
Multiply both sides by r: r2=2rcos(θ).
Step 3
Substitute r2 with x2+y2 and rcos(θ) with x: x2+y2=2x.
Step 4
Rearrange the equation: x2−2x+y2=0.
Step 5
Complete the square: (x−1)2+y2=1.
Final Answer:
The center of the circle is at (1, 0).
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| Question Text | The centre of circle r=2cos theta |
| Updated On | Nov 27, 2024 |
| Topic | All topics |
| Subject | Smart Solutions |
| Class | Class 12 |
| Answer Type | Text solution:1 |
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When a lessor accounts for a finance lease under IFRS 16, the lessor must derecognize the asset from its statement of financial position and record a lease receivable upon commencement of the lease. If the lessor is a manufacturer or dealer, then the lessor must also record sales revenue and cost of goods sold upon commencement of the lease.
The lessor will then recognize interest revenue throughout the lease term. At the end of the lease, the lessor will restore the asset to its statement of financial position. If the actual residual value turns out to be less than the guaranteed residual value, the lessor will receive cash from the lessee to cover the shortfall.
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let's do an example of how a lessor would account for a lease under ifrs 16 when there's a guaranteed residual value so let's say that the lessor is a manufacturer or a dealer and we've got a three year lease the lessor's implicit interest rate is 5 percent okay so that is what is going to be used for any discounting when we do our present value calculations remember when we did the accounting from the lessee's perspective and we said well if the lessee doesn't know the lessors implicit rate the lessee would use their incremental borrowing rate well we don't have to worry about the lessee's incremental borrowing rate at all because we are doing this from the perspective of the lessor and the lessor is going to know their own implicit interest rate so we're always going to use that so this would be five percent in this example to do the discounting now we have a guaranteed residual value of 20 000 that means the lessee has guaranteed the less ore that this asset will be worth 20 thousand dollars when they return it to the lessor at the end of the lease now there's gonna be three payments we've got three rental payments of a hundred thousand dollars so it's a hundred thousand dollars a year first payment made at the beginning of the year so day one of the lease uh we will have a payment of a hundred thousand dollars now the present value of the rental payments those three payments of a hundred thousand if you go three periods five percent hundred thousand dollar payment you get two hundred and eighty five thousand nine hundred and forty one dollars now the present value the guaranteed residual value that twenty thousand if you take the present value of that you get seventeen thousand two seventy seven why do these numbers matter if you add together the present value of the rental payments and the present value the guaranteed residual value that gives you the fair market value of the asset at the commencement of the lease three hundred and three thousand two hundred and eighteen and that also happens to be this is going to be the amount of our initial lease receivable okay i'm gonna show you a table where i'll show you how we calculate the interest revenue for the lessor and so forth and i'll show you the journal entries how we would go through this we are going to book so so the sales in this example because there's a guaranteed residual value the sales revenue is also going to be equal to this 303 218. so when the lessor at the commencement at least they're going to book a sale of 303 218 and they're also going to recognize cost of goods sold and that's going to be based on their cost basis which in this example say 200 000 so they're going to recognize 303 218 in sales revenue and 200 000 in cost of goods sold upon commencement of the lease okay so they're gonna have a selling profit of 103 218 and then they're going to recognize interest revenue throughout the lease i'm gonna show you how to do that okay now just uh we we had a video where i talked about at least uh classification for a lessor this is gonna be a finance lease if you haven't figured it out already from the details i'm gonna talk there's gonna be a finance lease i don't want to go too much in the details of that but just note that the present value the lease payments amounts to substantially all the fair market value the fair value of the asset so that would be that's one of the classification tests for whether it's a finance lease but you can also just take my word for it in this example let's treat this as a finance lease so i mentioned that we're going to have sales revenue 303 218 so we can see that here so 303 000 218 of sales revenue that's going to go on the company's income statement 200 000 of cost of goods sold okay now so we're gonna have a selling profit of 103 218. okay that's the difference between the sales and the cost of goods sold 103 218. now they also need to remove the inventory right from their statement of financial position and then record the lease receivable so let's think about what's happening here with this finance lease because this wouldn't happen if it was an operating lease it wouldn't go down this way what is happening is this piece of equipment or whatever it is that is is being rented to this lessee okay it was in the lessors inventory and we're saying okay let's take it out of the inventory let's take it off the statement of financial position but so so we're de-recognizing the asset from the lessors do you recognize in the asset but they're also recognized as least receivable which again the least receivable is the lessor is expecting receive rental payments and then the residual value whatever this thing's gonna be worth at the end of the lease term okay so that you add those together you discount them right using that five percent but you add them together and that's where we got this lease receivables they're expecting receive rental payments and the residual value okay so this is how this all ties together so the lessor get rid of inventory adding at least receivable and then there's going to be an effect on the income statement as well now the lease amortization table we're going to start with a lease receivable 303 to 18 which i already showed you how we calculated okay and that's that there so 303 218. now on january 1st 2022 so here's just to record the sale from the lessor but i'll show you another in a minute i'll show you they gotta make a journal entry to receive uh the hundred thousand dollar payment right because the lessee makes a payment so now the lease receivable and immediately on day one goes down to 203 218 because that's a hundred thousand is no longer receivable because the lessee just paid it so when we're calculating the interest revenue for the lessor we take this 203 218 times the five percent and that equals this ten thousand one sixty one so in terms of the income statement effect the lessor they're gonna have two hundred and three thousand and two eighteen dollars in selling profit and 10 161 in interest revenue for year one of the lease year two they're going to have just the interest revenue five thousand six sixty nine which is this number right here times five percent okay and uh so so basically they're not gonna have any selling profit in year two or year three of the lease um year three we got the 952 that's this number here times five percent i'm just trying to show you how all the calculations are made now if we look at the undiscounted lease payments here we see okay there's three payments of a hundred thousand this 20 000 at the end is the return that's the residual value right that's the guaranteed residual value that's the the lessee is returning the asset to the lessor and it's supposed to have a value of twenty thousand dollars now i mentioned so this lease payment i mentioned there was a journal entry here so basically the lessor would increase the cash account decrease the lease lease receivable just remember look the lease receivable it's eventually gonna go to zero because at the end of the lease term it's like okay the lessee gave the thing back to the lessor they don't have any more payments that they owe so this is representing the the payments they're going to be made plus the um the residual value what the assets can be worth at the end now i also mentioned the interest revenue being recorded okay interest revenue when that goes up the lease receivable is going to increase as well that's why this goes from 203 to 18 to 213 379 in case you're wondering like the math of how this all works out okay so i mentioned already the income statement effect we've got the 103 218 in selling profit and we have the 10 161 in interest revenue now this is assuming that the lessor is a manufacturer dealer and so they're selling something okay now let's say it's a piece of equipment or something like that but you could have an arrangement where it's just strictly the lease is a financing arrangement and it's not it's not really a sale it's not a manufacturer dealer let's say it's a bank for example and the lease is basically just strictly a financing arrangement and there is no selling profit there is no selling profit no sale recognized on the commencement lease that's possible okay that's normally called a direct financing lease and so if you have that situation uh basically there wouldn't be any uh sales revenue or cost of goods sold what would happen what would happen is basically so let's say it was a piece of equipment okay that the the bank was uh leasing out so they would reduce the equipment account okay and then they would inc they'd recognize the lease receivable but there wouldn't be any sales revenue or cost of goods sold or anything like that then the only profit that would recruited the bank as the lessor would be the interest okay but when it's when the lessor is a manufacturer or dealer they're getting the selling profit at the commencement lease and interest along the way when it's a direct financing lease like a bank and they're not they're not selling inventory then there's no sale recognized no selling profit the the bank is just the lesser is just getting interest revenue is is the profit along the way okay that's just a little side note now the statement of financial position aka balance sheet uh as of uh december 31st 2022 so i should i should have mentioned up here okay so i've got the dates here january 1st 2022 that's the first day of the lease so right here december 31st 2022 we see a lease receivable 213 379 okay that's these two added together now we've got a current portion and a non-current the current is the payment that's expected to be received in the next year okay so this is just the financial statement effects for the lessor like what their financials would look like now year two of the lease they receive another 100 000 payment lease receivable goes down pretty straightforward and then the interest revenue i already showed you how that was recognized so for year two at least this is the only thing that would affect the p l statement for the lessors you just have this 5669 of interest revenue okay there's no selling profit in year two year three we have another payment so we receive cash lease receivable goes down and then we have the um additional interest revenue okay so this would be the only income statement effect for year three for the lessor okay now if all goes well the lessee returns the asset at the end of the lease the lessor puts it back on their books so the lessor is gonna increase their inventory account when they get the asset back and then they're gonna basically zero out the lease receivable account okay now i said if all goes well because there's a guaranteed residual value of 20 000. so lessee had said look this is going to be worth 20 000 at the end of the lease but what if it's not what if it's not uh so just a little hypothetical scenario here let's say that the residual value is actually less than the amount was guaranteed so if we had so the guaranteed residual value was 20 000 but let's say the actual residual value like when the lessee gives the asset back to less or it's only worth 8 000 and they're like well sorry well hey you're gonna have to do more than sorry because you guaranteed that this thing would be worth twenty thousand so they're gonna have to give twelve thousand dollars cash okay to the lessor the less he's gonna have to pay out this cash so the lessor in that case right this is the hypothetical scenario right this is the final thing in the example i was covering but i'm saying if it wasn't worth 20 000 at the end then they put it back in their inventory at 8 000 so you increase the inventory account by 8 000 but also increase the cash count by 12 000 because the lessee's giving them 12 000 cash to compensate for that that shortfall and then they zero out the least receivable account |
16068 | https://www.cs.tufts.edu/r/geometry/pdf/haas04planar.pdf | Planar Minimally Rigid Graphs and Pseudo-Triangulations Ruth Haas a, David Orden b,1, G¨ unter Rote c,2, Francisco Santos d,1, Brigitte Servatius e, Herman Servatius e, Diane Souvaine f,3, Ileana Streinu g,4, Walter Whiteley h,5 aDepartment of Mathematics, Smith College, Northampton, MA 01063, USA.
bDepartamento de Matem´ aticas, Universidad de Alcal´ a de Henares, E-28871 Alcal´ a de Henares, Spain.
cInstitut f¨ ur Informatik, Freie Universit¨ at Berlin, Takustraße 9, D-14195 Berlin, Germany dDepartamento de Matem´ aticas, Estadistica y Computacion, Universidad de Cantabria, E-39005 Santander, Spain.
eMathematics Department, Worcester Polytechnic Institute, Worcester, MA 01609, USA.
fDepartment of Computer Science, Tufts University, Medford MA, USA.
gDeptartment of Computer Science, Smith College, Northampton, MA 01063, USA.
hDepartment of Mathematics and Statistics, York University, Toronto, Canada.
Abstract Pointed pseudo-triangulations are planar minimally rigid graphs embedded in the plane with pointed vertices (adjacent to an angle larger than π). In this paper we prove that the opposite statement is also true, namely that planar minimally rigid graphs always admit pointed embeddings, even under certain natural topolog-ical and combinatorial constraints. The proofs yield efficient embedding algorithms.
They also provide the first algorithmically effective result on graph embeddings with oriented matroid constraints other than convexity of faces. These constraints are described by combinatorial pseudo-triangulations, first defined and studied in this paper. Also of interest are our two proof techniques, one based on Henneberg inductive constructions from combinatorial rigidity theory, the other on a general-ization of Tutte’s barycentric embeddings to directed graphs.
Key words: Pseudotriangulations, rigidity, graph drawing.
Preprint submitted to Elsevier Science 28 May 2004 1 Introduction In this paper we bring together two classical topics in graph theory, planarity and rigidity, to answer the question (posed in ) of characterizing the class of planar graphs which admit pointed pseudo-triangular embeddings. Our main result is that this coincides with the class of all planar minimally rigid graphs (planar Laman graphs). Furthermore we extend the result in several directions, attacking the same type of question for other (not necessarily pointed) classes of pseudo-triangulations and for combinatorial pseudo-triangulations, a new class of objects first introduced and studied in this paper.
Novelty. As opposed to traditional planar graph embeddings, where all the faces are designed to be convex, ours have interior faces which are as non-convex as possible (pseudo-triangles). Planar graph embeddings with non-convex faces have not been systematically studied. Our result links them to rigidity theoretic and matroidal properties of planar graphs. We show how to adapt Tutte’s barycentric embeddings, designed for convex faces, to work for pointed pseudo-triangulations. To the best of our knowledge, this is the first result holding for an interesting family of graphs on algorithmically efficient graph embeddings with oriented matroid constraints other than convexity of faces. In contrast, the universality theorem for pseudo-line arrangements of Mn¨ ev implies that the general problem of embedding graphs with oriented matroid constraints is as hard as the existential theory of the reals.
Proof Techniques and Algorithmic Results. We present two proof tech-niques of independent interest. The first one is of a local nature, relying on incremental (inductive) constructions known in rigidity theory as Henneberg constructions. The second one is based on a global approach making use of a directed version of Tutte’s barycentric embeddings. Both proofs are construc-tive, yield efficient algorithms, emphasize distinct aspects of the result and Email addresses: rhaas@math.smith.edu. (Ruth Haas), david.orden@uah.es (David Orden), rote@inf.fu-berlin.de (G¨ unter Rote), santos@matesco.unican.es (Francisco Santos), bservat@math.wpi.edu (Brigitte Servatius), hservat@math.wpi.edu (Herman Servatius), dls@cs.tufts.edu (Diane Souvaine), streinu@cs.smith.edu (Ileana Streinu), whiteley@mathstat.yorku.ca (Walter Whiteley).
1 Supported by grant BFM2001-1153 of the Spanish Ministry of Science and Tech-nology.
2 Partly supported by the Deutsche Forschungsgemeinschaft (DFG) under grant RO 2338/2-1.
3 Supported by NSF Grant EIA-9996237.
4 Supported by NSF grants CCR-0105507 and CCR-0138374.
5 Supported by NSERC (Canada) and NIH (USA).
2 (a) ( b) ( c ) 2 3 4 5 1 6 8 7 2 3 4 5 1 6 7 8 2 3 4 5 1 8 6 7 Fig. 1. (a) A pseudo-triangulation (necessarily non-pointed, since the underlying graph is a circuit, not a Laman graph) and two embeddings of a planar Laman graph: (b) is a pointed pseudo-triangulation, (c) is not: the faces 2876 and 1548 are not pseudo-triangles and the vertices 6 and 8 are not pointed.
lead into new directions of further investigation: combinatorial versus geomet-ric embeddings, local versus global coordinate finding.
Laman Graphs and Pseudo-Triangulations. Let G = (V, E) be a graph with n vertices V = {1, 2, . . . , n} and m = |E| edges. G is a Laman graph if m = 2n −3 and every subset of k ≥2 vertices spans at most 2k −3 edges. 6 An embedding G(P) of the graph G on a set of points P = {p1, . . . , pn} ⊂R2 is a mapping of the vertices V to points in the Euclidean plane i 7→pi ∈P.
Throughout the paper, without any further explicit reference, all the geometric objects (embeddings, polygons, convex hulls) reside in the Euclidean plane R2.
In general, it is not required that the points be distinct, but in this paper we will work only with embeddings on sets of distinct points. The edges ij ∈E are mapped to straight line segments pipj. We say that the vertex i of the embedding G(P) is pointed if all its adjacent edges lie (strictly) on one side of some line through pi. Equivalently, some consecutive pair of edges adjacent to i (in the circular order around the vertex) spans a reflex angle (strictly larger than π). An embedding G(P) is non-crossing if no pair of segments pipj and pkpl corresponding to non-adjacent edges ij, kl ∈E, i, j ̸∈{k, l} have a point in common. A graph G is planar if it has a non-crossing embedding.
In a simple polygon, a vertex is convex if its interior angle is strictly between 0 and π and reflex if strictly between π and 2π. Throughout the paper, all the angles incident to polygon vertices will be convex or reflex (i.e., we will never encounter 0, π or 2π angles).
6 This is called the definition by counts of Laman graphs. An equivalent definition via Henneberg constructions will be given in Section 2.
3 A pseudo-triangle is a simple polygon with exactly three convex vertices (and all the others reflex). A pseudo-triangulation of a set of points in the plane is a non-crossing embedded graph G(P) whose outer face is the complement of the convex hull of the point set and whose interior faces are pseudo-triangles.
In a pointed pseudo-triangulation all the vertices are pointed. See Figure 1.
Historical Perspective. Techniques from Rigidity Theory have been re-cently applied to problems such as collision-free robot arm motion planning [10,45], molecular conformations [25,49,53] or sensor and network topologies with distance and angle constraints .
Laman graphs are the fundamental objects in 2-dimensional Rigidity Theory.
Also known as isostatic or generically minimally rigid graphs, they characterize combinatorially the property that a graph, embedded on a generic set of points in the plane, is infinitesimally rigid (with respect to the induced edge lengths).
See [20,27,54].
Pseudo-triangulations are relatively new objects, introduced and applied in Computational Geometry for problems such as visibility [36, 37, 43], kinetic data structures and motion planning for robot arms . They have rich combinatorial, rigidity theoretic and polyhedral properties [32, 41, 45], many of which have only recently started to be investigated [1–3,6,7,26,38].
In particular, the fact that they are Laman graphs which become expansive mechanisms when one of their convex hull edges is removed, has proven to be crucial in designing efficient motion planning algorithms for planar robot arms . Finding their 3-dimensional counterpart, which is perhaps the main open question about pseudo-triangulations and expansive motions, may lead to efficient motion planning algorithms for certain classes of 3-dimensional linkages, with potential impact on understanding protein folding processes.
Graph Drawing is a field with a distinguished history, and embeddings of planar graphs have received substantial attention in the literature [9, 13, 17, 19, 42, 50, 51]. Extensions of graph embeddings from straight-line to pseudo-line segments have been recently considered (see e.g. ). It is natural to ask which such embeddings are stretchable, i.e. whether they can be realized with straight-line segments while maintaining some desired combinatorial sub-structure. Indeed, the primordial planar graph embedding result, F´ ary’s The-orem , is just an instance of answering such a question. Graph embedding stretchability questions have usually ignored oriented matroidal constraints, allowing for the free reorientation of triplets of points when not violating other combinatorial conditions. The notable exception concerns the still widely open visibility graph recognition problem, approached in the context of pseudo-line arrangements (oriented matroids) by . In it is shown that it is not always possible to realize with straight-lines a pseudo-visibility graph, while 4 maintaining oriented matroidal constraints.
In contrast, this paper gives the first non-trivial stretchability result on a nat-ural graph embedding problem with oriented matroid constraints, other than convexity. It adds to the already rich body of surprisingly simple and ele-gant combinatorial properties of pointed pseudo-triangulations by proving a natural connection.
Main Result. We are interested in planar Laman graphs. Not all Laman graphs fall into this category. For example, K3,3 is Laman but not planar. But the underlying graphs of all pointed pseudo-triangulations are planar Laman.
We prove that the converse is always true: Theorem 1 (Main Theorem) Every planar Laman graph can be embedded as a pointed pseudo-triangulation.
The following characterization follows then from well known properties of Laman graphs: Corollary 1 Given a planar graph G, the following conditions are equivalent: (1) G is a Laman graph.
(2) Generically, G is minimally rigid.
(3) G can be embedded as a pointed pseudo-triangulation.
We prove in fact several stronger results, stated later in the paper after intro-ducing the appropriate definitions. They allow the a priori choice of the facial structure (Theorem 2) and even of the combinatorial information regarding which vertices are convex in each face (Theorem 6). This last result needs the apparatus of combinatorial pseudo-triangulations, first introduced and studied in this paper.
Finally, we answer a natural question related to the underlying matroidal structure of planar rigidity and extend the result to planar rigidity circuits, which are minimal dependent sets in the rigidity matroid, where the bases (maximally independent sets) are the Laman graphs. By adding edges to a pointed (minimum) pseudo-triangulation while maintaining planarity, the graph has increased dependency level (in the rigidity matroid) and can no longer be realized with all vertices pointed, but it can always be realized with straight edges. Our concern is to maintain the minimum number of non-pointed vertices, for the given edge count. For circuits, this number is one, and we show that it can be attained.
Overview. The paper is organized as follows. In Section 2 we give the basic 5 terminology and definitions needed for an independent reading of Section 3.
For increased readability, additional technical definitions are later included in the sections that use them. The first proof of the Main Theorem is pre-sented in Section 3, which is further devoted to all the proofs (combinato-rial or geometric) making use of the inductive Henneberg technique: planar Laman graphs, combinatorial pseudo-triangulations, pointed pseudo-triangu-lations and Laman-plus-one combinatorial and geometric pseudo-triangula-tions. Section 4 is devoted to combinatorial pseudo-triangulations and to the perfect matching technique for assigning combinatorial pseudo-triangular la-belings to plane graphs. Section 5 focuses on the second proof technique based on Tutte embeddings and contains our most general result on plane graph embeddings compatible with given combinatorial pseudo-triangulations. We conclude in Section 6 with a list of further directions of research and open questions.
2 Preliminaries For the standard graph and rigidity theoretical terminology used in this paper we refer the reader to and . For relevant facts about pointed pseudo-triangulations, see . In this section we continue what we started in the introduction and give most of the definitions needed for reading Sections 3 and 4. The technically denser Section 5 contains its own additional concepts.
Notation and Abbreviations. Throughout the paper we will abbreviate counter-clockwise as ccw. To emphasize that a graph has n vertices we may denote it by Gn. We will occasionally abbreviate combinatorial pseudo-tri-angulation as cpt and pointed combinatorial pseudo-triangulation as pointed cpt.
Plane Graphs. A non-crossing embedding of a connected planar graph G partitions the plane into faces (bounded or unbounded), edges and vertices.
Their incidences are fully captured by the vertex rotations: the ccw circular order of the edges incident to each vertex in the embedding. A spherical graph refers to a choice of a facial structure (and thus of a system of rotations) in a graph which has some planar embedding (not necessarily unique), and is oblivious of an “outer” face. It is well-known (Whitney ) that a 3-connected planar graph induces a unique facial structure (or set of rotations, modulo reorientation), but 2-connected ones may induce several. A plane graph is a spherical graph with a choice of a particular face as the outer face. Every simple plane graph can be realized with straight-line edges in the plane (F´ ary’s Theorem ).
6 (a) (b) 2 4 3 3 1 4 2 5 1 5 6 Fig. 2. Illustration of the two types of steps in a Henneberg sequence, with vertices labelled in the construction order. The shaded part is the old graph, to which the black vertex is added. (a) Henneberg I for vertex 5, connected to old vertices 2 and 4. (b) Henneberg II for vertex 6, connected to old vertices 1, 4 and 5.
A (combinatorial) angle (incident to a vertex or a face in a plane graph) is a pair of consecutive edges (consecutive in the order given by the rotations) incident to the vertex or face.
Pseudo-Triangulations. We have defined pseudo-triangles, pseudo-triangu-lations and pointed pseudo-triangulations in the introduction. In addition, we will use the following related concepts. The corners of a pseudo-triangle are its three convex angles, and its side chains are the pieces of the boundary between two corners (vertices and edges). The extreme edges of a pointed vertex are the two edges incident with its unique incident reflex angle.
Minimally Rigid (Laman) Graphs and Henneberg constructions. Be-sides the definition by counts given in the introduction, Laman graphs can be characterized in a variety of ways. In particular, a Laman graph on n vertices has an inductive construction as follows (see [23,54]). Start with an edge for n = 2. At each step, add a new vertex in one of the following two ways: • Henneberg I (vertex addition): the new vertex is connected via two new edges to two old vertices.
• Henneberg II (edge splitting): a new vertex is added on some edge (thus splitting the edge into two new edges) and then connected to a third vertex.
Equivalently, this can be seen as removing an edge, then adding a new vertex connected to its two endpoints and to some other vertex.
See Figure 2, where we show drawings with crossing edges, to emphasize that the Henneberg constructions work for general, not necessarily planar Laman graphs.
7 We will make heavy use of the following result, essentially stated by Henneberg , and of its proof, due to Tay and Whiteley .
Lemma 2 A graph is Laman if and only if it has a Henneberg construction.
The proof of Lemma 2 proceeds inductively to show that there always exists a vertex of degree 2 or 3 which can be removed in the reverse order of a Henneberg step while maintaining the Laman property. It is instructive to give a slightly more general proof. We will make use of it in Section 3.
Lemma 3 A Laman graph has a Henneberg construction starting from any prescribed subset of two vertices. Moreover, if there are three vertices of degree 3 mutually connected in a triangle, then we can prescribe them as the first three vertices of the Henneberg construction.
PROOF. Let Gn = (V, E) be a Laman graph on n = |V | vertices and let V2 ⊂V be any subset of two vertices. We show that as long as n > 2 we can always remove a vertex not in V2 in the opposite direction of a Henneberg step.
In the actual Henneberg construction this amounts to starting the induction from this prescribed pair.
Vertices of degree 0 or 1 do not exist in Laman graphs, otherwise the Laman property would be violated on a subset of n −1 vertices. Since Gn has 2n −3 edges, a simple count shows that there are at least three vertices of degree at most 3, hence either of degree 2 or 3. At least one of them (call it v) is not in V2. This will be the vertex we choose to remove, in a backwards application of a Henneberg step.
If v has degree 2, we remove v and its incident edges: the resulting graph on n −1 vertices and 2(n −1) −3 edges is clearly Laman. If v has degree 3, let its neighbors be v1, v2 and v3. The removal of v and of its three adjacent edges produces a graph G′ n−1 with a deficit of one edge: n −1 vertices but only 2(n −1) −4 edges. We must put back one edge joining one of the three pairs of vertices in v1, v2, v3. Consider the rigid components of G′ n−1: maximal subsets of some k vertices spanning 2k −3 edges. The three endpoints v1, v2 and v3 cannot belong to the same rigid component (otherwise the Laman count would be violated in Gn on the subset consisting of this component and v).
Two rigid components share at most one vertex, otherwise their union would be a larger Laman subgraph. Suppose that v1 and v2 are not in a common rigid component. Then adding an edge between v1 and v2 doesn’t violate the Laman condition on any subset and completes G′ n−1 to a Laman graph Gn−1.
If Gn contains a subset V3 of three degree 3 vertices connected in a triangle, then a similar counting argument shows that there is an additional vertex of degree at most 3. This fourth vertex can be removed in such a way that 8 the invariant (of having three vertices of degree 3 connected in a triangle) is maintained. Hence the vertices of V3 may be prescribed as the three starting vertices. To finish, let us show that the invariant is maintained. Let G′ be the subgraph induced on the vertices in V \ V3: if it contains more than two vertices, then it spans 2n −3 −6 = 2(n −3) −3 edges, hence it is Laman. Let N(V3) be the neighbors of V3 in Gn: none of these vertices can be of degree 2, otherwise G′ would not be Laman. If there is a vertex v′ of degree 3 in N(V3) which is removed at some Henneberg step, one must put back an edge incident to two of its neighbors, and one of them must be in V3: otherwise, the induced subgraph G′ on V \ V3 (after performing the reverse Henneberg step) would violate the Laman counts. Since at any reverse Henneberg step we remove either vertices of degree 2 (which are not incident to V3) or of degree 3, which do not change the degree of their neighbors in V3, it follows that the vertices in V3 maintain their degrees and the property of being connected in a triangle throughout the construction (in fact, until n = 5, from which point it is easy to see that V3 can still be prescribed).
2 Laman-plus-one Graphs and Rigidity Circuits. A Laman-plus-one graph is a Laman graph with one additional edge. It has 2n−2 edges and every sub-set of k edges spans at most 2k −2 edges. A rigidity circuit (shortly, a circuit) is a graph with the property that removing any edge produces a Laman graph.
It is therefore a special Laman-plus-one graph. In a rigidity circuit G with n vertices, the number n of vertices is at least 4, the number m of edges is 2n−2 and every subset of k < n vertices spans at most 2k −3 edges. Moreover, the minimum degree in a circuit is 3. It is straightforward to prove that a Laman-plus-one graph contains a unique rigidity circuit: take the maximal subgraph satisfying the circuit counts. It is unique, because otherwise the union of two circuits would violate the Laman-plus-one counts.
These concepts are motivated by the matroid view of Rigidity Theory, see : Laman graphs correspond to bases (maximal independent sets of edges) in the generic rigidity matroid, while the circuits are the minimally dependent sets.
It has been proven recently in that 3-connected rigidity circuits admit an inductive (Henneberg-type) construction (using only Henneberg II steps and starting from K4), where all intermediate graphs are themselves circuits. All rigidity circuits are 2-connected, hence they can be obtained by making use of Tutte’s Theorem on the structure of 2-connected graphs in terms of 3-blocks, see [15,52].
We show now (and use later) that Laman-plus-one graphs also admit a simple Henneberg construction. This type of inductive proof (much easier than because of a simpler inductive invariant) will be used in Section 3 to show stretchability of planar Laman-plus-one graphs, and thus of planar rigidity 9 circuits.
Lemma 4 A graph G is Laman-plus-one if and only if it has a Henneberg construction starting from a K4.
PROOF. The proof is similar to that of Lemma 2 and uses the 2n−2 counts.
Vertices of degree 2 are outside the circuit and are removed just as in the Laman case. Thus we assume for the remainder of the proof that there are no vertices of degree 2. Then there are at least four vertices of degree 3.
Let C = (Vc, Ec) be the unique induced subgraph which is the circuit of G, Vo = V \ Vc the vertices outside the circuit and Vb ⊂Vc the boundary of Vc, i.e. the set of vertices in Vc adjacent to a vertex in Vo.
Let v be a vertex of degree 3. If v ∈Vo, we show that we can always carry out a reverse Henneberg II step. The three neighbors of v cannot all belong to the circuit, because otherwise the subgraph induced on Vc ∪{v} would violate the 2n −2 counts. Remove (temporarily) an edge ab of G from inside C: the resulting graph is Laman, containing C without this edge as a rigid block (subset on which the Laman count is satisfied with equality). By Lemma 3 there is a well defined way of removing v and placing back an edge to perform a Henneberg II step in reverse: the added edge is not between two vertices of Vc. Therefore we can put back the temporarily removed edge ab to get a Laman-plus-one graph.
If v ∈Vc, note first that it cannot be on the boundary Vb, otherwise its degree in C would be at most 2, contradicting the fact that C is a circuit. So all its three neighbors v1, v2 and v3 are in Vc. Removing v and its incident edges produces a Laman graph. Either all of the edges v1v2, v1v3, v2v3 are present in G or not. If not (say, v1v2 is missing), then we add v1v2, get a Laman-plus-one graph and continue the induction. Otherwise, the circuit C is a K4. If there are no vertices outside the circuit, we are done. Otherwise, since the graph is connected, there must be some edge from Vc to Vo, increasing the degree of at least one vertex in the circuit K4 to at least 4. Since there are in total at least four vertices of degree 3, at least one of them must be in Vo. We will perform the Henneberg step on this vertex (and thus not touch K4 until the end).
2 Combinatorial Pseudo-Triangulations. Let G be a plane 2-connected graph 7 . A combinatorial pseudo-triangulation (cpt) of G is an assignment of labels big (or reflex) and small (or convex) to the angles of G such that: 7 The definition is valid in a more general setting than what we use in this paper, and works even with multiple edges, vertices of degree one and loops. In such a case, a vertex of degree one is incident to a unique angle, labeled big.
10 (1) Every face except the outer face gets three vertices marked small. These will be called the corners of the face.
(2) The outer face gets only big labels (has no corners).
(3) Each vertex is incident to at most one angle labeled big. If it is incident to a big angle, it is called pointed.
(4) A vertex of degree 2 is incident to one angle labeled big.
By analogy with pseudo-triangulations, we also define extreme edges, side chains and non-pointed vertices of combinatorial pseudo-triangulations.
Combinatorial pseudo-triangulations share many properties with pseudo-tri-angulations. The following lemma follows easily from the definition.
Lemma 5 A combinatorial pseudo-triangulation on n vertices has at least 2n−3 edges. If a cpt has m ≥2n−3 edges, then it contains exactly m−(2n−3) non-pointed vertices.
PROOF. Let VK be the set of non-pointed vertices. Let m be the number of edges, f the number of faces, k the size of VK and dv the degree of a vertex v. We count the number of small angles in two ways; summing over the faces we get 3(f −1), summing over vertices we get P v̸∈VK(dv −1) + P v∈VK dv = P v∈V dv −(n −k) = 2m −n + k. Applying Euler’s formula formula gives m = 2n −3 + k and proves the statement.
2 A cpt with exactly 2n −3 edges will have all vertices pointed: we’ll call it a pointed combinatorial pseudo-triangulation (pointed cpt). Another case of interest in this paper is when exactly one vertex is combinatorially non-pointed, i.e. has no incident big angle: we call it a pointed-plus-one cpt. Note that in this case the non-pointed vertex has degree at least 3, is interior, i.e. not incident to the outer face and that the cpt has 2n −2 edges.
In Section 3 we will prove that all planar Laman graphs and all planar Laman-plus-one graphs have cpt assignments. The previous lemma implies that such cpt assignments must be pointed for Laman graphs, resp. pointed-plus-one for Laman-plus-one graphs.
Pointed-plus-one and Circuit Pseudo-triangulations. A pointed-plus-one pseudo-triangulation is a pseudo-triangulation with precisely one non-pointed vertex. It is easy to see that it has 2n −2 edges, and is in fact just a planar Laman-plus-one graph embedded as a pseudo-triangulation. A pseudo-triangulation circuit is a planar rigidity circuit embedded as a pseudo-triangu-lation. A reminder that by a pseudo-triangulation we mean any decomposition 11 into pseudo-triangles, which may not be pointed. In fact, any such pseudo-tri-angulation with more than 2n−3 edges is necessarily non-pointed by Lemma 5, or, alternatively, because pointed pseudo-triangulations are maximal pointed sets of edges on any planar set of points and must have exactly 2n −3 edges (cf. ). See Figure 1 for an example.
3 Main result: Inductive Proof via Henneberg construction We are now ready to give our first proof of the Main Theorem, in the following slightly more general form, and extend it to planar Laman-plus-one graphs.
Theorem 2 Any plane Laman graph has a pointed pseudo-triangular embed-ding.
Theorem 3 Any plane Laman-plus-one graph has a pointed-plus-one pseudo-triangular embedding.
Both proofs have the same structure (based on an inductive Henneberg con-struction) and are divided into three steps: topological, combinatorial and geometric. To avoid a cluttered proof with too many details at once, we have chosen to present them as separate entities: once the reader has understood the Laman case, the Laman-plus-one follows naturally. The common theme of this section is centered around the Henneberg technique, and to empha-size it we have included two additional Lemmas (8 and 12) which will find applications in the next Section.
3.1 Pseudo-Triangular Embeddings of Plane Laman Graphs The proof of Theorem 2 is a consequence of the four lemmas stated and proven below. Lemma 6 reduces the construction to the case when the outer face is a triangle. Lemma 7 provides the framework for a Henneberg induction on plane graphs. This is then used in Lemma 8 to compute a combinatorial pseudo-tri-angulation assignment and in Lemma 9 to realize the same thing geometrically.
Theorem 2 follows from Lemma 9. We remark that Lemma 8 is not needed for the proof of Theorem 2. It is however natural to include it here because it makes use of the same Henneberg technique (ubiquitous in this section). It also gives a better intuition about the combinatorial structure of the many possibilities involved in a complete proof by case analysis of Lemma 9.
Lemma 6 (Fixing the Outer Face) Embedding a plane Laman graph as a pseudo-triangulation reduces to the case when the outer face is a triangle.
12 Fig. 3. Reducing to an embedding with a triangular outer face.
PROOF. Let G be a plane Laman graph with an outer face having more than three vertices. We construct another Laman graph G′ of n+3 vertices by adding 3 vertices in the outer face and connecting them to a triangle containing the original graph in its interior. Then we add an edge from each of the three new vertices to three distinct vertices on the exterior face of G. See Figure 3. We now realize G′ as a pseudo-triangulation with the new triangle as the outer face. The graph G, as a subgraph of G′, must be realized with its outer face convex by the following argument. The three new interior edges of G′ provide two corners each at their end-point incident to the outer face and at least one corner in the interior one. Since the three faces incident to them have nine corners in total, the boundary of G provides no corner to the three new interior faces of G′.
2 Note that a plane Laman graph (on n vertices) always has at least two tri-angular faces: the dual plane graph has n −1 vertices (including the vertex corresponding to the outer face) and 2(n−1)−1 edges, hence there are at least two of degree three. The construction in the previous lemma makes it possible to use the stronger invariant of the Henneberg construction from Lemma 3 and to start any geometric embedding with a triangular outer face, then to insert only on interior faces. This feature is not needed in the proof of the topological or combinatorial lemmas below.
Lemma 7 (The Topological Lemma) Every plane Laman graph has a plane Henneberg construction in which: (1) All intermediate graphs are plane.
(2) At each step, the topology is changed only on edges and faces involved in the Henneberg step: either a new vertex is added inside a face of the previous graph (Henneberg I), or inside a face obtained by removing an edge between two faces of the previous graph (Henneberg II).
In addition, if the outer face of the plane graph is a triangle, we may perform the Henneberg construction starting from that triangle. The Henneberg steps 13 Fig. 4. A plane Henneberg construction. Top row: Gn, to which a new vertex will be added. Middle row: Henneberg I on the outer, resp. an interior face. Bottom row: Henneberg II on the outer, resp. an interior face.
will never insert vertices on the outer face.
PROOF. We follow the structure of the basic Henneberg construction from Lemma 3. See Figure 4 for an illustration. Find an appropriate vertex of degree 2 or 3. Removing it, and its incident edges, merges two (resp. three) faces into one. The other endpoints of the removed edges are incident to this face, hence the added edge in the Henneberg II step simply splits this face and maintains the planarity of the embedding.
2 Lemma 8 (The Combinatorial Lemma) Every plane Laman graph admits a combinatorial pseudo-triangulation assignment.
PROOF. Let Gn be a plane Laman graph on n vertices. We may assume that the outer face is a triangle. We proceed with a plane Henneberg construction guaranteed by Lemma 7, which will insert only on interior 8 faces. The base case is a triangle and has a unique cpt labeling. At each step we have, by induction, a cpt labeling which we want to extend. The proof will guarantee that each one can be extended (so there is no need to backtrack).
In a Henneberg I step, the new vertex v is inserted on a face T (already labeled as a pseudo-triangle), and joined to two old vertices v1 and v2. The new edges 8 This is just a technical simplification reducing the size of our case analysis. The Henneberg steps would work just as well for insertions on the outer face.
14 Fig. 5. Extending a combinatorial pseudo-triangulation in a Henneberg I step. Top left: the combinatorial face is represented as a circle with its three corners, denoted by white vertices, marked small (a small black dot). Top right, a representative situation: the two endpoints of the newly added edges (in black) are distributed on two distinct side chains of the face. Two distinct labelings are possible in this case (bottom row): the newly created angles after the insertion of the new vertex (grey) are labeled with a small black dot for small (or convex) and with a large arc for big (or reflex).
Fig. 6. Merging two faces into one: a representative case for the analysis of a combi-natorial Henneberg II step. The markings of small and big angles follow the conven-tions from Figure 5. The black vertices are the endpoints of the removed edge v1v2.
vv1 and vv2 partition the face F and its three corners into two. The following cases may happen. If neither v1 nor v2 is a corner of F, the three corners can be split between the two new faces, either as as 2 + 1 or 3 + 0. It can happen that one of the corners of F is split by a new edge (say v1 is a corner) and the other two are either both in the same new face or are separated; otherwise, two corners are split, and the third corner is in one of the two newly created faces. In either case, the assignment of big and small labels is what one would expect: a small angle is split into two small angles, a big angle is split into a big and a small angle, and the new point gets exactly one big angle. We illustrate one representative case in Figure 5 and omit the rest of the straightforward details of this case analysis.
In a Henneberg II step, an edge v1v2 is first removed, merging two faces labeled as combinatorial pseudo-triangles into one face T. In this process, some angles are merged into one: their labels must be reassigned, but we make no changes 15 Fig. 7. Extending a combinatorial pseudo-triangulation in a Henneberg II step.
Top: a representative case of a combinatorial face with four corners. Left: two (out of three) possible placements of the third vertex. Right: the possible labelings of the induced faces as combinatorial pseudo-triangles.
to the labels of the other angles. The rules for assigning labels to merged angles are simple, mimicking what one would expect to happen in a straight-line situation: if one old angle was big, the merged angle is marked big, otherwise small. The face T thus gets exactly four small angles. Its boundary is separated by the vertices v1 and v2 into two chains: each contains at least one corner. Four cases may happen: v1 and v2 are both small (corners), separating the other two corners; only one is small (say, v1), and the other corners are distributed as 1-2 on the chains; or both v1 and v2 are big, and the four corners are distributed as either 2-2 or 1-3. See Figure 6 for a representative case. Note that it is impossible to have all four corners on only one chain induced by v1 and v2.
The new vertex v is now inserted inside this face T, and joined to the old vertices v1 and v2, and to some other vertex v3 on T. The new edges vv1, vv2 and vv3 partition the face T and its four corners into three parts, which can be assigned the labels in several ways. See Figure 7 for a representative case: the systematic verification of all the cases is straightforward.
2 Note that in general the pointed combinatorial pseudo-triangulation guaran-teed by Lemma 8 is not unique. The Lemma allows to generate many different cpts. We next prove that at least one of them is realizable with straight-lines via a similar Henneberg extension technique.
16 Lemma 9 (The Geometric Lemma) Every plane Laman graph G can be embedded as a pseudo-triangulation.
PROOF. Let Gn be a plane Laman graph on n vertices with a triangular outer face. Assume we have a plane Henneberg construction for Gn starting with the outer face and adding vertices only on interior faces. We basically follow the same analysis as in Lemma 8. This time, however, we will not choose the big/small labels of the angles, but rather show that there exists a way of placing a point pn inside a face which realizes a compatible partitioning of the face into pseudo-triangles as prescribed by the Henneberg step on the vertex vn of degree 2 or 3.
As in Lemma 8, the Henneberg I step is straightforward on an interior face (which is what we do here). As an exercise pointing out the difference between the combinatorial and the geometric case, we leave it to the reader to verify that this is not the case on an outer face, where the placement of a vertex at step n + 1 may be constrained by the realization up to step n, and thus may not directly allow an embedding with a certain prescribed outer face.
The analysis of a Henneberg II step is identical to that performed in the combinatorial lemma, and leads to several cases to be considered. We illustrate here only a representative case (but have verified them all). The important fact is that it is always possible to realize with straight-lines at least one of the possible Henneberg II combinatorial pseudo-triangular extensions.
Consider the (embedded) interior face F with four corners obtained by remov-ing an interior edge pipj, and let pk be a vertex on the boundary of F. We must show that there exists a point p inside F which, when connected to pi, pj and pk partitions it into three pseudo-triangles and is itself pointed. The three line segments ppi, ppj and ppk must be tangent to the side chains of F.
We define the feasibility region of an arbitrary point pa on the boundary of F as the (single or double) wedge-like region inside F from where tangents to the boundary of F at pa can be taken. The feasibility region of several points is the intersection of their feasibility regions. An important fact is that the feasibility region of pi and pj always contains the part of the supporting line of the removed edge pipj, and that the feasibility region of any other vertex pk cuts an open segment on it. In fact, the feasibility region of pk intersects the feasibility region of pi and pj in a non-empty feasible 2-dimensional region on one side or the other (or both) of this segment. One can easily see that not only is this region non-empty, but it contains a subregion where a placement of p as a pointed vertex is possible (we call that a pointed-feasible region). We skip the rest of the straightforward details. See Figure 8 for a representative case.
2 17 i k j Fig. 8. Henneberg II step on an interior face. One sees that the feasible region of pk (light grey) intersects the feasible region of the two endpoints pi and pj (dark grey) of the removed edge. We show the two feasible regions (of the pair pi, pj, resp. pk), the final pointed-feasible region and a placement of a pointed vertex p and its three tangents.
Proof of Theorem 2. Let G be a plane Laman graph. If its outer face is not a triangle, apply Lemma 6 to get a new graph Gn which will contain G in its embedding. Follow the plane geometric Henneberg construction described in Lemma 9 to embed Gn starting from a triangle and always inserting new vertices in some interior face.
2 Algorithmic analysis. The proof of Theorem 2 can be turned into an efficient polynomial time algorithm. Given a Laman graph, verifying its planarity and producing a plane embedding (stored as a quad-edge data structure with face information) can be done in linear time . One then chooses an outer face and in linear time one can perform the construction from Lemma 6 to get a triangular outer face. For producing a topological Henneberg construction, we’ll keep an additional field in the vertex data structure, storing the degree of the vertex. We will keep the vertices in a min-heap on the degree field.
To work out the Henneberg steps in reverse we need to do efficiently the following operations: a) detect a vertex of minimum degree (which will be 2 or 3), b) if the minimum degree is 3, corresponding to a vertex v, we must find an edge e that will be put back in after the removal of the neighbors of v, and c) restore the quad-edge data structure. Step a) can be done in O(log n) time. Step c) can be done in constant time. Step b) requires deciding which of the three possibilities for e among v1v2, v1v3 and v2v3 (where v1, v2 and v3 are the neighbors of the vertex removed in a reverse Henneberg II step) produces a Laman graph. Testing for the Laman condition on a graph with 2n −3 edges can be done by several algorithms (the algorithms of Imai and Sugihara, via reductions to network flow or bipartite matching, or via matroid (tree) decompositions, see and the references given there), and takes O(n2). Therefore the time for performing a reverse Henneberg step is dominated by b), which gives a total running time of O(n3).
18 The embedding is done now by performing the Henneberg steps, starting with the outer triangular face embedded on an arbitrary initial triple of points. It is straightforward to see that each step takes constant time to determine a position for the new vertex, and the whole embedding takes linear time once the Henneberg sequence is known.
3.2 Pseudo-Triangular Embeddings of Plane Laman-plus-one Graphs We now turn to a proof of Theorem 3 using Henneberg constructions for plane Laman-plus-one graphs. It is very similar to the proof of Theorem 2. It is instructive though to see the differences, which lie in the combinatorial (and hence also geometric) pseudo-triangulation assignment, where we must keep track of the non-pointed vertex. We have two items which may in principle be prescribed: the outer face and the vertex to become the unique non-pointed one. The non-pointed vertex may only be interior to the circuit. In a Henneberg construction, we will see that it is easy to prescribe either the outer face or the interior vertex to be non-pointed, but the analysis becomes more complicated for the prescription of both. In Section 4 we use a different, global argument to do the simultaneous prescription of the outer face and of the non-pointed vertex, in the case of a rigidity circuit.
The next two lemmas are straightforward extensions of the Laman case.
Lemma 10 (Fixing the Outer Face) Embedding a plane Laman-plus-one graph as a pseudo-triangulation reduces to the case when the outer face is a triangle.
2 One subtle but important difference between Laman-plus-one graphs and rigidity circuits is that for any degree three vertex v in a Laman-plus-one graph there is a Henneberg construction whose last step is the addition of v, while for circuits this need not be the case, i.e. performing a reverse Henneberg II step on a vertex of degree 3 of a circuit does, in general, not result in an-other circuit. This means that if we want to restrict our attention to circuits, our technique of not touching the outer triangular face during the Henneberg construction may not always work simultaneously with maintaining the circuit property.
Lemma 11 (The Topological Lemma) Every plane Laman-plus-one graph has a plane Henneberg construction.
2 A planar Laman-plus-one graph (on n vertices) always has at least two trian-gular faces: the dual planar graph has n vertices (including the vertex corre-sponding to the outer face) and 2n −2 edges, hence there are at least two of degree three. The previous construction allows us to prescribe the outer face 19 in a geometric embedding, should we want to do so, and is not needed in the proof of the combinatorial lemma below.
Lemma 12 (The Combinatorial Lemma) Every plane Laman-plus-one graph admits a pointed-plus-one combinatorial pseudo-triangulation assign-ment.
PROOF. The proof has the same basic structure (but more cases to analyze) as Lemma 8, and relies on the details of the Henneberg construction from Lemma 4. The base case is K4 which has a unique cpt assignment for a choice of an outer face. It is easy to see that Henneberg I steps cause no problem, and the Henneberg II steps work as before when the vertex of degree 3 is not inside the circuit, is not the pointed vertex, and it is not incident to it.
Let vivj be the removed edge and vk the third vertex involved in the Henneberg II step. The only problematic case is when the edge vivj is incident to the unique non-pointed vertex. In this case, the resulting face after the removal of vivj is a pseudo-triangle: it has three, not four corners. We must argue that in at least one combinatorial pseudo-triangulation compatible with the information so far, the three vertices vi, vj and vk cannot lie all three on the same side chain of this face, otherwise the extension to a cpt is impossible.
There is a way around this, which would guarantee that both the outer face and the non-pointed vertex could be prescribed. We will describe this, in a more general setting, in a forthcoming paper. For the time being, it suffices to notice that if this happens during the Henneberg construction, we can always pick up one of the other three vertices guaranteed to have degree three (when there are no degree two vertices), and continue from there. Note that this may change the outer face assignment, though.
2 Lemma 13 (The Geometric Lemma) Every plane Laman-plus-one graph G can be embedded as a pseudo-triangulation.
2 This proof, and the proof of Theorem 3 are now straightforward extensions of those done for the Laman case. Notice that it may not be possible in general to guarantee a certain outer face or non-pointed vertex.
Remarks. The inductive technique described in this section works in fact for rigid graphs on n vertices and fewer than 2n edges (i.e. Laman graphs with at most two extra edges). Indeed, the only ingredient that is needed is the existence of a vertex of degree at most 3 whose removal either leaves a rigid graph or a graph with one degree of freedom. Given a rigid graph with a combinatorial pseudo-triangulation assignment we can add edges and combinatorially assign the big and small labels to the angles, preserving at 20 each step the property that every face has exactly three small angles and every vertex has at most one big angle and the outside face has only big angles.
However, not all of these combinatorial angle assignments are geometrically realizable.
The algorithmic analysis is similar to the case of plane Laman graphs.
4 Combinatorial pseudo-triangulations In this section, we will extend the results from Section 3 on pointed and pointed-plus-one combinatorial pseudo-triangulations. We present a global, non-inductive technique for generating cpt assignments for planar Laman graphs and planar circuits. It is based on a reduction to finding perfect match-ings in a certain associated bipartite graph. By showing that Hall’s condition is satisfied, we are guaranteed to have a solution (and hence a cpt) for both plane Laman graphs and circuits. We also show that the existence of a pointed com-binatorial pseudo-triangulation assignment is not restricted to plane Laman graphs: it also works for certain (connected, multi-) graphs with 2n −3 edges.
Let G = (V, E, F) be a connected plane graph with vertices V , edges E and faces F. Assume |V | = n and |E| = 2n −3. Euler’s relation implies that |F| = n−1. Denote by F ′ the set of interior faces and by fo the outer face (with h vertices, possibly appearing with multiplicities), F = F ′ ∪{fo}. We define a bipartite graph H with the two sets of the bipartition labeled V and W. V stands for the set of vertices V of G and has n elements. The set of face nodes W corresponds to the faces F of G taken with certain multiplicities. For an interior face f ∈F ′ of degree (number of edges on the face, possibly appearing with multiplicities) df, we will put df −3 nodes in W. For the outer face fo we will put h = dfo nodes in W. The total number of elements in W is thus P f∈F ′(df−3)+h = P f∈F df−3|F ′| = 2|E|−3(n−2) = 2(2n−3)−3(n−2) = n.
1 2 3 4 5 A B C D E F A B C D E F 1a 1b 1c 4a 4b 5a Fig. 9. Left: a plane graph with n = 6 vertices (labeled A to F), 2n −3 = 9 edges and 5 faces (labeled 1 to 5). Right: the associated bipartite graph H.
A vertex v ∈V is connected in H to the nodes in W corresponding to the 21 interior faces f of degree larger than 3 to which it belongs in G, and to the nodes corresponding to the outer face (if it belongs to it). Hence if v belongs to the faces f1, f2, . . ., and these faces have multiplicities d1, d2, . . . in W, then v is connected to d1 copies of the node for f1, d2 copies for f2, etc. See Figure 9.
The 6 vertices and 5 faces of degrees 3 (outer face 1), 3 (faces 2 and 3), 4 (face 5) and 5 (face 4) lead to the bipartition sets V = {1, 2, 3, 4, 5, 6} and W = {1a, 1b, 1c, 4a, 4b, 5a}, connected by edges as in the figure.
The connections (edges) in the bipartite graph H represent potential assign-ments of big angles, where an angle is viewed as a pair (vertex,face). Since each vertex must receive a big angle, we want a perfect matching. Since each interior face receives all but three big angles, and the outer face receives all big angles, the choice of multiplicities reflects just that. These considerations lead to the following Lemma.
Lemma 14 There is a one-to-one correspondence between the combinatorial pseudo-triangulations of a plane graph G with n vertices and 2n−3 edges and the perfect matchings in the associated bipartite graph H.
2 (a) (b) 1 2 3 4 5 Fig. 10. (a) A plane graph with 2n −3 edges and no combinatorial pseudo-triangu-lation assignment. (b) A plane non-Laman graph with a cpt assignment.
In general, plane graphs satisfying the conditions of the previous Lemma may or may not have combinatorial pseudo-triangulation assignments. See Fig-ure 10 for examples. But for Laman graphs, we are guaranteed a solution.
The main result of this section is: Theorem 4 If G is a Laman graph, then H has a perfect matching. Hence G has a pointed combinatorial pseudo-triangulation.
PROOF. We will check Hall’s condition (see [14, Theorem 2.1.2, p. 31]) to guarantee the existence of a perfect matching. Let A ⊂V be a subset of vertices. Hall’s condition requires that the number of face nodes in the bipartite graph H which are adjacent to the nodes corresponding to A is at least |A|.
Let FA be the set of faces incident to the vertices in A, and let D = P f∈FA df.
We need to show that |A| ≤D −3|FA|. In fact, when the outer face belongs to FA, it would be sufficient to prove |A| ≤D −3(|FA| −1), but we will prove 22 the stronger inequality, unless FA contains all faces, in which case the desired relation |A| ≤D −3(|FA| −1) follows trivially from |A| ≤n.
Fig. 11. The analysis in the proof of Theorem 4: it suffices to analyze separately the face-connected components of FA (two in this case, shaded differently).
It suffices to carry out the analysis on different face-connected components of FA separately, see Figure 11. From now on, let us assume that FA is face-connected. We consider the plane graph GA consisting of all vertices and edges which are incident to faces of FA. Suppose GA has mI interior edges and mB boundary edges, being incident to faces of FA on both sides and on one side, respectively. We have D = 2mI + mB. Similarly, let nI and nB denote the number of interior vertices, which are completely surrounded by faces of FA, and of remaining (boundary) vertices, respectively. Since every vertex of A is an interior vertex, we have |A| ≤nI. We denote by fI = |FA| the number of “interior” faces of GA, and by fX ≥1 the number of remaining “exterior” faces, including the unbounded face if it does not belong to FA. Thus, in order to establish |A| ≤D −3|FA|, it is sufficient to prove nI + 3fI ≤D.
We apply Euler’s relation to the subgraph of boundary vertices and edges and obtain mB +(fX +1) ≥nB +2. (We only have inequality here, since the graph need not be connected.) Thus, fX −1 ≥nB −mB.
(1) Euler’s relation for the whole graph GA gives (nI + nB) + (fI + fX) = (mI + mB) + 2. Hence fI = mI −nI + 2 + mB −nB −fX.
Laman’s condition implies mI +mB ≤2(nI +nB)−3, hence mI ≤2nI +2nB − mB−3. Now to show nI+3fA ≤D we need nI+3(mI−nI+2+mB−nB−fX) ≤ 2mI + mB, i.e., mI ≤2nI −6 + 3nB −2mB + 3fX. Since we know mI ≤ 2nI +2nB −mB −3, it remains to show 2nB −mB −3 ≤3nB −2mB +3fX −6, i.e., mB −nB ≤3(fX −1), which follows directly from (1) since fX ≥1.
2 The result of Theorem 4 extends to the case of plane circuits. Moreover, we will be able to show in this case a more general version of Theorem 11, by 23 being able to prescribe both the outer face and the non-pointed vertex (which may be chosen as any vertex non-incident to the outer face). In this case, the associated bipartite graph is slightly different: the set V contains all vertices but one, namely the vertex prescribed to be the non-pointed one. The set W has the same description, but its size now is n −1 (because the number of faces is n).
Theorem 5 If G is a plane circuit, then H has a perfect matching. Hence G has a pointed-plus-one combinatorial pseudo-triangulation with a prescribed outer face and prescribed non-pointed vertex.
PROOF. The analysis from proof of Theorem 4 still holds, because in the case of a circuit, the condition on subsets of size k < n is exactly the same as for Laman graphs: they span at most 2k −3 edges. Therefore the analysis works whenever FA does not cover the whole polygon. Since at least one vertex is missing from A (the vertex prescribed to be non-pointed), this is always the case.
2 Algorithmic analysis. To check whether a graph admits a combinatorial pseudo-triangulation (and to compute one) we will use the O(n3/2) time algo-rithm for the maximum flow problem of Dinits (see ) to solve the bipartite matching problem described above.
Remark. This set of degree-constrained subgraphs of a bipartite graph can be modelled as a network flow problem. Thus the set of combinatorial pseudo-triangulations of a given graph (with a given planar embedding, including a specification of the outer face) is in one-to-one correspondence with the vertices of a polytope, given by the equations and inequalities of the network flow.
5 Stretching combinatorial pseudo-triangulations We have seen in the previous sections two proofs of the fact that every plane Laman graph can be assigned a combinatorial pseudo-triangulation labeling.
The technique from Section 3 does not realize geometrically every such possible combinatorial structure. In this section we give the strongest version of the main result by proving the following theorem.
Theorem 6 For any plane Laman graph G and for any of its combinatorial pseudo-triangular assignments, there is a compatible straight-line embedding, and it can be found efficiently. The same holds for plane circuit graphs.
24 The proof is a consequence of two general results of independent interest.
We first give in Theorem 7 two characterizations of stretchable combinatorial pseudo-triangulations. The stretchability proof relies on a directed version of Tutte’s Barycentric Embedding Theorem (Theorem 8). Finally, we show that the characterization in Theorem 7 is satisfied for Laman (Theorem 9) and circuit plane graphs (Theorem 10) with cpt assignments.
5.1 Two characterizations of stretchability In this section we give two combinatorial characterizations of stretchability of combinatorial pseudo-triangulations in terms of the number of corners of planar subcomplexes and in terms of 3-connectivity properties of an associated directed graph.
Let G = (V, E) be a plane graph with a combinatorial pseudo-triangulation la-beling. We do not impose any restrictions on its number of non-pointed vertices or rigidity properties. As a plane graph, every subgraph GS = (S, ES) induced by a subset of vertices S ⊂V has an induced plane embedding and a well-defined unbounded region. The boundary of the unbounded region consists of cycles of vertices and edges, with one cycle for each connected component of GS. Some edges and/or vertices may be repeated in these cycles. For example, if GS is a tree then every edge appears twice.
Corners of boundary cycles. We have defined corners in combinatorial pseudo-triangulations as being the angles marked small. We extend the con-cept to the vertices on boundary cycles of induced subgraphs GS by looking at the labels of angles in G incident to v on the outer face of GS. We call v a corner of type 1 if it contains a big label on the outer face, or a corner of type 2 when v is non-pointed in G but contains two consecutive small labels on the outer face.
The following simple counting lemma will be useful later.
Lemma 15 Let GS be a subgraph of a cpt induced by the subset S ⊂V .
Assume that GS is connected and that it contains all edges lying in the interior of its boundary cycle. Let GS have e edges, k pointed vertices, l non-pointed vertices, c1 corners of type 1 (with big angles in the outer boundary), and a boundary cycle of length b. Then, e = 2k + 3l −3 −b + c1.
In this statement a vertex in GS is called pointed if and only if it was pointed in G.
25 PROOF. Let f be the number of interior pseudo-triangles. The number of interior angles in GS is 3f + k −c1, because there are 3f small interior angles and k −c1 interior big angles. But the number of interior angles also equals 2e −b (since the total number of angles in any plane graph equals 2e). Hence, 2e −b = 3f + k −c1, or 3(e −f) = e + k + b −c1.
Finally, Euler’s formula applied to GS (as it contains all its interior edges) is (k+l)+(f +1) = e or e−f = k+l−1, which implies 3k+3l−3 = e+k+b−c1 and thus the desired statement.
2 The partially directed auxiliary graph D of a combinatorial pseudo-triangulation G. A partially directed graph D = (V, E, ⃗ E) is a graph (V, E) together with an assignment of directions to some of its edges, in such a way that edges are allowed to get two directions, one direction only, or remain undirected.
A plane embedding of a partially directed graph (V, E, ⃗ E) is 3-connected to the boundary if from every interior vertex p there are at least three vertex-disjoint directed paths in ⃗ E ending in three different boundary vertices. Equivalently, if for any interior vertex p and for any pair of forbidden vertices q and r there is a directed path from p to the boundary not passing through q or r.
Lemma 16 For every combinatorial pseudo-triangulation G, there is a par-tially directed graph D satisfying the following conditions: (1) D is planar and contains the underlying graph of G.
(2) The vertices on the outer face have no out-neighbors.
(3) Every interior vertex v which is pointed has three out-neighbors: its two neighbors in G along extreme edges and a neighbor along the interior of the pseudo-triangle containing the big angle at v.
(4) For every non-pointed vertex v of G its out-neighbors in D are exactly its neighbors in G.
PROOF. We extend the underlying graph of G to a (topological) triangula-tion by triangulating the pseudo-triangles of G with more than three vertices in such a way that every big angle of G is dissected by at least one new edge.
This can be achieved by recursively dissecting each face with an edge joining a pointed vertex on the face to the opposite corner. Then the edges are oriented as required by the statement. See Figure 12 for an illustration of how a face is triangulated and how the edges incident to big angles are oriented.
2 26 Fig. 12. Left: A combinatorial pseudo-triangular face, with a small black dot indi-cating a small angle (big angles are not marked). Middle: a compatible triangulation of the face. Right: the edges of the auxiliary directed graph.
The main result of this section can now be stated.
Theorem 7 For a combinatorial pseudo-triangulation G with non-degenerate (simple polygonal) faces the following are equivalent: (1) G can be stretched into a compatible pseudo-triangulation.
(2) Every subgraph of G with at least three vertices has at least three corners.
(3) Every partially directed graph satisfying the requirements of Lemma 16 is 3-connected to the boundary.
5.2 Proof of Theorem 7 The implication from part 1 to part 2 is trivial. If G is embedded as a pseudo-triangulation, there is no loss of generality in assuming that the embedding is in general position, so that every subgraph with at least three vertices has at least three convex hull vertices. And all convex hull vertices of a subgraph of G will be corners according to our definition.
Proof of 2 ⇒3 in Theorem 7. To prove 3-connectedness, we will show that from every interior vertex a there is a directed path in D going to the boundary and not passing through two arbitrary (but fixed) vertices b and c.
Let us consider A, the directed connected component of vertex a, defined as the set of all vertices and directed edges of D that can be reached from v without passing through b or c. We define this component as not containing the forbidden points b and c, but it may contain edges arriving at them. Our goal is to prove that A contains a vertex on the boundary of D. We argue by contradiction. Suppose that all the vertices of A are interior to D.
For each interior pointed vertex v, let Tv be the unique pseudo-triangle of G 27 containing the big angle at v. Thus v is in an edge-chain of Tv containing also the two extreme adjacent edges of v. Let GS = (S, ES) be the graph enclosing all the pseudo-triangles Tv associated to the pointed vertices v of A and all the pseudo-triangles incident to the non-pointed vertices. Clearly GS contains A: indeed, every directed edge of G is contained in a pseudo-triangle associated to its source vertex.
It is easy to prove that GS has at least three corners. Indeed, consider the original vertex v, which is certainly in A. If v is pointed, then GS contains at least the pseudo-triangle containing the big angle at v. If v is not pointed then GS contains at least v and all its neighbors in G. One has to prove that in the latter case v cannot have a single neighbor. If it did, consider the (unique) pseudo-triangle T containing v. As a subgraph of G it has at least three vertices, hence it has at least three corners. But its corners must be corners of T as a pseudo-triangle. This is impossible because v itself is one of the three corners of the pseudo-triangle and is not a corner of the subgraph.
We now use the fact that GS has at least three corners. We claim that at least one of them, d, belongs to A. This gives the contradiction, because then there is an edge of D \ G jumping out of that corner d (by the conditions imposed on the partial orientation in D), which means that the pseudo-triangle(s) corresponding to d should have been contained in GS and hence d is not a corner of GS anymore.
To prove the claim, let v1, . . . , vk be the corners of GS which are not in A. We want to prove that k ≤2. For this let T1, . . . , Tk be pseudo-triangles in GS, each having the corresponding vi as a corner (there may be more than one valid choice of the Ti’s; we just choose one). By definition, some non-corner pointed vertex or some corner non-pointed vertex of each Ti is in the component A.
Were there no forbidden points, from a non-corner pointed vertex we could arrive to the three corners of Ti by three disjoint paths: two of them along the concave chain containing the initial point and the third starting with an edge of D \ G. From a non-pointed corner vertex we could arrive to the other two corners by two disjoint paths: moving out from the vertex to the two neighbors in the incident side chains and then following along them. In particular, since vi is not in A, one of the forbidden points must obstruct one of these paths, which implies that either vi equals one of the forbidden points b and c or that Ti is the pseudo-triangle of one of the two forbidden points. And, clearly, each of the two forbidden points contributes to only one of the indices i (either as a corner of GS or via its associated pseudo-triangle if it is not a corner, but not both). This shows that k ≤2 and completes the proof.
2 Tutte’s equilibrium condition. To prove 3 ⇒1 we use a directed version of Tutte’s Theorem on barycentric embeddings of graphs.
28 An embedding D(P) of a partially directed graph D = (V, E, ⃗ E) on a set of points P = {p1, . . . , pn}, together with an assignment w: ⃗ E →R of weights to the directed edges is said to be in equilibrium at a vertex i ∈V if X (i,j)∈⃗ E wij(pi −pj) = 0.
(2) Theorem 8 (Directed Tutte Theorem) Let D = ({1, . . . , n}, E, ⃗ E) be a partially directed plane graph, 3-connected to the boundary, whose boundary cycle has no repeated vertices. Let (k + 1, . . . , n) be the ordered sequence of vertices in this boundary cycle and let pk+1, . . . , pn be the ordered vertices of a convex (n −k)-gon. Let w: ⃗ E′ →R be an assignment of positive weights to the internal directed edges. Then: (i) There are unique positions p1, . . . , pk ∈R2 for the interior vertices such that all of them are in equilibrium in the embedding D(P), P = {p1, . . . , pn}.
(ii) In this embedding, all cells of D are realized as non-overlapping convex polygons.
PROOF. Part (i) is essentially Lemma 9 in . Since the proof is not long, we reproduce it here. For simplicity, let us represent the position of each point as one complex number pi (instead of a pair of real coordinates). The equilibrium conditions become then a linear system of k equations in k complex unknowns, the positions p1, . . . , pk. We prove that the square matrix M of this system is non-singular. Hence the system has a unique solution for any choice of pk+1, . . . , pn. We will use only the fact that D is connected (not 3-connected) to the boundary.
Note that each diagonal entry mii in M equals the sum P j̸=i wij. Each non-diagonal entry mij equals the negative of the corresponding wij. In particular, if y = (y1, . . . , yk) is an element in the kernel of M, then for every i we have the equation: n X j=1 wij yi = k X j=1 wijyj In this equation we implicitly take wii = 0, as well as wij = 0 whenever ij is not a directed edge in D.
Consider an entry yi of y of maximum absolute value. We prove now that yi = 0, and hence y = (0, . . . , 0). If yi is not zero, the maximality of |yi| (together with the above equation) implies that wij = 0 for all j > k and yj = yi whenever wij ̸= 0. The second condition, recursively, implies that yj = yi (and hence yj has maximum absolute value, too) for every vertex j 29 that can be reached from i by a directed path in D. The first condition implies that no vertex whose coordinate in y has maximum absolute value is joined to the boundary. These two assertions contradict the fact that D is connected to the boundary.
For part (ii), the proof of Tutte’s Theorem given in [39, Theorem 12.2.2, pp. 123–132] works with only minor modifications. First, in the definition of good representation (Definition 12.2.6, p. 126), each point pi is required to be in the relative interior of its out-neighbors, since this is what the directed equilibrium condition gives. Second, Claim 1 on p. 126 proves that in a good representation it is not possible for a vertex p that p and all its neighbors lie in a certain line ℓ, using 3-connectedness. The proof can be adapted to use 3-connectedness to the boundary as follows: consider three vertex disjoint paths from p to the boundary. Call q any of the three end-points, assumed not to lie in the line ℓ. Complete the other two paths to end at q using boundary edges in opposite directions. This produces three vertex-disjoint paths from p to a vertex q not lying on ℓ. The rest needs no change.
2 Note that edges of E which don’t have a representative in ⃗ E have no weight w and don’t appear at all in the system (2). Their presence or absence has no influence on the locations of the points pi. Nevertheless, the whole graph D is planar. This is less surprising than it may seem at first sight, since the subgraph of edges in ⃗ E already has convex faces (by statement (ii) of the theorem), and therefore the additional edges can just be added into these faces.
Proof of 3 ⇒1 in Theorem 7. Construct an auxiliary partially directed graph D in the conditions of Lemma 16, choose arbitrary positive weights for its directed edges, and apply the Directed Tutte Theorem to it. Since all weights are positive, the equilibrium condition on an interior vertex p, together with the convexity of faces that comes from Tutte’s theorem, implies that every interior vertex is in the relative interior of the convex hull of its out-neighbors.
The conditions on D then imply that the straight-line embeddding of G so obtained has big and small angles distributed as desired.
2 Time Analysis. Suppose that we are given a cpt that can be stretched.
Tutte’s theorem actually gives an algorithm to find a stretching: construct the auxiliary graph D of Lemma 16, choose coordinates for the boundary cycle in convex position and arbitrary positive weights for the directed edges, and then compute the equilibrium positions.
Everything can be done in linear time, except for the computation of the 30 equilibrium position for the interior vertices. In this computation one writes a linear equation for each interior vertex, which says that the position of the vertex is the average of its (out-)neighbors. The position of the boundary vertices is fixed. It has been observed [8, Section 3.4] that the planar structure of this system of equations allows it to be solved in O(n3/2) time, using the √n-separator theorem for planar graphs in connection with the method of Generalized Nested Dissection (see [28,29] or [40, Section 2.1.3.4]), or even in time O(M(√n)), where M(n) = O(n2.375) is the time to multiply two n × n matrices.
The above complexity estimate assumes the unit-cost arithmetic model of computation. If we specialize and set all weights wij = 1, the algorithm is polynomial in the bit complexity model as well, because all input coefficients are small integers. Indeed, each row of the coefficient matrix of the system (2) above has a diagonal entry equal to the outdegree and at most one non-zero entry for each outgoing edge, which is equal to −1. The right-hand sides are the coordinates of the boundary vertices, which form a convex polygon. Setting pk+i = (i, i2), for example, yields coordinates between 0 and n2. Gaussian elimination is polynomial in the bit complexity model, and thus our algorithm is polynomial in the bit complexity model as well.
More specifically, since the matrix is diagonally dominant, the determinant cannot exceed the product of the diagonal entries, and hence it can be bounded by 6n, using the fact that the sum of the degrees is at most 6n (cf. Richter-Gebert for the corresponding calculation in the case of the original Tutte theorem). This determinant is a common denominator for all elements of the solution. Thus, if we scale the resulting embedding by this factor, the resulting pseudo-triangulation is embedded on an integer grid of side length n26n.
5.3 Laman and circuit combinatorial pseudo-triangulations can be stretched Not all combinatorial pseudo-triangulations can be stretched: see for instance the second example in Figure 10. Its non-stretchability can be proved either by showing that the graph is not Laman (while the graph of every pointed pseudo-triangulation must be so) or by applying the characterization given in Theorem 7 and finding a subgraph with fewer than three corners.
Our next goal is to prove that if the underlying graph of a combinatorial pseudo-triangulation is Laman (or is a rigidity circuit) then it can be stretched.
The proof uses the Laman counting condition to show that every subgraph has at least three corners. We recall that both Laman graphs and circuits have the property that a subgraph induced on a subset of k, 2 ≤k ≤n −1 vertices spans at most 2k −3 edges. The complementary set of n −k vertices 31 is therefore incident to at least m−(2k −3) edges, where m, the total number of edges, is m = 2n −3 for Laman graphs and m = 2n −2 for circuits. This is equivalent to saying that: • In a Laman graph with n vertices, every subset of k ≤n −2 vertices is incident to at least 2k edges.
• In a rigidity circuit graph with n vertices, every subset of k ≤n−2 vertices is incident to at least 2k + 1 edges.
We use this rephrased Laman property to prove the following theorem.
Theorem 9 Every subgraph GS of a Laman combinatorial pseudo-triangula-tion G has at least 3 corners. Therefore G can be stretched.
PROOF. We show first that there is no loss of generality in assuming that GS is simply connected (i.e. it is connected and contains all the edges of G interior to its contour) and that no edge appears twice in the boundary cycle.
If GS has an edge which appears twice on the boundary cycle, its removal does not change the number of corners; indeed, each end-point of such an edge that is a corner after the removal must be a corner before as well. If GS is not connected, either some connected component has at least three vertices or all the vertices of GS are corners. If GS is connected but not simply connected, then adding to GS the pseudo-triangles, edges and vertices of G that fill in the holes does not change the number of corners.
We now observe that since G is pointed, the equation in Lemma 15 becomes e = 2k −3 −b + c, if GS has e edges, k vertices, b boundary edges, and c corners. Let b0 be the number of boundary vertices of GS (which may be smaller than b, if a boundary vertex appears twice in the boundary cycle). We now consider the set of edges incident to vertices in the interior of GS. Since there are k −b0 interior vertices, the (rephrased) Laman property tells us that there are at least 2(k −b0) such edges. On the other hand, these edges are all interior to GS, and the total number of interior edges in GS is e −b. Hence: 2k −2b0 ≤e −b = 2k −3 −2b + c, which implies the desired relation c ≥3 + 2b −2b0 ≥3.
2 Theorem 10 Every subgraph GS of a rigidity circuit cpt G has at least 3 corners. Hence, G can be stretched.
32 PROOF. As in Theorem 9 we may assume without loss of generality that GS is connected, contains all the edges of G enclosed by its boundary cycle and its boundary cycle has no repeated edges.
Let GS have e edges, k pointed vertices, l non-pointed vertices, and b boundary edges. By Lemma 15 we have e = 2k + 3l + c1 −b −3.
(3) If GS has no interior edge then the statement is trivial: either GS contains no pseudo-triangle and then all its vertices are corners, or it contains only one pseudo-triangle and the three corners of it are corners of GS. We show that if GS has at least one interior edge then e ≥2k + 2l −b + 1.
(4) Indeed, let A ⊂S be the set of vertices interior to GS, so that its cardinality is k+l−b0, where b0 ≤b is the number of boundary vertices in GS. If A is empty then b0 = k +l ≤b and our inequality (4) is equivalent to e ≥b+1−2(b−b0), which holds by the existence of at least one interior edge. If A is not empty we apply the Laman condition to GA, which says that the number of interior edges of GS is at least 2(k + l −b0) + 1, hence the total number of edges in GS is at least 2(k + l −b0) + 1 + b ≥2(k + l −b) + 1 + b = 2k + 2l −b + 1.
Formulas (3–4) imply l + c1 ≥4 and, since l ≤1 (because there is only one non-pointed vertex in G), c1 ≥3.
2 With this, the proof of Theorem 6 has been completed.
2 (a) (b) (c) Fig. 13. (a) An embedding of the pseudo-triangle of Figure 12. (b) A different embedding of the same pseudo-triangle with the same weights for the interior edges.
(c) A pseudo-triangle which is not affinely equivalent to (a) and (b).
33 5.4 Specifying the Shape of Pseudo-Triangles When one analyzes the system of equilibrium equations (2) for our directed graph D, one sees that for the vertices interior to the side chains of some pseudo-triangle t, all their out-neighbors lie on t, see Figure 13a. The only way to “leave” t is by one of the corners. It follows that we can apply an affine transformation to t without destroying equilibrium in the vertices of the side chain, see Figure 13b.
This observation can be used to gain some control over the shape of pseudo-triangles. Suppose that we have some desired shape of a pseudo-triangle t.
After drawing t in the plane, we triangulate it and select the additional directed interior edges in the process of Lemma 16 in such a way that every vertex on a side chain lies in the convex hull of its three out-neighbors. Then we choose positive weights for the outgoing edges of every vertex on a side chain in such a way that equilibrium holds. (This choice is unique for every vertex up to a scalar factor; this scalar factor only multiplies one equation of the system (2) by a constant and hence does not change the solution.) It follows from these considerations that, for any choice of the remaining weights, the shape of t will be an affine image of the given shape: the ver-tices of the side chain will always have the same barycentric coordinates with respect to the three corners.
If we call an equivalence class of shapes under affine transformations an affine shape, we may say that we can independently control the affine shape of each pseudo-triangle. Of course, for a triangle, specifying the affine shape is no restriction: all triangles have the same affine shape. For a k-gon, specifying its affine shape reduces the degrees of freedom by 2(k −3). See Figure 13c for an example of different affine shape.
Together with the fact that the exterior polygon can be chosen freely, this gives the following strengthening of Theorem 6.
Theorem 11 Let G be a plane Laman graph or circuit graph with a combina-torial pseudo-triangular assignment. For any positions of the exterior vertices as a convex polygon and for any given set of pseudo-triangle shapes there is a straight-line embedding in which each pseudo-triangle is an affine image of the given shape.
2 This embedding is unique for Laman graphs (but not for circuit graphs). In fact, there even is a one-to-one correspondence between all embeddings of a given Laman graph with a cpt and a (n −k)-gon as an outer face and the set shape specifications. In this context, a shape specification consists of a convex (n−k)-gons in the plane together with an affine shape for each pseudo-triangle.
34 6 Conclusions and Open Problems We have shown that any combinatorial pseudo-triangulation of a plane Laman graph or of a plane rigidity circuit is stretchable. In the latter case, we may even prescribe the non-pointed vertex. In addition, Laman-plus-one graphs are also stretchable, although we may not in general be able to prescribe the outer face or the non-pointed vertices.
The Main Result stated in the introduction has thus been extended along several lines, leading to interesting combinatorial objects to study and several open questions, some solved in this paper, some left for the future. We end with a listing of the main directions for further investigations.
Embeddability of planar generically rigid graphs as pseudo-triangu-lations. The goal here is the clarification of the connection between minimum (pointed) pseudo-triangulations of a planar point set and triangulations (max-imal planar graphs embedded on the same point set). Triangles are pseudo-triangles, and every triangulation is a pseudo-triangulation, but some or all of the vertices of the embedding may not be pointed. All planar graphs con-taining a Laman graph are rigid (although not minimally so). Stratifying by the number of additional edges (besides a minimally rigid substructure) added to a Laman graph, we want to investigate realizability as triangulations with some prescribed number of non-pointed vertices. In this paper, we solved the case of one additional edge (via the special case of rigidity circuits). We leave open the question of completing the characterization for the whole hierarchy.
Such an investigation will shed light into new intrinsic properties of planar tri-angulations, some of the best studied and still elusive objects in Combinatorial Geometry. We make the following conjecture.
Conjecture 17 Given a plane graph G, the following conditions are equiva-lent: (1) G is generically rigid.
(2) G can be straightened as a pseudo-triangulation.
Combinatorial pseudo-triangulations and embeddings on oriented matroids (pseudolines). We have seen that not all planar graphs admit-ting combinatorial pointed pseudo-triangular labelings are Laman graphs. But those which are Laman also have straight-line realizations. A further direction of research emerging from our work is to study the connection between the combinatorial pseudo-triangulations and realizations in the oriented matroid sense (on pseudo-configurations of points).
35 Grid size of pseudo-triangular embeddings. Every planar graph can be embedded on a grid of size O(n) × O(n), see for example [18, 19, 42]. As mentioned at the end of Section 5.2, pseudo-triangulations can be embedded on a grid of exponential side length, roughly 6n. Here is a natural remaining problem.
Open Problem 1 Can a planar Laman graph be embedded as a pseudo tri-angulation on a O(nk) × O(nk) size grid? What is the smallest such k?
Reciprocal duals of pseudo-triangulations. Planar graphs have combina-torial duals, obtained by replacing faces with vertices and vice-versa. Moreover, when an embedded planar graph supports a self-stress, it has a geometric dual, the so-called reciprocal diagram of Maxwell . A natural question (which will be answered in a subsequent paper) concerns the connection between stressed pseudo-triangulations (necessarily not minimal) and the planarity of their re-ciprocal duals, see .
Algorithmic questions. We conjecture that our embedding algorithm from Section 5 can be improved from O(n3/2) to O(n log n) time. This includes the construction of a cpt for a given planar Laman graph (analyzed at the end of Section 4), and the stretchability via Tutte’s embeddings (analyzed at the end of Section 5).
The time complexity of the first embedding algorithm, presented in Section 3, would be improved by a positive answer to the following open questions: Open Problem 2 Is it possible to decide the Laman condition in linear time for a planar graph?
Open Problem 3 Is it possible to decide, faster than by testing the Laman condition, which edge to put back in a Henneberg II step for a planar graph?
For a combinatorial pseudo-triangulation?
Acknowledgements This research was initiated at the Workshop on Rigidity Theory and Scene Analysis organized by Ileana Streinu at the Bellairs Research Institute of McGill University in Barbados, Jan. 11–18, 2002, and partially supported by NSF grant CCR-0203224. A preliminary version of this paper appeared in .
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40 |
16069 | http://anslyn.cm.utexas.edu/courses/PhysOrg_Fall2021/Exams_files/Final_2013.pdf | a) Write the mechanism of Friedel-Crafts alkylation of ethyl benzene to give 1,4-diethylbenzene. Show all arrow pushing. b) Using resonance and inductive effects, explain why an ethyl group will direct a second ethyl group to the para position in Friedel-Crafts alkylation. Br AlBr3 c) For the Friedel-Crafts alkylation reaction shown below, there is one major product where all ethyls are meta. Explain why this product is preferred and why its regiochemistry is surprising. Br AlBr3 heat for a long time excess Identify any strains present in the following molecules in the exact conformation drawn. (5 pts) The relative SN1 solvolysis rates of two alkyl tosylates is shown below: a) Draw orbitals to show why the structure of the adamantyl ring prevents the resonance stabilization of the allylic carbenium ion intermediate formed from the second structure. (3 pts) a) Explain why the rate of the solvolysis of the second structure is so much slower. (2 pts) OTs OTs 1 105 1. (22 points) Let’s make meth(anol)! a) Draw the Walsh diagram for pyramidal CH3. You can draw the linear form if it helps you, but it is not necessary. Mix the two highest energy orbitals that have the same symmetry and populate the orbitals for methyl radical. (8 points) b) Draw the occupied orbitals from part a) on the left side of the diagram below. Consider the mixing of the orbitals as indicated and fill in the diagram with the appropriate amount of electrons. (14 points) c) The calculated wavefunctions for all of methanol’s occupied orbitals are shown below. Match each orbital to its correct energy level on the above diagram by filling in the appropriate letter in the boxes above. z x y Draw the CH3 orbitals here 4. (23 points) Four parts on stereochemistry: a) Circle any chiral molecules in the set below (6 points): b) Label the indicated groups as homotopic, enantiotopic, or diastereotopic (8 points): i. ii. iii. iv. O OH OH O Cl Cl O Cl Cl O TsO TsO faces of alkene O TsO TsO O OH OH c) For the following reactions, fill in the box with the appropriate term (enantiomers, diastereomers, same molecule) that defines the stereochemical relationship between the two indicated molecules (5 points). i. ii. iii. H-Br H Br H Br Br H Br H + + + O NaO H HO O H H H OH O H H H + 1.
2. H3O+ O TsO TsO Na O H O TsO O TsO + 1 equiv.
d) Define the sets of reactions below as stereospecific, stereoselective, or both stereoselective and stereospecific (4 points). i. ii. O NaO H HO O H H H OH O H H H + 1.
2. H3O+ O NaO H H OH O H H 1.
2. H3O+ 100% 60% 40% H-Br H Br H Br Br H Br H + + + 5% 5% 45% 45% 5. (24 points) Mannose is a diastereomer, specifically a C2 epimer, of glucose. In solution, the alpha anomer of D-mannose dominates over the beta form. This is in contrast to glucose, where the beta anomer is preferred. O H HO H HO OH OH H H H OH O H HO H HO OH H H H OH OH O H HO H HO H OH OH H H OH O H HO H HO H H OH H OH OH α-D-mannose 67% β-D-mannose 33% α-D-glucose 45% β-D-glucose 55% 5. (continued) Let’s explain why mannose has a smaller preference for the beta anomer (33% β) compared to glucose (55% β). a) Fill in the following Newman projections for α-D-mannose and β-D-mannose looking down the highlighted bonds on the previous page. The anomeric carbon should be on the left. Count the total number of gauche interactions in each, keeping in mind each gauche interaction with an atom in the ring counts for two gauche. (6 points) Hint: Once you have filled in one Newman projection, you should be able to quickly construct the next by only switching two groups of one stereocenter to create the epimer. b) Draw the Newman projections of α-D-glucose and β-D-glucose. Again, you should be able to quickly construct them by switching two groups at one stereocenter. Count the total number of gauche interactions. (6 points) C H O OH C H O OH C H O OH C H O OH β-D-mannose # gauche _ α-D-glucose # gauche _ β-D-glucose # gauche _ α-D-mannose # gauche _ c) Based on the Newman projections you have just drawn, circle the preferred anomer of each pair below and write in their energy differences, assuming each gauche is worth the same value. (4 points) α-D-mannose or β-D-mannose energy difference: _ kcal/mol α-D-glucose or β-D-glucose energy difference: _ kcal/mol d) Use the energy differences from part c) to estimate the ratios of the alpha and beta anomers of both sugars at equilibrium. Which sugar, mannose or glucose, has the higher preference for the beta anomer (prefers the alpha anomer to a lesser extent)? (5 points) α-D-mannose : β-D-mannose α-D-glucose : β-D-glucose e) Experimental evidence shows that the alpha anomer of D-mannose is preferred over the beta 67:33. Draw the relevant orbitals to help explain why α-D-mannose is preferred. Comment on how your diagram affects any gauche interactions you identified in the Newman projections. (3 points) 8. (10 points) The effect of varying the substituent, X, on the rate-determining chloride departure in the SN1 reaction of the starting materials below was measured using Hammett plots. a) Fill in the missing x-axis for the Hammett plot above. Which parameter did you choose, σ, σ+, or σ-, and why? (2 points) b) Why are the slopes in this plot all negative? (2 points) c) Assign the letters corresponding to the starting materials to each of the three lines in the plot. Justify your choice for each starting material. (6 points) X O X O O X Cl Cl Cl A B C log (kX') 9. (6 points, 2 each) Describe the kinetic isotope effects you would expect to observe for the following reactions using the terms primary, secondary, normal, and inverse. a) b) (D)H (D)H H(D) H(D) H(D)Br (D)H Br 2 HBr H(D) Br H H Br H(D) 13. (12 pts, 3 each) Draw the predominant product for each of the following reactions. State whether you expect the reaction to proceed by an SN1/E1, SN2, E2, or E1CB mechanism. a) b) c) d) OTs tBuOH (solvent) heat Cl NaSeH EtOH O TsO NaOEt H OTs H NaOH/H2O 14. (6 pts) The rate of SN2 reactions was found to reverse depending on the counter ion of the nucleophile used for the following SN2 reaction: a) Circle the halogen that will react the fastest when the counter ion is Li+. Explain your choice. (3 points) Cl- Br- I- b) Circle the halogen that will react the fastest when the counter ion is tetrabutylammonium. Explain your choice. (3 points) Cl- Br- I- Br O S O O O H3C CH3X Br O S O O O X + + 15. (10 points) In the following 4 + 2 cyclization, the endo product dominates. Let’s figure out why! a) Fill in the correct phasing for the HOMO of cyclobutadiene on the following diagram (2 points): b) Treat the dienophile, methyl vinyl ketone, as though it were butadiene. Draw the correct phasing for the LUMO on the following two diagrams that are simply different conformers of the same dienophile (2 points): O + O O endo exo + HOMO LUMO O LUMO O c) Use the phasing from parts a) and b) to fill in the orbital diagrams below to show the reaction is indeed allowed for both endo and exo. (4 points) d) Use the orbital diagrams from part c) to explain why the endo product is preferred for Diels-Alder cyclization. (2 points) O endo exo O 16. (5 points) Your TA Alex uses a variety of dyes to create her rainbow hair. One of these dyes is Brilliant Blue FCF, a dye approved for food, drug, and cosmetic use: Brilliant Blue FCF a) Given that the dye appears blue, use the color wheel to determine the wavelengths of light it absorbs. (2 points) b) From the choices below, circle the absorption spectrum that best corresponds to that of Brilliant Blue FCF. (1 point) The majority of Alex’s hair is dyed pink using Eosin B. This dye emits via fluorescence when exposed to UV radiation. Eosin B c) Use the color wheel and the given absorption/emission spectra of Eosin B to determine what color Alex’s hair appears when excited using a UV lamp. (2 points) 17. (10 points) This is simply an arrow-pushing problem. Starting from the imine, push arrows correctly to make the following indole. Make sure to draw all lone pairs, proton transfers, and relevant arrows and steps. There is one step that involves a [3,3] sigmatropic shift and one step that involves a 5-exo-trig cyclization. HN NH2 H O H2SO4/H2O HN + H2O + NH3 HN N imine 2. (10 points) a) Circle the strongest acid for the following pairs of compounds and briefly explain your choice. (6 pts) b) Indicate which of the following compounds would be the strongest acid in the gas phase. Explain your choice. (2 pts) c) In solution the order reverses. Briefly explain why. (2 pts) 3. (6 points) a) Circle the structure that has a higher hydride affinity value, and explain your choice. (3 pts) b) Circle the solvent that will give the slowest rate of the following SN2 reaction, and explain your choice. (3 pts) or 11. (18 points, 6 each) The following three reactions follow either specific base, general base, or no catalysis. a. Predict which corresponds to each mechanistic alternative. (1 pt) b. Draw the transition state of the rate-determining step for each reaction in the brackets given. (3 pts) c. Draw the corresponding reaction coordinate diagram for each mechanistic alternative, where you include all intermediates. (2 pts) a) Reaction 1:____ b) c) Reaction coordinate ΔG⁰ a) Reaction 2:___ b) c) Reaction coordinate ΔG⁰ a) Reaction 3:_____ b) c) Reaction coordinate ΔG⁰ 7. (12 points) The following reaction shows a 1,4-addition of a thiol to an enone in basic conditions. a) What type of catalysis does the reaction above show? (1 pt) b) For the above mechanism, write the rate law for the formation of “D” using the steady state approximation. Use the letters provided to write the rate law. (4 pts) c) Does this reaction have any kinetic dependence on ? Briefly explain why. (2 pts) d) Does this reaction have any kinetic dependence on the base (:B)? Briefly explain why. (2 pts) e) Given your answer in part (d), and the pKa of triethylamine, is this reaction going to be faster at pH 7 or 12? Briefly explain your answer. (3 pts) 10. (15 points) The decomposition of aryl-substituted acetophenone methyl hemiacetal was studied under base-catalyzed conditions with an added base (A:ϴ) and its conjugate acid (HA). To understand the transition state, the following Bronsted plot was generated by varying the pKa of HA and measuring the rate of decomposition of the hemiacetal group. a) What kind of catalysis will the kinetics of the above mechanism indicate? (2 pts) Hint: the rate determining step uses a general acid after prior equilibrium deprotonation. b) Does the Bronsted plot support your answer to part (a)? (1 pt) c) Now, we teach you that α + β = 1 in general catalyzed reactions. So given this information and β = 0.61, what is the α-value for the rds in the mechanism given above? (1 pt) d) For the rds, fill in the remaining corners and axes on the More O’Ferrall-Jencks plot with the correspondent species. (4 pts) e) Approximate the placement of the transition state using the answer that you found on part (c) and a βLG of -0.3, labeling this point as “E”. (1 pt) f) Starting from “E”, how would the transition state change using a stronger base? Label this new transition state as “F”. (2 pts) g) With a stronger base, how would the extent of protonation by its conjugate acid change in the rds? (2 pts) h) With a stronger base, how would the extent of βLG change in the rds? Why does this make sense? (2 pts) 12. (9 points)The following methylation reaction proceeds through a radical mechanism. This reaction produces two different products, where B is the favored product. a) Using the one conformer of A1 given above, draw all three conformers and label the two new ones as A2 and A3. (5 pts) b) Circle the lowest energy conformer in part (a), and briefly explain the type of strain that is minimized in this conformer. (2 pts) c) What is the postulate or principle that you should consider when attempting to explain the product ratios observed in the reaction? (2 pts) 6. (4 points)The following rearrangement can occur via radical or carbocation intermediates shown below. a) Propose two experiments to distinguish between these mechanisms. |
16070 | https://www.sciencedirect.com/science/article/abs/pii/S0720048X18301256 | Nonhypervascular pancreatic neuroendocrine tumors: Spectrum of MDCT imaging findings and differentiation from pancreatic ductal adenocarcinoma - ScienceDirect
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Cited by (37)
European Journal of Radiology
Volume 110, January 2019, Pages 66-73
Research article
Nonhypervascular pancreatic neuroendocrine tumors: Spectrum of MDCT imaging findings and differentiation from pancreatic ductal adenocarcinoma
Author links open overlay panel Grigory Karmazanovsky a b, Elena Belousova a b, Wolfgang Schima c, Andrei Glotov d, Dmitry Kalinin d, Andrei Kriger e
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Highlights
•Nonhypervascular PNETs are not rare and require differential diagnosis with PDAC.
•The most valuable contrast enhanced MDCT features to differentiate between nonhypervascular PNETs and PDACs are determined.
•Contrast enhanced MDCT information may be useful for therapeutic and surgery planning.
Abstract
Purpose
The purpose of our study was to determine contrast-enhanced MDCT features to differentiate nonhypervascular pancreatic neuroendocrine tumors (PNETs) from pancreatic ductal adenocarcinomas (PDACs).
Methods
and materials: We included 74 patients with PNETs and 80 patients with PDACs who underwent preoperative MDCT. Two radiologists evaluated the morphologic characteristic and enhancement patterns of the tumors. Quantitative and qualitative analysis was performed, including evaluation of tumor size, homogeneity, contrast enhancement pattern, presence of pancreatic duct dilatation and tumor invasion to the adjacent vessels and peripancreatic infiltration. Tumor-to-pancreas enhancement ratio was defined as the Hounsfield units (HU) value of the tumor divided by the HU value of the pancreas. the first group was hypervascular PNETs showing hyperenhancement on arterial phase images and nonhypervascular PNETs, showing iso- or hypoenhancement on arterial phase images. After that, two radiologists estimated the possibilities of PNET or PDAC were for nonhypervascular PNETs.
Results
On the basis of arterial enhancement, 43 PNETs were hypervascular and 31 were nonhypervascular. When compared to PDAC, nonhypervascular PNETs more frequently had well-defined tumor margins, intratumoral cystic components, calcifications and blood vessels and less frequently had main pancreatic duct dilatation, peripancreatic infiltration and vascular invasion (p < 0.01 for all). Nonhypervascular PNETs had higher tumor-to-pancreas enhancement ratio in venous phase (1.02 vs. 0.78, p = 0.012). Nonhypervascular PNETs more often had portal-venous hyperenhancement or persistent isoenhancement, while PDAC more often had persistent hypoenhancement or gradual delayed enhancement (p < 0.001). The absence of pancreatic duct dilatation and portal-venous hyperenhancement or persistent isoenhancement were the independent predictors for nonhypervascular PNETs.
(The most accurate MDCT-findings to predict nonhypervascular PNET were the absence of pancreatic duct dilatation and peripancreatic infiltration (79% and 92% accuracy), portal-venous phase hyperenhancement or persistent isoenhancement (77%), the presence of intratumoral blood vessels (77%) and relative enhancement intensity in venous phase >0.9 (76%). Using these criteria, the area under curve for differentiation of PNET from PDAC was 0.906–0.846.
Conclusion
Combined assessment of the enhancement and morphologic characteristics can improve the differentiation between nonhypervascular PNETs and PDAC at contrast-enhanced MDCT.
Introduction
Pancreatic neuroendocrine tumors (PNETs) are a heterogeneous group of tumors that arise from neuroendocrine cells of the pancreas. PNETs account for 2% of all pancreatic neoplasms and have an incidence of 1–2 per 100,000 persons per year [1,2]. Clinically, PNETs are divided into hyperfunctioning (functional) or non-hyperfunctioning (non-functional), depending on the presence of a specific clinical syndrome, caused by hormonal hypersecretion.
Among hyperfunctioning tumors, insulinomas and gastrinomas are the most common types. In most of the patients (up to 85%), however, the hormones are produced in small amounts or are functionally inactive, so these (non-hyperfunctioning) tumors are either discovered incidentally or due to local symptoms . According to the recent World Health Organization (WHO) 2017 classification system, pancreatic neuroendocrine neoplasms (PNENs) are divided into Well-differentiated PNENs: PNETs (grade 1, grade 2 and grade 3) and poorly differentiated PNENs: pancreatic neuroendocrine carcinomas .
Contrast enhanced CT and/or MRI are the initial imaging techniques for patients with suspected pancreatic tumor. Typically, PNETS present as small, well-demarcated, homogeneously enhanced hypervascular lesions, yet up to 41% of PNETs are hypovascular . Previous studies attributed tumor hypovascularity to lower microvessel count, extensive fibrosis or the presence of necrosis, that are more commonly encountered in less differentiated (grade 2 or 3) PNETs .
The differential diagnosis of PNETs and other hypervascular pancreatic lesions (solid-pseudopapillary tumor, pseudosolid serous cystadenoma, renal cancer metastases, intrapancreatic accessory spleen etc.) has been previously described in literature . However, only few studies focused on the differentiation between PNET and pancreatic ductal adenocarcinoma (PDAC) . Moreover, there has not been a direct comparison between MDCT imaging features of nonhypervascular PNETs and PDAC. Differential diagnosis of these two entities is of great importance, especially in the case of non-hyperfunctioning PNETs, since PNET and pancreatic adenocarcinoma have different prognosis and require different treatment strategies. The purpose of our study is to investigate the MDCT-features of hypovascular PNETs and to evaluate the MDCT performance for the differential diagnosis of PNET from PDAC.
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Study population
Institutional review board approval was obtained for this retrospective study, with informed consent being waived. A total of 89 patients with histologically proven PNETs were identified in our pathology database from the period between September 2011 and March 2017. Five patients who had a fine-needle aspiration biopsy and didn't undergo the surgery were excluded. 10 patients were excluded due to the lack of preoperative multiphasic MDCT-examination images. Finally, 74 patients with PNETs
Results
A total of 74 PNETs were analyzed (Table 3). Thirty-one (42%) tumors were classified as hypervascular and 43 (58%) tumors were classified as nonhypervascular. The majority (n = 32/43, 74%) of hypervascular tumors were grade 1 (Fig. 1) and the majority of nonhypervascular tumors (n = 23/31, 74%) were grades 2 and 3 (Fig. 2, Fig. 3), with a statistically significant difference between the two groups (p < 0.001). The mean MVD was significantly lower in nonhypervascular PNETs than in hypervascular
Discussion
Our results showed that the enhancement pattern and additional imaging characteristics of the PNETs help to discriminate them from PDACs. Using specific MDCT findings, the area under curve for differentiation of PNET from PDAC was 0.906 for reader 1 and 0.846 for reader 2.
According to the old WHO 2010 classification system, PNETs were divided into three tumor grades, based on the mitotic count and Ki-67 index: well differentiated neuroendocrine tumors – grade 1 and grade 2, and poorly
Conflicts of interest
None.
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Cited by (37)
Evaluation of contrast-enhanced computed tomography for the differential diagnosis of hypovascular pancreatic neuroendocrine tumors from chronic mass-forming pancreatitis
2020, European Journal of Radiology Citation Excerpt :
Mass contrast enhancement (CT attenuation values) measured in the arterial, portal, and delayed phases were recorded as ACE, PCE, and DCE. Mass-to-pancreas enhancement ratio was defined as HU values of mass divided by that of adjacent pancreatic parenchyma, measured in the arterial (AER), portal (PER), and delayed phases (DER), respectively [5,14,20]. We adopted Cohen’s d to calculate the effect size and statistical power with the aim of justifying sample size . Show abstract To assess the role of contrast-enhanced computed tomography (CECT) for differentiation of hypovascular pancreatic neuroendocrine tumors (hypo-PNETs) from chronic mass-forming pancreatitis (CMFP). A retrospective study of 59 patients (27 hypo-PNETs patients vs 32 CMFP patients) who underwent preoperative CECT between July 2012 and July 2019 was performed. Qualitative and quantitative analysis was performed, including mass location, size, margin, cystic changes, calcification, pancreatic or bile duct dilatation, pancreatic atrophy, local vessels involvement, mass contrast enhancement and mass-to-pancreas enhancement ratio. Multivariate logistic regression analyses were used to identify relevant CT imaging findings in differentiation between hypo-PNETs and CMFP. When compared to CMFP, hypo-PNETs more frequently had a well-defined margin and cystic changes and less frequently had a history of pancreatitis and calcification. CMFP had higher mass contrast enhancement and mass-to-pancreas enhancement ratio in the portal and delayed phases than hypo-PNETs. After multivariate logistic regression analyses, areas under the curve (AUCs) were 0.795 (95 % CI: 0.652−0.899), 0.752 (95 % CI: 0.604−0.866), 0.859 (95 % CI: 0.726−0.943), and 0.929 (95 % CI: 0.814−0.983) for Model 1(clinical factors), Model 2 (qualitative parameters), Model 3 (quantitative parameters), and their combinations, respectively. Combined assessment of clinical factors, qualitative, and quantitative imaging characteristics can improve the differentiation between hypo-PNETs and CMFP at CECT.
### The prognostic value of multidetector CT features in predicting overall survival outcomes in patients with pancreatic neuroendocrine tumors
2020, European Journal of Radiology Show abstract To assess the prognostic value of multidetector CT in predicting overall survival outcomes in patients with pancreatic neuroendocrine tumors (PNETs). Seventy-one patients pathologically diagnosed with PNETs were retrospectively included. The clinical and imaging information was evaluated by two radiologists. The difference between well-differentiated and poorly differentiated PNETs was analyzed. Cox proportional hazards models were created to determine the risk factors for overall survival. Kaplan-Meier survival analyses with log-rank tests were used among different subgroups of patients with PNETs. In the whole cohort, the median survival was 36 months, and the 5-year survival rate was 84.8 %. Patients with poorly differentiated PNETs were more likely to present with symptoms, abnormal tumor markers, larger diameters, irregular shapes, ill-defined margins, invasion into nearby tissues, liver and lymph node metastases, and lower enhancement ratio than those with well-differentiated PNETs (P < 0.05). In the multivariate analysis, lymph node metastases (hazard ratio: 21.52, P = 0.009) and a portal enhancement ratio less than 1.02 (hazard ratio: 30.89, P = 0.024) were significant factors for overall survival. Overall survival decreased with an ill-defined margin, irregular shape, poor differentiation, grade 3 disease, nonfunctional status, abnormal tumor marker levels, invasion into nearby tissues, lymph node and liver metastases, and lower enhancement ratio (log-rank P < 0.05). Poorly differentiated PNETs were more aggressiveness than well-differentiated PNETs. Lymph node metastases and a portal enhancement ratio < 1.02 were independent prognostic factors for worse overall survival outcomes in patients with PNETs.
### Differentiation of atypical non-functional pancreatic neuroendocrine tumor and pancreatic ductal adenocarcinoma using CT based radiomics
2019, European Journal of Radiology Citation Excerpt :
First, the patients enrolled in our study were nonfunctional pNET. Unlike F-pNET patients, their imaging features, like a kaleidoscope, overlap with those of PDAC patients; thus, it was more challenging to differentiate NF-pNET from PDAC . Second, we excluded 12 typical cases and included only NF-pNET case with malignant CT findings, thereby increasing the difficulty of differential diagnosis. Show abstract To develop and validate an effective model to differentiate NF-pNET from PDAC. Between July 2014 and December 2017, 147 patients (80 patients with PDAC and 67 patients with atypical NF-pNET) with pathology results and enhanced CT were consecutively enrolled and chronologically divided into primary and validation cohorts. Three models were built to differentiate atypical NF-pNET from PDAC, including a model based on radiomic signature alone, one based on clinicoradiological features alone and one that integrated the two. The diagnostic performance of the three models was estimated and compared with the area under the receiver operating characteristic curve (AUC) in the validation cohort. A nomogram was used to represent the model with the best performance, and the associated calibration was also assessed. In the validation cohort, the AUC for differential diagnosis was 0.884 with the integrated model, which was significantly improved over that of the model based on clinicoradiological features (AUC = 0.775, p value = 0.004) and was comparable to that of the model based on the radiomic signature (AUC = 0.873, p value = 0.512). The nomogram representing the integrated model achieved good discrimination performances in both the primary and validation cohorts, with C-indices of 0.960 and 0.884, respectively. The integrated model outperformed the model based on clinicoradiological features alone and was comparable to the model based on the radiomic signature alone with respect to the differential diagnosis of atypical NF-pNET and PDAC. The nomogram achieved an optimal preoperative, noninvasive differential diagnosis between atypical pNET and PDAC, which can better inform therapeutic choice in clinical practice.
### Machine intelligence in non-invasive endocrine cancer diagnostics
2022, Nature Reviews Endocrinology
### Computed tomography-based radiomics approach in pancreatic tumors characterization
2021, Radiologia Medica
### Evaluation of Spheroid 3D Culture Methods to Study a Pancreatic Neuroendocrine Neoplasm Cell Line
2019, Frontiers in Endocrinology
View all citing articles on Scopus
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16071 | https://www.quora.com/What-is-the-difference-between-every-second-day-and-every-other-day | What is the difference between “every second day” and “every other day”? - Quora
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What is the difference between “every second day” and “every other day”?
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Richard Lueger
Former editor, ESL teacher (Parliament & Gov't of Canada) · Author has 13.5K answers and 19.2M answer views
·2y
Originally Answered: Which one is correct: "every other day" or "every second day"? Why? ·
Both are correct and they often mean the same thing, but not always. They’d mean the same thing in “You should water this plant every other/second day,” but I couldn’t use ‘second’ in “Today has been just like every other day at work since the pandemic. Nothing special happened.”
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Do “every 2 days and every 2nd day” mean the same?
Why does the phrase “every other day” mean every second day instead of every day besides today?
What is the meaning of "every other day"?
Which one is correct: "every other day" or "every second day"? Why?
What is the difference between "every day" and "day by day"?
Nailah Gull Khan
masters in English (language)&International Relations, Convent of Jesus and Mary · Author has 6.3K answers and 4.3M answer views
·5y
Originally Answered: Why does the phrase “every other day” mean every second day instead of every day besides today? ·
phrase. If something happens every other day or every second day, for example, it happens one day, then does not happen the next day, then happens the day after that, and so on. You can also say that something happens every third week, every fourth year, and so on. I went home every other week.
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The phrases "every second day" and "every other day" are often used interchangeably, but they can have slightly different implications depending on context:
Every Second Day: This typically means an event occurs on one day and then skips the next day, occurring again on the following day. For example, if something happens on Monday, it would happen again on Wednesday, and then Friday, and so on.
Every Other Day: This phrase generally means the same as "every second day," indicating that something happens one day, skips the next, and then happens again on the following day.
In most contexts, they
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The phrases "every second day" and "every other day" are often used interchangeably, but they can have slightly different implications depending on context:
Every Second Day: This typically means an event occurs on one day and then skips the next day, occurring again on the following day. For example, if something happens on Monday, it would happen again on Wednesday, and then Friday, and so on.
Every Other Day: This phrase generally means the same as "every second day," indicating that something happens one day, skips the next, and then happens again on the following day.
In most contexts, they convey the same idea, but it’s always good to clarify if there's any potential for misunderstanding, as usage can vary by region or personal preference.
Upvote ·
John Roberts
Author has 62 answers and 43K answer views
·Mar 23
Originally Answered: Why does the phrase “every other day” mean every second day instead of every day besides today? ·
In Australia we tend to use “every second day” for alternate days so as not to confuse this with “every other day” meaning ‘every day except {a particular day}’, eg. ‘Dad goes to church and rests on Sunday; every other day he works very hard.’
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Is there a difference between “every day” and “everyday”?
When do you use "everyday" and "every day"?
Can you explain the difference between “every other day” and “two days a week”?
What is the difference between "on this day" and "on that day"?
What is the difference between every other day and every second day in English grammar? What about everyday vs. each day?
Paul Rastrick
Former Tax Inspector. · Author has 3.5K answers and 1.7M answer views
·6y
Originally Answered: Why does the phrase “every other day” mean every second day instead of every day besides today? ·
It is all down to context:-
“It happens every other day of the year except today.” - every day besides today.
“It happens every other day.” - every second day.
You cannot just quote without any context and expect a meaningful answer.
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Jamille Day
Former IT Tech at Centrica (company) · Author has 251 answers and 378K answer views
·6y
They can mean the same thing. “Every other day” always means that you skip a day, do a day, skip a day, do a day etc.
But “every second day”can be qualified. (I do my accounts every second day of the month”, meaning “I do my accounts on January 2, February 2, March 2…” etc)
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Richard P. Morrall
Teacher, Librarian, Coach retired. (1962–present) · Author has 8.6K answers and 19M answer views
·6y
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What is the difference between "every day" and "day by day"?
Every day is an adverbial phrase. It modifies (explains) a predicate, a verb. It tell your listener/reader that some action you are talking about occurs regularly, daily, each and every day. You can tell this because you may add another adverb, for example almost. Let me write a few examples. I drink coffee every day. Every day tells my readers that I drink coffee seven days a week, 364 days a year. I wash dishes almost every day. Almost every day tells my readers that while washing dishes happens most days, there are a few days when I do not do so. Perhaps I ate out or brought home fast food
Continue Reading
Every day is an adverbial phrase. It modifies (explains) a predicate, a verb. It tell your listener/reader that some action you are talking about occurs regularly, daily, each and every day. You can tell this because you may add another adverb, for example almost. Let me write a few examples. I drink coffee every day. Every day tells my readers that I drink coffee seven days a week, 364 days a year. I wash dishes almost every day. Almost every day tells my readers that while washing dishes happens most days, there are a few days when I do not do so. Perhaps I ate out or brought home fast food or was just too tired and I left a couple of items in the sink.
Every day also proceeds into the past and into the future. My reader may assume that I have been drinking coffee daily for a very long time and that I am going to continue drinking coffee every day. The same applies to my trying to get my dishes washed daily.
Day by day is also an adverbial phrase, but one can tell that it is somehow different from every day because one cannot add another adverb to it. Almost day by day doesn't sound right; it makes no sense. Day by day I will improve my running time. Day by day tells my readers that I am attempting to do something or that I have made a resolution to do something. It does nor proceed into the past. If I write Day by day I have tried to improve my running time, I am saying that the attempt to get better has been occuring in the past. It does not process into the future.
Ii would characterize every day as a positive statement of fact; day by day as a conditional statement of hope or intent.
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Colleen Kinerk
Raised 3 people to adulthood; no one died · Upvoted by
Nick Pharris
, Ph.D. Linguistics, University of Michigan (2006)
·5y
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What is the difference between "everyday" and "every day"?
Every day is the same as each day. I brusy my teeth every day. You say that to me every day.
Everyday is an adjective describing a noun, meaning nothing special, something that is very common. I don’t wear my everyday clothes to church. I ate my everyday lunch without interest.
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Landon Pasby
A bit of a language nerd. · Upvoted by
Roy Mitchell
, Ph.D. ABD Anthropology & Linguistics, University of California, Berkeley (1989) and
Charles Carlson
, Ph.D Uralic-Altaic & Linguistics, Indiana University Bloomington · Author has 276 answers and 4.3M answer views
·Updated 5y
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What is the difference between "everyday" and "every day"?
I’m glad you realize there is a difference, because many people don’t.
When you do something daily, let’s say, then you do it every day. Every. Day. Two words. Two separate words. “Every day” means, refers to, and includes all days, just like “every cantaloupe” means, refers to, and includes all cantaloupes. You get the idea. A few examples:
Ken and Terry go jogging every day.
Does she work in the office every day?
We make sure to eat a nutritious breakfast every day.
Every day is a winding road.
Every day at work, they tried not to notice the mob of demonstrators.
When something is commonplace, ordi
Continue Reading
I’m glad you realize there is a difference, because many people don’t.
When you do something daily, let’s say, then you do it every day. Every. Day. Two words. Two separate words. “Every day” means, refers to, and includes all days, just like “every cantaloupe” means, refers to, and includes all cantaloupes. You get the idea. A few examples:
Ken and Terry go jogging every day.
Does she work in the office every day?
We make sure to eat a nutritious breakfast every day.
Every day is a winding road.
Every day at work, they tried not to notice the mob of demonstrators.
When something is commonplace, ordinary, mundane, routine, average, run-of-the-mill, plain, or typical, then it can be accurately described as everyday. One word. One. “Everyday” is an adjective, a descriptor. For example:
After several months, the workers grew tired of the menial, everyday tasks and chores.
Both accountants think this quarter’s profits will far exceed anything resembling the everyday numbers the company is used to seeing.
I am everyday people.
Alex said Saturday’s banquet calls for formal attire, so it’s a good idea to forgo your everyday wardrobe.
That’s all there is to it. In sum, every day is an adverbial phrase meaning “each day”, while everyday is an adjective meaning “commonplace”, “ordinary”, or “routine”.
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Savit Dayneh
Studied at No Credentials, Just an Ordinary Knowitall · Author has 157 answers and 127K answer views
·6y
Originally Answered: Why does the phrase “every other day” mean every second day instead of every day besides today? ·
Because it is what it is, means exactly that; every other whatever you want. Otherwise it would be “any other” thing, and that would be a different story altogether.
Upvote ·
Mark Grinstein-Camacho
studied grammar in school · Author has 2.1K answers and 8.8M answer views
·8y
Related
Is there a difference between “every day” and “everyday”?
“Every day” refers to every day: Monday, Tuesday, etc.
Examples:
I study every day.
My friend works out every day.
Some people cook dinner every day.
“Everyday” refers to something that is normal or routine. It is an adjective that describes things that could happen any or every day.
Examples:
This recipe is fast enough for an everyday meal. (Note that this recipe is probably not going to be made every day, although that is possible; it is fast enough to cook any day.)
My everyday clothes are all white. My special-occasion clothes are all black.
Besides my everyday stretches, I go for a run twice a wee
Continue Reading
“Every day” refers to every day: Monday, Tuesday, etc.
Examples:
I study every day.
My friend works out every day.
Some people cook dinner every day.
“Everyday” refers to something that is normal or routine. It is an adjective that describes things that could happen any or every day.
Examples:
This recipe is fast enough for an everyday meal. (Note that this recipe is probably not going to be made every day, although that is possible; it is fast enough to cook any day.)
My everyday clothes are all white. My special-occasion clothes are all black.
Besides my everyday stretches, I go for a run twice a week. (Note that although the narrator probably does stretch every day, he describes that routine as “everyday stretches”. Were he to say that he stretches every day, the other form would be used: “I stretch every day.”
In short, “everyday” is an adjective. It describes something: an everyday meal, an everyday workout. “Every day” is a noun phrase, meaning “each day”: it might occur in “the workout you do every day,” or “the meal she eats every day.”
Also, the pronunciation is different: “Everyday” has one stress on the first syllable; “every day” has a stress on the first syllable of each word.
Please do not say something like “I work out everyday” or “I study everyday”. These are wrong. The correct form for these would be “I work out every day” or “I study every day”.
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Mary
12 years experience as an English L.Teacher in own home. · Author has 2.7K answers and 1.9M answer views
·2y
Related
Can you explain the difference between “every two days” and “twice a day”?
Water the plants every two days..(not every day) This means I only water them on Monday, Wednesday, Friday Sunday, Tuesday, Thursday etc ..
Take the medicine twice a day means two times …every day. I take my medicine both at Breakfast and at Dinner time on Monday, Tuesday, Wednesday, Thursday Friday, Saturday and Sunday . Over 7 days, I have taken 14 lots of medicine.
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Lou A.
Former Editor (2004–2012) · Author has 104 answers and 49.5K answer views
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Do “every 2 days and every 2nd day” mean the same?
Thats an interesting question. Every two days - could be every 48hrs.
Every 2nd day - depends where you start, mon- wed- Friday?
Lets try it in the same sentence.
Every 2 days of January I visit my great aunt in her care home.
Every 2nd day I visit my great aunt in her care home..
or
I visit my gran the same two days every week
I visit my gran every 2nd day.
They have different meanings depending on the context of the sentence.
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KEY DIFFERENCES>
DIFFERENCES BETWEEN PHARYNX AND LARYNX: A COMPREHENSIVE GUIDE
Difference Between Pharynx and Larynx
Written by Angesh singh December 12, 2023
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Written by Angesh singh December 12, 2023
Difference Between Pharynx and Larynx:The pharynx and larynx are essential components of the human respiratory and digestive systems, playing distinct roles in the processes of breathing, swallowing, and vocalization. These structures, located in the upper part of the throat, contribute to the intricate orchestration of air and food passage in the human body.
Pharynx:
The pharynx is a muscular tube situated behind the nasal cavity, mouth, and larynx.
It serves as a common pathway for both air and food, functioning in both the respiratory and digestive systems.
Divided into three sections: nasopharynx, oropharynx, and laryngopharynx, each serving specific functions in the overall process of ingestion and respiration.
Larynx:
Also known as the voice box, the larynx is located just below the pharynx and above the trachea.
It is primarily responsible for producing sound and facilitating speech.
Contains vocal cords, which vibrate when air passes through, producing various pitches and tones.
Plays a crucial role in preventing food and liquids from entering the trachea during swallowing, thanks to the epiglottis.
Difference Between Pharynx and Larynx
Here is a table highlighting the key differences between the Pharynx and Larynx:
CharacteristicPharynxLarynx
Location Behind the nasal cavity, mouth, and larynx Below the pharynx, above the trachea
Function Common pathway for air and food Voice production and sound modulation
Divisions Nasopharynx, oropharynx, laryngopharynx No distinct divisions, but includes vocal cords
Role in Respiration Contributes to the respiratory system Connects the pharynx to the trachea
Role in Digestion Participates in the digestive system No direct role in digestion
Structures Muscular tube Contains vocal cords and the epiglottis
Swallowing Assistance Assists in the initial stages of swallowing Prevents food and liquids from entering the trachea during swallowing
What is Pharynx?
The pharynx is a muscular tube-shaped structure located in the throat, behind the nasal cavity and mouth, and extending to the esophagus and larynx. It serves as a crucial anatomical component involved in both the respiratory and digestive systems. The pharynx plays a central role in the passage of air and food, connecting the nasal and oral cavities to the larynx and esophagus, respectively.
The pharynx is divided into three main regions:
Nasopharynx:The uppermost section of the pharynx, situated behind the nasal cavity. It functions primarily in the passage of air during breathing.
Oropharynx:Located behind the oral cavity (mouth), the oropharynx is responsible for the passage of both air and food. It serves as a common pathway for the respiratory and digestive systems.
Laryngopharynx: The lowermost portion of the pharynx, connecting to both the larynx and esophagus. It plays a crucial role in directing food and air to their respective pathways.
What is larynx?
The larynx, commonly known as the voice box, is a cartilaginous structure located in the neck, situated just below the pharynx and above the trachea. It plays a vital role in the production of sound and the modulation of pitch during speech. The primary functions of the larynx include vocalization and protecting the lower respiratory tract during swallowing.
Key features of the larynx include:
Vocal Cords: The larynx contains two pairs of vocal cords (or vocal folds), which are stretched across its interior. These cords vibrate as air passes through, generating sound waves that are shaped into speech sounds.
Epiglottis:A flap-like structure in the larynx that helps prevent food and liquids from entering the trachea during swallowing. It covers the trachea, directing substances toward the esophagus and away from the respiratory passages.
Cartilages: The larynx is composed of several cartilages, including the thyroid cartilage (Adam's apple) and the cricoid cartilage. These structures provide support and protection to the vocal cords.
Tracheal Connection:The larynx connects the pharynx to the trachea, serving as a transition point for air entering the respiratory system.
The larynx is a crucial anatomical structure that facilitates speech and protects the airway during the complex processes of breathing, swallowing, and vocalization.
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Similarity Between Pharynx and Larynx
The pharynx and larynx share several similarities, as both are integral components of the upper respiratory and digestive systems. Here are some key similarities:
Anatomical Location:
Both the pharynx and larynx are situated in the upper part of the throat.
Respiratory Function:
Both structures play roles in the respiratory system, contributing to the passage of air.
Muscular Composition:
Both the pharynx and larynx contain muscles that aid in their respective functions.
Connection to the Respiratory Tract:
They are interconnected, with the pharynx connecting to the larynx and then to the trachea, forming a continuous pathway for air.
Involvement in Swallowing:
Both structures are involved in the initial stages of swallowing. The pharynx directs food and liquids to the esophagus, while the larynx prevents their entry into the trachea.
Protection of Airway:
They contribute to the protection of the airway during swallowing. The epiglottis in the larynx covers the trachea to prevent aspiration of food and liquids into the respiratory passages.
Voice Production:
While the pharynx is not directly involved in voice production, it contributes to resonance. The larynx, on the other hand, is the primary structure responsible for voice production.
These shared characteristics highlight the interconnected and complementary roles of the pharynx and larynx in facilitating respiratory functions, swallowing, and, in the case of the larynx, vocalization.
FAQ's
What is the primary function of the pharynx and larynx?
The pharynx serves as a common pathway for air and food, contributing to both the respiratory and digestive systems. The larynx, on the other hand, is primarily responsible for voice production and modulation, as well as protecting the airway during swallowing.
How do the locations of the pharynx and larynx differ?
The pharynx is located behind the nasal cavity, mouth, and larynx, while the larynx is situated just below the pharynx and above the trachea in the neck.
Are there distinct divisions within the pharynx and larynx?
Yes, the pharynx is divided into three sections: nasopharynx, oropharynx, and laryngopharynx. The larynx does not have distinct divisions but includes structures such as vocal cords and the epiglottis.
What roles do the pharynx and larynx play in swallowing?
The pharynx assists in the initial stages of swallowing, directing food and liquids to the esophagus. The larynx protects the airway during swallowing, preventing the entry of substances into the trachea with the help of the epiglottis.
Do the pharynx and larynx contribute to the respiratory system?
Yes, both the pharynx and larynx play roles in the respiratory system. The pharynx allows the passage of air, while the larynx connects the pharynx to the trachea, facilitating the movement of air into the lower respiratory tract.
How do the structures differ between the pharynx and larynx?
The pharynx is a muscular tube, while the larynx is composed of cartilage and muscle tissue. The larynx includes vocal cords and the epiglottis, which are crucial for voice production and protecting the airway.
Is there a direct role of the larynx in digestion?
No, the larynx does not have a direct role in digestion. Its primary functions are related to voice production and protecting the airway during swallowing.
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16073 | https://artofproblemsolving.com/wiki/index.php/Ceva's_Theorem?srsltid=AfmBOopP0_J6-jZawinRNiYy2hID2NLJGYyOfChrfGNgG4KftT-PD4FJ | Art of Problem Solving
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Ceva's Theorem
Ceva's Theorem is a criterion for the concurrence of cevians in a triangle.
Contents
[hide]
1 Statement
2 Proof
3 Proof by Barycentric coordinates
4 Trigonometric Form
4.1 Proof
5 Problems
5.1 Introductory
5.2 Intermediate
6 See Also
Statement
Let be a triangle, and let be points on lines , respectively. Lines are concurrent if and only if where lengths are directed. This also works for the reciprocal of each of the ratios, as the reciprocal of is .
(Note that the cevians do not necessarily lie within the triangle, although they do in this diagram.)
The proof using Routh's Theorem is extremely trivial, so we will not include it.
Proof
We will use the notation to denote the area of a triangle with vertices .
First, suppose meet at a point . We note that triangles have the same altitude to line , but bases and . It follows that . The same is true for triangles , so (Note the result is just the area lemma) Similarly, and , so Now, suppose satisfy Ceva's criterion, and suppose intersect at . Suppose the line intersects line at . We have proven that must satisfy Ceva's criterion. This means that so and line concurs with and .
Proof by Barycentric coordinates
Since , we can write its coordinates as . The equation of line is then .
Similarly, since , and , we can see that the equations of and respectively are and
Multiplying the three together yields the solution to the equation:
Dividing by yields: , which is equivalent to Ceva's Theorem.
Trigonometric Form
The trigonometric form of Ceva's Theorem states that cevians concur if and only if
Proof
First, suppose concur at a point . We note that and similarly, It follows that
Here, the sign is irrelevant, as we may interpret the sines of directed angles mod to be either positive or negative.
The converse follows by an argument almost identical to that used for the first form of Ceva's theorem.
Problems
Introductory
Suppose , and have lengths , and , respectively. If and , find and . (Source)
Intermediate
In are concurrent lines. are points on such that are concurrent. Prove that (using plane geometry) are concurrent.
Let be the midpoint of side of triangle . Points and lie on line segments and , respectively, such that and are parallel. Point lies on line segment . Lines and intersect at and lines and meet at . Prove that are collinear. (Source)
See Also
Stewart's Theorem
Menelaus' Theorem
Routh's Theorem
Retrieved from "
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16074 | https://famus.org.uk/module/thoracic-ultrasound-theory/ | Thoracic Ultrasound Theory - FAMUS
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Thoracic Ultrasound Theory
Contents
Choice of probe
Anatomy review (brief!)
Lung sliding
Seashore sign
Stratosphere sign
Lung pulse
Normal lung base
A lines
B lines
Consolidation
Pleural effusion
Thoracic ultrasound has progressed from marking pleural procedures to an integral tool for the assessment of patients with respiratory failure. The protocol we endorse in FAMUS is theBLUE protocol, which was first published by Daniel Lichtenstein in Chest in 2008.
Choice of probe
For lung ultrasound two transducers are potentially needed. For assessment of the pleural line a high frequency linear transducer is preferable (e.g vascular probe). For obese patients the pleural line might only be seen with a curvilinear transducer due to the increased depth needed.
To assess pleural effusions, consolidation and for the presence of B-lines a curvilinear transducer (e.g abdominal probe) is the preferred option, however a phased array probe (e.g cardiac) might be used if no curvilinear transducer is available.
Anatomy review (brief!)
To understand lung ultrasound it is important to review some anatomy first. Rib shadows help with orientation. Ribs are high in calcium, so they will be seen as a very bright line on the anterior surface with a shadow cast below. If you move the transducer towards the cartilaginous part of the ribs you will then be able to look through the ribs and appreciate the pleura behind.
Video 1: normal appearances of ribs with shadow
Video 2: difference between rib and cartilage in the thorax
Lung sliding
In health, the visceral and parietal pleura are closely opposed, separated only by a minute layer of fluid which is not appreciable on ultrasound. This pleural line is seen as a bright white line just below the rib margin. During breathing (more specifically, ventilation) this fluid layer allows the pleura to slide across each other and this is appreciated on ultrasound as a shimmering appearance. This is usually described as lung sliding and is demonstrated in video 3. It is sometimes also referred to as ‘a marching of ants’.
Video 3: Lung sliding – the ‘shimmering’ effect of the parietal and visceral pleura sliding over each other. For an explanation of A lines, see below
ung sliding is an important concept to grasp. It’s presence means the parietal and visceral pleura are directly opposed and the lung is ventilating, and so rules out pleural effusion, pneumothorax and a host of conditions which dramatically reduce lung ventilation (eg very low tidal volume ventilation, ARDS, severe pneumonia, fibrosis). It’s often thought that a lack of lung sliding must mean pneumothorax, but as you will see this is not the case – it is one of the signs of pneumothorax but it is not diagnostic of it. See theBLUE protocol sectionfor more detail on how to diagnose the different pathologies.
Seashore sign
It can be difficult sometimes to clearly determine if lung sliding is present. In the first instance, make sure you are using a high frequency probes (eg vascular) and minimise your depth, as this will give maximum definition of the pleural line; the next option is to use M-mode across the pleural line.
In a patient with lung sliding, the pattern generated will be like ‘waves on a beach’, where the waves represent the intercostal tissues, the pleural line the edge of the sea and the beach is the scattered artefact from the lung below (Figure 1 below).
Figure 1: Waves on a beach appearance of normal lung when viewed through M mode – the ‘seashore’ sign
Another way of thinking about this is to imagine what happens when the ultrasound waves reach the pleural line. Normal lung has well aerated alveoli, which do not transmit sound very well (the impedance of air and lung tissue are very different), so the ultrasound waves are scattered all over once they pass through the visceral pleura which is moving with respiration (Figure 2). In an M-mode image this ‘scatter’ is represented by multiple individual pixels which together look like grains of sand.
Figure 2: US waves (red arrows) scatter when they meet aerated lung. (Note A lines marked by black arrows; see below)
Stratosphere sign
The converse of this is the appearance of the M mode image when there is a pneumothorax present. In this instance, there is no scatter of the ultrasound waves by the moving visceral pleura as it does not oppose the parietal pleura. Instead, all you see below the pleural line is reverberation artefact from the pleural line and tissues above. This has the appearance of a stratosphere with horizontal lines the whole way down the image, with the brightest usually being the pleural line (Figure 3). This is also referred to as the barcode sign.
Figure 3: M mode appearances of a pneumothorax – the stratosphere sign
The image below shows both the seashore and stratosphere signs in the same image, as is created by placing the M mode cursor across the lung point. It helps to highlight the difference between these two signs, which often can seem subtle in real time (Figure 4).
Figure 4: M mode through the lung point highlighting the seashore and stratosphere signs in the same image
Lung pulse
The third sign that may be elicited using M mode is called the lung pulse. This is created by the cardiac impulse being transmitted through the lungs to be picked up by the ultrasound probe, as occurs both in health and in many pathological conditions. It may therefore be seen in combination with both the ‘seashore’ and ‘stratosphere’ appearances. It is viewed on ultrasound as regular, subtle vertical lines (T lines) across an M mode trace (Figure 5).
Figure 5: the lung pulse – regular vertical lines (here indicated by green arrows) seen across an M mode trace
It is important to understand that for the lung pulse to be visible on M mode, the parietal and visceral pleura must be opposed. If there is air between the pleura then this would prevent transmission of the lung pulse from the lung to the probe (and you would see the stratosphere sign only). If there is fluid between the pleura it usually acts as a ‘shock absorber’ and so similarly prevents the lung pulse from being transmitted to the probe. However, it may often be seen in normal respiration with the seashore sign (particularly in health where tidal volumes are relatively low), and in combination with the stratosphere sign in states where the lung is expanded but not ventilating (for example one lung intubation, breath holding, severe pneumonia or inflammation, ARDS). It can therefore be understood that the presence of a lung pulse is one way of distinguishing pneumothorax from most other pathological states that generate a stratosphere sign (see table 1 below).
PathologySeashore/ stratosphereLung pulse
Normal Seashore Present
Pneumothorax Stratosphere Absent
One lung intubation, breath holding, pleurodesis Stratosphere Present
Low tidal volume ventilation, ARDS, severe pneumonia/inflammation/fibrosis Seashore/stratosphere Present
Table 1: difference in M mode appearances of various lung pathologies
Normal lung base
It is important to understand that in many instances, the image you see below the pleural line is all artefact, and is not the lung parenchyma that is being visualised. This helps in understanding why the appearance of both normal lung and pneumothorax is identical below the pleural line – because they both generate the same artefact on B mode imaging. In a patient with no pathology within the lung, the interface between solid abdominal organs and lung appears like a curtain coming in and out of the picture (see video 4).
Video 4: normal appearances of lung base where aerated lung casts a ‘curtain’ over the intra-abdominal organs
A lines
In videos 1 and 3 and figure 1 there are distinct horizontal bright lines visible within the aerated lung below the pleural line. They are the same distance from the pleural line as the pleura is from the top of the screen, and are a reverberation artefact.
They are generated when an ultrasound wave hits the pleural line, gets reflected back to the probe and then instead of being absorbed by the probe it reflects back down towards the lung. As this wave hits the pleura and reflects back to the probe it will have travelled from probe to pleura twice and so the machine will ‘paint’ a line that appears within the lung itself, twice the distance of the probe to pleura. This process repeats and so multiple A line artefacts may be generated each equidistant from the previous. As some of the ultrasound wave is absorbed at each step the amplitude of the signal reduces sequentially and so the A-lines get less pronounced the deeper they are shown (video 5).
Video 5: The generation of A lines
B lines
If there is increased amount of interstitial fluid (as in pulmonary oedema, contusion, infection or inflammation), or interstitial thickening (as in fibrosis), B lines will be seen. They arise from a marked difference in acoustic impedance between an object and its surroundings – in this case thought to be the aerated lung and the adjacent interlobular septae – creating a phenomenon of resonance (Figure 6).
Figure 6: The generation of a B line. Although they appear as a single line on screen, they are in fact thought to be hundreds of horizontal reverberation artefacts which coalesce as a single laser-like projection
B-lines start at the pleural surface and travel at least to a depth of 18cm. They will move with respiration. They will appear as a single line, however with more interstitial fluid they can coalesce and finally the lung will look bright with individual B lines difficult to separate. A feature of B lines is that they will obliterate A lines. All these features can be seen in video 6 below.
Video 6: B lines
Consolidation
Collapsed and consolidated lung can have the appearance of solid organs. In addition fully consolidated lung will show air bronchograms (look out for the white speckles within the consolidation). The periphery of consolidated lung will often have a coarse, irregular outline like the parenchyma has been ripped; this is the ‘shred sign’ of consolidation and is highlighted in the following video (video 7).
Video 7: lung consolidation
Pleural effusion
The vast majority of simple pleural effusions will be visible in the postero-lateral region of the supine patients (the PLAPS point of the BLUE protocol). Simple pleural fluid will usually be anechoic in appearance and, crucially, will allow the passage of ultrasound waves beyond the pleural line so you gn’).will start to see definition of the deep organs. In practice, this means you will often see the superficial diaphragm, the inferior border (right hand border on the screen) of the lung parenchyma and, if deep enough, the spinal column at the base of the picture (the ‘spine sign’).
Video 8 below shows examples of small basal effusions in both the left and right lung. Note how visible the diaphragm is due to the presence of the pleural fluid. Visibility of the superficial diaphragm is one of the features of a pleural effusion – to visualise the superficial diaphragm you must have pathology within (consolidation) or around (effusion) the lung base.
In addition, note the appearance of the segment of collapsed lung, thought to look like a jellyfish tentacle (the ‘jellyfish sign’).
Video 8: Small pleural effusions
In the following video there is a large pleural effusion causing lung collapse, and again the lung takes on a more solid-organ appearance.
Video 9: large pleural effusion with collapsed lung
There are more examples of pathology with explanations in thethoracic ultrasound pathology section.
For a detailed explanation of the BLUE protocol seehere.
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16075 | https://www.chegg.com/homework-help/questions-and-answers/find-equation-tangent-line-curve-given-point-y-sqrt-x-16-4-find-equation-line-need-slope-l-q23940398 | Solved Find an equation of the tangent line to the curve at | Chegg.com
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Find an equation of the tangent line to the curve at the given point. y = SQRT(x) , (16, 4) To find the equation of a line, we need the slope of the line and a point on the line. Since we are requested to find the equation of the tangent line at the point (16, 4), we know that (16, 4) is a point on the line. So we just need to find its slope. The slope of
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Question: Find an equation of the tangent line to the curve at the given point. y = SQRT(x) , (16, 4) To find the equation of a line, we need the slope of the line and a point on the line. Since we are requested to find the equation of the tangent line at the point (16, 4), we know that (16, 4) is a point on the line. So we just need to find its slope. The slope of
Find an equation of the tangent line to the curve at the given point.
y = SQRT(x) , (16, 4)
To find the equation of a line, we need the slope of the line and a point on the line.
Since we are requested to find the equation of the tangent line at the point (16, 4), we know that (16, 4) is a point on the line. So we just need to find its slope.
The slope of a tangent line to f(x) at x = a can be found using the formula
m tan = lim x→a f(x)-f(a)/ x-a
In this situation, the function is f(x) = _____
and a = ___
Consider the parabola y = 6 x − x 2.
(a) Find the slope of the tangent line to the parabola at the point (1, 5).
(b) Find an equation of the tangent line in part (a).
y =
If a ball is thrown into the air with a velocity of 50 ft/s, its height (in feet) after t seconds is given by y = 50 t− 16 t 2. Find the velocity when t = 2.
Let
h = s(t) = 50 t − 16 t 2
give the height of the ball at time t.
Then the ball's velocity at time t = a can be found by
v(a) = lim t→a s(t)-s(a) / t-a
We are requested to find the velocity at t = 2; therefore we use a =_____ and have
v(2)= Lim t→2 s(t)-s(2) / t-2
lim t→2 (50 t − 16 t^2) - (50(__________)-16(2)^2) / t-2
lim t→2 (50 t − 16 t^2) - (___________) / t-2
Find f '(a).
f(x) = 4 x 2 − 5 x + 4
f'(a)=_____
(a) Find the slope m of the tangent to the curve y = 7 + 4 x 2 − 2 x 3 at the point where x = a.
m = _____
(b) Find equations of the tangent lines at the points (1, 9) and (2, 7).
y(x)=(at the point (1, 9))
y(x)=(at the point (2, 7))
(c) Graph the curve and both tangents on a common screen.
Find f '(a).
f(t) = t 4 − 5 t
f '(a) = ______
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16076 | https://grammarphobia.com/blog/2012/04/back-to-back.html | Can three things be back to back?
Q: I bristle at the use of “back to back” for more than two things: “I’ve had over a dozen appointments today, all back to back.” Have you blessed this construction?
A: Some old familiar idioms lose their literal meaning over the years. This is the case with “back to back.”
In modern usage, this phrase is often used to describe not only physical objects alongside each other, but also events that come one after another.
As we all know, events come in threes and fours as well as twos, while logic would require that only two things can be “back to back.”
Besides, events don’t really have fronts and backs. And even if they did, they’d follow one another “front to back,” not “back to back.”
The point is that the phrase “back to back” has broken the bounds of logic.
There are several literal examples in the Oxford English Dictionary of the phrase used adverbially, including this one from a ballad version of Robin Hood (circa 1500): “And there they turnd them back to back.”
The OED’s earliest examples of the phrase used adjectivally (and hyphenated) are in 19th-century writings about houses. In those days, the phrase was meant literally.
The first quotation is from Dr. Lyon Playfair’s Report on the Sanitary Condition of the Large Towns in Lancashire (1845): “Back-to-back houses cannot be considered dwellings of proper construction.”
A century later, the phrase was used literally to describe paired fireplaces. This OED citation is from the sociologist Dennis Chapman’s The Home and Social Status (1954):
“The other living-room usually has a ‘back-to-back’ combination fireplace ‘shared’ with the kitchen.”
But when used to describe events, the OED says, the phrase means “following one upon another without a break, consecutive.” And by extension, it also means “full, crowded.”
The use of “back to back” for events is “chiefly” American, the OED says.
The dictionary’s earliest example is from a sports story that appeared in the New York Times on August 24, 1952:
“Back to back doubles by Gene Woodling and Joe Collins off Early Wynn in the fourth inning produced the only tally of the day.” (The Yankees beat the Indians, 1-0.)
And here’s an example of the phrase used in the sense of “crowded.” It’s from Lady Bird Johnson’s A White House Diary (1970): “Today was one of those full, back-to-back Washington days.”
Have we, you ask, blessed such constructions? We think the nonliteral use of “back to back” is now firmly established in English.
The American Heritage Dictionary of the English Language (5th ed.) defines the adverbial “back to back” as an idiom meaning “consecutively and without interruption: presented three speeches back to back.”
The dictionary defines the adjectival “back-to-back” as meaning “consecutive; successive: back-to-back performances; back-to-back home runs.”
American Heritage doesn’t even bother with the literal meaning. But Merriam-Webster’s Collegiate Dictionary (11th ed.) gives two definitions: (1) “facing in opposite directions and often touching,” and (2) “coming one after the other: consecutive.”
Apparently the only problem with this usage is what to do about the hyphens. Our advice is to use hyphens only when the phrase is used adjectivally before a noun (“back-to-back doubles”). Otherwise, drop the hyphens.
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16077 | https://e.math.cornell.edu/people/belk/measuretheory/Inequalities.pdf | Convexity, Inequalities, and Norms Convex Functions You are probably familiar with the notion of concavity of functions. Given a twice-differentiable function ϕ: R →R, • We say that ϕ is convex (or concave up) if ϕ′′(x) ≥0 for all x ∈R.
• We say that ϕ is concave (or concave down) if ϕ′′(x) ≤0 for all x ∈R.
For example, a quadratic function ϕ(x) = ax2 + bx + c is convex if a ≥0, and is concave if a ≤0.
Unfortunately, the definitions above are not sufficiently general, since they require ϕ to be twice differentiable. Instead, we will use the following definitions: Definition: Convex and Concave Functions Let −∞≤a < b ≤∞, and let ϕ: (a, b) →R be a function.
1. We say that ϕ is convex if ϕ (1 −λ)x + λy ≤(1 −λ)ϕ(x) + λϕ(y) for all x, y ∈(a, b) and λ ∈[0, 1].
2. We say that ϕ is concave if ϕ (1 −λ)x + λy ≥(1 −λ)ϕ(x) + λϕ(y) for all x, y ∈(a, b) and λ ∈[0, 1].
Convexity, Inequalities, and Norms 2 y jHxL Hx, jHxLL Hy, jHyLL Figure 1: For a convex function, every chord lies above the graph.
Geometrically, the function λ 7→ (1 −λ)x + λy, (1 −λ)ϕ(x) + λϕ(y) , 0 ≤λ ≤1 is a parametrization of a line segment in R2. This line segment has endpoints x, ϕ(x) and y, ϕ(y) , and is therefore a chord of the graph of ϕ (see figure 1). Thus our definitions of concave and convex can be interpreted as follows: • A function ϕ is convex if every chord lies above the graph of ϕ.
• A function ϕ is concave if every chord lies below the graph of ϕ.
Another fundamental geometric property of convex functions is that each tangent line lies entirely below the graph of the function. This statement can be made precise even for functions that are not differentiable: Theorem 1 Tangent Lines for Convex Functions Let ϕ: (a, b) →R be a convex function. Then for every point c ∈(a, b), there exists a line L in R2 with the following properties: 1.
L passes through the point c, ϕ(c) .
2.
The graph of ϕ lies entirely above L.
PROOF See exercise 1.
■ Convexity, Inequalities, and Norms 3 Figure 2: A tangent line to y = |x| at the point (0, 0).
We will refer to any line satisfying the conclusions of the above theorem as a tangent line for ϕ at c. If ϕ is not differentiable, then the slope of a tangent line may not be uniquely determined. For example, if ϕ(x) = |x|, then a tangent line for ϕ at 0 may have any slope between −1 and 1 (see figure 2).
We shall use the existence of tangent lines to provide a geometric proof of the continuity of convex functions: Theorem 2 Continuity of Convex Functions Every convex function is continuous.
PROOF Let ϕ: (a, b) →R be a convex function, and let c ∈(a, b). Let L be a linear function whose graph is a tangent line for ϕ at c, and let P be a piecewise-linear function consisting of two chords to the graph of ϕ meeting at c (see figure 3).
Then L ≤ϕ ≤P in a neighborhood of c, and L(c) = ϕ(c) = P(c). Since L and P are continuous at c, it follows from the Squeeze Theorem that ϕ is also continuous at c.
■ We now come to one of the most important inequalities in analysis: Convexity, Inequalities, and Norms 4 jHxL PHxL LHxL c Figure 3: Using the Squeeze Theorem to prove that ϕ(x) is continuous at c Theorem 3 Jensen’s Inequality (Finite Version) Let ϕ: (a, b) →R be a convex function, where −∞≤a < b ≤∞, and let x1, . . . , xn ∈(a, b). Then ϕ(λ1x1 + · · · + λnxn) ≤λ1ϕ(x1) + · · · + λnϕ(xn) for any λ1, . . . , λn ∈[0, 1] satisfying λ1 + · · · + λn = 1.
PROOF Let c = λ1x1 + · · · + λnxn, and let L be a linear function whose graph is a tangent line for ϕ at c. Since λ1 + · · · + λn = 1, we know that L(λ1x1 + · · · + λnxn) = λ1L(x1) + · · · + λnL(xn).
Since L ≤ϕ and L(c) = ϕ(c), we conclude that ϕ(c) = L(c) = L(λ1x1 + · · · + λnxn) = λ1L(x1) + · · · + λnL(xn) ≤λ1ϕ(x1) + · · · + λnϕ(xn).
■ This statement can be generalized from finite sums to integrals. Specifically, we can replace the points x1, . . . , xn by a function f : X →R, and we can replace the weights λ1, . . . , λn by a measure µ on X for which µ(X) = 1.
Convexity, Inequalities, and Norms 5 Theorem 4 Jensen’s Inequality (Integral Version) Let (X, µ) be a measure space with µ(X) = 1. Let ϕ: (a, b) →R be a convex function, where −∞≤a < b ≤∞, and let f : X →(a, b) be an L1 function.
Then ϕ Z X f dµ ≤ Z X (ϕ ◦f) dµ PROOF Let c = R X f dµ, and let L be a linear function whose graph is a tangent line for ϕ at c. Since µ(X) = 1, we know that L( R X f dµ) = R X(L ◦f) dµ. Since L(c) = ϕ(c) and L ≤ϕ, this gives ϕ(c) = L(c) = L Z X f dµ = Z X (L ◦f) dµ ≤ Z X (ϕ ◦f) dµ.
■ Means You are probably aware of the arithmetic mean and geometric mean of positive numbers: x1 + · · · + xn n and n √x1 · · · xn.
More generally, we can define weighted versions of these means.
Given positive weights λ1, . . . , λn satisfying λ1 +· · ·+λn = 1, the corresponding weighted arithmetic and geometric means are λ1x1 + · · · + λnxn and xλ1 1 · · · xλn n .
These reduce to the unweighted means in the case where λ1 = · · · = λn = 1/n.
Arithmetic and geometric means satisfy a famous inequality, namely that the geometric mean is always less than or equal to the arithmetic mean. This turns out to be a simple application of Jensen’s inequality: Theorem 5 AM–GM Inequality Let x1, . . . , xn > 0, and let λ1, . . . , λn ∈[0, 1] so that λ1 + · · · + λn = 1. Then xλ1 1 · · · xλn n ≤λ1x1 + · · · + λnxn.
Convexity, Inequalities, and Norms 6 y ex log a 1 2 Hlog a + log bL log b a b Ha + bL2 ab Figure 4: A visual proof that √ ab < (a + b)/2.
PROOF This theorem is equivalent to the convexity of the exponential function (see figure 4). Specifically, we know that eλ1t1+···λntn ≤λ1et1 + · · · λnetn for all t1, . . . , tn ∈R. Substituting xi = eti gives the desired result.
■ The following theorem generalizes this inequality to arbitrary measure spaces.
The proof is essentially the same as the proof of the previous theorem.
Theorem 6 Integral AM–GM Inequality Let (X, µ) be a measure space with µ(X) = 1, and let f : X →(0, ∞) be a measurable function. Then exp Z X log f dµ ≤ Z X f dµ PROOF Since the exponential function is convex, Jensen’s inequality gives exp Z X log f dµ ≤ Z X exp(log f) dµ = Z X f dµ.
■ By the way, we can rescale to get a version of this inequality that applies whenever Convexity, Inequalities, and Norms 7 µ(X) is finite and nonzero: exp 1 µ(X) Z X log f dµ ≤ 1 µ(X) Z X f dµ.
Note that the quantity on the right is simply the average value of f on X. The quantity on the left can be thought of as the (continuous) geometric mean of f.
p-Means There are many important means in mathematics and science, beyond just the arith-metic and geometric means. For example, the harmonic mean of positive numbers x1, . . . , xn is n 1/x1 + · · · + 1/xn .
This mean is used, for example, in calculating the average resistance of resistors in parallel. For another example, the Euclidean mean of x1, . . . , xn is r x2 1 + · · · + x2 n n .
This mean is used to average measurements taken for the standard deviation of a random variable.
The AM–GM inequality can be extended to cover both of these means. In partic-ular, the inequality n 1/x1 + · · · + 1/xn ≤ n √x1 · · · xn ≤x1 + · · · + xn n ≤ r x2 1 + · · · + x2 n n .
holds for all x1, . . . , xn ∈(0, ∞).
Both of these means are examples of p-means: Definition: p-Means Let x1, . . . , xn > 0. If p ∈R −{0}, the p-mean of x1, . . . , xn is xp 1 + · · · + xp n n 1/p .
For example: • The 2-mean is the same as the Euclidean mean.
• The 1-mean is the same as the arithmetic mean.
Convexity, Inequalities, and Norms 8 • The (−1)-mean is the same as the harmonic mean.
Though it may not be obvious, the geometric mean also fits into the family of p-means.
In particular, it is possible to show that lim p→0 xp 1 + · · · + xp n n 1/p = n √x1 · · · xn for any x1, . . . , xn ∈(0, ∞). Thus, we may think of the geometric mean as the 0-mean.
It is also possible to use limits to define means for ∞and −∞. It turns out the the ∞-mean of x1, . . . , xn is simply max(x1, . . . , xn), while the (−∞)-mean is min(x1, . . . , xn).
As with the arithmetic and geometric means, we can also define weighted versions of p-means.
Given positive weights λ1, . . . , λn satisfying λ1 + · · · + λn = 1, the corresponding weighted p-mean is λ1xp 1 + · · · + λnxp n 1/p.
As you may have guessed, the p-means satisfy a generalization of the AM-GM in-equality: Theorem 7 Generalized Mean Inequality Let x1, . . . , xn > 0, and let λ1, . . . , λn ∈[0, 1] so that λ1 + · · · + λn = 1. Then p ≤q ⇒ λ1xp 1 + · · · + λnxp n 1/p ≤ λ1xq 1 + · · · + λnxq n 1/q for all p, q ∈R −{0}.
PROOF If p = 1 and q > 1, this inequality takes the form λ1x1 + · · · + λnxn q ≤λ1xq 1 + · · · + λnxq n which follows immediately from the convexity of the function ϕ(x) = xq.
The case where 0 < p < q follows from this. Specifically, since q/p > 1, we have λ1(xp 1) + · · · + λn(xp n) q/p ≤λ1(xp 1)q/p + · · · + λn(xp n)q/p and the desired inequality follows. Cases involving negative values of p or q are left as an exercise to the reader.
■ Convexity, Inequalities, and Norms 9 Applying the same reasoning using the integral version of Jensen’s inequality gives p ≤q ⇒ Z X f p dµ 1/p ≤ Z X f q dµ 1/q for any L1 function f : X →(0, ∞), where (X, µ) is a measure space with a total measure of one.
Norms A norm is a function that measures the lengths of vectors in a vector space. The most familiar norm is the Euclidean norm on Rn, which is defined by the formula ∥(x1, . . . , xn)∥= q x2 1 + · · · + x2 n.
Definition: Norm on a Vector Space Let V be a vector space over R. A norm on V is a function ∥−∥: V →R, denoted v 7→∥v∥, with the following properties: 1. ∥v∥≥0 for all v ∈V , and ∥v∥= 0 if and only if v = 0.
2. ∥λv∥= |λ| ∥v∥for all λ ∈R and v ∈V .
3. ∥v + w∥≤∥v∥+ ∥w∥for all v, w ∈V .
For those familiar with topology, any norm ∥−∥on a vector space gives a metric d on the vector space defined by the formula d(v, w) = ∥v −w∥.
Thus any vector space with a norm can be thought of as a topological space.
The Euclidean norm on Rn can be generalized to the family of p-norms.
For any p ≥1, the p-norm on Rn is defined by the formula ∥(x1, . . . , xn)∥p = (|x1|p + · · · + |xn|p)1/p The usual Euclidean norm corresponds to the case where p = 2.
It is easy to see that the definition of the p-norm satisfies axioms (1) and (2) for a norm, but third axiom (which is known as the triangle inequality) is far from clear. The following theorem establishes the p-norm is in fact a norm.
Convexity, Inequalities, and Norms 10 Theorem 8 Minkowski’s Inequality If u, v ∈Rn and p ∈[1, ∞), then ∥u + v∥p ≤∥u∥p + ∥v∥p PROOF Since p ≥1, the function x 7→|x|p is convex. It follows that ∥(1 −λ)u + λv∥p p = n X i=1 |(1 −λ)ui + λvi|p ≤ n X i=1 (1 −λ)|ui|p + λ|vi|p = (1 −λ) ∥u∥p p + λ ∥v∥p p for all u and v and λ ∈[0, 1]. In particular, this proves that ∥(1 −λ)u + λv∥p ≤1 whenever ∥u∥p = ∥v∥p = 1.
From this Minkowski’s inequality follows. In particular, we may assume that u and v are nonzero. Then u/∥u∥p and v/∥v∥p are unit vectors, so ∥u + v∥p ∥u∥p + ∥v∥p = ∥u∥p ∥u∥p + ∥v∥p u ∥u∥p + ∥v∥p ∥u∥p + ∥v∥p v ∥v∥p p ≤1.
■ If ∥−∥is a norm on a vector space V , the unit ball in V is the set BV (1) = v ∈V ∥v∥≤1 .
For example, the unit ball in R2 with respect to the Euclidean norm is a round disk of radius 2 centered at the origin.
Figure 5 shows the unit ball in R2 with respect to various p-norms. Their shapes are all fairly similar, with the the unit ball being a diagonal square when p = 1, and the unit ball approaching a horizontal square as p →∞. All of these unit balls have a certain important geometric property.
Definition: Convex Set Let V is a vector space over R. If v and w are points in V , the line segment from v to w is the set L(v, w) = {λv + (1 −λ)w | 0 ≤λ ≤1}.
A subset S ⊆V is convex if L(v, w) ⊆S for all v, w ∈S.
The following theorem gives a geometric interpretation of Minkowski’s inequality.
Convexity, Inequalities, and Norms 11 p = 1 p = 1.5 p = 2 p = 3 p = 10 Figure 5: The shape of the unit ball in R2 for various p-norms.
Theorem 9 Shapes of Unit Balls Let V be a vector space over R, and let ∥−∥: V →R be a function satisfying conditions (1) and (2) for a norm. Then ∥−∥satisfies the triangle inequality if and only if the unit ball v ∈V ∥v∥≤1 is convex.
PROOF Suppose first that ∥−∥satisfies the triangle inequality. Let v and w be points in the unit ball, and let p = λv + (1 −λ)w be any point on the line segment from v to w. Then ∥p∥= ∥λv + (1 −λ)w∥≤∥λv∥+ ∥(1 −λ)w∥ = λ∥v∥+ (1 −λ)∥w∥≤λ(1) + (1 −λ)(1) = 1 so p lies in the unit ball as well.
For the converse, suppose that the unit ball is convex, and let v, w ∈V . If v or w is 0, then clearly ∥v + w∥≤∥v∥+ ∥w∥, so suppose they are both nonzero. Let ˆ v = v/∥v∥and ˆ w = w/∥w∥. Then ∥ˆ v∥= 1 ∥v∥v = 1 ∥v∥∥v∥= 1 and similarly ∥ˆ w∥= 1. Thus ˆ v and ˆ w both lie in the unit ball. Since ∥v∥ ∥v∥+ ∥w∥+ ∥w∥ ∥v∥+ ∥w∥= 1, the point v + w ∥v∥+ ∥w∥= ∥v∥ ∥v∥+ ∥w∥ˆ v + ∥w∥ ∥v∥+ ∥w∥ˆ w must lie in the unit ball as well, and it follows that ∥v + w∥≤∥v∥+ ∥w∥.
■ Convexity, Inequalities, and Norms 12 Finally, we should mention that there is an integral version of Minkowski’s in-equality. This involves the notion of a p-norm on a measure space.
Definition: p-Norm Let (X, µ) be a measure space, let f be a measurable function on X, and let p ∈[1, ∞). The p-norm of f is the quantity ∥f∥p = Z X |f|p dµ 1/p Note that ∥f∥p is always defined, since it involves the integral of a non-negative function, but it may be infinite. We can now state Minkowski’s inequality for arbitrary measure spaces.
Theorem 10 Minkowski’s Inequality (Integral Version) Let (X, µ) be a measure space, let f, g: X →R be measurable functions, and let p ∈[1, ∞). Then ∥f + g∥p ≤∥f∥p + ∥g∥p.
PROOF Since p ≥1, the function x 7→|x|p is convex. It follows that ∥(1 −λ)f + λg∥p p = Z X |(1 −λ)f + λg|p dµ ≤ Z X (1 −λ)|f|p + λ|g|p dµ = (1 −λ) ∥f∥p p + λ ∥g∥p p for any measurable functions f and g and any λ ∈[0, 1]. In particular, this proves that ∥(1 −λ)f + λg∥p ≤1 whenever ∥f∥p = ∥g∥p = 1.
From this Minkowski’s inequality follows. First, observe that if ∥f∥p = 0, then f = 0 almost everywhere, so Minkowski’s inequality follows in this case. A similar argument holds if ∥g∥p = 0, so suppose that ∥f∥p > 0 and ∥g∥p > 0. Let ˆ f = f/∥f∥p and ˆ g = g/∥g∥p, and note that ∥ˆ f∥p = ∥ˆ g∥p = 1. Then ∥f + g∥p ∥f∥p + ∥g∥p = ∥f∥p ∥f∥p + ∥g∥p ˆ f + ∥g∥p ∥f∥p + ∥g∥p ˆ g p ≤1.
■ Convexity, Inequalities, and Norms 13 H¨ older’s Inequality Recall that the Euclidean norm on Rn satisfies the Cauchy-Schwarz Inequality |u · v| ≤∥u∥2 ∥v∥2 for all u, v ∈Rn. Our next task is to prove a generalization of this known as H¨ older’s inequality.
Lemma 11 Young’s Inequality If x, y ∈[0, ∞) and p, q ∈(1, ∞) so that 1/p + 1/q = 1, then xy ≤xp p + yq q PROOF This can be written (xp)1/p(yq)1/q ≤1 p xp + 1 q yq which is an instance of the weighted AM–GM inequality.
■ Theorem 12 H¨ older’s Inequality Let u, v ∈Rn, and let p, q ∈(1, ∞) so that 1/p + 1/q = 1. Then |u · v| ≤∥u∥p ∥v∥q.
PROOF By Young’s inequality, |u · v| ≤|u1v1| + · · · + |unvn| ≤|u1|p + · · · + |un|p p + |v1|q + · · · + |vn|q q = ∥u∥p p p + ∥v∥q q q In particular, if ∥u∥p = ∥v∥q = 1, then |u · v| ≤1/p + 1/q = 1, which proves H¨ older’s inequality in this case.
For the general case, we may assume that u and v are nonzero. Then u/∥u∥p and v/∥v∥q are unit vectors for their respective norms, and therefore u ∥u∥p · v ∥v∥q ≤1.
Convexity, Inequalities, and Norms 14 Multiplying through by ∥u∥p ∥v∥q gives the desired result.
■ This inequality is not hard to generalize to integrals. We begin with the following definition.
Definition: Inner Product of Functions Let (X, µ) be a measure space, and let f and g be measurable functions on X. The L2 inner product of f and g is defined as follows: ⟨f, g⟩= Z X fg dµ.
Note that ⟨f, g⟩may be undefined if fg is not Lebesgue integrable on X. Note also that ⟨f, g⟩= ∥fg∥1 if f and g are nonnegative, but that in general ⟨f, g⟩≤∥fg∥1.
Theorem 13 H¨ older’s Inequality (Integral Version) Let (X, µ) be a measure space, let f and g be measurable functions on X, and let p, q ∈(1, ∞) so that 1/p + 1/q = 1. If ∥f∥p < ∞and ∥g∥p < ∞, then ⟨f, g⟩ is defined, and |⟨f, g⟩| ≤∥f∥p ∥g∥q PROOF Suppose first that ∥f∥p = ∥g∥q = 1. By Young’s inequality, ∥fg∥1 = Z X |fg| dµ ≤ Z X |f|p p + |g|q q dµ ≤∥f∥p p p + ∥g∥q q q = 1.
Since fg is L1, it follows that ⟨f, g⟩is defined. Then ⟨f, g⟩≤∥fg∥1 ≤1, which proves H¨ older’s inequality in this case.
For the general case, note first that if either ∥f∥p = 0 or ∥g∥q = 0, then either f = 0 almost everywhere or g = 0 almost everywhere, so H¨ older’s inequality holds in that case. Otherwise, let ˆ f = f/∥f∥p and ˆ g = g/∥g∥q. Then ∥ˆ f∥p = ∥ˆ g∥q = 1, so |⟨f, g⟩| = ∥f∥p ∥g∥p |⟨ˆ f, ˆ g⟩| ≤∥f∥p ∥g∥q ■ Convexity, Inequalities, and Norms 15 In the case where p = q = 2, this gives the following.
Corollary 14 Cauchy-Schwarz Inequality (Integral Version) Let (X, µ) be a measure space, and let f and g be measurable functions on X.
If ∥f∥2 < ∞and ∥g∥2 < ∞, then ⟨f, g⟩is defined, and |⟨f, g⟩| ≤∥f∥2 ∥g∥2.
Exercises 1.
a) Prove that a function ϕ: R →R is convex if and only if ϕ(y) −ϕ(x) y −x ≤ϕ(z) −ϕ(x) z −x ≤ϕ(z) −ϕ(y) z −y for all x, y, z ∈R with x < y < z.
b) Use this characterization of convex functions to prove Theorem 1 on the existence of tangent lines.
2. Let ϕ: (a, b) →R be a differentiable function. Prove that ϕ is convex if and only if ϕ′ is non-decreasing.
3. Let ϕ: R →R be a convex function. Prove that ϕ (1 −λ)x + λy ≥(1 −λ)ϕ(x) + λϕ(y) for λ ∈R −[0, 1]. Use this to provide an alternative proof that ϕ is continuous.
4. If x, y ≥0, prove that lim p→0 xp + yp 2 1/p = √xy and lim p→∞ xp + yp 2 1/p = max(x, y).
What is lim p→−∞ xp + yp 2 1/p ?
5.
a) If x1, . . . , xn, y1, . . . , yn ∈(0, ∞), prove that √x1y1 + · · · √xnyn n ≤ r x1 + · · · + xn n r y1 + · · · + yn n .
That is, the arithmetic mean of geometric means is less than or equal to the corresponding geometric mean of arithmetic means.
Convexity, Inequalities, and Norms 16 b) If λ, µ ∈[0, 1] and λ + µ = 1, prove that xλ 1yµ 1 + · · · xλ nyµ n n ≤ x1 + · · · + xn n λ y1 + · · · + yn n µ .
6. Prove the Generalized Mean Inequality (Theorem 6) in the case where p or q is negative.
7. Let f : [0, 1] →R be a bounded measurable function, and define ϕ: [1, ∞) →R by ϕ(p) = Z [0,1] f p dµ.
Prove that log ϕ is convex on [1, ∞).
8. Prove that 1 + x2y + x4y23 ≤ 1 + x3 + x62 1 + y3 + y6 for all x, y ∈(0, ∞).
9. Let (X, µ) be a measure space, let f, g: X →[0, ∞) be measurable functions, and let p, q, r ∈(1, ∞) so that 1/p + 1/q = 1/r. Prove that ∥fg∥r ≤∥f∥p ∥g∥q. |
16078 | https://www.quora.com/What-is-a-perfect-cube-What-are-some-examples-of-perfect-cubes-How-do-you-determine-if-something-is-a-perfect-cube-without-calculating-its-square-roots-first | What is a perfect cube? What are some examples of perfect cubes? How do you determine if something is a perfect cube without calculating its square roots first? - Quora
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What is a perfect cube? What are some examples of perfect cubes? How do you determine if something is a perfect cube without calculating its square roots first?
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Alexander Mathey
Former Chemical Engineer, retired, lives in Athens, GR · Author has 5.6K answers and 10.7M answer views
·1y
When speaking about numbers, a perfet cube is a natural number whose cubic root is also a natural number.
Examples:
2^3 = 8
3^3 = 27
4^3 = 64
5^3 = 125 and so on.
To determine if a number is a perfect cube:
Calculate, by some method, its cubic root: If it comes out an integer, then the number was a perfect cube.
An easy way to calculate the cubic root is by the Newton Raphson approximation:
1) Choose, by guessing, a first approximation, x for the cubic root of N.
2) Calculate the quantity (2x + N/x^2)/3
3) Make this the new x.
4) Go to (2).
Example 1: Find the cubic root of 12763.
Choose x(0) = 25 a
Continue Reading
When speaking about numbers, a perfet cube is a natural number whose cubic root is also a natural number.
Examples:
2^3 = 8
3^3 = 27
4^3 = 64
5^3 = 125 and so on.
To determine if a number is a perfect cube:
Calculate, by some method, its cubic root: If it comes out an integer, then the number was a perfect cube.
An easy way to calculate the cubic root is by the Newton Raphson approximation:
1) Choose, by guessing, a first approximation, x for the cubic root of N.
2) Calculate the quantity (2x + N/x^2)/3
3) Make this the new x.
4) Go to (2).
Example 1: Find the cubic root of 12763.
Choose x(0) = 25 as the first approximation.
x(1) = (225 + 12763/25^2)/3 = 23.4736
x(2) = (223.4736 + 12763/23.4736^2)/3 = 23.37004
x(3) = (223.37004 + 12763/23.37004^2)/3 = 23.36958 . . .
“12763” is not a perfect cube.
Example 2: Find the cubic root of 185193.
If it is a perfect cube, the last digit of its cubic root has to be 7. (7^3 = 343)
50^3 = 125000 < 185193
60^3 = 216000 > 185193
If the cubic root is an integer, it has to be 57.
Check: 57^3 = 185193
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How can I find whether a number is a perfect cube or not?
Who can find out if a number is a perfect cube or not?
Is 3 a perfect cube?
How do you find the square root of any perfect cube?
How do you find the square roots of cubes and the cubes of square roots?
Paul Snyderwine
Have learned to see unintended meanings in a question. · Author has 100 answers and 25.8K answer views
·1y
An object is either a cube, or not. If there is the slightest difference between one edge and any others, the object is not a cube. Your second question sounds like nonsense.
Or, maybe there is a textbook discussion of “perfect cube” and how that is different from an ordinary cube, and what squares have to do with it. The answers my friend would be blowing through the leaves of that book.
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Alon Amit
PhD in Mathematics; Mathcircler. · Upvoted by
Erik Bergland
, PhD Mathematics, Brown University (2024) and
Eknath Belbase
, PhD Mathematics & Statistics, Cornell University (1998) · Author has 8.7K answers and 171.7M answer views
·Updated 5y
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Is 64 the first perfect square and a perfect cube? Is it the only one?
Come, I want to show you something.
If x x is any integer, then x 3 x 3 is what you’d call a perfect cube. And if y y is any integer, then y 2 y 2 is what you’d call a perfect square. If a number is both a perfect square and a perfect cube, then this number is both y 2 y 2 and x 3 x 3, and since it’s the same number, we can write
y 2=x 3 y 2=x 3.
What’s the first solution? We can restrict our attention to nonnegative values of x x, since negative ones are clearly out of the question. The first possible candidate is x=0 x=0, and it kinda works: 0 2=0 3 0 2=0 3, and indeed zero is a perfect square and a perfect cube.
But tha
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Come, I want to show you something.
If x x is any integer, then x 3 x 3 is what you’d call a perfect cube. And if y y is any integer, then y 2 y 2 is what you’d call a perfect square. If a number is both a perfect square and a perfect cube, then this number is both y 2 y 2 and x 3 x 3, and since it’s the same number, we can write
y 2=x 3 y 2=x 3.
What’s the first solution? We can restrict our attention to nonnegative values of x x, since negative ones are clearly out of the question. The first possible candidate is x=0 x=0, and it kinda works: 0 2=0 3 0 2=0 3, and indeed zero is a perfect square and a perfect cube.
But that’s a little boring. We move on: how about x=1 x=1? Of course, 1 2=1 3 1 2=1 3 is also a perfect square which is a perfect cube. So 64 64 isn’t quite the first one, nor the only one, but perhaps it’s the first interesting one. It’s what we get when we pick x=4 x=4. What happened to x=2 x=2 and x=3 x=3? Of course, they don’t work: 8 8 and 27 27 aren’t squares. Why did 0,1 0,1 and 4 4 work for x x? Ah, of course: they worked because they are themselves squares. If we take x=9 x=9, for instance, then x=3 2 x=3 2, so x 3=3 6 x 3=3 6, and that’s a perfect square! It’s the square of y=3 3=27 y=3 3=27, and the number which is both a perfect square and a perfect cube is 3 6=729 3 6=729.
We can keep going, of course: the square-cube numbers go 0,1,64,729,4096,15625 0,1,64,729,4096,15625 and so on. More simply, they are 0 6,1 6,2 6,3 6,4 6,5 6 0 6,1 6,2 6,3 6,4 6,5 6 and so on. All the sixth powers.
Infinitely many.
So no, 64 64 isn’t the only one and isn’t one of a few. It’s one of infinitely many sixth powers, which are numbers that are simultaneously perfect squares and perfect cubes.
But this isn’t what I wanted to show you. What I wanted to show you is what happens when we go back to y 2=x 3 y 2=x 3 and forget that we are talking about integers. It’s a simple relation, and we can easily plot it: we let x,y x,y be real numbers and we plot the points (x,y)(x,y) with y 2=x 3 y 2=x 3.
What we get is mostly a nice, smooth curve, but there’s a particular point where it’s not so nice and smooth.
The point (0,0)(0,0) is called a cusp. It’s very obviously different from all other points, where the curve is as smooth as a baby’s cheek. It’s just this one, sharp, pointy cusp at (0,0)(0,0). And believe it or not, this cusp is the reason we have infinitely many perfect squares that are perfect cubes.
What do I mean by that?
Suppose we change the equation just slightly. Instead of y 2=x 3 y 2=x 3 we look at y 2=x 3+1 y 2=x 3+1, or y 2=x 3−1 y 2=x 3−1, or y 2=x 3+17 y 2=x 3+17, whatever. Here are a few of them:
In blue we have the original, spikey y 2=x 3 y 2=x 3. The red, green and orange curves are everywhere smooth as silk. And you know what else? The equations y 2=x 3−1 y 2=x 3−1, y 2=x 3+2 y 2=x 3+2 and y 2=x 3+17 y 2=x 3+17 don’t have infinitely many integer solutions. They only have finitely many (sometimes none at all).
And guess what else?
Proving this is hard. Very hard. Finding the solutions for each of those is hard. Sometimes just tricky, sometimes really hard, and sometimes exceedingly, insanely hard.
The equation y 2=x 3+k y 2=x 3+k is called Mordell’s equation, and it’s like a different episode for every k k. The only k k for which there’s a cusp is also the only k k for which there are infinitely many solutions, and also the only k k which is easy to handle. For every other k k, it’s hard. We’ve actually seen a few Mordells here on Quora, like this one.
To the untrained eye, the equations y 2=x 3 y 2=x 3 and y 2=x 3+1 y 2=x 3+1 seem almost identical. What’s the difference? We just added a silly +1+1. The real trouble here are those variables, and squares, and cubes, who cares about a measly +1+1? But imagine that: if we seek integer solutions, solving y 2=x 3 y 2=x 3 is an elementary exercise, and solving y 2=x 3+1 y 2=x 3+1 is, I would guess, beyond the abilities of most professional mathematicians, unless they bring in some cavalry for help – a number theorist colleague, or a few advanced papers and books.
Let me drive this home. Consider
y 2=x 3−3 x 2+3 y 2=x 3−3 x 2+4 y 2=x 3−3 x 2+3 y 2=x 3−3 x 2+4
Just by looking at the equations, can you guess which one is trivial and which one is very difficult to solve in integers? No? I wouldn’t either. But let’s look at their graphs:
See the difference? The blue one is smooth everywhere, and very difficult to solve in integers. The red one has a singularity: that intersection at (2,0)(2,0). And you can take it as a nice, fun exercise to show that it has infinitely many integer solutions, and they are very easy to describe. Find them!
Much more generally, cubic equations in two variables with integer coefficients are either smooth or they have singularities. When they are smooth, they are called elliptic curves, and then they can only have finitely many integer solutions, and proving this is very far from easy. On the other hand, when they are singular, they are much easier to handle. They are called “rational” in this case, for ancient reasons, and they can be parametrized very simply, like the parametrization (y=t 3,x=t 2)(y=t 3,x=t 2) we’ve found for the original question. And those parametrizations allow for infinitely many integer solutions.
This is one chapter of an epic tale, as deep as an ocean: the geometry of algebraic equations is connected to their arithmetic in wildly unexpected ways. The Weil conjectures, which shaped a huge amount of math in the 20th century, are another tome in this saga.
Where were we? Ah, yes: Is 64 the first perfect square and perfect cube? No. Is it the only one? No. Are there finitely many such numbers? No. And it’s all because of that tiny little pointy, spikey cusp on the graph of y 2=x 3 y 2=x 3.
The queen of mathematics is just full of surprises.
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Ellis Cave
40+ years as an Electrical Engineer · Author has 7.7K answers and 4.2M answer views
·May 27
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Why do the last digits of perfect cubes not repeat, and how does this help in finding cube roots quickly?
Generate the first 100 cubes, store them in cu, & list them::
]cu=.3^~>:i.100
1 8 27 64 125 216 343 512 729 1000 1331 1728 2197 2744 3375 4096 4913 5832 6859 8000 9261 10648 12167 13824 15625 17576 19683 21952 24389 27000 29791 32768 35937 39304 42875 46656 50653 54872 59319 64000 68921 74088 79507 85184 91125 97336 103823 110592 117649 125000 132651 140608 148877 157464 166375 175616 185193 195112 205379 216000 226981 238328 250047 262144 274625 287496 300763 314432 328509 343000 357911 373248 389017 405224 421875 438976 456533 474552 493039 512000 531441 551368 571787 592704 614125 636056 6585
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Generate the first 100 cubes, store them in cu, & list them::
]cu=.3^~>:i.100
1 8 27 64 125 216 343 512 729 1000 1331 1728 2197 2744 3375 4096 4913 5832 6859 8000 9261 10648 12167 13824 15625 17576 19683 21952 24389 27000 29791 32768 35937 39304 42875 46656 50653 54872 59319 64000 68921 74088 79507 85184 91125 97336 103823 110592 117649 125000 132651 140608 148877 157464 166375 175616 185193 195112 205379 216000 226981 238328 250047 262144 274625 287496 300763 314432 328509 343000 357911 373248 389017 405224 421875 438976 456533 474552 493039 512000 531441 551368 571787 592704 614125 636056 658503 681472 704969 729000 753571 778688 804357 830584 857375 884736 912673 941192 970299 1000000
Extract the last digit of each cube and list them:
{:"1 sep cu
1 8 7 4 5 6 3 2 9 0 1 8 7 4 5 6 3 2 9 0 1 8 7 4 5 6 3 2 9 0 1 8 7 4 5 6 3 2 9 0 1 8 7 4 5 6 3 2 9 0 1 8 7 4 5 6 3 2 9 0 1 8 7 4 5 6 3 2 9 0 1 8 7 4 5 6 3 2 9 0 1 8 7 4 5 6 3 2 9 0 1 8 7 4 5 6 3 2 9 0
So it looks like there is a 10-digit long repeating pattern: 1 8 7 4 5 6 3 2 9 0
10 10${:"1 sep cu
1 8 7 4 5 6 3 2 9 0
1 8 7 4 5 6 3 2 9 0
1 8 7 4 5 6 3 2 9 0
1 8 7 4 5 6 3 2 9 0
1 8 7 4 5 6 3 2 9 0
1 8 7 4 5 6 3 2 9 0
1 8 7 4 5 6 3 2 9 0
1 8 7 4 5 6 3 2 9 0
1 8 7 4 5 6 3 2 9 0
1 8 7 4 5 6 3 2 9 0
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Marc Hammes
Personal Finance Journalist at Almedia (2024–present)
·Fri
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After all, I wasn't losing anything by trying. So, I did…
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I used to scroll Reddit looking for ways to make extra cash – surveys, cashback apps, weird side hustles.
90% of it? Trash.
But then I found something that actually made me money while playing games.
Sounds fake?
That’s what I thought, too… until I cashed out $538 in my first two weeks with Freecash. Isn’t it mindblowing??
Frankly, what caught my attention was the $5 welcome bonus. No deposit, no catch, just for signing up.
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Related questions
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What is the first perfect square that is also a perfect cube?
Adding 1000 is a perfect cube. If true, then what is the perfect cube of 1000?
How can we find out the cube root of non perfect cube?
Which is the first even perfect cube natural number?
What are the first five perfect cubes?
Wayne VanWeerthuizen
Studied at Washington State University · Author has 123 answers and 107K answer views
·Updated 2y
Related
What is the definition of a perfect square? Is there any way to tell if an integer is a perfect square without finding its square root?
What is the definition of a perfect square?
Perfect squares are squares of natural numbers, that is, a natural number times itself.
The first 25 perfect squares are:
0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625.
The next 25 perfect squares are:
676, 729, 784, 841, 900, 961, 1024, 1089, 1156, 1225, 1296, 1369, 1444, 1521, 1600, 1681, 1764, 1849, 1936, 2025, 2116, 2209, 2304, 2401, 2500.
(I split it into two lists for this reason: compare the final two digits of the numbers in the second list with those in the first list. What do
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What is the definition of a perfect square?
Perfect squares are squares of natural numbers, that is, a natural number times itself.
The first 25 perfect squares are:
0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625.
The next 25 perfect squares are:
676, 729, 784, 841, 900, 961, 1024, 1089, 1156, 1225, 1296, 1369, 1444, 1521, 1600, 1681, 1764, 1849, 1936, 2025, 2116, 2209, 2304, 2401, 2500.
(I split it into two lists for this reason: compare the final two digits of the numbers in the second list with those in the first list. What do you notice?)
Is there any way to tell if an integer is a perfect square without finding its square root?
We can sometimes recognize at a glance that certain numbers (written in base ten) can’t be perfect squares.
No perfect square ends (rightmost digit) with the digits 3 or 7.
For example, 1234567 can’t be a perfect square because it ends with 7.
One or the other of the last two digits must be even.
Thus, 839 can’t be a perfect square, since both of it’s last two digits are odd.
A perfect square can’t be three more than a multiple of four.
For example, 843 can’t be a perfect square, since 840 is a multiple of four.
For a perfect square, the digit sum after casting out nines can only be 0, 1, 4, or 7.
For example, the digit sum of 208241 is 2+0+8+2+4+1 = 17, and repeating until we get down to a single digit, we have 1+7=8. Since this isn’t 0, 1, 4, or 7, we know 208241 isn’t a perfect square.
If a number is “too close” to another perfect square, it can’t be a perfect square.
For example, 425 can’t be a perfect square because it is too close to 400, which is 20². Starting at 400, the previous perfect square was 400–20–19 = 361, and the next will be 400+20+21 = 441. In general, the next square after n² is n²+n+(n+1).
There are many other rules similar to these which could help anyone doing fast mental arithmetic to quickly eliminate possibilities for perfect squares.
That said, I don’t know of any means to confirm that a number is a perfect square without finding its square root.
What about factoring the number into primes?
Yes, that works, and is often the best approach, at least for small numbers. Other times, though, factoring can be significantly harder than just using the square root algorithm. So it’s hit and miss.
If one has successfully factored the number and discovered that all the primes have even powers, one has only to divide the exponents in half to get a fully legitimate representation of the square root. I feel that is just too close to actually finding the square root to count as satisfying the “without finding its square root” component of the original question. Thus, I personally don’t consider “by using factoring” to qualify as a valid answer to the question.
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Lawrence Stewart
Research Computer Scientist · Author has 10.6K answers and 21.3M answer views
·Mar 4
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Can you provide examples of numbers that are not perfect squares but have perfect cube roots? What is the reason for this?
These are easy to find by considering the prime factors.
A number that has a perfect cube root is called a perfect cube
So we are looking for numbers that are not perfect squares, but are perfect cubes
A number that is a perfect cube must have prime factors with multiplicity 3, so for example 8 is 222 and 8 is a perfect cube.
A number that is a perfect square must have prime factors with multiplicty 2, so for example 4 is 22 and 4 is a perfect cube.
So we are looking for numbers whose prime factors have multiplicity 3 but not 2
8 (as above) is like this, as is 27
You can easily construct a number
Continue Reading
These are easy to find by considering the prime factors.
A number that has a perfect cube root is called a perfect cube
So we are looking for numbers that are not perfect squares, but are perfect cubes
A number that is a perfect cube must have prime factors with multiplicity 3, so for example 8 is 222 and 8 is a perfect cube.
A number that is a perfect square must have prime factors with multiplicty 2, so for example 4 is 22 and 4 is a perfect cube.
So we are looking for numbers whose prime factors have multiplicity 3 but not 2
8 (as above) is like this, as is 27
You can easily construct a number that is both a perfect square and a perfect cube by having prime factors where the multiplicity is both 2 and 3 (the number of copies of each prime factor is divisible by both 2 and 3).
So a constructed example is 2 2 2 2 2 2 = 64, which is both 4^3 and 8^2
You can make these as complicated as you want, provided that there are 6 or 12 or 18 copies of each prime factor.
Similary you can make numbers that are perfect cubes but not perfect squares by haing a prime factor that has multiplicity 3 but not 2, so 2^6 3^3 is one. Even though there are 6 two’s that isn’t good enough because there are only 3 3’s.
Now for extra credit, find me a number that is a perfect 11th power and a perfect cube.
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John K WilliamsSon
Accredited (MS Educ) nerd who loves talking about math · Author has 9K answers and 23M answer views
·1y
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What is the first perfect square that is also a perfect cube?
Let me ask you a few questions:
How do you write perfect squares of the number (n) ?
n 2 n 4 n 6 n 8 n 10 n 12 e t c.n 2 n 4 n 6 n 8 n 10 n 12 e t c.
Conclusion: If the exponent is even, it is a perfect square
How do you write perfect cubes of the number (n) ?
n 3 n 6 n 9 n 12 n 15 n 18 e t c.n 3 n 6 n 9 n 12 n 15 n 18 e t c.
Conclusion: If the exponent is a multiple of three, it is a perfect cube
What numbers are in both lists?
n 6 n 12 n 18 n 24 n 30 n 36 e t c.n 6 n 12 n 18 n 24 n 30 n 36 e t c.
Conclusion: If the exponent is a multiple of six, it is a perfect square and a p
Continue Reading
Let me ask you a few questions:
How do you write perfect squares of the number (n) ?
n 2 n 4 n 6 n 8 n 10 n 12 e t c.n 2 n 4 n 6 n 8 n 10 n 12 e t c.
Conclusion: If the exponent is even, it is a perfect square
How do you write perfect cubes of the number (n) ?
n 3 n 6 n 9 n 12 n 15 n 18 e t c.n 3 n 6 n 9 n 12 n 15 n 18 e t c.
Conclusion: If the exponent is a multiple of three, it is a perfect cube
What numbers are in both lists?
n 6 n 12 n 18 n 24 n 30 n 36 e t c.n 6 n 12 n 18 n 24 n 30 n 36 e t c.
Conclusion: If the exponent is a multiple of six, it is a perfect square and a perfect cube
Yes, this even applies to when the exponents are 0, -6, -12, -18, etc.
Here are some calculations using \left(\frac12\right)^2 to the powers -12, -6, 0, 6 and 12:
As you can see the square roots and the cube roots of the numbers {4096, 64, 1, 1/64 and 1/4096 are all rational numbers, so the original numbers are perfect squares and perfect cubes.
You’ll also note that these same numbers are also all perfect sixth powers.
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Alfred Shin
Former Professor at Humber College (1974–2004) · Author has 115 answers and 15.8K answer views
·1y
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What is the definition of a perfect square? What is the definition of a perfect cube? Are there any numbers that are both a perfect square and a perfect cube?
An integer n is a perfect square if there exists an integer m such that n = m^2.
For example, 4 and 9 are perfect squares because 4 = 2^2 and 9 =3^2
An integer n is a perfect cube if there exists an integer m such that n = m^3.
For example, 8 and 27 are perfect cubes because 8 = 2^3 and 27 =3^3
You will obtain numbers which are both perfect squares and cubes if you square the perfect cubes.
Squaring the perfect cube 8, you get 64. 64 is a perfect square of 8, and a perfect cube of 4.
Squaring the perfect cube 27, you get 729. 729 is a perfect square of 27, and a perfect cube of 9.
By the way, you can
Continue Reading
An integer n is a perfect square if there exists an integer m such that n = m^2.
For example, 4 and 9 are perfect squares because 4 = 2^2 and 9 =3^2
An integer n is a perfect cube if there exists an integer m such that n = m^3.
For example, 8 and 27 are perfect cubes because 8 = 2^3 and 27 =3^3
You will obtain numbers which are both perfect squares and cubes if you square the perfect cubes.
Squaring the perfect cube 8, you get 64. 64 is a perfect square of 8, and a perfect cube of 4.
Squaring the perfect cube 27, you get 729. 729 is a perfect square of 27, and a perfect cube of 9.
By the way, you can also obtain the same results by cubing the perfect squares.
Try cubing 4 and 9 and check for yourself.
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Scott
I write code · Author has 11.7K answers and 11.9M answer views
·2y
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Can 70 70..01 01 be a perfect cube?
No.
If a perfect cube ends in 1, the number that was cubed ends in 1.
Let’s assume we can cube a number and have it end in 1. So that number is of the form 10 k+1 10 k+1. Its cube would be 1000 k 3+300 k 2+30 k+1 1000 k 3+300 k 2+30 k+1. Since our perfect cube ends in 01, that means the 1s digit of k has to be 0, so the 10s digit of the number to be cubed is 0.
Okay, great. So we now know that our number must be of the form 100 k+1 100 k+1. The cube of that is 1000000 k 3+30000 k 2+300 k+1 1000000 k 3+30000 k 2+300 k+1. Since our perfect cube ends in 001, that means the 1s digit of k has to be 0.
Wash, rinse, repeat.
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Dave Reynolds
Former Software Developer · Author has 66 answers and 26.9K answer views
·May 28
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Why do the last digits of perfect cubes not repeat, and how does this help in finding cube roots quickly?
This statement is false.
Any number cubed ending in 0 will end in a 0.
Any number cubed ending in 1 will end in a 1.
Any number cubed ending in 2 will end in a 8.
Any number cubed ending in 3 will end in a 7.
Any number cubed ending in 4 will end in a 4.
Any number cubed ending in 5 will end in a 5.
Any number cubed ending in 6 will end in a 6.
Any number cubed ending in 7 will end in a 3.
Any number cubed ending in 8 will end in a 2.
Any number cubed ending in 9 will end in a 9.
Therefore, the periodicity is 10. The last digit will repeat every 10 numbers.
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Michiel Bogaert
Unable to carry all the maths books he studied · Author has 3.3K answers and 2.8M answer views
·5y
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Is every perfect cube the difference between two perfect squares?
hmmm … is it?
1³ = 1²-0²
2³ = 3²-1²
3³ = 6²-3²
4³ =10²-6²
5³= 15²-10²
… hmm … interesting.
the first number seems to be the sum up till A (for 4, it’s 1+2+3+4; while for 5 it’s 1+2+3+4+5)
1+2+3+…+A = A(A+1)/2
The second number seems to be the same, one lower, so that makes
(A-1)A/2
so … does
A³ = ( A(A+1)/2 ) ² - ( (A-1)A/2 )²
hold up? That’s easy to calculate:
4A³ = A²(A+1)²- (A-1)²A²
4A = (A+1)²- (A-1)²
4A = A² + 2A + 1 - ( A² -2A +1)
4A = 2A + 2A.
Yup !
A 3=(A(A+1)2)2−(A(A−1)2)2 A 3=(A(A+1)2)2−(A(A−1)2)2
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Charles Wolf
The more I learn about Math the more my ignorance is exposed · Author has 181 answers and 114.9K answer views
·1y
Related
What is the first perfect square that is also a perfect cube?
(n^3)^2, for all Natural Numbers, n>0 (Integers only assumed)
… since (n^3)^2 is equivalent to (n^2)^3
Note, (1^3)^2 = (1^2)^3
…so “1 is the first perfect square that is also a perfect cube”
64 is the “the next perfect square that is also a perfect cube”
…since (2^3)^2 = (2^2)^3, both expressions are 64
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Lukas Schmidinger
I have graduate CS and my studies included math courses. · Author has 27.5K answers and 14.7M answer views
·2y
Related
What is the definition of a perfect square? Is there any way to tell if an integer is a perfect square without finding its square root?
What is the definition of a perfect square?
A perfect square is an integer that has an integer as its square root. (You could extend this definition to rational numbers too, there check the numerator and denominator of the most simplefied fraction form, if they are perfect integer squares you have a perfect rational square otherwise not. With real numbers it is kind of a pointless as all positiv real numbers have a positiv real square root).
Is there any way to tell if an integer is a perfect square without finding its square root?
You could find its prime factorisation, if all exponents are an e
Continue Reading
What is the definition of a perfect square?
A perfect square is an integer that has an integer as its square root. (You could extend this definition to rational numbers too, there check the numerator and denominator of the most simplefied fraction form, if they are perfect integer squares you have a perfect rational square otherwise not. With real numbers it is kind of a pointless as all positiv real numbers have a positiv real square root).
Is there any way to tell if an integer is a perfect square without finding its square root?
You could find its prime factorisation, if all exponents are an even number it is square otherwise not. But that is tiresome and would also tell you the square root trivially. But there is a quiet simpler trick you can use: if you can show that n 2<z<(n+1)2 n 2<z<(n+1)2 for any n∈N 0 n∈N 0 than z z is not a perfect square. You can find this with a simple binary search.
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16079 | https://indico.ictp.it/event/a11187/session/11/contribution/7/material/0/0.pdf | P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem CMT I. Applications of the Rado-Hall Theorem Richard A. Brualdi University of Wisconsin-Madison, USA ICTP/IPM Workshop, 3-7 Sept 2012 September 4, 2012 P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem P. Hall’s Theorem A = (A1, A2, . . . , An), a family of n subsets of a finite set X.
A System of Representatives (SR) of A is a family of elements (x1, x2, . . . , xn) of X with x1 ∈A1, x2 ∈A2, . . . , xn ∈An.
If x1, x2, . . . , xn are all different, then (x1, x2, . . . , xn) is a System of Distinct Representatives (SDR) of A.
Example: X = {1, 2, 3, 4, 5, 6, 7} A = ({1, 3, 6, 7}, {2, 3, 5}, {1, 4, 6, 7}, {2, 5, 7}, {1, 3, 7}) Question: When does a family A have an SDR?
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem P. Hall’s Theorem A = (A1, A2, . . . , An), a family of n subsets of a finite set X.
A System of Representatives (SR) of A is a family of elements (x1, x2, . . . , xn) of X with x1 ∈A1, x2 ∈A2, . . . , xn ∈An.
If x1, x2, . . . , xn are all different, then (x1, x2, . . . , xn) is a System of Distinct Representatives (SDR) of A.
Example: X = {1, 2, 3, 4, 5, 6, 7} A = ({1, 3, 6, 7}, {2, 3, 5}, {1, 4, 6, 7}, {2, 5, 7}, {1, 3, 7}) Question: When does a family A have an SDR?
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem P. Hall’s Theorem A = (A1, A2, . . . , An), a family of n subsets of a finite set X.
A System of Representatives (SR) of A is a family of elements (x1, x2, . . . , xn) of X with x1 ∈A1, x2 ∈A2, . . . , xn ∈An.
If x1, x2, . . . , xn are all different, then (x1, x2, . . . , xn) is a System of Distinct Representatives (SDR) of A.
Example: X = {1, 2, 3, 4, 5, 6, 7} A = ({1, 3, 6, 7}, {2, 3, 5}, {1, 4, 6, 7}, {2, 5, 7}, {1, 3, 7}) Question: When does a family A have an SDR?
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem P. Hall’s Theorem continued Obvious necessary condition for A = (A1, A2, . . . , An) to have an SDR: (∗) | ∪i∈K Ai| ≥|K| K ⊆{1, 2, . . . , n}.
For instance, if 5 sets collectively contained only 4 elements then surely there cannot be an SDR.
Hall’s Theorem: Condition (∗) is both necessary and sufficient.
Induction can be used to prove Hall’s Theorem. My favorite proof is due to Richard Rado. It is based on the observation that if each of the sets A1, A2, . . . , An contain exactly one element: A = {x1}, A2 = {x2}, . . . , An = {xn}, then () implies that x1, x2, . . . , xn are all different and so (x1, x2, . . . , xn) is an SDR. So we try to reduce our problem to this situation.
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem P. Hall’s Theorem continued Obvious necessary condition for A = (A1, A2, . . . , An) to have an SDR: (∗) | ∪i∈K Ai| ≥|K| K ⊆{1, 2, . . . , n}.
For instance, if 5 sets collectively contained only 4 elements then surely there cannot be an SDR.
Hall’s Theorem: Condition (∗) is both necessary and sufficient.
Induction can be used to prove Hall’s Theorem. My favorite proof is due to Richard Rado. It is based on the observation that if each of the sets A1, A2, . . . , An contain exactly one element: A = {x1}, A2 = {x2}, . . . , An = {xn}, then () implies that x1, x2, . . . , xn are all different and so (x1, x2, . . . , xn) is an SDR. So we try to reduce our problem to this situation.
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem P. Hall’s Theorem continued Obvious necessary condition for A = (A1, A2, . . . , An) to have an SDR: (∗) | ∪i∈K Ai| ≥|K| K ⊆{1, 2, . . . , n}.
For instance, if 5 sets collectively contained only 4 elements then surely there cannot be an SDR.
Hall’s Theorem: Condition (∗) is both necessary and sufficient.
Induction can be used to prove Hall’s Theorem. My favorite proof is due to Richard Rado. It is based on the observation that if each of the sets A1, A2, . . . , An contain exactly one element: A = {x1}, A2 = {x2}, . . . , An = {xn}, then () implies that x1, x2, . . . , xn are all different and so (x1, x2, . . . , xn) is an SDR. So we try to reduce our problem to this situation.
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Rado’s proof of Hall’s Theorem (Keep in mind that for sets, |X1 ∪X2| + |X1 ∩X2| = |X1| + |X2|.) So suppose that one of the sets, say A1 contains (at least) two elements, a and b where a ̸= b.
Suppose that neither (A1 \ {a}, A2, . . . , An) nor (A1 \ {b}, A2, . . . , An) satisfy the () condition. Then there exists K1, K2 ⊆{2, . . . , n} such that |(A1 \ {a}) ∪∪i∈K1Ai| ≤|K1|, |(A1 \ {b}) ∪∪i∈K2Ai| ≤|K2|.
Then, |K1| + |K2| ≥ |(A1 \ {a}) ∪∪i∈K1Ai| + |(A1 \ {b}) ∪∪i∈K2Ai| ≥ |A1 ∪∪i∈K1∪K2Ai| + | ∪i∈K1∩K2 Ai| ≥ |K1 ∪K2| + 1 + |K1 ∩K2| = |K1| + |K2| + 1, a contradiction.
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Linear Independence of Vectors in a Vector Space Let X be a finite set of vectors in a vector space V , and let I be the collection of subsets of X that are linearly independent.
Then I satisfies the following familiar properties: ▶∅∈I, ▶A ∈I, A′ ⊆A implies A′ ∈I, ▶A1, A2 ∈I, |A1| < |A2| implies ∃x ∈A2 \ A1 such that A1 ∪{x} ∈I.
The rank ρ(Y ) of any subset Y of X is the dimension of the subspace they span: the maximum number of linearly independent vectors in Y . Recall that for subspaces U and W of V : dim(U + W ) + dim(U ∩W ) = dim(U) + dim(W ) gives, for subsets Y , Z of X: rank(Y ∪Z) + rank(Y ∩Z) ≤rank(Y ) + rank(Z).
These properties when viewed axiomatically are one way to define a matroid.
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Linear Independence of Vectors in a Vector Space Let X be a finite set of vectors in a vector space V , and let I be the collection of subsets of X that are linearly independent.
Then I satisfies the following familiar properties: ▶∅∈I, ▶A ∈I, A′ ⊆A implies A′ ∈I, ▶A1, A2 ∈I, |A1| < |A2| implies ∃x ∈A2 \ A1 such that A1 ∪{x} ∈I.
The rank ρ(Y ) of any subset Y of X is the dimension of the subspace they span: the maximum number of linearly independent vectors in Y . Recall that for subspaces U and W of V : dim(U + W ) + dim(U ∩W ) = dim(U) + dim(W ) gives, for subsets Y , Z of X: rank(Y ∪Z) + rank(Y ∩Z) ≤rank(Y ) + rank(Z).
These properties when viewed axiomatically are one way to define a matroid.
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Linear Independence of Vectors in a Vector Space Let X be a finite set of vectors in a vector space V , and let I be the collection of subsets of X that are linearly independent.
Then I satisfies the following familiar properties: ▶∅∈I, ▶A ∈I, A′ ⊆A implies A′ ∈I, ▶A1, A2 ∈I, |A1| < |A2| implies ∃x ∈A2 \ A1 such that A1 ∪{x} ∈I.
The rank ρ(Y ) of any subset Y of X is the dimension of the subspace they span: the maximum number of linearly independent vectors in Y . Recall that for subspaces U and W of V : dim(U + W ) + dim(U ∩W ) = dim(U) + dim(W ) gives, for subsets Y , Z of X: rank(Y ∪Z) + rank(Y ∩Z) ≤rank(Y ) + rank(Z).
These properties when viewed axiomatically are one way to define a matroid.
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Matroids X a finite set, I a collection of subsets of X called independent sets, satisfying: ▶∅∈I, ▶A ∈I, A′ ⊆A implies A′ ∈I, ▶A1, A2 ∈I, |A1| < |A2 implies ∃x ∈A2 \ A1 such that A1 ∪{x} ∈I.
These properties can be used to show the submodular inequality: For subsets Y and Z of X: ρ(Y ∪Z) + ρ(Y ∩Z) ≤ρ(Y ) + ρ(Z).
A matroid can also be defined in terms of circuits (minimal dependent subsets of X), or bases (maximal independent subsets of X), ... .
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Matroids X a finite set, I a collection of subsets of X called independent sets, satisfying: ▶∅∈I, ▶A ∈I, A′ ⊆A implies A′ ∈I, ▶A1, A2 ∈I, |A1| < |A2 implies ∃x ∈A2 \ A1 such that A1 ∪{x} ∈I.
These properties can be used to show the submodular inequality: For subsets Y and Z of X: ρ(Y ∪Z) + ρ(Y ∩Z) ≤ρ(Y ) + ρ(Z).
A matroid can also be defined in terms of circuits (minimal dependent subsets of X), or bases (maximal independent subsets of X), ... .
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Matroids X a finite set, I a collection of subsets of X called independent sets, satisfying: ▶∅∈I, ▶A ∈I, A′ ⊆A implies A′ ∈I, ▶A1, A2 ∈I, |A1| < |A2 implies ∃x ∈A2 \ A1 such that A1 ∪{x} ∈I.
These properties can be used to show the submodular inequality: For subsets Y and Z of X: ρ(Y ∪Z) + ρ(Y ∩Z) ≤ρ(Y ) + ρ(Z).
A matroid can also be defined in terms of circuits (minimal dependent subsets of X), or bases (maximal independent subsets of X), ... .
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Rado’s Theorem - the Hall-Rado Theorem A = (A1, A2, . . . , An): a family of n subsets of a finite set X.
I: the collection of independent sets of a matroid on X.
Then A has an SDR (x1, x2, . . . , xn), with x1 ∈A1, x2 ∈A2, . . . , xn ∈An, such that its set of elements {x1, x2, . . . , xn} is an independent set if and only if (∗) ρ(∪i∈KAi) ≥|K|, K ⊆{1, 2, . . . , n}.
Hall’s theorem is the special case where every subset of X is independent (a trivial matroid) and the rank function is ρ(Y ) = |Y |. The proof of Hall’s theorem can be adapted to give a proof of Rado’s theorem using the submodularity property of the rank function.
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Rado’s Theorem - the Hall-Rado Theorem A = (A1, A2, . . . , An): a family of n subsets of a finite set X.
I: the collection of independent sets of a matroid on X.
Then A has an SDR (x1, x2, . . . , xn), with x1 ∈A1, x2 ∈A2, . . . , xn ∈An, such that its set of elements {x1, x2, . . . , xn} is an independent set if and only if (∗) ρ(∪i∈KAi) ≥|K|, K ⊆{1, 2, . . . , n}.
Hall’s theorem is the special case where every subset of X is independent (a trivial matroid) and the rank function is ρ(Y ) = |Y |. The proof of Hall’s theorem can be adapted to give a proof of Rado’s theorem using the submodularity property of the rank function.
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Rado’s Theorem - the Hall-Rado Theorem A = (A1, A2, . . . , An): a family of n subsets of a finite set X.
I: the collection of independent sets of a matroid on X.
Then A has an SDR (x1, x2, . . . , xn), with x1 ∈A1, x2 ∈A2, . . . , xn ∈An, such that its set of elements {x1, x2, . . . , xn} is an independent set if and only if (∗) ρ(∪i∈KAi) ≥|K|, K ⊆{1, 2, . . . , n}.
Hall’s theorem is the special case where every subset of X is independent (a trivial matroid) and the rank function is ρ(Y ) = |Y |. The proof of Hall’s theorem can be adapted to give a proof of Rado’s theorem using the submodularity property of the rank function.
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem (0, 1)-matrices Let A = [aij] be an m × n (0, 1)-matrix. Let ri = the number of 1s in row i and sj = the number of 1s in column j.
Then R = (r1, r2, . . . , rm) is the row sum vector of A and S = (s1, s2, . . . , sn) is the column sum vector of A.
Example: A = 0 1 1 1 1 1 1 0 1 1 0 0 1 0 1 0 R = (3, 3, 2, 2), S = (3, 3, 3, 1) WLOG (by permuting rows and columns) we may assume that R and S are monotone: r1 ≥r2 ≥· · · ≥rm and s1 ≥s2 ≥· · · ≥sn.
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem (0, 1)-matrices Let A = [aij] be an m × n (0, 1)-matrix. Let ri = the number of 1s in row i and sj = the number of 1s in column j.
Then R = (r1, r2, . . . , rm) is the row sum vector of A and S = (s1, s2, . . . , sn) is the column sum vector of A.
Example: A = 0 1 1 1 1 1 1 0 1 1 0 0 1 0 1 0 R = (3, 3, 2, 2), S = (3, 3, 3, 1) WLOG (by permuting rows and columns) we may assume that R and S are monotone: r1 ≥r2 ≥· · · ≥rm and s1 ≥s2 ≥· · · ≥sn.
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem (0, 1)-matrices Let A = [aij] be an m × n (0, 1)-matrix. Let ri = the number of 1s in row i and sj = the number of 1s in column j.
Then R = (r1, r2, . . . , rm) is the row sum vector of A and S = (s1, s2, . . . , sn) is the column sum vector of A.
Example: A = 0 1 1 1 1 1 1 0 1 1 0 0 1 0 1 0 R = (3, 3, 2, 2), S = (3, 3, 3, 1) WLOG (by permuting rows and columns) we may assume that R and S are monotone: r1 ≥r2 ≥· · · ≥rm and s1 ≥s2 ≥· · · ≥sn.
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Vector Majorization/Dominance If X = (x1, x2, . . . , xn) and Y = (y1, y2, . . . , yn) are two real monotone vectors, then X ⪯Y (X is majorized by Y ) provided k X i=1 xi ≤ k X i=1 yi (1 ≤i ≤k) with equality when k = n. If Y is a nonnegative integral vector, then the conjugate of Y is the vector Y ∗= (y∗ 1 , y∗ 2 , y∗ 3 , . . .) where y∗ k = |{i : yi ≥k}|.
Example: Y = (4, 3, 1, 1), Y ∗= (4, 2, 2, 1): 1 1 1 1 4 1 1 1 3 1 1 1 1 4 2 2 1 9 P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Vector Majorization/Dominance If X = (x1, x2, . . . , xn) and Y = (y1, y2, . . . , yn) are two real monotone vectors, then X ⪯Y (X is majorized by Y ) provided k X i=1 xi ≤ k X i=1 yi (1 ≤i ≤k) with equality when k = n. If Y is a nonnegative integral vector, then the conjugate of Y is the vector Y ∗= (y∗ 1 , y∗ 2 , y∗ 3 , . . .) where y∗ k = |{i : yi ≥k}|.
Example: Y = (4, 3, 1, 1), Y ∗= (4, 2, 2, 1): 1 1 1 1 4 1 1 1 3 1 1 1 1 4 2 2 1 9 P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem The Inverse Problem: Gale-Ryser Theorem Let R = (r1, r2, . . . , rm) and S = (s1, s2, . . . , sn) be monotone nonnegative integral vectors. Then there exists an m × n (0, 1)-matrix A with row sum vector R and column sum vector S if and only if S ⪯R∗, that is, S is majorized by the conjugate of R.
The condition is satisfied if such a matrix A exists: Slide all the 1s of A to the left obtaining a matrix with column sum vector R∗.
E.G.
1 1 0 1 1 0 1 1 1 1 1 0 1 1 0 1 1 1 0 0 1 0 0 0 1 4 3 4 3 3 → 1 1 1 1 0 1 1 1 1 0 1 1 1 0 0 1 1 1 0 0 1 1 0 0 0 5 5 4 2 0 P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem The Inverse Problem: Gale-Ryser Theorem Let R = (r1, r2, . . . , rm) and S = (s1, s2, . . . , sn) be monotone nonnegative integral vectors. Then there exists an m × n (0, 1)-matrix A with row sum vector R and column sum vector S if and only if S ⪯R∗, that is, S is majorized by the conjugate of R.
The condition is satisfied if such a matrix A exists: Slide all the 1s of A to the left obtaining a matrix with column sum vector R∗.
E.G.
1 1 0 1 1 0 1 1 1 1 1 0 1 1 0 1 1 1 0 0 1 0 0 0 1 4 3 4 3 3 → 1 1 1 1 0 1 1 1 1 0 1 1 1 0 0 1 1 1 0 0 1 1 0 0 0 5 5 4 2 0 P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Gale-Ryser Theorem: Outline of Sufficiency Let X = {(i, j) : 1 ≤i ≤m, 1 ≤j ≤n}, the set of positions of an m × n matrix and let (X1, X2, . . . , Xm) where Xk is the set of positions (k, j) of the kth row. Define a matroid on X by: A set P of positions is independent if it has ≤rk positions from row k: |P ∩Xk| ≤rk (1 ≤k ≤m).
(Check this defines a matroid) Its rank function is: ρ(F) = m X k=1 min{|F ∩Xk|, rk} (F ⊆X).
Let A = (A1, A2, . . . , Ap) = (E1, . . . , E1 | {z } s1 , E2, . . . , E2 | {z } s2 , . . . , En, . . . , En | {z } sn ) where p = s1 + s2 + · · · + sn and Ej is the set of positions in column j. An independent SDR of A ‘is’ a (0, 1)-matrix with row sum vector R and column sum vector S.
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Gale-Ryser Theorem: Outline of Sufficiency Let X = {(i, j) : 1 ≤i ≤m, 1 ≤j ≤n}, the set of positions of an m × n matrix and let (X1, X2, . . . , Xm) where Xk is the set of positions (k, j) of the kth row. Define a matroid on X by: A set P of positions is independent if it has ≤rk positions from row k: |P ∩Xk| ≤rk (1 ≤k ≤m).
(Check this defines a matroid) Its rank function is: ρ(F) = m X k=1 min{|F ∩Xk|, rk} (F ⊆X).
Let A = (A1, A2, . . . , Ap) = (E1, . . . , E1 | {z } s1 , E2, . . . , E2 | {z } s2 , . . . , En, . . . , En | {z } sn ) where p = s1 + s2 + · · · + sn and Ej is the set of positions in column j. An independent SDR of A ‘is’ a (0, 1)-matrix with row sum vector R and column sum vector S.
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Gale-Ryser Theorem: Outline of Sufficiency Let X = {(i, j) : 1 ≤i ≤m, 1 ≤j ≤n}, the set of positions of an m × n matrix and let (X1, X2, . . . , Xm) where Xk is the set of positions (k, j) of the kth row. Define a matroid on X by: A set P of positions is independent if it has ≤rk positions from row k: |P ∩Xk| ≤rk (1 ≤k ≤m).
(Check this defines a matroid) Its rank function is: ρ(F) = m X k=1 min{|F ∩Xk|, rk} (F ⊆X).
Let A = (A1, A2, . . . , Ap) = (E1, . . . , E1 | {z } s1 , E2, . . . , E2 | {z } s2 , . . . , En, . . . , En | {z } sn ) where p = s1 + s2 + · · · + sn and Ej is the set of positions in column j. An independent SDR of A ‘is’ a (0, 1)-matrix with row sum vector R and column sum vector S.
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Gale-Ryser Theorem: Outline of Sufficiency cont.
Now we apply Rado’s Theorem. In checking Rado’s condition ρ(∪i∈KAi) ≥|K|, the left side depends only on whether at least one copy of Ej appears while the right side depends on how many copies appear. So it is equivalent to: ρ(∪j∈KEj) ≥P j∈K sj (K ⊆{1, 2, . . . , n}). ∪j∈KEj is the set of positions in the columns indexed by K, and so ρ(∪j∈KEj) = m X i=1 min{|K|, ri}, and so depends only on k = |K| and not K itself. By monotonicity, P j∈K sj is largest at Pk j=1 sj. Thus Rado’s condition reduces to k X i=1 r∗ i = ! m X i=1 min{k, ri} ≥ k X j=1 sj (1 ≤k ≤n).
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Tournaments (Tournament matrices) A tournament Tn of order n is an orientation of the complete graph Kn.
Example: K6 and an orientation of it giving a T6: 2 1 3 4 2 3 4, 3, 3, 2, 2, 1 is the score sequence P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Adjacency Matrix of T6 Often it is easier to visualize a tournament by constructing its adjacency matrix. The adjacency matrix of T6 with a particular ordering of its vertices is A6 = 0 1 1 1 0 1 0 0 0 1 1 1 0 1 0 1 1 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 1 0 1 0 .
All 0s on the main diagonal; of each pair of diagonally opposite entries one is a 1 and the other is a 0: aij + aji = 1 for i ̸= j. The scores of T6 are the row sums of A6: 4, 3, 3, 1, 2, 2.
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Adjacency Matrix of T6 Often it is easier to visualize a tournament by constructing its adjacency matrix. The adjacency matrix of T6 with a particular ordering of its vertices is A6 = 0 1 1 1 0 1 0 0 0 1 1 1 0 1 0 1 1 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 1 0 1 0 .
All 0s on the main diagonal; of each pair of diagonally opposite entries one is a 1 and the other is a 0: aij + aji = 1 for i ̸= j. The scores of T6 are the row sums of A6: 4, 3, 3, 1, 2, 2.
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Adjacency Matrix of T6 Often it is easier to visualize a tournament by constructing its adjacency matrix. The adjacency matrix of T6 with a particular ordering of its vertices is A6 = 0 1 1 1 0 1 0 0 0 1 1 1 0 1 0 1 1 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 1 0 1 0 .
All 0s on the main diagonal; of each pair of diagonally opposite entries one is a 1 and the other is a 0: aij + aji = 1 for i ̸= j. The scores of T6 are the row sums of A6: 4, 3, 3, 1, 2, 2.
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Transitive Tournaments Vertices can be ordered so that all arrows go down: 1 2 ↓ 3 ↓ .
.
.
n Scores are n −1, n −2, . . . , 2, 1, 0. In this case, it is absolutely clear who the best and worst players are!
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Adjacency Matrix of a Transitive Tournament of order 6 0 1 1 1 1 1 0 0 1 1 1 1 0 0 0 1 1 1 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 0 or 0 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 1 1 1 0 0 0 1 1 1 1 0 0 1 1 1 1 1 0 P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Landau’s Theorem If R = (r1, r2, . . . , rn) is a sequence of nonnegative integers, then R is the score sequence of a tournament of order n iff (∗) X i∈I ri ≥ |I| 2 , (I ⊆{1, 2, . . . , n}) with equality when I = {1, 2, . . . , n}.
Necessity of (∗) is clear: Any set I of vertices with |I| = k induces a tournament TI of order k; TI has k 2 edges; all of these edges, and more in general, are accounted for in the sum of their scores P i∈I ri. Thus P i∈I ri ≥ k 2 .
WLOG we assume monotonicity of the scores in the form:r1 ≤r2 ≤· · · ≤rn. Then (∗) is equivalent to k X i=1 ri ≥ k 2 , (k = 1, 2, . . . , n).
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Landau’s Theorem If R = (r1, r2, . . . , rn) is a sequence of nonnegative integers, then R is the score sequence of a tournament of order n iff (∗) X i∈I ri ≥ |I| 2 , (I ⊆{1, 2, . . . , n}) with equality when I = {1, 2, . . . , n}.
Necessity of (∗) is clear: Any set I of vertices with |I| = k induces a tournament TI of order k; TI has k 2 edges; all of these edges, and more in general, are accounted for in the sum of their scores P i∈I ri. Thus P i∈I ri ≥ k 2 .
WLOG we assume monotonicity of the scores in the form:r1 ≤r2 ≤· · · ≤rn. Then (∗) is equivalent to k X i=1 ri ≥ k 2 , (k = 1, 2, . . . , n).
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Landau’s Theorem If R = (r1, r2, . . . , rn) is a sequence of nonnegative integers, then R is the score sequence of a tournament of order n iff (∗) X i∈I ri ≥ |I| 2 , (I ⊆{1, 2, . . . , n}) with equality when I = {1, 2, . . . , n}.
Necessity of (∗) is clear: Any set I of vertices with |I| = k induces a tournament TI of order k; TI has k 2 edges; all of these edges, and more in general, are accounted for in the sum of their scores P i∈I ri. Thus P i∈I ri ≥ k 2 .
WLOG we assume monotonicity of the scores in the form:r1 ≤r2 ≤· · · ≤rn. Then (∗) is equivalent to k X i=1 ri ≥ k 2 , (k = 1, 2, . . . , n).
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Proof of Landau’s Theorem Several proofs exist: ▶A constructive proof of Ryser: Inductively construct row and column i for i = n, · · · , 2, 1 ▶Minimal counterexample proofs of Mahmoodian and Thomassen: Choose a counterexample with n minimum and then with minimum r1.
▶As a corollary of Rado’s theorem; see next slide.
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Landau’s Theorem from Rado’s Theorem Proof Outline: RAB and K. Kiernan 2009 ▶Define a matroid M on X = {(i, j); 1 ≤i, j ≤n, i ̸= j} whose independent sets are those subsets of X that do not contain a symmetric pair (i, j), (j, i) with i ̸= j.
▶Thus the minimal dependent sets - the circuits - are the n 2 disjoint sets {(i, j), (j, i)} of two pairs in X with i ̸= j.
▶We have ρ(X) = n 2 .
▶Let A = (A1, A2, . . . , An) where Ai = {(i, j) : 1 ≤j ≤n, j ̸= i} (i = 1, 2, . . . , n). Thus Ai consists of all the non-diagonal positions of row i.
▶There exists a tournament with score sequence r1, r2, . . . , rn if and only if there exists P1, P2, . . . , Pn, with Pi ⊆Ai and |Pi| = ri (1 ≤i ≤n), such that P = P1 ∪P2 ∪· · · ∪Pn is an independent set of M.
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Landau’s Theorem from Rado’s Theorem continued ▶There exists a tournament with score sequence r1, r2, . . . , rn if and only if there exists P1, P2, . . . , Pn, with Pi ⊆Ai and |Pi| = ri (1 ≤i ≤n), such that P = P1 ∪P2 ∪· · · ∪Pn is an independent set of M, ▶Equivalently, if and only if the family A′ = (A1, . . . , A1 | {z } r1 , A2, . . . , A2 | {z } r2 , . . . , An, . . . , An | {z } rn ) has an SDR whose elements form an independent set.
▶The desired tournament has vertices 1, 2, . . . , n and an edge from i to j if and only (i, j) is in Pi. The independence of P then implies that there is no edge from j to i.
▶Now one checks that Rado’s condition holds if the inequalities of Landau hold.
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Stronger Landau Inequalities RAB and J. Shen 2001: If R = (r1, r2, . . . , rn) is a sequence of nonnegative integers, then R is the score sequence of a tournament of order n if and only if (∗∗) X i∈I ri ≥1 2 X i∈I (i −1) + 1 2 |I| 2 (I ⊆{1, 2, . . . , n}) with equality when I = {1, 2, . . . , n}.
Since X i∈I (i −1) ≥ |I| X i=1 (i −1) = |I| 2 , the inequalities (∗∗) are individually, but not collectively, stronger than Landau’s inequalities (∗) X i∈I ri ≥ |I| 2 .
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem What good are these “stronger” inequalities?
Certainly not to show the existence of a tournament with specified scores — they only make it more difficult.
Answer: These stronger inequalities allow an especially simple proof of the wonderful theorem first proved by Ao and Hanson 1998 and, independently, Guiduli, Gy´ ar´ as, and Weidl 1998: that shows the existence of a very special tournament for each possible score vector: If there is a tournament with score sequence r1 ≤r2 ≤· · · ≤rn, then there is one such that the induced tournaments on the vertex sets {1, 3, 5, . . .} and {2, 4, 6, . . . , } are transitive tournaments, that is, there is the arc i →j whenever i > j and i ≡j mod 2.
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem What good are these “stronger” inequalities?
Certainly not to show the existence of a tournament with specified scores — they only make it more difficult.
Answer: These stronger inequalities allow an especially simple proof of the wonderful theorem first proved by Ao and Hanson 1998 and, independently, Guiduli, Gy´ ar´ as, and Weidl 1998: that shows the existence of a very special tournament for each possible score vector: If there is a tournament with score sequence r1 ≤r2 ≤· · · ≤rn, then there is one such that the induced tournaments on the vertex sets {1, 3, 5, . . .} and {2, 4, 6, . . . , } are transitive tournaments, that is, there is the arc i →j whenever i > j and i ≡j mod 2.
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem What good are these “stronger” inequalities?
Certainly not to show the existence of a tournament with specified scores — they only make it more difficult.
Answer: These stronger inequalities allow an especially simple proof of the wonderful theorem first proved by Ao and Hanson 1998 and, independently, Guiduli, Gy´ ar´ as, and Weidl 1998: that shows the existence of a very special tournament for each possible score vector: If there is a tournament with score sequence r1 ≤r2 ≤· · · ≤rn, then there is one such that the induced tournaments on the vertex sets {1, 3, 5, . . .} and {2, 4, 6, . . . , } are transitive tournaments, that is, there is the arc i →j whenever i > j and i ≡j mod 2.
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Example Thus to construct a tournament T10 with score sequence R = (1, 3, 4, 4, 4, 5, 5, 5, 7, 7), we have only to construct A below: 0 0 0 0 0 1 1 0 0 0 0 4 1 1 0 0 0 A 4 1 1 1 0 0 5 1 1 1 1 0 7 0 0 0 0 0 3 1 0 0 0 0 4 A′ 1 1 0 0 0 5 1 1 1 0 0 5 1 1 1 1 0 7 8 5 5 4 2 6 5 4 4 2 (A′ = J −At) with row and column sum vectors (1, 3, 2, 2, 3) and (2, 2, 2, 3, 2).
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Biclique Partitions A biclique of a graph G of order n is a complete bipartite subgraph Kr,s with r, s ≥1 (when r = s = 1 we just get one edge). The biclique partition number bp(G) of G is the smallest number of bicliques into which its edges can be partitioned. For example, Generalizing the left construction to arbitrary n we get bp(Kn) ≤n −1.
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Biclique Partitions A biclique of a graph G of order n is a complete bipartite subgraph Kr,s with r, s ≥1 (when r = s = 1 we just get one edge). The biclique partition number bp(G) of G is the smallest number of bicliques into which its edges can be partitioned. For example, Generalizing the left construction to arbitrary n we get bp(Kn) ≤n −1.
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Graham-Pollak Theorem I bp(Kn) = n −1: The proof uses elementary linear algebra.
A biclique partition of Kn into bicliques corresponds to a tournament of order n: Let Tn be the corresponding tournament matrix so that Tn + T t n = Jn −In. Let T ∗ n be the n × (n + 1) matrix obtaining by adjoining a column of all 1s to Tn, Let x be a vector in the left nullspace of T ∗ n : xtT ∗ n = 0. On the one hand, xt(Jn −In)x = xtJnx −xtx = −xtx.
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Graham-Pollak Theorem I bp(Kn) = n −1: The proof uses elementary linear algebra.
A biclique partition of Kn into bicliques corresponds to a tournament of order n: Let Tn be the corresponding tournament matrix so that Tn + T t n = Jn −In. Let T ∗ n be the n × (n + 1) matrix obtaining by adjoining a column of all 1s to Tn, Let x be a vector in the left nullspace of T ∗ n : xtT ∗ n = 0. On the one hand, xt(Jn −In)x = xtJnx −xtx = −xtx.
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Graham-Pollak Theorem I bp(Kn) = n −1: The proof uses elementary linear algebra.
A biclique partition of Kn into bicliques corresponds to a tournament of order n: Let Tn be the corresponding tournament matrix so that Tn + T t n = Jn −In. Let T ∗ n be the n × (n + 1) matrix obtaining by adjoining a column of all 1s to Tn, Let x be a vector in the left nullspace of T ∗ n : xtT ∗ n = 0. On the one hand, xt(Jn −In)x = xtJnx −xtx = −xtx.
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Graham-Pollak Theorem II xt(Jn −In)x = xtJnx −xtx = −xtx.
On the other hand, xt(Jn−In)x = xt(Tn+T t n)x = xtTnx+xtT t nx = (xtTn)x+xt(xtTn)t = 0 + 0 = 0 Thus xtx = 0 and so x = 0. Thus the rank of T ∗ n equals n and so the rank of Tn is ≥n −1.
But a decomposition of Kn into r bicliques expresses the matrix Jn −In as the sum Jn −In = Q + 2(A1 + A2 + · · · + Ar) where the Ai have rank 1 and Q is skew-symmetric. Thus In + Q = Jn −2(A1 + A2 + · · · + Ar) has eigenvalues with real part equal to 1 and is nonsingular. So r + 1 ≥n and r ≥n −1.
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Graham-Pollak Theorem II xt(Jn −In)x = xtJnx −xtx = −xtx.
On the other hand, xt(Jn−In)x = xt(Tn+T t n)x = xtTnx+xtT t nx = (xtTn)x+xt(xtTn)t = 0 + 0 = 0 Thus xtx = 0 and so x = 0. Thus the rank of T ∗ n equals n and so the rank of Tn is ≥n −1.
But a decomposition of Kn into r bicliques expresses the matrix Jn −In as the sum Jn −In = Q + 2(A1 + A2 + · · · + Ar) where the Ai have rank 1 and Q is skew-symmetric. Thus In + Q = Jn −2(A1 + A2 + · · · + Ar) has eigenvalues with real part equal to 1 and is nonsingular. So r + 1 ≥n and r ≥n −1.
P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Graham-Pollak Theorem III Example of a Q: 1 3 2 4 6 5 Q = 0 1 1 1 1 1 −1 0 1 1 1 1 −1 −1 0 1 1 1 −1 −1 −1 0 1 1 −1 −1 −1 −1 0 1 −1 −1 −1 −1 −1 0 P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Graham-Pollak Theorem III Example of a Q: 1 3 2 4 6 5 Q = 0 1 1 1 1 1 −1 0 1 1 1 1 −1 −1 0 1 1 1 −1 −1 −1 0 1 1 −1 −1 −1 −1 0 1 −1 −1 −1 −1 −1 0 P. Hall’s Theorem R. Rado’s Theorem Gale-Ryser Theorem Landau’s Theorem Graham-Pollak Theorem Conclusion The Rado-Hall Theorem is a wonderful theorem to have in ones’s mathematical toolchest! |
16080 | https://www.chegg.com/homework-help/questions-and-answers/lithium-chloride-solubility-55g-licl-100-g-h2o-25degrees-c-determine-following-mixtures-fo-q12711537 | Solved Lithium chloride has a solubility of 55g of LiCl in | Chegg.com
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Lithium chloride has a solubility of 55g of LiCl in 100 g of H2O at 25degrees C. determine if each of the following mixtures forms an unsaturated or saturated solution at 25 degrees C. a.) addding 10g of LiCl to 15g of H2O b.) adding 25g of LiCl to 50g of H2O C.) adding 75 g of LiCl to 150g of H2O
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Question: Lithium chloride has a solubility of 55g of LiCl in 100 g of H2O at 25degrees C. determine if each of the following mixtures forms an unsaturated or saturated solution at 25 degrees C. a.) addding 10g of LiCl to 15g of H2O b.) adding 25g of LiCl to 50g of H2O C.) adding 75 g of LiCl to 150g of H2O
Lithium chloride has a solubility of 55g of LiCl in 100 g of H2O at 25degrees C. determine if each of the following mixtures forms an unsaturated or saturated solution at 25 degrees C.
a.) addding 10g of LiCl to 15g of H2O
b.) adding 25g of LiCl to 50g of H2O
C.) adding 75 g of LiCl to 150g of H2O
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Calculate the solubility factor for LiCl by dividing the given mass of LiCl by the mass of water it dissolves in, resulting in 0.55 g m g m H 2 O. Given solubility of LiCl= 55g/100g =…
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16081 | https://www.quora.com/In-which-quadrant-does-0-lie-if-the-following-statements-are-true-cos-0-0-and-tan0-0 | In which quadrant does 0 lie if the following statements are true cos 0<0 and tan0<0? - Quora
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In which quadrant does 0 lie if the following statements are true cos 0<0 and tan0<0?
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Terry Moore
M.Sc. in Mathematics, University of Southampton (Graduated 1968) · Author has 16.6K answers and 29.4M answer views
·4y
In which quadrant does 0 lie if the following statements are true cos 0<0 and tan0<0?
Let’s use θ θ for the angle, shall we? Well, cos(θ)<0 cos(θ)<0 on the left of the y y axis and sin(θ)<0 sin(θ)<0 below the x x axis. Combine these two pieces of information with tan(θ)=sin(θ)cos(θ)tan(θ)=sin(θ)cos(θ) and you should have your answer.
You could just remember which functions are positive in which quadrants, but I’ve just given you a method to work it out for yourself.
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In which quadrant does sin = 0 and tan = 0 lie?
In which quadrant does (0,-0) lie?
What is your next number in this sequence: 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 0, 0, 4, 0, 1, 0, 0, 0, _?
If sin(t) >0 and cos(t<0), in which quadrant does the terminal point of angle t lie?
What is 0#0#?
Stuart Herring
Author has 11.7K answers and 8.2M answer views
·4y
In which quadrant does 0 lie if the following statements are true cos 0<0 and tan0<0?
(You mean “θ” there, not “0”. If you don’t have a θ symbol available, you could just say “theta” instead.)
cos(θ) is negative when the angle terminates in Quadrant II or Quadrant III, because it is related to the x-coordinate in the Cartesian system. When x<0, cos(x) < 0.
tan(θ) = sin(θ)/cos(θ), so it is negative if either sin(θ)<0 or cos(θ)<0.
sin(θ) is negative when the angle terminates in Quadrant III or Quadrant IV, because it is related to the y-coordinate in the Cartesian system. When y<0, sin(x)<0.
So tan
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In which quadrant does 0 lie if the following statements are true cos 0<0 and tan0<0?
(You mean “θ” there, not “0”. If you don’t have a θ symbol available, you could just say “theta” instead.)
cos(θ) is negative when the angle terminates in Quadrant II or Quadrant III, because it is related to the x-coordinate in the Cartesian system. When x<0, cos(x) < 0.
tan(θ) = sin(θ)/cos(θ), so it is negative if either sin(θ)<0 or cos(θ)<0.
sin(θ) is negative when the angle terminates in Quadrant III or Quadrant IV, because it is related to the y-coordinate in the Cartesian system. When y<0, sin(x)<0.
So tan(θ) is positive in Quadrant I because sin(θ)>0 and cos(θ)>0 there; it is negative in Quadrant II because sin(θ)>0 and cos(θ)<0 there; it is positive in Quadrant III because sin(θ)<0 and cos(θ)<0 there; it is negative in Quadrant IV because sin(θ)<0 and cos(θ)>0 there.
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To determine in which quadrant the angle 0 0 lies given the conditions cos(0)<0 cos(0)<0 and tan(0)<0 tan(0)<0, let's analyze the trigonometric functions:
Cosine: cos(θ)<0 cos(θ)<0 implies that the angle θ θ is in either the second quadrant (where cosine is negative) or the third quadrant (where cosine is also negative).
Tangent: tan(θ)<0 tan(θ)<0 implies that the angle θ θ is in either the second quadrant (where tangent is negative) or the fourth quadrant (where tangent is negative).
Now, let's summarize the quadrants based on these conditions:
Second Quadrant: cos<0 cos<0 and tan<0 tan<0 (Valid)
Third Quadrant:
Continue Reading
To determine in which quadrant the angle 0 0 lies given the conditions cos(0)<0 cos(0)<0 and tan(0)<0 tan(0)<0, let's analyze the trigonometric functions:
Cosine: cos(θ)<0 cos(θ)<0 implies that the angle θ θ is in either the second quadrant (where cosine is negative) or the third quadrant (where cosine is also negative).
Tangent: tan(θ)<0 tan(θ)<0 implies that the angle θ θ is in either the second quadrant (where tangent is negative) or the fourth quadrant (where tangent is negative).
Now, let's summarize the quadrants based on these conditions:
Second Quadrant: cos<0 cos<0 and tan<0 tan<0 (Valid)
Third Quadrant: cos<0 cos<0 and tan>0 tan>0 (Invalid)
Fourth Quadrant: cos>0 cos>0 and tan<0 tan<0 (Invalid)
Since the only quadrant that satisfies both conditions is the second quadrant, we conclude that 0 0 lies in the second quadrant.
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Jeffrey Stuart
PhD in Mathematics&İndustrial Engineering, University of Wisconsin - Madison (Graduated 1987) · Author has 2.9K answers and 1.3M answer views
·4y
Should we presume that some of your zeros are the number zero, and that some of your zeros are actually supposed to be the Greek letter theta? Not choosing a better symbol suggests that you are not just confused about trigonometry, you are confused about communicating, which is worse.
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Which quadrant would y lie if tan y is positive and sin y is negative?
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Celeste Martin
Former Maintenance Engineer. at Holiday Inn Express (2014–2015) · Author has 113 answers and 146.9K answer views
·5y
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If tan θ is positive and sin θ is negative, in which quadrant does θ lie?
Hi, thank you for your request.
I've noticed quite of few answers here, from just giving you the answer, to drilling the whole “All Students Take Calculus” type thing.
One thing you will need is the understanding that
C o s(θ)C o s(θ) does correspond with the x axis, and the
S i n(θ)S i n(θ) does correspond with the y axis. Also, the
T a n(θ)=s i n(θ)/c o s(θ)T a n(θ)=s i n(θ)/c o s(θ)
And right along the x axis is positive, and up along the y axis is positive, while left along the x axis is negative while down along the y axis is negative.
So, what I have shared was a picture from Google displaying a Unit circle, with a Triangle inside. A unit circle
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Hi, thank you for your request.
I've noticed quite of few answers here, from just giving you the answer, to drilling the whole “All Students Take Calculus” type thing.
One thing you will need is the understanding that
C o s(θ)C o s(θ) does correspond with the x axis, and the
S i n(θ)S i n(θ) does correspond with the y axis. Also, the
T a n(θ)=s i n(θ)/c o s(θ)T a n(θ)=s i n(θ)/c o s(θ)
And right along the x axis is positive, and up along the y axis is positive, while left along the x axis is negative while down along the y axis is negative.
So, what I have shared was a picture from Google displaying a Unit circle, with a Triangle inside. A unit circle is defined a circle with a radius of one unit in length. And when you place this on a graph you will notice that at every point along the perimeter has some coordinates in the form of (x,y). The equation for this is
x 2+y 2=1 x 2+y 2=1
This will actually draw out a circle on a grid. And the x is equal the cosine y is equal to the sine on a unit circle. Watch!
And if we were to calculate
±√(1−x 2)±√(1−x 2) then we would be able to find y. And conversely, if we were to calculate
±√(1−y 2)±√(1−y 2) then we would find x.
Remember squaring negative numbers does in fact make it positive, so you will have identify it's quadrant after calculations have been made.
If you know the slope of an angle we can calculate
(±1±m i)/(m 2+1)∗√(m 2+1)(±1±m i)/(m 2+1)∗√(m 2+1), where m=slope. However, this can be simplified even further. This equation is in complex form, and is equal to
C o s(θ)+s i n(θ)i C o s(θ)+s i n(θ)i, which can be written in the form of x+y i.x+y i. If, we were to replace the variables x and y with an arbitrary number, then all we would need to do divide that by the radius of the circle, or the displacement of the hypotenuse.
So let's assume 1+1 i 1+1 i and this gives us a slope 1 which is the same as 45° angle and a hypotenuse of √(2)√(2)
If we calculate (1+i)/√(2)(1+i)/√(2) we would get (1/√(2)+i/√(2)=(1/2+1/2 i)√(2)(1/√(2)+i/√(2)=(1/2+1/2 i)√(2).
Concurrently, we can find the displacement of the hypotenuse with the absolute value of x+yi, this equation then becomes (x+y i)/a b s(x+y i)(x+y i)/a b s(x+y i).
And in order to find the cosine, we just pull the real component from the complex number. And in order to find the sine, we just pull the imaginary number.
How cool is that?
What if we were to replace the variables x and y with cos(1) and the sin(1) respectively?
x+y i=c o s(1)+s i n(1)i x+y i=c o s(1)+s i n(1)i then raise this entire equation to the power of some angle?
Assume (c o s(1)+s i n(1)i)(30°)(c o s(1)+s i n(1)i)(30°)
Demoivre (French mathematician) has demonstrated that (cos(x)+sin(x)i)^k=cos(xk)+sin(xk)i
Therefore we can calculate (c o s(1)+s i n(1)i)(30°)=c o s(30°)+s i n(30°)i.(c o s(1)+s i n(1)i)(30°)=c o s(30°)+s i n(30°)i. Or (c o s(1°)+s i n(1°)i)(30)=c o s(30°)+s i n(30°)i(c o s(1°)+s i n(1°)i)(30)=c o s(30°)+s i n(30°)i.
This give us the basics of e(θ i)e(θ i) also known as Euler's formula, because e i=c o s(1)+s i n(1)i e i=c o s(1)+s i n(1)i.
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John K WilliamsSon
Accredited (MS Educ) nerd who loves talking about math · Author has 9K answers and 23.4M answer views
·6y
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If tan θ is positive and sin θ is negative, in which quadrant does θ lie?
Memorize this question about your teachers:
(Are)All Science Teachers Crazy?
or perhaps you prefer:
All Students Take Calculus
All Silly Tom Cats
Add Sugar To Coffee
All Stations to Central
All Statistics Too Cumbersome
My Trigonometry Professor taught us the All Students Take Calculus but I found I had trouble remembering it. Even today, I was wondering whether it started out “all calculus students”, so I decided to put All Science Teachers Crazy first?
What does this accomplish? Write these four letters, in order, in quadrants I, II, III and IV of your chart:
Next,
write all six trig functions in Quad
Continue Reading
Memorize this question about your teachers:
(Are)All Science Teachers Crazy?
or perhaps you prefer:
All Students Take Calculus
All Silly Tom Cats
Add Sugar To Coffee
All Stations to Central
All Statistics Too Cumbersome
My Trigonometry Professor taught us the All Students Take Calculus but I found I had trouble remembering it. Even today, I was wondering whether it started out “all calculus students”, so I decided to put All Science Teachers Crazy first?
What does this accomplish? Write these four letters, in order, in quadrants I, II, III and IV of your chart:
Next,
write all six trig functions in Quadrant One
write Sin and its inverse in Quadrant Two
write Tan and its inverse in Quadrant Three
write Cos and its inverse in Quadrant Four
All of these values are positive in the quadrants you wrote them in
Sine and Cosecant are positive in quadrants I and II
Tangent and Cotangent are positive in quadrants I and III
Cosine and Secant are positive in quadrants I and IV
The various trig values are negative in the quadrants where they are not written.
Let’s look at your question:
If tanø is positive and sin ø is negative; In which quadrant does ø lie?
Look at the chart:
Tangent is positive in quadrants I and III (as marked, All and Tan)
Sine is negative in quadrants III and IV (positive in All and Sine)
So, tan ø is positive, and sin ø is negative. What quadrant is ø in?
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Khenan Mak
Studied Engineering&Mathematics at MMU Cyberjaya (Graduated 2004) · Upvoted by
James McElhatton Ph.D. (Glasgow, 1976)
, B.Sc. Mathematics & Chemistry, University of Malta (1967) · Author has 2.6K answers and 1.2M answer views
·4y
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Given that for an angle θ θ, cos θ<0 cosθ<0 and sin θ>0 sinθ>0, what is the quadrant in which θ θ lies?
cos cos is -ve and sin sin is +ve. So quadrant II.
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Pinterest
cos cos is -ve and sin sin is +ve. So quadrant II.
Footnotes
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John K WilliamsSon
Accredited (MS Educ) nerd who loves talking about math · Author has 9K answers and 23.4M answer views
·2y
Related
In which quadrant lies when cos and tan?
Quora’s robot algorithm did a lousy job creating this random question. I will answer this question:
In which quadrant does the angle lie when you know whether cos and tan are negative and/or positive?
Step Zero: Get out your four-color pen and a piece of paper.
Step One: Draw an x-y coordinate graph showing all four quadrants (blue lines). Optional, label each multiple of 45° (red numbers), with 360° on the same line as 0°. You already know that both sine and tangent are positive in the first quadrant, so label that entire quadrant as +cos +tan (green)
Step Two: Make a chart of the values of cosin
Continue Reading
Quora’s robot algorithm did a lousy job creating this random question. I will answer this question:
In which quadrant does the angle lie when you know whether cos and tan are negative and/or positive?
Step Zero: Get out your four-color pen and a piece of paper.
Step One: Draw an x-y coordinate graph showing all four quadrants (blue lines). Optional, label each multiple of 45° (red numbers), with 360° on the same line as 0°. You already know that both sine and tangent are positive in the first quadrant, so label that entire quadrant as +cos +tan (green)
Step Two: Make a chart of the values of cosine and tangent for every multiple of 15, 30 or 45 degrees. Here is how I asked my TI-84+ CE Python graphing calculator to make a chart for every 45 degrees:
Step Three: Look at 135°. cos is negative, tan is negative, so label that entire quadrant as –cos –tan.
Step Four: Do the same thing for 225° and 315°
Step Five: Practice making this chart so that you can quickly sketch one on the back of your quiz/test paper during a trig quiz or test.
.
Do you appreciate the work I put into my many answers?
If you enjoy the quality and quantity of my answers, please consider joining the ranks of my generous Patreon supporters. Thank you! (patreon.com/John_K_WilliamsSon)
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Gordon M. Brown
Math Tutor at San Diego City College (2018-Present) · Author has 6.2K answers and 4.3M answer views
·1y
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Which quadrant would theta be in if sin theta < 0 and tan theta > 0?
sin θ < 0 if and only if the terminal side of θ is located in either the third quadrant, or the fourth. tan θ > 0 if and only if the terminal side of θ is located in either the first quadrant, or the third. Therefore, for both sin θ < 0 and tan θ > 0 to hold simultaneously, the terminal side of θ must be located in the third quadrant.
Even more important, in the larger scheme of things, is to understand why these facts obtain with regard to sin θ and tan θ. This understanding not only warrants, but mandates, a thoroughgoing understanding of the definitions of all six basic trig functions: sine,
Continue Reading
sin θ < 0 if and only if the terminal side of θ is located in either the third quadrant, or the fourth. tan θ > 0 if and only if the terminal side of θ is located in either the first quadrant, or the third. Therefore, for both sin θ < 0 and tan θ > 0 to hold simultaneously, the terminal side of θ must be located in the third quadrant.
Even more important, in the larger scheme of things, is to understand why these facts obtain with regard to sin θ and tan θ. This understanding not only warrants, but mandates, a thoroughgoing understanding of the definitions of all six basic trig functions: sine, cosine, tangent, cosecant, secant, and cotangent. Too many students whom I’ve tutored over the years rely too heavily on mnemonic devices (that is, certain “memory aids”) to keep track of properties of trig functions, without cultivating any real understanding of why these properties should obtain. My advice is to go for the understanding, and skip the mnemonics, the rubrics, and other suspect devices. You will be the much better student of trigonometry for it.
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Terry Moore
M.Sc. in Mathematics, University of Southampton (Graduated 1968) · Author has 16.6K answers and 29.4M answer views
·5y
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What quadrant does 'sin theta < 0 and tan theta is undefined' lie?
What quadrant does 'sin theta < 0 and tan theta is undefined' lie?
tan(θ)tan(θ) is undefined when θ θ is π/2 π/2 or 3 π/2 3 π/2. sin(θ)<0 sin(θ)<0 when π<θ<θ<2 π π<θ<θ<2 π.
So the answer is any quadrant θ 1<θ<θ 1+π/2 θ 1<θ<θ 1+π/2 where π<θ 1<θ<2 π π<θ 1<θ<2 π, e.g. 5 π/4<θ<7 π/4 5 π/4<θ<7 π/4.
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9 1
Dr-Michael-W-Ecker
PSU Math Professor, Ph.D. Math, Summa Cum Laude, 1978, CUNY · Author has 1.9K answers and 2.1M answer views
·7y
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If cos theta<0 and tan theta<0, in which quadrant does theta lie?
May I suggest something easier? I’ve never understood why anybody thinks that memorization is needed beyond just the definition, which I will give concisely now for a real number theta, but I will use t for theta and temporarily will pretend t > 0. (Note: For those of you stuck with angles, real number is same as number of radians.)
DEFINITION: You wind t units of arc length from (1, 0) along the unit circle in appropriate direction and you arrive at the point (x, y).
Then sin(t) = y, cos(t) = x, and tan(t) = y/x.
So, if cos(t)<0 then x<0.
From tan(t)=y/x<0, we get y>0.
OK… Look at coordinate axes
Continue Reading
May I suggest something easier? I’ve never understood why anybody thinks that memorization is needed beyond just the definition, which I will give concisely now for a real number theta, but I will use t for theta and temporarily will pretend t > 0. (Note: For those of you stuck with angles, real number is same as number of radians.)
DEFINITION: You wind t units of arc length from (1, 0) along the unit circle in appropriate direction and you arrive at the point (x, y).
Then sin(t) = y, cos(t) = x, and tan(t) = y/x.
So, if cos(t)<0 then x<0.
From tan(t)=y/x<0, we get y>0.
OK… Look at coordinate axes. Where is x<0 and y>0? Quad II.
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9 2
9 1
Al Cohen
Studied Electrical Engineering&English, Astronomy (Graduated 1900) · Author has 1.9K answers and 571.6K answer views
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In which quadrant does sin = 0 and tan = 0 lie?
Angles lie in quadrants. Their functions are just numbers, and don't lie in quadrants.
The angle (0) is on the boundary between Quadrants I and IV.
Its sine and tangent are both zero. Its cosine is 1.
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Dave Williamson
MathNerd/MS Ed/age 65/99th %ile GRE/seeks free tuition Univ adv.study.math study · Author has 1.7K answers and 12.3M answer views
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If cos theta<0 and tan theta<0, in which quadrant does theta lie?
Here’s an acronym that I sometimes use to remember which trig functions are positive in which quadrants?
ASTC: A ll S tudents T ake C alculus
Q1: All functions are positive
Q2: Sine (and its inverse, Cosecant) are positive
Q3: Tangent (and its inverse, Cotangent) are positive
Q4: Cosine (and its inverse, Secant) are positive
Graph a unit circle, a circle centered at (0,0) with a radius of 1 unit.
With a blue pencil, label all the quadrants in which a function is positive.
Q1: write down all six
Q2: write down sine and cosecant
Q3: write down tangent and cotangent
Q4: write down cosine and secant
With a red pe
Continue Reading
Here’s an acronym that I sometimes use to remember which trig functions are positive in which quadrants?
ASTC: A ll S tudents T ake C alculus
Q1: All functions are positive
Q2: Sine (and its inverse, Cosecant) are positive
Q3: Tangent (and its inverse, Cotangent) are positive
Q4: Cosine (and its inverse, Secant) are positive
Graph a unit circle, a circle centered at (0,0) with a radius of 1 unit.
With a blue pencil, label all the quadrants in which a function is positive.
Q1: write down all six
Q2: write down sine and cosecant
Q3: write down tangent and cotangent
Q4: write down cosine and secant
With a red pencil, label all the functions in the quadrants that you didn’t write them in blue.
Let’s look at your question:
If cos Θ<0 and tan Θ<0, in which quadrant does Θ lie?
Look at your diagram. What is the only quadrant in which you wrote cosine and tangent in red letters?
Upvote ·
99 10
9 1
Related questions
In which quadrant does sin = 0 and tan = 0 lie?
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What is your next number in this sequence: 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 0, 0, 4, 0, 1, 0, 0, 0, _?
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Given cot(0) = -1/2 and 0 belongs to Quadrant II, then what are the values of furthermore tan (0) and CSC (0)?
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Related questions
In which quadrant does sin = 0 and tan = 0 lie?
In which quadrant does (0,-0) lie?
What is your next number in this sequence: 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 0, 0, 4, 0, 1, 0, 0, 0, _?
If sin(t) >0 and cos(t<0), in which quadrant does the terminal point of angle t lie?
What is 0#0#?
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16083 | https://www.wpi.edu/sites/default/files/inline-image/Departments-Programs/Mathematical-Sciences/Marcel%20Blais%20Linear_Programming_Talk_MIST_2017.pdf | MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications Connecting Calculus to Linear Programming Marcel Y. Blais, Ph.D.
Worcester Polytechnic Institute Dept. of Mathematical Sciences July 2017 Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications Motivation Goal: To help students make connections between high school math and real world applications of mathematics.
Linear programming is based on a simple idea from calculus.
Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications Motivation Goal: To help students make connections between high school math and real world applications of mathematics.
Linear programming is based on a simple idea from calculus.
We can connect calculus to real-world industrial problems.
Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications Motivation Goal: To help students make connections between high school math and real world applications of mathematics.
Linear programming is based on a simple idea from calculus.
We can connect calculus to real-world industrial problems.
We’ll explore the some basic principles of linear programming and some modern-day applications.
Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications Continuous Functions on Closed Intervals x -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 y -2 -1 0 1 2 3 4 5 a b What can we say about a continuous function on a closed interval?
Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications Continuous Functions on Closed Intervals x -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 y -2 -1 0 1 2 3 4 5 a b Theorem If f is continuous on [a, b] then f attains both an absolute minimum value and an absolute maximum value on [a, b].
Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications Continuous Functions on Closed Intervals x -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 y -2 -1 0 1 2 3 4 5 a b Theorem If f is continuous on [a, b] then f attains both an absolute minimum value and an absolute maximum value on [a, b].
Further these occur at critical points of f or endpoints of [a, b].
Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications Linear Case Theorem If f is continuous linear on [a, b] then f attains both an absolute minimum value and an absolute maximum value on [a, b], and these occur at endpoints of [a, b].
x 0 1 2 3 4 5 6 7 8 9 10 y 0 5 10 15 20 25 30 35 a b Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications Higher-Dimensional Optimization 1 0.8 0.6 0.4 x 0.2 0 0 0.1 0.2 y 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -0.2 -0.5 -0.4 -0.3 0.2 -0.1 0 0.1 z What does this mean for a continuous function f (x, y) on a closed and bounded plane region R?
Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications Higher-Dimensional Optimization 1 0.8 0.6 0.4 x 0.2 0 0 0.1 0.2 y 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -0.2 -0.5 -0.4 -0.3 0.2 -0.1 0 0.1 z Theorem If f (x, y) is continuous on closed and bounded plane region R then f attains both an absolute minimum value and an absolute maximum value on R. Further, these values occur either at critical points of f in the interior of R or on the boundary of R.
Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications Higher-Dimensional Optimization 1 0.8 0.6 0.4 x 0.2 0 0 0.1 0.2 y 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -0.2 -0.5 -0.4 -0.3 0.2 -0.1 0 0.1 z Theorem If f (x1, x2, . . . , xn) is continuous on closed and bounded region R ⊂Rn then f attains both an absolute minimum value and an absolute maximum value on R. Further, these values occur either at critical points of f in the interior of R or on the boundary of R.
Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications Higher-Dimensional Linearity 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 x 0.2 0.1 0 0 0.2 y 0.4 0.6 0.8 4.5 0 0.5 1 5 1.5 2 2.5 3 3.5 4 1 z Theorem If f (x1, x2, . . . , xn) is linear on closed and bounded region R ⊂Rn defined by a system of linear constraints then f attains both an absolute minimum value and an absolute maximum value on R. Further, these values occur on the boundary of R.
Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications Higher-Dimensional Linearity 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 x 0.2 0.1 0 0 0.2 y 0.4 0.6 0.8 4.5 0 0.5 1 1.5 2 2.5 3 3.5 4 5 1 z Theorem If f (x1, x2, . . . , xn) is linear on closed and bounded region R ⊂Rn defined by a system of linear constraints then f attains both an absolute minimum value and an absolute maximum value on R. Further, these values occur on corner points of the boundary of R.
Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications Higher-Dimensional Linearity 1 0.8 0.6 0.4 x 0.2 0 0 0.1 0.2 0.3 y 0.4 0.5 0.6 0.7 0.8 0.9 0.1 0.6 0 1 0.9 0.8 0.7 0.2 0.5 0.4 0.3 1 z Example: A possible feasible region f (x, y, z) = ax + by + cz subjuect to 5 linear constraints.
A solution to minimize f over this region will occur at a corner.
Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications Linear Programming The Simplex Method was developed by George Dantzig in 1947.
“programming” synonymous with “optimization”.
Algorithm to traverse the corner points of the feasible polyhedron for a linear programming problem to find an optimal feasible solution.
Standard form for a linear programming problem: min xTc such that Ax ≤b and x ≥0.
(1) x, c ∈Rn, b ∈Rm, A ∈Rm×n - n variables and m constraints.
Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications Simplex Method min xTc such that Ax ≤b and x ≥0.
For each constraint (row i of A), add a new slack variable yi .
New system: min xTc such that [A + I] x y = b, x ≥0, y ≥0.
Underdetermined (more variables than equations). A + I The slack now in the system is the key.
Each slack variable is associated with a constraint boundary Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications Simplex Method Simplex algorithm: Split entries of x y into 2 subsets: Basic variables: xB ∈Rm (non-zero valued) Non-basic variables: xN ∈Rn (zero valued) Represents a corner point of the feasible region.
If not optimal, move to an adjacent corner point by swapping one entry in xB with one entry in xN and re-solving the system of equations.
Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications Simplex Geometry 1 0.8 0.6 0.4 x 0.2 0 0 0.1 0.2 0.3 y 0.4 0.5 0.6 0.7 0.8 0.9 0.1 0.6 0 1 0.9 0.8 0.7 0.2 0.5 0.4 0.3 1 z Figure: Feasible Region Example: min f (x1, x2, x3) = ax1 + bx2 + cx3 subject to 5 linear constraints.
Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications Simplex Geometry 1 0.8 0.6 0.4 x 0.2 0 0 0.1 0.2 0.3 y 0.4 0.5 0.6 0.7 0.8 0.9 0.1 0.6 0 1 0.9 0.8 0.7 0.2 0.5 0.4 0.3 1 z Example: min f (x1, x2, x3) = ax1 + bx2 + cx3 subject to 5 linear constraints.
Choose initial basis to correspond to the origin.
Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications Simplex Geometry 1 0.8 0.6 0.4 x 0.2 0 0 0.1 0.2 0.3 y 0.4 0.5 0.6 0.7 0.8 0.9 0.1 0.6 0 1 0.9 0.8 0.7 0.2 0.5 0.4 0.3 1 z Example: min f (x1, x2, x3) = ax1 + bx2 + cx3 subject to 5 linear constraints.
If the objective function is not optimal, move to a better adjacent corner.
Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications Simplex Geometry 1 0.8 0.6 0.4 x 0.2 0 0 0.1 0.2 0.3 y 0.4 0.5 0.6 0.7 0.8 0.9 0.1 0.6 0 1 0.9 0.8 0.7 0.2 0.5 0.4 0.3 1 z Example: min f (x1, x2, x3) = ax1 + bx2 + cx3 subject to 5 linear constraints.
Repeat until the objective function is minimized (no remaining adjacent corners will reduce it).
Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications Application: Shipping Network Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications Shipping Problem K warehouses: S1, S2, . . . , SK Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications Shipping Problem K warehouses: S1, S2, . . . , SK L retail locations: D1, D2, . . . , DL Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications Shipping Problem K warehouses: S1, S2, . . . , SK L retail locations: D1, D2, . . . , DL Cost to ship x amount of product from Si to Dj is cijx Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications Shipping Problem K warehouses: S1, S2, . . . , SK L retail locations: D1, D2, . . . , DL Cost to ship x amount of product from Si to Dj is cijx Warehouse Si can supply si amount of product Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications Shipping Problem K warehouses: S1, S2, . . . , SK L retail locations: D1, D2, . . . , DL Cost to ship x amount of product from Si to Dj is cijx Warehouse Si can supply si amount of product Retail location Dj demands dj amount of product Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications Shipping Problem K warehouses: S1, S2, . . . , SK L retail locations: D1, D2, . . . , DL Cost to ship x amount of product from Si to Dj is cijx Warehouse Si can supply si amount of product Retail location Dj demands dj amount of product Problem: Design a shipping schedule that satisfies demand at all retail locations at a minimal cost.
Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications Shipping Problem K warehouses: S1, S2, . . . , SK L retail locations: D1, D2, . . . , DL Cost to ship x amount of product from Si to Dj is cijx Warehouse Si can supply si amount of product Retail location Dj demands dj amount of product Problem: Design a shipping schedule that satisfies demand at all retail locations at a minimal cost.
Example: As of July 2017, Target Corp. has 37 distribution centers and 1, 802 stores.
Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications Shipping Network Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications Shipping Network Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications Shipping Problem Mathematical Model Constraints xij ≥0 - amount of product shipped from Si to Dj Objective Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications Shipping Problem Mathematical Model Constraints xij ≥0 - amount of product shipped from Si to Dj Amount leaving Si : xi1 + xi2 + · · · + xiL ≤si Objective Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications Shipping Problem Mathematical Model Constraints xij ≥0 - amount of product shipped from Si to Dj Amount leaving Si : xi1 + xi2 + · · · + xiL ≤si Amount entering Dj : x1j + x2j + · · · + xKj = dj Objective Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications Shipping Problem Mathematical Model Constraints xij ≥0 - amount of product shipped from Si to Dj Amount leaving Si : xi1 + xi2 + · · · + xiL ≤si Amount entering Dj : x1j + x2j + · · · + xKj = dj Objective Total cost of shipping schedule: K X i=1 L X j=1 cijxij Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications Shipping Problem Mathematical Model Constraints xij ≥0 - amount of product shipped from Si to Dj Amount leaving Si : xi1 + xi2 + · · · + xiL ≤si Amount entering Dj : x1j + x2j + · · · + xKj = dj Objective Total cost of shipping schedule: K X i=1 L X j=1 cijxij Minimize cost of shipping such that demand is satisfied Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications Shipping Problem Mathematical Model Constraints xij ≥0 - amount of product shipped from Si to Dj Amount leaving Si : xi1 + xi2 + · · · + xiL ≤si Amount entering Dj : x1j + x2j + · · · + xKj = dj Objective Total cost of shipping schedule: K X i=1 L X j=1 cijxij Minimize cost of shipping such that demand is satisfied Target example: 37 × 1, 802 = 66, 674 variables Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications Shipping Problem Mathematical Model Amount leaving Si : xi1 + xi2 + · · · + xiL ≤si A = 1 . . .
1 1 . . .
1 ...
1 . . .
1 I I I min K X i=1 L X j=1 cijxij such that Ax ≤ = s d , x ≥0.
Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications Shipping Problem Mathematical Model Amount leaving Si : xi1 + xi2 + · · · + xiL ≤si Amount entering Dj : x1j + x2j + · · · + xKj = dj A = 1 . . .
1 1 . . .
1 ...
1 . . .
1 I I I min K X i=1 L X j=1 cijxij such that Ax ≤ = s d , x ≥0.
Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications TV Project - MA 3231 Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications TV Project - MA 3231 Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications TV Project - MA 3231 Blais MIST 2017 MIST 2017 Blais Optimization Linearity Higher-Dimensional Optimization Higher-Dimensional Linearity Linear Programming Simplex Method Applications References 1 Blais, Marcel., MA 3231 Linear Programming Lecture Notes, WPI Department of Mathematical Sciecnes, 2016.
2 Griva, Igor, Stephen G. Nash, and Ariela Sofer., Linear and Nonlinear Optimization, Second Edition. SIAM, 2009.
3 Hillier, Frederick and Gerald Lieberman., Introduction to Operations Research, Tenth Edition. McGraw Hill Education, 2015.
Blais MIST 2017 |
16084 | https://sites.math.rutgers.edu/~ag930/Some%20Math/p-adic-constructions.pdf | CONSTRUCTIONS OF THE p-ADIC NUMBERS ALEJANDRO GINORY Contents 1.
Introduction 1 2.
p-adic Completion of Q 1 3.
Algebraic Construction 3 4.
Equivalence of Constructions 4 1. Introduction As with many students of mathematics, I learned to construct the p-adic numbers in two seemingly different ways.
The first way was as the fraction field of the projective limit of the family of commutative rings Z/pnZ under the canonical maps (Z/pn+1Z) →(Z/pnZ). The second was as the (metric) completion of the rational numbers under the p-adic metric. It is probably more ‘natural’ to learn it in the opposite order but my first exposure came from Jean-Pierre Serre’s wonderful book A Course in Arithmetic where he mentions the second construction in passing.
While the two constructions exhibit striking similarities, it is not immediately obvious that they yield the same field (up to natural isometric isomorphism). In this article, we work out their equivalence.
2. p-adic Completion of Q In this approach, we redefine the notion of ‘closeness’ to reflect divisibility by a prime number of interest. Let p be a prime and and note that every nonzero rational number can be written in the form pn(a/b) where a, b, n ∈Z and gcd(a, p) = gcd(b, p) = 1.
It is easy to see that n is uniquely determined, prompting the following definition.
Definition 2.1. Let x = pn(a/b) ∈Q −{0} where a, b, n ∈Z and gcd(a, p) = gcd(b, p) = 1 then the p-adic absolute value (sometimes referred to, somewhat inappropriately, as p-adic norm) of x, denoted |x|p, is p−n. We also define |0|p = 0.
Definition 2.2. The p-adic metric (or p-adic distance) on Q is defined dp(x, y) = |x −y|p.
Before we continue, it is good manners to show that the above definitions make sense. It is enough to show for the absolute value.
1 2 ALEJANDRO GINORY Definition 2.3. Let R be a ring then a function | · | : R →R is called an Archimedean absolute value on R if it satisfies: (i) (Positive Definite) For all x ∈R, |x| ≥0 with equality iffx = 0, (ii) (Multiplicative) For all x, y ∈R, |x · y| = |x| · |y|, (iii) (Triangle Inequality) For all x, y ∈R, |x + y| ≤|x| + |y|.
If property (iii) is replaced by the stronger property: (iii)∗For all x, y ∈R, |x + y| ≤max{|x|, |y|}, then | · | is called a non-Archimedean absolute value.
Proposition 2.4. The p-adic absolute value is a non-Archimedean absolute value on Q.
Proof. We run through the checklist. As noted in the first paragraph of this section, | · |p is positive definite. Second, let x = pm(a/b) and y = pn(c/d) where all of the variables are integers, cd ̸= 0, and p does not divide a, b, c, or d, then |x · y|p = pm+n ac bd p = p−(m+n) = |x|p · |y|p.
If y = 0, then clearly |x · 0|p = |x|p · |0|p. As for the non-Archimedean property |x + y|p ≤max{|x|p, |y|p}, if both are 0 then it is clear. Let x = pm(a/b) and y = pn(c/d) as before but where x ̸= 0 and n ≥m, then |x + y|p = pm a + pn−mc bd p ≤p−m = max{|x|p, |y|p}.
□ Recall that for any metric space M with metric dM a Cauchy sequence {xi}∞ i=1 is a sequence of points xi ∈M, i ≥1, so that for all ε > 0 we can find a large enough N so that for all i, j ≥N, dM(xi, xj) < ε. Let X is the set of all Cauchy sequences of M, the completion of M, denoted M , is the set M = X/ ∼ where {xi}∞ i=1 ∼{yi}∞ i=1 if and only if limi→∞dM(xi, yi) = 0. An analogous defini-tion applies to fields with absolute values, where we add and multiply (equivalence classes of) Cauchy sequences ‘component-wise’.
A complete metric space (resp.
field) is a metric space (resp. field) where all Cauchy sequences converge. Comple-tions of metric space and fields are complete.
Definition 2.5. The p-adic numbers, denoted Qp, is the completion of Q with respect to the p-adic distance.
Note that every element of Qp can be approximated by rational numbers ai/bi, i ≥1, so that aibj−ajbi bibj are highly divisible by p as i and j both tend to infinity. To make this definition less ‘wild’ seeming, consider the following lemma.
CONSTRUCTIONS OF THE p-ADIC NUMBERS 3 Lemma 2.6. For every Cauchy sequence {xi}∞ i=1 in Q, there exists a sequence {pnki}∞ i=1 such that {xi}∞ i=1 ∼{pnki}∞ i=1 where n ∈Z is fixed, ki ∈Z, and, when {xi}∞ i=1 does not converge to 0, lim i→∞|xi|p = p−n.
Moreover, when {xi}∞ i=1 does not converge to 0 we can choose the ki so that p ∤ki.
Proof. If {xi}∞ i=1 converges to 0, then the constant sequence 0, 0, . . . does the trick.
Suppose {xi}∞ i=1 does not converge to zero and has no zero terms.
Since non-Archimedean absolute values satisfy the triangle inequality, it follows that | |xi|p −|xj|p | ≤|xi −xj|p (where the outside absolute values on the left are the standard ones) and so the sequence {|xi|p}i converges in R. By the assumption that the sequence does not converge to zero, it converges to some p−n for n ∈Z, since the complement {0} ∪ {pi}i∈Z is open. This will be the n used in the desired sequence.
Choose N > 0 large enough so that |xi|p = p−n for all i ≥N. This means that we can write xi = pn(ai/bi) where p ∤aibi for all i ≥N. This means that [bi] ∈Z/piZ is invertible and so there is an integer ki, for i ≥N, so that kibi ≡ai (mod pi).
Note also that p ∤ki. It follows that the sequence {pnki}i∈N, where ki(i < N) are arbitrarily chosen non-multiples of p, is equivalent to {xi}i∈N and limi→∞|xi|p = p−n.
□ This little lemma will prove to be very useful in the upcoming discussion. Note, though, that the choice of the ki in the proof is only unique modulo pi as long as i is large enough. We will revisit this point of view after discussing the algebraic construction.
3. Algebraic Construction As before, fix a prime number p and consider the system of commutative rings {Z/pnZ}n≥1 along with the canonical maps σn : Z/pn+1Z →Z/pnZ [k] 7→[k], for k ∈Z.
It is clear that this is a well-defined homomorphism for each n ≥1. For the algebraic construction of the p-adic numbers, we require the notion of projective limit for rings.
Definition 3.1. Let (Λ, ⪯) be a partially ordered set and {Rλ}λ∈Λ be a family of rings along with homomorphisms ραβ :Rβ →Rα for α ⪯β satisfying ραα = idRα and ραβ ◦ρβγ = ραγ, for α ⪯β ⪯γ. The projective limit of {Rλ}λ∈Λ (and the homomorphisms) is lim ← − λ∈Λ Rλ := ( r ∈ Y λ∈Λ Rλ : ραβ(πβ(r)) = πα(r) for α ⪯β ) 4 ALEJANDRO GINORY where πλ are the natural projection maps.
Definition 3.2. Using the data in the first paragraph of this section (with the par-tially ordered set being the positive integers), the p-adic integers is the projective limit Zp := lim ← − Z/pnZ.
The p-adic numbers is the fraction field Qp := Frac(Zp).
We can represent elements in Zp as sequences ([k1], [k2], . . . , [kn], . . .) where ki ∈ Z and [ki] ∈Z/piZ with the property that for any i, j ≥1 such that i ≤j, kj ≡ki (mod pi).
Again, the situation may seem rather ‘wild’ but, as in the previous section, there is some order.
Lemma 3.3. Every element x ∈(Zp−{0}) can be uniquely factored x = pnu where u is invertible in Zp and n ∈N.
Proof. Let x = ([x1], . . . , [xi], . . .), then x ̸= 0 implies that [xi] ̸= 0 for some i. Let n be the unique integer so that [xi] = 0 for i ≤n and [xj] ̸= 0 for j > n. This means that pn|xj and pn+1 ∤|xj for all j > n, so we may write uj = xj/pn. For i ≤n, recursively find ui so that [ui] = σi([ui+1]) and, once this is done, notice that [uj] ̸= 0 for all j ≥1. It follows that [uj] has an inverse [vj] in Z/pjZ and so ([u1], . . . , [uj], . . .) · ([v1], . . . , [vj], . . .) = (, , . . .).
Setting u = ([u1], . . . , [uj], . . .) it is clear that x = pnu with u invertible.
Uniqueness follows from this computation pnu = pmv pnuv−1 = pm · (, , . . .), where u, v are invertible, since it implies that n = m (count the zeros) and that ui = for all i > n. The lower indexed terms are uniquely determined by the system’s homomorphisms.
□ Corollary 3.4. Every element x ∈(Qp −{0}) can be uniquely factored x = pnu where u is invertible in Zp and n ∈Z.
Using these facts, we can define an absolute value | · |p on Qp by |pnu| = p−n for u ∈Zp invertible and n ∈Z, along with |0|p = 0. The reader is encouraged to verify that this is a well-defined non-Archimedean absolute value.
4. Equivalence of Constructions It is natural to ask if the completion of Q with respect to the p-adic absolute value, which we will denote Q in this section, and the fraction field of the projective limit of the groups Z/pnZ, n ≥1, with the p-adic absolute value are ‘the same’. By the same, we mean a field isomorphism that preserves the absolute value, i.e., an isometric field isomorphism. The following theorem answers this in the affirmative.
Theorem 4.1. The fields Q and Frac lim ← − Z/pnZ , with their respective p-adic absolute values, are isometrically isomorphic as fields.
CONSTRUCTIONS OF THE p-ADIC NUMBERS 5 Proof. Write F = Frac lim ← − Z/pnZ for convenience. We will construct an isomor-phism ϕ: Q →F by systematically choosing a distinguished representative from each element of Q.
Let x ∈Q −{0}, then the lemma in the second section showed that x can be represented by a Cauchy sequence of the form {pnki}∞ i=1 for n, ki ∈Z, and where p ∤ki for all i ≥1. Since {pnki}∞ i=1 is a Cauchy sequence, we can find a subsequence {pnkN1, . . . , pnkN2, . . . , pnkNi, . . .} satisfying |kNi −kNj|p ≤p−i, for all j ≥i.
Set yi = kNi and notice that (∗) yj ≡yi (mod pi), j ≥i for all i ≥1. This assignment x 7→{pnyi}i is not unique, but it is well-behaved modulo powers of p in the following sense. If {pny′ i}i is another equivalent Cauchy sequence satisfying the property (∗) above, then for each i ≥1, yj −y′ j ≡yi −y′ i (mod pi), j ≥i, but, by their equivalence, the left hand side goes to 0, i.e., pi eventually divides the left side for all j > N for some N. This means that y′ i ≡yi (mod pi). We can make the choice unique by choosing yi ∈[1, . . . , pi −1].
Now we can define the isomorphism as ϕ(x) = pn([y1], [y2], . . . , [yi], . . .) where {pnyi}i ∈x is the distinguished representative discussed above and ϕ(0) = 0.
The map is well-defined since the yi satisfy (∗). The important thing to notice is that ([y1], [y2], . . .) is a unit since p ∤yi for i ≥1. The map is clearly surjective by fact that every element of F can be written pnu where u is invertible (shown in the last lemma of section three) and if u = ([u1], [u2], . . .), ui ∈Z, then {ui}i is a Cauchy sequence.
To show that it is injective, suppose ϕ(x) = 0, then [yn+1] = 0, [yn+2] = 0, . . . , which implies that x = 0. It remains to show that ϕ preserves the absolute value, but indeed if |x|p = p−n then |ϕ(x)|p = |pn([y1], . . .)|p = p−n since ([y1], . . .) is a unit. It follows that Q ∼ = F isometrically.
□ Department of Mathematics, Rutgers, The State University Of New Jersey, Piscat-away, NJ 08854 E-mail address: aginory@math.rutgers.edu URL: www.math.rutgers.edu/∼ag930 |
16085 | https://math.stackexchange.com/questions/4957450/how-to-combine-per-capita-data-across-multiple-years-for-the-same-region | statistics - How to combine per capita data across multiple years for the same region? - Mathematics Stack Exchange
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How to combine per capita data across multiple years for the same region?
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I have a decade's worth of yearly per capita data of various datapoints "per 1 million people" for several regions, and I want to get the per capita rate over the whole time period.
I understand that merging per capita data for different regions is wrong because you lose the weight of each dataset; i.e. you can't just add/average/etc. a per capita rate for California with the per capita rate for Wyoming, since California's population is an order of manitude higher, so it's not equivalent.
However, if I'm merely combining consecutive years of the same region, wherein each year's per capita rate was calculated using that year's population, is it valid to simply add the per capita values for all the years together, to get the per capita rate over the whole time period, even though the population is changing slightly from year to year?
For example, if I have some data point for a region whose yearly occurrences per 1 million people is: [1.1, 1.2, 1.3], is it accurate to say that the per capita rate over that 3 year period is 3.6 per million people?
Or, would I have to calculate the raw sum of occurrences, and then divide by the average population per year, or the median, or something?
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asked Aug 12, 2024 at 3:28
YurelleYurelle
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1 The same problem also exists when you obtain your data points: the population at which point of the year? If the population can change every month, how did you obtain your yearly occurrences per million people?peterwhy –peterwhy 2024-08-12 06:02:47 +00:00 Commented Aug 12, 2024 at 6:02
CDC projections based on the US Census. 1 record per year, per state.Yurelle –Yurelle 2024-08-12 06:04:18 +00:00 Commented Aug 12, 2024 at 6:04
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No. You have to be careful when dealing with averages because the denominator can change. Taking your example, when the occurrences per million people were [1.1,1.2,1.3][1.1,1.2,1.3] it would be correct to say the occurrences per million people over those three years were 3.6 3.6 if you had the same people in the region for all three years. On the other hand, suppose there were 10 10 million people in the region for the first two years and 1 1 billion people for the third year. The number of occurrences would be 11+12+1300=1323 11+12+1300=1323 and the number of people-years would be 1020 1020 million, so there are about 1.3 1.3 occurrences per million people-years. Multiplying by 3 3 you would have about 3.9 3.9 occurrences per million people in 3 3 years.
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answered Aug 12, 2024 at 4:30
Ross MillikanRoss Millikan
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Ok, that makes sense why I can't just add them together, but I don't quite understand what you're saying the correct math is. Are you saying to add the populations of each year together, then divide the raw total of occurrences over the same number of years by that mega-population? So, if the population for each year is: [11 million, 12 million, 13 million] for a total of 36 million, and the raw number of occurrences for those years are: [21, 22, 23] for a total of 66 occurrences; I divide 66 by 36 Million, then scale up by the per capita rate? So, 1.83 per 1 Million people?Yurelle –Yurelle 2024-08-12 06:13:13 +00:00 Commented Aug 12, 2024 at 6:13
1 You need to be clear what time period is involved. If you want occurrences per million people-years you total the occurrences, total the people and divide. If you want occurrences per million people sometime in the three years you average the number of people. This will be 3 3 times as high.Ross Millikan –Ross Millikan 2024-08-12 13:15:55 +00:00 Commented Aug 12, 2024 at 13:15
I don't understand. What's the difference is between "per million people-years" and "per million people sometime in the three years"? I'm not a math guy, so maybe I don't have the baseline statistics knowledge that you're assuming I do. :-)Yurelle –Yurelle 2024-08-12 18:59:01 +00:00 Commented Aug 12, 2024 at 18:59
1 If you have a million people for three years you have three million people-years. If you have one occurrence per million people-years you will have three occurrences. If you have one occurrence per million people sometime in the three years you will have one occurrence.Ross Millikan –Ross Millikan 2024-08-12 19:38:30 +00:00 Commented Aug 12, 2024 at 19:38
OH! That makes sense. Thank you, sir.Yurelle –Yurelle 2024-08-12 21:38:54 +00:00 Commented Aug 12, 2024 at 21:38
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16086 | https://innovint.com/wp-content/uploads/2024/05/Four-Minute-Intervals-The-Art-of-Problem-Solving-041613-1.pdf | 1 | P a g e FOUR-MINUTE INTERVALS THE ART OF PROBLEM SOLVING William A. Guillory, Ph.D. Innovations International, Inc. Salt Lake City, UT 84117 INTRODUCTION This approach to problem-solving is derived from the recent practice in coaching when a sports team is behind in points—particularly basketball. That is, to close the deficit by using four-minute intervals of stepwise success. The process requires refocusing from solving a “whole problem” to breaking down the process into four-minute intervals. These intervals include: 1. Defining a problem 2. Defining the source of a problem 3. Defining your role in the problem 4. Defining if the problem is cognitive, behavioral, or procedural 5. Defining a solution or resolution 6. Testing the permanency of the solution 7. Adoption of a personal methodology When you master this process, any problem you have can be solved in less than a half hour 1. DEFINING A PROBLEM—THE FIRST FOUR MINUTES A problem is an ongoing emotionally stressful pattern of incidents that is unfulfilling to an individual’s stated intention. For example, if you are evaluated as an “average performer” by your manager, and you feel you are outstanding, this disconnect is a problem. If you experience ongoing conflict in a relationship that is of value to you, the ongoing conflict is a problem. If an organization experiences a high level of turnover of 2 | P a g e its valued employees, relative to its industry standard, then the excessive loss of employees is a problem. The first observation we make is that a problem is ongoing with a variety of ways and reasons for its occurrence. In the case of the “average performer,” the evaluation represents a series of missed performance expectations, or at least a disconnect, between employee and manager. In the case of a conflicting valued relationship, we anticipate again a series of mutual missed expectations, emotional disappointment, or even conflicting verbal exchanges. And in the final example, a high level of turnover not only creates severe financial loss, but consequences that negatively impact internal morale, recruitment, and customer service. This series of events are indicators that a problem exists. We want to emphasize that the objective of this publication is not about value judgments such as “good or bad,” “right or wrong,” or even “fair or unfair.” It is simply to address an unfulfilling result with respect to an individual’s or an organization’s stated intention. In essence, a problem is present when you identify a series of unfulfilling incidents with respect to your physical, mental, emotional, and even spiritual well-being. From an organizational perspective, typical indicators of a problem are difficulty in hiring the best people, retaining the best people, poor individual and collective performance, average management and leadership, and lack of sufficient creativity and innovation to remain competitive or sustainable. The most obvious indicator of an organizational problem is loss of profitability. The consequences of the latter are obvious. The flipside of this situation is the assumption that if an organization is profitable, it does not have any significant problems. This is not necessarily accurate, but is the subject of another discussion. Action: List the series of indicators of the problem. “Where, there’s smoke; there’s fire.” 2. DEFINING THE SOURCE OF A PROBLEM—THE SECOND FOUR MINUTES The source of an ongoing problem is usually not readily recognized. It is usually subliminal, not something of which we are consciously aware. This is the most challenging four-minute interval of them all. Most of all, we commonly assume the source of a problem we are experiencing is external to us—even though our bodies are experiencing the physical, mental, emotional, and even spiritual distress. For example, I used to be Chairman of a Chemistry Department. In the initial stages of being chairman, I experienced a series of “conflicts” with a variety of faculty members. Then an advisor asked me a very simple question: “Bill, what do all of your conflicts have in common?” I thought for a minute and then came up with a brilliant response, “They are all faculty members!” “No,” he said, “You!” “What?” I replied. “You,” he repeated. It took three “Yous” for me to get it! However, that “aha” moment has always stayed with me. That’s 3 | P a g e an example of a transformational experience. The result is a realization. An awareness. A learning. An expansion of one’s consciousness or reality. More often than not, the source of a problem is a belief, perception, or even a value that is reflected as an expectation of others. For example, if we believe someone is inherently inferior by “group identification” with another culture, then our assignments, support, and evaluation of that individual will mimic our expectations. And the evaluation will be “average performance.” If we experience an occasional, but repeated conflict in a valued relationship, an unacknowledged disconnect involving a belief or perception is probably the source of the problem. Examples of such disconnects may include beliefs about control, trust, jealousy, money, religion, honesty, or commitment. It should be recognized that the lack of awareness that such a disconnect exists limits the level of honesty, trust, or intimacy available in a personal relationship. Most organizations have a history of beliefs and values that span many years. These beliefs and values are so inculcated into the culture that they are practically invisible. But they are reflected in the overall results the organization produces. The most common examples in Western-oriented organizations are the dominance of Eurocentric values—such as individualism, competition, task-orientation, and systems-thinking—in preference to non-Eurocentric values that represent the value, functioning, and well-being of people—such as inclusion, collaboration, empowerment, and work-life balance. The predictable results of using a one-dimensional system of operation with a multidimensional workforce are preferential advancement, limited engagement and performance, and systemic exclusion. A model that explains the source of a problem is the State of Mind diagram shown below: STATE OF MIND DIAGRAM Mind-set Attitude Beliefs Behaviors Processes Procedures Results First-Order Change (Behavioral Modification) Second-Order Change (Personal Transformation) Performance Functional Skills Cognitive Skills 4 | P a g e The source of most problems is programmed in an individual’s or an organization’s mind-set. This belief, perception, or value drives one’s behavior—overt or implicit—to produce the counterproductive results we observe. Action: Identify the source of a problem by reflecting on counterproductive perceptions and perceptions you may have of others, such as unconscious biases or characteristics that upset you most about others: “That which I dislike in others is a mirror-reflection of myself.” 3. DEFINING YOUR ROLE IN THE PROBLEM—THE THIRD FOUR MINUTES A basic assumption of this four-minute interval is that if your body or mind tells you it is experiencing discomfort or challenge, then the source of the problem you are experiencing is programmed within you! Sorry to be so direct, but we only have four minutes! This assumption is reflected by the Dynamic Relationship Model shown below for Person 1 and Person 2. DYNAMIC RELATIONSHIP MODEL If the underlying belief, perception, or value is programmed in your mind-set that and is both unconscious and unaware, then this “driving force” dictates your behavioral role in a problem. If we truly intend to solve a problem, this is where we need to focus our attention. P1 ← → P2 Manager Employee Underlying Issue Conscious Awareness Unconscious Unawareness 5 | P a g e In some cases Person 2 may have an underlying belief that is held in common by a number of employees. Such a belief is described as institutionalized. Examples include, teamwork is valued, but individual performance is rewarded; being aggressive is a good sign of leadership, for men, not women; information is power, reveal only what is necessary to others. Action: In this four-minute period, the critical decision that needs to be made is to take 100% responsibility for P1 or P2 and begin to explore, what “unreasonable” steps you need to take to ensure your performance is unquestionably outstanding—whether fair or unfair. In essence, what behaviors would have the power to invalidate your self-limiting belief? A similar approach would also be used for the other two situations of a personal relationship and organizational performance. The emphasis of this approach is a focus on problem resolution rather than the unfairness of the other person’s (or persons’) belief, perception, or values. In other words, “What actions must be taken, in spite of an unfair situation?” In essence, simply examine your repeated behavioral pattern (fight or flight) to a recurring incident. That behavior defines your role in maintaining the problem. Action: Focus on your counterproductive behaviors that sabotage your success: that defines your role in maintaining the problem. Then design a positive, proactive behavioral pattern that ensures your success. 4. DEFINING IF THE PROBLEM IS COGNITIVE, BEHAVIORAL, OR PROCEDURAL—THE FOURTH FOUR MINUTES If we make reference to the State of Mind Diagram above in Section 2, we notice above the Mind-Set box is Cognitive Skills and above Behaviors is Functional Skills. These classifications allow us to determine if we are dealing with beliefs, attitudes, or values (cognitive) or a behavior, process, or procedure (functional) in solving a problem. Obviously, functional skills are easier to “fix.” We simply need to deploy a new or improved behavior, process, or procedure. Examples include, mentoring, learning new ISO quality procedures, or conducting a meeting online. Identifying a cognitive perception that is counterproductive involves self-reflection, self-introspection, and self-exploration. It involves making a connection between undesirable results that are produced and the type of thinking that would produce that result. If I manage someone who I think is capable and yet his or her performance level is continually average. I might begin by asking myself, “How am I unsupportive of that individual’s growth and development?” “What project can I assign that would ‘push’ that individual to the next level of performance?” and finally, “How can I hold the individual responsible and accountable for producing at an outstanding level of performance?” When we take responsibility for assigning roles at an exceptional level for ourselves and others, it creates behaviors that transform our way of thinking—or more precisely, they 6 | P a g e invalidate, by experience, self-limitations. The real highly functioning individual who was “always” there begins to emerge. Action: First try a behavioral or procedural approach. If the problem sequence disappears, then there is no cognitive actin necessary. I the problem sequence persists, then use the next section for in-depth problem solving. We must honest with you, most persistent problems are cognitive. "Become an astronaut of inner space.” 5. DEFINING A SOLUTION OR RESOLUTION—THE FIFTH FOUR MINUTES The ending of the previous section initiated the discussion of problem solution or resolution. The distinction we make between a solution and a resolution is that a solution is a situation where we have 100% control of the outcome. Resolution refers to a situation where we experience personal transformation (detachment from a reactive, emotional response) and the situation outcome is not 100% in our control. For example, if I transform my potential for performance by adopting a mind-set of 100% responsibility and 100% accountability as ways of operating, then I have 100% control of my level of performance. If, on the other hand, I transform and dissociate myself from a conflicting pattern with someone in my life, I cannot control what the other person might decide to do, as a result of my change. However, whatever the ultimate resolution is, you will be able to handle or adapt to it with the least amount of emotional upheaval. The direct solution of personal transformation involves a three-step process: 1. Be willing to tell the truth about my own counterproductive belief, perception, or value. (This is the most difficult step.) 2. Create non-comfort zone behaviors that have the power to invalidate the counterproductive belief, perception, or value. 3. Put the behaviors into practice for four to five weeks until they become a “natural behavioral pattern.” The result is the acquisition of new skill. An example of a series of behaviors that have the potential to create personal transformation for the individual discussed in Section 1, in a managerial or supervisory role, is the following: i. Focus on the individual’s contribution, rather than your perception, and acknowledge the source where incorporated. ii. Commit to assigning the individual challenging, visible projects with coaching and mentoring by you. 7 | P a g e iii. Ensure the individual is part of succession planning and include them in leadership roles. iv. Ensure the individual has an approved career plan, the required skills for success, and a mentor/coach to support his or her success. In a like manner, according to the P1↔P2 diagram on page 4, the series of transformational behaviors for the underperforming individual includes the following: i. Adopt an empowered mind-set (100% responsibility and 100% accountability) ii. Ensure your performance is exceptional by going the extra mile on assigned projects; get feedback from your harshest critic. iii. Volunteer for visible opportunities and/or leadership roles; be insistent. iv. Create a career plan and acquire a mentor to oversee your progress. v. Document your career successes. Action: The key element solving or resolving a problem is to first take the appropriate action to transformation your own counterproductive pattern of thinking and behaving. Second is to deal with the other individual’s willingness to change, adapt to your change, adapt to his or her unwillingness to change, or sever your working relationship for resolution. The most powerful aspect of your transformation is that you are free to make choices in your best interest; where choice did not previously appear to exist. When resolving a situation involving an institutional belief held by the organization, create a challenging objective in the Results box of the model on page 3. Examples include an ambitious goal in terms of workforce representation in leadership, creation of an inclusive culture within an ambitious timeframe, or the creation of a transparent culture or professional relationship. Such goals will require new, ambitious, and personally-confronting behaviors that challenge the presently existing status quo. When implemented, with conviction and commitment, the mind-set of the organization transforms! The commitment of the CEO/President and leadership is critical for success. 6. TESTING THE PERMANCY OF THE SOULTION—THE SIXTH FOUR MINUTES The question in this four-minute interval is “Did the behaviors implemented in the previous discussion result in the permanent invalidation of the underlying issue in the model on page 3? Or did the culture transform by a comprehensive measurement instrument comprised of “significant” questions? (All survey questions are not created equal in terms of the conceptual and behavioral framework necessary for high 8 | P a g e performance) If so, then personal transformation and organizational transformation occurred. If you are still “triggered” by the same issue in challenging situations, then more work needs to be done—preferable with facilitative coaching. Facilitative coaching is the process of having someone assists you through an “irreversible change” in mind-set with respect to a self-limiting belief. Fortunately for sports teams, they get to practice new challenging behaviors every day with coaching provided. In the case of cultural transformation, most measurement instruments will reveal whether transformation occurred or not. In spite of the fact that instruments are only semi-quantitative, those of excellent design do a remarkable job of tracking an organization’s culture. If the issue has been resolved, a significantly expanded level of performance occurs as a result of a mind-set of success. It is also important to recognize that a series of stressful incidents have been eliminated, all resulting from the same underlying issue in different interactions or situations. Action: Use observations of behaviors, desired organizational outcomes, and organizational performance, engagement, and inclusion survey measurements to indicate the permanency of a solution. 7. ADOPTION OF A PERSONAL METHODOLOGY—THE SEVENTH FOUR MINUTES After practicing this sequence for problem solving, you will note that two or three steps might be skipped. Or, you may have friends or co-workers who can coach you through some of these questions. Whatever works best is the key. These steps should primarily be used as a guideline for the important checkpoints you want to make sure you cover. For example, we believe that defining the problem and accepting full responsibility are the two key elements in the entire process. It is not unusual to seek informal coaching until these two are mastered. We have also found that “direct confrontation” with a skillful coach commonly leads to a direct transformational experience—cognitive transformation. The most satisfying aspect of the process is that it can be applied to any problem you might want to solve or resolve in the future. Action: Design your personal process for problem solving, ensuring that you have included all six of the previous steps. 9 | P a g e SUMMARY This Four-Minute Intervals problem-solving method can be used for the majority of conflict situations you deal with—both in the workplace and in your personal lives. After mastering this technique, most of those situations can be handled in less than a half-hour. We highly recommend using this method for any undesirable situation you are presently dealing with and discover how effectively the Four-Minute Intervals works! |
16087 | https://econrsa.org/wp-content/uploads/2022/06/working_paper_345.pdf | Economic Research Southern Africa (ERSA) is a research programme funded by the National Treasury of South Africa. The views expressed are those of the author(s) and do not necessarily represent those of the funder, ERSA or the author’s affiliated institution(s). ERSA shall not be liable to any person for inaccurate information or opinions contained herein. Concentration measures as an element in testing the structure-conduct-performance paradigm Johann du Pisanie ERSA working paper 345 April 2013 Concentration measures as an element in testing the structure-conduct-performance paradigm Johann du Pisanie∗ April 23, 2013 Abstract The original structure-conduct-performance (SCP) paradigm, accord-ing to which market structure determines market conduct and market conduct determines market performance, underlies numerous competition policies. Since its development almost a century ago, the paradigm has been heavily criticised and numerous efforts have been made to test it by correlating measures of seller concentration with measures of market performance. The reliability of seller concentration measures that are fre-quently used, particularly in South Africa, was tested against the Hannah and Kay criteria, using hypothetical numbers of sellers and market shares.
The premise is that a concentration measure must be reliable in the sense that it should lead to a correct conclusion when the relevant concentra-tion curves do NOT cross. The following absolute concentration measures were found to meet the criteria: the Herfindahl-Hirschman index (HHI), the other Hannah and Kay indices [HKI(α)], the Rosenbluth index (RI), the numbers equivalent of the Hannah and Kay indices [HKIne(α)] and the entropy coefficient (EC). The discrete measures, concentration ratios (CRX) and the occupancy count (CRX%), do not always meet the crite-ria, nor do the relative concentration measures or measures of inequality, namely the Gini coefficient (GC), the variance of logarithms of market shares (VL) and the relative entropy coefficient (REC). The Horvath in-dex (HI), an absolute concentration measure, does not always meet the criteria. Studies that employed the unreliable measures should be disre-garded or reworked and students should be forewarned against the use of such measures.
Most competition policies are to some extent based on the original structure-conduct-performance (SCP) paradigm, according to which mar-ket structure determines market conduct and market conduct determines market performance. Since the development of the paradigm almost a cen-tury ago, numerous efforts have been made to test it by correlating mea-sures of seller concentration with measures of market performance. This paper shows that some of the seller concentration measures are unreliable and that research based on them should be disregarded or reworked.
∗J A du Pisanie, Emeritus Professor, Department of Economics, University of South Africa (Unisa).
1 1 THE ORIGINAL SCP PARADIGM The original structure-conduct-performance (SCP) paradigm follows from neo-classical microeconomic theory and, more specifically, the comparison between monopoly and perfect competition found in almost every introductory and in-termediate textbook in use in mainstream microeconomics courses. The basic postulate of the original SCP approach is that market structure determines market conduct, which in turn determines market performance.
Market structure consists of the relatively permanent (slow-changing) con-ditions within which the sellers operate; most importantly seller concentration (the number and size distribution of the sellers), barriers to entry to the market and the homogeneity of the sellers’ products.
Market conduct can be described as the efforts of suppliers to market their products and to limit competition between them — that is, marketing con-duct and competition-limiting conduct. Marketing conduct is the continuous formulation, implementation and reformulation of policies on pricing, product characteristics, marketing communication (mostly advertising) and distribution.
Competition-limiting conduct mainly consists of merging or collusion with com-petitors, coercion of competitors or would-be competitors, and the use or misuse of political participation to limit competition through government action.
Market performance is the degree to which the suppliers to a market con-tribute to the economic goals of society, such as efficiency in the use of resources, full employment of resources, equity in the distribution of income and wealth, stability in the general price level and progressiveness, including innovation (the introduction of new products and processes).
According to the original SCP paradigm, a market structure characterised by low seller concentration, a homogeneous product and free entry to and exit from the market (approaching the neoclassical model of perfect competition) leave firms little choice of market conduct. They are price takers, determine their output quantities by setting marginal cost equal to price, must produce efficiently and make only normal profits in the long run. By contrast, a market structure typified by high seller concentration, products that differ appreciably from those of competitors and limited entry to the market (approaching the monopoly model) leave the firm or firms in the market a greater choice of mar-ket conduct. Such a firm is regarded as a price maker who chooses the quantity supplied by equating marginal cost with marginal revenue and sets the corre-sponding market price. The price exceeds marginal cost and the firm may make abnormal profits in the long run, need not maximise profit and can choose not to produce efficiently (within limits). It is deduced that market performance in concentrated markets will be worse than in unconcentrated markets. The price will be higher, fewer units of the product will be supplied, allocative and possi-bly internal efficiency will be lower, buyers will have a limited choice of sources of the product, service is likely to be worse, and innovation will probably be low.
There are many criticisms of the original SCP paradigm, which have led to some adjustments. For example, some authors have added feedback effects to 2 the unidirectional flow of cause and effect from market structure to conduct to performance (Ferguson & Ferguson 1994:18-19) or even argue that the direction of causation runs from market performance to conduct to structure (Reekie 2000:36,67).
2 THE V ALIDITY OF THE ORIGINAL SCP PARADIGM The validity of the original SCP paradigm has been questioned, particularly since the 1970s (Shepherd 1997:6-7). These criticisms seem to be gaining mo-mentum.
There are two broad issues that affect the relevance or validity of the SCP paradigm, namely conceptual problems and the difficulty of empirically testing the paradigm.
2.1 Conceptual issues As indicated in section 1, the SCP paradigm is based on neoclassical microeco-nomic theory, in particular the comparison of perfect competition and monopoly.
However, neoclassical theories of oligopoly often fail to produce clear-cut con-clusions about the relationships between market structure, conduct and perfor-mance (Lipczynski et al 2009:16). For example, it is not clear whether a more concentrated oligopoly would result in worse market performance than a less concentrated oligopoly.
It is often difficult to decide if a variable belongs to market structure, conduct or performance (Lipczynski et al 2009:16). For example, "product differentia-tion" may refer to a condition existing in a market and then it would be an element of structure. However, it may also refer to the act of differentiating one’s product from those of competitors and then it would be a form of market conduct. When such a term is encountered, care must be taken to determine its meaning in the relevant context.
The original SCP approach has been criticised for being static in nature, instead of explaining the evolution of structure, conduct and performance over time (Lipczynski et al 2009:17). For example, the existing market structure is assumed to be given and market conduct and performance are deduced from that, while the market structure might be changing and the changes might lead to unexpected forms of market conduct and performance.
The original SCP approach in effect postulated what has become known as the "collusion hypothesis", according to which a positive relationship between seller concentration and the profitability of the firms in the market is interpreted as evidence that the firms are colluding or abusing market power in some other way to increase profits. Later, the Chicago school postulated the so-called "effi-ciency hypothesis", which states that a positive correlation between seller con-centration and profitability reflects that efficient firms are more profitable than 3 their rivals and eventually dominate their markets, meaning that both seller concentration and profitability would be high (Lipczynski et al 2009:17).
In addition, it is quite plausible that intelligent business people could decide for themselves that price-cutting is likely to lead to retaliation and consequently no appreciable increase in turnover — without discussing the matter with com-petitors. In other words, businesses could keep prices high, turnover low and profits high without collusion. The mere fact that prices and profits are high is no proof of collusion.
Furthermore, even if price fixing does occur in an industry, it might bring benefits to consumers. If price competition is excluded, competition may take other forms and result in high quality of goods, good service and innovation, which might not have occurred otherwise. In such cases, it could be argued that the collusion would not lead to a decline in societal welfare, i.e. to bad market performance.
The conclusion is that competition authorities should not rely on general-isations, but should judge each market on its own merits, based on the facts pertaining to that market.
2.2 Empirical tests Empirical tests of the SCP paradigm consist of measuring market structure, market conduct and market performance and trying to find statistically signif-icant relationships between them. The original SCP paradigm has not been proven or disproven, owing to problems with the definition of markets, the lim-itations of the statistical measures employed, data deficiencies, the weakness of statistical relationships found, and biased interpretation of the results.
Definition of markets It is often difficult to define the relevant market. If one considers a particular firm, its "direct" competitors have to be identified.
This group of firms may then be considered to form a market. According to neo-classical theory two firms are in the same market when the cross-price elasticity of demand for their products is "high". However, measuring the cross-price elasticity of demand is often difficult in practice, while choosing a threshold between "high" and "low" cross-price elasticity is an arbitrary act.
Measuring market structure Although market structure has a number of dimensions, in empirical work it has often been measured only by seller concen-tration. Suppose concentration is measured in two different markets using the same measure and the same numerical answer is found. If entry to one of the markets is difficult, but easy in the other case, it is reasonable to expect that prices and profits would be higher (relative to cost) in the market that is difficult to enter. Seller concentration and ease of entry should therefore be measured simultaneously to achieve meaningful results. By the same token, the degree of product differentiation should also be measured. The problem is that these other variables, apart from seller concentration, are more difficult to measure numerically.
4 Secondly, there are many measures of seller concentration and they yield different results. This is the main focus of this paper, considered in section 3.
Measuring market conduct It is clear from the definition of market conduct that this concept is very difficult to measure quantitatively.
Measuring market performance Market performance is the extent to which the suppliers to a market contribute to social welfare and can theoretically be measured by profits or price (Lipczynski et al 2009:283-289), among other things. Most people would intuitively associate high prices with bad market per-formance, but may not immediately realise that the SCP paradigm associates good profitability with bad market performance. This association results from the neoclassical view that monopolists may earn economic or abnormal prof-its in the long term, while perfect competitors cannot. The business manager would associate good profitability with good performance by the FIRM, but firm performance is not the same as MARKET performance, which focuses on the welfare of the society, not that of the firm. The Lerner index (Lipczynski et al 2009:62) may be used to measure market performance — the higher the index, the worse the market performance — but it is difficult to apply in practice. The price-cost margin (Lipczynski et al 2009:286) may be used instead.
Data deficiencies Apart from the problematic characteristics of some mea-sures, researchers have encountered data problems when trying to apply such measures, including those that are not inherently deficient.
Official organizations, such as competition commissions and courts, have the authority to demand information from firms in a market being investigated and can apply any measure of seller concentration. Other researchers often have to rely on statistics published by official compilers of statistics, for example, Statistics South Africa (Stats SA). Such statistics are subject to limitations, including: • Secrecy provisions in legislation preclude official organizations from pub-lishing information on individual firms; therefore they publish the infor-mation for groups of firms. Concentration measures that require the cal-culation of the market shares of individual firms cannot be calculated from such information. The organizations themselves may calculate and publish such concentration measures, but this is seldom done and the measures are subject to the limitations mentioned below.
• Official statistics are published in accordance with schemes for industry classification, such as the UK’s successive Standard Industrial Classifi-cations (SICs).
Stats SA and its predecessors devised a similar series of SICs based on the United Nations’ International Standard Industrial Classification (ISIC). These classifications rely on criteria that are seldom appropriate for identifying relevant markets where all sellers are in direct competition with one another. Another problem is that the information 5 is published for "establishments", not firms. In South Africa, this term could indicate a branch plant (e.g. a factory or shop) of a multiplant firm.
In other words, a firm with five factories or shops would be counted as five entities, whereas the firm, and not the establishment, is the decision maker of economic theory. The five units are managed by a single decision maker and therefore constitute a single competitor. Basing concentration measures on establishment data understates the degree of concentration.
Because of this problem, Stats SA has published some data for firms in-stead of establishments. These data are still subject to the problem that every registered company is regarded as a separate firm, whereas many are subsidiaries of other companies and are in effect branches of the holding or controlling company and are not decision makers in their own right.
Consequently, even measures of concentration based on the Stats SA data for firms would understate the degree of concentration.
• Establishments and firms often produce several types of products that are sold in different markets. Such entities are classified in SICs on the basis of their most important product and their total turnover is shown under that heading. A firm that produces products A and B might be classified in industry A and its total sales of A and B would then be published as if it pertained only to A. Sales of A would be overstated and sales of B understated.
• Turnover figures in official reports on, for example, manufacturing usually only cover sales by domestic producers, whereas imports may constitute an important source of supply to the relevant market and thus influence the nature of competition in that market. Similarly, a portion of a domestic producer’s production might be exported and this should be taken into account.
Relating market structure, conduct and performance to one another Numerous studies test simple correlations between some measure of market structure and some measure of market performance.
Partly because of the difficulty of measuring market conduct, no effort is made to test the intervening relationships between market structure and market conduct, or between market conduct and market performance. Furthermore, no effort is made to exclude the possible effects of variables other than market structure on the employed measure of market performance, say profits. Many variables influence profits, for example, GDP, employment, inflation, interest rates, exchange rates and tax.
Particularly when studying trends in seller concentration and performance, these variables should, for example, be included in multiple regressions, so that their influence may be excluded when determining the relationship between market structure and market performance. Some studies merely measure trends in seller concentration, suggesting that increases in seller concentration have negative consequences for society (without trying to correlate seller concentration with some measure of market performance).
6 Weak statistical relationships Numerous studies, which aim at determin-ing the relationship between market structure and profit, have been done in the US and the UK (Ferguson & Ferguson 1994:95-96). Where positive correlations were found, such correlations have hardly been statistically significant.
Interpreting results If a study finds no correlation between market struc-ture and market performance, opponents of the SCP paradigm may claim that the finding disproves the paradigm.
However, protagonists of the paradigm might claim that this is not the case, since there are problems with the market definition, the measures, the data or the statistical techniques employed.
Suppose a study does find a statistically significant relationship between a measure of market structure and one of market performance. Protagonists of the paradigm would probably claim that the study proves the paradigm. How-ever, even if possible disputes about the definition of the market, the measures, the data and the statistical techniques were ignored, the study would not nec-essarily prove the paradigm, because the direction of causality might run from performance to structure (Lipczynski et al 2009:287). The possibility of feed-back effects inhibits the empirical testing of the original paradigm, since the statistical tests do not indicate the direction of causation.
3 MEASURES OF SELLER CONCENTRATION The plethora of problems involved in empirical testing of the SCP paradigm may lead to the conclusion that all efforts at such testing should be abandoned.
However, many researchers persist in doing such empirical work and it is im-portant to evaluate the measures they use. In this paper, the spotlight falls on measures of seller concentration. Some measures increase when concentration rises — or are supposed to do so — while others decline — or are expected to decline — when concentration increases. The second category of concentration measures is sometimes called "inverse" measures. For want of a better term, the first category is referred to as "positive" measures in this paper.
For the purposes of this paper, concentration measures discussed in text-books (Lipczynski et al 2009:195-206, Ferguson & Ferguson 1994:39-43) plus additional ones used in South African literature (Leach 1992, Fourie 1996, Fed-derke & Szalontai 2009) are tested by means of hypothetical data. These tests are reported on in section 4. The measures are defined in the rest of this section.
3.1 Positive measures of concentration 3.1.1 X-firm concentration ratio Possibly the best known measures of seller concentration are the x-firm con-centration ratio and the Herfindahl-Hirschman index. The x-firm concentration ratio measures the market share of the largest x sellers in a market and is ab-7 breviated CRX, where X can be 3, 4, 5 or another number: CRX = n i=1 S2 i (1) where Si is the market share of the ith seller.
A variant of CRX is CRX%, the cumulative market share of the top X% of the sellers in a market (Fourie 1996:100). When there are small numbers of sellers, say 3, it might be difficult to determine, for example, CR5%. The top firm then represents 33% of the sellers. One encounters a similar problem in the case of CRX, when there are fewer than X sellers in a market.
3.1.2 Herfindahl-Hirschman index The Herfindahl-Hirschman index, abbreviated HHI, is the sum of the squared market shares of all the sellers in a market, where N is the total number of sellers: HHI = N i=1 S2 i (2) CRX is a so-called "discrete" measure, while HHI is classified as a "sum-mary" measure. The difference can be explained by means of a concentration curve.
To construct a concentration curve, the firms are ranked from largest to smallest, in terms of their market shares. The rank of each firm is plotted on the horizontal axis of a box diagram, up to N. The cumulative market shares are plotted on the vertical axis, the maximum value being 100% or 1 respectively, if the market shares are expressed as percentages or as ratios to 1. An example is given in Diagram 1, using ratios to 1. Hypothetical market shares are used, with N = 5.
CRX is known as a discrete measure because it represents only one point on the concentration curve (Marfels 1975:486). For example, if X=3, the combined market share of the largest three sellers is 0.85 and CR3 represents only point (3, 0.85) on the concentration curve in Diagram 1. To calculate CR3 at this point, one needs only the cumulative sales of the three largest sellers and the total sales of the N sellers involved. By contrast, HHI utilizes all the information represented by the concentration curve and is for this reason called a summary measure (Marfels 1975:488). The market share of each and every firm in the market is squared and added up to calculate the value of the index.
Since market shares are expressed as ratios to 1 in the formulas for most of the seller concentration measures, it is advantageous to use such ratios (instead of percentages) in graphs of concentration curves. Another advantage is that the surface area of the rectangle between (0, 0) and (N, 1) always equals N (base × height = N × 1). The larger N, the longer the concentration curve and the larger the area of the rectangle would be. Since a larger number of sellers is associated with more competition, an increase in the length of the concentration 8 curve indicates lower seller concentration. Conversely, a shorter concentration curve denotes fewer sellers and thus higher seller concentration.
The concentration curve also provides a visual depiction of the other ele-ment of seller concentration, namely the size distribution of the sellers. If the market shares of the sellers were equal, the concentration curve would coincide with the diagonal line from (0, 0) to (N, 1). The more unequal the distribution, the higher the curve would rise above the diagonal line before ending at (N, 1).
Consequently, the inequality can be measured by the area between the diagonal line and the concentration curve. To paraphrase Marfels (1975:486), a higher and shorter concentration curve depicts a higher level of seller concentration than a lower and longer curve. However, when two concentration curves cross, it is harder to state categorically which one depicts a higher level of seller con-centration. In such cases, one would have to rely on some concentration measure to make a judgement.
The premise of this paper is that such a concentration measure must be reliable in the sense that it should lead to a correct conclusion when the concen-tration curves do NOT cross. The reliability of various concentration measures is tested by using hypothetical numbers of sellers and market shares (sec 4).
3.1.3 Hanna and Kay indices While there is no need to attach weights to the market shares used in the CRX, it is necessary to use such weights in the case of summary measures, since the market shares of all the firms always add up to one. Marfels (1975:488) pointed out that the HHI weighs the market share of each seller by itself. Therefore, the largest seller’s market share is multiplied by the highest weight and the smallest market share by the lowest weight. A shift among the market shares of large sellers has a greater impact on the index than a similar shift among the shares of small sellers in a given market.
Hanna and Kay (1977:56) indicate that the HHI is just one of a numerous array of indices that are the sums of the market shares weighed by the shares themselves, raised to some power. In the HHI each weight is the market share raised to the power of one. One could also use weights equal to the market shares raised to the power of 0.5, 1.5, or literally any other power. (For the moment, powers of zero or less are disregarded.) Such an index is denoted HKI(α), where HKI = "Hanna and Kay Index" and α = the power to which the market share is raised.
Table 1 shows an example of such indices, using the market shares of the five hypothetical firms in Diagram 1. If weights of S0.5 i are used, the weighted values of the market shares will be S1 i × S0.5 i = S1.5 i , and so forth. The higher the power to which Si is raised (α), the lower the weights and the corresponding indices would be.
To aid in interpreting the size of a concentration index, the maximum and minimum values that it can take must be known. The maximum value of all the HKIs, including the HHI, is 1 and occurs in the case of monopoly, as shown in Table 1. The minimum values occur when all the sellers in a market have equal 9 market shares and their precise values depend on the number of sellers, N. Each seller’s market share would equal 1/N. The Herfindahl-Hirschman index, for example, would equal (1/N)2 × N = 1/N. HKI(2.5) would equal (1/N)2.5 × N = (1/N)1.5. This is shown in Table 1, along with some examples, with N equal to 5, 10 and 100 respectively. The minimum values of HKI(1.5) equal 0.4472, 0.3162 and 0.1000 respectively, for N equal to 5, 10 and 100. This proves that the index declines as N increases; correctly indicating that seller concentration declines.
Following Adelman (1969), Hannah and Kay (1977:54) themselves prefer the so-called "numbers equivalent" of the indices. This is revisited in section 3.2.2.
3.1.4 The Horvath index The Horvath index (HI) is a hybrid index in which the market share of the largest seller is not weighed (or weighed by 1), but those of the other sellers are squared, as in the Herfindahl-Hirschman index (HHI). In addition, the square of each market share (except S1) is "reinforced" by the multiplier (2 - Si) (Horvath 1970:446; Marfels 1975:490,500): HI = S1 + N i=2 S2 i (2 −Si) (3) Horvath (1970: 446) originally wrote the multiplier "(1 + [1 −Si])" (though with a different symbol) and stated that the multiplier reflects "the proportional size of the rest of the industry". For all sellers except the largest one "the square of the fraction of the industry it does have times one plus the fraction of the industry which it does not have" is calculated.
The weight of the largest firm’s market share is 1 and those of the other firms are Si(2 −Si), as shown in Table 2. Si obviously declines as firms get smaller, but (2 −Si) increases. In fact, since Si < 1(i = 2, ..., N), (2 −Si) > Si.
The second element of the weight does not "reinforce" the first, but more than counterbalances it. Since (2 −Si) includes the market share of the largest firm, the contribution of the largest firm to HI is reduced further than would be the case if it had merely not been weighed (or weighed by 1).
The maximum value of HI is one and occurs in the case of monopoly: S1 = 1 and the rest of the formula falls away. The minimum value occurs when the sellers have equal market shares and is "a decimal fraction which is higher than the dominant firm’s absolute percentage share" (Horvath 1970:448). The decimal fraction equals [(3N2 −3N + 1)/N3] (Marfels 1975:500). The minima are shown in Table 2 for N equal to 5, 10 and 100 respectively. They decline as N increases, correctly indicating the direction of change in seller concentration.
3.1.5 Rosenbluth index It is not always easy to determine the weights used in other measures of concen-tration, for example, the Rosenbluth index (RI). This index is defined as follows 10 (Marfels 1975: 500): RI = 1/ 2 N i=1 iSi −1 (4) The sellers are ranked by market share from highest to lowest, as before.
The term iSi indicates that the market share of each seller is multiplied by the rank of the seller. Ignoring for the moment the rest of the formula, it means that the rank of the seller is the weight of its market share. The relevant calculations appear in Table 3, using the same numbers as in the top part of Tables 1 and 2. The market share of the largest seller is weighed by 1, that of the second seller by 2, etc, meaning that the weights increase as the market shares decline.
This procedure differs from that utilised in the HHI and other HKIs, where the weights decline as the market shares decrease. Whereas the HHI and other HKIs emphasize the role of the larger sellers, the RI emphasizes that of the smaller sellers. But this is not where the matter ends.
Table 3 shows that the sum of the market shares weighed by rank is 2.2.
To calculate the RI, this number is multiplied by 2, yielding 4.4. The answer is reduced by 1, yielding 3.4. Finally, the inverse of the last number is taken, meaning that the RI equals 0.2941 (rounded). It is no longer obvious what the final weights are. In particular, inverting the expression in square brackets means that the weights might now decline when the market shares increase, instead of increasing too.
The RI can be related to the graph of the concentration curve. The expres-sion iSi represents an area on the graph. If i = 1, 1S1 equals the area OABC (= 0.35) in Diagram 2. Likewise, 2S2 equals the area CDEF (= 0.30), etc. The sum of these areas forms the area OABDEGHJKMNP. If the triangles OAB, BDE, EGH, HJK and KMN are removed from this area, the area above the concentration curve (OBEHKNP) remains. The area of each of these triangles equals (1/2) × base × height = (1/2)(1)Si. Consequently, the area above the concentration curve can be written as: N i=1 iSi − N i=1 1 2 Si (5) And simplified to: N i=1 iSi −Si 2 = N i=1 Si i −1 2 (6) The market shares are no longer weighed by the ranks of the sellers, but by the rank minus 1/2. If the weights are multiplied by 2, they become 2(i −1/2) = (2i −1). The relevant figures in the case of Diagram 2 are shown in Table 3, where N i=1 Si(2i −1) = 3.4 (7) 11 The reciprocal of the expression on the left side of the equation above pro-vides an alternative formula for the Rosenbluth index: RI = 1/ N i=1 Si(2i −1) (8) In this case, RI equals 0.2941, as calculated before. While the weights of the market shares are (2i −1) in 1/RI, it is not clear what they are once RI is calculated. If the inverse of the individual components of 1/RI are calculated and added up, their sum is not equal to 0.2941. In general, it is difficult to determine the actual weights of individual market shares when the sum of the weighted shares are manipulated after it has been calculated, for example when it is inverted or expressed as a ratio to something else.
The maximum value of RI occurs when N = 1: RI = 1/1(2 × 1 −1) = 1/1 = 1 (9) The minimum value of RI occurs when the sellers are of equal size and equals 1/N, which is equal to the market share of anyone of the sellers. RI declines as N increases, correctly indicating that seller concentration declines.
3.1.6 Gini coefficient The Gini coefficient (GC) is a relative concentration measure, as opposed to the HKIs, HI and RI, which are absolute concentration measures. Relative concen-tration measures indicate the inequality of firm sizes in a particular market and effectively ignore the number of firms present. Absolute concentration measures combine the number of firms and their size distribution in a single measure.
Marfels (1971:754) points out that an absolute concentration measure is the weighted sum of the market shares, while a relative concentration measure is their weighted average.
GC was originally devised to measure the degree of income inequality in a human population and is associated with the Lorenz curve. To construct a Lorenz curve, members of the population are ranked according to their income from lowest to highest. The cumulative percentages of the population are plotted on the horizontal axis of a box diagram, while the corresponding percentages of cumulative income are plotted on the vertical axis. If the income were equally distributed among the population, the Lorenz curve would be the straight line that connects the points (0, 0) and (100, 100). The area between this diagonal line and the actual Lorenz curve provides a visual impression of the inequality of income among the population. GC is the ratio between this area and the triangular area under the diagonal line.
If the Lorenz curve were used to indicate the size distribution of sellers in a market, the sellers would be ranked by their market shares from smallest to largest and the cumulative percentage of sellers would be plotted on the horizon-tal axis. The corresponding cumulative percentage shares of the market would be plotted on the vertical axis. Instead of percentages, one could use ratios, 12 as in the case of the concentration curves in Diagrams 1 and 2. Furthermore, the sellers may be ranked from largest to smallest, yielding a graph similar to a concentration curve instead of a Lorenz curve. The only difference would be that the cumulative proportions of the sellers would be plotted on the horizontal axis, instead of their ranks (i/N instead of i), as in Diagram 3. To emphasize the difference, such a curve may be called a "relative concentration curve". The total area of the box diagram equals 1, (instead of N as in Diagrams 1 and 2).
GC would now equal the area between the proportional concentration curve and the diagonal line, divided by the area of the triangle above the diagonal line. Since GC is the ratio of one area to another, the same answer would be derived from the graph of a concentration curve with the firms’ ranks on the horizontal axis.
There is a relationship between GC and RI, since both make use of surface areas in box diagrams of concentration curves. While RI uses the area above the concentration curve, GC uses the area between the concentration curve and the diagonal line. The latter is the complement of the area above the concentration curve (relative to the area of the triangle above the diagonal line). As shown in section 3.1.5, when the ranks of the sellers are plotted on the horizontal axis, the area above the concentration curve equals N i=1 Si i −1 2 (10) When ratios instead of ranks are used on the horizontal axis (i/N instead of i), the area above the concentration curve becomes: N i=1 Si i N −1 2N = N i=1 Si(2i −1) 2N (11) Furthermore, when ratios are used on both axes, the area of the box diagram equals one, and those of the triangles above and below the diagonal line equal 1/2 each. Therefore, the area between the concentration curve and the diagonal line can be written as 1 2 − N i=1 Si(2i −1) 2N (12) The formula for GC would then be: GC = 1 2 − N i=1 Si(2i −1) 2N /1 2 (13) which can be simplified to: GC = 1 − N i=1 Si (2i −1) N (14) 13 While the area above the concentration curve can be viewed as the sum of weighted market shares with the weights equal to (2i — 1)/N, the area between the concentration curve and the diagonal line cannot be viewed as such. The diagonal line is a different concentration curve that is calculated by means of different market shares (namely equal ones). Furthermore, once the sum has been calculated, it is subjected to further manipulation, obscuring the actual weights.
As Marfels (1971:756,759) points out, GC can be transformed into RI and vice versa, if the number of sellers, N, is known: RI = 1/[N(1 - GC)] or GC = 1 - 1/(N × RI).
The minimum value of GC is zero and occurs when the sellers have equal market shares. The concentration curve would then coincide with the diagonal line and the area between the concentration curve and the diagonal line would be zero. A numerical example is shown in Table 4.
Some sources state that the maximum value of GC occurs when there is only one seller (e.g. Ferguson & Ferguson 1994:43, Fedderke & Szalontai 2009:242).
However, this is not the case. When there is only one seller, the Gini coefficient equals zero. When there is only one firm, the relative concentration curve is the straight line from point (0, 0) to (1, 1) and coincides with the diagonal line.
Therefore, the area between the curve and the diagonal line is zero. The formula of GC also equals zero: GC = 1 — 1(2 × 1 — 1)/1 = 1 — 1 = 0. Furthermore, the formula for transforming RI into GC yields GC = 0 when N = 1 and RI = 1: GC = 1 - 1/(N × RI) = 1 - 1/(1 × 1) = 1 - 1 = 0.
This is reminiscent of an old joke — one person asks another: "What is the difference between a duck?" When the other person looks dumbfounded and says the question is incomplete, the questioner says something like: "Its beak is equally long!" When there is no one to share with, there is no inequality. The issue would not easily arise in the case of a human population because a human population would typically consist of numerous people. However, a single seller is possible in the case of a market.
Lipczynski et al (2009: 205) correctly state that the maximum possible value of GC "corresponds to the case of one dominant firm with a market share approaching one, and (N - 1) very small firms each with a negligible market share." In practice, a small firm would need some minimum market share to stay in business, say 0.05 (i.e. 5%). Table 4 shows three interesting possibilities.
If there were two firms, one with a market share of 0.95 and the other with 0.05, GC would be 0.45. If there were five firms, one with a market share of 0.8 and four with market shares of 0.05 each, most economists would agree that there is lower concentration and more competition in this market than in the first. However, GC would be higher in the second market (0.60 instead of 0.45), signifying higher concentration and less competition. GC would be 0.45 in a market consisting of 10 firms, one with a market share of 0.55 and nine with market shares of 0.05 each, correctly indicating lower concentration than in the second market. However, concentration as measured by GC is the same as in the first market. It is clear that the Gini coefficient sends unsatisfactory signals about seller concentration and competitive conditions in various markets.
14 3.1.7 Variance of natural logarithms of market shares Another relative concentration measure is the variance of the natural logarithms of the market shares (VL). Lipczynski et al (2009:203) put forward the following formula for this measure: V L = 1/N N i=1 [ln Si −1/N N i=1 ln Si]2 (15) The formula can also be written as (Ferguson & Ferguson 1994: 42): V L = 1/N N i=1 (lnSi)2 −1/N2( N i=1 lnSi)2 (16) where ln denotes the natural logarithm (log to the base e) and the other symbols have the same meaning as before.
The expression 1/N N i=1 ln Si is the arithmetic average of the natural log-arithms of the market shares. It is subtracted from each market share and this difference is squared.
The squared differences are added up and the sum is divided by N to calculate the arithmetic average of the squared differences.
Since the market shares are expressed as ratios to one, the natural logarithms of the market shares are negative (except for the case of monopoly where S1 = 1 and lnS1 = 0) and are therefore squared to yield positive numbers. Lipczynski et al (2009:203) point out that in statistics, a variance provides a standard measure of dispersion or inequality within any data set.
Once more, it is difficult to identify the precise weights of the market shares.
The minimum VL = 0 and occurs when all the sellers have equal market shares. As in the case of GC, VL also equals 0 in the case of monopoly, while the maximum possible value of VL corresponds to the case of one dominant firm with a market share approaching one, and (N-1) very small firms each with a negligible market share. The same figures as in Table 4 are used to calculate various maxima of VL in Table 5. In this case, the maximum can exceed 1, which makes the measure difficult to interpret. However, the maximum of VL declines consistently as more small firms are added to the market and the market share of the dominant seller is reduced accordingly.
3.2 Inverse measures of concentration The so-called inverse measures of concentration decline as concentration in-creases and vice versa. Whereas the positive measures of concentration have easily identifiable maxima (with the exception of GC and VL), the inverse mea-sures do not.
The inverse measures include the so-called "numbers equivalent" of some of the measures discussed above, namely the HKIs, including HHI. Other inverse measures include the occupancy count, the entropy coefficient and the relative entropy coefficient.
15 3.2.1 Occupancy count The X% occupancy count (OCX%) is the smallest number of sellers that have a combined market share of X%. Leach (1992:390), for example, uses the 80% oc-cupancy count. The lower OCX%, the higher the seller concentration would be.
This inverse measure of concentration is a discrete measure, since it represents a specific point on the concentration curve. As in the case of CRX and CRX%, it may be difficult to determine the number of sellers that have a combined market share of exactly X%. For example, the largest three sellers might have a combined market share of 70%, while that of the largest four sellers might be 90%. OC80% would then be 4, but OC90% would also be 4. Meaningful comparisons of different industries by means of OCX% might not be possible.
3.2.2 Numbers equivalent of the Hannah and Kay indices Suppose the value of an index has been determined for a particular market.
Then its numbers equivalent is the number of equal-sized sellers for which the index would have the same value. The numbers equivalent of the HKIs can be written as follows (Hannah & Kay 1977: 55, Lipczynski et al 2009: 200): n(α) = [ N i=1 Sα i ]1/(1−α) (17) The top part of Table 1 is reproduced in Table 6. The totals of the weighted values are the relevant HKIs for the hypothetical five-seller market. When α = 2, namely when HKI = HHI, the numbers equivalent is: n(2) = [ N i=1 S2 i ]1/(1−2) = [ N i=1 S2 i ]−1 = 1/[ N i=1 S2 i ] = 1/HHI (18) This is the reciprocal of HHI. In the table HHI is 0.2650 and its reciprocal is 3.7736, meaning that a market with 3.7736 equal-sized sellers would also have an HHI of 0.2650. Sellers are counted in whole numbers, so it is rather awkward to refer to 3.7736 sellers. One might interpret it such that an industry consisting of four sellers with equal market shares would have roughly the same HHI as the hypothetical five-seller market. Each of the four sellers has a market share of 0.025, which has a squared value of 0.0625. The sum of the squared market shares = 0.0625 × 4 = 0.2500, the HHI of the four-seller market, and is fairly close to the HHI of the five-seller market (0.2650).
When α = 3, the numbers equivalent is: n(3) = [ N i=1 S3 i ]1/(1−3) = [ N i=1 S3 i ]−1/2 = 1/[ N i=1 S3 i ]1/2 (19) For the five-seller market in Table 6 this is 3.5578. The table also indicates n(1.5) and n(2.5).
16 The minimum value that any of the numbers-equivalent HKIs can take occurs in the case of a monopoly and equals 1 for all values of α, except α=1, when n(α) is not defined. There is no upper bound, but given N sellers with equal market shares, the maximum value of any of the numbers-equivalent HKIs would simply be N, except when α=1. A few examples are shown in the bottom part of Table 6.
3.2.3 Entropy coefficient and relative entropy coefficient The entropy coefficient (EC) is a weighted-sum concentration measure. The weights are the natural logarithms of the reciprocals of the sellers’ market shares and are inversely related to the market shares (Lipczynski et al 2009: 202): EC = N i=1 Si ln 1 Si (20) The top part of Table 7 indicates EC for the hypothetical market shares used in Tables 1 to 6. The weights increase from 1.0498 to 2.9957 as the market shares decline from 0.35 to 0.05. EC equals 1.4306. To aid in interpreting this number, one needs to know the range of values EC can take. The minimum is zero and occurs in the case of monopoly. The maximum values occur when the market shares of the sellers are equal and these maximum values are equal to the respective weights, namely ln(1/Si).
The relative entropy coefficient (REC) is defined as EC/lnN: REC = 1 ln N N i=1 Si ln 1 Si (21) This can be rewritten as: REC = N i=1 Si[ln 1 Si ] 1 ln N (22) The latter expression indicates that REC is a weighted-sum concentration measure like EC, with weights of ln(1/Si). The top part of Table 8 indicates the calculation of REC for the hypothetical market shares used in Tables 1 to 7. The weights increase from 0.6523 to 1.8614 as the market shares decline from 0.35 to 0.05. REC equals 0.8889. REC is not defined in the case of monopoly. The maximum values occur when the market shares of the sellers are equal and these maximum values are equal to one, irrespective of the number of sellers. Since REC varies from zero to one, its value is easier to interpret than that of EC, the range of which does not have an absolute upper boundary.
However, the fact that REC does not vary with the number of sellers when the sizes of the sellers are equal, is a major drawback. It renders REC a measure of inequality, as in the case of GC and VL.
17 4 EV ALUATION OF MEASURES OF SELLER CONCENTRATION Hannah and Kay (1977:48-50) put forward a number of reasonable "axioms" or criteria which a concentration measure should always meet, namely (slightly rephrased): • the concentration curve ranking criterion: if firms are ranked from largest to smallest and plotted (on the horizontal axis) against their cumulative output (on the vertical axis) and the concentration curve of one market lies above that of another market at all points [except (0, 0) and, when N is equal in the two markets, (N, 1)], a positive concentration measure of the first market must be higher than that of the second market, while an inverse concentration measure must be lower • the sales transfer criterion: measured concentration should increase if cus-tomers switch from smaller to larger firms and vice versa • the entry criterion: if a new firm, smaller than the average size of existing firms, enters the market, measured concentration should decline (assuming that the relative market shares of the existing firms decline proportionately to accommodate the new firm) • the merger criterion: measured concentration must increase if existing firms merge The literature covering empirical tests of the SCP paradigm often make use of measures that do not meet the Hannah and Kay criteria, casting doubt on the validity of the tests.
In this section, 11 hypothetical markets are compared pairwise to weigh up the concentration measures defined in section 4 against the Hannah and Kay criteria. The market shares are chosen such that the concentration curves being compared do not cross. Thus the concentration curve ranking criterion is used in each case, along with one of the other criteria.
Two markets are compared in Diagram 4. Market 1 is a monopoly estab-lished by the producer of a new product. As patents run out, another seller enters and in time captures half the market. The new situation is called Market 2 and is compared with the original monopoly market. The cumulative market shares are shown below the diagram, as well as the values of the concentration measures. The first 11 measures (CR3 to VL) are positive measures and are expected to decline when new entry occurs. The last seven (OC80% to REC) are inverse measures, which are supposed to increase. The numbers that do not change in the expected direction are shaded. CR3, CR4 and CR5 do not decline when the new entrant arrives, because the number of sellers remains less than 3. GC and VL remain equal to zero, since the market shares of the sellers are equal in Market 2 and there is no firm to share with in Market 1, as explained 18 before. REC is not defined for the case of monopoly (Market 1) and equals 1 in Market 2. The other concentration measures change in the expected direction.
Diagram 5 depicts a market with 4 equally sized sellers (Market 3) into which a fifth seller enters and in time gains an equal portion of the market from each incumbent (Market 4). CR5, GC, VL, OC80% and REC remain unchanged, while the other measures change as expected.
Diagram 6 depicts a market with 5 equally sized sellers (Market 4). The distribution of market shares gradually changes to one where the largest firm captures 35% of the market, while the least successful seller is left with 5% (Market 5). These are the figures used in the top parts of tables 1 to 8. In the jargon of Hannah and Kay, sales transfers have taken place. Most economists would agree that concentration has increased and all but one of the concentra-tion measures move in the expected direction. The exception is CR5.
If the two smallest sellers in Market 5 merge, leading to the formation of Market 6 (Diagram 7), six concentration measures misbehave, namely CR3, CR5, GC, VL, OC80% and REC.
If the two largest sellers merge, resulting in Market 7, only one concentration measure misbehaves, namely CR5 (Diagram 8).
Diagram 9 depicts sales transfers from the third and fourth seller to the first and second sellers, while the market share of the smallest seller remains unaltered (Market 5 is transformed into Market 10). In this case, CR5 does not change in the expected direction, because there are only 5 sellers in the market.
The rest of the measures change as expected.
Diagram 10 starts offwith Market 2, a duopoly with equal-sized sellers. One of the sellers gradually captures 70% of the market, a situation depicted in Mar-ket 11. Since there are only two sellers, CR3, CR4 and CR5 are useless. The other discrete measure, OC80%, also remains constant. Perhaps surprisingly, one of the absolute concentration measures, HI, declines, instead of increas-ing. This is related to the inconsistent assigning of weights and the strange "reinforcement" of the weight of each market share (except S1) by the multi-plier (2 — Si). The weight of S1 remains the same, while that of S2 is reduced substantially. This causes a reduction in the weighted sum of the market shares.
Diagram 11 indicates the concentration curve of Market 5, with 5 sellers of unequal size, and that of Market 8, which results from the entry of 6 small firms, each gaining a market share of 1% at the expense of the incumbents. In this case all the concentration measures, except REC, correctly indicate that concentration has decreased.
Finally, Diagram 12 depicts the entry of a large seller into Market 5, which could for instance happen if an existing conglomerate decides to enter in a big way, with a sizeable factory, a well designed and packaged product and an exten-sive advertising campaign. It is assumed that the entrant quickly gains a mar-ket share of 32%, slightly above the share of 30% that the second-largest seller had. Whereas the largest seller loses 2% of its market share, the others shifts downwards in rank and collectively lose 30%. The entry of an additional firm lengthens the concentration curve, which signifies lower concentration. Since the sellers’ ranking changes, their size-distribution does not change much. The 19 identity of the sellers does not play a role in the ranking by market share. The bottom part of the new concentration curve (Market 9) largely coincides with the old one, but the top part lies slightly below the old one. The concentration measures should therefore indicate a decline in concentration. Four of them do not, namely GC, VL, OC80% and REC.
The above analysis leads to the conclusion that the discrete concentration measures, as well as the relative ones, do not yield reliable results and should not be used when measuring seller concentration. In addition, HI is less reliable than the other absolute measures and should best be set aside by researchers in this field.
These findings are nothing new. Based on a slightly different analysis, Han-nah and Kay (1977: 50-52) reached similar conclusions more than three decades ago. They found that the concentration ratios do "not necessarily react posi-tively to a merger or sales transfer", but still expressed the opinion they stand up "reasonably well" to the criteria, because they "will never be perverse in the direction of change". However, they concluded that a number of measures of inequality, including GC and VL, violate some of the axioms and should not be used to measure seller concentration. They emphasised "that inequality and concentration are not the same thing" and "that trends in one do not necessarily shed light on trends in the other", adding the following interesting remark: "It is tedious to labour what we hope the reader will find an obvious point.
Our only justification for doing so is that the arguments above were laid out with great clarity and lucidity by Adelman (1951) all of twenty-five years ago, and that nevertheless economists have regularly continued to make unwarranted inferences about changes in concentration on the basis of measurements of in-equality." It is alarming that these measures are still being used in studies that find their way into peer-reviewed journals and presented in textbooks as measures of seller concentration.
This issue was discussed in the South African Journal of Economics (SAJE) in the early 1990s, starting with Leach’s (1992) criticism of the work of Fourie and Smit (1989). Du Plessis (1978) had been the first to publish seller concen-tration measures in the SAJE, using data from the 1972 manufacturing census.
His article was based on work done for the Mouton Commission (1977) and his DComm thesis at the University of Stellenbosch. Thereafter Fourie published a series of seller concentration measures in the SAJE, either on his own or with co-authors, aiming to establish whether or not seller concentration levels in South Africa were high and whether or not they increased from one manufacturing census to the next.
Leach (1992: 386-387) criticized Fourie and Smit (1989) for using the Gini coefficient (GC) to measure seller concentration and reworked the figures for the years 1972 to 1985, using the Rosenbluth index (RI) and the 80% occupancy count (OC80%). Whereas Fourie and Smit found that seller concentration in manufacturing tended to increase from 1972 to 1982, Leach found no positive trend when using OC80% and a negative trend when using RI.
20 In his 1996 article Fourie did not accept the obvious deficiencies of GC and other indices of inequality and criticised the use of RI on the grounds that it is not used widely in industrial economics and "gives inordinate weight to an increase in the size of the fringe of an industry" and argues that the dominance of the n-factor in RI is extraordinary, since the fringe may increase without affecting the dominance of two or three large firms (Fourie 1996: 101-103).
Hannah and Kay (1977: 50) also list the Hall-Tideman index, which is exactly the same as RI, as one of the "unsatisfactory concentration indices".
According to them, this index does not meet either the entry or the merger criterion (the third and fourth axioms). The tests in this paper do not confirm this in respect of the direction of change in RI. It may be that Hannah and Kay thought it not sensitive enough to changes in the number of sellers (the opposite of Fourie’s argument).
RI is nevertheless a vast improvement on GC, since it is a measure of con-centration, not inequality. Probably the most important reason why many re-searchers have used GC is that it can be calculated from grouped data. The strength of RI is that it can be calculated from GC if the total number of sellers is known.
Fedderke and Szalontai (2009) use both GC and RI to indicate trends in concentration in South African three-digit manufacturing industries and to find correlations between concentration and certain measures of market performance, mainly because previous South African studies made use of GC and RI (Fedderke & Szalontai 2009: 242).
5 CONCLUSION The discrete measures, namely concentration ratios (CRX) and the occupancy count (CRX%), do not always meet the Hannah and Kay criteria, nor do the relative concentration measures or measures of inequality, namely the Gini coef-ficient (GC), the variance of logarithms of market shares (VL) and the relative entropy coefficient (REC). An absolute concentration measure that does not always meet the criteria is the Horvath index. The other absolute concentra-tion measures that have been tested, namely the Herfindahl-Hirschman index (HHI), the other Hannah and Kay indices [HKI(α)], the Rosenbluth index (RI), the numbers equivalent of the Hannah and Kay indices [HKIne(α)] and the entropy coefficient (EC) do meet the criteria.
Seeing that similar findings were made decades ago, it is alarming that mea-sures such as GC, VL and REC are still being used in studies that find their way into peer-reviewed journals. It is even more alarming that measures of inequality are still being put forward in textbooks on industrial organization as legitimate measures of seller concentration. Instead of learning from the mistakes of past generations, new generations of scholars are being set up for failure. Given the use of measures of inequality in the past, they need to be mentioned in text-books, but today’s students should be forewarned against their use as measures of seller concentration.
21 Another implication is that studies on seller concentration that were done in the past (e.g. those listed in Ferguson & Ferguson 1994:95-98) should be revisited and those that make use of GC, VL and REC should be disregarded, unless data is available to calculate RI from GC to reach a new conclusion in a particular case.
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Upper Saddle River, NJ: Prentice-Hall.
23 Table 1: Herfindahl-Hirschman and Hannah and Kay indices of concentration Seller Market share i Si Si 0.5 Si 1 Si 1.5 Si 2 Si 1.5 Si 2 Si 2.5 Si 3 1 0.3500 0.5916 0.3500 0.2071 0.1225 0.2071 0.1225 0.0725 0.0429 2 0.3000 0.5477 0.3000 0.1643 0.0900 0.1643 0.0900 0.0493 0.0270 3 0.2000 0.4472 0.2000 0.0894 0.0400 0.0894 0.0400 0.0179 0.0080 4 0.1000 0.3162 0.1000 0.0316 0.0100 0.0316 0.0100 0.0032 0.0010 5 0.0500 0.2236 0.0500 0.0112 0.0025 0.0112 0.0025 0.0006 0.0001 Total 1.0000 2.1264 1.0000 0.5036 0.2650 0.5036 0.2650 0.1434 0.0790 α 1.5 2 2.5 3 HKI(1.5) 0.5036 HHI = HKI(2) 0.2650 HKI(2.5) 0.1434 HKI(3) 0.0790 Maximum N=1 1.0000 1.0000 0.1000 1.0000 1.0000 1.0000 0.1000 1.0000 1.0000 Minimum 1/(N0.5) 1/N 1/(N1.5) 1/(N2) Equal sizes Per seller N=5 0.2000 0.4472 0.2000 0.0894 0.0400 0.4472 0.2000 0.0894 0.0400 N=10 0.1000 0.3162 0.1000 0.0316 0.0100 0.3162 0.1000 0.0316 0.0100 N=100 0.0100 0.1000 0.0100 0.0010 0.0001 0.1000 0.0100 0.0010 0.0001 Weights Weighted values Indices 24 Table 2: Horvath index of concentration Seller Market share Weights Weighted values i Si Si(2-Si), i=2 to 5 Si 2(2-Si), i=2 to 5 1 0.3500 1.0000 0.3500 2 0.3000 0.5100 0.1530 3 0.2000 0.3600 0.0720 4 0.1000 0.1900 0.0190 5 0.0500 0.0975 0.0049 Total 1.0000 2.1575 0.5989 Maximum Index N=1 1.0000 1.0000 1.0000 Minimum [(3N2–3N+1)/N3] Equal sizes Per seller i=2, …, N N=5 0.2000 0.3600 0.4880 N=10 0.1000 0.1900 0.2710 N=100 0.0100 0.0199 0.0297 Table 3: Rosenbluth index of concentration Seller Market share i Si i 2i -1 Sii Si(2i-1) 1 0.3500 1 1.0000 0.3500 0.3500 2 0.3000 2 3.0000 0.6000 0.9000 3 0.2000 3 5.0000 0.6000 1.0000 4 0.1000 4 7.0000 0.4000 0.7000 5 0.0500 5 9.0000 0.2500 0.4500 Total 1.0000 15.0000 25.0000 2.2000 3.4000 RI 0.2941 Maximum Index N=1 1.0000 1.0000 Minimum 1/N Equal sizes Per seller N=5 0.2000 0.2000 N=10 0.1000 0.1000 N=100 0.0100 0.0100 Weights Weighted values 25 Table 4: Gini coefficient Seller Market share Weights Weighted values i Si (2i-1)/N Si(2i-1)/N 1 0.3500 0.2000 0.0700 2 0.3000 0.6000 0.1800 3 0.2000 1.0000 0.2000 4 0.1000 1.4000 0.1400 5 0.0500 1.8000 0.0900 Total 1.0000 5.0000 0.6800 GC 0.3200 Maximum GC S1=0.95, S2=0.05 0.4500 S1=0.8, S2 to S5=0.05 0.6000 S1=0.55, S2 to S10=0.05 0.4500 Minimum Equal sizes Per seller N=1 1.0000 0.0000 N=5 0.2000 0.0000 N=10 0.1000 0.0000 N=100 0.0100 0.0000 26 Table 5: Variance of natural logarithms of market shares Seller Market share lnSi 1/N∑lnSi (lnSi - 1/N∑lnSi)2 i Si 1 0.3500 -1.0498 -1.8323 0.6123 2 0.3000 -1.2040 -1.8323 0.3948 3 0.2000 -1.6094 -1.8323 0.0497 4 0.1000 -2.3026 -1.8323 0.2212 5 0.0500 -2.9957 -1.8323 1.3536 Total 1.0000 -9.1616 -9.1616 2.6315 VL 0.5263 Maximum S1=0.95, S2=0.05 2.1674 S1=0.8, S2 to S5=0.05 1.2300 S1=0.55, S2 to S10=0.05 0.5175 Minimum Equal sizes Per seller N=1 1.0000 0.0000 0.0000 0.0000 N=5 0.2000 -1.6094 -1.6094 0.0000 N=10 0.1000 -2.3026 -2.3026 0.0000 N=100 0.0100 -4.6052 -4.6052 0.0000 27 Table 6: Numbers equivalent of Herfindahl-Hirschman and Hannah and Kay indices of concentration Seller Market share i Si Si 0.5 Si 1 Si 1.5 Si 2 Si 1.5 Si 2 Si 2.5 Si 3 1 0.3500 0.5916 0.3500 0.2071 0.1225 0.2071 0.1225 0.0725 0.0429 2 0.3000 0.5477 0.3000 0.1643 0.0900 0.1643 0.0900 0.0493 0.0270 3 0.2000 0.4472 0.2000 0.0894 0.0400 0.0894 0.0400 0.0179 0.0080 4 0.1000 0.3162 0.1000 0.0316 0.0100 0.0316 0.0100 0.0032 0.0010 5 0.0500 0.2236 0.0500 0.0112 0.0025 0.0112 0.0025 0.0006 0.0001 Total 1.0000 2.1264 1.0000 0.5036 0.2650 0.5036 0.2650 0.1434 0.0790 α 1.5 2 2.5 3 n(α)=Σsi α]1/(1-α) 3.9426 3.7736 3.6505 3.5578 Minimum N=1 1.0000 1.0000 1.0000 1.0000 1.0000 1.00 1.00 1.00 1.00 Maximum Per seller N N N N N=4 0.2500 0.5000 0.2500 0.1250 0.0625 4.00 4.00 4.00 4.00 N=5 0.2000 0.4472 0.2000 0.0894 0.0400 5.00 5.00 5.00 5.00 N=10 0.1000 0.3162 0.1000 0.0316 0.0100 10.00 10.00 10.00 10.00 N=100 0.0100 0.1000 0.0100 0.0010 0.0001 100.00 100.00 100.00 100.00 Weights Weighted values Numbers equivalent Table 7: Entropy coefficient Seller Market share Weights Weighted values i Si ln(1/Si) Siln(1/Si) 1 0.3500 1.0498 0.3674 2 0.3000 1.2040 0.3612 3 0.2000 1.6094 0.3219 4 0.1000 2.3026 0.2303 5 0.0500 2.9957 0.1498 Total 1.0000 9.1616 1.4306 EC Minimum N=1 1.0000 0.0000 0.0000 Maximum Per seller ln(1/Si) N=5 0.2000 1.6094 1.6094 N=10 0.1000 2.3026 2.3026 N=100 0.0100 4.6052 4.6052 28 Table 8: Relative entropy coefficient Seller Market share Weights Weighted values i Si [ln(1/Si)]/lnN Si[ln(1/Si)]/lnN 1 0.3500 0.6523 0.2283 2 0.3000 0.7481 0.2244 3 0.2000 1.0000 0.2000 4 0.1000 1.4307 0.1431 5 0.0500 1.8614 0.0931 Total 1.0000 5.6924 0.8889 REC Minimum N=1 1.0000 Not def Not def Maximum Per seller [ln(1/Si)]/lnN N=5 0.2000 1.0000 1.0000 N=10 0.1000 1.0000 1.0000 N=100 0.0100 1.0000 1.0000 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 0 1 2 3 4 5 Cumulative market share (Cum Si) Rank of seller Diagram 1 Concentration curve 29 0.00 0.20 0.40 0.60 0.80 1.00 0 1 2 3 4 5 Cumulative market share (Cum Si ) Rank of seller (i) N P Diagram 2 Area above the concentration curve O A B D E H C G 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 0.00 0.20 0.40 0.60 0.80 1.00 Cumulative market share (Cum Si) Cumulative proportion of sellers Diagram 3 Relative concentration curve 30 Rank of seller Market shares Measure Mkt 1 Mkt 2 % change Mkt 1 Mkt 2 CR3 1.000 1.000 0.0 1 1.00 0.50 CR4 1.000 1.000 0.0 2 0.50 CR5 1.000 1.000 0.0 HKI(1.5) 1.000 0.707 -29.3 HKI(2)=HHI 1.000 0.500 -50.0 HKI(2.5) 1.000 0.354 -64.6 HKI(3) 1.000 0.250 -75.0 HI 1.000 0.875 -12.5 RI 1.000 0.500 -50.0 GC 0.000 0.000 #DIV/0! VL 0.000 0.000 #DIV/0! OC80% 1.000 2.000 100.0 Rank of seller Cumulative market shares HKIne(1.5) 1.000 2.000 100.0 HKIne(2)=1/HHI 1.000 2.000 100.0 Mkt 1 Mkt 2 HKIne(2.5) 1.000 2.000 100.0 0 0.00 0.00 HKIne(3) 1.000 2.000 100.0 1 1.00 0.50 EC 0.000 0.693 #DIV/0! 2 1.00 REC #DIV/0! 1.000 #DIV/0! 0.00 0.20 0.40 0.60 0.80 1.00 0 1 2 Cumulative market share Rank of seller Diagram 4 Monopoly and entry Mkt 1 Mkt 2 31 Rank of seller Market shares Measures Mkt 3 Mkt 4 % change Mkt 3 Mkt 4 CR3 0.750 0.600 -20.0 1 0.25 0.20 CR4 1.000 0.800 -20.0 2 0.25 0.20 CR5 1.000 1.000 0.0 3 0.25 0.20 HKI(1.5) 0.500 0.447 -10.6 4 0.25 0.20 HKI(2)=HHI 0.250 0.200 -20.0 5 0.20 HKI(2.5) 0.125 0.089 -28.4 HKI(3) 0.063 0.040 -36.0 HI 0.578 0.488 -15.6 RI 0.250 0.200 -20.0 Rank of seller Cumulative market shares GC 0.000 0.000 #DIV/0! VL 0.000 0.000 #DIV/0! Mkt 3 Mkt 4 OC80% 4.000 4.000 0.0 0 0.00 0.00 HKIne(1.5) 4.000 5.000 25.0 1 0.25 0.20 HKIne(2)=1/HHI 4.000 5.000 25.0 2 0.50 0.40 HKIne(2.5) 4.000 5.000 25.0 3 0.75 0.60 HKIne(3) 4.000 5.000 25.0 4 1.00 0.80 EC 1.386 1.609 16.1 5 1.00 REC 1.000 1.000 0.0 0.00 0.20 0.40 0.60 0.80 1.00 0 1 2 3 4 5 Cumulative market share Rank of seller Diagram 5 Equal-size sellers and entry Mkt 3 Mkt 4 32 Rank of seller Market shares Measures Mkt 4 Mkt 5 % change Mkt 4 Mkt 5 CR3 0.600 0.850 41.7 1 0.20 0.35 CR4 0.800 0.950 18.8 2 0.20 0.30 CR5 1.000 1.000 0.0 3 0.20 0.20 HKI(1.5) 0.447 0.504 12.6 4 0.20 0.10 HKI(2)=HHI 0.200 0.265 32.5 5 0.20 0.05 HKI(2.5) 0.089 0.143 60.3 HKI(3) 0.040 0.079 97.5 HI 0.488 0.599 22.7 RI 0.200 0.294 47.1 Rank of seller Cumulative market shares GC 0.000 0.320 #DIV/0! VL 0.000 0.526 #DIV/0! Mkt 4 Mkt 5 OC80% 4.000 3.000 -25.0 0 0.00 0.00 HKIne(1.5) 5.000 3.943 -21.1 1 0.20 0.35 HKIne(2)=1/HHI 5.000 3.774 -24.5 2 0.40 0.65 HKIne(2.5) 5.000 3.650 -27.0 3 0.60 0.85 HKIne(3) 5.000 3.558 -28.8 4 0.80 0.95 EC 1.609 1.431 -11.1 5 1.00 1.00 REC 1.000 0.889 -11.1 0.00 0.20 0.40 0.60 0.80 1.00 0 1 2 3 4 5 Cumulative market share Rank of seller Diagram 6 From equal to unequal seller sizes Mkt 4 Mkt 5 33 Rank of seller Market shares Measures Mkt 5 Mkt 6 % change Mkt 5 Mkt 6 CR3 0.850 0.850 0.0 1 0.35 0.35 CR4 0.950 1.000 5.3 2 0.30 0.30 CR5 1.000 1.000 0.0 3 0.20 0.20 HKI(1.5) 0.504 0.519 3.0 4 0.10 0.15 HKI(2)=HHI 0.265 0.275 3.8 5 0.05 HKI(2.5) 0.143 0.148 3.5 HKI(3) 0.079 0.081 2.8 HI 0.599 0.617 3.0 RI 0.294 0.303 3.0 Rank of seller Cumulative market shares GC 0.320 0.175 -45.3 VL 0.526 0.111 -78.8 Mkt 5 Mkt 6 OC80% 3.000 3.000 0.0 0 0.00 0.00 HKIne(1.5) 3.943 3.714 -5.8 1 0.35 0.35 HKIne(2)=1/HHI 3.774 3.636 -3.6 2 0.65 0.65 HKIne(2.5) 3.650 3.568 -2.3 3 0.85 0.85 HKIne(3) 3.558 3.508 -1.4 4 0.95 1.00 EC 1.431 1.335 -6.7 5 1.00 REC 0.889 0.963 8.3 0.00 0.20 0.40 0.60 0.80 1.00 0 1 2 3 4 5 Cumulative market share Rank of seller Diagram 7 Merger of the two smallest sellers Mkt 5 Mkt 6 34 Rank of seller Market shares Measures Mkt 5 Mkt 7 % change Mkt 5 Mkt 7 CR3 0.850 0.950 11.8 1 0.35 0.65 CR4 0.950 1.000 5.3 2 0.30 0.20 CR5 1.000 1.000 0.0 3 0.20 0.10 HKI(1.5) 0.504 0.656 30.3 4 0.10 0.05 HKI(2)=HHI 0.265 0.475 79.2 5 0.05 HKI(2.5) 0.143 0.362 152.6 HKI(3) 0.079 0.284 259.2 HI 0.599 0.746 24.5 RI 0.294 0.476 61.9 Rank of seller Cumulative market shares GC 0.320 0.475 48.4 VL 0.526 0.897 70.5 Mkt 5 Mkt 7 OC80% 3.000 2.000 -33.3 0 0.00 0.00 HKIne(1.5) 3.943 2.322 -41.1 1 0.35 0.65 HKIne(2)=1/HHI 3.774 2.105 -44.2 2 0.65 0.85 HKIne(2.5) 3.650 1.968 -46.1 3 0.85 0.95 HKIne(3) 3.558 1.877 -47.2 4 0.95 1.00 EC 1.431 0.982 -31.4 5 1.00 REC 0.889 0.708 -20.3 0.00 0.20 0.40 0.60 0.80 1.00 0 1 2 3 4 5 Cumulative market share Rank of seller Diagram 8 Merger of the two largest sellers Mkt 5 Mkt 7 35 Rank of seller Market shares Measures Mkt 5 Mkt 10 % change Mkt 5 Mkt 10 CR3 0.850 0.870 2.4 1 0.35 0.40 CR4 0.950 0.950 0.0 2 0.30 0.35 CR5 1.000 1.000 0.0 3 0.20 0.12 HKI(1.5) 0.504 0.535 6.3 4 0.10 0.08 HKI(2)=HHI 0.265 0.306 15.4 5 0.05 0.05 HKI(2.5) 0.143 0.181 26.3 HKI(3) 0.079 0.109 38.3 HI 0.599 0.646 7.9 RI 0.294 0.327 11.1 Rank of seller Cumulative market shares GC 0.320 0.388 21.3 VL 0.526 0.666 26.5 Mkt 5 Mkt 10 OC80% 3.000 3.000 0.0 0 0.00 0.00 HKIne(1.5) 3.943 3.488 -11.5 1 0.35 0.40 HKIne(2)=1/HHI 3.774 3.270 -13.3 2 0.65 0.75 HKIne(2.5) 3.650 3.125 -14.4 3 0.85 0.87 HKIne(3) 3.558 3.026 -15.0 4 0.95 0.95 EC 1.431 1.340 -6.3 5 1.00 1.00 REC 0.889 0.833 -6.3 0.00 0.20 0.40 0.60 0.80 1.00 0 1 2 3 4 5 Cumulative market share Rank of seller Diagram 9 Sales transfer to largest sellers Mkt 5 Mkt 10 36 Rank of seller Market shares Measures Mkt 2 Mkt 11 % change Mkt 2 Mkt 11 CR3 1.000 1.000 0.0 1 0.50 0.70 CR4 1.000 1.000 0.0 2 0.50 0.30 CR5 1.000 1.000 0.0 HKI(1.5) 0.707 0.750 6.1 HKI(2)=HHI 0.500 0.580 16.0 HKI(2.5) 0.354 0.459 29.9 HKI(3) 0.250 0.370 48.0 HI 0.875 0.853 -2.5 RI 0.500 0.625 25.0 GC 0.000 0.200 #DIV/0! VL 0.000 0.179 #DIV/0! OC80% 2.000 2.000 0.0 Rank of seller Cumulative market shares HKIne(1.5) 2.000 1.778 -11.1 HKIne(2)=1/HHI 2.000 1.724 -13.8 Mkt 2 Mkt 11 HKIne(2.5) 2.000 1.680 -16.0 0 0.00 0.00 HKIne(3) 2.000 1.644 -17.8 1 0.50 0.70 EC 0.693 0.611 -11.9 2 1.00 1.00 REC 1.000 0.881 -11.9 0.00 0.20 0.40 0.60 0.80 1.00 0 1 2 Cumulative market share Rank of seller Diagram 10 Sales transfer to largest seller in duopoly Mkt 2 Mkt 11 37 Rank of seller Market shares Measures Mkt 5 Mkt 8 % change Mkt 5 Mkt 8 CR3 0.850 0.790 -7.1 1 0.35 0.34 CR4 0.950 0.900 -5.3 2 0.30 0.28 CR5 1.000 0.940 -6.0 3 0.20 0.17 HKI(1.5) 0.504 0.467 -7.3 4 0.10 0.11 HKI(2)=HHI 0.265 0.237 -10.5 5 0.05 0.04 HKI(2.5) 0.143 0.125 -12.7 6 0.01 HKI(3) 0.079 0.068 -14.5 7 0.01 HI 0.599 0.555 -7.3 8 0.01 RI 0.294 0.243 -17.5 9 0.01 GC 0.320 0.625 95.5 10 0.01 VL 0.526 2.066 292.6 11 0.01 OC80% 3.000 4.000 33.3 HKIne(1.5) 3.943 4.585 16.3 Rank of seller Cumulative market shares HKIne(2)=1/HHI 3.774 4.216 11.7 HKIne(2.5) 3.650 3.996 9.5 Mkt 5 Mkt 8 HKIne(3) 3.558 3.847 8.1 0 0.00 0.00 EC 1.431 1.672 16.9 1 0.35 0.34 REC 0.889 0.697 -21.5 2 0.65 0.62 3 0.85 0.79 4 0.95 0.90 5 1.00 0.94 6 0.95 7 0.96 8 0.97 9 0.98 10 0.99 11 1.00 0.00 0.20 0.40 0.60 0.80 1.00 0 1 2 3 4 5 6 7 8 9 10 11 Cumulative market share Rank of seller Diagram 11 Unequal seller sizes and entry of small sellers Mkt 5 Mkt 8 38 Rank of seller Market shares Measures Mkt 5 Mkt 9 % change Mkt 5 Mkt 9 CR3 0.850 0.810 -4.7 1 0.35 0.33 CR4 0.950 0.920 -3.2 2 0.30 0.32 CR5 1.000 0.970 -3.0 3 0.20 0.16 HKI(1.5) 0.504 0.487 -3.2 4 0.10 0.11 HKI(2)=HHI 0.265 0.252 -4.8 5 0.05 0.05 HKI(2.5) 0.143 0.135 -5.5 6 0.03 HKI(3) 0.079 0.074 -6.0 HI 0.599 0.579 -3.4 Rank of seller Cumulative market shares RI 0.294 0.275 -6.6 GC 0.320 0.393 22.9 Mkt 5 Mkt 9 VL 0.526 0.794 50.8 0 0.00 0.00 OC80% 3.000 3.000 0.0 1 0.35 0.33 HKIne(1.5) 3.943 4.209 6.7 2 0.65 0.65 HKIne(2)=1/HHI 3.774 3.962 5.0 3 0.85 0.81 HKIne(2.5) 3.650 3.791 3.9 4 0.95 0.92 HKIne(3) 3.558 3.669 3.1 5 1.00 0.97 EC 1.431 1.521 6.4 6 1.00 REC 0.889 0.849 -4.5 0.00 0.20 0.40 0.60 0.80 1.00 0 1 2 3 4 5 6 Cumulative market share Rank of seller Diagram 12 Unequal seller sizes and entry of a large seller Mkt 5 Mkt 9 39 |
16088 | https://par.nsf.gov/servlets/purl/10383950 | Foundations of Computational Mathematics
Spectral Graph Matching and Regularized Quadratic Relaxations II Erd ˝ os-Rényi Graphs and Universality
Zhou Fan 1 · Cheng Mao 2 · Yihong Wu 1 · Jiaming Xu 3
Received: 16 August 2019 / Revised: 23 September 2021 / Accepted: 15 March 2022 © SFoCM 2022
Abstract
We analyze a new spectral graph matching algorithm, GRAph Matching by Pair-wise eigen-Alignments (GRAMPA), for recovering the latent vertex correspondence between two unlabeled, edge-correlated weighted graphs. Extending the exact recov-ery guarantees established in a companion paper for Gaussian weights, in this work, we prove the universality of these guarantees for a general correlated Wigner model. In particular, for two Erd˝ os-Rényi graphs with edge correlation coefficient 1 − σ 2 and average degree at least polylog (n), we show that GRAMPA exactly recovers the latent vertex correspondence with high probability when σ 1/ polylog (n). Moreover, we establish a similar guarantee for a variant of GRAMPA, corresponding to a tighter quadratic programming relaxation of the quadratic assignment problem. Our analysis exploits a resolvent representation of the GRAMPA similarity matrix and local laws for the resolvents of sparse Wigner matrices. Communicated by Francis Bach. Z. Fan is supported in part by NSF Grant DMS-1916198. C. Mao is supported in part by NSF Grant DMS-2053333. Y. Wu is supported in part by the NSF Grants CCF-1527105, CCF-1900507, an NSF CAREER award CCF-1651588, and an Alfred Sloan fellowship. J. Xu is supported by the NSF Grants IIS-1838124, CCF-1850743, and CCF-1856424.
B Yihong Wu yihong.wu@yale.edu Zhou Fan zhou.fan@yale.edu Cheng Mao cheng.mao@math.gatech.edu Jiaming Xu jx77@duke.edu 1 Department of Statistics and Data Science, Yale University, New Haven, CT, USA 2 School of Mathematics, Georgia Institute of Technology, Atlanta, GA, USA 3 The Fuqua School of Business, Duke University, Durham, NC, USA
123 Foundations of Computational Mathematics
Keywords Graph matching · Quadratic assignment problem · Spectral methods ·
Random matrix theory · Erd˝ os-Rényi graph
Mathematics Subject Classification Primary 90C25 · Secondary 68Q87
1 Introduction
Given two (weighted) graphs, graph matching aims at finding a bijection between the vertex sets that maximizes the total edge weight correlation between the two graphs. It reduces to the graph isomorphism problem when the two graphs can be matched perfectly. Let A and B be the (weighted) adjacency matrices of the two graphs on n
vertices. Then, the graph matching problem can be formulated as solving the following
quadratic assignment problem (QAP) [5, 18]: max
Π∈Sn
〈A, Π BΠ 〉, (1) where Sn denotes the set of permutation matrices in Rn×n and 〈· , ·〉 denotes the matrix inner product. The QAP is NP-hard to solve or to approximate within a growing factor . In the companion paper , we proposed a computationally efficient spectral graph matching method, called GRAph Matching by Pairwise eigen-Alignments (GRAMPA). Let us write the spectral decompositions of A and B as
A = ∑
i
λi vi v
i
and B = ∑
j
μ j w j w
j
. (2) Given a tuning parameter η > 0, GRAMPA first constructs an n × n similarity matrix 1
X = ∑
i,j
η(λ i − μ j )2 + η2 vi v
i
Jw j w
j
, (3) where J is the n × n all-ones matrix. Then, it outputs a permutation matrix ̂ Π by “rounding” X to a permutation matrix, for example, by solving the following linear assignment problem (LAP) ̂
Π ∈ argmax
Π∈Sn
〈X, Π 〉. (4) Let Π∗ ∈ Sn be the latent true matching, and denote the entries of A and
Π∗ BΠ∗ as ai j and bπ∗(i)π ∗( j). A Gaussian Wigner model is studied in , where {(ai j , bπ∗(i)π ∗( j) )} are i.i.d. pairs of correlated Gaussian variables such that 1 In , X is defined without the factor η in the numerator. We include η here for convenience in the proof; this does not affect the algorithm as the rounded solution ̂ Π is invariant to rescaling X .
123 Foundations of Computational Mathematics
bπ∗(i)π ∗( j) = ai j + σ z i j for a noise level σ ≥ 0, and ai j and z i j are independent stan-dard Gaussian. It is shown that GRAMPA exactly recovers the vertex correspondence
Π∗ with high probability when σ = O(1/ log n). Simulation results in [14,Sect. 4.1] further show that the empirical performance of GRAMPA under the Gaussian Wigner model is very similar to that under the Erd˝ os-Rényi model where {(ai j , bπ∗(i)π ∗( j) )}
are i.i.d. pairs of correlated centered Bernoulli random variables, suggesting that the performance of GRAMPA enjoys universality. In this paper, we prove a universal exact-recovery guarantee for GRAMPA, under a general Wigner matrix model for the weighted adjacency matrix: Let A = (ai j )
be a symmetric random matrix in Rn×n , where the entries (ai j )i≤ j are independent. Suppose that
E [ai j
] = 0 for all i, j, E
[
a2
i j
]
= 1
n for all i = j, (5) and
E
[∣∣ai j
∣∣k ]
≤ C knd (k−2)/ 2 for all i, j and each k ∈
[
2, ( log n)10 log log n ]
, (6) where d ≡ d(n) is an n-dependent sparsity parameter and C is an absolute positive constant. Of particular interest are the following special cases: – Bounded case: The entries are bounded in magnitude by C√n . Then, (6) is fulfilled for d = n and all k.– Sub-Gaussian case: The sub-Gaussian norm of each entry satisfies
‖ai j ‖ψ2 sup
k≥1
k−1/2E
[∣∣ai j
∣∣k ]1/k
= O (1/√n) . (7) It is easily checked that (6) is satisfied for d = n/( log n)11 log log n and all large n.– Erd˝ os-Rényi graphs with edge probability p ≡ p(n). We may center and scale the adjacency matrix A such that ai j ∼ (Bern ( p) − p)/ √np (1 − p) for i = j, which satisfies (5) and (6) for d = np (1 − p) (cf. Lemma 1). With the moment conditions (5) and (6) specified, we are ready to introduce the correlated Wigner model, which encompasses the correlated Erd˝ os-Rényi graph model proposed in as a special case.
Definition 1 (Correlated Wigner model) Let n be a positive integer, σ ∈ [ 0, 1] an ( n-dependent) noise parameter, π∗ a latent permutation on [n], and Π∗ ∈ { 0, 1}n×n
the corresponding permutation matrix such that (Π ∗)iπ∗(i) = 1. Suppose that {(ai j , bπ∗(i)π ∗( j) ) : i ≤ j} are independent pairs of random variables such that both
A = (ai j ) and B = (bi j ) satisfy (5) and (6),
E [ai j bπ∗(i)π ∗( j)
] ≥ 1 − σ 2
n for all i = j, (8)
123 Foundations of Computational Mathematics
and for a constant C > 0, any D > 0, and all n ≥ n0(D),
P
{∥ ∥∥A − Π∗ BΠ∗
∥∥∥ ≤ Cσ
}
≥ 1 − n−D (9) where ‖ · ‖ denotes the spectral norm. The parameter σ measures the effective noise level in the model. In the special case of sparse Erd˝ os-Rényi model, A and B are the centered and normalized adjacency matrices of two Erd˝ os-Rényi graphs, which differ by a fraction 2 σ 2 of edges approx-imately. In this paper, we prove the following exact recovery guarantee for GRAMPA:
Theorem (Informal statement) For the correlated Wigner model, if d ≥ polylog (n)
and σ ≤ c (log n)−2κ for any fixed constant κ > 2 and a sufficiently small constant c > 0, then GRAMPA with η = 1/ polylog n recovers π∗ exactly with high probability for large n. If furthermore a i j and b i j are sub-Gaussian and satisfy (7) , then this holds with κ = 1.
This theorem generalizes the exact recovery guarantee for GRAMPA proved in for the Gaussian Wigner model, albeit at the expense of a slightly stronger requirement for σ than in the Gaussian case. The requirement that d ≥ polylog (n) and σ ≤
1/ polylog (n) is the state-of-the-art for polynomial time algorithms on sparse Erd˝ os-Rényi graphs , although we note that the recent work of provided an algorithm with super-polynomial runtime n O(log n) that achieves exact recovery when d ≥ n o(1)
under the much weaker condition of σ ≤ 1 − (log n)−o(1) (see the end of Sect. 2 for more detailed discussion). Numerical experiments in the companion paper suggest that the failure of GRAMPA occurs at σ = C/ log n for some constant C,indicating that our theoretical characterization of the performance of GRAMPA here is almost tight. In , we further demonstrate the superior empirical performance of GRAMPA on a variety of synthetic and real datasets, in terms of both statistical accuracy and computational efficiency. In the conference version , GRAMPA is also shown to improve existing shape matching algorithms on 3D deformable shape data. The analysis in relies heavily on the rotational invariance of Gaussian Wigner matrices, and does not extend to non-Gaussian models. Here, instead, our universality analysis uses a resolvent representation of the GRAMPA similarity matrix (3) via a contour integral (cf. Proposition 1). Capitalizing on local laws for the resolvent of sparse Wigner matrices [11, 12], we show that the similarity matrix (3) is with high probability diagonal dominant in the sense that min k X kπ∗(k) > max =π∗(k) X k . This enables rounding procedures as simple as thresholding to succeed. From an optimization point of view, GRAMPA can also be interpreted as solving a
regularized quadratic programming (QP) relaxation of the QAP. More precisely, the QAP (1) can be equivalently written as min
Π∈Sn
‖AΠ − Π B‖2
F
, (10)
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and the similarity matrix X in (3) is a positive scalar multiple of the solution ˜X to argmin
X∈Rn×n
‖AX − X B ‖2
F
η2‖X‖2
F
s.t. 1 X1 = n. (11) (See [14,Corollary 2.2].) This is a convex relaxation of the program (10) with an additional ridge regularization term. As a result, our analysis immediately yields the same exact recovery guarantees for algorithms that round the solution ˜X to (11) instead of X. In Sect. 6, we study a tighter relaxation of the QAP (10) that imposes row-sum constraints, and establish the same exact recovery guarantees (up to universal constants) by employing similar technical tools.
Organization The rest of the paper is organized as follows. In Sect. 2, we state the main exact recovery guarantees for GRAMPA under the correlated Wigner model, as well as the results specialized to the (sparse) Erd˝ os-Rényi model. We start the analysis by introducing the key resolvent representation of the GRAMPA similarity matrix in Sect. 3. As a preparation for the main proof, Sect. 4 provides the needed tools from random matrix theory. The proof of correctness for GRAMPA is then presented in Sect. 5. In Sect. 6, we extend the theoretical guarantees to a tighter QP relaxation. Finally, Sect. 7 is devoted to proving the resolvent bounds which form the main technical ingredient to our proofs.
Notation Let [n] {1, . . . , n}. Let i = √−1. In a Euclidean space Rn or Cn , let ei
be the i-th standard basis vector, and let 1 = 1n be the all-ones vector. Let J = Jn
denote the n × n all-ones matrix, and let I = In denote the n × n identity matrix. The subscripts are often omitted when there is no ambiguity. The inner product of u, v ∈ Cn is defined as 〈u, v 〉 = u∗v. Similarly, for matrices,
〈A, B〉 = Tr (A∗ B). Let ‖v‖ ≡ ‖ v‖2 = 〈 v, v 〉 and ‖v‖∞ = sup i |vi | for vectors. Let
‖M‖ ≡ ‖ M‖op = sup v:‖ v‖= 1 ‖Mv‖, ‖M‖2
F
= 〈 M, M〉, and ‖M‖∞ = sup i, j |M i j |
for matrices. Let x ∧ y = min (x, y) and x ∨ y = max (x, y). We use C, C′, c, c′, . . . to denote positive constants that may change at each appearance. For sequences of positive real numbers (a n )∞
n=1
and (bn )∞
n=1
, we write a n bn (resp. a n bn ) if there is a constant
C > 0 such that a n ≤ Cb n (resp. bn ≤ Ca n ) for all n ≥ 1, a n bn if both relations
a n bn and a n bn hold, and a n bn if a n /bn → 0 as n → ∞ . We write
a n = O(bn ) if |a n | bn and a n = o(bn ) if |a n | bn .
2 Exact Recovery Guarantees for GRAMPA
In this section, we state the exact recovery guarantees for GRAMPA, making the earlier informal statement precise.
Theorem 1 Fix constants a > 0 and κ > 2, and let η ∈ [ 1/( log n)a , 1]. Consider the correlated Wigner model with n ≥ d ≥ (log n)c0 where c 0 > max (32 + 4a, 4 + 7a).Then, there exist (a, κ) -dependent constants C 0, n0 > 0 and a deterministic quantity
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r(n) ≡ r(n, η, d, a) satisfying r (n) → 0 as n → ∞ , such that for all n ≥ n0, with probability at least 1 − n−10 , the matrix X in (3) satisfies
max
=π∗(k)
|X k | ≤ C0(log n) κ 1
√η ,
max
k
∣∣∣∣X kπ∗(k) − 1 − σ 2
η
∣∣∣∣ ≤ C0
(r(n)η + ση2 + (log n) κ 1
√η
)
. (12)
If there is a universal constant K for which a i j and b i j are sub-Gaussian with
‖ai j ‖ψ2 , ‖bi j ‖ψ2 ≤ K /√n, then the above holds also with κ = 1.
As an immediate corollary, we obtain the following exact recovery guarantee for GRAMPA.
Corollary 1 (Universal graph matching) Under the conditions of Theorem 1, there exist constants c , c′ > 0 such that for all n ≥ n0, if
(log n)−a ≤ η ≤ c(log n)−2κ and σ ≤ c′η, (13)
then with probability at least 1 − n−10 ,
min
k
X kπ∗(k) > max
=π∗(k)
X k , (14)
and hence ̂ Π that solves the linear assignment problem (4) equals Π∗.
Proof Let c = 1/( 64 C20 ) and c′ = 1/( 2C0), where C0 is the constant given in Theorem 1. Then under assumption (13), we have
C0(log n) κ √η ≤ C0(log n) κ
√c
(log n) κ = C0
√c ≤ 1/8,
so max =π∗(k) |X k | ≤ 1/( 8η) . We also have C0σ/η ≤ C0c′ = 1/2 and 1 − σ 2 > 7/8and C0r(n) < 1/8 for all large n, so that max k X kπ∗(k) > ( 7/8−1/8−1/2−1/8)/η =
1/( 8η) . This implies (14).
An important application of the above universality result is matching two correlated sparse Erd˝ os-Rényi graphs. Let G be an Erd˝ os-Rényi graph with n vertices and edge probability q, denoted by G ∼ G(n, q). Let A and B′ be two copies of Erd˝ os-Rényi graphs that are i.i.d. conditional on G, each of which is obtained from G by deleting every edge of G with probability 1 − s independently where s ∈ [ 0, 1]. Then, we have that A, B′ ∼ G(n, p) marginally where p qs . Equivalently, we may first sample an Erd˝ os-Rényi graph A ∼ G(n, p), and then define B′ by
B′
i j
∼
{Bern (s) if Ai j = 1
Bern
( p(1−s)
1−p
)
if Ai j = 0.
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Suppose that we observe a pair of graphs A and B = Π∗ B′Π∗, where Π∗ is an unknown permutation matrix. We then wish to recover the permutation matrix Π∗.We transform the adjacency matrices A and B so that they satisfy the moment conditions (5) and (6): Define the centered, rescaled versions of A and B by
A (np (1 − p)) −1/2(A − E[A]) and B (np (1 − p)) −1/2 (B − E[B]). (15) Then, (5) clearly holds, and we check the following additional properties.
Lemma 1 For all large n, the matrices A = (ai j ) and B = (bi j ) satisfy (6), (8), and
(9) with d = np (1 − p) and
σ 2 = max
( 1 − s
1 − p , (log n)7
d
)
.
Proof Assume without loss of generality that Π∗ is the identity matrix. For any k ≥ 2, we have
E
[∣∣ai j
∣∣k ]
= (np (1 − p)) −k/2 [
p (1 − p)k + (1 − p) p k ]
= (1 − p)k−1 + p k−1
nd (k−2)/ 2
≤ 1
nd (k−2)/ 2 .
Thus, the moment condition (6) is satisfied. In addition, we have that for all i < j,
E [ai j bi j
] = 1
d E [( Ai j − p) ( Bi j − p)] = 1
d
(
ps − p2)
= s − pn(1 − p) ≤ 1 − σ 2
n ,
where the last inequality holds by the choice of σ 2 . Thus, (8) is satisfied. Moreover, let Δi j = 1√2σ 2
(ai j − bi j
) . It follows that E [Δi j
] = 0 and
E
[∣∣Δi j
∣∣k ]
= 2 p(1 − s)
(2σ 2d)k/2 ≤ 1
n(2σ 2d) (k−2)/ 2where the last inequality is due to σ 2 ≥ 1−s
1−p
. Thus, by applying Lemma 3 and 2(log n)7 ≤ 2σ 2d ≤ n where the upper bound follows from p(1−s) ≤ s(1−s) ≤ 1/4, there exists a constant C > 0 such that for any D > 0, with probability at least 1 −n−D
for all n ≥ n0(D), we have ‖Δ‖ ≤ C and hence ‖A − B‖ ≤ √2Cσ . Thus, (9) is satisfied.
Combining Lemma 1 with Corollary 1 immediately yields a sufficient condition for GRAMPA to exactly recover Π∗ in the correlated Erd˝ os-Rényi graph model.
Corollary 2 (Erd˝ os-Rényi graph matching) Suppose that either
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(a) (dense case)
δ ≤ p ≤ 1 − δ, 1 − s
1 − p ≤ (log n)−c1
for constants δ ∈ (0, 1) and c 1 > 4, or (b) (sparse case) np (1 − p) ≥ (log n)c0 , 1 − s
1 − p ≤ (log n)−c1
for constants c 0 > 48 and c 1 > 8.There exist (δ, c0, c1)-dependent constants a , n0 > 0 such that if η = (log n)−a and n ≥ n0, then with probability at least 1 − n−10 ,
min
k
X kπ∗(k) > max
=π∗(k)
X k ,
and hence the solution ̂ Π to the linear assignment problem (4) coincides with Π∗.
Proof For (a), pick κ = 1 and any a such that c1/2 > a > 2κ = 2. For (b), pick any
a, κ such that c1/2 > a > 2κ > 4 and c0 > 32 + 4a > 4 + 7a. Then, all conditions of Theorem 1 and Corollary 1 are satisfied for large n, and the result follows.
Comparison to information-theoretic limits and existing algorithmic guarantees of exact recovery For the correlated Erd˝ os-Rényi graph model, exact recovery of the hidden vertex correspondence with high probability is shown to be information-theoretically possible if nps − log n → +∞ and p/s = O(log −3(n)) , and impossible if nps 2−log n = O(1) [7, 8]. From a computational perspective, recent work shows that degree matching can achieve exact recovery with high probability in polynomial time provided that np n4/5 log 7/5(n) and 1 − s p4/ log 6(n).This result is further improved to np = Ω( log 2(n)) and 1 − s ≤ O(log −2(n))
in by matching degree profiles (that is, empirical distributions of neighbors’ degrees). The performance guarantee of the proposed GRAMPA method matches the state of the art of polynomial-time algorithms up to polylogarithmic factors and holds for more general models of correlated matrices. It is worth noting that a quasi-polynomial time ( n O(log n)) algorithm is proposed in which succeeds when
np ∈ [n o(1) , n1/153 ] ∪ [ n2/3, n1− ] and s ≥ (log n)−o(1). However, it remains open whether exact recovery is achievable in polynomial time for any constant s bounded away from 1. It is conceivable that there exists a “hard regime” where exact recov-ery is information-theoretically possible but computationally intractable, resembling the conjectured computational hardness for the planted clique problem and the stochastic block model .
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3 Resolvent Representation
For a real symmetric matrix A with spectral decomposition (2), its resolvent is defined by
R A(z) (A − zI)−1 = ∑
i
1
λi − z vi v
i
for z ∈ C \ R. Then, we have the matrix symmetry R A(z) = R A(z), conjugate symmetry R A(z) = R A(¯z), and the following Ward identity.
Lemma 2 (Ward identity) For any z ∈ C \ R and any real symmetric matrix A, R A(z)R A(z) = Im R A(z)
Im z .
Proof By the definition of R(z) ≡ R A(z) and conjugate symmetry, it holds Im R(z)
Im z = R(z) − R(z)
z − ¯ z = (A − zI)−1 − (A − ¯ zI)−1
z − ¯ z = (A − zI)−1(A − ¯ zI)−1
= R(z)R(z).
The following resolvent representation of X is central to our analysis.
Proposition 1 Consider symmetric matrices A and B with spectral decomposi-tions (2) , and suppose that ‖A‖ ≤ 2.5. Then, the matrix X defined in (3) admits the following representation X = 12π Re
∮
Γ
R A(z)JR B (z + iη) dz , (16)
where
Γ = { z : | Re z| = 3 and | Im z| ≤ η/ 2 or | Im z| = η/ 2 and | Re z| ≤ 3} (17)
is the rectangular contour with vertices ±3 ± iη/ 2 (See Fig. 1 for an illustration).
Proof We have
X = η ∑
i,j
vi v
i
J w j w
j
(λ i − μ j )2 + η2
= η ∑
i
vi v
i
JR B (λ i + iη) R B (λ i − iη)
123 Foundations of Computational Mathematics
= Im ∑
i
vi v
i
JR B (λ i + iη) (18) by Lemma 2. Consider the function f : C → Cn×n defined by f (z) = JR B (z + iη) .Then, each entry f k is analytic in the region {z : Im z > −η}. Since Γ encloses each eigenvalue λi of A, the Cauchy integral formula yields entrywise equality
− 12πi
∮
Γ
f (z)λi − z dz = f (λ i ). (19) Substituting this into (18), we obtain
X = Im ∑
i
vi v
i
(
− 12πi
∮
Γ
f (z)λi − z dz
)
= 12π Re
∮
Γ
R A(z) f (z)dz , (20) which completes the proof in view of the definition of f .
4 Tools from Random Matrix Theory
Before proving our main results, we introduce the relevant tools from random matrix theory. In particular, the resolvent bounds in Theorem 2 constitute an important tech-nical ingredient in our analysis.
4.1 Concentration Inequalities
We start with some known concentration inequalities in the literature.
Lemma 3 (Norm bounds) For any constant ε > 0 and a universal constant c > 0, if n ≥ d ≥ (log n)6+6ε, then with probability at least 1 − e−c(log n)1+ε
,
‖A‖ ≤ 2 + (log n)1+ε
d1/4 .
Proof See [12,Lemma 4.3], where we fix the parameter ξ = 1 + ε in [12,Eq. (2.4)]. The notational identification is q ≡ √d.
Lemma 4 (Concentration inequalities) Let α, β ∈ Rn be independent random vectors with independent entries, satisfying
E[αi ] = E[βi ] = 0, E[α2
i
] = E[β2
i
] = 1
n ,
max (E[| αi |k ], E[| βi |k ]) ≤ 1
nd (k−2)/ 2 , for each k ∈ [ 2, ( log n)10 log log n ]. (21)
For any constant ε > 0 and universal constants C , c > 0, if n ≥ d ≥ (log n)6+6ε,then:
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(a) For each i ∈ [ n], with probability at least 1 − e−c(log n)1+ε
,
|αi | ≤ C
√d . (22) (b) For any deterministic vector v ∈ Cn , with probability at least 1 − e−c(log n)1+ε
,
∣∣∣vα
∣∣∣ ≤ (log n)1+ε
( ‖v‖∞√d + ‖v‖2
√n
)
. (23)
Furthermore, for any even integer p ∈ [ 2, ( log n)10 log log n ],
E
[∣ ∣∣vα
∣∣∣p]
≤ (C p )p
( ‖v‖∞√d + ‖v‖2
√n
)p
. (24) (c) For any deterministic matrix M ∈ Cn×n , with probability at least 1−e−c(log n)1+ε
,
∣∣∣∣α Mα − 1
n Tr M
∣∣∣∣ ≤ (log n)2+2ε
( 2‖M‖∞√d + ‖M‖Fn
)
(25)
and
∣∣∣α Mβ
∣∣∣ ≤ (log n)2+2ε
( 2‖M‖∞√d + ‖M‖Fn
)
. (26)
Proof See [12,Lemma 3.7, Lemma 3.8, and Lemma A.1(i)], where again we fix ξ =
1 + ε.
Next, based on the above lemma, we state concentration inequalities for a bilinear form that apply to our setting directly.
Lemma 5 (Concentration of bilinear form) Let α, β ∈ Rn be random vectors such that the pairs (α i , β i ) for i ∈ [ n] are independent, with
E[αi ] = E[βi ] = 0, E
[
α2
i
]
= E
[
β2
i
]
= 1
n , E [αi βi ] ≥ 1 − σ 2
n .
Let M ∈ Cn×n be any deterministic matrix.
(a) For any constant ε > 0, suppose (21) holds where n ≥ d ≥ (log n)6+6ε. Then, there are universal constants C , c > 0 such that with probability at least 1 −
e−c(log n)1+ε
,
∣∣∣∣α Mβ − 1 − σ 2
n Tr M
∣∣∣∣ ≤ C (log n)2+2ε
( 1
n ‖M‖F + 1
√d ‖M‖∞
)
. (27)
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(b) Suppose that αi , β i are sub-Gaussian with ‖αi ‖ψ2 = ‖ βi ‖ψ2 ≤ K√n for a constant K > 0. Then for any D > 0, there exists a constant C ≡ C K ,D only depending on K and D such that with probability at least 1 − n−D ,
∣∣∣∣α Mβ − 1 − σ 2
n Tr M
∣∣∣∣ ≤ C log nn ‖M‖F . (28)
Proof In view of the polarization identity
α Mβ = 14 (α + β) M(α + β) − 14 (α − β) M(α − β),
it suffices to analyze the two terms separately. Note that
E
[
(α + β) M(α + β)
]
= 4 − 2σ 2
n Tr M,
E
[
(α − β) M(α − β)
]
= 2σ 2
n Tr M,
which yields the desired expectation E[α Mβ] = 1−σ 2
n
Tr M. Thus, it remains to study the deviation. To prove the concentration bound (27), we obtain from (25) that, there is a universal constant c > 0 such that with probability at least 1 − e−c(log n)1+ε
,
∣∣∣(α ± β) M(α ± β) − E[(α ± β) M(α ± β) ]
∣∣∣ ≤ (log n)2+2ε
( 1
n ‖M‖F + 2
√d ‖M‖ ∞
)
,
from which (27) easily follows. The sub-Gaussian concentration bound (28) follows from the Hanson–Wright inequality [16, 20]. More precisely, note that max {‖ α + β‖ψ2 , ‖α − β‖ψ2 } ≤‖α‖ψ2 + ‖ β‖ψ2 ≤ 2K /√d, so taking δ = n−D /2 in [14,Lemma A.2] yields that with probability at least 1 − n−D ,
∣∣∣(α ± β) M(α ± β) − E
[
(α ± β) M(α ± β)
]∣ ∣∣ ≤ C K ,D
log nn ‖M‖F ,
which completes the proof.
4.2 The Stieltjes Transform
Denote the semicircle density and its Stieltjes transform by
ρ( x) = 12π
√
4 − x2 1{| x|≤ 2} and m0(z) =
∫ 1
x − z ρ( x)dx = −z + √z2 − 42 ,
(29)
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respectively, where m0(z) is defined for z /∈ [− 2, 2], and √z2 − 4 is defined with a branch cut on [− 2, 2] so that √z2 − 4 ∼ z as |z| → ∞ . We have the conjugate symmetry m0(z) = m0(¯z).We record the following basic facts about the Stieltjes transform.
Proposition 2 For each z ∈ C \ R, the Stieltjes transform m 0(z) is the unique value satisfying m0(z)2 + zm 0(z) + 1 = 0 and Im m0(z) · Im z > 0. (30)
Setting ζ ( z) min (| Re z − 2|, | Re z + 2|), uniformly over z ∈ C \ [− 2, 2] with
|z| ≤ 10 ,
|m0(z)| 1, | Im m0(z)| | Im z|, and | Im m0(z)|
{√ζ ( z) + | Im z| if | Re z| ≤ 2,
| Im z|/√ζ ( z) + | Im z| if | Re z| > 2. (31)
For x ∈ [− 2, 2], the continuous extensions m+
0
(x) lim
z→x:z∈C+
m0(z), m−
0
(x) lim
z→x:z∈C−
m0(z)
from C+ and C− both exist. For all x ∈ [− 2, 2], these satisfy m±
0
(x)2 + xm ±
0
(x) + 1 = 0, m+
0
(x) = m−
0
(x),
1
π Im m+
0
(x) = − 1
π Im m−
0
(x) = ρ( x), |m±
0
(x)| = 1. (32)
Proof (30) follows from the definition of m0 . (31) follows from [11,Lemma 4.3] and continuity and conjugate symmetry of m0 . For the existence of m+
0
(and hence also
m−
0
), see, e.g., the more general statement of [4,Corollary 1]. The first claim of (32) follows from continuity and (30), the second from conjugate symmetry, the third from the Stieltjes inversion formula, and the last from the fact that the two roots of (30) at z = x ∈ [− 2, 2] are m+
0
(x) and m−
0
(x) = m+
0
(x), so that 1 = m±
0
(x)m±
0
(x) =|m±
0
(x)|2 .
4.3 Resolvent Bounds
For a fixed constant a > 0 and all large n, we bound the resolvent R(z) = R A(z) over the spectral domain
D = D1 ∪ D2, where
D1 = { z ∈ C : Re z ∈ [− 3, 3], | Im z| ∈ [ 1/( log n)a , 1]} , and
D2 = { z ∈ C : | Re z| ∈ [ 2.6, 3], | Im z| ≤ 1/( log n)a }.
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Here, D1 is the union of two strips in the upper and lower half planes, and D2 is the union of two strips in the left and right half planes.
Theorem 2 (Resolvent bounds) Suppose A ∈ Rn×n has independent entries (ai j )i≤ jsatisfying (5) and (6) . Fix a constant a > 0 which defines the domain D, fix ε > 0,and set b = max (16 + 3ε + 2a, 3 + 3ε + 5a/2), b′ = max (16 + 4ε + 2a, 4 + 5ε + 6a).
Suppose n ≥ d ≥ (log n)b′
. Then for some constants C , c, n0 > 0 depending on a and ε, and for all n ≥ n0, with probability 1 − e−c(log n)( log log n), the following hold simultaneously for every z ∈ D:
(a) (Entrywise bound) For all j = k ∈ [ n],
|R jk (z)| ≤ C(log n)2+2ε+a
√d . (33)
For all j ∈ [ n],
|R j j (z) − m0(z)| ≤ C(log n)2+2ε+3a/2
√d . (34) (b) (Row sum bound) For all j ∈ [ n],
∣∣∣e
j
R(z)1
∣∣∣ ≤ C(log n)1+ε+a . (35) (c) (Total sum bound)
|1 R(z)1 − n · m0(z)| ≤ Cn (log n)b
√d . (36) The proof follows ideas of , and we defer this to Sect. 7. As the spectral param-eter z is allowed to converge to the interval [− 2, 2] with increasing n, this type of result is often called a “local law” in the random matrix theory literature. The focus of the above is a bit different from the results stated in , as we wish to obtain explicit logarithmic bounds for | Im z| 1/ polylog (n), rather than bounds for more local spectral parameters down to the scale of | Im z| polylog (n)/ n.
5 Proof of Correctness for GRAMPA
In this section, we prove Theorem 1. Note that the mapping B → Π∗ BΠ∗ for any permutation Π∗ induces w j → Π∗ w j and X → XΠ∗, since JΠ∗ = J. By virtue of this equivariance, throughout the proof, we may assume without loss of generality that
Π∗ = I, i.e., the underlying true permutation π∗ is the identity permutation. Then, we aim to show that X is diagonally dominant, in the sense that min k X kk > max k = X k .
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In view of Lemma 3, we have that ‖A‖ ≤ 2.5 holds with probability 1 − n−D for any D > 0 and all n ≥ n0(D). In the following, we assume that ‖A‖ ≤ 2.5 holds. On this event, by Proposition 1, we get that
X k = 12π Re
∮
Γ
(e
k
R A(z)1)( e R B (z + iη) 1)dz (37) Note that one may attempt to directly apply (35) to bound the row sums e
k
R A(z)1
and e R B (z + iη) 1. This would yield
∣∣∣(e
k
R A(z)1)( e R B (z + iη) 1)
∣∣∣ (log n)2+2ε+2a ,
and hence |X k | (log n)2+2ε+2a . However, this estimate is too crude to capture the differences between the diagonal and off-diagonal entries. In fact, the row sum
e
k
R A(z)1 does not concentrate on its mean, and the deviation e
k
R A(z)1 − m0(z) and
e R B (z + iη) 1 − m0(z) is uncorrelated for k = and positively correlated for k = .For this reason, the diagonal entries of (37) dominate the off-diagonals. Thus, it is crucial to gain a better understanding of the deviation terms. We do so by applying Schur complement decomposition.
5.1 Decomposition Via Schur Complement
We recall the classical Schur complement identity for the inverse of a block matrix.
Lemma 6 (Schur complement identity) For any invertible matrix M ∈ Cn×n and block decomposition M =
[ A B C D
]
,
if D is square and invertible, then M−1 =
[ S −S B D −1
−D−1C S D−1 + D−1C S B D −1
]
(38)
where S = (A − B D −1C)−1.
We decompose e
k
R A(z)1 and e R B (z + iη) 1 using this identity, focusing without loss of generality on (k, ) = (1, 2). Let R A,12 ∈ C2×2 be the upper-left 2 × 2 sub-matrix of R A, and let R(12 )
A
∈ C(n−2)×(n−2) be the resolvent of the (n − 2) × (n − 2)
minor of A with the first two rows and columns removed. Let a
1
and a
2
be the first two rows of A with first two entries removed, and let A
o
∈ R2×(n−2) be the stacking of a
1
and a
2
.The following deterministic lemma approximates e
1
R A(z)1 based on the Schur complement.
123 Foundations of Computational Mathematics
Lemma 7 Suppose |z| ≤ 10 , and
∥∥R A,12 (z) − m0(z)I∥∥ ≤ δ (39)
where 0 ≤ δ ≤ min z:| z|≤ 10 |m0(z)|/2. Then for a constant C > 0 and k = 1, 2
∣∣∣e
k
R A(z)1 − m0(z)
(
1 − a
k
R(12 )
A
(z)1n−2
)∣ ∣∣ ≤ Cδ ( 1 + ‖ R A(z)1‖∞) . (40)
Proof It suffices to consider k = 1. Applying the Schur complement identity (38), the first two rows of R A are given by
[
R A,12 −R A,12 A
o
R(12 )
A
]
. (41) Thus,
e
1
R A(z)1 = [1 0 ] [
R A,12 −R A,12 A
o
R(12 )
A
] [ 12
1n−2
]
= [1 0 ] R A,12
(
12 − A
o
R(12 )
A
1n−2
)
.
Denote ΔA R A,12 (z) − m0(z)I. Then,
e
1
R A(z)1 = [1 0 ] (m0(z)I + ΔA)
(
12 − A
o
R(12 )
A
1n−2
)
.
= m0(z)
(
1 − a
1
R(12 )
A
1n−2
)
[1 0 ] ΔA
(
12 − A
o
R(12 )
A
1n−2
)
.
= m0(z)
(
1 − a
1
R(12 )
A
1n−2
)
O
(
δ
(
1 +
∥∥∥A
o
R(12 )
A
1n−2
∥∥∥))
, (42) where the last equality applies (39). We next upper bound
∥∥∥A
o
R(12 )
A
1n−2
∥∥∥. In view of the fact that C ≥ | m0(z)| ≥ c for absolute constants c and C, the assumption (39) implies that R A,12 is invertible with ‖R−1
A,12
‖ 1. Using (41) again, we have
A
o
R(12 )
A
1n−2 = 12 − R−1
A,12
[e1 e2
] R A1n . (43) It follows that
∥∥∥A
o
R(12 )
A
1n−2
∥∥∥ 1 +
∣∣∣e
1
R A1n
∣∣∣ +
∣∣∣e
2
R A1n
∣∣∣ 1 + ‖ R A1n ‖ ∞ . (44) The desired bound (40) follows by combining (42) and (44).
123 Foundations of Computational Mathematics
5.2 Off-Diagonal Entries
Without loss of generality, we focus on the off-diagonal entry X12 :
X12 = 12π Re
∮
Γ
(
e
1
R A(z)1
) (
e
2
R B (z + iη) 1
)
dz .
For the given value a > 0 in Theorem 1, and for some small constant ε > 0, let b, b′
be as defined in Theorem 2. Under the given condition for c0 in Theorem 1, for ε > 0sufficiently small, we have c0 > b′ and c0 > 2b—thus d (log n)b′
so Theorem 2 applies, and also √d (log n)b. Fix the constant κ, where κ = 1 in the sub-Gaussian case where ‖ai j ‖ψ2 , ‖bi j ‖ψ2 1/√n, and κ > 2 otherwise. For ease of notation, we define
δ1 = (log n)2+2ε+3a/2
√d , δ2 = (log n)1+ε+a
√n , δ3 = (log n)b
√d , δ4 = (log n) κ/ 2
√n .
(45) Note that we have δi = o(1) for each i = 1, 2, 3, 4, and also δ1δ22 n = o(1).
5.2.1 Resolvent Approximation
Define an event E1 wherein the following hold simultaneously for all z ∈ Γ :
∥∥R A,12 (z) − m0(z)I∥∥ δ1 (46)
∥∥R B,12 (z + iη) − m0(z + iη) I∥∥ δ1 (47)
‖R A(z)1‖ ∞ δ2
√n (48)
‖R B (z + iη) 1‖∞ δ2
√n. (49) Applying the resolvent approximations given in Theorem 2, we have that
P {E1} ≥ 1 − e−c(log n)( log log n) .
In the following, we assume the event E1 holds. On E1 , by Lemma 7, we get that uniformly over z ∈ Γ ,
e
1
R A(z)1 = m0(z)
(
1 − a
1
R(12 )
A
1n−2
)
O (δ1δ2
√n) , (50)
e
2
R B (z + iη) 1 = m0(z + iη)
(
1 − b
2
R(12 )
B
1n−2
)
O (δ1δ2
√n) . (51) Each of (50) and (51) is itself O(δ 2
√n), by (48) and (49). Then multiplying the two, we have
123 Foundations of Computational Mathematics
[
e
1
R A(z)1
] [
e
2
R B (z + iη) 1
]
= m0(z)m0(z + iη)
(
1 − a
1
R(12 )
A
1n−2 − b
2
R(12 )
B
1n−2 + a
1
R(12 )
A
Jn−2 R(12 )
B
b2
)
O
(
δ1δ22 n
)
.
It follows that
∮
Γ
[
e
1
R A(z)1
] [
e
2
R B (z + iη) 1
]
dz
=
∮
Γ
m0(z)m0(z + iη) dz − a
1
g − b
2
h + a
1
Mb 2 + O
(
δ1δ22 n
)
, (52) where
g
∮
Γ
m0(z)m0(z + iη) R(12 )
A
(z)1n−2dz ,
h
∮
Γ
m0(z)m0(z + iη) R(12 )
B
(z + iη) 1n−2dz ,
M
∮
Γ
m0(z)m0(z + iη) R(12 )
A
(z)Jn−2 R(12 )
B
(z + iη) dz . (53)
5.2.2 Term-By-Term Analysis
Next, we bound the individual terms of (52). By the boundedness of m0(z), we have
∮
Γ
m0(z)m0(z + iη) dz = O(1). (54) Define the event E2 wherein the following hold simultaneously:
∣∣∣a
1
g
∣∣∣ +
∣∣∣b
2
h
∣∣∣ δ1 (‖g‖∞ + ‖ h‖∞) + δ4 (‖g‖2 + ‖ h‖2) (55)
∣∣∣a
1
Mb 2
∣∣∣ δ1‖M‖∞ + δ24 ‖M‖F . (56) Note that the triple (g, h, M) is independent of the pair (a1, b2) and a1 and b2 are independent. Hence, by first conditioning on (g, h, M) and then applying (23) and (26), we get that
P {E2} ≥ 1 − n−D
for any constant D > 0, 2 and all n ≥ n0(D), in both the sub-Gaussian ( κ = 1) and general ( κ > 2) cases. Henceforth, we assume E2 holds. It then remains to bound the
2
and ∞ norms of g, h, and M.2 The constant D can be made arbitrarily large by setting the hidden constants in (55) and (56) sufficiently large.
123 Foundations of Computational Mathematics
Fig. 1 Nested contours Γand Γ′
Recall that Γ is the rectangular contour with vertices ±3±i η
2
. Let us define another contour (to be used later) Γ ′ inside Γ , with vertices ±2.6 ± i η
4
, cf. Fig. 1. Define the event E3 wherein the following hold simultaneously for all z ∈ Γ ∪ Γ ′:
∥∥∥R(12 )
A
(z)1n−2
∥∥∥∞ δ2
√n, (57)
∥∥∥R(12 )
B
(z + iη) 1n−2
∥∥∥∞ δ2
√n, (58)
∣∣∣1
n−2
R(12 )
A
(z)1n−2 − m0(z)( n − 2)
∣∣∣ δ3n, (59)
∣∣∣1
n−2
R(12 )
B
(z + iη) 1n−2 − m0(z + iη)( n − 2)
∣∣∣ δ3n. (60) By Theorem 2, we have that P {E3} ≥ 1 − e−c(log n)( log log n). In the following, we assume the event E3 holds. Note that
‖g‖∞ sup
z∈Γ
‖R(12 )
A
(z)1n−2‖∞ δ2
√n, (61) where the second inequality holds in view of (57). Similarly, in view of (58), we have that ‖h‖∞ δ2
√n. Furthermore,
‖M‖∞ sup
z∈Γ
∥∥∥R(12 )
A
(z)Jn−2 R(12 )
B
(z + iη)
∥∥∥∞≤ sup
z∈Γ
∥∥∥R(12 )
A
(z)1n−2
∥∥∥∞
∥∥∥1
n−2
R(12 )
B
(z + iη)
∥∥∥∞ δ22 n. (62) The 2 bounds of g, h and M are deferred to Lemma 8. Applying (59), (60), and Lemma 8 with R A = R(12 )
A
and R B = R(12 )
B
, we get ‖g‖22 n log 1
η
, ‖h‖22 n log 1
η
and ‖M‖F n/√η.
Combining the above bounds on the norms of g, h, M with (55), (56), and (54), and plugging into (52), we conclude that on the event {‖ A‖ ≤ 2.5} ∩ E1 ∩ E2 ∩ E3 ,
123 Foundations of Computational Mathematics
|X12 | = 2π
∣∣∣∣∮
Γ
[
e
1
R A(z)1
] [
e
2
R B (z + iη) 1
]
dz
∣∣∣∣
1 + δ4
√
n log 1
η + δ24 n 1
√η + δ1δ22 n δ24 n 1
√η = (log n) κ 1
√η , (63) where in the third step we used δ1δ22 n = o(1) and η ≤ 1 so that δ4
√n = (log n) κ/ 2 √
η log 1
η
η1/4 .
5.2.3 Bounding the Norms of g, h and M
Lemma 8 Suppose ‖A‖ ≤ 2.5 and ∣∣1 R(z)1∣∣ n for all z ∈ Γ ∪ Γ ′ and both R(z) = R A(z) and R (z) = R B (z + iη) . Define g =
∮
Γ
m0(z)m0(z + iη) R A(z)1dz h =
∮
Γ
m0(z)m0(z + iη) R B (z + iη) 1dz M =
∮
Γ
m0(z)m0(z + iη) R A(z)JR B (z + iη) dz .
Then, ‖g‖2 n log 1
η
, ‖h‖2 n log 1
η
and ‖M‖2
F
n2
η
.
Proof Since ‖A‖ ≤ 2.5, the function m0(z)m0(z + iη) R A(z)1 is analytic in z in the region between Γ ′ and Γ . It follows that
g =
∮
Γ
m0(z)m0(z + iη) R A(z)1dz =
∮
Γ′
m0(w) m0(w + iη) R A(w) 1dw.
Thus,
‖g‖2 (a)
=
∮
Γ
dz
∮
Γ′
dw m0(z)m0(z + iη) m0( ¯w) m0( ¯w − iη) 1 R A( ¯w) R A(z)1
(b)
= −
∮
Γ
dz
∮
Γ′
dw m0(z)m0(z + iη) m0(w) m0(w − iη) 1 R A(w) R A(z)1
(c)
= −
∮
Γ
dz
∮
Γ′
dw m0(z)m0(z + iη) m0(w) m0(w − iη) 1 R A(z) − R A(w)
z − w 1
(d)
n
∮
Γ
dz
∮
Γ′
1
|z − w| (64) where (a) applies conjugation symmetry of m0 and R A; (b) changes variables w → ¯ w
which reverses the direction of integration along Γ ′; (c) follows from the identity
123 Foundations of Computational Mathematics
R A(z)R A(w) = (A − z)−1 (A − w) −1 = 1
z − w
[
(A − z)−1 − (A − w) −1]
= 1
z − w [R A(z) − R A(w) ] (65) and (d) holds because |m0(z)| 1 and ∣∣1 R A(z)1∣∣ n for all z ∈ Γ ∪ Γ ′ by assumption. For either z or w in the vertical strips of Γ ∪ Γ ′ of length O(η) , we apply simply |z − w| η. For both z and w in the horizontal strips, i.e., | Im z| = η/ 2 and
| Im w| = η/ 4, we apply |z − w| | Re (z) − Re (w) | + η. This gives
‖g‖2 n
(
1 +
∫ 3
−3
dx
∫ 2.6
−2.6
dy 1
|x − y| + η
)
n log 1
η .
For ‖h‖2 , we have similarly
‖h‖2 = −
∮
Γ
dz
∮
Γ′
dw m0(z)m0(z + iη) m0(w) m0(w − iη) 1
R B (z + iη) − R B (w − iη) (z + iη) − (w − iη) 1 n
∮
Γ
dz
∮
Γ′
1
|z − w + 2iη|.
We may again bound |z − w + 2iη| η if either z or w belongs to a vertical strip, or
|z − w + 2iη| | Re (z) − Re (w) | + η otherwise, to obtain ‖h‖2 n log (1/η) .Finally, we bound ‖M‖F . Since ‖A‖ ≤ 2.5, the function m0(z)m0(z +
iη) R A(z)JR B (z + iη) is analytic in z in the region between Γ ′ and Γ , so
M =
∮
Γ
m0(z)m0(z + iη) R A(z)JR B (z + iη) dz =
∮
Γ′
m0(w) m0(w + iη)
R A(w) JR B (w + iη) dw.
Consequently, by the same arguments that leads to (64),
‖M‖2
F
= Tr (M∗ M)
=
∮
Γ
dz
∮
Γ′
dw m0(z)m0(z + iη) m0(w) m0(w − iη) Tr
× [R A(z)11 R B (z + iη) R B (w − iη) 11 R A(w) ]
= −
∮
Γ
dz
∮
Γ′
dw m0(z)m0(z + iη) m0(w) m0(w − iη) 1 R A(w) R A(z)11
R B (z + iη) R B (w − iη) 1
= −
∮
Γ
dz
∮
Γ′
dw m0(z)m0(z + iη) m0(w) m0(w − iη)
1(R A(z) − R A(w)) 1
z − w
1(R B (z + iη) − R B (w − iη)) 1
z + iη − (w − iη)
123 Foundations of Computational Mathematics
n2
∮
Γ
dz
∮
Γ′
dw 1
|z − w|
1
|z − w + 2iη| .
If z or w belongs to a vertical strip of Γ ∪Γ ′, of length O(η) , then |z−w|·| z−w+2iη|
η2 ; otherwise, |z−w|·| z−w+2iη| (| Re (z)−Re (w) |+ η) 2 (Re (z)−Re (w)) 2+η2 .Then
‖M‖2
F
n2
( 1
η +
∫ 3
−3
dx
∫ 2.6
−2.6
dy 1
(x − y)2 + η2
)
n2
η .
5.3 Diagonal Entries
Without loss of generality, we consider the diagonal entry X11 :
X11 = 12π Re
∮
Γ
[
e
1
R A(z)1
] [
1 R B (z + iη) e1
]
dz .
By similar arguments as in the off-diagonal entry X12 that lead to (50) and (51), we obtain that for all z ∈ Γ ,
e
1
R A(z)1 = m0(z)
(
1 − a
1
R(1)
A
(z)1n−1
)
O (δ1δ2
√n)
e
1
R B (z + iη) 1 = m0(z + iη)
(
1 − b
1
R(1)
B
(z)1n−1
)
O (δ1δ2
√n) .
It follows that
[
e
1
R A(z)1
] [
1 R B (z + iη) e1
]
= m0(z)m0(z + iη)
(
1 − a
1
R(1)
A
1n−1 − 1
n−1
R(1)
B
b1 + a
1
R(1)
A
Jn−1 R(1)
B
b1
)
O
(
δ1δ22 n
)
,
where, respectively, a
1
and b
1
are the first rows of A and B with first entries removed; and R(1)
A
and R(1)
B
are the resolvents of the minors of A and B with first rows and columns removed. Thus, we get that
∮
Γ
[
e
1
R A(z)1
] [
1 R B (z + iη) e1
]
dz
=
∮
Γ
m0(z)m0(z + iη) dz − a
1
g − b
1
h + a
1
Mb 1 + O
(
δ1δ22 n
)
, (66) where
g
∮
Γ
m0(z)m0(z + iη) R(1)
A
(z)1dz ,
123 Foundations of Computational Mathematics
h
∮
Γ
m0(z)m0(z + iη) R(1)
B
(z + iη) 1dz ,
M
∮
Γ
m0(z)m0(z + iη) R(1)
A
(z)JR(1)
B
(z + iη) dz .
By the same argument as in the off-diagonal entry X12 , we can control each term above. The only difference is that for the bilinear form, instead of using (26), applying Lemma 5 to control a
1
Mb 1 gives an extra expectation term (1−σ 2)n−1 Tr M. There-fore, we obtain that for any fixed constant D > 0, with probability at least 1 − n−D ,for all sufficiently large n,
∣∣∣∣X11 − 1 − σ 22π Re Tr Mn
∣∣∣∣ (log n) κ 1
√η . (67) Denote by E4 the event where the following hold simultaneously for all z ∈ Γ :
‖A − B‖ σ
∣∣∣1
n−1
R(1)
A
(z)1n−1 − m0(z)n
∣∣∣ δ3n
∣∣∣1
n−1
R(1)
B
(z + iη) 1n−1 − m0(z + iη) n
∣∣∣ δ3n.
By the assumption (9) and Theorem 2, we have that P {E4} ≥ 1−n−D for any constant
D > 0 and all n ≥ n0(D).We defer the analysis of Tr M to Lemmas 9 and 10: Assuming E4 holds and applying Lemmas 9 and 10 with R A, R B replaced by R(1)
A
, R(1)
B
, respectively, we get 1
n Re Tr (M) = 2π + oη (1)η + O
( ση2 + δ3
η
)
. (68) Setting r(n) = oη (1) + δ3 , we get
∣∣∣∣X11 − 1 − σ 2
η
∣∣∣∣ r(n)η + ση2 + (log n) κ 1
√η .
5.3.1 Analyzing the Trace of M
Lemma 9 Suppose ‖A‖ ≤ 2.5 and ‖A − B‖ σ and
∣∣∣1 R A(z)1 − m0(z)n
∣∣∣ δ3n,
∣∣∣1 R B (z + iη) 1 − m0(z + iη) n
∣∣∣ δ3n, (69)
for all z ∈ Γ . Define M =
∮
Γ
m0(z)m0(z + iη) R A(z)JR B (z + iη) dz .
123 Foundations of Computational Mathematics
Then,
1
n Tr M = 1
iη
∮
Γ
m0(z)m0(z + iη)( m0(z + iη) − m0(z)) dz + O
( ση2 + δ3
η
)
.
Proof Applying the identity
R B (z + iη) − R A(z) = (B − (z + iη)) −1 − (A − z)−1 = R B (z + iη)( A − B + iη) R A(z),
we get R B (z + iη) R A(z) = 1
iη
(R B (z + iη) − R A(z) − R B (z + iη)( A − B)R A(z)) .Therefore Tr M =
∮
Γ
dz m 0(z)m0(z + iη) Tr [R A(z)JR B (z + iη) ]
=
∮
Γ
dz m 0(z)m0(z + iη) 1 R B (z + iη) R A(z)1
= 1
iη
∮
Γ
dz m 0(z)m0(z + iη) 1 (R B (z + iη) − R A(z)
−R B (z + iη)( A − B)R A(z)) 1. (70) To proceed, we use the following facts. First, it holds that
∣∣∣1 R B (z + iη)( A − B)R A(z)1
∣∣∣ ≤
∥∥∥1 R B (z + iη)
∥∥∥ ‖A − B‖ ‖ R A(z)1‖ .
For z ∈ Γ with Im z = ± η/ 2, in view of the Ward identity given in Lemma 2 and the assumption given in (69), we get that
‖R A(z)1‖2 = 1 R A(z)R A(z)1 = 2
η | Im 1 R A(z)1| n
η
For z ∈ Γ with Re z = ± 3, we have that ‖R A(z)1‖2 ≤ n ‖R A(z)‖2 n thanks to the assumption ‖A‖ ≤ 2.5. Similarly, we have ‖R B (z + iη) 1‖2 n/η . Combining these bounds with the assumption that ‖A − B‖ σ yields that
∣∣∣1 R B (z + iη)( A − B)R A(z)1
∣∣∣ nση .
Then applying |m0(z)| 1 and (69), we obtain 1
n Tr M = 1
iη
∮
Γ
m0(z)m0(z + iη)( m0(z + iη) − m0(z)) dz + O
( ση2 + δ3
η
)
.
123 Foundations of Computational Mathematics
Lemma 10 Let Γ be the rectangular contour with vertices ±3 ± iη/ 2. Then
Im
[∮
Γ
m0(z)m0(z + iη)( m0(z + iη) − m0(z)) dz
]
= 2π + oη (1).
Proof By Proposition 2, the integrand is analytic and bounded over
{z ∈ C : | z| ≤ 9, z /∈ [− 2, 2], z + iη / ∈ [− 2, 2]} .
Hence, we may deform Γ to the contour Γ with vertices ±(2+ε) ±iε, and take ε → 0(for fixed η). The portion of Γ where | Re z| > 2 has total length O(ε) , so the integral over this portion vanishes as ε → 0. We may apply the bounded convergence theorem for the remaining two horizontal strips of Γ to get (recall that contour integrals are evaluated counterclockwise):
∮
Γ
m0(z)m0(z + iη)( m0(z + iη) − m0(z)) dz
=
∫ −22
m+
0
(x)m0(x + iη)( m0(x + iη) − m+
0
(x)) dx
+
∫ 2
−2
m−
0
(x)m0(x + iη)( m0(x + iη) − m−
0
(x)) dx ,
where m+
0
and m−
0
are the limits from C+ and C− defined in Proposition 2. Now applying the bounded convergence theorem again to take η → 0, we have lim η→0 m0(x + iη) = m+
0
(x) and hence lim
η→0
∮
Γ
m0(z)m0(z + iη)( m0(z + iη) − m0(z)) dz
=
∫ 2
−2
m−
0
(x)m+
0
(x)( m+
0
(x) − m−
0
(x)) dx =
∫ 2
−2
|m+
0
(x)|2 · 2πiρ( x)dx = 2πi,
the last two steps applying (32). Thus, the imaginary part of the integral is 2 π + oη (1)
for small η.
6 A Tighter Regularized QP Relaxation
As discussed in the introduction, GRAMPA can be interpreted as solving the regular-ized QP relaxation (11) of the QAP (10). We further explore this optimization aspect in this section, and study a tighter regularized QP relaxation. Let us begin by recalling the following QP relaxation of the QAP (10) that replaces the feasible set of permutation matrices by its convex hull, the Birkhoff polytope consisting of all doubly stochastic matrices [1, 21]:
123 Foundations of Computational Mathematics
min
X∈Rn×n
‖AX − X B ‖2
F
s.t. X1 = 1, X1 = 1, X ≥ 0. (71) This program differs from the QP relaxation (11) that underlies GRAMPA in two aspects. First, the added ridge penalty η2‖X‖2
F
in (11) is crucial for ensuring the desired statistical property of the solution, 3 while for (71) there is no such need for regularization. Moreover, the Birkhoff polytope constraint, being the tightest possible convex relaxation, is significantly tighter than the constraint 1 X1 = n. Although it is much slower to solve (71) than to implement GRAMPA, the doubly stochastic relaxation achieves superior performance over the weaker program (11) as demon-strated by ample empirical evidence (cf. [10, 14]); nevertheless, a rigorous theoretical understanding is still lacking. As a further step toward understanding the relaxations, we analyze the following intermediate program between (71) and (11): min
X∈Rn×n
‖AX − X B ‖2
F
η2‖X‖2
F
s.t. X1 = 1, (72) where we enforce the sum of each row of X to be equal to one. The above program without the regularization term η2‖X‖2
F
has been studied in in a small noise regime. As we are analyzing the structure of the solution rather than the value of the program, the exact recovery guarantee for GRAMPA (and hence for (11)) does not automatically carries over to the tighter program (72). Fortunately, we are able to employ similar technical tools to analyze the solution to (72), denoted henceforth by Xc.The following result is the counterpart of Theorem 1 and Corollary 1:
Theorem 3 Fix constants a > 0 and κ > 2, and let η ∈ [ 1/( log n)a , 1].Consider the correlated Wigner model with n ≥ d ≥ (log n)c0 where c 0 >
max (34 + 11 a, 8 + 12 a). Then, there exist (α, κ) -dependent constants C , n0 > 0
and a deterministic quantity r (n) ≡ r(n, η, d, a) satisfying r (n) → 0 as n → ∞ ,such that for all n ≥ n0, with probability at least 1 − n−10 ,
max
π∗(k)=
∣∣n · Xc
k
∣∣ ≤ C(log n) κ 1
√η , (73) max
k
∣∣∣∣n · Xc
kπ∗(k)
− 4(1 − σ 2)πη
∣∣∣∣ ≤ C
(r(n)η + ση2 + (log n) κ 1
√η
)
. (74)
If ‖ai j ‖ψ2 , ‖bi j ‖ψ2 ≤ K /√n, then the above guarantees hold also for κ = 1, with constants possibly depending on K . Furthermore, there exist constants c , c′ > 0 such that for all n ≥ n0, if
(log n)−a ≤ η ≤ c(log n)−2κ and σ ≤ c′η, (75) 3 See [14,Sect. 1.3] for a more detailed discussion in this regard.
123 Foundations of Computational Mathematics
Fig. 2 Fraction of correctly matched pairs of vertices by GRAMPA and the tighter QP (72) (both followed by linear assignment rounding) on Erd˝ os-Rényi graphs with 1000 vertices and edge density 0.5, averaged over 10 repetitions
then with probability at least 1 − n−10 ,
min
k
X kπ∗(k) > max
π∗(k)=
X k . (76) Compared with Corollary 1, the theoretical guarantee for the tighter program (72) is similar to that for (11) and the GRAMPA method. In practice the performance of the former is slightly better (cf. Fig. 2). Furthermore, Theorem 3 applies verbatim to the solution of (72) with column-sum constraints X1 = 1 instead. This simply follows by replacing (A, B, X, Π ∗) with (B, A, X, Π ∗ ).
6.1 Structure of Solutions to QP Relaxations
Before proving Theorem 3, we first provide an overview of the structure of solutions to the QP relaxations (11), (72), and (71). Using the Karush–Kuhn–Tucker (KKT) conditions, the solution of (72) can be expressed as
Xc = ∑
i,j
〈vi , μ 〉〈 w j , 1〉
(λ i − μ j )2 + η2 vi w
j
, (77) where μ ∈ Rn is the dual variable corresponding to the row sum constraints, chosen so that Xc is feasible. Since
Xc1 = ∑
i,j
〈w j , 1〉2
(λ i − μ j )2 + η2 vi v
i
μ =
{∑
i
τi vi v
i
}
μ,
123 Foundations of Computational Mathematics
where
τi ∑
j
〈w j , 1〉2
(λ i − μ j )2 + η2 . (78) Solving Xc1 = 1 yields
μ = ∑
i
〈vi , 1〉
τi
vi , (79) so we obtain
Xc = ∑
i,j
1
(λ i − μ j )2 + η21
τi
vi v
i
Jw j w
j
. (80) Let us provide some heuristics regarding the solution Xc. As before we can express
τi via resolvents as follows:
τi = 1
η Im ∑
j
〈w j , 1〉2
μ j − (λ i + iη) = 1
η 1
[
Im ∑
j
1
μ j − (λ i + iη) w j w
j
]
1
= 1
η Im [1 R B (λ i + iη) 1]. (81) Invoking the resolvent bound (36), we expect τi ≈ n
η
Im [m0(λ i +iη) ], where, by prop-erties of the Stieltjes transform (cf. Proposition 2), Im [m0(λ i + iη) ] ≈ Im [m0(λ i )] =
πρ(λ i ) as η → 0. Thus, we have the approximation
Xc ≈ 1
πn
∑
i,j
η(λ i − μ j )2 + η21
ρ(λ i ) vi v
i
Jv j w
j
,
Compared with the unconstrained solution (3), apart from normalization, the only difference is the extra spectral weight 1
ρ(λ i)
according to the inverse semicircle density. The effect is that eigenvalues near the edge are upweighted while eigenvalues in the bulk are downweighted, the rationale being that eigenvectors corresponding to the extreme eigenvalues are more robust to noise perturbation.
Remark 1 (Structure of the QP solutions) Let us point out that solution of various QP relaxations, including (71), (72), and (11), is of the following common form:
X = ∑
i,j
η(λ i − μ j )2 + η2 vi v
i
Sw j w
j
, (82) where S is an n × n matrix that can depend on A and B. Specifically, from the loosest to the tightest relaxations, we have:
123 Foundations of Computational Mathematics
– For (11) with the total sum constraint, S = αJ, where the dual variable α > 0 is chosen for feasibility. Since scaling by α does not affect the subsequent rounding step, this is equivalent to (3) that we analyze. – For (72) with the row sum constraint, S = μ1 is rank-one with μ given in (79). – For (71) without the positivity constraint, S = μ1 + 1ν is rank-two. Unfortu-nately, the dual variables and the spectral structure of the optimal solution turn out to be difficult to analyze. – For (71) with the positivity constraint, S = μ1 + 1ν + H , where H ≥ 0 is the dual variable certifying the positivity of the solution and satisfies complementary slackness.
6.2 Proof of Theorem 3
We now apply the resolvent technique to analyze the behavior of the constrained solution Xc and establish its diagonal dominance.
6.2.1 Resolvent Representation of the Solution
We start by giving a resolvent representation of Xc via a contour integral.
Lemma 11 Consider symmetric matrices A and B with the spectral decomposi-tions (2) , and suppose that ‖A‖ ≤ 2.5. Then, the solution X c of the program (72)
admits the following representation Xc = 12π Re
∮
Γ
F(z)R A(z)JR B (z + iη), (83)
where Γ is defined by (17) and F(z) 2i1 R B (z + iη) 1 − 1 R B (z − iη) 1 . (84)
Proof By (81) we have τ −1
i
= ηF(λ i ). This leads to the following contour represen-tation of Xc analogous to (16) for the unconstrained solution:
Xc = η ∑
i
F(λ i )v i v
i
J
⎧⎨⎩∑
j
1
(λ i − μ j )2 + η2 w j w
j
⎫⎬⎭
(a)
= Im
[∑
i
F(λ i )v i v
i
JR B (λ i + iη)
]
(b)
= Im
[ 1
−2πi
∮
Γ
F(z)R A(z)JR B (z + iη)
]
= 12π Re
∮
Γ
F(z)R A(z)JR B (z + iη),
123 Foundations of Computational Mathematics
where (a) follows from the Ward identity (Lemma 2); (b) follows from Cauchy integral formula and the analyticity of F in the region enclosed by the contour Γ .
6.2.2 Entrywise Approximation
For some small constant ε > 0, let b, b′ be as defined in Theorem 2. Under the assumptions of Theorem 3, we have c0 > b′ for ε sufficiently small, so that Theorem 2 applies. Recall the notation δ1, . . . , δ 4 defined in (45). For sufficiently small ε > 0, we may also verify under the assumptions of Theorem 3 that δi = o(1) for each
i = 1, 2, 3, 4, and
δ1δ22 n
η ≤ 1, δ22 δ3n
η2 ≤ (log n) κ
√η , and δ3 ≤ η3. (85) We also assume throughout the proof that the high-probability event ‖A‖ ≤ 2.5 holds. Thanks to (36), we can approximate F(z) by
˜F(z) = 1
n
2i
m0(z + iη) − m0(z − iη) (86) and approximate Xc by
˜Xc = 12π Re
∮
Γ
˜F(z)R A(z)JR B (z + iη) (87)
= −1
πn Im
∮
Γ
1
m0(z + iη) − m0(z − iη) R A(z)JR B (z + iη). (88) The following lemma makes the approximation of Xc precise in the entrywise sense:
Lemma 12 Suppose (85) holds. On the high-probability event where Theorem 2 holds and also ‖A‖ ≤ 2.5,
‖˜Xc − Xc‖ ∞ δ22 δ3
η2 ≤ (log n) κ
n√η , (89)
where δ2, δ 3 are defined in (45) .
Proof For notational convenience, put G(z) = 2i/( n F (z)) and ˜G(z) = 2i/( n ˜F(z)) .Note that | Im (z)| ≤ η/ 2 for z ∈ Γ , and thus Im (z + iη) and Im (z − iη) have different signs. Therefore,
|˜G(z)| ≥ | Im ˜G(z)| = | Im m0(z + iη) | + | Im m0(z − iη) | η,
where the last step follows from (31). Furthermore, by (36), we have sup z∈Γ |G(z) −
˜G(z)| ≤ 2Cδ3 . In view of (85), δ3 η. Hence, we have |G(z)| η and sup
z∈Γ
|F(z) − ˜F(z)| 1
n
δ3
η2 .
123 Foundations of Computational Mathematics
Finally, by (83) and (87), we have
|(Xc − ˜Xc)k | ≤
∮
Γ
dz |F(z) − ˜F(z)|| e
k
R A(z)1|| e R B (z + iη) 1|.
By (35), for all k, , |e
k
R A(z)1| δ2
√n and |e R B (z + iη) 1| δ2
√n. Combining the last two displays yields the desired claim.
In view of the entrywise approximation, we may switch our attention to the approx-imate solution ˜Xc and establish its diagonal dominance, assuming without loss of generality π∗ is the identity permutation. The proof parallels the analysis in Sect. 5 so we focus on the differences. To make the scaling identical to the unconstrained case, define
Y n ˜Xc = 12π Re
∮
Γ
f (z)R A(z)JR B (z + iη), (90) with
f (z) 2i
m0(z + iη) − m0(z − iη) .
Compared with the unconstrained solution (16), the only difference is the weighting factor f (z).We aim to show that with probability at least 1 − n−D , for any constant D > 0, the following holds: 1. For off-diagonals, we have max
k=
|Yk | (log n) κ /√η. (91) 2. For diagonal entries, we have min
k
∣∣∣∣Ykk − 4(1 − σ 2)πη
∣∣∣∣ r(n)η + ση2 + (log n) κ 1
√η . (92) In view of Lemma 12, this implies the desired (73) and (74). Finally, analogous to Corollary 1, under the assumption (75) with constants c = 1/( 64 C2) and c′ = 1/( 2C),for all sufficiently large n,4(1 − σ 2)πη ≥ 78η > C
(r(n)η + ση2 + 2(log n) κ 1
√η
)
,
implying the diagonal dominance in (76).
123 Foundations of Computational Mathematics
6.2.3 Off-Diagonal Entries
Let us first consider Y12 . Recall that for z ∈ Γ , we have | Im (z + iη) | η, | Im (z −
iη) | η, and these imaginary parts have opposite signs. Then,
| f (z)| ≤ 2
| Im [m0(z + iη) − m0(z − iη) ]| = 2
| Im m0(z + iη) | + | Im m0(z − iη) | 1
η ,
(93) where the last step applies (31). Analogous to (52), we get 2πY12 = Re
(∮
Γ
f (z)
[
e
1
R A(z)1
] [
e
2
R B (z + iη) 1
]
dz
)
= Re
(
α − a
1
g − b
2
h + a
1
Mb 2
)
O
( δ1δ22 n
η
)
, (94) where
α
∮
Γ
f (z)m0(z)m0(z + iη) dz , (95)
g
∮
Γ
f (z)m0(z)m0(z + iη) R(12 )
A
(z)1n−2dz , (96)
h
∮
Γ
f (z)m0(z)m0(z + iη) R(12 )
B
(z + iη) 1n−2dz , (97)
M
∮
Γ
f (z)m0(z)m0(z + iη) R(12 )
A
(z)Jn−2 R(12 )
B
(z + iη) dz . (98) Here, the constant Re α is in fact equal to 2 π, which is consistent with the row-sum constraints. Indeed, opening up m0(z) and applying the Cauchy integral formula, we have Re α = Re
∮
dz 2i
m0(z + iη) − m0(z − iη) m0(z)m0(z + iη)
=
∫
ρ( x)dx Re
∮
dz 1
x − z
2i m0(z + iη)
m0(z + iη) − m0(z − iη)
=
∫
ρ( x)dx Re
[
(−2πi) 2i m0(x + iη)
m0(x + iη) − m0(x − iη)
]
= 2π
∫
ρ( x)dx Re
[ 2 m0(x + iη)
2i Im m0(x + iη)
]
= 2π
∫
ρ( x)dx = 2π. (99) As in Sect. 5.2.2, to bound the linear and bilinear terms, we need to bound the
∞
-norms and 2 -norms of g, h and M. Clearly, by (93), the ∞-norms are at most an O(1/η) factor of those obtained in (61) and (62), i.e., ‖g‖∞ δ2
√n/η and
‖M‖∞ δ22 n/η . The 2 -norms need to be bounded more carefully. The following result is the counterpart of Lemma 8:
123 Foundations of Computational Mathematics
Lemma 13 Assume the same setting of Lemma 8, and define M, g, and h as in (96–98)
with R A , R B in place of R (12 )
A
, R(12 )
B
. Then, ‖M‖2
F
n2/η , ‖g‖2 n log (1/η) , and
‖h‖2 n log (1/η) .
Proof We start with ‖M‖F , as the arguments for ‖g‖ and ‖h‖ are analogous and simpler. Recall the contour Γ ′ from Fig. 1. Proceeding as in the proof of Lemma 8, we have 1
n2‖M‖2
F=−
∮
Γ
dz
∮
Γ′dwm0(z)m0(z+iη) m0(w) m0(w −iη) f(z)f(w) ×
n−11(RA(z)−RA(w)) 1
z−w
n−11(RB(z+iη) −RB(w −iη)) 1
z+iη−(w −iη)
= −
∮
Γ
dz
∮
Γ′dwm0(z)m0(z+iη) m0(w) m0(w −iη) f(z)f(w) m0(z)−m0(w)
z−w
m0(z+iη) −m0(w −iη)
z+iη−(w −iη)
︸︷︷ ︸
(I)
+(II ),
where (II ) denotes the remainder term. Applying (36), (93), and the boundedness of
m0 , the residual term is bounded as
|(II )| δ3
∮
Γ
dz
∮
Γ′
dw| f (z)|| f (w) | 1
|z − w|
1
|z + iη − (w − iη) | δ3
η4 1
η .
(100) To control the leading term (I), let us define the auxiliary contours γ with vertices
±(2 + 2η) ± (η/ 2)i and γ ′ with vertices ±(2 + η) ± (η/ 4)i. By first deforming Γ ′
to γ ′ for each fixed z ∈ Γ , then deforming Γ to γ , and finally taking the complex modulus and applying |m0| 1, we get
|(I)|
∮
γ
dz
∮
γ′
dw | f (z)|| f (w) |
∣∣∣∣
m0(z) − m0(w)
z − w
∣∣∣∣∣∣∣∣
m0(z + iη) − m0(w − iη)
z + iη − (w − iη)
∣∣∣∣ .
The reason for performing these deformations is that for any z ∈ γ ∪ γ ′, since Re z ∈[− 2−2η, 2+2η], we have from (31) that Im m0(z+iη) √η + ζ ( z) and − Im m0(z−
iη) √η + ζ ( z), where ζ ( z) is as defined in Proposition 2. Then, we obtain from (93) the improved bound | f (z)| 1/√η + ζ ( z), and hence
|(I)|
∮
γ
dz
∮
γ′
dw 1
√η + ζ ( z)
1
√η + ζ (w)
∣∣∣∣
m0(z) − m0(w)
z − w
∣∣∣∣∣∣∣∣
m0(z + iη) − m0(w − iη)
z + iη − (w − iη)
∣∣∣∣ .
To bound the above integral, for a small constant c0 > 0, consider the two cases where |z − w| ≥ c0 and |z − w| < c0 . For the first case |z − w| ≥ c0 , we simply apply
|m0| 1 and √η + κ ≥ √η to get that
∮ ∮
|z−w|≥ c0
dz d w 1
√η + ζ ( z)
1
√η + ζ (w)
∣∣∣∣
m0(z) − m0(w)
z − w
∣∣∣∣ |
m0(z + iη) − m0(w − iη)
z + iη − (w − iη)
∣∣∣∣ 1
η . (101)
123 Foundations of Computational Mathematics
In the second case |z − w| < c0 , we claim that for c0 sufficiently small, we have
|m0(z) − m0(w) | √η + ζ ( z) + √η + ζ (w), (102)
|m0(z + iη) − m0(w − iη) | √η + ζ ( z) + √η + ζ (w). (103) Indeed, if ζ ( z) > c0 , then (102) and (103) hold because √η + ζ ( z)+√η + ζ (w) 1. If instead ζ ( z) ≤ c0 , say, Re z ≥ 2 − c0 , then from the explicit form (29) for m0(z)
we get 1 + m0(z) = 2−z+√z2−42 and hence
|1 + m0(z)| |z − 2| + √|z − 2|| z + 2| √|z − 2| √η + ζ ( z).
Furthermore, since Re w ≥ Re z − | z − w| ≥ 2 − 2c0 , we also have |1 + m0(w) | √η + ζ (w) . Then, (102) follows from the triangle inequality. The case of Re z ≤−2 + c0 , and the argument for (103), are analogous. Having established (102) and (103), we apply
(√η + ζ ( z) + √η + ζ (w) )2
√η + ζ ( z)√η + ζ (w)
√η + max (ζ ( z), ζ (w))
√η + min (ζ ( z), ζ (w))
≤√η + min (ζ ( z), ζ (w)) + √|ζ ( z) − ζ (w) |√η + min (ζ ( z), ζ (w)) ≤ 1 +√|z − w|√η
to get
∮ ∮
|z−w|<c0
dz d w 1
√η + ζ ( z)
1
√η + ζ (w)
∣∣∣∣
m0(z) − m0(w)
z − w
∣∣∣∣∣∣∣∣
m0(z + iη) − m0(w − iη)
z + iη − (w − iη)
∣∣∣∣
∮ ∮
|z−w|<c0
dz d w
(
1 +√|z − w|√η
) 1
|z − w|| z + iη − (w − iη) | .
Then divide this into the integrals where |z − w| < η and |z − w| ≥ η, applying
∮ ∮
|z−w|<η
dz d w 1
|z − w|| z + iη − (w − iη) |
∮ ∮
|z−w|<η
dz d w 1
η2 1
η
and
∮ ∮
η≤| z−w|<c0
dz d w
√|z − w|√η · 1
|z − w|| z + iη − (w − iη) |
1
√η
∮ ∮
η≤| z−w|<c0
dz d w 1
|z − w|3/2 1
√η
1
√η 1
η . (104) Combining with the first case (101), we get |(I)| 1/η . Finally, combining with (100), we get ‖M‖2
F
n2/η as desired.
123 Foundations of Computational Mathematics
Next we bound ‖g‖. Proceeding as in the proof of Lemma 8 and following the same argument as above, we get 1
n ‖g‖2
∮
γ
dz
∮
γ′
dw | f (z)|| f (w) | |m0(z) − m0(w) ||z − w| + O
( δ3
η3
)
∮
γ
dz
∮
γ′
dw 1
√η + ζ ( z)
1
√η + ζ (w)
|m0(z) − m0(w) ||z − w| + O
( δ3
η3
)
.
For |z − w| ≥ c0 , we have
∮ ∮
|z−w|≥ c0
dzd w 1
√η + ζ ( z)
1
√η + ζ (w)
|m0(z) − m0(w) ||z − w|
(∮ 1
√η + ζ ( z) dz
) (∮ 1
√η + ζ (w) dw
)
1.
For |z − w| < c0 , we apply |m0(z) − m0(w) | √η + ζ ( z) + √η + ζ (w) as above, so that
∮ ∮
|z−w|<c0
dzd w 1
√η + ζ ( z)
1
√η + ζ (w)
|m0(z) − m0(w) ||z − w|
∮
dz 1
√η + ζ ( z)
∮
dw 1
|z − w| +
∮
dw 1
√η + ζ (w)
∮
dz 1
|z − w|
log (1/η) ·
(∮
dz 1
√η + ζ ( z) +
∮
dw 1
√η + ζ (w)
)
log (1/η).
Combining the above yields ‖g‖2 n log (1/η) . The argument for ‖h‖2 is the same as that for ‖g‖2 .
Finally, proceeding as in (55)–(56) and using the preceding norm bounds, we obtain from (94):
|Y12 | 1 + δ4
√
n log 1
η + δ24 n
√η + δ1δ22 n
η δ24 n
√η = (log n) κ /√η,
with probability at least 1 − n−D , for any constant D. This implies the desired (91) by the union bound.
6.2.4 Diagonal Entries
We now consider Y11 . Following the derivation from (66) to (67) and using Lemma 13 in place of Lemma 8, we obtain, with probability at least 1 − n−D for any constant
D,
∣∣∣∣Y11 − 1 − σ 22π Re Tr (M)
n
∣∣∣∣ (log n) κ 1
√η , (105)
123 Foundations of Computational Mathematics
where
M ∮
Γ
f (z)m0(z)m0(z + iη) R(1)
A
(z)JR(1)
B
(z + iη) dz .
The trace is computed by the following result, which parallels Lemma 9 and Lemma 10:
Lemma 14 Suppose δ3 ≤ η2. Assume the setting of Lemma 9. Define M =
∮
Γ
f (z)m0(z)m0(z + iη) R A(z)JR B (z + iη) dz .
Then,
1
n Tr (M) = 8 + oη (1)η + O
( σ + δ3
η2
)
.
Proof Analogous to (70), we have 1
n
Tr (M) = (I) − (II ), where
(I) = 1
iη
∮
Γ
f (z)m0(z)m0(z + iη) 1
n 1(R B (z + iη) − R A(z)) 1dz
(II ) = 1
iη
∮
Γ
f (z)m0(z)m0(z + iη) 1
n 1 R B (z + iη)( A − B)R A(z)1 dz .
To bound (II), consider two cases: – For z ∈ Γ with | Im z| = η/ 2, by the Ward identity and (36), we have
‖R A(z)1‖2 = 2
η | Im 1 R A(z)1| n
η (| Im m0(z)| + O(δ 3)).
and similarly,
‖R B (z + iη) 1‖2 n
η (| Im m0(z + iη) | + O(δ 3)).
Thus, it holds that
∣∣∣1 R B (z + iη)( A − B)R A(z)1
∣∣∣ nση
(√ | Im m0(z) Im m0(z + iη) | + √δ3
)
.
Using (31) and (93), we conclude that
| f (z)|√| Im m0(z) Im m0(z + iη) | ≤ 2√| Im m0(z) Im m0(z + iη) || Im m0(z + iη) | + | Im m0(z − iη) | 1for all z ∈ Γ with | Im z| = η/ 2.
123 Foundations of Computational Mathematics
– For z ∈ Γ with Re z = ± 3, since ‖A‖ ≤ 2.5, ∣∣1 R B (z + iη)( A − B)R A(z)1∣∣
nσ .Furthermore, by (93), | f (z)| 1
η
for all z ∈ Γ . Combining the above two cases yields
|(II )| ση2
(
1 +√δ3
η
)
ση ση2 ,
since δ3 ≤ η2 by the assumption. For (I), applying (36) again and plugging the definition of f (z) yields
(I) = 2
η
∮
Γ
m0(z)m0(z + iη) m0(z + iη) − m0(z)
m0(z + iη) − m0(z − iη) dz + O
( δ3
η2
)
.
We now apply an argument similar to that of Lemma 10: Note that
|m0(z + iη) − m0(z − iη) | ≥ Im (m0(z + iη) − m0(z − iη)) η
by (31), so the integrand is bounded for fixed η. Then deforming Γ to Γ with vertices
±(2 + ε) ± iε, taking ε → 0 for fixed η, and applying the bounded convergence theorem, we have the equality
∮
Γ
m0(z)m0(z + iη) m0(z + iη) − m0(z)
m0(z + iη) − m0(z − iη) dz
=
∫ −22
m+
0
(x)m0(x + iη) m0(x + iη) − m+
0
(x)
m0(x + iη) − m0(x − iη) dx
+
∫ 2
−2
m−
0
(x)m0(x + iη) m0(x + iη) − m−
0
(x)
m0(x + iη) − m0(x − iη) dx . (106) We show that these integrands are uniformly bounded over small η: For any constant
δ > 0 and for |x| ≤ 2 − δ, we have the lower bound
|m0(x + iη) − m0(x − iη) | = 2 Im m0(x + iη) √ζ ( x) + η ≥ √δ. (107) Then, the above integrands are bounded by C/√δ for |x| ≤ 2 − δ. For |x| ∈ [ 2 − δ, 2],let us apply
|m0(x + iη) − m+
0
(x)| √ζ ( x) + η
as follows from (102) and taking the limit w ∈ C+ → x. We have also |m+
0
(x) −
m−
0
(x)| √ζ ( x) √ζ ( x) + η, so that
|m0(x + iη) − m−
0
(x)| √ζ ( x) + η.
Combining these cases with the first inequality of (107), we see that the integrands of (106) are uniformly bounded for all small η.
123 Foundations of Computational Mathematics
Now apply the bounded convergence theorem and take the limit η → 0, noting that lim η→0 m0(x + iη) = m+
0
(x) and lim η→0 m0(x − iη) = m−
0
(x). We get lim
η→0
∮
Γ
m0(z)m0(z + iη) m0(z + iη) − m0(z)
m0(z + iη) − m0(z − iη) dz
=
∫ 2
−2
m−
0
(x)m+
0
(x) m+
0
(x) − m−
0
(x)
m+
0
(x) − m−
0
(x) dx =
∫ 2
−2
|m+
0
(x)|2dx = 4.
This gives (I) = (8 + oη (1))/η + O(δ 3/η 2). Combining with the bound for (II ) yields the lemma.
Finally, combining (105) with Lemma 14 and δ3 η from (85), and applying a union bound yields the desired (92).
7 Proof of Resolvent Bounds
In this section, we prove Theorem 2. The entrywise bounds of part (a) are essentially the local semicircle law of [12,Theorem 2.8], restricted to the simpler domain {z :
dist (z, [− 2, 2]) ≥ (log n)−a } and with small modifications of the logarithmic factors. The bound in (b) follows from (a) using a straightforward Schur complement identity. The bound in (c) is more involved, and relies on the fluctuation averaging technique of [12,Sect. 5]. We provide a proof of all three statements using the tools of . For each statement, it suffices to establish the claim with the stated probability for each individual point z ∈ D. The uniform statement over z ∈ D then follows from a union bound over a sufficiently fine discretization of D (of cardinality an arbitrarily large polynomial in n) and standard Lipschitz bounds for m0 and R jk on the event of
‖A‖ ≤ 2.5—we omit these details for brevity.
7.1 Notation and Matrix Identities
In this section, for S ⊂ [ n], denote by A(S) ∈ Rn×n the matrix A with all elements in rows and columns belonging to S replaced by 0. Denote
R(S) (z) = (A(S) − zI)−1 ∈ Cn×n .
Note that R(S) (z) is block diagonal with respect to the block decomposition Cn =
CS ⊕ C[n]\ S , with S × S block equal to (−1/z)I|S| and ([n] \ S) × ([n] \ S) block equal to the resolvent of the corresponding minor of A. (We will typically only access elements of R(S) in this ([n]\ S)×([n]\ S) block, in which case R(S) may be understood as the resolvent of the minor of A.) For i ∈ [ n], we write as shorthand
i S = { i} ∪ S,(S)∑
k
= ∑
k∈[ n]\ S
.
123 Foundations of Computational Mathematics
We usually omit the spectral argument z for brevity.
Lemma 15 (Schur complement identities) For any j ∈ [ n],
1
R j j
= a j j − z −
(j)
∑
k,
a jk R( j)
k
a j . (108)
For any j = k ∈ [ n],R jk = − R j j
(j)
∑ a j R( j)
k
= R j j R( j)
kk
⎛⎝−a jk +
(jk )
∑
,m
a j R( jk )
m
a mk
⎞⎠ , (109)
e
k
R = e
k
R( j) + R k j R j j
· e
j
R, (110) 1
R kk
= 1
R( j)
kk
− (R k j )2
R( j)
kk
R j j R kk
. (111)
For any j , k, ∈ [ n] with j /∈ { k, },R k = R( j)
k
R k j R j
R j j
. (112)
These identities hold also for any S ⊂ [ n] with R replaced by R (S) and with j , k, ∈[n] \ S.
Proof For all but (110), see [11,Lemma 4.5] and [13,Lemma 4.2]. As for (110), it is equivalent to verify that (112) holds also for = j, which simply follows from
R( j)
k j
= 0, due to the block diagonal structure of R( j).
7.2 Entrywise Bound
We say an event occurs w.h.p. if its probability is at least 1 −e−c(log n)1+ε
for a universal constant c > 0. Let us show that (33) and (34) hold for z ∈ D w.h.p. We start with (34). Note that the jth row {a jk : k ∈ [ n]} is independent of A( j) and hence R( j). Applying (108), (22), and (25) conditional on A( j), w.h.p. for all j,
∣∣∣∣∣∣
1
R j j
z + 1
n
(j)
∑
k
R( j)
kk
∣∣∣∣∣∣
=
∣∣∣∣∣∣a j j −
(j)
∑
k,
a jk R( j)
k
a j + 1
n
(j)
∑
k
R( j)
kk
∣∣∣∣∣∣
≤ (log n)2+2ε
(
1
√d + 2‖R( j)‖∞√d + ‖R( j)‖Fn
)
.
123 Foundations of Computational Mathematics
Note that ‖R( j)‖∞ ≤ ‖ R( j)‖, ‖R( j)‖F ≤ √n‖R( j)‖, and d ≤ n. For z ∈ D1 and any
S ⊂ [ n], we have ‖R(S)‖ ≤ 1/| Im z| ≤ (log n)a . For z ∈ D2 , we have ‖R(S)‖ ≤ 10 on the event ‖A‖ ≤ 2.5, which occurs w.h.p. by Lemma 3. Then in both cases, we get
∣∣∣∣∣∣
1
R j j
z + 1
n
(j)
∑
k
R( j)
kk
∣∣∣∣∣∣ (log n)2+2ε+a
√d . (113) Since |z| ≤ 10, |R( j)
kk
| ≤ (log n)a , and d (log n)4+4ε, this implies 1 /|R j j |
(log n)a . Let m n (z) = n−1 Tr R(z) be the empirical Stieltjes transform. Then,
∣∣∣∣∣∣m n − 1
n
(j)
∑
k
R( j)
kk
∣∣∣∣∣∣
=
∣∣∣∣∣∣
1
n R j j + 1
n
(j)
∑
k
(
R kk − R( j)
kk
)∣∣∣∣∣∣
(
112 )
=
∣∣∣∣∣
1
n
∑
k
R2
k j
R j j
∣∣∣∣∣ = ‖e
j
R‖2
n|R j j | ≤ ‖R‖2
n|R j j | (log n)3an .
Using d ≤ n and combining with (113), w.h.p. for all j,
∣∣∣∣
1
R j j
z + m n
∣∣∣∣ (log n)2+2ε+a
√d . (114) Then by the triangle inequality, also w.h.p. for all j = k,
∣∣∣∣
1
R j j
− 1
R kk
∣∣∣∣ (log n)2+2ε+a
√d ,
so
∣∣∣∣
m nR j j
− 1
∣∣∣∣ =
∣∣∣∣∣n−1 ∑
k
R kk − R j j R j j
∣∣∣∣∣ ≤ max
k
∣∣∣∣
R kk − R j j R j j
∣∣∣∣
= max
k
|R kk |
∣∣∣∣
1
R j j
− 1
R kk
∣∣∣∣ (log n)2+2ε+2a
√d .
For d (log n)4+4ε+4a , this implies 32 |R j j | ≥ | m n | ≥ | R j j |/2 w.h.p. for all j. Then also
∣∣∣∣
1
R j j
− 1
m n
∣∣∣∣ = |R j j − m n ||R j j || m n | ≤ max
k
|R j j − R kk ||R j j || m n | ≤ max
k
2|R j j − R kk ||R j j || R kk |= 2 max
k
∣∣∣∣
1
R j j
− 1
R kk
∣∣∣∣ ,
123 Foundations of Computational Mathematics
so
∣∣∣∣
1
R j j
− 1
m n
∣∣∣∣ (log n)2+2ε+a
√d . (115) Combining with (114), w.h.p. we have 1
m n
z + m n = r n , |r n | (log n)2+2ε+a
√d (log n)−a .
Solving for m n yields
m n ∈ −z + r n ± √z2 − 4 − 2zr n + r2
n
2where the right side denotes the two complex square roots. Note that |z2 −4| = | z −2|· |z +2| (log n)−a |z| and |z| ≥ (log n)−a for all z ∈ D. Then, as (log n)−a | r n |, we have |z2 − 4| | zr n | | r n |2 . Letting m0 be the Stieltjes transform of the semicircle law, and letting ˜m0 = 1/m0 be the other root of the quadratic Eq. (30), we obtain by a Taylor expansion of the square root that min (|m n − m0|, |m n − ˜m0|) |r n |
(
1 + |z|
√|z2 − 4|
)
|r n |√ζ ( z) + | Im z| , (116) where ζ ( z) is as defined in Proposition 2. To argue that this bound holds for |m n − m0| rather than |m n − ˜m0|, consider first
z ∈ D1 with Im z > 0. In this case m n ∈ C+ and ˜m0 ∈ C−. Furthermore, note that (31) implies Im m0(z) ≥ (Im z)/ √ζ ( z) + Im z, and hence Im ˜m0 = − (Im m0)/ |m0|2 ≤−c(log n)−a /√ζ ( z) + Im z. Since Im m n > 0 and |r n | (log n)−a , (116) must hold for |m n −m0| rather than |m n −˜m0|. The same argument applies for z ∈ D1 with Im z <
For z ∈ D2 , we have || m0(z)|− 1| ≥ c and hence |m0(z)− ˜m0(z)| > c for a constant
c > 0. Consider the point z′ ∈ D1 ∩ D2 with Re z′ = Re z and Im z′ = (log n)−a .Note that for all z ∈ D2 , | ddz m0(z)| 1 and, on the event ‖A‖ ≤ 2.5, | ddz m n (z)| 1also. Thus |m0(z) − m0(z′)| ≤ C(log n)−a and |m n (z) − m n (z′)| ≤ C(log n)−a . Since we have already shown that (116) holds for |m n (z′) − m0(z′)| in the previous case, this implies also that (116) must hold for |m n − m0| rather than for |m n − ˜m0|.Applying | Im z| ≥ (log n)−a , (116) yields w.h.p.
|m n − m0| (log n)a/2|r n | (log n)2+2ε+3a/2
√d . (117) Recalling (115), |R j j | ≤ (log n)a and |m n | ≤ 32 |R j j |, we get
|R j j − m n | |R j j || m n | · (log n)2+2ε+a
√d (log n)2+2ε+3a
√d . (118)
123 Foundations of Computational Mathematics
Combining the last two displayed equations gives the weak estimate
|R j j − m0| (log n)2+2ε+3a
√d .
Since d (log n)4+4ε+6a by assumption, this and |m0(z)| 1 imply |R j j | 1w.h.p. Then, applying the last display and (117) to the first inequality of (118) yields the desired estimate
|R j j − m0| ≤ | R j j − m n | + | m n − m0| (log n)2+2ε+3a/2
√d .
To show (33) for the off-diagonals, we now apply (109), (22), (26) conditional on
R( jk ), |R j j | 1, |R( j)
kk
| 1, ‖R( jk )‖∞ ≤ (log n)a , ‖R( jk )‖F ≤ √n(log n)a , and
d ≤ n to get w.h.p.
|R jk | = | R j j || R( j)
kk
|
∣∣∣∣∣∣−a jk +
(jk )
∑
,m
a j R( jk )
m
a mk
∣∣∣∣∣∣
(log n)2+2ε
(
1
√d + 2‖R( jk )‖∞√d + ‖R( jk )‖Fn
)
(log n)2+2ε+a
√d .
7.3 Row Sum Bound
We now show that (35) holds for z ∈ D w.h.p. Set
Zi
(i)
∑
j,k
aik R(i)
k j
=
(i)
∑
k
aik
(
e
k
R(i)1
)
(119) where the last equality holds because R(i)
ki
= 0 for k = i. Applying (109),
e
i
R1 = ∑
j
R i j = R ii − R ii Zi .
Then applying (34), w.h.p. for every i ∈ [ n],
∣∣∣e
i
R1
∣∣∣ 1 + | Zi |. (120) Applying (23) conditional on A(i), w.h.p. for every i ∈ [ n],
|Zi | ≤ (log n)1+ε
⎛⎝ max k =i |e
k
R(i)1|√d +
√ ∑(i)
k
|e
k
R(i)1|2
n
⎞⎠ . (121)
123 Foundations of Computational Mathematics
For the second term above, we apply ‖R(i)‖ ≤ (log n)a w.h.p. to get
(i)
∑
k
∣∣∣e
k
R(i)1
∣∣∣2
≤ 1 R(i) R(i)1 ≤ (log n)2a n. (122) For the first term, we apply (110), (33), and (34) to get, w.h.p. for all k = i,
∣∣∣e
k
R(i)1
∣∣∣ =
∣∣∣∣e
k
R1 − R ki R ii
· e
i
R1
∣∣∣∣ ≤
∣∣∣e
k
R1
∣∣∣ + C(log n)2+2ε+a
√d
∣∣∣e
i
R1
∣∣∣ .
(123) Applying d (log n)4+4ε+2a and substituting (122) and (123) into (121) and then into (120), we get that
∣∣∣e
i
R1
∣∣∣ 1 + (log n)1+ε
( max k |e
k
R1|√d + (log n)a
)
(124) Taking the maximum over i and rearranging yields (35).
7.4 Total Sum Bound
Finally, we show that (36) holds with probability 1 − e−c(log n)( log log n) for z ∈ D. As above, we set
Zi =
(i)
∑
j,k
aik R(i)
k j
=
(i)
∑
k
aik
(
e
k
R(i)1
)
. (125) Note that if we apply (122), (123), and (35) to (121), we obtain w.h.p. that for every
i ∈ [ n],
|Zi | ≤ (log n)1+ε+a . (126) The main step of the proof of (36) is to use the weak dependence of Z1, . . . , Zn to obtain a bound on n−1 ∑
i
Zi that is better than (log n)1+ε+a . The idea is encapsulated by the following abstract lemma from .
Lemma 16 (Fluctuation averaging) Let Ξ be an event defined by A, let Z1, . . . , Zn be random variables which are functions of A, let p be an (n-dependent) even integer, and let x , y > 0 be deterministic positive quantities. Suppose there exist random variables
Z[U ]
i
, indexed by U ⊆ [ n] and i ∈ [ n] \ U , which satisfy Z[∅]
i
= Zi as well as the following conditions:
(i) Let a i denote the i th row of A. Then Z[U ]
i
is independent of {a j : j ∈ U }, and
Ei
[
Z[U ]
i
]
= 0 where Ei is the partial expectation over only a i .
123 Foundations of Computational Mathematics
(ii) For any U ⊆ S ⊂ [ n] with |S| ≤ p, and for any i /∈ S, denote u = | U | + 1 and
ZS,Ui = ∑
T:T⊆U
(−1)|T |Z[(S\U )∪T ]
i
. (127)
Then for a constant C > 0 and any integer r ∈ [ 0, p],
E
[
1{Ξ }
∣∣∣ZS,Ui
∣∣∣r ]
≤
(
y(C xu )u )r
.
Furthermore, x ≤ 1/( p5 log n).
(iii) Let A ⊂ Rn×n be the matrices satisfying Ξ , i.e., Ξ = { A ∈ A}. Let Ai = { B ∈
Rn×n : B(i) = A(i) for some A ∈ A}, and define the event Ξi = { A ∈ Ai }. For a constant C > 0 and any U , S, i as above, E
[
1{Ξi }
∣∣∣ZS,Ui
∣∣∣2]
≤ n C p .
(iv) For a constant C > 0 and any U ⊆ [ n], 1{Ξ }
∣∣∣Z[U ]
i
∣∣∣ ≤ yn C .
(v) For a constant ε > 0, P[Ξ ] ≥ 1 − e−c(log n)1+ε p .Then for constants C ′, n0 > 0 depending on C , ε above, and for all n ≥ n0,
P
[
1{Ξ }
∣∣∣∣∣n−1 ∑
i
Zi
∣∣∣∣∣ ≥ p12 y
(
x2 + n−1)]
≤ (C′/ p)p.
Proof See [12,Theorem 5.6]. (The theorem is stated for 1 + ε = 3/2 in condition (v), but the proof holds for any ε > 0.)
The important condition encapsulating weak dependence above is (ii). Applying (ii) with U = ∅ , the condition requires first that each |Z[S]
i
|, and in particular each
|Zi | = | Z[∅]
i
|, is of typical size C x y . In the application of this lemma, for S = U and
i /∈ U , we will define the variables Z[V ]
i
for ∅ ⊆ V ⊆ U such that the quantity ZU ,Ui
in (127) is the variable Zi with its dependence on all {a j : j ∈ U } projected out by an inclusion–exclusion procedure. Then, condition (ii) requires that Zi depends weakly on {a j : j ∈ U }, in the sense that |ZU ,Ui | is of typical size x|U |+ 1 y · (C(|U | + 1)) |U |+ 1 ,which is roughly smaller than |Zi | by a factor of x|U | for each element of U . Assuming 1/√n x p−12 , the above then estimates the average |n−1 ∑
i
Zi | to be of the smaller order p12 yx 2 x y . We refer the reader to the discussion in for additional details. We will check that the conditions of this lemma hold for Zi as defined by (125), with the appropriate construction of variables Z[U ]
i
. To this end, we first extend (33), (34), and (35) to R(S) for |S| ≤ log n in the following deterministic lemma:
123 Foundations of Computational Mathematics
Lemma 17 Suppose (33), (34), and (35) hold with the constant C ≡ C0 for a deter-ministic symmetric matrix A, some z ∈ D, and all j , k ∈ [ n]. Then for all S ⊂ [ n]
with |S| ≤ log n, and all j = k ∈ [ n] \ S,
|R(S)
j j
(z) − m0(z)| ≤ 2C0(log n)2+2ε+3a
√d , (128)
|R(S)
jk
(z)| ≤ 2C0(log n)2+2ε+a
√d , (129)
|e
j
R(S) (z)1| ≤ 2C0(log n)1+ε+a . (130)
Proof For integers s ≥ 0, let
Λds = max
{
|R(S)
j j
− m0 S| = s, j ∈ [ n] \ S
}
,Λos = max
{
|R(S)
jk
S| = s, j = k ∈ [ n] \ S
}
.
When (33) and (34) hold, we have that Λds ≤ C0(log n)2+2ε+3a /√d and Λos ≤
C0(log n)2+2ε+a /√d for s = 0. By (112), we have for each s ≥ 1 and ∗ ∈ { d, o} that
Λ∗
s+1
≤ Λ∗
s
(Λ os )2
|m0| − Λds
. (131) Assume inductively that for some s ≤ log n,
Λds ≤ C0(log n)2+2ε+3a
√d
(
1 + 4C0(log n)2+2ε+a
|m0|√d
)s
,Λos ≤ C0(log n)2+2ε+a
√d
(
1 + 4C0(log n)2+2ε+a
|m0|√d
)s
. (132) Applying d (log n)6+4ε+2a , |m0| ≥ c, and s ≤ log n, this implies in particular that
Λds ≤ 2C0(log n)2+2ε+3a
√d , Λos ≤ 2C0(log n)2+2ε+a
√d .
We then have |m0| − Λds ≥ | m0|/2 for d (log n)4+4ε+6a , so (131) yields
Λ∗
s+1
≤ max (Λ ∗
s
, Λ os )
(
1 + 2Λos
|m0|
)
≤ max (Λ ∗
s
, Λ os )
(
1 + 4C0(log n)2+2ε+a
|m0|√d
)
.
Thus, both bounds of (132) hold for s + 1, completing the induction. This establishes (128) and (129). To show (130), set
Γs = max {| e
j
R(S)1 S| = s, j /∈ S}.
123 Foundations of Computational Mathematics
When (35) holds, Γ0 ≤ C0(log n)1+ε+a . Applying (110) and the bound |m0| − Λds ≥|m0|/2, we have
Γs+1 ≤ (1 + 2Λos /|m0|)Γ s
(
129 )
≤
(
1 + 4C0(log n)2+2ε+3a
|m0|√d
)
Γs ,
Thus, Γs ≤ 2Γ0 for all s ≤ log n.
Lemma 18 Fix z ∈ D. Let Zi be defined in (125) . For U ⊂ [ n] not containing i, define
Z[U ]
i
=
(iU )
∑
j,k
aik R(iU )
k j
=
(iU )
∑
k
aik
(
e
k
R(iU )1
)
.
Let Ξ be the event where – (33), (34), and (35) all hold at z, for all distinct j , k ∈ [ n],– |ai j | ≤ 1 for all i , j ∈ [ n], and – ‖A‖ ≤ 2.5.Let p ∈ [ 2, ( log n) − 1] be an even integer, and set x = (log n)2+2ε+a
√d , y = C′ √d(log n)−ε
for a sufficiently large constant C ′ > 0. Then, all of the conditions of Lemma 16 are satisfied.
Proof Condition (i) is clear by definition, as row ai of A is independent of R(iU ).To check (ii), note first that the bound x ≤ 1/( p5 log n) follows from d ≥
(log n)16 +4ε+2a . For U ⊆ S and i /∈ S we write
ZS,Ui = ∑
T:T⊆U
(−1)|T |Z[(S\U )∪T ]
i
= ∑
T:T⊆U
(−1)|T |
(( i S \U)∪T)
∑
k
aik (e
k
R(( i S \U )∪T )1)
= ∑
k∈U
aik
⎛⎝ ∑
T:T⊆U{ k}
(−1)|T |(e
k
R(( i S \U )∪T )1)
⎞⎠
+
(i S )
∑
k
aik
⎛⎝ ∑
T:T⊆U
(−1)|T |(e
k
R(( i S \U )∪T )1)
⎞⎠
∑
k∈U
aik αk +
(i S )
∑
k
aik βk .
123 Foundations of Computational Mathematics
We claim that deterministically on the event Ξ , there is a constant C > 0 such that for any W , V ⊂ [ n] disjoint with |W ∪ V | ≤ log n, and any i /∈ W ∪ V , we have
∣∣∣∣∣∣∑
T:T⊆W
(−1)|T | (
e
i
R(V ∪T )1
)∣∣∣∣∣∣ ≤ ˜y(C x w) w , (133) where w = | W | + 1, x = (log n)2+2ε+a /√d, and ˜y = C√d(log n)−1−ε. We will verify this claim at the end of the proof. Assuming this claim, we apply it above with
V = i S \ U and either W = U or W = U \ { k}. Then setting u = | U | + 1 ≥ w, we have on Ξ that
|αk | ≤ ˜y(C xu )|U |, |βk | ≤ ˜y(C xu )|U |+ 1. (134) Let r be any even integer with r ≤ p ≤ (log n) − 1. As αk , β k are independent of row
ai of A by definition, we have for the partial expectation Ei over ai that
Ei
[
1{Ξ }
∣∣∣ZS,Ui
∣∣∣r ]
= Ei
⎡⎣1{Ξ }
∣∣∣∣∣∣∑
k∈U
aik αk +
(i S )
∑
k
aik βk
∣∣∣∣∣∣
r
⎤⎦
≤ 1{| αk | ≤ ˜y(C xu )|U | and |βk | ≤ ˜y(C xu )|U |+ 1 for all k}· Ei
⎡⎣∣∣∣∣∣∣∑
k∈U
aik αk +
(i S )
∑
k
aik βk
∣∣∣∣∣∣
r
⎤⎦ .
We apply (24) for the conditional expectation Ei , with v having entries vk = αk for
k ∈ U , vk = βk for k /∈ i S , and vk = 0 otherwise. Recall that w ≤ | U | ≤ | S| ≤ log n.Since C x w 1 and |U |(C x w) 2|U | (n − | U |)( C x w) 2|U |+ 2 by the definition of x
and d ≤ n, the bounds (134) imply
‖v‖∞ ≤ ˜y(C xu )|U |, ‖v‖2 ≤ √2n · ˜y(C x w) |U |+ 1.
Then for a constant C′ > 0, (24) gives
Ei
[
1{Ξ }
∣∣∣ZS,Ui
∣∣∣r ]
≤ (C′r˜y(C xu )u )r .
Then, taking the full expectation and setting y = C′(log n)˜y ≥ C′r˜y (since r ≤ p ≤
log n) yields condition (ii).
123 Foundations of Computational Mathematics
For condition (iii), we have
E
[
1{Ξi }
∣∣∣ZS,Ui
∣∣∣2]
≤ 2|U | ∑
T:T⊆U
E[1{Ξi }| Z[(S\U )∪T ]
i
|2]= 2|U | ∑
T:T⊆U
(( i S \U)∪T)
∑
k,k′
E[aik aik ′ ]E
[
1{Ξi }(e
k
R(( i S \U )∪T )1)( e
k′
R(( i S \U )∪T )1)
]
= 2|U | ∑
T:T⊆U
(( i S \U)∪T)
∑
k
E[a2
ik
]E
[
1{Ξi }
∣∣∣e
k
R(( i S \U )∪T )1
∣∣∣2]
,
where the second line applies the independence of ai and A(i). Note that on Ξi , we have
‖A(i)‖ ≤ 2.5. Then, applying |U | ≤ log n, the norm bound ‖R(( i S \U )∪T )‖ ≤ (log n)a
on Ξi , and E[a2
ik
] ≤ C2/n, we get (iii). For (iv), we apply the condition |aik | ≤ 1 by definition of Ξ , together with the bound ‖R(iU )‖ ≤ (log n)a on Ξ . Finally, (v) holds by the probability bound of 1 − e−c(log n)1+ε
established for (33), (34), (35), (22), and in Lemma 3. It remains to establish the claim (133). For W = ∅ , this follows from (35). Assume then that w ≥ 1, and write W = { j1, . . . , jw−1} (in any order). For a function f :
Rn×n → C and any index j ∈ [ n], define Q j f : Rn×n → C by
(Q j f )( A) = f (A) − f (A( j) ).
Note that if f is in fact a function of A(S), i.e., f (A) = f (A(S) ) for every matrix
A ∈ Rn×n , then Q j f (A) = f (A(S) ) − f (A( j S ) ). Fix i and V , and define f (A) =
e
i
R(V )1. This satisfies f (A) = f (A(V ) ) for every A. Then by inclusion–exclusion, the quantity to be bounded is equivalently written as
∑
T:T⊆W
(−1)|T | (
e
i
R(V ∪T )1
)
= (Q jw−1 . . . Q j2 Q j1 f ) (A).
We apply Schur complement identities to iteratively to expand Q jw−1 . . . Q j1 f :First applying (110), we get
Q j1 f (A) = e
i
R(V )1 − e
i
R( j1 V )1 = R(V )
i j 1
· 1
R(V )
j1j1
· e
j1
R(V )1.
Then applying (110), (111), and (112) to the three factors on the right side above, and using the identity
x yz − ˜x˜y˜z = x y (z −˜z) + x(˜y − y)˜z + (˜x − x)˜y˜z,
123 Foundations of Computational Mathematics
we get
Q j2 Q j1 f (A) = R(V )
i j 1
· 1
R(V )
j1j1
·
( R(V )
j1j2
R(V )
j2j2
· e
j2
R(V )1
)
R(V )
i j 1
·
⎛⎜⎝−
(
R(V )
j1j2
)2
R( j2 V )
j1j1
R(V )
j2j2
R(V )
j1j1
⎞⎟⎠ · e
j1
R( j2 V )1
R(V )
i j 2
R(V )
j2j1
R(V )
j2j2
· 1
R( j2 V )
j1j1
· e
j1
R( j2 V )1.
Applying (112), (111), and (110) to each factor of each summand above, and repeating iteratively, an induction argument verifies the following claims for each
t ∈ { 1, . . . , w − 1}:– Q jt . . . Q j1 f (A) is a sum of at most ∏t−1
s=1
4s summands (with the convention
∏0
s=1
4s = 1), where – Each summand is a product of at most 4 t factors, where – Each factor is one of the following three forms, for a set S ⊆ V ∪ W : R(S)
jk
for
j, k /∈ S distinct, or 1 /R(S)
j j
for j /∈ S, or e
j
R(S)1 for j /∈ S. Furthermore, – Each summand of Q jt . . . Q j1 f (A) satisfies: (a) It has exactly one factor of the form e
j
R(S)1. (b) The number of factors of the form 1 /R(S)
j j
is less than or equal to the number of factors of the form R(S)
jk
for j = k. (c) There are at least t factors of the form R(S)
jk
for j = k.Finally, we apply this with t = w − 1 and use the bound
t−1
∏
s=1
4s ≤ (4w) w .
By Lemma 17, since |W ∪ V | ≤ log n, we have |R(S)
jk
| ≤ C(log n)2+2ε+a /√d,
|R(S)
j j
| ≥ | m0|/2, and |e
j
R(S)1| ≤ C(log n)1+ε+a on the event Ξ . Thus, we get
|Q jw−1 . . . Q j1 f (A)| ≤ (4w) w ·
( C(log n)2+2ε+a
√d
)w−1
· C(log n)1+ε+a ≤˜y(C′xw) w
for x = (log n)2+2ε+a /√d and ˜y = C√d(log n)−1−ε, as claimed.
We now show (36) holds for z ∈ D with probability 1 − e−c(log n)( log log n). The diagonal bound (34) implies
| Tr R − n · m0| ≤ Cn (log n)2+2ε+3a/2
√d . (135)
123 Foundations of Computational Mathematics
To bound the sum of off-diagonal elements of R, we apply (109) to write
∑
i =kR ik = − ∑
iR ii Zi = − m0
∑
i
Zi − ∑
i
(R ii − m0)Zi . (136) Applying (34) and (126) yields
∑
i
|(R ii − m0)Zi | ≤ Cn (log n)3+3ε+5a/2
√d . (137) Then applying Lemma 16 with x, y, Ξ as defined in Lemma 18 and with p being the largest even integer less than (log n) − 1, we have
1{Ξ }
∣∣∣∣∣n−1 ∑
i
Zi
∣∣∣∣∣ ≤ C(log n)12 · √d(log n)−ε · (log n)4+4ε+2a /d ≤ C(log n)16 +3ε+2a
√d
(138) with probability 1 − e−c(log n)( log log n). Since 1 R1 = Tr R + ∑
i =k R ik , multiplying (138) by n · m0 and combining with (135)–(137) yields (36).
References
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Handbook of combinatorial optimization , vol. 3 (D.-Z. Du, P.M. Pardalos, eds.). Kluwer Academic Publishers, 1998, pp. 241–337. 6. Y. Chen and J. Xu. Statistical-computational tradeoffs in planted problems and submatrix local-ization with a growing number of clusters and submatrices. In Proceedings of ICML 2014 (Also arXiv:1402.1267), Feb 2014. 7. D. Cullina and N. Kiyavash. Improved achievability and converse bounds for Erdös-Rényi graph matching. In Proceedings of the 2016 ACM SIGMETRICS International Conference on Measurement and Modeling of Computer Science , pp 63–72. ACM, 2016. 8. D. Cullina and N. Kiyavash. Exact alignment recovery for correlated Erdös-Rényi graphs. arXiv preprint arXiv:1711.06783, 2017. 9. O. E. Dai, D. Cullina, N. Kiyavash, and M. Grossglauser. On the performance of a canonical labeling for matching correlated Erd˝ os-Rényi graphs. arXiv preprint arXiv:1804.09758, 2018. 10. J. Ding, Z. Ma, Y. Wu, and J. Xu. Efficient random graph matching via degree profiles. Probability Theory and Related Fields , pp 1–87, Sep 2020. 11. L. Erd˝ os, A. Knowles, H.-T. Yau, and J. Yin. The local semicircle law for a general class of random matrices. Electron. J. Probab , 18(59):1–58, 2013. 12. L. Erd˝ os, A. Knowles, H.-T. Yau, and J. Yin. Spectral statistics of Erd˝ os–Rényi graphs I: local semicircle law. The Annals of Probability , 41(3B):2279–2375, 2013.
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L. Erd˝ os, H.-T. Yau, and J. Yin. Bulk universality for generalized Wigner matrices. Probability Theory and Related Fields , 154(1-2):341–407, 2012. 14. Z. Fan, C. Mao, Y. Wu, and J. Xu. Spectral graph matching and regularized quadratic relaxations I: The Gaussian model. arxiv preprint arXiv:1907.08880, 2019. 15. Z. Fan, C. Mao, Y. Wu, and J. Xu. Spectral graph matching and regularized quadratic relaxations: Algorithm and theory. In International Conference on Machine Learning (ICML) , Jul 2020. 16. D. L. Hanson and F. T. Wright. A bound on tail probabilities for quadratic forms in independent random variables. Ann. Math. Statist. , 42:1079–1083, 1971. 17. K. Makarychev, R. Manokaran, and M. Sviridenko. Maximum quadratic assignment problem: Reduc-tion from maximum label cover and LP-based approximation algorithm. Automata, Languages and Programming , pp 594–604, 2010. 18. P. M. Pardalos, F. Rendl, and H. Wolkowicz. The quadratic assignment problem: A survey and recent developments. In In Proceedings of the DIMACS Workshop on Quadratic Assignment Problems, volume 16 of DIMACS Series in Discrete Mathematics and Theoretical Computer Science , pp 1–42. American Mathematical Society, 1994. 19. P. Pedarsani and M. Grossglauser. On the privacy of anonymized networks. In ACM SIGKDD Inter-national Conference on Knowledge Discovery and Data Mining , pp 1235–1243, 2011. 20. M. Rudelson and R. Vershynin. Hanson-Wright inequality and sub-Gaussian concentration. Electron. Commun. Probab. , 18(82):9, 2013. 21. M. Zaslavskiy, F. Bach, and J.-P. Vert. A path following algorithm for the graph matching problem.
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Publisher’s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
123 |
16089 | https://www.youtube.com/watch?v=vgrM9zS16no | Fun Desmos Classroom Slope Activity
Tim Brzezinski
13700 subscribers
4 likes
Description
493 views
Posted: 9 Jan 2023
Transcript:
algebra and geometry teachers if you want to actively engage your students more when it comes to working with slope and have them be less passive right then why not give them some Desmos classroom marble slides check this activity out right here tiny URL right there but here we have to actually take these number boxes to create a ramp with a given slope see how the slope of that ramp is three-sevenths right my rise over run isn't really that good there I can knock over one star but the goal is to knock out all the stars I got to collect them all so let's go back to the John board should I make my rise my rise bigger or smaller what should I do with my run right here students can come to play with the concept of slope conceptually and also very meaningfully as they build to create here still not good enough let's try this I think this Ah that's nice yeah look at that one two three we have a success this activity has a bunch of slides like this where kids can play have fun and engage and you can formatively assess at the end be sure to check it out foreign |
16090 | https://stats.stackexchange.com/questions/646259/asymptotic-unbiasedness-asymptotic-zero-variance-consistency | mathematical statistics - Asymptotic unbiasedness + asymptotic zero variance = consistency? - Cross Validated
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Asymptotic unbiasedness + asymptotic zero variance = consistency?
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Asked 1 year, 5 months ago
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Here, Ben shows that an unbiased estimator θ^θ^ of a parameter θ θ that has an asymptotic variance of zero converges in probability to θ θ. That is, θ^θ^ is a consistent estimator of θ θ.
It makes sense that we should be able to relax the condition to asymptotic unbiasedness: lim n→∞E[θ^n−θ]=0 lim n→∞E[θ^n−θ]=0. That seems to stay within the spirit of the estimator closing-in on the true parameter value...
...but math has surprised me before.
Can we relax the condition to asymptotic unbiasedness? What is the proof?
mathematical-statistics
convergence
estimators
asymptotics
consistency
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edited May 2, 2024 at 12:04
DaveDave
asked May 1, 2024 at 0:32
DaveDave
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2 Answers 2
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Your question may be restated as
Let ⟨θ^n⟩n∈N⟨θ^n⟩n∈N be a sequence of random variables such that
θ n:=E[θ^n]→θ Var(θ^n)→0 as n→∞.as n→∞.(1)(2)(1)θ n:=E[θ^n]→θ as n→∞.(2)Var(θ^n)→0 as n→∞.
Does (1)(1) and (2)(2) imply θ^n→p θ θ^n→p θ?
The answer is in the affirmative and the proof is similar to the original case (i.e., using Chebyshev's inequality, but with slightly more work to split the deviance |θ^n−θ||θ^n−θ|).
Given ε>0 ε>0, by (1)(1), there exists N N sufficiently large such that |θ n−θ|<ε/2|θ n−θ|<ε/2 for all n>N n>N, it then follows that for all n>N n>N,
=≤=≤P(|θ^n−θ|>ε)P(|θ^n−θ n+θ n−θ|>ε)P(|θ^n−θ n|>ε/2)+P(|θ n−θ|>ε/2)P(|θ^n−θ n|>ε/2)4 Var(θ^n)/ε 2.P(|θ^n−θ|>ε)=P(|θ^n−θ n+θ n−θ|>ε)≤P(|θ^n−θ n|>ε/2)+P(|θ n−θ|>ε/2)=P(|θ^n−θ n|>ε/2)≤4 Var(θ^n)/ε 2.
The right-hand side of the inequality converges to 0 0 as n→∞n→∞ by (2)(2), which implies θ^n→p θ θ^n→p θ. This completes the proof.
An equivalent way of seeing it is by decomposing θ^n−θ θ^n−θ as (θ^n−θ n)+(θ n−θ)(θ^n−θ n)+(θ n−θ) then applying the Slutsky's theorem (or merely the simple asymptotic fact that if X n→p X X n→p X and Y n→p Y Y n→p Y then X n+Y n→p X+Y X n+Y n→p X+Y) -- the first part θ^n−θ n→p 0 θ^n−θ n→p 0 is exactly the original case you linked, while the second part θ n−θ→0 θ n−θ→0 follows by the asymptotic unbiasedness condition (1)(1).
At my first reading, I mistakenly interpreted OP's problem as "Does asymptotic unbiasedness alone imply consistency?", which is certainly not true. One counterexample is: let θ=0 θ=0, θ^n θ^n is a Rademacher random variable, i.e., P(θ^n=1)=P(θ^n=−1)=1 2 P(θ^n=1)=P(θ^n=−1)=1 2 for all n n. It is easy to verify that E[θ^n]≡θ E[θ^n]≡θ while θ^n↛p θ θ^n↛p θ. Note that in this case Var(θ^n)≡1 Var(θ^n)≡1.
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edited May 1, 2024 at 12:07
Mahmoud
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answered May 1, 2024 at 1:13
ZhanxiongZhanxiong
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I thought the question was whether variance going to zero plus asymptotic unbiasedness was enough (as a slight relaxation of variance going to zero plus unbiasedness) [which is true]Thomas Lumley –Thomas Lumley 2024-05-01 01:14:18 +00:00 Commented May 1, 2024 at 1:14
2 I have stated a more general result, Zhanxiong.User1865345 –User1865345 2024-05-01 01:14:37 +00:00 Commented May 1, 2024 at 1:14
@ThomasLumley reading the body, I thought whether asymptotic unbiasedness was enough.User1865345 –User1865345 2024-05-01 01:16:21 +00:00 Commented May 1, 2024 at 1:16
1 @User1865345 OK, I restored it (btw, do you know how to make smaller fonts? I hope by doing so the appendix wouldn't distract OP's true question).Zhanxiong –Zhanxiong 2024-05-01 01:49:06 +00:00 Commented May 1, 2024 at 1:49
1 @mhdadk Appreciate it.Zhanxiong –Zhanxiong 2024-05-01 12:09:19 +00:00 Commented May 1, 2024 at 12:09
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The following general result would be relevant here:
_Given a measure space (Ω,A,μ),(Ω,A,μ), consider a sequence of measurable functions ⟨f n⟩n∈N⟨f n⟩n∈N where f n∈L p(Ω,A,μ),p∈[1,∞).f n∈L p(Ω,A,μ),p∈[1,∞). If the sequence converges in the L p L p norm to f,f, then
f n→μ f.f n→μ f._
To see that, note for an arbitrary ε>0,ε>0,
∥f n−f∥p p≥∫{Ω:|f n−f|≥ε}|f n−f|p d μ≥ε p μ{Ω:|f n−f|≥ε},‖f n−f‖p p≥∫{Ω:|f n−f|≥ε}|f n−f|p d μ≥ε p μ{Ω:|f n−f|≥ε},
but by assumption ∥f n−f∥p→0,‖f n−f‖p→0, so f n f n converges in measure to f f on Ω.Ω.
--
Reference:
[I][I] Real Analysis: Theory of Measure and Integration, J. Yeh, World Scientific, 2014,2014, Th. 16.25.16.25.
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edited May 1, 2024 at 1:30
answered May 1, 2024 at 1:03
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| Некоторые распространенные композиции. Рассмотрим теперь некоторые комбинации движений, используемые достаточно часто, но не уделяя им особого внимания. Композиции отражений в плоскости. Теорема 1. Движение пространства первого рода представимо в виде композиции двух или четырех отражений в плоскости. Движение пространства второго вида есть либо отражение в плоскости, либо представимо в виде композиции трех отражений в плоскости. Отсюда мы можем объяснить уже известные нам движения так: Композиция отражения в 2 параллельных плоскостях есть параллельный перенос. Композиция отражения в 2 пересекающихся плоскостях есть поворот вокруг прямой пересечения этих плоскостей. Центральная симметрия относительно данной точки является композицией 3 отражений относительно любых 3 взаимно перпендикулярных плоскостей, пересекающихся в этой точке. Винтовые движения. Определение. Винтовым движением называется композиция поворота и переноса на вектор, параллельный оси поворота. Представление о таком движении дает ввинчивающийся или вывинчивающийся винт. Теорема 2. Любое движение пространства первого рода - винтовое движение (в частности поворот вокруг прямой или перенос) . Зеркальный поворот. Определение. Зеркальным поворотом вокруг оси a на угол (называется композиция поворота вокруг оси a на угол (и отражения в плоскости, перпендикулярной оси поворота. Теорема 3. Любое движение пространства второго рода, имеющее неподвижную точку, является зеркальным поворотом, который, в частности, может быть центральной или зеркальной симметрией. Скользящие отражения. Определение. Скользящим отражением называется композиция отражения в некоей плоскости и переноса на вектор, параллельный этой плоскости. Теорема 4. Движение пространства второго рода, не имеющее неподвижных точек, есть скользящее отражение. Теорема Шаля. Движение плоскости первого рода является либо поворотом, либо параллельным переносом. Движение плоскости второго рода является скользящим отражением. | |
Некоторые распространенные композиции.
Рассмотрим теперь некоторые комбинации движений, используемые достаточно часто, но не уделяя им особого внимания.
Композиции отражений в плоскости.
Теорема 1. Движение пространства первого рода представимо в виде композиции двух или четырех отражений в плоскости.
Движение пространства второго вида есть либо отражение в плоскости, либо представимо в виде композиции трех отражений в плоскости.
Отсюда мы можем объяснить уже известные нам движения так: Композиция отражения в 2 параллельных плоскостях есть параллельный перенос.
Композиция отражения в 2 пересекающихся плоскостях есть поворот вокруг прямой пересечения этих плоскостей.
Центральная симметрия относительно данной точки является композицией 3 отражений относительно любых 3 взаимно перпендикулярных плоскостей, пересекающихся в этой точке.
Винтовые движения.
Определение. Винтовым движением называется композиция поворота и переноса на вектор, параллельный оси поворота. Представление о таком движении дает ввинчивающийся или вывинчивающийся винт.
Теорема 2. Любое движение пространства первого рода - винтовое движение (в частности поворот вокруг прямой или перенос) .
Зеркальный поворот.
Определение. Зеркальным поворотом вокруг оси a на угол (называется композиция поворота вокруг оси a на угол (и отражения в плоскости, перпендикулярной оси поворота.
Теорема 3. Любое движение пространства второго рода, имеющее неподвижную точку, является зеркальным поворотом, который, в частности, может быть центральной или зеркальной симметрией.
Скользящие отражения.
Определение. Скользящим отражением называется композиция отражения в некоей плоскости и переноса на вектор, параллельный этой плоскости.
Теорема 4. Движение пространства второго рода, не имеющее неподвижных точек, есть скользящее отражение.
Теорема Шаля. Движение плоскости первого рода является либо поворотом, либо параллельным переносом.
Движение плоскости второго рода является скользящим отражением.
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16092 | https://www.khanacademy.org/math/early-math/cc-early-math-add-sub-20/cc-early-math-add-20/e/add-within-20-visually | Add within 20 visually (practice) | Khan Academy
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Early math review
Course: Early math review>Unit 4
Lesson 1: Addition within 20
Adding within 20 using place value blocks
Adding within 20 using ten frames
Adding 7 + 6
Adding 8 + 7
Add within 20 visually
Add within 20
Adding 5 + 3 + 6
Add 3 numbers
Math>
Early math review>
Addition and subtraction within 20>
Addition within 20
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Add within 20 visually
CCSS.Math: 1.OA.C, 1.OA.C.6, 2.OA.B, 2.OA.B.2
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Hint: Use the number line to help solve 3+10.
=3+10
05101520+3+10
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Video 2 minutes 38 seconds 2:38 Adding within 20 using ten frames
Video 1 minute 24 seconds 1:24 Adding within 20 using place value blocks
Video 3 minutes 53 seconds 3:53 Adding 7 + 6
Video 3 minutes 19 seconds 3:19 Adding 8 + 7
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16093 | https://mathworld.wolfram.com/SingularMatrix.html | TOPICS
Singular Matrix
A square matrix that does not have a matrix inverse. A matrix is singular iff its determinant is 0. For example, there are 10 singular (0,1)-matrices:
The following table gives the numbers of singular matrices for certain matrix classes.
| | | |
---
| matrix type | OEIS | counts for , 2, ... |
| -matrices | A057981 | 1, 33, 7875, 15099201, ... |
| -matrices | A057982 | 0, 8, 320, 43264, ... |
| -matrices | A046747 | 1, 10, 338, 42976, ... |
See also
Determinant, Ill-Conditioned Matrix, Matrix Inverse, Nonsingular Matrix, Singular Value Decomposition
Explore with Wolfram|Alpha
More things to try:
null space calculator
[{{1, 2, 3}, 1, -2, 1}, {0, 0, 0}}
null space {{-1,2},{2,-4}}
References
Ayres, F. Jr. Schaum's Outline of Theory and Problems of Matrices. New York: Schaum, p. 39, 1962.Faddeeva, V. N. Computational Methods of Linear Algebra. New York: Dover, p. 11, 1958.Golub, G. H. and Van Loan, C. F. Matrix Computations, 3rd ed. Baltimore, MD: Johns Hopkins, p. 51, 1996.Kahn, J.; Komlós, J.; and Szemeredi, E. "On the Probability that a Random Matrix is Singular." J. Amer. Math. Soc. 8, 223-240, 1995.Komlós, J. "On the Determinant of -Matrices." Studia Math. Hungarica 2, 7-21 1967.Marcus, M. and Minc, H. Introduction to Linear Algebra. New York: Dover, p. 70, 1988.Marcus, M. and Minc, H. A Survey of Matrix Theory and Matrix Inequalities. New York: Dover, p. 3, 1992.Sloane, N. J. A. Sequences A046747, A057981, and A057982 in "The On-Line Encyclopedia of Integer Sequences."
Referenced on Wolfram|Alpha
Singular Matrix
Cite this as:
Weisstein, Eric W. "Singular Matrix." From MathWorld--A Wolfram Resource.
Subject classifications |
16094 | https://tasks.illustrativemathematics.org/content-standards/tasks/1123 | Typesetting math: 100%
Tile Patterns II: hexagons
No Tags
Alignments to Content Standards:
8.G.A.5
Student View
Task
A common tiling pattern with hexagons is pictured below:
A regular hexagon is a hexagon with 6 congruent sides and 6 congruent
interior angles. Find the measure of the interior angles in a regular hexagon.
Show that three equally sized regular hexagons sharing a common vertex can be arranged
in the configuration of the above picture.
Show that a single regular hexagon can be surrounding by six regular hexagons
with no spaces and no overlap, as in the picture below:
IM Commentary
Tile patterns will be familiar with students both from working with geometry tiles and from the many tiles they encounter in the world. Here one of the most important examples of a tiling, with regular hexagons, is studied in detail. This provides students an opportunity to use what they know about the sum of the angles in a triangle and also the sum of angles which make
a line. If hexagonal and equilateral triangular tiles are available, the teacher may wish to use them to help students visualize.
The pattern of equally sized regular hexagons fitting together to fill the
plane is familiar not only from tile but also in nature:
the molecular structure of graphene (a one atom thin ''sheet'' of graphite)
consists of carbon atoms placed at the vertices of regular hexagons,
convection patterns for thin layers of heated liquids,
beehives are often built in this pattern of hexagons.
This task is ideally suited for instruction purposes where students can take their time and develop several of the Mathematical Practice standards, as the mathematical content is directly related to, but somewhat exceeds, the content of standard 8.G.5 on sums of angles in triangles.
Careful analysis of the angles requires students to construct valid arguments (MP3) using abstract and quantitative reasoning (MP2). Producing the picture in part (c) helps students identify a common mathematical argument repeated multiple times (MP8).
If students use pattern blocks in order to develop the intuition for decomposing the hexagon into triangles, then this is also an example of MP5.
Solutions
Solution:
Measuring angles
One way to calculate the measure of the interior angles on the hexagon is
the following. Suppose you are
travelling counterclockwise along the hexagon. At each corner, you make a turn. After
6 of these turns you are heading in the same direction in which you started
and have made one full revolution of 360∘. Since the angles in the
regular hexagon are all congruent, each of these turns must be 360∘6=60∘. Since the sum of the interior angle and 60∘ is 180∘,
this means that the interior angle must measure 120∘:
In the picture, the interior angle is angle ABC while the 60∘ ''turn''
is angle PBC.
2. The picture below includes a second hexagon adjacent to the first:
We know that
m(∠BCD)+m(∠BCF)+m(∠DCF)=360.
Since m(∠BCF)=m(∠DCF)=120 from part (a) we conclude
that m(∠BCD)=120. This means, from part (a), that angle BCD
forms an interior angle of a regular hexagon congruent to the two hexagons
pictured in part (a). In other words, a third hexagon can be placed as in
the picture:
3. The method of part (b) needs to be applied four more times, inserting a new
hexagon in each open space, with one side coming from the central hexagon
and one side coming from one of the adjacent hexagons. Continuing to work
around clockwise, the hexagons below are labelled in the order in which
they are added to the initial hexagon:
At the last step, we have three sides of the sixth hexagon already in place: they
are all congruent and form two 120∘ angles so the sixth hexagon will
fit in place to complete the set of seven hexagons.
Solution:
2 Triangulating Hexagon (alternate solution for part a)
We can divide the hexagon into triangles by adding a vertex at the center of the hexagon
as pictured below:
Since there are 360∘ in a circle, each of the six interior angles for the
triangles measures 60∘. The six triangles are isosceles since the vertex is at the center of the regular hexagon. This means that the base angles of the six triangles also each measure 60∘. So these
six triangles are equilateral. Each interior angle of the hexagon is made up of two angles of equilateral triangles and so they must each measure 120∘. |
16095 | https://www.coursehero.com/file/39066247/Acid-Base-flowchartjpg/ | Acid-Base flowchart.jpg - acid -Pause Decision Tree Identity major hspecies in myits. heydeligals ADAPTED FROM . J. Chem. Edi. 2 007 84 p | Course Hero
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16096 | http://home.hiwaay.net/~jalison/hepta.html | Untitled Document
HEPTAGRAM CONSTRUCTION
It is supposedly impossible to construct an exact heptagram with only a ruler and a compass. Below is a virtually perfect heptagram construction based on the dimensions of the great pyramid.
Draw a circle with center at point A. Construct a vertical and horizontal axis. Mark point B at the intersection point of the vertical axis and the edge of the circle. Draw circle BA. Repeat this process until 11 segments equal in length to segment AB are constructed. Segment AC is seven times the length of segment AB. Construct a horizontal line through point C. Segment AD is 10 times the length of segment AB. Construct circle AD. Segment AE is 11 times the length of segment AB. Arc segment AE down to the point where it intersects the right horizontal axis (point F). Construct the midpoint of segment AF (point G). Draw a perpendicular line from point G, intersecting horizontal line C at point H. With a baselength of 11 and a height of 7, the angular dimensions of triangle AHF are the same as the great pyramid. Construct the midpoint of segment AH (point I). Construct the midpoint of segment AI (point J). The dark shaded triangle has the same angular dimensions as the larger triangle and the great pyramid. Arc segment AJ to the point where it intersects the lower vertical axis (point K). Construct a horizontal segment from point K, intersecting the large circle at points 3 and 6.
Having fixed the distance of the horizontal segment from the center of the circle through the above operation, construct a vesica pisces with points 3 and 6 as the centers of the large outer circles. Mark the points where these circles intersect the central circle (points 2 and 7). Construct a circle with a radius from point 3 to point 2. Construct a circle with a radius from point 6 to the point 7. Mark the two lower points where the smaller circles intersect the central circle (points 4 and 5). The first point of the heptagram is point D from the diagram above, where the vertical axis intersects the top of the central circle.
The acute angles at the points of an exact heptagram are 25° 42' 51" (180°/7). The acute angles formed by the construction above range from 25° 42' 14" to 25° 43' 05". Given a diameter of 20 for the circle around the heptagram, the length of each arm of a perfect heptagram is 18.019. The segments formed by the construction above range from 18.018 to 18.021.
Given a height of 280 cubits and a baselength of 440 cubits (7/11), the base angles of the great pyramid are 51° 50' 34". The ratio between the height of 7 and the half base of 5.5 is 1.272727. Beginning with an exact heptagram and performing the operation described above in reverse produces the following dimensions for the pyramid triangle. If segment AK is arced to the point it crosses the vertical axis of point J in the diagram above, the ratio between the height and the half base of the resultant triangle is 1.272401. This produces a base angle of 51° 50' 08". Given a baselength of 440 cubits, this base angle produces a height of 279.928 cubits. If segment AK is arced to the point it crosses the horizontal axis of point J in the diagram above, the ratio between the height and the half base of the resultant triangle is 1.273255. This produces a base angle of 51° 51' 15". Given a height of 280 cubits, this base angle produces a baselength of 439.817 cubits.
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16097 | https://www.ncbi.nlm.nih.gov/books/NBK430733/ | Anatomy, Abdomen and Pelvis: Genitofemoral Nerve - StatPearls - NCBI Bookshelf
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StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
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Anatomy, Abdomen and Pelvis: Genitofemoral Nerve
Marco Gupton; Matthew A. Varacallo.
Author Information and Affiliations
Authors
Marco Gupton 1; Matthew A. Varacallo 2.
Affiliations
1 Campbell University
2 Penn Highlands Healthcare System
Last Update: October 24, 2022.
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Introduction
The genitofemoral nerve arises from the lumbar plexus (see Image. Genitofemoral Nerve) and supplies sensation to the skin of the anterior scrotal area in males, mons pubis in women, and the upper segment of the anterior thigh in both males and females. While distinct from the femoral nerve, the genitofemoral nerve originates from the upper lumbar segments L1 to L2. The nerve then descends inferiorly, piercing the psoas major muscle before emerging on its anterior surface. The nerve then traverses the retroperitoneum, descending over the anterior surface of the psoas muscle. The nerve continues inferiorly, ultimately separating into two divisions- the femoral and the genital branches. When the genitofemoral nerve reaches the groin area, it enters the deep inguinal ring coursing through the inguinal canal.In men, the nerve continues inferiorly and supplies the skin of the scrotum. In women, the genitofemoral nerve runs alongside the round ligament of the uterus and terminates superior to the mons pubis and labia majora. Further, after crossing the inguinal ligament, the genitofemoral nerve travels adjacent to the external iliac artery, supplying sensation to the anterior upper thigh.
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Structure and Function
The genitofemoral nerve supplies sensation via the femoral branch and motor innervation via the genital branch. The cremasteric reflex is a function of genitofemoral nerve innervation,as it supplies sensation to the superior medial aspect of the thigh. In at least 50% of individuals, there is some variation in the course of the genitofemoral nerve as it travels within the retroperitoneum and ultimately enters into the inguinal canal. Variation is also found at the level of its bifurcation into genital and femoral branches.
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Blood Supply and Lymphatics
The genitofemoral nerve is formed in the psoas muscle's midsection by the branches' union from the anterior rami of L1 and L2 nerve roots. The nerve then courses inferiorly within the psoas muscle and finally emerges on the anterior surface of the muscle distally. The nerve then descends inferiorly in the retroperitoneum along the psoas major muscle and courses deeper and on the left to the gonadal vessels, ureter, and the ileocolic vein and artery. Just before the nerve reaches the inguinal ligament, the right and left genitofemoral nerves perforate the psoas fascia and divide into the femoral and genital branches.
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Nerves
The genital branch of the genitofemoral nerve enters the inguinal canal via the deep inguinal ring. In men, the genital branch supplies the scrotal skin and cremaster (see Image.Sacral and Coccygeal Nerves). In women, the genital branch is accompanied by the round ligament of the uterus and supplies sensation to the labia majora and mons pubis. The femoral branch of the genitofemoral nerve travels beneath the inguinal ligament in the lateral muscular compartment, where the femoral nerve courses through into the leg. The femoral branch enters the cribriform fascia of the greater saphenous vein opening and supplies the thigh's anterior, upper, and medial skin.
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Physiologic Variants
Cadaveric studies have reported significant variation in the course of the genitofemoral nerve in over half of the specimens. While most specimens exhibit a predictable bifurcation pattern of the nerve in the mid-substance of the psoas muscle, a recent study noted that the genitofemoral nerve prematurely bifurcates in the upper portion of the psoas muscle about 20% of the time. Also, the femoral and genital branches do not form a common trunk within the psoas muscle and remain as distinct nerves as they course in the retroperitoneum.
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Surgical Considerations
Inguinal hernia repair
During surgery to repair an inguinal hernia, one has to visualize the genitofemoral nerve, which is found within the deep inguinal ring. If damaged, the patient will complain of pain in the scrotal area and anterior thigh. The injury can result in genitofemoral neuralgia, causing morbidity as no good treatment currently exists. For exam purposes, one must differentiate between the genitofemoral and ilioinguinal nerves, which can be injured during surgery.
Lumbar spinal interbody fusion
During lumbar spine surgical procedures, the surgeon must know the anatomic relationships between the lumbar plexus and the intervertebral disc at all levels. This concept is particularly relevant when performing lateral interbody fusion by minimally invasive retroperitoneal trans-psoas approaches. Ventral nerve roots of the lumbar plexus are particularly vulnerable during psoas muscle retraction. Surgeons must be aware of the "safe working zone" about the particular level of interest regarding the level of pathology being treated. This "safe zone"is typically located between the lumbar nerve's anterior border and the sympathetic trunk's posterior border. The genitofemoral nerve appears to be solely responsible for narrowing this safe zone, with its most significant anatomic impact being recognized at the L2 to L3 level.
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Clinical Significance
The genitofemoral nerve can be injured during surgery in several ways:
This can be lacerated during injury to the groin and can occur with penetrating trauma, a knife wound, or surgical exploration of the groin.In some cases, during surgery on the saphenous vein for varicose vein ligation, retractors can cause abrasive injury.
The genitofemoral nerve can also be injured in the inguinal canal, usually during inguinal hernia repair. In most cases, the injury goes unnoticed during surgery and is diagnosed in the postoperative period when the patient complains of pain.
The genitofemoral nerve can also be injured during a motor vehicle accident. The lower back injury may result in an extension of the back while the patient is seated.
Lifting heavy objects can also cause injury to the genitofemoral nerve.
When there is spinal stenosis of the L1 or L2 segments, compression fracture, or metastatic lesions to the lumbar spine, the nerve roots, which give rise to the genitofemoral nerve, can be injured.
Psoas abscess which may occur after an open pelvic fracture or a retroperitoneal hematoma, is also known to cause irritation of the genitofemoral nerve.
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Other Issues
Genitofemoral neuralgia is a relatively common pain syndrome observed in men and women. The patient typically presents with unilateral pain in the lower abdomen. Rarely the pain may be bilateral. The pain is sometimes referred to as the groin area (because of the path of the genital nerve) and the upper medial thigh (because of the femoral branch). The pain varies in intensity from moderate to severe and is worsened by movements that cause lower back extension. In most patients, palpation of the lower abdomen in the inguinal region can reproduce the pain. In most cases, the cause is compression of the nerve in the inguinal canal, usually after open hernia surgery. The diagnosis is often difficult in patients who have not had surgery.
The prognosis of patients who suffer entrapment or injury to the genitofemoral nerve is guarded. If the nerve is entrapped in the inguinal canal after inguinal hernia surgery, one may try decompression, but often the scarring makes visualization of the nerve difficult. In patients with a lacerated nerve, many pain-relieving modalities have been used, but none works effectively. Some patients respond to nerve blocks, but most patients require long-term anticonvulsants and opiates for pain relief.
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Review Questions
Access free multiple choice questions on this topic.
Comment on this article.
Figure
Sacral and Coccygeal Nerves. The illustration depicts the sacral plexus of the right side, pelvic area, coccyx, dorsal nerve of the penis, nerve to bulb, and genitofemoral nerve. Henry Vandyke Carter, Public Domain, via Wikimedia Commons
Figure
Genitofemoral Nerve. The genitofemoral nerve arises from the lumbar plexus and supplies sensation to the thigh and genitalia. Illustration by E Gregory
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References
1.
Mahabadi N, Lew V, Kang M. StatPearls [Internet]. StatPearls Publishing; Treasure Island (FL): Apr 11, 2023. Anatomy, Abdomen and Pelvis: Femoral Sheath. [PubMed: 29494010]
2.
Benes J, Nádvornik P, Dolezel J. Abdominoinguinal pain syndrome treated by centrocentral anastomosis. Acta Neurochir (Wien). 2000;142(8):887-91. [PubMed: 11086827]
3.
Lee KS, Sin JM, Patil PP, Hanna AS, Greenberg JA, Zea RD, Brace CL. Ultrasound-Guided Microwave Ablation for the Management of Inguinal Neuralgia: A Preliminary Study with 1-Year Follow-up. J Vasc Interv Radiol. 2019 Feb;30(2):242-248. [PubMed: 30717957]
4.
Iwanaga J, Simonds E, Schumacher M, Kikuta S, Watanabe K, Tubbs RS. Revisiting the genital and femoral branches of the genitofemoral nerve: Suggestion for a more accurate terminology. Clin Anat. 2019 Apr;32(3):458-463. [PubMed: 30592097]
5.
Cirocchi R, Henry BM, Mercurio I, Tomaszewski KA, Palumbo P, Stabile A, Lancia M, Randolph J. Is it possible to identify the inguinal nerves during hernioplasty? A systematic review of the literature and meta-analysis of cadaveric and surgical studies. Hernia. 2019 Jun;23(3):569-581. [PMC free article: PMC6586705] [PubMed: 30570686]
6.
Mellick LB, Mowery ML, Al-Dhahir MA. StatPearls [Internet]. StatPearls Publishing; Treasure Island (FL): Apr 19, 2023. Cremasteric Reflex. [PubMed: 30020720]
7.
Zhu CL, Zhong H, Li CH. [Anatomic application of the genitofemoral nerve in uroandrological surgery]. Zhonghua Nan Ke Xue. 2017 Mar;23(3):276-279. [PubMed: 29706052]
8.
Narita M, Jikihara S, Hata H, Matsusue R, Yamaguchi T, Otani T, Ikai I. Surgical experience of laparoscopic retroperitoneal triple neurectomy for a patient with chronic neuropathic inguinodynia. Int J Surg Case Rep. 2017;40:80-84. [PMC free article: PMC5612803] [PubMed: 28942229]
Disclosure:Marco Gupton declares no relevant financial relationships with ineligible companies.
Disclosure:Matthew Varacallo declares no relevant financial relationships with ineligible companies.
Introduction
Structure and Function
Blood Supply and Lymphatics
Nerves
Physiologic Variants
Surgical Considerations
Clinical Significance
Other Issues
Review Questions
References
Copyright © 2025, StatPearls Publishing LLC.
This book is distributed under the terms of the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International (CC BY-NC-ND 4.0) ( which permits others to distribute the work, provided that the article is not altered or used commercially. You are not required to obtain permission to distribute this article, provided that you credit the author and journal.
Bookshelf ID: NBK430733 PMID: 28613484
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Introduction
Structure and Function
Blood Supply and Lymphatics
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Clinical Significance
Other Issues
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[Damage to the inguino-femoral nerves in the treatment of hernias. An anatomical hazard of traditional and laparoscopic techniques].[Ann Chir. 1996][Damage to the inguino-femoral nerves in the treatment of hernias. An anatomical hazard of traditional and laparoscopic techniques].Chevallier JM, Wind P, Lassau JP. Ann Chir. 1996; 50(9):767-75.
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16098 | https://turgor.ru/lktg/2012/1/1-1en.pdf | Amazing properties of binomial coeffcients 1 Amazing properties of binomial coefficients Several research topics will be set to you at the conference. Your aim is the maximal advance in one of these topics. You can co-operate in the solving of problems, arbitrary teams are allowed (i.e. the team may consist of participants from different cities). If you solve problems in different topics you may take part in different teams. The only thing you should avoid is to sign up the solutions of those problems that you really were not solving (this may happen if the team is too big and not all of its members solve the problems of some topic actively).
The following is the introductory set of problems about binomial coefficients. You may hand in the (written) solutions to Kokahs K. (coach 15, seat 17) In Teberda the set of problems will be enlarged a lot and you may hand in your solutions of this set of problems, except 1.2, in Teberda, too. You can hand in the solutions of the problem 1.2 in train only.
1 Problems for solving in train 1.1. Prove that a) p−1 k ≡(−1)k (mod p); b) 2n n ≡(−4)n p−1 2 n (mod p) при n ⩽p−1 2 .
1.2. Prove that the number of odd binomial coefficients in n-th row of Pascal triangle is equal to 2r, where r is the number of 1’s in the binary expansion of n.
1.3. Fix a positive integer m. By a m-arithmetical Pascal triangle we mean a triangle in which binomial coefficients are replaced by their residues modulo m. We will also consider similar triangles with the arbitrary residues a instead of 1’s along the lateral sides of the triangle. The operation of the multiplying by a number and addition of triangles of equal size are correctly defined. We will consider these operations modulo m.
a a a a 2a a a 3a 3a a b b b b 2b b b 3b 3b b + a+b a+b a+b a+b 2(a+b) a+b a+b 3(a+b) 3(a+b) a+b = x· a a a a 2a a a 3a 3a a = ax ax ax ax 2ax ax ax 3ax 3ax ax Let all the elements of s-th row of m-arithmetical Pascal triangle except the first and the last one be equal to 0. Prove that the triangle has a form depicted on fig. 1. Shaded triangles consist of zeroes, triangles ∆k n consist of s rows and satisfy the following relations 1) ∆k−1 n + ∆k n = ∆k n+1; 2) ∆k n = Ck n · ∆0 0 (mod m).
The well known puzzle Tower of Hanoi consists of three rods, and a number of disks of different sizes which can slide onto any rod. The puzzle starts with the disks in a neat stack in ascending order of size on one rod, the smallest at the top, thus making a conical shape. The objective of the puzzle is to move the entire stack to another rod, obeying the following rules: 1) оnly one disk may be moved at a time; 2) each move consists of taking the upper disk from one of the rods and sliding it onto another rod, on top of the other disks that may already be present on that rod; 3) no disk may be placed on top of a smaller disk.
Let n be the number of disks. Let THn be a graph, whose vertices are all possible correct placements of disks onto 3 rods and edges connect placements that can be obtained one from another by 1 move. Consider also graph Pn, whose vertices are 1’s located in the first 2n rows of the 2-arithmetical Pascal triangle and edges connect neighboring 1’s (i.e. two adjacent 1’s in the same row or neighboring 1’s by a diagonal in two adjacent rows ) 1.4. prove that graphs THn and Pn are isomorphic.
1.5. Prove that that first 106 rows of 2-arithmetical Pascal triangle contain less than 1 % of 1’s.
1.6. Prove that if n is divisible by p −1, then n p−1 + n 2(p−1) + n 3(p−1) + . . . + n n ≡1 (mod p). Or, even better prove the general statement: if 1 ⩽j, k ⩽p −1 и n ≡k (mod p −1), then n j + n (p −1) + j + n 2(p −1) + j + n 3(p −1) + j + . . . ≡ k j (mod p).
∆0 0 ∆0 1 ∆1 1 ∆0 2 ∆1 2 ∆2 2 ∆0 3 ∆1 3 ∆2 3 ∆3 3 ∆0 4 ∆1 4 ∆2 4 ∆3 4 ∆4 4 Рис. 1: Рис. 2: Amazing properties of binomial coeffcients 2 Amazing properties of binomial coefficients — 2 “The official theoretical source” for this set of problems is Vinberg’s article . Particularly the following theorems are considered to be known.
1. Wilson’s theorem. For any prime p (and for primes only) the equivalence holds (p −1)! ≡−1 (mod p).
2. Lukas’ theorem. Write the numbers n and k in base p: n = ndpd + nd−1pd−1 + . . . + n1p + n0, k = kdpd + kd−1pd−1 + . . . + k1p + k0.
(1) Then n k ≡ nd kd nd−1 kd−1 · . . . · n1 k1 n0 k0 (mod p) .
3. Kummer’s theorem. The exponent ordp n k is equal to the number of “carries” when we add k and ℓ= n −k in base p.
4. Wolstenholme’s theorem. If p ⩾5 then 2p p ≡2 (mod p3), or, that is the same, 2p−1 p−1 ≡1 (mod p3).
Remind that 0 0 = 1, n k = 0 for k > n and for k < 0 by definition.
We denote by p a prime number. For any natural n denote by (n!)p the product of all integers from 1 to n not divisible by p. If a number p is given the symbols ni, mi etc. denote the digits of numbers n, m etc. in base p.
2 Arithmetical triangle and divisibility 2.1. a) Prove that the first 3k rows of 3-arithmetical Pascal triangle contain 1 2(6k + 4k) residues “1” and 1 2(6k −4k) residues “2”.
b) Find the number of zero elements in the first 5k rows of 5-arithmetical Pascal triangle.
c) Find the number of non-zero elements in the first pk rows of p-arithmetical Pascal triangle.
2.2. Prove that the number of 1’s in the first m rows of 2-arithmetical Pascal triangle equals n−1 X i=0 mi · 2 n−1 k=i+1 mk · 3i.
If m = 2α1 +2α2 +. . . +2αr, where α1 > α2 > . . . > αr, then we can rewrite the last expression in the form 3α1 + 2 · 3α2 + 22 · 3α3 + . . . + 2r−1 · 3αr .
2.3. Consider n-th row of Pascal triangle modulo 2 as binary expansion of some integer Pn. Prove that Pn = Fi1 · . . . · Fis, where i1, . . . , is are numbers of positions where 1’s occur in the binary expansion of n, and Fi = 22i + 1 is i-th Fermat number.
2.4. Prove that the number of non-zero elements in n-th row of p-arithmetical Pascal triangle equals d Q i=0 (ni + 1).
2.5. a) All the binomial coefficients n k , where 0 < k < n, are divisible by p if and only if n is a power of p.
b) All the binomial coefficients n k , where 0 ⩽k ⩽n, are not divisible by p if and only if n + 1 is divisible by pd, in other words, all the digits of n, except the leftmost, in base p are equal to p −1.
2.6. Let 0 < k < n + 1. Prove that if n k−1 ̸ .
.
. p and n k ̸ .
.
. p, then n+1 k ̸ .
.
. p, except the case, when n + 1 is divisible by p.
Amazing properties of binomial coeffcients 3 3 Generalization of Wilson’s and Lukas’ theorems 3.1. Prove that ordp(n!) = n −(nd + . . . + n1 + n0) p −1 .
3.2. Prove the following generalizations of Wilson’s theorem.
a) (−1)n/pp ≡n0! (mod p); b) Prove that for p ⩾3 (pq!)p ≡−1 (mod pq) , and for p = 2, q ⩾3 (pq!)p ≡1 (mod pq).
c) n!
pµ ≡(−1)µn0!n1! . . . nd! (mod p), where µ = ordp(n!) 3.3. Generalized Lukas’ theorem. Let r = n −k, ℓ= ordp( n k ). Then 1 pℓ n k ≡(−1)ℓ n0!
k0!r0!
n1!
k1!r1!
. . .
nd!
kd!rd!
(mod p) 3.4. a) Prove that (1 + x)pd ≡1 + xpd (mod p) for all x = 0, 1, . . . , p −1.
b) Prove Laukas’ theorem algebraically.
3.5. a) Let m, n, k be nonnegative integers, and (n, k) = 1. Prove that Ck mn ≡0 (mod n).
b) Prove that if n .
.
. pk, m ̸ .
.
. p, then n m .
.
. pk.
3.6. Let fn,a = n P k=0 ( n k )a. Prove that fn,a ≡ d Q i=0 fni,a (mod p).
4 Variations on Wolstenholme’s theorem 4.1. Prove that 1 1 + 1 2 + . . . + 1 p −1 ≡0 (mod p2).
4.2. Let p = 4k + 3 be a prime number. Find 1 02 + 1 + 1 12 + 1 + . . . + 1 (p −1)2 + 1 (mod p).
4.3. a) Let k be a nonnegative integer such that for any prime divisor p of the number m k is not divisible by (p−1). Prove that 1 1k + 1 2k +. . . + 1 (p −1)k ≡0 (mod m) (summation over all fractions whose denominators are coprime to m).
b) Let k be odd and (k + 1) ̸ .
.
. (p −1). Prove that 1 1k + 1 2k + . . . + 1 (p −1)k ≡0 (mod p2).
4.4. Prove that the equivalence (12) from Vinberg’s article holds in fact modulo p4.
4.5. Prove that the following properties are equivalent 1) 2p−1 p−1 ≡1 (mod p4); 2) 1 1 + 1 2 + . . . + 1 p −1 ≡0 (mod p3); 3) 1 12 + 1 22 + . . . + 1 (p −1)2 ≡0 (mod p2).
4.6. a) Prove algebraically that for any prime p and arbitrary k and n ( pk pm − k m ) .
.
. p2. (In Vinberg’s article this fact is proven combinatorially.
b) Prove the statement (9) form Vinberg’s article: for any prime p ⩾5 and arbitrary k and n ( pk pm − k m ) .
.
. p3.
4.7. Let p ⩾5. Prove that a) p2 p ≡ p 1 (mod p5); b) ps+1 p ≡ps (mod p2s+3).
4.8. Prove that p3 p2 ≡ p2 p (mod p8).
Amazing properties of binomial coeffcients 4 Amazing properties of binomial coefficients — 3 Additional problems to previous topics 2.7. Prove that pn−1 k ≡(−1)Sk (mod p), where Sk is the sum of digits of k in base p.
2.8. Prove that if the binomial coefficient n k is odd i.e. ki ⩽ni for all i = 0, 1, . . . , d in the notations of (1), then n k ≡ d Y i=1 (−1)ki−1ni+kini−1 (mod 4).
2.9. Prove that if there are no two consecutive 1’s in the binary expansion of n then all the odd entries in n-th row ≡1 (mod 4), otherwise the number of entries ≡1 (mod 4) equals the number of entries ≡−1 (mod 4).
2.10. Prove that the number of 5’s in each row of 8-arithmetical Pascal triangle is a power of 2. Prove the same for 1’s, 3’s and 7’s.
2.11. Prove that if we consider all the elements of the two sets 2n −1 1 , 2n −1 3 , 2n −1 5 , . . . , 2n −1 2n −1 and {1, 3, 5, . . . , 2n −1} as a reminders modulo 2n, then these sets coincide.
2.12. Prove that elements of a row of Pascal triangle are not coprime in the following sence. For any ε > 0 there exists N, such that for all integer n > N and k1, k2, . . . , k100 < ε√n the numbers 2n n + k1 , 2n n + k2 , . . . , 2n n + k100 have a common divisor.
2.13. a) The non negative numbers m > 1, n, k are given. Prove that at least one of the numbers n k , n+1 k , . . . , n+k k is not divisible by m.
b) Prove that for each k there exist infinite set of numbers n, such that all the numbers n k , n+1 k , . . . , n+k−1 k are divisible by m.
4.9. Prove that for n > 1 2n+1 2n − 2n 2n−1 is divisible by 22n+2.
4.10. Prove that for p ⩾5 (−1) p−1 2 p−1 p−1 2 ≡4p−1 (mod p3).
Amazing properties of binomial coeffcients 5 Amazing properties of binomial coefficients — 4 Additional problems to previous topics 4.11. Let m be a non negative integer, p ⩾5 be a prime. Prove that 1 mp + 1 + 1 mp + 2 + · · · + 1 mp + (p −1) ≡0 (mod p2).
4.12. Let p and q be primes. Prove that 2pq−1 pq−1 ≡1 (mod pq) if and only if 2p−1 p−1 ≡1 (mod q) and 2q−1 q−1 ≡1 (mod p).
5 Sums of binomial coefficients 5.1. a) Prove that the sum 3a−1 P k=0 2k k is divisible by 3; b) is divisible by 3a.
5.2. Let Ck = 1 k+1 2k k be Catalan numbers. Prove that n P k=1 Ck ≡1 (mod 3) if and only if the number n+1 contains at least one digit “2” in base 3.
5.3. Let p ⩾3, k = [2p/3]. Prove that the sum p 1 + p 2 + . . . + p k is divisible by p2.
5.4. Let n .
.
. (p −1), where p is an odd prime. Prove that n p −1 + n 2(p −1) + n 3(p −1) + . . . ≡1 + p(n + 1) (mod p2).
5.5. Prove that if 0 ⩽j ⩽p −1 < n and q = n−1 p−1] then X m:m≡j (mod p) (−1)m n m ≡0 (mod pq).
5.6. Let p be an odd roime. Prove that n .
.
. (p + 1) if and only if n j − n j + (p −1) + n j + 2(p −1) − n j + 3(p −1) + . . . ≡0 (mod p) for all j = 1, 3, . . . , p −2.
Amazing properties of binomial coeffcients 6 Solutions 1 Problems for solving in train 1.1. a) S o l u t i o n 1.
p−1 k = (p −1)(p −2) . . . (p −k) 1 · 2 · · · k ≡(−1)(−2) . . . (−k) 1 · 2 · · · k ≡(−1)k (mod p).
S o l u t i o n 2. It is evident by the formula for binomial coefficients that p i is divisible by p when 1 ⩽i ⩽p−1. Since p−1 k−1 + p−1 k = p k and p−1 0 = 1 ≡1 (mod p), then ( p−1 0 + p−1 1 ) .
.
. p, and therefore p−1 1 ≡−1 (mod p). But p−1 1 + p−1 2 is divisible by p also, hence p−1 2 ≡1 (mod p) etc.
б) This problem is taken from [3, problem 162]. Since the fractions 2n+2 n+1 / 2n n and p−1 2 n+1 / p−1 2 n are highly reducible, the statement can be easily proven by induction. But we suggest a direct calculation from .
It easy to see that 2n n = 2n · 1 · 3 · · · (2n −1) n!
and 1 · 3 · · · (2n −1) = (−1)n(−1)(−3) · · · (−2n + 1) ≡(−1)n(p −1)(p −3) · · · (p −2n + 1) = = (−1)n2np −1 2 p −3 2 · · · p −2n + 1 2 = (−1)n2np −1 2 p −1 2 −1 · · · p −1 2 −n + 1 = = (−1)n2n (p−1 2 )!
(p−1 2 −n)!
(mod p).
Therefore 2n n ≡(−1)n4n (p−1 2 )!
n!(p−1 2 −n)!
= (−4)n p−1 2 n (mod p).
1.2. It follows directly from self-similar structure of an arithmetical Pascal triangle, that is described in the next problems. It follows from Lucas’ theorem also, you can read the proof in .
1.3. We restrict ourselves with small contemplation, the full solution can be found in [3, problem 133].
Since the s-th row contains a long sequence of zeroes, then below these zeroes in (s+1)-th row we have the sequence of zeroes, too, (it is one element shorter than the upper sequence); in (s + 2)-th row there are the sequence of zeroes also (it is one element shorter again) and so on. This explains the presence of the grey triangle below ∆0 0 (fig. 1).
Further, the non-zero elements of the s-th row are equal to 1, hence the numbers situated along the sloped sides of the grey triangle all are 1’s (due to the recurrence for binomial coefficients). So all the numbers along the sloped sides of the triangles ∆0 1 and ∆1 1 are 1’s, and therefore both triangles are identical to ∆0 0.
Now it is clear, what is the (2s)-th row of the triangle. The left- and the rightmost elements are 1’s, all other elements equal 0, except the central element that is equal to 2, because it is a sum of the two upper 1’s. Thus we obtain that two grey triangles are situated below 2s-th row, the triangles ∆0 2 and ∆2 2 to the left and to the right of them are identical to ∆0 0, and the triangle ∆1 2 with 2’s along its sloped sides is equal to 2 · ∆0 0.
And so on.
1.4. This statement we found in , several facts about binomial coefficients are proven there via Tower of Hanoi and the graph THn.
Let a be the diameter of the upper disc on the first rod, b be the diameter of the upper disc on the second rod and c be the diameter of the upper disc on the third rod. W.l.o.g. a < b < c, then we have 3 possible moves in this configuration: from a to b or c and from b to c, we analogously have 3 moves if one rod is without discs. If all the discs are placed on one rod then we have 2 possible moves only; let A1, A2, A3 denote the configurations of this type.
Observe that by the problem 1.2 all the elements of 2s-th row of Pascal triangle are 1’s. Therefore graph Pn has the rotational symmetry of the third order, because the recurrence n k−1 + n k = n+1 k , that allows us to construct the triangle from top to bottom, is equivalent in arithmetic modulo 2 to the recurrences n k−1 = n k + n+1 k and n k = n k−1 + n+1 k , that allows us to construct the triangle from the low left Amazing properties of binomial coeffcients 7 A1 A3 A2 A2 A1 A3 A1 A3 A2 2 1 3 Рис. 3: corner in the upper right direction and from the low right corner in the upper left direction. It follows also that the triangle of the double size contains 3 copies of the initial triangle.
Now let us prove by induction that there exists a bijection between THn and Pn, such that the vertices of the triangle Pn correspond to the configurations A1, A2, A3. The base n = 1 is evident.
Proof of the step of induction. Assume that the bijection between THn and Pn has been constructed.
The 2-arithmetical Pascal triangle with the side length 2n+1 contains 3 copies of the triangle with the side length 2n. Number the copies and mark its vertices as shown on fig.3. Consider all the configurations of the Tower of Hanoi for which the (n + 1)-th (biggest) disc is placed on rod i. If we fix the placement of this disc then displacements of other discs correspond to the graph that is isomorphic to TPn. By induction hypothesis we can choose a bijection between this graph and the graph Pn in the i-th copy of the triangle, such that the configurations Aj correspond to the vertices of the triangles with the same marks. When we move the biggest disc, say, from the first rod to the second, all other discs must be on the 3rd rod. This move correspond to the edge connecting two neighboring vertices A3 on the left sloped side of big triangle.
The same reasons concern other moves of the biggest disc. Therefore we obtain an isomorphism between TPn+1 and Pn.
1.5. The bijection with Tower of Hanoi gives us a formula (when the number of rows is a power of 2): the first 2k rows of 2-arithmetical Pascal triangle contain 3k 1’s. The formula can be also proved by induction via recurrence from the problem 1.3. Using this formula we can obtain an estimation. Since 106 < 220, the total number of elements in these rows equals 1 2 · 106(106 + 1), and the number of 1’s is at most 320. The proportion does not exceed 2·320 106(106+1) ≪0.01.
1.6. We found this statement in .
S o l u t i o n 1 ([CSTTVZ]). For p = 2 the statement can be easily checked. So we can assume that p is odd prime. Let n = x(p −1) + k. We use induction on x.
The base x = 0 is trivial: k j ≡ k j (mod p).
To prove the step of induction we need the following property of binomial coefficients: a + b s = X i a s −i b i (summation in natural bounds), both sides of which calculate in how many ways we can choose s balls in the box that contains a black and b white balls. Let n = m + (p −1). Observe that n ℓ(p−1) + j = m + (p−1) ℓ(p−1) + j = p−1 X i=0 m ℓ(p−1) + j −i p −1 i ≡ p−1 X i=0 (−1)i m ℓ(p−1) + j −i (mod p) (the last equivalence is due to problem 1.1 a). Remark that the sign of the first and last terms in the last Amazing properties of binomial coeffcients 8 sum is “plus” . Now transform the sum from the problem statement: X ℓ n ℓ(p −1) + j ≡ ≡ m j − m j −1 + . . .
+ m p −1 + j − m p −1 + j −1 +. . .+ m j + + m 2(p−1) + j − m 2(p−1) + j −1 +. . .+ m 2(p−1) + j + . . .
= m X i=0 (−1)i m i + X ℓ m ℓ(p−1) + j (mod p).
the first sum is equal to 0, the second sum is equivalent k j (mod p) by the induction hypothesis.
S o l u t i o n 2 ([J], [T]). Induction by n. The base n ⩽p −1 is trivial: both sides contain the same term. Prove the step of induction.
n j + n (p −1) + j + . . . = n −1 j + n −1 j −1 + n −1 (p −1) + j + n −1 (p −1) + j −1 + . . . = = n −1 j + n −1 (p −1) + j + . . .
+ n −1 j −1 + n −1 (p −1) + j −1 + . . .
≡ ≡ k −1 j + k −1 j −1 = k j (mod p).
But it should be accurate in cases when p −1 divides j or k, because the induction hypothesis does not hold for j = 0 or k = 0 (it uses the value p −1 instead of 0). Therefore we must consider more carefully the cases when j = 1 or k = 1. We restrict ourselves by consideration of one partial case only. Let p = 5, j = 1 and we fulfill step to n = 13. Then we have 1 1 ?
≡ 13 1 + 13 6 + 13 11 = 12 1 + 12 6 + 12 11 + 12 0 + 12 5 + 12 10 .
By induction hypothesis the sum in the first parentheses has a residue 4 1 (and not 0 1 as the previous calculation shows). In the second parentheses the induction hypothesis covers all the terms except the first one, so the sum has residue 12 0 + 4 0 . Writing p −1 instead of 4 for clarity, we obtain that the whole sum is equivalent to n−1 0 + p−1 1 + p−1 0 ≡ 1 1 (mod p), as required.
S o l u t i o n 3 (algebraical reasoning with Luka’s theorem, ). Induction by n. Base n ⩽p −1 is trivial. Now let n ⩾p, write all parameters in base p, let σp(m) denotes the sum of digits of m. It is clear that if m ≡j (mod p), then σp(m) ≡j (mod p). The sum under consideration is equal by Luka’s theorem to X n0 m0 n1 m1 . . .
nd md (mod p) , where the summation is over all m = md . . . m1m0 ⩽n, for which σp(m) ≡j (mod p). This sum is equal to the sum of coefficients of xj, xj+p−1, xj+2(p−1), . . . in the expression (1 + x)n0(1 + x)n1 . . . (1 + x)nd = (1 + x)σp(n) .
But it is evident that this sum of coefficients equals X 1⩽r⩽σp(n) r≡j (mod p−1) σp(n) r , which satisfy the induction hypothesis because 1 ⩽σp(n) ⩽n−1, and supply the desired equivalence since σp(n) ≡n ≡j (mod p).
S o l u t i o n 4 (linear algebra, [D]). The polynomials x, x2, . . . , xp−1 are linearly independent over Zp and form a basis in the space of functions f : Zp →Zp, f(0) = 0. By Fermat’s little theorem (1 + x)n ≡ (1 + x)k (mod p). Applying the relations xi+a(p−1) ≡xi to the left hand side, we obtain that our sum as an element of Zp is equal to the coefficient of xj in the right hand side, i. e.
k j .
Amazing properties of binomial coeffcients 9 2 Arithmetical triangle and divisibility 2.1. a) This result is due to Roberts . By ak denote the number of 1’s in the first 3k rows, and by bk denote the number of 2’s. Due to the recurrence from problem 1.3 we obtain ak+1 = 5ak + bk, bk+1 = 5bk + ak.
Now the statement of problem follows by induction.
b) A n s w e r: 1 2 · 5k(5k + 1) −15k. By ak denote the number of nonzero elements in the first 5k rows.
As in previous problem we have a recurrence ak+1 = 15ak + 10 · 5k(5k −1) 2 .
Since the whole triangle consists of 5k(5k+1) 2 elements, it is natural to change variables ak = 5k(5k+1) 2 −bk.
Then we can rewrite the previous relation in terms of bk as bk+1 = 15bk.
c) A n s w e r: p(p+1) 2 k. This is Fine’s result . It can be obtained by induction by means of recurrence of the problem 1.3.
2.2. S o l u t i o n 1. Induction by α1. The base α1 = 0, 1 can be easily checked. Let the statement has been proven for all α1 < a. Prove it for α1 = a. Evedently ˜ m −2α1 < 2α1. Let s = 2α1 (in notations of problem 1.3). Consider the ˜ m-th row in the triangle ∆0 0, where ˜ m = 2α2 + 2α3 + . . . + 2αr. By the induction hypothesis the number of 1’s in this row and above it equals 3α2 + 2 · 3α3 + . . . + 2r−2 · 3αr.
(2) Then for the number m = ˜ m + 2α1 we have a row that intersects the triangles ∆1 0 and ∆1 1 (due to 2-arithmetics they are both identical to triangle ∆0 0). The part of Pascal triangle from top to this row contains triangle ∆0 0 (containing 3α1 1’s by induction hypothesis) and partially triangles ∆1 0 and ∆1 1 (the number of 1’s in them is given by (2)). So the total number of 1’s is 3α1 + 2(3α2 + 2 · 3α3 + . . . + 2r−2 · 3αr).
S o l u t i o n 2 (combinatorial sense of coefficients, [T]).
L e m m a 1. Let the k-th row contains 2r 1’s (or, equivalently, k contains r 1’s in base 2) and let α1 > α2 > · · · > αm, 2αm > k. Then the row with number 2α1 + 2α2 + . . . + 2αm + k contains 2m+r 1’s.
P r o o f. It is clear that the number 2α1 + 2α2 + . . . + 2αm + k in base 2 contains m + r 1’s and hence the corresponding row contains 2m+r 1’s.
L e m m a 2. The rows with the following numbers 2α1 + 2α2 + . . . + 2αm−1, 2α1 + 2α2 + . . . + 2αm−1 + 1, . . . , 2α1 + 2α2 + . . . + 2αm−1 + 2αm −1, contain 2k3αm 1’s.
P r o o f. By lemma 1 the row with number 2α1 + 2α2 + . . . + 2αm−1 + i contains 2kxi 1’s, where xi is the number of 1’s in i-th row. Then the total number of 1’s in these rows equals 2k P xi. But P xi is the number of 1’s in the first 2αm −1 rows of Pascal triangle, this number is equal to 3αm (it is known, for example, by problem 1.4).
The statement of problem follows from lemma 2.
2.3. The problem is from , the solution is from . The problem statement follows from Luka’s theorem due to the following observation (it is also mentioned in ): a binomial coefficient n k is odd if and only if the set of 1’s in the binary expansion of k is the subset of the set of 1’s in the binary expansion of n.
Therefore Pn = P 2k, where the summation is over all k described in the previous phrase. For p = 2 let Sn = {i : ni = 1} in notations of formula (1). Then Pn = X I⊆Sn Y i∈I 22i = Y i∈Sn Fi.
Amazing properties of binomial coeffcients 10 2.4. This result of Fine (1947) is an easy corollary of Kummer’s theorem. If p does not divide n k , then there are no carries when we add k and n −k in base p. For a fixed n it means that we can choose i-th digit of k in base p by ni + 1 ways.
2.5. a) It follows from the formula proven in the previous problem because here we have a row with 2 elements only not divisible by p.
b) . If Если pd | (n + 1), then n = a(p −1)(p −1) . . . (p −1) in base p. Then for any k, 0 ⩽k ⩽n, each digit of k does not exceed the corresponding digit of n. Therefore all the binomial coefficients ni ki are not equal to 0 and ̸≡0 (mod p). By Lukas’ theorem n k is not divisible by p.
The reverse statement. Assume that all the coefficients n k are not divisible by p, but n is not the number of the form a(p −1)(p −1) . . . (p −1). Therefore one of its digits, say, ni is less than p −1. Choose k = (p −1) · pi. Then ki = p −1 and hence ni ki = 0, and p | n k by Lukas’ theorem. A contradicition.
2.6. This problem we found in .
S o l u t i o n 1. Assume that n k−1 ̸ .
.
. p and n k ̸ .
.
. p, but n+1 k = n k−1 + n k .
.
. p. Then n k ≡− n k−1 (mod p). Since both binomial coefficients are not divisible by p, we can reduce the equivalence and obtain n−k+1 k ≡−1 (mod p). Therefore n + 1 ≡0 (mod p).
S o l u t i o n 2 ([K]). Though the statement remind us the main recurrence for binomial coefficients, the part “ n k−1 ̸ .
.
. p” is unnecessary. Indeed, if (n+1) ̸ .
.
. p, then 0 ⩽n0 ⩽p−2. Since n k ̸ .
.
. p, then by Kummer’s theorem ki ⩽ni for all i But analogous inequalities hold also for the pair k and n + 1, because n and n + 1 have the same digits except the lower ones that differs by 1. Hence n+1 k ̸ .
.
. p.
2.7. . It follows from Lukas’ theorem and problem 1.1.a).
2.8. The problem is from . Induction by number of digits. The base is trivial. For the proof of induction step add one more digit to the rightmost position. Since the binomial coefficient is odd we have the inequalities ni ⩾ki. Now we will use the recurrence n k = n−1 k−1 + n−1 k and consider distinct variants of parity n и k. Applying Kummer’s theorem and the problem 4.6a) we will reduce the question to the induction hypothesis.
For example, let n = 2ℓ+ 1 be odd and k = 2m be even. Consider a subcase k1 = 1. Then we have binary representations k = . . . 10, n = . . . 11, k −1 = . . . 01 and n −k = . . . 01 (the latter because by Kummer’s theorem there are no carries when we add k and n −k). Now when we add k −1 and n −k we have 1 carry, i.e.
n−1 k−1 ≡2 (mod 4), and hence n k = n −1 k −1 + n −1 k ≡− n −1 k = − 2ℓ 2m ≡− ℓ m (mod 4) , the latter equivalence is by problem 4.6a). The minus sign in it corresponds to the multiplier (−1)k0n1+k1n0.
2.9. The problem is from . The statement follows from the previous problem. If the binary representation of n does not contains two consecutive 1’s, then for all k all the exponents ki−1ni+kini−1 are equal to 0 and all the binomial coefficients in n-th row have are equivalent 1 modulo 4. But if the binary representation of n contains several consecutive 1’s starting from nj = 1 then the one half of all coefficients have kj = 0, and one half of them have kj = 1. By the formula of previous problem these two halves differ by a sign.
2.10. Two articles in Monthly [19, 20] discuss this dark problem.
2.11. This is a problem of D. Dzhukich was presented at the olympiad of 239 school of St.-Petersburg, 2002, and after that appeared at short-list of IMO-2008.
All the binomial coefficients in the problem statement are odd by Lukas’ theorem, therefore, it is sufficient to check that all the numbers 2n−1 1 , 2n−1 3 , . . . , 2n−1 2n−1 have distinct reminders modulo 2n.
S o l u t i o n 1 ([D]). Assume by the contrary that 2n−1 k ≡ 2n−1 m (mod 2n) for odd k and m, k > m.
Observe that 2n −1 k = 2n k − 2n −1 k −1 = 2n k − 2n k −1 + 2n −1 k −2 = · · · = = 2n k − 2n k −1 + 2n k −2 −. . . − 2n m + 1 + 2n −1 m .
Amazing properties of binomial coeffcients 11 In particular 2n k − 2n k −1 + 2n k −2 −. . . − 2n m + 1 ≡0 (mod 2n) .
Calculate the exponent ord2 2n r by Kummer’s theorem. If ord2 r = a then we have n−a carries in addition r and 2n −r (it is clear by the standard algorithm of addition), hence ord2 2n r = n −a. In particular 2n | 2n r for odd r, that allows us to consider only one half of summands: 2n k −1 + 2n k −3 + . . . + 2n m + 1 ≡0 (mod 2n) .
Now all the 2n i in the left hand side have even parameter i, therefore ord2 2n x < n.
We will prove that this congruence is impossible and obtain a contradiction. Choose x with minimal ord2 2n x . Since ord2 2n x < n and the whole sum is divisible by 2n, there exists y, for which ord2 2n x = ord2 2n y . Then the binary representations of x and y end with equal number of 0’s, and hence there exists z between x and y which binary representation ends with bigger number of 0’s. Then ord2 2n z < ord2 2n x , a contradiction.
S o l u t i o n 2 ([CSTTVZ]). Induction by n. We prove the step of induction. Let the statement be proven for all numbers less than n. Assume by the contrary that there exist k and ℓ, k ̸= ℓ, 0 ⩽k, ℓ⩽2n −1, such that 2n−1 2k+1 ≡ 2n−1 2ℓ+1 mod 2n. Observe that 2n −1 2k + 1 = 2n 1 −1 2n 2 −1 . . .
2n 2k + 1 −1 = = 2n 1 −1 2n 3 −1 . . .
2n 2k + 1 −1 · 2n−1 1 −1 2n−1 2 −1 . . .
2n−1 k −1 = (3) = 2n 1 −1 2n 3 −1 . . .
2n 2k + 1 −1 · 2n−1 −1 k ≡ ≡(−1)k+1 2n−1 −1 k (mod 2n) and analogously 2n−1 2ℓ+1 ≡(−1)ℓ+1 2n−1−1 ℓ (mod 2n). It follows by induction hypothesis that both k and ℓcan not be odd. Besides, due to the symmetry 2n−1 r = 2n−1 2n−1−r the problem statement means that all the “even” binomial coefficients 2n−1 2r are pairwise distinct modulo 2n and form the same set of residues as “odd” binomial coefficients 2n−1 2r+1 . Therefore k and ℓcan not be even simultaneously.
It remains to consider a case when k and ℓhave distinct parity, say k = 2a + 1, ℓ= 2b. Then 2n−1 −1 2a + 1 + 2n−1 −1 2b ≡0 (mod 2n) .
If a = b the congruence is impossible because 2n−1−1 2a is odd and 2n−1 −1 2a + 1 + 2n−1 −1 2a = 2n−1 −1 2a 1+ 2n−1 −1 −2a 2a + 1 = 2n−1 −1 2a · 2n−1 2a + 1 ≡2n−1 (mod 2n) .
If b ̸= a, then 2n−1−1 2a ̸= 2n−1−1 2b by the induction hypothesis, Since 2n−1−1 2a + 2n−1−1 2a+1 is divisible by 2n−1, the sum 2n−1−1 2b + 2n−1−1 2a+1 can not be divisible by 2n−1.
2.12. The author of this problem is A. Belov. Observe that 2n n + k = 2n n · n(n −1) . . . (n −k + 1) (n + 1)(n + 2) . . . (n + k), and therefore 2n n+k have many common divisors with 2n n , because the denominator is not very big, more precisely, it does not exceed (2n)k. Write the analogous equalities for all binomial coefficients 2n n+k1 , Amazing properties of binomial coeffcients 12 2n n+k2 , . . . , 2n n+k100 . Then GCD of all denominators in the right hand sides of the equalities does not exceed (n+1)(n+2) . . .
n+[ε√n ] < (2n)ε√n. But for big n the binomial coefficient 2n n is much greater, so after reducing by GCD the quotient is very big, and it divides all 100 binomial coefficients.
Explain more accurate the last reasoning. Observe that 2n n = 2n n · 2n −1 n −1 . . . n + 1 1 > 2n and (2n)100ε√n = 2ε√n log2 n+ε√n.
For each ε there exists N such that for all n > N we have the equality n 2 > ε√n log2 n + ε√n. If we reduce 2n n by GCD for these n, the quotient is at least 2n/2.
2.13. a) The problem was presented at Leningrad olympiad, 1977.
S o l u t i o n 1 (without Kummer’s theorem). This is solution from the excellent book . Assume that all these numbers are divisible by m. Then the numbers n + k −1 k −1 = n + k k − n + k −1 k , n + k −2 k −1 = n + k −1 k − n + k −2 k , . . .
n k −1 = n + 1 k − n k are also divisible by m. Then analogously m divides all the numbers n+i j , where i ⩽j are arbitrary nonnegative integers. But n 0 (i = j = 0) is not divisible by m. A contradiction.
S o l u t i o n 2 (Kummer’s theorem). Let p be a prime divisor of m. Prove that at least one of the numbers n k , n+1 k , . . . , n+k k is not divisible by p. By Kummer’s theorem if we choose ℓ(n −k ⩽ℓ⩽n) such that the addition k + ℓfulfills in base p without carries then the binomial coefficient k+ℓ k is not divisible by p.
We will explain how to choose ℓby giving a concrete example. Let p = 7, k = 133. We will write all the numbers in base 7. Since we try to choose ℓin the set of k + 1 numbers, we can always choose ℓsuch that k + ℓto be one of the following numbers . . . 133, . . . 233, , . . . , . . . 633.
(Remind that 6 is the greatest digit in our example.) It is clear that the addition k + ℓfulfills without carries.
b) We found this problem in . It is not difficult to construct n by Kummer’s theorem. Let ordp m = s, and k have d+1 digits in base p. Let n .
.
. pd+s+1. Then the representations of numbers n−k, n−k +1, . . . , n −1 contain digits (p −1) in positions from (d + 2) to (d + s + 2). When we add k to these numbers we have carries in these positions. Therefore by Kummer’s theorem all the corresponding binomial coefficients are divisible by ps.
Since it is not difficult to combine our reasoning for distinct p, the statemetn is proven.
3 Generalizations of Wilson’s and Lukas’ theorems 3.1. It is well known that ordp(n!) = P k n pk . If n = ndpd + nd−1pd−1 + . . . + n1p + n0 (representation in base p), then n pk = ndpd−k + nd−1pd−k−1 + . . . + nk+1p + nk and we can rewrite the formula for ordp(n!) in the form ordp(n!) = d X k=1 d X i=k nipi−k !
= d X i=1 ni(pi−1 + pi−2 + . . . + p + 1) = d X i=1 ni pi −1 p −1 = d P i=0 nipi − d P i=0 ni p −1 .
This is exactly what we need.
Amazing properties of binomial coeffcients 13 3.2. a) Split the factors of n! on groups of (p −1) factors: (n!)p = [ n p ]−1 Y k=0 (kp+1)·(kp+2) · · · (kp+p−1) · [n p]p+1 [n p]p+2 . . .
[n p]p+n0 ≡(−1)[ n p ]n0!
(mod p) .
б) This statement can be found in Gauss works . The product (pq!)p contains factors in pairs: a factor and its inverse modulo pq, the product of each pair is 1 modulo pq. So we need to watch on those factors m which equals to its inverse, this factors satisfy the congruence m2 ≡1 (mod pq).
For odd prime p the congruence has 2 solutions: ±1. For p = 2, q ⩾3 the congruence has two more solutions: 2q−1 ± 1.
c) Since n! = (n!)p · p[ n p ] [n p] !, the statement can be proven by induction by means of the congruence of statement a) of this problem.
3.3. We found this problem on the web-page of A.Granville . It well known Legendre’s formula for the number ℓis that ℓ= ordp n k = hn p i − hk p i − hr p i + h n p2 i − h k p2 i − h r p2 i + . . .
(4) Denote ˜ n = [n/p] for brevity and so forth, and collect all terms divisible by p in the the formula for a binomial coefficient: n k = (n!)p (k!)p(r!)p · p[n/p] p[k/p] · p[r/p] · ˜ n!
˜ k! · ˜ r!
.
By generalized Wilson’s theorem (problem 3.2, b) the first fraction equals ± n0!
k0!r0! (mod p), the third fraction allows us to apply induction, and the middle fraction (together with the sign of the first fraction) supply all the expressions containing ℓby the formula (4).
3.4. a) Expand brackets in (1 + x)pd use the fact that p | pd k for 1 ⩽k ⩽pd −1 by Kummer’s theorem.
b) Let n = n′p + n0, k = k′p + k0. By the previous statement (1 + x)pn′ ≡(1 + xp)n′ (mod p). Then (1 + x)n = (1 + x)pn′(1 + x)n0 ≡(1 + xp)n′(1 + x)n0 (mod p).
This congruence means that we transform the coefficients of the polynomial modulo p. The coefficient of xk at the l.h.s. equals n k . All the exponents in the first brackets at the r.h.s. are divisible by p, hence the only way to obtain the term xpk′+k0 is multiplying the xpk′ from the first bracket and xk0 from the second.
Thus we obtain n′ k′ n0 k0 and so n k = n′ k′ n0 k0 . Now Lukas’ theorem follows by induction.
3.5. a, b) It follows from Kummer’s theorem.
3.6. . In the following calculation we use that ni ki = 0 for ki > ni; this allows us to apply Lukas’ theorem and truncate a lot of summands: fn,a = n X k=0 n k a ≡ nd X kd=0 nd−1 X kd−1=0 · · · n0 X k0=0 d Y i=0 ni ki a ≡ d Y i=0 ni X ki=0 ni ki a ≡ d Y i=0 fni,a (mod p).
4 Variations on Wolstenholme’s theorem 4.1. This is an exercise on reading an article. The statement is proven in article . Observe that 2 p−1 X i=1 1 i = p−1 X i=1 1 i + 1 p −i = p p−1 X i=1 1 i(p −i) .
Amazing properties of binomial coeffcients 14 Hence the sum under consideration is divisible by p. Since 1 i ≡−1 p−i (mod p), it remains to check that p−1 X i=1 1 i2 ≡0 (mod p).
But 1 12, 1 22 , . . . , 1 (p−1)2 modulo p is the same set as1, что 12, 22, . . . , (p −1)2. Therefore it is sufficient to prove that p−1 X i=1 i2 ≡0 (mod p).
(5) Let p−1 P i=1 i2 ≡s (mod p). It p > 5 we can always choose a, such that a2 ̸≡1 (mod p). Then the sets {1, 2, . . . , p −1} and {a, 2a, . . . , (p −1)a} coincide (the proof is the same as in the footnote) and s ≡ p−1 X i=1 i2 = p−1 X i=1 (ai)2 = a2 p−1 X i=1 i2 ≡a2s (mod p) .
Thus s ≡0 (mod p).
4.2. A n s w e r: 2k + 2. This problem of A. Golovanov was presented at Tuimaada-2012 olympiad. Observe that for p = 4k + 3 the equation x2 + 1 = 0 has no solutions in the set of residues modulo p, and hence the denominators of all fractions are non zero.
S o l u t i o n 1. Let ai = i2 + 1, i = 0, . . . , p −1. Then the expression equals σp−1(a0, a1, . . . , ap−1) σp(a0, a1, . . . , ap−1) , where σi is an elementary symmetrical polynomial of degree i. Find the polynomial for which the numbers ai are its roots: p−1 Y i=0 (x −1 −i2).
Change the variable x −1 = t2 and obtain p−1 Y i=0 (t2 −i2) = p−1 Y i=0 (t −i) p−1 Y i=0 (t + i) ≡(tp −t)(tp −t) = t2p −2tp+1 + t2.
Now apply the inverse change of variables and obtain for p = 4k + 3 p−1 Y i=0 (x −1 −i2) ≡(x −1)p −2(x −1) p+1 2 + (x −1) = xp + . . . + (p + 2 · p+1 2 + 1)x −4.
By Viete’s theorem σp ≡4 (mod p), σp−1 ≡2 (mod p), therefore σp−1 σp ≡1 2 ≡2k + 2 (mod p).
S o l u t i o n 2. Split all nonzero residues modulo p, except ±1, on pairs of reciprocal. We obtain 2k pairs and in each pair (i, j) ij ≡1 ⇔ i2j2 ≡1 ⇔ (ij)2 + i2 + j2 + 1 ≡i2 + j2 + 2 (mod p).
Therefore, 1 ≡(ij)2 + i2 + j2 + 1 (i2 + 1)(j2 + 1) ≡ i2 + j2 + 2 (i2 + 1)(j2 + 1) = 1 i2 + 1 + 1 j2 + 1 (mod p).
So, the sum is equal to 1 02+1 + 1 12+1 + 1 (−1)2+1 + 2k ≡2k + 2.
1 These sets coincide because they contain p −1 element each, and it is clear that all the reminders in each set are non zero and pairwise distinct.
Amazing properties of binomial coeffcients 15 S o l u t i o n 3. By Fermat’s little theorem the operations x 7→x−1 and x 7→xp−2 modulo p coincide.
So it is sufficient to calculate the sum p−1 X x=0 (x2 + 1)p−2 = p−1 X x=0 p−2 X m=0 p −2 m x2m = p−2 X m=0 p −2 m S2m, (6) where S2m = p−1 P x=0 x2m. Evidently S2m ≡−1 (mod p) for m = p−1 2 . Prove that S2m ≡0 (mod p) for all other m ⩽p −1. Indeed, for each m we can choose a non zero residue a such that a2m ̸≡1 (mod p) and after that we can reason as in (5). For the sum (6) we have p−2 X m=0 p −2 m S2m ≡− p −2 p−1 2 = − 4k + 1 2k + 1 = −(4k + 1) · 4k · . . . · (2k + 1) 1 · 2 · . . . · (2k + 1) ≡ ≡−(−2) · (−3) . . . (2k + 2) 1 · 2 · . . . · (2k + 1) ≡2k + 2 (mod p).
4.3. We found these statements in .
a) For each prime divisor p | m choose ap such that p ∤(ak p −1). By the Chinese reminder theorem choose a such that a ≡ap (mod p) for all p. Then the result can be proven by reasoning as in (5).
b) Observe that for odd k by the binomial formula we have ik + (p −ik) ≡kik−1p (mod p2). Then 2 p−1 X i=1 1 ik = p−1 X i=1 1 ik + 1 (p −i)k = p−1 X i=1 ik + (p −i)k ik(p −i)k ≡ p−1 X i=1 kik−1p ik(−i)k ≡−kp p−1 X i=1 1 ik+1 (mod p2).
The sum in the r.h.s is divisible by p by the statement a).
4.4. The congruence holds even modulo p7 (see ), but it goes a bit strong. We can reason as in , tracing all powers till p4, and obtain p −1 2p −1 = (2p −1)(2p −2) · . . . · (p + 1) p!
= 2p 1 −1 2p 2 −1 · . . . · 2p p −1 −1 ≡ ≡1 −2p p−1 X i=1 1 i +4p2 p−1 X i,j=1 i<j 1 ij −8p3 p−1 X i,j,k=1 i<j<k 1 ijk (mod p4).
(7) The last sum can be expressed via power sums: p−1 X i,j,k=1 i<j<k 1 ijk = S3 3 −S1S2 2 + S3 1 6 , where Sk = p−1 X i=1 1 ik .
We now that S1 and S3 are divisible by p2 (the latter due to problem 4.3b). Therefore the last term in the formula (7) can be omitted.
4.5. The problem is from , variations can be found in . Since 2 p−1 X k=1 1 k2 = p−1 X k=1 1 k2 + 1 (p −k)2 = p−1 X k=1 k2 + (p −k)2 k2(p −k)2 ≡−2 p−1 X k=1 1 k(p −k) (mod p2) , the statement 3) is equivalent to the congruence p−1 X k=1 1 k(p −k) ≡0 (mod p2). The statement 2) is equivalent to the same congruence, because 2 p−1 X k=1 1 k = 2 p−1 X k=1 1 k + 1 p −k = 2p p−1 X k=1 1 k(p −k). Finally we know from the previous problem that 2p −1 p −1 ≡1 −p2 p−1 X i=1 1 i(p −i) + 4p2 p−1 X i,j=1 i<j 1 ij (mod p4).
Amazing properties of binomial coeffcients 16 So the statement 1) is equivalent to the congruence p−1 X i=1 1 i(p −i) ≡4 p−1 X i,j=1 i<j 1 ij (mod p2).
(8) Rewrite the expression in the r.h.s.: 4 p−1 X i,j=1 i 1 every non block sample consists of at least 3 blocks, so in this case the statement is true. It remains to consider a case when k = 1 and we count the number of non block samples of p objects from the set of 2p objects. This number equals 2p p −2, by Wolstenholme’s theorem it is divisible by p3.
S o l u t i o n 2. In the formula a b = a(a−1)...(a−b+1) b(b−1)...1 split the numerator and the denominator onto blocks of p terms, reduce the first terms in each block, and collect the quotients in a separate expression: mp kp = m̸p · (mp −1) . . .
mp −(p−1) k̸p · (kp −1) . . .
kp −(p−1) · (m−1)̸p · (m−1)p −1 . . .
(m−1)p −(p−1) (k−1)̸p · (k−1)p −1 . . .
(k−1)p −(p−1) · . . . × × (m−k+1)̸p · (m−k+1)p −1 . . .
(m−k+1)p −(p−1) ̸p · (p −1) . . . 1 = = m k · (mp −1) . . .
mp −(p−1) (kp −1) . . .
kp −(p−1) · . . . · (m−k+1)p −1 . . .
(m−k+1)p −(p−1) (p −1) . . . 1 .
It remains to check that the product of fractions is congruent to 1 (mod p3). For this prove the congruence (np −1) . . .
np −(p−1) (rp −1) . . .
rp −(p−1) ≡1 (mod p3) or, even, it would be better to prove the following congruence (np −1) . . .
np −(p−1) (p −1)!
≡(rp −1) . . .
rp −(p−1) (p −1)!
(mod p3) .
This is true because both parts are congruent to 1 (mod p3), that can be shown analogously to the proof of Wolstenholme’s theorem.
4.7. a) [5, theorem 2.14]. Transform the difference p2 p − p 1 = p2(p2 −1) . . . (p2 −(p −1)) 1 · 2 · . . . · (p −1)p −p = p (p −1)!
(1−p2)(2−p2) . . . ((p−1)−p2)−1·2·. . .·(p−1) .
It remains to check that (1 −p2)(2 −p2) . . . ((p −1) −p2) ≡1 · 2 · . . . · (p −1) (mod p4).
Expand brackets in the l.h.s.: (1−p2)(2−p2) . . . ((p−1)−p2) = 1·2·. . .·(p−1)+p2 1 + 1 2 + . . . + 1 p −1 (p−1)!+terms divisible by p4.
By the problem 4.1 the second summand is divisible by p4.
b) Observe that ps+1 p = ps · ps+1−1 p−1 , hence it is sufficient to prove that ps+1−1 p−1 ≡1 (mod ps+3).
ps+1 −1 p −1 = (ps+1 −1)(ps+1 −2) . . . (ps+1 −(p −1)) 1 · 2 · · · (p −1) = ps+1 1 −1 ps+1 2 −1 . . .
ps+1 p −1 −1 ≡ ≡(−1)p−1 + ps+1 1 + 1 2 + . . . + 1 p −1 (mod ps+3).
Since (−1)p−1 = 1 and 1 + 1 2 + . . . + 1 p−1 ≡0 (mod p2) we are done.
Amazing properties of binomial coeffcients 18 4.8. The problem is from , we present solution [T].
p3 p2 − p2 p = p p3 −1 p2 −1 − p2 −1 p −1 = = p p3 1 −1 p3 2 −1 . . .
p3 p2 −1 −1 − p2 1 −1 p2 2 −1 . . .
p2 p −1 −1 = = p p2 1 −1 p2 2 −1 . . .
p2 p −1 −1 p2−1 Y k=1 p∤k p3 k −1 −1 .
It is sufficient to prove that the last bracket is divisible by p7. Transform the product: p2−1 Y k=1 p∤k p3 k −1 = p2−1 2 Y k=1 p∤k p3 k −1 p3 p2−k −1 = p2−1 2 Y k=1 p∤k p6 −p5 k(p2−k) +1 ≡1+p5(p−1) p2−1 2 X k=1 p∤k 1 k(p2−k) (mod p7).
Now we have to check that the last sum is divisible by p2. This is true because by problem 4.3a) p2−1 2 X k=1 p∤k 1 k(p2−k) ≡− p2−1 2 X k=1 p∤k 1 k2 ≡0 (mod p2).
4.9. The statement is taken from [6, theorem 5], its generalization can be found in .
S o l u t i o n 1 ([5, proposition 2.19]). Use the fact that the difference 2k+1 2k − 2k 2k−1 is equal to the coefficient of x2k in the polynomial (1 + x)2k+1 −(1 −x2)2k = (1 + x)2k (1 + x)2k −(1 −x)2k = = 1 + 2k 1 x + 2k 2 x2 + . . . + x2k !
· 2 2k 1 x + 2k 3 x3 + . . . + 2k 2k −1 x2k−1 !
.
Since the second polynomial contains odd exponents only, the coefficient of x2k in the product equals 2 2k 1 2k 2k −1 + 2k 3 2k 2k −3 + . . . + 2k 2k −1 2k 1 !
.
By problem 3.5 b) 2k divides each binomial coefficient in this expression, moreover each term occurs twice in the sum, and the sum itself is multiplied by 2. Thus all the expression is divisible by 22k+2.
S o l u t i o n 2 ([CSTTVZ]). Since 2n+1 2n = 2 2n+1−1 2n−1 , it is sufficient to prove that 2n+1 −1 2n −1 ≡ 2n −1 2n−1 −1 (mod 22n+1).
Similarly to (3) we obtain 2n+1 −1 2n −1 = 2n+1 1 −1 2n+1 3 −1 . . .
2n+1 2n −1 −1 · 2n −1 2n−1 −1 .
It is sufficient to prove that L = 2n+1 1 −1 2n+1 3 −1 . . .
2n+1 2n −1 −1 ≡1 (mod 22n+1) .
Amazing properties of binomial coeffcients 19 This is true because L ≡(−1)2n−1 −2n+1 1 1 + 1 3 + 1 5 + . . . + 1 2n −1 ≡ ≡1 −2n+1 2n 1 · (2n −1) + 2n 3 · (2n −3) + . . . + 2n (2n−1 −1)(2n−1 + 1) ≡1 (mod 22n+1) .
4.10. This is theorem of Morley .
S o l u t i o n 1 (author’s proof, 1895). It goes a bit beyond the school curriculum.
Take the formula which expresses cos2n+1 x via cosines of multiple angles,1 or, as they were saying in that times, write cos2n+1 x in the form handy for integrating: 22ncos2n+1x = cos(2n + 1)x+(2n + 1) cos(2n−1)x+(2n+1) · 2n 1 · 2 cos(2n−3)x+. . .+(2n+1) · 2n . . . (n+2) n!
cos x.
Now integrate it2 over the interval [0, π 2 ]: 22n Z cos2n+1 x dx = sin(2n + 1)x 2n + 1 + 2n + 1 2n −1 sin(2n −1)x + . . . , 22n π/2 Z 0 cos2n+1 x dx = (−)n 1 2n + 1 −2n + 1 2n −1 + . . .
.
Every first grade student of university knows that it is convenient to use integration by parts for calculating this integral: I2n+1 = π/2 Z 0 cos2n+1 x dx = π/2 Z 0 cos2n x cos x dx = cos2n x sin x π/2 0 +2n π/2 Z 0 cos2n−1 x sin2 x dx = = 0 + 2n π/2 Z 0 cos2n−1 x(1 −cos2 x) dx = 2n · I2n−1 −2n · I2n+1 , therefore I2n+1 = 2n 2n+1 · I2n−1. Since I1 = 1, we can apply the formula n times and obtain π/2 Z 0 cos2n+1 x dx = 2n · (2n −2) . . . 2 (2n + 1)(2n −1) . . . 3 .
Equating of these two results give us the formula 22n 2n · (2n −2) . . . 2 (2n + 1)(2n −1) . . . 3 = (−)n 1 2n + 1 −2n + 1 2n −1 + . . . + (2n+1) · 2n . . . (n+2) n!
.
Let p = 2n+1 be a prime number. We obtain the desired congruence by multiplying the last formula by p: 22n 2n · (2n −2) . . . 2 (2n −1)(2n −3) . . . 3 ≡(−)n (mod p2) .
S o l u t i o n 2 ([CSTTVZ]). We will use the following notations: A = p−1 2 X i=1 1 i , B = X 1⩽i 1 , f(n, 1) ≡f(n′, 1) + f(n′, p −2) .
Now the part “only if” of the problem statement follows from the induction hypothesis, and the part “if”, too: if f(n, j) ≡0 (mod p) for j = 1, 3, . . . , p −2, then f(n′, p −2) ≡f(n′, p −4) ≡. . . ≡f(n′, 1) ≡−f(n′, p −2) , from where f(n′, j) ≡0 (mod p) for all required j, and then n′ .
.
. (p + 1), hence n .
.
. (p + 1).
References The authors of many solutions are participants of the conference: [D] Didin Maxim; [К] Krekov Dmitri; [J] Jastin Lim Kai Ze; [T] Teh Zhao Yang Anzo; [CSTTVZ] ´ Cevid Domagoj, Stoki´ c Maksim, Tanasijevii´ c Ivan, Trifunovi´ c Petar, Vukorepa Borna, ˇ Zikeli´ c Ðorđe Список литературы Винберг Э. Б. Удивительные свойства биномиальных коэффициентов. / / Мат. просвещение. Третья серия. Вып. 12.
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Петербургские математические олимпиады, 1961–1993. СПб: Лань, 2007.
Табачников С.Л., Фукс Д.Б. Математический дивертисмент. 30 лекций по классической математике. М.: МЦНМО, 2011.
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Syntax - Sentence structure, or the branch of linguistics which studies the structure of sentences. (From Trask'sA Student’s Dictionary of Language and Linguistics.)
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The Blackwell Companion to Syntax by Martin Everaert (Editor); Henk C. Van Riemsdijk (Editor); Rob Goedemans (Editor); Bart Hollebrandse (Editor) Call Number: Ref P291 .B53 2006 Publication Date: 2006
The Routledge Handbook of Syntax by Andrew Carnie (Editor); Dan Siddiqi (Editor); Yosuke Sato (Editor) Call Number: P291 .R68 2014 Publication Date: 2014
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