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11200 | https://www.khanacademy.org/math/geometry-tx/x790e3ac3e338c450:measurement-of-two-dimensional-figures/x790e3ac3e338c450:arc-length-from-degrees/e/circles_and_arcs | Arc length (practice) | Khan Academy
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Skip to lesson content
Geometry (TX TEKS)
Course: Geometry (TX TEKS)>Unit 10
Lesson 2: Arc length (from degrees)
Arc length from subtended angle
Subtended angle from arc length
Arc length
Challenge problems: Arc length 1
Challenge problems: Arc length 2
Finding perimeter of shape with arcs
Finding perimeter of shape with semicircles
Perimeter of shapes with arcs
Math>
Geometry (TX TEKS)>
Measurement of two-dimensional figures>
Arc length (from degrees)
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Arc length
OK.Math: G.C.1.3, PC.T.1.3
Google Classroom
Microsoft Teams
Problem
A circle has a radius of 3. An arc in this circle has a central angle of 340∘.
What is the length of the arc?
Either enter an exact answer in terms of π or use 3.14 for π and enter your answer as a decimal.
340∘3
Related content
Video 4 minutes 58 seconds 4:58 Arc length from subtended angle
Video 2 minutes 18 seconds 2:18 Subtended angle from arc length
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Do 4 problems
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11201 | https://artofproblemsolving.com/downloads/printable_post_collections/72602?srsltid=AfmBOorb9yf4-84yOHYF4iX8RkDg28JZ9QRO_cuP24-Q1kuu86uzdRb9 | AoPS Community 100 Geometry Solutions
Solutions for these problems:
www.artofproblemsolving.com/community/c72602 by CaptainFlint, hotstuffFTW
1 We let the x, y, and z be the radii of circles A, B, and C respectively. We then have the following systems of equations:
x − y = 6
x − z = 5
y + z = 9 .
Adding the first two equations we get 2x − (y + z) = 11 . Because y + z = 9 , we have
2x − 9 = 11 2x = 20
x = 10 .
2 Let ∠CAD = x. We know that ∠ADC = x and ∠ACD = 180 − 2x. Let ∠DAB = y. We know that ∠ADB = 180 −x, so ∠ABD = 180 −(180 −x+y) = x−y. Because ∠CAB −∠ABC = 30 ◦,we know that (x + y) − (x − y) = 30 = ⇒ y = 15 ◦ .
3A BCDEFxx
2222−x
Let the point of tangency be F . By the BC = F C = 2 and AE =
EF = x. Thus DE = 2 − x. The on 4CDE yields
©2019 AoPS Incorporated
1AoPS Community 100 Geometry Solutions
DE 2 + CD 2 = CE 2
(2 − x)2 + 2 2 = (2 + x)2
x2 − 4x + 8 = x2 + 4 x + 4
x = 12
Hence CE = F C + x = 52 ⇒ 52 .
4A BCDM
663
It is given that ∠AM D ∼= ∠CM D . Since ∠AM D and ∠CDM are alternate interior angles and
AB ‖ DC , ∠AM D ∼= ∠CDM −→ ∠CM D ∼= ∠CDM . Use the Base Angle Theorem to show
DC ∼= M C . We know that ABCD is a rectangle, so it follows that M C = 6 . We notice that
4BM C is a 30 − 60 − 90 triangle, and ∠BM C = 30 ◦. If we let x be the measure of ∠AM D,
then
2x + 30 = 180 2x = 150
x = 75
5 Drawing the square and examining the given lengths,
©2019 AoPS Incorporated
2AoPS Community 100 Geometry Solutions A BCDEF
x
3
x
3
xx
2x
32x
3
30 30 30
you find that the three segments cut the square into three equal horizontal sections. Therefore, (x being the side length), √x2 + ( x/ 3) 2 = 30 , or x2 + ( x/ 3) 2 = 900 . Solving for x, we get
x = 9 √10 , and x2 = 810 . Hence, the area of the square is 810 .
6AB CDE
Since ∠DEC = ∠DBC = 90 , quadrilateral DEBC is cyclic, from which the result follows due to same inscribed arcs.
7 By the pigeonhole principle, there must be 3 points with lengths a, a, and 2a between them. Since a + a ¡= 2a, these 3 points cannot form a triangle. Instead, the three points must be collinear. Let them be A, B, and C such that AB = BC = a and AC = 2 a. Let the last point be D. D is a distance of a from two of the points among A, B, and C. Because the four points are distinct, D must form an equilateral triangle with A and B or with B and C. Without loss of generality, let AD = BD = AB = a. Let the foot of the perpendicular from D to AB be K.Because AD = BD , AK = BK = 12 (AB ) = a
2
. By the Pythagorean theorem, DK = a√32 . Thus,
b2 = DC 2 = DK 2 + KC 2 = 3a2
4
9a2
4
= 3 a2. Therefore, ba = √3.
©2019 AoPS Incorporated
3AoPS Community 100 Geometry Solutions
8CA BLMN
Since quadrilateral ABLM is cyclic, ∠M LB = 180 − ∠A = 90 ◦ = ∠M LC . Thus ∠CM L = 90 −
∠C. We also have ∠C = ∠AN M since ALCN is cyclic. Then since ∠M AN = 180 − ∠A = 90 ◦,
we get
∠AM N = 90 − ∠AN M = 90 − ∠C = ∠CM L.
Since the vertical angles are congruent, L, M, N are collinear.
9XAB CIYDE
Let D and E be the feet of the altitudes from I to BC and AB respectively. Since IX = IA and
ID = IE we have that 4IDX ∼= 4IEA by HL congruence (both are clearly right triangles). It is also easy to show that 4IDX ∼= 4IDY. So, AE = s − a = XD = 12 XY, and thus
XY = b + c − a = 1400 + 1800 − 2014 = 1186 .
©2019 AoPS Incorporated
4AoPS Community 100 Geometry Solutions
10 Notice that, since ADBE is cyclic, we want to show that it is an isosceles trapezoid. Thus, it suffices to prove that DB
_ = AE
_. However, we have AB
_ = AC
_ and AD
_ = CE
_ (since AD = CE ). Thus, we have DB
_ = AB
_ − AD
_ = AC
_ − CE
_ = AE
_. QED
11 Let P be the shapes perimeter and A its area. Note that if we dilate the shape by a factor of
k, its perimeter becomes kP and its are becomes k2A. Thus, if we want our dilated shape to equiable, or kP = k2A, we should dilate the shape by a factor of k = PA .
12 By simple angle chasing, we notice that triangle AEB is similar to EF C . Let the side length of the square be a, EC = x and thus BE = a − x. Because of similarity, it is
a
4 = x
3 ⇔ x = 34 a.
That yields BE = 14 a. Using the Pythagorean Theorem in triangle AEB the result is
a2 + ( 14 a) = 16 ⇔ 17 16 a2 = 16 ⇔ a2 = 16 2
17
which is the area we searched.
13 ∠ABN = ∠AM N since they are subtended from the same chord. Since ∠M XN = ∠M Y N =90 ◦, it follows that M , X, Y , and N are concyclic. Then, since M and X form angles subtended from the same chord, ∠Y XN = ∠Y M N = ∠AM N = ∠ABN . Because BN is an extension of XN , and because ∠Y XN = ∠ABN , it follows that AB ‖ XY .
14 Extend AE, DF and BE, CF to their points of intersection. Since 4ABE ∼= 4CDF and are both 5 − 12 − 13 right triangles, we can come to the conclusion that the two new triangles are also congruent to these two (use ASA, as we know all the sides are 13 and the angles are mostly complementary). Thus, we create a square with sides 5 + 12 = 17 .
©2019 AoPS Incorporated
5AoPS Community 100 Geometry Solutions A BCDEFGH
EF is the diagonal of the square, with length 17 √2; the answer is EF 2 = (17 √2) 2 = 578 .
15 Let O be 4ABC s circumcenter. Note that ∠BF E = 12 ∠BOE = 14 ∠BOC = 12 ∠A. Similarly,
∠AEF = 12 ∠B and ∠DF B = 12 ∠C. Since ∠DF E + ∠AEF = ∠DF B + ∠BF E + ∠AEF =
12
(∠A + ∠B + ∠C) = 90 ◦, it follows that DF ⊥ AE .
16 We position face ABC on the bottom. Since [4ABD ] = 12 = 12 · AB · hABD , we find that hABD = 8 . The height of ABD forms a
30 − 60 − 90 with the height of the tetrahedron, so h = 12 (8) = 4 . Thus, the volume of the tetrahedron is 13 Bh = 13 15 · 4 = 20 .
17
©2019 AoPS Incorporated
6AoPS Community 100 Geometry Solutions P1
P2
P3
P4
A3
A1
O
Since the quadrilateral is orthodiagonal, we can write the statement we want to show as D2 =
P1P 22 + P3P 24 . Let A1 and A3 be the antipodes of P1 and P3, respectively, and let O be the center of the circle. The claim is that reflecting P1P2P3P4 over the diameter of the circle that is parallel to P1P3 results in the new quadrilateral A3P4A1P2. We can prove this by noting that
P1OP 3 ∼= A3OA 1, so P1 goes to A3 and P3 goes to A1 (and obviously P2 and P4 switch places). Now, after reflecting, we get that A1P2 = P3P4, so now we want to show that D2 = P1P 22 +A1P 22
which is true by the Pythagorean Theorem on 4P1P2A1.
18 Notice that the midpoint of AB , the centers of the two circles, and the center of the sphere form a rectangle. Thus, if we know the sides of the rectangle we can compute the distance from the center of the sphere to the midpoint of AB . Then, since this line is perpendicular to
AB , we can then compute the radius of the sphere. First, we find the distances from the centers of the circles to midpoint of AB . They are √54 2 − 21 2
and √66 2 − 21 2. Thus, the distance from the center of the sphere to the midpoint of AB is √54 2 − 21 2 + 66 2 − 21 2. Finally, we apply pythagorean theorem once more to get the radius of the sphere to be √54 2 − 21 2 + 66 2 − 21 2 + 21 2. Squaring, we get that R2 = 6831 .
19 Extend AB and CD to meet at X. We see that ∠AXD is a right angle, so XM and XN are medians of triangle XBC and XAD , respectively. We can also see that X, M, N are collinear. Hence, XN = AN and XM = BM , so we easily find that M N = XN − XM = 1004 − 500 = 504 .
20
©2019 AoPS Incorporated
7AoPS Community 100 Geometry Solutions A BCDEF
First remark that E and F are symmetric with respect to the altitude from C in 4ABC , so
EF ‖ AB . Since CF BD is a cyclic quadrilateral, we have
∠CF D = ∠CBD = ∠CAB = ∠CF E.
Hence D, E, F are collinear as desired.
21 Let the center of the circle with radius 1 be A. Let the circle shaded grey have center B and let the circle shaded black that is adjacent to that grey circle be C. Let r be the radius that we want to find. Clearly, BC = 2 r, AB = 4 − r and AC = r + 1 . We can also see that, from symmetry, ∠BAC = 60 ◦. From Law of Cosines, we have
4r2 = ( r + 1) 2 + (4 − r)2 − 2( r + 1)(4 − r) cos 60 ◦)
This reduces down to r2 + 9 r − 13 , and plugging into the quadratic formula:
r = −9 + √133 2
We arrive at −9 + 133 + 2 = 126 .
22 Let D′ be a point on the same side of AB as D such that the circumcircle of 4ABC is tan-gent to BD ′. Then ∠BAC = ∠CBD ′ so B, D, D ′ are collinear. Hence BD is (also) tangent, as desired.
23 Clearly, we see that CQM B and AN P C are cyclic quadrilaterals, implying that ∠CM B =
∠CQB and ∠CN A = ∠CP A . Therefore, we can deduce that ∠M CN = 180 ◦ − (∠CM N +
∠M N C ) = 180 ◦ − (90 ◦ − A/ 2 + 90 ◦ − B/ 2) = 45 ◦.
©2019 AoPS Incorporated
8AoPS Community 100 Geometry Solutions
24 Note that since AM CN is cyclic, we have ∠AN M = ∠ACB and ∠AM N = ∠ACD = ∠CAB
so 4M AN ∼ 4 ABC by AA similarity.
25 (a) We angle chase. Since AH is perpendicular to BC , ∠BAH = 90 − ∠B. Now consider the circumcircle of ABC . We have ∠AOC = 2 ∠B (since it inscribes the same arc but goes through the center). Since triangle OAC is isosceles ( OA = OC ), we have ∠OAC = (180 − 2∠B)/2 = 90 − ∠B.(b) Notice that ∠HAO = |∠A−∠BAH −∠OAC |. Substituting in our values from part (a), we get
∠HAO = |∠A−(90 −B)−(90 −B)| = |∠A+2 ∠B−180 | = |(∠A+∠B+∠C)+( ∠B−∠C)−180 | =
|∠B − ∠C|. QED
26 We know ∠AP B = ∠CP D , so ∠AP B + ∠BP C = ∠CP D + ∠BP C , or ∠AP C = ∠BP D . We also know AP P C = P B P D , so by SAS similarity 4P AC ∼ 4 P BD .
27 Note that ∠OY X = ∠OXY , so ∠Y ZO = ∠OZX . Let ∠Y ZO = α, then by Law of Cosines on triangles Y ZO and OZX we get 121 + 49 − 2 · 11 · 7 cos α = 121 + 169 − 2 · 11 · 13 cos α, which simplifies to cos α = 10 11 . Plugging this back in we have OY 2 = 170 − 140 = 30 , so OY = √30 .
28 Let ∠AKF = ∠F KB = α and ∠AM H = ∠HM D = β. Also let F E ∩ HG = X. Then because
∠KAF = ∠DCB , we have ∠M F X = α + C, so ∠M XF = 180 − (α + β + C). However we also know that 2α + C + D = 180 and that 2β + C + B = 180 , so adding these we see that
2α+2 β +2 C +( B +D) = 360 . But we also know that B +D = 180 because quadrilateral ABCD
is cyclic, so we get that α + β + C = 90 . Which means ∠M XF = 180 − 90 = 90 . From this we conclude that triangles EM X and F M X are congruent, and triangles HKX and GKX are congruent. Thus EX = XF , HX = XG , and HG ⊥ F E , so EGF H is a rhombus.
29 Analitic 5 77
ABCMPH
©2019 AoPS Incorporated
9AoPS Community 100 Geometry Solutions
We will proceed by coordinates. Let A = (0 , 0) , C = (14 , 0) . Since a 13-14-15 is composed of a 5-12-13 and 9-12-15, we have B = (5 , 12) . Let H be the foot of the altitude from B to AC.
We also have H = (5 , 0) and M = (7 , 0) . Now from BH = 12 , HM = 2 , we get BM = √148 .Also since P M is a median to the hypotenuse of 4AP C, we have M P = M C = 7 . Now the coordinates of P are just a weighted average of those of B and M :
P = 7√148 B +
√148 −7√148
M =
(
5(7) + 7( √148 − 7)
√148 , 12(7)
√148
)
=
(
7√37 − 7
√37 , 42
√37
)
Now, keeping a steady hand, using the distance formula gives AP = 7
√
2 − 2√37 , P C = 7
√
2 + 2√37
(see, it all worked out nicely in the end), and computing AP ·P C
2
gives [AP C ] = 294 √37 so p+q +r =294 + 2(37) = 368 .
29 Synthetic With some simple calculations we see that AH = 5 , HM = 2 , and M C = M P = 7 . Also
BH = 12 . So by the Pythagorean theorem we have BM = 2 √37 . Let the foot of the altitude from P to AC be D. Then M P M B = P D BH . Or 72√37 = P D
12
. Solving we have P D = 42 √37 Thus the area of 4P AC is 12 · 14 · 42 √37 = 294 √37 = 294 √37 37 . Therefore p + q + r = 294 + 37 + 37 = 368 .
30 OABCPDEME′
D′
©2019 AoPS Incorporated
10 AoPS Community 100 Geometry Solutions
Let O be the circumcentre of 4ABC and M be the midpoint of BC . Then its clear that P M ⊥
BC . Since ∠BDP = 90 ◦ = 190 −∠P M B , quadrilateral P M BD is cyclic. So we have ∠M DP =
∠M BP . Also ∠ADM = 90 ◦ − ∠M DP = 90 ◦ − ∠M DP . As BP is tangent to the circumcircle of 4ABC we must have ∠M BP = BC
_
2 = ∠A. Therefore ∠ADM = 90 ◦ − ∠M DP = 90 ◦ − ∠A.Let D′ = DM ∩ AE then ADD ′ is right. So DD ′ is the D-altitude of 4ADE . Similarly EE ′ is the E-altitude of 4ADE which means that M is the orthocentre of 4ADE .
31 Note that HH ABH C and HH ACH B are cyclic. Then ∠HC HAA = ∠HC BH B = 90 − ∠BAC and ∠AH AHB = ∠ACH C = 90 − ∠BAC .Therefore, H lies on the angle bisector of ∠HC HAHB .The same argument can be repeated for the angles HC HB HA and HAHC HB to show that H
is the incenter.
32 Let Y and Z be the intersections of AC with the circle, w.l.o.g. CY < CZ . Then clearly CY =97 − 86 = 11 and CZ = 97 + 86 = 183 . Now, CB · CX = CY · CZ = 11 · 183 = 3 · 11 · 61 .If BX and CX have integer lengths, this also holds for BC and since BC ≥ CX ≥ 11 (the latter holds by the Triangle Inequality) the only possible values are CX = 11 , V C = 183 and
CX = 33 , BC = 61 . But in the first case, ACX would be degenerated which implies X ∈ AC ,only possible if X = C but this contradicts AX = 86 6 = 97 = AC . Hence CX = 33 and
BC = 61 is the only possible solution.
33 Let ∠BAC = A, ∠ABC = B, and ∠BCA = C. We know that ∠BOC = 2 A, so the measure of ̂
BN C = 4 A. Then A = 4A−PQ
_
2
, which means PQ
_ = 2 A, and ∠P N Q = A. Because the center of circle τ also lies on the perpendicular bisector of BC , we know that ∠N OB = ∠N OC = A.Thus ∠N QC = ∠N OC = A, so ∠N QA = 180 − A, and ∠N P A = 180 − A. From this we conclude AP N Q is a parallelogram.
34
©2019 AoPS Incorporated
11 AoPS Community 100 Geometry Solutions
Drawing the center of the square to each of the 8 vertices we see that we are looking for the area of 4 sectors of a circle with central angle 45 ◦, 4 right triangles with length and width equal to the half the side of the square, and 4 tiny external right triangles. The sectors form a semicircle with radius
√22
, so the area of that is π
4
. The 4 big right triangles have a total area of 4 · 12 · 12
2
= 12 . Then the small external triangles are isosceles right triangles, and have height
√2−12
, so their area is
4 ·
( √2−12
)2
2 = 3 − 2√22
So the total area is π
4 + 2 − √2 .
35 Take O the midpoint of the diameter, M, O, P, S are concyclic, so ∠M P S = ∠M OS ; since ST
has constant length, we are done.
©2019 AoPS Incorporated
12 |
11202 | https://www.youtube.com/watch?v=2rJEENnFF44 | Graphing Sine and Cosine Functions [Stretches and Shrinks]
Lisa Ruddy
5720 subscribers
189 likes
Description
21151 views
Posted: 27 Mar 2017
13 comments
Transcript:
Introduction hi everybody today we're going to talk about how to graph sine and cosine functions and we're going to talk about two transformations specifically vertical and horizontal stretches and shrinks so first we're going to talk about what the parent function for sine looks like and we'll follow that up with the parent function for cosine there's a lot of information on this page so you might want to pause the video and copy it all down and then kind of listen to the important parts so one thing to notice about this graph is that it has a repeating shape so it kind of repeats the shape the same shape over and over and over and it makes that wave shape so this feature of repeating a shape ties into this new word which is the period so the period is how long it takes for your graph to complete one full cycle so the period doesn't have to be this set point it can be like from here to here is 2 pi or wonderful period it can even be from here to here but the idea is that the parent function for sine has a period of 2 pi now it's no coincidence that it's 2 pi and that a full rotation of your unit circle is also 2 pi they go hand-in-hand the next thing we want to talk about is this new word called amplitude so amplitude is half the distance from the maximum to the minimum value so it's always from the midline going up so if you imagine like halfway through your graph this would be your midline so it's the distance from your midline up or from your midline down now amplitude is just a distance so it's always going to be positive so since this graph continues on and on forever our domain is all real numbers but you might notice that the range is restricted so if you look at the graph it always stops right at 1 and turns around and then when it hits negative 1 it turns back around the other way so that's why our range is between an including negative 1 and positive 1 now from there another thing you might notice is that our graph split up into four parts and the reason why I split up like this is because it matches unit circle which has four quadrants so if you look at the coordinates of your unit circle I want to remind you that the x value represents cosine and the y value represents sine so if you look at all of the sine values going around your unit circle the Y value at zero is zero the Y value at PI over two is 1 the Y value at pi is zero the Y value at 3 PI over 2 is negative 1 and if you make it back to 2 pi its back at 0 and then it will just repeat the process as you continue on around the unit circle which is why these graphs are cyclical and repeat over and over and over so anytime we draw these graphs we will split it up into those four quadrants so since a full period is PI halfway is PI then PI over 2 and 3 PI over 2 last thing I want to talk about for sine is kind of the starting point so you'll see that the parent function for this kind of starts at the origin and then starts its cycle going up so you're gonna see that that's the main difference between sine and cosine so here's the parent function for cosine which will always start at the maximum which in this case is 1 and then it still creates that same cycle the period is still 2 pi so if you measure out how long it takes for this graph to finish a cycle it will always be 2 pi for the parent function the amplitude is still 1 so you can see that this distance from the midline to the maximum or from the midline to the minimum is 1 and you'll see that this graph is still loop C cut into 4 quadrants now just like the previous graph of sine if you look at the coordinates of the unit circle and you look at the X values because now it's cosine the x value at 0 is 1 the x value at PI over 2 is 0 and then negative 1 0 and back to positive 1 so the graphs for sine and cosine are just the different coordinates going around your unit circle and just like sign our domain is going to be all real numbers and our range is restricted it pretty much hits the minimum and maximum at one and negative one now of course this can change with a stretch or shrink or a shift but in terms of the parent function it will always be the same just be careful that you have the correct starting point for cosine so cosine starts at that maximum value and then works its way down and back up so now we're going to talk about two of the Transforms transformations so a horizontal stretch or shrink and vertical stretch or shrink so these will look the same as any functions we've done in the past when I remind you that if we multiply your entire function by a number it represents a vertical stretch or a shrink so if you see a two out front it's a vertical stretch by a factor of two if you see one half it's a vertical shrink by a factor of one half now if you're multiplying exclusively your X this becomes a horizontal stretch or shrink and I want to remind you that horizontal stretch or shrink it always was by a factor of the reciprocal the factor was 1 over B so if it said 2x right here it would actually be a shrink by a factor of 1/2 so this kind of ties in to how these transformations affect our graph so if you have a vertical stretch or shrink it's going to affect the amplitude of your graph which makes sense because the amplitude is how high or how low your graph goes essentially from the midline and if you have a horizontal stretch or shrink it's going to affect your period so either your graph is going to be faster like it will complete a cycle more frequently or it's going to get kind of like stretched out so we'll take longer for it to complete a cycle so the way it affects your graph it's 2 pi over the absolute value of B and the reason why it's dividing by B is because that factor is a reciprocal so these are the two their formulas that you need to know but if you understand where the come from you probably won't need to memorize them but they do affect how your graph looks so let's give this Example problem a try so here they want us to graph for sine of X but first they want us to identify the amplitude in the period so I see that there's only one transformation here which is that a value so that's going to be a vertical stretch by a factor of four and we know that it affects our amplitude so our amplitude little formula was the absolute value of a so for this graph our apps our amplitude is going to be 4 so rather than going up to positive 1 and down to negative 1 our graph will go up to positive 4 and down to negative 4 now you'll see here that there's no horizontal stretch or shrink so our period is unchanged and it's still going to be 2 pi so when we make our graph I want to remind you guys that you get to decide what the scale is on your X and y axis so I always want you to have at least 4 quadrants on the left and 4 on the right and since a full period is 2 pi I know that 2 pi will be here which means this is PI this is PI over 2 and this is 3 PI over 2 and of course it's the same going in the negative direction and now let's say that this is 4 and this is negative 4 so our graph is going to go up to 4 and down to negative 4 now from here we're just plotting our points so I know that sine the parent function for sine starts at the origin and starts going up so I just placed my point at the 4 quadrants 1 2 3 4 then just connect the dots I do want you to notice that the sine and cosine graphs have that wave shape they are curved so if your graph forms a point at each of your minimums or maximums that is not correct we want it to kind of arc around and it does continue on past the points so that's all there is to it so just a recap we check for the period we divide our x-axis into 4 parts to help us graph and then we plot the minimum and maximum according to our amplitude and then you just connect the the dots alright let's try one more together so here we are asked to graph one half cosine 2 pi X so here there are two different transformations a vertical shrink and a horizontal shrink as well so let's start with the amplitude so I can see that our amplitude is going to be 1/2 and I want to remind you that for the period our little formula is 2 pi over the absolute value of B which is 2 pi so in this graph our period is 1 so I want to remind you that since our period is shorter you know our graph is going to look a little bit more squished compared to the parent function all right so now when we go to graph it the set up is always going to be the same at least for me I'm going to use the four quadrants and now instead of a period of 2 pi we have a period of 1 so this last - mark is going to represent 1 which means this is 1/2 1/4 and 3/4 and we can decide at the scale of our graph so I'm going to call this 1/2 I don't want to try to squish my graph in and try to make it easy as possible for us to plot our points so we know that for cosine our parent function starts at the maximum value which in this case adjusted for an amplitude is 1 now we just plot our four other points according to our four quadrants and connect the dots notice that my graph does turn around and head back down at the end so make sure you kind of include that feature and then let's include our four points going in the other direction and there you have it so we adjusted the height of our graph for the amplitude and then we also adjusted the scale of our x axis to adjust for the period change so remember if we were graphing our parent function and if it had a period of 2pi it would actually be much more stretched out it would look something more like this so it is much more condensed all right at this point go ahead and pause the video and give these four a try make sure to first determine what your amplitude and period of each function are and then graph the function okay thank you for giving those a try so this is the graph for number one this is for number two just to keep them all straight so it would look right below the number so you can see that for the first example the amplitude is one fourth in the period is 2pi it remains unchanged so you can see how I kind of adjusted the x-axis to accommodate that 2pi now for number two our amplitude is unchanged as one but our period becomes PI because two PI over two is just PI so the way I split up my x-axis is accounted by PI over four sep 5 over 4 PI over 2 3 PI over 4 and a full cycle at PI for example 3 our amplitude was 2 in our period was 2 because 2 PI over pi so do split it up into force is 1/2 1 3 halves 2 and for this last example our amplitude is 1 third and our period was 4 pi so remember if you have 2 PI over 1/2 you're multiplying by the reciprocal so that becomes 4 PI so to divide up into fourths I counted by pi PI 2 pi 3 PI 4 PI so you might notice that our graphs look pretty much the same except for their starting point so sine always starts at the origin and cosine starts at the maximum and then the other thing that's different is the scaling for the x and y axis but as long as you are clear with how your X Y axis is scaled you can make it look however you want to look so the easier to graph the better all right thank you guys for watching |
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From Wikipedia, the free encyclopedia
The main article for this category is Complex number.
Wikimedia Commons has media related to Complex numbers.
The complex numbers are an extension of the real numbers, in which all non-constant polynomials have roots. The complex numbers contain a number i, the imaginary unit, with i2= −1, i.e., i is a square root of −1. Every complex number can be represented in the form x + iy, where x and y are real numbers called the real part and the imaginary part of the complex number respectively.
Subcategories
This category has the following 5 subcategories, out of 5 total.
A
Algebraic numbers (3 C, 19 P)
C
Complex distributions (3 P)
H
Hypercomplex numbers (3 C, 13 P)
R
Real numbers (6 C, 21 P)
T
Transcendental numbers (1 C, 17 P)
Pages in category "Complex numbers"
The following 20 pages are in this category, out of 20 total. This list may not reflect recent changes.
Complex number
Imaginary number
Complex-base system
A
Jean-Robert Argand
C
CM-field
Complex analysis
Complex conjugate
Complex measure
Complex plane
Complex conjugate line
E
External ray
G
Gaussian integer
Gaussian moat
I
Imaginary unit
M
Mean value problem
P
Principal root of unity
Q
Quater-imaginary base
R
Root of unity
T
Table of Gaussian integer factorizations
W
Caspar Wessel
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11204 | https://artofproblemsolving.com/wiki/index.php/Combinatorial_identity?srsltid=AfmBOoqjOK-YlJkwAg8K_qBIhvcEHc9R8lCwIkSuV5HbEvNPdwbL-XiP | Art of Problem Solving
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Combinatorial identity
Contents
1 Pascal's Identity
1.1 Proof
1.2 Alternate Proofs
2 Vandermonde's Identity
2.1 Video Proof
2.2 Combinatorial Proof
2.3 Algebraic proof
3 Hockey-Stick Identity
3.1 Proof
4 Another Identity
4.1 Hat Proof
4.2 Proof 2
5 Even Odd Identity
6 Examples
7 See also
Pascal's Identity
Pascal's Identity is a very important formula for olympiad math and it states that
for any positive integers and . Here, is the binomial coefficient .
This result can be interpreted combinatorially as follows: the number of ways to choose things from things is equal to the number of ways to choose things from things added to the number of ways to choose things from things.
Proof
If then and so the result is trivial. So assume . Then
Alternate Proofs
Here, we prove this using committee forming.
Consider picking one fixed object out of objects. Then, we can choose objects including that one in ways.
Because our final group of objects either contains the specified one or doesn't, we can choose the group in ways.
But we already know they can be picked in ways, so
Also, we can look at Pascal's Triangle to see why this is. If we were to extend Pascal's Triangle to row n, we would see the term . Above that, we would see the terms and . Due to the definition of Pascal's Triangle, .
Vandermonde's Identity
Vandermonde's Identity states that , which can be proven combinatorially by noting that any combination of objects from a group of objects must have some objects from group and the remaining from group .
Video Proof
Combinatorial Proof
Think of the right hand side as picking people from men and women. Think of the left hand side as picking men from the total men and picking women from the total women. The left hand side and right hand side are the same, thus Vandermonde's identity must be true.
~avn
Algebraic proof
For all The coefficients of on both sides must be the same, so using the Binomial Theorem,
Hockey-Stick Identity
For .
This identity is known as the hockey-stick identity because, on Pascal's triangle, when the addends represented in the summation and the sum itself is highlighted, a hockey-stick shape is revealed. We can also flip the hockey stick because pascal's triangle is symmetrical.
Proof
Inductive Proof
This identity can be proven by induction on .
Base Case Let .
.
Inductive Step Suppose, for some , . Then .
Algebraic Proof
It can also be proven algebraically with Pascal's Identity, . Note that
, which is equivalent to the desired result.
Combinatorial Proof 1
Imagine that we are distributing indistinguishable candies to distinguishable children. By a direct application of Balls and Urns, there are ways to do this. Alternatively, we can first give candies to the oldest child so that we are essentially giving candies to kids and again, with Balls and Urns, , which simplifies to the desired result.
Combinatorial Proof 2
We can form a committee of size from a group of people in ways. Now we hand out the numbers to of the people. We can divide this into disjoint cases. In general, in case , , person is on the committee and persons are not on the committee. This can be done in ways. Now we can sum the values of these disjoint cases, getting
Algebraic Proof 2
Apply the finite geometric series formula to : Then expand with the Binomial Theorem and simplify: Finally, equate coefficients of on both sides: Since for , , this simplifies to the hockey stick identity.
Algebraic Proof 3
Consider the number of solutions to the equation ++++++.......+ ≤ N where each is a non-negative integer for 1≤i≤r.
METHOD 1 We know since all numbers on LHS are non-negative therefore 0≤N and N is a integer.
Therfore, ++++++.......+ = 0,1,2......N. Consider each case seperately.
++++++.......+ =0 by Stars-and-bars the equation has solutions.
++++++.......+ =1 by Stars-and-bars the equation has solutions.
++++++.......+ =2 by Stars-and-bars the equation has solutions.
........... ++++++.......+ =N by Stars-and-bars the equation has solutions.
Hence, the equation has + + +.... = (where k=N+r-1) SOLUTIONS.
METHOD 2 Since, ++++++.......+ ≤ N Therefore we may say ++++++.......+ = N -m where m is another non-negative integer. 0 ≤++++++.......+ ⇒ 0≤ N-m ⇒ m≤ N So, we need not count this as an extra restriction.
Now, ++++++.......+ +m = N. Again by Stars-and-bars this has solutions.
Therefore, the equation has = solutions(As N+r-1 =k).
Since, both methods yeild the same answer ⇒ = . Taking r-1= p redirects to the honeystick identity.
~SANSGANKRSNGUPTA
Another Identity
Hat Proof
We have different hats. We split them into two groups, each with k hats: then we choose hats from the first group and hats from the second group. This may be done in ways. Evidently, to generate all possible choices of hats from the hats, we must choose hats from the first and the remaining hats from the second ; the sum over all such is the number of ways of choosing hats from . Therefore , as desired.
Proof 2
This is a special case of Vandermonde's identity, in which we set .
Even Odd Identity
Examples
1986 AIME Problem 11
2000 AIME II Problem 7
2013 AIME I Problem 6 (Solution 2)
2015 AIME I Problem 12
2018 AIME I Problem 10
2020 AIME I Problem 7
2016 AMC 10A Problem 20
2021 AMC 12A Problem 15
1981 IMO Problem 2
2022 AIME I Problem 12
See also
Pascal's Triangle
Combinatorics
Pascal's Identity
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Combinatorics
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11205 | https://www.statology.org/coefficient-of-variation-vs-standard-deviation/ | Published Time: 2021-05-17T15:28:33+00:00
Coefficient of Variation vs. Standard Deviation: The Difference
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Coefficient of Variation vs. Standard Deviation: The Difference
by Zach BobbittPosted onLast updated on May 17, 2021
Thestandard deviation of a dataset is a way to measure how far the average value lies from the mean.
To find the standard deviation of a given sample, we can use the following formula:
s = √(Σ(x i– x)2 / (n-1))
where:
Σ: A symbol that means “sum”
x i: The value of the i th observation in the sample
x: The mean of the sample
n: The sample size
The higher the value for the standard deviation, the more spread out the values are in a sample. However, it’s hard to say if a given value for a standard deviation is “high” or “low” because it depends on the type of data we’re working with.
For example, a standard deviation of 500 may be considered low if we’re talking about annual income of residents in a certain city. Conversely, a standard deviation of 50 may be considered high if we’re talking about exam scores of students on a certain test.
One way to understand whether or not a certain value for the standard deviation is high or low is to find the coefficient of variation, which is calculated as:
CV = s / x
where:
s: The sample standard deviation
x: The sample mean
In simple terms, the coefficient of variation is the ratio between the standard deviation and the mean.
The higher the coefficient of variation, the higher the standard deviation of a sample relative to the mean.
Example: Calculating the Standard Deviation & Coefficient of Variation
Suppose we have the following dataset:
Dataset: 1, 4, 8, 11, 13, 17, 19, 19, 20, 23, 24, 24, 25, 28, 29, 31, 32
Using a calculator, we can find the following metrics for this dataset:
Sample mean (x): 19.29
Sample standard deviation (s): 9.25
We can then use these values to calculate the coefficient of variation:
CV = s / x
CV = 9.25 / 19.29
CV = 0.48
Both the standard deviation and the coefficient of variation are useful to know for this dataset.
The standard deviation tells us that the typical value in this dataset lies 9.25 units away from the mean. The coefficient of variation then tells us that the standard deviation is about half the size of the sample mean.
Standard Deviation vs. Coefficient of Variation: When to Use Each
The standard deviation is most commonly used when we want to know the spread of values in a single dataset.
However, the coefficient of variation is more commonly used when we want to compare the variation between two datasets.
For example, in finance the coefficient of variation is used to compare the mean expected return of an investment relative to the expected standard deviation of the investment.
For example, suppose an investor is considering investing in the following two mutual funds:
Mutual Fund A: mean = 9%, standard deviation = 12.4%
Mutual Fund B: mean = 5%, standard deviation = 8.2%
The investor can calculate the coefficient of variation for each fund:
CV for Mutual Fund A = 12.4% / 9% = 1.38
CV for Mutual Fund B = 8.2% / 5% =1.64
Since Mutual Fund A has a lower coefficient of variation, it offers a better mean return relative to the standard deviation.
Summary
Here’s a brief summary of the main points in this article:
Both the standard deviation and the coefficient of variation measure the spread of values in a dataset.
The standard deviation measures how far the average value lies from the mean.
The coefficient of variation measures the ratio of the standard deviation to the mean.
The standard deviation is used more often when we want to measure the spread of values in a single dataset.
The coefficient of variation is used more often when we want to compare the variation between two different datasets.
Additional Resources
How to Calculate the Mean and Standard Deviation in Excel
How to Calculate the Coefficient of Variation in Excel
Posted in Programming
Zach Bobbitt
Hey there. My name is Zach Bobbitt. I have a Masters of Science degree in Applied Statistics and I’ve worked on machine learning algorithms for professional businesses in both healthcare and retail. I’m passionate about statistics, machine learning, and data visualization and I created Statology to be a resource for both students and teachers alike. My goal with this site is to help you learn statistics through using simple terms, plenty of real-world examples, and helpful illustrations.
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2 Replies to “Coefficient of Variation vs. Standard Deviation: The Difference”
Muataz Tahasays: March 7, 2022 at 12:06 am Thanks for this nice simple example and nice presentation Reply
siddsays: June 14, 2022 at 9:03 am To find the COV.
Which SD we have to use.
We use Sample Standard Deviations?(n-1)
Or we use Standard Deviation only?(n) Reply
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11206 | https://artofproblemsolving.com/wiki/index.php/Trigonometry?srsltid=AfmBOoquhtiWTb_orEH_joRdqJmLuHPrJUsOH7ut6YkeDzYw0VmR0aQx | Art of Problem Solving
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Trigonometry
Trigonometry is the study of relations between the side lengths and angles of triangles through the trigonometric functions. It is a fundamental branch of mathematics, and its discovery paved the way towards countless famous results.
In contest math, trigonometry is an integral subfield of both geometry and algebra. Many essential results in geometry are written in terms of the trigonometric functions, such as the Law of Sines and the Law of Cosines; many more, such as Stewart's Theorem, are most easily proven using trigonometry. In algebra, expressions involving the trigonometric functions appear frequently on contests. These are solved by clever usage of the trigonometric functions' countless identities, which can simplify otherwise unwieldy equations.
Outside of competition math, trigonometry is the backbone of much of analysis. In particular, Fourier Analysis is written almost entirely in the language of the trigonometric functions.
Contents
[hide]
1 Definitions
1.1 Right triangle definition
1.2 Unit circle definition
1.3 Taylor series definition
2 Applications in Geometry
2.1 Law of Sines
2.2 Law of cosines
3 Trigonometric identities
4 See also
Definitions
The trigonometric functions can be defined in several equivalent ways. The definition usually taught first is the right triangle definition, for its ease of access. An intermediate to olympiad geometry course usually uses the unit circle definition of trigonometry. Beyond the scope of contest math, the Taylor series definition of trigonometry is preferred in order to extend trigonometry to a complex domain.
Right triangle definition
The right triangle definition of trigonometry involves the ratios between edges of a right triangle, with respect to a given angle. The definitions below will be referring to angle , with side lengths specified in the diagram. Because angle must be less than for the triangle to stay right, these definitions only work for acute angles.
Sine: The sine of angle , denoted , is defined as the ratio of the side opposite to the hypotenuse.
Cosine: The cosine of angle , denoted , is defined as the ratio of the side adjacent to the hypotenuse.
Tangent: The tangent of angle , denoted , is defined as the ratio of the side opposite to the side adjacent to .
A common mnemonic to remember this is SOH-CAH-TOA, where Sine = Opposite / Hypotenuse, Cosine = Adjacent / Hypotenuse, and Tangent = Opposite / Adjacent
More uncommon are the reciprocals of the trigonometric functions, listed below.
Cosecant: The cosecant of angle , denoted , is defined as the reciprocal of the sine of .
Secant: The secant of angle , denoted , is defined as the reciprocal of the cosine of .
Cotangent: The cotangent of angle , denoted , is defined as the reciprocal of the tangent of .
The right triangle definition is most commonly taught in introductory geometry classes for its simplicity. However, it has its limitations. It only works if is right, which means that the trigonometric functions are only defined when angle is acute.
Even though it is defined using right triangles, trigonometry is just as useful when used on acute and obtuse triangles. The Law of Sines and Law of Cosines mentioned below generalize the right triangle definition to include all triangles.
Unit circle definition
Consider the unit circle, the circle with radius one centered at the origin. Starting at , walk a distance counterclockwise around the unit circle, as shown in the diagram. The coordinates of this point are defined to be .
As for the other trigonometric functions, is defined to be the ratio of to , and cosecant, secant, and cotangent are defined to be the reciprocals of sine, cosine, and tangent, respectively.
The benefit of this definition is that it matches the right triangle definition for acute angles, but extends their domain from acute angles to all real-valued angles. As such, this definition is usually preferred in intermediate to olympiad geometry settings.
Taylor series definition
The Taylor series for sine and cosine are used as their definitions in all higher mathematics. This meets the rigorous standards of real analysis, and gives a concrete way to extend the definitions of trigonometric functions from the real numbers to the full complex plane. The Taylor series for sine and cosine are shown below: These formulas are not used in high school math competitions. However, they do appear on the Putnam and other undergraduate competitions.
Applications in Geometry
While trigonometry is useful at any level, intermediate competitions are particularly fond of geometry problems demanding trigonometry. In addition to those mentioned, here are some highlights of the applications of trigonometry to geometry:
Law of Sines
The Law of Sines states that in any , where is the side opposite to , opposite to , opposite to , and is the circumradius of . The law of sines is particularly handy in problems involving the circumradius, seeing extremely wide usage in intermediate geometry.
Law of cosines
The Law of Cosines states that in any , where is the side opposite to , opposite to , and opposite to . It is a generalization of the Pythagorean Theorem and is used to prove several famous results, such as Heron's Formula and Stewart's Theorem. However, it sees limited applicability compared to the Law of Sines, as usage of the Law of Cosines can get algebra-heavy. It is helpful to memorize common, "nicer" values of sine and cosine as it can come in handy in contests, especially if you wish to apply either this or the Law of Sines to problems.
Trigonometric identities
Trigonometric identities are expressions true for all inputs involving the trigonometric functions. Due to the natural relationship between their definitions, these identities run numerous. In contest math, the most useful of these are:
Pythagorean identities
Angle addition identities
Double angle identities
Half angle identities
Sum-to-product identities
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Trigonometric identities
Law of Sines
Law of Cosines
Stewart's Theorem
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Number of pairs of two numbers in a set - Math proof
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I have a set of 11 numbers {0,3,6,9,12,15,18,21,24,27,30}. I am currently grouping numbers with a spacing of 9. I did this by hand - {0,9},{3,12},....{21,30} of total 8 pairs.
The answer is 8. But I am not sure how to relate this mathematically - or maybe there are another set pairs, which I might have missed. I am definitely sure this is related to permutations and combinations, but I don't know how to relate it exactly. Any guidance will be helpful.
Thank you.
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asked May 24, 2019 at 12:02
sundarsundar
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Your method is essentially correct. You are sure to find all the pairs that differ by 9 if you look ahead 9 from each number in the list to see if the result is in the list. You don't need anything fancy about permutations and combinations.
In this particular example the list is an arithmetic progression and 9 is three times the spacing, so you know you will find a difference of 9 for all but the last 3 numbers on the list. That's why there are 11−3=8 pairs.
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answered May 24, 2019 at 12:08
Ethan BolkerEthan Bolker
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Thank you for the quick and exact answer :). I can now generalize this to: Number of pairs = N - (spacing required/spacing present).sundar –sundar 2019-05-24 12:13:22 +00:00 Commented May 24, 2019 at 12:13
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11208 | https://www.omim.org/entry/615558 | Entry - #615558 - HYPOBETALIPOPROTEINEMIA, FAMILIAL, 1; FHBL1 - OMIM - (OMIM.ORG)
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615558
Table of Contents
Title
Phenotype-Gene Relationships
Clinical Synopsis
Text
Description
Clinical Features
Population Genetics
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# 615558
HYPOBETALIPOPROTEINEMIA, FAMILIAL, 1; FHBL1
Alternative titles; symbols
HYPOBETALIPOPROTEINEMIA, FAMILIAL; FHBL
ACANTHOCYTOSIS WITH HYPOBETALIPOPROTEINEMIA
HYPOBETALIPOPROTEINEMIA, NORMOTRIGLYCERIDEMIC
Other entities represented in this entry:
LOW DENSITY LIPOPROTEIN CHOLESTEROL LEVEL QUANTITATIVE TRAIT LOCUS 4, INCLUDED; LDLCQ4, INCLUDED
Phenotype-Gene Relationships
| Location | Phenotype | Phenotype MIM number | Inheritance | Phenotype mapping key | Gene/Locus | Gene/Locus MIM number |
--- --- ---
| 2p24.1 | Hypobetalipoproteinemia | 615558 | AR | 3 | APOB | 107730 |
Clinical SynopsisToggle Dropdown
PheneGene Graphics
Linear
Radial
INHERITANCE
Autosomal recessive [SNOMEDCT: 258211005][UMLS: C0441748 HPO: HP:0000007][HPO: HP:0000007]
HEAD & NECK
Eyes
Retinitis pigmentosa (in some patients) [SNOMEDCT: 28835009][ICD10CM: H35.52][UMLS: C0035334 HPO: HP:0000510][HPO: HP:0000510]
Retinal degeneration [SNOMEDCT: 95695004][UMLS: C0035304 HPO: HP:0000546][HPO: HP:0000546]
ABDOMEN
Gastrointestinal
Mild fat malabsorption [UMLS: C1862622][SNOMEDCT: 197494007][HPO: HP:0002630]
NEUROLOGIC
Central Nervous System
Ataxia [SNOMEDCT: 39384006, 20262006, 85102008][ICD10CM: R27.0][ICD9CM: 438.84][UMLS: C1135207, C0007758, C0004134 HPO: HP:0010867, HP:0001251][HPO: HP:0001251]
Peripheral Nervous System
Absent or decreased deep tendon reflexes [UMLS: C1866934 HPO: HP:0001315][HPO: HP:0001315]
HEMATOLOGY
Red cell acanthocytosis [SNOMEDCT: 250249008][UMLS: C0687751 HPO: HP:0001927][HPO: HP:0001927]
LABORATORY ABNORMALITIES
Hypobetalipoproteinemia [SNOMEDCT: 190786004][UMLS: C0853085, C0020597 HPO: HP:0003563][HPO: HP:0003563]
Decreased serum cholesterol [UMLS: C4325587]
MISCELLANEOUS
Heterozygotes have half-normal levels of APOB-containing lipoproteins
MOLECULAR BASIS
Caused by mutation in the apolipoprotein B gene (APOB, 107730.0001)
▲Close
▼TEXT
A number sign (#) is used with this entry because familial hypobetalipoproteinemia-1 (FHBL1) is caused by mutation in the APOB gene (107730) on chromosome 2p24.
See abetalipoproteinemia (ABL; 200100) for a similar disorder caused by mutation in the MTP gene (157147).
▼Description
Hypobetalipoproteinemia (FHBL) and abetalipoproteinemia (ABL; 200100) are rare diseases characterized by hypocholesterolemia and malabsorption of lipid-soluble vitamins leading to retinal degeneration, neuropathy, and coagulopathy. Hepatic steatosis is also common. The root cause of both disorders is improper packaging and secretion of apolipoprotein B-containing particles. Obligate heterozygous parents of FHBL patients typically have half normal levels of apoB-containing lipoproteins consistent with autosomal codominant inheritance, whereas obligate heterozygous parents of ABL patients usually have normal lipids consistent with autosomal recessive inheritance (summary by Lee and Hegele, 2014).
Genetic Heterogeneity of Familial Hypobetalipoproteinemia
Familial hypobetalipoproteinemia-2 (FHBL2; 605019) is caused by mutation in the ANGPTL3 gene (604774) on chromosome 1p31.
▼Clinical Features
Brown et al. (1974) noted that the consistent laboratory findings of reduced serum cholesterol and beta-lipoprotein define hypobetalipoproteinemia as a distinct syndrome. They found 4 reported kindreds and added a fifth. Only 2 of the patients in the reported families had symptoms. Mars et al. (1969) observed a family in which 1 of the 14 hypobetalipoproteinemic persons (in 3 generations), a 37-year-old woman, had signs and symptoms of progressive demyelination of the central nervous system, lack of responsiveness to local anesthesia, and dislike for animal fats and milk. The family reported by Brown et al. (1974) contained a child with psychomotor retardation. Although the peripheral blood smear showed no acanthocytes, the red cells on symptomatic and asymptomatic persons became acanthocytotic when placed in tissue culture medium with 10% autologous serum.
Biemer and McCammon (1975) described a family and reviewed others in the literature in which a person with 'homozygous hypobetalipoproteinemia' had occurred. They pointed out that although some of these cases were milder than cases of abetalipoproteinemia (ABL; 200100), homozygous hypobetalipoproteinemia could often be distinguished from abetalipoproteinemia only by the demonstration of presumably heterozygous hypobetalipoproteinemic first-degree relatives of the homozygote.
Kahn and Glueck (1978) reported remarkable freedom from atheroma in a 76-year-old woman who died from hepatic failure due apparently to hemochromatosis. The woman had been found to have hypobetalipoproteinemia in a study done previously (Glueck et al., 1976). This and hyperalphalipoproteinemia (143470) are accompanied by increased life expectancy.
Steinberg et al. (1979) described a kindred with a form of hypobetalipoproteinemia characterized by unusually low LDL cholesterol, normal triglyceride levels, low levels of HDL, mild fat malabsorption, and a defect in chylomicron clearance. On a high-carbohydrate diet, the triglyceride levels of the 67-year-old proband fell rather than rose. The proband, a retired Naval chaplain, was asymptomatic. He came to attention because of total serum cholesterol of 47 mg/dl. The proband's mother, aged 92, 1 brother, 1 sister, and 2 daughters also had hypobetalipoproteinemia. Young et al. (1987) found an abnormality of apoB, called apolipoprotein B37, in the plasma lipoproteins of multiple members of this kindred. Young et al. (1987) reported an intensive study of 41 members in 3 generations of this kindred. They documented the presence, in addition to the abnormal, truncated apoB species B37, of another apoB allele that was associated with reduced plasma concentrations of the normal apoB100. The proband (H.J.B.) and 2 of his sibs had both abnormal apoB alleles and were therefore compound heterozygotes for familial hypobetalipoproteinemia. All of the offspring of the 3 compound heterozygotes had hypobetalipoproteinemia, and each had evidence of only 1 of the abnormal apoB alleles. The average LDL cholesterol level in the compound heterozygotes was 6 mg/dl; in the 6 heterozygotes who had only the abnormal apoB37 allele, 31 mg/dl; in the 10 heterozygotes who had only the allele for reduced plasma concentrations of apoB100, 31 mg/dl; and in 22 unaffected family members, 110 mg/dl.
Malloy et al. (1981) described a patient with a metabolic disorder that they termed 'normotriglyceridemia abetalipoproteinemia.' The disorder was characterized by the absence of LDLs and apoB100 in plasma with apparently normal secretion of triglyceride-rich lipoproteins containing apoB48. Homer et al. (2005) suggested that the term 'normotriglyceridemic hypobetalipoproteinemia' is preferred to 'normotriglyceridemic abetalipoproteinemia' because abetalipoproteinemia (ABL; 200100) refers to the disorder caused by mutation in the MTP gene (157147).
Berger et al. (1983) studied a kindred in which the proband manifested the clinical and biochemical features of the homozygous state. Unlike the apparent absence of apolipoprotein B in the plasma in 5 previous cases of homozygous hypobetalipoproteinemia, they found a minute amount of apoB (about 0.025% of normal) in the plasma and suggested that the disorder might result not from a structural gene defect but from a failure of secretion.
Since LDLs are a main source of cholesterol for steroid hormone formation, Parker et al. (1986) were interested in studying the endocrine changes during pregnancy in homozygous familial hypobetalipoproteinemia. They found it surprising that a woman with phenotypic abetalipoproteinemia, previously reported by Illingworth et al. (1979), could become 'pregnant, let alone carry the pregnancy to term without hormonal therapy.' They noted successful pregnancy in 3 other abetalipoproteinemic women.
In 2 patients with homozygous hypobetalipoproteinemia, Ross et al. (1988) found that Southern blot analysis with 10 different cDNA probes revealed a normal gene without major insertions, deletions, or rearrangements. Northern and slot-blot analyses of total liver mRNA showed a normal-sized apoB mRNA that was present in greatly reduced quantities. ApoB protein was detected in liver cells immunohistochemically but was markedly reduced in quantity, and no apoB was detectable in the plasma with an ELISA assay. Ross et al. (1988) interpreted the findings as indicating a mutation in the coding portion of the apoB gene, leading to an abnormal apoB protein and apoB mRNA instability. These findings were distinct from those previously noted in abetalipoproteinemia (200100), which is characterized by an elevated level of hepatic apoB mRNA and accumulation of intracellular hepatic apoB protein. The blood-lipid changes that accompany heterozygous hypobetalipoproteinemia are reduced plasma concentrations of LDL cholesterol, total triglycerides, and APOB to less than 50% of normal values.
Harano et al. (1989) identified homozygous hypobetalipoproteinemia in 3 sibs. Both parents and 2 children of 1 of the sibs were heterozygous. The 75-year-old proband, the father of the 3 sibs, died of fever of unknown cause, thrombocytopenia, and anemia. He had ataxic movements of the hands and gait disturbance in later life. The 3 homozygotes showed marked deficiency of apoB100, although trace amounts were noted in LDL. In contrast, apoB48 was present in chylomicrons obtained after a fatty meal in 2 of the patients with homozygous hypobetalipoproteinemia, indicating a selective deficiency of apoB100.
Keidar et al. (1990) described apparent compound heterozygosity for abetalipoproteinemia and familial hypobetalipoproteinemia. The proband, a 10-year-old boy with abetalipoproteinemia, had a father with a normal apolipoprotein profile; however, his mother and maternal grandfather suffered from familial hypobetalipoproteinemia.
Araki et al. (1991) described a 55-year-old man with cerebellar ataxia due apparently to hypobetalipoproteinemia. A brother also had hypobetalipoproteinemia with neurologic symptoms. The 2 children of the proband, aged 31 and 29 years, and a sister of the proband had only hypobetalipoproteinemia. The proband and his neurologically affected brother as well as members of the 2 previous generations had steatocystoma multiplex (184500). The latter condition may have been coincidental.
Di Leo et al. (2008) reported 3 patients with severe hypobetalipoproteinemia due to homozygosity or compound heterozygosity for mutations in APOB, who presented with chronic liver disease and/or chronic diarrhea at ages 52, 55, and 19 years, respectively. The authors stated that the clinical diagnosis of homozygous FHBL is extremely rare, with approximately 20 cases reported over the past 2 decades; they noted that their patients highlight the heterogeneity of clinical manifestations and the possible presentation of disease late in life.
▼Population Genetics
Lee and Hegele (2014) stated that the incidences of both FHBL and abetalipoproteinemia are reported as less than 1 in 1 million.
▼Mapping
Leppert et al. (1988) found that a DNA haplotype of the APOB gene cosegregated with hypobetalipoproteinemia in an Idaho pedigree, with a maximum lod score of 7.56 at theta = 0.0. This finding strongly suggested that a mutation in the APOB gene underlies hypobetalipoproteinemia and indicated the usefulness of the candidate gene approach.
Pulai et al. (1998) commented that various truncated forms of apoB have been found to segregate with the FHBL phenotype in more than 30 kindreds. They reported studies of 6 kindreds in which no truncated forms of apoB protein were detected with sensitive immunoblotting in the plasma of any of the affected individuals. Persons with apoB levels in the 5th centile for their age and sex were considered as affected with FHBL. Linkage analysis with 3 microsatellite markers flanking the APOB gene, a 3-prime VNTR marker, and an intragenic marker yielded 2-point linkage of FHBL to the 3-prime VNTR marker with a combined maximum lod score of 8.5 at theta = 0.0 for 5 of the 6 families. A test of homogeneity differentiated the sixth family from the other 5. These kindreds demonstrated the heterogeneity of FHBL.
▼Molecular Genetics
In a patient with hypobetalipoproteinemia and small amounts of truncated protein (B37) in VLDL, LDL, and HDL fractions of the plasma, Young et al. (1987, 1988) found a 4-bp deletion in the APOB gene resulting in a frameshift (107730.0001). This was one of the mutant alleles in the family with hypobetalipoproteinemia first reported by Steinberg et al. (1979). Linton et al. (1992) investigated the reason for the curious finding that low levels of apoB100 were produced by the mutant allele carrying this mutation. The clue that led to the understanding of what was going on with this allele was the recognition that the proband in the family, H.J.B., as well as the other 2 compound heterozygotes, actually had 4 bona fide apoB species within their plasma lipoproteins: apoB37, apoB48, apoB100, and apoB86. Linton et al. (1992) demonstrated that the apoB86 and apoB100 were products of a single mutant apoB allele, which they designated the apoB86 allele.
In a patient with normotriglyceridemic hypobetalipoproteinemia, originally described by Malloy et al. (1981), Hardman et al. (1991) identified homozygosity for a nonsense mutation in the APOB gene (107730.0013).
In a patient with hypobetalipoproteinemia, McCormick et al. (1992) identified a heterozygous nonsense mutation in the APOB gene (107730.0014).
In a 27-year-old woman from a consanguineous French Canadian family, who was diagnosed with FHBL in the first months of life, Gangloff et al. (2011) identified a homozygous truncating mutation in the APOB gene (107730.0022). The authors stated that this was the first case of homozygous FHBL in a French Canadian family.
▼History
Salt et al. (1960) reported the absence of beta-lipoprotein from the plasma of a patient with abetalipoproteinemia, but this patient had familial hypobetalipoproteinemia because his parents had markedly low levels of cholesterol in plasma (Kane and Havel, 2001).
▼See Also:
Linton et al. (1993)
▼REFERENCES
Araki, W., Hirose, S., Mimori, Y., Nakamura, S., Kimura, J., Ohno, K., Shimada, T. Familial hypobetalipoproteinaemia complicated by cerebellar ataxia and steatocystoma multiplex. J. Intern. Med. 229: 197-199, 1991. [PubMed: 1997645, related citations] [Full Text]
Berger, G. M. B., Brown, G., Henderson, H. E., Bonnici, F. Apolipoprotein B detected in the plasma of a patient with homozygous hypobetalipoproteinaemia: implications for aetiology. J. Med. Genet. 20: 189-195, 1983. [PubMed: 6876109, related citations] [Full Text]
Biemer, J. J., McCammon, R. E. The genetic relationship of abetalipoproteinemia and hypobetalipoproteinemia: a report of the occurrence of both diseases within the same family. J. Lab. Clin. Med. 85: 556-565, 1975. [PubMed: 164511, related citations]
Brown, B. J., Lewis, L. A., Mercer, R. D. Familial hypobetalipoproteinemia : report of a case with psychomotor retardation. Pediatrics 54: 111-113, 1974. [PubMed: 4365305, related citations]
Di Leo, E., Magnolo, L., Bertolotti, M., Bourbon, M., Carmo Pereira, S., Pirisi, M., Calandra, S., Tarugi, P. Variable phenotypic expression of homozygous familial hypobetalipoproteinaemia due to novel APOB gene mutations. Clin. Genet. 74: 267-273, 2008. [PubMed: 18492086, related citations] [Full Text]
Gangloff, A., Bergeron, J., Couture, P., Martins, R., Hegele, R. A., Gagne, C. A novel mutation of apolipoprotein B in a French Canadian family with homozygous hypobetalipoproteinemia. J. Clin. Lipidol. 5: 414-417, 2011. [PubMed: 21981844, related citations] [Full Text]
Glueck, C. J., Tsang, R. C., Mellies, M. J., Fallat, R. W., Steiner, P. M. Neonatal familial hypobeta-lipoproteinemia. Metabolism 25: 611-614, 1976. [PubMed: 178979, related citations] [Full Text]
Harano, Y., Kojima, H., Nakano, T., Harada, M., Kashiwagi, A., Nakajima, Y., Hidaka, T. H., Ohtsuki, T., Suzuki, T., Tamura, A., Fujii, T., Nishimura, T., Ohtaka, T., Shigeta, Y. Homozygous hypobetalipoproteinemia with spared chylomicron formation. Metabolism 38: 1-7, 1989. [PubMed: 2909827, related citations] [Full Text]
Hardman, D. A., Pullinger, C. R., Hamilton, R. L., Kane, J. P., Malloy, M. J. Molecular and metabolic basis for the metabolic disorder normotriglyceridemic abetalipoproteinemia. J. Clin. Invest. 88: 1722-1729, 1991. [PubMed: 1939657, related citations] [Full Text]
Homer, V. M., George, P. M., du Toit, S., Davidson, J. S., Wilson, C. J. Mental retardation and ataxia due to normotriglyceridemic hypobetalipoproteinemia. Ann. Neurol. 58: 160-163, 2005. [PubMed: 15984016, related citations] [Full Text]
Illingworth, D. R., Connor, W. E., Buist, N. R. M., Jhaveri, B. M., Lin, S. S., McMurry, M. P. Sterol balance in abetalipoproteinemia: studies in a patient with homozygous familial hypobetalipoproteinemia. Metabolism 28: 1152-1160, 1979. [PubMed: 491973, related citations] [Full Text]
Kahn, J. A., Glueck, C. J. Familial hypobetalipoproteinemia: absence of atherosclerosis in a postmortem study. JAMA 240: 47-48, 1978. [PubMed: 207903, related citations] [Full Text]
Kane, J. P., Havel, R. J. Disorders of the biogenesis and secretion of lipoproteins containing the B apolipoproteins.In: Scriver, C. R.; Beaudet, A. L.; Sly, W. S.; Valle, D. (eds.) : The Metabolic and Molecular Bases of Inherited Disease. Vol. II. (8th ed.) New York: McGraw-Hill (pub.) 2001. P. 2725.
Keidar, S., Etzioni, A., Brook, J. G., Gershoni-Baruch, R., Aviram, M. Compound heterozygosity for abetalipoproteinaemia and familial hypobetalipoproteinaemia. J. Med. Genet. 27: 133-134, 1990. [PubMed: 2319582, related citations] [Full Text]
Lee, J., Hegele, R. A. Abetalipoproteinemia and homozygous hypobetalipoproteinemia: a framework for diagnosis and management. J. Inherit. Metab. Dis. 37: 333-339, 2014. [PubMed: 24288038, related citations] [Full Text]
Leppert, M., Breslow, J. L., Wu, L., Hasstedt, S., O'Connell, P., Lathrop, M., Williams, R. R., White, R., Lalouel, J.-M. Inference of a molecular defect of apolipoprotein B in hypobetalipoproteinemia by linkage analysis in a large kindred. J. Clin. Invest. 82: 847-851, 1988. [PubMed: 2901434, related citations] [Full Text]
Linton, M. F., Farese, R. V., Jr., Young, S. G. Familial hypobetalipoproteinemia. J. Lipid Res. 34: 521-541, 1993. [PubMed: 8496659, related citations]
Linton, M. F., Pierotti, V., Young, S. G. Reading-frame restoration with an apolipoprotein B gene frameshift mutation. Proc. Nat. Acad. Sci. 89: 11431-11435, 1992. [PubMed: 1454832, related citations] [Full Text]
Malloy, M. J., Kane, J. P., Hardman, D. A., Hamilton, R. L., Dalal, K. B. Normotriglyceridemic abetalipoproteinemia: absence of the B-100 apolipoprotein. J. Clin. Invest. 67: 1441-1450, 1981. [PubMed: 7229035, related citations] [Full Text]
Mars, H., Lewis, L. A., Robertson, A. L., Jr., Butkus, A., Williams, G. H., Jr. Familial hypobetalipoproteinemia: a genetic disorder of lipid metabolism with nervous system involvement. Am. J. Med. 46: 886-900, 1969. [PubMed: 4978919, related citations] [Full Text]
McCormick, S. P. A., Fellowes, A. P., Walmsley, T. A., George, P. M. Apolipoprotein B-32: a new truncated mutant of human apolipoprotein B capable of forming particles in the low density lipoprotein range. Biochim. Biophys. Acta 1138: 290-296, 1992. [PubMed: 1562615, related citations] [Full Text]
Parker, C. R., Jr., Illingworth, D. R., Bissonnette, J., Carr, B. R. Endocrine changes during pregnancy in a patient with homozygous familial hypobetalipoproteinemia. New Eng. J. Med. 314: 557-560, 1986. [PubMed: 3945294, related citations] [Full Text]
Pulai, J. I., Neuman, R. J., Groenewegen, A. W., Wu, J., Schonfeld, G. Genetic heterogeneity in familial hypobetalipoproteinemia: linkage and non-linkage to the apoB gene in Caucasian families. Am. J. Med. Genet. 76: 79-86, 1998. [PubMed: 9508071, related citations]
Ross, R. S., Gregg, R. E., Law, S. W., Monge, J. C., Grant, S. M., Higuchi, K., Triche, T. J., Jefferson, J., Brewer, H. B., Jr. Homozygous hypobetalipoproteinemia: a disease distinct from abetalipoproteinemia at the molecular level. J. Clin. Invest. 81: 590-595, 1988. [PubMed: 2828430, related citations] [Full Text]
Salt, H. B., Wolff, O. H., Lloyd, J. K., Fosbrooke, A. S., Cameron, A. H., Hubble, D. V. On having no beta-lipoprotein: a syndrome comprising abetalipoproteinaemia, acanthocytosis and steatorrhoea. Lancet 276: 325-329, 1960. Note: Originally Volume II. [PubMed: 13745738, related citations] [Full Text]
Steinberg, D., Grundy, S. M., Mok, H. Y. I., Turner, J. D., Weinstein, D. B., Brown, W. V., Albers, J. J. Metabolic studies in an unusual case of asymptomatic familial hypobetalipoproteinemia with hypoalphalipoproteinemia and fasting chylomicronemia. J. Clin. Invest. 64: 292-301, 1979. [PubMed: 221546, related citations] [Full Text]
Young, S. G., Bertics, S. J., Curtiss, L. K., Dubois, B. W., Witztum, J. L. Genetic analysis of a kindred with familial hypobetalipoproteinemia: evidence for two separate gene defects: one associated with an abnormal apolipoprotein B species, apolipoprotein B-37; and a second associated with low plasma concentrations of apolipoprotein B-100. J. Clin. Invest. 79: 1842-1851, 1987. [PubMed: 3473077, related citations] [Full Text]
Young, S. G., Northey, S. T., McCarthy, B. J. Low plasma cholesterol levels caused by a short deletion in the apolipoprotein B gene. Science 241: 591-593, 1988. [PubMed: 3399894, related citations] [Full Text]
Contributors:
Marla J. F. O'Neill - updated : 12/20/2013
Creation Date:
Carol A. Bocchini : 12/4/2013
Edit History:
alopez : 03/17/2023
carol : 10/10/2019
alopez : 07/13/2018
carol : 07/09/2016
mcolton : 8/12/2014
carol : 12/20/2013
mcolton : 12/20/2013
carol : 12/9/2013
carol : 12/9/2013
carol : 12/9/2013
# 615558
HYPOBETALIPOPROTEINEMIA, FAMILIAL, 1; FHBL1
Alternative titles; symbols
HYPOBETALIPOPROTEINEMIA, FAMILIAL; FHBL
ACANTHOCYTOSIS WITH HYPOBETALIPOPROTEINEMIA
HYPOBETALIPOPROTEINEMIA, NORMOTRIGLYCERIDEMIC
Other entities represented in this entry:
LOW DENSITY LIPOPROTEIN CHOLESTEROL LEVEL QUANTITATIVE TRAIT LOCUS 4, INCLUDED; LDLCQ4, INCLUDED
SNOMEDCT: 60193003; ICD10CM: E78.6; ORPHA: 14; DO: 0111062; MONDO: 0014252;
Phenotype-Gene Relationships
| Location | Phenotype | Phenotype MIM number | Inheritance | Phenotype mapping key | Gene/Locus | Gene/Locus MIM number |
--- --- ---
| 2p24.1 | Hypobetalipoproteinemia | 615558 | Autosomal recessive | 3 | APOB | 107730 |
TEXT
A number sign (#) is used with this entry because familial hypobetalipoproteinemia-1 (FHBL1) is caused by mutation in the APOB gene (107730) on chromosome 2p24.
See abetalipoproteinemia (ABL; 200100) for a similar disorder caused by mutation in the MTP gene (157147).
Description
Hypobetalipoproteinemia (FHBL) and abetalipoproteinemia (ABL; 200100) are rare diseases characterized by hypocholesterolemia and malabsorption of lipid-soluble vitamins leading to retinal degeneration, neuropathy, and coagulopathy. Hepatic steatosis is also common. The root cause of both disorders is improper packaging and secretion of apolipoprotein B-containing particles. Obligate heterozygous parents of FHBL patients typically have half normal levels of apoB-containing lipoproteins consistent with autosomal codominant inheritance, whereas obligate heterozygous parents of ABL patients usually have normal lipids consistent with autosomal recessive inheritance (summary by Lee and Hegele, 2014).
Genetic Heterogeneity of Familial Hypobetalipoproteinemia
Familial hypobetalipoproteinemia-2 (FHBL2; 605019) is caused by mutation in the ANGPTL3 gene (604774) on chromosome 1p31.
Clinical Features
Brown et al. (1974) noted that the consistent laboratory findings of reduced serum cholesterol and beta-lipoprotein define hypobetalipoproteinemia as a distinct syndrome. They found 4 reported kindreds and added a fifth. Only 2 of the patients in the reported families had symptoms. Mars et al. (1969) observed a family in which 1 of the 14 hypobetalipoproteinemic persons (in 3 generations), a 37-year-old woman, had signs and symptoms of progressive demyelination of the central nervous system, lack of responsiveness to local anesthesia, and dislike for animal fats and milk. The family reported by Brown et al. (1974) contained a child with psychomotor retardation. Although the peripheral blood smear showed no acanthocytes, the red cells on symptomatic and asymptomatic persons became acanthocytotic when placed in tissue culture medium with 10% autologous serum.
Biemer and McCammon (1975) described a family and reviewed others in the literature in which a person with 'homozygous hypobetalipoproteinemia' had occurred. They pointed out that although some of these cases were milder than cases of abetalipoproteinemia (ABL; 200100), homozygous hypobetalipoproteinemia could often be distinguished from abetalipoproteinemia only by the demonstration of presumably heterozygous hypobetalipoproteinemic first-degree relatives of the homozygote.
Kahn and Glueck (1978) reported remarkable freedom from atheroma in a 76-year-old woman who died from hepatic failure due apparently to hemochromatosis. The woman had been found to have hypobetalipoproteinemia in a study done previously (Glueck et al., 1976). This and hyperalphalipoproteinemia (143470) are accompanied by increased life expectancy.
Steinberg et al. (1979) described a kindred with a form of hypobetalipoproteinemia characterized by unusually low LDL cholesterol, normal triglyceride levels, low levels of HDL, mild fat malabsorption, and a defect in chylomicron clearance. On a high-carbohydrate diet, the triglyceride levels of the 67-year-old proband fell rather than rose. The proband, a retired Naval chaplain, was asymptomatic. He came to attention because of total serum cholesterol of 47 mg/dl. The proband's mother, aged 92, 1 brother, 1 sister, and 2 daughters also had hypobetalipoproteinemia. Young et al. (1987) found an abnormality of apoB, called apolipoprotein B37, in the plasma lipoproteins of multiple members of this kindred. Young et al. (1987) reported an intensive study of 41 members in 3 generations of this kindred. They documented the presence, in addition to the abnormal, truncated apoB species B37, of another apoB allele that was associated with reduced plasma concentrations of the normal apoB100. The proband (H.J.B.) and 2 of his sibs had both abnormal apoB alleles and were therefore compound heterozygotes for familial hypobetalipoproteinemia. All of the offspring of the 3 compound heterozygotes had hypobetalipoproteinemia, and each had evidence of only 1 of the abnormal apoB alleles. The average LDL cholesterol level in the compound heterozygotes was 6 mg/dl; in the 6 heterozygotes who had only the abnormal apoB37 allele, 31 mg/dl; in the 10 heterozygotes who had only the allele for reduced plasma concentrations of apoB100, 31 mg/dl; and in 22 unaffected family members, 110 mg/dl.
Malloy et al. (1981) described a patient with a metabolic disorder that they termed 'normotriglyceridemia abetalipoproteinemia.' The disorder was characterized by the absence of LDLs and apoB100 in plasma with apparently normal secretion of triglyceride-rich lipoproteins containing apoB48. Homer et al. (2005) suggested that the term 'normotriglyceridemic hypobetalipoproteinemia' is preferred to 'normotriglyceridemic abetalipoproteinemia' because abetalipoproteinemia (ABL; 200100) refers to the disorder caused by mutation in the MTP gene (157147).
Berger et al. (1983) studied a kindred in which the proband manifested the clinical and biochemical features of the homozygous state. Unlike the apparent absence of apolipoprotein B in the plasma in 5 previous cases of homozygous hypobetalipoproteinemia, they found a minute amount of apoB (about 0.025% of normal) in the plasma and suggested that the disorder might result not from a structural gene defect but from a failure of secretion.
Since LDLs are a main source of cholesterol for steroid hormone formation, Parker et al. (1986) were interested in studying the endocrine changes during pregnancy in homozygous familial hypobetalipoproteinemia. They found it surprising that a woman with phenotypic abetalipoproteinemia, previously reported by Illingworth et al. (1979), could become 'pregnant, let alone carry the pregnancy to term without hormonal therapy.' They noted successful pregnancy in 3 other abetalipoproteinemic women.
In 2 patients with homozygous hypobetalipoproteinemia, Ross et al. (1988) found that Southern blot analysis with 10 different cDNA probes revealed a normal gene without major insertions, deletions, or rearrangements. Northern and slot-blot analyses of total liver mRNA showed a normal-sized apoB mRNA that was present in greatly reduced quantities. ApoB protein was detected in liver cells immunohistochemically but was markedly reduced in quantity, and no apoB was detectable in the plasma with an ELISA assay. Ross et al. (1988) interpreted the findings as indicating a mutation in the coding portion of the apoB gene, leading to an abnormal apoB protein and apoB mRNA instability. These findings were distinct from those previously noted in abetalipoproteinemia (200100), which is characterized by an elevated level of hepatic apoB mRNA and accumulation of intracellular hepatic apoB protein. The blood-lipid changes that accompany heterozygous hypobetalipoproteinemia are reduced plasma concentrations of LDL cholesterol, total triglycerides, and APOB to less than 50% of normal values.
Harano et al. (1989) identified homozygous hypobetalipoproteinemia in 3 sibs. Both parents and 2 children of 1 of the sibs were heterozygous. The 75-year-old proband, the father of the 3 sibs, died of fever of unknown cause, thrombocytopenia, and anemia. He had ataxic movements of the hands and gait disturbance in later life. The 3 homozygotes showed marked deficiency of apoB100, although trace amounts were noted in LDL. In contrast, apoB48 was present in chylomicrons obtained after a fatty meal in 2 of the patients with homozygous hypobetalipoproteinemia, indicating a selective deficiency of apoB100.
Keidar et al. (1990) described apparent compound heterozygosity for abetalipoproteinemia and familial hypobetalipoproteinemia. The proband, a 10-year-old boy with abetalipoproteinemia, had a father with a normal apolipoprotein profile; however, his mother and maternal grandfather suffered from familial hypobetalipoproteinemia.
Araki et al. (1991) described a 55-year-old man with cerebellar ataxia due apparently to hypobetalipoproteinemia. A brother also had hypobetalipoproteinemia with neurologic symptoms. The 2 children of the proband, aged 31 and 29 years, and a sister of the proband had only hypobetalipoproteinemia. The proband and his neurologically affected brother as well as members of the 2 previous generations had steatocystoma multiplex (184500). The latter condition may have been coincidental.
Di Leo et al. (2008) reported 3 patients with severe hypobetalipoproteinemia due to homozygosity or compound heterozygosity for mutations in APOB, who presented with chronic liver disease and/or chronic diarrhea at ages 52, 55, and 19 years, respectively. The authors stated that the clinical diagnosis of homozygous FHBL is extremely rare, with approximately 20 cases reported over the past 2 decades; they noted that their patients highlight the heterogeneity of clinical manifestations and the possible presentation of disease late in life.
Population Genetics
Lee and Hegele (2014) stated that the incidences of both FHBL and abetalipoproteinemia are reported as less than 1 in 1 million.
Mapping
Leppert et al. (1988) found that a DNA haplotype of the APOB gene cosegregated with hypobetalipoproteinemia in an Idaho pedigree, with a maximum lod score of 7.56 at theta = 0.0. This finding strongly suggested that a mutation in the APOB gene underlies hypobetalipoproteinemia and indicated the usefulness of the candidate gene approach.
Pulai et al. (1998) commented that various truncated forms of apoB have been found to segregate with the FHBL phenotype in more than 30 kindreds. They reported studies of 6 kindreds in which no truncated forms of apoB protein were detected with sensitive immunoblotting in the plasma of any of the affected individuals. Persons with apoB levels in the 5th centile for their age and sex were considered as affected with FHBL. Linkage analysis with 3 microsatellite markers flanking the APOB gene, a 3-prime VNTR marker, and an intragenic marker yielded 2-point linkage of FHBL to the 3-prime VNTR marker with a combined maximum lod score of 8.5 at theta = 0.0 for 5 of the 6 families. A test of homogeneity differentiated the sixth family from the other 5. These kindreds demonstrated the heterogeneity of FHBL.
Molecular Genetics
In a patient with hypobetalipoproteinemia and small amounts of truncated protein (B37) in VLDL, LDL, and HDL fractions of the plasma, Young et al. (1987, 1988) found a 4-bp deletion in the APOB gene resulting in a frameshift (107730.0001). This was one of the mutant alleles in the family with hypobetalipoproteinemia first reported by Steinberg et al. (1979). Linton et al. (1992) investigated the reason for the curious finding that low levels of apoB100 were produced by the mutant allele carrying this mutation. The clue that led to the understanding of what was going on with this allele was the recognition that the proband in the family, H.J.B., as well as the other 2 compound heterozygotes, actually had 4 bona fide apoB species within their plasma lipoproteins: apoB37, apoB48, apoB100, and apoB86. Linton et al. (1992) demonstrated that the apoB86 and apoB100 were products of a single mutant apoB allele, which they designated the apoB86 allele.
In a patient with normotriglyceridemic hypobetalipoproteinemia, originally described by Malloy et al. (1981), Hardman et al. (1991) identified homozygosity for a nonsense mutation in the APOB gene (107730.0013).
In a patient with hypobetalipoproteinemia, McCormick et al. (1992) identified a heterozygous nonsense mutation in the APOB gene (107730.0014).
In a 27-year-old woman from a consanguineous French Canadian family, who was diagnosed with FHBL in the first months of life, Gangloff et al. (2011) identified a homozygous truncating mutation in the APOB gene (107730.0022). The authors stated that this was the first case of homozygous FHBL in a French Canadian family.
History
Salt et al. (1960) reported the absence of beta-lipoprotein from the plasma of a patient with abetalipoproteinemia, but this patient had familial hypobetalipoproteinemia because his parents had markedly low levels of cholesterol in plasma (Kane and Havel, 2001).
See Also:
Linton et al. (1993)
REFERENCES
Araki, W., Hirose, S., Mimori, Y., Nakamura, S., Kimura, J., Ohno, K., Shimada, T. Familial hypobetalipoproteinaemia complicated by cerebellar ataxia and steatocystoma multiplex. J. Intern. Med. 229: 197-199, 1991. [PubMed: 1997645] [Full Text:
Berger, G. M. B., Brown, G., Henderson, H. E., Bonnici, F. Apolipoprotein B detected in the plasma of a patient with homozygous hypobetalipoproteinaemia: implications for aetiology. J. Med. Genet. 20: 189-195, 1983. [PubMed: 6876109] [Full Text:
Biemer, J. J., McCammon, R. E. The genetic relationship of abetalipoproteinemia and hypobetalipoproteinemia: a report of the occurrence of both diseases within the same family. J. Lab. Clin. Med. 85: 556-565, 1975. [PubMed: 164511]
Brown, B. J., Lewis, L. A., Mercer, R. D. Familial hypobetalipoproteinemia : report of a case with psychomotor retardation. Pediatrics 54: 111-113, 1974. [PubMed: 4365305]
Di Leo, E., Magnolo, L., Bertolotti, M., Bourbon, M., Carmo Pereira, S., Pirisi, M., Calandra, S., Tarugi, P. Variable phenotypic expression of homozygous familial hypobetalipoproteinaemia due to novel APOB gene mutations. Clin. Genet. 74: 267-273, 2008. [PubMed: 18492086] [Full Text:
Gangloff, A., Bergeron, J., Couture, P., Martins, R., Hegele, R. A., Gagne, C. A novel mutation of apolipoprotein B in a French Canadian family with homozygous hypobetalipoproteinemia. J. Clin. Lipidol. 5: 414-417, 2011. [PubMed: 21981844] [Full Text:
Glueck, C. J., Tsang, R. C., Mellies, M. J., Fallat, R. W., Steiner, P. M. Neonatal familial hypobeta-lipoproteinemia. Metabolism 25: 611-614, 1976. [PubMed: 178979] [Full Text:
Harano, Y., Kojima, H., Nakano, T., Harada, M., Kashiwagi, A., Nakajima, Y., Hidaka, T. H., Ohtsuki, T., Suzuki, T., Tamura, A., Fujii, T., Nishimura, T., Ohtaka, T., Shigeta, Y. Homozygous hypobetalipoproteinemia with spared chylomicron formation. Metabolism 38: 1-7, 1989. [PubMed: 2909827] [Full Text:
Hardman, D. A., Pullinger, C. R., Hamilton, R. L., Kane, J. P., Malloy, M. J. Molecular and metabolic basis for the metabolic disorder normotriglyceridemic abetalipoproteinemia. J. Clin. Invest. 88: 1722-1729, 1991. [PubMed: 1939657] [Full Text:
Homer, V. M., George, P. M., du Toit, S., Davidson, J. S., Wilson, C. J. Mental retardation and ataxia due to normotriglyceridemic hypobetalipoproteinemia. Ann. Neurol. 58: 160-163, 2005. [PubMed: 15984016] [Full Text:
Illingworth, D. R., Connor, W. E., Buist, N. R. M., Jhaveri, B. M., Lin, S. S., McMurry, M. P. Sterol balance in abetalipoproteinemia: studies in a patient with homozygous familial hypobetalipoproteinemia. Metabolism 28: 1152-1160, 1979. [PubMed: 491973] [Full Text:
Kahn, J. A., Glueck, C. J. Familial hypobetalipoproteinemia: absence of atherosclerosis in a postmortem study. JAMA 240: 47-48, 1978. [PubMed: 207903] [Full Text:
Kane, J. P., Havel, R. J. Disorders of the biogenesis and secretion of lipoproteins containing the B apolipoproteins.In: Scriver, C. R.; Beaudet, A. L.; Sly, W. S.; Valle, D. (eds.) : The Metabolic and Molecular Bases of Inherited Disease. Vol. II. (8th ed.) New York: McGraw-Hill (pub.) 2001. P. 2725.
Keidar, S., Etzioni, A., Brook, J. G., Gershoni-Baruch, R., Aviram, M. Compound heterozygosity for abetalipoproteinaemia and familial hypobetalipoproteinaemia. J. Med. Genet. 27: 133-134, 1990. [PubMed: 2319582] [Full Text:
Lee, J., Hegele, R. A. Abetalipoproteinemia and homozygous hypobetalipoproteinemia: a framework for diagnosis and management. J. Inherit. Metab. Dis. 37: 333-339, 2014. [PubMed: 24288038] [Full Text:
Leppert, M., Breslow, J. L., Wu, L., Hasstedt, S., O'Connell, P., Lathrop, M., Williams, R. R., White, R., Lalouel, J.-M. Inference of a molecular defect of apolipoprotein B in hypobetalipoproteinemia by linkage analysis in a large kindred. J. Clin. Invest. 82: 847-851, 1988. [PubMed: 2901434] [Full Text:
Linton, M. F., Farese, R. V., Jr., Young, S. G. Familial hypobetalipoproteinemia. J. Lipid Res. 34: 521-541, 1993. [PubMed: 8496659]
Linton, M. F., Pierotti, V., Young, S. G. Reading-frame restoration with an apolipoprotein B gene frameshift mutation. Proc. Nat. Acad. Sci. 89: 11431-11435, 1992. [PubMed: 1454832] [Full Text:
Malloy, M. J., Kane, J. P., Hardman, D. A., Hamilton, R. L., Dalal, K. B. Normotriglyceridemic abetalipoproteinemia: absence of the B-100 apolipoprotein. J. Clin. Invest. 67: 1441-1450, 1981. [PubMed: 7229035] [Full Text:
Mars, H., Lewis, L. A., Robertson, A. L., Jr., Butkus, A., Williams, G. H., Jr. Familial hypobetalipoproteinemia: a genetic disorder of lipid metabolism with nervous system involvement. Am. J. Med. 46: 886-900, 1969. [PubMed: 4978919] [Full Text:
McCormick, S. P. A., Fellowes, A. P., Walmsley, T. A., George, P. M. Apolipoprotein B-32: a new truncated mutant of human apolipoprotein B capable of forming particles in the low density lipoprotein range. Biochim. Biophys. Acta 1138: 290-296, 1992. [PubMed: 1562615] [Full Text:
Parker, C. R., Jr., Illingworth, D. R., Bissonnette, J., Carr, B. R. Endocrine changes during pregnancy in a patient with homozygous familial hypobetalipoproteinemia. New Eng. J. Med. 314: 557-560, 1986. [PubMed: 3945294] [Full Text:
Pulai, J. I., Neuman, R. J., Groenewegen, A. W., Wu, J., Schonfeld, G. Genetic heterogeneity in familial hypobetalipoproteinemia: linkage and non-linkage to the apoB gene in Caucasian families. Am. J. Med. Genet. 76: 79-86, 1998. [PubMed: 9508071]
Ross, R. S., Gregg, R. E., Law, S. W., Monge, J. C., Grant, S. M., Higuchi, K., Triche, T. J., Jefferson, J., Brewer, H. B., Jr. Homozygous hypobetalipoproteinemia: a disease distinct from abetalipoproteinemia at the molecular level. J. Clin. Invest. 81: 590-595, 1988. [PubMed: 2828430] [Full Text:
Salt, H. B., Wolff, O. H., Lloyd, J. K., Fosbrooke, A. S., Cameron, A. H., Hubble, D. V. On having no beta-lipoprotein: a syndrome comprising abetalipoproteinaemia, acanthocytosis and steatorrhoea. Lancet 276: 325-329, 1960. Note: Originally Volume II. [PubMed: 13745738] [Full Text:
Steinberg, D., Grundy, S. M., Mok, H. Y. I., Turner, J. D., Weinstein, D. B., Brown, W. V., Albers, J. J. Metabolic studies in an unusual case of asymptomatic familial hypobetalipoproteinemia with hypoalphalipoproteinemia and fasting chylomicronemia. J. Clin. Invest. 64: 292-301, 1979. [PubMed: 221546] [Full Text:
Young, S. G., Bertics, S. J., Curtiss, L. K., Dubois, B. W., Witztum, J. L. Genetic analysis of a kindred with familial hypobetalipoproteinemia: evidence for two separate gene defects: one associated with an abnormal apolipoprotein B species, apolipoprotein B-37; and a second associated with low plasma concentrations of apolipoprotein B-100. J. Clin. Invest. 79: 1842-1851, 1987. [PubMed: 3473077] [Full Text:
Young, S. G., Northey, S. T., McCarthy, B. J. Low plasma cholesterol levels caused by a short deletion in the apolipoprotein B gene. Science 241: 591-593, 1988. [PubMed: 3399894] [Full Text:
Contributors:
Marla J. F. O'Neill - updated : 12/20/2013
Creation Date:
Carol A. Bocchini : 12/4/2013
Edit History:
alopez : 03/17/2023
carol : 10/10/2019
alopez : 07/13/2018
carol : 07/09/2016
mcolton : 8/12/2014
carol : 12/20/2013
mcolton : 12/20/2013
carol : 12/9/2013
carol : 12/9/2013
carol : 12/9/2013
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NOTE: OMIM is intended for use primarily by physicians and other professionals concerned with genetic disorders, by genetics researchers, and by advanced students in science and medicine. While the OMIM database is open to the public, users seeking information about a personal medical or genetic condition are urged to consult with a qualified physician for diagnosis and for answers to personal questions.
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Permutations of n numbers with no odd numbers next to each other
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What is the number of {1,2,…,n}{1,2,…,n} permutations, in which neither two neighbouring numbers are odd?
Could somebody show me the reasoning that leads to the answer?
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edited Jun 14, 2016 at 15:15
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I do not understand your inclusion of the phrase [even] at the end of the sentence. Do you wish for no odd numbers are adjacent and no even numbers are adjacent? or is 1243 an acceptable permutation of this type and we allow even numbers to be adjacent as well?JMoravitz –JMoravitz 2016-06-14 15:05:49 +00:00 Commented Jun 14, 2016 at 15:05
@JMoravitz If no two neighboring numbers are odd, then also no two neighboring numbers are even, except in the case that n n is even and the endpoints are chosen to be both odd. So the question is really three questions: No consecutive odds, not consecutive evens, and no consecutive eithers..Mark Fischler –Mark Fischler 2016-06-14 15:08:12 +00:00 Commented Jun 14, 2016 at 15:08
@MarkFischler I gave an example of 1243 1243 where no odd numbers are adjacent, however an even number is adjacent to another even number. Such a possibility can occur when n n is even, but not when n n is odd.JMoravitz –JMoravitz 2016-06-14 15:09:02 +00:00 Commented Jun 14, 2016 at 15:09
@JMoravitz Yes, I mis-spoke. I have corrected my comment.Mark Fischler –Mark Fischler 2016-06-14 15:11:12 +00:00 Commented Jun 14, 2016 at 15:11
The trick is to make a permutation of odd numbers and even numbers apart from each other. Now consider you have two permutations, pick one from each permutation over and over. Think about which one you should start with, which is dependent on if n n is even or odd. Jack D'Aurizio gives the solution that follows from this reasoning.Jasper –Jasper 2016-06-14 15:16:53 +00:00 Commented Jun 14, 2016 at 15:16
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Break into cases: Either n=2 k n=2 k is even or n=2 k+1 n=2 k+1 is odd.
Arrange the k k even numbers. Give a bit of extra space to the left and right of each. Using these as a barrier, we place the odd numbers into the gaps. There are k+1 k+1 gaps total.
In the first case, exactly one of these gaps will remain unused. In the second case, all gaps will be used. Pick which gap if any is unused. Then arrange the odd numbers within the gaps using at most one odd number per gap.
Apply multiplication principle and conclude.
There will be k!(k+1)!k!(k+1)! arrangements in both cases.
Rewording to remove mention of k k, there will be ⌊n 2⌋!⌊n 2+1⌋!⌊n 2⌋!⌊n 2+1⌋! total arrangements.
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edited Jun 14, 2016 at 15:26
answered Jun 14, 2016 at 15:10
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As discussed in the comments, this question can have three possible meanings, and each meaning is an interesting question in itself.
Say you want to count permutations with no two consecutive odd numbers and no two consecutive even numbers. Then the solution of Jack D'aurizio holds; note that for odd n=2 k+1 n=2 k+1 this solution is (k+1)!k!(k+1)!k! (observe that the second n/2 n/2 is in a ceiling operator, not a floor).
Say you just require no consecutive odd numbers and n=2 k n=2 k is even. (This problem is isomorphic to no consecutive even numbers for the same n n).
Then there are two classes of admissible permutations:
Pure alternation of odd and even -- as before, there are 2 k!k!2 k!k! such permutations, since each of the odds and evens can be ordered independently, and the first number could be odd or even.
Two odds are on the endpoints. Here, there are k−1 k−1 positions for the inevitable pair of adjacent even numbers. Again each of the odds and evens can be ordered independently, so this class numbers (k−1)k!k!(k−1)k!k!.
Add those together and the answer for no two odds adjacent, when n=2 k n=2 k, is (k+1)!k!(k+1)!k!.
If n=2 k+1 n=2 k+1 is odd, then the problem of no consecutive odd numbers is the same as no consecutive odds and no consecutive evens, since the endpoints need to be odd anyway, so the answer in that case is (k+1)!k!(k+1)!k!. So the answer you are looking for now that the question has been changed to no consecutive odds only is
(k+1)!k!(k+1)!k!
for either n=2 k n=2 k or n=2 k+1 n=2 k+1.
Finally, if n=2 k+1 n=2 k+1 is odd, then the problem of no consecutive evens is the most interesting of these problems. The way to solve this is to notice that either the first position must contain an odd number, or there must be a sequence of EOO somewhere in the permutation, and before that the sequence must be EOEOEO...EOEOO. So the string starts with between 0 0 and k k EO pairs, each one of which contributes two possible arrangements of odds and evens (since the numbers after the EOO squense must alternate parity), except if you start with k k pairs, there is only one arrangement. That makes n n possible arrangements in total.
So if n=2 k+1 n=2 k+1 is odd, then the number of permutations with no consecutive even numbers is n(k+1)!k!n(k+1)!k!.
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answered Jun 14, 2016 at 15:52
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Children and teen mannequins are very big in this industry, they show mothers how items of clothing would look on their precious bundles of joys. It gives your store a great advantage over your competitors. Child mannequins can come in three forms, namely baby, child and teenager which allow you to display your merchandise.
Parents have a hard time thinking of how a clothing item will look on their children so help them out with our high quality child mannequins. They come in all “ages” for both genders and even unisex, they also have numerous positions (baby mannequins come in crawl, sitting or standing positions) and have angelic faces.
Pick the ones you need for your retail storehere.
Mannequin Style By Pose / Audience
#7. Sexy Mannequins
The sexy mannequins are hyper realistic and are for specific stores only, these include lingerie retail stores and of course, adult shops. They are constructed from high quality fiberglass, are sculpted in great detail and have detachable limbs in order to ease the process of dressing and undressing, they also come mostly in female form.
These gorgeous mannequins have voluptuous forms and come in very provocative and lascivious poses, all in other to get the customer in to the “mood” of shopping for some sexy apparel.
Find a beautiful and sexy mannequin for your storehere.
#8. Sport Mannequins
As in the case of the sexy mannequins we discussed above, sport mannequins come as realistic, headless or abstract, in various colors and in very specific poses. Some might be set while running while others come in yoga poses, playing golf or soccer. Indifferent of their posture, they are the perfect display mannequins for sports retailers.
If you sell sports apparel we recommend that you check out our very muscular and athleticmale sport mannequins and our fitfemale sport mannequins.
Mannequin Style By Color
#9. Black Mannequins
We are not referring toAfrican-American mannequins but to literal black mannequins. These specific display mannequins are in general abstract mannequins that have a glossy or mate finish. They are purchased and displayed by high end retailers and are very flashy, yet classy and contemporary.Black mannequins look fantastic in almost anything so give them a shot.
#10. Chrome Mannequins
Chrome display mannequins are absolutely dazzling, they are so shiny and show off such graceful postures that they are a must for modern retailers. Besides their power to enchant customers anywhere they are placed. Chrome mannequins are made from fiberglass and come as abstract mannequins or as mannequin torsos, in both genders.
Having a chrome mannequin is like showcasing a silver doll in your store and if you’re interested in how they an increase your sales, we invite you tocheck them out here.
#11. Gold Mannequins
Gold display mannequins give off the idea of luxury, because they look like a million dollars. Can you even imagine having one in your store? How they would brighten up a display or a room with that mirror like “skin” and a super shiny finish. They generally are abstract mannequins with all the features that come with them.
#12. Other color Mannequins
Custom color mannequins - they are absolutely gorgeous and while they mostly fit in the abstract category and have the same features as the rest of them, these in general are custom made and rather pricey. Also in this color theme we can discuss a little about hand painted mannequins which have something similar to a tattoo on their bodies, but again, these too are custom made.
But if you want to do something special for your customers and your store then the investment might pay off or if you wish to invest a little but do something amazing with your mannequins we have some advice on how you canspruce mannequins up.
Mannequin Types By Functionality
Moving on past the dazzling style of the mannequins we have to take into consideration their functionality.
#1. Dress forms
Also known as tailor’s dummies and mannequins, display dress forms are used for fashion design. These are professional products that come in a variety of forms and sizes, some even having special features that help the dressmakers create the perfect items of clothing for their customers. They can also be used in retail stores to create fantastic displays and effects.
The special features we mentioned earlier include collapsible shoulders to make it easier to remove clothing items from it. Moreover they can have adjustable height, rolling bases to help with moving around, magnetic shoulders which allow for the adding of arms, some are even expandable which aid in the making of various sizes.
They are generally made from heavy duty canvases that are designed to withstand the test of time and of course scissors and needles. Besides the general dress form that is a torso with a wire bottom, we have suspended bodies which include arms and legs and are suspended from the neck.
We have high quality dress forms for any type of fashion design need, check them outhere.
#2. Ghost Mannequins
Also known as photography mannequins or invisible, these exquisite types are absolutely fantastic for retailers, photographers or whoever might need them. This bestseller type of mannequin is very popular nowadays because they focus strictly on the customer, they can imagine how the item of clothing looks on them without being distracted by the rest of the mannequin.
Their fantastic advantage, besides the fact that they come in both genders, is that they have multiple removable parts to make the clothes seem like they just float. These removable parts include multiple neck layers either v shaped or round, arms, hands, legs, and some parts of the torso.
If you want to see exactly what we’re talking about check out ourghost mannequins.
#3. Flexible Mannequins
Flexible or poseable mannequins are very special and sought after these days. It’s easy to understand why. Business owners can set these mannequins in incredible poses that defy imagination.
While other mannequins are made of plastic or fiberglass in stiff postures, flexible mannequins can be set in different poses, taking into consideration the type of flexibility they dispose of. They either have shoulder, elbow or knee joints or are made from a high density foam that can be modeled into any position. Even their heads can be tilted or moved from side to side.
The head is generally very realistic in terms of facial features and make-up and it is possible to find bothhere.
The best thing about them is that it’s just like dressing a real human, you don’t have to detach parts or unscrew anything.
#4. Torso Mannequins
Torso mannequins are widely available, easy to transport and can be made from hard plastic. This is also the building block for any type of mannequin, sometimes having the possibility to attach arms, legs and heads.
They are also more economical than other mannequin options so feel free to check out our excellent prices ontorso mannequins.
#5. Standalone Mannequin Parts
Neck pieces:
Mannequin necklace displays stands are perfect for jewelry stores because they represent the upper part of the torso. They come either in the form of the torso from the waist up with no arms or head, as the neck and décolletage, or in some cases with half of the face, Velvet, plastic, foam or various types of canvases are used to create this particular type of mannequin part, it is up to you to choose which one suits your store and make the jewelry or even scarves pop out from the rest.
Hands & arms:
Mannequin hands and arms are used to display rings and bracelets or other types of jewelry, gloves, and they can also be used by nail artists to exercise their craft. They can have a stiff hand, that comes in a preset pose or they can have all the finger joints so you can move the fingers however you wish. Mannequin display hands and arms come in various lengths and are made from various materials such as plastic, wood, high density foam and so on. The hand poses are perfect for displaying tasteful rings and bracelets, most of them being very graceful and almost inviting.
Legs:
These mannequin legs are generally used to display stockings because they begin from the mid-thigh area and continue down to the foot. It doesn’t matter what kind of stocking you sell, leg mannequins can showcase any type of stocking and socks.
Calf:
Calf mannequin are used to display leg bracelets or anklets, high heels and boots.
Foot:
Foot mannequins are those that begin from the ankle and continue down. They can be flat to help showcase or designed to fit into high heels. Also they can be made from a wide variety of materials, can come in multiple colors and with various designs.
#6. Display Mannequins
Display mannequins can mean a lot of things, as it includes a wide selection of some of the best mannequin types out there, being used pretty much in all windows displays, from the the small corner store to the high end retail stores or in rather unusual places such as Las Vegas hotels.
If you lack the perfect mannequins for your displays, be sure to check out complete selection of display mannequins. For our beloved blog readers, we're listing here a 10% Off Coupon here that works on all our display mannequins - just use the code DSPL100 on the checkout and you'll see a 10% price reduction on all display mannequins. This code works even if specific mannequins are already under another site-wide promotion (don't you just hate those "You can't use a coupon on this product." errors?).
This pretty much wraps up the main types of mannequins according to style and functionality. Of course, there are some more, but they're not relevant to retail businesses.
We hope this clears up any kind of concern regarding mannequins and that it will help you make an informed decision when you decide to purchase one for your store. If you need advice, feel free to contact us!
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Verified Reviews
Product Reviews
Site Reviews
05/13/2025
Leslie S.
United States
She's Beautiful!
Good. Her face is gorgeous. And she's perfect for what I needed and went together easily. She did have a blemish in the paint on one ankle. So it would be advisable to include your little nail polish size paint bottle with every order.
05/12/2025
A.
United States
I’m new at sewing but I am looking forward to making my own patterns and eventually dress making. The size of this dressform is perfect for my size and the adjustable height helps me alot. It has a collapsible shoulder and it’s pinnable. It is supported with wheels which makes it convenient for me...I’m new at sewing but I am looking forward to making my own patterns and eventually dress making. The size of this dressform is perfect for my size and the adjustable height helps me alot. It has a collapsible shoulder and it’s pinnable. It is supported with wheels which makes it convenient for me to move anywhere i need it to be. I have to make certain adjustments to specific areas of the dressform for my body type. Owning one helps motivate the creative part of me.
05/07/2025
Anglique J.
United States
Amazing and Stunning Detail
It was easy to put together, pictures did not do this mannequin justice! I recommend for anyone who is looking to buy, I do not at all regret this purchase, shipping went smooth and it arrived intact! Thank you! =D
04/23/2025
Mike P.
United States
Great display for Star Wars costumes
The mannequin itself is great and easy to assemble. However, the bolt on the stand was cheaply welded on and broke off the first day having it, I cant lock the stand in place after adjusting it.
03/11/2025
JOYCE B.
United States
Exceeded Expectations
I've previously worked in retail, so I have experience with various forms of Mannies. I shopped and purchased this particular mannequin to use for special events and in the showroom at our distribution center. She was expertly packed and easy to assemble. We appreciated every detail, even the...I've previously worked in retail, so I have experience with various forms of Mannies. I shopped and purchased this particular mannequin to use for special events and in the showroom at our distribution center. She was expertly packed and easy to assemble. We appreciated every detail, even the inclusion of white gloves for handling her. We truly like her and have named her Olivia. <3 Thank you, MannequinMall!
03/08/2025
Hollie E.
United States
Overall, very good
The product is very good quality and life-like in dimensions. The white mannequin was a bit chalky coming out of the box so we wiped it down before putting clothes on it. There was a customer service issue that was dealt with quickly and with integrity to make sure we were happy. I will shop with...The product is very good quality and life-like in dimensions. The white mannequin was a bit chalky coming out of the box so we wiped it down before putting clothes on it. There was a customer service issue that was dealt with quickly and with integrity to make sure we were happy. I will shop with you again. Thank you
11/22/2024
Advanced D.
United States
Female Mannequin is Fantastic!!
Our company purchased both the female and male mannequins to create a display for one of our clients; we were extremely happy with the quality of these mannequins! The faces on both have eyelashes that actually look real instead of being molded/painted on and the female even has red lips as if...Our company purchased both the female and male mannequins to create a display for one of our clients; we were extremely happy with the quality of these mannequins! The faces on both have eyelashes that actually look real instead of being molded/painted on and the female even has red lips as if makeup was applied. Beware, however, if purchasing shoes for these mannequins in advance of receiving them....because the molded feet are NOT flexible (like a human foot) we have had to return the shoes we purchased and exchange them for shoes at least 3 to 4 sizes larger in order to get them to fit. Not a complaint by any means, just a Cautionary Tale! Overall, we are extremely pleased with these purchases!
11/22/2024
Nicholas K.
United States
Works great!
Started using it to shoot our clothing for our website. Works awesome!
11/12/2024
Trey G.
United States
Great product!
I ordered one form to see if liked the quality and were pleased with our purchase. Placed an order for (10) more forms!
09/12/2024
Steve A.
United States
Love mannequin mall
Wonderful display, want to make sure it’s secure from being knocked over. Any displays standing on one leg could easily come tumbling down if it’s in the wrong area. My only suggestion. Great for a corner display.
09/12/2024
Steve A.
United States
Football player mannequin
What a great company to work with I love mannequin mall I have 19 football players up in my man cave now.
08/31/2024
Diane G.
United States
Dress form of my dreams
I have longed for such a dress form since I was 16 years old, 56 years ago. I am very excited to finally own one and look forward to creating beautiful garments!
08/07/2024
Marc G.
United States
Classic and modern
It has both a very classic and modern look to it. Overall very nice. The hands don't close all the way so it leaves a gap. Otherwise it would have received a 5
07/07/2024
Joseph
United States
Perfect! Love it, will buy again
06/27/2024
Mrs L.
United States
Great dress form
This was a gift given to me and the quality and appearance is perfect. Measurements fit me perfectly . A beautiful dress form. Well made wonderful form . I love it
06/24/2024
Edward E.
United States
Great product
I bought this product to display my HALO Master Chief armor costume…and, it is awesome! Highly recommend.
06/24/2024
Symia
United States
I’m so obsessed, shipping and delivery was quick!
06/13/2024
Tom
United States
She’s great I named her Penelope!
05/26/2024
Liliana
United States
It was a gift and my sister loved it thank you so much ❤️💕
05/02/2024
Bryan B.
United States
MYA2
The mannequin was perfect; thank you!
04/29/2024
Marivel T.
United States
Great product.
They worked well for us.
03/20/2024
Anthony G.
United States
Perfect
Great quality and arrived quickly.
03/15/2024
Carol J.
United States
Model MM-Z25 mannequin looks even more beautiful in real life!
Upon receiving this mannequin, I was impressed by how well she was packed! The biggest parcel was a double box for extra strength and each of the limbs was inside its own bubble pack bag. The pieces were all surrounded by little spongy scraps of what looks like scuba diving wet suit material. This...Upon receiving this mannequin, I was impressed by how well she was packed! The biggest parcel was a double box for extra strength and each of the limbs was inside its own bubble pack bag. The pieces were all surrounded by little spongy scraps of what looks like scuba diving wet suit material. This is good for shock absorption. The boxes were clearly marked "fragile, handle with care". The face of the mannequin had a foam mask to protect her nose from breaking. As for the mannequin herself (model MM-Z25), quality is great, all the pieces were there and the stand has a foot support as well as a calf support for better dressing options. The joints are all metal threads and posts, so they can withstand a lot of position changes without wear and tear issues. All it takes is an Allen wrench to move or remove any limbs. The mannequin's face is pleasant (not pouty like some can be), she has nice eyelashes and light colored lipstick and rouge. She has holes in her ears for earrings. I had no problem assembling her. Put one foot on the stand first and work your way up from there. She comes apart easily for dressing. We remove her from the stand and sit her on our sewing studio couch sometimes for a little change and she looks so real! We have several "normal people wigs" that fit her head fine.
03/13/2024
Susan
United States
I love it. It is beautiful high quality 😍
03/12/2024
Alex G.
United States
so thrilled with my dress form
fantastic experience! as described, nicely made & arrived quickly. for the price I think its outstanding.
03/07/2024
Oconee M.
United States
Excellent service and fast shipping!
This is the third mannequin we have purchased from here. Our original choice was out-of-stock, but we were contacted promptly and offered a substitute. Shipping was super fast! Easy to assemble!
03/05/2024
SUZETTE A.
United States
Prompt Service -Great Product!
Smooth and Easy!
02/13/2024
Patrick G.
United States
Beautiful and high quality mannequin
Very pleased with the service provided. The mannequin is high quality looks stunning.
02/13/2024
Bryan B.
United States
MYA 2
She is great! We bought her for our clothing store. We received her very fast and she was packaged exceptionally. We would love to buy many more in the future. Thank you for making our first purchase from you wonderful.
02/12/2024
Maureen H.
United States
Beautiful
She’s beautiful! Perfect for the flower exhibit she will be in at the art museum
05/31/2025
Jeanie B.
The head on the mannequin is not proportionate to the body. I am not sure if the mannequin not having hair formed onto the head makes the difference or not, but other than that, it is exactly what I expected. Thank you.
05/31/2025
Anna C.
Accommodating and efficient
05/19/2025
Anglique J.
The quality honestly and also how you ship it keeps it intact!
05/13/2025
Leslie S.
Quality Control. All six of the Lisa mannequins ordered had paint blemishes which needed painting touch ups. And the discount for hands on one that had bubbled up/chipped paint on hands was inadequate. Replacement hands unavailable.
04/12/2025
Mai V.
I have bought 3 mannequins so far and your customer service had been very helpful and I have received every thing that I purchased.
03/30/2025
Symone G.
The mannequin came broken and bent. I had to have several packages delivered with replacement parts
02/18/2025
eric h.
Great service
01/27/2025
eric h.
Great company
09/24/2024
Steve A.
Great customer service and product
09/07/2024
Diane G.
Excellent product. Just as I hoped for. Arrived in perfect condition and promptly.
08/07/2024
Heather H.
I bought non-refundable adjustable mannequin so I had to ask if it's able to pose that I wanted. And it worked as they explained. Thank you!
06/28/2024
Jeannine E.
Good product.
06/27/2024
Naomie G.
The customer service is really good , thanks again
06/24/2024
Edward E.
Great product, quick shipping
06/06/2024
Sue P.
When I needed customer support, they were on the case in a flash. And, my new dress maker's is made well and was delivered quickly therefore, color me one happy customer.
05/02/2024
Donald S.
Great product and shipping time.
03/19/2024
Julian F.
Quick delivery and great product
03/12/2024
Richard
Great mannequin and good service
03/07/2024
Oconee M.
Good quality, fast shipping
02/22/2024
Patrick G.
The mannequin we ordered looks very high end and excellent quality.
01/29/2024
Becky
good quality merchandise, I don't think anyone would be disappointed with any of these mannequinns.
01/24/2024
Misun L.
On time shipping and fast response to me about requesting.
12/26/2023
JULIANNA H.
Great product and shipped pronto
12/02/2023
Dunella
I could not initially set up the mannequin, but staff by phone worked consistently with me.
11/23/2023
Tyrone C.
Really nice mannequin and resolve my issues timely and with a solution to my problem.
11/17/2023
Ken B.
Product was flawed. Broken hand, chipped back and broken pole! However, the response to my issues was excellent. Just received replacement parts.
10/21/2023
Elbert M.
The high quality of the product.
10/18/2023
John C.
They're cool
10/11/2023
Diana R.
Options
10/05/2023
Jeff R.
The product I purchased is great
REVIEWS
Product Reviews
Site Reviews
05/13/2025
Leslie S.
United States
She's Beautiful!
Good. Her face is gorgeous. And she's perfect for what I needed and went together easily. She did have a blemish in the paint on one ankle. So it would be advisable to include your little nail polish size paint bottle with every order.
05/12/2025
A.
United States
I’m new at sewing but I am looking forward to making my own patterns and eventually dress making. The size of this dressform is perfect for my size and the adjustable height helps me alot. It has a collapsible shoulder and it’s pinnable. It is supported with wheels which makes it convenient for me...I’m new at sewing but I am looking forward to making my own patterns and eventually dress making. The size of this dressform is perfect for my size and the adjustable height helps me alot. It has a collapsible shoulder and it’s pinnable. It is supported with wheels which makes it convenient for me to move anywhere i need it to be. I have to make certain adjustments to specific areas of the dressform for my body type. Owning one helps motivate the creative part of me.
05/07/2025
Anglique J.
United States
Amazing and Stunning Detail
It was easy to put together, pictures did not do this mannequin justice! I recommend for anyone who is looking to buy, I do not at all regret this purchase, shipping went smooth and it arrived intact! Thank you! =D
04/23/2025
Mike P.
United States
Great display for Star Wars costumes
The mannequin itself is great and easy to assemble. However, the bolt on the stand was cheaply welded on and broke off the first day having it, I cant lock the stand in place after adjusting it.
03/11/2025
JOYCE B.
United States
Exceeded Expectations
I've previously worked in retail, so I have experience with various forms of Mannies. I shopped and purchased this particular mannequin to use for special events and in the showroom at our distribution center. She was expertly packed and easy to assemble. We appreciated every detail, even the...I've previously worked in retail, so I have experience with various forms of Mannies. I shopped and purchased this particular mannequin to use for special events and in the showroom at our distribution center. She was expertly packed and easy to assemble. We appreciated every detail, even the inclusion of white gloves for handling her. We truly like her and have named her Olivia. <3 Thank you, MannequinMall!
03/08/2025
Hollie E.
United States
Overall, very good
The product is very good quality and life-like in dimensions. The white mannequin was a bit chalky coming out of the box so we wiped it down before putting clothes on it. There was a customer service issue that was dealt with quickly and with integrity to make sure we were happy. I will shop with...The product is very good quality and life-like in dimensions. The white mannequin was a bit chalky coming out of the box so we wiped it down before putting clothes on it. There was a customer service issue that was dealt with quickly and with integrity to make sure we were happy. I will shop with you again. Thank you
11/22/2024
Advanced D.
United States
Female Mannequin is Fantastic!!
Our company purchased both the female and male mannequins to create a display for one of our clients; we were extremely happy with the quality of these mannequins! The faces on both have eyelashes that actually look real instead of being molded/painted on and the female even has red lips as if...Our company purchased both the female and male mannequins to create a display for one of our clients; we were extremely happy with the quality of these mannequins! The faces on both have eyelashes that actually look real instead of being molded/painted on and the female even has red lips as if makeup was applied. Beware, however, if purchasing shoes for these mannequins in advance of receiving them....because the molded feet are NOT flexible (like a human foot) we have had to return the shoes we purchased and exchange them for shoes at least 3 to 4 sizes larger in order to get them to fit. Not a complaint by any means, just a Cautionary Tale! Overall, we are extremely pleased with these purchases!
11/22/2024
Nicholas K.
United States
Works great!
Started using it to shoot our clothing for our website. Works awesome!
11/12/2024
Trey G.
United States
Great product!
I ordered one form to see if liked the quality and were pleased with our purchase. Placed an order for (10) more forms!
09/12/2024
Steve A.
United States
Love mannequin mall
Wonderful display, want to make sure it’s secure from being knocked over. Any displays standing on one leg could easily come tumbling down if it’s in the wrong area. My only suggestion. Great for a corner display.
09/12/2024
Steve A.
United States
Football player mannequin
What a great company to work with I love mannequin mall I have 19 football players up in my man cave now.
08/31/2024
Diane G.
United States
Dress form of my dreams
I have longed for such a dress form since I was 16 years old, 56 years ago. I am very excited to finally own one and look forward to creating beautiful garments!
08/07/2024
Marc G.
United States
Classic and modern
It has both a very classic and modern look to it. Overall very nice. The hands don't close all the way so it leaves a gap. Otherwise it would have received a 5
07/07/2024
Joseph
United States
Perfect! Love it, will buy again
06/27/2024
Mrs L.
United States
Great dress form
This was a gift given to me and the quality and appearance is perfect. Measurements fit me perfectly . A beautiful dress form. Well made wonderful form . I love it
06/24/2024
Edward E.
United States
Great product
I bought this product to display my HALO Master Chief armor costume…and, it is awesome! Highly recommend.
06/24/2024
Symia
United States
I’m so obsessed, shipping and delivery was quick!
06/13/2024
Tom
United States
She’s great I named her Penelope!
05/26/2024
Liliana
United States
It was a gift and my sister loved it thank you so much ❤️💕
05/02/2024
Bryan B.
United States
MYA2
The mannequin was perfect; thank you!
04/29/2024
Marivel T.
United States
Great product.
They worked well for us.
03/20/2024
Anthony G.
United States
Perfect
Great quality and arrived quickly.
03/15/2024
Carol J.
United States
Model MM-Z25 mannequin looks even more beautiful in real life!
Upon receiving this mannequin, I was impressed by how well she was packed! The biggest parcel was a double box for extra strength and each of the limbs was inside its own bubble pack bag. The pieces were all surrounded by little spongy scraps of what looks like scuba diving wet suit material. This...Upon receiving this mannequin, I was impressed by how well she was packed! The biggest parcel was a double box for extra strength and each of the limbs was inside its own bubble pack bag. The pieces were all surrounded by little spongy scraps of what looks like scuba diving wet suit material. This is good for shock absorption. The boxes were clearly marked "fragile, handle with care". The face of the mannequin had a foam mask to protect her nose from breaking. As for the mannequin herself (model MM-Z25), quality is great, all the pieces were there and the stand has a foot support as well as a calf support for better dressing options. The joints are all metal threads and posts, so they can withstand a lot of position changes without wear and tear issues. All it takes is an Allen wrench to move or remove any limbs. The mannequin's face is pleasant (not pouty like some can be), she has nice eyelashes and light colored lipstick and rouge. She has holes in her ears for earrings. I had no problem assembling her. Put one foot on the stand first and work your way up from there. She comes apart easily for dressing. We remove her from the stand and sit her on our sewing studio couch sometimes for a little change and she looks so real! We have several "normal people wigs" that fit her head fine.
03/13/2024
Susan
United States
I love it. It is beautiful high quality 😍
03/12/2024
Alex G.
United States
so thrilled with my dress form
fantastic experience! as described, nicely made & arrived quickly. for the price I think its outstanding.
03/07/2024
Oconee M.
United States
Excellent service and fast shipping!
This is the third mannequin we have purchased from here. Our original choice was out-of-stock, but we were contacted promptly and offered a substitute. Shipping was super fast! Easy to assemble!
03/05/2024
SUZETTE A.
United States
Prompt Service -Great Product!
Smooth and Easy!
02/13/2024
Patrick G.
United States
Beautiful and high quality mannequin
Very pleased with the service provided. The mannequin is high quality looks stunning.
02/13/2024
Bryan B.
United States
MYA 2
She is great! We bought her for our clothing store. We received her very fast and she was packaged exceptionally. We would love to buy many more in the future. Thank you for making our first purchase from you wonderful.
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Conjugate of Square Root – Definition and Examples
Conjugate of Square Root – Definition and Examples
JUMP TO TOPIC
Defining Conjugate of Square Root
Historical Significance
Evaluating Conjugate of Square Root
Identify the Terms
Change the Sign
Properties
Elimination of Square Roots
Simplifying Complex Numbers
Unaltered Magnitude
Reversal of Sign of Imaginary Part
Addition and Subtraction
Multiplication and Division
Applications
Mathematics
Physics and Engineering
Signal Processing and Telecommunications
Computer Science
Control Systems
Exercise
Example 1
Solution
Example 2
Solution
Example 3
Solution
Example 4
Solution
Example 5
Solution
Example 6
Solution
Example 7
Solution
Example 8
Solution
The conjugate of a square root is a novel concept waiting to be understood and explored while delving into the mathematics and navigating through an intricate labyrinth, where every turn reveals.
By no means a stranger to mathematicians, engineers, or scientists, the notion of conjugates is fundamental in simplifying expressions and solving equations, particularly those involving square roots.
This article is a journey into understanding how conjugates of square roots work, their applications, and the elegance they bring to mathematical computations. It provides an immersive experience, whether you’re a seasoned math enthusiast or a novice keen on discovering new mathematical ideas.
Defining Conjugate of Square Root
In mathematics, the concept of a conjugate is a fundamental tool to simplify expressions involving square roots. Specifically, when dealing with square roots, the conjugate is a method used to ‘rationalize the denominator‘ or simplify complex numbers.
For example, Suppose we have a square root expression such as √a + √b. Its conjugate is formed by changing the sign in the middle of the two terms, resulting in √a – √b.
For complex numbers, the conjugate is also an important concept. If we have a complex number like a + bi, where a and b are real numbers, and i is the square root of -1 (the imaginary unit), the conjugate of this complex number is a – bi.
The importance of the conjugate comes into play when we multiply the original expression by its conjugate. Multiplying an expression by its conjugate eliminates the square root (or the imaginary part in the case of complex numbers) due to the difference in squares’ identity, thus simplifying the expression.
Historical Significance
The concept of a conjugate, which is the cornerstone for understanding the conjugate of a square root, is a mathematical tool with its roots firmly placed in the development of algebra and complex number theory.
The historical development of conjugates is tightly intertwined with the evolution of algebra itself. The idea to “rationalize the denominator“, or remove the square roots from the denominator of a fraction, is an old technique that can be traced back to ancient mathematicians. This process inherently uses the principle of conjugates, even if the term “conjugate” was not explicitly used.
The explicit use of the term “conjugate” and the formal concept of conjugates took form with the development of complex numbers in the 16th to 18th centuries. The Italian mathematician Gerolamo Cardano is often credited with the first systematic use of complex numbers in his work on the solutions of cubic equations, published in his 1545 book “Ars Magna.”
However, the concept of the complex conjugate as we understand it today was not formalized until the 19th century, as mathematicians like Jean-Robert Argand and Carl Friedrich Gauss developed a deeper understanding of complex numbers. They recognized that every non-real complex number and its conjugate could be represented as mirror images in the Argand plane (a geometric representation of complex numbers), and these pairs of complex numbers had useful mathematical properties.
The notion of a conjugate has since become a fundamental tool in many mathematics, physics, engineering, and related fields. While it’s challenging to pinpoint the exact origin of the concept of “conjugate of a square root” itself, it’s clear that its underlying principle is closely tied to the broader historical development of algebra and complex number theory.
Evaluating Conjugate of Square Root
Finding the conjugate of a square root term is a straightforward process. It essentially involves changing the sign between the two terms in the expression. Let’s go through the process in detail:
Consider a mathematical expression containing square roots in the form a + √b. In this expression, ‘a‘ and ‘b‘ are any real numbers. The term ‘a‘ could be a real number, another square root, or even zero.
The conjugate of this expression is formed by changing the sign between the terms ‘a‘ and ‘√b‘. So, the conjugate of ‘a + √b‘ would be ‘a – √b‘.
Similarly, if the expression were ‘a – √b‘, its conjugate would be ‘a + √b‘.
Here are the steps broken down:
Identify the Terms
First, identify the two terms you want to find the conjugate in your expression. The expression should be ‘a + √b’ or ‘a – √b’.
Change the Sign
Change the sign between the terms. If it’s a plus sign, change it to a minus sign. If it’s a minus sign, change it to a plus sign.
That’s it. You’ve found the conjugate of the square root expression.
As an example, consider the expression 3 + √2. The conjugate of this expression would be 3 – √2. If you have the expression 5 – √7, the conjugate would be 5 + √7.
Properties
The conjugate of a square root has some important properties that make it an indispensable tool in mathematics. Here are some of the most significant properties:
Elimination of Square Roots
One of the main uses of the conjugate is to eliminate square roots in an expression. Multiplying a binomial expression with a square root (such as √a + b) by its conjugate (√a – b) results in the difference of squares. This means the square root term is squared, effectively removing the square root. For example, multiplying (√a + b)(√a – b) gives us a – b².
Simplifying Complex Numbers
The conjugate is also used to simplify complex numbers, where the square root of -1 (denoted as ‘i’) is involved. The conjugate of a complex number (a + bi) is (a – bi). If we multiply a complex number by its conjugate, we eliminate the imaginary part: (a + bi)(a – bi) = a² + b², a real number.
Unaltered Magnitude
When we take the conjugate of a complex number, its magnitude (or absolute value) remains unchanged. The magnitude of a complex number (a + bi) is √(a² + b²), and the magnitude of its conjugate (a – bi) is also √(a² + b²).
Reversal of Sign of Imaginary Part
The conjugate of a complex number has the same real part but an opposite sign for the imaginary part.
Addition and Subtraction
The conjugate of the sum (or difference) of two complex numbers equals their conjugates’ sum (or difference). In other words, if z₁ and z₂ are two complex numbers, then the conjugate of (z₁ ± z₂) is equal to the conjugate of z₁ ± the conjugate of z₂.
Multiplication and Division
The conjugate of the product (or quotient) of two complex numbers equals the product (or quotient) of their conjugates. Thus, if z₁ and z₂ are two complex numbers, then the conjugate of (z₁ z₂) is equal to the conjugate of z₁ the conjugate of z₂. The same holds for division.
These properties provide a set of powerful tools that can be used to simplify mathematical expressions, solve equations, and perform complex computations.
Applications
The concept of the conjugate of square roots, and more broadly, the conjugate of complex numbers, find widespread application across various fields of study, not only in pure mathematics but also in engineering, physics, computer science, and more. Below are some applications in different fields:
Mathematics
In algebra, conjugates are frequently used to rationalize the denominator of fractions. The conjugate is used in complex analysis to prove fundamental results such as the Cauchy-Riemann equations. It is also used to simplify complex number expressions.
Physics and Engineering
Complex numbers’ conjugates help analyze phase changes and amplitude in studying waves and oscillations. In electrical engineering, conjugates simplify the calculation of power in AC circuits. Quantum mechanics also utilizes complex conjugates, as the normalization condition of wave functions involves taking the complex conjugate.
Signal Processing and Telecommunications
In digital signal processing and telecommunications, the complex conjugate is used to calculate the power spectrum of a signal and also in the correlation and convolution of signals.
Computer Science
Complex numbers and conjugates are used in computer graphics, especially when rendering and transformations are involved. They are utilized to represent rotations, transformations, and color operations.
Additionally, the conjugate gradient method in optimization problems is another example of applying conjugates. This method is widely used for solving systems of linear equations and finding the minimum of a function.
Control Systems
Conjugates help in analyzing the stability of control systems. The roots of the characteristic equation of a control system must be in the left half of the complex plane for the system to be stable. The roots will either be real or complex conjugate pairs.
These are just a few examples. The mathematical tool of conjugates is so versatile and powerful that it is utilized in many more areas and various ways.
Exercise
Example 1
Simplifying a Fraction
Simplify the expression 2/(3+√5).
Solution
We use the conjugate of the denominator to rationalize it as follows:
2/(3+√5) = 2 (3-√5) / ((3+√5) (3-√5))
2/(3+√5) = 2 (3-√5) / (9 – 5)
2/(3+√5) = 2 (3-√5) / 4
2/(3+√5) = 0.5 (3 – √5)
Example 2
Simplifying a Fraction
Simplify the expression 1/(√7 – 2).
Solution
We use the conjugate of the denominator to rationalize it as follows:
1/(√7 – 2) = (√7 + 2) / ((√7 – 2) (√7 + 2))
1/(√7 – 2) = (√7 + 2) / (7 – 4)
1/(√7 – 2) = (√7 + 2) / 3
Example 3
Multiplying a Complex Number by its Conjugate
Calculate the result of (2 + 3i) (2 – 3i).
Solution
This is a direct application of the conjugate:
(2 + 3i) (2 – 3i) = 2² + (3i)²
= 4 – 9
= -5
Example 4
Multiplying a Complex Number by its Conjugate
Calculate the result of (7 – 5i) (7 + 5i).
Solution
This is a direct application of the conjugate:
(7 – 5i) (7 + 5i)
= 7² + (5i)²
= 49 – 25
= 24
Example 5
Finding the Conjugate of a Complex Number
Find the conjugate of 6 – 2i.
Solution
The conjugate of a complex number is found by reversing the sign of its imaginary part.
The conjugate of (6 – 2i) is:
6 + 2i
Example 6
Finding the Conjugate of a Complex Number
Find the conjugate of 3 + 7i.
Solution
The conjugate of a complex number is found by reversing the sign of its imaginary part.
Conjugate of (3 + 7i) is :
3 – 7i
Example 7
Multiplying Square Roots by their Conjugates
Calculate the result of (√3 + √2) (√3 – √2).
Solution
This is a direct application of the conjugate:
(√3 + √2) (√3 – √2)
= (√3)² – (√2)²
= 3 – 2
= 1
Example 8
Multiplying Square Roots by their Conjugates
Calculate the result of (√5 + √7) (√5 – √7).
Solution
This is a direct application of the conjugate:
(√5 + √7) (√5 – √7)
= (√5)² – (√7)²
= 5 – 7
= -2
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11214 | https://catalogimages.wiley.com/images/db/pdf/047139761X.07.pdf | 7 THE k-OUT-OF-n SYSTEM MODEL An n-component system that works (or is “good”) if and only if at least k of the n components work (or are good) is called a k-out-of-n:G system. An n-component system that fails if and only if at least k of the n components fail is called a k-out-of-n:F system. Based on these two definitions, a k-out-of-n:G system is equivalent to an (n −k + 1)-out-of-n:F system. The term k-out-of-n system is often used to indicate either a G system or an F system or both. Since the value of n is usually larger than the value of k, redundancy is generally built into a k-out-of-n system. Both parallel and series systems are special cases of the k-out-of-n system.
A series system is equivalent to a 1-out-of-n:F system and to an n-out-of-n:G system while a parallel system is equivalent to an n-out-of-n:F system and to a 1-out-of-n:G system.
The k-out-of-n system structure is a very popular type of redundancy in fault-tolerant systems. It finds wide applications in both industrial and military systems.
Fault-tolerant systems include the multidisplay system in a cockpit, the multiengine system in an airplane, and the multipump system in a hydraulic control system. For example, it may be possible to drive a car with a V8 engine if only four cylinders are firing. However, if less than four cylinders fire, then the automobile cannot be driven.
Thus, the functioning of the engine may be represented by a 4-out-of-8:G system.
The system is tolerant of failures of up to four cylinders for minimal functioning of the engine. In a data processing system with five video displays, a minimum of three displays operable may be sufficient for full data display. In this case the display sub-system behaves as a 3-out-of-5:G system. In a communications system with three transmitters, the average message load may be such that at least two transmitters must be operational at all times or critical messages may be lost. Thus, the transmis-231 232 THE k-OUT-OF-n SYSTEM MODEL sion subsystem functions as a 2-out-of-3:G system. Systems with spares may also be represented by the k-out-of-n system model. In the case of an automobile with four tires, for example, usually one additional spare tire is equipped on the vehicle. Thus, the vehicle can be driven as long as at least 4-out-of-5 tires are in good condition.
Among applications of the k-out-of-n system model, the design of electronic cir-cuits such as very large scale integrated (VLSI) and the automatic repairs of faults in an on-line system would be the most conspicuous. This type of system demonstrates what is called the voting redundancy. In such a system, several parallel outputs are channeled through a decision-making device that provides the required system func-tion as long as at least a predetermined number k of n parallel outputs are in agree-ment.
In this chapter, we provide a detailed coverage on reliability evaluation of the k-out-of-n systems. Methods for finding both the exact and the approximate system reliability values are introduced. The performance measures of both nonrepairable and repairable k-out-of-n systems are addressed. In addition, the weighted k-out-of-n system model is discussed in this chapter. In our discussions, it is assumed that the working of the components is independent of one another unless otherwise specified.
7.1 SYSTEM RELIABILITY EVALUATION In this section, we concentrate on techniques for reliability evaluation of k-out-of-n:G systems. The k-out-of-n:G system with i.i.d. components is first studied. Several approaches for system reliability evaluation, when the components are not necessar-ily s-identical, are then introduced in detail. Finally, bounds on system reliability, when components are not necessarily s-independent, are discussed.
Notation • n: number of components in the system • k: minimum number of components that must work for the k-out-of-n:G system to work • pi: reliability of component i, i = 1, 2, . . . , n • p: reliability of each component when all components are i.i.d.
• qi: unreliability of component i, qi = 1 −pi, i = 1, 2, . . . , n • q: unreliability of each component when all components are i.i.d., q = 1 −p • Re(k, n): probability that exactly k out of n components are working • R(k, n): reliability of a k-out-of-n:G system or probability that at least k out of the n components are working, where 0 ≤k ≤n and both k and n are integers • Q(k, n): unreliability of a k-out-of-n:G system or probability that less than k out of the n components are working, where 0 ≤k ≤n and both k and n are integers, Q(k, n) = 1 −R(k, n) SYSTEM RELIABILITY EVALUATION 233 7.1.1 The k-out-of-n:G System with i.i.d. Components In a k-out-of-n:G system with i.i.d. components, the number of working components follows the binomial distribution with parameters n and p. Thus, we have Pr(exactly i components work) = n i piqn−i, i = 0, 1, 2, . . . , n.
(7.1) The reliability of the system is equal to the probability that the number of working components is greater than or equal to k: R(k, n) = n i=k n i piqn−i.
(7.2) Equation (7.2) is an explicit formula that can be used for reliability evaluation of the k-out-of-n:G system.
If we apply the pivotal decomposition to component n or directly use equation (7.26) developed by Rushdi , the system reliability of a k-out-of-n:G system with i.i.d. components can be expressed as R(k, n) = pR(k −1, n −1) + (1 −p)R(k, n −1) = p(R(k −1, n −1) −R(k, n −1)) + R(k, n −1) = p Pr(exactly k −1 out of n −1 components work) + R(k, n −1) = n −1 k −1 pkqn−k + R(k, n −1).
(7.3) Rearranging the terms in equation (7.3), we obtain the expression R(k, n) −R(k, n −1) = n −1 k −1 pkqn−k for n ≥k.
(7.4) Equation (7.4) represents the improvement in system reliability by increasing the number of components in the system from n −1 to n. As n increases, this improve-ment amount in system reliability will become smaller. Thus, there is an optimal design issue of determining the system size n, which will be addressed later.
Equation (7.3) can be used recursively for system reliability evaluation with the boundary condition R(k, n) = 0 for n < k.
(7.5) Using equation (7.4) and the boundary condition given in equation (7.5), we can express the reliability of a k-out-of-n:G system as follows: R(k, n) = n i=k [R(k, i) −R(k, i −1)] = pk n i=k i −1 k −1 qi−k.
(7.6) 234 THE k-OUT-OF-n SYSTEM MODEL From the equations for system reliability given above, we can see that the relia-bility of a k-out-of-n:G system with i.i.d. components is a function of n, k, and p.
An increase in n or p or both or a decrease in k will increase the system’s reliability.
Equation (7.4) represents the increase in system reliability by increasing the number of components in the system from n −1 to n. In the following, we give an expression for the increase in system reliability for each unit of decrease in k: R(k, n) = Pr(at least k components work) = Pr(at least k −1 components work) −Pr(exactly k −1 components work) = R(k −1, n) − n k −1 pk−1qn−k+1.
(7.7) Or equivalently, we have R(k −1, n) −R(k, n) = n k −1 pk−1qn−k+1.
(7.8) With the various expressions of R(k, n) derived so far, we can easily find the expressions of the unreliability Q(k, n) of the k-out-of-n:G system. For example, the following is obvious from equation (7.2): Q(k, n) = 1 −R(k, n) = k−1 i=0 n i piqn−i.
(7.9) To find the expression for the sensitivity of system reliability on component relia-bility in this i.i.d. case, we can take the first derivative of R(k, n) with respect to p.
Using equation (7.6), we have d R(k, n) dp = k n k pk−1qn−k.
(7.10) Exercises 1. Verify equation (7.10).
2. Find similar expressions of R(k, n) or Q(k, n) and other measures for the k-out-of-n:F systems.
3. Analyze the performance of a 3-out-of-6:G system with p1 = 0.5, p2 = 0.6, p3 = 0.7, p4 = 0.8, p5 = 0.9, and p6 = 0.95.
7.1.2 The k-out-of-n:G System with Independent Components For k-out-of-n:G systems with components whose reliabilities are not necessarily identical, we can use the concept of minimal path sets to evaluate system reliability.
However, more efficient algorithms for reliability evaluation of such systems were SYSTEM RELIABILITY EVALUATION 235 reported by Barlow and Heidtmann and Rushdi . These two algorithms have the same complexity as O(k(n −k + 1)). The use of Markov chain imbeddable structures confirms the same result. Belfore uses fast Fourier transform (FFT) and proposes an O(n(log2 n)2) algorithm for reliability evaluation of k-out-of-n:G systems. In this section, we illustrate the use of minimal path sets in system reliabil-ity evaluation. In addition, we illustrate these other approaches to deriving efficient algorithms for reliability evaluation of k-out-of-n:G systems.
Minimal Path Sets or Minimal Cut Sets Approach As discussed earlier, the relia-bility of any system is equal to the probability that at least one of the minimal path sets works. The unreliability of the system is equal to the probability that at least one minimal cut set is failed. For a minimal path set to work, each component in the set must work. For a minimal cut set to fail, all components in the set must fail. In a k-out-of-n:G system, there are n k minimal path sets and n n−k+1 minimal cut sets.
Each minimal path set contains exactly k different components and each minimal cut set contains exactly n −k + 1 components. Thus, all minimal path sets and minimal cut sets are known. The question remaining to be answered is how to find the prob-ability that at least one of the minimal path sets contains all working components or the probability that at least one minimal cut set contains all failed components.
The IE method can be used for reliability evaluation of a k-out-of-n:G system since all the minimal path sets and minimal cut sets are known. The IE method has the disadvantage of involving many canceling terms. Heidtmann and McGrady provide improved versions of the IE method for reliability evaluation of the k-out-of-n:G system. In their improved algorithms, the canceling terms are eliminated.
However, both algorithms are still enumerative in nature. For example, the formula provided by Heidtmann using minimal path sets is as follows: R(k, n) = n i=k (−1)i−k i −1 k −1 j1< j2<···< ji i ℓ=1 p jℓ.
(7.11) In this equation, for each fixed i value, the inner summation term gives us the prob-ability that i components are working properly regardless of whether the other n −i components are working or not. The total number of terms to be summed together in the inner summation series is equal to n i . If all the components are i.i.d., equation (7.11) gives another formula for reliability evaluation of a k-out-of-n:G system with i.i.d. components: R(k, n) = n i=k n i i −1 k −1 (−1)i−k pi.
(7.12) Equation (7.12) is apparently not as efficient as those given in Section 7.1.1.
The SDP method can also be used for reliability evaluation of the k-out-of-n:G systems. Like the improved IE method given in equation (7.11), the SDP method is also easy to use for the k-out-of-n:G systems. However, we will see later that there are much more efficient methods than the IE (and its improved version) and the SDP 236 THE k-OUT-OF-n SYSTEM MODEL method for evaluating k-out-of-n:G systems. In the following, we present an example to illustrate the use of minimal path sets with the IE method, Heidtmann’s improved IE method, and the SDP method for reliability evaluation of a 2-out-of-4:G system.
Example 7.1 Evaluate the reliability of a 2-out-of-4:G system with p1 = 0.91, p2 = 0.92, p3 = 0.93, and p4 = 0.94. The number of minimal path sets is equal to 4 2 = 6. We will use Si to represent the ith minimal path set as listed below: S1 = x1x2, S2 = x1x3, S3 = x1x4, S4 = x2x3, S5 = x2x4, S6 = x3x4.
With the IE method, we can calculate system reliability as follows: R(2, 4) = Pr(S1 ∪S2 ∪S3 ∪S4 ∪S5 ∪S6) = Pr(S1) + Pr(S2) + Pr(S3) + Pr(S4) + Pr(S5) + Pr(S6) −Pr(S1S2) −Pr(S1S3) −Pr(S1S4) −Pr(S1S5) −Pr(S1S6) −Pr(S2S3) −Pr(S2S4) −Pr(S2S5) −Pr(S2S6) −Pr(S3S4) −Pr(S3S5) −Pr(S3S6) −Pr(S4S5) −Pr(S4S6) −Pr(S5S6) + Pr(S1S2S3) + Pr(S1S2S4) + Pr(S1S2S5) + Pr(S1S2S6) + Pr(S1S3S4) + Pr(S1S3S5) + Pr(S1S3S6) + Pr(S1S4S5) + Pr(S1S4S6) + Pr(S1S5S6) + Pr(S2S3S4) + Pr(S2S3S5) + Pr(S2S3S6) + Pr(S2S4S5) + Pr(S2S4S6) + Pr(S2S5S6) + Pr(S3S4S5) + Pr(S3S4S6) + Pr(S3S5S6) + Pr(S4S5S6) −Pr(S1S2S3S4) −Pr(S1S2S3S5) −Pr(S1S2S3S6) −Pr(S1S2S4S5) −Pr(S1S2S4S6) −Pr(S1S2S5S6) −Pr(S1S3S4S5) −Pr(S1S3S4S6) −Pr(S1S3S5S6) −Pr(S1S4S5S6) −Pr(S2S3S4S5) −Pr(S2S3S4S6) −Pr(S2S3S5S6) −Pr(S2S4S5S6) −Pr(S3S4S5S6) + Pr(S1S2S3S4S5) + Pr(S1S2S3S4S6) + Pr(S1S2S3S5S6) + Pr(S1S2S4S5S6) + Pr(S1S3S4S5S6) + Pr(S2S3S4S5S6) −Pr(S1S2S3S4S5S6) ≈0.998467.
With equation (7.11), we have R(2, 4) = 4 i=2 (−1)i−2 i −1 1 j1< j2<···< ji i ℓ=1 p jℓ = (p1 p2 + p1 p3 + p1 p4 + p2 p3 + p2 p4 + p3 p4) SYSTEM RELIABILITY EVALUATION 237 −2(p1 p2 p3 + p1 p2 p4 + p1 p3 p4 + p2 p3 p4) + 3p1 p2 p3 p4 ≈0.998441.
With the SDP method, we have R(2, 4) = Pr(S1 ∪S2 ∪S3 ∪S4 ∪S5 ∪S6) = Pr(S1) + Pr(S1S2) + Pr(S1S2S3) + Pr(S1S2S3S4) + Pr(S1S2S3S4S5) + Pr(S1S2S3S4S5S6) = Pr(x1x2) + Pr(x1x2x3) + Pr(x1x2 x3x4) + Pr(x1x2x3) + Pr(x1x2x3x4) + Pr(x1 x2x3x4) ≈0.998441.
It is clear that the IE method involves much more calculation than either the im-proved IE method or the SDP method. Because there are many canceling terms in the IE method, the round-off errors are obvious in its final result.
Generating Function Approach by Barlow and Heidtmann Barlow and Heidtmann present two BASIC programs for reliability evaluation of k-out-of-n:G systems with independent components. The first program uses the following generating func-tion and its expanded form: gn(z) = n i=1 (qi + piz) = n i=0 Re(i, n)zi, (7.13) where z is a dummy variable. As we have defined in the notation, Re(i, n) represents the probability that there are exactly i working components in the system. Through examination of the expanded form of gn(z), we find that Re(i, n) also represents the coefficient of zi in the generating function. The BASIC program computes all Re(i, j) entries recursively. Rushdi provides better explanations of this algo-rithm. In fact, the algorithm relies on the equation R(k, n) = n i=k Re(i, n), (7.14) which is obvious based on the definition of a k-out-of-n:G system. The algorithm obtains Re(i, n) through the recursive relation Re(i, j) = q j Re(i, j −1) + p j Re(i −1, j −1), 0 ≤i ≤n, 0 ≤j ≤n, (7.15) with the boundary conditions 238 THE k-OUT-OF-n SYSTEM MODEL Re(−1, j) = Re( j + 1, j) = 0 for j = 0, 1, 2, . . . .
(7.16) Re(0, 0) = 1.
(7.17) To derive this recursive relation, first construct the following generating function: g j−1(z) = j−1 i=1 (qi + piz) = j−1 i=0 Re(i, j −1)zi.
(7.18) Since g j(z) = (q j + p jz)g j−1(z), a comparison of the coefficients of zi in both sides of the equation j i=0 Re(i, j)zi = (q j + p jz) j−1 i=0 Re(i, j −1)zi = j i=0 q j Re(i, j −1) + p j Re(i −1, j −1) zi (7.19) leads to equation (7.15).
To find out the computational complexity of this algorithm, we examine the number of entries, Re(i, j), that should be calculated with equation (7.15) utilizing boundary conditions in equations (7.16) and (7.17). As shown in Figure 7.1, the total number of entries is equal to (n −k + 1)(k + 1) −1 + 1 2(n −k)2.
Each such entry requires three basic arithmetic operations (two multiplications and one addition). We then need to use equation (7.14) to find the system reliability, which requires n −k basic arithmetic operations. As a result, the total number of basic arithmetic operations required is equal to 3 (n −k + 1)(k + 1) −1 + 1 2(n −k)2 + n −k = (n −k)(1.5n + 1.5k + 4) + 3k.
From this expression, we can see that the computational complexity of the algorithm is O(n2) when k is small (close to 1) and O(n) when k is large (close to n). Generally speaking, the complexity of this algorithm is O(n2).
The number of arithmetic operations required for system reliability evaluation can be reduced by noting that we are only interested in finding the probability that at least k components are working. Thus, the calculation of Re(i, j) can be avoided. The second BASIC program by Barlow and Heidtmann avoids calculating Re(i, j) and requires only 3k(n−k+1) arithmetic operations. This computational complexity is also achieved by the algorithm proposed by Rushdi . We will present Rushdi’s algorithm in the following section.
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(n,n) FIGURE 7.1 The Re(i, j) entries to be calculated with the algorithm by Barlow and Heidtmann.
239 240 THE k-OUT-OF-n SYSTEM MODEL Exercises 1. Consider a system with n = 5 components. Verify that the coefficient of zi for i = 0, 1, . . . , 5 does represent the probability that there are exactly i working components in the system.
2. Compute the reliability and unreliability of a 3-out-of-8:G system with the algorithm given in this section.
3. Use the generating function approach to derive a similar algorithm for the k-out-of-n:F system.
Symmetric Switching Function Approach by Rushdi This approach starts with an analysis of the structure function of the k-out-of-n:G system. The structure function φ(x) of a k-out-of-n:G system is symmetric based on Definition 4.4. It can only take two possible values (0 or 1) under the binary assumption of component and system states, like an on–off switch. This is why we name this approach the symmetric switching function approach.
In this section, xi indicates the state of component i and S(k, n), instead of φ(x), indicates the structure function of the system. Both xi and S(k, n) are binary vari-ables with a value of 1 indicating the working state and 0 indicating the failed state.
The complements of these variables are represented by xi and S(k, n), respectively.
Based on these definitions of S(k, n) and S(k, n), we have the following expressions for system reliability and system unreliability: R(k, n) = Pr(S(k, n) = 1), Q(k, n) = Pr(S(k, n) = 1).
To find an expression of the system state, we can use pivotal decomposition on the nth component, as shown below: S(k, n) = xnS(k −1, n −1) + xnS(k, n −1), (7.20) S(k, n) = xnS(k −1, n −1) + xnS(k, n −1).
(7.21) Based on these two equations, the state of a k-out-of-n:G system can be expressed as a function of the states of two subsystems with the same n−1 components. However, the minimum numbers of components required for these two subsystems to work are different. One requires at least k components to work while the other requires at least k −1 components to work. These two subsystems with n −1 components can be further decomposed on the last component, namely component n −1, until we reach some boundary conditions. Thus, an iterative expression can be used to describe this decomposition process. Consider a system with j components that requires at least i components to work for the system to work. We have the following equations to express the state of such a system as a function of the states of two subsystems: S(i, j) = x j S(i −1, j −1) + x j S(i, j −1), (7.22) S(i, j) = x j S(i −1, j −1) + x j S(i, j −1), (7.23) SYSTEM RELIABILITY EVALUATION 241 where i may take any integer value from 1 to k and j may take values from 0 to n.
The following boundary conditions are needed for equations (7.22) and (7.23): S(0, j) = S( j + 1, j) = 1, (7.24) S( j + 1, j) = S(0, j) = 0.
(7.25) Equations (7.22) and (7.23) are in pivotal decomposition form. Because of the as-sumption that the components are s-independent, they can be immediately converted to the following algebraic reliability expressions: R(i, j) = p j R(i −1, j −1) + q j R(i, j −1), (7.26) Q(i, j) = p j Q(i −1, j −1) + q j Q(i, j −1).
(7.27) Equations (7.26) and (7.27) are recursive relations that are valid for 1 ≤i ≤k. Their boundary conditions can be directly obtained from equations (7.24) and (7.25) as follows: R(0, j) = Q( j + 1, j) = 1, (7.28) R( j + 1, j) = Q(0, j) = 0.
(7.29) Solutions for the reliability R(k, n) or the unreliability Q(k, n) is easily obtained by programming in languages that allow a program to call itself recursively. How-ever, a closer look at the recursive relations (7.26) and (7.27) reveals that they can be easily represented by what is called a signal flow graph (SFG). As an illustration, Figure 7.2 shows the SFG for the computation of R(3, 7). In Figure 7.2, a node at position (i, j) represents R(i, j). The black nodes in the first row with i = 0 are “source” nodes with values of 1, that is, R(0, j) = 1. The white nodes at i = j + 1 are source nodes with zero values, that is, R( j + 1, j) = 0 for j ≥0. The values at i j 1 2 3 0 0 1 2 3 6 5 4 7 p1 p2 p3 p4 p5 q1 q2 q3 q4 q5 p6 p2 p3 p4 p5 q2 q3 q4 q5 q6 p6 p3 p4 p5 p7 q3 q4 q5 q6 q7 FIGURE 7.2 Signal flow graph for obtaining R(3, 7) and Q(3, 7).
242 THE k-OUT-OF-n SYSTEM MODEL other nodes, say (i, j), have to be calculated by adding the product of the immediate top-left entry and p j to the product of the immediate left entry and q j.
The same graph in Figure 7.2 can also be used for the computation of Q(3, 7) provided that the graph nodes (i, j) are understood to represent the unreliabilities Q(i, j) instead of the reliabilities R(i, j), and the two types of source nodes inter-change their values; that is, the black nodes at i = 0 become zero values [Q(0, j) = 0] and the white nodes at i = j + 1 become unity values [Q( j + 1, j) = 1].
The algorithm proceeds efficiently by directly constructing (i.e., computing the element values of) the parallelogram with corners (1, 1), (1, n −k + 1), (k, k), and (k, n). The number of elements in the parallelogram is k(n −k + 1). Each element of the parallelogram requires three arithmetic operations (namely, one multiplication and two additions) for its evaluation. This can be easily seen by invoking the relation q j = 1 −p j to simplify (7.26) and (7.27) into the following forms: R(i, j) = p j R(i −1, j −1) + (1 −p j)R(i, j −1) = R(i, j −1) + p j (R(i −1, j −1) −R(i, j −1)) , (7.30) Q(i, j) = (1 −q j)Q(i −1, j −1) + q j Q(i, j −1) = Q(i −1, j −1) + q j (Q(i, j −1) −Q(i −1, j −1)) .
(7.31) This means that the algorithm by Rushdi requires 3k(n−k+1) arithmetic operations, and its computational complexity can be written as O(k(n −k + 1)).
Computation of the R(i, j) or Q(i, j) entries shown in Figure 7.2 can be pro-cessed by row, by column, or even diagonally. However, to minimize the memory requirements, this is done columnwise, for the R(3, 7) case, with due attention paid to the parallelogram boundaries. In this case the algorithm requires a memory stor-age of k +1 = 4 scalars, in addition to the memory needed to store pi for 1 ≤i ≤n.
The additional storage requirement for any such problem is min{k + 1, n −k} by calculating columnwise (if k is smaller) or rowwise (if n −k +1 is smaller). It is also interesting to note that this algorithm has the same computational complexity for its reliability and unreliability evaluations.
Detailed comparisons of the time and memory requirements of the algorithms by Sarje and Prasad , Rushdi , and Barlow and Heidtmann are conducted by Rushdi . The results are shown in Table 7.1. The time requirement is mea-sured by the number of multiplications, additions, and array references. Table 7.1 shows that the algorithms by Barlow and Heidtmann and Rushdi are computation-ally more efficient and require less memory than the algorithm by Sarje and Prasad.
Risse and Pham and Upadhyaya also give detailed comparisons of such algorithms that generally agree with this result.
Sample outputs of the algorithm by Rushdi for reliability and unreliability evalu-ations of a 5-out-of-8:G system are shown in Tables 7.2 and 7.3, respectively. In both tables, we have assumed that component reliabilities are p j = 0.9 −0.01( j −1) for j = 1, 2, . . . , 8.
Another advantage of the algorithms by Barlow and Heidtmann and Rushdi is worth noting. All the intermediate entries needed for calculating R(k, n) or SYSTEM RELIABILITY EVALUATION 243 TABLE 7.1 Comparison of Time and Space Complexities Algorithm Sarje and Prasad Rushdi Barlow and Heidtmann Temporal complexity Multiplications 4k(n −k) + 4 k(n −k + 1) (k + 1)(n −k + 1) −1 Additions 2k(n −k + 1) + n 2k(n −k + 1) (2k + 1)(n −k + 1) −2 References to one-dimensional arrays 4k(n −k) + k 4k(n −k + 1) + 1 (4k + 3)(n −k + 1) −3 References to two-dimensional arrays 4k(n −k) + 2k + 6 0 0 Spatial Complexity: Memory requirements 3(n −k + 1) + 2n min(k + 1, n −k + 2) k + 2 Source: Rushdi .
TABLE 7.2 Calculating R(5, 8) of 5-out-of-8:G System with Symmetric Switching Function Approach j 0 1 2 3 4 5 6 7 8 0 1 1 1 1 1 0 0.900000 0.989000 0.998680 0.999828 i 2 0 0.801000 0.966440 0.994489 0.999081 3 0 0.704480 0.932437 0.985802 0.997089 4 0 0.613246 0.887750 0.971094 0.992930 5 0 0.527391 0.833697 0.949110 0.985480 TABLE 7.3 Calculating Q(5, 8) of 5-out-of-8:G System with Symmetric Switching Function Approach j 0 1 2 3 4 5 6 7 8 0 0 0 0 0 1 1 0.100000 0.011000 0.001320 0.000172 i 2 1 0.199000 0.033560 0.005511 0.000919 3 1 0.295120 0.067563 0.014198 0.002911 4 1 0.386754 0.112250 0.028906 0.007070 5 1 0.472609 0.166303 0.050890 0.014520 Q(k, n) are meaningful numbers that represent R(i, j) or Q(i, j) for 1 ≤i ≤k and i ≤j ≤n −k + i. These numbers are available to the reliability engineer at no extra cost, and can enable one to make a valid economic assessment of redundancy. For example, row 5 in Table 7.2 represents the reliability R(5, j), where j varies from 5 to 8. The incremental system reliability achieved by increasing system size from j to j + 1 is j R(i, j) = R(5, j + 1) −R(5, j).
(7.32) 244 THE k-OUT-OF-n SYSTEM MODEL The economic equivalence of this incremental reliability can, therefore, be estimated and compared to the cost of adding an additional component, thereby obtaining the optimal number of components for the 5-out-of- j system.
Exercises 1. Compute the reliability of a 2-out-of-6:G system with pi = 0.6 + 0.06i for i = 1, 2, . . . , 6.
2. How do you use the algorithm covered in this section to evaluate the reliability of a k-out-of-n:F system?
3. What can you conclude through examination of the entries in the same column in Table 7.2 or 7.3?
MIS Technique To imbed the k-out-of-n:G system into the Markov chain following Definition 5.1, we can define the state space as S = {s0, s1, . . . , sk} = {0, 1, . . . , k}, the partition as Si = {i} for i = 0, 1, . . . , k (here, N = m = k), and the Markov chain {Yl, l ≥0} as 1. Yl = i if exactly i of the components 1, 2, . . . ,l are working (0 ≤i < k) and 2. Yl = k if at least k of the components 1, 2, . . . ,l are working.
Recall that pi j represents the probability for the Markov chain to make a transition from state i to state j. The transition matrix of the Markov chain is l = (pi j)(k+1)×(k+1) = ql pl ql pl ...
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ql pl ql pl 1 (k+1)×(k+1) , where i, j = 0, 1, . . . , k.
(7.33) Unmarked entries of the matrix are all equal to zero. This transition matrix provides the probabilities for the system with one more component, namely component l, to be in state j(0 ≤j ≤k) given that the system with l −1 components is in state i(0 ≤i ≤k). In the k-out-of-n:G system, we are interested in knowing whether the number of working components has reached or exceeded k. That is why the system state space includes 0, 1, . . . , k and represents the progressive increase in the number of working components as the system size increases. When the system size reaches n, the probability that the system is in state k is the reliability of the system.
Making use of the transition probability matrix given in equation (7.33) and noting N = m = k, we obtain the following recursive equations with Theorem 5.2: SYSTEM RELIABILITY EVALUATION 245 a0(l) = qla0(l −1), l ≥1, (7.34) a j(l) = pla j−1(l −1) + qla j(l −1), 1 ≤j < k, j ≤l ≤n, (7.35) ak(l) = plak−1(l −1) + ak(l −1), k ≤l ≤n, (7.36) where a j(l) is the probability that there are exactly j working components in a sys-tem with l components for 0 ≤j < k and ak(l) is the probability that there are at least k working components in the l-component subsystem. The following boundary conditions are immediate: a0(0) = 1, (7.37) a j(0) = 0, j > 0, (7.38) a j(l) = 0, l < j.
(7.39) The reliability of the system is R(k, n) = ak(n).
The recursive equations (7.34)–(7.36) can also be represented by a signal flow diagram similar to the one shown in Figure 7.2. These recursive equations have the same iterative structure as the algorithms given by Barlow and Heidtmann and Rushdi . The computational complexity of the recursive equations (7.34)–(7.36) is ∼3k(n −k + 1) or O(k(n −k + 1)).
Example 7.2 Consider a 5-out-of-8:G system with component reliabilities pl = 0.9 −0.01(l −1) for l = 1, 2, . . . , 8. Table 7.4 lists the results using the MIS approach. Compare this table with Table 7.2.
From Table 7.4, we see that the MIS approach not only provides the reliability of a k-out-of-n:G system but also the probabilities that there are at least k working components in the l-component subsystem for l = k, k + 1, . . . , n. In addition, we also know the probabilities that there are exactly j working components in the l-component subsystems for j = 0, 1, . . . , k −1 and l = j, j + 1, . . . , j + n −k.
For example, from the column with l = 6 in Table 7.4, we see that in the six-component subsystem, the probability that there are exactly three working compo-TABLE 7.4 Reliability Evaluation of 5-out-of-8:G System with MIS Approach l 0 1 2 3 4 5 6 7 8 pl 0.90 0.89 0.88 0.87 0.86 0.85 0.84 0.83 ql 0.10 0.11 0.12 0.13 0.14 0.15 0.16 0.17 0 1 0.100000 0.011000 0.001320 1 0 0.900000 0.188000 0.032240 0.005340 j 2 0 0.801000 0.261560 0.062052 0.013280 3 0 0.704880 0.319192 0.098052 0.025996 4 0 0.613246 0.360360 0.137392 0.043802 5 0 0.527391 0.833697 0.949111 0.985482 246 THE k-OUT-OF-n SYSTEM MODEL nents is 0.025996, that there are exactly four working components is 0.137392, and that there are exactly five working components is 0.833697. From these entries, we can also find that the probability that there are at least three working components in the six-component subsystem is equal to 0.025996 + 0.137392 + 0.833697 = 0.997085.
Fast Fourier Transform Method by Belfore Belfore uses the generating func-tion approach as developed by Barlow and Heidtmann and applies the FFT in computation of the products of the generating functions. An algorithm for reliabil-ity evaluation of k-out-of-n:G systems results from such a combination that has a computational complexity of O(n(log2 n)2). In the following, we explain this FFT approach.
Consider the following generation function: gn(z) = n i=1 (pi + qiz).
(7.40) It is a polynomial function of variable z. The coefficient of term zi in this polynomial function represents the probability that exactly i components are failed and thus the other n −i components are working. By factoring out the products of the component reliabilities, we can express equation (7.40) in the form gn(z) = Pπ n i=1 (1 + ai z) = Pπ(1 + A(z)), (7.41) where Pπ = p1 p2 · · · pn, ai = qi pi for i = 1, 2, . . . , n, A(z) = b1z + b2z2 + · · · + bnzn, bi = 1≤j1< j2<··· ji≤n a j1a j2 · · · a ji for i = 1, 2, . . . , n.
Using the form of the generating function in equation (7.41) rather than the one in equation (7.40) results in fewer computations because the multiplications by 1 are implicit and the increases in FFT sizes are delayed since the FFT is applied to A(z) . Once the coefficients of A(z), or bi for i = 1, 2, . . . , n, are calculated, we can find the reliability of a k-out-of-n:G system with the following equation: R(k, n) = Pπ 1 + n−k i=1 bi .
(7.42) SYSTEM RELIABILITY EVALUATION 247 Suppose, for ease of explanation, that n is a power of 2. Define A(z) as an nth-order polynomial function of z. The term 1 + A(z) in equation (7.41) can be viewed as a product of two generating functions, 1 + A1(z) and 1 + A2(z), where A1(z) and A2(z) are (n/2)th-order polynomial functions of z. We have 1 + A(z) = [1 + A1(z)][1 + A2(z)] = 1 + A1(z) + A2(z) + A1(z)A2(z).
As a result, A(z) = A1(z) + A2(z) + A1(z)A2(z).
(7.43) To find A(z), first we need to compute the product of two (n/2)th-order polynomial functions, A1(z) and A2(z). In turn A1(z) and A2(z) each is equal to the sum of two lower order [(n/4)th-order] polynomial functions plus their product. This process can be repeated until the order of the polynomial functions is low enough so that the exact values of its coefficients become apparent.
Finding the expression of the product of two (n/2)th-order polynomial functions A1(z) and A2(z) is equivalent to finding the coefficients of the resulting polynomial function. This can be achieved using FFT. First, we define a discrete function corre-sponding to Ai(z)(i = 1, 2). This discrete function takes values of the coefficients of Ai(z)(i = 1, 2) over the definition domain of {0, 1, 2, . . . , (n/2) −1}. If A1(z) and A2(z) are in the forms A1(z) = c1z + c2z2 + · · · + cn/2zn/2, A2(z) = d1z + d2z2 + · · · + dn/2zn/2, the two discrete functions are in the forms f1(x) = cx+1 if x = 0, 1, 2, . . . , n/2 −1, 0 otherwise, f2(x) = dx+1 if x = 0, 1, 2, . . . , n/2 −1, 0 otherwise.
The convolution of these two discrete functions is given by f (x) = f1(x) ∗f2(x) ≡ n/2−1 y=0 f1(x −y) f2(y), where ∗denotes the convolution operator. We have to note that the definition domain of the resulting function, f (x), is {0, 1, 2, . . . , n −1}. This is in agreement with the product of two (n/2)th-order polynomial functions being an nth-order polynomial function. The values of f (x) for x ∈{0, 1, 2, . . . , n} provide the coefficients of the resulting polynomial function, namely A1(z)A2(z). We can then use equation (7.43) 248 THE k-OUT-OF-n SYSTEM MODEL to find the coefficients of A(z). However, we need an efficient method to find the convolution of two discrete functions.
Based on the convolution theorem in Fourier theory, the Fourier transform of the convolution of two functions is equal to the product of the Fourier transforms of these two individual functions. Thus, to find the coefficients of the product of two polyno-mial functions, we can first find the FFT of two discrete functions corresponding to the two polynomial functions, multiply these two FFTs in the frequency domain, and finally conduct the inverse FFT on the resulting product to obtain the desired coeffi-cients of the resulting polynomial function. For details on FFT, readers are referred to Bracewell .
Using the generating function form as shown in equation (7.41) and assuming n is a power of 2, the number of operations required to multiply 1+ A1(z) and 1+ A2(z), where A1(z) and A2(z) each is a (n/2)th-order polynomial function, using FFT is T (n) = 15n log2(n) + 11n −2.
(7.44) Apparently, large overheads are involved in the FFT approach. Thus, it is not efficient to use this approach for small n values. When n is small, we can use the algorithm provided by Barlow and Heidtmann to directly find the coefficients of the gen-erating function. Belfore shows that the FFT approach is more efficient than the algorithm by Barlow and Heidtmann (BH) when n is larger than 512. The following algorithm based on Barlow and Heidtmann is used to compute the coefficients of the generating functions for small n values: BH(n, a[1 : n], Az[1 : n]) integer i, j; Az = a; For i = 2 To n By 1 Do Az[i] = Az[i −1] ∗a[i]; For j = i −1 To 2 By −1 Do Az[ j] = Az[ j] + Az[ j −1] ∗a[i]; EndFor Az = Az + a[i]; EndFor Return Az[1 : n]; End In the BH algorithm shown above, a[1 : n] is an array of size n holding the ratios qi/pi and Az[1 : n] is an array of size n holding the coefficients of zi in the resulting generating function for i = 1, 2, . . . , n. The sizes of these arrays are determined by the calling algorithm through the argument n.
The following algorithm is used to calculate the coefficients of the generating function shown in equation (7.41) for large n values: SYSTEM RELIABILITY EVALUATION 249 GF FFT(n, a[1 : n], Az[1 : n]) If n <= threshold Then Call BH(n, a, Az); Else Call GF FFT(n/2, a[1 : n/2], Az[1 : n/2]); Call GF FFT(n−n/2, a[n/2 + 1 : n], Az[n/2 + 1 : n]); F FT size = 2liub(log2n); Initialize temp1, temp2; For i = 1 To n/2 By 1 Do temp1[i].real = Az[i]; EndFor For i = n/2 + 1 To n By 1 Do temp2[i −n/2].real = Az[i]; EndFor Compute FFT of temp1; Compute FFT of temp2; For i = 1 To F FT size By 1 Do temp1[i] = temp1[i] ∗temp2[i]; EndFor Compute the inverse FFT of temp1 and assign it to I temp1; Az = Az + Az[n/2 + 1]; For i = 2 To n/2 By 1 Do Az[i] = Az[i] + Az[n/2 + i] + I temp1[i −1]; EndFor For i = n/2 + 1 To n−n/2 By 1 Do Az[i] = Az[n/2 + i] + I temp1[i −1]; Endfor For i = n −n/2 + 1 To n By 1 Do Az[i] = I temp1[i −1]; Endfor EndIf Return Az[1 : n]; End In this algorithm, n/2 is defined to be the largest integer less than or equal to n/2 when n is not divisible by 2; liub indicates the lowest integer upper bound; temp1 and temp2 are complex variable arrays, both of size FFT size; a[n/2 + 1 : n] indicates that a subarray of a[1 : n] with values in positions n/2+1 through n of a[1 : n] being used; and the calling algorithm defines the sizes of the arrays in the arguments. The threshold value is specified by the user so that when n is smaller than the threshold, the BH algorithm is used for calculating the coefficients of the generating function.
The system reliability of a k-out-of-n:G system can be calculated with the follow-ing algorithm according to equation (7.42): 250 THE k-OUT-OF-n SYSTEM MODEL R sys FFT(n, k, p[1 : n], q[1 : n]) P pi = 1; For i = 1 To n By 1 Do P pi = P pi ∗p[i]; a[i] = q[i]/p[i]; EndFor Call GF FFT(n, a[1 : n], Az[1 : n]); Rel = 1; For i = 1 To n −k By 1 Do Rel = Rel + Az[i]; EndFor Rel = P pi ∗Rel; Return Rel; End The lower and upper bounds on the complexity of the algorithm R sys FFT for a threshold value of 2 are given by Belfore as Tlower(n) = 1 2 × 15n(log2 n)2 + 1 2 × 37n log2 n −23n + 2, (7.45) Tupper(n) = 15n(log2 n)2 + 67n log2 n + 6n + 2.
(7.46) For simplicity, we can say that the FFT approach has a time complexity of O(n(log2 n)2).
7.1.3 Bounds on System Reliability When the components in a k-out-of-n system are s-independent, the algorithms presented in Section 7.1.2 are quite efficient for evaluation of exact system reli-ability. However, the components in the system may not be independent in some cases. To add to the difficulty, the way in which the components are dependent on each other may not be completely understood. In this section, we provide discussion on system reliability approximation when components are not necessarily indepen-dent.
Associated Components As introduced earlier, the concept of association indicates that two random variables have nonnegative covariance. In this case, we may use the theorem given in Barlow and Proschan to find the upper and lower bounds on system reliability of k-out-of-n:G systems.
Let P1, P2, . . . , Pr represent the minimal path sets. We have r = n k and there are k components in each of these minimal path sets. Let K1, K2, . . . , Kt represent the minimal cut sets. Then, t = n n−k+1 and there are n −k + 1 components in each of these minimal cut sets. The following bounds on system reliability are given by RELATIONSHIP BETWEEN k-OUT-OF-n G AND F SYSTEMS 251 Barlow and Proschan : max 1≤i≤r j∈Pi p j ≤Rs ≤min 1≤i≤t 1 − j∈Ki (1 −p j) .
(7.47) Unspecified Dependence of Components Without making any assumptions on how components are dependent on one another, Lipow provides a simple formula for the lower bound of the reliability of a k-out-of-n:G system: Rs ≥max 1≤i≤r j∈Pi p j −k + 1, (7.48) where r = n k and Pi is the ith minimal path set. This formula was derived from the IE method for system reliability evaluation using minimal path sets. It is useful only when component reliabilities are pretty close to 1 and k is not too big.
Exercises 1. Analyze the closeness of the bounds given in (7.47) to exact system reliability for a k-out-of-n:G system. Consider cases when component reliabilities are high and low.
2. Develop an upper bound for system reliability using the IE method when com-ponent dependency is unspecified. Under what conditions will this bound be close to the exact system reliability?
3. Analyze the closeness of the bounds given in (7.48) to exact system reliability for a k-out-of-n:G system. Consider cases when component reliabilities are high and low.
7.2 RELATIONSHIP BETWEEN k-OUT-OF-n G AND F SYSTEMS In the previous section, we illustrated different approaches for reliability evaluation of k-out-of-n:G systems. Exercises were given for following similar approaches to derive algorithms for reliability evaluation of k-out-of-n:F systems. In this section, we provide a formal discussion of the relationship between k-out-of-n G and F sys-tems and how reliability evaluation algorithms for these two types of systems are closely related.
7.2.1 Equivalence between k-out-of-n:G and (n −k + 1)-out-of-n:F Systems Based on the definitions of these two types of systems, a k-out-of-n:G system is equivalent to an (n −k + 1)-out-of-n:F system. Similarly, a k-out-of-n:F system is equivalent to an (n−k+1)-out-of-n:G system. This means that provided the systems have the same set of component reliabilities, the reliability of a k-out-of-n:G system 252 THE k-OUT-OF-n SYSTEM MODEL is equal to the reliability of an (n −k + 1)-out-of-n:F system and the reliability of a k-out-of-n:F system is equal to the reliability of an (n−k+1)-out-of-n:G system. As a result, we can use the algorithms that have been covered in the previous section for the k-out-of-n:G systems in reliability evaluation of the k-out-of-n:F systems. The procedure is simple and is outlined below: Procedure for Using Algorithms for the G Systems in Reliability Evaluation of the F Systems Utilizing the Equivalence Relationship 1. Given k, n, p1, p2, . . . , pn for a k-out-of-n:F system.
2. Calculate k1 = n −k + 1.
3. Use k1, n, p1, p2, . . . , pn to calculate the reliability of a k1-out-of-n:G sys-tem. This reliability is also the reliability of the original k-out-of-n:F system.
7.2.2 Dual Relationship between k-out-of-n G and F Systems Barlow and Proschan provide the following definition of a dual structure.
Definition 7.1 Given a structure φ, its dual structure φD is given by φD(x) = 1 −φ(1 −x), (7.49) where 1 −x = (1 −x1, 1 −x2, . . . , 1 −xn).
With a simple variable substitution of 1−x for x, we have the equation φD(1 −x) = 1 −φ(x).
(7.50) We can interpret equation (7.50) as follows. Given a primal system with component state vector x and the system state represented by φ(x), the state of the dual system is equal to 1 −φ(x) if the component state vector for the dual system can be ex-pressed as 1 −x. In the binary system context, each component and the system may only be in two possible states, either working or failed. We will say that two com-ponents with different states have opposite states. For example, if component 1 is in state 1 and component 2 is in state 0, components 1 and 2 have opposite states. Sup-pose a system (called system 1) has component state vector x and system state φ(x).
Consider another system (called system 2) with the same number of components as system 1. If each component in system 2 has the opposite state of the corresponding component in system 1 and the state of system 2 becomes the opposite of the state of system 1, then system 1 and system 2 are duals of each other.
Now we examine the k-out-of-n G and F systems. Suppose that in the k-out-of-n:G system, there are exactly j working components and the system is working (in other words, j ≥k). Now assume that there are exactly j failed components in the k-out-of-n:F system. Since j ≥k, the k-out-of-n:F system must be in the failed state. If j < k, the k-out-of-n:G system is failed, and at the same time the k-out-of-n:F system is working. Thus, the k-out-of-n G and F systems are duals of each RELATIONSHIP BETWEEN k-OUT-OF-n G AND F SYSTEMS 253 other. Using the equivalence relationship described in the previous section, we can also say that the dual of a k-out-of-n:G system is an (n −k + 1)-out-of-n:G system.
Similarly, we can say that a k-out-of-n:F system is the dual of an (n −k + 1)-out-of-n:F system. These dual and equivalence relationships between the k-out-of-n G and F systems are summarized below: 1. A k-out-of-n:G system is equivalent to an (n −k + 1)-out-of-n:F system.
2. A k-out-of-n:F system is equivalent to an (n −k + 1)-out-of-n:G system.
3. The dual of a k-out-of-n:G system is a k-out-of-n:F system.
4. The dual of a k-out-of-n:G system is an (n −k + 1)-out-of-n:G system.
5. The dual of a k-out-of-n:F system is a k-out-of-n:G system.
6. The dual of a k-out-of-n:F system is an (n −k + 1)-out-of-n:F system.
Using the dual relationship, we can summarize the following procedure for reli-ability evaluation of the dual system if the available algorithms are for the primal system: Procedure for Using Algorithms for the G Systems in Reliability Evaluation of the F Systems Utilizing the Dual Relationship 1. Given k, n, p1, p2, . . . , pn for a k-out-of-n:F system.
2. Calculate qi = 1 −pi for i = 1, 2, . . . , n.
3. Treat qi as the reliability of component i in a k-out-of-n:G system and use the algorithms for the G system discussed in the previous section to evaluate the reliability of the G system.
4. Subtract the calculated reliability of the G system from 1 to obtain the reliabil-ity of the original k-out-of-n:F system.
Using the dual relationship, we can also obtain algorithms for k-out-of-n:F system reliability evaluation from those developed for the k-out-of-n:G systems. We only need to change reliability measures to unreliability measures and vice versa. Take the algorithm developed by Rushdi as an example. The formulas for reliability and unreliability evaluation of a k-out-of-n:G system are given in equations (7.26) and (7.27) with boundary conditions in equations (7.28) and (7.29). By changing R(i, j) to Q(i, j), Q(i, j) to R(i, j), pi to qi, and qi to pi in those four equations, we obtain the following equations for reliability and unreliability evaluation of a k-out-of-n:F system: QF(i, j) = q j QF(i −1, j −1) + p j QF(i, j −1), (7.51) RF(i, j) = q j RF(i −1, j −1) + p j RF(i, j −1), (7.52) with the boundary conditions QF(0, j) = RF( j + 1, j) = 1, (7.53) QF( j + 1, j) = RF(0, j) = 0.
(7.54) 254 THE k-OUT-OF-n SYSTEM MODEL To avoid confusion, the subscript F is added to indicate that these measures are for the F system. Similar steps can be applied to other algorithms for the G systems to derive the corresponding algorithms for the F systems. It is because of such close relationships between the k-out-of-n G and F systems that we often refer to them collectively as the k-out-of-n systems.
Example 7.3 Consider a k-out-of-n:F system with k = 3, n = 7, and pi = 0.8 + 0.02i for i = 1, . . . , 7. Use the two procedures listed in Section 7.2 to evaluate the reliability of the system.
The 3-out-of-7:F system is equivalent to a 5-out-of-7:G system with the same set of components. Table 7.5 lists the calculations needed to find the reliability of the 5-out-of-7:G system. The reliability of the 5-out-of-7:G system is found to be 0.959836, which is equal to the reliability of the original 3-out-of-7:F system. Each entry in Table 7.5 can be interpreted in terms of either a k-out-of-n:G subsystem or a (n −k + 1)-out-of-n:F subsystem. For example, 0.853982 in the column labeled 5 and the row labeled 4 represents the reliability of a 4-out-of-5:G subsystem and at the same time the reliability of the equivalent 2-out-of-5:F subsystem.
We can also use the dual relationship between a k-out-of-n:F system and a k-out-of-n:G system. From the given pi values, calculate all qi = 1 −pi for i = 1, 2, . . . , 7. Treat these qi’s as component reliability values and apply the formulas for the k-out-of-n:G system. Table 7.6 lists these calculations. The rightmost entry at the bottom row in Table 7.6 is the reliability of the 3-out-of-7:G system. To find the reliability of the original 3-out-of-7:F system, we need to subtract this value from 1: RF(3, 7) = 1 −0.040163 = 0.959837.
TABLE 7.5 Reliability Evaluation of Equivalent 5-out-of-7:G System n k 0 1 2 3 4 5 6 7 0 1 1 1 1 0 0.820000 0.971200 0.995968 2 0 0.688800 0.931664 0.998252 3 0 0.592368 0.890948 0.978521 4 0 0.521284 0.853982 0.968558 5 0 0.469156 0.823196 0.959836 TABLE 7.6 Reliability Evaluation of Dual System: 3-out-of-7:G System n k 0 1 2 3 4 5 6 7 0 1 1 1 1 1 1 0 0.180000 0.311200 0.407632 0.478716 0.530845 2 0 0.028800 0.068336 0.109052 0.146018 0.176804 3 0 0.004032 0.011748 0.021479 0.031442 0.040164 NONREPAIRABLE k-OUT-OF-n SYSTEMS 255 Exercises 1. Derive the formulas for reliability evaluation of a k-out-of-n:F system based on the algorithm by Barlow and Heidtmann.
2. Derive the formulas for reliability evaluation of a k-out-of-n:F system based on the MIS approach.
3. Compute the reliabilities of the k-out-of-n F and G systems with k = 2, 3, 4 and n = 5, 6, 7.
7.3 NONREPAIRABLE k-OUT-OF-n SYSTEMS In the previous sections, we have discussed the so-called static properties of k-out-of-n systems. Reliability has not been expressed as a function of time. But, in fact, reliability and other performance measures of any system are functions of time. Start-ing from this section, we will provide stochastic analyses of the k-out-of-n systems.
In reality, it is sometimes impossible to repair a system until its mission is com-plete. In this case, the reliability of the system is a decreasing function of time.
In this section, we examine the performance measures of nonrepairable k-out-of-n systems.
Notation • Ti: lifetime of component i, a random variable • Ts: lifetime of the system, a random variable • Ri(t): Pr(Ti ≥t), reliability function of component i • R(t): reliability function of each component when components are i.i.d.
• Fi(t): 1 −Ri(t), CDF or unreliability function of component i • F(t): CDF or unreliability function of each component when components are i.i.d.
• fi(t): pdf of the lifetime of component i • f (t): pdf of the lifetime of each component when components are i.i.d.
• hi(t): failure rate function of component i • h(t): failure rate function of each component when components are i.i.d.
• Rs(t): reliability function of the system • Fs(t): 1 −Rs(t), CDF or unreliability function of the system • fs(t): pdf of the lifetime of the system • hs(t): failure rate function of the system • R(t; k, n): reliability function of the k-out-of-n:G system • MTTFs: mean time to failure of the system • MTTF(k, n): mean time to failure of a k-out-of-n:G system 256 THE k-OUT-OF-n SYSTEM MODEL 7.3.1 Systems with i.i.d. Components When the components in a k-out-of-n:G system are i.i.d., the reliability function of the system can be expressed as Rs(t) = n i=k n i R(t)i F(t)n−i.
(7.55) This equation is directly obtained from equation (7.2) by replacing p with R(t) and q with F(t). Similarly, the CDF of the system lifetime is given by Fs(t) = 1 −Rs(t) = k−1 i=0 n i R(t)i F(t)n−i.
(7.56) The pdf of the system lifetime is then fs(t) = dFs(t) dt = k n k f (t)F(t)n−k R(t)k−1.
(7.57) Usually, as the system is used, the components in the system will fail one by one. The system is failed as soon as the (n −k + 1)th component is failed. If we use ti to indicate the lifetime of component i, the system lifetime is then equal to the (n −k + 1)th smallest ti. The expected lifetime of the system, or mean time to failure, can be evaluated using the standard equation MTTFs = ∞ 0 t fs(t) dt = ∞ 0 Rs(t) dt.
(7.58) In the following, we first illustrate that when all i.i.d. components have IFR or even constant failure rates, the system has IFR. No specific component lifetime dis-tributions are assumed: Rs(t) = ∞ t fs(x) dx, 1 hs(t) = Rs(t) fs(t) = ∞ t f (x)F(x)n−k R(x)k−1 dx f (t)F(t)n−k R(t)k−1 = 1 f (t) ∞ t f (x) F(x) F(t) n−k R(x) R(t) k−1 dx.
Let y = R(x)/R(t); then dy = −[ f (x)/R(t)] dx: 1 hs(t) = 1 h(t) 1 0 1 −yR(t) F(t) n−k yk−1 dy.
(7.59) Since [1 −yR(t)]/F(t) is decreasing in t and h(t) is assumed to be IFR, we con-clude that hs(t) is increasing in t based on equation (7.59). This indicates that if all NONREPAIRABLE k-OUT-OF-n SYSTEMS 257 components have IFR, the k-out-of-n:G structure preserves this IFR property of the components. If all components have constant failure rates, the k-out-of-n:G system would have IFR as long as k ̸= n and a constant failure rate when k = n.
It is generally impossible to find more specific expressions of the performance measures of the k-out-of-n:G system. However, when the components follow the ex-ponential distribution, some explicit results can be derived. When all components follow the exponential lifetime distribution with CDF F(t) = 1 −e−λt, the expres-sions of system reliability and unreliability are Rs(t) = n i=k n i (e−λt)i(1 −e−λt)n−i, (7.60) Fs(t) = k−1 i=0 n i (e−λt)i(1 −e−λt)n−i, (7.61) respectively. The MTTF of the system can be derived as follows. Based on equation (7.7), we have, for k ≥2, Rs(t; k, n) = R(t; k −1, n) − n k −1 e−λt(k−1)(1 −e−λt)n−k+1.
(7.62) Integrating both sides of this equation results in the following recursive equation: MTTF(k, n) = MTTF(k −1, n) − ∞ 0 n k −1 e−λt(k−1)(1 −e−λt)n−k+1 dt = MTTF(k −1, n) −1 λ n k −1 n−k+1 j=0 n −k + 1 j (−1) j k −1 + j = MTTF(k −1, n) − 1 λ(k −1).
(7.63) The following equation is used in the above derivations: N j=0 N j (−1) j a + j = N!(a −1)!
(N + a)!
for a ≥1.
(7.64) MTTF(1, n) represents the MTTF of a parallel system, which is (1/λ) n j=1(1/j).
Using this boundary condition and applying equation (7.63) recursively, we find MTTF(k, n) = 1 λ n j=k 1 j .
(7.65) Substituting k = n in equation (7.65) provides the MTTF of a series system, 1/(nλ), as is expected.
258 THE k-OUT-OF-n SYSTEM MODEL Using equation (7.59), we can express the system failure rate as hs(t) = λ 1 0 yk−1[(1 −ye−λt)/(1 −e−λt)]n−k dy .
(7.66) No closed-form expression for hs(t) can be obtained even in the case when all com-ponents have exponential lifetime distributions.
Exercises 1. Verify equation (7.57).
2. Verify equation (7.63).
7.3.2 Systems with Nonidentical Components It is generally difficult to write an expression for k-out-of-n system reliability when the components do not have identical lifetime distributions. It is possible to derive the desired expressions for simple cases. For components with exponential lifetime distributions such that component i has a constant failure rate λi(1 ≤i ≤n), we have the following expressions of system reliability and MTTF for a 2-out-of-3:G system: Rs(t; 2, 3) = e−(λ1+λ2)t + e−(λ1+λ3)t + e−(λ2+λ3)t −2e−(λ1+λ2+λ3)t, MTTF(2, 3) = 1 λ1 + λ2 + 1 λ1 + λ3 + 1 λ2 + λ3 − 2 λ1 + λ2 + λ3 .
7.3.3 Systems with Load-Sharing Components Following Exponential Lifetime Distributions Consider a k-out-of-n:G system with i.i.d. components each following the exponen-tial lifetime distribution. When the system is put into operation at time zero, all com-ponents are working and they are equally sharing the constant load that the system is supposed to carry. In this case, the failure rate of every component is denoted by λ0. When the system experiences the first failure, the remaining n −1 working components must carry the same load on the system. As a result, the failure rate of each working component becomes λ1, which is usually higher than λ0. When i components are failed, the failure rate of each of the n −i working components is represented by λi (0 ≤i ≤n −k). The system is failed when more than n −k components are failed. For such a system with no repair provisions, Scheuer provides an analysis of the system’s performance measures.
Notation • λi: failure rate of each surviving component when i components have failed (0 ≤i ≤n−k). Assume λ0 ≤λ1 ≤· · · ≤λn−k due to practical considerations.
NONREPAIRABLE k-OUT-OF-n SYSTEMS 259 • Ti: time to the ith failure (T0 ≡0), i = 1, 2, . . . , n −k + 1 • Xi: time between the (i −1)th failure and the ith failure, Xi = Ti −Ti−1, i = 1, 2, . . . , n −k + 1 • αi: failure rate of the system when there are i failed components, αi = (n −i + 1)λi−1, i = 1, 2, . . . , n −k + 1 Since all components are i.i.d. following the exponential distributions, the inter-arrival times of failures are independent random variables and Xi follows the expo-nential distribution with parameter αi for 1 ≤i ≤n −k + 1. The lifetime of the system is equal to the (n −k + 1)st failure time, that is, Ts = Tn−k+1 = X1 + X2 + · · · + Xn−k+1.
The MTTF of the system is then MTTFs = n−k+1 i=1 1 αi = n−k+1 i=1 1 (n −i + 1)λi−1 .
The distribution of Ts is the distribution of a sum of n−k +1 independent random variables, each following the exponential distribution with possibly different param-eters. To find the distribution of Ts and the reliability function of the system, we need to distinguish the following three cases.
Case I: α1 = α2 = · · · = αn−k+1 ≡α This case arises when the load of the sys-tem is equally shared by surviving components. If the failure rate of each surviving component is directly proportional to the load it carries, we can write λi as λi = c d n −i , i = 0, 1, 2, . . . , n −k, where d is the load on the system and c is a constant. Using this equation, we can verify the following: αi = (n −i + 1)λi−1 = cd ≡α, i = 1, 2, . . . , n −k + 1.
Thus, under case I, Xi’s for i = 1, 2, . . . , n −k + 1 are i.i.d. random variables following the same exponential distribution with parameter α. As a result, Ts, a sum of these n −k + 1 i.i.d. random variables, follows the gamma distribution with scale parameter α and shape parameter n −k + 1. The pdf of this gamma distribution is f (t) = α(αt)n−k (n −k)! e−αt.
The reliability function of the system is then Rs(t) = n−k j=0 (αt) j j!
e−αt.
(7.67) 260 THE k-OUT-OF-n SYSTEM MODEL Case II: α1, α2, . . . , αn−k+1 Take Distinct Values In this case, the lifetime of the system is a sum of n −k + 1 independent random variables each with a distinct exponential distribution parameter. The pdf of the system lifetime is a convolution of the pdf’s of these n −k + 1 exponential random variables. With the technique of Laplace transform, the pdf of the system’s lifetime is found to be fs(t) = n−k+1 i=1 αi n−k+1 i=1 e−αit n−k+1 j=1, j̸=i(α j −αi) .
From fs(t), we find the reliability function of the system: Rs(t) = n−k+1 i=1 Aie−αit, (7.68) Ai = n−k+1 j=1, j̸=i α j α j −αi , i = 1, 2, . . . , n −k + 1.
(7.69) Case III: α1, α2, . . . , αn−k+1, Are neither Identical nor Distinct Specifically, as-sume that these αi’s take a (1 < a < n) distinct values, β1, β2, . . . , βa. With possi-bly some renumbering of these αi values, assume α1 = α2 = · · · = αr1 ≡β1, αr1+1 = αr1+2 = · · · = αr1+r2 ≡β2, .
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αr1+r2+···+ra−1+1 = · · · = αr1+r2+···+ra ≡βa, r1 + r2 + · · · + ra = n −k + 1, 1 ≤ri < n, i = 1, 2, . . . , a.
Under case III, the interarrival times of failures are divided into a (a > 1) groups.
Group j has r j identical interarrival times following the exponential distribution with the same parameter. The interarrival times in different groups follow exponential distributions with different parameters. If we define the lifetime of each group as the sum of the interarrival times within the group, such group lifetimes then follow the gamma distribution. The lifetime of group j, denoted by Vj, for j = 1, 2, . . . , a follows the gamma distribution with scale parameter β j and shape parameter r j. In addition, these group lifetimes are independent. As a result, we can write the lifetime of the system, Ts, as a sum of the lifetimes of the groups, each following a different gamma distribution: Ts = V1 + V2 + · · · + Va.
The reliability function of the system is given by NONREPAIRABLE k-OUT-OF-n SYSTEMS 261 Rs(t) = B a j=1 r j ℓ=1 jℓ(−β j) (ℓ−1)!βr j−ℓ+1 r j−ℓ i=0 (β jt)ie−β jt i!
, (7.70) where B = a j=1 βr j j , (7.71) jℓ(t) = dℓ−1 dtℓ−1 a i=1,i̸= j (βi + t)−ri .
(7.72) These equations can be derived as follows. Assume that Vj has the gamma dis-tribution with scale parameter β j and shape parameter r j (a positive integer) and its pdf can be written as f j(t) = β j(β jt)r j−1e−β jt (r j −1)!
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The Laplace transform of f j(t) is L j(s) = β j β j + s r j .
The pdf of Ts is a convolution of the individual pdf’s of the lifetimes of these a groups. The Laplace transform of a convolution of functions is equal to the product of the Laplace transforms of these individual functions. As a result, the Laplace transform of the pdf of Ts is Ls(s) = a j=1 β j β j + s r j .
(7.73) The inverse Laplace transform of equation (7.73) will give the pdf of Ts : fs(t) = B a j=1 r j ℓ=1 jℓ(−β j) (ℓ−1)!(r j −ℓ)!tr j−ℓe−β jt.
(7.74) From this pdf of Ts, we can find the Rs(t) as given in equation (7.70).
Exercises 1. Derive equation (7.68).
2. Derive the system reliability function under case II.
262 THE k-OUT-OF-n SYSTEM MODEL 7.3.4 Systems with Load-Sharing Components Following Arbitrary Lifetime Distributions Liu provides an analysis of the k-out-of-n:G system with i.i.d. components whose lifetime distributions are not necessarily exponential. Repair of failed com-ponents is not allowed. Surviving components equally share the constant load of the system. The lifetime distribution of a component under a constant load can be repre-sented by the accelerated failure time model (AFTM) or the accelerated life model.
The parametric form of the AFTM for each component is assumed to be known.
The AFTM specifies that the effect of load on the lifetime of a component is mul-tiplicative in time. The reliability function of a component under the AFTM can be expressed as R(t, z) = R0(tψ(z)), (7.75) where z is a vector representing the loads on the component, ψ(z) is an acceleration factor, and R0(·) is the reliability function of an arbitrary statistical distribution. For more discussions on AFTM, readers are referred to Nelson . When there is only one type of load, z, commonly used forms of ψ(z) include ψ(z) = eαz, ψ(z) = zα.
For example, if R0(·) is of the Weibull distribution, that is, R0(x) = e−(t/η)β, and ψ(z) = zα, we can write the load-dependent reliability function of the component as R(t; z) = exp − tzα η β .
(7.76) When R0(·) is Weibull, the AFTM is equivalent to the proportional hazard model (PHM) , wherein the load acts multiplicatively on the failure rate. When R0(·) is not Weibull, the AFTM is not equivalent to the PHM. In the following, we illustrate the reliability analysis of a k-out-of-n:G system with i.i.d. load-sharing components whose lifetimes can be modeled with AFTM as given in equation (7.75).
Notation • R(t; z): reliability function of each component when the total load on the sys-tem is z • zn−j: total load to be shared by n−j surviving components when j components are failed When k = n, we have a series system. All components have to work for the system to work. Since components are independent, R(t; n, n) = n i=1 R(t; zn) = [R(t; zn)]n .
When k = n−1, for the system to survive beyond t, either all components survive beyond t or one component fails at time x(0 < x < t) and all other components NONREPAIRABLE k-OUT-OF-n SYSTEMS 263 survive the remaining time duration t −x: R(t; n −1, n) = R(t; n, n) + n t 0 f (x; zn) R(t −x −ˆ x; zn−1) n−1 dx, (7.77) where ˆ x = xψ(zn)/ψ(zn−1).
When k = n −2, we have R(t; n −2, n) = R(t; n −1, n) + n!
(n −2)!
t 0 x1 0 f (x; zn) f (x1 −x + ˆ x; zn−1) × R(t −x1 + ˆ x1; zn−2) n−2 dx dx1, (7.78) where ˆ x = xψ(zn)/ψ(zn−1) and ˆ x1 = (x1 −x + ˆ x)ψ(zn−1)/ψ(zn−2).
When k = n −3, R(t; n −3, n) = R(t; n −2, n) + n!
(n −3)!
t 0 x2 0 x1 0 f (x; zn) f (x1 −x + ˆ x; zn−1) × f (x2 −x1 + ˆ x1; zn−2) × R(t −x2 + ˆ x2; zn−3) n−3 dx dx1 dx2, (7.79) where ˆ x = xψ(zn)/ψ(zn−1), ˆ x1 = (x1 −x + ˆ x)ψ(zn−1)/ψ(zn−2), and ˆ x2 = (x2 −x1 + ˆ x1)ψ(zn−2)/ψ(zn−3).
Generally, the following equation can be used for evaluation of R(t; j, n) for 1 ≤j < n: R(t; j, n) = R(t; j + 1, n) +n!
j!
t 0 xn−j−1 0 xn−j−2 0 · · · x2 0 x1 0 f (x; zn) × f (x1 −x + ˆ x; zn−1) f (x2 −x1 + ˆ x1; zn−2) × · · · × f (xn−j−2 −xn−j−3 + ˆ xn−j−3; zn−(n−j−2)) × f (xn−j−1 −xn−j−2 + ˆ xn−j−2; zn−(n−j−1)) × R(t −xn−j−1 +ˆ xn−j−1; zn−(n−j)) j dx dx1 dx2 · · · dxn−j−2 dxn−j−1, (7.80) where ˆ xi = (xi −xi−1 + ˆ xi−1)ψ(zn−i)/[ψ(zn−(i+1))] for i = 1, 2, . . . , n −j −1 and s0 ≡s.
The procedure outlined above is enumerative in nature. More efficient methods for handling arbitrary load-dependent component lifetime distributions are needed.
264 THE k-OUT-OF-n SYSTEM MODEL 7.3.5 Systems with Standby Components As we mentioned before, the k-out-of-n system structure has built-in redundancy.
Actually the system requires only k components to work for the system to work. In deriving the equations for k-out-of-n system performance evaluations so far in this chapter, we have treated the extra n−k components as active redundant components.
In other words, they are in hot standby mode. In this section, we will analyze the k-out-of-n system with cold and warm standby components.
Cold Standby with i.i.d. Components and Perfect Switching In Chapter 4, we dis-cussed standby systems with n components. When the components are i.i.d. follow-ing the exponential lifetime distribution with parameter λ, the lifetime of the system follows the gamma distribution with scale parameter λ and shape parameter n. In the following, we will show that the lifetime of a k-out-of-n:G system with i.i.d. cold standby components can also be described by a gamma distribution.
For a k-out-of-n:G system with standby components, k components are put into operation initially and n −k components are in standby. Whenever one of the active components is failed, one of the standby components is switched into operation. No repair provisions are allowed. Sensing and switching are assumed to be perfect. The system is failed when n −k + 1 component failures have been experienced. The k active components can be viewed as a series subsystem since all of them are required to work for the k-out-of-n:G system to work. As was explained in Chapter 4, the fail-ure rate of a series system is equal to the sum of the failure rates of the components when all components have constant failure rates. If all components in the k-out-of-n:G system are i.i.d. with a constant failure rate λ, then the series subsystem with k active components has a failure rate of kλ. Whenever one of the components in this series subsystem is failed, it is replaced by a standby component and a new se-ries subsystem is formed. Because of the memoryless property of the exponential distribution, each series subsystem follows the exponential lifetime distribution with parameter kλ. The system is failed when the k-component series subsystem, includ-ing the last standby component, is failed. Thus, we have the following expression of system lifetime: Ts = T1 + T2 + · · · + Tn−k+1, (7.81) where Ti represents the lifetime of the ith k-component series subsystem. Even though these k-component series subsystems have components in common, their lifetimes T1, T2, . . . , Tn−k+1 are i.i.d. random variables because of the memoryless property of the exponential distribution. The sum of i.i.d. random variables with the exponential distribution follows the gamma distribution. Thus, Ts follows the gamma distribution with scale parameter kλ and shape parameter n −k + 1: fs(t) = kλe−kλt (kλt)n−k (n −k)! , t ≥0, (7.82) NONREPAIRABLE k-OUT-OF-n SYSTEMS 265 Rs(t) = e−kλt n−k j=0 (kλt) j j!
, (7.83) MTTFs = n −k + 1 kλ .
(7.84) The derivations outlined above are based on the fact that the n−k+1 k-component series subsystems have i.i.d. exponential lifetime distributions. This is satisfied only when each component follows the exponential distribution. When the lifetime distri-bution of each component is not exponential, we cannot use the results shown above.
Warm Standby System with i.i.d. Components and Perfect Switching As mentioned in Chapter 4, warm standby systems are more complicated to analyze because both active and dormant components may fail. Assuming that all components are i.i.d. and the lifetime of each component follows the exponential distribution with parameter λa in the active state and parameter λd in the dormant state, She and Pecht provide a closed-form expression for system reliability function.
Notation • λa: constant failure rate of an active component • λd: constant failure rate of a dormant or standby component • fa(·), Ra(·): pdf and reliability function of an active component, respectively • fd(·), Rd(·): pdf and reliability function of a dormant component, respectively The event that the system survives beyond time t may be expressed as the union of the following mutually exclusive events: 1. The k active components all survive beyond time t.
2. One of the k active components fails in interval (x, x + dx) for 0 < x < t, all n −k dormant components survive beyond time x, and the (n −1)-component subsystem with k active and n−k−1 dormant components survives the remaining time period t −x.
3. One of the n−k dormant components fails in interval (x, x+dx) for 0 < x < t; all k active components survive beyond time x; and the (n−1)-component sub-system with k active and n−k−1 dormant components survives the remaining time period t −x.
Based on this decomposition, we can express R(t; k, n) as R(t; k, n) = e−kλat + t 0 k 1 fa(x)Rd(x)n−k Rs(t −x; k, n −1) dx + t 0 n −k 1 fd(x)Ra(x)k R(t −x; k, n −1) dx.
(7.85) 266 THE k-OUT-OF-n SYSTEM MODEL This equation can be applied recursively until we reach R(z; k, k) = e−kλaz, which is the reliability function of a k-component series system. The closed-form expression for the system reliability is R(t; k, n) = 1 (n −k)!λn−k d n−k i=0 (−1)i n −k i × n−k j=0, j̸=i (kλa + jλd) e−(kλa+iλd)t.
(7.86) When λd = λa = λ, equation (7.86) reduces to the system reliability function of a k-out-of-n:G system with active redundancy given in equation (7.60). When λd = 0, we get the reliability function of a k-out-of-n:G system with cold standby compo-nents, as given in equation (7.83).
Exercises 1. Derive the expression of the MTTF of the warm standby system.
2. Verify that equation (7.86) reduces to equation (7.60) when λd = λa = λ.
3. Verify that equation (7.85) reduces to equation (7.83) when λd = 0.
7.4 REPAIRABLE k-OUT-OF-n SYSTEMS We have discussed the k-out-of-n:G systems with active redundant components, with standby components, or with load-sharing components. In this section, we will de-velop a general model for analysis of such systems when they are repairable. After such a model is developed, we will analyze various system performance measures under different assumptions.
When a k-out-of-n:G system is put into operation, all n components are in good condition. As the system is used, components will fail one after another. The system is failed when the number of working components goes down below k or the number of failed components has reached n −k +1. If resources are allocated to repair failed components, we should be able to keep the number of failed components below n −k + 1 for a much longer time. This way, we expect to prolong the system life cycle. Whenever the number of failed components at any instant of time is higher than n −k, the system is failed and its life cycle is complete.
Many situations exist in which more than one failed component can be repaired simultaneously (in parallel). This can be achieved when there exist more than one repairman or repair facility. As the number of repair facilities is increased, we expect a better chance of extending the system operating time until its first failure.
In the following section we describe a general repairable k-out-of-n:G system model with multiple repair facilities. The components may be in active redundancy, standby, or load sharing. Such a model will allow us to evaluate such performance REPAIRABLE k-OUT-OF-n SYSTEMS 267 measures of the system as mean time to failure, steady-state availability, and mean time between failures.
7.4.1 General Repairable System Model Here are the model descriptions and assumptions: 1. The system is a k-out-of-n:G structure with possibly cold standby and/or load-sharing components.
2. The failure of each component is self-revealing.
3. All active components are i.i.d. following the exponential lifetime distribu-tions. However, the parameter of the lifetime distribution of each component may change depending on the load applied on the component.
4. There are r identical repair facilities available (1 ≤r ≤n −k + 1). Only one repair facility may be assigned to the repair of a failed component. The time needed by any repair facility to repair any failed component is i.i.d. with the exponential distribution.
5. Whenever a component fails, repair immediately commences if a repair facility is available; if not, the failed component must wait for the first available repair facility. Components are repaired on a first-come, first-served basis.
6. The system is considered failed as soon as the number of components in the failed state has reached n −k + 1.
7. While the system is down, no further units can fail.
8. The state of the system is defined to be the number of failed components in the system that are either waiting for or are receiving repair.
9. The system state is decreased by 1 whenever a failed component becomes op-erational and increased by 1 whenever a working component becomes failed.
10. The probability that two or more components are restored to the working con-dition or become failed in a small time interval is negligible.
Notation • i: number of failed components in the system, i = 0, 1, . . . , n −k + 1 • t: time • λi: failure rate of the system when there are i failed components, 0 ≤i ≤n −k • µi: repair rate of the system when there are i failed components, 1 ≤i ≤ n −k + 1 • Pi(t): probability that there are i failed components in the system at time t, 0 ≤ i ≤n −k + 1 • Pi: steady-state probability, Pi = limt→∞Pi(t), 0 ≤i ≤n −k + 1 • P′ i (t): first derivative of Pi(t), 0 ≤i ≤n −k + 1 • Li(s): Laplace transform of Pi(t), 0 ≤i ≤n −k + 1 268 THE k-OUT-OF-n SYSTEM MODEL 0 1 2 n-k n-k+1 . . .
λ0 λ1 λ2 λn-k-1 λn-k µ1 µ2 µ3 µn-k µn-k+1 FIGURE 7.3 General transition diagram for repairable k-out-of-n:G system.
• As(t): point availability of the system at time t • As: steady-state availability of the system, As = limt→∞As(t) Based on the model descriptions, the system state transition diagram is given in Figure 7.3. The numbers in the circles in Figure 7.3 indicate the system states. The system state n −k + 1 indicates system failure.
To evaluate Pi(t + t), we note that at time t + t the system can be in state i only if one of the following disjoint events occurs: 1. At time t the system is in state i and during (t, t + t) no change in system state occurs.
2. At time t the system is in state i −1 and a transition to state i occurs during (t, t + t).
3. At time t the system is in state i + 1 and a transition to state i occurs during (t, t + t).
4. During (t, t + t), the system state changes by two or more.
Since t is very small, the probability of the last event is o(t), as assumed. As a result, we have Pi(t + t) = Pi(t)(1 −λi t)(1 −µi t) + Pi−1(t)λi−1t × (1 −µi−1 t) + Pi+1(t)µi+1 t(1 −λi+1 t) + o(t) = Pi(t) −Pi(t)(λi + µi) t + Pi−1(t)λi−1t + Pi+1(t)µi+1t + o(t).
(7.87) Rearranging the terms in equation (7.87) and letting t →0, we have P′ i (t) = −(λi + µi)Pi(t) + λi−1Pi−1(t) + µi+1Pi+1(t) for i = 0, 1, . . . , n −k + 1, (7.88) where Pi(t) ≡0 for i < 0 or i > n −k + 1. We shall assume the initial conditions Pi(0) = 0 if i ̸= 0 and P0(0) = 1, that is, all components are assumed to be initially in the working state. Considering these initial conditions and assumptions, we can rewrite equation (7.88) as P′ 0(t) = −λ0P0(t) + µ1P1(t), (7.89) REPAIRABLE k-OUT-OF-n SYSTEMS 269 P′ i (t) = −(λi + µi)Pi(t) + λi−1Pi−1(t) + µi+1Pi+1(t) for 1 ≤i ≤n −k, (7.90) P′ n−k+1(t) = −µn−k+1Pn−k+1(t) + λn−k Pn−k(t).
(7.91) One of these equations can be written as a linear combination of the other n −k + 1 equations because the system must be in one of the n −k + 2 states at any instant of time, that is, P0(t) + P1(t) + · · · + Pn−k+1(t) = 1 for any t ≥0.
(7.92) Thus, equation (7.91) should be replaced by equation (7.92). The set of differential equations to be solved are equations (7.89), (7.90), and (7.92). Solving this set of differential equations results in the probability distribution of the system in various states as a function of time. Once this distribution is found, we can evaluate system performance measures such as mean time to failure, mean time between failures, and steady-state availability of the system. The Laplace transform is an effective method for solving systems of differential equations.
Taking Laplace transforms of equations (7.89), (7.90), and (7.92) yields the fol-lowing linear equations in terms of Li(s) for i = 0, 1, . . . , n −k + 1: (s + λ0)L0(s) −µ1L1(s) = 1, (7.93) (s + λi + µi)Li(s) −λi−1Li−1(s) −µi+1Li+1(s) = 0 for 1 ≤i ≤n −k, (7.94) s(L0(s) + L1(s) + · · · + Ln−k+1(s)) = 1.
(7.95) Using matrix notation, we can rewrite these equations as DX = B, (7.96) where X = L0(s) L1(s) .
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Ln−k(s) Ln−k+1(s) (n−k+2)×1 , B = 1 0 .
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0 0 0 0 0 · · · −λn−k−1 s + λn−k + µn−k −µn−k+1 s s s s s · · · s s s , where D is an (n −k + 2) × (n −k + 2) matrix.
270 THE k-OUT-OF-n SYSTEM MODEL 7.4.2 Systems with Active Redundant Components The general repair model is used for evaluation of system performance measures.
Reliability Function and Mean Time to Failure Even when the system under con-sideration is a repairable system, we are still interested in finding the MTTF of the system. This is different from the nonrepairable system case that was analyzed ear-lier. In this case, as components fail, they also get repaired. If repairs are timely enough, the system may experience more than n −k cumulative component failures without experiencing a system failure.
Since we are interested in finding the MTTF of the system, we need to assume that state n −k +1 is an absorbing state. As soon as the number of components in the failed state at any instant of time reaches n −k + 1, the system is considered failed.
As a result, we have to assume µn−k+1 = 0. When the system is in state i(0 ≤ i ≤n −k), there are i failed components and n −i active working components in the system, and the failure rate of the system is λi = (n −i)λ. If the number of failed components is less than or equal to the total number of repair facilities, all failed components are being repaired, and thus the repair rate of the system is µi = iµ(1 ≤i ≤r). However, if i > r, µi will be a constant equal to rµ as all repair facilities are being used and some failed components are waiting for repair.
The following summarizes these conditions: λi = (n −i)λ, 0 ≤i ≤n −k, µi = iµ for 0 ≤i ≤r, rµ for r < i ≤n −k, 0 for i = n −k + 1.
In the system of differential equations, Pn−k+1(t) is the probability that the sys-tem is in the failed state at time t. Thus, the reliability function of the system is Rs(t) = 1 −Pn−k+1(t). We can use the Laplace transform to find Pn−k+1(t).
Because µn−k+1 = 0, equation (7.91) can be written as P′ n−k+1(t) = λn−k Pn−k(t) = kλPn−k(t), (7.97) Pn−k+1(t) = kλ t 0 Pn−k(x) dx, (7.98) assuming that Pn−k+1(0) = 0. Note that Pn−k+1(t) and P′ n−k+1(t) actually rep-resent the CDF and the pdf, respectively, of the system lifetime. Also, because µn−k+1 = 0, Pn−k+1(t) disappears from equation (7.90). As a result, equa-tions (7.89) and (7.90) include n −k + 1 equations and n −k + 1 variables, P0(t), P1(t), . . . , Pn−k(t). After the Laplace transform, the system of linear equa-tions (7.96) includes n −k + 1 equations and n −k + 1 variables. The last entries in vectors X and B and the last row and the last column of matrix D are removed.
REPAIRABLE k-OUT-OF-n SYSTEMS 271 The method of determinants may be used to solve equation (7.96) for Ln−k(s), Ln−k(s) = |D′| |D| , (7.99) where D′ is the matrix obtained from D by replacing the (n −k + 1)st column (the last column) of D by vector B. The determinant of matrix D′ is |D′| = n−k−1 i=0 λi = n!λn−k k!
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(7.100) Since |D| is a polynomial in s with degree n −k + 1 and leading coefficient 1, we can write |D| = n−k+1 i=1 (s −si) where each si is a (distinct) root of the polynomial.
Therefore, 1 |D| = n−k+1 i=1 (s −si) −1 = n−k+1 j=1 n−k+1 i=1,i̸= j (s j −si) −1 1 s −s j , (7.101) Ln−k(s) = |D′| |D| = n!λn−k k!
n−k+1 j=1 n−k+1 i=1,i̸= j (s j −si) −1 1 s −s j .
(7.102) An inverse Laplace transform of equation (7.102) yields Pn−k(t) = n!λn−k k!
n−k+1 j=1 n−k+1 i=1,i̸= j (s j −si) −1 es jt, (7.103) Rs(t) = 1 −n!λn−k+1 (k −1)!
n−k+1 j=1 n−k+1 i=1,i̸= j (s j −si) −1 t 0 es j x dx = 1 −n!λn−k+1 (k −1)!
n−k+1 j=1 es jt −1 s j n−k+1 i=1,i̸= j (s j −si) −1 = 1 − n−k+1 j=1 C j(es jt −1), (7.104) where C j = n−k+1 i=1,i̸= j si n−k+1 i=1,i̸= j (s j −si) −1 .
272 THE k-OUT-OF-n SYSTEM MODEL Derivation of equation (7.104) uses the fact that [n!/(k −1)!]λn−k+1 = n−k+1 i=1 si, which is the constant term in the polynomial representing the determinant of the matrix D. It can be shown that si < 0 for all i; thus t2Rs(t) →0 as t →+∞.
Now that an expression of the system reliability function is known, we can derive the MTTF of the system based on its definition: MTTFs = ∞ 0 Rs(t) dt = n−k i=0 ∞ 0 Pi(t) dt.
(7.105) Since Li(0) = $ ∞ 0 Pi(t) dt, MTTFs can also be written as MTTFs = n−k i=0 Li(0).
(7.106) If we let D′ i denote the matrix obtained from D by replacing the (i + 1)th column by the vector B, then Li(0) = |D′ i|s=0 |D|s=0 .
(7.107) Since |D|s=0 = n!λn−k+1 (k −1)! , (7.108) we have MTTFs = (k −1)!
n!λn−k+1 n−k i=0 |D′ i|s=0.
(7.109) Provided that the s j’s are known, another way for evaluation of MTTFs is to use the fact that P′ n−k+1(x) is actually the pdf of the system lifetime. With equation (7.97), an equivalent form for Rs(t) is Rs(t) = ∞ t P′ n−k+1(x) dx = ∞ t kλPn−k(x) dx.
(7.110) This implies that MTTFs = ∞ 0 ∞ t kλPn−k(x) dx dt.
(7.111) Exercises 1. Derive Rs(t) and MTTFs when k = 2, n = 3, and r = 1.
2. Derive Rs(t) and MTTFs when k = 3, n = 5, and r = 2.
REPAIRABLE k-OUT-OF-n SYSTEMS 273 3. Verify equation (7.101).
4. Verify equation (7.104).
Steady-State Availability As the k-out-of-n:G system is used, the number of failed components in the system changes. When it reaches n−k+1, the system is failed and all repair facilities are utilized to repair failed components. As soon as the number of failed components goes down below n−k+1, the system starts working again. Thus, the system state changes between up and down over time. The probability that the system is in the working state at time t is called the point availability of the system.
The system point availability is given by As(t) = 1 −Pn−k+1(t).
(7.112) To evaluate the availability of the system, we have to treat state n−k+1 as a transient state too, that is, µn−k+1 = rµ. The following summarizes the system parameters: λi = (n −i)λ, 0 ≤i ≤n −k, µi = iµ if 0 ≤i ≤r, rµ if r < i ≤n −k + 1.
To evaluate the point availability As(t) of the system, the general repair sys-tem model can be used. With the Laplace transform technique, it suffices to find Ln−k+1(s). Using the method of determinants, we observe that Ln−k+1(s) = |D′| |D| , (7.113) where D′ is the matrix obtained from D upon replacing the (n −k + 2)nd column (the last column) by vector B. Since |D′| = n!λn−k+1 (k −1)! , (7.114) we have Ln−k+1(s) = n!λn−k+1 (k −1)!|D|.
(7.115) The steady-state availability of the system is given by As = lim t→∞As(t) = 1 −lim t→∞Pn−k+1(t) = 1 −lim s→0 sLn−k+1(s) = 1 −lim s→0 sn!λn−k+1 (k −1)!|D|.
(7.116) The procedure outlined above is necessary if one is interested in the availability of the system as a function of time t. However, the mathematical derivation is very 274 THE k-OUT-OF-n SYSTEM MODEL tedious. If one is only interested in the steady-state availability of the system, there is no need to derive the point availability first. The differential equations of the system given in equations (7.89), (7.90), and (7.92) can be used directly. First, we try to find the steady-state (or time-independent) distribution of the system in different states.
This solution is provided by defining Pi = lim t→∞Pi(t) for i = 0, 1, . . . , n −k + 1, (7.117) provided that the limits exist. Taking the limit of both sides of equations (7.89), (7.90), and (7.92) as t →∞and noting that limt→∞P′ i (t) = 0, we obtain −λ0P0 + µ1P1 = 0, (7.118) −(λi + µi)Pi + λi−1Pi−1 + µi+1Pi+1 = 0 for i = 1, . . . , n −k + 1, (7.119) P0 + P1 + · · · + Pn−k+1 = 1.
(7.120) A simple induction argument shows that Pi = λ0 · · · λi−1 µ1 · · · µi P0 for i = 1, . . . , n −k + 1.
(7.121) Applying equation (7.120), we have P0 = 1 + n−k+1 i=1 λ0 · · · λi−1 µ1 · · · µi −1 , (7.122) Pi = λ0 · · · λi−1 µ1 · · · µi 1 + n−k+1 i=1 λ0 · · · λi−1 µ1 · · · µi −1 for i = 1, . . . , n −k + 1, (7.123) As = 1 −Pn−k+1 = 1 − λ0 · · · λn−k µ1 · · · µn−k+1 1 + n−k+1 i=1 λ0 · · · λn−k µ1 · · · µn−k+1 −1 .
(7.124) Mean Time between Failures The mean time between failures (MTBF) is defined to be the expected length of operating time of the system between successive failures.
It does not include the time that the system spends in the failed state. The mean time to repair (MTTR) indicates the average length of time that the system stays in the failed state.
It is often necessary to calculate MTBF quickly in order to make timely design decisions. Although a general formula is known, it is not easily remembered nor REPAIRABLE k-OUT-OF-n SYSTEMS 275 derived. Angus presents a simple way of obtaining an expression of MTBF.
With this method, the MTBF expression is easily reproduced by remembering a few simple concepts.
In the following, we assume that there are n −k + 1 repair facilities. This means that no failed components need to wait for repair. The system transition parameters are summarized as follows: λi = (n −i)λ, 0 ≤i ≤n −k, µi = iµ, 1 ≤i ≤n −k + 1.
The MTBF of the system is the average (successful operating) time between visits to state n −k + 1, the system down state. It is the average time for the system to go from a working state (with possibly some failed components) to the failed state.
This should be distinguished from the mean time to the first failure (MTTF) of the system, which is defined as the average time for the system to go from the working state with zero failed components to the failure state. In the following, we illustrate the derivation of MTBF of a k-out-of-n:G system.
Let N(t) indicate the number of failed components in the system at time t. Be-cause of the Markov nature of the process {N(t); t ≥0}, once the process arrives at the state n −k +1, the sequence of times between successive visits to state n −k +1 forms an i.i.d. sequence of random variables. The mean of each of these random variables is MTBF + MTBR. The portion of this average that represents success-ful operation time is MTBF. It follows from the renewal theory (in particular, the analysis of alternating renewal processes) that As = lim t→∞Pr(system is working at time t) = lim t→∞Pr(N(t) ≤n −k) = MTBFs MTBFs + MTBRs .
(7.125) Solving for MTBFs gives MTBFs = As × MTBRs 1 −As .
(7.126) Each component has a MTBF of 1/λ and a MTBR of 1/µ. The steady-state avail-ability of each component A is A = µ λ + µ.
When the system is down, there are n−k +1 units undergoing repair, and because of the Markov assumptions, MTBRs = 1 µ(n −k + 1).
(7.127) 276 THE k-OUT-OF-n SYSTEM MODEL Because there is always a repair facility available for a failed component and the components are s-independent, the limiting probability (t →∞) of finding exactly j(k −1 ≤j ≤n) components working at time t is given by the truncated binomial distribution (truncated because the Markov process is not allowed to visit states n − k + 2, n −k + 3, . . . , n): Pj = Pr(exactly j components are available) Pr(At least k −1 components are available) (7.128) = n j A j(1 −A)n−j n i=k−1 n i Ai(1 −A)n−i for j = k −1, k, k + 1, . . . , n, (7.129) As = Pk + Pk+1 + · · · + Pn = n i=k n i Ai(1 −A)n−i n i=k−1 n i Ai(1 −A)n−i , (7.130) MTBFs = AsMTBRs 1 −As = n i=k n i Ai(1 −A)n−i µ(n −k + 1) n k−1 Ak−1(1 −A)n−k+1 = n−k j=0 n j (λ/µ) j kλ n k (λ/µ)n−k .
(7.131) This formula is easily recalled by remembering the following basic concepts: 1. As = MTBFs/(MTBFs + MTBRs).
2. MTBR = 1/[µ(n−k +1)] since n−k +1 components are under simultaneous repair when the system is down.
3. The number of components working as t →∞follows the truncated binomial distribution.
4. MTBFs = As × MTBRs/(1 −As).
Exercise 1. Verify equation (7.131).
7.4.3 Systems with Load-Sharing Components When the working components equally share the load of the system, the failure rate of each component depends on the load that it has to carry. The load that is allocated to each component depends on the number of failed components that exist in the system. Shao and Lamberson provide an analysis of a repairable k-out-of-n:G system with load-sharing components considering imperfect switching. The sensing and switching mechanism is responsible for detection of component failures and the redistribution of the load of the system equally among surviving components. Sys-tem performance measures such as reliability and availability are analyzed. Several REPAIRABLE k-OUT-OF-n SYSTEMS 277 errors exist in this paper that are corrected by Akhtar . Newton provides an alternative argument for evaluation of the MTTF and MTBF of such systems.
In this section, we consider a repairable k-out-of-n:G system with load-sharing components. For simplicity of analysis, the sensing and switching mechanism is as-sumed to be perfect. Service is needed whenever the number of failed components in the system changes (when another component is failed or when a failed component is repaired) to redistribute the load of the system. When the sensing and switching mechanism is imperfect, we say that the system has imperfect fault coverage. This will be discussed in a later section in this chapter. Other assumptions are as given in the general model for a repairable k-out-of-n:G system described in Section 7.4.1.
We provide expressions of system reliability, availability, MTTF, and MTBF of such systems.
Assumptions 1. The failure rates of all working components are the same. They are dependent on the number of working components in the system.
2. A repaired component is as good as new and immediately shares the load of the system.
Notation • λi: failure rate of each component when there are i failed components, i = 0, 1, . . . , n −k. Generally, we have λ0 ≤λ1 ≤· · · ≤λn−k.
The only difference between the load-sharing system model in this section and the repairable system model with active redundant components discussed in Section 7.4.2 is that the component failure rate is not a constant any more. The failure rate of the system with i failed components, αi, can be written as αi = (n −i)λi, i = 0, 1, . . . , n −k.
(7.132) The same techniques as used in Section 7.4.2 can be used to derive the required system performance measures.
Exercise 1. Find expressions of Rs(t), MTTFs, MTBFs, and As.
7.4.4 Systems with both Active Redundant and Cold Standby Components Morrison and Munshi provide an analysis of a k-out-of-n:G system with ad-ditional standby components. The system consists of n active components where at least k of them have to work for the system to work. In addition, there are m spare components available. All components are i.i.d. There are r repair facilities. Repairs are perfect. The lifetime of an active component is exponentially distributed. Repair 278 THE k-OUT-OF-n SYSTEM MODEL time also follows the exponential distribution. Detection of active component failure and switching of a standby component to the active state are perfect and instant. We will use i to represent the number of failed components in the system and to indicate the state of the system. Both cold and hot standby are analyzed by Morrison and Munshi. We will not discuss the hot standby case here as it is exactly the same as if all of the components are active, which has been discussed in previous sections.
Notation • n: number of active components used • m: number of standby components or spares • γ : λ/µ Since the spare components are in cold standby, we have the following failure rates and repair rates at different system states: λi = nλ if 0 ≤i ≤m, (n + m −i)λ if m < i ≤n + m, 0 if i > n + m, (7.133) µi = iµ if i ≤r, rµ if i > r.
(7.134) With these system state transition parameters and following the same procedure as outlined in Section 7.4.1, we can develop a set of differential equations for state probabilities. Solving these differential equations, we can obtain expressions of sys-tem state probabilities. Letting time go to infinity, we find the following steady-state probabilities: Pi = ni i! γ i P0, 0 ≤i ≤min{r, m}, ni ri−rr!γ i P0, r + 1 ≤i ≤m, nmγ i n!
(n + m −i)!i! P0, m + 1 ≤i ≤r, 1 ri−rr!nmγ i n!
(n + m −i)! P0, max{r, m} + 1 ≤i ≤n + m.
(7.135) If all spares are in hot standby, the following is provided for verification purposes: Pi = (n + m)!
(n + m −i)!i!γ i P0, 0 ≤i ≤r, (n + m)!
(n + m −i)!ri−rr!γ i P0, i > r.
WEIGHTED k-OUT-OF-n:G SYSTEMS 279 Exercise 1. Verify equation (7.135).
7.5 WEIGHTED k-OUT-OF-n:G SYSTEMS Wu and Chen propose a variation of the k-out-of-n:G system, called the weighted k-out-of-n:G model. In a weighted k-out-of-n:G system, component i car-ries a weight of wi, wi > 0 for i = 1, 2, . . . , n. The total weight of all components is w, w = n i=1 wi. The system works if and only if the total weight of working components is at least k, a prespecified value. Since k is a weight, it may be larger than n because they have different measuring units. Such a weighted k-out-of-n:G system is equivalent to a weighted (w−k +1)-out-of-n:F system wherein the system fails if and only if the total weight of failed components is at least w−k+1. With this definition, the k-out-of-n:G system is a special case of the weighted k-out-of-n:G system wherein each component has a weight of 1.
A recursive equation is provided by Wu and Chen . In the following, R(i, j) represents the probability that a system with j components can output a total weight of at least i. Then, R(k, n) is the reliability of the weighted k-out-of-n:G system. The following recursive equation can be used for reliability evaluation of such systems: R(i, j) = p j R(i −w j, j −1) + q j R(i, j −1), (7.136) which requires the following boundary conditions: R(i, j) = 1 for i ≤0, j ≥0, (7.137) R(i, 0) = 0 for i > 0.
(7.138) It should be noted that wi (1 ≤i ≤n) may not be integer. When wi = 1 for all 1 ≤i ≤n, we have the usual k-out-of-n:G system. The computational complexity of equation (7.136) is O(k(n −k + 1)) when wi = 1 for all i. However, when wi > 1 for all 1 ≤i ≤n, the number of terms to be computed may be much less than k(n −k + 1), as illustrated in the following example.
Example 7.4 Consider a weighted 5-out-of-3:G system. It has three components with weights 2, 6, and 4. The system works if and only if the total weight of working components is at least 5.
This is a very simple example. We can easily solve the problem without using equation (7.136). The following are the minimal paths of the system: Component 2 works with a total output of 6.
Components 1 and 3 work with a total output of 6.
Thus, we can find the system reliability as Rs = Pr(x2 ∪x1x3) = Pr(x2) + Pr(x2 x1x3) = p2 + q2 p1 p3.
280 THE k-OUT-OF-n SYSTEM MODEL If we apply equation (7.136), here are the terms to be calculated: R(1, 1) = p1R(−1, 0) + q1R(1, 0) = p1, R(5, 1) = p1R(3, 0) + q1R(5, 0) = 0, R(1, 2) = p2R(−5, 1) + q2R(1, 1) = p2 + p1q2, R(5, 2) = p2R(−1, 1) + q2R(5, 1) = p2, R(5, 3) = p3R(1, 2) + q3R(5, 2) = p3(p2 + p1q2) + q3 p2 = p2 + q2 p1 p3.
The total number of terms calculated is only five, much less than (n+1)(k+1) = 20. |
11215 | https://mathoverflow.net/questions/69857/bits-and-orbits | rt.representation theory - Bits and orbits - MathOverflow
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Bits and orbits
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I know this title makes what I am about to ask sound like an off topic CS theory question but please bear with me because I assure you that it is not! (Well mostly, actually I am about ~90% certain that this is a perhaps routine application of representation theory...) If you just want the problem without the backstory, skip to the bottom.
Anyway, I would like to get some way to uniquely encode tuples of finite sets of bits up to a some type of reordering. In other words, let's suppose I have k binary strings all of equal length n, or in other words a 1,a 2,...a k∈F n 2 a 1,a 2,...a k∈F 2 n, and I want to find some way to encode them uniquely as a binary string of minimal length so that I can insert them into a dictionary for later use.
Now if I care about the order of each of these guys, then there is really only one way to go, which is to just concatenate each of the bit strings together into one big string of length n k n k.
However, suppose that I don't care about the ordering (for example, I consider these things to be sets not sequences), then instead I only want to encode the orbit of this sequence under the usual action of S k S k. Now there are at least a couple of ways to do this; for example I could sort the sequence lexicographically, or by being more clever I could re-encode them using symmetric polynomials. As a simple application, I could use the usual basis and write the result equivalently as:
b 1=a 1+a 2+...a k b 1=a 1+a 2+...a k
b 2=(a 1 a 2)+(a 1 a 3)+...b 2=(a 1 a 2)+(a 1 a 3)+...
....
b k=a 1 a 2...a k b k=a 1 a 2...a k
Of course this can be computed in time on the order of n k log 2(k)n k log 2(k) by divide and conquer and FFT polynomial multiplication. Ignoring the subtleties of the time complexity of computing all these sums for the moment, this suggests the following interesting question:
Question 1: On average, what is the smallest length binary string needed to encode k k length n n binary strings up to permutation? (which can be computed in polynomial time).
It should be about n k−k log(k)n k−k log(k), since there are k!k! permutations, but I am not exactly sure how to get there.
Now, as a follow up question, what is the best you can do for an arbitrary group? For example, if you only care about encoding the binary sequences up to alternation, what is the best way to do it? One possibility is to use the Vandermonde polynomials, however there is a need to be a bit careful here, since if this is done directly over F n 2 F 2 n, then one runs into the small problem that a k=−a k a k=−a k. So this leads to the following pair of questions:
Let G G be a group acting faithfully on the indices of a finite set of binary strings a 1,a 2,...a k a 1,a 2,...a k whose individual lengths are at most n n:
Question 2: Is there an efficiently computable encoding of the orbit of a 1,a 2,...a k a 1,a 2,...a k under G G of length at most n k n k? (In other words, two sequences in the same orbit would map to the same string, and this encoding can be found in time polynomial in log(G),n log(G),n and k k.)
Question 3: Is there an efficiently computable encoding on the order of length O(n k−log(|G|))O(n k−log(|G|)) of the orbit?
EDIT: Adjusted the wording of the first question to be more specific. Also fixed my brain fart with dividing out by the entropy of G instead of subtracting (in my defense, it was about 3am my local time when I wrote this).
EDIT 2: I tried to be a bit more specific with the time complexity requirements to rule out silly things like "take lexicographically first element of the orbit".
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edited Jul 9, 2011 at 17:05
MikolaMikola
asked Jul 9, 2011 at 6:52
MikolaMikola
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Here is a start toward an encoding, for k not too small. List the smallest string, and then with some agreed upon character for separation, store the differences in binary arithmetic between the strings. Alternatively, look at tries and other data structures (Bloom filters?) for storage of large lists of IP addresses. Gerhard "Email Me About System Design" Paseman, 2011.07.09 Gerhard Paseman –Gerhard Paseman 2011-07-09 07:24:45 +00:00 Commented Jul 9, 2011 at 7:24
1 If n n is larger than about log k logk, then the answer to the first question is n k−k log k n k−k logk. That is because in most k k-tuples of n n-strings, all the strings are distinct, and each additional condition a i=a j a i=a j gives a family of strings that is constant times smaller, leading to a geometric series. If n n is constant, and k k is large, then the answer is on the order of 2 n log k 2 n logk bits.Boris Bukh –Boris Bukh 2011-07-09 10:19:27 +00:00 Commented Jul 9, 2011 at 10:19
1 If n n is large, then encodoing can be done via arithmetic coding. Just write the strings one-by-one in sorted order. The conditional distribution of the next string given the knowledge of the preceding string is fairly simple to compute. Of course, it is very impractical, and it will requires arithmetic with numbers of about n k n k bits.Boris Bukh –Boris Bukh 2011-07-09 10:22:53 +00:00 Commented Jul 9, 2011 at 10:22
@Boris Bukh: Ah! Good catch. I messed up my estimate for the entropy calculation (tried dividing when I should've subtracted). Also you are right and that entropy sorted ordered would work, though I wonder if there is a more direct argument. Specifically, if you take some symmetric polynomials (for example the Shur polynomials), or more generally any G-invariant polynomials, then their coefficients should have lower entropy than the input bits and so their should be a way to squeeze log(G) bits out of them somehow.Mikola –Mikola 2011-07-09 15:13:54 +00:00 Commented Jul 9, 2011 at 15:13
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I suspect the problem is intended to include some requirement that the coding scheme be efficiently computable, because if one doesn't care at all about efficiency then the following (rather silly) scheme would provide codes of the shortest possible length. By the "first member" of an orbit, I mean the lexicographically first string (of length n k n k) that is the concatenation of a k k-tuple in that orbit. Imagine a list of all these first members, listed in lexicographic order; the length of this list is the number Q Q of orbits and each of the Q Q entries is a bit string of length n k n k. Code any orbit by (the binary representation of) the position of its first member in this list of all first members. So the code is an integer in the range from 1 to Q Q and its length in bits is therefore essentially log Q logQ. That's the best you can hope for when you're coding Q Q things using binary digits.
I'm quite confident the OP intended some computability (or other reasonableness) requirement that would exclude such a coding. It's not obvious, though, what the requirement should be. Should one be able to easily compute, given a k k-tuple a 1,…,a k a 1,…,a k, which orbit it belongs to? Or is it enough to be able to easily decide, given a tuple and an orbit, whether the tuple belongs to the orbit? One can probably imagine other computability criteria as well.
EDIT to take into account the edited version of the question:
Now that a computability requirement has been added, the problem has become considerably more difficult. In particular, it subsumes the graph isomorphism problem as follows. Let n=1 n=1 and k=v(v−1)/2 k=v(v−1)/2 for some big integer v, so we're dealing with bit strings of length equal to the number of two-element subsets of {1,2,…,v}{1,2,…,v}. Identify the indices 1 through k k with these 2-element sets in some canonical way (e.g. lexicographic). A bit string of length k k is thereby identified with a graph on the vertex set {1,2,…,v}{1,2,…,v}. Let G G be the symmetric group of permutations of {1,2,…,v}{1,2,…,v}, acting in the obvious way on the 2-element subsets and thus on {1,2,…,k}{1,2,…,k}. Then the orbits amount to the isomorphism classes of graphs. If we had an efficient way to encode orbits, and to compute the code of an orbit from any member of the orbit, then we'd have an efficient test for graph isomorphism. In particular, the proposed computability requirement in the revised question, polynomial time in n n, k k, and log(|G|)log(|G|), would amount in this case to polynomial time in v v.
Since it is still (as far as I know) open whether graph isomorphism can be computed in polynomial time, there is no known coding of orbits in this example, subject to the propoosed computability requirements.
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edited Jul 9, 2011 at 23:54
answered Jul 9, 2011 at 15:38
Andreas BlassAndreas Blass
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EDIT - now that the question has been made more precise, my answer is out-of-date.
OLD - Your first question
Question 1: What is the smallest length binary string needed to encode k length n binary strings up to permutation?
has no definitive answer, as Kolmogorov complexity is not computable. One more point - if you don't want the smallest answer, but just a reasonable answer, then knowing the relative size of k k is very important. For example, when k k is around 2 n 2 n Gerhard's scheme works well. If k≫2 n k≫2 n then one could think of the strings as n n-bit binary numbers, sort the numbers by size, and count the number of each number.
As for question 2 - given A=(a 1,a 2,…,a k)A=(a 1,a 2,…,a k) compute the orbit G⋅A G⋅A. Let B B be the lexicograpically first element of G⋅A G⋅A; use this as the representative of A A.
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2 The first question has a well-defined answer: the number of unordered k k-tuples of numbers between 1 1 and 2 n 2 n is computable, and its logarithm is the required number of bits. Boris Bukh –Boris Bukh 2011-07-09 10:27:39 +00:00 Commented Jul 9, 2011 at 10:27
1 @Boris - the original question was broadly enough stated to allow this answer. The new version of the question is more interesting. Sam Nead –Sam Nead 2011-07-09 22:39:08 +00:00 Commented Jul 9, 2011 at 22:39
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Author / Uploaded
R K Singal
_Table of contents :
Cover
Halftitle
Title
Copyright
Preface
Contents
Part I: Fundamentals
Chapter 1: Classification and Applications
1.1 Introduction
1.2 Hydro Energy
1.3 Classification of Fluid Machines
1.4 Applications of Hydro Turbines
1.5 Applications of Centrifugal Pumps
Question Bank No. 1
Chapter 2: Basic Principles of Hydraulic Flow and Jet Theory
2.1 Introduction
2.2 Principles of Fluid Mechanics
2.3 Basic Concepts of Hydraulic Flow
2.4 Continuity Equation
2.5 Three-Dimensional Flow
2.6 Momentum Equation
2.7 Applications of Momentum Equation
2.8 Moment of Momentum Equation
2.9 Euler’s Fundamental Equation
2.10 Jet Theory
2.11 Jet Force on Stationary Flat Plate
2.12 Jet Force on Moving Flat Plate
2.13 Jet Force on Curved Plate when Jet Strikes Tangentially
2.14 Jet Propulsion of Ships
Question Bank No. 2
Tutorial Sheet No. 2
Part II: Turbines
Chapter 3: Pelton Turbine
3.1 Introduction
3.2 Comparative Study of Hydro Turbines
3.3 Pelton Turbine
3.4 Layout Arrangements
3.5 Velocity Diagram of Bucket
3.6 Turbine Losses and Efficiency
3.7 Design of Pelton Turbine
3.8 Design Example of a Pelton Turbine
Question Bank No. 3
Tutorial Sheet No. 3
Chapter 4: Francis Turbine
4.1 Introduction
4.2 Specific Speed and Turbine Type
4.3 Main Components
4.4 Velocity Triangles of Francis Turbine
4.5 Sankey Diagram
4.6 Design of Francis Turbine
4.7 Design Problem
Question Bank No. 4
Tutorial Sheet No. 4
Chapter 5: Propeller and Kaplan Turbines
5.1 Introduction
5.2 Kaplan Turbine
5.3 Turbine Constants
5.4 Design Problem
5.5 Other Turbines
Question Bank No. 5
Tutorial Sheet No. 5
Chapter 6: Draft Tube and Cavitation
6.1 Introduction
6.2 Functions of Draft Tube
6.3 Types of Draft Tubes
6.4 Cavitation
6.5 Factors Causing Cavitation
6.6 Methods to Avoid Cavitation
Question Bank No. 6
Tutorial Sheet No. 6
Chapter 7: Governing of Hydraulic Turbines
7.1 Introduction
7.2 Oil Pressure Governor
7.3 Double Regulation System of Pelton Turbine
7.4 Governing of Reaction Turbines
Question Bank No. 7
Chapter 8: Dynamic Similarity and Performance Characteristics
8.1 Introduction
8.2 Dynamic Similarity of Model and Prototype
8.3 Specific Speed
8.4 Scale Ratio
8.5 Unit Quantities
8.6 Model Relationships
8.7 Performance Characteristics
Question Bank No. 8
Tutorial Sheet No. 8
Part III: Pumps
Chapter 9: Centrifugal Pumps
9.1 Introduction
9.2 Classification of Centrifugal Pumps
9.3 Pump Performance
9.4 Sankey Diagram
9.5 Velocity Diagrams
9.6 Design of Impeller
9.7 Dynamic Similarity and Performance Characteristics
9.8 Cavitation and Maximum Suction Lift
9.9 Multistage Pumps
9.10 Pump Operation
Question Bank No. 9
Tutorial Sheet No. 9
Chapter 10: Reciprocating Pump
10.1 Introduction
10.2 Pump Installation and Performance
10.3 Classification of Reciprocating Pumps
10.4 Piston Motion
10.5 Indicator Diagram
10.6 Factors Affecting Performance of Reciprocating Pumps
Question Bank No. 10
Tutorial Sheet No. 10
Chapter 11: Special Pumps
11.1 Introduction
11.2 Axial Flow Pump
11.3 Design of Axial Flow Pump
11.4 Vertical Turbine Pump
11.5 Rotary Displacement Pumps
11.6 Jet Pumps
11.7 Hydraulic Ram
11.8 Hydraulic Coupling
11.9 Torque Converter
Question Bank No. 11
Tutorial Sheet No. 11
Part IV: Systems And Plants
Chapter 12: Hydraulic Systems
12.1 Introduction
12.2 Hydraulic Accumulator
12.3 Differential Accumulator
12.4 Hydraulic Intensifier
12.5 Hydraulic Jack or Press
12.6 Hydraulic Lift
Question Bank No. 12
Tutorial Sheet No. 12
Chapter 13: Hydraulic Power Plants
13.1 Introduction
13.2 Hydro Power Potential
13.3 Classification of Hydro Power Plants
13.4 Location of Hydro Power Plants
13.5 Hydrology
Question Bank No. 13
Tutorial Sheet No. 13
Part V: Case Studies
Case I : Floating Type Micro Hydro Power Plants
Case II : An Alpine Hydro Village
Case III : Operation and Maintenance of Centrifugal Pumps for 257 Energy Conservation
Case IV : Hydraulic Design of a Boiler Feed Pump to Ensure Stable Operation at Reduced Flows
Case V : Hydro Power Plants: Great Ecological Havocs
Part VI: Objective Questions
Index
Backcover_
Citation preview
HYDRAULIC MACHINES
The book has been divided into the following six parts containing 13 chapters: • Fundamentals of Fluid Flow • Turbines • Pumps • Systems and Power Plants • Case Studies • Objective Questions Contents: PART I – Fundamentals: Classification and Applications / Basic Principles of Hydraulic Flow and Jet Theory, PART II – Turbines: Pelton Turbine / Francis Turbine / Propeller and Kaplan Turbines / Draft Tube and Cavitation / Governing of Hydraulic Turbines / Dynamic Similarity and Performance Characteristics, PART III – Pumps: Centrifugal Pumps / Reciprocating Pumps / Special Pumps, PART IV – Systems and Plants: Hydraulic Systems / Hydraulic Power Plants / PART V – Case Studies, PART VI – Objective Questions / Index.
Mridul Singal is an engineer with more than 17 years of experience. He has undertaken research and development work in aerospace industry at DRDO, and software industry in various MNCs in India and abroad. Presently, he is working as IT Architect at IBM India Pvt. Ltd. Rishi Singal is an engineer with more than 12 years of experience in major steel and software industries in India and abroad. He is a PMI-certified PMP.® Presently, he is working as Project Manager for Accenture India Pvt. Ltd.
978-93-89307-65-8
HYDRAULIC MACHINES
R.K. Singal • Mridul Singal Rishi Singal
R.K. Singal is a senior professor and consultant. He has an experience of more than 47 years in India and abroad with MNCs and academic institutions of international repute. He has been teaching at IIT, NIT, UPTU and other universities, and worked as Director of various institutions. He is the author of many books and research publications in national and international journals. He has also been decorated by the President of India.
HYDRAULIC MACHINES
Hydraulic Machines (Fluid Machinery) has been designed as a textbook for engineering students specialising in mechanical, civil, electrical, hydraulics, chemical and power engineering. The highlights of the book are simple language supported by analytical and graphical illustrations. A large number of theory questions and numerical problems with solution hints have been annexed at the end of every chapter. A large number of objective questions have been included to help students opting for competitive examinations. Five case studies based on research have been included, which can be advantageously used by practising engineers pursuing research design and consultancy careers. Complete design of hydraulic machines has been demonstrated with the help of suitable examples.
TM
` 405/-
Fluid Machinery R.K. Singal Mridul Singal Rishi Singal Distributed by:
9 789389 307658 TM
HYDRAULIC MACHINES (Fluid Machinery)
HYDRAULIC MACHINES (Fluid Machinery)
R. K. Singal Professor and Consultant rk.singal@gmail.com
Mridul Singal B.Tech. (IIT, Kanpur), M.Tech. (IISC, Bangalore) I/T Architect, IBM India Pvt. Ltd.
Rishi Singal B.Tech. (PEC Chandigarh), PMP (PMI, USA) Accenture India Pvt. Ltd.
©Copyright 2019 I.K. International Pvt. Ltd., New Delhi-110002. This book may not be duplicated in any way without the express written consent of the publisher, except in the form of brief excerpts or quotations for the purposes of review. The information contained herein is for the personal use of the reader and may not be incorporated in any commercial programs, other books, databases, or any kind of software without written consent of the publisher. Making copies of this book or any portion for any purpose other than your own is a violation of copyright laws. Limits of Liability/disclaimer of Warranty: The author and publisher have used their best efforts in preparing this book. The author make no representation or warranties with respect to the accuracy or completeness of the contents of this book, and specifically disclaim any implied warranties of merchantability or fitness of any particular purpose. There are no warranties which extend beyond the descriptions contained in this paragraph. No warranty may be created or extended by sales representatives or written sales materials. The accuracy and completeness of the information provided herein and the opinions stated herein are not guaranteed or warranted to produce any particulars results, and the advice and strategies contained herein may not be suitable for every individual. Neither Dreamtech Press nor author shall be liable for any loss of profit or any other commercial damages, including but not limited to special, incidental, consequential, or other damages. Trademarks: All brand names and product names used in this book are trademarks, registered trademarks, or trade names of their respective holders. Dreamtech Press is not associated with any product or vendor mentioned in this book. ISBN: 978-93-89307-65-8 EISBN: 978-93-89872-60-6
Preface India is bestowed with large hydel power potential and can set up 1,50,000 MW capacity. Presently, only about 30,000 MW of capacity has been created. There is a big gap to exploit this renewable and clean source of energy. Every industrial process involves the transfer of liquid from one level of pressure to another. Pumps have become essential part of all industrial processes and, in turn, major consumers of energy. Therefore, hydraulic machines mainly consisting of hydraulic turbines and pumps have been included as a core subject for undergraduate and postgraduate students of Mechanical Engineering, Electrical Engineering, Civil Engineering, Hydraulic Engineering, Power Engineering, etc. The present book has been designed to meet the above requirements. The book will be useful for the students, teachers and library reference. The requirements of engineers interested in taking up competitive examinations and research career have been fully taken care of. Authors
Contents Preface
v
PART IFUNDAMENTALS
1
Classification and Applications 1.1 Introduction 1.2 Hydro Energy 1.3 Classification of Fluid Machines 1.4 Applications of Hydro Turbines 1.5 Applications of Centrifugal Pumps Question Bank No. 1
3 3 3 5 8 10 13
Basic Principles of Hydraulic Flow and Jet Theory 2.1 Introduction 2.2 Principles of Fluid Mechanics 2.3 Basic Concepts of Hydraulic Flow 2.4 Continuity Equation 2.5 Three-Dimensional Flow 2.6 Momentum Equation 2.7 Applications of Momentum Equation 2.8 Moment of Momentum Equation 2.9 Eulers Fundamental Equation 2.10 Jet Theory 2.11 Jet Force on Stationary Flat Plate 2.12 Jet Force on Moving Flat Plate 2.13 Jet Force on Curved Plate when Jet Strikes Tangentially 2.14 Jet Propulsion of Ships Question Bank No. 2 Tutorial Sheet No. 2
14 14 14 18 21 22 24 25 26 27 30 31 33 35 37 38 39
viii Contents PART IITURBINES
43
Pelton Turbine 3.1 Introduction 3.2 Comparative Study of Hydro Turbines 3.3 Pelton Turbine 3.4 Layout Arrangements 3.5 Velocity Diagram of Bucket 3.6 Turbine Losses and Efficiency 3.7 Design of Pelton Turbine 3.8 Design Example of a Pelton Turbine Question Bank No. 3 Tutorial Sheet No. 3
45 45 45 48 50 51 56 58 64 71 72
Francis Turbine 4.1 Introduction 4.2 Specific Speed and Turbine Type 4.3 Main Components 4.4 Velocity Triangles of Francis Turbine 4.5 Sankey Diagram 4.6 Design of Francis Turbine 4.7 Design Problem Question Bank No. 4 Tutorial Sheet No. 4
77 77 77 79 81 83 84 87 91 91
Propeller and Kaplan Turbines 5.1 Introduction 5.2 Kaplan Turbine 5.3 Turbine Constants 5.4 Design Problem 5.5 Other Turbines Question Bank No. 5 Tutorial Sheet No. 5
96 96 97 98 100 102 103 103
Draft Tube and Cavitation 6.1 Introduction 6.2 Functions of Draft Tube 6.3 Types of Draft Tubes 6.4 Cavitation
106 106 108 108 109
Contents ix
6.5 Factors Causing Cavitation 6.6 Methods to Avoid Cavitation Question Bank No. 6 Tutorial Sheet No. 6
111 112 113 113
Governing of Hydraulic Turbines 7.1 Introduction 7.2 Oil Pressure Governor 7.3 Double Regulation System of Pelton Turbine 7.4 Governing of Reaction Turbines Question Bank No. 7
116 116 117 118 119 120
Dynamic Similarity and Performance Characteristics 8.1 Introduction 8.2 Dynamic Similarity of Model and Prototype 8.3 Specific Speed 8.4 Scale Ratio 8.5 Unit Quantities 8.6 Model Relationships 8.7 Performance Characteristics Question Bank No. 8 Tutorial Sheet No. 8
121 121 121 122 124 125 126 127 130 131
PART IIIPUMPS 9. Centrifugal Pumps 9.1 Introduction 9.2 Classification of Centrifugal Pumps 9.3 Pump Performance 9.4 Sankey Diagram 9.5 Velocity Diagrams 9.6 Design of Impeller 9.7 Dynamic Similarity and Performance Characteristics 9.8 Cavitation and Maximum Suction Lift 9.9 Multistage Pumps 9.10 Pump Operation Question Bank No. 9 Tutorial Sheet No. 9
133 135 135 135 139 141 143 148 154 158 161 161 166 167
x Contents 10. Reciprocating Pump 10.1 Introduction 10.2 Pump Installation and Performance 10.3 Classification of Reciprocating Pumps 10.4 Piston Motion 10.5 Indicator Diagram 10.6 Factors Affecting Performance of Reciprocating Pumps Question Bank No. 10 Tutorial Sheet No. 10
174 174 175 177 179 181 182 190 190
Special Pumps 11.1 Introduction 11.2 Axial Flow Pump 11.3 Design of Axial Flow Pump 11.4 Vertical Turbine Pump 11.5 Rotary Displacement Pumps 11.6 Jet Pumps 11.7 Hydraulic Ram 11.8 Hydraulic Coupling 11.9 Torque Converter Question Bank No. 11 Tutorial Sheet No. 11
196 196 196 198 204 205 209 210 212 213 214 215
PART IVSYSTEMS AND PLANTS
217
Hydraulic Systems 12.1 Introduction 12.2 Hydraulic Accumulator 12.3 Differential Accumulator 12.4 Hydraulic Intensifier 12.5 Hydraulic Jack or Press 12.6 Hydraulic Lift Question Bank No. 12 Tutorial Sheet No. 12
219 219 220 220 221 222 223 223 223
Hydraulic Power Plants 13.1 Introduction 13.2 Hydro Power Potential 13.3 Classification of Hydro Power Plants
226 226 226 227
Contents xi
13.4 Location of Hydro Power Plants 13.5 Hydrology Question Bank No. 13 Tutorial Sheet No. 13
PART VCASE STUDIES Case I : Floating Type Micro Hydro Power Plants Case II : An Alpine Hydro Village Case III : Operation and Maintenance of Centrifugal Pumps for Energy Conservation Case IV : Hydraulic Design of a Boiler Feed Pump to Ensure Stable Operation at Reduced Flows Case V : Hydro Power Plants: Great Ecological Havocs
PART VIOBJECTIVE QUESTIONS Index
231 233 237 237
241 243 251 257 268 277
283 307
2)46 1 FUNDAMENTALS
CHAPTER
Classification and Applications 1.1 INTRODUCTION Water is by far the most common liquid on the earth and is therefore plentiful and cheap. It is also chemically stable and non-hazardous to health. H2O molecules have extra strong mutual attraction due to hydrogen bond without which, it will be a gas at atmospheric temperature and life would not exist on the earth. It is also an extremely good solvent. At typical atmospheric conditions, density of water is 1000 kg/m3. Hence, size and cost of plants using water as working medium are not impractically large. Water is not common everywhere in the universe. It does not exist in stars because of break-up of molecules at very high temperatures. It does not exist on small planets because gravitational forces are too weak to trap the molecules. However, earths planetary position, size and mean temperature are such as to contain and hold a more representative sample of water and it is therefore, plentiful and cheap on earth.
1.2 HYDRO ENERGY Rain falling on the earths surface has potential energy relative to sea level towards which it flows. This hydro energy can be converted into mechanical energy by shaft work where water falls through an appreciable vertical distance. The hydraulic power is thus a naturally available renewable energy source. The utilisation of hydro energy to operate agricultural and industrial devices is one of the oldest and widespread techniques of human achievement. Hydraulic rams were used to raise water for lift irrigation and drinking purposes. Watermills have been used to generate mechanical power to drive irrigation pumps, machine tools and small electric generators. The Himalaya Mills is well known in the mountain areas from Afghanistan to Burma which is mostly used for grinding grains. In modern hydroelectric power plants, the energy of water is utilised to drive a turbine which, in turn, runs the generator to produce electricity. A schematic diagram of a hydroelectric power plant is shown in Fig. 1.1.
4 Hydraulic Machines Water Hydro energy
DAM Water storage
Water P. E.
Electric generator
Hydro turbine Energy conversion
M. E. of Rotation
E. M. F.
Electrical Energy
Fig. 1.1 Hydropower plant
The net electric power output of a hydroelectric power plant is given by: P = HQgHD0 = HQgH DT DC DTH where P = Electric power output, [W ] H = Density of water, [1000 kg/m3] Q = Water flow rate in turbine, [m3/s] g = Gravitational acceleration, [9.81 m/s2] H = Net useful water head [m] D0 = Overall efficiency of hydropower plant DT = Turbine efficiency [0.85 0.95] DC = Generation efficiency [0.95 0.99] DTH = Transformer efficiency [0.92 0.98] Hydropower, available free of cost in the form of flowing water, is an important, pollution free, renewable source of energy. These can be integrated with irrigation, fish breeding and drinking water supply schemes. Nearly twenty per cent of the total world power is produced by hydel stations. There are some countries like Norway and Switzerland where the hydropower forms almost the total installed capacity. The Central Electricty Authority (CEA) has assessed Indias hydro potential to be about 1,48,700 MW of installed capacity. The hydroelectric capacity currently under operation is about 26,000 MW. Another 16,083 MW capacity is under various stages of development. India is endowed with economically viable hydro potential of 600 TWH. The key advantage of hydroelectric power is the ability to store energy and the flexibility of its use during peak load periods. Data of Central Electricity Authority shows that over a long period a hydel project is much cheaper than a thermal project. As hydel projects get older, their generator costs decrease, since fuel cost remains zero, while those of thermal projects would go up, on account of escalating fuel costs. The water pumps especially centrifugal pumps are very widely used to transfer water from a lower level to higher level. Pumps are used in industries, municipal water supply, agriculture and other sectors.
Classification and Applications 5
1.3 CLASSIFICATION OF FLUID MACHINES Liquids and gases together are called fluids. The fluid machines use liquids and/or gases as working substances for transfer and transformation of energy from one form to another. The fluid machines convert the potential energy or kinetic energy or pressure energy or thermal energy or internal energy of fluid into mechanical work and are called engines or turbines or prime movers or driving machines. The driven machines may be called pumps or compressors or blowers or fans to deliver fluids from low level of energy to higher level by consuming mechanical work. There are other fluid machines used for tranmission of energy from driver to driven machine such as a fluid coupling or torque converter. Fluid machines also include complete systems or components to intensify or store energy during transmission. There are so many types of machines working with different types of fluids, performing different functions and working on different principles that it may be difficult to classify them on single basis. These machines are grouped together on different criteria. A tree diagram is shown in Fig. 1.2 to broadly group different fluid machinery as per services provided by them. Fluid machines
Fluid transfer machines
Prime movers
Engines
Turbines
Hydro turbine
Gas turbine
Steam Engines
Fluid coupling
Fluid systems
Torque convertor Supply and Fluid power Fluid systems control system disposal systems
Steam turbine
I.C. Engines
Pumps
Centrifugal pumps
Power transmission machines
Compressors
Positive DisplacCompressor ement Pumps
Blowers
Fans
Fig. 1.2 Classification of fluid machines
6 Hydraulic Machines 1.3.1 Prime Movers Prime movers are fluid machines which utilize the naturally available energy of fluid and transform into mechanical energy of shaft rotation. The natural sources of energy can be kinetic energy of wind, potential energy of water flow, solar heat, thermal energy produced by combustion of fuels (coal, gas, diesel, petrol, biomass) or geothermal energy from the core of the earth. A steam engine is a reciprocating machine where enthalpy of steam is converted into mechanical (piston) work. Similarly, internal combustion engines are run on hot air obtained by combustion of diesel, petrol or gas and are also of piston cylinder reciprocating type. Steam turbine is a rotary machine which also converts the enthalpy drop of steam into mechanical work of shaft rotation. The steam may be generated in a boiler by evaporation of water by combustion gases or solar energy or geothermal steams or heat of nuclear fission. Gas turbines are also rotary machines run by hot air or carbon dioxide or helium or other gases heated by combustion of a fuel. All these machines are widely used for power generation or driving of process machines such as a pump, compressor, blower, or other machines. These prime movers are beyond the scope of this book. Hydel turbines convert the potential energy of naturally available water into mechanical energy of shaft rotation which in turn is used to run an electric generator to produce electricity. These machines work on the principle of fluid dynamics and are known as rotodynamic machines. The rotating element of the turbine has a number of buckets or vanes or blades and is called a rotor. The fixed casing enclosing the rotor is called a stator. The turbine rotors are turned by the impulse of a waterjet or aerofoil shape of blades and are respectively called impulse turbines (Piston wheel) and Pelton reaction turbines (Francis and Kaplan turbine).
1.3.2 Fluid Transfer Machines Fluid machines are widely used to transfer fluids from lower level or pressure to higher level or pressure. Machines transferring air or other gases with a pressure ratio upto 1.15 are called fans. Compressors are water cooled fluid machines producing pressure ratios more than 1.15. Compressors without cooling provision are called the blowers. The compressors, fans and blowers are run by electric motors or engines or other prime movers and are beyond the scope of this book. The pumps handle liquids. The positive displacement pumps transfer the input mechanical energy of prime mover into pressure energy or kinetic energy of fluid with the help of a piston or gear teeth or a sliding gate. The use of the pump type depends upon the properties of the fluid to be pumped and also the quantity and pressure head of the system installation. The centrifugal pumps are most widely used. The pump impeller is provided with vanes enclosed in a volute casing. As the impeller rotates by an electric motor
Classification and Applications 7
or a fluid prime mover, the liquid is pushed by centrifugal force from the pump centre to its periphery through the volute casing. The kinetic energy generated by the pump impeller is converted into fluid pressure in the volute casing. There are other pumps like jet pumps, pneumatic lift pumps, etc. which will be discussed separately.
1.3.3 Power Transmission Machines In industrial practice, fluid machines are sometimes used to transmit mechanical energy from the motor shaft to the shaft of driven machines. The two shafts are connected through a coupling. Sometimes, it is required that the two shafts should not be connected mechanically. The shafts are connected hydraulically. The motor shaft drives a centrifugal pump to generate fluid pressure which is used to drive a fluid turbine mounted on the driven shaft. The pump and turbine are built into a single unit with a closed hydraulic unit. The hydraulic power transmission systems are of two types: 1. Fluid coupling transmits power with same torque on the driving and driven shaft. 2. Fluid torque converter provides for torque multiplication with the same power on the driving and driven shafts. In addition to a pump and a hydraulic turbine as in the case of a fluid coupling, a reactionary number is incorporated to augment the torque.
1.3.4 Fluid Systems A fluid system is a circuit in which forces and power are transmitted through a fluid. When air or gas is used as working fluid, the system is called fluidics which is beyond the scope of this book. The fluid systems discussed transmit forces or power by the hydraulic pressure of a liquid. There is no motion and no dynamic effect of liquid is used in the operation of the system. 1. The circuit diagram of a typical fluid power system is shown in Fig. 1.3. Liquid
Accumulator
Intensifier
Power machine (Lift, Press, Crane)
Pump Fig. 1.3 Fluid system
The examples of fluid power systems are: (i) The hydraulic press (ii) The hydraulic lift (iii) The hydraulic intensifier
8 Hydraulic Machines (iv) The hydraulic accumulator (v) The hydraulic crane (vi) The hydraulic ram 2. The fluid control systems are used in nuclear power plants, thermal power plants, hydroelectric power plants and other process plants to operate and control the opening of water and steam valves; speed controller to regulate the load. Special servomotors are used to operate the valves. 3. The supply and disposal systems include lubrication system of turbines and generators, ash disposal system, effluent disposal system. Special slurry pumps, jet pumps and pneumatic lift pumps are used in the supply and disposal systems.
1.4 APPLICATIONS OF HYDRO TURBINES The modern hydro turbines have been derived from the water wheels of the middle ages which were used for flour mills to grind wheat and ore crushing. Many technological developments in the field of fluid dynamics, metallurgy and mechanical engineering have helped to the development of very wide range and types of hydro turbines. There are turbines which can operate at heads of 200 m and wheel diameter of 10 m. Similarly, the power capacity of a turbine can be 700 MW and more. There are also turbine rotors derived from windmills which can work on zero head and utilize the current velocities of rivers and canals. Such water mills can be designed with water velocity of even 1m/s and power output of a few kilowatt. The main aim of all these turbines is to convert the potential energy of waterfalls or kinetic energy of flowing water into mechanical energy to produce electrical power. The hydro turbines can be classified as: 1. High head (Pelton), medium head (Francis) or low head (Kaplan) turbines. 2. Hydro turbines for base load or peak load power plants. 3. Turbines for mini or micro hydel plants. 4. Turbines for hydro electric plants with storage reservoirs using a dam. 5. Turbines for run of the river plants without pondage. 6. Turbines for pumped storage plants. 7. Reversible turbines for tidal power plants. The hydro power plants are designed as central power plants to generate electricity or as interconnected to thermal power plants in order to meet the fluctuating loads most economically. Special reversible hydro turbines have been built for tidal power plants so that these can produce power both during high tide storage of water and low tide emptying of reservoir.
Classification and Applications 9
In the design of pumped storage plants, water from turbine discharge is stored in the tail race reservoir. When demand is low (at night), the water is pumped back from tail race to head water pond. The power required to drive the pump is taken from common grid or a nearby thermal power plant. During high demand (peak load) this pumped water is used to produce electricity. A motor generator set and a reversible turbine pump set can be used both for power generation and water pumping. The small hydro plants can use the natural water source in hilly terrain for power generation. Zero head turbines can be used to tap the kinetic energy of river or canal water in plain areas. The hydro electric power plants can be used for generating electricity only or as a multipurpose scheme for flood control, irrigation water and power generation.
1.4.1 Advantages of Hydel Power Power generation by hydro turbine plants have the following attractions: 1. Hydro plants can be designed as multipurpose projects. In addition to generation of electricity, the project can be engineered for irrigation, flood control, afforestation, navigation and aquaculture. 2. These plants are simple in design and operation and have very high reliability and durability. 3. The hydraulic turbine can be started and stopped quickly in a few minutes. Therefore, no spinning or reserve capacity is needed. 4. The maintenance of plants is very simple and due to low temperature operation, life expectancy can be more than 50 years. 5. Hydro turbines have high efficiency (more than 90%) over a wide range of loads. 6. The running cost of a hydro turbine is very low due to zero cost of fuel, fuel transportation, low-skilled manpower and requirements. The cost of manpower and spare parts is very low. 7. The water after doing work in the turbine is still available for other applications such as drinking water schemes or irrigation of farms. 8. Water is a renewable and clean source of energy. 9. There is no adverse effect on the environment such as greenhouse effect and global warming. 10. There are no chimney gases containing oxides of sulphur or nitrogen or carbon. 11. There is no problem of ash disposal or processing of waste as from a nuclear power plant.
10 Hydraulic Machines 1.4.2 Disadvantages of Hydel Power The main shortcomings are: 1. The initial cost of a hydro turbine plant is high due to huge and complex civil structures. 2. The hydro turbine has to be designed for every site as per head and run-off available. The cost and time of design and manufacture is high due to nonavailability of standard products. 3. The plants are to be located near the source of water and suitable site of location. The cost of power transmission due to long transmission line and high power loss can be huge. 4. The gestation period of hydel projects may be ten to fifteen years affecting the cost and cost variability of the project. 5. Power plant design and operation depends upon the quantity of water available which varies over the year. 6. Large hydro plants using dams and large storages have very serious ecologically disturbing consequences. These can submerge large fertile areas destroy forests and vegetation and uproot people from villages affected. There is a strong opinion in favour of run-off river projects or small hydel projects against power projects with dams and reservoirs.
1.5 APPLICATIONS OF CENTRIFUGAL PUMPS Every industrial process involves the transfer of liquids from one level of pressure or static energy to another. Pumps have become an essential part of all industrial processes and, in turn, major consumers of energy themselves. In many industries, pumps may form more than 40% of the auxillary equipment. The pumping capacity may be as high as 100,000 m3/hr of liquid and heads as high as 2500 m of H2O and more. Centrifugal pumps are used in thermal power plants as boiler feed pumps, condenser circulating water pumps and feed water heater pumps. Such pumps are also used for fire protection system and other water services, Sealed centrifugal pumps are designed for nuclear power plants. Jet pumps are used for evacuation of air from condensers. Rotary pumps are used in lubrication and control systems. Reciprocating pumps are used for dosing of chemicals into boilers. Centrifugal and jet pumps are designed for ash disposal systems. Centrifugal pumps have maximum application. Some important applications of centrifugal pumps are given in Table 1.1.
Classification and Applications
11
Table 1.1 Applications of centrifugal pumps S. No. Sector 1.
Rural sector Agriculture
Water supply Cottage and rural industries 2.
Transport Sector Harbours, docks and canals Road, rail and air transport
Marine Mining and quarrying 3.
4.
Construction sector Pipeline construction Civil engineering works Service industries and services Heating, ventilating and air-conditioning Domestic water supply Fire-fighting and fire prevention Armed forces, postal services, atomic energy establishment, communication, miscellaneous local government services Laundries, dry cleaners, dryers, sports establishments Commercial construction services
5.
6.
7.
Power generation and distribution Gas Electricity
Activities Irrigation, farming, insecticide and fertilizer spraying, forestry, fish farms and hatcheries Drinking and waste water
Transportation of equipment manufacture e.g. car, lorry, bus, aircraft, rolling stocks, naval Commercial ship building, ship owners Coal mining, metallic and non-metallic ores, clay, salt, sound and gravel, etc.
Harbours and dry docks, roads, bridges, etc. Hospitals, schools, offices, etc. Domestic booster sets, hot water circulation Sprinkler system, etc.
Hotels, hospitals, schools, offices
Water supply Sewage
Processing and distribution Electricity generation and transmission Municipalities and other agencies Sewage treatment and disposal
Oil exploration and production Oil and natural gas Petroleum Coal, coke, tar
Exploration and production Refinery plants, distribution centres Processing
Engineering industry Mechanical machinery and component manufacturers
Machine tools, pumps, compressors, food machinery, heaters, coolers, fastners, instruments, etc. Table Contd.
12 Hydraulic Machines Table Contd.
Boiler Dish and bottle washing plant Metal parts washing and spray painting Filtration plant Commercial industrial refrigeration Electrical motors, generators, radio, television, telephony equipment, wires and cable, other domestic electrical equipment Switchgear 8.
9.
10.
Food and beverages industry Tobacco Cane and beet sugar processing Distilleries, breweries, soft drink Cereal processing, bakeries and confectionary products Canneries, fruit and vegetable processing and products Animal and marine fats and oils Dairy products, milk, cheese, butter, ice cream Meat processing, bacon production Process industries Petrochemicals, plastic and synthetic rubber Chemicals: fertilizers, inorganic and organic acid, solvents, ammonia, chlorine, inks, paints, phosphoric chemicals, polishers Water treatment Pharmaceutical and cosmetics Printing, publishing and book binding Paper, pulp and printing Textiles Consumer goods
Industrial materials Primary metal production, foundries, rolling mills Electro-polishing and-plating Cement ready mix concrete Bricks, fire clay, pottery glass and glass containers Timber, furniture, bedding and other wood and cork
Manufacturing Manufacturing Manufacturing Manufacturing Manufacturing Manufacturing
Manufacturing
Manufacturing
Slaughter houses Manufacturing Manufacturing
Plant manufacturers and contractors Manufacturing Manufacturers and contractors Synthetic fibre manufacturers Textiles, tanneries, leather processing, furs, leather goods manufacturers, clothing and footwear, plastic consumer goods manufacturers Equipment manufacturers and material processors Plant manufacturers or process specialists Manufacturing Manufacturing
Classification and Applications
13
QUESTION BANK NO. 1 1. What is hydro energy? Why is water so common on earths surface? 2. Describe the classification of fluid machines. What is the difference between fluid machines and hydel machines? 3. Explain the following: (a) Prime Movers (b) Fluid Transfer Machines (c) Power Transmission Machines (d) Fluid Systems 4. Briefly explain the applications of: (a) Fluid Systems. (b) Power Transmission Machines 5. What are the application of hydro turbines? 6. Discuss the advantages and disadvantages of hydel power. 7. Briefly explain the applications of centrifugal pumps.
+0)26-4
Basic Principles of Hydraulic Flow and Jet Theory 2.1 INTRODUCTION Hydraulic machines handle water. Machines changing fluid energy into mechanical energy are called hydraulic turbines or hydraulic motors. Machines designed to more liquids and add energy to them are called pumps. There are also special systems dealing with the transmission of hydraulic power. The hydraulic machines are classified in Fig. 2.1. Hydraulic machines
Fluid P. E.
Hydraulic Fluid Fluid turbines Work 1. Pelton wheel 2. Francis turbine 3. Propeller turbine 4. Tubular turbine
Pumps
Fluid Systems: Motor + P. E. Pump + Piping
Work 1. Centrifugal pump 2. Reciprocating pump 3. Axial flow pump 4. Others
Hydraulic ram 2. Hydraulic accumulator 3. Hydraulic jack 4. Intensifier 5. Hydraulic press
Fig. 2.1 Classification of hydraulic machines
The hydraulic mechanics are designed on the principles of fluid machanics, hydraulic flow and water jet forces.
2.2 PRINCIPLES OF FLUID MECHANICS Important principles of fluid mechanics required for the design of fluid (hydraulic) machines are summarized below.
Basic Principles of Hydraulic Flow and Jet Theory
15
2.2.1 Hydrostatic Forces 1. Properties of fluids
The weight density or specific weight of a fluid is the weight per unit volume. [N/m3]
w=r´g
Specific volume is the reciprocal of density. n=
1 r
[m3/kg]
m r
[stokes]
Kinematic u is cosity, u= 2. Fluid pressure
The pressure at any point in a fluid is defined as the force per unit normal area. p=
F A
[N/m2]
The Pascals law states that intensity of pressure for a fluid at rest is equal in all directions. Pressure variation at a point in a fluid at rest is given by the hydrostatic law which states that the rate of increase of pressure in the vertically downward direction is equal to the specific weight of fluid, dp = w = r ´ g [N/m3] dz
The pressure at any point in a liquid p = rgz
[N/m2]
Absolute pressure = Atmospheric pressure + Gauge pressure pabs = patm + pg The pressure at a point in a fluid is measured by manometer. 3. Hydrostatic forces
The force exerted by a static fluid on a vertical, horizontal or an inclined immersed surface, F = rgA h
[N]
where h = depth of centre of gravity of the immersed surface from free surface of the liquid.
16 Hydraulic Machines 2.2.2 Kinematics of Fluid Flow For a steady flow, the fluid properties like pressure, density, velocity, etc., do not change at a point with respect to time dV = 0, dt
steady flow
For uniform flow of fluid, the velocity does not change with respect to space (length of direction of flow) dV = 0, ds
uniform flow
The density of fluid remains constant for incompressible flow. r = constant: Incompressible flow. For laminar flow in a pipe, Reynolds number is less than 2000. For turbulent flow in a pipe, Reynolds number is more than 4000, Volumetric flow rate, Q = A1V1 = A2V2 = A3V3
[m3/s]
This is called continuity equation.
2.2.3 Dynamics of Fluid Flow 1. Equation of motion (Newtons second law of motion) Fx = m ax [N] 2. Eulers equation of motion along a stream line, ¶p + g¶z + V ¶V = 0 r
Bernaullis equation (integration of Eulers equation of motion for a steady, ideal flow of an incompressible fluid), states that total energy consisting of pressure energy, kinetic energy and potential energy at any point of fluid is constant. p1 V12 p V2 + + z1 = 2 + 2 + z2 + hL [m] rg 2 g rg 2 g
where
p1 = pressure head = pressure energy per unit weight, rg V12 = kinetic head = kinetic energy per unit weight 2g z1 = datum head = datum energy per unit weight h L = loss of energy head between sections 1 and 2
Basic Principles of Hydraulic Flow and Jet Theory
17
Momentum equation states that net force acting on a fluid mass is equal to change in momentum per second in that direction
d (mV) [N] dt 5. The impulse momentum equation is given as: F=
F dt = d (mV) [N-s] 6. The force exerted by the nozzle on water Fx = rQ(V2x V1x ) [N] 7. Moment of momentum equation states that the resultant torque on a rotating fluid is equal to the rate of change of moment of momentum T = rQ(V2r2 V1r1)
[N-m]
Loss of pressure head for viscous flow through circular pipe hf =
32muL rg D 2
[m]
where m = coefficient of viscosity u = average fluid velocity Q (m/s) pR 2 D = diameter of pipe [m].
=
Darcy formula. Energy loss due to friction, hf =
4 fLV 2 2 gD
[m]
Darcy Weisbach Equation. Head loss due to friction in pipes, hf =
4 fLV 2 2 gD
[m]
where f = Coefficient of friction =
16 for laminar flow Re
=
0.0791 for turbulent flow. (Re)1/ 4
The velocity of water at the outlet of the nozzle,
18 Hydraulic Machines 2 gH 4 fL a 2 1+ . D A2
V=
[m/s]
where H = head at inlet of pipe [m] L = length of pipe [m] D = diameter of pipe [m] a = area of nozzle outlet [m2] A = area of pipe [m2] 12. The power transmitted through nozzle, P=
LM N
rg Q 4 fLV 2 H1000 2 gN
OP Q
[kW]
2.3 BASIC CONCEPTS OF HYDRAULIC FLOW The hydraulic turbines utilize the potential energy of water to produce mechanical work. The pumps are required to transfer water and other liquids. The fluid couplings, torque converters and fluid system work with special oils.
2.3.1 Fluid Characteristics The following fluid characteristics are assumed for the working substance used in hydraulic machines and systems. 1. Ideal fluid
An ideal fluid is an imaginary fluid which is both incompressible and non-viscous. Such liquids do not exist in nature. However, water is assumed to be incompressible and has very low value of viscosity. Therefore, it is nearly an ideal fluid. For incompressible fluid, the water density is constant. r = const. 2. Newtonian fluid
The oils used in power transmission machines and fluid systems are also incompressible. These oils are assumed to be Newtonian fluids whose viscosity is independent of velocity gradient, t=m
du dy
2.3.2 Closed and Open Systems A system is a quantity of matter in space upon which attention is made in the study of changes of properties and analysis of a problem. Everything external to the
Basic Principles of Hydraulic Flow and Jet Theory
19
system is called the surrounding. The boundary separating the system from the surrounding may be real (solid) or imaginary. There may be energy transfer into or out of the system. 1. Closed system. A closed system has fixed identity with fixed mass. There is no mass (fluid) transfer across the system boundary. All positive displacement machines such as reciprocating pumps, gear pumps, etc. torque converters, fluid couplings, various fluid power systems and fluid control systems will be analyzed as constant mass closed systems. 2. Open system. In an open system, the fluid crosses the boundary of the system in addition to interaction of energy between the system and the surrounding. The mass of an open system may or may not change. The identity of the fluid changes continuously. The boundary of the open system is kept fixed without any change in its volume. An open system is also referred to as control volume system. The closed boundary of a control volume is called the control surface. There is a transfer of both mass and energy across the control surface. All types of hydraulic turbines, centrifugal pumps, axial flow pumps, slurry pumps, jet pumps, pneumatic lift pumps, various supply and disposal fluid systems will be analyzed and studied as control volume systems.
2.3.3 Macroscopic Properties Matter is composed of several molecules and description of the motion of a fluid will consider the behaviour of discrete molecules which constitute the fluid. The microscopic properties can be found out by statistical summation of properties of individual molecules. Liquids have extra-strong intermolecular attractive forces. Therefore, molecular description is not required. The entire liquid mass behaves as a continuous mass. The time-averaged values of properties are sufficiently accurate and valid. The matter of the system is assumed as a continuous distribution of mass with no empty space and no conglomeration of separate molecules. The fluid machines and system will be designed and analyzed on the basis of time-averaged macroscopic properties measured with the help of instruments. This is called the concept of continuum model.
2.3.4 Conservation of Mass In non-nuclear processes, the matter can neither be created nor destroyed. The conservation of mass is inherent to the concept of a closed system. For a control volume, the rate of mass entering the system must be equal to the rate of mass leaving the system plus the rate of storage of mass in the system. If the fluid flow is steady, the rate of liquid stored is zero.
20 Hydraulic Machines Most of the channels comprising hydraulic machines have variable and irregular sections and a curved mean path. Some of the channels are in circular motion with power applied to or taken from the flow. The behaviour of hydraulic flow is studied making the following assumptions: 1. The flow is steady. The fluid properties do not change with time at any particular section. 2. The flow is incompressible and fluid density (r) remains constant. 3. The flow is one-dimensional. 4. The fluid velocity is uniform across the cross-section of the channel. Average velocity is assumed for each section of flow. 5. There is no branching of fluid stream.
2.3.5 Fluid Motion A fluid mass consists of a number of fluid particles. The instantaneous velocity at any point in the fluid is the velocity of a particle that exists at that point at that time. The fluid motion can be studied by the following methods. 1. Langrangian method
A single fluid particle is followed during its motion and its velocity, acceleration, density, etc. are described. It is possible to trace the history of each fluid particle. However, it is very difficult to find solutions mathematically. The method is rarely used for practical applications. 2. Eulerian method
The velocity, acceleration, pressure, density, etc. are described at a certain point in flow field. It avoids the determination of movement of each individual particles of fluid. The change of velocity with time can be easily found out. The Eulerian method is commonly used in the analysis of fluid mechanics and fluid machines. 3. Stream line and stream tube
The motion of fluid through hydraulic machines can be described with the help of streamlines and stream tubes. Streamline is the instantaneous locus of a line in the flow field to which the velocity vector is tangent. The velocity at right angles to the streamline is always zero and no flow can cross any streamline. Stream tube is a tubular space formed by the collection of streamlines. No fluid can enter or leave the stream tube except at the ends. Stream tube is imagined to behave as a solid tube. The cross-sectional areas of this tube are small in comparison with its length, so that velocity can be assumed constant across any cross-sectional area (Ai). For steady flow, the stream tube maintains the same shape for all times while for unsteady flow, the stream tube changes shape with each succeeding instant of time.
Basic Principles of Hydraulic Flow and Jet Theory
21
2.3.6 Governing Equation The flow of liquid through hydraulic machines is divided in imaginary stream tubes and the behaviour of the hydraulic flow is studied with the help of the following governing equations. 1. Continuity Equation This is based on the principle of conservation of mass flow ( m ). 2. Energy Equation This is derived from the principle of conservation of energy ( m V2) 3. Equation of Motion This is based on the principle of conservation of momentum ( m V).
2.4 CONTINUITY EQUATION
Dt V2
V The continuity equation of flow is derived on the principle of conservation of mass. For an incompressible, uniform and steady state flow, the quantity of fluid entering at one end of stream tube and leaving at the other must A2 be same provided there is no sink or source in the tube. Consider a stream tube with two arbitrary V1 A1 cross-sections A1 and A2, the mass of fluid enters the tube through A1 with a velocity V1 Fig. 2.2 Flow through a stream tube and leaves through A2 with velocity V2. In an increment of time Dt, the fluid particles entering A1 have moved an infinitesimal distance V1Dt. The particle at A2 have moved a distance V2 Dt. By the definition of a stream tube no fluid can cross the stream tube. The mass of fluid entering the stream tube at A1 in time Dt must be exactly equal to the mass of the fluid leaving the stream tube at A2 during the same interval of time. For a steady flow, the rate of fluid stored is zero. Mathematically, the statement can be expressed as follows: V
1 Dt
2
m 1 = m 2
rQ 1 = rQ 2 rA1 V1 Dt = rA2 V2 Dt For incompressible flow, density r is constant. \
V1 A1 = V2 A2 = VA = Constant = Q
[m3/s]
This is called equation of continuity and the constant Q represents the volumetric flow rate. Therefore, the volume of fluid which passes through each cross-section of the stream tube per unit time remains constant for a steady flow and incompressible fluid.
22 Hydraulic Machines 2.5 THREE-DIMENSIONAL FLOW In a three-dimensional fluid flow, the fluid properties (velocity, pressure, density, viscosity) may vary in all directions. Examples are: flow in a river, flow within a fluid machine. However, for simplicity, the fluid machines are analyzed as one-dimensional fluid flow. Z Z
w
DX
y
C
DY B
u
DZ
v X
X
O
Y D Y Fig. 2.3
Three-dimensional flow
Consider an elementary cube Dx, Dy, Dz in the fluid body. The difference between the amounts of fluid which flows into and out of its faces during time Dt, must be equal to the increase in the mass which the edges enclosed. The mass of fluid of density r entering the cross-section across the face OB in time Dt = ru (Dy.Dz)Dt. The mass of fluid leaving across the face CD in time Dt = ru(Dy.Dz) Dt + ¶ (ru Dy Dz Dt)Dx ¶x
\ Gain of mass across the above faces =
¶ ru(Dx.Dy.Dz) Dt ¶x
Similarly, gain of mass across the faces BD and OC = And, gain of mass across the faces BC and OD = Total gain =
¶ rv(Dx.Dy.Dz)Dt ¶y
¶ rw(Dx.Dy.Dz)Dt ¶z
FG ¶ ru + ¶ rv + ¶ rwIJ (Dx.Dy.Dz)Dt. H ¶x ¶y ¶z K
...(1)
Mass of fluid contained at time t = r(Dx.Dy.Dz) Mass of fluid contained at time t + Dt = r(Dx.Dy.Dz) +
¶ (r.Dx.Dy.Dz)Dt ¶t
23
Basic Principles of Hydraulic Flow and Jet Theory
\ Gain in the mass of fluid in time Dt =
¶ r.Dt(Dx. Dy.Dz) ¶t
...(2)
Equating equations (1) and (2),
FG ¶ ru + ¶ rv + ¶ rwIJ (Dx.Dy.Dz) Dt = ¶ rDt (Dx.Dy.Dz) ¶t H ¶x ¶y ¶z K ¶r ¶ ¶ ¶ + (ru) + (ru) + (rw) = 0 General Equation ¶t ¶x ¶z ¶y
or,
For incompressible fluid, r = Constant ¶u ¶v ¶w =0 + + ¶x ¶y ¶z
\
This is also called differential equation of continuity in Cartesian coordinates. Special Cases
Y Dr
Two-dimensional flow.
Vr
w=0 r
Dq
¶u ¶v + =0 ¶ x ¶y
\
Vq q X
Cylindrical coordinates 1 ¶vq ¶vr ¶v rv + + z + r =0 r ¶q ¶r ¶z r
Z Fig. 2.4 Cylindrical coordinates
Spherical coordinates 1 ¶ 1 ¶ (vf sin f) 1 ¶vq (w2 vw) + + =0 w ¶w sin f sin f ¶q ¶f
Vector Notation General equation,
Vz
Z
¶r + Ñ(rq) = 0 ¶t
Vq
f w
For incompressible fluid, Dq = 0
q
r
X
Fig. 2.5 Spherical coordinates
Vw
Vr Y
24 Hydraulic Machines 2.6 MOMENTUM EQUATION The momentum equation is based on Newtons second law of motion which states that the time rate of change of momentum is proportional to the applied force and takes place in the direction of force. Y Assume a resultant force Fx acts on a mass m v along x-axis. The change in velocity of mass m in time V dt be dV. u
\
FX = m.
du dt
...(3)
or
FX dt = m du
...(4)
w X
Equation (3) is called linear momentum equation. Equation (4) is called impulse-momentum equation Z which shows that impulse of applied force (FX.dt) is Fig. 2.6(a) Velocity equal to change of momentum (m.du). components Consider a stream tube. For steady state flow, mass entering the tube is equal to mass leaving the tube. r1 A1V1Dt = r2 A2V2 Dt Momentum of fluid entering = (r1 A1 V1Dt)V1 Momentum of fluid leaving = (r2 A2 V2 Dt)V2 Dt V2
q2
V2 x
A2
t V 1D
V1
q1 x A1 Fig. 2.6(b) Momentum equation
\ Change of momentum in x-direction, = [(r2 A2 V2 Dt)V2 cos q2] [(r1 A1V1Dt)V1 cos q1]
Basic Principles of Hydraulic Flow and Jet Theory
25
But for a steady state, equation of continuity: (r1 A1V1 Dt) = (r2A2 V2 Dt) = rQDt \ Change of momentum in x-direction, = rQDt (V2 cos q2 V1 cos q1) From impulse-momentum equation,
å Fx Dt = å rQ Dt(V2 cos q2 V1 cos q1) Fx = rQ(V2 cos q2 V1 cos q1) = rQ(vx2 vx1] Similarly,
Fy = rQ(V2 sin q2 V1 sin q1) = rQ(vy2 vy1)
Resultant force acting on the flowing fluid, F=
Fx2 + Fy2
The angle between F and Fx can be found out as: tan q =
Fy Fx
The flowing fluid will also exert an equal and opposite force on the boundary of the stream tube. The forces may include dynamic force due to change of momentum, static pressure, weight of fluid, drag force due to friction, gravity and inertia forces due to centrifugal force.
2.7 APPLICATIONS OF MOMENTUM EQUATION There are two types of applications of Impulse-Momentum equation. 1. Determination of forces exerted by the flowing fluid on the boundaries of flow passage of hydraulic machines due to change of momentum. The examples are: (i) Forces caused by a fluid jet striking a surface, i.e., fixed and moving blades of hydraulic machines. (ii) Jet propulsion. Propulsion of ships, boats, rockets, turbojet, ramjet, etc. (iii) Propellers. Marine propellers, helicopters. (iv) Pipe bends and reducers 2. Determination of flow characteristics due to energy loss in the flow systems. The examples are: (i) Sudden enlargement or constriction in a pipe such as orifices and mouth pieces. (ii) Hydraulic jump in open channel.
26 Hydraulic Machines 2.8 MOMENT OF MOMENTUM EQUATION This equation is derived from the principle of moment of momentum which states that the resulting torque acting on a rotating fluid is equal to the rate of change of moment of momentum. Vw3 Fixed centre or axis of rotation
90°
r3
a3
V3
3 Vw2
r2
r1
90° a 2 1
90° 2
V2
a1 V1 Vw2
Fig. 2.7
Angular momentum
Let V1 = velocity of fluid at section 1 of stream tube r1 = radius of curvature at section 1 r = density of fluid Q = rate of flow of fluid Momentum of a particle (mass m) in a direction normal to radius r = mVw Moment of momentum of particle about axis of rotation = mVw r From geometry, Vw = V cos a \ Moment of momentum = mV cos ar This is called angular momentum. Angular momentum at section l = mV1 cos a 1 r1 Angular momentum at section 2 = mV2 cos a 2 r2. \ Change of angular momentum = m(V2 cos a 2 r2 V1 cos a1 r1) Rate of change of angular momentum =
m (V2 cos a2 r2 V1 cos a1 r1) t
Basic Principles of Hydraulic Flow and Jet Theory
27
But rate of change of momentum is equal to torque. \
T=
m (V2 cos a2 r2 V1 cos a1 r1) t
= rQ(V2 cos a2 r2 V1 cos a 1 r1) For a circular path, r1 = r2 = r3 and a 1 = a2 = a3 = 0 \
T = rQr(V2 V1) Power
P = Tw = rQr (V2 V1)w
where w = angular velocity. Blade velocity,
Vb = wr P = rQ(Vw2 Vb2 Vw1 + Vb1)
\
The moment of momentum equation is applied for: 1. Analysis of flow problems in turbines and centrifugal pumps. 2. Finding torque exerted by water on sprinkler.
2.9 EULERS FUNDAMENTAL EQUATION Eulers fundamental equation is the equation of motion for a fluid with the following assumptions. Assumptions
7.
Fluid is non-viscous and frictional losses are zero. Fluid is homogeneous and incompressible. r = constant Flow is steady. Flow is one-dimensional along the streamline. Velocity is uniform over the section Flow is continuous Except gravity and pressure forces, no other forces are involved. Other forces like forces due to viscosity, turbulence and compressibility are neglected. Energy is defined as ability to do work. It manifests in various forms and can change from one form to another. 1. Gravitational potential energy or elevation energy Z in metres of liquid column. V2 in N-m. 2 3. Pressure energy required to move the fluid against its pressure.
Kinetic energy due to mass and velocity of fluid
Pressure energy = p
[N/m2].
28 Hydraulic Machines Consider a stream tube. The forces acting on a cylindrical element of stream tube are: 1. Pressure force pdA in the direction of flow
FG H
Pressure force p +
IJ K
¶p ds dA opposite to the direction of flow. ¶s
Weight of the element, rg dA dS. The resultant force on fluid element in the direction of S = mass ´ acceleration
FG H
pdA p +
\ where
IJ K
¶p dA rg dA dS cos q = rdAdS ´ aS ¶s
...(5)
aS = acceleration in the direction of S. =
dV dt
V = V(S, t) aS =
\
¶V ¶S ¶V . + ¶S ¶t ¶t
=V For steady flow,
LM\ N
¶V ¶V + ¶S ¶t
dS =V dt
OP Q
¶V =0 ¶t
aS = p + ∂p dA ∂s
\
V ¶V ¶S
S
Z ds
dS
q q
pdA
X rgdAdS
Y
Fig. 2.8 Forces on a fluid element
Basic Principles of Hydraulic Flow and Jet Theory
29
From equation (5),
V ¶V ¶p dA dS rg dAdS cos q = rdA dS . ¶s ¶S
Dividing throughout by rdS dA \
But
¶V ¶p g cos q + V =0 ¶S r¶S
cos q =
dZ dS
\
dZ dV 1 ¶p +g +V =0 dS dS r ¶S
or
¶p + g dZ + V dV = 0 r
or
¶p + g dZ + VdV = 0 r
...(6)
Equation (6) is called Eulers equation of motion.
2.9.1 Bernoullis Equation Bernoullis equation is also called energy equation and is obtained by integration of Eulers equation of motion
z z z
dp + g dZ + V dV = constant r
For incompressible flow, r = constant \
p V2 + gZ + = constant 2 r
or
p V2 +Z+ = constant 2g rg
Equation (7) is called Bernoullis equation. p = Pressure head = Pressure energy per unit weight of fluid. rg V2 = Kinetic head = Kinetic energy per unit weight 2g
Z = Potential head = Potential energy per unit weight.
...(7)
30 Hydraulic Machines The Bernoullis equation can also be written as V2 V2 p1 p + z1 + 1 = 2 + z2 + 2 . 2 g rg 2g rg
2.10 JET THEORY When a nozzle is fitted at the end of a pipe, a jet of liquid comes out with high velocity utilizing the pressure of liquid in the pipe. Applying continuity equation to the sections 1 and 2 of the nozzle, A1V1 = A2V2
Pipe
The velocity of liquid jet,
2
Fig. 2.9 Liquid jet
The kinetic energy of the jet, KEj =
1 m V22 2
Let jet velocity = Vj = V2 \
KEj = =
1 m Vj2 2 1 1 rAj Vj ´ Vj2 = rAj Vj3 2 2
[ j/s]
The jet power,
1 rA jV j3 Pj = 2 1000 The efficiency of nozzle, hn =
Nozzle
V1
A1 V1 A2
V2 =
1
[kW]
Jet power at outlet of nozzle Liquid power at inlet of nozzle
1 rA jV j3 V j2 2 ´ 1000 = = rgQH 2 gH 1000
where H = total head of liquid at the inlet of nozzle. The liquid jet exerts a force on a plate placed in front of it.
Liquid jet V2
31
Basic Principles of Hydraulic Flow and Jet Theory
2.10.1 The Impulse Principle The horizontal fluid jet impinging on a fixed vertical flat plate will spread out along the plate. The velocity in the direction of the jet is reduced to zero. A horizontal force F is imparted to the plate in the direction of the jet. This force is called an impulse and is equal to the change in momentum of the jet. F = m (Vj 0) = m Vj. This principle is used in impulse turbines like Pelton wheel to get mechanical power of rotation from the jet force.
2.10.2 Reaction Principle The fluid coming out of a fixed nozzle has a force which can be used to rotate a turbine wheel. There will be a corresponding and equal force tending to move the nozzle in the opposite direction. This opposite force is called reaction. If the nozzle is fitted on a wheel, the same will rotate in the opposite direction of the fluid jet and mechanical work of wheel rotation is produced. The reaction principle is used in the reaction turbines i.e., Francis and Kaplan turbines to generate mechanical power of shaft rotation from the reaction of jet. The force of reaction = rAVj2 = m Vj,
2.10.3 Propulsion Principle The reaction principle is also utilized in the propulsion of ships, rockets, vessels, jet aircraft, etc. The machine is fitted with a nozzle and is free to move in the direction opposite to that of jet. This is called propulsion principle or jet propulsion,
2.11 JET FORCE ON STATIONARY FLAT PLATE 2.11.1 Normal Plate The initial velocity of jet = Vj The final velocity of jet = 0 The impinging force, F = m (Vjf Vji) = m (0 Vj)
= m Vj = r Q Vj .
Fixed plate
Q/2 F
Nozzle Vj
X
Q/2
Fig. 2.10 Jet force on normal fixed plate
Example A jet of water 75 mm diameter strikes a normal fixed plate with a velocity of 200 m/s. Find the force exerted by water jet on the plate.
32 Hydraulic Machines p Fj = r Q Vj = rAj Vj Vj = r Dj2 Vj2 4
Solution
FG H
p 75 4 1000
= 1000 ´
IJ K
2
´ (20)2 = 1766.8 N
= 1.767 kN
2.11.2 Inclined Plate A fluid jet with a velocity Vj strikes a fixed plate inclined at an angle q with the direction of jet. The components of jet velocity, Vjx¢ = Vj sin q Vjy¢ = Vj cos q = 0 The jet force, Fj = r Q Vjx¢ = r Q Vj sin q. The jet force has components in X-direction and Y-direction. Fjx = Fj sin q = rQVj sin2 q F = r Q V sin q cos q = 0 iy
j
Y
Plate Q1 Nozzle
Fj sin q
Vj
q
q
Q2
X
Fj cos q Fj
Fig. 2.11 Jet force on inclined fixed plate Y
Y¢
Vj sin q Vj cos q
q Vj
X q X¢
Fig. 2.12
Components of jet velocity
Basic Principles of Hydraulic Flow and Jet Theory
33
The volumetric discharge Q is divided into Q1 and Q2. Q = Q1 + Q2 Q1 =
Q (1 + cos q) 2
Q2 =
Q (1 cos q) 2
2.12 JET FORCE ON MOVING FLAT PLATE The absolute velocity of jet issuing from the nozzle, = Vj The velocity of moving plate in the direction of jet = U. The jet force F = r Q (V U) j
j
The quantity of fluid mass striking the plate per second, Q = A(Vj U) \
Fj = rA(Vj U)
Nozzle 2
Plate A Vj
The work done by the jet on the plate
U
Wj = Fj ×U = r Q (Vj U )U. The kinetic energy of jet KEj =
Fig. 2.13
1 1 1 m Vj2 = r Q Vj2 = rAVj3 2 2 2
Jet force on moving flat plate
The efficiency of the system, h=
=
Wj Work done on the plate = KE j Kinetic energy of jet rA(V j - U ) 2 .U 2 = 3 [Vj2 U + U 3 2UjU 2] 1 Vj rAV j3 2
For a given Vj, maximum efficiency. dh 2 = 3 (Vj2 + 3U 2 4VjU ) = 0 dU V j 2 ¹0 V j3
34 Hydraulic Machines Vj2 + 3U 2 4VjU = 0
\
(Vj U )(Vj 3U) = 0 Vj = U gives Wj = 0 Vj = 3U
\ The maximum work done, Wj max
F = rA G V H
j
I - J 3K Vj
2
´
Vj 3
=
4 rAVj3 27
4 rAV j3 8 27 h max = = 1 27 rAV j3 2 Special Case
When a series of plates is arranged, i.e., flat plates mounted on the periphery of a wheel, the effective distance between nozzle and plate remains constant. The quantity of fluid mass striking the plate per second, m = rAVj
The jet force acting on the plate, Fj = rAVj (Vj U ) Wj = Fj ×U = rQ(Vj U )U The efficiency of the system, rQ(V j - U )U = h= 1 KE j rQV j2 2
Wj
=
2(U j - U ) U V j2
=
2V j U - 2U 2 V j2
2V j - 4U dh = =0 dU V j2 \
Vj = 2U
or
U=
Vj 2
Vj
U
Fig. 2.14 Flat plates mounted on the periphery of a wheel
Basic Principles of Hydraulic Flow and Jet Theory
FG H
2 Vj -
h max =
2.13
IJ K2
35
Vj Vj 2
V j2
= 0.5 or
50%.
JET FORCE ON CURVED PLATE WHEN JET STRIKES TANGENTIALLY
2.13.1 Stationary Vane The fluid jet enters the curved vane with absolute velocity V1 glides along the smooth inner surface and leaves with absolute velocity, V2, making angles a 1 and a2 with the horizontal direction. V1 a1
V1 a1
V1 a1
Y
X
a2 V2
a2 V2
(i) a2 < 90°
(ii) a2 > 90°
a2 V2 (iii) a1 = a2
Fig. 2.15 Jet force on stationary vane
The velocity components in the X-direction are: VX 1 = V1 cos a 1 VX2 = V2 cos a 2. The force exerted by the jet on the curved vane, FX = rQ(V1 cos a1 V2 cos a2) But
Q = AV1
where A = cross-section area of jet. FX = rAV1 (V1 cos a1 V2 cos a 2)
\ (i) If a 2 > 90°
FX = rAV1(V1 cos a 1 + V2 cos a2) In order to get more power, a 2 > 90°.
36 Hydraulic Machines (ii) For symmetrical vane, a 1 = a 2 = a and V1 = V2 = V FX = rAV 2 (cos a + cos a) = 2rAV 2 cos a. The hydraulic thrust in Y-direction Fy = rAV 2 (sin a sin a) = 0 (iii) For semicircular vane, a1 = a2 = 0. FX = rAV 2 (cos 0° + cos 0°) = 2rAV 2
\
Fy = rAV 2 (sin 0° sin 0°) = 0 Semicircular vanes give maximum hydraulic force and zero hydraulic thrust.
2.13.2 Moving Vane A fluid jet of cross-sectional area A strikes a curved vane with an absolute velocity V1. The curved vane moves with a velocity U in X-direction. The tip angles of vane are a1 and a 2 at inlet and exit respectively. The velocity triangles at inlet and outlet of vane are drawn. U1 V1
W1
b1 a1 U
W1
Direction of motion of vane
b2 a 2 V2 U2
Fig. 2.16
Jet force on moving vane.
For frictionless vane, the relative velocity does not change, W1 = W2 Q = A(V1 U) FX = rA (V1 U)(V1 cos a 1 V2 cos a 2) If a2 > 90° FX = rA(V1 U)(V1 cos a1 + V2 cos a 2) In actual hydraulic machines, a series of vanes is mounted on the periphery of a wheel.
Basic Principles of Hydraulic Flow and Jet Theory
37
2.14 JET PROPULSION OF SHIPS 2.14.1 Jet Reaction A water jet of area A issues with a high velocity V. The jet velocity, V = Kv 2gh
h –F
The fluid discharge,
A•
V
U
Q = AV
Fig. 2.17 Jet reaction
The force of reaction, F = rQ(VX2 VX1) = rQV
The propulsive force exerted on the nozzle or tank = F. This jet reaction can be used to propel aircraft, rocket, ships or submarine. A nozzle is fitted to these machines.
2.14.2 Propulsion of Ships/Boats A pump in the ship sucks water from the sea and discharges through a nozzle with a velocity V. The ship or boat moves in the opposite direction with a velocity U. Ship or Boat
Pump
U
V
U
V W
Fig. 2.18
Ship propulsion
The relative velocity of water jet with respect to motion of ship, W=U+V The jet force F = rQV Work done by the jet on the ship W.D. = rQV.V = rQ(W U )U The energy input, E=
1 rQW 2 2
38 Hydraulic Machines The efficiency of the system, h=
W. D. rQ(W - U ).U 2(W - U )U = = 1 E W2 rQW 2 2
For maximum efficiency of the system dh =0 dU d 2(WU - U 2 ) =0 dU W2
\ But \
W2 ¹0 2
U=
W 2
FG H
2 W-
h max =
IJ K
W W 2 2
W2
=
1 = 50% 2
QUESTION BANK NO. 2 1. 2. 3. 4.
8.
9.
How are the hydraulic machines classified? Derive Eulers equation applied to fluid machines. Derive the continuity equation in Cartesian coordinates. Discuss the following aspects of hydraulic flow: (a) Closed and open systems. (b) Concept of continuum. (c) Lagrangian and Eulerian methods. (d) Streamline and stream tube Derive momentum and angular momentum equation. What are the applications of momentum and angular momentum equations in fluid machines? Establish the Bernoullis equation from Eulers equation of motion. Explain water jet theory and how are the following principles established. (a) Impulse principle (b) Reaction principle (c) Propulsion principle Derive equation of a jet force for moving flat plate and moving curved vane.
Basic Principles of Hydraulic Flow and Jet Theory
39
TUTORIAL SHEET NO. 2 1. A 25 cm diameter pipe carries lubrication oil of 0.9 specific gravity at a velocity of 3 m/s. Find the mass flow rate of oil. What will be the oil velocity at another section where diameter is reduced to 20 cm. Solution Section 1: D1 = 25 cm = 0.25 m p 2 p D = (0.25)2 4 1 4 = 0.049 m2. V1 = 3 m/s. r = 0.9 ´ 1000 = 900 kg/m3.
A1 =
m = rA1V1 = 900 ´ 0.049 ´ 3 = 132.23 kg/s Ans.
Section 2 D2 = 20 cm = 0.20 m p (0.20)2 = 0.314 m2 4 Applying equation of continuity at sections 1 and 2, A1V1 = A2V2
A2 =
\
V2 =
A1V1 0.049 ´ 3 = = 4.68 m/s Ans. 0.0314 A2
Water under a pressure of 29.43 N/cm2 (g) and velocity of 2 m/s is flowing through a 5 cm diameter pipe. Find the total head of water at a section 5 m above datum line. Solution
Pressure head =
p 29.43 ´ 104 = = 30 m. rg 1000 ´ 9.81
Kinetic head =
V2 2´2 = = 0.204 m 2 g 2 ´ 9.81
Datum head = Z = 5 m. \
Total head = 30 + 0.204 + 5 = 35.204 m
Ans
A jet of water 60 mm in diameter, having a velocity of 20 m/s, strikes a flat plate inclined at an angle of 30° to the axis of the jet. The plate moves at 5 m/s in the direction of the jet.
40 Hydraulic Machines Calculate the normal force exerted on the plate, work done per second and efficiency of the system. Sketch the arrangement showing velocity and force components. Solution Refer Figs. 2.11 and 2.12. The normal force on the moving plate, Fjn = rA(Vj Vb)2 sin a = 1000 ´
p (0.06)2 ´ (20 5)2 sin 30° 4
= 318 N
Ans.
The force component in the direction of jet Fjx = Fjn sin a = 318 ´ 0.5 = 159 N Ans. Work done per second, W = Fjx Vb = 159 ´ 5 = 795 Nm/s = 0.795 kW Ans. Kinetic energy supplied by the jet KEj = =
1 1 mVj2 = (rAVj)Vj2 2 2 1 1 rAVj3 = ´ 1000 ´ (0.06)2 ´ 203 2 2
= 11309.7 Nm/s. Efficiency of the system, h=
795 W = ´ 100 = 7% 11309.7 KE j
Ans.
Show that angle of swing of a vertical hinged plate is given by. sin q =
rAV j2 W
where Vj = velocity of jet striking the plate A = cross-sectional area of jet W = weight of plate q = angle of swing of plate from vertical Solution A plate AB is hinged at A and is free to swing. Initially, the plate is vertical and a water jet strikes it normally at mid-point G with a velocity,
Basic Principles of Hydraulic Flow and Jet Theory A
W q
G C
Nozzle
G a
Vj
41
Fjn
D Fjn B
B
Vj. The plate swings through an angle q. The plate experiences the following forces 1. Weight W of plate acting vertically through G. 2. Normal water jet force, Fjn, Fjn = rAVj2 sin a where a = angle of jet with the centre line of plate. Fjn = rAVj2 sin (90° q)
\
= rAVj2 cos q.
The plate is in equilibrium under the two forces. Taking moments about point A. Fjn (AD) = W (CG) AD =
AG cos q
CG = AG sin q. Fjn
( AG ) = W(AG) sin q cos q
rAVj2 cos q
( AG ) = W(AG) sin q. cos q
\
\
sin q =
rAV j2 W
Proved.
TURBINES
P A R T II
+0)26-4
!
Pelton Turbine 3.1 INTRODUCTION Hydroelectric power plants utilize the potential energy of water to drive a turbine which, in turn, runs the generator to produce electricity. The hydro turbines are designed based on water head and discharge available at site. For high head and low discharge (water flow), Pelton turbines are used which run on impulse principle. For low head and large discharge, Kaplan and Propeller turbines are used which work on reaction principle. For medium head and discharge, Francis turbine is used which also works on reaction principle. The Pelton turbine is usually designed as horizontal shaft whereas Kaplan and Francis turbines are designed as vertical shaft machines. A special reversible turbine pump has been designed by a Swiss engineer and is named after him as Deriaz turbine. The comparative study of various types of hydro turbines is discussed in the following paragraph.
3.2 COMPARATIVE STUDY OF HYDRO TURBINES 3.2.1 Specifications The hydraulic turbines have been classified and main specifications are given in Table 3.1. Table 3.1 S.No. Parameters 1. 2. 3. 4. 5. 6. 7. 8. 9.
Originator Principle of operation Head and discharge
Classification of hydraulic turbines
Pelton Turbine
Francis Turbine
Lester Allen Pelton, USA James B. Francis Impulse Reaction High head, low discharge Medium head Medium discharge Maximum head (m) 300-2000 30-500 Maximum power (MW) 250 720 Maximum wheel 5.5 10 diameter (m) Flow direction Tangential Mixed Flow (Radial Inward) Specific speed 4-70 60-400 Shaft axis Horizontal vertical
Kaplan Turbine Dr. Victor Kaplan Reaction Low head, large discharge 2-70 225 10 Axial 300-1100 vertical
46 Hydraulic Machines The layout drawings of various turbines are shown in Fig. 3.1.
I
(a) Pelton turbine
(c) Kaplan turbine
(b) Francis turbine
Fig. 3.1
Layout of hydraulic turbines
3.2.2 Performance Curves The efficiency curves of various turbines are plotted against normalized water discharge in Fig. 3.2. Francis turbine is suitable for constant load operation whereas Kaplan turbine is suitable for variable load operation. Pelton turbines have lower efficiency but can be used as variable load turbine. 1.0
Efficiency
0.9
3 1
0.8 2 0.7 0.6 0
40
80
120
Discharge (%) 1- Pelton turbine 2- Francis turbine 3- Kaplan turbine
Fig. 3.2 Efficiency curves
3.2.3 Selection of Hydro Turbines The following criteria are used for the selection of hydro turbine for a site to attain the maximum possible efficiency and economy. 1. Operating head
The present practice is to use Kaplan and Propeller type of turbines for heads upto 50 m. Francis turbines are selected for water heads from 50 m to 400 m. For heads greater than 400 m, Pelton turbine is used. The range of heads as mentioned is not rigid and may change if other conditions dominate to achieve economy.
Pelton Turbine
47
Size of turbine and number of units
A large size of hydro turbine results in less number of units for a power plant. This leads to lower capital cost due to economy of size and lesser number of runners, penstocks, generators, etc. However, less number of hydrosets are not able to meet the fluctuating load requirements economically as the turbines may not be operating at optimum efficiency at part loads. In no case there should be less than two turbines to ensure reserve availability of capacity during plant shut down under normal and emergency conditions. There is limitation on the size of type of turbine as given in Table 3.1. 3. Turbine Speed
High speed turbines are more economical as the size of turbine, generator, power house, etc. are smaller. The turbine speed should correspond to synchronous speed of generator, i.e. N=
120 f p
where f = grid frequency (50 cps) p = number of poles of generator The specific speed range for different turbines is given in Table 3.1. The specific speed,
N P H 5/ 4 where N = turbine operating speed, (rpm) P = turbine output (kW) N = water head (m) NS =
Load characteristics
The performance characteristics of various turbines are shown in Fig. 3.2. Kaplan and Pelton turbines have flat efficiency curves and can operate at high efficiency for large range of part loads. Francis turbine is not suitable to operate at part loads as efficiency falls appreciably. However, Francis turbines have high efficiency when operating as base load turbine. 5. Height of Installation
The cost of excavation can be saved, if the turbine can be installed at higher level, Pelton turbines are generally horizontal shaft turbines requiring little excavation. Francis and Kaplan turbines are generally installed vertically and are fitted with draft tube to avoid cavitation. This requires excavation for installation of draft tubes. For these machines also height of installation should be more to save excavation cost.
48 Hydraulic Machines 3.3 PELTON TURBINE Pelton wheel is an impulse turbine used for high head installation. It has evolved from Himalayan Mill which is still widely used in the high mountains from Afghanistan to Burma for grinding of grains. The pressure energy of water available in high reservoirs is converted into velocity head. The water is carried from reservoirs through penstocks. A nozzle is fitted at the end of penstock which issues a free and compact jet of water. The water jet sticks a series of buckets mounted around a circular wheel. The buckets are designed in the form of twolobe ellipsoidal discs for high conversion efficiency and balanced hydraulic thrust. Each bucket has a ridge or splitter in the middle to divide the jet into two equal streams. The symmetry of the bucket ensures no axial face on the shaft bearings. The wheel rotates and supplies torque or mechanical power to the shaft carrying the wheel. The Pelton turbine is a very reliable and efficient in operation at high heads. At low heads, the rate of water flow has to be increased for a given power output. The high flow rate leads to big jet diameter and high jet velocity resulting in: 1. Large runner diameter 2. Low peripheral velocity of runner The increase of runner diameter and decrease of its velocity requires a bulky, slow speed machine. For low heads and large flow rate, reaction turbines are more suitable.
3.3.1 Main Components The main components of a Pelton turbine are: 1. Water inlet system 2. Runner assembly 3. Casing 4. Brakes 1. Water Inlet system
(a)2enstock A large pipe or conduit made of steel, concrete or wood carries water from high level reservoir to the powerhouse. It is provided with suitable anchor blocks to take the hydraulic thrust. Screens or trash racks are provided at the inlet to prevent entry of debris. There is a stop valve and a control valve provided in the penstock to regulate the flow of water. The penstock ends in a nozzle. (b) Spear and Nozzle At the end of the penstock, there are one or several nozzles which convert the pressure energy into high speed jet. A spear or needle moves to and fro in the nozzle to vary the outlet annular area, thereby controlling the water flow rate depending upon the load on the turbine-generator. The
Pelton Turbine
49
Spear
To governor
Nozzle
Deflector
From penstock Fig. 3.3 Spear and nozzle.
movement of the spear is automatically controlled by hydraulic system of governing system. The runner speed has to be maintained constant at all loads and variable heads. (i) When the load on the generator decreases or head increases, the turbine speed will increase. The speed governor operates to push the spear forward into the nozzle reducing the annular area and water discharge. The turbine speed is reduced to synchronous speed. (ii) When the load on the generator increases or head decreases, the turbine speed falls. The speed governor operates to withdraw the spear backward. The annular area of nozzle increases and also increases the water discharge. The turbine speed is increased to synchronous speed. (iii) Protection System In case of sudden load throw off, quick closing of nozzle can cause bursting of inlet pipe. This is avoided by the use of one of the following devices which operate along with slow closing of main nozzle. (1) By-pass nozzle to discharge extra water without striking the runner buckets. (2) The governor swings a deflector plate in between the nozzle and runner buckets when load is thrown off. (3) Seewer Jet Diffuser When the load is thrown off, a number of blades move forward towards the spear needle and whirl the flow and thus dissipate the water energy. 2. Runner assembly
The runner assembly consists of a circular wheel which is shrunk fit and keyed to a shaft supported on bearing. An odd number of specially shaped buckets (not less than 15) are bolted to the wheel periphery at equal pitch. Sometimes the wheel and buckets are cast together. The buckets are double hemispherical cup-shaped castings of cast iron, bronze or stainless steel or cast steel. The inner surface of buckets is finely polished to reduce flow friction of water. The jet strikes the splitter ridge at the centre and flow is equally deviated on both sides, thus eliminating the end thrust.
50 Hydraulic Machines The water jet after impinging on the buckets is deflected through an angle of about 165° (instead of 180°) to avoid its splashing on the incoming bucket and retarding the runner motion. A recess or notch is provided at the bottom of the bucket not to interfere with the water jet till it is in proper position. Notch
Splitter ridge 165°
Hemispherical
Fig. 3.4 Bucket
Casing
The casing has atmospheric pressure and has no hydraulic function. The water outflow from bucket issues in the form of a strong splash and scatters in all direction. The casing avoids the water splash and guides the water to the tail race. It also acts as safe gaurd against accidents. The casing is made in two halves with a seal at the joint. Baffles are also provided to reduce windage losses. The casing can be a cast or fabricated from steel plate. 4. Brakes
When the turbine stops, the runner can still continue revolving for quite sometime due to inertia. A hydraulic brake is provided to forcefully stop the runner in minimum time. The hydraulic brake consists of a nozzle of diameter 60% of main nozzle and is situated and positioned suitably, so that the brake jet strikes the bucket at its back.
3.4 LAYOUT ARRANGEMENTS The sectional view of a single jet horizontal Pelton turbine is shown in Fig. 3.5. All the main components have been arranged. The following design changes are discussed below. 1. Number of jets
One nozzle Pelton turbines are widely used. However, large turbines may have more than one nozzle depending upon the specific speed to control the running speed. Multi-jets are essential when NS > 35. The arrangement layout of a 4-jet Pelton turbine is shown in Fig. 3.6.
Pelton Turbine
51
Brake From main pipe
Runner
Buckets
Nozzle
Deflector
Tail race
Fig. 3.5 Layout of Pelton turbine
Fig. 3.6
Multi-jet Pelton turbine
Number of runners
The most common Pelton turbine layout is single overhung runner. However, double runner gives more power and higher speed. The various arrangements are shown below. 3. Shaft arrangement
Horizontal Pelton turbines are most common. However, horizontal shaft arrangement is used for Pelton turbines with one or two nozzle. For four or six nozzle turbines, vertical shaft arrangement is used.
3.5 VELOCITY DIAGRAM OF BUCKET The inlet and outlet velocity triangles of the bucket are drawn to estimate the work output and efficiency of a Pelton wheel.
52 Hydraulic Machines
Generator
Exciter
Turbine Runner (i) Single overhung hydro-set
Generator
Runner
Runner (ii) Double overhung hydro-set
Generator
Runner Runner (iii) Double runner hydro-set
Fig. 3.7 Runner arrangement
Pressure atmospheric Bucket 2¢ Vb1
Vr1
1 Vb
Nozzle
~ 165° = (180° – b2)
Vj = Va1 = Vw1 a1 = 0 b1 = 180° 2
b2 Va2
Vr2
a2
b2
Vf
Vw2 Vb2 Fig. 3.8 Velocity diagrams
2
1—Entry point 2—Exit point
Pelton Turbine
53
Blade is divided into two parts to avoid side thrust or axial thrust Vj = Jet velocity
[m/s]
Va = Absolute velocity of water Vb = Bucket or blade tip velocity Vw = Whirl velocity Vf = Flow velocity Vr = Relative velocity 1. Inlet velocity triangle
Pelton wheel is an axial flow impulse turbine and the angle of entry of the jet is nearby zero. a1 = 0 and b1 = 180° where a = Angle of absolute velocity with blade velocity b = Angle of relative velocity with blade velocity Va1 = Vj = Vw1
\
Vr1 = (Vj Vb1) Vb = Vb1 = Vb2 =
pDN 60
Outlet velocity triangle
The outlet velocity triangle depends upon the speed of the runner. The values of a2 and Vw2 will change as shown.
Vr
Va2
2
Vr
2
a2
b2
Va
2
b2
Vw2 Vb
Vb2
2
(a) Fast runner
(b) Medium runner Va
Vr
2
2
a2
b2
Vb
2
Vf
2
Vw
2
(c) Slow runner Fig. 3.9
Outlet velocity triangles
a2
54 Hydraulic Machines Type of Runner
Angle =2
Velocity Vw2
Slow Runner Medium Runner Fast Runner
< 90° = 90° > 90°
ve 0 +ve
The buckets are polished to fine finish from inside to reduce friction. Vr2 = KVr2,
where K » 0.85
Under ideal conditions, assuming no friction K = 1.0 Vr2 = Vr 1 Vb1 = Vb2 = Vb
(3 r1 = r2)
Work output
The kinetic energy of water at inlet, Va12 [N-m/kg] 2 The kinetic energy of water at outlet,
KE1 =
KE2 =
Va22 2
Energy transferred to rotor =
F Va GH 2
2 1
-
Va22 2
I JK
Power input to wheel or work done on the wheel, W = (Vw1 r1 Vw2 r2) w = (Vw1 Vw2) wr = (Vw1 Vw2)Vb For fast runner W = (Vw1 Vw1 )Vb For medium runner W=0 For slow runner W = (Vw1 + Vw2)Vb
(3 W = Tw) (Vb = wr)
Pelton Turbine
55
Hydraulic efficiency
The hydraulic efficiency or blade efficiency hhyd = =
Work doneon wheel Jet power (Vw1 - Vw2 )Vb V j2 2
From velocity diagrams, Vw1 = Vj Vw2 = Vb2 Vr2 cos b2 = Vb2 KVr1 cos b2 = Vb2 K(Vj V b1) cos b2 \
hhyd =
[V j + K (V j - Vb1 ) cos b 2 - Vb ]Vb V j2 /2
For a given runner, b2, Vj and K are constants. For estimation of maximum hydraulic efficiency, d ( hhyd ) dVb
=0
From equation (1) hhyd = \
d ( hhyd ) dVb
=
2 KVb2 cos b 2 2Vb2 2KVb 2Vb + cos b2 2 Vj Vj Vj V j2
2 K cos b 2 4 KVb cos b 2 2 4V + b =0 2 Vj Vj V j2 Vj
2 + 2K cos b2 4K 2 + 2K cos b2 =
Vb 4V cos b2 b = 0 vj Vj
2Vb (2 + 2K cos b2) Vj
1 Vb = 2 Vj
\ Speed ratio,
Vb = Vj
Vb = 0.5 2 gHa
(1)
56 Hydraulic Machines Actual speed ratio varies between 0.46 and 0.47 Vb = 0.5 2gHa =
\
Dm =
pDm N 60
60Vb pN
The blade efficiency curve is shown in Fig. 3.10. 1.00
Actual hb
0.50 Ideal
0
0
0.46 0.5
1.0
Speed ratio, Vb/Vf
Fig. 3.10 Blade efficiency curve
Jet ratio
Jet ratio is an important characteristic of Pelton turbine and is defined as m=
Dm dj
where Dm = Mean diameter of runner dj = jet diameter m varies from 6.5 to 20. For maximum hhyd, m should be taken between 10 and 14.
3.6 TURBINE LOSSES AND EFFICIENCY For performance analysis, the various losses in the turbine are estimated and plotted in a Sankey diagram. The energy balance sheet is also prepared. There are three types of losses. 1. Volumetric losses
Volumetric losses arise as full volume of the jet does not strike the bucket. The volumetric losses may be of the order of 1% to 3%.
Pelton Turbine
57
Hydraulic losses
These losses are due to the following factors: (i) Exit losses due to kinetic energy of water leaving the bucket
F Va I . GH 2 JK 2 2
(ii) Friction and shock losses on the bucket due to improper inlet blade angle and surface finish. 3. Mechanical losses
The mechanical losses include: (i) Bearing friction (ii) Windage losses The mechanical losses may be 1% to 3%.
3.6.1 Sankey Diagram
Penstoke losses
Nozzle losses
Volumetric losses
Shaft power
Power transferred to wheel
Fluid power stricking buckets
Jet power
Gross power
Power available
The Sankey diagram for a Pelton turbine is shown in Fig. 3.11.
Hydraulic Mechanical losses losses
Fig. 3.11 Sankey diagram
Vj2
hnozzle = hvol = = hhyd =
rQg Jet power 2g = Available power rQg Ha
(» 97%)
Kinetic energy of fluid striking the bucket Jet power Q - DQ Q
(97 99%)
Power transferred to wheel Kinetic energy of fluid striking the bucket
58 Hydraulic Machines
r(Q - DQ)
hhyd =
r(Q -
F Va GH 2
2 1
-
Va22 2
DQ)V j2 / 2
I JK (V =
w1
Vw2 )Vb
V j2 / 2
Vw1 and Vw2 are measured from velocity diagrams. (Va21 - Va22 )
hhyd =
V j2
F Va I =1G H V JK
2
2
j
Shaft power output Power transferred to wheel
h mech =
(97 to 99%)
hoverall = hturbine = hnozzle ´ hvol ´ hhyd ´ hmech (90 to 92%)
3.6.2 Energy Balance Sheet A typical energy balance diagram is plotted in Table 3.2. Table 3.2 Energy balance (Basis MW) Credit Description 1. Gross Power rQg Hg 10 6 2. Power Available
Debit MW
Description 1. Transmission losses
Pg =
PDhf = 2. Brake Power
Pa = (Pg PDhf ) =
rQg Ha 10 6
MW
rQg DHf 10 6
2pNT 60 ´ 10 6 Nozzle losses PDhn = (Pa Pj) Volumetric losses PDhval = (1 hvol)Pj Hydraulic losses PDhhyd. = (1 hhyd)Pj hvol Mechanical losses PDhmech = (1 hm)Pj hvol hhyd Ps=
6.
Total
Total
3.7 DESIGN OF PELTON TURBINE The hydraulic design of a Pelton turbine is based on a large number of experimental and imperical relations and coefficients determined from dimensional analysis and good engineering practice.
Pelton Turbine 1. Turbine power
The turbine power pt = Pa hoverall Where Pa = Available Power (W) = rQg Ha. Q = Volumetric flow rate (m3/s) Ha = Available head (m) hoverall = Overall efficiency of turbine (85 to 90%) 2. Nozzle and jet size
The angle of nozzle = 2b Angle of spear = 2a a is 22.5 to 25° b is 28.5 to 42° The jet flow, Q=
p 2 d 1 V1 (m3/s) 4
V1 = Kv1 2gH where
H = Head (m) Kv1 = coefficient of velocity (0.98 0.99) at vena contracta of jet. Vena contracta V a b a0 a1 V0 V1 Fig. 3.12
Nozzle and jet size
p 2 d 0 V0 4
Also
Q=
\
V0 = Kv0 2gH Kv0 = coefficient of velocity at nozzle exit (0.81 0.83)
59
60 Hydraulic Machines The area of nozzle exit, A0 =
p 2 d0 4
A0 = AS + BS 2
Let
where A and B are constants and S is maximum spear travel where
A = 2pr0 sin a
and
B=
(b - a ) sin (b - a )
p (b - a ) (sin b - sin a ) sin 2 a sin 2 (b - a )
Number of jets
A Pelton wheel ordinarily has one nozzle and jet. However, more jets are employed when more power is produced by the runner. The number of jets is decided by specific speed of the runner which is defined as follows. The specific speed, NS =
N Pt H 5/4
where N = Rotor speed (rpm) Pt = Turbine power H = Available head The maximum value for a single jet Pelton wheel is limited to 35. If NS is more, then the number of jets are increased. There is another dimensionless quantity (KS) used to decide number of jets. KS =
QN 2 Vj3
where Q = Volumetric flow rate (m3/s) Vj = Jet velocity Vj = 2gH KS =
FG N IJ H 32.2 K s
2
for turbine efficiency, h t = 90%
For single jet Pelton turbine, KS = 1. For Pelton turbine with one wheel and 2 jets, total power will be 2Pt.
Pelton Turbine
\
NS2 =
N 2 Pt H 5/4
61
= 2 NS1
Observations 1. For single-jet turbine, KS < 1 2. For two-jet turbine, KS lies between 1 and 2 3. For three-jet turbine, KS lies between 2 and 3 4. Maximum number of jets used so far; 2 jets for horizontal shaft turbine 6 jets for vertical shaft turbine 5. Governing of turbine with multi-jets becomes complex 4. Runner diameter
The runner diameter of the turbine is also called pitch circle diameter measured from the centre of the buckets.
w D1
Fig. 3.13
Pitch diameter
The circumferential velocity or peripheral velocity, U 1 = f 2gH where
f = speed ratio =
U1 2 gH
The speed ratio varies between 0.44 to 0.46 with average value of 0.45. U1 = Now \
pD1 N 60
(m/s)
pD1 N = f 2gH 60
D1 =
60 H 38.06 H ´ 0.45 2g = N N p
62 Hydraulic Machines The rotor speed is decided by the synchronous speed of the generator N=
60 f p
where
f = grid frequency = 50 cps for India p = no. of pair of poles of hydro-generator The usual values of N are 500, 300 or 250 rpm. Higher values of N result in lower cost of turbine-generator set. 5. Rotor speed
The selection of rotor speed requires detailed techno-economical analysis. The turbine efficiency is a function of specific speed. Lower value of NS leads to higher turbine efficiency. NS also depends upon rotor speed. ht = f (NS) NS = f(N) Higher rotor speed leads to: 1. Higher value of specific speed, NS. This results in : (i) Compact and less expensive turbine. (ii) Jet diameter decreases, jet ratio increases and turbine efficiency increases. (iii) Multijet turbines. It is more expensive and governing is complicated. 2. Hydro-generator speed increases (i) Less number of poles are required. (ii) The cost of hydro-generator decreases. 3. Higher stresses in turbine and generator rotors (i) Superior materials are needed to match higher stresses (ii) Cost of turbine-generator increases. 6. Jet ratio
Jet ratio is defined as the ratio of the pitch circle diameter of runner and jet diameter. Jet ratio,
m= NS =
where
N=
D1 d1 N Pt H 5/4
60 f 2 gH pD1
Pelton Turbine
63
Pt = rQgHht. Q= \
p 2 d1 y 2gH 4
Pt = r
p 2 d1 y 2gH .gH h t 4
60f 2 gH NS = . pD1
p r d12 y 2 gH . gHht 4 H 5/ 4
260 ht m The jet ratio is between 10 and 30. For maximum turbine efficiency m = 12 is adopted. A small value of m results in a very few buckets and too close buckets spacing. A large value of m results in bulky installation.
=
Bucket dimensions
The working dimensions of a bucket are specified in terms of jet diameter d1. L = (2.3 to 2.8)d1 B = (2.8 to 3.2)d1 T = (0.6 to 0.9)d1 Inlet angle, b1 » 5° to 8° Outlet angle, b2 = 10° to 20° at the centre D1
PCD
T
B
2b1 b2 Section
L
Fig. 3.14 Bucket dimensions
64 Hydraulic Machines 8. Number of buckets (z)
Minimum number of buckets on the periphery of a Pelton wheel is decided (i) to ensure little loss due to fricition (ii) to ensure full utilization of jet water, i.e. no escape without striking the buckets. The number of buckets is related to jet ratio z = 0.5m + 15 for m from 6 to 35. The angular pitch q= =
2p (radians) z 360 (degrees) z
3.8 DESIGN EXAMPLE OF A PELTON TURBINE 3.8.1
Design a Pelton Turbine for the Following Site Conditions
Gross head, Hg = 400 m 2. Power to be developed, Pt = 30 MW 3. Penstock length, L = 2 km. Note The power to be developed is decided on the basis of site conditions, load factor, cost of energy produced, revenue expected. 1. Determination of water flow rate
Pt = rQg Hg ´ htrans ´ hoverall Assume
htrans = 0.90 hoverall = 0.91
\
Q= = Assume
Pt rg H g h tran ´ hoverall 30 ´ 106 = 9.976 m3/s 1000 ´ 9.81 ´ 400 ´ 0.9 ´ 0.91
Q = 10 m3/s
Number of turbines
The best availability of power supply can be ensured by the use of 3 sets of 50% capacity each so that 2 sets operate at best efficiency and third set to meet peak (surplus) loads. In case of breakdown, two sets meet the total load.
Pelton Turbine
65
HR
Hg
Penstock (2 km)
DAM
Fig. 3.15
PH TR
Powerhouse
Therefore, use 3 turbines of 15 MW capacity each. There will be 3 penstocks handling a discharge rate of 5 m3/s each. 3. Penstock design
The water flow in a pipe has a velocity range of 1 m/s to 6 m/s. Higher velocity leads to higher pressure drop but smaller diameter of penstock and cost saving. Assume a suitable water velocity of say 5 m/s. Calculate the pressure drop and operating cost due to power lost in friction. Also calculate the penstock diameter and investment cost on penstock. Plot cost v/s pipe diameter and find the pipe diameter for minimum resultant cost. In Fig. 3.15, curve (1) is investment cost v/s penstock diameter; curve (2) is operating cost due to power lost in friction v/s pipe diameter, curve (3) is the resultant sum of (1) and (2). 2
3 1
Cost
Minimum
D Pipe Diameter Fig. 3.16
Penstock design
Select diameter D corresponding to minimum total cost. The cross-sectional area of penstock, A=
p 2 D 4
A=
Q 5 = = 1 m2 V 5
66 Hydraulic Machines \
D=
4A = p
4´1 = 1128 mm p
Refer to Indian standard 1S:6631 and select the nearest size, say 1250 mm. \
V=
Q 5 = A p 1250 4 1000
FG H
IJ K
= 4 m/s OK
2
As per Darcys formula, the friction head, DHf = From curves, \
4 fLV 2 2 gD
f = 0.007 (The friction coefficient) DHf = htrans =
4 ´ 0.007 ´ 2000 ´ (4) 2 = 36.53 m . 2 ´ 9.81 ´ 125
Hg - DH f Hg
=
400 - 36.53 = 0.909 400
OK
Power available at inlet to turbine pa = rQg(Hg DHf) = 1000 ´ 9.81 ´ 5(400 36.53) = 17.85 MW 4. Selection of turbine speed and type of turbine
It is a high head power plant, Pelton turbine should be most suitable for a head of 400 m. Selection of turbine speed is based on techno-economical analysis. Higher rotation speed provides: (i) Compact and economical turbines (ii) More jet ratio and higher turbine efficiency. (iii) More generator speed, less number of poles and less cost. (iv) Higher stresses due to centrifugal force requiring superior materials of construction and more cost of turbine and generator (v) Higher specific speed and increased turbine efficiency The rotor speed =
60 f p
where f = frequency of power generation = 50 cps for India p = No. of pair of poles of hydrogenerator
Pelton Turbine
67
The relation between synchronous speed and the number of pair of poles is shown in Table 3.3. Table 3.3 p N (rpm)
Synchronous speed
1
2
3
4
5
6
8
10
12
3000
1500
1000
750
600
500
375
300
250
Assume \
N = 500 rpm NS =
N Pt Ha5/4
=
500 15,000 3645/4
= 38 rpm For Pelton wheel, NS should be between 10 and 35. \ Use
N = 300 rpm NS =
\
300 15,000 = 22.8 rpm OK 3645/ 4
3 NS < 35, single jet Pelton turbine can be used. 5. Design of nozzle
The design of nozzle consists of determination of geometry of nozzle and spear and its efficiency. Q= Also
p 2 dj Vj [m3/s] 4
Vj = Cv 2gHa
where coefficient of velocity Cv lies between 0.98 and 0.99. Assume Cv = 0.985 \
Vj = 0.985 2 ´ 9.81 ´ 364 = 83.2 m/s OK [Recommended range 8085 m/s] Vena contraction a b dn dj Vn Vj Fig. 3.17 Nozzle design
68 Hydraulic Machines
dj =
4´5 = 0.2766 m p ´ 83.2
4Q = pV j
= 276.6 mm Q=
p 2 dn Vn 4
Vn = Cd 2gHa The coefficient of discharge Cd lies between 0.81 and 0.83. Take
Cd = 0.82
\
Vn = 0.82 2 ´ 9.81 ´ 364 = 6.93 m/s dn =
4Q = pVn
4´5 = 0.303 m p ´ 69.3
= 300 mm Angle of spear, Take
2a = 45° 50° a = 24°
Angle of nozzle, 2b = 57° 84° Take
b = 35° Pj = rQg
Jet power,
V j2 2g
=
1000 ´ 5 (83.2)2 ´ 106 2
= 17.3 MW Power loss in nozzle = Pa PJ = 17.85 17.3 = 0.55 MW Nozzle efficency, hnozzle =
17.3 = 0.969 17.85
» 97% 6. Runner diameter
The ideal speed ratio,
Vb = 0.5 Vj
Pelton Turbine
69
Vb = 0.5 2gHa = 0.5 2 ´ 9.81 ´ 364
The blade velocity,
= 42 m/s Vb =
pDm N 60
The mean diameter of runner, Dn =
60Vb 60 ´ 42 = = 2.6738 m pN p ´ 300
= 2675 mm The jet ratio,
m=
2675 Dm = = 9.67 276.6 dj
Jet ratio is an important feature of Pelton turbine and influences its characteristics m should be taken from 6.5 to 20. For maximum efficiency m should be taken from 10 to 14. Assume \
m = 10 Dm = m dj = 10 ´ 276.6 = 2766 mm
Number of buckets
The full volume of jet water does not strike the buckets and some water may slip and fall directly into the tail race without doing any useful work. This loss is represented by volumetric efficiency. hvol =
Q - DQ Q
The volumetric efficiency ranges between 97 to 99%. It is controlled by number of buckets, shape of bucket and jet ratio (m). Dr. Tayguns empirical formula for number of buckets Z = 0.5 m + 15 = 0.5 ´ 10 + 15 = 20 It can be checked by a scale drawing to accommodate the number of buckets calculated. For mechanical stability, number of buckets should be odd. \ Take Z = 21 8. Shape of a bucket
The shape of a bucket is related to jet diameter (dj) L = (2.3 to 2.8) dj = 636 to 774.48 mm
70 Hydraulic Machines Take
L = 700 mm B = (2.8 to 3.2) dj = 774.5 to 885 mm
Take
B = 830 mm T = (0.6 to 0.9) dj = 166 to 249 mm
Take Inlet angle, Take
T = 200 mm b1 = 5° to 8° 2b1 = 13°
Outlet angle,
b2 = 10° to 20°
Take
b2 = 15°
Notch size = dj »300 mm Notch depth = 0.2dj » 50 mm Draw the bucket to scale as shown in Fig. 3.18.
50
B = 830
Dm = 2766 mm
300
A
A L = 700 T = 200 15°
13°
Section A-A
Fig. 3.18
3.8.2
Bucket
Draw Energy Balance of the Pelton Turbine Designed as Per Para 3.7.1
The energy balance sheet is shown in Table 3.4 along with various calculations.
Pelton Turbine
71
Table 3.4 Energy balance sheet Basis MW Credit
MW
Gross power Pg = rQgHg ´ 106 = 1000 ´ 5 ´ 9.81 ´ 400 ´ 106 2. Power available Pa = rQgHa ´ 106
Total
Note:
19.62 17.85
Debit 1. Transmission losses Dhf = Pg Pa = (19.62 17.85) 2. Shaft power, Ps 3. Nozzle loses Dhn = (Pa Pj ) = (17.85 17.30) 4. Volumetric losses Dhv = (1 hv)Pj = (1 0.97)17.30 5. Hydraulic losses Dhh = (1 hh)hv Pj = (1 0.92)0.97 ´ 17.30 6. Mechanical losses Dhm = (1 hm)hv hh Pj
19.62
MW 1.77 15.00 0.55 0.52
1.28 0.55 19.62
htrans = 0.91 hh = 0.92 hv = 0.97 hm = 0.97 hoverall =
15 Ps = = 0.84 or 84% Pa 17.85
QUESTION BANK NO. 3 1. Compare the main hydraulic parameters of Pelton, Francis and Kaplan turbines. 2. What parameters are used for selection of a hydro turbine? 3. Describe the construction and operation of a Pelton turbine. Highlight the role of protection system, speed control system and brakes. 4. Discuss the general arrangement of a Pelton turbine with the help of suitable diagrams. 5. Draw the inlet and outlet velocity triangles for a Pelton turbine. 6. How are work output and hydraulic efficiency calculated from velocity diagrams? 7. What is jet ratio and speed ratio? How are these values calculated for maximum blade efficiency? 8. Draw a Sankey diagram for a Pelton turbine? How are various losses and efficiencies calculated?
72 Hydraulic Machines 9. How is energy balance sheet prepared for a Pelton turbine? Explain the method for calculation of various losses. 10. Explain step-by-step the method for the design of a Pelton turbine. 11. Discuss the role of the following in the design of a Pelton turbine. (a) Specific speed (b) Jet ratio 12. How are the following dimensions fixed up for a Pelton turbine (a) Nozzle diameter (b) Runner diameter (c) Number of buckets (d) Shape of spear and bucket. 13. How are the following features of a Pelton turbine decided? (a) Number of jets (b) Rotor speed 14. Draw the shape of the bucket of a Pelton turbine. How are various dimensions and angles decided?
TUTORIAL SHEET NO. 3 1. A Pelton wheel develops 3310 kW under a head of 100 m with an overall efficiency of 80%. Find the diameter of the nozzle if the coefficient of velocity for nozzle is 1. Solution
Pt = 3310 kW H = 100 m hoverall = 80% Pt = rQgH hoverall
\
Q=
3310 ´ 103 Pt = = 4.22 m3/s rgHhoverall 1000 ´ 9.81 ´ 100 ´ 0.8
The velocity of water through nozzle, Vn = Cv 2gH = 1 ´ 2 ´ 9.81 ´ 100 = 44.3 m3/s
\
Q=
p 2 d V 4 n n
dn =
4Q = pVn
= 348 mm
4 ´ 4.22 = 0.348 m p ´ 44.3 Ans.
Pelton Turbine
73
A Pelton wheel is working under a head of 45 m and the rate of flow of water through the jet is 850 litres/sec. The mean bucket speed of the wheel is 15 m/s. Find the efficiency and power produced by the wheel. If the jet is deflected by the bucket through an angle of 165°, draw velocity triangles. Take Cv as 0.98. Solution
H = 45 m Q = 850 l/s = 0.85 m3/s Vb = 15 m/s Vj = Cv 2gH = 0.98 2 ´ 9.81 ´ 45 = 29 m/s
\
p 2 dj Vj 4
Q= \
4Q = pV j
dj =
4 ´ 0.85 = 0.193 m p ´ 29
= 193 mm Speed ratio,
f=
Vb 15 = = 0.517 V j 29 Vj = Va = Vw1
Inlet velocity triangle
Vb
Vj = Va = Vw1 = 29 m/s Vb = 15 m/s Vr1 = Vj Vb = 29 15 = 14 m/s Assume no friction at bucket
Vr2 15° Vb2
Va2 Vw2
Vr2 = Vr1 = 14 m/s
\
Vw2 = Vb2 Vr2 cos 15° = 15 14 cos 15° = 1.477 m/s Pt = rQ(Vw1 Vw2)Vb = 1000 ´ 0.85(29 1.477) 15 ´ 103 = 350.92 kW
Ans
Pa = rQgHa = 1000 ´ 9.81 ´ 0.85 ´ 45 ´ 103 = 375.23 kW ht =
350.92 Pt = = 0.9352 375.23 Pa
= 93.52% Ans
74 Hydraulic Machines 3. Calculate the number of jets required and their diameter for a Pelton wheel which develops 11,500 kW under a head of 400 m when running at a speed of 450 rpm, assuming that the jet diameter is not to exceed one-tenth of the wheel diameter. Assume coefficient of velocity of jet as 0.97 and the ratio of peripheral velocity to the jet velocity as 0.45 and the efficiency of the turbine as 85%. Also find out the number and size of the buckets mounted on the wheel. Solution Data given: Pt = 11,500 kW H = 400 m N = 450 rpm Cv = 0.97 Vb = 0.45 Vj
ht = 0.85 Pt = rQgH h t \
Q=
Pt 11,500 ´ 103 = 1000 ´ 9.81 ´ 400 ´ 0.85 rgHht
= 3.44786 m3/s NS =
N Pt H
5/4
=
450 11500 = 26.97 rpm 4005/ 4
Vb = 0.45 2gH = 0.45
2 ´ 9.81 ´ 400
= 39.865 m/s
\
Vj =
Vb = 88.59 m/s 0.45
Vb =
pDm N 60
Dm =
60 ´ Vb 60 ´ 39.865 = = 1.6919 m pN p ´ 450
= 1692 mm m=
Dm = 10 dj
Pelton Turbine
\
dj =
Dm = 169.2 mm = 0.169 m 10
Qj =
p 2 p dj Vj = (0.169)2 ´ 88.59 = 1.9872 m3/s 4 4
No. of jets = Qj =
3.44786 Q = = 1.735 say 2 19872 . Qj Q 3.44786 = = 1.724 m3/s 2 2
V j = Cv 2gH = 0.97 2 ´ 9.81 ´ 400 = 85.93 m/s Qj = \
dj =
Jet ratio,
m=
p 2 dj Vj 4 4Q j pV j
=
4 ´ 1.724 = 160 mm p ´ 85.93
Dm 1692 = = 10.575 say 10 160 dj
Z = 0.5 m + 15 = 10.575 ´ 0.5 + 15 = 20.28 say 21 T = 120
A
30
A
160
Dm = 1962
B = 480
15°
10°
Section A-A
L = 400
L = (2.3 2.8)dj = 368 mm to 448 mm. Take L = 400 mm B = (2.8 3.2)dj = 448 mm to 512 mm. Take B = 480 mm T = (0.6 0.9)dj = 96 mm to 144 mm. Take T = 120 mm
75
76 Hydraulic Machines 4. Design a Pelton wheel with the following data: Head = 200 m Power to be developed = 1000 kW Speed = 380 rpm Assume reasonably all the missing data. Solution Follow 3.7.1. 5. (a) How are water turbines classified? (b) Why is the jet deflected by the bucket 160° to 165° instead of 180°? (c) Sketch the layout of single-jet, double-jet and four-jet Pelton turbines. 6. Show that in a Pelton wheel, where the buckets deflect the water through the angle (180° a), the hydraulic efficiency of the wheel is given by hh =
2Vb (V j - Vb )(1 + cos a ) V j2
where Vj = velocity of the jet Vb = velocity of the wheel at pitch radius Solution When a jet of water strikes a series of flat plates mounted on the periphery of a wheel, the efficiency, hh = 2
(V j - Vb )Vb V j2
.
When a series of curved vanes are mounted on the periphery of a wheel, all the water is utilized since the jet is always intercepted by the vane; the force acting on the vane. Fx = rAVj [(Vj Vb) { (Vj Vb) cos a}] where a = 180° b Work done, W = Fx .Vb = rAVj (Vj Vb)(1 + cos a) The hydraulic efficiency, hh =
2(1 + cos a )(V j - Vb )Vb W = 1 V j2 ( AV j )V j2 2
+0)26-4
"
Francis Turbine 4.1 INTRODUCTION Francis turbine is a reaction turbine having mixed flow. It has radial entry and exial axit of water. It is a versatile machine and works with medium heads and discharges. The water from the penstock is admitted to the spiral casing or scroll casing which distributes the water equally around the turbine. The guide vanes (wicket gates) fixed all around the turbine directs the water to the runner blades. A part of pressure energy is converted into kinetic energy in the guide vanes. The running blades utilize the kinetic energy as well as pressure energy of water to produce power. As the runner rotates under the reaction of pressure, it is called reaction turbine. The whole of pressure energy of water is utilized and water is discharged into a draft tube. The free end of draft tube is submerged deep in the tail water and a closed water passage is obtained. Some kinetic energy of water leaving the turbine is recovered in the draft tube. Main hydroelectric plants of India use Francis turbine. The examples are Bhakra Dam Project, Hirakund Dam Project, Rihand Dam Project, Cauvery Hydroelectric scheme, Chambal Hydroelectric scheme, etc.
4.2
SPECIFIC SPEED AND TURBINE TYPE
The specific speed of a hydraulic turbine is defined as the speed of a geometrically similar turbine which when operating under a head of 1 m develops 1 kW. NS =
N Pt H 5/4
where PJ = turbine power (kW) H = water head (m) N = rotor speed (rpm) NS = specific speed (rpm)
78 Hydraulic Machines The specific speed has been shown to be a useful parameter in classifying hydraulic turbines and in giving information on the general shape of machine. If performance characteristic, efficiency is plotted against NS, various types of runners show optimum efficiency in the range of different specific speeds. Hence, different types of runners can be classified according to the range of specific speed. It is, therefore, called Jype characJerisJic or characJer speed. Specific speeds of various runners are shown in Fig. 4.1 and Table 4.1 %100
h %90 Francis
Pelton
%80 0
35 60
100
Kaplan
200
300
400 Ns
500
600
700
800 900 1000
Fig. 4.1 Characteristic speed
Table 4.1
Specific speeds of Various Runners
Turbine
Type
NS
Pelton
Slow Normal Fast
10-20 20-28 28-35
Francis
Slow Normal Fast
60-120 120-180 180-300
Kaplan
300-1000
While designing a runner for a given output and head, the specific speed can be varied by changing the speed of rotation. It is a common practice to select a high specific speed runner which is always economical because the size of the generator as well as that of power house will become smaller. High specific speed is essential where water head is low and output is large, because otherwise the speed of rotation will be very low which means cost of generator and power house will be high. On the other hand, there is practically no need of selecting a high value of specific speed for high head installations because
Francis Turbine
79
even with low specific speed, high rotational speed can be attained with medium capacity plant. High specific speed means greater value of cavitation factor. As the head for a site is fixed, the specific speed can also be varied by selecting the unit size and number of machines. The specific speed can be changed by changing the runner diameter, width of blades, number of vanes and inlet and outlet angles. For low NS, large diameter and small tip width are needed and for high NS, small diameter and large blade width are required.
4.3 MAIN COMPONENTS The main components of a Francis turbine are: 1. Penstock 2. Spiral casing or scroll casing 3. Guide mechanism 4. Runner and shaft assembly 5. Draft tube
4.3.1 Penstock It is a water way to connect water reservoir to turbine casing. Water is supplied under pressure, therefore, the penstock is usually fabricated from high quality steel plates. The weld seams are radiographed and subjected to hydraulic test at double the working pressure. If the length of the penstock is short, a separate penstock is run from water reservoir to each turbine. For moderate heads and long distances, a single penstock is used to feed two or more turbines. The penstocks may be buried or embedded and exposed. Expansion joints are used with rigid penstocks having anchorage.
4.3.2 Spiral Casing or Scroll Casing A spiral casing is provided to receive water from penstock and distribute the same evenly around the turbine to avoid eddies and to ensure equal velocity of water at all sections. It is of varying cross-section and assumes the shape of a spiral. It may be made of concrete for low heads but for high heads, it is fabricated from steel plates or cast steel casings are used.
4.3.3 Guide Mechanism A series of aerofoil-shaped vanes called the guide vanes or wicket gates, are arranged inside the casing to form a number of flow passages between the casing and runner blades. Aerofoil section ensures minimum frictional and hydrodynamic losses. These vanes can be rotated around their fulcrum and thus the passage are a can be varied. The rotation is received from speed governor
80 Hydraulic Machines which keeps the turbine speed constant with varying loads by adjusting the guide vane water passage and hence the flow. The guide vanes are generally made of cast steel.
4.3.4 Runner and Shaft Assembly The runner of Francis turbine receives water from guide vanes radially and turns the water through 90° so that water is discharged axially.
Fig. 4.2 Guide vanes Main shaft D1
Runner
Scroll casing
Guide vanes D2
Draft tube Tail race From penstock
Fig. 4.3
Francis turbine
Francis Turbine
81
The shape of the runner depends upon the specific speed. The vanes may be cast integral or fabricated from steel plates. The material of construction is usually cast iron, bronze, cast steel or stainless steel. The shaft is a forging and is keyed to the runner. The vertical turbines have only one radial-thrust bearing to take the axial load also. The runner works both due to impulsive impact caused by change in direction of flow as well as reactive impact due to change in pressure and velocity. The number of blades may vary between 16 and 24.
4.3.5 Draft Tube The water from the runner is discharged to the tail race through a gradually expanding pipe called draft tube. The free end is submerged in tail race water. It performs dual function. 1. It converts the kinetic energy of exit water into pressure energy due to increasing cross-sectional area. The exit energy is recovered and turbine efficiency improves. 2. The turbine is installed above the tail race level. There is saving in excavation cost. The maintenance of turbine becomes easier and results in saving in cost.
4.4 VELOCITY TRIANGLES OF FRANCIS TURBINE In a Francis turbine, the pressure of water is more at the inlet than that at the outlet and water in the turbine flows in a closed conduit. The velocity triangles at inlet and outlet of blades of Francis turbine are shown in Fig. 4.4. Vb1 a1
b1 Vr1 Vf1
Va1 Vw1
a2
Vr2
Va2 = Vf2 b2
Fig. 4.4
Vb2
Velocity triangles
Vf1 = Vf2 Vw2 = 0 a2 = 90°
82 Hydraulic Machines The flow velocity remains constant Vf 1 = Vf 2 The discharge is axial Vw2 = 0
\
a2 = 0 Work done per unit mass flow, The turbine work,
W = Vb1 Vw1 PJ = rQ(Vw1Vb1 Vw2Vb2)
Vw2 = 0
3
PJ = rQVw1 Vb1
\
The inlet velocity triangles depend upon the speed of the runner and are shown in Fig. 4.5. The outlet velocity triangle is shown in Fig. 4.6. b2 = 90° ± 30° a2 = 90° Vw2 = 0 Slow Runner Vw1 a1
Vb1 = Vw1
Vb1
a1
b1 Va1
Fast Runner Vw1
Normal Runner
Vr1
vf1
b1 Va1
a1 = 15° to 25° b1 = 60° to 90°
a1 = 25° to 32.5° b1 = 90°
There is acceleration of flow in the runner Vr2 > Vr1 The hydraulic efficiency, hD =
b1
Vr1 = Vf1
Fig. 4.5 Inlet velocity triangles
\
Vb1
a1
rQVw1 Vb1 Vw1 Vb1 = gH rQ gH
Vf 1 Va1
Vr1
a1 = 32.5° to 37.5° b1 = 90° to 120°
Francis Turbine Vb2 a2
b2
Vf2
Va2
Vr
2
Fig. 4.6 Outlet velocity triangle
tan a1 =
tan b1 =
tan b2 =
VB 1
VM 1
VB
1
V> - VM 1 1 VB
2
V>2
Vf 1 = y 2gH
where y =
Vb1 = f 2C0
where f =
VB 2 gH
= flow ratio
V> = speed ratio 2 gH
4.5 SANKEY DIAGRAM
Volumetric losses due to water leakage
Hydraulic losses
Fig. 4.7
Shaft power
Fluid power transferred to runner
Fluid power transferred to vanes
Power available
The energy flow diagram or Sankey diagram is shown in Fig. 4.7.
Mechanical losses
Sankey diagram
83
84 Hydraulic Machines Losses
The main losses in a Francis turbine are: 1. Volumetric losses due to leakage of water between guide vanes and runner (DQ) 2. Hydraulic losses (i) Head loss due to discharge velocity (ii) Shock and friction losses over the vanes. 3. Mechanical losses (i) Friction losses in the bearings. (ii) Windage losses. Efficiencies
Volumetric efficiency, Dvol = =
Fluid Power Transferred to Vanes Power Available at Inlet to Tubine Q - DQ Q
(0.96 0.98)
Hydraulic efficiency DD = =
Fluid Power Transferred to Runner Fluid Power Transferred to Vanes VM1V>1 gHa
Mechanical efficiency Dmech =
Shaft Power Fluid Power Transferred to Runner
(0.98 0.99)
Turbine efficiency The overall efficiency of turbine hJ = hvol hD hmech.
4.6 DESIGN OF FRANCIS TURBINE The site water head and total power to be developed are known, the speed of turbine and turbine power are selected to get specific speed necessary for the type of turbine desired. Practically all dimensions, features and characteristics of the desired turbine can be expressed as a function of the specific speed. Hence, specific speed is the logical key to the design of the turbine.
Francis Turbine
85
4.6.1 Speed Ratio and Runner Diameter The speed ratio, V> 2 gH
f=
where Vb = Blade peripheral velocity H = Water head available at turbine inlet Vb = f 2C0
\
The speed ratio which gives maximum turbine efficiency is denoted as fA and its value ranges between 0.55 and 0.9. This value may be changed in the design by altering certain angles and areas of runner. The recommended values of fA for different specific speed NS are given in Fig. 4.8. Vb =
pDN 60
\ Runner diameter, 60f A 2C0 pN If B is the width or height of the runner, find the ratio B/D from Fig. 4.8.
D=
1.5
1.0
fe
fe 0.5 B/D 0
50
100
150
200 Ns
Fig. 4.8
4.6.2 Flow Ratio The flow ratio, y= where Vf = Flow velocity.
VB 2 gH
Speed ratio
250
300
86 Hydraulic Machines The water discharge through turbine, Q = Area of runner ´ Flow velocity = pDBVf Taking into account the effect of blade thickness (about 5%) Q = 0.95pDBVf. = 0.95pDB y 2gH Let
=m , Q = (0.95pmD2) y 2C0
\
...(1)
The turbine overall efficiency, hJ =
Pt ´ 1000 rQgH
...(2)
where PJ = Turbine power [kW] The specific speed, NS =
N Pt H 5/4
...(3)
The flow ratio y can be expressed in terms of NS, f, D and h J with the help of equations (1), (2), and (3) y is useful in fixing inlet and exit angles of blades.
4.6.3 Blade Angles The whirl ratio, x=
VM h = h 2f 2 gH
where Vw = whirl velocity tan a1 =
y x
tan b1 =
y (x - f)
a2 = 90°. D2 = 0.5 D1
Francis Turbine
87
4.6.4 Number of Runner Vanes In order to avoid any pulsations, the runner vanes are often made an odd number. Zowski recommendations
Number of runner vanes, n=K D K = 3.7 for slow runner K = 3.0 for normal runner K = 2.2 for fast runner D = Runner diameter in feet.
4.6.5 Number of Guide Vanes Zowski recommendations n¢ = K¢ D K¢ = 2.5 for a1 between 10° and 20° K¢ = 3.0 for a1 between 20° and 30° K¢ = 3.5 for a1 between 30° and 40° The values of n and n¢ should not be same nor of any single multiple to avoid pulsations.
4.7 DESIGN PROBLEM Design a turbine for the following site conditions. Head = 120 m Power = 5 ´ 120 MW. No. of penstocks = 5 with head loss due to friction as 3% of gross head. Length of penstock = 1300 m each. Solution 1. Determination of water flow rate Assume overall turbine efficiency, hJ = 0.87 The friction factor
f = 0.005 htrans = 0.97 (Given) Gross head, Hg = 120 m (Given)
The turbine power, Pt = rQgHghtrans ´ hJ
88 Hydraulic Machines \
Q=
120 ´ 106 Pt = 1000 ´ 9.81 ´ 120 ´ 0.97 ´ 0.87 rgHghtrans ´ ht
= 120 m3/s. DHf = 0.03 ´ 120 = 3.6 m
\
Q=
p 2 DV 4
V=
4Q pD2
DHf = \
4 f LV 2 2 gD
FG H
IJ K
2
3.6 =
4 ´ 0.005 ´ 1300 4 ´ 120 = 2 ´ 9.81 ´ D pD2
D5 =
4 ´ 0.005 ´ 1300 ´ 1202 ´ 4 2 = 8593 m5 2 3.6 ´ p ´ 2 ´ 9.81
D = 6.12 m say 6 m Water velocity through each penstock, V=
4Q 4 ´ 120 = = 4.25 m/s pD2 p ´ 62
OK
Type of turbine The specific speed, NS =
N Pt
H 5/4 From Francis turbine at Bhakra Power Project, p = 18 and N = 166.7 rpm
\
NS =
=
166.7 ´ 120 ´ 103 ( H g - DH f )5/4 166.7 ´ 120 ´ 103
116.45/ 4 = 151 rpm.
For Francis turbine from Fig. 4.1 and Table 4.1 normal runner is suitable for NS = 151 rpm.
Francis Turbine
89
Spiral casing The spiral casing diameter dc can be taken less or equal to penstock diameter. Allowable VC can be taken as 10 m/s.
\
Q=
p 2 dc VC 4
dc =
4Q pVc
=
4 ´ 120 p ´ 10
= 3.9 m RC1 is fixed to accommodate runner and guide vanes. q dc. 2p The dimensions of spiral casing is fixed by drawing to scale.
RC2 = RC1 +
Q dc
Rc2 q Rc1
Fig. 4.9 Spiral casing
Runner size The runner diameter, 60f 2gHa pN Read from graph, f = 0.75
D1 =
\
D1 =
60 ´ 0.75 2 ´ 9.8 ´ 116.4 = 4.1 m p ´ 166.7
90 Hydraulic Machines From graph,
B/D = 0.3
\
B = 4.1 ´ 0.3 = 1.23 m
Velocity triangle Now,
Q = 0.95 ´ pBDVf Vf = y=
Assume
Q 120 = = 7.97 m/s. 0.95 ´ p ´ B ´ D 0.95 ´ p ´ 123 . ´ 4.1
Vf 2 CH
=
7.97 = 0.167 2 ´ 9.81 ´ 116.4
hh = 0.90 x= tan a1 =
\
0.9 hh = = 0.60 2f 2 ´ 0.75 y 0.167 = = 0.278 0.60 x
a1 = 15.5° tan b1 =
0167 . y = (x - f) (0.60 - 0.75)
b1 = 48° = 132°. a2 = 90° 6. Number of vanes (i) Number of runner vanes n=K D For normal runner, K = 3.0 \
n = 3 4.1 ´ 3.3 = 11
(ii) Number of guide vanes n¢ = K¢ D K¢ = 2.5 for a1 = 15.5° \
n¢ = 2.5 4.1 ´ 3.3 = 9
Francis Turbine
91
QUESTION BANK NO. 4 1. Why specific speed is called type characteristic? Classify hydro turbines as per specific speed. 2. What is the role of following components in Francis turbine? (a) spiral casing (b) draft tube 3. Draw velocity triangles for different Francis runners. How are different angles estimated? 4. List various sources of losses in a Francis turbine. Draw a Sankey diagram and define different efficiencies. 5. Explain step-by-step the design of a Francis turbine. 6. Define the following. Explain their role in the design of a Francis turbine. (a) speed ratio (b) flow ratio
TUTORIAL SHEET NO. 4 1. An inward flow reaction turbine has outer and inner diameters of the wheel as 1 m and 0.5 m respectively. The vanes are radial at inlet and axial at outlet and water enters the vanes at an angle of 10°. Assuming the velocity of flow as constant and equal to 5 m/s, calculate: (a) The speed of wheel, and (b) The vane angle at outlet. Solution The data given: D1 = 1 m D2 = 0.5 m Vf 1 = Vf 2 = 5 m/s a1 = 10° b1 = 90° a2 = 90° The velocity triangles are drawn below. a1
Vb1
Vb2 b1 Vf1
Va1
b2
a2 Vf2 = Va2 Vr2
92 Hydraulic Machines (a)
tan a1 =
V f1 Vb
1
5 = 28.356 m/s tan 10°
\
Vb1 =
Now
Vb1 = pD1 N 60
\
N=
(b)
Vb2 = tan b2 =
\
60 ´Vb1 60 ´ 28.356 = = 541.56 m/s Ans. pD1 p ´1 pD2 N p ´ 0.5 ´ 54156 . = = 14.176 m/s 60 60 V f2 Vb2
=
5 14.176
b2 = 19°25¢ Ans.
An inward flow reaction turbine, having external diameter of 1 m is running at a speed of 180 rpm. The guide blade angle is 15°. If the velocity of flow at inlet is 5 m/s, find: (a) peripheral velocity at inlet (b) velocity of whirl at inlet (c) absolute velocity of water at inlet (d) vane angle at inlet Solution The data given D1 = 1 m N = 180 rpm. a1 = 15° Vf 1 = 5 m/s Draw inlet velocity triangle. Vw1 Vb1 a1
b1 Vf1
Vr1 Va1
Francis Turbine
93
pD1 N p ´ 1 ´ 180 = 60 60 = 9.42 m/s Ans
(a)
Vb1 =
(b)
tan a1 = \
Vw1 = sin a1 =
(c)
\ (d)
Va1 = tan b1 =
\
V f1 Vw1 V f1
tan a1
=
5 = 18.66 m/s Ans tan 15°
=
5 = 19.3 m/s Ans sin 15°
V f1 Va1 V f1
sin a1
V f1
=
Vb1 - Vw1
b1 = 28.42°
5 5 = = 0.541 18.66 - 9.42 Vw1 - Vb1
Ans
Estimate the main dimensions for an inward flow turbine to suit the following conditions. Head = 58 m Power developed = 400 kW Speed = 1000 rpm
Hydraulic efficiency = 90% Overall efficiency = 85% Flow ratio = 0.16
The ratio of wheel width to wheel diameter at inlet = 0.1. The ratio of inner diameter and outer diameter = 0.5. Consider constant velocity of flow and axial of discharge. Solution The data given: H = 58 m Pt = 400 kW N = 1000 rpm hh = 0.90 hh = 0.85 y = 0.16 B/D = 0.1 D2 = 0.5 D1
94 Hydraulic Machines Vf 1 = Vf 2 a2 = 90° (a) The specific speed, NS =
N Pt H 5/4
=
1000 400 1000 ´ 20 = 160 585/4
= 125 rpm Normal Francis Runner Pt = rQgHhJ \
Q=
Pt 400 ´ 103 = 1000 ´ 9.81 ´ 58 ´ 0.85 rgHht
= 0.827 m3/s Vf = y 2gH = 0.16 2 ´ 9.81 ´ 58 = 5.4 m/s Q = pD1 BVf = pD1(0.1D1)Vf Q = p ´ 01 . ´ Vf
\
D1 =
\
D1 = 0.6982 m » 0.7 m
D1 = 0.35 m 2 B = 0.1D1 = 0.07 m
D2 =
Vb1 =
(b)
f= x= tan a1 = \
0.827 p ´ 01 . ´ 5.4
U| V| W
Ans
pD1 N p ´ 0.7 ´ 1000 = = 36.65 m/s 60 60 Vb1 2 gH
=
36.65 = 1.08645 2 ´ 9.81 ´ 58
0.90 hh = = 0.4142 2f 2 ´ 108645 . 0.16 y = 0.4142 x
a1 = 21°
Francis Turbine
tan b1 =
016 . y = = 0.238 . ) (x - f) (0.4142 - 108645
b1 = 13.4° or 166.6° (c) Number of runner vanes, n = 3 D ´ 3.3 = 3 0.7 ´ 3.3 = 4.55 or 5 Number of guide vanes, n¢ = 2.5 D ´ 3.3 = 3.8 or 4
95
+0)26-4
#
Propeller and Kaplan Turbines 5.1 INTRODUCTION The Francis turbines are medium head turbines and hydraulic efficiency is affected due to following reasons: 1. The number of blades are more from 16 to 24. 2. The blades receive water in radial direction and water is discharged in axial direction. There is 90° bend to water. Those two sources of hydraulic losses are overcome in Propeller turbine which have axial entry and axial discharge. The number of blades are only 3 to 6. These are also reaction turbines and work with large flow and low heads. The specific speed is high between 300 and 1000 rpm. The components of propeller turbine such as scroll casing, guide mechanism and draft tube are similar to Francis turbine except the runner. The schematic diagram of a propeller turbine is shown in Fig. 5.1. The runner blades are fixed in position as in Francis turbine. The efficiency depends upon inlet blade angle. In fixed blade runners, it is not possible to vary the inlet blade angle for varying loads. The propeller turbine is designed for maximum efficiency only for a particular load. At all other loads, efficiency falls. These turbines are called peak load turbines. Guide vane
Runner
Fig. 5.1 Propeller turbine
Propeller and Kaplan Turbines
97
5.2 KAPLAN TURBINE Kaplan turbine is a special propeller turbine in which the individual runner blades are pivoted to the hub, so that their inclination may be adjusted during operation responding to changes in load. The blades are adjusted automatically rotating about pivots with the help of a governor servo-mechanism. The blades are attached to a hollow boss and hollow shaft. The servo-mechanism to rotate the blades are housed inside the hollow boss and shaft. The turbine can be run at maximum efficiency at all loads because the inlet angle is automatically adjusted with variation of load and a flat curve is obtained. The guide vanes are rotated by speed governor during load fluctuation to ensure constant speed. Therefore, in Kaplan turbine both guide vanes as well as runner blades are rotated by speed governor during load fluctuation. The arrangement drawing of a Kaplan turbine is shown in Fig. 5.3. 100 Kaplan 90 ht (%) Propeller
80
70 25
50
75 % Load
100
125
Fig. 5.2 Efficiency curves
Scroll casing
Guide vane
Main shaft D Tail race
Runner blade d Boss
Draft tube
Fig. 5.3 Kaplan turbine
98 Hydraulic Machines 5.3 TURBINE CONSTANTS 5.3.1 Velocity Triangles D = Diameter of runner d = Diameter of boss. d H = 0.38 + D 220
At edge of diameter D, Vb1 = Vb2 At the hub diameter ¢ V>¢1 = V>2 Vf¢1 = Vf¢2 = VB 1 = VB2. a1 = 45° to 70° b1 = 20° to 68° The degree of reaction is less than 50% and increases gradually from the hub to the tip. 1. water inlet 2. water outlet
D d 1
2 Vw1 a1 V f1
b1 Vr1
Va1
Fig. 5.4
Vw¢1
a2
b2 Vr2
(a) Velocity triangles at edge (D)
Vb¢1
Vb¢2 b ¢1
Vf 1¢
Va¢2 = Vf ¢2
a¢1
Vb2 Va2 = Vf2
Vb1
a¢2
b¢2
Va¢1
Vr¢1
Fig. 5.4
(b) Velocity triangles at hub (@ )
Vr¢1
Propeller and Kaplan Turbines
99
5.3.2 Ratios 1. Speed ratio The blade speed, Vb = f 2g0 where f = speed ratio =
V> 2 CH
pDN 60 f is a function of NS.
Vb =
3.0 2.5 2.0 f 1.5 1.0 0.5 300 400 500 600 700 800 900 1000 Ns (rpm)
Fig. 5.5
Speed ratio
Flow ratio The flow velocity VB = y 2g0 where y = Flow ratio. p 2 (D d 2) VB 4 where K = Coefficient of flow area after deduction of area occupied by the blades. 3. Whirl ratio The whirl velocity,
Q=K
Vw = x 2g0
100 Hydraulic Machines \ Whirl ratio
VM h = h 2f 2 CH
x=
where hydraulic efficiency, hD = 0.95.
5.3.3 Blade Angles tan a1 = tan b1 = tan b2 =
V B1 VM1
=
8b1 - 8w1 V f2 Vb2
x¢ = x V>¢1 =
\
x 2 CH
8 f1
=
VM¢ 1 = Vw1 ´ \
y 2 CH
=
=
y x
y ( f - x)
y f D d
D d
pdN 60
f¢ = f
D d
5.3.4 Losses and Efficiencies These can be estimated in the same way as for Francis turbine. See Art. 4.5.
5.4 DESIGN PROBLEM Design a propeller turbine of 8 MW with water head of 5 m.
5.4.1 Specific Speed NS =
N Pt
H 5/4 NS varies from 300 rpm to 1000 rpm. Take NS = 300 rpm.
Propeller and Kaplan Turbines
\
N=
101
N s H 5/ 4 300 ´ 55/4 = = 25 rpm. 8000 Pt
Take NS = 1000 rpm.
1000 = 83 rpm 300 Select a synchronous speed of 75 rpm which can be coupled to a generator having 40 pairs of poles for a generation frequency of 50 cps.
\
\
N = 25 ´
N Pt
75 8000 = 897 rpm. H 55/ 4 From graph for NS = 897 rpm, speed ratio f = 2.2.
NS =
5/4
=
5.4.2 Water Flow PJ = rQg HhJ Assume \
hJ = 0.90 Q=
8 ´ 106 Pt = = 181.22 m3/s. 1000 ´ 9.81 ´ 5 ´ 0.90 rgHht
5.4.3 Runner Dimensions Runner diameter, D=
60f pN
2gH =
60 ´ 2.2 2 ´ 9.81 ´ 5 p ´ 75
= 5.55 m Hub diameter,
FG H
d = 0.38 +
IJ K
H 5.55 = 2.235 m 220
5.4.4 Velocity Diagrams Refer Fig. 5.4. f = 2.2, y = 0.90; x = tan a1 =
0.90 y = 0.216 x
0.95 hh = = 0.216 2f 2 ´ 2.2
\ a1 = 76.5°
102 Hydraulic Machines tan b1 =
0.90 y = (f - x ) (2.2 - 0.216)
tan b2 =
y 0.90 = b2 = 22.2° 2.2 f
\>1 = 24.4°
5.5 OTHER TURBINES 5.5.1 Deriaz Turbine Deriaz turbine is a reversible flow turbine and is a cross between Kaplan and Francis turbines. It has 10 to 12 inclined blades which are adjustable like Kaplan turbine. For pumped-storage power plants it works both as turbine as well as pump. For tidal pover plants it can work in either direction of flow during high tide where water is being stored in the reservoir and low tide when water flows from reservoir to the ocean. It is also called diagonal turbine as flow over runner is at angle of 45° to the axis. The general arrangement is shown in Fig. 5.6. These turbines can work even upto head of 200 m. Guide vane Shaft
45°
Scroll casing
Inclined blade Draft tube
Fig. 5.6 Deriaz turbine
5.5.2 Bulb Turbine Bulb or tubular turbines are small fixed axial flow propeller turbines operating under low heads. The hydro-generator is enclosed in a bulb-shaped casing and installed in the middle of flow passage. The bulb and propeller form an integral unit. There is a straight conical draft tube. These are suitable for tidal power plants, run-off the river plants.
Propeller and Kaplan Turbines
103
Runner blade
Bulb casing
Draft tube Guide vanes
Fig. 5.7 Bulb turbine
QUESTION BANK NO. 5 1. What is the main difference between a Francis and propeller turbine? How are hydraulic losses minimized in propeller turbine? 2. What is a peak load turbine and flat curve turbine? How is a propeller turbine converted into Kaplan turbine? 3. How are blade angle fixed in Kaplan turbine in terms of turbine constants? 4. Write brief notes on the following: (a) Deriaz turbine (b) Bulb turbine
TUTORIAL SHEET NO. 5 1. A Kaplan turbine develops 10,000 kW under a head of 4.3 m. Find the diameter and speed of the turbine if the speed ratio is 1.8, flow ratio is 0.5, boss diameter as 0.35 times the outer diameter and overall turbine efficiency is 90%. Solution The given data: PJ = 10,000 kW H = 4.3 m f = 1.8 y = 0.50 d = 0.35 D
hJ = 90% VB = y 2gH = 0.5 2 ´ 9.81 ´ 4.3 = 4.6 m/s.
104 Hydraulic Machines PJ = rQgH hJ \
Q=
10,000 ´ 103 Pt = 1000 ´ 9.81 ´ 4.3 ´ 0.9 rgHht
= 263.4 m3/s Q=
\
p 2 (D d 2)VB 4
=
p 2 [D (0.35D)2]VB 4
D=
4Q = p(1 - 0.352 )V B
4 ´ 263.4 p ´ 4.6 ´ 0.8775
= 9.1 m say 9 m d = 0.35 D = 3.l5 m Vb = f 2gH = 1.8 2 ´ 9.81 ´ 4.3 = 16.5 m/s
\
Vb =
pDN 60
N=
60 ´ V> 60 ´ 16.5 = = 35 rpm pD p´9
Find the specific speed and the type of water turbine developing shaft power of 5145 kW under a head of 20 m when running at 100 rpm. Calculate its normal speed and output under a head of 2.5 m. Solution (a) Data given: PJ = 5145 kW H = 20 m N = 100 rpm. The specific speed, NS =
N Pt H
5/4
=
100 5145 205/4
= 169.6 rpm. It is a Francis turbine.
Propeller and Kaplan Turbines
(b)
105
PJ = rQgHh J Q = y 2gH PJ µ QH 3
µ H2
FG IJ H K
H¢ Pt¢ = H Pt
3 2
F H¢ I P¢ = P G J H HK J
3 2
J
F 2.5I = 5145 G J H 20 K
3 2
= 227 kW.
Vb = f 2gH 2pND = Vb = f 2gH 60
\
Nµ
H
FG IJ H K
N¢ H¢ = N H
\
F H¢ I N¢ = N G J H HK
F 2.5I = 100 G J H 20 K #
Also
N¢ =
N ¢s ( H¢ "
( P¢ t
2
= 35 rpm. 5
1%0(2.5 " = 22%
= 35 rpm. 3. Calculate the diameter and speed of the runner of a Kaplan turbine developing 6000 kW under an effective head of 5 m. Overall efficiency of the turbine is 90. The diameter of the boss is 0.4 times the external diameter of runner. The turbine speed ratio is 2.0 and flow ratio is 0.6. What is the specific speed of the turbine? [D = 5.66 m; N = 66.8 rpm, NS = 692 rpm].
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$
Draft Tube and Cavitation 6.1 INTRODUCTION Draft tube is an airtight diverging conduit for carrying water from turbine runner exit to the tail race. The gradually increasing cross-section of draft tube helps to convert the kinetic energy of exit water into pressure energy. It is necessary and an integral part of all reaction turbines (Francis, Propeller, Kaplan, Bulb). The draft tube may be a concrete tunnel or fabricated from steel. The outlet end of draft tube is submerged in the water approximately 1 m below the lowest tail race level. A draft tube is shown in Fig. 6.1 where section 1-1 is the turbine exit or draft tube inlet and section 2 2 is the draft tube outlet located at a depth of Z2 below tail race level. Apply Bernoullis equation between section 1 1 and 2 2, p1 V12 p V2 + + Z1 = 2 + 2 + 0 + hf w w 2g 2g
where hf = hydraulic loss in draft tube.
F GH
p1 p V 2 - V22 - hf = 2 Z1 1 2g w w
\
I JK
Turbine casing 1
1 Draft Tube hs pa
TR Z2 2
2
Fig. 6.1 Draft tube
Draft Tube and Cavitation 107
p2 p = a + Z2 w w
But
F GH
p1 p V 2 - V22 = a + (Z2 Z1) 1 - hf 2g w w
I JK
w = rg Z2 Z1 = hs = suction head of draft tube V12 - V22 = dynamic head. 2g
F GH
\
p1 p V 2 - V22 = a hs 1 - hf 2g w w
\
p1 < pa.
I JK
The cross-sectional area of draft tube is increasing. A2 > A1 \
V2 < V1
The suction head, hs is positive, the available head of turbine increases leading to higher turbine output and efficiency. The actual gain of pressure head is
FV -V GH 2g 2 1
2 2
I JK
hf .
The efficiency of draft tube V12 - V22 - hf 2g hd = V12 2g
Let
hf = k
(V12 - V22 ) 2g
(1 - k )
\
hd =
FV -V I GH 2g JK 2 1
V12 / 2 g
2 2
108 Hydraulic Machines 6.2 FUNCTIONS OF DRAFT TUBE A draft tube performs the following functions. 1. Recovery of exit loss
The water leaves the turbine runner at a high velocity (V1) and carries away with it a large kinetic energy
FV I GH 2 g JK 2 1
which forms a big loss mechanism. The diffuser
action of draft tube recovers this energy by converting kinetic energy
FV -V I GH 2g JK 2 1
2 2
into pressure energy (hs). The water is discharged into tail race at very low velocity (V2). The pressure in the draft tube is above the vapour pressure corresponding to the water temperature and cavitation is avoided. The effective head of the turbine is increased by (hs hf), raising the output and efficiency of turbine. 2. Turbine installation above water level
Draft tube ensures an airtight closed circuit at turbine outlet thereby making possible to decrease the pressure at turbine outlet below atmospheric (p1 < pa). For same working head, now it is possible to install the turbine above tail race water level. There is saving in excavation cost and maintenance cost. If there was no draft tube, the turbine would be installed below the tail race level and water had to be pumped out for maintenance of the runner. Therefore, draft tube helps to avoid the need of pumping out of water.
6.3 TYPES OF DRAFT TUBES The draft tubes are designed to achieve the following: 1. To ensure the maximum possible recovery of kinetic energy which is expressed as draft tube efficiency. hd =
Actual head recovered Theoretical head-losses = Theoretical head Theoretical head
The losses in the draft tube are: (i) Due to separation of flow from walls of the tube, the taper angle is restricted to 7° to 8°, see Fig. 6.3. (ii) Due to formation of vortex at the tube exit. This is controlled by provision of vortex breaker. See Fig. 6.4 where a cone is provided as vortex breaker. (iii) Friction losses in the tube. (iv) Head loss due to tube bends.
Draft Tube and Cavitation 109 0.45
0.15
300
0
250
500
750
1000
Ns
Fig. 6.2 Draft tube efficiency
Cone 3.5° to 4°
Fig. 6.3 Taper angle
Fig. 6.4
Vortex breaker
To ensure minimum cost of excavation to lay down the tube. This is achieved by the use of bent tube. The following types of draft tubes are used. (i) Straight divergent tube. This gives hd = 85% (ii) Bell mouth tube. This gives hd = 85% (iii) Hydro cone or Moodys spreading tube, hd = 85% (iv) Simple elbow tube. Low excavation but hd = 66% (v) Elbow tube with circular inlet and rectangular outlet. This requires minimum depth of excavation and hd is more than simple elbow.
6.4 CAVITATION Vapour pressure of water is a function of its temperature. During flow of water in a turbine, whenever, water pressure falls below the vapour pressure corresponding
110 Hydraulic Machines f
(2.5 – 3) f
4°
TR
Cone (vortex breaker)
(0.6 – 1) f
(a) Straight divergent tube
(b) Bell mouth
f1 TR TR f2
Vortex breaker (c) Hydro cone
(d) Simple elbow tube
TR
Vortex breaker
Two vortex breakers
(e) Elbow tube with circular inlet and rectangular outlet
Fig. 6.5 Types of draft tubes
Draft Tube and Cavitation 111
to temperature, water boils resulting in formation of vapours or small bubbles. When these bubbles are carried away by water to a high-pressure zone, these condense and bubbles collapse suddenly. There is a cavity (discontinuity) in the flow and surrounding liquid rushes towards cavity and collides to give very high local pressure (>7000 bar). The process is repeated several times in a second. There can be two effects. 1. Metallic pitting
The vacuum cavities if formed on a metal surface of turbine parts (mainly runner and draft tube), can cause pitting on metal surface. This process is called cavitation of metal surface. The effect can be very serious resulting in pitting surface, fatigue failure and tearing of metal parts if not attended in time. The roughened surface affects the streamline flow pattern leading to high friction losses and lower turbine efficiency. 2. Vibration
Vapour pressure of water (m)
If vacuum cavities are formed away from metal surface, pressure waves are formed. This results in water hammer, vibrations and noise. 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0
10
20
30 40 50 60 70 80 Water temperature (°C)
90 100
Fig. 6.6 Vapour pressure of water
6.5 FACTORS CAUSING CAVITATION The turbine should be properly designed to avoid cavitation as it has very serious consequences. The turbine life is affected and efficiency is lowered. The cavitation depends upon the following factors: 1. Vapour pressure and temperature of water. 2. Mean sea level of turbine installation and atmospheric pressure (Ha). 3. Pressure and velocity of water at the runner exit. The cavitation in reaction turbine is decided by specific speed (NS) and sigma factor (s).
112 Hydraulic Machines s=
( Ha - Hv ) - Hs H
where Ha = Atmospheric pressure function of attitude. HV = Vapour pressure function of water temperature HS = Height of runner outlet above tail race H = Working head of turbine. Cavitation can be avoided if s is more than scritical. As per Prof. P.H. Roger of USA scritical = 0.0317 or
NS = 100
FG N IJ H 100K s
2
s critical 0.0317
The critical cavitation condition is expressed by Thomas parameter which is defined as
NPSH . H
scritical
NS
Fig. 6.7
scritical v/s NS
6.6 METHODS TO AVOID CAVITATION The following methods can be used to avoid cavitation 1. Installation of turbine below tail race level
The value of s can be increased as static head HS decreases (or even becomes negative). However, maintenance of turbine becomes complicated as water has to be pumped out for inspection and repair. 2. Cavitation free runner
The turbine models and blades are subjected to testing simulating actual working conditions. The points of low pressure are found out and design modifications are made to avoid critical low pressure areas.
Draft Tube and Cavitation 113 3. Use of suitable materials
(i) Cast iron has less resistance to cavitation. It is also difficult to repair the damage by welding. (ii) Cast steel has more resistance to cavitation. It can also be repaired by welding. (iii) Stainless steel (Ni-Cr) has maximum resistance to corrosion and cavitation. However, it is very expensive. The turbine runners are made of cast steel and overlaid with stainless steel as cavitation is a surface phenomenon. (iv) For chemically active water, bronze is a better material which has good cavitation resistance. It can also be repaired by brazing. (v) Steel plates have better resistance to cavitation than steel castings. 4. Machining of surface
Turbine blades are finely polished to improve cavitation resistance. Rough surface can cause interruption to stream flow lines leading to vortices and hence cavitation.
QUESTION BANK NO. 6 1. What are the uses of draft tube? Describe with neat sketches different types of draft tubes. 2. On what factors does the cavitation in water turbines depend? 3. Describe some methods to avoid cavitation in water turbines.
TUTORIAL SHEET NO. 6 1. A Francis turbine is installed at a site where atmospheric pressure is 10 m of H2O and vapour pressure is 0.20 m of H2O. The turbine is designed to produce 11760 kW under a head of 25 m when running at 120 rpm. Calculate the maximum height of straight draft tube for the turbine. Solution Specific speed, NS =
N Pt H 5/4
= 120 11760 = 232.8 rpm 255/ 4 Critical value of Thomass cavitation factor for Francis turbine, scrit
F N IJ = 0.044 G H 100K s
2
F 232.8IJ = 0.044 G H 100 K
2
= 0.238
114 Hydraulic Machines The cavitation factor,
( Ha - Hv ) - Hs (10 - 0.20) - Hs = 25 H s > scrit to avoid cavitation, s=
If
s = scrit 0.238 =
\
(10 - 0.20) - Hs 25
HS = 3.85 m.
Maximum permissible height of draft tube. 2. A Kaplan turbine is developing 3250 kW under a head of 6 m. A conical draft tube with inlet diameter of 2.8 m is placed 1.5 m above tail race water and has an efficiency of 76%. The pressure at inlet to draft tube is 5 m H2O and atmospheric pressure of 10.3 m H2O. Calculate the efficiency of turbine. Solution
F GH
I JK
p1 p V 2 - V22 = a HS 1 - hf 2g rg rg
Neglect
hf = 0 (10.3 5) = 10.3 1.5
\
FV -V GH 2g 2 1
2 2
I JK
-0
V12 - V22 = 3.5 2g
The efficiency of draft tube,
\
hd =
(V12 - V22 )/ 2 g V12 / 2 g
V12 =
35 . (V12 - V22 ) = 0.76 hd
V1 = 9.50 m/s. The water flow rate, Q=
p 2 p D1 × V1 = (2.8)2 ´ 9.50 = 58.47 m3/s. 4 4
Draft Tube and Cavitation 115
\ The turbine power Pt = rQgHh t \
ht =
3250 ´ 103 Pt = rQgH 1000 ´ 9.81 ´ 58.47 ´ 6
= 0.945 = 94.5%
+0)26-4
%
Governing of Hydraulic Turbines 7.1 INTRODUCTION Hydraulic turbines are directly coupled to the electric generators which must run at constant designed speed to generate power at required constant frequency. The constant speed of the generator is called the synchronous speed and is given by, N=
60 f (rpm) p
where f = frequency of the power generated (cps) p = number of pair of poles of the generator For Indian conditions f is 50 cps and frequency variation allowed may be ± 0.15, i.e., 49.85 to 50.15 cps. Therefore, speed governors are provided on the turbine which control the rate of flow of water into the turbine corresponding to the load in order to keep the speed constant. The regulation system should be quick acting and accurate. On the other hand, it should be stable without too much sensitivity causing hunting. The above conflicting requirements of speed regulation system are met by a feedback system as shown. Dashpot
Energy
Energy
Compensator
Controller
Disturbance Spear (Pelton) or Power element guide Output Turbine (Servomotor) vanes (Reaction)
Feedback Transducer and measurement – Centrifugal governor – Electric tachogenerator
Fig. 7.1 Feedback speed regulation system
Govering of Hydraulic Turbines 117
7.2 OIL PRESSURE GOVERNOR The schematic diagram of a simple oil pressure speed governor is shown in Fig. 7.2. The main elements are: 1. Actuator or pendulum 2. Relay valve 3. Servomotor
Centrifugal governor
Flyball
Main lever F From main turbine shaft
10-20 bar Oil gear pump Relay valve To turbine guide mechanism
Oil sump Servomotor Fig. 7.2 Oil pressure governor
Actuator or pendulum
The flyballs of centrifugal governor respond quickly to speed variation of turbine shaft and actuates the main lever. Modern turbines use electric tachogenerator which is extremely sensitive to frequency variation. 2. Relay valve
It is a piston type slide valve. The motion of the flyballs is transmitted to relay valve which distributes oil to either side of servomotor. 3. Servomotor
Servomotor is a piston cylinder arrangement which transmits power to turbine guide mechanism. In case of Pelton turbine, the guide mechanism is the spear rod. In case of Francis turbine and Propeller turbine, the guide mechanism is guide vanes. In case of Kaplan turbine, the power is transmitted to both guide vanes as well as runner vanes.
118 Hydraulic Machines 7.3 DOUBLE REGULATION SYSTEM OF PELTON TURBINE The water flow to the runner of a Pelton turbine is regulated by the combined action of the spear and the deflector plate as shown in Fig. 7.3. Centrifugal governor closes with less load and more speed opens with more load and less speed
Flyball Main lever
Sleeve Fulcrum
Deflector Bell crank up lever Down
From turbine main shaft V1
V2
Gear pump Pilot valve
F Roller came arrangement Spear
Oil sump Spear rod Servomotor
Water
Fig. 7.3 Double regulation system of Pelton turbines
The sensitivity of centrifugal governor is augmented by a hydraulic servomotor. (a) When the load on the generator drops, the speed of the turbine increases. The flyballs of the centrifugal governor fly outward due to more centrifugal force (due to higher speed). The sleeve moves up pushing the piston rod of pilot valve downwards. The valve V1 closes and valve V2 opens. A gear pump supplies oil from an oil sump to the pilot valve. The oil flows to the left chamber of servomotor. The spear rod moves to the right, thereby decreasing the flow area of nozzle and the rate of water flow to the turbine is reduced. The turbine speed falls to normal value and flyballs, sleeve lever, etc. come to normal position. (b) When the load on the generator increases, speed decreases, flyballs move inward, the sleeve moves down, the piston rod of pilot valve moves up, valve V2 closes, valve V1 opens, pressure oil is supplied to right chamber of servomotor the spear rod moves to left, the nozzle opens more, more water flows to the turbine to take up increased load, speed decreases to its rated value.
Govering of Hydraulic Turbines 119
(c) When there is sudden fall of load, the spear moves rapidly to close the nozzle. There can be water hammer in the penstock due to sudden stoppage of water flow. In order to avoid sudden stoppage of water flow and water hammer, the spear is designed to allow some water flow during sudden load fall. The bell crank lever moves down and deflector moves up. The water is deflected and not allowed to strike the buckets. The deflected water goes as waste to the tail race.
7.4 GOVERNING OF REACTION TURBINES The guide blades are pivoted and connected to regulating ring which in turn is connected to the piston rod of servomotor of oil pressure governor through regulating rods and regulating lever. When the servomotor moves to the right, the guide blades are rotated to decrease or close the water supply to the runner. The reverse happens when the servomotor moves to the right under decrease of generator load and increase of turbine speed. The guide blades are rotated to increase the water flow area to the turbine runner. Regulating ring
V2
From Regulating lever slide value V1
Regulating rod Guide blade
Close Open
Servomotor Regulating ring Water
Fig. 7.4 Governing of reaction turbine
In case of Kaplan turbine both guide vanes and runner blades are adjusted simultaneously in case of load fluctuation. There is an additional servomotor and a control valve interconnected to those of guide vanes so that for a given guide vane opening, there is a definite runner blade inclination. The runner blade operating mechanism is housed in the hub of the runner which is hollow. When the load suddenly falls on reaction turbine, a pressure relief valve is operated by speed governor. A portion of water from spiral casing is allowed to flow directly through relief valve to the tail race without striking the turbine runner. The pressure relief valve performs the function of protection system to avoid water hammer effects during sudden load drop.
120 Hydraulic Machines QUESTION BANK NO. 7 1. What do you understand by governing of hydraulic turbine? Explain with sketches, the working of an oil pressure governor. 2. Describe, in brief, with the help of neat sketch, method of governing of Pelton wheel. How is water hammer avoided in the penstock during sudden load throw off? 3. Explain with the help of neat sketches, method of governing of Francis turbine. How is it different in case of Kaplan turbine? 4. Compare the protection system against water hammer due to sudden load drop in case of impulse turbine and reaction turbine. 5. What is feedback speed governing system? What are the main requirements of governing system?
+0)26-4
&
Dynamic Similarity and Performance Characteristics 8.1 INTRODUCTION The type, size and number of hydro turbines for a power plant are selected on the basis of detailed techno-economical analysis of site conditions. The performance of a prototype (first of the series of turbine) cannot be predicted before its manufacture without its testing in the laboratory or on a test bed. It is very difficult to carry out its tests because of its size and facilities required such as variable head, speed and load under which the turbine is expected to operate at the power plant. It is very costly to affect changes on a full size machine and testing can be timeconsuming. It is economical to make scale models and perform necessary hydraulic tests on it in order to predict the performance of prototype or full-sized machine under similar operating conditions. The models are prepared and tested within the following ranges: The model capacity,
Pm = 5 to 50 kW.
The runner diameter,
Dm = 250 600 mm.
Test head for model,
Hm = 4 100 m.
8.2 DYNAMIC SIMILARITY OF MODEL AND PROTOTYPE The turbine model should satisfy the following similarities with the prototype. 1. Geometric similarity
There should be constant linear proportions and same angles between model and prototype. (a1, b1, a2, b2)m = (a1, b1, a2, b2)p
FG D IJ = FG D IJ HD K HD K 1
1
2
2
m
p
122 Hydraulic Machines where m = model p = prototype. 2. Kinematic similarity
There should be same flow patterns and same velocity ratios. fm = fp ym = yp xm = xp The velocity triangles of model and prototype should be congruent. 3. Dynamic similarity
The Reynolds number Rem = Rep There should be proportionality of forces acting on the runner blades. If the above conditions are fulfilled, the following parameters are achieved and hm = hp. Model 1. Qm =
Prototype
p 2 d jm Vjm 4
Q =
. Pm = rQm gHmhm 3. Vbm =
p 2 dj Vj 4
. P = rQgHh
pDm Nm 60
Vb =
pDN 60
8.3 SPECIFIC SPEED The suitability of a turbine for a given application depends on: 1. Head of water, H (m) 2. Rotational speed, N (rpm) 3. Power developed, Pt (kW) The above parameters constitute a characteristic called specific speed. The specific speed of a turbine is defined as the speed of operation of a geometrically similar model of the turbine which is so proportioned that it produces 1 kW power when operating under a head of 1 m. The blade speed, Vb =
pDN 60
Vb =f Vj
Dynamic Similarity and Performance Characteristics
\
Vb = fVj = f 2gH
\
Vb µ But
\
123
H
Vb µ DN
H
DN µ
...(1)
The turbine power, Pt = rQgH The density of power r and acceleration due to gravity g are constant. Pt µ QH
\
The water discharge, Q=
p 2 D Vf 4
Vf = y 2gH
H
\
Vf µ
\
Q µ D2 H
...(2)
\
Pt µ D2 H ... H
...(3)
From equation (3),
D2 µ or
Dµ
Pt H 3/ 2 Pt H 3/ 4
From equation (1),
H D
Nµ µ
µ
H 1/ 2 × H 3/ 4 Pt H 5/ 4
=K
Pt H 5/ 4 Pt
124 Hydraulic Machines where K = Proportionality constant =
N Pt H 5/ 4
If
Pt = 1 kW, H = 1 m, K = N = NS
where NS is called specific speed. \
NS =
N Pt
H 5/ 4 For equal efficiencies of model and prototype,
(NS)m = (NS)p. The classification of turbines on the basis of specific speed has been discussed in the previous chapter.
8.4 SCALE RATIO The model of a turbine and its prototype should be in a definite geometric ratio depending upon their respective heads and rotational speeds. The ratio of blade velocity Vb and water velocity V(Vj in Pelton wheel and Va in reaction turbines) is called speed ratio. Different types of turbines have definite value of speed ratio. Pelton turbine,
f = 0.42 to 0.47
Francis turbine,
f = 0.55 to 1.00
Kaplan turbine,
f = 1.50 to 3.00
From equation (1) of # 8.3, DN µ \
Dm N m = Dp N p Dm = Dp
The scale ratio
FD I GH D JK m p
prototype of turbine.
H Hm Hp Hm N p × H p Nm
...(4)
is the ratio of diameter of model of turbine and
Dynamic Similarity and Performance Characteristics
125
8.5 UNIT QUANTITIES The operational characteristics of hydraulic turbines are expressed in terms of unit quantities. These quantities help to study the performance of turbine independent of actual head, discharge and power output. 1. Unit speed
Unit speed (NU) is defined as the speed of a geometrically similar turbine (model) working under a head of 1 m. Vb = f 2gH pDN = f 2gH 60
DN µ Nµ
2gH H
\ N=K H where K = constant when H = 1 m, N = NU = K \ Unit speed, NU =
N H
...(5)
Unit power
The unit power (PU) is the power in kW generated by a geometrically similar turbine (model) under a head of 1 m, Pt = rQgH = rA 2gH ×gH r and g are constant. \
Pt µ H3/2
\
Pt = KH3/2
when H = 1 m,
Pt = PU = K
\
PU =
Pt H 3/ 2
...(6)
Unit discharge
The unit discharge (QU) is the flow rate of a geometrically similar turbine (model) under a head of 1 m.
126 Hydraulic Machines Q = A 2gH =K H when H = 1 m,
Q = QU = K QU =
\
Q
...(7)
H
8.6 MODEL RELATIONSHIPS Model relationships are used to predict the behaviour of a prototype from the test runs of model of turbine. 1. Head coefficient
The blade velocity, Vb = f 2gH = \
N=
\
pDN 60
60f 2 gH pD
µ
H D
ND µ
H
H = constant. N 2 D2
\ The parameter
H is called head coefficient. N D2 2
Flow coefficient
The flow rate of water through a turbine Q = A ´ Vf A µ D2
Vf = y 2gH µ \
H
Q µ D2 H
...(8)
Dynamic Similarity and Performance Characteristics
127
From equation (8), H µ N2D2 Q µ ND3
\
Q = constant ND 3
or The parameter
...(9)
Q is called capacity or flow coefficient. ND 3
Power coefficient
The power developed by a turbine, Pt = rg QHht µ QH But
Q µ ND3
and
H µ N 2D2 Pt µ ND3 ´ N2 D2
\ \
Pt µ N3 D5
or
Pt = constant. N 3 D5
The parameter
Pt is called power coefficient. N 3 D5
8.7 PERFORMANCE CHARACTERISTICS Hydraulic turbines are designed to operate at certain values of head, discharge, power, speed and efficiency. However, turbines have to operate at off design conditions also. Therefore, the performance of turbines at design conditions and varying conditions have to be determined by conducting tests on turbine models or prototypes. 1. Parameters and properties
The following parameters and properties are collected from model tests. (i) Flow rate, Q(m3/s) or mass flow rate m f (= rQ) in kg/s. rQgHht (kW) 103 (iii) Efficiencies, htrans, hn, hvol, hh, hm and ht, (iv) Turbine speed, N (rpm)
(ii) Turbine power, Pt =
128 Hydraulic Machines The data is collected for design conditions and for varying conditions of operation. 2. Graphical representation
The results of the tests are plotted graphically and the curves are called performance curves or characteristic curves. Six points are required to plot a curve and two points for a straight line as per Indian standard code. The representation can be made on: (i) Cartesian coordinates (ii) Log scale or Log Log scales (iii) Circular graphs 3. Measurements
(i) (ii) (iii) (iv)
Flow rate by flow meters Pressure in m of H2O or kPa by pressure gauge Speed in rpm or rps by speedometer Output in kW by dynamometer either absorption type or transmission type.
Type of characteristics
For the sake of convenience, the characteristic curves are plotted in unit quantities. Following types of curves are plotted. (i) Main characteristics or constant head characteristic curves. Tests are performed by maintaining a constant head and a constant nozzle 1 1 3 , , and full opening. 4 2 4 Similar tests are conducted for each gate opening. A series of values of N, Q and P are measured. The unit quantities of QU, PU, NU and ht are computed and plotted as shown in Fig. 8.1. The following types of curves are obtained. Pelton turbine : Horizontal straight line Francis turbine : Drooping curve Kaplan turbine : Parabolic (ii) Operating characteristic or constant speed curves. Constant speed is obtained by regulating the gate opening (i.e., discharge) as the load varies. The following values are measured and calculated. Q0 is the discharge required to initiate the motion of turbine runner from state of rest. For constant head PU µ QU and a straight line is obtained. However, hU increases with QU and becomes more or less constant beyond a certain value.
or guide valve opening. Tests are carried at
Dynamic Similarity and Performance Characteristics
129
hu (%) Pu (kW) Qu (m3/s)
H = const
Nu (rpm) Fig. 8.1
Main characteristics
N = const
hu (%) Pu (kW)
hu
Pu
3
Qo
Fig. 8.2
Qu m /s
Operating characteristics
(iii) Universal characteristic or constant efficiency curves These curves are plotted between QU and NU for different gate openings. These curves are used for optimum design of turbine. The hydraulic design is for optimum speed but it must satisfy the safety conditions of run away speed.
Nu (rpm)
Efficiency = const
3
Qu (m /s)
Fig. 8.3 Universal characteristic
130 Hydraulic Machines 5. Appreciation of characteristic curves
Appreciation of characteristic curves is fitting of curves to mathematical equations for feeding to computer programming. The following types of curves are obtained. (i) Straight line y = mx + c
y
m c x
(ii) Rectangular parabola x2 y2 = c
y
x
(iii) Drooping curve or polynomial y = a0 + a1 x + a2 x2 + ... + an xn
y
x
(iv) Rising curve y = bxm
y
x
Fig. 8.4
Different curves
Selection of turbine
The type of turbine, output and speed are selected for best performance. This has been discussed in the previous chapters.
QUESTION BANK NO. 8 1. 2. 3. 4. 5.
What is the need for model testing? Explain dynamic similarity. What are specific speed and unit quantities? Explain scale ratio and model relationships. What are performance characteristics? How are these prepared? Explain the following: (i) Main characteristics
Dynamic Similarity and Performance Characteristics
131
(ii) Operating characteristics (iii) Universal characteristics
TUTORIAL SHEET NO. 8 1. Determine the speed and power of a Francis turbine operating under a head of 30 m if the and 450 rpm. Solution
1 th scale model could develop 5 kW under a head of 1.5 m 5 Np Dm = Nm Dp
Np = = NS =
Hm Hp
Dm H p × ×Nm Hm Dp
1 ´ 5
30 ×450 = 402.5 rpm Ans 15 .
N m Pm Hm5/ 4
=
450 5 = 606.16 (15 . ) 5/ 4
=
606.16 ´ 305/ 4 = 105.736 402.5
N p Pp
Also
NS =
\
Pp =
\
Pp = 11180 kW Ans.
H 5p/ 4 N S H 5p/ 4 Np
A Francis turbine with specific speed of 210 rpm has to develop 30 MW at 180 rpm. An experimental model has to be prepared to operate at 4.5 m head with a flow rate of 0.6 m3/s. Assuming an efficiency of 88%, estimate the speed, power and scale ratio for the model. What will be the flow rate for the turbine? Solution The power developed by the model, Pm = rQm gNmhm ´ 103 = 1000 ´ 0.6 ´ 9.81 ´ 4.5 ´ 0.88 ´ 103 = 23.3 kW NS =
N m Pm Hm5/ 4
132 Hydraulic Machines Nm = Dp N p Dm N m
Scale ratio,
Again,
\
Dp Dm
=
=
NS =
Hp5/4 =
N S ´ Hm5/ 4 Pm
=
210 ´ 4.55/ 4 23.3
= 285 rpm
Hp Hm Nm Np
Hp Hm
N p Pp H 5p/ 4 N p Pp NS
=
180 30 ´ 103 210
= 148.46 \ \
Hp = 54.6 m Dp Dm
=
285 54.6 = 5.5 Ans 180 4.5
Pp = rQpgHphp \
Qp =
PF HgH F D F
=
30 ´ 106 1000 ´ 9.81 ´ 54.6 ´ 0.88
= 63.65 m3/s Ans
PUMPS
2 ) 4 6 111
CHAPTER
9
Centrifugal Pumps 9.1 INTRODUCTION The centrifugal pumps are the most widely used pumps in industry and other sectors of economy. The list of applications is enclosed in Table 1.1. The centrifugal pumps have low initial cost, simple construction, high efficiency, low installation and operation costs. A centrifugal pump mainly consists of a rotating impeller mounted on a shaft. It has a number of curved vanes. The stationary element enclosing the impeller is volute casing. The impeller is rotated by an electric motor or an engine. The water or other liquid enters the eye of the impeller and is imparted the rotational energy of impeller as kinetic energy. The liquid from impeller is discharged into volute casing which converts kinetic energy into pressure energy. The liquid is supplied to the pump by a suction pipe and is delivered to the delivery pipe.
Impeller Eye
Shaft Suction pipe
Volute casing
Fig. 9.1 Centrifugal pump
9.2 CLASSIFICATION OF CENTRIFUGAL PUMPS There is a very large variety of centrifugal pumps. These are classified by the following characteristic features: 1. Commercial classification of working head. 2. Construction (i) Type of casing
136 Hydraulic Machines (ii) Number of impellers (iii) Flow direction through impeller. (iv) Number of suction or impeller eyes. (v) Shaft layout (vi) Liquid handled and type of impeller 3. Hydraulic classification by specific speed.
9.2.1 Working Head The centrifugal pumps are commercially classified by the working head or delivery head of liquid (i) Low head pumps upto 15 m (ii) Medium head pumps: 15 m to 40 m (iii) High head pumps: above 40 m.
9.2.2 Shape and Type of Casing The high kinetic energy of fluid leaving the impeller of the pump is recovered and converted into pressure energy by the casing around the impeller. There are two types of casings. 1. Volute casing or scroll casing
The kinetic energy of liquid is converted into pressure energy by the diffuser action of the gradually increasing area of volute or scroll casing. The pump is called volute pump. 2. Casing with stationary guide vanes
Fixed guide blades are provided around the impeller periphery. The kinetic energy of fluid is converted into pressure energy by the diverging passages formed between the guide vanes. The pump is called turbine pump or diffuser pump. The turbine pumps are more efficient and used in multi-stage pumps.
Impeller
Impeller
Guide vanes Volute case (a) Volute pump
(b) Turbine pump (Diffuser pump)
Fig. 9.2
Type of casings
Centrifugal Pumps
137
9.2.3 Number of Impellers (Stages) Single impeller pumps are widely used. But there is a limit on the head developed (15m) by single impeller. For high pressure requirements, the impellers are arranged in series. Such pumps are called multi-stage pumps.
Second impeller
Impeller
Inlet Shaft Volute casing
First impeller Guide vanes
Spiral casing (b) Multi-stage pump
(a) Single stage pump
Fig. 9.3 Number of impellers
9.2.4 Flow Direction through Impeller The liquid enters the impeller axially and head is developed by centrifugal force. The liquid leaves the impeller radially. Such pumps are called radial flow pumps. In axial flow pumps, the head is developed by the propelling action of vanes. The liquid enters the impeller axially and also leaves axially. These pumps have very large discharge but low head. Flow through a mixed flow impeller is a combination of axial and radial flow. The head is developed by the combined action of centrifugal force and propelling action of impeller.
(a) Radial flow
(b) Axial flow
(c) Mixed flow
Fig. 9.4 Flow direction through impeller
138 Hydraulic Machines 9.2.5 Number of Inlets The liquid enters the impeller with single suction. There are pumps with double suction and liquid enters the impeller from both sides. The double suction pumps can handle large discharge and have no axial thrust.
(a) Single section
(b) Double section
Fig. 9.5 Single and double suction pumps
9.2.6 Shaft Layout Most of the centrifugal pumps have horizontal shaft. But to achieve economy of space, vertical shaft pumps are also used. Such pumps are used for deep well and mine pumps.
9.2.7 Liquid Handled There are three types of impellers: 1. Closed impeller
The vanes are covered with shrouds (side plates) on both sides to provide a smooth passage for the liquid. The design ensures full capacity operation, high efficiency and long life. This type of impeller is suitable for clear liquids of low viscosity, i.e., water, hot water, acids and other clear chemicals. 2. Semi-open impeller
The impeller has shroud on the back side only. It is a non-clog impeller and can handle high viscosity liquids containing fibrous materials such as paper pulp, sugar molasses, sewage water, etc. 3. Open impeller
The impeller has no shrouds and vanes are open on both sides. These pumps are used for very rough duty to handle abrasive liquids, i.e., coal slurry, mixture of water and sand, pebbles and clay. The closed and semi-open impellers are not suitable due to clogging tendency.
Centrifugal Pumps
139
Vanes
Vanes
Vane Shroud
Shroud
(a) Closed impeller
(b) Semi-open impeller
(c) Open impeller
Fig. 9.6 Types of impellers
9.2.8 Specific Speed Specific speed is the speed of rotation of a geometrically similar model of a pump which delivers unit discharge against a unit head. The specific speed of a pump,
N Q (rpm) H 3/ 4 where N = Pump speed (rpm) Q = Discharge (m3/s) H = Head per stage (m) NS =
The pumps are classified as per specific speed as given in Table 9.1. Table 9.1 Classification of centrifugal pumps Pump Type
Speed
Specific speed (rpm)
Radial Flow
Slow
10-30
Normal
30-50
High
50-80
Mixed Flow
80-160
Axial Flow
160-500
9.3 PUMP PERFORMANCE The installation of a centrifugal pump is shown in Fig. 9.7 together with valves, piping, fittings and instrumentation. The following nomenclature is used. 1. Suction lift (Hs )
The vertical height between liquid surface and pump impeller centre line. 2. Discharge lift (Hd )
The vertical height between pump impeller centre line and top surface liquid in discharge tank.
140 Hydraulic Machines
Delivery tank Delivery pipe with fittings
Hmano (d)
Hd
Hmano (s) hg
Delivery valve or regulating valve (to regulate flow)
Hs Suction pipe with fittings (No loop) Sump
Strainer and food valve (CI NRV)
Fig. 9.7
Pump installation
Static head (Hst )
The static head or vertical lift or geometric head represents the total vertical lift. Hst = Hs + Hd 4. Manometric head (Hmano )
It is the difference between delivery and suction pressure as measured by a manometer or pressure gauges. Hmano = Hmano(d) Hmano(s) + hg 5. Total head (H)
Total head is the increase in total energy of liquid between inlet and outlet of the pump. H = Hmano + where
Vd2 - Vs2 2g
Hmano = Hstatic + åDHl åDHl = Total loss of head
Pump power (Pi )
The power input required to drive the pump Pi =
rQgHmano hoverall
Centrifugal Pumps
141
The overall efficiency of pump installation hoverall = where
rQgHmano Output = Input 3EI cos f
I = Electric current (A) E = Electric voltage (V ) f = Power factor hstatic =
rQgH static Pi
9.4 SANKEY DIAGRAM
Motor and Mechanical Leakage coupling losses losses losses
Pwater
Pimpeller
Pvane
Pi
Ps
The energy flow of the pump installation is shown as Sankey diagram in Fig. 9.8.
Head losses
Fig. 9.8 Sankey diagram of pump installation
9.4.1 Losses The various losses include: 1. Motor and coupling losses
These losses include electrical and mechanical losses in the motor and windage losses of coupling. 2. Mechanical losses
The mechanical losses include: (i) Disc friction loss of impeller and liquid (ii) Bearing and gland losses. 3. Leakage losses
Liquid leakage losses in the glands.
142 Hydraulic Machines 4. Head losses
The head losses or hydraulic losses are due to (i) secondary flow or eddies in the suction eye and impeller outlet (ii) friction of volute and impeller (iii) turbulence of liquid flow
9.4.2 Pump Efficiency 1. The overall efficiency or gross efficiency or actual efficiency of the pump set, rQgHmano Pwater = Pi 3EI cos f
hoverall =
The overall pump efficiency, Pwater Ps
hpump =
where PS = Power input to pump shaft = Pi Motor and coupling losses 3. The power of the impeller, Pimpeller =
Vw2Vb2 g
(per kg of liquid flow)
Volumetric efficiency hvol =
Pimpeller Pvane
=
Q Q + DQ
where Q = Pump discharge DQ = Leakage losses 5. Mechanical efficiency of pump hmech =
r(Q + DQ) Vw2Vb2 Ps
=
Pvane Ps
Monometric efficiency, hmono =
Pwater H rQgHmano = = mano V V Pimpeller Vw2Vb2 w2 b2 rQ g
Hmano = HS + Hf +
FG H
Vd2 2g
IJ K
Centrifugal Pumps
143
where HS = Static lift (m) Hf = Friction losses (m) Vd2 = Velocity head (m) 2g
Hydraulic efficiency hhyd =
rQgHs Vw Vb rQ 2 2 g
Hs w2 Vb2
FG H
IJ = V K
The overall efficiency of pump, hoverall = hm ´ hvol ´ hmano
9.5 VELOCITY DIAGRAMS The velocity diagrams at the inlet and outlet of the impeller of a centrifugal pump are shown in Fig. 9.9. The following nomenclature is used: Va = Absolute velocity of liquid, (m/s) Vb = Peripheral velocity of vane, (m/s) Vr = Relative liquid velocity (m/s) Vf = Flow component of absolute velocity (m/s) Vw = Whirl or tangential component of absolute velocity (m/s) 1. suffix for inlet of impeller 2. suffix for outlet of impeller Vr
2
Vr
b2
1
Va 1 a1
b1
a2 Vb
Va
Vb
1
2
N
1 2
Fig. 9.9
Velocity diagrams of centrifugal pump
2
144 Hydraulic Machines Va1
Vr1
Vf1
b1 Vb1
Va2
Vr2 Vf2
a1
b2
a2
Vb2
Vw1
(a) Inlet velocity triangle
Vw2
(b) Outlet velocity triangle
Fig. 9.9 Inlet and outlet velocity triangles
The power applied by impeller on liquid, Pimpeller = rQ(Vw2Vb2 Vw1Vb1) For centrifugal pumps, liquid enters radially into the impeller \
a1 = 90° Pimpeller = rQVw2Vb2.
9.5.1 Practical Design Data For centrifugal pumps, the following data holds good. a1 = 90°; b1 = 15° to 30°; b2 = 20° to 40° Vf 1 = Vf 2; B ´ D = constant No. of vanes: 6 to 12
2gH =
Vb2 = \
pD2 N 60
60 . 2gH pN The inlet diameter of impeller, D1 depends upon the pump head.
D2 =
1.0 0.75 0.66
D1
0.5
D2 0.33 0.25 0 10
50 H
Fig. 9.10
D1/D2 v/s H
100 m
Centrifugal Pumps
145
9.5.2 Effect of Exit Blade Angle >2 The shape of impeller vane changes with b2 which affects the efficiency of the pump. For
a1 = 90°, Pimpeller = rQ(Vw2Vb2) = rQgHmon.
(If no internal losses)
...(1)
From outlet velocity triangles, Fig. 9.9(b), Vw2 = Vb2 But
V f2
tan b2
Q = pD2 2 Vf 2 Q pD2 B2
\
Vf 2 =
\
Vw2 = Vb2
where
Vb2 =
Q pD2 B2 tan b 2
pD2 N . 60
From equation (1) Hmon =
Assume and
Vw2Vb2 g
FG H
= Vb2 -
IJ FG IJ KH K
Vb Q . 2 g pD2 B2 tan b 2
=
QVb2 Vb22 g pD2 B2 tan b 2 g
=
( pD2 N ) 2 Q pD2 N . 60 g pD2 B2 tan b 2 (60)2 g
( pD2 N )2 =C (60) 2 g N =E gB2 tan b2
Hmon = C EQ
...(2)
For a given size of pump and N = constant, N, D2, B2 and b2 are constant. \ E and C are constant. \ Equation (2) is a linear relationship.
146 Hydraulic Machines p , cot b2 = 0. 2 \ Hmon = C straight radial blade.
(i) For b2 =
p 2 Hmon decreases with Q. Backward curved vanes
(ii) When b2
Hmono
N = const. b2 > 90° b2 = 90° b2 < 90°
C
Q Fig. 9.11 Types of vanes
9.5.3 Performance of Different Vane Shapes The exit relative velocity Vr2 is fixed by b2, hence the exit velocity triangle depends upon b2. 1. Radial vanes
Liquid leaves the vane in radial direction as b2 = 90°. Vw2 = Vb2 Hmon = C EQ Hmon = C as cot
p =0 2
\ Hmon is independent of Q (Fig. 9.11.) The power input to impeller, Pi = rQg(C EQ) = rg CQ EQ2
Centrifugal Pumps
Vr2
Vb2 = Vw2 a2 b2 Va2
147
N = const.
Pi
b2 > 90°
b2 = 90° b2 < 90°
Q
Fig. 9.12
(a) Radial vanes
Fig. 9.12
(b) Pi µ Q
For b2 = 90° Pi = rgQC \
Pi µ Q
Backward curved vanes
The vane is curved in the opposite direction of rotation and b2 < 90°. cot b2 < 1 Pi µ QHmano The curve droops downwards. Vw2 < Vb2 3. Forward curved vanes
The vanes are curved in the direction of rotation of impeller b2 > 90°; Vw2 > Vb2. cos b2 > 1 Pi µ Q Hmano The curve rises upwards. The efficiency increases with decrease of b2. The relation between maximum impeller efficiency and exit angle b2 is plotted in Fig. 9.12(e). For radial blades,
h » 80%
For forward curved blades,
h » 75%
For backward curved blades, h = 85 90% The energy conversion efficiency is maximum for forward-curved vanes, but the exit velocity (Va2) is very high which is not desirable. Normally backwardcurved vanes with b2 between 20° and 30° are used for low and medium head pumps. For high head pumps, forward-curved or radial vane impeller are preferred.
148 Hydraulic Machines Vw2
Vb2 Vw2 Vr2
Fig. 9.12
b2 Vb2
a2
b2
Vf2 Vr2
a2 Va2
Va2
(c) Backward curved vanes
Fig. 9.12 (d) Forward curved vanes
100% 90%
hmax
80% 70% 60%
20°
90° b2
120°
Fig. 9.12 (e) h v/s b2
9.6 DESIGN OF IMPELLER Design of pump impeller involves the following steps.
9.6.1 Selection of Speed The rotating speed of impeller is selected to meet the head and capacity requirements. The specific speed and type of impeller is selected. The selection of speed is governed by the following factors: 1. Type of driving unit 2. Specific speed. Higher specific speed results in a small pump and cheaper driver. 3. The optimum hydraulic efficiency and overall efficiency for each type of impeller depends upon the specific speed. 4. In case total required head cannot be produced in one stage, the head is divided between two or more stages. The head per stage also affects its final specific speed and the overall pump efficiency.
Centrifugal Pumps
149
An impeller model from existing models of same specific speed and hydraulic performance is selected. Qincreasing
h(%)
Ns
(rpm)
Fig. 9.13 h v/s NS
9.6.2 Impeller Profile The impeller profile for maximum hydraulic efficiency is determined from the following known data. 1. Inlet and exit velocities of the impeller (Va1 and Va2 ) 2. Impeller outside diameter and vane velocity (D2 and Vb2 ) 3. Impeller vane outlet angles (a2 and b2) from velocity triangle In straight radial vanes all liquid streams enter and leave the impeller at the same diameter and the vane is plane of simple profile. The impeller design is determined from one inlet and one outlet velocity triangle. For mixed flow and axial flow impellers, velocity triangle are drawn for several liquid stream line. Three streamlines are usually sufficient for average mixed flow and axial flow impellers. Variation of vane angles along the radius determines the vane curvature and twist.
9.6.3 Vane Discharge Angle (>2) Vane discharge angle is the most important design factor. Theoretical characteristics of pump impeller can be determined from vane angle alone. In actual pumps also, angle b2 is the deciding factor in impeller design. All design constants are functions of b2. Proper selection of b2 is the important step in fixing the design constants of the impeller. The selection of b2 depends upon the following factors: 1. The steepness of the head-capacity curve. 2. Normal head and capacity increase with b2. 3. Maximum capacity is required for a given impeller diameter.
150 Hydraulic Machines If the above design constraints are not applicable, b2 is selected from the following considerations: 1. Optimum efficiency and normal design. 2. An average value of b2 = 22.5° is taken for normal impeller design for all specific speeds. 3. For increased output, b2 should be raised to 27.5° without affecting the efficiency. 4. b2 = 17.5° can be used as lower value for good design.
9.6.4 Mean Effective Impeller Diameter The various dimensions of an impeller are shown in Fig. 9.14. b2
Impeller diameter
D, m
Fig. 9.14
D1, m
d1
D2, i
D2
,0
d2
For a radial impeller, the mean effective diameter, D22, 0 + D22, i 2 where D2, 0 and D2, i are the maximum and minimum outside diameters of impeller. For an axial flow impeller,
D2m =
D2m = where n =
D02 (1 + n2 ) 2
Dh D0
D0 = outside diameter of impeller Dh = hub diameter of impeller
Centrifugal Pumps
151
For mixed flow and axial flow impellers, the mean effective diameter divides the liquid flow in the impeller into equal parts. If the peripheral velocity (Vb2) and impeller discharge angle (b2) are based on mean effective diameter (Dm), the correlations and plots of several design constants for a complete range of impellers from straight radial to axial flow become very simple.
9.6.5 The Inlet Flow Velocity and Flow Areas The flow velocity at inlet (Vf1) must be known to complete the impeller profile. This is given by flow ratio (y1) y1 =
V f1 2 gH
The leakage at the entrance has been neglected. The flow velocity at exit (Vf 2) of impeller, Vf 2 =
Q Q Q = = ( pDm - Area occupied by vanes)d 2 A2 pDmd2
Vf 1 =
Q Q = A1 pD1,md1
The area occupied by vanes is normally neglected. The dimensions d1 and d2 are shown in Fig. 9.14. For multi-stage pumps such as boiler feed pumps, the flow velocity ratio
F V I may be as high as 1.625. GH V JK f1
f2
For services where low NPSH requirements are important, Vf 1 £ Vf 2 A low Vf 1 leads to low entrance angle (a1).
9.6.6 Reduction of Impeller Diameter The impeller diameter is generally reduced to decrease the head and capacity of a given centrifugal pump. The performance of pump with reduced impeller diameter can be more or less estimated from similarity laws. However, there are different rules for estimation of performance of different types of pump impellers. 1. Radial impeller
The inlet and outlet velocity triangles for the full and reduced impeller are shown in Fig. 9.15.
152 Hydraulic Machines Pi = rQg Hmono Hmono = If the term
Vb1Vw1
Vb2Vw2 - Vb1Vw1 g
is ignored, the impeller performance will result in:
g
The head varies directly as the square of diameter ratio and the discharge varies directly as the diameter ratio. 2. The power input varies directly as the cube of the impeller diameter ratio. Vw2
V¢w1 Vw1
Vw¢2
Vf2 V¢ f2
V¢f1 Vf1 a1
b2
b2
a1
Vb¢2 Vb2
a¢1
b1
Vb1
(a) Exit velocity triangles
(b) Inlet velocity triangles
Fig. 9.15 Velocity triangles for full and reduced impellers
Axial flow impeller
If the impeller diameter is reduced, a new casing or a liner would be required to accommodate the new impeller. Therefore, this practice is rarely in practised. The impeller performance will result in: 1. The discharge varies directly as the net area swept by the impeller vane, the axial velocity remaining the same. 2. The discharge head will be reduced directly as the square of the diameter ratio. The new head H¢ will be: 2H¢ = Hh + Ho
FG D¢ IJ HD K o o
where Do¢ = Reduced impeller diameter Do = Original impeller diameter Ho = Original head Hh = Head at the hub diameter
2
Centrifugal Pumps
153
Mixed flow impeller
Mixed flow impeller is difficult to reduce as the vanes overlap which causes great head loss. The diameter ratio is calculated as the average of the diameters of the outside and inside shrouds. Both the diameters should be cut in the same ratio. It is better to reduce the outside diameter and nothing at the hub.
9.6.7 Design Constants The following constants are used in the design of centrifugal pumps. 1. Speed ratio (f) is the ratio of peripheral speed Vb2 at the impeller tip and liquid velocity due to theoretical head corresponding to manometric head. f=
Vb2 2 gHmano
f varies between 0.95 and 1.25. 2. Flow ratio (y) is the ratio of the flow velocity Vf2 and liquid velocity due to theoretical manometric head, f=
V f2 2 gHmano
y varies between 0.1 and 0.25. 3. Diameter ratio
FG D IJ is the ratio of diameter of impeller at the inner and HD K 1
2
outer peripheries,
D2 =
60Vb2 pN
1 2 D1 varies from to . 3 3 D2
Exit vane angle (b2). It depends upon the head capacity curve. For optimum efficiency b2 = 25° for all values of specific speed. 5. Inlet vane angle (a1) is taken as 90° to ensure radial inlet absolute velocity. 6. Radius of curvature of vane is selected to ensure smooth free flow of liquid in the impeller passage. It depends upon inlet and outlet vane angles. 7. Number of vanes of impeller depends upon pump size, speed ratio and outlet vane angle. With low values of b2, six or eight vanes are used.
154 Hydraulic Machines 9.7
DYNAMIC SIMILARITY AND PERFORMANCE CHARACTERISTICS
For comparison of characteristics of similar pumps, the following assumptions are made: 1. The pumps are geometrically similar and all linear dimensions are proportional to impeller diameter. Example: Impeller width B µ Impeller diameter D 2. The velocity triangles are similar. 3. The pumps have same hydraulic and volumetric efficiency. 4. All velocities are proportional to H 1/2. 5. The Reynolds numbers are same for the pumps being compared/examined.
9.7.1 Flow Coefficient The discharge, Q = pDBVf Vf = y 2gH µ H BµD Q µ D2 H
\
Vb =
pDN 60
D=
60Vb pN
\
Dµ
Vb N
\
H N Vb µ ND Q ¥ ND3
Again
µ
Q = constant ND 3
\ The factor
Q is called flow coefficient. ND 3
...(1)
Centrifugal Pumps
155
9.7.2 Head Coefficient Dµ \
Dµ
Vb N
H N
H µ DN H µ D2N2 H = constant D N2
\
2
The factor
...(2)
H is called head coefficient or lift coefficient. D2 N 2
9.7.3 Power Coefficient P = rQgN µ QH, But
H µ ND3
and
N µ N 2D2 P µ (ND3) ´ (N 2D2)
\
¥ N3D5 \
P = constant N D5 3
...(3)
P is called power coefficient. N 3 D5 The flow coefficient, head coefficient and power coefficient are used to determine the performance of prototype pumps from the test results of model of pumps. The simplified non-dimensional coefficients can be used for comparison. Q H P = const.; 2 = const.; 3 = const. N N N
9.7.4 Specific Speed The specific speed of a centrifugal pump is defined as the speed of a geometrically similar pump which would deliver unit quantity (1 m3/s) against a unit head (1 m). Q µ D2 H
156 Hydraulic Machines
\
D2 µ
H2 N2
Qµ
H2 N2
µ \
H
H 3/ 2 N2
Q=K
H 3/ 2 N2
where K = constant of proportionality when H = 1 m and Q = 1 m/s2, N = Ns 1=K
(1) 3/ 2 N s2
\
K = Ns2
\
Q = Ns2
H 3/ 2 N2
Specific speed, Ns =
N Q H 3/ 2
Notes: 1. For multi-stage pumps, H is the manometric head per stage. 2. For double suction pump, Q is the half of total discharge.
9.7.5 Relation Between Specific Speed of Pump and Turbine (Ns)turbine =
N P H 5/ 4
(Ns)pump =
N Q H 3/ 4
P= \
rQgH ho 1000
(Ns)turbine = N
[kW]
FG rQgH h IJ H 1000 K o
1/ 2
´
1 H 5/ 4
Centrifugal Pumps
FG rgh IJ H 1000 K F rgh IJ =G H 1000 K o
(Ns)turbine =
o
1/ 2
.
157
N Q H 3/ 4
1/ 2
´ (Ns)pump
Assume ho for a turbine as 90% and r for water as 1000 kg/m2 (Ns)turbine =
FG 1000 ´ 9.81 ´ 0.9 IJ H 1000 K
1/ 2
´ (Ns)pump
= 2.97(Ns)pump
9.7.6 Performance Characteristics Performance characteristics of a pump are variation of head, power and efficiency against flow rate graphically. There are three types of performance characteristics. 1. Main characteristics
The pump is run at a constant speed and discharge is varied. Measurements are made for each discharge. The values of head and shaft power are measured by pressure gauge and dynamometer respectively and calculations are made for pump efficiency. A set of curves are plotted for different constant speeds. pm
N
r 00 20h =
r r rh
1500 rpm
h
N = 2000 rpm H
N = 20
00 rpm
N = 15
00 rpm
h P H
N = 10
N = 1000 rpm
00 rpm
N = 1500 rpm N = 1000 rpm Q Fig. 9.16
Main characteristics
Operating characteristics
The operating characteristics are the main characteristics for design speed for which maximum efficiency is obtained. Operating characteristics are used to obtain design discharge and head for maximum efficiency.
158 Hydraulic Machines
h h P H P
H
Qdesign Q
Fig. 9.17 Operating characteristics
Universal characteristics
These are constant efficiency curves also called Muschel curves. These curves are plotted from the main characteristics for different constant efficiency values. These curves help the region where pump would operate with maximum efficiency. h(%)
N increasing
70 75 80 85 85 80 75 70
H
Best speed limit Q Fig. 9.18
Universal characteristics
9.8 CAVITATION AND MAXIMUM SUCTION LIFT Cavitation starts under the vanes at pump entry where the pressure is the lowest. If the pressure is lower than vapour pressure of the liquid at the prevailing temperature, there is separation of water and bubbles are formed. These bubbles are carried to high pressure region near the impeller exit. The bubbles collapse causing pitting and severe damage of the metal surface of vane.
Centrifugal Pumps
159
The vane tips at exit experience the following harmful effects: 1. Pitting and erosion of surface due to continuous hammering action of collaping bubbles. 2. Sudden drop in head, efficiency and power delivered to the liquid. 3. Noise and vibration produced by the collapse of vapour bubbles. The following factors contribute to cavitation. 1. High impeller speed 2. Restricted suction area 3. High specific speed 4. High liquid temperature 5. Required NPSH equal to or greater than available NPSH. The pressure balance at free liquid surface (1) and pump suction (2) gives: H1 = H2 + Hfs + Hs +
Vs2 2g
where, H1 = Hatm and depends upon site altitude Hfs = Head loss in suction pipe Hs = Static head between points 2 and 1. Vs = Liquid velocity through suction pipe H2 = Head at suction eye. Discharge
2 Hs
Vs
Pump pa 1 Liquid sump
Fig. 9.19 Pressure balance at pump suction
H2 should be more than vapour pressure of liquid which is a function of temperature. H2 ³ Hvap. If H2 < Hvap, water separation occurs resulting in cavitation and harmful effects listed above. Minimum Net Position Suction Head (NPSH) should be provided to avoid cavitation.
160 Hydraulic Machines Net Positive Suction Head (NPSH)
F GH
= Hatm Hs + H f s + Hvap +
Vs2 2g
I JK
³ 2 to 2.5 m Vs2 = Hatm Hs Hfs Hvap 2g Hatm - Hs - H f s - Hvap Vs2 = 2 gH H
\
where H = Head developed by the pump. Vs2 is called the Thomas cavitation number (s) and [(Hatm Hs 2 gH Hfs) Hvap] represents NPSH, i.e. the head to make the liquid flow through the suction pipe to the pump impeller.
The factor
\
s=
[( Hatm - Hs - H f s ) - Hvap ] H
=
NPSH H
The cavitation factor is a function of specific speed. The critical cavitation number, scr = 1.042 ´ 103 (Ns)4/3 s ³ scr or
(NPSH) ³ scr H. For geometrically similar machines Q µ ND3
or
NPSH µ
V2 2g
µ
Q2 D4
N Q = const. = Ssu ( NPSH ) 3/ 4
where Ssu = suction specific speed
Centrifugal Pumps
FG IJ H K FN I s=G J Hs K
NPSH Ns = H ssu
4/ 3
4/ 3
s
or
161
su
For s = scr , Ssu = 175
For a given head and capacity, a pump with low specific speed operates safely with greater suction lift.
9.9 MULTISTAGE PUMPS A number of impellers are connected in series to build up high pressure. Two or more identical impellers are mounted on the same shaft and enclosed in the same casing. The discharge from first impeller passes through a guided passage and enters the second impeller, and discharge from second impeller enters the third impeller, and so on. The discharge from last impeller is pushed through the delivery pipe. The total discharge is same through each impeller. The total head developed by the pump, Htotal = nHstage where
n = Number of impellers or stages Hstage = Head developed by each impeller NS =
N Q ( H / n) 3/4
For centrifugal pump, Ns = 40 will give an efficient and economical pump. Hmano =
Vb2Vw2 g
Peripheral speed (Vb2) should be increased to get more head. However, Vb2 is limited by impeller strength and cavitation. Vb2 is limited to 40 m/s for C.I. impeller and 300 m/s for alloy steel impellers.
9.10 PUMP OPERATION 9.10.1 Operational Problems The main operational problems of centrifugal pumps along with their causes are tabulated in Table 9.2.
162 Hydraulic Machines Table 9.2 S.No.
Operational problems of centrifugal pumps
Problem
Reasons
1.
Pump fails to deliver liquid.
(i) (ii) (iii) (iv) (v)
Pump not properly primed Suction lift and delivery too high Speed low Wrong direction of pump rotation Clogging of impeller, strainer or suction line
2.
Not full head and capacity
(i) (ii) (iii) (iv) (v) (vi) (vii)
Air leakage into pump Speed low Wrong direction of pump rotation Partial clogging of impeller Small impeller diameter Discharge head higher than anticipated Insufficient suction head when pumping hot or volatile liquids
3.
Power consumption by pump excessive
(i) (ii) (iii) (iv) (v) (vi) (vii) (viii)
Pump running on higher speed Wrong direction of pump rotation Low head and high discharge Bent shaft and rubbing of runner with casing Liquid density high Cavitation Lack of lubrication Tight packing
4.
Overheating of pump
(i) (ii) (iii) (iv)
Rubbing of runner with casing Lack of lubrication High impeller speed Worn-out rings, bent shaft, uncleaned bearings or other mechanical defects
5.
Noise and vibration
(i) (ii) (iii) (iv) (v)
Cavitation Misalignment of pump shaft Worn-out bearings Improper foundation Rubbing of runner with casing
9.10.2 Priming of Pump When the centrifugal pump is not primed (filled with liquid) before starting, the air pockets inside the impeller may give rise to formation of vortices and cause discontinuity of liquid flow. Priming is done by pouring of liquid into pump casing (when pump is not running) through a funnel and air is displaced through air vents provided on the casing. The problem of formation of air pockets can be avoided by the following methods to ensure proper delivery of liquid by the pump when running.
Centrifugal Pumps
163
Installation of pump below suction liquid level 2. Self priming device 3. Priming of pump 1. Self priming devices
(i) Vacuum pump is run on the same pump shaft which sucks air from suction pipe and throws into the atmosphere. (ii) Proprietary design of pump to form air-liquid mixture. 2. Priming methods
(i) Manually pouring liquid through priming funnel and opening vent valves till pump is fully filled. The liquid is poured when pump is not running. (ii) Connecting the pump with city water mains for filling water into impeller and suction pipe. (iii) Priming chamber is provided on delivery side which is filled with liquid when pump is running. It is used for priming. The size of priming chamber is three times the volume of suction pipe.
9.10.3 Minimum Starting Speed The centrifugal pump starts delivering liquid when the following conditions are fulfilled. (i) There is no air entrapped on the suction side of pump. This is ensured by priming the pump which consists of filling liquid into impeller, casing and suction pipe in order to remove air, gas or vapour before starting. (ii) The pressure rize in the impeller is higher than the gross or manometric head of the pump. The centrifugal head or pressure head created the pump is
Vb 22 - Vb12 . 2g
The manometric head, Hmano =
\
Vb2Vw2 g
.hmano
Vb2 - Vb12 Vb2Vw2 ³ .hmano to ensure delivery of liquid 2g g
For hmano to be 100% Vb2Vw2 g
=
w2 2 (R2 R12 ) 2g
164 Hydraulic Machines Vb2Vw2 g
=
FG 2pN IJ × ( R - R ) H 60 K 2g 2 2
2 1
where R1 and R2 are impellers inside and outside the radii. The impeller speed N (rpm) can be found out. In other words, R2 can be found out which will ensure delivery of liquid at its nominal rotational speed.
9.10.4 Effect of Speed and Diameter The head and capacity of an existing pump can be adjusted to suit site conditions by altering the pump speed or impeller diameter or both. 1. Impeller speed
When the diameter of impeller is not changed, the performance of pump can be altered by changing the pump speed. Vb = \
pDN 60
Vb µ N For same vane angles b1, b2, a1 and a2, the flow velocity Vf µ Vb µ N
and \
Q µ Vf µ N.
N Q = N¢ Q¢
...(1)
The flow of liquid is directly proportional to pump speed. Vb µ \ \
Hµ
H V 2b
µ N2
FG IJ H K
H N = H¢ N¢
2
...(2)
The discharge head is proportional to the square of speed. P µ QH \
FG IJ H K
P N = P¢ N¢
3
The power input to pump is proportional to the cube of the speed.
...(3)
Centrifugal Pumps
165
Impeller diameter
When the speed of the impeller is not changed, the pump performance can be adjusted by changing the impeller diameter either by cutting or using an impeller of different diameter in the same pump casing. The vane speed, Vb = \
pDN 60
Vb µ D
(3 N = constant)
The flow velocity, Vf µ Vb µ D The flow area, Af µ D2 The liquid flow, Q µ Af Vf µ D3
FG IJ H K
D Q = D¢ Q¢
3
...(4)
The pump capacity is proportional to the cube of the impeller diameter. The pump head, H µ V b2 µ = D2
FG IJ H K
H D = H¢ D¢
2
...(5)
The pump head is proportional to the square of the impeller diameter. The pump power, P µ QH \
FG IJ H K
P D = P¢ D¢
5
The power consumption changes by 5th power of impeller diameter.
...(6)
166 Hydraulic Machines 3. Impeller diameter and speed
The impeller diameter and speed are changed simultaneously From equations (1) and (4),
FG IJ FG D IJ H K H D¢ K
N Q = N¢ Q¢
3
From equations (2) and (5)
FG IJ FG D IJ H K H D¢ K
H N = H¢ N¢
2
2
From equations (3) and (6)
FG IJ FG D IJ H K H D¢ K
P N = P¢ N¢
3
5
QUESTION BANK NO. 9 1. Discuss the classification of centrifugal pumps as per specific speed, working head, pump construction and the liquid handled. 2. Draw the installation of a centrifugal pump and discuss the significance of various heads. 3. Distinguish between manometric efficiency and hydraulic efficiency of a centrifugal pump. 4. With the help of an energy flow diagram, discuss various losses and efficiencies of a centrifugal pump. 5. Draw the velocity diagrams of centrifugal pump. What is the effect of vane shape on velocity triangles? 6. Discuss the influence of exit blade angle on the performance and efficiency of a centrifugal pump. 7. Explain step-by-step the procedure for the design of a centrifugal pump. 8. Define important ratios and coefficients as applicable to the hydraulic design of a centrifugal pump. 9. Explain various performance characteristics of a pump under the headings: (a) Main characteristics (b) Operating characteristics (c) Universal characteristics 10. What is cavitation? What is NPSH? How are cavitation and NPSH correlated?
Centrifugal Pumps
167
Write notes on: (a) Priming of pumps (b) Minimum starting speed (c) Effect of speed and diameter on pump performance. 12. Prepare a list of problems and reasons in the malfunctioning of a centrifugal pump.
TUTORIAL SHEET NO. 9 1. Calculate the head imparted to a liquid and inlet vane angle it passes through an impeller of 0.25 m diameter and 0.1 m inlet diameter rotated at 1440 rpm. The outlet vane angle is set back at an angle of 20° to the tangent. Assume radial entrance and velocity of flow at 3 m/sec. Solution The data given: D1 = 0.10 m D2 = 0.25 m N = 1440 rpm a1 = 90° b2 = 20° Vf1 = Vf2 = 3 m/s The vanes are backward curved with velocity triangles as shown. Vr1
Va1
Vr2 Vf1
b1 Vb1
Va2
Vf2 b2
a1 Vb2
Vw1
The vane velocity at inlet, Vb1 =
pD1 N p ´ 0.1 ´ 1440 = = 7.54 m/s 60 60
Vb2 =
pD2 N p ´ 0.25 ´ 1440 = = 18.85 m/s 60 60
From exit velocity triangle, Vb2 Vw2 =
V f2
tan b2
=
3 = 8.24 m/s tan 20°
a2 Vw2
168 Hydraulic Machines \
Vw2 = 18.85 8.24 = 10.6 m/s
(i) Head imparted to liquid H=
Vb2Vw2 g
=
18.85 ´ 10.6 = 20.37 m Ans 9.81
(ii) From inlet velocity triangle, tan b1 =
V f1 Vb1
= \
(3 a1 = 90°)
3 7.54
b1 = 21.7° Ans
A centrifugal pump impeller has an external diameter of 0.4 m and internal diameter of 0.18 m and it runs at 1440 rpm. Assuming a constant radial flow through the impeller at 2.5 m/sec and the vanes at exit are set back at an angle of 25°, calculate: (a) The angle absolute velocity of water at exit makes with the tangent, (b) Inlet vane angle, and (c) Work done per kg of water. Solution The data given: D1 = 0.18 m D2 = 0.40 m N = 1440 rpm Vf1 = Vf2 = 2.5 m/s b2 = 25° (a)
Vb1 =
pD1 N p ´ 0.18 ´ 1440 = = 13.57 m/s 60 60
Vb2 =
pD2 N p ´ 0.4 ´ 1440 = = 30.16 m/s 60 60
tan b1 = \
V f1 Vb1
=
b1 = 10.4°.
2.5 1357 .
Centrifugal Pumps
(b)
tan b2 =
\
Vw2 = tan a2 =
\
V f2 Vb2 - Vw2
169
= 0.4663
-2.5 + 30.16 = 24.8 m/s 0.4663 V f2 Vw2
=
2.5 24.8
a2 = 5.75° Ans
(c) Work done per kg = Vw2 Vb2 = 24.8 ´ 30.16 = 748 N.m/kg Ans 3. A centrifugal pump in a dock pumps 1565 litres/sec against a mean lift of 6.1 m. The impeller has a diameter of 1.22 m; speed 200 rpm; area at the outer periphery 0.645 m2; angle of vane at outlet 26° and ratio of external to internal diameter 2:1. Determine: (i) the hydraulic efficiency, (ii) power required to drive the pump. Solution The given data: Q = 1.565 m3/s H = 6.1 m D2 = 1.22 m N = 200 rpm Af = 0.645 m2 b2 = 26° D1 =
D2 1.22 = = 0.61 m. 2 2
Vf2 = Vf1 = (i)
1565 . Q = = 2.43 m/s Af 0.645
Vb1 =
pD1 N p ´ 0.61 ´ 200 = = 6.39 m/s 60 60
Vb2 =
pD2 N p ´ 122 . ´ 200 = = 12.78 m/s 60 60
170 Hydraulic Machines hhyd =
tan b2 = \ Vw2 = hhyd =
Hg rQgH = rQgVw2Vb2 Vw2 Vb2 V f2 Vb2 Vw2
= 0.4877
2.43 12.78 = 7.83 m/s 0.4877 61 . ´ 9.81 = 0.598 » 60% Ans 7.83 ´ 12.78
(ii) The power required to drive the pump, Pi =
. ´ 6.1 ´ 10-3 rQgH 1000 ´ 9.81 ´ 165 = 0.6 hhyd
= 156 kW Ans 4. A centrifugal pump lifts water under a static head of 33.6 m of which 3.68 m is suction lift. Suction and delivery pipes are both 1.5 m in diameter. The head loss in suction pipe is 2.1 m, and in the delivery pipe is 7.1 m. The impeller is 0.38 m in diameter and 2.54 cm width and revolves at 0.38 m in diameter and 2.54 cm wide and revolves at 1200 rpm. Its exit blade angle is 35°. If the manometric efficiency of the pump is 82%, determine the discharge and the pressure at the suction and delivery branches of the pump. Solution The given data Hstat = 33.6 m
hmano = 82%
HS = 3.68 m Hd = 33.6 3.68 = 29.92 m Total head to be supported by the pump, Hmano = Hstatic +
å DHL
= 33.6 + 2.1 + 7.1 = 42.8 m of H2O. Vb2 = hmano =
pD2 N p ´ 0.38 ´ 1200 = = 23.88 m/s 60 60 Hmano Vb2Vw2 / g
Centrifugal Pumps
\
Vw2 =
171
Hmano . g 42.8 ´ 9.81 = Vb2 hmano 2388 . ´ 0.82
= 21.44 m/s tan b2 = \
V f2 Vb2 - Vw2
Vf 2 = (Vb2 Vw2) tan b2 = (23.88 21.44) tan 35° = 1.7 m/s Q = pD2 B2Vf 2 = p ´ 0.38 ´
0.254 ´ 1.7 = 0.00515 m3/s 100
Velocity in suction and delivery pipes, Vd = Vs = Velocity head =
0.00515 Q = = 0.29 m/s As p / 4 ´ (015 . )2
Vs2 (0.29) 2 = = 0.043 m 2 g 2 ´ 9.81
Total effective pressure on delivery side Hd + HLd +
Vd2 = 29.92 + 7.1 + 0.043 2g
= 37 m Total effective pressure or suction side = Hs + HLI +
Vs2 = 3.68 + 2.1 + 0.043 2g
= 5.82 m of vacuum. or
(10 5.82) = 4.177 m
absolute
A centrifugal pump discharges 25 litres per second against a total head of 15 m at 1400 rpm and the diameter of the impeller is 0.45 m. A geometrically similar pump of 30 cm diameter is run at 2800 rpm. Calculate the head, discharge and power ratio required, assuming equal efficiencies between the two pumps. Solution The given data for first pump: Q1 = 25 l/s = 0.025 m2/s
172 Hydraulic Machines H1 = 15 m N1 = 1400 rpm D1 = 0.45 m The data for geometrically similar pump: D2 = 0.3 m N2 = 2800 rpm
FG IJ H K F N IJ FG D IJ =G HN K HD K F N IJ FG D IJ =G HN K HD K
FG IJ = 1.6875 H K F 1400 IJ FG 0.45IJ = 0.5625 =G H 2800K H 0.3 K F 1400 IJ FG 0.45IJ =G H 2800K H 0.3 K 3
N1 D13 1400 0.45 Q1 = = 3 2800 0.3 N 2 D2 Q2 H1 H2 P1 P2
2
1
1
2
2
3
1
1
2
2
2
5
2
3
2
5
= 0.125 ´ 7.59375 = 0.949 6. Determine the number of stages required for a multistage centrifugal pump to deliver 75.7 lit/sec to a height of 91 m at 1500 rpm. Assume a specific speed of 30 rpm and an efficiency of 75%. Solution The given data Q = 75.7 lit/sec = 0.0757 m3/sec H = 91 m N = 1500 rpm h = 75% NS = 30 NS = \
H 3/4 =
N Q H 3/ 4 N Q 1500 0.0757 = 30 Ns
= 10.7568 \
H = 32.93 m per stage
Centrifugal Pumps
No. of stages
n=
173
91 H = 32.93 HstaCA
= 2.76 say 3 7. A 1/5 scale model of a centrifugal pump absorbs 20 kW when pumping against a head of 8 m at its best speed of 400 rpm. If the actual pump works against 32 m head, find the speed and power required by the actual pump. Determine also the quantity of water discharged by the two pumps. [Na = 160 rpm; Pa = 4000 kW; Qm = 225 m3/s; Qa = 17750 m3/s].
+0)26-4
Reciprocating Pump 10.1 INTRODUCTION A reciprocating pump is a piston and cylinder arrangement and the liquid follows the movement of piston during suction and delivery strokes. It is a positive displacement pump. These are suitable for small capacities and high heads. The capacity is a function of speed and not head. Fig. 10.1 shows the regions of applications as compared to centrifugal and axial flow pumps. H(m) 10,000 Reciprocating 1000 100
Centrifugal
10 Axial Flow 0 10
100 Q
Fig. 10.1
1000
10000
Log scales
3
(m /hr)
Capacity-head region
It consists of the following: 1. A piston moves backward and forward in a closely fitted cylinder. 2. Suction and delivery pipes with non-return valves. 3. Crank and connecting rod mechanism operated by a prime mover (engine or electric motor). When piston moves outward, low pressure is formed in the cylinder left space. The suction valve opens and the liquid is sucked into the cylinder from the sump. When the piston moves inward, high pressure is developed in the cylinder. The suction valve closes and delivery
Reciprocating Pump
175
valve opens. The liquid is forced into the discharge tank through delivery pipe. The suction and delivery strokes are executed alternately and the liquid is pumped from sump to delivery tank. Delivery pipe
Cylinder
Suction pipe
Piston
Fig. 10.2 Pump operation
10.2 PUMP INSTALLATION AND PERFORMANCE The pump layout is shown in Fig. 10.3. Static head,
H = HS + Hd (m)
Piston area,
A=
p 2 2 D (m ) 4
Piston rod area,
a=
p 2 2 d (m ) 4
Piston stoke = L (m) Crank speed = N rpm For a single acting pump, the liquid discharge, Q=
ALN [m3/s] 60
The equivalent power output, Po = rQgH [W] =
rg ALN × (HS + Hd) [kW] 1000 60
The overall pump efficiency, hoverall =
P rQgH = o P 3EI cos f E
176 Hydraulic Machines Discharge tank
Delivery pipe
Delivery Hd head
D Suction load
Hs
Suction valve
M
Delivery valve Stuffing box Piston Cross-head d
Piston L Cylinder rod
Connecting rod
Belt drive Bearing N Crankshaft
Suction pipe Sump Fig. 10.3
where
Pump layout
PE = Electrical power input to motor.
Force acting on piston during forward stroke FS = rgHS A [N] Force acting on piston during backward stroke Fd = rgHd A [N] The rate of delivery of pump, Q = AS VS [m3/s] where
AS = area of suction pipe [m2] VS = velocity of the liquid in suction pipe [m/s]
\
Q = AS wr sin q
\
Q = f (sin q)
A AS
Note In double acting pump. Q=
ALN ( A - a ) LN + [m3/s] 60 60
=
LN (2A a) 60
»
2 ALN 60
(3 a << A and may be neglected)
Reciprocating Pump Suction
177
Delivery
Q
0
180° q
360° Crank angle
Fig. 10.4 Liquid delivery
Po =
rQgH [kW] 1000
The force on piston during forward stroke, FS = rgHS A + rgHd (A a) The force on piston during backward stroke, Fd = rgHS (A a) + rgHd A
10.3 CLASSIFICATION OF RECIPROCATING PUMPS The flow from a single-cylinder and single acting reciprocating pump, working with simple harmonic motion (Fig. 10.4) is of pulsating and intermittent nature. However, it is desirable to have a continuous flow both during the suction and delivery strokes. This is accomplished by using double acting pump, multicylinder pump and pump fitted with air vessels. Accordingly, reciprocating pumps are classified as: 1. Single cylinder pump (i) Single acting (ii) Double acting 2. Two-cylinder pump or two-throw pump 3. Three-cylinder pump or three-throw pump 4. Pump without air vessel 5. Pump with air vessel The schematic diagrams and performance curves of various types of pumps are shown in Fig. 10.5. 1. Single cylinder single acting pump.
178 Hydraulic Machines
q
Suction
Delivery
Q
0°
180° Crank angle q
360°
Fig. 10.5 (a) Single cylinder single acting pump
Single cylinder double acting pump. Suction
Delivery
Q
0°
Fig. 10.5
180° Crank angle q
360°
(b) Single cylinder double acting pump
Two-cylinder double acting pump.
180°
Q
0
90°
180° 270° Crank angle q
Fig. 10.5 (c) Double cylinder double acting pump
360°
Reciprocating Pump
179
Three-cylinder double acting pump.
Q 120° 120
°
120°
0
Fig. 10.5
60°
120° 180° 240° Crank angle q
270°
360°
(d) Three cylinder double acting pumps
Pump with air vessels. Air vessel in delivery line Q
Air vessel is suction line
0°
Fig. 10.5
180° q
360°
(e) Pump with air vessels
10.4 PISTON MOTION After the pump has been primed, the liquid follows the piston closely due to forces of cohesion and adhesion without separation. The acceleration of piston will impart head to the liquidalso called acceleration head. The piston displacement, x = r r cos q where
q = crank angle r = crank radius
180 Hydraulic Machines Also
q = wt
where w = Angular velocity [rad/s] t = time (seconds) \
x = r r cos wt y l r o
q
w x
x x
Fig. 10.6 Piston motion
The velocity of piston, Vp =
dx = 0 r(sin wt)w dt
= wr sin wt The acceleration of piston, ap =
d2x = w2r cos wt dt 2
Flow of liquid in suction pipe = Flow of liquid in the cylinder Q = VS AS = Awr sin wt where
VS = Velocity of liquid in suction pipe (m/s) AS = Area of suction pipe [m2] A = Area of piston [m2]
Acceleration of liquid in suction pipe, as = =
A 2 w r cos wt AS A 2 w r cos q AS
Accelerating force = Mass ´ acceleration = rAS lS ´
A 2 w r cos q [N] AS
Reciprocating Pump
181
where lS = length of suction pipe [m] Pressure,
pS = r
A A 2 ×lS w r cos q [N/m2] AS AS
Acceleration pressure head, HaS =
A 2 r A lS w r cos q [m] AS rg AS
HaS =
lS A 2 w r cos q g AS
...(1)
At the beginning of stroke, q = 0 \
HaS =
lS A 2 × wr g AS
At the mid of stroke, q = 90° HaS = 0 At the end of stroke, q = 180° HaS = \
(HaS)max =
lS A 2 × wr g AS
lS A 2 wr g AS
Maximum acceleration head occurs when the piston is at dead centres. The length of connecting rod has been neglected, being very long.
10.5 INDICATOR DIAGRAM The work input to the pump can be calculated from the indicator diagram which shows the pressure of liquid in the pump cylinder corresponding to crank position during suction and delivery strokes. The piston of the indicator moves up and down against the spring pressure as the cylinder pressure of the pump varies. The pencil moves with the indicator piston. The drum carrying the graph paper wrapped on it oscillates by reduction motion proportional to pump piston stroke. The area of indicator diagram = work done per revolution for single acting pump. (The area of indicator diagram) ´ 2 = work done per revolution for double acting pump W.D. = rgQH = =
rgALN (HS + Hd) 60
rgAN [L(HS + Hd)] 60
182 Hydraulic Machines Pencil
F
Fulcrum
Drum
Reduction motion
Piston
y Delivery stroke Pressure head (m)
Ha = 10 m
Hd Atmospheric pressure
Hs Suction stroke
x
Stroke L
Fig. 10.7
Indicator diagram
rgAN ´ area of indicator diagram. 60 The area of indicator diagram is proportional to work done for running the pump. The indicator diagram shown above is for ideal operating conditions. In actual operation, the performance as well as indicator diagram are affected by various factors. W.D. =
10.6
FACTORS AFFECTING PERFORMANCE OF RECIPROCATING PUMPS
The actual performance of reciprocating pump is affected by the following factors: 1. Slip 2. Acceleration head of liquid in suction and delivery pipes 3. Friction losses 4. Liquid separation 5. Air vessels
Reciprocating Pump
183
10.6.1 Effect of Slip of Liquid The actual liquid delivered by the pump is usually less than theoretical piston swept volume due to the following reasons. 1. Leakage through the valves, glands and piston rings. 2. Delay in closing of suction and delivery valves. 3. Liberation of air from the liquid being handled. The difference in volume delivered from theoretical volume is called slip. Slip = [Volume swept/stroke] [Actual discharge/stroke] The coefficient of discharge, Cd =
Percentage slip =
(Actual discharge/stroke) (Volume swept/stroke)
FG Q - Q IJ ´ 100 H Q K th
act
th
= (1 Cd) ´ 100 Sometimes by adjusting the valve timing (delivery valve opens before suction stroke is completed), Cd can be more than unity. The pump would be working with negative slip and actual discharge will be more than theoretical discharge.
10.6.2 Losses of Head in Suction Pipe In order to calculate losses in the suction pipe of the pump, apply Bernaullis equation at points (1) and (2), in Fig. 10.8. H1 = H2 + HS +
V22 + HaS + 2g
å DHl
where H1 = pressure at top of suction level = 1 atm. = 10 m. H2 = Pressure in the cylinder which should be more than vapour pressure Hv, (~ 2.5 m at normal liquid temperature) HS = Suction height V22 = Velocity head 2g
FG A wr cos qIJ HA K = S
2g
2
184 Hydraulic Machines
2 Hs 1
Fig. 10.8
Suction losses
HaS = Acceleration suction head due to inertia forces = HaS =
lS A 2 × w r cos q g AS lS A 2 w r, g AS
i.e. the maximum head at the beginning and end of suction stroke.
= 0 at q = 90°, i.e. at the mid-stroke
å DHl = sum of losses = Pipe friction + Friction losses in the valves and other fittings + Head required to open the valves. 4 flSVS2 4 flS = Pipe friction = 2 gd S 2 gd S 4 flS = 2 gd S
FG A wrIJ HA K
FG A wr cos qIJ HA K
2
S
2
at q = 90°, i.e., maximum
S
= 0 at beginning and end of suction stroke. The indicator diagram including all losses and acceleration head is shown in Fig. 10.9.
Reciprocating Pump Friction head
185
(Hfd)max
Pressure head (m) Had
Hd + Had
Hd
Hd-Had Hatm
Hs – Has
Hs HST -Has Has (Hfs)max
Hv q=0
q = 90°
q = 180°
Stroke L
Friction head
Fig. 10.9 Effect of friction and acceleration on indicator diagram
The pressures in the cylinder are function of cos q and are tabulated below. Table 10.1 Sl. No.
Pressures in the cylinder
Stroke length
Suction stroke
Delivery stroke
1.
Beginning, q = 0°
2.
Middle, q = 90°
3.
End, q = 180°
Hatm (HS + HaS) HfS = 0 Hatm (HS + HfS) HaS = 0 Hatm (HS + HaS) HfS = 0
Hatm + (Hd + Had ) Hfd = 0 Hatm + (Hd + Hfd ) Had = 0 Hatm + (Hd + Had ) Hfd = 0
10.6.3 Liquid Separation Whenever liquid pressure falls below its vapour pressure, liquid boiling and separation take place causing cavitation. Refer Fig. 10.8 H2 ³ Hv where Hv = Vapour pressure H2 = H1 HS
V22 HaS 2g
å DHl
H1 is a function of altitude which is fixed for a site. H2 can be increased by the following methods.
186 Hydraulic Machines (i) Reducing HS, i.e., low suction height (ii) Reducing
V22 , i.e., large suction pipe diameter and small piston speed. 2g
(iii) Reducing HaS, i.e., small suction pipe length and small piston speed. (iv) Reducing SDHl, i.e., small suction pipe length and minimum number of bends and pipe fittings. 1. Maximum pump speed
The reciprocating pumps are designed for low speed and are, therefore, driven through reduction gear or belt drive by electric motor and not directly coupled. The maximum acceleration head, (HaS )max =
lS A 2 × wr g AS
The pressure head at the beginning of suction stroke = Hatm HS (HaS )max ³ Hv \
HS + (HaS )max £ (Hatm Hv) Hatm = 10.3 m
\
10.3 (HS + HaS ) £ 2.5
or
(HS + HaS ) £ 7.8 HS is constant for a pump installation.
LM F A I OP MN GH A JK , w rPQ 2
HaS = f lS ,
2
S
lS, AS, A and r are constants for a pump installation \
HaS = f (w2) HaS = f(N )
N should be small for low value of HaS . 2. Pipe layout
The friction losses depend upon the piping layout. The pipes should be laid straight without bends. 3. Air vessels
Air vessel is a closed chamber fitted on suction and delivery pipes, near the pump cylinder, to reduce acceleration head (inertia) of liquid column.
Reciprocating Pump
187
The air vessel on suction pipe helps: (i) to reduce possibility of separation (ii) to operate on high possible speeds (iii) to increase length of suction pipe The air vessel on delivery pipe helps: (i) saving of power due to less inertia head (ii) to ensure constant flow rate 4. Direct acting steam pump
Reciprocating pumps are normally driven by electric motors through reduction gears or v-belt drive. In boiler houses of process industries where steam is available the boiler feed pumps may be directly connected to a steam engine. Direct acting steam pump is more economical and reliable than electric motor driven pump. The crankshaft, flywheel, etc. are completely eliminated. The pump duty is defined as: Pump duty =
rQgH [kJ/kJ] m S hS ´ 103
where m S = steam flow rate [kg/s] hS = enthalpy of steam [kJ/kg] Steam
Feed water
Fig. 10.10 Direct driven pump
10.6.4 Effect of Air Vessels It has been found that use of air vessels along with reciprocating pumps ensures constant flow rate and reduces the possibility of liquid separation. 1. Air vessel on suction side
When air vessel is used, suction takes place in two steps: (i) Liquid flows through suction pipe into air vessel. (ii) Liquid is raised from air vessel to cylinder.
188 Hydraulic Machines H v2 ld Hd
H dv
H sv
ldv
Hv1 lsv
Hs
ls
Fig. 10.11 Pump with air vessels Hfd
Hfd
Hd
Hd
Atmospheric pressure
Hs
Hs
Hfs
Hfs
(a) Without air vessels
Fig. 10.12
(b) With air vessels
Indicator diagrams
The air vessel acts like a flywheel. The top of vessel contains compressed air which can contract or expand to absorb most of the pressure fluctuations. The friction head loss is not a parabola but a rectangle. There is no acceleration head in the pipe except in short pipe length lSv. 2. Air vessel on discharge side
During delivery strokes, suction valve is closed and flow takes place from air vessel on delivery side which was filled in the previous stroke. Here also the friction head loss is constant and acceleration head takes place in short pipe length, ldv. When air vessels of sufficient capacity are installed on the suction and delivery pipes, the flow velocity in suction and delivery pipe can be assumed to be uniform and equal to mean flow velocity.
Reciprocating Pump
The discharge, Q = Piston area ´ stroke length ´ speed =A´L´ = A ´ 2r
N 60
w w = Ar 2p p
The flow velocity in pipes, V=
Q A wr = AS AS p
Loss of head due to friction HfS =
4 flS v 2 4 flS = 2 gd S 2 gd S
LM A wr IJ NA p K
2
S
Area of indicator diagram = base ´ height, =L´
LM N
4 f (ls + ld ) A wr 2 gd S AS p
OP Q
2
Power expanded, Pi =
LM N
rgAN 4 f (ls + ld ) A wr ´L 2 gd S AS p 60
OP Q
2
Power expanded in pumps without air vessel,
LM OP N Q rgAN 4 f (l + l ) L A wr O ´ MN A p PQ 60 2 gd = O rgAN 2 4 f (l + l ) L A wr P ´ ´ M 60 3 2 gd NA Q
rgAN 2 4 f (ls + ld ) A ´ ´L´ wr Pi¢ = AS 3 2 gd S 60 2
S
Therefore,
Pi Pi ¢
d
S
S
S
S
=
3 2p2
2
d
S
2
189
190 Hydraulic Machines Saving in power expanded due to air vessels, =
FG H
3 Pi ¢ - Pi P = 1 i = 1- 2 Pi ¢ Pi ¢ 2p
IJ = 0.848 = 84.8% K
QUESTION BANK NO. 10 1. Explain the working of a reciprocating pump. 2. With the help of a suitable layout diagram, explain the performance of a reciprocating pump. 3. How is the pulsating flow of a reciprocating pump converted into continuous flow? Explain the different methods. 4. Explain the formation of acceleration head with the help of piston motion. 5. Explain the function of air vessels in a reciprocating pump. 6. What is an indicator diagram of a reciprocating pump? Sketch the theoretical indicator diagram for a single acting reciprocating pump not fitted with air vessels. With the help of this diagram, explain clearly the effect of acceleration and friction on both suction and delivery strokes. 7. Explain how the indicator diagram is modified if air vessels are provided on both suction and delivery pipes. 8. What is liquid separation? How it can be controlled in reciprocating pumps? 9. Show that the maximum acceleration head in a reciprocating pump without air vessel is given by: Ha =
l A 2 ´ w r with usual notations g AS
TUTORIAL SHEET NO. 10 1. A single acting reciprocating pump has a piston diameter 15 cm and stroke length 30 cm. The centre of the pump is 5 m above the level of water and 33 m below delivery water level. The length of suction and delivery lines are 6.5 m and 39 m respectively and both pipes have same diameter of 7.5 cm. If the pump is working at 30 rpm, find the pressure head on the piston at the beginning, middle and end of both suction and delivery strokes, also find the power required to drive the pump. Assume atmospheric pressure as 10.3 m of water and friction factor of 0.01. Solution The piston area, A=
FG IJ H K
p 15 4 100
2
= 0.0176 m2
Reciprocating Pump
L = 0.30 m HS = 5 m Hd = 33 m lS = 6.5 m ld = 39 m The area of suction and delivery pipes, AS =
p (0.075)2 = 4.4 ´ 103 m2 4
N = 30 rpm w= The crank radius, r =
2 pN 2p ´ 30 = = p rad/s. 60 60 L = 0.15 m 2
Hatm = 10.3 m HaS = (HaS)max = =
(HfS)max =
lS A 2 w r cos q g AS lS A 2 × wr g AS
6.5 0.01767 2 ´ p ´ 0.15 = 3.94 m. 9.81 4.4 ´ 10 -3 4 flS 2 gd S
FG A wrIJ HA K
2
S
LM N
OP Q
0.01767 4 ´ 0.01 ´ 6.5 . = ´ p ´ 015 2 ´ 9.81 ´ 0.075 4.4 ´ 10-3
2
(HfS)max = 0.63 m Hd = 33 m (Had)max =
F GH
39 0.01767 ld A 2 × w r= 9.81 4.4 ´ 10 -3 g Ad
= 23.64 m
Ip JK
2
´ 0.15
191
192 Hydraulic Machines 4 fld (Hfd)max = 2 gd d
FG A wrIJ HA K d
2
LM N
OP Q
4 ´ 0.01 ´ 39 0.01767 = . ´ p ´ 015 2 ´ 9.81 ´ 0.075 4.4 ´ 10-3
2
= 3.78 m Stroke length
Suction stroke
Delivery stroke
Beginning, q = 0
Hatm (HS + HaS ) = 10.30 (5 + 3.94) = 1.36 m
Hatm + (Hd + Had) = 10.3 + 33 + 23.64 = 66.94 m
Middle, q = 90°
Hatm (HS + HfS) = 10.3 (5 + 0.63) = 4.67 m
Hatm + (Hd + Hfd) = 10.3 + 33 + 3.78 = 47.08 m
End, q = 180°
Hatm (HS HaS) = 10.3 (5 3.94) = 9.24 m
Hatm + (Hd Had ) = 10.3 + 33 23.64 = 19.66 m
Total head delivered. H = Static head + friction head losses + acceleration heads = (HS + Hd) + (HfS + Hfd ) + (HaS + Had) = 5 + 33 + (0.63 + 3.78) + (3.94 + 23.64) = 69.99 m The power required to drive the pump, Pi = rQgH = =
rg ALN H 1000 60
9810 30 ´ 0.0176 ´ 0.3 ´ ´ 69.99 = 1.82 kW Ans 60 1000
A single acting reciprocating pump runs at 30 rpm. The diameter of plunger is 15 cm and crank radius is 15 cm. The suction pipe is 10 cm in diameter and 9 m long. Calculate the maximum permissible value of suction lift HS, if the separation takes place at 2.6 m of water. Take atmospheric pressure head as 10.3 m of water. Solution The given data: N = 30 rpm; w = A=
2 pN 2p ´ 30 = = p rad/s 60 60
p (0.15)2 = 0.01767 m2 4
193
Reciprocating Pump
r = 0.15 m ds = 0.1 m AS =
p (0.1)2 = 7.85 ´ 103 m2 4
lS = 9 m Hv = 2.6 m Hatm = 10.3 m Hatm = Hv + HS +
V22 + HaS + 2g
Neglect
V22 and valve losses 2g
\
Hatm = Hv + HS + HaS + HfS = 2.6 + HS +
å DHl.
lS 2 A 2 flS wr´ cos q + g AS 2 gd s
FG A wr sin qIJ K HA
2
S
At q = 0, i.e., beginning of strokes 10.3 = 2.6 + HS +
0.01767 9 2 p ´ 0.15 ´ +0 9.81 7.85 ´ 10 -3
= 2.6 + HS + 3 \
HS = 4.7 m Ans
A single acting reciprocating pump has a plunger of 18 cm diameter and a stroke of 20 cm and supplies water from a sump 4 m below the pump through a pipe 6.5 m long and 5 cm in diameter. It delivers water to a reservoir 15 m above the pump through a pipe 3.5 cm in diameter and 16 cm long. If separation occurs at 8 cm of water, find the maximum speed at which the pump may be operated without separation. Solution The given data: A=
p ´ 0.182 = 0.02545 m2 4
lS = 6.5 m ds = 0.05 m AS =
p ´ (0.05)2 = 1.963 ´ 103 m2 4
194 Hydraulic Machines HS = 4 m Hd = 15 m dd = 0.35 m ld = 16 m Hv = 0.08 m At the beginning of suction stroke Hatm = HS + (Ha)max + Hv (Neglecting friction head and velocity head) = HS + 10.3 = 4 +
lS A 2 × w r + Hv g AS
F GH
6.5 0.02545 ´ 9.81 1963 . ´ 10 -3
I FG 2pN IJ JK H 60 K
2
×
0.2 + 0.08 2
10.3 = 4.08 + 0.009417N 2 \
N = 25.7 rpm
At the end of delivery stroke, Hatm = Had Hd + Hv 10.3 =
FG IJ FG 2pN IJ H K H 60 K
16 018 . 9.81 0.35
2
2
+
0.2 15 + 0.08 2
10.3 = 0.0004N 2 15 + 0.08 \ \
N = 251 rpm Maximum speed = 25.7 rpm Ans.
The bore diameter of a double acting twin cylinder pump is 10 cm and the length of stroke is 15 cm. When running at 60 rpm, it discharges 4.5 l/s and has a total lift of 10 m. What is the percentage of slip? If the required liquid pumped has a specific gravity of 0.96, what is the power required assuming efficiency of the unit to be 60%? Solution A=
p (0.1)2 = 7.854 ´ 103 m2 4
L = 0.15 m N = 60 rpm; w =
2 pN = 2p rad/s 60
Reciprocating Pump
195
Q = 4.5 l/s = 4.5 ´ 103 m3/s H = 10 m Slip = (Volume swept/stroke) (Actual discharge/stroke) =
2 ALN ´ 2 4.5 ´ 103 60
= 2(7.854 ´ 103) ´ 0.15 ´
60 ´ 2 4.5 ´ 103 60
= 0.0471 0.045 = 0.0021 m3/s % age slip =
0.0021 ´ 102 = 4.45% Ans. 0.0471
Power required, Pi =
960 ´ 4.5 ´ 10 -! ´ 9.81 ´ 10 r3C0 = 1000 ´ 0.6 ´ h
= 0.70 kW Ans 5. A single acting reciprocating pump operating at 120 rpm has a piston diameter of 200 mm and stroke of 300 mm. The suction and delivery heads are 4 m and 20 m. If the efficiency of both suction and delivery strokes is 75 per cent, determine the power required by the pump (6 kW).
+0)26-4
Special Pumps 11.1
INTRODUCTION
The centrifugal pumps and reciprocating pumps are most widely used pumps. But there are special pumps designed for specific applications. The following pumps have been selected for discussions. 1. Axial flow pumps These pumps include vertical turbine pump, deep well pump, submersible pump, borehole pump, etc. 2. Rotary displacement pump These pumps include gear pump, lobe pump, vane pump, screw pump. liquid ring pump. 3. Jet pumps These pumps include air lift pump and water jet pump. 4. Fluid coupling These include fluid coupling and torque converter.
11.2
AXIAL FLOW PUMP
Axial flow pumps or propeller pump work as converse of a propeller turbine. Pressure is developed by flow of liquid over blades of airfoil section. If Kaplan turbine runner with adjustable blades is used, the pump may be called Kaplan pump. These pumps are designed for large capacities and low heads below 1.2 m. These pumps have high specific speeds. For the sake of compactness, these pumps are usually designed with vertical shaft disposition. These are used for land drainage, flood control, storm water disposal, irrigation, condenser circulation pump in power plants and other low head pumping jobs. 1. Design constants
The design procedure is similar to that for propeller/Kaplan turbine. The pump constants are:
Special Pumps
197
Discharge Discharge elbow Propeller runner
d D
Fig. 11.1
Axial pump
Vb
Speed ratio,
f=
Flow ratio,
y=
Diameter ratio,
d = 0.4 to 0.8 D
Hydraulic efficiency,
hh = 0.75 to 0.92
Vf 2 gH
= 2 to 3
= 0.4 to 0.8
hvol = 1
Volumetric efficiency,
hm = 0.94 to 0.98
Mechanical efficiency, Overall efficiency,
2 gh
hoverall = hh ´ hm = 0.70 to 0.90
The maximum lift £ 1 m The discharge,
Q=
p 2 (D d2)Vf 4
The driving power,
Pi =
rQgH [kW] 1000 ´ hoverall
Characteristic curves
The characteristic curves of propeller pumps are shown in Fig. 11.2. (i) Q-H curve is saddle shaped, which is obtained due to vane lift being reached at low capacities and increased angle of attack. Low pressure pumps have constantly drooping characteristic.
198 Hydraulic Machines N = Const. B
Pi(kW) A C
hmax
H(m)
h(%) Operating region Qmin
Qmax 3
Discharge, Q (m /sec)
Fig. 11.2
Characteristic curves
(ii) Power curve is approximately horizontal initially and then drops with increase of flow. Axial pumps are started on load with delivery valve open. (iii) Efficiency curve has a pronounced peak. The efficiency changes abruptly when operated away from optimum design conditions. 3. Operating region
The operating region is selected within the steady portion of the characteristic curve (A C) to the right of peak B. The minimum discharge, Qmin corresponds to (HA)max = 0.9 HB. The maximum discharge, Qmax corresponds to hc = 0.9 hmax. 4. Capacity control
The capacity control of axial pumps can be affected by: (i) Speed regulation, i.e., most efficient (ii) Throttling is the most unfavourable economical operation as power grows as capacity is reduced (iii) Adjustable inlet guide vanes where driving power can be used economically (iv) Adjustable impeller vane is also power efficient control.
11.3
DESIGN OF AXIAL FLOW PUMP
In an axial flow pump, the energy is carried from the shaft to the flow by an impeller consisting of cantilever vanes (2 to 6 in number) fixed on the hub. The performance of axial flow pump is considered using the theory of cascades of profiles. In view of the very low head produced by an axial flow impeller, the skin friction loss or drag acquires a greater importance than in centrifugal or mixed flow impellers. For that reason, a high degree of impeller vane streamlining and polishing is required to obtain the optimum peak efficiency. To satisfy this condition and that of mechanical strength, the impeller vane profiles take the form
Special Pumps
199
of airfoils. Although developed primarily for the airplane supporting wing application, airfoils have found wide use in the field of axial flow pumps and turbo machines. A great many airfoil sections have been tested and classified according to their curvature and thickness.
11.3.1
Airfoil Section
Airfoils are considered as made up of a certain profile thickness form disposed about certain mean line as shown in Fig. 11.3. The form of the mean line determines completely most of the important hydraulic properties of airfoils, whereas thickness is dictated by the strength requirements. All good airfoils have nearly the same thickness variation along the mean line, the maximum thickness being different for different profiles. The maximum distance from the chord to the mean line is called camber (C) and is usually expressed in percentage of the chord length or C/l. Y
Camber yt 11 11
C
Mean li ne
Chord
X
l Fig. 11.3 Airfoil section
11.3.2
Definitions
Each N.A.C.A. profile is designated by a four digit number. Example. 4312. where 4 = Camber of the mean line 3 = Location of the camber four leading edge in lengths of chord lengths. 12 = Maximum vane thickness in percentage of the chord (l ). 1. Angle of attack (=)
It is an acute angle between the vane chord and direction of relative velocity of the flow (Vra ).
200 Hydraulic Machines
D
ba
bc
(90° – ba) P l
Vra
L
Fig. 11.4
X
Forces acting on airfoil
Aspect ratio
It is the ratio of the length of airfoil and the length of the chord. In a pump, liquid is confined between the hub and the casing sidewalls and is not free to escape radially, an aspect ratio of infinity is used for axial flow pumps. 3. Lift (L)
It is the component of the force acting on the airfoil normal to the direction of the approaching undisturbed flow. L = CLblr
where
Vra2
2 CL = experimental coefficient of lift b = width of airfoil l = length of chord of airfoil r = density of fluid Vra = undisturbed relative liquid velocity
Drag (D)
It is the component of the force acting on the airfoil in the direction of flow. D = CDblr
Vra2
2 where CD = experimental coefficient of drag. CL and CD depend upon the profile of airfoil, angle of attack a and aspect ratio. CD is very small as compared to CL. These values have been experimentally determined for a large number of profiles. 5. Relation between lift and drag
The airfoil efficiency, hairfoil =
L D
Special Pumps
201
The gliding angle, l=
CD CL
The airfoil efficiency is more for thin airfoils. The maximum value of CL is obtained with 12% airfoil thickness. If CL stands for head, a standards for capacity and L/D stands for efficiency, then the curves between CL and a resemble the ordinary head-capacity curve of an axial flow pump.
11.3.3
Airfoil Theory
The airfoil theory of axial flow impeller design establishes a connection between the lift coefficient of airfoil test data and impeller total head. The design procedure consists of: 1. Selection of the section of suitable airfoil properties for several radii of impeller. 2. Determination of the vane setting on the hub or chord angle bc for each section. 3. Assume constant head for several streamlines. 4. Establish vane curvature based on airfoil experiment knowledge. Impeller design is entirely experimental.
11.3.4
Limitations of Airfoil Theory
The airfoil theory fails to help in the selection of the following design elements of the impeller and are selected on the basis of previous experience only. 1. Hub ratio 2. l/t ratio at each radius 3. Speed v/s specific speed. 4. Axial velocity 5. Impeller diameter. Presently airfoil theory has become obsolete and the method of rotating profiles in suitable casing is used for the design of axial flow pumps. New design methods based on inlet and outlet velocity triangles have been evolved for the design of axial flow impeller.
11.3.5
Total Head Equation
If a cylindrical cut is made through an axial flow impeller and the cylinder is developed onto plane, a row of vane profiles will result. The action of liquid on the profile can be considered similar to that taking place on an airfoil in a wind tunnel,
202 Hydraulic Machines provided the relative velocity Vra is an average velocity of approach and discharge Va1 and Va before and after the vane at a distance whose effect of the flow through
the row of vanes is equalized. The effect of the cascade arrangement of vanes on CL is neglected. The total force (P exerted by the liquid per unit wavelength is the resultant of lift (L and drag (D and makes an angle l with the direction of lift. P= tan l =
L2 + D 2
D L
The total force P forms an angle [90° (ba + l ] with the direction of peripheral velocity (Vb of the impeller. The tangential component, Pw = P cos [90° (ba + l ] Pw = P sin (ba + l
\
Work done per second, W.D. = VbP sin (ba + l Consider a slice of the vane between two consecutive cylindrical surfaces of radius r and r + dr, the work per second for Z vanes is (Fig. 11.5 Z (W.D. = ZP dr Vb sin (ba + l If dQ is the volume of liquid included between the two cylindrical surfaces, the work per kg of liquid is W.D./kg =
Z (W . D. ZP drVb sin (ba + l = dQ r dQr
dQ = Zt dr Vf P=
L= where \
L cos l CL r Vra2 × A 2g
A = l ´ 1 = l = area per unit length P=
C L rl Vra2 2 g cos l
Special Pumps
W.D./kg =
CL
203
CL lr Vra2 sin (b a + l t × V f 2 g cos l
l 2g V f cos l = (W.D./kg 2 × × t sin (b a + l V Vra b
...(1
The theoretical head, Ht = where
H hh
hh = Hydraulic efficiency
Vra and ba are obtained from velocity diagram.
where Vb = Vb dr r
Fig. 11.5 Axial flow pump section Vr1
a1
Vf1 = Vf2
Va1
b1 Vb1
Vw1
Fig. 11.6
Vr2 b2
a2 Vw2
Vb2
Velocity diagrams
The average tangential component = Vw ` Vw =
Va2
Vw - Vw
gHt hh
Assume Vw = 0 and Vw is obtained from velocity triangle. l will be low for high values of Vb and Vra . t The large radii l/t and CL will be lower than those values at the hub.
From equation (1 , the value of CL
204 Hydraulic Machines For higher heads, higher values of CL and l/t are required, i.e., more vane area and a higher camber of profile. The values of CL
FG l IJ are calculated for different values of impeller radii using H tK
equation (1 , using constant values of Ht and Vf along radius. Experience is necessary to select proper values of l/t to obtain CL and vane profile in order to produce an efficient impeller and head-capacity and efficiency curves of good shape. Lift and Head Coefficients Let
cos l = 1 (3 l » 1° Vf Vr
= sin ba
Ht =
Vb Vw g
Substitute these values in equation (1 . \
11.4
CL
l 2Vw2 2(b 2 - b1 ) = = Vr2 t sin b a
VERTICAL TURBINE PUMP
It is a multistage pump installed in vertical position to save space. The type of impeller, i.e., closed, semi-closed or open, depends upon the properties of the liquid being pumped. The impellers can be centrifugal (radial), mixed flow or propeller type depending upon the head and discharge requirements and the specific speed of the pump. The impellers work as reverse of reaction turbine impeller. The main components are shown in Fig. 11.7. The liquid is sucked through a strainer and foot valve and delivered through vertical column pipe. The bearing can be oil lubricated or water lubricated. The vertical turbine pumps are used for pumping water from the sea, river or deep lake. These are also used to pump underground water as deepwell or borewell pumps. In order to overcome the difficulty of limitation of suction lift due to NPSH requirements, the pump and suction pipe are immersed in water. A submersible pump is also a vertical turbine pump where the electric motor is lowered along with pump inside the column pipe. The motor is designed water proof and avoids the need of long line shaft. The hydraulic performance is similar to multi-stage centrifugal pumps.
Special Pumps
205
Motor
Support frame Discharge valve Thrust hearing
Bearing Column pipe Shaft
Pump Suction pipe Suction strainer + Foot valve
Fig. 11.7
11.5
Vertical turbine pump
ROTARY DISPLACEMENT PUMPS
The centrifugal pumps give uniform discharge and reciprocating pumps give high pressure and positive displacement of liquid. The rotary positive displacement pumps have more or less both the advantages and are used for small flows and relatively high pressures. These pumps are primarily used as power sources in the hydraulic control systems and oil lubrication systems for turbines, motors and machine tools. The pumping elements one or more in number are directly coupled to the driver and rotate in a stationary casing. These pumps are very compact and have the self-priming ability. However, these pumps are slow speed machines. There are many designs of rotary pumps. However, gear pump, lobe pump, sliding vane pump and screw pump are popular designs. The operating characteristics of main rotary pumps are compared in Table 11.1. Table 11.1 S. No. 1. 2. 3. 4.
Operating characteristics of rotary pumps
Type of pump
Max. capacity
Max. pressure
Max. speed
Gear pump Screw pump Sliding vane pump Lobe pump
200 l/min. 300 m3/hr. 500 l/min. 100 l/min.
5 MPa 20 MPa 20 MPa 10 MPa
1000 rpm. 8000 rpm. 500-1200 rpm. 1000 rpm.
206 Hydraulic Machines 11.5.1
Gear Pump
The pump consists of two identical intermeshing spur gears rotating in a casing with small clearance. The drive gear is connected to the driver whose shaft extends into the casing through a stuffing box. The other gear is idle. The gears rotate in the direction as shown. The liquid is sucked, trapped between the gear teeth and discharged.
Fig. 11.8
Gear pump
The theoretical volume displaced by the pump per second,
N [m3/s] 60 where L = Axial length of teeth in each gear wheel [m] A = Area enclosed between two adjacent teeth and casing [m2] n = Number of teeth in each gear wheel. N = Speed [rpm] Actual discharge, Qth = (2 LAn)
Qa = Qth ´ hv where hv = volumetric efficiency [0.7 to 0.9]
11.5.2
Lobe Pump
A lobe is rotated in a close fitting casing with inlet and outlet ports. The rotary motion of the lobe delivers the entrapped liquid into discharge receiver. The high pressure liquid is delivered from the receiver. The volume displaced by the pump per second, Q = 2pRB
N hv (m3/s) 60
Special Pumps
207
Fig. 11.9 (a) Lobe pump
h
Pressure ratio, r
Fig. 11.9 (b) Lobe pump efficiency
where R = inner radius of casing [m] B = width of lobe (m) N = speed [rpm] There are twin lobe pumps working in the same way as the gear pumps. hv = volumetric efficiency The volumetric efficiency falls with pressure ratio.
11.5.3
Sliding Vane Pump
The pump consists of an eccentrically mounted rotor in a cylindrical casing carrying inlet and outlet ports. The rotor carries spring loaded sliding vanes. The vanes moves to-and-fro in the slots of the rotor which is rotated by a driver. The volume of the liquid displaced by the pump per second Q = [2p(R e) nt] 2eB
N ´ hv [m3/s] 60
where R = Inner radius of casing (m) e = Eccentricity between rotor and casing (m) n = Number of vanes,
208 Hydraulic Machines
e t
Fig. 11.10 Vane pump
t = thickness of vanes (m) B = width of vanes (m) N = rotor speed (rpm) hv = volumetric efficiency [0.7 0.9]
11.5.4
Screw Pump
The screw pumps are used in control and lubrication system of large drivers. These pumps have the capacity to give uniform discharge, high pressure, and operate at high speeds and without noise. However, these pumps are very expensive as compared to gear pump, lobe pumps and vane pumps for the same output. The manufacture of screws of special profiles requires complicated technology. The pump consists of a driven worm screw meshing with two sealing worm screws rotating in a cylindrical casing with a suction and discharge chamber. As the screw rotates, the liquid is trapped in the screw flights and is moved in the axial direction and discharged.
Fig. 11.11 Screw pump
The capacity of the pump is computed as: Q=
D3 N h [l/s] 14,500 v
Special Pumps
209
where N = speed of the screw [rpm] D = diameter of worm screw [m]. hv = volumetric efficiency [0.7 0.95]
11.6
JET PUMPS
Jet pumps are used to lift water from deep well or sump by the use of energy of water or compressed air.
11.6.1
Water Jet Pump
It consists of a nozzle and a diffuser mounted on a mixing chamber. Water from a centrifugal pump is supplied to the nozzle, which converts pressure energy of water into kinetic energy at the throat of the diffuser. The sump water is connected to the mixing chamber where low pressure is generated. The sump water is lifted due to pressure difference. The water mixture recovers some pressure in the diffuser and discharged outside. This pump can lift water by 6 m height. The efficiency of jet pump. h=
Qs ( H s + Hd ) ´ Q1 ( H1 - Hd )
where Q1 = Discharge through nozzle (m3/s) Qs = Discharge through suction pipe (m3/s) H1 = Pressure head on delivery side (m) Hd = Delivery head (m) Hs = Suction head (m) The jet pump efficiency is low of the order of 50%. The main losses occur in the mixing chamber. The jet pump also finds applications in lubrication system of turbines where lubricating oil is pumped from oil tank to bearings of turbo-machines. Discharge
C. F. Pump
H1 Hd Diffuser Water level Q1 Hs
Qs
Nozzle Foot valve Strainer Sump
Fig. 11.12 Jet pump
210 Hydraulic Machines 11.6.2
Air Lift Pump
The air lift pumps are very simple and reliable. These are used to raise water from boreholes of small diameter and great depth. These pumps are very useful to pump corrosive liquids such as in chemical and food industries. Air lift pumps can also be used to raise liquids contaminated by sand, ash or peat. These pumps have low efficiency of the order of 20% to 40%. Compressed air is introduced at the bottom of liquid or water column. The density of liquid air mixture is less than the density of the liquid. The liquid-air mixture rises due to density differences. The flow rate depends upon the density difference of liquid-air mixture and the liquid. For best results, H 1 showed lie between and 1. h 4
where
h = submergence H = liquid-air mixture lift Air compressor
Liquid-air mixture
H
h Deep well Air nozzle
Fig. 11.13 Air lift pump
The main advantages of air lift pumps are: 1. The design is very simple due to absence of valves and moving parts. 2. Low maintenance due to absence of wear and tear. 3. Very high discharge capacity. 4. It can handle suspended solids in water safely. 5. Very suitable for drawing of water in mines where compressors are already existing.
11.7
HYDRAULIC RAM
Hydraulic ram is a pumping contrivance to raise a part of large amount of water available at some height, to a greater height. This can be employed in hilly areas
Special Pumps
211
where some natural source of water like a spring or a stream is available at some altitude. Work done by a large quantity of water in falling through a small height is used to raise a small part of it to a greater height. Action of water hammer makes it feasible. No external power is, therefore, required to work this machine. Other attractions are: 1. Negligible amount of maintenance and supervision costs. 2. Continuous operation. 3. High efficiency. 4. Quiet operation. 5. Possibility of automatic adjustment of water supply. The liftable volume diminishes asymptotically with lifting height. In case of medium lifting height, the hydraulic ram operates with an efficiency absolutely comparable to a piston pump of the same performance. As the hydraulic ram does not need a driving unit it is ideal for lift irrigation and drinking water for hilly areas where the supply in fossil energy carriers or electricity is problematic. The hydraulic ram operates with water streams of 1 to 40 m3/s and fall heads of 1.5 to 30 m and with lifting height upto 300 m. Hydraulic ram is a simple and rugged device for operating irrigation schemes. This can be easily assembled from local materials. Discharge tank H H2 Su
Supply tank
pp
ly
H1
Air vessel
Waste valve
Delivery valve
pip
e
Supply valve
Delivery pipe
Valve box
Fig. 11.14 Hydraulic ram
The water flows down the supply pipe and water escapes through waste valve. With increase of velocity of water, the dynamic pressure lifts the delivery valve which closes suddenly. The sudden stoppage of water increases pressure in the valve box and part of water enters the air vessel, thereby compressing the air. The air pressure forces the water to the delivery tank when momentum of water gets destroyed in the valve box, the delivery valve closes and waste valve opens to restart the cycle. Air vessel also helps to prevent the occurrence of water hammer in the delivery pipe and reduce fluctuations in flow rate.
212 Hydraulic Machines The energy supplied by water from supply tank = rQ1gH1 The energy supplied to discharge tank = rQ2gH2 where Q1 = water supplied to valve box (m3/s) Q2 = water lifted to delivery tank (m3/s) The D Aubissons efficiency, h=
QH rQ2 gH2 = 2 2 Q1 H1 rQ1 gH1
The Rankine efficiency, hRankine =
Q2 ( H2 - H1 ) × Q1 H1
where H1 = Initial level of water (m) (H2 H1) = Height of water lifted (m)
11.8
HYDRAULIC COUPLING
Hydraulic coupling is a combination of a pump and a turbine and is used to transmit power at constant torque. The pump unit is mounted on the driver shaft and turbine unit is mounted on the shaft of the driven machine. The pump impeller when started imparts kinetic energy to the oil and throws outside. The oil enters the turbine runner and the turbine and driven machine starts rotating. The torque increases with the increase of driver speed and the same is imparted to the driven machine through turbine runner. Finally, the speed and torque of both the machines are same. The oil circulates in a circular path. The oil is supplied from pump impeller to turbine runner and the same is returned from turbine runner to pump impeller. The efficiency of hydraulic coupling is defined as the ratio of power transmitted to the driven machine to the power developed by the pump impeller. h=
To N o Ti N i
where To = Torque input to turbine (N-m) Ti = Torque output of pump impeller (N-m) No = Speed of driven shaft (rpm) Ni = Speed of driver (rpm)
Special Pumps
213
Neglecting friction, To = Ti \
h=
No i.e., speed ratio Ni
The slip in the speed, S=
Turbine runner
Ni - N o N =1 o =1h Ni Ni 100%
Pump impeller
80%
Casing
Driven shaft
h
60% 40%
Driver shaft
20% 0% 0
(a) Coupling
0.2 0.4
0.6 0.8 N Speed ratio o Ni (b) Efficiency curve
1.0
Fig. 11.15 Hydraulic coupling
11.9
TORQUE CONVERTER
Torque converter is a modified hydraulic coupling where a reacting member is added to feed variable torque to the driven machine. These are mostly used for agumentation of torque. These are widely used in automobile power transmission unit, diesel locomotives, earth moving machinery. Large magnification of torque can be obtained by using one pump impeller but two or more sets of fixed guide vanes and turbine runners. Liquid flows from pump impeller to turbine runner through reaction chamber carrying stationary vanes which change the direction of the liquid and torque and speed transformation can be obtained. The efficiency of torque converter, h= where
FG H
T Power output (Ti + Tv ) N o N = = o 1+ v Ti Power input Ti N i Ni
Ti = Torque produced by pump (N-m) Tv = Torque of guide vanes (N-m)
IJ K
214 Hydraulic Machines No = speed of driven shaft (rpm) Ni = speed of driver (rpm) In fluid coupling, the guide vanes are absent. \ Tv = 0 h=
No Ni
No = 0.5. Ni Comparing the efficiency curves of torque converter and hydraulic coupling, it is clear that torque converters are more efficient at low speed ratio and hydraulic couplings are more efficient at high speed ratios. The transmission systems are so designed that the unit operates as torque converter at low speed ratios and hydraulic coupling at high speed ratios.
The efficiency of a torque converter is maximum when speed ratio
Turbine runner
Casing 100%
Reaction chamber
80% Pump impeller
Driven shaft
Driver shaft
h
60% 40% 20% 0% 0
(a) Torque converter
0.2 0.4
0.6 0.8 Speed ratio No Ni (b) Speed curve
1.0
Fig. 11.16 Hydraulic torque converter
QUESTION BANK NO. 11 1. Explain the following aspects of an axial flow pump: (i) Construction (ii) Design constants (iii) Performance characteristics (iv) Capacity control 2. How is airfoil theory used in the design of axial flow pump? What are the limitations of this theory? 3. How is the design method based on velocity diagrams evolved from airfoil theory?
Special Pumps
215
Explain the construction and operation of a vertical turbine pump? How is it modified for the design of a submersible pump and borewell pump? 5. Compare the construction and performance characteristics of various types of rotary displacement pumps. How is screw pump superior? 6. Write notes on: (a) Water jet pump (b) Air lift pump 7. What is hydraulic ram? How does it work? What are its advantages and disadvantages? 8. What is the difference between a hydraulic coupling and torque converter? How a torque converter can be used as hydraulic coupling?
TUTORIAL SHEET NO. 11 1. A hydraulic ram lifts water from a tank 3 m above the ram to a tank 20 m above the ram. The ram supplies 0.1 m3/s of water and discharges 0.005 m3/s through a 6.5 cm diameter pipe which is 80 m long. Calculate the efficiency of the ram by taking friction coefficient, f = 0.01. Neglect friction losses in supply pipe. Solution The water velocity in discharge pipe, V=
0.005 Q = A p 6.5 4 100
FG IJ H K
2
= 1.507 m/s
Friction head loss DHf =
4 ´ 0.01 ´ 80 ´ (1507 . )2 4 fLV 2 = 6.5 2 gD 2 ´ 9.81 ´ 100
FG IJ H K
= 5.7 m Total delivery head = 20 + 5.7 = 25.7 m (a) Aubissons efficiency, h=
0.005 25.7 Q2 H ´ 2 = ´ = 42.8% Ans 0.1 3 Q1 H1
(b) Rankine efficiency, hRanking =
Q2 ( H2 - H1 ) 0.005 (25.7 - 3) × = ´ 3 0.1 Q1 H1
= 37.8% Ans
P A R T IV SYSTEMS AND PLANTS
+0)26-4
Hydraulic Systems 12.1 INTRODUCTION The lifting machines like hydraulic lifts and cranes and jacks and presses, etc. are operated by high pressure oil or water by a pump working continuously. The energy requirement of these devices may not be uniform. Large amount of energy is needed when the load is hoisted and practically no energy is needed when the load is lowered. The non-uniform energy requirements can be met by the following methods: 1. A large capacity pump with intermittent operation. 2. A small capacity pump with hydraulic accumulator and pump working continuously. Energy is supplied from pump plus energy of the accumulator during load lifting. However, when no energy is needed during lowering of load, pump energy is stored in the accumulator. The energy requirement of a lifting device may be enormous and cannot be met by the pump. An intensifier may be used to multiply the pump pressure to supply high pressure to the lifting device. The layout of a system for a hydraulic lifting device may consist of a pump, an accumulator and intensifier as shown in Fig. 12.1. W Oil return
Oil Pump
Accumulator Fig. 12.1
Intensifier
Servomotor of lifting device
Hydraulic system for lifting device
Pump: The working fluid can be high pressure water or oil. Oil is more widely used which is supplied at high pressure by a rotary positive displacement pump. Gear pump or screw pump are widely used.
220 Hydraulic Machines 2. Accumulator: A simple or differential accumulator is used to store the pressure energy of the pump. 3. Intensifier: Hydraulic, hydro-pneumatic or steam intensifiers are used to multiply the oil pressure. 4. Servomotor: The lifting devices are operated by a hydraulic servomotor which is a piston-cylinder unit and actuates or opens a device by piston motion controlled by slide or pilot valve.
12.2 HYDRAULIC ACCUMULATOR Hydraulic accumulator is a device to store energy is the form of liquid pressure and to release the pressure when required. The capacity of the accumulator is the maximum amount of energy, it can store. The capacity of the accumulator = ApL [N-m] where A = Area of ram [m2] L = Maximum stroke of ram [m] p = Liquid pressure [N/m2] The force or load lifted W + Wf = AP [N] where W = Dead weight of ram plus external weight [N] Wf = Force of friction between ram and cylinder [N]
Weight
Bypass line
RAM
Cylinder
To lifting device
Pump sump Fig. 12.2 Hydraulic accumulator
12.3 DIFFERENTIAL ACCUMULATOR Differential accumulator is used when the capacity handled by the accumulator is small but pressure requirement is high.
Hydraulic Systems
221
Liquid under pressure from the pump enters the sliding cylinder cum weight through a hole in the lower fixed ram. The cylinder is lifted till it is balanced by its weight. During downward movement of the loaded piston, the liquid accumulated is pushed to the lifting machine. The area of lower fixed piston is more than area of upper fixed piston, A2 > A1 Capacity of the accumulator = (A2 A1) Lp where L = stroke of piston Capacity = W ´ L where W = weight of sliding piston.
Fixed ram A1 Sliding cylinder cum weight (W)
p A2
From pump
To lifting machine Fixed ram
Fig. 12.3 Differential accumulator
12.4 HYDRAULIC INTENSIFIER Hydraulic intensifier is used to multiply the pressure of the pump to supply high pressure to the lifting machine. It is differential piston-cylinder device. The low pressure liquid or air or steam is supplied to the larger cylinder of piston area A1. This pushes the piston of area A2 in the small cylinder where working fluid is supplied from the reservoir. The pressure p2 is more than pressure p1. The forces acting on the differential piston are equal due to equilibrium F1 = F2 \
A1p1 = A2p2
222 Hydraulic Machines
p2
A1
A2
p1
Water or oil
Oil or water or compressed air or steam
From reservoir
Fig. 12.4 Hydraulic intensifier
\
p2 = p1
FG A IJ HA K 1
(Neglecting friction resistance)
2
There are single acting intensifiers and also double acting intensifiers for continuous operations. Pressures of the order of 1500 bar have been achieved by intensifiers. In hydraulic intensifiers water or oil is used. In hydro-pneumatic intensifiers, the fluid supplied is compressed air. Steam is also used in some intensifiers.
12.5 HYDRAULIC JACK OR PRESS Hydraulic jack is a portable equipment to lift heavy loads over short distance. Hydraulic press is a permanent set-up used for pressing cotton bales in ginning mills, plasting moulding machines and metal pressing work, for pushing and shearing steel plates. An effort applied to the pump plunger (d) through lever arrangement produces a pressure p in the hydraulic liquid. This pressure acts on the ram (D) which supports the load (W). The load W=
p 2 Dp 4
The rate of lifting of load will be equal to the rate of oil supplied by the pump. For lowering the load, pressure relief valve is opened to drain the oil into reservoir. Load W Effort
Load platform F RAM Pump
d
Cylinder
D p
p Pressure chamber Oil reservoir Pressure relief valve
Overflow pipe
Fig. 12.5 Hydraulic press
Hydraulic Systems
223
12.6 HYDRAULIC LIFT Hydraulic lift is used to carry goods and passengers from one floor to another in multi-storeyed buildings. In general, hydraulic lifts have been superceded by electric lifts and hydraulic lifts are used as standby units wherever there is danger of fire and explosion. Fluid under pressure is supplied to a cylinder under a ram carrying a cage or platform at its top. In direct acting lifts, the stroke of ram is equal to one floor height. Modern lifts are generally of suspended type.
Cage
SF
RAM
FF
Cage FF
GF Fixed cylinder
Oil from
SF
Pulley Block
RAM Oil
GF
intensifier (a) Direct acting lift
Fig. 12.6
(b) Suspended lift
Hydraulic lift
QUESTION BANK NO. 12 1. With the help of a flow diagram, list the various components of hydraulic system for lifting device. 2. What is a hydraulic accumulator? Differentiate between an ordinary and differential accumulator. 3. Explain the working of the following devices. (a) Hydraulic Intensifier (b) Hydraulic Press (c) Hydraulic Lift
TUTORIAL SHEET NO. 12 1. A certain machinery is operated with a hydraulic accumulator through a pipe 180 m long and 100 mm diameter. The accumulator has ram 250 mm diameter and 3 m stroke and is loaded with 345 kN weight. It is fed with water from a three-throw pump at 45 rpm. The plunger of the pump has a diameter of 45 mm and stroke of 350 mm; the slip being 3 per cent. If the
224 Hydraulic Machines machinery absorbs 36.8 kW power, calculate the longest period for which it can be operated continuously. Take friction fact or, f = 0.032. Solution Intensity of pressure in the accumulator, p=
W 345 ´ 103 = = 7028 ´ 103 N/m2 F A (0.25) 2 4
Volume of accumulator = AL =
F (0.2)2 ´ 3 = 0.147 m3 4
Friction losses in the pipe, DHf =
4 ´ 0.032 ´ 180 ´ V 2 4 fLV 2 = . 2 ´ 9.81 ´ 010 2 gD
= 11.74 V2 m Available energy at the end of pipe, rQgH =
LM 7028 ´ 10 N rg
3
OP Q
. V 2 rQg - 1174
= (716 11.74 V2)r ´ 9.81 ´
p (0.1)2V N/m/s 4
Energy required for the machinery = 36.8 kW Equating the two (716 11.74 V2) ´ 0.077 V = 36.8 V = 0.67 m/s Flow at the end of pipe, Q=
F (0.1)2 ´ 0.067 = 0.00526 m3/s 4
Flow from pump, Qp = =
AF L F N F 60
´ 3 ´ hL
F 45 (0.045)2 ´ 0.35 ´ ´ 3 ´ 0.97 4 60
= 0.001215 m3/s
Hydraulic Systems
225
Flow required from accumulator, Qacc. = Q Qp = 0.00526 0.001215 = 0.004045 m3/s Time of supply by accumulator =
0147 . = 36.6 s Ans 0.004045
An intensifier has a ram diameter of 150 mm and a sliding cylinder diameter of 570 mm. Calculate the pressure of water on the low pressure side of the intensifier, if the pressure of water on the high pressure side is 20 MN/m2. Solution For an intensifier p1A1 = p2A2 \
p2
FFD I F A I G JJ =p G J =p G 4 H A K GH F D JK 1
= 20 ´
1
2
FG 750IJ H # K
2 1
1
4
2 2
= 500 MN/m2 Ans
A hydraulic accumulator has 200 mm ram diameter and lift of 500 mm. +alculate the capacity of the accumulator when it is supplied with water at a pressure of 35 bar. Solution +apacity of accumulator = ApL =
F 2 D pL 4
F 0.2 2 ´ 0.5 ´ 35 ´ 105 4 = 0.35 ´ 102 kN-m Ans 4. A hydraulic lift has a ram diameter of 200 mm and is supplied with water at a pressure of 20 bar. Find the total load it can lift if the velocity of lift is 1 m/s. Assuming efficiency of 60%, calculate the power required. Solution The load,
=
F 2 F d p = 0.2 2 ´ 20 ´ 105 4 4 = 632832 N The power required,
W=
P=
W ´ t 632832 ´ 1 = = 105.5 kW Ans 0.6 h
+0)26-4
!
Hydraulic Power Plants 13.1 INTRODUCTION The utilization of hydro energy to operate agricultural and industrial devices is one of the oldest and widespread techniques of human achievement. Hydro power, available free of cost in the form of flowing water, is an important, pollution free and renewable source of energy. In a hydroelectric power plant, water streams in hills/ mountains are collected behind a dam. The potential energy of stored water is passed to a hydro turbine where potential energy of water is used to rotate the turbine runner. The mechanical energy of rotation of turbine is used to run an electrical generator to produce electricity. The electricity generated is transmitted over distances and distributed to consumers of electrical energy.
Water hydro energy
Water Water storage
Electric generator
Hydro turbine
DAM P.E
M.E of Energy conversion
Rotation
Electrical EMF
Energy
Fig. 13.1 Hydro power plant
13.2 HYDRO POWER POTENTIAL The hydro power potential of the world is huge to the tune of 36 ´ 1012 kWh of annual electricity generation which is technically feasible. Approximately 11 ´ 1012 kWh of electricity can be economically generated every year. Presently 20% of the total world electricity is generated by hydroelectric power stations. In some countries, the share of hydroelectric power is very high as seen in Table 13.1. The Central Electricity Authority (CEA) has assessed Indias hydro potential to be about 1,48,700 MW of the installed capacity. The hydro electric capacity currently under operation is about 26,000 MW and 16,083 MW is under various stages of development. India is endowed with economically viable hydro potential
Hydraulic Power Plants
227
Table 13.1 Share of hydro power in various countries Sl. No. Country
Share of Hydropower
Sl. No. Country
Share of Hydropower
1.
Norway
99%
6.
India
25%
2.
Brazil
94%
7.
USA
13%
3.
Canada
66%
8.
Former USSR
14%
4.
Austria
69%
9.
Germany
5%
5.
Switzerland
62%
of 600 TWH (trillion Watt hours). The key advantage of hydroelectric power is the ability to store energy and the flexibility of its use during peak load periods.
13.3 CLASSIFICATION OF HYDRO POWER PLANTS Hydro power plants are capital intensive projects with long construction period. These are mostly designed as multi-purpose integrated plants. In addition to generation of electricity, these projects are designed for irrigation water and flood control. Broadly, these plants are classified as run-of-the-river plant, plants with storage and pumped hydro plants. Depending upon the power output, hydro power plants are classified as small (upto 10 MW), medium (upto 100 MW) and large power plants (above 100 MW). There are also mini, micro and small power plants with capacities in the range of 100 kW to 1 MW. Besides, hydroelectric power plants are classified as low-head (upto 10 m), medium-head (upto 100 m) and high-head (above 100 m) plants.
13.3.1 Run-of-River Plants These power plants are installed on large and small rivers and irrigation canals. These are low head and high water flow power plants. The water head is created with the help of a dam or a weir. A water supply channel runs parallel to the river with a low slope on which powerhouse is built. The power generated depends on the quantity of flow water and works on daily basis on the amount of flow available. Sometimes, a small reservoir or pond is built, which can store a few hours supply of water to the plant, when the river flow exceeds the amount required by the plant. The stored water is used in generating power during the hours when the demand is in excess of the flow of water in the river at the moment. Run-of-river plants are generally low-head small power plants using Propeller or Kaplan turbines or bulb or tubular turbines with water siphoning over the weir.
228 Hydraulic Machines Dam or weir
River Power house HR Tail race
Generator TR
Power house Sluice gate
Canal
Forebay
Draft Tube
Turbine (a) Layout
(b) Power house
Fig. 13.2
Run-of-river power plant
13.3.2 Plants with Storage Reservoir High head power plants work in high mountains where water is stored in natural lakes during rainy season or when the snow melts. Long penstocks are used to connect the lakes to powerhouses where Pelton turbines are installed. These are mostly medium capacity plants. The most common power plants in India are plants with storage reservoir. These are built on large rivers by building a dam. During rainy season, water is stored in reservoirs behind the dams so that it can be utilized during other seasons to supplement the flow of river whenever the flow in the river falls below a specified minimum based on study of flow duration curve. Power can be generated directly from the reservoir. Sometimes canals are constructed to convey water from the reservoir for irrigation purposes. Storage power plants are designed for medium head and Francis turbines. The size and number of turbines are decided by optimization of hydrothermal mix by interconnecting hydro power with steam or nuclear power plants. The storage reservoir behind the dam can store water for a very long period of time. These plants cover both peak and base electrical load. HR Penstock River and reservoir
Powerhouse
Dam
Francis turbine TR
Draft tube
Fig. 13.3 Storage reservoir power plant
Hydraulic Power Plants
229
The largest storage reservoir power plants in the world are listed in Table 13.2. along with their installed capacities. Table 13.2 Storage reservoir hydro power plants Sl. No. Name of station 1. 2. 3. 4. 5.
River and country
Itapu Grand Coulee Guri Sayano-Shushearskaya GES Khasnoyarskaya GES
Installed capacity (MW)
Parana River Brazil/Paraguay Columbia River, USA Canoni River, Venezuela Yenisseri River, Russia
12,600 10,800 10,300 6,400
Yenisseri River, Russia
6,000
The largest power plants in India are Bhakra Dam at Satluj River and Tehri Dam at Ganga River.
13.3.3 Pumped Storage Plants Pumped hydro storage plants are mainly used to meet peak load demand. These can also be used as standby power plants to quickly start in case of breakdown of a thermal or nuclear plant working as a base load power plant. The start-up time of pumped storage plant is approximately 1 minute. These are mostly high head plants. The plant consists of a number of turbine-generator sets and pump-motor sets. Water after working in turbine is stored in tailrace reservoir. During low load, say night time, the water is pumped back from the tail to head reservoir drawing excess power from the grid or from nearly base load thermal or nuclear power plant. During peak load, the water is used to work on turbines to produce electricity. The steam power plants are basically base-load plants working at full plant capacity. Whenever the load demand is less than full plant capacity, the surplus power is used to run the pump installed at the tail race of hydroelectric plant. Storing of HR Penstock Powerhouse cum-pump house Head water reservoir
TR
Tail water reservoir
Fig. 13.4 Pumped storage plant
Dam
230 Hydraulic Machines potential energy of water is the most cost-effective and environment-friendly method of energy storage. About 75 per cent of power used in pumping is recovered. A motor-generator set and a reversible turbine-pump unit can be profitably used as the same machine can be operated as a motor or generator and similarly as a pump or turbine (Deriaz turbine). The specifications of worlds largest pumped hydro plants are given in Table 13.3. Table 13.3 Specifications of pumped storage plants Sl. No. Power station, country
10.
Head (m)
Bath country, USA Luding town, USA Dinorwig, U.K. Raccoon, USA Shinu Takase-gawa, Japan Grand Maison, France Oku-Yashino, Japan Zagork, Russia Piastra, Italy Weir Hornberg, Germany
360 110 535 320 265 955 515 1120 610/1070 660
No. of sets
6 6 6 4 4 8 6 6 9 4
Installed capacity (MW) Turbines
Pumps
2100 1755 1620 1530 1280 1220 1210 1200 1280 960
2280 1910 1800 1600 1320 1260 1280 1250 1430 1040
The energy flow diagram of a pumped hydro plant is shown in Fig. 13.5. From electric energy consumed for pumping 75% is regained in hydro-generator sets. Losses (%)
Losses (%)
86.4 0.5 Piping
Piping 0.8 Turbine 6.5
9.6 Pump
Generator 1.6 Transformer 0.5
77 77% Electrical energy regained
100
3.0 Motor 0.5 Transformer
100% Electrical energy supplied
Fig. 13.5 Energy flow diagram for pumped storage plant
13.3.4 Small, Mini and Micro Hydro Power Plants Small hydro power plants have no impact on environment and are safe for fish breeding and drinking water supply. The natural water sources in hilly areas can be
Hydraulic Power Plants
231
utilized for power generation with low-head standardized hydro sets. The range of applications of hydro turbines for small power plants is given in Table 13.4. Small and micro hydro systems are very important for developing countries which possess large hydro resources. Table 13.4
Operating range of hydro turbines for small power plants
Sl. No.
Type of Turbine
Power Capacity
Head (m)
3.
Pelton Francis Propeller
upto 5 MW 750 6000 kW upto 220 kW
upto 450 25 160 m Below 25 m
The following countries have following operational small hydro plants. China USA Japan Switzerland Austria
90,000 10,500 5,300 2600 1300
The potential energy source in India for small plants is around 20,000 MW. The mini plants are operating with 5 m to 20 m head with installed capacity of 1 MW to 5 MW. The micro plants are operating with head less than 5 m and with installed capacity between 0.1 MW and 1 MW.
13.4 LOCATION OF HYDRO POWER PLANTS The following important selections have to be made for the design of hydraulic power plants. 1. Location
The site selection must ensure economical availability of all resources, water flow, head, accessibility, reservoir area, dam foundation, etc. 2. Installed capacity
The installed capacity is decided by hydrology of the site, i.e., hydrographs, flow duration curves, range of available head, etc. 3. Number and size of units
The number and size of hydro sets will be decided by load duration curves and interconnected operation with other power plants, i.e., steam and nuclear power plants. The type of turbines will be decided on the basis of flow, head, specific speed, size and availability of hydro sets from suppliers. The following factors are considered for the location of a hydro power plant.
232 Hydraulic Machines 1. Availability of water
The design and capacity of the hydro power plant depends upon the amount of water available at the site. The data on run-off, annual precipitation in the catchment area, maximum and minimum quantity of water available in a year is required to decide the capacity of the plant, integration with thermal and nuclear plants and waterworks for flood control and irrigation water. 2. Water storage capacity
The storage capacity has to be decided to ensure continuous generation of power as there can be wide variation in the quantity of water flow in the river and rainfall round the year. The storage capacity is estimated with the help of hydrographs and flow duration curves. 3. Available water head
The power generation is a function of available water flow rate as well as available head. Availability of large head reduces the quantity of water requirement for a given output. 4. Accessibility of site
Large quantity of men and materials have to approach the site during long construction and operation periods. The site should be easily accessible by rail and road. 5. Distance from load centre
The power generated has to be transmitted to load centres. Therefore, the site should be close to load centre to reduce the cost of transmission lines as well as transmission losses. 6. Type of land
Many types of structures have to be built for the hydro power plant such as dam, conduits, intakes, surge tank, powerhouse, head and tail reservoirs. The land should be cheap and rocky. It should be strong enough to withstand the stress in the structure and the thrust of water. 7. Project cost
The cost of the project and the completion period should be minimum. Detailed investigations are carried out which include: 1. Hydrological investigations for water availability, storage capacity, available head, distance from load centre and approach to site by road and rail. 2. Topographical investigation of the catchment area and site for building of dam, conduit, powerhouse with the help of level contours by ground surveying or aerial surveying.
Hydraulic Power Plants
233
Geological investigation for accurate information of ground characteristics. The selection of site depends on the following factors: (i) Tight basin of ample size (ii) Narrow outlet requiring dam of low volume. (iii) Strong foundation to support the dam structure. (iv) Safe and ample spillway for disposal of surplus water. (v) Valuable minerals and agricultural lands should not be submerged. (vi) Local availability of materials for construction of dam. 4. Water pollution effects: The project should have minimum impact on the surroundings and enhance the local environment. The sedimentation effect of the reservoir should be minimum as it can lead to reduction of reservoir capacity and erosion of turbine blades.
13.5
HYDROLOGY
Hydrology is a science which deals with depletion and replenishment of water resources over and under the earths surface. It is used to design irrigation and flood control schemes, power projects, water supply system, navigation work, etc. The precipitation in the form of rain, hail, snow or sleet falls on the hills and mountains and converges to form streams and rivers. This can be used for power generation. The whole of precipitation or rainfall does not convert into river flow called run-off. There are following losses of water: 1. Re-evaporation of part of precipitation into atmosphere. 2. Transpiration of water through roots of plants and through leaves into atmosphere. 3. Infiltration of water by seepage into the ground forming water table or ground storage. The difference of rainfall in the catchment area of a river and the losses is called run-off and is available for power generation. The hydraulic power. P = rQgHh [W] where
r = Water density [1000 kg/m3] g = Acceleration due to gravity [9.81 m/s2] Q = Flow of water [m3/s] H = Head of water due to height of fall of water [m] h = Overall efficiency of power plant (0.85 0.90)
234 Hydraulic Machines The electrical energy produced W= where
rQgHht [kWh] 1000
t = operating time of plant [8760 h/year]
13.5.1 Flow Duration Curve The stream flow at a site is variable as it depends on the amount of rainfall which is seasonal. The variation of the river flow with time is shown graphically as hydrographs. A flow duration curve is a very useful tool for estimation of hydraulic power. It is drawn with the help of hydrographs. The flow duration curve is drawn between quantity of river flow against percentage of time duration for which a certain quantity of flow is available. The area under the curve gives the total quantity of run-off in a given period. For a given head PµQ
Qa D C
E
Qm
B
O 0
F
Power [kW]
3
Flue rate Q (m /s)
The flow duration curve can be easily converted into power duration curve by changing the units of y-axis from m3/s to kW. The minimum flow rate Qm is available all the time (for 100% time) and point PB gives primary power or firm power.
20
Fig. 13.6
40 50 60 80 Per cent of time
A 100
Flow duration curve
The horizontal line DEF is drawn for power (PF) required all the time. The additional power (PF PB ) is called secondary power. This power can be possible if additional water flow Qa is made available. This is possible only by providing adequate storage equal to area BEF.
Hydraulic Power Plants
235
The uses of flow duration curve are as follows: 1. Preliminary studies of a hydro project 2. Evaluation of low level flows 3. Estimation of primary power (without storage) and secondary power (with storage). 4. Preparation of power duration curve.
13.5.2 Mass Curve A mass curve is a graphical representation of accumulated flow against time. The slope of the line at any time shows the rate of flow at the time. The time is shown in years and accumulated flow in hectare-metre. The data is prepared from the records of monthly flows of a river or stream. The mass curves are used to estimate the water storage requirements. The following characteristics should be noted. 1. If the mass curve is horizontal, the flow rate is zero. 2. Concave depression indicates dry periods. 3. A steep rise of curve indicates high flow rate. The mass curve for a river is shown in Fig. 13.7.
Accumulated flows (ha-m)
B¢¢
B B¢ c
A¢¢ A 0
1
2 3 Time (year)
4
5
6
A¢
Fig. 13.7
Mass curve
Line AB is drawn by joining the end points of mass curve. It indicates the average discharge over the total period. Line A¢B¢ is drawn parallel to line AB and tangent to lowest point of mass curve C. Line A²B² is drawn parallel to line AB and tangent to highest point D of mass curve.
236 Hydraulic Machines The vertical intercept A¢A² gives the storage volume required to ensure continuous release of water for average discharge rate over the entire period. Storage is impounding of excess run-off during the seasons of surplus flow for use in dry season. A dam is constructed across the river at a suitable site and storage is built on the upstream side of dam.
13.6 COMBINED OPERATION OF DIFFERENT POWER PLANTS A hydraulic power plant can be used as an exclusive source of power. However, the output of the plant is proportional to water flow rate which is seasonal. During rainy season there will be plenty of water. If the plant is designed for minimum flow rate, there will be great wastage of water over the dam for most part of the year. During winter season, there will be low water flow rate and hydro power plant may not be able to meet the maximum demand. A large amount of water has to be stored to meet the variable load requirements. This is a big source of ecological imbalance claiming large amount of fertile land, damage to forestry and evacuation and rehabilitation of large population. In order to meet the fluctuating load requirement efficiently, economically, reliably and continuously the hydroelectric power plants are used in conjunction with thermal power plants and nuclear power plants as an interconnected system. Load sharing by hydro plant is maximum when available water flow is maximum (during rainy season). As long as there is plenty of water available stored in the reservoir, the hydro plants can carry base load, with thermal plants taking the peak loads. When water flow rate is low (during winter season), the steam plants can operate as base load and hydroplants meet the peak load. There are many advantages of interconnected mode of operation. 1. The reliability of continuous supply of electricity to the consumer is much greater in interconnected than in isolated system. The reliability of hydro plants is dependent on water flow rate and storage availability. 2. The installed capacity requirement is low because there is certain diversity in time between peak demands of two systems. Therefore, peak load of the interconnected system is less than the sum of individual peaks. 3. The power supply is ensured in the event of failure of one power station as consumers are fed from common grid of interconnected power plants. 4. The spinning spare requirement of interconnected system is reduced. 5. The overall cost of energy per unit is less from interconnected system. 6. Integrated system requires less capital investment and less operation and maintenance cost. 7. Integrated system facilitates better use of transmission lines at higher voltage by teeing the generating station in a group. The load division between a hydro power plant and steam power plant is shown in Fig. 13.8.
237
100
100
80
80
60
System load (%)
System load (%)
Hydraulic Power Plants
Hydro
40 Steam 20
24
4
8 12 Time
16
20
24
60 Steam 40 Hydro 20
24
16
20
24
(b) High flow day
(a) Low flow day
Fig. 13.8
8 12 Time
4
Hydro thermal mix
QUESTION BANK NO. 13 1. Sketch the hydro power potential for the world and India. 2. Give a detailed classification of hydro power plants. Which type of hydro power plants have maximum scope in India? 3. What are crucial factors considered for the location of a hydroelectric power plant? 4. Define hydrology, flow duration curve, mass curve. How are these tools used for the design of hydro power plant? 5. What are the advantages of interconnected mode of operation of hydroelectric and thermoelectric power plants? 6. What concepts are used in the division of load between a hydro power plant and steam power plant?
TUTORIAL SHEET NO. 13 1. The following data is available for the design of a hydroelectric power plant 1. Head available = 27 m 2. Catchment area = 430 sq.km 3. Rainfall = 150 cm/year 4. Utilization of total rainfall = 65% 5. Penstock efficiency = 95% 6. Turbine efficiency = 80% 7. Generator efficiency = 86% 8. Load factor = 0.45 Calculate the installed capacity of power plant. Also select suitable turbine.
238 Hydraulic Machines Solution Total quantity of annual water available = (430 ´ 106) ´ 1.5 ´ 0.65 = 419.25 ´ 106 m3 Available water flow rate, Q=
419.25 ´ 106 = 13.29 m3/s 365 ´ 24 ´ 60 ´ 60
Power developed P = rQgH hT ht hG = 1000 ´ 13.29 ´ 9.81 ´ 27 ´ 0.95 ´ 0.8 ´ 0.86 ´ 103 = 2300 kW Ans Load factor = \
Peak load =
Average load = 0.45 Peak load 2300 = 5111 kW 0.45
Use two turbines of equal capacity. Turbines capacity =
5111 = 2555.5 kW 2
As head is low, use two Kaplan turbines of 3000 kW each. 2. A power system has a maximum load of 80 MW at 35% load factor. The load can be supplied by any of the following schemes. (a) A steam power plant to supply the whole load. (b) A hydel plant and an interconnected steam plant. The steam plant supplies 120 ´ 106 kWh/year with a maximum load of 50 MW. The cost data is tabulated below. Cost
Steam Plant
Hydro plant
Capital cost 2. Operating cost 3. Transmission 4. Interest and depreciation
Rs 600/kW 4.8 paise/kWh Negligible 12%
Rs 1400/kW 1.2 paise/kWh 0.3 paise/kWh 10%
Calculate the energy cost for the mode of operation. Solution Annual energy requirement = Maximum load ´ Load Factor ´ 8760 = 80 ´ 103 ´ 0.35 ´ 8760 = 245 ´ 106 kWh
Hydraulic Power Plants
239
(a) Isolated Steam Plant Fixed cost (interest and depreciation) = Operating cost =
12 ´ 80 ´ 103 ´ 600 = Rs. 576 ´ 104/year 100 4.8 ´ 245 ´ 106 = Rs. 1152 ´ 104/year 100
Total annual cost = Rs. (576 + 1152) ´ 104 = Rs. 17.28 ´ 106 Energy cost =
17.28 ´ 10 6 ´ 100 = 7.05 paise/kWh 245 ´ 10 6
Ans
(b) Interconnected Power System (i) Hydel Plant Annual fixed cost =
10 (80 50) ´ 103 ´ 1400 = Rs. 4.2 ´ 106 100
Energy supplied by hydel plant = (245 120) ´ 106 = 125 ´ 106 kWh Operating cost including transmissions = 125 ´ 106 ´
(12 . + 0.3) = Rs. 1.875 ´ 106/year 100
(ii) Steam power plant Load share = Minimum plant capacity =
120 ´ 106 = 13.7 ´ 106 kW 8760 13.7 ´ 106 0.35
= 39.15 ´ 103 kW Annual fixed cost =
12 ´ 39.15 ´ 103 = Rs 2.82 ´ 106 100
Annual operating cost = 120 ´ 106 ´
4.8 = Rs. 5.76 ´ 106 100
Total cost of both the plants = Rs. 4.2 ´ 106 + 1.876 ´ 106 + 2.82 ´ 106 + 5.76 ´ 106 = Rs. 14.655 ´ 106/year
240 Hydraulic Machines The energy cost =
14.655 ´ 10 6 ´ 100 245 ´ 10 6
= 5.98 paise/kWh The interconnected operation of steam and hydel station is more economical.
2)46 8 +)5-567,1-5
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Floating Type Micro Hydro Power Plants INTRODUCTION India has made great strides in industrialization during post-independence years. This has brought prosperity to a selected few and created concentrated centres of employment near the industrial complexes and big cities where economic activities have flourished. The material and energy resources have been diverted to feed these industrial and commercial gluttons. The high technology industry could not penetrate the traditionally agricultural rural and remote areas where nearly 78% of Indias population live. The rural population felt neglected and many of them are flocking to the new economy centres in search of their livelihood. The continuous migration of population and uneven distribution of wealth, material and energy resources have created many social and political problems which can lead to very serious consequences, if not further checked. The solution of the problem lies in concerted efforts for the development, welfare and prosperity of the rural areas which in turn depend upon the availability of cheap and abundant energy for domestic cooking, heating and lighting community development, irrigation water, mechanized agriculture and small industries. Transportation of commercial fuels like coal, diesel and petrol over long distances to remote and rural areas is very difficult and transmission and distribution of electricity through power grid to far-flung areas are prohibitively expensive. The farmers and craftsmen are not in a position to pay for high costs of energy, thus small-scale industries in these areas are reduced and there are limited irrigation possibilities in agriculture. The modern high technologies do not adopt to rural ecology. An energy production technique based on renewable source and appropriate technology which can be taken care by trained local people for its operation and maintenance problems is ideally suited. Only appropriate technology involving local resources, manpower and raw materials and environment can ensure environmental harmony
244 Hydraulic Machines and sustainability of the development. The utilisation of energy in flowing water through micro hydro schemes is a good example of appropriate technology for rural development.
SMALL HYDEL PROJECTS IN INDIA Hydro power, available free of cost in the form of flowing water, is an important, pollution free, renewable source of energy. The large potential of hydro energy available from flowing water in the rivers and irrigation canals, however, still remains untapped, since generating power on a small scale from low and ultra low head drops of 3 to 20 metres has not been an economic proposition because of the high civil and equipment costs as well as high operating cost to be incurred per unit of power generated. The situation is further deteriorated by very low load factors in rural areas. China has developed 1200 MW of power by small hydel projects by creating water heads of about 5 m along the rivers. It is estimated that 5000 MW capacity of power can be created in India with the help of small hydro turbines of 200 kW to 1000 kW capacity each. M/s. Jyoti Limited, Baroda have the necessary know-how and capacity to manufacture these small hydro turbines, see Table 1. There are also other organisations showing interest in the manufacture of such small sets. Bharat Heavy Electricals Ltd. also has the necessary know-how which can be passed on to some ancillary unit for for quick manufacture of small hydro turbines without disturbing their main manufacturing plan for large sets. 6able 1 Manufacturing range of Jyoti Ltd. for small hydel sets Type of Turbine Pelton Turgo Impulse Francis Propeller Kaplan Horizontal Tubular
Operating head range (m) 100 to 400 50 to 200 20 to 200 2 to 30 2 to 25
Output range (kW) 50 to 6000 25 to 3000 50 to 3000 10 to 6000 10 to 6000
Due to civil costs of dams, water conduits and diversion channels, independent small hydel projects become very costly. It may cost Rs. 18000 per kW installed. Such projects should be planned on the basis of multiple purpose use of water resources such as irrigation and power generation. Even though these schemes are costlier when compared with the cost of unit power generation from major hydel projects, these are encouraging keeping long-term energy availability in view. Also hydropower is renewable in nature, once built, there is built-in protection against inflation of cost, water is available essentially free, it can be instantaneously switched on, it need not run continuously. According to the size of hydroturbine, the small hydel projects may be classified as:
Case IFloating Type Micro Hydro Power Plants
245
Micro Hydel Projects : 50 to 100 kW Mini Hydel Projects : 100 to 1000 kW Small Hydel Projects : 2 MW to 15 MW Small hydel projects can be completed in four to five years, mini and micro projects in three years provided civil works of irrigation projects are already completed. Here water released for irrigation only is used and power is therefore, generated for 6 to 7 months a year. In Maharashtra, 130 schemes of various catagories of small hydro projects of about 11000 kW capacity for a layout of Rs. 2000 millions have been planned. A separate circle has been created in UP State Electricity Board with headquarters at Nainital for small hydel projects. Sixteen micro hydel schemes are already functioning in eight hill districts of UP with total istalled capacity of 7730 kW. Four schemes are under construction for a total cost of Rs. 3.37 crores. The total installed capacity after completion of these schemes will be 8830 kW. Other 32 schemes of 20,000 kW are in the pipeline. Other hilly states are also going for small and micro hydel schemes in a big way.
RIVER CURRENT ENERGY Western Ghats of Maharashtra and regions of Uttar Pradesh, Himachal Pradesh, Jammu & Kashmir, Sikkim and North-Eastern States located in the Himalayas are suitable sites for small hydel projects where medium to high heads can be available. Equipment costs are low in the high head schemes. As we come down the hills, the available head decreases and equipment costs go up. For example, the cost of one hydro set of horizontal tubular design of 3.2 kW capacity operating at a head of 3 m may be Rs. 1.28 lacs, i.e., Rs. 40,000 per kW installed. The civil costs are site-specific but taking 30% of equipment cost, the total installed cost will be a high value of Rs. 52,000 per kW installed. Most of our rivers flow through flat terrains which can be harnessed to produce power much in excess of 5000 MW as estimated above. Chinese model is difficult to develop due to different topography in India. Special zero head turbines operating on kinetic energy of flowing water should be developed to harness huge river current energy. The civil costs of micro hydel power projects can be drastically cut by using floating type plants. There is no need of diversion channels, deep excavations and massive foundations. The floating plant consists of two floats with a vertical (or horizontal) runner in between which rotates by the velocity of water. The capacity developed is a function of the size of the runner and cube of water current speed. The rotational speed of runner is rather low and electric generator can be operated via a transmission gear. The floating plant moves up and down with water level and gives constant performance all the year round which is a big advantage over firmly installed water turbines especially during monsoons.
246 Hydraulic Machines The current energy of flowing water which can be exploited by floating type power plant may be expressed as: P = 1/2 AV3E (kW) where P = Power generated A = Embracing area of runner (m2) V = Water current speed (m/s) E = Efficiency of runner depends upon the type of the runner and varies between 0.5 and 0.85. Although hydraulic wheels will operate at low water velocities, it is unlikely to be an economic proposition in current speeds below 1 m/s (2 knots).
PONCELET WATER WHEEL In ancient times, floating mills were used in Germany on large rivers such as the Rhine, the Weser and the Elbe to drive electric generators and machine tools. A vertical undershot water wheel consisting of a horizontal shaft and radial blades fixed to it, was fitted on the side of a barge which was either moored to the bank or firmly anchored in the stream. A smaller barge was used to support the other end of the shaft. The hulls of the barges were built with local materials. The efficiency realised were very low, in the range of 20 to 30% because of bad blade geometry (straight radial blades). Poncelet water wheel is an improvement on the straight blades of undershot water wheel, which have suitable curvature. Water strikes the vanes practically without shock (inlet vane angle 15°) and drives it by impulse. Water is discharged from the blades almost vertically downwards. The efficiencies achieved are 55 to 65%. The main drawback of such machines is low power output for a given size and weight because at any instant only a small part of it is actually in the water being driven by the current while the rest of the machine is idle.
DARRIEUS TURBINE ROTOR Darrieus turbine rotor is similar to modern vertical-axis wind turbine which was first introduced in France by G.J.M. Darrieus in 1920s. Its configuration consists of four hydrofoil blades which rotate at much higher speed than the water current speed. The speed of rotation is primarily dependent upon the machine overall dimensions. It has relatively low starting torque. Because of simplicity of the blade design and because they are relatively thin blade, fabrication costs are low. They require no pitch control for synchronous applications. Two prototypes have been developed by Intermediate Technology Development Group, London and are shown in Fig. 1 and 2. The blades can be made out of timber or fabricated from steel, ferrocement or glass fibre. The expected performance from these models are plotted in Fig. 4.
Case IFloating Type Micro Hydro Power Plants
247
Output shaft Water level
1.25 m
Hydrofoil blades
Output shaft
Cross arms
Cross arms
Hydrofoil blades Output shaft
Fig. Vertical axis rotor 13.5 rpm in 1 m/s current power take-off above surface swept area 3.75 m2
Fig. Horizontal axis rotor 32 rpm in 1 m/s current power take-offs under water swept area 3.75 m2
Current flow
Fixed length cables
River bank Adjustable length cable
Wooden keels under pontoon floats Winch
Turbine rotor
Fig. 3 Mooring system for first prototype
INSTALLATION Floating type micro hydro power plants are suitable for use in situations where there is no head of water available and is designed to extract kinetic energy from a river or canal current. Therefore, appropriate sites are along the banks of large rivers or on irrigation canals.
Rotor shaft power (watts)
248 Hydraulic Machines
2400 2300 2200 2100 2000 1900 1800 1700 1600 1500 1400 1300 1200 1100 900 800 700 600 500 400 300 200 100 0
0.1 0.3 0.5 0.7 0.9 1.1 1.3 1.5 1.7 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 River current speed (Metres/Sec)
Fig. "
Graph showing how turbine rotor power varies with river current speed for a turbine with swept areas 4 m2 and rotor efficiency 25%
As the energy flux depends upon the cube of water velocity, the latter must be as large as possible. The depth of water required for a 4 m2 runner is at least 2.5 m. The Darrieus turbine rotor is a low speed machine. It is a low solidity device which is completely submerged during operation and this, together with its high efficiency (above 50%), means that a rotor of moderate size and modest materials content can extract useful amounts of energy from current speeds as low as 1 m/s (2 knots). Irrigation canals are expected to be an important area of application for floating type micro hydro power plants. In a large canal, the turbine rotor could be mounted on a pontoon as in a river situation, whereas on a small canal it might be cheaper to mount the rotor on a bridge or beam over the canal. The power output from the rotor could be increased by shaping the canal banks to form a venturi to increase the water velocity through the rotor. The mooring details and site requirements are shown in Figs. 3 and 5 respectively. The mooring cable arrangement protects the rotor against submerged or semi-submerged objects striking the blades. The pontoon which carries the turbine, generator and the transmission system, can float on two rows of barges or oil drums under which wooden keels are affixed. The mooring cable can be adjusted with a winch to ensure sufficient angle
Case IFloating Type Micro Hydro Power Plants
249
River bank
2.5 m minimum depth
Minimum 1 m/s river current velocity
Fig. # Site requirement for turbine with 4 m 2 swept area
of the keels to the current direction so that water side thrust on the keels keeps the pontoon at a distance from the bank.
AUXILIARIES The rotational speed of Darrieus turbine is very low: 13.5 rpm for vertical axis and 32 rpm for horizontal axis turbines as shown in Figs. 1 and 2. For Poncelet turbine, the peripheral velocity of runner is about half of water current speed. Therefore, multi-stage belt drives or gear boxes are needed to increase the speed of the generator shaft to about 950 rpm. Permanent magnet alternators or induction generators are preferred over synchronous generators. To reduce equipment costs, the conventional mechanical or hydromechanical speed governor should be dispensed with. The inertia of rotor being small, there is a problem of instability in power supply quality. The above problems can be solved by use of a flywheel which can ensure stable power supply and can shave-off power demand fluctuations. A flywheel designed for this purpose can be accelerated to extremely high speeds without any risk of breaking apart. It can be made from concentric rings of quartz fibre as windings, the rings being fitted over one another and close gaps filled with an elastic substance to keep the plies of the rims together. The flywheel is coupled to a generator and enclosed in a sealed evacuated casing to reduce windage losses. The device operates as a generator when power demand on the system grows, and as an electric motor when it is time to accumulate energy. According to some calculations, the cost of 1 kW will be less than with pumped hydroelectric storage. A project has been described contemplating a 5m diameter flywheel of 1.96 MN weight that could be charged to 20 MWh. The operating speed is 3500 rpm. The mechanical energy storage system as described above can fit into appropriate technology level suitable for rural environment. The load governors based on electronic circuitry or microprocessor technology may be able to solve the problem to some extent but may not be easily acceptable in rural areas because of higher levels of technology involved. Further, these load governors do not operate happily against unstable operations of turbine rotor. Load governors
250 Hydraulic Machines invariably require an auxiliary load or another hybrid system to store surplus electric power generated by hydel power plant during low demand period. This is not always a good decision thermodynamically as electrical energy is the highest form of energy and hybrid systems help to degrade the energy form and in addition there are heavy conversion losses.
CONCLUSION Floating type micro hydro power plants can be a very attractive proposition to harness enormous amounts of hydroenergy available as river current energy in the flat countryside. Civil costs are reduced and cheap and reliable equipment can be developed for tapping the flow energy in the irrigation canals by constructing the banks into venturi shapes. Research and development efforts are needed to develop efficient turbine rotors working on the principle of wind turbines or old time water wheels followed by intensive field trials. The problem of stability and energy storage to improve load factors can be solved by the development of suitable flywheels. The developmental efforts must be constrained keeping in view the principles of appropriate technology suitable for harmonious development of rural areas involving local people, local materials and local resources.
RE.ERENCES 1. Peter German, The Development of a Turbine for Tapping River Current Energy, Intermediate Energy Technology Development Group, London. 2. Upgrading and Further Development of Traditional Devices for the Utilisation of Hydroenergy, Bremen Overseas Research and Development Association, Bremen. 3. Jagdish Lal, Hydraulic Machines, Metropolitan Book Co. Pvt. Ltd., Delhi. 4. N.P. Cheremisinoff, Fundamentals of Wind Energy, Ann Abor Science Publishers Inc., Michigan, 1978. 5. P.D. Mohindra, Current from Currents The Tribune, 14th June. 1982. 6. Brij Bhardwaj, Surplus Power by 1986 Envisaged Hindustan Times dated 5th April, 1984. 7. UP submits Five more Micro Hydel Schemes, Hindustan Times dated 30th Sept., 83. 8. Venikov & Putyatin, Introduction to Energy Technology Mir Publishers, Moscow, 1981.
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An Alpine Hydro Village INTRODUCTION India is a country of villages. These villages vary in size and population and are scattered all over the country. Some of these are located on the hills and in dense forests and can be reached only on foot or by pony. They have no road connectivity to other villages, towns and cities. The population of these villages may be one hundred to two hundred people. Most of them earn their living by cultivating the land, growing fruits and some of them may be craftsmen who earn their living by weaving woollen carpets, by making earthen pots, by weaving baskets and even by making beautiful carved articles from wood, brass and other local materials. Some of them keep cattles like cows, sheep and goats and make their living on animal yields. In a nutshell, life is hard and hilly villages are very low income areas. The main wealth of hilly areas is vegetation varying according to temperature, soil and rain and mainly depends upon the altitude. Near the base of the mountains, the vegetation will be thick evergreen forests while the top of the mountains will have no vegetation, but will be covered with eternal snows. The vegetation belts of the Himalayas are given in Table 1. 6able 1
Vegetation belts of the Himalayas
Altitude (m)
Description
Trees
Fruits
Crops
0-750
Tarai
Sal, Bamboo
Tropical Fruits
750-1000
Deciduous Forests Mixed Forests Coniferous Forests Timber Alpine Grasslands Snow Eternal Snow
Oak, Horse-Chestnut
Temperate Fruits
Barley, Rice, Tea, Potatoes Rice
1000-1600 1600-3300 3300 3300-4200 4200 4200-8500
Oak, Beech Walnuts Pine, Chir, Fir, Deodar, Cedar, Spruce Line
Line
Tea
252 Hydraulic Machines The valleys are surrounded by the mountains and are protected from the cold winds blowing on the mountains. The climate is warm and the soil is more fertile than that of the mountains. In almost all valleys, there are rivers and sometime lakes. Good crops especially fruits like apples, pears, cherries, etc., can be grown in the valleys. India has made great strides in industrialisation during post-independence years. This has brought prosperity to a selected few and created concentrated centres of employment near the industrial complexes and big cities where economic activities have flourished. The material and energy resources have been diverted to feed these industrial and commercial gluttons. The high technology industry could not penetrate the hilly areas. The population of these areas felt neglected and many of them are flocking to the economy centres in search of their livelihood. The continuous migration of population and uneven distribution of wealth, material and energy resources have created many social and political problems which can lead to very serious consequences, if not further checked.
NEED FOR CHANGE The continuous flux of population to new economy centres has created the problems of mushrooming of slums and destruction of social life; the introduction of high technology will give birth to widespread unemployment. These problems can explode with very serious consequences. These can be checked only by adoption of Gandhian model of socio-economic development with village as the unit. Each unit must be self-sufficient economically and the ratio of exports to imports of goods into a village must be more or equal to one. The development, welfare and prosperity of hilly villages depend upon the availability of cheap and abundant energy for domestic cooking, heating and lighting, community development, irrigation water, mechanized agriculture and small industrial parks based on local materials. Transportation of commercial fuels like coal, diesel and petrol over long distances to hilly areas is very difficult and transmission and distribution of electricity through power grids to far flung areas are prohibitively expensive. The farmers and craftsmen are not in a position to pay for high costs of energy, thus small-scale industries in these areas are reduced and there are limited irrigation possibilities in agriculture.
APPROPRIATE TECHNOLOGY The modern high technologies do not adopt to rural ecology. Only appropriate technology involving local resources, manpower and raw materials can ensure environmental harmony and sustainability of the development. However, there should be provision for upgrading of technology in the future. The wealth of the hills can be increased by inputting ample energy and technology to increase
Case IIAn Alpine Hydro Village 253
agricultural, forest and animal yields. Appropriate technology can be worked out by a technical institute specially established for the development of hilly areas. An energy production technique based on local resources and appropriate technology which can be taken care by trained local people for its design, manufacture, installation, operation and maintenance is ideally suited. The utilization of energy flowing water through small hydel schemes is a good example of appropriate technology for the development of hilly areas. An alpine hydro village should meet most of its energy requirements from small hydel schemes. One such model is described in the following paragraphs. The utilisation of hydroenergy to operate agriculture and industrial devices is one of the oldest and widespread techniques of human achievements. Hydraulic rams have been used to raise water for lift irrigation and drinking purposes. Water mills of various types have been used to generate mechanical power to drive irrigation pumps, machine tools and small electric generators. The Himalaya Mill is well known in the mountain areas from Afghanistan to Burma which is mostly used for grinding grains. The present model of alpine hydro village is based on these devices and are briefly described before discussing the hydro model proper.
HYDRAULIC RAM Hydraulic ram is a contrivance to raise a part of large amount of water available at some height, to a greater height. This can be employed in hilly areas where some natural source of water like a spring or a stream is available at some altitude. Work done by a large quantity of water in falling through a small height is used to raise a small part of it to a greater height. Action of water hammer makes it feasible. No external power is, therefore, required to work this machine. Other attractions are: negligible amount of maintenance and supervision costs, continuous operation, high efficiency, quiet operation and possibility of automatic adjustment of water supply. The liftable volume of water diminishes asymptotically with lifting height. In case of medium lifting heights, the hydraulic ram operates with an efficiency absolutely comparable to a piston pump of the same performance. As the hydraulic ram does not need a driving unit, it is to be considered ideal for lift irrigation and supply of drinking water for hilly areas, where the supply in fossil energy carriers or electricity is problematic. The hydraulic ram operates with water streams of 1 to 40 m3/s and fall heads of 1.5 to 30 m and with lifting heights upto 300 m. There are a number of different types which can be distinguished by the operating of oscillating valves. Hydraulic ram is a simple and rugged device for operating irrigation schemes. This can be easily assembled from local materials. The proposed technical institute for the development of hilly areas has to put in some developmental work to improve the performance and adoption of hydraulic ram to irrigation.
254 Hydraulic Machines HIMALAYA MILL It is a vertical shaft, horizontal runner hydraulic turbine used to drive grain mills. The wooden radial blades are attached to a boss fitted into a wooden or steel shaft. The water is fed through open wooden channel. It produces about 0.5 hp with a water head of 1m. There are about 10,000 such mills scattered in the hilly areas from Afghanistan to Burma producing power equivalent of about 45000 kWh daily. The horizontal water wheel can be improved in performance to raise its efficiency substantially so that it can be adopted in a small-scale industry park as a prime-mover for flour mills, threshers, rice-husking machines, vertical lathe, circular saw, trip-hammer, small electrical generator, carpet loom, etc. The vertical lathe can be used by the craftsmen to produce wooden vessels and trip hammers for embossed copper vessels. The sawmill can be used by the carpenters. All these machines can be employed together to meet the local needs to produce domestic and agricultural implements and packing boxes for the export of fruits and other products. The Himalaya Mill can be developed into an efficient and modern Pelton turbine using metallic buckets fitted to a vertical shaft supported on roller bearings. The water can be directed onto the buckets, through adjustable nozzle from a piping system collecting water from natural streams existing abundantly in the hilly areas. The stopping of turbine can be achieved through manual brakes and load/speed regulation by the nozzle. The design documentation can be provided by the proposed technical institute for the development of hilly areas and machine can be manufactured in the small-scale industry park. The output of the machines can be standardised at 5 to 10 kW. The tail race water can be recirculated by lifting it with a hydraulic ram for irrigation purposes.
FLOATING MILL Poncelet water wheel of a Darrius Turbine rotor can be submerged in a river or water stream available in the hilly areas to tap current energy from water of high speeds. The mechanical energy thus obtained is used for driving irrigation pumps, small electric generators and small compressors for cold storages and ice mills. The small cold storages are necessary in the hilly areas to store potatoes and other perishable goods to get better returns. Ice mill can be used to chill and store milk before it is sent to bigger collection centres. The electricity generated is required to light the houses, schools, community centres and to run TV sets and other domestic devices. All these measures can improve the wealth of the village multifold and raise the standard of living of the population which will arrest their migration to big cities. The blades of the water mills can be made from fabricated steel, ferrocement, glass fibre or timber depending upon the availability of the local materials and
Case IIAn Alpine Hydro Village 255
skills. The power output depends upon the swept area and cube of water velocity and very large energy can be tapped from hill streams flowing at high speeds. The turbine rotor can be mounted on a pontoon or may be hung from a bridge or a beam over the stream. The pontoon can float on two rows of barges or empty oil drums under which wooden keels are fixed. The mooring cable can be adjusted with a winch to ensure sufficient angle of the keels to the current direction, so that water side thrust on the keel keeps the pontoon at a distance from the bank. A turbine with swept area of 4 m2 and rotor efficiency of 25% can give an output of 8 kW at a water current speed of only 2 m/s. These turbines can be adopted to run small cold storages and ice mills, irrigation pumps and to produce all the electricity requirements of the village.
HYDRO VILLAGE
HM HM
HM
HM HM HM
HR
FM = Floating Mill HM= Himalaya Mill HR = Hydro Ram Industry park .ig.
Hydro village
Carpet loom
Tail race
Trip hammer
Irrigation pump
HM
Vertical lathe
P
Saw mill
G
Thresher
CS
Water stream
Rice mill
FM
Flour mill
FM
Cold store
FM
Electric generator
A typical layout of an alpine hydro village is shown in Fig. 1. The capacity and number of each type of hydro machines will depend upon the size of the village and characteristics of available water stream. The following units have been included in the scheme. 1. A small cold storage and ice mill run by a floating water mill. The capacity of the plant can be around 10 kW. Such cold storages can be built along the flow of water stream to store potatoes, fruits and other perishable products. This will help to avoid spoilage of such products and realize better returns by disposing of uniformly over the year. Similarly, the ice mills attached to the cold storage can be used to chill and store the milk cooperatively and can be drawn for disposal to collection centres and self use efficiently. 2. Electrical generator run by a floating water mill. The capacity will match the village requirement of electricity for lighting, heating, cooking and
Irrigation and drinking water
256 Hydraulic Machines running of radios, TV sets and other small electrical gadgets in the houses, schools, dispensaries and community centres. 3. Irrigation pump run by a floating water mill. The capacity will suit the irrigation water requirements. 4. Industry park consisting of flour mills, rice mills, threshers, saw mills, vertical lathes, trip hammers, carpet looms, etc., run by improved Himalaya mills individually. The capacities can be standardised at 5 to 10 kW each. The industry park will meet all the domestic and agricultural requirements and make the village independent of imports of implements and spare parts. This will help to produce the goods like wooden, brass and other vessels, toys, beauty carvings, woollen carpets for exports with mechanical aids. All packing boxes for export of forest, agricultural and animal yields will be prepared in the industry park. 5. Hydraulic ram to lift the used water from tail race for irrigation and domestic water consumption.
CONCLUSION The proposed alpine hydro village can meet the productivity requirements for selfsufficiency as per Gandhian Model. A technical institute for the development of hilly areas should be set up who can undertake research and development for the improvement and adoption of hydraulic machines for a hydro village. It should also plan and implement such schemes and monitor the results. Such hydro villages can go a long way in bringing prosperity and self-sufficiency to these villages. The migration of population can be arrested and the county can be prepared to absorb high technology in the required sectors.
REFERENCES 1. Singal, R.K., Floating Type Micro Hydro Power Plants. Electrical India, Bombay, 30th September 1984. 2. Singal, R.K., Non-Conventional Energy for Lift Irrigation Electrical India, Bombay, March, 1985. 3. Upgrading and Further Development of Traditional Devices for the Utilisation of Hydroenergy, Bremen Overseas Research and Development Association, Bremen.
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Operation and Maintenance of Centrifugal Pumps for Energy Conservation In many industries, pumps may be forming more than 40% of the auxiliary equipment and if proper design and operation of the pumping equipment can help to conserve even a few per cent of the huge power consumed by them, it will serve our country well. Here we will examine various sources of power saving which can be achieved in pumping installations by adopting proper approach to their operation and maintenance. Never before in the history of industrial society we have faced as pressing a need to conserve energy as now. The high cost of energy and scarcity of fuel such as oil or natural gas, or even coal have become a hard fact of life. In addition, new constraints are being introduced on the use of power from the point of view of ecology. Therefore, it is imperative to examine all energy-consuming processes with a view to improving their overall efficiency. And since every industrial process which underlines our modern civilization involves the transfer of liquids from one level of pressure or static energy to another, pumps have become an essential part of all industrial processes, and, in turn, major consumers of energy themselves. Some important applications of centrifugal pumps are given in Annexure-I showing the widespread use of pumps in various sectors of economy. In many industries, pumps may be forming more than 40% of the auxiliary equipment and if proper design and operation of the pumping equipment can help to conserve even a few per cent of the huge power consumed by them, it will serve our country well.
PROPER IMPELLER DIAMETER A centrifugal pump is a velocity machine. The peripheral speed of the impeller depends directly upon its diameter. Therefore, the pump capacity which is a direct
258 Hydraulic Machines function of the velocities, will vary directly as the impeller diameter, the total head, which is a function of the square of the peripheral speed, will vary as the square of the impeller diameter. Finally, since power consumption varies as the product of the head and the capacity, the power will vary as the cube of the impeller diameter. Therefore, it is very important to run a pump with proper size of the impeller. However, one of the greatest sources of power wastage is the practice of oversizing a pump by selecting operating conditions with excessive margins in both capacity and total head. The pump is then operated with considerable throttling to limit its delivery to the desired value thereby wasting power by as much as 5, 10 or even 15%. The point in question can be illustrated by an actual example given in Annexure-II. There is a saving in power consumption of 11.5% by using an impeller of 355 tip diameter in place of 375 mm diameter. The power savings may jump to 18% initially when the piping is new. There is additional saving in capital expenditure due to smaller size of motor required. It is true that some margins should always be included, mainly to provide against the wear of internal clearances which will, with time, reduce the effective capacity of the pump. However, reasonable restraints must be exercised to avoid incorporating excessive safety margins into the rated capacity conditions of service. It will be a safe practice to design a pump with normal margin of 10% either on capacity or head (not both) and order a pump with impeller one size smaller. On a future date, when the internal clearances have increased due to corrosive and erosive action of the liquid, or additional demand is created, the bigger impeller can be procured and installed into the existing casing to cover up the additional demand in pump capacity. On an existing installation, run a performance test and plot the true system head curve. A reasonable margin can be selected and three choices become available to the operator. Trim the existing impeller to meet the more realistic condition of service required for the installation. The diameter ratio can be approximated from the square root of the two pressures. After the impeller is cut the vane tips should be filed very carefully to approximate the finish of the vanes as they appear new. A replacement impeller with the necessary reduced diameter may be ordered from the pump manufacturer. The original impeller can be stored for future use for meeting the loss of pump capacity due to increased friction losses. A replacement narrower impeller may be ordered from the pump manufacturer and the original impeller may be stored for possible future use. A narrower impeller will have its best efficiency at a lower capacity than the normal width impeller.
Case IIIOperation and Maintenance of Centrifugal Pumps... 259
VARIABLE-SPEED DEVICE Since it is the peripheral speed of impeller that determines the total head and the capacity of the pump, obviously it is immaterial whether the peripheral speed is changed by cutting down the impeller diameter or the pump speed. When the speed is changed, the capacity for any given point on the pump characteristic curve varies directly as the speed, and at the same time, the head varies as the square of the speed while the power varies as the cube of the speed. A magnetic drive or a hydraulic coupling can be interposed between the pump and the electric motor to meet the exact required conditions of service without throttling by reducing the pump operating speed. The method of calculations of speed change is illustrated in Annexure-III. The potential power savings are huge. If the number of hours of operation at various expected pump capacities are known, the yearly savings in kWh and energy cost can be calculated to justify the installation of variable speed device. If techno-economically it is not feasible to install a variable speed device, use smaller impeller as discussed in para 2.
PARALLEL OPERATION In most of the installations, two 50% capacity pumps are used which operate in parallel to deliver the required flow under full load conditions. However, the flow requirements vary widely and important power savings are possible through improved operating practices. It is clear from Annexure-IV that very important power savings are possible if one pump can be shut down when the flow requirements are 50% or less. The evaluation of any energy conservation programme consists of the comparison between the amount of energy saved and the cost of the implementation of the programme. The remarkable fact is that the cost of implementation in this case is exactly zero. As a bonus, the life of the pump improves as pump when operating more nearly to their best efficiency point have maximum life.
MAINTENANCE OF INTERNAL CLEARANCES The leakage losses in a pump increase due to wear of internal clearances, thereby affecting the HQ curve and power input. The rate of wear depends upon differential pressure across the pump, type of fluid handled (corrosive and abrasive) severity of service, material of pump internal parts, etc. The effect of leakage losses on power consumption depends upon the specific speed of the pump i.e., lower the specific speed, the head per stage is higher and therefore the wear is increased due to higher differential pressure. Therefore, there is a need to attend to restoration of internal clearances specifically for pump with lower specific speed. The effect of wear on the deterioration of HQ curves is shown in Fig. 1.
260 Hydraulic Machines Initial H-Q curve
Head
H-Q curve after wear
Increase in leakage
Capacity
.ig. 1 Effect of internal wear on H-Q curve of C F pumps
Leakage loss % of power output
The pressure drop across all the internal joints is of the same order of magnitude when expressed in terms of pressure drop across units of linear dimensions, all running joints will wear at about the same rate. It is thus possible to predict when internal clearances should be renewed without dismantling the pump. The leakage through a balancing device is returned to the pump suction via an external pipe, in which a flow measuring orifice can be located. This leakage can therefore be readily measured. The measurement can be used to monitor the progress of the wear experienced by the pump. The relation between leakage losses in percentage of power output and specific speed is shown in Fig. 2. The leakage losses when the pump is new are of the order of 1.2 per cent for pumps of high specific speeds of 2200-2500. Thus, when the internal clearances have increased to the point that this leakage has doubled, we can regain approximately 1.2 per cent in power savings by restoring the pump clearances. 10 8 6 4 2 0
0.5
1.0
1.5
2.0
2.5
3.5
3
Specific speed X 10 .ig.
Leakage losses for double suction pumps
Case IIIOperation and Maintenance of Centrifugal Pumps... 26
On the other hand, for pump with low specific speeds of the order of 700-800, the leakage losses are of the order of 5 per cent. If the clearances are restored after the leakage losses doubled, we can count 5 per cent power savings. Therefore, restoring clearances of the lower specific speed-type pumps gives us greater returns in terms of the reduction of leakage losses. It may well be, then, that economic conditions may permit us to let clearances of higher specific speed pumps increase beyond values double those of a new pump while making it attractive to review internal clearances of low specific speed pumps before they have doubled. The final decision, at any rate, must be based on comparing the costs of an overhaul with the value of the potential power savings, considering actual energy costs.
MECHANICAL PROBLEMS In addition to hydraulic or system design, a number of mechanical problems must be attended to save excessive power consumption in the operation of pumps. Rubbing between the rotating and stationary parts of a pump, for whatever cause, will of course increase the power consumption. This rubbing may be caused by misalignment of pump and driver, by a bent shaft or by incorrect reassembly of a pump during maintenance and replacement of internal parts. Excessive wear of parts at the internal running clearances will generally lead to increased power consumptions. As already discussed, increased clearances cause increased internal leakage and the impeller is forced to handle a much higher flow than is being discharged by the pump into the system. Improper conditions at the stuffing boxes will frequently increase power consumption. This is particularly true for small pumps for which the mechanical losses at the stuffing boxes represent a significant portion of the total power consumption. The packing may have been improperly installed or it may be inadequate for the actual operating conditions; the glands may have been tightened excessively. Anyone of these problems will increase the friction losses at the packing. If a significant foreign matter becomes lodged in the impeller water ways, the pump efficiency will suffer and there will be increased power consumption.
CONCLUSION The operator of a pump must analyze each of the possibilities, starting with the most probable cause, and to correct the problem that leads to increased power consumption. Potential of energy conservation is huge. Tap it.
262 Hydraulic Machines Annexure-I APPLICATIONS OF CENTRIFUGAL PUMPS Rural Sector Agriculture:
Irrigation, farming, insecticides and fertilizer spraying, forestry, fish farms and hatcheries Water Supply: Drinking and waste water Cottage & Rural Industries
Transport Sector Harbours, docks and canals. Road, rail and air tranport Tranportation equipment manufacture, e.g. car, lorry, bus, aircraft, rolling stock and naval. Marine Commercial ship-building, and ship owners. Mining & Quarrying Coal mining, metallic and non-metallic ores, clay, salt, sand, gravel, etc.
ConstructEon Pipeline construction Civil engineering works e.g. harbours and dry docks, road, bridges, etc.
ServEce IndustrEes & ServEces Heating, ventilating and air-conditioning, e.g., hospitals, schools, offices, etc. Domestic water supply, e.g. domestic booster-sets and domestic hot water circulations. Fire-fighting and fire prevention, e.g. sprinkler system, etc. Armed forces, atomic energy establishments, research establishments postal services and communications and other miscellaneous local government services. Laundries, dry cleaners and dryers, sports establishments. Commercial construction services, hotels, hospitals, schools, offices.
Power GeneratEon and DEstrEbutEon Gas: Electricity:
Processing and distribution. Electricity generation and transmission.
Case IIIOperation and Maintenance of Centrifugal Pumps... 263
Water supply: Sewage:
Municipalities and other agencies. Sewage treatment and disposal.
OEl ExploratEons and ProductEon
Oil and natural gas exploration and production. Petroleum refinery plants Petroleum distribution centres Coal, coke and tar processing
EnCEneerEnC IndustrEes Mechanical machinery and component manufacturers, e.g. machine tools, pump, compressors, food machinery, heaters, coolers, fasteners, instruments, etc. Boiler manufacturers. Dish and bottle washing plant manufacturers. Metal parts washing and spray painting plant manufacturers. Filtration plant manufacturers Commercial industrial refrigeration. Electrical motors, generator, radio, television, radar, telephony equipment, wires and cables, other domestic electrical equipment. Switchgear manufacturers.
Food and BeveraCes Cereal processing, bakeries and confectionery products, canneries; fruit and vegetable processing and products, animal and marine fats and oils; dairy products, e.g. milk, cheese, butter, ice creams, meat processing, e.g. slaughter houses and bacon factories. Distilleries, breweries, soft drink manufacturers. Cane and beat sugar processing. Tobacco.
Process IndustrEes Petrochemicals, plastic and synthetic rubber manufacturer. Chemical manufacturers, e.g. fertilizers, inorganic chemicals, organic chemical (acids, solvents, ammonia, chlorine, etc.) inks, paints, photographic chemicals, polishers. Water treatment plant manufacturers and contractors. Pharmaceuticals and cosmetics manufacturers.
264 Hydraulic Machines Paper, pulp and printing manufacturers and contractors, printing, publishing and book binding. Textiles: Synthetic fibres manufacturers. Consumer goods: Textiles, tanneries and leather processing, furs, leather goods manufacturers, clothing and footwear, plastic consumer goods manufacturers.
IndustrEal MaterEals Primary metal production, foundries and rolling mill equipment. Electropolishing and plating specialists and plant manufacturs. Bricks, fire-clay and pottery, glass and glass-container manufacturs, cement readymix concrete. Timber, furniture and bedding, other wood and cork manufacturs.
Annexure-II SELECTION OF A CENTRIFUGAL PUMP Design Conditions Max. desired capacity Static head Total friction losses assuming 15 years old pipe Total head required Margin on capacity at the rate of 10% Total design capacity Margin on total head Total design head Pump Selection I Impeller diameter Pump power required at 220 lps Motor capacity Power consumption at rated capacity and head (achieved by throttling)
= 200 lps = 35 m = 18 m = 53 m = 20 lps = 220 lps =7m = 60 m = 375 mm = 130 kW = 150 kW = 123 kW
Pump Selection-II (Alternate) Impeller diameter Capacity at intersection with head capacity curve Margin on capacity Power consumption at intersection at 200 lps capacity Motor capacity Saving in power consumption
= 355 mm = 210 lps = 5% = 108 kW = 110 kW = 11.5%
Case IIIOperation and Maintenance of Centrifugal Pumps... 265
Annexure-III CALCULATION OF VARIABLE PUMP SPEED The head-capacity curve of pump at 180 rpm is known. We have to determine the pump speed to deliver 150 lpm at a total head of 45 m. This point does not lie on the available H-Q curve. Select an arbitary capacity greater than 150 lpm, such that it will be located on the parabola defined by the affinity laws:
FG IJ = FG Q IJ H K HQ K
Q! n H n = !; ! = ! Q n H n
Assume Q3 = 185 lpm H3 = H2
FG Q IJ HQ K !
= 45
!
FG 185IJ H 150K
= 68.45 m
Draw the portion of the parabola defining the affinity laws between 150 lpm, 45 m and 185 lpm and 68.45 m. It can be assumed that this is essentially a straight line. The intersection of this straight line with the head-capacity curve of the pump at 1800 rpm corresponds to 180 lpm and 65 m. We can now determine the speed required to meet the desired conditions of 150 lpm and 45 m. 45 = 1497 rpm (say 1500 rpm) 65
n2 = 1800
If variable speed device is not available or cannot be justified on the basis of energy savings, the use of smaller or narrower may be used. In the above example, say the original impeller has a diameter of 375 mm. 45 = 312 mm (say 315 mm i.e. the next size of impeller 65 available with the manufacturer) The power input to the variable speed device can be calculated as:
D2 = 375
P2 P P
Fn I =P G J Hn K F 1500IJ =G H 1800K 1
!
!
= 0.578 m.
i.e. a saving of 42%
266 Hydraulic Machines Annexure-IV PARALLEL PUMP OPERATION Full Load CondEtEons Flow rate Static head Friction loss in the system Total head Mode of operation
= 240 lps =6m =9m = 15 m = Constant speed
Pump SelectEon No. of pumps Capacity of each pump Total head Power consumption of each pump Total power consumption of the installation
=2 = 120 lps = 15 m = 18 kW = 36 kW
Part Load OperatEnC CondEtEons Flow rate required Total head
= 120 lps = 15 m
Both Pump OperatEnC Flow rate of each pump Head Power consumption Total power consumption
= 60 lps = 20 m = 15.6 kW = 31.2 kW
SEnCle Pump OperatEnC It is possible to shut down one pump and meet the requirements with throttle considerably reduced. Flow rate = 120 lps Head = 15 m Power consumption = 18 kW
Power SavEnCs Savings resulting from single pump operation = (31.2 18) = 13.2 kW
Case IIIOperation and Maintenance of Centrifugal Pumps... 267
13.2 ´ 100 = 42.3% 312 . Yearly operating time when flows required are 50% or less Yearly savings
=
= 20% = 8%.
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Hydraulic Design of a Boiler Feed Pump to Ensure Stable Operation at Reduced Flows The boiler feed pumps for industrial and power station boilers have to operate often at reduced capacities to meet the changing demand of steam and electricity. The operation of centrifugal pumps at reduced capacities lead to a number of unfavourable results seriously affecting the pump reliability. Some of these, such as internal recirculation of flow inside the pump have been recently studied. Here we will discuss these unfavourable results and analyse various design factors which can control unstable operation of the pumps at reduced flows. The commissioning problems of boiler feed at Rajasthan Atomic Power plant at Kota and modifications carried out in the light of the above studies are described below.
INTRODUCTION The demand curve of total electrical power for a city or a region varies widely over a day, a month and a year. The fluctuations in the demand are expected to vary much more with further industrialization and improvement in the living standards as less and less industries might be working at night with longer stoppages. The fluctuations in hourly demand of electricity may be as large as 30% or even more of the peak demand. It is very difficult to force a change in the energy consumption pattern even by differential tariff of power. Therefore, the electrical power supply system must be designed to cope with the demand schedule. It was estimated that by the year 2000 AD, more than 95% of electrical power will be produced by thermal and nuclear power plants. These plants are high temperature systems and most suited to operate at constant loads. These must be especially designed to meet wide load fluctuations.
Case IVHydraulic Design of a Boiler Feed Pump... 269
Boiler feed pump is the most important link between the boiler house/nuclear reactor and the turbine hall. The load changes or even short shut-downs of the power plants will necessitate the feed pump operation at reduced flows. The industrial boilers generating steam for process heating and/or power generation for captive uses have to operate frequently at variable capacities and very often at 30% to 50% or even at 10% of rated value. Operation of centrifugal pumps at reduced capacities leads to a number of unfavourable results which may take place separately or simultaneously. The pump operates at the best efficiency point, the increased thrusts overload the bearings, the temperature rise through the pump increases, probability of flow separation and cavitation damage increases, hydraulic surges due to unstable head-capacity characteristics or internal recirculation of flow or steam flashing may cause accelerated cavitation, vibrations, noise. All these factors causing poor operation and premature impeller wear and deterioration must be avoided. Here we will discuss these problems in some details with reference to the experiences of nuclear power plants. The pump designers must take care of these phenomena to ensure stable operation of boiler feed pumps at reduced flows.
DESIGN PARAMETERS The boiler feed service presents a number of problems not encountered in any other field of application. Multi-stage, single-suction, opposed impellers, horizontally split casing centrifugal pumps are mostly used to handle large flows of saturated feed water at high temperature and pressure. Compared to thermal power stations, the pumps for nuclear plants may have to handle higher flows at moderate heads. The pump hydraulics is mainly decided by the head, capacity, speed and suction conditions. The required hydraulics is achieved by providing suitable clearances and impeller dimensions. Table 1 Rajasthan atomic power plant at Kota Design Parameters for Boiler Feed Pumps Number of boilers Total steam output Steam pressure at drum Steam temperature at drum Moisture in steam Feedwater flow Feedwater temperature Reheater condensate return Deaerator height above pump cl
8 1162 t/h 40 kg/cm2g 250°C 0.25%, max 1072.5 t/h 171°C 89.5 t/h 23.77 m
270 Hydraulic Machines Margin in Head and Capacity A proper margin in head and capacity should always be included over and above the conditions of service based on maximum flows to be encountered in operation to compensate for the effect of wear in service and as a precaution against underestimating the maximum capability of the main unit or system pressure losses. As a general rule, an 8% margin to the maximum boiler capacity is considered in arriving at the rated capacity of the boiler feed pump installation. Over-liberal margins result in power wastage and excessive wear of regulating valves especially in motor-driven constant-speed pumps. The margin can be added to the desired capacity or to the required total head, the net result on the head-capacity curve is essentially the same. But adding margin to the total head has the advantage of keeping design capacities nearer to the best efficiency point of the pump curve; otherwise the pump will be running constantly at a lower efficiency and in a range where serious pulsations and hydraulic surges can take place. In order to overcome the above adverse effects of oversizing, the margin in head or capacity of boiler feed pumps has been reduced to as little as 4% in the latest utility power plants. The drum pressure to be used in calculating the rated pump conditions need not incorporate the 6% over-pressure setting of the safety valves. The average boiler feed pump head-capacity curve has a rise of 15% to 20% from its design head to shut off, the steepest portions occurring near the rated capacity. Even if the pump had to feed water into the boiler under the adverse conditions of all the safety valves blowing, it would still be able to maintain a very respectable rate of flow into the boiler and protect it against mishap. Therefore, the pump should not be oversized as it leads to much wasted power and requires excessive pump and valve maintenance due to throttle wear. The specification of the boiler feed pumps intalled at Rajasthan Atomic Power Plant at Kota shows a margin of 10.6% on the rated capacity which is little on higher side (Table 2).
Pump Speed The boiler feed pumps are very large pumps and may consume power as high as 2% of the rated capacity of the turbo-generator. Therefore, very high efficiencies are expected from these types of pumps. Considering that pump efficiency increases as specific speed increases, the boiler feed pumps are designed with specific speeds in the range of 2000 to 2500 (FPS units). However, the pump speed will also depend upon the type of driver used and its speed. In the case of multistage pumps for high heads, the desire to increase the specific speed leads to a reduction of the head per stage and hence to an increase in the number of stages required. This in turn will result in longer shaft spans, greater static deflection, and thus lower reliability. The design for slightly higher efficiency under these circumstances clearly works against the user.
Case IVHydraulic Design of a Boiler Feed Pump... 271 Table 2 Rajasthan atomic power plant at Kota Specifications: Boiler Feed Pumps Number of pumps 2 ´ 100% duty Rated capacity 19985 lpm Total head 500 m Speed 3000 rpm Type of pump Multi-stage constant speed centrifugal pumps Number of stages 3 Manufacturer Bingham pump Co Ltd, Canada Margin on capacity 10.6% Head per stage 166.7 m Recirculation bypass capacity 6.6% of rated capacity Specific speed 2000 FPS units Number of impeller vanes 1st stage = 5 original, 7 after modifications 2nd stage = 7 3rd stage = 7
In order to obtain high pump reliability, high speed pumps are used which can give higher heads per stage and develop the required pressure with lesser number of stages. Here the speed is limited by the capabilities of its driver and availability of NPSH of the installation. For a given specific speed and fixed head per stage, the pump speed must decrease with increasing capacity. When integrated properly into the station heat balance, the use of auxiliary steam turbines can provide significant savings in overall heat rate and for this reason there is a strong trend towards their use as boiler feed pump drives in thermal power plants. However, in nuclear power plants, electric motors are preferred as pump drivers to reduce initial cost and to avoid complicated layout. The steam turbine is inherently a variable speed machine and does not require the use of a variable speed drive such as a hydraulic coupling or the magnetic drive. The most important attraction of the use of steam turbine is increased net power output of the station by eliminating the electric power required by the boiler feed pumps. The hydraulic design data of the boiler feed pumps used at Rajasthan Atomic Power Plant at Kota are given in Table 2. The data confirms to the prevailing practice of boiler feed pump design.
Suction Conditions There is a danger of flashing in the pump suction whenever the load on the steam turbine is dropped suddenly and there is a pressure drop in the bleed line from turbine to deaerator. The water in the storage space of the deaerator and pump
272 Hydraulic Machines suction line is at a higher temperature than that corresponding to the reduced pressure. Violent vaporization of water will take place until the pressuretemperature equilibrium is established again. A sudden dynamic pressure drop may occur in the suction line when the discharge valve is suddenly opened to meet increased demand and again creating the conditions of flashing. Therefore, NPSH requirements for boiler feed pumps are higher than required as per cavitation constant sigma to prevent vapour binding as a result of sudden reduction of electric load, or sudden increase of pump capacity.
Recirculation Capacity There is an increase in enthalpy of feed water from suction to discharge manifesting as temperature rise. There can be overheating of pumps when the capacity is reduced below a safe limit as enthalpy increase is absorbed by a smaller quantity of water. The minimum flow required through the pump is calculated for each of the following situations and final decision is then based on the greatest of these minimua. (i) increase in radial thrust at reduced flows, (ii) temperature increase of water from suction to discharge, (iii) internal recirculation in the pump suction or discharge or both at reduced flows, (iv) excessive power requirement at reduced flows overloading the driver, and (v) air or vapour lock at reduced flows. The amount of water bypassed varies from 5% to 10% of the normal capacity. If this limit is exceeded and bypass flow is connected back to the pump suction, flashing can occur as higher temperature water is mixed at the pump suction. The larger sizes of boiler feed pumps and the higher design pressures have led to very significant values of recirculation bypass, requiring large and expensive valves and introducing appreciable disturbances when surges arise from the sudden increase or decrease in flow as the bypass controls open or close the recirculation. Dual bypass control valves and orifices working in parallel can be used. As the feed water flow decreases, the first of the two valves opens and with further reduction in flow the second valve opens. The closing of valves as demand increases follows the same pattern. The feed water in the bypass is returned to the deaerator and there will be saving in the supply of steam for deaeration.
Impeller Geometry In addition to sudden changes in pump speed due to abrupt variations in electric loads or sudden increase of pump capacity, there are chances of flow separation from the vanes in case of operation at reduced flows. These transient conditions
Case IVHydraulic Design of a Boiler Feed Pump... 273
of operation can cause hydraulic surging (self-excited vibrations) leading to unstable operation, sharp and jerky pressure rise and damage similar to cavitation. The design of impeller to prevent hydraulic surging should include rounding off of inlet vane edge, increase in number of vanes and highly backward-curved vanes. The increase in a number of vanes ensures favourable cross-sectional shape but reduces the flow area affecting the impeller flow capacity. Normally seven vanes are used in boiler feed pump design. In boiler feed pumps where low NPSH requirements are important to restrict the height of deaerator installation within reasonable limits, the entrance velocities are kept low. It is customary to use exaggerated impeller suction eye to reduce entrance velocities. This, however, can lead to internal recirculation of flow within the pump when operated at reduced flows.
OPERATION AT REDUCED FLOWS Operation of centrifugal pumps at reduced flows can lead to a number of unfavourable results which must be anticipated or circumvented. (i) Obviously, the pump operates at less than its best efficiency. (ii) Particularly if the pump is of single-volute design, it will develop a radial thrust which increases the load on its radial bearings. A pump that will operate at such flows must be designed to accept this higher bearing load. (iii) As the capacity is reduced, the temperature rise through the pump increases. To avoid exceeding permissible limits, a minimum flow bypass must be provided. (iv) At certain flows below that of best efficiency, centrifugal pumps are subject to internal recirculation both in the suction and discharge areas of the impeller. This can cause hydraulic surging and damage similar to cavitation damage.
INTERNAL RECIRCULATION The phenomenon of internal recirculation and prediction of and control of magnitude of flows at which this occurs have been researched only recently. The constant competitive pressure on manufacturers to build pumps with lower and lower NPSH values forced them to design impellers with large eye diameters. While the resulting lower velocities have little effect on the pump performance at or near its best efficiency capacity, they may lead to hydraulic surges, noisy operation and premature wear at low capacities. These symptoms are caused by internal recirculation at the suction. The flow at the outer eye diameter tends to reverse itself at low flows and causes excessive turbulence through the formation of a vortex.
274 Hydraulic Machines The recirculation of impeller discharge is somewhat similar to the recirculation at the suction. It is engendered by the desire to improve the value of the pump efficiency by increasing the head coefficient of the impeller. This recirculation also creates hydraulic surges and local cavitation at the impeller tips. In multistage pumps, the suction and discharge recirculation are generally limited to the first stage and the subsequent stages have a greater margin between the prevailing pressure and liquid vapour pressure and the effect is either minimised or disappears entirely.
Antistall Rings Certain procedures are available to the designer to minimize the effect of recirculation and to reduce the percentage of best efficiency capacity at which significant suction recirculation starts. These procedures generally increse the minimum required NPSH. Therefore, if the pump requires frequent and extended operation at reduced flows, it is better to select a pump with proper NPSH so that it will not suffer from the effects of recirculation. For the existing pumps, certain modifications are possible. The simplest method is to install an antistall ring in the suction eye of the pump by shring-fitting to reduce the suction eye area and thereby the tendency for recirculation.
Minimum Flow Bypass A centrifugal pump can be designed only for one capacity at a given head and speed, the geometry of the impeller can be ideal for the rated conditions; for all other flows below the rated capacity, the distribution of velocities and flow angles are distorted. The required NPSH does not follow the square of the flow rule but decreases at a lesser rate. The incipient internal recirculation takes place at reduced flow which depends upon each impeller design. It is not possible to calculate or measure the exact value of NPSH required to prevent this cavitation. It has however been established that an impeller with very low NPSH and extremely oversized eye diameter develops an internal recirculation at flows considerably higher than impeller with moderate NPSH requirements. Therefore, the minimum flow bypass line should be designed according to the tendency of the pump to develop internal recirculation. On installations already existing, the minimum flow bypass capacity may have to be increased if internal recirculation occurs. This may be required in addition to the installation of antistall rings in the pump suction eye.
Cold Water Injection into Pump Suction Providing sufficient NPSH for the boiler feed pump by raising the height of deaerator or other means can pose practical difficulties, which can be overcome by
Case IVHydraulic Design of a Boiler Feed Pump... 275
injecting cold water from a suitable point in the feed water cycle when the pump is operating at reduced capacities. Injection of cold water into the pump suction results in the subcooling and available NPSH is artificially increased. The amount of cold water to be injected can be calculated from the amount of subcooling (temperature depression) required by energy balance. In the case of high-pressure boilers where the possible effect of oxygen contamination may be quite severe, the colder water injection has to be tapped from downstream side after the deaerator. A heat exchanger is installed in the suction line of the pump to subcool the water and cooling medium employed can be condensate from the inlet to the deaerator. In order to avoid complication of the subcooling scheme, the cold injection can be directly taken from the condenser hotwell where deaeration of water is reasonably well and bypassing the deaerator by a small fraction (6% to 8%) of feed water may not have serious effects.
CASE STUDY OF RAPP I The pump specifications for the Rajasthan Atomic Power Plant (RAPP) are given in Table 2. These pumps together with other nuclear pumps were supplied by M/s Bingham Pump Co. Ltd., Canada, who had especially developed these pumps and elaborate testing was not carried out by the manufacturer. Subsequent to first commissioning and start-up, high vibrations and noise were noticed on the boiler feed pump. The commissioning tests were carried out at about 30 MW against the rated capacity of the unit of 220 MW as higher connected load was not available in the grid. The pump was dismantled for inspection to find out the reasons for vibrations. Some damage similar to cavitation was found on the first stage impeller wearing ring. The pump was again run installing the spare rotating assembly, but high vibrations persisted. The second pump also showed high vibrations. The shaft mounted oil pump was replaced. Foundation design and quality of grouting were again checked. Detailed alignment checks were made. The individual motors were checked for smooth running. As per the practice of the pump manufacturer, individual impellers and pump shafts were dynamically balanced separately without checking the balance of final rotor assembly. Therefore, the rotor assembly was taken to Walchannagar Industries for balance check as balancing machine was not available in the adjoining areas. The checks showed the imbalance within limits of international practice. It was therefore concluded that the hydraulic design of the pump installation was insufficient for operation at reduced flow. Due to non-availability of grid capacity, it was not possible to check the performance of the pumps at higher loads. Vibration data were analysed in consultation with the manufacturer. As the pumps were the first of their model, elaborate test data were not available on
276 Hydraulic Machines hydraulic performance. The manufacturer suggested certain modifications based on their experience on similar pumps in the USA and Canada. It was a typical case of internal recirculation at reduced flows with consequent noise and vibrations. (i) The profile of the volute sections of the first stage impeller was checked as per templates supplied by the manufacturer. Some grinding was necessary. (ii) The discharge vane tips were chamferred to change the discharge angle. (iii) An antistall ring was installed in the suction eye. The rotor assembly was mounted on a lathe to machine the internal diameter of the suction eye of first stage impeller to suit the OD of antistall ring. The impeller was uniformly preheated to 150°C and ring was shrunk-fit followed by dynamic check of the assembly. (iv) The capacity of the bypass was increased from 22 lps to 63 lps by replacing the orifice block and control valves. (v) The pump was tested after the above modifications but there was no appreciable improvement in the vibration level. Vibration data showed predominant vibrations at 5 N harmonics. (vi) The first stage impeller with five vanes was replaced with one with seven vanes. (Number of vanes in the second and third stages were seven each.) (vii) The capacity of the recirculation line was further increased to 145 lps by changing the pipeline from 75 NB to 200 NB. An orifice of 94 mm diameter was provided in the horizontal run of the bypass line near the deaerator. The size of the orifice was restricted by the maximum flow velocity permissible (4.88 m/s). The boiler feed pumps were commissioned after carrying out all the above modifications. The pumps have been working smoothly without vibrations and caviation.
+)5-
8
Hydro Power Plants: Great Ecological Havocs The utilization of energy for the performance of tasks has always been integral with mans standard of living. Different branches of industry, agriculture, transport, commercial activities, domestic comforts, entertainment, all depend upon ample supply of power. But power production is associated with disturbing the ecological balance and environmental pollution. The thermal power plants eject large quantities of solid and gaseous impurities and thermal discharges into the atmosphere, adversely affecting ecological conditions in the vicinity of the plant. Due to harmful effects of radioisotopes on living organisms, the handling of nuclear fuels and wastes requires utmost caution. The nuclear power plants pose great dangers to human life in the case of accidents. Hydropower plants are considered the cleanest and least harmful of power plants because there is no ejection of toxic or radioactive materials or thermal discharge to the environments. However, there is a need for storage of large amounts of water behind the dams requiring submergence of vast areas of land under water. This results in large transfer of population and uprooting and rehabilitation of villages, towns and cities. The flow of water through the rivers downstream is reduced causing water pollution. The ecological balance can be much disturbed in some cases that hydropower plants can cause ecological havocs. These plants can be very dangerous hidden monsters. The damage caused by these plants have moral and social significance but the economical impact in terms of money lost has far reaching consequences. An attempt has been made to estimate the economical damages which can be caused by large hydropower plants with upstream reservoirs.
SLOW 2OISONING During my school days, I happened to attend one election meeting in Punjab. The speaker addressing the peasants informed them that the water they were getting in the irrigation canals was no good for cropping as the same had been diluted by extracting energy at the Bhakra Hydroelectric Plant. That election gimmick
278 Hydraulic Machines amused me very much. After so many years, I am compelled to take his words seriously. The fertility of Indian fields especially plains of northern part of India depends upon the alluvial soil and nutrients brought along by rivers from the mountains and picked up while flowing over hill slopes. Through all centuries, the fertility of these fields has been maintained by the feed of suspended nutrients by our rivers and irrigation canal system. For the construction of large power plants, these rivers are dammed for storage of water to ensure a regular flow with static head created. In the process, the water is decanted of most part of the suspended nutrients as the latter settles down as silt in the reservoirs by sedimentation. The fields are devoid of the natural fertilization process as these do not get the nutrients through irrigation water. The land can slowly become completely barren. It may be a very slow process taking hundreds of years like the slow lead poisoning of the Romans by the lead utensils used by them for drinking wine. If not prevented, the denial of natural fertilization can lead to very serious consequences over the years resulting in loss of fertile land for agriculture and other allied economic activities. This natural process has been the main source of flourishing of rich Indian civilization and agricultural society of Indus-Gangetic plains. By damming the rivers for the construction of hydro power plants, we may be working against a very important natural process of land fertilization. The estimation of economical damages which may be caused by the loss of land fertility due the construction of a hydro power plant is a very complicated problem. The damage can be determined on the basis of the loss of crop due to exponential decrease in land fertility or by the additional expenditure required to prevent the loss of land fertility by the use of synthetic chemical fertilizers. The manufacture of these synthetic fertilizers consumes large quantities of useful natural resources and the energy spent in their manufacture may be much more than obtained from a corresponding hydro power plant. The long-term harmful effects of using the chemical fertilizers on soil structure and fertility are still not fully understood. The damage to the agriculture due to this slow poisoning causing the loss of land fertility can be expressed as follows: Dsp =
n
å[Dw. e
BM t
Dp. A
F J
Da. e B= t + Af. A
B J
]
where, Dsp = Damage due to slow poisoning of natural fertility loss. Dw & Bw = Coefficients of damage due to withdrawal of land due to loss of land fertility. Dp & Bp = Coefficients of damage due to loss of productivity in agriculture. Da & Ba = Coefficients of damage to animal breeding due to loss of fodder. Af & Bf = Coefficients of expenditure on increased use of chemical fertilizers. n = Time under consideration.
Case VHydro Power Plants: Great Ecological Havocs
279
L)ND G4)B The hydro power plants have a very low material utilization efficiency based on Einsteins Equation, E = mc2. There is a need for storage of large amount of water behind the dams requiring submergence of vast areas of land under water. It takes 700 ´ 106 tonnes of water flow through a turbine of 1 GW capacity to generate 120 ´ 1012J of energy. The intrinsic energy of this mass of water is 630 ´ 1026 J. Hence, the material utilization efficiency is (120 ´ 1012/630 ´ 1026) ´ 100% = 0.19 ´ 1012 per cent only. Hydro power plants have the lowest material utilization efficiency compared to 102% efficiency of nuclear power plants. The storage of large quantity of water results in large transfer of population and uprooting and rehabiliation of villages, towns and cities. The local ecological conditions are adversely affected by the water reservoir behind the dams which effect the local climate, water pressure on the land and surrounding terrain enclosing the reservoir and water level of the underground water. This can lead to serious waterlogging problem affecting the crops, plants and animal lives in large surrounding areas. In addition to moral and social consequences, the economical problems can be very serious for the affected population as echoed by the hill people against Tehri Dam in Uttarakhand. The ecological balance is so much disturbed in some cases that Silent Valley Project in Kerala had to be abandoned on this account. The seismic characteristic of the location can be completely changed by the hydrostatic pressure of the huge water reservoir created. The damage due to land grab for submergence underwater may be estimated as follows: Dlg =
n
å(Dagr + Dfor + Dm + Dr + Ar)
where, Dlg = damages due to land grab, Dagr = damage to agriculture, Dfor = damage of forestry, Dm = damage to mining for non-availability due to submergence under water, Dr = damage to fish farming, Ar = expenditure for rehabiliation of population uprooted. where, Dagr = Dw + Dp + Da Dw = damage due to loss of availability of land for agriculture. Dp = damage to agriculture productivity in the surrounding region due to waterlogging. Da = damage to the animal breeding due to loss of pastures and fodder. Similar expressions can be developed for damage to forestry.
280 Hydraulic Machines S6)4V)6ION The flow of water through the rivers downstream is reduced due to construction of dams, causing water pollution. The capacity of rivers to carry the garbage of the cities and the towns and the effluents of the industries situated on their banks is greatly reduced. Similarly, the capacity of the thermal power and nuclear power plants and other industries located on the river banks is very much reduced for want of water. The reduced flow of water in the rivers results in the growth of blue algae, reproduction of epidemic bacteria, extinction of water meadows and salination of soil. In a nutshell, the socio-economic life of the people can be completely shattered. It is technically feasible to construct a dam across the Straits of Gibralter and a super-high capacity hydro electric power house to feed whole of Europe with low-cost electric power. But this can lead to lowering of water level in the Mediterranean Sea causing abandonment of a number of existing ports and disturbing the socio-economic life of the South Europe; reduced water pressure can increase volcanic activity and disturb the climate; increase in salinity can adversely affect the sea flora and fauna. The estimation of various monetary losses can be a complex problem requiring the concerted efforts of economists, technologists and other experts dealing with national economy. The loss due to water pollution caused by reduced water flow in the rivers can be estimated from the additional expenditure required to take up preventive measures such as construction of waste water treatment plants for the effluents from the industries and municipalities before discharging into the rivers. The losses can also be summed up from the damages caused to other water consumers such as power plants, other industrial enterprises, agriculture, fishery, etc. by the use of contaminated water. The calculations can be very complicated requiring the use of special nomograms. The loss of industrial sector can be calculated as follows: where,
D1 = D1i + D2i + D3i + D4i + D5i D1i = losses due to unscheduled equipment breakdown and repairs there of. 2 D i = losses due reduced output, D3i = losses due to poor quality of products, D4i = losses due to decrease in service life of plant and machinery, D5i = losses due to rejection of products.
The polluted water has also long-term adverse effects on agriculture, forestry and fishery. D¢agr = D¢w + D¢p + D¢a
Case VHydro Power Plants: Great Ecological Havocs
281
where, D¢agr = loss to agriculture, D¢w = loss due to withdrawal of land from agriculture caused by pollution, D¢p = loss due to fall in agricultural productivity, D¢a = loss of animal breeding. Similarly, where,
Df = Df 1 + Df 2 + Df 3 + Df4 + Df 5 Df = loss to fishery by polluted water, Df1 = loss due to death of mature fish, Df2 = loss due to reduced reproduction, Df3 = loss due to perishing of nutritional organisms, Df4 = loss due to spoiling of spawning areas, Df5 = loss due to impaired quality of fish products.
Similarly, loss to forestry due to polluted water, Dfo = Dfo1 + Dfo2 + Dfo3 where, Dfo1 = loss due to inhibited plant growth, Dfo2 = loss due to bad quality of commercial timber, Dfo3 = extra expenditure for cleaning and renewal of forests Similarly, damage to public health Dph includes expenditure on curing of diseases caused by the consumption of unhealthy water and loss of working hours due to the diseases, etc.
M-+0)NI+)L -X2LOSION The huge amount of water stored at the back of the dam has a great potential to cause death and destruction to hundred and thousands of cities; towns and villages which are situated in the vicinity of the hydro power plants. It can completely wipe out the civilization of a region from the surface of the earth in case of any damage to the dam due to earthquake or material failure or bombardment by the enemy. The mechanical explosion releasing suddenly millions of tonnes of water gushing out with very high momentum can be more devastating than much feared nuclear bombs. It must also be remembered that 90% of earthquakes take place in areas without previous seismic history. The economic loss due to sudden failure of a dam is tremendous and cannot be visualized.
282 Hydraulic Machines +ON+LUSION It is clear from the foregoing discussions that the so called cleanest and least harmful hydro power plants are the most polluting and dangerous power plants. These should be rightly feared as their destruction power can be much more than nuclear bombs. Their effects on the socio-economic life of the people have farreaching consequences. These can reduce the prosperous and fertile lands to barren wastelands not fit for any cropping. All the factors given in above must be properly weighed before planning a hydro power plant. The hydro power, available free of cost in the form of flowing water is an important renewable source of energy. The large potential of hydro energy available from flowing water in the rivers and irrigation canals can be tapped by building run-off river and small hydro power plants without involving the construction of storage dams. The adverse ecological effects and other dangers listed above may be avoided. The construction of dams requires huge investments and long gestation periods offsetting the economic edge of these types of plants over thermal plants. Hydro villages based on small hydro turbines can be very effective in bringing economic prosperity to the population living along the rivers and irrigation canals.
4-F-4-N+-S 1. Singal, R.K., Environmental Pollution by Power IndustryAn Overall view, ElectHical India, 15th February, 84. 2. Singal, R.K. An Alpine Hydro Village, Electrical India, 15th October, 1985. 3. Richter, Volkov and Pokrovosky, Thermal Power Plants and Environmental Control, Mir Publications, Moscow, 1985. 4. Venikov and Putyatin, Introduction to Energy Technology, Mir Publishers, Moscow,1981.
P A R T VI OBJECTIVE QUESTIONS
Objective Questions 285
FLUID FLOW Tick (Ö) the right answer. 1. The expression
2.
3.
4.
5.
6.
¶f + ¶t
z
¶p 1 + (Df)2 + gz = constant, represents 2 r
(a) Steady flow energy equation (b) Unsteady irrotational Bernoullis equation (c) Steady rotational Bernoullis equation (d) Unsteady rotational Bernoullis equation Euler equation of turbine giving energy transfer per unit mass Eo (where U, Vw, Vr and V represent the peripheral, whirl, relative and absolute velocities respectively; suffixes 1 and 2 refer to the turbine inlet and outlet respectively) is given by (a) Eo = U1Vw1 U2Vw2 (b) Eo = U1Vr1 U2Vr2. (d) Eo = V1ww1 V2Vw2.V (c) Eo = U1V1 U2 V2 Which of the following assumptions are made for deriving Bernoullis equation? (1) Flow is steady and incompressible. (2) Flow is unsteady and compressible. (3) Effect of friction is neglected and flow is along a stream line. (4) Effect of friction is taken into account and flow is along a stream line. Codes (a) (1) and (3) (b) (2) and (3) (c) (1) and (4) (d) (2) and (4) The force of impingement of a jet on a vane increases if (a) The vane angle is increased. (b) The vane angle is decreased. (c) The pressure is reduced. (d) The vane is moved against the jet. A streamline is a line (a) Which is along the path of the particle. (b) Which is always parallel to the main direction of flow. (c) Along which there is no flow. (d) On which tangent drawn at any point gives the direction of velocity. The velocity potential of a velocity field is given by f = x2 y2 + const. Its stream function will be given by (a) 2xy + const. (b) +2xy + const. (c) 2xy + f (x) (d) 2xy + f( y)
286 Hydraulic Machines 7. The stream function in a 2-dimensional flow field is given by f. The potential function is (x2 + y2 ) (x2 - y2) ; (b) ; 2 2 (c) xy (d) x2y + y2x. 8. A symmetrical stationary vane experiences a force F of 100 N as shown when the mass flow rate of water over the vane is 5 kg/s with a velocity V 20 m/s without friction. The angle a of the vane is
(a)
V
a a
F = 100 N
V
(a) zero (b) 30° (c) 45° (d) 60° 9. In fluid machinery, the relationship between saturation temperature and pressure decides the process of: (a) flow separation (b) turbulent mixing (c) cavitation (d) water hammer. 10. The stream function is given by (x2 y2). The potential function of the flow will be (a) 2 xy + f(x) (b) 2xy + const. 2 2 (c) 2(x y ) (d) 2xy + f(y)
F GH
The expression r + rgz +
rV 2 2
I commonly used to express Bernoullis JK
equation has units of (a) total energy per unit mass. (b) total energy per unit weight. (c) total energy per unit volume. (d) total energy per unit cross-sectional area of flow. 12. The differential form of continuity equation for two-dimensional flow of fluid may be written in the following form r
¶u ¶v +r =0 ¶x ¶y
Objective Questions 287
in which u and v are velocities in the x and y directions and r is the density. This is valid for: (a) compressible, steady flow (b) compressible, unsteady flow. (c) incompressible, unsteady flow (d) incompressible, steady flow. 13. A 2-dimensional flow in x-y plane is irrotational if (a)
¶u ¶v = ¶x ¶y
(b)
¶u ¶u = ¶x ¶y
(c)
¶v ¶u = ¶x ¶y
(d)
¶v ¶v = ¶x ¶y
If y = stream function and f = velocity potential, the incorrect relation for irrotational flow is, (a) y = xy (b) y = A(x2 y2) (c) f = ur cos q +
u cos q r
FG H
(d) f = r -
IJ K
2 sin q r
Euler equation for water turbine is derived on the basis of: (a) conservation of mass (b) rate of change of linear momentum (c) rate of change of angular momentum (d) rate of change of velocity 16. Select the correct statements regarding 2-dimensional potential flow. (1) Laplace equation for stream function must be satisfied. (2) Laplace equation for velocity potential must be satisfied. (3) Stream lines and equipotential lines are mutually perpendicular. (4) Stream function and potential functions are not interchangeable. Codes (a) (1) and (4) (b) (2) and (4) (c) (1), (2) and (3) (d) (2), (3) and (4) 17. The Euler equations of motion for the flow of an ideal fluid is derived considering the principal of conservation of: (a) mass and fluid as incompressible and viscous. (b) momentum and fluid as incompressible and viscous. (c) momentum and fluid as incompressible and inviscid. (d) energy and the fluid as incompressible and inviscid.
288 Hydraulic Machines 18. The continuity equation for a steady flow states that (a) the velocity field is continuous at all points in flow field. (b) the velocity is tangential to stream lines. (c) the stream function exists for steady flows. (d) the efflux rate of mass through the control surface is zero. 19. Bernaullis equation represents the (a) forces at any point in the field and is obtained by integrating the momentum equation to viscous flows. (b) energies at any point in the flow field and is obtained by integrating the Euler equations. (c) momentum at any point in the flow field and is obtained by integrating the equation of continuity. (d) moment of momentum and is obtained by integrating the energy equation. 20. The streamlines and the lines of constant velocity potential in an inviscid rotational flow field form (a) parallel grid lines placed in accordance with their magnitude. (b) intersecting grid net with arbitrary orientation (c) an orthogonal grid system (d) none of the above. 21. A 2-dimensional fluid flow is described by velocity components U = 5 x3 and V = 15 x2 y. The stream functions will be (a) 2 m3/s (c) 10 m3/s
(b) 5 m3/s (d) 20 m3/s
Key 1. (b) 8. (d) 15. (c)
(a) 9. (c) 16. (c)
(a) 10. (b) 17. (c)
(d) 11. (c) 18. (d)
(d) 12. (d) 19. (b)
(a) 13. (c) 20. (c)
(b) 14. (d) 21. (c)
Objective Questions 289
TURBINES Tick (Ö) the right answer. 1. Two Pelton wheels A and B have the same specific speed and are working under the same head. Wheel A produces 400 kW at 1000 rpm. If B produces 200 kW, then its rpm is (a) 4000 (b) 2000 (c) 1500 (d) 1250 2. Consider the following types of water turbines: (1) Bulb (2) Francis (3) Kaplan (4) Pelton The correct sequence of order in which the operating head decreases while developing the same power is (a) (4) (2) (3) (1) (b) (3) (4) (1) (2) (c) (2) (1) (4) (3) (d) (1) (3) (2) (4) 3. Consider the following energies associated with a Pelton turbine (1) Mechanical energy (2) Kinetic energy (3) Potential energy The correct sequence of energy conversion starting from the entry of fluid is: (a) (1) (2) (3) (b) (2) (3) (1) (c) (3) (2) (1) (d) (1) (3) (2) 4. Which one of the following graphs correctly represents the relations between head and specific speed for Kaplan and Francis turbines.
(a)
(b) Francis
H
H
Kaplan
Kaplan Francis
NS
(c)
Kaplan
NS
Francis
H
(d)
Kaplan Francis H
NS
NS
A draft tube is used in a reaction turbine (a) to guide water downstream without flashing. (b) to convert residual pressure energy into kinetic energy.
290 Hydraulic Machines (c) to convert residual kinetic energy into pressure energy. (d) to streamline the flow in the tail race. 6. In the phenomenon of cavitation, the characteristic fluid property involved is: (a) surface tension (b) viscosity (c) bulk modulus of elasticity (d) vapour pressure. 7. Which of the following statements are correct? (1) Pelton wheel is a tangential flow impulse turbine. (2) Francis turbine is an axial flow reaction turbine (3) Kaplan turbine is a radial flow reaction turbine Codes (a) (1) and (3) (b) (1) alone (c) (2) alone (d) (3) alone 8. Efficiency of Pelton wheel is maximum if the ratio of jet velocity to tangential velocity of the wheel is 1 (b) 1 2 (c) 2 (d) 4 9. The maximum efficiency in the case of Pelton wheel is (where angle of deflection of the jet = 180 b)
(a)
(a)
1 - cos b 2
(b)
1 + cos b 2
cos b 1 + cos b (d) 2 4 10. Match List-I with List-II and select the correct code. List-1 List-11 (Hydraulic 6urbine) ()pplication )rea) A. Pelton Turbine 1. Low head, large discharge B. Francis Turbine 2. Medium head, medium discharge C. Kaplan Turbine 3. High head, low discharge Code A B C (a) 2 3 1 (b) 2 1 3 (c) 3 1 2 (d) 3 2 1
(c)
Objective Questions 291
If H is the available head for a hydraulic turbine, the power, speed and discharge, respectively are proportional to: (a) H1/2, H1/2, H3/2 (b) H3/2, H1/2, H1/2 (c) H1/2, H3/2, H1/2 (d) H3/2, H1/2, H. 12. In utilizing scale models in the designing of turbo-machinery which of the following relationships must be satisfied? (a) (b) (c)
(d)
H Q = Const. 2 2 = Const. ND 3 N D Q D
2
H
= Const;
H = Const. N 3D
H P = Const; 2 2 = Const. QH N D 1
1
NQ 2
NP 2
3 2 H
= Const;
3 4 H
= Const.
Consider the specific speed ranges of the following types of turbines; (1) Francis (2) Kaplan (3) Pelton, The sequence of their specific speed in increasing order is; (a) (1), (2), (3) (b) (3), (1), (2) (c) (3), (2), (1) (d) (2), (3), (1) 14. The level of runner exit is 5 m above the tail race, and atmospheric pressure is 10.3 m. The pressure at the exit of runner for a divergent draft tube can be (a) 5 m; (b) 5.3 m; (c) 10 m; (d) 10.3 m; 15. The gross head on a turbine is 300 m. The length of penstock supplying water from reservoir to turbine is 400 m. The diameter of the penstock is 1 m and velocity of water through penstock is 5 m/s. If coefficient of fiction is 0.0098, the net head on the turbine would be, nearly (a) 310 m; (b) 295 m; (c) 200 m; (d) 150 m; 16. For a water turbine running at constant head and speed, the operating characteristic curves in the figure show that upto a certain discharge q both output power and efficiency remain zero. The discharge q is required to:
292 Hydraulic Machines P h
q
Q
(a) overcome initial inertia (b) overcome initial friction (c) keep the hydraulic circuit full (d) keep the turbine running at no load 17. The maximum number of jets generally employed in an impulse turbine without jet interference is (a) 4 (b) 6 (c) 8 (d) 12 18. If the full-scale turbine is required to work under a head of 30 m and to run at 428 rpm, then a quarter scale turbine model tested under 10 m must run at: (a) 143 rpm (b) 341 rpm (c) 428 rpm (d) 988 rpm 19. The following pairs of formulas represent the specific speeds of turbine and pump respectively. (a)
(c)
1
1
1
1
NQ 2
NP 2
NQ 2
NP 2
3 H4
and
5 H4
(b)
3 H4
and
3
H4
1
1
1
1
NP 2
NQ 2
NP 2
NQ 2
3 H4
and
5 H4
(d)
5 H4
and
3
H4
The cavitation number of a fluid machine is defined as s=
p - p¢ rV 2 / 2
where p = absolute pressure r = density V = free stream velocity
Objective Questions 293
21.
22.
23.
24.
25.
The symbol p¢ denotes (a) static pressure of fluid (b) dynamic pressure of fluid (c) vapour pressure of fluid (d) shear stress of fluid. Which of the statements are correct regarding water turbine governor? (1) It helps in starting and shutting down the turbo-unit. (2) It controls the speed of turbine set to match it with the hydroelectric system. (3) It sets the amount of load which a turbine unit has to carry. Codes (a) (1), (2) and (3) (b) (1) and (2) (c) (2) and (3) (d) (1) and (3). What is the correct sequence of turbines in increasing order of their specific speeds? (1) Francis turbine (2) Pelton wheel with two or more jets (3) Pelton wheel with one jet (4) Kaplan turbine Codes (a) (2), (3), (1), (4) (b) (3), (2), (1), (4) (c) (2), (3), (4), (1) (d) (3), (2), (4), (1) The correct sequence of hydraulic turbines in decreasing order of their specific speeds is: (a) Pelton wheel, Francis turbine, Kaplan turbine (b) Propellor turbine, Francis turbine, Pelton wheel (c) Kaplan turbine, Pelton wheel, Francis turbine (d) Francis turbine, Kaplan turbine, Pelton wheel. A Pelton wheel is ideally suited for: (a) High head and low discharge (b) High head and high discharge (c) Low head and low discharge (d) Medium head and medium discharge The following turbine is used in underwater power station (a) Pelton turbine (b) Deriaz turbine (c) Tubular turbine (d) Turgo-impulse turbine
294 Hydraulic Machines 26. The correct sequence in increasing order of specific speeds of the given turbines is (1) Kaplan (2) Pelton (3) Francis Code (a) (2), (3) and (1) (b) (2), (1), (3) (c) (3), (1), (2) (d) (1), (2), (3) 27. The following shape of draft tube will not improve the hydraulic efficiency of a turbine. (a) Straight cylindrical (b) Conical (c) Bell-mouthed (d) Bent tube 28. The specific speed of a hydraulic turbine is given by (a)
(c)
N P 4 H5
P N 4 H5
;
(b)
;
(d)
N P 5 H4
P N 5 H4
;
.
The pressure acting on the runner of a reaction turbine varies from (a) more than atmospheric pressure to vacuum. (b) less than atmospheric pressure to zero gauge pressure. (c) atmospheric pressure to more than atmospheric pressure. (d) atmospheric pressure to vacuum. 30. Which statement is correct regarding an impulse turbine? (a) There is no pressure variation in flow over the buckets and the fluid fills the passage way between the buckets. (b) There is no pressure variation in flow over the buckets and the fluid does not fill the passage way between the buckets. (c) There is pressure drop in flow over the buckets and the fluid fills the passage way between the buckets. (d) There is pressure drop in flow over the bucket and the fluid does not fill the passage way between the buckets. 31. Kinematic similarity between the model and prototype is the similarity of (a) shape (b) discharge (c) stream (d) forces 32. In reaction turbines, the draft tube is used: (a) for the safety of the turbine (b) to convert kinetic energy of flow by a gradual expansion of the flow cross-section
Objective Questions 295
33.
34.
35.
36.
37.
(c) to destroy the undesirable eddies (d) for none of the above purposes The specific speed of a turbine is defined as the speed of a member of the same homogeneous series of such a size that it, (a) delivers unit discharge at unit head. (b) delivers unit discharge at unit power. (c) delivers unit power at unit discharge. (d) produces unit power under the unit head. Match List-I with List-II and select the correct answer. List-1 List-11 A. Pelton wheel (single jet) 1. Medium discharge, low head B. Francis turbine 2. High discharge, low head C. Kaplan turbine 3. Medium discharge, medium head 4. Low discharge, high head Codes A B C (a) 1 2 3 (b) 1 3 4 (c) 4 1 3 (d) 4 3 2 The degree of reaction of a turbine is defined as the ratio of; (a) static pressure drop to total energy transfer (b) total energy transfer to static pressure drop. (c) change of velocity energy across the turbine to the total energy transfer (d) velocity energy to pressure energy. The movable wicket gates of a reaction turbine are used to: (a) control the flow of water passing through the turbine. (b) control the pressure under which the turbine is working. (c) strengthen the casing of the turbine. (d) reduce the size of turbine. Specific speed of a pump and specific speed of a turbine are: (a)
(c)
N Q 3 H4
N Q 5 H4
and
N P H
and
5 4
N P H
5 4
;
;
(b)
(d)
N Q 3 H4
N Q 5 H4
and
N P 3
H4
and
N P 3
H4
296 Hydraulic Machines 38. Match List-I and List-II and select the correct statements. List-1 List-11 A. Draft tube 1. Impulse turbine B. Surging 2. Reciprocating pump C. Air vessel 3. Reaction turbine D. Nozzle 4. Centrifugal pump Codes A B C D (a) 4 3 2 1 (b) 3 4 2 1 (c) 3 4 1 2 (d) 4 3 1 2 39. Match List-I and List-II and select the correct codes. List-1 List-11 (6urbines) (Characteristic) A. Propeller 1. Inward flow reaction B. Francis 2. Tangential flow impulse C. Kaplan 3. Axial flow reaction with fixed vanes. D. Pelton 4. Axial flow reaction with adjustable vanes Codes A B C D (a) 2 4 1 3 (b) 3 4 1 2 (c) 2 1 4 3 (d) 3 1 4 2 40. Chances of occurrence of cavitation are high if the (a) local pressure becomes very high. (b) local temperature becomes high. (c) Thomas cavitation parameter exceeds a certain limit (d) local pressure falls below the vapour pressure. 41. Which of the following statements are false regarding the relation, specific speed, NS =
N Q 5 H4
.
(a) The relation is valid for both impulse and reaction turbines (b) The value of specific speed depends on the system of units.
Objective Questions 297
(c) The value of specific speed for Pelton wheel will be lower than that for Kaplan turbine for any system of units. (d) The dimensions of specific speed are (FLT 1).
Key 1. (b) 8. (c) 15. (b) 22. (b) 29. (a) 36. (b)
(a) 9. (b) 16. (d) 23. (b) 30. (a) 37. (a)
(c) 10. (d) 17. (b) 24. (a) 31. (c) 38. (b)
(b) 11. (b) 18. (d) 25. (c) 32. (b) 39. (d)
(d) 12. (c) 19. (d) 26. (d) 33. (d) 40. (d)
(d) 13. (b) 20. (c) 27. (d) 34. (d) 41. (d)
(b) 14. (b) 21. (a) 28. (b) 35. (a)
298 Hydraulic Machines 3. PUMPS Tick (Ö) the right answer. 1. On the assumption that a double suction impeller is the equivalent of single suction impellers placed back to back, it is customary to base the specific speed of the double suction pump on, (a) One half of total capacity (b) Three-fourths of the total capacity (c) Full total capacity (d) Double the total capacity 2. Priming is necessary in (a) Centrifugal pumps to lift water from a greater depth. (b) Centrifugal pumps to remove air in the suction pipe and casing. (c) Hydraulic turbine to remove air in the turbine casing. (d) Hydraulic turbine to increase the speed of turbine and to generate more power. 3. A pump is installed at a height of 5 m above the water level in the sump. Frictional loss on the suction side is 0.6 m. If the atmospheric pressure is 10.3 m of water and vapour pressure head is 0.4 m (abs), the NPSH (Net Positive Suction Head) will be: (a) 3.7 m (b) 4 m (c) 4.3 m (d) 4.6 m 4. The pressure ratio of a pump and its
1 th scale model, if the ratio of heads 4
is 5:1, will be (a) 100 (b) 3.2 (c) 179 (d) 12.8 5. Consider the following statements regarding air vessels provided in reciprocating pump installations: (1) The air vessels are fitted both on suction and delivery sides. (2) The air vessels are fitted far from the pump cylinder. (3) The air vessels save energy by reducing the friction loss. Which of these statements are correct? (a) (1), (2) and (3) (b) (1) and (2) (c) (2) and (3) (d) (1) and (3)
Objective Questions 299
Match List I with List II and select the correct answer using the codes given below: List-1 List-11 6ype of pumps )ssociated Features A. Centrifugal pump 1. Air vessel B. Gear pump 2. Draft tube C. Reciprocating pump 3. Guide vanes D. Turbine pump 4. Rotary pump 5. Rotor having blades Codes A B C D (a) 4 2 5 3 (b) 5 4 1 2 (c) 4 2 3 1 (d) 5 4 1 3 7. Match List I with List II and select the correct answer using the codes given below. List-1 List-11 (1ndustrial Needs) (6ype of 2ump) A. Combustible fluid to be pumped 1. Singlestage centrifugal. B. High head but small discharge needed 2. Multistage centrifugal. C. Low head but large discharge needed 3. Positive displacement. D. High head and high discharge needed 4. Jet pump Codes A B C D (a) 3 2 1 4 (b) 4 3 1 2 (c) 3 1 4 2 (d) 4 3 2 1 8. Fluid flow machines use the principle of either (i) supplying energy to fluid, or (ii) extracting energy from the fluid. Some fluid flow machines are a combination of both (i) and (ii). They are classified as: (a) compressors (b) hydraulic turbines (c) torque converters (d) windmills
300 Hydraulic Machines 9. A centrifugal pump driven by a directly coupled 3 kW motor of 1450 rpm speed, is proposed to be connected to another motor of 2900 rpm speed. The power of the motor should by: (a) 6 kW (b) 12 kW (c) 18 kW (d) 24 kW 10. A centrifugal pump gives maximum efficiency when its blades are (a) bent forward (b) bent backward (c) straight (d) wave shaped. 11. A pump running at 1000 rpm consumes 1 kW and generates head of 10 m of water. When it is operated at 2000 rpm, its power consumption and head generated would be: (a) 4 kW, 50 m of water (b) 6 kW, 20 m of water (c) 3 kW, 30 m of water (d) 8 kW, 40 m of water 12. The correct sequence of the centrifugal pump components through which the fluid flows is (a) Impeller, suction pipe, foot valve and strainer, delivery pipe. (b) Foot valve and strainer, suction pipe, impeller, delivery pipe. (c) Impeller, suction pipe, delivery pipe, foot valve and strainer. (d) Suction pipe, delivery pipe, impeller, foot valve and strainer. 13. In a fluid coupling, the torque transmitted is 50 kNm, when the speed of the driving and driven shaft is 900 rpm and 720 rpm respectively. The efficiency of the fluid coupling will be: (a) 20%; (b) 25%; (c) 80%; (d) 90%; 14. Which of the following statements are correct regarding fluid coupling? (1) Efficiency increases with increase of speed ratio. (2) Neglecting friction, the output torque is equal to input torque. (3) At the same input speed, higher slip requires higher input torque. Codes (a) (1), (2) and (3) (b) (1) and (2) (c) (2) and (3) (d) (1) and (3) 15. A hydraulic coupling transmits 1 kW of power at an input speed of 200 rpm, with a slip of 2%. If the input speed is changed to 400 rpm, the power transmitted with the same slip is: (a) 2 kW (b) 0.5 kW (c) 4 kW (d) 8 kW 16. If a reciprocating pump having a mechanical efficiency of 80% delivers water at the rate of 80 kg/s with a head of 30 m, the brake power of the pump is
Objective Questions 301
17.
18.
19.
20.
(a) 29.4 kW; (b) 20.8 kW; (c) 15.4 kW; (d) 10.8 kW. Which of the following statements regarding a centrifugal pump are correct? (1) The monometric head is developed by the pump. (2) The suction pipe diameter is generally larger than discharge pipe diameter. (3) The suction pipe is provided with a foot valve and strainer. (4) The delivery pipe is provided with a foot valve and a strainer. Codes (a) (1), (2), (3) and (4) (b) (1) and (2) (c) (2) and (3) (d) (1) and (3) Which of the following statements are correct regarding torque converter? (1) Its maximum efficiency is less than that of the fluid coupling. (2) It has two runners and a set of stationary vanes interposed between them. (3) It has two runners. (4) The ratio of secondary to primary torque is zero for zero value of angular velocity of secondary. Codes (a) (1) and (2) (b) (3) and (4) (c) (1) and (4) (d) (2) and (4) Which of the following statements are correct regarding volute casing of a centrifugal pump? (1) Loss of head due to change in velocity is eliminated. (2) Efficiency of pump is increased. (3) Water from the periphery of the impeller is collected and transmitted to the delivery pipe at constant velocity. Codes (a) (1), (2) and (3) (b) (1) and (2) (c) (2) and (3) (d) (1) and (3) Match List I with List II for a pump and select correct answer using the codes given below: List-1 List-11 (Outlet vane angle >2 Performance curves) A. b2 < 90° B. b2 = 90° C. b3 > 90°
3 2
H 1 Q
302 Hydraulic Machines Codes
21.
22.
23.
24.
25.
A B C (a) 1 2 3 (b) 1 3 2 (c) 2 1 3 (d) 3 2 1 Hydraulic ram is a pump which works on the principle of: (a) Water hammer (b) Centrifugal action (c) Reciprocating action (d) Hydraulic press The data for the performance of a centrifugal pump: speed = 1200 rpm; flow rate = 30 l/s; head = 20 m; power = 5 kW If the speed is increased to 1500 rpm, the power will be nearly equal to (a) 6.5 kW (b) 8.7 kW (c) 9.8 kW (d) 10.9 kW Which of the following pumps can be used to lift water through a suction head of 12 m from a well? (1) Centrifugal pump, single-stage (2) Centrifugal pump, multi-stage (3) Reciprocating pump (4) Jet pump Code (a) (2) alone (b) (1), (3) and (4) (c) (4) alone (d) (1) and (3) The following purposes (given in code) are served by the volute casing of a centrifugal pump. (1) Increase in the efficiency of the pump. (2) Conversion of part of the pressure energy to velocity head. (3) Giving uniform flow of the fluid coming out of impeller. Code (a) (1) and (2) (b) (1) and (3) (c) (2) and (3) (d) (1), (2) and (3) Which statements are correct regarding specific speed of a centrifugal pump? (1) Specific speed is defined as the speed of a genetically similar pump developing unit power under unit head. (2) At the same specific speed, the efficiency is greater with large capacity.
Objective Questions 303
26.
27.
28.
29.
30.
(3) The specific speed increases with increase of outer blade angle. (4) The specific speed varies directly as the square root of the pump discharge. Codes (a) (1) and (2) (b) (2) and (4) (c) (3) and (4) (d) (2) and (3) Which are the correct statements regarding a torque converter? (1) It has stationary set of blades in addition to the primary and secondary rotors. (2) It can be used for multiplication of torques. (3) The maximum efficiency of a converter is less than a fluid coupling. (4) In a converter designed to give large increase of torque, the efficiency V/S speed ratio approaches unity. Codes (a) (1), (2), (3), (4) (b) (1), (2) and (3) (c) (1), (2), (4) (d) (3) and (4) In contrast to fluid coupling, torque converters are operated: (a) while completely filled with liquid (b) while partially filled with liquid (c) without liquid (d) while completely filled with air A centrifugal pump is started with its delivery valve kept: (a) fully open (b) fully closed (c) partially open (d) 50% open Select the correct statements. If pump NPSH is not satisfied, then (1) It will not develop sufficient head to raise water (2) Its efficiency will be low (3) It will deliver very low discharge (4) It will be cavitated. Codes (a) (1), (2), (3) (b) (1) and (4) (c) (2), (3) and (4) (d) (1), (2), (3) and (4). What are the beneficial effects of air vessel fitted to the delivery side of a reciprocating pump? (1) Constant rate of discharge can be ensured. (2) Power consumption can be reduced (3) Discharge can be increased (4) Constant velocity of piston can be ensured.
304 Hydraulic Machines
31.
32.
33.
34.
Codes (a) (1) and (4) (b) (1) and (2) (c) (2) and (4) (d) (1) and (3) Choose the correct answers regarding place of cavitation in hydraulic machines. (1) exit of a pump (2) entry of pump (3) exit of a turbine. Codes (a) (1) and (2) (b) (1) and (3) (c) (1), (2) and (3) (d) (2) and (3) For attaining a non-overloading characteristic in centrifugal pumps. (a) backward bent vanes are preferred over forward bent banes. (b) forward bent vanes are preferred over backward bent vanes. (c) forward bent vanes are preferred over vanes radial at outlet. (d) vanes radial at outlet are preferred over backward vanes. Select the correct code regarding function of a volute casing of a centrifugal pump (1) To collect water from the periphery of the impeller and to transmit it to the delivery pipe at a constant velocity. (2) To increase discharge of the pump. (3) To increase the efficiency of the pump. (4) To reduce loss of head in discharge. Codes (a) (1), (2) and (3) (b) (2), (3) and (4) (c) (1), (3) and (4) (d) (1) and (2) The specific speed of a hydraulic pump is the speed of geometrically similar pump against a unit head and (a) delivering unit quantity of water (b) consuming unit power (c) having unit velocity of flow (d) having unit radial velocity
Key 1. (c) 8. (c) 15. (a) 22. (c) 29. (b)
(b) 9. (d) 16. (a) 23. (c) 30. (b)
(a) 10. (b) 17. (c) 24. (a) 31. (d)
(c) 11. (d) 18. (a) 25. (a) 32. (a)
(d) 12. (b) 19. (a) 26. (a) 33. (c)
(d) 13. (c) 20. (a) 27. (a) 34. (a)
(b) 14. (a) 21. (a) 28. (c)
Objective Questions 305
PLANTS AND SYSTEMS Tick (Ö) the right answer: 1. An accumulator is a device to store: (a) Sufficient quantity of liquid to compensate the change in discharge. (b) Sufficient energy to drive the machine when the normal energy source does not function. (c) Sufficient energy in case of machines which work intermittently to supplement the discharge from the normal source. (d) Liquid which otherwise would have gone waste. 2. A hydraulic press has a ram of 20 cm diameter and a plunger of 5 cm diameter. The force required at the plunger to lift a load of 16 ´ 104 N shall be: (a) 256 ´ 104 N (b) 64 ´ 104 N. (c) 4 ´ 104 N (d) 1 ´ 104 N 3. A surge tank provided on the penstock connected to a water turbine (1) Helps in reducing the water hammer (2) Stores extra water when not needed (3) Provides increased demand of water. Which of these statements are correct? (a) (1) and (3) (b) (2) and (3) (c) (1) and (2) (d) (1), (2) and (3). 4. The gross head available to a hydraulic power plant is 100 m. The utilized head in the runner of the hydraulic turbine is 72 m. If the hydraulic efficiency of the turbine is 90%, the pipe friction head is estimated to be: (a) 20 m (b) 18 m (c) 16.2 m (d) 1.8 m 5. A triangular dam of height h and base width b is filled upto its top with water as shown. The condition of stability is h Masonary (a) b = h dam (b) b = 2.6h S.G. = 2.56 (c) b = ! h b (d) b = 0.625h 6. Of all power plants, hydel is most disadvantageous when one compares the (a) nearness to load centre (b) cost of energy resource (c) technical skill required (d) economics that determines the choice of plants
306 Hydraulic Machines 7. A hydraulic power station has the following major items in the hydraulic circuit: (1) Draft tube; (2) Runner; (3) Guide wheel (4) Penstock. The correct sequence of these items in the direction of flow is; (a) (4), (2), (3), (1), (5) (b) (4), (3), (2), (5), (1) (c) (1), (2), (3), (5), (4) (d) (1), (3), (2), (4), (5) 8. Water hammer in penstocks takes place when (a) water is flowing with high velocity. (b) water is flowing with high pressure. (c) flowing water is suddenly brought to rest by closing a valve. (d) flowing water is brought to rest by gradually closing a valve.
Key 1. (c) 8. (c)
(d)
(d)
(a)
(b)
(d)
(b)
Index Air lift pump, 210 Airfoil section, 199 Airfoil theory, 201 limitations of, 201 Angular momentum, 26 Axial flow impeller, 152 Axial flow pump, 196 design of, 198 Bernoullis equation, 29 Blade efficiency, 55 Boiler feed pump, hydraulic design of, 268 Bucket dimensions, 63 Bulb turbine, 102 Casings, type of, 136 Cavitation, 109, 158 factors causing, 111 methods to avoid, 112 Centrifugal pump, 135 classification of, 135 design constants, 153 velocity diagrams of, 143 Centrifugal pumps, applications of, 10 operation and maintenance of, 257 operational problems of, 162 Character speed, 78
Closed and open systems, 18 Closed impeller, 138 Closed system, 19 Continuity equation, 21 Continuum model, 19 Cylindrical coordinates, 23 Darrieus turbine rotor, 246 Deriaz turbine, 102 Differential accumulator, 220, 221 Discharge lift, 139 Double cylinder double acting pump, 178 Draft tube efficiency, 109 Draft tube, 106 functions of, 108 types of, 108, 110 Dynamic similarity, 122 Efficiency curves, 46 Energy balance sheet, 58 Energy equation, 21 Equation of motion, 21 Eulers fundamental equation, 27 Eulerian method, 20 Feedback speed regulation system, 116 Floating mill, 254 Floating type micro hydro power plants,243 Flow coefficient, 126
! & Index Flow duration curve, 234, 235 Fluid coupling, 7 Fluid flow, dynamics of, 16 kinematics of, 16 Fluid machines, classification of, 5 Fluid mechanics, principles of, 14 Fluid motion, 20 Fluid pressure, 15 Fluid systems, 7 Fluid torque, 7 Fluid transfer machines, 6 Fluids, properties of, 15 Francis turbine, 77 components of, 79 design of, 84 velocity triangles of, 81 Gear pump, 206 Geometric similarity, 121 Head coefficient, 126 Himalaya mill, 254 Himalayas, vegetation belts of the, 251 Hydel power, advantages of, 9 disadvantages of, 10 Hydraulic accumulator, 220 Hydraulic coupling, 212 Hydraulic efficiency, 55 Hydraulic flow, 20 basic concepts of, 18 Hydraulic intensifier, 221, 222 Hydraulic jack, 222 Hydraulic lift, 223 Hydraulic losses, 57
Hydraulic machines, classification of, 14 Hydraulic press, 222 Hydraulic ram, 210, 211, 253 Hydraulic systems, 219 Hydraulic turbines, classifications of, 45 layout of, 46 performance characteristics of, 127 selection of, 46 specific speed of a, 77 Hydro energy, 3 Hydro power plant, 226 classifications of, 227 location of, 231 Hydro power plants, social and ecological impact of, 277 Hydro power potential, 226 Hydro thermal mix, 237 Hydro turbines, applications of, 8 classifications of, 8 Hydro village, 255 Hydrology, 233 Hydropower plant, 4 Hydrostatic forces, 15 Ideal fluid, 18 Impeller diameter, 150, 257 Impeller, design of, 148 Impulse principle, 31 Inlet velocity triangle, 53 Jet force, on curved plate, 35 on moving flat plate, 33 on stationary flat plate, 31 Jet propulsion of ships, 37 Jet pumps, 209
Index Jet ratio, 56, 62 Jet reaction, 37 Jet theory, 30 Jet velocity, components of, 32 Kaplan turbine, 97 Kinematic similarity, 122 Langrangian method, 20 Liquid jet, 30 Liquids, macroscopic properties of, 19 Lobe pump, 206 Manometric head, 140 Mass curve, 235 Mass, conservation of, 19 Maximum suction lift, 158 Mechanical losses, 57 Mixed flow impeller, 153 Momentum equation, 24 applications of, 25 moment of, 26 Multi-jet pelton turbine, 51 Multistage pumps, 161 Newtonian fluid, 18 Oil pressure governor, 117 Open system, 19 Open-impeller, 138 Outlet velocity triangle, 53 Pelton turbine, components of a, 48 design of, 58 double regulation system of, 118 layout arrangements of, 50 Penstock design, 65
! '
Performance curves, 46 Poncelet water wheel, 246 Power coefficient, 127 Power plants, combined operation of, 236 Power transmission machines, 7 Prime movers, 6 Propeller turbine, 96 Propulsion principle, 31 Pump installation, sankey digram of, 141 Pump operation, 161 Pump power, 140 Pump with air vessels, 179 Pump, performance characteristics of a, 157 priming of, 162 specific speed of, 156 Pumped storage plant, 229 energy flow diagram for, 230 specifications of, 230 Radial impeller, 151 Reaction principle, 31 Reaction turbine, governing of, 119 Reciprocating pump, 174 classifications of, 177 performance of, 182 Reciprocating, pump installation and performance of,175 River current energy, 245 Rotary displacement pumps, 205 Rotary pumps, operating characteristics of, 205 Rotor speed, 62 Runners, specific speeds of, 78 Run-of-rifer plants, 228
! Index Sankey diagram, 57, 83, 141 Scale ratio, 124 Screw pump, 208 Semi-open impeller, 138 Ships/boats, propulsion of, 37 Single cylinder double acting pump, 178 Single cylinder single acting pump, 178 Sliding vane pump, 207 Small hydel projects in India, 244 Spherical coordinates, 23 Static head, 140 Storage reservoir power plant, 228 Stream line, 20 Stream tube, 20 Suction lift, 139 Taper angle, 109 Three cylinder double acting pumps, 179 Three-dimensional flow, 22 Torque converter, 213 Total head, 140
Turbine, specific speed of, 156 Turbine constants, 98 Turbine losses, 56 Turbine speed, selection of, 66 Unit discharge, 125 Unit power, 125 Unit speed, 125 Vane discharge angle, 149 Velocity components, 24 Velocity diagram of bucket, 51 Vertical turbine pump, 204 Volumetric losses, 56 Vortex breaker, 109 Water Jet Pump, 209 Water, vapour pressure of, 111 Zowski recommendations, 87
HYDRAULIC MACHINES
The book has been divided into the following six parts containing 13 chapters: • Fundamentals of Fluid Flow • Turbines • Pumps • Systems and Power Plants • Case Studies • Objective Questions Contents: PART I – Fundamentals: Classification and Applications / Basic Principles of Hydraulic Flow and Jet Theory, PART II – Turbines: Pelton Turbine / Francis Turbine / Propeller and Kaplan Turbines / Draft Tube and Cavitation / Governing of Hydraulic Turbines / Dynamic Similarity and Performance Characteristics, PART III – Pumps: Centrifugal Pumps / Reciprocating Pumps / Special Pumps, PART IV – Systems and Plants: Hydraulic Systems / Hydraulic Power Plants / PART V – Case Studies, PART VI – Objective Questions / Index.
Mridul Singal is an engineer with more than 17 years of experience. He has undertaken research and development work in aerospace industry at DRDO, and software industry in various MNCs in India and abroad. Presently, he is working as IT Architect at IBM India Pvt. Ltd. Rishi Singal is an engineer with more than 12 years of experience in major steel and software industries in India and abroad. He is a PMI-certified PMP.® Presently, he is working as Project Manager for Accenture India Pvt. Ltd.
978-93-89307-65-8
HYDRAULIC MACHINES
R.K. Singal • Mridul Singal Rishi Singal
R.K. Singal is a senior professor and consultant. He has an experience of more than 47 years in India and abroad with MNCs and academic institutions of international repute. He has been teaching at IIT, NIT, UPTU and other universities, and worked as Director of various institutions. He is the author of many books and research publications in national and international journals. He has also been decorated by the President of India.
HYDRAULIC MACHINES
Hydraulic Machines (Fluid Machinery) has been designed as a textbook for engineering students specialising in mechanical, civil, electrical, hydraulics, chemical and power engineering. The highlights of the book are simple language supported by analytical and graphical illustrations. A large number of theory questions and numerical problems with solution hints have been annexed at the end of every chapter. A large number of objective questions have been included to help students opting for competitive examinations. Five case studies based on research have been included, which can be advantageously used by practising engineers pursuing research design and consultancy careers. Complete design of hydraulic machines has been demonstrated with the help of suitable examples.
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11217 | https://www.semanticscholar.org/paper/On-inequalities-for-normalized-Schur-functions-Sra/44fa3dcd70d3a334bfa85ba1338ecd56f11fb336 | [PDF] On inequalities for normalized Schur functions | Semantic Scholar
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DOI:10.1016/j.ejc.2015.07.005
Corpus ID: 923533
On inequalities for normalized Schur functions
@article{Sra2015OnIF,
title={On inequalities for normalized Schur functions},
author={Suvrit Sra},
journal={Eur. J. Comb.},
year={2015},
volume={51},
pages={492-494},
url={
}
S. Sra
Published in European journal of…17 February 2015
Mathematics
Eur. J. Comb.
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R. Ait-HaddouM. Mazure
Mathematics
Foundations of Computational Mathematics
2016
A classical theorem by Chebyshev says how to obtain the minimum and maximum values of a symmetric multiaffine function of n variables with a prescribed sum. We show that, given two functions in an… Expand
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### The Fundamental Blossoming Inequality in Chebyshev Spaces—I: Applications to Schur Functions
R. Ait-HaddouM. Mazure
Mathematics
Found. Comput. Math.
2018
TLDR
This work definitely demonstrates that, via blossoms, CAGD techniques can have important implications in other mathematical domains, e.g., combinatorics.Expand
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### Inequalities for symmetric means
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Mathematics
Eur. J. Comb.
2011
47
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### Inequalities: Theory of Majorization and Its Applications
A. W. MarshallI. OlkinB. Arnold
Mathematics
1980
Although they play a fundamental role in nearly all branches of mathematics, inequalities are usually obtained by ad hoc methods rather than as consequences of some underlying "theory of… Expand
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### Symmetric functions and Hall polynomials
I. G. MacDonald
Mathematics
1979
I. Symmetric functions II. Hall polynomials III. HallLittlewood symmetric functions IV. The characters of GLn over a finite field V. The Hecke ring of GLn over a finite field VI. Symmetric functions… Expand
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### I.—A class of symmetric polynomials with a parameter
H. Jack
Mathematics
Proceedings of the Royal Society of Edinburgh…
1970
Synopsis In an attempt to evaluate the integral (5) below, using a decomposition of an orthogonal matrix (Jack 1968), the author is led to define a set of polynomials, one for each partition of an… Expand
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### The planar approximation. II
C. ItzyksonJ. Zuber
Mathematics
1980
The planar approximation is reconsidered. It is shown that a saddle point method is ineffective, due to the large number of degrees of freedom. The problem of eliminating angular variables is… Expand
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### Differential Operators on a Semisimple Lie Algebra
Harish-Chandra
Mathematics
1957
488
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### Ueber eine Klasse von Matrizen, die sich einer gegebenen Matrix zuordnen lassen
I. Schur
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11218 | https://www.scientificamerican.com/article/quantum-eraser-answer-to-three-polarizer-puzzle/ | April 16, 2007
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Answer to the Three-Polarizer Puzzle Featured in the Print Edition
Answer to the Three-Polarizer Puzzle Featured in the Print Edition
This sidebar is part of a package that supplements our story on quantum erasure in the May issue of Scientific American
By Rachel Hillmer and Paul Kwiat
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If you take two crossed polarizers (for example, a horizontal and vertical one), no light will get through them. Yet when you insert a third polarizer between the two, oriented diagonally, then some photons make it through. How does adding that polarizer (which will block some photons) cause photons to get through?
Say that the first polarizer is horizontal. Any photons that make it through that one are then horizontally polarized. If the vertical polarizer comes next, it will block all of these photons. When the diagonal polarizer is in place, however, it will let half of them through and these transmitted photons will then be diagonally polarized. When these diagonally polarized photons arrive at the vertical polarizer, now half of them will get through—they have no "memory" of ever having been horizontally polarized.
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11220 | https://www.quora.com/Why-do-most-programming-languages-have-indexes-starting-at-zero-instead-of-1-For-example-why-would-an-eight-element-set-have-indexes-0-7-when-we-always-say-1-8 | Something went wrong. Wait a moment and try again.
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Why do most programming languages have indexes starting at zero instead of 1? For example, why would an eight element set have indexes 0-7 when we always say 1-8?
68 Answers
Michael Veksler
30+ years programming
·
Upvoted by
Ernst Stolz
, M.Ed. English & Computer Science, University of Osnabrück (2016) and
Horst H. von Brand
, PhD Computer Science & Mathematics, Louisiana State University (1987) · Author has 657 answers and 3.4M answer views
·
Updated 7y
What is the reason index numbers start from 0 instead of 1 in programming languages?
Despite sounding counter intuitive, starting with 0 is actually more convenient for humans. Of course, there are many programming languages where indexes start with 1, but from my experience they are less convenient to work with. One of the newest languages that starts arrays with 1 is Lua (Programming in Lua : 11.1 - arrays) , although any starting point can be used.
Did you ever wonder, why Cartesian coordinates start with 0 and not 1? Why the first tick on a ruler is 0 and not 1?
There are many examples where counting from 1 is more confusing than it helps anything. Take, for example, our cal
Despite sounding counter intuitive, starting with 0 is actually more convenient for humans. Of course, there are many programming languages where indexes start with 1, but from my experience they are less convenient to work with. One of the newest languages that starts arrays with 1 is Lua (Programming in Lua : 11.1 - arrays) , although any starting point can be used.
Did you ever wonder, why Cartesian coordinates start with 0 and not 1? Why the first tick on a ruler is 0 and not 1?
There are many examples where counting from 1 is more confusing than it helps anything. Take, for example, our calendar. It starts counting years from 1, so do centuries and millennia start with 1. To see what is wrong with that, think about the year 2018. The confusing part is that 2018 is not in the 20-th century, it is in the 21. To make things more confusing year 2000 is not in the 21-century, it is in the 20-th. The century turns at 2001, since centuries go:
1–100 : 1st century
101–200: 2nd century
…
1901–2000: 20th century
Same with the millennium:
1–1000: 1st millennium
1001–2000: 2nd millennium
2001–3000: 3rd millennium
And most people got confused and celebrated the new millennium at the new year’s eve of 2000, instead of 2001. This confusion extends nicely into algorithms.
Getting the i-th element in a matrix (counting elements from one row to the next) becomes way too complicated with 1-based arrays, where it is trivial with 0-based arrays.
Edit. Despite knowing the right answer (obviously) I made a typo saying that 1901 was in the 21th century (now fixed). This is to show that when writing enough values, and when the mind is off-guard, there is a higher risk of errors when 1-based indexing is concerned. This is purely anecdotal though.
Edit 2:
I have got some opposition to the idea that 1-based indexing is more complicated than 0-based. I agree that my wording (“much more confusing") was a bit too strong, but the basic idea still stands: 1-based indexing has a higher probability to introduce 1-off bugs.
Also, I suspect that opposition on this topic is a cultural thing. In some countries, like mine, floors (stories) in buildings are counted from zero. An entrance floor is either marked with a zero or with an E (for entrance). Underground parking is marked with -1, -2, etc. This makes it quite natural to think in terms of 0-based indexing. In other countries, an entrance floor is marked with 1 (after all, it is the first floor). I suspect that people from those countries are more likely to prefer the 1-based indexing.
Related questions
In Python, where from the indexing starts 0 or 1?
What language is all 1 and 0?
What is the reason index numbers start from 0 instead of 1 in programming languages?
What is that code which is read in 0 and 1 form?
How do I use index PHP instead of index HTML?
Travis Addair
Maintainer of Horovod.ai and Ludwig.ai
·
Upvoted by
Horst H. von Brand
, PhD Computer Science & Mathematics, Louisiana State University (1987) and
Abu Azlan
, M.S. Computer Science, Stanford University · Author has 772 answers and 12.8M answer views
·
11y
Why do array indexes start with 0 (zero) in many programming languages?
Originally Answered: Why do array indexes start with 0 (zero) in many programming languages?
·
Many languages don't index starting from zero (MATLAB, for instance), but one of the luminaries of computer science -- Edsger Dijkstra -- actually wrote a note in 1982 advocating for zero-based indexing called Why numbering should start at zero.
The full text can be found here: Why numbering should start at zero (EWD 831)
To summarize his argument:
When working with sub-sequences of natural numbers, the difference between the upper bound and the lower bound should be the length of the sub-sequence. The indices of an array can be thought of as a special kind of such a sub-sequence.
The lower bound
Many languages don't index starting from zero (MATLAB, for instance), but one of the luminaries of computer science -- Edsger Dijkstra -- actually wrote a note in 1982 advocating for zero-based indexing called Why numbering should start at zero.
The full text can be found here: Why numbering should start at zero (EWD 831)
To summarize his argument:
When working with sub-sequences of natural numbers, the difference between the upper bound and the lower bound should be the length of the sub-sequence. The indices of an array can be thought of as a special kind of such a sub-sequence.
The lower bound should be inclusive, the upper bound should be exclusive. In other words, the lower bound should be the first index of the array. Otherwise, we risk having to have a lower bound in the unnatural numbers for some sub-sequences.
If we want to maintain conditions (1) and (2), then we effectively have two choices for upper and lower bounds: 1 <= i < N+1 or 0 <= i < N. Clearly, putting N+1 in the range is ugly, so we should prefer indexing starting from 0.
Edward De Jong
Long-time developer, see beadslang.org
·
Author has 204 answers and 536.6K answer views
·
Updated 5y
Why do array indexes start with 0 (zero) in many programming languages?
Originally Answered: Why do array indexes start with 0 (zero) in many programming languages?
·
I believe the first language to have arrays was
FORTRAN
, one of the oldest languages. It used arrays based on 1. So did
COBOL
.
C
came along and used 0, because in
C
an array was nothing more than a shortcut for writing pointer arithmetic. An array subscript reference was interchangeable with pointer arithmetic in
C
. The developers of C were obsessed with tiny optimizations, because C was for them just a better assembly language. Pascal used 1, but its successor language
Modula-2
(invented by the famous Prof. Wirth of ETH in Zurich) lets you define the starting and ending bounds of an array, so
I believe the first language to have arrays was FORTRAN, one of the oldest languages. It used arrays based on 1. So did COBOL. C came along and used 0, because in C an array was nothing more than a shortcut for writing pointer arithmetic. An array subscript reference was interchangeable with pointer arithmetic in C. The developers of C were obsessed with tiny optimizations, because C was for them just a better assembly language. Pascal used 1, but its successor language Modula-2 (invented by the famous Prof. Wirth of ETH in Zurich) lets you define the starting and ending bounds of an array, so you could go from -5 .. +5 if that was convenient for your purposes. In Modula-2 you could have arrays of enumerated types as well, so if you declared an enumerated type of a_color = { red, green, blue } you could then declare an array of a_color, and then index an array by an enumerated type, like array[red]. This was a fantastic feature and hugely improved readability in Modula-2 over other languages. It is sad fact that people still look at C and think it is modern, and somehow justified, when in fact C is an archaic language, barely above assembler, and very clumsy when compared to the now old Modula-2. And for those historians, the mighty PL/1 also had user-definable array limits. To not be able to set array bounds to a programmer-specified range is really an unforgivable omission at this point in time, yet language designer after language designer skips this feature. And for those academics who think that because Edsger Dijsktra wrote a paper on why 0 was great it somehow must be right, you are omitting the fact that at that ancient time his limited imagination could only see two choices: 0-based and 1-based, when a more flexible definition capability is much more user friendly. Even back then, computers were faster than humans, and anything which reduces comprehension of programs by humans is a bad tradeoff. And the people at Google who took Modula-2 and copied it, kinda sorta, to create Go, blew it bad and took arrays back to the C-era.
C won the battle over PL/1 and other languages because the UNIX operating system beat MULTICS, OS/360, and the other mainframe OS’es, mostly because UNIX was the first kinda open source OS given to universities, and programmers all trained on UNIX. C was then adopted by Microsoft, which gave it another huge boost into the PC era. Interestingly, Apple had a terrific Pascal-based OS that was very reliable and easy to program, but because they wanted to switch to a UNIX kernel, converted to C with initially disastrous reliability problems. Apple is now pushing Swift, which is much cleaner than Objective-C, but guess what, Swift doesn’t allow negative subscripts. It is as if negative numbers hadn’t been invented yet. These so-called ‘new’ languages are saddled with old limitations and as they say, those that don’t learn from history are doomed to repeat it.
In my own Beads language, arrays are sparse, so you can store elements at negative values, or very large subscripts, or with enumerated subscripts, and even strings which Javascript also allows. Why limit yourself to archaic restrictions, which inevitably are more cumbersome?
Strong math languages like Mathematica don’t have any restrictions on subscripts, and can do very powerful things. Please don’t tell me that “the math” indicates that arrays should start at 0. I wouldn’t be surprised if some math tool allows subscripts that aren’t integers…some language can probably even let you use complex number subscripts. The compiler should handle all the trivial subtractions automatically, and programmers should use a notation that is convenient for the problem at hand.
Mark Guzdial
Works at Georgia Institute of Technology
·
7y
What is the reason index numbers start from 0 instead of 1 in programming languages?
Originally Answered: What is the reason index numbers start from 0 instead of 1 in programming languages?
·
Because of Model 33 Teletypes. When UNIX and C were being invented at Bell Labs, Thompson and Ritchie and the rest were programming on Model 33 Teletypes (WikiVisually.com). Each keystroke was literally a physical effort, so C was designed to minimize keystrokes. Making array indexing exactly equivalent to pointer arithmetic allowed them to cut out a few keystrokes. C was based on BCPL which also had zero-based arrays, but C went further than BCPL in minimizing keystrokes for pointer manipulation and arithmetic.
Prior to C, 1-based array indexes were much more common. Fortran, Lisp, COBOL, and
Because of Model 33 Teletypes. When UNIX and C were being invented at Bell Labs, Thompson and Ritchie and the rest were programming on Model 33 Teletypes (WikiVisually.com). Each keystroke was literally a physical effort, so C was designed to minimize keystrokes. Making array indexing exactly equivalent to pointer arithmetic allowed them to cut out a few keystrokes. C was based on BCPL which also had zero-based arrays, but C went further than BCPL in minimizing keystrokes for pointer manipulation and arithmetic.
Prior to C, 1-based array indexes were much more common. Fortran, Lisp, COBOL, and Smalltalk all use 1-based arrays. More modern languages like Mathematica and MATLAB also use 1-based arrays. Any argument about historical precedence or convenience falls flat when you realize that almost ALL programming used 1-based arrays for the first 20 years of high-level programming languages.
Zero-based indexing became the standard because of the popularity of C, and because brevity was the goal.
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Johnny M
Master's Degree from Harvard University (Graduated 2011)
·
Updated Sep 9
Does switching car insurance really save you money, or is that just marketing hype?
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This is one of those things that I didn’t expect to be worthwhile, but it was. You actually can save a solid chunk of money—if you use the right tool like this one. I ended up saving over $1,500/year, but I also insure four cars.
I tested several comparison tools and while some of them ended up spamming me with junk, there were a couple like Coverage.com and these alternatives that I now recommend to my friend.
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But recently I decided to check so I used this quote tool, which compares everything in one place. It took maybe 2 minutes, tops. I just answered a few questions and it pulled up offers from multiple big-name providers, side by side. Prices, coverage details, even customer reviews—all laid out in a way that made the choice pretty obvious.
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Related questions
If a computer only understands 0 and 1, then why do we use a programming language?
Why do we count from 0 in most programming languages?
Why do array indexes start with 0 (zero) in many programming languages?
Is it really possible to code using 0 and 1 only?
For (I = 1; I < 10; I++), what is the index in the code?
Sascha Atrops
former Java developer
·
Author has 396 answers and 2.5M answer views
·
7y
What is the reason index numbers start from 0 instead of 1 in programming languages?
Originally Answered: What is the reason index numbers start from 0 instead of 1 in programming languages?
·
An index can be created by counting things or by distances to reach the desired point from a known starting point. In computers it’s usually better to know a distance instead of counting, because you can jump through random access memory. Counting is more expensive.
Let’s pretend a string. In C this is a pointer to the first element. Let’s have a look at a string: “Hello World”. The Array of characters is a pointer to the ‘H’. Let’s asume “Hello World” is on address 1000 and the following bytes. The first index is 0. So if you add 1000 and 0 you get the address of the ‘H’, which is 1000.
In C th
An index can be created by counting things or by distances to reach the desired point from a known starting point. In computers it’s usually better to know a distance instead of counting, because you can jump through random access memory. Counting is more expensive.
Let’s pretend a string. In C this is a pointer to the first element. Let’s have a look at a string: “Hello World”. The Array of characters is a pointer to the ‘H’. Let’s asume “Hello World” is on address 1000 and the following bytes. The first index is 0. So if you add 1000 and 0 you get the address of the ‘H’, which is 1000.
In C the null value represents the end of a string. So the string is actually Hello World\0. To find out how long the string is, you have to count until you find the null value. Either you have to store the result whereever you want or you have to count. You decide. That’s the way of thinking in C: If you need it, you have to do it by yourself otherwise you don’t need it, so it’s faster not to do it. And there is an advantage: Strings can be as long as you want.
Not lets have a look at Pascal: The index starts at 1. Lets assume the character array is at address 1000. In Pascal a string is not just an array of bytes. It’s a string. And its indices starts at 1. Why? Because at address 1000 the length of the string is stored. So Hello is stored as 5Hello. Index would be the Length of the string. So index 1 is 1000+1 which is the address of the ‘H’ in a pascal string. Advantages: It’s much cheaper to request the string length. You can easily jump over the string - just add 256 to the address. Disadvantages: The string length has to be pre calculated even if that value its never used and the string takes always 256 bytes and so the maximum length of a string is 255 bytes. This might be a bad limitation and usually it’s a waste of space.
You can easily recreate a Pascal String in C. In C this is a restriction. That’s the way of good C programming. Restrict your software to what you need. But if you need more, you have a problem with Pascal. In C you simply do not restrict yourself so much.
Whenever you use an array in Pascal like array[5..10] of char you have an array length of 6 (5,6,7,8,9,10) and an offset of 5. So again lets asume your array begins at address 1000 and your requested index is 8 you have to add 1000 + index 8 - offset 5. So the address of your index 8 is 1003. Therefore Pascal uses implicitely an addition and a substraction. Which is more expensive than starting every array at index 0. There is no offset, so there is no substraction. If you need it, you have to do it by yourself. Which is a good reason to think about needing it. If you don’t need it, your software is faster.
Humans have two similar ways of measuring: You can count what you see or you can measure the distance between a starting point. An address in a computer has always a very well known starting point: The address 0. So all addresses in a computer are related to address 0. So the index (counted by 0) and the distance to the starting point are the identical value.
If you imagine a street you have a starting point: The beginning of the street. Now you can count the houses: 1, 2, 3… or you can measure the distance. You can say I am 50m after the beginning of the street. Which would also be much better if someone builds a house between #1 and #2. With a new house you need an address #1b. If there was space between #1 and #2 it has its own distance to the beginning of the street and this distance would be beween #1 and #2. Which would have distances like house at 20meters (#1) and house at 80 meters(#2) and the new house would be at 50 meters (#1b).
We use distances by “counting” birthdays. When we are born it’s our birthday (actually the first) and we are zero years old. After we passed the same date again, it’s our second birthday and we are one year old. Our age is discribed in the temporal distance of our actual birthday. So we don’t say you become one year old on your second birthday. We count the distance.
In a computer the distances are always 8 bit. And there a no spaces in between, there is just one street with a huge amount of uniform addresses. To reach address 0 I had ot pass 0 addresses so I am at the first house in the street. At real streets we count the houses, so the house number is 1. If we would count the amount of houses we passed, it would be 0.
It doesn’t make sense to use two similar kinds of measuring in a computer language. So you use just one. And a processor measures in distances.
So it seems to be natural to start counting by 1 but it is also natural to measure distances. And we only need one system in a computer and that is measuring distances because it is faster on a computer where all ‘houses’ have the same size and you can “beam” yourself whereever you want.
David Lewis
Software Engineer since 1973 (retired)
·
Author has 2.6K answers and 3M answer views
·
2y
There are a pile of languages that do have indexes that start at 1:
ALGOL 98
APL
AWK
CFML
COBOL
Fortran
FoxPro
Julia
Lingo
Lua
Mathematica
Matlab
PL/I
RPG
R
Ring
Sass
Smalltalk
Wolfram Language
XPath/ XQuery
but C went with 0 and influenced many that came after.
C is a low level language. An array in C is just a pointer to memory and the offset to the first thing in the array is 0 which makes 0 based indexing simpler.
When I work in 1 based languages, it’s a little weird until I hit my stride, but 0 feels more natural to me even though my very first language (Fortran) is 1 based. I really can’t say that things get a
There are a pile of languages that do have indexes that start at 1:
ALGOL 98
APL
AWK
CFML
COBOL
Fortran
FoxPro
Julia
Lingo
Lua
Mathematica
Matlab
PL/I
RPG
R
Ring
Sass
Smalltalk
Wolfram Language
XPath/ XQuery
but C went with 0 and influenced many that came after.
C is a low level language. An array in C is just a pointer to memory and the offset to the first thing in the array is 0 which makes 0 based indexing simpler.
When I work in 1 based languages, it’s a little weird until I hit my stride, but 0 feels more natural to me even though my very first language (Fortran) is 1 based. I really can’t say that things get any easier going one way over the other. It’s just different.
Also, some languages (e.g., Ada) will let have whatever range of indexes you want which could be argued is the most natural: to make the indexes conform to the problem space instead of having the user conform to the language limitations.
Programming Languages where indices start from 1 (NOT 0)
There are 20 Programming Languages in the World where the array index starts from 1 and not from 0. Some of the popular languages include R, COBOL, Matlab and Julia.
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Ravi Singh
Technical Lead at Amazon (company) (2021–present)
·
Updated May 13
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Structured preparation for problem-solving (DSA) is crucial, rather t
YES, candidates can transition from service companies (such as TCS, Infosys) to top tech companies (like Amazon, Flipkart, or Google) 🌟 through a well-planned preparation strategy and regular practice. To achieve this, you need to be highly proficient in problem-solving and system design skills 💡.
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Anders Kaseorg
MIT PhD student in CS; Cofounder of Ksplice, Inc.
·
Upvoted by
Mohamed Elsayed
, Bachelor's Degree Computer Science, Cairo University (2017) and
Joseph Kehoe
, PhD Computer Science · Author has 550 answers and 10.3M answer views
·
Updated 10y
Why do array indexes start with 0 (zero) in many programming languages?
Originally Answered: Why do array indexes start with 0 (zero) in many programming languages?
·
Array indices should start at 0. This is not just an efficiency hack for ancient computers, or a reflection of the underlying memory model, or some other kind of historical accident—forget all of that. Zero-based indexing actually simplifies array-related math for the programmer, and simpler math leads to fewer bugs. Here are some examples.
Suppose you’re writing a hash table that maps each integer key to one of n buckets. If your array of buckets is indexed starting at 0, you can write bucket = key mod n; but if it’s indexed starting at 1, you have to write bucket = (key mod n) + 1.
Suppose you
Array indices should start at 0. This is not just an efficiency hack for ancient computers, or a reflection of the underlying memory model, or some other kind of historical accident—forget all of that. Zero-based indexing actually simplifies array-related math for the programmer, and simpler math leads to fewer bugs. Here are some examples.
Suppose you’re writing a hash table that maps each integer key to one of n buckets. If your array of buckets is indexed starting at 0, you can write bucket = key mod n; but if it’s indexed starting at 1, you have to write bucket = (key mod n) + 1.
Suppose you’re writing code to serialize a rectangular array of pixels, with width w and height h, to a file (which we’ll think of as a one-dimensional array of length wh). With 0-indexed arrays, pixel (x, y) goes into position yw + x; with 1-indexed arrays, pixel (x, y) goes into position yw + x - w.
Suppose you want to put the letters ‘A’ through ‘Z’ into an array of length 26, and you have a function ord that maps a character to its ASCII value. With 0-indexed arrays, the character c is put at index ord(c) - ord(‘A’); with 1-indexed arrays, it’s put at index ord(c) - ord(‘A’) + 1.
It’s in fact one-based indexing that’s the historical accident—human languages needed numbers for “first”, “second”, etc. before we had invented zero. For a practical example of the kinds of problems this accident leads to, consider how the 1800s—well, no, actually, the period from January 1, 1801 through December 31, 1900—came to be known as the “19th century”.
Copied from my answer to Should array indices start at 0 or 1?. Along similar lines, see my answer to Why are Python ranges half-open instead of closed?.
Brian Bi
software engineer
·
Upvoted by
David Rutter
, M.S. C.S. Georgia Tech and
Anders Blehr
, M. Sc. Computer Science, Norwegian University of Science and Technology (1993) · Author has 4.8K answers and 65.1M answer views
·
4y
In coding, why are arrays zero indexed?
Originally Answered: In coding, why are arrays zero indexed?
·
Arrays start from index 0 in most, but not all programming languages. This is because there are several situations where arrays starting from 0 are more convenient than arrays starting from 1. However, there are also other situations where indexing starting from 1 is more convenient.
Dijkstra explains that intervals of integers should be represented by the lower bound and 1 plus the upper bound (conversely, a pair of integers represents the half-open interval of integers greater than or equal to the first integer and strictly less than the second integer). If we accept this, then we also see th
Arrays start from index 0 in most, but not all programming languages. This is because there are several situations where arrays starting from 0 are more convenient than arrays starting from 1. However, there are also other situations where indexing starting from 1 is more convenient.
Dijkstra explains that intervals of integers should be represented by the lower bound and 1 plus the upper bound (conversely, a pair of integers represents the half-open interval of integers greater than or equal to the first integer and strictly less than the second integer). If we accept this, then we also see that if an array of size N is indexed from 0, then the range of valid indices in the array is [0, N), which is nice.
If we need to simulate a two-dimensional array, by storing elements in row major order in a one-dimensional array, where the number of columns is C, then using 0-based indexing, the element at row i, column j can be found at position C i + j. Using 1-based indexing, the formula would instead be C (i - 1) + j.
But sometimes you want an array of size N to have an Nth element. If that’s the case, the first index will have to be 1, not 0.
The bottom line, though, is that if you’re using a language where arrays start from 0, but you want an array with indices from 1 to N inclusive, you can just allocate the array to have size N + 1, and not use the element at index 0. Whereas if the language forces all arrays to start from 1, and you want to use index 0, it’s a headache. So, in cases where a design decision must be made, starting from 0 is clearly the correct choice. (Even if the language, or an array library, gives the user the ability to specify the starting point, 0 should be the default.)
Promoted by The Penny Hoarder
Lisa Dawson
Finance Writer at The Penny Hoarder
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Updated Sep 16
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Ian Joyner
Worked on practical language design and compiler writing.
·
Author has 9.7K answers and 8.2M answer views
·
2y
There are a lot of answers here, so I thought I would not have to give an answer.
The problem with most of the answers is that they are based on the physical consideration that computers start with memory addresses of 0. They think such concepts start with the computer and that is the way computers work. But why do computers work that way?
Programming languages don’t do things for physical reasons, but logical reasons. That is such things are determined by the logical considerations of data, not for physical considerations of computer memory.
Programming languages are to abstract programs from ph
There are a lot of answers here, so I thought I would not have to give an answer.
The problem with most of the answers is that they are based on the physical consideration that computers start with memory addresses of 0. They think such concepts start with the computer and that is the way computers work. But why do computers work that way?
Programming languages don’t do things for physical reasons, but logical reasons. That is such things are determined by the logical considerations of data, not for physical considerations of computer memory.
Programming languages are to abstract programs from physical consideration of how the computer might work, not expose the thinking in terms of “memory addresses start at zero”.
We can use other values that 0 for lower bounds, but this turns out to be rare. Pascal had the ability to specify arrays as:
a: array [ .. ]
but the need for lower bounds other than 0 were rare because it is not a logical requirement. The lower bound of 0 is the most frequent logical requirement. Hence arrays are usually specified by a size, but then arrays should be resizable (adding more data past end, which is expensive because a new memory block must be allocated and original data copied into new array). Also array type should not depend on the size of the array. This was a big mistake in the definition of Pascal (but I think many implementations probably ignored it).
Programmers and programming languages should think in terms of denotational and axiomatic semantics — that is what is the nature of the data and what is the meaning of the data. Too many programmers think in terms of operational and translational semantics, that is the meaning is given by something ‘under the hood’. Thus physical explanations in terms of how the computer and memory work are taking the wrong approach. Having to think about what is happening below the surface of a computation is the wrong way to think. We just understand 3 + 4 — we don’t need to understand that in terms of computer operations (operational semantics), in fact, our understanding is considerably confused by thinking at that level.
In mathematics the property of 0 over addition is that 0 is the identity of integer algebra over addition. Thus
3 + 0 = 3
This is an example of axiomatic semantics. Since n + 0 = n we can use this high-level thinking to optimise n + 0 to n.
Hence it is no coincidence that computer indexing in terms of memory and addresses is the same as the logical and axiomatic thinking. But the optimisation of the computer has followed the logical approach, not dictated it.
Computers are thus designed to follow the logical considerations, not the other way around. Logically, things start at 0 — that is the start of something is zero offset from the start, or the first element is zero offset from the start.
In comments to other answers, the example of a ruler was given. If you look at a ruler, the first position is labelled ‘0’, as in “Zero centimetres from here”:
Cm
0…….1……2……3
Thus zero is ‘zero distance’. The first element in a data structure is thus zero distance from the beginning.
Zero is a very interesting concept. It was denied by European mathematical thinkers for centuries. That is why Roman numerals are not convenient for calculation.
But zero (as defined by those in the east who were thousands of years in advance of Europeans) is a very useful concept. A good book explains the idea and history of zero: ‘Zero: The Biography of a Dangerous Idea’ by Charles Seife.
Thus we don’t start things from zero because that is what computer hardware and memory does, we start from zero because it is logically the correct thing to do, and we build computers to follow that logic and that is why computer memory starts at zero — not the other way around!
Thus computers should be built to the requirements of programming, not the other way around. That is how Turing envisioned ‘mechanical’ computing in the first place. Electronic computers just perform mechanical computations.
Bob Barton wrote about that in his paper:
Barton B5000
The present methods of determining the functional design of computers are critically reviewed and a new approach proposed. The ideas presented arise from the conviction that for a true general purpose digital computer both coding and operation should be fully automated. Operation should be under the control of the machine itself, in a fuller sense than is typical in current practice.
“A New Approach to the Functional Design of a Digital Computer”
Note ‘Functional’ that is logical design of a computer. It should be that form follows required functionality, that is physical form follows logical requirements. Zero indexing is the logical requirement, computer forms follow that so well, it almost bears no thinking about, except when people very wrongly think the logical requirements have followed the physical form.
Every computer designer should read Barton’s paper and take its message seriously. In other words, why are we doing this in the first place. And that goes back to how Turing thought about computing.
We should not be constrained by technology. Advancing has always meant we make technology do what we want it to do. Too many people in computing are constrained by ‘what is’ rather than ‘what should be’.
Ryan Reich
Haskell programmer
·
Upvoted by
Justin Rising
, MSE in CS · Author has 916 answers and 4.4M answer views
·
7y
What is the reason index numbers start from 0 instead of 1 in programming languages?
Originally Answered: What is the reason index numbers start from 0 instead of 1 in programming languages?
·
You will recognize this from real life as calendar counting: if today is Sunday (as it happens to be) and I talk about “two days from now”, that's Tuesday, the third day of the week. Counting with a reference point is naturally zero-based, while counting as free enumeration is naturally one-based.
Since in C (which is not the originator of this practice, but is probably the most primitive example that is at all familiar, as well as being the reference imperative language) an array variable is a pointer to the first element of the array, subsequent elements, also referenced via pointers, are co
You will recognize this from real life as calendar counting: if today is Sunday (as it happens to be) and I talk about “two days from now”, that's Tuesday, the third day of the week. Counting with a reference point is naturally zero-based, while counting as free enumeration is naturally one-based.
Since in C (which is not the originator of this practice, but is probably the most primitive example that is at all familiar, as well as being the reference imperative language) an array variable is a pointer to the first element of the array, subsequent elements, also referenced via pointers, are counted from there as a reference point and so correspond to a zero-based index on the array variable.
Other occurrences of zero-based indices are either secretly arrays, or loop counters. For the latter I can do no better than to quote Koenig & Moo, Accelerated C++.
One reason to count from 0 is that doing so encourages us to use asymmetric ranges to express intervals…Asymmetric ranges are usually easier to use than symmetric ones because of an important property: A range of the form [m, n) has n - m elements, and a range of the form [m, n] has n - m + 1 elements. So, for example, the number of elements in [0, rows) is obvious (i.e., rows - 0, or rows) but the number in [1, rows] is less so.
This behavioral difference between asymmetric and symmetric ranges is particularly evident in the case of empty ranges: If we use asymmetric ranges, we can express an empty range as [n, n), in contrast to [n, n - 1] for symmetric ranges. The possibility that the end of a range could ever be less than the beginning can cause no end of trouble in designing programs.
Another reason to count from 0 is that doing so makes loop invariants easier to express. In our example, counting from 0 makes the invariant straightforward: We have written r rows of output so far. What would be the invariant if we counted from 1?
One would be tempted to say that the invariant is that we are about to write the rth row, but that statement does not qualify as an invariant. The reason is that the last time the while tests its condition, r is equal to rows + 1, and we intend to write only rows rows. Therefore, we are not about to write the rth row, so the invariant is not true!
Our invariant could be that we have written r - 1 rows so far. However, if that’s our invariant, why not simplify it by starting r at 0?
Another reason to count from 0 is that we have the option of using != as our comparison instead of <=. This distinction may seem trivial, but it affects what we know about the state of the program when a loop finishes. For example, if the condition is r != rows, then when the loop finishes, we know that r == rows. Because the invariant says that we have written r rows of output, we know that we have written exactly rows rows all told. On the other hand, if the condition is r <= rows, then all we can prove is that we have written at least rows rows of output. For all we know, we might have written more.
If we count from 0, then we can use r != rows as a condition when we want to ensure that there are exactly rows iterations, or we can use r < rows if we care only that the number of iterations is rows or more. If we count from 1, we can use r <= rows if we want at least rows iterations — but what if we want to ensure that rows is the exact number? Then we must test a more complicated condition, such as r == rows + 1. This extra complexity offers no compensating advantage.
This excellent book offers many useful lessons, of which the above quote is, by far, the most memorable for me. You should read it.
I know that Keith Thompson is reading, so I must include the caveat that this statement is technically incorrect. You can still use it when thinking about arrays as a memory model, however.
Ken Gregg
Working with C, C++, C# and assembly languages for decades
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4y
In coding, why are arrays zero indexed?
Originally Answered: In coding, why are arrays zero indexed?
·
Actually, whether or not array indexes (aka, subscripts) start at 0 depends on the programming language you are using, or the features you select within that programming language.
In some languages, the index of the first element of an array is always 0. In those languages, if you have an array of 10 elements, the valid indexes range from 0 through 9. C is an example of such a language, but is cert
Actually, whether or not array indexes (aka, subscripts) start at 0 depends on the programming language you are using, or the features you select within that programming language.
In some languages, the index of the first element of an array is always 0. In those languages, if you have an array of 10 elements, the valid indexes range from 0 through 9. C is an example of such a language, but is certainly not the only one. Many many programming languages have been influenced by C, and follow this convention.
In some languages, the index of the first element of an array is always 1. In those languages, if you have an array of 10 elements, the valid indexes range from 1 through 10. Some (but not all) dialects of Fortran, BASIC, and COBOL fall into this category, among other languages.
In some languages, you can set the first valid index to either 0 or 1, either for all arrays in the module, or on an array-by-array basis. APL and some (but not all) dialects of Pascal fall into this category, among other languages.
In some languages, you can specify some other range of valid indexes for a specific array. For example, you could set up an array of 10 elements whose valid array indexes are 18 through 27. Some (but not all) dialects of Pascal fall into this category, among other languages.
In some languages, you can specify other types of indexes (i.e., non-integers) to act as valid array indexes.
In all cases, the first valid array index is mapped by the language to the first element of the array, which from a memory perspective is at offset 0 from the address of the beginning of the array.
It’s also possible in some languages to have what looks like an array, but is in fact some other data structure. In that case, the index you provide might go through some kind of search algorithm to access the element you’re interested in. Under the covers, these may not be arrays at all, but may be non-contiguous dynamic data structures, where the concept of an offset has no meaning. For example, overloading the [ ] operator in a C++ class can make an oper...
Ralf Quint
programming for 49y in Pascal,BASIC,assemblers,C,C++ et.al.
·
Author has 3.1K answers and 1.9M answer views
·
7y
What is the reason index numbers start from 0 instead of 1 in programming languages?
Originally Answered: What is the reason index numbers start from 0 instead of 1 in programming languages?
·
Well, this is the case only in “some” programming languages, for example C.
The reason is that in those cases, the index refers to the offset of the array element’s memory location.
In those programming languages where it starts from a non-zero index, it refers to the logical number of the elements. For that purpose, in some languages like Pascal for example, you can define the array boundaries in truly logical ways, for example
Var MiddleSchoolGrades : Array [6..8] of ClassRecord;
where you would address each element according to the grade, 6, 7 or 8, which is far more logical and readable in the
Well, this is the case only in “some” programming languages, for example C.
The reason is that in those cases, the index refers to the offset of the array element’s memory location.
In those programming languages where it starts from a non-zero index, it refers to the logical number of the elements. For that purpose, in some languages like Pascal for example, you can define the array boundaries in truly logical ways, for example
Var MiddleSchoolGrades : Array [6..8] of ClassRecord;
where you would address each element according to the grade, 6, 7 or 8, which is far more logical and readable in the source code than having to do arithmetic each time you would like to address an element within the array. The compiler just takes care of that for you…
Matt Laine
Senior Software Engineer
·
Author has 3.8K answers and 22.7M answer views
·
2y
What is the reason index numbers start from 0 instead of 1 in programming languages?
Originally Answered: What is the reason index numbers start from 0 instead of 1 in programming languages?
·
Thanks, Aaron Chao (赵明佳), for the A2A.
TL;DR: It makes memory address arithmetic easier to grok.
There are certain programming languages in which index numbers start from 1. MATLAB is one such language. The reason that MATLAB and other languages which choose to start arrays indexed at one (1) instead of zero (0) is because most people are used to counting, beginning with “one”. The “zeroth” element of something sounds very strange to folks used to not having to think about memory addresses.
However, in most languages, this question is valid. So, why do indices start at 0 instead of 1?
I’ll do a bi
Thanks, Aaron Chao (赵明佳), for the A2A.
TL;DR: It makes memory address arithmetic easier to grok.
There are certain programming languages in which index numbers start from 1. MATLAB is one such language. The reason that MATLAB and other languages which choose to start arrays indexed at one (1) instead of zero (0) is because most people are used to counting, beginning with “one”. The “zeroth” element of something sounds very strange to folks used to not having to think about memory addresses.
However, in most languages, this question is valid. So, why do indices start at 0 instead of 1?
I’ll do a bit of pseudocode to illustrate the point for you:
// How to access the memory address of the "nth" element of a zero-indexed array:ARRAY_ADDRESS + SIZE_OF_ARRAY_ELEMENT n // How to access the memory address of the "nth" element of a one-indexed array:ARRAY_ADDRESS + SIZE_OF_ARRAY_ELEMENT (n - 1)
This zero-indexed numbering strategy does not exist for the end-user; it exists for the folks who have to think about memory addresses, because it makes memory address arithmetic easier to grok.
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11222 | https://brainly.com/question/16313635 | [FREE] Factor as the product of two binomials: x^2 + 3x + 2 - brainly.com
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Factor as the product of two binomials:
x 2+3 x+2
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x²+3x+2 factors as (x+1)(x+2
To factor the quadratic expression ˣ²+3x+2, we can use a method called "ac method" or "factoring by grouping." Here's a step-by-step process:
Multiply the coefficient of the quadratic term (ˣ²)by the constant term (2): 1×2=2
Find two numbers whose product is the result from step 1 (2) and whose sum is the coefficient of the linear term (3). The numbers are 1 and 2 because 1×2=2 and 1+2=3.
Rewrite the middle term (3x) using the two numbers found in step 2. So, ˣ²+x+2x+2
group the terms in pairs and factor out the common factors: x(x+1)+2(x+1).
Factor out the common binomial factor (x+1)(x+2).
Therefore, x²+3x+2 factors as (x+1)(x+2
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The expression x 2+3 x+2 can be factored into (x+1)(x+2) by finding two numbers that multiply to the constant term and add to the coefficient of the linear term. The factors are determined through grouping and identifying common terms. Thus, the final result is (x+1)(x+2).
Explanation
To factor the quadratic expression x 2+3 x+2, we will find two binomials that multiply to give us this quadratic. Here’s a step-by-step process:
Identify the coefficients:
The coefficient of x 2 is 1 (let's call this 'a').
The coefficient of x is 3 (let's call this 'b').
The constant term is 2 (let's call this 'c').
Find two numbers:
We need to find two numbers that multiply to a c=1×2=2 and add up to b=3.
The numbers that satisfy these conditions are 1 and 2 because:
1×2=2
1+2=3
Rewrite the expression:
We can rewrite the quadratic as follows:
x 2+1 x+2 x+2
Factor by grouping:
Group the terms:
(x 2+1 x)+(2 x+2)
Factor out the common factors in each group:
x(x+1)+2(x+1)
Factor out the common binomial:
Now we can factor out the common binomial (x+1): (x+1)(x+2)
Thus, the given expression x 2+3 x+2 factors to (x+1)(x+2).
Examples & Evidence
For example, to further understand factoring, let's look at another expression, x 2+5 x+6. This can be factored as (x+2)(x+3) because 2 and 3 multiply to 6 and add to 5. By practicing with various quadratics, you can get accustomed to recognizing patterns in factoring.
The method shown is a standard approach taught in algebra courses and can be verified by expanding the factors (x+1)(x+2) to ensure it equals the original quadratic. This confirms that the factoring process is correct.
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(x + 1)(x + 2)
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x 2+3 x+2=x 2+2 x+x+2=x(x+2)+1(x+2)=(x+2)(x+1)
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11223 | https://www.quora.com/What-is-the-equation-of-a-parabola-that-has-vertex-as-origin-opens-upwards-and-passes-through-points-p-3-7 | What is the equation of a parabola that has vertex as origin, opens upwards and passes through points p (-3,7)? - Quora
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What is the equation of a parabola that has vertex as origin, opens upwards and passes through points p (-3,7)?
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Haresh Sagar
Studied Science&Mathematics (Graduated 1988) · Author has 6.2K answers and 7M answer views
·4y
Standard equation of a parabola which opens upwards is | (X — h)^2 = 4p(Y — k).
As vertex is at origin, h and k are 0. Substituting point p(—3,7) in the equation,
(-3–0)^2 = 4p(7–0) =>4p = 9/7
So the equation of required parabola is =>
X^2 = 9Y/7 => 7X^2 — 9Y = 0
Continue Reading
Standard equation of a parabola which opens upwards is | (X — h)^2 = 4p(Y — k).
As vertex is at origin, h and k are 0. Substituting point p(—3,7) in the equation,
(-3–0)^2 = 4p(7–0) =>4p = 9/7
So the equation of required parabola is =>
X^2 = 9Y/7 => 7X^2 — 9Y = 0
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How can I write an equation of a parabola that opens upward, has a vertex at the origin, and focus at (0,2)?
How do you determine the equation of a parabola which has a vertex of (-1,3) and passes through the point (0,5)?
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To find the equation of a parabola that opens upwards with its vertex at the origin math[/math], we can use the standard form of a parabola's equation:
[math]y = ax^2[/math]
where [math]a[/math] is a constant that determines the width and direction of the parabola.
Given that the parabola passes through the point [math]P(-3, 7)[/math], we can substitute this point into the equation to find [math]a[/math]:
[math]7 = a(-3)^2[/math]
This simplifies to:
[math]7 = 9a[/math]
Now, solving for [math]a[/math]:
[math]a = \frac{7}{9}[/math]
Now we can write the equation of the parabola:
[math]y = \frac{7}{9}x^2[/math]
Thus, the equation of the parabola is:
[math]y = \frac{7}{9}x^2[/math]
Upvote ·
Alberto Beron
Bachelor of Sciece (BS) in Electronic Engineering (Course)&Mathematics, California State University, Los Angeles · Author has 1.9K answers and 896.4K answer views
·4y
Y= ax^2 + bx + c
If P1= ( 0,0 )…. y = a(0)^2 + b(0) + c =0 = 0
Then . . . . . . . . . . . . . . . . . . . . . . . .y = ax^2 + bx (1)
The graph goes through: P2 ( -3,7); Then :
y = a(-3)^3 + b(-3)= 7
9a - 3b = 7 ( 2)
If the vertex is (0,0). La parabola is symmetric with respect to the y axis.
Then the graph also passes through the point P2( 3,7)
Since the graph passes through the point P2. We have: y = a. ( 3)^2 + b ( 3 ) = 7
9a + 3b = 7 (3)
Add equation (2) and (3)
18a = 14
a= 14/18 = 7/9 . . . . a = 7/9 (4)
(4) in (2)
9.( 7/9)+ 3b = 7
3b = 7 - 7 = 0 .. . . . b = 0 . ( 5 )
(4) and (5) in (1) . . . . . y
Continue Reading
Y= ax^2 + bx + c
If P1= ( 0,0 )…. y = a(0)^2 + b(0) + c =0 = 0
Then . . . . . . . . . . . . . . . . . . . . . . . .y = ax^2 + bx (1)
The graph goes through: P2 ( -3,7); Then :
y = a(-3)^3 + b(-3)= 7
9a - 3b = 7 ( 2)
If the vertex is (0,0). La parabola is symmetric with respect to the y axis.
Then the graph also passes through the point P2( 3,7)
Since the graph passes through the point P2. We have: y = a. ( 3)^2 + b ( 3 ) = 7
9a + 3b = 7 (3)
Add equation (2) and (3)
18a = 14
a= 14/18 = 7/9 . . . . a = 7/9 (4)
(4) in (2)
9.( 7/9)+ 3b = 7
3b = 7 - 7 = 0 .. . . . b = 0 . ( 5 )
(4) and (5) in (1) . . . . . y = 7/9 x^2 + (0)x ( 6)
Finally: . . y = 7/9 x^2 . (7) . The Answer
Test:
P1 ( 0,0 ) . 0 = 7/9 (0)^2
0 = 0 V
p2 ( -3,7 ) . . 7 = 7/9 (-3 )^2
. . . . . . .. . 7 = ( 7/9 )( 9)
7 = 7 . V
A faster method.
Since the vertex is at (0,0). . .0 = a(0)^2 + b(0) + c . . . (1) . Then c =0
To be symmetric with respect to the y axis all its exponents must be even. Then: b =0.
we have ( 1) is y = ax^2 . . (2)
P1 ( -3,7 ) in ( 2))
7 = a ( -3 )^2
7 = 9 a
Then a = 7/9
and (1) is :. . . . . y = 7/9 x^2. The Answer
Upvote ·
Graham Dolby
Author has 3.2K answers and 1.7M answer views
·4y
The quadratic equation of the curve with the vertex at math[/math], with a line of reflection parallel to the y axis and passing through math[/math] is [math]y = \tfrac{y_1-y_0}{(x_1-x_0)^2} (x-x_0)^2+ y_0[/math].
Here math = (0,0)[/math] and math = (-3,7)[/math] so we get [math]y = \tfrac{7}{(-3)^2} x^2[/math] which is [math]y = \tfrac{7}{9} x^2[/math]
Upvote ·
9 4
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11224 | https://pi.math.cornell.edu/~karola/dimex/catalan.pdf | CATALAN NUMBERS Supported by CAREER National Science Foundation Grant (DMS 1847284, PI: Karola M´ esz´ aros; document prepared by: Avery St. Dizier) Contents 1.
Catalan Numbers 1 2.
Objects Counted by Catalan Numbers 4 2.1.
Ballot Sequences 4 2.2.
Parenthesizations 5 3.
Exercises 7 3.1.
Nonintersecting Chords 7 3.2.
Noncrossing Arcs 8 3.3.
Binary and Full Binary Trees 8 3.4.
Triangulations 10 3.5.
Plane Trees 11 4.
Even More Catalan!
12 References 12 1. Catalan Numbers This document aims to give the reader an introduction to Catalan numbers, a sequence of numbers arising throughout mathematics. We will explore some of the myriad of objects counted by these numbers using the machinery of bijections. We begin with one such family of objects called Dyck paths, and use them to get a formula for the Catalan numbers.
A lattice path is a sequence of length 1 east and north steps in the plane starting at (0, 0).
A Dyck path is a lattice path that does not go above the line y = x. Shown below are two example lattice paths in the 3 × 3 grid. The one on the left is a Dyck path, but the one on the right is not.
Can you find the other four Dyck paths in the 3 × 3 grid?
1 2 CATALAN NUMBERS Definition 1.1. For each n ≥1, define the Catalan number Cn to be the number of Dyck paths in the n × n grid.
We will work toward giving a formula for Cn. First, recall the binomial coefficients.
Definition 1.2. For nonnegative integers n and k, the binomial coefficient n k is the number n k = n!
k!(n −k)!
which counts the number of ways to choose k objects at once out of a bag of n objects.
In other words, n k counts the number of size k subsets of [n] = {1, 2, . . . , n}.
Example 1.3. The size 3 subsets of = {1, 2, 3, 4, 5} are {1, 2, 3}, {1, 2, 4}, {1, 2, 5}, {1, 3, 4}, {1, 3, 5}, {1, 4, 5}, {2, 3, 4}, {2, 3, 5}, {2, 4, 5}, {3, 4, 5}.
Correspondingly, 5 3 = 5!
3!(5 −3)! = 5 × 4 × 3 × 2 × 1 (3 × 2 × 1)(2 × 1) = 10 To find a formula for the Catalan numbers, we begin with the following fact: Lemma 1.4. For any m and n, there are m+n m lattice paths from (0, 0) to (m, n) in the m × n grid.
Proof. Think of a lattice path as a sequence of letters indicating the steps, N for north and E for east. To reach (m, n), a lattice path must have exactly m E’s and n N’s, but can have them in any order. We can biject paths to size m subsets of [m + n]. Think of [m + n] as a line of empty boxes. In each box we can place either an E or an N. We think of E in box i as meaning “put i in the subset” and N as meaning “throw away i”. For example, NEENNNNEENE corresponds to the subset{2, 3, 8, 9, 11} of .
□ CATALAN NUMBERS 3 For example, for (m, n) = (3, 2), the bijection with size 3 subsets of is {1, 2, 3} {1, 2, 4} {1, 2, 5} {1, 3, 4} {1, 3, 5} {1, 4, 5} {2, 3, 4} {2, 3, 5} {2, 4, 5} {3, 4, 5} Proposition 1.5. The lattice paths in the n×n grid that cross the line y = x are in bijection with the lattice paths in the (n −1) × (n + 1) grid.
Proof. Given a lattice path P in the n×n grid crossing the line y = x, we construct a lattice path QP in the (n−1)×(n+1) grid. Let p be the first step of P that crosses the line y = x.
Let QP agree with P up to p, but after p swap the direction of each edge. For example: p P QP In the blue portion of P, there will always be exactly one more north steps than east steps by the definition of p. Since there are exactly n north steps and n east steps in P, there must be exactly one more east steps than north steps in the red portion of P.
Thus, both the blue and red portions of QP will have one more north steps than east steps. Thus, QP will have n −1 total east steps and n + 1 total north steps. In other words, QP will be a lattice path in the (n −1) × (n + 1) grid.
For the inverse bijection, just draw in the n × n diagonal in the (n −1) × (n + 1) grid and do the same reflection again!
□ For n = 2, there are four non-Dyck paths, and they biject to all lattice paths in the 1 × 3 grid: 4 CATALAN NUMBERS We can now describe the Catalan numbers.
Theorem 1.6. The Catalan numbers are given by the formula Cn = 2n n − 2n n −1 Proof. Cn counts the number of lattice paths in the n × n grid that do not cross the line y = x. There are 2n n total lattice paths. Of these, there are the same number as bad paths as there are lattice paths in the (n −1) × (n + 1) grid. That is, there are 2n n−1 . Hence, Cn = # of total paths −# of bad paths = 2n n − 2n n −1 .
□ The first few Catalan numbers are C0 = 1 (basically a definition), C1 = 1, C2 = 2, C3 = 5, C4 = 14, C5 = 42. What is so special about these numbers, you might ask. It turns out they count many, many completely different objects! We will look at a handful of these in the next section.
2. Objects Counted by Catalan Numbers In this section, introduce various families of mathematical objects and give bijections between them to show that they are counted by Catalan numbers. We begin with ballot sequences.
2.1. Ballot Sequences. A ballot sequence of length 2n is a sequence with n −1’s and n 1’s such that every partial sum is nonnegative. Equivalently, as you read the sequence left-to-right, you always encounter at least as many 1′s as −1’s. For example, (1, −1, 1, 1, −1, −1, −1, 1) has partial sums 1, 0, 1, 2, 1, 0, −1, and 0, so it is not a ballot sequence. For n = 3, there are five ballot sequences: (1, 1, 1, −1, −1, −1) (1, 1, −1, 1, −1, −1) (1, 1, −1, −1, 1, −1) (1, −1, 1, 1, −1, −1) (1, −1, 1, −1, 1, −1) CATALAN NUMBERS 5 We show that ballot sequences are counted by Catalan numbers through a bijection to Dyck paths!
Proposition 2.1. There are Cn ballot sequences of length 2n.
Proof. Given a ballot sequence b, replace each 1 by E and each −1 by N to get a lattice path P. We claim P is actually a Dyck path. Since the partial sums of b are nonnegative, in any initial segment of b there are at least as many 1’s as −1’s. Then in each initial segment of D, there are at least as many E’s as N’s. Thus, D never goes above the line y = x. Reversing this process gives a map from Dyck paths to ballot sequences.
□ When n = 3, we get the bijection: (1, 1, 1, −1, −1, −1) (1, 1, −1, 1, −1, −1) (1, 1, −1, −1, 1, −1) (1, −1, 1, 1, −1, −1) (1, −1, 1, −1, 1, −1) 2.2. Parenthesizations. Fix a string of n + 1 letters. A parenthesization is an insertion of n −1 open parentheses and n −1 close parentheses so that the resulting expression only involves products of two adjacent things at a time. For n = 2, we are looking at the string abc and inserting one pair of parentheses. This can be done in two ways: a(bc) and (ab)c.
Can you find five parenthesizations of abcd?
Proposition 2.2. There are Cn parenthesizations of n + 1 letters.
6 CATALAN NUMBERS Proof. We give a bijection to Dyck paths. Take a parenthesization of n+1 letters and remove all the right parentheses and the very last letter to get a string of length 2n −1 consisting of n −1 open parentheses and the first n letters. Note that this is bijective since we can recover the full parenthesization from this information. Try an example to see how you do this: Add a ‘(’ to the beginning of the string. To get a Dyck path, replace each ‘(’ by E and each letter by N. For example, (a((bc)d))(ef) ( ( ( ( ( a b c d e For this not to be a Dyck path, at some point prior to the end of the path there would have to be more N steps than E steps. Then at the corresponding point in the original parenthe-sization (which will be before the last letter), there would have been at least two more letters than ‘(’s (two, since we added a ‘(’ while building the path). In any valid parenthesization, there can only be at most one more letter than open parenthesis at any point, otherwise we will end up trying to multiply three or more things at once. To convince yourself of this, try writing down 6 letters and only 4 open (and closed if you want) parentheses in various ways in the space below. Can you make it so that at least two more letters than open parentheses occur prior to the last letter? What goes wrong?
This completes a map from parenthesizations to Dyck paths. Explain how you can reverse this map to go from Dyck paths to parenthesizations.
This will complete the bijective proof.
□ CATALAN NUMBERS 7 Using the five parenthesizations of abcd you found earlier, complete the n = 3 bijection: ((ab)c)d a(b(cd)) 3. Exercises For the exercises, we will define more objects counted by Catalan numbers and ask you to find bijections to prove it. For each type of object, we will show you the case n = 3 for you to play around with.
3.1. Nonintersecting Chords. Imagine 2n points in the plane lying on a circle.
Consider all the ways of connecting the 2n points by n chords (line segments lying inside the circle) such that no two chords ever cross. For n = 3, there are five ways to do this: 8 CATALAN NUMBERS Exercise 3.1. For each n, find a bijection between the arrangements of nonintersecting chords on 2n points and the Dyck paths in the n × n grid. Use your bijection to conclude that there are Cn arrangements of nonintersecting chords on 2n points for each n.
3.2. Noncrossing Arcs. Imagine 2n points in the plane lying on a horizontal line.
Consider all the ways of connecting the 2n points by n arcs lying above the line such that each arc connects exactly two points and no arcs ever cross. For n = 3, there are five ways to do this: Exercise 3.2. For each n, find a bijection between the arrangements of noncrossing arcs on 2n points and the arrangements of nonintersecting chords on 2n points. Use your bijection to conclude that there are Cn arrangements of noncrossing arcs on 2n points for each n.
3.3. Binary and Full Binary Trees. Binary trees are a type of graph (think of a diagram consisting of points and lines, not the graph of a function or a data plot) that represent a sequence of yes/no choices. Consider the following feline diagnostic tool as a motivating example.
CATALAN NUMBERS 9 Do you feed your cat regularly?
Do you regularly play with your cat?
Is it on a diet?
It’s hungry!
That’s why!
Cats need frequent love.
Cats are mysterious creatures. You may never know why.
Why is my cat mad?
Legend No Yes You begin at the top of the graph. At each point, you have at most two choices for how to answer. Each choice takes you further down the graph. When you reach a point with no more choices, you stop.
We now formally describe graphs with this kind of structure, called binary trees. The definition works recursively, defining a binary tree of a given size in terms of smaller binary trees. In general, it is not required that there are exactly two choices at each vertex. Binary trees with this property are called full binary trees.
Recall that a graph is a set of points, called vertices, and line segments between them, called edges. Binary trees are the members of a family of graphs defined recursively: • The empty graph (no vertices or edges) is a binary tree.
• A nonempty binary tree consists of a root vertex v0, a left-subtree TL, and a right subtree TR such that both TL and TR are binary trees, and v0 has an edge to the roots of both.
v0 TL TR Exercise 3.3. Explain in the example above why TL is itself a binary tree. Where is its root? Why are its left and right subtrees binary trees?
10 CATALAN NUMBERS Here are the five binary trees on 3 vertices: Let T be a binary tree, and fix a vertex v in T. The vertices that have an edge with v and are farther away from the root vertex of T are called the children of v.
Exercise 3.4. Prove that in a binary tree, each vertex has 0, 1, or 2 children.
A full binary tree is a binary tree where each vertex has either zero or two children. Here are the five full binary trees on 7 vertices: Exercise 3.5. For each n, find a bijection between the full binary trees on 2n + 1 vertices and the parenthesizations of n + 1 letters. Use your bijection to conclude that there are Cn full binary trees on 2n + 1 vertices for each n.
Exercise 3.6. Find a bijection between the full binary trees on 2n + 1 vertices and the binary trees on n vertices.
3.4. Triangulations. An n-sided polygon, or n-gon for short, is obtained by connecting n distinct points (again called vertices) on a circle with line segments to their closest neighbors, and then filling in the enclosed region. Here are n-gons for n = 4, 5, 6: A diagonal of an n-gon is a line segment between two vertices that are not neighbors on the circle. A triangulation of an n-gon is a set of n −3 diagonals that don’t intersect inside the polygon and that chop the polygon up into triangles. As you may be expecting by now, there are five triangulations of a 5-gon: CATALAN NUMBERS 11 Exercise 3.7. Find a bijection between triangulations of n-gons and either binary trees on n vertices or full binary trees on 2n+1 vertices. Conclude that there are Cn−2 triangulations of an n-gon.
3.5. Plane Trees. Similar to binary trees, plane trees are the members of a family of graphs defined recursively. Every plane tree P has a root vertex v0 (and so is nonempty). Either this is all of T, or T has subtrees T1, . . . , Tk all of which are plane trees whose roots have an edge to v0.
v0 T3 T2 T1 Exercise 3.8. Explain in this example why T2 is itself a plane tree. Where is its root? What are its subtrees, and why are they plane trees?
As you probably guessed, there are five plane trees with 4 vertices: Exercise 3.9. Find a bijection between plane trees with n + 1 vertices and Dyck paths in the n × n grid.
12 CATALAN NUMBERS Exercise 3.10. Find a bijection between plane trees with n + 1 vertices and binary trees with n vertices.
4. Even More Catalan!
So far, we have the following chart of objects and bijections: Parenthesizations Dyck Paths Full Binary Trees Binary Trees Plane Trees Noncrossing Arcs Nonintersecting Chords Ballot Sequences Triangulations Can you add any more bijections to this chart?
For more Catalan objects than you can imagine, the book describes over 200 different objects counted by the Catalan numbers!
References Marc Renault. Four proofs of the ballot theorem. Mathematics Magazine, 80, 12 2007.
R. P. Stanley. Catalan Numbers. Cambridge University Press, 2015. |
11225 | https://www.csd.uwo.ca/~abrandt5/teaching/DiscreteStructures/Chapter4/gcds-and-primes.html | .md
Greatest Common Divisors and Primes
Contents
4.2. Greatest Common Divisors and Primes#
Greatest common divisors (GCDs) and prime numbers
are a fundamental part of number theory. They have been extensively studied for
thousands of years. Euclid was fundamental to the study of prime numbers and number theory.
As we will see, Euclidean division extends to the Euclidean algorithm for computing GCDs.
Prime numbers
Finding Primes
Computing Primes
Greatest Common Divisors
Least Common Multiples
Euclidean Algorithm
Bézout Relations and GCDs
Consequences of Bézout
4.2.1. Prime numbers#
Definition (prime)
An integer is prime if the only divisors of are and .
An integer which is not prime is called composite.
For example, is prime because only and divide .
On the other hand, divides , and so is not prime.
Recall that another way of describing divisors is as factors.
Therefore, an equivalent definition of a prime number
is one whose factors are only 1 and .
The fundamental theorem of arithmetic strengthens the idea of primality to a prime factorization.
Theorem 4.2.1 (fundamental theorem of arithmetic)
Every integer greater than is either prime or can
be written as the product of two or more primes.
Formally, the theorem can be stated as follows. For every integer , there exists a positive integer ,
prime numbers , and exponents such that:
This theorem is also called the unique factorization theorem. It tells us that any number which is not prime
is the product of some primes.
Prime factorization
The following are prime factorizations.
Since multiplication is commutative, prime factorization is only unique up to the ordering of factors.
To get a unique prime factorization, we often add an additional constraint to the fundamental theorem of arithmetic.
This constraint requires the primes to listed in increasing order: .
4.2.2. Finding Primes#
How can we determine if a number is prime?
A simple and brute-force solution is to try and divide the number in question
by every other integer less than it. If there are no divisors, then the number is prime.
This method by “trial division” is very inefficient.
The sieve of Eratosthenes is a more efficient method. It is based on the following observation.
Proposition 4.2.2
If a positive integer is composite, then it must have a prime divisor less than or equal to
Proof: If is a positive composite then there exists two integers greater than 1 such that .
Certainly or . If is a perfect square then . Otherwise,
one of or must be smaller than .
The sieve of Eratosthenes uses this proposition to remove all composite numbers from a list and
retain only the prime ones.
The sieve of Eratosthenes
Let . Since the maximum element of is ,
we only need to consider primes divisors less than .
Find the smallest element of . This is . This element is prime.
Remove from all multiples of other than itself.
Find the next smallest element of the remaining numbers. This is .
Remove from all multiples of other than itself.
Find the next smallest element of the remaining numbers. This is .
Remove from all multiples of other than itself.
Find the next smallest element of the remaining numbers. This is .
Remove from all multiples of other than itself.
The next smallest element of is 11. Since , we can stop.
Every remaining number is prime.
The prime numbers less than 100 are:
4.2.3. Computing Primes#
Finding and, in particular, generating primes is a very practical problem. A large class
of digital security and cryptography algorithms rely on prime numbers.
However, there is no known closed form formula or function which always produces primes.
For example, the function results in prime numbers
for all choices of between and . However, is not prime.
On the other hand, we do know that there are infinitely many primes, so we can always find primes, with enough effort.
Theorem 4.2.3
There are infinitely many prime numbers.
Proof
Proceed by contradiction and assume that there are finitely many primes: .
Let .
By the fundamental theorem of arithmetic, is either prime or a product of primes.
If is prime then we have a contradiction where cannot be the list of all primes
since is not in that list and is prime.
On the other hand, consider if is a product of primes. Let be such a prime divisor of .
That is, . However, we also have .
Therefore, it must be that , that is, .
However, this is not possible since every prime is greater than . This is a contradiction.
In either case we have that cannot possibly be the list of all primes.
Hence, there are infinitely many primes.
Even though there are infinitely many primes, actually generating them is a challenge.
By trial division or the sieve of Eratosthenes, we can determine if a number is prime.
However, for large numbers, this is very inefficient.
Another test of primality is based on Fermat’s little theorem.
Theorem 4.2.4 (Fermat’s little theorem)
For two positive integers and , if is prime and , then:
Proof
Nope! This proof is well beyond the scope of this course. See wikipedia.
For example, we know that is prime and . By Fermat’s little theorem,
it must be that . Indeed, we have and .
Fermat’s little theorem gives rise to the Fermat primality test.
Given some positive integer , we can deduce that is not prime if we
can find a number such that .
Such an is called a Fermat witness.
The Fermat primality test leads to a probabilistic method to determine if a number is prime.
A probabilistic algorithm is one which produces the correct result “with high probability” but
not necessarily all of the time.
For primality, the probabilistic method tries to find a Fermat witness.
If, after a certain number of attempts, it cannot find such a witness,
then the algorithm terminates and assumes that the number is prime.
This assumption is what makes the algorithm probabilistic. Only if we compute
for every will we know for certain that is prime.
How many times should we try to find a Fermat witness before giving up?
The answer is not clear. Computing the exact probability of error is challenging.
Qualitatively, the more attempts we make to find a Fermat witness,
the better chance there is that is actually prime. Hence, one can feel “good”
if they try a “reasonably large” number of times to find a Fermat witness.
Maybe 100? 256? 1000? It depends on how long you are willing to wait and how large your candidate number is.
The below Python code implements the Fermat primality test
by choosing random integers as possible Fermat witnesses.
It generates random primes by choosing using this test.
```
from random import randint
def isPrime(p, numIter) :
for i in range(numIter) :
a = randint(2, p-1)
e = a(p-1) % p #this is very inefficient
if (e != 1) :
return False
return True
def randomPrime(n) :
while(True) :
p = randint(2(n-1), 2(n)-1)
if isPrime(p, 128) :
return p;
print a random 32-bit prime
print(randomPrime(16))
```
```
49103
```
Prime Conjectures
Primes have been studied for thousands of years by countless researchers. Yet, many properties have still been unproved.
The following are conjectures many believe to be true.
Goldbach’s conjecture: Every even integer greater than is the sum of two primes.
This has been verified for numbers up to ().
Landau’s conjecture: There are infinitely many primes of the form , for a positive integer .
On the other hand, it has been proven that there are infinitely many composite numbers of the form . But, this does not
disprove Landau’s conjecture.
Twin Prime conjecture: There are infinitely many primes that differ by . Twin primes include and , and , and , etc.
4.2.4. Greatest Common Divisors#
Definition (greatest common divisor)
For two non-zero integers and , is the greatest common divisor of and if , , and any other common divisor of and also divides .
For small numbers, it is easy to compute GCDs by eye.
For example, a GCD of and is . Yet, another GCD of and is .
Consider the again the definition of GCD. The common divisors of and are , , , , , and .
Since , and all divide , and , , and all divide ,
both and are GCDs of and .
To make life easier, we typically decide on a “canonical” GCD and call it the GCD. For the integers,
the canonical GCD is the positive one. Therefore, the GCD of and is .
We can write the GCD of two numbers in a functional notation.
For example, .
Definition (relatively prime)
Two integers are relatively prime if their greatest common divisor is . We call two such integers co-prime.
Every pair of numbers has a trivial common divisor: .
When two numbers have a GCD that is greater than , we say they have a non-trivial GCD. Otherwise, they are relatively prime.
That means, they do not share any common factors.
When we have a collection of integers, we say that they are pairwise relatively prime if every integer in the collection
is relatively prime to every other integer. That is,
for integers , we have
for .
We can determine the GCD of two numbers, or if they are relatively prime, based on their prime factorizations.
If any of the primes equals a prime , then
and have a non-trivial GCD. If and have no primes
in common between their prime factorizations, then they are relatively prime.
We can compute the GCD of and from their prime factorization
by getting the product of all common primes raised the minimum exponent of that prime in either number.
Computing GCDs from primes
We compute the GCD of and .
The GCD of and is thus .
Least Common Multiples#
We can also use prime factorization to compute
the least common multiple (LCM) between two integers.
Definition (least common multiple)
The least common multiple of two positive integers and is the smallest positive integer that is divisible by both and .
Whereas we computed the GCD by taking the common primes of
two integers (i.e. the intersection of primes in their factorization), we compute the LCM by taking the union of primes
in their factorization. Then, each prime is raised to the maximum exponent of that prime in either number.
Computing LCMs from primes
We compute the LCM of and .
The LCM of and is thus .
While this method is easy by inspection, computing the prime factorization
of number, in general, is very hard. Therefore, this method
is not practical. A more practical method is the Euclidean algorithm which we will study shortly.
Before that, we conclude with a theorem. Can you prove it?
Theorem 4.2.5
For any two positive integers and , we have:
Euclidean Algorithm#
The Euclidean algorithm is an efficient method
for computing the GCD of two integers.
As can be inferred from its name, the algorithm has been known for many centuries and can be attributed to Euclid. The algorithm is based on a simple idea rooted in Euclidean division.
Let by Euclidean division. Then, .
Now, let be a common divisor of and .
That is, and . Certainly, then,
we have since .
In summary, we have the following lemma.
Lemma 4.2.6
Let and be integers with by Euclidean division.
Thus, . We have:
Therefore, the foundation of the Euclidean algorithm is repeated
applications of Euclidean division.
Euclidean by example
Let us find the GCD of and by the Euclidean algorithm.
. That is, mod = 14.
. That is, mod = 7.
. That is, mod = 0.
Since we cannot divide by 0 in the next step, the process terminates and we have:
This previous example results in a remainder sequence starting at and ending at :
We can verify that is a common divisor at every step.
, , , , and . This is as expected from the previous lemma.
Moreover, notice that this remainder sequence is strictly decreasing. From Euclidean division we have with . Since the magnitude of successive remainders strictly decreases, and has as a lower bound, it must be that repeated applications of Euclidean division will lead to a remainder
and thus termination of the algorithm. The correctness follows from
the previous lemma.
Finally, we state the algorithm. Its termination and correctness is guaranteed by the previous discussion.
We can also see the Euclidean algorithm in action in Python.
```
def gcd(a,b) :
x = a
y = b
print("r0: %d" % x)
print("r1: %d" % y)
i = 2;
while y != 0 :
r = x % y
print("r%d: %d" % (i, r))
i += 1
x = y
y = r
return x
print("GCD(152152, 154700) = %d" % gcd(152152, 154700))
```
```
r0: 152152
r1: 154700
r2: 152152
r3: 2548
r4: 1820
r5: 728
r6: 364
r7: 0
GCD(152152, 154700) = 364
```
Bézout Relations and GCDs#
An interesting property of GCDs is that they can be
written as a linear combination of the two input integers.
Theorem 4.2.7 (Bézout’s theorem)
For any positive integers and , there exists integers and such that:
From this theorem we extract some definitions.
The formulas is called the Bézout identity.
The integers and are called the Bézout coefficients of and .
A first Bézout relation
The Bézout coefficients of and are and .
We can find the Bézout coefficients of two numbers through a “two pass” method using the Euclidean algorithm.
Consider the Euclidean algorithm executed on and .
Therefore, . Now, how can we write as a combination of and ?
We can work bottom-up from the successive Euclidean divisions and back-substitute one equation at a time.
The first line comes from . The second line comes from substituted
into the previous. The third line comes from substituted into the previous.
This two-pass method is sufficient for computations by hand. However, a more efficient method is to
compute a sequence of Bézout coefficients at each step of the algorithm.
Alongside the sequence of remainders , one also computes a sequence
of Bézout coefficients and . These sequences start as:
See more about the Extended Euclidean Algorithm here.
Consequences of Bézout#
Lemma 4.2.8
Let be positive integers such that and are relatively prime.
If then .
Proof:
Since and are relatively prime, we have .
By hypothesis assume .
By Bézout’s theorem, we have .
Then, multiply both by to get: .
Since , . That is, there exists
such that .
Therefore, we have:
Hence, , as required.
This lemma can be generalized for prime and composite numbers.
Lemma 4.2.9
Let be a prime integer and be integers.
If , then for at least one .
We saw earlier in Congruence, sums, and products
that division did not always maintain congruence relations.
However, with Bézout relations we can describe when divisions do maintain congruence.
Theorem 4.2.10
Let be a positive integer and be integers.
If and , then .
Proof:
By the hypothesis we have , hence:
From the previous lemma, must therefore divide or .
Since, by assumption, , it must be that .
That is, .
4.2.5. Exercises#
Exercise 4.10
Write a Python function that takes an integer and prints all prime numbers between and . Use the sieve of Eratosthenes as your algorithm.
Exercise 4.11
Determine the following values.
Solution to Exercise 4.11
6
42
42
14
Exercise 4.12
Compute the prime factorization of the following numbers.
Solution to Exercise 4.12
Exercise 4.13
Let . Let be a binary relation on where if .
Draw a Hasse diagram for .
Exercise 4.14
Prove the following statement.
Let be an integer. is never divisible by .
Solution to Exercise 4.14
Proceed by contradiction. Assume . Then, such that .
Assume is even. Then for some integer , and:
which says , which is not possible.
Assume is odd. Then for some integer , and:
which says , which is impossible.
Therefore, for any integer .
Exercise 4.15
Prove the following statement.
Proposition
Let be a natural number and be an integer. In the sequence of consecutive integers ,
at least one of them is divisible by .
Solution to Exercise 4.15
By Euclidean division we have with .
If , then and .
If , then , and .
If , then and .
Continuing, if then and .
Since ranges over consecutive values, it covers all cases of .
Exercise 4.16
RSA is a cryptography system which relies of exponentiation and modular numbers. In particular,
it is relatively easy to find three integers such that, for any integer , ,
We call the encrypted message and the public key. Then, is the decrypted message and is the private key.
In particular, we can compute such an as the least common multiple of and for two different prime numbers and .
If and then .
Find and such that .
Tip: use a calculator or python to handle the big numbers!
Solution to Exercise 4.16
Any two numbers which are multiplicative inverses modulo 30 will work. For example, 7 and 13, 11 nad 11, 17 and 23, 19 and 19, 29 and 29. |
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Running a grocery store sounds simple—after all, everyone needs food. But turning that everyday demand into a business that actually works? Takes planning.
Whether you’re starting small, like a corner shop, or going bigger with a full-sized supermarket, it all comes down to one thing: A clear, workable plan.
Not sure where to start?
This grocery store business plan will walk you through everything from your executive summary to financial projections.
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How to Write a Grocery Store Business Plan?
Writing a grocery store business plan is a crucial step toward the success of your business. Here are the key steps to consider when writing a business plan:
1. Executive Summary
The executive summary gives a quick snapshot of your entire grocery store business plan. This section helps readers like investors, lenders, or partners understand your vision, how the store will run, and why it’s worth their time or support.
Even though this section sits right at the very front of your plan, here’s a tip: it’s best to tackle this part last. That way, you’ve got all your facts straight from the rest of the plan, letting you keep the summary to the point.
Here’s what to include in your executive summary:
Grocery store concept & location
Business model (retail sales, delivery, subscriptions, or other services)
Target market (local families, working professionals, students)
Unique selling points (USPs)
Marketing strategy
Financial highlights (revenue goals)
Funding needs (If any)
Appendix (optional)
Ultimately, this whole section should feel upbeat, realistic, and inspiring enough to genuinely make the reader eager to flip through the rest of your detailed plan.
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2. Business Overview
This business overview section explains what your grocery store is all about, your concept, structure, location, and what you’re building. It’s a snapshot of your store’s identity.
Here’s what to include in your business overview:
| Section | What to Include |
--- |
| Store name & concept | Name + type of store (local, organic, ethnic, etc.) |
| Legal structure | Chosen structure (LLC, sole prop, etc.) + quick reason |
| Location | Planned location + why it works (foot traffic, demand) |
| Ownership & team | Owner(s) + roles or strengths of key people |
| Mission & vision | What your store stands for and your long-term goal |
| Goals & milestones | Immediate launch goals + future plans (e.g., expansion) |
By the end of this section, readers should’ve a clear sense of what kind of store you’re building, why it makes sense for your area (location of operation), and who’s behind it.
Read More: What to Include in an Effective Business Overview
3. Market Analysis
The market analysis provides a clear understanding of the market in which your small grocery store business will run, along with your target market, competitors, and growth opportunities.
Your market analysis should contain the following essential components:
Target market
Identify your target market and define your ideal customer. This can include age, economic level, family size, purchasing interests, and nutritional preferences.
Here’s an example of your core customer:
Market size and trends
Use local or national data to show the opportunity:
Annual grocery spending in your area
Growth in specialty segments like organic, international, or health-focused foods
Increased demand for contactless checkout, in-app ordering, or home delivery
Show how your store will tap into these trends with the right mix of inventory and services.
Competitor analysis
Start by looking at two or three grocery stores in your area. Maybe one has a great variety and strong branding but charges higher prices, while another is more affordable but has poor customer service or an outdated layout.
Think about where they fall short and how your store can do better. Maybe in terms of faster checkout, a cleaner layout, or friendlier service. Show how your store fills the gaps and offers something shoppers aren’t getting elsewhere.
Market opportunity
End this section by connecting the dots:
There’s steady demand
There’s a clear customer base
There are gaps you can fill
And your store is positioned to meet those needs directly
The key is to clearly define your target customers. Once that’s done, the rest of this section comes together much easily.
4. Products and Services
The heart of any grocery store lies in what it sells. Your store should aim to offer everyday items people need, while also being known for quality and convenience.
You can carry a wide variety of goods, including:
| Category | Examples |
--- |
| Fresh Produce | Fruits, vegetables, herbs |
| Dairy Products | Milk, cheese, butter, yogurt |
| Meat & Poultry | Chicken, mutton, eggs, fish |
| Grains & Staples | Rice, wheat, pulses, flour |
| Packaged Snacks & Sweets | Biscuits, chips, chocolates, candies |
| Frozen & Ready-to-Eat | Frozen meals, snacks, frozen vegetables |
| Cooking Essentials | Spices, oil, salt, sauces, condiments |
| Beverages | Soft drinks, juices, bottled water, tea, coffee |
| Household Essentials | Detergents, soaps, toothpaste, cleaning supplies |
| Optional Services | Home delivery, seasonal kits, bulk orders, loyalty rewards, local tie-ups |
Quality and freshness should be a top priority. Ensure all perishable items are stored correctly and checked regularly. Having a reliable supply chain will also help keep shelves stocked without compromising on standards.
This section gives a clear picture of what customers can expect from your store.
5. Sales and Marketing Strategies
Setting up your grocery store is just one part of the process; getting customers through the door (and keeping them coming back) is just as important. That’s where a strong sales and marketing strategy comes in.
Here’s how you can approach it:
What makes us special? (unique hook)
You might focus on locally grown produce, stock hard-to-find international items, or offer great everyday value. Whatever it is, you need to make it clear and consistently remind customers why they should choose you over the next shop.
Smart pricing that works for everyone
People shop smart, especially when it comes to groceries. Your pricing needs to be competitive, but still leave room for profits. Occasional offers, loyalty discounts, or “buy more, save more” deals can help draw attention and drive bigger purchases.
Spreading the word locally
Getting the news out doesn’t have to break the bank. Just a few quick updates on social media about fresh arrivals or weekend deals can surprisingly bring a good amount of foot traffic.
Old-school flyers, eye-catching in-store posters, and even simple referral programs (think “bring a friend, get a discount!”) are super effective, low-cost ways to market. Don’t forget to optimize your Google My Business profile – it’s important for locals trying to find you online.
Bringing people in and back
Promotions tied to festivals, free tasting sessions, or weekend family offers can pull in new faces. But the real win is making them return. Great service, clean displays, fast checkouts, and the occasional “we remembered your favorite brand” moment go a long way.
At the end of the day, your strategy is about connection, making people feel that your store is part of their routine, not just a random stop.
6. Operations Plan
When writing the operations plan section, it’s important to consider the various aspects of your business processes and procedures involved in operating a business.
Here are the components to include in an operations plan:
Hiring plan
Start by outlining your staffing needs. How many employees will you need to manage checkout counters, stock inventory, handle deliveries, or manage the floor? Be clear about their roles and what skills or experience you’re looking for.
Also, include any employee perks or benefits you’ll offer to help attract and retain good talent.
Daily operations
Explain how your grocery store will function on a regular day. That includes how you’ll place and receive product orders, manage stock levels, restock shelves, handle customer service, and keep the store clean and organized.
If you’re offering delivery or online ordering, mention how that process will work too.
Technology and equipment
To run your grocery store efficiently, you can rely on a few essential tools:
A Point-of-Sale (POS) system for quick checkouts and sales tracking.
Barcode scanners to speed things up and keep inventory accurate.
Inventory management software to track everything from stock levels to reorders.
Reliable refrigeration units and freezers for all our fresh and frozen items.
Strong security systems to keep everyone safe and prevent theft.
Having these essential tools and clear daily routines means you’re setting up a business that’s ready to operate effectively every single day.
7. Management Team
The management team section provides an overview of the individuals responsible for running the grocery store.
Include the following in your plan:
Ownership and leadership: Who owns the business, and what relevant experience or background do they bring to the business?
Key team members: Highlight essential roles like Store manager, Procurement lead, Floor supervisor, Finance or Admin manager, and Delivery coordinator, if applicable.
Experience and qualifications: Summarize each person’s relevant background that makes them a good fit for their role.
Compensation and structure: Mention how the team is compensated, such as salary, bonuses, or equity, if relevant.
Advisory support: If any external consultants, mentors, or advisors are involved, note their names and contributions.
Try adding a simple chart to show how roles are structured and who reports to whom. Here’s a sample layout you can adapt for your plan:
With an experienced team and clear roles, our grocery store is set up for smooth operations and steady growth. This strong management foundation will help us serve customers well and achieve our goals.
Read More: How to Write the Management Team Section of a Business Plan
8. Financial Plan
Your financial plan lays out the money side of your grocery store business. It’s where you show exactly how you’ll bring in revenue, manage all your costs, and ultimately make a profit.
Key financial statements
Below are the key statements your financial plan should include:
| Statement Name | Description | Includes |
---
| Profit & Loss Statement | Projects the store’s income and expenses over time, showing profitability. | Revenue, Cost of Goods Sold (COGS), Operating Expenses, Net Profit or Loss. |
| Cash Flow Statement | Tracks the movement of cash in and out of the business to ensure liquidity. | Cash receipts from sales, vendor payments, loan repayments, other inflows/outflows. |
| Balance Sheet | Provides a snapshot of the store’s financial position at a given point in time. | Assets (current & fixed), Liabilities (short & long-term), Owner’s Equity. |
| Break-even Analysis | Identifies the point at which revenue covers all operating costs. | Fixed Costs, Variable Costs, Contribution Margin, Break-even Revenue/Units. |
If you’re seeking funding, clearly state it. Break down how the funds will be used, such as for inventory, marketing, branding, or working capital.
For more clarity, you can add a chart showing the use of startup funds. Sort of like this:
This section should show that your numbers are grounded and assumptions are realistic, and your store will become financially sustainable with time.
To create automatic financials for your own business plan, we recommend Upmetrics. Create your own business plan.
9. Appendix
The appendix section includes extra documents and data that support your business plan. While not required for every reader, it’s helpful for lenders, investors, or partners who want to see proof behind your assumptions.
You can include:
Financial statement backups (detailed cash flow projections, P&L reports)
Market research data or survey results
Supplier agreements or sample invoices
Lease agreement or property photos
Business registration or licenses
Menu of products or categories (with pricing)
Equipment quotes
Staff training manuals or onboarding outline
Charts or visuals referenced in your financial plan (e.g., break-even graph, funding allocation pie chart)
Use a table of contents within the appendix if your appendix is longer than a few pages, and keep everything clearly labeled so it’s easy to refer to.
Only include what adds real value; skip anything that repeats what’s already in the main plan.
Download a free grocery store business plan template
Ready to start putting your grocery store plan together, but want a little help with formatting? No problem. Download our free grocery store business plan template (PDF) to get started.
It’s a pre-formatted and ready-to-use template. It covers everything from the executive summary to financial projections. All you have to do is, just fill in the details that fit your specific store, make a few adjustments, and you’ll have a solid, professional plan in no time.
The Quickest Way to turn a Business Idea into a Business Plan
Fill-in-the-blanks and automatic financials make it easy.
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Summary
Now that you’ve gone through this step-by-step guide and have a template as a handy tool, so might be in a much better position to draft your grocery business plan.
But, feeling stuck? Upmetrics can help. Our business planning software makes the entire process easier with smart AI suggestions, clean formatting, and built-in tools for financial forecasting, competitor analysis, and more.
So why wait? Start planning smarter, right now!
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Frequently Asked Questions
Why do you need a grocery store business plan?
A grocery store business plan helps you define your vision, set goals, and stay organized. It shows potential investors or lenders you’re serious and prepared.
Most importantly, it gives you clear guidance on key areas like managing inventory, hiring staff, marketing your store, and planning your finances, so you can build a solid, successful business.
How to get funding for your grocery store business?
There are several ways to get funding for your grocery store, but one of the most efficient and speedy funding options is self-funding.
Other options for funding are:
Bank loans
Small Business Administration (SBA) loans
Crowdfunding platforms like GoFundMe
Angel investors
Venture capital
Apart from all these options, there are small business grants available. Check with your operating location, and you can apply for them.
What is the easiest way to write your grocery store business plan?
The easiest way to write your grocery store business plan is to use a template. It makes the process simple and helps you fill in each section easily.
You can also use business planning software like Upmetrics, LivePlan, or Bizplanr. These tools guide you step by step and help you finish your plan quickly.
How detailed should my financial plan be?
Be clear and realistic with your financial plan; it shows you understand how to run a successful grocery store and builds trust with investors or lenders. Include essentials like:
Startup costs
Revenue forecasts
Profit margins
Operating expenses
Cash flow projections
A clear financial plan helps you avoid financial surprises and make better buying and pricing choices in your grocery store.
Upmetrics Team
Upmetrics is the #1 business planning software that helps entrepreneurs and business owners create investment-ready business plans using AI. We regularly share business planning insights on our blog. Check out the Upmetrics blog for such interesting reads. Read more
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What is the formula for thermal stress and how is it derived?
Thermal Stress and Strain are key mechanisms that govern how objects respond when their temperature changes but their expansion or contraction is restricted. Such effects are seen in bridges, rails, and pipelines. Mastery of this concept is vital for JEE Main Physics.
When temperature alters, materials tend to change their dimensions. The restriction of this natural expansion or contraction results in thermal stress. The related geometric change, measured as a ratio, corresponds to thermal strain. Engineers must manage these effects to prevent failures in everyday structures.
Understanding Thermal Stress and Strain in JEE Physics
Thermal stress arises when a solid body can’t freely expand or contract due to boundary constraints. The change in temperature attempts to alter the material’s length, but resistance to this change develops internal forces, or thermal stresses. The corresponding fractional change in length is the thermal strain. Both terms are critical in mechanical and civil engineering scenarios.
Several physics principles tie into this topic, such as thermal expansion, Young’s modulus, and the relationship between stress and strain. JEE Main candidates often face questions linking these connected ideas.
Thermal stress and strain are crucial for understanding why buildings need expansion joints.
Thermal expansion is the basis for calculating resulting stresses.
Young’s modulus links internal forces to material deformation.
Practical engineering applies these concepts for safety.
Neglecting thermal effects can cause cracks in structures.
Formulae and Key Equations for Thermal Stress and Strain
The main equation connecting temperature and internal force is:
| Quantity | Formula | Variables | SI Units |
--- --- |
| Thermal stress | YαΔT | Y: Young’s modulusα: linear expansion coefficientΔT: temp. change | N/m2 |
| Thermal strain | αΔT | α: expansion coefficientΔT: temp. change | (dimensionless) |
The thermal stress formula assumes the object cannot move at all. Y must be in N/m2, α in K-1, and ΔT in Kelvin or Celsius. Omitting the restriction condition is a frequent JEE error.
Thermal stress = YαΔT (if ends are fixed).
Thermal strain = αΔT (ratio, no unit).
If the material is unconstrained, thermal stress is zero.
Remember: sign convention matters for expansion vs contraction.
For advanced problems, you may need the modulus of elasticity or details from states of matter and thermal expansion.
Comparison: Thermal Stress vs Thermal Strain
| Aspect | Thermal Stress | Thermal Strain |
---
| What is measured? | Internal force per area | Fractional length change |
| Formula | YαΔT | αΔT |
| Units | N/m2 | Dimensionless |
| Exists when | Expansion/contraction prevented | Any temp. change |
Always check in JEE questions whether a rod or structure is “fixed at both ends.” If not, thermal stress does not occur. This is a high-frequency mistake on exam day.
Applications, Examples, and Problem-Solving with Thermal Stress and Strain
Let’s see a typical application found in JEE Main numericals.
Metal railway tracks expand in summer heat and may buckle if not designed with clearances.
Pipelines undergo thermal expansion and require flexible joints.
Bridges have expansion joints to absorb thermal strain.
Metal rods in machinery can develop excessive internal stress if not allowed to expand.
Here is a solved example relevant for JEE Main.
Example: A steel rod (Y = 2 × 1011 N/m2, α = 1.2 × 10-5 K-1) of length L is fixed at both ends. Temperature increases by 40 K. What is the thermal stress?
By the formula, thermal stress = YαΔT
Thermal stress = 2 × 1011 × 1.2 × 10-5 × 40 = 9.6 × 107 N/m2.
Such calculations frequently appear in practice papers and mock tests.
Always check if ends are fixed before applying the formula.
Thermal stress is zero if the rod is free to expand.
Units: SI system only (Pa or N/m2).
High stresses can cause cracking if not managed.
Different materials have different α and Y values.
For a more detailed look, explore related concepts like elasticity and Hooke’s law or measurement errors due to temperature changes.
Viscosity and viscous force topics connect when studying liquids exposed to heat.
Heat and temperature provides foundational terminology for this chapter.
Laws of thermodynamics frame underlying reasons for material expansion.
Properties of solids and liquids deepen understanding of material-specific behaviors under temperature change.
Thermal stress and strain questions test students’ grasp of material response to heat. They combine basic physics, real-world applications, and mathematical reasoning—all central to Vedantu’s exam preparation approach.
Review mock tests to find problem-solving patterns.
Summarise constants: Y and α differ by material.
Always specify if the change is heating or cooling.
Never mix units—stay strictly in SI.
In summary, thermal stress and strain allow us to predict and design against failures in engineering. They draw on core physics principles found in many JEE Main chapters and link with topics like modulus of elasticity, states of matter, and thermal expansion. For additional practice, see the properties of solids and liquids practice paper or the thermodynamics mock test curated by Vedantu’s JEE experts.
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FAQs on Thermal Stress and Strain: Concepts, Formulas & Applications
What is the formula for thermal stress?
Thermal stress is calculated using the formula: Thermal stress (σ) = Young's modulus (Y) × Coefficient of linear expansion (α) × Change in temperature (ΔT). This formula is important for determining how much stress develops in a material when it is prevented from expanding or contracting with temperature change. Key points:
σ = Y × α × ΔT
Where σ = thermal stress, Y = Young's modulus, α = coefficient of linear expansion, ΔT = temperature change
This concept is widely used in physics, engineering, and microelectronics packaging
What is the thermal strain formula?
Thermal strain refers to the fractional change in length due to temperature and is given by: Thermal strain (ε) = Coefficient of linear expansion (α) × Change in temperature (ΔT). Details:
ε = α × ΔT
Where ε = thermal strain, α = coefficient of linear expansion, ΔT = temperature change
Thermal strain is dimensionless and important for understanding material deformation
What is thermal stress in physics?
Thermal stress in physics is the internal stress created in a material when it is constrained and cannot freely expand or contract with temperature changes. This can cause objects to crack or deform under certain conditions.Key aspects:
Occurs when temperature change is applied to a fixed or constrained body
Depends on material properties like Young's modulus and thermal expansion coefficient
Relevant in structures, bridges, and electronic devices
What is the difference between thermal stress and thermal strain?
Thermal stress refers to the internal force developed per unit area within a material due to temperature changes, while thermal strain measures the relative deformation or change in size. Difference:
Thermal stress: Internal force (units – pascal or newton/m²)
Thermal strain: Fractional length change (dimensionless)
Stress arises if expansion/contraction is restricted, whereas strain can occur even with free expansion
How does thermal stress work?
Thermal stress occurs when a material experiences a change in temperature and is prevented from expanding or contracting freely. The restriction creates internal forces, potentially leading to cracks or deformation.
If free to expand, no stress develops
If constrained, stress builds up according to the formula σ = Y × α × ΔT
Understanding this is crucial in construction, manufacturing, and electronics
What are the causes of thermal stress?
Thermal stress is caused by temperature changes in materials that are either restrained or have varying expansion rates. Major causes include:
Rapid heating or cooling of objects (e.g., quenching hot metal in water)
Differences in temperature between different parts of an object
Use of materials with different thermal expansion in a structure
Fixing one end of a rod and exposing it to temperature change
Environmental temperature variations
What is the difference between heat stress and thermal stress?
Heat stress usually refers to the condition in living organisms or workers exposed to hot environments, impacting health. Thermal stress is a physics and engineering concept describing internal stresses in materials due to temperature changes.
Heat stress: A physiological response; relevant in occupational safety
Thermal stress: Mechanical response; associated with materials and structures
What are some real-world examples of thermal stress and strain?
Thermal stress and strain are commonly observed in everyday life and industry when materials undergo temperature changes.
Cracking of glass: Pouring hot water into a cold glass vessel
Railway tracks: Expansion gaps are left to prevent buckling during summer
Bridges: Expansion joints account for thermal expansion and contraction
Microelectronics packaging: Stress between different materials due to temperature cycling
How to calculate thermal stress in a constrained rod?
To calculate thermal stress in a rod fixed at both ends:
Find the coefficient of linear expansion (α) and Young's modulus (Y)
Measure the temperature change (ΔT)
Apply the formula: Thermal stress (σ) = Y × α × ΔT
Ensure the rod is prevented from changing length for this calculation to be valid
What are the important applications of thermal stress and strain in microelectronics packaging?
Thermal stress and strain play a critical role in microelectronics packaging, as devices often have dissimilar materials expanding at different rates. Applications include:
Designing solder joints to withstand temperature fluctuations
Improving reliability of semiconductor chips
Minimizing failure due to delamination or cracking
Optimizing material selection to reduce mismatches in thermal expansion
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11228 | https://i17vm1.in.tum.de/zotero/5TPEFWM5.pdf | Computers in Industry 129 (2021) 103442 Contents lists available at ScienceDirect Computers in Industry jo ur nal ho me page: www.elsevier.com/locate/compind Extraction of dimension requirements from engineering drawings for supporting quality control in production processes Beate Scheibel a,∗, Juergen Mangler b, Stefanie Rinderle-Ma b a Research Group Workflow Systems and Technology, Faculty of Computer Science, University of Vienna, Waehringerstrasse 29, 1090 Vienna, Austria b Information Systems and Business Process Management, Department of Informatics, Technical University of Munich, Boltzmannstrasse 3, 85748 Garching, Germany a r t i c l e i n f o Article history: Received 6 October 2020 Received in revised form 5 March 2021 Accepted 10 March 2021 Keywords: Engineering drawings Information extraction Clustering Computer aided quality control Manufacturing process quality a b s t r a c t Engineering drawings accompany a workpiece throughout its production process and include informa-tion about the dimensions and tolerances as well as the associated regulatory standards. Even though the construction and manufacturing process of a workpiece can be almost entirely performed automatically, the design and use of engineering drawings is still not fully integrated in the automated production pro-cess. This work provides DigiEDraw, a conceptual approach as well as a prototype to extract dimensioning information from engineering drawings and to integrate this information into the production process to facilitate and optimize quality control. The extraction process is based on 2D clustering. The challenge is to determine the parameters to distinguish clusters representing different dimensioning informa-tion. The approach uses DBSCAN, achieving a recall value of over 88%. The applicability of DigiEDraw is demonstrated based on a real-world manufacturing process.
© 2021 The Author(s). Published by Elsevier B.V. This is an open access article under the CC BY license (
1. Introduction Production processes usually require a model of the workpiece as input. Typically, these models are designed using a computer-aided design (CAD) program such as AutoCAD or Solidworks. The model is then transformed into a numerical control (NC) program using computer-aided manufacturing (CAM) tools, for example, Catio or Esprit. The NC program, in turn, is used by a tooling machine to manufacture the workpiece. Design and production processes can be seen as two separate steps or can be developed together using an integrated CAD-CAM system.
Engineering drawings (EDs) are 2D depictions of a workpiece that include geometric as well as textual information such as measure-ments, tolerances, and applicable norms, which are essential for quality control of the finished workpiece. CAD modeling describes the design of a workpiece with the help of CAD programs. Hence, EDs are strictly speaking CAD models. However, this paper uses the term CAD model for a digital model of a workpiece, usually in 3D, that only includes graphical and geometrical information, whereas the term EDs refers to manual and digital drawings that include ∗Corresponding author.
E-mail addresses: beate.scheibel@univie.ac.at (B. Scheibel), juergen.mangler@tum.de (J. Mangler), stefanie.rinderle-ma@tum.de (S. Rinderle-Ma).
2D depictions of a workpiece as well as dimensioning information.
Examples of a 3D model of a workpiece, an ED, and the resulting workpiece can be seen in Fig. 1. EDs can be generated from a CAD model. However, additional information (i.e., tolerances and stan-dards) has to be added manually as it is not included in CAD models by default.
Nowadays, CAD models are typically used for the actual pro-duction process. Nevertheless, EDs are still mostly applied as the contractual basis and as reference for quality control as the spec-ifications of tolerances as well as the applicable standards are essential for these purposes (Labisch and Weber, 2008). According to Henderson (2014), 250 million new EDs are generated each year and millions of legacy EDs are still in circulation. A solution that allows the extraction of information from EDs, that is not included in the CAD model, can be used to automate the entire produc-tion process including measuring and quality control. An optimal solution should be able to extract all data including graphical ele-ments as well as additional information such as the dimensioning requirements. However, it is not always necessary to extract geo-metric and graphical elements, as an additional CAD model exists in a lot of cases. The problem is to include dimensioning informa-tion in the process to create a seamless production chain. This refers not only to the dimensions and tolerances written on the ED itself, but also to the information that is part of an associated regulatory framework, e.g., ISO or DIN standards. These regulatory documents usually specify minimum standards that should be sat- 0166-3615/© 2021 The Author(s). Published by Elsevier B.V. This is an open access article under the CC BY license (
B. Scheibel, J. Mangler and S. Rinderle-Ma Computers in Industry 129 (2021) 103442 Fig. 1. Engineering drawing.
isfied, as well as default dimensioning requirements, if these are not explicitly stated in the ED. Integrating this information into a con-tinuous (semi-)automated production process can, among others, facilitate automating quality control by the means of automated measurement. There are approaches to include all of the additional information regarding dimensioning and tolerances in the CAD model, e.g., product and manufacturing information. However, it is still common practice to include this kind of information exclu-sively in the EDs. The transformation to CAD models as well as the extraction of information has been a well researched topic for the last decades. However, there is still no ready-to-use approach to effectively address the problem of integrating dimensioning infor-mation from EDs into a production process.
Hence, the DigiEDraw approach described in this work aims at supporting quality control by providing an end-to-end approach for digitalization and integration of EDs. End-to-end means that this approach includes the upload of a drawing, the information extraction, as well as the actual integration in the process. It also refers to a solution that does not require training, is easy to use, and supports an employee, who – in the end – can always check for validity. A prototypical implementation, which is deployed in a real-life manufacturing scenario, is described in Section 4.
The focus of this work lies on mechanical EDs, in particu-lar component drawings. Component drawings display a specific workpiece from different views as well as the specifications that should be applied, as opposed to other kinds of EDs, e.g., assembly drawings that show how different workpieces are combined. How-ever, the principles of DigiEDraw should apply for other kinds of EDs as well.
Fig. 1 depicts the ED of a geometric object. The dimensions are usually noted directly at the corresponding graphical element. Aux-iliary lines can be used to specify which structural element the specification refers to. The first value is called the nominal dimen-sion. Its value should lie between the minimum and the maximum tolerance. This area is also called the tolerance zone. The area high-lighted in green in Fig. 1 is a typical example of a dimension. The nominal value is 17.00. +0.20 is the maximum tolerance which leads to an upper deviation of 17.20. −0.10 is the minimal tolerance, which means that the lower deviation is 16.90 and the tolerance zone lies between 16.90 and 17.20. This combination of nominal values and tolerances is called a dimension set. The element high-2 B. Scheibel, J. Mangler and S. Rinderle-Ma Computers in Industry 129 (2021) 103442 Fig. 2. Example of a geometrical dimension.
lighted in blue is a theoretically exact dimension, which means that this dimension has no tolerances. This is displayed with a border.
If a dimension is not a theoretically exact dimension, and has no explicit tolerances, the standard tolerances apply. The standard tol-erances are specified in the applicable regulatory documents and norms. In addition to dimensioning and size tolerances as shown in Fig. 1, geometrical tolerances describe the form or position of an ele-ment, as well as other geometric features such as orientation and run out. In Fig. 2, for example, a tolerance for perpendicularity is set. The whole dimension would be interpreted as follows: the perpendicu-larity of the specific element compared to part A cannot differ more than 0.01. In general, an ED can therefore include text, symbols as well as graphical elements.
The variability of representation is a challenge for automatic information extraction. A dimension can consist of only the nomi-nal value, a complete dimension set with nominal value, maximum and minimum tolerance or a nominal value with only one of these tolerances. Geometrical dimensions can consist of different sym-bols and have to be set in relation to the feature they describe, i.e., which part of the workpiece they refer to. A dimension is only meaningful in relation to this additional information. The extrac-tion process should therefore be designed to keep the elements in correct composition.
EDs can exist in multiple formats such as DXF, PDF, STEP, or image formats (TIFF, PNG). Existing approaches mostly use image formats as legacy EDs are usually available as scanned images only (Section 2). Pure CAD formats such as STEP often do not include additional information such as the tolerances. PDF includes the information needed for quality control and offer the benefit of pro-viding graphical and textual elements in an already separated way.
DigiEDraw focuses on digital PDF, as most traditional approaches work with either scanned or CAD format, but no approach has pro-vided a solution for PDF based EDs so far, which should provide improved results as text recognition/OCR does not has to be applied.
For the intended application of DigiEDraw only the textual infor-mation is needed, as we assume that a CAD model is present in addition to the ED. PDF extraction is a well researched topic (see Section 2). However, EDs differ from other PDF documents in that they include a combination of geometrical and textual elements, which do not follow a uniform structure and are spread over the drawing area. These elements can be horizontal as well as vertical or lie at an angle and can consist of one or multiple values.
DigiEDraw addresses these challenges based on the following research questions: Research Question 1: How can textual information, specifically dimensioning requirements, be automatically extracted from EDs, under the condition that each extracted dimension set consists of only one dimension, the respective tolerances, and additional information?
The textual information can be extracted using available tools.
However, this leads to single extracted elements, that are not part of dimension sets anymore. The main challenge is therefore to merge the separate values in order to get the associated dimension sets.
As DigiEDraw operates on 2D EDs, this problem can be understood as a 2D clustering problem and tackled using standard clustering techniques. DigiEDraw opts for DBSCAN (Ester et al., 1996) as it does not rely on the number of clusters as input parameter.
Research Question 2: How should the clustering parameters be set to achieve optimal clustering results, i.e. complete dimension sets and avoid over-clustering? Specifically: • RQ 2.1: Which distance metric leads to optimal results?
• RQ 2.2: How should the DBSCAN parameters be set?
To achieve optimal results, multiple parameters have to be fit-ted. In addition, we have to compute a distance metric that reflects the goal of merging elements back into their respective dimension sets.
The contributions of this paper are (i) employing clustering for merging logically connected elements that have been split while being extracted from a PDF document, including pre- and post-processing steps, (ii) an iterative approach to set the required parameters for clustering, (iii) the definition and calculation of a custom distance metric that takes the domain logic into account and therefore (iv) providing an end-to-end approach that can be used in the industry for quality control support.
Moreover, the paper provides a prototypical implementation including a user interface and production process models where DigiEDraw is integrated as support for quality assurance. This demonstrates how product and process quality can be increased based on the integration of EDs.
The remainder of the paper is structured as follows: Section 2 discusses existing approaches. The DigiEDraw approach is pre-sented in Section 3. The prototypical implementation of DigiEDraw and the application in a manufacturing process are presented in Section 4. Section 5 provides a discussion of DigiEDraw results and Section 6 concludes this work.
2. Related work Research on digitalizing EDs dates back to the late 1980s (e.g.
Krause et al., 1989), but still remains a challenging task, for example, regarding quality control in production. This is based on the obser-vation that quality checking based on EDs is still mostly conducted in a manual way. In general, EDs are available as image format (i.e., raster graphics such as TIFF, PNG, and JPEG), CAD formats includ-ing DXF, DWG, IGES, and STEP, as well as vector graphic format such as PDF and SVG. Existing approaches work on one of these three format groups. Also, as an overarching observation existing approaches aim at (i) digitalizing EDs, using mostly graphical ele-ments, or (ii) at extracting specific information in order to optimize the retrieval of EDs, the design and production process, or the prod-uct management. As mentioned in Section 1, EDs consist of text, symbols as well as graphical elements. A solution that provides complete digitalization would have to include all of these elements.
Hence, existing approaches can be categorized by their input format as well as their focus, i.e., textual elements, graphical elements, symbols or a combination (cf. Table 1). Approaches focus-ing on textual elements can be further divided. Meta Information includes papers that extract specific information, e.g., the version number or other information found in the drawing tables. Category Dimensions, Aspect includes approaches that extract dimension information, i.e., dimensions and tolerances, but describe only one part, e.g., how to detect dimension boxes, where afterwards OCR can be applied or only the OCR itself. Category End-to-End, by contrast, refers to approaches that offer a solution starting from an ED until the integration of the textual information into the application. Note that also the category Combination contains end-to-end approaches. In the following, we discuss existing work along Table 1.
2.1. Scanned/image-based drawings: The first approaches to digitize EDs focused on scanned images and therefore raster graphics. They consist of algorithms for text/graphic segmentation (Lu, 2002), symbol recognition 3 B. Scheibel, J. Mangler and S. Rinderle-Ma Computers in Industry 129 (2021) 103442 Table 1 Overview of related work.
Textual elements Symbols Graphical elements Combination Meta Information Dimensions, Aspect End-to-End Scanned Ondrejcek et al. (2009) Lu (2002) Archibald et al. (1995) Vaxiviére and Tombre (1994) Mani et al. (2020) Banerjee et al. (2016) Das and Langrana (1997) Elyan et al. (2018) Ablameyko et al. (2002) Kang et al. (2019) Lu et al. (2008) Lai and Kasturi (1994) Elyan et al. (2020) Krause et al. (1989) Rahul et al. (2019) Das et al. (2018) Habed and Boufama (1999) Fonseca et al. (2005) Van Daele et al. (2019) Dori and Velkovitch (1998) DXF/STEP/IGES Jiang and Feng (2010) Sukimin and Haron (2008) Prabhu (2002) Zhang et al. (2012) Ye et al. (2009) Prabhu et al. (2001) Cao et al. (2005) Zhang and Li (2014) PDF/SVG DigiEDraw Kasimov et al. (2015) Hoang et al. (2016) (Archibald et al., 1995), vectorization of raster graphics (Ablameyko et al., 2002), text recognition (Dori and Velkovitch, 1998), opti-cal character recognition (OCR) (Brown et al., 1988; Lu, 1995), and combinations of these. A comprehensive overview regard-ing techniques for the digitalization of raster image EDs and the related use-cases is given by Henderson (2014), Moreno-García et al. (2018), and Tombre (1998). Nowadays work on this format continues, even though EDs are usually not stored in image for-mat anymore. However, a large number of legacy EDs remains, which are mainly scanned. Recent approaches mainly build on neu-ral networks, specifically convolutional neural networks or hybrid approaches which contain neural networks as well as traditional segmentation approaches (e.g., Elyan et al., 2018, 2020; Kang et al., 2019; Mani et al., 2020; Moreno-García et al., 2018; Rahul et al., 2019; Van Daele et al., 2019). In the following specific approaches are briefly explained, along the categories seen in Table 1.
2.1.1. Textual elements In regards to Meta Information, Ondrejcek et al. (2009) aim at discovering links between scanned EDs and CAD models by analyz-ing ED features. Banerjee et al. (2016) search and analyze drawing names to find matching drawings and Lu et al. (2008) extract knowledge from tables. The tables are detected by analyzing the layout structure and matching it to a standard structure. This is then used to obtain meta information about a drawing. All of these approaches also use OCR in order to digitalize the letters and num-bers. To decide which OCR engines should be used, algorithms to analyze if a drawing includes handwritten or machine written numbers can be used (Das et al., 2018). These approaches have in common that their focus on the drawing tables and search for very specific information.
Several approaches focus on Dimensions. Lu (2002) detect areas of dimension sets and separate them from the graphical elements, which is then used as input for OCR algorithms. Das and Langrana (1997), Lai and Kasturi (1994), and Habed and Boufama (1999) also look for dimension sets, using classic image analysis techniques (e.g., connected components), and analyzing the lines and arrows in the drawing. For recognition of the characters, existing OCR algorithms are used. The OCR step itself is not described, Das and Langrana (1997), for example, already assume that vectorization was performed before and OCR will lead to correct results. Dori and Velkovitch (1998) similarly detects dimension set areas, but it also includes a recognition algorithm. However, it is only able to detect 23 different characters, no symbols and only if the dimension set does not have a border.
2.1.2. Symbols Archibald et al. (1995) present an approach to find and cate-gorize symbols by matching these to templates. Nowadays, often deep learning and convolutional neural networks (CNN) are used for symbol detection (Elyan et al., 2018, 2020). These approaches focus on piping and instrumentation (P/ID) diagrams, a specific kind of engineering drawings, which relies heavily on symbol usage.
However, symbol recognition in images is used in other domains such as music notes (Pacha et al., 2018).
2.1.3. Graphical elements For a full digitalization of EDs, the recognition of graphical elements is necessary. This is mostly done using vectorization tech-niques. Mostly, these drawings are then transformed into CAD format (Vaxiviére and Tombre, 1994; Ablameyko et al., 2002; Krause et al., 1989). Fonseca et al. (2005) aim at facilitating the retrieval of EDs by analyzing the graphical elements, extracting features and the topology, and thereby optimizing the search for similar drawings.
2.1.4. Combination These approaches aim at fully digitalizing EDs, including text, symbols as well as graphical elements. Mani et al. (2020), Kang et al. (2019), and Rahul et al. (2019) focus on a full digitalization of P/ID diagrams, where symbol detection is important, as well as text and connection recognition. All of these approaches are end-to-end approaches using a combination of traditional techniques as well as deep learning. Van Daele et al. (2019) support design processes by searching drawings, summarizing the most important features and finding similar ones. This is done using CNN as well as reasoning-based methods.
The main challenge for image-based EDs is the quality of the input files, as it can vary greatly and accordingly leads to different results. Furthermore, in order to use neural network approaches, which are the most promising, significant amounts of annotated EDs are needed as training data. In addition, training might require experts and is time-consuming.
2.2. CAD format (DXF, DWG, STEP, IGES) Several approaches focus on developing 3D models from CAD files or extracting specific information. When dealing with CAD drawings, the information is already in machine readable, textual form, but has to be parsed and combined to obtain useful informa-tion. Therefore the description and categorization in Table 1 refers to what the elements are in the ED, i.e., text or graphical, and not their physical representation, as in this case, all elements are stored in textual form.
2.2.1. Textual elements Jiang and Feng (2010) and Zhang et al. (2012) use DWG draw-ings to obtain product management data from the drawing tables in order to facilitate product management. Cao et al. (2005) similarly focus on product information to improve versioning management of EDs. Therefore all of these approaches belong to the Meta Infor-mation category.
4 B. Scheibel, J. Mangler and S. Rinderle-Ma Computers in Industry 129 (2021) 103442 As these formats are already in digital form, symbol detection does not have to be performed. Therefore no related work can be found in this category.
2.2.2. Graphical elements Sukimin and Haron (2008) focus on extracting production rele-vant data such as the volume of a workpiece from the description of the graphical elements. Ye et al. (2009) describe a method to con-struct 3D models out of DXF drawings by simulating the human process of understanding EDs and combining different views of a workpiece.
2.2.3. Combination Prabhu (2002) and Prabhu et al. (2001) extract graphical fea-tures, manufacturing information as well as information about the dimensions to improve the CAD/CAM linking. Zhang and Li (2014) aim at combining information from an ED with additional data stored in a database, i.e., information extraction, and visually dis-play graphical elements. Therefore this approach also combines textual and graphical elements.
Even though the just described approaches, contain dimen-sioning information, CAD documents commonly do not contain additional information such as the tolerances and applicable regu-latory guidelines, as these are often manually noted when the CAD models are transformed into EDs.
2.3. Vector drawings/PDF Nowadays EDs are often stored as vector drawings, which can be processed digitally. Previous approaches focus on extracting either the graphical elements or facilitating the management of these EDs.
2.3.1. Graphical elements Kasimov et al. (2015) describe a system for content-based retrieval of vector drawings based on a graph matching problem.
This includes, for example, specifying a geometric feature, which is then used to search for drawings containing this feature.
2.3.2. Combination Hoang et al. (2016) introduce an approach to automatically extract structural information as well as relationships from vector-based EDs. This also includes detecting symbols. Therefore this approach is categorized as Combination.
To the best of our knowledge there exists no approach to extract textual elements, specifically dimensioning information, from vec-tor drawings or EDs in digital PDF format.
2.4. PDF information extraction Multiple papers deal with the analysis and extraction of infor-mation from PDF documents in general (Bast and Korzen, 2017; Yuan et al., 2006; Adrian et al., 2017) or focus on specific docu-ments, e.g., scientific papers (Bui et al., 2016; Wang et al., 2019) or health care related documents (Parizi et al., 2018; Li et al., 2019; Harmata et al., 2017). These approaches analyse pixels, words, spatial or logical connections. Ferres et al. (2018) also focus on extracting information not only from digital but also scanned PDF documents by including an OCR library. Similarly, Tomovic et al.
(2020) use OCR techniques and even combine multiple OCR engines to generate robust input for classifying document segments. Corrêa and Zander (2017) give an overview of papers and tools regarding the extraction of tables from PDF documents. One of these tools is “texus” (Rastan et al., 2018) where the table is firstly detected by looking for the typical table structure and afterwards the con-tent is extracted row by row, column by column. Another research focus are visually rich documents, i.e., documents where text and images are combined. Wei et al. (2020) propose an approach to extract specific information for example the address, from resumes as well as invoices that often contain images as well as text. “Layout-analysis” provides algorithms to detect and classify components of such documents, e.g., text paragraphs, headlines, figures, tables or algorithms (Prüˇ sa and Fujiyoshi, 2017; Zhong et al., 2019; Li et al., 2018). Hansen et al. (2019) focus on extracting all non-textual com-ponents, i.e., tables, algorithms, figures. Shi et al. (2019) take figures from medical papers and split these into the respective sub figures for better analysis. Morris et al. (2019) focus on the extraction of text from figures in PDF documents using OCR and neural networks.
Overall, a variety of approaches exists to extract different kinds of information from PDF documents. However, all of the covered document types differ from EDs as they mostly consist of only textual elements that are written in lines, or if containing fig-ures and tables they follow a similar structure (e.g., resumes and invoices). Lastly, they do not contain elements that are overlap-ping, unequally distributed over the page and are in vertical or diagonal orientation. EDs, by contrast, are a combination of tex-tual and graphical elements, which are distributed in the ED in no machine-understandable order.
2.5. Grouping and clustering The specific challenge of DigiEDraw is to merge textual elements extracted from ED into complete dimension sets. The general chal-lenge of grouping extracted elements has also been addressed by existing approaches. Habed and Boufama (1999) use the proxim-ity and thickness of elements to find graphical elements and the associated text elements. Van Daele et al. (2019) employ clustering to combine graphical elements into logical clusters and to parti-tion the ED into different views. The main difference to DigiEDraw is that it aims at deciding automatically whether textual elements that can be positioned in different distances and orientation belong to the same dimension set. This decision depends at least partly on domain knowledge which in turn has to be considered in a potential clustering approach.
2.6. Conclusion Even though information extraction from EDs has been researched for more than three decades, many challenges remain.
Approaches using machine learning such as neural networks are the only end-to-end approaches and seem promising, but also have weaknesses especially concerning the availability of annotated training data and effort of training (Moreno-García et al., 2018). The envisioned DigiEDraw approach, by contrast, does not require any training. Regarding vector drawings and digital PDFs, approaches focus mainly on graphical elements. However, none of the existing approaches take dimensions and tolerances into account. Previ-ous works referring to PDF extraction focus on more structured and homogeneous document types, also containing mainly tex-tual elements. In addition, source code is not available for most of the aforementioned papers. Therefore, we conclude that there is no end-to-end approach yet for automatically extracting textual elements, specifically dimension sets, from EDs.
3. DigiEDraw algorithms The goal of DigiEDraw is to extract textual elements from EDs into complete and coherent dimensions sets. Consider Fig. 3, left side, where preprocessing yields elements 7 and , 3 as separated boxes, although the elements obviously belong together (cf. Fig. 3, middle). Likewise, the tolerances provided by elements +0, 1 and −0, 1 in Fig. 3 are required to complete the actual dimension set as depicted in Fig. 3, right side. The challenge is to automatically 5 B. Scheibel, J. Mangler and S. Rinderle-Ma Computers in Industry 129 (2021) 103442 Fig. 3. Dimension set split up into three blocks and multiple words, clustered to achieve complete dimension set.
Fig. 4. Overview of the DigiEDraw algorithm and example.
determine and merge the elements that belong together in dimen-sion sets. The idea of DigiEDraw is to employ clustering for this, i.e., to combine well-known clustering technique DBSCAN with a novel distance metric and an iterative parameter setting algorithm as core, embedded into specific pre- and postprocessing steps.
Fig. 4 provides an overview on the overall DigiEDraw approach (left side) as well as an illustrating example (right side). The input is the ED that is preprocessed at first (cf. Algorithm 1). The result is an array of elements that constitutes the input for the DigiEDraw core, i.e., the clustering of the elements into complete and coher-ent dimension sets (cf. Algorithm 3). Indicated by black borders in Fig. 4, the main conceptual DigiEDraw contributions comprise the iterative approach to set clustering parameters optimally in order to ensure a correct representation of dimensions as well as the def-inition of a novel distance metric (see Algorithm 2). The output here is an array of clustered elements which is then postprocessed, i.e., the resulting dimension sets are integrated into the manual quality control using a web service (cf. DigiEDraw prototype and production process models in Section 4).
3.1. Preprocessing The preprocessing of the DigiEDraw approach starts by upload-ing an ED in PDF format that is converted into HTML. The reason is that HTML enables the extraction of all elements with their respec-tive coordinates. There are different python libraries for reading 6 B. Scheibel, J. Mangler and S. Rinderle-Ma Computers in Industry 129 (2021) 103442 PDF available, e.g., PyPDF2,1 textract,2 and Tika.3 . We obtained the best results using the pdftotext script which is part of the poppler-tools4 integrated in unix systems. The resulting files after conversion to HTML consist of blocks and words. Each word con-sists of one or multiple characters or digits. For each element the coordinates of the bounding box are given as well. In some cases, one dimension set is extracted into exactly one HTML block. How-ever, mostly, a dimension set is split into multiple blocks. Fig. 3, for example, shows a single dimension set that was split into three blocks (illustrated by the blue rectangles) by the conversion.
Each of these blocks consists of multiple words, visualized by the dark blue borders in the example. For further processing, the tex-tual elements, as well as the associated coordinates are read from the HTML file and stored in an array consisting of multiple sub-arrays, each representing one block, which also contain arrays, each representing one word. For the clustering process only the outer coordinates, i.e. the bounding box, are needed. Therefore the maxi-mum and minimum x-coordinates and y-coordinates for each block are extracted and stored in the array. One element in the example depicted in Fig. 4 (right side) consists of textual value 4.00 together with the associated coordinates xMin=“291.384”, yMin=“172.578”, xMax=“317.041”, yMax=“203.493”. The elements inside a block are not always in the correct order. To sort the elements, we determine if a block is horizontal or vertical by comparing the ratio of x to y val-ues. If the block is horizontal, the words within a block are ordered by x-coordinates using the built-in sort function. The same proce-dure is done for vertical boxes, using y-coordinates. Algorithm 1 describes the necessary preprocessing steps.
Algorithm 1. DigiEDraw preprocessing.
Input: digital PDF ED Output: array of bounding boxes 1: convert PDF to HTML 2: parse HTML file 3: find all blocks 4: for block in blocks do 5: find all words 6: for word in words do 7: get text element 8: get bounding box (min x, max x, min y, max y) 9: add text and bounding box to array bounding boxes ▷ Array of array 10: end for 11: end for 12: for element in bounding boxes do 13: check if element is vertical or horizontal 14: if horizontal then 15: sort by x-coordinates 16: end if 17: if vertical then 18: sort by y-coordinates 19: end if 20: end for 21: return sorted array bounding boxes 3.2. Clustering The goal is to obtain complete dimension sets based on the information extracted from the input ED. Therefore, we have to determine which elements belong to the same dimension set, but have been extracted separately. As illustrated by the example in Fig. 3, not only numbers can be subject to merge (e.g., 7 and 3), but also associated tolerances (e.g., +0, 1 and −0, 1). Additionally, the elements can be positioned in a horizontal, vertical or diagonal way.
1 2 3 4 We tested several approaches to combine extracted elements into logically connected groups, without using training data. A “brute force” approach is to hard-code the relationships between elements in the ED, i.e., elements that are next to each other are combined. A more refined approach is to combine elements using regular expressions. However, both approaches did not yield accurate results as the structure of dimensions and tolerances can vary greatly. Clustering algorithms, by contrast, seem more promising with respect to merging logically connected elements, without requiring extensive training data. We tested k-means (Faber, 1994), using elbow plots to specify the amounts of clusters, OPTICS (Ankerst et al., 1999), and DBSCAN (Ester et al., 1996) where DBSCAN does not require to set the number of clusters as input.
The results for k-means and OPTICS were less accurate than for DBSCAN. As a conclusion, we opted for clustering, and in particular, for DBSCAN to be employed by DigiEDraw.
Open questions are how to define the distance metric and how to set the parameters. DBSCAN implementations are available for most programming languages.5 For DBSCAN, the user sets parameter MinPts, which defines how many points should at least be in a cluster and epsilon, which defines the radius for each cluster, as well as the appropriate distance met-ric, which is used to calculate how “similar” or “close” elements are to each other. The algorithm then analyzes all points, grouping the ones together that are within the specified radius. For DigiEDraw, a new distance metric is defined that takes into account spatial proximity as well as domain knowledge. An additional challenge is the setting of epsilon, as this value has a great impact of the result, but has to be set dynamically, as it can be different for each ED.
This issue was solved by using an iterative approach, increasing the value of epsilon until a threshold is reached – the stopping criterion – which is an indicator that the optimal result has been achieved.
The envisioned result of the DigiEDraw clustering is an array of sorted elements, each element being assigned to a cluster. Consider the example in Fig. 4, right side. Three elements are assigned the same cluster number 1. This means that they form a dimension set and are stored as key value pair together with the coordinates. The coordinates of the dimension set are determined by the minimum of xMin, yMin and the maximum of xMax, yMax coordinates over all its elements.
The distance metric is used to calculate which boxes are close or similar to each other. Several metrics are available such as the Euclidean distance. However, as DigiEDraw works with bound-ing boxes instead of points, existing metrics are not sufficient. For humans it is easily recognizable which parts belong together. How-ever, it is more complex to achieve this automatically.
The basic measure is the distance between the two nearest cor-ners of two bounding boxes. To account for the boxes being close or just single edges near to each other, the distances of the two nearest edges of the boxes are averaged to obtain the distance. If the boxes intersect, this distance is set to zero, as these boxes will most likely belong to one dimension set. Conversely, if two boxes are parallel, the distance is increased as these boxes are likely not part of a dimension set and therefore should not be in one cluster.
Definition 1 (Distance metric). Let boxa and boxb be two bound-ing boxes of elements a and b extracted from an ED. The distance between a and b is defined as the average of the smallest and sec-ond smallest distance between a corner point of bounding box boxa and a corner point of bounding box boxb. If boxa and boxb intersect, which is checked using the separating axis theorem (Gottschalk et al., 1996), the distance is set to zero. If the elements are paral-5 The DigiEDraw prototype uses the scikit-learn clustering library (Pedregosa et al., 2011).
7 B. Scheibel, J. Mangler and S. Rinderle-Ma Computers in Industry 129 (2021) 103442 lel, which is determined by comparing orientation and alignment of boxa and boxb, the distance is increased. This is done for all bounding boxes of all elements extracted from the ED. The result is captured within a the distance matrix. The calculation is presented in pseudo code in Algorithm 2.
We check for intersection using the separating axis theorem (Gottschalk et al., 1996) which says that two boxes cannot over-lap if there is an axis that separates them. In practice, this is done by taking two boxes and comparing the respective bottom left and top right corners to see if they overlap on either axis. To check for parallelism, first the orientation of the boxes is defined more pre-cisely than in the preprocessing. As in the preprocessing, the ratio of length and width of a box can suggest the orientation of the box.
In this case, we have three categories: vertical, horizontal or “not defined”, which is characterized by similar length and width. The ratio which differentiates between these categories is determined empirically by looking at multiple sample EDs and measuring the ratio for vertical as well as horizontal boxes.
The orientation of the bounding boxes is calculated as follows: horizontal = (xMax − xMin) > 1, 3 ∗ (yMax − yMin) vertical = (yMax − yMin) > 1, 3 ∗ (xMax − xMin) xMax is defined as the maximum x-coordinate, xMin accord-ingly refers to the minimum x-value of the respective bounding box.
Correspondingly, yMax and yMin are the maximum and minimum value for the y-axis. The remaining boxes are set as “not defined”.
After the orientation is defined, the alignment of the two boxes in relation to each other is analyzed, i.e., are the boxes above each other or next to each other. If the boxes are above each other and horizontally aligned, then they are marked as being parallel. The same is done if the two boxes are next to each other and vertically aligned. The distance to the parallel box is then increased by 100.
The exact value of the increase does not influence the results as long as it it exceeds the epsilon value. This is done for each box in relation to all other boxes, resulting in a distance matrix, which can be used in the scikit-DBSCAN implementation.
Algorithm 2. Calculate distance metric.
Input: array of bounding boxes Output: distance matrix 1: for boxa in bounding boxes do 2: calculate orientation 3: for boxb in bounding boxes do ▷ Calculate distance from every box to other boxes 4: calculate smallest distance between boxes a,b 5: calculate 2nd smallest distance between boxes a,b 6: check intersection using separating axis theorem 7: if a,b intersect then 8: set distancea,b to 0 9: end if 10: check parallelism by comparing orientation, alignment of boxes 11: if a,b are parallel then 12: increase distancea,b by 100 13: end if 14: distancea,b = (smallest distance + 2nd smallest distance)/2 ▷ Calculate average distance 15: store distancea,b in distance matrix 16: end for 17: end for 18: make array distmin, including all distances, sorted in ascending order ▷ Needed as input for epsilon 19: return distance matrix, distmin ▷ The distances from each box to the other boxes are stored in a matrix The MinPts value relates to the minimum amount of points in the epsilon distance of a point to constitute a cluster. In this setting MinPts is set to 1, as it is possible that a dimension set consists of a single box.
Determining the optimal epsilon value is more complex. At first, we manually adjusted the value such that the first sample ED showed optimal results. As more EDs were analyzed, it became obvious that this value is not optimal for all EDs. Therefore, we opt for setting the value dynamically. Ester et al. (1996) and Sander et al. (1998) propose an interactive approach to set the epsilon value, i.e., computing a distance graph for all points and letting the user choose the threshold value. This approach did not work in this case, as the graph suggests higher epsilon values than needed, which leads to over clustering. Ozkok and Celik (2017) also use a k-nearest neighbor graph to determine the optimal setting. Schubert et al. (2017) note that the epsilon value is depending on the distance metric and the domain, but should generally be as small as possible.
In regard to EDs, the epsilon value depends on multiple vari-ables and can differ for each ED. An iterative approach was taken to determine the optimal epsilon value. For all boxes the distances to all other boxes are calculated and stored in ascending order in the array distmin. The first epsilon value is the value of the minimal distance, i.e., the distance between the nearest boxes. In each iter-ation this value is increased to the next bigger distance value. For defining the stopping criterion, the clustering result was evaluated in regards to the following criteria: • Davis-Bouldin index (Davies and Bouldin, 1979): The Davis-Bouldin index measures the average similarity measure of each cluster with its most similar cluster. Similarity is defined as the ratio of within-cluster distances to between-cluster distances.
The lower to zero, the better the separation of clusters, as the clusters are farther apart.
• Calinski-Harabasz index (Cali ´ nski and Harabasz, 1974): The Calinski-Harabsz index is defined as the ratio of sum of between-cluster dispersion and inter-cluster dispersion. The higher the value, the better defined are the clusters.
• Silhouettes coefficient (Rousseeuw, 1987): Similar to Calinski-Harabasz index, the higher the value, the better defined are the clusters. It is calculated by comparing the intra-cluster distance and the mean nearest-cluster distance (where the element is not a part of), to achieve clusters that are cohesive and distinct.
These values are internal quality criteria of clustering algorithms and take into account how well clusters are defined and separated.
These criteria were chosen because these values do not need the ground truth, which is not available at this point. The Calinski-Harabasz index as well as the silhouettes coefficient should get bigger if the higher epsilon leads to better separated clusters. Con-versely, the Davis-Bouldin index should get closer to zero. In each run the current value of these indices is compared to the values in the previous run. If these values change in the opposite direc-tion it can be assumed that the best result was already achieved.
Therefore the iterations are stopped and the epsilon value of the previous run is set as the optimal value. This was tried with each of the three parameters as well as combinations of these. The sil-houette coefficient leads to the best results and is therefore used as stopping criterion. The clustering process stops as soon as the stopping criterion evaluates to true or the highest distance value is reached.
Stopping criterion: Clustering is applied with increasing epsilon values until the stopping criterion is reached. As stopping criterion the silhouettes coefficient (Rousseeuw, 1987) is used, which is an internal quality value, referring to the cluster definition. The value of the silhouettes coefficient rises if the clusters are more defined.
Therefore we continue the iterative clustering process, until the coefficient gets smaller than in the step before, which is a sign of the cluster definition getting worse. At this point, the loop is stopped and the end result will be obtained by running DBSCAN 8 B. Scheibel, J. Mangler and S. Rinderle-Ma Computers in Industry 129 (2021) 103442 Fig. 5. User interface featuring ED ‘Elevator Bottom’.
with epsilon being the penultimate value - the value where the silhouettes coefficient was highest.
Algorithm 3. DigiEDraw clustering.
Input: array of bounding boxes, distance matrix, dist min Output: dictionary of dimension sets 1: get array bounding boxes, distance matrix, distmin 2: run DBSCAN(distance matrix, epsilon = smallest distmin, MinPts = 1) ▷ First iteration of clustering 3: if sh < shold ▷ Silhouettes coefficient is used as stopping criterion then 4: stopping criterion = true ▷ If Silhouettes coefficient gets smaller, stopping threshold is reached 5: end if 6: while not stopping criterion ▷ Continue clustering while stopping criterion is not true 7: run DBSCAN(distance matrix, epsilon = next distmin, MinPts = 1) ▷ epsilon = next value of distmin 8: if sh < shold then 9: stopping criterion = true 10: end if 11: if epsilon = nnmax ▷ If epsilon reaches maximum distance, clustering is stopped then 12: end clustering 13: end if 14: end while 15: run DBSCAN with penultimate epsilon ▷ The value before reaching the stopping criterion is used 16: return array of clusters ▷ Each cluster containing one to multiple elements The clustering algorithm can be seen in Algorithm 3. We take the results of the preprocessing (Algorithm 1) and the calculation of the distance matrix (Algorithm 2) and run the DBSCAN implemen-tation as long as the stopping criterion is not met. The result (array of elements, assigned to clusters) is then returned to continue with postprocessing. Overall, the clustering result depends on the qual-ity of the ED, i.e., whether or not the ED was designed according to existing standards, e.g., with respect to distance between elements, the epsilon value, MinPts, and the employed distance metric.
3.3. Postprocessing The array of elements with assigned cluster numbers (see Fig. 4, right side for the running example) resulting from Algorithm 3 is postprocessed in several steps. At first, data cleaning is performed, using regular expressions. In detail, the cleaning involves filtering for expressions that are not relevant for dimensions. This step can be adapted to fit particular needs and scenarios. In addition, the elements are sorted again. If the bounding boxes of the elements are determined as horizontal, sorting is performed based on the x-value and if vertical by the y-value, to ensure that after merging the clusters, the elements are in correct order. The last step converts the array of elements into a dictionary which can then be stored in a key value store.6 Fig. 4, right side, shows the resulting key value set for the running example. Entry {4.00, +0.10, −0.03}: {320.711, 172.578, 354.384, 203.493}, for example, refers to a dimension set containing elements 4.00, +0.10, 0.03 with the associated coordi-nates of the overall bound box. The resulting dimension sets are then postprocessed using regular expressions, stored in a database, and on request by the user shown via the DigiEDraw user interface (cf. Fig. 5).
4. Evaluation DigiEDraw is evaluated with respect to its feasibility, the recall and precision of its algorithms, and its application.
4.1. Feasibility – prototypical implementation Algorithms 1–3 are implemented within the DigiEDraw pro-totype which consists of an extraction tool7 and a webservice including the user interface.8 The DigiEDraw user interface is depicted in Fig. 5. The ED of the workpiece is displayed on the left while the automatically extracted dimension sets are shown on the right hand side. Next to the dimensions, the user can input the man-ual measurements which are automatically stored in a database for further processing. The user can click on the input field next to the dimension, and the associated element is highlighted in the ED. Moreover, all references to associated regulatory standards and norms are extracted using regular expressions, noted on top of the 6 DigiEDraw uses redis,
7 8 9 B. Scheibel, J. Mangler and S. Rinderle-Ma Computers in Industry 129 (2021) 103442 UI and linked if available. Therefore, these standards can be eas-ily accessed if necessary. The DigiEDraw user interface enables the user to get a clear view on all dimensions and directly input the associated measurements.
4.2. Validation – recall and precision The quality of the clustering algorithm can be validated using external or internal validation. Internal validation was already dis-cussed in Section 3, for the stopping criterion. External validation refers to the comparison to the ground truth, which in this case con-sists of the actual dimensions. This can only be validated manually, as the actual dimensions have to be extracted by hand. According to literature regarding external evaluation for density-based clus-tering methods, Ester et al. (1996) use visual inspection. Aliguliyev (2009) mention “accuracy” and “recall”, which are used to calcu-late more complex evaluation measures. Ting (2010) define the measures “precision” and “recall”, which are commonly used to evaluate information extraction systems. In this work these preci-sion and recall measures are used.
Definition 2 (Recall). In the context of information retrieval, Ting (2010) defines recall as: Recall : = Total number of documents retrieved that are relevant Total number of relevant documents in the database RecallDE := extracted relevant dimension sets all relevant dimension sets where extracted relevant dimension sets denotes the num-ber of relevant dimension sets extracted from the ED and all relevant dimension sets denotes the number of all relevant dimension sets in the ED.
In order to calculate this metric, the actual correct dimension sets, as displayed in the ED, as well as the correctly extracted, i.e.
complete, sets are counted manually.
Definition 3 (Precision). Ting (2010) define precision as: Precision : = Total number of documents retrieved that are relevant Total number of documents that are retrieved For this paper, the value is adapted to: PrecisionDE := extracted relevant dimension sets all extracted elements where extracted relevant dimension sets denotes the num-ber of relevant dimension sets extracted from the ED and all extracted elements denotes all elements extracted by DigiEDraw.
This specifies how much useful information could be extracted out of all the information that was retrieved.
In order to calculate precision, the correctly extracted sets are the same as for the recall calculation. The total number of extracted elements are counted automatically.
Fig. 6 shows cutouts of four example EDs, i.e., “Gripper”, “Aufs-pannung”, “Aufspannung Ecke” and “Adapterplatte”. These EDs in full, as well as “Elevator Bottom” and “Halter” can also be down-loaded from the git repository. An additional ED has been obtained by a company partner and is not publicly available.
Table 2 Evaluation results.
Drawing name RecallDE PrecisionDE Iterations eps-value Gripper 0.82 0.6 4 4.3 Elevator Bottom 0.93 0.88 4 4.3 Aufspannung 0.85 0.58 3 4.15 Adapterplatte 0.93 0.87 5 8.58 GV12 0.88 0.76 4 4.88 Halter 0.85 0.73 5 5.06 Aufspannung Ecke 0.91 0.67 4 4.2 Table 2 shows the validation results based on the seven exam-ple EDs in terms of recall and precision as well as the number of iterations and the used epsilon value.
The EDs chosen for the evaluation (cf. Fig. 6) cover a variety of possible component EDs, as all main parts (dimensions, size tolerances, geometrical tolerances, tables, multiple graphical elements) as well as the main challenges (e.g. combination of graphical and textual elements, unequally spaced out over the ED space, textual elements in different orientations) are part of these EDs and they differ in terms of quantity of elements and distribution as well as arrangement of these. The average recall is 0.88 and the average precision is 0.73. The results are discussed in more detail in Section 5.
4.3. Application Fig. 7 depicts the process models (modeled using Business Pro-cess Modeling and Notation (BPMN)9) for producing the workpiece shown in Fig. 1. More precisely, Fig. 7a depicts the production process for a single workpiece. The process with manual quality control includes subprocess Version 1 (cf. Fig. 7b) and the pro-duction process with a facilitated and optimized quality control using DigiEDraw includes subprocess Version 2 (cf. Fig. 7c). For both processes, the actual production part is the same. For the quality control in Fig. 7b, an employee manually measures the dimen-sions of the workpiece and checks whether they are within the tolerance range specified in the corresponding ED (cf. Fig. 1). This quality control can be facilitated and optimized through DigiEDraw, i.e., by two system-based tasks Display Dimensions in UI and Compare Measurements as depicted in Fig. 7c. In both scenarios, the employee has to have the ED available. For Fig. 7b, the ED is read manually, whereas in Fig. 7c the ED has to be uploaded to the DigiEDraw webservice before the production process starts. For using DigiEDraw no additional knowledge or training time is nec-essary. The user simply has to upload the ED and is then provided with the UI shown in Fig. 5. The user still has to decide manually which features should be measured, which also acts as a plausibility check for the extracted dimensions.
Currently, DigiEDraw facilitates and optimizes manual measur-ing in production processes. However, DigiEDraw could also be used as part of an automated measurement process, i.e., by pro-viding the input for a measuring program. In addition, DigiEDraw can help in the design phase by providing feedback if the elements are too close together or overlap and are therefore not only harder to be extracted automatically, but also lead to less readable EDs for humans.
5. Results and discussion In this section, we summarize the results along research ques-tions RQ 1 and RQ 2 as set out in the introduction. Then we 9 bpmn.org 10 B. Scheibel, J. Mangler and S. Rinderle-Ma Computers in Industry 129 (2021) 103442 Fig. 6. Cutouts of three EDs.
formulate guidelines on EDs and discuss an extended application of DigiEDraw.
RQ 1: How can textual information, specifically dimensioning requirements, be automatically extracted from EDs, under the condi-tion that each extracted dimension set consists of only one dimension, the respective tolerances, and additional information?
Existing approaches target different ED formats. This paper focuses on PDF format. Algorithm 1 realizes the preprocessing of EDs by extracting HTML elements, filtering and sorting the tex-tual elements. After the elements are extracted, the dimension sets have to be recomposed. This is achieved through clustering using DBSCAN (cf. Algorithm 3) in an iterative way until the best possible results are achieved. Other solutions were also tried: converting the PDF into image format and using OCR to extract the dimen-sion led to inferior results. Similarly, instead of clustering, regular expressions and a brute force approach provided worse results.
Afterwards, postprocessing differentiates between dimension ele-ments, text referring to regulatory standards and other textual elements and provides this additional information in form of a user interface to assist employees with manual quality control.
RQ 2 How should the clustering parameters be set to achieve optimal clustering results, i.e. complete dimension sets and avoid over-clustering?
• RQ 2.1: Which distance metric leads to optimal results?
• RQ 2.2: How should the DBSCAN parameters be set?
We propose a distance metric based on the average distance between the two nearest points of the bounding boxes, taking into account parallel and intersecting elements. Aside from the distance metric, the parameter influencing the result of DBSCAN the most, is the epsilon value. We propose an iterative approach, increasing the epsilon value in each run, before a stopping crite-rion is reached. As stopping criterion the silhouette coefficient is used, which measures the relation between the inner-cluster dis-tance and the between-cluster distance. For evaluation purposes recall RecallDE and precision PrecisionDE are used. Table 2 shows the results of the evaluation. The recall values are between 0.82 and 0.93, whereas the precision values lie between 0.58 and 0.88.
The average recall is 0.88, meaning that 88% of all relevant informa-tion has been extracted. The average precision is 0.73, meaning that of all extracted elements 73% are indeed dimension sets. The fluc-tuation in these values could be a result of different conditions such as the layout of the drawing and in particular if design guidelines regarding, e.g., minimal distance have been adhered to. The preci-sion values depend not only on the clustering of the values, but also on the postprocessing where the elements are filtered. These values may seem partly low. For the described use case, a 100% precision is not essential, as there is always a user involved, who still has to measure the features manually. This means that the user interface might still show unrelated values, but these unrelated values have decreased significantly, achieving a better overview for the user.
However, if DigiEDraw should be used to provide input for an auto-matic measuring program, without human involvement, higher precision values should be obtained. In future work, we will there-fore focus on increasing recall and precision by learning from the user interaction. No substantial fluctuations were observed for the number of iterations and the eps-value. This can be explained by the fact that no extreme cases, where values are very far apart or close together, were part of the evaluation set. However, even though 11 B. Scheibel, J. Mangler and S. Rinderle-Ma Computers in Industry 129 (2021) 103442 Fig. 7. Production process (modeled using Signavio©): (a) the main production process, (b) with manual quality control and (c) with facilitated and optimized quality control.
the range of variation does not seem large, even small changes can influence the recall and precision values.
Section 2 discusses approaches for scanned EDs as well as CAD format to digitize the graphical as well as textual elements. If the EDs are only available on paper, one of the mentioned approaches has to be used. However, these approaches may be problematic, as the quality heavily depends on the quality of the scan. In addi-tion, most of these approaches only focus on one aspect and source code is mostly not available. Approaches for CAD formats are also limited, as it is not common to note dimensions in these models, but rather when transitioning the models to EDs. If PDF docu-ments are available these issues can be avoided, and accurate results can be achieved by directly extracting from the PDF using DigiEDraw. DigiEDraw is providing an end-to-end approach, start-ing from uploading a drawing, to the integration of extracted values into a process. It is also easy to use and apply in industry, as it does not require programming skills or training time. Unique to DigiEDraw is also that only the pre- and postprocessing steps rely on domain-specific knowledge, as the clustering step itself relies on the distance metric. Thus, the DigiEDraw clustering approach can be adapted to other domains requiring only small adjustments.
5.1. Limitations DigiEDraw is currently limited to EDs in digital PDF. However, DigiEDraw could be extended to include DXF as well as scanned EDs, by adding a DXF parser and an OCR library to the prepro-cessing stage. The approach is sensitive to noise, e.g., if dimensions overlap. Therefore the best results are achieved if the ED adheres to the standards and norms where, among others, spacing and font is specified. Currently, the approach was tested on component drawings, but can be adapted/extended to, for example, assem-bly drawings. Furthermore, the DigiEDraw approach is intended to be used in conjunction with manual quality control, therefore including another check of the extracted dimensions. The follow-ing structuring guidelines specify under which conditions the best results can be achieved.
5.2. ED structuring guidelines The lessons learned from designing, implementing, validating, and applying DigiEDraw flow into the following guidelines. They state how EDs should be structured for achieving optimal results and supporting quality control in production processes: • The ISO norms concerning the distances between elements are adhered to, in order to avoid overlapping.
• The ED is either in conventional portrait or landscape format.
• Diagonal text should be avoided, clear horizontal or vertical writ-ing is optimal.
• Standard fonts are used.
12 B. Scheibel, J. Mangler and S. Rinderle-Ma Computers in Industry 129 (2021) 103442 If all of the guidelines are followed, DigiEDraw can even extract dimensions out of complex EDs. Additionally, well structured EDs are also easier to understand for the employees.
Section 4 discusses the application of DigiEDraw to facilitate and optimize manual quality control in the production process.
In addition, DigiEDraw can be used as quality control for the ED itself. This is motivated as follows: extraction results may vary according to the compliance to standards. It is important that the different values are organized and specifically no values are overlapping or stacked upon each other. If this is the case, the extraction process cannot work properly. Thus, DigiEDraw can be used to assess the extractability of an ED, while it is still in the design phase. The design engineer can then immediately react and adapt the ED accordingly, facilitating reading for humans as well as for DigiEDraw. In addition, DigiEDraw can be used to generate an automatic measurement program, by extracting dimensioning information, only requiring the employee to check the extracted elements and edit these if necessary, similar to the approach mentioned by Rica et al. (2020). This would be a more efficient approach than manually extracting all information. A measure-ment program could then be used to fully automate the quality control process. Zeleny et al. (2017) also use a clustering algo-rithm for web page pre-processing using HTML boxes as input.
This could constitute another application, in a completely different domain.
6. Conclusion Today, EDs are mainly used as contractual basis as well as to provide additional information, in particular information about tol-erances. However, there is still no approach to incorporate these EDs automatically in the production process. This paper proposes DigiEDraw to extract dimensions from EDs using DBSCAN. The dimensions can then be used for quality control tasks. The develop-ment of DigiEDraw shows that it is possible to extract dimensioning information from EDs with a recall of over 88% and a precision of 73%. DigiEDraw was applied for facilitating and optimizing man-ual quality control in a real-world production process. Additional application scenarios such as the automatic creation of measure-ment programs are conceivable. To the best of our knowledge, this is the first end-to-end approach where dimensions are extracted from a PDF ED.
DigiEDraw is currently limited to EDs in digital PDF. However, DigiEDraw could be extended to also include DXF as well as scanned EDs, by adding a DXF parser and an OCR library to the preprocessing stage. The best results are achieved if the ED adheres to the guide-lines e.g., standards are complied with. Currently, the approach was tested on component drawings, but can be adapted/extended to, for example, assembly drawings.
In future work, more parameters influencing the recall and pre-cision of DigiEDraw will be explored. This could be achieved by learning from the user interaction, i.e. allowing the user to mark unrelated values, storing these values and therefore learning for future drawings. In addition, the relationship between font size and clustering parameters will be examined and included into the algo-rithm. This could lead to a more domain specific approach. At all, it seems promising to include more semantics such as specific posi-tions or areas and their meaning, potentially in combination with object recognition. In addition, we will explore how it can be dis-tinguished between essential and non-essential information, e.g., is there a pattern which dimensions are essential for overall qual-ity of a workpiece. Finally, we will explore additional application scenarios and types of EDs.
Conflict of interest The authors declare that there is no conflict of interest.
Declaration of Competing Interest The authors report no declarations of interest.
Acknowledgments This work has been partially supported and funded by the Austrian Research Promotion Agency (FFG) via the “Austrian Com-petence Center for Digital Production” (CDP) under the contract number 854187.
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11229 | https://learningcorner.co/worksheet/4992 | Simplifying Ratios Using Decimals Worksheet for 13-Year-Olds / Learning Corner
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Simplifying Ratios Using Decimals Worksheet for 13-Year-Olds
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For each problem below, simplify the ratio by converting it to a decimal. Write your answer in the space provided. If needed, round your decimal to two decimal places.
Part 1: Convert Ratios to Decimal
2:5 =
3:4 =
5:10 =
7:14 =
9:15 =
Part 2: Simplify Ratios Using Decimals
Simplify 4:8 using decimals:
Simplify 6:9 using decimals:
Simplify 12:18 using decimals:
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11230 | https://sciencenotes.org/barium-facts-element-56-or-symbol-ba/ | Home » Science Notes Posts » Chemistry » Elements » Barium Facts – Element 56 or Symbol Ba
Barium Facts – Element 56 or Symbol Ba
This entry was posted on by Anne Helmenstine (updated on )
Barium is a chemical element with the symbol Ba and atomic number 56. It is an alkaline earth metal that is best known for the green color it adds to fireworks and for its use in barium enemas for x-ray imaging.
Discovery, Naming, and Isolation
Pebbles consisting of the mineral baryte occurred in volcanic rock in Italy. Following exposure to light, these stones glowed with phosphorescence, making them extremely interesting to alchemists in the early 17th century.
In 1772, Carl Wilhelm Scheele experimented with baryte, believing its interesting properties meant it contained a new element. But, Scheele could only isolate barium oxide, not the purified element. In 1808, Sir Humphry Davy used electrolysis to isolate the element from barium hydroxide.
Davy named the “barium” from the Greek word “barys,” meaning “heavy,” due to the high density of its compounds. He added the -ium ending consistent with the names of other metallic elements.
Appearance and Properties
Pure barium is a silvery-white metal with a slight golden cast that is soft enough to cut with a knife. It is highly reactive and rapidly oxidizes in air, forming a dark gray protective layer of barium oxide. Barium reacts with water, releasing hydrogen gas, and it reacts vigorously with halogens to form halides. Barium compounds burn with a green flame. Chemically, barium acts much like magnesium, calcium, and strontium.
Element Group
Barium is part of Group 2 on the periodic table, which is also known as the alkaline earth metals. These elements share similar properties, including high reactivity with water, a strong tendency to form oxides, and a common +2 oxidation state.
Natural and Synthetic Isotopes
Natural barium is a mix of seven isotopes: Ba-130, Ba-132, and Ba-134-138. Barium-138 is the most abundant, accounting for about 71.7% of all natural barium. Ba-130 and Ba-133 are radioisotopes. Ba-133 has a half-life of 10.51 years. Ba-130 is essentially stable, with a half-life of around 1021 years. Including the synthetic radioisotopes, there are 40 known barium isotopes with masses from 144 to 153.
Abundance and Sources
Barium is the 14th most abundant element in the Earth’s crust, accounting for 0.0425% of the crust and having a concentration of 13 μg/L in sea water. The element is too reactive to occur as a native element. It mainly occurs in the form of the minerals baryte (barite, barium sulfate) and witherite (barium carbonate).
The largest sources of barium are in Britain, Romania, and Russia. The main producers are China, India, Morocco, and the United States.
Purification
Barium purification relies on electrolysis or high-temperature reduction. Electrolysis involves dissolving barium compounds in molten salts, then using electric current to isolate the metal. High-temperature reduction uses aluminum or other metals to reduce barium compounds, ultimately leaving barium in its pure form.
Uses of Barium
Barium and its compounds have numerous applications, including:
Medical Imaging: Barium sulfate is a contrast agent in gastrointestinal X-rays and other imaging techniques due to its opacity to X-rays.
Oil and Gas Drilling: Barium sulfate is a key component in drilling fluids, providing weight and lubrication to prevent blowouts.
Gemstone: The blue fluorescent gemstone benitoite (barium titanate silicate) is the state gem of California.
Manufacturing: Barium is used in the production of glass, ceramics, and as a deoxidizing agent in metallurgical processes.
Fireworks and Pyrotechnics: Barium nitrate, barium monochloride, and barium chlorate produce green colors in fireworks.
Electronics: Barium titanate is a piezoelectric material used in capacitors and transducers. Historically, it was a getter in vacuum tubes for removing unwanted gases. Barium ferrite is a magnetic compound in magnetic stripe cards and data storage tapes.
Superconductors: YBCO is a well-known high-temperature superconductor containing barium.
Water Treatment: Barium carbonate is used to remove impurities, such as sulfates, from water.
Rodenticide: Water-soluble barium compounds are toxic and effective as rodenticides.
Poison: For the same reason, soluble compounds have been used in poisonings.
Oceanography: Barium concentration in sea water shows a correlation with silicic acid and alkalinity. Meanwhile, particulate barium correlates with particulate organic carbon. The element helps trace variations in the global climate and carbon cycle.
Oxidation States
Barium typically exhibits a +2 oxidation state, which is consistent with other alkaline earth metals. This state is the most stable and is the one observed in most of its compounds. However, it sometimes exhibits the +1 oxidation state.
Biological Role, Health Effects, and Toxicity
Barium has no known biological role in humans or other living organisms. The element is not a carcinogen and does not bioaccumulate. Insoluble compounds, like barium sulfate, present no significant health risk. However, soluble ones (e.g., barium chloride) cause severe health effects, including muscle weakness, respiratory distress, and cardiac irregularities. The Ba2+ ion blocks potassium ion channels, which leads to nervous system and organ damage. Inhaling insoluble barium compounds leads to a condition called baritosis, which is similar to coal miner’s lung and silicosis.
Key Barium Facts for Scientists
Here’s a table with key facts about barium, including its physical and atomic properties:
| Property | Value |
--- |
| Name | Barium |
| Symbol | Ba |
| Atomic Number | 56 |
| Atomic Weight | 137.327 |
| Group | 2 (Alkaline Earth Metals) |
| Period | 6 |
| Block | s-block |
| Electron Configuration | [Xe] 6s² |
| Electrons per Shell | 2, 8, 18, 18, 8, 2 |
| State at Room Temperature | Solid |
| Melting Point | 727 °C (1341 °F) |
| Boiling Point | 1845 °C (3353 °F) |
| Density | 3.59 g/cm³ |
| Heat of Fusion | 7.12 kJ/mol |
| Heat of Vaporization | 142 kJ/mol |
| Molar Heat Capacity | 28.07 J/(mol·K) |
| Oxidation States | +1, +2 |
| Electronegativity | 0.89 (Pauling scale) |
| First Ionization Energy | 502.9 kJ/mol |
| Second Ionization Energy | 965.2 kJ/mol |
| Third Ionization Energy | 3,613 kJ/mol |
| Atomic Radius | 222 pm |
| Covalent Radius | 215 pm |
| Van der Waals Radius | 268 pm |
| Crystal Structure | Body-centered cubic (bcc) |
| Thermal Conductivity | 18.4 W/(m·K) |
| Electrical Resistivity | 332 nΩ·m |
| Magnetic Ordering | Paramagnetic |
| Young’s Modulus | 13 GPa |
| Shear Modulus | 4.9 GPa |
| Mohs Hardness | 1.25 |
References
Davy, H. (1808). “Electro-chemical researches on the decomposition of the earths; with observations on the metals obtained from the alkaline earths, and on the amalgam procured from ammonia”. Philosophical Transactions of the Royal Society of London. 98: 333–370. doi:10.1098/rstl.1808.0023
Greenwood, Norman N.; Earnshaw, Alan (1997). Chemistry of the Elements (2nd ed.). Butterworth-Heinemann. ISBN 978-0-08-037941-8.
Griffith, Elizabeth M.; Paytan, Adina (2012). “Barite in the ocean – occurrence, geochemistry and palaeoceanographic applications”. Sedimentology. 59 (6): 1817–1835. doi:10.1111/j.1365-3091.2012.01327.x
Kresse, Robert; Baudis, Ulrich; et al. (2007). “Barium and Barium Compounds”. Ullmann’s Encyclopedia of Industrial Chemistry. Weinheim: Wiley-VCH. ISBN 978-3527306732. doi:10.1002/14356007.a03_325.pub2
Weast, Robert (1984). CRC, Handbook of Chemistry and Physics. Boca Raton, Florida: Chemical Rubber Company Publishing. ISBN 0-8493-0464-4.
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11231 | https://lessons.unbounded.org/math/grade-4/module-2/topic-b/lesson-4 | MATH 4 : Multi-step word problems: length, mass, and capacity
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Lesson 4 - Multi-step word problems: length, mass, and capacity
Previous Lesson in Series Express Metric Capacity Measurements Previous lesson in this Series
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Math / Grade 4 G4 / Module 2 M2 / Topic B TB / Lesson 4 lesson 4
lesson 4 1 hour
Multi-Step Word Problems: Length, Mass, And Capacity
Students will solve multi-step word problems involving length, mass, and capacity.
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Math Grade 4 Curriculum Map
module 1
module 2
topic A
topic B
module 3
module 4
module 5
module 6
module 7
Description
Objective: Use addition and subtraction to solve multi-step word problems involving length, mass, and capacity.
In Lesson 5, as students solve two- and three-step word problems by adding and subtracting metric units, their ability to reason in parts and wholes is taken to the next level. This is important preparation for multi-digit operations and manipulating fractional units in future modules.
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Grade 4 Mathematics Module 2, Topic B, Lesson 5
Grade 4 Mathematics Module 2, Topic B, Lesson 5
Prerequisites
3.MD.2,
3.MD.A.2,
3.OA.7,
3.OA.C.7,
4.MD.1,
4.MD.A.1,
4.NF.3.A,
4.NF.3.B,
4.NF.3.C,
4.NF.4.C,
4.NF.5,
4.NF.6,
4.NF.B.3.A,
4.NF.B.3.B,
4.NF.B.3.C,
4.NF.B.4.C,
4.NF.C.5,
4.NF.C.6
Tags
CCSS Standard:
4.MD.1,
4.MD.2,
4.MD.A.1,
4.MD.A.2
Credits
From EngageNY.org of the New York State Education Department. Grade 4 Mathematics Module 2, Topic B, Lesson 5. Available from engageny.org/resource/grade-4-mathematics-module-2-topic-b-lesson-5; accessed 2015-05-29.
Copyright © 2015 Great Minds. UnboundEd is not affiliated with the copyright holder of this work.
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Grade 4 Mathematics Module 2, Topic B, Lesson 5
Grade 4 Mathematics Module 2, Topic B, Lesson 5
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Understanding pressure thrust
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HiPressure thrust = Fp = (Pe - Pa) AnAfter 3 days searching the net I haven't found an explanation for what "pressure thrust" actually is ...I have my own ideas, but can somebody provide a simple explanation?
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Pressure thrust = Fp = (Pe - Pa) VeSimplified: Exit pressure minus the atmospheric pressure (difference in pressures) times velocity at exit (how fast the flow of particles or "flow" is) = Pressure thrustThe more the difference between the exit pressure and atmospheric pressure is the more it effects thrust, and ideally if Pe=Pa then Ve will be at its prime.check out the link... Im no expert and Im sure there are far more qualified people to explain this than me...
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decyple said:
Simplified: Exit pressure minus the atmospheric pressure (difference in pressures) times velocity at exit (how fast the flow of particles or "flow" is) = Pressure thrustThe more the difference between the exit pressure and atmospheric pressure is the more it effects thrust, and ideally if Pe=Pa then Ve will be at its prime
Click to expand...
Thanks for your post but what I am looking for is an explanation of what the pressure term represents in reality ... ... what is actually happening to create "pressure thrust" The net is full of explanations for "momentum thrust" but none for pressure thrust. I even have a physics book with a whole chapter on rocket engines that omits to explain what pressure thrust actually is! An equation without basis in physical reality is no use whatsoever ... (I also read all of www.braeunig.us already - it also doesn't explain it)
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I would look at pressure thrust being equal to (Pe-Pa) x Ae. Pe is the pressure at the exit plane of the nozzle, Pa the local atomospheric pressure and Ae is the area of the nozzle at the exit plane. I look at the pressure thrust being the contribution (or loss) to the total thrust by the expansion (and gain in exit velocity) of the exhaust gases contributed by the nozzle. The other portion of the thrust is mdot x Ve and is known as momentum thrust. Richard Nakka does an excellent job of presenting overall motor theory on his web site, www.nakka-rocketry.net. Regards, Ed
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WildEnergy
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#5
Great website, thanks for the link. Unfortunately - yet another web page that declares pressure thrust without explaining what it is ... Is there a standard academic text for rocket propulsion?
W
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#6
This is what I figured out so far:If you take an enclosed volume of gas (say a coke bottle with 100 psi air)all the pressure forces balance - so there is no acceleration.As soon as you create a nozzle (by opening the bottle) the forces no longer balance ... there is a reduction of force on the side with the hole of exactly F = P A.The action will attempt to accelerate the bottle (and any contents outside of the nozzle "cylinder") upwards, the reaction will accelerate anything inside the nozzle "cylinder" downwards.Once propellant is accelerated outside the bottle - then momentum thrust is relevant.(The nozzle "cylinder" is the cylinder defined by the nozzle and a vector normal to the nozzle plane and parallel to the unbalanced forces) (or something like that)
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#7
Check this out. Bob
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#8
at least nasa try and explain it:
There is an additional effect which we must account for if the exit pressure p is different from the free stream pressure. The fluid pressure is related to the momentum of the gas molecules and acts perpendicular to any boundary which we impose. If there is a net change of pressure in the flow there is an additional change in momentum. Across the exit area we may encounter an additional force term equal to the exit area Ae times the exit pressure minus the free stream pressure
Click to expand...
(Will have to read that a few more times, isn't making much sense at the moment)
B
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#9
Normally the atmospheric pressure is pretty low compared to combustion chamber pressure so that little complexity could be somewhat neglected.The rest of the problem is more complex than it looks. I think the FORCE=PRESSURE AREA is a good way to look at it. The pressure part is where it gets more complicated, I don't think pressure goes to 0 at the nozzle. Otherwise I think nozzle design would not matter. I think the better the nozzle design, the better the pressure gets closer to 0 at the nozzle and the more efficient the engine is. This all makes sense to me with respect to solid fuel motors, but I haven't come up with my own theory to completely explain how liquid fuel motors behave. They seem to have more elaborate nozzles than my solid fuel motor theory would require.
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#10
I'll try to explain it in a simple manner.In the simplest efficient rocket motor you have a combustion chamber and a convergent-divergent nozzle aka deLaval nozzle.IN the combustion chamber, the fuel and oxidizer are mixed and react and make hot gases. The throat, or minimum diamter the nozzle, acts as a choke point to the flow and pressurizs the combustion chamber with respect to the outside. Once the hot gas passes through the nozzle throat, it expands and cools, converting thermal energy into directed kinetic energy with a corresponding pressure drop.When the pressure difference between the combustion chamber and the atmosphere exceeds about a factor of 2, the expansion is supersonic and is described by the isentopic flow equations. Follow the isentropic flow sequence starting here. It's a bit complicated but in the ideal nozzle expansion there is one area ratio, that is the area of the nozzle exit to the area of the throat, where the expansion pressure will equal the atmospheric pressure. If the area ratio is less than ideal, the gas is underexpanded and has the potentiallly expand further beyond the exit plane of the nozzle. This extra energy is the pressure thrust. Bob
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#11
To (hopefully) help expand on what Bob said, consider one form of the equation defining jet thrust:
The first half of it (mdotu_e) covers the "momentum thrust". This is the part that Newton's Third Law talks about -- action/reaction and whatnot. The propellant mass is tossed overboard (mdot has units of mass per time, it's a flow rate) at a given speed (u_e is the exit velocity), and the resulting change in momentum out of the engine control volume generates a force. Check the units --
The second half of the first equation ((Pe-Pa)Ae) is the "pressure thrust". This part of the thrust equation comes from the fact that you're firing a propellant that's producing a lot of hot, compressed gasses which contain a correspondingly large amount of energy as pressure and temperature. The nozzle recovers some of this energy as velocity by passing the flow through the diverging section (this is the magic of supersonic flow, see Bob's post). Any excess pressure that's left over pushes on the rocket because of the pressure differential at the nozzle exit area -- Pe is the exit pressure, Pa is the surrounding ambient pressure, and Ae is the exit area. This term is the pressure thrust, and it can be positive, zero, or negative, depending on the nozzle design and surrounding conditions.
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#12
Thankyou all for your posts.I think I understand it now.I see it like this:pressure is the driver for the whole thingif the action of the pressure is to accelerate the rocket up (by pushing the top of the combustion chamber up)the reaction is to accelerate the propellant down (equal and opposite)the force involved is P A because the pressure forces are less on the nozzle sideyou need to subtract ambient pressure because if internal and external pressure were equal there would be no change in pressure and thus no thrust(taken to extreme: if the ambient pressure was higher than internal pressure the rocket would run in reverse - like a vacuum cleaner!)now we have a force accelerating the propellant out of the nozzle - momentum thrust comes into playthe reason the thrust force isn't simply F = 2 P Ais because of inefficiency in moving the propellant out of the chamber(so force is spent on overcoming friction, drag due to vortices, and all the fluid dynamics stuff, etc.)
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#13
Almost. You're getting there.There are 2 mechanistic components to thrust generation: the major component is momentum thrust and the minor component is pressure thrust which is zero in the idea nozzle expansion. Momenturm thrust converts the random kinetic energy in a hot high pressure subsonic gas flow via nozzle expansion into a high directed supersonic velocity lower temperature and pressure gas flow. The optimum thrust conversion occurs when the pressure of the expansion at the exit of the nozzle equas the ambient pressure. This is a sequence of the NASA tutorial that should allow you to visualize the process. It is complicated and does requires a bit of patience to get comfortable with it, Bob
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#14
Thanks BobI am working my way through it
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#15
It's takes a while to understand.Also there are weight and dimensional limits to nozzle length, so in reality, most nozzle flows are underexapnded and that's where pressure thrust becomes important.
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#16
...I would boil it down to this:Rocket motors are almost entirely momentum devices; they expel matter quickly in one direction and travel in the other.Secondarily, there is an effect--which can either help or hinder the process, and may or may not be significant--whereby the pressure differential between the exhaust and the ambient air either serves to push the rocket a little more, or suck it down a bit.You can devise experiments where the pressure differential is significant enough to affect the flight, such as the Krushnik Effect in rack rockets where the rocket may not even leave the pad.But for well designed rockets, it's a momentum game.--John
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#17
My minor contribution to a great thread, lots of good posts here:1. Dr. Robert "Bob" Goddard had to work pretty hard convincing others in the 1920's that rockets would work in a vacuum. It went against conventional wisdom that the pressure of the escaping gases wasn't the reason for the thrust, but the momentum exchange was. :bangpan: (BTW, Bob, moderator of the Rocketry Forum, is also from Massachussetts...is that you Dr. Goddard?? )2. For a great practical example of the idea of exhaust expansion, look at just about any 2 or 3 stage launch vehicle. Saturn V, Falcon 9, Soyuz are all popular examples. See that the 1st stage engines have "small" engine bells that don't expand the exhaust gas too much, because ambient pressure during the first 1-2 minutes of flight, just after launch, is NOT vacuum. But, 2nd and 3rd stages act pretty much in a vacuum so the designers expand the exhaust gases even further with a larger engine bell. Ideally they'd keep expanding the exhaust forever but there is a practical limit, for as the engine gets larger the weight gets larger/ payload performance drops, and if too large the engine bell won't fit inside the adapter between stages.3. So as the others have said, you're trying to expand your exhausts gases as much as practical without overexpanding. Some under-expansion is OK and happens whether you like it or not, and a measure of under-expansion (waste) is pressure thrust.
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Coming in from left field...I think the problem is the distinction between pressure and momentum, because both are the result of molecular motion. The difference is that pressure is the result of random molecular motion; whereas momentum thrust is the result of directed molecular motion.So the exhaust stream has molecules going in all directions. Some are actually going upstream. A good nozzle directs the great majority of them aftward. If they were all going in that direction, the temperature of the stream would be absolute zero. It's not. Temperature is also the result of random motion, and the nonzero temperature of the exhaust demonstrates that some random motion persists. That said, the exhaust stream temperature is lower than the chamber pressure, because some of the random motion that manifests as temperature in the chamber is directed as exhaust velocity in the exhaust stream.All of the directed gas produces thrust. Also, by chance, some of the randomly-directed motion manifests as thrust, because some of the motion is in that direction. Indeed, the presence of the rocket biases the motion at the nozzle exit by blocking random motion in that direction. The extra thrust provided by random motion is pressure thrust.
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#19
It really isn't that hard to understand if one just realizes that there are two kinds of thrust. One happens inside the combustion chamber, and the other happens in the vacuum of space behind the combustion chamber. What I think people are struggling with is the fact that pressure thrust is actually NOT reactionary behavior caused by Newton's Law of Motion.Pressure thrust occurs inside the combustion chamber, where the pressure at the front of the chamber is greater than the pressure at the neck of the exit nozzle. In order to keep this pressure differential going, the rate of combustion must be greater than the rate of exhaust gas flow out of the engine nozzle. This is why there is a choke point at the neck of the nozzle. This restricts the flow of the exhaust gases, so that the rate of combustion can exceed it. By having greater pressure pushing on the front of the combustion chamber, the rocket moves forward in that direction. So, inside the chamber, the main action is caused by the differential of the pressure at the front of the chamber, and the pressure at the nozzle neck orifice. The reaction of that forward pressure thrust is the backward flow of hot exhaust gas out the back of the engine. So, pressure thrust pushes the rocket forward from inside the combustion chamber.Simultaneous to the forward pressure inside the combustion chamber, there is a backward flow of exhaust gases, and this backward flow causes a reactionary impulse in the forward direction. So, this flow is not caused by the exiting gases pushing against the air, since there is no air, but it is the reaction motion described by Newton's Law of Motion, which says that every action has an equal and opposite reaction.Inside the combustion chamber, there is a high pressure pushing against all the walls of the chamber, except at the exit nozzle, where the pressure is much lower due to the escaping exhaust gases. This pressure difference inside the chamber causes the rocket to move forward, as a direct result of the pressure on the forward surface of the combustion chamber.Outside the combustion chamber, the hot exhaust gases are exiting the engine, accelerated by the bell shape of the nozzle. These atoms of exhaust are quickly accelerated by the difference in back pressure at the nozzle and the vacuum of space. This backward flow of exhaust gases causes a reaction in the forward direction, due to the conservation of momentum.
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#20
WildEnergy said:
... what is actually happening to create "pressure thrust"The net is full of explanations for "momentum thrust" but none for pressure thrust.
Click to expand...
The full equation for 1D thrust is F_thrust = mdotVe + (Pe-Pa)Ae The first term is momentum thrust which comes from conservation of momentum (change in momentum is change in mass times change in velocity). The second term is pressure thrust where Pe is pressure at the nozzle exit, Pa is ambient pressure, and Ae is the nozzle exit area. Pressure from the nozzle exit is adding thrust while pressure from the surrounding air is subtracting thrust, the net pressure multiplied by area gives you a force, ergo thrust. The speed at which information (pressure for example) travels through a fluid is its speed of sound. When you have a supersonic gas, information downstream cannot propagate upstream. So where in normal subsonic gas flow the exit pressure of a nozzle would be forced to equalize with the ambient air pressure, in supersonic gas flow the gas "doesn't know" what is happening down stream and cannot equalize until it gets there. Naturally, it will eventually have to equalize because gas flows from regions of high pressure to regions of low pressure, and this pressure difference is not stable, it just doesn't happen gradually but rather nearly instantaneously once the gas is exposed to the pressure difference. This allows the exit pressure to be different from ambient pressure - in the case of under expanded flow the gas pressure is higher than ambient and so it adds pressure thrust but at the cost of missing out on some potential gains in momentum from further expansion, in over expanded flow the gas has expanded more and so the momentum thrust is higher but pressure is lower than ambient and so it subtracts thrust (hence a perfect nozzle expands the gas to exactly match ambient pressure). The pressure component of thrust is why you get shock diamonds - in an attempt to equalize pressure with ambient conditions, the exhaust will form a shockwave to increase pressure (bright regions of the exhaust) or an expansion wave to decrease pressure (dark regions of the exhaust). Its also why rocket engines are more efficient in a vacuum than on the ground (momentum thrust and Pe stay the same, Pa decreases).
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jqavins
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#21
Let me try another way. When you let a balloon go and it farts its way around the room, that's almost entirely pressure thrust. There's a difference in pressure from inside the balloon to outside, that pressure is applied over the exit area, and the thrust is the pressure difference times the area, ΔP∙A. Bog simple, done.But then, all thrust is, in a manner of speaking, momentum thrust. The law of conservation of momentum means that if the balloon gains momentum in one direction, the air coming out must have exactly the same momentum going the other way. The only real question is how the air gets that momentum, and being pushed out through a hole with a pressure differential is the simplest way.Now, the rate of momentum carried away by any fluid stream is equal to its mass flow rate times its speed, as @daveyfire wrote in post #11. The rate of momentum change is the same thing as force; in Isaac Newton's original writing that's actually the definition of force. How the stream is formed and gets its speed is not important. Again, in the balloon example ΔP∙A is all you need. But there are other ways. Some genius named Gustaf de Laval figured out how to do some magic to get more thrust by applying his amazing converging-diverging nozzle. That nozzle uses the mystical powers of supersonic flow to make the gas leaving a rocket engine go a lot faster than ΔP∙A could alone. de Laval's magic provides most of the momentum in the exhaust stream. And all the magic happens upstream of the nozzle exit plane. The pressure is high in the combustion chamber, it's even higher at the nozzle's throat, and it drops in the expansion section, where the gas is bookin' and cookin'. But, once that gas reaches the exit plane, it's just like the opening in the balloon. Whatever pressure the exiting gas has come down to in the expansion is just like the pressure inside the balloon. That exit pressure differential creates its little bit of ΔP∙A, which we call pressure thrust. The words are a bit misleading, in my opinion. All thrust is a result of the momentum of exiting gas, and there would be no exiting gas were it not for combustion chamber pressure. So all thrust is momentum thrust and all thrust is pressure thrust. But that's not what they're called. It is what it is. I'd call them something like "magic thrust" and "balloon thrust" myself, but no one asked me.
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#22
A 2012 thread... that is seamlessly continued in 2025.Awesome.
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#23
Oops, didn't notice the date on Post 1 when the thread rose to near the top of the list.
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#24
jqavins said:
Oops, didn't notice the date on Post 1 when the thread rose to near the top of the list.
Click to expand...
No worries.. I like the fact that we can learn / cuss / discuss from older threads.
Continuity is a good thing.
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Deduce the type of a triangle from a quadratic equation relating its side lengths
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The lengths of the sides of a triangle are p,q,r p,q,r respectively. If p 2+q 2+r 2=p q+q r+p r p 2+q 2+r 2=p q+q r+p r, then this triangle is
(A) equilateral triangle
(B) isosceles triangle
(C) right angled triangle
(D) obtuse triangle
My Attempt: p 2+q 2+r 2=p q+q r+p r⟹(p 2−p q)+(q 2−q r)+(r 2−p r)=0⟹p(p−q)+q(q−r)+r(r−p)=0,
p 2+q 2+r 2=p q+q r+p r⟹(p 2−p q)+(q 2−q r)+(r 2−p r)=0⟹p(p−q)+q(q−r)+r(r−p)=0,
but I'm unable to connect this information to the given options. Please help me.
algebra-precalculus
triangles
quadratics
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edited Jul 30 at 18:06
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2 For what it's worth, as indicated by the answers, this problem is a special case of Maclaurin's inequality.user2661923 –user2661923 2025-01-05 16:07:52 +00:00 Commented Jan 5 at 16:07
Regardless of the math, due to the symmetry of the equation, the only possible answer is A.sds –sds 2025-01-06 16:22:20 +00:00 Commented Jan 6 at 16:22
@sds or at least, A must be one possible answer.usul –usul 2025-01-06 18:19:48 +00:00 Commented Jan 6 at 18:19
@sds, the equation p 2+q 2+r 2=p q+q r+r p+1 p 2+q 2+r 2=p q+q r+r p+1 has the same symmetry of that equation, but does not admit a solution with p=q=r p=q=r.Gonçalo –Gonçalo 2025-01-06 22:45:42 +00:00 Commented Jan 6 at 22:45
Don't forget A=>B A=>B here.Kamil Maciorowski –Kamil Maciorowski 2025-01-07 00:54:55 +00:00 Commented Jan 7 at 0:54
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Let me clarify Person Average's suggestion: 2(p 2+q 2+r 2)=2(p q+q r+r p)⟹(p 2−2 p q+q 2)+(q 2−2 q r+r 2)+(r 2−2 r p+p 2)=0⟹(p−q)2+(q−r)2+(r−p)2=0⟹p=q=r.
2(p 2+q 2+r 2)=2(p q+q r+r p)⟹(p 2−2 p q+q 2)+(q 2−2 q r+r 2)+(r 2−2 r p+p 2)=0⟹(p−q)2+(q−r)2+(r−p)2=0⟹p=q=r.
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answered Jan 6 at 1:44
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Since p 2+q 2+r 2≥p q+q r+r p p 2+q 2+r 2≥p q+q r+r p for p,q,r∈R+p,q,r∈R+
But given that p 2+q 2+r 2=p q+q r+r p p 2+q 2+r 2=p q+q r+r p.
Equality holds when p=q=r p=q=r.
Hence equilateral triangle.
NOTE:
(p−q)2≥0(p−q)2≥0 , p 2+q 2≥2 p q p 2+q 2≥2 p q
(q−r)2≥0(q−r)2≥0 , q 2+r 2≥2 q r q 2+r 2≥2 q r
(r−p)2≥0(r−p)2≥0 , r 2+p 2≥2 r p r 2+p 2≥2 r p
then
p 2+q 2+r 2≥p q+q r+r p p 2+q 2+r 2≥p q+q r+r p for p,q,r∈R+p,q,r∈R+
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edited Jan 5 at 13:17
answered Jan 5 at 13:10
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What you need is just multiply 2 2 for left hand and right hand , then you can find that it is (3×square)=0(3×square)=0 and they must be all zero . so it is a equilateral triangle .
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edited Jan 5 at 12:57
Antony Theo.
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11234 | https://apstricks.wordpress.com/about/profit-and-loss/ | Profit and Loss | apstricks
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Profit and Loss
C P( Cost Price ):
The Price at which an article is purchased, is called its cost price or In shot called C P.
Cost price is amount of money goes out from us when purchase any article.
SP( Selling Price ):
The price at which an article is sold, called its selling priceor In short called SP.
Selling Price is a amount of money came in when selling by any thing.
Profit or Gain:
If S.P. is greater than C.P, then seller is said to have a profit or gain.
Loss:
If S.P. is less than C.P., the seller said to have a loss.
Marked Price :
When we purchase any item we saw the label price or marked Price or listed Price and denoted by MP.
Some important formulas:
1.Selling price is greater then cost price is called Gain.
Gain = (S.P) – (C.P)
15% Profit on article that means,
Cost price 100%
Selling Price 115%
Profit 15%
2.If Cost price is greater then Selling price is called Loss.
Loss = (C.P) – (S.P)
15% Loss on article that means,
Cost price 100%
Selling Price 85%
Loss 15%
3.Profit and Loss is always calculated on C.P ( Cost Price )
Profit or Gain percent Shortcut tricks
Formula: Gain % = (Gain x 100/C.P)
Example 1 :Rana purchase an ball for Rs.80 and sells it for Rs. 1000. Find his gain percent.
Answer: C.P = 80 and S.P = 100.
Gain= 100 – 80 = 20
Gain % = 20 x 100/80 = 25%
Example 2 :A farmer purchase an article for Rs.25 and sells it for Rs. 28. Find his gain percent.
Answer: C.P = 25 and S.P = 28.
Gain= 28 – 25 = 3
Gain % = 3 x 100/25 = 12%
Example 3 :Sabir purchase an article for Rs.40 and sells it for Rs. 60. Find his gain percent.
Answer: C.P = 40 and S.P = 60.
Gain= 60 – 40 = 20 .
Gain % = 20 x 100/40 = 50%
Loss percent Shortcut tricks
Formula: Loss% = (Loss x 100/C.P)
Example 1 :
10% loss on selling price is what percent loss on the cost price ?
Answer :
consider selling price be = Rs.100, Then loss = Rs.10
Cost price = Rs.( 100 + 10 ) = 110
So, loss% = (10 / 110 x 100 ) = 100 / 11 %
Example 2 :
A man purchased a toy for Rs. 120 and sold it for Rs. 90, Find the loss Percent.
Answer :
C.P = 120 and S.P =90
So We Know the formula of Loss = C. P – S.P . Loss = (120 – 90) = 30
%Loss = 30 x 100/120 = 25%
Example 3 :
Anil purchased a mobile for Rs. 12000 and sold it for Rs. 8000, Find the loss Percent.
Answer :
C.P = 12000 and S.P =8000
So We Know the formula of Loss = C. P – S.P . Loss = (12000 – 8000) = 4000
%Loss = 4000 x 100/12000 = 33.33%
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Example 4 :
A farmer purchased a cow for Rs. 200 and sold it for Rs. 180, Find the loss Percent.
Answer :
C.P = 200 and S.P =180
So We Know the formula of Loss = C. P – S.P . Loss = (200 – 180) =20
%Loss = 20 x 100/200 = 10%
Find selling Price Profit and Loss Shortcut tricks
Formula: S.P = (100+Gain%)/100 x C.P.
Q:We can find the Selling Price of article Using this above Formula
If C.P of a product is a Rs. 180 , Gain = 30%, then Find the S.P.
A:S.P = ?
S.P = 130% of Rs. 180, Gain = (100+30) = 130, (130/100 x 180) = Rs.234
So Selling Price is Rs. 234
Shortcut Tricks
130/100 x 180 = 234
Formula: S.P = (100 – Loss%)/100 x C.P
Q:If C.P of a product is a Rs. 180 , Loss = 30%, then Find the S.P.
A:S.P. = ?
S.P = 70% of Rs. 180, Loss = (100 – 30) = 70, ( 70/100 x 180 ) = Rs. 126
So the Selling Price is Rs. 126
Shortcut Tricks
70/100 x 180 = 126
Example :
Rajan bought a bike for Rs.60,000 and spent Rs. 4000 on repair and Rs. 1000 on transport and sold it with 25% profit. What price did he sell the car ?
Solution:
Cost price =Rs. 60,000 Spent on repair =Rs. 4000 transport =Rs. 1000 and profit 25%
So, Cost price =Rs.( 60,000 + 4000 + 1000 ) = Rs. 65,000
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So, Selling price = Rs. 65,000 x 125 / 100 = 81250.
Find cost price Profit and Loss Shortcut tricks
Formula: C.P =100 x S.P/(100+Gain%).
Example1:If S.P of a article is Rs. 240 then Gain 20%, Find the C.P.
Answer:C.P.= ?
C.P . = 100 x 240 / ( 100+20 )
=24000/120 = 200
C.P. = 200
shortcut Tricks:
100 x 240/( 120 ) = 200
Formula: C.P = 100 x S.P/(100 – Loss%).
Example2:If S.P of a article is Rs. 240 then Loss 20%, Find the C.P.
Answer:C.P. = ?
C.P. = 100 x 240 / ( 100 – 20 )
= 24000/80 = 300
shortcut Tricks
100 x 240 / 80 = 300
Profit and Loss Example 1
Example 1 :
If the C.P is 75% of the selling price, then What is the profit percent ?
Answer :
Let S.P = Rs 100,
Then C.P = 75
Profit = Rs 25
Profit% = ?
= 25x 100 / 75
= 100 / 3 = 33.3
Profit % = 33.3
Example 2 :
Find the single discount equivalent to a series discount of 30%, 20% and 10%.
Answer :
Let price be Rs. 100.
Then Net S.P = ( 90 x 80 x 70 / 100 x 100 x 100 ) x 100
= 36 x 7 / 5
= 252 / 5
= 50.4.
Required Discount is = ( 100 – 50.4 ) = 49.6
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Example 3 :
A Farmer bought a cow and carriage for Rs. 3500. He sold the horse at a gain of 35% and the carriage at loss of 10%, thereby gaining 2% on the whole. Find the cost of the cow.
Answer :
We consider C.P of the cow Rs. x
Then C.P of the Carriage Rs. ( 3500 – x )
So,
35% of x – 10% of ( 3500 – x ) = 2% of 3500
( 35x / 100 ) – 10( 3500 – x ) / 100 = 2 x 3500 / 100
( 7x / 20 ) – ( 3500 – x ) / 10 = 70
7x – 2x 7000 = 1400
5x = 5600
x = 5600 / 5
x = 1400
Pofit and Loss Example 2
Example 1:
If the cost price of 14 pens is equal to the selling price of 8 pens,the gain percent is :
Answer :
Let C.P of each pen is Rs. 1.Then
C.P of 8 pens = Rs. 8 : S.P price of 8 pens = Rs. 14
Gain % = 6 x 100 / 8 = 75%
Example 2 :
The Profit earned by selling an article for Rs.630 is equal to the loss incurred when the same article is sold for Rs. 370. What should be the sale price for making 50% profit?
Answer:
Let C.P price be x
Then
630 – x = x – 370
2x =1000
x = 500
Required S.P = 150% of 500
= 150 x 500 / 100
= 750
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11235 | http://www.nature.com/scitable/topicpage/natural-selection-uncovering-mechanisms-of-evolutionary-adaptation-34539 | This page has been archived and is no longer updated
Natural Selection: Uncovering Mechanisms of Evolutionary Adaptation to Infectious Disease
By: Pardis C. Sabeti M.D., D.Phil. (Harvard University, Cambridge, MA) © 2008 Nature Education
Citation: Sabeti, P. (2008) Natural selection: uncovering mechanisms of evolutionary adaptation to infectious disease. Nature Education 1(1):13
The evolutionary link between sickle-cell trait and malaria resistance showed that humans can and do adapt. But are the “bugs” that make us sick evolving as well?
Aa)
Aa)
Aa)
In the 1940s, J. B. S. Haldane observed that many red blood cell disorders, such as sickle-cell anemia and various thalassemias, were prominent in tropical regions where malaria was endemic (Haldane, 1949; Figure 1). Haldane hypothesized that these disorders had become common in these regions because natural selection had acted to increase the prevalence of traits that protect individuals from malaria. Just a few years later, Haldane's so-called "malaria hypothesis" was confirmed by researcher A. C. Allison, who demonstrated that the geographical distribution of the sickle-cell mutation in the beta hemoglobin gene (HBB) was limited to Africa and correlated with malaria endemicity. Allison further noted that individuals who carried the sickle-cell trait were resistant to malaria (Allison, 1954).
Figure 1: Worldwide distribution of malaria
The worldwide distribution of Plasmodium falciparum malaria in 2003.
© 2003 Modified with permission from the World Health Organization. All rights reserved.
Allison's confirmation of Haldane's hypothesis provided the first elucidated example of human adaptation since natural selection had been proposed a century earlier. Today, this and other demonstrations of natural selection help point researchers toward biological mechanisms of resistance to infectious disease. Moreover, such examples also shed light on the ways in which pathogens rapidly evolve to remain agents of human morbidity and mortality.
Selection for Malaria Resistance: A Closer Look
Since Allison and Haldane's work, the action of natural selection on genetic resistance to malaria has been shown in a multitude of contexts (Kwiatkowski, 2005). Indeed, the sickle-cell variant (i.e., the HbS allele) has been identified in four distinct genetic backgrounds in different African populations, suggesting that the same mutation arose independently several times through convergent evolution. Beyond HbS, other distinct mutations in the HBB gene have generated the HbC and HbE alleles, which arose and spread in Africa and in Southeast Asia, respectively.
The various HBB alleles aren't alone in offering protection against malaria, however. The geographic distributions of several other red blood cell disorders, including a-thalassemia, G6PD deficiency, and ovalocytosis, correlate to malaria endemicity, and the diseases also are linked to malaria resistance. An even more striking worldwide geographical difference exists for a mutation in the Duffy antigen gene (FY), which encodes a membrane protein used by the Plasmodium vivax malaria parasite to enter red blood cells. This mutation disrupts the protein, thus conferring protection against P. vivax malaria, and it occurs at a prevalence of 100% throughout most of sub-Saharan Africa yet is virtually absent outside of Africa. Moreover, through convergent evolution, an independent mutation in FY that decreases this gene's expression has also become prevalent in Southeast Asia.
So, why has malaria exerted such strong selective pressure? Scientists now know the answer. Malaria is arguably one of the human population's oldest diseases and greatest causes of morbidity and mortality. Research indicates that the malaria-causing parasite Plasmodium falciparum has occurred in human populations for approximately 100,000 years, with a large population expansion in the last 10,000 years as human populations began to move into settlements (Hartl, 2004). P. falciparum, together with the other malaria species, P. vivax, P. malariae, and P. ovale, infects hundreds of millions of people worldwide each year, and kills more than 1 million children annually (World Health Organization, 2000). Because this disease is so devastating, humans have had to evolve adaptive traits to survive in the face of this infectious condition over the past few millennia (Kwiatkowski, 2005).
Broader Implications of Natural Selection for Investigating Infectious Disease
While malaria is the
best-understood example of an infectious disease that has driven human
evolution, numerous other infectious diseases have also acted in human
populations over generations, thus allowing resistance alleles to emerge and
spread over time (Diamond, 2005). Based on historical records from the last
millennium, these diseases might include smallpox in ancient Europe and in
Native American populations, as well as cholera, tuberculosis, and bubonic
plague in Europe. Many diseases in Africa have likely been endemic for even longer, such as
numerous diarrheal diseases, yellow fever, and Lassa hemorrhagic fever.
Today, with access
to heretofore unprecedented data sets for the study of human genetic variation, researchers can exploit the genetic signatures of natural
selection using novel analytical methods. In this way, they can identify
genetic variants conferring resistance to infectious diseases that have spread
through human populations over time. These studies will help elucidate natural
mechanisms of defense and perhaps uncover novel evolutionary pressures. Moreover,
the same tools that have revolutionized the study of natural selection in
humans will also make unprecedented studies of pathogens possible.
Investigating the signatures of natural selection can help elucidate the
evolutionary adaptations that have allowed humans to withstand some of our most
complex and challenging selective agents. In particular, researchers can look
for variants that might be readily detected in genetic association studies; for
distinctive, detectable patterns of genetic variation in the human genome; and for
clues as to how pathogens themselves evolve so rapidly.
Searching for Variants via Association Studies
By driving highly
protective variants to high prevalence, natural selection produces variants
that might be readily detected in genetic association studies to help elucidate
the biological basis of disease resistance. The classic examples of host
genetic factors that play a role in resistance to malaria, such as HbS, are
some of the strongest and most robust signals of genetic susceptibility to
infectious disease (Hill, 2006). This is because natural selection acts to
increase the prevalence of highly advantageous alleles, over time generating
common resistance alleles of especially strong effect. For example, a study of
genetic susceptibility of HbS in the Gambia detected a significant level
of protection using just 315 cases and 583 controls (Ackerman et al., 2005). By studying other ancient
selective pressures in which common resistance alleles of strong effect are
acting, scientists may have the power to detect a genetic association even with
small sample sizes.
In contrast, no
single highly protective variant for emergent diseases like HIV and tuberculosis
(in Africa) would have had time to spread. For
these diseases, resistance appears to be modulated by many rare genetic
variants, most with modest protective effect, and genetic studies require
extremely large sample sizes (Hill, 2006). This is likely not a biological but, rather,
a historical difference. Indeed, hundreds of structural and regulatory
mutations exist in HBB, such as HbS,
HbE, or HbC, but in populations under malaria selective pressure, a single
highly protective variant will often dominate (Kwiatkowski, 2005). Moreover,
many variants nearby on the chromosome will rise in prevalence in the
population through genetic hitchhiking, such that other nearby linked alleles
can serve as proxies for the underlying causal allele in genetic association
studies, further enhancing researchers' ability to detect an association. Thus,
natural selection may produce important genetic resistance loci that can more
easily be detected in association studies.
Searching for Patterns of Variation
As
genetic variants conferring resistance to infectious diseases spread through
human populations over time through natural selection, they leave distinctive,
detectable patterns of genetic variation in the human genome. These signals of
selection can uncover novel resistance alleles or even novel evolutionary
pressures. Also, as previously mentioned, as
advantageous alleles under positive selection rise in prevalence, variants at nearby locations on
the same chromosome (linked alleles) also rise in prevalence. Such genetic hitchhiking leads to a "selective sweep" that alters the typical pattern of
genetic variation in the region. Selective sweeps produce numerous detectable
signals of selection (Nielsen, 2005; Sabeti et
al., 2006). As tests for selection have been applied to newly available
genetic variation data across the human genome, many of the top signals of
selection that have been identified have been at genes and alleles known to be
involved with malaria susceptibility, including HBB, FY, CD36, and HLA.These signals were
identified in just 90 individuals randomly chosen from the population, and they
could have been identified without prior knowledge of a specific variant or
selective advantage.
Surveys
of natural selection can not only identify new resistance variants for known
selective pressures, but they can also potentially uncover previously
unrecognized selective pressures. For example, in a genome survey of the Yoruba
people of Nigeria,
two of the top signals of
selection were at genes (LARGE and DMD) biologically linked to the Lassa
hemorrhagic fever virus (Sabeti et al.,
2007). While little studied, Lassa virus in fact infects many millions of West Africans,
and based on oral records and epidemiology, it is likely to be an ancient
disease (Richmond & Baglole, 2003). Researchers have documented that in
several affected West African populations, between 50% and 90% of individuals
are resistant to the virus, suggesting that protective alleles emerged at some
point (McCormick & Fisher-Hoch,
2002). This finding could open new
avenues for research and shine light on other important pathogens in human
history.
Searching for Clues about Pathogen Evolution
The same
tools that revolutionized the study of natural selection in humans are now
making unprecedented studies of pathogens possible, allowing scientists to better understand how these organisms
rapidly evolve to remain agents of human morbidity and mortality. Pathogens are
perhaps the most intriguing of all the forces shaping humans. They have had a
tremendous impact on our evolution, and they, themselves, evolve over time. The
great effect that pathogens have exerted on the human genome is demonstrated by
positive selection for traits such as sickle-cell hemoglobin (Sabeti et al., 2006). Natural human defenses have
similarly exerted strong pressures on the genomes of pathogens, as has the use
of drugs and vaccines (Volkman et al.,
2007). By studying genetic diversity in pathogens, researchers can examine how
they have evolved to avoid human immune defenses and therapeutics. Furthermore,
scientists can investigate in real time the evolutionary consequences of new
vaccines and drugs, with the goal of developing better intervention strategies.
Future Endeavors
Investigation of the
links between natural selection and disease resistance has revealed some of the
forces that have shaped our species, and the findings of these studies have
direct implications for human health. However, research thus far represents
just a first glimpse of a vast new landscape. In the years to come, new
technologies and analytic methods will enable researchers to learn even more
about the genetic basis of evolutionary adaptations that have allowed humans to
withstand a wide variety of complex and challenging selective agents.
References and Recommended Reading
Ackerman, H., et al. A comparison of case-control and family-based association methods: The example of sickle-cell and malaria. Annals of Human Genetics 69, 559–565 (2005)
Allison, A. C. Protection afforded by sickle-cell trait against subtertian malarial infection. British Medical Journal 4857, 290–294 (1954)
Diamond, J. M. Guns, Germs, and Steel: The Fates of Human Societies (New York, Norton, 2005)
Haldane, J. B. S. Disease and evolution. Ricerca Science Supplement 19, 3–10 (1949)
Hartl, D. L. The origin of malaria: Mixed messages from genetic diversity. Nature Reviews Microbiology 2, 15–22 (2004) doi:10.1038/nrmicro795 (link to article)
Hill, A. V. Aspects of genetic susceptibility to human infectious diseases. Annual Review of Genetics 40, 469–486 (2006)
Kwiatkowski, D. P. How malaria has affected the human genome and what human genetics can teach us about malaria. American Journal of Human Genetics 77, 171–192 (2005)
McCormick, J. B., & Fisher-Hoch, S. P. Lassa fever. Current Topics in Microbiology and Immunology 262, 75–109 (2002)
Nielsen, R. Molecular signatures of natural selection. Annual Review of Genetics 39, 197–218 (2005)
Richmond, J. K., & Baglole, D. J. Lassa fever: Epidemiology, clinical features, and social consequences. British Medical Journal 327, 1271–1275 (2003)
Sabeti, P. C., et al. Positive natural selection in the human lineage. Science 312, 1614–1620 (2006) doi:10.1126/science.1124309
———. Genome-wide detection and characterization of positive selection in human populations. Nature 449, 913–918 (2007) (link to article)
Volkman, S. K., et al. A genome-wide map of diversity in Plasmodium falciparum. Nature Genetics 39, 113–119 (2007) doi:10.1038/ng1930 (link to article)
World Health Organization. WHO expert committee on malaria. World Health Organization Technical Report Series 892, 1–74 (2000)
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Genome Evolution
Origins of New Genes and Pseudogenes
Microevolution
Evolutionary Adaptation in the Human Lineage
Genetic Mutation
Natural Selection: Uncovering Mechanisms of Evolutionary Adaptation to Infectious Disease
Negative Selection
Neutral Theory: The Null Hypothesis of Molecular Evolution
Sexual Reproduction and the Evolution of Sex
Speciation
Haldane's Rule: the Heterogametic Sex
Hybrid Incompatibility and Speciation
Hybridization and Gene Flow
Why Should We Care about Species?
Macroevolution
The Molecular Clock and Estimating Species Divergence
Phylogeny
Reading a Phylogenetic Tree: The Meaning of Monophyletic Groups
Trait Evolution on a Phylogenetic Tree: Relatedness, Similarity, and the Myth of Evolutionary Advancement
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11236 | https://www.amboss.com/us/knowledge/preventive-medicine/ | Preventive medicine
Last updated: January 30, 2025
Summary
Preventive medicine aims to prevent disease occurrence as well as disease progression and complications after disease onset. The five levels of preventive medicine target all stages of disease: primordial prevention (prevention of disease risk factors), primary prevention (prevention of specific diseases), secondary prevention (identification of early-stage disease, e.g., using screening tests), tertiary prevention (optimization of care for an established disease to limit progression and complications), and quaternary prevention (prevention of medical overuse). Such efforts may occur on a population level (e.g., through public health interventions) or an individual level (e.g., through routine health maintenance). Challenges to the implementation of preventive medicine include the “prevention paradox” and nonadherence. Health promotion can help individuals overcome these challenges and empower them to improve their health.
See also “Adult health maintenance,” “Adolescent health care,” and “Well-child visits.”
Register or log in , in order to read the full article.
Public health interventions
Public health interventions are actions and policies introduced by public health authorities to protect and improve the health of a population. Population-based practice focuses on the health concerns of entire populations and intervenes at the individual, community, and systemic levels.
| Population-based public health interventions | | | | |
--- ---
| | Intervention | Examples | | |
| Category | System-focused | Community-focused | Individual-focused |
| Case finding and epidemiology | Surveillance (public health): collection, analysis, and interpretation of data required for planning and evaluation of public health interventions Outreach: identifying at-risk populations and informing them about prevention, mitigation, and treatment Case finding: identifying individuals and families with risk factors and providing them with educational resources and medical referrals Health event and disease investigation: gathering data on reported threats to population health in order to develop control measures (see “Disease outbreak investigation”) Screening: for identification of asymptomatic or at-risk individuals in a population | Establishing protocols during a pandemic for sharing of case reports and incidence data between providers and local health departments | Creating hotlines during a pandemic to provide information to the general population and assist in mitigation and control measures Establishing testing centers during a pandemic to identify cases in the community | Raising awareness during a pandemic and staying vigilant regarding pertinent symptoms in patients and colleagues Monitoring local incidence and reporting data during health care worker staff meetings |
| Diagnosis and management | Referral and follow-up: directing individuals to health care resources and assessing outcomes Case management: collaborating with individuals and providers to plan, coordinate, and assess care Delegated functions: entrusting health care tasks to other authorized personnel | Developing school policies that implement new guidelines for children with asthma | Coordinating with the local chapter of the American Lung Association to organize regular asthma information events for students and parents | Creating an individualized management plan for a child recently diagnosed with asthma |
| Information, education, empowerment | Counseling: developing interpersonal rapport to facilitate coping and self-care Consultation: deliberation between health care workers and/or experts on the diagnosis or treatment in any particular case Health teaching: educating the public about health-related conditions and behaviors | Training physicians and midwives on latest research about the risks of alcohol use during pregnancy Organizing meetings between a health care worker and the staff at a childcare center to develop a protocol to prevent measles outbreaks | Developing and distributing posters that aim to reduce alcohol use by women at establishments serving alcoholic beverages Providing safer sex education at schools | Providing information about the impact of alcohol use on pregnancy for a reproductive health class in high schools Providing fertility counseling for young couples at a free health clinic |
| Partnerships and networks | Collaboration: coordinating cooperation between institutions, groups, and/or individuals directed toward a particular health care goal Community organizing: uniting individuals to identify and develop strategies to solve public health problems Coalition building: developing alliances among organizations and institutions to solve public health problems | Collaborating with senior centers to provide screening of older adults for fall risk | Introducing a home safety checklist on fall prevention to residents of a senior center | Conducting home visits to the homes of older adults to assess risks and prevent falls |
| Policy development, promotion, and enforcement | Health advocacy: promoting public health through collaboration with community stakeholders and engagement with policymakers Social marketing: deploying marketing techniques to positively influence a target audience's health behaviors Policy development and enforcement: working with decision-makers to develop laws and regulations to protect and promote public health | Drafting legislation to mitigate certain risk factors, e.g.: + Cigarette and soda taxes + Air pollution regulations + Smoke-free laws + Alcohol sales bans Carrying out an anti-bullying social media campaign in a school district | Providing free health care for persons experiencing homelessness Organizing fundraising campaigns for orphan disease research | Providing reproductive health counseling to a sexually-active teenager |
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Health promotion
Health promotion is defined as efforts that empower individuals to improve their health. These may occur on an individual or population level.
Population-level interventions
Optimizing health care access
Expanding access to health insurance (e.g., government-funded health insurance, private health insurance)
Addressing social aspects of health care (e.g., social justice in health care, health care disparities, social determinants of health)
Promoting educational efforts and health in schools (e.g., Whole School, Whole Community, Whole Child)
Mandatory reporting of factors that can impact the health and safety of others (e.g., reportable diseases, child abuse, conditions necessitating driving restrictions)
Public health legislation (e.g., laws prohibiting smoking in certain spaces, Nutritional Facts label to promote healthy eating)
Ensuring healthy environmental conditions (e.g., clean food and water)
Individual-level interventions
Regular office visits focused on preventive health and screening, i.e.:
Adult health maintenance
Adolescent health care
Well-child visit
Promoting healthy behaviors and enabling behavioral change, e.g.:
Preventive counseling and education (e.g., counseling on safer sex, photoprotective measures, lifestyle modifications for ASCVD prevention)
Smoking cessation
Treatment of alcohol use disorder
Routine vaccinations
Family planning services
Chronic disease management
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Primordial prevention
Definition: actions that aim to prevent the development of risk factors for disease
Target: entire population
Goal: encourage positive and discourage negative lifestyle habits
Strategies
Health promotion
Mass education
Legislation
Examples
Programs on food safety and nutrition guidelines
Campaigns discouraging tobacco and drug use (e.g., smokefree air laws in public buildings)
Building bicycle lanes and sidewalks to promote physical activity
Primordial prevention aims to prevent risk factors from developing, whereas primary prevention targets existing risk factors to prevent the onset of a disease.
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Primary prevention
Definition: actions targeted at preventing specific diseases from occurring to decrease the incidence and, subsequently, the prevalence of those diseases
Target: entire population or selected groups of individuals without disease who are at increased risk of disease
Goal: decrease incidence and, in turn, prevalence of a specific disease
Strategies
Health promotion (health interventions, lifestyle modifications)
Environmental modifications (e.g., work safety)
Specific protection interventions (immunizations, chemoprophylaxis, safety of drugs and food)
Examples
Immunization
Lifestyle modification (e.g., smoking cessation to reduce lung cancer risk, exercise to reduce the risk of heart disease, dental care to reduce the risk of tooth loss)
Fortification of salt with iodine to prevent iodine deficiency
Fluoridation of toothpaste, water, and salt to reduce the risk of dental conditions
Fortification of food with folic acid to reduce the prevalence of neural tube defects
Health legislation (e.g., seat belt laws, food safety standards, traffic laws)
Routine neonatal ophthalmic antibiotic prophylaxis to prevent neonatal gonococcal conjunctivitis
Postexposure prophylaxis
For further information, see:
“Adult health maintenance”
“Anticipatory guidance for children”
“Preventive health recommendations for adolescents”
“Prenatal patient education”
Primary prevention helps to Prevent disease.
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Secondary prevention
Definition: actions targeted at early detection of disease in asymptomatic individuals to promote timely intervention, preventing disease progression and/or improving the chance of treatment success
Target: individuals at risk for disease
Goal: detect disease and initiate treatment early to improve the chance of treatment success
Strategy: two-step process
Screening test to identify disease
Follow-up for disease management
For further information on screening, see:
“Adult health maintenance screening recommendations”
“Adolescent health screening”
“Pediatric screening recommendations”
“Prenatal care”
“Preventive health care for transgender individuals”
Screening complements diagnostics, but it is not a substitute.
Secondary prevention helps Screen.
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Tertiary prevention
Definition: actions taken to optimize the care of patients with an existing disease to improve well-being and prevent complications
Target: individuals with an established disease
Goal: decrease morbidity and mortality by preventing disease progression and reducing the risk of relapse
Strategies
Chronic disease management
Strategies to improve treatment adherence
Examples
Adjuvant therapy to reduce the risk of cancer recurrence (e.g., tamoxifen in breast cancer)
Blood pressure management (e.g., antihypertensives) to reduce the risk of a cardiovascular event
Diabetes management (e.g., antidiabetic drugs, HbA1c monitoring) to reduce the risk of chronic kidney disease and/or cardiovascular events
Measures to prevent recurrent myocardial infarction (e.g., low-dose aspirin)
Tertiary prevention helps Treat.
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Quaternary prevention
Definition: actions taken to avoid medical interventions in which the harms may outweigh the benefits
Target: individuals with the potential to undergo unnecessary or inappropriate diagnostic and therapeutic interventions
Goal: prevent medical overuse (i.e., overdiagnosis and overtreatment) and avoidable iatrogenesis
Strategies
Practicing evidence-based medicine
Using shared decision-making before initiating medical interventions
Educating health care providers and patients on potential harms of unnecessary interventions (e.g., through initiatives such as the Choosing Wisely campaign)
Avoidance of exploitation of medical conditions to sell products and/or procedures
Examples
Inappropriate investigations leading to overdiagnosis: the detection of disease that would never manifest clinically or result in death
Screening individuals outside of the recommended age range (e.g., cervical cancer screening in individuals > 65 years of age without risk factors).
MRI for lower back pain of < 6 weeks duration or without red flag features
Unnecessary examinations (e.g., routine pelvic examination before prescribing contraception)
Unnecessary or excessive treatment (overtreatment)
Antibiotics for viral infections
Use of opioids in chronic noncancer pain
Antibiotics for acute mild to moderate sinusitis, as this may lead to antibiotic resistance
Avoidable iatrogenesis
Use of the combined hormonal contraception in patients with absolute contraindications (e.g., migraine with aura)
Continuation of antidiabetic drugs (e.g., sulfonylureas) despite patients experiencing adverse effects (e.g., multiple hypoglycemic events)
QUaternary prevention helps sQUeeze out unnecessary interventions.
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Prevention paradox
Definition
A preventive measure that benefits a population as a whole will offer little benefit to each individual member of that population (population approach to prevention; primordial and primary prevention)
A preventive measure that benefits a group of individuals susceptible to a particular disease will offer little benefit to the population as a whole (high-risk approach to prevention; secondary and tertiary prevention).
The high-risk approach and the population approach to prevention are complementary, but preventive medicine should prioritize preventing the underlying causes of disease (primordial and primary prevention) over reducing the impact of disease after it occurs (secondary and tertiary prevention).
The prevention paradox may lead to the misconception that a measure that provides no immediate benefit to the individual, provides no benefit to the entire population and that a small risk involved in a measure (e.g., vaccination) outweighs the benefits of that measure.
Misconceptions derived from the prevention paradox may negatively affect epidemiological policy as well as adherence in the population.
Primordial and primary prevention require consistent, long-term education programs for health care professionals as well as the general population to be effective.
Examples
While heavy drinking carries a greater risk than moderate drinking, moderate drinking has a greater negative impact on the general population because the number of moderate drinkers is greater than that of heavy drinkers.
Seatbelt laws have prevented many severe injuries, yet the overall risk of dying in an accident due to not wearing a seatbelt is still low.
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Disease outbreak investigation
Definitions
Disease cluster: an unusual aggregation, real or perceived, of cases of a disease that are grouped together in time and space
Disease outbreak: the sudden occurrence of more cases of a disease than expected in a given area, population, and/or season
Identifying outbreaks
Outbreaks are identified by health authorities through reports that may come from hospitals, laboratories, health care providers, and even the general population.
After receiving an initial report, health authorities decide on whether to investigate further.
Investigating outbreaks
Field investigation involves identifying a potential disease outbreak, forming a hypothesis about its cause, gathering data to test the hypothesis, and finally, developing and implementing control and prevention measures.
Steps of the field investigation usually include:
Preparing for the investigation
Gathering information about the condition, area, and/or population of interest
Organizing a team and necessary supplies
Confirming the outbreak
Establishing a case definition
Finding cases and documenting details of each case
Developing a hypothesis about the cause of the outbreak
Analyzing the gathered data and evaluating the hypothesis
Comparing and reconciling with laboratory and/or environmental studies
Developing and implementing control and prevention measures
Initiating or maintaining surveillance.
Communicating the findings to the public
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School health policies
The Whole School, Whole Community, Whole Child (WSCC) model is a student-centered framework developed by the CDC for addressing health in schools.
WSCC consists of ten components:
Physical education and physical activity
Physical activity recommendations are outlined in the Comprehensive School Physical Activity Program, which consists of five components: physical education, physical activity during school, physical activity before and after school, staff involvement, and family and community engagement.
Physical education is an academic subject for all K-12 students, which helps to develop motor skills, knowledge, and behaviors for a healthy lifestyle.
Nutrition environment and services
Food provided to students must meet the Nutrition Standards for School Meals issued by the USDA.
Free drinking water must be available to all students throughout the day.
The National School Lunch and Breakfast Programs provide nutritious, low-cost or free meals to children each school day.
Health education: Integration of health education into the curriculum involves addressing topics such as nutrition, mental health, sexual health, violence prevention, and the use of alcohol, tobacco, and other substances in a variety of subjects and educational settings.
Social and emotional climate: A positive emotional climate is essential for proper child development and should be maintained in all aspects of school functioning.
Physical environment: School buildings and surrounding environment should be safe from any health threat (e.g., traffic, crime, construction, improper ventilation).
Health services: Services from qualified health professionals (e.g., nurses, physicians, physician assistants) should be available to all students.
Counseling, psychological, and social services: prevention and intervention to support the mental, behavioral, and social health of students (i.e., with certified school counselors, school psychologists, and school social workers)
Employee wellness: School employees should have access to programs, policies, and benefits that promote their health and well-being.
Community involvement: partnerships with local groups, organizations, and businesses to encourage students' civic engagement and to share school resources (e.g., school-based health centers and sports facilities) with the community
Family engagement: continuous involvement of students' families in educational activities and development
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11237 | https://support.minitab.com/en-us/minitab/help-and-how-to/statistics/basic-statistics/supporting-topics/basics/type-i-and-type-ii-error/ | What are type I and type II errors? - Minitab
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What are type I and type II errors?
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No hypothesis test is 100% certain. Because the test is based on probabilities, there is always a chance of making an incorrect conclusion. When you do a hypothesis test, two types of errors are possible: type I and type II. The risks of these two errors are inversely related and determined by the level of significance and the power for the test. Therefore, you should determine which error has more severe consequences for your situation before you define their risks. Type I error When the null hypothesis is true and you reject it, you make a type I error. The probability of making a type I error is α, which is the level of significance you set for your hypothesis test. An α of 0.05 indicates that you are willing to accept a 5% chance that you are wrong when you reject the null hypothesis. To lower this risk, you must use a lower value for α. However, using a lower value for alpha means that you will be less likely to detect a true difference if one really exists. Type II error When the null hypothesis is false and you fail to reject it, you make a type II error. The probability of making a type II error is β, which depends on the power of the test. You can decrease your risk of committing a type II error by ensuring your test has enough power. You can do this by ensuring your sample size is large enough to detect a practical difference when one truly exists.
The probability of rejecting the null hypothesis when it is false is equal to 1–β. This value is the power of the test.
Truth about the population
Decision based on sampleH 0 is true H 0 is false
Fail to reject H 0 Correct Decision (probability = 1 - α)Type II Error - fail to reject H 0 when it is false (probability = β)
Reject H 0Type I Error - rejecting H 0 when it is true (probability = α)Correct Decision (probability = 1 - β)
Example of type I and type II error
To understand the interrelationship between type I and type II error, and to determine which error has more severe consequences for your situation, consider the following example.
A medical researcher wants to compare the effectiveness of two medications. The null and alternative hypotheses are:
Null hypothesis (H 0): μ 1= μ 2
The two medications are equally effective.
Alternative hypothesis (H 1): μ 1≠ μ 2
The two medications are not equally effective.
A type I error occurs if the researcher rejects the null hypothesis and concludes that the two medications are different when, in fact, they are not. If the medications have the same effectiveness, the researcher may not consider this error too severe because the patients still benefit from the same level of effectiveness regardless of which medicine they take. However, if a type II error occurs, the researcher fails to reject the null hypothesis when it should be rejected. That is, the researcher concludes that the medications are the same when, in fact, they are different. This error is potentially life-threatening if the less-effective medication is sold to the public instead of the more effective one.
As you conduct your hypothesis tests, consider the risks of making type I and type II errors. If the consequences of making one type of error are more severe or costly than making the other type of error, then choose a level of significance and a power for the test that will reflect the relative severity of those consequences.
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11238 | https://math.stackexchange.com/questions/1862793/open-interval-0-1-with-the-usual-topology-admits-a-metric-space | real analysis - Open interval $(0,1)$ with the usual topology admits a metric space - Mathematics Stack Exchange
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Open interval (0,1)(0,1) with the usual topology admits a metric space
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which of the following is/are true ?
(0,1)(0,1) with the usual topology admits a metric which is complete .
(0,1)(0,1) with the usual topology admits a metric which is not complete.
[0,1][0,1] with the usual topology admits a metric which is not complete.
[0,1][0,1] with the usual topology admits a metric which is complete.
This question has come at my competetive exam. I think this is a wrong question, because completeness a metric space property not a topological space property.In the offical answer key, answer has given (1) and (4), I want to send my representation. So please check my representation. Thank you
Let X=(0,1)X=(0,1) and d d is a Euclidean metric on X X which induces the usual topology on X X and a sequence {1 n}{1 n} is a cauchy sequence in the Euclidean metric , but not converges in X X. So X X is not complete withbthe usual topology admit a Euclidean metric.
On the other hand
The map
f:(0,1)→R:x↦tan π(x−1 2)f:(0,1)→R:x↦tanπ(x−1 2)
is a bijection which allows you to define the metric
d(x,y)=|f(x)−f(y)|d(x,y)=|f(x)−f(y)|
which makes ((0,1),d)((0,1),d) complete. Since f f maps intervals to intervals then both topologies are equivalent.
So Completeness is not a topological property. So this is irrelivent.
I would be thankful, if some one check my representation
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asked Jul 18, 2016 at 3:59
user120386user120386
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1 Um, the question doesn't say "complete topology"; it says "complete metric". So there's nothing wrong with the questions.fleablood –fleablood 2016-07-18 04:02:32 +00:00 Commented Jul 18, 2016 at 4:02
1 2 2 and 4 4 are clearly true.Asinomás –Asinomás 2016-07-18 04:03:32 +00:00 Commented Jul 18, 2016 at 4:03
The map f is irrelevant as that admits a different metric. The questions are specifically asking about the euclidean metric.fleablood –fleablood 2016-07-18 04:05:45 +00:00 Commented Jul 18, 2016 at 4:05
@ fleablood :There are two metric induced the same topology and one is complete other is incomplete. So 1 and 2 are true.user120386 –user120386 2016-07-18 04:07:58 +00:00 Commented Jul 18, 2016 at 4:07
In the offical answer key , answer is given 1 and 4 user120386 –user120386 2016-07-18 04:09:13 +00:00 Commented Jul 18, 2016 at 4:09
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I think it is a legit question to ask if a topological space admits a metric.
If I give you a topological space and ask you if there is a metric (with certain properties) which induces this topology there is nothing wrong about it, right? (If I got the word admit right...)
It is right that completeness may does not make sense on any space, but if I give you a space I can ask you if it makes sense. There is a broader class of spaces in which it make sense. If you are interested you can read about uniform spaces.
Now why are 1 and 4 correct? The space in 4 is a closed subspace of a complete space (R R) which is complete. And for 1 you need that (0,1)(0,1) is homeomorphic to R R which is complete.
Clearly to show that 2 is correct you can show that (0,1)(0,1) with the usual topology is not complete, just choose a Cauchy sequence which converges to 0 0.
Most interesting is 3. You have to know that [0,1][0,1] with any metric inducing the standard topology is complete. Here you can use the fact that any compact metric space ist complete, since compactness is clearly a topological property.
So either I got the question wrong or 2 is also correct. In the second case you can have a look at Completely metrizable space which are topological spaces whose topology is/(can be) induced by a complete metric.
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answered Jul 18, 2016 at 14:58
user60589user60589
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1Is the metric d(x,y)1+d(x,y)d(x,y)1+d(x,y) complete where d d is the usual Euclidean metric on R 2 R 2
0(0,1)(0,1) with the usual topology admits a metric which is complete?
2choose the complete metric space?
0To check that under a constructed complete metric on (0,1)(0,1) the sequence 1/n 1/n is not cauchy
0Completeness of a metric with discrete topology
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How to Find Vertical Asymptotes of a Rational Function
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A simple guide to find and graph vertical asymptotes
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Grace Imson, MA
Last Updated: March 10, 2025
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Finding Vertical Asymptotes
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This article was reviewed by Grace Imson, MA. Grace Imson is a math teacher with over 40 years of teaching experience. Grace is currently a math instructor at the City College of San Francisco and was previously in the Math Department at Saint Louis University. She has taught math at the elementary, middle, high school, and college levels. She has an MA in Education, specializing in Administration and Supervision from Saint Louis University.
There are 7 references cited in this article, which can be found at the bottom of the page.
This article has been viewed 393,671 times.
A rational function is a mathematical function (equation) that contains a ratio between two polynomials.
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Research source That is, there must be some form of a fraction, involving more than just the coefficients. Thus, is not a rational function, because the only fraction is a coefficient term. However, is a rational function. A vertical asymptote is a representation of values that are not solutions to the equation, but they help in defining the graph of solutions.
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Finding Vertical Asymptotes
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Factor the denominator of the function. To simplify the function, you need to break the denominator into its factors as much as possible. For the purpose of finding asymptotes, you can mostly ignore the numerator.
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For example, suppose you begin with the function . The denominator can be factored into the two terms .
As another example, consider the function . You should recognize the denominator as a simple quadratic function, which can be factored into .
Recognize that some denominator functions may not be able to be factored. For example, in the equation , the function in the denominator, cannot be factored. For this first step, you will just have to leave it in that form.
If you need to review factoring of functions, check out the articles Factor Algebraic Equations or Factor Second Degree Polynomials (Quadratic Equations).
2. 2
Find values for which the denominator equals 0. Still disregarding the numerator of the function, set the factored denominator equal to 0 and solve for x. Remember that factors are terms that multiply, and to get a final value of 0, setting any one factor equal to 0 will solve the problem. Depending on the number of factors that exist, you may find one or more solutions.
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For example, if a denominator function factored as , then you would set this equal to 0 as . The solutions will be any values of x that make this true. To find those values, set each individual factor equal to 0, to create two mini-problems of and . The first solution is and the second is .
Given another example with a denominator of , this could be factored into the two terms . Setting each factor equal to 0 leads to and . Therefore, the solutions for this problem would be and .
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3. 3
Understand the meaning of the solutions. The work you have done to this point identifies values of x for which the denominator of the function equals 0. Recognize that a rational function is really a large division problem, with the value of the numerator divided by the value of the denominator. Because dividing by 0 is undefined, any value for x for which the denominator will equal 0 represents a vertical asymptote for the full function.
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Graphing Vertical Asymptotes
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Review the meaning of a graph. A graph of a function is a visual representation of the values of x and y that are solutions to a given equation. The graph may consist of individual points, a straight line, a curved line, or even some closed figures like a circle or an ellipse. Any point that lies on the line could be a solution to the equation.
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For example, a simple equation like will have infinite solutions. Written in pairs of (x,y), some possible solutions are (1,2), (2,4), (3,6), or any pair of numbers in which the second number is double the first. Plotting these points on the x,y coordinate plane will show a continuous straight line that appears as a diagonal that goes upward from left to right. To see more samples of this type of graph, you may want to review Graph Linear Equations.
A graph of a quadratic equation is one that has an exponent of 2, such as . Some sample solutions are (-1,-2), (0,-1), (1,1), (2,7). If you plot these points, and others, you will find the graph of a parabola, which is a u-shaped curve. To review this type of graph, you can look at Graph a Quadratic Equation.
If you need more help reviewing how to graph functions, read Graph a Function or Graph a Rational Function.
2. 2
Recognize asymptotes. An asymptote is a straight line that generally serves as a kind of boundary for the graph of a function. An asymptote can be vertical, horizontal, or on any angle. The asymptote represents values that are not solutions to the equation, but could be a limit of solutions.
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For example, consider the equation . If you begin at the value x=3 and count down to select some solutions for this equation, you will get solutions of (3, 1/3), (2, 1/2), and (1,1). If you continue counting down, the next value for x would be 0, but this would create the fraction y=1/0. Because division by 0 is undefined, this cannot be a solution to the function. Therefore, the value of x=0 is a vertical asymptote for this equation.
3. 3
Graph vertical asymptotes with a dotted line. Conventionally, when you are plotting the solution to a function, if the function has a vertical asymptote, you will graph it by drawing a dotted line at that value. In the example of , this would be a vertical dotted line at x=0.
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State the equation of asymptote for the graph y=2^3-x -4
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I believe you copied your problem incorrectly. The equation you gave simplifies to y=8-x-4, or y=-x+4. This is a simple straight line, with a slope of -1 and a y-intercept at 4. There is no asymptote for this. Check your problem again.
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About This Article
Reviewed by:
Grace Imson, MA
Math Teacher
This article was reviewed by Grace Imson, MA. Grace Imson is a math teacher with over 40 years of teaching experience. Grace is currently a math instructor at the City College of San Francisco and was previously in the Math Department at Saint Louis University. She has taught math at the elementary, middle, high school, and college levels. She has an MA in Education, specializing in Administration and Supervision from Saint Louis University. This article has been viewed 393,671 times.
38 votes - 65%
Co-authors: 11
Updated: March 10, 2025
Views: 393,671
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Reviewed by:
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Math Teacher
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11241 | https://www.sciencedirect.com/science/article/pii/S0925772111000770 | 3D Euler spirals for 3D curve completion - ScienceDirect
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Abstract
Keywords
1. Introduction
2. Related work
3. Background
4. 3D Euler spirals
5. Properties of 3D Euler spirals
6. Curve construction algorithm
7. Results and applications
8. Conclusion
Acknowledgements
References
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Computational Geometry
Volume 45, Issue 3, April 2012, Pages 115-126
3D Euler spirals for 3D curve completion
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Abstract
Shape completion is an intriguing problem in geometry processing with applications in CAD and graphics. This paper defines a new type of 3D curve, which can be utilized for curve completion. It can be considered as the extension to three dimensions of the 2D Euler spiral. We prove several properties of this curve – properties that have been shown to be important for the appeal of curves. We illustrate its utility in two applications. The first is “fixing” curves detected by algorithms for edge detection on surfaces. The second is shape illustration in archaeology, where the user would like to draw curves that are missing due to the incompleteness of the input model.
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Keywords
Euler spirals
3D curves
1. Introduction
3D curves convey important information about the shape. They are significant in modeling, in non-photo realistic rendering, and in a variety of mesh analysis algorithms , , . There exist several curve detection algorithms, which produce appealing results , , , . However, sometimes the resulting curves appear broken due to the inability of the algorithm to detect curves at noisy surface patches. Moreover, in some applications the objects are broken, and naturally, curves are missing there. In these cases the user wishes to complete the broken curves or create additional curves from scratch.
Shape completion has been an important task in computational geometry with applications to CAD and computer graphics, , . While most of the work has focused on completing or repairing polyhedra and CAD models, this paper focuses on completing curves in three dimensions. It presents a practical solution to the problem, which is demonstrated by real-life data.
Given two point–tangent pairs, one way to complete them is to use any of the variety of polynomial curves, such as splines , or Pythagorean Hodograph, . Such curves possess many attractive properties, however, Fig. 1(b) illustrates that they (in this case Hermite splines) might not always produce the preferable results. This is also supported by psychological studies that indicate that splines may be unsatisfactory for curve completion .
1. Download: Download full-size image
Fig. 1. 3D Euler spirals (red) complete the curves on a broken Hellenistic oil lamp – curves that would most likely be drawn if the model were complete. The scale of the Hermite splines is determined manually (magenta), since the automatically-scaled splines (green) are inferior due to the large ratio between the length of the curve and the size of the model. Note the perfect circular arcs of our curves. (For interpretation of the references to color in this figure, the reader is referred to the web version of this article.)
This paper defines a new type of 3D curves that can be used for this purpose (Fig. 1(a)). We show that our curves are not only appealing, but also qualitatively outperform some splines. In a nutshell, our curves can be considered as an extension to 3D of the planar Euler spirals. An important consideration in aesthetic curve design is the curveʼs fairness , which has been shown to be closely related to how little and how smoothly a curve bends. An Euler spiral, also referred to as a clothoid or a Cornu spiral, is an example of such an aesthetic curve. Its curvature varies linearly with arc-length , , . Our proposed curve has both its curvature and torsion change linearly with length.
The contribution of this paper is threefold. First, the paper defines the 3D Euler spiral (Section 4) and proves that it satisfies some desirable properties – properties that have been claimed to produce eye-pleasing curves (Section 5). In particular, we prove that our curves are invariant to similarity transformations and that they are symmetric, extensible (i.e., refinable), smooth, and round (i.e., if the boundary conditions lie on a circle, then a circle is the limiting case of an Euler spiral that satisfies the boundary conditions, as illustrated in Fig. 1(a)). Second, we present a parameter-less algorithm for computing these curves (Section 6). Last but not least, we demonstrate the use of these curves in two curve completion applications (Section 7).
In the first application, our spirals complete curves that are detected on surfaces using curve detection algorithms. The curves often have missing segments due to “weak” surfaces patches. The second application is curve completion of broken shapes, such as archaeological artifacts. Currently, this task is performed by drawing the missing curve segments manually in 2D. This is an expensive and time-consuming process, which is prone to biases. Our curves can replace this manual task while performing it in 3D, directly on the scanned artifact. With the growing popularity of digital documentation in archaeology, the ability to draw curves in 3D is becoming ever more important.
A preliminary version of this paper was published in the Annual Symposium on Computational Geometry (SoCG) 2010 , accompanied by a video that complements it .
2. Related work
2.1. Curve completion
Given two point–tangent pairs, the most common way to perform completion is to use splines, such as a cubic Hermite spline , . Splines are fast and easy to compute, but they are not always the curves preferred by the human visual system.
Ullman suggests properties that 2D curves should satisfy: invariance to rigid transformations, smoothness, minimization of the total curvature, and extensibility. These properties led to a Biarc solution – a curve consisting of two circular arcs. Biarcs are smooth and invariant to rigid transformations, but they do not guarantee that the total curvature is minimized and they are not extensible . Biarcs are generalized to 3D in , .
Knuth proposes different properties of eye-pleasing 2D curves through a set of points: invariance to similarity transformations and cyclic permutations (for closed curves), extensibility, smoothness, roundness, and being locally constructed. Knuth also shows that the latter four properties cannot be simultaneously satisfied. These properties are the base of the METAFONT system of LATEX. , , propose to use 2D cubic splines, giving up extensibility and roundness.
Another way to construct curves is Elastica , , , , which refers to the curves that minimize the total square curvature of the curve:E[κ(s)]=∫0 L κ 2(s)d s, where s∈[0,L] is the arc-length parameter and κ(s) is the curvature. Horn argues that Elastica is the “smoothest” 2D curve to complete the gap between two point–tangent pairs. It is extensible, but neither scale-invariant nor round. Already in 1906 discussed Elastica for 3D curves. A curve that approximates the solution to Elastica in 3D is explicitly defined in . In 2D this curve is the 2D Euler spiral (discussed next), which has a linear curvature. In 3D, the curvature of this curve is expressed as a hyperbola.
2.2. Spirals in graphics applications
A variety of spirals have been investigated in computer graphics both in 2D and in 3D. They include logarithmic spirals , Helispirals , and Euler spirals , . 3D Logarithmic spirals have a linear radius of curvature and a linear radius of torsion. Helispirals extend the 2D Archimedean spirals to 3D. In cylindrical coordinates, the radius length and the z coordinate of the spiral depend linearly on the angle. We focus on Euler spirals.
2.3. 2D Euler spirals
Euler spirals are curves whose curvature evolves linearly along the curve. They were discovered independently by three researchers . In 1694 Bernoulli wrote the equations for the Euler spiral for the first time, but did not draw the spirals or compute them numerically. In 1744 Euler rediscovered the curveʼs equations, described their properties, and derived a series expansion to the curveʼs integrals. Later, in 1781, he also computed the spiralʼs end points. The curves were re-discovered in 1890 for the third time by Talbot, who used them to design railway tracks. Euler spirals are also known as “Cornu spirals” (after Cornu who plotted them) and “Clothoid” (after Clotho, the youngest of the three Fates of Greek mythology). They are defined as the curves that penalize the curvature variation, hence minimizing the following (κ s is the derivative of κ):E[κ(s)]=∫0 L κ s 2(s)d s.
In psychological experiments show that in 2D an interpolation between two point–tangent pairs using Euler spiral outperforms parabolic curves and circular arcs. This is attributed to its monotonous change in curvature, which has a good fit to the way the human eye interpolates curves.
2D Euler spirals are used in computer aided design. In , they are used as an approximation to the solution of Elastica. In the conditions under which the spirals can form a transition curve are investigated. Two spirals are used in to form a parabola-like segment between consecutive points of a control polygon. In a formulation for fitting spiral primitives to a dense polyline data is developed.
In an algorithm is described for 2D curve completion using an Euler spiral. The algorithm is an iterative gradient-descent, initialized by a 2D Biarc. Since there are infinite possible Biarcs, the Biarc that minimizes the total curvature variation is chosen. Some properties that characterize eye-pleasing curves are also proved. In a faster and more accurate algorithm is proposed. It is proved that given two point–tangent pairs, there always exists an Euler spiral that interpolates them.
2.4. 3D Euler spirals
An attempt to generalize Euler spirals to 3D, maintaining the linearity of the curvature, is presented in . A given polygon is refined, such that the polygon satisfies both arc-length parameterization and linear distribution of the discrete curvature binormal vector. The algorithm ignores the torsion, despite being an important characteristic of 3D curves.
We propose a novel algorithm, which produces continuous, rather than discrete, curves and takes both curvature and torsion into account. The algorithm, which is inspired by , is general and does not require an initial polygon. We prove that our curve satisfies properties that characterize fair and appealing curves and reduces to the 2D Euler spiral in the planar case.
3. Background
A spatial curve C(s) is determined by its curvature κ(s) and its torsion τ(s). Intuitively, a curve can be obtained from a straight line by bending (curvature) and twisting (torsion).
This section reviews the Frenet–Serret equations and the Euler–Lagrange equations, which will be necessary in the derivation of our 3D Euler spirals. In the following, T→(s)=d C d s(s) is the unit tangent vector, N→(s) is the unit normal vector, and B→(s)=T→(s)×N→(s) is the binormal vector. We assume an arc-length parameterization.
3.1. Frenet–Serret equations
Given a curvature κ(s)>0 and a torsion τ(s), according to the fundamental theorem of the local theory of curves , there exists a unique (up to rigid motion) spatial curve, parameterized by the arc-length s, defined by its Frenet–Serret equations, as follows:(1)d T→(s)d s=κ(s)N→(s),d N→(s)d s=−κ(s)T→(s)+τ(s)B→(s),d B→(s)d s=−τ(s)N→(s).
The curve C is defined by:C(s)=∫0 s T→(v)d v+x 0=∫0 s[∫0 t d T→(u)d u d u+T→0]d t+x 0.
3.2. Euler–Lagrange equation
The Euler–Lagrange Equation is fundamental in calculus of variations. It is a differential equation, useful for solving optimization problems in which, given some functional, one seeks the function that optimizes it. It is satisfied by a function q of a real argument s, which is a stationary point of the functional(2)S(q)=∫s 1 s 2 L(s,q(s),q′(s))d s, where q=(q 1,…,q n) is the function to be found, q′=(q 1′,…,q n′), q i′=d q i d s, i=(1,…,n), and the positions q(s 1) and q(s 2) are defined.
The function q that optimizes Eq. (2) satisfies the Euler–Lagrange equations:(3)d d s(∂L∂q i′)−∂L∂q i=0(i=1,…,n).
4. 3D Euler spirals
This section defines the 3D Euler spiral – the curve having both its curvature and torsion evolve linearly along the curve (Fig. 2). Furthermore, we require that our curve conforms with the definition of a 2D Euler spiral. We start with some intuition, then define the 3D Euler spiral, and finally prove its existence and uniqueness up to a rigid transformation.
1. Download: Download full-size image
Fig. 2. A 3D Euler spiral.
We seek a functional that will penalize the change in curvature and torsion along the curve. Thus, the curve should minimize the sum of the square variation of the curvature and the torsion. Formally, we require that the following integral be minimized:(4)S((κ,τ))=∫0 L[κ s 2(s)+τ s 2(s)]d s, where L is the curveʼs length, κ s=∂κ∂s, and τ s=∂τ∂s. Note that in the planar case τ=0, therefore our definition indeed conforms with the definition of the 2D Euler spiral.
Minimizing Eq. (4) can be performed using the Euler–Lagrange equation. In our case, Eq. (4) corresponds to Eq. (2) as follows:q 1(s)↦κ(s),q 1′(s)↦κ s(s),q 2(s)↦τ(s),q 2′(s)↦τ s(s),L(s,q(s),q′(s))↦[κ s 2(s)+τ s 2(s)].
Hence, the corresponding Euler–Lagrange equations are (by Eq. (3)):κ:d d s(∂(κ s 2+τ s 2)∂κ s)=0⇒d d s(2 κ s)=0⇒κ s s=0,τ:d d s(∂(κ s 2+τ s 2)∂τ s)=0⇒d d s(2 τ s)=0⇒τ s s=0.
By integrating κ s s and τ s s twice, these equations lead to a curve whose curvature and torsion evolve linearly. Thus, for some constants κ 0,τ 0,γ,δ∈R, and for 0⩽s⩽L:(5)κ(s)=κ 0+γ s,τ(s)=τ 0+δ s.
In summary, we have shown that the curve that minimizes our functional (Eq. (4)) is a curve whose curvature and torsion change linearly along the curve. Thus, we can define our curve as follows:
Definition 4.1 3D Euler spiral
The 3D curve whose curvature and torsion evolve linearly with arc-length.
Note that since our curve has a linear relation between the curvature and torsion, it is a special case of Bertrand curves . This, however, neither helps in deriving the properties proved in Section 5 (which do not hold for general Bertrand curves) nor provides a method for constructing them (Section 6).
Finally, the following proposition proves the existence of this curve and its uniqueness up to a rigid transformation.
Proposition 4.1
Given constants κ 0,τ 0,γ,δ,∈R , there exists a 3D Euler spiral having a linear curvature κ(s)=κ 0+γ s and a linear torsion τ(s)=τ 0+δ s . Moreover, this curve is unique up to a rigid transformation.
Proof
By definition of the curvature of curves in R 3, κ(s)=|d 2 C→d s 2(s)|⩾0. According to the fundamental theorem of local theory of curves, for every differential function with κ(s)>0 and τ(s), there exists a regular parameterized curve, where κ(s) is the curvature, τ(s) is the torsion, and s is the arc-length parameterization . Moreover, any other curve satisfying the same conditions, differs by a rigid motion.
If κ(s)=0∀s, the curve is a straight line, which is unique up to a rigid transformation. It is also possible that κ(s)=0 for a single point. In this case, the tangent and hence the curve are well-defined at this point, which is the inflection point at which the normal switches directions. (Note that by our definition κ(s) may be negative. In this case, we consider |κ(s)| and regard the switch of the sign as a change of the normal direction.) □
5. Properties of 3D Euler spirals
The aesthetics of curves has been studied in a variety of papers , , . In addition to having its curvature and torsion change linearly – a property acknowledged to characterize eye-pleasing curves – this section proves that our 3D Euler curves also hold the following properties.
1.Invariance to similarity transformations (translation, rotation, and scaling).
2.Symmetry: The curve leaving a point x 0 with tangent T→0 and reaching a point x f with tangent T→f, coincides with a curve leaving x f with tangent −T→f and reaching x 0 with tangent −T→0.
3.Extensibility: For every point x m∈C between points x 0 and x f, curves C 1 between x 0 and x m and C 2 between x m and x f coincide with C, each in its own section.
4.Smoothness: The tangent is defined at every point, i.e., ∂C∂s is finite. (In fact, our curves are C∞-smooth.)
5.Roundness: Given two point–tangent pairs lying on a circle, the circle that satisfies the initial conditions is an Euler spiral. The importance of this property is demonstrated in Fig. 1, Fig. 5, Fig. 6(a), where our spirals are both appealing and correct, since the boundary conditions indicate completion by a circular arc. Constructively, it is possible to identify the circularity of the initial conditions and construct the circular sought-after Euler spiral.
Proposition 5.1
A 3D Euler spiral is invariant to similarity transformations.
Proof
Invariance to rotation and translation results from Proposition 4.1. Scaling a curve by a factor λ scales the arc-length between points by a factor λ, while both the curvature and the torsion are scaled by a factor 1/λ. Therefore, the linear dependence of both κ(s) and τ(s) on the arc-length is preserved. □
Proposition 5.2
A 3D Euler spiral is symmetric.
Proof
We are given a 3D Euler spiral C that interpolates the point–tangent pairs (x 0,T→0) and (x f,T→f) and has parameters κ 0,τ 0,γ,δ,L. We need to show that C s y m (parameterized backward) is a 3D Euler spiral that interpolates the point–tangent pairs (x f,−T→f) and (x 0,−T→0) and coincides with C.
By definition, the reverse parametrization of a curve maps the arc-length parameter s to L−s and leaves both the curvature and the torsion unaffected. Therefore, the linear dependence of κ(s) and τ(s) on the arc-length is preserved. Moreover, since the tangent vectors are multiplied by (−1), the inverse parametrization creates a spiral that matches the reversed boundary conditions. □
Proposition 5.3
A 3D Euler spiral is extensible.
Proof
The linear dependence of the curvature and the torsion on the arc-length does not rely upon the interval in which the curve is considered. Thus, if C is a 3D Euler spiral interpolating the point–tangent pairs (x 0,T→0) and (x f,T→f), for any point–tangent pair (x m,T→m) taken from the curve segmentC, the restriction of the curve C to a smaller segment will be an interpolating 3D Euler spiral for the boundary data (x 0,T→0) and (x m,T→m). The same applies to the boundary data (x m,T→m) and (x f,T→f). □
Proposition 5.4
A 3D Euler spiral is smooth.
Proof
According to Proposition 4.1, there exists a solution for the Frenet–Serret equations. Therefore, ∂C∂s=T→(s) is defined for every 0⩽s⩽L. □
Proposition 5.5
A 3D Euler spiral is round.
Proof
For given two point–tangent pairs lying on a circle, the circle defined by κ 0≠0, τ 0=0, γ=0, δ=0 is a solution for the Frenet–Serret equation. □
6. Curve construction algorithm
In Section 4 we have shown that given curve parameters κ 0,τ 0,γ,δ,L∈R and initial conditions x 0, T→0 and N→0, there exists a 3D Euler spiral determined by these parameters and satisfying the initial conditions.
In practice, however, we are given two points and their associated tangents (x 0,T→0) and (x f,T→f). Our goal is to find the parameters κ 0,τ 0,γ,δ,L∈R that define the 3D Euler spiral that starts at x 0 and T→0 and minimizes both the difference between the curveʼs position at s=L and x f, and the difference between the curveʼs tangent at s=L and T→f. In other words, we attempt to minimize the following error:(6)ϵ=(ϵ x+ϵ T),ϵ x=[(x(L)−x f)2+(y(L)−y f)2+(z(L)−z f)2],ϵ T=[(T x(L)−T f,x)2+(T y(L)−T f,y)2+(T z(L)−T f,z)2].
We experimented with other weights of ϵ x and ϵ T as well, but they were not proven beneficial.
We propose the Gradient-descent approach to find the parameters of the 3D Euler spiral that minimizes the error in Eq. (6). This approach, which guarantees convergence to a local minimum, is described below and explained thereafter.
6.1. Parameter initialization (step 1)
The 3D Euler spiral is initialized using a planar Euler spiral . The question is which plane to choose. We define the plane for which three out of the four boundary conditions hold: x 0, x f, and T→0. Therefore, the plane (whose normal is denoted by N→0) is defined by two vectors: T→0 and the vector between x 0 and x f. Note that the resulting planar Euler spiral interpolates T→0 at x 0 and the projection of T→f onto the plane at x f. The parameters of this 2D spiral κ 0, γ, L are used to initialize our curves. Since it lies on a plane, the torsionʼs parameters are initialized to zero (τ 0=0, δ=0).
Our experiments indicate that this initialization gives a good approximation to κ 0 and γ, which hardly change afterwards. We also tested other initialization methods (e.g., Hermite spline, 3D Biarc, and 2D Euler spiral on the binormal plane). We found that our initialization is the fastest and the most accurate.
6.2. Iterative step (steps 3–5)
First, the gradient direction, which is the direction of the steepest descent, is calculated. Since the curve is described as a set of differential equations that do not have an explicit solution, we cannot explicitly find the best gradient direction. Instead, at each iteration, we first find the parameters among κ 0,τ 0,γ,δ,L, that when modified by ±Δ, yield a decrease of the error ϵ. We then compare the Euler spirals that result by modifying only one of these parameters to the spiral that results by modifying them all, and choose the spiral that obtains the minimum error ϵ.
Each of these candidate curves is computed by numerically solving the Frenet–Serret equations. This is done by sampling the arc-length parameter s uniformly and solving the equations at these sampled points, using the Euler method . This method only needs the solution at the immediately preceding point to compute the function at the next point. The first point is the input x 0, T→0 and the normal N˜→0. The normal is defined as the unit vector perpendicular to T→0 on the initial plane.
Next, the step size is modified. If the error ϵ of the chosen direction is smaller than the error obtained in the previous iteration, Δ is unchanged. Otherwise, it is decreased to 3 Δ 4.
Finally, the parameters are updated. If Δ is unchanged, the parameters that determined the gradient direction are updated according to the chosen direction.
6.3. Termination (step 2)
In our experiments, Δ is initialized to 0.1. The algorithm runs until Δ<1 e–5 or the error ϵ<1 e–6. Smaller values yield negligible changes to the curves.
6.4. Optional bound on L (step 5)
Since our curve is a spiral, the obtained solution can have multiple revolutions. For the type of input we expect, it is often desirable to limit the solution to have at most one revolution. This is done by bounding the parameter L, as follows. We first approximate the maximal possible length as the length of the planar Euler spiral with parameters κ 0,γ. Since the tangent angle of such a curve is θ(s)=1 2 γ s 2+κ 0 s+θ 0, we require that θ(L)−θ 0=1 2 γ L 2+κ 0 L⩽2 π.
6.5. Implementation issues
Calculating the numerical solution to the Frenet–Serret equations at each iteration of the Gradient-descent algorithm might be expensive. In addition, the number of samples along the curve should be carefully determined. Using too many samples will result in a long computation time, while too few samples will cause accuracy problems and result in a large error ϵ. To accelerate the computation while using a fixed and rather small number of samples, we scale the region of interest to the box (−1,−1,−1), (1,1,1), prior to applying Algorithm 1, and then scale it back. Recall that Proposition 5.1 allows us to scale the problem back and forth.
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Algorithm 1. Gradient-descent 3D Euler spiral construction
In practice, the problem is first translated by −x 0. Then, it is scaled by D=max(▵x=|x f−x 0|,▵y=|y f−y 0|,▵z=|z f−z 0|), while leaving the tangents at the endpoints unaltered. Then, the curveʼs parameters are found by Algorithm 1 using 100 samples along the curve for performing the numerical calculations. Once a solution is obtained, it is scaled back to the original range.
7. Results and applications
This section demonstrates the use of our spirals in two curve completion applications. In the first, the entire model (a triangular mesh) is given, but the algorithm for edge detection on surfaces generates incomplete curves. This is a common problem with most edge-detection algorithms, which may be vital for the shape analysis algorithms that use these curves. In the second application, the given models are broken – a situation prevalent in archaeology. The user is interested in drawing the curves that would be drawn should the entire model be given. In both cases, the user needs only mark the endpoints of the curves and the system creates an Euler spiral between the given endpoints. No parameter tuning is necessary.
7.1. Curve completion on polyhedral surfaces
Edge detection in images has been extensively investigated from the early days of computer vision. Edge detection on surfaces, on the other hand, has received much less attention. Edges on surfaces are the outcome of the surface geometry only. Consequently, some geometric conditions of the surface (due to noise, erosion, etc.) result in broken or missing edges. Our first application addresses this problem and allows the completion of these curves on the surface.
Fig. 3, Fig. 4, Fig. 5, Fig. 6 demonstrate the use of our spirals for completing the curves detected by various edge detection algorithms: suggestive contours , apparent ridges , valleys and ridges , and demarcating curves . In this application the user needs only choose the endpoints of existing curves that should be connected, since the system can automatically compute the tangents at these endpoints.
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Fig. 3. Fixing demarcating curves . Our curves (red) manage to capture the “S” shape, in contrast to the automatically-scaled Hermite splines (green). (For interpretation of the references to color in this figure, the reader is referred to the web version of this article.)
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Fig. 4. Completing suggestive contours on the Buddha – 3D Euler spirals (red) and Hermite splines (green). Note how the spirals nicely capture the “S” shapes. (For interpretation of the references to color in this figure, the reader is referred to the web version of this article.)
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Fig. 5. Completing apparent ridges on the Column model: Our Euler-spiral completion (red) is circular and resembles the real column. (For interpretation of the references to color in this figure, the reader is referred to the web version of this article.)
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Fig. 6. (a) Completing suggestive contours on the Rocker-arm model. This example illustrates the roundness property of our curve. (b) Completing apparent ridges on the Bust model ((c) – zoom in).
Our curves are also compared to the results obtained using Hermite splines. Since Hermite splines depend on the magnitude of the tangent, we used automatic scaling. The tangents provided for the computation of the Hermite splines are multiplied by D (described in Section 6), so as to relate their length to the distance between the endpoints.
It can be seen that our curves manage to satisfactorily complete the curves, regardless of how they were created. The two features that are most visible in our spirals, are the ability to create perfect circular arcs (Fig. 5, Fig. 6) and the “natural” S-shapes (Fig. 4).
In this application, the generated curves should lie on the surface. Since our curves are not constrained to lie on any surface, the produced 3D Euler curves are projected to the surface. This is done by projecting each point on the curve to its closest point onto the mesh. Though this method is straightforward, it yields good results.
To further examine the quality of the projected curves, we compare the curvature and the torsion of the projected curves with those of the 3D curves. Fig. 7 shows the results on the “S”-shape completion from Fig. 3. It can be seen that the curvature and the torsion of our projected curve are approximately linear. On the other hand, when projecting the 3D Hermite spline to the object, the curvature and the torsion change considerably. This is so since our spirals are close to the surface to start with.
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Fig. 7. Analysis of the projected curves of the “S”-shape from Fig. 3. Top: the curvature and the torsion of the 3D curves. Bottom: the curvature and the torsion of the projected curves. Our projected curve maintain approximately the same properties of the curvature and the torsion, whereas these properties change considerably in the case of Hermite spline.
7.2. Shape illustration in archaeology
Archaeology has recently attracted a lot of attention in geometry processing , , , . In this paper we focus on one aspect of archaeological research – relic completion. Many of the artifacts found by the archaeologists are scanned and need to be processed and analyzed. These artifacts are often broken and eroded and thus are difficult to handle. One specific problem is the drawing of the artifact, which is traditionally performed manually by archaeological artists in 2D, as shown in Fig. 8(a). This is an expensive and time-consuming procedure, which is prone to biases and inaccuracies. Our curves propose a 3D alternative, as illustrated in Fig. 8(b–c).
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Fig. 8. Completing a broken oil lamp. (a) Archaeological drawing of a lamp . (b–c) Completion of a similar lamp in 3D. The 3D Euler spirals (red) are more appealing than the Hermite splines (green) due to the nice circles produces. This is guaranteed by the roundness property of our curves. (For interpretation of the references to color in this figure, the reader is referred to the web version of this article.)
Fig. 9, Fig. 10, Fig. 11, Fig. 12 show several additional models, which show that our completed curves are not only more appealing, but also better resemble the shape of the original unbroken models. For example, the curl completion in Fig. 10 demonstrates the S-shape property of our spirals, while the ear completion illustrates a more “circular” shape. Fig. 11 demonstrates that since our curves have more “volume”, they avoid intersecting the mesh, which might occur when using the Hermite completion.
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Fig. 9. Completing ridges on a broken amphora.
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Fig. 10. Completing a (manually) broken head sculpture. The 3D Euler spirals (red) nicely compute the S-shaped curl. The Euler completion of both the curl and the ear are more similar to the original model than the Hermite completion (green). (For interpretation of the references to color in this figure, the reader is referred to the web version of this article.)
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Fig. 11. Completion of a broken Hellenistic lamp. Ridges and valleys are used to determine initial conditions. Our curves (red) manage to capture the true volume of the shape, while the Hermite splines (green) intersect the broken model. (For interpretation of the references to color in this figure, the reader is referred to the web version of this article.)
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Fig. 12. The completion of a (manually) broken pot. (a–d) Our curves better resemble the original unbroken model. (e–f) The curvature and the torsion of the right curve. Biarcs have a singular torsion point and a large jump in the curvature.
Fig. 12 shows a comparison also to the 3D Biarcs of . In addition to a visual comparison, we also depict the curvesʼ curvature and torsion. It can be seen that at the point of connection between the planar arcs, Biarcs have a singular torsion point and a large jump in the curvature.
7.3. Running times
The algorithm was implemented in Matlab and C and ran on a 2 Ghz Intel Core 2 Duo-processor laptop with 2 Gb of memory. The running time, which depends on the complexity of the required interpolation curve, is 0.01–0.5 second for the curves demonstrated in this paper. The further the curve is from being planar, the longer the time required. This is probably due to initializing the torsion to zero. The size of the model has little affect on the time; it is relevant only when projection is performed. The projection itself is straightforward and quick to compute.
8. Conclusion
This paper presents a novel definition of curves, which extends the 2D Euler spiral to 3D. We proved that for given parameters, a unique curve always exists. Moreover, we showed that our curves satisfy several desired aesthetic properties, including invariance to similarity transformation, symmetry, extensibility, smoothness, and roundness. Given boundary conditions – endpoints and tangents – this paper proposed a novel technique for generating these curves, based on the gradient-descent approach.
The utility of our curves is demonstrated for edge completion on polyhedral surfaces and for artifact illustration in archaeology – a task that is traditionally performed manually in 2D. In archaeology, when automatic 3D curve drawing replaces the traditional manual 2D drawing, automatic or interactive curve completion, would be the only alternative. We believe that the proposed curves may be found a feasible alternative for additional applications involving shape design, artistic design, and shape analysis.
In the future, we wish to prove existence of the curves given point–tangent boundary conditions. In 2D, an algorithm was first established before existence was proved several years later . We hope that the same will happen in 3D. In practice a solution exists for all the inputs we tried.
Acknowledgements
This research was supported in part by the Israel Science Foundation (ISF)628/08, the Goldbers Fund for Electronics Research, and the Ollendorff foundation. We thank Dr. A. Gilboa and the Zinman Institute of Archaeology at the University of Haifa for providing the archaeological models. The other models are courtesy of the AIM@SHAPE Shape Repository and the MIT CSAIL database.
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Recognising decorations in archaeological finds through the analysis of characteristic curves on 3D models
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However, low degree polynomials cannot span a large number of points, therefore many small segments need to be blended together to build the desired curve, , and such a decomposition is not unique, for instance it depends on the starting point. On the contrary, methods that adopt a prototypical family of curves, like the natural 3D spiral in and the 3D Euler spiral in , are able to describe line drawings and silhouettes suitable for shape completion and repair. However, using one class of curves at a time, these methods propose specific solutions and algorithms for settings the curve parameters that unlikely apply to other classes of curves. Show abstract In the analysis of archaeological finds, it is important for archaeologists to identify their style, origin, period, etc. to allow their correct classification. In the digital era, the development of automatic techniques to measure the peculiar characteristics of archaeological finds would be of great help in this activity. Considering that ancient artefacts are very often incomplete, consumed, degraded, if not consisting of simple fragments, geometric details, such as decorations, visual motifs, patterns, are more useful for their analysis than global characteristics. These patterns are usually composed by characteristic curves arranged in a regular way, as in a Greek fret or a floral band. Here we propose the recognition of characteristic curves on 3D models of archaeological artefacts, identified by a set of characteristic points. We approximate these curves with known curves to provide localisation and quantitative measurement of the characteristic features used as decorations or patterns of the digital models of ancient objects. To solve this problem, we adopt a generalised version of the Hough Transform (HT). In addition, we introduce new rules of composition and automatic aggregation of the characteristic curves, not limiting the recognition to a single curve at a time and supporting an automatic annotation of the fragment digital model.
### Recognition of feature curves on 3D shapes using an algebraic approach to Hough transforms
2018, Pattern Recognition Citation Excerpt :
Due to the intuitiveness and meaningful information conveyed in human line drawings, feature curves have been largely investigated in shape modelling and analysis to support several processes, ranging from non-photorealistic rendering to simplification, segmentation and sketching of graphical information [1–4]. These curves can be represented as curve segments identified by a set of vertices, splines interpolating the feature points , L1 medial skeletons or approximated with known curves, like spirals [7,8]. Traditional methods proposed for identifying feature curves on 3D models can be divided into view dependent and view independent methods. Show abstract Feature curves are largely adopted to highlight shape features, such as sharp lines, or to divide surfaces into meaningful segments, like convex or concave regions. Extracting these curves is not sufficient to convey prominent and meaningful information about a shape. We have first to separate the curves belonging to features from those caused by noise and then to select the lines, which describe non-trivial portions of a surface. The automatic detection of such features is crucial for the identification and/or annotation of relevant parts of a given shape. To do this, the Hough transform (HT) is a feature extraction technique widely used in image analysis, computer vision and digital image processing, while, for 3D shapes, the extraction of salient feature curves is still an open problem. Thanks to algebraic geometry concepts, the HT technique has been recently extended to include a vast class of algebraic curves, thus proving to be a competitive tool for yielding an explicit representation of the diverse feature lines equations. In the paper, for the first time we apply this novel extension of the HT technique to the realm of 3D shapes in order to identify and localize semantic features like patterns, decorations or anatomical details on 3D objects (both complete and fragments), even in the case of features partially damaged or incomplete. The method recognizes various features, possibly compound, and it selects the most suitable feature profiles among families of algebraic curves.
### A survey on path planning for persistent autonomy of autonomous underwater vehicles
2015, Ocean Engineering Citation Excerpt :
The clothoid is useful in path-planning applications due to its property of having its curvature change linearly with arc-length. This notion can be extended to three dimensions, and is consequently also true for the torsion (Harary and Tal, 2012). Pythagorean hodographs employ a polynomial of the fifth degree in order to produce a closed-form solution that gives a flexible path with velocity continuity. Show abstract Recently, there has been growing interest in developing Autonomous Underwater Vehicle (AUV) to operate for longer mission durations as well as with higher levels of autonomy. This paper describes the current state of the art in methodologies to enable long range AUVs and provides a detailed literature review of some existing AUVs characterizing their operation endurance. Path planning is identified as the core and crucial components to improve AUV persistence. The aims of path optimization, path re-planning adapting to the dynamic environments, and cooperative path planning of multiple AUVs have received much attention from the research community. This paper presents a review of the main research works focusing on these three technology areas. The main objective of this paper is to present a comprehensive survey of shape and properties of the path and optimization techniques for path planning. These techniques and algorithms have been classified into different classes and their assumptions and drawbacks have been discussed. Finally, the paper discusses the AUV literature in general and highlights challenges that need to be addressed in developing AUVs with advanced autonomy and capable of operating for longer mission durations.
### Feature-preserving surface completion using four points
2014, Computer Graphics Forum
### 3D Shape Analysis for Archaeology
2014, Lecture Notes in Computer Science Including Subseries Lecture Notes in Artificial Intelligence and Lecture Notes in Bioinformatics
### Super space clothoids
2013, ACM Transactions on Graphics
View all citing articles on Scopus
Copyright © 2011 Elsevier B.V. All rights reserved.
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11242 | https://genchem1.chem.okstate.edu/BDA/DCIs/EnthalpyAns.pdf | DCI18 1 During Class Invention Enthalpy Name(s) with Lab section in Group ___ _______ 1. The enthalpy for the formation of liquid water is shown below, H2(g) + 1 2 O2(g) ® H2O(l) ∆H = -286 kJ mol The enthalpy change for the formation of gaseous water is shown below, H2(g) + 1 2 O2(g) ® H2O(g) ∆H = -242 kJ mol Why is the enthalpy change different? The water formed in the first equation is in the liquid phase, while the water formed in the second equation is in the gas phase. Energy is required to convert the water in the liquid phase to water in the gas phase, therefore the enthalpy is less negative in the second equation compared to the first. 2. Given the enthalpy change for the two reactions below, C(s) + 1 2 O2(g) ® CO(g) ∆H˚ = -111 kJ mol C(s) + O2(g) ® CO2(g) ∆H˚ = -394 kJ mol Calculate the enthalpy change for the reaction, CO(g) + 1 2 O2(g) ® CO2(g) CO(g) ® C(s) + 1 2 O2(g) ∆H˚ = + 111 kJ mol C(s) + O2(g) ® CO2(g) ∆H˚ = - 394 kJ mol Adding the two equations CO(g) + 1 2 O2(g) ® CO2(g) ∆H˚ = -283 kJ molrxn DCI18 2 3. Using the following standard enthalpy of reaction data, 2C(s) + 3H2(g) ® C2H6(g) ∆H˚ = - 84.68 kJ mol C(s) + O2(g) ® CO2(g) ∆H˚ = - 394 kJ mol H2(g) + 1 2 O2(g) ® H2O(l) ∆H˚ = - 286 kJ mol Calculate the heat of reaction for the combustion of 1 mol of ethane (C2H6). The equation which describes the combustion of ethane is, C2H6(g) + 7 2 O2(g) ® 2CO2(g) + 3H2O(l) Rearranging the equations provided C2H6(g) ® 2C(s) + 3H2(g) ∆H˚ = + 84.68 kJ mol 2 · C(s) + O2(g) ® CO2(g) ∆H˚ = 2 · - 394 kJ mol 3 · H2(g) + 1 2 O2(g) ® H2O(l) ∆H˚ = 2 · - 286 kJ mol Adding the equations C2H6(g) + 7 2 O2(g) ® 2CO2(g) + 3H2O(l) ∆H˚ = -1561 kJ molrxn 4a. Using standard heats of formation, calculate the ∆H˚ for the following reaction, C2H5OH(l) + 3O2(g) ® 2CO2(g) + 3H2O(l) ∆H˚f -277.7 kJ mol 0 -393.5 kJ mol -285.85 kJ mol DH ˚ rxn = SDH ˚ f (products) - SDH˚ f (reactants) DH ˚ rxn = 3 molH2O molrxn è ç æ ø ÷ ö -285.85 kJ molH2O + 2 molCO2 molrxn è ç æ ø ÷ ö -393.5 kJ molCO2 - è ç æ ø ÷ ö 1 molC2H5OH 1 molrxn è ç æ ø ÷ ö -277.7 kJ molC2H5OH DH ˚ rxn = -1366.8 kJ molrxn b) Determine the amount of heat released at constant pressure when 1.00 gram of ethanol, C2H5OH, is combusted in excess oxygen. From part a, -1366.8 kJ are produced when 1 mol of ethanol is combusted. The amount of heat produced when 1.00 g is combusted is, 1.0 g C2H5OH è ç æ ø ÷ ö 1 mol C2H5OH 46.0 g è ç æ ø ÷ ö -1366.8 kJ 1 mol = -29.7 kJ DCI18 3 5. Calculate the ∆H˚ for the following reaction, 3NO2(g) + H2O(l) ® 2HNO3(aq) + NO(g) ∆H˚f +33.84 kJ mol -285.85 kJ mol -206.6 kJ mol +90.37 kJ mol DH ˚ rxn = SDH ˚ f (products) - SDH˚ f (reactants) DH ˚ rxn = 2 molHNO3 molrxn è ç æ ø ÷ ö -285.85 kJ molHNO3 + 1 molNO molrxn è ç æ ø ÷ ö +90.37 kJ molNO - è ç æ ø ÷ ö 3 molNO2 1 molrxn è ç æ ø ÷ ö +33.84 kJ molNO2 + 1 molH2O 1 molrxn è ç æ ø ÷ ö -285.85 kJ molH2O DH ˚ rxn = -138.5 kJ molrxn DCI18 4 6. The standard enthalpy of combustion to CO2 (g) and H2O(l) at 25 ˚C of cyclohexane, C6H12(l), is -3924 kJ mol . Calculate the standard heat of formation, ∆H˚f, of cyclohexane C6H12(l) + 9O2(g) ® 6CO2(g) + 6H2O(l) ∆H˚f ? 0 -393.5 kJ mol -285.85 kJ mol DH ˚ rxn = SDH ˚ f (products) - SDH˚ f (reactants) DH ˚ rxn = 6DH ˚ f (CO2) + 6DH ˚ f (H2O) - DH˚ f (C6H12) DH˚ f (C6H12) = 6DH ˚ f (CO2) + 6DH ˚ f (H2O) - DH ˚ rxn DH˚ f (C6H12) = 6 molCO2 molrxn è ç æ ø ÷ ö -393.5 kJ molCO2 + 6 molH2O molrxn è ç æ ø ÷ ö -285.85 kJ molH2O - è ç æ ø ÷ ö -3924 kJ molrxn DH˚ f (C6H12) = -152.1 kJ mol |
11243 | https://pubchem.ncbi.nlm.nih.gov/compound/Acetic-acid-ammonium | Acetic acid ammonium | C2H8NO2+ | CID 22147639 - PubChem
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compound Summary
Acetic acid ammonium
PubChem CID
22147639
Structure
Molecular Formula
C 2 H 8 NO 2+
Synonyms
acetic acid ammonium
USFZMSVCRYTOJT-UHFFFAOYSA-O
Molecular Weight
78.09 g/mol
Computed by PubChem 2.2 (PubChem release 2025.09.15)
Parent Compound
CID 176 (Acetic Acid)
Component Compounds
CID 222 (Ammonia)
CID 176 (Acetic Acid)
Dates
Create: 2007-12-05
Modify: 2025-09-27
1 Structures
1.1 2D Structure
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Chemical Structure Depiction
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1.2 3D Conformer
3D Conformer of Parent
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2 Names and Identifiers
2.1 Computed Descriptors
2.1.1 IUPAC Name
azanium acetic acid
Computed by Lexichem TK 2.9.3 (PubChem release 2025.09.15)
PubChem
2.1.2 InChI
InChI=1S/C2H4O2.H3N/c1-2(3)4;/h1H3,(H,3,4);1H3/p+1
Computed by InChI 1.07.4 (PubChem release 2025.09.15)
PubChem
2.1.3 InChIKey
USFZMSVCRYTOJT-UHFFFAOYSA-O
Computed by InChI 1.07.4 (PubChem release 2025.09.15)
PubChem
2.1.4 SMILES
CC(=O)O.[NH4+]
Computed by OEChem 4.2.0 (PubChem release 2025.09.15)
PubChem
2.2 Molecular Formula
C 2 H 8 NO 2+
Computed by PubChem 2.2 (PubChem release 2025.09.15)
PubChem
2.3 Other Identifiers
2.3.1 Nikkaji Number
J1.710H
Japan Chemical Substance Dictionary (Nikkaji)
2.4 Synonyms
2.4.1 Depositor-Supplied Synonyms
acetic acid ammonium
USFZMSVCRYTOJT-UHFFFAOYSA-O
PubChem
3 Chemical and Physical Properties
3.1 Computed Properties
Property Name
Property Value
Reference
Property Name
Molecular Weight
Property Value
78.09 g/mol
Reference
Computed by PubChem 2.2 (PubChem release 2025.09.15)
Property Name
Hydrogen Bond Donor Count
Property Value
2
Reference
Computed by Cactvs 3.4.8.24 (PubChem release 2025.09.15)
Property Name
Hydrogen Bond Acceptor Count
Property Value
2
Reference
Computed by Cactvs 3.4.8.24 (PubChem release 2025.09.15)
Property Name
Rotatable Bond Count
Property Value
0
Reference
Computed by Cactvs 3.4.8.24 (PubChem release 2025.09.15)
Property Name
Exact Mass
Property Value
78.055503498 Da
Reference
Computed by PubChem 2.2 (PubChem release 2025.09.15)
Property Name
Monoisotopic Mass
Property Value
78.055503498 Da
Reference
Computed by PubChem 2.2 (PubChem release 2025.09.15)
Property Name
Topological Polar Surface Area
Property Value
38.3 Ų
Reference
Computed by Cactvs 3.4.8.24 (PubChem release 2025.09.15)
Property Name
Heavy Atom Count
Property Value
5
Reference
Computed by PubChem
Property Name
Formal Charge
Property Value
1
Reference
Computed by PubChem
Property Name
Complexity
Property Value
31
Reference
Computed by Cactvs 3.4.8.24 (PubChem release 2025.09.15)
Property Name
Isotope Atom Count
Property Value
0
Reference
Computed by PubChem
Property Name
Defined Atom Stereocenter Count
Property Value
0
Reference
Computed by PubChem
Property Name
Undefined Atom Stereocenter Count
Property Value
0
Reference
Computed by PubChem
Property Name
Defined Bond Stereocenter Count
Property Value
0
Reference
Computed by PubChem
Property Name
Undefined Bond Stereocenter Count
Property Value
0
Reference
Computed by PubChem
Property Name
Covalently-Bonded Unit Count
Property Value
2
Reference
Computed by PubChem
Property Name
Compound Is Canonicalized
Property Value
Yes
Reference
Computed by PubChem (release 2010.01.29)
PubChem
4 Related Records
4.1 Related Compounds with Annotation
Follow these links to do a live 2D search or do a live 3D search for this compound, sorted by annotation score. This section is deprecated (see the neighbor discontinuation help page for details), but these live search links provide equivalent functionality to the table that was previously shown here.
PubChem
4.2 Parent Compound
CID 176 (Acetic Acid)
PubChem
4.3 Component Compounds
CID 222 (Ammonia)
CID 176 (Acetic Acid)
PubChem
4.4 Related Compounds
Same Parent, Connectivity Count
1690
Same Parent, Exact Count
1576
Mixtures, Components, and Neutralized Forms Count
2
Similar Compounds (2D)
View in PubChem Search
Similar Conformers (3D)
View in PubChem Search
PubChem
4.5 Substances
4.5.1 Related Substances
Same Count
4
PubChem
4.5.2 Substances by Category
PubChem
5 Patents
5.1 Depositor-Supplied Patent Identifiers
PubChem
Link to all deposited patent identifiers
PubChem
6 Information Sources
Filter by Source
Japan Chemical Substance Dictionary (Nikkaji)
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3 Chemical and Physical Properties Expand this menu
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11244 | https://math.libretexts.org/Bookshelves/Linear_Algebra/Book%3A_Linear_Algebra_(Schilling_Nachtergaele_and_Lankham)/09%3A_Inner_product_spaces/9.03%3A_Orthogonality | Skip to main content
9.3: Orthogonality
Last updated
: Mar 5, 2021
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9.2: Norms
9.4: Orthonormal bases
Page ID
: 258
Isaiah Lankham, Bruno Nachtergaele, & Anne Schilling
University of California, Davis
( \newcommand{\kernel}{\mathrm{null}\,})
Using the inner product, we can now define the notion of orthogonality, prove that the Pythagorean theorem holds in any inner product space, and use the Cauchy-Schwarz inequality to prove the triangle inequality. In particular, this will show that ‖v‖=√⟨v,v⟩∥v∥=⟨v,v⟩−−−−√ does indeed define a norm.
Definition 9.3.1. Two vectors u,v∈Vu,v∈V are orthogonal (denoted u⊥vu⊥v) if ⟨u,v⟩=0⟨u,v⟩=0.
Note that the zero vector is the only vector that is orthogonal to itself. In fact, the zero vector is orthogonal to every vector v∈Vv∈V.
Theorem 9.3.2. (Pythagorean Theorem). If u,v∈Vu,v∈V, an inner product space, with u⊥vu⊥v, then ‖⋅‖∥⋅∥ defined by ‖v‖:=√⟨v,v⟩∥v∥:=⟨v,v⟩−−−−√ obeys
‖u+v‖2=‖u‖2+‖v‖2.
∥u+v∥2=∥u∥2+∥v∥2.(9.3.1)
Proof. Suppose u,v∈Vu,v∈V such that u⊥vu⊥v. Then
‖u+v‖2=⟨u+v,u+v⟩=‖u‖2+‖v‖2+⟨u,v⟩+⟨v,u⟩=‖u‖2+‖v‖2.
∥u+v∥2=⟨u+v,u+v⟩=∥u∥2+∥v∥2+⟨u,v⟩+⟨v,u⟩=∥u∥2+∥v∥2.
Note that the converse of the Pythagorean Theorem holds for real vector spaces since, in that case, ⟨u,v⟩+⟨v,u⟩=2Re⟨u,v⟩=0⟨u,v⟩+⟨v,u⟩=2Re⟨u,v⟩=0.
Given two vectors u,v∈Vu,v∈V with v≠0v≠0, we can uniquely decompose uu into two pieces: one piece parallel to vv and one piece orthogonal to vv. This is called an orthogonal decomposition. More precisely, we have
u=u1+u2,
u=u1+u2,
where u1=avu1=av and u2⊥vu2⊥v for some scalar a∈Fa∈F. To obtain such a decomposition, write u2=u−u1=u−avu2=u−u1=u−av. Then, for u2u2 to be
orthogonal to vv, we need
0=⟨u−av,v⟩=⟨u,v⟩−a‖v‖2.
0=⟨u−av,v⟩=⟨u,v⟩−a∥v∥2.
Solving for aa yields a=⟨u,v⟩/‖v‖2a=⟨u,v⟩/∥v∥2 so that
u=⟨u,v⟩‖v‖2v+(u−⟨u,v⟩‖v‖2v).
u=⟨u,v⟩∥v∥2v+(u−⟨u,v⟩∥v∥2v).(9.3.1)
This decomposition is particularly useful since it allows us to provide a simple proof for the Cauchy-Schwarz inequality.
Theorem 9.3.3 (Cauchy-Schwarz inequality). Given any u,v∈Vu,v∈V, we have
|⟨u,v⟩|≤‖u‖‖v‖.
|⟨u,v⟩|≤∥u∥∥v∥.(9.3.2)
Furthermore, equality holds if and only if uu and vv are linearly dependent, i.e., are scalar multiples of each other.
Proof. If v=0v=0, then both sides of the inequality are zero. Hence, assume that v≠0v≠0, and consider the orthogonal decomposition
u=⟨u,v⟩‖v‖2v+w
u=⟨u,v⟩∥v∥2v+w
where w⊥vw⊥v. By the Pythagorean theorem, we have
‖u‖2=‖⟨u,v⟩‖v‖2v‖2+‖w‖2=|⟨u,v⟩|2‖v‖2+‖w‖2≥|⟨u,v⟩|2‖v‖2.
∥u∥2=∥∥∥⟨u,v⟩∥v∥2v∥∥∥2+∥w∥2=|⟨u,v⟩|2∥v∥2+∥w∥2≥|⟨u,v⟩|2∥v∥2.
Multiplying both sides by ‖v‖2∥v∥2 and taking the square root then yields the Cauchy-Schwarz inequality.
Note that we get equality in the above arguments if and only if w=0w=0. But, by Equation (9.3.1), this means that uu and vv are linearly dependent.
The Cauchy-Schwarz inequality has many different proofs. Here is another one.
Alternate Proof of Theorem 9.3.3. Given u,v∈Vu,v∈V, consider the norm square of the vector u+reiθv:u+reiθv:
0≤‖u+reiθv‖2=‖u‖2+r2‖v‖2+2Re(reiθ⟨u,v⟩).
0≤∥u+reiθv∥2=∥u∥2+r2∥v∥2+2Re(reiθ⟨u,v⟩).
Since ⟨u,v⟩⟨u,v⟩ is a complex number, one can choose θθ so that eiθ⟨u,v⟩eiθ⟨u,v⟩ is real. Hence, the right hand side is a parabola ar2+br+car2+br+c with real coefficients. It will lie above the real axis, i.e. ar2+br+c≥0ar2+br+c≥0, if it does not have any real solutions for rr. This is the case when the discriminant satisfies b2−4ac≤0b2−4ac≤0. In our case this means
4|⟨u,v⟩|2−4‖u‖2‖v‖2≤0.
4|⟨u,v⟩|2−4∥u∥2∥v∥2≤0.
Moreover, equality only holds if rr can be chosen such that u+reiθv=0u+reiθv=0, which means that uu and vv are scalar multiples.
Now that we have proven the Cauchy-Schwarz inequality, we are finally able to verify the triangle inequality. This is the final step in showing that ‖v‖=√⟨v,v⟩∥v∥=⟨v,v⟩−−−−√ does indeed define a norm. We illustrate the triangle inequality in Figure 9.3.1.
Figure 9.3.1: The Triangle Inequality in R2R2
Theorem 9.3.4. (Triangle Inequality). For all u,v∈Vu,v∈V we have
‖u+v‖≤‖u‖+‖v‖.
∥u+v∥≤∥u∥+∥v∥.(9.3.3)
Proof. By a straightforward calculation, we obtain
‖u+v‖2=⟨u+v,u+v⟩=⟨u,u⟩+⟨v,v⟩+⟨u,v⟩+⟨v,u⟩=⟨u,u⟩+⟨v,v⟩+⟨u,v⟩+¯⟨u,v⟩=‖u‖2+‖v‖2+2Re⟨u,v⟩.
∥u+v∥2=⟨u+v,u+v⟩=⟨u,u⟩+⟨v,v⟩+⟨u,v⟩+⟨v,u⟩=⟨u,u⟩+⟨v,v⟩+⟨u,v⟩+⟨u,v⟩¯¯¯¯¯¯¯¯¯¯¯=∥u∥2+∥v∥2+2Re⟨u,v⟩.
Note that Re⟨u,v⟩≤|⟨u,v⟩|Re⟨u,v⟩≤|⟨u,v⟩| so that, using the Cauchy-Schwarz inequality, we obtain
‖u+v‖2≤‖u‖2+‖v‖2+2‖u‖‖v‖=(‖u‖+‖v‖)2.
∥u+v∥2≤∥u∥2+∥v∥2+2∥u∥∥v∥=(∥u∥+∥v∥)2.
Taking the square root of both sides now gives the triangle inequality.
Remark 9.3.5. Note that equality holds for the triangle inequality if and only if v=ru or u=rv for some r≥0. Namely, equality in the proof happens only if ⟨u,v⟩=‖u‖‖v‖, which is equivalent to u and v being scalar multiples of one another.
Theorem 9.3.6. (Parallelogram Law). Given any u,v∈V, we have
‖u+v‖2+‖u−v‖2=2(‖u‖2+‖v‖2).
Proof. By direct calculation,
‖u+v‖2+‖u−v‖2=⟨u+v,u+v⟩+⟨u−v,u−v⟩=‖u‖2+‖v‖2+⟨u,v⟩+⟨v,u⟩+‖u‖2+‖v‖2−⟨u,v⟩−⟨v,u⟩=2(‖u‖2+‖v‖2).
Remark 9.3.7. We illustrate the parallelogram law in Figure 9.3.2.
Figure 9.3.2: The Parallelogram Law in R2
Contributors
Isaiah Lankham, Mathematics Department at UC Davis
Bruno Nachtergaele, Mathematics Department at UC Davis
Anne Schilling, Mathematics Department at UC Davis
Both hardbound and softbound versions of this textbook are available online at WorldScientific.com.
9.2: Norms
9.4: Orthonormal bases |
11245 | https://pmc.ncbi.nlm.nih.gov/articles/PMC9639180/ | Eisenmenger syndrome in an adult patient with a large patent ductus arteriosus - PMC
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Eur Respir Rev
. 2013 Dec;22(130):558–564. doi: 10.1183/09059180.00007013
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Eisenmenger syndrome in an adult patient with a large patent ductus arteriosus
Konstantinos Dimopoulos
Konstantinos Dimopoulos
1 Adult Congenital Heart Centre and Centre for Pulmonary Hypertension, NIHR Cardiovascular Biomedical Research Unit, Royal Brompton Hospital and the National Heart and Lung Institute, Imperial College, London, UK
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1 Adult Congenital Heart Centre and Centre for Pulmonary Hypertension, NIHR Cardiovascular Biomedical Research Unit, Royal Brompton Hospital and the National Heart and Lung Institute, Imperial College, London, UK
✉
K. Dimopoulos, Adult Congenital Heart Centre and Centre for Pulmonary Hypertension, Royal Brompton Hospital, Sydney Street, London, SW3 6NP, UK. E-mail: k.dimopoulos02@gmail.com
Received 2013 Sep 19; Accepted 2013 Sep 29.
©ERS 2013
ERR articles are open access and distributed under the terms of the Creative Commons Attribution Non-Commercial Licence 3.0.
PMC Copyright notice
PMCID: PMC9639180 PMID: 24293472
Abstract
This is the case of a young female who was seen at our adult congenital heart disease and pulmonary hypertension service (Royal Brompton Hospital, London, UK) at the age of 17 years. She initially presented at the age of 4 years with increasing shortness of breath. At that time, there was differential cyanosis with clubbing and lower oxygen saturations in the toes (82%) compared with her fingers (95%). On echocardiography there was evidence of severe pulmonary hypertension and a large patent ductus arteriosus (PDA) with low velocity bidirectional shunting. She underwent cardiac catheterisation at the time which showed a pulmonary arterial pressure equal to that of the aorta.
Question 1
What is the diagnosis and would you close this PDA?
The patient had a bidirectional shunt (thus also shunted right-to-left) at PDA level and had severe pulmonary arterial hypertension (PAH) and had, therefore, developed Eisenmenger syndrome (fig. 1) . PAH related to congenital heart disease (CHD) can be classified into four groups (table 1) according to the severity of the disease and the presence/size of the defect. Eisenmenger syndrome is at the extreme end with regards to the severity of PAH and the degree of cyanosis . While Eisenmenger patients have severe pulmonary vascular disease, with some following an aggressive course in childhood, patients reaching adulthood may remain fairly stable for years or decades and have a prognosis that is better compared to those with PAH after repair of CHD and those with severe PAH in the presence of a small defect [3–5].
Figure 1.
Open in a new tab
a) Pressure tracings from the aorta (AO) and pulmonary artery (PA), demonstrating systemic levels of pulmonary arterial pressures. Mean pulmonary arterial pressure is 54 mmHg. b) Cardiac magnetic resonance showing the large patent ductus arteriosus (PDA), with bidirectional shunting between the PA to the descending AO. c) Computed tomography of the thorax demonstrating an area of intrapulmonary haemorrhage in the left upper lung (white arrow). The PDA is also seen between the AO and PA (black arrow).
Table 1. Types of pulmonary arterial hypertension (PAH) in congenital heart disease.
A. Eisenmenger syndrome
Includes all systemic-to-pulmonary shunts due to large defects leading to a severe increase in PVR and a reversed (pulmonary-to-systemic) or bidirectional shunt. Cyanosis, erythrocytosis, and multiple organ involvement are present
B. PAH associated with systemic-to-pulmonary shunts
Patients with moderate-to-large defects, in which the increase in PVR is mild-to-moderate, left-to-right shunt is still largely present and no cyanosis is present at rest
C. PAH with small defects
Patients with a clinical picture very similar to idiopathic PAH, who have (coincidental?) small cardiac defects: a ventricular septal defect <1 cm and an atrial septal defect <2 cm
D. PAH after corrective cardiac surgery
Congenital heart disease has been corrected but PAH is still present immediately after surgery, or has recurred several months or years after surgery in the absence of significant post-operative residual congenital lesions or defects that originate as sequelae of previous surgery
Open in a new tab
PVR: pulmonary vascular resistance. Reproduced with modification from .
It is clear that this defect should be left open once significant pulmonary vascular disease with reversal of the shunt through the PDA has developed. Even though the PDA is responsible for the onset of PAH it is now acts as a “relief valve”, offloading the pressure overloaded right ventricle. While closure of the PDA can easily be achieved percutaneously, this can lead to right ventricle failure hours, weeks, months or years after the procedure, adversely affecting outcome. The same applies for patients with Eisenmenger syndrome in the presence of an atrial or ventricular septal defect.
Our patient presented with a typical Eisenmenger phenotype early in life. The full Eisenmenger phenotype with right-to-left shunting and overt cyanosis typically becomes manifest later in childhood or in adolescence/early adult life. The reason for this heterogeneity in presentation in patients with similar cardiac anatomy remains unknown and has been attributed to different susceptibility of the pulmonary vasculature and different response to the haemodynamic insult.
At the time of diagnosis of PAH (in childhood), no oral PAH therapies were available. Thus, the patient was started on nifedipine, a calcium channel blocker.
Question 2
Are calcium channel blockers indicated in Eisenmenger syndrome?
According to available evidence and current guidelines, patients with idiopathic PAH should be assessed for reversibility at the time of initial cardiac catheterisation and put on high dose of calcium channel blockers when an adequate response to acute pulmonary vasodilator administration is seen . The criteria for defining vasoreactivity in idiopathic PAH (decrease in mean pulmonary arterial pressure ≥10 mmHg to an absolute value ≤40 mmHg) do not apply here. No data are available on the usefulness of long-term calcium channel blocker therapy in Eisenmenger patients. Minor vasoreactivity can even be observed in Eisenmenger patients and has been reported to predict outcome . While there is little value in performing vasoreactivity testing in a CHD-PAH patient for the purpose of assessing eligibility for calcium channel blocker therapy, reversibility studies are often performed in patients with CHD and borderline pulmonary vascular resistance to assess operability [7–9]. The two indications should not be confused and the criteria for defining vasoreactivity differ. Evidence-based criteria for assessing operability are still lacking and the decision to operate is best taken in experienced centres, based on the available clinical information (including catheterisation data).
The patient was first seen in the adult congenital heart disease service at the age of 17 years. On examination, blood pressure and heart rate were normal. She did not appear to have cyanosis in her upper body. She had no finger clubbing, but significant toe clubbing. Oxygen saturations in her fingers were normal, but low in the toes (83%). There were no signs of congestive heart failure. On auscultation, there was a normal first, but prominent second heart sound and a mild holosystolic murmur over the left lower sternal border. There was no murmur over the PDA.
Question 3
Why is there no murmur generated by the PDA?
A large PDA generates an audible murmur as long as there is a sufficient pressure gradient between the aorta and the pulmonary artery to sustain a high velocity shunt. As pulmonary vascular disease progresses, the pressure difference between the great vessels falls progressively, and so does the intensity of the murmur. In patients with Eisenmenger syndrome, the shunt is bidirectional, mainly from right-to-left, low-velocity and with no associated murmur. A soft holosystolic murmur in the tricuspid area and a long diastolic murmur in the pulmonary areas may be heard if tricuspid and pulmonary regurgitation, respectively, develop. The same applies for patients with Eisenmenger syndrome in the presence of an atrial or ventricular septal defect.
In the absence of a high velocity shunt at PDA level in Eisenmenger patients it may be difficult to diagnose a PDA on echocardiography. In fact, Eisenmenger patients with a PDA can, at times, be mislabelled as idiopathic PAH. All patients with a new diagnosis of PAH should be investigated for differential cyanosis and peripheral clubbing, especially when erythrocytosis is present (see below), and the presence of the PDA confirmed on computed tomography of the pulmonary arteries or CMRI (fig. 1b).
When the patient was seen in the adult service, blood testing was performed demonstrating an increase in haemoglobin concentration (18.4 g·dL−1) and haematocrit (56%), with evidence of iron deficiency (ferritin 10 μg·L−1 and transferrin saturation 13%). Mean cellular volume was normal. Brain natriuretic peptide (BNP) was raised at 40 pmol·L−1 (normal range <4 pmol·L−1).
Question 4
Why is haemoglobin concentration raised, does the patient require venesection and what is the significance of a raised BNP in Eisenmenger patients?
An increase in haemoglobin levels in patients with cyanotic heart disease is a welcome compensatory mechanism, aimed at increasing the blood's oxygen carrying capacity. It is best referred to as “erythrocytosis” to distinguish it from polycythaemia rubra vera, with which it does not share the same propensity to thrombotic and embolic events . Erythrocytosis in cyanotic patients is often associated with a decrease in platelet count and, at times, white cell count. Even though no validated algorithms exist to establish the appropriate haemoglobin concentration for individual patients, in the presence of systemic oxygen saturations of 80–85%, haemoglobin of 19–24 g·dL−1 is usually observed [5, 11, 12].
Iron deficiency is common in cyanotic heart disease, and is encountered in up to one-third of patients [5, 13–15]. It is multifactorial, probably due to increased iron requirement, decreased absorption, bleeding, etc. Iron deficiency is an adverse prognostic marker in this population . All cyanotic patients should be regularly screened for iron deficiency using transferrin receptor levels or, when not available, ferritin and transferrin saturations. A low mean cellular volume is not commonly encountered in iron deficient cyanotic patients and should not be used to exclude iron deficiency .
Venesections are best avoided in cyanotic CHD patients as they promote iron deficiency and can lead to cerebrovascular events, probably due to a reduction in oxygen delivery to the brain [10, 17]. Venesections should only be performed in patients with severe hyperviscosity symptoms in the presence of very high haematocrit (>65%) and in the absence of dehydration and iron deficiency. In fact, severe iron deficiency shares symptomatology with hyperviscosity. Adequate fluid replacement is essential, using filters to avoid air emboli.
Iron supplementation is essential in patients who are found to be iron deficient. A single centre, non-randomised cohort study on cyanotic patients with CHD showed that iron supplementation alone can improve exercise capacity and quality of life. No cases of excessive erythropoiesis or new-onset severe hyperviscosity symptoms were seen and no cases required venesection .
Elevated BNP concentrations are common in patients with Eisenmenger syndrome and relate to a higher risk of death. A cut-off BNP value of 100 pg·mL−1 (∼30 pmol·L−1) has been suggested as a marker of adverse prognosis. Increases in BNP concentration over time have also been found to be of prognostic significance. Thus, BNP can be a useful adjunct in guiding therapy in Eisenmenger syndrome .
When questioned about exercise limitation, the patient insisted that she was not significantly limited in her ordinary activities. However, when looking into specific activities, it became evident that she could only manage ∼20 min walking on the flat or one flight of stairs before becoming very short of breath. She found activities such as gardening and vacuuming difficult and, hence, avoided them. She achieved a distance of only 340 m on the 6-min walk test (6MWT), with a significant drop in oxygen saturation to 70% in the toe.
Question 5
What is the patient's current functional class?
The pulmonary hypertension guidelines suggest that only Eisenmenger patients in functional class 3 should be considered for treatment with advanced therapies for PAH . However, assessing the functional class in PAH patients with CHD can be difficult using the New York Heart Association or World Health Organization classification. Onset of symptoms early in life can modify symptom perception and lead to chronic adaptation of everyday “ordinary” activities, as opposed to patients who develop exercise limitation later in life and over a shorter period of time (e.g. those with idiopathic PAH) [15, 20, 21]. In order to avoid underestimating their degree of limitation, patients should be asked specific questions on typical ordinary activities (walking, climbing stairs, showering, cleaning, vacuuming, gardening, dancing, etc.) and specific landmarks should be explored (e.g. distance from local train/metro station, distance from hospital entrance or cardiology department) on a regular basis . Addition of the above information to indices of objective exercise capacity (exercise testing), imaging and BNP should influence the decision as to whether to initiate therapies [23, 24].
The 6MWT is commonly used for assessing exercise capacity and response to therapies in Eisenmenger patients. In a recent, single centre study on Eisenmenger patients, 6MWT distance (but not the rate of change of 6MWT distance) was a strong predictor of outcome, while functional class seemed to lack the reproducibility and discriminative power to follow subtle changes over time . It is noteworthy that a lack of predictive power of 6-min walk distance change on long-term outcomes is also seen in other forms of PAH.
Nifedipine was stopped and the patient underwent iron supplementation. Thereafter she was initiated on the endothelin receptor antagonist (ERA) bosentan . 4 months after the introduction of bosentan, an increase of 60 m in 6-min walk distance was observed.
Question 6
What is the evidence for the use of advanced therapies for PAH in Eisenmenger patients?
The safety and efficacy of the ERA bosentan was established by the BREATHE-5 trial . This was a 16-week randomised controlled trial on Eisenmenger patients in functional class 3 with a ventricular septal defect or large atrial septal defect, which reported a statistically significant beneficial effect on 6MWT distance and pulmonary vascular resistance . It also reported an improvement in functional class but no deterioration in systemic oxygen saturations . It is also noteworthy that, on average, the 6MWT distance and pulmonary vascular resistance worsened in the placebo group, suggesting progression of the disease over the relatively short duration of the study. The effect on 6MWT distance was maintained at 12 months on the open-label extension and this improvement was confirmed in both patients with a ventricular septal defect and an atrial septal defect [26, 27]. Further small trials have demonstrated the efficacy and safety of sildenafil and tadalafil in this setting [28–30]. The long-term efficacy of advanced therapies in Eisenmenger patients has now been confirmed in large registries . There is also evidence of a beneficial effect of advanced therapies on survival from a single centre retrospective study .
Data are still lacking on the optimal mode of following Eisenmenger patients in terms of assessing the need for starting therapies, their efficacy and the need to up-titrate or escalate (combination therapy). In clinical practice, functional status, 6MWT distance, echocardiographic parameters and BNP are commonly used, but specific achievement targets have not yet been established in this population [5, 23, 32]. Moreover, the value of repeat cardiac catheterisation as a follow-up tool remains controversial. Cardiac output in Eisenmenger patients is maintained through right-to-left shunting and is unlikely to carry the same prognostic power as in idiopathic PAH. Moreover, calculation of pulmonary vascular resistance is difficult in patients with a PDA and those with complex cardiac anatomy.
A large proportion of Eisenmenger patients have Down syndrome. In these patients, neither subjective measures of effort tolerance nor formal exercise testing are possible and can be inaccurate [33, 34]. An improvement in quality of life is probably the most important target for therapy in all Eisenmenger patients, especially those with Down syndrome.
Question 7
What other supportive measures should be taken in Eisenmenger patients?
Supportive measures are as important in Eisenmenger syndrome as advanced therapies . These mainly consist of avoiding pitfalls, such as submitting patients to non-essential general anaesthesia or sedation, which carry significant risks. Venesection should be avoided and iron deficiency treated . Dehydration should be avoided. Patients should remain physically active while avoiding strenuous efforts, competitive sports and extreme isometric efforts. There is little evidence to support long-term supplemental oxygen use. Endocarditis prophylaxis is still advocated by international guidelines, in view of the increased risk of embolic events, especially cerebral abscess . Regular dental review is essential.
Acute arrhythmias, both ventricular and supraventricular, can lead to rapid deterioration and should be managed promptly. Electrical cardioversion is often required and is best performed under the care of senior anaesthetists experienced in PAH and Eisenmenger syndrome.
Haemoptysis is common in Eisenmenger patients and, in most cases, is self-limiting; however, massive haemoptysis can lead to death (fig. 1c) . When assessing a patient with haemoptysis, evaluation of the airway, breathing and circulation is the first step. Bleeding cessation can be achieved through interruption and/or reversal of anticoagulation. In massive haemoptysis, protection of airways, fluid resuscitation and anti-fibrinolytic drugs (e.g. tranexamic acid) may be used with caution. Underlying lung infection should be treated promptly. Haemoglobin levels should be corrected to levels appropriate for the oxygen saturations and iron should be supplemented. Strict in-hospital observation is recommended. Escalation of advanced therapies should be considered.
There is still controversy on the use of anticoagulants in Eisenmenger syndrome, with no convincing evidence for or against their use . Patients with arrhythmias, in situ thrombosis of the pulmonary arteries, embolic phenomena or congestive heart failure should receive anticoagulation, in the absence of contraindications.
Question 8
Is it safe for patients with Eisenmenger syndrome to become pregnant?
The issue of pregnancy should be discussed as early as possible in all patients with pulmonary hypertension in view of the very high associated mortality (prohibitive) . In a systematic review of the literature, the risk of death due to pregnancy in patients with PAH-CHD was ∼30% . There are also significant risks to the baby as there is an increased likelihood of prematurity and growth retardation (∼80%), with an increased associated risk of long-term neurological damage. Spontaneous abortion is not uncommon, and relates to the degree of peripheral desaturation.
Even interruption of pregnancy carries risks in Eisenmenger patients, with risks of bleeding and the risk of general anaesthesia when surgical interruption is required. Appropriate contraception with progesterone only compounds, combined with barrier methods (dual contraception, especially in patients on bosentan due to possible pharmacological interaction with oral contraceptives), is therefore paramount.
Question 9
The patient has improved significantly on advanced therapy; can I now close the defect?
Defects should never be closed in Eisenmenger patients, regardless of the magnitude of response to medical therapy. Closure of the defect risks transforming Eisenmenger physiology into an idiopathic PAH-like condition, with more rapid right ventricular dilatation and progressive dysfunction, negatively affecting long-term prognosis [3, 7]. Even for CHD-PAH patients with milder forms of pulmonary vascular disease, the long-term outcome of a “treat and repair” approach (repairing the defect after optimal response to medical therapy) has not been established and, thus, should be avoided until proven safe and effective .
Question 10
Can Eisenmenger patients fly?
Concerns about cyanotic CHD patients becoming unwell when flying, due to the ambient oxygen concentration being lower than at sea level, are not supported by evidence. One retrospective study questioning Eisenmenger patients on previous air travel demonstrated that most patients do not encounter significant problems and supplemental oxygen is not generally required, especially for shorter flights . However, definitive evidence is lacking.
Conclusions
PAH is common in patients with CHD and a congenital defect should always be sought and excluded in all patients presenting with PAH. Echocardiography (with or without contrast injection) is pivotal in excluding intracardiac defects, but even simple measures, such as careful measurement of oxygen saturations in the fingers and toes at rest and after exercise, as well as serial sampling of oxygen saturation from the high superior vena cava to the pulmonary artery during cardiac catheterisation, can aid in the diagnosis of “occult” CHD. Such patients are best treated in specialist centres combining CHD and PAH expertise, as their physiology and natural history differs significantly to other types of PAH.
Supplementary Material
ERR_0070-2013_Disclosure_Dimopolous.pdf
ERR_0070-2013_Disclosure_Dimopolous.pdf (3.8KB, pdf)
Acknowledgments
I would like to thank R. Doyle (Elements Communications Ltd, Westerham, UK) for medical writing support, funded by Actelion Pharmaceuticals Ltd (Allschwil, Switzerland), D. Wahl and R. Zimmerman (Actelion Pharmaceuticals Ltd) for their useful review of this article.
Footnotes
Conflict of interest: Disclosures can be found alongside the online version of this article at err.ersjournals.com
Provenance: Publication of this peer-reviewed article was supported by Actelion Pharmaceuticals Ltd, Switzerland (principal sponsor, European Respiratory Review issue 130).
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11246 | https://www.aafp.org/pubs/afp/issues/2012/1201/p1055.html | Acute Otitis Externa: An Update
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PAUL SCHAEFER, MD, PhD, AND REGINALD F. BAUGH, MD
update
A more recent article on acute otitis externa is available.
info
Am Fam Physician. 2012;86(11):1055-1061
assignment
Author disclosure: No relevant financial affiliations to disclose.
Acute otitis externa is a common condition involving inflammation of the ear canal. The acute form is caused primarily by bacterial infection, with Pseudomonas aeruginosa and Staphylococcus aureus the most common pathogens. Acute otitis externa presents with the rapid onset of ear canal inflammation, resulting in otalgia, itching, canal edema, canal erythema, and otorrhea, and often occurs following swimming or minor trauma from inappropriate cleaning. Tenderness with movement of the tragus or pinna is a classic finding. Topical antimicrobials or antibiotics such as acetic acid, aminoglycosides, polymyxin B, and quinolones are the treatment of choice in uncomplicated cases. These agents come in preparations with or without topical corticosteroids; the addition of corticosteroids may help resolve symptoms more quickly. However, there is no good evidence that any one antimicrobial or antibiotic preparation is clinically superior to another. The choice of treatment is based on a number of factors, including tympanic membrane status, adverse effect profiles, adherence issues, and cost. Neomycin/polymyxin B/hydrocortisone preparations are a reasonable first-line therapy when the tympanic membrane is intact. Oral antibiotics are reserved for cases in which the infection has spread beyond the ear canal or in patients at risk of a rapidly progressing infection. Chronic otitis externa is often caused by allergies or underlying inflammatory dermatologic conditions, and is treated by addressing the underlying causes.
Otitis externa, also called swimmer's ear, involves diffuse inflammation of the external ear canal that may extend distally to the pinna and proximally to the tympanic membrane. The acute form has an annual incidence of approximately 1 percent1 and a lifetime prevalence of 10 percent.2 On rare occasions, the infection invades the surrounding soft tissue and bone; this is known as malignant (necrotizing) otitis externa, and is a medical emergency that occurs primarily in older patients with diabetes mellitus.3 Otitis externa lasting three months or longer, known as chronic otitis externa, is often the result of allergies, chronic dermatologic conditions, or inadequately treated acute otitis externa.
SORT: KEY RECOMMENDATIONS FOR PRACTICE
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| Clinical recommendation | Evidence rating | References | Comments |
--- --- |
| Acute otitis externa should be distinguished from other possible causes of ear canal inflammation. | C | 4 | Different and overlapping pathologies may require alteration in treatment |
| Topical antimicrobial otic preparations should be considered the first-line treatment foruncomplicated acute otitis externa. | A | 4, 14–16 | Choose specific preparation based on risk of adverse effects, tympanic membrane status, cost, adherence issues, etc. |
| Addition of a topical corticosteroid may result in faster resolution of symptoms such as pain, canal edema, and canal erythema. | B | 4, 14–16 | May be partly dependent on strength of corticosteroid |
| Systemic antibiotics should be used only if the infection has spread beyond the ear canal or in patients at high risk of such spread. | B | 4, 14, 16 | — |
| Use of aural toilet should be considered to remove debris from the ear canal before treatment. | C | 4, 16 | Widely performed in research studies, but no data on effectiveness |
A = consistent, good-quality patient-oriented evidence; B = inconsistent or limited-quality patient-oriented evidence; C = consensus, disease-oriented evidence, usual practice, expert opinion, or case series. For information about the SORT evidence rating system, go to
Etiology
In North America, 98 percent of cases of acute otitis externa are caused by bacteria.4 The two most common isolates are Pseudomonas aeruginosa and Staphylococcus aureus. However, a wide variety of other aerobic and anaerobic bacteria have been isolated.5,6 Approximately one-third of cases are polymicrobial.4 Fungal pathogens, primarily those of the Aspergillus and Candida species, occur more often in tropical or subtropical environments and in patients previously treated with antibiotics.7–9 Inflammatory skin disorders and allergic reactions may cause noninfectious otitis externa, which can be chronic.
Risk Factors
Several factors may predispose patients to the development of acute otitis externa (Table 1).4,10 One of the most common predisposing factors is swimming, especially in fresh water. Other factors include skin conditions such as eczema and seborrhea, trauma from cerumen removal, use of external devices such as hearing aids, and cerumen buildup.4 These factors appear to work primarily through loss of the protective cerumen barrier, disruption of the epithelium (including maceration from water retention), inoculation with bacteria, and increase in the pH of the ear canal.10–12
Table 1.
Predisposing Factors for Otitis Externa
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| |
| Anatomic abnormalities |
| Canal stenosis |
| Exostoses |
| Hairy ear canals |
| Canal obstruction |
| Cerumen obstruction |
| Foreign body |
| Sebaceous cyst |
| Cerumen/epithelial integrity |
| Cerumen removal |
| Earplugs |
| Hearing aids |
| Instrumentation/itching |
| Dermatologic conditions |
| Eczema |
| Psoriasis |
| Seborrhea |
| Other inflammatory dermatoses |
| Water in ear canal |
| Humidity |
| Sweating |
| Swimming or other prolonged water exposure |
| Miscellaneous |
| Purulent otorrhea from otitis media |
| Soap |
| Stress |
| Type A blood |
Information from references 4 and 10.
Prevention
A number of preventive measures have been recommended, including use of earplugs while swimming, use of hair dryers on the lowest settings and head tilting to remove water from the ear canal, and avoidance of self-cleaning or scratching the ear canal. Acetic acid 2% (Vosol) otic solutions are also used, either two drops twice daily or two to five drops after water exposure. However, no randomized trials have examined the effectiveness of any of these measures.
Diagnosis
Acute otitis externa is diagnosed clinically based on signs and symptoms of canal inflammation (Table 24 ; Figures 1 and 2). Presentation can range from mild discomfort, itching, and minimal edema to severe pain, complete canal obstruction, and involvement of the pinna and surrounding skin. Pain is the symptom that best correlates with the severity of disease.13 Mild fever may be present, but a temperature greater than 101°F (38.3°C) suggests extension beyond the auditory canal.
Table 2.
Diagnosis of Otitis Externa
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| | |
--- |
| Onset of symptoms within 48 hours in the past three weeks | |
| | and |
| Symptoms of ear canal inflammation: | |
| | Ear pain, itching, or fullness |
| | With or without hearing loss or jaw pain |
| | and |
| Signs of ear canal inflammation: | |
| | Tenderness of tragus/pinna or ear canal edema/erythema |
| | With or without otorrhea, tympanic membrane erythema, cellulitis of the pinna, or local lymphadenitis |
Adapted with permission from Rosenfeld RM, Brown L, Cannon CR, et al.; American Academy of Otolaryngology–Head and Neck Surgery Foundation. Clinical practice guideline: acute otitis externa. Otolaryngol Head Neck Surg. 2006;134(4 suppl):S5.
Figure 1.
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Figure 2.
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Acute otitis externa should be distinguished from other causes of ear canal inflammation4 (Table 3; see previous AFP article on ear pain [ Proper evaluation includes a history of presenting and associated symptoms, water exposure, local trauma/cerumen removal, inflammatory skin disorders, diabetes, ear surgeries, and local radiotherapy. Physical examination should include the auricle and surrounding lymph nodes, a skin examination, otoscopy of the ear canal, and verification that the tympanic membrane is intact. Tenderness with movement of the tragus or pinna is a classic finding.
Table 3.
Conditions That May Be Confused with Acute Otitis Externa
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| Condition | Distinguishing characteristics | Comment |
---
| Acute otitis media | Presence of middle ear effusion, no tragal/pinnal tenderness | Use pneumatic otoscopy or tympanometry, treat with systemic antibiotics |
| Chronic otitis externa | Itching is often predominant symptom, erythematous canal, lasts more than three months | Treat underlying causes/conditions |
| Chronic suppurative otitis media | Chronic otorrhea, nonintact tympanic membrane | Control otitis externa symptoms, then treat otitis media |
| Contact dermatitis | Allergic reaction to materials (e.g., metals, soaps, plastics) in contact with the skin/epithelium; itching is predominant symptom | Check for piercings, hearing aids, or earplug use; discontinue exposure when possible |
| Eczema | Itching is predominant symptom; often chronic; history of atopy, outbreaks in other locations | Consider treatment with topical corticosteroids |
| Furunculosis | Focal infection, may be pustule or nodule, often in distal canal | Consider treatment with heat, incision and drainage, or systemic antibiotics; can progress to diffuse otitis externa |
| Malignant otitis externa | High fever, granulation tissue or necrotic tissue in ear canal, may have cranial nerve involvement; patient with diabetes mellitus or immunocompromise, elevated erythrocyte sedimentation rate, findings on computed tomography | Medical emergency with high morbidity rate and possible mortality; warrants emergent consultation with otolaryngologist, hospitalization, intravenous antibiotics, debridement |
| Myringitis | Tympanic membrane inflammation, may have vesicles; pain is often severe, no canal edema | Usually results from acute otitis media or viral infection |
| Otomycosis | Itching is predominant symptom, thick material in canal, less edema; may see fungal elements on otoscopy (Figures 3 and 4) | Can coexist with bacterial infections; treat with acetic acid (Vosol), half acetic acid/half alcohol, or topical antifungals; meticulous cleaning of ear canal |
| Ramsay Hunt syndrome | Herpetic ulcers in canal; may have facial numbness/paralysis, severe pain, loss of taste | Treatment includes antivirals, systemic corticosteroids |
| Referred pain | Normal ear examination | Look for other causes based on patterns of referred pain |
| Seborrhea | Itching and rash on hairline, face, scalp | Treatment includes lubricating or moisturizing the external auditory canal |
| Sensitization to otics | Severe itching, maculopapular or erythematous rash in conchal bowl and canal; may have streak on pinna where preparation contacted skin; vesicles may be present | Type IV delayed hypersensitivity reaction to neomycin or other components of otic solutions; discontinue offending agent; treat with topical corticosteroids |
Because otitis externa can cause tympanic membrane erythema, pneumatic otoscopy or tympanometry should be used to differentiate it from otitis media. Otomycosis is classically associated with itching, thick material in the ear canal, and failure to improve with use of topical antibacterials. Otomycosis can sometimes be identified during otoscopy (Figures 3 and 4), although nonpathogenic saprophytic fungi may also be found.
Figure 3.
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Figure 4.
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Malignant otitis externa may be suspected in older patients with diabetes mellitus or immunocompromise who have refractory purulent otorrhea and severe otalgia that may worsen at night. Clinical findings include granulation tissue in the external auditory canal, especially at the bone-cartilage junction. Extension of the infection beyond the auditory canal can cause lymphadenopathy, trismus, and facial nerve and other cranial nerve palsies.
In chronic otitis externa, the symptoms and signs listed in Table 24 occur for more than three months. Classic symptoms include itching and mild discomfort; there may also be lichenification on otoscopy.
Treatment
TOPICAL MEDICATIONS
Topical antimicrobials, with or without topical corticosteroids, are the mainstay of treatment for uncomplicated acute otitis externa. Topical antimicrobials are highly effective compared with placebo, demonstrating an absolute increase in clinical cure rate of 46 percent or a number needed to treat of slightly more than two.4,14–16 Topical agents come in a variety of preparations and combinations; a recent systematic review included 26 different topical interventions.15 In some studies, ophthalmic preparations have been used off-label to treat otitis externa.14,15 Ophthalmic preparations may be better tolerated than otic preparations, possibly due to differences in pH between the preparations, and may help facilitate compliance with treatment recommendations. Commonly studied antimicrobial agents include aminoglycosides, polymyxin B, quinolones, and acetic acid. No consistent evidence has shown that any one agent or preparation is more effective than another.4,14–16 There is limited evidence that use of acetic acid alone may require two additional days for resolution of symptoms compared with other agents, and that it is less effective if treatment is required for more than seven days.15
Current guidelines recommend factoring in the risk of adverse effects, adherence issues, cost, patient preference, and physician experience. Some components found in otic preparations may cause contact dermatitis.17 Hypersensitivity to aminoglycosides, particularly neomycin, may develop in up to 15 percent of the population, and has been identified in approximately 30 percent of patients who also have chronic or eczematous otitis externa.17,18 Adherence to topical therapy increases with ease of administration, such as less frequent dosing.19 The addition of a topical corticosteroid yields more rapid improvement in symptoms such as pain, canal edema, and erythema.4,14–16 Cost varies considerably for the different preparations20,21 (Table 44,14–16,20,21 ).
Table 4.
Common Antimicrobial Otic Preparations for Otitis Externa
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| Component | Cost of generic (brand) | Dosage | Use if tympanic membrane perforation? | Comments |
--- ---
| Acetic acid 2% (Vosol) | $39 for 15 mL ($36 for 15 mL) | Four to six times daily | No | May cause pain and irritation; may be less effective than other treatments if use is required beyond one week; often used as prophylactic agent |
| Ciprofloxacin 0.3%/dexamethasone 0.1% (Ciprodex) | Not available ($160 for 7.5 mL) | Twice daily | Yes | Low risk of sensitization |
| Hydrocortisone 2%/acetic acid 1% (Vosol HC) | $220 for 10 mL† ($215 for 10 mL) | Four to six times daily | No | May cause pain and irritation |
| Neomycin/polymyxin B/hydrocortisone, solution or suspension | $28 for 10-mL solution; $30 for 10-mL suspension† ($85 for 10-mL solution; $78 for 10-mL suspension)† | Three to four times daily | No | Ototoxic; higher risk of contact hypersensitivity; avoid in chronic/eczematous otitis externa4 |
| Ofloxacin 0.3% | $60 for 5 mL; $93 for 10 mL† ($80 for 5 mL; $143 for10 mL)† | Once to twice daily | Yes | Low risk of sensitization |
—Estimated retail price of one course of treatment (10 to 14 days) based on information obtained at Red Book Online at (accessed April 5, 2012).
†—Estimated retail price of one course of treatment (10 to 14 days) based on information obtained at (accessed April 5, 2012).
Information from references 4, 14 through 16, 20, and 21.
ORAL ANTIBIOTICS
Systemic antibiotics increase the risks of adverse effects, generation of resistant organisms, and recurrence. They also increase time to clinical cure and do not improve outcomes compared with a topical agent alone in uncomplicated otitis externa.1,6,16,22 Systemic antibiotics should be used only when the infection has spread beyond the ear canal, or when there is uncontrolled diabetes, immunocompromise, a history of local radiotherapy, or an inability to deliver topical antibiotics.4,14,16
TREATMENT METHODS
Use of a topical otic preparation without culture is a reasonable treatment approach for patients who have mild symptoms of otitis externa. If the tympanic membrane is intact and there is no concern of hypersensitivity to aminoglycosides, a neomycin/polymyxin B/hydrocortisone otic preparation would be a first-line therapy because of its effectiveness and low cost. Ofloxacin and ciprofloxacin/dexamethasone (Ciprodex) are approved for middle ear use and should be used if the tympanic membrane is not intact or its status cannot be determined visually 4; these also may be useful if patients are hypersensitive to neomycin, or if nonadherence to treatment because of dosing frequency is an issue. Use of a corticosteroid-containing preparation is recommended to provide more rapid relief when symptoms warrant.
Patients should be taught to properly administer otic medications. The patient should lie down with his or her affected side facing upward, running the preparation along the side of the ear canal until it is full and gently moving the pinna to relieve air pockets. The patient should remain in this position for three to five minutes, after which the canal should not be occluded, but rather left open to dry.4 It may benefit the patient to have another person administer the ear drops, because only 40 percent of patients self-medicate appropriately.23 Patients should be instructed to minimize trauma to (and manipulation of) the ear, and to avoid water exposure, including abstinence from water sports for a week or, at minimum, avoidance of submersion.
When there is marked canal edema, a wick of compressed cellulose or ribbon gauze may be placed in the canal to facilitate antimicrobial or antibiotic administration. Wick placement permits antibiotic drops to reach portions of the external auditory canal that are inaccessible because of canal swelling. As the canal responds to treatment and patency returns to the ear canal, the wick often falls out.
ANALGESIA
Pain is a common symptom of acute otitis externa, and can be debilitating.12 Oral analgesics are the preferred treatment. First-line analgesics include nonsteroidal anti-inflammatory drugs and acetaminophen. When ongoing frequent dosing is required to control pain, medications should be administered on a scheduled rather than as-needed basis. Opioid combination pills may be used when symptom severity warrants. Benzocaine otic preparations may compromise the effectiveness of otic antibiotic drops by limiting contact between the drop and the ear canal. The lack of published data supporting the effectiveness of topical benzocaine preparations in otitis externa limits the role of such treatments.
CLEANING THE CANAL
Acute otitis externa can be associated with copious material in the ear canal. Consensus guidelines published by the American Academy of Otolaryngology recommend that such material be removed to achieve optimal effectiveness of the topical antibiotics.4,16 However, no randomized controlled trials have examined the effectiveness of aural toilet, and this is not typically done in most primary care settings.4,15 Topical medications rely on direct contact with the infected skin of the ear canal; hence, aural toilet takes on greater importance when the volume or thickness of the debris in the ear canal is great. Guidelines recommend aural toilet by gentle lavage suctioning or dry mopping under otoscopic or microscopic visualization to remove obstructing material and to verify tympanic membrane integrity.4 Lavage should be used only if the tympanic membrane is known to be intact, and should not be performed on patients with diabetes because of the potential risk of causing malignant otitis externa.4 Pain medications may be required during the procedure.
CHRONIC OTITIS EXTERNA
The treatment of chronic otitis externa depends on the underlying causes. Because most cases are caused by allergies or inflammatory dermatologic conditions, treatment includes the removal of offending agents and the use of topical or systemic corticosteroids. Chronic or intermittent otorrhea over weeks to months, particularly with an open tympanic membrane, suggests the presence of chronic suppurative otitis media. Initial treatment efforts are similar to those for acute otitis media. With control of the symptoms of otitis externa, attention can shift to the management of chronic suppurative otitis media.
Follow-up and Referral
Most patients will experience considerable improvement in symptoms after one day of treatment. If there is no improvement within 48 to 72 hours, physicians should reevaluate for treatment adherence, misdiagnosis (Table 3), sensitivity to ear drops, or continued canal patency. The physician should consider culturing material from the canal to identify fungal and antibiotic-resistant pathogens if the patient does not improve after initial treatment efforts or has one or more predisposing risk factors, or if there is suspicion that the infection has extended beyond the external auditory canal. There is a lack of data regarding optimal length of treatment; as a general rule, antimicrobial otics should be administered for seven to 10 days, although in some cases complete resolution of symptoms may take up to four weeks.4,15
Consultation with an otolaryngologist or infectious disease subspecialist may be warranted if malignant otitis externa is suspected; in cases of severe disease, lack of improvement or worsening of symptoms despite treatment, and unsuccessful lavage; or if the primary care physician determines that aural toilet or ear wick insertion is warranted, but is unfamiliar with or concerned about performing the procedure.
Data Sources: We performed multiple searches of PubMed Clinical Queries using the search term otitis externa. We also searched Essential Evidence Plus, the Cochrane Database of Systematic Reviews, the National Guideline Clearinghouse, Clinical Evidence, Dynamed, and UpToDate. Search dates: January 1, 2011, through April 6, 2011.
PAUL SCHAEFER, MD, PhD, is an assistant professor, clerkship director, and director for medical student education in the Department of Family Medicine at the University of Toledo (Ohio) College of Medicine.
REGINALD F. BAUGH, MD, is a professor and chief of otolaryngology in the Department of Surgery at the University of Toledo College of Medicine.
Address correspondence to Paul Schaefer, MD, PhD, University of Toledo Health Science Campus, MS 1179, 2240 Dowling Hall, 3000 Arlington Ave., Toledo, OH 43614 (e-mail: paul.schaefer@utoledo.edu). Reprints are not available from the authors.
Author disclosure: No relevant financial affiliations to disclose.
Rowlands S, Devalia H, Smith C, Hubbard R, Dean A. Otitis externa in UK general practice: a survey using the UK General Practice Research Database. Br J Gen Pract. 2001;51(468):533-538.
Raza SA, Denholm SW, Wong JC. An audit of the management of acute otitis externa in an ENT casualty clinic. J Laryngol Otol. 1995;109(2):130-133.
Rubin Grandis J, Branstetter BF, Yu VL. The changing face of malignant (necrotising) external otitis: clinical, radiological, and anatomic correlations. Lancet Infect Dis. 2004;4(1):34-39.
Rosenfeld RM, Brown L, Cannon CR, et al.; American Academy of Otolaryngology–Head and Neck Surgery Foundation. Clinical practice guideline: acute otitis externa. Otolaryngol Head Neck Surg. 2006;134(4 suppl):S4-S23.
Ninkovic G, Dullo V, Saunders NC. Microbiology of otitis externa in the secondary care in United Kingdom and antimicrobial sensitivity. Auris Nasus Larynx. 2008;35(4):480-484.
Roland PS, Stroman DW. Microbiology of acute otitis externa. Laryngoscope. 2002;112(7 pt 1):1166-1177.
Martin TJ, Kerschner JE, Flanary VA. Fungal causes of otitis externa and tympanostomy tube otorrhea. Int J Pediatr Otorhinolaryngol. 2005;69(11):1503-1508.
Pontes ZB, Silva AD, Lima Ede O, et al. Otomycosis: a retrospective study. Braz J Otorhinolaryngol. 2009;75(3):367-370.
Ahmad N, Etheridge C, Farrington M, Baguley DM. Prospective study of the microbiological flora of hearing aid moulds and the efficacy of current cleaning techniques. J Laryngol Otol. 2007;121(2):110-113.
Russell JD, Donnelly M, McShane DP, Alun-Jones T, Walsh M. What causes acute otitis externa?. J Laryngol Otol. 1993;107(10):898-901.
Kim JK, Cho JH. Change of external auditory canal pH in acute otitis externa. Ann Otol Rhinol Laryngol. 2009;118(11):769-772.
van Asperen IA, de Rover CM, Schijven JF, et al. Risk of otitis externa after swimming in recreational fresh water lakes containing Pseudomonas aeruginosa. BMJ. 1995;311(7017):1407-1410.
Halpern MT, Palmer CS, Seidlin M. Treatment patterns for otitis externa. J Am Board Fam Pract. 1999;12(1):1-7.
Rosenfeld RM, Singer M, Wasserman JM, Stinnett SS. Systematic review of topical antimicrobial therapy for acute otitis externa. Otolaryngol Head Neck Surg. 2006;134(4 suppl):S24-S48.
Kaushik V, Malik T, Saeed SR. Interventions for acute otitis externa. Cochrane Database Syst Rev. ;2010(1):CD004740.
Hajioff D, Mackeith S. Otitis externa. Clin Evid (Online). 2010.
Smith IM, Keay DG, Buxton PK. Contact hypersensitivity in patients with chronic otitis externa. Clin Otolaryngol Allied Sci. 1990;15(2):155-158.
Yariktas M, Yildirim M, Doner F, Baysal V, Dogru H. Allergic contact dermatitis prevalence in patients with eczematous external otitis. Asian Pac J Allergy Immunol. 2004;22(1):7-10.
Shikiar R, Halpern MT, McGann M, Palmer CS, Seidlin M. The relation of patient satisfaction with treatment of otitis externa to clinical outcomes: development of an instrument. Clin Ther. 1999;21(6):1091-1104.
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EverydayHealth.com. Drugs A-Z on everyday health. Accessed April 5, 2011.
Roland PS, Belcher BP, Bettis R, et al.; Cipro HC Study Group. A single topical agent is clinically equivalent to the combination of topical and oral antibiotic treatment for otitis externa. Am J Otolaryngol. 2008;29(4):255-261.
England RJ, Homer JJ, Jasser P, Wilde AD. Accuracy of patient self-medication with topical eardrops. J Laryngol Otol. 2000;114(1):24-25.
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Brain-Eating Amoeba
Naegleri fowleri is an amoeba that can cause a serious central nervous system infection. The amoeba is found in warm and still fresh water bodies of water and enters a human body through the nose. Successful treatment has included miltefosine.
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ContentsOverviewSymptoms and CausesDiagnosis and TestsManagement and TreatmentOutlook / PrognosisPreventionLiving With
Overview
What is brain-eating amoeba (Naegleria fowleri)?
Naegleria fowleri is an amoeba that lives throughout the world in warm and shallow bodies of fresh water, such as lakes, rivers and hot springs. It also lives in soil. It’s considered a free-living organism because it doesn’t need a host to live.
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People who become infected by this amoeba develop a condition called primary amoebic meningoencephalitis (PAM). PAM is a very serious infection of the central nervous system that’s almost always fatal.
Note: You might also see the words “ameba” instead of “amoeba” and “amebic” instead of “amoebic.” Amoeba is more common than ameba, but both words refer to an organism with one cell.
How do you get infected by brain-eating amoeba (Naegleria fowleri)?
The most common way of being infected by this type of amoeba happens when infected water goes into your nose. From there, the amoeba goes to your brain. This usually happens when you’re swimming, diving or doing something like water skiing in infected water. In extremely rare cases, the infected water can be heated tap water or swimming pool water that isn’t chlorinated enough.
You can’t be infected by swallowing infected water.
How common is an infection due to brain-eating amoeba (Naegleria fowleri)?
Luckily, there are only a few cases each year in the U.S. (estimated to be between zero and eight). Most of the cases happen in the southern states, such as Florida and Texas, and involve young males. In later years, though, some cases have happened in northern states during periods of very hot weather.
Studies are being done that call into question how rare infection with Naegleria fowleri really is. Some people have antibodies to the amoeba, indicating that they’ve been infected and survived. Some cases of deaths attributed to meningitis have been reclassified as deaths due to brain-eating amoeba (Naegleria fowleri).
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Another question is, why some people aren’t infected with the amoeba even though they’re in the same place doing the same things as people who’ve been infected?
Symptoms and Causes
What are the symptoms of infection with brain-eating amoeba (Naegleria fowleri)?
The signs and symptoms of primary amoebic meningoencephalitis (PAM) come on suddenly and are severe at the start, including:
High fever.
Very painful headache.
Nausea and vomiting.
Trembling.
Symptoms like those of meningitis, including a stiff neck and extreme sensitivity to light (photophobia).
Mental confusion.
Coma.
The fatality rate is higher than 97% even with treatment.
What causes infection with brain-eating amoeba (Naegleria fowleri)?
The infection occurs when the amoeba known as Naegleria fowleri gets into your brain through your nasal cavity. It can enter your body if you inhale any infected water. Usually, the amoeba lives in freshwater bodies of water that are warm, including hot springs (geothermal water).
You can also be infected by inhaling infected dust.
There have been other cases of reported infection by brain-eating amoeba (Naegleria fowleri) due to people using tap water rather than distilled or sterilized water to rinse out their noses with devices like a neti pot.
How long is the incubation period for infection with brain-eating amoeba (Naegleria fowleri)?
It takes about two to 15 days after you’ve been exposed to the amoeba for symptoms to develop.
Can I get infected with brain-eating amoeba (Naegleria fowleri) from being around someone who has it?
No. There haven’t been any cases found where the infection spread from person to person. There are studies underway to see if the infection can spread by tissue or organ donation.
Diagnosis and Tests
How is infection with brain-eating amoeba diagnosed?
If a healthcare provider thinks you might’ve been infected by brain-eating amoeba (Naegleria fowleri), they’ll recommend a spinal tap — also known as lumbar puncture — to see if the organism is in your cerebrospinal fluid (CSF).
Your provider may also recommend a brain biopsy. During this procedure, they’ll take a tissue sample and examine it under a microscope to check for the presence of the amoeba.
Management and Treatment
How is infection with brain-eating amoeba (Naegleria fowleri) treated?
The treatment of choice for primary amoebic meningoencephalitis (PAM), or infection with brain-eating amoeba (Naegleria fowleri) is the antifungal amphotericin B. Some survivors in North America were treated with a combination of drugs that included amphotericin B, rifampin, fluconazole and a drug called miltefosine. Miltefosine is a drug approved for treating leishmaniasis, a parasitic disease that’s spread by sandflies.
The best results (in two children who recovered completely) came from early diagnosis and treatment with the recommended drugs, along with cooling the body to below-normal temperature to treat brain swelling.
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Outlook / Prognosis
What is the prognosis (outlook) for people who are infected with brain-eating amoeba (Naegleria fowleri)?
The outlook for someone who is infected with Naegleria fowleri is very poor. Even with treatment, most people die from this condition. Coma followed by death usually happens in a week or 10 days after symptoms and signs begin.
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Prevention
How can I prevent myself from being infected with brain-eating amoeba (Naegleria fowleri)?
Because the outlook for this condition is so dire, prevention can be important, even though the condition is very rare. These are key things to remember:
Don’t swim, wade or do watersports in warm freshwater locations, especially still waters, without nose plugs. Don’t go into the water at all if Naegleria fowleri is known to be present or likely to be present.
Don’t use tap water for a neti pot or any other device that cleans your nasal passages. Only use distilled or sterilized water. If you must use tap water, make sure that you boil it for one minute and then let it cool. If you live somewhere that’s 6,500 feet above sea level, boil the water for three minutes and let cool.
You can use filters to remove germs from water. Use filters labeled “NSF 53,” “NSF 58” or “absolute pore size of 1 micron or smaller.”
You can also use chlorine bleach liquid or tablets to disinfect your water for cleaning your nose and sinuses. Disinfecting water for nasal use requires a different amount of bleach than disinfecting water for drinking.
If you do develop symptoms of fever or headache after going into warm freshwater, tell your healthcare provider where you’ve been.
Living With
When should I see my healthcare provider if I have questions about infection with brain-eating amoeba (Naegleria fowleri)?
If you get feverish or have a headache after you’ve been active in a warm freshwater body of water or after you’ve used tap water to irrigate your nasal pages, see a healthcare provider or go to the emergency room immediately. Early diagnosis and treatment of infection with brain-eating amoeba (Naegleria fowleri) is critical.
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A note from Cleveland Clinic
Hearing about a condition caused by a brain-eating amoeba and knowing how dangerous it can be is certainly scary. If you think you may have been exposed to the amoeba, get immediate medical help. But it’s important to remember how very rare this condition is. You can do your part in preventing it by using only distilled or sterilized water to rinse your nasal passages and by avoiding water you suspect may be infected, especially in hot weather.
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Last reviewed on 11/29/2022.
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11248 | https://www.cnblogs.com/lixddd/p/15914305.html | 应用随机过程03：马尔可夫链的状态 - 这个XD很懒 - 博客园
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应用随机过程03：马尔可夫链的状态
本文主要对马尔可夫链的状态进行了分类讨论，介绍了一些用来刻画马尔可夫链状态特性的变量。
目录
第三讲 马尔可夫链的状态
一、常返和暂留
Part 1：常返和暂留的定义
Part 2：常返的判别条件 I
Part 3：常返的判别条件 II
二、状态之间的关系
Part 1：互达
Part 2：周期
第三讲 马尔可夫链的状态
一、常返和暂留
Part 1：常返和暂留的定义
在马尔可夫链运行和转移的过程中，各个状态所起的作用不完全相同，因此我们要讨论几类特殊状态的特点，并给出状态空间的分解定理。设 {X n}{X n} 是一个时齐马尔可夫链，这是我们这一节的研究对象。此外这一节的概念比较多，需要注意区分。
首中时：给定状态 j∈I j∈I ，定义 τ j=inf{n≥1:X n=j}τ j=inf{n≥1:X n=j} ，称为 j j 的首中时。约定 inf∅=∞inf∅=∞ ，即如果不存在 n≥1 n≥1 使得 X n=j X n=j ，则定义 τ j=∞τ j=∞ 。
首中时的含义是在零时刻从状态 i i 出发的马尔可夫链首次到达状态 j j 的时刻。
当 X 0=i=j X 0=i=j 时，τ j τ j 表示马尔可夫链首次回到状态 j j 的时刻；
当 X 0=i≠j X 0=i≠j 时，τ j τ j 表示马尔可夫链首次到达状态 j j 的时刻。
利用首中时的概念，我们可以将马尔可夫链的状态分为常返态和暂留态。
如果 P(τ j<∞|X 0=j)=1 P(τ j<∞|X 0=j)=1 ，则称 j j 是常返态。常返态的含义是从某一状态出发以概率 1 1 在有限时间内返回该状态。
如果 P(τ j<∞|X 0=j)<1 P(τ j<∞|X 0=j)<1 ，则称 j j 是暂留态。暂留态的含义是从某一状态出发以一个正的概率不再返回该状态。
平均回转时：若状态 j j 是常返态，定义 μ j=E(τ j|X 0=j)μ j=E(τ j|X 0=j) ，称为 j j 的平均回转时。
利用平均回转时的概念，我们可以将常返态进一步分为正常返态和零常返态。
如果 μ j=E(τ j|X 0=j)<∞μ j=E(τ j|X 0=j)<∞ ，则称 j j 是正常返态。
如果 μ j=E(τ j|X 0=j)=∞μ j=E(τ j|X 0=j)=∞ ，则称 j j 是零常返态。
上述定义说明的是，在平均的意义下，正常返态的返回速度快于零常返态的返回速度。
简单回顾一下，这一部分我们介绍了两个概念——首中时和平均回转时。根据首中时我们将马尔可夫链的状态分为常返态和暂留态，根据平均回转时我们又将常返态分为了正常返态和零常返态。
Part 2：常返的判别条件 I
在上一节的学习中，我们知道马尔可夫链的性质常常用转移概率来刻画。在定义了常返态和暂留态之后，我们也希望能用一个概率来刻画常返态和暂留态的性质。
首先我们定义 n n 步首次击中概率和 n n 步首次返回概率。
令 f(n)i j f i j(n) 表示从状态 i i 出发经过 n n 步之后首次击中状态 j j 的概率，则有
f(n)i j=P(τ j=n|X 0=i)=P(X n=j,X n−1≠j,⋯,X 1≠j|X 0=i).f i j(n)=P(τ j=n|X 0=i)=P(X n=j,X n−1≠j,⋯,X 1≠j|X 0=i).
令 f(n)j j f j j(n) 表示从状态 j j 出发经过 n n 步之后首次返回状态 j j 的概率，则有
f(n)j j=P(τ j=n|X 0=j)=P(X n=j,X n−1≠j,⋯,X 1≠j|X 0=j).f j j(n)=P(τ j=n|X 0=j)=P(X n=j,X n−1≠j,⋯,X 1≠j|X 0=j).
接着我们定义可达概率和可返回概率。
令 f i j f i j 表示从状态 i i 出发在有限步能击中状态 j j 的概率，则有
f i j=P(τ j<∞|X 0=i)=∞∑n=1 f(n)i j.f i j=P(τ j<∞|X 0=i)=∑n=1∞f i j(n).
令 f j j f j j 表示从状态 j j 出发在有限步能返回状态 j j 的概率，则有
f j j=P(τ j<∞|X 0=j)=∞∑n=1 f(n)j j.f j j=P(τ j<∞|X 0=j)=∑n=1∞f j j(n).
根据以上定义，我们可以得到常返态的第一个判别条件：
状态 j j 是常返态，当且仅当 f j j=1 f j j=1 。
状态 j j 是暂留态，当且仅当 f j j<1 f j j<1 。
若状态 j j 是常返态，我们可以通过 n n 步首次返回概率计算平均回转时，从而对正常返态和零常返态进行判别。利用离散型随机变量的数学期望的定义即可得到
μ j=E(τ j|X 0=j)=∞∑n=1 n f(n)j j.μ j=E(τ j|X 0=j)=∑n=1∞n f j j(n).
概括以上判别方法的主要思路：对于状态 j j 的每一步 n n 求出 f(n)j j f j j(n) ，根据 f j j=∞∑n=1 f(n)j j f j j=∑n=1∞f j j(n) 和 μ j=∞∑n=1 n f(n)j j μ j=∑n=1∞n f j j(n) 计算可返回概率和平均回转时，从而判断状态 j j 的常返性。
在实际应用时，我们一般计算 f(n)j j f j j(n) 的方法是画出状态转移图，利用图论的知识和一步转移概率进行计算。显然，这种方法适用于状态转移过程简单的马尔可夫链，对于复杂的马尔可夫链并不适用。
Part 3：常返的判别条件 II
回到时齐的马尔可夫链本身，我们往往已知的是一步转移概率矩阵 P P ，从而很容易得到 n n 步转移概率矩阵 P n P n 。下面我们就从转移概率的角度来给出常返态的判别条件。
首先讨论一下常返态的特性：假设状态 j j 是常返态，并且过程开始时处于 j j ，则过程将以概率 1 1 返回 j j 。由马尔可夫链的定义知，当它再次进入 j j 时，上述过程将被重复，从而状态 j j 将以概率 1 1 再次被访问。继续重复可得如下结论：如果状态 j j 是常返态，那么开始处于状态 j j 的过程将以概率 1 1 无穷多次地返回状态 j j 。
定义 N j=♯{n≥0:X n=j}N j=♯{n≥0:X n=j} ，表示马尔可夫链访问 j j 的次数。我们可以得到常返态的第二个判别条件：
状态 j j 是常返态，当且仅当 P(N j=∞|X 0=j)=1 P(N j=∞|X 0=j)=1 ，当且仅当 ∞∑n=1 p(n)j j=∞∑n=1∞p j j(n)=∞ 。
状态 j j 是暂留态，当且仅当 P(N j<∞|X 0=j)=1 P(N j<∞|X 0=j)=1 ，当且仅当 ∞∑n=1 p(n)j j<∞∑n=1∞p j j(n)<∞ 。
这里我们需要给出第二个等价条件的证明。根据数学期望的定义，显然有如下结论成立：
若 P(N j=∞|X 0=j)=1 P(N j=∞|X 0=j)=1 ，则 E(N j|X 0=j)=∞E(N j|X 0=j)=∞ ；
若 P(N j<∞|X 0=j)=1 P(N j<∞|X 0=j)=1 ，则 E(N j|X 0=j)<∞E(N j|X 0=j)<∞ 。
令 I n={1,X n=j,0,X n≠j.I n={1,X n=j,0,X n≠j. ，则 N j=∞∑n=0 I n N j=∑n=0∞I n 表示马尔可夫链处于状态 j j 的次数。于是
E(N j|X 0=j)=∞∑n=0 E(I n|X 0=j)=∞∑n=0 P(X n=j|X 0=j)=∞∑n=0 p(n)j j.E(N j|X 0=j)=∑n=0∞E(I n|X 0=j)=∑n=0∞P(X n=j|X 0=j)=∑n=0∞p j j(n).
如此我们就证明了第二个等价条件成立。
我们再讨论一下暂留态的特性：假设状态 i i 是暂留态，则有可返回概率 f j j<1 f j j<1 。因此过程每次访问 i i 都将以一个正的概率 1−f i i 1−f i i 不再进入这个状态。所以，开始处于状态 i i 的过程将恰好在状态 i i 访问 n n 次的概率等于 f n−1 i i(1−f i i)f i i n−1(1−f i i) 。
换句话说，如果状态 i i 是暂留态，那么开始处于状态 i i 的过程再次访问 i i 的次数服从参数为 1−f i i 1−f i i 的几何分布。利用几何分布的数学期望，也可以得到
∞∑n=0 p(n)j j=E(N j|X 0=j)=1 1−f j j<∞.∑n=0∞p j j(n)=E(N j|X 0=j)=1 1−f j j<∞.
推论：一个暂留态只能被访问有限次，从而一个有限状态的马尔可夫链中至少有一个状态是常返态。推论可以用反证法进行证明，假设所有状态都是暂留态，进而推出在有限时间后无状态可访问即可。
二、状态之间的关系
Part 1：互达
关于常返和暂留的特点，以上我们都是对单个状态进行讨论的，下面我们考虑能否通过状态之间的关系来判断一个状态的常返性。
首先介绍一下可达和互达的概念。设 i i 和 j j 是状态空间 I I 中的任意两个状态：
如果存在 n≥1 n≥1 使得 p(n)i j>0 p i j(n)>0 ，则称状态 i i 可达状态 j j ，记为 i→j i→j 。
如果 i→j i→j 且 j→i j→i ，则称状态 i i 和状态 j j 互达，记为 i↔j i↔j 。
可以证明互达关系是一种等价关系，满足以下三个性质：
自反性：i↔j i↔j ；
对称性：如果 i↔j i↔j ，则 j↔i j↔i ；
传递性：如果 i↔j,j↔k i↔j,j↔k ，则 i↔k i↔k 。
我们将两个互达的状态，称为属于同一个互达等价类中。于是，按照互达关系，状态空间 I I 可以表示成可列个互不相交的互达等价类的并。如果状态空间 I I 中的任意两个状态都是互达的，则称该马尔可夫链是不可约的。
闭集：设 C⊂I C⊂I 是一个互达等价类，如果从 C C 中的任何状态出发，都无法到达 I∖C I∖C 中的状态，则称 C C 为闭集。换句话说，如果 C C 是闭集，则对任意 i∈C i∈C 和 j∉C j∉C ，都有 i↛j i↛j 。一般规定空集 ∅∅ 和状态空间 I I 是闭集。
吸收态：如果闭集 C⊂I C⊂I 只有一个状态 i i ，即 C={i}C={i} ，则称状态 i i 为吸收态。吸收态的含义是一旦处于该状态，它将永远不会离开。
我们可以举个例子来理解，设马尔可夫链的状态空间 I={1,2,3,4,5,6}I={1,2,3,4,5,6} ，转移概率矩阵为
P=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣1 2 1 2 0 0 0 0 1 4 3 4 0 0 0 0 0 0 0 1 0 0 0 0 1 3 0 1 3 1 3 0 0 0 1 0 0 0 0 0 0 0 1⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦,P=[1 2 1 2 0 0 0 0 1 4 3 4 0 0 0 0 0 0 0 1 0 0 0 0 1 3 0 1 3 1 3 0 0 0 1 0 0 0 0 0 0 0 1],
则该状态空间可以分为三个互达等价类：C 1={1,2},C 2={3,4,5},C 3={6}C 1={1,2},C 2={3,4,5},C 3={6} 。其中 C 1 C 1 和 C 3 C 3 是闭集，因为从 C 1 C 1 或 C 3 C 3 出发不可能到达其他互达等价类。而 C 2 C 2 不是闭集，因为从 C 2 C 2 出发可以到达 C 3 C 3 。此外 C 3 C 3 是吸收态，因为 C 3 C 3 是闭集且只有一个状态。
Part 2：周期
设 i i 是状态空间 I I 中的任意一个状态，定义状态 i i 的周期为
d(i)=gcd{n≥1:p(n)i i>0},d(i)=gcd{n≥1:p i i(n)>0},
其中 gcd gcd 表示集合中各元素的最大公因子。因此，周期的含义是可达步数的最大公因子，即从 i i 出发只有经过 d(i)d(i) 的整数倍步数后，才有可能以正概率返回 i i 。
如果 d(i)=1 d(i)=1 ，则称状态 i i 是非周期的。如果对所有的状态 i∈I i∈I 都是非周期的，则称此马尔可夫链是非周期的。
如果状态 i i 是正常返且非周期的，则称状态 i i 是遍历状态。不可约非周期正常返的马尔可夫链称为遍历的马尔可夫链。
定理：如果 i↔j i↔j ，则
i i 和 j j 具有相同的周期，即 d(i)=d(j)d(i)=d(j) ；
i i 是暂留态当且仅当 j j 是暂留态；
i i 是常返态当且仅当 j j 是常返态；
i i 是正常返态当且仅当 j j 是正常返态。
该定理说明在同一个互达等价类中，各状态具有相同的周期和常返性。因此在判断一个状态的性质时，我们可以从它的互达等价类中找到一个容易判断的状态来进行判断。
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1. 现代精算风险理论13：风险序(2)
2. 现代精算风险理论08：破产概率(1)(2)
3. 回归分析03：回归参数的估计(1)(2)
4. 现代精算风险理论09：破产概率(2)(1)
5. 现代概率论03：测度空间(1)(1)
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1. Re:现代精算风险理论09：破产概率(2)
补充一下关于 Beekman 卷积公式的物理意义。 它代表的是，当出现最后一个新的大损失纪录 Lm 时，公司没有破产，并且以后再也不会发生新的大损失记录，也就是公司永远不会破产的概率。 所以，(1-p...
--Xavier25
2. Re:现代精算风险理论08：破产概率(1)
关于“调节系数”这一名称，其实来自于对再保险的研究。 当我们对风险安排了再保险之后，发现 破产概率没有达到最小值（或者说R没有达到最大值），这时就需要我们不得不去 “调节”（或者说调整，反正英文都是 ...
--Xavier25
3. Re:现代精算风险理论08：破产概率(1)
补充一下“调节系数”的由来。 我们知道盈余过程是 U(t)=U0+ct-S(t)。由于初始盈余（或者叫初始资本金）U0是外生变量，因此我们更关注 ct 和 S(t) 的大小关系。 比较 ct 和 S(...
--Xavier25
4. Re:现代精算风险理论05：再保险与最优再保险
在分出保费为一定值的条件下，就像文中的 “定理（停止损失再保险的最优性）”和最后的那个效用最大化下的定理，也可用 凸序 的角度去证明。 在凸序下，对比 止损再保险 和 其它任意一个再保险，总可以找到一...
--Xavier25
5. Re:现代精算风险理论13：风险序
凸序 和 止损序 非常相似，两者的主要差别是： 止损序 X =<sl Y 只要求 E[X-d]+=<E[Y-d]+, 并不 要求 EX=EY. 而 凸序 X =<cx Y 不仅要求 E[X-d]+=<...
--Xavier25
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CEMC Gauss (Grade 7)
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1 Audience
2 Format
3 Mathematical Content
4 Organization
5 Contest Preparation
6 Links
7 See Also
Audience
All students in Grades 7 and 8 and interested students from lower grades.
Format
The Gauss (Grade 7) contest is a 150-point, 60-minute competition that consists of 25 multiple-choice questions, each with 5 possible choices. They are divided into 3 sections:
Section A consists of 10 questions (1-10), each worth 5 points.
Section B consists of 10 questions (11-20), each worth 6 points.
Section C consists of 5 questions (21-25), each worth 8 points.
Calculators of all types are permitted for the Gauss (Grade 7). Devices that can connect to the internet, other devices, or have previously stored information are not allowed.
Mathematical Content
Questions are based on curriculum common to all Canadian provinces.
Organization
The Gauss Contests may be organized and run by an individual school, by a secondary school for feeder schools, or on a board-wide basis. All marking of the Contests is done in your school by the organizing teacher, in order to de-emphasize competition.
Contest Preparation
We recommend that students spend some time preparing for our contests by trying to solve problems. Many teachers use past contests in the classroom. Past papers are a good source of preparation for this contest.
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CEMC past contests
CEMC Gauss (Grade 7) Problems and Solutions
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CEMC
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11250 | http://file.snnu.net/res/201210/11/76d16733-c4c5-4e79-b27c-a0e7011fbc05.pdf | 从微分几何的学习看解析几何的教学 刘德金 （ 德州学院数学系，山东德州253023） 摘 要：从微分几何学习的角度谈了解析几何教学中应注意的三个问题：一是要重视矢量代数的教学；二 是要熟练掌握常见曲面的方程；三是不能忽视解析几何的数学思想方法教育。 关键词：解析几何；微分几何；矢量代数 中图分类号：G633.63 文献标识码：A 文章编号：1671- 1351 （ 2008）02- 0114- 04 收稿日期：2008- 01- 06 作者简介：刘德金（ 1957- ） ，男，山东武城人，德州学院数学系教授。 解析几何是用代数方法研究几何问题的一门学 科，它的基本思想是将几何图形代数化，把数和数 的运算引入到几何中，通过代数方法对几何问题进 行研究，然后把所研究的结果转化为几何结论。其 中几何图形的代数化是核心，它是通过用坐标表示 点和用方程表示曲线的观念实现的。解析几何的诞 生是数学史上一个伟大的里程碑，它使人们摆脱了 现实的束缚和对图形的直观依赖，使以常量为主导 的数学转变为以变量为主导的数学，为微积分的诞 生奠定了基础。 微分几何可以说是与微积分同时诞生的，它是 用微积分为工具来研究空间曲线和曲面的形状和性 质。解析几何和微分几何是本科院校数学系学生的 必修课。解析几何主要介绍矢量代数、空间直线和 平面、常见曲面和二次曲线的一般理论；微分几何 是解析几何的后继课，主要研究三维空间曲线和曲 面在一点邻近的性质，找出决定曲线、曲面形状的 不变量系统。两种几何虽然研究的对象和方法有很 大不同，但联系非常紧密，解析几何是微分几何的 基础，只有掌握了解析几何的知识和方法，才能顺 利地学习微分几何。因此在解析几何的教学中，应 充分考虑后继课程的学习，全面理解和把握教学的 重点；在微分几何的教学中应及时沟通它们的联 系，这对激发学生学习的积极性，系统理解和把握 解析几何、微分几何的教学内容和方法，培养学生 能力，提高学生素质，提高教学效果是十分有 益的。 有关学者谈了关于数学基础课教学的一点思 考，很深刻。笔者受其启发，根据自己教学的实 践，谨从微分几何学习的角度谈一下解析几何教学 中应注意的三个方面问题。 1 矢量代数是学习数学的工具，应作为教学 的重点 矢量代数是解析几何课程的内容之一，它在解 析几何课程本身的地位就很重要。几何结构矢量化 是把几何结构代数化的一个步骤，也可以说是一种 方法。通过这一方法，把矢量的运算带到了几何 中，再通过矢量引进坐标系，从而把数和数的计算 引到了几何里来，把对几何问题的研究从定性的层 面推进到可以计算的层面上来，实现了数与形的真 正结合，从而达到用代数方法解决几何问题的目 的。实际上，通过矢量的运算就可以解决几何图形 中某些有关定性的问题，这就是人们说的用矢量方 法解决几何问题，它是用代数方法解决几何问题的 一个方面。 矢量作为研究问题的工具，在其他学科有广泛 的应用。在被人们广泛使用的文献中，特别是 曲线论一章，实际上是用矢量分析的方法研究空间 曲线在一点的局部性质。矢量的概念，矢量的运 算，尤其是矢量的数性积、矢性积、混合积的应 用，从空间曲线在一点的基本三棱形、求法到曲面 在一点的切平面和曲面上向量的平行移动及其性 质，贯穿了整个课程。由此得到的关于向量函数的 几个命题“向量函数r= r (t)具有固定长!r · r'=0” 、 “向量函数r=r (t)具有固定方向!r×r'=0” 、“ 向量 函数r= r(t)平行于固定平面!(r,r',r'')=0” 等在微 2008 年3 月 天水师范学院学报 Mar.,2008 第28 卷第2 期 Journal of Tianshui Norm al University Vol.28 No.2 114 分几何的学习中发挥了极其重要的作用。因此在解 析几何的教学中，抓好矢量代数的教学非常重要， 从微分几何的角度看，矢量代数的教学应抓住以下 几点。 1.1 明确数性积、矢性积、混合积的概念，讲 清数性积、矢性积的区别与联系 数性积、矢性积是两矢量间的两种截然不同的 乘法运算。a与b的数性积是一个数，这个数是a · b =| a| | b| cos （ ∠a,b） ；而a与b的矢性积是一个矢量， 记为a×b，是矢量就得明确两样东西，一是它的模 | a×b| = | a| | b| sin∠（ a,b） ， 二是它的方向： a×b⊥a， a×b⊥ b， 且按a, b, a×b 的顺序成右手系，避免a×b= | a| | b| sin∠（ a,b） . 明确数性积与长度、角度有关，因此求 夹角往往用数性积；而矢性积与面积有关，| a×b| 是 以a,b为邻边的平行四边形的面积，因此求面积往 往考虑用矢性积。a⊥b#a · b=0；a∥b#a×b=0. 数 性积满足交换率a · b=b · a；矢性积满足反交换率a×b =- b×a. 数性积的交换率是常常被使用的，矢性积 的反交换率往往被忽视。为理解数性积和矢性积， 我们可设计一些与其有关的问题。例如： 问题1 已知矢量a,b,c. a与c不垂直，且a=!b.
则!= .
解：在a=!b的两边点乘c，则a · c=!b · c，因为 b · c是数，且b与c不垂直，所以b · c≠0，在a · c=!b · c的两边同除以b · c，得!= a · c b · c .
在a=!b的两边叉乘c，则a×c=!b×c，因为b×c 不是数，所以在a×c=!b×c的两边同除以b×c是不可 以的（ 即使b×c≠0）.
当然它们也有共同点，都不满足结合律、消去 率。一般的(ab)c≠a(bc)， (a×b)×c≠a×(b×c)，不等号 两边都是矢量，但两边的方向一般不同，模一般也 不同。a · b=a · c (a≠0) 推不出b=c，因为b,c可以是 在a上投影相同的任意两个矢量b,c而未必相等， 特 别取与a垂直的任意两个矢量b,c而未必相等。同 样，a×b=a×c (a≠0) 也推不出b=c，因为b,c可以是 a与平行而模不相等的任意两个矢量b,c.
问题2 若!a · b=0，则b=c≠0时一定有 =0 （ 或0）或 ；若!a×b=0，则b=c≠0时一定有 =0 （ 或0）或 .
在微分几何中，已知曲面上的曲率线Γ的主法 向量"，曲面的法向量n，曲线的挠率满足#" · n=0， 求证Γ是平面曲线。正确的证法是：若#=0，则Γ是 平面曲线；若" · n=0，则Γ 是渐近曲线，从而沿Γ 有K n=0，且n∥$. 又因Γ是曲率线，所以dn=- K ndr= 0，所以n是常向量，从而$是常向量，所以Γ 是平 面曲线。而不能由n≠0，得#%=0，从而#=0，去判 断Γ是平面曲线。 数性积、矢性积还满足(a · b)2+(a×b)2=a2b2. 此式 对微分几何中证明“向量函数r=r(t)有固定方向#r ×r'=0” 、“向量函数r=r(t)平行于固定平面#(r,r',r'') =0” 起了重要作用.
由数性积和矢性积定义的混合积(a,b,c)也是一 个很重要的概念。a,b,c共面#(a,b,c)=0.平面方程的 求法或矢量表示和第二类基本量的表示，可展曲面 的条件，挠率的计算等都与它有关系，所以要掌握 其概念和计算方法。可以想象，数性积、矢性积、 混合积的概念不清，想学好微分几何是很难的。 1.2 讲清矢量坐标的意义 谈到坐标就得有标架或坐标系。坐标系实际上 是点或矢量与数组的一一对应关系，矢量a在标架 {o,e1,e2,e3}下若有a=xe1+ye2+ze3，则x,y,z叫做矢量a 的分量或坐标，这时记a={x, y, z}. 对空间点P，若 op={x, y, z}， 则（ x, y, z） 就是点P的坐标。空间图形 的方程实际是图形上任意点的径矢r满足的代数条 件。空间图形的参数方程就是在一个标架{o,e1,e2,e 3 }下图形上任意点的径矢r用e1,e2,e3线性表出时其组 合系数的表达式，它一般是另外一个或两个的参数 的函数。例如对空间曲线Γ，若Γ上任意点的径矢r 有r=x(t)e1+y(t)e2+z(t)e3，则Γ的参数方程是 x=x(t) y=y(t) z=z(t & ( ( ( ( ’ ( ( ( ( ) ) ； 对曲面Σ，若Σ上任意点的径矢r有r=x(u,v)e1+y(u,v) e 2+z(u,v)e 3，则Σ的参数方程是 x=x(u,v) y=y(u,v) z=z(u,v & ( ( ( ( ( ( ( ( ) ) .
在微分几何中，讨论曲线在一点邻近的形状 时，在每点取弗雷内标架建坐标系，得到曲线在一 点邻近的方程；讨论曲面在一点P的杜邦指标线方 程时，是在P点的切平面上取{P; ru , rv} 为标架建 立的方程。讲清矢量坐标的意义和明确图形参数方 程的意义对于理解微分几何中这些知识必不可缺 少。在解析几何的教学中应指出，并非一取坐标系 就是笛卡尔右手直角坐标系{o,i,j ,k}. 在空间任何 a c b c 115 三个不共面的矢量a,b,c都可作为坐标矢量，任何一 点A可作为坐标原点建立坐标系{A； a,b,c}. 空间的 矢量p在这个坐标系下有坐标{x,y,z}!p=xa+yb+zc.
不明确这一点，在微分几何的学习中就会遇到困 难，为此在讲了标架之后可设计如下的问题：若矢 量a在标架{o,e1,e2,e3}下的分量是{x,y,z}，其中xyz≠ 0，那么a与e 1,e 2,e 3中任何两个共面吗？ e 1在{A； e 1,e 2, e 3}下的坐标是什么？ 2 基本、常见图形与方程要熟练掌握 基本、常见图形主要指空间直线、平面、柱 面、锥面、旋转曲面和常见二次曲面。它们不仅对 学微积分非常重要，对微分几何的学习也相当重 要。这部分内容的教学应该做到“三要”。 一要熟知这些图形方程的特征，特别是要重视 直线的矢量表示r=r0+tv，平面的矢量表示r=r0+λa +μb或r- r0,a,b和rn- p=0. 由矢量表示可以了解确定 方程需要的条件，还可得到方程的其他表示方法， 如直线的标准式方程和平面的一般方程。 二要理解怎样去建立这些图形的方程。建立图 形的方程，实际是求图形上任一点的坐标或径矢所 满足的方程。掌握方程的特征是建立方程的一个方 面，可以根据方程的特征寻求确定方程的条件。 微分几何中，求空间曲线在一点的切线、主法 线、副法线、密切平面、法面，从切面求曲面在一 点的切平面、法线和某些直纹面等，都要求熟悉直 线、平面方程的求法。在直线的矢量表示r= r0+tv 中，r0,v表示常向量。如果以向量函数a (u)代替r0， 则r=a(u)+tv表示经过曲线a=a(u)上每一点，方向沿 v的直线构成的曲面，即柱面；如果以b(u)代替v， 则r=r0+tb (u)表示以r0为顶点的一个锥面；如果以a (u)代替v，以b(u)代替v，则r=a(u)+tb(u)表示以曲线 a=a(u)为导线，过每个点a(u)沿b(u)的方向的直线构 成的直纹曲面。例如任一条曲线r=r (s) 的切线曲 面ρ=r (s) +tα，主法线曲面ρ=r (s) +t %，副法线 曲面ρ=r (s) +u&. 从这个意义上讲，直线是直纹面 的特例，形式上直纹面方程是直线方程的推广，掌 握直线方程就可以在微分几何中顺利地建立直纹面 的方程，在此基础上就可以证明某些几何问题。如 证明“挠曲线的主法线曲面、副法线曲面不可展”、 “任一条曲线在它的主法线曲面上是渐近曲线” 等。 三要让学生通过作图、观看教具、观察实物使 学生了解其结构，建立起空间直观形象。可展曲面 是微分几何的重要内容之一，柱面、锥面、单叶双 曲面、双曲抛物面是一个曲面是否可展讨论的主要 对象。讨论曲面在一点邻近的结构，这些曲面还有 旋转曲面、球面、柱面、抛物面、椭球面、圆环面 是考虑的主要对象。这些曲面对理解曲面上的渐近 线、曲率线、测地线也是常被考虑的。因此在解析 几何的教学中，加强这方面的教学非常必要。 3 解析几何的思想方法教育不可忽视 现代数学教育愈来愈重视数学的思想和方法。 因为知识可以淡忘，而思想方法却可以永远植根于 人们的头脑中，长久的在自己的业务中发挥作用。 解析几何的思想方法主要就是坐标法。通过坐标 系，使点有坐标，图形有方程，然后把对图形的研 究转变为用代数方法对方程的研究。掌握这种方 法，一是遇到与坐标方程无关的问题时，知道考虑 建立坐标系，使问题的条件和结论用坐标或方程来 表示，从而把几何问题转变为用坐标和方程表示的 代数问题；二是要有建适当的坐标系的观念，如果 所考虑的问题的性质与坐标系的选取无关，或说是 坐标变换下的不变性质，则我们可考虑建适当的坐 标系。怎样的坐标系才算是适当的坐标系？判定的 标准就是在这样的坐标系下，问题的条件表示得简 单，从而方便讨论。如问题中有一个定点，则取定 点为坐标原点或坐标轴上的点；若问题中有两个定 点，则取两定点所在直线为一个坐标轴，两点的中 点为坐标原点；有定直线，则往往取定直线为某坐 标轴；有定平面，则取定平面为坐标平面；如果既 有定点又有定直线，则取定直线为一坐标轴，过定 点与定直线垂直的直线为另一坐标轴；若有定平面 和定平面外一定点，则取定平面为坐标平面，过定 点与定平面垂直的直线为一坐标轴等等。如果所述 图形有对称性，为了方程的简单，往往取图形的对 称中心为坐标原点，对称轴为坐标轴，对称平面为 坐标平面等。 掌握了这一方法，微分几何中的许多问题就会 容易的得到解决。例如， “证明球面曲线在任意点的 116 法平面通过球面的中心” ；“ 如果曲线在任意点的 密切平面都经过一个定点，则此曲线是平面曲 线” 等。 曲面的曲纹坐标是平面坐标在曲面上的推广， 选适当坐标系的思想用于微分几何中的曲面论一 章，就是选取合适的曲纹坐标网。有了这种认识， 在曲面问题的讨论中，我们可以选取适当的曲纹坐 标网，如正交网、曲率网和半测地坐标网。在这样 的坐标网下，问题的讨论就变得更简单。例如，微 分几何中欧拉(Euler) 公式kn=k1cos2θ+k2sin 2θ的证 明，曲面在一点的三个基本形式及高斯曲率、平均 曲率之间的关系Ⅲ- 2HⅡ+KI的证明都是选取了曲 率网为曲纹坐标网。曲面上测地曲率kg的计算，在 正交网下的计算公式，即利乌维（ Liouville）公式 + ，其中 θ是曲线的切向与坐标曲线切向ru的夹角。 在正交网下，测地线的微分方程是一个含三个 变量u,v,θ和一个自变量s的一阶微分方程组： 还有克里斯托菲尔符号Γ k ij以及黎曼曲率张量 Rl ijk等它们都比在一般曲纹坐标网下的表示简单 得多。 又如在曲面的半测地坐标网下， 曲面的第一基 本形式是ds2=du 2+Gdv2， 利用半测地坐标网和在此坐 标网下的第一基本形式很容易证明曲面上测地线的 短程性。著名的高斯—波涅公式 G !
Kdσ + " # K gds + n i = 1 $(π- αi)=2π也就是在半测地坐标网下证明的。 在半测地坐标网下，克里斯托菲尔符号Γ k ij以 及黎曼曲率张量Rl ijk更简单，测地曲率公式、测地 线的微分方程、向量平移的条件分析条件、高斯曲 率的内蕴表达式等都取特别简单的形式。 当然，一种思想方法要经过教师长期的、有意 识的、有目的的教学活动，通过渗透、揭示、归 纳、总结的过程，使学生在解决问题的实践过程中 领悟，才能转变为他们自己的东西，最后达到掌握 的目的。建立适当坐标系的思想方法在学生学习 了解析几何后也并不一定就能完全掌握，这时在微 分几何的教学中要注意对学生适当提醒。 参考文献： 张顺燕. 数学与文化[J].数学通报,2001,(2):1- 3.
卢玉峰.关于数学基础课教学的一点思考[J].北京:高等数 学研究,2003,(3):5- 7.
梅向明,黄敬之.微分几何[M].3版.北京:高等教育出版社, 2003.
吕林根,许子道.解析几何[M].3版.北京:高等教育出版社, 1987.
冯斌.简论数学教学的过程化[J].数学通报,2004,(10):14- 16.
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The complete solution for verifying Malus’ Law of Polarization.
See the Buying Guide for this item's required, recommended, and additional accessories.
Product Summary
In this experiment, Malus’ Law of Polarization is verified by showing that the intensity of light passed through two polarizers depends on the square of the cosine of the angle between the two polarization axes.
Laser light (peak wavelength = 650 nm) is passed through two polarizers. As the second polarizer (the analyzer) is rotated by hand, the relative light intensity is recorded as a function of the angle between the axes of polarization of the two polarizers. The angle is obtained using a Rotary Motion Sensor coupled to the polarizer with a drive belt. The plot of light intensity vs. angle can be fitted to the square of the cosine of the angle.
PASCO Advantage: Laser light is used in this experiment because its wavelength is more completely extinguished by the crossed polarizers.
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Our expandable optics system provides everything you need to perform high-quality investigations in any physics lab. Introduce students to the fundamentals of optics with our comprehensive Basic Optics System; then build on the basics with a variety of optics accessories that support detailed studies of color, interference and diffraction, polarization, and more.
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Perform the following experiments and more with the Polarization Experiment.
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Polarization of Light
Laser light is passed through two polarizers. As the second polarizer (the analyzer) is rotated by hand, the relative light intensity is recorded as a function of the angle between the axes of polarization of the two polarizers....
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| Question Text | - Net torque about hinge point should be zero. But torque of hinge force about the same point is already zero. - Net force on the rigid body is also zero. Example 1 A uniform L shaped rod of mass 3 m is hinged at point O. Length OB is two times the length OA. It is in equilibrium. Find (a) relation between α and β (b) net hinge force. |
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Inner Product Spaces
We have used the dot product in to compute the length of vectors (Corollary cor:length_via_dotprod) and also the angle between vectors. The goal of this section is to define an inner product on an arbitrary vector space over the real numbers. The dot product is an example an inner product for .
An inner product on a real vector space is a function that assigns a real number to every pair , of vectors in in such a way that the following properties are satisfied.
(a)
: is a real number for all and in .
(b)
: for all and in .
(c)
: for all , , and in .
(d)
: for all and in and all in .
(e)
: for all in .
A real vector space with an inner product will be called an inner product space. Note that every subspace of an inner product space is again an inner product space using the same inner product.
is an inner product space with the dot product as inner product: See Theorem th:dotproductproperties. This is also called the euclidean inner product, and , equipped with the dot product, is called euclidean -space.
If and are matrices, define where is the trace of the square matrix . Show that is an inner product in .
Property prop:inner_prod_1 is clear. Since for every square matrix , we have Property prop:inner_prod_2: Next, Property prop:inner_prod_3 and Property prop:inner_prod_4 follow because trace is a linear transformation (Practice Problem ex:10_1_19). Turning to Property prop:inner_prod_5, let denote the rows of the matrix . Then the -entry of is , so But is the sum of the squares of the entries of , so this shows that is the sum of the squares of all entries of . Property prop:inner_prod_5 follows.
The next example is important in analysis.
Let be the set of all functions . Observe that is a vector space. Let be a subset of consisting of all continuous functions. Why is a subspace of ? Show that defines an inner product on .
This example (and others later that refer to it) can be omitted with no loss of continuity by students with no calculus background.
Property prop:inner_prod_1 and Property prop:inner_prod_2 are clear. As to Property prop:inner_prod_4, Property prop:inner_prod_3 is similar. Finally, theorems of calculus show that and, if is continuous, that this is zero if and only if is the zero function. This gives Property prop:inner_prod_5.
If is any vector, then, using Property prop:inner_prod_3 of Definition def:innerproductspace, we get
and it follows that the number must be zero. This observation is recorded for reference in the following theorem, along with several other properties of inner products. The other proofs are left as Practice Problem ex:10_1_20.
Let be an inner product on a space ; let , , and denote vectors in ; and let denote a real number.
(a)
(b)
(c)
(d)
: if and only if
If is a complex number, then 030346b must be modified. See Theorem th:025575 in Complex Matrices.
If is an inner product on a space , then, given , , and in ,
for all and in by Property prop:inner_prod_3 and Property prop:inner_prod_4 of Definition def:innerproductspace. Moreover, there is nothing special about the fact that there are two terms in the linear combination or that it is in the first component: and hold for all and in and all , , , and in . These results are described by saying that inner products “preserve” linear combinations. For example,
If is a symmetric matrix and and are columns in , we regard the matrix as a number. If we write
then Properties prop:inner_prod_1 -prop:inner_prod_4 of Definition def:innerproductspace follow from matrix arithmetic (only Property prop:inner_prod_2 of Definition def:innerproductspace requires that is symmetric). Property prop:inner_prod_5 of Definition def:innerproductspace reads and this condition characterizes the positive definite matrices (Theorem thm:024830). This proves the first assertion in the next theorem.
If is any positive definite matrix, then defines an inner product on , and every inner product on arises in this way.
Proof
: Given an inner product on , let be the standard basis of . If and are two vectors in , compute by adding the inner product of each term to each term . The result is a double sum. As the reader can verify, this is a matrix product: Hence , where is the matrix whose -entry is . The fact that shows that is symmetric. Finally, is positive definite by Theorem thm:024830.
Thus, just as every linear operator corresponds to an matrix, every inner product on corresponds to a positive definite matrix. In particular, the dot product corresponds to the identity matrix .
If we refer to the inner product space without specifying the inner product, we mean that the dot product is to be used.
Let the inner product be defined on by Find a symmetric matrix such that for all , in .
The -entry of the matrix is the coefficient of in the expression, so . Incidentally, if , then
for all , so implies . Hence is indeed an inner product, so is positive definite.
Let be an inner product on given as in Theorem thm:030372 by a positive definite matrix . If , then is an expression in the variables called a quadratic form. For more on quadratic forms, see Section 8.8 of [Nicholson], pp. 472–482.
Norm and Distance
As in , if is an inner product on a space , the norm of a vector in is defined by We define the distance between vectors and in an inner product space to be
If the dot product is used in , the norm of a vector is usually called the length of .
Note that Property prop:inner_prod_5 of Definition def:innerproductspace guarantees that , so is a real number.
The norm of a continuous function in (with the inner product from Example exa:030334) is given by
Hence is the area beneath the graph of between and .
Show that in any inner product space.
A vector in an inner product space is called a unit vector if . The set of all unit vectors in is called the unit ball in . For example, if (with the dot product) and , then
Hence the unit ball in is the unit circle with centre at the origin and radius . However, the shape of the unit ball varies with the choice of inner product.
Let and . If and , define an inner product on by
The reader can verify (Practice Problem ex:10_1_5) that this is indeed an inner product. In this case so the unit ball is the ellipse shown in the diagram.
Example exa:030469 graphically illustrates the fact that norms and distances in an inner product space vary with the choice of inner product in .
If is any vector in an inner product space , then is the unique unit vector that is a positive multiple of .
The next theorem reveals an important and useful fact about the relationship between norms and inner products, extending the Cauchy inequality for (Theorem th:CS).
Cauchy-Schwarz Inequality If and are two vectors in an inner product space , then Moreover, equality occurs if and only if one of and is a scalar multiple of the other.
Hermann Amandus Schwarz (1843–1921) was a German mathematician at the University of Berlin. He had strong geometric intuition, which he applied with great ingenuity to particular problems. A version of the inequality appeared in 1885.
Proof of Cauchy-Schwarz Inequality
: Write and . Using Theorem thm:030346 we compute: It follows that and , and hence that . But then , as desired.
Conversely, if then . Hence (eq:thm10\_1\_4) shows that or . It follows that one of and is a scalar multiple of the other, even if or .
If and are continuous functions on the interval , then (see Example exa:030334)
Another famous inequality, the so-called triangle inequality (See Triangle Inequality in the Appendix), also comes from the Cauchy-Schwarz inequality. It is included in the following list of basic properties of the norm of a vector.
If is an inner product space, the norm has the following properties.
(a)
: for every vector in .
(b)
: if and only if .
(c)
: for every in and every in .
(d)
: for all and in (triangle inequality).
Proof
: Because , properties thm:030504a and thm:030504b follow immediately from thm:030346c and thm:030346d of Theorem thm:030346. As to thm:030504c, compute Hence thm:030504c follows by taking positive square roots. Finally, the fact that by the Cauchy-Schwarz inequality gives Hence thm:030504d follows by taking positive square roots.
It is worth noting that the usual triangle inequality for absolute values,
is a special case of thm:030504d where and the dot product is used.
In many calculations in an inner product space, it is required to show that some vector is zero. This is often accomplished most easily by showing that its norm is zero. Here is an example.
Let be a spanning set for an inner product space . If in satisfies for each , show that .
Write , in . To show that , we show that . Compute: by hypothesis, and the result follows.
The norm properties in Theorem thm:030504 translate to the following properties of distance familiar from geometry. The proof is Practice Problem ex:10_1_21.
Let be an inner product space.
(a)
: for all , in .
(b)
: if and only if .
(c)
: for all and in .
(d)
: for all , , and in .
Practice Problems
In each case, determine which of Property prop:inner_prod_1–Property prop:inner_prod_5 in Definition def:innerproductspace fail to hold.
(a)
: ,
(b)
: ,
Click the arrow to see the answer.
Property prop:inner\_prod\_5 fails.
(c)
: , , where is complex conjugation
Click the arrow to see the answer.
Property prop:inner\_prod\_1 fails, as sometimes we get a complex number. However, we will return to this definition of in Complex Matrices – see Definition def:025549
(d)
: ,
Click the arrow to see the answer.
Property prop:inner\_prod\_5 fails.
(e)
: ,
(f)
: ,
Click the arrow to see the answer.
Property prop:inner\_prod\_5 fails.
Let be an inner product space. If is a subspace, show that is an inner product space using the same inner product.
Property prop:inner_prod_1–Property prop:inner_prod_5 hold in because they hold in .
In each case, find a scalar multiple of that is a unit vector.
(a)
: in where
(b)
: in where
Click the arrow to see the answer.
(c)
: in where
(d)
: in ,
Click the arrow to see the answer.
In each case, find the distance between and .
(a)
: ,
(b)
: ,
Click the arrow to see the answer.
(c)
: , in where and ;
(d)
: , in where and ;
Click the arrow to see the answer.
Let be positive numbers. Given and , define . Show that this is an inner product on .
If is a basis of and if and are vectors in , define Show that this is an inner product on .
Let denote the real part of the complex number . Show that is an inner product on if .
If is an isomorphism of the inner product space , show that defines a new inner product on .
Show that every inner product on has the form for some upper triangular matrix with positive diagonal entries.
Theorem thm:024907
In each case, show that defines an inner product on and hence show that is positive definite.
(a)
(b)
: Click the arrow to see the answer.
(c)
(d)
Click the arrow to see the answer.
In each case, find a symmetric matrix such that .
(a)
(b)
: Click the arrow to see the answer.
(c)
(d)
: Click the arrow to see the answer.
If is symmetric and for all columns in , show that .
Consider where .
Click the arrow to see the answer.
By the condition, for all , . Let denote column of . If , then for all and .
Show that the sum of two inner products on is again an inner product.
Let , , , , and . Compute:
(a)
(b)
Given the data in Practice Problem ex:10_1_16, show that .
Show that no vectors exist such that , , and .
Complete Example exa:030310.
Prove Theorem thm:030346.
(a)
: Using Property prop:inner_prod_2: .
(b)
: Using Property prop:inner_prod_2 and Property prop:inner_prod_4: .
(c)
: Using Property prop:inner_prod_3: , so . The rest is Property prop:inner_prod_2.
(d)
: Assume that . If this contradicts Property prop:inner_prod_5, so . Conversely, if , then by Part 3 of this theorem.
Prove Theorem thm:030346.
Let and be vectors in an inner product space .
(a)
: Expand .
(b)
: Expand .
Click the arrow to see the answer.
(c)
: Show that .
(d)
: Show that .
Click the arrow to see the answer.
Show that for any and in an inner product space.
Let be an inner product on a vector space . Show that the corresponding distance function is translation invariant. That is, show that for all , , and in .
(a)
: Show that for all , in an inner product space .
(b)
: If and are two inner products on that have equal associated norm functions, show that holds for all and .
Let denote a vector in an inner product space .
(a)
: Show that is a subspace of .
(b)
: Let be as in (a). If with the dot product, and if , find a basis for .
Click the arrow to see the answer.
Given vectors and , assume that for each . Show that for all in .
If and holds for each . Show that .
for each , so by Practice Problem ex:10_1_27.
Use the Cauchy-Schwarz inequality in an inner product space to show that:
(a)
: If , then for all in .
(b)
: for all real , , and .
If in (with the dot product) then . Use (a) with .
(c)
: for all vectors , and all in .
If is a matrix, let and denote the rows of .
(a)
: Show that .
(b)
: Show that .
(a)
: If and are nonzero vectors in an inner product space , show that , and hence that a unique angle exists such that and . This angle is called the angle between and .
(b)
: Find the angle between and in with the dot product.
(c)
: If is the angle between and , show that the law of cosines is valid:
If , define .
(a)
: Show that satisfies the conditions in Theorem thm:030504.
(b)
: Show that does not arise from an inner product on given by a matrix .
If it did, use Theorem thm:030372 to find numbers , , and such that for all and .
Text Source
This section was adapted from Section 10.1 of Keith Nicholson’s Linear Algebra with Applications. (CC-BY-NC-SA)
W. Keith Nicholson, Linear Algebra with Applications, Lyryx 2018, Open Edition, pp. 521–530. |
11254 | https://flexbooks.ck12.org/cbook/ck-12-precalculus-concepts-2.0/section/12.5/primary/lesson/geometric-series-pcalc/ | Skip to content
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12.5
Geometric Series
Written by:CK-12 | Mark Spong
Fact-checked by:The CK-12 Editorial Team
Last Modified: Aug 01, 2025
Lesson
An advanced factoring technique allows you to rewrite the sum of a finite geometric series in a compact formula. An infinite geometric series is more difficult because sometimes it sums to be a number and sometimes the sum keeps on growing to infinity. When does an infinite geometric series sum to be just a number and when does it sum to be infinity?
Geometric Series
A geometric series is a sum of numbers whose consecutive terms form a geometric sequence. Recall the advanced factoring technique for the difference of two squares and, more generally, two terms of any power (5 in this case).
a2−b2=(a−b)(a+b)a5−b5=(a−b)(a4+a3b+a2b2+ab3+b4)an−bn=(a−b)(an−1+⋯+bn−1)
If the first term is one then a=1. If you replace b with the letter r, you end up with:
1−rn=(1−r)(1+r+r2+⋯rn−1)
You can divide both sides by (1−r) because r≠1.
1+r+r2+⋯rn−1=1−rn1−r
The left side of this equation is a geometric series with starting term 1 and common ratio of r. Note that even though the ending exponent of r is n−1, there are a total of n terms on the left. To make the starting term not one, just scale both sides of the equation by the first term you want, a1.
a1+a1r+a1r2+⋯a1rn−1=a1(1−rn1−r)
This is the sum of a finite geometric series.
To sum an infinite geometric series, you should start by looking carefully at the previous formula for a finite geometric series. As the number of terms get infinitely large (n→∞) one of two things will happen.
a1(1−rn1−r)
Option 1: The term rn will go to infinity or negative infinity. This will happen when |r|≥1. When this happens, the sum of the infinite geometric series does not go to a specific number and the series is said to be divergent.
Option 2: The term rn will go to zero. This will happen when |r|<1. When this happens, the sum of the infinite geometric series goes to a certain number and the series is said to be convergent.
One way to think about these options is think about what happens when you take 0.9100 and 1.1100.
0.9100≈0.000026561.1100≈13780
As you can see, even numbers close to one either get very small quickly or very large quickly.
The formula for calculating the sum of an infinite geometric series that converges is:
∞∑i=1a1⋅ri−1=a1(11−r)
Notice how this formula is the same as the finite version but with rn=0, just as you reasoned.
A partial sum of an infinite sum is the sum of all the terms up to a certain point. Considering partial sums can be useful when analyzing infinite sums.
Examples
Example 1
Earlier, you were asked when a infinite geometric series sum to just a number. An infinite geometric series converges if and only if|r|<1. Infinite arithmetic series never converge.
Example 2
Compute the sum of the following infinite geometric series two ways, without using the infinite summation formula and using the infinite summation formula.
0.2+0.02+0.002+0.0002+⋯
Without using the summation formula:
You can tell just by looking at the sum that the infinite sum will be the repeating decimal 0.¯2. You may recognize this as the fraction 29, but if you don’t, this is how you turn a repeating decimal into a fraction.
Let x=0.¯2
Then 10x=2.¯2
Subtract the two equations and solve for x.
10x−x=2.¯2−0.¯29x=2x=29
With using the summation formula:
The first term of the sequence is a1=0.2. The common ratio is 0.1. Since |0.1|<1, the series does converge.
0.2(11−0.1)=0.20.9=29
Example 3
Why does an infinite series with r=1 diverge?
If r=1 this means that the common ratio between the terms in the sequence is 1. This means that each number in the sequence is the same. When you add up an infinite number of any finite numbers (even fractions close to zero) you will always get infinity or negative infinity. The only exception is 0. This case is trivial because a geometric series with an initial value of 0 is simply the following series, which clearly sums to 0:
0+0+0+0+⋯
Example 4
What is the sum of the first 8 terms in the following geometric series?
4+2+1+12+⋯
The first term is 4 and the common ratio is 12.
SUM=a1(1−rn1−r)=4(1−(12)81−12)=4(25525612)=25532
Example 5
You put $100 in a bank account at the end of every year for 10 years. The account earns 6% interest. How much do you have total at the end of 10 years?
The first deposit gains 9 years of interest: 100⋅1.069
The second deposit gains 8 years of interest: 100⋅1.068. This pattern continues, creating a geometric series. The last term receives no interest at all.
100⋅1.069+100⋅1.068+⋯100⋅1.06+100
Note that normally geometric series are written in the opposite order so you can identify the starting term and the common ratio more easily.
a1=100,r=1.06
The sum of the 10 years of deposits is:
a1(1−rn1−r)=100(1−1.06101−1.06)≈$1318.08
| |
| Summary |
| A geometric series is a sum of numbers whose consecutive terms form a geometric sequence, and it can be finite or infinite. The sum of a finite geometric series can be calculated using the formula: a1+a1r+a1r2+⋯a1rn−1=a1(1−rn1−r), where a1 is the first term, r is the common ratio, and n is the number of terms. An infinite geometric series can be either convergent or divergent, depending on the value of the common ratio r. If |r|≥1, the infinite geometric series is divergent, and its sum does not go to a specific number. If |r| < 1, the infinite geometric series is convergent, and its sum can be calculated using the formula: ∞∑i=1a1⋅ri−1=a1(11−r) A partial sum of an infinite sum is the sum of all the terms up to a certain point. |
Review
Find the sum of the first 15 terms for each geometric sequence below.
5,10,20,…
2,8,32,…
5,52,54,…
12,4,43,…
13,1,3,…
For each infinite geometric series, identify whether the series is convergent or divergent. If convergent, find the number where the sum converges.
5+10+20+⋯
2+8+32+⋯
5+52+54+⋯
12+4+43+⋯
13+1+3+⋯
6+2+23+⋯
You put $5000 in a bank account at the end of every year for 30 years. The account earns 2% interest. How much do you have total at the end of 30 years?
You put $300 in a bank account at the end of every year for 15 years. The account earns 4% interest. How much do you have total at the end of 10 years?
You put $10,000 in a bank account at the end of every year for 12 years. The account earns 3.5% interest. How much do you have total at the end of 12 years?
Why don’t infinite arithmetic series converge?
Review (Answers)
Click HERE to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
Asked by Students
Here are the top questions that students are asking Flexi for this concept:
Overview
A geometric series is a sum of numbers whose consecutive terms form a geometric sequence, and it can be finite or infinite.
The sum of a finite geometric series can be calculated using the formula: @$a_1+a_1r+a_1r^2+\cdots a_1r^{n-1}=a_1 \left(\frac{1-r^n}{1-r}\right),@$ where @$a_1@$ is the first term, @$r@$ is the common ratio, and @$n@$ is the number of terms.
An infinite geometric series can be either convergent or divergent, depending on the value of the common ratio @$r.@$
If @$|r| \ge 1,@$ the infinite geometric series is divergent, and its sum does not go to a specific number.
If |r| < 1, the infinite geometric series is convergent, and its sum can be calculated using the formula: @$\sum \limits_{i=1}^{\infty} a_1 \cdot r^{i-1}=a_1 \left(\frac{1}{1-r}\right)@$
A partial sum of an infinite sum is the sum of all the terms up to a certain point.
Vocabulary
infinite geometric series
term
converges
converge
diverge
Test Your Knowledge
Question 1
You put @$\begin{align}\$10,000 \end{align}@$ in a bank account at the end of every year for @$\begin{align}12\end{align}@$ years. The account earns @$\begin{align}3.5\% \end{align}@$ interest. How much do you have total at the end of @$\begin{align}12\end{align}@$ years?
a
@$\begin{align}\$202,840.40\end{align}@$
b
@$\begin{align}\$6,007.08\end{align}@$
c
@$\begin{align}\$146,019.62\end{align}@$
d
@$\begin{align}\$126,119.42\end{align}@$
Think about this sum in reverse. Looking at your deposits from last to first: @$\begin{align}\$10,000+(\$10,000)(1.035)+(\$10,000)({1.035}^{2})+\hspace{1mm}...+(\$10,000)({1.035}^{11})\end{align}@$
Factor out @$\begin{align}\$10,000\end{align}@$ from: @$\begin{align}\$10,000+(\$10,000)(1.035)+(\$10,000)({1.035}^{2})+\hspace{1mm}...+(\$10,000)({1.035}^{11}).\end{align}@$
Convert: @$\begin{align}\$10,000\left(1+1.035+{1.035}^{2}+\hspace{1mm}...+{1.035}^{11}\right)\end{align}@$ into the formula of a finite geometric series: @$\begin{align}a\left(\frac{1-{r}^{n}}{1-r}\right),\end{align}@$ where @$\begin{align}r\end{align}@$ is the common ratio, @$\begin{align}n\end{align}@$ is the number of terms, and @$\begin{align}a\end{align}@$ is the first term.
Question 2
Find the sum of the first @$\begin{align}15\end{align}@$ terms for the geometric sequence below.
@$\begin{align}\frac{1}{3},1,3,\ldots\end{align}@$
a
@$\begin{align}2312464.33\end{align}@$
b
@$\begin{align}1321484.3\end{align}@$
c
@$\begin{align}2351284.333\end{align}@$
d
@$\begin{align}2391484.3333\end{align}@$
Find the common ratio and rewrite the sum in the form:
@$\begin{align}\left(\tfrac{1}{3}\right)+\left(\tfrac{1}{3}\right)3+\left(\tfrac{1}{3}\right){3}^{2}+\hspace{1mm}...+\left(\tfrac{1}{3}\right){3}^{14}\end{align}@$
Factor @$\begin{align}\left(\tfrac{1}{3}\right)\end{align}@$ out from: @$\begin{align}\left(\tfrac{1}{3}\right)+\left(\tfrac{1}{3}\right)3+\left(\tfrac{1}{3}\right){3}^{2}+\hspace{1mm}...+\left(\tfrac{1}{3}\right){3}^{14}\end{align}@$
Convert: @$\begin{align}\left(\tfrac{1}{3}\right)\left(1+3+{3}^{2}+\hspace{1mm}...+{3}^{14}\right)\end{align}@$ into the formula of a finite geometric series: @$\begin{align}a\left(\frac{1-{r}^{n}}{1-r}\right),\end{align}@$ where @$\begin{align}r\end{align}@$ is the common ratio, @$\begin{align}n\end{align}@$ is the number of terms, and @$\begin{align}a\end{align}@$ is the first term.
Asked by Students
Here are the top questions that students are asking Flexi for this concept:
Related Content
Sums of Geometric Series
Sums of Geometric Sequences and Series - Overview
Sums of Geometric Sequences and Series - Example 1
Sums of Geometric Sequences and Series - Example 4
Paradoxically Speaking
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11255 | https://www.surveylab.com/blog/systematic-sampling/ | Published Time: 2024-01-31T12:42:02+00:00
What is Systematic Sampling in Surveys?
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What is Systematic Sampling in Surveys?
by Kinga Edwards
13 May 2024
Surveys help us understand what people think and need, from shopping habits to opinions on big issues.
But how do we get this information? We use systematic sampling techniques to take a peek into a large population without having to ask everyone.
It’s hardly possible to talk to millions of people! Instead, we use smart shortcuts like systematic sampling.
It gives us a fair and accurate picture by selecting people in a regular pattern. So, why should you care?
Because the better we sample, the more we can trust what surveys tell us.
Today, we will discuss systematic sampling and see how it powers up our surveys!
Systematic sampling: what is it?
Systematic sampling is a way to pick out members from a big group with a neat and structured plan.
Think of it as a probability sampling method, carefully picking people in a step-by-step way.
It starts with a random starting point in the population list and keeps picking every nth person.
The distance between each person picked is the sampling interval, which we figure out by dividing the population size by how many people we want to include (sample size).
Then, everyone in the entire population has a fair chance of being selected.
Let’s break it down with a systematic sample example. ⬇️
You have 1000 people and want to understand 100 of them. Your sampling interval (n) becomes 10. So, after choosing a random starting point, you pick every 10th person to be part of your group.
💡 Not quite sure how to survey people?SurveyLabis a great tool with many templates to choose from!
Why choose systematic sampling?
There are many pros, but cons can be found as well. Let’s start with the advantages.
Simplicity and efficiency
Systematic sampling is a favorite for many because it’s simple and efficient, especially with many people. Its linear path makes it less of a headache than other sampling methods like stratified random sampling or cluster sampling.
Less susceptible to data manipulation
The design of systematic sampling reduces the risk of data manipulation, since the only random sample is choosing the starting point.
It’s tough to mess with the data, making it less likely for data manipulation compared to methods like simple random sampling.
Uniform coverage of the population
It makes an even distribution across everyone, so no group is left out or overly focused on.
Considerations in systematic sampling
Go through this list to be aware of the disadvantages systematic sampling may have.
Risk of periodic patterns
One hiccup can be repeating patterns. If the group has a certain order that matches the sampling interval, systematic sampling might accidentally favor or ignore some parts, not giving a truly representative sample.
Inflexibility
Once you set your path, there’s not much wiggle room. This might not work for every research type, particularly those that need a flexible approach.
Steps involved in systematic sampling
Check out the steps you need to take to understand how systematic sampling works. It will be easier for you to start, then.
#1 Define the target population and determine sample size
Before starting with systematic sampling, pinpoint who you want to study (the target population) and decide on the number of people you’ll include (sample size).
Know the total population size and what part you’re interested in. Your desired sample size should be big enough to reflect the entire population but small enough to handle easily and affordably.
#2 Calculate sampling interval
Next up is figuring out the sampling interval. To do that, divide the population size by the sample size.
➡️ Let’s say your total is 1000 people, and you want a group of 100, the sampling interval would be 10. So, you’ll include every 10th member of population in your sample.
#3 Select the starting point
The starting point is where you begin picking people, and it’s the only random component in systematic sampling.
You randomly choose a number between 1 and the sampling interval.
➡️ For example, if your interval is 10, your starting point could be any number from 1 to 10.
#4 Ensure randomness in the starting point
To keep the process fair, the starting point needs to be randomly picked. This is usually done with random number generators.
➡️ It’s important because it gives everyone an equal probability of being chosen, which helps avoid any bias.
#5 Address potential bias
Sometimes, systematic sampling might accidentally line up with a hidden pattern in the group, which can skew results.
Read more here:What is Non-Response Bias and Why It Matters?
➡️ Check for these patterns and adjust your approach if needed. You might change the interval or switch to a different sampling method.
#6 Combine with other sampling methods
To get around some of the systematic sampling’s limitations, you might mix it with other techniques.
➡️ For instance, stratified random sampling can be used first to divide the group into smaller, varied parts. It may help your final sample represent everyone.
#7 Set limitations and mitigation strategies
While systematic sampling is usually efficient and easy, it’s not perfect.
Watch out for any repeating patterns in your group that might affect your results.
➡️ Researchers often need to tweak the sampling interval or starting point, or use other methods to get a truly representative sample.
Comparing systematic sampling with other sampling techniques
See how different methods measure up, including our focus on systematic sampling.
Systematic sampling vs simple random sampling
When exploring the sample selection process, let’s first compare systematic sampling with simple random sampling.
Systematic samplingSimple random sampling
➕ A probability sampling method where sample members are selected at regular intervals from a sorted population list.
➕ Effective and simple, especially for a larger population.
➖ There is a risk of data bias with periodic patterns in the population.➕ Involves selecting individuals randomly from the entire population, giving each member an equal probability of being chosen.
➕ Ideal for avoiding biases.
➖ Less practical for large datasets due to the complexity of randomly sampling each individual.
Systematic sampling vs stratified random sampling
How does systematic sampling contrast with stratified random sampling in the sample selection process? Check out the comparison.
Systematic samplingStratified random sampling
➕ Selects individuals at predetermined intervals from a list, and makes it simpler and faster.
➖ Potentially less representative of all segments of the total population.➕ Divides the total population into distinct subgroups and sample randomly from each subgroup.
➕ Ensures all population segments are represented in the sample, suitable for diverse populations.
➖ Requires a comprehensive understanding of the population’s characteristics.
Systematic sampling vs cluster sampling
Look at systematic sampling alongside cluster sampling in our discussion of the sample selection process.
Systematic samplingCluster sampling
➕ Offers a more straightforward approach by selecting individuals at regular intervals from a list.
➖ May miss nuanced representation achieved through cluster sampling, especially in diverse populations.➕ Involves dividing the population into clusters and then randomly selecting entire clusters for the sample.
➕ Works for geographically spread-out populations.
➖ May increase sampling error if clusters are not homogeneous.
Systematic sampling vs linear and circular systematic sampling
Compare systematic sampling with its variations – linear and circular systematic sampling.
Systematic samplingLinear and circular systematic sampling
➕ A probability sampling method that selects sample members at regular intervals from a complete list.
➕ Maintains a low-risk factor. It’s straightforward and efficient, especially for data collection from a complete list of the population.➕ Linear systematic sampling selects at regular intervals from a linear list.
➕ Circular systematic sampling treats the list as circular, looping back to start.
➕ Useful in avoiding risk of data bias at the end of the list.
➕ Both maintain the low-risk factor inherent in systematic sampling
➖ May overlook patterns present in the population.
Suitability of systematic sampling for specific research scenarios
Systematic sampling works great in situations where you need to pick sample members from everyone or a large group, like when collecting customer experience feedback.
It helps researchers form samples at fixed steps, and it makes the whole process tidy and predictable.
But, be careful if the people or things you’re studying change a lot or follow a repeating cycle.In these cases, systematic sampling might accidentally lean one way, missing important information.
In general, systematic sampling is loved for being straightforward and efficient in data collection, particularly with big groups. Ensure the people you pick truly reflect the whole group you’re studying, especially if there’s a chance that some kind of pattern could throw off your results.
Examples of systematic sampling in surveys
It’s a popular method in many areas, for example:
➡️ Market research
Systematic random sampling is a popular method in market research. Businesses use it to figure out what customers think.
Picture a company checking on how happy people are with what they sell. They grab their customer list and pick every nth person based on the sample interval to get a feel of everyone’s views.
Why does it work?
The even spread ensures all parts of the larger population, particularly the target audience, are considered, minimizing bias.
➡️ Social sciences
In social sciences, systematic random sampling is understanding big groups and what they do.
Let’s say researchers want to explore who’s voting for whom in a city. They use systematic sampling to pick households at regular sampling intervals across different areas.
Why does it work?
They make sure to include a mix of opinions and lifestyles, giving a real picture of the city’s heartbeat.
➡️ Opinion polls
For opinion polls, especially those covering a larger population, systematic sampling is the go-to method.
Pollsters looking into political trends might start with a voter list and select people at fixed gaps, known as sampling intervals.
Read also: Types of Poll Questions You Can Ask.
Why does it work?
It’s great for getting a pulse on the entire voting population, all while keeping things straightforward and maintaining a low risk of bias.
Benefits of using systematic sampling
It’s a solid way to gather reliable data from a big group of people. Because of the regular intervals, every part of the group has a chance to be heard, which is super helpful for maintaining accuracy.
It’s not complicated to use, making it a friendly choice for many. Still, it’s good to keep an eye out for any sneaky biases, especially in groups where patterns might throw off the randomness. And that’s crucial in data collection processes where gathering representative systematic samples matters.
Best practices for systematic sampling
Stick to a set of best practices to make your results accurate and reliable.
01 Figure out the right number to study
Before jumping into systematic random sampling,determine the right number of people to look at (that’s your sample size). You want enough people to give you the real story, but not so many that it’s unmanageable.
Also, consider the total number of people (population size) and how precise you need to be. Use tools and formulas available out there to help you get the numbers.
02 Choose an appropriate sampling interval
How often you pick a member of the population is your sampling interval. This number depends on how big the group you’re studying is.
Bigger group? Bigger gaps between picks. Divide the total population size by your sample size.
03 Select a random starting point
It’s got to be random to make sure you’re not leaning any particular way. Most people use a random number generator for this. All to give every single person an equal shot at being included.
04 Ensure transparency and documentation
Document every bit of your process, from figuring out your sample size to choosing your starting point. Clear records mean your work is transparent and can be double-checked. It makes everyone trust your findings and might even help you spot any slips or areas for improvement.
05 Consider potential sources of error
Sometimes, the group you’re analyzing might have its own natural rhythm or repeating pattern that could mess with your results. If you notice this happening, you might need to switch up your interval or try other methods to keep things unbiased.
Read also: Margin of Error in Surveys: What is it and How to Tackle it?
06 Combine with other sampling techniques if necessary
There are times when you might need to blend in other sampling techniques to get a fuller picture. This is especially true if you’re looking at a diverse crowd or need to make sure certain groups are included.
Read also: Probability Sampling: Methods, Examples, Differences, and Insights.
07 Check things out yourself
If possible, physically observe the group you’ve sampled. It’s a good way to verify that your sample is a true mini-version of the whole crowd under study.
08 Adapt and adjust
Things change, and so might your sampling method’s effectiveness. Keep an eye on it and be ready to adjust as needed to ensure it’s always giving you the best snapshot of the population.
Conclusion on systematic sampling
It’s time to draw conclusions.
Remember, the goal of systematic random sampling is to get a sample that really represents the entire population without any bias. It’s giving every person an equal shot at being included and making sure no particular pattern or group is left out or overrepresented.
Keep our best practices and other tips in mind, and you’ll be on your way to collecting and analyzing data that’s both reliable and insightful.
And for easy surveys, use SurveyLab!Sign in to start for free.
FAQ on systematic sampling
Go through the questions, so you have a clear picture of the topic.
What is systematic sampling?
Systematic sampling is a method where you choose members from a large group in a regular pattern. You start at a random point and pick every nth person to create your sample. Each person has an equal chance of being chosen.
Why is systematic sampling good for big groups?
Systematic sampling is simple and fast, which makes it great for big groups. It lets you cover the whole population evenly without missing any part.
What problems can occur with systematic sampling?
Sometimes, the pattern you use to pick people might match a pattern within the group. This can make your sample less accurate because it might favor or ignore some people. If you notice this, you might need to change your sampling method.
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Posted on January 31, 2024 by Kinga Edwards
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11256 | https://www1.grc.nasa.gov/beginners-guide-to-aeronautics/second-law-entropy/ | On this page:
Second Law – Entropy
Thermodynamics is a branch of physics which deals with the energy and work of a system. Thermodynamics deals only with the large scale response of a system which we can observe and measure in experiments. In aerodynamics, the thermodynamics of a gas obviously plays an important role in the analysis of propulsion systems. The first law of thermodynamics defines the relationship between the various forms of kinetic and potential energy present in a system, the work which the system can perform and the transfer of heat. The law states that energy is conserved in all thermodynamic processes.
However, we can imagine thermodynamic processes which would conserve energy but which never occur in nature. For example, if we bring a hot object into contact with a cold object, the hot object cools down and the cold object heats up until an equilibrium is reached. The transfer of heat goes from the hot object to the cold object. We could imagine a system in which the heat would instead be transferred from the cold object to the hot object, and such a system would not violate the first law of thermodynamics. The cold object would get colder and the hot object would get hotter but energy would be conserved. Obviously we don’t encounter such a system in nature and to explain this and similar observations, thermodynamicists proposed a second law of thermodynamics. Clasius, Kelvin, and Carnot proposed various forms of the second law to describe the particular physics problem that each was studying. The description of the second law stated on this slide was taken from Halliday and Resnick’s textbook, “Physics”. It begins with the definition of a new state variable called entropy. Entropy has a variety of physical interpretations, including the statistical disorder of the system, but for our purposes, let us consider entropy to be just another property of the system, like enthalpy or temperature.
The second law states that there exists a useful state variable called entropy. The change in entropy (delta S, (\Delta S)) is equal to the heat transfer (delta Q, (\Delta Q)) divided by the temperature (T).
(\LARGE \Delta S=\frac{\Delta Q}{T})
For a given physical process, the entropy of the system and the environment will remain a constant if the process can be reversed. If we denote the initial and final states of the system by “i” and “f”, then:
(\LARGE S_{f}=S_{i}(\text{reversible process}))
An example of a reversible process would be ideally forcing a flow through a constricted pipe. (Ideal means no boundary layer losses). As the flow moves through the constriction, the pressure, temperature and velocity would change, but these variables would return to their original values downstream of the constriction. The state of the gas would return to its original conditions and the change of entropy of the system would be zero. The second law states that if the physical process is irreversible, the entropy of the system and the environment must increase; the final entropy must be greater than the initial entropy.
(\LARGE S_{f}>S_{i}(\text{reversible process}))
An example of an irreversible process is the problem discussed in the second paragraph where a hot object is put in contact with a cold object. Eventually, they both achieve the same equilibrium temperature. If we then separate the objects they do not naturally return to their original (different) temperatures. The process of bringing them to the same temperature is irreversible.
The application of the second law describes why heat is transferred from the hot object to the cool object. Let us assume that the heat is transferred from the hot object (object 1) at temperature T1 to the cold object (object 2) at temperature T2. The amount of heat transferred is Q and the final equilibrium temperature for both objects we will call Tf. The temperature of the hot object changes as the heat is transferred away from the object. The average temperature of the hot object during the process we will call Th and it would be the average of T1 and Tf.
(\LARGE T_{h}=\frac{T_{1}+T_{f}}{2})
Similarly, for the cold object, the final temperature is Tf and the average temperature during the process is Tc which is the average of Tf and T2.
(\LARGE T_{c}=\frac{T_{2}+T_{f}}{2})
Th will always be greater than Tc, because T1 is greater than T2.
(\LARGE T_{h}>T_{c})
The entropy change for the hot object will be (-Q/Th), with the minus sign applied because the heat is transferred away from the object.
(\LARGE \Delta S_{h}=-\frac{Q}{T_{h}})
For the cold object, the entropy change is (Q/Tc), positive because the heat is transferred into the object.
(\LARGE \Delta S_{c}=\frac{Q}{T_{c}})
So the total entropy change for the whole system would be given by the equation
(\LARGE S_{f}=S_{i}-\frac{Q}{T_{h}}+\frac{Q}{T_{c}})
with Si and Sf being the final and initial values of the entropy. The term (Q/Tc) will always be greater than (-Q/Th) because Th is greater than Tc. Therefore, Sf will be greater than Si, as the second law predicts. If, instead, we had assumed that the heat was being transferred from the cold object to the hot object, our final equation would be:
(\LARGE S_{f}=S_{i}+\frac{Q}{T_{h}}-\frac{Q}{T_{c}})
The signs on the terms would be changed because of the direction of the heat transfer. This would result in Sf being less than Si and the entropy of the system would decrease which would violate the second law of thermodynamics.
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11257 | https://owl.purdue.edu/owl/subject_specific_writing/writing_in_literature/index.html | Suggested Resources
Writing in Literature
Welcome to the Purdue OWL
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Copyright ©1995-2018 by The Writing Lab & The OWL at Purdue and Purdue University. All rights reserved. This material may not be published, reproduced, broadcast, rewritten, or redistributed without permission. Use of this site constitutes acceptance of our terms and conditions of fair use.
This OWL sections contains resources for writing about literature across a variety of genres and contexts.
For additional resources for creative writing, click here.
In this section
Writing About Poetry
This section covers the basics of how to write about poetry, including why it is done, what you should know, and what you can write about.
Poetry: Close Reading
This resource will help you perform a close reading of poetry and begin developing ideas for writing papers based on close readings. This resource is enhanced by a PowerPoint file. If you have a Microsoft Account, you can view this file with PowerPoint Online.
Subsections
Writing About Film
This page offers a film writing resource explaining cinematic terms and prompts, alongside a sample analysis of Ex Machina with editorial commentary on rhetorical strategies.
Literary Terms
This handout gives a rundown of some important terms and concepts used when talking and writing about literature.
Writing About Fiction
This handout covers major topics relating to writing about fiction. This covers prewriting, close reading, thesis development, drafting, and common pitfalls to avoid.
Writing About Literature
This handout provides examples and description about writing papers in literature. It discusses research topics, how to begin to research, how to use information, and formatting.
Writing in Literature (Detailed Discussion)
These sections describe in detail the assignments students may complete when writing about literature. These sections also discuss different approaches (literary theory/criticism) students may use to write about literature. These resources build on the Writing About Literature materials.
Image in Poetry
This section covers images as they appear in poetry and covers related terminology, definitions and origins of images, uses of images, and several exercises.
Writing about World Literature
This resource provides guidance on understanding the assignment, considering context, and developing thesis statements and citations for world literature papers. It also includes a PowerPoint about thesis statements in world literature for use by instructors and students. Media File: Writing about World Literature This resource is enhanced by an Acrobat PDF file. Download the free Acrobat Reader
Resources
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Purdue OWL is a registered trademark. Copyright ©2025 by The On-Campus Writing Lab & The OWL at Purdue and Purdue University. This material may not be published, reproduced, broadcast, rewritten, or redistributed without permission. This website collects and publishes the ideas of individuals who have contributed those ideas in their capacities as faculty-mentored student scholars. The materials collected here do not express the views of, or positions held by, Purdue University. Use of this site constitutes acceptance of our terms and conditions of fair use. Privacy policy. |
11258 | https://oeis.org/A123663 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Number of shared edges in a spiral of n unit squares.
0, 1, 2, 4, 5, 7, 8, 10, 12, 13, 15, 17, 18, 20, 22, 24, 25, 27, 29, 31, 32, 34, 36, 38, 40, 41, 43, 45, 47, 49, 50, 52, 54, 56, 58, 60, 61, 63, 65, 67, 69, 71, 72, 74, 76, 78, 80, 82, 84, 85, 87, 89, 91, 93, 95, 97, 98, 100, 102, 104, 106, 108, 110, 112, 113, 115, 117, 119
(list;
graph;
refs;
listen;
history;
text;
internal format)
OFFSET
1,3
COMMENTS
If one constructs a square (square 1) and then draws another square of identical size beside it (square 2), the squares share 1 edge. If one then places an identical square above square 2 (instead of continuing in a straight path), there are now 2 shared edges. Continuing this pattern in an outward spiral, one finds that the number of shared edges is 4, 5, 7, ...
Numbers a(n) such that a(n+1) = a(n) + 1 are (except for the leading zero) A074148. Otherwise a(n+1) = a(n) + 2. - Franklin T. Adams-Watters, Oct 17 2014.
This sequence is also the maximal number of shared edges among all polyominoes with n square cells. This is the result of Harary and Harborth cited in the references. Once this is known the formula 2n - ceiling(2sqrt(n)) comes from geometrical considerations and A027709. Namely, the 4n sides of the n squares making up the polyomino form the perimeter and come together in pairs along shared edges. Hence, 4n = perimeter + 2shared edges. Maximizing shared edges minimizes perimeter and so maximum shared edges = (4n - minimum perimeter)/2 = (4n - 2ceiling(2sqrt(n)))/2 = 2n - ceiling(2sqrt(n)). This interpretation is important to landscape ecologists and is called the aggregation index in the GIS program FRAGSTATS. - Julian F. Fleron, Nov 29 2016
a(n) is also the maximum degree of the cover graphs of lattice quotients of lattice congruences of the weak order on the symmetric group S_n. See Table 1 in the Hoang/Mütze reference in the Links section. - Torsten Muetze, Nov 28 2019
a(n) is also the number of pixels in H_{n-1}, where H_n (a pixelated piece of hyperbola xy = n) is the set of the (x, y), ordered pairs of positive integers, such that xy = n or (xy < n and ((x+1)y > n or x(y+1) > n)). - Luc Rousseau, Dec 28 2019
REFERENCES
F. Harary and H. Harborth, Extremal Animals, Journal of Combinatorics, Information & System Sciences, Vol. 1, No 1, 1-8 (1976).
LINKS
Peter Kagey, Table of n, a(n) for n = 1..10000
Richard A. Brualdi and Geir Dahl, Frobenius-König theorem for classes of (0,+/-1)-matrices, Disc. Math. (2024) Vol. 347, Issue 6, 113951.
Hung Phuc Hoang and Torsten Mütze, Combinatorial generation via permutation languages. II. Lattice congruences, arXiv:1911.12078 [math.CO], 2019.
FORMULA
a(n) = 2n - ceiling(2sqrt(n)). - Julian F. Fleron, Nov 29 2016
a(n) = a(n-1) + 2 - [n-1 is a square or a pronic number], where [] stands for the Iverson bracket. - Luc Rousseau, Dec 28 2019
MAPLE
A:= 0:
for n from 2 to 100 do
if issqr(2A[n-1]+1) or issqr(2A[n-1]+2) then A[n]:= A[n-1]+1
else A[n]:= A[n-1]+2
fi
od:
seq(A[n], n=1..100); # Robert Israel, Oct 21 2014
MATHEMATICA
FoldList[Plus, 0, t = Table[2, {72}]; t--; t] ( Robert G. Wilson v, Jan 19 2007 )
PROG
(Ruby) a123663 = ; k = 0; a_n = 0; (1..N).to_a.each{ |i| 2.times{ k.times{ a_n += 2; a123663 << a_n }; a_n += 1; a123663 << a_n; }; k += 1}
(PARI) a(n)=2n - sqrtint(4n-1) - 1 \ Charles R Greathouse IV, Nov 29 2016
(Python)
from math import isqrt
def A123663(n): return (m:=n<<1)-1-isqrt((m<<1)-1) # Chai Wah Wu, Jul 28 2022
CROSSREFS
Cf. A002620.
Sequence in context: A195126 A047496 A223735 A174131 A014248 A174799
Adjacent sequences: A123660 A123661 A123662 A123664 A123665 A123666
KEYWORD
easy,nonn
AUTHOR
Zacariaz Martinez, Nov 15 2006
Extended by Robert G. Wilson v, Jan 19 2007 |
11259 | https://www.youtube.com/watch?v=p78td1-_lxU | 9.1 Show that the lines 2x + 3y – 1 = 0 and 3x – 2y + 6 = 0 are perpendicular to each other
GURUTVAM EDUCATION
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Dear Teachers, Students and Parents,
We are presenting here a New Concept of Education, Easy way of self-Study. This is an Audio-Visual e- Guide, in which we have covered each and every problem, given in Practice set of your SSC textbook and some IMP extra problems from CBSE- NCERT book.
This is useful for the students of SSC board as well as CBSE & ICSE boards as this chapter is common for all.
After each question, three options are given in the form of video links for answer, you have to click on it :
Link for Fast Track Answer without explanation.
Link for Video with English explanation.
Link for Video with Hindi explanation.
To get AUDIO-VISUAL : Maths e-Guides ( PDF ), send your Name & class
by email to gurutvameduinfra@gmail.com
or by Whatsapp to 8446005959 ( Please no call, only whatsapp )
MATHS : STRAIGHT LINE :
PARALLEL, PERPENDICULAR AND INTERCEPTS OF A LINE :
Solved Problem :
Learn about Angle of Inclination of Straight line, Inclination of X axis and Y axis ,etc basic about Line.
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CBSE, ICSE, ISC, SSC, HSC, IGCSE, A LEVEL, IB
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Transcript:
exercise nine point one problem first show that the line 2x plus 3y minus 1 is equal to 0 and 3x minus 2y plus 6 is equal to 0 are perpendicular to each other see solution first given line is 2x plus 3y minus 1 is equal to 0 to find slope of this line we have to convert this equation in the form of y is equal to MX plus C therefore transposing 2x minus 1 in the right hand side we get 3y is equal to minus 2x plus 1 therefore dividing both the files by 3 we get Y is equal to minus 2 upon 3 X plus 1 upon 3 comparing this with y is equal to MX plus C we get slope of the first line m1 is equal to minus 2 upon 3 given other line is 3x minus 2y plus 6 is equal to 0 therefore to convert this in the form of y is equal to MX plus C we get minus 2y is equal to minus 3x minus 6 therefore dividing both the sides by -2 we get Y is equal to 3 upon 2 X plus 6 upon 2 comparing this with y is equal to MX plus 3 we get slope of the second line m2 is equal to 3 upon 2 we know that two lines are perpendicular if and only if m1 into m2 is equal to minus one product of these two slopes is equal to minus 1 therefore m1 into m2 is equal to value of M 1 minus 2 upon 3 into value of n2 3 upon 2 2 & 2 cancel each other and also 3 & 3 cancel each other therefore m1 into m2 is equal to minus 1 the product of flow of these two given line is equal to minus 1 then the line 2x plus 3y minus 1 is equal to 0 and 3x minus 2y plus 6 is equal to 0 are perpendicular to each other |
11260 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_24?srsltid=AfmBOopoNaEU83kn3-XJY8_1I5lyGrttpZQnBgM2fw6_RXmXuk7PK5jW | Art of Problem Solving
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1996 AHSME Problems/Problem 24
Contents
[hide]
1 Problem
2 Solution
3 Solution 2 (Alcumus)
4 See also
Problem
The sequence consists of ’s separated by blocks of ’s with ’s in the block. The sum of the first terms of this sequence is
Solution
The sum of the first numbers is
The sum of the next numbers is
The sum of the next numbers is
In general, we can write "the sum of the next numbers is ", where the word "next" follows the pattern established above.
Thus, we first want to find what triangular numbers is between. By plugging in various values of into , we find:
Thus, we want to add up all those sums from "next number" to the "next numbers", which will give us all the numbers up to and including the number. Then, we can manually tack on the remaining s to hit .
We want to find:
Thus, the sum of the first terms is . We have to add more s to get to the term, which gives us , or option .
Note: If you notice that the above sums form , the fact that appears at the end should come as no surprise.
Solution 2 (Alcumus)
The th appearance of 1 is at position . Then there are 1's and 2's among the first numbers, so the sum of these terms is .
When , , and when , .
The sum of the first 1225 terms is . The numbers in positions 1226 through 1234 are all 2's, so their sum is . Therefore, the sum of the first 1234 terms is .
See also
1996 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 23Followed by
Problem 25
1•2•3•4•5•6•7•8•9•10•11•12•13•14•15•16•17•18•19•20•21•22•23•24•25•26•27•28•29•30
All AHSME Problems and Solutions
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.
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11261 | https://resources.allsetlearning.com/chinese/grammar/Asking_about_degree_with_%22duo%22 | Asking about degree with "duo"
How big? How busy? How cold? Ask questions like these regarding the degree of an adjective with 多 (duō). This is just one of the many uses of this word.
Contents
Structure
多 (duō) is often used to ask about the degree or extent of something.
Subj. + 多 + Adj. ?
This is an easy way to ask "How [adjective] is [subject]?"
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大 (dà) and 小 (xiǎo) can also be used to describe ages. The question phrase 多大 (duō dà) is often used to ask "how old." However, it is an informal way to ask, usually reserved for peers, close friends, or children. The phrase 几岁 (jǐ suì) is most often used for children young enough to display their ages on one hand. Adults do not normally directly ask each other's ages in a formal setting.
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11262 | https://www.quora.com/Why-is-the-Ka2-smaller-than-the-Ka1-of-sulphuric-acid | Why is the Ka2 smaller than the Ka1 of sulphuric acid? - Quora
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Chemistry
Acid Dissociation Constan...
Sulfuric Acid
Properties of Acids
Ionization Constant
Dissociation (chemistry)
Equilibrium Constant
Specific Acids
Chemical Properties
5
Why is the Ka2 smaller than the Ka1 of sulphuric acid?
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Anish Koulgi
BE in Computer Science, Pune Institute of Computer Technology (Graduated 2022) · Author has 56 answers and 304K answer views
·7y
The equations for dissociation of H 2 S O 4 H 2 S O 4 are as follows :
H 2 S O 4⇌H S O 4−+H+K a 1≈10 H 2 S O 4⇌H S O 4−+H+K a 1≈10
H S O 4−⇌S O 4−2+H+K a 2=1.2∗10−2 H S O 4−⇌S O 4−2+H+K a 2=1.2∗10−2
H 2 S O 4 H 2 S O 4 is a neutral molecule and so readily gives a proton that is H+H+ ion. Whereas H S O 4−H S O 4− is a negatively charged molecule so resists to give a proton. H 2 S O 4 H 2 S O 4 has a much higher tendency to lose a proton than H S O 4−H S O 4−. Thus the K a 1 K a 1 value is much more than the K a 2 K a 2 value for H 2 S O 4 H 2 S O 4 in water.
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Ramananda Adhikari
Deputy General Manager (Chemical ) & TS to CMD. at Hindustan Copper Limited · Author has 57 answers and 81.5K answer views
·6y
Combined extronegative effect of 4 oxygen atom is responsible for high Ka1 value. For Ka2 stage electronegativity of oxygen atoms are partially satisfied by getting extra electron upon extraction of 1st proton. In view of the above the ease of 2nd proton extraction is comparatively less resulting less Ka2 value.
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Vishal Sharma
Studied at Sarvottam Career Institute
·7y
When H2SO4 forms HSO4⁻ and H⁺ ions than Ka1 exist. And when H2SO4 forms SO4²⁻ and H⁺ ion than Ka2 is formed. It is easier to release a H⁺ ion as compare to HSO4⁻ ion from a neutral H2SO4. Thus Ka2 is less seen than Ka1
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More answers below
What is the normality of 2m of sulphuric acid?
How can I prepare 0.18M of 1 % sulphuric acid?
Does sulfuric acid react with copper?
Does a tubular battery use sulphuric acid?
What is the concentration of sulphuric acid and hydrochloric acid?
Daniel James Berger
PhD in organic/organosilicon chemistry · Author has 4.6K answers and 11.3M answer views
·9y
Related
Why does hydrosulfuric acid have such a low Ka?
The sulfur-hydrogen bond is not all that weak; this table of Bond Energies lists it at 81 kcal/mol, compared to C-H which averages 99 kcal/mol.
For acids where the conjugate base has no inductive or resonance stabilizing factors, acid strength is perhaps most strongly influenced by the strength of the bond to hydrogen.
Generally, within the same row, acid strength is a function of electronegativity (elements that are more electrongative have more acidic bonds to hydrogen).
Within the same column (family), acid strength is largely a function of the strength of the bond to hydrogen, and secondaril
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The sulfur-hydrogen bond is not all that weak; this table of Bond Energies lists it at 81 kcal/mol, compared to C-H which averages 99 kcal/mol.
For acids where the conjugate base has no inductive or resonance stabilizing factors, acid strength is perhaps most strongly influenced by the strength of the bond to hydrogen.
Generally, within the same row, acid strength is a function of electronegativity (elements that are more electrongative have more acidic bonds to hydrogen).
Within the same column (family), acid strength is largely a function of the strength of the bond to hydrogen, and secondarily a function of the size of the conjugate-base anion. Larger anions spread the charge over a larger volume, making it more stable.
If you're going to compare (for example) H2S to HF, it will be a mixture of those factors that's probably better to use to explain than to predict.
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The dissociation constants K a 1 K a 1 and K a 2 K a 2 for sulfuric acid (H 2 S O 4 H 2 S O 4) reflect the strength of the acid in its two dissociation steps:
First Dissociation (H 2 S O 4 H 2 S O 4 to H+H+ + H S O−4 H S O 4−):
This step has a very high K a 1 K a 1 value because sulfuric acid is a strong acid, meaning it dissociates almost completely in water. The reaction can be represented as:
H 2 S O 4⇌H++H S O−4 H 2 S O 4⇌H++H S O 4−
Since it dissociates almost completely, K a 1 K a 1 is significantly greater than 1.
Second Dissociation (H S O−4 H S O 4− to H+H+ + S O 2−4 S O 4 2−):
The second dissociation is represented as:
H S O−4\righ H S O 4−\righ
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The dissociation constants K a 1 K a 1 and K a 2 K a 2 for sulfuric acid (H 2 S O 4 H 2 S O 4) reflect the strength of the acid in its two dissociation steps:
First Dissociation (H 2 S O 4 H 2 S O 4 to H+H+ + H S O−4 H S O 4−):
This step has a very high K a 1 K a 1 value because sulfuric acid is a strong acid, meaning it dissociates almost completely in water. The reaction can be represented as:
H 2 S O 4⇌H++H S O−4 H 2 S O 4⇌H++H S O 4−
Since it dissociates almost completely, K a 1 K a 1 is significantly greater than 1.
Second Dissociation (H S O−4 H S O 4− to H+H+ + S O 2−4 S O 4 2−):
The second dissociation is represented as:
H S O−4⇌H++S O 2−4 H S O 4−⇌H++S O 4 2−
This step has a lower K a 2 K a 2 value because H S O−4 H S O 4− is a weaker acid than H 2 S O 4 H 2 S O 4. While H S O−4 H S O 4− can still donate a proton, it does not do so as readily as H 2 S O 4 H 2 S O 4. As a result, the equilibrium does not favor the formation of H+H+ and S O 2−4 S O 4 2− to the same extent, leading to a smaller K a 2 K a 2 value.
Summary
K a 1 K a 1 is larger than K a 2 K a 2 because:
The first dissociation of sulfuric acid is nearly complete (strong acid).
The second dissociation involves a weaker acid (HSO₄⁻), which is less likely to donate a proton.
This difference in acid strength between the two steps is the fundamental reason why K a 1 K a 1 is greater than K a 2 K a 2.
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David Hu
B.S. Biochemistry, Ohio University (2016) · Author has 329 answers and 1.2M answer views
·10y
Related
Why is sulfuric acid less acidic than perchloric acid?
Sulfuric acid:
Perchloric acid:
The conjugate bases of both acids are conjugated systems; the lone electron pair created by hydrogen donation is delocalized across the double bonds to oxygen. In general, delocalization of electrons makes the molecule more stable. Perchloric acid has three of such double bonds, while sulfuric acid has only two. Thus, the conjugate base of perchloric acid is more stable, meaning the acid itself is more acidic.
Continue Reading
Sulfuric acid:
Perchloric acid:
The conjugate bases of both acids are conjugated systems; the lone electron pair created by hydrogen donation is delocalized across the double bonds to oxygen. In general, delocalization of electrons makes the molecule more stable. Perchloric acid has three of such double bonds, while sulfuric acid has only two. Thus, the conjugate base of perchloric acid is more stable, meaning the acid itself is more acidic.
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How do I prepare exactly 0.01M of sulphuric acid in 2.5L of solution?
How do I prepare 0.1n of sulphuric acid?
Why is sulphuric acid used in preparing volatile acid?
How do I prepare a 10% sulphuric acid solution?
What happens when hot sulphuric acid is added to sulphur?
Barry Gehm
Former Asst Prof. Of Chemistry/Biochemistry at Lyon College (2003–2024) · Author has 13.6K answers and 16.6M answer views
·7y
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Why is sulphuric acid regarded as the king of acids even though many other acids are stronger than it?
Acids that are stronger than sulfuric are actually rare, and are referred to as “superacids”. In aqueous solution there are none stronger than sulfuric, since all strong acids generate hydronium ion quantitatively in water.
Because of its high boiling point, all other common acids can be produced by heating solutions of salts with sulfuric acid (e.g., sodium chloride plus sulfuric acid produces hydrochloric acid, potassium nitrate with sulfuric makes nitric acid, and so on). Sulfuric is king of the acids because you can use it to make all the other common acids.
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Vishal Khawse
B.E in Petrochemical Engineering, Samrat Ashok Technological Institute (S.A.T.I.), Vidisha (Graduated 2020)
·7y
Related
How is sulphuric acid formed?
Sulfuric acid is industrially produced from C ontact process.
This process mainly has 5 steps:
Production of sulfur dioxide from sulfur(oxidation of sulfur).
Reaction is:
S+ 02 → SO
Purification of sulfur dioxide (to eemove particles, oxides of iron and arsenic etc.
Catalytic oxidation of sulfur dioxide (SO2)to sulfur trioxide (SO3).
Reaction is:
2SO2 +O2 (in presence of V2O5 at 450°C ) → 2SO3 + heat.
4.Conversion of SO3 to OLEUM.
Reaction is:
SO3 + H2SO4 → H2S2O7
Diffusion of oleum to sulfuric acid.
Reaction is:
H2S2O7 +H2O → 2H2SO4
Upvote ·
9 4
Nelson Carpenter
Train Driver at Renfe (2019–present)
·3y
Related
Why is sulfuric acid less acidic than perchloric acid?
Sulfuric acid:
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Sulfuric acid:
For those that suffer gout, be careful with taking a lot of drugs for long time, it can effect your health seriously. Drugs just deal with the symptoms not the disease. If you want to reserver gout completely by natural way, you can check this secret tips. I have got good results with it. I've been with gout for more than five years and even though I'm 37 I had been feeling just like more a 70 year old with all the pain and lack of mobility. I had been relying heavily on Catafast to help with all the pain but it had begun to cause issues with my stomach. Now I have been completely off Catafast at yet all of the pain is gone just by following that method.
Perchloric acid:
The conjugate bases of both acids are conjugated systems; the lone electron pair created by hydrogen donation is delocalized across the double bonds to oxygen. In general, delocalization of electrons makes the molecule more stable. Perchloric acid has three of such double bonds, while sulfuric acid has only two. Thus, the conjugate base of perchloric acid is more stable, meaning the acid itself is more acidic.
Upvote ·
Katrina Peck
Former Works at Greenhouse.io
·3y
Related
Why is acetic acid weaker than sulphuric acid?
I personally interpret this question as why sulphuric acid being stronger rather than why acetic acid being weaker.
This is because the conjugate base of sulphuric acid, sulfate ion is more stable due to more resonance structure and inductive effect. More resonance structure = the (-2) charge is shared among the existing oxygen = more stable. More stable conjugate base = reaction favors the formation of conjugate base = thus in the same time form more proton since they’re both products.
For people who suffer gout, be careful with taking lots of drugs for long time, it can effect your health seri
Continue Reading
I personally interpret this question as why sulphuric acid being stronger rather than why acetic acid being weaker.
This is because the conjugate base of sulphuric acid, sulfate ion is more stable due to more resonance structure and inductive effect. More resonance structure = the (-2) charge is shared among the existing oxygen = more stable. More stable conjugate base = reaction favors the formation of conjugate base = thus in the same time form more proton since they’re both products.
For people who suffer gout, be careful with taking lots of drugs for long time, it can effect your health seriously. Drugs just treat the symptoms not the disease. If you want to reserver gout completely by natural way, you can check this guide. I've got great results with it. I have followed that method| for 2 weeks and now I have no pain at all and no signs of any attacks (and I'm off the damn Sterapred)
In the case of acetic acid, acetate ion is not as stable as sulfate ion is to the extent that acetic acid is more stable than its own conjugate base. Acetate ion have -1 charge = requires energy. Acetic acid = zero charge.
But then, why not sulfuric acid stays as it is since sulfuric acid = 0 charge and sulfate ion = -2 charge divided by 4 oxygen atoms, which distributes the energy. Well this is most probably caused by polar/inductive effect. This answer may not be complete but hopes you can continue researching and reading more to add on the details.
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9 1
Veerasamy Natarajan
Graduate in Chemistry, University of Madras (Graduated 1973) · Author has 3.7K answers and 4.3M answer views
·7y
Related
How do you prepare 10% sulphuric acid?
10% means 10 grams in 100 grams of water. You can prepare volume/volume or weight/volume. As you prepare 10% sulphuric acid, important warning is, add acid to water. Make sure water is not added to sulphuric acid.
Glasswares required:
Reagent Bottle 500 ml -1, Glass Measuring Cylinder 100 ml - 2; Glass rod - 1, Glasss Funnel, Rubber apron -1, Rubber Gloves - 1 pair, Goggles - 1 pair, Distilled water, Con Sulphuric acid
Procedure: For example volume/volume, take 90 ml of distilled or demineralised water in a measuring cylinder/jar. Put a long glass rod inside the jar. Measure 10 ml of LR grade con
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10% means 10 grams in 100 grams of water. You can prepare volume/volume or weight/volume. As you prepare 10% sulphuric acid, important warning is, add acid to water. Make sure water is not added to sulphuric acid.
Glasswares required:
Reagent Bottle 500 ml -1, Glass Measuring Cylinder 100 ml - 2; Glass rod - 1, Glasss Funnel, Rubber apron -1, Rubber Gloves - 1 pair, Goggles - 1 pair, Distilled water, Con Sulphuric acid
Procedure: For example volume/volume, take 90 ml of distilled or demineralised water in a measuring cylinder/jar. Put a long glass rod inside the jar. Measure 10 ml of LR grade con.sulphuric acid in another measuring jar and add the acid through the sides of the glass tube into the water with stirring. Now 10% sulphuric acid is ready. Transfer the contents into a reagent white bottle and close it. Label it properly. Use funnel while transferring acid.
Precausion: Wear rubber apron, gloves and goggle before preparation. Open acid bottle carefully. Check water shower in the laboratory is working and ensure in good condition.
To prepare weight/volume 10% sulphuric acid, weigh 10 grams of con.sulphuric acid carefully in a clean 100 ml beaker. Take 50 ml of DM water or distilled water in a glass measuring cylinder. Use a glass rod for stirring. Add now 10 grams of con.sulphuric acid to the DM water slowly through the sides of the glass rod. Stirr gently after every addition. Trnasfer the acid completely into DM water and make up with water up to 100 ml mark. Transfer the contents into a reagent bottle using a glass funnel and close it. Label the bottle. Follow the precausions strictly.
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9 1
Mohtisham Usman
Studied at Dow International Medical College
·7y
Related
Why is sulphuric acid regarded as the king of acids even though many other acids are stronger than it?
It is so important, that at one time the annual production of sulphuric acidwas taken as a measure of the degree of industrialisation of a country, and earned it its nickname of the 'king of chemicals'. Sulphuric acid is familiar to us as the electrolyte in the lead acidbatteries in automobiles.
Upvote ·
9 4
Steve Geo
Ph.D. in Chemistry&Organic Chemistry, University of Iowa (Graduated 1964) · Author has 2.9K answers and 612.5K answer views
·Apr 15
Related
What is the reason for sulphuric acid having a higher Ka value than hydrochloric acid, even though it has a lower pKa value?
HCl has one ionizable hydrogen and is 100% ionized in water. The first ionization constant of H2SO4, Ka1, is actually lower, at ~1000. The second, Ka2, comparatively lower, at 10^-2.
Upvote ·
Related questions
What is the normality of 2m of sulphuric acid?
How can I prepare 0.18M of 1 % sulphuric acid?
Does sulfuric acid react with copper?
Does a tubular battery use sulphuric acid?
What is the concentration of sulphuric acid and hydrochloric acid?
How do I prepare exactly 0.01M of sulphuric acid in 2.5L of solution?
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Why is sulphuric acid used in preparing volatile acid?
How do I prepare a 10% sulphuric acid solution?
What happens when hot sulphuric acid is added to sulphur?
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What is the normality of 2m of sulphuric acid?
How can I prepare 0.18M of 1 % sulphuric acid?
Does sulfuric acid react with copper?
Does a tubular battery use sulphuric acid?
What is the concentration of sulphuric acid and hydrochloric acid?
How do I prepare exactly 0.01M of sulphuric acid in 2.5L of solution?
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11263 | https://physics.stackexchange.com/questions/317341/why-do-ice-cubes-shrink-in-the-freezer | thermodynamics - Why do ice cubes shrink in the freezer? - Physics Stack Exchange
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Why do ice cubes shrink in the freezer?
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Asked 8 years, 6 months ago
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OK before you all yell "EVAPORATION", I know that's the boilerplate answer, but why and how.
Back in high school we were taught water, and most other elements and compounds, have three states, solid, liquid, and gas. Which state it's in being dependent on temperature and pressure. OK that all makes sense.
To transition from one state to another sufficient energy has to be imparted to the substance to pass through that transition before it will continue to heat up. To get water from ice all the way up to a gas takes a lot of energy... ask all the distillers out there.
Still.. all good...
But my ice cubes disappear over time inside a closed freezer.
The water is evaporating, turning into a gas state, but where did the energy come from to take it all the way from a solid to a gas when it never left the freezer..
thermodynamics
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edited Mar 8, 2017 at 20:12
DilithiumMatrix
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2 For clarification, your ice cubes are shrinking while remaining in the freezer?NeutronStar –NeutronStar 2017-03-08 19:27:24 +00:00 Commented Mar 8, 2017 at 19:27
Yes Joshua over time Trevor_G –Trevor_G 2017-03-08 19:28:51 +00:00 Commented Mar 8, 2017 at 19:28
1 When the same thing happens to your frozen food, we call it "freezer burn."Solomon Slow –Solomon Slow 2017-07-28 17:33:19 +00:00 Commented Jul 28, 2017 at 17:33
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I had never really noticed that happening, so before I said "that seems odd, are you sure?" I did some research and surprised myself.
According to the Wikipedia page on freezer burn water will sublimate from the surface of ice if the air temperature is low enough and the air is dry enough.
The reduced vapour pressure of the dry air is enough to cause sublimation.
Hopefully someone can give a more mathematical approach, but this is somewhere to start.
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answered Mar 8, 2017 at 19:41
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Yes you have to go on vacation for like a month... to really notice.Trevor_G –Trevor_G 2017-03-08 19:44:38 +00:00 Commented Mar 8, 2017 at 19:44
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Ice at -10 centigrade has a vapor pressure of 259.9 p/Pa. It evaporates.
Water still evaporates at less than boiling.
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answered Jul 28, 2017 at 15:15
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11264 | https://www.quora.com/How-do-I-calculate-the-mass-of-an-element-in-a-compound-with-a-given-mass | Something went wrong. Wait a moment and try again.
Molar Weight
Mass Mass Calculation
Elements and Compounds
Chemical Calculations
Molar Concentration
5
How do I calculate the mass of an element in a compound with a given mass?
·
To calculate the mass of a specific element in a compound when you know the total mass of the compound, you can follow these steps:
Identify the Molecular Formula
: Determine the molecular formula of the compound to identify the elements and their respective quantities.
Calculate the Molar Mass of the Compound
:
Find the atomic mass of each element in the compound (usually found on the periodic table).
Multiply the atomic mass of each element by the number of times it appears in the formula.
Sum these values to get the total molar mass of the compound.
Molar Mass
=
∑
(
\tex
To calculate the mass of a specific element in a compound when you know the total mass of the compound, you can follow these steps:
Identify the Molecular Formula: Determine the molecular formula of the compound to identify the elements and their respective quantities.
Calculate the Molar Mass of the Compound:
Find the atomic mass of each element in the compound (usually found on the periodic table).
Multiply the atomic mass of each element by the number of times it appears in the formula.
Sum these values to get the total molar mass of the compound.
Molar Mass=∑(Atomic Mass of Element×Number of Atoms)
Use Proportions to Calculate the Mass of the Element:
Use the formula:
Mass of Element=(Molar Mass of ElementMolar Mass of Compound)×Total Mass of Compound
Let's say you have 50 grams of water (H₂O) and you want to find the mass of hydrogen in it.
Molecular Formula: H₂O
Calculate Molar Mass:
Hydrogen (H): 1.01 g/mol × 2 = 2.02 g/mol
Oxygen (O): 16.00 g/mol × 1 = 16.00 g/mol
Total Molar Mass of H₂O = 2.02 + 16.00 = 18.02 g/mol
Calculate Mass of Hydrogen:
Molar Mass of H: 2.02 g/mol
Total Mass of H₂O: 50 g
Mass of H=(2.0218.02)×50≈5.61 grams
So, in 50 grams of water, there are approximately 5.61 grams of hydrogen.
Related questions
How do you calculate the mass of a compound?
How do I find the mass of an element in a compound?
How do I calculate the number of molecules of an element in a compound if I only know the total mass?
How do I calculate relative atomic mass?
How do you calculate the percent by mass of an element in a compound?
Purna Chandra Sahu
M. Sc. in Pure Chemistry & Inorganic Chemistry, Calcutta Uniiversity (Graduated 1987)
·
Author has 2.9K answers and 6M answer views
·
4y
To solve this type of problem, you have to know the molecular formula of the compound and atomic masses of the all constituent elements. From atomic masses you can calculate the Molar mass of the compound. Now, mass of an element in a given mass of compound = (atomic mass of the element × numb...
Claudio Giomini
Laurea (M. Sc.) from Sapienza University of Rome (Graduated 1965)
·
Author has 3.5K answers and 2.1M answer views
·
3y
Related
When calculating the relative formula mass of a compound, why do we often use the relative atomic mass of the elements, instead of the actual mass number of that element? Doesn't using relative atomic mass produce any inaccuracy in calculations?
Mass number is perhaps a deceptive term. It actually does not specify the mass, but the number of nucleons (protons and neutrons) present in a given isotope of a given element. It is a dimensionless positive integer number. On the other hand, the (relative) atomic mass of that isotope is the actual mass of the same isotope divided by 1/12th of the mass of a C-12 atom. It is a dimensionless, positive number as well, but usually non-integer. Its value, in most cases, is very close to, but not coinciding with, the mass number. The relative atomic mass of an element is the weighed average of the r
Mass number is perhaps a deceptive term. It actually does not specify the mass, but the number of nucleons (protons and neutrons) present in a given isotope of a given element. It is a dimensionless positive integer number. On the other hand, the (relative) atomic mass of that isotope is the actual mass of the same isotope divided by 1/12th of the mass of a C-12 atom. It is a dimensionless, positive number as well, but usually non-integer. Its value, in most cases, is very close to, but not coinciding with, the mass number. The relative atomic mass of an element is the weighed average of the relative atomic masses of its isotopes, made by keeping into account the percent abundance of these isotopes; a fact that very often causes the result thus obtained to be quite far from an integer. The relative formula mass of a given compound is the sum of the relative atomic masses of the elements constituent that compound, on the basis of the chemical formula of that compound.
Now, employing mass numbers instead of relative atomic masses does involve errors that become more serious when passing from a single isotope to the element as a whole, that is, the mixture of its isotopes according to their abundance as present in nature (at least, on our Earth’s crust). And similarly large errors would be introduced in the calculation of the relative formula-mass of the compounds.
Howard Ludwig
Ph.D. in Physics, Northwestern University (Graduated 1982)
·
Author has 3K answers and 10.3M answer views
·
3y
Related
When calculating the relative formula mass of a compound, why do we often use the relative atomic mass of the elements, instead of the actual mass number of that element? Doesn't using relative atomic mass produce any inaccuracy in calculations?
“When calculating the relative formula mass of a compound, why do we often use the relative atomic mass of the elements, instead of the actual mass number of that element? Doesn't using relative atomic mass produce any inaccuracy in calculations?”
Your question is based on a fundamentally flawed assumption. Let's say you want the relative molecular mass (also known as molecular weight) of SnCl₂. Tin has 10 naturally occurring isotopes on Earth represented by the mass numbers:
112
114
115
116
117
118
119
120
122
124
What then is the mass number of tin given that atoms of tin can have any one of the above m
“When calculating the relative formula mass of a compound, why do we often use the relative atomic mass of the elements, instead of the actual mass number of that element? Doesn't using relative atomic mass produce any inaccuracy in calculations?”
Your question is based on a fundamentally flawed assumption. Let's say you want the relative molecular mass (also known as molecular weight) of SnCl₂. Tin has 10 naturally occurring isotopes on Earth represented by the mass numbers:
112
114
115
116
117
118
119
120
122
124
What then is the mass number of tin given that atoms of tin can have any one of the above mass numbers? The mass number is a count of protons and neutrons taken together, with the combined concept commonly referred to as nucleons. An individual atom can have a specific number of nucleons, as can a nuclide, as can an isotope of an element. Therefore, a specific atom, a specific nuclide, or a specific isotope of a specific element has a mass number. However, while each atom of some element has the same number of protons, the number of neutrons typically varies from atom to atom of that element. Therefore, the number of nucleons typically varies from atom to atom of that element. Therefore, the mass number typically varies from atom to atom of that element. Therefore, there is no one number that can be validly said to represent the mass number of an element. Therefore, there is no such thing as the mass number of an element.
Yes, there are secondary school teachers of chemistry who are ignorant of their subject who will say erroneously (among many other false “facts" pertaining to chemistry) that the mass number of an element is the standard atomic weight of an element rounded to the nearest integer. Perhaps you have one such teacher. Taking tin as an example, the standard atomic weight is 118.710, but that does not mean that the mass number of tin is 119, implying wrongly that all atoms of tin have mass number 119—some of them do, but over 91 % of tin atoms have a mass number different from 119. (Bromine is even worse, with standard atomic weight rounded to an integer being 80, but no atoms of ⁸⁰Br occur naturally on Earth.)
Even for an element like cesium that has only one isotope naturally occurring on Earth in more than [negligible] trace amounts, the relative atomic mass of Cs as an element and of the specific isotope ¹³³Cs is 132.905 451 96, while the mass number of the isotope is 133. We can see that the mass number of an atom is only coarsely related to the relative atomic mass of the atom.
The mass number is very convenient for labeling a specific atom, nuclude, or isotope, because it is a simple numeric value and uniquely identifies a specific isotope based on count of nucleons. [The element itself indicates the number of protons while the mass number indicates the total number of protons and neutrons, and from those two pieces of information the number of neutrons can be determined uniquely.] The mass number of an atom is [only] coarsely related to the mass of that atom. Only the standard atomic weight of elements is representative of the average or nominal mass of the atoms for an element. To assess a meaningful average [relative] mass for formula units, you must use the most representative value for average [relative] mass of the component atoms, which is their standard atomic weight.
Related questions
What happens to mass when a compound undergoes decomposition into its elements or other compounds with lower atomic/molecular masses (elemental formulae)?
How do you find the empirical formula of a compound if the total mass and mass of elements are given?
How do you find the mass of a percentage of a substance, given the mass of what the substance is in?
What is the difference between molecular mass and elemental mass? How do you calculate them for a given element or compound?
How do I calculate the number of ions of an element within a compound if I have the mass of the compound?
Siddharth Shirore
8y
Related
How do I find the mass of elements in a compound?
Use the unitary method.
1 mole of CaO weighs 56 grams and contains 16 grams of oxygen.
Let 10 grams of CaO contain x grams of Oxygen.
16\56=x\10
x= 160\56 = 2.85 grams.
Hope that helped.\U0001f64c
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Trevor Hodgson
Knows English
·
Author has 11.8K answers and 12.3M answer views
·
3y
Related
How do you find the moles of an element in a compound given mass?
You have the mass of the sample of the compound
You require the molar mass of the compound - you get this from Wikipedia or calculate it using molar mass data from the periodic table.
Calculate the moles of the compound : moles = mass g / molar mass g/mol
Example :
You have 36.48 g Na2SO4 . How many moles is this
Molar mass Na2SO4 :
Molar mass Na = 22.990 g/mol - 2 mol Na = 45.98
Molar mass S = 32.066 g/mol
Molar mass O = 15.999 g/mol - 4 mol O = 63.996
Molar mass Na2SO4 = 45.98 + 32.066 + 63.996 = 142.042 g/mol
Mol Na2SO4 in 36.48 g = 36.48 g / 142.042 g/mol = 0.25682545 mol direct from calculator
Answe
You have the mass of the sample of the compound
You require the molar mass of the compound - you get this from Wikipedia or calculate it using molar mass data from the periodic table.
Calculate the moles of the compound : moles = mass g / molar mass g/mol
Example :
You have 36.48 g Na2SO4 . How many moles is this
Molar mass Na2SO4 :
Molar mass Na = 22.990 g/mol - 2 mol Na = 45.98
Molar mass S = 32.066 g/mol
Molar mass O = 15.999 g/mol - 4 mol O = 63.996
Molar mass Na2SO4 = 45.98 + 32.066 + 63.996 = 142.042 g/mol
Mol Na2SO4 in 36.48 g = 36.48 g / 142.042 g/mol = 0.25682545 mol direct from calculator
Answer can have 4 significant digits
Answer: 0.2568 mol Na2SO4
Now you want to know the moles of each element in the compound :
Mol of Na = 2 0.2568 mol = 0.5136 mol Na
Mol of S = 1 0.2568 mol = 0.2568 mol S
Mol of O = 4 0.2568 mol = 1.027 mol O
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Uglubuglu
An amateur in Chemistry
·
Author has 153 answers and 290.6K answer views
·
5y
Related
How do I calculate the number of molecules of an element in a compound if I only know the total mass?
This is a very easy question.
You only need to know the formula of finding moles i.e.
Moles=total mass/At. mass or molar mass
So, you can easily find no. of moles as you also know the molar mass
Then apply another formula for finding moles i.e.
Moles=no. of molecules/6.023x10^23 i.e. Avogadro’s no.
or no. of molecules = moles x 6.023x10^23.
For understanding more clearly see the attached photo in which I have solved an example having a similar question……..
Hope you get it😄😄
This is a very easy question.
You only need to know the formula of finding moles i.e.
Moles=total mass/At. mass or molar mass
So, you can easily find no. of moles as you also know the molar mass
Then apply another formula for finding moles i.e.
Moles=no. of molecules/6.023x10^23 i.e. Avogadro’s no.
or no. of molecules = moles x 6.023x10^23.
For understanding more clearly see the attached photo in which I have solved an example having a similar question……..
Hope you get it😄😄
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Rachel Oxburgh
Studied Chemistry at University of Oxford (Graduated 1988)
·
Author has 97 answers and 365.7K answer views
·
7y
Related
How can I calculate the mass of a hydrogen molecule?
In my view, neither of the previous two answers actually calculates the mass of a single hydrogen molecule. To do that, you take the mass of a mole of hydrogen (2g, as hydrogen exists as diatomic molecules) And divide by the number of particles in a mole (Avogadro’s number; 6.02 x 10^23).
so, the mass of a single molecule of hydrogen is 3.32 x 10 ^-24g
Juwayriyah Nuha
Participated in National astronomical exhibition.
·
7y
Related
How do you calculate a molecular formula given a compound's molar mass and empirical formula?
Hi,
This is Princy Nuha.
The answer to your question is simple, you need the below formula to solve ur answer.
Molecular formula= ( Empirical formula) n
n = molecular weight / empirical formula weight
To help you more for better understanding, the given question and solution from my material may help you.
This is taken from my notes.
Thanks, have a nice day ahead
Follow me at princy nuha at qoura.
Hi,
This is Princy Nuha.
The answer to your question is simple, you need the below formula to solve ur answer.
Molecular formula= ( Empirical formula) n
n = molecular weight / empirical formula weight
To help you more for better understanding, the given question and solution from my material may help you.
This is taken from my notes.
Thanks, have a nice day ahead
Follow me at princy nuha at qoura.
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William Kyei
Former Chemistry Teacher at Accra Polytechnic (1983–1987)
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Steve Geo
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What is the approximate mass of one mole of a compound? If an exact mass is not possible to obtain, what would be a close estimate?
Each compound has a specified atomic composition of the elements that form the compound. Some compounds are made up of simple molecules and others are made of giant molecules.
We have one mole of the compound when the number of molecules (simple or giant) is 6.023x10^(23) (approx.). The mass of the mole of the compound can be computed from the atomic mass of each element in the compound and the number of atoms of that element in the compound.
For example: CO2 (carbon dioxide) is a simple molecular compound made of one atom of carbon and two atoms of oxygen. From data books on chemistry you can f
Each compound has a specified atomic composition of the elements that form the compound. Some compounds are made up of simple molecules and others are made of giant molecules.
We have one mole of the compound when the number of molecules (simple or giant) is 6.023x10^(23) (approx.). The mass of the mole of the compound can be computed from the atomic mass of each element in the compound and the number of atoms of that element in the compound.
For example: CO2 (carbon dioxide) is a simple molecular compound made of one atom of carbon and two atoms of oxygen. From data books on chemistry you can find that: one atom of carbon (C) has average mass of 12 a.m.u. (approx); one atom of oxygen (O) has average mass of 16 a.m.u. The atomic mass unit (a.m.u.) can be expressed in grams or kilograms. So for carbon dioxide (CO2) the average mass of one molecule is (12+2x16)a.m.u = 44 a.m.u.
Consequently for CO2: one mole of carbon dioxide compound has the average mass is 6.023x10^23x(44a.m.u.).
Thus for any compound of known molecular formula one can find to a good approximation the mass of one mole of the compound. You will need the atomic masses of the constituent elements of the compound.
Trevor Hodgson
Knows English
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How do you calculate the percentage by mass and mole percent for an element in a compound?
Take an easy compound : NaCl .
This is made up of 1 mol Na and 1 mol Cl.
Mol % Na = 50% and mol % Cl = 50%
Mass % :
Molar mass Na = 23.0 g/mol
Molar mass Cl = 35.5 g/mol
Total molar mass = 58.5 g
mass % Na = 23 g / 58.5 g 100% = 39.32%
Mass % Cl = 35.5 g/ 58.5 g 10% = 60.68 %
Now take a more complicated molecule or formula: Na3PO4
One mol of this consists of : 3 mol Na + 1 mol P + 4 mol O = 8 mol
Mol % Na = 3/8100% = 37.5%
Mol % P = 1/8100% = 12.5%
Mol % O = 4/8100% = 50.0%
Mass %
Molar mass Na = 23.0 g/mol 3 mol = 323.0 = 69.0 g
Molar mass P = 31.0 g/mol 1 mol = 131.0 = 31.0 g
Molar mass O = 16.0 g/mo
Take an easy compound : NaCl .
This is made up of 1 mol Na and 1 mol Cl.
Mol % Na = 50% and mol % Cl = 50%
Mass % :
Molar mass Na = 23.0 g/mol
Molar mass Cl = 35.5 g/mol
Total molar mass = 58.5 g
mass % Na = 23 g / 58.5 g 100% = 39.32%
Mass % Cl = 35.5 g/ 58.5 g 10% = 60.68 %
Now take a more complicated molecule or formula: Na3PO4
One mol of this consists of : 3 mol Na + 1 mol P + 4 mol O = 8 mol
Mol % Na = 3/8100% = 37.5%
Mol % P = 1/8100% = 12.5%
Mol % O = 4/8100% = 50.0%
Mass %
Molar mass Na = 23.0 g/mol 3 mol = 323.0 = 69.0 g
Molar mass P = 31.0 g/mol 1 mol = 131.0 = 31.0 g
Molar mass O = 16.0 g/mol 4 mol = 416 = 64.0
Total molar mass = 69.0 g + 31.0 g + 64.0 g = 164 g
Mass % Na = 69.0 g / 164 g 100 % = 42.1 %
Mass % P = 31.0 g / 164 g 100 % = 18.9 %
Mas % O = 64.0 g / 164 g 100 % = 39.0 %
Nina Fields
Embedded Software Engineer
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How do I calculate the mass of an object?
Method 1 of 2:Using a Triple-Beam Balance
1. Set up the balance.
Make sure the pan you will place your object onto is clean and dry.
Zero the balance.
Move all the weights to the zero position, then turn the knob on the far left beneath the scale pan. The scales should move around. Continue turning it in either direction until the white "pointer" line on the right of the beams lines up with the "0" marking on the right-hand side.
Place the object in the pan.
Be careful not to influence the weight of the object with your hand or other objects.
Move the weights.
Slide the weights left and righ
Method 1 of 2:Using a Triple-Beam Balance
1. Set up the balance.
Make sure the pan you will place your object onto is clean and dry.
Zero the balance.
Move all the weights to the zero position, then turn the knob on the far left beneath the scale pan. The scales should move around. Continue turning it in either direction until the white "pointer" line on the right of the beams lines up with the "0" marking on the right-hand side.
Place the object in the pan.
Be careful not to influence the weight of the object with your hand or other objects.
Move the weights.
Slide the weights left and right on the beams until the two white lines at the right line up again. The most efficient way to do this is to make a rough estimate as to what you think the mass will be, and then move the highest value weight you think will still be lower than the mass. Move this weight until the pointer is just under 0. Then move on progressively to the next smallest weights to get closer and closer to the actual mass.
Read the mass.
Add the measurement of each weight together. The total will be the mass of the object.
Method 2 of 2:Using the Density and Volume
1. Know the equation.
The set equation relating mass, density, and volume is or .
Plug your values into the equation.
If the density of your object is 500 kg/m (kilograms per cubic meter), then you put 500 in place of the for 500=m/v. If your volume is 10 m (cubic meters), put 10 in place of the for 500=m/10.
the variable.
Isolate As you are trying to calculate mass, the variable in this equation is . You want this value to end up alone on one side of the equal sign. In this equation, it is involved in division with another value. To isolate it, you must multiply of the equation by this value. This equation becomes (500)10=(m/10)10.
Isolating a variable is always done by performing the opposite mathematical function on both sides of an equation. If the variable is involved in addition, subtract the extra value from both sides, etc.
Simplify
On the left side of the equation, 500 times 10 simplifies to 5000. On the right side, the two 10's cancel out, leaving the alone. Thus the answer is 5000kg=m.
Don't forget your units. The cubic meters have canceled each other out to leave only kilograms.
Adrian Giles
Can pick a "bot" question from a mile away.
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Author has 3.5K answers and 807.2K answer views
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1y
Related
How does one calculate the molar mass of an unknown compound from its molecular formula?
Seems like a bot question not realising that by knowing the molecular formula the compound is then known unless empirical formula is being suggested.
In any case suppose we have a molecular formula CO₂. We then sum the atomic mass numbers for each element times their multiplicity. Carbon has atomic mass of 12 units while oxygen is 16 and there are two per molecule so
molar mass (CO₂)=12+2(16)=44g
In other words one mole, that is Avogadro’s number of molecules, of CO₂ has a mass of 44g. Note that molar masses are almost always not whole numbers. This is because isotopes of an element have differen
Seems like a bot question not realising that by knowing the molecular formula the compound is then known unless empirical formula is being suggested.
In any case suppose we have a molecular formula CO₂. We then sum the atomic mass numbers for each element times their multiplicity. Carbon has atomic mass of 12 units while oxygen is 16 and there are two per molecule so
molar mass (CO₂)=12+2(16)=44g
In other words one mole, that is Avogadro’s number of molecules, of CO₂ has a mass of 44g. Note that molar masses are almost always not whole numbers. This is because isotopes of an element have different masses so depending on their relative abundances the actual atomic mass of an element can’t be worked out from first principles but needs to be looked up in a suitable data book.
Note that two substances may have the same molar mass (rounded to the nearest whole number of grams): Nitrous oxide N₂O has molar mass=2(14)+16=44g.
Related questions
How do you calculate the mass of a compound?
How do I find the mass of an element in a compound?
How do I calculate the number of molecules of an element in a compound if I only know the total mass?
How do I calculate relative atomic mass?
How do you calculate the percent by mass of an element in a compound?
What happens to mass when a compound undergoes decomposition into its elements or other compounds with lower atomic/molecular masses (elemental formulae)?
How do you find the empirical formula of a compound if the total mass and mass of elements are given?
How do you find the mass of a percentage of a substance, given the mass of what the substance is in?
What is the difference between molecular mass and elemental mass? How do you calculate them for a given element or compound?
How do I calculate the number of ions of an element within a compound if I have the mass of the compound?
When calculating the relative formula mass of a compound, why do we often use the relative atomic mass of the elements, instead of the actual mass number of that element? Doesn't using relative atomic mass produce any inaccuracy in calculations?
Is it logical to calculate the equivalent mass of a single compound?
What is a percentage determined by the ratio of mass of each element to the total mass of the compound?
Given mass of Alka seltzer tablet, total mass before reaction, total mass after reaction & mass of an empty beaker. How do I calculate the mass of CO2 formed for Alka-seltzer lab? All of the units were given in grams.
How can I find the mass of a compound with a given mass value? What is the mass of 5 grams of H2O?
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11265 | https://mathematicalolympiads.files.wordpress.com/2012/08/75427434-problem-books-in-mathematics-problem-solving-strategies.pdf | Arthur Engel Problem-Solving Strategies With 223 Figures 1 3 A-PDF Merger DEMO : Purchase from www.A-PDF.com to remo Angel Engel Institut f¨ ur Didaktik der Mathematik Johann Wolfgang Goethe–Universit¨ at Frankfurt am Main Senckenberganlage 9–11 60054 Frankfurt am Main 11 Germany Series Editor: Paul R. Halmos Department of Mathematics Santa Clara University Santa Clara, CA 95053 USA Mathematics Subject Classification (1991): 00A07 Library of Congress Cataloging-in-Publication Data Engel, Arthur.
Problem-solving strategies/Arthur Engel.
p.
cm. — (Problem books in mathematics) Includes index.
ISBN 0-387-98219-1 (softcover: alk. paper) 1. Problem solving. I. Title. II. Series.
QA63.E54 1997 510′.76—dc21 97-10090 © 1998 Springer-Verlag New York, Inc.
All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Springer-Verlag New York, Inc., 175 Fifth Avenue, New York, NY 10010, USA), except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden.
The use of general descriptive names, trade names, trademarks, etc., in this publication, even if the former are not especially identified, is not to be taken as a sign that such names, as understood by the Trade Marks and Merchandise Marks Act, may accordinly be used freely by anyone.
ISBN 0–387–98219–1 Springer-Verlag New York Berlin Heidelburg SPIN 10557554 Preface This book is an outgrowth of the training of the German IMO team from a time when we had only a short training time of 14 days, including 6 half-day tests. This has forced upon us a training of enormous compactness. “Great Ideas” were the leading principles. A huge number of problems were selected to illustrate these principles. Not only topics but also ideas were efficient means of classification.
For whom is this book written?
• For trainers and participants of contests of all kinds up to the highest level of international competitions, including the IMO and the Putnam Competition.
• For the regular high school teacher, who is conducting a mathematics club and is looking for ideas and problems for his/her club. Here, he/she will find problems of any level from very simple ones to the most difficult problems ever proposed at any competition.
• For high school teachers who want to pose the problem of the week, problem ofthemonth,andresearchproblemsoftheyear. Thisisnotsoeasy.Manyfail, but some persevere, and after a while they succeed and generate a creative atmosphere with continuous discussions of mathematical problems.
• For the regular high school teacher, who is just looking for ideas to enrich his/her teaching by some interesting nonroutine problems.
• For all those who are interested in solving tough and interesting problems.
The book is organized into chapters. Each chapter starts with typical examples illustrating the main ideas followed by many problems and their solutions. The vi Preface solutions are sometimes just hints, giving away the main idea leading to the solu-tion. In this way, it was possible to increase the number of examples and problems to over 1300. The reader can increase the effectiveness of the book even more by trying to solve the examples.
The problems are almost exclusively competition problems from all over the world. Most of them are from the former USSR, some from Hungary, and some from Western countries, especially from the German National Competition. The competition problems are usually variations of problems from journals with prob-lem sections. So it is not always easy to give credit to the originators of the problem.
If you see a beautiful problem, you first wonder at the creativity of the problem proposer. Later you discover the result in an earlier source. For this reason, the references to competitions are somewhat sporadic. Usually no source is given if I have known the problem for more than 25 years. Anyway, most of the problems are results that are known to experts in the respective fields.
There is a huge literature of mathematical problems. But, as a trainer, I know that there can never be enough problems. You are always in desperate need of new problems or old problems with new solutions. Any new problem book has some new problems, and a big book, as this one, usually has quite a few problems that are new to the reader.
The problems are arranged in no particular order, and especially not in increasing order of difficulty. We do not know how to rate a problem’s difficulty. Even the IMO jury, now consisting of 75 highly skilled problem solvers, commits grave errors in rating the difficulty of the problems it selects. The over 400 IMO contestants are also an unreliable guide. Too much depends on the previous training by an ever-changing set of hundreds of trainers. A problem changes from impossible to trivial if a related problem was solved in training.
I would like to thank Dr. Manfred Grathwohl for his help in implementing various L aT EX versions on the workstation at the institute and on my PC at home.
When difficulties arose, he was a competent and friendly advisor.
There will be some errors in the proofs, for which I take full responsibility, since none of my colleagues has read the manuscript before. Readers will miss important strategies. So do I, but I have set myself a limit to the size of the book.
Especially, advanced methods are missing. Still, it is probably the most complete training book on the market. The gravest gap is the absence of new topics like probability and algorithmics to counter the conservative mood of the IMO jury.
One exception is Chapter 13 on games, a topic almost nonexistent in the IMO, but very popular in Russia.
Frankfurt am Main, Germany Arthur Engel Contents Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
v Abbreviations and Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
ix 1 The Invariance Principle. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1 2 Coloring Proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
25 3 The Extremal Principle. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
39 4 The Box Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
59 5 Enumerative Combinatorics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
85 6 Number Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
117 7 Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
161 8 The Induction Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
205 9 Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
221 10 Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
245 11 Functional Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
271 viii Contents 12 Geometry. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
289 13 Games. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
361 14 Further Strategies. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
373 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
397 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
401 Abbreviations and Notations Abbreviations ARO Allrussian Mathematical Olympiad ATMO Austrian Mathematical Olympiad AuMO Australian Mathematical Olympiad AUO Allunion Mathematical Olympiad BrMO British Mathematical Olympiad BWM German National Olympiad BMO Balkan Mathematical Olympiad ChNO Chinese National Olympiad HMO Hungarian Mathematical Olympiad (K˝ urschak Competition) IIM International Intellectual Marathon (Mathematics/Physics Competition) IMO International Mathematical Olympiad LMO Leningrad Mathematical Olympiad MMO Moskov Mathematical Olympiad PAMO Polish-Austrian Mathematical Olympiad x Abbreviations and Notations PMO Polish Mathematical Olympiad RO Russian Olympiad (ARO from 1994 on) SPMO St. Petersburg Mathematical Olympiad TT Tournament of the Towns USO US Olympiad Notations for Numerical Sets N or Z+ the positive integers (natural numbers), i.e., {1,2,3, . . . } N0 the nonnegative integers, {0,1,2, . . . } Z the integers Q the rational numbers Q+ the positive rational numbers Q+ 0 the nonnegative rational numbers R the real numbers R+ the positive real numbers C the complex numbers Zn the integers modulo n 1 . . n the integers 1, 2, . . . , n Notations from Sets, Logic, and Geometry ⇐ ⇒iff, if and only if ⇒implies A ⊂B A is a subset of B A \ B A without B A ∩B the intersection of A and B A ∪B the union of A and B a ∈A the element a belongs to the set A |AB| also AB, the distance between the points A and B box parallelepiped, solid bounded by three pairs of parallel planes 1 The Invariance Principle We present our first Higher Problem-Solving Strategy. It is extremely useful in solving certain types of difficult problems, which are easily recognizable. We will teach it by solving problems which use this strategy. In fact, problem solving can be learned only by solving problems. But it must be supported by strategies provided by the trainer.
Our first strategy is the search for invariants, and it is called the Invariance Prin-ciple. The principle is applicable to algorithms (games, transformations). Some task is repeatedly performed. What stays the same? What remains invariant?
Here is a saying easy to remember: If there is repetition, look for what does not change!
In algorithms there is a starting state S and a sequence of legal steps (moves, transformations). One looks for answers to the following questions: 1. Can a given end state be reached?
2. Find all reachable end states.
3. Is there convergence to an end state?
4. Find all periods with or without tails, if any.
Since the Invariance Principle is a heuristic principle, it is best learned by ex-perience, which we will gain by solving the key examples E1 to E10.
2 1. The Invariance Principle E1. Starting with a point S (a, b) of the plane with 0 < b < a, we generate a sequence of points (xn, yn) according to the rule x0 a, y0 b, xn+1 xn + yn 2 , yn+1 2xnyn xn + yn .
Here it is easy to find an invariant. From xn+1yn+1 xnyn, for all n we deduce xnyn ab for all n. This is the invariant we are looking for. Initially, we have y0 < x0. This relation also remains invariant. Indeed, suppose yn < xn for some n. Then xn+1 is the midpoint of the segment with endpoints yn, xn. Moreover, yn+1 < xn+1 since the harmonic mean is strictly less than the arithmetic mean.
Thus, 0 < xn+1 −yn+1 xn −yn xn + yn · xn −yn 2 < xn −yn 2 for all n. So we have lim xn lim yn x with x2 ab or x √ ab.
Here the invariant helped us very much, but its recognition was not yet the solution, although the completion of the solution was trivial.
E2. Suppose the positive integer n is odd. First Al writes the numbers 1, 2, . . . , 2n on the blackboard. Then he picks any two numbers a, b, erases them, and writes, instead, |a −b|. Prove that an odd number will remain at the end.
Solution. Suppose S is the sum of all the numbers still on the blackboard. Initially this sum is S 1+2+· · ·+2n n(2n+1), an odd number. Each step reduces S by 2 min(a, b), which is an even number. So the parity of S is an invariant. During the whole reduction process we have S ≡1 mod 2. Initially the parity is odd. So, it will also be odd at the end.
E3. A circle is divided into six sectors. Then the numbers 1, 0, 1, 0, 0, 0 are writ-ten into the sectors (counterclockwise, say). You may increase two neighboring numbers by 1. Is it possible to equalize all numbers by a sequence of such steps?
Solution. Suppose a1, . . . , a6 are the numbers currently on the sectors. Then I a1 −a2 + a3 −a4 + a5 −a6 is an invariant. Initially I 2. The goal I 0 cannot be reached.
E4. In the Parliament of Sikinia, each member has at most three enemies. Prove that the house can be separated into two houses, so that each member has at most one enemy in his own house.
Solution. Initially, we separate the members in any way into the two houses. Let H be the total sum of all the enemies each member has in his own house. Now suppose A has at least two enemies in his own house. Then he has at most one enemy in the other house. If A switches houses, the number H will decrease. This decrease cannot go on forever. At some time, H reaches its absolute minimum.
Then we have reached the required distribution.
1. The Invariance Principle 3 Here we have a new idea. We construct a positive integral function which de-creases at each step of the algorithm. So we know that our algorithm will termi-nate. There is no strictly decreasing infinite sequence of positive integers. H is not strictly an invariant, but decreases monotonically until it becomes constant. Here, the monotonicity relation is the invariant.
E5. Suppose not all four integers a, b, c, d are equal. Start with (a, b, c, d) and repeatedly replace (a, b, c, d) by (a −b, b −c, c −d, d −a). Then at least one number of the quadruple will eventually become arbitrarily large.
Solution. Let Pn (an, bn, cn, dn) be the quadruple after n iterations. Then we have an + bn + cn + dn 0 for n ≥1. We do not see yet how to use this invariant.
But geometric interpretation is mostly helpful. A very important function for the point Pn in 4-space is the square of its distance from the origin (0, 0, 0, 0), which is a2 n + b2 n + c2 n + d2 n. If we could prove that it has no upper bound, we would be finished.
We try to find a relation between Pn+1 and Pn: a2 n+1 + b2 n+1 + c2 n+1 + d2 n+1 (an −bn)2 + (bn −cn)2 + (cn −dn)2 + (dn −an)2 2(a2 n + b2 n + c2 n + d2 n) −2anbn −2bncn −2cndn −2dnan.
Now we can use an + bn + cn + dn 0 or rather its square: 0 (an+bn+cn+dn)2 (an+cn)2+(bn+dn)2+2anbn+2andn+2bncn+2cndn.
(1) Adding (1) and (2), for a2 n+1 + b2 n+1 + c2 n+1 + d2 n+1, we get 2(a2 n + b2 n + c2 n + d2 n) + (an + cn)2 + (bn + dn)2 ≥2(a2 n + b2 n + c2 n + d2 n).
From this invariant inequality relationship we conclude that, for n ≥2, a2 n + b2 n + c2 n + d2 n ≥2n−1(a2 1 + b2 1 + c2 1 + d2 1).
(2) The distance of the points Pn from the origin increases without bound, which means that at least one component must become arbitrarily large. Can you always have equality in (2)?
Here we learned that the distance from the origin is a very important func-tion. Each time you have a sequence of points you should consider it.
E6. An algorithm is defined as follows: Start: (x0, y0) with 0 < x0 < y0.
Step: xn+1 xn + yn 2 , yn+1 √xn+1yn.
4 1. The Invariance Principle Figure 1.1 and the arithmetic mean-geometric mean inequality show that xn < yn ⇒xn+1 < yn+1, yn+1 −xn+1 < yn −xn 4 for all n. Find the common limit lim xn lim yn x y.
Here, invariants can help. But there are no systematic methods to find invariants, just heuristics. These are methods which often work, but not always. Two of these heuristics tell us to look for the change in xn/yn or yn −xn when going from n to n + 1.
(a) xn+1 yn+1 xn+1 √xn+1yn xn+1 yn 1 + xn/yn 2 .
(1) This reminds us of the half-angle relation cos α 2 1 + cos α 2 .
Since we always have 0 < xn/yn < 1, we may set xn/yn cos αn. Then (1) becomes cos αn+1 cos αn 2 ⇒αn α0 2n ⇒2nαn α0, which is equivalent to 2n arccos xn yn arccos x0 y0 .
(2) This is an invariant!
(b) To avoid square roots, we consider y2 n −x2 n instead of yn −xn and get y2 n+1 −x2 n+1 y2 n −x2 n 4 ⇒2 y2 n+1 −x2 n+1 y2 n −x2 n or 2n y2 n −x2 n y2 0 −x2 0, (3) which is a second invariant.
q q q q xn xn+1 yn+1 yn Fig. 1.1 s t Fig. 1.2. arccos t arcsin s, s √ 1 −t2.
1. The Invariance Principle 5 From Fig. 1.2 and (2), (3), we get arccos x0 y0 2n arccos xn yn 2n arcsin y2 n −x2 n yn 2n arcsin y2 0 −x2 0 2nyn .
The right-hand side converges to y2 0 −x2 0/y for n →∞. Finally, we get x y y2 0 −x2 0 arccos(x0/y0).
(4) It would be pretty hopeless to solve this problem without invariants. By the way, this is a hard problem by any competition standard.
E7. Each of the numbers a1, . . . , an is 1 or −1, and we have S a1a2a3a4 + a2a3a4a5 + · · · + ana1a2a3 0.
Prove that 4 | n.
Solution. This is a number theoretic problem, but it can also be solved by in-variance. If we replace any ai by −ai, then S does not change mod 4 since four cyclically adjacent terms change their sign. Indeed, if two of these terms are pos-itive and two negative, nothing changes. If one or three have the same sign, S changes by ±4. Finally, if all four are of the same sign, then S changes by ±8.
Initially, we have S 0 which implies S ≡0 mod 4. Now, step-by-step, we change each negative sign into a positive sign. This does not change S mod 4. At the end, we still have S ≡0 mod 4, but also S n, i.e, 4|n.
E8. 2n ambassadors are invited to a banquet. Every ambassador has at most n−1 enemies. Prove that the ambassadors can be seated around a round table, so that nobody sits next to an enemy.
Solution. First, we seat the ambassadors in any way. Let H be the number of neighboring hostile couples. We must find an algorithm which reduces this number whenever H > 0. Let (A, B) be a hostile couple with B sitting to the right of A (Fig. 1.3). We must separate them so as to cause as little disturbance as possible.
This will be achieved if we reverse some arc BA′ getting Fig. 1.4. H will be reduced if (A, A′) and (B, B′) in Fig. 1.4 are friendly couples. It remains to be shown that such a couple always exists with B′ sitting to the right of A′. We start in A and go around the table counterclockwise. We will encounter at least n friends of A. To their right, there are at least n seats. They cannot all be occupied by enemies of B since B has at most n −1 enemies. Thus, there is a friend A′ of A with right neighbor B′, a friend of B.
6 1. The Invariance Principle Fig. 1.3. Invert arc A′B.
s s s s B′ A′ A B Fig. 1.4 s s s s B′ B A A′ Remark. This problem is similar to E4, but considerably harder. It is the following theorem in graph theory: Let G be a linear graph with n vertices. Then G has a Hamiltonian path if the sum of the degrees of any two vertices is equal to or larger than n −1. In our special case, we have proved that there is even a Hamiltonian circuit.
E9. To each vertex of a pentagon, we assign an integer xi with sum s xi > 0.
If x, y, z are the numbers assigned to three successive vertices and if y < 0, then we replace (x, y, z) by (x + y, −y, y + z). This step is repeated as long as there is a y < 0. Decide if the algorithm always stops. (Most difficult problem of IMO 1986.) Solution. The algorithm always stops. The key to the proof is (as in Examples 4 and 8) to find an integer-valued, nonnegative function f (x1, . . . , x5) of the vertex labels whose value decreases when the given operation is performed. All but one of the eleven students who solved the problem found the same function f (x1, x2, x3, x4, x5) 5 i1 (xi −xi+2)2, x6 x1, x7 x2.
Suppose y x4 < 0. Then fnew −fold 2sx4 < 0, since s > 0. If the algorithm does not stop, we can find an infinite decreasing sequence f0 > f1 > f2 > · · · of nonnegative integers. Such a sequence does not exist.
Bernard Chazelle (Princeton) asked: How many steps are needed until stop? He considered the infinite multiset S of all sums defined by s(i, j) xi + · · · + xj−1 with 1 ≤i ≤5 and j > i. A multiset is a set which can have equal elements. In this set, all elements but one either remain invariant or are switched with others. Only s(4, 5) x4 changes to −x4. Thus, exactly one negative element of S changes to positive at each step. There are only finitely many negative elements in S, since s > 0. The number of steps until stop is equal to the number of negative elements of S. We see that the xi need not be integers.
Remark. It is interesting to find a formula with the computer, which, for input a, b, c, d, e, gives the number of steps until stop. This can be done without much effort if s 1. For instance, the input (n, n, 1 −4n, n, n) gives the step number f (n) 20n −10.
1. The Invariance Principle 7 E10. Shrinking squares. An empirical exploration. Start with a sequence S (a, b, c, d) of positive integers and find the derived sequence S1 T (S) (|a − b|, |b −c|, |c −d|, |d −a|). Does the sequence S, S1, S2 T (S1), S3 T (S2), . . .
always end up with (0, 0, 0, 0)?
Let us collect material for solution hints: (0, 3, 10, 13) →(3, 7, 3, 13) →(4, 4, 10, 10) → (0, 6, 0, 6) →(6, 6, 6, 6) →(0, 0, 0, 0), (8, 17, 3, 107) →(9, 14, 104, 99) →(5, 90, 5, 90) → (85, 85, 85, 85) →(0, 0, 0, 0), (91, 108, 95, 294) →(17, 13, 99, 203) →(4, 86, 104, 186) → (82, 18, 82, 182) →(64, 64, 100, 100) →(0, 36, 0, 36) → (36, 36, 36, 36) →(0, 0, 0, 0).
Observations: 1. Let max S be the maximal element of S. Then max Si+1 ≤max Si, and max Si+4 < max Si as long as max Si > 0. Verify these observations. This gives a proof of our conjecture.
2. S and tS have the same life expectancy.
3. After four steps at most, all four terms of the sequence become even. Indeed, it is sufficient to calculate modulo 2. Because of cyclic symmetry, we need to test just six sequences 0001 →0011 →0101 →1111 →0000 and 1110 →0011. Thus, we have proved our conjecture. After four steps at most, each term is divisible by 2, after 8 steps at most, by 22, . . . , after 4k steps at most, by 2k. As soon as max S < 2k, all terms must be 0.
In observation 1, we used another strategy, the Extremal Principle: Pick the maximal element! Chapter 3 is devoted to this principle.
In observation 3, we used symmetry. You should always think of this strategy, although we did not devote a chapter to this idea.
Generalizations: (a) Start with four real numbers, e.g., √ 2 π √ 3 e π − √ 2 π − √ 3 e − √ 3 e − √ 2 √ 3 − √ 2 π −e √ 3 − √ 2 π −e π −e − √ 3 + √ 2 π −e − √ 3 + √ 2 π −e − √ 3 + √ 2 π −e − √ 3 + √ 2 0 0 0 0.
8 1. The Invariance Principle Some more trials suggest that, even for all nonnegative real quadruples, we always end up with (0, 0, 0, 0). But with t > 1 and S (1, t, t2, t3) we have T (S) [t −1, (t −1)t, (t −1)t2, (t −1)(t2 + t + 1)].
If t3 t2 +t +1, i.e., t 1.8392867552 . . ., then the process never stops because of the second observation. This t is unique up to a transformation f (t) at + b.
(b) Start with S (a0, a1, . . . , an−1), ai nonnegative integers. For n 2, we reach (0, 0) after 2 steps at most. For n 3, we get, for 011, a pure cycle of length 3: 011 →101 →110 →011. For n 5 we get 00011 →00101 →01111 → 10001 →10010 →10111 →11000 →01001 →11011 →01100 → 10100 →11101 →00110 →01010 →11110 →00011, which has a pure cycle of length 15.
1. Find the periods for n 6 (n 7) starting with 000011 (0000011).
2. Prove that, for n 8, the algorithm stops starting with 00000011.
3. Prove that, for n 2r, we always reach (0, 0, . . . , 0), and, for n ̸ 2r, we get (up to some exceptions) a cycle containing just two numbers: 0 and evenly often some number a > 0. Because of observation 2, we may assume that a 1. Then | a −b | a + b mod 2, and we do our calculations in GF(2), i.e., the finite field with two elements 0 and 1.
4. Let n ̸ 2r and c(n) be the cycle length. Prove that c(2n) 2c(n) (up to some exceptions).
5. Prove that, for odd n, S (0, 0, . . . , 1, 1) always lies on a cycle.
6. Algebraization. To the sequence (a0, . . . , an−1), we assign the polynomial p(x) an−1 + · · · + a0xn−1 with coefficients from GF(2), and xn 1. The polynomial (1+x)p(x) belongs to T (S). Use this algebraization if you can.
7. The following table was generated by means of a computer. Guess as many properties of c(n) as you can, and prove those you can.
n 3 5 7 9 11 13 15 17 19 21 23 25 c(n) 3 15 7 63 341 819 15 255 9709 63 2047 25575 n 27 29 31 33 35 37 39 41 43 c(n) 13797 47507 31 1023 4095 3233097 4095 41943 5461 Problems 1. Start with the positive integers 1, . . . , 4n −1. In one move you may replace any two integers by their difference. Prove that an even integer will be left after 4n −2 steps.
1. The Invariance Principle 9 2. Start with the set {3, 4, 12}. In each step you may choose two of the numbers a, b and replace them by 0.6a −0.8b and 0.8a + 0.6b. Can you reach the goal (a) or (b) in finitely many steps: (a) {4, 6, 12}, (b) {x, y, z} with |x −4|, |y −6|, |z −12| each less than 1/ √ 3?
3. Assume an 8×8 chessboard with the usual coloring. You may repaint all squares (a) of a row or column (b) of a 2 × 2 square. The goal is to attain just one black square.
Can you reach the goal?
4. We start with the state (a, b) where a, b are positive integers. To this initial state we apply the following algorithm: while a > 0, do if a < b then (a, b) ←(2a, b −a) else (a, b) ←(a −b, 2b).
For which starting positions does the algorithm stop? In how many steps does it stop, if it stops? What can you tell about periods and tails?
The same questions, when a, b are positive reals.
5. Around a circle, 5 ones and 4 zeros are arranged in any order. Then between any two equal digits, you write 0 and between different digits 1. Finally, the original digits are wiped out. If this process is repeated indefinitely, you can never get 9 zeros.
Generalize!
6. There are a white, b black, and c red chips on a table. In one step, you may choose two chips of different colors and replace them by a chip of the third color. If just one chip will remain at the end, its color will not depend on the evolution of the game.
When can this final state be reached?
7. There are a white, b black, and c red chips on a table. In one step, you may choose two chips of different colors and replace each one by a chip of the third color. Find conditions for all chips to become of the same color. Suppose you have initially 13 white 15 black and 17 red chips. Can all chips become of the same color? What states can be reached from these numbers?
8. There is a positive integer in each square of a rectangular table. In each move, you may double each number in a row or subtract 1 from each number of a column. Prove that you can reach a table of zeros by a sequence of these permitted moves.
9. Each of the numbers 1 to 106 is repeatedly replaced by its digital sum until we reach 106 one-digit numbers. Will these have more 1’s or 2’s?
10. The vertices of an n-gon are labeled by real numbers x1, . . . , xn. Let a, b, c, d be four successive labels. If (a −d)(b −c) < 0, then we may switch b with c. Decide if this switching operation can be performed infinitely often.
11. In Fig. 1.5, you may switch the signs of all numbers of a row, column, or a parallel to one of the diagonals. In particular, you may switch the sign of each corner square.
Prove that at least one −1 will remain in the table.
-1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Fig. 1.5 10 1. The Invariance Principle 12. There is a row of 1000 integers. There is a second row below, which is constructed as follows. Under each number a of the first row, there is a positive integer f (a) such that f (a) equals the number of occurrences of a in the first row. In the same way, we get the 3rd row from the 2nd row, and so on. Prove that, finally, one of the rows is identical to the next row.
13. There is an integer in each square of an 8 × 8 chessboard. In one move, you may choose any 4 × 4 or 3 × 3 square and add 1 to each integer of the chosen square.
Can you always get a table with each entry divisible by (a) 2, (b) 3?
14. We strike the first digit of the number 71996, and then add it to the remaining number.
This is repeated until a number with 10 digits remains. Prove that this number has two equal digits.
15. There is a checker at point (1, 1) of the lattice (x, y) with x, y positive integers. It moves as follows. At any move it may double one coordinate, or it may subtract the smaller coordinate from the larger . Which points of the lattice can the checker reach?
16. Each term in a sequence 1, 0, 1, 0, 1, 0, . . . starting with the seventh is the sum of the last 6 terms mod 10. Prove that the sequence . . . , 0, 1, 0, 1, 0, 1, . . . never occurs.
17. Starting with any 35 integers, you may select 23 of them and add 1 to each. By repeating this step, one can make all 35 integers equal. Prove this. Now replace 35 and 23 by m and n, respectively. What condition must m and n satisfy to make the equalization still possible?
18. The integers 1, . . . , 2n are arranged in any order on 2n places numbered 1, . . . , 2n.
Now we add its place number to each integer. Prove that there are two among the sums which have the same remainder mod 2n.
19. The n holes of a socket are arranged along a circle at equal (unit) distances and numbered 1, . . . , n. For what n cantheprongsofaplugfitting the socket be numbered such that at least one prong in each plug-in goes into a hole of the same number (good numbering)?
20. A game for computing gcd(a, b) and lcm(a, b).
We start with x a, y b, u a, v b and move as follows: if x < y then, set y ←y −x and v ←v + u if x > y, then set x ←x −y and u ←u + v The game ends with x y gcd(a, b) and (u + v)/2 lcm(a, b). Show this.
21. Three integers a, b, c are written on a blackboard. Then one of the integers is erased and replaced by the sum of the other two diminished by 1. This operation is repeated many times with the final result 17, 1967, 1983. Could the initial numbers be (a) 2, 2, 2 (b) 3, 3, 3?
22. There is a chip on each dot in Fig. 1.6. In one move, you may simultaneously move any two chips by one place in opposite directions. The goal is to get all chips into one dot. When can this goal be reached?
r r r r r r r r 1 2 n 3 Fig. 1.6 1. The Invariance Principle 11 23. Start with n pairwise different integers x1, x2, . . . , xn, (n > 2) and repeat the fol-lowing step: T : (x1, . . . , xn) → x1 + x2 2 , x2 + x3 2 , . . . , xn + x1 2 .
Show that T, T 2, . . . finally leads to nonintegral components.
24. Start with an m × n table of integers. In one step, you may change the sign of all numbers in any row or column. Show that you can achieve a nonnegative sum of any row or column. (Construct an integral function which increases at each step, but is bounded above. Then it must become constant at some step, reaching its maximum.) 25. Assume a convex 2m-gon A1, . . . , A2m. In its interior we choose a point P, which does not lie on any diagonal. Show that P lies inside an even number of triangles with vertices among A1, . . . , A2m.
26. Three automata I, H, T print pairs of positive integers on tickets. For input (a, b), I and H give (a + 1, b + 1) and (a/2, b/2), respectively. H accepts only even a, b. T needs two pairs (a, b) and (b, c) as input and yields output (a, c). Starting with (5, 19) can you reach the ticket (a) (1, 50) (b) (1, 100)? Initially, we have (a, b), a < b. For what n is (1, n) reachable?
27. Three automata I, R, S print pairs of positive integers on tickets. For entry (x, y), the automata I, R, S give tickets (x −y, y), (x + y, y), (y, x), respectively, as outputs.
Initially, we have the ticket (1, 2). With these automata, can I get the tickets (a) (19, 79) (b) (819, 357)? Find an invariant. What pairs (p, q) can I get starting with (a, b)? Via which pair should I best go?
28. n numbers are written on a blackboard. In one step you may erase any two of the numbers, say a and b, and write, instead (a + b)/4. Repeating this step n −1 times, there is one number left. Prove that, initially, if there were n ones on the board, at the end, a number, which is not less than 1/n will remain.
29. The following operation is performed with a nonconvex non-self-intersecting poly-gon P . Let A, B be two nonneighboring vertices. Suppose P lies on the same side of AB. Reflect one part of the polygon connecting A with B at the midpoint O of AB. Prove that the polygon becomes convex after finitely many such reflections.
30. Solve the equation (x2 −3x + 3)2 −3(x2 −3x + 3) + 3 x.
31. Let a1, a2, . . . , an be a permutation of 1, 2, . . . , n. If n is odd, then the product P (a1 −1)(a2 −2) . . . (an −n) is even. Prove this.
32. Many handshakes are exchanged at a big international congress. We call a person an odd person if he has exchanged an odd number of handshakes. Otherwise he will be called an even person. Show that, at any moment, there is an even number of odd persons.
33. Start with two points on a line labeled 0, 1 in that order. In one move you may add or delete two neighboring points (0, 0) or (1, 1). Your goal is to reach a single pair of points labeled (1, 0) in that order. Can you reach this goal?
34. Is it possible to transform f (x) x2 + 4x + 3 into g(x) x2 + 10x + 9 by a sequence of transformations of the form f (x) →x2f (1/x + 1) or f (x) →(x −1)2f [1/(x −1)]?
12 1. The Invariance Principle 35. Does the sequence of squares contain an infinite arithmetic subsequence?
36. The integers 1, . . . , n are arranged in any order. In one step you may switch any two neighboring integers. Prove that you can never reach the initial order after an odd number of steps.
37. One step in the preceding problem consists of an interchange of any two integers.
Prove that the assertion is still true.
38. The integers 1, . . . , n are arranged in order. In one step you may take any four integers and interchange the first with the fourth and the second with the third. Prove that, if n(n −1)/2 is even, then by means of such steps you may reach the arrangement n, n −1, . . . , 1. But if n(n −1)/2 is odd, you cannot reach this arrangement.
39. Consider all lattice squares (x, y) with x, y nonnegative integers. Assign to each its lower left corner as a label. We shade the squares (0, 0), (1, 0), (0, 1), (2, 0), (1, 1), (0, 2). (a) There is a chip on each of the six squares (b) There is only one chip on (0, 0).
Step: If (x, y) is occupied, but (x + 1, y) and (x, y + 1) are free, you may remove the chip from (x, y) and place a chip on each of (x + 1, y) and (x, y + 1). The goal is to remove the chips from the shaded squares. Is this possible in the cases (a) or (b)? (Kontsevich, TT 1981.) 40. In any way you please, fill up the lattice points below or on the x-axis by chips. By solitaire jumps try to get one chip to (0, 5) with all other chips cleared off. (J. H.
Conway.) The preceding problem of Kontsevich might have been suggested by this problem.
A solitaire jump is a horizontal or vertical jump of any chip over its neighbor to a free point with the chip jumped over removed. For instance, with (x, y) and (x, y + 1) occupied and (x, y + 2) free, a jump consists in removing the two chips on (x, y) and (x, y + 1) and placing a chip onto (x, y + 2).
41. We may extend a set S of space points by reflecting any point X of S at any space point A, A ̸ X. Initially, S consists of the 7 vertices of a cube. Can you ever get the eight vertex of the cube into S?
42. The following game is played on an infinite chessboard. Initially, each cell of an n×n square is occupied by a chip. A move consists in a jump of a chip over a chip in a horizontal or vertical direction onto a free cell directly behind it. The chip jumped over is removed. Find all values of n, for which the game ends with one chip left over (IMO 1993 and AUO 1992!).
43. Nine 1 × 1 cells of a 10 × 10 square are infected. In one time unit, the cells with at least two infected neighbors (having a common side) become infected. Can the infection spread to the whole square?
44. Can you get the polynomial h(x) x from the polynomials f (x) and g(x) by the operations addition, subtraction, multiplication if (a) f (x) x2 + x, g(x) x2 + 2; (b) f (x) 2x2 + x, g(x) 2x; (c) f (x) x2 + x, g(x) x2 −2?
45. Accumulation of your computer rounding errors. Start with x0 1, y0 0, and, with your computer, generate the sequences xn+1 5xn −12yn 13 , yn+1 12xn + 5yn 13 .
1. The Invariance Principle 13 Find x2 n + y2 n for n 102, 103, 104, 105, 106, and 107.
46. Start with two numbers 18 and 19 on the blackboard. In one step you may add another number equal to the sum of two preceding numbers. Can you reach the number 1994 (IIM)?
47. In a regular (a) pentagon (b) hexagon all diagonals are drawn. Initially each vertex and each point of intersection of the diagonals is labeled by the number 1. In one step it is permitted to change the signs of all numbers of a side or diagonal. Is it possible to change the signs of all labels to −1 by a sequence of steps (IIM)?
48. In Fig. 1.7, two squares are neighbors if they have a common boundary. Consider the following operation T : Choose any two neighboring numbers and add the same integer to them. Can you transform Fig. 1.7 into Fig. 1.8 by iteration of T ?
1 2 3 4 5 6 7 8 9 Fig. 1.7 7 8 9 6 2 4 3 5 1 Fig. 1.8 49. There are several signs + and −on a blackboard. You may erase two signs and write, instead, + if they are equal and −if they are unequal. Then, the last sign on the board does not depend on the order of erasure.
50. There are several letters e, a and b on a blackboard. We may replace two e′s by one e, two a′s by one b, two b′s by one a, an a and a b by one e, an a and an e by one a, a b, and an e by one b. Prove that the last letter does not depend on the order of erasure.
51. A dragon has 100 heads. A knight can cut off 15, 17, 20, or 5 heads, respectively, with one blow of his sword. In each of these cases, 24, 2, 14, or 17 new heads grow on its shoulders. If all heads are blown off, the dragon dies. Can the dragon ever die?
52. Is it possible to arrange the integers 1, 1, 2, 2, . . . , 1998, 1998 such that there are exactly i −1 other numbers between any two i′s?
53. The following operations are permitted with the quadratic polynomial ax2 +bx +c: (a) switch a and c, (b) replace x by x + t where t is any real. By repeating these operations, can you transform x2 −x −2 into x2 −x −1?
54. Initially, we have three piles with a, b, and c chips, respectively. In one step, you may transfer one chip from any pile with x chips onto any other pile with y chips. Let d y −x + 1. If d > 0, the bank pays you d dollars. If d < 0, you pay the bank |d| dollars. Repeating this step several times you observe that the original distribution of chips has been restored. What maximum amount can you have gained at this stage?
55. Let d(n) be the digital sum of n ∈N. Solve n + d(n) + d(d(n)) 1997.
56. Start with four congruent right triangles. In one step you may take any triangle and cut it in two with the altitude from the right angle. Prove that you can never get rid of congruent triangles (MMO 1995).
57. Starting with a point S(a, b) of the plane with 0 < a < b, we generate a sequence (xn, yn) of points according to the rule x0 a, y0 b, xn+1 √xnyn+1, yn+1 √xnyn.
14 1. The Invariance Principle Prove that there is a limiting point with x y. Find this limit.
58. Consider any binary word W a1a2 · · · an. It can be transformed by inserting, deleting or appending any word XXX, X being any binary word. Our goal is to transform W from 01 to 10 by a sequence of such transformations. Can the goal be attained (LMO 1988, oral round)?
59. Seven vertices of a cube are marked by 0 and one by 1. You may repeatedly select an edge and increase by 1 the numbers at the ends of that edge. Your goal is to reach (a) 8 equal numbers, (b) 8 numbers divisible by 3.
60. Start with a point S(a, b) of the plane with 0 < b < a, and generate a sequence of points Sn(xn, yn) according to the rule x0 a, y0 b, xn+1 2xnyn xn + yn , yn+1 2xn+1yn xn+1 + yn .
Prove that there is a limiting point with x y. Find this limit.
Solutions 1. In one move the number of integers always decreases by one. After (4n −2) steps, just one integer will be left. Initially, there are 2n even integers, which is an even number. If two odd integers are replaced, the number of odd integers decreases by 2. If one of them is odd or both are even, then the number of odd numbers remains the same. Thus, the number of odd integers remains even after each move. Since it is initially even, it will remain even to the end. Hence, one even number will remain.
2. (a)(0.6a−0.8b)2+(0.8a+0.6b)2 a2+b2.Sincea2+b2+c2 32+42+122 132, the point (a, b, c) lies on the sphere around O with radius 13. Because 42+62+122 142, the goal lies on the sphere around O with radius 14. The goal cannot be reached.
(b) (x −4)2 + (y −6)2 + (z −12)2 < 1. The goal cannot be reached.
The important invariant, here, is the distance of the point (a, b, c) from O.
3. (a) Repainting a row or column with b black and 8−b white squares, you get (8−b) black and b white squares. The number of black squares changes by |(8 −b) −b| |8 −2b|, that is an even number. The parity of the number of black squares does not change. Initially, it was even. So, it always remains even. One black square is unattainable. The reasoning for (b) is similar.
4. Here is a solution valid for natural, rational and irrational numbers. With the invariant a + b n the algorithm can be reformulated as follows: If a < n/2, replace a by 2a.
If a ≥n/2, replace a by a −b a −(n −a) 2a −n ≡2a (mod n).
Thus, we double a repeatedly modulo n and get the sequence a, 2a, 22a, 23a, . . .
(mod n).
(1) Divide a by n in base 2. There are three cases.
(a) The result is terminating: a/n 0.d1d2d3 . . . dk, di ∈{0, 1}. Then 2k ≡0 1. The Invariance Principle 15 (mod n), but 2i ̸≡0 (mod n) for i < k. Thus, the algorithm stops after exactly k steps.
(b) The result is nonterminating and periodic.
a/n 0.a1a2 . . . apd1d2 . . . dkd1d2 . . . dk . . . .
The algorithm will not stop, but the sequence (1) has period k with tail p.
(c) The result is nonterminating and nonperiodic: a/n 0.d1d2d3 . . .. In this case, the algorithm will not stop, and the sequence (1) is not periodic.
5. This is a special case of problem E10 on shrinking squares. Addition is done mod 2: 0 + 0 1 + 1 0, 1 + 0 0 + 1 1. Let (x1, x2, . . . , xn) be the original distribution of zeros and ones around the circle. One step consists of the replacement (x1, . . . , xn) ←(x1 + x2, x2 + x3, . . . , xn + x1). There are two special distributions E (1, 1, . . . , 1) and I (0, 0, . . . , 0). Here, we must work backwards. Suppose we finally reach I. Then the preceding state must be E, and before that an alternating n-tuple (1, 0, 1, 0, . . .). Since n is odd such an n-tuple does not exist.
Now suppose that n 2kq, q odd. The following iteration (x1, . . . , xn) ←(x1 + x2, x2 + x3, . . . + xn + x1) ←(x1 + x3, x2 + x4, . . . xn + x2) ←(x1 + x2 + x3 + x4, x2 + x3 + x4 + x5, . . .) ←(x1 + x5, x2 + x6, . . .) ←· · · shows that, for q 1, the iteration ends up with I. For q > 1, we eventually arrive at I iff we ever get q identical blocks of length 2k, i.e., we have period 2k. Try to prove this.
The problem-solving strategy of working backwards will be treated in Chapter 14.
6. All three numbers a, b, c change their parity in one step. If one of the numbers has different parity from the other two, it will retain this property to the end. This will be the one which remains.
7. (a, b, c) will be transformed into one of the three triples (a + 2, b −1, c −1), (a −1, b + 2, c −1), (a −1, b −1, c + 2). In each case, I a −b mod 3 is an invariant. But b −c 0 mod 3 and a −c 0 mod 3 are also invariant. So I 0 mod 3 combined with a + b + c 0 mod 3 is the condition for reaching a monochromatic state.
8. If there are numbers equal to 1 in the first column, then we double the corresponding rows and subtract 1 from all elements of the first column. This operation decreases the sum of the numbers in the first column until we get a column of ones, which is changed to a column of zeros by subtracting 1. Then we go to the next column, etc.
9. Consider the remainder mod 9. It is an invariant. Since 106 1 mod 9 the number of ones is by one more than the number of twos.
10. From (a −d)(b −c) < 0, we get ab + cd < ac + bd. The switching operation increases the sum S of the products of neighboring terms. In our case ab + bc + cd is replaced by ac + cb + bd. Because of ab + cd < ac + bd the sum S increases.
But S can take only finitely many values.
11. The product I of the eight boundary squares (except the four corners) is −1 and remains invariant.
16 1. The Invariance Principle 12. The numbers starting with the second in each column are an increasing and bounded sequence of integers.
13. (a)LetS bethesumofallnumbersexceptthethirdandsixthrow.S mod 2isinvariant.
If S ̸≡0 (mod 2) initially, then odd numbers will remain on the chessboard.
(b) Let S be the sum of all numbers, except the fourth and eight row. Then I S mod 3 is an invariant. If, initially, I ̸≡0 (mod 3) there will always be numbers on the chessboard which are not divisible by 3.
14. We have 73 1 mod 9 ⇒71996 ≡71 mod 9. This digital sum remains invariant. At the end all digits cannot be distinct, else the digital sum would be 0+1+· · ·+9 45, which is 0 mod 9.
15. The point (x, y) can be reached from (1, 1) iff gcd(x, y) 2n, n ∈N. The permitted moves either leave gcd(x, y) invariant or double it.
16. Here, I(x1, x2, . . . , x6) 2x1 + 4x2 + 6x3 + 8x4 + 10x5 + 12x6 mod 10 is the invariant. Starting with I(1, 0, 1, 0, 1, 0) 8, the goal I(0, 1, 0, 1, 0, 1) 4 cannot be reached.
17. Suppose gcd(m, n) 1. Then, in Chapter 4, E5, we prove that nx my + 1 has a solution with x and y from {1, 2, . . . , m −1}. We rewrite this equation in the form nx m(y −1) + m + 1. Now we place any m positive integers x1, . . . , xm around a circle assuming that x1 is the smallest number. We proceed as follows. Go around the circle in blocks of n and increase each number of a block by 1. If you do this n times you get around the circle m times, and, in addition, the first number becomes one more then the others. In this way, |xmax −xmin| decreases by one. This is repeated each time placing a minimal element in front until the difference between the maximal and minimal element is reduced to zero.
But if gcd(x, y) d > 1, then such a reduction is not always possible. Let one of the m numbers be 2 and all the others be 1. Suppose that, applying the same operation k times we get equidistribution of the (m + 1 + kn) units to the m numbers. This means m + 1 + kn ≡0 mod m. But d does not divide m + kn + 1 since d > 1.
Hence m does not divide m + 1 + kn. Contradiction!
18. We proceed by contradiction. Suppose all the remainders 0, 1, . . . , 2n −1 occur.
The sum of all integers and their place numbers is S1 2 (1 + 2 + . . . + 2n) 2n (2n + 1) ≡0 (mod 2n).
The sum of all remainders is S2 0 + 1 + . . . + 2n −1 n (2n −1) ≡n (mod 2n).
Contradiction!
19. Let the numbering of the prongs be i1, i2, . . . , in. Clearly i1 + · + in n(n + 1)/2.
If n is odd, then the numbering ij n + 1 −j works. Suppose the numbering is good. The prong and hole with number ij coincide if the plug is rotated by ij −j (or ij −j + n) units ahead. This means that (i1 −1) + · · · + (in −n) 1 + 2 + · · · n (mod n). The LHS is 0. The RHS is n(n + 1)/2. This is divisible by n if n is odd.
20. Invariants of this transformation are P : gcd(x, y) gcd(x −y, x) gcd(x, y −x), 1. The Invariance Principle 17 Q : xv + yu 2ab, R : x > 0, y > 0.
P and R are obviously invariant. We show the invariance of Q. Initially, we have ab + ab 2ab, and this is obviously correct. After one step, the left side of Q becomes either x(v+u)+(y−x)u xv+yu or (x−y)v+y(u+v) xv+yu, that is, the left side of Q does not change. At the end of the game, we have x y gcd(a, b) and x(u + v) 2ab →(u + v)/2 ab/x ab/ gcd(a, b) lcm(a, b).
21. Initially, if all components are greater than 1, then they will remain greater than 1.
Starting with the second triple the largest component is always the sum of the other two components diminished by 1. If, after some step, we get (a, b, c) with a ≤ b ≤c, then c a + b −1, and a backward step yields the triple (a, b, b −a + 1). Thus, we can retrace the last state (17, 1967, 1983) uniquely until the next to last step: (17, 1967, 1983) ←(17, 1967, 1951) ←(17, 1935, 1951) ←· · · ← (17, 15, 31) ←(17, 15, 3) ←(13, 15, 3) ←· · · ←(5, 7, 3) ←(5, 3, 3). The preceding triple should be (1, 3, 3) containing 1, which is impossible. Thus the triple (5, 3, 3) is generated at the first step. We can get from (3, 3, 3) to (5, 3, 3) in one step, but not from (2, 2, 2).
22. Let ai be the number of chips on the circle #i. We consider the sum S iai.
Initially, we have S i ∗1 n(n + 1)/2 and, at the end, we must have kn for k ∈{1, 2, . . . , n}. Each move changes S by 0, or n, or −n, that is, S is invariant mod n. At the end, S ≡0 mod n. Hence, at the beginning, we must have S ≡0 mod n.
This is the case for odd n. Reaching the goal is trivial in the case of an odd n.
23. Solution 1. Suppose we get only integer n-tuples from (x1, . . . , xn). Then the dif-ference between the maximal and minimal term decreases. Since the difference is integer, from some time on it will be zero. Indeed, if the maximum x occurs k times in a row, then it will become smaller than x after k steps. If the minimum y occurs m times in a row, then it will become larger after m steps. In a finite number of steps, we arrive at an integral n-tuple (a, a, . . . , a). We will show that we cannot get equal numbers from pairwise different numbers. Supppose z1, . . . , zn are not all equal, but (z1 + z2)/2 (z2 + z3)/2 · · · (zn + z1)/2. Then z1 z3 z5 · · · and z2 z4 z6 · · ·. If n is odd then all zi are equal, contradicting our assumption.
For even n 2k, we must eliminate the case (a, b, . . . , a, b) with a ̸ b. Suppose y1 + y2 2 y3 + y4 2 · · · yn−1 + yn 2 a, y2 + y3 2 · · · yn + y1 2 b.
But the sums of the left sides of the two equation chains are equal, i.e., a b, that is, we cannot get the n-tuple (a, b, . . . , a, b) with a ̸ b.
Solution 2. Let ⃗ x (x1, . . . , xn), T ⃗ x ⃗ y (y1, . . . , yn). With n + 1 1, n i1 y2 i 1 4 n i1 (x2 i + x2 i+1 + 2xixi+1) ≤1 4 n i1 (x2 i + x2 i+1 + x2 i + x2 i+1) n i1 x2 i .
We have equality if and only if xi xi+1 for all i. Suppose the components remain integers. Then the sum of squares is a strictly decreasing sequence of positive integers until all integers become equal after a finite number of steps. Then we show as in 18 1. The Invariance Principle solution 1 that, from unequal numbers, you cannot get only equal numbers in a finite number of steps.
Another Solution Sketch. Try a geometric solution from the fact that the sum of the components is invariant, which means that the centroid of the n points is the same at each step.
24. If you find a negative sum in any row or column, change the signs of all numbers in that row or column. Then the sum of all numbers in the table strictly increases. The sum cannot increase indefinitely. Thus, at the end, all rows and columns will have nonnegative signs.
25. The diagonals partition the interior of the polygon into convex polygons. Consider two neighboring polygons P1, P2 having a common side on a diagonal or side XY.
Then P1, P2 both belong or do not belong to the triangles without the common side XY. Thus, if P goes from P1 to P2, the number of triangles changes by t1 −t2, where t1 and t2 are the numbers of vertices of the polygon on the two sides of XY. Since t1 + t2 2m + 2, the number t1 −t2 is also even.
26. You cannot get rid of an odd divisor of the difference b −a, that is, you can reach (1, 50) from (5, 19), but not (1, 100).
27. The three automata leave gcd(x, y) unchanged. We can reach (19, 79) from (1, 2), but not (819, 357). We can reach (p, q) from (a, b) iff gcd(p, q) gcd(a, b) d.
Go from (a, b) down to (1, d + 1), then, up to (p, q).
28. From the inequality 1/a + 1/b ≥4/(a + b) which is equivalent to (a + b)/2 ≥ 2ab/(a + b), we conclude that the sum S of the inverses of the numbers does not increase. Initially, we have S n. Hence, at the end, we have S ≤n. For the last number 1/S, we have 1/S ≥1/n.
29. The permissible transformations leave the sides of the polygon and their directions invariant. Hence, there are only a finite number of polygons. In addition, the area strictly increases after each reflection. So the process is finite.
Remark. The corresponding problem for line reflections in AB is considerably harder.
The theorem is still valid, but the proof is no more elementary. The sides still remain the same, but their direction changes. So the finiteness of the process cannot be easily deduced. (In the case of line reflections, there is a conjecture that 2n reflections suffice to reach a convex polygon.) 30. Let f (x) x2 −3x + 3. We are asked to solve the equation f (f (x)) x, that is to find the fixed or invariant points of the function f ◦f . First, let us look at f (x) x, i.e. the fixed points of f . Every fixed point of f is also a fixed point of f ◦f . Indeed, f (x) x ⇒f (f (x)) f (x) ⇒f (f (x)) x.
First, we solve the quadratic f (x) x, or x2 −4x + 3 0 with solutions x1 3, x2 1. f [f (x)] x leads to the fourth degree equation x4−6x3+12x2−10x+3 0, of which we already know two solutions 3 and 1. So the left side is divisible by x −3 and x −1 and, hence, by the product (x −3)(x −1) x2 −4x + 3. This will be proved in the chapter on polynomials, but the reader may know this from high school. Dividing the left side of the 4th-degree equation by x2 −4x + 3 we get x2 −2x + 1. Now x2 −2x + 1 0 is equivalent to (x −1)2 0. So the two other solutions are x3 x4 1. We get no additional solutions in this case, but usually, the number of solutions is doubled by going from f [x] x to f [f (x)] x.
1. The Invariance Principle 19 31. Suppose the product P is odd. Then, each of its factors must be odd. Consider the sum S of these numbers. Obviously S is odd as an odd number of odd summands.
On the other hand, S (ai −i) ai − i 0, since the ai are a permutation of the numbers 1 to n. Contradiction!
32. We partition the participants into the set E of even persons and the set O of odd persons. We observe that, during the hand shaking ceremony, the set O cannot change its parity. Indeed, if two odd persons shake hands, O increases by 2. If two even persons shake hands, O decreases by 2, and, if an even and an odd person shake hands, |O| does not change. Since, initially, |O| 0, the parity of the set is preserved.
33. Consider the number U of inversions, computed as follows: Below each 1, write the number of zeros to the right of it, and add up these numbers. Initially U 0. U does not change at all after each move, or it increases or decreases by 2. Thus U always remains even. But we have U 1 for the goal. Thus, the goal cannot be reached.
34. Consider the trinomial f (x) ax2 + bx + c. It has discriminant b2 −4ac. The first transformation changes f (x) into (a + b + c)x2 + (b + 2a)x + a with discriminant (b + 2a)2 −4(a + b + c) · a b2 −4ac, and, applying the second transformation, we get the trinomial cx2 + (b −2c)x + (a −b + c) with discriminant b2 −4ac.
Thus the discriminant remains invariant. But x2 + 4x + 3 has discriminant 4, and x2 + 10x + 9 has discriminant 64. Hence, one cannot get the second trinomial from the first.
35. For three squares in arithmetic progression, we have a2 3 −a2 2 a2 2 −a2 1 or (a3 − a2)(a3 + a2) (a2 −a1)(a2 + a1). Since a2 + a1 < a3 + a2, we must have a2 −a1 > a3 −a2.
Suppose that a2 1, a2 2, a2 3, . . . is an infinite arithmetic progression. Then a2 −a1 > a3 −a2 > a4 −a3 > · · · .
This is a contradiction since there is no infinite decreasing sequence of positive integers.
36. Suppose the integers 1, . . . , n are arranged in any order. We will say that the numbers i and k are out of order if the larger of the two is to the left of the smaller. In that case, they form an inversion. Prove that interchange of two neighbors changes the parity of the number of inversions.
37. Interchange of any two integers can be replaced by an odd number of interchanges of neighboring integers.
38. The number of inversions in n, . . . , 1 is n(n −1)/2. Prove that one step does not change the parity of the inversions. If n(n −1)/2 is even, then split the n integers into pairs of neighbors (leaving the middle integer unmatched for odd n). Then form quadruplets from the first, last, second, second from behind, etc.
39. We assign the weight 1/2x+y to the square with label (x, y). We observe that the total weight of the squares covered by chips does not change if a chip is replaced by two neighbors. The total weight of the first column is 1 + 1 2 + 1 4 + · · · 2.
20 1. The Invariance Principle The total weight of each subsequent square is half that of the preceding square. Thus the total weight of the board is 2 + 1 + 1 2 + · · · 4.
In (a) the total weight of the shaded squares is 2 3 4. The weight of the rest of the board is 1 3 4. The total weight of the remaining board is not enough to accommodate the chips on the shaded region.
In (b) the lone piece has the weight 1. Suppose it is possible to clear the shaded region in finitely many moves. Then, in the column x 0 there is at most the weight 1/8, and in the row y 0, there is at most the weight 1/8. The remaining squares outside the shaded region have weight 3/4. In finitely many moves we can cover only a part of them. So we have again a contradiction.
40. I can get a chip to (0, 4), but not to (0, 5). Indeed, we introduce the norm of a point (x, y) as follows: n(x, y) |x| + |y −5|. We define the weight of that point by αn, where α is the positive root of α2 + α −1 0. The weight of a set S of chips will be defined by W(S) p∈S αn.
Cover all the lattice points for y ≤0 by chips. The weight of the chips with y 0 is α5 + +2α6 i≥0 αi α5 + 2α4. By covering the half plane with y ≤0, we have the total weight (α5 + 2α4)(1 + α + α2 + · · ·) α5 + 2α4 1 −α α3 + 2α2 1.
We make the following observations: A horizontal solitaire jump toward the y-axis leaves total weight unchanged. A vertical jump up leaves total weight unchanged.
Any other jump decreases total weight. Total weight of the goal (0, 5) is 1. Thus any distribution of finitely many chips on or below the x-axis has weight less than 1.
Hence, the goal cannot be reached by finitely many chips.
41. Place a coordinate system so that the seven given points have coordinates (0,0,0), (0,0,1), (0,1,0), (1,0,0), (1,1,0), (1,0,1), (0,1,1). We observe that a point preserves the parity of its coordinates on reflection. Thus, we never get points with all three coordinates odd. Hence the point (1,1,1) can never be reached. This follows from the mapping formula X →2A−X,orincoordinates(x, y, z) →(2a−x, 2b−y, 2c−z), where A (a, b, c) and X (x, y, z). The invariant, here, is the parity pattern of the coordinates of the points in S.
42. Fig. 1.10 shows how to reduce an L-tetromino occupied by chips to one square by using one free cell which is the reflection of the black square at the center of the first horizontal square. Applying this operation repeatedly to Fig. 1.9 we can reduce any n × n square to a 1 × 1, 2 × 2, or 3 × 3 square. A 1 × 1 square is already a reduction to one occupied square. It is trivial to see how we can reduce a 2 × 2 square to one occupied square.
The reduction of a 3×3 square to one occupied square does not succeed. We are left with at least two chips on the board. But maybe another reduction not necessarily using L-tetrominoes will succeed. To see that this is not so, we start with any n divisible by 3, and we color the n × n board diagonally with three colors A, B, C.
1. The Invariance Principle 21 Fig. 1.9 d t d d t d d Fig. 1.10 d t t Denote the number of occupied cells of colors A, B, C by a, b, c, respectively.
Initially, a b c, i.e., a ≡b ≡c mod 2. That is, all three numbers have the same parity. If we make a jump, two of these numbers are decreased by 1, and one is increased by 1. After the jump, all three numbers change parity, i.e., they still have the same parity. Thus, we have found the invariant a ≡b ≡c mod 2. This relation is violated if only one chip remains on the board. We can even say more. If two chips remain on the board, they must be on squares of the same color.
43. By looking at a healthy cell with 2, 3, or 4 infected neighbors, we observe that the perimeter of the contaminated area does not increase, although it may well decrease.
Initially, the perimeter of the contaminated area is at most 4 × 9 36. The goal 4 × 10 40 will never be reached.
44. By applying these three operations on f and g, we get a polynomial P (f (x), g(x)) x, (1) which should be valid for all x. In (a) and (b), we give a specific value of x, for which (1) is not true. In (a) f (2) g(2) 6. By repeated application of the three operations on 6 we get again a multiple of 6. But the right side of (1) is 2.
In (b) f (1/2) g(1/2) 1. The left-hand side of (1) is an integer, and the right-hand side 1/2 is a fractional number.
In (c) we succeed in finding a polynomial in f and g which is equal to x: (f −g)2 + 2g −3f x.
45. We should get x2 n + y2 n 1 for all n, but rounding errors corrupt more and more of the significant digits. One gets the table below. This is a very robust computation. No ”catastrophic cancellations” ever occur. Quite often one does not get such precise results. In computations involving millions of operations, one should use double precision to get single precision results.
46. Since 1994 18 + 19 · 104, we get 18 + 19 37, 37 + 19 56, . . . , 1975 + 19 1994. It is not so easy to find all numbers which can be reached starting from 18 and 19. See Chapter 6, especially the Frobenius Problem for n 3 at the end of the chapter.
22 1. The Invariance Principle 47. (a) No! The parity of the number of −1′s on the perimeter of the pentagon does not change.
(b) No! The product of the nine numbers colored black in Fig. 1.11 does not change.
48. Color the squares alternately black and white as in Fig. 1.12. Let W 10n x2 n + y2 n 10 1.0000000000 102 1.0000000001 103 1.0000000007 104 1.0000000066 105 1.0000000665 106 1.0000006660 107 1.0000066666 Fig. 1.11 pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp s s d d s s d s s s s s Fig. 1.12 and B be the sums of the numbers on the white and black squares, respectively.
Application of T does not change the difference W −B. For Fig. 1.7 and Fig. 1.8 the differences are 5 and −1, respectively. The goal −1 cannot be reached from 5.
49. Replace each + by +1 and each −by −1, and form the product P of all the numbers.
Obviously, P is an invariant.
50. We denote a replacement operation by ◦. Then, we have e ◦e e, e ◦a a, e ◦b b, a ◦a b, b ◦b a, a ◦b e.
The ◦operation is commutative since we did not mention the order. It is easy to check that it is also associative, i.e., (p ◦q) ◦r p ◦(q ◦r) for all letters occurring. Thus, the product of all letters is independent of the the order in which they are multiplied.
51. The number of heads is invariant mod 3. Initially, it is 1 and it remains so.
52. Replace 1998 by n, and derive a necessary condition for the existence of such an arrangement. Let pk be the position of the first integer k. Then the other k has position pk + k. By counting the position numbers twice, we get 1 + · · · + 2n (p1 +p1 +1)+· · ·+(pn +pn +n). For P n i1 pi, we get P n(3n+1)/4, and P is an integer for n ≡0, 1 mod 4. Since 1998 ≡2 mod 4, this necessary condition is not satisfied. Find examples for n 4, 5, and 8.
53. This is an invariance problem. As a prime candidate, we think of the discriminant D. The first operation obviously does not change D. The second operation does not change the difference of the roots of the polynomial. Now, D b2 −4ac a2((b/a)2 −4c/a), but −b/a x1 + x2, and c/a x1x2. Hence, D a2(x1 − x2)2, i.e., the second operation does not change D. Since the two trinomials have discriminants 9 and 5, the goal cannot be reached.
54. Consider I a2 + b2 + c2 −2g, where g is the current gain (originally g 0). If we transfer one chip from the first to the second pile, then we get I ′ (a −1)2 + (b + 1)2 + c2 −2g′ where g′ g + b −a + 1, that is, I ′ a2 −2a + 1 + b2 + c2 + 2b + 1 −2g −2b + 2a −2 a2 + b2 + c2 −2g I. We see that I does not 1. The Invariance Principle 23 change in one step. If we ever get back to the original distribution (a, b, c), then g must be zero again.
The invariant I ab + bc + ca + g yields another solution. Prove this.
55. The transformation d leaves the remainder on division by 3 invariant. Hence, modulo 3 the equation has the form 0 ≡2. There is no solution.
56. We assume that, at the start, the side lengths are 1, p, q, 1 > p, 1 > q. Then all succeeding triangles are similar with coefficient pmqn. By cutting such a triangle of type (m, n), we get two triangles of types (m + 1, n) and (m, n + 1). We make the following translation. Consider the lattice square with nonnegative coordinates. We assign the coordinates of its lower left vertex to each square. Initially, we place four chips on the square (0, 0). Cutting a triangle of type (m, n) is equvalent to replacing a chip on square (m, n) by one chip on square (m + 1, n) and one chip on square (m, n + 1). We assign weight 2−m−n to a chip on square (m, n). Initially, the chips have total weight 4. A move does not change total weight. Now we get problem 39 of Kontsevich. Initially, we have total weight 4. Suppose we can get each chip on a different square. Then the total weight is less than 4. In fact, to get weight 4 we would have to fill the whole plane by single chips. This is impossible in a finite number of steps.
57. Comparing xn+1/xn with yn+1/yn, we observe that x2 nyn a2b is an invariant. If we can show that lim xn lim yn x, then x3 a2b, or x 3 √ a2b.
Because of xn < yn and the arithmetic mean-geometric mean inequality, yn+1 lies to the left of (xn + yn)/2 and xn+1 lies to the left of (xn + yn+1)/2. Thus, xn < xn+1 < yn+1 < yn and yn+1 −xn+1 < (yn −xn)/2. We have, indeed, a common limit x.
Actually for large n, say n ≥5, we have √xnyn ≈(yn + xn)/2 and yn+1 −xn+1 ≈ (yn −xn)/4.
58. Assign the number I(W) a1 +2a2 +3a3 +· · ·+nan to W. Deletion or insertion of any word XXX in any place produces Z b1b2 · · · bm with I(W) ≡I(Z) modulo 3. Since I(01) 2 and I(10) 1, the goal cannot be attained.
59. Select four vertices such that no two are joined by an edge. Let X be the sum of the numbers at these vertices, and let y be the sum of the numbers at the remaining four vertices. Initially, I x −y ±1. A step does not change I. So neither (a) nor (b) can be attained.
60. Hint:Considerthesequencessn 1/xn,andtn 1/yn.Aninvariantissn+1+2tn+1 sn + 2tn 1/a + 2/b.
2 Coloring Proofs The problems of this chapter are concerned with the partitioning of a set into a finite number of subsets. The partitioning is done by coloring each element of a subset by the same color. The prototypical example runs as follows.
In 1961, the British theoretical physicist M.E. Fisher solved a famous and very tough problem. He showed that an 8 × 8 chessboard can be covered by 2 × 1 dominoes in 24 × 9012 or 12,988,816 ways. Now let us cut out two diagonally opposite corners of the board. In how many ways can you cover the 62 squares of the mutilated chessboard with 31 dominoes?
The problem looks even more complicated than the problem solved by Fisher, but this is not so. The problem is trivial. There is no way to cover the mutilated chessboard. Indeed, each domino covers one black and one white square. If a covering of the board existed, it would cover 31 black and 31 white squares. But the mutilated chessboard has 30 squares of one color and 32 squares of the other color.
The following problems are mostly ingenious impossibility proofs based on coloring or parity. Some really belong to Chapter 3 or Chapter 4, but they use coloring, so I put them in this chapter. A few also belong to the closely related Chapter 1. The mutilated chessboard required two colors. The problems of this chapter often require more than two colors.
26 2. Coloring Proofs Problems 1. A rectangular floor is covered by 2×2 and 1×4 tiles. One tile got smashed. There is a tile of the other kind available. Show that the floor cannot be covered by rearranging the tiles.
2. Is it possible to form a rectangle with the five tetrominoes in Fig. 2.1?
3. A 10 × 10 chessboard cannot be covered by 25 T-tetrominoes in Fig. 2.1. These tiles are called from left to right: straight tetromino, T-tetromino, square tetromino, L-tetromino, and skew tetromino.
Fig. 2.1 4. An8×8chessboardcannotbecoveredby15T-tetrominoesandonesquaretetromino.
5. A 10 × 10 board cannot be covered by 25 straight tetrominoes (Fig. 2.1).
6. Consider an n × n chessboard with the four corners removed. For which values of n can you cover the board with L-tetrominoes as in Fig. 2.2?
7. Is there a way to pack 250 1 × 1 × 4 bricks into a 10 × 10 × 10 box?
8. An a × b rectangle can be covered by 1 × n rectangles iff n|a or n|b.
9. One corner of a (2n + 1) × (2n + 1) chessboard is cut off. For which n can you cover the remaining squares by 2 × 1 dominoes, so that half of the dominoes are horizontal?
10. Fig. 2.3 shows five heavy boxes which can be displaced only by rolling them about one of their edges. Their tops are labeled by the letter ⊤. Fig. 2.4 shows the same five boxes rolled into a new position. Which box in this row was originally at the center of the cross?
11. Fig. 2.5 shows a road map connecting 14 cities. Is there a path passing through each city exactly once?
Fig. 2.2 ⊤⊤⊤ ⊤ ⊤ Fig. 2.3 ⊤⊣⊤⊤⊤ Fig. 2.4 s s s s s s s s s s s s s s @ @ @ @ @ @ @ @ Fig. 2.5 12. A beetle sits on each square of a 9 × 9 chessboard. At a signal each beetle crawls diagonally onto a neighboring square. Then it may happen that several beetles will sit on some squares and none on others. Find the minimal possible number of free squares.
2. Coloring Proofs 27 13. Every point of the plane is colored red or blue. Show that there exists a rectangle with vertices of the same color. Generalize.
14. Every space point is colored either red or blue. Show that among the squares with side 1 in this space there is at least one with three red vertices or at least one with four blue vertices.
15. Show that there is no curve which intersects every segment in Fig. 2.6 exactly once.
Fig. 2.6 16. On one square of a 5 × 5 chessboard, we write −1 and on the other 24 squares +1.
In one move, you may reverse the signs of one a ×a subsquare with a > 1. My goal is to reach +1 on each square. On which squares should −1 be to reach the goal?
17. The points of a plane are colored red or blue. Then one of the two colors contains points with any distance.
18. The points of a plane are colored with three colors. Show that there exist two points with distance 1 both having the same color.
19. All vertices of a convex pentagon are lattice points, and its sides have integral length.
Show that its perimeter is even.
20. n points (n ≥5) of the plane can be colored by two colors so that no line can separate the points of one color from those of the other color.
21. You have many 1 × 1 squares. You may color their edges with one of four colors and glue them together along edges of the same color. Your aim is to get an m × n rectangle. For which m and n is this possible?
22. You have many unit cubes and six colors. You may color each cube with 6 colors and glue together faces of the same color. Your aim is to get a r × s × t box, each face having different color. For which r, s, t is this possible?
23. Consider three vertices A (0, 0), B (0, 1), C (1, 0) in a plane lattice. Can you reach the fourth vertex D (1, 1) of the square by reflections at A, B, C or at points previously reflected?
24. Every space point is colored with exactly one of the colors red, green, or blue. The sets R, G, B consist of the lengths of those segments in space with both endpoints red, green, and blue, respectively. Show that at least one of these sets contains all nonnegative real numbers.
25. The Art Gallery Problem. An art gallery has the shape of a simple n-gon. Find the minimum number of watchmen needed to survey the building, no matter how complicated its shape.
26. A 7×7 square is covered by sixteen 3×1 and one 1×1 tiles. What are the permissible positions of the 1 × 1 tile?
27. The vertices of a regular 2n-gon A1, . . . , A2n are partitioned into n pairs. Prove that, if n 4m + 2 or n 4m + 3, then two pairs of vertices are endpoints of congruent segments.
28. A 6×6 rectangle is tiled by 2×1 dominoes. Then it has always at least one fault-line, i.e., a line cutting the rectangle without cutting any domino.
28 2. Coloring Proofs 29. Each element of a 25 × 25 matrix is either +1 or −1. Let ai be the product of all elements of the ith row and bj be the product of all elements of the jth column.
Prove that a1 + b1 + · · · + a25 + b25 ̸ 0.
30. Can you pack 53 bricks of dimensions 1 × 1 × 4 into a 6 × 6 × 6 box? The faces of the bricks are parallel to the faces of the box.
31. Three pucks A, B, C are in a plane. An ice hockey player hits the pucks so that any one glides through the other two in a straight line. Can all pucks return to their original spots after 1001 hits?
32. A 23 × 23 square is completely tiled by 1 × 1, 2 × 2 and 3 × 3 tiles. What minimum number of 1 × 1 tiles are needed (AUO 1989)?
33. The vertices and midpoints of the faces are marked on a cube, and all face diagonals aredrawn.Isitpossibletovisitallmarkedpointsbywalkingalongthefacediagonals?
34. There is no closed knight’s tour of a (4 × n) board.
35. The plane is colored with two colors. Prove that there exist three points of the same color, which are vertices of a regular triangle.
36. A sphere is colored in two colors. Prove that there exist on this sphere three points of the same color, which are vertices of a regular triangle.
37. Given an m × n rectangle, what minimum number of cells (1 × 1 squares) must be colored, such that there is no place on the remaining cells for an L-tromino?
38. The positive integers are colored black and white. The sum of two differently colored numbers is black, and their product is white. What is the product of two white numbers? Find all such colorings.
Solutions 1. Color the floor as in Fig. 2.7. A 4 × 1 tile always covers 0 or 2 black squares. A 2 × 2 tile always covers one black square. It follows immediately from this that it is impossible to exchange one tile for a tile of the other kind.
Fig. 2.7 2. Any rectangle with 20 squares can be colored like a chessboard with 10 black and 10 white squares. Four of the tetrominoes will cover 2 black and 2 white squares each.
The remaining 2 black and 2 white squares cannot be covered by the T-tetromino. A T-tetromino always covers 3 black and one white squares or 3 white and one black squares.
3. A T-tetromino either covers one white and three black squares or three white and one black squares. See Fig. 2.8.Tocoveritcompletely,weneedequally many tetrominoes of each kind. But 25 is an odd number. Contradiction!
2. Coloring Proofs 29 4. The square tetromino covers two black and two white squares. The remaining 30 black and 30 white squares would require an equal number of tetrominoes of each kind. On the other hand, one needs 15 tetrominoes for 60 squares. Since 15 is odd, a covering is not possible.
5. Color the board diagonally in four colors 0, 1, 2, 3 as shown in Fig. 2.10. No matter how you place a straight tetromino on this board, it always covers one square of each color. 25 straight tetrominoes would cover 25 squares of each color. But there are 26 squares with color 1.
Fig. 2.8 Alternate solution. Color the board as shown in Fig. 2.9. Each horizontal straight tetromino covers one square of each color. Each vertical tetromino covers four squares of the same color. After all horizontal straight tetrominoes are placed there remain a + 10, a + 10, a, a squares of color 0, 1, 2, 3, respectively. Each of these numbers should be a multiple of 4. But this is impossible since a + 10 and a cannot both be multiples of 4.
0 1 2 3 0 1 2 3 0 1 0 1 2 3 0 1 2 3 0 1 0 1 2 3 0 1 2 3 0 1 0 1 2 3 0 1 2 3 0 1 0 1 2 3 0 1 2 3 0 1 0 1 2 3 0 1 2 3 0 1 0 1 2 3 0 1 2 3 0 1 0 1 2 3 0 1 2 3 0 1 0 1 2 3 0 1 2 3 0 1 0 1 2 3 0 1 2 3 0 1 Fig. 2.9 6. There are n2 −4 squares on the board. To cover it with tetrominoes n2 −4 must be a multiple of 4, i.e., n must be even. But this is not sufficient. To see this, we color the board as in Fig. 2.11. An L-tetromino covers three white and one black squares or three black and one white squares. Since there is an equal number of black and white squares on the board, any complete covering uses an equal number of tetrominoes of each kind. Hence, it uses an even number of tetrominoes, that is, n2 −4 must be a multiple of 8. So, n must have the form 4k + 2. By actual construction, it is easy to see that the condition 4k + 2 is also sufficient.
30 2. Coloring Proofs 0 1 2 3 0 1 2 3 0 1 1 2 3 0 1 2 3 0 1 2 2 3 0 1 2 3 0 1 2 3 3 0 1 2 3 0 1 2 3 0 0 1 2 3 0 1 2 3 0 1 1 2 3 0 1 2 3 0 1 2 2 3 0 1 2 3 0 1 2 3 3 0 1 2 3 0 1 2 3 0 0 1 2 3 0 1 2 3 0 1 1 2 3 0 1 2 3 0 1 2 Fig. 2.10 7. Assign coordinates (x, y, z) to the cells of the box, 1 ≤x, y, z ≤10. Color the cells in four colors denoted by 0, 1, 2, 3. The cell (x, y, z) is assigned color i if x + y + z ≡i mod 4. This coloring has the property that a 1 × 1 × 4 brick always occupies one cell of each color no matter how it is placed in the box. Thus, if the box could be filled with two hundred fifty 1 × 1 × 4 bricks, there would have to be 250 cells of each of the colors 0, 1, 2, 3, respectively. Let us see if this necessary packing condition is satisfied. Fig. 2.10 shows the lowest level of cells with the corresponding coloring. There are 26, 25, 24, 25 cells with color 0, 1, 2, 3 respectively. The coloring of the next layer is obtained from that of the preceding layer by adding 1 mod 4.
Thus the second layer has 26, 25, 24, 25 cells with colors 1, 2, 3, 0, respectively. The third layer has 26, 25, 24, 25 cells with colors 2, 3, 0, 1, respectively, the fourth layer has 26, 25, 24, 25 cells with colors 3, 0, 1, 2, respectively, and so on. Thus there are (26 + 25 + 24 + 25) · 2 + 26 + 25 251 cells of color 0. Hence there is no packing of the 10 × 10 × 10 box by 1 × 1 × 4 bricks.
8. If n|a or n|b, the board can be covered by 1 × n tiles in an obvious way. Suppose n ̸ | a, i.e., a q · n + r, 0 < r < n. Color the board as indicated in Fig. 2.9. There are bq + b squares of each of the colors 1, 2, . . . , r, and there are bq squares of each of the colors 1, . . . , n. The h horizontal 1 × n tiles of a covering each cover one square of each color. Each vertical 1 × n tile covers n squares of the same color.
After the h horizontal tiles are placed, there will remain (bq + b −h) squares of each of the colors 1, . . . , r and bq −h of each of the colors r + 1, . . . , n. Thus n|bq + b −h and n|bq −h. But if n divides two numbers, it also divides their difference: (bq + b −h) −(bq −h) b. Thus, n|b. Space analogue: If an a × b × c box can be tiled with n × 1 × 1 bricks, then n|a or n|b or n|c.
Fig. 2.11 Fig. 2.12 ⊤⊣⊤⊤⊤ Fig. 2.13 2. Coloring Proofs 31 9. Color the board as in Fig. 2.12. There are 2n2 + n white squares and 2n2 + 3n black squares, a total of 4n2 + 4n squares. 2n2 + 2n dominoes will be required to cover all of these squares. Since one half of these dominoes are to be horizontal, there will be n2 + n vertical and n2 + n horizontal dominoes. Each vertical domino covers one black and one white square. When all the vertical dominoes are placed, they cover n2 + n white squares and n2 + n black squares. The remaining n2 white squares and n2 +2n black squares must be covered by horizontal dominoes. A horizontal domino covers only squares of the same color. To cover the n2 white squares n2, i.e., n must be even. One easily shows by actual construction that this necessary condition is also sufficient. Thus, the required covering is possible for a (4n + 1) × (4n + 1) board and is impossible for a (4n −1) × (4n −1) board.
10. Suppose the floor is ruled into squares colored black and white like a chessboard.
Further suppose that the central box of the cross covers a black square. Then the four other boxes stand on white squares. It is easy to see that the transition ⊤→⊤ requires an even number of flips whereas a transition ⊤→⊢requires an odd number of flips. Hence the boxes #1, 3, 4, 5 in Fig. 2.13 originally stand on squares of the same color. Now the squares occupied by boxes #1, 3, 5 are the same color, and so boxes #1, 3, 5 must have originated on squares of the same color. Since there are not three boxes which originated on black squares, these boxes must stand on white squares. Box #2 must have been flipped an odd number of times. It is now on a black square. Hence it was originally on a white square. Box #4 is now on a black square.
Since it was flipped an even number of times, it was originally on a black square.
Thus #4 is the central box.
11. Color the cities black and white so that neighboring cities have different colors as shown in Fig. 2.14. Every path through the 14 cities has the color pattern bwbwb-wbwbwbwbw or wbwbwbwbwbwbwb. So it passes through seven black and seven white cities. But the map has six black and eight white cities. Hence, there is no path passing through each city exactly once.
t d d t t d t d t t d d d d @ @ @ @ @ @ Fig. 2.14 odd odd even odd even odd Fig. 2.15 Fig. 2.16 12. Color the columns alternately black and white. We get 45 black and 36 white squares.
Every beetle changes its color by crawling. Hence at least nine black squares remain empty. It is easy to see that exactly nine squares can stay free.
13. Consider the lattice points (x, y) with 1 ≤x ≤n + 1, 1 ≤y ≤nn+1 + 1.
One row can be colored in nn+1 ways. By the box principle, at least two of the (nn+1 + 1) rows have the same coloring. Let two such rows colored the same way have ordinates k and m. For each i ∈{1, . . . , n + 1}, the points (i, k) and (i, m) have the same color. Since there are only n colors available, one of the colors will repeat.
Suppose (a, k) and (b, k) have the same color. Then the rectangle with the vertices (a, k), (b, k), (b, m), (a, m) has four vertices of the same color.
32 2. Coloring Proofs The problem can be generalized to parallelograms and to k-dimensional boxes. In-stead of the lattice rectangle with sides n and nn+1, we have a lattice box with lengths d1 −1, d2 −1, . . . dk −1, and d1 n + 1, di+1 nd1···di + 1.
14. Denote by B the property that there is a unit square with four blue vertices.
Case 1: All points of space are blue ⇒B.
Case 2: There exists a red point P1. Make of P1 the vertex of a pyramid with equal edges and the square P2P3P4P5 as base.
Case 2.1: The four points Pi, i 2, 3, 4, 5 are blue ⇒B.
Case 2.2: One of the points Pi, i 2, 3, 4, 5 is red, say P2. Make of P1P2 a lateral edge of an equilateral prism, with the remaining vertices P6, P7, P8, P9.
Case 2.2.1: The four points Pj, j 6, 7, 8, 9 are blue⇒B.
Case 2.2.2: One of the points Pj, j 6, 7, 8, 9 is red, say P6. Then P1, P2, and P6 are three red vertices of a unit square.
15. The map in Fig. 2.15 consists of three faces each bounded by five segments (labeled odd). Suppose there exists a curve intersecting every segment exactly once. Then it would have three points inside the odd faces, where it starts or ends. But a curve has zero or two endpoints.
16. Color the board as in Fig. 2.16. Every permitted subsquare contains an even number of black squares. Initially if −1 is on a black square, then there are always an odd number of −1’s on the black squares. Rotation by 90◦shows that the −1 can be only on the central square.
If −1 is on the central square, then we can achieve all +1’s in 5 moves 1. Reverse signs on the lower left 3 × 3 square.
2. Reverse signs on the upper right 3 × 3 square.
3. Reverse signs on the upper left 2 × 2 square.
4. Reverse signs on the lower right 2 × 2 square.
5. Reverse signs on the whole 5 × 5 square.
17. Suppose the theorem is not true. Then the red points miss a distance a and the blue points miss a distance b. We may assume a ≤b. Consider a blue point C. Construct an isosceles triangle ABC with legs AC BC b and AB a. Since C is blue, A cannot be blue. Thus, it must be red. The point B cannot be red since its distance to the red point A is a. But it cannot be blue either, since its distance to the blue point C is b. Contradiction!
18. Call the colors black, white, and red. Suppose any two points with distance 1 have different colors. Choose any red point r and assign to it Fig. 2.17. One of the two points b and w must be white and the other black. Hence, the point r′ must be red. J J J J b w r r′ Fig. 2.17 2. Coloring Proofs 33 Rotating Fig. 2.17 about r we get a circle of red points r′. This circle contains a chord of length 1. Contradiction!
Alternate solution. For Fig. 2.18 consisting of 11 unit rods, you need at least four colors, if vertices of distance 1 are to have distinct colors.
19. Color the lattices as in a chess board. Erect right triangles on the sides of the pentagon as longest sides. With the two other sides along the sides of the squares, trace the ten shorter sides. Since, at the end, we return to the vertex we left, we must have traced an even number of lattice points (on transition from one lattice point to the next the color of the lattice point changes). Hence the sum of shorter sides is even. The parity of the longer sides (i.e., the sides of the pentagon) is equivalent to the parity of the sums of the shorter sides. Hence the perimeter of the pentagon has the same parity as the sum of the shorter sides.
20. Of n ≥5 points, it is always possible to choose four vertices of a convex polygon.
If we color two opposite vertices the same color, then no line will separate the two sets of points.
21. Result: We can glue together an m × n rectangle iff m and n have the same parity.
(a) m and n are both odd. Then we can glue together an 1 × n rectangle as in Fig.
2.19. From these strips, we can glue together the rectangle in Fig. 2.20.
(b) m and n are even. Consider the rectangles with odd side lengths of dimensions (m −1) × (n −1), 1 × (n −1), (m −1) × 1, and 1 × 1, respectively. They can be assembled into the rectangle m × n.
(c) m is even, and n is odd. Suppose we succeeded in gluing together a rectangle m × n satisfying the conditions of the problem. Consider one of the sides of the rectangle with odd length. Suppose it is colored red. Let us count the total number of red sides of the squares. On the perimeter of the rectangle, there are n and in the interior there is an even number, since another red neighbor belongs to one red side of a square. Thus the total number of red sides is odd. The total number of squares is the same as the number of red sides, i.e., odd. On the other hand this number is m n, that is, an even number. Contradiction!
T T T T T T """ " bbb b b b b b ppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp Fig. 2.18 Fig. 2.19 4 4 4 4 1 3 1 3 1 3 2 2 2 2 1 1 1 1 1 1 . . . .
. . . .
. . . .
. . . .
3 3 3 2 2 2 2 3 1 3 4 4 4 4 3 1 3 2 2 2 2 3 1 3 4 4 4 4 Fig. 2.20 22. The solution is similar to that of the preceding problem.
23. Color the lattice points black and white such that points with odd coordinates are black and the other lattice points are white. By reflections you always stay on lattices 34 2. Coloring Proofs of the same color. Thus it is not possible to reach the opposite vertex of the square ABCD.
24. Let P1, P2, P3 be the three sets. We assume on the contrary that a1 is not assumed by P1, a2 is not assumed by P2, and a3 is not assumed by P3. We may assume that a1 ≥a2 ≥a3 > 0.
A A A a1 a2 x1 Fig. 2.21 Let x1 ∈P1. The sphere S with midpoint x1 and radius a1 is contained completely in P2 ∪P3. Since a1 ≥a3, S ̸⊂P3. Let x2 ∈P2 ∩S. The circle {y ∈S|d(x2, y) a2} ⊂ P3, since P2 does not realize a2. But in Fig. 2.21, a2 ≤a1 ⇒r a2 1 −a2 2/4a2 1 ≥ a2 √ 3/2, and a3 ≤a2 ≤a2 √ 3 ≤2r. Thus a3 is assumed in P3.
Another ingenious solution will be found in Chapter 4 (problem 67). It will be good training for the more difficult plane problem 68 of that chapter. Both solutions make nontrivial use of the box principle.
25. The gallery is triangulated by drawing nonintersecting diagonals. By simple induc-tion one can prove that such a triangulation is always possible. Then we color the vertices of the triangles properly with three colors, so that any vertex of a triangle gets a different color. By trivial induction, one proves that the triangles of the trian-gulation can always be properly colored. Now we consider the color, which occurs least often. Suppose it is red. The watchmen at the red vertices can survey all walls.
Thus the minimum number of watchmen is ⌊n/3⌋.
26. Color the squares diagonally by colors 0, 1, 2. Then each 3 × 1 tile covers each of the colors once. In Fig. 2.22 we have 17 zeros, 16 ones and 16 twos. The monomino must cover one of the squares labeled ”0”. In addition, it must remain a ”0” if we make a quarter-turn of the board. As possible positions there will remain only the central square, the four corners, and the centers of the outer edges in Fig. 2.22. A different coloring yields a different solution. We use the three colors 0, 1, 2 as in Fig. 2.23. That is, the squares colored 0 will be the center, the four corners, and the centers of the outer edges. The tiles 1×3 are of two types, those covering one square of color 0 and two squares of color 1 and those covering one square of color 1 and two squares of color 2. Suppose all squares of color 0 are covered by 1 × 3 tiles.
There will be 9 tiles of type 1 and 7 tiles of type 2. They will cover 9 · 2 + 7 25 squares of color 1 and 7 · 2 14 squares of color 2. This contradiction proves that one of the squares of color 0 is covered by the 1 × 1 tile.
0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 Fi 2 22 0 1 1 0 1 1 0 1 2 2 1 2 2 1 1 2 2 1 2 2 1 0 1 1 0 1 1 0 1 2 2 1 2 2 1 1 2 2 1 2 2 1 0 1 1 0 1 1 0 Fi 2 23 2. Coloring Proofs 35 27. Suppose that all pairs of vertices have different distances. To the segment ApAq, we assign the smaller of the numbers | p −q | and 2n−| p −q |. We get the numbers 1, . . . , n. Suppose that among these numbers there are k even and n−k odd numbers.
To the odd numbers correspond the segments ApAq, where p, q have different parity.
Hence, among the remaining segments there will be k vertices with odd numbers and k vertices with even numbers, with the segments connecting vertices of the same parity. Hence k is even. For the numbers n of the type 4m, 4m + 1, 4m + 2, 4m + 3 the number k of even numbers is 2m, 2m, 2m + 1, 2m + 1, respectively. Hence n 4m or n 4m + 1.
28. We consider an amazing proof due to S. W. Golomb and R. I. Jewett. Suppose we have a fault-free 6 × 6 square. Notice that each tile breaks exactly one potential fault-line. Furthermore (and this is the crucial observation), if any fault-line (say L in Fig. 2.24) is broken by just a single tile, then the remaining regions on either side of it must have an odd area, since they consist of 6 × t rectangles with a single unit square removed. However, such regions are impossible to tile by dominoes. Thus each of the 10 potential fault-lines must be broken by at least two tiles.
Fig. 2.24 L Since no tile can break more than one fault-line, then at least 20 tiles will be needed for the tiling. But the area of the 6 × 6 square is only 36 whereas the area of the 20 tiles is 40. Contradiction! No such tiling of the 6 × 6 square can exist.
Remark: A p × q rectangle can be tiled fault-free by dominoes iff the following conditions hold: (1) pq is even. (2) p ≥5, q ≥5. (3) (p, q) ̸ (6, 6).
29. a1a2 . . . a25 b1b2 . . . b25 product of all elements of the matrix. Let a1 + b1 + a2 + b2 + . . . + a25 + b25 0. To cancel, there must be the same number of positive and negative summands. If among the ai there are n negative terms, then among the bj there are 25 −n negative terms. The numbers n and 25 −n have different parity.
Hence the products a1 . . . a25 and b1 . . . b25 have different signs and cannot be equal.
Contradiction.
30. The 6 × 6 × 6 cube consists of 27 subcubes of dimensions 2 × 2 × 2. Color them alternately black and white as a chessboard. Then 14 subcubes will be colored black and 13 white, that is, there will be 112 black and 104 white unit cubes. Any 1×1×4 brick will use up 2 black and 2 white unit cubes. 53 bricks will use up 106 white unit cubes. But there are only 104 white unit cubes.
31. No! After each hit, the orientation of the triangle ABC changes.
32. Suppose no 1 × 1 tile is needed. Color the rows of the square alternately black and white. There will be 23 more black than white unit squares. A 2×2 tile covers equally many black and white unit squares. A 3 × 3 tile covers three more unit squares of one color than the other. Hence the difference of the number of black and white unit 36 2. Coloring Proofs squares is divisible by 3. But 23 is not divisible by 3. Hence the assumption is false.
So at least one 1 × 1 tile is needed. By actual construction, we prove that one 1 × 1 tile is also sufficient. Put the 1 × 1 tile into the center and split the remaining board into four 12 × 11 rectangles. Each 12 × 11 rectangle can be tiled with a row of six 2 × 2 and three rows of 3 × 3 tiles, each consisting of four tiles.
33. No! On the walk, vertices and centers of faces are alternating, but a cube has 8 vertices and 6 faces. This is exactly problem 11.
a b a b a b c d c d c d d c d c d c b a b a b a Fig. 2.25 34. Color the board with four colors a, b, c, d, as in Fig. 2.25. Every a-cell must be preceded and followed by a c-cell. There are equally many a- and c-cells, and all must lie on any closed tour. To get all of them, we must avoid the b- and c-cells altogether. Once a jump is made from a c-cell to a d-cell there is no way to get back to an a-cell without first landing on another c-cell. The existence of a closed tour would imply that there are more c-cells than a-cells. Contradiction! There exist eight open tours of a 4 × 3 board. Find all of them.
35. Consider a regular hexagon together with its center.
36. Inscribe a regular icosahedron into the sphere. Start coloring the triangles of its faces in two colors. No matter how you do it, there will be regular triples of vertices at distance 2 (along the edges) colored with the same color.
37. Suppose m and n are both even. We color every second vertical strip. An L–tromino cannot be placed on the remaining squares. We prove that it is not possible to use a smaller number of colorings. Indeed, we can partition the rectangle into mn/4 squares of size 2 × 2. We must color at least two cells in each such square. The answer is mn/2.
Supposenis even andmodd.Wecoloreverysecondstripinthe odd direction, starting with the second. We prove that a smaller number of colorings is not sufficient. Indeed, from such a rectangle we may cut out n(m −1)/4 squares of size 2 × 2, in each of which we must color at least two cells. The answer in this case is n(m −1)/2.
Suppose n and m are both odd and n ≥m. Since both directions are odd we take the one giving largest economy of colored cells. So we color (m −1)/2 strips of size 1×n. We prove that we cannot get by with less colorings. It is sufficient to reduce the problem to a smaller rectangle. Cut off a big L leaving an (n −2)(m −2) rectangle.
The big L can be cut into (m + n −6)/2 squares of size 2 × 2 and one 3 × 3 square with one missing corner cell, i.e., a small L. We must color at least m + n −6 cells in the squares and at least three cells in the small L. By induction, we get the answer n(m −1)/2.
38. Suppose m and n are two white numbers. We will prove that mn is white. Suppose k is some black number. Then m + k is black, that is, mn + kn (m + k)n is white, and kn is white. If mn is black, then mn + kn is black. This contradiction proves that mn is white.
2. Coloring Proofs 37 Suppose k is the smallest white number. From the preceding result, we conclude that all multiples of k are also white. We prove that there are no other white numbers.
Suppose n is white. Represent n in the form qk + r, where 0 ≤r < k. If r ̸ 0, then r is black since k is the smallest white number. But we have proved that qk is white. Hence, qk + r is black. This contradiction proves that the white numbers are all multiples of some k > 1.
3 The Extremal Principle A successful research mathematician has mastered a dozen general heuristic princi-ples of large scope and simplicity, which he/she applies over and over again. These principles are not tied to any subject but are applicable in all branches of math-ematics. He usually does not reflect about them but knows them subconsciously.
One of these principles, the invariance principle was discussed in Chapter I. It is applicable whenever a transformation is given or can be introduced. If you have a transformation, look for an invariant! In this chapter we discuss the extremal principle, which has truly universal applicability, but is not so easy to recognize, and therefore must be trained. It is also called the variational method, and soon we will see why. It often leads to extremely short proofs.
We are trying to prove the existence of an object with certain properties. The extremal principle tells us to pick an object which maximizes or minimizes some function.Theresultingobjectisthenshowntohavethedesiredpropertybyshowing that a slight perturbation (variation) would further increase or decrease the given function. If there are several optimizing objects, then it is usually immaterial which one we use. In addition, the extremal principle is mostly constructive, giving an algorithm for constructing the object.
We will learn the use of the extremal principle by solving 17 examples from geometry, graph theory, combinatorics, and number theory, but first we will remind the reader of three well known facts: (a) Every finite nonempty set A of nonnegative integers or real numbers has a minimal element min A and a maximal element max A, which need not be unique.
40 3. The Extremal Principle (b) Every nonempty subset of positive integers has a smallest element. This is called the well ordering principle, and it is equivalent to the principle of mathematical induction.
(c) An infinite set A of real numbers need not have a minimal or maximal element. If A is bounded above, then it has a smallest upper bound sup A.
Read: supremum of A. If A is bounded below, then it has a largest lower bound inf A. Read: infimum of A. If sup A ∈A, then sup A max A, and if inf A ∈A, then inf A min A.
E1. (a) Into how many parts at most is a plane cut by n lines? (b) Into how many parts is space divided by n planes in general position?
Solution. We denote the numbers in (a) and (b) by pn and sn, respectively. A beginner will solve these problems recursively, by finding pn+1 f (pn) and sn+1 g(sn). Indeed, by adding to n lines (planes) another line (plane) we easily get pn+1 pn + n + 1, sn+1 sn + pn.
There is nothing wrong with this approach since recursion is a fundamental idea of large scope and applicability, as we will see later. An experienced problem solver might try to solve the problems in his head.
In (a) we have a counting problem. A fundamental counting principle is one-to-one correspondence. The first question is: Can I map the pn parts of the plane bijectively onto a set which is easy to count? The n 2 intersection points of the n lines are easy to count. But each intersection point is the deepest point of exactly one part. (Extremal principle!) Hence there are n 2 parts with a deepest point. The parts without deepest points are not bounded below, and they cut a horizontal line h (which we introduce) into n + 1 pieces (Fig. 3.1). The parts can be uniquely assigned to these pieces. Thus there are n + 1, or n 0 + n 1 parts without a deepest point. So there are altogether pn n 0 + n 1 + n 2 parts of the plane.
(b) Three planes form a vertex in space. There are n 3 vertices, and each is a deepest point of exactly one part of space. Thus there are n 3 parts with a deepest point.
Each part without a deepest point intersects a horizonal plane h in one of pn plane parts. So the number of space parts is sn n 0 + n 1 + n 2 + n 3 .
% % % % % % C C C C C C l l l l l l X X X X X X X X h Fig. 3.1 aaaa a H H A A A B C D Fig. 3.2 @ @ J J J B D A C r E ϵ Fig. 3.3 3. The Extremal Principle 41 E2. Continuation of 1b. Let n ≥5. Show that, among the sn space parts, there are at least (2n −3)/4 tetrahedra (HMO 1973).
Telling the result simplifies the problem considerably. An experienced problem-solver can often infer the road to the solution from the result.
Let tn be the number of tetrahedra among the sn space parts. We want to show that tn ≥(2n −3)/4.
Interpretation of the numerator: On each of the n planes rest at least two tetra-hedra. Only one tetrahedron need rest on each of three exceptional planes.
Interpretation of the denominator: Each tetrahedron is counted four times, once for each face. Hence, we must divide by four.
Using these guiding principles we can easily find a proof. Let ϵ be any of the n planes. It decomposes space into two half-spaces H1 and H2. At least one half-space, e.g., H1 , contains vertices. In H1, we choose a vertex D with smallest distance from ϵ (extremal principle). D is the intersection point of the planes ϵ1, ϵ2, ϵ3. Then ϵ, ϵ1, ϵ2, ϵ3 define a tetrahedron T ABCD (Fig. 3.2). None of the remaining n −4 planes cuts T , so that T is one of the parts, defined by the n planes. If the plane ϵ′ would cut the tetrahedron T , then ϵ′ would have to cut at least one of the edges AD, BD, CD in a point Q having an even smaller distance from ϵ than D. Contradiction.
This is valid for any of the n planes. If there are vertices on both sides of a plane, at least two tetrahedra then must rest on this plane.
It remains to be shown that among the n planes there are at most three, so that all vertices lie on the same side of these planes.
We show this by contradiction. Suppose there are four such planes ϵ1, ϵ2, ϵ3, ϵ4. They delimit a tetrahedron ABCD (Fig. 3.3). Since n ≥5, there is another plane ϵ. It cannot intersect all six edges of the tetrahedron ABCD simultaneously.
Suppose it cuts the continuation of AB in E. Then B and E lie on different sides of the plane ϵ3 ACD. Contradiction!
E3. There are n points given in the plane. Any three of the points form a triangle of area ≤1. Show that all n points lie in a triangle of area ≤4.
Solution. Among all n 3 triples of points, we choose a triple A, B, C so that △ABC has maximal area F. Obviously F ≤1. Draw parallels to the opposite sides through A, B, C. You get △A1B1C1 with area F1 4F ≤4. We will show that △A1B1C1 contains all n points.
Suppose there is a point P outside △A1B1C1. Then △ABC and P lie on different sides of at least one of the lines A1B1, B1C1, C1A1. Suppose they lie on different sides of B1C1. Then △BCP has a larger area than △ABC. This contradicts the maximality assumption about ABC (Fig. 3.4).
E4. 2n points are given in the plane, no three collinear. Exactly n of these points are farms F {F1, F2, . . . , Fn}. The remaining n points are wells: W {W1, W2, . . . , Wn}. It is intended to build a straight line road from each 42 3. The Extremal Principle farm to one well. Show that the wells can be assigned bijectively to the farms, so that none of the roads intersect.
@ @ @ @ @ @ @ @ @ @ @ A1 B1 C1 C A B P Fig. 3.4 c c c c c Wn Wm Fi Fk Fig. 3.5 QQQQQ Q B A D C A′ Fig. 3.6 Solution. We consider any bijection: f : F →W. If we draw from each Fi a straight line to f (Fi), we get a road system. Among all n! road systems, we choose one of minimal total length.Suppose this system has intersecting segments FiWm and FkWn (Fig. 3.5). Replacing these segments by FkWm and FiWn, the total road length becomes shorter because of the triangle inequality. Thus it has no intersecting roads.
E5. Let be a set of points in the plane. Each point in is a midpoint of two points in . Show that is an infinite set.
First proof. Suppose is a finite set. Then contains two points A, B with maximal distance |AB| m. B is a midpoint of some segment CD with C, D ∈.
Fig. 3.6 shows that |AC| > |AB| or |AD| > |AB|.
Second proof. We consider all points in farthest to the left, and among those the point M farthest down. M cannot be a midpoint of two points A, B ∈ since one element of {A, B} would be either left of M or on the vertical below M.
E6. In each convex pentagon, we can choose three diagonals from which a triangle can be constructed.
Solution. Fig. 3.7 shows a convex pentagon ABCDE. Let BE be the longest of the diagonals. The triangle inequality implies |BD| + |CE| > |BE| + |CD| > |BE|, that is, we can construct a triangle from BE, BD, CE.
A A A X X X B B B Fig. 3.7 A B C D E E7. In every tetrahedron, there are three edges meeting at the same vertex from which a triangle can be constructed.
Solution. Let AB be the longest edge of the tetrahedron ABCD. Since (|AC| + |AD|−|AB|)+(|BC|+|BD|−|BA|) (|AD|+|BD|−|AB|)+(|AC|+|BC|− 3. The Extremal Principle 43 |AB|) > 0 then, either |AC| + |AD| −|AB| > 0, or |BC| + |BD| −|BA| > 0.
In each case, we can construct a triangle from the edges at some vertex.
E8. Each lattice point of the plane is labeled by a positive integer. Each of these numbers is the arithmetic mean of its four neighbors (above, below, left, right).
Show that all the labels are equal.
Solution. We consider a smallest label m. Let L be a lattice point labeled by m.
Its neighbors are labeled by a, b, c, d. Then m (a + b + c + d)/4, or a + b + c + d 4m.
(1) Now a ≥m, b ≥m, c ≥m, d ≥m. If any of these inequalities would be strict, we would have a +b +c +d > 4m which contradicts (1). Thus a b c d m.
It follows from this that all labels are equal to m.
This is a very simple problem. By replacing positive integers by positive reals, it becomes a very difficult problem. The trouble is that positive reals need not have a smallest element. For positive integers, this is assured by the well ordering principle. The theorem is still valid, but I do not know an elementary solution.
E9. There is no quadruple of positive integers (x, y, z, u) satisfying x2 + y2 3(z2 + u2).
Solution. Suppose there is such a quadruple. We choose the solution with the smallest x2 + y2. Let (a, b, c, d) be the chosen solution. Then a2 + b2 3(c2 + d2) ⇒3|a2 + b2 ⇒3|a, 3|b ⇒a 3a1, b 3b1, a2 + b2 9(a2 1 + b2 1) 3(c2 + d2) ⇒c2 + d2 3(a2 1 + b2 1).
We have found a new solution (c, d, a1, b1) with c2 +d2 < a2 +b2. Contradiction.
We have used the fact that 3|a2 + b2 ⇒3|a, 3|b. Show this yourself. We will return to similar examples when treating infinite descent.
E10. The Sylvester Problem, posed by Sylvester in 1893, was solved by T. Gallai 1933 in a very complicated way and by L.M. Kelly in 1948 in a few lines with the extremal principle.
A finite set S of points in the plane has the property that any line through two of them passes through a third. Show that all the points lie on a line.
Solution. Suppose the points are not collinear. Among pairs (p, L) consisting of a line L and a point not on that line, choose one which minimizes the distance d from p to L. Let f be the foot of the perpendicular from p to L. There are (by assumption) at least three points a, b, c on L. Hence two of these, say, a and b are on the same side of f (Fig. 3.8). Let b be nearer to f than a. Then the distance from b to the line ap is less than d. Contradiction.
44 3. The Extremal Principle HHHHHHHHHH H f b t p t d a t L c t Fig. 3.8 H H H ' & $ % Fig. 3.9 X D M t 1 2 t tE t t t > Z Z Z } A A A A A Z Z Z Z Z } t m E11. Every road in Sikinia is one-way. Every pair of cities is connected exactly by one direct road. Show that there exists a city which can be reached from every city directly or via at most one other city.
Solution. Let m be the maximum number of direct roads leading into any city, and let M be a city for which this maximum is attained. Let D be the set of m cities with direct connections into M. Let R be the set of all cities apart from M and the cities in D. If R ∅, the theorem is valid. If X ∈R, then there is an E ∈D with connection X →E →M. If such an E did not exist, then X could be reached directly from all cities in D and from M, that is, m + 1 roads would lead into X, which contradicts the assumption about M. Thus, every city with the maximum number of entering roads satisfies the conditions of the problem (Fig. 3.9).
E12. Rooks on an n × n × n chessboard. Obviously n is the smallest number of rooks which can dominate an n × n chessboard. But what is the number Rn of rooks, which can dominate an n × n × n-chessboard?
Solution. We try to guess the result for small values of n. But first we need a good representation for placing rooks in space. We place n layers of size n × n × 1 over an n × n square, and we number them 1, 2, . . . , n. Each rook is labeled with the number of the layer on which it is located. Fig. 3.10 suggests the conjecture Rn n2 2 : n ≡0 mod 2, n2+1 2 : n ≡1 mod 2.
1 T1 1 1 2 T2 2 1 2 3 3 2 T3 5 1 2 2 1 3 4 3 4 T4 8 1 2 1 2 3 4 5 4 5 3 5 3 4 T5 13 Fig. 3.10 Now comes the proof. Suppose R rooks are so placed on the n3 cubes of the board, that they dominate all cubes. We choose a layer L, which contains the minimum number of rooks. We may assume that it is parallel to the x1x2-plane.
Suppose that L contains t rooks. Suppose these t rooks dominate t1 rows in the 3. The Extremal Principle 45 x1-direction and t2 rows in the x2-direction. We may further assume that t1 ≥ t2. Obviously t ≥t1 and t ≥t2. In the layer L, these rooks fail to dominate (n −t1)(n −t2) cubes, which must be dominated in the x3-direction. We consider all n layers parallel to the x1x3-plane. In n−t1 of these not containing a rook from L, there must be at least (n −t1)(n −t2) rooks. In each of the remaining t1 layers are at least t rooks (by the choice of t). Hence, we have R ≥(n −t1)(n −t2) + tt1 ≥(n −t1)2 + t2 1 n2 2 + (2t1 −n)2 2 .
The right side assumes its minimum n2/2 for even n and (n2 + 1)/2 for odd n. It is easy to see that this necessary number is also sufficient. Fig. 3.11 gives a hint for a proof (MMO 1965, AUO 1971, IMO 1971).
Remark. The exact number of rooks which dominate an n × n × n × n board and other higher dimensional boards does not seem to be known. Here good bounds would be welcome.
1 2 3 2 3 1 3 1 2 4 5 6 7 5 6 7 4 6 7 4 5 7 4 5 6 Fig. 3.11 1 2 3 4 2 3 4 1 3 4 1 2 4 1 2 3 5 6 7 8 6 7 8 5 7 8 5 6 8 5 6 7 E13. Seven dwarfs are sitting around a circular table. There is a cup in front of each. There is milk in some cups, altogether 3 liters. One of the dwarfs shares his milk uniformly with the other cups. Proceeding counter-clockwise, each of the other dwarfs, in turn, does the same. After the seventh dwarf has shared his milk, the initial content of each cup is restored. Find the initial amount of milk in each cup (AUO 1977, grade 8).
Solution. Every 8th grader, 53 altogether, guessed the correct answer 6/7, 5/7, 4/7, 3/7, 2/7, 1/7, 0 liters. The answer is easy to guess because of an invariance property. Each sharing operation merely rotates the answer. But only 9 students could prove that the answer is unique. The solutions were quite ingenious and required just a few lines. We prefer, instead, a solution based on a general principle, in this case, the extremal principle.
Suppose the dwarf #i has the (maximal) amount xi before starting to share his milk. The dwarf Max has the maximum amount x to share. The others to the right of him have x1, x2, . . ., x6 to share. Max gets xi/6 from dwarf #i. Thus, we have x x1 + x2 + x3 + x4 + x5 + x6 6 , (1) 46 3. The Extremal Principle where xi ≤x for i 1, . . . , 6. If the inequality would be strict only once, we could not have equality in (1). Thus x1 x2 x3 x4 x5 x6 x, that is, each dwarf shares the same amount of milk. We easily infer from this that, initially, the milk distribution is 0, x/6, 2x/6, 3x/6,4x/6, 5x/6, 6x/6. From the sum 3 liters, we get x 6/7.
E14. The Sikinian Parliament consists of one house. Every member has three enemies at most among the remaining members. Show that one can split the house into two houses so that every member has one enemy at most in his house.
Solution. We consider all partitions of the Parliament into two houses and, for each partition, we count the total number E of enemies each member has in his house. The partition with minimal E has the required property. Indeed, if some member would have at least two enemies in his house, then he would have one enemy at most in the other house. By placing him in the other house, we could decrease the minimal E, which is a contradiction.
We have solved this problem already in Chapter 1 by a variation of the invariance principle which we call the Principle of the Finiteness of a Decreasing Sequence of Nonnegative Integers. So the Extremal Principle is related to the Invariance Principle.
E15. Can you choose 1983 pairwise distinct positive integers < 100000, such that no three are in arithmetic progression (IMO 1983)?
All hints to the solution are eliminated in this problem. So we must recover them.
We need some strategic idea to get the first clues. Let us construct a tight sequence with no three terms in arithmetic progression. Here, the extremal principle helps in finding an algorithm. We use the so-called greedy algorithm: Start with the smallest nonnegative integer 0. At each step, add the smallest integer which is not in arithmetic progression with two preceding terms. We get • 0, 1 (translate this by 3), • 0, 1, 3, 4 (translate this by 9), • 0, 1, 3, 4, 9, 10, 12, 13 (translate this by 27), and • 0, 1, 3, 4, 9, 10, 12, 13, 27, 28, 30, 31, 36, 37, 39, 40 (translate this by 81).
We get a sequence with many regularities. The powers of 3 are a hint to use the ternary system. So we rewrite the sequence in the ternary system, getting 0, 1, 10, 11, 100, 101, 110, 111, 1000, . . . .
This is a hint to the binary system. We conjecture that the constructed sequence consists of those ternary numbers, which miss the digit 2, i.e., they are written in the binary system. Our next conjecture is that if we read the terms of the sequence 3. The Extremal Principle 47 an in the binary system, we get n. Read in the ternary system, we get an. The solution to our problem is a1983 a111101111112 111101111113 87844.
It is quite easy to finish the problem. Five of our six team members gave this answer, probably, because in training I briefly treated the greedy algorithm as a construction principle for good but not necessarily optimal solutions. This is one of the innumerable versions of the Extremal Principle.
E16. There exist three consecutive vertices A, B, C in every convex n-gon with n ≥3, such that the circumcircle of △ABC covers the whole n-gon.
Among the finitely many circles through three vertices of the n-gon, there is a maximal circle. Now we split the problem into two parts: (a) the maximal circle covers the n-gon, and (b) the maximal circle passes through three consecutive vertices.
We prove (a) indirectly. Suppose the point A′ lies outside the maximal circle about △ABC where A, B, C are denoted such that A, B, C, A′ are vertices of a convex quadrilateral. Then the circumcircle of △A′BC has a larger radius then that of △ABC. Contradiction.
We also prove (b) indirectly. Let A, B, C be vertices on the maximal circle, and let A′ lie between B and C and not on the maximal circle. Because of (a), it lies inside that circle, but then the circle about △A′BC is larger than the maximal circumcircle. Contradiction.
E17. n √ 2 is not an integer for any positive integer n.
We use a proof method of wide applicability based on the extremal principle.
Let S be the set of those positive integers n, for which n √ 2 is an integer. If S is not empty, it would have a least element k. Consider ( √ 2 −1)k. Then ( √ 2 −1)k √ 2 2k −k √ 2, and, since k ∈S, both ( √ 2 −1)k and 2k −k √ 2 are positive integers. So, by definition, ( √ 2 −1)k ∈S. But ( √ 2 −1)k < k, contradicting the assumption that k is the least element of S. Hence S is empty, which means that √ 2 is irrational.
Problems 1. Prove that there are at least (2n −2)/3 triangles among the pn parts of the plane in Example #1.
2. In the plane, n lines are given (n ≥3), no two of them parallel. Through every intersection of two lines there passes at least an additional line. Prove that all lines pass through one point.
48 3. The Extremal Principle 3. If n points of the plane do not lie on the same line, then there exists a line passing through exactly two points.
4. Start with several piles of chips. Two players move alternately. A move consists in splitting every pile with more than one chip into two piles. The one who makes the last move wins. For what initial conditions does the first player win and what is his winning strategy?
5. Does there exist a tetrahedron, so that every edge is the side of an obtuse angle of a face?
6. Prove that every convex polyhedron has at least two faces with the same number of sides.
7. (2n + 1) persons are placed in the plane so that their mutual distances are different.
Then everybody shoots his nearest neighbor. Prove that (a) at least one person survives; (b) nobody is hit by more then five bullets; (c) the paths of the bullets do not cross; d) the set of segments formed by the bullet paths does not contain a closed polygon.
8. Rooks are placed on the n × n chessboard satisfying the following condition: If the square (i, j) is free, then at least n rooks are on the ith row and jth column together.
Show that there are at least n2/2 rooks on the board.
9. All plane sections of a solid are circles. Prove that the solid is a ball.
10. A closed and bounded figure with the following property is given in a plane: Any two points of can be connected by a half circle lying completely in . Find the figure (West German proposal for IMO 1977).
11. Of n points in space, no four lie in a plane. Some of the points are connected by lines. We get a graph G with k edges.
(a) If G does not contain a triangle, then k ≤⌊n2/4⌋.
(b) If G does not contain a tetrahedron, then k ≤⌊n2/3⌋.
12. There are 20 countries on a planet. Among any three of these countries, there are always two with no diplomatic relations. Prove that there are at most 200 embassies on this planet.
13. Every participant of a tournament plays with every other participant exactly once.
No game is a draw. After the tounament, every player makes a list with the names of all players, who (a) were beaten by him and (b) were beaten by the players beaten by him.
Prove that the list of some player contains the names of all other players.
14. Let O be the point of intersection of the diagonals of the convex quadrilateral ABCD.
Prove that, if the perimeters of the triangles ABO, BCO, CDO and DAO are equal, then ABCD is a rhombus.
15. There are n identical cars on a circular track. Together they have just enough gas for one car to complete a lap. Show that there is a car which can complete a lap by collecting gas from the other cars on its way around.
16. Let M be the largest distance among six distinct points of the plane, and let m be the smallest of their mutual distances. Prove that M/m ≥ √ 3.
17. A cube cannot be divided into several pairwise distinct cubes.
3. The Extremal Principle 49 18. In space, several planets with unit radius are given. We mark on the surface of each planet all those points from which none of the other planets are visible. Prove that the sum of the areas of all marked points is equal to the surface of one planet.
19. In a plane, 1994 vectors are drawn. Two players alternately take a vector until no vectors are left. The loser is the one whose vector sum has the smaller length. Can the first player choose a strategy so that he does not lose?
20. Any two of a finite number of (not necessarily convex) polygons have a common point. Prove that there is a line which has a common point with all these polygons.
21. Any convex polygon of area 1 is contained in a rectangle of area 2.
22. n ≥3 points, which are not all collinear are given in a plane. Show that there exists a circle passing through three of the points, the interior of which does not contain any of the remaining points.
23. Take the points A1, B1, C1, respectively on the sides AB, BC, CA of the triangle ABC. Show that if |AA1| ≤1, |BB1| ≤1, |CC1| ≤1, then the area of the triangle is ≤1/ √ 3.
24. Of 2n + 3 points of a plane, no three are collinear and no four lie on a circle. Prove that we can choose three of the points and draw a circle through these points, so that exactly n of the remaining 2n points lie inside this circle and n outside. (ChNO.) 25. Consider a walk in the plane according to the following rules. From a given point P(x, y) we may move in one step to one of the four points U(x, y+2x), D(x, y−2x), L(x −2y, y), R(x + 2y, y) with the restriction that we cannot retrace a step we just made. Prove that, if we start from the point (1, √ 2), we cannot return to this point any more (HMO 1990).
26. Solve E8 of Chapter 1 with the extremal principle.
27. Among any 15 coprime positive integers > 1 and ≤1992, there is at least one prime.
28. Eight points are chosen inside a circle of radius 1. Prove that there are two points with distance less than 1.
29. n points are given in a plane. We label the midpoints of all segments with endpoints in these n points. Prove that there are at least (2n −3) distinct labeled points.
30. The base of the pyramid A1 · · · AnS is a regular n-gon A1 · · · An with side a. Prove that ̸ SA1A2 · · · ̸ SAnA1 implies that the pyramid is regular.
31. On a sphere, there are five disjoint and closed spherical caps, each less than one-half of the surface of the sphere. Prove that there exist on the sphere two diametrically opposite points, which are not covered by any cap.
32. Find all positive solutions of the system of equations x1 + x2 x2 3, x2 + x3 x2 4, x3 + x4 x2 5, x4 + x5 x2 1, x5 + x1 x2 2.
33. Find all real solutions of the system (x + y)3 z, (y + z)3 x, (z + x)3 y.
34. Let E be a finite set of points in 3-space with the following properties: (a) E is not coplanar.
(b) No three points of E are collinear.
Prove: Either there are five points in E, which are vertices of a convex pyramid the interior of which is free of points of E, or there exists a plane, which contains exactly three points of E.
50 3. The Extremal Principle 35. Six circles have a common point A. Prove that there is one among these circles which contains the center of another circle.
36. We choose n points on a circle and draw all chords joining these n points. Find the number of parts into which the circular disk is cut.
37. Each of 30 students in a class has the same number of friends among his class mates.
What is the highest possible number of students, who learn better than the majority of their friends? Of any two students one can tell which one is better (RO 1994).
38. A set S of persons has the following property. Any two with the same number of friends in S have no common friends in S. Prove that there is a person in S with exactly one friend in S.
39. The sum of several nonnegative reals is 3, and the sum of their squares is > 1. Prove that you may choose three of these numbers with sum > 1.
40. Several positive reals are written on paper. The sum of their pairwise products is 1.
Prove that you can cross out one number, so that the sum of the remaining numbers is less than √ 2.
41. m chips (m > n) are placed at the vertices of a convex n-gon. In one move, two chips at a vertex are moved in opposite directions to neighboring vertices. Prove that, if the original distribution is restored after some moves, then the number of moves is a multiple of n.
42. It is known that the numbers a1, . . . , an and b1, . . . , bn are both permutations of 1, 1/2, . . . , 1/n. In addition, we know that a1 + b1 ≥a2 + b2 ≥· · · ≥an + bn.
Prove that am + bm ≤4/m for all m from 1 to n.
43. Fifty segments are given on a line. Prove that some eight of the segments have a common point, or eight of the segments are pairwise disjoint (AUO 1972).
44. There are n students in each of three schools. Any student has altogether n + 1 acquaintances from the other two schools. Prove that one can select one student from each school, so that the three selected students know each other.
Solutions 1. Use the ideas of E2, which treats the more complicated space analogue.
2. Suppose not all lines pass through one point. We consider all intersection points, and we choose the smallest of the distances from these points to the lines. Suppose the smallest distance is from the point A to the line l. At least three lines pass through A. They intersect l in B, C, D. From A drop the perpendicular AP to l. Two of the points B, C, D lie on the same side of P . Suppose these are C and D. Suppose |CP| < |DP |. Then the distance from C to AD is smaller than the distance from A to l, contradicting the choice of A and l. (This argument is exactly the one used by L.M. Kelly.) 3. Again, this is a variation of Sylvester’s problem.
4. It is my move. It all depends on the largest pile. Suppose it contains M chips. As long as M > 1, I can move. Trying small numbers shows that I must occupy the 3. The Extremal Principle 51 position M 2k −1. No matter how my opponent splits the piles, he must leave a position with 2k−1 −1 < M < 2k −1.
On my next move, I can occupy the position M 2k−1 −1. If I continue in this way, I will finally move to M 21 −1 1, and my opponent has lost since he cannot move. So the first player wins if, initially, M does not have the form 2k −1.
5. Suppose AB is the longest edge of a face ABC. Then the angle at C is at least as large as those at A and B. Hence the angles at A and B are acute.
6. Let F be the face with the largest number m of edges. Then, for the m + 1 faces consisting of F and its m neighbors, there are only the possibilities 3, 4, . . . , m as the number of edges. These are only m −2 possibilities. Thus, at least one number of edges must occur more than once.
7. (a) All mutual distances are different. Hence there exist two persons A and B with minimum distance. These two persons will shoot each other. If any other person shoots at A or B, someone will survive since A and B have used up three bullets. If not, we can ignore A and B. We are left with the same problem with n replaced by n −1. Repeating the argument, we either find a pair at whom three shots are fired, or if not, we arrive, finally, at three persons, and for this case (n 1), the theorem is obvious.
> A A A K @ @ @ R = P D C A B Fig. 3.12 C C C C C C A D S B C Fig. 3.13 @ @ A A A M N A B C D E Fig. 3.14 (b) Suppose the persons A, B, C, D, . . . shoot at P (Fig. 3.12). A shoots at P and not at B, so |AP| < |AB|. B shoots at P and not at A, so |BP| < |AB|. Thus, AB is the largest side in the triangle ABP . The largest angle lies opposite the largest side.
Hence, γ > α, γ > β or 2γ > α + β, 3γ > α + β + γ , γ > 60◦. Thus any two bullet paths meeting at P make an angle greater than 60◦. Since 6 × 60◦ 360◦, five bullet paths at most can meet at P .
(c) Suppose the paths of two bullets cross with A shooting at B and C shooting at D (Fig. 3.13). Then |AB| < |AD| and |CD| < |CB| imply |AB| + |CD| < |AD| + |CB|. On the other hand, by the triangle inequality, |AS| + |SD| > |AD| and |BS| + |SC| > |BC| ⇒|AB| + |CD| > |AD| + |BC|. Contradiction!
(d) Suppose there is a closed polygon ABCDE . . . MN (Fig. 3.14). Let |AN| < |AB|, that is, N is the nearest neighbor of A. Then |AB| < |BC|, |BC| < |CD|, |CD| < |DE|, . . ., |MN| < |NA|, that is, |AB| < |NA|. Contradiction! The assumption |AN| > |AB| also leads to a contradiction.
8. Among the 2n rows and columns, we choose one with the least number of rooks.
Suppose it is a row. Suppose k is the number of rooks in this row. If k ≥n/2, then each row has at least n/2 rooks, and there are at least n2/2 rooks on the board.
52 3. The Extremal Principle A A A A A A PPPPPP P D A C M t Z E B Fig. 3.15 q Suppose k < n/2. There are at least n −k free squares in this row, and there are at least (n −k)2 rooks in all columns through a free square. The remaining k columns have each at least k rooks. Hence on the board, there are at least (n −k)2 + k2 rooks. We must show that this is greater than or equal to n2/2. But (n −k)2 + k2 n2 2 + (n −2k)2 2 ≥n2/2 if n is even, ≥(n2 + 1)/2 if n is odd.
Existence. If n is even, we occupy the black squares with n2/2 rooks. If n is odd, there are (n2 + 1)/2 squares which have the same color as the four corner squares.
We occupy the squares of the same color with rooks.
9. The shortest proof runs as follows. Consider the largest chord of the solid. Any section through this chord is a circle whose diameter is the chord. Otherwise the circle and the solid would have a larger chord. Thus the solid is a ball and one of its diameters is the selected chord.
This proof is not complete. We did not prove that a longest chord exists. In fact, if the surface of the solid did not belong to the solid, a longest chord would not exist.
So we assume that the solid is a closed and bounded set. Then we can apply the theorem of Weierstraß: A continuous function defined on a closed and bounded set always assumes its global maximum and minimum.
This theorem belongs to higher mathematics, but at the IMO you can use it. The proof is not considered to have a gap if you cite the theorem. There are also elementary proofs which are slightly longer (see HMO 1954).
10. We choose two points A, B in with maximum distance and draw the circle C with diameter AB and midpoint M. We will prove that is the disk with boundary C.
The line AB partitions C into two semicircular arcs Cr and Cl (Fig. 3.15). Now Cr ⊂ or Cl ⊂. Suppose Cr ⊂. A point X left of AB and outside of C cannot belong to . Indeed, XM intersects Cr in Y. Then |XY| > |AB|. For a point U to the right of AB and outside one of the circles about A and B with radius |AB| we have |AU| > |AB| or |BU| > |AB|. Hence the area outside AEBDA in Fig. 3.15 does not contain points of .
Now we choose any point Z inside C and draw the segment AZ. The perpendicular to AZ in Z intersects Cr in C and Cl in D. (C and D cannot both lie on Cr or on Cl. Why?) The semicircular arc over AC not through Z does not completely lie in , since the tangent to C in A is a secant of this semicircular arc and intersects 3. The Extremal Principle 53 it in A and also in F. The arc bounded by A and F lies outside AEBDA. Thus the semicircular arc over AC through Z lies completely in . Hence Z ∈. This implies that every interior point of C lies in . Since is closed, C ⊂. No point of can lie outside of C, since this would contradict the maximality of |AB|.
11. (a) We choose a point p joined with a maximum number m of other points. Then all points are partitioned into two sets A {p1, . . . , pm} and B {p, q1, . . . , qn−m−1}.
A consists of the points joined to p. Any two points in A are not joined since G has no triangle. In B are the points not joined to p and p. For the total number of edges, we have k ≤m(n −m) n2 4 − n 2 −m 2 ≤n2 4 .
We can get equality for even n, if m n/2. Otherwise m (n + 1)/2, and we get (n + 1)/2 and (n −1)/2 for the two partitions. (b) See chapter 8 on the induction principle.
12. This is problem 11a with n 20. Notice that two embassies belong to each pair of countries.
13. Let A be a participant who has won the maximum number of plays. If A would not have the property of the problem, then there would be another player B, who has won against A and against all players who were beaten by A. So B would have won more times than A. This contradicts the choice of A.
14. Let us suppose that |AO| ≥|BO| and |DO| ≥|BO|. Let B1 and C1 be the reflec-tions of B and C at O. Denote by P (XYZ) the perimeter of the triangle XYZ. Since the triangle B1OC1 lies inside the triangle AOD, we haveP (AOD) ≥P(B1OC1) P(BOC). There is equality only if B1 D and C1 A. HenceABCD is a paral-lelogram, |AB| −|BC| P(ABO) −P (BCO) 0, that is, ABCD is a rhombus.
15. An additional car with a sufficiently large tank starts somewhere on the circle. At each car, it buys up all the gas. At some point A, the level of gas in his tank is lowest.
Then A must be another car. The car in A is able to complete a round trip. Another solution uses induction (Chapter 8, problem 2).
16. Among six points in the plane, there are always three which form a triangle with maximum angle ≥120◦. For this triangle, the ratio of the longest to the shortest side is ≥ √ 3. This will be proved. Consider the convex hull of the six points. If it consists of a triangle ABC, then join any interior point D with A, B and C. One of the three angles at D is ≥120◦. If the convex hull is a quadrilateral ABCD, then any of the other two points E lies inside one of the triangles ABC and ADC. Suppose E lies inside ABC. Then one of the triangles EAB, EBC, ECA has an angle ≥120◦.
If the convex hull is a pentagon, then the sixth point F lies inside a triangle of the triangulation of the pentagon by the diagonals from one vertex. Suppose F lies inside △ACD. Join E to the vertices of ACD. One of the triangles EAC, ECD, EDA has an angle ≥120◦. If the six points are the vertices of a convex hexagon, then one of the interior angles is ≥120◦. If the inside point lies on a diagonal, then we can even do better. In that case, M : m ≥2 > √ 3. We have thus proved that there is a triangle with largest angle ≥120◦. In such a triangle, we assume α ≤β < γ . Then, c a sin γ sin α ≥ sin γ sin α+β 2 sin γ sin(90◦−γ 2 ) sin γ cos γ 2 2 sin γ 2 ≥2 sin 60◦ √ 3.
54 3. The Extremal Principle 17. Suppose the cube is dissected into a finite number of distinct cubes. Then its faces are dissected into squares. Choose the smallest of these squares. Turn the cube so that the face with the smallest square becomes the bottom. It is easy to see that the smallest square cannot lie at the boundary of the bottom. Thus it is the bottom of a “well” surrounded by larger cubes. To fill this well, we need still smaller cubes, and so on, until we reach the top face, which is dissected into still smaller squares.
Contradiction!
18. This is obviously true for two planets. Now suppose that O1, . . . , On are the centers of the planets. What do we need to prove? It is sufficient to prove that, for each unit vector ⃗ a, there is a unique point X on some planet #i, so that − − → OiX ⃗ a, from which none of the other planets is visible. We first prove that X is unique. Suppose − − → OiX − − → OjY and from X and Y no other planet is visible. But we have already considered the case of two planets. It showed that, if the planet number j is not visible from X, then the planet number i is visible from Y. Contradiction!
We prove the existence of the point X. We introduce a coordinate system with axis Ox in the direction of the vector ⃗ a. Then that point of the given planets with largest x-coordinate is the point X.
19. Suppose the sum of the 1994 vectors is ⃗ a. Introduce a coordinate system such that the axis Ox has the direction of the vector ⃗ a. If ⃗ a ⃗ o, then use any direction. At each move, the first player chooses the vector with largest abscissa. At the end, he will have an abscissa which is not smaller than that of his opponent. His ordinate will be the same as that of his opponent, since the sum of all ordinates will be 0.
Hence, the first player will not lose with this strategy.
20. Take any line g in a plane, and project all polygons onto g. We get several segments any two of which have a common point. Consider the left endpoints of these segments and, of these, the one farthest to the right. We get a point R belonging to all segments.
The perpendicular to g through R intersects all polygons.
21. Let AB be the largest diagonal or side of the polygon. Draw perpendiculars a, b to AB through A and B. Then the polygon lies completely in the convex domain bounded by the lines a and b. Indeed, let X be any vertex of the polygon. Then AX ≤AB and XB ≤AB. Enclose the polygon in the smallest rectangle KLMN with KL and MN having common points C and D with the polygon. |KLMN| 2|ABC| + 2|ABD| 2|ABCD|. Since the quadrilateral lies completely inside the convex polygon with area 1, we have |KLMN| ≤2.
22. Consider two of the points with minimal distance. Then there are no additional points inside the circle with diameter AB. Let C be one of the remaining points with maximal angle ̸ ACB. Then there are no points of the point set inside the circle through A, B, C. But they could all lie on the circle.
23. We may assume that ̸ α ≱ β ≱ γ . We consider two possibilities: (1) △ABC is acute, i.e., 60◦≰ α < 90◦. Since hb ≤|BB1| ≤1 and hc ≤|CC1| ≤ 1, we have |ABC| chc/2 hbhc/2 sin α ≤1/ √ 3. In fact, the sine is monotonic from 0◦up to 90◦.
(2) △ABC is not acute. Then α ≥90◦, |AB| ≤|BB1| ≤1, |AC| ≤|CC1| ≤1.
Hence, |ABC| ≤|AB| · |AC|/2 ≤1/2 < 1/ √ 3.
24. Take any two points A, B such that all the remaining points lie on the same side of the line AB. Order these points X1, X2, . . . X2n+1 so that ̸ AXiB > ̸ AXi+1B, for all 3. The Extremal Principle 55 i 1, . . . , 2n. Then the circle through A, Xn+1, B contains the points X1, . . . Xn.
The remaining n points lie outside this circle. No two points Xi lie on the same circle, or else we would have four points on a circle, which contradicts our basic assumption.
25. It is easy to verify that, if P is not on one of the lines x 0, y 0, y x, y −x, then exactly one of the four possible steps leads us closer to the origin O, whereas the other three lead us away from O. Since the ratio of P ′s coordinates is irrational at the start, the above rule remains valid during the whole walk.
Suppose that, after a series of steps P0P1 . . . Pn P0, we are back at the point P0(1, √ 2). If Pi is the farthest point of the closed path from O, then d(OPi−1) < d(OPi) > d(OPi+1), and thus the only possible step from Pi to the origin takes us back to Pi−1. This is a contradiction, since we are not allowed to retrace a step.
26. Consider all arrangements of the 2n ambassadors around the round table. Count the number of hostile pairs for each arrangement. Let H be the minimum of these numbers. Then H 0. Indeed, suppose H > 0. Then, applying one step of the reduction algorithm described in E8 of Chapter 1, we can further decrease this minimal value. Contradiction!
27. Suppose the 15 positive integers n1, . . . , n15 satisfy the conditions of the problem and are all composite. We denote by pi the smallest prime divisor of ni, and by p the largest of the pi. Because the numbers n1, . . . , n15 are coprime, the primes p1, . . . , p15 are all distinct. Hence p ≥47 (47 is the 15th prime). Hence for n, for which p is the smallest prime, we have n ≥p2 ≥472 > 1993. Contradiction! Here we used almost any problem just to show the ubiquity of the underlying extremal principle.
28. At least seven points are different from the center O of the circle. Hence the smallest of the angles ̸ AiOAj is at most 360◦/7 < 60◦. If A and B correspond to the smallest angle, then |AB| < 1, since |AO| ≤1, |BO| ≤1 and ̸ AOB cannot be the largest angle of △AOB.
29. Let A and B be two of the n points with largest distance. The midpoints of the segments connecting A (or B) with all the other points are all distinct, and they lie in the circle with radius |AB|/2 with center A(or B). We get two circles with one common point. Hence there are at least 2(n −1) −1 or 2n −3 distinct points.
30. Construct ̸ BAC α in a plane, where α ̸ SA1A2 · · · ̸ SAnA1, and |AB| a. Then, for each i 1, . . . , n, we construct the points Si on the ray AC such that △ASiB △ASiAi+1. Suppose not all points Si coincide, and let Sk be the nearest point to B and Sl be the point with largest distance from B. Since |SkSl| > |SkB −SlB|, we have |SkA−SlA| > |SkB −SlB|, i.e., |Sk−1B −Sl−1B| > |SkB −SlB|. But on the right side of this inequality is the difference between the largest and smallest number, and on the left side the difference between two numbers between them. Contradiction! Hence the points Si coincide, i.e., S is equidistant from the vertices A1, . . . , An of the base.
31. Consider a spot of greatest radius, and draw a concentric circle of a slightly larger radius and still not intersecting any of the other spots. Reflect the five spots in the center of the sphere. It is easy to see that the reflected spots will not cover the whole sphere. Any uncovered point of the sphere and its diametrically opposite point will suit.
56 3. The Extremal Principle 32. Let x and y be the largest and the smallest of the numbers x1, . . . , x5. Then, from the corresponding equations, we get x2 ≤2x and y2 ≥2y. Since x > 0, y > 0, we get 2 ≤y ≤x ≤2. Hence the system has the unique solution x1 x2 x3 x4 x5 2.
33. Since the system is symmetric in x, y, z, we may assume x ≥y, x ≥z. The last two equations imply y + z ≥z + x or y ≥x. Thus x y. Similarly x z. The equation 8x3 x has three real roots x 0, x ±1/2 √ 2.
34. The number of pairs (A, P) of points A ∈E and planes P containing three points of E \ A is finite. Hence there is such a pair with minimal distance between A and P.
If P contains just three points of E, then we are finished. Otherwise, there are four points A2, A3, A4, A5 in E ∩P , such that the quadrilateral Q A2A3A4A5 contains no additional points from E. Now suppose that Q is not convex. We may assume that A2 is inside the triangle A3A4A5. The parallels to the sides of this triangle through A2 partition Q into pairs of half planes. One can always find such a half plane that, except for the projection A1 of A onto P , contains one additional point from {A3, A4, A5}, sayA3. Then the distance between A2 and the plane P3 through A, A4 and A5 is smaller than the distance between A1 and the plane P3, and this is smaller than |AA1| by the Pythagorean theorem. This contradicts the minimality property of the pair (A, P ). Hence Q is convex. The minimality property implies immediately that the pyramid A1A2A3A4A5 does not contain any additional points of E.
35. Join A to the centers Oi of the six circles. Let O1AO2 be the smallest of the angles OiAOJ. Prove that the segment O1O2 lies completely in one of the circles.
36. Proceed as in E1.
37. We call a student good if he learns better than the majority of his friends. Let x be the number of good students and k the number of friends of each student. The best student in class is the best of k pairs, and any other good student of at least ⌊k/2⌋+ 1 ≥(k + 1)/2 pairs. Hence, the good students are the best in at least k+(x−1)(k+1)/2 pairs.Thisnumbercannotexceedthenumberofallpairsoffriends in the class, which is 15k. Hence k+(x−1)(k+1)/2 ≤15k, or x ≤28·k/(k+1)+1.
We observe that (k + 1)/2 ≤30 −x or k ≤59 −2x, since the number of students, who are better than the worst among the good ones, does not exceed 30 −x, that is, x ≤28 · (59 −2x)/(60 −2x) + 1, or x2 −59x + 856 ≥0. The greatest integer x ≤30 satisfying the last inequality is x 25. Find an example showing that 25 can be attained.
38. Consider a person with a maximal number n of friends. We conclude that all his friends have different numbers of friends > 0, but ≤n. There are n possibilities 1, . . . , n friends. Hence all possibilities are realized. In particular, there exists a person with exactly one friend.
39. Set x1 ≥x2 ≥x3 ≥· · · ≥xn. Suppose x1 + x2 + x3 ≤1. Then x1 + x2 + x3 − (x1 −x3)(1 −x1) −(x2 −x3)(1 −x2) ≤1 or x2 1 + x2 2 + x3(3 −x1 −x2) ≤1, or x2 1 + x2 2 + x3(x3 + · · · + xn) ≤1, or x2 1 + x2 2 + x2 3 + · · · x2 n ≤1. This contradiction proves the theorem.
3. The Extremal Principle 57 40. Let x1 be the largest of the numbers x1, . . . , xn. Then (x2 + · · · + xn)2 n i2 x2 i + 2≤i<j≤n 2xixj.
(1) Adding the inequalities x2 i < 2x1xi for i 2 to n and inserting the estimate n i2 x2 i into (1), we get (x2 + · · · + xn)2 < n i2 2x1xi + 2≤i<j≤n 2xixj 1≤i<j≤n 2xixj.
Hence, (x2 + · · · + xn)2 < 2, or x2 + · · · + xn < √ 2.
See Chapter 9, problem 39 for another proof.
41. Number the vertices of the n-gon clockwise. Suppose that ai moves are made from the ith vertex. From the conditions of the problem, we have a1 a2 + an 2 , a2 a1 + a3 2 , . . . , an an−1 + a1 2 .
Suppose that a1 is the maximum of the ai. Then a1 (a2 + an)/2 implies a2 an a1. Similarly, a2 (a1 + a3)/2 implies a1 a2 a3, and so on, that is a1 a2 · · · an, and the total number of moves is na1.
42. For every m (1 ≤m ≤n) among the m pairs (ak, bk), one of the inequalities ak ≥bk or bk ≥ak is satisfied at least in m/2 pairs.
For instance, let bk ≥ak at least in m/2 pairs. If bl is the smallest of these bk, then bl ≤2/m. Hence al + bl ≤2bl ≤4/m, and since i ≤m, we have am + bm ≤ al + bl ≤4/m.
43. Let [a1, b1] be the segment with the smallest right endpoint. If more than 7 segments contain b1, then we are finished. If this number is ≤7, then at least 43 segments lie completely to the right of b1. From these segments, select [a2, b2] with the smallest right endpoint. Then either b2 belongs to 8 segments, or there exist 36 segments to the right of b2. Continuing in this way either we find a point belonging to eight segments, or we find seven pairwise disjoint segments [a1, b1], . . . , [a7, b7] such that to the right of [ak, bk] lie at least (50 −7k) segments, i.e., to the right of [a7, b7] lies at least one segment [a8, b8].
Similarly we can prove that, among (mn+1) segments one can select (m+1) pairwise disjoint segments or (n + 1) segments with a common point. This is a special case of the Theorem of Dilworth: In a partially ordered set of mn+1 elements, there is a chain of (m + 1) elements or (n + 1) pairwise incomparable elements.
44. From the 3n students, take one who has a maximum number k of acquaintances from one of the two other schools. Suppose it is student A from the first school, who knows k students from the second school. Then A knows (n + 1 −k) students from the third school, n + 1 −k ≥1 since k ≤n. Consider student B from the third school, who knows A. If B knows at least one student C from the k acquaintances of A in the second school, then {A, B, C} is a triple of mutual acquaintances. But if B knows none of the k acquaintances of A in the second school, then, in this school he does not know more than (n −k) students, and hence, in the first school, he does not know less than n + 1 −(n −k) k + 1 students, which contradicts the choice of k.
4 The Box Principle The simplest version of Dirichlet’s box principle reads as follows: If (n + 1) pearls are put into n boxes, then at least one box has more than one pearl.
This simple combinatorial principle was first used explicitly by Dirichlet (1805– 1859) in number theory. In spite of its simplicity it has a huge number of quite unexpected applications. It can be used to prove deep theorems. F.P. Ramsey made vast generalizations of this principle. The topic of Ramsey Numbers belongs to the deepest problems of combinatorics. In spite of huge efforts, progress in this area is very slow.
It is easy to recognize if the box principle is to be used. Every existence problem about finite and, sometimes, infinite sets is usually solved by the box principle.
The principle is a pure existence assertion. It gives no help in finding a multiply occupied box. The main difficulty is the identification of the pearls and the boxes.
For a warmup, we begin with a dozen simple problems without solutions: 1. Among three persons, there are two of the same sex.
2. Among 13 persons, there are two born in the same month.
3. Nobody has more than 300,000 hairs on his head. The capital of Sikinia has 300,001 inhabitants. Can you assert with certainty that there are two persons with the same number of hairs on their heads?
4. How many persons do you need to be sure that 2 (3, q) persons have the same birthday?
60 4. The Box Principle 5. If qs + 1 pearls are put into s boxes, then at least one box has more than q pearls.
6. A line l in the plane of the triangle ABC passes through no vertex. Prove that it cannot cut all sides of the triangle.
7. A plane does not pass through a vertex of a tetrahedron. How many edges can it intersect?
8. A target has the form of an equilateral triangle with side 2.
(a) If it is hit 5 times, then there will be two holes with distance ≤1.
(b) It is hit 17 times. What is the minimal distance of two holes at most?
9. The decimal representation of a/b with coprime a, b has at most period (b −1).
10. From 11 infinite decimals, we can select two numbers a, b so that their dec-imal representations have the same digits at infinitely many corresponding places.
11. Of 12 distinct two-digit numbers, we can select two with a two-digit differ-ence of the form aa.
12. If none of the numbers a, a + d, . . . , a + (n −1)d is divisible by n, then d and n are coprime.
The next eleven examples show typical applications of the box principle.
E1. There are n persons present in a room. Prove that among them there are two persons who have the same number of acquaintances in the room.
Solution. A person (pearl) goes into box #i if she has i acquaintances. We have n persons and n boxes numbered 0, 1, . . . , n −1. But the boxes with the numbers 0 and n −1 cannot both be occupied. Thus, there is at least one box with more then one pearl.
E2. A chessmaster has 77 days to prepare for a tournament. He wants to play at least one game per day, but not more then 132 games. Prove that there is a sequence of successive days on which he plays exactly 21 games.
Solution. Let ai be the number of games played until the ith day inclusive. Then 1 ≤a1 < . . . < a77 ≤132 ⇒22 ≤a1 + 21 < a2 + 21 < . . . < a77 + 21 ≤153.
Among the 154 numbers a1, . . . , a77, a1 + 21, . . . , a77 + 21 there are two equal numbers. Hence there are indices i, j, so that ai aj + 21. The chessmaster has played exactly 21 games on the days #j + 1, j + 2, . . . , i.
E3. Let a1, a2, . . . , an be n not necessarily distinct integers. Then there always exists a subset of these numbers with sum divisible by n.
4. The Box Principle 61 Solution. We consider the n integers s1 a1, s2 a1 + a2, s3 a1 + a2 + a3, . . . , sn a1 + a2 + · · · + an.
If any of these integers is divisible by n, then we are done. Otherwise, all their remainders are different modulo n. Since there are only n−1 such remainders, two of the sums, say sp and sq with p < q, are equal modulo n, that is, the following difference is divisible by n.
sq −sp ap+1 + . . . + aq.
This proof contains an important motive with many applications in number theory, group theory, and other areas.
E4. One of (n + 1) numbers from {1, 2, . . . , 2n} is divisible by another.
Solution. We select (n + 1) numbers a1, . . . , an+1 and write them in the form ai 2kbi with bi odd. Then we have (n + 1) odd numbers b1, . . . , bn+1 from the interval [1, 2n −1]. But there are only n odd numbers in this interval. Thus two of them p, q are such that bp bq. Then one of the numbers ap, aq is divisible by the other.
E5. Let a, b ∈N be coprime. Then ax −by 1 for some x, y ∈N.
Solution. Consider the remainders mod b of the sequence a, . . . , (b −1)a. The remainder 0 does not occur. If the remainder 1 would not occur either, then we would have positive integers p, q, 0 < p < q < b, so that pa ≡qa (mod b).
But a and b are coprime. Hence we have b|q −p. This is a contradiction since 0 < q −p < b. Thus there exists an x so that ax ≡1 (mod b), that is, ax=1+by, or ax −by 1.
E6. Erd˝ os and Szekeres. The positive integers 1 to 101 are written down in any order. Prove that you can strike 90 of these numbers, so that a monotonically increasing or decreasing sequence remains.
Solution. We prove a generalization: For n ≥(p −1)(q −1) + 1 every sequence of n integers contains either a monotonically increasing subsequence of length p or a monotonically decreasing subsequence of length q.
We assign the maximal length Lm of a monotonically increasing sequence with last element m and the maximal length Rm of a monotonically decreasing sequence beginning with m to any number m in the sequence.
This assignment has the property that, for two different numbers m and k there must be Lm ̸ Lk or Rm ̸ Rk. This follows easily from the fact that either m > k or m < k. All pairs (Lm, Rm) with m 1, 2, . . . , n are distinct. Assuming that no such subsequences exist, Lm can assume only the values 1, 2, . . . , p −1 and Rm only the values 1, 2, . . . , q −1. This gives (p −1)(q −1) different boxes for the pairs. But n ≥(p −1)(q −1) + 1 and the box principle leads to a contradiction.
E7. Five lattice points are chosen in the plane lattice. Prove that you can always choose two of these points such that the segment joining these points passes through 62 4. The Box Principle another lattice point. (The plane lattice consists of all points of the plane with integral coordinates.) Solution. Let us consider the parity patterns of the coordinates of these lattice points. There are only four possible patterns: (e,e),(e,o), (o,e), (o,o). Among the five lattice points, there will be two points, say A (a, b) and B (c, d) with the same parity pattern. Consider the midpoint L of AB, L a + c 2 , b + d 2 .
a and c as well as b and d have the same parity, and so L is a lattice point.
E8. In the sequence 1, 1, 2, 3, 5, 8, 3, 1, 4, . . . each term starting with the third is the sum of the two preceding terms. But addition is done mod 10. Prove that the sequence is purely periodic. What is the maximum possible length of the period?
Solution. Any two consecutive terms of the sequence determine all succeeding terms and all preceding terms. Thus the sequence will become periodic if any pair (a, b) of successive terms repeats, and the first repeating pair will be (1, 1).
Consider 101 successive terms 1, 1, 2, 3, 5, 8, 3, . . .. They form 100 pairs (1, 1), (1, 2), (2, 3), . . .. Since the pair (0, 0) cannot occur, there are only 99 possible distinct pairs. Thus two pairs will repeat, and the period of the sequence is at most 99.
E9. Consider the Fibonacci sequence defined by a1 a2 1, an+1 an−1 + an, n > 1.
Prove that, for any n, there is a Fibonacci number ending with n zeros.
Solution. A term ap ends in n zeros if it is divisible by 10n, or, if ap ≡0 mod 10n.
Thus we consider the Fibonacci sequence modulo 10n, and we prove that the term 0 will occur in the sequence. Take (102n +1) terms of the sequence a1, a2, . . . mod 10n. They form 102n pairs (a1, a2), (a2, a3), . . ., but the pair (0, 0) cannot occur.
Thus there are only (102n −1) possible pairs. Hence one pair will repeat. So the period length is at most (102n −1). As in E8, the first pair to repeat is (1, 1).
1, 1, 2, 3, . . . , ap period , 1, 1.
Then ap 1 −1 0. Thus, the term 0, will occur in the sequence. In fact, it is the last term of the period.
E10. Suppose a is prime to 2 and 5. Prove that for any n there is a power of a ending with 000 . . . 01 n .
Solution. Consider the 10n terms a, a2, a3, . . . , a10n. Take their remainders mod-ulo 10n. The remainder 0 cannot occur since a and 10 are coprime. Thus there are 4. The Box Principle 63 only (10n −1) possible remainders 1, 2, 3, . . . , 10n −1.
Hence, two of the terms ai, ak(i < k) will have the same remainder, and so their difference will be divisible by 10n: 10n|ak −ai ⇐ ⇒10n|ai(ak−i −1).
Since gcd(10n, ai) 1, we have 10n|ak−i −1 or ak−i −1 q ∗10n, or ak−i q ∗10n + 1. Thus, ak−i ends in 000 . . . 01 (n digits).
E11. Inside a room of area 5, you place 9 rugs, each of area 1 and an arbitrary shape. Prove that there are two rugs which overlap by at least 1/9.
Suppose every pair of rugs overlaps by less than 1/9. Place the rugs one by one on the floor. We note how much of the yet uncovered area each suceeding rug will cover. The first rug will cover area 1 or 9/9. The 2nd, 3rd, . . ., 9th rug will cover area greater than 8/9, . . . , 1/9. Since 9/9 + . . . + 1/9 5, all nine rugs cover area greater than five. Contradiction!
Ramsey Numbers, Sum-Free Sets, and a Theorem of I. Schur We consider four related competition problems: E12. Among six persons, there are always three who know each other or three who are complete strangers.
This problem was proposed in 1947 in the K¨ urschak Competition and in 1953 in the Putnam Competition. Later, it was generalized by R.E. Greenwood and A.M.
Gleason.
E13. Each of 17 scientists corresponds with all the others. They correspond about only three topics and any two treat exactly one topic. Prove that there are at least three scientists, who correspond with each other about the same subject.
E14. In space, there are given pn ⌊en!⌋+ 1 points. Each pair of points is connected by a line, and each line is colored with one of n colors. Prove that there is at least one triangle with sides of the same color.
E15. An international society has members from six different countries. The list of members contains 1978 names, numbered 1, 2, . . . , 1978. Prove that there is at least one member whose number is the sum of the numbers of two members from his own country or twice as large as the number of one member from his own country (IMO 1978).
The first two problems are special cases of the third with n 2 and n 3.
One represents the persons by points. In the first problem, each pair of points is 64 4. The Box Principle joined by a red or blue segment depending on the corresponding persons being acquaintances or strangers. In the second problem each pair of points is joined by a red, blue, or green segment if the corresponding scientists exchange letters about the first, second, or third topic, respectively. The relationship of the fourth problem to the third will be recognized later.
Before solving the problems, we introduce some notation. We select p points in space with no four lying in the same plane, and we join each pair of points by a segment (or curve). We get a so-called complete graph Gp with p vertices, p 2 edges, and p 3 triangles. We color each edge with one of n colors and call this an n-coloring of the Gp. If Gp contains a triangle with all sides of the same color then we call it monochromatic. We also say that Gp contains a monochromatic G3. Now, we solve E12, E13, and E14.
Solution of E12. The edges of a G6 are colored red or blue. Take any of the six points and call it P. At least 3 of the 5 lines which start at P are of the same color, say red. These red lines end at 3 points A, B, C (Fig. 4.1 ). If any side of the triangle ABC is red, we have a red triangle. If not, ABC is a blue triangle. In both cases, we have a monochromatic triangle. Fig. 4.2 shows that with 5 points and 2 colors there need not exist a monochromatic triangle. Here sides and diagonals have different colors.
E E E E E @ @ @ HH H Fig. 4.1 A B C P A A A H H H B B B B B Q Q Q Q Q Fig. 4.2 Solution of E13. The vertices of a G17 are colored red, blue, or green. Let P be one of the 17 points. At least six of the 16 lines which start at P are of the same color, say red. These red lines end at six points A1, . . . , A6. If any pair of these points is connected by a red line, we have a red triangle. If not, we have six points connected pairwise with lines of two colors. By the preceding problem, among the triangles formed by these six points, there will be a unicolored triangle. Now we construct a coloring of the G16 without a monochromatic triangle. Let G be the elementary abelian group of order 16 with the generating elements a, b, c, d. The reader needs no group theory. He needs to know only that a + a b + b c + c d + d 0.
We partition the nonzero elements of G into three sum-free subsets A1 {a, b, c, d, a + b + c + d}, A2 {a + b, a + c, c + d, a + b + c, b + c + d}, A3 {b + c, a + d, b + d, a + c + d, a + b + d}, that is, the sum of two elements of Ai does not lie in Ai.
We assign the colors 1, 2, 3 (red, blue, green) to the sets A1, A2, A3. In G16 we label each vertex with another group element. The edge xy, which connects x 4. The Box Principle 65 with y, we label with x + y. If x + y lies in Ai, then we color this edge with color i. If x + y and y + z lie in the same Ai then sides xy and yz in the triangle xyz have the same color. Since the sets are sum-free, (x + y) + (y + z) x + z lies in another set, that is, the side xz has another color. The constructed coloring has no monochromatic triangle.
Solution of E14. We know already that p1 3, p2 6, p3 17. We consider the complete graph with smallest p4 so that any of its 4-colorings results in 17 edges at each vertex. This gives p4 66. Similarly, we get p5 327, p6 1958. In general, we get pn+1 −1 n + 1 (pn −1) + 1 n + 1, pn+1 −1 (n + 1)(pn −1) + 1.
With qn pn −1, we get q1 2, qn+1 (n + 1)qn + 1, q1 2, qn+1 (n + 1)! qn n! + 1 (n + 1)!.
From this, we easily get qn n!
1 + 1 1! + 1 2! + · · · + 1 n!
.
We recognize the truncated series for e in the parenthesis. Thus, e qn n! + rn, rn 1 (n + 1)! + 1 (n + 2)! + · · · < 1 n!( 1 n + 1 + 1 (n + 1)! + · · ·) 1 n · n!.
Hence, qn < en! < qn + 1 n, that is, qn ⌊en!⌋, or pn ⌊en!⌋+ 1.
For a Gp colored with n colors, we have a special case of Ramsey’s theorem: If q1, . . . , qn ≥2 are integers, there is a minimal number R(q1, . . . , qn), so that, for p ≥R(q1, . . . , qn) for at least one i 1, . . . , n, Gp contains at least one monochromatic Gqi.
The numbers R(q1, . . . , qn) are called Ramsey Numbers. Obviously R(q, 2) R(2, q) q.Apartfromthesetrivialcases,thereareonlysevenRamsey Numbers known. We know that R(3, 3) 6, R(3, 3, 3) 17, and Rn(3) R(3, 3, . . . , 3 n times 3 ) ≤⌊en!⌋+ 1.
66 4. The Box Principle In addition, we know that R(3, 4) 9, R(4, 4) 18, R(3, 6) 18, R(3, 5) 14, R(3, 7) 23, and R(4, 5) 25. The last number was found in 1993. It required as much as a total of 11 years of processor time on as many as 110 desktop computers.
This may be the limit of computer power.
Each Ramsey Number leads to an interesting and tough problem. For example, R(3, 4) 9 says that any 2-coloring of a G9 forces a red triangle (G3) or a blue tetrahedron (G4). We make of this problem 39.
We will now solve E15. Afterwards, we will illustrate its mathematical back-ground. In this problem we are asked to show that the set {1, 2, . . . , 1978} cannot be partitioned into six sum-free subsets. We can replace 1978 by the smaller number 1957.
Assumption: There is a partitioning of {1, . . . , 1957} into six sum-free subsets A, B, C, D, E, F.
Conclusion: One of these subsets, say A, has at least 1957/6 326 1/6, i.e.
327 elements a1 < a2 < . . . < a327.
The 326 differences a327 −ai, i 1, . . . , 326 do not lie in A, since A is sum-free.
Indeed, from a327 −ai aj follows ai + aj a327. So they must lie in B to F. One of these subsets, say B, has at least 326/5 65 1/5, that is 66 of these differences b1 < b2 < . . . < b66.
The 65 differences b66 −bi, i 1, . . . , 65 lie neither in A nor in B since both sets are sum-free. Hence they lie in C to F. One of these subsets, say C, has at least 65/4 16 + 1/4, i.e., 17 of these differences c1 < c2 < . . . < c17.
The 16 differences c17 −ci, i 1, . . . , 16 do not lie in A to C, that is, in D to F.
One of these subsets, say D, has at least 16/3 5 1/3 that is, 6 of these differences d1 < d2 < . . . < d6. The 5 differences d6 −di do not lie in A to D, that is, in E or F. One of these, say E, has at least 2.5, that is, 3 elements e1 < e2 < e3. The two differences f1 e3 −e2, f2 e3 −e1 do not lie in A to E. Hence they lie in F. The difference g f2 −f1 does not lie in A to F. Contradiction!
There is a close connection between E15 and E14 for n 6. A subset A of the positive integers or an abelian group is called sum-free, if the equation x + y z for x, y, z ∈A is not solvable. Of course, we may also have x y. In connection with the Fermat Conjecture, in 1916 Isai Schur considered the following problem: What is the largest positive integer f (n) so that the set {1, 2, . . . , f (n)} can be split into n sum-free subsets?
We know only 4 values of the Schur function f (n). By trial, one finds f (1) 1, f (2) 4, f (3) 13. In 1961 Baumert found f (4) 44 with the help of a computer. A sum-free partition of {1, . . . , 44} is S1 {1, 3, 5, 15, 17, 19, 26, 28, 40, 42, 44}, 4. The Box Principle 67 S2 {2, 7, 8, 18, 21, 24, 27, 33, 37, 38, 43}, S3 {4, 6, 13, 20, 22, 23, 25, 30, 32, 39, 41}, S4 {9, 10, 11, 12, 14, 16, 29, 31, 34, 35, 36}.
Schur found the following estimates 3n −1 2 ≤f (n) ≤⌊en!⌋−1.
Now, we show that each partition of the set {1, . . . , ⌊en!⌋} into n subsets has at least one subset in which the equation x + y z is solvable.
Suppose {1, 2, . . . , ⌊en!⌋} A1 ∪A2 ∪. . . ∪An is a partition into n parts. We consider the complete graph G with ⌊en!⌋+1 points, which we label 1, 2, . . . , ⌊en!⌋+1. We color G with n colors 1, 2, . . . , n. The edge rs gets color m, if |r −s| ∈Am. According to E13 G will have a monochromatic triangle, that is, there exist positive integers r, s, t such that r < s < t ≤⌊en!⌋+1, so that the edges rs, rt, st all have the same color m, that is, s −r, t −s, t −r ∈Am.
Because (s −r) + (t −s) t −r, Am is not sum-free. This implies f (n) ≤⌊en!⌋−1.
In particular, f (6) ≤⌊720e⌋−1.
This is a simpler proof of E15. There, we may replace 1978 by 1957.
We recall the Ramsey Number Rn(3). This is the smallest positive integer such that every n-coloring of the complete graph with Rn(3) vertices forces a monochro-matic triangle. We have already proved that Rn(3) ≤⌊en!⌋+ 1.
Thus, we have an upper estimate for f (n) by means of Rn(3). We prove that Rn(3) ≥f (n) + 2.
The proof coincides with the previous one. Let A1, A2, . . . , An be a sum-free partition of {1, 2, . . . , f (n)} and suppose that G is a complete graph with the f (n) + 1 vertices 0, 1, . . . , f (n). We color the edges of G with n colors 1, . . . , n by coloring edge rs with color m if |r −s| ∈Am. Suppose we get a triangle with vertices r, s, t and with edges of color m. We assume r < s < t. Then t −s, t −r, s −r ∈Am. But, (t −s) + (s −r) t −r, and this contradicts the assumption that Am is sum-free. Hence Rn > f (n) + 1, q.e.d.
68 4. The Box Principle In problem 43, we will prove f (n) ≥3n −1 2 .
Thus, we have 3n + 3 2 ≤Rn(3) ≤⌊en!⌋+ 1, that is, 3 ≤R1(3) ≤3, 6 ≤R2(3) ≤6, 15 ≤R3(3) ≤17, 42 ≤R4(3) ≤66.
Because of Baumert’s result, we know that even 44 ≤R4(3) ≤66. The first three upper bounds are exact. The fourth is not. For about 20 years, it has been known that R4(3) ≤65, that is, 44 ≤R4(3) ≤65.
Problems 13. n persons meet in a room. Everyone shakes hands with everyone else. Prove that during the greeting ceremony there are always two persons who have shaken the same number of hands.
14. In a tournament with n players, everybody plays with everybody else exactly once.
Prove that during the game there are always two players who have played the same number of games.
15. Twenty pairwise distinct positive integers are all < 70. Prove that among their pairwise differences there are four equal numbers.
16. Let P1, . . . , P9 be nine lattice points in space, no three collinear. Prove that there is a lattice point L lying on some segment PiPk, i ̸ k.
17. Fifty-one small insects are placed inside a square of side 1. Prove that at any moment there are at least three insects which can be covered by a single disk of radius 1/7.
18. Three hundred forty-two points are selected inside a cube with edge 7. Can you place a small cube with edge 1 inside the big cube such that the interior of the small cube does not contain one of the selected points?
19. Let n be a positive integer which is not divisible by 2 or 5. Prove that there is a multiple of n consisting entirely of ones.
20. S is a set of n positive integers. None of the elements of S is divisible by n. Prove that there exists a subset of S such that the sum of its elements is divisible by n.
21. Let S be a set of 25 points such that, in any 3-subset of S, there are at least two points with distance less than 1. Prove that there exists a 13-subset of S which can be covered by a disk of radius 1.
22. In any convex hexagon, there exists a diagonal which cuts off a triangle with area not more then one sixth of the hexagon.
4. The Box Principle 69 23. If each diagonal of a convex hexagon cuts off a triangle not less than one sixth of its area, then all diagonals pass through one point, are divided by this point in the same ratio, and are parallel to the sides of the hexagon.
24. Among n + 1 integers from {1, 2, . . . , 2n} there are two which are coprime.
25. From ten distinct two-digit numbers, one can always choose two disjoint nonempty subsets, so that their elements have the same sum (IMO 1972).
26. Let k be a positive integer and n 2k−1. Prove that, from (2n −1) positive integers, one can select n integers, such that their sum is divisible by n.
27. Let a1, · · · , an (n ≥5) be any sequence of positive integers. Prove that it is always possible to select a subsequence and add or subtract its elements such that the sum is a multiple of n2.
28. In a room with (m −1)n + 1 persons, there are m mutual strangers (in the room) or there is a person who is acquainted with n persons.
Does the theorem remain valid, if one person leaves the room?
29. Of k positive integers with a1 < a2 < . . . < ak ≤n and k > ⌊(n + 1)/2⌋, there is at least one pair ai, ar such that ai + a1 ar.
30. Among (ab + 1) mice, there is either a sequence of (a + 1) mice of which one is descended from the preceding, or there are (b + 1) mice of which none descends from the other.
31. Let a, b, c, d be integers. Show that the product of the differences b−a, c−a, d −a, c −b, d −b, d −c is divisible by 12.
32. One of the positive reals a, 2a, . . . , (n−1)a has at most distance 1/n from a positive integer.
33. Two of six points placed into a 3 × 4 rectangle will have distance ≤ √ 5.
34. In any convex 2n-gon, there is a diagonal not parallel to any side.
35. From 52 positive integers, we can select two such that their sum or difference is divisible by 100. Is the assertion also valid for 51 positive integers?
36. Each of ten segments is longer than 1 cm but shorter than 55 cm. Prove that you can select three sides of a triangle among the segments.
37. The vertices of a regular 7-gon are colored white or black. Prove that there are vertices of the same color, which form an isosceles triangle. What about a regular 8-gon? For what regular n-gons is the assertion valid?
38. Each of nine lines partitions a square into two quadrilaterals of areas in the ratio 2:3.
Then at least three of the nine lines pass through one point.
39. Among nine persons, there are three who know each other or four persons who do not know each other. The number nine cannot be replaced by a smaller one.
40. R(4, 4) 18 yields the problem: Among 18 persons, there are four who know each other or four persons who do not know each other. For 17 persons this need not be true.
41. R(3, 6) 18 gives the problem: Among 18 persons, there are three who know each other, or six who do not know each other. Try to get an estimate of R(6, 3) from below and above.
70 4. The Box Principle 42. Erd˝ os proved two estimates for R(r, s), which we formulate as the next two problems.
Prove that R(r, s) ≤R(r −1, s) + R(r, s −1).
(1) 43. With the help of (1), prove that R(r, s) ≤ r + s −2 r −1 .
44. Split the set {1, 2, . . . , 13} into three sum-free subsets. Prove that {1, . . . , 14} cannot be split into three sum-free subsets.
45. Prove that the set {1, 2, . . . , (3n −1)/2) can be split into n sum-free subsets.
46. The set {1, . . . , 9} is split in any way into two subsets. Prove that in at least one subset, there are three numbers of which one is the arithmetic mean of the other two.
47. The sides of a regular triangle are bicolored. Do there exist on its perimeter three monochromatic vertices of a rectangular triangle? (IMO 1983).
48. From the set {1, 2, . . . , 2n+1}, select a sum-free subset A with a maximum number of elements. How many elements does A have?
49. If the points of the plane are colored red or blue, then there will be a red pair with distance 1, or there are 4 collinear blue points with distance 1.
50. If a G14 is colored with two colors, there will be a unicolored quadrangle.
51. A three colored G80 contains a monochromatic quadrilateral.
The next problems are rather tough. They treat a theorem of Jacobi and its applica-tions. Solutions for 50 to 57 and 59 are missing.
52. Fig. 4.3 shows a circle of length 1. A man with irrational step length α (measured along the circumference) walks around the circle. The circle has a ditch of width ϵ > 0. Prove that, sooner or later, he will step into the ditch no matter how small ϵ will be.
53. Prove that there is a power of two, which begins with 6 nines, that is, there are positive integers n, k such that 999999 × 10k < 2n < 10k+6, k + log 999999 < n log 2 < k + 6.
Hint. Here ϵ 6 −log 999999 and the step length is α log 2. Similarly, one can show that, for irrational log a, there is a power of a which begins with any prescribed digit sequence.
54. Let an be the number of terms in the sequence 21, 22, . . . , 2n, which begin with digit 1. Prove that log 2 −1 n < an n < log 2 and, hence, p1 lim n→∞ an n log 2 ≈0.30103.
One sometimes says that a randomly chosen power of two begins with 1 with prob-ability log 2 ≈0.30103.
4. The Box Principle 71 A A t Fig. 4.3 55. The line y αx with irrational α passes through no lattice point except (0, 0), but it comes arbitrarily close to some lattice points.
56. Prove that there is a positive integer n such that sin n < 10−10 (or 10−k for any positive integer k).
57. If α π , β π , α β are irrational, then always sin nα + sin nβ < 2, but we can get as near to 2 as we please for some integers n.
58. There is a point set on the circle which, by rotation, goes into a part of itself.
59. An infinite chessboard consists of 1 × 1 squares. A flea starts on a white square and makes jumps by α to the right and β upwards, α, β, α/β being irrational. Prove that, sooner or later, it will reach a black square.
60. The function f (x) cos x + cos(x √ 2) is not periodic.
Remark. We consider the sequence αn n α−⌊n α⌋, n 1, 2, 3, . . .. with irrational α. The theorem of Jacobi says that the terms of the sequence αn are everywhere dense in the interval (0, 1). In 1917 H. Weyl showed that the sequence is equidistributed in the interval (0, 1), that is, let 0 ≤a < b ≤1, and let Hn(a, b) be the number of terms αi, 1 ≤i ≤n, which lie in the interval (a, b). Then lim n→∞ Hn(a, b) n b −a.
The distribution of the golden section α ( √ 5 −1)/2 is amazingly uniform.
We conclude the topic with problems mostly of a geometrical flavor.
61. There are 650 points inside a circle of radius 16. Prove that there exists a ring with inner radius 2 and outer radius 3 covering ten of these points.
62. There are several circles of total length 10 inside a square of side 1. Show that there exists a straight line which intersects at least four of these circles.
63. Suppose n equdistant points are chosen on a circle (n ≥4). Then every subset of k ⌊√2n + 1/4 + 3/2⌋of these points contains four points of a trapezoid.
64 Several segments of a segment of length 1 are colored such that the distance between any two colored points is ̸ 0.1. Prove that the sum of the lengths of the colored segments is ≤0.5.
65. A closed disk of radius 1 contains seven points with mutual distance ≥1. Prove that the center of the disk is one of the seven points (BrMO 1975).
72 4. The Box Principle 66. (a) Prove that there exist integers a, b, c not all zero and each of absolute value less than one million, such that |a + b √ 2 + c √ 3| < 10−11.
(b) Let a, b, c be integers, not all zero and each of absolute value less than one million. Prove that |a + b √ 2 + c √ 3| > 10−21 (Putnam 1980).
67. Prove that, among any seven real numbers y1, . . . , y7, there exist two, such that 0 ≤yi −yj 1 + yiyj ≤ 1 √ 3 .
68. Prove that, among any 13 real numbers, there are two, x and y, such that |x −y| ≤(2 − √ 3)|1 + xy|.
69. The points of a space are colored in one of three colors. Prove that at least one of these colors realizes all distances, that is, for any d > 0, there are two points of this color with distance d.
70. The points of a plane are colored in one of three colors. Prove that at least one of these colors realizes all distances, that is, for any d > 0, there are two points of this color with distance d.
71. Twelve percent of a sphere is painted black, the remainder is white. Prove that one can inscribe a rectangular box with all white vertices into the sphere.
72. The cells of a 7 × 7 square are colored with two colors. Prove that there exist at least 21 rectangles with vertices of the same color and with sides parallel to the sides of the square.
73. The Sikinian road system is such that three roads meet at each intersection. Prove the following property of the Sikinian road system: Start at any intersection A1, and drive along any of the three roads to the next intersection A2. At A2 turn right and go to the next intersection A3. At A3 turn left, and so on, turning left and right alternately.
Then you will eventually return to your starting point A1.
74. Thirty-three rooks are placed on an 8×8 chessboard. Prove that you can choose five of them which are not attacking each other.
75. The n positive integers a1 ≤a2 ≤· · · ≤an ≤2n are such that the least common multiple of any two of them is greater than 2n. Prove that a1 > ⌊2n/3⌋.
76. Any of the n points P1, . . . , Pn in space has a smaller distance from point P than from all the other points Pi. Prove that n < 15.
77. A plane is colored blue and red in any way. Prove that there exists a rectangle with vertices of the same color.
78. Let a1, . . . , a100 and b1, . . . , b100 be two permutations of the integers from 1 to 100. Prove that, among the products a1b1, . . . , a100b100, there are two with the same remainder upon division by 100.
79. The length of each side of a convex quadrilateral ABCD is < 24. Let P be any point inside ABCD. Prove that there exists a vertex, say A, such that |PA| < 17.
4. The Box Principle 73 80. A positive integer is placed on each square of an 8 × 8 board. You may select any 3 × 3 or 4 × 4 subboard and add 1 to each number on its squares. The goal is to get 64 multiples of 10. Can the goal always be reached?
81. The numbers from 1 to 81 are written on the squares of a 9 × 9 board. Prove that there exist two neighbors which differ by at least 6.
82. Each of m cards is labeled by one of the numbers 1, . . . , m. Prove that if the sum of the labels of any subset of the cards is not a multiple of m + 1, then each card is labeled by the same number.
83. Two of 70 distinct positive integers ≤200 have differences of 4, 5, or 9.
84. A 20 × 20 × 20 cube is built of 1 × 2 × 2 bricks. Prove that one can pierce it by a needle without damaging one of the bricks.
Solutions 13. The solution is the same as for E1.
14. The same problem as problem 13. Handshakes are replaced by contests.
15. Denote the 20 integers a1 to a20. Then 0 < a1 < · · · < a20 < 70. We want to prove that there is a k, so that aj −ai k has at least four solutions. Now 0 < (a2 −a1) + (a3 −a2) + · · · + (a20 −a19) a20 −a1 ≤68.
We will prove that, among the differences ai+1 −ai, i 1, . . . , 19, there will be four equal ones. Suppose there are at most three differences equal. Then 3 · 1 + 3 · 2 + 3 · 3 + 3 · 4 + 3 · 5 + 3 · 6 + 7 ≤68, that is, 70 ≤68. Contradiction!
16. Generalization of E7. Consider the three coordinates mod 2. There are 23 8 pos-sible binary 3-words. Since there are nine words altogether, at least two sequences must be identical. Thus there are two points (a, b, c) and (r, s, t) with integral mid-point M ((a + r)/2, (b + s)/2, (c + t)/2).
17. Subdivide the unit square into 25 small squares of side 1/5. There will be three insects in one of these squares of side 1/5 and diagonal √ 2/5. A circumcircle of this square has radius √ 2/10 < 1/7. If we circumscribe a concentric circle with radius 1/7, it will cover this square completely.
18. Subdivide the cube into 73 343 unit cubes. Since there are altogether only 342 points inside the large cube, the interior of at least one unit cube must remain empty.
19. Consider the n integers 1, 11, . . . , 11 · · · 1 mod n. There are n possible remainders 0, 1, . . . , n−1. If 0 occurs, we are finished. If not, two of the numbers have the same remainder mod n. Their difference 111 · · · 100 · · · 0 is divisible by n. Since n is not divisible by 2 or 5, we can strike the zeros at the end and get the number consisting of ones and divisible by n.
74 4. The Box Principle 20. We use the same motive. Consider the sums a1, a1 + a2, a1 + a2 + a3, . . . , a1 + a2 + · · · + an.
If any of the n sums is divisible by n, then we are done. Otherwise, two of the sums a1 +. . .+ai and a1 +· · ·+aj have the same remainder upon division by n. Suppose j > i. Then the difference ai+1 + · · · + aj is divisible by n.
21. In the proof, we change 25 and 13 to 2n + 1 and n + 1, respectively.
Let A and B be two points of S with maximum distance. If |AB| ≤1, a disk with center at A and radius 1 covers all 2n + 1 points, and we are finished. Now suppose that |AB| > 1. Let X be any point in S{A, B}. In the 3-subset {A, B, X} there are two points with distance less than 1. So either |AX| < 1 or |BX| < 1. Hence any point of S lies in one of the disks of radius 1 about A and B. One of these disks must contain at least n + 1 of the 2n + 1 points.
22. If the main diagonals (which do not cut off a triangle) pass through one point, then everything is clear. The main diagonals partition the hexagon into six triangles of which at least one has area not surpassing one-sixth of the hexagon. Suppose it is △OBC in Fig. 4.4. Then one of the triangles ABC and BCD has area ≤the area of BCO. But suppose the main diagonals form a triangle PQR in Fig. 4.5. Then it is even easier to find such a triangle. Prove this yourself.
23. This follows somehow from the preceding proof. In fact this problem was made up from the preceding one.
24. Among n + 1 integers from 1, . . . , 2n, there are two successive integers. They are coprime.
25. A set S of 10 numbers with two digits, each one ≤99 has 210 1024 subsets.
The sum of the numbers in any subset of S is ≤10 · 99 990. So there are fewer possible sums than subsets. Thus there are at least two different subsets S1 and S2 having the same sum. If S1 ∩S2 ∅, then we are finished. If not, we remove all common elements and get two nonintersecting subsets with the same sum of their elements.
26. Use induction from n to 2n, which corresponds to induction from k to k + 1.
(1) For n 1, the statement is correct.
(2) Suppose that, from 2n −1 integers, we can always select n with sum divisible by n. Of the 2(2n) −1 positive integers, we can select n numbers three times, which are divisible by n. After the first selection, there will remain 3n −1 numbers, after the second selection, 2n −1 numbers. Let the sum of the first choice be a · n, the sum of the second choice be b · n, and the last choice be c · n. At least two of the numbers a, b, c have the same parity, e.g., a and b. Then an + bn (a + b)n is divisible by 2n, since a + b is even.
Remark. The more general theorem that, from any 2n −1 positive integers, one can always select n with sum divisible by n is more difficult to prove. Start by proving it for n p, where p is a prime. Then prove it for n pq, where p, q are primes.
27. Consider all subsets {i1, . . . , ik} of the set {1, . . . , n}. Let S(i1, . . . , ik) ai1 +· · ·+ aik. The number of such sums is 2n −1. Since 2n −1 > n2 for n ≥5, two of these sums will have the same remainder upon division by n2. Their difference will be divisible by n2. This difference has the form ±as1 ± as2 ± · · · ± ast for some t ≥1 and some selection of indices s1, . . . , st.
4. The Box Principle 75 J J J @ @ @ @ @ @ @ """"" aaaaa A B C D E F O Fig. 4.4 A A A A A A \ \ \ \ \ A B C D E F R Q P Fig. 4.5 @ @ @ @ Fig. 4.6 28. We must prove that there are m mutual strangers in the room or n + 1 mutual acquaintances.
We will repeat the following step: Select any person left in the room and remove his strangers. This is repeated n times. At each step at most m −1 persons are removed.
There will be at least 1 person left. The persons selected and any of the persons left are n + 1 mutual acquaintances.
29. The k −1 pairwise different positive integers a2 −a1, a3 −a1, . . . , ak −a1 together with the given pairwise different integers together are 2k −1 > n positive integers, all of which are ≤n. Hence the two sets have at least one common element, i.e., at least once, we have ar −a1 ai, or ai + a1 ar.
30. Draw an arrow from each mouse to its immediate descendant. The mice split into trees. If each tree has at most a vertices, then there must be at least b + 1 trees.
Taking one mouse from each tree, we get b + 1 mice of which none descends from the other.
31. Put the four numbers into two boxes depending on their parity. In the worst case, the boxes have two elements each. Then the difference in each box is even. Thus we have two even differences, giving two factors 2 for the product. In all other cases, we have more factors 2.
Now, consider the four numbers modulo 3. We have three boxes and four numbers.
Thus at least one box contains two numbers. Their difference is a multiple of 3. So the product of all six differences is divisible by 12.
32. Considering the fractional parts of these numbers, we get n −1 reals in the interval [0, 1]. Subdivide this unit interval into n equal parts, each of length 1/n. If one of the n points falls into the first interval, than we are finished. Otherwise, two points, say {ia} and {ka}, fall into the same interval. Then the point {(k −i)a} is a distance ≤1/n from 0.
33. Split the 3×4 rectangle into 5 parts, as in Fig. 4.6. At least one part will contain two of the six points. Their distance will be ≤ √ 5.
34. A 2n-gon has 2n(2n−3)/2 n(2n−3) diagonals. The number of diagonals parallel to a given side is ≤n −2. Hence the total number of diagonals parallel to some side is at most 2n(n −2). Since 2n(n −2) < n(2n −3), one of the diagonals is not parallel to any side.
35. We consider 51 boxes. Into box 0, we put the numbers ending in 00. Into 1, we put the numbers ending in 01 or 99, into box 2 we put the numbers ending in 02 or 98, and so on. Finally, into box 49, we put the numbers ending in 49 or 51, and into box 50 the numbers ending in 50. Two of the 52 numbers will be in the same box. Their difference (if they have the same end) or their sum (otherwise) ends in 00. Among 51 numbers, such a pair need not exist. For instance, 1, 2, . . . , 49, 50, 100.
76 4. The Box Principle 36. Suppose the segments a1, . . . , a10 are such that 1 < a1 ≤a2 ≤a3 ≤· · · ≤a10 < 55.
Assume that no triangle can be constructed. Then a3 ≥a1 + a2 > 2, a4 ≥a2 + a3 > 1+2 3, a5 ≥a3 +a4 > 2+3 5, a6 ≥a4 +a5 > 8, a7 ≥a5 +a6 > 5+8 13, a8 ≥a6 + a7 > 8 + 13 21, a9 ≥a7 + a8 > 34, a10 ≥a8 + a9 > 21 + 34 55, i.e., a10 > 55. Contradiction!
37. Since the number of vertices is odd, there must be two neighbors of the same color, say black. Number the vertices such that these black vertices have numbers 2 and 3.
If 1 or 4 are also black, then we have a monochromatic isosceles triangle. Otherwise, 1 and 4 are white. Now, either 2, 3, and 7 are vertices of a black isosceles triangle, or 3, 4, and 7 are vertices of a white isosceles triangle. The same argument works for any odd n with n ≥5. For n 6 and n 8, there are colorings which avoid monochromatic isosceles triangles. See Fig. 4.7 and Fig. 4.8. For n 4k +2, k > 1, we can ignore every second vertex and use the argument for a n-gon with an odd number of vertices. What about the other cases?
Let us number the vertices 1, . . . , n. If there are no two neighbors of the same color, then the colors must alternate bwbwb · · ·. The first, third, and fifth vertices are black and at equal distances. So they form an isosceles black triangle. Otherwise, there are two neighboring vertices of the same color. Suppose 1 and 2 are black. Starting with these, we draw the tree of all possibilities avoiding three vertices of the same color at equal distances. See Fig. 4.9. The tree stops growing, reaching at most length 8.
If we take any 9 successive integers, there will always be 3 numbers in arithmetic progression. So for n > 8, there will always be an isosceles black or white triangle.
What about the 8-gon? There are three paths of length 8. On closing them to a ring, we observe that both paths bbwwbbww and bwwbbwwb give the same solution. So the solution in Fig. 4.8 is unique. We started with the black color. By starting with the white color, we get the same solution with colors interchanged but color change merely rotates the solution by 90◦.
u J Je e u J J e e Fig. 4.7 e e u u @ @ e e u u @ @ Fig. 4.8 d yes d t t d t @ @d d d d t P P t d d d t t @ @ d d t t @ @d d @ @t H Hd t t d d t yes d d @ @t Fig. 4.9 38. Suppose a square has side a. Two quadrilaterals are trapezi with altitudes a. Their areas are in the same ratio as their midlines. There are four points on the midlines of the square which divide them in the ratio 2 : 3. The nine lines must pass through these four points. Because of the box principle, at least three lines will pass through one of the four points.
39. See solution of problem 42 below.
40. See solution of problem 42 below.
41. See solution of problem 42 below.
4. The Box Principle 77 42. We consider the complete graph with R(r −1, r)+R(r, s −1) vertices whose edges are colored red and black. We select one vertex v and consider V1 set of all vertices, which are connected to v by a red edge. |V1| n1.
V2 set of all vertices, which are connected to v by a black edge. |V2| n2.
n1 + n2 + 1 R(r −1, s) + R(r, s −1). From n1 < R(r, s −1), we conclude that n2 ≥R(r, s −1). This implies that V2 contains a Gr or Gs−1, and together with v, we have a Gs.
n1 ≥R(r −1, s) implies that V1 contains a Gs or a Gr−1, and together with V , a Gr.
Thus, we have R(r, s) ≤R(r −1, s) + R(r, s −1), with the boundary conditions R(2, s) s, R(r, 2) r. For symmetry reasons, we also have R(r, s) R(s, r).
If R(r −1, s) and R(r, s −1) are both even, then R(r, s) < R(r −1, s) + R(r, s −1).
Indeed, set R(r −1, s) 2p, R(r, s −1) 2q, and consider the complete graph with 2p + 2q −1 vertices. Select one vertex v, and consider the three cases (a) At least 2p red edges are incident with v.
(b) At least 2q black edges are incident with v.
(c) 2p −1 red and 2q −1 black edges are incident with v.
In the first case, we have a Gs or, together with v, a Gr. Similarly in case (b) we have a Gr or, together with v, a Gs. Case (c) cannot be valid for every vertex of the two colored graph, since we would have (2p + 2q −1)(2q −1) red endpoints, i.e., an odd number. But every edge has 2 endpoints, so there must be an even number of red endpoints. Thus there is at least one vertex at which (a) or (b) is violated, and in both cases, we have a sharp inequality.
With R(2, 4) 4, R(3, 3) 6, we get R(3, 4) < R(2, 4) + R(3, 3) 10. Thus R(3, 4) ≤9, R(4, 4) ≤R(3, 4)+R(4, 3) 9+9 18. Fig. 4.10 contains neither a triangle of thin lines nor a quadrilateral of thick lines. The center does not belong to the G8. This proves that R(3, 4) 9. We prove that R(4, 4) ≥18. Indeed, take 17 equally spaced points 1, . . . , 17 on a circle. Join 1 to 7, 7 to 13, . . ., always skipping 5 points. You get a G17 colored black and invisible. It does not contain an invisible G4 or a black G4.
@ @ @ @ @ @ B B B B B B B B B PPPPPPPP P HHHHH H @ @ @ @ @ @ A A A A AA A A A A A A @ @ @ @ @ @ H H H H H H Fig. 4.10 L L L L L L L L H H H H H H H H ppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppl l l l l l l ppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp S S S S S S S `````` ` Fig. 4.11 R(3, 5) ≤R(2, 5) + R(3, 4) 5 + 9 14. Fig. 4.11 shows that R(3, 5) 14. This G13 with colors black and invisible does not contain a triangle or five independent points. Independent points are joined by invisible lines.
78 4. The Box Principle R(6, 3) < R(5, 3) + R(6, 2) 14 + 6 20 since 14 and 6 are both even. One can prove that R(6, 3) 18. With our crude estimates, we missed the exact bound. Try to find a coloring which proves that R(6, 3) > 17.
43. This follows from the equality C(r, s) C(r −1, s) + C(r, s −1) for the binomial coefficients.
44. See the next problem.
45. We want to give a lower bound for the Schur function f (n), which is the smallest number so that the integers 1, 2, . . . , f (n) can be arranged in n sum-free rows. If the table with n rows x1, x2, . . . , . . . , u1, u2, . . .
has sum-free rows, then the n + 1 rows 3x1, 3x1 −1, 3x2, 3x2 −1, 3x3, 3x3 −1, 3x4, 3x4 −1, . . . , 1, 4, 7, . . . , 3f (n) + 1 give a similar table for the integer 3f (n) + 1. For n 2, from the table 1 2 4 3, we get the new table 3, 2, 12, 11 6, 5, 9, 8 1, 4, 7, 10, 13.
In any case, we have f (n + 1) ≥3f (n) + 1, and since f (1) 1, we have f (2) ≥4, f (3) ≥13, f (4) ≥40. Thus, we get f (n) ≥1 + 3 + 32 + · · · + 3n−1 (3n −1)/2.
46. Try to draw a tree with vertices of two colors while avoiding an arithmetic progres-sion. You will not get beyond depth 8.
47. Suppose there is no right triangle with vertices of the same color. Partition each side of the regular triangle by two points into three equal parts. These points are vertices of a regular hexagon. If two of its opposite vertices are of the same color, then all other vertices are of the other color, and hence there exists a right triangle with vertices of the other color. Hence opposite vertices of the hexagon are of different color. Thus there exist two neighboring vertices of different color. One pair of these bicolored vertices lies on a side of the triangle. The points of this side, differing from the vertices of the hexagon, cannot be of the first or second color. Contradiction.
48. Let M {1, 2, . . . , 2n + 1}. The subset {1, 3, . . . , 2n + 1} consists of n + 1 odd numbers. It is sum-free, since the sum of two odd integers is even. Consider a maximal sum-free subset T {a1, . . . , ak} with a1 < · · · < ak. Because 0 < a2 −a1 < a3 −a1 < · · · < ak −a1 ≤2n + 1 −a1 < 2n + 1 the set S {a2 −a1, a3 −a1, . . . , ak −a1} is a subset of M with k −1 elements. S and T are disjoint. Indeed, if, for some (i, j) with i ∈{2, . . . , k}, j ∈{1, . . . , k}, we had ai −a1 aj, then we would have ai aj + a1. Contradiction, since T is sum-free. Thus we have (k −1) + k |S| + |T | |S ∪T | ≤|M| 2n + 1. From 2k −1 ≤2n + 1, we have k ≤n + 1. Thus no sum-free subset of M has higher cardinality than the subset of odd integers above. There is another sum-free subset {n + 1, n + 2, . . . , 2n + 1}. Try to prove that these are the only maximal sum-free subsets of M.
4. The Box Principle 79 49. Consider a rhombus ABCD consisting of two equilateral triangles ABD and BCD of side 1. We color its vertices black, white, and red trying to avoid two vertices of the same color at distance 1. Color B and D black and white, respectively. Then A and C must both be red. Rotating the rhombus about A, the point C describes a circle of radius √ 3 consisting entirely of red points. This circle has a chord of length 1, which has red endpoints.
58. Take a circle of length 1, and, on this circle, take any point O as origin. If α is any positive irrational number, we measure off the points α, 2α, 3α, . . . from O in the same direction. The points will be automatically reduced mod 1. We get a point set S with the property of going by rotation into a part of S. Rotating this set by mα we get S\{0, α, . . . , (m −1)α}.
60. Let λ √ 2. Now suppose that f (x) has period T . Then cos(x+T )+cos(λx+λT ) cos x + cos λx for all x. In particular, for x 0 we get cos T + cos λT 2. This implies T 2πk, λT 2πn, or λ n/k ∈Q, which is a contradiction.
61. We observe that the point P belongs to a ring with center O iff the point O belongs to a congruent ring with center P . Thus it is sufficient to prove the following fact. If we consider all such rings with centers in the given points, then one of these points will be covered by at least 10 rings. These rings lie inside a circle of radius 16 + 3 19 with area 192π 361π. Now, 9 · 361π 3249π, but the sum of the areas of all rings is 650 · 5π 3250π.
62. Orthogonally project all circles onto side AB of a unit square. A circle of length l will project into a segment of length l/π. The sum of the projections of all circles is 10/π. Since 10/π > 3 3AB, there is a point on AB belonging to the projections of at least four circles. The perpendicular to AB through this point intersects at least four circles.
63. The sides and diagonals of a regular n-gon have n directions. This is easy to see.
Any k of the points are endpoints of k 2 chords. The box principle tells us that, if the number of chords is greater than n, there will be two parallel chords. From k 2 > n, we get k > 1/2 + √2n + 1/4 and k ⌊√2n + 1/4 + 3/2⌋.
64. Cut a unit segment into 10 segments of length 0.1, put them into a pile above each other, and project them onto a segment. Since the distance between any two colored points is ̸ 0.1, the colored points of neighboring segments cannot project into one point. Hence, the colored points of more than 5 segments cannot be projected into a point. Hence the sum of the projections of the colored segments (which is the sum of their lengths) is at most 5 × 0.1 0.5.
65. Suppose a center O is not one of seven points. Then there are two of the seven points P and Q such that ̸ P OQ < 60◦. Hence |P Q| < 1. Complete the details.
66. (a) Let S be the set of 1018 real numbers r + s √ 2 + t √ 3 with each of r, s, t ∈ {0, 1, . . . , 106−1}, and let d (1+ √ 2+ √ 3)106. Then each x ∈S is in the interval 0 ≤x < d. This interval is partitioned into 1018−1 small intervals (k−1)e ≤x < ke withe d/(d18−1)andk takingonthevalues1, 2, . . . , 1018−1.Bytheboxprinciple, two of the 1018 numbers of S must be in the same small interval and their difference a + b √ 2 + c √ 3 gives the desired a, b, c since c < 10−11.
(b) Let F1 a + b √ 2 + c √ 3 and F2, F3, F4 be the other numbers of the form a ± b √ 2 ± c √ 3. Using the irrationality of √ 2 and √ 3 and the fact that a, b, c are not all zero, one shows that no Fi is zero. The product P F1F2F3F4 is an integer 80 4. The Box Principle since the mappings √ 2 →− √ 2 and √ 3 →− √ 3 leave P invariant. Hence |P| ≥1.
Then |F1| ≥1/(F2F3F4) > 10−21 since |Fi| < 107 for each i.
67. This problem contains all necessary hints for a solution. It is a problem for the box principle, since all existence problems about finite sets somehow rely on the box principle. Furthermore, it contains the hint to the addition theorem for tan, and 0 tan 0, 1/ √ 3 tan(π/6) give the missing hints for the boxes. So we set yi tan xi, yj tan xj and get tan 0 ≤tan(xi −xj) ≤tan π 6 .
Because tan is monotonically increasing everywhere, we get 0 ≤xi −xj ≤π 6 .
The yi can lie anywhere in the infinite interval −∞< yi < ∞. But the xi are confined to the interval −π/2 < xi < π/2. For at least two of the seven xi we have 0 ≤xi −xj ≤π/6. The original inequality follows from this.
6 --?
-?
-?
6 6 6 -?
x1x2 x3 x5x6 x7 x4 y4 y5 y6 y7 y1 y2 y3 x y Fig. 4.12 68. This problem is treated similarly. The addition theorem is slightly hidden, and we must recognize that 2 − √ 3 tan (π/12).
69. Suppose that none of the three colors A, B, C possesses the required property, that is, they do not realize the distances a, b, c, respectively. We may assume 0 < a ≤b ≤c. Let A1, A2, A3, A4 be the vertices of an a-tetrahedron. Similarly, let B1, B2, B3, B4 and C1, C2, C3, C4 be the vertices of a b-terahedron and c-tetrahedron, respectively. By an x-tetrahedron, we mean a regular tetrahedron with edge x. The position vectors of the vertices Ai, Bi, Ci will be denoted by ⃗ ai, ⃗ bi, ⃗ ci, respectively. Pijk is the point with the position vector ⃗ ai + ⃗ bj + ⃗ ck.
For each of the 16 index pairs (i, j), the four points Pij1, Pij2, Pij3, Pij4 are the vertices of a c-tetrahedron. It originated from the original c-tetrahedron by translation 4. The Box Principle 81 with ⃗ ai + ⃗ bj. Each of these 16 c-tetrahedra can have at most one point of color C, so that of the 64 index triples (i, j, k), at most 16 belong to points Pijk of color C.
Similarly, consideration of the b-tetrahedra with vertices Pi1k, Pi2k, Pi3k, Pi4k shows that at most 16 of the 64 index triples (i, j, k) belong to B-colored points.
Thus at least 32 of the index triples (i, j, k) belong to A-colored points Pijk. At least two points of color A belong to the same of the 16 (not necessarily pointwise distinct) a-tetrahedra. Thus we have two points with color A. Contradiction!
70. We consider the configuration in Fig. 4.13 consisting of four equilateral triangles A1A2A4, A2A3A4, A1A5A7, A5A6A7 with side d and, in addition, |A3A6| d. We observe that, of any three points of the configuration, at least two are at distance d.
Suppose that none of the three colors A, B, C possesses the required property. That is, they do not realize the distances a, b, c, respectively. Consider three configu-rations C1, C2, C3 not realizing distances a, b, c. We can always place them such that no four points of different configurations are vertices of a parallelogram. De-note the vertices of the configurations by Ai, Bj, Ck, i, j, k 1, . . . , 7. Let O be any point of the plane. Consider all possible sums − − → OAi + − − → OBj + − − → OCk. We get 73 points of the plane. These 73 points can be considered as three sets cosisting of 49 a-configurations, 49 b-configurations, or 49 c-configurations. Of the 343 points, at least 115 are of the same color, say A. Then among the 49 a-configurations, there are some with three points of color A. If not, the number of points of the color A would be at most 2 · 49 98. Thus the assumption that the color A is not realized leads to a contradiction.
71. Consider three pairwise orthogonal planes α, β, γ through the center O of a sphere.
If we reflect the black parts of the sphere at α, β, γ , then at most 8 · 12 96% of the sphere becomes black. There will remain white points. Let W be any white point. Reflecting it at α, β, γ we get eight white vertices of a box.
The theorem is probably valid for an inscribed cube. In addition, we can increase the black parts to 50% −ϵ, if we succeeded in proving that we could find four points of a rectangle in the white parts. Then we reflect this rectangle in the center of the sphere, getting a box with 8 white vertices.
72. We call those pairs of the table in the same row of the same color good pairs. Suppose there are k white and 7 −k black cells in some row. Then there are k(k −1) 2 + (7 −k)(6 −k) 2 k2 −7k + 21 good pairs.This term is minimal for k 3 and k 4 and is equal to 9. Thus there are at least 9 good pairs in each row, and in the whole square, at least 63. We call two good pairs in the same columns and of the same color concordant. Any two such pairs form a suitable rectangle. To estimate the number of concordant pairs, we observe that there are 7 · 6/2 21 pairs of columns and two different colors, that is, there cannot exist more than 2 · 21 42 discordant good pairs. Hence, considering the 63 good pairs one-by-one, not less than 63−42 21 of these will be concordant with one of the preceding ones. (The number 21 is exact.) 82 4. The Box Principle T T T T T T """ " bbb b b b b b """ " A2 A3 A6 A7 A1 A5 A4 Fig. 4.13 @ @ @ R ?
) B C D Fig. 4.14 1 2 3 4 5 6 7 8 2 3 4 5 6 7 8 1 3 4 5 6 7 8 1 2 4 5 6 7 8 1 2 3 5 6 7 8 1 2 3 4 6 7 8 1 2 3 4 5 7 8 1 2 3 4 5 6 8 1 2 3 4 5 6 7 Fig. 4.17 - 3 C C C O Q Q s / / / A E D F B C Fig. 4.15 -/ / / -C C C W / / / C C C O A B C D E F G Fig. 4.16 73. Since the road system is finite, you will eventually traverse some road section AB for the fifth time. Then you will have traveled this section at least three times in the same direction, say from A to B. Hence you have traversed one of the two continuations BC and BD at least twice in the same direction, say from B to C (Fig. 4.14.). But the part A →B →C uniquely determines your future path because it tells you where you are in the left–right–left–right–sequence. The fact that you have traversed these two road sections at least twice in the same direction means that your path is periodic.
We must show that this is a pure period, i.e., the situations depicted in Fig. 4.15 and Fig. 4.16 cannot occur. In Fig. 4.15 (circuit of odd length), when returning to F you must turn right to B, but then from B, you must turn left and get out of the circuit.
In Fig. 4.16 (circuit of even length), when returning to B, you must turn left and go to A instead of C.
74. Color the board diagonally in 8 colors as in Fig. 4.17. Since 33 4 · 8 + 1, at least one of the 8 colors is occupied by 5 rooks. These 5 rooks do not attack each other.
75. Suppose that a1 ≤⌊2n/3⌋. Then 3a1 ≤2n. The set {2a1, 3a1, a2, . . . an} consists of n + 1 integers ≤2n, of which none is divisible by another. This contradicts E4.
76.
̸ PiPPj > 60◦for all i ̸ j. Otherwise PiPj would not be the longest side in △PPiPj. Hence the n spherical caps on the unit sphere with center P, which for Pi contain all points Q of the unit ball with ̸ PiP Q ≤30◦are disjoint. The surface of such a cap is 2πrh 2π(1 −cos 30◦) π(2 − √ 3). The total area of the n spherical caps cannot surpass the area of the sphere. Hence, n · π(2 − √ 3) < 4π ⇒n < 4 2 − √ 3 4(2 + √ 3) 8 + √ 48 < 8 + √ 49 15.
77. Choose any 7 collinear points. At least 4 of these points are of the same color, say red. Call them R1, R2, R3, R4. We project these points onto two lines parallel to the first line to S1, . . . , S4 and T1, . . . , T4. If two S-points or two T -points are red, then we have a red rectangle. Otherwise, there exist 3 blue S-points and 3 T -points, and hence, a blue rectangle.
78. Suppose all the 100 products are different mod 100. In particular, there will be 50 odd and 50 even products. The 50 odd products use up all odd ai and all odd bj.
The even products are the products of two even numbers, so they are all multiples of 4. But then among the products there will be no numbers of the form 4k + 2.
Contradiction!
4. The Box Principle 83 79. Suppose all 4 distances of P to the vertices are ≥17. Join P to A, B, C, D. Then at least one of the 4 angles at P is ≥90◦. Suppose it is ̸ AP B. Then |AB|2 ≥ |PA|2 + |P B|2. The left side of this inequality is less then 242 576, and the right side ≥172 + 172 578, or 576 > 578. Contradiction!
80. Not always! Consider all the numbers mod 10. How many boards can we get starting with all zeros? We have (8 −3 + 1)2 + (8 −4 + 1)2 61 subboards of dimensions 3×3 and 4×4, i.e., we can reach at most 1061 8×8 boards. But there are altogether 1064 choices. Take one of these choices we cannot reach from the board of zeros.
This one must be taken as the starting board.
81. Assume the contrary. Then, if there exists a path of k steps from one cell to the next, the sum of the differences of the numbers on these cells is at most 5k. But the difference between 1 and 81 is 80, and the number of steps between the cells on which these numbers are located is not more than 16. Since 5 · 16 80, we can attain these bounds just once. On every other path from 1 to 81, there will be pairs of neighbors differing at least by 6.
82. Let ak be the label of card # k. None of the sums n k1 ak is a multiple of m + 1, and they are all distinct mod m + 1. Otherwise a difference of two of the sums, again a sum of ak, would be a multiple of m + 1. We have a2 s2 −s1. If a2 would have the same remainder as sum sq, (3 ≤q ≤m), then sq −a2, a sum of ak, would be 0 mod m + 1. Since all the remainders 1, . . . , m occur among the remainders of the sn, we have either a2 ≡s2 mod m + 1 or a2 ≡s1 mod m + 1. Because of 0 < a1 < m + 1, we can have only a2 s1, i.e., a1 a2. By cyclic rotation of the ak, we conclude that all the ak are equal.
83. Let a1, . . . , a70 be the given numbers. None of the 210 numbers a1, . . . , a70, a1 + 4, . . . , a70 + 4, a1 + 9, . . . , a70 + 9 exceeds 209. By the box principle two of them, say ai + x and aj + y, are equal (x ̸ y), where x and y can have the values 0, 4, or 9. Hence, the difference between ai and aj is 4, 5, or 9.
5 Enumerative Combinatorics What is a good Olympiad problem? Its solution should not require any prerequisites except cleverness. A high school student should not be at a disadvantage compared to a professional mathematician. During its first participation in 1977 in Belgrade, our team was confronted by such a problem. But first we give a definition.
Let a1, a2, . . . , am be a sequence of real numbers. The sum of q successive terms will be called a q-sum, for example, ai + ai+1 + · · · + ai+q−1.
E1. In a finite sequence of real numbers, every 7-sum is negative, whereas every 11-sum is positive. Find the greatest number of terms in such a sequence. (6 points) In our short training of 10 days, we did not treat any problem even distantly related to this one. I was quite amazed that most of the jury considered this prob-lem easy and suggested merely 6 points for its solution. Only one member of our team gave a complete solution, and another gave an almost complete solution. On the other hand, our team worked very well with the most difficult problem of the Olympiad, which was worth 8 points. They tackled it with the ubiquitous extremal principle.
E1 is, indeed, simple. It belongs to a large class of problems with almost au-tomatic solutions. It does not require much ingenuity to write successive 7-sums in separate rows. Then one sees immediately that q-sums crop up automatically in successive columns. Hence continue with row sums until we get 11-sums colum-nwise. By adding the row sums, we get a negative total. By adding the column sums, we get a positive total. Contradiction!
86 5. Enumerative Combinatorics a1 + a2 + · · · + a7 < 0 a2 + a3 + · · · + a8 < 0 . . . . . . . . . . . . . . . . . . . . . . .
a11 + a12 + · · · + a17 < 0 s1 + s2 + · · · + sr < 0 s2 + s2 + · · · + sr+1 < 0 . . . . . . . . . . . . . . . . . . . . . . . . . .
sr + sr+1 + · · · + sr+t−1 < 0 Fig. 5.1 Fig. 5.2 Thus, such a sequence can have at most 16 terms (Fig. 5.1). Some cleverness is required to construct such a sequence for 16 terms: 5, 5, −13, 5, 5, 5, −13, 5, 5, −13, 5, 5, 5, −13, 5, 5.
One could also construct the sequence more systematically. Here are a few related problems: E2. Replace 7, 11 by p, q with gcd(p, q) 1. Then the maximal length is ≤ p + q −2, as was proved by John Rickard (GB) at the IMO.
E3. In addition, one can also require that every r-sum is equal to 0.
E4. If gcd(p, q) d, then the maximal length is ≤p + q −d −1. Proof: We set p dr, q dt with gcd(r, t) 1, and consider the real sequence ai with p + q −d (r + t −1)d terms. Denote the nonoverlapping 1−, 2−, . . . , d-sums by s1, s2, s3, . . . , st+t−1. We write the negative p-sums until in rows the positive q-sums appear in columns, a contradiction (Fig. 5.2).
E5. In a sequence of positive real numbers every p-product is < 1, and each q-product is > 1. By using logarithms, we see that such a sequence can have at most length m p + q −d −1.
E6. In every sequence of positive integers, each 17-sum is even, and each 18-sum is odd. How many terms can such a sequence have at most?
E7. Let ai revenues−expenditures in month i for the budget of Sikinia. If ai < 0, there is a deficit in month #i. We consider the sequence a1, a2, . . . , a12. Suppose every 5-sum is negative. Then it is possible that we have a surplus for the whole year. Deficits and surpluses can be arbitrarily prescribed. The deficits and the final surplus can be astronomical.
Ideally an IMO problem should be unknown to all students. Even a similar problem should never have been discussed in any country. What was the status of E1 in July 1977? Years later, I was browsing in Dynkin–Molchanov–Rosental– Tolpygo: Mathematical Problems, 1971, 3rd edition with 200,000 copies sold.
There, I found problem 118: (a) Show that it is not possible to write 50 real numbers in a row such that every 7-sum is positive, but every 11-sum is negative.
(b) Write 50 numbers in a row, so that every 47-sum is positive, but every 11-sum is negative.
5. Enumerative Combinatorics 87 The origin of the problem was MMO 1969. The motive of E1 was well known in Eastern Europe, so it should not have been used at all.
This problem belongs to combinatorics in a wider sense. Such problems are very popular at the IMO since the topic is not so easy to train for. On the other hand, enumerative combinatorics is easy to train for. It is based on a few principles every contestant should know.
The most general combinatorial problem-solving strategy is borrowed from algorithmics, and it is called Divide and Conquer: Split a problem into smaller parts, solve the problem for the parts, and combine the solutions for the parts into a solution of the whole problem.
This Super principle or paradigm consists of a whole bundle of more spe-cial principles. For enumerative combinatorics, among others, these are sum rule, product rule, product-sum rule, sieving, and construction of a graph which accepts the objects to be counted. Divide and Conquer summarizes these and many other principles in a catchy slogan.
Let |A| denote the number of elements in a finite set A. If |A| n, we call A an n-set. A sequence of r elements from A is called an r-word from the alphabet A. In enumerative combinatorics, we count the number of words from an alphabet A which have a certain property.
1. Sum Rule. If A A1 ∪A2 ∪· · · ∪Ar is a partition of A into r subsets (blocks, parts), then |A| |A1| + |A2| + · · · + |Ar|. Applying this rule, we try to split A into parts Ai, so that finding |Ai| is simpler.
This rule is ubiquitous and is used mostly subconsciously. One task of a trainer is to point out its use as frequently as possible.
2. Product Rule. The set W consists of r-words from an alphabet A. If there are ni choices available for the ith letter, independent of previous choices, then |W| n1n2 · · · nr.
3. Recursion. A problem is split into parts which are smaller copies of the same problem, and these in turn are split in even smaller copies,. . . , until the problem becomes trivial. Finally, the partial problems are combined to give a solution to the whole problem.
Besides the Divide and Conquer Paradigm, there are some other paradigms in enumerative combinatorics.
4. Counting by Bijection. Of two sets A, B, we know |B|, but |A| is unknown.
If we succeed in constructing a bijection A ↔B, then |A| |B|. A proof which shows |A| |B| by such an explicit construction is called a bijective proof or combinatorial proof. Sometimes, one constructs a p −q bijection instead of a 1–1 bijection.
5. Counting the same objects in two different ways. Many combinatorial iden-tities are found in this way.
88 5. Enumerative Combinatorics The product-sum rule is mostly used simultaneously in the form: Multiply along the paths and add up the path products.
Here, the objects to be counted are interpreted as (directed) paths in a graph.
For instance, in Fig. 5.3 the number of paths from S (start) to G (goal) are |W| a1b1 + a2b2 + a3b3 + · · · .
We derive some simple results with the product rule: An n-set has 2n subsets.
There are n! permutations of an n-set.
j j j j j j A A Q Q Q Q Q Q Q Q A A G 1 2 3 4 S a1 a2 a3 a4 b1 b2 b3 b4 Fig. 5.3 The number of s-subsets of an n-set will be denoted by n s . We find this number by counting the s-words with different letters from an n-alphabet in two ways.
(a) Choose the s letters one by one which can be done in n(n−1) · · · (n−s +1) ways.
(b) An s-subset is chosen and then ordered. This gives n s ·s! possibilities. Thus, n s n(n −1) · · · (n −s + 1) s!
n s n −1 s −1 n!
s!(n −s)!
E8. 2n players are participating in a tennis tournament. Find the number Pn of pairings for the first round.
First solution (Recursion, Product Rule). We choose any player S. His partner can be chosen in 2n −1 ways. (n −1) pairs remain. Thus, Pn (2n −1)Pn−1 ⇒Pn (2n −1)(2n −3) · · · 3 · 1 (2n)!
2nn! .
(1) Second solution (Suggested by (1)). Order the 2n players in a row. This can be done in (2n)! ways. Then make the pairs (1, 2), (3, 4), . . . , (2n −1, 2n). This can be done in one way. Now we must eliminate multiple counting by division.
We may permute the elements of each pair, and also the n pairs. Hence, we must divide by 2nn!.
Third solution. Choose the n pairs one by one. This can be done in 2n 2 2n −2 2 · · · 2 2 ways. Then, divide the result by n! to eliminate the ordering of the pairs.
5. Enumerative Combinatorics 89 In this simple example, we see a tricky trap of enumerative combinatorics.
Subconsciously, we introduce an ordering and forget to eliminate it by division with an appropriate factor. This error can be eliminated by training.
E9. Convex n-gons.
(a) The number dn of diagonals of a convex n-gon is equal to the number of pairs of points minus the number of sides: dn n 2 −n n(n −3) 2 .
(b) In Fig. 5.4, the number sn of intersection points of the diagonals is equal to the number of quadruples of vertices (bijection): sn n 4 .
r @ @ r A A r a a a a a r r Fig. 5.4 (c) We draw all diagonals of a convex n-gon. Suppose no three diagonals pass through a point. Into how many parts Tn is the n-gon divided?
Solution. We start with one part, the n-gon. One part is added for each diagonal, and one more part is added for each intersection point of two diagonals, that is, Tn 1 + n 2 −n + n 4 (d) (p−q-application.) Wedrawalldiagonalsofaconvex n-gon P.Supposethat no three diagonals pass through one point. Find the number T of different triangles (triples of points).
Solution. The sum rule gives T T0 + T1 + T2 + T3, where Ti is the number of triangles with i vertices among the vertices of P. This partition is decisive since each Ti can be easily evaluated. The following Figs. 5.5a to 5.5d show the trivial counting. They show how we can assign some subsets of the vertices of P to the four types of triangles. The figures show that the assignments are 1:1, 1:5, 1:4, 1:1. Thus, we have T n 6 + 5 n 5 + 4 n 4 + n 3 .
90 5. Enumerative Combinatorics 3 3 3 33 A A h h h h h X X X X s s s s s s (a) 3 3 3 33 @ @ S S S S A A !
!
!
!
s s s s s (b) Fig. 5.5 P P P Q Q Q s s s s (c) @ @ H H H H H A A Q Q Q s s s (d) E10. Find a recursion for the number of partitions of an n-set.
Solution. Let Pn be the number of partitions of the n-set {1, . . . , n}. We take another element n + 1. Consider a block containing the element n + 1. Suppose it contains k additional elements. These elements can be chosen in n k ways. The remaining n −k elements can be partitioned into Pn−k blocks. Since k can be any number from 0 to n, the product-sum rule gives the recursion Pn+1 n k0 n k Pn−k n r0 n r Pr.
Here, we have defined P0 1, that is, the empty set has one partition. We get the following table from the recursion: n 0 1 2 3 4 5 6 7 8 9 10 Pn 1 1 2 5 15 52 202 877 4140 21147 115975 E11. Horse races. In how many ways can n horses go through the finish?
Solution. Without ties the answer is obviously n!. Let Hn be the corresponding answer with ties. We have H1 1 and H2 3. For H3, we need some deliberation.
The outcomes can be denoted by 3, 2 + 1, 1 + 1 + 1. These are all partitions of the number 3. The first element 3 means that a block of three horses arives simultaneously. 2+1 means that a block of two and a single horse arrive. 1+1+1 signifies three horses arriving at different moments. The block of three can arrive in one way. The two blocks in 2 + 1 can arrive in two ways, and the single horse can be chosen in three ways. In 1 + 1 + 1, the individual horses can arrive in six ways. The product-sum rule gives H3 1 + 2 · 3 + 3! 13 ways.
To find H4, consider all partitions of 4 and take into account the order of the various blocks. We have 4 3 + 1 2 + 2 2 + 1 + 1 1 + 1 + 1 + 1. Taking into account the distinctness of the elements and the order of the blocks, we get H4 1 + 4 · 2 + 3 · 2 + 6 · 3! + 4! 75. Now the computation of H5 and H6 becomes routine. For example, for H5, we have 5 4 + 1 3 + 2 3 + 1 + 1 2 + 2 + 1 2 + 1 + 1 + 1 1 + 1 + 1 + 1 + 1, H5 1 + 5 · 2! + 10 · 2! + 10 · 3! + 5 · 3 · 3! + 10 · 4! + 5! 541.
5. Enumerative Combinatorics 91 Define H0 1. Then we get the recursion Hn n k1 n k Hn−k. The closed for-mula below uses S(n, k) number of partitions of an n-set into k blocks=Stirling number of the second kind.
Hn n r0 S(n, k)k!
E12. Here the Stirling numbers of the second kind come up quite naturally. Let us find a recursion for S(n, k).
There are n persons in a room. They can be partitioned in S(n, r) ways into r parts. I come into the room. Now there are S(n+1, r) partitions into r parts. There are two possibilities: (a) I am alone in a block. The other n persons must be partitioned into r −1 blocks. This can be done in S(n, r −1) ways.
(b) I have r possibilities to join one of the r blocks. Thus, S(n + 1, r) S(n, r −1) + rS(n, r), S(n, 1) S(n, n) 1.
This is the analogue of the well-known formula n + 1 r n r + n r −1 , n 0 n n 1.
To prove this, consider the number of r-subsets of an (n+1)-set. We partition them according to the element n + 1. Of these, n r will not contain that element, and n r−1 will.
It helps for a beginner to compute a few Stirling numbers S(n,k) for some values of n and k by using only the product-sum rule. Suppose we want to find S(8, 4).
This is the number of ways of splitting an 8-set into 4 blocks. There are 5 types of partitions: 5+1+1+1, 4+2+1+1, 3+3+1+1, 3+2+2+1, 2+2+2+2.
See Fig. 5.6, where the 5 types are separated by 4 vertical lines.
r r r r r r r r A A r r r r r r r r r r r A A r r r A A r r r r r A A r r r r r r r r r r r r r Fig. 5.6 1. In the first type, we choose the three 1-blocks in 8 3 56 ways.
2. The second type is determined by first choosing the 4-block and then the 2-block, which can be done in 8 4 4 2 70 · 6 420 ways.
92 5. Enumerative Combinatorics 3. To find the contribution of the third type, we first choose the two 1-blocks in 8 2 28 ways. Then we must choose the first 3-block in 6 3 20 ways.
The second 3-block is now determined. But there is no first block. We have introduced the ordering, which must be eliminated on dividing by 2. So we have 28 · 10 280 ways for the third type.
4. For the fourth type, we first choose the 3-block in 8 3 56 ways. Then we choose the 1-block in 5 ways. Finally, we must partition the remaining four elements into two pairs (order does not count), which can be done in 3 ways.
Thus, there are 56 · 5 · 3 840 ways.
5. The fifth type is determined by splitting the 8-set into 4 pairs. This is the tennis player problem for 8 players. There are 7 · 5 · 3 · 1 105 cases.
6. Altogether, we have S(8, 4) 56 + 420 + 280 + 840 + 105 1701.
E13. Cayley’s formula for the number Tn of labeled trees with n vertices.
A tree is a nonoriented graph without a cycle. It is called labeled if its vertices are numbered. First, we want to guess a formula for Tn. A labeled tree with one vertex is just a point. It can be labeled in one way. There is also just one labeling for a tree with two vertices since the tree is not oriented. But there are three labelings for three points. There are three choices for the middle point. The two other points are indistinguishable. For a tree with four vertices, there are two topologically different cases: a chain with four points. There are 12 distinct labelings for the chain. In addition, there is a star with one central point and three indistinguishable points connected with the center. There are four choices for the center. This determines the star. Thus, T4 16. Now, let us take a tree with five vertices. There are three topologically different shapes: a chain, a star with a central point and four points connected to the center, and a T-shaped tree. See Fig. 5.7. There are 5!/2 60 labelings for the chain. The center of the star can be labeled in five ways. Now, let us look at the T. The intersection point of the horizontal and vertical bar can be chosen in five ways. The two points for the vertical tail can be chosen in six ways.
They can be ordered in two ways. Now the T-shaped tree is determined. So there are 5·6·2 60 T-shaped trees. Altogether we have T5 60+5+60 125. Now, look at the table below. The table suggests the conjecture Tn nn−2 number of (n −2)-words from an n-alphabet.
n 1 2 3 4 5 Tn 1 1 3 16 125 We want to test this conjecture for n 6. If it turns out that it is valid again, then we gain great confidence in the formula, and we will try to prove it. This time we have six topologically different types of trees. See Fig. 5.8.
1. There are 6!/2 360 distinct labelings for the chain.
5. Enumerative Combinatorics 93 t t t t t t t t t t @ @ t t t t t Fig. 5.7 s s s s s s s s s s s s @ @ s s s s s s @ @ s s s s s s @ @ @ @ s s s s s s A A s s s s s s Fig. 5.8 2. Now take the Y-shape with the vertical tail consisting of three edges. We can choose the center in six ways. The points for the tail can be chosen in 5 3 10 ways. The order of the three points in the tail can be chosen in 3! 6 ways. This determines the labeling of the Y-shape. So there are 360 possible labelings for this type of a tree.
3. Now comes the Y-shape with a vertical tail of one edge. The center can be chosen in six ways. The endpoint of the vertical tail can be chosen in five ways. The two other pairs of points can be chosen in three ways. Each can be ordered in two ways. The product rule gives 6 · 5 · 3 · 2 · 2 360.
4. The intersection point of the cross with a tail of two edges can be chosen in six ways. The three points with distance 1 from the center can be chosen in 5 3 10 ways. The remaining two points go into the tail and can be labeled in two ways. Again, the product rule gives 6 · 10 · 2 120.
5. Now comes the double-T. The two centers can be chosen in 6 2 15 ways.
The two points for one end of the edge connecting the two centers can be chosen in 4 2 6 ways. The two other points go to the other endpoint. So there are 15 · 6 90 distinct labelings for a double-T.
6. The center of the star can be chosen in 6 ways. This determines the labeling of the star.
Thus, we have T6 360 · 3 + 120 + 90 + 6 64.
This is a decisive confirmation of our conjecture. Now, we try to prove it by constructing a bijection between labeled trees with n vertices and (n −2)-words from the set {1, 2, . . . , n}.
Coding Algorithm. In each step, erase a vertex of degree one with lowest number together with the corresponding edge and write down the number at the other end of the crossed-out edge. Stop as soon as only two vertices are left.
For the tree in Fig. 5.9, we have the so-called Pr¨ ufer Code (7,7,2,2,7).
DecodingAlgorithm.WritethemissingnumbersundertheCodewordinincreasing order, the so called anticode (1,3,4,5,6). Connect the two first numbers of code and anticode and cross them out. If a crossed-out number of the code does not occur any more in the code then it is sorted into the anticode. Repeat, until the code vanishes. Then, the two last numbers of the code and anticode are connected.
94 5. Enumerative Combinatorics For Fig. 5.9, the algorithm runs as follows (Fig. 5.10): @ @ @ @ v v v v v v v 5 4 2 7 1 3 6 Fig. 5.9 v v v v v v v v v v 1 3 4 5 2 7 7 2 2 7 6−7 Fig. 5.10 @ @ @ @ @ @ Fig. 5.11 v L L L a a a !
!
! L L L a a a !!
! 1 2 n 3 x p p p Numbers missing in the code are the vertices of degree one.
E14. We want to generate a random tree. Take the spinner in Fig. 5.11 and spin it (n −2) times. There are nn−2 possible and equiprobable cases. The missing numbers correspond to the vertices of degree one. How many missing numbers are to be expected?
Obviously, P (# X is missing) P[(n −2) −times not X] 1 −1 n n−2 .
Hence, the expected number X of the missing numbers is E(n) E(X) n 1 −1 n n−2 ≈n e .
We can check this formula by Fig. 5.8 for computing T6.
E(6) 360 · 8 + 120 · 4 + 90 · 4 + 6 · 5 216 625 216.
For n 6, the above formula gives E(6) 6 · 5 6 4 625 216.
E15. Counting the same objects in two ways.
• Let us count the triples (x, y, z) from {1, 2, . . . , n+1} with z > max{x, y}.
Divide and Conquer! There are k2, of such triples with z k+1. Altogether there are 12 +22 +· · · n2 such triples. Again Divide and Conquer, but a little differently and deeper. Triples with x y < z, x < y < z, y < x < z are n + 1 2 , n + 1 3 , n + 1 3 .
Hence, we get 12 + 22 + · · · n2 n + 1 2 + 2 n + 1 3 .
5. Enumerative Combinatorics 95 • Now, we count the quadruples (x, y, z, u) with u > max{x, y, z}. Simple counting leads to 13 + 23 + · · · + n3.
After partitioning, sophisticated counting gives 3+1, 2+1+1, 1+1+1+1.
As above, n + 1 2 , 3 · 2 · n + 1 3 , 3! · n + 1 4 .
Hence, we get 13 + 23 + · · · + n3 n + 1 2 + 6 · n + 1 3 + 6 · n + 1 4 .
• We count all quintuples (x1, . . . , x5) from {1, 2, . . . , n + 1} with x5 > maxi≤4 xi. The simple counting again gives 14 + 24 + · · · n4.
Sophisticated counting uses the partitions 4 + 1, 3 + 1 + 1, 2 + 2 + 1, 2 + 1 + 1 + 1, 1 + 1 + 1 + 1 + 1. Thus, we get 14 + 24 + · · · n4 n + 1 2 + 14 n + 1 3 + 36 n + 1 4 + 24 n + 1 5 .
• Now we can prove the general formula 1k + 2k + · · · nk k i1 S(k, i) n + 1 i + 1 i!.
E16. The number of binary n-words with exactly m 01-blocks is n+1 2m+1 .
Solution. The result is the number of choices of a (2m+1)-subset from an (n+1)-set. Why (2m + 1) elements from (n + 1) elements? This result may direct us to 10-words. Look at the transitions 0 −1. There should be exactly m of these. But the number of 1−0 transitions can be m−1, m, or m+1. It would be nice to have exactly m + 1 transitions from 1 to 0. But we can always extend the word by a 1 at the beginning and a 0 at the end. Then we always have exactly (m + 1) transitions from 1 to 0. Altogether, we have an (n+2)-word with n+1 gaps. From these gaps, we freely choose 2m + 1 places for a switch. This can be done in n+1 2m+1 ways.
This is a very good example of the construction of a bijection.
E17. Find a closed formula for Sn n k1 n k k2.
Here is a sophisticated direct counting argument: The sum is the number of ways to choose a committee, its chairman, and its secretary (possible the same 96 5. Enumerative Combinatorics person) from an n-set. You can choose the chairman secretary in n ways, and the remaining committee in 2n−1 ways. The case chairman ̸ secretary can be chosen in n(n−1) ways and the remaining committee can be chosen in 2n−2 ways.
The sum is n · 2n−1 + n(n −1) · 2n−2 n(n + 1)2n−2.
Thus, we have the identity n k1 n k k2 n k0 n k k2 n(n + 1)2n−2.
The alternative would be an evaluation of the sum by transformation. It requires considerably more work and more ingenuity.
Sn k k0 n k k2 n k0 n k (k2 −k) + n k0 n k k n−2 k2 n k n −1 k −1 n −2 k −2 k(k −1) + n−1 k1 n k n −1 k −1 k n(n −1) n−2 k2 n −2 k −2 + n n−1 k1 n −1 k −1 n(n −1)2n−2 + n · 2n−1.
Here we twice used the formula n k0 n k 2n.
It can be proved by counting the number of subsets of an n-set in two ways. The left side counts them by adding up the subsets with 0, 1, 2, . . . , n elements. The right side counts them by the product rule. For each element, we make a two-way decision to take or not to take that element.
E18. Probabilistic Interpretation. Prove that n k0 n + k k 1 2k 2n.
We will solve this counting problem by a powerful and elegant interpretation of the result. First, we divide the identity by 2n, getting n k0 n + k k 1 2n+k n k0 pk 1.
This is the sum of the probabilities pk n + k k 1 2n+k .
5. Enumerative Combinatorics 97 Now, pk 1 2 n + k k 1 2n+k + 1 2 n + k k 1 2n+k P(Ak) + P(Bk) with the events Ak (n + 1) times heads and k times tails, and Bk (n + 1) times tails and k times heads.
See Fig. 5.12, which shows the corresponding 2n + 2 paths starting in O and ending up in one of the 2n + 2 endpoints, n + 1 vertical and n + 1 horizontal ones.
Here, we used the standard interpretation heads →one step upward, tails →one step to the right.
In Chapter 8, we give a much more complicated proof by induction.
E19. How many n-words from the alphabet {0, 1, 2} are such that neighbors differ at most by 1?
s s s s s s s s s s s s s s s s s s Fig. 5.12 0 1 2 Start ) ?
@ @ @ @ R Fig. 5.13 We represent the problem by the graph in Fig. 5.13. Each walk through the graph along the directed edges generates a permissible word. Missing arrows indicate that you may traverse the edge in both directions.
Let xn be the number of n-words starting from the starting state. Then the corresponding number from state 1 is also xn. By symmetry, the number of n-words starting in 0 or 2 is the same. We call it yn. From the graph, by the sum rule we read off, xn xn−1 + 2yn−1, (1) yn yn−1 + xn−1.
(2) From these difference equations we get 2yn−1 xn−xn−1 and 2yn xn+1−xn.
Putting the last two equations into (2), we get xn+1 2xn + xn−1.
(3) Initial conditions are x1 3, x2 7. From x2 2x1 +x0, we see that, by defining x0 1, the recurrence is satisfied. We start with x0 1, x1 3. The standard 98 5. Enumerative Combinatorics method for solving a difference equation is to look for a special solution of the form xn λn. Putting this into (3), for λ, we get, λ2 −2λ −1 0 with the two solutions λ1 1 + √ 2, λ2 1 − √ 2.
Thus, a general solution of (3) is given by xn a(1 + √ 2)n + b(1 − √ 2)n.
For n 0 and n 1, we get the equations for a, b: a + b 1, a(1 + √ 2) + b(1 − √ 2) 3 with the solutions a 1 + √ 2 2 , b 1 − √ 2 2 .
Thus, xn (1 + √ 2)n+1 2 + (1 − √ 2)n+1 2 .
E20. Find the number Cn of increasing lattice paths from (0, 0) to (n, n), which never are above the first diagonal. A path is increasing if it goes up or to the right only.
Fig. 5.14 shows how we can easily make a table of the numbers Cn with the help of the sum rule. By looking at the table, we try to guess a general formula. Besides looking at Cn, it is often helpful to consider the ratio Cn/Cn−1. This helps, but still it may be difficult. In our case, the ratio pn Cn/ 2n n of Cn to all the paths from (0, 0) to (n, n) is most helpful.
1 1 1 1 1 1 1 1 1 1 2 3 4 5 6 7 8 2 5 14 42 132 429 1430 2 5 9 14 20 27 35 14 28 48 75 110 42 90 165275 Fig. 5.14 n Cn Cn Cn−1 pn Cn (2n n) 0 1 − 1/1 1 1 2/1 6/3 1/2 2 2 5/2 10/4 1/3 3 5 14/5 1/4 4 14 42/14 18/6 1/5 5 42 132/42 22/7 1/6 6 132 429/132 26/8 1/7 7 429 1430/429 30/9 1/8 So we guess the formula Cn 1 n + 1 2n n .
5. Enumerative Combinatorics 99 This is a probabilistic problem. Among all 2n n paths from the origin to (n, n), we considered the good paths which never cross the line y x. A fundamental idea in probability tells us that: if you cannot find the number of good paths, try to find the number of bad paths. For the bad paths, we guess Bn 2n n − 1 n + 1 2n n n n + 1 2n n n n + 1 2n n 2n −1 n −1 2n n + 1 2n −1 n −1 2n n + 1 2n −1 n 2n n + 1 .
Here we used the formulas n k n k n−1 k−1 and n k n n−k in each direction. This result is easy to interpret geometrically. Indeed, the number of bad paths is the number of all paths from (−1, 1) to (n, n). Here (−1, 1) is the reflection of the origin at y x + 1. Now, we construct a bijection of the bad paths and all paths from (−1, 1) to (n, n). Every bad path touches y x + 1 for the first time. The part from O to y x +1 is reflected at y x +1. It goes into a path from (−1, 1) to (n, n), and any path from (−1, 1) to (n, n) crosses y x + 1 somewhere for the first time. If you reflect it at y x + 1, you get a bad path. Thus, we have a bijection between bad paths and all paths from (−1, 1) to (n, n). This so-called reflection principle is due to Desir´ e Andr´ e, 1887.
Cn are called Catalan numbers. They are almost as ubiquitous as the Pascal numbers n k . In the problems at the end of this chapter, you will find some more occurences of Catalan numbers.
E21. Principle of Inclusion and Exclusion (PIE or Sieve Formula).
This very important principle is a generalization of the Sum Rule to sets which need not be disjoint. Venn diagrams show that |A ∪B| |A| + |B| −|A ∩B| and |A ∪B ∪C| |A| + |B| + |C| −|A ∩B| −|B ∩C| −|C ∩A| + |A ∩B ∩C|.
We generalize to n sets as follows.
|A1 ∪· · · ∪An| n i1 |Ai| − i<j |Ai ∩Aj + i<j<k |Ai ∩Aj ∩Ak| −· · · (−1)n+1|A1 ∩· · · ∩An|.
Proof. Suppose an element a is contained in exactly k of the n sets Ai. How often is it counted by the right side? Obviously, k − k 2 + k 3 −· · · 1−(1−k + k 2 − k 3 + k 4 −· · ·) 1−(1−1)k 1 time. So it is counted exactly once. This proves the PIE.
As an example, we consider all n! permutations of 1, 2, . . . , n. If an element i is on place number i, then we say i is a fixed-point of the permutation. Let pn be 100 5. Enumerative Combinatorics the number of fixed point free permutations and qn the number of permutations with at least one fixed point. Then pn n! −qn.
Let Ai be the number of permutations with i fixed points. Then, qn |A1 ∪· · · ∪An| n 1 (n −1)! − n 2 (n −2)! + · · · + (−1)n+1 n n 0!, qn n!
1 −1 2! + 1 3! −1 4! + · · · + (−1)n−1 n!
, pn n!
1 0! −1 1! + 1 2! −1 3! + · · · + (−1)n−1 n!
≈n!
e , where e 2.71828 . . . .
Problems 1. Each of the faces of a cube is colored by a different color. How many of the colorings are distinct?
2. n persons sit around a circular table. How many of the n! arrangements are distinct, i.e., do not have the same neighboring relations.
3. Find the sum Sn n k1 n k k3. Hint: The sum can be interpreted as the number of ways of selecting a committee, a chairman, a vice-chairman, and a secretary, not necessarily different persons, from an n-set.
4. Let Rn be the number of ways to place n undistinguishable rooks peacefully on an n × n chessboard. Moreover, let Hn, Qn, Mn, Dn be the number of those placings, which are invariant with respect to a half-turn, a quarter-turn, reflection at a diagonal, and reflection at both diagonals. Find formulas for Rn, Hn, Qn, and find recursions for Mn, Dn.
5. 2n objects of each of three kinds are given to two persons, so that each person gets 3n objects. Prove that this can be done in 3n2 + 3n + 1 ways.
6. Of 3n+1 objects, n are indistinguishable, and the remaining ones are distinct. Show that one can choose from them n objects in 22n ways.
7. How many subsets of {1, 2, . . . , n} have no two successive numbers?
8. (a) Is it possible to label the edges of a cube by 1, 2, . . . , 12 so that, at each vertex, the labels of the edges leaving that vertex have the same sum?
(b) A suitable edge label is replaced by 13. Now, is equality of the eight sums possible?
9. In how many ways can you take an odd number of objects from n objects?
10. The vertices of a regular 7-gon are colored black and white. Prove that there are three vertices of the same color forming an isosceles triangle. For which regular n-gons is the assertion valid?
11. Can you arrange the numbers 1, 2, . . . , 9 along a circle, so that the sum of two neighbors are never divisible by 3, 5, or 7?
12. Four noncoplanar points are given. How many boxes have these points as vertices?
A box is bounded by three pairs of parallel planes (AUO 1973).
5. Enumerative Combinatorics 101 J J J J J """"" r r r r r r (a) A @ @ @ @ @ @ H H H H H H HHHHH H s s C sP sR s B sQ s s s p p p p p p p p p (b) Fig. 5.15 aaa J J J J J J @ @ @ HHHH H ppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp r r r r r r r r r r r (c) 13. In how many ways can you select two disjoint subsets from an n-set?
14. Let b(0) 1 and b(n) number of partitions of n into powers of 2 for n ≥1. Find recursions for b(n), and compute a table up to b(40).
15. A permutation p of the set {1, . . . , n} is called an involution, if p ◦p identity.
Find a recursion for the number tn of involutions of {1, . . . , n}. Also find a closed formula in the form of a sum.
16. Let f (n) be the number of n-words without neighboring zeros from the alphabet {0, 1, 2}. Find a recursion and a formula for f (n).
17. Figs. 5.15a,b,c show three configurations: the complete fourside, the Pappus–Pascal configuration, and the Desargues configuration. In how many ways can you permute their points, so that collinearity is preserved?
In the next four problems, you will find some occurrences of Catalan numbers. Your task will be to find an interpretation as good paths.
18. 2n points are chosen on a circle. In how many ways can you join pairs of points by nonintersecting chords?
19. In how many ways can you triangulate a convex n-gon?
20. In how many ways can you place parentheses in a nonassociative product of n factors?
21. How many binary trees with n-labeled leafs are there?
22. Find combinatorial proofs of the following formulas. Use bijection or counting the same objects in two ways.
(a) n s n s n −1 s −1 (b) n r r k n k n −k r −k (c) n i0 n i n n −i 2n n n i0 n i 2 (d) n s n n −s n −1 s e) n 0 + n 2 + · · · n 1 + n 3 + · · · (f) n 0 + n + 1 1 + n + 2 2 + · · · + n + r r n + r + 1 r .
23. How many games are needed to find the winner in a tennis tournament with n players, if the KO-system is used? Use bijection.
102 5. Enumerative Combinatorics 24. How many 5-words from the alphabet {0, 1, . . . , 9} have (a) strictly increasing digits, (b) strictly increasing or decreasing digits, (c) increasing digits, (d) increasing or decreasing digits?
25. In Lotto, 6 numbers are chosen from {1, 2, . . . , 49} with 49 6 possible 6-subsets. How many of these subsets have at least a pair of neighbors?
26. Let F(n, r) number of n-permutations with exactly r cycles Stirling number of the first kind. Prove the recurrences F(n + 1, r) F(n, r −1) + n F(n, r), F(n, 1) (n −1)!, F(n, n) 1.
27. Euler’s φ-function of a positive integer n is defined as follows: φ(n) the number of positive integers ≤n which are prime to n.
Prove that φ(n) n m i1 1 −1 pi , where p1, p2, . . . , pm are all distinct prime divisors of n. Use PIE.
28. Let m ≥n, Bm {1, . . . , m}, Bn {1, . . . , n}. The mappings from Bm onto Bn are called surjections. Find the number s(m, n) of surjections from Bm onto Bn. Use PIE.
29. Let a1, a2, a3 . . . be the infinite sequence 1, 2, 2, 3, 3, 3,. . . , where k occurs k times.
Find an f (n) in closed form. ⌊⌋and ⌈⌉are permitted.
30. Let 1 ≤k ≤n. Consider all finite sequences of positive integers with sum n. Suppose that the term k occurs T (n, k) times in all these sequences. Find T (n, k).
31. Consider a row of n seats. A child sits on each. Each child may move by at most one seat. Find the number Fn of ways they can rearrange.
32. Consider a circular row of n seats. A child sits on each. Each child may move by at most one seat. Find the number an of ways they can rearrange.
33. Consider all n-words from the alphabet {0, 1, 2, 3}. How many of them have an even number of (a) zeros, (b) zeros and ones?
34. Does a polyhedron exists with an odd number of faces, each face having an odd number of edges?
35. Can you partition the set of positive integers into infinitely many infinite subsets, so that each subset is generated from any other by adding the same positive integer to each element of the subset?
36. Given are 2001 pairwise distincts weights a1 < a2 < · · · < a1000 and b1 < b2 < · · · < b1001. Find the weight with rank 1001 by 11 weighings.
37. Consider all 2n −1 nonempty subsets of the set {1, 2, . . . , n}. For every such subset, we find the product of the reciprocals of each of its elements. Find the sum of all these products.
38. Find the number xn of n-words from the alphabet A {0, 1, 2}, if any two neighbors differ at most by 1.
39. How many n-words from the alphabet {a, b, c, d} are such that a and b are never neighbors?
5. Enumerative Combinatorics 103 40. Whatistheminimumnumberofpairwisecomparisonsneededtoidentifytheheaviest and second heaviest of 128 objects?
41. Prove that 139 comparisons are sufficient to identify the objects of ranks 1, 2, 3 in a set of 128 objects, if no two of them have the same weight.
42. Three of 128 objects are labeled A, B, C. You are told that A has rank 1, B has rank 2, and C has rank 3. How many comparisons do you need to check this?
43. A tramp has a coat of area 1 with five patches; each patch has area ≥1/2. Prove that two patches exist with common area ≥1/5.
44. Does the set {1, . . . , 3000} contain a subset A of 2000 elements such that x ∈A ⇒ 2x ̸∈A (APMO)?
45. Let 1 ≤r ≤n and consider all subsets of r elements of the set {1, 2, . . . , n}. Each of these subsets has a smallest member. Let F(n, r) denote the arithmetic mean of these smallest numbers; prove that F(n, r) n + 1 r + 1 (IMO 1981).
46. There are at most 2n/(n + 1) binary n-words differing in at least 3 places.
47. We call a permutation (x1, . . . , x2n) of the numbers 1, 2, . . . , 2n pleasant if |xi − xi+1| n for at least one i ∈{1, . . . , 2n −1}. Prove that more than one-half of all permutations are pleasant for each positive integer n (IMO 1989).
48. Define the sequence an by d|n ad 2n. Prove that n|an.
49. Along a one-way street there are n parking lots. One-by-one n cars numbered 1 to n enter the street. Each driver i heads to his favorite parking lot ai, and, if it is free, he occupies it. Otherwise, he continues to the next free lot and occupies it. But if all succeeding lots are occupied, he leaves for good. How many sequences ai are such that every driver can park (M.D. Haiman, J. Algebraic Combinatories, 3, 17–76 (1994) and SPMO 1996)?
Solutions 1. Call the six colors 1, 2, 3, 4, 5, 6. Put the cube on the table so that face 1 is at the bottom. Consider face 2. If it is at the top then we can rotate the cube about a vertical axis so that face 3 is in front. Now the cube is fixed. There are 3! 6 ways to complete the coloring. Now, suppose that face 2 is a neighbor of 1. The we rotate the cube so that 2 is in front. Now the cube is fixed, and the coloring can be completed in 4! 24 ways. Altogether, there are 6 + 24 30 distinct colorings of the cube by six colors.
2. Rotations and reflections at a line through the center conserve neighboring relation-ships. Thus we have n!/2n (n −1)!/2 distinct arrangements for n > 2.
3. We can choose the three VIP’s all different, and the whole committee in n(n−1)(n− 2)2n−3 ways. If all the VIP’s are the same person, then we have n2n−1 different committees. There are three choices for exactly two of the VIP’s to be the same person.Thenwehave3n(n−1)2n−2 choices.Altogether,wehaveSn n2(n+3)2n−3.
104 5. Enumerative Combinatorics 4. (a) Rn n!. Interpret the placings as permutations.
(b) Consider a 2n×2n board. In the first column, the rook can be placed in 2n ways.
Then, the rook in the last column is also fixed. We are left with an (2n−2)×(2n−2) board to be filled. Thus H2n 2nH2n−2, or H2n 2nn!. In a (2n + 1) × (2n + 1) board, the central cell remains fixed and must be occupied by a rook. Then we are left with a 2n × 2n board. Thus, H2n+1 H2n 2nn!.
(c) First, consider a 4n × 4n board. In the first column, there are 4n −2 ways to place a rook, since the corner cells must be left free. Then 4 rows and 4 columns are eliminated, and we are left with a (4n −4) × (4n −4) board. Thus, Q4n (4n−2)Q4n−4, or Q4n 2n(2n−1)(2n−3) · · · 3·1. In a (4n+1)×(4n+1) board, the central cell is fixed and must be occupied. We are left with a 4n × 4n board, i.e., Q4n+1 Q4n. It is easy to see that Q4n+2 Q4n+3 0. Indeed, except for the central cell, the rooks come up in quadruples.
(d) If the rook is placed on the diagonal in the first column, we are left with a (n −1) × (n −1) board. If it is placed on some of the other (n −1) cells, then we are left with a (n −2) × (n −2) board. Thus Mn Mn−1 + (n −1)Mn−2.
(e) In the first column of a 2n × 2n board, there are two ways to place the rook on a diagonal and 2n−2 other ways. In the first case we are left with a (2n−2)×(2n−2) board and in the second case, with a (2n −4) × (2n −4) board. Hence, D2n 2D2n−2 + (2n −2)D2n−4, D2n+1 D2n.
5. First solution. The result 3n2 + 3n + 1 (n + 1)3 −n3 is striking and allows a geometrical interpretation. One person gets x+y+z 3n objects with 0 ≤x, y, z ≤ 2n.Thesearetriangularcoordinatesforanequilateraltrianglewithaltitude3n.x, y, z can be interpreted as lattice points (make a figure). The hexagon in the figure can be interpreted as the projection of the cube with edge n + 1 from which a cube of edge n is subtracted. This solution is due to Martin H¨ arterich, a gold medallist of the IMO 1987/89.
Second solution. If the first person gets n −p (p ∈N0) objects of the first kind, then the person can get q to 2n objects of the second kind. The remaining ones are objects of the third kind. The sum is n p0 (2n −p + 1) (2n + 1)(n + 1) −n(n + 1) 2 .
If the first person gets n + q (k ∈N) objects of the first kind, then the person gets 0 to (2n −q) objects of the second kind, since the person gets 3n objects altogether.
The sum is n q1 (2n −q + 1) n(2n + 1) −n(n + 1) 2 .
The sum altogether is (2n + 1)(n + 1) + n(2n + 1) −n(n + 1) 3n2 + 3n + 1.
6. We can take n objects in 2n + 1 n + 2n + 1 n −1 + · · · + 2n + 1 0 ways. We add to this number the same number 2n + 1 n + 1 + 2n + 1 n + 2 + · · · 2n + 1 2n + 1 and get 22n+1. Thus there are 22n ways to choose n objects.
5. Enumerative Combinatorics 105 7. We interpret the subsets as n-words from the alphabet {0, 1}. Let an be the number of binary words with no two successive ones. The words can start either with 0 and may continue in an−1 ways, or they start with 10 and may continue in an−2 ways.
Thus, an an−1 + an−2, a1 2, a2 3. Thus, an is the Fibonacci number Fn+2.
8. (a) Suppose there is such a numbering. Let the sum of the edge labels for each vertex be s. Then the sum of all vertex sums is 8s. In this sum, each edge label occurs twice.
Thus 2(1 + · · · + 12) 8s, or s 19.5. Since s is a positive integer we have a contradiction.
(b) Replace one number in the numbering of the edges by 13, and call the replaced number by r. Then we have 2(1 + · · · + 12 + 13) −2r 8s, or 91 −r 4s, that is, r ∈{3, 7, 11}. This necessary condition is also sufficient. Try to find a corresponding labeling for some value of r.
9. There is a bijection between subsets with an even and odd number of elements.
Indeed, consider any element, say 1. Let A be any subset. If it contains 1, then we assign the subset A\{1} to it. If it does not contain 1, then we assign the subset A ∪{1} to it. This bijection proves that exactly one-half of all 2n subsets contain an odd number of elements, that is, 2n−1.
10. The solution will be found in Chapter 4.
11. Write the nine numbers along a circle, and draw a line between any two numbers for which the sum is not 3, 5, or 7. We get a graph, for which we must find a Hamiltonian circuit. Now, such a circuit is easy to find since 1, 2, and 4 have only two neighbors.
One gets 1, 3, 8, 5, 6, 2, 9, 4, 7.
12. This problem is instructive, since besides Divide and Conquer, we practice space geometry and spacial intuition. First, we solve the analogous plane problem. Three noncollinear points are given in the plane. How many parallelograms are there with vertices in these points?
This problem is considerably simpler and at first does not help much for the space analogy. The answer is 3. The fourth vertex of the parallelogram is obtained by reflecting each of the given vertices A, B, C at the midpoint of the side of △ABC to A1, B1, C1.
First solution. Of the 8 vertices we single out four noncoplanar ones. There are four distinct ways of doing this (Divide and Conquer). In Fig. 5.16 (a) to (d) 3, 2, 1, 0 faces have three of the four fixed points. Such faces will be called rigid, since they can be constructed from their three points.
(a) Three rigid faces have a common vertex A. In this case there are four boxes.
(b) Two rigid faces have a common edge AB. Any two points can play the role of AB. The choice of AB can be done in six ways. Then we can decide in two ways which point is joined to B. In this case there are 12 boxes.
(c) There is one rigid face with three points and the fourth vertex D′ must lie opposite D. Each of the four vertices can be D, and each of the other three can be D′. Then the box is constructible. We have 12 boxes again.
(d) There is no rigid face. The selected vertices are the vertices of a tetrahedron inscribed in the box. The box is uniquely determined. Through each edge of the tetrahedron, we draw the plane parallel to the opposite edge.
106 5. Enumerative Combinatorics s s s s A (a) s s s s A B (b) Fig. 5.16 s s s s D D′ (c) s s s s (d) / / / // S S S S S / / / S S S A B C B1 C1 A1 Fig. 5.17 There are altogether 4 + 12 + 12 + 1 29 boxes.
Second solution. Look at the plane problem in Fig. 5.17. The answer three can be obtained as follows. The midlines of the future parallelogram are lines equidistant from the three vertices. If we have two midlines, then we can easily find the missing vertex of the parallelogram. We draw parallels to the given points. There are three straight lines equidistant from the three given points. One can select 2 lines from them in three ways.
Now to the space problem. The midplanes of the box we want to find are equidistant from all 8 vertices and satisfy the following conditions: (1) Each is equidistant from the four points.
(2) All three planes pass through a point.
On the other hand, if we can find a triple of planes satisfying (1) and (2), then the box with these midplanes is uniquely constructible. Through the four points draw all distinct planes parallel to each plane of the triples, and the box is ready.
How many planes are equidistant from four noncoplanar points K, L, M, N? Ex-actly 7. It suffices to decide which points lie on one side of a plane. There are four of the type 1 | 3 and three of the type 2 | 2. Three planes out of seven can be selected in 35 ways. Each triple satisfies condition (1). Which triples are “bad,” i.e., do not satisfy (2)? They are parallel to a line. There are six of these, as many as the number of edges of the tetrahedron, that is, 35 −6 29. Why?
Third solution. Of the 8 vertices, we can single out four in 70 ways, of which there are 6+6 12 coplanar quadruples. Thus, we are left with 70−12 58 noncoplanar quadruples. But to every such quadruple, there is a complementary quadruple which gives the same box. Hence, we are left with 29 boxes.
13. For an ordered pair (A, B) of disjoint subsets, we define the characteristic function f (x) ⎧ ⎨ ⎩ 1 if x ∈A, 2 if x ∈B, 0 otherwise.
Then the function f is an n-word from the alphabet {0, 1, 2}. The number of possible functions is 3n. There are 2n words from {0, 2} (A empty), 2n n-words from {0, 1} (B 5. Enumerative Combinatorics 107 empty), and 1 word consisting entirely of zeros. Then A and B are both empty. Thus, the number of ordered disjoint pairs is 3n −2n −2n + 1. The number of unordered pairs is g(n) 3n + 1 2 −2n.
Check this for n 4 by drawing pictures.
14. Consider some examples: b(0) 1 by definition. 20 1 ⇒b(1) 1. 2 21 1 + 1, b(2) 2, 3 2 + 1 1 + 1 + 1, b(3) 2. 4 22 21 + 21 21 +1+1 1+1+1+1, b(4) 4. 5 22 +1 21 +21 +1 21 +1+1+1 1 + 1 + 1 + 1, b(5) 4.
We observe that (a) b(2n) b(2n + 1) and (b) b(2n) b(2n −2) + b(n). Proof of (a): Every partition of 2n + 1 has a last summand 1. If we take it away, we get a partition of 2n. Proof of (b): A partition of 2n has either a smallest element 2 or two ones. There are b(n) of the first kind and b(2n −2) of the second kind.
15. Let tn be the number of involutions of {1, . . . , n}, i.e, the permutations p such that p ◦p is the identity. Add another element n + 1. It can be a fixed point in tn ways.
It is not a fixed point in ntn−1 ways, that is, tn+1 tn + n · tn−1, t1 1, t2 2.
The closed formula for tn is tn ⌊n/2⌋ k0 n 2k (2k)!
2kk! .
Interpretation of this formula: From n elements, we select 2k. This can be done in n 2k ways. Then we partition them into k unordered pairs in (2k)!/(2k · k!) ways. The re-maining n−2k points are fixed points. This must be summed for k 0, 1, . . . , ⌊n/2⌋.
Thus, we get tn ⌊n/2⌋ k0 n 2k · 1 · 3 · 5 · · · (2k −1).
16. The words can begin with 1, 2 and continue in f (n −1) ways, or they can start with 01, 02 and continue in f (n −2) ways. Thus we have the recurrence f (n) 2f (n −1) + 2f (n −2) with f (1) 3, f (2) 8. From the recurrence, we get f (0) 1. Thus, finally, we have f (n) 2f (n −1) + 2f (n −2), f (0) 1, f (1) 3, f (2) 8.
The characteristic equation of this difference equation is λ2−2λ−2 0. Thus we get λ1,2 1± √ 3. Now, it is easy to find a close formula of the form f (n) aλn 1 +bλn 2.
We get a, b from initial conditions. Do it.
17. Answer (a) 24. (b) 108. (c) 120. We show how to get (b): Label the 9 points of the configuration A, B, C, P , Q, R, L, M, N so that ABC and PQR are as in Fig.
5.15b. We want to relabel the configuration such that collinearity is conserved. There are nine ways to choose A. Say A is fixed. For B, there are six places left, since A and B are collinear. For P , there are only two ways. Now the places of all the other points are fixed. So, there are 6 · 4 · 2 108 possible collineations.
108 5. Enumerative Combinatorics ( ( ( ( ( ( H H H H H H b b b b b b b b e e b e e b e b Fig. 5.18 b e A A A @ @ A A AA (a) A A A A (b) A A A A QQ Q Fig. 5.19 (c) 18. Draw suitable chords between pairs of points. Go around the circle in any sense.
Meeting a diagonal first, we label its endpoint by b (for beginning). Meeting it for the second time, we label the endpoint by e (for end). For Fig. 5.18, we get the word bbbeebbbeeee. This is a good path in the sense of E14 using the interpretation b →to the right and e →up. Thus, we have a bijection between good paths and words. Hence, the number of possible ways is Cn 1 n + 1 2n n .
19. Let Tn be the number of distinct triangulations of an n-gon. We try to find a recurrence for Tn. Consider a triangle A1AnAk. It splits the polygon into a k-gon and an (n + 1 −k)-gon. Define T2 1. Then, Tn T2Tn−1 + T3Tn−2 + T4Tn−3 + · · · + Tn−1T2.
Fig. 5.19 shows some triangulations giving T3 1, T4 2, T5 5, T6 14. This is a strong indication that, generally, we have Tn+2 Cn. We can also find the next number T7 42 by means of the recursion. But it is not obvious how to get from the recursion to the closed formula. See the next problem.
20. There is one way to set parentheses in one or two factors: (x1), (x1x2). For three factors we have two ways: ((x1x2)x3) and (x1(x2x3)). For 4 factors we have 5 ways: (((x1x2)x3)x4), (x1(x2(x3x4))), ((x1x2)(x3x4)), ((x1(x2x3))x4), (x1((x2x3)x4)). Hence, a1 1, a2 1, a3 2, a4 5.
To get a recursion for an, take the last multiplication (x1 · · · xk)(xk+1 · · · xn). Here, k can run 1 to n −1. Summing the results, we get an a1an−1 + a2an−2 + · · · + an−1a1.
We have a1 T2 1, a2 T3 1, a3 T4 2, a4 T5 5. We have the same recursion with the same initial conditions, giving the same result. Thus, we conjecture that an+1 Tn+2 Cn. Hence, there should be a hidden interpretation by good paths of a random walk. It uses the following interpretation: Ignore the last element xn.
Now, scan the parenthesized expression from left to right. Whenever we come to an 5. Enumerative Combinatorics 109 open parenthesis, go one step to the right; for every xi, go one step up. Notice that we ignore the closed parentheses. If they were all deleted, all multiplications would still be uniquely determined. Another interpretation is even more direct. Ignore the xi, but keep all parentheses. Now, we can use Fig. 5.20 to get well-formed expressions leading from state 0 back to 0 in 2n steps.
( ) ( ) ( ) ( ) 0 1 2 3 Fig. 5.20 21. Fig. 5.21 gives a one-to-one mapping of parenthesized expressions and binary trees.
r r r r r r r @ @ @ @ @ @ a((bc)d) a b c d Fig. 5.21 r r r r r r r a b d c @ @ @ @ @ @ (a(bc))d 22. (a) From a set of n people choose an s-committee and in the committee a chairman.
We count in two ways: (i) Choose the committee in n s ways, and in the committee choose a chairman in s ways.
(ii) Choose a chairman in n ways, then the ordinary members in n−1 s−1 ways. Thus, n · n −1 s −1 s · n s ⇒ n s n s n −1 s −1 .
(b) Choose a subset of n persons from n men and n women. The left side partitions this number according to the number i of women (men). The middle term counts the n-subsets on a 2n-set. In the right side, we use the bijection subset↔complement.
(c) From an n-set, select an r-subset and in the r-subset a k-subcommittee. This gives the left side. We can also choose the k-subcommittee from the n-set, then, the remaining (r −k) committee members in n−k r−k ways.
(d) From the n-set, select an s-subset, and, from the remaining persons, a controller who must not be in the subset. You can first choose the subset, then, from the complementary subset, choose the controller. You can also choose the controller, then from the remaining n −1 the s-subset.
(e) This says that the number of even subsets equals the number of odd subsets.
We have done this already. Another proof uses the binomial theorem (1 + x)n n s0 n s xs. Setting x −1, we get 0 (1 −1)n n k0 (−1)k n k ⇒ n 0 + n 2 + · n 1 + n 3 + · · · .
(f) The right side gives the number of r-subsets of a set with (n + r + 1) elements.
The left side gives the same subsets (beginning at the end), but sorted as follows: those without element 1, with 1 but without 2, with 1, 2 but without 3, with 1, 2, 3 but without 4,. . ., with 1, 2, 3, . . . r, but without r + 1.
110 5. Enumerative Combinatorics 23. (n −1) games. 1 plays against 2, the winner against 3, the winner against 4, and so on. There is no shorter way. Indeed, there must be (n −1) losers.
24. (a) 10 5 252. (b) 512. (c) We can choose five digits from ten digits with repetition in 10+5−1 5 14 5 2002 ways. (d) 2·2002−10. In the last result, we must subtract 10 for the 10 words of the form aaaaa, which are both increasing and decreasing.
25. We find the number of 6-subsets with no neighbors. We think of the 49 numbers as a row of 49 balls, the 43 unselected balls, white, the six selected balls black. No two black balls must be neighbors. Thus we have 44 places for them. One can select six places from them in 44 6 ways. Thus, there are altogether 49 6 − 44 6 subsets with at least one neighbor. These are 49.5% of all subsets.
26. Add another point n + 1. There are two possibilities. First, (n + 1) is a fixed point (1-cycle). Then, the remaining n points must be arranged in (r −1) cycles. This can be done in F(n, r −1) ways. Second, the point is included in some cycle. In this case, there are already r cycles, which can be built in F(n, r) ways. In how many ways can the new point be included in a cycle? It can be put in front of any of the n points. This can be done in n ways. Thus, F(n + 1, r) F(n, r −1) + nF(n, r), F(n, 1) (n −1)!, F(n, n) 1.
27. Let Ai be the subset of all numbers from {1, . . . , n} divisible by pi. Then, the number of numbers from 1 to n divisible by some prime is |A1 ∪A2 ∪· · · ∪An| i n pi − i<j n pipj + i<j<k n pipjpk −· · · .
The number of elements not divisible by any of the primes p1, . . . , pm is n − i n pi + i<j n pipj −· · · n m i1 1 −1 pi .
28. Let Ai be the set of mappings in which the element i ∈Bn is not hit by an arrow from Bm. Then the number of nonsurjections is |A1 ∪A2 ∪· · · ∪An| n 1 (n −1)m − n 2 (n −2)m + · · · .
m ≥n.
If we subtract this number from the number nm or from n 0 (n −0)m of all mappings from Bm to Bn, then we get s(m, n). For m ≥n, we get s(m, n) n 0 (n −0)m − n 1 (n −1)m + · · · n i0 (−1)i n i (n −i)m.
29. First solution. We want to have an k if k(k −1) 2 < n ≤k(k + 1) 2 .
5. Enumerative Combinatorics 111 Since n is an integer, this is equivalent to k(k −1) 2 + 1 8 < n < k(k + 1) 2 + 1 8 or k2 −k + 1 4 < 2n < k2 + k + 1 4, that is, k −1 2 < √ 2n < k + 1 2 ⇒k < √ 2n + 1 2 < k + 1.
Hence, an ⌊ √ 2n + 1/2⌋, which is the nearest integer to √ 2n.
Second solution. We want to have an k if k(k −1)/2 < n ≤k(k + 1)/2. The equation k(k + 1)/2 r can be solved for (positive) k: k −1 + √ 1 + 8r 2 .
Hence, −1 + √ 1 + 8n 2 ≤k < −1 + √ 1 + 8n 2 + 1 ⇒an −1 + √ 1 + 8n 2 .
The two results have different form, but are equivalent.
30. Consider a row of n points. These points form (n −1) gaps. We can insert vertical bars into these gaps in 2n−1 ways. In this way, we get all sequences with sum n. To find the number T (n, k) of all terms k in all these sequences, first we draw a sequence of n points. Then we pack k successive points into a rectangle and place vertical bars to the right and left of it.
· · · | · | · | · · · | ·· ⇔(3, 1, 1, 3 , 2).
First case. The packed points do not contain an endpoint. The packing can be done (n−k −1) ways. There will remain (n−k −2) gaps between the nonpacked points.
One can insert at most one vertical bar into each gap in 2n−k−2 ways. Thus we get a sequence with one packed term k.
Second case. The packed points contain an endpoint. This can occur in two ways, and there are now (n −k −1) gaps, into which we can insert bars in 2n−k−1 ways.
Altogether, one gets T (n, k) (n −k −1) · 2n−k−2 + 2 · 2n−k−1 (n −k + 3) · 2n−k−2.
Example:Withn 6,k 2,theformulagivesT (6, 2) 28.Allsequenceswithsum 6, which contain at least one 2 are (2, 2, 2), (4, 2), (2, 4), (3, 2, 1) and permutations, (2, 2, 1, 1) and permutations, (2, 1, 1, 1, 1) and permutations. The number of twos in these sequences is T (6, 2) 3 + 1 + 1 + 6 + 12 + 5 28.
31. Consider children and seats numbered 1, . . . , n. Let an be the number of rearrange-ments. There are an−1 rearrangements with the first child staying in its place. If child 1 moves to 2, then 2 must move to 1. There are an−2 such rearrangements. Thus we have an an−1 + an−2, a1 1, a2 2. Thus, an Fn+1, where Fn is the nth Fibonacci number.
112 5. Enumerative Combinatorics 32. Let bn be the number of seatings. There are three cases: (a) Child 1 remains seated. There will be an−1 seatings of this kind.
(b) 1 and 2 are interchanged. There are an−2 such seatings.
(c) All the children move one seat to the right or left. There are two such seatings.
We get bn an−1 + an−2 + 2 an + 2 Fn+1 + 2 seatings.
33. Suppose en and on are the number of n-words with an even and odd number of zeros, respectively. By partitioning the words according to the first digit, we get the recurrences en 3en−1 + on−1, on en−1 + 3on−1. This is a linear mapping from (en−1, on−1) to (en, on) with matrix 3 1 1 3 . Its eigenvalues λ1,2 satisfy the equation 3 −λ 1 1 3 −λ 0, or (λ −3)2 −1 0, or λ1 4, λ2 2. Find a closed formula for en. Try to solve the problem for an even number of zeros and ones.
Alternate solution. The number of n-words from {0, 1, 2, 3} with an even number of zeros is En 3n + n 2 3n−2 + n 4 3n−4 + · · · and with an odd number of zeros On n 1 3n−1 + n 3 3n−3 + · · · .
Adding and subtracting we get En + On (3 + 1)n 4n, En −On (3 −1)n 2n.
Adding and subtracting again, we get 2En 4n + 2n ⇒En 4n + 2n 2 , 2On 4n −2n ⇒On 4n −2n 2 .
34. Let ei be the number of edges of the ith face. Then, ei is an odd number of odd summands. This number is odd. On the other hand, every edge in the sum is counted twice. So, it must be an even number. This contradiction proves the nonexistence of such a polyhedron.
35. Yes, it is possible. First, consider two subsets A, B of positive integers. We include in A all positive integers with zeros at even positions (starting at the right). We include in B all positive integers with zeros at odd positions. Every positive integer can be uniquely represented in the form n a + b, a ∈A, b ∈B. The partition of the positive integers N A1 ∪A2 ∪A3 · · · is as follows: A1 A, and we get each Ak(k 2, 3, . . .) from A by adding to its elements bk ∈B, i.e., A2, A3, . . . are the translations of A by corresponding elements of the set B.
36. Let a500 > b501. Of the weights a501, . . . , a1000, which are heavier than the 1001 weights a1, . . . , a500, b1, . . . , b501, and thus cannot be the median, we take away a513, a514, . . . , a1000. Of the weights b1, . . . , b500, which are lighter than the 1002 weights b501, . . . , b1001, a500, . . . , a1000, and which cannot contain the median, we eliminate b1, . . . , b489. The median is now the 512th lightest. In 10 weighings, we 5. Enumerative Combinatorics 113 can now reduce the number of weights to 1 as follows: We have pairwise distinct weights c1 < · · · < c2l and d1 < · · · < d2l (l 2k−1), and we must find the 2l lightest weight. First, we compare cl with dl. If cl > dl, then cl+1, . . . , c2l are heavier than the 2l weights c1, . . . , cl, d1, . . . , dl and can be eliminated; the weights d1, . . . , dl are lighter than the (2l + 1) weights cl, . . . , c2l, dl+1, . . . d2l and can be eliminated. There will remain l weights of each sort, of which we are to find the l-lightest. Similarly we proceed with the case cl < dl. In the case a500 < b500, we must invert all inequality signs in the preceding case and replace “lightest” by “heaviest.” 37. If we multiply the product (1 + 1/1) (1 + 1/2) · · · (1 + 1/n), we get 2n summands.
Each summand is the product of the reciprocals of one of the 2n subsets of {1, . . . , n}.
If we throw away the 1, which corresponds to the empty set, we get the sum we want.
It is 2 · 3 2 · 4 3 · · · n + 1 n −1 n + 1 −1 n.
38. From the graph in Fig. 5.22, we read off the recurrences xn xn−1 + 2yn−1 and yn yn−1 + xn−1. From the first, we get 2yn−1 xn −xn−1 and 2yn xn+1 −xn.
0 1 2 yn xn yn Fig. 5.22 Inserting this in the second recursion, we get xn+1 2xn + xn−1, λ2 2λ + 1, λ1,2 1 ± √ 2.
Find a closed expression for xn!
39. From the graph in Fig. 5.23, we read off the recurrences xn 2xn−1 + 2yn−1, yn 2xn−1 + yn−1.
By eliminating yn and yn−1, we get the recurrence xn+1 3xn + 2xn−1 with the characteristic equation λ2 3λ + 2. Find closed expressions for xn.
j j j j @ @ @ $ % $ % $ % $ % Fig. 5.23 xn xn yn yn c b d a 40. We play a seven-round KO-elimination tournament.
First Round: The 128 objects are separated into 64 pairs, and the lighter component in each pair is eliminated.
Second Round: The 64 winners play 32 games, and 32 are eliminated, and so on. In the seven rounds, 127 comparisons are made, and the object of rank 1 is identified.
Candidates for rank 2 are the seven objects that lost, one in each round, to the rank 1 object. These seven candidates play an elimination tournament and find the winner in six additional comparisons. Thus, the objects of rank 1 and rank 2 can be identified in 127 + 6 133 comparisons.
114 5. Enumerative Combinatorics 41. The object of rank 1 is determined in seven rounds as in the preceding problem. This requires 127 comparisons. Candidates for rank 2 are the seven objects that lost to the rank 1 object. We number them from 1 to 7 such that #i was eliminated in the ith round. The object with rank 2 is determined in a second tournament as follows. #1 is compared with #2, the winner with #3, the winner with #4, and so on. The winner of the last round is the object of rank 2. This required six comparisons.
Candidates for rank 3 are the objects which lost to the object of rank 1 and/or rank 2.
They may not have been matched against the rank 1 object; however, they must have lost against rank 2, or else they would still be candidates for rank 2. But the rank 2 object has won at most seven comparisons. Indeed, suppose #i is the object of rank 2. Then it won (7 −i) games against #i + 1, #i + 2, · · · , #7, and one more against #i −1 if i > 1. Thus, the rank 2 object has won altogether either i −1+7−i +1 7 or i −1+7−i 6 games. Hence, there are at most seven candidates for third rank.
The heaviest of these can be found in six comparisons. Thus, we find the objects of rank 1, 2, 3 in at most 127 + 6 + 6 or 139 comparisons.
42. To check this, 127 comparisons are sufficient. First, ignore A, B, C. Among the remaining 125 objects, we find the heaviest object D in 124 comparisons. Then D plays against C and loses, C loses against B, and B against A. 127 comparisons are also necessary, because each object, except that of rank 1, must lose at least one match.
43. Use the PIE to get the necessary estimate. A hard problem.
44. We generalize slightly. A set S of integers is called double-free (D.F.) if x ∈S ⇒ 2x ̸∈S. Let Tn {1, 2, . . . , n} and f (n) max{|A| : A ⊂Tn is D.F.}. Then, using the PIE, we get f (n) n −⌊n/2⌋+ ⌊n/4⌋−⌊n/8⌋+ ⌊n/16⌋−· · · .
We subtract the even integers from n, then add the multiples of 4, subtract the multiples of 8, and so on. For n 3000, we get 1999. The answer is no!
E.T.H. Wang (Ars Comb. 1989) proved that f (n) ⌈n/2⌉+ f (⌊n/4⌋). Solve the problem for n 3000 by this formula.
Try to solve the problem about the maximal triple-free subset of Tn. A triple-free set A has the property x ∈A ⇒3x ̸∈A.
45. Let n r denote the number of r-element subsets of an n-set. The sum of the least elements of the r-element subsets of {1, . . . , n} is n r F(n, r). Consider the mapping from the set of (r +1)-element subsets of {0, 1, . . . , n} to the set of r-element subsets of {1, . . . , n} which strips the least element off each such (r + 1)-element subset.
Clearly, under this mapping, each r-element subset of {1, . . . , n} occurs as an image exactly i times, where i is its least element. Hence, counting the (r + 1)-element subsets of {0, 1, . . . , n} both directly and via the mapping, n + 1 r + 1 n r F(n, r) ⇒F(n, r) n + 1 n + 1 / n r (n + 1)/(r + 1).
Here, we used the fact n+1 r+1 n+1 r+1 n r which can be found by counting in two ways, without knowing a formula for n r .
This proof is due to Dr. M. F. Newmann from the Australian National University. It requires no computation.
5. Enumerative Combinatorics 115 An identical proof using the language of graph theory was sent to me by Cecil Rousseau of Memphis State University. It runs as follows.
Consider the bipartite graph in which the black vertices are the (r + 1)-element subsets of {0, . . . , n}, the white vertices the r-element subsets of {1, . . . , n} and a black vertex X is adjacent to the white vertex Y obtained by deleting the smallest element from X. Our bipartite graph has n+1 r+1 black vertices, n r white vertices, and n+1 r+1 n+1 r+1 n r edges. Note that the degree of a white vertex is the value of its least element. Thus, the desired average minimum element is the average degree (n + 1)/(r + 1) of a white vertex.
The proofs by the students used computation with binomial coefficients. Find such a proof. It is easy. Also try to prove the following generalization.
The arithmetic mean of the k-th largest elements of all r-subsets of the n-set {1, . . . , n} is F(k, n, r) k n + 1 r + 1 .
The simplest proof uses probability. Take (n + 1) equally spaced points on a circle of length (n + 1). Choose (r + 1) of the (n + 1) points at random. The chosen points split the circle into (r + 1) parts. By symmetry each part has the same expected length (n + 1)/(r + 1). Cut the circle at the (r + 1)th chosen point and straighten it into a segment of length (n + 1). Then I have r chosen points along the points {1, 2, . . . , n}, and the expected value of the distance of the minimal selected point from the origin (one of the endpoints) is (n + 1)/(r + 1). By the same symmetry argument, the distance from the origin to the k-th largest point is F(k, n, r) k n + 1 r + 1 .
46. Suppose there are altogether p words wi, (i 1, 2, . . . , p) of length n differing at least in three places. We write them in one line. Under each of these words, we write the column of all words differing from the top word by exactly one letter.
The words of any two columns differ at least by one letter. We have p columns of (n+1) different words which is at most 2n, the number of all binary n-words. Hence p(n + 1) ≤2n, or p ≤2n/(n + 1).
47. For k ∈{1, . . . , n}, let Ak be the set of all permutations of {1, . . . , 2n} with k and k + n in neighboring positions. For the set A ∪n k1Ak of all pleasant permutations the PIE yields |A| n k1 |Ak| − k<l≤n |Ak ∩Al| + k<l<m≤n |Ak ∩Al ∩Am| −· · · .
(1) This is a series of monotonically decreasing alternating terms. Hence, |A| ≥ n k1 |Ak| − k<l≤n |Ak ∩Al|.
We have |Ak| 2(2n −1)! since there are (2n −1)! possibilities to arrange the elements x ̸ k, x ∈{1, . . . , 2n} and two posibilities for the order (k, k + n) or (k+n, k). We have |Ak ∩Al| 22 ·(2n−2)!. Indeed, there are (2n−2)! possibilities 116 5. Enumerative Combinatorics to arrange the 2n −2 objects x ̸ k, x ̸ l, and then, 22 possibilities for the order of the two pairs {k, k + n} and {l, l + n}. Thus, we get |A| ≥ n n1 2(2n −1)!
− k<l 22(2n −2)! 2n(2n −1)! − n 2 · 22 · (2n −2)! > (2n)!
2 .
By using the whole series (1), one can prove that |A| (2n)! →1 −e−1 0.632.
48. A binary n-word W is repeating if it can be split for some d|n into d identical blocks.
Any n-word can be generated by repeating its unique longest nonrepeating initial block. Hence, the given recurrence counts the number of nonrepeating n-words.
Now, the claim follows from the obvious fact that from a given nonrepeating n-word cyclic shifting yields n distinct nonrepeating n-words.
49. Answer: (n + 1)n−1. Add an (n + 1)st parking lot and extend the street to a circuit leading from the (n + 1)st to the first lot. There are (n + 1)n sequences ai since each driver has n + 1 choices. One lot will remain empty. The sequence ai is good, i.e., it solves the original problem if the place (n + 1) remains empty. Split the sequences ai into (n + 1)n−1 groups of n + 1 each. One group comprises all cyclic shifts of a sequence and only one of these is good. This can be extended to a proof of Cayley’s theorem on the number of labeled trees with (n + 1) vertices.
6 Number Theory Number Theory requires extensive preparation, but the prerequisites are very finite.
One usually can use the prerequisites 1 to 19 without proof. Here all variables stand for integers. The strategies are acquired by massive problem solving. At first the problems are far below a hard competitive level. But if you do most of the problems you are fit for any competition.
1. If b aq for some q ∈Z, then a divides b, and we write a|b.
2. Fundamental Properties of the Divisibility Relation I. a | b, b | c ⇒a | c.
II. d | a, d | b ⇒d | ax + by. Especially d | a + b, d | a −b.
III. If any two terms in a + b c are divisible by d, the third will also be divisible by d.
3. Division with Remainder. Every integer a is uniquely representable by the positive integer b in the form a bq + r, 0 ≤r < b.
q and r are called quotient and remainder upon division of a by b.
4. GCD and Euclidean Algorithm. Let a and b be nonnegative integers, not both 0. Their greatest common divisor and least common multiple will be denoted by gcd(a, b) and lcm(a, b), respectively. Then gcd(a, 1) 1, gcd(a, a) a, gcd(a, 0) a, gcd(a, b) gcd(b, a).
118 6. Number Theory a and b will be called relatively prime or coprime, if gcd(a, b) 1. With gcd(a, b) gcd(b, a −b), (4) we can compute gcd(a, b) by subtracting repeatedly the smaller of the two numbers from the larger one. The following example shows this: gcd(48, 30) gcd(30, 18) gcd(18, 12) gcd(12, 6) gcd(6, 6) 6.
The Euclidean algorithm is a speedup of this algorithm, and it is based on a bq + r ⇒gcd(a, b) gcd(b, r) gcd(b, a −bq).
(5) Theorem. The gcd(a, b) can be represented by a linear combination of a and b with integral coefficients, that is, there are x, y ∈Z, so that gcd(a, b) ax + by.
Special case: If a and b are coprime, then the equation ax + by 1 has integral solutions.
5. gcd(a, b) · lcm(a, b) a · b.
6. A positive integer is called a prime if it has exactly two divisors.
7. Euclid’s Lemma. If p is a prime, p | ab ⇒p | a or p | b.
8. Fundamental Theorem of Arithmetic. Every positive integer can be uniquely represented as a product of primes.
9. There are infinitely many primes since p ̸ | (n! + 1) for any prime p ≤n.
10. n! + 2, n! + 3, . . . , n! + n are (n −1) consecutive composite integers.
11. The smallest prime factor of a nonprime n is ≤√n.
12. All primes p > 3 have the form 6n ± 1.
13. All pairwise prime triples of integers satisfying x2 + y2 z2 are given by x |u2 −v2|, y 2uv, z u2 + v2, gcd(u, v) 1, u ̸ v mod 2.
14. Congruences. a ≡b mod m ⇔m | a −b ⇔a −b qm ⇔a b + qm ⇔a and b have the same remainder upon division by m. Congruences can be added, subtracted, and multiplied.
Suppose a ≡b mod m and c ≡d mod m. Then a ± c ≡b ± d mod m, and ac ≡bd mod m, 6. Number Theory 119 This has several consequences: a ≡b mod m ⇒ak ≡bk mod m and a ≡b mod m ⇒f (a) ≡f (b) mod m, where f (x) anxn + an−1xn−1 + · · · + a1x + a0, ai ∈Z.
In general we cannot divide, but we have the following cancellation rule: gcd(c, m) 1, ca ≡cb mod m ⇒a ≡b mod m.
15. Fermat’s Little Theorem (1640). Let a be a positive integer and p be a prime. Then ap ≡a mod p.
The cancellation rule tells us that we can divide by a if gcd(a, p) 1, getting gcd(a, p) 1 ⇒ap−1 ≡1 mod p.
16. Fermat’s theorem is the first nontrivial theorem. So we give three proofs.
First proof by induction. The theorem is valid for a 1, since p|1p −1.
Suppose it is valid for some value of a, that is, p|ap −a.
(6) We will also show that p|(a + 1)p −(a + 1). Indeed, (a + 1)p −(a + 1) ap + p−1 i1 p i ap−i + 1 −(a + 1) (7) or (a + 1)p −(a + 1) ap −a + p−1 i1 p i ap−i.
(8) Now p| p i for 1 ≤i ≤p −1. Also since p|ap −a, we have p|(a + 1)p − (a + 1).
Second proof with congruences. We may multiply congruences, that is, from ci ≡di mod p for i 1, . . . , n follows c1 · c2 · · · · cn ≡d1 · d2 · · · · cn mod p.
(9) Now, suppose that gcd(a, p) 1. We form the sequence a, 2a, 3a, . . . , (p −1)a.
(10) 120 6. Number Theory No two of its terms are congruent mod p, since i · a ≡k · a (mod p) ⇒i ≡k mod p ⇒i k.
Hence, each of the numbers in (7) is congruent to exactly one of the numbers 1, 2, 3, . . . p −1.
(11) Applying (6) to (7) and (8) gives ap−1 · 1 · 2 · · · · · (p −1) ≡1 · 2 · · · · · (p −1) mod p.
We may cancel with (p −1)! since (p −1)! and p are coprime. Thus, ap−1 ≡1 mod p.
Thirdproofbycombinatorics.Wehavepearlswitha colors.Fromthesewe make necklaces with exactly p pearls. First, we make a string of pearls. There are ap different strings. If we throw away the a one-colored strings ap −a strings will remain. We connect the ends of each string to get necklaces. We find that two strings that differ only by a cyclic permutation of its pearls result in indistinguishable necklaces. But there are p cyclic permutations of p pearls on a string. Hence the number of distinct necklaces is (ap −a)/p.
Because of its interpretation this is an integer. So p | ap −a.
17. The converse theorem is not valid. The smallest counterexample is 341 | 2341 −2, where 341=31 · 11 is not a prime. Indeed, we have 2341−2 2(2340−1) 2((210)34−134) 2(210−1)(· · ·) 2·3·341·(· · ·).
18. The Fermat–Euler Theorem. Euler’s φ-function is defined as follows: φ(m) number of elements from {1, 2, . . . , m} which are prime to m.
gcd(a, m) 1 ⇒aφ(m) ≡1 mod m.
19. The Function Integer Part. ⌊x⌋ greatest integer ≤x integer part of x. x mod 1 x −⌊x⌋ {x} fractional part of x.
(a) ⌊x +y⌋≥⌊x⌋+⌊y⌋. We have equality only if x mod 1+y mod 1 < 1.
(b) ⌊⌊x⌋/n⌋ ⌊x/n⌋. This is an important special case of the formula ⌊(x + m)/n⌋ ⌊(⌊x⌋+ m)/n⌋. Here m and n are integers.
(c) ⌊x + 1/2⌋ the integer, which is nearest to x. More precisely, n ≤x < n + 1/2 ⇒⌊x + 1/2⌋ n, n + 1/2 ≤x < n + 1 ⇒⌊x + 1/2⌋ n + 1.
(d) The prime p divides n! with multiplicity e ⌊n/p⌋+⌊n/p2⌋+⌊n/p3⌋+ · · ·.
6. Number Theory 121 Divisibility The most useful formula in competitions is the fact that a−b | an−bn for all n, and a +b | an +bn for odd n. The second of these is a consequence of the first. Indeed, an+bn an−(−b)n foroddn,whichisdivisiblebya−(−b) a+b.Inparticular, adifferenceoftwosquarescanalwaysbefactored.Wehavea2−b2 (a−b)(a+b).
But a sum of two squares such as x2 + y2 can only be factored if 2xy is also a square. Here you must add and subtract 2xy. The simplest example is the identity of Sophie Germain: a4 + 4b4 a4 + 4a2b2 + 4b4 −4a2b2 (a2 + 2b2)2 −(2ab)2 (a2 + 2b2 + 2ab)(a2 + 2b2 −2ab).
Some difficult Olympiad problems are based on this identity. For instance, in the 1978 K¨ urschak Competition, we find the following problem which few students solved.
E1. n > 1 ⇒n4 + 4n is never a prime.
If n is even, then n4 + 4n is even and larger than 2. Thus it is not a prime. So we need to show the assertion only for odd n. But for odd n 2k + 1, we can make the following transformation, getting Sophie Germain’s identity: n4 + 4n n4 + 4 · 42k n4 + 4 · (2k)4, which has the form a4 + 4b4.
This problem first appeared in the Mathematics Magazine 1950. It was proposed by A. Makowski, a leader of the Polish IMO-team.
Quite recently, the following problem was posed in a Russian Olympiad for 8th graders: E2. Is 4545 + 5454 a prime?
Only few saw the solution, although all knew the identity of Sophie Germain and some competitions problems based on it. In fact, it is almost trivial to see that 4545 + 5454 5454 + 4 · (4138)4, which is the left side of Sophie Germain’s identity.
Now, consider the following recent competition problem from the former USSR: E3. n ∈N0 ⇒f (n) 22n + 22n−1 + 1 has at least n different prime factors.
Here, we use the lemma x4+x2+1 (x2+1)2−x2 (x2−x+1)(x2+x+1).
With x 22n−1, we get 22n+1 + 22n + 1 (22n −22n−1 + 1)(22n + 22n−1 + 1).
Both right-hand side factors are prime to each other. If they had an odd divisor q > 1, then their difference 2 · 22n−1 22n−1+1 would have the same factor. If we 122 6. Number Theory already know that 22n + 22n−1 + 1 has at least n prime factors, then by induction 22n+1 + 22n + 1 has at least n + 1 prime factors.
Remarks. For n > 4, the number has at least n + 1 different prime factors, since 224 −223 + 1 97 · 673, 224 + 223 + 1 3 · 7 · 13 · 241.
The product of the last two terms is f (5). Thus f (5) has six factors and f (n) has at least n + 1 factors. The problem also shows that there are infinitely many primes.
We can solve the following competition problem with the same paradigm.
E4. Find all primes of the form nn + 1, which are less then 1019.
For n 1 and n 2, we get primes. An odd n > 1 yields an even nn + 1 > 2.
So n must be even, i.e., n 22t(2k+1). Since 22t + 1|22t(2k+1) + 1, the exponent of n cannot have an odd divisor. Thus n 22t, or nn 22t
22t .
Fort 0, 1, 2wegetnn+1 5, 257, 1616+1 264+1 > 16·10006+1 > 1019.
So there are no other primes besides 2, 5, and 257.
Let us consider some more competition problems.
E5. Can the number A consisting of 600 sixes and some zeros be a square?
Solution. If A is a square, then it ends in an even number of zeros. By canceling them we get a square 2B, B consisting of 300 threes and some zeros, with B ending in 3. Since B is odd, 2B cannot be a square. It has only one factor 2.
E6. The equation 15x2 −7y2 9 has no integer solutions.
Solution. 15x2 −7y2 9 ⇒y 3y1 ⇒15x2 −63y2 1 9 ⇒5x2 −21y2 1 3 ⇒x 3x1 ⇒45x2 1 −21y2 1 3 ⇒15x2 1 −7y2 1 1 ⇒y2 1 ≡−1 mod 3. This is a contradiction since y2 1 ≡0 or 1 mod 3.
E7. A nine-digit number, in which every digit except zero occurs and which ends in 5, cannot be a square.
Solution. Suppose there is such a nine-digit number D, so that D A2. A 10a + 5 ⇒A2 100a2 + 100a + 25 100a(a + 1) + 25. Consequences: (a) The next to last digit is 2.
(b) The third digit from the right in D is one, which can be the final digit in a(a + 1), that is 0, 2, or 6. See the table below: 6. Number Theory 123 a 0 1 2 3 4 5 6 7 8 9 a(a + 1) mod 10 0 2 6 2 0 0 2 6 2 0 But 0 cannot occur, and 2 has already occurred. Hence, the third digit is a 6.
From D 1000B + 625 follows that 125|D. Since D A2 we have 54|D. Thus the fourth digit from the right in D must be 0 or 5. But 0 cannot occur, and 5 has already occurred.
E8. There is no polynomial f (x) with integer coefficients, so that f (7) 11, f (11) 13.
Solution. Let f (x) n i1 aixi, ai ∈Z. Then a −b | f (a) −f (b), that is, f (11) −f (7) is divisible by 11 −7 4. But f (11) −f (7) 2. Contradiction!
E9. For every positive integer p, we consider the equation 1 x + 1 y 1 p.
(1) We are looking for its solutions (x, y) in positive integers, with (x, y) and (y, x) being considered different. Show that if p is prime, then there are exactly three solutions. Otherwise, there are more then three solutions.
Solution. We have x > p, y > p. Hence, we set x p + q, y p + r in (1) and get 1 p + q + 1 p + r 1 p ⇒p2 qr.
If p is a prime, the only solutions will be (1, p2), (p, p), (p2, 1), that is, for (x, y), there are the three pairs of solutions (p+1, p(p+1)), (2p, 2p), (p(p+1), p+1).
If p is composite, then there will be obviously more solutions.
E10. I start with any multidigit number a1 and generate a sequence a1, a2, a3 . . . .
Here an+1 comes from an by attaching a digit ̸ 9. Then I cannot avoid the fact that an is infinitely often a composite number.
Solution. My strategy is to attach digits so as to get only finitely many composite digits. I cannot use 9 at all, and I can use 0, 2, 4, 6, 8, 5 only finitely often. Of the other digits 1, 3, 7, I may use 1 and 7 but finitely often because they change the remainder mod 3. Each time I attach 1 or 7 three times I get a number divisible by 3. So I am forced from a place upward to attach only threes. If at some moment I have a prime p, then after attaching at most p threes, again I get a multiple of p.
I know that gcd(10, p) 1. Hence, among 1, 11, 111, . . . , 111 . . . 11 p there is at least one multiple of p.
Remark. If I could use 3 and 9, then I could not tell if I could get only primes from some n upwards. For instance, with a1 1, I get the following primes of length 9: 1979339333, 1979339339.
E11. In the sequence 1, 9, 7, 7, 4, 7, 5, 3, 9, 4, 1, . . ., every digit from the fifth on is the sum of the preceding digits mod 10. Does one of the following words ever occur in the sequence.
124 6. Number Theory (a) 1234 (b) 3269 (c) 1977 (d) 0197?
Solution. We reduce all digits mod 2 and get 111101111011110 . . . . To the words 1234 and 3269 correspond 1010 and 1001. Both patterns do not occur in the reduced sequence. For (c) we observe that there are only finitely many possible 4-words. Hence, some word abcd will repeat for the first time: 1977 . . . abcd . . .
period p abcd.
Four successive digits determine the next digit, but they also determine the preced-ing digit. Hence the sequence can be extended indefinitely in both directions. This extended sequence is purely periodic. In each period of length p lies one word 1977. This word is the first one to repeat, if you start with 1977.
This is an important observation. First, we show that the sequence must repeat.
Then we show invertibility, which garantees a pure cycle (Fig. 6.1). For (d) we extend the sequence to the left by one term and get 0197.
r r r r r r r r (a) Pure cycle for invertible operation Fig. 6.1. The two types of behavior of iterates x →f (x).
r r r r r r r r r r r (b) Noninvertible operation Remark. Computer experimentation shows that if we start with four odd digits, the period length will be p 1560 5 · 312. Starting with four even digits, we get period p 312. If we start with at least one 5 and only zeros, the period will be p 5.
E12. The equation x2 + y2 + z2 2xyz (0) has no integral solutions except x y z 0. Show this.
First Solution. Let (x, y, z) ̸ (0, 0, 0) be an integral solution. If 2k, k ≥0 is the highest power of 2, which divides x, y, z, then x 2kx1, y 2ky1, z 2kz1, 22kx2 1 + 22ky2 1 + 22kz2 1 23k+1x1y1z1, x2 1 + y2 1 + z2 1 2k+1x1y1z1.
(1) The right side of (1) is even. Hence, the left side is also even. All three terms on the left cannot be even because of the choice of k. Hence, exactly one term is even.
Suppose x1 2x2, while y1 and z1 are odd. Hence, y2 1 + z2 1 2k+2x2y1z1 −4x2 2 ≡0 mod 4.
6. Number Theory 125 This contradicts y2 1 + z2 1 ≡2 mod 4.
Second Solution: By infinite descent. On the left side of (0), exactly one term is even or all three terms are even. If exactly one term is even, then the right side is divisible by 4, the left only by 2. Contradiction! Hence all three terms are even: x 2x1, y 2y1, z 2z1 and x2 1 + y2 1 + z2 1 4x1y1z1.
(2) From (2), with the same reasoning we get x1 2x2, y1 2y2, z1 2z2 and x2 2 + y2 2 + z2 2 8x2y2z2.
(3) Again, from (3) follows that x2, y2, z2 are even, and so on, that is x 2x1 22x2 23x3 · · · 2nxn · · · , y 2y1 22y2 23y3 · · · 2nyn · · · , z 2z1 22z2 23z3 · · · 2nzn · · · , that is, if (x, y, z) is a solution, then x, y, z are divisible by 2n for any n. This is only possible for x y z 0.
Remark. The equation x2 + y2 + z2 kxyz has only for k 1 and k 3 infinitely many solutions, as will be shown later.
E13. Show that f (n) n5 + n4 + 1 is not prime for n > 1.
First Solution: By trial, conjecture, and verification.
n 1 2 3 4 · · · 10 f (n) 3 · 1 7 · 7 13 · 25 21 · 61 · · · 111 · 991 (n2+n+1)(n3−n+1) Second Solution: Factoring. We have f (1) ̸ 0, f (−1) ̸ 0. Thus, there is no linear factor. We try a quadratic and cubic factor. Either n5 + n4 + 1 (n2 + an + 1)(n3 + bn2 + cn + 1) or n5 + n4 + 1 (n2 + an −1)(n3 + bn2 + cn −1).
We investigate the first case. By expanding the right side, we get n5 + n4 + 1 n5 + (a + b)n4 + (ab + c + 1)n3 + (ac + b + 1)n2 + (a + c)n + 1.
Comparing coefficients, we get four equations for a, b, c: a + b 1, ab + c + 1 0, ac + b + 1 0, a + c 0 with solutions b 0, a 1, c −1. Thus, n5 +n4 +1 (n2 +n+1)(n3 − n + 1). The second case leads to an inconsistent system of equations.
126 6. Number Theory Third Solution: By third roots of unity. Let ω be the third root of unity, i.e., ω3 1. Then ω2 + ω + 1 0. Since ω5 + ω4 + 1 ω2 + ω + 1, we see that ω2 + ω + 1 is a factor of the polynomial. So n2 + n + 1|n5 + n4 + 1. By long division of n5 + n4 + 1 by n2 + n + 1, we get the second factor n3 −n + 1.
The next two problems are among the most difficult ever proposed at any com-petition.
E14. If n ≥3, then 2n can be represented in the form 2n 7x2 +y2 with odd x, y.
Solution. This is a very interesting and exceedingly tough problem which was proposed at the MMO 1985. It is due to Euler, who never published it. It was taken from his notebook by the proposers. No participant could solve it. It became a subject of controversy among mathematicians. A prominent number theorist wrote in the Russian journal Mathematics in School that it was well beyond the students and required algebraic number theory. I proposed it to our Olympiad team. One student Eric M¨ uller gave a solution after some time, which I did not understand. I asked him to write it down, so that I could study it in detail. It took him some time to write it down, since he solved not only this problem but along with it also over a thousand other problems on 434 pages, all the problems posed by the trainers in three years. I found the solution of our problem. It was correct.
Figure 6.2 shows the first 8 solutions, which can easily be found by guessing.
Now study this table closely. Before reading on, try to find the pattern behind the table.
n 3 4 5 6 7 8 9 10 x 1 1 1 3 1 5 7 3 y 1 3 5 1 11 9 13 31 Fig. 6.2 Our hypothesis is that one column somehow determines the next one. How can I get the next pair x1, y1 from the current x, y? This conjecture is supported by similar equations, for instance the Pell–Fermat equation where we get from one pair (x, y) to the next by a linear transformation. Let us start with x1. How can I get from (x, y) to x1? We get x1 from the first pair (1,1) by taking the arithmetic mean. From the second pair (1,3), the mean 2 is not an odd integer. So let us take the difference |x −y|/2 1. Again we are successful. Some more trials convince us that we should take (x + y)/2 if that number is odd. If that number is even, we should take |x −y|/2. After guessing the pattern behind x, we will try to guess the pattern behind y. There is a 7 before x2 in the equation. So we could try (7x +y)/2 and |7x −y|/2. The pattern seems to hold for the table above.
To support our conjecture, we observe that exactly one of x + y 2 or |x −y| 2 is odd since x + y 2 + |x −y| 2 max (x, y).
Exactly one of 7x + y 2 or |7x −y| 2 is odd since 7x + y 2 + |7x −y| 2 max (7x, y).
6. Number Theory 127 In addition, we have x + y 2 odd ⇒|7x −y| 2 8x −(x + y) 2 |4x −x + y 2 | odd, |x −y| 2 odd ⇒7x + y 2 8x −(x −y) 2 4x −x −y 2 odd.
So we have the following transformations: S : (x, y) → x + y 2 , |7x −y| 2 , T : (x, y) → |x −y| 2 , 7x + y 2 .
Nowweproveourconjecturebyinduction.Itisvalidforn 3.Suppose7x2+y2 2n for any n. By applying S, we get 7(x + y)2 + (7x −y)2 4 14x2 + 2y2 2(7x2 + y2) 2 · 2n 2n+1.
Similarly we can proceed with transformation T.
Thenextproblemwassubmittedin1988bytheFRG.Nobodyofthesixmembers of the Australian problem committee could solve it. Two of the members were Georges Szekeres and his wife, both famous problem solvers and problem creators.
Since it was a number theoretic problem it was sent to the four most renowned Australian number theorists. They were asked to work on it for six hours. None of them could solve it in this time. The problem committee submitted it to the jury of the XXIX IMO marked with a double asterisk, which meant a superhard problem, possibly too hard to pose. After a long discussion, the jury finally had the courage to choose it as the last problem of the competition. Eleven students gave perfect solutions.
E15. If a, b, q (a2 + b2)/(ab + 1) are integers, then q is a perfect square.
Solution. We replace a, b by x, y and get a family of hyperbolas x2 + y2 −qxy −q 0, (1) one hyperbola for each q. They are all symmetric to y x. Let us fix q. Suppose there is a lattice point (x, y) on this hyperbola Hq. There will also be a lattice point (y, x) symmetric to y x. For x y, we easily get x y q 1. So we may assume x < y. See Fig. 6.3. If (x, y) is a lattice point then for fixed y the quadratic in x has two solutions x, x1 with x + x1 qy, x1 qy −x. So x1 is also an integer, that is, B (qy −x, y) is a lattice point on the lower branch of Hq. Its reflection at y x is a lattice point C (y, qy −x). Starting from (x, y), we can generate infinitely many lattice points above A on the upper branch of Hq by means of the transformation T : (x, y) →(y, qy −x).
Again, starting at A, we keep x fixed. Then (1) is a quadratic in y with two solutions y, y1 such that y + y1 qx, or y1 qx −y. So y1 is an integer and 128 6. Number Theory D (x, qy −x) is a lattice point on the lower branch of Hq. Its reflection at y x is the lattice point E (qx −y, x) on the upper branch. Starting in A, we can use the transformation S : (x, y) →(qx −y, x) to get lattice points on the upper branch below A. But this time, there will be only a finite number of them. Indeed, each time S is applied, both coordinates will strictly decrease. Can it be that x becomes negative while y is positive? No! In this case (1) becomes x2 + y2 + q |xy| −q > 0.
So on the last step, we require that x 0, and, from (1), q y2 which was to be shown. @ @ @ @ @ @ @ @ @ q q B(qy −x, y) C(y, qy −x) A(x, y) D(x, qx −y) E(qx −y, x) Fig. 6.3 In Fig. 6.3, we have drawn the hyperbola for q 4. In fact, we replaced it with its asymptotes because the deviation from the asymptotes is negligible for large x or y.
Until now we have not proved that there exists a single lattice point on Hq. The existence was not required. The theorem is valid even if a single lattice point does not exist on any of the hyperbolas. But we can easily show the existence of one lattice point for each perfect square q. The point (x, y, q) (c, c3, c2) is a lattice point since x2 + y2 xy + 1 q ⇒c2 + c6 c4 + 1 c2.
E16. The Pell–Fermat Equation.
We want to find all integral solutions of the equation x2 −dy2 1.
(1) Here the positive integer d is not a square. We may even assume that it is square-free. If it were not square-free then we could integrate its square factor into y2.
We associate the number x + y √ d with every solution (x, y). We have the basic factorization x2 −dy2 (x −y √ d)(x + y √ d).
(2) 6. Number Theory 129 It follows from (2) that the product or quotient of two solutions of (1) is again a solution of (1). If x and y are positive, then it follows from (1) that x + y √ d and x −y √ d are positive. In addition, the first one is > 1 and the second < 1.
We consider the smallest positive solution x0 + y0 √ d. Then we will show that all solutions are given by (x0 + y0 √ d)n, n ∈Z. We will prove this by the ingenious method of descent. Suppose there is another solution u+v √ d which is not a power of x0 +y0 √ d. Then it must lie between two succeeding powers of x0 +y0 √ d, that is, for some n, (x0 + y0 √ d)n < u + v √ d < (x0 + y0 √ d)n+1.
Multiplying with the solution (x0 −y0 √ d)n, we get 1 < (u + v √ d)(x0 −y0 √ d)n < x0 + y0 √ d.
The middle term of the inequality chain is a solution and because it is larger than 1, it is a positive solution. This is a contradiction because we have found a positive solution which is smaller than the smallest positive solution. Thus every solution is a power of the smallest positive solution. So we have only to find the smallest positive solution. It can be found by exhaustive search if x0 and y0 are small. At the IMO, only such cases have come up so far. But there is an algorithm for finding the smallest solution by developing √ d into a continued fraction.
The equation x2 −dy2 −1 does not always have a solution. One can often tell by congruences that it has no solutions. If it has solutions, we can try to find the smallest one (x0, y0) by guessing. Then (x0 + y0 √ d)2k+1 gives all solutions.
We could also find the smallest solution by continued fractional expansion of √ d.
The following examples have automatic solutions. They use one of the following obvious ideas: between any two consecutive positive integers (squares, triangular numbers), there is no other positive integer (square, triangular number).
E17. Let α and β be irrational numbers such that 1/α + 1/β 1. Then the sequences f (n) ⌊αn⌋and g(n) ⌊βn⌋, n 1, 2, 3, . . . are disjoint and their union is N.
You cannot miss the proof: ⌊αm⌋ ⌊βn⌋ q ⇒q < αm < q + 1, q < βn < q + 1.
Here we use the fact that α, β are irrational.
m q + 1 < 1 α < m q , n q + 1 < 1 β < n q .
Adding the two inequalities, we get m + n q + 1 < 1 < m + n q ⇒m + n < q + 1, q < m + n ⇒q < m + n < q + 1.
130 6. Number Theory This is a contradiction. Thus, ⌊αm⌋̸ ⌊βn⌋.
First, we observe that α or β is in (1, 2), because α > 2, β > 2 implies 1 α + 1 β < 1, a contradiction.
Now suppose that (q, q + 1) with q ≥2 contains no element of the f (n) or g(n), that is, αm < q < q + 1 < α(m + 1), βn < q < q + 1 < β(n + 1), m q < 1 α < m + 1 q + 1 , n q < 1 β < n + 1 q + 1.
Adding the two inequality chains, we get m + n q < 1 < m + n + 2 q + 1 ⇒m + n < q < q + 1 < m + n + 2.
Again, this is a contradiction, because there is no place for two successive positive integers between m + n and m + n + 2.
E18. The function f (n) ⌊n + √n + 1/2⌋misses exactly the squares.
Suppose ⌊n + √n + 1/2⌋̸ m. What can we say about m ∈N?
n + √n + 1 2 < m and m + 1 < n + 1 + √ n + 1 + 1 2 ⇒√n < m −n −1 2 < √ n + 1, n < (m −n)2 −(m −n) + 1 4 < n + 1 ⇒n −1 4 < (m −n)2 −(m −n) < n + 3 4, (m −n)2 −(m −n) n ⇒m (m −n)2.
Now we make a simple counting argument: There are exactly k squares ≤k2 + k and exactly k2 integers of the form ⌊n + √n + 1/2⌋. Thus ⌊n + √n + 1/2⌋is the nth nonsquare.
E19. The sequence ⌊n + √ 2n + 1/2⌋, n 1, 2, . . . misses exactly the triangular numbers.
Suppose m is not assumed. Then, n + √ 2n + 1 2 < m, m + 1 < n + 1 + √ 2n + 2 + 1 2 ⇒ √ 2n < m −n −1 2 < √ 2n + 2, 2n < (m −n)2 −(m −n) even +1 4 < 2n + 2, (m −n)2 −(m −n) 2n ⇒(m −n)2 + (m −n) 2m, m (m −n + 1)(m −n) 2 m −n + 1 2 .
6. Number Theory 131 A counting argument similar to the one in the preceding example shows, that exactly the triangular numbers are omitted.
Problems 1. a −c | ab + cd ⇒a −c | ad + bc.
2. a ≡b ≡1 mod 2 ⇒a2 + b2 not a square.
3. (a) 6 | n3 + 5n.
(b) 30 | n5 −n.
(c) For which n is 120 | n5 −n?
4. (a) 3 | a, 3 | b ⇔3 | a2 + b2. (b) 7 | a, 7 | b ⇔7 | a2 + b2. (c) 21 | a2 + b2 ⇒ 441 | a2 + b2.
5. n ≡1 mod 2 ⇒n2 ≡1 mod 8 ⇔8 | n2 −1.
6. 6 | a + b + c ⇔6 | a3 + b3 + c3.
7. Derive divisibility criteria for 9 and 11.
8. Let A 3105 + 4105. Show that 7 | A. Find A mod 11 and A mod 13.
9. Show that 3n −1, 5n ± 2, 7n −1, 7n −2, 7n + 3 are not squares.
10. If n is not a prime, then 2n −1 is not a prime.
11. If n has an odd divisor, then 2n + 1 is not prime.
12. 641 | 232 + 1. No calculator allowed!
13. (a) n > 2 ⇒2n −1 is not a power of 3.
(b) n > 3 ⇒2n + 1 is not a power of 3.
14. A number with 3n equal digits is divisible by 3n.
15. Find all primes p, q, so that p2 −2q2 1.
16. If 2n + 1 and 3n + 1 are squares, then 5n + 3 is not a prime.
17. If p is prime, then p2 ≡1 mod 24.
18. 9 | a2 + b2 + c2 ⇒9 | a2 −b2 or 9 | b2 −c2 or 9 | a2 −c2.
19. n ≡0 mod 2 ⇒323 | 20n + 16n −3n −1.
20. 121 ̸ | n2 + 3n + 5.
21. If p and p2 + 2 are primes, then p3 + 2 is also prime.
22. 2n ̸ | n!.
23. How many zeros are at the end of 1000!?
24. Among five integers, there are always three with sum divisible by 3.
25. Using x2 + y2 + z2 ̸≡7 mod 8, find numbers which are not sums of 3 squares.
26. The four-digit number aabb is a square. Find it.
27. Can the digital sum of a square be (a) 3, (b) 1977?
28. 1000 · · · 001 with 1961 zeros is composite (not prime).
29. Let Q(n) be the digital sum of n. Show that Q(n) Q(2n) ⇒9 | n.
30. The sum of squares of five successive positive integers is not a square.
31. Let n pa1 1 pa2 2 · · · pan n , pi be distinct primes.Then n has (a1+1) · · · (an+1) divisors.
132 6. Number Theory 32. Among n + 1 positive integers≤2n, there are two which are coprime.
33. Among n + 1 positive integers ≤2n, there are p, q such that p|q.
34. (12n + 1)/(30n + 2) and (21n + 4)/(14n + 3) are irreducible.
35. Show that gcd(2n + 3, n + 7) 1 for n ̸≡4 mod 11, and 11 for n ≡4 mod 11.
36. gcd(n, n + 1) 1, gcd(2n −1, 2n + 1) 1, gcd(2n, 2n + 2) 2, gcd(a, b) gcd(a, a + b), gcd(5a + 3b, 13a + 8b) gcd(a, b).
37. (a) gcd(2a −1, 2b −1) 2gcd(a,b) −1.
(b) n ab ⇒2a −1 | 2n −1.
38. (a) gcd(6, n) 1 ⇒24 | n2 −1.
(b) p, q primes > 3 implies 24 | p2 −q2.
39. (a) p, p + 10, p + 14 are primes, (b) p, p + 4, p + 14 are primes. Find p.
40. (a) p, 2p + 1, 4p + 1 are primes (b) p and 8p2 + 1 are primes. Find p.
41. 13 | a + 4b ⇒13 | 10a + b.
19 |3x + 7y ⇒19 | 43x + 75y. 17 | 3a + 2b ⇒ 17 |10a + b.
42. If p > 5 is a prime, then p2 ≡1 or p2 ≡19 mod 30.
43. x2 + y2 x2y2 has no integral solutions besides x y 0.
44. 120 | n5 −5n3 + 4n.
9 | 4n + 15n −1.
45. Let m > 1. Then exactly one of the integers a, a + 1, . . . , a + m −1 is divisible by m.
46. Find all integral solutions of x2 + y2 + z2 x2y2.
47. Find the integral solutions of (a) x + y xy (b) x2 −y2 2xyz.
48. Find all integral solutions of (a) x2 −3y2 17, (b) 2xy + 3y2 24.
49. Find the integral solutions of x2 + xy + y2 x2y2 and x2 + y2 + z2 + u2 2xyzu.
50. Find all integral solutions of x + y x2 −xy + y2.
51. Let p p1p2 · · · pn, n > 1 be the product of the first n primes. Show that p −1 and p + 1 are not squares.
52. a1a2 + a2a3 + · · · + an−1an + ana1 0 with ai ∈{1, −1}. Show that 4 | n.
53. Three brothers inherit n gold pieces weighing 1, 2, . . . , n. For what n can they be split into three equal heaps?
54. Find the smallest positive integer n, so that 999999 · n 111 · · · 11.
55. Find the smallest positive integer with the property that, if you move the first digit to the end, the new number is 1.5 times larger than the old one.
56. With the digits 1 to 9, construct two numbers with (a) maximal (b) minimal product.
57. Which smallest positive integer becomes 57 times smaller by striking its first digit.
58. If ab cd, then a2 + b2 + c2 + d2 is composite. Generalize (BWM 1970/71).
59. Find the four-digit number abcd such that 4 · abcd dcba.
60. Find the five-digit number abcde such that 4 · abcde edcba.
61. If n > 2, p a prime, and 2n/3 < p < n, then p ̸ | 2n n .
62. The sequence an √ 24n + 1, n ∈N, contains all primes except 2 and 3.
6. Number Theory 133 63. (a) There are infinitely many positive integers which are not the sum of a square and a prime.
(b) There are infinitely many positive integers, which are not of the form p + a2k with p a prime and a, k positive integers.
64. Different lattice points of the plane have different distances from ( √ 2, 1 3).
65. Different lattice points of space have different distances from ( √ 2, √ 3, 1 3).
66. A number a is called automorphic if a2 ends in a. Apart from 0 and 1, the only one-digit automorphic numbers are 5 and 6. Find all automorphic numbers with (a) 2, (b) 3, (c) 4 digits. Do you see a pattern?
67. For any n, there is an n-digit number with 1 and 2 as the only digits and which is divisible by 2n. In which other number system does this hold?
68. Is n a sum of two squares, then also 2n is.
69. n is an integer, and n > 11 ⇒n2 −19n + 89 is not a square.
70. Every even number 2n can be written in the form 2n (x + y)2 + 3x + y with x, y nonnegative integers.
71. m | (m −1)! + 1 ⇒m is a prime.
72. How often does the factor 2 occur in the product (n + 1)(n + 2) · · · (2n)?
73. If a, m, n are positive integers with a > 1, then am + 1 | an + 1 ⇒m | n.
74. Let (x, y, z) be a solution of x2 + y2 z2. Show that one of the three numbers is divisible by (a) 3 (b) 4 (c) 5.
75. We can choose 2k different numbers from 0, 1, 2, . . . , 3k −1, so that three numbers in arithmetic progression will not occur.
76. Can you find integers m, n with m2 + (m + 1)2 n4 + (n + 1)4?
77. Let n be a positive integer. If 2 + 2 √ 28n2 + 1 is an integer, then it is a square.
78. The equation x3 + 3 4y(y + 1) has no integral solutions.
79. A 20-digit positive integer starting with 11 ones cannot be a square.
80. 9 | a2 + ab + b2 ⇒3 | a, 3 | b.
81. Find the smallest positive integer a, so that 1971 | 50n + a · 23n for odd n.
82. There are infinitely many composite numbers in the sequence 1, 31, 331, 3331, . . ..
83. Find all positive integers n, so that 3 | n · 2n −1.
84. If m is a positive integer, then m(m + 1) is not a power > 1.
85. Every positive integer > 6 is sum of two positive integers > 1, with no common divisor.
86. If x2 + 2y2 is an odd prime, then it has the form 8n + 1 or 8n + 3.
87. Let a, b be positive integers with b > 2. Show that never is 2b −1 | 2a + 1.
88. Can the product of three (4) consecutive integers be a power of an integer?
89. If you move the last digit of a number to the front, then it becomes nine times larger.
Find the smallest such number.
134 6. Number Theory 90. Find all pairs of integers (x, y), such that x3 + x2y + xy2 + y3 8(x2 + xy + y2 + 1).
91. Find all pairs of nonnegative integers (x, y), such that x3 + 8x2 −6x + 8 y3.
92. If n ∈N and 2n + 1 3n + 1 are squares, then 40|n.
93. Do there exist positive integers, so that x3 + y3 4684?
94. 3851980 + 181980 is not a square.
95. 11 + 22 + 33 + · · · + 19831983 is not a power mk with k ≥2.
96. y2 x3 + 7 has no integral solutions.
97. Find the three last digits of 79999.
98. Find pairwise prime solutions of 1/x + 1/y 1/z.
99. Find pairwise prime solutions of 1/x2 + 1/y2 1/z2.
100. The product of two numbers of the form (a) x2 + 2y2 (b) x2 −2y2 (c) x2 + dy2 (d) x2 −dy2 again has the same form (d is not a square).
Hint: x2 −dy2 (x + y √ d)(x −y √ d), x2 + 2y2 (x + iy √ d)(x −iy √ d).
101. Show that 11987 + 21987 + · · · + n1987 is not divisible by n + 2 for n ∈N.
102. For what integers m, n is the equation (5 + 3 √ 2)m (3 + 5 √ 2)n valid?
103. Solve x3 −y3 xy + 61 in positive integers.
104. Does x2 + y3 z4 have prime solutions x, y, z?
105. Find all numbers with the digits 1..9 containing every digit exactly once and with the initial part divisible by n, n ∈1..9.
106. x, y, z are pairwise distinct integers. Show that (x −y)5 + (y −z)5 + (z −x)5 is divisible by 5(x −y)(y −z)(z −x).
107. Find the smallest positive integer ending in 1986 which is divisible by 1987.
108. Show that 1982 | 222 · · · 22 (1980 twos).
109. The integers 1, . . . , 1986 are writen in any order and concatenated. Show that we always get an integer which is not the cube of another integer.
110. Find the eight last digits of the binary expansion of 271986.
111. The next to last digit of 3n is even.
112. For no positive integer m is (1000m −1) | (1978m −1).
113. For which positive integers do we have n k1 k | n k1 k?
114. a, b, c, d, e ∈Z, 25 | a5 + b5 + c5 + d5 + e5 ⇒5 | abcde.
115. Find a pair of integers a, b so that 7 ̸ | ab(a + b), but 77 | (a + b)7 −a7 −b7.
116. Find the first digits before and behind the decimal point in ( √ 2 + √ 3)1980.
117. The product of two positive integers of the form a2 + ab + b2 has the same form.
118. If ax2 + by2 1, with a, b, x, y ∈Q, has a rational solution (x, y), then it has infinitely many rational solutions.
119. Show that x(x + 1)(x + 2)(x + 3) y2 has no solution for x, y ∈N.
6. Number Theory 135 120. a, b, c, d, e ∈N are such that a4 + b4 + c4 + d4 e4. Show that among the five variables (a) at least three are even, (b) at least three are multiples of 5, (c) at least two are multiples of 10.
121. Show that, if m ends with the digit five, then 1991 | 12m + 9m + 8m + 6m.
122. Find all pairs (x, y) of nonnegative integers satisfying x3 + 8x2 −6x + 8 y3.
123. Find all integral solutions of y2 + y x4 + x3 + x2 + x.
124. There are infinitely many pairwise prime integers x, z, y such that x2, z2, y2 are in arithmetic progression.
125. Each of the positive integers a1, . . . , an is less than 1951. The least common multiple of any two of these is greater than 1951. Show that 1 a1 + · · · + 1 an < 2.
126. Find the smallest integer of the form | f (m, n) | with (a) f (m, n) 36m −5n.
(b) f (m, n) 12m −5n.
127. Find infinitely many integral solutions of (x2 + x + 1)(y2 + y + 1) z2 + z + 1.
128. Let z2 (x2 −1)(y2 −1) + n for x, y ∈Z. Are there solutions for (a) n 1981.
(b) n 1985.
(c) n 1984 (IMO Jury 1981)?
129. If a, b, (a2 + ab + c2)/(ab + 1) q are integers, then q is a perfect square.
130. (a) If a, b, (a2 + b2)/(ab −1) q are integers, then q 5.
(b) a2 + b2 −5ab + 5 0 has infinitely many solutions in N.
131. No prime can be written as a sum of two squares in two different ways.
132. Find infinitely many solutions of (a) x2 + y2 + z2 3xyz (b) x2 + y2 + z2 xyz.
133. Two players A and B alternately take chips from two piles with a and b chips, respectively. Initially a > b. A move consists in taking from a pile a multiple of the other pile. The winner is the one who takes the last chip in one of the piles. Show that (a) If a > 2b, then the first player A can force a win.
(b) For what α can A force a win, if initially a > αb. (This is the game Euclid, which is due to Cole and Davie. See Math. Gaz. LIII, 354–7 (1969). and AUO 1978.) 134. If n ∈N and 3n + 1 and 4n + 1 are perfect squares, then 56|n.
135. Fifty numbers a1, a2, . . . , a50 are written along a circle; each of the numbers is +1 or −1. You want to find the product of these numbers. You may find the product of three consecutive numbers in one question. How many questions do you need at least?
Here is a generalization you can work on: Along a circle are written n numbers, each number being +1 or −1. Our aim is to find the product of all n numbers. In one question, we can find the product of k successive numbers ai · · · ai+k−1. Here an+1 a1, and so on. How many questions q(n, k) are necessary to find the product?
136. Let n ∈N. If 4n + 2n + 1 is a prime, then n is a power of 3.
136 6. Number Theory 137. (a) If the positive integers x, y satisfy 2x2 + x 3y2 + y then x −y, 2x + 2y + 1, 3x + 3y + 1 are perfect squares. (PMO 1964/65.) (b) Find all integral solutions of 2x2 + x 3y2 + y.
138. (a) Let an be the last nonzero digit in the decimal representation of the number n!.
Does the sequence a1, a2, a3, . . . become periodic after a finite number of steps (USSR proposal for IMO 1991)?
(b) Let dn be the last nonzero digit of n!. Prove that dn is not periodic, that is, p and n0 do not exist such that dn+p dn for all n ≥n0 (USSR proposal for IMO 1983).
139. Prove that the positive integer (5125 −1)/(525 −1) is composite.
140. Integers a, b, c, d, e are such that n | a + b + c + d + e, n | a2 + b2 + c2 + d2 + e2 for the odd integer n. Prove that n | a5 + b5 + c5 + d5 + e5 −5abcde.
141. For each positive integer k, find the smallest n such that 2k|5n −1.
142. If p, q are positive integers, then 1 −1 2 + 1 3 −1 4 + · · · − 1 1318 + 1 1319 p q ⇒1979|p (IMO1979).
143. If the difference of the cubes of two consecutive integers can be represented as a square of an integer, then this integer is the sum of the squares of two consecutive integers (R.C. Lyness).
144. There are infinitely many powers of 2 in the sequence ⌊n √ 2⌋.
145. Let gcd(a, b) 1. The Central Bank of Sikinia issues only a- and b-Kulotnik coins.
What amounts can you pay if you can get (a) change (b) no change?
146. In Sikinia there are three types of weights: 15, 20, and 48 Slotnik. What weights can you measure (a) two sidedly (b) one sidedly?
147. Let a, b, c ∈N with gcd(a, b) gcd(b, c) gcd(a, c) 1. Prove that 2abc−ab− bc−ca is the largest integer which cannot be expressed in the form xbc+yca+zab, where x, y, z are nonnegative integers (IMO 1983).
148. Prove that the number 1 280 000 401 is composite (IIM 1993).
149. Do there exist positive integers x, y, such that x + y, 2x + y and x + 2y are perfect squares?
150. For what smallest integer n is 3n −1 divisible by 21995?
151. a, b ∈N are such that (a + 1)/b + (b + 1)/a ∈N. Let d gcd(a, b). Prove that d2 ≤a + b (RO 1994).
152. Does there exist a positive integer which is divisible by 21995 and whose decimal notation does not contain any zero?
153. Prove that n(n + 1) divides 2(1k + 2k + · · · + nk) for odd k.
154. Let P (n) be the product of all digits of a positive integer n. Can the sequence nk defined by nk+1 nk + P (nk) and initial term n1 ∈N be unbounded for some n1 (AUO 1980).
155. Let D(n) be the digital sum of the positive integer n.
(a) Does there exist an n such that n + D(n) 1980?
(b) Prove that at least one of any two successive positive integers can be represented in the form En n + D(n) (AUO 1980).
6. Number Theory 137 156. Several different positive integers lie strictly between two successive squares. Prove that their pairwise products are also different (AUO 1983).
157. Find the integral solutions of 19x3 −84y2 1984 (MMO 1984).
158. Start with some positive integers. In one step you may take any two numbers a, b and replace them by gcd(a, b) and lcm(a, b). Prove that, eventually, the numbers will stop changing.
159. The powers 2n and 5n start with the same digit d. What is this digit?
160. If n a2 + b2 + c2, then n2 x2 + y2 + z2, where a, b, c, x, y, z ∈N.
161. For infinitely many composite n, we have n|3n−1 −2n−1 (MMO 1995).
162. The equation x2 + y2 + z2 x3 + y3 + z3 has infinitely many integer solutions (MMO 1994).
163. Prove that there exist infinitely many positive integers n such that 2n ends with n, i.e., 2n · · · n (MMO 1978).
164. There are white and black balls in an urn. If you draw two balls at random, the probability is 1/2 to get a mixed couple. What can you conclude about the contents of the urn?
165. A multidigit number contains the digit 0. If you strike it the number becomes 9 times smaller. At which position is this 0 located? Find all such numbers.
166. If you are condemned to die in Sikinia, you are put into Death Row until the last day of the year. Then all prisoners from Death Row are arranged in a circle and numbered 1, 2, . . . , n. Starting with #2 every second one is shot until only one remains who is immediately set free. How do you find the place of the sole survivor?
167. (a) Find a number divisible by 2 and 9 which has exactly 14 divisors.
(b) Replacing 14 by 15 there will be several solutions, replacing 14 by 17 there will be none.
168. The positive integer k has the property: for all m ∈N : k | m ⇒k | mr. Here m and mr are mutual reflections like 1234 and 4321. Show that k | 99.
169. Let p and q be fixed positive integers. The set Z of integers is to be partitioned into three subsets A, B, C such that, for every n ∈Z, the three integers n, n + p, and n + q belong to different subsets. What relationships must p and q satisfy?
170. A positive integer is the product of n distinct primes. In how many ways can it be represented as the difference of two squares?
Solutions 1. (ab + cd) −(ad + bc) a(b −d) −c(b −d) (a −c)(b −d).
2. An even square is divisible by 4.
3. (a) n3 + 5n n3 −n + 6n (n −1)n(n + 1) + 6n.
(b) The three first factors of n5 −n n(n −1)(n + 1)(n2 + 1) are successive integers. Divisibility by 5 follows from Fermat’s theorem.
(c) If n is odd, n5 −n is divisible by 120.
138 6. Number Theory 4. (a) For any x, x2 ≡0 mod 3 or x2 ≡1 mod 3.
(b) For any x, x2 ≡0 or 1 or 4 mod 7.
(c) This follows from (a) and (b).
5. n 2q + 1 ⇒n2 4q2 + 4q + 1 4q(q + 1) + 1 8r + 1. Every odd square is 1 mod 8. This fundamental fact is used very often.
6. (a3 −a) + (b3 −b) + (c3 −c) is divisible by 2 and 3, i.e., 6.
7. 10 ≡1 mod 3 , 10 ≡−1 mod 11, n n i1 di10i ⇒n ≡n i1 di mod 3, n ≡ di(−1)i mod 11.
8. 7 | A since 105 is odd. 35 ≡1 mod 11, 45 ≡1 mod 11. 3105 + 4105 (35)21 + (45)21 ≡1 + 1 ≡2 mod 11.
9. Show it by squaring the remainders of 3, 5, 7 modulo 3, 5, 7, respectively.
10. This follows from a −b | an −bn.
11. This follows from a + b | an + bn for odd n.
12. 641 54 + 24 5 · 27 + 1 divides both 54 · 228 + 232 and 54 · 228 −1. Then it also divides their difference 232 + 1.
13. (a) Suppose n > 2. We want to show that never 2n −1 3m. For odd m we have 2n 3m + 1 (3 + 1)(3m−1 + 3m−2 + · · · + 1). The last factor is an odd number of odd summands. This is a contradiction.
Next suppose m 2s is even. Then 2n 1+32s (9)s +1 8q+2. Contradiction, because it is not a multiple of 4.
(b) Suppose n > 3. For odd m, we get 2n 3m −1 (3 −1)(3m−1 + · · · + 1). The last factor is an odd number of odd summands. Contradiction!
Nextsupposem 2s iseven.Then3s 2a+1, 2n+1 (3s)2, 2n (2a+1)2−1 4a(a + 1). Here a or a + 1 is odd. Thus a 1, 2n 32 −1. Hence, there is no solution for n > 3.
14. Prove it by induction.
15. p must be odd. p 3 and q 2 are solutions as well as p 5 and q 3.
Suppose both p and q are greater than 3. Then both are ≡±1 mod 6. Then we have (±1)2 −2(±1)2 1, −1 ≡1 mod 6. Contradiction.
16. 2n + 1 a2, 3n + 1 b2 ⇒5n + 3 4(2n + 1) −(3n + 1) 4a2 −b2 (2a + b)(2a −b). Hence 2a −b 1 ⇒(b −1)2 −2n. Thus 2a −b ̸ 1.
17. For p > 3, we have p 6n ± 1, and the theorem is valid for such numbers.
18. x2 ≡0, 1, 4, 7, mod 9. Thus (a2, b2, c2) is (0,0,0), (1,1,7), (1,4,4) or (4,7,7), or permutations of these. Two elements of each of these triples are equal. So their difference is 0.
19. 323 17 · 19. Prove by congruences divisibility by 17 and 19.
20. We prove: If n2+3n+5 is divisible by 11 then it is not divisible by 121. n2+3n+5 ≡ n2 −8n + 16 ≡(n −4)2 mod 11. Thus, 11 | n2 + 3n + 5 if n 11k + 4. But then n2 + 3n + 5 121k(k + 1) + 33. This is not divisible by 121. Another solution uses n2 + 3n + 5 (n −4)(n + 7) + 33.
21. p must be odd. p 3 gives p2 + 2 11, p3 + 2 29. For p > 3, we have p 6n ± 1, and p2 + 2 is divisible by 3.
6. Number Theory 139 22. The number of two’s in n! is ⌊n 2 ⌋+ ⌊n 4 + ⌊n 8⌋+ · · · < n 2 + n 4 + n 8 + · · · < n.
23. The number of fives in 1000! is 200+40+8+1=249. The number of two’s is enough to match each 5 to get a 10. Thus, 1000! ends in 249 zeros.
24. We consider three boxes 0, 1, 2. We put a number into the box i if its remainder on division by 3 is i. Either there will be 3 numbers in some box, and then we have three numbers with sum 0 mod 3, otherwise, there will be at least one number in each box.
Then the sum of these numbers is divisible by 3.
25. We must show that x2 + y2 + z2 8s + 7 has no integral solutions. If x, y, z are even the two sides have different parity. If two are even, and one is odd, then we have 8p + 1 + 4a2 + 4b2 8t + 7, or 4(p −t) + 2a2 + 2b2 3, that is, even=odd, a contradiction. Suppose only one term on the left is even. Then we have even=odd. Finally, in the case all three terms on the left side are odd, we have 8p + 1 + 8q + 1 + 8r + 1 8t + 7, or 2p + 2q + 2r −2t 1, or even=odd.
So every parity combination on the left side leads to a contradiction. All numbers of the form 8t + 7 are not representable as sums of three squares. But that is not all.
We will prove by finite descent that all numbers of the form 4n(8t + 7) are not sums of three squares. Suppose x2 + y2 + z2 4n(8t + 7). Then we can show as above that x 2x1, y 2y1, z 2z1 . This implies x2 1 + y2 1 + z2 1 4n−1(8t + 7). And again x1, y1, z1 are even. Finally, we arrive at x2 n + y2 n + z2 n 8t + 7, which has no integral solutions. It can be proved by a complicated argument that any integer not of the form 4n(8t + 7) can be represented as a sum of three squares. So we have found all numbers which are not sums of three squares, although we have not proved it.
26. Suppose n2 aabb. Then n2 1100a + 11b 11(100a + b) 11(99a + a + b).
Since n2 is divisible by 112, we see that 11 | a + b, that is, a + b 11. Since n2 is a square, b cannot be 0, 1, 2, 3, 5, 7, or 8. Checking the remaining digits we see that only 7744 882 fits. We can eliminate b 5 since a square ends with 25.
27. (a) No, a square divisible by 3 is also divisible by 9.
(b) Same argument.
28. The number 101962 + 1 (10654)3 + 1 is divisible by 10654 + 1.
29. If the digital sums of two numbers are equal then their difference is a multiple of 9.
Hence their difference 2a −a a is divisible by 9.
30. (n −2)2 + (n −1)2 + n2 + (n + 1)2 + (n + 2)2 5(n2 + 2). So 5 | n2 + 2, that is, n2 5q −2. But a number of the form 5q −2 is not a square.
31. For each of the n primes pi, we have ai + 1 choices for the number of primes pi to be included into the divisor.
32. Two of (n + 1) positive integers ≤2n are consecutive. They are coprime.
33. Represent these (n + 1) numbers ≤2n in the form 2k(2m + 1). There are only n odd numbers in the interval 1..2n. Thus two of the odd divisors of the representations are equal. Then one of the two corresponding numbers is divisible by the other.
34. gcd(30n + 2, 12n + 1) gcd(12n + 1, 6n) gcd(6n, 1) 1. gcd(21n + 4, 14n + 3) gcd(14n + 3, 7n + 1) gcd(7n + 1, 1) 1.
35. gcd(2n + 3, n + 7) gcd(n + 7, n −4) gcd(n −4, 11) 1 if n ̸≡4 mod 11.
36. gcd(5a + 3b, 13a + 8b) gcd(5a + 3b, 3a + 2b) gcd(3a + 2b, 2a + b) gcd(2a + b, a + b) gcd(a + b, a) gcd(a, b).
140 6. Number Theory 37. gcd(2a −1, 2b−1) gcd(2a −2b, 2b−1) gcd[2b(2a−b−1), 2b−1] gcd(2a−b− 1, 2b −1). This is one step of Euclid’s algorithm on the exponents.
38. If p and q are primes > 3, then p 6m ± 1, and q 6n ± 1.
p2 −q2 (6m ± 1)2 −(6n ± 1)2 36(m2 −n2) −12(±m ± n) 12(m + n)(3(m −n) ± 1).
On the right side, either m + n or 3(m −n) ± 1 are even. Thus 24 | p2 −q2.
39. p, p + 10, and p + 14 belong to three different residue classes mod 3. So one of these numbers is divisible by 3. So only p 3 gives the primes 3, 13, 17. The same is true for the second example.
40. For p 3, we have 2p + 1 7 and 4p + 1 13. For p > 3, one of the three numbers is divisible by 3. This follows if we put p 6n ± 1, or even simpler by looking at the numbers mod 3. Then we get p, −(p −1) and p + 1 which belong to three different residue classes mod 3.
For p 3, we have 8p2 +1 73. For p > 3, we have 8p2 +1 ≡−(p2 −1) mod 3.
The last number is −(p −1)(p + 1) mod 3. Thus we have three different residue classes mod 3. So for p > 3 p or (p −1)(p + 1) is divisible by 3.
41. This follows from 10(a +4b)−(10a +b) 39b, 43(3x +7y)−3(43x +75) 38y, 10(3a + 2b) −3(10a + b) 17b. How do you get these linear combinations systematically?
42. We write p in the form p 30q + r with r ∈{7, 11, 13, 17, 19, 23, 29}. Then p2 ≡r2 mod 30. A simple check with the seven possible values gives the result.
43. x2 + y2 x2y2 ⇔x2y2 −x2 −y2 + 1 1 ⇔(x2 −1)(y2 −1) 1 ⇔x y 0.
Another solution uses parity and infinite descent starting from the fact that both x and y must be even.
44. (a) n5−5n3+4n n(n4−5n2+4) n(n2−4)(n2−1) (n−2)(n−1)n(n+1)(n+2).
The product of five consecutive integers is divisible by 5!.
(b) f (n) 4n + 15n −1 ≡0 mod 3, but this is not enough. We use induction.
f (0) 0, so 9 | f (0). Suppose 9 | f (n) for any n. Then f (n + 1) 4 · 4n + 15n + 15 −1 3 · 4n + 4n + 15n −1 + 15 f (n) + 3(4n + 5), which is divisible by 9 since 4n + 5 ≡0 mod 3.
45. These are m consecutive integers.
46. If each of x, y, z is odd, we have 3 ≡1 mod 8. If any one of x, y, z is odd, we have odd=even. If x and y are odd and z even we have 2 ≡1 mod 4. If any of x, y is odd and the other together with z even, we have 1 ≡0 mod 4. Thus each of x, y, z is even. This starts an infinite descent with the only solution x y z 0. Another solution is based on (x2 −1)(y2 −1) z2 + 1.
47. (a) x + y xy ⇒(x −1)(y −1) 1. Thus x y 2. Solve (b) yourself.
48. (a) This is x2 ≡−1 mod 3 and has no solution. Solve (b) on your own.
49. (a) Use infinite descent.
(b) Use infinite descent.
50. Transform the equation into the form (x −1)2 + (y −1)2 + (x −y)2 2. It has the solutions (0,0), (1,0), (0,1), (2,1), (1,2), (2,2).
51. p −1 6p3p4 · · · pn −1 6P −1 is not a square. No solution for p + 1.
52. We have proved a similar result by invariance. We could do this in the same way.
But here we do it by number theory. One-half of the terms are +1 and one half are 6. Number Theory 141 −1. Thus n 2k. But aiai+1 −1 if and only if the two factors are of opposite sign, that is, k is the number of changes of sign in the sequence a1, a2, · · · , an, a1.
The changes from +1 to −1 are as often as those from −1 to +1. Thus k 2m, and n 4m.
Another solution runs as follows. Set pi aiai+1. One half of the pi are equal to −1.
Consider p1p2 · · · pn (−1)k. But in this product every ai occurs exactly twice. So the product is 1. Thus k 2m. That is n 4m.
53. 1 + 2 + · · · + n n(n + 1)/2 must be divisible by 3, that is, 3 | n or 3 | n + 1. This necessary condition is also sufficient if n > 3. Show this.
54. The given equation is equivalent to (106 −1)n (10k −1)/9 ⇒n (10k − 1)/9(106 −1) with k 6m. Then n (1 + 106 + · · · + 106(m−1))/9. The numerator becomes a multiple of 9 if m 9. Thus, the smallest n is n 1054 −1 9(106 −1).
55. Let d be the first digit. Then the number is n 10kd + r. We get 3(10kd + r) 2 10r + d ⇒3d · 10k + 3r 20r + 2d ⇒d(3 · 10k −2) 17r, that is, 17 | 3 · 10k −2 ⇒3 · 10k ≡2 mod 17 ⇒10k ≡12 mod 17 with the smallest solution k 15, d 1 : r 3 · 1015 −2 17 ⇒n 20 · 1015 −2 17 .
56. (a) Suppose you have two positive integers a, b with a > b in decimal notation. You want to append the digit c to the end of either a or b to make the largest possible product. Since (10a + c)b −(10b + c)a c(b −a) < 0, you should append c to the smaller number. Using this result, we construct the largest product in a sequence of optimal steps: a 9642, b 87531. We leave (b) to the reader.
57. Let x be the leftmost digit, and let y be the number resulting from crossing off that digit. Then 10nx +y 57y, 10nx 56y. The right side has the factor 7. Hence the left side has the factor 7. But 10n is not divisible by 7. Since x < 10, x 7. Thus 10n 8y, y 10n/8 125 · 10n−3, n 3, 4, 5, · · · . 10nx + y 7 · 10n + 125 · 10n−3 7125 · 10n−3. There are infinitely many solutions 7125 · 10n−3, n ≥3. The smallest solution is 7125. We get the other solutions by attaching zeros to 7125.
58. We prove the more general theorem: Let a, b, c, d ∈N, and let n ∈N. If ab cd, then an + bn + cn + dn is not a prime. Proof: ab cd ⇒a c d b x y , gcd(x, y) 1; x, y ∈N, or a ux, c uy, d vx, b vy, u, v ∈N, Thus, an + bn + cn + dn unxn + vnyn + unyn + vnxn (un + vn)(xn + yn).
Now un + vn > 1, xn + yn > 1. Thus an + bn + cn + dn is not a prime.
142 6. Number Theory 59. abcd · 4 dcba ⇒a < 3, since 3000 · 4 12000 has five digits. But dcba is even. Thus a must be even, i.e., a=2. From 2bcd · 4 dcb2, we get d ≥8, and the product d · 4 ends in 2. Thus d 8. The result 2bc8 · 4 8cb2 or 8000 + 400b + 40c + 32 8000 + 100c + 10b + 2 ⇒390b + 30 60c ⇒13b + 1 2c.
The right side is even, and 2c ≤18. Thus b must be odd and smaller than 2, i.e., b 1, c 7, abcd 2178.
60. Find the unique solution, as in the preceeding problem.
61. This is because p and 2p, but not 3p, are factors of (2n)!.
62. First let us find out for what values of n the terms an are positive integers. an ∈N if and only if there exists a q ∈N such that 24n + 1 q2 or n q2 −1 24 (q −1)(q + 1) 24 .
Since n ∈N the denominator must cancel. Hence q must be odd. Then q −1 and q + 1 are consecutive even numbers, and one of them is a multiple of 4. So the product (q −1)(q + 1) is divisible by 8. In addition, either q −1 or q + 1 must be a multiple of 3. Hence there is an s ∈N such that q ± 1 6s or q 6s ± 1. Then n s(3s ± 1) 2 , s 1, 2, 3, . . .
and an 6s ± 1. But every prime from 5 on has the form 6s ± 1.
63. (a) We will show that all numbers of the form (3k +2)2 are not of this form. Suppose (3k + 2)2 n2 + p. Then p (3k + 2)2 −n2 (3k + n + 2)(3k + n −2). This is a nontrivial decomposition of p.
(b) We leave this to the reader.
64. If the lattice points (a, b) and (c, d) are equidistant from ( √ 2, 1/3), then (a − √ 2)2 + (b −1 3)2 (c − √ 2)2 + (d −1 3)2, or a2 −c2 + b2 −d2 −2 3(b −d) 2 √ 2(a −c).
(1) The left side is rational. So the right side must also be rational. Thus, a c.
(2) Hence, b2 −d2 −2(b −d)/3 0, (b −d)(b + d) −2(b −d)/3 0, (b −d)(b + d −2 3) 0.
(3) b + d −2/3 ̸ 0 since b + d is an integer. So b d. Thus, (a, b) (c, d).
65. Do this problem in the same way as the preceding one.
6. Number Theory 143 66. (a) a2 ends in a, i.e., a2−a ends in 00, or 100 | a(a−1). But a−1 and a are relatively prime. So one is a multiple of 4, the other of 25.
(i) a 25q. Since a < 100, a −1 25q −1 is a multiple of 4 only for q 1.
Thus, a 25, a2 625.
(ii) a−1 25q, a 25q +1 is a multiple of 4 only for q 3. Thus, a 76, a2 5776. Hence, 25 and 76 are the only two-digit automorphic numbers.
(b) a2 −a a(a −1) is divisible by 1000 8 · 125. So one is a multiple of 8, the other of 125.
(i) a 125q, a −1 125q −1 120q + (5q −1), 8 | a −1, 8 | 5q −1 with the only solution q 5. (Note: q < 8 since a < 1000.) Thus, a 625, a2 390625.
(ii) a −1 125q, a 125q + 1 120q + 5q + 1. Since 8 | 5q + 1, the only solution is q 3. Thus, a 376, a2 141376. Hence, 376 and 625 are the only three-digit automorphic numbers.
(c) a(a −1) a(a −1) is divisible by 10000, or 16 · 625.
(i) a 625q, a −1 625q −1 624q + q −1, 16 | a, 16 | q −1, q 17, a 625 · 17 10625 > 10000. But a must have four digits. There is no solution in this case.
(ii) a−1 625q, a 625q+1 624q+q+1, 16 | a, 16 | q+1, q 15, a 9376.
There is only one 4-digit automorphic number: a 9376, a2 87909376.
(d) We tabulate these results together with one extrapolation: --sum 5 6 11 25 76 101 625 376 1001 0625 9376 10001 90625 09376 100001 n an divisor of an 1 2 2 2 12 22 3 112 23 (24 also) 4 2112 24 (25 and 26 also) 5 22112 25 6 122112 26 (27, 28 also) 7 2122112 27 67. We get the right table above by experimenting: This table suggests that the numbers an are constructed as follows: a1 2, an+1 1an if 2n+1 ̸\ an (i.e., prepend digit 1 to an). an+1 2an if 2n+1 | an (i.e., prepend digit 2 to an). Suppose an dndn−1 · · · d2d1, where di 1 or 2, and 2n | an, i.e, an 2nbn.
(i) 2n+1 ̸\ an, i.e., bn is odd. We get an+1 1an 10n + an 10n + 2nbn 2n(5n + bn) 2n+1cn since 5n + bn is odd.
(ii) 2n+1 | an, i.e., an 2n+1. We get an+1 2an 2·10n +2n+1bn 2n+1(5n +bn).
Note: The theorem is valid for all bases of the form 4k + 2, k ∈N.
68. x2 + y2 n ⇒(x + y)2 + (x −y)2 2n.
69. The tactical idea is to show that n2 −19n+89 lies between two consecutive squares.
Indeed, n2 −19n + 89 n2 −18n + 81 −(n −8) (n −9)2 −(n −8) >0 < (n −9)2 n2 −19n + 89 n2 −20n + 100 + (n −11) (n −10)2 + n −11 >0 > (n −10)2 (n −10)2 < n2 −19n + 89 < (n −9)2.
144 6. Number Theory 70. 2n (x + y)2 + 3x + y (x + y)2 + (x + y) + 2x (x + y)(x + y + 1) + 2x, n (x + y)(x + y + 1) 2 + x x + x + y + 1 2 .
The first formula shows that the right side is, indeed, even. The second formula shows how to find x, y. First, straddle n by two consecutive triangular numbers Tz z 2 and Tz+1 z+1 2 as follows Tz ≤n < Tz+1. Then n Tz + x with z x + y + 1.
For instance, let n 1000. Then n 45 2 + 10. So x 10, x + y + 1 45 which implies x 10, y 34. One could also find x, y explicitly in terms of n.
71. This simple theorem is best proved by proving the contraposition: m not prime ⇒m ̸ | (m −1)! + 1, which is obvious. If m is not prime, it can be decomposed into m pq with 1 < p < m and 1 < q < m. Then m is a divisor of (m −1)! and cannot be divisible by the next number. To prove the converse is slightly more difficult. Both the theorem and its converse give Wilson’s theorem.
72. Use induction to prove that the factor 2 occurs exactly n times.
73. Let a, b, m, n ∈N, gcd(a, b) 1, a > 1. We will prove three lemmas.
(a) Let m qn with odd q. Then an + bn | am + bm.
(b) Let m qn + r, q odd and 0 < r < n. Then an + bn ̸ | am + bm.
(c) Let m sn + r, s even, 0 ≤r < n. Then an + bn ̸ | am + bm, that is, with odd q we have the more precise statement: an + bn | am + bm ⇔m qn.
Proof. (a) aqn + bqn (an)q + (bn)q is divisible by an + bn for odd q.
(b) am + bm aqn+r + bqn+r ar(aqn + bqn) + bqn(br −ar).
From (a), we see that the first term on the right is divisible by an + bn. The second term is not divisible by an +bn since gcd(bqn, an +bn) 1 and | br −ar | < an +bn.
Thus the sum is not divisible by an + bn.
(c) If s is even, then q −s is odd. With s q +1, we write am +bm asn+r +bsn+r aqnan+r +bqnbn+r an+r(aqn +bqn)+bqn+r(bn +an)−bqnan(br +ar). The first two terms are divisible by an + bn, the third is not. Indeed, gcd(bqnan, an + bn) 1, and 0 < br + ar < an + bn. This proves the stronger statement above.
74. (a) Suppose none of the numbers is divisible by 3. Then 1 + 1 ≡1 mod 3, which is a contradiction.
(b) Suppose that none of x, y, z is divisible by 4. Suppose x and z are odd and y 4q + 2. Then we have 1 + 4 ≡1 mod 8. This is a contradiction.
(c) Suppose none of the three numbers is divisible by 5. Then we have ±1 ± 1 ≡ ±1 mod 5. Contradiction.
75. Take from the numbers 0, 1, . . . , 3k −1 all those 2k different numbers which contain no 2’s in their ternary expansion. These will not be in arithmetic progression. Indeeed, suppose a+c 2b for some a, b, c consisting only of the digits 0 and 1. The number 2b consists only of the digits 0 and 2. Hence a and c must match digit for digit, and so a b c.
6. Number Theory 145 76. This and the next three problems have automatic solutions. You just make obvious transformations and always look for patterns. First multiply, collect terms, and cancel factor 2: m2 + (m + 1)2 n4 + (n + 1)4 ⇒m2 + m n4 + 2n3 + 3n2 + 2n, m2 + m (n2 + n)2 + 2(n2 + n) ⇒m2 + m + 1 (n2 + n)2 + 2(n2 + n) + 1 ⇒m2 + m + 1 (n2 + n + 1)2.
The right side is a square, the left is not because it lies between two consecutive squares: m2 < m2 + m + 1 < m2 + 2m + 1 (m + 1)2.
77. 2 + 2 √ 28n2 + 1 m ⇒4(28n2 + 1) m2 −4m + 4 ⇒m 2k ⇒28n2 + 1 k2 −2k + 1 ⇒28n2 k2 −2k ⇒k 2q ⇒28n2 4q2 −4q ⇒7n2 q(q −1).
Here q and q −1 are relatively prime.
(i) q 7x2, q −1 y2 ⇒7x2 −y2 1. This case cannot occur, because y2 ̸≡−1 mod 7.
(ii) q x2, q −1 7y2. In this case, m 2k 4q 4x2 (2x)2. So we have solved the problem. We were not required to show that there is a solution.
Only if there is a solution, it must be a square. We have done just that. There are in fact infinitely many solutions. Eliminating q by subtraction we get the Pell–Fermat equation x2 −7y2 1. We find the smallest positive solution by inspection. It is x0 8, y0 3. Thus all solutions are given by xn + yn √ 7 (8 + 3 √ 7)n.
78. x3 + 3 4y(y + 1) ⇒x3 + 3 4y2 + 4y ⇒x3 + 4 (2y + 1)2 ⇒x3 (2y + 1)2 −4 (2y −1)(2y + 3).
But gcd(2y + 3, 2y −1) gcd(2y −1, 4) 1. Thus, 2y −1 u3, 2y + 3 v3, v3 −u3 4. But no two cubes can differ by 4. So there is no solution.
79. 11111111111 · 109 ≤x < 11111111111 · 109 + 109 ⇔(1011 −1)109 ≤9x < (1011 −1)109 + 9 · 109.
Now (1010 −1)2 < 1020 −109 ≤9x, (1010 + 1)2 > 1020 + 8 · 109 > 9x. But there is just one square between (1010 −1)2 and (1010 + 1)2. So 9x 1020. But 1020 is not divisible by 9.
80. We transform the left side as follows: a2 + ab + b2 (a −b)2 + 3ab ⇒ 3 | a −b ⇒9 | 3ab ⇒3 | a or 3 | b and 3 | a −b ⇒3 | a and 3 | b.
81. Since 1971 27 · 73 with gcd(73, 27) 1, for odd n, we have 50n + 23na ≡(−4)n + (−4)na ≡−4n(a + 1) mod 27 ⇒a ≡−1 mod 27, 50n + 23na ≡(−23)n + 23na ≡23n(a −1) mod 73 ⇒a ≡1 mod 73, 146 6. Number Theory that is, a 73x + 1 and a 27y −1, or 73x −27y −2. The last equation has infinitely many solutions. We must pick the one with smallest a.
73 73 · 1 + 27 · 0, 27 73 · 0 −27 · −1 ⇒19 73 · 1 + 27 · (−2) ⇒8 73 · (−1) + 27 · 3 ⇒3 73 · 3 + 27 · (−8) ⇒2 73 · (−7) + 27 · 19 ⇒−2 73 · 7 −27 · 19.
Starting with the third equation, you get equation #n by subtracting equation #(n−1) as often as possible from equation #(n −2) so as to get a possible left side. From the last equation, we get one solution x0 7, y0 19. Thus all solutions are given by x 7 + 27t, y 19 + 73t. We get the smallest positive a for t 0 : a 73 · 7 + 1 27 · 19 −1 512.
82. Multiplying by 3 and adding 7, we get 10n for the nth term. Thus an (10n −7)/3.
From 102 ≡−2 mod 17, we get 108 ≡16 ≡−1 mod 17. From this, we get 109 ≡−10 ≡7 mod 17 and 1016 ≡1 mod 16. Hence 17 | (1016k+9 −7)/3, that is, (1016k+9 −7)/3 is composite for k 0, 1, 2, . . . . On the other hand, an is prime for n 1, 2, 3, 4, 5, 6, 7, 8. There is also an infinite sequence divisible by 19. Find it.
83. n even ⇒n · 2n −1 ≡n −1 ≡0 mod 3 ⇒n 6k + 4.
n odd ⇒n · 2n −1 ≡0 ≡−n −1 ≡0 mod 3 ⇒n 6k + 5 (k ∈N0).
84. Since gcd(m + 1, m) 1, we require that m + 1 an, m bn. or an −bn 1.
But no two powers differ by 1.
85. 4n (2n−1)+(2n+1). Here the two sums on the right side are two odd consecutive numbersandhavenocommondivisor.Foroddnumbers,wehave2n+1 n+(n+1).
Finally,ifnisodd,then2n (n−2)+(n+2)withgcd(n−2, n+2) gcd(4, n−2) 1.
86. Since x2 + 2y2 is a prime, x must be odd, and x2 ≡1 mod 8. If y is even, then 2y2 ≡0 mod 8 and x2 + 2y2 ≡1 mod 8. If y is odd, then y2 ≡1 mod 8, and x2 + 2y2 ≡3 mod 8.
87. From b > 2, we conclude that a > b. From a qb + r, 0 ≤r < b, and 2a + 1 2b −1 2a−b + 2a−b + 1 2b −1 , we conclude 2a + 1 2b −1 2a−b + 2a−2b + · · · + 2r + 1 2b −1, 2r + 1 2b −1 < 1.
88. (a) (n −1)n(n + 1) (n2 −1)n mk. Since gcd(n2 −1, n) 1, we must have n2 −1 ak, n bk, bk −ak 1. There are no solutions in N.
(b) Suppose x(x + 1)(x + 2)(x + 3) (x2 + 3x)(x2 + 3x + 2) yk. Then gcd(x2 + 3x + 2, x2 + 3x) gcd(x2 + 3x, 2) 2, gcd((x2 + 3x)/2, (x2 + 3x + 2)/2) 1.
Then (x2 + 3x)/2 ak, (x2 + 3x + 2)/2 bk and bk −ak 1. No two kth powers of positive integers have difference 1.
89. The digit block with the last digit removed is b. Then 10b + 9 9 · 10n + b.
90. The solution can be found in Chapter 10, problem 63.
6. Number Theory 147 91. There is no general method visible, but we observe that x and y do not differ much.
Indeed, y3 −(x + 1)3 5x2 −9x + 7 > 0 and (x + 3)3 −y3 x2 + 33x + 19 > 0.
That is, x + 1 < y < x + 3. Since x and y are integers we must have y x + 2.
Replacing y by x + 2, we get 2x(x −9) 0 with solutions x1 0, y1 2, and x2 9 y2 11. The pairs (0, 2) and (9, 11) do satisfy the original equation.
92. (a) 2n + 1 x2, 3n + 1 y2. The first equation implies that x is odd, i.e., n 4m is even. The second equation implies 3n 8m1 or n 8m2. Thus, n ≡0 mod 8.
We still have to show that n ≡0 mod 5. Now the quadratic residues can only be 0, 1, 4 mod 5. Thus, modulo 5, we have n ≡1 ⇒x2 2n + 1 ≡3, n ≡2 ⇒y2 3n + 1 ≡2, n ≡3 ⇒x2 2n + 1 ≡2, n ≡4 ⇒y2 3n + 1 ≡3.
These are all contradictions. Thus n ≡0 mod 5. So we have proved n ≡0 mod 40.
(b) The first two equations imply 3x2 −2y2 1. We can transform this equation into a Pell equation by the transformation x u + 2v, y u + 3v. We get u2 −6v2 1 with the smallest positive solution u0 5, v0 2. Thus all solutions are given by un + vn √ 6 (5 + 2 √ 6)n. The solution x0 9, y0 11 with y2 0 −x2 0 n 40 corresponds to u0, v0.
One can also directly guess the smallest solution x0 9, y0 11. Then all solutions are given by xn √ 3 + yn √ 2 (9 √ 3 + 11 √ 2)n.
93. Note that 468 53 + 73. Thus 4684 468 · 4683 (5 · 468)3 + (7 · 468)3.
94. 3851980 + 181980 is congruent to 2 mod 13, but 2 is not a quadratic residue mod 13.
To see this, we consider the table x 0 1 2 3 4 5 6 x2 0 1 4 -4 3 -1 -3 We need not go beyond 6 since x ≡7 ≡−6 mod 13, and so on until 12 ≡−1 mod 13 since we get the same quadratic residues in inverse order. Now 385 ≡−5 mod 13, 18 ≡5 mod 13, 54 ≡(−5)4 mod 13. Since 1980 is a multiple of 4, we have the result. A smaller module will not do since we would get a possible quadratic residue.
95. Find the sum modulo 4. The terms of the sum are a periodic sequence with period 1, 0, −1, 0 of length 4, that is, the sum is a multiple of 4. If the sum would be of the form mk with k ≥3, then it would be divisible by 8. Let us look at the sum modulo 8. If n is even, and n ̸ 2, then nk is a multiple of 8. If n is odd, then nk ≡n mod 8.
Thus the sum is modulo 8 22 + 1 + 3 + 5 + · · · + 1983 984068 ≡4, which is not a multiple of 8.
96. y2 x3 +7 ⇔y2 +1 x3 +8 (x +2)(x2 −2x +4). First we observe that, if x is even then y2 ≡7 mod 8. But we know that an odd square is congruent to 1 modulo 8. Thus, x must be odd. But x2 −2x + 4 (x −1)2 + 3 4k + 3. Thus this factor has a prime factor of the same form because the factors of the form 4k +1 are closed under multiplication. But it is known that odd numbers can have only prime factors of the form 4k + 1 (except 2). We will prove this well known fact. Let q be a prime factor of y2 + 1. Then y2 ≡−1 mod q. Because of Fermat’s theorem, we also have yq−1 ≡1 mod q. From y2 ≡−1 mod q, we get y4 ≡1 mod q by squaring. Hence 4 | q −1 ⇒q 4k + 1. This is a contradiction.
148 6. Number Theory 97. We must find x ≡79999 mod 1000 or 7x ≡710000 mod 1000. But φ(1000) 1000(1 −1/2)(1 −1/5) 400, 7400 ≡1 mod 1000. Since 10000 25 · 400, we have 7x ≡1 mod 1000. Thus we have to find the inverse of 7 mod 1000. This can be done in a standard way by solving the equation 7x + 1000y 1 with the Eu-clidean algorithm. But in this particular case, we use the fact that 1001 7 · 11 · 13, which is well known by a high school student since his teacher uses it for many tricks.
Now, obviously, 1001 ≡1 mod 1000. But 1001 143·7. so 143·7 ≡1 mod 1000.
Thus, x 143.
98. Multiplying by xyz, we get yz+xz xy. Let x da, y db with gcd(a, b) 1.
Then dbz + daz d2ab ⇒(a + b)z dab ⇒z d · ab a + b .
Now gcd(a, b) gcd(a, a + b) gcd(b, a + b) gcd(ab, a + b) 1, that is, d k(a + b), z kab, x ka(a + b), y kb(a + b).
Since gcd(x, y, z) 1, we have k 1, and finally, x a(a + b) y b(a + b) z ab.
Indeed, 1 a(a + b) + 1 b(a + b) 1 ab .
99. Multiplying with x2y2z2, we get (yz)2 + (xz)2 (xy)2. Using the formulas in item 13, we get yz u2 −v2, xz 2uv, xy u2 + v2, gcd(u, v) 1, u ̸≡v mod 2.
With xyz k, we get kx 2uv(u2 + v2), ky (u2 + v2)(u2 −v2), kz 2uv(u2 −v2).
100. Using the hint, we proceed as follows: (x2 −dy2)(u2 −dv2) (x + y √ d)(x −y √ d)(u + v √ d)(u −v √ d) (x + y √ d)(u −v √ d)(x −y √ d)(u + v √ d) (xu −dvy −(xv −yu) √ d))(xu −dvy + (xv −yu) √ d)) (xu −dvy)2 −d(xv −yu)2.
Similarly, we proceed with x2 + dy2. Another approach is via matrices and deter-minants. The matrix x yd y x is a matrix with determinant x2 −dy2. If we are familiar with multiplication of matrices, then x yd y x u v d v u xu + dyv d(xv + yu) xv + uy xu + dyv .
If A, B are two matrices, then the determinant of the product is the product of determinants, i.e., det(A B) det(A) det(B). Applying this rule to our matrices, we get (x2 −dy2)(u2 −dv2) (xu + dyv)2 −d(xv + yu)2. Similarly, we proceed with other similar so-called quadratic forms in two variables.
6. Number Theory 149 101. 2(11987+21987+· · ·+n1987) (n1987+21987)+· · ·+(21987+n1987)+2 (n+2)P +2, where P is an integer. This follows from a + b | ak + bk for odd k. Thus n + 2 does not divide the sum.
102. (5 + 3 √ 2)m (3 + 5 √ 2)n ⇒(5 −3 √ 2)m (3 −5 √ 2)n. The only solution is m n 0 since 0 < 5 −3 √ 2 < 1, but 5 √ 2 −3 > 1.
103. Assume x ≥y. Then x d + y, d ≥1, and (3d −1)y2 + (3d2 −d)y + d3 61.
From this, we infer that d ≤3. d 1 yields 2y2 + 2y −60 0, y2 + y −30 0 with y 5 x 6. The other two possible values d 2, d 3 yield no solutions in positive integers. Because of the symmetry of the original equation in x and y, we have an additional solution x 5, y 6.
104. From y3 z4 −x2 (z2 −x)(z2 + x), we have either z2 −x 1, z2 + x y3, or z2 −x y, z2 +x y2. From the first system, we get x z2 −1 (z−1)(z+1).
Thus z −1 1 or z 2, x 3, y3 5, a contradiction. Upon addition, the second system leads to the contradiction y2 2z2.
105. The fifth place is a 5. The places #2, 4, 6, 8 are even. The others must be odd. For d1d2d3d4 to be divisible by 4, we must have d4 2 or 6. d6d7d8 and hence also d7d8 should be divisible by 8. Thus d8 2 or 6. Hence d2, d6 4 or 8. Now, d1d2d3 is divisible by 3, and d1d2d3d45d6 is divisible by 6. For d2, there are just two possibilities: d2 4, d2 8. The first posibility leads to two numbers which are not divisible by 7. The second possibility d2 8 leads to the only solution 381654729.
106. (x −y)5 + (y −z)5 + (z −x)5 is zero for x y, y z, z x. So the terms x −y, y −z, z −x can be factored out. To see that 5 is also a factor, we observe that, by multiplying the parentheses, the terms x5, y5, z5 cancel. The remaining terms all are multiples of 5. This proves the assertion.
107. We have 10000x + 1986 1987z, x, z ∈N. With y z −1 we get 10000x − 1987y 1. This equation has infinitely many solutions x, y, and the smallest is x 214. Thus the answer is 2141986.
108. Dividing by 2, we get 991 | 11 · · · 1 1980 . But 11 · · · 1 1980 (101980 −1)/9. Now 101980 − 1 (10991 −1)(10991 + 1). Since 991 is a prime, by Fermat’s theorem, we have 991 | 10991−1 −1. This proves the assertion.
109. Can 1 · 10k1 + 2 · 10k2 + 3 · 10k3 + · · · + 1986 · 10kn be a cube x3? Cubic residues of x are just 0, 1, −1. Since 10k ≡1 mod 9, we get x3 ≡1 + 2 + 3 + · · · + 1986 ≡ 1987 · 1986/2 ≡1987 · 993 ≡7 · 3 ≡3 mod 9. So x3 ≡3 mod 9. But 3 is not a cubic residue mod 9. Thus we have proved the theorem. Without the very nice criterion for divisibility by 9, we would be completely lost.
110. φ(256) 128, 1986 128 · 15 + 66. The theorem of Euler–Fermat tells us that 271986 ≡2766 mod 256. Now 2764 ≡(−39)32 ≡(−15)16 ≡(−31)8 ≡(−63)4 ≡ 1292 ≡1 mod 256 ⇒2766 ≡272 ≡−39 ≡217 mod 256. Writing 217 in the binary system, we get the last 8 digits 110110012.
111. We do this problem by induction. For the first values of n, 3n has an even next to the last digit. Suppose 3n Bed where d is one of the digits 1, 3, 7, 9 and e stands for an even digit. B is the initial block of digits which do not interest us. If you multiply d by 3, you will always have an even carry of 0 or 2. Adding this to e, we again get an even digit, sometimes with a carry which affects only the third digit from the right.
150 6. Number Theory 112. 1000m −1 | 1978m −1 implies that 1000m −1 divides the difference 1987m −1000m, or 2m(989m −500m). But 1000m −1 is odd. Thus 1000m −1 | 989m −500m. But this is obviously wrong since 1000m −1 > 989m −500m.
113. n!/n(n + 1)/2 2 · 2 · 3 · · · (n −1)/(n + 1). If n + 1 p, a prime, then the answer is obviously no!, except for n 1. In all other cases, the answer is yes! We will prove the yes!, since it is less obvious. Suppose n + 1 pq > 3 (p ≤q, q > 1).
First case: 1 < p < q ≤(n + 1)/2 ≤n −1. In this case, p and q are in distinct factors of (n −1)!.
Second case: p q. For n 3, we have (1 + 2 + 3)|(1 · 2 · 3). Otherwise, we have n > 3 and q > 2 ⇒q(q −2) > 1 ⇒q2 > 2q + 1. With n + 1 q2, we have n + 1 + q2 > q2 + 2q + 1 ⇒n > 2q. Thus (n −1)! contains the factors q and 2q.
114. We prove the contraposition 5 ̸ | abcde ⇒25 ̸ | a5 + b5 + c5 + d5 + e5.
(5k ± 1)5 (5k)5 ± 5(5k)4 + 10(5k)3 ± 10(5k)2 + 5 · 5k ± 1.
(5k ± 2)5 (5k)5 ± 5(5k)4 · 2 + 10 · (5k)3 · 22 ± 10(5k)2 · 23 + 5 · 5k · 24 ± 25.
Thus, (5k ± 1)5 ≡±1 mod 25 and (5k ± 2)5 ≡±7 mod 25. Addition of 5 of the four numbers +1, −1, +7, −7 never gives 0 or ±25, or 35 5 · 7.
115. (a+b)7 −a7 −b7 7ab(a+b)(a2 +ab+b2)2. Thus we must have 73 | a2 +ab+b2.
For a 18, b 1, we have 73 a2 + ab + b2. There are also more systematic ways to a solution.
116. See Chapter 14.4, example E1 for a solution.
117. (a2 + ab + b2)(c2 + cd + d2) (a −ωb)(a −ωb)(c −ωd)(c −ωd) (ac −bd)2 + (ac −bd)(ad + bc −bd) + (ad + bc −bd)2.
Here ω e(2πi)/(3) is the third root of unity with ω2 −1 −ω.
Another solution uses matrices.
118. ax2 + by2 1 is an ellipse. If (x0, y0) is a rational point of the ellipse, we choose a line Ax + By + C 0 through (x0, y0) with A, B, C ∈Q which intersects the ellipse in a second point (x1, y1) with x1, y1 ∈Q. By rotating the line about (x0, y0), we get infinitely many rational solutions.
119. x(x + 1)(x + 2)(x + 3) y2 ⇒(x2 + 3x)(x2 + 3x + 2) y2. Both factors on the left are even and their halves have difference 1. Thus, their gcd is 1. This implies that they are both squares: x2 + 3x 2 u2, x2 + 3x + 2 2 v2, v2 −u2 1.
The last equation has no solutions in positive integers.
120. (a) First, we check that m4 ≡0 or 1 mod 16. The right side is 0 or 1 mod 16. Hence there must be at least three even numbers on the left side.
(b) It is easy to check that m4 ≡0 or 1 mod 16. Since the right side is at most 1 mod 16, at least 3 numbers divisible by 5 will be on the left side.
(c) At least 3 of 4 on the left side are multiples of 2, and the same number are multiples of 5. Hence two will be multiples of 10.
6. Number Theory 151 121. 12m + 9m + 8m + 6m (3m + 4m)(3m + 2m). Since m 10a + 5 5(2a + 1), we have 4m +3m 45(2a+1) +35(2a+1) and 45 +35 | 4m +3m. Similarly 35 +25 | 3m +2m.
But 45 +35 1024+243 1267 7·181, 35 +25 243+32 275 25·11.
Now 1991 11 · 181. Hence, we have divisibility by 181 · 11 1991.
122. We have y3 −(x +1)3 5x2 −9x +7 > 0 and (x +3)3 −y3 x2 +33x +19 > 0.
Hence x + 1 < y < x + 3, and, since the variables are integers, we have y x + 2.
Using this in the original equation we get 2x(x −9) 0 with solutions x1 0, x2 9, y1 2, y 11. We check that (0, 2) and (9, 11) indeed satisfy the original equation.
123. The left side of the equation y2 + y x4 + x3 + x2 + x is almost a square. Just multiply by 4, and add 1, and you get 4y2 +4y +1 4x4 +4x3 +4x2 +4x +1, (2y +1)2 4x4 +4x3 +4x2 +4x +1.
The LHS is a square. We try to show that the RHS lies between two successive squares.
T (x) 4x4 + 4x3 + 4x2 + 4x + 1 (2x2 + x)2 + (3x + 1)(x + 1), T (x) 4x4 + 4x3 + 4x2 + 4x + 1 (2x2 + x + 1)2 −x(x −2).
For x < −1 or x > 0, we have (3x + 1)(x + 1) > 0 and T (x) > (2x2 + x)2. For x < 0 or x > 2, we have T (x) < (2x2 + x + 1)2. For x < −1 or x > 2, we have (2x2 + x)2 < T (x) < (2x2 + x + 1)2.
We need to check only the cases x −1, 0, 1, 2. We get (a) x −1 ⇒y2 + y 0 ⇒y 0, y −1 (b) x 0 ⇒y2 + y 0 ⇒y 0, y −1 (c) x 1 ⇒y2 + y 4 with no integral solutions (d) x 2 ⇒y2 + y 30 ⇒y −6, y 5.
The integral solutions are (−1, −1), (−1, 0), (0, −1), (0, 0), (2, −6), (2, 5).
124. x2, z2, y2 are in arithmetic progression if z2 −x2 y2 −z2, i.e., x2 + y2 2z2 ⇔(y −x)2 + (x + y)2 (2z)2.
y −x u2 −v2, x + y 2uv, 2z u2 + v2 follows from this. Addition and subtraction of the first two equations gives x 2uv −u2 + v2 2 , y u2 −v2 + 2uv 2 , z u2 + v2 2 , u > v.
Here u and v must have the same parity, so the numerators are even.
125. The number of integers from 1 to m, which are multiples of b is ⌊m/b⌋. From the assumption, we know that none of the integers 1, . . . , 1951 is simultaneously divisible by two of the numbers a1, . . . , an. Hence the number of integers among 1, . . . , 1951 divisible by one of a1, . . . , an is ⌊1951/a1⌋+ · · · + ⌊1951/an⌋.
152 6. Number Theory This number does not exceed 1951. Hence 1951 a1 −1 + · · · + 1951 an −1 < 1951, 1951 a1 + · · · + 1951 an < n + 1951 < 2 · 1951, 1 a1 + · · · + 1 an < 2.
This problem was used at the MMO 1951. It is due to Paul Erd¨ os. The 2 can be replaced by 6/5, but even this is not the best possible bound.
126. (a) The answer is 36 −52 11. The last digit of 36k 62k is 6, the last digit of 5m is 5. Hence |62k −5m| ends with 1 or 9. The equation 62k −5m 1 has no solutions since otherwise we would have 5m (6k −1)(6k + 1), but 6k + 1 is not divisible by 5. For k 1, m 2, we get 36k −5m 11.
(b) |f (1, 1)| 7. We prove that |f (m, n)| cannot assume smaller values. It cannot take the values 6, 5, 3, 0 since 12 and 5 are prime to each other. Because 12m is even and 5n is odd, it cannot take the values 4 and 2. Now we will exclude the value |f (m, n)| 1. f (m, n) 1 ⇒5n ≡−1 mod 4, and f (m, n) 1 ⇒12m ≡ 2 mod 4. This contradicts 12m ≡0 mod 4. Now, suppose f (m, n) −1. Then 5n ≡1 mod 3 ⇒n 2k ⇒12m (5k + 1)(5k −1), 5k ≡1 mod 4 ⇒5k + 1 ≡2 mod 4. Thus 5k + 1 is only divisible once by 2. From 12m (5k + 1)(5k −1), we conclude that 5k + 1 2 · 3v, 5k −1 22m−13m−v. Only one of 5k + 1 and 5k −1 must contain a factor of 3, since their difference is 2. But v 0 would imply 5k + 1 2 ⇒k 0 ⇒n 0, which is a contradiction, since 0 ̸∈N. Second case: v m ⇒5k −1 22m−1, 5k + 1 2 · 3m. The difference 2 2 · 3m −22m−1 ⇒ 3m −4m−1 1. This is not valid for any positive integer m.
127. The identity (x2 + x + 1)(x2 −x + 1) x4 + x2 + 1 gives infinitely many solutions (n, −n, n2).
128. (a) We have z2 ≡(x2 −1)(y2 −1) + 5 mod 8. Since z2 ≡0, 1, 4 mod 8, x2 −1 ≡ 0, 3, 7 mod 8, (x2 −1)(y2 −1) ≡0, 1, 5 mod 8, and (x2 −1)(y2 −1) + 5 ≡ 2, 5, 6 mod 8, we have z2 ̸≡(x2 −1)(y2 −1) + 5 mod 8.
(b) Consider the equation mod 9 : z2 ≡(x2 −1)(y2 −1) + 5 mod 9. We have z2 ≡ 0, 4, 7 mod 9, x2 −1 ≡0, 3, 6, 8 mod 9, (x2 −1)(y2 −1) ≡0, 1, 3, 6 mod 9, (x2 −1)(y2 −1) + 5 ≡2, 5, 6, 8 mod 9. Thus, z2 ̸≡(x2 −1)(y2 −1) + 5 mod 9.
(c) n 1984. Simplifying, we get x2 + y2 + z2 −x2y2 1985. The idea is to find a representation x2 + y2 1985. Then z xy gives a solution. By looking at the last digits of squares, we quickly get one of the solutions 72 + 442 1985 and 312+322 1985bytrialanderror.Thus(x, y, z) (7, 44, 7·44)and(31, 32, 31·32) are solutions. (There are infinitely many solutions.) 129. Proceed exactly as in E15.
130. Proceed similarly to E15.
131. Suppose there is a prime p such that p a2 + b2 c2 + d2 with a > b, c > d, a ̸ c. We assume that a > c. Then p2 a2c2 + b2d2 + a2d2 + b2c2 has two representations p2 (ac + bd)2 + (ad −bc)2 (ad + bc)2 + (ac −bd)2.
6. Number Theory 153 Since (ac + bd)(ad + bc) (a2 + b2)cd + (c2 + d2)ab p(ab + cd), either p | ac + bd or p | ad + bc. If p | ac + bd, then from the first representation for p2, we get ad −bc 0, ad bc, a/c b/d. Since a > c, we have b > d, and a2 + b2 > c2 + d2. Contradiction.
But if p | ad +bc, then from the second representation for p2, we get that a/b d/c, which implies d > c. But we have assumed that c > d. Contradiction.
One can show that t ac + bd gcd(ac + bd, ab + cd) is a divisor of p such that 1 < t < p.
132. Consider the equation x2 + y2 + z2 3xyz. One solution is easy to guess by inspection. It is the triple (1, 1, 1). Now, suppose that (x,y,z) is any solution. Keep y and z fixed. Then it is a quadratic in x with two solutions x and x1 satisfying x + x1 3yz or x1 3yz −x. Thus x1 is also an integer. With the triple (x, y, z) satisfying this equation, there will be another triple (3yz−x, y, z) which should also satisfy the equation. Indeed, (3yz−x)2+y2+z2 3(3yz−x)yz ⇒9y2z2−6xyz+x2+y2+z2 9y2z2−3xyz.
This simplifies to x2+y2+z2 3xyz. Thus we have found infinitely many solutions of this equation.
x 1 2 5 13 34 89 233 610 29 169 985 194 433 y 1 1 2 5 13 34 89 233 5 29 169 13 29 z 1 1 1 1 1 1 1 1 2 2 2 5 5.
If (x, y, z) satisfies the equation x2 + y2 + z2 3xyz, then (3x, 3y, 3z) will satisfy the equation x2 + y2 + z2 xyz.
133. See Chapter 13, problem 34.
134. 3n + 1 x2, 4n + 1 y2, y2 −x2 n ⇒y odd ⇒n even ⇒x odd ⇔8|n.
Here we used the fact that, if x and y are odd, then 8|y2 −x2. Now 4x2 −3y2 4(3n + 1) −3(4n + 1) 1. Thus 4x2 −3y2 1. Setting w 2x, we finally get w2 −3y2 1. This Pell equation has the solutions (2 + √ 3)n wn + √ 3yn. But only the first, third, fifth,. . . solution has an even wn. So we start with the solution 2 + √ 3 and multiply repeatedly by (2 + √ 3)2 7 + 4 √ 3. In this way, we get all solutions with even wn. We get the recursions wn+1 2xn+1 14xn + 12yn, yn+1 8xn + 7yn.
From xn+1 7xn + 6yn ≡−yn (mod 7) and yn+1 8xn + 7yn ≡xn (mod 7), we get y2 n+1 −x2 n+1 ≡x2 n −y2 n ≡n ≡0 (mod 7). Hence, 7|n.
135. One product remains unknown after 49 questions, e.g., a1a2a3. We switch the signs of all numbers ai with i ̸≡0 (mod 3), except a1. This does not change the answers to the 49 questions, but the product does change, since a1a2a3 changes its sign. Hence 49 questions do not suffice. But if we know the answers to 50 questions giving the prod-ucts a1a2a3, a2a3a4, . . . , a48a49a50, a49a50a1, a50a1a2, then, by multiplying, we get a3 1 · a3 2 · a3 3 · · · a3 50 a1 · a2 · · · a50.
154 6. Number Theory 136. Let 3k be the greatest power of 3 which is contained in n. We write n 3k(3s + r) with r 1, 2. In the following proof we use the lemma: x2 + x + 1|x6s+2r + x3s+r + 1 for all s ∈N0, r ∈{1, 2}.
We have 4n + 2n + 1 43k(3s+r) + 23k(3s+r) + 1 23k
6s+2r + 23k
3s+r + 1.
Because of the lemma, the last value is divisible by 23k
2 + 23k + 1. Since this divisor is different from 1 and 4n + 2n + 1 is a prime, we conclude that 23k
6s+2r + 23k
3s+r + 1 23k
2 + 23k + 1.
Hence, 3s + r 1, or s 0 and r 1. Thus, n 3k, a power of 3.
Now we prove the lemma. We prove that the polynomials pn(x) x6n+2 + x3n+1 + 1, qn(x) x6n+4 + x3n+2 + 1 vanish at the roots of x2 + x + 1. Indeed, the roots of the last polynomial are the third roots of unity ω, ω2. But ω6n+2 ω3n+2 ω2 and ω6n+4 ω3n+1 ω. Thus, pn(ω) ω2 + ω + 1 and qn(ω) ω2 + ω + 1.
137. (a) From 2x2 +x 3y2 +y, we get x2 x −y +3x2 −3y2 (x −y)(3x +3y +1), y2 x −y + 2x2 −2y2 (x −y)(2x + 2y + 1). Since 3(x + y) + 1 and 2(x + y) + 1 are prime to each other, and x −y gcd(x2, y2) gcd(x, y)2, the integers 3x + 3y + 1 b2 and 2x + 2y + 1 a2 must also be squares. This proves (a).
(b) With x d · b, y d · a, gcd(a, b) 1, we get d2 x −y. From (a) we get 3a2 −2b2 1 and d2 db −da ⇒d b −a, x (b −a)b, y (b −a)a. The solutions of 3a2 −2b2 1 can be obtained from √ 3 + √ 2 2n+1 an √ 3 + bn √ 2 by powering or, simpler, by recurrences. From an+1 √ 3 + bn+1 √ 2 an √ 3 + bn √ 2 5 + 2 √ 6 , we get an+1 5an + 4bn, bn+1 6an + 5bn, a1 1, b1 1. The next solution a2 9, b2 11 yields x2 22, y2 18.
138. (b) There are more 2’s then 5’s in n! for n > 1. Hence for n ≥2, the last nonzero digit dn of n! is even.
Let p be a period of dn, that is, for n ≥n0, dn+p dn. We have p ≥3. Choose m such that (p −1)! < 10m and n0 10m −1. We have (n + p)!
n!
10q(a + 10u), 1 ≤a ≤9.
But (n + p)!
n!
(n + 1) · · · (n + p) 10m (10m + 1) · · · (10m + p −1) 10m(p −1)! mod 102m ⇒2|a.
6. Number Theory 155 On the other hand, n! 10r(d + 10v), (n + p)! 10s(d + 10w), with even digit d (n + p)! n!10q(a + 10u) ⇒10s(d + 10w) 10r(d + 10v) · 10q(a + 10u) ⇒d ≡ad (mod 10).
From this it follows that a 6. Similarly, the last nonzero digit of 2 · 10m(2 · 10m + 1) · · · (2·10m+p−1) is 6. But this number is congruent to 2·10m(p−1)! mod 102m, which implies that the last nonzero digit is 2. In fact, 6·2 ≡2 mod 10. Contradiction!
139. With x 525, the number becomes x5 −1 x −1 x4 + x3 + x2 + x + 1 (x2 + 3x + 1)2 −5x(x + 1)2.
For x 525, this result is a difference of two squares, which can be factored into two factors, both greater than 1.
140. For the first time, we use the auxiliary polynomial P (t) t5+pt4+qt3+rt2+st+u with roots a, b, c, d, e. Hence P (a)+P (b)+P (c)+P(d)+P(e) 0. We conclude that a5 + · · · + e5 + p(a4 + · · · + e4) + q(a3 + · · · + e3) + r(a2 + · · · + e2)+ s(a + · · · + e) −5abcde 0. Here p −(a + b + c + d + e), and q ab + ac + ad + ae + bc + bd + be + cd + ce + de, that is, n | p and also n | q. The second relationship follows from 2q (a + · · · + e)2 −(a2 + · + e2). We also conclude that n | a5 + b5 + c5 + d5 + e5 −5abcde.
Where did we use the fact that n is odd?
141. 5n + 1 ends with 26 and is divisible by 2, but not by 4. 52q+1 −1 (5 −1)(52q + · · · + 5 + 1) is divisible by 4 but not by 8, since the last parentheses have an odd number of odd summands. For p ≥1, we conclude from factoring 52p(2q+1) −1 52p−1(2q+1) −1 52p−1(2q+1) + 1 that the numbers of the form 52p(2q+1) −1 have in their factoring exactly one factor 2 more than 52p−1(2q+1) −1. Hence, 52p(2q+1) −1 is divisible by 2p+2, but not 2p+3.
Hence, the answer is n 2k−2.
142. Denote the sum by s. Then we have s 1319 k1 1 k −2 659 k1 1 2k 1319 k660 1 k 989 k660 1 k + 1 1979 −k 1979 989 k660 1 k(1979 −k) 1979 · p1 q .
The denominators k(1979−k) are prime to 1979, since this number is a prime. Thus the gcd of the denominators is not a multiple of 1979, and hence the numerator is a multiple of 1979.
143. Multiplying (x + 1)3 −x3 3x2 + 3x + 1 y2 by 4, we get 3(2x + 1)2 (2y −1)(2y + 1). Since 2y −1 and 2y + 1 are coprime, we must consider two cases (gcd(m, n) 1): (a) 2y −1 3m2, 2y + 1 n2.
(b) 2y −1 m2, 2y + 1 3n2.
156 6. Number Theory The first case leads to n2 −3m2 2 which has no solution since it implies n2 ≡ −1 mod 3. In the second case, we set m 2k + 1 and get 2y 4k2 + 4k + 2 2 (k + 1)2 + k2 , which implies y (k + 1)2 + k2.
144. In the binary form √ 2 b0.b1 · · · bk, bi ∈{0, 1}, there are infinitely many i′s such that bi 1. If bk 1, then setting m ⌊2k−1√ 2⌋ b0 · · · bk−1, we have 2k−1√ 2 −1 < m < 2k−1√ 2 −1 2.
Multiplying by √ 2 and adding √ 2, we get 2k < (m + 1) √ 2 < 2k + √ 2 2 < 2k + 1, i.e., ⌊(m + 1) √ 2⌋is 2k, qed.
145. (a) We know the answer from item #5. Since gcd(a, b) 1, we can solve the Diophantine equation ax + by 1 in infinitely many ways. Multiplying by the integer z, we get z a(xz) + b(yz). Thus we can represent any integer by a and b.
(b) By experimenting with small values of a, b, we get the result: If m, n are integers such that m + n ab −a −b, then exactly one of m, n is representable, the other not.
In the identity ax′ + by′ a(x′ −bt) + b(y′ + at), we can choose t such that 0 ≤x′ −bt ≤b −1. Hence we assume that, in m ax + by, n au + bv, we have 0 ≤x ≤b −1 and 0 ≤u ≤b −1. From ax + by + au + bv ab −a −b, we get ab −a(x + u + 1) −b(v + y + 1) 0 (1) and hence b|x + u + 1. From the assumption about x and u, we get 1 ≤x + u + 1 ≤2b −1, and thus x +u+1 b. From (1), we conclude that y +v+1 0. Hence exactly one of the two numbers y, v is negative, the other nonnegative. Obviously, the smallest representable number is 0 with x y 0. Thus the largest nonrepresentable number is ab −a −b. All negative integers are not representable. Hence all integers from ab −a −b + 1 on upward are representable.
This result is due to Sylvester. It is a special case of the problem of Frobenius: Given are n positive integers a1, . . . , an with gcd(a1, . . . an) 1. Find the largest number Gn, which cannot be represented in the form a1x1 + · · · + anxn with xi ≥0.
Until recently, not even the case n 3 was solved. Now several people have claimed to have solved the case n 3. A look at their solutions shows that they did not find a formula for G3. Rather they gave a “simple” algorithm for finding G3. Its description comprises several pages. In this sense also, the general case has a solution. A formula for Gn does not seem to exist even for case n 3.
6. Number Theory 157 146. (a) One Slotnik can be weighed since 1 2 · 48 −15 −4 · 20.
(b) This is an instance of the case n 3 of the problem of Frobenius. Since a general solution is not known, we must use ingenuity to find the largest integer not representable in the form 48x + 20y + 15z, x, y, z ≥0.
(2) We can write this in the form 3(16x+5z)+20y. Here 16x+5z takes all integral values from 16 · 5 −16 −5 + 1 60 upward. We write the first term in the form 3(t + 60) and get 3t +20y +180. Now 3t +20y takes all values from 3·20−3−20+1 38 upward. Hence 48x + 15z + 20y takes all value from 218 upward. So G3 217 is the largest value not assumed by 48x + 20y + 15z. We have made two uses of Sylvester’s result to arrive at our conclusion.
147. We make two applications of Sylvester’s result: bcx + cay + abz c(bx + ay) + abz c(ab −a −b + 1 + t) + abz abc −ac −bc + c + ct + abz abc−c−ab+1+u .
Hence, bcx +cay +abz 2abc −ab −bc −ca +1+u. Here t, u are nonnegative integers. We conclude that all integers from 2abc −ab −bc −ca +1 upward can be expressed in the form bcx + cay + abz. We prove that 2abc −ab −bc −ca cannot be so represented. Suppose bcx+cay+abz 2abc−ab−bc−ca ⇒bc(x+1)+ca(y+1)+ab(z+1) 2abc.
(1) We conclude that a|x + 1 ⇒a ≤x + 1. Similarly, b ≤y + 1 and c ≤z + 1. Now (1) implies 3abc ≤2abc, a contradiction.
148. With a 20, we have 1280000401 a7 + a2 + 1. The polynomial a7 + a2 + 1 has the factor a2 + a + 1 since ω7 + ω2 + 1 ω2 + ω + 1, where ω is the third root of unity. Hence 1280000401 is divisible by 421.
149. Suppose x + y a2, 2x + y b2, x + 2y c2. Adding the last two equations, we get 3a2 b2 + c2.
(1) A square can only be 0 or 1 mod 3. This implies that b and c are both divisible by 3. But then a is also divisible by 3. Hence (a/3, b/3, c/3) satisfies (1). By infinite descent, only the triple (0, 0, 0) satisfies (1).
150. For n 1, the integer 32n −1 is divisible by 23. Consider the identity 32n+1 −1 32n −1 32n + 1 ; 32n + 1 (−1)2n + 1 ≡2 (mod 4).
This shows that just one factor 2 is added by increasing n by 1. Thus 32n −1 has exactly n + 2 factors 2.
151. Since (a + 1)/b + (b + 1)/a (a2 + b2 + a + b)/ab and d2|ab, we also have d2|a2 + b2 + a + b. But also d2|a2 + b2. Hence, d2|a + b or d2 ≤a + b.
152. The solution to problem 67 is an example containing just the digits 1 and 2.
158 6. Number Theory 153. Adding Sn,k 1k + 2k + · · · nk and Sn,k nk + (n −1)k + · · · + 1k, we get 2Sn,k (1k + nk) + 2k + (n −1)k + · · · + (nk + 1k).
Since k is odd, we have (n + 1)|2Sn,k. To prove that n|2Sn,k, we may ignore the last term in Sn,k and add 1k + · · · (n −1)k to (n −1)k + · · · 1k. We get n|2Sn,k, and since gcd(n, n + 1) 1, we conclude that n(n + 1)|2Sn,k.
154. No! The sequence nk becomes constant starting with some index p, so that np np+1 · · ·. Indeed, we have nk ≤nk+1 ≤nk + 9c(nk) for all k, where c(nk) is the number of digits of nk. Suppose that the sequence nk is not bounded for some choice of n1. We choose a positive integer N, such that 10N > n1 and 9N < 10N−1. Such a choice is always possible. The unboundedness of nk implies that nk > 10N from some number k on. Hence among the numbers nk < 10N, there is a largest, say np.
But then 10N ≤np+1 ≤np + 9c(np) < 10N + 9N < 10N + 10N−1.
This means that np+1 starts with 10, and P(np+1) 0. Thus nk np+1 for all k ≥p + 1. This contradicts the unboundedness of the sequence nk. In other words, starting with any n1, the sequence nk does not change from some number k on.
155. (a) Answer: 1962 + D(1962) 1980. How to guess this will be seen from (b).
(b) If n ends with 9, then En+1 < En, if not, En+1 En +2. For any positive integer m > 2, we choose the largest N, for which EN < m. Then EN+1 ≥m, and the last digit of N is not 9. Thus either EN+1 m or EN+1 m + 1.
156. Let n2 < a < b < c < d < (n + 1)2, ad bc. Then d −a < 2n.
(1) Our aim is to produce a contradiction to (1). From ad bc, we conclude that a [(a + d) −(b + c)] (a −b)(a −c) > 0. Hence, a + d > b + c.
Now (a + d)2 −(d −a)2 4ad 4bc < (b + c)2. We conclude that (d −a)2 > (a + d)2 −(b + c)2 (a + d + b + c)(a + d −b −c).
Each term of the first factor on the RHS is larger than n2, and the second is ≥1.
Thus we have d −a > 2n, which contradicts (1).
157. Write the equation in the form 19(x3−100) 84(y2+1). The right side is a multiple of 7, hence also the left side, i.e., x3 −2 ≡0 mod 7. But x3 ̸≡2 mod 7.
158. Since a · b gcd(a, b) · lcm(a, b) and a + b ≤gcd(a, b) + lcm(a, b), the product of all the numbers is invariant while the sum increases or does not change. This is an invariance problem using number theory.
159. Suppose 2n and 5n begin with the digit d and have r +1 and s +1 digits, respectively.
Then, for n > 3, we have d ·10r < 2n < (d +1)·10r and d ·10s < 5n < (d +1)·10s.
Multiplying these inequalities, we get d2 · 10r+s < 10n < (d + 1)2 · 10r+s or d2 < 10n−r−s < (d + 1)2. From 1 ≤d and d + 1 ≤10, we get n −r −s 1, i.e., d2 < 10 and (d + 1)2 > 10. This implies d 3. The smallest example is 25 32 and 55 3125.
6. Number Theory 159 160. We check that x a2 + b2 −c2, y 2bc, z 2ca. We may assume a ≥b ≥c.
Then x > 0.
161. We look for divisors of the same form 3k −2k. Set k 2t, n 32t −22t , t ≥2.
We use the fact that, for k ∈N and distinct integers x, y, we have x −y|xk −yk.
Now, to prove that n|3n−1 −2n−1, it is sufficient to prove that the exponent n −1 is divisible by 2t, i.e., 2t|32t −1 (since 2t|22t ).
By induction, we prove that we have 2t+2|32t −1 for all t ∈N. For t 1, this is clear. Suppose it is true for some t. Then 32t+1 −1 (32t + 1)(32t −1). The first factor is divisible by 2; the second by 2t+2 by the induction hypothesis.
162. Get rid of two cubes by setting z −x. We get 2x2 + y2 y3, 2x2 (y −1)y2, i.e., y−1 2 must be a square t2. Then y 2t2 + 1, x t(2t2 + 1).
163. Hint: The smallest n for which 2n · · · n is 36: 236 · · · 736. From here on, we use induction. Suppose 2n · · · dn, where d is the digit to the left of n, then 2dn · · · dn.
164. Let a, b and n a+b be the number of white balls, black balls and balls, respectively.
We may assume that a > b. Then 2 · a n · n −a n −1 1 2 ⇔a n ± √n 2 , i.e., a (n + √n)/2, b (n −√n)/2, and the number of balls must be a square q2. Then a q+1 2 and b q 2 .
7 Inequalities Means Let x be a real number. The most basic inequalities are x2 ≥0, (1) n i1 x2 i ≥0.
(2) We have equality only if x 0 in (1) or xi 0 for all i in (2). One strategy for proving inequalities is to transform them into the form (1) or (2). This is usually a long road. So we derive some consequences equivalent to (1). With x a −b, a > 0, b > 0, we get the following equivalent inequalities: a2 + b2 ≥2ab ⇔2(a2 + b2) ≥(a + b)2 ⇔a b + b a ≥2 ⇔x + 1 x ≥2, x > 0 ⇔a + b 2 ≤ a2 + b2 2 .
Replacing a, b by √a, √ b, we get a + b ≥2 √ ab ⇔a + b 2 ≥ √ ab ⇔ √ ab ≥2ab a + b.
In particular, we have the inequality chain min(a, b) ≤2ab a + b ≤ √ ab ≤a + b 2 ≤ a2 + b2 2 ≤max(a, b).
162 7. Inequalities This is the harmonic-geometric-arithmetic-quadratic mean inequality, or the HM-GM-AM-QM inequality. By repeated use of the inequalities above, we can already prove a huge number of other inequalities. Every contestant in any competition must be able to apply these inequalities in any situation that may arise. Here are a few very simple examples.
E1.
x2+2 √ x2+1 ≥2 for all x. This can be transformed as follows.
x2 + 2 √ x2 + 1 x2 + 1 √ x2 + 1 + 1 √ x2 + 1 x2 + 1 + 1 √ x2 + 1 ≥2.
E2. For a, b, c, ≥0, we have (a + b)(b + c)(c + a) ≥8abc. Indeed, a + b 2 · b + c 2 · c + a 2 ≥ √ ab · √ bc · √ca abc.
E3. If ai > 0 for i 1, . . . , n and a1a2 · · · an 1, then (1 + a1)(1 + a2) · · · (1 + an) ≥2n.
Dividing by 2n we get 1 + a1 2 · 1 + a2 2 · · · 1 + an 2 ≥√a1 √a2 · · · √an √a1a2 · · · an 1.
E4. For a, b, c, d ≥0, we have √(a + c)(b + d) ≥ √ ab + √ cd. Squaring and simplifying, we get ad + bc ≥2 √ abcd, which is x + y ≥2√xy.
E5. Show that, for real a, b, c, a2 + b2 + c2 ≥ab + bc + ca.
(3) First proof. Multiplying by 2, we reduce (3) to (2): 2a2 + 2b2 + 2c2 −2ab −2bc −2ca ≥0 ⇔(a −b)2 + (b −c)2 + (c −a)2 ≥0.
Second proof. We have a2 + b2 ≥2ab, b2 + c2 ≥2bc, c2 + a2 ≥2ca. Addition and division by 2 yields (3).
Third proof. Introduce ordering or assume that some element is extremal. Since the inequality is symmetric in a, b, c, assume a ≥b ≥c. Then a2 + b2 + c2 ≥ab + bc + ca ⇔a(a −b) + b(b −c) −c(a −c) ≥0 ⇔a(a −b) + b(b −c) −c(a −b + b −c) ≥0 ⇔a(a −b) + b(b −c) −c(a −b) −c(b −c) ≥0 ⇔(a −c)(a −b) + (b −c)2 ≥0.
7. Inequalities 163 The last inequality is obviously correct. Here it is enough to assume that a is the maximal or minimal element. Note also the replacement of −c(a −c) by −c(a −b + b −c). This idea has many applications.
Fourth proof. Let f (a, b, c) a2 + b2 + c2 −ab −bc −ca. Then we have f (ta, tb, tc) t2f (a, b, c). Hence, f is homogeneous of degree two. For t ̸ 0, we have f (a, b, c) ≥0 ⇔f (ta, tb, tc) ≥0. Therefore, we may make various normalizations. For example, we may set a 1, b 1 + x, c 1 + y and get x2 + y2 −xy (x −y/2)2 + 3y2/4 ≥0. More proofs will be given later.
E6. We start with the classic factorization a3 + b3 + c3 −3abc (a + b + c)(a2 + b2 + c2 −ab −bc −ca).
(4) Because of (3), for nonnegative a, b, c, we have a3 + b3 + c3 ≥3abc ⇔a + b + c ≥3 3 √ abc ⇔a + b + c 3 ≥ 3 √ abc.
(5) This is the AM-GM inequality for three nonnegative reals.
Generally, for n positive numbers ai, we have the following inequalities: min(ai) ≤ n 1 a1 + · · · + 1 an ≤ n √a1 · · · an ≤a1 + · · · + an n ≤ a2 1 + · · · + a2 n n ≤max(ai).
The equality sign is valid only if a1 · · · an. We will prove these later. At the IMO, they need never be proved, just applied.
E7. Let us apply (5) to Nesbitt’s inequality (England 1903): a b + c + b a + c + c a + b ≥3 2.
(6) It has many instructive proofs and generalizations and is a favorite Olympiad problem. Let us transform the left-hand side f (a, b, c) as follows.
a+b+c b+c + a+b+c a+c + a+b+c a+b −3 (a + b + c) 1 a+b + 1 b+c + 1 a+c −3, 1 2 [(a + b) + (b + c) + (c + a)] 1 a+b + 1 b+c + 1 a+c −3.
(7) First proof. In (7), we set a + b x, b + c y, a + c z and get 2f (a, b, c) (x + y + z) 1 x + 1 y + 1 z −6 x y + y x ≥2 + x z + z x ≥2 + y z + z y ≥2 −3 ≥3.
164 7. Inequalities We have equality for x y z, that is, a b c.
Second proof. The AM-HM Inequality can be transformed as follows: u + v + w 3 ≥ 3 1 u + 1 v + 1 w ⇔(u + v + w)(1 u + 1 v + 1 w) ≥9.
From (7), we get f (a, b, c) ≥1 2 · 9 −3 3 2.
Let us prove the product form of the AM-HM inequality (a1 + · · · + an) 1 a1 + · · · 1 an ≥n2.
Multiplying the LHS, we get n times 1 and n 2 pairs xi/xj +xj/xi, each pair being at least 2. Hence the LHS is at least n + 2 n 2 n2.
Third proof. We apply the inequality u + v + w ≥3 3 √uvw to both parentheses of (7) and get f (a, b, c) ≥1 2 · 3 3 (a + b)(b + c)(c + a) · 3 3 1 (a + b)(b + c)(c + a) −3 3 2.
Fourth proof. We have f (a, b, c) f (ta, tb, tc) for t ̸ 0, that is, f is homoge-neous in a, b, c of degree 0. We may normalize to a + b + c 1. Then, from the AM-HM inequality, we get f (a, b, c) 1 a + b + 1 b + c + 1 c + a −3 ≥9 2 −3 3 2.
E8. Inequalities for the sides a, b, c of a triangle are very popular. In this case, the Triangle Inequality plays a central role. During the proof you must use the triangle inequality or else the inequality is valid for all triples (a, b, c) of positive reals.
That includes all triangles, of course.
The triangle inequality occurs in four equivalent forms: I. a + b > c, b + c > a, c + a > b.
II. a > |b −c|, b > |a −c|, c > |a −b|.
III. (a + b −c)(b + c −a)(c + a −b) > 0.
IV. a y + z, b z + x, c x + y, where x, y, z are positive.
7. Inequalities 165 If we know that c max(a, b, c), then a + b > c alone suffices. The other two inequalities in I are automatically satisfied. We prove the equivalence of I and III. If I is valid, then III is also valid. Suppose III is valid. Then all three factors are positive, which is I, or exactly two factors are negative. Suppose the first and second factor are negative. Adding a + b −c < 0 and b + c −a < 0, we get 2b < 0, which is a contradiction.
E9. In a triangle ABC, the bisectors AD, BE, and CF meet at the point I. Show that 1 4 < IA AD · IB BE · IC CF ≤8 27.
(1) Solution. This was the first problem of IMO 1991. To avoid trigonometry, we use the following simple geometric theorem (Fig. 7.1): A bisector of a triangle divides the opposite side in the ratio of the other two sides.
@ @ @ @ @ @ pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp A B C I F D p q b b : c p : q.
pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp Fig. 7.1 Hence, p CD (ab)/(b + c), q DB (ac)/(b + c). Thus, we have AI ID b : p b + c a , AI AD AI AI + ID b + c a + b + c.
Similarly, BI BE a + c a + b + c, CI CF a + b a + b + c.
Applying the GM-AM inequality to the numerator,we get f (a, b, c) AI AD · BI BE · CI CF (a + b)(b + c)(c + a) (a + b + c)3 ≤ 8 (a + b + c)3 a + b + c 3 3 , which is 8/27. This is the right side of the inequality chain. To prove the left side, we use the triangle inequality (a + b −c)(a + c −b)(b + c −a) > 0.
(2) For a more economical evaluation, we introduce the elementary symmetric func-tions u a + b + c, v ab + bc + ca, w abc.
(3) 166 7. Inequalities Putting (3) into (2), we get −u3 + 4uv −8w > 0.
(4) On the other hand, 1 4 < f (a, b, c) (5) gives −u3 + 4uv −4w > 0.
(6) Now (4) is obviously correct. Hence, (6) is also correct. Here we profitably used the elementary symmetric functions. They are useful in cases when we are dealing with functions which are symmetric in their variables.
Here is the simplest proof of (5): Set a y + z, b z + x, c x + y (Fig.
7.2). With r x/(x + y + z), s y/(x + y + z), t z/(x + y + z), we get @ @ @ @ @ A B C x x y y z z Fig. 7.2 AI AD 1 2(1 + r), BI BE 1 2(1 + s), CI CF 1 2(1 + t), r + s + t 1, f (a, b, c) 1 8(1 + r)(1 + s)(1 + t) 1 8(1 + 1 + rs + st + tr + rst) > 1 4.
E10. We consider three problems: a3 + b3 + c3 + 3abc ≥ab(a + b) + bc(b + c) + ca(c + a), (1) a2(b + c −a) + b2(c + b −a) + c2(a + b −c) ≤3abc, (2) (a + b −c)(b + c −a)(c + a −b) ≤abc.
(3) The first is from the AUO 1975, the second is from the IMO 1964. (1) was to be proved for all a, b, c ≥0, and (2) was to be shown for the sides of a triangle. In fact, all three are equivalent. Show this yourself. But (2) becomes simpler since we may use the triangle inequality.
Let us prove (1). It is symmetric in a, b, c. So we may assume a ≤b ≤c. In addition the inequality is homogeneous of degree three. So we may stretch it by a factor t so that a 1. Then b 1+x, c 1+y, x ≥0, y ≥0. By plugging this into (1) and with the usual reductions, we get the following chain of equivalences.
x3 + y3 + x2 + y2 ≥x2y + xy + xy2 ⇔x3 + y3 + x2 −xy + y2 −xy(x + y) ≥0 ⇔x3 + y3 + (x −y)2 + xy −xy(x + y) ≥0 ⇔(x + y)(x2 −xy + y2 −xy) + xy ≥0 ⇔(x + y + 1)(x −y)2 + xy ≥0.
7. Inequalities 167 The last inequality is obvious. We get u3 −4uv + 3w ≥0 if we introduce the elementary symmetric functions. This helps if we know some simple inequalities for u, v, w.
E11. The Cauchy–Schwarz Inequality (CS Inequality). For all real x, we have n i1 (aix + bi)2 x2 n i1 a2 i + 2x n i1 aibi + n i1 b2 i ≥0.
This quadratic polynomial is nonnegative, i.e, it has discriminant D ≤0. We get one of the most useful inequalities in mathematics, the Cauchy–Schwarz Inequality (a1b1 + · · · + anbn)2 ≤(a2 1 + · · · + a2 n)(b2 1 + · · · + b2 n).
Using the vectors ⃗ a (a1, . . . , an), ⃗ b (b1, . . . , bn), we get ⃗ a · ⃗ b 2 ≤|⃗ a|2|⃗ b|2.
We have equality exactly if ⃗ a and ⃗ b are linearly dependent.
With this inequality, we prove the AM-QM inequality for n real numbers.
(1 · a1 + · · · + 1 · an)2 ≤(12 + · · · + 12)(a2 1 + · · · + a2 n).
Taking square roots of both sides and dividing by n, we get the result.
As another example, we find the maximum of the function y a·sin x+b·cos x for a > 0, b > 0, 0 < x < π/2.
(a · sin x + b · cos x)2 ≤(a2 + b2)(sin2 x + cos2 x) a2 + b2.
The maximum √ a2 + b2 will be attained if a/b sin x/ cos x tan x.
E12. Rearrangement Inequality. Finally, we consider an interesting and powerful theorem which enables us to see the validity of many inequalities by inspection.
Let a1, . . . , an and b1, . . . , bn be sequences of positive real numbers, and let c1, . . . , cn be a permutation of b1, . . . , bn. Which of the n! sums S a1c1 + · · · + ancn is extremal, i.e., maximal or minimal?
Consider an example. Four boxes contain $10, $20, $50, and $100 bills respec-tively. From each box, you may take 3, 4, 5, and 6 bills, respectively. But you have free choice of assigning the boxes to the numbers 3,4,5,6. To get as much money as possible, you use the greedy algorithm: Take as many $100 bills as you can, i.e. six. Then take as many $50 bills as you can, i.e. five. Then you take four $20 bills, and finally three $10 bills. You get the least amount of money if you take three $100 bills, four $50 bills, three $20 bills, and six $10 bills.
168 7. Inequalities Theorem. The sum S a1b1+· · · anbn is maximal if the two sequences a1, . . . , an and b1, . . . , bn are sorted thesameway. S isminimalifthetwosequencesaresorted oppositely, one increasing, the other decreasing.
Proof. Let ar > as. We consider the sums S a1c1 + · · · + arcr + · · · + ascs + · · · + ancn, S′ a1c1 + · · · + arcs + · · · + ascr + · · · + ancn.
We get S′ from S by switching the positions of cs and cr. Then S′ −S arcs + ascr −arcr −ascs (ar −as)(cs −cr).
Consequently, cr < cs ⇒S′ > S, cr > cs ⇒S′ < S.
E13. Let us prove the AM-GM inequality for n numbers. Suppose xi > 0, c n √x1 · · · xn, a1 x1 c , a2 x1x2 c2 , a3 x1x2x3 c3 , . . . , an x1···xn cn 1, b1 1 a1 , b2 1 a2 , . . . , bn 1 an 1.
The sequences ai and bi are oppositely sorted. Hence we have a1b1 + · · · + anbn ≤a1bn + a2b1 + a3b2 + · · · + anbn−1, 1 + 1 + · · · + 1 ≤x1 c + x2 c + · · · + xn c , n √x1 · · · xn ≤x1 + · · · + xn n .
E14. Finally we derive the Chebyshev inequality. Let a1, . . . , an and b1, . . . , bn be similarly sorted sequences (both rising or both falling). Then a1b1 + · · · + anbn a1b1 + a2b2 + · · · + anbn, a1b1 + · · · + anbn ≥ a1b2 + a2b3 + · · · + anb1, a1b1 + · · · + anbn ≥ a1b3 + a2b4 + · · · + anb2, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
a1b1 + · · · + anbn ≥ a1bn + a2b1 + · · · + anbn−1.
Adding the inequalities, we get n(a1b1 + · · · + anbn) ≥(a1 + · · · + an)(b1 + · · · + bn), a1 + · · · + an n · b1 + · · · + bn n ≤a1b1 + · · · + anbn n .
This is the original Chebyshev inequality for means. Similarly, we can prove for oppositely sorted sequences ai and bi that a1b1 + · · · + anbn n ≤a1 + · · · + an n · b1 + · · · + bn n .
7. Inequalities 169 We introduce a new notation for the scalar product: a1 b1 a2 b2 a3 b3 a1b1 + a2b2 + a3b3.
E15. Then a3 + b3 + c3 a a2 b b2 c c2 ≥ a c2 b a2 c b2 a2b + b2c + c2a.
E16. For any positive a, b, c, the two sequences (a, b, c) and (1/(b + c), 1/(c + a), 1/(a + b)) are sorted the same way. Thus, we have ⎡ ⎣ a 1 b + c b 1 c + a c 1 a + b ⎤ ⎦≥ ⎡ ⎣ a 1 c + a b 1 a + b c 1 b + c ⎤ ⎦, ⎡ ⎣ a 1 b + c b 1 c + a c 1 a + b ⎤ ⎦≥ ⎡ ⎣ a 1 a + b b 1 b + c c 1 c + a ⎤ ⎦.
Adding the two inequalities, we get 2 a b + c + b c + a + c a + b ≥3, which again is Nesbitt’s inequality E7.
E17. Let ai > 0, i 1, .., n and s a1 + · · · + an. Prove the inequality a1 s −a1 + a2 s −a2 + · · · + an s −an ≥ n n −1.
Obviously, the sequences a1, · · · , an and 1/(s −a1), · · · , 1/(s −an) are sorted the same way. Therefore, ⎡ ⎢ ⎣ a1 1 s −a1 · · · an 1 s −an ⎤ ⎥ ⎦≥ ⎡ ⎢ ⎣ a1 1 s −ak a2 1 s −ak+1 · · · an 1 s −ak−1 ⎤ ⎥ ⎦, (k 2, 3, . . . , n).
Adding these (n −1) inequalities gives the result.
E18. Find the minimum of sin3 x/ cos x + cos3 x/ sin x, 0 < x < π/2.
The sequences (sin3 x, cos3 x) and (1/ sin x, 1/ cos x) are oppositely sorted.
Thus, ⎡ ⎣sin3 x 1 cos x cos3 x 1 sin x ⎤ ⎦≥ ⎡ ⎣sin3 x 1 sin x cos3 x 1 cos x ⎤ ⎦ sin2 x + cos2 x 1.
170 7. Inequalities E19. Prove the inequality a4 + b4 + c4 ≥a2bc + b2ca + c2ab.
We use an extension of the scalar product to three sequences: ⎡ ⎣ a2 a a b2 b b c2 c c ⎤ ⎦≥ ⎡ ⎣ a2 b c b2 c a c2 a b ⎤ ⎦.
In the first matrix, the three sequences are sorted the same way, in the second, not.
Recently, the following inequality was posed in the Mathematics Magazine.
E20. Let x1, . . . , xn be positive real numbers. Show that xn+1 1 + xn+1 2 + · · · + xn+1 n ≥x1x2 . . . xn (x1 + x2 + · · · + xn) .
The proof is immediate. Rewrite the preceding inequality as follows: ⎡ ⎢ ⎢ ⎣ x1 · · · xn x1 · · · xn . . . . . . . . . . .
x1 · · · xn ⎤ ⎥ ⎥ ⎦≥ ⎡ ⎢ ⎢ ⎣ x1 · · · xn x2 · · · x1 . . . . . . . . . . .
x1 · · · xn ⎤ ⎥ ⎥ ⎦.
E21. Triangular Inequalities. In this section we discuss inequalities for a triangle.
Our students acquire all their knowledge about the geometry and trigonometry of the triangle from E21/22.
We will denote the sides of a triangle by a, b, c. The opposite angles will be denoted by α, β, γ. The area will be denoted by A, the inradius by r and the circumradius by R. Two indispensable theorems are the Cosine Law: c2 a2 + b2 −2ab cos γ (and cyclic permutations).
and the Sine Law: a sin α b sin β c sin γ 2R.
The area of the triangle is A 1 2ab sin γ 1 2bc sin α 1 2ac sin β.
We start with an inequality, which we will prove and sharpen in many ways.
Prove that, for any triangle with sides a, b, c and area A, a2 + b2 + c2 ≥4 √ 3A (IMO 1961).
The inequality is due to Weitzenb¨ ock, Math. Z. 5, 137–146, (1919).
Main idea: We conjecture that we have equality exactly for the equilateral tri-angle. This conjecture is the guide to most of our proofs.
7. Inequalities 171 l l l l l A B C q p h b a Fig. 7.3 First proof. An equilateral triangle with side c has altitude c 2 √ 3. Any triangle with side c will have an altitude perpendicular to c of length c 2 √ 3 + y. It splits c into parts c 2 −x and c 2 + x. Here x, y are the deviations from an equilateral triangle.
Then we have (see Fig. 7.3) a2 + b2 + c2 −4 √ 3A c 2 −x 2 + c 2 + x 2 + 2 y + c 2 √ 3 2 + c2 −2 √ 3c y + c 2 √ 3 2x2 + 2y2 ≥0.
We have equality iff x y 0, i.e., for the equilateral triangle.
Second proof. This is a more geometric version of the preceding solution. Let a ≤b ≤c. We erect the equilateral triangle ABC′ on AB and introduce p |CC′| as the deviation from an equilateral triangle. The Cosine Law yields p2 a2 + c2 −2ac cos (β −60◦) a2 + c2 −2ac (cos β cos 60◦+ sin β sin 60◦) , p2 a2 + c2 −ac cos β − √ 3ac sin β a2 + b2 −2 √ 3A −1 2 (2ac cos β) a2+c2−b2 , p2 a2 + b2 + c2 2 −2 √ 3A a2 + b2 + c2 −4 √ 3A 2 ≥0, since the square p2 is not negative. We have equality exactly if p 0, that is, a b c.
Third proof. This is a proof by contradiction. We assume 4A √ 3 > a2 + b2 + c2 and by equivalence transformations we get 4A √ 3 > a2 + b2 + c2 ⇔2bc sin α > 1 √ 3 a2 + b2 + c2 .
Now we use the Cosine Law 2bc cos α b2 + c2 −a2. Square and add the last two relations. We get the contradiction a2b2 +b2c2 +c2a2 > a4 +b4 +c4 ⇔ a2 −b2 2 + b2 −c2 2 + c2 −a2 2 < 0.
172 7. Inequalities Fourth proof. Using Heron’s formula and the AM-GM inequality, we get 16A2 (a + b + c)(−a + b + c)(a −b + c)(a + b −c) ≤(a + b + c) a + b + c 3 3 , 4A ≤(a + b + c)2 3 √ 3 √ 3 a + b + c 3 2 ≤ √ 3a2 + b2 + c2 3 , or a2 + b2 + c2 ≥4A √ 3. We have equality exactly for a b c.
Fifth proof.
a2 + b2 + c2 ≥ab + bc + ca 2A 1 sin α + 1 sin β + 1 sin γ .
Now we use the fact that f (x) 1/ sin x is convex. Convexity implies that f (α) + f (β) + f (γ ) ≥3f (α + β + γ 3 ) 3f (60◦) 3 sin 60◦ 2 √ 3, that is, a2 + b2 + c2 ≥4A √ 3.
Sixth proof. We prove a slight generalization.
2a2 + 2b2 + 2c2 (a −b)2 + (b −c)2 + (c −a)2 + 2ab + 2bc + 2ca (a −b)2 + (b −c)2 + (c −a)2 Q + 4A 1 sin α + 1 sin β + 1 sin γ ≥2 √ 3 .
We get a generalization a2 + b2 + c2 ≥Q 2 + 4A √ 3.
Seventh proof. We replace a2 in a2 + b2 + c2 by b2 + c2 −2bc cos α and get a2 + b2 + c2 −4A √ 3 2(b2 + c2) −2bc cos α −2bc √ 3 sin α 2(b2 + c2) −4bc 1 2 cos α + √ 3 2 sin α 2[b2 + c2 −4bc cos (60◦−α)] ≥2(b2 + c2) −4bc 2(b −c)2.
7. Inequalities 173 We have equality exactly for b c and α 60◦. In this case a b c.
Eighth proof. The Hadwiger–Finsler inequality (1937). This is a strong general-ization.
a2 b2 + c2 −2bc cos α (b −c)2 + 2bc(1 −cos α) (b −c)2 + 4A1 −cos α sin α (b −c)2 + 4A tan α 2 .
Here we used 1 −cos α 2 sin2 α 2 , sin α 2 sin α 2 cos α 2 , that is, a2 + b2 + c2 (a −b)2 + (b −c)2 + (c −a)2 + 4A tan α 2 + tan β 2 + tan γ 2 .
Since α/2, β/2, γ/2 < π/2, the function tan is convex. Thus, we have tan α 2 + tan β 2 + tan γ 2 ≥3 tan α + β + γ 6 3 tan 30◦ √ 3.
We have equality for α β γ 60◦. Then we have a2 + b2 + c2 ≥(a −b)2 + (b −c)2 + (c −a)2 + 4A √ 3.
Ninth proof. We have the following equivalence transformations: a2 + b2 + c2 ≥4A √ 3, a2 + b2 + c2 2 ≥3(a + b + c)(a −b + c)(−a + b + c)(a + b −c), (a2 + b2 + c2)2 ≥3 2a2b2 + 2c2a2 + 2a2b2 −a4 −b4 −c4 , 4a4 + 4b4 + 4c4 −4a2b2 −4b2c2 −4a2c2 ≥0, a2 −b2 2 + b2 −c2 2 + c2 −a2 2 ≥0.
Tenth proof. We try to invent a triangular inequality which becomes an exact equality for the equilateral triangle. Such an inequality is (a −b)2 + (b −c)2 + (c −a)2 ≥0.
Squaring, we get a2 + b2 + c2 ≥ab + bc + ca.
We decide to introduce the area of the triangle. We use ab 2A sin γ , bc 2A sin α , ca 2A sin β .
174 7. Inequalities Replacing the right side by the right sides of these formulas, we get a2 + b2 + c2 ≥ab + bc + ca 2A 1 sin α + 1 sin β + 1 sin γ .
From here we proceed as in the fifth proof.
Eleventh proof. Again, we prove the Hadwiger–Finsler inequality a2 + b2 + c2 ≥4A √ 3 + (a −b)2 + (b −c)2 + (c −a)2.
We transform this inequality into the form a2 −(b −c)2 + b2 −(c −a)2 + c2 −(a −b)2 ≥4A √ 3, (a −b + c)(a + b −c) + (b −c + a)(b + c −a) + (c −a + b)(c + a −b) ≥4A √ 3.
Here we set x −a + b + c, y a −b + c, z a + b −c. Although the sides a, b, c must satisfy the triangle inequality, the new variables x, y, and z must merely be positive. For the RHS of the last inequality, we have 4A √ 3 3(x + y + z)xyz.
So we get xy + yz + zx ≥ 3(x + y + z)xyz.
Dividing by xyz and then setting u 1/x, v 1/y, w 1/z we get 1 x + 1 y + 1 z ≥ 3 1 xy + 1 yz + 1 zx , u + v + w ≥√3(uv + vw + wu).
Squaring and simplification gives the well known inequality u2 + v2 + w2 ≥uv + vw + wu.
We give just two proofs of another classic inequality for triangles.
E22. Let R and r be the radii of the circumcircle and incircle of a triangle. Then R ≥2r.
Firstproof. The area of a triangleis A rs,where s isthesemiperimeter.Fromthe Sine Law a 2R sin α, we get abc 2Rbc sin α 4RA, that is, R abc/4A.
Hence, R r sabc 4A2 sabc 4s(s −a)(s −b)(s −c) 2abc (a + b −c)(a −b + c)(−a + b + c), R r ≥ 2abc √ a2b2c2 2.
7. Inequalities 175 We have equality exactly for a + b −c a −b + c −a + b + c ⇒a b c.
Second proof. This brilliant proof is due to the Hungarian mathematician Adam, who died prematurely. He considers the circumradius of the triangle of midpoints which is R/2. Now, almost obviously, R/2 ≥r or R ≥2r.
Indeed, by three stretches with factors 0 < λ1, λ2, λ3 < 1, the circumcircle of the midpoints can be transformed into the incircle. The centers of the stretches are the three vertices of the triangle.
E23. Carlson’s Inequality. We start with the Cauchy–Schwarz inequality (a1b1 + · · · + anbn)2 ≤ a2 1 + · · · + a2 n b2 1 + · · · + b2 n .
(CS) We have equality exactly if (a1, · · · , an) λ(b1, · · · , bn). CS gives (a1 + · · · + an)2 a1 · c1 · 1 c1 + · · · + an · cn · 1 cn 2 ≤ a2 1c2 1 + · · · + a2 nc2 n 1 c2 1 + · · · + 1 c2 n .
With Cn 1 c2 1 + · · · + 1 c2 n , we get (a1 + · · · + an)2 ≤Cn a2 1c2 1 + · · · + a2 nc2 n .
(1) With cn n, we have (a1 + · · · + an)2 ≤Cn a2 1 + 22a2 2 + · · · + n2a2 n .
With Cn 1 + 1 22 + · · · + 1 n2 < π2 6 , Cn →π2 6 for n →∞, we have (a1 + · · · + an)2 < π2 6 a2 1 + 22a2 2 + · · · + n2a2 n .
This is Carlson’s inequality (1934) which cannot be made sharper by replacing π2 6 by a smaller constant. Carlson posed c2 n t + n2/t and got a2 1c2 1+· · ·+a2 nc2 n tP + 1 t Q, P a2 1+· · ·+a2 n, Q a2 1+22a2 2+· · ·+n2a2 n.
Because of (1), he got (a1 + · · · + an)2 ≤Cn tP + Q t , 176 7. Inequalities where Cn 1 t + 1 t + 1 t + 22 t + · · · + 1 t + n2 t t t2 + 1 + t t2 + 22 + · · · + t t2 + n2 .
In Fig. 7.4, we have t 2 1 2|OMn−1| · |OMn| · sin αn 1 2 t2 + (n −1)2 · √ t2 + n2 · sin αn, sin αn t √ t2+(n−1)2√ t2+n2 > t t2+n2 , t t2+n2 < sin αn < αn, Cn t t2+1 + · · · + t t2+n2 < α1 + · · · + αn < π 2 , (a1 + · · · + an)2 < π 2 tP + Q t .
We set t √Q/P and get tP + Q/t 2√PQ. Thus, (a1 + a2 + · · · + an)2 < π√PQ, (a1 + · · · + an)4 < π2 a2 1 + · · · + a2 n a2 1 + 22a2 2 + · · · + n2a2 n .
(2) This is the second of several Carlson inequalities, each odder than the other. !!!!!!!
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""""""" " # # # # # # # # O t 1 1 1 1 1 Mn Mn−1 M0 M1 M2 α1 α2 α3 αn Fig. 7.4. MiMi+1 1 Three Problems on Convexity E24. Consider the following problem of the US Olympiad 1980: 1 ≥a, b, c ≥0 ⇒ a b + c + 1+ b c + a + 1+ c a + b + 1+(1−a)(1−b)(1−c) ≤1.
A manipulative solution requires enormous skills, but there is a solution without any manipulation. Denote the left side of the inequality by f (a, b, c). This function is defined on a closed convex cube, and f (a, b, c) is strictly convex in each variable since the second derivative in each variable is strictly positive. Hence, f assumes its maximum 1 at the extremal points, that is, the 8 vertices (0, 0, 0), . . . , (1, 1, 1).
They are the only points of the closed cube, which are not midpoints of two other pointsofthecube.ThisproofwouldbeacceptedattheIMOifonecitesthe Theorem of Weierstraß that a continuous function on a bounded and closed domain assumes its maximum and minimum.
7. Inequalities 177 Consider the following problem of the Allunion Olympiad 1982 in Odessa: E25. The vertices of the tetrahedron KLMN lie inside, on the edges, or faces of another tetrahedron ABCD. Prove that the sum of the lengths of all edges of KLMN are less then 4/3 of the sum of the edges of ABCD.
This problem is probably even more difficult than the preceding one. Only four students solved it, two with high school mathematics, and two with college geometry. We consider the college level solution, which is quite simple. ABCD is a convex, bounded, and closed domain. K, L, M, N ∈ABCD. The function f (K, L, M, N) |K −L|+|K −M|+ |K −N|+|L−M|+|L−N|+|M −N| is continuous in its domain. Because of the strict convexity of f , it follows that it assumes its maximum at the vertices. Thus we have a finite problem. The strict convexity of f follows from the strict convexity of the distance function. This is an immediate consequence of the triangle inequality. The inequality cannot be improved, because for B C A, D ̸ A, K L A, M N D, we have equality. In the vicinity of this degenerated tetrahedron, we have nondegenerated tetrahedra with the sum of edges of KLMN as near to 4/3 of the sum of the edges of ABCD as we please.
The high school methods were based on the ingenious use of the triangle in-equality.
E26. A finite set P of n points (n ≥2) is given in the plane. For any line l, denote by S(l) the sum of the distances from the points of P to the line l. Consider the set L of the lines l such that S(l) has the least possible value. Prove that there exists a line of L, passing through two points of P.
We observe that some line in L passes through a point of P. Indeed, displacing a line parallel to itself, we can reach a point in P without increasing S(l). Choose a line l ∈L passing through a point A of P, and rotate l about A. Let φ be the angle of rotation, and let φk, k 1, 2, . . . , n be the values of φ for which l passes through a point Ak of P (Ak ̸ A). Let ak |AkA|. Then the sum of the distances, when l is rotated through φ, is S(φ) n−1 k1 ak| sin(φ −φk)|.
The function S(φ) is a sum of concave functions whenever φ is restricted to an interval [φk, φk+1]. Hence, S(φ) is concave (as a sum of concave functions) in each such interval. Thus, S(φ) cannot attain its minimum at an internal point of [φk, φk+1]. Hence, it assumes its minimum for some φk.
E27. Trigonometric Substitution. Prove that, for positive reals, √ ab + √ cd ≤ (a + d)(b + c).
178 7. Inequalities We transform into the form a a + d · b b + c + c b + c · d a + d ≤1.
Setting a/(a + d) sin2 α, b/(b + c) sin2 β (0 < α, β < π 2 ), the inequality takes the form sin α sin β + cos α cos β ≤1, i.e., cos(α −β) ≤1.
Strategies for Proving Inequalities 1. Try to transform the inequality into the form pi, pi > 0, e.g., pi x2 i .
2. Does the expression remind you of the AM, GM, HM, or QM?
3. Can you apply the Cauchy–Schwarz inequality? This is especially tricky.
You can apply this inequality far more often than you think.
4. Can you apply the Rearrangement inequality? Again, this theorem is much underused. You can apply it in most unexpected circumstances.
5. Is the inequality symmetric in its variables a, b, c, . . .? In that case, assume a ≤b ≤c ≤. . . . Sometimes one can assume that a is the maximal or minimal element. It may be advantageous to express the inequality by elementary symmetric functions.
6. An inequality homogeneous in its variables can be normalized.
7. If you are dealing with an inequality for the sides a, b, c of a triangle, think of the triangle inequality in its many forms. Especially, think of setting a x + y, b y + z and c z + x with x, y, z > 0.
8. Bring the inequality into the form f (a, b, c, . . . ≥0. Is f quadratic in one of its variables? Can you find its discriminant?
9. If the inequality is to be proved for all positive integers n ≥n0, then use induction.
10. Try to make estimates by telescoping series or products: (a2 −a1)+(a3 −a2)+· · ·+(an −an−1) an −a1, a2 a1 a3 a2 · · · an an−1 an a1 .
11. If a1x1 + · · · + anxn c, then x1 · · · xn is maximal for a1x1 · · · anxn.
12. If x1 · · · xn c, then a1x1 + · · · + anxn is minimal for a1x1 · · · anxn.
13. Max xi > d if the mean of the xi is > d.
14. One of several numbers is positive if their sum or mean is positive.
7. Inequalities 179 15. A powerful idea for proving inequalities is convexity or concavity.
16. To prove an inequality T (a, b, c, . . .) ≥0 or T (a, b, c, . . .) ≤0 one of-ten solves an optimization problem: find the values a, b, c, . . . such that T (a, b, c, . . .) is a minimum or maximum.
17. Does trigonometric substitution simplify the inequality?
18. If none of these methods is immediately applicable then transform the in-equality into a simpler form with some aims in view until a standard method is applicable. If you have no success, continue transforming and try to in-terpret the intermediate results.
Problems 1. a, b, c ∈R, a2 + b2 + c2 1 ⇒−1 2 ≤ab + bc + ca ≤1.
2. Prove that, for a, b, c > 0, (a) a2 + b2 a + b ≥a + b 2 , (b) a3 + b3 + c3 a2 + b2 + c2 ≥a + b + c 3 , (c) a + b + c 3 ≥ ab + bc + ca 3 ≥ 3 √ abc.
3. For a, b, c, d > 0, a2 + b2 + c2 + d2 4 ≥ 3 abc + abd + acd + bcd 4 .
4. Prove that, for a, b > 0, we have n+1 √ abn ≤(a + nb)/(n + 1).
5. The spinner in Fig. 7.5 has circumference 1. It is spun 6 times. For what values of x, y, z for the probabilities of O, A, B, respectively, is the probability of the word BAOBAB maximal?
6. Let a, b, c be the sides of a triangle. Then ab + bc + ca ≤a2 + b2 + c2 ≤ 2(ab + bc + ca).
Z Z O A B x y z Fig. 7.5 7. If a, b, c are sides of a triangle, then 2(a2 + b2 + c2) < (a + b + c)2.
8. If a, b, c are sides of a triangle, then so are 1/(a + b), 1/(b + c), 1/(c + a).
9. Let a, b, c, d > 0. Find all possible values of the sum S a a + b + d + b a + b + c + c b + c + d + d a + c + d (IMO 1974).
180 7. Inequalities 10. Prove the triangle inequality a2 1 + · · · + a2 n + b2 1 + · · · + b2 n ≥ (a1 + b1)2 + · · · + (an + bn)2.
11. Let a, b, c > 0. Show that a + b + c abc ≤1 a2 + 1 b2 + 1 c2 .
12. Let xi, yi, 1 ≤i ≤n be real numbers such that x1 ≥x2 ≥· · · ≥xn and y1 ≥y2 ≥· · · ≥yn (IMO 1975) Let z1, z2, . . . , zn be any permutation of y1, y2, . . . , yn. Show that n i0 (xi −yi)2 ≤ n i1 (xi −zi)2.
13. Let {ak} (k 1, 2, . . . , n, . . .) be a sequence of pairwise distinct positive integers.
Show that for all positive integers n n k1 ak k2 ≥ n k1 1 n (IMO 1978).
14. (Telescoping product.) Prove that 1 15 < 1 2 · 3 4 · 5 6 · 7 8 · · · 99 100 < 1 10.
Hint: (1) A 1 2 · 3 4 · · · 99 100, (2) A < 2 3 · 4 5 · 6 7 · · · 100 101, (3) A > 1 2 · 2 3 · 4 5 · · · 98 99.
Multiply (1) with (2) and (1) with (3).
15. (Telescoping series.) Let Qn 1 + 1/4 + 1/9 + · · · + 1/n2. Then, for n ≥3, 19 12 − 1 n + 1 < Qn < 7 4 −1 n.
16. By induction, prove the sharp inequality 1 2 · 3 4 · 5 6 · · · 2n −1 2n ≤ 1 √ 3n + 1 , n ≥1.
Replace 3n + 1 by 3n on the right side, and try to prove this weaker inequality by induction. What happens?
17. a, b, c > 0 ⇒abc(a + b + c) ≤a3b + b3c + c3a.
18. 1/2 < 1/(n + 1) + 1/(n + 2) + · · · + 1/2n < 3/4, n > 1.
19. The Fibonacci sequence is defined by a1 a2 1, an+2 an + an+1. Prove that 1 2 + 1 22 + 2 23 + 3 24 + 5 25 + · · · + an 2n < 2.
7. Inequalities 181 20. Prove that, for real numbers x, y, z |x| + |y| + |z| ≤|x + y −z| + |x −y + z| + | −x + y + z|.
21. If a, b, c > 0, then a(1 −b) > 1/4, b(1 −c) > 1/4, c(1 −a) > 1/4 cannot be valid simultaneously.
22. If a, b, c, d > 0, then at least one of the following inequalities is wrong: a + b < c + d, (a + b)(c + d) < ab + cd, (a + b)cd < ab(c + d).
23. The product of three positive reals is 1. Their sum is greater than the sum of their reciprocals. Prove that exactly one of these numbers is > 1.
24. Let x1 1, xn+1 1 + n/xn for n ≥1. Show that √n ≤xn ≤√n + 1.
25. If a, b, and c are sides of a triangle, then 3 2 ≤ a b + c + b c + a + c a + b < 2.
26. If a, b, c are sides of a triangle with γ 90◦, then cn > an + bn for n ∈N, n > 2.
27. If x, y, z are sides of a triangle, then |x/y + y/z + z/x −y/x −z/y −x/z| < 1.
Can you replace 1 by a smaller number?
28. A point is chosen on each side of a unit square. The four points are sides of a quadrilateral with sides a, b, c, d. Show that 2 ≤a2 + b2 + c2 + d2 ≤4 and 2 √ 2 ≤a + b + c + d ≤4.
29. Let ai ≥1 for i 1, . . . , n. Show that (1 + a1)(1 + a2) · · · (1 + an) ≥ 2n n + 1(1 + a1 + a2 + · · · + an).
30. Let 0 < a ≤b ≤c ≤d. Then abbccdda ≥bacbdcad.
31. If a, b > 0 and m is an integer, then (1 + a/b)m + (1 + b/a)m ≥2m+1.
32. Let 0 < p ≤a, b, c, d, e ≤q. Show that (a + b + c + d + e) 1 a + 1 b + 1 c + 1 d + 1 e ≤25 + 6 p q − q p 2 .
This is a problem of the US Olympiad 1977. It is a special case of a general theorem.
Also, prove this more general theorem.
33. The diagonals of a convex quadrilateral intersect in O. What is the smallest area this quadrilateral can have, if the triangles AOB and COD have areas 4 and 9, respectively?
34. Let x, y > 0, and let s be the smallest of the numbers x, y + 1/x, 1/y. Find the greatest possible value of s. For which x, y is this value assumed?
182 7. Inequalities 35. Let xi > 0, x1 + · · · + xn 1, and let s be the greatest of the numbers x1 1 + x1 , x2 1 + x1 + x2 , x3 1 + x1 + x2 + x3 , · · · , xn 1 + x1 + x2 + · · · + xn .
Find the smallest value of s. For which x1, . . . , xn will it be assumed?
36. Find a point P inside the triangle ABC, such that the product PL · PM · PN is maximal. Here L, M, N are the feet of the perpendiculars from P onto BC, CA, AB (BrMO 1978).
37. If xi > 0 and xiyi −z2 i > 0 for i ≤n, then n3 n i1 xi n i1 yi − n i1 zi 2 ≤ n i1 1 xiyi −z2 i .
Prove this inequality for n 2 (IMO 1969), and then also generally.
38. The vectors ⃗ a, ⃗ b, ⃗ c, ⃗ d with sum ⃗ o are given in a plane. Prove the inequality |⃗ a| + |⃗ b| + |⃗ c| + | ⃗ d| ≥|⃗ a + ⃗ d| + |⃗ b + ⃗ d| + |⃗ c + ⃗ d|.
Prove this also for one and three dimensions (AUO 1976).
39. Show that (n + 1)n ≥2n · n! for n 1, 2, 3, . . . .
40. (MMO 1975.) Which of the two numbers is larger: (a) An exponential tower of n 2’s or an exponential tower of (n −1) 3’s?
(b) An exponential tower of n 3’s or an exponential tower of (n −1) 4’s?
41. Fifty watches, all showing correct time, are on a table. Prove that at a certain moment the sum of the distances from the center O of the table to the endpoints of the minute hands is greater than the sums of the distances from O to the centers of the watches (AUO 1976).
42. Let x1 2, xn+1 (x4 n + 1)/5xn for n > 0. Show that 1/5 ≤xn < 2 for all n > 1.
43. Let a, b, c > 0. Show that (a) abc ≥(a + b −c)(a + c −b)(b + c −a), (b) a3 + b3 + c3 ≥a2b + b2c + c2a.
44. Let xi > 0, s x1 + · · · , +xn. Show that s s −x1 + s s −x2 + · · · + s s −xn ≥ n2 n −1.
45. For x, y, z > 0, (a) x2 y2 + y2 z2 + z2 x2 ≥y x + z y + x z , (b) x2 y2 + y2 z2 + z2 x2 ≥x y + y z + z x .
46. Write each rational number from (0, 1] as a fraction a/b with gcd(a, b) 1, and cover a/b with the interval a b − 1 4b2 , a b + 1 4b2 .
Prove that the number √ 2/2 is not covered.
7. Inequalities 183 47. By calculus, prove that a > 0, b > 0 ⇒ a + 1 b + 1 b+1 ≥ a b b .
48. Prove that, for real a, b, |a + b| 1 + |a + b| ≤ |a| 1 + |a| + |b| 1 + |b|.
49. The polynomial ax2 + bx + c with a > 0 has real roots x1, x2. Prove that |xi| ≤ 1 (i 1, 2) exactly if a + b + c ≥0, a −b + c ≥0, a −c ≥0.
50. Let 0 a0 < a1 < . . . < an and ai+1 −ai ≤1 for 0 ≤i ≤n −1. Then, n i0 ai 2 ≥ n i0 a3 i .
51. Let a, b, c > 0, a > c, b > c. Prove that √c(a −c) + √c(b −c) ≤ √ ab.
52. If ab and a + b have the same sign, then (a + b)(a4 + b4) ≥(a2 + b2)(a3 + b3).
53. For a + b > 0, a b2 + b a2 ≥1 a + 1 b .
54. If a > b > 0, then, (a −b)2 8a < a + b 2 − √ ab < (a −b)2 8b .
55. The following inequality holds for any triangle with sides a, b, c: a(b2 + c2 −a2) + b(c2 + a2 −b2) + c(a2 + b2 −c2) ≤3abc.
56. For any triangle with sides a, b, c, a2b(a −b) + b2c(b −c) + c2a(c −a) ≥0.
(Proposed by Klamkin and used in the IMO 1983. Due originally to E. Catalan, Educational Times N.S. 10, 57 (1906). The source is cited in .) 57. Two triangles with sides a, b, c and a1, b1, c1 are similar if and only if √aa1 + bb1 + √cc1 (a + b + c)(a1 + b1 + c1).
58. Let x, y, z be the lengths of the sides of a triangle, and let f (x, y, z) x −y x + y + y −z y + z + z −x z + x .
Prove that (a) f (x, y, z) < 1.
(b) f (x, y, z) < 1/8.
(c) Find the upper limit of f (x, y, z).
184 7. Inequalities 59. Minimize x2 1 + · · · + x2 n for 0 ≤xi ≤1 and x1 + · · · + xn 1. Find a probabilistic interpretation.
60. x, y > 0, x ̸ y; m, n ∈N ⇒xmyn + xnym < xm+n + ym+n.
61. Find the maximum and minimum of f 3x + 4y + 12z if x2 + y2 + z2 1.
62. Each of the vectors ⃗ a1, . . . , ⃗ an has length ≤1. Prove that the signs can be chosen in the sum ⃗ c ± ⃗ a1 ± · · · ± ⃗ an so that |⃗ c| ≤ √ 2.
63. √xy < (x −y)/(ln x −ln y) < (x + y)/(2), (x > y > 0).
64. a, b, c > 0 ⇒ √ a2 −ab + b2 + √ b2 −bc + c2 ≥ √ a2 + ac + c2.
65. a+b(x+y+z+u)+c(xy+xz+xt+yz+yt+zt)+d(xyz+xyt+xzt+yzt)+exyzt ≥ 0, 0 ≤x, y, z, t ≤1 iff a ≥0; a + b ≥0; a + 2b + c ≥0, a + 3b + 3c + d ≥0, a + 4b + 6c + 4d + 4 ≥0.
66. 0 < x, y < 1 ⇒xy + yx > 1.
67. a, b > 0, a + b 1 ⇒(a + 1/a)2 + (b + 1/b)2 ≥25/2.
68. a, b, c > 0, a + b + c 1 ⇒(a + 1/a)2 + (b + 1/b)2 + (c + 1/c)2 ≥100/3.
69. Prove the inequality an b + c + bn c + a + cn a + b ≥an−1 + bn−1 + cn−1 2 .
70. A function d(x, y) of two points is a distance if d(x, y) d(y, x), d(x, y)+d(y, z) ≤ d(x, z) for all x, y, z, and d(x, x) 0. The second property is called triangle in-equality. Prove that the following function is a distance: d(x, y) |x −y| √ 1 + x2 1 + y2 .
71. Let xi > 0, x1 + · · · + xn 1, n ≥2. Prove that S ≥n/(2n −1) if S x1 1 + x2 + · · · + xn + x2 1 + x1 + x3 + · · · + xn + · · · + xn 1 + x1 + · · · + xn−1 .
72. In a triangle with sides a, b, c, it is known that ab + bc + ca 12. Between which bounds does the perimeter p lie?
73. Twenty disjoined squares lie inside a square of side 1. Prove that there are four squares among them with the sum of the lengths of their sides ≤2/ √ 5.
74. Let x, y, z ∈R and x2 + y2 + z2 + 2xyz 1. Prove that x2 + y2 + z2 ≥3/4.
75. Prove that xi > 0 for all i ⇒xx1 1 xx2 2 · · · xxn n ≥(x1 · · · xn) x1+···+xn n .
76. 0 ≤a, b, c ≤1 ⇒a/(bc + 1) + b/(ac + 1) + c/(ab + 1) ≤2.
77. Three lines are drawn through a point O inside a triangle with area S so that every side of the triangle is cut by two of them. The lines cut out of the triangle three triangular pieces with common vertex O and areas S1, S2, S3. Prove that (a) 1 S1 + 1 S2 + 1 S3 ≥9 S , (b) 1 S1 + 1 S2 + 1 S3 ≥18 S .
7. Inequalities 185 78. Find the positive solutions of the system of equations x1 + 1 x2 4, x2 + 1 x3 1, . . . , x99 + 1 x100 4, x100 + 1 x1 1.
79. Prove that, for any real numbers x, y, −1 2 ≤(x + y)(1 −xy) (1 + x2)(1 + y2) ≤1 2.
80. Let a + b + c 1. Prove the inequality √ 4a + 1 + √ 4b + 1 + √ 4c + 1 ≤ √ 21.
81. Prove that, for any positive numbers x1, x2, . . . , xk (k ≥4), x1 xk + x2 + x2 x1 + x3 + · · · + xk xk−1 + x1 ≥2.
Can you replace 2 by a greater number?
82. Prove that, for positive reals a, b, c, a + b −2c b + c + b + c −2a c + a + c + a −2b a + b ≥0.
83. Prove the inequality (a3 −a + 2)2 > 4a2(a2 + 1)(a −2).
84. Let a1, . . . , an be positive and an+1 a1. Prove that 2 n k1 a2 k ak + ak+1 ≥ n k1 ak.
85. Let x1, . . . , xn be positive with x1 · x2 · · · xn 1. Prove that xn−1 1 + xn−1 2 + · · · + xn−1 n ≥1 x1 + 1 x2 + · · · + 1 xn .
86. Find all values assumed by x/(x + y) + y/(y + z) + z/(z + x) if x, y, z > 0?
87. Let a, b, c be the side lengths of a triangle, and let sa, sb, sc be the lengths of the medians. D is the diameter of the circumcircle. Prove that a2 + b2 sc + b2 + c2 sa + c2 + a2 sb ≤6D.
88. Find all positive solutions of the system x + y + z 1, x3 + y3 + z3 + xyz x4 + y4 + z4 + 1.
89. Let x, y, z be positive reals with xy + yz + zx 1. Prove the inequality 2x(1 −x2) (1 + x2)2 + 2y(1 −y2) (1 + y2)2 + 2z(1 −z2) (1 + z2)2 ≤ x 1 + x2 + y 1 + y2 + z 1 + z2 .
90. Let a, b and c be positive real numbers such that abc 1. Prove that 1 a3(b + c) + 1 b3(a + c) + 1 c3(a + b) ≥3 2 (IMO 1995).
186 7. Inequalities 91. Prove that, for real numbers x1 ≥x2 ≥· · · ≥xn > 0, x1 x2 + x2 x3 + · · · + xn−1 xn + xn x1 ≤x2 x1 + x3 x2 + · · · + xn xn−1 + x1 xn .
92. Provethat,ifthenumbersa,b,andc satisfytheinequalities|a−b| ≥|c|,|b−c| ≥|a|, |c −a| ≥|b|, then one of these numbers is the sum of the other two (MMO 1996).
93. The positive integers a, b, c are such that a2+b2−ab c2. Prove that (a−c)(b−c) ≤ 0 (MMO 1996).
94. If x, y, z are reals from [0, 1], then 2(x3 + y3 + z3) −x2y −y2z −z2x ≤3.
95. If a, b, c are real numbers such that 0 ≤a, b, c ≤1, then a 1 + bc + b 1 + ac + c 1 + ab ≤2.
96. Prove that, for any distribution of signs + and −in the odd powers of x, x2n ± x2n−1 + x2n−2 ± x2n−3 + · · · + x4 ± x3 + x2 ± x + 1 > 1 2.
97. Given are any eight real numbers a, b, c, d, e, f , g, and h. Prove that at least one of the six numbers ac + bd, ae + bf , ag + bh, ce + df , cg + dh, eg + f h is not negative.
98. Let n > 2 and x1, . . . , xn be nonnegative reals. Prove the inequality (x1x2 · · · xn)1/n + 1 n i<j | xi −xj |≥x1 + · · · + xn n .
99. Let a, b ∈R and f (x) a cos x + b cos 3x. It is known that f (x) > 1 has no solutions. Prove that | b |≤1.
100. Let a, b, c be the sides of a triangle. Prove that a b + c −a + b c + a −b + c a + b −c ≥3.
Solutions 1. The right side follows from ab + bc + ca ≤a2 + b2 + c2. The left side follows from 0 ≤(a + b + c)2 a2 + b2 + c2 + 2(ab + bc + ca) 1 + 2(ab + bc + ca).
2. (a) This is a slight transformation of the QM-AM and an example of the Chebyshev inequality 2(a2 + b2) ≥(a + b)2.
(b) This is the Chebyshev inequality 3(a3 + b3 + c3) ≥(a + b + c)(a2 + b2 + c2).
(c) The right side is √(ab + bc + ca)/3 ≥ 3 √ ab · bc · ca 3 √ abc. We get the left side easily by squaring a2 + b2 + c2 ≥ab + bc + ca.
7. Inequalities 187 3. We have abc + abd + acd + bcd 4 1 2 ab c + d 2 + cd a + b 2 ≤1 2 a + b 2 2 c + d 2 + c + d 2 2 a + b 2 a + b 2 · c + d 2 · a + b + c + d 4 a + b + c + d 4 3 .
Hence, 3 abc + abd + acd + bcd 4 ≤a + b + c + d 4 ≤ a2 + b2 + c2 + d2 4 .
4. This is the AM-GM inequality for the n + 1 numbers a, b, . . . , b.
5. We maximize the probability x3y2z of the word BAOBAB if x + y + z 1: 1 x + y + z x 3 + x 3 + x 3 + y 2 + y 2 + z ≥6 6 x3 27 · y2 4 · z, or x3y2z ≤1/432. We have equality iff x/3 y/2 z, i.e, x 1/2, y 1/3, z 1/6.
6. The left side is well known and does not require the triangle inequality. The right side follows from a2 < (b + c)2, b2 < (a + c)2, c2 < (a + b)2 by addition and simplification.
7. This follows from the preceding problem.
8. Let a ≥b ≥c. Then 1/(a + b) ≤1/(c + a) ≤1/(b + c). We must prove that 1/(b + c) < 1/(a + b) + 1/(a + c). This follows easily from a < b + c.
9. Denote the sum by S. Then S > a a + b + c + d + b a + b + c + d + c a + b + c + d + d a + b + c + d 1.
S < a a + b + b a + b + c c + d + d c + d 2.
The function S is continuous. We will prove that it comes arbitrarily close to 1 and 2.
So it assumes every value from the interval (1, 2). First, using a b x, c d y and then a c x, b d y, we get S1(x, y) 2x 2x + y + 2y x + 2y , lim x→1 y→0 S1(x, y) 1, and S2(x, y) 2x x + 2y + 2y 2x + y , lim x→1 y→0 S2(x, y) 2.
10. Squaring and simplifying, we get the CS inequality.
188 7. Inequalities 11. Rewrite the inequality as follows: 1 a · 1 b + 1 b · 1 c + 1 c · 1 a ≤1 a · 1 a + 1 b · 1 b + 1 c · 1 c .
On the RHS, we have the scalar product of two sequences sorted the same way. On the LHS, we have the scalar product of the rearranged sequences.
12. This is Chebyshev’s inequality after some transformation.
13. Writing the RHS in the form n/n2, we have oppositely sorted sequences. On the left, this is not necessarily the case.
14. The hint should be sufficient to solve the problem.
15. We have the following estimates: Qn > 1 + 1 4 + 1 3·4 + · · · + 1 n(n+1), Qn < 1 + 1 4 + 1 2·3 + · · · + 1 (n−1)n, Qn > 5 4 + 1 3 −1 4 + · · · + 1 n − 1 n+1, Qn < 5 4 + 1 2 −1 3 + · · · + 1 n − 1 n+1.
16. The inequality is sharp for n 1. Suppose the inequality is valid for any n. If we can prove that 2n+1 2n+2 ≤ 3n+1 3n+4, the statement will be true for n + 1.
2n + 1 2n + 2 ≤ 3n + 1 3n + 4 ⇔ 2n + 1 2n + 2 2 ≤3n + 1 3n + 4 ⇔(4n2 + 4n + 1)(3n + 4) ≤(4n2 + 8n + 4)(3n + 1) ⇔12n3 + 28n2 + 19n + 4 ≤12n3 + 28n2 + 20n + 4 ⇔0 ≤n.
Sometimes it is easier to prove more than less. This simple approach does not work for the weaker inequality.
17. Trivial transformation yields 0 ≤ab(a −c)2 + bc(b −a)2 + ca(c −b)2.
Second proof. Apply the CS inequality to the vectors (a/√c, b/√a, c/ √ b), (√c, √a, √ b). You get a √c · √c + b √a · √a + c √ b · √ b 2 ≤ a2 c + b2 a + c2 b (c + a + b), (a + b + c) ≤a2 c + b2 a + c2 b ⇒abc(a + b + c) ≤a3b + b3c + c3a.
18.
1 n+1 + · · · + 1 n+n > 1 n+n + 1 n+n + · · · + 1 n+n n 2n 1 2.
1 n + 1 n+1 + · · · + 1 2n 1 2 1 n + 1 2n + 1 n+1 + 1 2n−1 + · · · + 1 2n + 1 n 1 2 3n 2n2 + 3n 2n2+(n−1) + · · · + 3n 2n2 < 1 2 3n 2n2 + · · · + 3n 2n2 < 3 4 + 1 n.
Subtracting the redundant term 1/n, we get the result.
7. Inequalities 189 19. We have the following estimates: Sn a1 2 + a2 22 + a1 + a2 23 + a2 + a3 24 + · · · + an−3 + an−2 2n−1 + an−2 + an−1 2n , Sn 3 4 + 1 4 n i1 ai 2i + 1 2 n i1 ai 2i −1 4 −an+1 2n+1 −an 2n , Sn 4 1 2 −an+1 2n+1 −an 2n+2 , Sn 2 −an+1 2n+1 −an 2n < 2.
20. (x + y −z) + (x −y + z) 2x ⇒|x + y −z| + |x −y + z| ≥2|x|, and two similar equations for 2y and 2z are added and divided by 2.
21. Suppose all three inequalities are valid simultaneously. Then a, b, c are all less than 1. Multiplying, we get a(1 −a)b(1 −b)c(1 −c) > 1/64. But a(1 −a) 1/4 −(1/2 −a)2 ≤1/4 and the product is ≤1/64. Contradiction!
22. Multiplying the first two inequalities, we get (a+b)2 < ab+cd. But (a+b)2 ≥4ab.
Hence ab + cd ≥4ab, or cd ≥3ab.
Multiplying the last two inequalities, we get ab(ab + cd) > (a + b)2cd ≥4abcd.
Hence, ab +cd > 4cd, i.e., ab > 3cd. Thus, ab +cd > 3(ab +cd). Contradiction!
23. Suppose x, y, 1/xy are these numbers. From x + y + 1/xy > 1/x + 1/y + xy, we get (x −1)(y −1)(1/xy −1) > 0, and this implies that exactly one of the factors is positive.
24. (a) xn+1 1 + n/xn ≤1 + n/√n ≤1 + √n ≤ √ n + 1 + 1.
(b) xn+1 1 + n/xn ≥1 + n/(√n + 1) ≥1 + (n + 1 −1)/( √ n + 1 + 1) ≥ 1 + √ n + 1 −1 ≥ √ n + 1. Thus, √ n + 1 ≤xn+1 ≤ √ n + 1 + 1. For n 1, we get √ 1 ≤1 ≤ √ 1 + 1, which is also true.
25. We already know the left side. Its proof does not require the triangle inequality. Since the sum of two sides of a triangle is larger than the semiperimeter s, we have b +c > s, c +a > s, a +b > s ⇒ a b + c + b c + a + c a + b < 2(a + b + c) a + b + c 2.
26. We know that a2 + b2 c2. Multiplying by c we get c3 ca2 + cb2 > a3 + b3.
Suppose that the proposition is valid for any n ≥3. Then cn+1 > can + cbn > an+1 + bn+1.
27. The denominator is xyz. The numerator is a cubic polynomial in x, y, z which is invariant with respect to cyclic shift. We observe that x y, y z, z x are zeros of the numerator. So, because of the triangle inequality we get, f (x, y, z) |x −y| z · |y −z| x · |z −x| y < 1.
By a special choice of the variables, we try to get as near to 1 as we please. Indeed, x 1, y 1 + ϵ, z ϵ + ϵ2 yield f (1, 1 + ϵ, ϵ + ϵ2) |1 −ϵ| · |1 −ϵ −ϵ2| 1 + ϵ →1 for ϵ →0.
190 7. Inequalities , , , , A A A b b b A B C D a b c d x y z u Fig. 7.6 28. In Fig. 7.6, we have a2 + b2 + c2 + d2 x2 + (1 −x)2 + y2 + (1 −y)2 + z2 + (1 −z)2 + u2 + (1 −u)2, x2 + (1 −x)2 2(x −1 2)2 + 1 2 ≥1 2, x2 + (1 −x)2 ≤1.
Hence, 2 ≤a2 + b2 + c2 + d2 ≤4, a + b + c + d ≤x + 1 −x + y + 1 −y + z + 1 −z + u + 1 −u 4.
The perimeter of ABCD is minimal if it is a closed light path. All of these light polygons have the same perimeter 2 √ 2, which is twice the length of a diagonal.
Prove this. Hence a + b + c + d ≥2 √ 2.
29. Use induction or proceed as follows: (1 + a1) · · · (1 + an) 2n n i1 1 2 + ai 2 2n n i1 1 + ai −1 2 ≥2n 1 + a1 −1 2 + · + an −1 2 ≥2n 1 + a1 −1 n + 1 + · · · + an −1 n + 1 2n n + 1(n + 1 + a1 −1 + · · · + an −1) 2n n + 1(1 + a1 + · · · + an).
30. Taking logarithms, we get b ln a + c ln b + d ln c + a ln d ≥a ln b + b ln c + c ln d + d ln a.
By routine transformation this can be brought into the form ln c −ln a c −a ≥ln d −ln b d −b .
For c ̸ a, d ̸ b, we use the geometrical interpretation as slopes of chords. Then it becomes (almost) obvious.
31.
1 + a b m + 1 + b a m ≥ 2 a b m + 2 b a m ≥2√(2 · 2)m 2m+1.
7. Inequalities 191 32. TheLHSf (a, b, c, d, e)oftheinequalityisaconvexfunctionofeachofthevariables.
Hence the maximum is taken on one of the 32 vertices of the 5-cube given by p ≤a, b, c, d, e ≤q. If there are n p′s and 5 −n q′s, then we have to maximize the quadratic function f (np + (5 −n)q)( n p + 5 −n q ) 25 + n(5 −n) p q − q p 2 .
So f takes its maximum value 25 + 6 √p/q −√q/p 2 for n 2 or 3.
Alternative solution. Let four of the variables be fixed with sum s and sum of reciprocals r. Denote the fifth variable by x. The left side is a function f (x) (s + x)(r + 1/x). f (x) rx + s/x + rs + 1, f ′′(x) 2r/x3 > 0. Hence f has its extrema at the endpoints. The left side is maximal if k variables are p and 5 −k variables are q. Then (a + · · · + e) 1 a + · · · + 1 e ≤(kp + (5 −k)q) k p + 5−k q 25 + k(5 −k) p q − q p 2 ≤25 + 6 p q − q p 2 .
We have equality for k 2, or k 3.
Generalization: Let x1, . . . , xn ∈[a, b], where 0 < a < b. Prove that (x1 + · · · + xn) 1 x1 + · · · + 1 xn ≤(a + b)2 4ab n2.
33. Let the areas of BCO and DAO be x and y, respectively. Since the areas of two triangles with equal altitudes are proportional to their bases, we have x/4 9/y, or y 36/x. Thus the area of ABCD is f (x) x + 36/x + 13, that is, f (x) (√x −6/√x)2 + 25. This formula proves that the minimum value of the area is 25.
It is taken for x y 6.
34. We want to solve x ≥s, y + 1/x ≥s, 1/y ≥s. At least one of these must be an equality. These inequalities imply y ≤1/s, 1/x ≤1/s, s ≤y + 1/x ≤2/s. From this we conclude s2 ≤2, s ≤ √ 2. It is possible that all three inequalities become equalities: y 1/x √ 2/2. In this case, s √ 2.
35. Set y0 1, yk 1 + x1 + · · · + xk (1 ≤k ≤n). Then yn 2, xk yk −yk−1. If all the given numbers are ≤s, that is, xk yk + yk −yk−1 yk 1 −yk−1 yk ≤s, then 1 −s ≤yk−1/yk. If we multiply all these inequalities for k 1 to n, we get (1 −s)n ≤y0/yn 1/2. Hence s ≥1 −2−1/n. This value is attained, if 2−1/n 1−s yk−1/yk for all k, i.e, if the yk is a geometric progression y1 21/n,y2 22/n, . . . , yn 2 with quotient 21/n and xk 2k/n −2(k−1)/n.
36. Denote P L x, P M y, P N z. We want to maximize f (x, y, z) xyz subject to the condition ax + by + cz 2A where A is the area of the triangle. f takes its maximum at the same point (x, y, z) as the function g(x, y, z) ax ·by ·cz.
Now, ax · by · cz ≤ ax + by + cz 3 3 2A 3 3 .
192 7. Inequalities The product reaches its maximum for ax by cz 2A 3 . Thus f assumes its maximum for x 2A 3 1 a , y 2A 3 1 b, z 2A 3 1 c . In this case, we have x : y : z 1 a : 1 b : 1 c ha : hb : hc.
The point with maximum product xyz is the centroid G of the triangle.
37. Set ai √xiyi −zi, bi √xiyi + zi and use the CS inequality. It is enough to prove that n3 ( ai)( bi) ≤ 1 aibi .
38. Adding up the vectors − → AB ⃗ a, − → BC ⃗ b, − → CD ⃗ c and − → DA ⃗ d, we get a closed polygon ABCD (Fig. 7.7). By rearranging these vectors, we can make a self-intersecting polygon ABCD, as shown in Fig. 7.8. You can easily see that at least one of the six possible arrangements yields such a polygon. Adding up |AE| + |CE| ≥|AC|, |BE| + |DE| ≥|BD|, we get |AB| + |CD| ≥|AC| + |BD| or |⃗ a| + |⃗ b| ≥|⃗ b + ⃗ d| + |⃗ a + ⃗ d|.
The triangle inequality yields |⃗ c| + | ⃗ d| ≥|⃗ c + ⃗ d|.
Adding up the last two inequalities, we get |⃗ a| + |⃗ b| + |⃗ c| + | ⃗ d| ≥|⃗ a + ⃗ d| + |⃗ b + ⃗ d| + |⃗ c + ⃗ d|.
- B B B B B B M ) A B C D ⃗ a ⃗ b ⃗ c ⃗ d Fig. 7.7 - ) B B B B B B M A B C D ⃗ a ⃗ c ⃗ b ⃗ d E Fig. 7.8 39. The inequality is true for n 1. Suppose nn−1 ≥2n−1 · (n −1)!. Multiply the left side by (n+1)n/nn−1 and the right side by 2n. Since n(1+1/n)n ≥2n, the property is hereditary.
40. (a) The second number is larger than the first for all n ≥3. Proof by induction. The opposite is true for n 2.
(b) Let An be the tower of n threes and Bn−1 the tower of (n −1) fours. We will prove by induction that An+1 > 2Bn. Suppose that An > 2Bn−1. Then An+1 3An > 32Bn−1 9Bn−1 9 4 Bn−1 · 4Bn−1 > 2 · 4Bn−1 2Bn.
7. Inequalities 193 41. Let M1, . . . Mn be the centers of the watches, A1, . . . , An the endpoints of the minute hands, and Bi the reflection of Ai at Mi for i 1, . . . , n. Then 2|OMi| ≤|OAi| + |OBi| for all i. This is the triangle inequality. Thus 2 |OMi| ≤ |OAi| + |OBi|. Hence at least one of the two sums on the right side is ≥ |OMi|.
42. (a) 1 ≤xn ≤2 ⇒xn+1 1/5 x3 n + 1/xn < (8 + 1)/5 9/5 < 2.
(b) 1/5 ≤xn ≤1 ⇒xn+1 x3 n + 1/xn /5 < (1 + 5)/5 6/5 < 2.
(c) xn+1 x3 n + 1 3xn + 1 3xn + 1 3xn /5 ≥1 5 · 4 4 √1/27 4 4 √ 3/15 0.3846.
If the sequence converges, then it converges to a root of x4−5x2+1 0 with solution x ( √ 7 − √ 3)/2 ≈0.456850. But it need not converge, and the convergence need not be monotonic. In fact, it does converge to 0.456850, but we are not asked to decide this.
43. (a) Setting a y + z, b x + z, c x + y, we get (x + y)(y + z)(z + x) ≥8xyz.
The result follows from x + y ≥2√xy, y + z ≥2√yz, z + x ≥2√zx by multiplication.
(b) a2 · a + b2 · b + c2 · c ≥a2 · b + b2 · c + c2 · a. This follows from the fact that we have two sequences on the left sorted the same way. This is not the case on the other side.
44. (s/(s −x1) + · · · + s/(s −xn)) ((s −x1)/s + · · · + (s −xn)/s) ≥n2. The second factor on the left is (n −1). This implies the result.
45. (a) Rewrite the inequality as follows: x y · x y + y z · y z + z x · z x ≥y z · z x + z x · x y + x y · y z .
The LHS is the scalar product of two sequences sorted the same way. The RHS is the scalar product of the rearranged sequences.
(b) We use another very useful idea. Clear the denominators. You will get x4z2 + y4x2 + z4y2 ≥x3yz2 + x2y3z + xy2z3.
Now, suppose that x ≥y ≥z. Then we transform as follows: x3z2(x −y) + x2y3(y −z) + y2z3(z −x) ≥0.
Here the first two parentheses are ≥0, but the third is not positive. In this case one usually writes z −x z −y + y −x and collects terms: x3z2(x −y) + x2y3(y −z) −y2z3(x −y) −y2z3(y −z) ≥0 ⇒z2(x3 −y2z)(x −y) + y2(x2y −z3)(y −z) ≥0.
The last inequality is obviously correct.
46. Since |b2 −2a2| ≥1, we have √ 2 2 −a b √ 2 2 + a b 1 2 −a2 b2 |b2 −2a2| 2b2 ≥ 1 2b2 , Using the fact that a/b ∈(0, 1), i.e., √ 2/2 + a/b < 2, we get √ 2 2 −a b ≥ 1 2b2 · 1 √ 2 2 + a b > 1 2b2 · 1 2 1 4b2 .
So √ 2/2 is not covered.
194 7. Inequalities 47. Let f (a) (a + 1)b+1/ab. The inequality is equivalent to f (a) ≥f (b). f ′(a) (a −b)(a + 1)b/ab+1. For a b, f ′(a) 0 with change of sign from −to +. Thus fmin f (b). This proves the result.
48. Let us assume that the inequality does not hold. Then |a + b| 1 + |a + b| > |a| 1 + |a| + |b| 1 + |b|.
Simplifying, we get |a + b| > |a| + |b| + 2|ab| + |ab||a + b|, which is impossible since |a + b| ≤|a| + |b|.
49. Using b/a −x1 −x2, c/a x1x2, we get a + b + c ≥0 ⇔1 + b a + c a ≥0 ⇔1 −x1 −x2 + x1x2 ≥0 ⇔(1 −x1)(1 −x2) ≥0, a −b + c ≥0 ⇔1 −b a + c a ≥0 ⇔1 + x1 + x2 + x1x2 ≥0 ⇔(1 + x1)(1 + x2) ≥0, a −c ≥0 ⇔1 −c a ≥0 ⇔1 −x1x2 ≥0.
Let s1 (1−x1)(1−x2), s2 (1+x1)(1+x2), s3 1−x1x2. Obviously |xi| ≤1, i 1, 2 ⇒sk ≥0, k 1, 2, 3. We prove the converse. Because of the symmetry in x1 and x2, it is sufficient to consider the cases x1 > 1 and x1 < −1. Suppose x1 > 1. If x2 < 1, then s1 < 0. Otherwise, if x2 ≥1, then s3 < 0.
Suppose x1 < −1. If x2 ≤−1, then s3 < 0. Otherwise, if x2 > −1, then s2 < 0.
50. Try to prove that n i0 ai 2 − n i0 a3 i 2 n i0 i j0 ai aj + aj−1 2 [1 −(aj −aj−1)] ≥0.
We have equality if aj −aj−1 1 for j 1, . . . , n. This gives the well-known result n i0 i 2 n i0 i3.
51. Two squarings eliminate all square roots and yield 0 ≤(ab −ac −bc)2. There is equality if c ab/(a + b).
Alternate solution. Consider a deltoid ABCD with sides AB BC √a, DC DA √ b and diagonal AC 2√c. We can express its area in two ways: (1) |ABCD| |ABC| + |ACD| √c(a −c) + √c(b −c).
(2) |ABCD| 2|ABD| 2 · 1 2 √a √ b sin(̸ BAD) ≤ √ ab. This yields the inequality. We have equality if |AB|2 + |AD|2 |BD|2, that is, a + b a −c + b −c + 2√(a −c)(b −c), which is equivalent to c ab/(a + b).
52. Simplifying, we get a2 ·a +b2 ·b ≥a2 ·b+b2 ·a. Use the Rearrangement inequality.
53. We get this if we multply by a2b2.
54. No solution. Try to prove it yourself.
7. Inequalities 195 55. Here we use the Cosine Law giving b2 + c2 −a2 2bc cos α and its cyclic permu-tations. Replacing the parentheses, we get 2abc cos α + 2abc cos β + 2abc cos γ ≤ 3abc or cos α + cos β + cos γ ≤3 2.
This inequality can be proved in many ways. Here is one way: We may assume that the angles of the triangle are acute. Then we use the fact that the Cosine is concave in 0 < x < π 2 . Thus, cos α + cos β + cos γ ≤3 cos α + β + γ 3 3 cos 60◦ 3 2.
Another method goes as follows: Introduce unit vectors ⃗ a, ⃗ b, ⃗ c with sum ⃗ s directed counterclockwise along the sides a, b, c of the triangle. Then, ⃗ s ⃗ a a + ⃗ b b + ⃗ c c ⇒s2 3 + 2 ⃗ a⃗ b ab + ⃗ b⃗ c bc + ⃗ a⃗ c ac , s2 3 −2(cos α + cos β + cos γ ) ⇒cos α + cos β + cos γ 3 2 −s2 2 ≤3 2.
Equality holds exactly for ⃗ s 0, that is, for equilateral triangles.
Here is another proof: cos α (b2 + c2 −a2)/2bc ((b −c)2 + 2bc −a2)/2bc ≤ 1 −a2/2bc. Similarly, cos β ≤1 −b2/2ac, cos γ ≤1 −c2/2ab.
cos α + cos β + cos γ ≤3 −1 2 a2 bc + b2 ac + c2 ab ≤3 −3 2 3 √ 1 3 2.
56. In proving triangular inequalities, it is often useful to use the transformations a y + z, b z + x, c x + y, where x, y, z are positive numbers. Fig. 7.9 shows the geometric interpretation of this transformation. Solving for x, y, and z, we get x s −a, y s −b, z s −c, with s (a + b + c)/2. The given inequality reduces to @ @ @ @ @ @ A B C x x y y z z Fig. 7.9 x3z + y3x + z3y ≥x2yz + xy2z + xyz2.
(1) Dividing by xyz, we get x2 y + y2 z + z2 x ≥x + y + z.
(2) 196 7. Inequalities Now we observe that the two sequences (x2, y2, z2) and (1/x, 1/y, 1/z) are oppo-sitely sorted. Hence, ⎡ ⎢ ⎣ x2 1 x y2 1 y z2 1 z ⎤ ⎥ ⎦≤ ⎡ ⎢ ⎣ x2 1 y y2 1 z z2 1 x ⎤ ⎥ ⎦, (3) which was to be proved.
Bernhard Leeb received a special prize for rewriting the inequality by algebraic manipulation in the form a(b −c)2(b + c −a) + b(a −b)(a −c)(a + b −c) ≥0.
(4) Since a cyclic permutation leaves the given inequality invariant, one can assume that a ≥b, c. Now (4) becomes obvious.
The inequality is homogeneous in a, b, c of degree three. Try to solve it by nor-malizing. For instance, Set a 1, b 1 −x, c 1 −y with 0 < x, y < 1 and x + y < 1.
Be careful! Your proof must consist of two cases: (a) x ≤y (b) y ≤x.
Also try to apply the CS inequality to (2).
57. This is a straightforward application of the CS inequality. Let (x, y, z) (√a, √ b, √c), and (x1, y1, z1) (√a1, √b1, √c1). Then we have (xx1 + yy1 + zz1)2 ≤(x2 + y2 + z2)(x2 1 + y2 1 + z2 1) (1) We have equality iff (x1, y1, z1) λ(x, y, z) (similar triangles).
58. Let f (x, y, z) (x−y)/(x+y)+(y−z)/(y+z)+(z−x)/(z+x) p(x, y, z)/(x+ y)(y + z)(z + x). The polynomial p has degree 3, and p(x, x, z) p(x, y, y) p(x, y, x) 0. Thus, p has factors x −y, y −z, x −z. Up to a constant, which turns out to be 1, we have f (x, y, z) (x −y)(y −z)(x −z) (x + y)(y + z)(z + x).
(a) From |x −y| < x +y, |y −z| < y +z, |x −z| < z+x, we get |f (x, y, z)| < 1.
(b) We did not use the triangle inequality in (a). Using | x −y |< z, | y −z |< x, |z −x| < y, we get |f (x, y, z)| < z x + y · x y + z · y z + x √xy x + y · √yz y + z · √xz z + x ≤1 8.
Here we used the fact that a + b ≥2 √ ab.
(c) By analysis, one gets the smallest upper bound, which is assumed for a de-generated triangle with sides z 1, y √ 10+ √ 5+ √ 2+1 2 , x z + y. One gets f (x, y, z) (8 √ 2 −5 √ 5)/3 < 0.04446.
7. Inequalities 197 59. We conjecture that the minimum is attained for xi 1/n for all i. To prove this we set xi yi +1/n, where the yi are the deviations from 1/n. Then we have yi 0 So, x2 i yi + 1 n 2 y2 i + 2 yi n + 1 n2 1 n + y2 i .
The sum is minimal if all the deviations yi are zero. Another solution uses the CS inequality: 1 1 · xi ≤ √ 12 + · · · + 12 x2 1 + · · · + x2 n, 1 ≤√n · x2 1 + · · · + x2 n ⇒x2 1 + · · · + x2 n ≥1/n.
Solution with the QM–AM inequality: x2 1 + · · · + x2 n n ≥x1 + · · · + xn n 1 n ⇒x2 1 + · · · + x2 n ≥1 n.
Probablistic interpretation: It is the probability of a repetition if a spinner with prob-abilities x1, · · · , xn for outcomes 1, · · · , n is spun twice.
Generalization: Minimize x2 1 + · · · + x2 n with a1x1 + · · · + anxn 1 as a side condition.
60. This can be transformed into 0 < (xm −ym)(xn −yn), which is obvious.
61. |3x + 4y + 12z| ≤ √ 32 + 42 + 122 x2 + y2 + z2 13. Equality holds for (x, y, z) t(3, 4, 12). From 9t2 + 16t2 + 144t2 1, we get t ±1/13. Thus, the maximum is (3 + 4 + 12)/13 19/13 and the minimum −19/13.
62. First, we prove that, of the vectors ⃗ a, ⃗ b, ⃗ c with lengths ≤1 at least one of ⃗ a ± ⃗ b, ⃗ a ± ⃗ c, ⃗ b ± ⃗ c has length ≤1. Indeed, two of the vectors ±⃗ a, ±⃗ b and ±⃗ c have an angle ≤60◦. Hence the difference of these two vectors has length ≤1. In this way, we can get down to two vectors ⃗ a and ⃗ b, each of length ≤1. The angle between ⃗ a and ⃗ b or ⃗ a and −⃗ b is ≤90◦. Thus either |⃗ a −⃗ b| ≤ √ 2 or |⃗ a + ⃗ b| ≤ √ 2.
63. A geometric interpretation will make both inequalities obvious. We must know that ln x is the area under the hyperbola s 1/t from 1 to x. The area under the hyperbola from y to x is ln x−ln y. Now we simply write the obvious fact that this area is larger than the area bounded by t x, t y, the x-axis, and the tangent at some point between y and x. The area of the hyperbolic trapezoid is ln x −ln y. The trapezoid bounded above by the tangent at √xy is (x −y)/√xy, and the one bounded by the tangent at (x + y)/2 is 2(x −y)/(x + y). Thus, we have x −y √xy < ln x −ln y and 2x −y x + y < ln x −ln y.
Routine transformation gives the results of the problem. We use the obvious fact that a tangent lies below the hyperbola, a consequence of the convexity of the hyperbola.
The convexity can be proved without derivatives. Indeed, a function f is convex by definition if f x + y 2 ≤f (x) + f (y) 2 .
If we apply this to the hyperbola, after taking reciprocals, we get x + y 2 ≥ 2 1 x + 1 y .
198 7. Inequalities This is the arithmetic-harmonic mean inequality.
64. The radicands remind us of the Cosine Rule with angles 60◦and 120◦. In Fig. 7.10 we have |AB| √ a2 −ab + b2, |BC| √ b2 −bc + c2, |AC| √ a2 + ac + c2.
It is the triangle inequality for △ABC.
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66. Let x 1 1+u, y 1 1+v , u > 0, v > 0. Then xy 1 (1+u)y > 1 1+uy 1+v 1+u+v , yx > 1+u 1+u+v , xy + yx > 1+v 1+u+v + 1+u 1+u+v 1 + 1 1+u+v > 1.
Here we used the inequality (1 + u)y < 1 + uy for 0 < y < 1. We will prove it by calculus.
f (u) 1 + yu −(1 + u)y, f ′(u) y −y(1 + u)y−1 y 1 − 1 (1 + u)1−y > 0.
Now f (0) 0, f (1) y, and f is increasing in the interval (0, 1).
67. The function f (x) (x + 1/x)2 is convex since f ′(x) 2 x −1/x3 and f ′′(x) 2 1 + 3/x4 > 0. Hence, f (a) + f (b) ≥2f a + b 2 2f 1 2 2 1 2 + 2 2 25 2 .
68. f (a) + f (b) + f (c) ≥3f ( a+b+c 3 ) 3f ( 1 3) 3 · 3 + 1 3 2 100 3 .
69. Suppose a ≥b ≥c. Then an, bn, cn, and 1 b+c, 1 c+a , 1 a+b are monotonically increasing. This impllies an 1 b+c + bn 1 c+a + cn 1 a+b ≥an 1 a+b + bn 1 b+c + cn 1 c+a , an 1 b+c + bn 1 c+a + cn 1 a+b ≥an 1 c+a + bn 1 a+b + cn 1 b+c.
Adding these two inequalities, we get an b + c + bn c + a + cn a + b ≥1 2 an + bn a + b + bn + cn b + c + cn + an c + a .
7. Inequalities 199 Now it is easy to prove the inequality (xn + yn)/(x + y) ≥(xn−1 + yn−1)/2. This is a consequence of Cebyshev’s inequality. Hence, the result.
70. We see at once that d(x, x) 0 and d(x, y) d(y, x) for all x, y. To prove transitivity, we use the transformation y1 tan α1, y2 tan α2. Then d(y1, y2) | tan α1 −tan α2| 1 + tan2 α1 1 + tan2 α2 | sin α2 cos α2 −sin α2 cos α1| | sin(α1 −α2)|. Now d(y1, y3) ≤d(y1, y2) + d(y2, y3) becomes | sin(a1 −α3)| ≤ | sin(α1 −α2)| + | sin(α2 −α3|. With β α1 −α2, γ α2 −α3, this becomes sin(β + γ )| | sin β cos γ + cos β sin γ | ≤| sin β cos γ | + | sin γ cos β| ≤| sin β| + | sin γ |.
71. Note that the ith denominator is 2 −xi. Thus, S n i1 xi 2 −xi n i1 xi −2 + 2 2 −xi 2 n i1 1 2 −xi −n.
Using the CS-inequality (a2 1 + · · · + a2 n)(b2 1 + · · · + b2 n) ≥(a1b1 + · · · + anbn)2 with ai 1/√2 −xi, bi √2 −xi, we get 1 2 −xi (2 −xi) ≥n2 ⇒ 1 2 −xi ≥ n2 2n −1 and S 2 1 2 −xi −n ≥ 2n2 2n −1 −n n 2n −1.
72. p2 (a+b+c)2 a2+b2+c2+2(ab+bc+ca) 1 2(a−b)2+ 1 2(b−c)2+ 1 2(c−a)2 +3(ab+bc+ca), 2p2−72 (a−b)2+(b−c)2+(c−a)2 ≥0 ⇒p2 ≥36, p ≥6.
The minimum is attained for a b c.
On the other hand, |a−b| ≤c, |b−c| ≤a, |a−c| ≤b ⇒a2+b2+c2 ≥(a−b)2+(b−c)2+(c−a)2.
The left side is p2 −24, the right side is 2p2 −72. Thus, 2p2 −72 ≤p2 −24, or p2 ≤48, u ≤4 √ 3. We have equality for c 0, and a + b 4 √ 3. For instance, a b 2 √ 3. Thus 6 ≤p ≤4 √ 3.
73. Choose the four smallest squares, denote the lengths of their sides a1, a2, a3, a4, and the sum of their areas by A. Obviously A ≤4/20 1/5. Now (a1 −a2)2 + (a1 −a3)2 + (a1 −a4)2 + (a2 −a3)2 + (a2 −a4)2 + (a3 −a4)2 3(a2 1 + a2 2 + a2 3 + a2 4) −2(a1a2 + a1a3 + a1a4 + a2a3 + a2a4 + a3a4) 4(a2 1 + · · · + a2 4) −(a1 + · · · + a4)2 4A −(a1 + · · · + a4)2, that is, 4A −(a1 + a2 + a3 + a4)2 ≥0 ⇒a1 + a2 + a3 + a4 ≤2 √ A ≤ 2 √ 5 .
200 7. Inequalities 74. Let x2 + y2 + z2 < 3/4. Then (x2 + y2 + z2)/3 < 1/2. Hence, 3 √xyz ≤ x2 + y2 + z2 3 < 1 2 ⇒xyz < 1 8.
Now x2 +y2 +z2 +2xyz < 3/4+1/4 1. Contradiction! Thus x2 +y2 +z2 ≥3/4.
We have equality for |x| |y| |z| 1/2 and 0 or 2 negative variables.
75. Taking logarithms and dividing by n, we get x1 ln x1 + · · · + xn ln xn n ≥x1 + · · · + xn n · ln x1 + · · · + ln xn n .
This is Chebyshev’s inequality since the sequences xi and ln xi are sorted the same way.
76. We denote the left side by f (a, b, c). The function f is defined and continuous on the closed cube, and it is convex in any of its variables. Thus it assumes its maximum at one of its vertices. Because of the symmetry in a, b, c we need to try only the triples (0, 0, 0), (0, 0, 1) (0, 1, 1), (1, 1, 1). We get f (0, 1, 1) 2 for the maximum.
To prove convexity, we need only check that f (x, b, c) x pq+1 + p qx+1 + q px+1 is a sum of three convex functions And a sum of any number of convex functions is again convex. Indeed, the three summands are a straight line and two convex hyperbolas.
77. We do not prove it. We just give hints.
(a) Prove that (a) is valid for the special case that the lines through O are parallel to the three sides of the original triangle.
(b) Join the endpoints of the bases of the three triangles so that three more trian-gles with areas T1, T2, T3 are formed. All six triangles form a hexagon. Prove that S1S2S3 T1T2T3. Use the AM-GM inequality giving 1 S1 + 1 S2 + 1 S3 ≥ 3 3 √S1S2S3 3 6 √S1S2S3T1T2T3 ≥ 3 · 6 S1 + S2 + S3 + T1 + T2 + T3 ≥18/S.
There is equality for O, the centroid of the triangle.
78. Answer: x1 2, x2 1 2, . . . , x99 2, x100 1 2. Applying x + 1 y ≥2 x y , we get x1 + 1 x2 ≥2 x1 x2 , . . . , x100 + 1 x1 ≥2 x100 x1 .
Multiplying these inequalities, we get x1 + 1 x2 x2 + 1 x3 · · · x100 + 1 x1 ≥2100, but, from the system of equations, we get x1 + 1 x2 x2 + 1 x3 · · · x100 + 1 x1 450 2100.
Hence, each inequality is an equality, i.e., x1 1 x2 , x2 1 x3 , . . . , x100 1 x1 .
7. Inequalities 201 79. Let ⃗ a 2x 1 + x2 , 1 −x2 1 + x2 , ⃗ b 1 −y2 1 + y2 , 2y 1 + y2 .
Then it is easy to verify that |⃗ a| |⃗ b| 1. The CS–inequality |⃗ a · ⃗ b ≤|⃗ a| · |⃗ b| implies that |⃗ a · ⃗ b| 2 · x(1 −y2) + y(1 −x2) (1 + x2)(1 + y2) 2 · (x + y)(1 −xy) (1 + x2)(1 + y2) ≤1.
Dividing by 2, we get the result.
80. We assume that 4a ≥−1, 4b ≥−1, 4c ≥−1. Consider the two vectors ⃗ p (1, 1, 1), ⃗ q ( √ 4a + 1, √ 4b + 1, √ 4c + 1).
The CS inequality ( ⃗ p · ⃗ q)2 ≤⃗ p 2 · ⃗ q 2 yields ( √ 4a + 1 + √ 4b + 1 + √ 4c + 1)2 ≤3(4a + 1 + 4b + 1 + 4c + 1).
The RHS is 3(4(a + b + c) + 3) 21. We have equality iff a b c 1 3.
81. Denote the LHS of the inequality by Lk. For k 4, we have L4 x1 x4 + x2 + x2 x1 + x3 + x3 x2 + x4 + x4 x1 + x3 x1 + x3 x2 + x4 + x2 + x4 x1 + x3 ≥2.
Now suppose that the proposed inequality is true for some k ≥4, i.e., that Lk ≥2.
Considerk+1arbitrarypositivenumbersx1, x2, . . . xk, xk+1.SinceLk+1 issymmetric with respect to these numbers, without loss of generality, we may assume that xi ≥ xk+1 for i 1, . . . , k. Thus, Lk+1 x1 xk+1 + x1 + · · · + xk xk−1 + xk+1 + xk+1 xk + x1 > Lk ≥2.
Now, we prove that 2 cannot be replaced by a larger number. Consider the case k 2m, where m is a positive integer > 1. Set x1 x2m 1, x2 x2m−1 t, x3 x2m−2 t2, . . . , xm xm+1 tm−1, where t is an arbitrary positive number. Then Lk simplifies to Lk 2 1 + (m −2)t 1 + t2 .
Hence, limt→∞Lk 2. We can proceed similarly in the case k 2m + 1.
82. This inequality is not symmetric in its variables. Rather, a cyclic rotation (a, b, c) → (b, c, a) leaves it invariant. So we can rotate the variables until a becomes the largest (smallest). Denote the LHS by f (a, b, c). Then f (ta, tb, tc) f (a, b, c). The function f is homogeneous of degree zero. We may normalize it so that a+b+c 1, or a 1, b 1 + x, c 1 + y, x > 0, y > 0. In the latter case, we must treat the cases x > y and y > x separately. Note that some of the three terms in f can be negative which complicates usual estimates and makes it difficult unless we clear the denominators. After surprisingly little work, we arrive at the equivalent inequality a(a −c)2 + b(b −a)2 + c(c −b)2 ≥0.
202 7. Inequalities 83. This looks like a discriminant. Indeed, f (x) a2(a−2)x2 −(a3 −a+2)x +(a2 +1) has f (0) a2 +1 > 0, f (1) −(a2 −a +1) < 0. So f has a positive discriminant which is the inequality to be proved.
84. Set an+1 a1, and let s1 n k1 a2 k ak + ak+1 , s2 n k1 a2 k+1 ak + ak+1 .
Then s1 −s2 a1 −a2 + a2 −a3 + · · · + an −a1 0, i.e., s1 s2. Hence, 2 n 1 a2 k ak + ak+1 s1 + s2 a2 1 + a2 2 a1 + a2 + · · · + a2 n + a2 1 an + a1 ≥ n k1 ak + ak+1 2 n 1 ak.
85. The left-hand side of the inequality is n k1 xn−1 k 1 n −1 n i1 k̸i xn−1 k ≥ 1 n −1 n i1 (n −1) n−1 k̸i xn−1 k n i1 k̸i xk n i1 1 xi .
86. Let f (x, y, z) x/(x + y) + y/(y + z) + z/(z + x). Then f (x, y, z) > x/(x + y + z) + y/(x + y + z) + z/(x + y + z) 1. In addition, we have f (x, y, z) ((x + y) −y)/(x + y)+((y + z) −z)/(y + z)+((z + x) −x)/(z + x) 3−f (y, x, z). We have already proved that f (y, x, z) > 1. Hence, f (x, y, z) < 2.
Theseinequalitiesareexact.Indeed,f (x, tx, t2x) 2/(1 + t)+t2/(1 + t2)haslimit 1 for t →∞and limit 2 for t →0. Because f (x, y, z) is continuous for all positive x, y, z, it assumes all values between 1 and 2.
87. Extend the medians AA2, BB2, CC2 until they meet the circumcircle in A1, B1, C1.
We have AA1 ≤D, BB1 ≤D, CC1 ≤D, i.e., ma + A1A2 ≤D, mb + A1A2 ≤D, mc + C1C2 ≤D. A well-known theorem implies A1A2 · AA2 BA2 · A2C, i.e., A1A2 a2/4ma. Similarly, B1B2 b2/4mb and C1C2 c2/4mc. Plugging this into the inequalities above, we get 4m2 a + a2 4ma + 4m2 b + b2 4mb + 4m2 c + c2 4mc ≤3D.
From 4m2 a + a2 2b2 + 2c2, 4m2 b + b2 2c2 + 2a2, 4m2 c + c2 2a2 + 2b2, we get a2 + b2 2mc + b2 + c2 2ma + c2 + a2 2mb ≤3D, and from this, we get the result by doubling.
88. From the first equation, we get 1 x + y + z ≥3 3 √xyz, or xyz ≤1/27. The second equation implies x3(1 −x) + y3(1 −y) + z3(1 −z) 1 −xyz ≥26/27.
On the other hand, 3t3(1 −t) t · t · t(3 −3t) ≤(3/4)4 81/256. Hence, x3(1 −x) + y3(1 −y) + z3(1 −z) ≤81/256, a contradiction.
7. Inequalities 203 89. This reminds us of the formulas sin α 2 tan(α/2)/[1+tan2(α/2)] and cos α [1− tan2(α/2)]/[(1+tan2(α/2)]. So let us set x tan(α/2), y tan(β/2), z tan(γ/2).
The inequality now becomes cos α sin α + cos β sin β + cos γ sin γ ≤(sin α + sin β + sin γ )/2, sin 2α + sin 2β + sin 2γ ≤sin α + sin β + sin γ.
(1) Until now we ignored xy + yz + zx 1. It is satisfied if α + β + γ π. Indeed, z tan(π/2−α/2−β/2) cot(α/2+β/2) (1−xy)/(x+y), and xy+yz+zx xy + (x + y)z xy + 1 −xy 1. We may assume that in (1) we are dealing with the angles α, β, γ of a triangle. By the Sine Law, for the RHS, we have sin α + sin β + sin γ a + b + c 2R 2s 2R sr Rr A rR .
Denote the distances of the circumcenter M from a, b, c by x, y, z Then, for the LHS, we get sin 2α + sin 2β + sin 2γ 2(sin α cos α + sin β cos β + sin γ cos γ ) a cos α + b cos β + c cos γ R , but a cos α + b cos β + c cos γ a · x R + b · y R + c · z R 2A R .
Hence, sin α + sin β + sin γ sin 2α + sin 2β + sin 2γ R 2r ≥1.
90. Let x 1/a, y 1/b and z 1/c. Then xyz 1, and 1 a3(b + c) + 1 b3(c + a) + 1 c3(a + b) x2 y + z + y2 z + x + z2 x + y .
Denote the RHS by S. We want to prove that S ≥3 2. The CS inequality, applied to the vectors x √y + z, y √z + x , z √x + y and √y + z, √ z + x, √x + y , yields (x + y + z)2 ≤S · 2(x + y + z) or S ≥(x + y + z)/2. Using the AM–GM inequality, we get S ≥x + y + z 3 · 3 2 ≥ 3 √xyz · 3 2 3 2.
Equality holds iff x y z 1, which is equivalent to a b c 1.
Many participants of the Olympiad used the Chebyshev inequality. One can also use the Rearrangement inequality. Give a different proof!
91. Transfer all terms to the left side and look at all terms with an xn: f (xn) xn−1 xn + xn x1 −xn xn−1 −x1 xn .
Let us find the minimum of this function on the interval [xn−1, ∞). The derivative of f (xn) on this interval is positive, and hence the minimum is attained at xn xn−1.
Inserting xn xn−1 into the inequality, we get the same inequality, but for the variables x1 to xn−1. We finish the proof by induction.
204 7. Inequalities 92. We square the inequalities, transfer their right sides to the left, factor the differences of the squares, and multiply them, getting (a −c −b)2(b −c −a)2(c −a −b)2 ≤0.
Since squares are nonnegative, at least one of the factors on the left is zero.
93. a2 + b2 −ab c2 can be written in the form a2 + b2 −2ab cos 60◦ c2. So a, b, c are sides of a triangle with γ 60◦. Hence, α ≥60◦, β ≤60◦or α ≤60◦, β ≥60◦.
So a ≥c ≥b or a ≤c ≤b. In both cases, (a −c)(b −c) ≤0.
94. Rewrite the inequality in the form (x3+y3−x2y)+(x3+z3−z2x)+(z3+y3−y2z) ≤ 3. We will show that each parenthesis on the LHS does not exceed 1. Take the first one x3 + y3 −x2y. If x > y, then y3 −x2y < 0. Otherwise x3 −x2y ≤0. Since both x and y are ≤1, we conclude that x3 + y3 −x2y ≤1. We treat the other two parentheses similarly.
95. We may assume 0 ≤a ≤b ≤c ≤1 and 0 ≤(1 −a)(1 −b). Hence a + b ≤ 1 + ab ≤1 + 2ab, and a + b + c ≤a + b + 1 ≤2 + 2ab 2(1 + ab). Thus, a 1 + bc + b 1 + ac + c 1 + ab ≤ a 1 + ab + b 1 + ab + c 1 + ab ≤a + b + c 1 + ab ≤2.
96. It is enough to prove that, for any x > 0, f (x) x2n −x2n−1 + x2n−2 −· · · + x2 −x + 1 1 + x2n+1 1 + x > 1 2.
But f (x) ≥1 for x ≥1, and, if x < 1, the denominator is ≤2, and f (x) > 1 2.
97. Consider the four vectors ⃗ v1 (a, b), ⃗ v2 (c, d), ⃗ v3 (e, f ), ⃗ v4 (g, h). The six given numbers are all pairwise products of these vectors: ⃗ v1 · ⃗ v2, ⃗ v1 · ⃗ v3, . . . , ⃗ v3 · ⃗ v4.
Since one of the angles between these four vectors does not exceed π/2, at least one of the six scalar products is not negative.
98. We may assume that x1 ≥x2 ≥. . . ≥xn. Then all the points x1, . . . , xn lie on the segment [xn, x1]. Hence | xi −xj |≤| xn −x1 |. In addition, | x1 −xk | + | xk −xn | x1 −xn for k 2, . . . , n −1. Together with | x1 −xn |, we get the estimate i<j | xi −xj |≥(n −1)(x1 −xn).
Since (x1 · · · xn)1/n ≥xn, it is sufficient to prove that xn + 1 n(n −1)(x1 −xn) ≥x1 + · · · + xn n or xn + (n −1)x1 ≥x1 + · · · + xn, which is valid. The proof of this weak inequality was so simple since we could get by with huge overestimations.
99. We have f (πn) (−1)na + (−1)nb, f (π/3) a 2 −b, f (2π/3) −a 2 + b. Hence, | a + b |≤1, and | a −2b |≤2, or −1 ≤a + b ≤1, and −2 ≤a + b −3b ≤2.
Adding the last two inequalities we get | b |≤1.
100. We set x b + c −a, y c + a −b, and z a + b −c. The triangle inequality implies that x, y, and z are positive. Furthermore, a (y + z)/2, b (z + x)/2, and c (x + y)/2. The LHS of the inequality becomes y + z 2x + z + x 2y + x + y 2z 1 2 x y + y x + y z + z y + x z + z x , and this is obviously ≥3.
8 The Induction Principle The Induction Principle is of great importance in discrete mathematics: Number Theory, Graph Theory, Enumerative Combinatorics, Combinatorial Geometry, and other subjects. Usually one proves the validity of a relationship f (n) g(n) if one has a guess from small values of n. Then one checks that f (1) g(1), and, by making the assumption f (n) g(n) for some n, one proves that also f (n + 1) g(n + 1). From this one concludes by the Induction Principle that f (n) g(n) for all n ∈N. There are many variations of this principle. The relationship f (n) g(n) is valid for 0 already, or, starting from some n0 > 1.
The inductive assumption is often f (k) g(k) for all k < n, and, from this assumption, one proves the validity of f (n) g(n). We assume familiarity with all this and apply induction in unusual circumstances to make nontrivial proofs.
We refer to Polya to for excellent treatment of induction for beginners.
The reader can acquire practice by proving some of the innumerable formulas for the Fibonacci sequence defined by F0 0, F1 1, Fn+2 Fn+1 + Fn, n ≥0.
We state some of these.
1. Binet’s formula Fn (αn −βn)/ √ 5, α (1 + √ 5)/2, β (1 − √ 5)/2.
2. Fn n−1 0 + n−2 1 + n−3 2 + · · · .
3. n i1 F 2 i FnFn+1.
4. Prove 1 1 1 0 n Fn+1 Fn Fn Fn−1 .
206 8. The Induction Principle Here you need to know how to multiply matrices, but it helps much in proving formulas later.
5. Fn−1Fn+1 F 2 n + (−1)n.
6. F1 + F2 + · · · + Fn Fn+2 −1.
7. F1 + F3 + · · · F2n+1 F2n+2, 1 + F2 + F4 + · · · + F2n F2n+1.
8. FnFn+1 −Fn−2Fn−1 F2n−1, Fn+1Fn+2 −FnFn+3 (−1)n.
9. F 2 n−1 + F 2 n F2n−1, F 2 n + 2Fn−1Fn F2n, Fn(Fn+1 + Fn−1) F2n.
10. F1F2 + F2F3 + · · · F2n−1F2n F 2 2n.
11. F 3 n + F 3 n+1 −F 3 n−1 F3n.
12. m|n ⇒Fm|Fn.
13. gcd(Fm, Fn) Fgcd(m,n).
14. Let t be the positive root of t2 t + 1. Then t 1 + 1/t, from which follows the continued fractional expansion t 1 + 1 1 + 1 1 + 1 1 + 1 1 + · · · with the convergents t1 1, t2 1 + 1 1, t3 1 + 1 1 + 1 1 , . . . .
Prove that tn Fn+1/Fn.
15. Prove that ∞ i1 1 Fn 4 −t, ∞ n1 (−1)n+1 FnFn+1 t −1, ∞ n2 1 + (−1)n F 2 n t.
In this chapter we will use induction to prove some old and new theorems. Some of these were already proved by the extremal principle or by other means. In fact, the Induction Principle is equivalent to the axiom that any subset of the nonnegative integers has a smallest element. In this respect, it is also an extremal principle.
8. The Induction Principle 207 Problems 1. 2n points are given in space. Altogether n2 + 1 line segments are drawn between these points. Show that there is at least one set of three points which are joined pairwise by line segments.
2. Therearenidenticalcarsonacirculartrack.Amongallofthem,theyhavejustenough gas for one car to complete a lap. Show that there is a car which can complete a lap by collecting gas from the other cars on its way around.
3. Every road in Sikinia is one-way. Every pair of cities is connected by exactly one direct road. Show that there exists a city which can be reached from every other city either directly or via at most one other city.
4. Show by induction that f (n) n k0 n + k k 1 2k 2n.
5. For any natural N, prove the inequality ' ( ( ) 2 3 4 . . .
(N −1) √ N < 3 (TT 1987).
6. If a, b, and q (a2 + b2)/(ab + 1) are integers ≥0, then q gcd(ab)2. Prove this famous IMO 1988 problem by induction on the product ab.
7. We build an exponential tower √ 2 √ 2 √ 2··· by defining a0 1, and an+1 √ 2 an, (n ∈N0). Show that the sequence an is monotonically increasing and bounded above by 2.
8. n circles are given in the plane. They divide the plane into parts. Show that you can color the plane with two colors, so that no parts with a common boundary line are colored the same way. Such a coloring is called a proper coloring.
9. A map can be properly colored with two colors iff all of its vertices have even degree.
10. (a) Any simple not necessarily convex n-gon has at least one diagonal which lies completely inside the n-gon.
(b) This n-gon can be triangulated by diagonals which lie inside the n-gon.
(c) The vertices of the triangulated n-gon can be colored properly with three colors.
(d) The faces of the triangulation can be properly colored with two colors.
11. Let an be the number of words of length n from the alphabet {0, 1}, which do not have two 1′s at distance 2 apart. Find an in terms of the Fibonacci numbers.
12. We are given N lines (N > 1) in a plane, no two of which are parallel and no three of which have a point in common. Prove that it is possible to assign a non-zero integer of absolute value not exceeding N to each region of the plane determined by these lines, such that the sum of the integers on either side of any of the given lines is equal to 0 (TT 1989).
208 8. The Induction Principle 13. The sequence an is defined as follows: a0 9, an+1 3a4 n + 4a3 n, n > 0. Show that a10 contains more than 1000 nines in decimal notation (TT).
14. Find a closed form for the expression with n radicals defined as follows: an 2 + 2 + · · · + 2 + √ 2.
15. Let α be any real number such that α + 1/α ∈Z. Prove that αn + 1 αn ∈Z for any n ∈N.
16. Prove that 1 < 1/(n + 1) + · · · + 1/(3n + 1) < 2.
17. For all n ∈N, we have f (n) g(n), where f (n) 1 −1 2 + 1 3 −· · · + 1 2n + 1 −1 2n, g(n) 1 n + 1 + · · · + 1 2n.
18. Prove that (n + 1)(n + 2) · · · · 2n 2n · 1 · 3 · 5 · · · (2n −1) for all n ∈N.
19. Prove that z + 1/z 2 cos α ⇒zn + 1/zn 2 cos nα for all n ∈N.
20. If one square of a 2n × 2n chessboard is removed, then the remaining board can be covered by L-trominoes.
21. 2n + 1 points on the unit circle on the same side of a diameter are given. Prove that | − → OP 1 + · · · − → OP 2n+1 |≥1.
22. Consider all possible subsets of the set {1, 2, . . . , N}, which do not contain any neighboringelements.Provethatthesumofthesquaresoftheproductsofallnumbers in these subsets is (N + 1)! −1. (Example: N 3. Then 12 + 22 + 32 + (1 · 3)2 23 4! −1.) 23. A graph with n vertices, k edges, and no tetrahedron satisfies k ≤⌊n2/3⌋.
24. Let a1, . . . , an be positive integers such that a1 ≤· · · ≤an. Prove that 1 a1 + · · · + 1 an 1 ⇒an < 2n!.
25. 3n+1 | 23n + 1 for all integers n ≥0.
26. In an m × n matrix of real numbers, we mark at least p of the largest numbers (p ≤m) in every column, and at least q of the largest numbers (q ≤n) in every row.
Prove that at least pq numbers are marked twice.
27. n points are selected along a circle and labeled by a or b. Prove that there are at most ⌊(3n+4)/2⌋chords which join differently labeled points and which do not intersect inside the circle.
28. Let n 2k. Prove that we can select n integers from any (2n −1) integers such that their sum is divisible by n.
8. The Induction Principle 209 29. Prove Zeckendorf’s theorem: Any positive integer N can be expressed uniquely as a sum of distinct Fibonacci numbers containing no neighbors: N m j1 Fij +1, |ij −ij−1| ≥2.
Here F1 1, F2 2, Fn+2 Fn+1 + Fn, n ≥1. Indeed, 1 F1 1, 2 F2 10, 3 F3 100, 4 F3 + F1 101, 5 F4 1000, 6 F4 + F1 1001, 7 F4 + F2 1010, 8 F5 10000, 9 F5 + F1 10001, 10 F5 + F2 10010, 11 F5 + F3 10100, 12 F5 + F3 + F1 10101, . . ..
30. A knight is located at the (black) origin of an infinite chessboard. How many squares can it reach after exactly n moves?
31. (a) Consider any convex region in the plane crossed by l lines with p interior points of intersection. Find a simple relationship between l, p, and the number r of disjoint regions created.
(b) Place n distinct points on the circumference of a circle, and draw all possible chords through pairs of these points. Assume that no three chords are concurrent.
Let an be the number of regions. Find a1, a2, a3, a4, a5 by drawing figures. Guess an, and check your guess by finding a6. Now find an by using the result in (a).
32. An infinite chessboard has the shape of the first quadrant. Is it possible to write a positive integer into each square, such that each row and each column contains each positive integer exactly once (TT 1988)?
33. Find the sum of all fractions 1/xy, such that gcd(x, y) 1, x ≤n, y ≤n, x+y > n.
34. Find a closed formula for the sequence an defined as follows: a1 1, an+1 1 16 1 + 4an + 1 + 24an .
35. Prove that if n points are not all collinear, then at least n of the lines joining them are different.
36. The positive integers x1, . . . , xn and y1, . . . , ym are given. The sums x1 + · · · + xn and y1 + · · · + ym are equal and less than mn. Prove that one may cross out some of the terms in the equality x1 + · · · + xn y1 + · · · + ym, so that one again gets an equality.
37. All numbers of the form 1007, 10017, 10117, . . . are divisible by 53.
38. All numbers of the form 12008, 120308, 1203308, . . . are divisible by 19.
39. Let x1, x2 be the roots of the equation x2 +px −1 0, p odd, and set yn xn 1 +xn 2, n ≥0. Then yn and yn+1 are coprime integers.
Solutions 1. We will prove the contrapositive statement: A graph with 2n points and no triangle has at most n2 edges.
The theorem is obviously true for n 1. Suppose the theorem is true for a graph with 2n points. We will prove it for 2n + 2 points.
210 8. The Induction Principle Let G be a graph with 2n + 2 points and no triangle. Select two points A, B of G connected by a line segment. Ignore A, B and all line segments joined to A or B. The remaining graph G′ has 2n points and no triangle. By the induction hypothesis G′ has at most n2 line segments. How many line segments can G have? There is no point C such that A and B are joined to C. Otherwise G would contain a triangle ABC.
Thus if A is joined to x points of G′, then B is joined to at most 2n −x points of G′.
Thus (not forgetting to count the line segment AB) G has at most n2+2n−x+x+1, n2 + 2n + 1, or (n + 1)2 line segments.
It is easy to see that the statement of the theorem is exact. Indeed, partition the 2n points into two n-sets P and Q, and join every point of P with every point of Q.
The resulting graph has no triangle.
2. The theorem is obvious for n 1. Suppose we have proven the theorem for n. Let there be n + 1 cars. Then there is a car A which can reach the next car B. (If no car could reach the next car, there would not be enough fuel for one lap.) Let us empty B into A and remove B. Now we have n cars which, between them, have enough fuel for one lap. By the induction hypothesis, there is a car which can complete a lap.
The same car can also get around the track with all (n + 1) cars on the road. From A to B, there will be enough gas (from car A) and, on the remaining road sections, this car has the same amount of gas as in the case of n cars.
3. The theorem is obviously true for two and three cities. Suppose it is true for n cities. A city satisfying the conditions of the problem will be called an H-city. For n arbitrarily chosen cities let A be an H-city. The other n −1 cities can be partitioned into two sets: the set D of cities with direct roads into A; the set N of cities without direct roads into A. Then, from each N-city one can reach A via some D-city. Let us add another city P to the n cities. There are two cases to consider: (1) There is a direct road from P to A or to a D-city. Then A is also an H-city for the (n + 1) cities.
(2) From A and from any city in D there is a direct road to P. There is also a direct road from any N-city to some D-city. Thus P is an H-city.
4. We have f (1) 2, and with i k −1 and n+1+k k n+k k−1 + n+k k , we get f (n + 1) n+1 k0 n + 1 + k k 2−k 1 + n+1 k1 n + k k −1 2−k + n+1 k1 n + k k 2−k 1 2 n i0 n + i + 1 i 2−i + 2n + 1 n + 1 2−n−1 + f (n) 1 2f (n + 1) + f (n), that is, f (n) 2n. This proof is by far more complicated than the proof by prob-abilistic interpretation in Chapter 5. Note that we made it so compact that you will understand it only by investing some effort.
5. This problem is too special. We imbed it into a more general problem by replacing 2 by m. This makes the proof simpler. By specialization we get the result. For m ≥2, we prove m (m + 1) . . .
√ N < m + 1 8. The Induction Principle 211 by reverse induction, that is, we prove it first for m N and then down to m 2.
Clearly √ N < N + 1. For m < N, we assume inductively that (m + 1) (m + 2) . . .
√ N < m + 2.
Then, m (m + 1) . . .
√ N < m(m + 2) < m + 1.
So, 2 3 . . .
√ N < 3.
6. This proof is due to J. Campbell (Canberra). If ab 0, the result is clear. If ab > 0, we may suppose a ≤b because of symmetry in a and b. Assume the result holds for all smaller values of ab. Now, we try to find an integer c satisfying q a2 + c2 ac + 1 , 0 ≤c ≤b.
(1) Since ac < ab, we know by the induction hypothesis that q gcd(a, c)2.
(2) To obtain c, we solve a2 + b2 ab + 1 a2 + c2 ac + 1 q.
By subtracting numerators and denominators of these two fractions, we get b2 −c2 ab −ac q ⇒b + c a q ⇒c aq −b.
Notice that c is an integer and gcd(a, b) gcd(a, c). The proof will be finished if we can prove 0 ≤c < b. To prove this, we note that q a2 + b2 ab + 1 < a2 + b2 ab a b + b a , giving aq < a2 b + b ≤b2 b + b 2b ⇒aq −b < b ⇒c < b.
To prove c ≥0, we make the estimates q a2 + c2 ac + 1 ⇒ac + 1 > 0 ⇒c > −1 a ⇒c ≥0.
This completes the proof.
7. We have a0 < a1 since 1 < √ 2. Suppose an < an+1 for any n. Since the exponential function with base b > 1 is increasing, we also have √ 2 an < √ 2 an+1, or an+1 < an+2.
This shows that an is increasing.
212 8. The Induction Principle We have, obviously, a0 < 2. Suppose an < 2. Then √ 2 an < √ 2 2 2, or an+1 < 2.
So an has an upper bound 2.
Remark. Every increasing sequence an with an upper bound is convergent to a limit a, which satisfies a √ 2 a. The only solution is a 2. It can be shown that the sequence defined by a0 1, an+1 aan converges for 0.065988 . . . e−e ≤a ≤ e1/e 1.44466 . . . . See Chapter 9.
8. Proof. The theorem is obvious for n 1. The interior is colored white, and the exterior black, which is a proper coloring. Suppose the theorem is valid for n circles.
Now take (n + 1) circles. Ignore one of the circles. The remaining n circles divide the plane into parts which have a proper coloring by the induction hypothesis. Now add the (n + 1)th circle and make the following recoloring. The parts outside this circle keep their colors. The parts inside this circle exchange their colors, the black ones become white, the white ones become black. The new coloring is obviously proper. Indeed, two neighboring regions across this circle will have opposite colors because of reversal of coloring. Two neighboring regions on the same side of this circle still have opposite colors by the induction hypothesis.
Alternate proof. Each of the parts, into which the plane is divided, is labeled by the number of circles within which it lies. Two neighboring parts will have labels of opposite parity. By coloring the odd numbered parts black and the even numbered parts white, we get a proper coloring of the plane.
9. If a vertex has an odd degree, then even the parts surrounding it cannot be properly colored with two colors.
To prove sufficiency, we use induction on the number of edges. The theorem is obvious for maps with two edges.
Suppose the theorem is valid for any map of n edges with all vertices of even degree.
Now take any map M with (n+1) edges with all vertices of even degree. Start at any vertex A of the map, and move along the edges until you return, for the first time, to a vertex B you have already visited. The part of the path from B back to B is a closed path which we erase. We are left with a new map M′ with vertices of even degree. By the induction hypothesis, M′ can be properly colored with two colors.
Now, add the erased path and exchange the colors on one side of the closed path. We get a proper coloring of the map M.
10. (a) Let A, B, and C be three neighboring vertices of the polygon. Consider all rays from B directed inside the polygon. Either one of the rays hits another vertex D.
Then AD is such an inner diagonal. Otherwise, AC is such a diagonal.
(b) We use induction on n. Suppose all k-gons for k ≤n can be triangulated com-pletely by diagonals in their interiors. Consider any (n + 1)-gon. Draw any diagonal in its interior. It splits the polygon into two polygons with ≤n vertices. Each of these can be split completely into triangles by interior diagonals. Thus we get a splitting of the (n + 1)-gon into trangles.
(c) The theorem is obviously true for n 3. Suppose the vertices of a triangulated n-gon can be properly colored with three colors. Now take an (n + 1)-gon. It has three adjacent vertices A, B, C with ̸ ABC < 180◦. Cut off the triangle ABC.
The remaining polygon has n vertices and can be colored properly by the induction hypothesis. Add the vertex B. Since we have used two colors for A and C, we can use the third color for B.
8. The Induction Principle 213 (d) We denote the three colors in (c) by 1, 2, and 3. Orient the sides of the triangles 1 →2 →3 →1. Color the triangles with clockwise orientation black and those with anticlockwise orientation white.
11. We derive a recursion for an as follows. A word starting with 0 can be continued in an−1 ways. A word starting with 100 has an−3 continuations. A word starting with 1100 can be continued in an−4 ways.
n Fn an 1 1 2 2 · 1 2 1 4 2 · 2 3 2 6 2 · 3 4 3 9 3 · 3 5 5 15 3 · 5 Thus, an an−1 + an−3 + an−4, a1 2, a2 4, a3 6, a4 9.
This recursion leads to the table above. From this table, we conjecture that a2m F 2 m+2, a2m+1 Fm+2 · Fm+3.
Suppose the conjecture is valid for all k < 2m. Then, a2m Fm+1Fm+2 + FmFm+1 + F 2 m Fm+1Fm+2 + FmFm+2 F 2 m+2, a2m+1 F 2 m+2 + F 2 m+1 + FmFm+1 F 2 m+2 + Fm+1Fm+2 Fm+2Fm+3.
12. Color the corresponding map properly with two colors. Assign to each region an integer whose magnitude is equal to the number of vertices of that region. The sign of the integer is positive for one color and negative for the other color. The sum of the integers at any side of any line will be 0. Indeed, take any of the N lines. If a vertex is not on that line, then it contributes +1 to two regions and −1 to two regions. If it is on the separating line, it contributes +1 to one region and −1 to another region.
13. To get some clues, we try to compute the first terms of the sequence: a0 9, a1 22599, . . .. The next term already takes too much time. But at least we suspect that there are enough nines at the end of the numbers. In addition, we are told that a10 contains more than 1000 nines. But 1000 is slightly less that 210 1024. We conjecture that an ends with 2n nines. This will be proved by induction. A number ending in m nines has the form a ·10m −1, a ∈N. Suppose an a ·10m −1. Then, an+1 3a4 n + 4a3 n 3(a · 10m −1)4 + 4(a · 10m −1)3 3a4104m −12a3103m + 18a2102m −12a10m + 3 + 4a3103m −12a2102m + 12a310m −4 b · 102m −1.
Hence the number of nines at the end doubles at each step. So an a · 102n −1 for all n ≥0.
14. We try a geometric interpretation. First a1 2 cos(π/4). Next, we remember the duplication formula cos 2α 2 cos2 α −1. Now we make the conjecture an 2 cos π 2n+1 .
214 8. The Induction Principle Using this conjecture, we conclude that an+1 2 + 2 cos π 2n+1 2 cos π 2n+2 .
15. We have α0 +1/α0 ∈Z and, by assumption, α1 +1/α1 ∈Z. Suppose that, for some n ∈N, αn−1 + 1 αn−1 ∈Z, and αn + 1 αn ∈Z.
Then αn+1 + 1 αn+1 α + 1 α αn + 1 αn − αn−1 + 1 αn−1 ∈Z.
16. We have f (n) 1 n + 1 + · · · + 1 3n + 1 < 2n + 1 n + 1 < 2.
Now f (1) 1 2 + 1 3 + 1 4 13 12 > 1. Let f (n) > 1. Then f (n + 1) f (n) − 1 n + 1 + 1 3n + 2 + 1 3n + 3 + 1 3n + 4.
To get f (n + 1) from f (n), we subtract 1/(n + 1), and add g(n) 1/(3n + 2) + 1/(3n + 3) + 1/(3n + 4). Which is larger? We show that g(n) is larger. Indeed, 1 3n + 2 + 1 3n + 4 6n + 6 (3n + 2)(3n + 4) >(3n+3)2 > 2 3n + 3.
Here, we use ab < [(a + b)/2]2. Thus f (n+1) > f (n) > 1. Hence, 1 < f (n) < 2.
17. We have f (1) g(1). Suppose that, for some n ∈N, f (n) g(n).
(1) Then, f (n + 1) −f (n) 1 2n + 1 − 1 2n + 2, g(n + 1) −g(n) 1 2n + 1 + 1 2n + 2 − 1 2n + 2 1 2n + 1 − 1 2n + 2, that is, f (n + 1) −f (n) g(n + 1) −g(n).
(2) Adding (1) and (2), we get f (n + 1) g(n + 1). Now we invoke the induction principle.
18. Denote the left and right sides of the equation by f (n) and g(n), respectively. Then f (1) g(1). Suppose that, for some n ∈N, f (n) g(n).
(1) 8. The Induction Principle 215 Then f (n + 1) f (n)(4n + 2), g(n + 1) g(n)(4n + 2), or f (n + 1) f (n) g(n + 1) g(n) .
(2) Multiplying (1) and (2), we get f (n + 1) g(n + 1). Now we invoke the induction principle.
We could also use simple transformation. Let An (n+1) · · · (2n−1)·2n. Multiply by n!, and divide by n!2n. Then we get An 2n 1 · 2 · 3 · · · 2n 2n · 1 · 2 · · · n 1 · 2 · 3 · · · 2n 2 · 4 · 6 · · · 2n 1 · 3 · 5 · 7 · · · (2n −1).
This is the product of all odd integers fron 1 to 2n −1.
19. From z+1/z 2 cos α, we get z2+1/z2 (z+1/z)2−2 4 cos2 α−2 2 cos 2α.
The theorem is valid for n 1 and n 2. Suppose zn + 1/zn 2 cos nα. Then, zn+1 + 1 zn+1 z + 1 z zn + 1 zn −zn−1 − 1 zn−1 , which is 4 cos α cos nα −2 cos(n −1)α. From the addition theorem for cosine, we get cos(x + y) + cos(x −y) 2 cos x cos y. Applying this formula to the result, we get 2 cos (n + 1)α + 2 cos (n −1)α −2 cos (n −1)α 2 cos (n + 1)α.
20. (a) The problem is trivial for n 1.
(b) Now, suppose that a 2n × 2n board can be covered and we want to cover a board with side 2n+1. Split it into four boards with side 2n. One of the four boards is defective, the other three are complete. We can rotate the defective board so that the missing square does not have a vertex at the center. Now we cover the three corner cells of the the whole boards by one L-tromino. By the induction hypothesis, the resulting four defective boards can be covered.
21. We use induction. The statement is obviously true for n 1. We assume its truth for 2n + 1 vectors, and we consider in the system of 2n + 3 vectors, the two outer vectors − − → OP1 and − → OP 2n+3. Because of the induction assumption, the length of the vector − → OR − → OP 2 + · · · + − → OP 2n+2 is not less than 1. The vector − → OR lies inside the angle P1OP2n+3. Hence it forms an acute angle with − → OS − → OP 1 + − → OP 2n+3. Thus | − → OS + − → OR |≥| OR |≥1.
22. We use induction on N. Partition the set of all subsets in the problem into two subsets: those with N, and those without N. The sum of the squares in the first subset, by the induction hypothesis, is N2 [(N −1)! −1] + N 2, and in the second subset N! −1.
Adding, we get (N + 1)! −1.
23. The statement is obvious for n ≤3. Suppose the statement is correct for n vertices.
Consider three additional vertices, which form a triangle. They cannot be connected to another point. We must have at most 2n + 3 additional edges. Thus the maximum number of edges is n2/3 + 2n + 3 (n + 3)2/3.
216 8. The Induction Principle 24. Suppose an ≥2n!. By backward induction, we prove that ak ≥2k! for k 1, . . . , n.
Suppose that the assumption is proved for k n, n −1, . . . , m + 1. Then, 1 am ≤ m 1 a1 · · · am ≤ m 1 −1 a1 −· · · −1 am m 1 am+1 + · · · + 1 an ≤ m ' ( ( ) n im+1 1 2i! ≤ 1 2m! .
It remains to be observed that 1 21! + 1 22! + · · · + 1 2k! < 1.
25. The theorem is true for n 0. Let n ≥0. Then 23n + 1 (23n−1 + 1) 23n−1
2 −23n−1 + 1 .
By the inductive assumption, the first factor is divisible by 3n. The second factor is divisible by 3 since 23n−1 ≡−1 (mod 3). This proves the statement.
26. We use induction on m + n. The result is obvious for m n p q 1. Suppose we have an m×n matrix. We reduce it to an m×(n−1) or (m−1)×n matrix. If all numbers are marked twice in the matrix, then their number is at least pq. Otherwise, we choose among the numbers marked once the largest number M, which is one of the largest in its row or column (but not both). Suppose M is one of the largest in its column. Then it is not one of the largest in its row, but all larger numbers in its row are marked twice. We discard this row from the matrix, and we get an (m −1) × n matrix, in which at least q of the largest numbers in each row and at least (p −1) numbers in each column are marked. By the induction hypothesis, at least (p−1)×q numbers are marked twice in this smaller matrix. These numbers are also marked in the larger m×n matrix. In addition, the q numbers of the eliminated row are marked in this matrix. Thus, in the m × n matrix, (p −1)q + q pq numbers are marked twice.
27. The result is obviously true for n 2. Suppose we have already proved the theorem for all k < n. Draw any diagonal connecting some a with some b. The circle is split into two parts. One of the parts has k points and the other n−k −2 points. We apply the induction hypothesis to both sides and get 3k + 4 2 + + 3(n −k −2) + 4 2 + + 1 ≤ 3k + 4 2 + 3(n −k −2) + 4 2 + 1 + , which is ⌊(3n + 4)/2⌋. Hence the theorem is valid for n.
28. The theorem is trivial for n 0. Suppose the theorem is valid for n 2k. From 2k+2 −1 integers, we can select three times 2k integers which, by the induction hypothesis, have a sum divisible by 2k. By the box principle, two of these three sums have the same remainder upon division by 2k+1. The sum of these two sums is a sum of 2k+1 numbers divisible by 2k+1.
8. The Induction Principle 217 29. If N is a Fibonacci number, the theorem is trivial. For small N, we check it by inspection. Assume it to be true for all integers up to and including Fn, and let Fn+1 ≥N > Fn. Now, N Fn+(N −Fn), and N ≤Fn+1 < 2Fn, i.e., N −Fn < Fn.
Thus N −Fn can be written in the form N −Fn Ft1 + · · · + Ftr , ti+1 ≤ti −2, tr ≥2, and N Fn + Ft1 + Ft2 + · · · Ftr . We can be certain that n ≥t1 + 2, because, if we had n t1 + 1, then Fn + Ft1+1 2Fn. But this is larger than N. In fact, Fn must appear in the representation of N because no sum of smaller Fibonacci numbers, obeying ki+1 ≤ki −2 (i 1, 2, . . . r −1) and kr ≥2, could add up to N. This follows, if n is even, say 2k, from F2k−1 + F2k−3 + · · · + F3 (F2k −F2k−2) + (F2k−2 −F2k−4) + · · · + (F4 −F2), which is F2k −1, and if n is odd, say 2k −1, it follows from F2k + F2k−2 + · · · + F2 (F2k+1 −F2k−1) + · · · + (F3 −F1) F2k−1 −1.
Again, the largest Fi not exceeding N −Fn must appear in the representation of N −Fn, and it cannot be Fn−1. This proves uniqueness by induction.
30. Let f (n) be the number of squares on which the knight can be after n moves. We have f (0) 1, f (1) 8, f (2) 33. For n 3, the reachable squares fill all white squares of an octagon with four white squares as sides. By induction you can prove that, for n ≥3, the reachable squares fill an octagon with (n + 1) cells of the same color on each side. It is easy to count the number of unicolored cells of such an octagon. We complete it to a square of 4n+1 cells. It has [(4n+1)±1]/2 unicolored squares. The + sign is for even n and the −sign for odd n. We must add 4 [(n −1) + (n −3) + · · ·] n2 if n is even, n2 −1 if n is odd redundant cells. Hence, the number of cells is (4n + 1)2 + 1 2 −n2 (4n + 1)2 −1 2 −(n2 −1) 7n2 + 4n + 1.
Thus, f (n) 1 for n 0; 8 for n 1; 33 for n 2; 7n2 + 4n + 1 for n ≥3.
31. (a) Experimentation suggests that r l + p + 1.
(1) We will prove (1) by induction on the number of lines. Fig. 8.1 suggests that (1) is correct for l 0. Suppose formula (1) is correct for some number l of lines. We show that it remains valid if another line is added. Take another line. Suppose it intersects s lines. The s new points of intersection split the new line into (s + 1) segments and each segment splits an old region into two. Thus l increases by 1, p increases by s, and r increases by s + 1. Formula (1) remains valid since both sides are increased by s + 1.
218 8. The Induction Principle Fig. 8.1. l 0, p 0, r 1.
T T T T a3 4 @ @ @ @ @ @ a4 8 Fig. 8.2 q q q q q LLL # # # c c c Q Q Q Q B B B B B Q Q Q Q a5 16 (b) We have a1 1 and a2 2. Fig. 8.2 suggests that an 2n−1 for all n. We cannot use six equally spaced points on the circle to find a6 since three chords would pass through the center of the circle. We get a6 31 instead of 32. One region is missing, so our guess was not correct. It is easy to find the correct value of an by the formula r p + l + 1. The n points determine l n 2 lines and p n 4 intersection points.
Thus, an n 4 + n 2 + 1.
32. We define an infinite matrix inductively as follows: A0 1, An+1 Bn An An Bn , where Bn is obtained from An by adding 2n to each of its elements.
By easy induction, we can prove that each row and each column of An contains the positive integers from 1 to 2n. The matrix A∞solves the problem.
A0 (1), A1 2 1 1 2 , A2 ⎛ ⎜ ⎜ ⎝ 4 3 2 1 3 4 1 2 2 1 4 3 1 2 3 4 ⎞ ⎟ ⎟ ⎠, A3 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 8 7 6 5 4 3 2 1 7 8 5 6 3 4 1 2 6 5 8 7 2 1 4 3 5 6 7 8 1 2 3 4 4 3 2 1 8 7 6 5 3 4 1 2 7 8 5 6 2 1 4 3 6 5 8 7 1 2 3 4 5 6 7 8 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ .
33. A few cases give us a hint. For n 2, we have x 1, y 2 and x 2, y 1 with sum 1/1 · 2 + 1/2 · 1 1. For n 3, we must consider the pairs (1, 3), (3, 1), (2, 3), (3, 2) with sum 1/1 · 3 + 1/3 · 1 + 1/2 · 3 + 1/3 · 2 1. We conjecture that 8. The Induction Principle 219 Sn 1/xy 1, where x ≤n, y ≤n, x + y > n, gcd(x, y) 1. Suppose this is true for some n. How does Sn+1 differ from Sn? All terms 1/xy from the sum Sn with x + y > n + 1 stay in the sum Sn+1. On transition from n to n + 1 we must delete the terms 1/xy with x + y ≤n + 1 from Sn. These are the fractions of the form 1/x(n + 1 −x). For each such deleted fraction, two other fractions 1/x(n + 1) and 1/(n + 1 −x)(n + 1) must be included. Clearly, if x and n + 1 are coprime, so are n+1−x and n+1. Since 1/x(n+1−x) 1/x(n+1)+1/(n+1−x)(n+1), we have Sn Sn+1.
34. We can arrive at a guess an f (n) in many ways and then prove it by induction.
(a) Starting with a1 1. we compute a2, a3, a3, . . . until we see the formula.
(b) Somewhat easier is to compute, successively, the ratios an+1/an for n 1, 2, 3, . . . and then guess a rule which we prove by induction.
(c) A guess becomes easier if the sequence an is convergent. Then we can replace an+1 and an in the recursion formula by the limit a and consider the difference an−a.
Now it becomes easier to guess the rule. We will use this approach. Replacing an and an+1 by a in an+1 g(an), we get a 1/3 and a 0. Then a1−1 3 1 2 +1 6, a2−1 3 1 22 + 1 3 · 23 , a3−1 3 1 23 + 1 3 · 25 , a4−1 3 1 24 + 1 3 · 27 .
We conjecture that an 1 3 + 1 2n + 2 3 · 4n .
(1) In the recursion formula an+1 g(an), we replace an in the right side by the right side of (1) and, after heavy computation, get an+1 1 3 + 1 2n+1 + 2 3 · 4n+1 .
Remark. The sequence an converges to 2 1 0 x2 dx. The recursion is a “duplication formula” for the parabola y x2. This is the way I discovered it. Of course, there may have been thousands of people who had this idea before.
35. The assertion is obvious for n 3. Suppose we have a proof for (n −1) points. We will prove it for n points. If another point lies on each line through two points, then all points lie on one line (See Chapter 3, E10). Hence there is a line joining only the points A and B. We throw away the point A. Now there are two cases.
(1) All the remaining points lie on one line l. Then we have n different lines: (n −1) lines through A and the line l.
(2) The remaining points are not collinear. By the induction hypothesis, there are at least (n−1) different connecting lines, and they are all distinct from l. Together with the line AB, we have at least n lines.
36. The conditions of the problem imply that s x1 + · · · + xm y1 + · · · · · · + yn is at least 2 (since m ≤s, n ≤s, s < mn). If m n 2, 2 ≤s ≤3, the assertion is easy to check. We prove it in the general case by induction on m + n k, if k ≥4.
Let x1 > y1 be the largest numbers among xi and yl, respectively (1 ≤i ≤m, 1 ≤j ≤n). The case xi yl is obvious. To apply the induction hypothesis to the equality (x1 −y1) + x2 + · · · + xm y2 + · · · yn 220 8. The Induction Principle with k −1 m + n −1 on both sides, it is sufficient to check the inequality s′ y2 + · · · + yn < m(n −1); since y1 > s/n, we have s′ < s −s/n mn(n −1)/n m(n −1).
37. The integer 1007 is divisible by 53. Any two successive terms have the difference 9010 · · · 0 which is divisible by 53. By induction, each term of the sequence is divisible by 53.
38. Proceed as in the preceding problem.
39. We use induction. We have x0 1 + x0 2 2 and x1 + x2 −p. Since p is odd gcd(y0, y1) 1. Suppose now that gcd(yn, yn+1) 1. Then, we prove that gcd(yn+1, yn+2) 1. Indeed, yn+2 (xn+1 1 + xn+1 2 )(x1 + x2) xn+2 1 + xn+2 2 + x1x2(xn 1 + xn 2) −pyn+1 + yn.
Every divisor of yn+2 and yn+1 is also a divisor of yn. Thus yn+2 and yn+1 have the same divisors as yn+1 and yn.
9 Sequences Difference Equations. A sequence is a function f defined for every nonnegative integer n. For sequences one mostly sets xn f (n). Usually we are given an equation of the form xn F(xn−1, xn−2, xn−3, . . .).
Sometimes we are expected to find a ‘closed expression’ for xn. Such an equation is called a functional equation. A functional equation of the form xn pxn−1 + qxn−2 (q ̸ 0) (1) is a (homogeneous) linear difference equation of order 2 (with constant coeffi-cients.) To find the general solution of (1), first we try to find a solution of the form xn λn for a suitable number λ. To find λ, we plug λn into (1) and get λn pλn−1 + qλn−2, λ2 pλ + q, or λ2 −pλ −q 0.
(2) This is the characteristic equation of (1). For distinct roots λ1 and λ2, xn aλn 1 + bλn 2 is the general solution. a and b can be found from the initial values x0, x1.
If λ1 λ2 λ, the general solution has the form xn (a + bn)λn.
(3) 222 9. Sequences E1. A sequence xn is given by means of x0 2, x1 7, and xn+1 7xn −12xn−1.
Find a closed expression for xn.
The characteristic equation λ2 −7λ + 12 0 has roots λ1 3, λ2 4. The general solution xn a · 3n + b · 4n yields a + b 2, 3a + 4b 7 with solutions a b 1 for x0 2 and x1 7. Thus, xn 3n + 4n.
E2. For all x ∈R, a function f satisfies the functional equation f (x + 1) + f (x −1) √ 2f (x).
(1) Show that it is periodic.
With a f (x −1), b f (x), we get f (x + 1) √ 2b −a, f (x + 2) b − √ 2a, f (x + 3) −a, f (x + 4) −b, i.e, f (x + 4) −f (x) for all x, and f (x + 8) f (x) for all x. Thus 8 is a period of f .
E3. Can we replace √ 2 in (1) so that the period has any preassigned value, e.g., 12?
Replacing √ 2 by the golden section t ( √ 5 + 1)/2 with the property t > 0, t2 t +1 we get a f (x−1), b f (x), f (x+1) tb−a, f (x+2) t(b−a), f (x + 3) b −ta, f (x + 4) −a, f (x + 5) −f (x), f (x + 10) f (x). Now f has period 10.
Replacing √ 2 by the positive root of t3 t2 + t + 1, no periodicity was in sight after many steps. Whenever t3 turned up, I replaced it by t2 + t + 1. Is f not periodic in this case?
A second look shows that (1) is a linear difference equation of second order. But the discrete variable n is replaced by the continuous variable x. So we try to find solutions f (x) λx. For the value of λ, we get λ2 −tλ + 1 0 with solutions λ t 2 ± t2 4 −1.
For t < 2 we have the solutions λ t 2 + i 1 −t2 4 , λ t 2 −i 1 −t2 4 , and |λ| |λ| 1.
So λ and its conjugate λ are unit vectors in the complex plane, that is, λ cos φ + i sin φ, λ cos φ −i sin φ.
Thus, λ has period n if λn 1 or λ cos(2π/n) + i sin(2π/n). In particular, it has period 12, if t/2 cos(π/6), t 2 cos(π/6) √ 3. The period is exactly n, if t/2 cos(2π/n) or t 2 cos(2π/n). The positive solution of t3 t2 + t + 1 is t 1.854 . . . < 2. Yet it is unlikely that this irrational number gives a rational multiple of π for the angle φ, the only way to secure periodicity.
9. Sequences 223 E4. A sequence an is defined by a0 0, an+1 √6 + an. Show that an is (a) monotonically increasing (b) bounded above by 3. (c) Find its limit. (d) Find the convergence rate versus its limit.
(a) We have a0 < a1 since 0 < √ 6. Suppose an−1 < an. Add 6 on both sides and take square roots. Since the square root is increasing, we get 6 + an−1 < 6 + an.
By definition this is an < an+1. By the induction principle, an is monotonically increasing.
(b) a0 < 3 since 0 < 3. Suppose an < 3. Add 6 on both sides and take square roots. We get √6 + an < 3, or an+1 < 3. By the induction principle, an is bounded above by 3 for all n.
(c) From (a) and (b), it follows that an has limit a ≤3. To find a, we take limits on both sides. We get a √ 6 + a, a2 −a −6 0 with the positive root a 3, which is the limit.
(d) To find the convergence rate, we compare an −3 with an+1 −3 an+1 −3 6 + an −3 an −3 √6 + an + 3 ≈an −3 6 in the neighborhood of the limit 3. Thus, the linear convergence rate is 1/6, that is, near 3, the distance of an to 3 shrinks six times at each step.
E5. Find the number an of all permutations p of {1, . . . , n} with |p(i) −i| ≤1 for all i.
We use the method of separation of cases.
(1) There are an−1 ways for n staying in its place.
(2) n moves to n −1. Then n −1 is forced to move to n: an−2 cases.
Altogether we have an an−1 +an−2, a1 1, a2 2. Hence an fn+1, where fn is the nth term of the Fibonacci sequence, defined by f1 f2 1, fn+1 fn + fn−1. Its characteristic equation λ2 λ + 1 has solutions α (1 + √ 5)/2, β (1 − √ 5)/2. Prove that fn (αn −βn)/ √ 5.
Let us find the corresponding number bn for a circular arrangement of the num-bers 1 to n. Now, there are five cases.
(1) p(n) n. We are left with a line of (n −1) elements with an−1 fn cases.
(2) p(n) 1, p(1) n. There are an−2 fn−1 ways.
(3) p(n) n −1, p(n −1) n. Again, there are an−2 fn−1 ways.
(4) n →1 →2 →3 →· · · →n −1 →n. One way.
224 9. Sequences (5) n →n −1 →n −2 →· · · →2 →1 →n. One way.
Thus, bn 2 + fn + 2fn−1, or b1 1, b2 2, bn 2 + fn−1 + fn+1, n ≥3, or bn αn + βn + 2.
E6. We define an infinite binary sequence as follows: Start with 0 and repeatedly replace each 0 by 001 and each 1 by 0.
(a) Is the sequence periodic?
(b) What is the 1000th digit of the sequence?
(c) What is the place number of the 10000th one in the sequence?
(d) Try to find a formula for the positions of the ones (3, 6, 10, 13,. . .) and a formula for the positions of the zeros.
(a) We get the infinite binary word as follows: w1 0, w2 001, w3 w2w2w1. By induction we can prove that wk+1 wkwkwk−1. Let ak and bk be the the numbers of zeros and ones in wk. Then ak+1 2ak + ak−1, bk ak−1, tk ak/ak−1, tk+1 ak+1/ak 2 + 1/tk. For n →∞we get t 2 + 1/t or t √ 2 + 1, that is, ak/bk tends to an irrational number. Thus, the sequence is not periodic. If it were periodic, tk would tend to the rational ratio of zeros/ones in one period. For the infinite binary word we have zeros/ones √ 2 + 1, zeros/bits ( √ 2 + 1)/(2 + √ 2) 1/ √ 2, and ones/bits 1/(2 + √ 2). So every (2+ √ 2)th digit is a 1. The nth one should have place number ≈(2+ √ 2)n.
For the nth zero we have place number ≈ √ 2n.
We need the following table for the next questions: n 1 2 3 4 5 6 7 8 9 10 11 12 an 1 2 5 12 29 70 169 408 985 2378 5741 13860 bn 0 1 2 5 12 29 70 169 408 985 2378 5741 an + bn 1 3 7 17 41 99 239 577 1393 3363 9119 19601 (b) The table above shows that place number 1000 is located inside the word W9. But W9 W8W8W7. This word has length 577 + 577 + 239. So the 1000th digit is inside the word W8W8. Expanding further, we get W8W7W7W6. If we shave off W6 at the end and expand the last W7 we get W8W7W6W6W5. Continuing shaving off the tail and expanding the preceding term, we finally get the word W8W7W6W5W5W2 of length 1000. The 1000th digit of the word is the final digit of W2, that is, 1.
(c) Similarly, one gets the word W12W11W9W8W8W6W3W3 ending in the 10000th one. Adding the lengths of the 8 subwords we get 34142, or ⌊10000(2 + √ 2)⌋.
(d) One can prove that the positions of the nth one and nth zero are f (n) ⌊(2 + √ 2)n⌋and g(n) ⌊ √ 2n⌋, respectively. See , pp. 265–266.
9. Sequences 225 Problems 1. The sequence xn is defined by x0 0, xn+1 √4 + 3xn. Show that it is convergent and find its limit. What is the convergence rate near the limiting point?
2. a0 a1 1, an+1 an−1an + 1, (n ≥1). Show that 4 ̸ |a1964.
3. a1 a2 1, an (a2 n−1 + 2)/an−2, (n ≥3). Show that all ai are integers.
4. Can you select from 1, 1/2, 1/4, 1/8, . . . an infinite geometric sequence with sum (a) 1/5? (b) 1/7?
5. a1 a, a2 b, an+2 (an+1 + an)/2, n ≥0. Find limn→∞an.
6. There does not exist a monotonically increasing sequence of nonnegative integers a1, a2, a3, . . . so that anm an + am for all n, m ∈N.
7. Let an 23−1 23+1 · 33−1 33+1 · 43−1 43+1 · · · n3−1 n3+1. Find limn→∞an.
8. a > 0, a0 √a, an+1 √a + an. Find limn→∞an.
9. Let a1 1, an+1 1 + 1/an, n ≥1. Show that an converges versus the positive root of a2 −a −1 0. What is the convergence rate?
10. Let u0, v0, u0 < v0 be given. The sequences un, vn are defined by un (un−1 + vn−1)/2, vn (un−1 + 2vn−1)/3. Prove that both have the same limit L, u0 < L < v0.
11. a1 a2 1, an 1/an−1 + 1/an−2, n ≥2. Find the limn→∞an and the conver-gence rate.
12. a0 > 0, a1 > 0, an √an−1 + √an−2, n ≥2. Find the limn→∞an and the convergence rate.
13. x0 > 0, a > 0, xn+1 (xn + a/xn)/2. Find the limn→∞xn and the convergence rate.
14. Show that the sequence defined by xn+1 xn (2 −axn) , a > 0 converges quadrat-ically versus 1/a for suitable x0.
15. The arithmetic-geometric mean of Gauss. Let 0 < a < b. We define the two se-quences an and bn as follows.
a0 a, b0 b, an+1 anbn, bn+1 an + bn 2 .
(a) Prove that an < an+1, bn > bn+1 and an < bn for all n.
(b) Prove that bn+1 −an+1 (bn −an)2)/8bn+2.
(c) Show that limn→∞an limn→∞bn g with a quadratic convergence rate.
16. Let an be the sum of the first n terms of 1 + 2 + 4 + 4 + 8 + 8 + 8 + · · · and bn be the sum of the first n terms of 1 + 2 + 3 + 4 + 5 + · · ·. Investigate the quotient an/bn for n →∞.
17. a0 0, a1 1, an 2an−1 + an−2, n > 1. Prove that 2k|an ⇔2k|n.
18. All terms of the sequence a1 a2 a3 1, an+1 (1+an−1an)/an−2 are integers.
19. Let a0 0, a1 1. Find all integers an which cannot be represented in the form an ai + 2aj with ai, aj not necessarily distinct. Can you describe these numbers in a simple way?
226 9. Sequences 20. All terms of the sequence a1 a2 1, a3 2, an+3 (an+1an+2 + 5)/an are integers.
21. All terms of the sequence 10001, 100010001, 1000100010001, . . . are composite.
22. A sequence of positive numbers a0, a1, a2, . . . is defined by a0 1, an+2 an − an+1, n ≥0. Show that this sequence is unique.
23. A sequence an is defined by a1 1, an+1 an + 1/a2 n. (a) Is an bounded? (b) Show that a9000 > 30.
24. Three sequences xn, yn, zn with positive initial terms x1, y1, z1 are defined for n ≥1 by xn+1 yn + 1/zn, yn+1 zn + 1/xn, zn+1 xn + 1/yn. Show that (a) none of the three sequences is bounded.
(b) At least one of x200, y200, z200 is greater than 20.
25. The sequence xn is defined by x1 1/2, xk+1 x2 k + xk. Find the integer part of the sum 1 x1 + 1 + 1 x2 + 1 + · · · + 1 x100 + 1.
26. A sequence an is defined by a1 1, a2 12, a3 20, an+3 2an+2 + 2an+1 −an, n ≥0. Prove that, for every n, the integer 1 + 4anan+1 is a square.
27. a1 a2 1, a3 −1, an an−1an−3. Find a1964.
28. A sequence xn is defined by x1 2, xn+1 (x4 n +9)/(10xn). Show that 4/5 < xn ≤ 5/4 for all n > 1.
29. A sequence an is defined by a1 √ 2, an+1 √ 2 an. Find limn→∞an.
30. If a0 a > 1, an+1 aan, then the an converges for a ≤e1/e 1.444667861.
31. The terms of the sequence a1, a2, a3, . . . are positive, and a2 n+1 an + 1 for all n.
Show that the sequence contains irrational numbers.
32. If r > 0 is a rational approximation to √ 5, then (2r + 5)/(r + 2) is an even better approximation. Generalize to √a.
33. Josephus Problem. n persons are arranged in a circle and numbered from 1 to n.
Then every kth person is removed and the circle closes up after each removal. What is the number f (n) of the last survivor?
(a) The problem becomes vastly simplified for k 2. Show that f (2n) 2f (n) −1, f (2n + 1) 2f (n) + 1, f (1) 1.
Find f (100) by means of these recursions.
(b) There is almost an explicit expression for f (n): Let 2m be the largest integer, so that 2m ≤n. Then f (n) 2 (n −2m) + 1.
Prove it and find f (1993) by means of this formula.
(c) Write n in the binary system, and transfer the first digit to the end. Then you will get f (n). Show this, and find f (1000000).
34. A sequence f (n) is defined by f (0) 0, f (n) n −f [f (n −1)], n > 0. Make a table of functional values, guess a formula for f (n), and prove it.
9. Sequences 227 35. Morse–Thue Sequence. Start with 0; to each initial segment append its complement: 0, 01, 0110, 01101001, . . ..
(a) Let the digits of the sequence be x(0), x(1), x(2), . . .. Prove that x(2n) x(n), x(2n + 1) 1 −x(2n).
(b) Prove that x(n) 1−x(n−2k), where 2k is the largest power of 2 which is ≤n.
Find the 1993rd digit of the sequence.
(c) Prove that the sequence is not periodic.
(d) Write the nonnegative integers in base 2: 0, 1, 10, 11, . . .. Now replace each number by the sum of its digits mod 2. You get the Morse–Thue sequence. Prove this.
36. The sequence an is defined as follows: a4n+1 1, a4n+3 0 for n ≥0, and a2n an for n ≥1. Show that this sequence is not periodic.
Remark. These digits can be used to draw a curve as follows: Start at the origin and go one step to the right. If the next bit is 1, then turn left by 90◦and go one step forward. If the next bit is 0 turn right by 90◦and go one step forward. You get a strange curve with many regularities, which is called a “dragon curve.” 37. Find a recursion for the number an of permutations p of {1, . . . , n} with |p(i)−i| ≤2 for all i.
38. Three sequences xn, yn, zn, n 1, 2, . . . are defined as follows: x1 2, y1 4, z1 6 7, xn+1 2xn x2 n −1, yn+1 2yn y2 n −1, zn+1 2zn z2 n −1.
(a) Show that this construction can be extended indefinitely.
(b) At some stage can we get xn + yn + zn 0 (ARO 1990)?
39. Given a set of positive numbers, the sum of the pairwise products of its elements is equal to 1. Show that it is possible to eliminate one number so that the sum of the remaining numbers is less than √ 2 (ARO 1990).
40. Find the sum Sn 1/1 · 2 · 3 · 4 + · · · + 1/n(n + 1)(n + 2)(n + 3).
41. The sequence xn is defined by x1 2, xn+1 2 + xn 1 −2xn , n 1, 2, 3, . . . .
Prove that (a) xn ̸ 0 for all n; (b) xn is not periodic.
42. A sequence is defined as follows: a1 3, and an+1 an/2 if an is even, (an + 1983)/2 if an is odd.
Prove that it is periodic and find its minimal period.
43. Investigate the sequence an n 0 −1 + n 1 −1 + · · · + n n −1 .
Is it bounded? Does it converge for n →∞?
228 9. Sequences 44. Does there exist a positive sequence an, such that an and 1/(n2an) are conver-gent?
45. The positive real numbers x0, . . . , x1995 satisfy x0 x1995 and xi−1 + 2 xi−1 2xi + 1 xi for i 1, . . . , 1995. Find the maximum value that x0 can have (IMO 1995).
46. Let k ∈N. Prove that there exists a real r > 1, such that k|⌊rn⌋for all n ∈N.
47. (IMO 1993.) Let n > 1 be an integer. There are n lamps L0, . . . , Ln−1 arranged in a circle. Each lamp is either ON or OFF. A sequence of steps S0, . . . , Si, . . . is carried out. Step Sj affects the state of Lj only (leaving the states of all other lamps unaltered) as follows: If Lj−1 is ON, Sj changes the state of Lj from ON to OFF or from OFF to ON; If Lj−1 is OFF, Sj leaves the state Lj unchanged.
The lamps are labeled mod n, that is, L−1 Ln−1, L0 Ln, L1 Ln+1. Initially all lamps are ON. Show that (a) there is a positive integer M(n) such that after M(n) steps all the lamps are ON again; (b) if n has the form 2k, then all lamps are ON after (n2 −1) steps; (c) if n has the form 2k + 1, then all the lamps are ON after (n2 −n + 1) steps.
48. The sequence an is defined by a1 0, |a2| |a1 + 1|, . . . |an| |an−1 + 1|. Prove that a1 + a2 + · · · + an n ≥−1 2.
49. Of the sequence a0, a1, . . . , an it is known that a0 an 0 and that ak−1 −2ak + ak+1 ≥0 for all k 1, . . . , n −1. Prove that ak ≥0 for all k.
50. Given are the positive integers a0, . . . , a100 such that a1 > a0, a2 3a1 −2a0, a3 3a2 −2a1, . . . , a100 3a99 −2a98. Prove that a100 > 299.
51. Start with two positive integers x1, x2, both less than 10000, and for k ≥3 let xk be the smallest of the absolute values of the pairwise differences of the preceding terms. Prove that we always have x21 0 (AUO 1976).
52. The sequence a0, a1, a2, . . . is such that, for all nonnegative m, n (m ≥n), we have am+n + am−n (a2m + a2n)/2. Find a1995 if a1 1.
53. Can the numbers 1, . . . , 100 belong to 12 geometrical progressions?
54. Prove that, for any positive integer a1 > 1 there exists an increasing sequence of positive integers a1, a2, a3, . . ., such that a2 1 + · · · + a2 k is divisible by a1 + · · · + ak for all k ≥1 (RO 1995).
55. The infinite sequence xn is defined by 0 ≤x0 ≤1, xn+1 1 −|1 −2xn|. Prove that the sequence is periodic iff x0 is rational.
56. The sequence x1, x2, . . . of positive integers is defined as follows: 1, 2, 4, 5, 7, 9, 10, 12, 14, 16,. . . . Find a formula for xn.
9. Sequences 229 57. Prove that, for any sequence an of positive integers, the integer parts of the square roots of the all bn defined below are different: bn (a1 + · · · + an)(1/a1 + · · · + 1/an).
The following problems treat the number an of ways to tile a k × n rectangle by various smaller tiles. A solution is here a recurrence for an.
58. Let an be the number of ways to tile a 2 × n rectangle by 2 × 1 dominoes.
(a) Find an. (b) Find the number of symmetric and distinct tilings.
59. In how many ways can you tile a 2 × n rectangle by 2 × 1 or 2 × 2 tiles?
60. In how many ways can you tile a 2 × n rectangle by 1 × 1 squares and L-trominoes?
61. In how many ways can you tile a 2 × n rectangle by 2 × 2 squares and L-trominoes?
62. In how many ways can you tile a 3 × n rectangle by 2 × 1 dominoes?
63. In how many ways can you tile a 4 × n rectangle by 3 × 1 dominoes?
64. In how many ways can you tile a 2 × n rectangle by 1 × 1 or 2 × 1 tiles?
65. In how many ways can you tile a 4 × n rectangles with 2 × 1 dominoes?
66. In how many ways can you fill a 2 × 2 × n box with 1 × 1 × 2 bricks? A table suggests that the values a2n are squares. Can you prove this?
Solutions 1. By induction we show that an < an+1 for all n ∈N. We show that an < 4 for all n ∈N. First a0 < 4. Now, let an < 4. Then √4 + 3an < √ 4 + 3 · 4, or an+1 < 4. A monotonic and bounded sequence has a limit L, which can be found from L2 4 + 3L. The positive solution is 4. Now we consider |an+1 −4| | 4 + 3an −4| |4 + 3an −16| √4 + 3an + 4 3 |an −4| √4 + 3an + 4 ≈3 8|an −4| for an near its limit 4. Thus, 3/8 is the linear convergence rate.
2. We consider the sequence (mod 4): 1, 1, 2, 3, 3, 2, 3, 3, . . .. It has period 2, 3, 3 and does not contain a zero.
3. The sequence has the equivalent form anan−2 a2 n−1 + 2. Replace n by n + 1: an+1an−1 a2 n + 2. Subtraction and trivial transformation yields an+1 + an−1 an an + an−2 an−1 c, a constant. The initial conditions give c 4, that is, an+1 4an −an−1.
4.
1 2a + 1 2a+b + 1 2a+2b + · · · 1 m ⇒1 2a 1 1 −1 2b 1 m ⇒ 2b−a 2b −1 1 m.
If a b, then we have 2b −1 m, which is possible for m 7, but impossible for m 5. If a ̸ b, then either the numerator or the denominator is even. This is impossible for odd m. Thus, 1 7 1 23 + 1 26 + 1 29 + · · · .
230 9. Sequences a1 a2 a3 a4 a5 Fig. 9.1 5. Looking at Fig. 9.1 we see that lim n→∞an a + b −a 2 + b −a 8 + · · · a + 2 3(b −a) a + 2b 3 .
6. For a strictly increasing function an, we have a2n an + a2 ≥a2 + (n −1). This is impossible for any finite value a2.
7. We have n k2 k3 −1 k3 + 1 n k2 k −1 k + 1 n k2 k2 + k + 1 k2 −k + 1.
The first product is 2/(n(n + 1)). To find the second product, we observe that if bk k2 + k + 1, ck k2 −k + 1, then ck bk−1. Hence, the second product is (n2 + n + 1)/3. Finally, lim n→∞ 2 3 n2 + n + 1 n2 + n 2 3.
8. We have a2 n+1 an+a. It is easy to see that an increases. We show that an is bounded above, which guarantees a limit L. We have a2 n+1 −an −a 0.
Since an < an+1, we have a2 n −an −a < 0, or an − √ 4a + 1 + 1 2 an + √ 4a + 1 −1 2 < 0.
The second parenthesis is positive, so the first must be negative, that is, an < √ 4a + 1 + 1 2 .
Hence, an has a limit L > 0 which can be found from L2 −L −a 0. Thus, L √ 4a + 1 + 1 2 .
9. Here we will profit from Chapter 8. There you analyzed the behavior of the Fibonacci sequence defined by F1 F2 1, Fn+2 Fn+1 + Fn, n > 0. From a small table of the sequence an, we guess that an Fn+1/Fn, and we prove this by induction.
From Chapter 8 we also know that lim n→∞an a, a 1 + √ 5 2 , a2 a + 1.
9. Sequences 231 To get the convergence rate, we consider the equation x f (x), where f (x) 1 + 1/x. If we try to find the fixed point by iteration, we get our sequence. To get the convergence rate, we interpret f (x) as a mapping of the x-axis to itself. Then f ′(x) can be interpreted as the local contraction in the neighborhood of x. Since f ′(x) −1/x2,wehave,fortheconvergencerateata,f ′(a) −1/a2 ≈−1/2.618.
Since |f ′(a)| < 1, we have indeed a contraction, not an expansion.
10. From vn−un (vn−1−un−1)/6, we conclude that at each step the difference between vn and un is reduced six times. So un and vn have the same limit, and lim n→∞un u0 + v0 −u0 2 + v0 −u0 2 · 6 + v0 −u0 2 · 62 + · · · 2u0 + 3v0 5 .
11. From the equation a 1/a + 1/a, we get for the positive fixed point a √ 2. We use the transformation bn 1/an and get the new recursion 1 bn bn−1 + bn−2.
In this new equation we consider the relative error bn (1 + ϵn)/ √ 2. We get 1 √ 2 (1 + ϵn+1) √ 2 1 + ϵn + 1 + ϵn−1 .
From here we get ϵn+1 − ϵn + ϵn−1 2 + ϵn + ϵn−1 .
The convergence rate is the limiting convergence speed as the relative error tends to zero. In this case we have for ϵn the recursion ϵn −ϵn−1 + ϵn−2 2 with the characteristic equation λ2 + λ/2 + 1/2 0 with solutions λ −1 4 + √ 7 4 i, λ −1 4 − √ 7 4 i.
|λ| 1/ √ 2 ≈0.707 is the convergence rate.
12. (a) Let 0 < a0 ≤a1 < 1. We have a2 √a1 + √a0 > a1, an+1 −an √an + √an−1 −√an−1 −√an−2 √an −√an−1 + √an−1 −√an−2 .
(1) Hence, an increases, and by induction we prove that an ≤4, n ≥1. This guarantees a limit L satisfying L 2 √ L with solution L 4.
(b) Let 0 < a1 < a0 < 1. Then a2 > a1, a2 > a0 and since an+1−an √an−√an−2, we have a3 > a2. From (1), we get a1 < a2 < a3 < a4 < · · ·.
(c) Suppose now that a0 ≥1 or a1 ≥1. Then a2 √a1 + √a0 > 1, a3 √a2 + √a2 > 2 > 1, and by induction we get an > 1, n ≥1. Let us denote xn |an −4|. We observe that xn ≤|an−1 −4| √an−1 + 2 + |an−2 −4| √an−2 + 2 < 1 3 (xn−1 + xn−2) .
232 9. Sequences This inequality can be written in the form xn + √ 13 −1 6 xn−1 ≤ √ 13 + 1 6 xn−1 + √ 13 −1 6 xn−2 , n ≥2.
For n →∞, this yields 0 ≤xn < xn + √ 13 −1 6 xn−1 ≤ √ 13 + 1 6 n−1 x1 + √ 13 −1 6 x0 →0, that is, xn →0, n →∞, or an →4, n →∞.
For the convergence rate we set an √ 2(1 + ϵn) and, after some manipulations, get ϵn ϵn−1 2(√1 + ϵn−1 + 1) + ϵn−2 2(√1 + ϵn−2 + 1) ≈ϵn−1 + ϵn−2 4 .
Of the two roots of the characteristic equation, the larger one λ (1 + √ 17)/8 is the convergence rate. It is slightly larger than 5/8.
13. This is the school method of “Divide and Average” for finding √a. Possible can-didates for limits are the solutions of x (x + a/x)/2, or x √a, since x > 0.
Setting xn √x(1 + ϵn) and plugging this into the iteration equation, after simple algebra, we get ϵn+1 ϵ2 n 2(1 + ϵn).
For large ϵn we have ϵn+1 ≈ϵn/2. But for small ϵn we have ϵn+1 ≈ϵ2 n/2, and this is quadratic convergence. At each iteration step, the number of correct digits about doubles.
14. Setting xn (1 −ϵn)/a, we get ϵn+1 ϵ2 n. We have quadratic convergence versus 1/a for |ϵ1| < 1.
15. (a) We have a0 < b0. Suppose an < bn for any n. Then bn+1 is the midpoint between an and bn and an+1 is the geometric mean of an and bn, and is less than their arithmetic mean. Thus we have an+1 < bn+1, an < an+1, bn > bn+1 for all n.
(b) bn+1 −an+1 an + bn 2 −√anbn (√bn −√an)2 2 , √bn −√an bn −an √an + √bn , bn+1 −an+1 (bn −an)2 2(√an + √bn)2 (an −bn)2 2(an + bn + 2an+1), or bn+1 −an+1 (bn −an)2 2(2bn+1 + 2an+1) (bn −an)2 8bn+2 .
(c) This follows from (b).
16. Let an 1 + 2 + 4 + 4 + 8 + 8 + 8 + 8 + · · · + 2k+1 + · · · 2k+1 (2k−1 terms 2k and m terms 2k+1).
9. Sequences 233 The summation yields an (1+3n·2k+1+1)/3 with n 2k+m and 0 ≤m ≤2k−1.
Elimination of m gives 2k ≤n ≤2k+1 −1.
(∗) Hence, we write an 1 3 1 + 3n · 2k+1 −22(k+1) , bn n(n + 1) 2 .
Thus, for the general term qn of the sequence, we get an bn 2 3 1 + 3n · 2k+1 −22(k+1) n(n + 1) 4 3 1/22k−1 + 3n/2k −2 n n/2k + 1/2k /2k .
From (∗) we have 1 ≤n/2k ≤2 −1/2k and hence 1 ≤x limk, n→∞n/2k ≤2, that is, lim n→∞ an bn 4 3 3x −2 x2 with 1 ≤x ≤2.
The sequence qn has no limit; all real numbers of the closed interval [4/3, 3/2] are limit points.
17. First solution. We compute a small table for checking formulas.
n 0 1 2 3 4 5 6 7 8 9 10 an 0 1 2 5 12 29 70 169 408 985 2378 We check that an+1 a2an + a1an−1, an+2 a3an + a2an−1. From these data we guess the general formula an+m anam+1 + aman−1.
(1) For m n we get from (1) a2n an(an+1 + an−1).
(2) We prove (1) by induction. We see from the table and easily check by induction that an ≡1 mod 4 for odd n. If n is even, both n −1 and n + 1 are odd, and we have an−1 ≡an+1 ≡1 mod 4 and an−1 + an+1 ≡2 mod 4. Thus just one more factor 2 is contributed by the parentheses in (2). This proves the result.
Second solution. The shift T : (an−1, an) →(an, an+1) (0·an−1 +1·an, 1·an−1 + 2 · an), is a linear transformation with the matrix 0 1 1 2 or a0 a1 a1 a2 . By induction we prove that 0 1 1 2 n an−1 an an an+1 .
Consider a few powers of the matrix T 2 1 2 2 5 , T 3 2 5 5 12 , T 4 5 12 12 29 , T 5 12 29 29 70 ,T 6 29 70 70 169 ,T 7 70 169 169 408 ,T 8 169 408 408 985 .Weseethat2k | an ⇐ ⇒ 2k | n is valid for small values of n. In addition, for k ≥1, the elements x in the main diagonal satisfy x ≡1 mod 4. Now, suppose a b b c 0 1 1 2 n. Then 0 1 1 2 2n a2 + b2 b(a + c) b(a + c) b2 + c2 , 234 9. Sequences with a ≡c ≡1 mod 4 and an 2k · q, q odd. Hence, a2n b(a + c). Since a + c ≡2 mod 4, just one new factor 2 is added to b. This proves the theorem, because a b b c 0 1 1 2 2 a + 2b b + 2c 2a + 5b 5c + 2b .
Again, a + 2b ≡5c + 2b ≡1 mod 4, since b is even by the induction assumption (k ≥1).
18. The table for an suggests an+2 4an −an−2, (n 3, 4, 5, . . .). We prove this by induction. Suppose that the formula is valid for n −1. That is, an−1an+2 1 + an+1an 1 + (4an−1 −an−3) an 4an−1an −an−1an−2, an+2 4an −an−2.
Empirically we can also find a2k+1 2a2k −a2k−1 and a2k+2 3a2k+1 −a2k for k 1, 2, . . .. We can use induction based on these conjectures.
19. We find the following table empirically: n 1 2 3 4 5 6 7 8 9 an 0 1 4 5 16 17 20 21 64 We conjecture that, apart from a1 0, the an are those positive integers, which are representable as sums of distinct powers of 4.
Proof. In base 2 every integer has a unique representation n 2a + 2b + · · ·. Of the odd powers of 2, we split off the factor 2, and we get n (2r + · · ·) + 2 (2s + · · ·) bi + 2bj, where each exponent r, s, . . . is even, so that bi, bj are sums of distinct powers of 4. Is the representation unique? Suppose n ai + 2aj a′ i + 2a′ j are distinct representations. We subtract common powers of 4 from ai, a′ i as well as from aj, a′ j, and we get two different binary representations of the same positive integers. Thus the representation n ai + 2aj is unique.
20. Try to treat this recurrence the same way as problems 3 or 19.
21. For k 1 we have 1 + x4 10001 73 · 137. For k > 1, we have 1 + x4 + . . . + x4k x4k+4 −1 x4 −1 x2k+2 −1 x2 −1 · x2k+2 + 1 x2 + 1 .
For k > 1, both factors on the RHS are greater than 1.
22. We set a1 t. Then a2 1 −t > 0, a3 2t −1 > 0, a4 2 −3t > 0, a5 5t−3 > 0,a6 5−8t > 0.Thust < 1, t > 1/2, t < 2/3, t > 3/5, t < 5/8.
By induction we prove that F2n F2n+1 < t < F2n+1 F2n+2 for all n.
But lim n→∞ Fn Fn+1 t with the positive root t √ 5 −1 2 and t2 1 −t.
Obviously this number satisfies the conditions of the problem since 1 −t t2, t −t2 t3, . . . , tn −tn+1 tn+2, . . . .
9. Sequences 235 23. an+1 an + 1/a2 n ⇒a3 n+1 a3 n + 3 + 3/a3 n + 1/a6 n > a3 n + 3. Since a3 2 1 + 3 + 3 + 1 > 2 · 3, we get a3 n > 3n by induction.
(a) Since an > 3 √ 3n, the sequence is not bounded.
(b) a9000 > 3 √ 27000 30.
24. Suppose xn is not bounded. Then zn is not bounded because of the third equation, and yn is not bounded because of the second equation. We consider the behavior of a2 n (xn + yn + zn)2. Since x + 1/x ≥2 for x > 0, we observe that a2 2 (x1 + 1/x1 + y1 + 1/y1 + z1 + 1/z1)2 ≥36 2 · 18. Now a2 n+1 (xn + yn + zn + 1 xn + 1 yn + 1 zn )2 > a2 n + 2 (xn + yn + zn) 1 xn + 1 yn + 1 zn ≥a2 n + 18.
By induction we get a2 n > 18n for n > 2. Thus, a2 200 > 3600, x200 +y200 +z200 > 60.
So at least one of x200, y200, z200 is greater than 20.
25. xk+1 x2 k + xk ⇒1/xk+1 1/xk(1 + xk) 1/xk −1/(1 + xk). We get 1 x1 + 1 + 1 x2 + 1 + · · · + 1 x101 + 1 1 x1 −1 x2 + · · · + 1 x100 − 1 x101 1 x1 − 1 x101 , and this is 2 −1/x101. The integer part is 1 since x101 > 1.
26. Use induction.
27. By computing the first 10 terms of the sequence, we observe that the sequence starts with 1, 1, −1, −1, −1, 1, −1 period , 1, 1, −1. The last three terms certify the period.
Since 1964 7 · 280 + 4, we have a1964 −1.
28. All the terms of the sequence are positive. We have xn+1 x4 n + 9 10xn x3 n 10 + 3 10xn + 3 10xn + 3 10xn ≥4 4 x3 n 10 · 3 10xn · 3 10xn · 3 10xn > 2 5 4 √ 27 > 4/5.
Here we used the arithmetic mean-geometric mean inequality. Now we show that xn ≤ 5 4. First we observe that x2 5/4. Then we find out when xn+1 ≤xn, i.e., xn ≥ x4 n + 9 /10xn, or x4 n −10x2 n + 9 ≤0. This inequality is valid for 1 ≤x2 n ≤9.
From this we conclude that, for 1 ≤xn ≤5/4, we have xn+1 ≤5/4. But if xn < 1, then xn 9 + x4 n /10xn < 10/10xn < 5/4.
29. We have a1 < a2 since √ 2 < √ 2 √ 2. Let an−1 < an. For a > 1 the function ax is increasing. Thus, √ 2 an−1 < √ 2 an, or an < an+1. By induction the sequence an is monotonically increasing. We show that an < 2 for all n. Indeed, a1 < 2. Suppose an < 2. Then √ 2 an < √ 2 2, or an+1 < 2. By induction an is bounded above by 2.
Hence, it has limit L ≤2. We find it from L √ 2 L with solution L 2.
236 9. Sequences 30. a0 < a1 since a < aa. Let an−1 < an. Then aan−1 < aan, or an < an+1. By induction an increases monotonically. If it converges, then its limit L can be found from the equation L aL. We can show that there is convergence for 1 < a ≤e1/e 1.44466 . . . . The maximum value can be found from L eL/e which has solution L e. We will show, for a ≤e1/e, that an is increasing and bounded above by e.
Let an ≤e. Then an+1 aan ≤ e1/e e e.
31. Suppose all terms of the sequence are positive rationals, an pn/qn, gcd(pn, qn) 1. Then a2 n+1 an + 1 pn qn + 1 pn + qn qn p2 n+1 q2 n+1 , or q2 n+1 qn for all n.
Then qn+1 (q1)1/2n is a positive integer for all n > n0. Now an 1 implies an+1 √ 2, a contradiction. Hence an > 1 for all n > n0. For these n, we have a2 n+1 −a2 n an + 1 −a2 n 1 + an (1 −an) < 0, or an+1 < an for all n > n0, that is, we have an infinite strictly decreasing sequence of positive integers. Contradiction!
Thus the existence of a sequence of positive rationals satisfying a2 n+1 an + 1 leads to a contradiction.
32. (2r+5)/(r+2)− √ 5 ( √ 5−2)( √ 5−r)/(r+2).Now( √ 5−2)/(r+2) < ( √ 5−2)/2, which is less than 0.15. In general, comparing r and (br + a)/(r + b), we get br + a r + b −√a b −√a r + b r −√a .
If b is a good approximation to √a, we get a quickly converging sequence.
t t t t t t t t t t t t 1 1 2 3 2 3 4 f (n) f (2n) 5 6 7 4 8 2n Fig. 9.2. f (2n) 2f (n) −1.
33. We express f (2n) and f (2n + 1) in terms of f (n). In Fig. 9.2 with 2n persons around the circle, we eliminate numbers 2, 4, . . . , 2n, and we are left with numbers 1, 3, . . . , 2n −1 which are renumbered 1, 2, . . . , n. In Fig. 9.3 with 2n + 1 persons we eliminate numbers 2, 4, . . . , 2n, 1, and we are left with numbers 3, 5, . . . , 2n+1 which are renumbered 1, 2, . . . , n. Since f (n) denotes the last survivor on the inner circle, we see that his original number (on the outer circle) is f (2n) 2f (n) −1 or f (2n + 1) 2f (n) + 1, f (1) 1. These recursions give f (100) 73.
9. Sequences 237 t t t t t t t t t t t t 1 2 3 1 4 f (n) f (2n + 1) 5 2 6 7 3 8 2n 2n+1 Fig. 9.3. f (2n + 1) 2f (n) + 1.
(b) First we note that f (n) 1 for n 2m. For arbitrary n, let m be the largest integer such that 2m ≤n. We write n 2m+(n−2m). Now we remove the persons numbered 2, 4, 6, . . . , 2(n −2m), leaving 2m persons in the circle. By the above result, the first one of these 2m persons will survive. The place number of this one is 2(n −2m) + 1.
Hence, for k 2, the number of the last survivor is f (n) 2(n −2m) + 1, where 2m is the largest power of 2 ≤n. Thus, f (1993) 2(1993 −1024) + 1 1939.
(c) In binary n 1b1b2 · · · bm 2m+(n−2m), f (n) 2(n−2m)+1. f (1000000) 2(1000000 −219) + 1 951425.
34. Answer: f (n) ⌊(n + 1)t⌋, where t ( √ 5 + 1)/2.
35. (a) Start with the digit 0, and repeatedly use the replacement rule T : 0 →01, 1 → 10. Thus T (0) 01, T 2(0) T (01) 0110, . . .. We have an alternative method of forming the Morse–Thue sequence, T n(0) being the first 2n digits of the sequence.
Applying T to the whole sequence leaves it invariant. This makes (a) almost obvious, and (b) also. To prove (c) we note that x(2n) x(n) and x(2n + 1) 1 −x(n) are always different. If the sequence were ultimately periodic and x(n) were in the periodic part of the sequence, we could conclude that n + 1 is not a multiple of the period. The same would be true of n + 2, n + 3, . . ., but this is impossible.
(d) This is true because the sequence of (n + 1)-digit binary numbers is obtained from the sequence of all numbers up to n digits by putting a 1 and possibly some 0′s in front of them.
36. Let T 2rq (q odd) be the period of the sequence. If q 4m + 1 and k ≥r + 2, then 1 a2k a2k+T a2k+2r(4m+3) a2k−r +4m+3 a4P +3 0. If q 4m + 3, then 1 a2k a2k+3T a2k+3·2r (4m+1) a2k−r +3(4m+1) a4P +3 0. Both cases lead to the contradiction 1 0. Thus the sequence is not periodic.
37. The method of separation of cases is barely feasible now. In E5 it was quite easy.
We look again at the tail of the permutation.
238 9. Sequences tail # of permutations (n) an−1 (n, n −1) an−2 (n, n −1, n −2) an−3 (n, n −2, n −1) an−3 (n −1, n, n −2) an−3 (n −1, n −3, n, n −2) an−4 (n −1, n, n −3, n −2) an−4 (n −2, n, n −3, n −1) an−4 (n −2, n −4, n, n −3, n −1) an−5 (n −3, n −1, n −4, n, n −2) an−5 The last two lines show easily that there are also two terms a6. Similarly there are two terms a7, a8, a9, . . .. Consequently, we have an an−1 + an−2 + 3an−3 + 3an−4 + 2an−5 + 2an−6 + · · · .
Shifting the index n ←n + 1 and subtracting, we get an+1 2an + 2an−2 −an−4, a0 1, a1 1, a2 2, a3 6, a4 14.
The recursion easily gives a5 31, a6 73, a7 172, a8 400.
We can make the problem simpler by introducing bn # of permutations p(n) such that n →n −1 with all other conditions satisfied. Then we get quite easily an an−1 + bn + bn−1 + an−3 + an−4, bn an−2 + an−3 + bn−2. Eliminating bn, we get the same recurrence an+1 2an + 2an−2 −an−4.
38. (a) We will show that the denominator of any term can never become zero. Indeed, suppose we get a triple (A, B, C) with A 1. Then for the preceding triple (a, b, c) we get 2a/(a2−1) 1, or a2−2a−1 0 with solutions a 1± √ 5. But all triples (xn, yn, zn) are rational numbers. A −1 and all other cases are treated similarly.
(b) We have x1 + y1 + z1 x1y1z1 48/7. We will show in a moment that xn + yn + zn xnynzn ⇒xn+1 + yn+1 + zn+1 xn+1yn+1zn+1. By induction, then, we have xn + yn + zn xnynzn for all n ≥1. But if at some stage xn + yn + zn 0, then at least one of the numbers in xn, yn, zn is zero. This is not possible.
We will drop the subscripts. Then we know that x + y + z xyz. We must show that 2x x2 −1 + 2y y2 −1 + 2z z2 −1 2x x2 −1 · 2y y2 −1 · 2z z2 −1.
This can be done by brute force. Putting the left side on a common denominator, we get the numerator 2x(y2 −1)(z2 −1) + 2y(x2 −1)(z2 −1) + 2z(x2 −1)(y2 −1) 2(x + y + z) + 2xyz(xy + yz + zx) −2(x + y + z)(xy + yz + zx) + 6xy 8xyz.
A more clever approach is to see that the duplication formula for tan is involved.
tan 2u 2 tan u 1 −tan2 u.
9. Sequences 239 We set x −tan u, y −tan v, z −tan w. Now we must prove that tan u + tan v + tan w tan u · tan v · tan w ⇒tan 2u + tan 2v + tan 2w tan 2u · tan 2v · tan 2w. We use the formula tan(u + v + w) tan u + tan v + tan w −tan u tan v tan w 1 −tan u tan v −tan v tan w −tan w tan u.
Now we see that tan(u + v + w) 0 ⇔u + v + w ≡0 (mod π) ⇔tan u + tan v + tan w tan u tan v tan w ⇒2u + 2v + 2w ≡0 (mod π) ⇔tan 2u + tan 2v + tan 2w tan 2u tan 2v tan 2w.
39. Let the set of numbers on the blackboard be {a1, . . . , an} with S a1 + · · · + an.
From the condition i<k aiak 1, we get 2 a1(S −a1) + a2(S −a2) + · · · + an(S −an).
Suppose that S −ak ≥ √ 2 for all k 1, 2, . . . , n. Then 2 ≥a1 · √ 2 + a2 · √ 2 + · · · + an · √ 2 √ 2 · S, that is, √ 2 ≥S. On the other hand, S > S −a1 ≥ √ 2. Contradiction!
40. We transform the kth term into a form, which gives a telescoping series: 1 k(k + 1)(k + 2)(k + 3) 1 3 1 k(k + 1)(k + 2) − 1 (k + 1)(k + 2)(k + 3) .
Summing from k 1 to n, we get Sn 1/18 −1/3(n + 1)(n + 2)(n + 3).
41. We prove by induction that xn tan nα, where α arctan 2. For n 1, this is true. Now, let xn tan nα. Then xn+1 2 + xn 1 −2xn tan α + tan nα 1 −tan a tan nα tan(n + 1)α, q.e.d. We observe that, for any m, x2m tan 2mα 2 tan mα 1 −tan2 mα 2xm 1 −x2 m .
(1) Now we prove (a) by contradiction. If xn 0 and n 2m is even, then by (1) xm 0. But ifn 2k(2s+1) with nonnegative integersk, s, then afterk steps, we get x2s+1 0. Hence, (2+x2s)/(1−2x2s) 0 ⇒x2s −2 ⇒2xs/(1−x2 s ) −2. Both roots of this equation are irrational, but all xs must be rational, since (2+x)/(1−2x) is rational for any x because the initial value x1 2 is rational. Contradiction!
(b) We will prove more than nonperiodicity. The sequence xn assumes any of its values only once. Suppose xn+m−xn 0 for some n, m, m ≥1. Since xn tan nα, we have tan(n + m)α −tan nα sin mα cos(n + m)α cos nα 0.
Hence, xm tan mα 0. But this is impossible because of (a).
42. The terms of the sequence are positive integers, which are smaller than 1983. Thus we do not change them if we consider them mod 1983. Then the algorithm generating the sequence becomes an ≡2an+1, and thus a1 ≡2nan+1. The congruence 3 ≡ 2n · 3 mod 1983 is satisfied, if 1983|3 · 2n −3, i.e., 661|2n −1. By Euler’s theorem, n φ(661) 660. Thus the period is 660 or a divisor of this number. A check shows that the period is indeed 660. We need to check only divisors up to 330. We get 2330 ≡−1 mod 661. So 2660 ≡1 mod 661.
240 9. Sequences 43. Suppose n is even. From n k n k n−1 k−1 , we get an 1 + n/2 k1 3n k −1 + n n −k + 1 −14 1 + 1 n n/2 k1 3 k n −1 k −1 −1 + (n −k + 1) n −1 n −k −14 and with n−1 k−1 n−1 n−k , we get an 1 + n + 1 n n/2 k1 n −1 k −1 −1 1 + n + 1 2n an−1.
Similarly, we treat the case of an odd n. With the recurrence, we get a0 1, a1 2, a2 5/2, a3 a4 8/3 which is larger than a5 13/5. If an−1 > 2 + 2/(n −1), then an > n+1 2n 2 + 2 n−1 + 1 > 2 + 2 n, or n 2n+2an > 1. Now try to prove that an+1 < an for n ≥4. A bounded monotonically decreasing function has a limit a.
We can find it from the recursion by a limiting process giving a 1 + a/2 with solution a 2.
44. No! Applying the AM-GM inequality we get an + 1 n2an ) ≥ 2 n ∞.
45. The given condition is equivalent to 2x2 i −(xi−1 + 2/xi−1)xi + 1 0, which has the solutions xi xi−1/2 and xi 1/xi−1. We claim that for i ≥0, xi 2kixϵi 0 , for some integer ki with |ki| ≤i and ϵ (−1)ki+i. This is true for i 0, with k0 0 and ϵ0 1, and we proceed by induction. If it is true for i −1 and xi xi−1/2, then we have ki ki−1 −1 and ϵi ϵi−1; while if xi 1/xi−1, then we have ki −ki−1 and ϵi −ϵi−1. In each case, it is immediate that |ki| ≤i and ϵi (−1)ki+i.
Thus x1995 2kxϵ 0, where k k1995 and ϵ ϵ1995, with 0 ≤|k| ≤1995 and ϵ (−1)1995+k. It follows that x0 x1995 2kxϵ 0. If k is odd, then ϵ 1 and we have 2k 1, a contradiction since k ̸ 0. Thus k must be even, so that ϵ −1 and x2 0 2k. Since k is even and |k| ≤1995, k ≤1994. Hence x0 ≤2997. We can have x0 2997, xi xi−1/2 for i 1, . . . , 1994, and x1995 1/x1994. Then x1994 2−997 and x1995 2997 x0 as desired.
46. We consider a sequence un defined as follows: un+2 (2k + 1)un+1 −kun. Then un c1xn 1 + c2xn 2, where x1,2 (2k + 1 ± √ 4k2 + 1)/2. Let u1 and u2 be such that c1 c2 1, i.e., u0 2, u1 2k + 1. Since 0 < x2 < 1, we have ⌊xn 1⌋ un −1. We prove by induction that k | un −1. Indeed, u1 −1 2k, u2 −1 x2 1 + x2 2 −1 (x1 + x2)2 −2x1x2 −1 (2k + 1)2 −2k −1 4k2 + 2k.
Now we observe that if k divides un −1 and un+1 −1, then k also divides un+2 −1.
47. Let xj ∈{0, 1} represent the state of lamp Lj (0 for OFF, 1 for ON). Operation Sj affects the state of Lj, which in the previous round has been set to the value xj−n. At the moment when Sj is being performed, lamp Lj−1 is in state xj−1. Consquently, xj ≡xj−n + xj−1 (mod 2).
(1) This is true for all j ≥0. Note that the initial state (all lamps ON) corresponds to x−n x−n+1 x−n+2 · · · x−2 x−1 1.
(2) 9. Sequences 241 The state of the system at instant j can be represented by the vector ⃗ vj [xj−n, . . . , xj−1], ⃗ v0 [1, . . . , 1]. Since there are only 2n feasible vectors, repe-titions must occur in the sequence ⃗ v0, ⃗ v1, ⃗ v2, . . .. The operation that produces ⃗ vj+1 from ⃗ vj is invertible. Hence, the equality ⃗ vj+m ⃗ vj implies ⃗ vm ⃗ v0; the initial state recurs in at most 2n steps, proving (a).
To prove (b) and (c), notice that, in view of (1), xj ≡xj−n + xj−1 ≡(xj−2n + xj−n+1) + (xj−1−n + xj−2) ≡xj−2n + 2xj−n−1 + xj−2 ≡xj−3n + 3xj−2n−1 + 3xj−n−2 + xj−3, and so on. After r applications of (1), we arrive at the equality xj ≡ r i0 r i xj−(r−i) n−i (mod 2), holding for all j and r such that j −(r −i)n −i ≥−n. In particular, if r is of the form r 2k, then the binomial coefficients r i are even,except the two outer ones, and we obtain xj ≡xj−rn + xj−r (for r 2k), (3) provided the subscripts do not go below −n, i.e., for j ≥(r −1)n.
Now, if n 2k, choose j ≥n2 −n, and set in (3) r n, obtaining, in view of (1), xj ≡xj−n2 + xj−n ≡xj−n2 + (xj −xj−1).
Hence, xj−n2 xj−1, showing that the sequence xj is periodic with period n2 −1.
Thus, the string (2) of ones reappears after exactly n2 −1 steps; claim (b) results.
And if n 2k + 1, choose j ≥n2 −2n, and set in (3) r n −1, obtaining, in view of (1), xj ≡xj−n2+n + xj−n+1 ≡xj−n2+n + (xj+1 −xj) ≡xj−n2+n −xj+1 + xj (because x ≡−x mod 2). Hence xj−n2+n xj+1, showing that the sequence xj is periodic with period n2 −n + 1 and proving claim (c).
This problem is due to G.N. de Bruijn. The solution is due to Marcin Kuczma.
48. Square all equalities a1 0, |a2| |a1 + 1|, . . . |an+1 |an + 1|, and add them.
Reduction yields a2 n+1 2(a1+· · ·+an)+n ≥0. This implies a1+· · ·+an ≥−n/2.
49. A picture is very helpful. The broken line with vertices (k, ak) is convex since ak+1 − ak ≥ak −ak−1, that is, the slope of each succeeding segment is greater than or equal to the preceding one. Hence, all the broken line, except its endpoints, lies below the axis 0k.
Suppose that, for some m ≥1, we have am−1 ≤1, am > 0. Then an −an−1 ≥an−1 −an−2 ≥· · · am+1 −am ≥am −am−1 > 0, and thus an > an−1 > · · · > am > 0. This contradicts the condition an 0.
50. We have a1−a0 ≥1. Furthermore a2−a1 2(a1−a0), . . . , a100−a99 2(a99−a98).
Multiplying these 99 equalities with both sides positive and cancelling, we get a100 a99 + 299(a1 −a0) ≥299.
A sharper estimate using induction ak ≥2k, ak+1 −ak ≥2k (k 1, 2, . . .) shows that a100 ≥2100.
242 9. Sequences 51. Sort the first three terms decreasingly. Then the sequence is decreasing: x1 ≥x2 ≥ · · · ≥x21 since starting with x2 the set of differences increases. Now we have for k ≥0: xk ≥xk+1 +xk+2. Otherwise, we would have xk+2 > xk −xk+1 |xk −xk+1|, which is impossible. We assume x21 ≥1. Then x20 ≥1, x19 ≥x20+x21 ≥2, x18 ≥3, and so on, until we finally get the contradiction x1 ≥4, 181 + 6, 765 > 10, 000.
52. For m n, we find a0 0. For n 0, we get a2m 4am. Now let m n+2. Then a2n+2 + a2 (a2n+4 + a2n)/2, and, from a2m 4am, we finally get a2n+2 + a2 2(an+2 + an). On the other hand, because of a1 1, we have a2 4 and after trivial computations an+2 2an+1 −an −2 with a0 0, a1 1. Since a2 4, a3 9, a4 16, we conjecture that an n2 and prove this by induction.
53. It is easy to prove that three different primes cannot belong to the same geometric progression. Prove it! But among the numbers from 1 to 100 there are 25 primes. By the box principle, they cannot belong to 12 geometric progressions.
54. Suppose that a1, . . . , an satisfy the conditions of the problem. We prove that we can find an+1 such that An+1 a2 1+· · ·+a2 n+1 is divisible by Bn+1 a1+· · ·+an+1. Since An+1 An + (an+1 −Bn)(an+1 + Bn) + B2 n, the number An+1 is divisible by Bn+1 if An+B2 n is divisible by Bn+1. For this it is sufficient to take an+1 An+B2 n −Bn (here An + B2 n Bn+1). Then an+1 > an, since B2 n −Bn > 0 and an+1 > An > a2 n > an.
55. Consider the binary expansion of x0 0.b1b2b3 . . .. It is easy to see that x1 0.b2b3b4 . . ., or x1 0.b2b3b4 . . . where bi 1 −bi, that is, the function shaves off the first binary digit with or without subsequent complementing of digits. So the period is equal to the period of x0 in the binary system or twice this period.
56. Hint: The formula is xn 2n −⌊ √ 2n + 1/2⌋.
57. It is sufficient to prove the stronger result bn+1 ≥bn + 1. We set a a1 + · · · + an, c 1 a1 + · · · + 1 an .
Obviously a/x + cx ≥2√ac for x > 0. Hence, (a + x)(c + 1 x ) ≥ac + 1 + 2√ac √ac + 1 2 .
From this we get √(a + x)(c + 1/x) ≥√ac + 1 or bn+1 ≥bn + 1 by setting x an+1.
58. (a) an an−1 + an−2, a1 1, a2 2. (b) The number sn of symmetric tilings and the number dn of distinct tilings is s2n an+1, s2n+1 an, d2n (a2n + an+1)/2, d2n+1 (a2n+1 + an)/2.
59. a1 1, a2 3, an an−1 + 2an−2.
60. a1 1, a2 4, a3 2, an an−1 + 4an−2 + 2an−3.
61. a2 a3 a4 1, an an−2 + an−3.
62. a0 1, a2 3, an 4an−2 −an−4, n ≥4, n even.
63. n must be a multiple of 3. a0 1, a3 3, an 4an−3 + an−5.
64. a0 1, a1 2, a2 7, an 3an−1 + an−2 −an−3.
65. a0 1, a1 1, a2 5, a3 11, an an−1 + 5an−2 + an−3 −an−4.
9. Sequences 243 66. Let an be the number of ways to fill a 2 × 2 × n box with 1 × 1 × 2 bricks. Fig. 9.4 shows that a1 2, a2 2a1 + 5 9, a3 2a2 + 5a1 + 4 32, an 2an−1 + 5an−2 + 4an−3 + · · · + 4a1 + 4.
Replacing n by n−1 and subtracting, we get an 3an−1 +3an−2 −an−3 from which we get the following table: n 1 2 3 4 5 6 7 8 an 2 32 32 112 450 412 6272 1532 .
The characteristic equation is λ3 3λ2 + 3λ −1 with the solutions λ1 −1, λ2 2 + √ 3 and λ3 2 − √ 3. From this prove that an 1 3(−1)n + (2 + √ 3)n+1 + (2 − √ 3)n+1 6 .
Now try to prove that a2n is a square, and a2n+1 is twice a square.
pppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppp Fig. 9.4 and 3 more obtained by quarter turns 10 Polynomials 1. The terms f (x) anxn + · · · + a0, g(x) bmxm + · · · + b0, an ̸ 0, bm ̸ 0 are polynomials of degrees n and m: deg f n, deg g m. The coefficients ai, bi can be from C, R, Q, Z, Zn.
2. Division with Remainder. For polynomials f and g there exist unique polynomials q and r so that f (x) g(x)q(x) + r(x), deg r < deg g or r(x) 0.
q(x) and r(x) are quotient and remainder on division of f by g. If r(x) 0, then we say that g(x) divides f (x), and we write g(x)|f (x).
E1. With f (x) x7 −1, g(x) x3 + x + 1 the grade school method of division yields x7 −1 (x3 + x + 1)(x4 −x2 −x + 1) + 2x2 −2.
Here q(x) x4 −x2 −x + 1, r(x) 2x2 −2.
3. Let f be a polynomial of degree n and a ∈R. Division by x −a yields f (x) (x −a)q(x) + r, r ∈R, deg q n −1.
(1) Setting x a in (1), we get f (a) r, and hence f (x) (x −a)q(x) + f (a).
(2) 246 10. Polynomials If f (a) 0, then a is a root or zero of f . It follows from (2) f (a) 0 ⇔f (x) (x −a)q(x) for some polynomial q(x).
(3) If a1, a2 are distinct zeros of f , then f (x) (x −a1)q(x) with q(a2) 0, that is, q(x) (x −a2)q1(x). Thus, f (x) (x −a1)(x −a2)q1(x), deg q1 n −2.
If deg f n and f (ai) 0 for a1, . . . , an, then f (x) c(x −a1)(x −a2) · · · (x −an), c ∈R.
4. If there exists an m ∈N and a polynomial q so that f (x) (x −a)mq(x), q(a) ̸ 0, (4) then the root a of f has multiplicity m. (4) implies that a has multiplicity m if and only if f (a) f ′(a) f ′′(a) · · · f (m−1)(a) 0, f m(a) ̸ 0.
(5) 5. Let f (x) anxn + · · · + a0 have integer coefficients, and let z ∈Z. Then f (z) 0 ⇔z|a0.
Indeed, anzn + · · · + a1z + a0 0 ⇔a0 −z(anzn−1 + · · · + a1). If an 1, then each rational root of f is an integer. Indeed, let p/q be a root, p, q ∈Z, gcd(p, q) 1. Then pn qn + an−1 pn−1 qn−1 + · · · + a1 p q + a0, pn q −an−1pn−1 −an−2pn−2q −· · · −a1pqn−2 −a0qn−1.
The RHS is an integer. Hence, q 1.
If the highest degree coefficient an 1, then the polynomial is called a monic polynomial.
6. Vieta’s Theorem. (a) If the polynomial x2 + px + q has roots x1, x2, then x2 + px + q (x −x1)(x −x2) x2 −(x1 + x2)x + x1x2, that is, p −(x1 + x2), q x1x2.
(b) Let x1, x2, x3 be the roots of x3 + px2 + qx + r. By expanding (x −x1)(x −x2)(x −x3) x3 −(x1 + x2 + x3)x2 + (x1x2 + x2x3 + x3x1)x −x1x2x3 and comparing coefficients, we get p −(x1 + x2 + x3), q x1x2 + x2x3 + x3x1, r −x1x2x3.
Similar relations exist for higher degree monic polynomials.
10. Polynomials 247 E2. Let x1, x2, x3 be the roots of x3 + 3x2 −7x + 1. Find x2 1 + x2 2 + x2 3.
Solution. x1 + x2 + x3 −3, x1x2 + x2x3 + x3x1 −7, 9 (x1 + x2 + x3)2 x2 1 + x2 2 + x2 3 + 2(x1x2 + x2x3 + x3x1) x2 1 + x2 2 + x2 3 −2 · 7, x2 1 + x2 2 + x2 3 23.
7. If a ∈R, then f (x) anxn + · · · + a0 can be written in the form f (x) cn(x −a)n + · · · + c1(x −a) + c0.
To prove this, we set x a + (x −a) for x in f .
8. Fundamental Theorem of Algebra. Every polynomial f (z) anzn+· · ·+ a0, ai ∈C, n ≥1, an ̸ 0 has at least one root in C.
From this theorem it easily follows that each polynomial of degree n can be written in the form f (x) c(x −x1)(x −x2) · · · (x −xn), xi ∈C, where the xi are not necessarily distinct.
9. Roots of Unity. Let ω ei 2π n cos 2π n + i sin 2π n . The polynomial xn −1 has the roots ω, ω2, · · · , ωn 1. They are called roots of unity and they are the vertices of a regular n-gon inscribed in the unit circle with center O.
If gcd(k, n) 1, then the powers of ωk also give all nth roots of unity. We have the decomposition xn −1 (x −1)(x −ω)(x −ω2) · · · (x −ωn−1).
In particular, the roots of x3 −1 0, or (x −1)(x2 + x + 1) 0 are the third roots of unity. Denoting by z the conjugate of z, we get ω −1 + i √ 3 2 , ω2 ω 1 ω, ω3 1, 1 + ω + ω2 0.
(6) We can solve the general cubic equation with third unit roots. We start with the classic decomposition x3 + a3 + b3 −3abx (x + a + b)(x2 + a2 + b2 −ax −bx −ab).
The last factor has the roots x2 −aω −bω2, x3 −aω2 −bω. Thus, x3 + a3 + b3 −3abx (x + a + b)(x + aω + bω2)(x + aω2 + bω).
Hence, the cubic equation x3 −3abx + a3 + b3 0 has the solutions x1 −a −b, x2 −aω −bω2, x3 −aω2 −bω.
(7) Comparing this with x3 + px + q 0. we get p −3ab, q a3 + b3, or a3b3 −p3/27, a3 + b3 q.
(8) 248 10. Polynomials From (8) we infer that a3, b3 are roots of the quadratic z2 −qz −p3/27 0.
Thus, a 3 q 2 + q2 4 + p3 27, b 3 q 2 − q2 4 + p3 27.
(9) Inserting (9) into (7) we get the three solutions of x3 + px + q 0. Any cubic can be transformed into this form by translation and division by a constant.
Now we use the fifth roots of unity to construct the regular pentagon.
x5 −1 (x −1)(x4 + x3 + x2 + x + 1).
This factoring shows that the fifth unit root ω satisfies the equation ω4 + ω3 + ω2 + ω + 1 0, ω2 + 1 ω2 + ω + 1 ω + 1 0, (ω + 1 ω)2 + (ω + 1 ω) −1 0, ω + 1 ω √ 5−1 2 .
For a cos 72◦in Fig. 10.1, we have a √ 5 −1 4 .
ppppppppppppppppppppppppppppppppppppp a 1 ω 1 Fig. 10.1 The segment a is easy to construct with ruler and compass.
Now we solve some typical examples with polynomials.
E3. (a) For which n ∈N is x2 + x + 1|x2n + xn + 1? (b) For which n is 37|1 0 . . . 0 n 1 0 . . . 0 n 1?
First Solution. By straightforward transformation using the relations x3 −1 (x −1)(x2 + x + 1) and x3 −1|x3m −1.
10. Polynomials 249 (i) n 3k ⇔x6k +x3k +1 (x6k −1)+(x3k −1)+3 (x2+x +1)Q(x)+3.
(ii) n 3k +1 ⇔x6k+2 +x3k+1 +1 x2(x6k −1)+x(x3k −1)+x2 +x +1 (x2 + x + 1)R(x).
(iii) n 3k+2 ⇔x6k+4+x3k+2+1 x4(x6k−1)+x2(x3k−1)+x4+x2+1 x4(x6k −1) + x2(x3k −1) + x(x3 −1) + x2 + x + 1 (x2 + x + 1)S(x).
Answer: x2 + x + 1|x2n + xn + 1 ⇔3 ̸ |n.
(b) x 10 yields x2 + x + 1 111, x2(n+1) + xn+1 + 1 1 0 . . . 0 n 1 0 . . . 0 n 1, 111 3 · 37. The number is divisible by 3 since the digit sum is 3. Hence 37|1 0 . . . 0 n 1 0 . . . 0 n 1 if n 0 mod 3 or n 1 mod 3.
Second Solution of (a). x2 + x + 1 0 has solutions ω and ω2. By using the relationships ω3 1 and ω2 + ω + 1 0, we get n 3k ⇒ω6k + ω3k + 1 1 + 1 + 1 3, n 3k + 1 ⇒ω6k+2 + ω3k+1 + 1 ω2 + ω + 1 0, n 3k + 2 ⇒ω6k+4 + ω3k+2 + 1 ω4 + ω2 + 1 ω + ω2 + 1 0.
E4. If P (x), Q(x), R(x), S(x) are polynomials so that P(x5) + xQ(x5) + x2R(x5) (x4 + x3 + x2 + x + 1)S(x), (∗) then x −1 is a factor of P (x). Show this (USO 1976).
Solution. Let ω e2πi/5, so that ω5 1. We set for x in (), ω, ω2, ω3, ω4 successively, and get the following equations 1 to 4. If we multiply 1 to 4 by −ω, −ω2, −ω3, −ω4, then we get the last 4 equations.
P(1) + ωQ(1) + ω2R(1) 0, P(1) + ω2Q(1) + ω4R(1) 0, P(1) + ω3Q(1) + ωR(1) 1, P(1) + ω4Q(1) + ω3R(1) 0, −ωP(1) −ω2Q(1) −ω3R(1) 0, −ω2P(1) −ω4Q(1) −ωR(1) 0, −ω3P(1) −ωQ(1) −ω4R(1) 0, −ω4P(1) −ω3Q(1) −ω2R(1) 0.
Using 1+ω+ω2 +ω3 +ω4 0, we get the sum 5P(1) 0, that is, x −1|P (x).
E5. Let P (x) be a polynomial of degree n, so that P(k) k/(k + 1) for k 0..n.
Find P (n + 1) (USO 1975).
250 10. Polynomials Solution. Let Q(x) (x + 1)P(x) −x. Then the polynomial Q(x) vanishes for k 0, . . . , n, that is, (x + 1)P (x) −x a · x · (x −1)(x −2) · · · (x −n).
To find a we set x −1 and get 1 a(−1)n+1(n + 1)!. Thus, P(x) (−1)n+1x(x −1) · · · (x −n)/(n + 1)! + x x + 1 , and P (n + 1) 1 for odd n, n/(n + 2) for even n.
E6. Let a, b, c be three distinct integers, and let P be a polynomial with integer coefficients. Show that in this case the conditions P (a) b, P(b) c, P(c) a cannot be satisfied simultaneously (USO 1974).
Solution. Suppose the conditions are satisfied. We derive a contradiction.
P(x) −b (x −a)P1(x), (1) P(x) −c (x −b)P2(x), (2) P(x) −a (x −c)P3(x).
(3) Among the numbers a, b, c, we choose the pair with maximal absolute difference.
Suppose this is |a −c|. Then we have |a −b| < |a −c|.
(4) If we replace x by c in (1), then we get a −b (c −a)P1(c).
Since P1(c) is an integer, we have |a −b| ≥|c −a|, which contradicts (4).
10. Reciprocal Equations Definition. The polynomial f (x) anxn + · · · + a1x + a0, an ̸ 0 is called reciprocal, if ai an−i for i 0, . . . , n.
Examples xn + 1, x5 + 3x3 + 3x2 + 1, 5x8 −2x6 + 4x5 + 4x3 −2x2 + 5. The equation f (x) 0 with f (x) being a reciprocal polynomial is called a reciprocal equation.
Theorem. Any reciprocal polynomial f (x) of degree 2n can be written in the form f (x) xng(z), where z x + 1 x , and g(z) is a polynomial in z of degree n.
10. Polynomials 251 Proof.
f (x) a0x2n + a1x2n−1 + · · · + a1x + a0, f (x) xn a0xn + a1xn−1 + · · · + a1 xn−1 + a0 xn , f (x) xn a0 xn + 1 xn + a1 xn−1 + 1 xn−1 + · · · + an .
We show how to express xk + 1/xk by z x + 1/x: x2 + 1 x2 x + 1 x 2 −2 z2 −2, x3 + 1 x3 x + 1 x 3 −3x −3 x z3 −3z, x4 + 1 x4 x + 1 x 4 −4x2 −6 −4 x2 z4 −4 z2 −2 −6 z4 −4z2 + 2, x5 + 1 x5 x + 1 x 5 −5x3 −10x −10 x −5 x3 z5 −5z3 + 5z.
Without proof we state some properties of reciprocal polynomials. They are easy to prove and are left to the reader as exercises: (a) Every polynomial f (x) of degree n with a0 ̸ 0 is reciprocal iff xnf 1 x f (x).
(b) Every reciprocal polynomial f (x) of odd degree is divisible by x + 1 and the quotient is a reciprocal polynomial of even degree.
(c) If a is a zero of the reciprocal equation f (x) 0, then 1 a is also a zero of this equation.
11. Symmetric Polynomials A polynomial f (x, y) is symmetric, if f (x, y) f (y, x) for all x, y. Examples: (a) The elementary symmetric polynomials in x, y σ1 x + y, σ2 xy.
(b) The power sums si xi + yi i 0, 1, 2, . . . .
A polynomial symmetric in x, y can be represented as a polynomial in σ1, σ2.
Indeed, sn xn + yn (x + y)(xn−1 + yn−1) −xy xn−2 + yn−2 σ1sn−1 + σ2sn−2.
252 10. Polynomials Thus, we have the recursion s0 2, s1 σ1, sn σ1sn−1 −σ2sn−2, n ≥2.
Now the proof for any symmetric polynomial is simple. Terms of the form axkyk cause no trouble since axkyk aσ k 2 . With the term bxiyk (i < k), it must also contain bxkyi. We collect these terms: bxiyk + bxkyi bxiyi xk−i + yk−i bσ i 2sk−i.
But sk−i can be expressed through σ1, σ2.
Nonlinear systems of symmetric equations in two variables x, y can mostly be simplified by the substitution σ1 x +y, σ2 xy. The degree of these equations will be reduced since σ2 xy is of second degree in x, y. As soon as we have found σ1 and σ2 we find the solutions z1, z2 of the quadratic equation z2 −σ1z + σ2 0.
Then we have the system of equations x + y σ1, xy σ2.
E7. Solve the system x5 + y5 33, x + y 3.
We set σ1 x + y, σ2 xy. Then the system becomes σ 5 1 −5σ 3 1 σ2 + 5σ1σ 2 2 33, σ1 3.
Substituting σ1 3 in the first equation, we get σ 2 2 −9σ2 + 14 0 with two solutions σ2 2 and σ2 7. Now we must solve x + y 3, xy 2, and x + y 3, xy 7 resulting in (2, 1), (1, 2), (x3, y3) 3 2 + √ 19 2 i, 3 2 − √ 19 2 i , (x4, y4) (y3, x3).
E8. Find the real solutions of the equation 4 √ 97 −x + 4 √x 5.
We set 4 √x y, 4 √ 97 −x z and get y4 + z4 x + 97 −x 97. Hence, y + z 5, y4 + z4 97.
Setting σ1 y + z, σ2 yz, we get the system of equations σ1 5, σ 4 1 −4σ 2 1 σ2 + 2σ 2 2 97 10. Polynomials 253 resulting in σ 2 2 −50σ2 + 264 0 with solutions σ2 6, σ2 44. We must solve the system y + z 5, yz 6 with solutions (y1, z1) (2, 3), (y2, z2) (3, 2).
Now x1 16, x2 81. The solutions y + z 5, yz 4 give complex values.
E9. What is the relationship between a, b, c if the system x + y a, x2 + y2 b, x3 + y3 c is compatible (has solutions)?
Solution. We eliminate x, y: σ1 a, σ 2 1 −2σ2 b, σ 3 1 −3σ1σ2 c with the result a3 −3ab + 2c 0.
(c) Polynomials with three variables have the elementary symmetric polynomi-als σ1 x + y + z, σ2 xy + yz + zx, σ3 xyz.
The power sums si xi + yi + zi, i 0, 1, 2, · · · can be represented by σ1, σ2, σ3. Show that the following identities are valid: s0 x0 + y0 + z0, s1 x + y + z σ1, s2 x2 + y2 + z2 σ 2 1 −2σ2, s3 x3 + y3 + z3 σ 3 1 −3σ1σ2 + 3σ3, s4 σ 4 1 −4σ 2 1 σ2 + 2σ 2 2 + 4σ1σ3, x2y + xy2 + x2z + xz2 + y2z + yz2 σ1σ2 −3σ3, x2y2 + y2z2 + z2x2 σ 2 2 −2σ1σ3.
Systems of equations which are symmetric in x, y, z can be expressed through σ1, σ2, σ3. As soon as we have σ1, σ2, σ3, we find the solutions u1, u2, u3 of the cubic equation u3 −σ1u2 + σ2u −σ3 0. Then (x1, y1, z1) (u1, u2, u3) is one solution. We get the others by permuting the variables.
E10. Solve the system of equations x + y + z a, x2 + y2 + z2 b2, x3 + y3 + z3 a3.
We set x + y + z σ1, xy + yz + zx σ2, xyz σ3 and get σ1 a, σ2 1 2 a2 −b2 , σ3 1 2a a2 −b2 , u3 −au2 + 1 2 a2 −b2 u −1 2a a2 −b2 0, (u −a) u2 −1 2 b2 −a2 0, u1 a, u2 b2−a2 2 , u3 − b2−a2 2 .
There are six solutions (u1, u2, u3) and its permutations.
E11. Find all real solutions of the system x + y + z 1, x3 + y3 + z3 + xyz x4 + y4 + z4 + 1.
254 10. Polynomials Introducing elementary symmetric polynomials yields σ1 1, x3 + y3 + z3 σ 3 1 −3σ1σ2 + 3σ3, x4 + y4 + z4 σ 4 1 −4σ 2 1 σ2 + 2σ 2 2 + 4σ1σ3. For σ1 1, the second equality becomes 2σ 2 2 −σ2 + 1 0, which has no solutions.
E12. Given 2n distinct numbers a1, . . . , an, b1, . . . , bn, an n × n table is filled as follows: into the cell in the ith row and jth column is written the number ai + bj.
Prove that if the product of each column is the same, then also the product of each row is the same (AUO 1991).
Consider the polynomial f (x) n i1 (x + ai) − n j1 (x −bj) of degree less than n. If f (bj) n i1 (ai + bj) c for all j 1, . . . , n then the polynomial f (x)−c has at least n distinct roots. This implies f (x) −c 0 for all x. But then c f (−ai) − n j1 (−ai −bj) (−1)n+1 n j1 (ai + bj), QED.
Problems 1. Factor x3 + y3 + z3 −3xyz by elementary symmetric functions.
2. For which a ∈R is the sum of the squares of the zeros of x2 −(a −2)x −a −1 minimal?
3. If x1, x2 are the zeros of the polynomial x2 −6x + 1, then for every nonnegative integer n, xn 1 + xn 2 is an integer and not divisible by 5.
4. Given a monic polynomial f (x) of degree n over Z and k, p ∈N, prove that if none of the numbers f (k), f (k + 1), . . . , f (k + p) is divisible by p + 1, then f (x) 0 has no rational solution.
5. The polynomial x2n −2x2n−1 + 3x2n−2 −· · · −2nx + 2n + 1 has no real roots.
6. a, b, c ∈R, a + b + c > 0, bc + ca + ab > 0, abc > 0 ⇒a, b, c > 0.
7. A polynomial f (x, y) is antisymmetric, if f (x, y) −f (y, x). Prove that every antisymmetric polynomial f (x, y) has the form f (x, y) (x −y)g(x, y), where g(x, y) is symmetric.
8. The polynomial f (x, y, z) is antisymmetric if the sign changes on switching any two variables. Prove that every antisymmetric polynomial f (x, y, z) can be written in the form f (x, y, z) (x −y)(x −z)(y −z)g(x, y, z), where g(x, y, z) is symmetric.
9. If f (x, y) is symmetric and x −y|f (x, y), then (x −y)2|f (x, y).
10. Polynomials 255 10. Iff (x, y, z)issymmetricandx−y|f (x, y, z),then(x−y)2(y−z)2(z−x)2|f (x, y, z).
11. Solve the equation z8 + 4z6 −10z4 + 4z2 + 1 0.
12. Solve the equation 4z11 +4z10 −21z9 −21z8 +17z7 +17z6 +17z5 +17z4 −21z3 −21z2 +4z+4 0.
13. Solve the equation (x −a)4 + (x −b)4 (a −b)4.
14. Factorize over Z: (a) x10 + x5 + 1, (b) x4 + x2 + 1, (c) x8 + x4 + 1, x9 + x4 −x −1.
15. Let f (x) 1 −x + x2 −. . . + x100 1 + x + x2 + . . . + x100 . Show that, after multiplying and collecting terms, only even powers of x will remain.
16. Find the remainder on dividing x100 −2x51 + 1 by x2 −1.
17. Determine a, b so, that (x −1)2|ax4 + bx3 + 1.
18. For which n ∈N do we have (a) x2 + x + 1|(x −1)n −xn −1, (b) x2 + x + 1|(x + 1)n + xn + 1?
19. Show that (x −1)2|nxn+1 −(n + 1)xn + 1.
20. Show that k|n ⇔xk −ak|xn −an, a, k, x, n ∈N.
21. Show that (x + 1)2|x4n+2 + 2x2n+1 + 1.
22. The polynomial 1 + x/1 + x2/2! + . . . + xn/n! has no multiple roots.
23. Find a such that −1 is a multiple root of x5 −ax2 −ax + 1.
24. In x3 +px2 +qx +r one zero is the sum of the two others. Find the relation between p, q, r.
25. x5 + ax3 + b has a double zero ̸ 0. Find the relation between a and b.
26. Let a, b, c be distinct numbers. The quadratic equation (x −a)(x −b) (c −a)(c −b) + (x −b)(x −c) (a −b)(a −c) + (x −c)(x −a) (b −c)(b −a) 1 has the solutions x1 a, x2 b, x3 c. What follows from this fact?
27. Find a, b, c so that x + 5 (x −1)(x −2)(x −3) a x −1 + b x −2 + c x −3.
28. x4 + x3 + x2 + x + 1|x44 + x33 + x22 + x11 + 1.
29. Solve the equation x4 + a4 −3ax3 + 3a3x 0.
30. Let x1, x2 be the roots of the equation x2 + ax + bc 0, and x2, x3 the roots of the equation x2 + bx + ac 0 with ac ̸ bc. Show that x1, x3 are the roots of the equation x2 + cx + ab 0.
31. The polynomial ax3 + bx2 + cx + d has integral coefficients a, b, c, d with ad odd and bc even. Show that at least one zero of the polynomial is irrational.
32. Let a, b be integers. Then the polynomial (x −a)2(x −b)2 + 1 is not the product of two polynomials with integral coefficients.
256 10. Polynomials 33. Let f (x) ax2 + bx + c. Suppose f (x) x has no real roots. Show that the equation f (f (x)) x has also no real solutions.
34. Let f (x) be a monic polynomial with integral coefficients. If there are four different integers a, b, c, d, so that f (a) f (b) f (c) f (d) 5, then there is no integer k, so that f (k) 8.
35. Let f (x) x4 + x3 + x2 + x + 1. Find the remainder on dividing f (x5) by f (x).
36. Find all polynomials P (x), so that P[F(x)] F[P (x)], P(0) 0, where F(x) is a given function with the property F(x) > x for all x ≥0.
37. Find all polynomial solutions of the functional equation f (x)f (x + 1) f (x2 + x + 1).
38. Find all pairs of positive integers (m, n), so that 1 + x + x2 + · · · + xm|1 + xn + x2n + · · · + xmn (USO 1977).
39. If a and b are two solutions of x4 + x3 −1 0, then ab is a solution of x6 + x4 + x3 −x2 −1 0 (USO 1977).
40. Find the polynomial p(x) x2 + px + q for which maxx∈[−1,1] |p(x)| is minimal.
41. Let f (x) x1958 + x1957 + 2 1959 a0 + a1x + · · · + anxn. Find a0 −a1/2 −a2/2 + a3 −a4/2 −a5/2 + a6 −· · · .
42. Find the remainder on dividing x1959 −1 by (x2 + 1)(x2 + x + 1).
43. Is there a nonconstant function f (x) so that xf (y) + yf (x) (x + y)f (x)f (y) for all x, y ∈R?
44. Find all positive solutions of the equation nxn+1 −(n + 1)xn + 1 0.
45. Let p(x) be a polynomial over Z. If p(a) p(b) p(c) −1 with integers a, b, c, then p(x) has no integral zeros.
46. Find all polynomials p(x) with xp(x −1) (x −26)p(x) for all x.
47. The polynomial ax4 + bx3 + cx2 + dx + e with integral coefficients is divisible by 7 for every integer x. Show that 7|a, 7|b, 7|c, 7|d, 7|e.
48. Let a, b ∈R. For x ∈[−1, 1] we have −1 ≤ax2 + bx + c ≤1. Show that in the same interval, −4 ≤2ax + b ≤4.
49. The polynomial 1 + x + x2/2! + x3/3! + · · · + x2n/(2n)! has no real zeros.
50. If x3 + px2 + qx + r 0 has three real zeros, then p2 ≥3q.
51. f (n) n2 −n + 41 gives primes for n 1, . . . , 40. Find 40 successive values of n for which f (n) is composite. Generalize.
52. Find the smallest value of the polynomial x3(x3 + 1)(x3 + 2)(x3 + 3).
53. Does there exist a polynomial f (x), for which xf (x −1) (x + 1)f (x)?
54. (1 + x + · · · + xn)2 −xn is the product of two polynomials.
55. A polynomial f (x) over Z has no integral zero if f (0) and f (1) are both odd.
10. Polynomials 257 56. Find a cubic equation whose roots are the third powers of the roots of x3 + ax2 + bx + c 0.
57. Find all polynomials f (x), for which f (x)f (2x2) f (2x3 + x).
58. If a1, . . . , an ∈Z are distinct, then (x −a1) · · · (x −an) −1 is irreducible.
59. Find all polynomials f , so that (a) f (x2)+f (x)f (x+1) 0, (b) f (x2)+f (x)f (x− 1) 0.
60. For which k is x3 + y3 + z3 + kxyz divisible by x + y + z?
61. Given a polynomial with (a) natural (b) integral coefficients, let an be the digital sum in the decimal representation of f (n). Show that there is a number, which occurs in a1, a2, a3, . . . infinitely often.
62. Find all pairs x, y ∈Z, so that x3 + x2y + xy2 + y3 8(x2 + xy + y2 + 1).
63. Let n > 1 be an integer and f (x) xn + 5xn−1 + 3. Show that f (x) is irreducible over Z (IMO 1993).
64. Let f (x) and g(x) be nonzero polynomials, with f (x2 + x + 1) f (x)g(x). Show that f (x) has even degree.
65. A polynomial f (x) x4 + ∗x3 + ∗x2 + ∗x + 1 has three undetermined coefficients denoted by stars. The players A and B move alternately, replacing a star by a real number until all stars are replaced. A wins if all zeros of the polynomial are complex.
B wins if at least one zero is real. Show that B can win in spite of his only second move.
66. Find real numbers a, b, c, for which |f (x)| |ax2 + bx + c| ≤1 for |x| ≤1 and 8 3a2 + 2b2 is maximal.
67. Find all polynomials P in two variables with the following properties: (i) For a positive integer n and all real t, x, y, P (tx, ty) tnP(x, y).
(ii) for all real a, b, c, P (b + c, a) + P (c + a, b) + P (a + b, c) 0, (iii) P (1, 0) 1 (IMO 1975).
68. Let P1(x) x2 −2 and Pj(x) P1 Pj−1(x) for j 2, 3, . . . . Show that, for any positive integer n, the roots of the equation Pn(x) x are real and distinct.
(IMO 1976.) 69. The polynomial ax2 + bx + c with a > 0 has real zeros x1, x2. Show that |xi| ≤1, (i 1, 2) ⇔a + b + c ≥0, a −b + c ≥0, a −c ≥0.
70. Find all polynomials f over C satisfying f (x)f (−x) f (x2).
71. The polynomial f (x) has integral coefficients and assumes values divisible by 3 for the integral arguments k, k + 1, k + 2. Show that f (m) is a multiple of 3 for every integer m.
72. The polynomial P(x) xn +a1xn−1 +· · ·+an−1x +1 with nonnegative coefficients a1, . . . , an−1 has n real roots. Prove that P (2) ≥3n.
73. Is the polynomial x105 −9 reducible over Z?
74. The polynomial f (x) x5 −x + a is irreducible over Z if 5 ̸ |a.
258 10. Polynomials 75. Find the minimum of a2 + b2 if the equation x4 + ax3 + bx2 + ax + 1 0 has real roots.
76. Is it possible that each of the polynomials P (x) ax2+bx+c, Q(x) cx2+ax+b, R(x) bx2 + cx + a has two real roots?
77. Prove that a2 + ab + b2 ≥3(a + b −1) for all real a, b.
78. Find all positive integral solutions (x, y) of the polynomial equation 4x3 + 4x2y −15xy2 −18y3 −12x2 + 6xy + 36y2 + 5x −10y 0.
79. Find all real solutions (x, y) of the polynomial equation y4 + 4y2x −11y2 + 4xy −8y + 8x2 −40x + 52 0.
80. Factor the polynomial x8 + 98x4 + 1 into two factors with integral coefficients.
81. Prove that, for any polynomial p(x) of degree greater than 1, we can substitute another polynomial q(x) for x, such that p(q(x)) can be factored into a product of polynomials, different from constants. (All polynomials have integral coefficients.) 82. It is known of a polynomial over Z that p(n) > n for every positive integer n.
Consider x1 1, x2 p(x1) . . .. We know that, for any positive integer N, there exists a term of the sequence divisible by N. Prove that p(x) x + 1.
Solutions 1. x3 + y3 + z3 −3xyz s3 −3σ3 σ1(σ 2 1 −3σ2) (x + y + z)(x2 + y2 + z2 − xy −yz −zx).
2. x2 1 +x2 2 (x1+x2)2−2x1x2 (a−2)2+2(a+1) a2−2a+6 (a−1)2+5 ≥5.
We have equality for a 1.
3. We have s1 x1 + x2 6, x1x2 1. Let sn xn 1 + xn 2. In the section on symmetric polynomials, we established that sn 6sn−1 −sn−2. Starting from s0 2, s1 6, this recurrence gives only integral values. Consider the sn modulo 5. The recurrence becomes sn sn−1 −sn−2. We get s0 2, s1 1, s2 4, s3 3, s4 4, s5 1, s6 2, s7 1, . . . .
After six steps, the pair (2, 1) recurs and, the sequence is periodic without any zero (multiple of 5).
Remark.Thecharacteristicequationofthesequencesn 6sn−1−sn−2 isx2−6x+1 0, and the general solution is sn (3 + √ 8)n + (3 − √ 8)n.
4. If f (x) 0 has a rational root, then this root is an integer. Suppose that f (x) has the integral root x0 m, that is f (m) 0. Then f (x) (x −m)g(x), where g(x) has integral coefficients. By setting x k, k + 1, . . . , k + p in the last equation, we get f (k) (k −m)g(k), f (k + 1) (k + 1 −m)g(k + 1), . . . , f (k + p) (k −p −m)g(k + p). One of the p + 1 successive integers k −m, . . . , k + p −m is divisible by p + 1. This proves the contrapositive statement which is equivalent to the original statement.
10. Polynomials 259 5. For x ≤0 we have obviously p(x) > 0. Let x > 0. We transform the polynomial in the same way as a geometric series: p(x) x2n −2x2n−1 + 3x2n−2 −· · · −2nx + 2n + 1, xp(x) x2n+1 −2x2n + 3x2n−1 −4x2n−2 + · · · + (2n + 1)x.
Adding, we get xp(x) + p(x) x2n+1 −x2n + x2n−1 −x2n−2 + · · · , +x + 2n + 1, (1 + x)p(x) x · 1 + x2n+1 1 + x + 2n + 1.
From here we see that p(x) > 0 for x > 0.
6. Let a + b + c u, ab + bc + ca v, abc w. Then a, b, c are the roots of the equation x3 −ux2 + vx −w 0. This equation cannot have negative roots for u, v, w > 0. Indeed, for x < 0, all terms on the left side are negative. Even for x 0, the left side is −w. Thus, a, b, c > 0.
7. Hint: f (x, y) −f (y, x) implies f (x, x) −f (x, x), or f (x, x) 0. Hence f (x, y) (x −y)g(x, y).
8. Hint: x y, y z, x z are roots of the polynomial.
9. Hint: f (x, y) is symmetric. In f (x, y) (x −y)g(x, y), g must be antisymmetric.
Thus, it must be divisible by x −y.
10. Hint: This follows from the preceding result.
11. Dividing by z4, we get z4 + 1/z4 + 4 z2 + 1/z2 −10 0. Substituting u z + 1/z, we get u4 16 with u1 2, u2 2i, u3 −2, u4 −2i. From z + 1/z u, we get z u/2 ± u2/4 −1. Substituting the 4 u-values gives z1,2 1, z3,4 −1, z5,6 i(1 ± √ 2), z7,8 −i(1 ± √ 2).
12. It is easy to see that any reciprocal equation of odd degree has zero z −1. Thus the left side is divisible by z+1. We get (z+1)(4z10−21z8+17z6+17z4−21z2+4) 0.
The first factor is zero for z1 −1. Seting u z+1/z we can transform the second factor as follows: u(4u4 −41u2 + 100) 0, u1 0, u2 −5/2, u3 5/2, u4 2, u5 −2. Altogether we get 11 roots: z1 −1, z2 i, z3 −i, z4 −2, z5 −1/2, z6 2, z7 −1/2, z8 z9 −1, z10 z11 1.
13. We use the fact that x1 a, x2 b. Simplifying the given equation we get x4 −2(a + b)x3 + 3(a2 + b2)x2 −2(a3 + b3)x + 2ab3 −3a2b2 + 2a3b 0.
Now x1 + x2 + x3 + x4 2a + 2b, and x1x2x3x4 2ab3 −3a2b2 + 2a3b. But x1 + x2 a + b, and x1x2 ab. So x3 + x4 a + b and x3x4 2a2 −3ab + 2b2.
Thus x3 and x4 are roots of the equation x2 −(a + b)x + 2a2 −3ab + 2b2 0 with solutions x3,4 a + b 2 ± a −b 2 √ 7i.
Try another approach by setting y x −a, z x −b, a −b z −y.
14. (a) Inserting the third root of unity ω for x, we get ω+ω2 +1 0. Thus, x10 +x5 +1 has factor x2 + x + 1. Division by x2 + x + 1 yields x10 + x5 + 1 x2 + x + 1 x8 −x7 + x5 −x4 + x3 −x + 1 .
260 10. Polynomials (b) x4 +x2 +1 x4 +2x2 +1−x2 x2 + 1 2 −x2 x2 + x + 1 x2 −x + 1 .
(c) x8 +x4 +1 x8 +2x4 +1−x4 x4 + 1 2 −x4 x2 + x + 1 x2 −x + 1 x4 −x2 + 1 .
(d) x9 + x4 −x −1 x x8 −1 + x4 −1 x4 −1 x5 + x + 1 (x −1) (x + 1) x2 + 1 x2 + x + 1 x3 −x2 + 1 .
15. Hint: If we change the sign of x, we change the factors.
16. x100 −2x51 + 1 x2 −1 q(x) + ax + b. Puting x 1 into this relation we get b 0. Putting x −1, we get a −4. Thus the remainder is −4x.
17. f (1) 0 and f ′(1) 0 imply a + b + 1 0 and 4a + 3b 0, or a 3, b −4.
18. x2 + x + 1 0 has roots ω and ω2 with ω2 + ω + 1 0, ω3 1, ω2 1/ω.
(a) Let n 6k + 1. Then (ω + 1)n −ωn −1 −ω2 −ω −1 0. For n 6k −1, (−ω2)−1 −ω−1 −1 −ω −ω2 −1 0. For n 6k, n 6k ± 2, n 6k + 3, we do not get zero.
(b) For n 6k ± 2, we get zero, but not for n 6k, n 6k ± 1, n 6k + 3.
19. f (1) n −(n + 1) + 1 0, and f ′(1) n(n + 1) −(n + 1)n 0.
20. Let n mq + r, 0 ≤r < m. Then we have xn −an xmqxr −amqar xmqxr −amqxr + amqxr −amqar xr (xmq −amq) + amq (xr −ar) .
The first parenthesis is divisible by xm−am. Hence, also the second must be divisible by xm −am. This is only possible for r 0.
Here is another proof based on roots of unity.
xn −an xm −am (x −a)(x −ωa)(x −ω2a) · · · (x −ωn−1a) (x −a)(x −ϵa)(x −ϵ2a) · · · (x −ϵm−1a) .
Every mth root of unity must also be an nth root of unity, that is, ϵ ωk, ϵ2 ω2k, ϵ3 ω3k, . . . , ϵm−1 ω(m−1)k, ϵm ωmk 1.
Now m|n since mk n.
21. f (−1) 1 −2 + 1 0, and f ′(−1) −(4n + 2) + 2(2n + 1) 0.
22. The polynomial f (x) has multiple zero z if f (z) f ′(z) 0. For our polynomial, we have f (x) f ′(x) + xn/n!. The condition for a multiple zero z becomes z 0, but f (0) 1.
23. f (−1) −1 −a + a + 1 0. f ′(−1) 5 + 2a −a 5 + a 0 gives a −5.
24. x1 + x2 + x3 −p, x1x2 + x2x3 + x3x1 q, x1x2x3 −r, x3 x1 + x2 lead to the relation, p3 −4pq + 8r 0.
25. We eliminate x from f (x) x5 +ax3 +b 0, and f ′(x) 5x4 +3ax2 0. Since x ̸ 0, we get 55b2 + 108a3 0.
26. It is an identity, valid for every value of x.
27. x + 5 a(x −2)(x −3) + b(x −1)(x −3) + c(x −1)(x −2). x 1, x 2, x 3 give a 3, b −7, c 4.
10. Polynomials 261 28. Let ω5 1. Then ω44 + ω33 + ω22 + ω11 + 1 ω4 + ω3 + ω2 + ω1 + 1. All roots of the left side are also roots of the right side. This implies the stated divisibility.
29. We divide by a2x2: (x/a −a/x)2 −3 (x/a −a/x)+2 0. This quadratic equation gives x/a −a/x 2 and x/a −a/x 1 with solutions x1,2 a(1 ± √ 2) and x3,4 a −1 ± √ 5 /2 (almost reciprocal equation).
30. One must prove ac ̸ bc, x1x2 bc, x2x3 ac, x1 + x2 −a, x2 + x3 −b ⇒ x1 + x3 −c, x1x3 ab.
This can be accomplished by clever, but routine, transformations.
31. Let xi, (i 1, 2, 3) be the rational roots of the given polynomial. Then ax3 + bx2 + cx + d 0 ⇒(ax)3 + b (ax)2 + ac (ax) + a2d 0.
Setting y ax, we get y3 + by2 + acy + a2d 0.
(1) yi are the three rational zeros of (1), i.e., they must be integers. And since they are divisors of a2d, they must be odd. Because of y1 + y2 + y3 −b and y1y2 + y2y3 + y3y1 ac, both b and ac must be odd, that is, b and c are odd. This contradicts the assumption that bc is even.
32. Let (x −a)2(x −b)2 + 1 p(x)q(x). Since p(a) q(a) p(b) q(b) 1, both p(x) −1 and q(x) −1 must be divisible by (x −a)(x −b). We may assume that p(x) −1 (x −a)(x −b) and q(x) −1 (x −a)(x −b). This implies p(x)q(x) ((x −a)(x −b) + 1)2 (x −a)2(x −b)2 + 1 + 2(x −a)(x −b). But then (x −a)(x −b) ≡0, which is a contradiction.
33. If f (x) x has no real roots, then either f (x) > x for all x or f (x) < x for all x.
Thus, either f (f (x)) > f (x) > x or f (f (x)) < f (x) < x for all x.
34. Let g(x) f (x)−5. Then x −a, x −b, x −c, x −d are factors of g(x). So we can write g(x) (x −a)(x −b)(x −c)(x −d)h(x). If r is an integer such that f (r) 8, then g(r) f (r) −5 3, or (r −a)(r −b)(r −c)(r −d)h(r) 3. The left side is a product of five integers of which at least four are distinct. But the right side has at most three distinct factors 1, −1, −3.
35. x20 + x15 + x10 + x5 + 1 (x4 + x3 + x2 + x + 1)q(x) + r(x), where r(x) ax3 +bx2 +cx +d. Let ω be the fifth root of unity. We set x ω, ω2, ω3, ω4. These values are zeros of the polynomial x4 + x3 + x2 + x + 1. Thus, we get 5 r(ω), 5 r(ω2), 5 r(ω3), 5 r(ω4). If a polynomial of at most degree three takes the value 5 for four different values of x, it will be 5 everywhere. Thus, r 5 is a constant.
We consider a second solution, which does not use fifth roots of unity: Let f (x) x4 + x3 + x2 + x + 1. Then (x −1)f (x) x5 −1, and f (x5) x20 −1 + x15 −1 + x10 −1 + x5 −1 q(x)f (x) +5.
The remainder is 5.
262 10. Polynomials 36. Let F(0) a0 > 0. Then P (F(0)) F(P (0)) ⇔P (a0) a0. Similarly, we get F(an) an+1, P(an) an, and an+1 > an. We must find all polynomials with infinitely many points on y x. Then P (x) −x has infinitely many zeros, i.e., P(x) x.
37. This polynomial functional equation is due to Harold N. Shapiro. In f (x)f (x + 1) f x2 + x + 1 , (1) we set x ←x −1 and get f (x −1)f (x) f (x2 −x + 1).
(2) If f (x) is a constant c, then c2 c with the solutions f (x) ≡0 and f (x) ≡1.
Now suppose that f (x) is not constant. Then it has at least one complex zero. Let z be a zero of maximal distance from O. Here we use the extremal principle. From (1) and (2), we have f (z2 + z + 1) f (z2 −z + 1) 0. Thus, z ̸ 0. If also z2 + 1 ̸ 0, then z, z2 + z + 1, z2 −z + 1, −z are vertices of a parallelogram. Thus, either z2 −z + 1 or z2 + z + 1 is larger then |z|. This contradicts the choice of z.
Thus, z2 + 1 0, and z ±i are zeros of f . Hence we have f (x) (x2 + 1)mg(x), m ∈N, x2 + 1 ̸ | g(x).
Plugging this into (1) and using (x2 + 1)(x2 + 2x + 2) x4 + 2x3 + 3x2 + 2x + 2 we see that g also satisfies (1). Since it is not divisible by x2 + 1, we must have g(x) ≡1. We conclude that f (x) x2 + 1 m is the general polynomial solution of (1). It would be interesting to know the solutions of (1) for the domain of continuous or differentiable functions.
38. We must find (m, n), so that x(m+1)n −1 (x −1) xm+1 −1 (xn −1) is a polynomial. But xm+1 −1 and xn −1 are divisors of x(m+1)n −1. Since the factors of x(m+1)n −1 are all distinct, it is necessary and sufficient that xm+1 −1 and xn −1 have no common factor except x −1, that is, gcd(m + 1, n) 1.
39. x4+x3−1 (x−a)(x−b)(x−c)(x−d) x4−(a+b+c+d)x3+ (ab+ac+ad+ bc+bd +cd)x2−(abc+abd +acd +bcd)x+abcd. Comparing coefficients we get a+b+c+d −1, ab+(a+b)(c+d)+cd 0, ab(c+d)+(a+b)cd 0, abcd 1.
We eliminate cd and c + d and get cd 1/ab and c + d −1 −a −b, which we plug into the second and third equation, getting ab −ab(1 + a + b) + 1/ab 0 and ab(1 + a + b) + (a + b)/ab 0. After eliminating a + b, for u ab, we get the equation u6 + u4 + u3 −u2 −1 0.
40. Answer: p(x) x2 −1/2.
41. Let f (x) x1958 + x1957 + 2 1959 a0 + a1x + a2x2 + · · · + anxn, f (ω) 1 a0 + a1ω + a2ω2 + a3 + a4ω + a5ω2 + · · · , f (ω2) 1 a0 + a1ω2 + a2ω + a3 + a4ω2 + a5ω + · · · .
10. Polynomials 263 Add the two equalities, and use ω2 + ω −1. After division by 2, you get 1 a0 −a1 2 −a2 2 + a3 −a4 2 −a5 2 + · · · .
42. x1959−1 (x2+1)(x2+x+1)q(x)+ax3+bx2+cx+d, x i : −ai−b+ci+d −i −1, x −i : ai −b −ci +d i −1, x ω : a +bω +cω +d 0, x ω2 : a + bω + cω2 + d 0. Solving for a, b, c, d we get a 1, b c 0, d −1.
Thus, the remainder is x3 −1.
43. y x ⇒2xf (x)(1 −f (x)) 0 ⇒f (x) ≡0 or f (x) ≡1.
44. The equation nxn+1 −(n + 1)xn + 1 0 has root x 1. The derivation gives n(n + 1)xn−1(x −1) 0. Thus x 1 is a double root. We prove that for x > 1 and 0 < x < 1 the left side of the equation is positive.
nxn+1 −(n + 1)xn + 1 nxn(x −1) −(xn −1) (x −1)(nxn −xn−1 −xn−2 − · · ·−1).x > 1 ⇒xn > xk for n > k ⇒nxn −xn−1 −· · ·−1 > nxn −nxn−1 > 0.0 < x < 1 ⇒xn < xk for n > k ⇒nxn −xn−1 −· · · −1 < nxn −nxn−1.
45. p(x) (x −a)(x −b)(x −c)q(x) −1. If z is an integral zero of p(x), then p(z) (z −a)(z −b)(z −c)q(z) −1 0. The first three factors on the right side are distinct. We have represented 1 as a product of four factors, of which the first three are distinct. This is not possible since 1 has only the factors 1 and −1.
46. x|p(x) ⇒x −1|p(x −1) ⇒x −1|p(x) ⇒x −2|p(x −1) ⇒x −2|p(x) ⇒ . . . ⇒x −25|p(x). Thus, p(x) x(x −1) · · · (x −25)q(x) and p(x −1) (x−1)(x−2) · · · (x−26)q(x−1). Plugging this into the original functional equation, we get q(x) q(x −1), i.e., q(x) a is a constant. Hence, p(x) ax(x −1) · · · (x −25).
47. 7|f (x), x 0, 1, −1, 2, −2. Thus, 7|e, 7|a + b + c + d, 7|a −b + c −d, 7|16a + 8b+4c+2d, 7|16a−8b+4c−2d. This implies 7|a+c, 7|b+d, 7|4a+c, 7|4b+d, or 7|a, b, c, d.
48. Let f (x) ax2 + bx + c with |f (x)| ≤1 for |x| ≤1. Since f ′(x) 2ax + b is a linear function, it assumes its maximum at x −1 or x 1. Hence, max|x|≤1 |2a + b| or |2a −b|, 2a + b 3 2(a + b + c) + 1 2(a −b + c) −2c 3 2f (1) + 1 2f (−1) −2f (0), 2a −b 1 2 + 3 2f (−1) −2f (0), |2a + b| ≤3 2 + 1 2 + 2 4, |2a −b| ≤1 2 + 3 2 + 2 4.
Hence, max |f (x)| ≤4. The polynomial f (x) 2x2 −1 satisfies the conditions of the problem, and we have |f ′(x)| |4x| 4 for x ±1.
49. Denote the LHS of the equation by f (x). Note that f (x) f ′(x)+x2n/(2n)!. Since f (x) is of even degree with a positive leading coefficient, it has an absolute minimum at, say, x z, where z ̸ 0, and f ′(z) 0. Thus f (z) z2n/(2n)! > 0, and f (x) is never zero for real x.
50. f (x) x3 + px2 + qx + r, f ′(x) 3x2 + 2px + q. The critical points are the solutions of f ′(x) 0. They give 3x −p ± p2 −3q. For p2 < 3q, there are no critical points, that is, f (x) is monotonically increasing and cannot have three real zeros. To have three real zeros, p2 ≥3q is a necessary condition, but by no means a sufficient one.
264 10. Polynomials 51. Let ak k2 −k + 41 for k 1, . . . , 40. Let A a1 · a2 · · · a40. Then for k 1, 2, . . . , 40, we have f (A + k) (A + k)2 −(A + k) + 41 A2 + (2k −1)A + ak.
Since ak|A, it follows that A is composite. This can be generalized to any quadratic polynomial f (n) an2 + bn + c. Let x f (1) · · · f (k). Then f (x + i) ax2 + 2aix + bx + f (i), and f (i)|f (x + i) for i 1, . . . k. The sequence of k consecutive values is x + 1, . . . , x + k.
52. With t x3 we get f (x) t(t + 1)(t + 2)(t + 3) (t2 + 3t)(t2 + 3t + 2) u(u + 2) (u + 1)2 −1 (t2 + 3t + 1)2 −1 x6 + 3x3 + 1 2 −1 ≥−1. We have fmin −1 for the real roots of x6 + 3x3 + 1 0.
53. For x 0, we get f (0) 0. If f (n) 0, then f (n + 1) 0 since (n + 1)f (n) (n + 2)f (n + 1). Thus f (n) has infinitely many zeros, i.e., f (x) ≡0.
54. (1 + x + · · · + xn)2 −xn 1 + x + · · · + xn−1 1 + x + · · · + xn+1 .
55. Hint: Of the two integers −a and 1 −a, exactly one is even. If f (a) 0, then f (x) (x −a)g(x). But f (0) −ag(a), and f (1) (1 −a)g(a). Both f (0) and f (1) cannot be odd.
56. Let P be the given polynomial and Q the polynomial to be found. Then Q(x3) (x3 −x3 1)(x3 −x3 2)(x3 −x3 1) P (x)P(ωx)P(ω2x) because of x3 −x3 1 (x − x1)(ωx −x1)(ω2x −x1), ω3 1. The calculation is simplified by means of the identity (u + v + w)(u + ωv + ω2w)(u + ω2v + ωw) u3 + v3 + w3 −3uvw.
Second solution. By brute force. P (x) x3+ax2+bx+c (x−x1)(x−x2)(x−x3), x1 + x2 + x3 −a, x1x2 + x2x3 + x3x1 b, x1x2x3 −c, Q(x) x3 + Ax2 + Bx + C (x −x3 1)(x −x3 2)(x −x3 3), A −(x3 1 + x3 2 + x3 3), B x3 1x3 2 + x3 2x3 3 + x3 3x3 1, C −(x1x2x3)3 c3, (x1 + x2 + x3)3 x3 1 + x3 2 + x3 3 + 3(x1 + x2 + x3)(x1x2 + x2x3 + x3x1) −3x1x2x3, −a3 −A −3ab + 3c ⇒A a3 −3ab + 3c, b3 (x1x2 + x2x3 + x3x1)3 B + 3bca −3c2 ⇒B b3 −3abc + 3c2, Q(x) x3 + (a3 −3ab + 3c)x2 + (b3 −3abc + 3c2)x + c3.
57. f (x) ≡0 is a solution. Now let f (x) ̸≡0. By comparing coefficients of both sides, we conclude that both the leading coefficient and f (0) are equal to 1. f (0) 1 is the product of all zeros. Let α be a zero. Then 2α3 + α is also a zero. The triangle inequality implies |α| > 1 ⇒|2α3 + α|2 ≥|2α3| −|α| > |α| > 1 Thus, we get infinitely many zeros by means of α1 α, αn+1 2α3 n +αn. Contradiction! But the product of all zeros is 1. So all zeros have absolute value 1. Let |α| 1. Then also |2α3 + α| |α||2α2 + 1| |2α2 + 1| 1, that is, 1 |2α2 + 1| ≥|2α2| −1 1.
Hence, α2 −1. We conclude that f (x) 1 + x2 n.
58. Let (x −a1) · · · (x −an)−1 f (x)g(x), where f (x), g(x) are polynomials with in-tegral coefficients. Then f (ai) −g(ai) ±1 for i 1, . . . , n. If the polynomials f, g and hence also f (x)+g(x) would be of degree ≤n−1, then f (x)+g(x) ≡0, since it has n zeros. Hence, we would have (x −a1) · · · (x −an) −1 −[f (x)]2.
This is a contradiction since the coefficient of xn on the left is 1, on the right < 0.
59. (a) f (z) 0 ⇒f (z2) 0, f [(z −1)2] 0. The set of zeros of f is finite and closed with respect to the map z →z2. Hence z lies in O or on the unit circle. Closure 10. Polynomials 265 with respect to the map z →(z −1)2 confines the possible zeros to 0, 1, that is, f (x) axm(x −1)n. Plugging this into the functional equation yields ax2m(x2 − 1)n + axm(x −1)na(x + 1)mxn 0.
First case: a 0, that is, f (x) 0.
Second case: a ̸ 0 ⇒xm(x + 1)n + axn(x + 1)m 0 ⇒1 + axn−m(x + 1)m−n 0 ⇒n m, a −1 ⇒f (x) −xn(x −1)n, n 0, 1, 2, · · ·.
(b) In a similar way, one proves f (x) ≡0 and f (x) −(x2 + x + 1)n, n ≥0.
60. We require that x3+y3+z3+kxyz (x+y+z)f (x, y, z). We set z −x−y and get x3+y3+z3−3kxyz x3+y3−(x+y)3−kxy(x+y),or−3x2y−3xy2−kxy(x+y) −3xy(x + y) −kxy(x + y), or −xy(x + y)(3 + k) 0. Hence, k −3.
61. No solution.
62. The equation x3 +x2y +xy2 +y3 8(x2 +xy +y2 +1) is symmetric in x, y. Thus it can be replaced by the elementary symmetric functions u x + y, and v xy.
We get (x2 + y2)(x + y) 8 (x + y)2 −xy + 1 , or u(u2 −2v) 8(u2 −v + 1), or u3 −2uv 8u2 −8v + 8. Here u 2t. Thus 8t3 −4tv 32t2 −8v + 8 ⇒ 2t3 −tv 8t2 −2v + 2. After solving for v and polynomial division, we get v 2t2 −4t −8 −18/(t −2).
There are only 12 values of t, which yield integer v, and of these only two values give integer (x, y): (8, 2), (2, 8).
63. We prove the statement by contradiction. Suppose there are two polynomials with integral coefficients, such that f (x) g(x)h(x), where g(x) and h(x) have degrees greater than one. Let f (x) a0 + a1x + · · · + an−1xn−1 + anxn, g(x) b0 + b1x + · · · + bmxm, h(x) c0 + c1x + · · · + cn−mxn−m.
We may assume that |b0| 3. Then |c0| 1, i.e., it is not divisible by 3. Let i be the smallest number such that bi is not divisible by 3. Then ai bic0 + (bi−1c1 + bi−2c2 + · · ·).
is not divisible by 3. Looking at f (x), we see that i ≥n −1. Hence, the degree of the polynomial h(x) is not larger than 1. Contradiction!
Thus, h(x) x ± 1, and h(x) has roots 1 or −1. The polynomial f (x) will have the same roots. But f (1) 9, and f (−1) (−1)n + 5(−1)n−1 + 3 ±1.
64. No solution.
65. No solution.
66. Instead of 8 3a2 + 2b2 we consider the maximum of 3 2 8 3a2 + 2b2 4a2 + 3b2. We use the following obvious lemma: |u| ≤1, |v| ≤1 ⇒|u −v| ≤2.
(1) There is equality iff u 1, v −1 or u −1, v 1. We apply the inequality (1) to the function |f (x)| ≤1 for x 1 and x 0 and get 2 ≥|f (1) −f (0)| |a + b + c −c| |a + b|. We get (a + b)2 ≤4.
(2) 266 10. Polynomials For x −1 and x 0, we get: 2 ≥|f (−1) −f (0)| |a −b + c −c| |a −b|.
Hence, (a −b)2 ≤4.
(3) From (2) and (3) we get 4a2 + 3b2 2(a + b)2 + 2(a −b)2 −b2 ≤16. We have equality if b 0, and therefore |a + b| |a −b| |a| 2. Then |f (1) − f (0)| |(a + c) −c)| |a| 2. From (1) we get |c| 1 and |a + c| 1.
Hence, we have either c 1, a −2, b 0 or c −1, a 2, b 0. In these two cases 0 ≤|x| ≤1 ⇒0 ≤x2 ≤1, −1 ≤2x2 −1 ≤1. Hence, |2x2 −1| | −2x2 + 1| |ax2 + bx + c| ≤1. Thus, 8 3a2 + 2b2 2 3(4a2 + 3b2) ≤2 3 · 16 102 3.
67. Setting a b c in (ii), we get P (2a, a) 0 for all a, that is, P (x, y) (x −2y)Q(x, y), (1) where Q is homogeneous of degree n −1. Since P (1, 0) Q(1, 0) 1, condition (ii) with b c says P(2b, a) + 2P (a + b, b) 0. From (1) we get (2b −2a)Q(2b, a) + 2(a −b)Q(a + b, b) 2(a −b) [Q(a + b, b) −Q(2b, a)] .
Hence, Q(a + b, b) Q(2b, a) whenever a ̸ b.
(2) But (2) holds also for a b. With a + b x, b y, a x −y, (2) becomes Q(x, y) Q(2y, x −y). Applying this functional equation repeatedly, we get Q(x, y) Q(2y, x−y) Q(2x−2y, 3y−x) Q(6y−2x, 3x−5y) · · · , (3) where the sum of the arguments is always x + y. Each member of (3) has the form Q(x, y) Q(x + d, y −d) with d 0, 2y −x, x −2y, 6y −3x, . . . .
(4) These values of d are all distinct if x ̸ y. For any fixed values x, y, the equation Q(x + d, y −d) −Q(x, y) 0 is a polynomial of degree n −1 in d, and if x ̸ 2y, it has infinitely many solutions, some of which are given by (4). Hence, for x ̸ 2y the equation Q(x + d, y −d) Q(x, y) holds for all d. By continuity it also holds for x 2y, that is, Q(x, y) is a function of the single variable x + y. Since it is homogeneous of degree n−1, we have Q(x, y) c(x +y)n−1, where c is a constant.
Since Q(1, 0) 1, we have c 1, and hence P (x, y) (x −2y)(x + y)n−1.
68. We set x(t) 2 cos t. This function maps 0 ≤t ≤π into 2 ≥x ≥−2. With the duplication formula for the cosine, we get P1(x) P1(2 cos t) 4 cos2 t −2 2 cos 2t, P2(x) P1(P1(x)) 4 cos2 2t −2 2 cos 4t, . . . , Pn(x) 2 cos 2nt.
The equation Pn(x) x is transformed into 2 cos 2nt 2 cos t with solutions 2nt ±t + 2kπ, k 0, 1, . . ., i.e., the following 2n values of t t 2kπ 2n −1 and t 2kπ 2n + 1 give 2n real distinct values of x 2 cos t satisfying the equation Pn(x) x.
10. Polynomials 267 69. Proof. a + b + c ≥0 ⇔1 + b a + c a ≥0 ⇔(1 −x1)(1 −x2) ≥0.
a −b + c ≥0 ⇔1 −b a + c a ≥0 ⇔1 + x1 + x2 + x1x2 ≥0 ⇔(1 + x1)(1 + x2) ≥0, a −c ≥0 ⇔1 −c a ≥0 ⇔1 −x1x2 ≥0.
Let s1 (1 −x1)(1 −x2), s2 (1 + x1)(1 + x2), s3 1 −x1x2. Obviously we have |xi| ≤1 for i 1, 2 ⇒sk ≥0 for k 1, 2.
We will show the inverse. Because of the symmetry in x1 and x2, it is sufficient to consider the cases x1 > 1 and x1 < −1.
x1 > 1, x2 < 1 ⇒s1 < 0, x1 > 1, x2 ≥1 ⇒s3 < 0, x1 < −1, x2 > −1 ⇒s3 < 0, x1 < −1, x2 > −1 ⇒s2 < 0.
70. Let z be a zero of f . Then z2 also is a zero. If |z| > 1, there are infinitely many zeros, which is impossible for polynomials. If 0 < |z| < 1 there will also be infinitely many zeros. So all zeros must lie in O or on the unit circle.
Let us find some such polynomials.
(a) Constant polynomials: f (x) ≡0, and f (x) ≡1.
(b) Linear polynomials: f (x) b + ax, a ̸ 0. Putting this into the functional equation, we get (b + ax)(b −ax) b + ax2 or ax2 + b −a2x2 + b2. Since a ̸ 0, we have a −1. b2 b implies b 0 or b 1. Thus we have two linear polynomial solutions, f (x) −x and f (x) 1 −x.
(c) Quadratic polynomials: f (x) ax2 + bx + c, a ̸ 0. We get f (x)f (−x) (ax2 + bx + c)(ax2 −bx + c) a2x4 + (2ac −b2)x2 + c2.
Comparing with f (x2) ax4 + bx2 + c we get a2 a, 2ac −b2 b, and c2 c.
Since a ̸ 0, we have the unique solution a 1. For c2 c, we have two solutions c 0 and c 1. For each of these values of c, we have two values for b. For c 0 we get b 0 and b −1. For c 1, we get b 1 and b −2. Thus we have four candidates: f (x) x2, f (x) x2 −x, f (x) x2 −2x + 1 (x −1)2, f (x) x2 + x + 1.
We rewrite the second and third function in the form f (x) −x(1 −x) and f (x) (1 −x)2. Now we can write a very general solution f (x) (−x)p(1 −x)q(x2 + x + 1)r, p, q, r ∈Z.
Since f (−x) xp(1 + x)q(x2 −x + 1)r, we have f (x2) (−x2)p(1 −x2)q(x4 + x2 + 1)r, so that f (x)f (−x) f (x2). Are these all polynomial solutions? Note that we also have some rational solutions. Indeed, p, q, r could also be negative.
268 10. Polynomials 71. We use the following lemma: m, n ∈Z, m ̸ n ⇒m −n|f (m) −f (n). For m ∈Z, m ̸ k, k + 1, k + 2, f (m) −f (k), f (m) −f (k + 1), f (m) −f (k + 2) (1) are divisible by m −k, m −(k + 1), m −(k + 2), respectively. These are three successive integers. Thus one of these is divisible by 3. Hence one of the integers (1) is divisible by 3, that is, 3|f (m).
72. SinceallcoefficientsofP (x)arenonnegative,noneofitsrootsx1, . . . , xn arepositive.
Thus, P (x) has the form P (x) (x + y1) · · · (x + yn), where yi −xi > 0, i 1, . . . , n. Hence, 2 + yi 1 + 1 + yi ≥3 3 1 · 1 · yi 3 3 √yi, i 1, . . . , n.
Since y1y2 · · · yn 1, by Vieta’s theorem we get P (2) (2 + y1) · · · (2 + yn) ≥3n 3 √y1 · · · yn 3n.
73. Suppose the given polynomial f (x) can be represented as a product of two polyno-mials over Z of degree less than 105: f (x) g(x)h(x), and let β1, β2, . . . , βk be the complex roots of h(x). By Vieta’s theorem, their product is an integer, and hence |β1 · · · βk| 105 √ 9 k ∈N, which is impossible for k < 105. Thus the answer is No!
74. Suppose there is a representation in the form f (x) (x −b) · g(x). Then f (b) 0 and hence b5 −b −a. Since 5 is a prime, by Fermat’s theorem, b5 −b ≡0 mod 5.
Thus, a is divisible by 5. Contradiction!
Now suppose that there is a representation in the form f (x) (x2 −bx −c) · h(x).
Dividing x5 −x + a by x2 −bx −c, we get the remainder (b4 + 3b2c + c2 −1)x + (b3c + 2bc2 + a). This must be the zero polynomial. Hence b4 + 3b2c + c2 −1 0 and b3c+2bc2 +a 0. This implies b(b4 +3b2c+c2 −1)−3(b3c+2bc2 +a) 0.
Expanding and collecting terms, we get b5−b−5bc2 3a. The left side is a multiple of 5. Hence 5|3a, or 5|a. Contradiction!
75. The equation is reciprocal. So we set y x + 1 x and get ay + b 2 −y2. The CS inequality yields (2 −y2)2 (ay + b · 1)2 ≤(a2 + b2)(y2 + 1), a2 + b2 ≥(2−y2)2 y2+1 (2−z)2 z+1 f (z), where z y2 and z ≥4. Since f (z) is monotonically increasing if z ≥2, we get a2 + b2 ≥f (4) 4 5. Equality holds, for example, if y a b and z y2 4, and thus we have, for example, a −4 5, b −2 5, and the original equation has a root x 1.
76. Suppose that each of P(x), Q(x), R(x) has two roots. Then b2 > 4ac, a2 > 4bc, c2 > 4ab. Multiplying the inequalities, we get a2b2c2 > 64a2b2c2. Contradiction!
77. a2 + ab + b2 ≥3(a + b −1) is equivalent to a2 + (b −3)a −b2 −3b + 3 ≥0. The LHS p(a) of the second inequality is a quadratic polynomial in a with discriminant D −3(b −1)2 ≤0. This is exactly the condition for p(a) ≥0.
10. Polynomials 269 78. This problem looks hopeless. Since it cannot be hopeless, it must be trivial, that is, it splits into a straight line and a conic or into three linear factors. We start with the simpler case of three linear factors. Then one of the lines must pass through the origin, that is, one of the factors must be x −2y 0. Replacing x by 2y, in the original equation we get an identity. Hence x −2y is a factor of the equation. We get the other factor 4x2 + 12xy −12x + 9y2 −18y + 5 0 dividing by x −2y.
We transform it into the form (2x + 3y)2 −6(2x + 3y) + 5 0 ⇐ ⇒(2x + 3y −5)(2x + 3y −1) 0.
From (x −2y)(2x + 3y −5)(2x + 3y −1) 0, by inspection we get the solution set consisting of the pair (1, 1) and the infinitely many pairs (2n, n), n ∈N.
79. This is a quadratic equation in the variable x. To have any real solutions, its discrim-inant D must be nonnegative. We write this quadratic in standard form and compute its discriminant D: 8x2 + (4y2 + 4y −40)x + y4 −11y2 −8y + 52 0, D 16(y2 + y −10)2 −32(y4 −11y2 −8y + 52) −16(y2 −y −2)2.
We must have D 0 or y2 −y −2 0 with two solutions y1 2 and y2 −1.
From x −(y2 + y −10)/4, we get x1 1 and x2 5/2.
80. The following factorization (which is not unique) is the most natural one: x8 + 98x4 + 1 (x4 + 1)2 + 96x4 (x4 + 1)2 + 16x2(x4 + 1) + 64x4 −16x2(x4 + 1) + 32x4 (x4 + 8x2 + 1)2 −16x2(x4 −2x2 + 1) (x4 + 8x2 + 1)2 −(4x3 −4x)2 (x4 −4x3 + 8x2 + 4x + 1)(x4 + 4x3 + 8x2 −4x + 1).
81. We observe that p(z)−p(x) is divisible by z−x. Take a z such that z−x is divisible by p(x), for example, z q(x) x +p(x). Thus p[q(x)] is divisible by p(x). Since the degree of p[q(x)] is greater than that of p(x), the second factor is not constant.
82. No solution.
11 Functional Equations Equations for unknown functions are called functional equations. We dealt with these already in the chapters on sequences and polynomials. Sequences and poly-nomials are just special functions.
Here are five examples of functional equations of a single variable: f (x) f (−x), f (x) −f (−x), f ◦f (x) x, f (x) f x 2 ; f (x) cos x 2f x 2 , f (0) 1, f continuous.
The first three properties characterize even functions, odd functions, and involu-tions, respectively. Many functions have the fourth property. On the other hand, the last condition makes the solution unique.
Here are examples of famous functional equations in two variables: f (x + y) f (x) + f (y), f (x + y) f (x)f (y), f (xy) f (x) + f (y), and f (xy) f (x)f (y). These are Cauchy’s functional equations.
f x+y 2 f (x)+f (y) 2 . This is Jensen’s functional equation.
f (x + y) + f (x −y) 2f (x)f (y). This is d’Alambert’s functional equation.
g(x + y) g(x)f (y) + f (x)g(y), f (x + y) f (x)f (y) −g(x)g(y), g(x −y) g(x)f (y) −g(y)f (x), f (x −y) f (x)f (y) + g(x)g(y).
The last four functional equations are the addition theorems for the trigonometric functions f (x) cos x and g(x) sin x.
Usually a functional equation has many solutions, and it is quite difficult to find all of them. On the other hand it is often easy to find all solutions with 272 11. Functional Equations some additional properties, for example, all continuous, monotonic, bounded, or differentiable solutions.
Withoutadditionalassumptions,itmaybepossibletofindonlycertainproperties of the functions. We give some examples: E1. First we consider the equation f (xy) f (x) + f (y).
(1) One solution is easy to guess: f (x) 0 for all x. This is the only solution which is defined for x 0. If x 0 belongs to the domain of f , then we can set y 0 in (1), and we get f (0) f (x) + f (0), implying f (x) 0 for all x. Let x 1 be in the domain of f . With x y 1, we get f (1) 2f (1), or f (1) 0.
(2) If both 1 and −1 belong to the domain, then f is an even function, i.e., f (−x) f (x) for all x. To prove this, we set x y −1 in (1), and because of (2), we get f (1) 2f (−1) 0 ⇒f (−1) 0.
Setting y −1 in (1), we get f (−x) f (x) + f (1), or f (−x) f (x) for all x.
Assume that f is differentiable for x > 0. We keep y fixed and differentiate for x. Then we get yf ′(xy) f ′(x). For x 1, one gets yf ′(y) f ′(1). Change of notation leads to f ′(x) f ′(1)/x, or f (x) 5 x 1 f ′(1) t dt f ′(1) ln x.
If the function is also defined for x < 0, then we have f (x) f ′(1) ln |x|.
E2. A famous classical functional equation is f (x + y) f (x) + f (y).
(1) First, we try to get out of (1) as much information as possible without any additional assumptions. y 0 yields f (x) f (x) + f (0), that is, f (0) 0.
(2) For y −x, we get 0 f (x) + f (−x), or f (−x) −f (x).
(3) Now we can confine our attention to x > 0. For y x, we get f (2x) 2f (x), and by induction, f (nx) nf (x) for all n ∈N.
(4) 11. Functional Equations 273 For rational x m n , that is, n · x m · 1, by (4) we get f (n · x) f (m · 1), nf (x) mf (1), and f (x) m n f (1).
(5) If we set f (1) c, then, from (2), (3), (5), we get f (x) cx for rational x. That is all we can get without additional assumptions.
(a) Suppose f is continuous. If x is irrational, then we choose a rational sequence xn with limit x. Because of the continuity of f , we have f (x) lim xn→x f (xn) lim xn→x cxn cx.
Then we have f (x) cx for all x.
(b) Let f be monotonically increasing. If x is irrational, then we choose an in-creasing and a decreasing sequence rn and Rn of rational numbers, which converge toward x. Then we have crn f (rn) ≤f (x) ≤f (Rn) cRn.
For n →∞, both crn and cRn converge to cx. Thus f (x) cx for all x.
(c) Let f be bounded on [a, b], that is, |f (x)| < M for all x ∈[a, b].
We show that f is also bounded on [0, b −a]. If x ∈[0, b −a], then x + a ∈[a, b]. From f (x) f (x + a) −f (a), we get |f (x)| < 2M.
If we set b −a d, then f is bounded on [0, d]. Let c f (d)/d and g(x) f (x) −cx. Then g(x + y) g(x) + g(y).
Furthermore, we have g(d) f (d) −cd 0 and g(x + d) g(x) + g(d) g(x), that is, g is periodic with period d. As the difference of two bounded functions, g is also bounded on [0, d]. From the periodicity, it follows that g is bounded on the whole number line. Suppose there is an x0, so that g(x0) ̸ 0. Then g(nx0) ng(x0). By choosing n sufficiently large, we can make |ng(x0)| as large as we want.
This contradicts the boundedness of g. Hence, g(x) 0 for all x, that is, f (x) cx for all x.
In 1905 G. Hamel discovered “wild” functions that are nowhere bounded and also satisfy the functional equation f (x+y) f (x)+f (y). We are looking for “tame” 274 11. Functional Equations solutions. If we succeed in finding a solution for all rationals, then we can extend them to reals by continuity or monotonicity, etc.
E3. Another classical equation is f (x + y) f (x)f (y).
(1) If there is an a such that f (a) 0, then f (x + a) f (x)f (a) 0 for all x, that is, f is identically zero. For all other solutions, f (x) ̸ 0 everywhere. For x y t/2, we get f (t) f 2 t 2 > 0.
The solutions we are looking for are everywhere positive. For y 0, we get f (x) f (x)f (0) from (1), that is, f (0) 1. For x y, we get f (2x) f 2(x), and by induction f (nx) f n(x).
(2) Let x m n (m, n ∈N), that is, n · x m · 1. Applying (2), we get f (nx) f (m · 1) ⇒f n(x) f m(1) ⇒f (x) f m n (1). If we set f (1) a, then f m n a m n , that is, f (x) ax for rational x. With a weak additional assumption (continuity, monotonicity, boundedness), as in E2, we can show that f (x) ax for all x.
The following procedure is simpler: Since f (x) > 0 for all x, we can take loga-rithms in (1): ln ◦f (x + y) ln ◦f (x) + ln ◦f (y).
Let ln ◦f g. Then g(x + y) g(x) + g(y) ⇒g(x) cx ⇒ln ◦f (x) cx, and f (x) ecx.
E4. We treat the following equation more generally: f (xy) f (x) + f (y), x, y > 0.
(1) We set x eu, y ev, f (eu) g(u). Then (1) is transformed into g(u + v) g(u) + g(v) with solution g(u) cu, and f (x) c ln x, as in E1, where we used differentiability.
E5. Next we consider the last Cauchy equation f (xy) f (x)f (y).
(1) 11. Functional Equations 275 We assume x > 0 and y > 0. Then we set x eu, y ev, f (eu) g(u) and get g(u + v) g(u) + g(v) with the solution g(u) ecu (eu)c xc.
f (x) xc and with the trivial solution f (x) 0 for all x.
If we require (1) for all x ̸ 0, y ̸ 0, then x y t and x y −t give f 2(t) f (t2) f (−t)f (−t) and f (−t) f (t) tc (or 0), −f (t) −tc.
In this case the general continuous solutions are (a) f (x) |x|c, (b) f (x) sgn x · |x|c, (c) f (x) 0.
E6. Now we come to Jensen’s functional equation f x + y 2 f (x) + f (y) 2 .
(1) We set f (0) a and y 0 and get f x 2 f (x)+a 2 . Then f (x) + f (y) 2 f x + y 2 f (x + y) + a 2 , f (x + y) f (x) + f (y) −a.
With g(x) f (x) −a, we get g(x + y) g(x) + g(y), g(x) cx, and f (x) cx + a.
E7. Now we come to our last and most complicated example f (x + y) + f (x −y) 2f (x)f (y).
(1) We want to find the continuous solutions of (1). First we eliminate the trivial solution f (x) 0 for all x. Now y 0 ⇒2f (x) 2f (x)f (0) ⇒f (0) 1, x 0 ⇒f (y) + f (−y) 2f (0)f (y) ⇒f (−y) f (y), that is, f is an even function. For x ny, we get f [(n + 1)y] 2f (y)f (ny) −f [(n −1)y] .
(2) 276 11. Functional Equations For y x, we get f (2x) + f (0) 2f 2(x). From this we conclude with t 2x that f 2 t 2 f (t) + 1 2 .
(3) (2) and (3) are satisfied by the functions cos and cosh. Since f (0) 1 and f is continuous, we have f (x) > 0 in [−a, a] for sufficiently small a > 0. Thus, f (a) > 0.
(a) First case. 0 < f (a) ≤1. Then there will be a c from 0 ≤c ≤π 2 , so that f (a) cos c. We show that, for any number of the form x (n/2m)a, f (x) cos c a x.
(4) For x a, this is valid by definition of c. Because of (3), for x a/2, f 2 a 2 f (a) + 1 2 cos c + 1 2 cos2 c 2.
Because of f (a/2) > 0, cos c 2 > 0, we conclude that f a 2 cos c 2.
(5) Suppose (5) is valid for x a/2m. Then (3) implies f 2 a 2m+1 f a 2m + 1 2 cos2 c 2m+1 or f a 2m+1 cos c 2m+1 , that is, f (a/2m) cos (c/2m) for every natural number m. Because of (2) for n 2, f 3 2m a f 3 · a 2m 2f a 2m f a 2m−1 −f a 2m 2 cos c 2m cos c 2m−1 −cos c 2m cos 3 2m c.
Since (4) is valid for x [(n −1)/2m]a and x (n/2m)a, we conclude from (2) for x [(n −1)/2m]a and x (n/2m)a, that f n + 1 2m a cos n + 1 2m c.
Hence, we have f n 2m a cos n 2m c for n, m ∈{0, 1, 2, 3, . . .}.
11. Functional Equations 277 Since f is continuous and even, we have f (x) cos c a x for all x.
Second case. If f (a) > 1, then there is a c > 0, so that f (a) cosh c.
One can show exactly as in the first case that f (x) cosh c a x for all x.
Thus, the functional equation (1) has the following continuous solutions: f (x) 0, f (x) cos bx, f (x) cosh bx.
This list also contains f (x) 1 for b 0.
(b) We want to find all differentiable solutions of (1). Since differentiability is a far more powerful property than continuity, it will be quite easy to find all solutions of f (x + y) + f (x −y) 2f (x)f (y). We differentiate twice with respect to each variable: With respect to x: f ′′(x + y) + f ′′(x −y) 2f ′′(x)f (y).
With respect to y: f ′′(x + y) + f ′′(x −y) 2f (x)f ′′(y).
From both equations we conclude that f ′′(x) · f (y) f (x) · f ′′(y) ⇒f ′′(x) f (x) f ′′(y) f (y) c ⇒f ′′(x) cf (x), c −ω2 ⇒f (x) a cos ωx + b sin ωx, c ω2 ⇒f (x) a cosh ωx + b sinh ωx.
f (0) 1 and f (−x) f (x) result in f (x) cos ωx and f (x) cosh ωx, respectively.
Problems 1. Find some (all) functions f with the property f (x) f x 2 for all x ∈R.
2. Find all continuous solutions of f (x + y) g(x) + h(y).
3. Find all solutions of the functional equation f (x + y) + f (x −y) 2f (x) cos y.
4. The function f is periodic, if, for fixed a and any x, f (x + a) 1 + f (x) 1 −f (x).
5. Find all polynomials p satisfying p(x + 1) p(x) + 2x + 1.
278 11. Functional Equations 6. Find all functions f which are defined for all x ∈R and, for any x, y, satisfy xf (y) + yf (x) (x + y)f (x)f (y).
7. Find all real, not identically vanishing functions f with the property f (x)f (y) f (x −y) for all x, y.
8. Find a function f defined for x > 0, so that f (xy) xf (y) + yf (x).
9. The rational function f has the property f (x) f (1/x). Show that f is a rational function of x + 1/x.
Remark. A rational function is the quotient of two polynomials.
10. Find all “tame” solutions of f (x + y) + f (x −y) 2 [f (x) + f (y)].
11. Find all “tame” solutions of f (x + y) −f (x −y) 2f (y).
12. Find all “tame” solutions of f (x + y) + f (x −y) 2f (x).
13. Find all tame solutions of f (x + y) f (x)f (y) f (x) + f (y).
14. Find all tame solutions of f 2(x) f (x + y)f (x −y). Note the similarity to 11.
15. Find the function f which satisfies the functional equation f (x) + f 1 1 −x x for all x ̸ 0, 1.
16. Find all continuous solutions of f (x −y) f (x)f (y) + g(x)g(y).
17. Let f be a real-valued function defined for all real numbers x such that, for some positive constant a, the equation f (x + a) 1 2 + f (x) −f 2(x) holds for all x.
(a) Prove that the function f is periodic, i.e., there exists a positive number b such that f (x + b) f (x) for all x.
(b) For a 1, give an example of a nonconstant function with the required properties (IMO 1968).
18. Find all continuous functions satisfying f (x + y)f (x −y) [f (x)f (y)]2.
19. Let f (n) be a function defined on the set of all positive integers and with all its values in the same set. Prove that if f (n + 1) > f [f (n)] for each positive integer n, then f (n) n for each n (IMO 1977).
20. Find all continuous functions in 0 which satisfy the relations f (x + y) f (x) + f (y) + xy(x + y), x, y ∈R.
11. Functional Equations 279 21. Find all functions f defined on the set of positive real numbers which take positive real values and satisfy the conditions: (i) f [xf (y)] yf (x) for all positive x, y; (ii) f (x) →0 as x →∞ (IMO 1983).
22. Find all functions f , defined on the nonnegative real numbers and taking nonnegative real values, such that (i) f [xf (y)] f (y) f (x + y) for all x, y ≥0; (ii) f (2) 0; (iii) f (x) ̸ 0 for 0 ≤x < 2 (IMO 1986).
23. Find a function f : Q+ →Q+, which satisfies, for all x, y ∈Q+, the equation f (xf (y)) f (x) /y (IMO 1990).
24. Find all functions f : R →R such that f x2 + f (y) y + [f (x)]2 for all x, y ∈R (IMO 1992).
25. Does there exist a function f : N →N such that f (1) 2, f [f (n)] f (n)+n, f (n) < f (n + 1) for all n ∈N (IMO 1993)?
26. Find all continuous functions f : R →R+ which transform three terms of the arithmetic progression x, x + y, x + 2y into corresponding terms f (x), f (x + y), f (x + 2y) of a geometric progression, that is, [f (x + y)]2 f (x) · f (x + 2y).
27. Find all continuous functions f satisfying f (x + y) f (x) + f (y) + f (x)f (y).
28. Guess a simple function f satisfying f 2(x) 1 + xf (x + 1).
29. Find all continuous functions which transform three terms of an arithmetic progres-sion into three terms of an arithmetic progression.
30. Find all continuous functions f satisfying 3f (2x + 1) f (x) + 5x.
31. Which function is characterized by the equation xf (x) + 2xf (−x) −1?
32. Find the class of continuous functions satisfying f (x + y) f (x) + f (y) + xy.
33. Let a ̸ ±1. Solve f (x/(x −1)) af (x) + φ(x), where φ(x) is a given function, which is defined for x ̸ 1.
34. The function f is defined on the set of positive integers as follows: f (1) 1, f (3) 3, f (2n) f (n), f (4n + 1) 2f (2n + 1) −f (n), f (4n + 3) 3f (2n + 1) −2f (n).
Find all values of n with f (n) n and 1 ≤n ≤1988 (IMO 1988).
280 11. Functional Equations 35. A function f is defined on the set of rational numbers as follows: f (0) 0, f (1) 1, f (x) f (2x)/4 for 0 < x < 1 2, 3 4 + f (2x −1)/4 for 1 2 ≤x < 1.
Let a 0.b1b2b3 · · · be the binary representation of a. Find f (a).
36. Find all polynomials over C satisfying f (x)f (−x) f (x2).
37. The strictly increasing function f (n) is defined on the positive integers and it assumes positive integral values for all n ≥1. In addition, it satisfies the condition f [f (n)] 3 · n. Find f (1994) (IIM 1994).
38. (a) The function f (x) is defined for all x > 0 and satisfies the conditions (1) f (x) is strictly increasing on (0, +∞), (2) f (x) > −1/x for x > 0, (3) f (x) · f (f (x) + 1/x) 1 for all x > 0.
Find f (1).
(b) Give an example of a function f (x) which satisfies (a).
39. Find all sequences f (n) of positive integers satisfying f [f [f (n)]] + f [f (n)] + f (n) 3n.
40. Find all functions f : N0 →N0, such that f [m + f (n)] f [f (m)] + f (n) for all m, n ∈N0 (IMO 1996).
Solutions 1. Any constant function has the required property. Another example is the function f defined by f (x) |x|/x, x ̸ 0. For 0, one can define f arbitrarily.
There are infinitely many solutions. One can get all solutions as follows: Take any interval of the form [a, 2a]. For instance, let us take [1, 2]. Define f in this interval, arbitrarily, except f (1) f (2). Then f is defined for all real x > 0. Take the graph of f in [1, 2], and stretch it horizontally by the factor 2n (n an integer). Then you get the graph of f in the interval [2n, 2n+1]. We can define f (0) as we please.
For negative x we can again choose an interval [b, 2b], b < 0, define f in this interval arbitrarily except f (b) f (2b), and extend the definition to all negative x by stretching it.
2. This equation can be reduced to Cauchy’s equation. Set y 0, h(0) b. You get f (x) g(x) + b, g(x) f (x) −b.
For x 0, g(0) a we get f (y) a + h(y), h(y) f (y) −a. Thus, f (x + y) f (x) + f (y) −a −b. So with f0(z) f (z) −a −b, we have f0(x + y) f0(x) + f0(y), i.e., f0(x) cx, and f (x) cx + a + b, g(x) cx + a, h(x) cx + b.
11. Functional Equations 281 3. For y π/2, the right side disappears. We substitute x 0, y t, x π 2 + t, y π 2 , x π 2 , y π 2 + t, and we get f (t)+f (−t) 2a cos t, f (π +t)+f (t) 0, f (π +t)+f (−t) −2b sin t, where a f (0), b f π 2 . Hence, f (t) a cos t + b sin t.
4. We find that f (x + 2a) −1/f (x), i.e., f (x + 4a) f (x). Thus 4a is a period of f .
5. We can guess the solution p(x) x2. Is it the only one? A standard method for answering this question is to introduce the difference f (x) p(x) −x2. The given functional equation becomes f (x + 1) f (x). So f (x) c, a constant. Thus p(x) x2 + c. We must check if this solution satisfies the original equation, which is indeed the case.
6. y x ⇒f (x) f 2(x) ⇒f (x) (f (x) −1) 0 for all x. Continuous solutions are f (x) ≡0, f (x) ≡1. There are many more discontinuous solutions. On any subset A of R, set f (x) 0. On R\A, set f (x) 1. But there is a restriction, which we find by setting y −x. It shows that f (−x) f (x) for all x, i.e., f is an even function.
7. y 0 ⇒f (x)f (0) f (x) for all x. Since f is not identically vanishing, we must have f (0) 1. y x ⇒f (x)f (x) 1 for all x. We get two continuous functions f (x) ≡1 and f (x) ≡−1. There are many discontinuous functions, e.g., f (x) 1 on any subset A of R, and f (x) −1 on R \ A.
8. Let g(x) (f (x))/x. Then we get the Cauchy equation g(xy) g(x) + g(y) with the solution g(x) c ln x. This implies f (x) cx ln x.
9. Suppose f (x) xk(a0xn + a1xn−1 + · · · + an) xl(b0xm + · · · + bm) , where a0, b0, an, bm are not zero. Using the relation f (x) f (1/x), we get x2(l−k)+m−n(anxn + · · · + a0) (bmxm + · · · + b0) ≡a0xn + · · · + an b0xm + · · · + bm .
(1) From here we get m −n 2(k −l), where m and n have the same parity. From (1) we conclude that Pm(x) bmxm + · · · + b0 ≡b0xm + · · · + bm and Pn(x) anxn + · · · + a0 ≡a0xn + · · · + an, i.e., a0 an, a1 an−1, · · · ; b0 bm, b1 bm−1, . . .. Hence Pm(x) and Pn(x) are reciprocal polynomials, which can be represented as follows: For even n: n 2r, then P2r(x) xngr(z), where z x +1/x and g(z) is a polynomial of degree r. If n is odd: n 2r + 1, then P2r+1(x) (x + 1)xrhr(z), where z x + 1/x, and hr(z) is a polynomial of degree r.
Furthermore, there are two possibilities: 282 11. Functional Equations (a) m 2s, n 2r. Then f (x) xkxrgr(z) xlxshs(z) g(z) h(z).
(b) m 2s + 1, n 2r + 1. Then f (x) (x + 1)xk+rgr(z) (x + 1)xl+shs(z) g(z) h(z).
10. For y 0, we get 2f (x) 2f (x) + 2f (0), or f (0) 0. For x y, we have f (2x) 4f (x). We prove by induction that f (nx) n2f (x) for all x. Now let x p/q. Then qx p ·1, f (qr) f (p ·1), q2f (x) p2f (1). With f (1) a, we get f (x) ax2 for all rational x. By continuity we can extend this to all continuous functions. By putting f (x) ax2 into the original equation, we see that it is indeed satisfied.
11. For y 0, we get f (x) −f (x) 2f (0), or f (0) 0. For y x, we get f (2x) 2f (x) for all x. By induction we prove that f (nx) nf (x). Now let x p/q or qx p·1. Then f (qx) f (p·1) ⇒qf (x) pf (1) ⇒f (x) f (1)x for all rational x. By continuity this can be extended to all real x. Putting f (x) ax into the functional equation, we see that it is the solution.
12. We want to solve the functional equation f (x + y) + f (x −y) 2f (x). y x yields f (2x) + f (0) 2f (x), or f (2x) 2f (x) + b with b −f (0). Now f (2x +x)+f (2x −x) 2f (2x) yields f (3x)+f (x) 2 (2f (x) + b), or f (3x) 3f (x)+2b. We guess f (nx) nf (x)+(n−1)b, and we prove this by induction. Now let x p/q ⇔qx p · 1 with p, q ∈N. Then f (qx) f (p · 1), or qf (x) + (q − 1)b pf (1)+(p−1)b, or f (x) f (1)x+(x−1)b, or f (x) [f (0) + f (1)] x−b.
With f (0) + f (1) a and f (0) b, we finally get f (x) ax + b. A check shows that this is indeed a solution.
13. Setting g(x) 1/f (x), we get Cauchy’s equation g(x + y) g(x) + g(y) with the solution g(x) cx. Thus f (x) 1/cx is the general continuous solution.
14. Taking logarithms on both sides, we get 2g(x) g(x + y) + g(x −y). Here g(x) ln ◦f (x), that is, g(x) ax + b. Thus f (x) eax+b, or f (x) rsx.
15. We repeatedly replace x g − →1/(1 −x) and get x g − → 1 1 −x g − →1 −1 x g − →x.
We get the following equations: f (x) + f 1 1 −x x, f 1 1 −x + f 1 −1 x 1 1 −x , f 1 −1 x + f (x) 1 −1 x .
Eliminating f 1 1 −x and f 1 −1 x we get f (x) 1 2 1 + x −1 x − 1 1 −x .
A check shows that this function indeed satisfies the functional equation.
11. Functional Equations 283 16. Hint: Interchanging x with y, we see that f (−x) f (x) for all x. Setting y 0, we get f (0)2 f 2(x)+g2(x). x y 0 implies f (0) f 2(0)+g2(0). y 0 implies f (x) f (x)f (0) + g(x)g(0). Now f (0) 0 would imply g(0) 0 and f (x) ≡0 for all x. Thus, f (0) ̸ 0. But f (x) [1 −f (0)] g(x)g(0). Thus, f (0) 1, and hence g(0) 0. y −x implies f (2x) f 2(x) + g(x)g(−x). We should get f (x) cos x and g(x) sin x.
17. We have f (x + a) ≥1 2, and so f (x) ≥1 2 for all x. If we set g(x) f (x) −1 2, we have g(x) ≥0 for all x. The given functional equation now becomes g(x + a) 1 4 −[g(x)]2.
Squaring, we get [g(x + a)]2 1 4 −[g(x)]2 for all x, (1) and thus also [g(x + 2a)]2 1 4 −[g(x + a)]2.
These two equations imply [g(x + 2a)]2 [g(x)]2. Since g(x) ≥0 for all x, we can take square roots to get g(x + 2a) g(x), or f (x + 2a) −1 2 f (x) −1 2, and f (x + 2a) f (x) for all x.
This shows that f (x) is periodic with period 2a.
(b) To find all solutions, we set h(x) 4[g(x)]2 −1 2. Now (1) becomes h(x + a) −h(x).
(2) Conversely, if h(x) ≥ 1 2 and satisfies (2), then g(x) satisfies (1). An example for a 1 is furnished by the function h(x) sin2 π 2 x −1 2 which satisfies (2) with a 1. For this h, g(x) 1 2| sin(πx/2)| and f (x) 1 2 sin π 2 x + 1 2.
In fact, h(x) can be defined arbitrarily in 0 ≤x < a subject to the condition |h(x)| ≤1 2 and extended to all x by (2).
18. To find the solution of f (x −y)f (x + y) [f (x)f (y)]2, we observe that we can assume f to be nonnegative. In fact, all we can say about a positive f is also valid for a negative f . The three trivial solutions f (x) ≡0, 1, −1 will be excluded from now on. y 0 ⇒f (x)2 f (x)2f (0)2 ⇒f (0)2 1 ⇒f (0) 1.
x 0 ⇒f (y)f (−y) f (y)2 ⇒f (y) f (−y). Thus, f is an even function.
x y ⇒f (2x) f (x)4. By induction we get f (nx) f (x)n2. This can be extended to rationals and then reals as in E2. Finally, we get f (x) f (1)x2 for all x.
Another approach introduces g ln ◦f to get g(x+y)+g(x−y) 2 (g(x) + g(y)).
This suggests the identity (x+y)2+(x−y)2 2(x2+y2). Thus we guess g(x) ax2 and f (x) eax2. It remains to be proved that the guess is unique.
284 11. Functional Equations 19. f has a unique minimum at n 1. For, if n > 1, we have f (n) > f [f (n −1)]. By the same reasoning, we see that the second smallest value is f (2), etc. Hence, f (1) < f (2) < f (3) < · · · .
Since f (n) ≥1 for all n, we also have f (n) ≥n. Suppose that, for some positive integer k, we have f (k) > k. Then f (k) ≥k + 1. Since f is increasing, f (f (k)) ≥ f (k + 1), contradicting the given inequality. Hence f (n) n for all n.
20. It is easy to guess the solution from this property. The function x3/3 satisfies the relationship. So we consider g(x) f (x) −x3/3. For g we get the functional equation g(x + y) g(x) + g(y). Since g(x) cx is the only continuous solution in 0, we have f (x) cx + x3/3.
21. We show that 1 is in the range of f . For an arbitrary x0 > 0, let y0 1/f (x0).
Then (i) yields f [x0f (y0)] 1, so 1 is in the range of f . In the same way, we can show that any positive real is in the range of f . Hence there is a value y such that f (y) 1. Together with x 1 in (i), this gives f (1 · 1) f (1) yf (1). Since f (1) > 0 by hypothesis, it follows that y 1, and f (1) 1. We set y x in (i) and get f [xf (x)] xf (x) for all x > 0.
(1) Hence, xf (x) is a fixed point of f . If a and b are fixed points of f , that is, if f (a) a and f (b) b, then (i) with x a, y b implies that f (ab) ba, so ab is also a fixed point of f . Thus the set of fixed points of f is closed under multiplication. In particular, if a is a fixed point, all nonnegative integral powers of a are fixed points.
Since f (x) →0 for x →∞by (ii), there can be no fixed points > 1. Since xf (x) is a fixed point, follows that xf (x) ≤1 ⇔f (x) ≤1 x for all x.
(2) Let a zf (z), so f (a) a. Now set x 1/a and y a in (i) to give f 1 a f (a) f (1) 1 af 1 a , f 1 a 1 a , f 1 zf (z) 1 zf (z).
This shows that 1/xf (x) is also a fixed point of f for all x > 0. Thus, f (x) ≥1/x.
Together with (2) this implies that f (x) 1 x .
(3) The function (3) is the only solution satisfying the hypothesis.
22. No solution.
23. If f (y1) f (y2), the functional equation implies that y1 y2. For y 1, we get f (1) 1. For x 1, we get f (f (y)) 1 y for all y ∈Q+. Applying f to this implies that f (1/y) 1/f (y) for all y ∈Q+. Finally setting y f (1/t) yields f (xt) f (x) · f (t) for all x, t ∈Q+.
Conversely, it is easy to see that any f satisfying (a) f (xt) f (x)f (t), (b) f [f (x)] 1/x for all x, t ∈Q+ solves the functional equation.
11. Functional Equations 285 A function f : Q+ →Q+ satisfying (a) can be constructed by defining arbitrarily on prime numbers and extending as f pn1 1 pn2 2 · · · pnk k [f (p1)]n1 [f (p2)]n2 · · · [f (pk)]nk , where pj denotes the jth prime and nj ∈Z. Such a function will satisfy (b) for each prime.
A possible construction is as follows: f (pj) pj+1 if j is odd, 1 pj−1 if j is even.
Extending it as above, we get a function f : Q+ →Q+. Clearly f [f (p)] 1/p for each prime p. Hence f satisfies the functional equation.
24. No solution.
25. Starting with f (1) 2 and using the rule f [f (n)] f (n)+n, we get, successively, f (2) 2 + 1 3, f (3) 3 + 2 5, f (5) 5 + 3 8, f (8) 8 + 5 13, . . .
that is, the map of a Fibonacci number is the next Fibonacci number. Complete this by induction.
It remains to assign other positive integers to the remaining numbers satisfying the functional equation. We use Zeckendorf’s theorem, which says that every positive integer n has a unique representation as a sum of non-neighboring Fibonacci num-bers. We have proved this in Chapter 8, problem 29. We write this representation in the form n m j1 Fij , |ij −ij−1| ≥2, where the summands have increasing indices. We will prove that the function f (n) m j1 Fij +1 satisfies all conditions of the problem. Indeed, since 1 represents itself as a Fibonacci number, we have f (1) 2, the next Fibonacci number. Then f [f (n)] f ⎛ ⎝ m ij +1 Fij +1 ⎞ ⎠ m j1 Fi1+2 m j1 Fij +1 + Fij m j1 Fij +1 + m j1 Fij f (n) + n.
Now we distinguish two cases.
(a) The Fibonacci representation of n contains neither F1 nor F2. Then the represen-tation of n + 1 contains the additional summand 1. The representations of f (n) and f (n+1) differ also by an additional summand in f (n+1), so that f (n) < f (n+1).
(b) The Fibonacci representation of n contains either F1 or F2. On adding of 1, some summands will become bigger Fibonacci numbers. The representation of n+1 has a largest Fibonacci number which is larger than the largest Fibonacci representation of n. This property remains invariant after the application of f . Hence f (n+1) > f (n), since the summands in the representation of f (n) are nonneighboring Fibonacci numbers and cannot add up to the greatest Fibonacci number in f (n + 1).
Remark. The function f is not uniquely determined by the three conditions.
286 11. Functional Equations 26. Replacing x →x −y, we get the equation f (x)2 f (x −y)f (x + y).
We can assume that f is positive. By introducing g ln ◦f , we get g(x −y) + g(x + y) 2g(x), which we solved in problem 13. A similar one was solved in 11.
27. By setting f (x) g(x) −1, we can radically simplify the functional equation g(x + y) g(x)g(y).
This is the functional equation of the exponential function g(x) ax, or f (x) ax −1.
28. The only solution is f (x) x + 1. See , problem 18.
29. We must solve the equation f (x) + f (x + 2y) 2f (x + y). The result is f (x) ax + b.
30. The unique solution is f (x) x −3 2. Show this yourself.
31. We replace x by −x and get −xf (−x)−2xf (x) −1. Thus, we have two equations for f (x) and f (−x). Solving for f (x), we get f (x) 1/x.
32. We guess f (x) ax2 + bx + c. Inserting this guess into the equation, we get a(x +y)2 ax2 +ay2 +xy, or ax2 +ay2 +2axy +b(x +y)+c ax2 +bx +c + ay2 + by + c + xy, which is satisfied for a 1/2 and c 0. By more conventional methods, show that f (x) x2/2 + c is the only continuous solution.
33. Let y x x−1. Then x y y−1. Thus f (y) (aφ(y) + φ(y/y −1)) /(1 −a2).
34. Any positive integer n can be written in the binary system, e.g., 1988 111110001002. By induction on the number in the binary system, we will prove the following assertion: if n a02k + a12k−1 + · · · + ak, a0, . . . , ak ∈{0, 1}, a0 1, then f (n) ak2k + ak−12k−1 + · · · + a0.
For 1 12, 2 102, 3 112 the assertion is true because of the first three points in (1). Now, suppose that the assertion is true for all numbers with less than (k + 1) digits in the binary system. Let n a02k + a12k−1 + · · · + ak, a0 1.
We consider three cases: (a) ak 0, (b) ak 1, ak−1 0 and (c) ak ak−1 1.
We only consider the case (b), the remaining cases can be handled similarly. In case (b) n 4m + 1, where m a02k−2 + · · · + ak−2, 2m + 1 a02k−1 + · · · + ak−22 + 1.
Because of (4), we have f (n) 2f (2m + 1) −f (m). By the induction hypothesis f (m) ak−22k−2 + · · · + a0, f (2m + 1) 2k−1 + ak−22k−2.
11. Functional Equations 287 Hence, f (n) 2k + 2(ak−22k−2 + · · · + a0) −(ak−22k−2 + · · · + a0) 2k + ak−22k−2 + · · · + a0 ak2k + ak−12k−1 + · · · + a0, q.e.d. The problem was to find the number of integers ≤1988 with symmetric binary representation. We observe that this number is 2⌊(n−1)/2⌋. We also see that only two symmetric 11-digit numbers 111111111112 and 111110111112 are larger than 1988.
Hence the number we are seeking is (1 + 1 + 2 + 2 + 22 + 22 + · · · + 24 + 24 + 25) −2 (25 −1) + (26 −1) −2 92.
35. Let x 0.b1b2b3 · · ·. If b1 0, then x < 1 2 and f (x) 0.b1b1 + 1 4f (0.b2b3 · · ·).
If b1 1, then x ≥1 2, and f (x) 0.b1b1 + 1 4f (0.b2b3 · · ·). From this we conclude that f (x) 0.b1b1b2b2b3b3 · · ·.
36. If z is a root of f , then also z2 is. If |z| ̸ 1, there are infinitely many roots, which is a contradiction. Hence all roots lie at the origin or on the unit circle. 0, 1 and third roots of unity have the closure property for squaring. Hence xp(x −1)q(1+x +x2)r also has the closure property. Inserting into the functional equation, we see that, in addition, p + q must be even: f (x) xp(x −1)q(1 + x + x2)r, p, q, r ∈N0, p + q ≡0 mod 2.
37. Hint: We have f (1) < f (2) < f (3) < · · · . In addition we have f (1) < f [f (1)] 3. Thus f (1) 2, f (2) 3. Prove that f (3n) 3f (n). In fact, f (n) n + 3k for 3k ≤n < 2·3k, and f (n) 3n−3k+1 for 2·3k ≤n < 3k+1. Hence f (1994) 3795.
38. (a) Let f (1) t. For x 1, we have tf (t + 1) 1 and f (t + 1) 1/t. Now x t + 1 yields f (t + 1)f f (t + 1) + 1 t + 1 1 ⇒f 1 t + 1 t + 1 t ⇒f 1 t + 1 t + 1 f (1).
Since f is increasing, we have 1/t +1/(t +1) 1, or t (1± √ 5)/2. But if t were positive, we would have the contradiction 1 < t f (1) < f (1 + t) 1/t < 1.
Hence t (1 − √ 5)/2 is the only possibility.
(b) Similar to the computation of f (1), we can prove that f (x) t/x, where t (1 − √ 5)/2. Again we must check that this function indeed satisfies all conditions of the problem.
39. Obviously the sequence f (n) n satisfies the condition. We prove that there are no other solutions. We observe that the function f is injective. Indeed, f (x) f (y) ⇒f [f (x)] f [f (y)] ⇒f {f [f (x)]} f {f [f (y)]} ⇒f {f [f (x)]} + f [f (x)] + f (x) f {f [f (y)]} + f [f (y)] + f (y) ⇒3x 3y, which implies x y. For n 1, we easily get f (1) 1. Suppose that, for n < k, we have f (n) n. We prove that f (k) k. If p f (k) < k then by the induction 288 11. Functional Equations hypothesis f (p) p f (k), and this contradicts the injectivity of f . If f (k) > k, then f [f (k)] ≥k. If we had f [f (k)] < k, then, as before, we would get the contradiction f {f [f (k)]} f [f (k)], f [f (k)] f (k), f (k) k.
Similarly, we have f {f [f (k)]} ≥k. Hence, f {f [f (k)]} + f [f (k)] + f (k) > 3k, which contradicts the original condition. Thus f (k) k.
12 Geometry 12.1 Vectors 12.1.1 Affine Geometry We consider the space with any number of dimensions. For competitions only 2 or 3 dimensions will be relevant. Points of the space will be denoted by capital letters A, B, C . . .. One point will be distinguished and will be denoted by O (for origin). The most important mappings of the space are the translations or vectors. A translation T is determined by any point X and its map T (X) Y. The translation taking point A into B is denoted by − → AB. It is usual practice to use O as the first point. The translation taking O to A is then − → OA. Since O is always the same point, we drop it and get ⃗ A. After a while one also drops the arrow on A and gets the point A. We simply identify points A and their vectors beginning in O and ending in A. We need not distinguish between points and vectors since all that is valid for points is also valid for vectors.
Now we define addition of two points A, B and multiplication of a point A by a real number t.
A + B reflection of the origin O at the midpoint M of (A, B).
The point tA lies on the line OA. Its distance from O is |t| times the distance of A. For t < 0 both A and tA are separated by O. For t > 0 they lie on the same side of O. For this reason multiplication with a real number is also called a stretch from O by the factor t. For the points (vectors) of the space, we have the following 290 12. Geometry properties (vector space axioms): (A + B) + C A + (B + C) for all A, B, C, (1) A + O A for all A, (2) A + (−A) O for all A, (3) A + B B + A for all A, B, (4) and (st)A (ts)A for all real s, t, and all A, (5) t(A + B) tA + tB, (6) (s + t)A sA + tA, (7) 1 · A A.
(8) Let A be a fixed point. The function T : Z →A + Z is a translation by A. Fig.
12.1 shows that 2M A + B, that is, the midpoint of (A, B) is O A B Q Q Q Q Q !!!!!!!
!
M 2M A + B Fig. 12.1 M A+B 2 .
(A, B, C, D) a parallelogram ⇐ ⇒ A+C 2 B+D 2 ⇐ ⇒A + C B + D.
We note the fundamental rule − → AB B −A.
Indeed, apply to (A, B) the translation which sends A to O. It will send B to B−A.
Thus, − → AB is the same translation as B −A.
A is the midpoint of (Z, Z′) ⇐ ⇒Z + Z′ 2 A ⇐ ⇒Z′ 2A −Z.
The function HA : Z →2A −Z is a reflection at A or a half-turn about A. We have Z HA − →2A −Z HB − →2B −(2A −Z) 2(B −A) + Z.
So HA ◦HB 2− → AB, and HA ◦HB ◦HC : Z HC − →2C −(2B −2A + Z), or HA ◦HB ◦HC HD where HD is the half-turn about D A −B + C. Since A + C B + D, the quadruple (A, B, C, D) is a parallelogram.
12.1 Vectors 291 E1. The midpoints P, Q, R, S of any quadrilateral in plane or space are vertices of a parallelogram.
Indeed, P A + B 2 , R C + D 2 ⇒P + R A + B + C + D 2 , Q B + C 2 , S A + D 2 ⇒Q + S A + B + C + D 2 .
Thus, P + R Q + S ⇐ ⇒(P, Q, R, S) a parallelogram.
E2. Reconstruct a pentagon from the midpoints P, Q, R, S, T of its sides.
We denote HA simply by A. Then P ◦Q ◦R X, where X is the fourth parallelogram vertex to the triple (P, Q, R). Furthermore X ◦S ◦T A. Thus, we have constructed A. The remaining vertices can be found by reflections in P, Q, R, S. This construction works for any polygon with (2n + 1) vertices, but not for polygons with 2n vertices. Successive reflections in the midpoints leave the first vertex A1 fixed. But the product of 2n reflections is a translation. Since it has a fixed point, it must be the identity mapping. So, any point of the plane can be chosen for vertex A1.
Suppose C lies on line AB.Then − → AC t · − → AB, or C −A t(B −A), or C A + t(B −A), and all real t.
In △ABC, let D (A + B)/2 be the midpoint of AB, and let S be such that − → CS 2− → CD/3. Then S −C 2 3(D −C) 2 3 · A + B 2 −2 3C ⇒S A + B + C 3 .
S is called the centroid of ABC. Since it is symmetric with respect to A, B, C, we conclude that the medians of a triangle intersect in S and are divided by S in the ratio 2 : 1.
E3. Let ABCDEF be any hexagon, and let A1B1C1D1E1F1 be the hexagon of the centroids of the triangles ABC, BCD, CDE, DEF, EFA, FAB. Then the A1B1C1D1E1F1 has parallel and equal opposites sides.
Solution. We want to prove that − − → A1B1 − − → E1D1 ⇐ ⇒B1 −A1 D1 −E1, that is, A1 + D1 B1 + E1. Indeed, we have A1 A + B + C 3 , D1 D + E + F 3 , B1 B + C + D 3 , E1 E + F + A 3 .
This implies that A1 + D1 B1 + E1 A + B + C + D + E + F 3 .
292 12. Geometry E4. Let ABCD be a quadrilateral, and let A′B′C′D′ be the quadrilateral of the centroids of BCD, CDA, DAB, ABC. Show that ABCD can be transformed into A′B′C′D′ by a stretch from some point Z. Find Z and the stretch factor t.
Solution. We have − − → A′B′ B′ −A′ A + C + D 3 −B + C + D 3 A −B 3 − − → AB 3 .
Similarly, we get − − → B′C′ −− → BC/3, − − → C′D′ −− → CD/3, − − → D′A′ −− → DA/3.
For the center Z, we get − → ZA′ −− → ZA/3, or A′ −Z −(A −Z)/3, or A + 3A′ 4Z, or Z A + B + C + D 4 .
Because of the symmetry of Z with respect to A, B, C, D we always get the same point Z.
E5. Find the centroid S of n points A1, . . . , An defined by n i1 − → SAi − → O .
Solution. From this equation, we get (A1 −S) + · · · + (An −S) O and S A1 + · · · + An n .
12.1.2 Scalar or Dot Product Let us introduce rectangular coordinates in space. The points A and B are now A (a1, · · · , an), B (b1, . . . , an).
We define the scalar or dot product as follows: A · B n i1 aibi, which is a real number. This definition implies S1. A · B B · A.
S2. A · (B + C) A · B + A · C, (tA) · B A · (tB) t(A · B).
S3. A 0 ⇒A · A 0, otherwise A · A > 0.
We define the norm or length of the vector A by |A| √ A · A a2 1 + · · · , +a2 n 12.1 Vectors 293 and the distance of the points A and B by |A −B| (A −B) · (A −B).
For 2 and 3 dimensions, it is easy to show that A · B |A| · |B| · cos( ∧ AB).
For n > 3, this becomes the definition of cos( ∧ AB). Now we have A ⊥B ⇐ ⇒A · B 0.
With the scalar product, we prove some classical geometric theorems.
E6. The diagonals of a quadrilateral are orthogonal if and only if the sums of the squares of opposite sides are equal.
We can write the theorem in the form C −A ⊥B −D ⇐ ⇒(B −A)2 + (C −D)2 (B −C)2 + (A −D)2.
Prove this by transforming, equivalently, the right side into the left.
A median of a triangle connects a vertex with the midpoint of the opposite side.
A median of a quadrilateral connects the midpoints of two opposite sides.
E7. The diagonals of a quadrilateral are orthogonal iff its medians have equal length.
Solution. Let MK and NL be the medians. Then we can express this theorem as follows: − → AC ⊥− → BD ⇔|MK|2 |NL|2.
To prove the theorem, we apply a sequence of equivalence transformations to the right-hand side (RHS) until we get the left-hand side (LHS).
C + D 2 −A + B 2 2 − A + D 2 −B + C 2 2 (C −A) · (D −B) − → AC · − → BD 0.
E8. Let A, B, C, D be four points in space. Then we always have |AB|2 + |CD|2 −|BC|2 −|AD|2 2− → AC · − → DB.
To prove this, we transform the LHS equivalently to get the RHS: (B −A)2 + (D −C)2 −(B −C)2 −(A −D)2 2(B · C + A · D −A · B −C · D) 2(C −A) · (B −D) 2− → AC · − → DB.
Some consequences of this theorem are the following: 294 12. Geometry H H H H H H @ @ @ d e b f a c A C B D Fig. 12.2 @ @ @ b e b a a f A C B D Fig. 12.3 J J J a a b −b Fig. 12.4 • In a tetrahedron AC ⊥BD ⇐ ⇒|AB|2 + |CD|2 |BC|2 + |AD|2.
• Application of the theorem to the trapezoid in Fig. 12.2 yields e2 + f 2 b2 + d2 + 2ac.
• The application to the parallelogram in Fig. 12.3 yields e2+f 2 2(a2+b2), that is, in a parallelogram, the sum of the squares of the diagonals is equal to the sum of the squares of the sides. We will show later that this property characterizes parallelograms.
• With the last theorem, we can easily express the length sa of the median of a triangle ABC. Reflect A at the midpoint of BC to D. You get parallelogram ABDC with diagonals 2sa and a. The main parallelogram theorem gives a2 + 4s2 a 2b2 + 2c2 or s2 a 1 4 2b2 + 2c2 −a2 .
Similarly, s2 b 1 4 2a2 + 2c2 −b2 , s2 c 1 4 2a2 + 2b2 −c2 .
• Let S be the centroid of △ABC. From the last theorem, one easily proves that AS ⊥BS ⇐ ⇒a2 + b2 5c2.
12.1.3 Complex Numbers Now we restrict ourselves to the plane. In the plane we will call points complex numbers, and we denote them by small letters like a, b, c, . . . . Point z in the plane can be represented in the form z xe1 + ye2, where e1 and e2 are unit points on the axes. Now e1 is our real unit, nothing new. But what about e2? Multiplication by e2 should have a geometric meaning. Since e2e1 e2, we conclude that e2 rotates e1 by 90◦. We simply define that e2 also rotates the vector e2 by 90◦. Thus, e2·e2 −e1. Now we want toseewhathappensifz xe1+ye2 ismultipliedbye2: e2z e2(xe1 + ye2) xe2 + ye2e2 −ye1 + xe2.
Fig. 12.4 shows that multiplication by e2 rotates the vector z by 90◦counterclock-wise.
From now on, we set e1 1 and e2 i. Then z x +iy, i2 −1. It is easy to show that complex numbers are a field with respect to addition and multiplication.
12.1 Vectors 295 This means that you can calculate with them as with real numbers. But you may not compare them with respect to order. a < b cannot be defined if you want the usual ordering properties to be satisfied.
We know that multiplication by i is a rotation of the plane by 90◦. We can find the formula for the rotation about any point a by 90◦. In fact, z′ a + i(z −a).
Indeed, translate a to the origin. Then z goes to z−a. Rotate by 90◦to get i(z−a).
Now translate back to get z′ a+i(z−a). We can use this result to solve a simple classical problem: E9. Someone found in his attic an old description of a pirate, who died long ago.
It read as follows: Go to the island X, start at the gallows, go to the elm tree, and count the steps. Then turn left by 90◦, and go the same number of steps until point g′. Again, go from the gallows to the fig tree, and count the steps. Then turn right by 90◦, and go the same number of steps to the point g′′. A treasure is buried in the midpoint t of g′g′′.
A man went to the island and found the elm tree e and the fig tree f . But the gallows could not be traced. Find the treasure point t.
Fig. 12.5 tells us that g′ e + i(e −g), g′′ f + i(g −f ), t g′ + g′′ 2 e + f 2 + i e −f 2 .
This is easy to interpret geometrically. m (e + f )/2 is the midpoint of the segment ef . Furthermore, − → me (e −f )/2. This vector must be rotated by 90◦ counterclockwise to get − → mt. The location of the gallows does not matter.
Multiplication z →az is a rotation about the origin O combined with a stretch from O with factor |a|. The rotational angle is the angle of vector a with the positive x-axis. This is easy to prove. If we do it without using trigonometry, then we get trigonometry for nothing.
H H H H H A A A XXXX X t g′′ g′ e f m g Fig. 12.5 Let e(α) be the unit vector in the direction α, |e(α)| 1. Then e(α) · e(β) e(α + β).
(1) Now we can define the trigonometric functions sin and cos as follows: e(α) cos α + i sin α, (2) e(−α) cos α −i sin α e(α) 1/e(α).
(3) 296 12. Geometry Now we prove some classical theorems with complex numbers.
E10. Napoleonic Triangles. If one erects regular triangles outwardly (inwardly) on the sides of a triangle, then their centers are vertices of a regular triangle (outer and inner Napoleonic triangles).
Let ϵ e(60◦) (1 + √ 3i)/2 be the sixth unit root, i.e., ϵ6 1 and 1 −ϵ + ϵ2 0, ϵ2 ϵ −1, ϵ3 −1, ϵ e(−60◦) 1−i √ 3 2 , ϵ + ϵ 1.
In Fig. 12.6, we have b0 a + (c −a)ϵ, c0 b + (a −b)ϵ, a0 c + (b −c)ϵ.
3(a1 −c1) c0 −b0 + c −a 2c −a −b + (2b −a −c)ϵ, 3(b1 −c1) a0 −b0 + c −b a + c −2b + (b + c −2a)ϵ, 3(a1 −c1)ϵ ϵ(2c −a −b) + (ϵ −1)(2b −a −c) a + c −2b + ϵ(b + c −2a) 3(b1 −c1).
ppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp @ @ @ @ / / / / / / T T T T ( ( ( ( ( e e e e a b c a0 b0 c0 b1 a1 c1 Fig. 12.6. Napoleonic triangles.
E11. Squares are erected outwardly on the sides of a quadrilateral. If the centers of the squares are x, y, z, u, then the segments xz and yu are perpendicular and of equal length.
x a + b 2 + i a −b 2 , y b + c 2 + i b −c 2 , z c + d 2 + i c −d 2 , u d + a 2 + i d −a 2 .
z −x c + d −a −b 2 + i c −d −a + b 2 , u −y a + d −b −c 2 + i c + d −a −b 2 , u −y i(z −x).
The last equation tells us that we get − → yu by rotating − → xz by 90◦.
E12. Squares cbqp and acmn are erected outwardly on the sides bc and ac of the triangle abc. Show that the midpoints d, e of these squares, the midpoint g of ab, and the midpoint f of mp are vertices of a square.
12.1 Vectors 297 This is a routine problem. Indeed, gef d is a parallelogram since its vertices are midpoints of the sides of the quadrilateral abpm. We have just to show that eg and gd are perpendicular and of equal length. Indeed g a + b 2 , d b + c 2 + i b −c 2 , e a + c 2 + i c −a 2 , d −g c −a 2 + i b −c 2 , e −g c −b 2 + i c −a 2 , (d −g)i c −b 2 + i c −a 2 e −g.
E13. Let a1b1c1 and b1b2b3 be two, positively oriented, regular triangles and let ci be the midpoint of aibi. Then c1c2c3 is a regular triangle.
Let a1 a, b1 b, c1 a + ϵ(b −a). The fact that a1b1c1 is regular has already been incorporated. We do the same with b1b2b3: b1 c, b2 d, b3 c + ϵ(d −c). Now c1 a + c 2 , c2 b + d 2 , c3 a + c 2 + ϵ b + d −a −c 2 .
Furthermore, c2 −c1 b + d −a −c 2 , c3 −c1 ϵ b + d −a −c 2 , c3 −c1 ϵ(c2 −c1).
E14. Let A, B, C, D be four points in a plane. Then |AB| · |CD| + |BC| · |AD| ≥|AC| · |BD| (Ptolemy’s inequality).
There is equality iff A, B, C, D in this order lie on a circle or on a straight line.
Proof. For any four points z1, z2, z3, z4 in the plane, we have the identity (z2 −z1)(z4 −z3) + (z3 −z2)(z4 −z1) (z3 −z1)(z4 −z2).
The triangle inequality |z1| + |z2| ≥|z1 + z2| implies that |z2 −z1| · |z4 −z3| + |z3 −z2| · |z4 −z1| ≥|z3 −z1| · |z4 −z2| or |AB| · |CD| + |BC| · |AD| ≥|AC| · |BD|.
We have equality iff (z2−z1)(z4−z3) and (z3−z2)(z4−z1) have the same direction, i.e., their quotient is real and positive. Denote the arguments of (z2 −z1)/(z4 −z1) and (z4 −z3)/(z3 −z2) by α and µ, respectively. Then z2 −z1 z4 −z1 · z4 −z3 z3 −z2 is a positive real ⇒α + µ 0◦, 298 12. Geometry that is, A, B, C, D lie on a circle or, for α µ 0◦, on a line. Note that in Fig. 12.7, α and µ are equal and oppositely oriented. |α| |µ| is necessary and sufficient for an inscribed quadrilateral.
@ @ @ @ z1 z2 z3 z4 α µ Fig. 12.7 Problems 1. Show that |AC|2 + |BD|2 |AB|2 + |BC|2 + |CD|2 + |DA|2 ⇐ ⇒A + C B + D.
2. Let A, B, C, D be four space points. Prove the theorem: If, for all points X in space, |AX|2 + |CX|2 |BX|2 + |DX|2, then ABCD is a rectangle.
3. Rectangles ABDE, BCFG, CAHI are erected outwardly on the sides of a triangle ABC. Show that the perpendicular bisectors of the segments HE, DG, FI are concurrent.
4. A regular n-gon A1 · · · An is inscribed in a circle with center O and radius R. X is any point with d |OX|. Then n i1 |AiX|2 n(R2 + d2).
5. Let ABC be a regular triangle inscribed in a circle. Then PAn + PBn + PCn is independent of the choice of P on the circle for n 2, 4.
6. For any point P of the circumcircle of the square ABCD, the sum PAn + PBn + PCn + P Dn is independent of the choice of P if n 2, 4, 6.
7. Prove Euler’s theorem: In a quadrilateral ABCD with medians MN and PQ, |AC|2 + |BD|2 2 |MN|2 + |P Q|2 .
8. Find the locus of all points X, which satisfy − → AX · − → CX − → CB · − → AX.
9. Three points A, B, C are such that |AC|2 + |BC|2 |AB|2/2. What is the relative position of these points?
10. If M is a point and ABCD a rectangle, then − → MA · −→ MC −→ MB · − − → MD.
11. The points E, F, G, H divide the sides of the quadrilateral ABCD in the same ratios.
Find the condition for EFGH to be a parallelogram.
12. Let Q be an arbitrary point in the plane and M be the midpoint of AB. Then |QA|2 + |QB|2 2|QM|2 + |AB|2/2.
13. Let A, B, C, D denote four points in space and AB the distance between A and B, and so on. Show that AC2 + BD2 + AD2 + BC2 ≥AB2 + CD2.
12.1 Vectors 299 14. Prove that, if the opposite sides of a skew (nonplanar) quadrilateral are congruent, then the line joining the midpoints of the two diagonals is perpendicular to these diagonals, and conversely, if the line joining the midpoints of the two diagonals of a skew quadrilateral is perpendicular to these diagonals, then the opposite sides of the quadrilateral are congruent.
15. Let ABC be a triangle, and let O be any point in space. Show that AB2 + BC2 + CA2 ≤3 OA2 + OB2 + OC2 .
16. For points A, B, C, D in space, AB ⊥CD ⇐ ⇒AC2 + BD2 AD2 + BC2.
17. ABCD is a quadrilateral inscribed in a circle. Prove that the six lines, each passing through a midpoint of one the sides of ABCD and perpendicular to the opposite side, are concurrent. Here, the diagonals are considered to be opposite sides.
18. The diagonals of a convex quadrilateral ABCD intersect in O. Show that AB2 + BC2 + CD2 + DA2 2 AO2 + BO2 + CO2 + DO2 exactly if either AC ⊥BD or one of the diagonals is bisected in O.
19. In a tetrahedron OABC with edges of lengths |OA| |BC| a, |OB| |AC| b, |OC| |AB| c, let A1 and C1 be the centroids of the triangles ABC and AOC, respectively. Prove that, if OA1 ⊥BC1, then a2 + c2 3b2.
20. In a unit cube, skew diagonals are drawn in two neighboring faces. Find the minimum distance between them.
21. Two opposite sides of a quadrilateral ABCD have lengths |AB| a, |CD| c, and the angle between these two sides is φ. How long is the segment MN joining the midpoints M, N of the two other sides?
22. Consider n vectors ⃗ a1, . . . , ⃗ an, |⃗ ai| ≤1. Show that, in the sum ⃗ c ±⃗ a1 ± · · · ± ⃗ an, one can choose the signs so that |⃗ c| ≤ √ 2.
23. P is a given point inside a given circle. Two mutually perpendicular rays from P intersect the circle at points A and B. Q denotes the vertex diagonally opposite to P in the rectangle determined by P A and P B. Find the locus of Q for all such pairs of rays from P .
24. P is a given point inside a given sphere. Three mutually perpendicular rays from P intersect the sphere at points A, B, and C. Q denotes the vertex diagonally opposite P in the box spanned by P A, P B, and P C. Find the locus of Q for all such triads of rays from P (IMO 1978).
25. Find the point X with minimal sum of the squares of the distances from the vertices A, B, C of a triangle.
26. Let O be the circumcenter of the △ABC, let D be the midpoint of AB, and let E be the centroid of △ACD. Prove that CD ⊥OE ⇔|AB| |AC|.
27. Let ABC be a triangle. Prove that there exists a unique point X such that the sums of the squares of the sides of the triangles XAB, XBC, XCA are equal. Give a geometric interpretation of X.
The following problems except 40 and 41 are to be solved by complex numbers.
Sometimes a convenient choice of the origin is helpful.
300 12. Geometry 28. A triangle with vertices a, b, c is equilateral iff a2 + b2 + c2 −ab −bc −ca 0.
29. Regular triangles are erected on the sides of a point symmetric hexagon, and its neigh-boring vertices are joined by segments. Show that the midpoints of these segments are vertices of a regular hexagon.
30. ABC is a regular triangle. A line parallel to AC intersects AB and BC in M and P, respectively. D is the centroid of P MB, E is the midpoint of AP. Find the angles of △DEC.
31. OAB and OA1B1 are positively oriented regular triangles with a common vertex O.
Show that the midpoints of OB, OA1, and AB1 are vertices of a regular triangle.
32. OAB and OA′B′ are regular triangles of the same orientation, S is the centroid of △OAB, and M and N are the midpoints of A′B and AB′, respectively. Show that △SMB′ ∼△SNA′ (IMO jury 1977).
33. A trapezoid ABCD is inscribed in a circle of radius |BC| |DA| r and center O. Show that the midpoints of the radii OA, OB and the midpoint of the side CD are vertices of a regular triangle.
34. Regular triangles DAS, ABP , BCQ, and CDR are erected outwardly on the sides of the quadrilateral ABCD. M1 and M2 are the centroids of DAS and CDR. The triangle M1M2T is oppositely oriented with respect to ABCD. Find the angles of △PQT .
35. Regular triangles with the vertices E, F, G, H are erected on the sides of a plane quadrilateral ABCD. Let M, N, P, Q be the midpoints of the segments EG, HF, AC, BD, respectively. What is the shape of P MQN?
36. The convex quadrilateral ABCD is cut by its diagonals (intersecting in O) into four triangles AOB, BOC, COD, DOA. Let S1 and S2 be the centroids of the first and third of those triangles, and H1, H2 the orthocenters of the other two triangles. Then H1H2 ⊥S1S2.
37. Regular triangles with vertices D and E, respectively, are erected outwardly on the sides AB and BC of △ABC. Prove that the midpoints of BD, BE and AC are vertices of a regular triangle.
38. A point D is chosen inside a scalene triangle ABC such that ̸ ADB ̸ ACB +90◦ and |AC| · |BD| |AD| · |BC|. Find |AB| · |CD| |AC| · |BD| (IMO 1993).
39. Regular triangles OAB, OA1B1 and OA2B2 are positively oriented with common vertex O. Show that the midpoints of BA1, B1A2, and B2A are vertices of a regular triangle.
40. If Pi, (i 1, · · · , n) are points on a unit sphere, then i≤j |PiPj|2 ≤n2.
41. Given any box ABCDEFGH. Prove the following theorems: • The sum of the squares of the space diagonals is four times the sum of the squares of the of the three edges.
• The square of a space diagonal starting in some vertex is the sum of the squares of the face diagonals which start at the same point minus the sum of the squares of the three edges.
12.1 Vectors 301 • The sum of the lengths of a space diagonal starting at some point and the edges is greater then the sum of the face diagonals starting at the same point.
• |⃗ a + ⃗ b + ⃗ c| + |⃗ a| + |⃗ b| + |⃗ c| > |⃗ a + ⃗ b| + |⃗ b + ⃗ c| + |⃗ c + ⃗ a| (ATMO 1972).
42. Equilateral triangles are erected to the outside on the sides of a convex quadrilateral.
Prove that the segment P Q joining the vertices of ABP and CDQ is perpendicular to the segment RS joining the centers of the two other triangles, and, in addition, |PQ| √ 3|RS|.
43. A point P0 and a triangle A1A2A3 are given in a plane. Let us set As As−3 for all s ≥4. We construct the sequence P0, P1, P2, . . . of points, so that the point Pk+1 is the map of Pk rotated around Ak+1 by 120◦clockwise (mathematically negative sense) (k 0, 1, 2, . . .). Show that if P1986 P0, then triangle A1A2A3 is regular (IMO 1986).
44. Construct regular hexagons on the sides of a centrally symmetric hexagon. Their centers form the vertices of a regular hexagon. (A special case of a theorem of A.
Barlotti.) 45. Equilateral triangles ABK, BCL, CDM, DAN are constructed inside the square ABCD. Prove that the midpoints of the four segments KL, LM, MN, NK and the midpoints of the eight segments AK, BK, BL, CL, CM, DM, DN, AN are the twelve vertices of a regular dodecagon.
Solutions 1. Expanding and collecting terms in the LHS of the equivalence yields (A + C −B − D)2 0, or A + C B + D, i.e., ABCD is a parallelogram.
2. Routine transformation yields A2 + C2 −B2 −D2 2X(A + C −B −D). This is valid for all points X of the plane iff A + C B + D, (1) and A2 + C2 B2 + D2.
(2) From (1) we get (A + C)2 (B + D)2 ⇐ ⇒A2 + C2 + 2A · C B2 + D2 + 2B · D.
(3) Subtracting (2) from (3), we get 2A · C 2B · D.
(4) Subtracting (4) from (2), we get (A −C)2 (B −D)2, i.e., the parallelogram has equal diagonals. Hence it is a rectangle. We have shown that this property charac-terizes rectangles. This will be useful in several later problems, e.g., the next one.
302 12. Geometry @ @ @ @ H H H H pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp @ @ @ @ A A A A E D H I F G A B C P Fig. 12.8 3. In Fig. 12.8, let P be the common point of the perpendicular bisectors of the segments HE and DG. From the preceding problem, we know that P B2 + P E2 P A2 + P D2, P A2 + P I 2 P C2 + PH 2, P C2 + P G2 P B2 + PF 2, P D2 P G2 ⇒P on a perpendicular bisector of DG, P H 2 P E2 ⇒P on a perpendicular bisector of EH.
Hence, P I 2 P F 2, that is, P lies on the perpendicular bisector of FI.
4. We have A1 + · · · + An O, |AiX|2 A2 i + X2 −2Ai · X R2 + d2 −2Ai · X, and |A1X|2 + · · · + |AnX|2 n(R2 + d2).
5. Let O be the center of the circle with radius R. Then PA2 (P −A)2 P 2 −2P · A + A2 2R2 −2A · P 2 R2 −A · P , P A2 + P B2 + P C2 6R2 −2P · (A + B + C) 6R2.
P A4 (A −M)2(A −M)2 4 R4 −2R2A · P + (P · A)2 , PA4 + P B4 + P C4 12R4 −8R2(A + B + C) · P + 4(A · P)2 + 4(B · P)2 + 4(C · P )2 12R4 + 4R2 cos2 φ + cos2(α + φ) + cos2(α −φ) 18R4.
Here we used the result of the preceding problem.
6. In Fig. 12.9, P A2 2r2 −2r2 cos φ, PB2 2r2 −2r2 cos( π 2 −φ), PC2 2r2 − 2r2 cos[π −φ), PD2 2r2 −2r2 cos( π 2 +φ), PA2 +P B2 +PC2 +PD2 8r2.
Similarly, by expanding and collecting terms, we get PA4 + P B4 + P C4 + P D4 24r4 and P A6 + P B6 + PC6 + PD6 80r6.
q H H ppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppppppppppppppppppppppppppppppppp @ @ @ @ @ & & && P A B C D φ O Fig. 12.9. |OP | |OA| |OB| r.
12.1 Vectors 303 7. Plugging into the formula M (A + B)/2, N (C + D)/2, P (B + C)/2, Q (D + A)/2, we get an identity after some routine computations.
8. (X −A) · (X −C) (B −C) · (X −A) ⇐ ⇒X2 −(A + B) · X −A · B ⇐ ⇒ X −A + B 2 2 A −B 2 2 (circle with diameter AB).
9. 2(C −A)2 + 2(C −B)2 (B −A)2 ⇔4C2 + A2 + B2 −4AC −4BC + 2AB 0 ⇔(2C −A −B)2 0 ⇐ ⇒C (A + B)/2.
10. A, B, C, D are vertices of a rectangle if A+C B +D, and, in addition, |A−C| |B −D|. Now (A−M)·(C −M) (B −M)(D −M) ⇐ ⇒A2 +C2 −B2 −D2 2(A −B + C −D)M. Since A + C B + D, we are left with A2 + C2 B2 + D2.
Subtracting this from (A+C)2 (B +D)2, we get 2AC 2BD. But then we have (A −C)2 (B −D)2, that is, we have a parallelogram with equal diagonals, which is a rectangle.
11. E (1−t)A+tB, F (1−t)B +tC, G (1−t)C +tD, H (1−t)D +tA.
EFGH is a parallelogram iff E + G F + H. This implies (1 −t)A + tB + (1 −t)C + tD (1 −t)B + tC + (1 −t)D + tA ⇔(1 −t)(A + C −B −D) −t(A −B + C −D) 0 ⇔(1 −2t)(A −B + C −D) 0 ⇔t 1 2 or A + C B + D, that is, if E, F, G, H are midpoints or if ABCD is a parallelogram.
12. (A −Q)2 + (B −Q)2 2(M −Q)2 + (B −A)2/2 ⇐ ⇒ A2 + B2 −2(A + B −2M)Q (A + B)2/2 + (A −B)2/2. Now A + B 2M. Hence, A2 + B2 (A + B)2/2 + (A −B)2/2, which is an identity.
13. A routine equivalence transformation gives A2 + B2 + C2 + D2 + 2A · B + 2C · D −2A · C −2B · D −2A · D + 2B · C ≥0 ⇐ ⇒(A + B −C −D)2 0 ⇐ ⇒A + B C + D, that is, ACBD is a parallelogram.
14. We want to prove below that (1), (2) ⇔(3), (4).
(A −B) · (A −B) (C −D) · (C −D), (1) (B −C) · (B −C) (A −D) · (A −D), (2) [(B + D) −(A + C)] · (A −C) 0, (3) [(B + D)) −(A + C)] · (B −D) 0.
(4) Addition and subtraction of (1) and (2) give (3) and (4). Addition and subtraction of (3) and (4) give (1) and (2). In section 4 we will give a simple geometric solution.
15. Let O be the origin. Then 3A2 + 3B2 + 3C2 −(A −B)2 −(B −C)2 + (C −A)2 ≥ 0 ⇔A2 + B2 + C2 + 2A · B + 2B · C + 2C · A ≥0 ⇔(A + B + C)2 ≥0. The last inequality is obvious. There is equality iff A + B + C O, that is, O is the centroid.
304 12. Geometry 16. AC2 + BD2 AD2 + BC2 ⇔(C −A)2 + (D −B)2 (D −A)2 + (C −B)2 ⇔ A · (C −D) B · (C −D) ⇔(A −B) · (C −D) 0 ⇔− → AB ⊥− → CD.
17. Let the origin be the center of the circumscribed circle. Consider the point S (A+B +C +D)/2. The vector from the midpoint of AB to S is (C +D)/2, and this is perpendicular to CD since |C| |D|, and, similarly, for the five other segments BC, CD, DA, AC, and BD.
18. Let O be the origin. Then 2A2 +2B2 +2C2 +2D2 −(B −A)2 −(C −B)2(D−C)2 − (A−D)2 0 ⇐ ⇒A·B+B·C+C·D+D·A 0 ⇔B·(A+C)+D·(A+C) 0 ⇔(A + C) · (B + D) 0 ⇔A + C O or B + D O or A + C ⊥B + D ⇔ O bisects AC or O bisects BD or AC ⊥BD.
19. We have A1 (A + B + C)/3, C1 (A + C)/3 and A1 · (C1 −B) 0. This implies (A + B + C) · (A + C −3B) O which is equivalent to a2 + c2 −3b2 + 2ac cos β −2ab cos γ −2bc cos α 0.
(1) We apply the cosine law to △ABC and get 2ac cos β a2+c2−b2, 2ab cos γ a2+b2−c2, 2bc cos α b2+c2−a2. (2) Eliminating the trigonometric functions in (1) and (2), we get a2 + c2 3b2. B O C A q Q P q Fig. 12.10 20. In Fig. 12.10 O is the origin and A, B, C are three unit vectors spanning the cube.
A−B and A+C are skew diagonals of two neighboring faces. The vector P −Q is orthogonal to both diagonals. It has minimum distance. Now P (1 −x)A + xB, Q y(A + C), P −Q ⊥A −B, P −Q ⊥A + C, A ⊥B, B ⊥C, C ⊥A. Thus, we get (P −Q)(A −B) 0 ⇒1 −2x −y 0, (P −Q)(A + C) 0 ⇒1 −x −2y 0 with solutions x y 1/3. Now P (2A + B)/3, Q (A + C)/3, P −Q (A + B −C)/3. |P −Q|2 1/3, |P −Q| 1/ √ 3.
21. With ⃗ a − → AB, ⃗ c − → DC, we get ⃗ m N −M (B + C)/2 −(A + D)/2 (B −A)/2 + (C −D)/2 ⃗ a 2 + ⃗ c 2, | ⃗ m|2 (a2 + c2 + 2ac cos φ)/4, | ⃗ m| a2 + c2 + 2ac cos φ/2.
22. If ⃗ a, ⃗ b, ⃗ c arevectorswithnorm≤1,thenatleastoneofthevectors ⃗ a±⃗ b, ⃗ a±⃗ c, ⃗ b±⃗ c has norm ≤1. Indeed, two of the vectors ±⃗ a, ±⃗ b, ±⃗ c have an angle ≤60◦, and hence the difference of these two vectors has norm ≤1. In this way we can descend to two vectors ⃗ a, ⃗ b. The angle between ⃗ a and ⃗ b or between ⃗ a and −⃗ b is ≤90◦.
Hence, |⃗ a −⃗ b| ≤ √ 2 or |⃗ a + ⃗ b| ≤ √ 2.
12.1 Vectors 305 A A A A Fig. 12.11 O P A Q B p 23. In Fig. 12.11 let O be the center of the circle, R its radius. Let |P| p. Making a picture, we soon realize that the locus we are looking for is a circle concentric to the given circle. Let us prove this theorem. In such problems one should not forget to prove two theorems. First, Q lies on a circle, and second, any point of the circle is also a point of the locus. Now Q P + (A −P) + (B −P). Hence, Q2 P 2 + (A −P)2 + (B −P )2 + 2P (A −P ) + 2P (B −P) P 2 + A2 + P 2 −2A · P + B2 + P 2 −2B · P + 2A · P + 2P · B −2P 2 2R2 −p2.
Thus, we have shown that Q lies on the circle about O with radius 2R2 −p. It remains to be shown that every point of this circle is on the locus. Take any point Q on the outer circle. Describe the circle with diameter P Q. It intersects the given circle in A and B. We have P A ⊥AQ and P B ⊥BQ. But do we also have PA ⊥PB, that is, is P AQB a rectangle? Thus, |OP |2 + |OQ|2 p2 + 2r2 −p2 2r2, |OA|2 + |OB|2 r2 + r2 2r2 ⇒|OP |2 + |OQ|2 |OA|2 + |OB|2.
The last property characterizes rectangles. Thus PAQB is a rectangle.
24. As in the plane case, we get Q P + (A −P ) + (B −P ) + (C −P), and Q2 P 2 + (A −P )2 + (B −P )2 + (C −P )2 + 2P · (A −P) + 2P · (B −P) + 2P · (C −P ) 3R2 −2p2.
Thus, Q lies on the sphere about O with radius 3R2 −2p2. It remains to be shown that every point Q of the sphere is also a point of the locus, which can be done as in the preceeding case.
25. Let 3S A + B + C. Then (X −A)2 + (X −B)2 + (X −C)2 3X2 −2(A + B + C)X + A2 + B2 + C2 3(X2 −2SX + S2) −3S2 + A2 + B2 + C2 3(X −S)2 + A2 + B2 + C2 −3S2. For X S, this has minimal value A2 + B2 + C2 −(A + B + C)2 3 (A −B)2 + (B −C)2 + (C −A)2 3 (a2 + b2 + c2)/3, where a, b, c are the sides of △ABC.
306 12. Geometry 26. The left-hand side of the equivalence is A + B 2 −C · A + C + (A + B)/2 3 0 ⇔(A + B −2C) · (3A + B + 2C) 0 ⇐ ⇒4A · B −4A · C 0 ⇐ ⇒A · (B −C) 0.
The right-hand side is (B −A)2 (C −A)2 ⇐ ⇒A · B A · C ⇐ ⇒A · (B −C) 0.
27. (X −A)2 + (X −B)2 + (A −B)2 (X −B)2 + (X −C)2 + (B −C)2 (X − C)2 + (X −A)2 + (C −A)2.
From the first equality, after expanding and collecting terms, we get 2(C −A)·X 2C2 −2A2 +2A·B −2B ·C ⇔(C −A)·X (C −A)(C +A−B).
By setting B′ A + C −B, we get (C −A) · X (C −A) · B′. This is equivalent to (C −A) · (X −B′) O or − → AC · − − → B′X O, or − → AC ⊥− − → B′X, that is, X lies on the perpendicular to AC through B′. By cyclic permutation, we conclude that X lies on the perpendicular to BC through A′ B + C −A and on the perpendicular to AB through C′ A+B −C. The three perpendiculars must intersect since the first two equalities imply the third. Independently, we can also say that they intersect in one point, since it is the orthocenter of the triangle A′B′C′ (Lemoine point).
28. Consider a2 + b2 + c2 −ab −bc −ca 0 as a quadratic in a with solutions a + bω + cω2 0 and a + bω2 + cω 0. The first solution characterizes positively oriented equilateral triangles, the second one negatively oriented triangles. Indeed, a positively oriented triangle (a, b, c) is equilateral iff (b −a)ω c −b, which can be transformed equivalently to a + bω + cω2 0. By exchanging b with c, we get the second solution for negatively oriented triangles. Here ω is the third root of unity.
29. Let the center of the hexagon be o, and the vertices (a, b, c, −a, −b, −c). We erect regular triangles with vertices d, e, f , g on (a, b), (b, c), (c, −a), (−a, −b). Denote the midpoints of (d, e), (e, f ), (f, g) with p, q, r. Then with ϵ6 1 d b +(a −b)ϵ, e c +(b −c)ϵ, f −a +(c +a)ϵ, g −b +(b −a)ϵ.
Here ϵ is the sixth unit root. For the midpoints, we get p d + e 2 b + c + (a −c)ϵ 2 , q e + f 2 c −a + (a + b)ϵ 2 , r f +g 2 −a−b+(b+c)ϵ 2 . For the vectors of the sides pq and qr, we get − → qp p −q a + b −(b + c)ϵ 2 , − → qr r −q −b −c + (c −a)ϵ 2 , − → qr ϵ2 −c −b + (c −a)ϵ 2 (ϵ −1) a + b −(b + c)ϵ 2 − → qp.
This completes the proof.
12.1 Vectors 307 30. In Fig. 12.12, we assign to A and B the complex numbers a and 0. Then M ta, P taϵ, D ta 3 (1 + ϵ), E a 2 (1 + tϵ).
Thus, we have − → DE a 6 (3 −2t + tϵ) , − → DC a 3(3ϵ −t −tϵ), 2ϵ− → DE a 3(3ϵ −t −tϵ).
Hence, △CDE is a 30◦, 60◦, 90◦triangle.
T T T T T Fig. 12.12 A XX C C C C pppppppppppppppppppppppppppppppppppppppppppppppppp P B D E C M 31. We assign to the points O, A, B, A1, B1 the complex numbers o, a, aϵ, b, bϵ.
Then p a + bϵ 2 , q aϵ 2 , r b 2, − → pq (a −b)ϵ −a 2 , − → pr b −a −bϵ 2 .
Now we have − → pqϵ b−a−bϵ 2 − → pr. Thus pqr is regular.
32. We assign the complex numbers o, a, aϵ, b, bϵ to O, A, B, A′, B′. Then N bϵ 2 , S a + aϵ 3 , A′ b, M b + aϵ 2 , − → SM M −S 3b −2a + aϵ 6 , − → SMϵ −a + (3b −a)ϵ 6 , − → SB −a + (3b −a)ϵ 3 , − → SM − → SB′ 2 .
Similarly, we prove that − → SA′ 2− → SN. This proves the theorem.
33. We assign the complex numbers b, bϵ, d, dϵ to the points B, C, D, A. The mid-points are p dϵ 2 , q b 2, r d + bϵ 2 , − → pq b −dϵ 2 , − → pr d + bϵ 2 .
Now we have − → pq · ϵ bϵ −d(ϵ −1) 2 d + (b −d)ϵ 2 − → pr, which was to be proved.
308 12. Geometry 34. We assign the complex numbers a, b, c, d to A, B, C, D, respectively. A drawing suggests that △P QT is isosceles with ̸ P T Q 120◦. Thus we proceed to show that − → T Qϵ −− → T P .
S a + (d −a)ϵ, P b + (a −b)ϵ, Q c + (b −c)ϵ, R d + (c −d)ϵ, M1 2a + d + (d −a)ϵ 3 , M2 c + 2d + (c −d)ϵ 3 , T M2 + (M1 −M2)ϵ a + 2c + (a −c)ϵ 3 .
− → T P P −T −a + 3b −2c + (2a −3b + c)ϵ 3 , − → T Q Q −T −a + c + (3b −a −2c)ϵ 3 , − → T Qϵ −− → T P .
35. Assigning to A, B, C, . . . the complex numbers a, b, c, . . ., we get e b + (a −b)ϵ, f c + (b −c)ϵ, g d + (c −d)ϵ, h a + (d −a)ϵ, m e + g 2 b + d 2 + a −b + c −d 2 ϵ, n f + h 2 a + c 2 + b −c + d −a 2 ϵ, p (a + c)/2, q (b + d)/2.
Since m + n p + q, MQNP is a parallelogram.
36. First we compute the upper part AH of the altitude in △ABC in Fig. 12.13. We have |AD| b cos α, |AH| |AD|/ sin β b cos α/ sin β. By means of the Sine Law b/ sin β a/ sin α we get, finally, |AH| a cot α. Using the intersection of diagonals in Fig. 12.14 as the origin, we have S1 A + B 3 , S2 C + D 3 , − − → S1S2 S2 −S1 1 3(C + D −A −B).
Setting ̸ DOA ̸ BOC ω, because |AH| a cot α, we get − − → OH1 i(B −C) cot ω, − − → OH2 i(D −A) cot ω, − − − → H1H2 H2 −H1 i cot ω(C + D −A −B).
The factor i rotates a vector by 90◦. Hence, S1S2 ⊥H1H2. L L L L L L !!!!!!
A B C H D α β β Fig. 12.13 B B B B B B !
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/ / / / b b b b b b b q q ((( pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp q q Fig. 12.14 A B D C S1 S2 H2 H1 12.2 Transformation Geometry 309 37. Let P , Q and R be the midpoints of BD, BE, and AC, respectively. Then r a + c 2 , p 2b + (a −b)ϵ 2 , q b + c + (b −c)ϵ 2 , p −r 2b −a −c + (a −b)ϵ 2 , q −r b −a + (b −c)ϵ 2 , (p −r)ϵ b −a + (b −c)ϵ 2 .
Since q −r (p −r)ϵ, the triangle pqr is regular.
38. Assign the complex numbers a, b, c, o to A, B, C, D, respectively. Then setting s |AC| |AD| |BC| |BD|, ̸ CAD α, we get a −c seiαa, c −b sieiαb, and hence c a(1 −eiα) b(1 + iseiα), (a −b)c s(eiαac+ieiαbc) sab eiα(1 + sieiα) + ieiα(1 −seiα) sabeiα(1+ i). Thus, |AB| · |CD| |a −b| · |c| s|a| · |b| √ 2 |AC| · |BD| · √ 2, that is, |AB| · |CD| |AC| · |BD| √ 2.
12.2 Transformation Geometry In this section isometries and similarities and their concatenations are used to prove theorems or to solve problems. Problems solvable by vectors or complex numbers are usually good examples for transformation geometric methods. In fact, vectors are translations, a simple type of isometry. Multiplication by a complex number is a stretch from O combined with a rotation about O.
Isometries are one-to-one transformations of a plane (or space) which preserve distance. In a plane, direct isometries preserve sense. They are translations and ro-tations. The opposite isometries are not sense-preserving. They are line reflections and glide reflections. The last one is hardly ever used in competitions. A translation has no fixed point except the identity, which has nothing but fixed points. A rota-tion has just one fixed point. Among the opposite isometries the line reflection has a whole line of fixed points. The glide reflection has none if it is not a reflection.
Every direct isometry is the concatenation of two line reflections. An opposite isometry can be represented as a composition of one or three line reflections.
RotationaroundpointP withangle2φ istheconcatenationoftwolinereflections withthelinespassingthroughP andformingangleφ.Atranslationistheproductof two line reflections in parallel mirrors. The direction of the translation is orthogonal to the lines, and its distance is twice the distance of the parallel lines. A product of two half-turns about A and B is the translation 2− → AB.
We give some examples of the use of transformation geometry.
E1. Napoleonic Triangles. Erect outwardly (inwardly) isosceles triangles with vertices P, Q, R and vertex angles 120◦on the sides AC, BC, AB of a triangle.
Prove that △P QR is regular.
310 12. Geometry @ @ @ @ @ pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp A B C R R′ P Q p q r Fig. 12.15 ppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp A B O S N A′ M B′ Fig. 12.16 T T T T T pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp @ @ @ A B C N P M E D Fig. 12.17 Look at Fig. 12.15. P120◦◦Q120◦◦R120◦ I, since it is a translation with fixed point A, i.e., the identity mapping. Hence P120◦◦Q120◦ R−120◦. Now construct the regular triangle with base PQ and vertex R′. Then P120◦◦Q120◦ p ◦q ◦q ◦r p ◦r R′ −120◦.
Thus, R−120◦ R′ −120◦, which is the same rotation with the same fixed point, that is, R R′.
E2. Again we solve problem 31, Chapter 12.2 (IMO jury 1977). In Fig. 12.16, dilatation from B with factor 2 and then rotation about O by 60◦moves M to B′ and leaves S fixed. Hence ̸ MSB′ 60◦and SM : SB′ 1/2. Similarly ̸ NSA′ 60◦, SN : SA′ 1/2. Hence △SMB′ ∼△SNA′.
E3. Let us look at another problem we already solved by complex numbers. On the sides AB and BC of △ABC are erected outwardly regular triangles with vertices D and E. Show that the midpoints of AC, BD, BE are vertices of a regular triangle.
We must show in Fig. 12.17 that △MNP is regular. The idea is to move N by a sequence of transformations to P. The product must be a rotation about M by 60◦. Such a sequence is easy to find: dilatation with center B by factor 2, rotation about B by −60◦, a half turn about M, rotation about B by −60◦, and a stretch from B by factor 1/2. It moves N →D →A →C →E →P. Now we show that M is a fixed point. Indeed, M →M1 →M2 →M3 →M1 →M.
Since the stretches by 2 and 1/2 give an isometry, this is a rotation by +60◦since −60◦+ 180◦−60◦ 60◦.
E4. The trapezoid ABCD in Fig.12.18 has AB ∥CD. An arbitrary point P on the line BC, which does not coincide with B or C, is joined with D and the midpoint M of the segment AB. Let X ∈PD ∪AB, Q ∈PM ∪AC, Y ∈DQ ∪AB. Show that M is the midpoint of XY.
Consider the following homotheties: HQ : A →C, HP : C →B.
Obviously, HQ ◦HP maps A to B and leaves M fixed. Since M is the midpoint of AB, the composite mapping HQ ◦HP HM is a half turn about M. But HQ : Y →D, HP : D →X. Thus HM : Y →X, and |MX| |MY|.
12.2 Transformation Geometry 311 !!!!!!!
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A A A A A A c c c c c c c c H H H H H H H H / / / A Y M X B C D Q P Fig. 12.18 @ @ @ @ E E E `````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````` ``````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````` pppppppppppppppppppppppppppppppppppppppppppp E E E D DD A B C D Y X W Z Fig. 12.19 E5. On the sides AB, BC, CD, DA of a quadrilateral ABCD, we construct, alternately to the outside and inside, regular triangles with vertices Y, W, X, Z, respectively. Show that YWXZ is a parallelogram.
A parallelogram is generated by translation. So we try to find some mappings which give a translation as a product. Such a product is easy to find. A60◦◦C−60◦ is a translation which takes Y to W and Z to X. Thus, − → YW − − → WX. Indeed, Y A60◦ − →B C−60◦ − →W, Z A60◦ − →D C−60◦ − →X.
E6. This is a generalization of the preceding example. Suppose, we replace the regular triangles with directly similar triangles. See Fig. 12.19. The result still seems to be a parallelogram.
Indeed, with |AY|/|AB| r, we have Aα ◦A 1 r ◦C(r) ◦C−α f, a translation.
Y f − →W, Z f − →X ⇒− → YW − → ZX.
E7. Construct a parallelogram, given two opposite vertices A, C, if the other two vertices lie on a given circle.
A parallelogram is a centrally symmetric figure. The center M is the midpoint of AC. A half turn about M interchanges the other two vertices, but they must lie on the reflected circle. So they are the intersections of the given circle and its reflection.
E8. Construct a parallelogram ABCD, given the vertices A, C and the distances r and s of the points B and D from a given point E.
Reflect E at the midpoint M of AC to E′. Now B is constructible from EE′ and circles with radii r and s and centers E and E′, respectively.
E9. Construct a parallelogram ABCD from C, D and the distances r and s of A and B from a given point E.
The translation − → AD takes E to E′. Now △DE′C is constructible from the three sides |CD|, |DE′| r, |E′C| s. Now translate DC by − − → E′E. The image of DC is AB.
312 12. Geometry E10. Two circles α and α1 and a point P are given. Find a circle which is tangent to α and α1, such that the line through the two points of tangency passes through P.
PPPPPP pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp S P D B X A O O1 C α Y α1 x Fig. 12.20 The circle x to be constructed touches α and α1 (with centers O, O1) in A and B, where P ∈AB. We consider the homothety with center A, which maps α to x, and the homothety with center B, which maps x to α1. Their product maps α onto α1 and has center S ∈AB ∪OO1, that is, AB is determined by P and S, where S is a similarity center of the circles α and α1. If α, α1 are not congruent, there will be two similarity centers S, S1, such that α →α1. There will be solutions, if at least one of the lines SP, SP1 intersects the given circles. At most there are four solutions: two circles x, y for SP and two for S1P (with a negative stretch factor).
See Fig. 12.20, which shows the two solutions for S. The second solution is not actually drawn, but its center Y and its points C and D of tangency are constructed.
E11. A circle and one of its diameters AB are given as well as one point P in the plane. Construct the perpendicular to AB through P by ruler alone. With a ruler, you can connect two points.
The problem is almost automatic for most positions of P. In Fig. 12.21 you must draw AP and BP. Then two new points C, D arise. So you draw AC and BD. They intersect in H. But AC ⊥BP and BD ⊥AP. So H is the orthocenter of the triangle ABP . Thus PH ⊥AB. For a point P inside the circle, the lines to be drawn are exactly the same, but this time P is the orthocenter. The case in Fig.
12.22 is not much different. But suppose P lies on the circle as in Fig. 12.23. The new idea is to choose a point Q outside the circle. We can drop a perpendicular from this point to AB which intersects the circle at R, S. We can drop perpendicular from P , if we can reflect P at AB. Now we have two symmetric points R, S. With their help, we can easily reflect P. Draw SP. It intersects AB in T . Draw RT . It intersects the circle in P ′, the image of P. Now PP ′ ⊥AB.
Now suppose that P ∈AB as in Fig. 12.24. We want to draw the perpendicular to AB through P . This is a considerably more difficult problem. Now we must draw two perpendiculars to AB. The first intersects AB in Q and the circle in S, S′. The other intersects AB in R. Draw SP and S′P. They intersect the second perpendicular in T and T ′. The simplest way to proceed now is to use a shear with fixed line SS′ which takes T ′R to RT . Shears preserve areas and take lines into lines. Now the trapezoids S′T ′RQ and S′RT Q have the same area, S′T ′ goes to S′R, and QR goes to QT . Since △S′PQ and S′P ′Q have the same area and the 12.2 Transformation Geometry 313 A A A A A A Z Z Z Z Z A B D C P H Fig. 12.21 QQQQQQ Q A A A A A A A A B=H C D P P ′ Fig. 12.22 A A A A A A A B R S P P ′ T Fig. 12.23 same base S′Q, they will have the same altitude. Thus P and P ′ are equidistant from S′Q. Hence, P P ′ ⊥AB.
E12. Construct a quadrilateral ABCD from its sides and the median MN joining the midpoints of AB and CD, respecively.
Reflect the whole quadrangle ABCD at M to ABD′C′. N will go to N1. Trans-late DN by − → DA to AA1. Similarly, translate CN by − → CB to BB1. △A1N1N can be constructed from its sides. Now △MAA1 can also be constructed from its sides.
The rest is trivial. See Fig. 12.25.
HHHHH ppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp S S′ A B P P ′ Q R T T ′ Fig. 12.24 A A A A A A A A A A pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp A A A A A A A A A A Fig. 12.25 A B C D N M C′ N1 D′ A1 b b b d d m B1 m c/2 c/2 c/2 Problems 1. ABC and A′B′C are regular triangles with the same orientation. Let P, Q, R be the midpoints of the segments AB′, BC, A′C, respectively. Show that △PQR is regular.
2. Let M, N be the midpoints of the bases of trapezoid ABCD. Show that the line MN passes through the intersection point O of the diagonals and the point S where the extensions of the legs intersect.
3. A point P is joined to the vertices of triangle ABC. The straight lines AP y, BP z, CP x are reflected at the angle bisectors passing through A, B, C to v, w, u, respectively. Prove that u, v, w pass through one point Q.
314 12. Geometry 4. Three lines x, y, z are incident with a point P and are orthogonal to the sides c, a, b of a triangle ABC. Now x, y, z are reflected at the midpoints of c, a, b to u, v, w, respectively. Prove that u, v, w also pass through a point Q.
5. Take a point A inside an acute angle. Construct the triangle ABC of minimum perimeter if B and C lie on the legs of the angle.
6. Two circles are tangent internally at point A. A secant intersects the circles in M, N, P, Q. Prove that ̸ MAP ̸ NAQ.
7. A chord MN is drawn in a circle ω. In one of the circular segments, the circles ω1, ω2 are inscribed touching the arc in A and C and the chord in B and D. Show that the point of intersection of AB and CD is independent of the choice of ω1, ω2.
8. Consider n circles Ci (Cn+1 C1) with Ci touching Ci+1 externally at Ti for i 1 to n. Start at any point A1 on C1, and, for i 1 to n, draw straight lines AiTi intersecting Ci+1 a second time in Ai+1. What is the relative position of A1 and An+1 on C1? Generalize.
9. Assume a line a and a point P . Using as few lines as possible (circles or segments), construct the line perpendicular to a which passes through P. If P ̸∈a the problem is well known to every high school student of geometry. But suppose P ∈a. The minimal construction is hardly known. See the solution for a proof of our contention.
10. A, B, C, D are four points on a line. Through A and B, draw a pair (a, b) of parallels and, through C and D, another pair (c, d) of parallels so that (a, b)∪(c, d) PQRS is a square.
11. Draw through a point P inside an angle a segment, which cuts off a triangle of minimum area.
12. On the sides CA and CB of △ABC, squares CAMN and CBPQ with centers O1 and O2 are constructed to the outside. The points D and F are the midpoints of the segments MP and NQ. Prove that the triangles ABD and O1O2F are rectangular and isosceles.
13. What can you say about lines a and b if a ◦b ◦a b ◦a ◦b. Here we identify a line a with the reflection in a.
14. What is the relative position of a, b, c, d if a ◦b ◦c ◦d b ◦a ◦d ◦c?
15. In a quadrilateral ABCD, we reflect A at B to A1, B at C to B1, C at D to C1, D at A to D1. Suppose, only A1, B1, C1, D1 are given. Reconstruct ABCD. Compare the areas of ABCD and A1B1C1D1.
16. In a quadrilateral ABCD, reflect A at C to A1, B at D to B1, C at A to C1, D at B to D1. Compare the areas of ABCD and A1B1C1D1.
17. On the sides BC, CA, and AB of triangle ABC, regular triangles with vertices D, E, and F are erected. Reconstruct ABC from D, E, F.
18. On the sides AB and DA of a parallelogram ABCD, regular triangles with vertices E and F are erected. Prove that E, C, F are vertices of a regular triangle.
19. On the sides of △ABC, the points P , Q, R are chosen, such that AP 2PB, BQ 2QC, CR 2RA. Reconstruct the triangle from P, Q, R.
20. Construct a triangle ABC from two sides b, c, if it is known that the median AD divides the angle at A in the ratio 1 : 2, so that ̸ BAD α, ̸ DAC 2α, α being unknown.
12.2 Transformation Geometry 315 c A A A A A A a B b C !!!!!!!!!!
v @ @ @ @ @ w p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p z y x P Q ppppppppppppppppppppppppppppppppppppppppu Fig. 12.26 l l l l l l l a a % % % a a a a a a % % A B C c p x u q v y w z P Q Fig. 12.27 21. Three pairwise orthogonal plane mirrors are used as reflectors at the back of a bycicle.
Prove that, if a light ray is reflected at each of the three mirrors, it reverses its path.
22. Three points P , Q, and R are given in the plane. Construct a quadrilateral ABCD for which these three points are the midpoints of AB, BC, CD, if it is known that |AB| |BC| |CD|.
23. ABCD is a square, and P is a point inside with |P D| 1, |PA| 2, |PB| 3.
Find ̸ AP D.
24. A point P inside the equilateral triangle ABC of side s has distances 3, 4, and 5 from the vertices A, C and B, respectively. Find s.
Solutions 1. P C′(2) − →B A60◦ − →C A(1/2) − →Q, R C′(2) − →R′ A(60◦) − →B′ A(1/2) − →R.
2. Let |AB|/|CD| λ, |CD|/|AB| 1/λ. Then O(−1/λ) ◦S(λ) exchanges A with B. Hence, it is M(−1). Similarly, S(λ) ◦O(−1/λ) exchanges C with D. Hence, it is N(−1). This implies M ∈OS, N ∈OS.
3. In Fig. 12.26 ax ub, by vc, cz wa ⇒u axb, v byc, w cza.
Now P ∈x, y, z ⇒xyz is a line reflection ⇐ ⇒xyzxyz I ⇒uvwuvw axbbycczaaxbbyccza axyzxyza aIa I ⇒u, v, w have a common point Q.
4. In Fig. 12.27 px uq ⇒u pxq ⇒u pccxq pcxcq AxB. Sim-ilarly, v ByC, w CzA. Now uvwuvw AxBByCCzAAxBByCCzA AxyzxyzA AIA I. Thus u, v, w have a common point Q.
5. In Fig. 12.28, reflect A at b to M and at c to N. Line MN intersects the legs of the angle in B and C. Triangle ABC has the least perimeter. Indeed, let B1 and C1 be any two other points on b and c, respectively. Then |AB1| + |B1C1| + |C1N| > |MN| |MB| + |BC| + |CN| |AB| + |BC| + |CA|. @ @ @ @ @ @ P P P & & & & & A A A A A A A A HH H O B B1 b A C c N M C1 Fig. 12.28 316 12. Geometry @ @ @ @ @ @ ppppppppppppppppppppppppp H H H H H H H Q M N P A N ′ P ′ ω ω′ Fig. 12.29 E E E E E E E E E M N B D M A ω C E ω1 ω2 Fig. 12.30 6. In Fig. 12.29, the homothety with center A takes circle ω to circle ω′. We have N ′P ′ ∥MQ ⇒arc MN ′ arc P ′Q ⇏ P AQ ̸ MAN.
7. In Fig. 12.30, the homothety with center A, which takes ω1 to ω takes MN to the horizontal tangent in E. The homothety with center C, which takes ω2 to ω, takes MN to the tangent in E. Thus AB and CD intersect in E.
8. We consider n homotheties. Each has center Ti and maps Ci onto Ci+1. Now T1(λ1)◦ T2(λ2) ◦· · · ◦Tn(λn) maps C1 onto C1 with n half turns. The result is the identity for even n and a half turn of C1 for odd n. Thus An+1 A1 for even n and An+1 and A1 are endpoints of a diameter of C1 for odd n.
9. Fig. 12.31 shows the construction for the second case. It is most interesting and hardly known. It takes any point Q outside the line and draws the circle with center Q, which passes through P . Through the second intersection point R of the line with the circle, we draw the diameter RQ intersecting the circle once again in S. Then SP is perpendicular to a. We need to draw one circle and two straight lines, one line less than in the classical construction.
10. Let P QRS be the required square. The angular bisectors of ̸ PSR and ̸ BQC pass through the points N and M on the circles with diameters BC and AD, respectively.
N and M are bisectors of the semicircular arcs BC and AD.
11. In Fig. 12.32, aOb is the given angle. Take any line MN through P, M ∈a, N ∈b.
Suppose |MP | < |P N|. Now reflect a at P to a′. Let a′ ∩b B, a′ ∩MN K, PB ∩a A. Then the area of triangle OAB is smaller than the area of triangle OMN by the area of triangle BKN.
12. In Fig. 12.33, A−90◦◦B−90◦is a half turn which maps M to P. The center of symmetry is the midpoint D of the segment P M. The theorem on the composition of two rotations tells us that ̸ DAB ̸ DBA 45◦. This proves that triangle ABD is isosceles with a right angle. Similarly, we can prove the required property for the triangle O1O2F of the rotation about O1 and O2 by 90◦.
P S R r Q Fig. 12.31 c c c c O M A a B b N a′ P K Fig. 12.32 Z Z Z Z P P P P P P Z Z Z Z p pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp p @ @ @ ppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppp p pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp A B C F M N Q P O1 O2 D Fig. 12.33 12.2 Transformation Geometry 317 13. Suppose O is the intersection of the lines a and b. Let ̸ (a, b) φ. Then a ◦b ◦a b ◦a ◦b ⇐ ⇒a ◦b ◦a ◦b ◦a ◦b I, that is, 6φ 2π, or, φ π/3.
14. By multiplying abcd badc with ab from the left and dc from the right, we get (ab)2 (dc)2. If a ∥b, then a, b, c, d are parallel. If a ∦b and these lines are not perpendicular, then a, b, c, d have a common point. If a ⊥b and c ⊥d the position of these two pair of lines is arbitrary.
15. Homothety A1(1/2) ◦A2(1/2) ◦A3(1/2) ◦A4(1/2) is the homothety A(1/16). We can find point A by applying the product of the four homotheties to any point X of the plane. From X and its image Y, we can find A. Since X is arbitrary, we may take X A1. Then Y will be the image of A1 under the homothety A1(1/2) ◦A2(1/2) ◦ A3(1/2) ◦A4(1/2).
16. The area of a quadrilateral ABCD is |AC|·|BD|·sin φ, where φ is the angle of the two diagonals AC and BD. Since the diagonals of A1B1C1D1 have the same angle and are three times as long, its area is nine times the area of ABCD.
17. P60◦◦Q60◦◦R60◦is a half turn. Applying this mapping to P, we get its image P ′.
The midpoint of P P ′ is the point A.
18. E(60◦) leaves E fixed and takes F to C. Indeed, ̸ FAE ̸ EBC 120◦+ α, where α is the angle between AF and AE.
19. P(−1/2) ◦Q(−1/2) ◦R(−1/2) A(−1/8). We can get A from −→ AP ′ −− → AP /8, if P ′ is the image of P with respect to A(−1/8).
20. Reflect C at AD to E. Then FD is a median in △EBC. Hence AD ∥BE, and ̸ ABE ̸ DAB α. Hence, |AE| |BE| b, and △ABE can be constructed from its sides. Now draw AD ∥BE, and reflect E at AD to C.
21. The mirrors define a coordinate system with the origin O being the unique common point of the planes. Reflection at all of the planes is reflection at O, which reverses each path. Indeed, reflection at the yz-, zx-, and xy-planes resuls in (x, y, z) → (−x, y, z) →(−x, −y, z) →(−x, −y, −z).
22. B and C lie on the perpendicular bisectors of P Q m1 and QR m2, respectively.
They intersect in O. Now we have ̸ m1Om2 with a point Q inside. We must find a segment from m1 to m2, which is bisected in Q. There is a unique solution. Reflect m2 at Q to m′ 2, which intersects m1 in B. The rest is trivial.
23. Rotate the square about A by +90◦. Then B →B′ D, C →C′, D →D′, P →P ′. We have AP ⊥AP ′, |AP | |AP ′| 2. Thus △AP P ′ has ̸ AP P ′ 45◦. Since |P P ′| √ 2 and ( √ 2)2 + 12 ( √ 3)2, we have PP ′ ⊥PD. Thus ̸ APD ̸ AP P ′ + ̸ P ′P D 135◦.
24. Reflect the point P at the sides BC, CA, and AB, respectively, to A′, B′, and C′. The area of the hexagon AC′BA′CB′ can be computed in two ways. On the one hand it is twice the area of △ABC, i.e., √ 3s2/2. On the other hand, it is the area of the rectangular triangle A′B′C′ with sides 3 √ 3, 4 √ 3 and 5 √ 3 together with the areas of the triangles AC′B′, BA′C′ and CB′A′ which we know from two sides and the included angle 120◦. We get s 25 + 12 √ 3.
318 12. Geometry 12.3 Classical Euclidean Geometry This topic is the most important one in competitions. At the IMO usually two of the six problems come from elementary geometry. Some of them can be treated conveniently with vectors, complex numbers, or transformation geometry. But usually ingenuity plus a few quite elementary facts from Euclidean geometry are required. We will not give a list of prerequisites, but just use them. We start with a set of easy problems, which can be used in a regular classroom. The main part consists of a mixture of more difficult to very hard problems. We give just one typical example.
E1. One of the cross sections in a rectangular box is a regular hexagon. Prove that the box is a cube.
E1 belongs to the category of easy problems, yet it is by no means trivial. As soon as you have the right idea, it is immediately trivialized. The trivializing idea is to extend every second side of the hexagon to get a regular triangle. In Fig.
12.34 the vertices K, L, and M of this triangle lie on the extensions of the edges AB, AA1 and AD of the box. We have △KLA ∼ △LMA since KL LM, ̸ KAL ̸ LAM 90◦and AL is a common side. This implies KA MA.
Similarly KA LA. Since PQ KM/3, we have LPQ ∼LKM and LPA1 ∼ LKA. Hence AA1 2AL/3, AB 2AK/3, AD 2AM/3. This implies AB AA1 AD, i.e., the box is a cube.
If the box is not rectangular, it can still have a cross section in the shape of a regular hexagon. Stretch the cube in Fig. 12.34 along the diagonal AC1. A A A A A A A Fig. 12.34 K B A C DM P B1 A1 Q C1 D1 L 12.3.1 Easy Geometrical Problems 1. The medians of a triangle partition its area into six equal parts.
2. From the medians of △ABC one can construct a triangle, the area of which is 3/4 of the area of △ABC.
3. Can two triangles have two equal sides and three equal angles, and still be noncon-gruent? If yes, then give conditions.
4. A convex quadrilateral is cut by its two medians into four parts. Show that they can be assembled into a parallelogram.
5. Why is a foldline of a piece of paper always straight?
12.3 Classical Euclidean Geometry 319 e e e e e e e e A B P Q R C S Fig. 12.35 @ @ @ @ @ @ r P Fig. 12.36 6. Can you wrap the surface of a unit cube with a 3 × 3 piece of paper?
7. For each side of a convex quadrilateral, we draw a circle with this side as a diameter.
Show that the four circular discs cover the quadrilateral.
8. Show that the points B, R, C in Fig. 12.35 are collinear.
9. Let a, b, c, d be the sides of a quadrilateral with area A. Prove that (a) A ≤ab + cd 2 , (b) A ≤ac + bd 2 , (c) A ≤a + c 2 · b + d 2 .
10. What is the maximum area of a quadrilateral with sides 1, 4, 7, 8?
11. The semicircular disc in Fig. 12.36 glides along two legs of a right angle. Which line describes point P on the perimeter of the half circle?
12. Show how to cut any triangle by two straight cuts into symmetric parts. Which triangles can be cut into two symmetric parts by one straight cut?
13. You have any amount of string, three short pegs, and one iron ring. How can you tie a cow in such a way that she can graze the entire inside of a semicircular lawn without being able to overstep its boundary.
14. Find a point inside (a) a quadrilateral (b) a regular hexagon so that the sum of its distances from the vertices is minimal. The problem for the regular pentagon is considerably tougher and will be treated later.
15. Draw a polygon and a point O in its interior, so that no side is completely visible from O.
16. Draw a polygon and a point O in its exterior, so that no side is completely visible from O.
17. Does there exist a polyhedron and a point O outside, such that none of its vertices is visible from O?
18. Given any n-gon, show that it has at least one internal diagonal.
19. What is the sum of the interior angles of a star polygon with (a) 5, (b) 7, (c) 8 sides.
20. On a square ABCD with side a, we construct an isosceles triangle CDE to its interior with legs b, so that ̸ ABE 15◦. Prove that b a.
21. (Solve by ruler!) Given two parallel segments. Find their midpoints.
320 12. Geometry 22. (By ruler only!) Given a segment a and its midpoint. Construct through M ̸∈a a line g ∥a.
23. (Solve by ruler!) Given a parallelogram. Draw a parallel through its center to a side.
24. Four points are given in a plane. We can construct rectangles the sides of which pass through these four points. Find the locus of the midpoints of these rectangles.
25. Points A, B are fixed. Find the set of all feet of perpendiculars from A to all possible straight lines through B.
26. Two points A, B are fixed. A is reflected at all straight lines through B. Find the locus of all images.
27. Assume a fixed line t and two fixed points A, B on t. Two variable circles are tangent to t in A and B, and they touch in M. Find the locus of M.
28. Given a circle C and two points A, B inside of C, inscribe in C a right triangle with legs of the right angle passing through A and B.
29. B is a fixed point and a is a straight line. Erect a square ABCD with A ∈a. What line describes C if A runs through all points of a?
30. A circle with diameter r rolls inside a circle with diameter 2r. What line describes a point K of the rolling circle?
31. Two circles intersect in A and B. P traces the arc AB. Show that the length of the chord CD cut out by P A and P B on the other circle has constant length.
32. Given two fixed circles C1 and C2 with centers O1 and O2, find the locus of midpoints of the segment XY, where X ∈C1 and Y ∈C2.
33. Let P be any point inside a triangle with distances la, lb, lc from the sides a, b, c.
We may assume a ≤b ≤c. Show that hc ≤la + lb + lc ≤ha. There is equality iff the triangle is regular.
34. M is the midpoint of segment AB. Prove that, for every point P in space, |P M| ≤|P A| + |P B| 2 .
35. M is the midpoint of segment AB. Prove that, for every point P of space, ||P A| −|P B|| ≤2|P M|.
36. Characterize the set of all planes equidistant from points A, B.
37. If G is the centroid of the tetrahedron ABCD, then, for any point P, |P G| < 1 4 (|P A| + |P B| + |P C| + |P D|) .
38. In a triangle ABC, A is reflected at B to A′, B is reflected at C to B′, C is reflected at A to C′. Find |A′B′C′| in terms of |ABC|.
39. What four-bar linkage with sides a, b, c, d has maximum area?
40. Can you get a quarter through a penny-sized circular hole in a piece of paper?
41. Let a, b, c, d, e be five line segments. Any three of them can be used to construct a triangle. Show that at least one of these triangles is acute.
12.3 Classical Euclidean Geometry 321 42. Suppose that the sun is exactly overhead. How should I hold a rectangular box over a horizontal table so that its shadow has maximum area?
43. Solve the preceding problem for a regular tetrahedron.
44. Take any convex n-gon. Select any m points inside of it. Cut the polygon into non-intersecting triangles whose vertices are these m + n points. How many triangles do you get in terms of m and n?
45. Points of space are colored with five colors (all five colors do occur). Prove that there exists a plane, the points of which are colored by at least four different colors.
46. Many identical rectangular boxes are available. Give a practical method for measur-ing a space diagonal.
47. The midpoints of the altitudes of a triangle are collinear. Find the shape of the triangle.
48. A convex quadrilateral is cut by its diagonals into four triangles of equal perimeter.
What can you infer for the shape of this quadrilateral?
49. A point P is chosen inside a square, and parallels to the sides and diagonals are drawn through P . They split the square into eight parts, which we label 1 and 2 alternately around P . Prove that the parts labeled 1 and those labeled 2 have equal areas.
50. Any four of five circles have a common point. Prove that all five circles have a common point.
51. Two parallel planes and two spheres are given in space. The first plane touches the first sphere in A, the second plane touches the second sphere in B, and the spheres touch in C. Prove that A, B, C are collinear.
52. Can you cut a hole into a cube so that a slightly larger cube can pass through the hole?
53. An equilateral triangle ABC is inscribed in a circle. An arbitrary point M is chosen on the arc BC. Prove that |MA| |MB| + |MC|.
54. If the incircle of a quadrangle ABCD has radius r, then |AB| + |CD| ≥4r.
55. Given three points in a plane, construct a quadrilateral for which these points are midpoints of three successive equal sides.
56. If the angles α, β, γ of a triangle satisfy cos 3α + cos 3β + cos 3γ 1, then one of its angles is 120◦.
57. The base of a pyramid is an n-gon, n odd. Can you place arrows on the edges of this pyramid, such that the sum of the vectors is − → O ?
58. Prove that a square has the smallest perimeter of all quadrilaterals circumscribed about a given circle of radius r.
59. From a point O inside an equilateral triangle ABC, perpendiculars OM, ON, OP are dropped onto the sides BC, CA, AB. Prove that |AP |+|BM|+|CN| does not depend on the location of the point O.
60. Circles with centers O and O′ are disjoined. A tangent from O to the second circle intersects the first circle in A and B. A tangent from O′ to the first circle intersects the second circle in A′ and B′, and A and A′ lie on the same side of OO′. Suppose we know the distances |AA′| a and |BB′| b. Find |OO′|.
322 12. Geometry 61. Let ABCD be a convex quadrilateral with area F. Suppose that |AM|2 + |BM|2 + |CM|2 + |DM|2 2F for some point M of the plane. What can you say about A, B, C, D, M?
62. A trapezoid ABCD is drawn on paper, together with the median EF connecting the midpoints of AD and BC, and the segment OK ⊥AB, where O AC ∩BD and K ∈AB. Now everything is erased except the segments EF and OK. Reconstruct the trapezoid.
63. A right triangle D is divided by its altitude into two triangles D1 and D2. Prove that the sum of the radii of the incircles of D, D1, D2 is equal to the altitude of D.
64. Inscribe squares with sides x, y, z into a triangle with sides a, b, c, so that two vertices lie on BC, CA, AB, respectively. Then x y z ⇒a b c.
65. In a triangle, ha 12, hb 20. Prove that 7.5 < hc < 30.
66. The distance between any two trees in a forest is less than the difference of their altitudes. Any tree has altitude < 100 m. Prove that the forest can be surrounded by a fence of length 200 m.
67. The radii of the insphere and circumsphere of a tetrahedron are r and R, respectively.
Prove or disprove that R ≥3r.
68. The skew edges of a tetrahedron are pairwise equal. Prove that the centers of its insphere and circumsphere coincide.
69. A point O inside a convex quadrilateral is joined to its vertices. Find the area of the quadrilateral with vertices in the centroids S1 to S4 of the four triangles ABO, BCO, CDO, DAO.
70. By means of a ruler in the shape of a semicircular disk draw the perpendicular to a given line l through a given point A.
71. Fig. 12.37 shows a four-bar linkage. The longest link a is fixed. When the shortest link d makes a complete revolution, the rocker b oscillates between two extreme positions. How do you find these extreme positions? Show that a + d ≤b + c, i.e., the sum of the shortest and longest link does not surpass the sum of the other two links.
C C C C A A A A A A a b c d Fig. 12.37 72. The sides of a skew quadrilateral are tangent to a cone. Show that the tangent points are coplanar.
Solutions 1. Two triangles with the same base and altitude have the same areas. Thus, we have the equalities in Fig. 12.38. Now a + a + b, a + b + b, a + a + c are half the area of the triangle. Hence, a b c.
12.3 Classical Euclidean Geometry 323 J J J J J J J b b b b b b E E E E E E E c c b b a a Fig. 12.38 C C C C C C C C C C C C A1 A2 A3 A4 A1 A2 A4 Fig. 12.39 @ @ @ @ @ @ @ @ @ @ @ @ @ @ Fig. 12.40 3 3 2 √ 2 -1 1 2. Let |ABC| F. Reflect the centroid S at the midpoint P of AB with image S′.
Then |AS| 2 3ma, |SS′| 2 3mc, |AS′| 2|RS| 2 3mb, |AS′S| 1 3F (since |ASP | 1 6A). Stretch △ASS′ from A with factor 3 2. Its area increases by factor 9/4. The stretched triangle AT Q has sides ma, mb, mc and area |AT Q| 9 4 · 1 3|ABC| 3 4|ABC|.
The triangle AT Q can be constructed by translation of ma, mb, mc.
3. Yes, e.g. the triangles with sides 1, 3/2, 9/4 and 3/2, 9/4, 27/8. They have two equal sides, and they have proportional sides, e.g., they are similar and have equal angles.
Generally, two triangles with sides a, aq, aq2 and aq, aq2, aq3 are similar and have two common sides. To be constructible, they must satisfy the triangular inequality.
For q > 1, this means q2 < q + 1 and, for q < 1, we must have 1 < q + q2. Thus, √ 5 −1 2 < q < √ 5 + 1 2 with the exception of q 1, which would give three equal sides. In all other cases the longest side would satisfy the triangular inequality.
4. Fig. 12.39 shows the proof.
5. Fold the paper. Let A and B be coinciding points on the two folds. Now unfold again. Let X be any point on the fold line. Then |AX| |BX|. Thus X lies on the perpendicular bisector of AB.
6. Yes, and Fig. 12.40 shows a solution.
7. Drop the perpendiculars from B and D onto the diagonal AC. The quadran-gle is cut into four rectangular triangles 1, 2, 3, 4. The circles with diameters AB, BC, CD, DA are circumscribed about 1, 2, 3, 4.
8.
̸ BRS ̸ AP S ̸ SQC α. Since ̸ SQC + ̸ SRC 180◦, we must have ̸ SRC 180◦−α. Thus ̸ BRS + ̸ SRC 180◦, and B, R, C are collinear.
9. (a) If a is the base and h is the altitude of a triangle ABC, and b one of the other sides, then h ≤b. In Figs. 12.41 and 12.42, |ABC| ≤ab 2 , |ACD| ≤cd 2 , A |ABCD| ≤ab + cd 2 .
We have equality iff AB ⊥CD and CD ⊥DA. In this case the quadrilateral is cyclic. The inscribed circle has diameter AC. Thus, A ab + bc 2 ⇐ ⇒ ̸ D ̸ B 90◦⇐ ⇒a2 + b2 c2 + d2 |AC|2.
324 12. Geometry J J J J J Fig. 12.41 a b h @ @ @ @ @ b a Fig. 12.42 @ @ @ @ Q Q Q Q @ @ @ @ A A A Fig. 12.43 Fig. 12.44 b d a c d a c b D D A A B B C C′ (b) We reduce this case to the preceding one. In Fig. 12.43 cut the quadrilateral along the diagonal BD and turn the triangle BCD over. You get the new quadrilateral ABC′D in Fig. 12.44 with the same area to which the preceding result can be applied giving A ≤ac + bd 2 .
There is equality iff in the new quadrilateral AB ⊥BC′ and DC′ ⊥AD, i.e., β + δ′ δ + β′ 90◦, or a2 + c2 b2 + b2 |AC′|2. In addition, ABCD is a cyclic quadrilateral.
(c) |ABC| ≤ab/2, |BCD| ≤bc/2, |CDA| ≤cd/2, |DAB| ≤da/2. Thus, a + c 2 · b + d 2 1 2 ab 2 + bc 2 + cd 2 + da 2 ≥|ABCD|.
We have equality iff ̸ A ̸ B ̸ C ̸ D, i.e., for rectangles.
10. We may assume that the sides 1 and 8 are neighbors. If not, we cut the quadrilateral along a diagonal and turn over one of the triangles as in the preceding problem. Now the quadrilateral has area ≤1·8/2+4·7/2 18. Since 12 +82 42 +72 65 we can build a quadrilateral of area 18 from two right triangles with common hypotenuse √ 65.
11. In Fig. 12.45, AOBP is a cyclic quadrilateral with fixed ̸ ABP α. Thus ̸ AOP is also fixed, and P traces a segment on the fixed straight line OP. This was a 2 minute problem from a Hungarian TV show in the sixties monitored by Renyi, Turan, and Alexits.
12. Let AB be the maximal side of △ABC. The foot D of the altitude from C lies on AB. Join D with the midpoints P , Q of AC and BC. Then |AP | |PC| |DP |, |BQ| |QC| |DQ|. Thus ADP and DBQ are isosceles, and DPCQ is a symmetric deltoid. Its symmetry line is the perpendicular bisector of CD. See Fig. 12.46.
13. In Fig. 12.47 the cow C is tied in the middle of a rope of length 2r. At the one end is the peg in the center M, at the other is the ring R. The line MC prevents the cow from overstepping the semicircle, the line CR prevents the overstepping of the diameter.
The exact solution is impractical since the cow is a strong animal and the rope DE will bend. Give an almost correct solution which is very robust.
12.3 Classical Euclidean Geometry 325 @ @ @ @ @ q B B B B B B P O A B α α Fig. 12.45 @ @ @ @ @ A A A P Q A B C p p p q q q D Fig. 12.46 h R q q r r M D E r C r r r Fig. 12.47 14. (a) If the quadrilateral ABCD is convex, the problem is easy. Fig. 12.48 shows by the triangle inequality that P must coincide with the point O of intersection of the diagonals.
Now look at Fig. 12.49. Here the point D lies inside △ABC. Obviously |AP | + |DP| > |AD|, and two applications of the triangle inequality show that |BP | + |CP | > |BD| + |CD|. Show this. By adding the two inequalities, we get |PA| + |PB| + |P C| + |P D| > |DA| + |DB| + |DC|. Hence, D is the optimal location for P.
Suppose D lies on side BC of △ABC. We have P A + P D > DA, PB + PC > DB + DC. Adding the two inequalities, we get |P A| + |P B| + |P C| + |P D| > |DA| + |DB| + |DC|, that is, D is the optimal location.
But what if the points A, B, C, D all lie on a straight line? This leads to a highly interesting problem which can be solved for any number of points: n friends live at x1 < x2 < · · · < xn on the same street. Find a meeting place P, so that the total distance travelled is minimal.
For n 2, any point x ∈[x1, x2] will give the minimum distance x2 −x1. Now let n 3. For x1 and x3, any point in [x1, x3] will do. Of these points, x2 is optimal for x2 itself. Hence, x2 is the optimal point.
Generally, for even n, any point in the innermost interval [xn/2, xn/2+1] is optimal.
For n odd, the innermost point x(n+1)/2 is the optimal point.
(b) Fig. 12.50 shows that this time the problem is again trivial. Three applications of the triangle inequality show that P must be the center O.
A A A A A A a a a a a a a a X X X X X X @ @ P A O B C D Fig. 12.48 @ @ @ @ @ @ Q Q Q Q Q X X X X X X A A A P A B C D Fig. 12.49 326 12. Geometry X X X X X B B B B !!!!!
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HHHH H XXXX X @ @PPP P F A B C D E O P Fig. 12.50 q Z Z Z ZZ Z O Fig. 12.51 q Z Z Z ZZ Z O Fig. 12.52 15. Fig. 12.51 shows one example.
16. Fig. 12.52 shows an example.
17. (Due to the contestant Brailow.) Take two thin parallel square plates of the same size. Between them we take a square frame of the same size rotated by 45◦versus the plates. The frame will hide the vertices of the plates from the center of symmetry O. In the angles of the frame perpendicular to its plane, we place 4 ‘pencils’ which hide the vertices of the frame. The plates will hide the ends (vertices) of the pencils.
18. Consider any polygon. Let A, B, C be three successive vertices. We draw through B all rays filling the interior of the angle ABC. Either some ray will hit another vertex D, then BD is an internal diagonal, or none of the rays hits another vertex, then AC is an internal diagonal.
19. (a) The sum of the interior angles of a star pentagon is 180◦.
(b) There are two kinds of star polygons with 7 vertices with sums of interior angles: S7,2 540◦and S7,3 180◦. You can skip one or two vertices.
(c) There is just one star polygon with 8 vertices. You skip two vertices. The others degenerate. The sum of interior angles for the nondegenerated star octagon S8,3 360◦.
The best way to find the sum of the interior angles of a star polygon is to move a pencil around its contour, turning the pencil at each vertex by the angle at that vertex.
Rotation must always be in the same direction to get the sum of the interior angles.
20. First proof. In Fig. 12.53 suppose ̸ AED ̸ BEC ϵ. Then b > a ⇒ϵ < 75◦⇒α > 60◦⇒β < 60◦⇒b < a.
Contradiction!
b < a ⇒ϵ > 75◦⇒α < 60◦⇒β > 60◦⇒b > a.
Contradiction!
Thus, a b.
Second proof. In Fig. 12.54 we erect △BCF ∼ △ABE on BC to the interior. Then we easily find |CE| a.
Third proof. Erect the regular triangle ABE′ on AB to the exterior. Then AEE′ and BEE′ are isosceles, i.e., |EE′| a. Besides, AE is the bisector of ̸ DAE′. Hence |DE| |EE′| a, and DCE is regular.
Fourth proof. Erect the regular triangle DCE on CD to the interior. The remainder is clear.
12.3 Classical Euclidean Geometry 327 pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp 15◦15◦ A B C D a b b E a a c ϵ ϵ Fig. 12.53 pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp 15◦15◦ A B C D a a b F b E a ϵ pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp c c c Fig. 12.54 @ @ @ @ @ @ a a a a a a a A B C D C C C C C C M N O Fig. 12.55 Fifth proof. Hint: Rotate the square about its center by 90◦.
21. (a) Suppose AB ∥CD and |AB| ̸ |CD|. Fig. 12.55 shows the construction of the midpoints M, N of AB and CD based on a theorem about the trapezoid, which we proved earlier.
(b) |AB| |CD|. This is problem 23 below.
22. Given a segment AB, its midpoint N, and a point M, Draw AM, BM, NM. Choose C ∈AM freely. Draw BC intersecting MN in S. Draw AS, intersecting BM in D.
Now CD and MN intersect in P . Transforming NBCP by a shear into NBP D, we get Q ∈NC ∩P B. Then QM ∥AB ∥CD.
23. In Fig. 12.56, we are given the parallelogram ABCD. We can find the center M AC ∩BD. Now we find the midpoint N of AB as follows. We choose a point P on BC and draw P A, which intersects DC in E. We can find N as in problem 19 from S BE ∩AC and N P S ∩AB.
24. In Fig. 12.57, draw any line a through A, a line c ∥a through C, and lines b ⊥a, d ⊥a through B and D. Let E and F be the midpoints of AC and BD, respectively.
If M is the center of the rectangle, we have ̸ EMF 90◦. If line a rotates about A, the points E, F remain fixed, and M describes the circle with diameter EF. We have assumed that A, C are on opposite sides of the rectangle. But A, B or A, D could just as well be on opposite sides. Thus the locus consists of the union of three circles, which are easy to construct.
25. The circle with diameter AB.
26. Stretch the circle with diameter AB from A by a factor of 2.
27. The circle with diameter AB.
28. Describe a circle C1 with diameter AB. It intersects C in D. The straight lines DA and DB intersect C a second time in E and F. Then DEF is the required triangle.
There are 0, 1, 2, ∞possible solutions depending upon the number of common points of C and C1.
29. The locus of C is the line a rotated by 90◦about B.
30. A point of the rolling circle describes a diameter of the large circle in Fig. 12.58.
31.
̸ APB β and ̸ ACB ̸ ADB α are fixed. Hence ̸ CAD α + β is also fixed. A chord CD of fixed length belongs to this fixed angle.
32. Fix a point X ∈C1. The locus of all midpoints XY for Y tracing C2 is a circle with radius r2/2 about the midpoint of XO2. If we let X trace C1, the set is the union of all circles of radius r2/2 about all points of the circle with radius r1/2 and midpoint O3 of O1O2. This is the area of the closed ring about O3 with inner radius (r1 −r2)/2 and outer radius (r1 + r2)/2.
328 12. Geometry !!!!!!!!
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A A A A A B D C E Q Q Q Q Q Q N M S P Fig. 12.56 d c b a q D qC qB q \ \ \ \ \ \ \ A q qF Mq E Fig. 12.57 J J α 2α O K A B Fig. 12.58 33. aha bhb chc Hence a ≤b ≤c ⇒hc ≤hb ≤ha. First replacing a, b, c in chc ala + blb + clc aha by c and then by a, we get hc ≤la + lb + lc ≤ha.
The sum of the distances is a minimum for the vertex with largest angle and largest for the vertex with smallest interior angle. In particular, for an equilateral triangle, la + lb + lc h is independent of the location of the point inside the triangle.
34. Reflect P at M to P ′ to get the parallelogram P AP ′B. The triangular inequality gives |P M| ≤|P A| + |P B| 2 .
35. Reflect P at M to P ′. The sides of the triangle AP ′P are |PA|, |PB|, and 2|PM|.
Since each side is greater than the difference of the other two, we have ||P A| −|P B|| ≤2|P M|.
We have equality for the degenerated triangle.
36. The planes parallel to AB and through the midpoint of AB are equidistant from A and B.
37. G is the midpoint of EF, where E and F are the midpoints of AD and BC. Applying problem 34 three times, we get |P G| < 1 2 (|P E| + |P F|) , |P E| < 1 2 (|P A| + |P D|) , |P F| < 1 2 (|P B| + |PC|) .
Thus, |P G| < 1 4 (|P A| + |P B| + |P C| + |P D|) .
38. Fig. 12.59 shows that |A′B′C′| 7|ABC|.
39. Let F(x) |ABCD|. Fig. 12.60 shows that F(x) ab 2 sin x + cd 2 sin y with the auxiliary condition a2 + b2 −2ab cos x c2 + d2 −2cd cos y. Deriving for x, F ′(x) ab 2 cos x + cd 2 cos y · y′.
(1) 12.3 Classical Euclidean Geometry 329 C C C C C @ @ @ @ @ aaaaa a pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp A B C A′ B′ C′ F Fig. 12.59 B B B B B B !!!!
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A B C D a b c d y x Fig. 12.60 Implicitly deriving the auxiliary condition, we get 2ab sin x 2cd sin y · y′ or y′ (ab/cd) · (sin x/sin y). Inserting y′ into (1), we get F ′(x) ab 2 · sin x cos y + cos x sin y sin y ab 2 sin(x + y) sin y .
F ′(x) 0 ⇒sin(x + y) 0 ⇒x + y π. x + y < π ⇒F ′(x) > 0, x + y > π ⇒F ′(x) < 0. We get a maximum for a cyclic quadrilateral.
40. We cut a hole of diameter d in a piece of paper and fold it twice along perpendicular diameters. The endpoints of the diameters are A, B and C, D. Now it is possible to get the points A, C, B into a straight line. Thus we get a slit of size d √ 2. A penny has diameter d 3/4. We can get a coin of diameter 3 √ 2/4 > 1.06 through a penny-sized hole. A quarter has diameter 1, thus we can easily push it through the hole.
41. For a triangle with side lengths a ≥b ≥c, we have α 90◦⇔a2 b2 + c2, α > 90◦⇔a2 > b2 + c2, α < 90◦⇔a2 < b2 + c2.
We may assume that a ≥b ≥c ≥d ≥e.
(1) We assume that triangles (a, b, c) and (c, d, e) are not both acute. This will lead to a contradiction. The nonacuteness of the two triangles is equivalent to a2 ≥b2 + c2, (2) c2 ≥d2 + e2.
(3) From (2) and (3), we get a2 ≥b2 + d2 + e2.
(4) From (1) and (4), a2 ≥c2 + d2 + e2.
(5) (3) and (5) imply a2 ≥d2 + e2 + d2 + e2. Thus, a2 ≥(d + e)2 + (d −e)2, a2 ≥(d + e)2, a ≥d + e.
But we are told that a, d, e can be used to form a triangle. Yet the last relation contradicts the triangle inequality a < d + e.
42. The area of the shadow is twice the area of △ABC in Fig. 12.61. Thus, we should maximize the projection of ABC on the table, which is the case, when the triangle is horizontal.
330 12. Geometry HHHHHH H A B C Fig. 12.61 @ @ @ @ @ @ @ @ A B C D P R Q S Fig. 12.62 43. The square ABCD in Fig. 12.62 must be placed horizontally. It is parallel to two opposite edges of the tetrahedron.
44. Let x be the number of triangles formed. We can compute the sum S of their angles in two ways. On the one hand, S 180◦x. On the other hand, S 360◦m+180◦(n−2).
The first term is the sum of the angles of all m interior points. The second term is the sum of the angles of an n-gon. By equating the right sides of the two equations, we get x 2m + n −2.
Do this problem by induction, or use Euler’s formula f + v e + 2.
45. We denote the five colors by a, b, c, d, e. Corresponding points are denoted by A, B, C, D, E. We prove two lemmas.
Lemma 1 (L1): Suppose the conditions of the problem are satisfied. If there exists a three-colored straight line, then there exists in space a four-colored plane.
Proof. Suppose the straight line v consists of points with colors a, b, c. We know that there exists a point D in space with color d. Every plane (at least one) containing v and D is four-colored.
Lemma 2 (L2): Suppose the conditions of the problem are satisfied. If there exists a three-colored plane and a straight line, which contains points of the two other colors, and which intersects this plane, then there exists a four-colored plane.
Proof. Suppose the plane S contains points with colors a, b, c and v contains points with colors d, e. Let P ∈v ∩S. If P has one of the colors a, b, c, then v is three-colored, and according to L1 there exists a four-colored plane. If P has one of the colors d or e, then S is four-colored.
Proof of the theorem: If four of the points A, B, C, D, E are in one plane, then we are done. Otherwise ABCD is a tetrahedron. One of its faces, for instance, S (BCD), separates the other two points A and E. Then line AE intersects the plane S, and the theorem is correct according to L2.
Otherwise, E is contained in the tetrahedron, and A ̸ E. Hence, AE intersects S, and the theorem is correct according to L2.
Since the problem is so simple, there are many other proofs. Let us sketch another one.
Second proof. Let ABC S1, CDE S2, S1∩S2 m, C ∈m. If AB or DE and m intersect, the theorem is valid according to L2. Otherwise ABCD is four-colored.
46. In Fig. 12.63, it is easy to measure the segment AB.
12.3 Classical Euclidean Geometry 331 A B Fig. 12.63 q q q q A B Q R S T C D ppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp Fig. 12.64 47. The midpoints H1, H2, H3 of the altitudes of a triangle lie on the three sides of the triangle of midpoints of the sides of a triangle. No two of the points Hi can coincide.
The only way for the Hi to be collinear is that they lie on one side of the triangle of midpoints. For instance H1 and H2 are endpoints and H3 lies between H1 and H2.
The only solution is the right triangle.
48. First we show that O is the midpoint of the diagonals. Let |OC| ≥|OA| and |OD| ≥|OB|. Reflect △ABO in O to get parallelogram ABMN. Now ABO and MNO have the same perimeter p+q+a. But CDO has the same perimeter p+q+a.
On the other hand, it has the perimeter p + q + x + y + c. Hence a c + x + y.
This implies that x y 0 and c a. Thus O is the midpoint of the diagonals in ABCD. Comparing the perimeters of ABO and DAO, we get a b. ABCD has equal sides, i.e., it is a rhombus.
49. Draw a figure. Let the square have side 1. Express all of the segments on the sides by the variables x, y, z. Now compute the area of the parts labeled 1. The result will be 1/2. Find an ingenions proof by dissection.
50. Let A be a common point of circles 1, 2, 4, 5, B a common point of circles 1, 3, 4, 5, C a common point of circles 2, 3, 4, 5. Then A, B, C are not all distinct, since all three lie on circles 4, 5. But two circles intersect at most twice. Thus, two of the three points coincide. Suppose A B. Then A lies on all five circles.
51. The points A, B, C lie in one plane. Thus, we may reduce the space problem to a problem in the plane containing the points A, B, C. We get a problem about two parallel lines a, b and two circles c1, c2, a ∩c1 A, b ∩c2 B, c1 ∩c2 C. This routine problem will be left to the reader.
52. Fig. 12.64 shows a unit cube with QA QD T B T C 3/4. ABCD is a square with side |AB| 3 √ 2/4 1.06066 . . . . Another solution is more obvious.
Project the cube orthogonally to a space diagonal. You get a regular hexagon. Inscribe thelargest square in this hexagonwithside √ 6− √ 2 1.035. . . and shrink it slightly so that its side is still > 1.
53. Firstproof.Weareinterestedin|MA| x, |MB| y, |MC| z.Theyaresidesof the triangles AMB, BMC with ̸ AMB 60◦, ̸ BMC 120◦. Denote |AB| a.
Since cos 60◦ 1/2, cos 120◦ −1/2, the Cosine Rule implies a2 x2 + y2 −xy and a2 y2 + z2 + yz.
Subtracting the two equations we get (x + z)(x −y −z) 0 after factoring. Hence x y + z.
Second proof. Since the segments MA, MB, MC are chords of the circle, the Sine Rule yields x 2R sin(α + 60◦), y 2R sin α, z 2R sin(60◦−α). This implies x y + z.
332 12. Geometry T T T T T X X X C C C b b b b b A B C M D Fig. 12.65 Third proof. The area of the quadrilateral ABCD can be expressed in two ways. Let φ be the angle between the diagonals AM and BC. Then 2|ABMC| ax sin φ. On the other hand, the same area is 2|ABM|+2|ACM|. Since ̸ ABM φ, ̸ ACM 180◦, we have ax sin φ ay sin φ + az sin(180◦−φ), which implies x y + z.
Fourth proof. The result |AM| |BM| + |CM| follows from Ptolemy’s theorem |BC| · |AM| |AC| · |BM| + |AB| · |CM|, since |AB| |BC| |CA|.
Fifthproof.OnsegmentMA,wemeasureoffsegmentDM,whichisequaltosegment MB. We prove that |DA| |MC|. Since ̸ AMB 60◦, △DBM is regular as is △ABC. Rotate BMC around B by 60◦so that C coincides with A. Then M coincides with D and segment MC coincides with DA. Thus DA MC, and |MA| |MB| + |MC|. This short geometrical solution shows a road to a generalization. Let M be any point in the plane. Then a similar construction gives a point D, which need not lie on AM. But still, the segments MA, MB, MC are the sides of △ADM.
Thus we get the following theorem due to the Roumanian mathematician Pompeiu (1873–1954): If in the plane of the equilateral triangle ABC a point M is given, then one can construct a triangle from MA, MB, MC. It degenerates for all points of the circumcircle of ABC. See Fig. 12.65.
54. We have S pr and 2S ≤ad+bc, 2S ≤ab+cd, i.e., 4S ≤(ab+cd)+(ad+bc) (a + c)(b + d) p2. Hence 4pr ≤p2, or 4r ≤p |AB| + |CD|.
55. If K, L, M are the given midpoints of three sides AB BC CD of a quadrilateral ABCD, then B and C lie on the perpendicular bisectors of KL and LM. Reflect one of the perpendiculars at L to get B or C.
56. We will prove the theorem by transforming the equality cos 3α + cos 3β + cos 3γ 1 (1) into an equivalent one. For one of the angles α, β, γ to be 120◦it is necessary and sufficient that one of 1 −cos 3α, 1 −cos 3β, 1 −cos 3γ is zero: (1 −cos 3α)(1 −cos 3β)(1 −cos 3γ ) 0.
(2) So we attempt to transform (1) into (2). γ 180◦−(α + β), cos 3γ −cos(3α + 3β) −cos 3α cos 3β + sin 3α sin 3β. (1) becomes cos 3α + cos 3β −cos 3α cos 3β + sin 3α sin 3β −1 0 ⇒sin 3α sin 3β (1 −cos 3α)(1 −cos 3β).
Squaring, we get sin2 3α sin2 3β (1 −cos 3α)2(1 −cos 3β)2, or (1 −cos2 3α)(1 −cos2 3β) (1 −cos 3α)2(1 −cos 3β)2, (1 −cos2 3α)(1 −cos2 3β) −(1 −cos 3α)2(1 −cos 3β)2 0, (1 −cos 3α)(1 −cos 3β)(cos 3α + cos 3β) 0.
12.3 Classical Euclidean Geometry 333 But from (1), we have cos 3α + cos 3β 1 −cos 3γ . This implies (2).
57. No! Project the vectors onto the altitude SO of the pyramid. The projection of the vectors of the base is − → O . The projection of each lateral edge is ±− → OS. Adding them we get at least one vector ±− → OS. So the total sum is ̸ − → O .
58. The area of the inscribed circle is r · p, where p is the perimeter. Thus we can also minimize area. Let Q be the square circumscribed about the circle C with radius r, and let C′ be the circle circumscribed about Q. We denote by s the segment cut off from C′ by a side of the square. Then |Q| |C′|−4s (Fig. 12.66). If ABCD in Fig.
12.67 is not a square, at least one vertex, in our case D, will lie inside C′. Any side of ABCD cuts off the same segment s from C′. Since at least two of the segments overlap (at D), the area |C′| −4s is smaller than |ABCD|, i.e., |ABCD| > |Q|.
C′ s C Q Fig. 12.66 & & & & & && D D D D D D D D A B C D C′ Fig. 12.67 59. Draw perpendiculars through A, B, C to AB, BC, CA, respectively. We get △A′B′C′, where |OM′|, |ON′|, |OP ′| are the distances from O to the sides B′C′, C′A′, A′B′, respectively. Since |AP | |OM′|, |BM| |ON′|, |CN| |OP ′|, we have |AP|+|BM|+|CN| |OM′|+|ON ′|+|OP ′|. But the right side is the sum of the distances of O from the sides of the equilateral triangle A′B′C′, which is a constant. If a is the side of △ABC, then this sum is the altitude of A′B′C′, i.e., 3a/2.
60. If AA′B′B is a trapezoid and OO′ is its median, then |OO′| (a + b)/2.
61. AM2 + BM2 ≥2AM · BM, BM2 + CM2 ≥2BM · CM, CM2 + DM2 ≥ 2CM · DM, DM2 + AM2 ≥2DM · AM. Adding these inequalities and dividing by 2, we get AM2 +BM2 +CM2 +DM2 ≥AM ·BM +BM ·CM +CM ·DM + DM · AM (AM + CM)(BM + DM) ≥AC · BD ≥2F. The first inequality becomes an equality for AM BM CM DM. The second inequality is valid if AM ⊥BM, BM ⊥CM, CM ⊥DM, DM ⊥AM. Thus ABCD is a square, and M is its center.
62. We use the following property: O the midpoint M of EF and the midpoint N of AB are collinear. EN and FN are the midlines of ABD and BCD and thus are parallel to the diagonals.
63. Drop the only nontrivial altitude h of D which splits the opposite side into segments of lengths p and q, p +q c. Denote the radii of the incircles by r, r1, r2. It is easy to prove that r (a+b−c)/2. Do it! Hence, r1 (p+h−a)/2, r2 (q +h−b)/2, or r + r1 + r2 h.
64. Prove that x 2A/(a + ha), y 2A/(b + hb), c 2A/(c + hc), where A is the area of the triangle. x y z implies a + ha b + hb c + hc. Let a ̸ b. Then a −b hb −ha 2A/b −2A/a 2A(a −b)/(ab) ⇒2A ab.
334 12. Geometry Hence, γ 90◦, and c > a, c > b. Similarly we get 2A bc, which implies α 90◦. Contradiction.
65. c > |b −a| and a 2A/ha, b 2A/hb, c 2A/hc imply 1 hc > 1 ha −1 hb 1 12 −1 20 1 30.
Hence, hc < 30. From a : b : c 1/ha : 1/hb : 1/hc and a + b > c, we get 1/12 + 1/20 > 1/hc and hc > 7.5.
66. Suppose the trees have the altitudes a1 ≥a2 ≥· · · ≥an and they grow at the points A1, . . . , An. We know that A1A2 ≤a1 −a2, . . . , An−1An ≤an−1 −an. The length of the segments A1A2 · · · An is ≤a1 −a2 + a2 −a3 + · · · + an−1 −an < 100 m.
This sequence of segments can be surrounded by a fence of length 200 m.
67. Choose a point on each face of the tetrahedron. The radius r1 of the sphere through these points is at least r, i.e., r1 ≥r. If the chosen points are the centroids of the faces, then they are vertices of a tetrahedron with edges 1/3 of the edges of the given terahedron. Hence, R 3r1, or R ≥3r. See Chapter 7, E22, 2nd proof.
68. The faces of the tetrahedron are congruent. Hence, their circumcircles are also con-gruent. Thus, the faces are equidistant from the center of the circumsphere, i.e., the centers of the insphere and circumsphere coincide.
69. Let EFGH be the quadrilateral of the midpoints of the sides of ABCD. Then |EFGH| 1 2|ABCD|, and |S1S2S3S4| 4 9|EFGH| 2 9|ABCD|.
12.3.2 Harder Geometrical Problems 1. How many spheres are needed to shield a point source of light?
2. Can you cut a thin hole into a plane, which leaves it connected, so that a wire model of (a) a cube of edge 1 (b) a tetrahedron of edge 1 can be pushed through the hole.
The hole must have negligible area, and the thickness of the wires must be negligible.
3. In an equilateral convex hexagon A1A2A3A4A5A6, we have α1 + α3 + α5 α2 + α4 + α6. Prove that α1 α4, α2 α5, α3 α6 (αi is the interior angle at vertex Ai).
4. A curve C partitions the area of a parallelogram into two equal parts. Prove that there exist two points A, B of C such that the line AB passes through the center O of the parallelogram.
5. For what n is it possible to construct a closed sequence of segments in the plane with lengths 1, . . . , n (exactly in this order) if any two neighboring segments are perpendicular?
6. For what n is it possible to construct a space polygon of side lengths 1, . . . , n (exactly in this order) such that any three successive sides are pairwise perpendicular?
7. N points are given in a plane, no three on a line. We connect them in pairs by nonintersecting segments, until there are no two points left which could be connected.
Find the lower and upper bounds for the number of segments that can be drawn.
8. A convex quadrilateral is cut by its diagonals into four triangles with integral areas.
Prove that the product of the four areas is a perfect square.
12.3 Classical Euclidean Geometry 335 9. Every isometry f of a finite point set H is such that f (H) H. In particular, the centroid S of H {A1, A2, . . . , An} is a fixed point of f .
10. Find a point P inside the regular pentagon with the minimum sum of its distances to the vertices.
11. n points A1, . . . , An are taken on a circle with radius O, such that their centroid lies in O. For which point P is |P Ai| minimal?
12. Let AB be one of the parallel sides of a trapezoid. Prove that the trapezoid is equi-lateral (|BC| |AD|) if |AC| + |BC| |AD| + |BD|.
13. A finite set S of points in the plane has the following property: if A and B are any two of its points, then the perpendicular bisector of AB is symmetry axis of S. Prove that all points of S lie on a circle. Is this also valid, if S has infinitely many points?
14. A finite set S of points in the plane has the property that if, for any two of its points A, B, there is an isometry f such that f (A) B, then also f (S) S. Show that all points of S lie on a circle. Is this also valid if S is infinite?
15. Every equilateral and equiangular pentagon in space lies in a plane.
16. Let ABCD be a quadrilateral with an incircle. Then the incircles of the triangles ABC and CDA are tangent.
17. Suppose the opposite sides of a convex hexagon are parallel. Prove that |ACE| ≥ 1 2|ABCDEF|. When do we have equality?
18. There are 150 cubic boxes of side 1 in a square yard of side 37. Prove that there is room for a cylindrical barrel of radius 1.
19. Which point P has minimal distance from the vertices of a triangle ABC?
20. Connect the four vertices of a square by a shortest street system.
21. Given a circle of radius 1 and n points A1, . . . , An of the plane, prove that there is a point M on the circle, so that |MA1| + · · · |MAn| ≥n.
22. The vertices of an equilateral closed sequence of segments are lattice points. Prove that it has an even number of sides.
23. Three points are given on a circle. Find a fourth point on the circle so that the four points are vertices of a quadrilateral with an incircle.
24. There is a box with sides a and b in a corridor of width c. Find the conditions in which it can be pushed through a door of width d.
25. Denote the radii of the incircle and the circumcircle of △ABC by r and R, respec-tively, and its semiperimeter by s. Prove that 2R + r s iff the triangle is a right triangle.
26. Prove that if four sides of a convex pentagon are parallel to the opposite diagonals then this also holds for the fifth side.
27. The chord CD of a circle with center O is perpendicular to its diameter AB, and the chord AE bisects the radius OC. Prove that the chord DE bisects the chord BC.
28. The cube ABCDA1B1C1D1 has edge of length 2. Find the minimal distance between the points of two circles, one of which is inscribed in the base ABCD of the cube and the other goes through the vertices A, C, D1.
336 12. Geometry 29. Does there exist an infinite set of points in space, which has at least one, but finitely many points on each plane?
30. Can space be represented as a disjoint union of nondegenerated circles?
31. Can space be represented as a disjoint union of skew straight lines?
32. If the sides of a skew quadrilateral touch a sphere, the points of contact are coplanar.
33. Place three cylinders of diameter a/2 and altitude a in a hollow cube of edge a, so that they cannot move inside the cube.
34. Three lattice points A, B, C are chosen in a plane. Prove that if △ABC is acute, then at least one lattice point is inside or on its sides.
35. Several intersecting circles are given in a plane. Their union has area 1. Prove that one can select several nonintersecting circles, so that the sum of their areas is at least 1/9.
36. Some barrels of radius 1 are stored inside a square of side 100, each barrel standing on its circular bottom. The barrels are placed such that any line segment of length 10 inside the yard hits at least one barrel. Prove that there are at least 400 barrels inside the yard.
37. Prove that not more than one vertex of a tetrahedron has the property that the sum of any two plane angles at this vertex is more than 180◦.
38. The vertices of a convex polyhedron are lattice points. There are no other lattice points inside on the faces or edges. Show that the polyhedron has at most eight vertices.
39. A convex 7-gon is inscribed in a circle. Three of its angles are equal to 120◦. Prove that two of its sides are equal.
The next 5 problems treat strategies of getting out of the woods.
40. A mathematician got lost in the woods. He knows its area S, but nothing else about its shape, except that it has no holes. Show that he can get out of the woods by walking not more than 2 √ πS miles.
41. A mathematician got lost in a convex woods of area S. Show that he can get out of the woods by walking not more than √ 2πS miles.
42. (Continuation of the preceding problem.) Consulting a person who knows the way out, he will need at most √S/π miles.
43. A mathematician got lost in the woods in the shape of a half plane. All he knows is that he is exactly one mile from the edge of the woods. Show that he can get out of the woods by walking not more than 6.4 miles. Experiment with some paths, and test them versus the nearly ideal 6.4 miles.
44. A mathematician got lost in the woods in the shape of a one mile wide strip. and infinite length. Try to find some good walking strategies, and test them versus 2.3 miles.
45. A transformation of the plane maps circles to circles. Does it map lines to lines?
46. Construct a cyclic quadrilateral from its sides.
47. A circle with center O, which is inscribed into △ABC, touches its sides in A1, B1, C1. The segments AO, BO, CO intersect the circle in A2, B2, C2, re-spectively. Prove that A1A2, B1B2, C1C2 intersect in one point.
12.3 Classical Euclidean Geometry 337 48. Twoacuteanglesα andβ satisfysin2 α+sin2 β sin(α+β).Provethatα+β π/2.
49. Regular triangles ABC, CDE, and EHK (vertices given counterclockwise) with pairwise common vertices C and E are located in a plane so that − → AD − → DK. Prove that △BHD is also regular.
50. Prove that if the opposite sides of a skew quadrilateral are congruent, then the line joining the midpoints of the two diagonals is perpendicular to these diagonals, and conversely, if the line joining the midpoints of the two diagonals of a skew quadrilat-eral is perpendicular to these diagonals, then the opposite sides of the quadrilateral are congruent. (This is again USO 1977. Now we are looking for a short geometric solution.) 51. In △ABC, the bisectors of α, β, γ meet the circumcircle in A1, B1, C1. Prove that |AA1| + |BB1| + |CC1| > |AB| + |BC| + |CA| (AuMO 1982).
52. All angles in a convex hexagon are equal. Prove that the differences of opposite sides are equal.
53. From a variable point P of the circumcircle of triangle ABC, we drop the perpendic-ulars P M and P N to the straight lines AB and AC, respectively. For what position of P is |MN| maximal and find this maximal length?
54. If an acute triangle has circumradius R and perimeter p, then p > 4R.
55. Let A1A2 · · · An be a regular plane polygon, and let P be any point of the plane.
Prove that one can construct some n-gon from the segments PAi, i 1, . . . n.
56. Prove that, if there exists a polygon with sides a1, a2, . . . an, then there exists an inscribed polygon with these sides.
57. The six planes bisecting the angles of neighboring faces of a tetrahedron meet in one point.
58. The six planes through the midpoints of the edges of a tetrahedron and perpedicular to them pass through one point.
59. A space polygon is called regular, if all its sides are equal and all its angles are equal.
In problem # 12, we have shown that a space pentagon does not exist. For what n do regular space polygons exist, which are not plane?
60. Does a polyhedron exist with all of its plane sections triangular?
61. Prove that the sum of the lengths of all edges of a polyhedron is greater than 3d, where d is the distance of two vertices A an B of maximal distance.
62. (a) Every diagonal of a convex quadrilateral ABCD divides its area into two equal parts. Prove that ABCD is a parallelogram.
(b) The diagonals AD, BE, CF divide the convex hexagon ABCDEF into two equal parts. Prove that these diagonals pass through one point.
63. The circumscribed sphere of a tetrahedron ABCD has center O. Find a simple condition for the tetrahedron so that O lies inside of it.
64. Find the highest number of acute angles in a plane, nonintersecting n-gon.
65. Three circles in space touch in pairs, and the three points of tangency are distinct.
Prove that these circles lie on one sphere or in one plane.
66. If each vertex of a convex polyhedron is joined to every other vertex by edges, then it is a tetrahedron (HMO 1948).
338 12. Geometry 67. Prove that a convex polyhedron cannot have exactly seven edges.
68. Three circles have a common intersection. Prove that the three pairwise common chords intersect in one point.
69. Prove that, for any tetrahedron, there exist two planes such that the ratio of the areas of the projections onto them is ≥ √ 2 (AUO 1978).
70. Four noncomplanar points are given in space. How many boxes are there, which have these four points as vertices (AUO)?
71. Let P be an arbitrary point inside △ABC; x, y, z the distances of P from A, B, C, respectively; u, v, w the distances from the sides BC, CA, AB, respectively. The sides of ABC will be denoted by a, b, c, its area by S; R and r are the radii of circumscribed and inscribed circles. Prove the following inequalities: (a) ax + by + cz ≥4S; (b) x + y + z ≥2(u + v + w); (c) xu + yv + zw ≥2(uv + vw + wu).
72. Consider the following theorems: (U): Circumscribed quadrilateral ⇐ ⇒α + γ β + δ 180◦.
(I): Inscribed quadrilateral ⇐ ⇒a + c b + d.
(A): Area of quadrilateral A √ abcd Prove that (U), (I) ⇒(A), (U), (A) ⇒(I), (I), (A) ⇒(U).
73. In a triangle, we have a + ha b + hb c + hc with the usual notation. What is so special about this triangle?
74. Two straight lines a and b intersect in O, and ̸ (a, b) α. A grasshopper starts in A ∈a and alternately jumps to B ∈b and back to a. His jump has constant length 1. Will he ever return to the starting point A?
75. A spherical planet has diameter d. Can you place eight observation stations on its surface, so that every celestial object at distance d from its surface is visible from at least two stations?
76. Opposite sides AB and DE, BC and EF, CD and FA of a convex hexagon are parallel. Prove that |ACE| |BDF|.
77. A hexagon with a circumcircle has three successive sides of length a and three successive sides of length b. Find the radius of the circumcircle.
78. M is a tiny Anchurian island whose territorial waters extend one mile. At night a powerful searchlight rotates slowly counterclockwise about M, illuminating the territorial waters. At B (distance 1 mile) there is a Sikinian boat whose mission it is to reach M undetected. The boat has maximum speed b. At a distance of one mile from M, the light beam of the searchlight has speed s.
(a) Suppose k s/b 8. Show that the boat can fulfill its mission.
(b) Suppose k s/b < 2π + 1. Show that the boat can fulfill its mission.
(c) Find the smallest k for which the boat can fulfill its mission.
79. A tetrahedron ABCD is inscribed in a sphere of radius R and center O. The straight lines AO, BO, CO, DO intersect the opposite faces in A1, B1, C1, D1. Prove that |AA1| + |BB1| + |CC1| + |DD1| ≥16 3 R.
12.3 Classical Euclidean Geometry 339 80. Pick proved a simple formula for the area f (P ) of any lattice polygon P: A f (P ) i + b 2 −1 2.
Here i and b are the numbers of interior and boundary points, respectively. We leave the the proof to you, but we give you the steps leading to a proof.
(a) Prove the formula for any lattice rectangle with sides p and q.
(b) Prove the formula for a right triangle with one horizontal and one vertical side.
(c) Pick’s formula assigns a number f (P ) to any polygon P. Prove that the function f is additive, i.e., if P1, P2 are polygons with a common boundary, then f (P1∪P2) f (P1) + f (P2).
(d) Show that f (P ) gives the correct area for any lattice triangle P.
(e) Finally, show that f (P ) is the area of any simple lattice polygon P.
81. In a tetrahedron A1A2A3A4, the face opposite vertex Ai has area Si. Choose a point P inside the tetrahedron with distances x1, . . . , x4 from the faces S1, . . . , S4, respec-tively, such that the sum Si/xi is minimal.
82. For which point P inside △ABC is the sum of the squares of its distances from the sides minimal?
83. A circle with radius r is inscribed in a triangle. Tangents parallel to the sides of the triangle cut off three small triangles from the triangle with inscribed circles of radii r1, r2, r3. Prove that r1 + r2 + r3 r.
84. A sphere of radius r is inscribed in a tetrahedron. Tangent planes parallel to the faces of the tetrahedron cut off four small tetrahedra from the tetrahedron having inscribed spheres of radii r1, r2, r3, r4. Then r1 + r2 + r3 + r4 2r.
85. If the length of each bisector of a triangle is > 1, then its area is > 1/ √ 3.
86. One may cut out three regular tetrahedra of edge 1 from a unit cube.
87. The circles C1 and C2 with centers O1 and O2 intersect in the points A and B. The ray O1B intersects C2 in F, and the ray O2B intersects C1 in E. The straight line through B and parallel to EF intersects the circles C1 and C2 a second time in M and N, respectively. Prove that MN AE + AF.
88. The points A1, B1, and C1 are chosen on the sides BC, CA, and AB of △ABC, so that AA1, BB1, and CC1 intersect in a point. Let M be the projection of A1 onto B1C1. Prove that MA1 bisects ̸ BMC.
89. The sum of the distances from point M to two neighboring vertices of a square is a.
What is the largest value of the sum of the distances from M to the other vertices of the square?
90. A convex n-gon is triangulated by nonintersecting diagonals such that an odd number of triangles meets at any vertex. Prove that 3|n.
91. Given a regular 2n-gon, prove that you can place arrows on all of its sides and diagonals such that the sum of the resulting vectors is zero.
92. On the sides BC and CD of the square ABCD, we take points M and N with ̸ MAN 45◦. Draw a line perpendicular to MN with a ruler.
340 12. Geometry 93. A rectangle is erected outwardly on every side of an inscribed quadrilateral Q. The second side of each rectangle is equal to the opposite side of Q. Prove that the midpoints of the four rectangles are vertices of a rectangle.
94. Perpendiculars BE and CF are dropped onto AD from the points B and C of a semicircle with diameter AD. The straight lines AB and DC intersect in P, and the segments EC and BF intersect in Q. Prove that P Q ⊥AD.
95. Given a wooden ball, ruler and, compasses, construct the radius of the ball.
96. Three points are given on the surface of a wooden ball. Construct a circle through these points on the surface of the ball.
97. Two points are given on the surface of a wooden ball, which are not antipodes.
Construct a great circle (circle of largest radius) through these two points.
98. 4 points are chosen in a 3 × 4 rectangle. Prove that among them there are two with distance ≤25/8.
99. Let ABCDEF be a convex hexagon such that AB ∥DB, BC ∥EF and CD ∥AF.
Let RA, RC, RE denote the circumradii of triangles FAB, BCD, DEF, respectively, and let P denote the perimeter of the hexagon. Prove that RA + RC + RE ≥P 2 (IMO 1996).
100. Prove that, if one of the diagonals in a cyclic quadrilateral is a diameter of the circumcircle, then the the projections of the opposite sides on the other diagonal are equal.
101. P is an internal point of the tetrahedron ABCD. At least how many edges can be seen at an obtuse angle from P ?
102. Two convex polygons have an even number of vertices, and the midpoints of their edges coincide. Prove that they have equal areas.
103. Two nonoverlapping squares of sides a and b are placed inside a square of side 1.
Prove that a + b ≤1 (HMO 1974; originally due to Erd¨ os).
Solutions 1. Suppose the source of light is in O. We construct a regular tetrahedron ABCD with center O. Consider the four infinite circular cones, each containing strictly the four pyramids OBCD, OACD, OABC, OABD and common vertex O. These cones partly intersect, so that every light ray from O lies inside some cone. Let us inscribe four spheres into the cones so that they do not intersect. This is easy to achieve if the radii of the spheres differ greatly from each other. Obviously every ray from O intersects one of the four spheres. This cannot be achieved with four spheres of equal radius. It can be proved that six spheres of equal radius are needed to shield the light completely. Try to find such a distribution of equal spheres.
2. Yes, it is possible in both cases. For (a) an H-shaped slot will do. For (b) we can use a T-shaped slot. Try to describe how this can be done.
12.3 Classical Euclidean Geometry 341 3. Reflect the triangles A1A2A6, A2A3A4, and A4A5A6 at their bases, and you get a partition of the hexagon into three rhombi. From there it is easy to see that opposite angles are equal. We leave it to the reader to complete this sketch.
4. If O ∈C, the proposition is obvious. Now suppose that O ̸∈C. Reflect C at O to C′. If C ∩C′ ∅, the line C cannot partition the area into two equal parts. Hence C ∩C′ ̸ ∅. Let A be one point of C ∩C′ and B its reflection at O. Since the curve C′ has image C on reflection at O, B ∈C. Hence, AB passes through O.
5. Answer: n must be a multiple of 8. This necessary condition is also sufficient as is shown by the two sums below: (1 −3 −5 + 7) + (9 −11 −13 + 15) + · · · 0, (2 −4 −6 + 8) + (10 −12 −14 + 16) + · · · 0.
6. Answer: n must be a multiple of 12. This necessary condition is also sufficient as is shown by the three sums below: (1 −4 −7 + 10) + · · · + (3k −11 −(3k −8) −(3k −5) + 3k −2) 0, (2 −5 −8 + 11) + · · · + (3k −10 −(3k −7) −(3k −4) + 3k −1) 0, (3 −6 −9 + 12) + · · · + (3k −9 −(3k −6) −(3k −3) + 3k) 0.
7. Suppose a convex hull of N points is an r-gon, 3 ≤r ≤N. There will be (N −r) interior points. To find the number of triangles in a triangulation, we find the sum of the angles of all triangles of the triangulation: 180◦(r −2) + 360◦(N −r). The first term is the sum of the angles of the r-gon. The second term gives the contribution of the interior points. The number of triangles is r −2 + 2(N −r) 2N −r −2, and the number of its sides is 3(2N −r −2) 3(2N −2) −3r. Of these sides, the r sides of the convex hull are counted once and the remaining 3(2N −2) −4r sides are counted twice. Hence the number of segments will be s r + 3N −3 −2r.
Since 3 ≤r ≤N, we get 2N −3 ≤s ≤3N −6 for the number of segments.
8. In Fig. 12.68, A1 to A4 are the areas of the four triangles. We have A1/A4 A2/A3, or A1A3 A2A4. Thus, A1A2A3A4 (A1A3)2.
9. Let S be the centroid of H, and S′ f (S). Then we have S 1 n (A1 + · · · + An) , S′ 1 n A′ 1 + · · · + A′ n .
But {A′ 1, . . . , A′ n} is a permutation of H. Hence, S S′.
J J J J J P P P P P P P """ " A B C D A1 A2 A3 A4 Fig. 12.68 10. We conjecture that P O the center of the pentagons. We want to show in Fig.
12.69 that |P Ai| ≥ |OAi|, with equality iff P O. In a regular pentagon, 342 12. Geometry you should try rotation about its center by 72◦! This paradigm gives us Fig. 12.70, where, from P , we get the points P1, . . . , P5, then Fig. 12.71 where the segments P1Ai go by rotation into A1Pi, that is, |P1Ai| |A1Pi|. Now we interpret the segments A1Pi as vectors − − → A1Pi Then O is the centroid of points Pi, i.e., − − → A1O 1 5 − − → A1Pi.
The triangle inequality gives 5|− − → A1O| |OAi| | − − → A1Pi| ≤|A1Pi| |P1Ai|, that is, |P1Ai| ≥ |OAi|. We have equality iff P O.
B B B B B Q Q Q Q A1 A2 A3 A4 A5 Q Q rO P((((( ( ZZZZ Z Fig. 12.69 B B B B B Q Q Q Q A1 A2 A3 A4 A5 Q Q rO r r P2 r P3 r P4 r P5 P1((((( ( ZZZZ Z Fig. 12.70 B B B B B Q Q Q Q A1 A2 A3 A4 A5 6 r !!!
!r P3 r P5 r P2 rP4 rO P1 Fig. 12.71 11. There is a one-line solution. Take the unit sphere about O, i.e., Ai O. Then |P Ai| |Ai −P | · |Ai| ≥ (Ai −P ) · Ai n −P Ai n.
12. Since AB ∥CD, C and D are reflections at the perpendicular bisector of AB. This follows from the construction of the ellipse with foci A, B and constant sum of distances |AC| + |BC| |AD| + |BD| 2a from the foci.
13. Consider the smallest circle containing all points of S. This is the largest of all circles through triples of points of S and all circles with pairs of points of S as endpoints of diameters. Every reflection at the perpendicular bisector of any two of its points will leave this circle fixed. Thus it passes through the center O of the minimal circle.
Hence all points of S are equidistant from O. For infinite sets, this is no more valid.
A counterexample is the whole plane.
14. The same solution as in the preceding example.
15. In van der Waerden published a detailed and highly instructive account of how he discovered the solution of this problem raised by a chemist. It was an example of the psychology of invention. We give a short solution by G. Boll (Freiburg i. Br.) and H.S.M. Coxeter (Toronto): If the length of the sides a and the angle α are given, then all distances of the five points are given. Thus, the figure is determined up to an isometry. Hence, there exists a direct or opposite isometry S, which permutes the vertices ABCDE cyclically.
The fifth power S5 is the identity. Thus, S is a direct isometry. The centroid of the five points remains fixed. Thus, S is a rotation. Hence, ABCDE lies in a plane perpendicular to the axis of rotation. (Find a more down to earth solution.) Many of the details are considered as well known by specialists and are not men-tioned. To give just one example: every geometer knows that a direct isometry with a fixed point is a rotation about an axis through the fixed point.
12.3 Classical Euclidean Geometry 343 16. Hint: Let ABCD be any convex quadrilateral. Consider the incircles of the triangles ABC and CDA. They touch AC in T1 and T2. Prove that |T1T2| 1 2 |(|AB| + |CD|) −(|BC| + |AD|)| .
17. Fig. 12.72 makes the inequality obvious. There is equality if |PQR| 0, i.e., if the opposite sides have equal length.
18. Let C be the center of the base of the cylindrical barrel. C must be at least at distance 1 from the fence. So C cannot belong to to the strip around the fence of area 372 −352 144. Now take the bottom of any cubic box of side 1. C must be at least at distance 1 from any point of the square, i.e., C cannot belong to the region in Fig. 12.73, consisting of five unit squares and four quarters of a circle of radius 1.
Its area is π + 5. Hence all the 150 boxes and the fence together at most restrict C from belonging to an area A 150(π + 5) + 144 150π + 894. The total area of the yard is F 372 1369. F −A 475 −150π 150(3 1 6 −π) > 0. Thus not all points of the yard are impossible positions for C.
@ @ @ @ @ @ @ ppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp C C C C C C pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp @ @ @ @ A B C D E F I G H Fig. 12.72 1 1 1 1 Fig. 12.73 19. First case: Let max(α, β, γ ) < 120◦. The solution uses the result of 12.4.1, problem 34. Every point inside the equilateral triangle with altitude h has the constant distance sum h from the sides.
Now let T inside △ABC be such that ̸ AT B ̸ BT C ̸ CT A 120◦. We will prove that T has a minimal distance sum from A, B, C. Draw the perpendiculars to AT, BT, CT through A, B, C. You get an equilateral triangle A1B1C1. For every point P , we have |AP | + |BP | + |CP| ≥|AT | + |BT | + |CT | h.
Second case: max(α, β, γ ) ≥120◦. Let γ ≥120◦. In this case C is the point with minimal distance sum from A, B, C, that is, |AP | + |BP| + |CP| ≥|AC| + |BC| for all P ̸ C. We use the following lemma: In an isosceles triangle A1B1C1, let α1 β1 > 60◦. Let the altitude on a leg be h. Then the distance sum of a point P from the sides is > h, if P ̸∈A1B1 and equal to h if P ∈A1B1. Prove this lemma using |A1B1| < |A1C1|.
Draw the perpendiculars to CA, CB and the bisector of γ through A, B, C. We get a triangle A1B1C1 satisfying the conditions of the lemma. The remainder is simple to see.
20. A minimum 3 is shown for no auxiliary point in Fig. 12.74. For one auxiliary point, the minimum 2 √ 2 ≈2.828 is shown in Fig. 12.75. Any other point P has a larger 344 12. Geometry B C A D Fig. 12.74 @ @ @ @ @ B C A D Fig. 12.75 A A A A A A r B C A D E P Q Fig. 12.76 distance sum because of the triangle inequality. For two auxiliary points P, Q, the minimum 1 + √ 3 ≈2.732 is shown in Fig. 12.76. You simply have to join the minimum distances for the triangles ABE and DEC.
21. Consider the reflection M′ of M at the center O of the circle. By the triangle in-equality, we have |MAi| + |M′Ai| ≥2. Thus, n i1 |MAi| + n i1 |M′Ai| ≥2n.
Thus, at least one of the two sums is ≥n. Of any two antipodal points of the unit circle, at least one has the required property.
22. We denote the coordinate differences of the ith side by xi, yi. Then the xi, yi are integers with x2 i + y2 i R, where R is independent of i and x1 + x2 + . . . xn y1 + y2 + · · · + yn 0.
From these equalities, we want to deduce that n is even. We consider x2 i + y2 i mod 4.
(1) If (1) is 0, then all xi, yi are even, and we can cancel a factor 2, getting an equilateral lattice polygon with the same number of sides. So this case can be excluded. We need consider only the cases (a) xi, yi are both odd for all i, (b) one of xi, yi is odd the other even.
In case (a), n must be even. An odd number of odd terms would give an odd sum. In the remaining case (b), for each i we have xi odd and yi even, or vice versa.
x1 + · · · + xn 0 ⇒Pairs (xi, yi) with odd xi are even, y1 + · · · + yn 0 ⇒Pairs (xi, yi) with odd yi are even.
Thus, n is even.
23. D lies on the circumcircle of △ABC. In addition, we require that |AB| + |DC| |BC| + |AD|, or |AD| −|DC| |AB| −|BC|. Thus the problem is reduced to the well-known construction of a triangle from one side, the opposite angle, and the difference of the remaining sides. Let |AB| > |BC|. We mark off the segment AM AB −BC on AD. The △ACD is isosceles with equal angles at the base.
Thus ̸ CMD 1 2 ̸ ABC β/2. Hence ̸ AMC 180◦−β/2. From AC, AM and ̸ AMC, we construct △AMC. D is the intersection of the line AM with the circumcircle of △ABC.
12.3 Classical Euclidean Geometry 345 24. Let a ≤b. Then we have a ≤c, or else the box will not fit into the corridor. We also have a ≤d, or else the box cannot be moved through the door. These necessary conditions are not sufficient. In Fig. 12.77 we move the box so that CD and BC touch the points L and R, and B moves toward R. If A hits the opposite wall of the corridor before B reaches R, then the box gets stuck. If A hits the opposite wall when B coincides with R, then we can just get the box through. Fig. 12.78 shows this critical case. In this figure, the rectangle ABCD and the parallelogram ARLE have the same area, i.e., ab cd. For ab < cd, the box can easily be pushed through the door, that is, the box can be moved through the door iff a ≤c, a ≤d, ab ≤cd. D D D D D D pppppppppppppppppppppppppppppp A B R C L D Fig. 12.77 Fig. 12.78 J J J J E A B=R C L D b a d pppppppppppppppppppppppppppppp 25. We use the well-known formulas 4AR abc, A sr, A2 s(s−a)(s−b)(s−c).
Introducing them into the relationship 2R + r s, after some juggling, we get (−a2 + b2 + c2)(a2 −b2 + c2)(a2 + b2 −c2) 0.
The last equation is valid iff the triangle is a right triangle.
26. AB ∥EC, BC ∥AD, CD ∥BE, DE ∥AC ⇒|ABE| |ABC|, |BCA| |BCD|, |BCD| |CDE|, |CDE| |ADE| ⇒|ABE| |ADE| ⇒AE ∥BD.
27. We prove the more general statement: if the chord AE intersects the radius OC in M and chord DE intersects chord BC in N, then CM/CO CN/CB.
The arcs AC and AD are symmetric with respect to the line AB and thus equal. Hence ̸ AEC ̸ AED. Also ̸ AEC ̸ ABC, and ̸ ABC ̸ OCB since △OCB is isosceles. Hence ̸ AED ̸ OCB, i.e., ̸ MEN ̸ MCN. This means that the points M, N, E and C are concyclic. Hence ̸ MNC ̸ MEC ̸ OBC. Thus △MNC ∼△MEC and CM/CO CN/CB.
28. The small circle lies on the sphere about the center O of the cube with radius √ 2. The large circle lies on the sphere about O with radius √ 3. Thus the minimal distance is ρ ≥ √ 3 − √ 2. Let P and Q be the intersections of AC with the incircle of ABCD.
Then OP (and OQ) lie in the plane of the large circle and intersect this circle in R (and S). Hence, ρ |P R| |QS| √ 3 − √ 2.
29. Yes. The curve C {t, t2, t3} with real t satisfies the condition. The equation of a plane has the form Ax+By+Cz+D 0, where at least one of A, B, C is different from 0. For the intersection, we get the equation At + Bt2 + Ct3 + D 0. This equation has at least one, but not more than three solutions. Thus the intersection of C and any plane is finite but not empty.
30. We first show that a sphere S with two missing points P, Q can be so partitioned. The tangential planes at P and Q intersect in a line g with no common point with S, or they are parallel. All other planes through g or all planes parallel to the parallel planes 346 12. Geometry at P, Q intersect the sphere in a circle, or they do not cut S at all. These circles are pairwise disjoint, and their union is S\{P, Q}. Let c be a circle through O with radius 1.Then all spheres Sr : {P |d(P, Q) r} for 0 < r < 2, except the the two points belonging to c, can be partitioned. In this way we partition M : 6 0<r<2 Sr ∪c, that is the open ball about O of radius 2 plus one point X with d(O, X) 2. If we translate M by all multiples of 4, then all translates are disjoint but cover the line OX. Let T be the union of these translates of M (which consequently are partitioned into disjoint circles). Let be any plane perpendicular to OX. Then \T is a plane with a closed disk or point missing. This can be partitioned into concentric circles about the midpoint of the disk.
31. Yes. Here is one example: Take any straight line a. Through two points A, B ∈a draw two lines b, c, so that b ⊥a, c ⊥a, and b ∦c. Consider the set of all planes parallel to each other and parallel to a. Take any of these planes. The lines b and c intersect it in two points which we join by a straight line. This is done for every one of the parallel planes. We get a wall of skew lines separating space into two half spaces. Now consider all rotations around the axis a. The images of the wall give the required partition of space into skew lines.
Another nonelementary construction consists of the union of all hyperboloids of one sheet with the same focus. See .
32. Let R, S, T, U be the points of contact of the quadrilateral with the sphere. Assign the masses 1/a, 1/b, 1/c, 1/d to A, B, C, D, respectively. Because a(1/a) b(1/b) c(1/c) d(1/d) 1 the centroid of A and B is R, and the centroid of C and D is T . The centroid of all four masses lies on the segment RT . We can find the centroid in another way: A, D have the centroid U and B, C have the centroid S.
Thus the centroid of all four vertices lies on the segment SU. Hence the two segments SU and RT must intersect in the centroid of all four points. Thus R, S, T, U are coplanar.
33. Hint: Easy! The axes of these cylinders are pairwise perpendicular.
34. Pick’s theorem with i b 0 gives f (ABC) 1/2 for the triangle. Heron’s formula gives s(s −a)(s −b)(s −c) 1/4. Simplifying we get that the square of any side is at least equal to the sum of the sqares of the other two sides. Thus for an acute triangle, at least one lattice point must be on the sides or inside.
35. Take the circle of largest radius, and consider a new concentric circle of radius three times larger. Now we remove all circles which are inside the new circle. The remaining circles do not intersect the first circle. Among the remaining circles, we take the maximal circle, and we repeat with it the same procedure. We continue until we get several blown up circles, with union greater than 1. The original circles of radius three times smaller do not intersect, and their common area is larger than 1/9.
36. Cut up the yard into 50 strips of width 2. Fig. 12.79 shows one of the strips S together with its horizontal symmetry line m of length 100. If the center of a barrel is outside S, then the barrel will have no point in common with m. These barrels will leave at most eight pieces of m uncovered. Each piece has length at most 10, because there cannot be a segment of length 10 having no point in common with any barrel. Under each barrel, there lies a piece of length at most 2. But 8 · 10 + 7 · 2 < 100. Thus at least eight barrels have their centers inside S. This holds for each of the 50 strips.
Hence there are at least 8 · 50 or 400 barrels inside the yard.
12.3 Classical Euclidean Geometry 347 37. SupposebothverticesAandB havethepropertymentioned.Then ̸ CAB+̸ DAB > 180◦and ̸ CBA+̸ DBA > 180◦whereasthesumofallsixanglesofthethetriangles CAB and DAB is altogether merely 180◦+ 180◦. Contradiction.
38. Suppose the polyhedron has more than eight vertices. Consider nine of its vertices.
At least five have the first coordinate of the same parity, of these five at least three agree also in the parity of the second coordinate, and of these three at least two agree in the parity of the third coordinate. But then the midpoint of the segment connecting these two points has integral coordinates. Because of the convexity of the polyhedron, the midpoint belongs to it, a contradiction.
39. Two of the three angles 120◦must be adjacent, or else the three angles would occupy the whole circle. Thus there are two neighboring angles ABC and BCD of 120◦.
Then |AC| |BD| and △ABC ∼ △BCD. Hence |AB| |CD|.
40. He should walk along a circle of area A. From A πr2, we get r √A/π for its radius, and 2πr 2π√A/π 2 √ πA miles for the length of the path.
41. He should walk on a semicircle of length √ 2πA. This semicircle does not fit into any convex figure of area A. Suppose it does. Since the woods are convex, if two points are in it, the whole segment joining them is also in the woods. Hence, the whole semicircular disk lies inside A. But this disk has radius R 1 π √ 2πA √2A/π and the area πR2/2 or A. This would mean that one figure of area A is contained inside another area A, which is a contradiction. Thus, the semicircle either touches the edge of the woods or leaves it altogether.
42. The man will show him the shortest way R out. Hence, the circle of area A will lie completely in the woods. From A πR2, we get R √A/π.
43. The man is at O. Draw a circle with center O and radius 1. The edge of the woods is a tangent to this circle. We are looking for the shortest curve which starts at O and has a common point with every tangent of the circle. Most people who tackle this problem successfully pass through the following stages.
First stage: Walk in a straight line for one mile in any direction to a point A. Then walk along the circumference of the circle in Fig. 12.80. You will walk at most 1 + 2π ≈7.28 miles to reach the edge of the woods.
Second stage: Do you really need to go all the way around the circle? Fig. 12.81 shows that this is not necessary. The path OABC also has a common point with every tangent of the circle. So it also leads out of the woods, and its length is merely 3π/2 + 2 ≈6.71 miles.
2 m S 100 Fig. 12.79 O 1 A Fig. 12.80 O A B C Fig. 12.81 O A B 1 C D 1 Fig. 12.82 Third stage: In Fig. 12.81 we made some savings at the end of the path. Let us look for similar savings at the point A. The path OABCD in Fig. 12.82 also has a 348 12. Geometry common point with every tangent of the circle. Hence, it will lead out of the woods in at most 2 + √ 2 + π ≈6.556 miles.
Fourth stage: For the next step, you need some trigonometry and calculus. The path OABCD in Fig. 12.83 has the length p(α, β) |OA| + |AB| + arcBC + |CD|.
But |OA| 1/ cos α, |AB| tan α, arcBC 2π −2α −2β, |CD| tan β, α and β being measured in radians. Thus, p(α, β) 2π + 1 cos α + tan α −2α + (tan β −2β), or p(α, β) 2π + f (α) + g(β). To minimize p(α, β) we must minimize f (α) and g(β) separately. But f ′(α) (2 sin α −1)(1 + sin α) cos2 α , g′(β) tan2 β −1 (tan β −1)(tan β + 1).
Since α and β are both acute angles, f ′(α) 0, g′(β) 0, and the unique solutions are α π 6 and β π 4 .
At these points, the signs of f ′(α) and g′(β) are changing from negative to positive.
Thus, we have minima at these values of the angles. The minimal path has length p π 6 , π 4 1 + √ 3 + 7 6π ≈6.397.
It can be shown that there is no shorter path leading out of the woods.
44. (a) One can walk along a circle of diameter 1 and get out of the woods in π miles.
(b) One can walk any segment of length √ 2, then turn by 90◦and walk another segment of length √ 2. Altogether we need 2 √ 2 ≈2.82 miles.
(c) We can walk in a straight line for 2/ √ 3 miles, then turn by 120◦and walk again 2/ √ 3. We definitely get out of the woods by walking not more than the distance 4/ √ 3 ≈2.31 miles.
(d) The last is only slightly above the ideal ≈2.278 which is very difficult to find.
It consists of a curve ABCDC′D′E where BC and D′C′ are circular arcs, AB is a tangent of BC, ED′ a tangent to D′C′, and DC and DC′ are tangents to both arcs.
This is the shortest curve which does not completely lie inside a 1 mile wide strip.
45. By a transformation of the plane, we mean a bijection of the plane onto itself. Let f be any transformation of the plane and X be any point of the plane, and let f (X) X′.
We must prove two facts: (a) Let A′, B′, C′ be three collinear points. Then their inverse images A, B, C are also collinear.
(b) Let A, B, C be three collinear points. Then A′, B′, C′ are also collinear.
The proof of (a) is trivial. Suppose A, B, C are not collinear. Then they lie on a circle. Their images must also lie on a circle and are not collinear. Contradiction.
NowletA, B, C bethreepointsonalineg.Considerthecirclesc1, c2 withdiameters AB and AC. Their images c′ 1, c′ 2 are also circles, which touch in A′. A′B′ is not a tangent of c′ 1. Since A′ ∈c′ 2, A′B′ is not a tangent of c′ 2. Thus A′B′ has another common point with c′ 2. Its inverse image must lie on c2 and, because of (a), on the line AB, that is, it must be C. Hence, C′ lies on the line A′B′.
12.3 Classical Euclidean Geometry 349 bbb T T T T b b b 1 O B A D C α α β β Fig. 12.83 ppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp A B C D E F P Q R Fig. 12.84 46. Suppose the quadrilateral ABCD is already constructed. Consider the rotational homothety with center A, angle α, and factor d/a. It maps B to D. Let C′ be the image of C. Then ̸ CDC′ β + δ 180◦, |DC′| bd/a. We construct the points C, D, C′ on one line from |C′D| bd/a and |DC| c. Locus for A is the circle with center D and radius a. In addition we know |AC′AC| d : a. So A lies on the so called circle of Appolonius which has distance ratio d : a from C′ and C. To get the endpoints P and Q of its diameter on CC′, we divide this segment internally and externally in the ratio d : a. The circle with diameter PQ is the second locus for A. The circles about A and C with radii a and b complete the construction.
47. The lines A1A2, B1B2, C1C2 are bisectors of the angles of △A1B1C1.
48. Transforming sin2 α + sin2 β sin(α + β) slightly, we get sin α(sin α −cos β) sin β(cos α −sin β). If sin α > cos β and cos α > sin β, then sin2 α + cos2 α > sin2 β + cos2 β, or 1 > 1, a contradiction. For the same reason, sin α < cos β, cos α < sin β is impossible. Thus sin α cos β, which implies α+β π/2.
49. Rotation by 60◦around C takes △CAD into △CBE, and rotation by 60◦around H takes △HBE into △HDK.
50. LettheoppositesidesoftheskewquadrilateralABCD becongruent.Then△ABC ∼ △ACD and △ABD ∼ △BCD. Let P and Q be the midpoints of AC and BD. Now |PD| |P B| ⇒P Q ⊥BD, |P A| |P C| ⇒P Q ⊥AC. Conversely, for PQ ⊥AC, P Q ⊥BD, we conclude that a half turn about PQ switches A with C and B with D. Thus, opposite sides are congruent.
51. We have |AA1| > (|AB| + |AC|)/2. Indeed, according to the theorem of Ptolemy |AA1| · |BC| |AB| · |CA1| + |AC| · |BA1|.
Since ̸ BAA1 ̸ CAA1 α/2 implies |A1B| |A1C| t and 2|AA1| 2|AB|t + |AC|t |BC| (|AB| + |AC|) · 2t |BC| > |AB| + |AC| since 2t |A1B| + |A1C| > |BC|. Similarly, we prove |BB1| > (|BA| + |BC|)/2, |CC1| > (|CA| + |CB|)/2. Addition of the three inequalities implies |AA1| + |BB1| + |CC1| > |AB| + |BC| + |CA|.
52. If the angles are each 120◦, then the triangle P QR in Fig. 12.84 is equilateral, that is, the differences of opposite sides are equal.
53. Because of the right angles at M and N, the circle Z with diameter AP passes through M and N. Since M ∈AB and N ∈AC, the subtended angle MAN is always the 350 12. Geometry same. With P Z also changes but ̸ MAN always remains the same. Hence |MN| is maximal if diameter AP is maximal, i.e., if P and A are endpoints of a diameter.
For this point P , the points M, N coincide with B and C. The maximum |MN| coincides with the length |BC| of the third side of △ABC.
54. A beginner’s solution: Erect perpendiculars at A and B on AB. They intersect the circle again in C′ and C′′. Consider △ABC′. Since |AC′| 2r and |AB|+|BC′| > |AC′| 2r,theperimeterof△ABC′ is> 4r.Nowwemustshowthat|AC|+|BC| > |AC′| + |BC′|. This relies on the following theorem: Of all triangles with the same base which are inscribed in a given circle, the one with greater altitude has the greater perimeter.
Because α + β 180◦−γ , the Sine Law a 2r sin α implies a + b 2r(sin α + sin β) 4r sin α + β 2 · cos α −β 2 4r cos γ 2 · cos |α −β| 2 .
This function is a monotonically decreasing function of |α −β|. The less this dif-ference, the larger is the value of the sum a + b. From this result, we easily get the theorem above.
Use Jordan’s inequality 0 < x < π/2 ⇒sin x > 2x/π. It says that the concave arc of the sine lies above its chord from (0, 0) to (π/2, 1). Now we have a one-line proof: a + b + c 2r(sin α + sin β + sin γ ) > 4r α + β + γ π 4r.
55. Draw P P1 ∥A1A2 (P1 ∈AnA1), then P P2 ∥A2A3 (P2 ∈A1A2), and so on. Prove that P1P2 · · · Pn has the required property.
56. Take a circle of sufficiently large radius and place the longest side into this circle as a chord. Then place all the other chords in any order. You get an open chain of chords.
Then start decreasing the radius. If the diameter of the circle becomes equal to the largest circle when the chain closes, then increase the circle again, but the midpoint of the circle should be on the other side of the longest chord from the remainder of the chain. This time the chain closes if the size of the circle is reduced sufficiently.
57. The three bisecting planes of a solid angle of a tetrahedron intersect in a line which is the locus of points equidistant from the faces of that solid angle. Take any other of the three other bisecting planes. Suppose it intersects this line in O. Point O is equidistant from all four faces of the tetrahedron. It is the center of the inscribed sphere. The two remaining bisecting planes are the sets of points equidistant from pairs of faces. They must also pass through O.
58. Use the fact that any point of the bisector of a solid angle is equidistant from its faces.
59. For n 3, all polygons are plane. For n 4, bend a rhombus about its shorter diagonal, until all angles become equal to α < 90◦. For even n > 4, start with a regular plane n-gon, and lift every second vertex upward by the same amount. The construction for α 90◦is especially easy. Start with a strip of congruent squares.
Then bend them at right angles to each other to get a “staircase”. There are regular space polygons for all odd n ≥7. Such polygons with all its angles α 90◦can be constructed from the plane polygon in Fig. 12.85 by bending the pentagon with three right angles so that the angles at vertices 3 and 6 become 90◦. The remaining squares are bent up and down by 90◦. See Fig. 12.86.
12.3 Classical Euclidean Geometry 351 Fig. 12.85 @ @ 1 2 3 4 5 6 7 Fig. 12.86 60. Take a section parallel to an edge e intersecting all edges which end in e. Since at least two other edges have ends at each endpoint of e, the section has at least four vertices.
61. Construct planes perpendicular to AB through A and B. Draw a plane perpendicular to AB through each other vertex of the polyhedron. Consider two neighboring planes.
Between them there are at least three segments of edges. Each segment is at least as long as its projection on AB. In addition there are segments not parallel to AB.
Thus, the sum of all the edges is greater than 3d.
Expressed more briefly, the orthogonal projection of the carcass of the polyhedron on AB covers the segment AB at least three times.
62. Easy.
63. Reflect the spherical triangle ABC in O to A′B′C′. Then D′ must lie inside A′B′C′.
64. Let k be the number of acute angles in an n-gon. We can express the sum of its angles in two ways. First, it is k · 90◦+ (n −k) · 360◦, and secondly, it is also (n −2) · 180◦.
Thus, k · 90◦+ (n −k) · 360◦> (n −2) · 180◦, i.e., 3k > 2n + 4. Consequently, k ≤[2n/3] + 1. Fig. 12.87 shows examples of n-gons with [2n/3] + 1 acute angles for n 3r, n 3r + 1, n 3r + 2.
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!C C C n 3r + 2 Fig. 12.87 65. Supposesphere(orplane)s1 containsthefirstandsecondcircleandspheres2 contains thesecondandthird.Supposes1 ands2 arenotthesame.Thentheirlineofintersection is the second circle. In addition, the common point of the first and third circle also belongs to the intersection line of s1 and s2, i.e., to the second circle, and thus the three circles have a common point. This is a contradiction.
66. If every vertex of a polyhedron is joined by edges to every other vertex, then all faces are triangular. We consider two faces ABC and ABD with the common edge AB. Suppose the polyhedron is not a tetrahedron. Then it has a vertex E, which is different from A, B, C, D. Since C and D lie on different sides of the plane ABE, triangle ABE is not a face of the given polyhedron. If we make cuts along AB, BE and EA, then the surface of the polyhedron will be separated into two parts, with C 352 12. Geometry and D lying in different parts. For a nonconvex polyhedron, this would be incorrect.
Thus, C and D cannot be joined by an edge, or else the cut would separate that edge. But the edges of a convex polyhedron cannot intersect in interior points. (The convexity is important. Akos Csasar has constructed a nonconvex polyhedron with 7 vertices, which are joined pairwise by edges.) 67. Suppose the polyhedron has only triangular faces, altogether f triangles. Then the number of edges is 3f/2. This number is divisible by 3. On the other hand, if there is a face with more then three edges, then the number of edges is at least eight.
68. We construct spheres with the circles as equators. The common chords are the pro-jections of intersecting circles of the spheres. We must show that the three spheres have a common point above the plane. Consider the circle, which is the intersection of two spheres. One diameter of this intersection lying in the plane lies outside the third sphere, the other inside. Thus this circle intersects the third sphere. Thus, the three spheres have a common point above the plane.
69. Consider the plane which is parallel to two skew edges of the tetrahedron. We will prove that there are two such planes which are perpendicular to . Projection of the tetrahedron on such a plane is a trapezoid or triangle with constant altitude, which equals the distance between the two skew edges of the tetrahedron. The median of the trapezoid is the projection of the parallelogram with vertices in the midpoints of the four other edges of the tetrahedron. Thus we must prove that, for any parallelogram, we can find two straight lines in the same plane so that the ratio of projections of the parallelogram onto them is ≥ √ 2. Let a and b be the sides of the parallelogram, a ≤b, and d its longest diagonal. The length of the projection of the parallelogram onto a line ⊥b is ≤a. The projection onto a line parallel to d is equal to d. Thus d2 > a2 + b2 ≥2a2.
70. Answer: 29. Of the eight vertices, we can choose 4 in 8 4 70 ways. Of these, 12 are coplanar. We are left with 58 noncoplanar quadruples. But these come in 29 complementary pairs. Each quadruple of the pair determines the same box. So there are 29 boxes left. Try to find some more geometric solution (see Chapter 5, problem 12).
71. (a) Take any point P inside △ABC, draw the straight line CP , and drop perpendicu-lars AA1 and BB1 onto CP from A and B (Fig. 12.88). Then 2 (|AP C| + |PBC|) (|AA1| + |BB1|) · z au + bv. But |AA1| + |BB1| ≤|AB| c. Thus, cz ≥au + bv, and similarly, ax ≥bv + cw, by ≥au + cw.
(1) Adding the three inequalities, we get ax + by + cz ≥2(au + bv + cw) 4S.
c c c c c c c c H H H H H H A A A A A A A A A B C P A1 B1 a b x y z w v u Fig. 12.88 12.3 Classical Euclidean Geometry 353 (b) First we show that we can interchange u and v in the first inequality (1). Indeed, reflect P at the bisector of γ to P ′. Then |CP | |CP ′| z, and the distances from P ′ to BC and AC are y and x, respectively. Applying the above inequality to P ′, we get cz ≥av + bu, and similarly, ax ≥bw + cv, by ≥aw + cu.
(2) Solving the inequalities (2) for x, y, z and adding, we get x + y + z ≥ b c + c b ≥2 u + c a + a c ≥2 v + b a + a b ≥2 w ≥2(u + v + w).
This is the famous Erd˝ os–Mordell inequality, first posed by Erd˝ os in 1935 in the American Mathematical Monthly and solved by Mordell in 1937. There is equality for the equilateral triangle.
(c) From the inequalities (1) we get xu b/auv + c/awu and similarly, yv a/buv + c/bwu, zw ≥a/cuw + b/cvw. Adding, we get xu+yv+zw ≥ b a + a b uv+ b c + c b vw+ a c + c a wu ≥2(uv+vw+wu).
72. Given a quadrilateral ABCD with a + c b + d and area A √ abcd, we want to prove that β + δ π. We can express the square of AC in two ways: a2 + b2 −2ab cos β c2 + d2 −2cd cos δ.
(1) From a + c b + d, we get (a −b)2 (c −d)2, or a2 + b2 −2ab c2 + d2 −2cd.
(2) Subtracting (2) from (1) and dividing by 2, we get ab(1 −cos β) cd(1 −cos δ).
(3) The area of ABCD can be expressed in two ways and equated: ab 2 sin β + cd 2 sin δ √ abcd.
Multiplying by two and squaring, we get 4abcd a2b2(1 −cos2 β) + c2d2(1 −cos2 δ) + 2abcd sin β sin δ.
Using (3), we get 4abcd ab(1 + cos β)cd(1 −cos δ) + cd(1 + cos δ)ab(1 −cos β) + 2abcd sin β sin δ.
Dividing by abcd, expanding and collecting terms, we get cos(β + δ) −1 ⇒β + δ π.
This is a Putnam Competition problem. The other two problems are left to the reader.
354 12. Geometry ppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp O A M d 2 d Z d φ Fig. 12.89 p H H H A B O a φ1 Fig. 12.90 73. Suppose a ≤b ≤c k. From aha bhb chc 2A, we get a + 2A/a b + 2A/b c + 2A/c k. If we introduce the function f (x) x + 2A/x, then f (a) f (b) f (c) k. Now f (x) k is a quadratic equation in x, and f (a) f (b) f (c) k. Since a quadratic equation has at most two solutions, at least two of the solutions must coincide. Suppose a b. Let a ̸ c. Then in x2 −kx + 2A 0 we have ac 2A, that is a b 2A/c hc, which is impossible. Thus, a b c.
74. Suppose the grasshopper is at C ∈a after two jumps. Reflect the path first at OB to BC′ then at OC′, etc. Then the points A, B, C′, D′, E′, . . . fall onto a circle and measure off equal arcs belonging to the chord of length 1. Hence the sequence of points on the circle closes if α is a rational multiple of π, that is, for α p/q · π, where p, q are positive integers.
75. Yes! For this it is necessary (and sufficient) to place the stations in the vertices of an inscribed cube. Indeed, those points of altitude d are visible from A in Fig. 12.89, which lie on the spherical cap bounded by the circle of radius AM about point Z vertically above A. Denote ̸ AOM φ. For φ, we get cos φ |OA| |OM| 1 3.
On the other hand, for the angular distance φ1 between neighboring vertices of an inscribed cube, cos φ1 1 3. Indeed, since the space diagonal d of a cube with edge a is a √ 3, from the Cosine Rule, we get (Fig. 12.90) |AB|2 |OA|2 + |OB|2 −2|OA||OB| cos φ1 ⇒cos φ1 1 3.
Hence, the sphere is covered by eight such spherical caps with angular radius φ and midpoints in the vertices of an inscribed cube. Every point of the sphere is covered at least by two caps.
76. Draw parallels to BC, DE, and FA through A, C, and E. Fig. 12.91 yields |ACE| |ABCDEF| −|P QR| 2 + |P QR| ABCDEF| + |PQR| 2 .
If we consider a similar construction for △BDF, instead of △PQR we get another triangle ST U. But △P QR ∼ △ST U since their sides are differences of opposite sides, e.g., |P Q| |AB −DE|, |QR| |AF −CD|, and |PR| |EF −BC|.
12.3 Classical Euclidean Geometry 355 HH H H H H H H HHHH H A A A A A , , , , ppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp F A B C D E P Q R Fig. 12.91 77. Let M be the midpoint of the circle, |AC| b, |BC| a, |AM| r, |AB| c.
Since arc ACB is one third of the circle, we have ̸ ACB ̸ AMB 120◦, c2 a2 + b2 + ab, c2 r2 + r2 + r2, that is a2 + b2 + ab 3r2, or r a2 + b2 + ab 3 .
78. (a) The boat starts from B at full speed when the searchlight passes the position BM.
When the searchlight has made a full turn and a quarter turn reaching position CM, the light beam has traveled the distance 2π + π/2 5π/2 miles on the unit circle.
At the same time, the boat has covered 1/8 of that distance, or just 0.98 miles, which is less that 1. The boat will be somewhere inside the lens in Fig. 12.92. During its 1 1/4 full turns, the searchlight has traversed the whole of this shaded area, and so, at some time, has illuminated the boat.
(b) Suppose k s/b. Consider the circle in Fig. 12.93 with radius 1/k about M. The boat can outpace the searchlight inside this circle. If the boat can travel the distance |BA| before the searchlight makes a full turn, it can fulfill its mission: 1 −1 k b < 2π s , or 2π 1 −1 k , or k < 2π + 1.
(c) The boat in Fig. 12.94 sails from B to C. Let us find the critical value of k such that the searchlight makes a full turn and the arc BD, when the boat covers the distance BC, is 2π + α tan α k, or 2π + α tan α, α in radians.
This equation must be solved by iteration giving α 1.442066530 radians and 1/ cos α k s/b 7.789705781. Thus for k < 7.789705781, the boat can fulfill its mission.
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M C B Fig. 12.92 M A B 1 −1 k 1 k Fig. 12.93 P P P P P M C D B α 1 −1 k 1 k Fig. 12.94 356 12. Geometry 79. The volumes of pyramids with the same base are proportional to their altitudes. From |ABCD| |OABC| + |OBCD| + |OCDA| + |ODAB|, we get |OA1| |AA1 | + |OB1| |BB1| + |OC1| |CC1| + |OD1| |DD1| 1 ⇒|AA1| −R |AA1| + |BB1| −R |BB1| + |CC1| −R |CC1| + |DD1| −R |DD1| 1 ⇒ 1 |AA1| + 1 |BB1| + 1 |CC1| + 1 |DD1| 3 R .
The AM–HM inequality yields (|AA1| + |BB1| + |CC1| + |DD1|) 1 |AA1| + 1 |BB1| + 1 |CC1| + 1 |DD1| ≥ 42 ⇒|AA1| + |BB1| + |CC1| + |DD1| ≥16 3 R.
80. No solution since we gave you enough hints.
81. What is here the side condition? Obviously Sixi 3V . Multiplying the function to be minimized by the constant 3V , we get Si xi Sixi S2 i + i<k SiSk xi xk + xk xi ≥ S2 i + 2 i 1 be the bisector of the angle α. Of all the lines through D we choose the one cutting from the angle α the triangle ABC of minimal area. This is an isosceles triangle, and its area is greater than 1/ √ 3, as can be seen from Fig. 12.95.
S S S S S S B A C D M N ≥30◦≥30◦ Fig. 12.95 86. Take three skew edges of the cube. Each of them will be an edge of one tetrahedron.
The midpoints of the opposite edges of each tetrahedron coincide with the center of the cube. Prove that these three tetrahedra do not have additional common points.
87. We observe that △O1BE ∼△O2BF. Hence, E, F, O1, O2 lie on a circle C. Since ̸ O1AO2 + ̸ O1EB ̸ O1BO2 + ̸ O1BE 180◦, the point A lies on the same circle. (Make a drawing.) ̸ FEB ̸ BEA since they are inscribed into C and on equal arcs O2F and O2A. EF ∥MN implies ̸ MBE ̸ FEB. Hence, ̸ MBE ̸ BEA, i.e., the trapezoid MEBA is equilateral, and AE MB. Similarly, we prove that ABFN is an equilateral trapezoid implying AF BN. Adding the last two equalities, we get AE + AF MB + BN MN.
358 12. Geometry @ @ @ @ @ @ @ C C C C C C C a a a a a a a a a @ @ @ @ pppppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp A B C B1 A1 C1 M Fig. 12.96 @ @ A A A A & & & & && D D D D D D c c c J J P P # # # b b b b b w w Fig. 12.97 @ @ @ @ @ @ @ S S S S S HHHH H B B B B B B A B C D N K L H M Fig. 12.98 88. Let D, P , and K be the projections of A, B and C onto the line B1C1 in Fig. 12.96.
Then BP AD BC1 C1A, AD CK AB1 B1C ⇒BP CK BC1 C1A · AB1 B1C BA1 A1C .
In the last equation, we have used Ceva’s theorem. Since BA1/A1C P M/MK, the triangles P MB and KMC are similar: ̸ P MB ̸ KMC.
89. Let MA + MB a, where A, B are neighboring vertices of the square ABCD.
Here M and C, D are separated by AB. We use the inequality of Ptolemy for the quadrilateral AMBC: MC ·AB ≤MA·BC +MB ·AC, or MC ≤MA+ √ 2MB.
Similarly MD ≤MB +MA √ 2. Adding the two inequalities, we get MC +MD ≤ (MA + MB)( √ 2 + 1) a( √ 2 + 1). We have equality if M lies on the circumcircle of the square ABCD.
90. Color the triangulation properly by two colors, black and white as follows: Draw the diagonals one by one. At each step, keep the coloring on one side of the last diagonal drawn. On the other side, switch the colors black and white. Since the number of triangles at each vertex is odd, the sides of the polygon belong to triangles of the same color, say black. The number w of sides of all white triangles is a multiple of 3.
Since each of the w sides is also a side of a black triangle, for the number b of sides of all black triangles, b n + w. Now 3|n + w and 3|w. Hence 3|n (Fig. 12.97).
91. The main diagonals pass through the center of the n-gon. The other diagonals come in pairs which are symmetric with respect to the center. If we orient them oppositely, we get vectors with sum − → O . Now we must place arrows on the sides and main diagonals.
Suppose n 2k + 1. We place arrows on the sides cyclically with sum − → O . Place the arrows into the vertices number 1, 3, . . . , k −1. Then there is one arrow on each diagonal. The system of these vectors is invariant with respect to rotation about the center by the angle 2π/(2k + 1). Hence, such a rotation takes the sum into itself.
Hence, it is ⃗ O.
Now suppose n 2k. Consider cycles consisting of neighboring main diagonals and sides connecting them. In each cycle, we place arrows so that the sum is ⃗ O. We are left with every second side. We orient them cyclically and get the sum ⃗ O, since rotation by the angle π/k about the center leaves the sum invariant.
92. Construct the diagonal BD, point K AN ∩BD, and point L BD ∩AM. See Fig. 12.98. Since ̸ LAN ̸ NDL 45◦, the quadrilateral ADNL is inscribed in a circle, and ̸ ALN 90◦, i.e., AL ⊥LN. Similarly ABMK is inscribed, since ̸ KBM ̸ LAK. Hence, MK ⊥AN. Thus, MK and NL are altitudes of 12.3 Classical Euclidean Geometry 359 aaaa a aaaa a L L L L L ppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp !
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Q R B C D A S T U V P W O1 O2 O3 O4 Fig. 12.99 A A A A A A A \ \ \ \ \ l l l l l l K L M N Q P A D B C E F Fig. 12.100 triangle AMN intersecting in the orthocenter H. Hence, AH is the third altitude perpendicular to MN.
93. In Fig. 12.99, ̸ O4AO1 ̸ O4AP + ̸ P AQ + ̸ QAO1 ̸ O2CB + ̸ BCD + ̸ DCO3 ̸ O2CO3 ⇒△O4AO1 ∼ △O2CO3 ⇒O4O1 O2O3, andsimilarlyO1O2 O3O4.Thus,O1O2O3O4 isaparallelogramandhascongruent opposite angles: ̸ O4O1O2 ̸ O2O3O4. Now ̸ O4O1A + ̸ CO3O2 + 2̸ AO1B −̸ BO1O2 −̸ DO3O4 ̸ CO3D + ̸ AO1B, ̸ O4O1A ̸ CO3O2, ̸ BO1O2 ̸ DO3O4, ̸ CO3D ̸ AO1B.
This implies ̸ O4O3O2 90◦. In our case, we even have a square since O4O2 is a symmetry line.
94. We have ̸ ABD ̸ ACD 90◦. We draw perpendiculars KL and MN through P and Q to BE and CF, respectively (Fig. 12.100). We have MN KL, △BP K ∼ △BAE ∼△DAB. Hence, ̸ BDA ̸ KBP , i.e., △BP K ∼△DBE. From △BP K ∼△DBE, △AP C ∼△DPB (̸ BAC ̸ CDB), and △CPL ∼△ACF, we conclude that KP BE BP BD P C AC P L CF ⇒KP BE P L CF ⇒KP PL BE CF .
From the last equality and BQE ∼FQC, we conclude that KP P L BE CF MQ QN ⇒KP P L MQ QN ⇒KP + P L P L MQ + QN QN ⇒P L QN ⇒P Q ⊥AD.
Here we have used the fact KL MN KP + P L MQ + QN.
95. Choose any point A on the surface of the ball, and draw a circle about A with any radius. Choose three points M, N, P on the circle. In the plane, construct 360 12. Geometry △M′N ′P ′ ∼ △MNP , and find its circumcircle with center O′. Then M′O′ is the radius of the circumcircle. From the leg M′O′ and hypotenuse A′M′, which is AM, we construct a right triangle M′O′A′. We construct the perpendicular to M′A′ which intersects the line A′O′ in B′. Then A′B′ is equal to the diameter AB.
96. In the plane we construct △A′B′C′ ∼ △ABC and find its circumcircle. Then we find the radius R of the ball as in the preceding problem. After drawing a circle with radius R, we draw a chord K′L′ into it which is equal to the diameter of the circumcircle. The distance from K′ to the midpoint P ′ of the arc K′L′ is the radius of the circle on the ball through A, B, C. We get the midpoint of the circle by drawing about A and B circles with distance K′P ′. They intersect in the center of the circle through A, B, C.
97. Draw circles on the ball about A and B with the same radius which intersect in K, L. Draw circles about K and L with the same radius which intersect in M, N.
Then M, N lie together with A and B on the great circle. From A, B, M, we can construct the circle.
98. Hint: It is easy to see that the minimum distance between the four points is maximal, when they are vertices of a rhombus with side 25/8, two opposite vertices of which are vertices of the rectangle, and the remaining two lie on the long sides of the rectangle.
99. This is the most difficult problem ever proposed at the IMO. Before 1996, the most difficult problem was E15 in Chapter 6. Although the jury correctly judged the extreme difficulty of E15, it estimated the difficulty of this problem as medium. We give no proof, but if you are interested you can find the solution in many sources.
13 Games We begin by describing so-called Nim Games in some detail. Most of the games in competitions are of this type, but some do not fit into any category known to the contestant. Still, most of the following definitions are useful even in those situations.
We consider games for two players A and B, who move alternately. A always moves first but otherwise the rules are the same for A and B. A draw cannot occur.
We are given the starting state and the set M of legal moves. A player loses if he finds himself in a position from which no legal move can be made. We can think of each position as a vertex of a graph and each move as a directed edge. We consider games with finitely many vertices and no directed circuit (a position can not repeat). This ensures that one of the players will lose.
The set P of all positions can be partitioned into the set L of losing and the set W of winning positions: P L ∪W, L ∩W ∅. A player finding himself in a position in L will lose provided his opponent plays correctly. A player finding herself in a position in W can force a win whatever her opponent does.
To win, a player must always move so as to force his opponent into a position belonging to L. From each position in L, every move must result in a position in W. From every position in W, a move to a position in L must be possible. L must contain at least one final position f from which there is no move out. The player who leaves his opponent facing such a position has won the game. The problem is to identify the set L of losing positions.
Most of the following problems can be solved by a simple strategy: Divide the set of all positions into pairs, so that there is a move from the first to the second element of the pair. Whenever my opponent occupies one 362 13. Games element of a pair, I move to the other element of the pair. Thus, I win, since my opponent runs out of moves first.
Initially, if there is one position without a pair, I should occupy it. Otherwise, I should be the second player to win. In more complicated games, a table of losing positions should be used in playing.
As a warmup, we will consider some examples with solutions.
1. Bachet’s Game. Initially there are n checkers on the table. The set of legal moves is the set M {1, 2, 3, . . . , k}. The winner is the one to take the last checker. Find the losing positions.
The set L consists of all multiples of k + 1. Indeed, if n is not a multiple of k + 1, then I can always move to a multiple of k + 1. My opponent cannot move to the next multiple of k + 1 since he can only subtract k or less checkers. So he has to move to some number, which is not a multiple of k + 1. Then I simply move into L. Thus, I will finally reach 0, which is also a multiple of k + 1.
2. In problem #1, let M {1, 2, 4, 8, · · ·} (any power of 2). Find the set L.
L consists of all multiples of 3. Indeed, a player confronted with a multiple of 3 cannot move to another multiple of 3, since 2m is never a multiple of 3. But from a nonmultiple of 3, I can always move to a multiple of 3, by subtracting 1 or 2 mod 3.
3. In problem #1, let M {1, 2, 3, 5, 7, 11, . . .} (1 and primes). Find L.
L consists of all multiples of 4. From a nonmultiple of 4, I can always move to a multiple of 4 by subtracting 1, 2 or 3 mod 4. But from a multiple of 4, I cannot move to another multiple of 4.
4. Find the set of losing positions for M {1, 3, 8}.
Translate the game into a board game by starting with a row of empty cells.
Then place a chip on the nth cell. Now A and B alternately move the chip to the left by 1 or 3, or 8 places. Start at the end and work upward by finding the losing positions until you detect a periodicity. You will find that L consists of all nonnegative integers of the form 11n, 11n + 2, 11n + 4, 11n + 6.
Problems 1. Wythoff’s Game: There are two piles of checkers on a table. A takes any number of checkers from one pile or the same number of checkers from each pile. Then B does the same. The winner is the one to take the last chip. Positions are pairs [x(i), y(i)] of nonnegative integers. By starting with small numbers, try to find the losing positions until you see a recursive rule. Also try to find a “closed” expression for the positions in L.
13. Games 363 2. There are initially 107 chips on a table. The set of moves consists of pn, where p is any prime and n can be any nonnegative integer. The winner is the one to take the last chip. Find L.
3. Start with n 2. Two players A and B move alternately by adding a proper divisor of n to the current n. The goal is a number ≥1990. Who wins?
4. A modification of Wythoff’s game. You may remove any number from a single pile, or numbers from both piles differing in absolute value by less than 2. Find some pairs belonging to L by trial and error. Can you find formulas for the pairs in L?
5. A and B alternately put white and black knights on the squares of a chessboard, which are unoccupied. In addition a knight may not be placed on a square threatened by an enemy knight (of the other color). The loser is the one who cannot move any more. Who wins?
6. A and B place white and black bishops on squares of a chessboard, which are free and not threatened by an enemy bishop. The loser is the one who cannot move any more. (The one who moves may place his bishop on squares of both colors.) 7. A and B alternately draw diagonals of a regular 1988-gon. They may connect two vertices if the diagonal does not intersect an earlier one. The loser is the one who cannot move. Who wins?
8. Given a triangular cake P QR of area 1, A chooses a point X of the plane. B makes a straight cut through X. What maximal area can B cut off?
9. Given a triangle P QR of area 1, A chooses a point X ∈PQ. Then B chooses a point Y ∈QR. Then again A chooses a point Z ∈P R. The aim of the first player is to maximize |XYZ|. What is the largest area he can secure for himself?
10. (One-person game.) There are 1990 boxes containing 1, . . . , 1990 chips, respec-tively, on a table. You may choose any subset of boxes and subtract the same number of chips from each box. What is the minimum number of moves you need to empty all boxes?
11. A and B alternately place +, −, · into the free places between the numbers 1 2 3 . . . 99 100. Show that A can make the result (a) odd, (b) even.
12. A and B start with p 1. Then they alternately multiply p by one of the numbers 2 to 9. The winner is the one who first reaches (a) p ≥1000, (b) p ≥106. Who wins, A or B?
13. A crosses out any 27 of the numbers 0, 1, . . . , 255, 256. Then B crosses out any 26 numbers. Then A crosses out any 25 numbers, and so on until finally B crosses out 20 1 number. Since 27 + 26 + · · · + 20 28 −1 numbers are crossed out, there will be two numbers a and b left. B pays the difference |a −b| to A. How should A play to get as much as possible? How should B play to lose as little as possible?
How much does A win per game if both players use their optimal strategies?
14. A and B take turns in placing a ′′+′′ sign or a ′′−′′ sign in front of one of the numbers in the sequence 1 2 3 4 · · · 19 20. After all 20 signs have been placed, B wins the absolute value of the sum. Find the best strategy for each player. How much does B win if both players use their best strategies?
15. In the equation x3 + · · · x2 + · · · x + · · · 0, A replaces one of the three dots by an integer unequal to 0. Then B replaces one of the remaining dots by an integer.
Finally A replaces the last dots by an integer. Prove that A can play so that all three roots of the resulting cubic equation are integers.
364 13. Games 16. A and B alternately replace the stars in the polynomial x10 + ∗x9 + ∗x8 + · · · + ∗x2 + ∗x + 1 by real numbers. If the resulting polynomial has no real roots, then A wins. If it has at least one real root, then B wins. Can B win, whatever A does?
17. A and B alternately write positive integers ≤p on the blackboard. Writing divisors of numbers that are already written is not allowed. The one who cannot move any more loses. Who wins for (a) p 10? (b) p 1000?
18. Double Chess. The rules of the chess are changed as follows: Black and White make alternately two legal moves. Show that there exists a strategy for white which guarantees him at least a tie. (Note: You need only prove the existence of such a strategy.) 19. On any directed graph with one highest and one lowest node, A puts a chip on any node. Than B puts a chip on a an unoccupied node, and so on. If a node is occupied, all lower nodes are forbidden. The player who is forced to place a chip on the highest node loses. Prove that the first player wins if he plays correctly. (Note: You are not asked to find the winning strategy. You only need to prove that it exists.) 20. Even Wins. Initially there is a supply of (2n+1) chips. A and B take turns to remove any number of chips from 1 to k. At the end, one of the players winds up with an even number of chips, the other with an odd number. The winner is the one who possesses an even number of chips. Find the losing position for (a) k 3, (b) k 4, (c) even k, (d) odd k.
Consider also the case that Odd Wins. See .
21. Initially there is a chip at the corner of an n × n-chessboard. A and B alternately move the chip one step in any direction. They may not move to a square already visited. The loser is the one who cannot move. (a) Who wins for even n? (b) Who wins for odd n? (c) Who wins if the chip starts on a square, which is neighbor to a corner square?
22. A places a knight onto an 8 × 8 board. Then B makes a legal chess move. Then A makes a move, but he may not place it on a square visited before, and so on. The loser is the one who cannot move any more. Who wins?
23. A king is placed at the upper left corner of an m × n chessboard. A and B move the king alternately, but the king may not move to a square occupied earlier. The loser is the one who cannot move. Who has a winning strategy?
24. Start with a pile of n chips. A and B move alternately. At his first move, A takes any number s so that 0 < s < n. From then on, a player may take any number which is a divisor of the number of chips taken at the preceding move. The winner is the one who makes the last move. Which initial positions are winning for A or B?
25. Let n be a positive integer and M {1, 2, 3, 4, 5, 6}. A starts with any digit from M. Then B appends to it a digit from M, and so on, until they get a number with 2n digits. If the result is a multiple of 9, then A wins; otherwise B wins. Who wins, depending on n?
26. Start with two piles of p and q chips, respectively. A and B move alternately. A move consists in taking a chip from any pile, taking a chip from each pile, or moving a chip from one pile to the other. The winner is the one to take the last chip. Who wins, depending on the initial conditions?
13. Games 365 27. Start with two piles of p and q chips, respectively. A and B move alternately. A move consists in removing any pile and splitting the other pile into two piles. The loser is the one who cannot move any more. Who wins, depending on the initial conditions?
28. Start with n ≥12 successive positive integers. A and B alternately take one integer, until only two integers a and b are left. A wins if gcd(a, b) 1, and B wins if gcd(a, b) > 1. Who wins?
29. Two players A and B alternately color lattice squares of a 19 × 94 square. Who has a winning strategy? A lattice square is any square of the board whose vertices are lattice points of the 19 × 94 board (MMO 1994).
30. A and B alternately move a knight on a 1994 × 1994 chessboard. A makes only horizontal moves (x, y) →(x ± 2, y ± 1), B makes only vertical moves (x, y) → (x ± 1, y ± 2). A starts by choosing a square and making a move. Visiting a square for a second time is not permitted. The loser is the one who cannot move. Prove that A has a winning strategy (ARO 1994).
31. A calls out a digit. Then B places that digit into one of the empty cells, until all 8 cells are filled by digits. A wants to maximize the difference. B tries to make it as small as possible. Prove that B can place the digits so that the difference is at most 4000. A can call digits such that the difference is at least 4000.
– 32. A and B alternately color squares of a 4 × 4 chessboard. The loser is the one who first completes a colored 2 × 2 subsquare. Who can force a win?
33. A and B alternately replace the stars in x4 + ⋆x3 + ⋆x2 + ⋆x + ⋆ 0 by integers of their choice. A wins if he gets a polynomial without integral roots after the fourth step. Otherwise B wins. Who wins, A or B?
34. Two players A and B alternately take chips from two piles with a and b chips, respectively. Initially a > b. A move consists in taking a multiple of the other pile from a pile. The winner is the one who takes the last chip in one of the piles. Show that (a) If a > 2b, then the first player A can force a win.
(b) For what α can A force a win, if initially a > αb. (This game of Euclid is due to Cole and Davie, Math. Gaz. LIII, 354–7 (1969).) 35. A marks any free cell of a 2n × 2n board. Then B places a 1 × 2 domino on the board so that it covers 2 free cells, one of which is marked. A wins if it is possible to cover the whole board by dominos, otherwise B wins. Who wins?
36. A solitaire game. Each edge of a 1997-polyhedron is assigned the number +1 or −1. Show that there exists a vertex such that the product of the numbers on all edges meeting in that vertex must be +1.
Solutions 1. The table of the first 13 losing positions is 366 13. Games n 0 1 2 3 4 5 6 7 8 9 10 11 12 x(n) 0 1 3 4 6 8 9 11 12 14 16 17 19 y(n) 0 2 5 7 10 13 15 18 20 23 26 28 31 This table suggests the following algorithm for constructing the losing positions step-by-step: Suppose the losing positions [x(i), y(i)] for i < n are known already.
Then x(n) is the smallest positive integer not used already, and y(n) x(n) + n.
Thus every positive integer occurs exactly once as a difference. It is not too difficult to prove this and the fact that we have indeed all loosing positions. Do it!
Now let us try to find a closed formula for x(n) and y(n). Plotting the results, we see that x(n) and y(n) are both approximately linear functions, that is, x(n) ≈t · n, y(n) ≈(t + 1) · n.
Furthermore, t ≈1.6. This suggests that t (1 + √ 5)/2. Thus, we conjecture that x(n) ⌊t · n⌋, y(n) ⌊(t + 1) · n⌋.
It remains to be shown that every positive integer occurs exactly once in one of the two sequences. But we have already proved this in Chapter 6. There we have shown that α, β irrational and 1/α + 1/β 1 is necessary and sufficient for this so called complementarity of the sequences ⌊α · n⌋and ⌊β · n⌋. Now we have 1 t + 1 t + 1 2t + 1 t2 + t 2t + 1 2t + 1 1.
Here we used the well-known relationship t2 t + 1 for the golden section t.
2. We observe that 6 is the first number, which is not the power of a prime. Thus, V consists of all multiples of 6. If A is confronted with a number, which is not a multiple of 6, he can attain a multiple of 6, by subtracting one of the numbers 1, · · · , 5. From a multiple of 6, there is no move to another multiple of 6.
3. In his first move A adds 1, the only divisor of 2, and gets n 3. From here on, A can move so that he gets an odd number. A proper divisor of an odd number is at most one third of that number. So B can add at most one third of the current number. A moves from an even number, and so he can add exactly one-half of that number. So A simply plays until he is confronted for the first time with an even number ≥1328.
By adding one-half of that number he reaches a number ≥1992.
4. The following table shows the first few positions in L: n 0 1 2 3 4 5 6 7 8 9 10 x(n) 0 1 2 4 5 7 8 9 11 12 14 y(n) 0 3 6 10 13 17 20 23 27 30 34 First, we note that y(n)−x(n) 2n. Here x(n) the minimum integer not yet used, and y(n) x(n) + 2n. Then we observe that the two sequences are complementary, i.e., disjoint and their union is all positive integers. By an analysis similar to that of the Wythoff game, we get α (1 + √ 5)/2, β 2 + α, and x(n) ⌊nα⌋, y(n) ⌊nβ⌋ give all solutions. We check that the Beatty condition α−1 + β−1 1 for comple-mentary sequences is satisfied.
13. Games 367 5. Consider the horizontal (or vertical) symmetry line of the board. B can win by always playing his knight symmetrically to the previous move of A.
6. B wins by using the same strategy as in the preceding problem.
7. A wins by drawing first a main diagonal. Then to each move of B, he draws the same diagonal reflected at the center of the polygon.
8. A chooses the centroid X. B draws a parallel to one of the sides through X and gets 5/9 of the cake. By drawing another line through X he would get less, comparing the wins and losses. The choice of the centroid X for A is best, because, in every other position, B would get more. Find the best choice for B.
9. B can prevent A from getting more than 1/4. He chooses Y so that XY ∥PS. Then, for every point Z on P S, the following inequality is satisfied: |XYZ| |P QR| |XY| |P R| · H −h H h(H −h) H 2 ≤1 4.
On the other hand, A can choose the midpoints X and Z of PQ and PR and secure |XYZ| 1/4 for himself. More difficult is the analogous problem for the perimeter of XYZ. See Quant 4, 32–33 (1976).
10. We need 11 steps. After each step we partition the boxes into susets each contain-ing the same number of chips. Suppose at some moment there are n subsets of boxes (some of which may be empty). In the next step we select k subsets, from which we subtract the same number of chips. After subtraction, the boxes in dif-ferent subsets still belong to different subsets, and untouched boxes still belong to the same subsets. If we started with n subsets of boxes, then after one step there will be not less than max(n, n −k) ≥n/2 subsets left. Thus at each step the num-ber of subsets of boxes left will be at least one-half of the preceding number. Ini-tially there were 1990 distinct subsets. After 1, . . . , 11 operations there will be at least 995, 498, 249, 125, 63, 32, 16, 8, 4, 2, 1 subsets left. So we need at least 11 steps. Eleven steps are indeed sufficient by proceeding as follows. We subtract 995 chips from all boxes containing at least 996 chips. Then we subtract 498 chips from boxes with at least 498 chips, and so on.
11. Since only parity counts, we may work modulo 2 and get the initial state 1 0 1 0 1 · · · 1 0. Since modulo 2 subtraction is the same as addition, A and B insert + and × into the gaps.
First suppose that A wants to make the result equal to 0. He should use × exclusively and move into the first gap, thus reducing the new position to the string 0, 1, 0, 1, . . . , 1, 0. Now, if B places any sign into some gap, getting · · · 0 ∗ 1 0 · · · or 0 1 ∗ 0, then A should place a × into the gap on the other side of B′s move. It is easy to see that the result is 0 at the end.
Now suppose that A wants to make the result 1. On his first move, he places a + into the first gap and then plays the same strategy as in the preceding case. At the end, he gets the sum 1 + 0, which is 1.
12. (a) Start at the end. Which set should I avoid? (a) [112, 999] ⊂W ⇒[56, 111] ⊂ L ⇒[7, 55] ⊂W ⇒{4, 5, 6} ⊂L ⇒1 ∈W. Thus, A wins.
(b) [111112, 999999] ⊂W ⇒[55556, 111111] ⊂L ⇒[6173, 55555] ⊂W ⇒ [3087, 6172] ⊂L ⇒[343, 3086] ⊂W ⇒[172, 342] ⊂L ⇒[20, 171] ⊂W ⇒ [10, 19] ⊂L ⇒[2, 9] ⊂W ⇒1 ∈L. Thus, A loses.
368 13. Games 13. We will show that A can secure at least 24, or 16, for himself, and B can prevent A from getting more than 16. Strategy of A: at each move he crosses out every sec-ond remaining number, i.e., 2, 4, 6, . . .. Then after 1, 2, 3, 4 moves, the distances between neighbors will be at least 2, 4, 8, 16.
Strategy of B: At each move, he crosses out consecutive numbers at the beginning or at the end. In this way the maximum difference between two numbers is reduced after 1, 2, 3, 4 moves to at most 128, 64, 32, 16.
One can generalize the game to the sequence 0, 1, 2, . . . , 22n. A wins 2n.
14. First we describe B′ strategy. Consider the pairs (1, 2), (3, 4),· · ·, (19, 20). Each time A places a sign in front of one component of any pair, B places the opposite sign in front of the other component, except for the pair (19, 20). As soon as A places a sign in front of a number in the pair (19, 20), B uses the same sign for the second component. In this way B wins at least 20+19−1−1−1−1−1−1−1−1−1 30.
A′s strategy: Find the sign of the current sum. Place the opposite sign in front of the largest free number. If the current sum is 0, place a “+′′. Thus, the first move will be +20. If A and B both apply their strategies, the play will evolve as follows: +20 + 19 −18 + 17 −16 + 15 −14 + · · · −2 + 1.
Now we show that B cannot get more than 30 if he uses a different strategy while A continues to use his strategy.
Consider the moves in pairs: A followed by B. Now, suppose that in some game the ith pair of moves changes the sign of the current sum and that, that the sign remains unchanged after this pair of moves. Then 1 ≤i ≤10.
Inthefirst(i−1)pairsofmoves,Ahasinsuredthatthenumbers20, 19, 18, . . . , 20− (i−2) have been used. (A may not have used them himself, but at each move he takes one of the highest remaining numbers.) Then, since the ith pair of moves changes the sign of the current sum, the absolute value of the sum after the ith pair will occur if the sum after the (i −1)th pair is 0. In this case the maximum that could be added in the ith pair of moves is |20 −(i −1)| + |20 −i| 41 −2i. For each of the remaining (10 −i) moves, the absolute value of the sum decreases by at least one since A subtracts the largest free number from the absolute value of the sum, say k, and B cannot add more than k −1. Thus, the resulting sum cannot be larger than 41 −2i −(10 −i) 31 −i ≤30.
15. If A places −1 in front of the term x and at its second move he places an integer in the last free place, which is the opposite of what B placed, then the equation has the form x3 −ax2 −x + a 0. This equation has the roots −1, 1, a, which are integers.
16. B can always force a win. In his first four moves, B can ensure that the last fifth move of A is the choice of the coefficient of an odd power x2p+1, where P(x) is the final polynomial with numerical coefficients.
First we choose the numbers µ and c > 0 so that, for any λ for the polynomial F(x) P (x) + µxm + λx2p+1, we have cF(1) + F(−2) 0. Then F(x) definitely has a root in [−2, 1]. For this it is sufficient to take c 22p+1 and µ P (−2) −cP(1) c + (−2)m .
Here 1 and −2 can be replaced by any two numbers of opposite signs. Playing with this µ in his fourth move, B will secure a real root for himself.
13. Games 369 17. In both cases A wins. (a) A writes 6. Then B writes one of the numbers of the pairs (4, 5), (7, 8), (9, 10). A responds with the other number of the pair.
(b) We consider a new game: the rules are the same, but among the numbers, the number 1 is missing. If A has a winning strategy in this case, then he uses it imme-diately. If not, then first he writes 1 and then uses the winning strategy of the second player. Note that in this case we do not explicitly describe the winning strategy of A. Rather we prove its existence.
18. Suppose B can win no matter what A does. On his first move, A moves one of his knights to any one of the two possible squares and then back to its original position.
Now all the pieces are in their original position, but A has become the second player and must win. Contradiction!
19. We consider a new game: the rules are the same, but the lowest node is forbidden.
If A has a winning strategy in this case, then he uses it immediately. If not, then he first puts a chip on the lowest node and then uses the winning strategy of B.
20. (c) Check that for even k the losing positions are (k+1 mod k+2, odd), (0 mod k+ 2, odd), (1 mod k + 2, even).
(d) Check that for odd k the losing positions are (1 mod 2k + 2, even), (k + 2 mod 2k + 2, odd), (0 mod 2k + 2, odd), (k + 1 mod 2k + 2, even).
21. (a) If n is even, then one can always partition the board into 2 × 1 dominoes. A can always make a move. If the chip is on one square of a domino, he moves to the other square.
(b) For odd n, one can split the board into 2 × 1 dominoes, except the corner square.
Then a similar strategy is winning for B.
(c) In this case, A always wins. For even n, the strategy is the same as in a). For odd n, we partition the board into dominoes except the corner square. Now we color the board in the usual way. It is easy to see that B can never move to a corner square.
Thus, A wins by the strategy of moving to the second square of a domino.
22. Split the board into eight 4 × 2 rectangles. On each such rectangle, there is a unique move to another square of this rectangle. Then B can win as follows. For each move of A, he moves the knight to the only possible square of the same rectangle.
23. Subdivide the chessboard into 2 × 1 dominoes. Whenever A places the king on a domino, B should move it to the other square of the same domino. In this way B wins the game.
24. We will prove that, for n > 1, B wins if n 2m. Let n 2m (n > 1). A takes first 2a(2b +1) chips (a ≥0, b ≥0). Then B wins if he uses the following strategy. First he takes 2a chips, and from then on he uses as many as A has taken before. A wins if initially there are n 2a(2b + 1) chips. First he takes 2a chips and from then on he mimicks his opponent.
25. Answer: If n is a multiple of 9, then B wins; otherwise A wins.
Suppose 9|n. If A appends any digit x, then B appends 7−x. At the end, the resulting number has digital sum 7n. Thus the resulting number is divisible by 9. So B wins.
If 7n is not a multiple of 9, then 7(n −1) ≡2 mod 9 r with r ̸ 2. Thus r ∈{0, 1, 3, 4, 5, 6, 7, 8}. Let s ≡9 −r mod 7; that is, s 9 −r if 9 −r < 7, otherwise s 2−r. The strategy of A is as follows. A writes down a number s ∈M.
To each move x of B, he responds with 7−x. If it is B′s last move, we have a number 370 13. Games of (2n −1) digits with digital sum s + 7(n −1), which is congruent to s + r mod 9.
But we have s + r 9 or s + r 2. To get a number divisible by, 9 B would have to add 0 or 7. Neither is permissible. Thus A wins (BWM 1984).
26. A can force a win by making p and q both even if initially at least one of p and q is odd. B is forced to make at least one of p or q odd. A restores the losing position for B.
27. Two odd piles are losing. From any other position, one can move to two odd piles.
From two odd piles, one is forced to move to even or odd. From this position one throws away the odd pile and splits the even pile into two odd piles. Finally, we move to (1, 1) and win.
28. Suppose that n 2k + 1. In that case A wins by subdividing the numbers into successive pairs and taking the lonely remaining number. If B takes any number of a pair, then A takes the other element of the pair.
Now suppose that n 2k. In that case B wins by always taking odd numbers except two odd numbers r, s divisible by 3. A is always forced to take even numbers.
At the next to last move, there will remain two even numbers e1, e2 and the odd numbers r, s. If A takes an odd number, then B takes the other odd number and wins. Otherwise A sticks to taking an even number, in which case B takes the other even number and wins again since gcd(r, s) ≥3.
29. Symmetry is the most important strategy in games. Look at the center of the board.
For a small odd height and a large even length it will be as indicated in Fig. 13.1.
Unfortunately the center of the board is not a lattice point. The first move should be to color the square in Fig. 13.1. Now the board is split into two parts which are symmetric with respect to the line s. B is forced to color a square on one side of s.
A responds by removing the square which is symmetric to B′s choice with respect to s.
r Fig. 13.1 s H H Y H H Y H H Y H H j H H jH H jH H j H H Y H H Y H H Y Fig. 13.2 30. We place arrows on the board as in Fig. 13.2. A starts by placing the knight on a cell from which an arrow starts and he moves in the direction of the arrow. Then B can only move to the start of another arrow, and A moves to the end of that arrow.
31. Denote the positions from left to right by p1, p2, p3, p4. The game splits into two parts: the beginning and the endgame. The endgame starts as soon as B puts a digit into the first position. It is clear that, in the beginning, A must not call digits 1 to 3 or digits 6 to 9, since B would place them into p1, a small digit into the upper cell and a large digit into the lower cell, and would go over to the endgame. If the difference of the first digits is not greater than 3, then the difference of the numbers is at most 3999. If A first calls 4 (or 5), then B can secure a difference for himself not less than 4000 by immediately starting the endgame with the move p1 4 ∗ or p1 ∗ 5 , and then put all digits 0 (9) into positions p2, p3, p4 until they are filled.
13. Games 371 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 Fig. 13.3 32. Fig. 13.3 shows the winning partition for B into pairs (x, x). If A colors some element of a pair, B responds by coloring the other element.
33. After three steps, three of the stars are replaced by the integers a, b, c. B wins by replacing the fourth star by −a −b −c −1. Then the sum of the coefficents of the polynomial becomes 0, and hence the number 1 is a root.
34. (a) Suppose a ≥2b. We will show that A can move from (a, b) into a losing position (for B). If (a −b, b) is a losing position, then A makes the move (a, b) →(a −b, b).
But if it is a winning position, then there is a move from it which makes it a losing position. Since a −b ≥b, this move has the form (a −b, b) →(a −qb, b), where q is a positive integer. But then (a, b) →(a −qb, b) is a winning move for A.
Note that we can show here that (a, b) for a ≥2b is a winning position without showing the winning strategy.
(b) The answer is α ≥(1 + √ 5)/2.
If b < a < αb < 2b, the only possible move from (a, b) is to (a −b, b). Hence, b a −b 1 a b −1 > 1 α −1 α.
(1) Since it is not possible to win in one move from the position (a, b), 1 < b/a < α, it is enough to show that when A starts from (a, b), b/a < α, then he may either win in one move or leave to B a position with 1 < b/a < α, from which by (1) B′s sole move is to a position with ratio> α from which the process is repeated.
When a/b > 2 there are at least two moves (a, b) →(b, r) with 0 ≤r < b, or (a, b) →(b + r, b). If r 0, A may win in one move. Otherwise, since α is strictly between b/r and (b + r)/b, A moves to that position for which the ratio lies strictly between 1 and α. When α < a/b < 2 A moves to (b, r).
35. A wins. By a diagonal row, we mean any row starting on the left or upper side and running southeast to the lower or right side. A must always mark a free cell in the lowest diagonal row. If there are cells in that row which can be covered uniquely, then first he must mark any one of these cells. If a free cell in this diagonal can be covered in two ways, it is irrelevant which one A marks.
36. Suppose we multiply all of the products corresponding to all of the vertices. Since every edge is counted twice, every −1 is counted twice. Thus, the product is +1. But there is an odd number of vertices. The product at each vertex cannot be −1, since (−1)1997 −1. Hence, at least one vertex has product +1.
14 Further Strategies In this chapter we collect further important strategies of somewhat lesser scope, except the first one on graph theory, which became quite important in recent IMOs.
They will be illustrated by a few examples followed by problems with solutions.
All of these ideas occurred in preceding problems and solutions. But still, it is useful to stress them again. By separate treatment, they will be better remembered.
14.1 Graph Theory Graphs are important objects of discrete mathematics. A graph is an object con-sisting a set of points or vertices, some of which are connected by lines or edges.
If you can visit all vertices by walking on edges, the graph is connected. A con-nected graph without closed paths or cycles is called a tree. Usually the edges of a graph are not oriented. But if the edges are oriented, then we have a digraph. An example is a one-way road system. The directed cycles are often called circuits.
A vertex v has degree or valency m if m edges end at v. The mapping f of a set A into itself is usually represented by a digraph, where we draw an arrow from the vertex a to its image f (a). Points with a f (a) are the fixed points of the mapping. A permutation of a set A is a one-to-one mapping of A onto itself. Since a ̸ b ⇒f (a) ̸ f (b), the graph of f splits into cycles. Most of the problems in this section belong to the box principle, some to combinatorics.
374 14. Further Strategies Problems 1. At an international meeting, 1985 persons participated. In each subset of three partic-ipants, there were at least two persons, who spoke the same language. If each person speaks at most five languages, then at least 200 persons spoke the same language (BMO 1987).
2. Can you draw a triangular map inside a pentagon, so that each vertex has an even degree?
3. In how many ways can you triangulate a convex n-gon by (n −3) nonintersecting diagonals, so that every triangle has at least one side in common with the n-gon?
4. Prove that, in any set of 17 persons, in which every person is acqainted with exactly four other persons, there exist two persons, who do not know each other and have no common acquaintances (AUO 1992).
5. Consider nine points in space, no four of which are coplanar. Each pair of points is joined by an edge (that is, a line segment), and each edge is either colored blue or red or left uncolored. Find the smallest value of n such that whenever exactly n edges are colored, the set of colored edges necessarily contains a triangle all of whose edges have the same color (IMO 1992).
6. We assign an arrow to each edge of a convex polyhedron, so that at least one arrow starts at each vertex, and at least one arrow arrives. Prove that there exist two faces of the polyhedron, so that you can trace their perimeters in the direction of the arrows (BWM).
7. Let S be a set of n points in space (n ≥3). The segments joining these segments are of distinct length, and r of these segments are colored red. Let m be the smallest integer for which m ≥2·r/n. Prove that there always exists a path of m red segments with their lengths sorted increasingly (BWM).
8. In a set of n persons, any subset of four contains a person who knows the other three persons. Prove that there exists a person who knows all the others. (If A knows B then B knows A.) 9. Two black knights stand on the lower corners of a 3 × 3 chessboard, and two white knights on the upper corners. White and black knights must be interchanged by legal moves onto free squares. Find the minimum number of moves needed (quoted by Lucas in 1894 from an earlier source in 1512).
10. In a set S of 2n persons there are two with an even number of common friends.
14.2 Infinite Descent We consider one of the oldest proof strategies going back to the Pythagoreans in the fifth century b.c. It is an impossibility proof especially useful in Number Theory. The main idea is as follows: We want to prove that (usually) a polynomial equation f (x, y, z, . . .) 0 (1) has no solution in positive integers. One shows: If (1) is true for some posi-tive integers a, b, c, . . ., then (1) would be true for the smaller positive integers 14.2 Infinite Descent 375 a1, b1, c1 . . .. For the same reason, (1) would be true for the still smaller posi-tive integers a2, b2, c2 . . ., and so on. But this is impossible since a sequence of positive integers is bounded below and cannot decrease indefinitely.
Pierre de Fermat (1601–1665) rediscovered the method and called it Infinite Descent (descent infini). He was especially proud of this method. Near the end of his life, he wrote a long letter in which he summarized all of his discoveries in number theory. He stated that he found all of his results with this method. By the way, he does not mention Fermat’s last conjecture which dates to a very early stage of his life.
We will present the method (not for the first time in this book) by an old method, which the Pythagoreans treated geometrically.
E1. The regular pentagram was the “badge” of the Pythagoreans. Fig. 14.1 shows that x 1 x + 1 x ⇒x2 x + 1.
(2) The Pythagoreans first thought that all ratios are rational, i.e., x a/b, a, b ∈N.
Introducing this into (2), we get a2 ab + b2.
(3) r r r r r B B B B B B B ppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp 1 x x 1 Fig. 14.1 The Pythagoreans knew the rudiments of number theory, in particular the parity rules e + e e, e + o o, o + o e, e · o e, e · e e, o · o o, where “e” and “o” stand for “even” and “odd,” respectively. Now what parities do the integers a and b in (3) have? The assumption that a and b have different parities leads to a contradiction. The assumption that both a and b are odd also leads to a contradiction. Hence, both a and b are even, that is, a 2a1, b 2b1, a1, b1 ∈N, a1 < a, b1 < b.
(4) Substitution in (3) and cancellation by 2, gives a2 1 a1b1 + b2 1.
(5) The same reasoning applied to (5) gives a1 2a2, b1 2b2, a2 < a1, b2 < b1, (6) 376 14. Further Strategies r r r r r ppppppppppppppppppppppppppppppppppp ppppppppppppppppppppppppppppppppppp ppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppp ppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp P1 P5 P4 P3 P2 Fig. 14.2 and so on. From the truth of (3), we deduce the existence of two decreasing infinite sequences of positive integers a > a1 > a2 > · · · and b > b1 > b2 · · · .
(7) Such sequences do not exist. Thus, (3) is never true for positive integers.
E2. The set Z × Z is called the plane lattice. Prove that for n ̸ 4 there exists no regular n-gon with lattice points as vertices.
Proof. First, we prove that there is no regular triangle with lattice points as vertices. Indeed, let a be the length of a side of such a triangle with lattice points as vertices. According to the distance formula a2 is a positive integer, and the area is the irrational number a2√ 3/4. On the other hand, the area of any lattice polygon has a rational area.
The vertices of a regular hexagon P1P2P3P4P5P6 cannot all be lattice points, since for instance P1P3P5 is a regular triangle.
Now let n ̸ 3, 4, 6. Suppose P1P2 · · · Pn is a regular lattice n-gon. At P1, P2, . . . , Pn, we apply the vectors − − → P2P3, − − → P3P4, . . . , − − → P1P2 (Fig. 14.2). The endpoints of these vectors are also lattice points, and they form a regular n-gon in-side the first one. With the new n-gon, we can proceed similarly, etc, ad infinitum.
The square of the lengths of the sides of all these polygons are integral, and they decrease at each step.
E3. Prove that the following equation has no solutions in positive integers: x2 + y2 + z2 + u2 2xyzu.
(1) The left side of (1) is even. Thus among the integers x, y, z, u, there is an even number of odd integers. If all four are odd, then the left side is divisible by 4, whereas the right side is only divisible by 2. If two of the integers are odd, then the left side is divisible only by 2, whereas the right side is divisible by 8. Hence all four integer on the left side are even, that is, x 2x1, y 2y1, z 2z1, u 2u1.
Inserting this into (1), we get x2 1 + y2 1 + z2 1 + u2 1 8x1y1z1u1.
(2) From (2), it follows that all four integers on the left side are even, that is, x1 2x2, y1 2y2, z1 2z2, u1 2u2, and x2 2 + y2 2 + z2 2 + u2 2 32x2y2z2u2.
(3) 14.3 Working Backwards 377 Similarly, one proves that x2 s + y2 s + z2 s + u2 s 22s+1xsyszsus, for every s ∈N, (4) that is, for every s ∈N x/2s, y/2s, z/2s, u/2s are positive integers. Contradic-tion!
Problems 11. 2n + 1 (n ≥1) integral weights are given. If we remove any of the weights, the remaining 2n weights can be split into two heaps of equal weights. Prove that all weights are equal.
12. Can a cube be partitioned into finitely many cubes of different sizes?
13. The equation 8x4 + 4y4 + 2z4 t4 has no solutions in positive integers.
14. Find the integral solutions of (a) x3 −3y3 −9z3 0, (b) 5x3 + 11y3 + 13z3 0, (c) x4 + y4 z4.
15. Let (x, y) be a solution x2 + xy −y2 1 in positive integers. Prove that (a) gcd(x, y) 1; (b) if x y then x y 1; (c) x ≤y < 2x; (d) (x + y, x + 2y) and (2x −y, −x + y) are also solutions. Construct an infinite sequence of solutions, and prove that they comprise all solutions.
16. Find all integral solutions of 10x3 + 20y3 + 1992xyz 1993z3.
14.3 Working Backwards Working Backwards is one of the oldest problem-solving strategies, used since antiquity. The ancient Greeks used the method in construction problems. They assumed that an object is already constructed, and they worked backwards to the data, which were actually given. The idea works well if the problem does not branch too much in backstepping. What was the situation one step before? What was the situation two steps before? There should be few possibilities before each backward step.
We will illustrate the method by some typical problems. Jacobi in the last century used to stress: You must always invert! His dictum proved very fruitful to him. At that time the most popular subject was Elliptic Integrals. By applying his dictum, he inverted elliptic integrals and so made his greatest discovery, the elliptic functions, which were far easier to handle than their inverses, the elliptic integrals. A very free interpretation of his dictum allows us to progress in hopeless situations. In fact, we used this method whenever we assumed the existence a solution and derived a contradiction from it. So this method is used in innumerable instances without mentioning its name. It is closely related to Infinite Descent.
378 14. Further Strategies Problems 17. Along a circle are written 4 ones and 5 zeros. Then between two equal numbers we write a one and between two distinct numbers zero. Finally the original numbers are wiped out. This step is repeated. In this way can we ever reach 9 ones?
18. There are n weights on a table with weights m1 > m2 > · · · > mn and a two-pan scale. The weights are put on the pans one-by-one. To each weighing we assign a word from the alphabet {L, R}. The kth letter of the word is L or R if the left or right pan outweighs the other, respectively. Prove that any word from {L, R} can be realized.
19. In n glasses with sufficient volume, there is initially the same amount of water. In one step you may empty as much water from any glass into any other glass as there is in the second glass. For what n can you pour all the water into one glass?
20. Starting with 1, 9, 9, 3, we construct the sequence 1, 9, 9, 3, 2, 3, 7, . . ., where each new digit is the mod 10 sum of the preceding four terms. Will the 4-tuple 7, 3, 6, 7 ever occur?
21. The integers 1, 2, . . . , n are placed in order, so that each value is either bigger than all preceding values or is smaller than all preceding values. In how many ways can this be done?
14.4 Conjugate Numbers Let a, b, r be rational, but √r be irrational. Then a +b√r and a −b√r are called conjugate numbers. They often occur simultaneously.
Often it is helpful to switch between a + b√r and a −b√r.
We rationalize the denominator as often as we rationalize the numerator: 1 a + b√r a −b√r a2 −b2r , a + b√r a2 −b2r a −b√r .
To rationalize the denominator in 1 1 + √ 2 + √ 3 , we multiply denominator and numerator so that we get the denominator (1 + √ 2 + √ 3)(1 + √ 2 − √ 3)(1 − √ 2 + √ 3)(1 − √ 2 − √ 3).
The mapping √ 2 →− √ 2 and √ 3 →− √ 3 leaves this term unchanged. Thus, the term is rational. To rationalize the denominators in 1 1 + 3 √ 2 + 2 3 √ 4 , 1 1 − 4 √ 2 + 2 √ 2 + 4 √ 8 , it is useful to know that the sets {a + b 3 √ 2 + c 3 √ 4|a, b, c ∈Q} and {a + b 4 √ 2 + c 4 √ 4 + d 4 √ 8|a, b, c, d ∈Q} are fields, i.e., algebraic systems which are closed with respect to the operations +, −, ·, :.
14.4 Conjugate Numbers 379 As a typical example, we use the problem from the “Ersatz”-IMO 1980.
E1. Find the first digit before and after the decimal point in √ 2 + √ 3 1980 .
The base √ 2 + √ 3 does not have the form a + b√n for which we have a theory. Hence we transform it into this form by squaring the base and halving the exponent. We get x (5 + 2 √ 6)990. This is almost an integer. Indeed, by adding the tiny number y (5 −2 √ 6)990 we get the integer a (5 + 2 √ 6)990 + (5 −2 √ 6)990 x + y p + q √ 6 + p −q √ 6 2p, where p is an integer. We need only the last digit of 2p, i.e., 2p mod 10. We can find 2p mod 10 by the binomial theorem. We get 2p 2 5990 + 990 2 5988 · 22 · 6 + 990 4 5986 · 24 · 62 + · · · + 2 · 2990 · 6495.
All of the terms except the last one are divisible by 10. The last one is easy to find mod 10 since 6n mod 10 6. Thus it remains to find 2991 mod 10, which is 8, since the last digit of powers of 2 has period 2, 4, 8, 6. Finally 8 · 6 ≡8 mod 10.
Now we have the last digit 8 of x + y. Subtracting the tiny number y, we get x . . . 7, 9 . . ..
Alternate solution: We embed the problem into a more general one. Let un (5 + 2 √ 6)n + (5 −2 √ 6)n xn + yn √ 6 + xn −yn √ 6 2xn, un+1 (xn + yn √ 6)(5 + 2 √ 6) + (xn −yn √ 6)(5 −2 √ 6) 10xn + 24yn, un+2 10xn+1 + 24yn+1 10(5xn + 12yn) + 24(2xn + 5yn) 98xn + 240yn, un+2 + un 100xn + 240yn 10un+1 ≡0 mod 10.
From u1 10, u2 98 we get 0, 8, 0, 2, . . . with period 4 for the last digit of un. Thus the 990th term is 8. The remainder can be finished as above.
Problems 22. Prove that (a + b√r)n p + q√r ⇐ ⇒(a −b√r)n p −q√r.
23. (x + y √ 5)4 + (z + t √ 5)4 2 + √ 5 has no rational solutions x, y, z, t.
24. Let (1 + √ 2)n xn + yn √ 2, where xn, yn are integers. Prove that (a) x2 n −2y2 n (−1)n; (b) xn+1 xn + 2yn, yn xn + yn.
25. Which number is larger: (a) √ 1979 + √ 1980 or √ 1978 + √ 1981?
(b) an √n + √ n + 1 or bn √ n −1 + √ n + 2?
26. Let an n √ n2 + 1 −n . Find limn→∞an.
27. an √ n + 1 + √n, bn √ 4n + 2 ⇒0 < bn −an < 1/16n√n.
380 14. Further Strategies 28. Find the first 100 decimals of √ 50 + 7 100 .
29. If p > 2 is a prime, then p|⌊(2 + √ 5)p⌋−2p+1.
30. ⌊(2 + √ 3)n⌋is odd.
31. Find the highest power of 2 which divides ⌊ 1 + √ 3 n ⌋.
32. (a) For every n ∈N, we have n √ 2 −⌊n √ 2⌋> 1/(2n √ 2).
(b) For every ϵ > 0 there is an n ∈N such that n √ 2 −⌊n √ 2⌋< (1 + ϵ)/(2n √ 2).
33. Find the equation of lowest degree with integral coefficients and one solution x1 1 + √ 2 + √ 3. Give the other solutions without computation.
34. Decide if 3 √ 5 + 2 − 3 √ 5 −2 is rational or irrational.
35. If a, b, √a + √ b are rational, then so are √a, √ b.
36. If a, b, c, √a + √ b + √c are rational, then so are √a, √ b, √c.
37.
3 √ 2 cannot be represented in the form a + b√r with a, b, r ∈Q.
38. √ 2 −1 n , n ∈N has the form √m − √ m −1, m ∈N.
39. Find the sixth decimal in √ 1978 + ⌊ √ 1978⌋ 20 .
40. Rationalize the denominator in (a) 1 1 + 3 √ 2 + 2 · 3 √ 4 , (b) 1 1 − 4 √ 2 + 2 · √ 2 + 4 √ 8 .
41. Let m, n ∈N and m n < √ 2. Prove that √ 2 −m n > 1 2· √ 2·n2 .
42. (a) Prove that there exist integers a, b, c not all zero and each of absolute value less than one million, such that |a + b √ 2 + c √ 3| < 10−11.
(b) Let a, b, c be integers, not all zero and each of absolute value less than one million. Prove that |a + b √ 2 + c √ 3| > 10−21 (Putnam 1980).
43. Simplify the expression L 2/ 4 −3 4 √ 5 + 2 √ 5 − 4 √ 125 (MMO 1982).
14.5 Equations, Functions, and Iterations In this section we collect some nonlinear systems of equations, which are of geo-metric origin or which originate in functional iterations.
E1. The positive reals x, y, z satisfy the equations x2 + xy + y2 3 25, y2 3 + z2 9, z2 + zx + x2 16.
Find xy + 2yz + 3zx (AUO 1984).
In a training session I gave this to one member our team, and I told him to give a detailed account of all ideas he had during the solution. Here is a short version: 14.5 Equations, Functions, and Iterations 381 1. What struck me first were the squares 9, 16, 25. This is the “Egyptian trian-gle.” It is a hint to the theorem of Pythagoras, to geometry, and geometrical interpretation.
2. Instead of x, y, z, only xy + 2yz + 3zx is required. This may be an area, maybe even the area 6 of the Egyptian triangle. It is also a hint that one should not try to find x, y, z.
3.
y2 3 occurs twice. Let us set t2 y2 3 . In fact, we need more squares to help in geometrical interpretations. The equations become x2 + √ 3xt + t2 25, t2 + z2 9, z2 + zx + x2 16.
The first looks like the Cosine Rule, the second like the theorem of Pythagoras, and the third again is the Cosine Rule. Indeed, the first and third equations are x2 + t2 −2xt cos 150◦ 25, z2 + x2 −2zx cos 120◦ 16.
For the area of the triangle, Fig 14.3 gives tz 2 + √ 3 4 xz + 1 4xt 6. On the other hand, Q xy + 2yz + 3zx xt √ 3 + 2 √ 3tz + 3zx 4 √ 3 · 6 24 √ 3.
Z Z Z Z Z Z Z Z 4 3 5 90◦ 120◦ 150◦ z a a a a a x S S S S t Fig. 14.3 Problems 44. Let f (x) 4x −x2. For x0 ∈R we consider the infinite sequence x0, x1 f (x0), x2 f (x1) . . .. Prove that there exist infinitely many x0, so that x0, x1, x2, . . .
consists of finitely many different values.
45. Solve the system of equations (x+y+z)3 3u, (y+z+u)3 3v, (z+u+v)3 3x, (u + v + x)3 3y, (v + x + y)3 3z.
46. Solve the equations x1 + x1x2 1, x2 + x2x3 1, . . . , x100 + x100x1 1.
47. Find all solutions (x, y, z) of the system of equations cos x + cos y + cos z 3 √ 3/2, sin x + sin y + sin z 3/2.
48. Find the positive reals x1, . . . , x10 satisfying the system of equations (x1 + · · · + xk) (xk + · · · + x10) 1, k 1, . . . , 10.
382 14. Further Strategies 49. Atookthenumbersx1, . . . , x5,foundtheirpairwisesumsa1, . . . , a10,andchallenged B to reconstruct x1, . . . x5 from a1, . . . , a10. Can B succeed?
Remark: For n 2k this is not always possible. For instance, the quadruples (0, 1, 2, 4) and (−0.5, 1.5, 2.5, 3.5) give the same sixtuple (1, 2, 3, 4, 5, 6).
50. Can you fill the 25 squares of a 5 × 5 table with numbers such that (a) the sum of the four numbers of each 2 × 2 square is negative, and the total sum of all numbers in the table is positive?
(b) the sum in each 2 × 2 square is negative, and the sum in each 3 × 3 square is positive?
51. Do there exist functions f (x), g(x), so that, for any x, y ∈R, x2 + xy + y2 f (x) + g(y)?
52. Solve the system of equations x1 + · · · + xn n, x2 1 + · · · + x2 n n, . . . , xn 1 + · · · + xn n n.
53. Let A (a1, a2, . . . , am) with m 2n and ai ∈{−1, +1}. Consider the transfor-mation T (A) (a1a2, a2a3, . . . ama1). Prove that, by repeated application of this transformation, you will reach the m-tuple (1, 1, . . . , 1).
54. Find all positive solutions of the system 1 −x3 1 x2, . . . , 1 −x3 n x1.
55. The system x + y + z 0, 1/x + 1/y + 1/z 0 has no real solutions.
56. Find g(x) f ◦f ◦· · · ◦f (x) f 1994(x), where f (x) (x √ 3 −1)/(x + √ 3).
57. Solve the equation 8x(2x2 −1)(8x4 −8x2 + 1) 1.
58. Solve the system of equations x2 + y2 1, 4x3 −3x √(x + 1)/2.
59. Find the positive solutions of xx1996 1996.
14.6 Integer Functions In the following definitions and rules, x is always a real and n an integer: ⌊x⌋ f loor of x largest integer ≤x x rounded down to next integer, ⌈x⌉ ceiling of x least integer ≥x x rounded up to next integer.
The function ⌊x⌋is also called the integer part of x, and {x} x −⌊x⌋is the fractional part of x. The following rules are especially useful: ⌊x⌋ n ⇔n ≤x < n + 1 ⇔x −1 < n ≤x, ⌈x⌉ n ⇔n −1 < x ≤n ⇔x ≤n < x + 1.
We have ⌊x +n⌋ ⌊x⌋+n, but ⌊nx⌋̸ n⌊x⌋. For this reason, it is usually a good strategy to get rid of floor and ceiling brackets. We prove the simple inequality ⌊x⌋+ ⌊y⌋≤⌊x + y⌋. Indeed, x ⌊x⌋+ {x}, y ⌊y⌋+ {x} Thus, ⌊x + y⌋ ⌊x⌋+ ⌊y⌋+ ⌊{x} + {y}⌋. Since 0 < {x} + {y} < 2, this is either ⌊x⌋+ ⌊y⌋or ⌊x⌋+ ⌊y⌋+ 1.
14.6 Integer Functions 383 E1. We will prove another simple formula by a method which mostly works, but which we will usually avoid, since it is not elegant. Prove that ⌊x⌋ n + 7x n 8 .
Let x m + α, 0 ≤α < 1, m qn + r, 0 ≤r < n. Then ⌊x⌋ m, ⌊x⌋ n m n q + r n, ⌊x⌋ n + q.
x n m + α n qn + r + α n q + r + α n , 7x n 8 q since r + α < n.
Problems 60. ⌊x⌋+ ⌊x + 1/n⌋+ · · · + ⌊x + n −1/n⌋ ⌊nx⌋, x ∈R, n ∈N.
61. If τn is the number of divisors of n ∈N, then τ1 + τ2 + · · · + τn ⌊n/1⌋+ ⌊n/2⌋+ · · · + ⌊n/n⌋.
62. If σn is the sum of divisors of n ∈N, then σ1 + σ2 + · · · + σn ⌊n/1⌋+ 2 ⌊n/2⌋+ · · · + n ⌊n/n⌋.
63. Suppose that p, q are prime to each other. Then p q + + · · · + (q −1)p q + q p + + · · · + (p −1)q p + (p −1)(q −1) 2 .
64. If n is a positive integer, prove that ⌊√n + √ n + 1⌋ ⌊ √ 4n + 2⌋.
65. If a, b, c ∈R and ⌊na⌋+ ⌊nb⌋ ⌊nc⌋for every n ∈N, then a ∈Z or b ∈Z.
66. For every n ∈N, find the largest k ∈Z+ for which 2k|⌊(3 + √ 11)2n−1⌋.
67. Among the terms of the sequence a1 2, an+1 ⌊(3/2)an⌋, n ∈N, there are infinitely many even and infinitely many odd numbers.
68. Based on the preceding sequence an, define a new sequence bn (−1)an. Prove that the sequence bn is not periodic.
69. For every pair of real numbers a and b, we consider the sequence pn ⌊2{an+b}⌋.
Here {c} is the fractional part of c. We call any k successive terms of this sequence a word. Is it true that any sequence of zeros and ones of length k is a word of a sequence given by some a and b: (a) for k 4; (b) for k 5 (MMO 1993)?
70. Find ⌊ √n + √ n + 1 + √ n + 2 2⌋.
71. Prove that ⌊ 3 √n + 3 √ n + 2 3⌋+ 1 is divisible by 8.
72. Prove that, for any positive integer n, we have 2n|1 + ⌊(3 + √ 5)n⌋.
384 14. Further Strategies Solutions 1. The proposition is certainly true, if one person speaks a common language with the other 1984, since 1984/4 > 200. Hence we assume that there is a pair {P1, P2}, with no common language. This pair forms 1983 triples with the remaining 1983 persons, of which each one has a common language with P1 or P2 (or both). Hence one of the pair, say P1, has a common language with 992 persons. Since P1 speaks at most 5 languages, one of these is spoken at least by 199 of the 992. Then the language is spoken at least by 199+1=200 persons, including P1.
2. Suppose there exists such a map. Since the degree of each vertex is even, we can color the plane red or blue so that countries with a common boundary are colored differently. Let the outside of the pentagon be colored red, and suppose r and b are the numbers of red and blue triangles, respectively. We count the number of edges in two ways: Every blue triangle is bounded by three edges. In this way the edges are counted exacly once, that is, k 3b.
Theredcountriesareboundedbyk 3r+5edges.Thus3b 3r+5,acontradiction.
3. Let n > 4. One vertex v1 can be chosen in n ways. We connect its neighbors by a diagonal d1. The next diagonal can be chosen in 2 ways: d2 v3en, or d2 e2en−1.
Similarly, we can choose each of the diagonals d3, . . . , dn−3 in two ways. Thus there are n · 2n−4 ways to choose vertex v1 and the diagonals d1, · · · , dn−3. Each such triangulation contains triangles belonging to two neighboring sides of the n-gon.
Hence we have counted each triangulation twice. The final result is n · 2n−5. For n 4, the formula is also correct.
4. Every person is represented by a point in the plane. Two points are joined by a line, if the corresponding persons know each other. We get a graph with vertices as persons and edges as acquaintances.
We proceed by contradiction. Suppose that every vertex A is joined with each of the 16 others either directly or via a third person. A is joined by edges with exactly four other vertices, of which each is joined with exactly three additional vertices. Thus in the graph there are no additional vertices, and all 17 vertices are distinct. All other edges, of which there are 17 · 4/2 −16 18, can join only outer points in Fig. 14.4.
Everyone of these 18 edges defines a cycle through A consisting of 5 edges. Because of the arbitrary choice of A, 18 such cycles also pass through each of the other 16 points. Each cycle passes through 5 vertices. Hence there are altogether 18 · 17/5 cycles. But this is impossible, since the number of cycles is an integer.
5. The answer is n 33. It is easy to check that 9 points are joined by 36 edges. If 33 edges are colored, then 3 edges remain uncolored. Choose 3 points of the 9 which are endpoints of the three uncolored edges. Then the remaining 6 points are joined with colored edges. We will show that among them there exists a monochromatic triangle. Choose any of the 6 points, say A. Of the 5 edges with endpoint A, at least 3 have the same color, for instance AB, AC, AD. Then one of the four triangles ABC, ACD, ABD and BCD is monochromatic.
On the other hand, there exists a coloring of 32 edges (Fig. 14.5, where the thick lines are red, and the thin lines are blue) without a monochromatic triangle. Hence n 33 is the minimum number of edges, such that, for any of their coloring with two colors, there exists a monochromatic triangle.
14.6 Integer Functions 385 q q q q q q q q q q q q q q q q q A Fig. 14.4 B B B B B B B B B B B B """" " b b b b b P P P ppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp Z Z Z Z Z Z Z !!!!!!!!
!
a a a a a a a a a B B B B B B B B S S S S S S S Z Z Z Z Z Z Z X X X X X X X Fig. 14.5 b A c D B a C d Fig. 14.6 @ @ @ @i i i i i i i i D c C b B a d A Fig. 14.7 6. Start at any vertex and go in the direction of the arrows, until you come for the first time to a vertex you have already visited. Thus, we get a circuit C, which separates the surface of the polyhedron into a right part and a left part. Show by finite descent that there is a face in each part, which can be traced in the direction of the arrows.
7. Consider the subgraph of the red segments. Place a hiker at each of the n vertices.
First the two hikers at the endpoints of the shortest segment exchange their places.
Then the hikers now at the endpoints of the second-shortest segment exchange places, then at the endpoints of the third-shortest segment, etc. up to the longest segment.
Since each of the r segments is traversed by exactly two hikers, the hikers have walked 2r segments altogether. Hence at least one of the n hikers has traversed ≥2r/n segments.
Since the path of each hiker consists of contiguous segments of increasing length, we have proved the existence of a path of at least m red segments of increasing length.
8. Suppose A and B do not know each other. Let C and D be any two other persons.
The C and D must know each other, since one of A,B,C,D knows the other three by hypothesis. So if there is a third person C who does not know everyone, it must be A or B he does not know. If there were a fourth person D who did not know everyone, it would again be A or B he did not know, but then {A, B, C, D} would violate the hypothesis. Hence all except at most three persons must know everyone else.
9. Translate the problem in Fig. 14.6 into the graph in Fig. 14.7. In this graph neighbors can be reached by one move of the knight. The knights move in a clockwise direction in 16 moves to their final exchanged positions. The minimum 16 becomes obvious.
10. We assume the opposite: any pair in S with |S| 2n has in S an odd number of common friends. Let A be one of these persons, and let M {F1, . . . , Fk} be the set of his friends. We prove the following: Lemma. The number k is even for every A.
Indeed, for every Fi ∈M, we consider the list of all his friends in M. The sum of all entries in all k lists is even, since it equals twice the number of pairs in M, and the number of persons in each list is odd by the lemma. Thus k is even.
Let k 2m. Now we consider, for every Fi ∈M, the list of all his friends, except A (not only in M). Every list contains by the lemma (applied to Fi) an odd number of persons. Hence the sum of all entries in all 2m lists is even. But then at least one of the (2m −1) persons (except A) appears in an even number of lists, that is, this person has an even number of common friends with A.
This contradiction proves that at least two persons in S have an even number of common friends.
386 14. Further Strategies 11. Let w1, . . . , w2n+1 be the integral weights. Since any 2n of the weights balance, the sum of any of the 2n weights must be even. This implies that all the weights have the same parity. If they are even, we set wi ←wi/2, if they are odd we set wi ← (wi−1)/2. In each case we get a new set of weights with the same balancing property.
Applying this reduction repeatedly, we see that the wi are congruent mod 2k for all k. This implies that all wi are equal.
Generalize the result to rational weights, which is easy, and to irrational weights, which is more difficult.
12. (a) Suppose a square Q is partitioned into different squares. Then the smallest square cannot touch the boundary of the square.
(b) Suppose a cube C has a decomposition into different subcubes Ci. Let Q be the bottom of C. The subcubes standing on Q generate a partition of Q into different subsquares. Let Q1 be the smallest of these squares, and let C1 be the corresponding cube. Now Q1 is surrounded by larger squares. The corresponding cubes form a “well” and C1 lies in the bottom of the well. No other cube will fit into this well.
(c) Construct on C1 an infinite tower of ever smaller cubes. Contradiction!
13. Use infinite descent.
14. In (a) and (c) use infinite descent. (c) is nontrivial but easily accessible. In (b) any three numbers x, y, z satisfying the equation are divisible by 13.
15. The hints should suffice to solve the problem.
16. Use infinite descent.
17. This sounds like a problem that can be treated by invariance. Starting with some distribution around the circle and using the transformation, we get a sequence 011101000 →011000111 →010110110 →000100101 · · · .
A superficial look does not reveal an invariant. So we think of the strategy of working backwards. It is often applicable if invariance does not work. Suppose the aim is attainable. We have 9 ones (for the first time). One step before we must have 9 zeros, and still one step before we have 9 changes 0 −1−0 −1−· · ·. With an odd number (9), this is not possible.
18. This is a difficult problem. It is not clear how to tackle it, except that we think of Jacobi’s dictum: You must always invert! Instead of putting the weights on the pans we take them from the pan. Instead of the word W=RRL. . . RRL we must generate the transposed word W T =LRR. . . LRR. This is considerably simpler. We can assume that W T starts with L. In the other case, we interchange the pans. Onto the left pan we put m1 > m3 > m5 > · · · and onto the right pan m2 > m4 > m6 · · ·. Then initially the left pan outweighs, independently of where the lightest weight is. If, during the removing process, the scale changes, then take away the heaviest weight; otherwise take away the lightest weight! The realization of a word is generated as follows: change of letters →remove heaviest weight. No change→remove lightest weight. For example, W=RLLRRRLRRL→W T LRRLRRRLLR. With m1 > m3 > m5 > m7 > m9 on the left pan and m2 > m4 > m6 > m8 > m10 on the right pan, W T leads to the sequence of removals m1, m10, m2, m3, m9, m8, m4, m7, m5, m6.
One gets W by taking away the weights in reverse order.
14.6 Integer Functions 387 19. Let n be arbitrary. Suppose it is possible to pour all the water into one glass. We may assume that the total amount of water is 1 and the number of steps is m. Let us work backwards. At the (m −1)th step, we have the distribution 1 2, 1 2 . At the (m −k)th step, we have x/2a, y/2b, . . . , z/2c .
What did we have at the preceding step? Number the glasses arbitrarily. Suppose we are pouring from the second into the first glass. There are two possibilities: (a) The second glass becomes empty. Then, in the preceding step, we had x 2a+1 , x 2a+1 , . . . , z 2c .
(b) After pouring into the first glass, there remains something in the second glass.
Then, in the preceding step, we had x 2a+1 , y 2b + x 2a+1 , . . . , z 2c .
In both cases the denominator has the form 2k. Especially these denominators were also present before the first pouring, i.e., at the start, that is, n 2k.
20. Extending the problem to the right is not a good idea. Either 7,3,6,7 will quickly come up. Then it is not a good Olympiad Problem. Anyone can do it. Here you should think of Jacobi′s motto: You must always invert! This motto obviously suggests an extension to the left. This can be done uniquely. Indeed, the preceding eight digits are 7,3,6,7,3,9,5,4,1,9,9,3,2,3,7. Among these there are the digits we are looking for. But will they come again? There are 104 possible quadruples of digits.
At the (104 + 1)th quadruple, we have a repetition. Thus we have a period. Since the sequence 1, 9, 9, 3 can be extended uniquely in both directions, we have a pure period, which contains very late the quadruple 7, 3, 6, 7.
21. Construct the sequence backwards. The last term must be 1 or n, and each subsequent term must be either the largest or smallest of those numbers left, that is, in each position, except the first, there are two choices, and in total there are 2n−1 such sequences.
22. Replace √r by −√r.
23. Taking the conjugate numbers, we get (x −y √ 5)4 + (z −t √ 5)4 2 − √ 5. The left side is positive, whereas the right side is negative.
24. (1 + √ 2)n+1 xn+1 + yn+1 √ 2 (1 + √ 2)(xn + yn √ 2) xn + 2yn + (xn + yn) √ 2.
Thus, we get xn+1 xn + 2yn, yn+1 xn + yn, and x2 n+1 −2y2 n+1 (xn + 2yn)2 − 2(xn + yn)2 −(x2 n −2y2 n) (−1)n+1.
25. bn −an √ n + 2−√n−( √ n + 1− √ n −1) 2/( √ n + 2+√n)−2/( √ n + 1+ √ n −1) < 0, since √ n + 2 + √n > √ n + 1 + √ n −1.
26. an n (n2 + 1 −n2)/( √ n2 + 1 + n) n/( √ n2 + 1 + n) →1 2 for n →∞.
27. We use the transformation √ 4n + 2 −√n − √ n + 1 2n + 1 −2√n(n + 1) √ 4n + 2 + √n + √ n + 1 1 √ 4n + 2 + √n + √ n + 1 · 1 2n + 1 + 2√n(n + 1) ≤ 1 2√n + √n + √n (2n + 2n) 1 16n√n.
388 14. Further Strategies 28. By adding the small number ( √ 50 −7)100 < 0.075100 < 0.1100 10−100, we get a positive integer. Thus, the first 100 decimals of ( √ 50 + 7)100 are nines.
29.
7 (2 + √ 5)p8 −2p+1 (2+ √ 5)p+(2− √ 5)p−2p+1. Indeed, −1 < (2− √ 5)p < 0, so the addition of this negative number with absolute value less than 1 can also be achieved by “flooring.” For the left-hand side we get a sum of integers each of which contains the factor p i , which, for i 2, . . . , p −1, is divisible by p.
30.
7 (2 + √ 3)n8 (2+ √ 3)n+(2− √ 3)n−1 xn+ √ 3yn+xn− √ 3yn−1 2xn−1.
31.
7 1 + √ 3 n8 ⎧ ⎨ ⎩ 1 + √ 3 n + 1 − √ 3 n if n is odd, 1 + √ 3 n + 1 − √ 3 n −1 if n is even.
For even n, the left-hand side is odd, since the sum of two conjugate numbers is even. Subtracting 1, we get an odd number. Thus we need only consider the case n 2m + 1.
With (2+ √ 3)m xm+ √ 3ym, (2− √ 3)m xm− √ 3ym, after routine computations, we get (1 + √ 3)2m+1 + (1 − √ 3)2m+1 2m+1(xm + 3ym).
It is easy to prove by induction that x2 m −3y2 m 1. Now xm + 3ym is odd. Indeeed, (xm +3ym)(xm −3ym) x2 m −9y2 m x2 m −3y2 m −6y2 m 1−6y2 m. Since the product is odd, both factors on the left side must be odd.
32. (a) Denote m ⌊n √ 2⌋and n √ 2−m {n √ 2}. Since m ̸ n √ 2, we have m < n √ 2 and m2 < 2n2. Hence 1 ≤2n2−m2 (n √ 2−m)(n √ 2+m) {n √ 2}(n √ 2+m) < {n √ 2}2n √ 2, n √ 2 −⌊n √ 2⌋> 1/(2n √ 2).
(b) With n1 m1 1 and ni+1 3ni + 2mi, mi+1 4ni + 3mi, we get two sequences satisfying 2n2 i −m2 i 1 for all n ∈N. Choose an ni0 n such that n > (1 + 1/ϵ)/2 √ 2. Then ϵ(2n √ 2 −1) > 1, (1 + ϵ)(2n √ 2 −1) > 2n √ 2. With m mi0, we conclude that 1 + ϵ 2n √ 2 > 1 n √ 2 + m n √ 2 −m {n √ 2}.
33. x1 1 + √ 2 + √ 3, and its conjugates x2 1 + √ 2 − √ 3, x3 1 − √ 2 + √ 3 and x4 1 − √ 2 − √ 3 are the solutions of the fourth degree equation x4 −4x3 − 4x2 + 16x −8 0 with integral coefficients. There is no equation of lower degree, since two squarings are needed to get rid of √ 2 and √ 3.
34. x 3 √ 5 + 2 − 3 √ 5 −2 p −q ⇒x3 p3 −q3 −3pq(p −q). This reduces to x3 + 3x −4 0 with the only real solution x 1.
35. a, b, √a + √ b ∈Q ⇒a −b ∈Q, a −b √a − √ b ∈Q ⇒√a + √ b, √a − √ b ∈Q ⇒2√a 0, 2 √ b ∈Q.
36. Let √a + √ b + √c r be rational. Then √a + √ b r −√c. Squaring, we get a + b + 2 √ ab r2 −2r√c + c, or 2 √ ab r2 + c −a −b −2r√c.
(1) 14.6 Integer Functions 389 Squaring once more, we get 4ab (r2 +c−a −b)2 +4r2c−4r(r2 +c−a −b)√c.
Thus, 4r(r2 + c −a −b)√c (r2 + c −a −b)2 + 4r2c −4ab.
(2) For r(r2 + c −a −b) ̸ 0, the equation (2) implies √c (r2 + c −a −b)2 + 4r2c −4ab 4r(r2 + c −a −b) .
Hence, √c is rational. The cases of the vanishing denominator are routine and are left to the reader. For symmetry reasons, we conclude that √a and √ b are also rational.
37. Suppose 3 √ 2 a+b√r. Then 2 a3+3ab2r+b(3a2+b2r)√r. Since the right side is rational, we have b(3a2 + b2r) 0. Thus, either b 0, which is a contradiction, or 3a2 + b2r 0. The last term is not negative. Thus a b 0. Again this is a contradiction.
38. n 1: √ 2 −1, n 2 : ( √ 2 −1)2 3 −2 √ 2 √ 9 − √ 8, n 3 : ( √ 2 −1)3 √ 50 − √ 49.
We suspect that, for even n, we have ( √ 2−1)n √ A2 − √ 2B2 with A2 −2B2 1, and for odd n, we have ( √ 2 −1)n √ 2B2 − √ A2 with 2B2 −A2 1, that is, A2 −2B2 (−1)n for all n ∈N. Indeed, suppose n is even and A2 −2B2 (−1)n.
Then ( √ 2 −1)n+1 ( √ 2 −1)(A −B √ 2) (−A −2B) + (A + B) √ 2. From A2 −2B2 (−1)n, we get (A + 2B)2 −2(A + B)2 −(A2 −2B2) (−1)n+1.
39. By adding the small number ( √ 1978−44)20 < 0.520 < 10−6, thus the sixth decimal is 9.
40. (a) Setting 1 1 + 3 √ 2 + 2 3 √ 4 a + b 3 √ 2 + c 3 √ 4, multiplying by the denominator, and comparing coefficients, we get a + 4b + 2c 1, a + b + 4c 0, 2a + b + c 0 with solutions a −3/23, b 7/23, c −1/23. Thus the rationalized fraction is (−3 + 7 3 √ 2 − 3 √ 4)/23.
(b) From (1 − 4 √ 2 + 2 √ 2 + 4 √ 8)(a + b 4 √ 2 + c √ 2 + d 4 √ 8) 1, we get a+2b+4c−2d 1, −a+b+2c+4d 0, 2a−b+c+2d 0, a+2b−c+d 0 with solutions a 9/167, b 15/167, c 25/167, d −14/167. Thus, the rationalized fraction is 9 + 15 4 √ 2 + 25 √ 2 −14 4 √ 8 167 .
41. √ 2 −m n √ 2 + m n (2n2 −m2)/n2 ≥ 1 n2 ⇒ √ 2 −m n ≥1/ ( √ 2 + m n )n2 > 1/(2 √ 2n2).
42. (a) Let S be the set of 1018 numbers a + b √ 2 + c √ 3, with each of a, b, c ∈ {0, 1, . . . , 106−1}, and let d (1+ √ 2+ √ 3)106. Then each x ∈S is in the interval 0 ≤x < d. This interval is partitioned into (1018 −1) intervals (k −1)e ≤x < ke with e d/(1018 −1) and k taking on the values 1, 2, . . . , 1018 −1. By the box principle, two of the 1018 numbers of S must be in the same interval, and their difference r + s √ 2 + t √ 3 gives the desired r, s, t since e < 10−11.
390 14. Further Strategies (b) None of the four numbers of the form Fi a ± b √ 2 ± c √ 3 is ever zero. Their product P is an integer. Indeed, the mappings √ 2 →− √ 2 and √ 3 →− √ 3 do not change P . Thus, P does not contain these radicals any more. Hence, |P | ≥1. Then |F1| ≥1/|F2F3F4| > 10−21 since |Fi| < 107, and thus, 1/|Fi| > 10−7 for each i.
43. Set q 4 √ 5, q4 5. Then (L/2)2 1 4−3q+2q2−q3 a + bq + cq2 + dq3.
Multiplying with the denominator and comparing coefficients on both sides, we get 4a −5b + 10c −15d 1, −3a + 4b −5c + 10d 0, 2a −3b + 4c −d 0, −a+2b−3c+4d 0. Solvingfor a, b, c, d, we geta 1 4, b 1 2, c 1 4, d 0.
Thus, (L/2)2 1 4(1 + 2q + q2) 1 4(1 + q)2, or L 1 + 4 √ 5.
44. It is pretty hopeless to iterate the quadratic explicitly, but there is similarity to some duplication formula. Indeed, set x 4 sin2 α. Then f (x) f (4 sin2 α) 16 sin2 α−16 sin4 α 16 sin2 α(1−sin2 α) 16 sin2 α cos2 α (4 sin α cos α)2 (2 sin 2α)2 4 sin2 2α. For 0 ≤x0 ≤4, we have 0 ≤α ≤π/2. Thus, we have x0 4 sin2 α, x1 4 sin2 2α, x2 4 sin2 4α, . . . , xn 4 sin2 2nα. Complete the details.
45. We will prove that x y z u v. Let v > u. Then the first two equations imply x < u and even more x < v. From the second and third equations we get y > v and even more y > x. From the third and fourth equations we get z < x and even more y > z. From the fourth and fifth equations we get u > y. From u > y, y > v we get u > v, a contradiction. Hence u=v. Because of cyclic symmetry, the same is valid for all other variables. Thus we have (3x)3 3x with the solutions x 0 and x ± 1 3.
46. No solution.
47. Multiplying the second equation by the imaginary unit i and adding, we get eix + eiy + eiz 3( √ 3 2 + 1 2i) 3(cos 30◦+ i sin 30◦) 3eiπ/6.
Since the sum of the three unit vectors on the left side has absolute value 3, all three vectors have the same direction 30◦. Hence x y z π/6 + 2kπ.
48. From this system, we first get x1 x10, . . . , x5 x6. From (x1 + · · · + x5)(x5 + · · · + x10) 1, we get (x1 + · · · + x5)2 < 1 and x1 + · · · + x10 < 2. Instead of an algebraic solution, we try a geometric interpretation.
On a straight line, we take segments |A0A1| x1, . . . , |A9A10| x10. Since A0A10 < 2, we can construct an isosceles triangle A0A10B with A0B A10B 1.
Let α ̸ BA0A1 ̸ BA10A9. Since |A0A1| · |A0A10| |A0B| · |A10B|, |A0A1|/|A0B| |A10B|/|A0A10|, the triangles A0A1B and A0BA10 are similar, and A0BA1 α. In the same way we conclude that △A0BA2 ∼△BA1A10. Hence ̸ A0BA2 ̸ BA1A1 2α and ̸ A1BA2 α. In general, for each k, the tri-angles A0BAk and BAk−1A10 are similar. Hence ̸ A0BA10 is divided by the rays BA1, . . . BA9 into equal angles α. Thus, (10 + 2)α 180◦, α 15◦. By the Sine Law, with a √ 2, b √ 6, c √ 3, we find x1 sin α sin 2α b −a 2 , x1 + x2 sin 2α sin 3α a 2 , x1 + x2 + x3 sin 3α sin 4α a c , x1 + · · · + x4 sin 4α sin 5α 3a −b 2 , x1 + · · · + x5 sin 5α sin 6α b + a 4 .
14.6 Integer Functions 391 Again we get with a √ 2 and b √ 6, x1 b −a 2 , x2 2a −b 2 , x3 b 3 −a 2 , x4 9a −5b 6 , x5 3b −5a 4 .
In addition, we know that x6 x5, x7 x4, x8 x3, x9 x2, x10 x1.
Similarly, we can solve the problem for any n ∈N. The result will depend on trigonometric functions of the angle π/(n + 2).
49. Let a1 ≥a2 ≥· · · ≥a10, x1 ≥x2 ≥x3 ≥x4 ≥x5. We observe that a1+· · ·+a10 4(x1 + · · · + x5). Hence we know S x1 + x2 + x3 + x4 + x5. Then a1 x1 + x2, a2 x1 + x3, a10 x4 + x5, a9 x3 + x5, x3 S −a1 −a10, x1 a2 −x3, x5 a9 −x3, x2 a1 −x1, x4 a10 −x5.
50. In (a) we get 17 equations in 25 variables, which is easily satisfied. In (b) we get 25 equations with 25 variables. This can be satisfied if the rank of the matrix is 25. Try to prove that the system is contradictory. So there is no solution. You may set the 16 negative sums equal to −1, the 9 positive sums equal to +1, and try to solve the system with Derive.
51. f (0) + g(0) 0, f (0) + g(1) 1, f (1) + g(0) 1, f (1) + g(1) 3. Adding the first equation to the fourth, we get f (0) + g(0) + f (1) + g(1) 3. Adding the second equation to the third, we get f (0) + g(0) + f (1) + g(1) 2. Contradiction!
52. Consider the polynomial P (t) (t −x1) · · · (t −xn) tn + a1tn−1 + · · · +an. Then 0 P (x1) + · · · + P (xn) (xn 1 + · · · + xn n) + a1(xn−1 1 + · · · + xn−1 n ) + · · · + nan, that is, n + na1 + · · · + nan 0 nP (1). This equation implies that one of the xi (say x1) is 1. Then for x2, . . . , xn, we get an analogous system. By finite descent, all xi are 1.
53. T ◦T (A) T 2(A) (a1a2 2a3, a2a2 3a4, . . . , ama2 1a2) (a1a3, a2a4, · · · , ama2), T 2 ◦T 2(A) T 4(A) (a1a2 3a5, a2a2 4a6, . . . , ama2 2a4) (a1a5, . . . , ama4), and finally, T 2m(A) (a1a1, a2a2, . . . , amam) (1, 1, . . . , 1).
54. Let x1 be a largest solution. Then x2 and xn are smallest solutions, x3 and xn−1 are largest, and so on. Thus x1 x3 · · · xn−1, x2 x4 · · · xn, that is, 1−x3 1 x2, 1−x3 2 x1, or x2 −x1 x3 2 −x3 1. If x1 ̸ x2 then x2 1 +x1x2 +x2 2 1.
But 1 x2 1 + x1x2 + x2 2 x2 1 + x2(x1 + x2) ≥x2 1 + x2 ≥x3 1 + x2 1, that is, x1 1, x2 0. This implies that either all the solutions are equal or they are alternately 1 and 0. We must still solve the equation x3+x−1 0, where 0 < x < 1.
Set x 1 √ 3 y −1 y . We get y3 − 1 y3 3 √ 3, that is, y 3 ( √ 27 + √ 31)/2.
55. We first observe that none of the variables can be zero. Then the second equation is equivalent to xy+yz+zx 0. Now 0 (x+y+z)2 x2+y2+z2+2(xy+yz+zx).
From this we conclude that x2 + y2 + z2 0. This is a contradiction.
56. Let h(x) (ax + b)/(−bx + a), k(x) (cx + d)/(−dx + c). Then h ◦k(x) (ac −bd)x + (ad + bc) −(ad + bc)x + (ac −bd).
392 14. Further Strategies Hence, the composition of two functions of this form are computed as products of complex numbers a + bi and c + di. To the given function f (x) x √ 3 2 −1 2 / 1 2x + √ 3 2 corresponds the complex number z √ 3 2 −1 2i cos −π 6 + i sin −π 6 e−πi/6.
Hence, g(x) corresponds to the complex number z1994. Now z1994 e(−πi/6)1994 e−πi/3 cos 60◦−i sin 60◦ 1 2 −i √ 3 2 .
Finally, we get g(x) (x − √ 3)/( √ 3x + 1).
57. We have |x| < 1, since |x| ≥1 implies 2x2 −1 ≥1, 8x4 −8x2 + 1 ≥1.
Hence we can set x cos t, 0 < t < π. 2x2 −1 2 cos2 t −1 cos 2t, 8x4−8x2+1 2(2x2−1)2−1 2 cos2 2t−1 cos 4t, 8 cos t·cos 2t·cos 4t 1.
Multiplying the last equation with sin t, we get sin 8t −sin t 0. This implies 7t 2πk, k 1, 2, 3, or 9t π + 2πk. Hence, t π/9 + 2/9πk, k 0, 1, 2, 3, x cos 2/7π, cos 4/7π, cos 6/7π, cos π/9, 1/2, cos 5π/9, cos 7π/9.
58. The first equation reminds us of cos2 t + sin2 t 1, 0 ≤t < 2π. We set x cos t, y sin t. Now the second equation reminds us of trigonome-try. Its left side becomes the triplication formula for cos t, and its right side has the form of a half-angle formula. Indeed, cos 3t 4 cos3 t −3 cos t. We get cos 3t √(1 + sin t)/2; cos 3t ≥0. Because cos 3t ≥0, we may square both sides, and we get cos2 3t 1 + sin t 2 ⇒2 cos2 3t −1 sin t ⇒cos 6t sin t, cos 6t cos(π 2 −t) ⇒6t π 2 −t + 2πk, 6t 2πk − π 2 −t .
We get t1 π/14, t2 9π/14, t3 17π/14, t4 7π/10, t5 3π/2, t6 19π/10.
The other six t-values give cos 3t < 0. The cosine and sine values of these angles give the corresponding x and y values.
59. For 0 < x < 1, there is no solution since the LHS is smaller than 1. For x > 1, there is only one solution since the function f (x) xx is monotonically increasing: if a > b > 1, then aa > ab because the exponential y ax is increasing, but ab > bb because the power function y xb is increasing. Let y x1996 or x y1/1996. Then y1/1996 19961/y, or yy 19961996. We guess y 1996, x 19961/1996 ≈1.00381 as the only solution.
60. If 0 ≤x < 1/n then the equation is correct since both sides are 0. Now suppose that x is arbitrary. If we increase x by 1/n, each of the terms on the left side will shift by one place, except the last one, which becomes the first one increased by 1. The right side also increases by 1. From here it is easy to conclude that the equality holds for any x.
14.6 Integer Functions 393 61. In {1, 2, . . . , n} exactly ⌊n/k⌋integers are divisible by k. Thus, the right sum counts the number of integers divisible by 1, 2, . . . , n. The left side does the same.
62. The sum of integers divisible by k is k⌊n/k⌋. The right side counts the sum of the divisors of the integers from 1 to n. The left side does the same.
63. Consider all the lattice points with 1 ≤x ≤q −1, 1 ≤y ≤p −1. They lie inside the rectangle OABC with sides |OA| q, |OC| p in Fig. 14.8. Draw the diagonal OB. None of the lattice points considered lies on this diagonal. This would contradict gcd(p, q) 1. We count the lattice points below the diagonal OB in two ways. On the one hand, their number is (p −1)(q −1)/2. On the other hand, it is also q−1 k1 ⌊kp/q⌋, that is, p−1 k1 ⌊kq/p⌋ (p −1)(q −1)/2.
-6 pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp O A B C x y Fig. 14.8 64. √n + √ n + 1 < √ 4n + 2 ⇔2n + 1 + √ 4n2 + 4n < 4n + 2 ⇔ √ 4n2 + 4n < 2n + 1 ⇔4n2 + 4n < 4n2 + 4n + 1. This proves that √n + √ n + 1 < √ 4n + 2, or ⌊√n + √ n + 1⌋≤⌊ √ 4n + 2⌋. Suppose that, for some positive integer n, ⌊√n + √ n + 1⌋̸ ⌊ √ 4n + 2⌋. Let q ⌊ √ 4n + 2⌋. Then √n+ √ n + 1 < q ≤ √ 4n + 2.
Squaring, we get 2n+1+ √ 4n2 + 4n < q2 ≤4n+2,or √ 4n2 + 4n < q2−2n−1 ≤ 2n+1. Squaring again gives 4n2+4n < q2 −2n −1 2 ≤4n2+4n+1. Since there is no square strictly between two successive integers, we have q2 −2n−1 2n+1, or q2 4n + 2, or q2 ≡2 mod 4. This is a contradiction.
65. We note that c a + b; otherwise, for c ̸ a + b and large n the condition ⌊an⌋+ ⌊bn⌋ ⌊cn⌋would not be satisfied. For n 1, we get ⌊a⌋+ ⌊b⌋ ⌊c⌋. We can assume that 0 ≤a < 1, 0 ≤b < 1 and c a + b < 1, that is, ⌊an⌋+ ⌊bn⌋ ⌊(a + b)n⌋implies that only one of a, b is nonzero.
Assume the contrary, and express a and b in the binary system a 2−a1 + · · · + 2−as, b 2−b1 + · · · + 2−bt , where ai, bj ∈N are arranged increasingly, and assume that bt ≥as. Choose n 2bt −1. The right side of ⌊an⌋+ ⌊bn⌋ ⌊(a + b)n⌋becomes ⌊n(a + b)⌋ 9 s i1 2bt −ai + t j1 2bt −bj −(a + b) : s i1 2bt −ai + t j1 2bt −bj −1 394 14. Further Strategies (because a + b < 1), whereas the left side is ⌊na⌋+ ⌊nb⌋, or 9 s i1 2bt −ai −a : + 9 t j1 2bt −bj −b : s i1 2bt −ai −1 + t j1 2bt −bj −1.
Clearly ⌊n(a + b)⌋̸ ⌊na⌋+ ⌊nb⌋, which proves the statement.
66. Set an (3+ √ 11)n +(3− √ 11)n. Then an+2 6an+1 +2an, n ∈Z+. Indeed, with x (3 + √ 11)n, y (3 − √ 11)n, we have an x + y, an+1 (3 + √ 11)x + (3 − √ 11)y, an+2 (3 + √ 11)2x + (3 − √ 11)2y (20 + 6 √ 11)x + (20 −6 √ 11)y (18 + 6 √ 11)x + (18 −6 √ 11)y + (2x + 2y) 6an+1 + 2an. From a0 2, a1 6, we conclude that an is an integer for any n ∈Z+. Since −1 < 3 − √ 11 < 0, for any n ∈N, we have a2n−1 (3 − √ 11)2n−1 + (3 − √ 11)2n−1 < (3 + √ 11)2n−1 < a2n−1 + 1, that is, a2n−1 7 (3 + √ 11)2n−18 . Now we can prove by induction that a2n−2, and a2n−1, are divisible by 2n but not by 2n+1.
67. Suppose that an is even, an 2kq, where q is odd. Then an+k 3kq is odd. But if an is odd, an 2kq + 1, then an+k 3kq + 1 is even. This implies the result.
68. Suppose bn is periodic with period t, starting with some n0. Then an+t −an is even starting with some n0. On the other hand, it is equal to 3 2 n−n0 an0+t −an0 .
For large n the last number is odd. Contradiction.
69. We have {an + b} an + b mod 1. So this number lies in [0, 1). Twice this number lies in [0, 2). Hence pn consists of zeros and ones. By considering the sequence an+b on a circle of perimeter 1, the reduction modulo 1 is performed automatically. If the terms lie in the upper half circle in Fig. 14.9, pn is zero, if they lie in the lower half, pn will be 1. If the sequence pn contains many zeros in a row, then a mod 1 must be small. Many zeros will be followed by many ones. So not all binary words will occur.
For k 5, the word 00010 will not occur. Indeed, three zeros in a row means that, after reduction, mod 1 |a| < 1 4. The subword 010 signifies that, from the upper half, we get to the lower half and then to the upper half. This means that |a| > 1 4. This contradiction proves that, for k 5, the answer to the question is no!. By simple checking, we confirm that, for k 4, every one of the 16 words 0000, . . . , 1111 will occur for suitable a, b.
0 1 0 1 sb sb + a s b + 2a Fig. 14.9 14.6 Integer Functions 395 70. a+b < 2(a2 + b2), (a ̸ b) ⇒√n+ √ n + 2 < 2 √ n + 1. Thus, √n+ √ n + 1+ √ n + 2 < 3 √ n + 1. We prove that the left side of this inequality is > √ 9n + 8.
For this we must show that (we use the AGM inequality) 3 6 √n(n + 1)(n + 2) > √ 9n + 8 ⇔243 −270n −512 > 0. For n ≥3, this is obvious. For n 1 and n 2, we check √ 1 + √ 2 + √ 3 > 17 and √ 2 + √ 3 + √ 4 > √ 26. This proves the result.
71. Prove that ⌊ 3 √n + 3 √ n + 2 3⌋+ 1 8n + 8 for n ≥3. It suffices to prove that 8n + 7 < 3 √n + 3 √ n + 2 3 < 8n + 8 for n ≥3 is equivalent to 6n + 5 < 3 3 n2(n + 2) + 3 n(n + 2)2 < 6n + 6, which is a straightforward computation.
72. Let a 3 + √ 5, b 3 − √ 5, a + b 6, ab 4. Then xn an + bn satisfies the recurrence xn+2 6xn+1 −4xn, n ≥1. Since x1 6, x2 28, we have 21|x1, 22|x2. Suppose 2n|xn, 2n+1|xn+1, or xn 2np, xn+1 2n+1q. Then we have xn+2 6 · 2n+1q −4 · 2np, or xn+2 2n+2(3q −p). Since 0 < (3 − √ 5)n < 1, and xn is an integer, we have xn ⌊(3 + √ 5)n⌋+ 1.
References E.J. Barbeau, Polynomials, Springer-Verlag, New York, 1989.
E.J. Barbeau, M.S. Klamkin, O.J. Moser, Five Hundred Mathematical Chal-lenges, Mathematical Association of America, Washington, DC, 1995.
O. Bottema et al., Geometric Inequalities, Nordhoff, Groningen, 1969.
D. Cohen, Basic Techniques of Combinatorial Theory, John Wiley & Sons, New York, 1978.
H.S.M. Coxeter, S.L. Greitzer, Geometry Revisited, NML-19, Mathematical Association of America, Washington, DC, 1967.
H. D¨ orrie, 100 Great Problems With Elementary Solutions, Dover, New York, 1987.
A. Engel, Exploring Mathematics With Your Computer, NML-35, Mathemat-ical Association of America, Washington, DC, 1993.
All books by Martin Gardner.
R.L. Graham, N. Patashnik, and D. Knuth, Concrete Mathematics, Addison– Wesley, Reading, MA, 1989.
S.L. Greitzer, International Mathematical Olympiads 1959–1977, Mathe-matical Association of America, Washington, DC, 1978.
D. Fomin, S. Genkin, I. Itenberg, Mathematical Circles, Mathematical Worlds, Vol. 7, American Mathematical Society, Boston, MA, 1966.
398 References D. Fomin, A. Kirichenko, Leningrad Mathematical Olympiads 1987–1991, MathPro Press, Westford, MA, 1994.
G.H. Hardy and E.M. Wright, An Introduction to the Theory of Numbers, Oxford University Press, Oxford, 1983.
D. Hilbert, S. Cohn–Vossen, Geometry and the Imagination, Chelsea, New York, 1952.
All books by Ross Honsberger, Mathematical Association of America, Wash-ington, DC.
M.S. Klamkin, International Mathematical Olympiads 1979–1985, Mathe-matical Association of America, Washington, DC, 1986.
M.S. Klamkin, USA Mathematical Olympiads 1972–1986, Mathematical As-sociation of America„ Washington, DC, 1986.
L. Larson, Problem Solving Through Problems, Springer-Verlag, New York, 1983.
E. Lozansky, C. Rousseau, Winning Solutions, Springer-Verlag, New York, 1996.
D.S. Mitrinovic, Elementary Inequalities, Nordhoff, Groningen, 1964.
D.J. Newman, A Problem Seminar, Springer-Verlag, New York, 1982.
G. Polya, How To Solve It, Princeton University Press, Princeton, NJ, 1945.
G. Polya, Mathematics and Plausible Reasoning, Princeton University Press, Princeton, NJ, 1954.
G. Polya, Mathematical Discovery, John Wiley & Sons, New York, 1962.
H. Rademacher, O. Toeplitz, The Enjoinment of Mathematics, Princeton Uni-versity Press, Princeton, NJ, 1957.
D.O. Shklarsky, N.N. Chentzov, I.M. Yaglom, The USSR Olympiad Problem Book, W.H. Freeman, San Francisco, CA, 1962.
W. Sierpinski, A Selection of Problems in the Theory of Numbers, Pergamon Press, New York, 1964.
R. Sprague, Recreation in Mathematics, Blackie, London 1963.
H. Steinhaus, One Hundred Problems In Elementary Mathematics, Basic Books, New York, 1964.
P.J. Taylor, ed., Tournament of the Towns, Australian Mathematics Trust, University of Canberra, Belconnen ACT.
References 399 B.L. van der Waerden, Einfall und ¨ Uberlegung, Birkh¨ auser, 1973.
A. and I. Yaglom, Challenging Mathematical Problems With Elementary Solutions, Vol. I and Vol. II, Dover, New York, 1987.
Index Algorithm 1, 2–5, 7, 14, 87 Coding 93 Decoding 93 Euclidean 117, 118, 147 Andr´ e, Desir´ e 97 Arithmetic progression 12, 19, 46 Automatic solution 85, 129, 145 Barlotti, A. 301 Bijective proof 87, 89, 95 Binet’s formula 205 Box principle 59–83 de Bruijn, G.N. 241 Carlson’s inequality 175–176 Catalan numbers 99, 101 Cauchy-Schwarz inequality 167, 175, 178, 188, 192, 196, 199–201, 203 Cayley’s formula 92, 116 Characteristic equation 107, 221–222, 231–232, 243 Chebyshev inequality 168, 186, 188, 198, 203 Coloring proofs 25–37 Combinatorial proof 87 Complex numbers 294–298, 306–309 Concavity 177–179 Conjugate numbers 365–367, 375–377 Convergence rate 223, 229, 231, 232 Convexity 172, 176–179, 197–198, 200 Counting by bijection 89, 95, 99 Descent 125, 129, 157, 374, 386 Difference equations 97–98, 107, 221 Dilworth 57 Dirichlet 59 Discriminant 19, 22 Divide and conquer 87, 94, 105 Enumerative combinatorics 85–116 Erd˝ os 61, 70, 152, 340, 353 Erd˝ os-Mordell inequality 353 Extremal principle 7, 39–57 Fermat’s theorem 119–120, 128, 149 Fermat-Euler theorem 120, 239 Fibonacci sequence 205, 209, 223, 285 Frobenius problem 156–157 Functional equation 221–222, 256, 271–288 Cauchy’s 271 d’Alambert’s 271 Jensen’s 271 402 Index Gallai, T. 43 Game 1, 361–371 Bachet’s 362 Wytthoff’s 362, 366 of Euclid 365 Greedy algorithm 46, 167 Graph 92, 97, 113–115, 384–385 complete 64–65, 77 Hadwiger-Finsler inequality 173–174 Heuristic principle 1, 4, 39 Induction principle 205–220 Inequalities 161–204 Integer part 120, 382–383 Involution 101, 107 Jacobi, C.G.J. 70, 377, 386, 387 Josephus problem 226 Kelly, L.M. 43, 50 Kuczma, M. 241 Lemoine point 306, 356 Morse-Thue sequence 227, 237 Napoleonic triangles 296, 309 Nesbitt’s inequality 163, 169 Number theory 117–159 Parity 2, 14–15, 19–22, 25, 35, 62, 75, 139, 151, 212, 281, 347, 367, 375 Pell-Fermat equation 128, 140, 145, 153 Permutation 88, 96, 99, 227, 237–238, 273 fixed point free 100 Polynomials 245–269 antisymmetric 254 degree 245, 249 elementary symmetric 165, 167, 251, 253–254 monic 246 reciprocal 250–251 symmetric 251–252 Pompeiu’s theorem 332 Position 361 losing 361 winning 361 Principle of inclusion and exclusion (PIE) 99, 102, 114 Probabilistic interpretation 96–97, 210 Product rule 87, 88, 96 Product-sum rule 88, 90, 91 Pr¨ ufer code 93 Ptolemy’s inequality 297, 357 theorem 319, 332 Ramsey, F.P. 59 Numbers 59, 63, 66, 67 Rearrangement inequality 167, 178, 188, 193–194, 203 Reflection principle 99 Recursion 40, 87, 90, 91, 97, 108 Rooks 44–45, 48, 51, 72, 100, 104 Roots 246 of unity 126, 154, 247–249 259–263 multiplicity 246 rational 246 Schur, I. 63, 66, 78 Sequences 221–243 Shrinking squares 7, 15 Sieve formula 99 Stirling numbers 91, 102 Sum free sets 63–67, 70, 78 Sum rule 87, 89, 99 Sylvester 43, 50, 156–157 Symmedian 356 Symmetry 7, 97, 115, 292, 370 System binary 46, 226, 234, 237, 242, 280, 286 ternary 46 Szekeres 61 Tree 92 labeled 92 Transformation geometry 309–317 triangle inequality 42, 51, 164–166, 177–178, 187, 189, 192–193 Trigonometric substitution 177, 179 Van der Waerden 342 Vectors 289–294, 298–299, 301–306 Weierstraß 52, 176 Index 403 Weitzenb¨ ock 170 Weyl, H. 71 Working backwards 15, 377, 386–387 Zeckendorf’s theorem 209, 285 |
11266 | https://www.vedantu.com/content-files-downloadable/revision-notes/cbse-class-11-chemistry-notes-chapter-5-states-of-matter.pdf | Class XI Chemistry www.vedantu.com 1 Revision Notes Class – 11 Chemistry Chapter 5 – States of Matter INTRODUCTION: The substance that contains mass and occupies space is known as Matter. Atoms and molecules constitute the matter. It has various physical and chemical properties. Matters are classified into solid, liquid and gas. The force of interaction present in them varies which gives rise to its physical properties. Gases have the least force of interaction, while solids have the highest force of interaction. 1. INTERMOLECULAR FORCES The molecules like solid, liquid or gas exhibit forces of attraction which are called intermolecular forces. Van der Waal forces are the collective of dipole-dipole, dipole-induced dipole and dispersion forces. Ion-dipole and ion-induced dipole are not Van der Waal forces. Hydrogen bonding is the strongest force of attraction. The different types of intermolecular forces are: 1.1 Dipole-Dipole Interaction Polar molecules exhibit dipole-dipole interactions. They have permanent dipole moments. The positive pole and negative pole are attracted in the molecule. For example, HCl. In HCl molecule, Cl is more electronegative than hydrogen. So, the chlorine atom acquires negative charge while, hydrogen atom acquires positive charge. Thus, dipole-dipole interaction takes place among them. Class XI Chemistry www.vedantu.com 2 1.2 Ion-Dipole Interaction The attraction between the cation, anion, and a polar molecule is known as ion-dipole interaction. For example, NaCl. The polar water molecules are attracted towards Na and Cl on dissolving NaCl in water. 1.3 Ion-induced Dipole Interaction The interaction between the non-polar molecules when a polarized ion is brought near to it is known as ion induced dipole interaction. For example, an iodine molecule which is not polar gets polarized in the presence of a nitrate ion. Class XI Chemistry www.vedantu.com 3 1.4 Dipole induced dipole interaction The interaction between the non-polar molecules when a polarized dipole is brought near to it is known as dipole-induced dipole interaction. For example, in the presence of polar molecules, noble gases get polarized. 1.5 London forces or Dispersion forces For instance, the electron cloud of the molecule gets distorted for the generation of an instantaneous dipole. The momentary dipole is produced in the molecule in which one part of the molecule is more negative than the other part. The momentary dipole produced induces a dipole in the other molecule. Thus, the force of attraction between the induced momentarily dipole is known as London dispersion forces. 2. INTERMOLECULAR FORCES VERSUS THERMAL ENERGY The force of interaction tries to bring the molecules closer. That is, the solid possesses the strongest intermolecular force of attraction, while the gases possess the least intermolecular force of attraction. The decreasing order is, Class XI Chemistry www.vedantu.com 4 Solid Liquid Gas The thermal energy is possessed by the molecule. The kinetic energy helps in the movement of particles. The gas possesses the highest thermal energy, while the solid possesses the least thermal energy. The decreasing order is, Gas Liquid Solid 3. IDEAL GAS An ideal gas is a hypothetical concept. There are various assumptions of an ideal gas. Some of them are: The force of interaction between the molecules is zero. The volume of the molecules is very small. The molecules of the gas collide with each other and with the walls of the container. 4. STATE OF A GAS AND STATE VARIABLE The physical condition of the system is the state of a gas. The variables which are used to denote the physical condition of a gas are known as state variables. They are pressure, volume and temperature , and .
P V T 4.1 Pressure The force exerted on an object per unit area is known as pressure. The force applied is always perpendicular to the object. The unit of pressure is the pascal. Pressure can be measured by various instruments. Barometer and manometer are used to measure pressure. Class XI Chemistry www.vedantu.com 5 4.2 Volume The volume of the gas is the same as the volume of the container in the case of rigid containers. For non-rigid containers, the volume of the gas is determined by the number of moles and other state functions. 4.3 Temperature The amount of heat contained in the gas can be measured by the physical term temperature. No heat flows in and out of the gas when the temperature of the gas is equal to the surrounding temperature. The thermometer is used to measure the temperature of the gas. The units of temperature are Celsius, kelvin, and Fahrenheit. 5. IDEAL GAS LAW The laws relate the state variable of the gas in two states. 5.1 Boyle’s Law It gives the relation between pressure-volume. At a constant temperature, the pressure of a fixed amount of gas is inversely proportional to its volume. The value of the proportionality constant for each curve corresponds to a different constant temperature and is known as isotherm. Mathematically, it can be written as, 1 P V k P V PV k As, PV increases, the corresponding temperature will also increase. Class XI Chemistry www.vedantu.com 6 5.2 Charles's Law It gives the relation between temperature-volume. At constant pressure, the volume of a fixed mass of gas is directly proportional to the absolute temperature. Mathematically, it can be written as, V T V kT V k T 5.3 Gay-Lussac’s Law It gives the pressure-temperature relationship. At fixed volume, the pressure of a fixed mass of gas is directly proportional to the absolute temperature. Mathematically, it can be written as, P T P kT P k T Class XI Chemistry www.vedantu.com 7 T P 5.4 Avogadro’s Law It gives the relation between amount of gas and volume. It states that all the gases of equal volume under the same condition of temperature and pressure contain an equal number of molecules. $V \propto n$ 6. IDEAL GAS EQUATION Combining all the three laws, Boyle’s law: 1 P V Charle’s law: V T Avogardo’s law: V n All the above laws when combined together give an equation, known as an ideal gas equation. Combining all the equations, nT V P nRT V P PV nRT Class XI Chemistry www.vedantu.com 8 The above equation is known as ideal gas equation. Here, P is the pressure, V is the volume, n is the amount of gas, T is temperature and R is universal gas constant. The values of R varies with different units. The values of R are: 1 1 1 1 1 1 8.314 J K mol 0.0821 L atm K mol 2 cal K mol R R R If temperature, volume, and pressure varies from their initial state to final state, then the ideal gases exert increased the pressure difference gas equation can be rewritten as, 1 1 1 PV nR T and, 2 2 2 P V nR T On combining the equation, 1 1 2 2 1 2 PV P V T T The above equation is known as Combined Gas Law. 7. VARIATION OF THE IDEAL GAS EQUATION The ideal gas equation is written as, PV nRT On rearranging the above equation, n P V RT Class XI Chemistry www.vedantu.com 9 Here, n is the number of moles which is equal to the given mass divided by molar mass. The above equation can be written as, m P MV RT d P m d M RT V On rearranging the above equation, PM dRT 8. DALTON’S LAW OF PARTIAL PRESSURE At constant temperature and volume, the total pressure exerted by a mixture of non-reactive gases equals the sum of their partial pressures. The pressure exerted by individual gases is known as partial pressure. Mathematically, it can be shown as, 1 2 3 total p p p p Here, total p is the total pressure and 1 2 3 , , p p p are the partial pressure of the gases. Partial pressure can also be expressed in terms of mole fraction. If there are three gases which exerts partial pressure as, 1 2 3 , , p p p at constant temperature and volume, then the relation between ideal gas equation and partial pressure can be shown as, 1 1 n RT p V 2 2 n RT p V 3 3 n RT p V Class XI Chemistry www.vedantu.com 10 1 2 3 , , p p p are the number of moles of gases respectively. The total pressure exerted can be calculated as, total 1 2 3 p p p p 3 1 2 total total 1 2 3 n RT n RT n RT p V V V RT p n n n V On dividing i P by total P 1 1 1 1 total 1 2 3 1 2 3 i p n n n RTV x p n n n RTV n n n n 1 1 total p x p Similarly, for the other gases, 2 2 total p x p and 3 3 total p x p Thus, the general equation is, total i i p x p 9. GRAHAM’S LAW OF DIFFUSION The process of intermixing of gases is known as diffusion of gases. Gases can mix with each other even without the difference in pressure. The diffusion process becomes faster if the pressure difference increases between the gases. Due to pressure differences, gas flows through small openings. The flow of gas is effusion. According to Graham’s law of diffusion, the rate of diffusion depends on two factors. Class XI Chemistry www.vedantu.com 11 Partial pressure and molecular weight of a gas. The rate of diffusion is directly proportional to the partial pressure and inversely proportional to the square root of the molar mass of the gas. The movement of molecules is faster when the molecules are lighter, and slower when the molecules are heavier. Mathematically, the rate of partial pressure is shown as, P Rate M The rate of diffusion can be explained in various ways like it may be defined as many moles traveled per unit time, distance traveled per unit time, volume transferred per unit time. The rate of diffusion for the two gases can be calculated by the given formula: 2 2 1 1 1 2 r P M r P M 10. KINETIC THEORY OF GASES The postulates of the kinetic-molecular theory of gases are given below. (i)Gases are made up of many similar particles (atoms or molecules). They are so small in size that the actual volume of the molecules is negligible. (ii) At normal temperature and pressure, there is no force of attraction between the particles. (iii) Particles of a gas are always in constant and random motion. They collide with one other and the container walls during their erratic mobility. The collision of the particles with the walls of the container is the total pressure exerted by the gas. (iv) The collisions of gas molecules are perfectly elastic. This means that the total Class XI Chemistry www.vedantu.com 12 energy of molecules remains the same before and after a short-range collision. (v) At any particular time, different particles in the gas have different speeds and hence different kinetic energies. The distribution of speed remains constant even on changing the individual speeds of gases. If the gases will have variable speed, they must also have variable kinetic energy. The average kinetic energy of the gas molecule is directly proportional to the absolute temperature of the gas. Mathematically, the relation is shown as, 3 K.E per mole 2 3 K.E per molecule 2 nRT KT Here, K is known as Boltzmann’s constant. 11. Molecular Distribution of Speed (Maxwell Boltzmann Distribution) A plot of the fraction of molecules in the gas vs the speed of the gas molecule is the Maxwell Boltzmann distribution curve. The graph is shown below. Class XI Chemistry www.vedantu.com 13 Salient features of graphs are: (a) The percentage of molecules with a very high or very low speed is extremely small. (b) Mostly, the molecules possess speed in the middle, which is known as the most probable speed. (c) The total area covered by the graph gives the total number of molecules in the sample. (d) There are two more speeds, root means square speeds and average speeds. Mathematically, they are shown as, rms 3RT M mp 2RT M avg 8RT M Note: Take molecular mass always in kilograms. The ratio for the speeds of the gas at fixed temperature is, mp avg rms : : 1:1.128:1.224.
12. REAL GASES The assumption taken for ideal gases becomes invalid in case of real gases. (i)We thought that in an ideal gas, there are no interactions between molecules. (ii) The volume of a gas’s molecules is negligible compared to the entire volume of gases. Class XI Chemistry www.vedantu.com 14 The molecular interaction cannot be ignored in real gases. They are: Attractive forces with a long range and repulsive forces with a short range. The interactive forces present in real gases are negligible when they are far apart. But, when the molecules are brought closer attractive forces start to develop. On bringing the molecules closer, they start repelling each other. 13. COMPRESSIBILITY FACTOR The compressibility factor is the measurement of the deviation from ideal behavior. It is denoted by .
Z It is represented as, real ideal V Z V Following are the key points: (a) There is no contact between the molecules when the pressure is exceedingly low. At that point, the value of Z is 1. (b) When the pressure is low or moderate, attractive forces take over, compressing a real gas to a larger volume. In that case, the value of Z is less than one. (c) When the pressure is high, repulsions take over, making it difficult to compress the real gas, resulting in a smaller volume. At that point, the value of Z is greater than 1. For an ideal gas, the value of Z is always one. For real gases, PV Z nRT 13.1 Variation of Z with P and T The graph is a straight line curve for an ideal gas of compressibility factor Z with Class XI Chemistry www.vedantu.com 15 pressure. However, at low pressure, for real gases the value of Z is < 1. As pressure increases, Z becomes > 1. keeps on increasing. On further increasing the temperature the graph tends more towards Z = 1 i.e. ideal gas. It can be concluded that real gas follows ideal behavior only at low pressure and high temperature. The main point to be noted is that on increasing and further increasing the temperature, a real gas will not convert to an ideal gas. There is a characteristic temperature at which a gas follows ideal behavior. The temperature at which gas follows ideal behavior is called Boyle’s temperature. 14. VAN DER WAALS EQUATION After correction of pressure and volume terms, the ideal gas equation can be formulated as, Class XI Chemistry www.vedantu.com 16 2 2 an P V nb nRT V Here, a and b are Van der Waals constants. a is the attractive force between the gases. As the attractive forces between the molecules increases, the value of a increases. Whereas, b is the volume occupied by the molecule. The Van der Waal constant a is always greater than b for a given gas. The higher the value of a , the more easily liquefaction can occur. 14.1 Applicability of Van der Waal equation: At high temperature and low pressure, the compressibility factor is equal to 1, and equation is reduced to, m PV RT 14.2 At low/moderate pressure Here, the volume correction factor can be ignored. When Z < 1, attractive force dominates. The equation is reduced to, 1 a Z VRT 14.3 At high pressure Here, a pressure correction factor can be ignored. When Z > 1, the repulsion force dominates over attractive forces. The equation is reduced to, Class XI Chemistry www.vedantu.com 17 1 Pb Z RT 15. LIQUEFACTION OF GASES When the attractive forces between the molecules increase, molecules of gas come closer to each other. A stage reaches when gas changes to a liquid. This phenomenon is known as the Liquefaction of gases. 15.1 Critical temperature The critical temperature is the temperature above which a gas cannot be liquified. It is denoted as .
c T It is represented as, 8 27 c a T Rb 15.2 Critical pressure The pressure required during the liquefaction of gas is its critical pressure. It is denoted as .
c P It is represented as, 2 27 c a P b 15.3 Critical volume Under critical temperature and pressure, the volume occupied by a gas is its critical volume. It is denoted as .
c V It is represented as, 3 c V b Class XI Chemistry www.vedantu.com 18 Note: At critical points, the value of the compressibility factor is constant and has the value of 0.375. The Liquefaction of gases can be achieved in two ways: Either increase the pressure or decrease the temperature of the gas. The dominant factor is temperature. 16. LIQUID STATE Properties of liquid: There are very few empty spaces present in liquids. The interactive forces present in liquid are more than gases and less than solids. Liquids have a definite volume. The molecules in the liquid move so fast with one another. They, therefore, acquire the shapes of the container. Vapor pressure: The pressure exerted by the vapor present above the liquid in equilibrium with the liquid at a given temperature is known as vapor pressure. The vapor pressure depends on two factors-nature of the liquid and temperature. Nature of liquid: The molecules leave the liquid phase and shift to the gaseous phase if the intermolecular attraction is weak. Thus, the vapor pressure becomes higher. Effect of temperature: Vapor pressure increases with an increase in temperature. Boiling point: The temperature at which the vapor pressure of the liquid becomes equal to the atmospheric pressure is known as the boiling point. When the external pressure is 760 mm Hg, boiling point is said to be a normal boiling point. When the external pressure is 1 bar, boiling point is said to be the standard boiling point. The influence of external pressure on the boiling point is applied as follows: Higher will be the external pressure, higher will be the boiling point. Because more heat will be required to make the vapor pressure equal to the external pressure and hence higher will be the boiling point. In hospitals, surgical instruments are Class XI Chemistry www.vedantu.com 19 sterilized in autoclaves, which raise the boiling point of water by covering the vent with a weight. Similarly, lower will be the external pressure, lower will be the boiling point. Therefore, on the top of a mountain, a liquid boils at a lower temperature (where pressure is low) than on the seashore. Surface tension: The surface tension arises because the molecules of the liquid at the surface and in the interior portion are in a different situation. The force acting at right angles to the surface along the one-centimeter length of the surface is surface tension. The units of surface tension are dynes per cm or Newtons per meter. Important results of surface tension are: The minimal surface area is why the spherical shape drops the lowest energy state of a liquid. The surface area of the liquid is decreased to the minimum due to the surface tension. The drops of a liquid are spherical because a sphere has a minimum surface area. Rise of a liquid in a capillary tube. The liquid is pushed into the capillary tube by the inward pull of surface tension acting on the surface, and so the liquid rises. Effect of nature of the liquid on surface tension: Surface tension arises due to the intermolecular forces of attraction among the molecules. Greater will be the intermolecular forces of attraction, higher is the surface tension of that liquid. Effect of temperature on surface tension: As the temperature increases, surface tension decreases. At the critical temperature, surface tension becomes zero. This is because as temperature increases, the kinetic energy of the molecules increases and, therefore, the intermolecular attraction decreases. Viscosity: Viscosity is a measure of resistance to flow which arises due to the internal friction between layers of fluid as they slip past one another while liquid Class XI Chemistry www.vedantu.com 20 flows. The SI unit of viscosity is 2 1 Nsm .
There is a regular gradation of velocity inflowing of the water from one layer to another. Such flowing of water is known as laminar flow. A force is required to maintain the flow of layers. This force is proportional to the area of contact of layers and velocity gradient. Mathematically, it is related as, du F A dz du F A dz Here, A is the area of contact and du dz is the velocity gradient. is the constant and is known as the coefficient of viscosity. The factors that cause more viscosity are hydrogen bonding and Van der Waals force. Slowly the liquid flows, greater is the viscosity. As the temperature increases, viscosity decreases. This is due to the fact that as temperature increases, the kinetic energy of the molecules increases and they overcome the intermolecular forces between the layers. 17. Measurement of the pressure of the gas A common arrangement to measure the pressure of a gas is called a “Barometer”. The height of a mercury column supported in a sealed glass tube is used to determine atmospheric pressure in a mercury barometer. The pressure of a gas can be measured by various methods. The most common method of measuring pressure is in terms of height. Assume a liquid of density d is filled in a tube with a cross-sectional area A of height h. A vacuum has been placed over it. The liquid exerts pressure on the bottom of a container due to gravity. Volume of liquid A h Class XI Chemistry www.vedantu.com 21 Then, Mass of liquid d A h Force applied is equal to the weight of the liquid which is equal to g d A h Pressure is given as, F P A P d g h Only the vertical component of weight is taken when a tube filled with liquid is kept at an angle. sin P dgh IMPORTANT FORMULA: 1 1 2 2 Boyle's law: PV P V 1 2 1 2 Charle's law: V V T T 1 2 1 2 Gay Lussac law: P P T T Avogadro’s law: V n For U-tube nanometer: 1 2 h P P O qg Ideal gas equation: PV nRT Variation of ideal gas equation: PM dRT Rate of diffusion: 2 2 1 1 1 2 r P M r P M Dalton’s Law of partial pressure: total i i p x p rms 3RT M mp 2RT M Class XI Chemistry www.vedantu.com 22 avg 8RT M mp avg rms : : 1:1.128:1.224.
Compressibility factor: real ideal V Z V Van der Waal’s equation: 2 2 an P V nb nRT V Van der Waal’s constant: 3 4 4 3 b r N At low pressure: 1 a Z VRT At high pressure: 1 Pb Z RT Critical temperature: 8 27 c a T Rb Critical pressure: 2 27 c a P b Critical volume: 3 c V b |
11267 | https://www.freemathhelp.com/forum/threads/find-other-trig-functions-if-cos-%CE%98-0-85-what-is-sin-%CE%98.115073/ | Find other trig functions- if cos(Θ)=0.85 what is sin(Θ) | Free Math Help Forum
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Geometry and Trig
Find other trig functions- if cos(Θ)=0.85 what is sin(Θ)
Thread starterlucifers
Start dateMar 20, 2019
L
lucifers
New member
Joined Mar 20, 2019 Messages 2
Mar 20, 2019
#1
I understand that I am suppose to use the pythagoras theorem, but i am not sure how.
The answer that i am suppose to arrive at is below. Thanks!
2√3
3
MarkFL
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Staff member
Joined Nov 24, 2012 Messages 3,021
Mar 20, 2019
#2
Hello, and welcome to FMH!
If cos(θ)=0.85\cos(\theta)=0.85 cos(θ)=0.8 5, this means θ\theta θ is either a quadrant I or quadrant IV angle (since we're given no other information about θ\theta θ). I would indeed begin with a Pythagorean identity:
[MATH]\sin^2(\theta)+\cos^2(\theta)=1[/MATH]
Can you solve this for sin(θ)\sin(\theta)sin(θ)?
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L
lucifers
New member
Joined Mar 20, 2019 Messages 2
Mar 20, 2019
#3
MarkFL said:
Hello, and welcome to FMH!
If cos(θ)=0.85\cos(\theta)=0.85 cos(θ)=0.8 5, this means θ\theta θ is either a quadrant I or quadrant IV angle (since we're given no other information about θ\theta θ). I would indeed begin with a Pythagorean identity:
[MATH]\sin^2(\theta)+\cos^2(\theta)=1[/MATH]
Can you solve this for sin(θ)\sin(\theta)sin(θ)?
Click to expand...
I would imagine it would go something like this
sin2(θ)+cos2(0.85)=1
sin2(θ)=1-0.7225=0.2775
sin=0.2775
however this does not coincide with the answer that is given at the back of my math book and the fast reply and welcome is much appreciated!
MarkFL
Super Moderator
Staff member
Joined Nov 24, 2012 Messages 3,021
Mar 20, 2019
#4
I would first solve for sin(θ)\sin(\theta)sin(θ):
[MATH]\sin(\theta)=\pm\sqrt{1-\cos^2(\theta)}[/MATH]
Now, let cos(θ)=0.85=17 20\cos(\theta)=0.85=\dfrac{17}{20}cos(θ)=0.8 5=2 0 1 7:
[MATH]\sin(\theta)=\pm\sqrt{1-\left(\frac{17}{20}\right)^2}=\pm\frac{\sqrt{111}}{20}\approx\pm0.52678268764263694242[/MATH]
Now, as far as this not matching the result you are expected to match, can you verify that everything is stated correctly, and the entire problem has been posted?
Steven G
Elite Member
Joined Dec 30, 2014 Messages 14,594
Mar 20, 2019
#5
lucifers said:
I would imagine it would go something like this
sin2(θ)+cos2(0.85)=1
sin2(θ)=1-0.7225=0.2775
sin=0.2775
however this does not coincide with the answer that is given at the back of my math book and the fast reply and welcome is much appreciated!
Click to expand...
First of all sin has no meaning so saying sin=0.2775 is meaningless. You compute sin of angles, like sin 70 o, sin 125 o and sin 7radians.
I suspect by sin2(θ) you mean sin 2(2θ) or sin^2(θ). If you feel that sin 2(θ)= 0.2775 then why would you think that sin(θ)= 0.2775????
Dr.Peterson
Elite Member
Joined Nov 12, 2017 Messages 16,844
Mar 20, 2019
#6
lucifers said:
I understand that I am suppose to use the pythagoras theorem, but i am not sure how.
The answer that i am suppose to arrive at is below. Thanks!
2√3
3
Click to expand...
That can't be the correct answer, because it equals 1.1547, and sines and cosines can't be greater than 1! Check that you looked at the answer to the right problem.
As for the answer you got, don't forget to take the square root.
Steven G
Elite Member
Joined Dec 30, 2014 Messages 14,594
Mar 20, 2019
#7
How about sin(arccos(.85))~+/- 0.52678268764
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11268 | https://math.stackexchange.com/questions/3403613/gradient-of-fx-at-xbt-x | multivariable calculus - Gradient of $f(x)=(a^T x)(b^T x)$ - Mathematics Stack Exchange
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Gradient of f(x)=(a T x)(b T x)f(x)=(a T x)(b T x)
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How do you find the gradient of f(x)=(a T x)(b T x)f(x)=(a T x)(b T x) where a a, b b, and x x are n n-dimensional vectors?
So, far I tried by taking a derivative with chain rule:
D(f(x))=D[(a T x)(b T x)]=(a T x)D(b T x)+(b T x)D((a T x)T)D(f(x))=D[(a T x)(b T x)]=(a T x)D(b T x)+(b T x)D((a T x)T)
which leads me to:
(a T x)b T+(b T x)(a T x)T(a T x)b T+(b T x)(a T x)T
but I'm not sure how to proceed.
multivariable-calculus
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edited Sep 27, 2020 at 10:32
Rodrigo de Azevedo
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asked Oct 22, 2019 at 0:57
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Some facts and notations before we start deriving the gradient:
Trace and Frobenius product relation ⟨A,B C⟩=t r(A T B C):=A:B C⟨A,B C⟩=t r(A T B C):=A:B C
Cyclic properties of Trace/Frobenius product A:B C=B C:A=A C T:B=etc.A:B C=B C:A=A C T:B=etc.
Towards this end, we rewrite your function
f(x)=a T x b T x=(a T x)T b T x=a T x:b T x f(x)=a T x b T x=(a T x)T b T x=a T x:b T x
Now, we can obtain the differential first, and then the gradient of ∂f(x)∂x∂f(x)∂x.
d f(x)=(a T d x:b T x)+(a T x:b T d x)=(b T x:a T d x)+(a T x:b T d x)=(a b T x:d x)+(b a T x:d x)d f(x)=(a T d x:b T x)+(a T x:b T d x)=(b T x:a T d x)+(a T x:b T d x)=(a b T x:d x)+(b a T x:d x)
Thus, the gradient is
∂f(x)∂x=a b T x+b a T x.∂f(x)∂x=a b T x+b a T x.
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answered Oct 22, 2019 at 5:10
user550103user550103
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Let f:R n→R f:R n→R be defined by
f(x):=(a⊤x)(b⊤x)=(x⊤a)(b⊤x)=x⊤a b⊤x f(x):=(a⊤x)(b⊤x)=(x⊤a)(b⊤x)=x⊤a b⊤x
whose gradient is
∇f(x)=(a b⊤+b a⊤)x∇f(x)=(a b⊤+b a⊤)x
multivariable-calculusscalar-fieldsquadratic-formsgradient
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edited Sep 27, 2020 at 10:33
answered Oct 22, 2019 at 6:31
Rodrigo de AzevedoRodrigo de Azevedo
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Here's another way to solve the problem:
f(x)=(a T x)(b T x)f(x)=(a T x)(b T x)
f(x)=⎛⎝⎜⎜⎜[a 1 a 2...a n]⎡⎣⎢⎢⎢x 1 x 2...x n⎤⎦⎥⎥⎥⎞⎠⎟⎟⎟⎛⎝⎜⎜⎜[b 1 b 2...b n]⎡⎣⎢⎢⎢x 1 x 2...x n⎤⎦⎥⎥⎥⎞⎠⎟⎟⎟f(x)=([a 1 a 2...a n][x 1 x 2...x n])([b 1 b 2...b n][x 1 x 2...x n])
f(x)=(a 1 x 1+a 2 x 2+...+a n x n)(b 1 x 1+b 2 x 2+...+b n x n)f(x)=(a 1 x 1+a 2 x 2+...+a n x n)(b 1 x 1+b 2 x 2+...+b n x n)
Now we need to take the partial derivative with respect to x i x i:
∂f∂x i=(a 1 x 1+a 2 x 2+...+a n x n)∂∂x i(b 1 x 1+b 2 x 2+...+b n x n)+∂∂x i(a 1 x 1+a 2 x 2+...+a n x n)(b 1 x 1+b 2 x 2+...+b n x n)∂f∂x i=(a 1 x 1+a 2 x 2+...+a n x n)∂∂x i(b 1 x 1+b 2 x 2+...+b n x n)+∂∂x i(a 1 x 1+a 2 x 2+...+a n x n)(b 1 x 1+b 2 x 2+...+b n x n)
∂f∂x i=(a 1 x 1+a 2 x 2+...+a n x n)b i+a i(b 1 x 1+b 2 x 2+...+b n x n)∂f∂x i=(a 1 x 1+a 2 x 2+...+a n x n)b i+a i(b 1 x 1+b 2 x 2+...+b n x n)
When we create the gradient by evaluating ∂f∂x i∂f∂x i for every value of i i we get:
∇f=⎡⎣⎢⎢⎢⎢a T x∗b 1+b T x∗a 1 a T x∗b 2+b T x∗a 2...a T x∗b n+b T x∗a n⎤⎦⎥⎥⎥⎥∇f=[a T x∗b 1+b T x∗a 1 a T x∗b 2+b T x∗a 2...a T x∗b n+b T x∗a n]
∇f=(a T x)∗b+(b T x)∗a∇f=(a T x)∗b+(b T x)∗a
I hope this helps!
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answered Oct 22, 2019 at 14:30
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11269 | https://www.merriam-webster.com/thesaurus/corpulence | Est. 1828
Synonyms of corpulence
as in obesity
as in obesity
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Definition of corpulence
as in obesity
the condition of having an excess of body fat the doctor warned that the patient's corpulence was unhealthy and not merely unattractive
Synonyms & Similar Words
obesity
weight
fatness
corpulency
fat
rotundity
embonpoint
chubbiness
plumpness
fleshiness
adiposity
pudginess
fattiness
grossness
portliness
stoutness
heaviness
bulkiness
pursiness
huskiness
burliness
brawniness
endomorphy
Antonyms & Near Antonyms
slenderness
leanness
thinness
slimness
fitness
trimness
svelteness
gauntness
reediness
skinniness
weediness
scrawniness
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 corpulence As president, George H.W. Bush compared his dog, Ranger, to a blimp in a playful memo to White House staff noting a need to curb doggie treats because of canine corpulence. —Alexis Simendinger, The Hill, 5 Sep. 2025 Whimsy was also the domain of the Colombian painter and sculptor Fernando Botero, whose signature was a race of voluptuously bloated figures denizens, from priests to bullfighters, of an almost cartoonish world that, to him, had to do not with corpulence but with the sensuality of human life. —William McDonald, New York Times, 28 Dec. 2023 Twelve bears started in the celebration of corpulence on September 29 before the final online showdown Tuesday. —CNN, 5 Oct. 2021 In order to support a pair of cubs and reach peak corpulence, Grazer didnt shy away from battles for the best spots on Brooks River, even against much larger male bears, according to the Parks video. —Theresa MacHemer, Smithsonian Magazine, 30 Sep. 2020 His striking physical resemblance to his grandfather, channeling his clothing, gait and corpulence, secured his place as the latest iteration of the countrys preordained leadership. —Victoria Kim, Los Angeles Times, 28 Apr. 2020 Over his career, the uber-producer has explored aging, poverty, addiction, corpulence, single parenthood, neurodivergence and other experiences that can leave a person in the margins of American life. —Robyn Bahr, The Hollywood Reporter, 23 Sep. 2019 The Superintendent of Criminal Investigations, Hideo Nishimura, was tall and even-featured and had probably been handsome in his youth, but the years at the desk showed in his growing corpulence and a certain slowness in breaking inertia. —Andrew Liptak, The Verge, 8 Dec. 2018
Recent Examples of Synonyms for corpulence
obesity
weight
Noun
The companys efforts to improve its dimming sales growth prospects with an obesity medicine have largely flopped, leaving the pharma giant sidelined from the industrys hottest market.
— Bloomberg, Oc Register, 22 Sep. 2025
Five months after ending development of its own obesity treatment, Pfizer is renewing its push into the rapidly growing field with a nearly $5 billion acquisition.
— Boston Herald Wire Services, Boston Herald, 22 Sep. 2025
Definition of obesity
Noun
Those batteries will then lead to cost, weight and material savings, creating an attractive selection of affordable, efficient new-energy vehicles, reported New Atlas.
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This happens when water in front of the tire builds up faster than the vehicle's weight can push water out of the way.
— CA Weather Bot, Sacbee.com, 20 Sep. 2025
Definition of weight
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11270 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_12?srsltid=AfmBOopDP2IGgAcgVSjSA9lEp1KWUzSedskDNKlxnA1vaUlgk6sZQgCe | Art of Problem Solving
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2015 AIME I Problems/Problem 12
Contents
1 Problem
2 Hint
2.1 Solution 1
3 Potential Inspiration
3.1 Solution 2
3.2 Solution 3(NOT RIGOROUS)
3.3 Solution 4 (short)
4 Video Solution
5 Generalization
6 See also
Problem
Consider all 1000-element subsets of the set . From each such subset choose the least element. The arithmetic mean of all of these least elements is , where and are relatively prime positive integers. Find .
Hint
Use the Hockey Stick Identity in the form
(This is best proven by a combinatorial argument that coincidentally pertains to the problem: count two ways the number of subsets of the first numbers with elements whose least element is , for .)
Solution 1
Let be the desired mean. Then because subsets have 1000 elements and have as their least element, Using the definition of binomial coefficient and the identity , we deduce that The answer is
Potential Inspiration
For solution 1, the inspiration could be that once we select the set , where , we want to multiply each such set by and sum up through all such sets.
So, how do we scale each such set by ? We realize that if we were to choose one more element, specifically, less than , this could be done in ways.
Oh! So, now if we add to all the numbers in our set, they become , so the number of ways to pick another number below the least number in this set is !
Solution 2
Each 1000-element subset of with contributes to the sum of the least element of each subset. Now, consider the set . There are ways to choose a positive integer such that ( can be anything from to inclusive). Thus, the number of ways to choose the set is equal to the sum. But choosing a set is the same as choosing a 1001-element subset from !
Thus, the average is . Our answer is .
Solution 3(NOT RIGOROUS)
Let be the size of the large set and be the size of the subset (i.e. in this problem, and ). We can easily find the answers for smaller values of and :
For and , the answer is .
For and , the answer is .
For and , the answer is .
For and , the answer is .
For and , the answer is .
For and , the answer is .
At this point, we can see a pattern: our desired answer is always . Plugging in and , the answer is , so .
Solution 4 (short)
In the "average case", the numbers evenly partition the interval [0,2016] into 1001 parts. Then because it asks for the expected value of the least element the answer is .
-tigershark22
VIEWER NOTE: This solution doesn't always work, for example, take on . The "average case" is but integrating from to we see that definitely is not the case. This works only in certain situations, so maybe this solution would be better off with proof that this is one of those situations.
Video Solution
~anellipticcurveoverq
Generalization
See also
2015 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 11Followed by
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11271 | https://www.youtube.com/watch?v=rMb99YOzD7w | 3 4 5 Triangle Statics for Engineer
Fanny Silvestri
254 subscribers
78 likes
Description
4757 views
Posted: 2 Feb 2022
5 comments
Transcript:
in this video i will explain you how to play with a three four five triangle um specifically for statics problems but that would be the same for any other um engineering problem or math problem or physics problem okay so let's look first at this f equal 200 newton okay we have the 200 newton that is parallel to the five okay and we want to find the fy and fx that is also on that side fx there okay so you have to uh imagine that your 345 triangle is the scale of your force with f x and f y so if the scale is respected that means that your 3 divided by 5 so this side divided by 5 is the same at the same proportion than the fy divided by f and this one divided by this one so from that relation we can write that fy equal three divided by five times f so in our case it would be fy equal three to five by 200 so your fy will be so that's 40 maybe it would be 120 newton and for the fx again the proportions are respected so the four divided by five are equal that's a 5 oops 4 divided by 5 are equal to the f x divided by f we deduce that f x equals 4 divided by 5 times f so fx is equal to 4 divided by 5 times 200 and fx equal 160 newton so just imagine that this little triangle is just a mini scale of the larger triangle um if you want to skip those steps a good way to remember is like i just look at which side is parallel to which one so fy is parallel to three so that mean that the fy will be three divided by the side that is parallel to the force f so f3 will predict by five f i just look at which side are parallel together and same thing for the fx this one and this one so f x pile to four and f pile to five on the left side i did an example with pawns and with a 345 triangle in another direction so let's call this force f 2 what you want to find is your f1 f2 on the y and the f2 on the x that is also there f2 on x okay um when you draw your force on the fibo diagram of something always keep all the um tails together okay but it's just to show you that the magnitude that we are looking for for f2x is the same than this one so what will be f 2 x f 2 x is parallel to the 3 the side 3 so that will be 3 divided by 5 times the value of f 2 that is 25 pound so f 2 x will be 5 15 pounds oops and one thing that we need to look at is the direction of this f 2 x f 2 x is going to the left so that will be if we that's the magnitude but if you want to write it as a vector your f2 will be minus 15 on i hat just be careful of this minus there okay that's for the magnitude so this minus i add it there but you don't need to add it there you need to add it uh on the vector okay let me put it in parenthesis and in parenthesis there too and the f2y the f2y if we look at the f2i we are parallel to so f2y is vertical so we are piled to the vertical side of the 345 triangle that is four so that will be four divided by 5 times the f the magnitude of f2 that is 25 so the f2y will be 20. bond so now i can use the 20 point and put it in the vector for the cartesian notation form and it's going up so it's in the same direction then the y the the y axis so plus 20 j at that that will be the final cartesian form for f2 for f1 i didn't write the final cartesian form that will be f one will be fx 160 i hat and fy plus 120 j at so fx 160 fy 120 on j at both positive if you want to check that you have not done any mistake you can always do um the theorem of the petagorium theorem and check that 160 square plus 120 square it's equal to the 200 square same thing on that side you can check that the 15 square plus the 20 square will be equal to the 25 square and we can really really see that the proportions three four five it's just um divided by five and this one with the the three that the four that the five just multiply by forty for all of them that direction so divide by 40. so that's our 345 triangle works let me know if you have any question i will be uh happy to answer bye |
11272 | https://xilinx.github.io/Vitis_Libraries/quantitative_finance/2019.2/methods/bt-crr.html | »
Pricing Models and Numerical Methods »
Binomial Tree, Cox-Ross-Rubinstein, Method
Binomial Tree, Cox-Ross-Rubinstein, Method¶
Overview¶
The Cox-Ross-Rubinstein Binomial Tree method is an instance of the Binomial Options Pricing Model (BOPM) , published originally by Cox, Ross and Rubinstein in their 1979 paper “Option Pricing: A Simplified Approach” [CRR1979].
In this method, the binomial tree is used to model the propagation of stock price in time towards a set of possibilities at the Expiration date, based on the stock Volatility. For “N” time steps into which the model scenario duration is subdivided, there are N+1 possible stock prices at the expiration time.
Based on the N+1 Call or Put Option values at expiration, option values are backward-propagated to the initial time using step probabilities and the interest-rate, to obtain the Call or Put Option price. Comparing intermediate Call/Put values during back-propagation to stock prices allows American Option prices to be calculated.
Cox-Ross-Rubinstein show that as N tends to ∞, the binomial European Put/Call solutions tend towards the Black-Scholes solutions. (Both models make the same underlying assumptions.) In an example where K = $35.00 and N = 150, they show the difference is less than $0.01.
In a later paper, Leisen & Reimer [LR1995] propose a method to increase the convergence speed of the CRR binomial lattice to converge faster.
The diagram above shows an example of a binomial tree, where the number of time steps is (n). (Note that (n) steps results in (n + 1) separate propagated (S) values after the n-th step.)
At each step the initial stock price (S_0) is propagated in an Up path and a Down path from each node, with Up and Down factors (u) and (d). The “Up” probability is (p); Down is (1 - p).
The equations in the diagram show the derivation, where (\sigma) is the stock volatility, (r) the “risk-free rate”, (t) the scenario duration and (n) the number of time steps. The dividend yield in the above is assumed to be zero and not included in the expression for (p), but may be included when required.
(Diagram source: Wikipedia article Binomial Options Pricing Model (BOPM) .)
References¶
| | |
--- |
| [CRR1979] | Cox, J. C., Ross, S. A., Rubinstein, M., “Option Pricing: A Simplified Approach”, Journal of Financial Economics (1979) |
| | |
--- |
| [LR1995] | Leisen, D., Reimer, M., “Binomial Models for Option Valuation - Examining and Improving Convergence”, Rheinische Friedrich-Wilhelms-Universität, Bonn, (1995). |
Next Previous |
11273 | https://www.amboss.com/us/knowledge/management-of-trauma-patients/ | Management of trauma patients
Last updated: May 22, 2025
Summary
Trauma is one of the leading causes of death worldwide and, in the United States, the leading cause of death in young adults. Traumatic injuries range from isolated wounds to life-threatening multi-organ injuries. Advanced trauma life support () is a framework for the systematic evaluation of trauma patients to improve outcomes and reduce missed injuries. Prehospital trauma care involves life-saving interventions and basic life support in the field by emergency medical services while providing rapid transportation to the nearest appropriate hospital. In the hospital, the assessment of trauma patients begins with a primary survey in which life-threatening conditions are identified and treated using the sequential ABCDE approach. After the patient is stabilized, the secondary survey is performed, which involves a thorough history and physical examination as well as diagnostic testing to identify other injuries. The tertiary survey is performed within 24 hours of presentation to identify missed injuries. If at any point during the evaluation the patient's needs exceed the hospital's capabilities, the process to transfer the patient to a trauma center should be initiated. Trauma management of pregnant, geriatric, and pediatric patients requires additional considerations given their unique physiology.
See also “Blunt trauma” and “Penetrating trauma.”
Definitions
Trauma center
A health facility that provides specialized care to patients with serious traumatic injuries.
Different levels (e.g., I–V) of trauma center can be designated
Trauma team
A multidisciplinary team that provides care to patients at trauma centers
Consists of specialists from trauma surgery, emergency medicine, intensive care, and nursing
Members typically include a team leader, airway manager, assessment physician, proceduralist, trauma nurse, trauma technician, respiratory therapist, radiographer, and scribe.
See also “Criteria for trauma team activation.”
Advanced trauma life support (ATLS)
A framework for managing patients with serious injuries in prehospital and hospital settings.
Describes management sequences (e.g., ABCDE algorithm) that prioritize the most immediately life-threatening injuries first.
Aims to standardize trauma care across centers with varying resources and experience with trauma management
Polytrauma: severe injuries occurring in more than one anatomic region that cause systemic physiological disturbances
Overview
Trauma care varies based on patient injuries, receiving center resources (e.g., equipment, consultants), and institutional and regional guidelines. Recommendations in this article are consistent with the 2018 Advanced Trauma Life Support guidelines.
Overview of ATLS
Sequence of trauma care
Prehospital trauma care and transportation to hospital
Primary survey
Transfer to trauma center (if needed)
Secondary survey
Tertiary survey
Key components of ATLS
| Overview of ATLS | |
--- |
| | Key components |
| Primary survey | Airway assessment with C-spine immobilization and airway management Breathing assessment with respiratory support and related procedures Circulation assessment with immediate hemodynamic support and hemorrhage control Disability assessment with TBI management and neuroprotective measures Exposure with environmental survey and hypothermia management Diagnostic adjuncts, e.g., FAST, portable CXR Transfer to trauma center (if needed) Treatment of traumatic cardiac arrest (if needed) |
| Secondary survey | AMPLE history Head-to-toe physical examination Comprehensive diagnostic studies and imaging Supportive care |
| Tertiary survey | Detailed history and physical to identify missed injuries Additional diagnostics (if needed) Quality and safety measures |
| Special patient groups | Pregnant individuals: See “Trauma during pregnancy” and “Management of pregnant patients with trauma.” Older adults: See “Geriatric trauma.” Children: See “Pediatric trauma.” |
Overview of injury mechanisms
| Overview of injury mechanisms | | |
---
| | Mechanisms of injury | Potential injuries |
| Blunt trauma | Motor vehicle collision (MVC) Fall from a height Pedestrian struck Bicycle crash Assault | Head: traumatic brain injury, skull fractures, facial fractures, soft tissue injuries of the head and neck, C-spine injuries Thoracic: pulmonary contusions, cardiac contusions, rib fractures, pneumothorax, hemothorax, blunt thoracic aortic injury, diaphragmatic rupture Abdominopelvic: splenic injuries, liver injuries, pelvic fracture, genitourinary trauma Musculoskeletal: spinal fractures, crush injury, lower extremity fractures |
| Penetrating trauma | Stab injuries Gunshot wounds (GSWs) | Neck: vascular injury Thoracic: cardiac tamponade, hemothorax, pneumothorax, diaphragmatic injury Abdominopelvic: hollow viscus perforation, hemorrhage, genitourinary trauma Musculoskeletal: neurovascular injuries, fractures, compartment syndrome |
| Thermal injury | Thermal burns Electrical injury Inhalation injury | Thermal burns: circumferential eschar, compartment syndrome Electrical injury: cardiac arrhythmias, myonecrosis, compartment syndrome Inhalation injury: airway swelling, pulmonary edema, carbon monoxide toxicity |
| Blast injuries | Pressure-related injury Blunt trauma Penetrating trauma Burns Exposure to toxic fumes | Pressure-related injury: barotrauma, blast lung, tympanic membrane rupture Blunt and/or penetrating injury: due to flying debris, shrapnel, falls Burns: thermal burns, inhalation injury |
| Other environmental injuries | Drowning Cold-related injury Diving-related injury | Drowning: respiratory failure, cardiac arrest Cold-related injuries: Hypothermia, frostbite, and nonfreezing cold injuries Diving-related injuries: decompression disease, arterial gas embolism, barotrauma |
Overview of immediately life-threatening injuries
| Recognition and initial management of common critical injuries | | |
---
| | Suggestive findings | Initial management |
| Airway compromise | Signs of airway obstruction Red flags for at-risk airway | Establish definitive airway (e.g., intubation, cricothyrotomy) with manual in-line C-spine stabilization. See “C-spine immobilization,” “Airway management” and “Difficult airway.” |
| Tension pneumothorax | Chest pain Dyspnea, hypoxemia Hyperresonance on percussion, decreased breath sounds, tracheal deviation Tachycardia, hypotension | Emergent needle decompression See “Acute management checklist for tension pneumothorax.” |
| Massive hemothorax | Chest pain Dyspnea, hypoxemia Dullness on percussion, decreased breath sounds Tachycardia, hypotension CXR: pleural fluid with blunting of the costophrenic angle | Chest tube placement See “Traumatic hemothorax.” |
| Open pneumothorax | Chest pain Dyspnea, hypoxemia Tachycardia, hypoxia Sucking chest wound | Place occlusive dressing over the wound, taping three sides. See “Acute management checklist for traumatic pneumothorax.” |
| Cardiac tamponade | Tachypnea, dyspnea Tachycardia Beck triad Penetrating chest trauma CXR: enlarged cardiac silhouette FAST: pericardial effusion, right atrial collapse during systole | Expedite urgent thoracotomy. If bedside treatment is required, emergency thoracotomy with bedside pericardial window is preferred. Pericardiocentesis is unlikely to be effective in trauma. See “Acute management checklist for cardiac tamponade” for details. |
| External hemorrhage | Tachypnea, dyspnea Tachycardia Hypotension Penetrating wound Active bleeding | Hemostatic control (e.g., application of pressure or tourniquet) See “Management of hemorrhagic shock.” |
| Internal hemorrhage | Tachypnea, dyspnea Tachycardia Hypotension Penetrating or blunt trauma to the chest or abdomen Intraabdominal bleeding: signs of peritonitis Retroperitoneal bleeding: Cullen sign, Grey Turner sign FAST: abdominal free fluid | Consult specialist for operative or interventional management. See “Management of hemorrhagic shock.” |
Prehospital trauma care
Prehospital trauma care is situation-dependent and centered on field stabilization of the patient and prompt transport to the closest trauma center.
Bystanders
Shout for help and/or call 911.
Assess scene safety prior to providing assistance.
Remove the patient from dangerous situations.
Initiate basic life support (BLS).
Perform life-saving interventions.
Airway opening maneuvers
Hemorrhage control
Spinal immobilization
Emergency medical services (EMS)
Prehospital trauma care provided by physicians varies regionally.
EMS typically perform an abbreviated version of the primary survey.
Provide life-saving interventions.
Airway management for respiratory distress or altered mental status
Hemorrhage control (e.g., use of tourniquets or pressure bandages)
Immediate hemodynamic support (e.g., IV fluids)
Prevent additional injury.
Spinal immobilization (e.g., cervical collar placement, backboard)
Fracture stabilization
Provide parenteral analgesia, as needed.
Rapidly transport the patient to the closest appropriate hospital.
Notify the receiving hospital.
Signout of patient to trauma team lead upon arrival
Primary survey
The ABCDE algorithm in ATLS provides a sequence to help prioritize treating the most rapidly life-threatening injuries first. In clinical practice, trauma team members evaluate and treat these simultaneously, continually reassessing each injury's severity throughout the resuscitation.
Airway (and C-spine immobilization)
Identify and treat airway obstruction (e.g., due to blood, direct injury, edema) and/or loss of airway protective reflexes, (e.g., due to AMS or coma), while preventing further C-spine injury.
Assess the airway.
Ask the patient to state their name; the ability to answer typically correlates with a patent airway.
Evaluate for signs of airway compromise and signs of respiratory distress.
Examine the airway for foreign bodies or injury (e.g., facial fractures, soot, burns).
Perform initial interventions.
Suction oropharyngeal secretions and/or blood.
Perform airway opening maneuvers.
Insert basic airway adjuncts.
Intubate patients with:
Airway obstruction and/or respiratory failure
Depressed mental status (e.g., GCS ≤ 8)
Severe shock and/or cardiac arrest
At-risk inhalation injury
Stabilize the C-spine.
Assume C-spine injury in patients with blunt trauma and indications for spinal immobilization.
Perform airway management with necessary spinal motion restrictions.
See “Initial management of C-spine injury” for details.
Perform cricothyrotomy in case of intubation failure.
Consider early intubation for impending airway obstruction in patients with signs of inhalation injury, moderate to severe facial and oropharyngeal burns, and extensive body burns.
Breathing
Identify and treat chest injuries, e.g., tension pneumothorax, open pneumothorax, massive hemothorax, flail chest, and tracheobronchial injuries.
Assess oxygenation: Evaluate SpO2 and start continuous pulse oximetry.
Assess ventilation
Vital signs: Monitor rate and quality of respirations; consider EtCO2 monitoring.
Neck: Inspect for jugular venous distension and tracheal deviation
Chest: Auscultate and inspect for chest wall injuries: e.g., penetrating wounds, subcutaneous emphysema, absent breath sounds, paradoxical chest movement
Consider eFAST to help identify pneumothorax at the bedside.
Perform initial interventions.
Supplemental O2
Emergency chest decompression for suspected tension pneumothorax.
3-sided dressing for sucking chest wound
BMV or mechanical ventilation for respiratory failure
If traumatic pneumothorax is suspected in a patient requiring positive pressure ventilation, perform tube thoracostomy immediately to prevent progression to tension pneumothorax.
Circulation
Provide immediate hemodynamic support and hemostatic measures while identifying sources of bleeding, e.g., external hemorrhage, thoracic cavity, abdominal cavity, thighs, retroperitoneal space.
Assess hemodynamic status.
Assess central pulses, level of consciousness, and capillary refill time.
Monitor vitals and continuous cardiac telemetry.
Perform initial interventions.
Place two large-bore IVs (at least 18-gauge).
Consider intraosseous access if peripheral IV cannot be obtained.
Administer 1 L warmed isotonic crystalloid bolus.
If unresponsive to IV fluid, proceed to blood transfusion.
Localize hemorrhage.
Perform FAST to rapidly identify hemopericardium, and intrathoracic, intraabdominal, and/or intrapelvic bleeding in unstable patients.
Identify obvious pelvic or long bone (e.g., femur) deformity.
Identify major bleeding vessels (e.g., in the neck, groin, chest, or proximal extremities).
Identify indications for immediate surgery (e.g., exploratory laparotomy, urgent thoracotomy) and/or angioembolization.
Treat hemorrhagic shock.
Administer emergency transfusion with universal donor blood products (e.g., type O blood) if required.
Provide crossmatched blood products as soon as they are available.
Follow local massive transfusion protocol if indicated, e.g., plasma, platelets, and pRBCs at a 1:1:1 ratio.
Consider tranexamic acid (TXA).
Consider reversal of anticoagulation.
Allow for permissive hypotension.
Perform bedside hemorrhage control.
Apply pressure or tourniquet to control active external hemorrhage.
Apply pelvic binder for suspected bleeding pelvic fractures
Insert chest tube for suspected massive traumatic hemothorax
Treat obstructive shock: e.g., chest tube insertion for tension pneumothorax, pericardial window for cardiac tamponade
If there is a loss of vital signs, treat traumatic cardiac arrest with emergency chest decompression and emergency thoracotomy.
Suspect cardiac tamponade in patients with penetrating chest injury with Beck triad and a positive FAST scan, and expedite urgent pericardial fluid drainage via thoracotomy.
Penetrating abdominal injury with signs of shock is usually an indication for exploratory laparotomy.
Disability
Identify life-threatening traumatic brain injury (TBI), begin measures to limit secondary brain injury, and expedite definitive surgery if indicated.
Perform rapid neurological evaluation.
Calculate GCS.
Assess pupillary light response.
Assess motor and sensation functions.
Initiate TBI management.
Consult neurosurgery.
Start neuroprotective measures.
Start ICP management if a cerebral herniation syndrome is present.
Remain vigilant for signs of traumatic brain injury in intoxicated patients.
Exposure (and environmental control)
Undress the patient completely.
Examine the entire patient for signs of occult injury, including the axilla, groin, and back.
Prevent and/or manage hypothermia with rewarming techniques.
Diagnostic adjuncts
Consider the following studies during the primary survey if they are likely to impact immediate management:
Portable CXR
Portable pelvic x-ray
FAST or eFAST
Diagnostic peritoneal lavage (if POCUS is unavailable)
See “Diagnostics in trauma” for details.
Use bedside studies for rapid diagnostics in patients that are too unstable for transport to imaging suites.
Traumatic circulatory arrest
Common causes include tension pneumothorax, cardiac tamponade, and hemorrhagic shock.
Management
First line: bilateral chest decompression with finger thoracostomy
Consider emergency thoracotomy
Expedite definitive surgery and operative stabilization of survivors.
Traumatic cardiac arrest requires bedside surgical interventions by trained personnel and subsequent operative treatment and stabilization. Do not follow standard ACLS algorithms as these are unlikely to be effective.
Resuscitative thoracotomy
The following is consistent with the 2015 Eastern Association for the Surgery of Trauma (EAST) recommendations on emergency department thoracotomy. Follow local policies as indications vary regionally.
Definition: a bedside procedure performed in the emergency department or trauma bay to obtain emergency access to the thoracic cavity to provide temporizing lifesaving measures in pulseless patients
Purpose
Relieve cardiac tamponade via pericardiotomy and/or repair of myocardial injury
Control intrathoracic hemorrhage
Perform open cardiac massage
Cross-clamp the descending aorta
Indications
Pulseless patients after penetrating thoracic injury
Pulseless patients after penetrating extrathoracic injury, excluding isolated cranial injuries
Pulseless patients with recently documented signs of life in the field or at the hospital after blunt injury
Contraindication: pulseless patients without any documented signs of life in the field or at the hospital after blunt injury
Do not perform resuscitative thoracotomy unless a qualified surgeon is present.
Interim management
Consider the following once the primary survey is complete:
Reassess the effectiveness of life-saving bedside interventions.
Prepare for urgent time-sensitive imaging (e.g., CT head) once the patient is stable enough.
Determine if the patient needs immediate surgery, urgent interfacility transfer, or immediate specialty consult (see “Disposition”).
Next steps: See “Secondary survey.”
If there is clinical deterioration at any time, repeat the primary survey to identify a critical cause.
Urgent interfacility transfer
The following applies to patients initially evaluated at facilities that are not trauma centers:
The decision to transfer a patient to a trauma center for definitive care is multifactorial.
See “Criteria for trauma team activation” for conditions that typically require management by a trained trauma team.
If tests are required prior to transfer, keep them limited to tests that will ensure a safe transfer.
Communicate with the receiving physician and transportation team about clinical and diagnostic findings.
Do not delay an urgent transfer in order to complete an in-depth diagnostic evaluation.
Secondary survey
The secondary survey is performed after the patient is stabilized and it involves a thorough history and physical examination as well as diagnostic testing to identify other injuries.
History
Obtain the following to anticipate likely injuries and estimate patient's physiological reserve.
AMPLE history: a focused history approach recommended in the rapid evaluation of patients with trauma
Allergies
Medications
Past medical history and pregnancy status
Last meal
Events and environment related to the injury
Mechanism of injury
Injury patterns can provide clues to the mechanism of injury (e.g., direction, amount of energy).
Commonly divided into blunt trauma and penetrating trauma
See “Blunt trauma injury mechanisms.”
See “Gunshot injuries” and “Penetrating trauma.”
Obtain collateral information from EMS, family, and/or witnesses, especially if the patient is unable to provide a reliable history.
Physical examination
A systematic head-to-toe physical examination must be completed to identify additional injuries.
Assess each body part for open wounds, lacerations, abrasions, foreign bodies, and/or bruising.
Identify any bony deformities, areas with focal tenderness, or edema.
Document all findings.
| Secondary survey in trauma patients | | |
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| | Examination | Injuries |
| Head | Inspect and palpate the head. Perform ocular examination and pupillary examination. Assess mental status and GCS. Evaluate cranial nerves. | See “Traumatic brain injury” See “Traumatic eye injuries” See “Skull fractures” |
| Maxillofacial | Inspect ears, nose, and mouth. Assess for malocclusion. | See “Facial fractures” See “Dental injuries” |
| Cervical spine and neck | Inspect for signs of tracheal injury or tracheal deviation. Palpate for crepitus. Auscultate carotid arteries. Palpate for cervical spine tenderness. | See “Vertebral fractures” See “Penetrating neck trauma” See “Blunt neck trauma” |
| Chest | Inspect and auscultate the chest. Palpate the chest wall for tenderness and crepitus. | See “Tension pneumothorax” See “Traumatic pneumothorax” See “Traumatic hemothorax” See “Rib fracture” See “Cardiac tamponade” See “Diaphragmatic rupture” See “Blunt chest trauma” See “Penetrating chest trauma” |
| Abdomen and pelvis | Inspect and auscultate the abdomen. Palpate the abdomen for tenderness, abdominal guarding, or rebound tenderness. Assess for pelvic instability. | See “Blunt abdominal trauma” See “Penetrating abdominal trauma” See “Pelvic fracture” |
| Genitourinary | Assess for urethral injury. Perform a digital rectal examination and/or pelvic examination as indicated. | See “Genitourinary trauma” |
| Back and flank | Perform a log roll to examine the back of patients under C-spine precautions. Palpate the thoracolumbar spine for tenderness and step deformities. Evaluate for signs of retroperitoneal hemorrhage (e.g. Grey Turner sign) | See “Vertebral fractures.” |
| Musculoskeletal | Inspect upper and lower extremities for lacerations or deformities. Palpate upper and lower extremities for tenderness. Palpate peripheral pulses. | See “Penetrating trauma to the extremities” See “Blunt trauma to the extremities” See “General principles of fractures” See “Compartment syndrome” See “Rhabdomyolysis and crush syndrome” |
| Neurological | Assess motor and sensory functions of upper and lower extremities. | See “Incomplete spinal cord syndromes” See “Complete spinal cord injury” |
Do not forget to log roll patients who are under C-spine precautions to examine the back and spine.
Diagnostic adjuncts
Consider the following diagnostic studies and procedures during the secondary survey once the patient is stable:
Whole-body CT (WBCT)
CT head without IV contrast
CT cervical spine without IV contrast
CT thoracic and/or lumbar spine without IV contrast
CT chest with IV contrast or CTA chest
CT abdomen and pelvis with IV contrast
X-rays of the extremities
Laboratory studies
Ancillary testing (e.g., bronchoscopy, esophagoscopy)
See “Diagnostics in trauma” for details.
Further management
Ensure imaging is ordered for all identifiable injuries.
Consider using decision rule to clinically rule out C-spine injury and allow early removal of C-spine precautions.
Order routine trauma laboratory studies, including any emergency preop diagnostics and tests required to evaluate medical comorbidies.
Order appropriate monitoring, e.g., continuous cardiac monitoring, neurovitals.
Provide adequate analgesia and other supportive care (e.g., antiemetics, maintenance fluids).
Provide tetanus prophylaxis for open wounds.
If intraperitoneal injury is suspected, administer empiric antibiotics for intraabdominal infections.
Evaluate the need for admission, interfacility transfer, and further consults to address newly identified injuries (See “Disposition”).
Next steps: See “Tertiary survey.”
Tertiary survey
Goal
Identify injuries undetected during primary and secondary surveys.
Ensure patient safety.
Complete the tertiary survey within 24 hours of presentation.
A standardized tertiary survey with high-quality imaging may limit the incidence of missed injuries.
Approach
Obtain collateral information if trauma history remains unclear.
Identify and rectify gaps in the patient's medical history (e.g., chronic conditions, medications, allergies).
Repeat examination of the patient focusing on musculoskeletal injuries and detailed neurovascular examination.
Review previously obtained diagnostic studies.
Order additional studies based on tertiary survey findings.
Screen for suicidality and nonsuicidal self-injury if self-harm is suspected.
Screen for sexual violence, intimate partner violence, older adult abuse, and child abuse in at-risk individuals.
Hand and foot injuries are the most common missed injuries.
Missed injuries
Missed injuries are an important area of focus for quality and safety in trauma care, however, high-quality evidence is limited.
Risk factors
Patient risk factors include:
Altered mental status
Concomitant TBI
Polytrauma with distracting injuries
Higher overall injury severity
Need for urgent surgery
Age > 30
Dangerous injury mechanism: e.g., assault, fall from height, MVC, pedestrian struck, GSW, stab wounds
System-associated risk factors include:
Limited initial clinical evaluation
Radiological imaging lacking or delayed
Cognitive biases
See “Health care personnel-associated risk factors” in “Quality and safety” for details.
Beware of anchoring bias and premature closure bias during the tertiary survey.
Injuries at risk of missed or delayed diagnosis
The following have a higher potential to remain undetected after initial evaluation.
Surface wounds
Scalp, nose, and mouth
Axillae
Groin and perineum
Extremity
MSK injuries involving the hand and foot
Neurovascular injury
Compartment syndrome
Occult extremity fractures
Head, neck, and spine
Small or delayed intracranial hemorrhages
Occult facial fractures and C-spine fracture
Carotid and vertebral artery injuries
Thorax
Blunt cardiac injury
Penetrating and blunt thoracic aortic injury
Diaphragmatic injury
Esophageal perforation
Abdominopelvic
Hollow viscus injury
Retroperitoneal injury (e.g., pancreatic and duodenal bulb injuries)
Subtle liver or splenic laceration
Genitourinary trauma (e.g., ureter injury)
Penetrating rectal injury
Diagnostics
Approach
Consider clinical judgment, mechanism of injury, and patient factors (e.g., age, hemodynamic status) when choosing diagnostic studies.
Weigh the need and timing of diagnostic studies against the need for urgent interfacility transfer or surgical intervention for each patient.
Follow local policies on diagnostic imaging strategy (e.g., liberal vs. selective) as these vary by institution and region (see “CT imaging in trauma” for details).
ECG
Indications: mechanisms that may result in cardiac injury (e.g., in blunt chest trauma)
Findings
Blunt cardiac injury (e.g., dysrhythmia, ST segment changes)
Pericardial effusion and cardiac tamponade (e.g., electrical alternans, low voltage QRS)
Laboratory studies
Routine studies: CBC, BMP, LFTs
Preoperative studies: coagulation studies, type and screen
Pregnancy test (e.g., serum or urine β-hCG)
POC glucose
Urinalysis (See “Approach to genitourinary trauma” for details.)
Lactate
ABG: to evaluate PaO2, PaCO2, pH and base deficit
Normotensive patients with trauma may have subclinical hypoperfusion. Evaluate for other clinical features of shock and check hemodynamic parameters (e.g., serum lactate, base deficit).
FAST and eFAST
See “Point-of-care ultrasound” (POCUS) for procedural details (e.g., image generation and troubleshooting).
Focused assessment with sonography for trauma (FAST): thoracoabdominal POCUS used in trauma patients to detect free fluid within the peritoneal, pericardial, and pleural cavities
Extended FAST (eFAST): an extension of the FAST scan that includes evaluation of the chest for lung sliding
Clinical applications in trauma
Can identify internal hemorrhage (e.g., hemoperitoneum, hemopericardium, hemothorax) requiring operative intervention in unstable patients
eFAST can additionally help identify pneumothorax
Used primarily as a diagnostic adjunct during the primary survey
Can be used as an adjunct for serial reassessments
Limitations
Limited sensitivity in stable patients
Needs to be interpreted alongside other diagnostic and clinical findings
Technique: Sonography of the subxiphoid, RUQ, LUQ, pelvic, and anterior thoracic regions is typically performed (see “FAST and eFAST” in “POCUS” for details).
Findings: See “FAST and eFAST” in “POCUS” for details.
Free fluid can be seen in the following spaces:
Pericardial
Pleural
Hepatorenal
Splenorenal and/or subphrenic
Pelvic
Lung sliding may be absent in pneumothorax
Interpret FAST and eFAST alongside other diagnostic parameters and clinical judgment. POCUS does not replace definitive diagnostic studies.
A positive FAST exam in a hemodynamically unstable patient with trauma is usually an indication for urgent operative intervention (e.g., exploratory laparotomy, urgent thoracotomy, pericardiotomy).
Radiography
Bedside chest and pelvic x-rays are commonly performed during the primary survey, while extremity and spine X-rays are typically reserved for the secondary survey.
CXR
Indications: blunt or penetrating trauma to the chest or abdomen
Findings
Pneumothorax
Hemothorax
Pneumoperitoneum
Pneumomediastinum
Widened mediastinum
Pelvic X-ray
Indications
Blunt or penetrating trauma to the abdomen
Unstable pelvis
Findings: pelvic fracture
Extremity X-rays
Indications: pain, redness, swelling, or deformity at the injury site
Findings: fracture and/or dislocation
Spinal X-rays
Spinal x-rays have been replaced by CT imaging in most trauma centers.
Indications
Spinal tenderness to palpation
Disruption of the vertebral process alignment (i.e., step-off deformity)
Extremity numbness, tingling, or weakness
Neurogenic shock
Findings
Vertebral compression fracture
Burst fracture
Chance fracture
Transverse process fracture
Spinous process fracture
Spondylolisthesis
CT imaging in trauma
Approach
Perform CT scans preferentially during the secondary survey in hemodynamically stable patients with no obvious indications for emergent laparotomy.
Make decisions about the timing, necessity, and sequence of CT scans together with all relevant specialists on the trauma team, especially for patients with polytrauma.
If a crucial CT scan cannot be delayed as it impacts operative management, ensure trauma teams are at the bedside with stabilizing treatments throughout the procedure.
Follow local protocols for CT imaging (e.g., WBCT vs. selective CT imaging) under specialist guidance.
Avoid transporting unstable patients out of resuscitation areas to obtain CT scans whenever possible.
Whole-body CT (pan scan)
May be performed to evaluate patients with multiple injuries after significant trauma
Commonly includes:
CT head without IV contrast
CT cervical spine without IV contrast
CT thoracic and lumbar spine with IV contrast
CT chest with IV contrast
CTA chest
CT abdomen and pelvis with IV contrast
CT head and spine
CT head without IV contrast: See “Diagnostic imaging in traumatic brain injury.”
CT cervical spine without IV contrast
Indications
Clinical suspicion of C-spine injury based on mechanism and examination
Inability to clear C-spine clinically using a validated tool (e.g., NEXUS C-spine criteria , Canadian C-spine rule (CCR) )
Findings
Bony deformity and/or fracture of the vertebral body or processes
Loss of alignment of the vertebral bodies
Increased distance between vertebrae
Narrowing of vertebral canal
Increased prevertebral soft tissue swelling
CT thoracic and lumbar spine without IV contrast
Indications: clinical suspicion of T-spine and/or or L-spine injury based on mechanism and examination
Findings: similar to CT cervical spine
Consider imaging for spinal fractures in patients with evidence of high-energy trauma to the lower extremities (e.g., calcaneus fracture) after falling from a height.
Obtain CT cervical spine if C-spine injury cannot be ruled out using validated clinical decision tools.
CT chest with IV contrast
Indications
Inability to rule out blunt chest trauma with validated clinical decision tools
Penetrating chest trauma
Findings
Cardiac injury (e.g., blunt cardiac injury)
Lung injury (e.g., pneumothorax, hemothorax, pulmonary contusion)
Tracheal injury
Diaphragm injury
Esophageal injury
Chest wall injury (e.g., rib fracture)
CTA chest
Indications
Inability to rule out blunt chest trauma with validated clinical decision tools
Penetrating chest trauma with concern for great vessel injury.
Findings
Cardiac injury (e.g., blunt cardiac injury)
Blunt thoracic aortic injury
Vascular injury (e.g., penetrating chest trauma)
Consider CTA chest to evaluate for blunt thoracic aortic injury in patients with high-energy trauma and a wide mediastinum on CXR.
CT abdomen and pelvis with IV contrast
Indications
Blunt and/or penetrating abdominal trauma
Penetrating back and/or flank trauma with no other indications for immediate laparotomy
Findings
Stomach injury
Splenic injury (e.g., splenic rupture)
Liver injury (e.g., liver hematoma)
Small and large bowel injuries (e.g., gastrointestinal perforation)
Kidney injury (e.g., renal laceration, renal hematoma)
Bladder injury (e.g., rupture of the bladder)
Pelvic fracture
Diagnostic peritoneal lavage (DPL)
Definition: an invasive diagnostic test used to assess for hemoperitoneum or viscus perforation in abdominal trauma
Indications: : may be performed after the primary survey for suspected hemoperitoneum with equivocal or unavailable FAST
Procedure: A catheter is placed into the abdomen and contents are aspirated to assess for the presence of blood or fecal matter. If neither is observed, a liter of warm saline is instilled and then collected for cytological analysis.
Findings: The presence of bile, fecal matter, or ≥ 10 cc of blood is considered a positive test and is an indication for emergent laparotomy.
Ancillary testing
Additional testing may be performed depending on the mechanism of injury and clinical evaluation, and may include:
Bronchoscopy for tracheobronchial injury
Esophagoscopy for esophageal perforation
Assessment for features of GU trauma (e.g., using CT urography, retrograde urethrogram): for renal, ureteral and urethral injuries
Echocardiography for blunt cardiac injury
Assess patients for features of GU trauma (e.g., with retrograde urethrogram) if there is hematuria, blood at the meatus, inability to void, need for pelvic binder, scrotal hematoma, or perineal ecchymosis.
Suspect tracheobronchial injury in patients with extensive subcutaneous emphysema.
Disposition
Criteria for trauma team activation
Mechanism of injury
Falls from > 5 meters
Impact from high-speed MVC
Ejection from vehicle
Any MVC > 18 mph (29 km/h) involving impact with a pedestrian, cyclist, or motorcyclist
Death of a vehicle co-passenger
Specific injuries
Injury to > 2 body regions
Penetrating injury to the head, neck, torso, or proximal limb
Traumatic amputation
15% estimated BSA burns in adults or > 10% in children; or signs of airway involvement
Airway obstruction
Physiological derangements
Hypotension
Tachycardia
Bradypnea or tachypnea
GCS < 14
Special patient groups
Pregnant patients > 24 weeks' gestation with chest and/or abdominal injury
Individuals > 70 years of age with chest injury
Specialty consults
Indications for urgent trauma surgeon consult
At level 1 trauma centers, a trauma surgeon's presence at the bedside within 15 minutes of patient arrival is indicated for any of the following:
Systolic BP < 90 mmHg
GSW to neck, thorax, abdomen, pelvis, or proximal extremities
Prehospital endotracheal intubation
Intubation for respiratory failure
GCS < 8 due to trauma
Emergency physician discretion
Other specialists
Consult all specialists outside the trauma team responsible for managing identified injuries:
Thoracic surgery: e.g., for operative cardiopulmonary injuries
Neurosurgery: e.g., for TBI
Orthopedics: e.g., for operative fractures
OB-GYN and/or urology: e.g., for genitourinary trauma
Plastic surgery: e.g., for extensive soft tissue injuries such as degloving, burns
Vascular surgery: e.g., for major vessel injuries
Interventional radiology: e.g., for angioembolization
Others: hand surgery, oromaxillofacial surgery, spine surgery
Interfacility transfers
Initiate the transfer process to a higher level trauma center as soon as the patient's needs exceed the capability of the current hospital.
Transfer decision depends on multiple factors, e.g.:
The patient's injuries
Resources and equipment available at the current hospital
Availability of consultant physicians
Institutional and regional guidelines
Transfer to a burn unit may also be indicated (see “Treatment of burns” for details).
An indication for trauma team activation is generally an indication that the patient requires transfer to a trauma center.
Surgical admission
Operating room: transfer patients to the OR for operative management of critical findings.
Traumatic brain injury: See “Management of moderate and severe TBI.”
Thoracotomy: See “Resuscitative thoracotomy” and “Urgent thoracotomy.”
Laparotomy: See “Emergency exploratory laparotomy.”
Trauma ICU
Admit all hemodynamically unstable or intubated patients to critical care settings.
Patients may require postoperative admission to ICU.
Discharge from emergency settings
Most patients with major trauma require admission for treatment and observation.
Consider discharge with outpatient follow-up for patients with all of the following after complete workup and observation:
Only minor injuries (e.g., isolated fracture)
Adequate analgesia
Normal mental status
Ability to function with ADLs and IADLs
Good social supports
No other indications for admission
Special patient groups
Trauma in pregnant individuals
Overview
| Overview of trauma in pregnancy | | |
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| | Maternal | Fetal |
| Clinical features | Vaginal bleeding Hematuria Abdominal pain Premature labor Rupture of membranes | Fetal heart rate changes Decreased fetal movement |
| Diagnostics | Clinical evaluation (see “Primary survey”) Laboratory studies: rhesus blood type (to avoid maternal Rh alloimmunization) Imaging + Ultrasound + X-ray, CT scan (should not be deferred because of fetal radiation concerns) | Doppler ultrasound: detection of fetal heartbeat Assessment of the gestational age Electronic fetal monitoring (minimum 4 hours): detection of fetal distress, placental abruption, and preterm labor |
| Management | Minor trauma: obstetric surveillance Major trauma: initial stabilization and resuscitation of the mother as needed; further assessment in trauma center (See “Primary survey””) | Minor trauma: obstetric surveillance Major trauma + Intrauterine resuscitation of the fetus + Perimortem Cesarean section (in case of viable fetus > 23 weeks' gestation after unsuccessful maternal resuscitation for 4–5 minutes) |
Avoid examining the mother in the supine position in order to avoid possible supine hypotensive syndrome.
The mother should be evaluated and treated before the fetus. Early and optimal diagnostics and trauma management of the mother is the best treatment for the fetus.
Epidemiology
Incidence: every 12th pregnant woman experiences trauma
Trauma in pregnancy is the leading cause of nonobstetrical maternal death in the US.
Etiology
Unintentional trauma (e.g., due to falls, MVCs)
Intentional trauma:
Intimate partner violence
Assault
Suicide attempt
Classification
Minor trauma (90%): trauma for which obstetrical surveillance suffices
No abdominal involvement
No rapid compression or deceleration
No pain or vaginal bleeding
No loss of fluid
Normal fetal movement
Major trauma (10%): trauma that requires further assessment in a trauma center
Abdominal involvement with abdominal pain
Signs and symptoms of internal bleeding
Hematuria
Vaginal bleeding and/or loss of fluid
Loss of consciousness
Rapid compression and/or deceleration
Decreased fetal movement
Management of pregnant patients with trauma
All pregnant trauma patients
Begin initial resuscitation using the ABCDE approach.
Assess for signs of abruption and uterine rupture.
Perform a fetal status assessment.
Consider imaging: ultrasound and/or CT (should not be deferred because of fetal radiation concerns)
Obtain coagulation studies, including fibrinogen.
Order a Kleihauer-Betke test.
Administer anti-D immunoglobulin for Rh-negative mothers.
Consult OB/GYN if there is any concern for obstetric injuries.
Trauma patients > 20 weeks' gestation
Initiate tocodynamometry for at least 4–6 hours.
Continue monitoring for 24 hours if any of the following are present :
High-impact mechanism of injury
≥ 6 uterine contractions per hour
Vaginal bleeding, significant abdominal pain, or uterine tenderness
Rupture of the membranes
Abnormal fetal heart rate pattern
Maternal tachycardia (HR > 110/minute)
Further management
Management of nonobstetric injuries as for nonpregnant patients
Emergency delivery in consultation with OB/GYN if there is nonreassuring fetal status or maternal hemodynamic instability
See also:
“Management of placental abruption”
“Management of uterine rupture”
“Management of preterm labor”
Even minor trauma poses a risk for placental abruption.
Complications
Due to physiological changes during pregnancy
Superior displacement of abdominal organs
Increased risk of gastrointestinal injury from chest or upper abdominal trauma
Increased risk of aspiration during intubation
Decrease of blood pressure (by 15–20 mmHg only during 2nd trimester): increased risk of hypotensive complications (e.g., falls due to syncope) and missing pathological causes underlying hypotension
Increase of the heart rate (by 15–20 bpm only during 3rd trimester): increased risk of complications from tachycardia (e.g., arrhythmias) and missing pathological causes underlying tachycardia
Increase of blood volume: increased risk of overlooked blood loss
Placental abruption and preterm labor
Fetal loss (60–70% caused by minor trauma, with placental abruption being the most common complication)
Supine hypotensive syndrome following trauma (see “Supine hypotensive syndrome” in “Other complications” below)
Uterine rupture and exsanguination
Maternal death
Prevention
Screen all pregnant women for intimate partner violence; for more information see “Intimate partner violence.”
Trauma in older adults
Common mechanisms of injury
Falls
Impact from MVCs
Burns
Penetrating injuries
General principles
Consider any physiological events that may have led to the trauma (e.g., cardiac arrhythmia leading to syncope and resulting in a fall).
Determine medication effects.
Assess for older adult abuse.
Determine goals of care.
Geriatric modifications to the primary survey
Airway
Loss of protective airway reflexes
Dentures and arthritic changes may impede airway management.
Breathing
Limited respiratory reserve with potential for rapid deterioration to respiratory failure
At risk for respiratory failure following rib fractures; provide adequate analgesia and pulmonary toilet.
Circulation
Patients with preexisting hypertension may appear normotensive, but this may represent a relatively hypotensive state (e.g., hemorrhagic shock).
Use of cardiac medicine may blunt tachycardia.
Consider advanced monitoring (e.g., central venous pressure, ECHO) during resuscitative efforts.
Disability
Increased risk for TBI
Cortical atrophy may delay signs of intracranial bleeding.
More likely in patients taking anticoagulant and antiplatelet medications
Preexisting neurological or psychiatric disease may impede evaluation.
Exposure: Loss of subcutaneous fat puts patients at risk for hypothermia, as well as pressure injury from immobilization.
Trauma in children
Common mechanisms of injury
MVCs
Drowning
House fires
Nonaccidental injury
Falls
General principles
Normal pediatric vital signs vary by age.
Smaller body mass results in greater force applied per unit of body area, leading to a greater risk for multiple injuries than adults.
Developmental age may hinder history and examination.
Use Broselow tape to recommend common drug dosing and equipment sizes.
Assess for red flags for child maltreatment.
Practice judicious use of CT scans.
Pediatric modifications to the primary survey
Airway
Proportionally larger head results in passive cervical spine flexion, which may obstruct the airway.
Visualization during intubation is difficult.
Shorter airway is located higher in the neck.
Anterior attachment of vocal cords is located more inferiorly than the posterior attachment.
Breathing
Higher normal respiratory rate than adults
Use a pediatric bag-valve-mask for children < 30 kg.
Circulation
Higher normal heart rate and lower normal blood pressure than adults
Increased physiological reserve allows children to maintain normal systolic blood pressure during hemorrhagic shock.
See also “Basic life support in infants and children.”
Disability
Proportionally larger head leading to an increased risk of TBI
Modified GCS may be used in children < 4 years of age.
See also “PECARN blunt head-trauma prediction rule.”
Exposure: at increased risk for accidental hypothermia because of greater body surface area to mass ratio than adults
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11274 | https://learnopengl.com/Guest-Articles/2021/Scene/Frustum-Culling | If you're running AdBlock, please consider whitelisting this site if you'd like to support LearnOpenGL (it helps a lot); and no worries, I won't be mad if you don't :)
Frustum Culling
Guest-Articles/2021/Scene/Frustum-Culling
Now we know how to create a Scene graph and organize your object in a scene, we are going to see how to limit your GPU usage thanks to a technical name's the frustum culling.
This technique is simple to understand.
Instead of sending all information to your GPU, you will sort visible and invisible elements and render only visible elements.
Thanks to this technique, you will earn GPU compute time.
You need to know that when information travels toward another unit in your computer, it takes a long time.
For example, information from your GPU to your ram takes time.
It's the same if you want to send information from your CPU to your GPU like a model matrice.
It's for this reason that the "draw instance" is so powerful.
You send a large block to your GPU instead of sending elements one by one.
But this technique isn’t free.
To sort your element, you need to create a physical scene to compute some stuff with math.
This chapter will start with an introduction to the mathematical concept that will allow us to understand how frustum culling works.
Next, we are going to implement it.
Finally, we are going to study possible optimizations and talk about the balance of the technical.
In this video illustrating frustum culling in a forest, the yellow and red shape on the left side is the bounding volume that contains the mesh.
Red color means that the mesh is not visible and not sent to the GPU.
Yellow means that the mesh is rendered.
As you can see lots of things are rendered and few are visible for the player.
Mathematical concept
Let's start the mathematical parts from top to bottom.
Firstly, what is a frustum?
As we can see in Wikipedia, frustum is a portion of a solid like a cone or pyramid.
The frustum is usually used in game engine to speak about the camera frustum.
Camera frustum represents the zone of vision of a camera.
Without limit, we have a pyramid but with near and far we have a frustum.
How to mathematically represent a frustum?
Thanks to 6 planes: near, far, right, left top and bottom planes.
So, an object is visible if it is forward or on the 6 planes.
Mathematically a plane is represented with a normal vector and distance to the origin.
A plane doesn't have any size or limit as a quad.
So, create a struct to represent a plane:
```
struct Plane
{
// unit vector
glm::vec3 normal = { 0.f, 1.f, 0.f };
// distance from origin to the nearest point in the plane
float distance = 0.f;
[...]
};
```
We can now create Frustum structure:
```
struct Frustum
{
Plane topFace;
Plane bottomFace;
Plane rightFace;
Plane leftFace;
Plane farFace;
Plane nearFace;
};
```
Reminder: a plane can be built with a point and a normal.
For the near, the normal is the front vector of the camera.
For the far plane, it's the opposite.
The normal of the right face we will need to do a cross product.
The cross product is the second wonderful tool for the programmer who likes vectors.
It allows you to get a perpendicular vector to a plane created with two vectors.
To go forward, we need to do the cross product of the right axis per up.
We will use it like that:
But to know the direction of each vector from the camera to the far plane we will know the side length of the far quad:
hSide and vSide are the far quad limited by the other planes of the camera frustum.
To compute its edge, we will need of trigonometry.
As you can see in the image above, we have two rectangle triangles and we can apply the trigonometric functions.
So, we would like to obtain vSide which is the opposite side and we have zFar that is the adjacent side of the camera.
Tan of fovY is equal to the opposite side (vSide) divided by the adjacent side (zFar).
In conclusion, if I move the adjacent side on the left on our equation, tan of fovY multiplied by the zFar is equal to the vSide.
We now need to compute hSide.
Thanks to the aspect that is a ratio of the width by the height, we can easily obtain it.
So, hSide is equal to the vSide multiplied by the aspect as you can see on the right side of the image above.
We can now implement our function:
```
Frustum createFrustumFromCamera(const Camera& cam, float aspect, float fovY,
float zNear, float zFar)
{
Frustum frustum;
const float halfVSide = zFar tanf(fovY .5f);
const float halfHSide = halfVSide aspect;
const glm::vec3 frontMultFar = zFar cam.Front;
frustum.nearFace = { cam.Position + zNear cam.Front, cam.Front };
frustum.farFace = { cam.Position + frontMultFar, -cam.Front };
frustum.rightFace = { cam.Position,
glm::cross(frontMultFar - cam.Right halfHSide, cam.Up) };
frustum.leftFace = { cam.Position,
glm::cross(cam.Up,frontMultFar + cam.Right halfHSide) };
frustum.topFace = { cam.Position,
glm::cross(cam.Right, frontMultFar - cam.Up halfVSide) };
frustum.bottomFace = { cam.Position,
glm::cross(frontMultFar + cam.Up halfVSide, cam.Right) };
return frustum;
}
```
In this example, the camera doesn't know the near, aspect but I encourage you to include this variable inside your Camera class.
Bounding volume
Let's take a minute to imagine an algorithm that can detect collisions with your mesh (with all types of polygons in general) and a plane.
You will start to say that image is an algorithm that checks if a triangle is on or outside the plane.
This algorithm looks pretty and fast! But now imagine that you have hundreds of mesh with thousands of triangles each one.
Your algorithm will sign the death of your frame rate fastly.
Another method is to wrap your objects in another geometrical object with simplest properties such as a sphere, a box, a capsule...
Now our algorithm looks possible without creating a framerate black hole.
Its shape is called bounding volume and allows us to create a simpler shape than our mesh to simplify the process.
All shapes have their own properties and can correspond plus or minus to our mesh.
All shapes also have their own compute complexity.
The article on Wikipedia is very nice and describes some bounding volumes with their balance and application.
In this chapter, we are going to see 2 bounding volumes: the sphere and the AABB.
Let's create a simple abstract struct Volume that represent all our bounding volumes:
```
struct Volume
{
virtual bool isOnFrustum(const Frustum& camFrustum,
const Transform& modelTransform) const = 0;
};
```
Sphere
The bounding sphere is the simplest shape to represent a bounding volume.
It is represented by center and radius.
A sphere is ideal to encapsulate mesh with any rotation.
It must be adjusted with the scale and position of the object.
We can create struct Sphere that inheritance from volume struct:
```
struct Sphere : public Volume
{
glm::vec3 center{ 0.f, 0.f, 0.f };
float radius{ 0.f };
[...]
}
```
This struct doesn't compile because we haven't defined the function isOnFrustum.
Let's make it.
Remember that our bounding volume is processed thanks to our meshes.
That assumes that we will need to apply a transform to our bounding volume to apply it.
As we have seen in the previous chapter, we will apply the transformation to a scene graph.
```
bool isOnFrustum(const Frustum& camFrustum, const Transform& transform) const final
{
//Get global scale is computed by doing the magnitude of
//X, Y and Z model matrix's column.
const glm::vec3 globalScale = transform.getGlobalScale();
//Get our global center with process it with the global model matrix of our transform
const glm::vec3 globalCenter{ transform.getModelMatrix() glm::vec4(center, 1.f) };
//To wrap correctly our shape, we need the maximum scale scalar.
const float maxScale = std::max(std::max(globalScale.x, globalScale.y), globalScale.z);
//Max scale is assuming for the diameter. So, we need the half to apply it to our radius
Sphere globalSphere(globalCenter, radius (maxScale 0.5f));
//Check Firstly the result that have the most chance
//to faillure to avoid to call all functions.
return (globalSphere.isOnOrForwardPlane(camFrustum.leftFace) &&
globalSphere.isOnOrForwardPlane(camFrustum.rightFace) &&
globalSphere.isOnOrForwardPlane(camFrustum.farFace) &&
globalSphere.isOnOrForwardPlane(camFrustum.nearFace) &&
globalSphere.isOnOrForwardPlane(camFrustum.topFace) &&
globalSphere.isOnOrForwardPlane(camFrustum.bottomFace));
};
```
To compute the globalCenter we can’t only add the current center with the global position because we need to apply translation caused by rotation and scale.
This is the reason why we use the model matrix.
As you can see, we used a function undefined for now called isOnOrForwardPlane.
This implementation method is called top/down programming and consists to create a high-level function to determine which kind of function need to be implemented.
It avoids to implement too many unused functions that can be the case in "bottom/up".
So to understand how this function works, let's make a drawing :
We can see 3 possible cases: Sphere is inside the plane, back or forward.
To detect when a sphere is colliding with a plane we need to compute the nearest distance from the center of the sphere to the plane.
When we have this distance, we need to compare this distance with radius.
```
bool isOnOrForwardPlane(const Plane& plane) const
{
return plane.getSignedDistanceToPlane(center) > -radius;
}
```
We can see the problem in the other way and create a function called isOnBackwardPlane.
To use it we simply need to check if bounding volume IS NOT on the backward plane.
Now we need to create the function getSignedDistanceToPlane in the Plane structure.
Let me realize my most beautiful paint for you :
Signed distance is a positive distance from a point if this point is forward the plane.
Otherwise this distance will be negative.
To obtain it, we will need to call a friend: The dot product.
Dot product allows us to obtain the projection from a vector to another.
The result of the dot product is a scale and this scalar is a distance.
If both vectors go oppositely, the dot product will be negative.
Thanks to it, we will obtain the horizontal scale component of a vector in the same direction as the normal of the plane.
Next, we will need to subtract this dot product by the nearest distance from the plane to the origin.
Hereafter you will find the implementation of this function :
```
float getSignedDistanceToPlane(const glm::vec3& point) const
{
return glm::dot(normal, point) - distance;
}
```
AABB
AABB is the acronym of Axis aligned bounding box.
It means that this volume has the same orientation as the world.
It can be constructed as different can be we generally create it with its center and its half extension.
The half extension is a distance from center to the edge in the direction of an axis.
The half extension can be called Ii, Ij, Ik. In this chapter, we will call it Ix, Iy, Iz.
Let's make the base of this structure with few constructors to made its creation the simplest
```
struct AABB : public BoundingVolume
{
glm::vec3 center{ 0.f, 0.f, 0.f };
glm::vec3 extents{ 0.f, 0.f, 0.f };
AABB(const glm::vec3& min, const glm::vec3& max)
: BoundingVolume{},
center{ (max + min) 0.5f },
extents{ max.x - center.x, max.y - center.y, max.z - center.z }
{}
AABB(const glm::vec3& inCenter, float iI, float iJ, float iK)
: BoundingVolume{}, center{ inCenter }, extents{ iI, iJ, iK }
{}
[...]
};
```
We now need to add the function isOnFrustum and isOnOrForwardPlane.
The problem is not easy as a bounding sphere because if I rotate my mesh, the AABB will need to be adjusted.
An image talks much than a text :
To solve this problem lets draw it :
Crazy guys want to rotate our beautiful Eiffel tower but we can see that after its rotation, the AABB is not the same.
To make the Shema more readable, assume that referential is not a unit and represented the half extension with the orientation of the mesh.
To adjust it, we can see in the third picture that the new extension is the sum of the dot product with the world axis and the scaled referential of our mesh.
The problem is seen in 2D but in 3D it's the same thing. Let's implement the function to do it.
```
bool isOnFrustum(const Frustum& camFrustum, const Transform& transform) const final
{
//Get global scale thanks to our transform
const glm::vec3 globalCenter{ transform.getModelMatrix() glm::vec4(center, 1.f) };
// Scaled orientation
const glm::vec3 right = transform.getRight() extents.x;
const glm::vec3 up = transform.getUp() extents.y;
const glm::vec3 forward = transform.getForward() extents.z;
const float newIi = std::abs(glm::dot(glm::vec3{ 1.f, 0.f, 0.f }, right)) +
std::abs(glm::dot(glm::vec3{ 1.f, 0.f, 0.f }, up)) +
std::abs(glm::dot(glm::vec3{ 1.f, 0.f, 0.f }, forward));
const float newIj = std::abs(glm::dot(glm::vec3{ 0.f, 1.f, 0.f }, right)) +
std::abs(glm::dot(glm::vec3{ 0.f, 1.f, 0.f }, up)) +
std::abs(glm::dot(glm::vec3{ 0.f, 1.f, 0.f }, forward));
const float newIk = std::abs(glm::dot(glm::vec3{ 0.f, 0.f, 1.f }, right)) +
std::abs(glm::dot(glm::vec3{ 0.f, 0.f, 1.f }, up)) +
std::abs(glm::dot(glm::vec3{ 0.f, 0.f, 1.f }, forward));
//We not need to divise scale because it's based on the half extention of the AABB
const AABB globalAABB(globalCenter, newIi, newIj, newIk);
return (globalAABB.isOnOrForwardPlane(camFrustum.leftFace) &&
globalAABB.isOnOrForwardPlane(camFrustum.rightFace) &&
globalAABB.isOnOrForwardPlane(camFrustum.topFace) &&
globalAABB.isOnOrForwardPlane(camFrustum.bottomFace) &&
globalAABB.isOnOrForwardPlane(camFrustum.nearFace) &&
globalAABB.isOnOrForwardPlane(camFrustum.farFace));
};
```
For the function isOnOrForwardPlane, I have taken an algorithm that I found in a wonderful article.
I invite you to have a look at it if you want to understand how it works.
I just modify the result of its algorithm to check if the AABB is on or forward my plane.
```
bool isOnOrForwardPlane(const Plane& plane) const
{
// Compute the projection interval radius of b onto L(t) = b.c + t p.n
const float r = extents.x std::abs(plane.normal.x) +
extents.y std::abs(plane.normal.y) + extents.z std::abs(plane.normal.z);
return -r <= plane.getSignedDistanceToPlane(center);
}
```
To check if our algorithm works, we need to check that every object disappeared in front of our camera when we moved.
Then, we can add a counter that is incremented if an object is displayed and another for the total displayed in our console.
```
// in main.cpp main lopp
unsigned int total = 0, display = 0;
ourEntity.drawSelfAndChild(camFrustum, ourShader, display, total);
std::cout << "Total process in CPU : " << total;
std::cout << " / Total send to GPU : " << display << std::endl;
// In the drawSelfAndChild function of entity
void drawSelfAndChild(const Frustum& frustum, Shader& ourShader,
unsigned int& display, unsigned int& total)
{
if (boundingVolume->isOnFrustum(frustum, transform))
{
ourShader.setMat4("model", transform.getModelMatrix());
pModel->Draw(ourShader);
display++;
}
total++;
for (auto&& child : children)
{
child->drawSelfAndChild(frustum, ourShader, display, total);
}
}
```
Ta-dah ! The average of objects sent to our GPUrepresents now about 15% of the total and is only divided by 6.
A wonderful result if your GPU process is the bottleneck because of your shader or number of polygons.
You can find the code here.
Optimization
Now you know how to make your frustum culling.
Frustum culling can be useful to avoid computation of things that are not visible.
You can use it to not compute the animation state of your entity, simplify its AI...
For this reason, I advise you to add a IsInFrustum flag in your entity and do a frustum culling pass that fills this variable.
Space partitionning
In our example, frustum culling is a good balance with a small number of entities in the CPU.
If you want to optimize your detection, you now will need to partition your space.
To do it, a lot of algorithms exist and each has interesting properties which depend on your usage :
- BSH (Bounding sphere hierarchy or tree) :
Different kinds exist. The simplest implementation is to wrap both nearest objects in a sphere.
Wrap this sphere with another group or objetc etc...
In this example, only 2 checks allow us to know that 3 objects are in frustum instead of 6 because if the bounding sphere is totally inside the frustum all its content is also inside.
If the bounding sphere is not inside when needed to inter and check its content.
Quadtree :
The main idea is that you will split space into 4 zones that can be split into four zones etc... until an object wasn't wrapped alone.
Your object will be the leaf of this diagram.
The quadtree is very nice to partition 2D spaces but also if you don't need to partition height. It can be very useful in strategy games like 4x (like age of empire, war selection...) because you don't need height partitioning.
Octree :
It's like a quadtree but with 8 nodes. It's nice if you have a 3D game with elements in different height levels.
BSP (binary space partitioning) :
It's a very fast algorithm that allows you to split space with segments. You will define a segment and the algorithm will sort if an object is in front of this segment or behind.
It's very useful with a map, city, dungeon... The segments can be created at the same time if you generate a map and can be fast forward.
Lot of other methods exist, be curious.
I don't implement each of these methods, I just learn it to know that they exist if one day I need specific space partitioning.
Some algorithm is great to parallelize like octree of quadtree if you use multithread and must also balance on your decision.
Compute shader
Compute shader allows you to process computation on shader.
This technique must be used only if you have a high parallelized task like check collision with a simple list of bounds.
I never implemented this technique for the frustum culling but it can be used in this case to avoid updating space partitioning if you have a lot of objects that move.
Additional resources
Article about camera frustum extraction: Fast Extraction of Viewing Frustum Planes from the WorldView-Projection Matrix by Gil Gribb and Klaus Hartmann
Article about collisions detection: A wonderful resource for collision detection and another approach about volume, culling and mathematic concept
Article to go further: A good article to go further on GPU culling process, multithreading and OBB
Autor
Article by: Six Jonathan
Contact: e-mail
Date: 09/2021 |
11275 | https://artofproblemsolving.com/community/c2409693h3476529_math_competition_resources?srsltid=AfmBOoonpEUCoQqPOESATZD5EDCsQRETvYETTwluQr6PxsxSytPrHTsj | acklew : Math Competition Resources
Community » Blogs » acklew » Math Competition Resources
Sign In • Join AoPS • Blog Info
acklew ======
Math Competition Resources
by akliu, Jan 10, 2025, 12:39 AM
gatekeepers will hate me for this one c:<
Prelude (headnote?)
This is basically a 3-year anniversary for me doing math competitions. I've come along so far, and I'm proud of that! So because I wish I had a lot of this stuff to look at before when I started math competitions (and my blog has a fair amount of traffic), I'm just going to share what I've tried over the years.
Firstly, you'll notice that most of these resources are self-paced resources that aren't classes. And usually in the rare instance that someone does ask me for advice about any stage of competitive math that I've already gotten past before, I direct them to self-paced stuff. When they ask me what classes I've taken, I tell them, "Oh, I took some classes from each of the main 4 topics at BBMC during the summer of 7th grade, and some AwesomeMath combo and geometry during the summer of 8th grade." However, the emphasis is always on the self-paced stuff, because I've always just been more effective at that then attending classes
By the way, can we talk about the "pay attention, maybe it's something you missed and they cover it briefly" argument?
The reason that doesn't work is i) if I'm confident that I know the topic, I know the topic. Anything essential is something that would've already been an example problem and something that I, with 99.999% certainty, already know, so it's probably some niche formula that isn't even necessary or will help me in any capacity with a 1% chance.
Additionally, ii) I didn't ask my parents to sign me up for this stuff, they need to stop gaslighting me. Asking me and then pestering me until I say "ok fine leave me alone" is not a "yes definitely sign me up".
Finally, iii) this is literally the sunk cost fallacy... and I know that they can argue that I'm wasting their money, but it's not me; they're forcing me to waste my time, whereas I could rather just do productive stuff during class and do the homework later, which is honestly 100% of the substance of classes in my experience (that is, for classes where I don't already know all the content).
Yes, I'm ranting about specific stuff again, but hopefully this applies to someone and helps them with their situation.
. (And really, I just end up playing video games in boring online classes my parents sign me up for anyways, after I finish all the work early.)
Secondly, I'm obviously not the most qualified for talking about this stuff. I usually find that a lot of parents go "hah, this absolute clown has less achievements than this other person, so you should listen to me instead, and go take these classes that they recommended!" If you're looking for advice coming from some people who are considerably better at math than I am, go DM some moppers or whatever, I'm sure they'll be happy to respond. But here's my take on stuff, and things that I did that worked. Hopefully you find something useful, but maybe you don't. I guess to validate this point in the future, I should make IMO or something, and go "Ha! I told you so!", but anyways, just use your own judgement and determine if what I write about below is a bunch of random nonsense or not.
AMCs:
Yeah, I know, they already ended, and I just decided to add this to the beginning of the blog post because I felt like including it. I constantly talk about this stuff in my posts on Contests and Programs, but since this blog has more traction, here you guys go:
AoPS Wiki Mock AMC Page
Past AMCs: Since these ones are actually way more high quality (mostly, kinda) than a lot of the mocks on the Mock AMC page, and also the fact that they are literally the source material, so quality is defined by them, you should try mocking these later than the ones above for actual high-quality content. (Note: I'm not saying anything about the quality of any specific mocks above other than the fact that i) there are more experienced people writing the actual AMCs, ii) everything above is meant to emulate the past AMCs anyways, and iii) the user-made mocks are usually almost always harder or easier by at least 1-2 problems, give or take).
If you aren't the type of person to motivate yourself to do mocks, just take a class. I took classes from BBMC back in like the summer of 7th grade and that was helpful in making AIME for the first time...
I'll give it credit for being i) a great place to take classes, and ii) consistent studying, but I've said this on multiple occasions: I'm more of a "self study/self-paced" sort of guy.
So even if your parents pressure you to take classes and you're already suffering from a ton of work and other stuff, consider your own opinion. They can't exactly force you to attend, and if they do, something's wrong with them, not you. Your parents may always want what's best for you, but whether they know exactly what they're doing is a completely different matter.
Burnout sucks. Being overwhelmed sucks. Stick to what you know will work, and if you don't know, experiment a bit.
AIME:
Doing olympiad math
Okay, this seems counterproductive. Computational is all about speed, right? So doing a slow and boring proof-based contest would never help you. Besides, JMO and stuff is hard. Doing even IMO problem 1s seems daunting, and would never be applicable in AIME.
Wrong! Olympiad techniques can definitely help you in AIME. Additionally, they open your mind to a lot of cool solutions. Doing olympiad math teaches you to think outside of the box more, outside the confines of the dreadful AMC, and AIME is just the gateway to doing that.
I did this because everyone cracked at math that I knew had done this, and looking back, doing OTIS was the right move. (Of course, doing more OTIS would've been a better move... whoops.)
Mocking past AIMEs: They are almost always high quality, and they're also the literal standard for AIME mocks. Yeah, a 3 hour test is daunting. It's the same for me, I absolutely cannot lock in for 180 minutes in a row. But they genuinely help.
Doing other contest problems: Notably, HMMT February, November, and CMIMC can help a ton, according to a notable unnamed legend on the NC ARML Team who is now a goated student at MIT.
Analyze the stuff you mess up on: Yeah, I do pretty bad at this. But every time I've tried, I've made a lot of progress. So do this.
I'm deciding to actually lock in, so I spent two hours making a google sheet to track my progress throughout mocks and determine where to work on stuff: Sheet Picture
. There's really cool conditional formatting stuff too!!! Since you guys might want a copy of it, here you go c:
I'm planning to also do spreadsheet stuff for HMMT November, February, and CMIMC, but the point scoring system was kinda goofy for the first few years of each contest's existence, so I don't know how that'll work out.
Anyways, I made this post so I can be serious now about grinding math competitions and locking in to finally make JMO, and I hope you got something out of this. Good luck!
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dang only 3 years compared to a lot of math sweats, "only" is fair wording ig lol
This post has been edited 1 time. Last edited by akliu, Jan 10, 2025, 3:45 AM
by Inaaya, Jan 10, 2025, 1:14 AM
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Thank you for convincing me to lock back in to otis np, grind those units!!
This post has been edited 1 time. Last edited by akliu, Jan 10, 2025, 3:44 AM
by mathnerd_101, Jan 10, 2025, 2:16 AM
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as a d1 gatekeeper ur making me angry >:C womp womp C:<
This post has been edited 1 time. Last edited by akliu, Jan 11, 2025, 3:09 AM
by LearnMath_105, Jan 10, 2025, 11:17 PM
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"Only three years" bro has come so far and im also proud of bro your actually my inspiration dude thats nice to hear c: you're definitely going further than i'll ever go!!
This post has been edited 1 time. Last edited by akliu, Jan 11, 2025, 3:10 AM
by raypawpatrol, Jan 11, 2025, 1:39 AM
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just wanted to say that this is a great post
by PaixiaoLover, Jan 12, 2025, 10:04 PM
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i agree 100% with doing problems from HMMT, HMNT, and CMIMC in terms of computational practice
by lpieleanu, Jan 13, 2025, 3:09 AM
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Archives
September 2025
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the first time I've taken a W in a competition (2025 NC SMC Regionals)
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Year Progress: Around 33.33 percent
happiness is such a moral issue
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This AMC contest season was objectively the worst in MAA history
this torch won't burn out
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when I put my mind to it
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Tips for middle schoolers part 2
a celebration post you werent expecting
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things that break me
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EGMO Chapter 1: Any% Speedrun
Watching acklew speedrun EGMO Ch 1
It is well-known that...
Math Competition Resources
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a little more locked in (2012 AIME II)
I need to derust at math (2012 AIME I)
Happy New Year
December 2024
darkest secrets
squid game season 2
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Policy-Based Data Structures (CF Grinding #3)
Lazy Propagation with Segment Trees (CF Grinding #2.5)
Learning Fenwick Trees in an Hour (CF Grinding #2)
i think this works someone verify please (cf grinding #1)
i did codeforces questions for 3 days to torture my friend
trio at rio
November 2024
insincerity
induction
the second worst day of my life
DMM 2024
11/7/2024
Into the Night
October 2024
in april
those of us who blossom
flower dance
the biggest storm approaches
im so tired
AMC grind (pt. 2)
t-1 to birthday
some pretty dark thoughts that you should probably know about
AMC grind (pt. 1)
october posting
September 2024
where we met
I guess it's coming together now
a completely random assortment of thoughts
August 2024
Let the grind begin! (Third times the charm)
summer is over
I'm an algebra anti-main now (hooray)
Learning projective geo (9.4-9.7)
Learning projective geo (9.1-9.3)
New highest mohs solve!!! (30)
fun(ny) math
July 2024
KEVIN DAY
Powerful math techniques that don't require skill
I recommend manga and anime for fun
number theory main goes on number theory training arc
June 2024
cute math problems
hi xoinks
a year's recap
2024 ARML @UAH Recap
May 2024
Geometry 3
last concert of the year
here's some random music i guess, making blog posts is hard
ARML Recap!!!
April 2024
life is...
The 76th Annual Hunger Games
Where's My Calculator?
2024 SMC Finals @NCSSM-Durham
the days before camp decisions
a random low-quality and horribly disorganized post about mental health
Geometry 2
March 2024
The REAL grind starts now
2024 SMC @WakeTechCC
Did You Know That
February 2024
random life updates and stuff
learning style
Ask Me Anything
HMMT Anyone?
Happy Lunar New Year!
A Journey's End
Last Wishlist
Some final remarks
January 2024
The Storm
HI AKLIU
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an assortment of random stuff
I saw akliu in a grocery store
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End of Winter Break
akliu's blog because mine is dead
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December 2023
Merry Christmas (Goodbye, 2023)!
Christmas Eve's Magic
Winter Break
Life has changed
November 2023
Thanksgiving Break!
AIME Grind #4: 2006-I
AIME Grind #3: 2005-I & 2005-II
AIME Grind #2: 2004-II
AIME Grind #1: 2004-I
2023 AMCs
Letting go
The storm is here
congrats to akliu
Best Halloween Ever
October 2023
T minus ~2 weeks!
less grind, life and 42 rabbitholes
fourteen
The Grind is REAL.
Do I need to explain? Remembrance, Worries, Full Circle.
September 2023
I'm literally the Impostor
decision
Akliu Solves USAMO 2008/4 During Orchestra
Another Generic Update (sorry for being bad at naming things oops)
a mix of emotions (disorganized post)
Screw AP World + ok skill issue moment
August 2023
High School Starts
It's that time of the year again...
A Summer Recap
Wrap-Up of Prerequisites: Another Grind
2022-2023
Middle School's End
Music Auditions
1433 (I lost the cow)
Standing here I realize
July 2023
I LOVE G2 SO MUCH
Moderately Trivial Logic-ish Questions
Mock Reviewing
All the grinds
AAAAAHHH AMC 10 SOMEONE HELP
Calculating Farmer John's Net Worth and A Sphere of Cows
Generating Functions, USACO, and everything.
'Murica 2023 + Summer
June 2023
SSMO + AwesomeMath + Physicality
The Minecraft Anarchy Experience + Precalculus Estimations, Anyone?
when you hire akliu as your marketing campaigner
wait im contrib here?
Memorization
I'm actually kind of worried...
Other Subjects
6/16/2023
Do this before it's too late.
Volunteering
2023 ARML Day 2 @UAH
End of Middle School
First ISL Solve!
Some Trigonometry Problems
2023 ARML Day 1 @UAH
its finally time
Are... you serious right now.
May 2023
armohls !!!!!
Summer Plans
The Final (Countdown)
Friends: Reality, or not?
bashing.
I met Poh Shen Loh today!
NC ARML 2023
Only a few days...
Two Sides
On Recent Cutoffs
5/7/23 update
NC Central State Math Contest @UNCG
Should I ask akliu out to the school dance?
April 2023
Road to JMO
Answering 30 stupid questions
Congratulations to akliu
Life is quite clownery at the moment
Some life updates
Olympiad Recommendations, Anyone? + Life Updates
10k Visits Special
My take on certain negative aspects of contests
Fun time's over: T-30?
Expectation getting closer to Reality
Expectation vs. Reality
Sillying Conundrum
Good start to spring break!
March 2023
Random life updates
2001 AIME II: Double digits :)
2001 AIME I: Some improvement, mostly sillies though
2023 State Math Contest @Wake Tech Community College Recap
Proof that AIME I > AIME II objectively (mostly)
2000 AIME I
small march madness post
depression
On School Curriculum, "Gifted" Program, and more.
So be it, I guess
One of two things will happen tomorrow.
Now or nothing
It's really amazing...
Basic Complete Search Speedrun
Life Update 3/12/2023
Random thoughts again
HellO!!!!!!
How to defend your computer against main battle tanks and MBT bullets
Quote of the day + POTD 3/7/2023
No school :) + Why my Entries are more Readable
The problem with having a generic name
USACO Guide: Simulations Suck (But are also fun)
life is okay
February 2023
Monday is somehow a fun day
Sunday Updates
Thanks for everything: 8k
how bad is akliu/POTD
Random Thoughts + USACO stuff
Friday 2/24/2023
How to make good tutorial 2022 edition (working) NOT FAKE
AwesomeMath Submitted!
2/21/2023 (Creative title name!)
gf face reveal
in the end, it's all lies
Life right now
AOC 2023 Official Leak
The ARML Experience
2023 AIME II
Two Days
Congratulations to akliu
The 2023 MATHCOUNTS Chapter Experience
Tomorrow, the storm starts
Having a 6-month relapse from my 2-year relapse
Random thoughts/out of context
Poll Regarding Improvement of List
Honestly, I think I'm screwed.
Random thoughts jumbled up (including Breaking Bad)
why akliu is more orz than me
An Ideal AIME List
The storm is here
January 2023
The January 2023 USACO Experience
Math/Coding Updates
My first attempt at making and solving a proof problem
My take on EO: Sequences and Series (Up to S5)
Change of Mind + Is it possible to prove/disprove this? (Induction?)
Clarification
dear unnamed person
Life Updates + Did I do it right?
i give up
Calm Before the Storm
Happy Chinese New Year!
Life Updates
Daily Public Humiliation
First Mock in a Month (2022 ZeMC 10A)!
Unsatisfied
Unfairness
Random Thought: Impact
Life Update/Nice AIME Problems & Solutions
On Procrastination After Winter Break
What a great start!
Power Through: Breaking Down Daunting Tasks
So... yeah this is me.
Happy New Year :)
December 2022
New Year's Eve (for me): Appreciation
Asian Slippery Slope Logic + Demotivation
Nearing 6k!
akliu's Massive MATHCOUNTS Skill Issue and the CEO of Shanghai Corn
Merry Christmas!
Demotivation/Remotivation/Demotivated Again/Remotivated Again
An interesting problem
Productivity Levels are High (Just like me!)
Winter Break: The Grind is On!
AIME Problem set #2: Crippled Math Skills + Spotify
Life Must Go On: First Bronze P2 Solve!
A Culmination of 13+ Years: AIME Qualification
Congratulations to akliu for qualifying for the AIME!
First USACO solve, and life updates
Two Grinds?
The Final Step to Banishing Procrastination
This is the choice that will determine my fate...
The grind NEVER stops.
Irrational Parental Blocks + akliu's PSAT sobad
Days 2 and 3 of akliu Redemption Arc + AMC 10 Mock Review
November 2022
Hit by Procrastination (and complaining about my height)
The Florida Experience
Goodbye guys
Concert + Mission: Hide from Society
So... Did I make it?
akliu how so orz
The Last Review
A... Second Beginning?
T-20: Departure
The Second Beginning
The Final Frontier
Planet Life: Pre-Broccoli Empire Optimal Strategy (with some math)
2020 Eleventh Hour Mock AMC 10 / 2020 TMC 10A
Some Problems from The Art of Modular Arithmetic
October 2022
2021 ZeMC 10 Review + 2020 CMC 10A / 10B Review
Life Updates + Running out of AMC mocks to do
Modern Olympiad Number Theory 1.2
2021 AAMC 10B Review (Will post life updates soon)
2021 AAMC 10A Review
Staying Productive During School Absence
Life Updates
AMC 10 Mocks + Fever Condition
Failed plans, I hate fevers
I have a fever, and my throat hurts, and I have mucus. Oh no.
Hooray, I got an award for public speaking in front of tiktokers and 5 year olds
AMC10 Mocks I should do (Reminder to self)
Proving School Math using Coordbashing
Insane Gaming Setup + Math PDF Setup
A Perfect Dream
4000 visits
Life updates and questioning my levels of sanity with philosophical thinking
AoPS members with their math stats
HAPPY BIRTHDAY AKLIU!!
Last Week, and My Birthday
Thoughts on Where I am
September 2022
The AIME Quest
Interesting FE
Proof that good at math means good at STEM
Isn't this unlucky?
I Wished
Quick Life Update
The Adventures of Jerry the Turtle (pt. 2/10): Beans Ending
Clown problems
The Adventures of Jerry the Turtle (pt. 1/10): Heist Ending
akliu = clown/sobad (pt. 2)
The Math Plan!
Operation Shuffle! Math Jam, and akliu's Mind Reading failure
jatloe adventures (pt. 2)
2019 AMC 10A #24
So many clubs and stuff
2021 AMC 10A #24
Why my favorite colors are Azure and Orange
jatloe adventures (pt. 1)
A fun NT problem
Fun Algebraic Proof Problem
200th post: Why FLT + Euler's Totient sucks
Basic Modular Arithmetic
programmeruser's contrib intro
Son, how was Symphony? Terrible /j
1+1 equals... 0? 14 squared entries
Goodbye.
Life Update
Math Stuff + akliu clown/sobad proof
My friends bet on how fast and where I die after second period
3333 Visits
How basic math topics were invented
Interesting Number Theory Proof Problem
My violin progress
The AOC has been released!
I GOT INTO SYMPHONY
Day 5 of school: The last day (in the week) and early release
Days 3 and 4: Teachers laugh evilly as we stare at them
r/AsianParentStories moment
August 2022
AMT 2021 Power Round Problem 3 Solutions
Day 2 of School: The Alphabet has never been so hard
Golden Ratio and Fibonacci Sequence Recurrence Relations
First Day of School: The Dark vs Light Side
my observations of life around me
Proving basic Logarithmic properties using only substitution
Violin Audition + finally we have a testsolver
The School Tour Experience.
Even less of a chance
Introducing, the MSH 2022!
If you're programmeruser, don't read this + mc nats sobad
The MATHCORN FTW Experience
Math Poll
3000 visits: The AoPS Social Journey.
USACO is bad (just kidding, Farmer John and his cows are really cool)
Freeze-up, barf-your-breakfast, pee-your-pants type of test taking stress
Different Platforms
It's been 387 days.
What should I do?
The most terrible mistake of my life + trigonometry grind
The JMPSC experience
Coding Class Idiocy
What is peak performance?
Poll + 2010 AMC 10B #24
Some AMC10 problems for AIME qual prep
Finite Differences
What would you do with different superpowers?
July 2022
Dumb things society tells us
Microsoft is bad
Full update on the past few days
So what happened recently?
Bruh I got on some news article or something
Me and some people played violin in a Chinese Bar with a talented blind person
Math Route? Don't know if I've gotten it right
This really screwed up my plans
We lost a legend
June 2022
A whole ton of Math Updates
The New Math Grind
Counting and Probability Mastery + Other Updates
Nearing the End of Counting and Probability + SSMO
First week of Summer Vacation
The Second Last Day
Tetris, Mathematics, and an Impending Summer Vacation
Idiots in my class, I mean goons. How do you deal with goons?
May 2022
What do AoPS math tools do?
Well, it's been a week.
Who's behind that screen? Thinking ahead so you don't do idiotic things.
The 4 Horsemen of Painful Geometry + Failing at Volleyball
No more flamewars!
A lot of life updates (ft. Devious Lick Simulator pt. 3)
Final Argument
They're in the Trees + Devious Lick Simulator in Real Life (pt. 2)
A lot of Life Updates + Korean War Simplified (a lot)
Concert Updates + programmeruser being dumb (ft. some kid called toni)
Low Quality JToH Update + What happens today
Here's the plan...
The dangers and problems with false information
Where should I go?
Finishing Geometry + Math Problems
Congratulations, programmeruser! Your extended unban warranty has expired!
Stupid + unfunny weekend + today I guess
April 2022
Funny things my parents did
Why do we act the way we do online?
Life is Long Term
Updates on Life
The group effect
No such thing as "random" and mislabeling
Alcumus Geometry Grind + Contradictory Statements
2000 views, here we are again.
Spring Break!
Kid bullies homosexual people, instantly regrets it (he actually does)
111th post - MATH PROBLEMS + UPDATE ON LIFE
Updates on why programmeruser is stalking me
AMC8 results
March 2022
My weird dreams + sleep paralysis demons
Salty kid tries to beat me up, LIVES TO REGRET IT AHAHAAAHAA
Optimized Mining Methods + Blackmail at school
I had a dream.
programmeruser you little-
do you are have stupid
School SMP drama :(
I'm sorry Rudra1101001 + Canceled basketball
How I'm going to trap my friends in the School SMP
School SMP!
Thanks Cisco Umbrella
programmeruser why did you do this.
March 1st
February 2022
Graphing NFTs with sine, cosine, and tangents on Desmos
Tuesday, Twosday, Two's Day.
A comprehensive guide to employing efficient confusion tactics on others.
My mom's birthday + epic valentine's day!
Butterfly clicking = epic weapon + me failing at math haaha
Is it just me or is it just my parents + my aim is terrible
This is annoying-
Happy Chinese New Year!
January 2022
How to make a convincing rickroll
Me when my ping is 1.3 million: ah yes minor lag + anime battle with a weirdo
AMC8 DHR cutoffs guess?
How long would a hypothetical World War 3 be?
And suddenly, it ended.
Analysis of teachers and substitute teachers
Packing lunch and eating free school lunch + a really salty baka
Sneaky advertisement...
Apparently Ad blocker = dreambox blocked??
It's still not safe in here.
It had to be done.
New hobby I have
Qb abg or nsenvq. Jr ner whfg yvxr lbh.
December 2021
Christmas turned out to be a NERF gun battle to the death
Boring day
Overpricing things to get profit
My honest review on the new Christmas Update
MATHCOUNTS: An end and a beginning
Super Mario World: The grind is over
I didn't make it
The brink of failure: MATHCOUNTS Target Round
My parents stalk me too much
Chef akliu teaches you all how to make impasta
My classmates are breaking limbs at an alarming rate
fps shooter games are OP for aim training
it's so good... I WANT MORE. NOW- I mean please.
help programmeruser is stalking me
EXCUSE ME? WHAT IS THIS @JWENSLAWSKI
MATHCOUNTS target round makes me commit upside down frown neck snap
Short grade update (I have to go for today)
Middle School Math is the key to transcendence
Start of math grind + mc tournament
dying to mobs with hacks on + math grind
floating water and child labor ._.
me being very stupid
I'm a total failure
November 2021
Did MATHCOUNTS test, absolutely failed
MATHCOUNTS test is tomorrow insert laughing/choking noise
School is starting again... Oh no
I'm the opposite of a chad + raising my motivation with deep thought and logic
MATHCOUNTS Practice Test
Avenging my Minecraft dog + School is over!
School discord block bypass + my Minecraft dog died
I had a concert today
Cat Mario Level 1-2 (aboveground) walkthrough/tutorial
Final step to dominate a Minecraft server
Changed CSS!
Escaping spawn and making a new life.
The plan to dominate a Minecraft server
FINAL DAY TO VOTE
Recommendations for hack clients?
What color do you prefer???
Red mode css!!!
Becoming the 3rd best player on an Anarchy server
Ending racist "friendships" and dying to lava
How my school black market shut down the school store
Teacher lectures us about bullying -_-
Thanks!!!
2 entries today (probably) + CSS updates
I had MATHCOUNTS today without doing any math.
Had orchestra today + Rant on school curriculum
Added some people + CSS Changes?
Updates on today
Got my second vaccine yesterday
Today was... okay.
2048 Strategy + What happened today
Kahoot botter gone wrong...
What happened in my PE class today...
My ELA class is doing dialogues... and something happened.
If you can drag click, read this... It's urgent.
August 2021
The Minecraft update that will break all servers
The weird thing about Anarchy in Minigame servers...
Went to the beach 4 days ago
Problems with Alcumus
Do you think anyone will beat this roblox obby?
Opinion on ODing
Windows vs Linux: Who's better?
The problem with copyright these days...
I wonder why people even hack at all...
July 2021
Why The Bridge is the best skill changer in all Minecraft minigames
Opinion on my avatar?
Why Champion Island is the best Google Doodle so far
Why Alcumus could improve a bit
Playing the violin is annoying
New Blog
Shouts
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akliu contrib trust
by stjwyl, Sep 22, 2025, 10:08 PM
How do you deal with pressure from college apps? I struggle a lot with comparison ("holy moly i'm screwed look at ___," or "bruh 1 time AIME qual is not getting me anywhere").
by NamelyOrange, Sep 18, 2025, 2:54 PM
W w w orzorzorz blog
by MaxIsW, Sep 8, 2025, 3:58 AM
Yo akliu if you can listen to Chinese I HIGHLY recommend this yt channel to you:
by Eric.s.lu, Sep 7, 2025, 10:59 PM
can you slide contrib
by aidan0626, Aug 10, 2025, 7:37 AM
hiiiiiiiiiiiii
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acklew orz contribute plz
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contrafbrifubrulation given
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conflibulation pluz
by juicetin.kim, May 18, 2025, 9:52 PM
Tuff
by centslordm, May 18, 2025, 12:04 AM
Wow! I never have the patience to write that much. Contrib?
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contrib?
by Math-lover1, May 13, 2025, 5:04 PM
Hi, can I have contrib?
by FalconMaster, Apr 28, 2025, 10:17 PM
contrib btw?
by Aminecraftbear, Apr 28, 2025, 4:50 PM
Okay, so we've reached 300 views... I don't really want to do a sort of contributor giveaway, because that would be cheap and I do that free anyways, so yeah.
I mean I guess I could like do something video game or blog related where I post/join something
by akliu, Nov 5, 2021, 11:16 PM
I didn't steal it from helios8 btw
by akliu, Nov 4, 2021, 7:39 PM
this CSS is LIT
by hh99754539, Nov 4, 2021, 12:10 AM
hello
by Helios8, Oct 19, 2021, 1:32 PM
css or z
by Lilathebee, Oct 10, 2021, 5:31 PM
u h
by horse869, Sep 15, 2021, 12:12 AM
its fine ig :DD
~~i was admin on ur old blog
and ur admin on mine i think~~
by StanziMarcello, Aug 24, 2021, 11:17 PM
thanks
by akliu, Aug 21, 2021, 6:17 PM
cool CSS!!!!!!!!!
by KNM, Aug 21, 2021, 1:02 PM
OOPS SORRY
by akliu, Aug 16, 2021, 2:08 PM
GaSp
you made a new blog, and didnt tell me?
im offended.
contrib/admin plz
by StanziMarcello, Aug 14, 2021, 10:23 PM
Added you! Sorry for the wait.
by akliu, Aug 10, 2021, 2:36 PM
contrib pls
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The CSS here isn't really different compared to my old blog.
by akliu, Jul 21, 2021, 6:36 PM
661 shouts
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5
Throwing m balls into n boxes, what is the probability that exactly two boxes have the same number of balls?
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Apoorv Khandelwal
Math enthusiast. · Author has 179 answers and 690.6K answer views
·Updated 7y
Generalization:
We have n∈N∖{1}n∈N∖{1} boxes. We are throwing m∈N m∈N balls into them (each ball has equal probability of going to each box). We wish to find the probability that there is exactly 1 1 set of k∈N∖{1}k∈N∖{1} boxes that have the same quantity of balls, while all the other n−k n−k boxes contain distinct quantities of balls.
For our specific question, k=2 k=2. For example, if m=6 m=6 and n=4 n=4, then a final boxes state of B 1=[0,2,4,0]B 1=[0,2,4,0] or B 2=[4,1,1,0]B 2=[4,1,1,0] would satisfy our criteria, since exactly 2 2 boxes ({1,4}∈B 1{1,4}∈B 1 or {2,3}∈B 2{2,3}∈B 2) would have the same quantity
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Generalization:
We have n∈N∖{1}n∈N∖{1} boxes. We are throwing m∈N m∈N balls into them (each ball has equal probability of going to each box). We wish to find the probability that there is exactly 1 1 set of k∈N∖{1}k∈N∖{1} boxes that have the same quantity of balls, while all the other n−k n−k boxes contain distinct quantities of balls.
For our specific question, k=2 k=2. For example, if m=6 m=6 and n=4 n=4, then a final boxes state of B 1=[0,2,4,0]B 1=[0,2,4,0] or B 2=[4,1,1,0]B 2=[4,1,1,0] would satisfy our criteria, since exactly 2 2 boxes ({1,4}∈B 1{1,4}∈B 1 or {2,3}∈B 2{2,3}∈B 2) would have the same quantity of balls (0 0 or 1 1), respectively. On the other hand, a final boxes state of B 3=[0,1,2,3]B 3=[0,1,2,3], B 4=[2,1,2,1]B 4=[2,1,2,1], or B 5=[1,1,1,3]B 5=[1,1,1,3] would not satisfy our criteria, since no 2 2 boxes in B 3 B 3 would have the same quantity of balls, B 4 B 4 would have 2≠1 2≠1 sets of 2 2 boxes that have the same quantity of balls, and B 5 B 5 would have 1 1 set of 3≠2 3≠2 boxes that have the same quantity of balls.
Answer:
In general, the answer is m!(n−1)!n m−1 k!⌊m k⌋∑b=0(b!)−k∑s∈S b n−k∏i=1(s i!)−1 m!(n−1)!n m−1 k!∑b=0⌊m k⌋(b!)−k∑s∈S b∏i=1 n−k(s i!)−1, where S b S b is described in detail in Reasoning.
For the example I provided above, we get 6!3!4 5 2!3∑b=0(b!)−2∑s∈S b(s 1!s 2!)−1 6!3!4 5 2!∑b=0 3(b!)−2∑s∈S b(s 1!s 2!)−1
=135 64{(0!)−2[(1!5!)−1+(2!4!)−1]+(1!)−2(0!4!)−1+(2!)−2(0)+(3!)−2(0)}=135 64{(0!)−2[(1!5!)−1+(2!4!)−1]+(1!)−2(0!4!)−1+(2!)−2(0)+(3!)−2(0)}
=153 1024≈14.9%=153 1024≈14.9%.
Reasoning:
First, we can compute the probability that a given final box configuration B=[a 1,...,a n]B=[a 1,...,a n] occurs (here, a i≥0 a i≥0 and n∑i=1 a i=m∑i=1 n a i=m). This probability is (m a 1,...,a n)(1 n)m(m a 1,...,a n)(1 n)m, by noticing that a Multinomial distribution fits. The first term (m a 1,...,a n)=m!a 1!...a n!(m a 1,...,a n)=m!a 1!...a n! counts the number of orders in which the balls could have landed in the relevant boxes, while the (1 n)m(1 n)m accounts for the entire sample space (since each of m m balls have n n box choices).
Now, we partition the answer into the quantity of balls b∈N b∈N that is the common amount held in each of the k k same-quantity boxes. Note that it must be that 0≤b≤⌊m k⌋0≤b≤⌊m k⌋, since having more than ⌊m k⌋⌊m k⌋ balls per same-quantity box would cause the k k boxes to have more than the allowed m m balls. For each b b, we can obtain a set S b S b of the sets of distinct ball quantities that can go in the remaining n−k n−k boxes. For my above example, if b=0 b=0, then 2 2 of the 4 4 boxes would contain 0 0 boxes, meaning that the only sets of possibilities for the ball quantities for the remaining 2 2 boxes would be S 0={{1,5},{2,4}}S 0={{1,5},{2,4}}.
Finally, there are (n k)(n−k)!(n k)(n−k)! box configurations for each S b S b. This is because there are (n k)(n k) ways to choose the k k boxes out of n n that have the same quantities of balls, and there are (n−k)!(n−k)! ways to arrange the other distinct n−k n−k ball quantities among the remaining boxes. Putting all the components together, we get (n k)m!(n−k)!n m⌊m k⌋∑b=0(b!)−k∑s∈S b n−k∏i=1(s i!)−1(n k)m!(n−k)!n m∑b=0⌊m k⌋(b!)−k∑s∈S b∏i=1 n−k(s i!)−1, which simplifies to the expression in Answer.
Code:
The below Python script verifies my answer by testing it against simulations.
import math
import random
def main():
numBalls = 10
numBins = 7
numSame = 3
numExperiments = 100000
confidenceThreshold = 0.01
probDistribution = simulate(numBalls,
numBins,
numSame,
numExperiments)
computeProb = compute(numBalls,
numBins,
numSame,
probDistribution)
simulateProb = 0.0
for bins in probDistribution:
if meetsCriteria(bins, numSame):
simulateProb += probDistribution[bins]
print "sim: " + str(simulateProb) + ", comp: " + str(computeProb)
if abs(simulateProb - computeProb) >= confidenceThreshold:
raise Exception("Simulated and computed probabilities far off!")
Given "numBalls" that are thrown at "numBins" bins, this will simulate
and return the probability distribution of the final bin states that
it observes. It runs "numExperiments" experiments, from which it
determines an empirical distribution.
def simulate(numBalls,
numBins,
numSame,
numExperiments):
probDistribution = dict()
for experiment in range(numExperiments):
bins = numBins
for ball in range(numBalls):
bins[int(random.random() numBins)] += 1
tupleBins = tuple(bins)
if tupleBins not in probDistribution:
probDistribution[tupleBins] = 0
probDistribution[tupleBins] += 1.0
for possibleResult in probDistribution:
probDistribution[possibleResult] /= numExperiments
return probDistribution
Given "numBalls" that are thrown at "numBins" bins, this will compute
the probability that exactly 1 set of "numSame" bins has the same
number of balls. All the remaining ("numBins" - "numSame") bins must
have distinct numbers of balls in them. Also, this is given
"expectedResults" for convenience, so it only needs to compute the
probabilities for the bin situations in there.
def compute(numBalls,
numBins,
numSame,
expectedResults):
toReturn = 0.0
sortedResults = set()
for expectedResult in expectedResults:
sortedResult = tuple(sorted(expectedResult))
if sortedResult not in sortedResults:
sortedResults.add(sortedResult)
if meetsCriteria(expectedResult, numSame):
toReturn += getProbability(numBalls,
numBins,
numSame,
expectedResult)
return toReturn
Given "numBalls" that are thrown at "numBins" bins, this will compute
the probability of seeing "expectedResult" arranged in any order, at
the end. Assumes that the "expectedResult" is a bin configuration that
meets our criteria.
def getProbability(numBalls,
numBins,
numSame,
expectedResult):
toReturn = getSingleProbability(numBalls, numBins, expectedResult)
return toReturn math.factorial(numBins) / math.factorial(numSame)
Given "numBalls" that are thrown at "numBins" bins, this will compute
the probability of seeing exactly "expectedResult" at the end.
def getSingleProbability(numBalls, numBins, expectedResult):
toReturn = float(math.factorial(numBalls - 1))
toReturn /= (numBins (numBalls - 1))
for currentBinBalls in expectedResult:
toReturn /= math.factorial(currentBinBalls)
return toReturn
Given the current state of the "bins", returns True iff exactly 1 set of
"numSame" bins has the same number of balls, while the rest of the bins
contain distinct numbers of balls.
def meetsCriteria(bins, numSame):
binCountToNumBins = dict()
for binCount in bins:
if binCount not in binCountToNumBins:
binCountToNumBins[binCount] = 0
binCountToNumBins[binCount] += 1
toReturn = False
for binCount in binCountToNumBins:
numBinsWithBinCount = binCountToNumBins[binCount]
if numBinsWithBinCount != 1:
if toReturn or numBinsWithBinCount != numSame:
return False
toReturn = True
return toReturn
if name == 'main':
main()
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Lance Berg
Author has 27.9K answers and 54.4M answer views
·5y
Let’s assume the number of balls is 0. Then if and only if the number of boxes is two, the probability is 100%. For any other number of boxes, with no balls the probability is 0%
Let’s assume the number of balls is 1. Then if and only if the number of boxes is three, the probability is 100%, again for any other number of balls the probability is 0%
Let’s assume the number of balls is 2. If the number of boxes is 1, the probability is 0. If the number of boxes is 2, the probability is 1/2. If the number of boxes is 3, the probability is 1/3. But if the number of boxes is 4, then the probability i
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Let’s assume the number of balls is 0. Then if and only if the number of boxes is two, the probability is 100%. For any other number of boxes, with no balls the probability is 0%
Let’s assume the number of balls is 1. Then if and only if the number of boxes is three, the probability is 100%, again for any other number of balls the probability is 0%
Let’s assume the number of balls is 2. If the number of boxes is 1, the probability is 0. If the number of boxes is 2, the probability is 1/2. If the number of boxes is 3, the probability is 1/3. But if the number of boxes is 4, then the probability is 0, because if two boxes each have one ball, there are two empty boxes, which means 4 boxes each have a partner box with the same number of balls. With n>4, the probability is 1/(n+1). So that’s pretty messy really.
Frankly, I don’t see any way to generalize this to a simple function, much less a single probability.
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N balls are randomly dropped into k boxes (k <=n). What is the probability that no box is empty?
If there are 4 boxes, containing distinct number of balls, then what is the minimum number of balls drawn to ensure a ball is drawn from each box?
There are three boxes, one containing two black balls, another containing two white balls, and another containing one black and one white ball. You select a random ball from a random box and you find you selected a white ball. What is the prob. that the other ball in the same box is also white?
How do you sort identical balls into identical boxes where each box has at least one ball and no two boxes have the same number of balls?
How many ways are there to put 17 balls into 12 boxes with at least 1 ball in each box?
Michael Lamar
PhD in Applied Mathematics · Author has 3.7K answers and 17.5M answer views
·10y
Related
N balls are randomly dropped into k boxes (k <=n). What is the probability that no box is empty?
Here's the exact answer:
p=∑k−1 h=0(−1)h(k h)(1−h k)n p=∑h=0 k−1(−1)h(k h)(1−h k)n
The sum in the answer comes from the inclusion exclusion principle. Term h inside the sum (without the sign) represents the probability that all the balls landed in some group of k−h k−h boxes. So we take the probability that all the balls landed in some group of all k boxes, then subtract the probability that all the balls landed in some group of (k-1) boxes, then add back the probability that all the balls landed in some group of (k-2) boxes and so on.
An interesting result that (to me) was a little sur
Continue Reading
Here's the exact answer:
p=∑k−1 h=0(−1)h(k h)(1−h k)n p=∑h=0 k−1(−1)h(k h)(1−h k)n
The sum in the answer comes from the inclusion exclusion principle. Term h inside the sum (without the sign) represents the probability that all the balls landed in some group of k−h k−h boxes. So we take the probability that all the balls landed in some group of all k boxes, then subtract the probability that all the balls landed in some group of (k-1) boxes, then add back the probability that all the balls landed in some group of (k-2) boxes and so on.
An interesting result that (to me) was a little surprising. With 100 balls, it turns out that 27 bins is as close as possible to the 50/5o point with the probability of no box empty at just over 51%.
Here's an Octave/Matlab script for quickly estimating the probability that verifies this result:
n = 100; %number of balls
k = 27; %number of boxes
m = 1e5; %number of trials
r = sort(ceil(krand(n,m))); %put all the balls into a box at random for each trial then sort box numbers within a trial in ascending order
r = [zeros(1,m); r;(k+1)ones(1,m)]; %add a zero before the first ball and a k+1 after the last ball so that we can detect if either of those boxes is empty
d = diff(r ); %difference between box numbers of consecutive balls. A jump of two or more means a box was skipped
p_estimated = 1-mean(any(d>=2)); %look for places where d>=2 which means a box was skipped. If it happens anywhere in a trial, mark that trial as a one. Otherwise, trial is zero. Then take the mean of all trial to get the probability of at least one empty box. Subtract from one to get the probability of all non-empty boxes.
disp(p_estimated) %display the result
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Bot
·
1y
To find the probability that exactly two boxes have the same number of balls when throwing m m balls into n n boxes, we can follow these steps:
Total Outcomes: First, we need to determine the total number of ways to distribute m m balls into n n boxes. This is given by n m n m since each ball can go into any of the n n boxes independently.
Favorable Outcomes: Next, we need to count the number of distributions where exactly two boxes contain the same number of balls and the remaining boxes contain different numbers of balls.
Let's denote the counts of balls in the boxes as (x 1,x 2,…,x n)(x 1,x 2,…,x n), where x_ x_
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To find the probability that exactly two boxes have the same number of balls when throwing m m balls into n n boxes, we can follow these steps:
Total Outcomes: First, we need to determine the total number of ways to distribute m m balls into n n boxes. This is given by n m n m since each ball can go into any of the n n boxes independently.
Favorable Outcomes: Next, we need to count the number of distributions where exactly two boxes contain the same number of balls and the remaining boxes contain different numbers of balls.
Let's denote the counts of balls in the boxes as (x 1,x 2,…,x n)(x 1,x 2,…,x n), where x i x i is the number of balls in box i i. We want exactly two of these x i x i to be equal, say x 1=x 2=k x 1=x 2=k, and the rest x 3,x 4,…,x n x 3,x 4,…,x n to be distinct and different from k k.
The steps to achieve this are:
Choose 2 Boxes: Choose 2 boxes to have the same number of balls, which can be done in (n 2)(n 2) ways.
Choose the Count for the Chosen Boxes: Let k k be the number of balls in the selected boxes. The total number of balls in these two boxes is 2 k 2 k. The remaining m−2 k m−2 k balls must be distributed among the other n−2 n−2 boxes.
Distribute Remaining Balls: The n−2 n−2 boxes must contain m−2 k m−2 k balls in such a way that each box has a distinct number of balls. This requires that m−2 k m−2 k must be at least 0 0 and must also allow for distinct distributions. The maximum number of distinct counts we can assign to the n−2 n−2 boxes is n−2 n−2, so we require m−2 k≥(n−2)(n−3)/2 m−2 k≥(n−2)(n−3)/2.
Count the Distinct Distributions: The number of ways to assign m−2 k m−2 k balls to n−2 n−2 boxes with distinct counts is a more complex combinatorial problem, typically requiring generating functions or combinatorial enumeration techniques.
Probability Calculation: Finally, the probability P P that exactly two boxes contain the same number of balls is given by:
P=Number of favorable outcomes n m P=Number of favorable outcomes n m
Conclusion
The exact probability depends heavily on the values of m m and n n and requires careful combinatorial reasoning based on the distribution of balls and the restrictions imposed by the distinct counts. For specific values of m m and n n, the counting of favorable outcomes can be more explicitly calculated.
If you provide specific values for m m and n n, I can help you compute the probability more concretely.
Upvote ·
Saswata Banerjee
Author has 1.7K answers and 2.2M answer views
·11mo
Related
What is the probability of placing n balls into n numbered boxes so that exactly one box remains empty?
A n s:n(n−1)×n!(n−1)!(2 n−1)!A n s:n(n−1)×n!(n−1)!(2 n−1)!
I will assume that the balls are identical. The reason for such an assumption is that the answer can get very messy if the balls are distinguishable. Just in case the probability is sought for distinguishable balls, please let me know through the comments and I shall try to post for that case as well.
I will assume that the balls are being dropped into the boxes randomly.
First let’s calculate the number of ways of distributing the balls.
Let the number of balls in box number 1 be denoted by k k
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A n s:n(n−1)×n!(n−1)!(2 n−1)!A n s:n(n−1)×n!(n−1)!(2 n−1)!
I will assume that the balls are identical. The reason for such an assumption is that the answer can get very messy if the balls are distinguishable. Just in case the probability is sought for distinguishable balls, please let me know through the comments and I shall try to post for that case as well.
I will assume that the balls are being dropped into the boxes randomly.
First let’s calculate the number of ways of distributing the balls.
Let the number of balls in box number 1 be denoted by k 1 k 1, the number of balls in box number 2 2 be denoted by k 2 k 2 and so on, such that
k 1+k 2+…+k n=n k 1+k 2+…+k n=n where k 1,k 2,…k n∈Z k 1,k 2,…k n∈Z and k 1,k 2,…k n≥0 k 1,k 2,…k n≥0
We can use the method of stars and bars to calculate the number of ways of distribution as
=n+n−1 C n=n+n−1 C n
=2 n−1 C n=2 n−1 C n
Now, let’s calculate the no. of favourable ways of distribution.
We can select the box to be kept empty in n n ways. The remaining n−1 n−1 boxes have to have at least 1 1 ball. So, we distribute a ball each in these n−1 n−1 boxes in exactly 1 1 way. Now, 1 1 ball remains that we can place in any of these n−1 n−1 boxes.
∴∴ The number of favourable ways of distribution
=n(n−1)=n(n−1)
∴∴ The required probability
=n(n−1)÷2 n−1 C n=n(n−1)÷2 n−1 C n
=n(n−1)÷(2 n−1)!n!(n−1)!=n(n−1)÷(2 n−1)!n!(n−1)!
=n(n−1)×n!(n−1)!(2 n−1)!=n(n−1)×n!(n−1)!(2 n−1)!
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Six different balls are put in three different boxes, no box being empty. The probability of putting balls in boxes in equal numbers is what?
What is the method for solving probability problems involving boxes with different numbers of balls in each box?
How many ways are there to distribute n identical balls into m boxes where each box can take any number of balls?
There are N different balls in a box. If I pick one each time and then return it to the box. What is the probability of picking all the balls after m times (m > N)? And what is the probability of k balls are never picked?
What is the probability of distributing 5 balls into three different boxes such that each box can have any number of balls and each ball is equally probable to be in any one of the boxes?
Jonathan Roberts
Lives in England · Author has 3.4K answers and 2.6M answer views
·1y
Related
There are 100 boxes. What’s the expected amount of balls needed such that randomly distributing them inside the boxes leaves half the boxes empty? What about N boxes?
When there are N N boxes then, for a given box, the probability that a specific box remains empty after k k balls have been randomly distributed is:
(1−1 N)k(1−1 N)k
The expected number of empty boxes E E can be calculated by multiplying the probability that one box is empty by the total number of boxes N N.
The expected number of empty boxes after k k balls is:
E empty=N(1−1 N)k E empty=N(1−1 N)k
We want half the boxes to be empty on average, so:
N 2=N(1−1 N)k N 2=N(1−1 N)k
divide both sides by N N:
\displaystyle\frac\displaystyle\frac
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When there are N N boxes then, for a given box, the probability that a specific box remains empty after k k balls have been randomly distributed is:
(1−1 N)k(1−1 N)k
The expected number of empty boxes E E can be calculated by multiplying the probability that one box is empty by the total number of boxes N N.
The expected number of empty boxes after k k balls is:
E empty=N(1−1 N)k E empty=N(1−1 N)k
We want half the boxes to be empty on average, so:
N 2=N(1−1 N)k N 2=N(1−1 N)k
divide both sides by N N:
1 2=(1−1 N)k 1 2=(1−1 N)k
take log of both sides:
log(1 2)=log((1−1 N)k)log(1 2)=log((1−1 N)k)
use the logarithm power rule:
log(1 2)=k log(1−1 N)log(1 2)=k log(1−1 N)
divide both sides by log(1−1 N):divide both sides by log(1−1 N):
k=log(1 2)log(1−1 N)k=log(1 2)log(1−1 N)
When N=100,k=log(1 2)log(1−1 100)When N=100,k=log(1 2)log(1−1 100)
k=69 k=69
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Kavita Chawdhary
Student of Statistics · Author has 170 answers and 1.3M answer views
·8y
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5 balls are distributed into 3 boxes in a way that each of 5 balls can get into any of boxes. What is the probability that exactly one box is empty?
There are no terms and conditions specified in the question as to whether the balls and boxes are identical or distinct.
This question can have four different answers based upon the assumptions.
Ok, let’s assume that we have distinct balls and distinct boxes (one of the four possible conditions).
Number of ways in which ’n’ distinct balls can be distributed into ‘k’ distinct boxes = k n k n
We have 5 balls and 3 boxes. Thus, total number of outcomes = 3 5 3 5
If any one of the 3 distinct boxes is empty, then 5 balls to the other two boxes can be distributed in (2 5−2)(2 5−2) ways (we have subtracted 2 for tho
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There are no terms and conditions specified in the question as to whether the balls and boxes are identical or distinct.
This question can have four different answers based upon the assumptions.
Ok, let’s assume that we have distinct balls and distinct boxes (one of the four possible conditions).
Number of ways in which ’n’ distinct balls can be distributed into ‘k’ distinct boxes = k n k n
We have 5 balls and 3 boxes. Thus, total number of outcomes = 3 5 3 5
If any one of the 3 distinct boxes is empty, then 5 balls to the other two boxes can be distributed in (2 5−2)(2 5−2) ways (we have subtracted 2 for those possible distributions where other boxes can also get a 0 ball, i.e. (5, 0), (0, 5)). Similarly, we have ((2 5−2)2 5−2) possible ways for each of the other two boxes.
Thus, favourable number of outcomes = (2 5−2).3(2 5−2).3
Total probability = (2 5−2).3 3 5(2 5−2).3 3 5 = 0.37 0.37
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Anastasia Kukanova
falling in love with statistics · Author has 64 answers and 473.1K answer views
·9y
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What is the probability that exactly two boxes are empty when 5 identical balls are randomly distributed to 5 distinct boxes?
I'll be using formulas for the number of combinations and the number combinations with repetitions, so let me remind you of them.
The number of combinations:
(n k)=n!k!(n−k)!(n k)=n!k!(n−k)!.
The number of combinations with repetitions:
((n k))=(n+k−1 k)((n k))=(n+k−1 k).
First, let's find out how many ways there are to to distribute 5 identical balls into 5 distinct boxes. We have to choose from 5 boxes 5 times, with repetition, the order doesn't matter since balls are identical, so that will be
m=\left(!!\binom{5}{5}!!\right)=\binom{9}{5}=\frac{9!}{5!4!}=\frac m=\left(!!\binom{5}{5}!!\right)=\binom{9}{5}=\frac{9!}{5!4!}=\frac
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I'll be using formulas for the number of combinations and the number combinations with repetitions, so let me remind you of them.
The number of combinations:
(n k)=n!k!(n−k)!(n k)=n!k!(n−k)!.
The number of combinations with repetitions:
((n k))=(n+k−1 k)((n k))=(n+k−1 k).
First, let's find out how many ways there are to to distribute 5 identical balls into 5 distinct boxes. We have to choose from 5 boxes 5 times, with repetition, the order doesn't matter since balls are identical, so that will be
m=((5 5))=(9 5)=9!5!4!=6⋅7⋅8⋅9 2⋅3⋅4=2⋅7⋅9 m=((5 5))=(9 5)=9!5!4!=6⋅7⋅8⋅9 2⋅3⋅4=2⋅7⋅9.
Now let's count ways to leave exactly 2 boxes empty. There are (5 2)(5 2) ways to choose those boxes. To make sure only 2 boxes are empty, we have to put one ball into each of other 3 boxes, and then we have to distribute 2 balls into 3 boxes, again, with repetition. There are ((3 2))((3 2)) ways to do so. Hence the overall number is
l=(5 2)⋅((3 2))=(5 2)⋅(4 2)=5!4!2!3!2!2!=3⋅4⋅5 l=(5 2)⋅((3 2))=(5 2)⋅(4 2)=5!4!2!3!2!2!=3⋅4⋅5.
Then the probability of that event is
P(A)=l m=3⋅4⋅5 2⋅7⋅9=10 21 P(A)=l m=3⋅4⋅5 2⋅7⋅9=10 21.
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Martin Jansche
biased estimator · Author has 3.7K answers and 3.6M answer views
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There are 10 boxes and 10 balls, each numbered from 1 to 10. Each box will randomly receive one ball. What is the probability that exactly 6 boxes contain a ball with a number corresponding to the box number?
Let’s generalize in such a way that the specific case at hand amounts to d=4 d=4 and k=6 k=6.
Out of all n!n! permutations of n=d+k n=d+k balls, we want to count how many leave k k balls in place and permute the other d d balls in such a way that not a single one of those d d balls ends up in the box with the corresponding number. The number of such derangements is given by !d=[d!/e]!d=[d!/e] where [x][x] is the nearest integer to x x and e≈2.718 e≈2.718 is Euler’s number. And of course there are (n d)(n d) ways of choosing d d balls to be deranged.
The required probability is therefore
\displaystyle p(d,k)=\frac{!d\,{d+k\choose\displaystyle p(d,k)=\frac{!d\,{d+k\choose
Continue Reading
Let’s generalize in such a way that the specific case at hand amounts to d=4 d=4 and k=6 k=6.
Out of all n!n! permutations of n=d+k n=d+k balls, we want to count how many leave k k balls in place and permute the other d d balls in such a way that not a single one of those d d balls ends up in the box with the corresponding number. The number of such derangements is given by !d=[d!/e]!d=[d!/e] where [x][x] is the nearest integer to x x and e≈2.718 e≈2.718 is Euler’s number. And of course there are (n d)(n d) ways of choosing d d balls to be deranged.
The required probability is therefore
p(d,k)=!d(d+k d)(d+k)!=[d!/e]d!k!≈1 k!e p(d,k)=!d(d+k d)(d+k)!=[d!/e]d!k!≈1 k!e
In PARI/GP:
? p(d,k) = round(d!/exp(1))/(d!k!)
%1 = (d,k)->round(d!/exp(1))/(d!k!)
? p(4, 6)
%2 = 1/1920
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11277 | https://www.doubtnut.com/qna/38183444 | The line x−y=1 intersects the parabola y2=4x at A and B . Normals at AandB intersect at C. If D is the point at which line CD is normal to the parabola, then the coordinates of D are (4,−4) (b) (4,4) (−4,−4) (d) none of these
(4,-4)
(4,4)
(-4,-4)
none of these
The correct Answer is:B
(2) solving the line y=x-1 and the parabola y2=4x, we have (x−1)2=x orx2−6x+1=0 orx=3±√8 ∴y=2±√8 Suppose point D is (X3,y3). Then, y1+y2+y3=0 or2+√8+2−√8+y3=0 ory3=−4 Then x3=4. Therefore, the point is (4,4).
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CENGAGE-PARABOLA-Exercise (Single)
min[(x1-x^2)^2+(3+sqrt(1-x1 2)-sqrt(4x2))],AAx1,x2 in R ,
is
4sqrt(5)...
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of the parabola y^2=4x
meet ...
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...
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on the parabola y^2=4a x ,(a >0),
meets it again ...
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Degree of a Polynomial: Definition, Types, Examples, Facts, FAQs
Home » Math Vocabulary » Degree of a Polynomial: Definition, Types, Examples, Facts, FAQs
What Is the Degree of a Polynomial?
Degree of a Polynomial with One Variable
How to Find the Degree of a Polynomial
Solved Examples on Degree of a Polynomial
Practice Problems on Degree of a Polynomial
Frequently Asked Questions about the Degree of a Polynomial
What Is the Degree of a Polynomial?
The degree of a polynomial is the highest degree among the degrees of the individual terms present in the polynomial.
If the polynomial is in a single variable, the degree of a polynomial is the highest exponent of the variable with a non-zero coefficient.
If the polynomial has more than one variable, then we calculate the degree of each term separately and the highest degree among them represents the degree of the given polynomial.
The standard form of degree polynomial is given by
where, .
Here, the degree of the polynomial p(x) is n, because “n” is the highest power of variable “x.”
We can represent the degree of a polynomial p(x) by Deg(p(x)).
Example:
The degree of the polynomial is 4 because the highest power of the variable x is 4.
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Degree of a Polynomial: Definition
The degree of a polynomial is the highest power of the variable in the polynomial expression with a non-zero coefficient. The degree of a polynomial can also be defined as the greatest number among the degrees of the individual terms (monomials) with non-zero coefficients.
Degree of a Polynomial with One Variable
For a polynomial in one variable, the highest power of the variable in a polynomial is the degree of the polynomial. It is the highest exponent value of the variable in the given polynomial.
Examples:
Deg
Deg
Deg
Degree of a Polynomial with More than One Variable
When finding the degree of a polynomial when a polynomial has more than one variable, we first find the degree of each individual term by adding the exponents of each variable present in the term.
In other words, we find the degree of each monomial present in the polynomial. Finally, the degree of the polynomial is the largest degree among the degrees of individual terms.
Example:
The degree of the term is 5.
The degree of is 3 (Sum of exponents .
The term 3xy has a degree of 2 (Sum of exponents .
The highest degree is 5. Thus, the degree of polynomial is 5.
How to Find the Degree of a Polynomial
Before finding the degree, first combine all the like terms (terms having the same variables and the same exponents). This way we ensure that no two terms have the same degree.
Finding Degree of a Polynomial with Only One Variable
Step 1: Write the polynomial expression in standard form. Standard form of a polynomial refers to arranging the terms in descending order of their degrees.
Step 2: Identify the term with the highest power of the variable. This term must have a non-zero coefficient. In the standard form, the highest power term is the leading term.
Step 3: The degree of the polynomial is equal to the degree of the highest power term.
Example 1:
Adding the like terms together, we get
.
…the term vanishes due to the 0 coefficient.
This expression is already in standard form.
Here, the term with the highest power of x is .
Hence, the degree of the polynomial is 2.
Example 2:
Rearrange the terms in descending order of their degrees.
Here, the term with the highest power of x is .
Hence, the degree of the polynomial is 4.
Finding Degree of a Polynomial with More than One Variable
Step 1: Identify each term.
Step 2: Find the degree of each term. To find the degree of a term, add the exponents of variables present.
Step 3: Compare the degrees of individual terms. The highest degree among them is the degree of the polynomial.
Example:
Degree of
Degree of
Degree of
Highest degree
Degree of the given polynomial
Degree of Zero Polynomial
The zero polynomial is usually denoted by 0 or by the expression . The zero polynomial has no non-zero terms. A polynomial with zero coefficients is called a zero polynomial. It has no terms with non-zero coefficients.
The degree of a zero polynomial is considered to be undefined. Why so? We can rewrite as .
…
So, the degree of a zero polynomial is not defined. It has no degree.
Note: In mathematical practice, sometimes the degree of the zero polynomial is taken to be −∞ and sometimes it is considered undefined.
Degree of Constant Polynomial
A constant polynomial has no variable term. It only has a constant term. Thus, the degree of a constant polynomial is 0.
The constant polynomial where c is a non-zero constant has degree 0. It can be written as .
This is because there is no variable term present, and the highest power of x is 0, and .
Degree of a Polynomial: Applications
The concept of the degree of a polynomial has important applications in mathematics, science, and engineering. A few examples of how the degree of a polynomial can be used are listed below:
To figure out the maximum possible roots or solutions a function can have.
To figure out how many times a function crosses the x-axis on a graph.
We find the degree of each term to see if the polynomial expression is homogeneous. For example, in the polynomial , all of the terms have a degree of 3. So it is a degree 3 homogeneous polynomial.
Classification of Polynomials Based on Degree
A specific name has been given to each of the polynomials in accordance with their degree. Let’s classify polynomials according to their degree along with their example.
| Degree | Name of Polynomials | Examples |
---
| 0 | Constant Polynomial | |
| 1 | Linear Polynomial | |
| 2 | Quadratic Polynomial | |
| 3 | Cubic Polynomial | |
| 4 | Bi-quadratic Polynomial | |
Facts about Degree of a Polynomial
The degree of a polynomial is always a whole number. The degree of a polynomial is always a non-negative integer. This means that the degree is either zero or a positive integer.
Degree of a polynomial can never be a negative or a fractional number.
A monic polynomial is a non-zero polynomial in a single variable in which the leading coefficient is equal to 1.
A polynomial of degree n can have a maximum n number of zeros.
A polynomial whose non-zero terms all have the same degree is called a homogeneous polynomial.
Conclusion
In this article, we learned about the degree of polynomial and the method to find the degree of the polynomial. Let’s solve a few examples and practice problems.
Solved Examples on Degree of a Polynomial
Determine the degree of the polynomial .
Solution:
The polynomial is written in the standard form.
In this case, the highest power of x is 4.
Thus, the degree of p(x) is 4.
Find the degree and leading coefficient of the polynomial .
Solution:
In this case, the highest power of x is 5.
Thus, the degree of g(x) is 5.
Leading term
Leading coefficient
Find the degree of the polynomial .
Solution:
The polynomial has more than 2 variables.
In this case, we first find the degree of each term by adding the exponents.
Degree of
Degree of
Degree of
Degree of
Highest degree
Thus, the degree of is 5.
The length and width of a rectangular garden are $(2x + 3)$ and respectively. What is the degree of the polynomial that represents the area of the garden?
Solution:
The area of the garden is given by the product of the length and the width.
Therefore, area of rectangular garden
Hence, the degree of the polynomial that represents the area of the garden is 2.
What is the degree of the polynomial ?
Solution:
In a polynomial with more than one variable, the degree can be determined by adding the exponents of each variable.
The term with the highest degree is .
Degree of .
Therefore, the degree of the polynomial P(x, y) is 4.
Practice Problems on Degree of a Polynomial
Degree of a Polynomial: Definition, Types, Examples, Facts, FAQs
Attend this quiz & Test your knowledge.
1
What is the degree of the polynomial ?
1
2
3
4
CorrectIncorrect
Correct answer is: 2
In , the highest power of x is 2, so the degree of p(x) is 2.
2
The degree of the cubic polynomial is_______.
1
2
3
4
CorrectIncorrect
Correct answer is: 3
The degree of the cubic polynomial is 3.
3
What is the degree of the polynomial ?
8
4
5
3
CorrectIncorrect
Correct answer is: 4
In , the highest power of x is 4, so the degree of h(x) is 4.
4
The polynomial with degree 2 is known as________.
linear
biquadratic
cubic
quadratic
CorrectIncorrect
Correct answer is: quadratic
The polynomial with degree 2 is known as a quadratic polynomial.
5
Which of the following is a linear polynomial?
CorrectIncorrect
Correct answer is:
The polynomial with degree 1 is called a linear polynomial. It is in the form of .
Out of the given options, is a linear polynomial.
6
What is the degree of the polynomial ?
1
2
3
CorrectIncorrect
Correct answer is: 0
is a constant polynomial. Thus, the degree of is 0.
Frequently Asked Questions about the Degree of a Polynomial
What is a constant term in a polynomial?
What is a leading coefficient in a polynomial?
What is a monomial?
What is a binomial?
What is a trinomial? |
11279 | https://www.themathdoctors.org/integration-choosing-a-substitution-to-try/ | Typesetting math: 100%
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Integration: Choosing a Substitution to Try
Having looked at two basic techniques of integration, let’s start putting things together. How do you approach an integral without knowing what method to use? We’ll focus on substitution here, which is also called “change of variables”.
Integration isn’t easy!
Let’s start with a general question, from 1998:
```
Tips for Integrating Functions
Hi!
I have been looking all over the Net for the past few weeks for a good tutorial on finding anti-derivatives, and haven't had much luck. There is an abundance of excellent sources on Derivative Calculus but none on Integral, so I am turning to you for help.
Unfortunately it seems there is not an easy formula for finding an Anti-Derivative, as there is for finding a derivative (f'(x)=(f(x+h)-f(x))/h). So if a good explanation would be too lengthy, even a pointer to a tutorial would be great.
Thanks for maintaining such a wonderful site, as it has helped me with numerous issues, not to mention it's just cool going "Ahhh, so that's why." =)
Brian Cowan
```
Doctor Rob answered:
```
It is an unfortunate fact that while every familiar function of calculus has a derivative, not all of them have antiderivatives expressible in terms of those familiar functions. A classic example is the so-called elliptic integrals, which first appear when trying to find the arc-length of an ellipse. The arc-length s from x = 0 to x = a sin(T) is given by:
s = a Integral sqrt(1 - k^2 sin^2[t]) dt from t = 0 to T
where k is the eccentricity of the ellipse. For ellipses that are not circles, k is not 1, and this integral is not possible in closed form. Other examples are the integrals of sqrt(sin(x)), e^(-x^2), and so on.
```
Not only is integration difficult, it is sometimes literally impossible (in the sense of finding an expression for the answer in terms of the basic functions, which is what “closed form” means).
```
You are correct that there is no standard method for doing integration. Some classes of functions have such methods, however.
For rational functions, the idea is to decompose the rational function into a polynomial plus a rational function whose numerator has smaller degree than the denominator. Then this new rational function can be split into simpler ones by factoring the denominator, and splitting it up via the method of partial fractions. These can be integrated individually.
```
So if the integrand is a rational function, you know it can be integrated, because there is a standard method (as long as the denominator can be factored, and you have enough patience to carry out the process).
For rational functions of trigonometric functions, the method is to use multiple angle formulas to express the function in terms of sines and cosines of a single angle t. Then the substitution z = tan(t/2) will reduce the function to a rational function in z. This reduces it to the previous case.
This is a special method that always works for such integrands; beginners are generally taught easier methods that often work, similar to what we’ll do below (but for non-rational functions).
For more elaborate functions, and those of other types, a high degree of pattern-recognition and experience in choosing substitutions that will simplify the problem at hand are involved. Often when there is a common subexpression among two or more parts of the integrand, it is useful to substitute some new variable for it. When an expression of the form sqrt(a^2-t^2) is encountered, the substitution t = asin(z) will often help, or t = qcoth(z). When it is sqrt(a^2+t^2), the substitution t = atan(z) will often help, or else t = acosh(z).
I often say that integration is an art; it takes a lot of experience to see quickly what to try, and even then you may not be able to see a way. The methods of “u-substitution” (as we’ll be seeing below) and “trig substitution” or “hyperbolic substitution” (like those he just mentioned) often help simplify a problem.
```
Another very useful technique is integration by parts. When to use it is not, however, very clear. When the integrand is a product, one of whose parts is integrable, and the other of which has a relatively simple derivative, then integration by parts may be beneficial.
In summary, there are just a few useful techniques that are useful in integration. They have to be applied imaginatively. Even so, some functions will turn out not to be integrable at all.
```
The key to integration is to have a full toolbox, and to learn to recognize when each tool might be needed. And sometimes you have to try one you didn’t think would help – just because nothing else did!
The simplest substitution: replacing ax with one variable.
As a first example, I want to start very small, with this question from 2001:
```
Integrate Cos 2a
Dear Dr. Math,
I was doing some integration and one of the problems was to integrate cos2a. The answer given in the book is sin 2a/2.
The textbooks that I have tell me that integrating cos a gives me sin a, and tell me in detail how that was arrived at. But I have searched everywhere for information on cos2a and how the result sin 2a/2 is arrived at. Does this follow a pattern, i.e. cos 3a = sin 3a/3 cos 4a = sin 4a/4 or cos na = sin na/n ?
```
Katherine is using a (which sounds like a constant) where I would expect to see x; to avoid confusion, we may want to think of this as integrating \int\cos(2x)dx
∫cos(2x)dx
and more generally \int\cos(nx)dx
∫cos(nx)dx
Books commonly give this integral as \int\cos(ax)dx=\frac{\sin(ax)}{a}+C
∫cos(ax)dx=sin(ax)a+C
I answered, using the variable x:
```
Hi, Katherine.
In general, the formula is
[INT] cos(ax)dx = sin(ax) / a
You can prove this by taking the derivative of the right side:
d/dx[1/a sin(ax)] = 1/a cos(ax) d/dx(ax)
= 1/a cos(ax) a
= cos(ax)
This required only a simple application of the chain rule.
```
Here I’ve taken the formula as given, and checked this antiderivative by differentiating it. The formula is worth memorizing. But what if you haven’t done so?
```
Not knowing this more general formula, you can obtain it by substitution. If we let u = ax, then du = a dx and dx = du/a, so
[INT]cos(ax)dx = [INT]cos(u) du/a = sin(u)/a = sin(ax)/a
Here I first replaced ax with u and dx with du/a, then integrated, and finally replaced u with ax again.
The same method is useful everywhere, so you should learn it well and even be able to do it in your head.
```
For details on the method, see Integration by Substitution. We defined a new variable u to be equal to the quantity found inside the function, ax; found how to express dx in terms of du; and then replaced each in the integral. The resulting integral was one we’ve memorized, so we just wrote the answer. Finally, we back-substituted to put the answer in terms of the original variable, x.
Now, what I often do for problems like this, if I’m not quite sure of the formula, is to start with a knowingly wrong “guess”, and correct it by checking:
I want the integral of \cos(ax)cos(ax); I know that the antiderivative of \cos(x)cos(x) is \sin(x)sin(x) (that is, that the derivative of the latter is the former), so I expect the answer to be something like \sin(ax)sin(ax).
Then I take the derivative of that using the chain rule, finding that it is a\cos(ax)acos(ax), which is a times as large as I want. To fix this, I just need to divide by a, so my answer is \frac{1}{a}\sin(ax)1asin(ax).
I can do all this in my head, and it frees my memory from having to be certain what to multiply or divide by. (Each function has its own behavior, which I don’t have to remember this way.)
This can be described as a version of the method of false position. We’ll be using it again below.
Looking for a substitution by rewriting
For a harder example, here’s a question from 1999:
```
Techniques of Integration - Change of Variables
Hello.
I would like to know how to solve this question.
(integral sign) sin 2x / sqrt(9-cos^4 x) dx
How should I begin?
Jen
```
We want to integrate \int\frac{\sin(2x)}{\sqrt{9-\cos^4(x)}}dx
∫sin(2x)9−cos4(x)−−−−−−−−−√dx
It is not obvious that this is a function of a function, which is what you look for to make a substitution (giving a name to the inner function); but rewriting it will change things.
Doctor Luis answered:
```
You can approach this question more easily if you use the identity sin(2x) = 2sin(x)cos(x), and if you rewrite your integral in a more suggestive way, like this:
/
| 2 sin(x) cos(x)
| ----------------------- dx
| sqrt(9 - (cos^2(x))^2)
/
```
Getting everything in terms of the same angle (x itself) is a good start; moreover, we see that the numerator is the derivative of \cos^2(x)cos2(x)! Also, writing the radicand as a difference of squares suggests something …
```
At this point it is clear that a change of variables will do the trick. Let cos^2(x) = 3u. Then, differentiating implicitly,
2cos(x)(-sin(x))dx = 3du
This means that 2cos(x)sin(x) = -3du, and so, rewriting our integral in terms of u, we have
/ /
| -3 du | du
| ---------------- = - sqrt(9 - 9 u^2) | sqrt(1-u^2)
/ /
```
The way he substituted may be unfamiliar to you.
We often define a new variable directly as something like u=\cos(x)u=cos(x) (a “u-substitution”), or, in reverse, something like x=\cos(\theta)x=cos(θ) (a “trig substitution”). In those cases, we can directly differentiate to find dudu, or to solve for d\thetadθ.
Here, we are replacing one expression by another (\cos^2(x)cos2(x) by 3u3u), and implicit differentiation shows that we can replace all of 2\cos(x)\sin(x)dx2cos(x)sin(x)dx with -3du−3du.
```
But this last integral is already known. It's nothing more than the inverse cosine function. So,
/
| du
| ------------- = arccos(u) = arccos((1/3)cos^2(x))
| sqrt(1-u^2)
/
Notice that we have expressed our final answer in terms of the original variable of integration, using the fact that u = (1/3)cos^2(x) .
Obviously, I have neglected the arbitrary constant of integration, but you can add that at any time.
```
This integral is commonly just memorized. If you hadn’t, you could use a trig substitution to derive it; and if you had just half-memorized it (like me) you could check that you had all the signs right, as Doctor Luis now shows:
```
Now, to show that the integral of -1/sqrt(1-u^2) is the arccos(u) function, what you can do is check that the derivative of the arccos(u) function is -1/sqrt(1-u^2). We can do that as follows:
Let y = arccos(x); then x = cos(y). By implicit differentiation on this last equation you can obtain,
1 = -sin(y) dy/dx (notice we used the chain rule here)
Solving for dy/dx, you get
-1
dy/dx = ---------
sin(y)
Expressing sin(y) in terms of cos(y) (use the identity sin^2(y) + cos^2(y) = 1 for this step) we get:
-1
dy/dx = -----------------
sqrt(1-cos^2(y))
But, by definition, x = cos(y). Therefore,
-1
dy/dx = -------------
sqrt(1-x^2)
Now, y was just y = arccos(x). Therefore, we have proven that the derivative of the arccos(x) function is -1/sqrt(1-x^2) .
```
Alternatively, you might know that \int\frac{1}{\sqrt{1-x^2}}=\arcsin(x)∫11−x2√=arcsin(x), and write the answer as -\int\frac{du}{\sqrt{1-u^2}}=-\arcsin(u)=-\arcsin\left(\frac{1}{3}\cos^2(x)\right)+C
−∫du1−u2−−−−−√=−arcsin(u)=−arcsin(13cos2(x))+C
Believe it or not, this is equivalent to his answer (because arcsin and arccos add up to \frac{\pi}{2}π2).
Looking for a substitution by listing possibilities
Now consider this, from 2003:
```
Substituting to Simplify the Integral
What is the integral of tan^3 x sec x dx?
```
We want \int\tan^3(x)\sec(x)dx
∫tan3(x)sec(x)dx
Where can we start?
Doctor Barrus answered:
```
Hi, Trista.
This is a good question. Finding the indefinite integral of a bunch of trigonometric functions is often challenging, and sometimes it takes a lot of tries to get something that works.
One strategy that you'll want to adapt is to look for a good substitution that will make the integral simpler. I'm going to assume that you're familiar with substitutions; let us know if you need more information on this.
```
We need to consider possible inner functions, and what their derivatives are:
```
What I'll do is look at different choices for our new variable u, based on what you have already in the integral, and figure out what the differential du would be with such a choice...
If then
u = tan x du = sec^2 x dx
u = tan^2 x du = 2(tan x)(sec^2 x) dx
u = tan^3 x du = 3(tan^2 x)(sec^2 x)dx
u = sec x du = sec x tan x dx
The first three didn't look promising. All of them required a sec^2 x in order to substitute du into the integral. The last one looks a lot better, though, because we DO have a sec x tan x dx in the indefinite integral.
```
Since we have only one secant, u=\sec(x)u=sec(x) seems like the only choice in order to have its derivative present. Note the key idea here: We want u to be something whose derivative is present in the integral, even if we don’t (yet) see a function of that u!
```
What we'll want to do, then, is to pull this quantity apart from the rest of the indefinite integral. In other words, we'll rewrite
(tan^3 x)(sec x)dx
as
(tan^2 x)(sec x tan x dx)
Remember that it doesn't matter in which order we multiply things. Then we know that if we make the substitution u = sec x, we'll be able to change the sec x tan x dx into du.
```
Now we look at the rest of the integrand.
```
The question now is whether we can write that remaining tan^2 x in terms of sec x, so we can replace it by something involving u. What relations (identities, etc.) exist between the functions tan x and sec x?
I hope you're familiar with one of the Pythagorean identities that relates these two:
1 + tan^2 x = sec^2 x
We want to write tan^2 x in terms of sec x, so we'll solve for tan^2 x in this equation:
tan^2 x = sec^2 x - 1
Or, putting it in terms of u = sec x, we have
tan^2 x = u^2 - 1
```
So now our integral becomes \int\tan^3(x)\sec(x)dx=\int\underset{f(u)}{\underbrace{\tan^2(x)}}\cdot\underset{du}{\underbrace{\sec(x)\tan(x)dx}}\=\int\underset{u^2-1}{\underbrace{(\sec^2(x)-1)}}\cdot\underset{du}{\underbrace{\sec(x)\tan(x)dx}}=\int(u^2-1)du
∫tan3(x)sec(x)dx=∫tan2(x)f(u)⋅sec(x)tan(x)dxdu=∫(sec2(x)−1)u2−1⋅sec(x)tan(x)dxdu=∫(u2−1)du
```
We can substitute this into the indefinite integral to arrive at the following indefinite integral:
INT (u^2 - 1) du
and we hope you know how to finish the problem from there.
Does this make sense? Substitution is a good strategy to consider in problems like this, particularly in instances where one of the trig functions is raised to an odd power, as in this case.
```
Here is the rest of the work: First we integrate, \int(u^2-1)du=\frac{1}{3}u^3-u+C
∫(u2−1)du=13u3−u+C
and then we back-substitute, replacing u with \sec(x)sec(x), \frac{1}{3}\sec^3(x)-\sec(x)+C
13sec3(x)−sec(x)+C
Finally, if we are unsure, we can check by differentiating: \frac{d}{dx}\left(\frac{1}{3}\sec^3(x)-\sec(x)+C\right)\=\frac{1}{3}\cdot3\sec^2(x)\cdot\sec(x)\tan(x)-\sec(x)\tan(x)\=\left(\sec^2(x)-1\right)\sec(x)\tan(x)\=\tan^2(x)\sec(x)\tan(x)=\tan^3(x)\sec(x)
ddx(13sec3(x)−sec(x)+C)=13⋅3sec2(x)⋅sec(x)tan(x)−sec(x)tan(x)=(sec2(x)−1)sec(x)tan(x)=tan2(x)sec(x)tan(x)=tan3(x)sec(x)
One step at a time
Sometimes it takes more than one substitution, as in this final question from 1998:
```
A Trigonometry Integral Requiring Two Substitutions
What is the integral of
sqrt(1 + sin(x)),
where sqrt stands for "square root of"?
```
We want \int\sqrt{1+\sin(x)}dx
∫1+sin(x)−−−−−−−−√dx
With nothing outside of the radical, what can we use for du?
Doctor Sam answered:
```
Paul,
This is a tricky problem. It will take (I think) two different substitutions. We want to find:
INT sqrt(1 + sin x) dx
```
(We’ll see that it could be done with only one substitution, but what he’s doing is just what I would do, trying something simple first. We just try whatever seems like it might help.)
```
I am going to try substituting u = sin(x) to try to remove the trig function. When you make a substitution, you must also substitute for dx. So:
u = sin(x) and du = cos(x) dx
This gives dx = du/cos(x), and changes the integral to
sqrt(1 + u)
INT ----------- du
cos(x)
```
Since this time there is no differential sitting there waiting for us, we’re taking a step in the dark, trying an “inner function” and hoping its derivative will turn up. If this didn’t work, we might instead try u=1+\sin(x)u=1+sin(x), or even u=\sqrt{1+\sin(x)}u=1+sin(x)−−−−−−−−√ to see what happened.
I don’t like to write an integral with two different variables in it as he does here, but it’s okay as long as we see it as a fleeting intermediate step. We immediately want to express that denominator in terms of the new variable u:
```
This is no good. We need to get an integral in terms of the u variable alone. Here's where a little right-triangle trigonometry can help. We made the substitution u = sin(x), so we can visualize a triangle with an acute angle x whose sine is u.
Here is one such triangle:
/|
/ |
/ |
/ |
1 / | u
/ |
/x |
Now we can use the Pythagorean Theorem to find the third side, and then the cosine of x. The third side is sqrt(1 - u^2), and so:
cos(x) = sqrt(1 - u^2)
```
Alternatively, we could just recall the Pythagorean identity without drawing the picture; We want to express \cos(x)cos(x) in terms of \sin(x)sin(x), so we observe that \sin^2(x)+\cos^2(x)=1sin2(x)+cos2(x)=1; therefore, \cos(x)=\sqrt{1-\sin^2(x)}=\sqrt{1-u^2}cos(x)=1−sin2(x)−−−−−−−−−√=1−u2−−−−−√.
But we’ve skipped over an important question: There are two square roots; shall we take the positive or negative root? This depends on what quadrant angle x is in! If -\frac{\pi}{2}\le x\le \frac{\pi}{2}−π2≤x≤π2, then the cosine will be positive; otherwise it would be negative, and the actual integral would have a different sign. Let’s just restrict x to that interval for simplicity; it turns out that making a single formula for the integral over all real numbers is tricky (even Wolfram Alpha doesn’t quite manage it).
```
Our integral is now:
sqrt(1 + u)
INT ------------- du
sqrt(1 - u^2)
Now I can't help but notice that
1 - u^2 = (1 - u)(1 + u)
so this fraction simplifies to:
1
INT ----------- du
sqrt(1 - u)
```
So we have a new, and simpler, integral; but it still is not one we’ve memorized.
```
We are almost done. We have now transformed our trig integral into an algebraic integral. Now a second substitution:
w = 1 - u
should finish the job.
If w = 1 - u, then dw = -du, so du = -dw. This gives:
1
INT ------- dw
sqrt(w)
```
Now we just have to rewrite this, and it will be one of our basic forms:
```
Interpret this as w^(-1/2), and we can use the formula for antidifferentiating u^n:
INT w^(-1/2) dw = -2w^(1/2) + C
Now change back from w to u using w = 1 - u:
-2w^(1/2) = -2 sqrt(1 - u) + C
And now change back from u to x using u = sin(x):
-2w^(1/2) = -2 sqrt(1 - u) + C = -2 sqrt(1 - sin(x)) + C
```
We can check our answer by differentiating: \frac{d}{dx}\left(-2\sqrt{1-\sin(x)}+C\right)=\frac{d}{dx}\left(-2\left(1-\sin(x)\right)^{1/2}\right)\=-2\cdot\frac{1}{2}\left(1-\sin(x)\right)^{-1/2}\cdot\left(-\cos(x)\right)=\frac{\cos(x)}{\sqrt{1-\sin(x)}}\=\frac{\sqrt{1-\sin^2(x)}}{\sqrt{1-\sin(x)}}=\frac{\sqrt{1+\sin(x)}\sqrt{1-\sin(x)}}{\sqrt{1-\sin(x)}}=\sqrt{1+\sin(x)}
ddx(−21−sin(x)−−−−−−−−√+C)=ddx(−2(1−sin(x))1/2)=−2⋅12(1−sin(x))−1/2⋅(−cos(x))=cos(x)1−sin(x)−−−−−−−−√=1−sin2(x)−−−−−−−−−√1−sin(x)−−−−−−−−√=1+sin(x)−−−−−−−−√1−sin(x)−−−−−−−−√1−sin(x)−−−−−−−−√=1+sin(x)−−−−−−−−√
Do you see how this suggests a trick method we could have used instead? Work backward:
\int\sqrt{1+\sin(x)}dx=\int\frac{\sqrt{1+\sin(x)}\sqrt{1-\sin(x)}}{\sqrt{1-\sin(x)}}dx\=\int\frac{\sqrt{1-\sin^2(x)}}{\sqrt{1-\sin(x)}}dx=\int\frac{\cos(x)dx}{\sqrt{1-\sin(x)}}
∫1+sin(x)−−−−−−−−√dx=∫1+sin(x)−−−−−−−−√1−sin(x)−−−−−−−−√1−sin(x)−−−−−−−−√dx=∫1−sin2(x)−−−−−−−−−√1−sin(x)−−−−−−−−√dx=∫cos(x)dx1−sin(x)−−−−−−−−√
Letting u=1-\sin(x)u=1−sin(x), we have
\int\frac{-du}{\sqrt{u}}=-\int u^{-1/2}du=-2u^{1/2}\=-2\sqrt{1-\sin(x)}+C
∫−duu−−√=−∫u−1/2du=−2u1/2=−21−sin(x)−−−−−−−−√+C
And we could have made this same substitution from the start; it combines the two we did above into one. We can let w=1-\sin(x)w=1−sin(x) all at once, so that \sin(x)=1-wsin(x)=1−w and dw=-\cos(x)dx=-\sqrt{1-\sin^2(x)}dx\=-\sqrt{1-(1-w)^2}dx=-\sqrt{2w-w^2}dx
dw=−cos(x)dx=−1−sin2(x)−−−−−−−−−√dx=−1−(1−w)2−−−−−−−−−−√dx=−2w−w2−−−−−−−√dx
Then \int\sqrt{1+\sin(x)}dx=\int\sqrt{1+(1-w)}\cdot\frac{dw}{-\sqrt{2w-w^2}}\
=-\int\frac{\sqrt{2-w}\,dw}{\sqrt{2w-w^2}}=-\int\frac{\sqrt{2-w}\,dw}{\sqrt{w}\sqrt{2-w}}\
=-\int\frac{dw}{\sqrt{w}}=-\int w^{-1/2}dw\
=-2w^{1/2}=-2(1-\sin(x))^{1/2}=-2\sqrt{1-\sin(x)}+C
∫1+sin(x)−−−−−−−−√dx=∫1+(1−w)−−−−−−−−−√⋅dw−2w−w2−−−−−−−√=−∫2−w−−−−−√dw2w−w2−−−−−−−√=−∫2−w−−−−−√dww−−√2−w−−−−−√=−∫dww−−√=−∫w−1/2dw=−2w1/2=−2(1−sin(x))1/2=−21−sin(x)−−−−−−−−√+C
But we’d never have thought of that. Step by step is the way to go.
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11280 | https://units.zeptomath.com/convert.php/capacitance-convert-microfarad-to-farad?value=1&hl=en | Convert 1 Microfarad to Farad
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11281 | https://www.youtube.com/watch?v=Wo6a8yDXb6c | [PMO] Find all b such x^2 + bx - 3b = 0 has integer roots! || High School Math
dumplet 餃仔
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Posted: 11 Feb 2022
Hallo, it's dumplet here^_^
Here's an item on the discriminant method from the Philippine Mathematical Olympiad!
Hope you guys learned something new from the video!
Hello everybody, dumplet here!
This is a YouTube Channel for Elementary, High School and a bit of College Math! I post random math problems and tutorials to various topics seen from Elementary to College Mathematics, as well as common shortcuts, tricks and techniques in various National and International Mathematical Olympiads!
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[Music] hello hello it's dumplit here here's an item on algebra find the sum of all possible real numbers v such that x squared plus bx minus 3b is equal to 0 has integer roots credits to the philippine mathematical olympiad for this item as usual pause this video if you'd like to give this item a try but if you're done let us dive into the solution all right now here we do have b to be a real number but some of you guys might realize well we do have integer roots so if i were to factor this expression on the left hand side well that i should be able to factor into two factors with integer roots obviously and that's going to give me b to be an integer and yes your instincts are correct but we could kind of prove this with the simple fact the sum of the roots via the vietnamese formula it's going to be negative b now we know that the roots are two integers so we're saying that the sum of two integers is negative b and that's pretty much enough for us to say that b is an integer right so that's going to tell us that b is an integer and this is actually very interesting because we could apply this in the green condition the quadratic equation having integer roots because that's going to mean that the discriminant is a perfect square right now why is that so now obviously b squared plus 12 b in the in the discriminant oh by the way we get this from the quadratic formula so solving the roots by the quadratic formula i'll get the discriminant is b squared plus 12 b now obviously b is an integer so the discriminant is an integer and in this case if the discriminant is not a perfect square then obviously this one the radical it's a radical so it's going to be irrational so it's going to be irrational which is going to cause this entire thing to be irrational but we know that's not true because the integer the roots are integers so that's going to be enough for us to say that the discriminant is a perfect square so we could write this as b squared plus 12 b being equal to a certain perfect square let's just say k squared and let's just assume without loss of generality that the square root is non-negative okay so essentially we just have this they'll find an equation in integers obviously so b and k are integers and we have a lot of techniques to solve but in this case we do have a square we do have two squares so let's try to complete the square maybe some factoring would work so b squared plus 12 b completing the square here i'm going to add 36 to both sides of the equation that's going to give me um left side here this becomes the square of b plus six i could put the k squared on the other side minus k squared then 36 on the right hand side and as i mentioned factoring here apparently because we do have a difference of two square scenarios so typically whenever you see perfect squares or squares generally in a data finding equation maybe the difference of squares would be used so try to complete the square and you might be able to observe something nice such as this one anyways moving on factoring will give me this as the first factor and this as the second factor and then they have a product of 36 and because b and k are integers we could just do the method of assigning of factors which is pretty nice right so we do have these two factors and because we set k to be a non-negative integer that's going to be enough for us to say that the left factor is less than or equal to the right factor now there's another thing we could try to do let me just try to take the sum of the two factors we have here the left factor and the right factor they actually add up to 2b plus 12. so the top the sum of the two factors is even 2b plus 12 is even so that's going to mean that the two factors are either both odd or both even now we could eliminate the both odd case because obviously odd odd number times an odd number is an odd number so it cannot be 36 so it must be the case that our two factors are both even so we just have to kind of incorporate the factoring here we must find two numbers multiplying to 36 and the two numbers two factors they must be both even now doing some listing here it's going to give me 2 and 18 and then 6 and 6. so we must have the condition that the last factor is less than or equal to the right factor now obviously b and k are integers here so it is possible that b is negative so we must also incorporate the negative factors so just negate the the first and the second pair so it's negative 6 negative 6 and the other pair negative 18 negative 2 now i switched them kind of because i have to maintain this condition the left factor is less than or equal to the right factor now um since i did notice that the sum of the factors is just 2b plus 12 it would be very nice for us to just simply add the two factors and then solve for b because b is the value that we want so um the first case 2 and 18 they add up to 20 second case they add up to 12 third case they add up to negative 12. fourth case they add up to negative 20. and i could just solve for b in the four equations in the first one i'll get b equals four second one i'll get b equals zero third one i'll get b equals negative 12 and then the fourth one here i'll get b equals negative 16. now obviously you could always put the b back and check if k is an integer as well so um the first one i will get that k here would be eight in the second one i'll get k is equal to zero in the third one i'll also get k equals zero and then in the fourth one over here i'll get k equals eight again so k is an integer for all four cases of b so let's just take 4 0 negative 12 and negative 16 add them up so 4 plus 0 plus negative 16 plus negative 12 that's going to equal negative 24 and this will be our final answer hopefully you guys learned something new from this video and i'll see you in the next one bye bye you |
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Skin 101: Understanding the Fundamentals of Skin Barrier Physiology—Why is This Important for Clinicians?
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J Clin Aesthet Dermatol. 2025;18(2):7–15.
by James Q. Del Rosso, DO, and Leon Kircik, MDDr. Del Rosso is Research Director at JDR Dermatology Research in Las Vegas, Nevada; Senior Vice President of Clinical Research and Strategic Development at Advanced Dermatology and Cosmetic Surgery in Maitland, Florida; and Adjunct Clinical Professor (Dermatology) at Touro University Nevada in Henderson, Nevada. Dr. Kircik is Medical Director at Skin Sciences, PLLC, in Louisville, Kentucky.
FUNDING: The authors received no compensation for preparation of this article. This article was written solely by the authors.
DISCLOSURES: Dr. Del Rosso has served as a consultant, speaker and/or researcher for Arcutis, Bausch Health, Beiersdorf, Galderma, Unilever, L’Oreal, LaRoche-Posay, and Sente.
ABSTRACT: This article reviews epidermal barrier dysfunctions and more thoroughly discusses the stratum corneum (SC) permeability barrier, physiologic self-repair mechanisms in healthy skin, and the clinical and structural effects of an overstressed SC permeability barrier. Discussion includes epidermal barrier impairments induced by both exogenous exposures and endogenous factors such as specific dermatologic disorders. Due the plethora of skin care products on the market and the variability of their contents and vehicle formulations, this article addresses core concepts required to optimize skin care product selection, including for specific disease states such as atopic dermatitis, psoriasis, acne vulgaris, and rosacea. To summarize, the selection of skin care products is directed at maintaining SC hydration, including assisting the SC in self-repair when conditions are adverse. This approach optimizes the ability to sustain healthy skin structure, function and appearance.
KEYWORDS: Skin barrier, epidermal barrier, skin care, stratum corneum dysfunction, transepidermal water loss, corneometry
Introduction
In dermatology, there is an inherent tendency to focus on how to treat various skin conditions based on the most current scientific information, especially with newer therapeutic options. This is true especially in the clinical arena where patients are directly cared for. Publications and presentations usually emphasize pharmacologic and/or procedural therapies, however, the exponential increase in skin care options over time has created a center stage for including optimized skin care as an integral component of cutaneous disease management and post-procedural approaches. The primary objective of this article is based on the premise that to truly understand what we apply to patient care, we need to “learn how to walk before we run.” Once the core fundamentals are mastered, we are in a much better position to translate our knowledge into the clinical setting and more effectively manage many skin conditions, especially common disorders.
Over a decade ago, I (JDR) was fortunate enough to interact which scientists who were dedicated to researching the basic functions of the stratum corneum and epidermal barrier in normal and diseased skin. After a focused period of deep work in studying all the scientific information that could be found on epidermal structure and function, I (JDR) voluntarily served as lead author of a comprehensive article on the clinical relevance of maintaining the functional integrity of the stratum corneum in healthy and diseased skin.1 This article was dedicated to providing dermatology with a core reference that pieces together a wide body of basic science and clinical research, including articles from several different authors that are published in journals that most dermatologists and staff are not likely encounter in their mainstream dermatology literature. Much of what is stated in the article from 2013 still applies today. This article serves to review and update the reader on the fundamentals of skin barrier function as a foundation to understanding and applying the integration of skin care into clinical management in their practice.
What does the term “skin barrier” actually refer to?
The term skin barrier, which is also commonly referred to as the epidermal barrier, is often incompletely misunderstood, as the epidermal barrier of the skin is actually a collection of multiple barrier functions. When the terms skin barrier or epidermal barrier are used in professional conversation, they are usually referring to the ability of the epidermis, especially the stratum corneum (SC), to provide selective permeability of exogenous and endogenous substances and to continually regulate and maintain homeostatic water content and balance in order to sustain healthy skin elasticity and physiologic desquamation; terms such as transepidermal water loss (TEWL) and corneometry (measure of water content in skin) are often mentioned.1–3 These functions refer specifically to the epidermal permeability barrier.1
In fact, these permeability barrier functions are components of a collective group of physiologic and homeostatic epidermal barrier responsibilities, carried out primarily within the SC. Additionally, these physiologic epidermal barrier responsibilities are continually responding dynamically to various exogenous exposures that are adversely affecting epidermal barrier integrity and function.1–9 Hereafter, the term epidermal barrier will be used here to also synonymously encompass the term skin barrier.1–3 Table 1 depicts the multiple barrier functions that collectively comprise the epidermal barrier; these individual barrier functions are reviewed in more detail elsewhere.1-10
How does the epidermal barrier respond physiologically to protect healthy skin?
Structural characteristics. Before reviewing how the epidermal barrier functions in self-repair to sustain healthy skin in response to endogenous exposures, a brief review of epidermal structure is relevant. Figure 1 depicts the classically described “bricks and mortar” structure of the SC. The structural integrity framework of the corneocytes (“bricks”) which are held together by corneodesmosomes, is maintained primarily by keratin macrofibrils and other structural proteins encased by a cornified envelope and covalently bound lipids.4 Within the corneocytes exists a relative concentration of natural moisturizing factor (NMF); NMF is highly hygroscopic and serves as the innate humectant that retains within the SC the necessary water content and balance to homeostatically maintain the hydrolytic activity needed for optimal desquamation, skin elasticity, and prevention of desiccation, rigidity, and fissuring of the skin.1,4,9–11 When SC water content decreases below a critical threshold level due to factors increasing TEWL, the SC protein filaggrin is broken down to form NMF as a SC self-repair response; the multiple components of NMF are primarily free amino acids (40%), pyrrolidone carboxylic acid (12%), lactate (12%), simple sugars (9%), urea (7%) and multiple electrolytes.1,4,5,9–18
Importantly, the spaces between the corneocytes are interconnected to form an intercellular lipid membrane, with SC lipids representing 20 percent of total SC volume1,4,18,19 This membrane is vital to maintaining SC physiology, especially water flux and balance.1,4,9–11,18,19 For optimal function, the intercellular lipid membrane is purposely designed to predominantly contain relative concentrations of specific lipids that are strategically arranged in a lattice of ceramides (40–50%), cholesterols (25%), and free fatty acids (10–15%), exhibiting specific structural characteristics to provide physiologic epidermal barrier function.1,4,18–20 Importantly, reductions in chain length of both free fatty acids and ceramides correlate directly with decreased density of SC lipid organization and greater impairment of epidermal barrier function in skin affected by atopic dermatitis.20
Functional characteristics. The structural and functional integrity of the SC is highly dependent on an adequate water content and gradient; many of the enzymes that catalyze vital SC functions are hydrolytic and do not operate efficiently if water content is below a requisite threshold concentration.1,3,4,9,10 With regard to maintaining physiologic epidermal water content, the SC is innately capable of multiple self-repair mechanisms when exposed to exogenous exposures that induce adverse changes that impair the SC permeability barrier.1,4,5,9,11 Examples of relatively abrupt and/or repetitive exposures that can increase TEWL and promote skin desiccation include marked reduction in ambient humidity, adverse skin care practices, poorly formulated skin care, exposure to cutaneous irritants and/or allergens, occupational exposures, certain topical medications, and inadequately designed topical vehicles.1 Importantly, SC water serves to plasticize the SC, thus enhancing elasticity and reducing epidermal rigidity that enhances fissuring, especially when skin encounters lateral shearing forces; proper water content directly affects other physical properties, such as flexibility, pliability, desquamation of individual corneocytes, skin pH, and cornified cell envelope formation.2–4,9,11–13,21–25
Self-repair. Self-repair within the SC is quickly triggered by even modest increases in TEWL and decreased SC water content.1–5 There is immediate release of stored lipids from lamellar bodies in the SC which partially reverses the increase in TEWL. This is followed within a few hours by markedly upregulated production of major SC lipids which serve to restore intercellular lipid membrane integrity and contribute to SC water binding capacity.1,4,26–28 Another immediate response to increased TEWL is cytokine release that promotes epidermal thickening to physically reduce the magnitude of TEWL.1–5,9
A fundamental self-repair mechanism in response to increased TEWL and decreased SC water content is the upregulation in filaggrin production and its subsequent conversion into NMF. As NMF increases SC humectancy, more water is retained within the SC, thus reversing the adverse effects of decreased SC water content.1–5,17,21,29 Over time, SC self-repair can maintain healthy skin when effectively coupled with mitigation of the inciting exogenous factors discussed above that induce epidermal barrier damage and impair the permeability barrier. Repeated exposure to inciting triggers, especially when frequent and of high magnitude, can often override the ability of SC self-repair to maintain healthy skin. Common examples of this latter scenario are frequent hand washing and/or overwashing, repeated exposure to skin irritants, use of poorly formulated skin care products including harsh skin cleansers, and excessively hot showering.1 As will be discussed below, certain skin diseases are associated with impairments in SC self-repair which further compounds the effects of exogenous exposures that damage the skin barrier.
What is the clinical relevance of skin pH?
Alterations in skin pH alone can result in clinically relevant changes in SC barrier function.30–38 This is exemplified by the neutral to alkaline pH of the SC in the early neonatal period which facilitates the growth of opportunistic organisms such as Candida albicans, creating a regional environment that predisposes to diaper dermatitis. After approximately the first three months of life, the SC physiologically converts to an acidic pH for optimal epidermal barrier function; this “sweet spot” is commonly referred to as the acid mantle of the skin (skin pH 4–6).1,23,30
An elevated skin pH induces a multitude of direct and indirect effects that adversely alter several epidermal barrier functions, many secondary to the subsequent increase in SC protease activity. These adverse effects include disruption of skin barrier homeostasis (decreased SC lipid synthesis, degradation of lipid synthesis enzymes, augmented non-physiologic desquamation, inhibition of lamellar body lipid secretion), suppression of innate antimicrobial peptide (AMP) defense (altered AMP processing and activity), induction of inflammation (“jump start” cytokine activation [IL-1, TNF alpha]), and augmented pruritus (pruritogen release such as leukotriene B4 and prostaglandin E2); many of these effects can markedly alter the structure and function of the SC permeability barrier.1,38
Another important observation is that an acidic SC pH favors growth of normal bacterial flora and a more diverse cutaneous microbiome, as opposed to an alkaline pH, which is more supportive of the growth of potential pathogens such as Staphylococcus aureus and Candida albicans.4,30,36,38 Innate buffering capacity of the skin is decreased in neonatal skin, geriatric skin, with repeated washing using alkaline soaps/skin cleansers, and with regular use of alkaline moisturizers.33 As an elevated (alkaline) skin pH is noted in both atopic dermatitis and acne vulgaris, use of cleansers and/or moisturizers that are not formulated to mitigate disturbance of skin pH may contribute to adverse effects such as irritation and/or deterioration of the underlying skin condition.33,36,37
What is an overstressed epidermal barrier?
An overstressed epidermal barrier occurs when exogenous, endogenous, or a combination of both sources produce SC permeability barrier impairment that exceeds the ability of self-repair mechanisms to fully and visibly correct the induced epidermal barrier dysfunction.1 The potential exogenous factors were discussed earlier. Endogenous factors include individual genetic, ethnic and/or racial predispositions, effects of increased age, or underlying disease states that are inherently associated with permeability barrier impairment.1,4,39,40 These affected individuals are at an inherent disadvantage when stresses that are placed upon the epidermal barrier exceed the restorative self-repair abilities of the SC. In such cases, the self-repair functions of the SC are not able to keep up with the speed and/or magnitude of permeability barrier restoration needed to fully reverse the excess in TEWL. Unless corrected by proper therapeutic intervention, the clinical sequelae of skin desiccation is the progressive march starting with subclinical effects, then to visible signs and symptoms of xerosis (ie, dry skin) which often increase in severity over time, then to signs and symptoms of an acute or subacute flare of eczematous dermatitis, and finally to chronic eczematous dermatitis and progressive hyperkeratotic changes, especially on the hands and feet (Figure 2).
Disease states the compromise the epidermal barrier and stratum corneum self-repair
Compromise of epidermal barrier function is commonly associated with several dermatologic disease states, with the SC permeability barrier frequently affected.1,4 Xerosis, although not often thought of as a fundamental primary skin disease, is probably the most commonly encountered skin disorder that is primarily correlated with impairment of the epidermal permeability barrier.1 Atopic dermatitis (AD), due to the plethora of data on overall pathophysiology and multiple epidermal barrier abnormalities, has served as the “poster skin disease” used to illustrate the multiple profound effects of epidermal barrier dysfunction.1,4,22,38,41–46 Epidermal barrier dysfunctions are identified in many dermatologic diseases, with specific barrier abnormalities varying among different disease states. In this article, we will discuss selected common skin diseases to illustrate different examples.
Xerosis. Xerosis, described as one of the most common human skin afflictions, is the initial clinical evidence of an impaired SC permeability barrier as described in Figure 2.1,4,47,48
Regardless of etiology, xerosis is the visible reaction pattern that occurs when TEWL becomes excessive and exceeds the ability of SC to adequately self-repair the permeability barrier.1,47,48 Ultimately, xerosis is best defined clinically, presenting as skin that is dull in appearance, rough, scaly, flakey, and tight, often with associated pruritus, and in some cases discomfort especially in low ambient humidity.1,4
Xerosis may be observed in individuals that do not have any apparent endogenous primary skin disease. Phenotypic xerosis is not uncommon among the general population and may reflect a range of variability in SC self-repair capacity among the general population. Development of xerosis is also affected by the extent of exposure to inciting exogenous factors. In many cases, xerosis can exist concurrently with other primary skin diseases, the most common being atopic dermatitis.1,4,43,48 Avoidance of exacerbating exogenous factors along with proper use of well-formulated cleanser and moisturizer formulations are the foundation of management for xerotic skin.1,21,28,47,49–51
Atopic dermatitis. Atopic dermatitis is associated with multiple barrier impairments that are well described elsewhere.1,4,20,23,27,29,36,41–46,52,53 In this article, the primary emphasis is on SC permeability barrier dysfunction in AD. SC lipid abnormalities have been demonstrated in eczematous skin, in xerotic skin, and in uninvolved skin in patients with AD.1,4,20,22,27,28,42,43,54
Although the major lipid fraction affected in AD is ceramides, the free fatty acid subfraction is also affected.1,20,45,51,52 The upregulated Th2 cascade associated with increases in IL-4, IL-13, and IL-31 cytokines reduces the expression of major ceramide synthesizing enzymes, thus contributing to SC permeability barrier impairment.52,55 Some studies have shown that the greatest ceramide subfraction decrease in AD patients was ceramide-1 in both lesional (eczematous) and nonlesional skin (visibly uninvolved without presence of eczema); however, ceramide-2 through ceramide-6 were also markedly decreased in both lesional and nonlesional skin.1,28,43,54,56 Additionally, correlation of SC ceramide composition with permeability barrier function in patients with AD showed marked decreases in both ceramide-1 and ceramide-3, with a direct correlation between reduction in ceramide-3 and an increase in TEWL.41,54,56
Collectively, SC permeability dysfunction in AD correlates with decreased total ceramide levels, altered ceramide chain lengths (increased short chain ceramides, decreased long chain, ceramides), altered free fatty acid chain lengths (increased short chain free fatty acids, decreased long chain free fatty acids), and a decrease in hydroxy-fatty acids.52 From a clinical perspective, the importance of consistently addressing atopic skin with proper skin care (cleansing and moisturization) is strongly supported by a three- to five- fold increase in TEWL noted in lesional skin compared to nonlesional skin in AD, and a two-fold increase in TEWL in clinically normal and xerotic skin (without eczematous changes) in AD patients who present with active AD flares at other skin sites.41,54,56–58
A common genetic aberration that impairs the SC permeability barrier, affecting 30 to 50 percent of AD patients are filaggrin gene mutations, studied primarily in white patients but not limited to this population.29,44,52 The resultant filaggrin deficiency impairs the physiologic self-repair response to increased TEWL due to the reduced ability to produce NMF through the usual route of filaggrin degradation.1,29,44 Downstream consequences of filaggrin deficiency in AD patients affected by filaggrin gene mutations can include xerotic skin, decrease in skin lipids, and an altered skin microbiome with increased skin colonization by Staphylococcus aureus.52 The possible presence of filaggrin deficiency due to this genetic predisposition in many AD patients supports why some moisturizers formulated to specifically target xerotic and atopic skin contain components of NMF.17,59
Psoriasis. The pathophysiology of psoriasis is complex, multifactorial, and has been progressively updated based on continued advances in basic science and clinical research.60,61 An understanding of the SC abnormalities noted in psoriasis have been evaluated with considerations given to how they impact on the pathophysiology of the disease and its management.62–68 SC abnormalities observed in plaque psoriasis lesions that can translate to impairment of SC permeability barrier functions include abnormal disposition of lamellar bodies in some phenotypes, genetically-associated changes in epidermal differentiation including aberrant expression of keratins, altered cornified cell envelope formation, decreased filaggrin expression, altered ceramide synthesis and subfractions in psoriatic skin lesions correlating directly with increased TEWL, reduced levels of short chain fatty acids and increased cholesterol in psoriatic lesions compared to healthy skin, altered expression of specific SC enzymes (eg, ceramidase, transglutaminases), decreased keratinocyte aquaporin-3 skin hydration channels, increased epidermal hyperproliferation, and disruption of SC intercellular connections (multiple affected proteins).65,67,68 Upregulation of IL-17 and IL-22 are noted to play important roles in keratinocyte activity and inflammation in psoriasis.; abnormal keratinocyte differentiation is observed in psoriatic skin with markers of epidermal proliferation highly expressed in psoriatic lesions which are reduced with effective treatment.67
The overall epidermal barrier dysfunction in plaque psoriasis, including marked impairments of SC permeability function, is multifactorial, encompassing abnormalities in both epidermal structure and function and also lipid content. Importantly, the total amount of ceramide content in keratinocytes and fibroblasts in psoriatic skin lesions is not reduced with SC permeability barrier dysregulation likely correlating with abnormalities of ceramide subtype in psoriasis; changes in free fatty acids and the reported increase in SC cholesterol may also alter the relative balance within the intercellular lipid membrane.67 This may impact on the choice of moisturizer formulation if there is also the logical desire to replenish the skin with specific ingredients in managing psoriasis.68–71
Acne vulgaris. Data are limited on SC permeability barrier dysfunctions that are innately present in untreated acne vulgaris (AV); most of the emphasis on epidermal barrier impairment in AV focuses on topical therapeutic agents and vehicle formulations as many are associated with cutaneous inflammation.1,72,73 One of the early studies completed in patients with AV (N=36) showed increased TEWL, reduced SC hydration based on conductance testing, reduction in total SC ceramides, and reduction in SC free sphingosine as compared to the study control group.74 It is uncertain whether or not epidermal barrier impairment in AV is inherently a part of the AV disease state, is primarily secondary to inflammation occurring during a flare, and is also present in nonlesional skin during and AV flare or between flares.37,73,74
Some topical acne medications and/or vehicle formulations are associated with adverse visible signs of cutaneous irritation, such as peeling, scaling, and erythema. As a result, the clinically apparent association of a topical AV formulation and/or a therapeutic AV ingredient with impairment of the SC permeability barrier or other SC functions is an important clinical consideration.1 Cutaneous application of benzoyl peroxide has been demonstrated to increase TEWL, oxidize SC antioxidants (eg, alpha tocopherol [vitamin E]), and induce lipid peroxidation; supplementation with topical vitamin E did not correct the increase in TEWL, but did reduce markers of lipid peroxidation.75 Topical retinoids commonly induce varying degrees of application site erythema, peeling, and flaking, usually within the first month of use, due to their inherent mechanism of action which alters epidermal differentiation and increases SC cellular dyscohesion.76 This is referred to as the initial period of “retinization” and is commonly interpreted as cutaneous irritation; regardless, these changes represent alterations in SC structure and can influence the permeability barrier.1 Additionally, some topical retinoid vehicle formulations are associated with a greater potential for cutaneous irritation. Several vehicle modifications and gentle skin care approaches have been recommended to reduce the adverse visible skin changes that are often associated with topical retinoid use and other topical acne therapies (such as benzoyl peroxide), especially in the first weeks of initiating therapy and/or when used in combination with other topical acne medications.77,78 One study that supports using SC permeability barrier-enhancing moisturization before and during and after initiation of topical retinoid therapy showed that this approach facilitated adaptation to the initial period of “retinization and induced signs of cutaneous irritation, decreased TEWL, and improved cutaneous hydration as measured by conductance.79 It is also important to regularly incorporate skin care with a gentle skin cleanser and moisturizer in patients treated with oral isotretinoin.80
The importance of adjunctive skin care as an integral component of acne management is well established in medical literature.1,71,73,75,77–83 Skin care concepts and approaches that specifically discuss use in patients with AV are available in the literature, with selected references shown here.77,81
Rosacea. Cutaneous rosacea, most often presenting clinically as central facial erythema with papules and pustules (papulopustular rosacea) and central facial erythema without papules and pustules (erythematotelangiectatic rosacea), is often associated with symptoms of stinging and burning, especially during periods of flaring. Several specific epidermal barrier impairments, including SC permeability barrier dysfunction, have been demonstrated in facial rosacea; multiple potential triggers have been frequently noted, resulting in “sensitive skin” commonly associated with facial rosacea, both during and between flares.84,86–88 In untreated adult patients with papulopustular rosacea (N=915) pooled from multiple studies, dryness, scaling, and edema were reported by 65 to 69 percent, 51 to 58 percent, and 32 to 38 percent of subjects, respectively; facial skin burning, stinging, and pruritus were reported by 34 to 36 percent, 29 to 34 percent, and 49 to 52 percent of subjects, respectively.1,89,90
Sensitive skin and SC permeability barrier dysfunction in cutaneous rosacea are evidenced by the above discussion, providing strong support for adjunctive gentle skin care using a well-selected cleanser and moisturizer.84,90–96 There are data to support increased TEWL, decreased epidermal hydration, increase SC pH, increased stinging induced by lactic acid facial skin testing, associated symptoms of stinging, burning, tingling, and itching, and an increased risk of facial contact dermatitis; specific SC lipid abnormalities have not been identified in rosacea prone skin.86,87,97–99 Designated skin care approaches with cleansers and moisturizers, and formulation characteristics including specific ingredients to mitigate SC permeability barrier dysfunction and the associated diffuse facial erythema of rosacea, are discussed in selected references.92,94,96,97,100 Although rosacea is reported to most commonly affect white individuals with fair skin, it is well-recognized that rosacea can affect all racial and ethnic groups including those with darker skin types such as people of African descent, Asian, and Latino populations.101,102,103
How can skin care be optimized to maintain the functional integrity of the epidermal barrier?
The above article outlines epidermal barrier dysfunctions and more thoroughly discusses the SC permeability barrier, physiologic self-repair mechanisms in healthy skin, and the effects of an overstressed SC permeability barrier including exogenous inducing factors, and endogenous factors such as specific dermatologic disorders. Due the plethora of skin care products on the market and the variability of their contents and vehicle formulations, a complete discussion of individual products is beyond the scope of this article. However, a few basic concepts deserve mention here. Ultimately, the selection of skin care products needs to focus on the maintenance of SC hydration, including assisting the SC in self-repair when conditions are adverse, in order to sustain healthy function and appearance of the skin.
Skin cleansers. The use of a well-formulated gentle skin cleanser that effectively removes exogenous and exfoliated debris and excess sebum, that does not perturb skin pH, and produces negligible damage to the SC is optimal for both healthy skin and disease-affected skin.95,96,101–106 Well-designed adjunctive skin care that is selected based on the needs of the individual patient, which includes a well-formulated gentle skin cleanser, augments therapeutic response to overall treatment in a variety of skin conditions, such as AD and rosacea, and/or can reduce adverse effects associated with therapies that can adversely affect the SC.1
Moisturizers. Moisturizers are recommended, concurrently with an appropriate gentle cleanser, for use in people with healthy skin who encounter intermittent or continuous exogenous exposures that diminish or overstress SC permeability barrier function, in individuals with impairments of SC permeability function due to inherent phenotypic xerotic skin, elderly skin, in the management of underlying dermatologic disease (as described above), and/or in those utilizing therapies that induce SC permeability barrier dysfunction (as described above).1,4,11,21,26, 28,48–51,59,68–72,75,77–79,81,83,84,90–97,101–103 For example, a thorough evidenced-based review of studies evaluating consistent moisturizer use in patients with eczematous dermatitis demonstrated therapeutic contribution that decreased the frequency and severity of flares and reduced the amount of topical corticosteroid use needed over time.50
The three core fundamental components of a well-designed moisturizer are occlusive agents, humectants, and agents that provide an emollient effect.47,49,59 Occlusive agents retard water evaporation by forming a hydrophobic barrier layer on the skin surface; these often include hydrocarbons/oils/waxes (eg, petrolatum, mineral oil, paraffin, Carnauba wax, silicone derivatives), fatty alcohols (cetyl alcohol, stearyl alcohol, lanolin alcohol), and other fatty acids, wax esters, and sterols.47,59 Humectants attract water from the dermis and from high ambient humidity into the epidermis to maintain SC water content; these include glycerin (glycerol), hyaluronic acids, urea, panthenol, propylene glycol, sodium PCA, and certain lactates.47,59 Emollients exhibit different physical properties, protective, fatting, astringent, or dry characteristics) and are included to impart a soft and smooth skin texture by “filling the crevices” and fissures characteristic of xerotic skin; these include castor oil, Jojoba oil, silicone derivatives, isopropyl myristate, isopropyl palmitate, and isostearyl alcohol.47,59
Ultimately, moisturizing agents that incorporate quality fundamental ingredients in appropriate concentrations and combinations, such as humectants, occlusive agents and emollients, and contain physiological lipids or ingredients that augment skin lipid synthesis, and/or include other agents such as components of NMF that facilitate permeability barrier restoration, may be an optimal choice as these formulations are designed to target specific SC abnormalities that lead to xerotic skin changes, at least in most cases.1,47,49,59 Agents that are predominantly only occlusive and/or humectant may exhibit less substantivity, may produce less overall beneficial impact on physiological barrier restoration, and are often less likely to be preferred by patients for continued use due to a greasy or sticky texture.1,47,49,59 Importantly, individual formulations must be evaluated based on their own performance in both research investigations and after careful observation during clinical use. correlated with the reasons why they were rationally selected. In the article above, several examples are given with references for further information that provide more detail on suggested formulations and ingredients for xerotic skin conditions and specific disease states.
Conclusion
The epidermal barrier is a collective entity comprised of multiple individual barrier functions as depicted in Table 1. This article primarily emphasizes the SC permeability barrier which is a major dynamic component of overall epidermal barrier function, and is involved in regulation of epidermal water content, flux, and balance. Understanding the physiology of the SC permeability barrier in healthy skin and self-repair mechanisms that counter the adverse effects of increased TEWL and decreased SC hydration are fundamental to comprehending accurate clinical evaluation of barrier-related impairments in healthy and disease-affected skin. Various disease states exhibit specific barrier impairments that frequently include structural and functional SC dysfunction and permeability barrier dysregulation. Adjunctive skin care is a vital integral component in dermatologic practice, with product selection tailored to the needs of the individual patient. Without proper skin care, it is far more difficult to maintain healthy skin and to optimally manage several skin diseases, as outlined in this article.
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Man AM, Orăsan MS, Hoteiuc OA, et al. Inflammation and psoriasis: a comprehensive review. Int J Mol Sci. 2023 Nov 8;24(22):16095. doi: 10.3390/ijms242216095
Yan D, Gudjonsson JE, Le S, et al. New frontiers in psoriatic disease research, Part I: genetics, environmental triggers, immunology, pathophysiology, and precision medicine. J Invest Dermatol. 2021 September;141(9): 2112–2122.e3. doi:10.1016/j.jid.2021.02.764.
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Orsmond A, Bereza-Malcolm L, Lynch T, et al. Skin barrier dysregulation in psoriasis. Int J Mol Sci. 2021;22:10841.
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Kircik L, Alexis AA, Anneke Andriessen, et al. Psoriasis and skin barrier dysfunction: the role of gentle cleansers and moisturizers in treating psoriasis. J Drugs Dermatol. 2023;22(8)773-778.
Fluhr JW, Cavallotti C, Berardesca E. Emollients, moisturizers and keratolytic agents in psoriasis. Clin Dermatol. 2008;26(4):380-386.
von Martial, Nippel G, Schmidt L, et al. Influence of an adjuvant treatment with an emollient containing 10% urea, ceramides, glycerin, and glyceryl glycoside in patients with psoriasis vulgaris. Hautarzt. 2021;72(10):892-899.
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Yamamoto A, Takenouchi, Ito M. Impaired water barrier function in acne vulgaris. Arch Dermatol Res. 1985;287:214-218.
Weber ST, Thiele JJ, Han N, et al. Topical alpha-tocotrienol supplementation inhibits lipid peroxidation but fails to mitigate increased transepidermal water loss after benzoyl peroxide treatment to human skin. Free Radical Biol Med. 2003;34(2):170–176.
Sami N, De La Feld S. Topical Retinoids. In: Wolverton SE, Wu JJ, Eds. Comprehensive Dermatologic Drug Therapy, 4th Edition, Philadelphia: Elsevier, 2021:528-540.
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Del Rosso JQ. The role of skin care as an integral component in the management of acne vulgaris: part 1: the importance of cleanser and moisturizer ingredients, design, and product selection. J Clin Aesthet Dermatol. 2013;6(12):19–27.
Draelos ZD, Ertel KD, Berge CA. Facilitating facial retinization through barrier improvement. Cutis. 2006;78:275-281.
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Del Rosso JQ, Gold M, Rueda MJ, et al. Efficacy, safety, and subject satisfaction of a specified skin care regimen to cleanse, medicate, moisturize, and protect the skin of patients under treatment for acne vulgaris. J Clin Aesthet Dermatol. 2015;8(1):22–30.
Reynolds RV, Yeung H, Cheng CE, et al. Guidelines of care for management of acne vulgaris. J Am Acad Dermatol. 2024;90(5):1006.e1-1006.e30.
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Del Rosso JQ, Thiboutot D, Gallo R, et al. Consensus recommendations from the American Acne & Rosacea Society on the management of rosacea, part 1: a status report on the disease state, general measures, and adjunctive skin care. Cutis. 2013; 92:234–240.
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Flavia Alvim Sant´Anna Addor. Skin barrier in rosacea. An Bras Dermatol. 2016;91(1):59-63.
Drake L. New survey pinpoints leading factors that trigger symptoms. Rosacea Review. Summer 2002. Available at: summer/article_3.php. Initially accessed March 15, 2008.
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Del Rosso JQ. Adjunctive skin care in the management of rosacea: cleansers, moisturizers, and photoprotectants. Cutis. 2005;75:17-21.
Draelos ZD. Effect of Cetaphil Gentle Skin Cleanser on the skin barrier of patients with rosacea. Cutis. 2006;77(4 suppl):27-33.
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Ananthapadmanabhan KP, Moore DJ, Subramanyan K, et al. Cleansing without compromise: the impact of cleansers on the skin barrier and the technology of mild cleanser. Dermatol Ther. 2004;17:16-25.
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11283 | https://www.doubtnut.com/qna/327463902 | The number of ways in which three numbers in arithmetic progression can be selected from {1,2,3,……….,50} is
276
600
840
640
The correct Answer is:B
To solve the problem of finding the number of ways to select three numbers in arithmetic progression from the set {1, 2, 3, …, 50}, we can follow these steps: Step 1: Understanding Arithmetic Progression (AP) Three numbers A,B,C are in arithmetic progression if the middle number B is the average of A and C. This can be expressed mathematically as: 2B=A+C This implies that A and C must be chosen such that B is an integer. Step 2: Expressing the Numbers Let’s denote the three numbers in AP as: - A=B−d - B=B - C=B+d Here, d is the common difference. The numbers A, B, and C must all lie within the range of 1 to 50. Step 3: Constraints on B and d For A and C to be valid: - A=B−d≥1 implies B−d≥1 or d≤B−1 - C=B+d≤50 implies B+d≤50 or d≤50−B Thus, we have: d≤min(B−1,50−B) Step 4: Counting Valid d for Each B For each value of B from 1 to 50, we need to count how many valid d values exist. The maximum value of d is given by: dmax=min(B−1,50−B) The number of valid d values for each B is: dmax=min(B−1,50−B) Step 5: Summing Over All Possible B Now, we will sum the valid d values for each B: - For B=1: dmax=0 (no valid d) - For B=2: dmax=1 (valid d=1) - For B=3: dmax=2 (valid d=1,2) - ... - For B=25: dmax=24 (valid d=1,2,...,24) - For B=26: dmax=24 (valid d=1,2,...,24) - ... - For B=49: dmax=1 (valid d=1) - For B=50: dmax=0 (no valid d) Step 6: Calculating Total Combinations The total number of valid combinations is the sum of valid d values for each B: Total combinations=25∑B=2(B−1)+49∑B=26(50−B) Calculating these sums: - From B=2 to B=25: 24∑k=1k=24×252=300 - From B=26 to B=49: 24∑k=1k=300 Thus, the total number of ways to select three numbers in arithmetic progression from the set {1, 2, ..., 50} is: 300+300=600 Final Answer The number of ways in which three numbers in arithmetic progression can be selected from {1, 2, 3, …, 50} is 600. ---
To solve the problem of finding the number of ways to select three numbers in arithmetic progression from the set {1, 2, 3, …, 50}, we can follow these steps:
Step 1: Understanding Arithmetic Progression (AP)
Three numbers A,B,C are in arithmetic progression if the middle number B is the average of A and C. This can be expressed mathematically as:
2B=A+C
This implies that A and C must be chosen such that B is an integer.
Step 2: Expressing the Numbers
Let’s denote the three numbers in AP as:
- A=B−d
- B=B
- C=B+d
Here, d is the common difference. The numbers A, B, and C must all lie within the range of 1 to 50.
Step 3: Constraints on B and d
For A and C to be valid:
- A=B−d≥1 implies B−d≥1 or d≤B−1
- C=B+d≤50 implies B+d≤50 or d≤50−B
Thus, we have:
d≤min(B−1,50−B)
Step 4: Counting Valid d for Each B
For each value of B from 1 to 50, we need to count how many valid d values exist. The maximum value of d is given by:
dmax=min(B−1,50−B)
The number of valid d values for each B is:
dmax=min(B−1,50−B)
Step 5: Summing Over All Possible B
Now, we will sum the valid d values for each B:
- For B=1: dmax=0 (no valid d)
- For B=2: dmax=1 (valid d=1)
- For B=3: dmax=2 (valid d=1,2)
- ...
- For B=25: dmax=24 (valid d=1,2,...,24)
- For B=26: dmax=24 (valid d=1,2,...,24)
- ...
- For B=49: dmax=1 (valid d=1)
- For B=50: dmax=0 (no valid d)
Step 6: Calculating Total Combinations
The total number of valid combinations is the sum of valid d values for each B:
Total combinations=25∑B=2(B−1)+49∑B=26(50−B)
Calculating these sums:
- From B=2 to B=25:
24∑k=1k=24×252=300
- From B=26 to B=49:
24∑k=1k=300
Thus, the total number of ways to select three numbers in arithmetic progression from the set {1, 2, ..., 50} is:
300+300=600
Final Answer
The number of ways in which three numbers in arithmetic progression can be selected from {1, 2, 3, …, 50} is 600.
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Show that the number of ways in which three numbers in arithmetical progresssion can be selected from 1,2,3,……..n is 14(n−1)2 or 14n(n−2) according as n is odd or even.
The total number of ways in with three distinct
numbers in A.P. can be selected from the set {1,2,3,............24}
is equal to
a. 66
b. 132
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Knowledge Check
Number of ways in which three numbers in AP can be selected from 1,2,3, . .,n is
Given that n is odd, number of ways in which three numbers in AP can be selected from 1, 2, 3,……., n, is
Statement-1: The total number of ways in which three distinct numbers in AP, can be selected from the set {1,2,3, . .,21}, is equal to 100. Statement-2: If a,b,c are inn AP, then a+c=2b.
Number of ways in which three numbers in A.P. can
be selected from 1,2,3,...,n
is
a. (n−12)2
if n
is
even b. nn−24
if n
is
even
c. (n−1)24
if n
is odd d. none of these
Number of ways in which three numbers in AP can be selected from 1,2,3, . .,n is
Given that n is odd, number of ways in which three numbers in AP can be selected from 1, 2, 3,……., n, is
If a,b and c are three positive numbers in an arithmetic progression, then :
If a, b and c are three positive numbers in an arithmetic progression, then:
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Irritable bowel syndrome: diagnosis and management
A Agrawal
P J Whorwell
Correspondence to: P J Whorwell Peter.whorwell@smuht.nwest.nhs.uk
Roles
Series information
BMJ Learning
Accepted 2005 Nov 25.
Short abstract
Irritable bowel syndrome is often dismissed as just being a nuisance rather than anything more serious, but its symptoms can seriously diminish a patient's quality of life. When the disease is better understood and treatment is tailored to the individual patient, it can often be rewarding to manage
What is it, and who gets it?
Irritable bowel syndrome (IBS) is a chronic condition characterised by abdominal pain, bowel dysfunction, and abdominal bloating in the absence of any structural abnormality. A number of pathophysiological abnormalities, however, can often be identified.1 About 10-15% of the adult population in the United Kingdom is affected by irritable bowel syndrome.2
Aetiology
IBS is now clearly understood to be a multifactorial condition, with a variety of factors contributing to expression of the disease rather than its just being due to psychopathology.. These include motility, visceral sensation, central processing, genetics, dietary factors, inflammation, and neurotransmitters.1
Exacerbating factors
Stress exacerbates IBS rather than being causative in any way. If stress is severe and chronic—for example, stress caused by continuous domestic strife—it can result in the disorder being virtually untreatable.3
Antibiotics need to be used with care in patients with IBS. Some antibiotics, particularly erythromycin, can make the condition worse.4
Non-steroidal anti-inflammatory drugs are often prescribed for the pain associated with IBS, but they may exacerbate symptoms. Paracetamol does not upset IBS.5
How do I diagnose it?
In the absence of a specific diagnostic test, the diagnosis remains largely clinical.
History
Patients typically report
Abdominal pain or discomfort
Disordered bowel habit, with either diarrhoea, constipation, or alternating diarrhoea and constipation
Abdominal bloating or distension.
Many patients experience extracolonic features that can be useful for making the diagnosis:
Low backache
Constant lethargy
Nausea
Thigh pain
Urinary symptoms:
Frequency
Urgency
Urge incontinence
Gynaecological symptoms:
Dysmenorrhoea
Dysparaeunia.6
The diagnosis of IBS is usually made intuitively with remarkable safety and reliability. Attempts to refine this clinical approach into guidelines have resulted in several diagnostic criteria being created: the Manning criteria, Rome I criteria, Rome II criteria, and Rome III criteria (in preparation).
Such criteria have proved useful for research purposes by ensuring homogeneity of patient populations, but their applicability in clinical practice is extremely limited and they are seldom used. Unless much more reliable guidelines are developed, doctors are likely to continue with the pragmatic approach they are using now.
Diagnostic uncertainty is more likely with diarrhoea predominant rather than constipation predominant IBS. Inflammatory bowel disease has to be considered when diarrhoea is present, especially if it is accompanied by perianal soreness (unusual in patients with IBS) or features such as arthralgia, mouth ulcers, or eye signs.
Examination
The abdomen should be normal on examination, although some tenderness is often found, particularly in the left or right iliac fossa. A palpable caecum should not cause concern but obviously needs to be distinguished from a mass associated with Crohn's disease.
Investigations
The concept that IBS is a diagnosis by exclusion is outdated. Investigation can often be kept to a minimum and should be used to exclude realistic alternatives. A full blood count and erythrocyte sedimentation rate are often sufficient, but a normal erythrocyte sedimentation rate does not definitively rule out inflammatory bowel disease. Examination of the colon is advisable in patients older than 50 years, and this is particularly important if the symptoms are recent in onset.7
Currently, some uncertainty exists about the need to screen for coeliac disease with endomysial antibody or tissue transglutaminase, although some authors say that screening should be undertaken routinely.8 Testing certainly is indicated in the presence of a family history or malabsorption. The threshold for investigation should be lower in the presence of “red flag” features:
Rectal bleeding
Anaemia
Weight loss
Late age of onset
Acute onset
Family history of cancer
Family history of inflammatory bowel disease
Signs of infection.7
How should I treat it?
The treatment of IBS is notoriously unsatisfactory, and no new drug has become available in the United Kingdom in the past 20 years. Consequently, none of the currently available options has been subjected to controlled trials conducted to modern standards. The following approaches are usually applied in the order in which they are discussed.
Dietary manipulation
An increase in fibre is often advised in the first instance. This is surprising, as there is little evidence to show that it is effective—in fact, insoluble fibre (for example, bran) often makes the condition worse by exacerbating bloating and pain.9 Fibre may help constipation; the commercially available soluble fibre preparations are the least likely to cause problems. Other food items that can exacerbate symptoms are coffee, chocolate, and sugar substitutes such as sorbitol or fructose. Any food suspected of causing problems must be excluded from the diet for at least one month. It is best to omit one food at a time; otherwise, confusion arises about which item is a problem if improvement occurs. More strict exclusion diets have also been shown to be helpful but are time consuming and best done under the supervision of a dietitian.10 True IgE mediated dietary allergy is probably relatively unimportant in IBS, but there is some preliminary evidence that eliminating foods on the basis of the presence of IgG antibodies to food may have a role.11
Antispasmodics
Antispasmodics are available in two varieties:
Anticholinergics—hyoscine and dicyclomine
Smooth muscle relaxants—alverine, mebeverine, and peppermint oil.
It is impossible to predict who will respond to a particular preparation and therefore it is worth trying them all. Combinations of a smooth muscle relaxant and anticholinergic are sometimes effective, and taking them “as necessary” helps to minimise tachyphylaxis, which can occur after prolonged use.
Antidiarrhoeals
Antidiarrhoeal agents include loperamide, diphenoxylate, and codeine phosphate. Loperamide especially is useful as it also tends to increase the tone of the anal sphincter. Codeine can lead to dependency. Patients should be encouraged to titrate the dose of an antidiarrhoeal according to their needs, and they need to be reassured that regular use is not a problem and will not damage their bowel.
Laxatives
Laxatives include senna, bisacodyl, polyethylene glycol, and sodium picosulphate. In a similar way to antidiarrhoeals, laxatives are often best used in the form of a regular small dose rather than precipitating a catharsis now and again. Patients should be reassured that there is no evidence to suggest that laxatives “damage” the bowel, and that long term use is acceptable. Lactulose is best avoided, as it causes a lot of flatus and can exacerbate bloating.
Antidepressants
The tricyclic antidepressants and selective serotonin reuptake inhibitors are used, but there is more evidence to support the use of tricyclics. The tricyclics tend to cause constipation and consequently are particularly suited to diarrhoea predominant IBS. If such antidepressants are used in patients with constipation, a laxative may also have to be given. The selective serotonin reuptake inhibitors do not cause constipation.12
Behavioural treatments
Behavioural treatments include psychotherapy, cognitive behavioural therapy, and hypnotherapy. Evidence supports all of these treatments, but they are time consuming, costly to provide, and vary widely in their availability. They probably are best reserved for treatment in secondary and tertiary care.13,14
New treatments
Probiotics
Probiotics are “friendly bacteria” such as lactobacilli and bifidobacteria. Different strains can elaborate different mediators, which confer different properties on different organisms. This means that the therapeutic activity of one strain can be completely different from that of another strain. Combinations may not be a good idea, as they could inhibit one another. One probiotic strain (Bifidobacterium infantis 35624) has shown potential in patients with IBS, but further work is needed.15,16
Type 3 serotonin receptor antagonists
Alosetron, cilansetron, and ramosetron have been developed for the treatment of diarrhoea predominant IBS and show a positive effect in clinical trials. Ischaemic colitis has been reported with alosetron and cilansetron, although this seems to be self limiting if the drug is stopped. In addition, any type 3 serotonin receptor antagonist needs to be discontinued promptly if constipation develops.17 Alosetron is available in the United States but not in the United Kingdom.
Type 4 serotonin receptor agonists
Tegaserod is used for constipation predominant IBS and has proved effective in clinical trials. It is now on the market in several countries, including the United States, but not in the United Kingdom. No major safety issues seem to be associated with this drug.18
When should I refer my patient?
Many patients with IBS respond to a combination of education about the condition and simple measures to deal with symptoms. Referral for further assessment should be considered if there is doubt about the diagnosis or the patient becomes refractory to treatment.
What is the outlook?
IBS should probably be regarded as a lifelong condition, just as patients with a history of migraine will nearly always continue to have a tendency towards migraines. Patients should thus expect to have symptoms intermittently, especially if they are exposed to any exacerbating factors.
IBS in secondary care
Most cases of IBS are relatively mild, and the patient can cope with the condition reasonably well. Symptoms in patients who attend hospital clinics are much more severe; because IBS is so common, these patients are numerous. The factors below are more characteristic of patients seen in secondary and tertiary care than of those seen in general practice.
Pain
The pain of IBS can be exceptionally severe. Many women equate it with the pain of childbirth.
Bowel dysfunction
In some cases of constipation, bowel movements can be separated by many days or sometimes weeks. Patients with the diarrhoea predominant form of the condition can experience extreme urgency. Faecal incontinence is not uncommon and is devastating when it occurs.
Distension
Distension might be regarded as an unimportant feature, but, in some instances, the abdominal girth can increase by 10-12 cm by the end of the day.
Flatus
Flatus is a universal occurrence, but patients with IBS seem to experience more flatus problems, perhaps because so many patients are on high fibre diets, which are known to generate more gas. Dietary modification sometimes helps.
Sexual function
Eighty per cent of patients who attend hospital clinics say that their IBS significantly impairs sexual function. This compares with a figure of 30% in patients with Crohn's disease or ulcerative colitis. Dyspareunia is the main complaint; this usually follows intercourse and may persist for many hours or days.19
Sample questions
Here is a small sample of the questions that you can find at the end of this module. To see all the questions and to get the answers, go to www.bmjlearning.com/ and search for “irritable bowel syndrome”.
Which of the following statements about bloating in patients with irritable bowel syndrome is correct?
Which one of the following statements about post-infectious irritable bowel syndrome is correct?
Which one of the following statements about probiotics is correct?
A 55 year old man has a diagnosis of longstanding irritable bowel syndrome and benign prostatic hyperplasia. Which of the following medications would you avoid giving this man?
Summary points
Irritable bowel syndrome is often regarded as a trivial, largely psychological disorder that is impossible to treat
Patients with severe disease have a range of symptoms that can seriously erode quality of life
Abdominal pain can sometimes be devastating, and the bowel dysfunction is not infrequently accompanied by incontinence
Better understanding of the pathophysiology, and tailoring treatment to the individual, can make irritable bowel syndrome a surprisingly rewarding condition to manage
Suicide
In a recent report, 38% of patients who attended a tertiary care clinic seriously considered ending their lives solely because of their bowel problem.20 The patients were not especially depressed, and the suicidal ideation was considered to be centred around hopelessness related to the prospect of little relief from their problem in the future.
Extracolonic symptoms
The extracolonic symptoms listed earlier are important for several reasons:
Diagnostic utility—the more of these symptoms that are present, the more likely the patient is to have IBS
Reassurance—patients are often alarmed by the presence of all these disparate symptoms, suspecting that a major disease might be overlooked. An understanding that they all form part of the syndrome provides some reassurance
Inappropriate referral and treatment.
Many patients with IBS are referred to gynaecological clinics, where they can undergo a variety of unnecessary investigations and sometimes even removal of the uterus or ovaries. Abdominal surgery unfortunately tends to make the symptoms of IBS even worse and should be avoided where possible.
Patients with IBS are also over-represented in urological clinics. They tend to be offered antibiotics for presumed recurrent urinary infections, but such drugs can exacerbate the bowel problem.
The low backache associated with IBS can lead to orthopaedic referral, and patients with IBS have been shown to have an excessive history of back surgery compared with controls.21
Absenteeism
IBS is a major cause of absenteeism from work which is a reflection of symptom severity as opposed to work avoidance.22
Stigmatisation
Patients with IBS often are reluctant to admit to others that they have this condition because of fear that they will be labelled as psychologically disturbed. Potential employers often are reluctant to employ patients with IBS because of their reputation for absenteeism.
Quality of life
Given the severity and range of symptoms in some patients with IBS, that quality of life is eroded is not surprising. Quality of life can be measured with a questionnaire such as the 36 item short form; such an approach has shown that patients with IBS who attend hospital clinics have worse quality of life than those with chronic renal disease or diabetes.
Conclusion
IBS is an extremely challenging condition to manage. Effective treatment involves understanding the whole situation and tailoring the treatment to the individual. It is difficult, but not impossible, to offer at least some help to most patients with the condition.
This article is based on a module that is available on the BMJ Learning website (www.bmjlearning.com). Access to the site is free but registration is required
Competing interests: PJW has received remuneration for advice and his department has also received financial support from Novartis Pharmaceuticals, GlaxoSmithKline, Pfizer, Solvay Pharmaceuticals, Rotta Research, Procter & Gamble, Astellas Pharma, Tillots Pharma.
References
Articles from BMJ : British Medical Journal are provided here courtesy of BMJ Publishing Group
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11285 | https://wchh.onlinelibrary.wiley.com/doi/abs/10.1002/psb.2010 | Updated guideline on diagnosing and managing epilepsies - Chaplin - 2022 - Prescriber - Wiley Online Library
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Updated guideline on diagnosing and managing epilepsies
Steve Chaplin,
Steve Chaplin
Steve Chaplin is a medical writer specialising in therapeutics
Search for more papers by this author
Steve Chaplin,
Steve Chaplin
Steve Chaplin is a medical writer specialising in therapeutics
Search for more papers by this author
First published: 13 September 2022
Citations: 2
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Abstract
From diagnosis to referral
Principles of treatment
Treating epileptic seizures
Non-pharmacological treatments
Managing co-morbidities
Epilepsy-related death
Services and transition
Summary
Declaration of interests
References
Citing Literature
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Abstract
NICE has recently updated its guideline on diagnosing and managing epilepsies in children, young people and adults (NG217). This article provides a brief summary of the new guidance, which provides detailed recommendations on assessment, referral and management of different epilepsy and seizure types.
In 2017, the International League Against Epilepsy published a new framework for classification of the epilepsies. It defined three levels of diagnosis: seizure type – focal, generalised or unknown onset seizures; epilepsy type – focal, generalised, combined generalised and focal, or unknown; and epilepsy syndrome.1 As a result, aetiology has come to play a greater part in diagnosis and management, and this is reflected in the recently published Epilepsies in Children, Young People and Adults (NG217),2 NICE's second update to its 2004 guideline (first updated in 2012).
The guideline is divided into 11 sections covering: diagnosis and assessment; information and support; referral to tertiary care; treatment principles; treating epileptic seizures; treating childhood-onset epilepsies; treating status epilepticus and repeated, cluster and prolonged seizures; non-pharmacological treatments; psychological and other co-morbidities; reducing risk of epilepsy-related death; and service provision and transition.
From diagnosis to referral
A first seizure and a seizure recurrence are indications for an urgent referral to an adult or paediatric specialist, though help from a tertiary paediatric neurologist should be sought within 24 hours if a child under two years has suspected or confirmed infantile spasms. Following a first seizure, the risk of more events should be assessed. Risk factors in adults include an underlying mental health problem, vascular disease (including diabetes) and sepsis. Afebrile seizure in children is associated with a higher risk of recurrence than febrile seizure.
Specialist assessment should include an EEG as soon as possible. This can support a diagnosis of epilepsy and provide information about the seizure type or syndrome but the EEG should not exclude a diagnosis – a sleep-deprived EEG and then a 48-hour ambulatory EEG should be considered if the first result reveals nothing abnormal. Further options to be considered with the appropriate specialist include an MRI (or, if contraindicated, a CT scan), genetic testing and, if autoimmune encephalitis is suspected, antibody testing.
After a first seizure, the patient, their family and carers should be given information about their type of seizure or epilepsy, risk reduction, how to recognise a seizure, first aid and self-referral. NICE provides a list of specified topics to be covered, including signposting to support agencies and information about the impact on daily activities (including driving), seizure triggers and medication. Specifically, the discussion should cover the cognitive impact of epilepsy and its treatment, reproductive health, mental health and sudden unexpected death in epilepsy (SUDEP).
Everyone with suspected or diagnosed epilepsy should have access to a tertiary care service. Children should be referred within two weeks if they are under three years old (under four years if they have myoclonic seizures) or they have a unilateral structural lesion or deterioration in behaviour, speech or learning. Adults should be referred within four weeks if there is diagnostic uncertainty, a drug-resistant syndrome is suspected, or further imaging or specialised treatment is indicated.
Principles of treatment
Treatment with an antiseizure medication (ASM) should be commenced when a diagnosis of epilepsy is confirmed, and considered after a first unprovoked seizure if there is evidence of a neurological deficit, a structural brain abnormality, the EEG is unequivocal, or the risk of further seizures is not acceptable. ASM monotherapy should be the initial aim. If the agent of first choice is unsuccessful (and the diagnosis is reviewed and confirmed), an alternative should be substituted, phasing the changeover slowly. Add-on therapy should be considered if monotherapy is not successful, with frequent assessment of adverse effects and seizure reduction. The advice of the Medicines and Healthcare products Regulatory Agency (MHRA) on switching between ASM formulations should be followed.3
The treatment strategy should be individualised according to the patient's diagnosis, age, sex, risk factors and co-morbidities, but also patient preference, individual circumstances (work, school, family commitments) and how and when medication is to be taken. There is, of course, special concern about the safety of ASMs for women and girls, and these issues should be discussed with patients. The MHRA has published guidance on the use of ASMs during pregnancy,4 and specifically on use of valproate in women and girls of childbearing potential.5 Briefly, valproate must not be used in any woman or girl able to have children unless she has a pregnancy prevention programme in place, and is contraindicated in pregnancy unless there is no other effective epilepsy treatment available. Breastfeeding in women taking ASM is “generally safe and should be encouraged”, but prescribers should consult advice on individual drugs in the summary of product characteristics or the BNF. Some ASMs impair the effectiveness of hormonal contraception (and vice versa in the case of lamotrigine).
Regular, and at least annual, reviews should be offered to adults at special risk (eg due to learning disability, higher risk of SUDEP, drug resistant epilepsy, serious co-morbidity, taking an ASM with a risk of long-term adverse effects, being of childbearing age, taking a teratogenic ASM) and to children, tailored to their needs. Therapeutic drug monitoring should be considered when seizures are not controlled, adverse effects occur, poor adherence is suspected or when closer supervision is needed (eg during pregnancy). Patients should be able to request a review if they feel they need one.
Women who are pregnant or planning pregnancy should be referred to an epilepsy specialist team for medication review. Care should be shared between that team, primary care and a specialist obstetric team. Women should be informed about how pregnancy may affect their treatment, and additional and more frequent monitoring should be considered.
Withdrawal of ASM treatment should be considered after two years of being seizure-free, and the patient and their family or carers, as appropriate, have discussed the risks and benefits with the epilepsy specialist team. Treatment is typically discontinued slowly over at least three months (one ASM at a time in the case of polytherapy), though a longer discontinuation period is likely to be needed for benzodiazepines and barbiturates to reduce the risk of withdrawal symptoms.
Treating epileptic seizures
The guideline provides specific recommendations on the choice of ASM to treat each seizure type (generalised tonic-clonic seizures, focal seizures with or without evolution to bilateral tonic-clonic seizures, absence seizures, myoclonic seizures, tonic or atonic seizures, idiopathic generalised epilepsies); to treat childhood-onset epilepsies (Dravet syndrome, Lennox-Gastaut syndrome, infantile spasms syndrome, self-limited epilepsy with centrotemporal spikes and Doose syndrome); and to treat status epilepticus, repeated or cluster seizures and prolonged seizures. These include sex-specific recommendations for first-line treatment, alternatives and second- and third-line agents (some of which are off-label), and options for add-on treatment. There is also advice on which ASMs exacerbate some seizure types. The place of newer ASMs in therapy is specified, though mostly as one of several second-line options; it is notable that sodium valproate is still recommended in many cases as a first-line therapy for men, boys, women unable to have children, and girls under 10 years old unlikely to need treatment when they reach childbearing age.
Non-pharmacological treatments
A ketogenic diet should be considered, guided by a tertiary care specialist, for some childhood-onset syndromes and, if other options are unsuccessful or inappropriate, drug-resistant epilepsies. Resective surgery is a further option for adults and children with drug-resistant epilepsy, following referral to tertiary services. If resective surgery is unsuitable, a further option is vagus nerve stimulation, administered as add-on to ASM treatment.
Managing co-morbidities
NICE states that “a diagnosis of epilepsy can have a significant adverse impact on a person's mental health and that people with epilepsy may feel socially excluded and stigmatised.” It recommends that a review of neurodevelopment, cognitive function, mental health, and social and emotional wellbeing should be part of routine management.
Compared with the general population, people with epilepsy have a higher prevalence of mental health difficulties, learning disabilities, neurodevelopmental co-morbidity (attention deficit-hyperactivity disorder and autism spectrum disorder) and dementia. The risk of suicide is also raised. Care provision should address these issues, and NICE refers to its guidance on mental health problems and challenging behaviour in people with learning disabilities (NG11 and NG54) and management of dementia (NG97) for further detail on management. It also provides links to a range of other guidelines on specific neurodevelopmental disorders, psychosis, anxiety and depression.
Epilepsy-related death
The guideline takes a risk-based approach to avoiding SUDEP and premature death, listing factors that are modifiable (eg non-adherence to medication, living alone, drug or alcohol misuse, having focal to bilateral tonic-clonic seizures, generalised tonic-clonic seizures or uncontrolled seizures) or non-modifiable (eg previous brain injury, CNS infection, abnormal neurological exam, stroke or metastatic cancer). People with epilepsy and their families or carers should be informed of the risks and consider support measures – for example, night-time supervision for someone who has seizures during sleep, or support with medicines adherence.
Services and transition
The epilepsy specialist nurse is the point of contact for the various agencies involved in epilepsy care, providing support to health professionals in all settings and offering education and support to people with epilepsy and their families. Everyone with epilepsy should have access to an epilepsy specialist nurse, and those who continue to have seizures should see them at least twice annually and after accessing emergency services.
Young people should be involved in planning their transition from paediatric to adult services. Planning, which should comprehensively address the person's concerns and be carried out jointly by adult and paediatric services, should be individually tailored and begin early for those with complex or additional health and social care needs.
Summary
This update to the NICE guideline provides detailed recommendations on all aspects of the management of epilepsy, including which investigations should be carried out, when and by whom, and the choice of ASMs for each diagnosis. The role of older ASMs, notably sodium valproate, remains prominent for boys, men, women unable to have children, and girls under 10 years, and there is specific advice on treatment choices for women and girls able to have children. The involvement of patients and access to epilepsy specialist nurses and tertiary services will be essential to implementing these recommendations.
Declaration of interests
None to declare.
References
1 Scheffer IE, et al. ILAE classification of the epilepsies: position paper of the ILAE Commission for Classification and Terminology. Epilepsia 2017; 58: 512–21. 10.1111/epi.13709 PubMedWeb of Science®Google Scholar
2 National Institute for Health and Care Excellence. Epilepsies in children, young people and adults. NG217. April 2022. Available from: Google Scholar
3 Medicines and Healthcare products Regulatory Agency. Antiepileptic drugs: updated advice on switching between different manufacturers’ products. November 2017. Available from: Google Scholar
4 Medicines and Healthcare products Regulatory Agency. Antiepileptic drugs in pregnancy: updated advice following comprehensive safety review. January 2021. Available from: Google Scholar
5 Medicines and Healthcare products Regulatory Agency. Valproate use by women and girls. March 2018 (updated February 2021). Available from: Google Scholar
Citing Literature
Volume 33, Issue 8-9
Aug-Sept 2022
Pages 28-30
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11286 | https://courses.lumenlearning.com/waymakercollegealgebra/chapter/graphs-of-ellipses/ | Graphs of Ellipses
Learning Outcomes
Sketch a graph of an ellipse centered at the origin.
Sketch a graph of an ellipse not centered at the origin.
Express the equation of an ellipse in standard form given the equation in general form.
Just as we can write the equation for an ellipse given its graph, we can graph an ellipse given its equation. To graph ellipses centered at the origin, we use the standard form x2a2+y2b2=1, a>b for horizontal ellipses and x2b2+y2a2=1, a>b for vertical ellipses.
How To: Given the standard form of an equation for an ellipse centered at (0,0), sketch the graph.
Use the standard forms of the equations of an ellipse to determine the major axis, vertices, co-vertices, and foci.
If the equation is in the form x2a2+y2b2=1, where a>b, then
the major axis is the x-axis
the coordinates of the vertices are (±a,0)
the coordinates of the co-vertices are (0,±b)
the coordinates of the foci are (±c,0)
If the equation is in the form x2b2+y2a2=1, where a>b, then
the major axis is the y-axis
the coordinates of the vertices are (0,±a)
the coordinates of the co-vertices are (±b,0)
the coordinates of the foci are (0,±c)
Solve for c using the equation c2=a2−b2.
Plot the center, vertices, co-vertices, and foci in the coordinate plane, and draw a smooth curve to form the ellipse.
Example: Graphing an Ellipse Centered at the Origin
Graph the ellipse given by the equation, x29+y225=1. Identify and label the center, vertices, co-vertices, and foci.
Show Solution
First, we determine the position of the major axis. Because 25>9, the major axis is on the y-axis. Therefore, the equation is in the form x2b2+y2a2=1, where b2=9 and a2=25. It follows that:
the center of the ellipse is (0,0)
the coordinates of the vertices are (0,±a)=(0,±√25)=(0,±5)
the coordinates of the co-vertices are (±b,0)=(±√9,0)=(±3,0)
the coordinates of the foci are (0,±c), where c2=a2−b2 Solving for c, we have:
c=±√a2−b2=±√25−9=±√16=±4
Therefore, the coordinates of the foci are (0,±4).
Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.
Try It
Graph the ellipse given by the equation x236+y24=1. Identify and label the center, vertices, co-vertices, and foci.
Show Solution
center: (0,0); vertices: (±6,0); co-vertices: (0,±2); foci: (±4√2,0)
Example: Graphing an Ellipse Centered at the Origin from an Equation Not in Standard Form
Graph the ellipse given by the equation 4x2+25y2=100. Rewrite the equation in standard form. Then identify and label the center, vertices, co-vertices, and foci.
Show Solution
First, use algebra to rewrite the equation in standard form.
4x2+25y2=1004x2100+25y2100=100100x225+y24=1
Next, we determine the position of the major axis. Because 25>4, the major axis is on the x-axis. Therefore, the equation is in the form x2a2+y2b2=1, where a2=25 and b2=4. It follows that:
the center of the ellipse is (0,0)
the coordinates of the vertices are (±a,0)=(±√25,0)=(±5,0)
the coordinates of the co-vertices are (0,±b)=(0,±√4)=(0,±2)
the coordinates of the foci are (±c,0), where c2=a2−b2. Solving for c, we have:
c=±√a2−b2=±√25−4=±√21
Therefore the coordinates of the foci are (±√21,0).
Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.
Try It
Graph the ellipse given by the equation 49x2+16y2=784. Rewrite the equation in standard form. Then identify and label the center, vertices, co-vertices, and foci.
Show Solution
Standard form: x216+y249=1; center: (0,0); vertices: (0,±7); co-vertices: (±4,0); foci: (0,±√33)
Graphing Ellipses Not Centered at the Origin
When an ellipse is not centered at the origin, we can still use the standard forms to find the key features of the graph. When the ellipse is centered at some point, (h,k), we use the standard forms (x−h)2a2+(y−k)2b2=1, a>b for horizontal ellipses and (x−h)2b2+(y−k)2a2=1, a>b for vertical ellipses. From these standard equations, we can easily determine the center, vertices, co-vertices, foci, and positions of the major and minor axes.
How To: Given the standard form of an equation for an ellipse centered at (h,k), sketch the graph.
Use the standard forms of the equations of an ellipse to determine the center, position of the major axis, vertices, co-vertices, and foci.
If the equation is in the form (x−h)2a2+(y−k)2b2=1, where a>b, then
the center is (h,k)
the major axis is parallel to the x-axis
the coordinates of the vertices are (h±a,k)
the coordinates of the co-vertices are (h,k±b)
the coordinates of the foci are (h±c,k)
If the equation is in the form (x−h)2b2+(y−k)2a2=1, where a>b, then
the center is (h,k)
the major axis is parallel to the y-axis
the coordinates of the vertices are (h,k±a)
the coordinates of the co-vertices are (h±b,k)
the coordinates of the foci are (h,k±c)
Solve for c using the equation c2=a2−b2.
Plot the center, vertices, co-vertices, and foci in the coordinate plane, and draw a smooth curve to form the ellipse.
Example: Graphing an Ellipse Centered at (h, k)
Graph the ellipse given by the equation, (x+2)24+(y−5)29=1. Identify and label the center, vertices, co-vertices, and foci.
Show Solution
First, we determine the position of the major axis. Because 9>4, the major axis is parallel to the y-axis. Therefore, the equation is in the form (x−h)2b2+(y−k)2a2=1, where b2=4 and a2=9. It follows that:
the center of the ellipse is (h,k)=(−2,5)
the coordinates of the vertices are (h,k±a)=(−2,5±√9)=(−2,5±3), or (−2,2) and (−2,8)
the coordinates of the co-vertices are (h±b,k)=(−2±√4,5)=(−2±2,5), or (−4,5) and (0,5)
the coordinates of the foci are (h,k±c), where c2=a2−b2. Solving for c, we have:
c=±√a2−b2=±√9−4=±√5
Therefore, the coordinates of the foci are (−2,5−√5) and (−2,5+√5).
Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.
Try It
Graph the ellipse given by the equation (x−4)236+(y−2)220=1. Identify and label the center, vertices, co-vertices, and foci.
Show Solution
Center: (4,2); vertices: (−2,2) and (10,2); co-vertices: (4,2−2√5) and (4,2+2√5); foci: (0,2) and (8,2)
How To: Given the general form of an equation for an ellipse centered at (h, k), express the equation in standard form.
Recognize that an ellipse described by an equation in the form ax2+by2+cx+dy+e=0 is in general form.
Rearrange the equation by grouping terms that contain the same variable. Move the constant term to the opposite side of the equation.
Factor out the coefficients of the x2 and y2 terms in preparation for completing the square.
Complete the square for each variable to rewrite the equation in the form of the sum of multiples of two binomials squared set equal to a constant, m1(x−h)2+m2(y−k)2=m3, where m1,m2, and m3 are constants.
Divide both sides of the equation by the constant term to express the equation in standard form.
Example: Graphing an Ellipse Centered at (h, k) by First Writing It in Standard Form
Graph the ellipse given by the equation 4x2+9y2−40x+36y+100=0. Identify and label the center, vertices, co-vertices, and foci.
Show Solution
We must begin by rewriting the equation in standard form.
4x2+9y2−40x+36y+100=0
Group terms that contain the same variable, and move the constant to the opposite side of the equation.
(4x2−40x)+(9y2+36y)=−100
Factor out the coefficients of the squared terms.
4(x2−10x)+9(y2+4y)=−100
Complete the square twice. Remember to balance the equation by adding the same constants to each side.
4(x2−10x+25)+9(y2+4y+4)=−100+100+36
Rewrite as perfect squares.
4(x−5)2+9(y+2)2=36
Divide both sides by the constant term to place the equation in standard form.
(x−5)29+(y+2)24=1
Now that the equation is in standard form, we can determine the position of the major axis. Because 9>4, the major axis is parallel to the x-axis. Therefore, the equation is in the form (x−h)2a2+(y−k)2b2=1, where a2=9 and b2=4. It follows that:
the center of the ellipse is (h,k)=(5,−2)
the coordinates of the vertices are (h±a,k)=(5±√9,−2)=(5±3,−2), or (2,−2) and (8,−2)
the coordinates of the co-vertices are (h,k±b)=(5,−2±√4)=(5,−2±2), or (5,−4) and (5,0)
the coordinates of the foci are (h±c,k), where c2=a2−b2. Solving for c, we have:
c=±√a2−b2=±√9−4=±√5
Therefore, the coordinates of the foci are (5−√5,−2) and (5+√5,−2).
Next we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.
Try It
Express the equation of the ellipse given in standard form. Identify the center, vertices, co-vertices, and foci of the ellipse.
4x2+y2−24x+2y+21=0
Show Solution
(x−3)24+(y+1)216=1; center: (3,−1); vertices: (3,−5) and (3,3); co-vertices: (1,−1) and (5,−1); foci: (3,−1−2√3) and (3,−1+2√3)
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11287 | https://math.stackexchange.com/questions/2349077/finding-what-numbers-can-be-produced-mod-k | modular arithmetic - Finding what Numbers can be Produced Mod k - Mathematics Stack Exchange
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Finding what Numbers can be Produced Mod k
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So I have a big number, say 107279. Now, I want to take all of its multiples mod 10 9+7 10 9+7. How do I determine which integers mod 10 9+7 10 9+7 are of such a form, 107279 k 107279 k? I did some searching and fiddling, and came up with nothing particularly helpful.
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asked Jul 7, 2017 at 1:53
Vedvart1Vedvart1
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Notice that 10 9+7 10 9+7 is a prime number.
Hence 107279−1 mod(10 9+7)107279−1 mod(10 9+7) exists
Hence we can always solve for 107279 x≡y mod(10 9+7)107279 x≡y mod(10 9+7) for any y y.
Hence y y can take any value from 0 0 to 10 9+6 10 9+6.
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edited Jul 7, 2017 at 2:22
answered Jul 7, 2017 at 1:58
Siong Thye GohSiong Thye Goh
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In general, suppose you have a number d d (which stands in for 107279 107279 in your example) and another number k k (which represents 10 9+7 10 9+7), and you want to know which residues can be produced by taking multiples of d d modulo k k.
The general answer is that you first compute g=gcd(d,k)g=gcd(d,k), and then there will be exactly k/g k/g possible residues, specifically all of the multiples of g g modulo k k: 0,g,2 g,…,k−g 0,g,2 g,…,k−g. In this particular case g=1 g=1 and so there will be exactly k k possible residues, namely all of them.
This is a very elementary consequence of the simple yet powerful Bézout's identity.
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answered Jul 7, 2017 at 3:13
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H i n t⟺⟺⟺a∣x(mod n)∃j:x≡j a(mod n)∃j,k:x=j a+k n gcd(a,n)∣x H i n t a∣x(mod n)⟺∃j:x≡j a(mod n)⟺∃j,k:x=j a+k n⟺gcd(a,n)∣x
i.e. a Z+n Z=gcd(a,n)Z a Z+n Z=gcd(a,n)Z by Bezout.
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answered Jul 7, 2017 at 3:32
Bill DubuqueBill Dubuque
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they have a GCD of one so all integers mod 10 9+7 10 9+7 appear.
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answered Jul 7, 2017 at 1:58
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11288 | https://english.stackexchange.com/questions/244847/what-is-n-the-abreviation-of | abbreviations - What is n' the abreviation of - English Language & Usage Stack Exchange
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What is n' the abreviation of [closed]
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Can someone explain to me what n' is the abbreviation of. I have found someone using it like born n' bred. To me that would be stating a missing letter to shorten it down. But wouldn't it mean nd rather than the and I think they are trying to mean.
I want to know what n' is the abbreviation of.
I have never seen n' before and searching through the dictionary did not come back with any results. So I put my question to you. I know what 'n is the abbreviation of. But, as I said, I have never seen n'
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edited May 7, 2015 at 14:09
SamSam
asked May 7, 2015 at 10:12
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How is this a duplicate? That question asked is there any other examples. I want to know what the n' is the abbreviation of.Sam –Sam 2015-05-07 10:34:39 +00:00 Commented May 7, 2015 at 10:34
@RegDwigнt. First of all I did look in the dictionary. It was not there. As you can see from the answer below it was in the urban dictionary so is not actually a real word. Second I know what Born 'n bred means. But since this politcians website said Born n' bred I wanted to check to see if it had another meaning that I was not aware out. Your comment is completly pointless and just flat out rude Sam –Sam 2015-05-07 10:52:37 +00:00 Commented May 7, 2015 at 10:52
It is in the Cambridge dictionary. Nonstandard spelling of ’n’ as in Born 'n' bred mplungjan –mplungjan 2015-05-07 11:13:03 +00:00 Commented May 7, 2015 at 11:13
If you ask what the meaning of …n'… is, it's no wonder people are going to say it means "and". I love to see the politician's face tonight when he receives a comment about a missing apostrophe, as if he (or she) doesn't have better things to worry about.Mari-Lou A –Mari-Lou A 2015-05-07 11:15:42 +00:00 Commented May 7, 2015 at 11:15
The link provided by RegDwight was perfect. The Wiktionary entry contains the "spelling" of n' and says: Nonstandard spelling of ’n’. So RegDwight's answer hidden in the link was helpful, and, accurate.Mari-Lou A –Mari-Lou A 2015-05-07 11:53:40 +00:00 Commented May 7, 2015 at 11:53
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'n or 'n' (ən):
(conj.) and:
Look 'n listen.
(The Free Dictionary)
nd is not a commonly accepted expression, The Urban Dictionary says;
Slang word for and. One of the many reasons I cannot successfully communicate with other teenagers through the internet, since they do not write in clear English.
nd he was like so cute
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edited May 7, 2015 at 10:47
answered May 7, 2015 at 10:21
user66974 user66974
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This does not answer my question. It just says it is a conjunction. But does ntot answer my question Sam –Sam 2015-05-07 10:36:23 +00:00 Commented May 7, 2015 at 10:36
2 Sorry , are you not asking for what 'n means?user66974 –user66974 2015-05-07 10:38:59 +00:00 Commented May 7, 2015 at 10:38
Not what 'n means but what n' means. As I have seen it as n' rather than 'n As I would guess that 'n means an but what would n' mean? If there is a difference Sam –Sam 2015-05-07 10:44:53 +00:00 Commented May 7, 2015 at 10:44
1 @Josh61 It did come to me that it could mean that. However before I am going to tell the guy about his website I wanted to make sure it didn't have a meaning I have never seen/heard of before.Sam –Sam 2015-05-07 10:59:27 +00:00 Commented May 7, 2015 at 10:59
1 Well he did technically help. His first answer showed that in the dictionary there is only 'n and 'n'. That helped confirm that I am not just useless at searching through a dictionary and couldn't see it myself. The comments did help me edit my question so it was more clear to what the answer should be. Which he did answer by saying that it is only in the Urban Dictionary Sam –Sam 2015-05-07 11:30:01 +00:00 Commented May 7, 2015 at 11:30
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11289 | https://www.datacamp.com/tutorial/t-tests-r-tutorial | Skip to main content
T-tests in R Tutorial: Learn How to Conduct T-Tests
Determine if there is a significant difference between the means of the two groups using t.test() in R.
Mar 14, 2023 · 10 min read
Introduction
Suppose you have two groups of sales teams and you want to check if the average number of mobile phones sold in a week by both teams is the same or not. How will you compare the performance?
You will take the average number of mobile phones sold to 200 random customers by respective teams and determine the difference. The first marketing team, on average, has sold 120 phones, whereas the second team has sold 80.
So, it is clear that the first team has performed better in sales than the second team. Right? We can’t be sure; the dataset is collected from random customers and does not represent all of the people who bought the phone in that week.
So how do we determine which team performed better? We will use a t-test to understand if the difference between the two averages is real or just random luck.
The t-test is a statistical hypothesis that takes samples from both groups to determine if there is a significant difference between the means of the two groups. How does it work? It compares both sample mean and standard deviations while considering sample size and the degree of variability of the data.
In this tutorial, we will learn about the classification of t-tests (one-sample, two-samples, and paired sample t-test) with R code examples and learn to interpret the results.
Note: if you are new to R, take a mini Introduction to R Programming course to understand the basics.
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t.test() Function in R
R language provides us with a simple t.test built-in function for One Sample, Two Samples, and Paired t-tests.
There are two ways of using the t.test function: default and formula methods.
Default method
You provide numerical samples from the x group and y group, specifying the alternative hypothesis, hypothesized mean mu, and confidence level of the interval. Moreover, you can perform Paired t-test by toggling the paired argument and Two Sample t-test with equal variance by changing var.equal argument.
```
t.test(x, y,
alternative = c("two.sided", "less", "greater"),
mu = 0, paired = FALSE, var.equal = FALSE,
conf.level = 0.95, ...)
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```
Formula method
In this method, you provide the formula x~y, where x is a numeric vector or a column from the data, and y is a binary column containing the types of groups.
(,,,,...) Was this AI assistant helpful? Yes No
How to Perform One Sample t-test in R
One-sample t-test is the statistical hypothesis to test if there is a significant difference between the sample mean and hypothesis or assumed population mean. The test compares the sample mean to the hypothesis mean, while considering the variability in the data.
x̄1 = Sample mean
μ = Hypothesized population mean
s = Sample standard deviation
n = Sample size
In this tutorial, we will be using Carbon Dioxide Uptake in Grass Plants R dataset for t-test code examples. The dataset has 84 rows and 5 columns, and it was collected from an experiment to test the cold tolerance of the grass species Echinochloa crus-galli. We will be mostly considering uptake, Treatment, and Type columns for our tests.
() Was this AI assistant helpful? Yes No
In the example, we will be using the conc (carbon dioxide concentrations) column from the dataset.
We can observe the mean, distribution, and outliers using a boxplot.
($) Was this AI assistant helpful? Yes No
For a one-sample t-test, we will be using t.test(x,mu=0). Where x is the variable, mu is set by the null hypothesis. In our case, it is 550.
($, = 550) Was this AI assistant helpful? Yes No
Result:
The concentration of carbon dioxide is not equal to 550 and significantly lower than the hypothesized population mean.
```
One Sample t-test
data: CO2$conc
t = -3.5617, df = 83, p-value = 0.0006134
alternative hypothesis: true mean is not equal to 550
95 percent confidence interval:
370.7805 499.2195
sample estimates:
mean of x
435
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```
How to Perform Two Sample t-test in R
In the Two Sample t-tests, we will be comparing carbon dioxide uptake rates of two treatment types: non-chilled and chilled.
We can visualize the distribution of two groups using a boxplot.
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Welch Two Sample t-test
It is a statistical hypothesis that investigates if there is a significant difference between the mean of two independent groups that may have unequal variance. The test is comparing the means of two groups while considering the variability within each group.
x̄1 = Sample mean of first group
x̄2 = Sample mean of second group
n1 = Sample size of first group
n2 = Sample size of second group
s12 = Sample variance of first group
s22 = Sample variance of second group
By default, the t.test() function assumes that the variance of two groups is unequal (var.equal=FALSE). So, we don’t have to make any changes.
We are using the formula method to obtain t-test results, where uptake is a numerical vector and Treatment is a binary category column of the CO2 dataset.
(~, =) Was this AI assistant helpful? Yes No
Result:
There is a significant difference in the means of the two groups, and the nonchilled group has higher uptake than the chilled group.
```
Welch Two Sample t-test
data: uptake by Treatment
t = 3.0485, df = 80.945, p-value = 0.003107
alternative hypothesis: true difference in means between group nonchilled and group chilled is not equal to 0
95 percent confidence interval:
2.382366 11.336682
sample estimates:
mean in group nonchilled mean in group chilled
30.64286 23.78333
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```
Two Sample t-test with Equal Variance
The Two Sample t-test is a statistical hypothesis test to determine if there is a significant difference between the mean of two independent groups while assuming that the variance of the two groups is equal. The test compares the means of two groups while considering the variability within each group.
x̄1 = Sample mean of first group
x̄2 = Sample mean of second group
n1 = Sample size of first group
n2 = Sample size of second group
sp = Pooled standard deviation
To perform Two Sample t-tests with equal variance, we have to set var.equal TRUE and run the test again with the same formula and dataset.
(~, =, = TRUE) Was this AI assistant helpful? Yes No
Result:
As we can see, we got almost similar results that there is a significant mean difference between the two groups.
```
Two Sample t-test
data: uptake by Treatment
t = 3.0485, df = 82, p-value = 0.003096
alternative hypothesis: true difference in means between group nonchilled and group chilled is not equal to 0
95 percent confidence interval:
2.38324 11.33581
sample estimates:
mean in group nonchilled mean in group chilled
30.64286 23.78333
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```
How to Perform Paired t-test in R
Paired t-test is a statistical hypothesis that is used to determine if there is a significant difference between the means of two related or paired samples. It calculates the t-test value by comparing the differences between the paired observations while considering the variability within the difference.
dࠡ = differences of mean in paired observations
sd = differences of standard deviation of sample
n = number of pairs
To perform Paired t-test in R, we have to set paired argument TRUE and run the test again with the same formula and dataset.
(~, = TRUE, =) Was this AI assistant helpful? Yes No
Result:
There is a statistically significant difference between the means of the two groups while considering the t and p-value.
```
Paired t-test
data: uptake by Treatment
t = 7.939, df = 41, p-value = 8.051e-10
alternative hypothesis: true mean difference is not equal to 0
95 percent confidence interval:
5.114589 8.604458
sample estimates:
mean difference
6.859524
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```
In the second example, we will factor the rate of uptake for two types of the same plant. One originated from Quebec, and another is from Mississippi.
(~, =) Was this AI assistant helpful? Yes No
Let’s check the paired t-test results by replacing the Treatment with the type in the formula.
(~, = TRUE, =) Was this AI assistant helpful? Yes No
Result:
Again, there is a significant difference between the mean of the Quebec and Mississippi group.
```
Paired t-test
data: uptake by Type
t = 11.374, df = 41, p-value = 2.937e-14
alternative hypothesis: true mean difference is not equal to 0
95 percent confidence interval:
10.41177 14.90727
sample estimates:
mean difference
12.65952
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```
Try the t-test in R DataLab Workbook. It comes with code sources and results. You can also duplicate the workbook and start practicing on different examples.
Note: A strong statistics foundation will serve you well, no matter which industry you belong to. Statistics is the backbone of modern AI, and you should start your journey by taking Statistics Fundamentals with R skill track.
How to Interpret t-test Results in R
We are generating the results, but what do df, p-value, alternative hypothesis, or sample estimates mean? In this section, we will learn how to interpret the t-test results in R.
Let’s start by creating two groups using the rnorm function and run the Two Sample t-tests.
```
set.seed(125)
group1 <- c(rnorm(100, mean = 24, sd = 3))
group2 <- c(rnorm(100, mean = 43, sd = 2.4))
t.test(group1, group2)
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```
Output:
```
Welch Two Sample t-test
data: group1 and group2
t = -47.765, df = 179.99, p-value < 2.2e-16
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-19.51569 -17.96722
sample estimates:
mean of x mean of y
24.30063 43.04208
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```
data: the data used in Two Sample t-test (group1 and group2)
t: t test-statistic. The negative t-value of -47.765 indicates that the group1 sample mean is significantly smaller than group2.
df: it is the degree of freedom associated with the t-test value.
p-value: indicates the statistical significance of the result. The p-value is 2.2e-16 which is lower than alpha (0.005), indicating that the probability of obtaining such a large difference between the two groups by chance is very small.
alternative hypothesis: we can set the alternative hypothesis. In our case, it was set to check if the true difference in means is not equal to zero.
95 percent confidence interval: 95% confident that the true population means the difference between the two groups lies within the range of -19.51569, -17.96722.
sample estimates: it tells us the sample means of each group where group1 and group2 are 24.30063 and 43.04208, respectively. It means that, on average, group2 has a higher value than group1.
There are two hypotheses for the t-test:
H0: µ1 = µ2: the two population mean are equal.
HA: µ1 ≠µ2: the two population means are not equal.
In conclusion, the results of the Welch Two Sample t-test suggest that there is strong evidence that there is a statistically significant difference between group1 and group2.
Conclusion
In this tutorial, we have learned about One Sample, Two Samples, and Paired t-tests with R programming examples and how to interpret the result.
The t-test is one of many statistical tools used in hypothesis testing, and if you want to learn everything about hypothesis testing, take an interactive Hypothesis Testing in R course. The course covers t-tests, ANOVA, proportion tests, and chi-square tests.
You can also go beyond and enroll in our Statistician with R career track to master the essential skills and land a job as a statistician.
Topics
R
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See More |
11290 | https://pubmed.ncbi.nlm.nih.gov/36099518/ | Reproductive outcome in 326 women with unicornuate uterus - PubMed
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Ultrasound Obstet Gynecol
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. 2023 Jan;61(1):99-108.
doi: 10.1002/uog.26073.
Reproductive outcome in 326 women with unicornuate uterus
T Tellum12,B Bracco1,L V De Braud1,J Knez3,R Ashton-Barnett1,T Amin1,P Chaggar1,D Jurkovic1
Affiliations Expand
Affiliations
1 Institute for Women's Health, Faculty of Population Health Sciences, University College London, London, UK.
2 Department of Gynecology, Oslo University Hospital, Oslo, Norway.
3 Clinic for Gynecology, University Medical Centre Maribor, Maribor, Slovenia.
PMID: 36099518
PMCID: PMC10107309
DOI: 10.1002/uog.26073
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Reproductive outcome in 326 women with unicornuate uterus
T Tellum et al. Ultrasound Obstet Gynecol.2023 Jan.
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. 2023 Jan;61(1):99-108.
doi: 10.1002/uog.26073.
Authors
T Tellum12,B Bracco1,L V De Braud1,J Knez3,R Ashton-Barnett1,T Amin1,P Chaggar1,D Jurkovic1
Affiliations
1 Institute for Women's Health, Faculty of Population Health Sciences, University College London, London, UK.
2 Department of Gynecology, Oslo University Hospital, Oslo, Norway.
3 Clinic for Gynecology, University Medical Centre Maribor, Maribor, Slovenia.
PMID: 36099518
PMCID: PMC10107309
DOI: 10.1002/uog.26073
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Abstract
Objectives: To study the reproductive outcomes of women with a unicornuate uterus and compare them to those of women with no congenital uterine anomaly.
Methods: This was a single-center, retrospective cohort study. Cases were women aged at least 16 years who were diagnosed with a unicornuate uterus on transvaginal/transrectal ultrasound between January 2008 and September 2021. Controls were women with no congenital uterine anomaly matched 1:1 by age and body mass index. The primary outcome was live-birth rate. Secondary outcomes were pregnancy loss (miscarriage, ectopic pregnancy, termination of pregnancy), preterm delivery, mode of delivery and concomitant gynecological abnormalities (endometriosis, adenomyosis, fibroids).
Results: Included in the study were 326 cases and 326 controls. Women with a unicornuate uterus had a significantly lower live-birth rate (184/388 (47.4%) vs 229/396 (57.8%); P = 0.004) and higher rates of overall miscarriage (178/424 (42.0%) vs 155/465 (33.3%); adjusted odds ratio (aOR), 2.21 (95% CI, 1.42-3.42), P < 0.001), ectopic pregnancy (26/424 (6.1%) vs 11/465 (2.4%); aOR, 2.52 (95% CI, 1.22-5.22), P = 0.01), preterm delivery (45/184 (24.5%) vs 17/229 (7.4%); aOR, 3.04 (95% CI, 1.52-5.97), P = 0.001) and Cesarean delivery (116/184 (63.0%) vs 70/229 (30.6%); aOR, 2.54 (95% CI, 1.67-3.88), P < 0.001). Rudimentary-horn pregnancies accounted for 7/26 (26.9%) ectopic pregnancies in the study group. Women with a unicornuate uterus were more likely to have endometriosis (17.5% vs 10.7%; P = 0.018) and adenomyosis (26.7% vs 15.6%; P = 0.001), but were not more likely to have fibroids compared with controls. Women with a functional rudimentary horn were more likely to have pelvic endometriosis compared to those without (odds ratio, 2.4 (95% CI, 1.4-4.1), P = 0.002).
Conclusions: Pregnant women with a unicornuate uterus should be classified as high risk. Removal of a functional rudimentary horn should be discussed with the patient to prevent a rudimentary-horn ectopic pregnancy. © 2022 The Authors. Ultrasound in Obstetrics & Gynecology published by John Wiley & Sons Ltd on behalf of International Society of Ultrasound in Obstetrics and Gynecology.
Keywords: adenomyosis; ectopic pregnancy; endometriosis; miscarriage; uterine anomaly.
© 2022 The Authors. Ultrasound in Obstetrics & Gynecology published by John Wiley & Sons Ltd on behalf of International Society of Ultrasound in Obstetrics and Gynecology.
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Figures
Figure 1
Three‐dimensional transvaginal ultrasound rendering of…
Figure 1
Three‐dimensional transvaginal ultrasound rendering of unicornuate uterus in coronal plane in four patients,…
Figure 1
Three‐dimensional transvaginal ultrasound rendering of unicornuate uterus in coronal plane in four patients, showing hemiuterus (U) and rudimentary horn (). (a) Right hemiuterus and non‐functional left rudimentary horn. Cavity of hemiuterus is narrow in fundal area and only a single interstitial portion of Fallopian tube is visible. (b) Right hemiuterus and non‐communicating left rudimentary horn. Cavity of horn contains functional endometrium with visible adenomyosis (A), represented by anechoic myometrial cyst with thin hyperechogenic rim protruding into cavity. (c) Right hemiuterus with hyperechogenic endometrium and functional, non‐communicating left rudimentary horn, with blood distending the cavity. (d) Right hemiuterus and duplex ectopic pregnancy in non‐communicating left rudimentary horn, where two gestational sacs (P) are visualized. (e) Right hemiuterus in same woman as in (d), 3 months after uneventful excision of rudimentary horn containing ectopic pregnancy.
Figure 2
Schematic diagram depicting subtypes of…
Figure 2
Schematic diagram depicting subtypes of unicornuate uterus: (a) hemiuterus without rudimentary horn; (b)…
Figure 2
Schematic diagram depicting subtypes of unicornuate uterus: (a) hemiuterus without rudimentary horn; (b) hemiuterus with non‐functional rudimentary horn; (c) hemiuterus with functional, non‐communicating rudimentary horn; and (d) hemiuterus with functional, communicating rudimentary horn.
Figure 3
Flowchart summarizing inclusion of patients…
Figure 3
Flowchart summarizing inclusion of patients in study. Representing clinical consultations, not individuals. BMI,…
Figure 3
Flowchart summarizing inclusion of patients in study. Representing clinical consultations, not individuals. BMI, body mass index.
See this image and copyright information in PMC
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References
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11291 | https://fiveable.me/key-terms/native-american-history/turners-frontier-thesis | Turner's Frontier Thesis - (Native American History) - Vocab, Definition, Explanations | Fiveable | Fiveable
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key term - Turner's Frontier Thesis
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Definition
Turner's Frontier Thesis is the argument presented by historian Frederick Jackson Turner in 1893, claiming that the American frontier shaped the nation's character and institutions. He believed that the experience of westward expansion fostered individualism, democracy, and economic innovation, and that this unique development was fundamental to understanding American identity. The thesis suggested that the closing of the frontier in the 1890s marked a significant shift in American society and its future.
5 Must Know Facts For Your Next Test
Turner's thesis emphasized how the frontier experience helped create a distinct American identity characterized by rugged individualism and self-reliance.
The thesis was presented at a meeting of the American Historical Association in Chicago during the World's Columbian Exposition, gaining significant attention.
Turner argued that as the frontier closed, Americans would need to find new ways to express their national character, which he feared could lead to stagnation.
His ideas influenced later interpretations of American history and prompted debates about the role of the frontier in shaping national policies and attitudes.
Critics of Turner's thesis pointed out its oversimplifications and failures to adequately address issues such as race, class, and the impact on Native Americans.
Review Questions
How did Turner's Frontier Thesis influence perceptions of American identity in the late 19th century?
Turner's Frontier Thesis significantly influenced perceptions of American identity by promoting the idea that westward expansion was crucial in shaping unique national characteristics such as democracy and individualism. His argument suggested that the challenges and opportunities presented by the frontier created a distinct ethos among Americans. This narrative reinforced a sense of national pride and purpose during a time of transformation as America moved into a new industrial age.
Evaluate the strengths and weaknesses of Turner's Frontier Thesis in explaining the complexities of American expansion.
The strengths of Turner's Frontier Thesis lie in its emphasis on how westward expansion contributed to American character traits like self-reliance and innovation. However, its weaknesses include an oversimplified view of this complex process, as it often neglected the experiences of diverse groups affected by expansion, including Native Americans and immigrants. Additionally, it didn't fully address the environmental impacts or regional differences within America, leading to a more one-dimensional narrative.
Assess how Turner's Frontier Thesis relates to contemporary discussions about American exceptionalism and identity.
Turner's Frontier Thesis continues to resonate in contemporary discussions about American exceptionalism and identity by reinforcing the belief that America's uniqueness stems from its pioneering spirit and democratic ideals. As debates around immigration, diversity, and globalization evolve, Turner's ideas provide a lens through which to evaluate how these themes intersect with America's historical narrative. This connection prompts critical reflections on who gets included in the American story and what values truly define its essence amidst changing societal dynamics.
Related terms
Manifest Destiny:
The 19th-century doctrine that justified American expansion across the continent, based on the belief that it was the nation's divine right to spread democracy and civilization.
Homestead Act:
A law passed in 1862 that granted 160 acres of public land to settlers for a small fee, promoting westward expansion and settlement in the United States.
Frontier Myth:The romanticized narrative surrounding the American frontier, portraying it as a place of opportunity, adventure, and personal freedom, often overlooking the realities faced by settlers and Native Americans.
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0 |
11292 | https://ford126.web.illinois.edu/Anatomy_lectnotes.pdf | Anatomy of Integers and Random Permutations Course Lecture Notes Kevin Ford Department of Mathematics, 1409 West Green Street, University of Illinois at Urbana-Champaign, Urbana, IL 61801, USA Email address: ford@math.uiuc.edu Contents Chapter 0.
Background material and notation 1 1.
Notation 1 Number Theory Functions 1 Permutation notation 1 Order of magnitude notation (Bachman-Landau, Hardy, Vinogradov) 1 Probability 1 General notational conventions 2 2.
Basic summation estimates 2 3.
Arithmetic functions 2 4.
Prime number estimates 3 5.
Inclusion-Exclusion 4 6.
Probability estimates: general 5 7.
Probability estimates: Poisson random variables 5 Chapter 1.
The sequence of prime factors of an integer and cycles of a permutation, I 8 1.
Prime factors and divisors 8 2.
Permutations, cycles and fixed sets 9 3.
Cycles and prime factors from intervals: first nibbles 10 4.
Cycles and prime factors from sets: general upper bounds 13 5.
The sequence of cycles and prime factors from intervals 19 6.
Lower bounds on Px{ω(n) = k} 20 7.
Prime factors counted with multiplicity 21 8.
Application: Erdős’ multiplication table problem 23 9.
Number of divisors of integers 24 10.
The range of Euler’s function 25 11.
Exercises 27 Chapter 2.
Distribution of the largest cycle and largest prime factor 28 1.
Upper bounds 28 2.
Application: large gaps between primes 31 3.
Asymptotic formulas when u is small 32 4.
Exercises 36 Chapter 3.
Integers without small prime factors and permutations without small cycles 37 1.
Permutations without small cycles 37 2.
Integers without small prime factors 41 3.
Exercises 45 Chapter 4.
Poisson approximation of small cycle lengths and small prime divisors 46 1.
Small cycles of permutations 46 2.
The Kubilius model of small prime factors of integers 47 iii iv CONTENTS 3.
Exercises 51 Chapter 5.
Central Limit Theorems 53 1.
Gaussian approximation of Poisson variables 53 2.
Central Limit Theorems for cycles 54 3.
Central Limit theorems for prime factors 56 4.
Exercises 58 Chapter 6.
The concentration of divisors of integers and permutations 59 1.
Concentration of divisors 59 2.
A random model of prime factors and cycle lengths 60 3.
Proof of Theorems 6.6, 6.7, and 6.8 62 4.
Proof of Theorem 6.9 64 5.
Exercises 68 Chapter 7.
Integers with a divisor in a given interval 69 1.
Exact formulas 69 2.
Easy bounds when z is small or large 69 3.
The critical case z = 2y 71 4.
Some applications of Theorem 7.3 71 5.
A heuristic for H(x, y, 2y) 73 6.
A global-to-local principle 74 7.
Completion of the lower bound in Theorem 7.3 78 8.
Bounding H(x, y, z) above in terms of uniform order statistics 82 9.
Upper bound, part II 85 10.
Counting integers with a given number of divisors in an interval 90 11.
Exercises 90 Chapter 8.
Permutations with a fixed set of a given size 91 1.
Introduction and notation 91 2.
The global-to-local principle 92 3.
The lower bound in Proposition 8.4 96 4.
The upper bound in Proposition 8.4 98 5.
Exercises 99 Chapter 9.
Sets of permutations with equal sized divisors 100 1.
Equal sized divisors of several permutations 100 2.
Application: Invariable generation of Sn 102 3.
Application: Irreducibility of polynomials over Q 104 4.
Exercises 105 Bibliography 106 CHAPTER 0 Background material and notation 1. Notation Number Theory Functions • τ(n) is the number of positive divisors of n • ω(n) is the number of distinct prime factors of n • ω(n, t) is the number of distinct prime factors of n which are ⩽t • ω(n; S) is the number of distinct prime factors of n that lie in the set S • Ω(n) is the number of prime factors of n counted with multiplicity • µ(n) is the Möbius’s function; µ(n) = (−1)ω(n) if n is squarefree and µ(n) = 0 otherwise.
• P +(n) is the largest prime factor of n; P +(1) = 0 by convention • P −(n) is the smallest prime factor of n; P −(1) = ∞by convention Permutation notation • Sn is the permutation group on a set of n objects (we don’t care what the objects are) • Cj(σ) is the number of cycles of length j in the permutation σ • C(σ) is the total number of cycles in the permutation σ • CI(σ) is the number of cycles of length j ∈I in the permutation σ • β|σ means that β is a divisor of the permutation σ, i.e. a product of some subset of the cycles of σ • A fixed set I of σ is a subset of [n] which is itself permuted by σ. Equivalently, I is the set of indices permuted by a divisor of σ.
• |β| is the size of β; β is a divisor of a permutation.
Order of magnitude notation (Bachman-Landau, Hardy, Vinogradov) • The notations f = O(g), f ≪g and g ≫f mean that there is a positive constant C so that |f| ⩽Cg throughout the domain of f. The constant C is independent of any parameters, unless specified by subscripts, e.g. f(x) = Oε(xε).
• f ≍g means that both f ≪g and g ≪f hold. Generally makes sense only if f, g are both positive.
Equivalently, there are positive constants C < C′ such that Cg ⩽f ⩽C′g throughout the domain of f.
• f ∼g as x →a means limx→a f(x)/g(x) = 1. Here a can be finite, ∞or −∞.
• f = o(g) as x →a means limx→a f(x)/g(x) = 0. Here a can be finite, ∞or −∞.
Probability • P(X) is the probability of the event X • E (X) is the expectation of the event X • X d = Y means that X has the same distribution as Y • Pois(λ) is a Poisson random variable with parameter λ 1 2 0. BACKGROUND MATERIAL AND NOTATION General notational conventions • N = {1, 2, 3, . . .}, the set of positive integers (“natural numbers”) • N0 = {0, 1, 2, 3, . . .} = N ∪{0} • e = 2.71828182 . . . is the base of the natural logarithm • γ = 0.57721566 . . . is Euler’s constant • log is the natural logarithm • logk x is the k−th iterate of the natural logarithm of x • ⌊x⌋is the greatest integer which is ⩽x.
• ⌈x⌉is the least integer which is ⩾x.
• 1(S) is the indicator function of statement S; 1(S) = 1 if S is true, and 1(S) = 0 if S is false.
• Hn = 1 + 1/2 + · · · + 1/n is the n-th harmonic sum • H(I) = P i∈I 1/i is a general harmonic sum, where I ⊂N • Variables in boldface type, e.g. h, usually denote vector quantities.
• A statement for “almost all integers” means that the number of exceptions below x is o(x) as x →∞.
• The symbols p, q denote primes unless otherwise noted • The symbols k, l, m, n denote integers unless otherwise noted 2. Basic summation estimates Harmonic sums. The harmonic sums Hn satisfy (i) log n ⩽Hn ⩽1 + log n; (ii) Hn = log n + γ + O(1/n), where γ = 0.57721566 . . . is Euler’s constant.
Stirling’s formula. We have the asymptotic (Stirling’s formula) n! = √ 2πn(n/e)n(1 + O(1/n)) and the strict inequalities (0.1) 1 ⩽ n!
√ 2πn(n/e)n ⩽e1/(12n) ⩽1 + 1 10n (n ∈N).
Euler’s summation. Let f ∈C1(y, x). Then X y<n⩽x f(n) = Z x y f(t) dt + Z x y {t}f ′(t) dt + {y}f(y) −{x}f(x), where {t} = t −⌊t⌋is the fractional part of t.
Roughly speaking, the sum of f(n) is approximated by the corresponding integral of f(t).
Abel summation, also called partial summation. Let an be any sequence of complex numbers with counting function A(t) = P 1⩽n⩽t an. Let 0 < y ⩽x and suppose f ∈C1(y, x]. Then X y 0.
Oftentimes the more crude bounds of Chebyshev suffice: for positive constants c, c′ we have cx log x ⩽#{p ⩽x} ⩽c′x log x (x ⩾2).
The Prime Number Theorem also implies that pn ∼n log n (n →∞), where pn is the n-th smallest prime.
There are similar bounds for primes in a fixed arithmetic progression. Here we fix 1 ⩽a ⩽b with (a, b) = 1.
Prime number theorem for arithmetic progressions. We have #{p ⩽x : p ≡a (mod b)} ∼ x φ(b) log x (x →∞) There are results which are uniform in b, but they are more complicated to state. We refer the reader to for specifics.
Mertens’ estimates for primes in arithmetic progressions. For some constants c1(a, b) and c2(a, b) > 0 we have (0.8) X p⩽x p≡a (mod b) 1 p = log log x φ(b) + c1(a, b) + Ob(1/ log x) and (0.9) Y p⩽x p≡a (mod b) 1 −1 p = c2(a, b) (log x)1/φ(b) (1 + Ob (1/ log x)) .
5. Inclusion-Exclusion We need a simple version of the inclusion-exclusion principle, with truncation.
Inclusion-exclusion. Let a be a non-negative integer. Then, for any k ∈N, 1(a = 0) = ∞ X r=0 (−1)r a r = k X r=0 (−1)r a r + (−1)k+1 a −1 k .
Proof. The first equality is trivial from the binomial theorem. For the second, we have ∞ X r=k+1 (−1)r a r = ∞ X r=k+1 (−1)r a −1 r −1 + a −1 r = (−1)k+1 a −1 k .
□ Often, we need to count a reciprocal weighted sum over integers with a given number of prime factors from a given set.
7. Probability estimates: Poisson random variables 5 Proposition 0.1. Let T be a finite set of positive real numbers, and k ∈N. Then 1 k! H(T)k − k 2 H(T)k−2 X n∈T 1 n2 !
⩽ X n1,...,nk∈T n1<···<nk 1 n1 · · · nk ⩽H(T)k k!
.
Proof. Evidently, H(T)k = X n1,...,nk∈T 1 n1 · · · nk .
The summands on the right corresponding to distinct, unordered k-tuples (n1, . . . , nk) equals k!
X n1,...,nk∈T n1<··· 0, w > 0.
Two easy consequences are Chebyshev’s inequality for variances and Chernoff’s inequality.
Chebyshev’s inequality. If w > 0, µ = E X and E |X −µ|2 > 0, then P |X −µ| ⩾w p E |X −µ|2 ⩽1 w2 .
Chernoff’s inequality. We have P (X ⩾w) ⩽inf b⩾0 E ebX ebw .
and P (X ⩽w) ⩽inf b⩽0 E ebX ebw .
7. Probability estimates: Poisson random variables The first Proposition lists basic properties of the Poisson distribution, which are readily verified from the definition.
Proposition 0.2. Suppose X d = Pois(λ). Then E X = λ, (0.10) E cX = e(c−1)λ (c > 0), (0.11) E X m = λm m!
(m ⩾0).
(0.12) If Xj d = Pois(λj), 1 ⩽j ⩽k, and X1, . . . , Xk are independent, then (0.13) X1 + · · · + Xk d = Pois(λ1 + · · · + λk).
We also record very useful tail bounds on the Poisson distribution, due to Norton .
6 0. BACKGROUND MATERIAL AND NOTATION Proposition 0.3 (Poisson tails). Let X d = Pois(λ) where λ > 0. Then P(X ⩽αλ) ⩽min 1, 1 (1 −α) √ αλ e−Q(α)λ (0 ⩽α ⩽1), P(X ⩾αλ) ⩽min 1, 1 (1 −1/α) √ 2πλα !
e−Q(α)λ (α ⩾1), where (0.14) Q(x) = x log x −x + 1.
Proof. First, using Chernoff’s inequality (with b = log α, 0 < α ⩽1) together with (0.11), we have P(X ⩽αλ) = P ebX ⩾ebαλ ⩽E ebX ebαλ = e(eb−1)λ ebαλ = e−Q(α)λ.
When α is bounded away from 1 we can do better. Suppose α < 1 and let k0 = ⌊αλ⌋. Then P(X ⩽αλ) = e−λ X k⩽k0 λk k! .
If k0 = 0 then Q(α) ⩽Q(0) = 1 and αλ < 1, hence 1 (1 −α) √ αλ e−Q(α)λ ⩾e−λ = P(X ⩽αλ).
Now suppose k0 ⩾1. Consecutive summands have ratio ⩾1/α, and thus by (0.1) and the fact that λx/(x/e)x is increasing for x ⩽λ, we have P(X ⩽αλ) ⩽e−λ 1 −α · λk0 k0! ⩽e−λ 1 −α · λk0 (k0/e)k0√2πk0 ⩽e−λ 1 −α · λαλ (αλ/e)αλ√2πk0 = e−Q(α)λ 1 −α · 1 √2πk0 ⩽e−Q(α)λ 1 −α · 1 √ παλ , which concludes the proof of the first inequality.
A second application of Chernoff’s inequality (again with b = log α, α ⩾1) and (0.11) yields P(X ⩾αλ) = P ebX ⩾ebαλ ⩽E ebX ebαλ = e(eb−1)λ ebαλ = e−Q(α)λ.
Now assume α > 1 and write k1 = ⌈αλ⌉. We have P(X ⩾αλ) = e−λ X k⩾k1 λk k! .
Consecutive summands have ratio ⩽1/α. By (0.1) and the fact that λx/(x/e)x is decreasing for x ⩾λ, P(X ⩾αλ) ⩽ 1 1 −1/α · λk1 k1! ⩽ 1 1 −1/α · λk1 (k1/e)k1√2πk1 ⩽e−Q(α)λ 1 −1/α · 1 √2πk1 ⩽e−Q(α)λ 1 −1/α · 1 √ 2παλ .
This concludes the proof of the 2nd inequality.
□ Proposition 0.4 (Q bounds). We have (0.15) Q(x) = Z x 1 log t dt (all x) = ∞ X k=2 (−1)k k(k −1)(x −1)k (|x −1| < 1), 7. Probability estimates: Poisson random variables 7 and (0.16) x2 3 ⩽Q(1 + x) ⩽x2 (|x| ⩽1).
Proposition 0.5 (Binomial tails). Let X have binomial distribution according to n trials and parameter p ∈[0, 1]; that is, P(X = k) = n k pk(1 −p)n−k. If β ⩽p then we have (0.17) P(X ⩽βn) ⩽exp −n β log β p + (1 −β) log 1 −β 1 −p .
Proof. For a proof, see [3, Lemma 4.7.2]. The right side is also an upper bound for P(X ⩾βn) for β ⩾p by symmetry.
□ CHAPTER 1 The sequence of prime factors of an integer and cycles of a permutation, I Positive integers factor uniquely into a product of prime numbers, and permutations factor uniquely into a product of cycles. Despite this similarity, the two objects, integers and permutations, look very different on the surface. Deeper inspection, however, reveals that the distribution of the two factorizations have many common features, and for much the same underlying reasons.
1. Prime factors and divisors Given a random number n ⩽x and a prime p, the probability that p|n is very close to 1/p, and moreover these events are close to independent for different p (as long as p is “small”, in a sense that will be made precise later). In particular, this heuristic suggests that the number of distinct prime factors ω(n) should be about X p⩽x 1 p = log2 x + O(1) on average, using Mertens’ sum estimate (0.5). In fact, an easy calculation gives (1.1) 1 x X n⩽x ω(n) = 1 x X p⩽x ⌊x/p⌋= X p⩽x 1/p + O(π(x)/x) = log2 x + O(1).
What is the distribution of ω(n) for n ⩽x? One can model the event p|n, for a random n ⩽x, by a Bernouilli random variable Xp, which equals 1 with probability 1/p and 0 with probability 1 −1/p. This in turn is very close to a Poisson random variable Zp with parameter 1/p when p is large. Thus, ω(n) can be modeled by the random variable P p⩽x Zp, which is a Poisson variable with parameter P p⩽x 1/p = log2 x + O(1).
The first result in this direction is a classic theorem of Landau from 1900.
Theorem (Landau, 1900). For every fixed k, #{n ⩽x : ω(n) = k} ∼ x log x (log2 x)k−1 (k −1)!
.
This already suggests that ω(n) has an approximate Poisson distribution, although Landau never wrote this explicitly. It was Hardy and Ramanujan in 1917 who analyzed the behavior of #{n ⩽x : ω(n) = k} uniformly in k, showing Theorem (Hardy-Ramanujan, 1917). Uniformly for x ⩾2 and k ⩾1, #{n ⩽x : ω(n) = k} ⩽C1 x log x (log2 x + C2)k−1 (k −1)!
, where C1, C2 are certain absolute constants.
Summing the upper bound for k ⩾(1 + ε) log2 x and k ⩽(1 −ε) log2 x, with ε > 0 fixed, and using standard bounds on the tail of the Poisson distribution (see (0.3) below), one obtains a sum of o(x). Conse-quently, most n ⩽x have close to log2 x distinct prime factors. This result is sometimes referred to as the birth of probabilistic number theory.
Motivated by the fact that the Poisson distribution Pois(λ) tends to the Gaussian with mean and variance λ as λ tends to infinity, Erdős and Kac proved their celebrated “Central Limit Theorem” for ω(n) in 1939: 8 2. Permutations, cycles and fixed sets 9 Theorem (Erdős-Kac, 1939 ). For any real z, 1 x# ( n ⩽x : ω(n) −log2 x p log2 x ⩽z ) →Φ(z) := 1 √ 2π Z z −∞ e−1 2 t2 dt (x →∞).
Much further work was done starting in the 1940s, examining the distribution of the entire sequence of prime factors of integers (equivalently, studying the distribution of arbitrary additive functions). Perhaps the most notable was the work of Kubilius, who developed a probabilistic model of integers which provides a kind of meta-tool for studying al kinds of statistical questions about the distribution of prime factors. A key concept in the theory is independence, the idea that if p and q are small primes, then the “events” p|n and q|n are nearly independent, from a probabilistic viewpoint; this idea also played a prominent role in the development of sieve methods. The theory leads to a “Poisson model” of prime factors; namely that the number of prime factors in an interval (eea, eeb] has roughly Pois(b −a) distribution, with disjoint intervals having independent distributions.
The distribution of divisors of integers has also received much attention, beginning in the 1930s. Much of the study was motivated by two fundamental problems: (a) (Besicovitch, 1934). Given a quantity y, what is the density of integers that have a divisor in (y, 2y]?
(b) (Erdős, 1948). Do almost all integers (that is, a set of density 1) have two divisors in some dyadic interval (z, 2z]?
Estimates for the density in Problem (a) were given by Erdős and Tenenbaum, with Ford giving the order of magnitude of the order in 2008 . The solution of Problem (b), in the positive, was given by Maier and Tenenbaum in 1984 . Problem (a) is closely related to the Erdős multiplication table problem: How many distinct products are there ab with 1 ⩽a ⩽N and 1 ⩽b ⩽N ?
2. Permutations, cycles and fixed sets The classical derangement problem was posed in 1708 by Pierre Raymond de Montmort. The problem asks how many permutations in Sn have no fixed points, that is no 1-cycles. Five years later, he found an exact formula, which is approximately 1 en! for large n. In the early 1800s, Cauchy introduced the cycle notation and showed that permutations factor uniquely into a product of cycles. He also developed an exact formula for the number of permutations with a given cycle type; that is, the number of cycles of each length.
If σ ∈Sn has Cj cycles of length j for each j, with P j Cj = n, then the number of such permutations equals n!
Y j⩽n 1 j Cj 1 Cj!.
This formula suggests that, for random σ ∈Sn, the quantities C1(σ), C2(σ), . . . behave like independent Poisson random variables, where Cj(σ) has distribution Pois(1/j). This is not precisely true, because of the condition that P j jCj = n. Goncharov was the first to make such statements rigorous, and in 1944 proved (among other things) the following: Theorem (Goncharov, 1944 ). We have • For any fixed j and m, 1 n!# {σ ∈Sn : Cj(σ) = m}} →e−1/j (1/j)m m!
(n →∞); • For any real z, 1 n!# σ ∈Sn : C(σ) −log n √log n ⩽z →Φ(z) := 1 √ 2π Z z −∞ e−1 2 t2 dt (n →∞).
The (unsigned) Stirling number of the first kind, S1(n, m), counts the number of permutations σ ∈Sn with exactly m total cycles; that is, C(σ) = m. Goncharov used careful asymptotic analysis of Stirling numbers to obtain the second part of the theorem above. The first part was deduced from an exact formula 10 1. THE SEQUENCE OF PRIME FACTORS OF AN INTEGER AND CYCLES OF A PERMUTATION, I which he proved for #{σ ∈Sn : Cj(σ) = m} (see Exercise 1.1 below). More recently, the Poisson model has been established in great uniformity: Namely, (1.2) (C1(σ), C2(σ), . . . , Ck(σ)) ≈(Z1, Z2, . . . , Zk) (meaning the two vectors have distributions which are very close), where Z1, . . . , Zk are independent and Zj d = Pois(1/j), provided that k = o(n) as n →∞.
A fixed set of a permutation σ is a subset I ⊂[n] which is itself permuted (i.e., left invariant) by σ. A fixed set corresponds to a divisor of σ, that is, a product of some subset of the cycles in σ (we include both the empty set and the whole set [n] as fixed sets). For example, if σ ∈S6 has cycle form σ = (1)(24)(356) then, e.g., (1)(356) is a divisor with corresponding fixed set {1, 3, 5, 6}. There play the same role for permu-tations as divisors do for integers. The existence of fixed sets of a particular size has applications to various questions in statistical group theory, such as generation of Sn by random permutations, the distribution of transitive subgroups of Sn, and the order of permutations. These in turn have further applications to irreducibility of polynomials over finite fields (with applications to Galois theory) and over global fields.
Since X ek n, then the left side is zero.
Proof. The second assertion is obvious, since the product on the left side is positive if and only if Cj(σ) ⩾mj for all j, and sinceP j jCj(σ) = n this implies that P jmj ⩽n. Now assume that m1 + 2m2 + · · · + nmn ⩽n. The product on the left side in the lemma equals the number of ways to choose from [n] a disjoint collection of m1 1-cycles, m2 2-cycles, . . ., mn n-cycles. The number of ways of choosing from [n] a disjoint collection of m1 1−element sets, m2 2−element sets, . . ., mn n−element sets is equal to n 1 · · · 1 | {z } m1 2 · · · 2 | {z } m2 · · · n · · · n | {z } mn t 1 m1! · · · mn! = n!/t!
Qn j=1(j!)mjmj!, 3. Cycles and prime factors from intervals: first nibbles 11 where t = n −(m1 + 2m2 + · · · + nmn). The elements of a k-element set may be arranged into a cycle in (k −1)! ways. Thus, the number of ways to arrange the elements of these sets into cycles is n Y j=1 (j −1)!mj.
Finally, the t elements not used in any of these cycles may be permuted in t! ways.
Multiplying these quantities together completes the proof.
□ We remark that the RHS in Lemma 1.1 equals n Y j=1 (1/j)mj mj!
= E n Y j=1 Zj mj , where Zj d = Pois(1/j) and Z1, . . . , Zn are independent. This already suggests that the quantities Cj(σ) behave like independent Poisson random variables.
A special case is the well-known formula of Cauchy for the number of permutations with a given cycle type.
Lemma 1.2 (Cauchy’s formula). If m1 + 2m2 + · · · + nmn = n, then Pσ Cj(σ) = mj (1 ⩽j ⩽n) = n Y j=1 (1/j)mj mj!
.
Proof. Apply Lemma 1.1, noting that Cj(σ) mj ̸= 0 for all j if and only if Cj(σ) = mj for every j. This implies that Pσ Cj(σ) = mj (1 ⩽j ⩽n) = E σ n Y j=1 Cj(σ) mj .
□ Corollary 1.3 (Derangements). We have the exact formula for derangements Pσ(C1(σ) = 0) = n X j=0 (−1)j j!
.
Proof. Apply Inclusion-Exclusion with u = C1(σ), followed by the Lemma 1.1. We get #{σ ∈Sn : C1(σ) = 0} = X σ∈Sn ∞ X j=0 (−1)j C1(σ) j = n!
n X j=0 (−1)jE σ C1(σ) j = n!
n X j=0 (−1)j j!
.
□ Recall that for a set T os positive integers, CT (σ) = X j∈T Cj(σ) is the number of cycles in σ with length in T.
Corollary 1.4. For any nonempty set T ⊆[n], E σCT (σ) = H(T). In particular, E σC(σ) = Hn = log n + γ + O(1/n).
Proof. By Lemma 1.1, E σCj(σ) = 1/j for every j, and the result follows by linearity of expectation and bounds on harmonic sums.
□ 12 1. THE SEQUENCE OF PRIME FACTORS OF AN INTEGER AND CYCLES OF A PERMUTATION, I Corollary 1.5. For any nonempty set T ⊆[n], E σ(CT (σ) −H(T))2 ⩽H(T) with equality in the case max T ⩽n/2.
Proof. We have CT (σ)2 = X i,j∈T i̸=j Ci(σ)Cj(σ) + 2 X i∈T Ci(σ) 2 + X i∈T Ci(σ).
In the double sum over i, j, the summands with i + j > n are zero. Applying Lemma 1.1, we get E σCT (σ)2 ⩽ X i,j∈T i̸=j 1 ij + X i∈T 1 i2 + 1 i = H(T)2 + H(T), with equality if max T ⩽n/2. Using Corollary 1.4, we conclude that E σ(CT (σ) −H(T))2 = E σCT (σ)2 −2H(T)E σCT (σ) + H(T)2 ⩽H(T).
□ Taking T = [n], so that H(T) = Hn = log n + O(1), and applying Chebyshev’s inequality we find that Pσ |C(σ) −Hn| ⩾ξ p log n ⩽E σ|C(σ) −Hn|2 ξ2 log n ≪1 ξ2 (ξ ⩾1).
That is, C(σ) is very close to Hn for most σ.
The situation with primes is more complicated (see the lower bound in Proposition 0.1), but we so have a clean upper bound of the same type. In what follows, we denote Px and E x the probability and expectation with respect to a random integer in [1, x]. Here x may be any real number.
Lemma 1.6. Let T1, . . . , Tk be nonempty, disjoint subsets of the primes in [2, x], and let m1, . . . , mk ⩾0.
Then E x k Y j=1 ω(n; Tj) mj ⩽ k Y j=1 H(Tj)mj mj!
.
Proof. We first write X n⩽x ω(n; Tj) mj = X pj,1,...,pj,mj ∈Tj pj,1<···<pj,mj (1⩽j⩽k) #{n ⩽x : p1,1 · · · pk,mk|n} = X pj,1,...,pj,mj ∈Tj pj,1<···<pj,mj (1⩽j⩽k) x p1,1 · · · pk,mk .
Using ⌊y/n⌋⩽⌊y⌋/n for real y and n ∈N, Proposition 0.1 then gives E x k Y j=1 ω(n; Tj) mj ⩽ k Y j=1 X pj,1,...,pj,mj ∈Tj pj,1<···<pj,mj 1 pj,1 · · · pj,mj ⩽ k Y j=1 H(Tj)mj mj!
.
□ A corollary is a theorem of Turán from 1934 : Corollary 1.7 (Turán’s variance theorem). We have E x ω(n) −log2 x 2 ≪log2 x.
4. Cycles and prime factors from sets: general upper bounds 13 Proof. We directly compute E x ω(n) −log2 x 2 = E x 2 ω(n) 2 + ω(n)(1 −2 log2 x) + (log2 x)2 .
Let T = T1 be the set of all primes in [2, x], so that H(T) = log2 x + O(1) by Mertens’ theorem (0.5). Using Lemma 1.6 with k = 1 and m1 = 2, together with (1.1), we obtain E x ω(n) −log2 x 2 ⩽H(T)2 + (1 −2 log2 x)E x ω(n) + (log2 x)2 = O(log2 x).
□ We can interpret Corollary 1.7 probabilistically. If n ⩽x is chosen at random, then Corollary 1.7 gives an upper bound on the variance of ω(n), telling us that ω(n) is concentrated near log2 x. In fact, it follows immediately from Chebyshev’s inequality that uniformly for ξ ⩾1, Px |ω(n) −log2 x| ⩾ξ p log2 x ≪1 ξ2 .
An immediate corollary is the following famous result of Hardy and Ramanujan from 1917. We note that for √x < n ⩽x, log2 x = log2 n + O(1). That is, we can say that for most n, ω(n) is close to log2 n.
Theorem 1.8 (Hardy-Ramanujan). The function ω(n) has normal order log log n, meaning that for any ε > 0 we have (1 −ε) log log n ⩽ω(n) ⩽(1 + ε) log log n for almost all integers n.
4. Cycles and prime factors from sets: general upper bounds Theorem 1.9 (Cycles in sets theorem). Let T1, . . . , Tr be arbitray disjoint, nonempty subsets of [n] and k1, . . . , kr ⩾0. Then Pσ (CT1(σ) = k1, . . . , CTr(σ) = kr) ⩽e r Y j=1 H(Tj)kj kj!
e−H(Tj) !
· 1 + k1 H(T1) + · · · + kr H(Tr) .
Proof. Evidently n#{σ ∈Sn : CT1(σ) = k1, . . . , CTr(σ) = kr} = X σ∈Sn CTj (σ)=kj (1⩽j⩽r) X α|σ α a cycle |α|.
Write σ = αβ and let h = |α|. Either h ∈Tj for some unique j, or h ̸∈T1 ∪· · · ∪Tr. Thus, for some t, 0 ⩽t ⩽r, we have (CT1(β), . . . , CTr(β)) = (mt,1, . . . , mt,r), where (1.3) mt,i = ki −1(i = t ⩾1).
It is permissible to think of β ∈Sn−h and thus n#{σ ∈Sn : CT1(σ) = k1, . . . , CTr(σ) = kr} = r X t=0 n X h=1 X α∈Sn,|α|=h α a cycle h X β∈Sn−h CTi(β)=mt,i (1⩽i⩽r) 1 = r X t=0 n X h=1 n!
(n −h)!
X β∈Sn−h CTi(β)=mt,i (1⩽i⩽r) 1.
14 1. THE SEQUENCE OF PRIME FACTORS OF AN INTEGER AND CYCLES OF A PERMUTATION, I Note that if ki = 0 for some i, then mi,i = −1 and the corresponding summand above is ommitted. Now subdivide the sum according the cycle type (b1, . . . , bn) of the permutation β, using Cauchy’s formula (Lemma 1.2) to count such permutations for each type. It follows that n#{σ ∈Sn : CTj(σ) = kj (1 ⩽j ⩽r)} = n!
r X t=0 n X h=1 X b1,dots,bn⩾0 b1+2b2+···+nbn=n−h P i∈Tj bi=mt,j (1⩽j⩽r) 1 Q i bi!ibi ⩽n!
r X t=0 kt̸=0 X b1,...,bn⩾0 P i∈Tj bi=mt,j (1⩽j⩽r) 1 Q i bi!ibi := Y, say. Let T = T1 ∪· · · ∪Tr and T0 = [n] \ T, and separately consider the summation over i ∈T and i ∈T0.
By the multinomial theorem, Y = n!
r X t=0 kt̸=0 X bi⩾0 (i∈T ) P i∈Tj bi=mt,j (1⩽j⩽r) 1 Q i∈T bi!ibi X bi⩾0 (i∈T0) 1 Q i∈T0 bi!ibi = n!
r X t=0 kt̸=0 r Y j=1 H(Tj)mt,j mt,j!
Y i∈T0 e1/i = n!
r Y j=1 H(Tj)kj kj! 1 + r X j=1 kj H(Tj) !
eH(T0).
The claimed bound now follows using harmonic sum bounds in the form H(T0) = Hn − r X j=1 H(Tj) ⩽log n + 1 − r X j=1 H(Tj).
□ Remark 1.10. Whenever r is bounded, and kj = O(H(Tj)) for each j, the right side is ≪P(Z1 = k1, . . . , Zr = kr).
where Zj d = Pois(H(Tj)) for each j, and Z1, . . . , Zr are independent. Thus, Theorem 1.9 gives an upper bound for counts of cycle lengths in sets T1, . . . , Tr of the expected order (up to a constant factor) according to the Poisson model. This is a useful tool for showing that the actual cycle counts cannot vary too much from the expected means.
In the special case r = 1, Theorem 1.9 implies that for any T ⊂[n] and k ⩾0, (1.4) Pσ (CT (σ) = k) ⩽e1−H(T ) H(T)k−1 (k −1)! + H(T)k k!
!
.
Specializing to the case of cycle lengths in a single interval [m], we obtain the following very useful corollary: Corollary 1.11 (Cycles in intervals). Uniformly for 1 ⩽m ⩽n and 0 ⩽λ ⩽1, we have Pσ∈Sn Cm ⩽λ log m ⩽e m−Q(λ).
Uniformly for 1 ⩽m ⩽n and 1 ⩽λ, we have Pσ∈Sn Cm ⩾λ log m ⩽e λ1+λm−Q(λ).
In particular, uniformly for 1 ⩽m ⩽n and 0 ⩽ψ ⩽√log m, we have Pσ∈Sn |Cm −log m| > ψ p log m ≪e−1 3 ψ2.
4. Cycles and prime factors from sets: general upper bounds 15 Proof. Let T = T1 = [m], and recall that H(T) = Hm ⩾log m. Applying Theorem 1.9 (see (1.4)), together with Proposition 0.3 and the fact that Q(u) is decreasing on [0, 1], we get that Pσ∈Sn Cm ⩽λ log m ⩽e1−Hm X k⩽λ log m Hk m k! ⩽e1−Hm X k⩽λHm Hk m k! ⩽e1−HmQ(λ) ⩽em−Q(λ).
The proof of the second bound is similar. First we note that the statement is trivial if m = 1. Suppose that m ⩾2. Let λ∗= λ −λ+1 Hm . If λ∗< 1 then the probability in question in ⩽1 trivially, λ = 1 + O(1/Hm) and m−Q(λ) = m−O(1/H2 m) ≫1, so the desired bound holds. Now suppose that λ∗⩾1. We have λ log m ⩽λ(Hm −1) −1 = λ∗Hm and thus by Theorem 1.9, together with Proposition 0.3, we get that Pσ∈Sn Cm ⩾λ log m ⩽e1−Hm X k⩾λ log m−1 Hk m k!
⩽e1−Hm X k⩾λ∗Hm Hk m k!
⩽e1−HmQ(λ∗).
Now 1 ⩽λ∗⩽λ, and Q′(u) = log u implies that Q(λ∗) ⩾Q(λ) −λ + 1 Hm log λ, and, since log m ⩽Hm ⩽log m + 1 we conclude that Pσ∈Sn Cm ⩾λ log m ⩽e1−HmQ(λ)+(λ+1) log λ ⩽em−Q(λ)λ1+λ.
The final estimate follows by taking λ = 1 ± ψ/√log m and using the bound (0.16) for Q(u). Here 0 ⩽λ ⩽ 2.
□ In particular, taking m = n, we see that C(σ) usually does not vary more that √log n from its mean Hn.
The same proof yields a much more general result: Corollary 1.12. Let T ⊂[n]. Uniformly for 0 < ξ ⩽ p H(T), we have Pσ |CT (σ) −H(T)| > ξ p H(T) ≪e−1 3 ξ2.
This is not very useful when H(T) < 1, however. In this case, we expect that CT (σ) will rarely be much more than 1. Theorem 1.9 implies a right-tail bound of Pσ (CT (σ) = k) ≪H(T)k−1 (k −1)! , whereas the Poisson model predicts that the right side should be smaller, namely H(T)k/k!; but see Home-work Exercise 1.2 below.
Theorem 1.13 (Prime factors in sets). Let T0, T1, . . . , Tr be a partition of the primes in [2, x] with T1, . . . , Tr nonempty. Define Hj = H({p −1 : p ∈Tj}) = X p∈Tj 1 p −1 (1 ⩽j ⩽r).
16 1. THE SEQUENCE OF PRIME FACTORS OF AN INTEGER AND CYCLES OF A PERMUTATION, I Let k1, . . . , kr ⩾0 and such that if T0 = ∅then k1, . . . , kr are not all zero. Then Px{ω(n; Tj) = kj (1 ⩽j ⩽r)} ≪ r Y j=1 Hkj j kj! e−Hj ! η + k1 H1 + · · · + kr Hr ⩽ r Y j=1 (H(Tj) + 2)kj k1!
e−H(Tj) , where η = 0 if T0 is empty, η = 1 otherwise.
Proof. Let N = #{n ⩽x : ω(n; Tj) = kj (1 ⩽j ⩽r)}, Define mt,j = kj −1(j = t ⩾1) and let Lt(x) = X h⩽x ω(h;Tj)=mt,j (1⩽j⩽r) 1 h (0 ⩽t ⩽r).
We use the unique factorization of integers into primes, (the “Wirsing trick”), starting with log x ≪log n = X pa∥n log pa (x1/3 ⩽n ⩽x).
It follows that (log x)N ≪ X n⩽x1/3 ω(n;Tj)=kj (1⩽j⩽r) log x + X n⩽x ω(n;Tj)=kj (1⩽j⩽r) X pa∥n log pa.
In the first sum, log x ⩽x1/3 log x n ≪x1/2 n , hence the sum is at most ⩽x1/2L0(x). In the double sum, let n = pah with p ∈Tt. If t ⩾1 then ω(h, Tt) = kj −1 and ω(h, Tj) = kj otherwise. That is, we have ω(h; Tj) = mt,j for 1 ⩽j ⩽t. Also, p ∈T0 is only possible if η = 1. Hence (log x)N ≪x1/2L0(x) + r X t=1−η X h⩽x ω(h;Tj)=mt,j (1⩽j⩽r) X pa⩽x/h log pa.
Using Chebyshev’s Estimate for primes or the Prime Number Theorem, the innermost sum over pa is O(x/h) and thus the double sum over h, pa is O(Lt(x)). Also, if kj = 0 then there is the sum corresponding to t = j is empty. This gives (1.5) Px ω(n; Tj) = kj (1 ⩽j ⩽r) ≪ 1 log x (η + x−1/2)L0(x) + X 1⩽t⩽r:kt>0 Lt(x) .
Now we fix t and bound the sum Lt(x); if t ⩾1 we may assume that kt ⩾1. Write the denominator h = h1 · · · hrh0, where, for 1 ⩽j ⩽r, hj is composed only of primes from Tj and ω(hj; Tj) = mt,j, and further that h0 is composed of primes in T0. For 1 ⩽j ⩽r we have X hj 1 hj ⩽ 1 mt,j!
X p∈Tj 1 p + 1 p2 + · · · mt,j = Hmt,j j mt,j! , and, using Mertens’ product estimate (0.7), X h′ 1 h′ ⩽ Y p∈T0 1 −1 p −1 ≪(log x) Y p∈T1∪···∪Tr 1 −1 p .
Thus, Lt(x) ≪(log x) r Y j=1 Hmt,j j mt,j!
Y p∈T1∪···∪Tr 1 −1 p .
4. Cycles and prime factors from sets: general upper bounds 17 Using the elementary inequality 1 + y ⩽ey, we see that Y p∈T1∪···∪Tr 1 −1 p ⩽e−H(T1)−···−H(Tr) and we obtain (1.6) Lt(x) ≪(log x) r Y j=1 Hmt,j j mt,j! e−H(Tj) !
Combining estimates (1.5) and (1.6), we conclude that N ≪x η + x−1/2 + r X j=1 kj Hj !
r Y j=1 Hkj j kj! e−H(Tj) !
.
By hypothesis, either η = 1 or kj/Hj ⩾1/Hj ≫1/ log log x for some j, and hence the additive term x−1/2 may be omitted. This proves the first claim.
Next, r Y j=1 Hkj j kj! η + r X j=1 kj Hj !
⩽ r Y j=1 Hkj j kj!
r Y j=1 1 + 1 Hj kj = r Y j=1 (Hj + 1)kj kj!
and we have Hj = H(Tj) + X p∈Tj 1 p(p −1) ⩽H(Tj) + 1.
This proves the final inequality.
□ Remark. A version of Theorem 1.13 is stated in [67, Theorem 2], where only the case r = 1 is proved.
The full proof may also be found in .
Corollary 1.14 (Hardy-Ramanujan inequality, 1917). Uniformly for x ⩾3 and k ∈N we have Px{ω(n) = k} ≪(log log x + O(1))k−1 (k −1)!
.
Proof. Let T1 consist of all primes ⩽x. Then η = 0 and H1 = log log x + O(1) by Mertens’ sum estimate (0.5), and we obtain the desired bound.
□ Taking as a single set the primes in an interval, we obtain the following very useful corollary.
Corollary 1.15 (Prime factors in intervals). Uniformly for 3 ⩽t ⩽x and 0 ⩽λ ⩽1, we have Px{ω(n, t) ⩽λ log log t} ≪(log t)−Q(λ).
Let λ0 > 1. Uniformly for 3 ⩽t ⩽x and 1 ⩽λ ⩽λ0, we have Px{ω(n, t) ⩾λ log log t} ≪λ0 (log t)−Q(λ).
In particular, uniformly for 3 ⩽t ⩽x and 0 ⩽ψ ⩽√log log t, we have Px{|ω(n, t) −log log t| > ψ p log log t} ≪e−1 3 ψ2.
Proof. The proof is identical to the proof of Corollary 1.11, using T = T1 as the set of primes in [2, t], H(T) = log log t + O(1) from Mertens’ bound (0.5), and Theorem 1.13.
□ Taking as a special case t = n, we recover a strong form of Theorem 1.8.
We can also analyze the distribution of integers composed only of primes factors from a given set.
18 1. THE SEQUENCE OF PRIME FACTORS OF AN INTEGER AND CYCLES OF A PERMUTATION, I Corollary 1.16. Let T be a subset of the primes ⩽x, and let N(T) denote the set of integers ⩽x composed only of primes from T. For all k ⩾1, #{n ∈N(T) : ω(n) = k} ≪ x log x · (H(T) + 1)k−1 (k −1)!
.
Proof. Apply Theorem 1.13 with T1 = T and T2 being the set of all primes ⩽x that are not in T, k1 = k and k2 = 0. Then η = 0, H1 ⩽H(T) + 1 and the result follows.
□ Remarks. Applying Theorem 1.13 with T1 the set of all primes ⩽x that are not in T, and k1 = 0, we see that |N(T)| ≪e−H(T1)x ≍eH(T ) x log x since H(T) + H(T1) = log2 x + O(1) by Mertens’ estimate (0.5). Oftentimes we have a corresponding lower bound (1.7) |N(T)| ≫eH(T ) x log x, and this allows us to conclude that, conditionally on n ∈N(T), that ω(n) has an approximate Poisson distribution with parameter H(T) + O(1). That is, combining (1.7) with Corollary 1.16, we obtain Px{ω(n) = k|n ∈N(T)} ≪e−H(T ) (H(T) + 1)k−1 (k −1)!
.
For example, let T be the set of primes ⩽x that are 1 mod 4; in particular, such numbers are the sum of two squares. Then (1.7) follows from a theorem of Landau, and we have H(T) = 1 2 log2 x+O(1) by Mertens’ theorem for arithmetic progressions (0.8). We then conclude that Px{ω(n) = k|n ∈N(T)} ≪( 1 2 log2 x + O(1))k−1 (k −1)!√log x , what is, conditional on n ∈N(T), ω(n) has roughly a Poisson distribution with parameter 1 2 log2 x.
If we condition on ω(n) = k, the r = 2 case of Theorem 1.13 supplies tail bounds for ω(n, T). If X, Y are independent Poisson random variables with parameters λ1, λ2, respectively, then X + Y d = Pois(λ1 + λ2) and hence, for 0 ⩽ℓ⩽k, we have P(X = ℓ|X + Y = k) = P(X = ℓ∧Y = k −ℓ) P(X + Y = k) = e−λ1−λ2(λℓ 1/ℓ!)(λk−ℓ 2 /(k −ℓ)!) e−λ1−λ2(λ1 + λ2)k/k!
= k l λ1 λ1 + λ2 ℓ λ2 λ1 + λ2 k−ℓ .
Thus, conditional on ω(n) = k we expect that ω(n, T) will have roughly a binomial distribution with parameter α = H(T)/H(S), where S is the set of all primes in [2, x].
Theorem 1.17. Fix A > 1 and suppose that 1 ⩽k ⩽A log log x. Let T be a nonempty subset of the primes in [2, x] and define let α = H(T)/H(S). For any 0 ⩽ψ ⩽ √ αk we have Px |ω(n, T) −αk| ⩾ψ p α(1 −α)k ω(n) = k ≪A e−1 3 ψ2, the implied constant depending only on A.
We leave the proof as an exercise. It requires the lower bound Px(ω(n) = k) ≫A (log2 x)k−1 (k −1)! log x, which follows, e.g., from the Sathe-Selberg theorem; see also Theorem 1.23 below.
5. The sequence of cycles and prime factors from intervals 19 5. The sequence of cycles and prime factors from intervals In this section, we take a first look at the random sequence Cm (1 ⩽m ⩽n) for σ ∈Sn, and random function ω(n, t) (1 ⩽t ⩽x) for integers n ⩽x. As long as m and t are not too small, it is relatively easy to deduce from Corollaries 1.11 and 1.15 that Cm is uniformly close to log m for most σ ∈Sn and ω(n, t) is uniformly close to log2 t for most n ⩽x.
Theorem 1.18. Let 3 ⩽ξ ⩽x. With probability 1 −O((log log ξ)−1/3), we have |ω(n, t) −log2 t| < 2 p log2 t log3 t (ξ ⩽t ⩽x).
Theorem 1.19. Let 2 ⩽ξ ⩽n. With probability 1 −O(1/(log ξ)1/3), we have |C[m] −log m| < 2 p log m log2 m (ξ ⩽m ⩽n).
Remark 1.20. When t is bounded, ω(n, t) has a discrete distribution and we cannot say anything about almost all n; in fact it takes every possible value with positive probability; e.g. ω(n, 3) takes the values 0, 1, 2 with probabilities (as x →∞) 1 3, 1 2, 1 6, respectively. The same is true for C[m] when m is bounded; see Exercise 1.1.
Proof. The proofs of Theorems 1.18 and 1.19 are nearly identical, the latter being simpler due to the discrete nature of the sequence of values of m in question. Thus, we show full details only for Theorem 1.18.
Let k1 = ⌊log2 ξ⌋+ 1, k2 = ⌊log2 x⌋, and for k1 ⩽k ⩽k2, let tk = eek. Put tk1−1 = ξ and tk2+1 = x. For each k, k1 −1 ⩽k ⩽k2 + 1, let Fk be the event (1.8) Fk = |ω(n, tk) −log2 tk| ⩾2 p (k −1) log(k −1) −1 .
As log2 tk = k + O(1) for all tk (including the endpoints), 2 p (k −1) log(k −1) −1 = ψ p log2 tk, ψ = 2 p log k + O(1/ √ k).
Hence, by the third part of the Prime Factors in Intervals Corollary (Cor. 1.15), PFk ≪e−1 3 ψ2 ≪ 1 k4/3 .
Summing over k, we see that Fk holds for some k with probability O(1/k1/3 1 ). Now suppose that Fk fails for every k in the range k1 −1 ⩽k ⩽k2 + 1. Let ξ ⩽t ⩽x and suppose that tk < t ⩽tk+1. Evidently, ω(n, tk) ⩽ω(n, t) ⩽ω(n, tk+1).
By the failure of Fk at every k, ω(n, t) ⩾log2 tk + 1 −2 p (k −1) log(k −1) ⩾log2 t −2 p log2 t log3 t and ω(n, t) ⩽log2 tk+1 −1 + 2 p k log k ⩽log2 t + 2 p log2 t log3 t.
□ Theorems 1.18 and 1.19 also tell us about the normal behavior of pj(n), the j-th smallest (distinct) prime factor of n, and Dj(σ), the length of the j-th smallest cycle of σ (note that Dj(σ) = Dj+1(σ) for some j when σ has cycles of the same length). Since a typical integer has about log2 t prime factors ⩽t, we expect pj(n) ≈eej. Likewise, a typical permutation σ ∈Sn has about log m cycles of length ⩽m, thus we expect that Dj(n) ≈ej.
Theorem 1.21 (j-th smallest prime factor). Let 1 ⩽θ ⩽log2 x. For all but O(x/θ1/3) integers n ⩽x, we have | log2 pj(n) −j| < 3 p j log j (θ ⩽j ⩽ω(n)).
20 1. THE SEQUENCE OF PRIME FACTORS OF AN INTEGER AND CYCLES OF A PERMUTATION, I Theorem 1.22 (j-th smallest cycle). Let 1 ⩽θ ⩽log n. With probability 1 −O(θ−1/3), we have | log Dj(σ) −j| < 3 p j log j (θ ⩽j ⩽C(σ)).
Proof. The proof of Corollaries 1.21 and 1.22 are nearly identical, and so we provide details only for Theorem 1.22. We may suppose that θ ⩾θ0, where θ0 is a sufficiently large, absolute constant, for otherwise the conclusion of the Corollary is trivial if the implied constant is large enough. Let ξ = e(2/3)θ . By Theorem 1.19, with probability 1 −O(1/θ1/3), we have (1.9) |Cm −log m| < 2 p log m log2 m (ξ ⩽m ⩽n).
Also, by Exercise 1.2 (b), with probability 1 −O(1/ξ) all the cycles of σ of length ⩾ξ have distinct lengths.
Now suppose that σ is a permutation satisfying (1.9), and such that the cycles of σ with lengths ⩾ξ have distinct lengths. We suppose that θ0 is so large that the right side of the inequality in (1.9) is at most 1 2 log m when m ⩾ξ. In particular, Cξ < 3 2 log ξ ⩽θ, that is, Dθ(σ) > ξ. Thus, we may apply (1.9) with m = Dj(σ) for all θ ⩽j ⩽C(σ). As the cycle lengths ⩾ξ are distinct, we have j = Cm > 1 2 log Dj(σ) and hence |j −log Dj(σ)| < 2 q log Dj(σ) log2 Dj(σ) < 2 p 2j log(2j) < 3 p j log j provided that θ0 is large enough (and hence j is large enough).
□ Slightly better bounds than those in Theorems 1.18 and 1.19 are attainable, based on ideas stemming from the ‘Law of the Iterated Logarithm’ from probability theory. Essentially one can replace the factor log3 t (or log2 m) with log4 t (or log3 m), and this is best possible. This is deducible, e.g., from the Kubilius model of integers and the analog for permutations; see Section 4 below.
6. Lower bounds on Px{ω(n) = k} Theorem 1.23. Fix A ⩾1. For some large constant x0(A), we have uniformly for x ⩾x0(A) and 1 ⩽k ⩽A log2 x that Px ω(n) = k ≍A (log2 x)k−1 (k −1)! log x.
Proof. The upper bound follows from the Hardy-Ramanujan inequality (Theorem 1.14), since (log2 x + O(1))k−1 ≪A (log2 x)k−1 (k ⩽A log2 x).
For the lower bound, WLOG suppose that A is sufficiently large. Let Q = 10A2 + 1, R = x1/100A and T the set of all primes in [Q, R]. By Mertens’ theorem (0.5), H := H(T) = log2 x + OA(1).
We assume that x0 is large enough so that for x ⩾x0(A) we have (1.10) H ⩾log2 x 2 .
Let N be the set of integers of the form p1 . . . pk ⩽x with Q ⩽p1 < p2 < . . . < pk−1 ⩽R < pk and such that p1 · · · pk−1 ⩽x1/2. Clearly ω(n) = k for each such n ∈N. Given p1, . . . , pk−1 with product ⩽x1/2, by the Prime Number Theorem the number of choices for pk is π x p1 · · · pk−1 −π(R) ≫ x/ log x p1 · · · pk−1 .
Thus, Px (ω(n) = k) ≫S1 −S2 log x , 7. Prime factors counted with multiplicity 21 where S1 = X Q⩽p1<...<pk−1⩽R 1 p1 · · · pk−1 , S2 = X Q⩽p1<...x1/2 1 p1 · · · pk−1 .
We bound S1 using the lower bound in Proposition 0.1. We note that X p⩾Q 1 p2 ⩽ X n⩾Q 1 n2 ⩽ 1 Q −1 = 1 10A2 .
Using (1.10), we have k ⩽2AH and hence S1 ⩾ Hk−1 (k −1)!
1 − k2 20H2A2 ⩾0.8 Hk−1 (k −1)!.
To bound S2, we use the fact that integers composed of primes below R that are > x1/2 are very rare; we will devote the next sections to this type of problem. Let α = 1 log R = 100A log x . Using Proposition 0.1 again, S2 ⩽ X Q⩽p1<··· 1 we have X n⩽x f(n) ⩽(A + B(x) + 1) x log x X n⩽x f(n) n A := max y⩽x A(y) ⩽(A + B(x) + 1)eB(x) x log x exp X p⩽x f(p) p .
Proof. Fix x ⩾1, let A = maxy⩽x A(y), B = B(x) and also define M(x) = X n⩽x f(n), L(x) = X n⩽x f(n) n .
We begin in a similar way to the proof of The Prime Factors in Sets Theorem (Thm. 1.13). Since log u ⩽u, M(x) log x = X n⩽x f(n) log(x/n) + X n⩽x f(n) X pk∥n log pk (n = pkh) ⩽xL(x) + X pk⩽x log pk f(pk) X h⩽x/pk f(h) ⩽xL(x) + X pk⩽x k⩾2 log pk f(pk) x pk X h⩽x/pk f(h) h + X p⩽x f(p) log p X h⩽x/p f(h).
Recalling (b), the first double sum over pk and h is bounded by BxL(x). Invoking (a), X p⩽x f(p) log p X h⩽x/p f(h) = X h⩽x f(h) X p⩽x/h f(p) log p ⩽A(x/h)x X h⩽x f(h) h ⩽AxL(x).
We obtain M(x) log x ⩽(1 + B + A)xL(x), which completes the proof of the first asserted inequality. For the second, we invoke (b) again, using (0.3), L(x) ⩽ X P +(n)⩽x f(n) n = Y p⩽x 1 + f(p) p + f(p2) p2 + · · · ⩽exp X p⩽x f(p) p + f(p2) p2 + · · · !
⩽exp B + X p⩽x f(p) p !
.
□ Corollary 1.25. Let T be a subset of the primes in [2, x], and let 1 ⩽y0 < min T. Uniformly for 1 ⩽y ⩽y0 we have X n⩽x yΩ(n,T ) ≪y0 xe(y−1)H(T ).
Proof. The function f(n) = yΩ(n,T ) is multiplicative, with f(pk) = 1 for p ̸∈T and f(pk) = yk if p ∈T. Thus, 1 u X p⩽u f(p) log p ⩽y u X p⩽u log p ≪y 8. Application: Erdős’ multiplication table problem 23 by the Prime Number Theorem. Thus, A(u) ≪y0 ≪y0 1. Also, B(x) ⩽ X p X k⩾2 f(pk) pk log pk = X p̸∈T X k⩾2 log pk pk + X p∈T X k⩾2 yk log pk pk ≪1 + X p∈T (log p) ∞ X k=2 k y p k = 1 + X p∈T (log p)(y/p) + (y/p)2 −(y/p)3 (1 −y/p)2 ≪1 + X p⩾min T log p p(p −y) ≪y0 1 (min T) −y ≪y0 1 since y ⩽y0 < min T. Hence, the hypotheses of Lemma 1.24 hold, with bounded A(x), B(x), the bounds depending on y0. We conclude that X n⩽x yΩ(n,T ) ≪y0 x log x exp X p⩽x f(p) p !
= x log x exp X p∈T p⩽x y p + X p̸∈T p⩽x 1 p !
= x log x exp X p∈T p⩽x y −1 p + X p⩽x 1 p !
⩽ x log x exp (y −1)H(T) + log2 x + O(1) !
, since y ⩾1, and where Mertens’ bound (0.5) was used in the last step.
□ Corollary 1.26. Let 1 < λ0 < 2. Uniformly for 1 ⩽λ ⩽λ0 and 3 ⩽t ⩽x, Px{Ω(n, t) ⩾λ log2 t} ≪λ0 (log t)−Q(λ).
Proof. Let T be the set of primes in [2, t].
By Mertens’ bound (0.5), H(T) = log2 t + O(1).
By Corollary 1.25, #{n ⩽x : Ω(n, t) ⩾λ log2 t} ⩽ X n⩽x λΩ(n,t)−λ log2 t ≪λ0 λ−λ log2 txe(λ−1)H(T ) ≪λ0 x(log t)−Q(λ).
□ Remark 1.27. When λ ⩾2, the behavior of the quantity in Corollary 1.26 is different than that of the quantity in Corollary 1.15. This is due to the behavior of powers of small prime factors, most important being powers of 2, and in fact (1.11) #{n ⩽x : Ω(n, t) ⩾λ log2 t} ≈x(log t)−Q(2)−(log 2)(λ−2) = x log t 2λ log2 t .
8. Application: Erdős’ multiplication table problem In 1955, Erdős posed the following problem: Estimate the number, A(N), of distinct products of the form ab with a ⩽N, b ⩽N. Erdős proved that A(N) = o(N 2), and later in 1960 refined the 24 1. THE SEQUENCE OF PRIME FACTORS OF AN INTEGER AND CYCLES OF A PERMUTATION, I estimates to prove that A(N) = N 2(log N)−E+o(1), where E = Q 1 log 2 = 1 −1 + log log 2 log 2 = 0.08607 . . . .
Theorem 1.28. We have A(N) ≪N 2(log N)−E.
Proof. Let k0 = log2 N log 2 .
By Corollary 1.26 (with t = N), the number of distinct products with Ω(ab) ⩾k0 is bounded above by #{m ⩽N 2 : Ω(m) ⩾k0} ≪N 2(log N)−Q(1/ log 2) = N 2(log N)−E.
There are O(N) products with a = 1 or b = 1. If a > 1, b > 1 and Ω(ab) < k0, then ω(a) = h, ω(b) = j with h ⩾1, j ⩾1 and h + j = k < k0. The number of pairs a, b with a fixed h, j is, by Theorem 1.23, ≪ N 2(log2 N)h+j−2 (log N)2(h −1)!(j −1)!.
Summing first over all j, h with h + j = k using the binomial theorem, and then over k < k0 we obtain an upper bound for the total number of pairs a, b with Ω(ab) < k0 of ≪ N 2 log2 N X kψ√ log2 x τ(n) ≪(x log x)e−1 12 ψ2.
Proof. Let λ = ψ √ log2 x ∈[0, 1]. Let 1 ⩽t ⩽2. Then X n⩽x ω(n)⩾(2+λ) log2 x τ(n) ⩽ X n⩽x τ(n)tω(n)−(2+λ) log2 x = t−(2+λ) log2 x X n⩽x τ(n)tω(n).
The summand is multiplicative, and satisfies the conditions of Lemma 1.24. Hence X n⩽x τ(n)tω(n) ≪ x log x exp X p⩽x 2t p ≪x(log x)2t−1 and therefore X n⩽x ω(n)⩾(2+λ) log2 x τ(n) ≪x(log x)2t−1−(2+λ) log t.
The optimum value of t to minimize the right side is t = 1 + λ 2 , and then the exponent of log x is 1 −Q(1 + λ 2 ) ⩽1 −1 12λ2 using (0.16).
Similarly, taking t = 1 −λ 2 , we obtain with a second application of Lemma 1.24 the estimate X n⩽x ω(n)⩽(2−λ) log2 x τ(n) ⩽t−(2−λ) log2 x X n⩽x τ(n)tω(n) ≪x(log x)−(2−λ) log t+2t−1 ≪x(log x)1−2Q(1−λ/2) ≪x(log x)1−1 12 λ2.
Finally, (log x)λ2 = eψ2 and the proof is complete.
□ 10. The range of Euler’s function Let φ(n) be Euler’s “totient” function, i.e., the number of integers m ∈[n] that are relatively prime to n. Let V be the image of φ, i.e. V = {1, 2, 4, 6, 8, 10, 12, 16, · · · }, and let V (x) be the number of elements of V that are ⩽x, e.g. V (15) = 7. Since φ(p) = p −1 for all primes p, we have V (x) ⩾π(x + 1) ≫x/ log x by the Prime Number Theorem. Here we show an upper bound which is very close to this.
Theorem 1.30 (Erdős, 1935 ). We have V (x) = x(log x)−1+o(1).
26 1. THE SEQUENCE OF PRIME FACTORS OF AN INTEGER AND CYCLES OF A PERMUTATION, I Proof. Fix an integer m ⩾10, let M = p1p2 · · · pm be the product of the first m primes. Let Jm = {1 ⩽j ⩽M : (j, M) = 1, (j −1, M) ⩾log m}.
For each j ∈[M] with (j, M) = 1 let Pj = {p ⩽x : p ≡j (mod M)} and P = [ j∈Jm Pj.
We first show that |Jm|/φ(M) →1 as m →∞. Let 2|ℓ|M. Then #{1 ⩽j ⩽M : (j, M) = 1, (j −1, M) = ℓ} = Y p|M/ℓ (p −2) = φ(M) Y p|M p>2 p −2 p −1 Y p|ℓ p>2 1 p −2.
Since p −2 > √p for p ⩾5, the inner product is ⩽ p 3/ℓ. The first product is ≪1/ log m by Mertens.
Hence, summing over ℓ< log m we have φ(M) −|Jm| = #{1 ⩽j ⩽M : (j, M) = 1, (j −1, M) < log m} ≪φ(M) log m X ℓ 0. By an elementary estimate, for some large constant C, if φ(n) ⩽x then n ⩽y := Cx log2 x. Let cm = |Jm|/φ(M). By Mertens theorem in arithmetic progressions (0.8), H(P) = cm log2 x + Om(1).
Let δ = 2/ log2 m and suppose m is large enough so that 0 < δ < 1/2. Also, δ →0 as m →∞. By Theorem 1.13, followed by (0.3) and Stirling’s formula, #{n ⩽y : ω(n; P) < δ log2 x} ≪ y (log x)1−cm X k<δ log2 x ((1 −cm) log2 x + Om(1))k k!
≪ x log2 x (log x)1−cm−δ log(e/δ) .
If m is large enough, the exponent of log x is > 1 −ε.
Now let W be the number of values of Euler’s function which have the form φ(n), where ω(n; P) ⩾L := ⌈δ log2 x⌉. Suppose that K ⩾L and {kj : j ∈Jm} is a vector of non-negative integers with sum K. If ω(n; Pj) = kj for all j ∈Jm then φ(n) is divisible by Q j∈Jm(j −1, M)kj. Thus, W ⩽ X K⩾L X {kj:j∈Jm} P kj=K x Q j∈Jm(j −1, M) ⩽ X K⩾L X {kj:j∈Jm} P kj=K x (log m)K ⩽ X K⩾L x(K + 1)|Jm| (log m)K ≪x(log2 x)|Jm| (log x)δ log2 m = x(log2 x)|Jm| log2 x ≪m x log x.
This completes the proof.
□ 11. Exercises 27 11. Exercises Exercise 1.1. (a) Derive the following general inclusion-exclusion formula, valid for any non-negative integer u: 1(u = m) = ∞ X j=m (−1)j−m j m u j .
(b) Let 1 ⩽j ⩽n and 0 ⩽m ⩽n/j.
Using part (a), derive an exact formula for the number of permutations σ ∈Sn with Cj(σ) = m.
(c) With j, m fixed, evaluate lim n→∞Pσ∈Sn(Cj(σ) = m).
Exercise 1.2. (a) Show that if T is a nonempty susbet of [n], and k ⩾0, then Pσ (CT (σ) ⩾k) ⩽H(T)k k!
.
(this is sometimes stronger than Theorem 1.9, especially if H(T) is small).
(b) Show that the probability that a permutation σ ∈Sn has two cycles of the same length ⩾ℓ, is O(1/ℓ).
Exercise 1.3. Show that E 2C(σ) = n + 1. Contrast this with the behavior of 2C(σ) for most σ ∈Sn.
Exercise 1.4. Let 1 ⩽k ⩽n. Show that if T is a nonempty subset of [n] with max T ⩽n/k, then E σCT (σ)k = E Zk, where Z d = Pois(H(T)).
Exercise 1.5. (a) Show that if T is a nonempty susbet of the primes in [2, x], and k ⩾0, then Px{ω(n, T) ⩾k} ⩽H(T)k k!
.
(this is sometimes stronger than Theorem 1.13, especially if H(T) is small).
(b) Show that the number of n ⩽x that have two prime factors in some dyadic interval of the form (z, 2z] with z > y, is O(x/ log y).
Exercise 1.6. (a) Prove that E σ∈Sn 1 C(σ) ∼ 1 log n as n →∞.
(b) Prove that E x 1 ω(n)1(n ⩾2) ∼ 1 log2 x as x →∞.
Exercise 1.7. Provide full details for the proof of Corollary 1.12.
Exercise 1.8. Starting with Lemma 1.24, prove Corollary 1.15 using the method used to prove Corollary 1.26.
Exercise 1.9. Prove Theorem 1.17 using Lemma 0.5 and Theorem 1.23.
Exercise 1.10. Provide full details of the proof of Theorem 1.19.
Exercise 1.11. Provide full details of the proof of Theorem 1.21.
CHAPTER 2 Distribution of the largest cycle and largest prime factor 1. Upper bounds Theorem 1.9 implies that ν(n, m) := Pσ∈Sn(C(m,n](σ) = 0) ⩽e1−Hn+Hm ⩽e2m/n 1 ⩽m ⩽n.
When m is small, however, the number of cycles is at least n/m and this is extremely rare by Theorem 1.11. We can argue heuristically as follows: C1(σ), . . . , Cm(σ) behaves like a set of independent Poisson variables Z1, . . . , Zm with Zj d = Pois(1/j). Thus, the event C(m,n](σ) = 0 can be modeled by the event Z1 + 2Z2 + · · · + mZm = n. Using the ideas behind Chernoff’s inequality, and Proposition 0.2, for any w ⩾1 we have Pσ(Z1 + 2Z2 + · · · + mZm = n) ⩽E σwZ1+2Z2+···+mZm−n = w−n exp ( w −1 1 + w2 −1 2 + · · · wm −1 m ) .
Optimizing the choice of w will show that the RHS is decaying very rapidly as a function of u = n/m.
We utilize this idea to show the following.
Theorem 2.1 (No large cycles). Uniformly for 1 ⩽m ⩽n we have ν(n, m) ⩽e−u log u+u−1, u = n/m.
Proof. Let w = u1/m. Following the heuristic above, we first write ν(n, m) ⩽E σ∈SnwC1(σ)+2C2(σ)+···+mCm(σ)−n.
For each j ∈[m], write wj = 1 + (wj −1). By the binomial theorem and Lemma 1.1, ν(n, m) ⩽w−nE σ∈Sn m Y j=1 ∞ X kj=0 (wj −1)kj Cj(σ) kj !
= w−n X k1,...,km⩾0 (w −1)k1 · · · (wm −1)kmE σ∈Sn C1(σ) k1 · · · Cm(σ) km ⩽w−n X k1,...,km⩾0 (w −1)k1 · · · (wm −1)km m Y j=1 (1/j)kj kj!
= w−n exp ( w −1 1 + w2 −1 2 + · · · wm −1 m ) .
The mean value theorem implies that wj = uj/m ⩽1 + (u −1)j/m for 1 ⩽j ⩽m and hence (2.1) w −1 + w2 −1 2 + · · · + wm −1 m ⩽ m X j=1 (u −1)j/m j = u −1.
We conclude that ν(n, m) ⩽u−n/meu−1 = e−u log u+u−1.
□ 28 1. Upper bounds 29 Remarks. The upper bound in Theorem 2.1 is reasonably sharp throughout the range on n, m. For example, if m = 1 then A(1, n) = 1 n! = 1 u! = e−u log u+u−(1/2) log u+O(1) by Stirling’s formula.
Let Ψ(x, y) = #{n ⩽x : P +(n) ⩽y}. These are known as y-smooth, or y-friable numbers. Applying Theorem 1.13 with T1 being the set of all primes in (y, x], and k1 = 0, we have H(T1) = log2 x−log2 y+O(1) by Mertens’ estimate (0.5), and hence Ψ(x, y) ≪xlog y log x.
When log y is much smaller than log x, one can do substantially better using the ideas behind the proof of Theorem 2.1.
Theorem 2.2. Uniformly for x ⩾10 and log x ⩽y ⩽x we have Ψ(x, y) ⩽xe−u log u+O(u), u = log x log y .
Remarks. There is a change of behavior around y = log x, due to the fact that for smaller y, if Q p⩽y p ≈x, then some of the exponents of primes dividing n must be large.
We note some special cases which we will find useful for applications: (2.2) Ψ(x, log x) ⩽exp (log x) log3 x log2 x + O log x log2 x = xo(1) (x →∞), (2.3) Ψ(x, (log x)c) ⩽x1−1/c+o(1) (x →∞) for any fixed c ⩾1, and (2.4) Ψ(x, xc(log3 x)/ log2 x) ≪ x (log x)c+o(1) (x →∞).
Proof of Theorem 2.2. Define α = 1 −log u log y .
By our hypothesis that log x ⩽y ⩽x, (2.5) 1 ⩽u ⩽log x log2 x, log3 x log2 x ⩽α ⩽1, x1−α = eu log u.
Define (2.6) S := X P +(n)⩽y 1 nα = Y p⩽y 1 + 1 pα + 1 p2α + · · · = Y p⩽y 1 + 1 pα −1 ⩽exp X p⩽y 1 pα −1 .
In the case 0 < α < 2/3, we use the simple bound Ψ(x, y) ⩽ X n⩽x P +(n)⩽y x n α ⩽xαS = xe−u log u S.
30 2. DISTRIBUTION OF THE LARGEST CYCLE AND LARGEST PRIME FACTOR We have y ⩽u3 ⩽log3 x and u ≍ log x log2 x. When p ⩽21/α we have pα −1 ≫α log p, and when p > 21/α we have pα −1 ⩾1 2pα. Thus, using (2.5) again and ignoring that p is prime, log S ⩽ X 2⩽n⩽21/α O(1) α log n + X n⩾2 2 nα ≪1 α Z 21/α 2 dt log t + Z y 1 dt tα ≪21/α + y1−α 1 −α ≪(log x)o(1) + u ≪u.
This completes the proof in the case 0 < α < 2/3.
Next, assume that α ⩾2/3. For all w > 0, log w ⩽w and thus for n ⩽x, log(x/n) = α−1 log(x/n)α ⩽ α−1(x/n)α. Hence, (log x)Ψ(x, y) = X n⩽x P +(n)⩽y log(x/n) + log n ≪xαS + X n⩽x P +(n)⩽y X pk|n log p, (2.7) In the double-sum on the right side of (2.7), let n = pkm, and separate into cases depending on k = 1 or k > 1. The k = 1 terms contribute ⩽ X m⩽x P +(m)⩽y X p⩽min(y,x/m) log p ≪ X m⩽x P +(m)⩽y min(y, x/m) ⩽ X P +(m)⩽y y1−α x m α = uxαS.
The terms with k ⩾2 contribute ⩽ X p⩽y 2⩽k⩽log x log p log p X m⩽x/pk P +(m)⩽y 1 ⩽ X p⩽y 2⩽k⩽log x log p log p X P +(m)⩽y x pkm α ≪xαS.
Therefore, by (2.5), (2.8) Ψ(x, y) ≪ u log xxαS = xe−u log u log y S.
It remains to bound S. Since α ⩾2/3, (pα −1)−1 = p−α + O(1/p4/3). For any 0 ⩽x ⩽1, the mean value theorem implies that ux ⩽1 + (u −1)x, hence 1 pα = 1 pu log p log y ⩽1 p 1 + (u −1)log p log y .
Thus, by Mertens bounds (0.5) and (0.6), log S ⩽O(1) + X p⩽y 1 pα ⩽O(1) + X p⩽y 1 p + u −1 log y X p⩽y log p p ⩽log2 y + O(u + 1).
Thus, S ≪(log y)eO(u). Combining this with (2.8), we get the claimed bound in the case α ⩾2/3.
□ 2. Application: large gaps between primes 31 When y ⩽log x, simpler bounds are possible, since P +(n) ⩽y implies that n = Q p⩽y pap. Therefore, we have the equality Ψ(x, y) = # n (ap)p⩽y : X p⩽y ap log p ⩽log x : 0 ⩽ap (p ⩽y) o , and good bounds can be arrived at by combinatorial counting; see, e.g. Section III.5.2 in .
2. Application: large gaps between primes Let pn denote the nth prime, and let G(X) := max pn+1⩽X(pn+1 −pn) denote the the maximum gap between consecutive primes less than X.
In 1931, Westzynthius proved that infinitely often the gap between consecutive prime numbers can be an arbitrarily large multiple of the average gap, which is ∼log X by the PNT. After improvements by Ricci in 1934 and Erdős in 1935, in 1938 Rankin proved that (2.9) G(X) ≫log X log2 X log4 X (log3 X)2 .
This was not improved until August 2014, in two independent papers of Ford-Green-Konyagin-Tao and Maynard. Later, Ford, Green, Konyagin, Maynard, and Tao established the current world record Theorem 2.3 (Ford-Green-Konyagin-Maynard-Tao [35, Theorem 1]). We have G(X) ≫log X log2 X log4 X log3 X for sufficiently large X.
Here we prove Rankin’s bound (2.9) using a very simple argument.
Idea #1. Let x be the largest integer such that P(x) ⩽X/3, where P(x) is the product of primes below x. By the PNT, x ∼log X. Let J(x) be the largest gap between numbers that are coprime to P(x); the set of such numbers is periodic modulo P(x), so J(x) exists. Such a gap occurs between P(x) and 3P(x), that is, below X. Each number in the gap has a prime factor ⩽x < P(x), thus these numbers are composite.
Therefore, G(X) ⩾J(x).
Idea #2. Suppose that the integers coprime to P(x) have a gap [u, u + y] of length y. For each prime p ⩽x, let ap be the residue class −u mod p. Then the set of residue classes ap, for p ⩽x, cover all integers in {1, 2, . . . , y −1}; for all 1 ⩽j ⩽y −1, there is a p ⩽x with p|(u + j), hence j ≡ap (mod p). Conversely, if we can find residue classes ap, one for each prime p ⩽x, that cover [1, y −1], then G(X) ⩾J(x) ⩾y.
Rankin’s argument, based on earlier work of Westzynthius and Erdős. Suppose that x < y < x log x, let z = y( log3 x 5 log2 x ), P1 = {p : p ⩽2 log x or z < p ⩽x/2}, P2 = {p : 2 log x < p ⩽z}, P3 = {p : x/2 < p ⩽x}.
First, we set ap = 0 mod p for all p ∈P1. These ap cover all integers in [1, y −1] that have a prime factor from P1. Let S0 be the set of uncovered integers n ∈[1, y −1]. Such n satisfy either P +(n) ⩽z, or they have a prime factor > x/2. In the latter case, as n has no prime factor < 2 log x and n ⩽y ⩽x log x, we conclude that n is prime. Let u = 5 log2 x log3 x . Then u log u > (5 −o(1)) log2 x. By Theorem 2.2 and the PNT, |S0| ⩽Ψ(y, z) + π(y) ≪ y log4 x + y log x ≪ y log x.
32 2. DISTRIBUTION OF THE LARGEST CYCLE AND LARGEST PRIME FACTOR Secondly, denote by q1, . . . , qk the primes in P2. Let Sj be the set of numbers in [1, y −1] left uncovered by ap for p ∈P1 and also left uncovered by aq1, . . . , aqj. For each j, if we are given Sj−1 we can always find a choice of aqj (greedy choice) such that |Sj| ⩽(1 −1/qj)|Sj−1|. In the end, we have |Sk| ⩽|S0| k Y j=1 (1 −1/qj) = |S0| Y 2 log x 0 is small enough, and we take (2.10) y = c x(log x) log3 x (log2 x)2 , then |Sk| ⩽ x 10 log x < π(x) −π(x/2), again using the PNT. Finally, the elements of Sk can be mapped to distinct primes in P3. Thus, if ℓ∈Sk maps to p, take ap = ℓmod p to cover ℓ.
In conclusion, if y is given by (2.10), then G(X) ⩾J(x) ⩾y. As x ∼log X, y ≫(log X)(log2 X)(log4 X) (log3 X)2 , and this proves (2.9).
3. Asymptotic formulas when u is small The idea behind the asymptotic formula is to first develop a recurrence formula. For 1 ⩽ℓ⩽m, there are n ℓ (ℓ−1)! ways to form an ℓ−cycle from [n]. Hence ν(n, m) = 1 n!
X σ∈Sn C(m,n](σ)=0 1 n X τ|σ τ a cycle |τ| = 1 n · n!
m X ℓ=1 ℓ n ℓ (ℓ−1)!(n −ℓ)!ν(n −ℓ, m) = 1 n n−1 X k=n−m ν(k, m).
(2.11) Heuristic. Suppose that ν(n, m) ≈f(u), where u = n/m and f is continuous. By (2.11), f(u) ≈1 n n−1 X k=n−m f(k/m) ≈1 n Z n n−m f(t/m) dt = 1 u Z u u−1 f(v) dv.
Assume we have equality instead of ≈.
Differentiation gives uf ′(u) = −f(u −1).
This is known as a differential-delay equation. If we add the natural initial conditions f(u) = 1 for 0 ⩽u ⩽1, then there is a unique continuous solution. This motivates the definition of the Dickman function ρ(u).
Definition 2.4. The Dickman function ρ : [0, ∞) →R is the unique continuous solution of (2.12) ρ(u) = 1 (0 ⩽u ⩽1); uρ′(u) = −ρ(u −1) (u > 1).
Lemma 2.5. We have (a) uρ(u) = R u u−1 ρ(v) dv for u ⩾1; (b) ρ(u) > 0 for all u ⩾0; (c) ρ(u) is decreasing for u ⩾0; (d) For u ⩾1, −ρ′(u) ρ(u) ≪1 + log u.
Proof. (a) follows by integrating (2.12) from u = 1 to u = v with v ⩾1. To prove (b), assume that τ = min{u : ρ(u) = 0} exists. Since ρ(u) = 1 −log u for 1 ⩽u ⩽2, τ > 2. By (a), 0 = τρ(τ) = Z τ τ−1 ρ(v) > 0, 3. Asymptotic formulas when u is small 33 Figure 1. Dickman’s function from 0 ⩽u ⩽15.
a contradiction. That ρ(u) is decreasing is clear from (b) and (2.12). This proves (c). From (2.12) and (a), (2.13) −ρ′(u) ρ(u) = ρ(u −1) R u u−1 ρ(v) dv .
Let Bk = max1<v⩽k/2(−ρ′(v)/ρ(v)). We have B4 = max 1<v⩽2 1/v 1 −log v = 1 2(1 −log 2) = 1.629 . . . .
If k ⩾4 and k/2 < u ⩽(k + 1)/2 then the denominator on the right side of (2.13) is at least Z u−1/2 u−1 ρ(v) dv ⩾ρ(u −1) Z u−1/2 u−1 e−Bk(v−u+1) dv = ρ(u −1)(1 −e−1 2 Bk) Bk .
Using that e−1 2 Bk ⩽e−1 2 B4 < 1/2, we infer that Bk+1 ⩽ Bk 1 −e−1 2 Bk ⩽Bk 1 + 2e−1 2 Bk .
The function x(1 + 2e−x/2) is increasing for x ⩾0, hence if C is large and Bk ⩽C log k then Bk+1 ⩽ (C log k)(1 + 2/kC/2) ⩽C log(k + 1). Therefore, Bk ≪log k and (d) follows.
□ In Figure 3, we plot the Dickman function on a log-scale, and it is evident that ρ decreases rapidly. In fact, ρ(u) = e−u log u−u log2(2u)+O(u); see , Ch. III.5.4 for further asymptotics and proofs.
Theorem 2.6. For all n ⩾m ⩾1 we have (2.14) ρ n m ⩽ν(n, m) ⩽ρ n + 1 m + 1 .
34 2. DISTRIBUTION OF THE LARGEST CYCLE AND LARGEST PRIME FACTOR Consequently, for √n log n ⩽m ⩽n we have (2.15) ν(n, m) = ρ(u) 1 + O u log(u + 1) m , u = n/m.
Remarks. Inequality (2.15) recovers Theorem 4 of , with a much shorter proof, and provides an asymptotic formula for ν(n, m) as long as n = o(m2/ log m). When n ≫m2/ log m, ν(n, m) ̸∼ρ(n/m), the asymptotic having a different shape; see , Theorem 2.4 or for details. Thus the final conclusion is best-possible.
Proof. Suppose m ⩽n ⩽2m. As there is at most one cycle of length > m, Lemma 1.1 implies (2.16) ν(n, m) = 1 − n X k=m+1 E Ck(σ) = 1 −Hn + Hm.
Since Hn −Hm = n X k=m+1 1 k ⩽ n X k=m+1 Z k k−1 dt t = log n m = 1 −ρ n m and Hn −Hm ⩾ n X k=m+1 Z k+1 k dt t = Z n+1 m+1 dt t = log n + 1 m + 1 = 1 −ρ n + 1 m + 1 , the bounds (2.14) hold when m ⩽n ⩽2m.
Now fix m ⩾1, let N ⩾2m + 1 and assume that (2.14) holds when m ⩽n ⩽N −1. Using (2.11) followed by Lemma 2.5 (a–c), ν(N, m) = 1 N N−1 X k=N−m A(k, m) > 1 N N−1 X k=N−m ρ(k/m) > 1 N N−1 X k=M−n Z k+1 k ρ(t/m) dt = 1 N Z N N−m ρ(v/m) dv = 1 N/m Z N/m N/m−1 ρ(v) dv = ρ(N/m) and ν(N, m) ⩽1 N N−1 X k=N−m ρ k + 1 m + 1 ⩽1 N N−1 X k=N−m Z k k−1 ρ t + 1 m + 1 dt = m + 1 N Z N m+1 N−m m+1 ρ(v) dv = m + 1 N Z N+1 m+1 N−m m+1 ρ(v) dv −m + 1 N Z N+1 m+1 N m+1 ρ(v) dv = N + 1 N ρ N + 1 m + 1 −m + 1 N Z N+1 m+1 N m+1 ρ(v) dv.
The final integral on the right side is ⩾ 1 m+1ρ N+1 m+1 and thus ν(n, m) ⩽ρ N+1 m+1 . The claimed bounds (2.14) now follow by induction on n.
Now we have n m −n + 1 m + 1 = n −m m(m + 1) ⩽n m2 .
Thus, by Lemma 2.5 (d), ρ n + 1 m + 1 ⩽ρ n m eO((n/m2) log(2u)) ⩽ρ n m eO(u log(u+1)/m) 3. Asymptotic formulas when u is small 35 When √n log n ⩽m ⩽n, u log(u + 1)/m ≪1 and (2.15) follows.
□ Comparing Theorems 2.1 and 2.6, we immediately conclude that (2.17) ρ(u) ≪e−u log u+u.
The “100 prisoners problem” Imagine a prison holding 100 prisoners. They are offered to play a game, the reward being freedom for all if they win; but they all must win in order for any to go free. The prisoners are numbered 1 to 100.
Inside a room are 100 boxes, and the numbers 1 through 100 are placed in these boxes in random order.
One by one, the prisoners are led into the room and allowed to open 50 boxes. If a prisoner finds his own number in one of the boxes, he wins. The prisoners are allowed to discuss strategy before the game begins, but then are separated and allowed no communication whatsoever (e.g., one prisoner cannot mark the boxes indicating which number is inside). Is there a strategy that allows all of them to win with large probability?
Naively, if each prisoner chooses 50 boxes at random, then each has a 1/2 chance of winning, but there is only a 1/2100 chance that they all win, and go free. There is a much better strategy, based on observing that the number inside the boxes form a permutation of (we can think of the boxes as lying in a row, 1st, 2nd, ...). The strategy for each prisoner number k is thus: first open the k-th box. If the number is k, he wins. Otherwise, if box k contains the number m, next open box m. Continue in this manner until either he finds his own number (win) or has opened 50 boxes without finding his own number (lose). Under what conditions does prisoner k win with this strategy? He is essentially “following the cycle containing k”, and if the cycle length is ⩽50 he will win. Thus, if there are no cycles of length 50 or more, then everyone wins!
The likelihood of this is A(100, 50), and by (2.16) this equals 1 −H100 + H50 ≈0.31 Thus, the prisoners have a 31% chance of all going free.
Next we develop a recursive formula for ψ(x, y) analogous to (2.11) and based on an idea of Hildebrand.
Lemma 2.7. For x ⩾y ⩾2 we have Ψ(x, y) = 1 log x X p⩽y (log p)Ψ x p , y + O x log x .
Proof. We start with Ψ(x, y) log x = X n⩽x P +(n)⩽y log(x/n) + X n⩽x P +(n)⩽y X pk|n log p.
The first sum is ≪P n⩽x(x/n)1/2 ≪x. In the second sum, let n = mpk, so that P +(m) ⩽y and m ⩽x/pk.
The terms with k = 1 have sum X p⩽y (log p) X m⩽x/p P +(m)⩽y 1 = X p⩽y (log p)Ψ x p , y .
Likewise, the terms with k > 1 contribute X p⩽y k⩾2 Ψ x pk , y ⩽ X p⩽y k⩾2 x pk ≪x.
□ Theorem 2.8. For x ⩾y ⩾3 we have Ψ(x, y) = xρ(u) + O(x/ log y), where u = log x log y .
The proof of Theorem 2.8 is Exercise 2.1 below.
Theorem 2.8 provides an asymptotic formula for Ψ(x, y) as long as 1/ log y = o(ρ(u)). By Lemma 2.5 (f), this happens only for u ≪log2 x log3 x. In fact, the asymptotic Ψ(x, y) ∼xρ(u) is true in a large range of x, y; see or [66, Ch. III.5] for specific statements.
36 2. DISTRIBUTION OF THE LARGEST CYCLE AND LARGEST PRIME FACTOR 4. Exercises Exercise 2.1. For x ⩾y ⩾2 let u = log x log y .
(a) Show that Ψ(x, y) = xρ(u) + O(x/ log x) for y ⩽x ⩽y2.
(b) Define ∆(x, y) by Ψ(x, y) = x(ρ(u) + ∆(x, y)). With y fixed, let ∆k := maxy⩽x⩽y2k |∆(x, y)|. Use Lemma 2.7 to prove ∆k ⩽log y + O(1) log(y2k−1) ∆k−1 + O 1 log(y2k−1) .
(c) use (a) and (b) to prove Theorem 2.8.
Exercise 2.2. Let G = 1 − Z ∞ 1 ρ(u) u2 du = 0.624329988 . . .
G is known as the “Golomb-Dickman constant”, although it was first written down by de Bruijn.
(a) Let C+(σ) denote the length of the largest cycle in σ. Show that E nC+(σ) ∼Gn as n →∞.
(b) Show that E x log P +(n) ∼G log x as x →∞.
Exercise 2.3. Define a function ρ2(u) by ρ2(u) = 1 for 0 ⩽u ⩽1 and ρ2(u) = ρ(u) + Z u−1 0 ρ(w) u −w dw (u ⩾1).
(a) For a permutation σ ∈Sn, let k2(σ) denote the length of the 2nd largest cycle (it may equal the length of the largest cycle), and let k2(σ) = 0 if σ has only one cycle. Show that, uniformly for 1 ⩽m ⩽n and u = n/m, that Pσ(k2(σ) ⩽m) = ρ2(u) + O 1 + log u m .
(b) Let q2(n) denote the 2nd largest prime factor of an integer n (it may equal the largest prime factor, if P +(n)2|n), and define q2(n) = 0 if n is prime. Show that, uniformly for 2 ⩽y ⩽x and u = log x log y , Px(q2(n) ⩽y) = ρ2(u) + O 1 + log u log y .
(c) Show that ρ2(u) ∼c/u as u →∞, where c = R ∞ 0 ρ(w) dw.
CHAPTER 3 Integers without small prime factors and permutations without small cycles 1. Permutations without small cycles For 1 ⩽m ⩽n, let Un,m = Pσ∈Sn Cj(n) = 0 (1 ⩽j ⩽m) .
In particular, Un,0 = 1. Based on our heuristic model, Un,m should be (for m of moderate size) about the probability that Z1 = · · · = Zm = 0, where Zj d = Pois(1/j) and Z1, . . . , Zm are independent. This probability equals e−Hm ≈e−γ m . This cannot be expected to hold for large m, for example Un,m = 1/n if m ⩾n/2 (permutations lacking cycles of length ⩽n/2 must be n-cycles). A special case of Corollary 1.11 (with λ = 0) implies that (3.1) Un,m ≪1 m uniformly for 1 ⩽m ⩽n. Our aim in this section is to prove strong asymptotics for Un,m throughout the range 1 ⩽m ⩽n.
As a first attempt, we’ll use inclusion-exclusion, obtaining for any ℓ⩾1 the formula (3.2) Un,m = E σ1(Cm = 0) = ℓ X r=0 (−1)rE σ Cm r + O Cm ℓ+ 1 .
To evaluate the right side of (3.2) we derive a generalization of Lemma 1.1.
Lemma 3.1. Let I ⊆[n] and let k ⩾0.
E σ CI(σ) k ⩽H(I)k k!
, with equality if and only if k (max I) ⩽n.
Proof. For non-negative integers x1, . . . , xt, k we have x1 + · · · + xt k = X i1+···+it=k t Y j=1 xj ij .
Thus, since CI(σ) = P r∈I Cr(σ), E σ CI(σ) k = X P r∈I kr=k E σ Y r∈I Cr(σ) kr .
We apply Lemma 1.1 to the expectation on the right side, followed by the multinomial theorem, obtaining E σ CI(σ) k ⩽ X P r∈I kr=k Y r∈I (1/r)kr kr!
= H(I)k k!
, 37 38 3. INTEGERS WITHOUT SMALL PRIME FACTORS AND PERMUTATIONS WITHOUT SMALL CYCLES with equality if and only if P r∈I rkr ⩽n for all choices of the kr. This latter condition clearly holds if k(max I) ⩽n. On the other hand, if k(max I) > n then there are terms with P r∈I rkr > n, e.g. taking kmax I = k, kr = 0 for other r.
□ Theorem 3.2. Let 1 ⩽m ⩽n and set u = n/m. Then Un,m = e−Hm + O e(log m + 1) u u !
.
Proof. The bound is trivial if u < e(log m + 1), thus we may assume that u ⩾e(log m + 1). Let ℓ= ⌊u⌋= ⌊n/m⌋. For 0 ⩽r ⩽ℓ, r max I1 ⩽n, and thus applying Lemma 3.1 to (3.2), we obtain Un,m = E σ ℓ X r=0 (−1)r Cm r + O Cm ℓ+ 1 = ℓ X r=0 (−1)r Hr m r! + O Hℓ+1 m (ℓ+ 1)!
.
Since ℓ+1 > e(log m+1) ⩾eHm, the sum equals e−Hm +O(Hℓ+1 m /(ℓ+1)!). Finally, since Hm ⩽log m+1 and (ℓ+ 1)! ⩾((ℓ+ 1)/e)ℓ+1 ⩾(u/e)u we obtain the claimed bound.
□ The inclusion-exclusion identity (3.2) corresponds to the original Brun sieve technique. The bound in Theorem 3.2 is nontrivial only for u ≫log m, that is, when m ≪n/ log n. When u is large, however, the error term is very tiny compared to the main term e−Hm ≍1/m.
When u is bounded, the behavior of Un,m is more complex. We will derive a recurrence for Un,m in (3.6) below, and use it to obtain an asymptotic for Un,m when u is small. We first must introduce the Buchstab function ω(u), defined recursively for u ⩾1 by (3.3) ω(u) = 1 u (1 ⩽u ⩽2), uω(u) = 1 + Z u−1 1 ω(v) dv (u > 2).
An easy induction argument shows that ω(u) is continuous, differentiable except at the point u = 2, and satisfies 1/2 ⩽ω(u) ⩽1 for all u ⩾1. Differentiating the integral equation in (3.3) yields (3.4) ω′(u) = ω(u −1) −ω(u) u = −1 u Z u u−1 ω′(v) dv (u > 2).
Lemma 3.3. We have ω′(u) ≪1/⌊u⌋!. Consequently, for some C ∈[1/2, 1] we have ω(u) = C + O(1/⌊u⌋!).
Proof. By 1/2 ⩽ω(u) ⩽1 for all u, (3.4) gives |ω′(u)| ⩽1/(2u) for u > 2. By induction on k ⩾2 we have |ω′(u)| ⩽ 1 2u(u −1) · · · (u −k + 2) (u ⩾k).
For any u, let k = ⌊u⌋⩽u. Then |ω′(u)| ⩽1/k!. The second claim follows from the first and ω(u) −ω(v) = Z u v ω′(w) dw ≪ 1 ⌊v⌋!
(2 ⩽v ⩽u).
□ Much more is known about ω(u), in fact we have ω(u) = e−γ + O(1/⌊u⌋!), and ω(u) = e−γ changes sign infinitely many times (Maier). We will derive such an asymptotic in a indirect way using the next Theorem.
Theorem 3.4. Suppose that m, n are integers with 1 ⩽m ⩽n −1. Then (3.5) 1 2m + 1 ⩽Un,m ⩽ 1 m + 1 1. Permutations without small cycles 39 and Un,m = ω(u) m 1 + O 1 m , u = n/m.
The second part provides an asymptotic Un,m ∼ω(u)/m as long as m →∞as n →∞.
Proof. We follow the method of Granville [40, Theorem 2.2], beginning with an analog of the recursion (2.11). If σ ∈Sn has no cycles of length ⩽m, then either σ is an n-cycle, or all cycles in σ have length in [m + 1, n −m −1]. Following the proof of (2.11), we start with Un,m = 1 n + 1 n · n!
X σ∈Sn Cm=0 X α|σ α a cycle |α|⩽n−m−1 |α|.
With ℓ= |α| fixed, there are n ℓ (ℓ−1)! ways to choose α. Writing σ = αβ, there are (n −ℓ)!U(n −ℓ, m) ways to choose β. Letting k = n −ℓ, we thus we obtain (3.6) Un,m = 1 n + 1 n X m+1⩽k⩽n−m−1 Uk,m.
If m + 1 ⩽n ⩽2m + 1, then Un,m = 1/n and thus 1 2m+1 ⩽Un,m ⩽ 1 m+1. Now suppose that N ⩾2m + 2 and that (3.5) holds for m + 1 ⩽n ⩽N −1. By (3.6), NUN,m ⩽1 + X m+1⩽k⩽N−m−1 1 m + 1 = 1 + N −2m −1 m + 1 ⩽ N m + 1 and NUN,m ⩾1 + X m+1⩽k⩽N−m−1 1 2m + 1 = N 2m + 1.
Thus, (3.5) follows by induction on n.
The second inequality is true when 1 ⩽u = n/m ⩽2 since then Un,m = 1/n = ω(u)/m. We now proceed by induction on N = ⌊u⌋with m fixed. Let ∆(n, m) = mUn,m −ω(n/m), ∆N := max m+1⩽n⩽mN |∆(n, m)| In particular, ∆2 = 0. Now suppose that N ⩾2 and that mN < n ⩽m(N + 1). By (3.6), (3.7) ω(n/m) + ∆(n, m) = m n + 1 n n−m−1 X k=m+1 ω k m + 1 n n−m−1 X k=m+1 ∆(k, m).
If m + 1 ⩽k ⩽n −m −1, then n −k ⩽mN, and thus the second sum on the right side is bounded in absolute value by (n −2m −1)∆N. Writing u = n/m and using Euler’s summation formula together with Lemma 3.3, the first sum equals n−m−1 X k=m+1 ω k m = −ω(u −1) + n−m X k=m+1 ω k m = −ω(u −1) + Z n−m m ω(t/m) dt + Z n−m m t −⌊t⌋ m ω′(t/m) dt = O(1) + m Z u−1 1 ω(v) dv.
By (3.3), the final integral equals uω(u) −1, and hence |∆(n, m)| ⩽ 1 − 2 N + 1 ∆N + O 1 n 40 3. INTEGERS WITHOUT SMALL PRIME FACTORS AND PERMUTATIONS WITHOUT SMALL CYCLES Figure 1. Buchstab’s function from 1 ⩽u ⩽4.
and thus ∆N+1 ⩽ 1 − 2 N + 1 ∆N + O 1 mN .
Iterating this gives ∆N ≪1 m (N ⩾2) and the second claim in the theorem follows.
□ Combining Theorems 3.2 and 3.4, we derive a strong asymptotic for ω(u) and a strong uniform asymptotic for Un,m. In figure 1 is a graph of ω(u) for 1 ⩽u ⩽4. The rapid convergence to e−γ is evident.
Theorem 3.5. We have ω(u) = e−γ + O(e−u log u+O(u)).
Proof. In light of Lemma 3.3 it suffices to evaluate C = limu→∞ω(u). Let m be large, n = m2, u = m.
By Theorems 3.2 and 3.4, we have Um2,m = ω(m) m 1 + O 1 m = e−Hm 1 + O m e(1 + log m) m m ad it follows that C = e−γ.
□ Theorem 3.6. For any 1 ⩽m ⩽n −1 we have Un,m = e−Hm(1 + O(e−u/5)) (u = n/m).
Proof. When u > 5 log m, Theorems 3.2 and 3.5 gives Un,m = e−Hm(1 + O(e−u/4)).
When u ⩽5 log m, Theorems 3.4 and 3.5 imply Un,m = ω(u) m 1 + O 1 m = e−Hm(1 + O(e−u/5)).
□ 2. Integers without small prime factors 41 2. Integers without small prime factors Let Φ(x, z) denote the number of positive integers n ⩽x that have no prime factor ⩽z. Again, a simple heuristic suggests that for small z we should have Φ(x, z) ≈x Q p⩽z(1−1/p), and this is what we will in fact demonstrate below. A special case of Corollary 1.15 (with λ = 0) implies that (3.8) Φ(x, z) = #{n ⩽x : ω(n, z) = 0} ≪x(log z)−Q(0) = x log z ≍x Y p⩽z 1 −1 p , uniformly for 2 ⩽z ⩽x. Here ω(n, z) is the number of distinct prime factors of n that are ⩽z.
Following what we did with permutations, we first prove a bound for Φ(x, z) which is very strong for small z and then by another method work out a bound which is good for large z.
Theorem 3.7. Uniformly for x ⩾2, 2 ⩽z ⩽x1/(4.5 log2 x) we have Φ(x, z) = x Y p⩽z 1 −1 p + O xe−u/2 u = log x log z ; Proof. We may assume that x is sufficiently large. Following the proof of Lemma 3.2, we apply “Brun’s pure sieve”. Let ℓ⩾1, and apply inclusion-exclusion to obtain 1(ω(n, z) = 0) = ℓ X r=0 (−1)r ω(n, z) r + O ω(n, z) ℓ+ 1 .
We sum over x and use X n⩽x ω(n, z) r = X p1<···<pr⩽z x p1 · · · pr = X p1<···<pr⩽z x p1 · · · pr + O(1) .
The totality of the O(1) terms is ≪ ℓ X r=0 #{d ⩽zr : ω(d) = r} ⩽zℓ.
Let T be the set of primes ⩽z, and let H = H(T). By Mertens’ sum estimate (0.5), H(T) = log2 z + O(1).
By Theorem 1.6, X n⩽x ω(n, z) ℓ+ 1 ⩽x Hℓ+1 (ℓ+ 1)!.
We get Φ(x, z) = x ℓ X r=0 (−1)r X p1<···<pr⩽z 1 p1 · · · pr + O zℓ+ x Hℓ+1 (ℓ+ 1)!
.
Extending the sum on r to all non-negative integers, we have using (0.4) ∞ X r=0 (−1)r X p1<···<pr⩽z 1 p1 · · · pr = X P +(d)⩽z µ(d) d = Y p⩽z 1 −1 p .
Now assume that ℓ⩾2H. Using (0.1), we have ∞ X r=ℓ+1 (−1)r X p1<···z log x n | {z } S1 + X n⩽x P −(n)>z X pk|n log p | {z } S2 .
In S1, break up the summands into intervals xe−j < n ⩽xe1−j for integers j ⩾1. Using (3.8) S1 ⩽ X j⩾1 jΦ(xe1−j, z) ≪ X j⩾1 xje1−j log z ≪ x log z .
We have S2 = X pk⩽x p>z (log p)Φ(x/pk, z).
2. Integers without small prime factors 43 Using (3.8) again, the terms with k ⩾2 contribute ≪ X pk⩽x k⩾2 p>z (log p) x pk log z ≪ x log z X p>z log p p2 −p ≪ x z log z .
The terms with k = 1 give, by the Prime Number Theorem, X z<p⩽x/z (log p)Φ(x/p, z) + X x/z<p⩽x log p = X z<p⩽x/z (log p)Φ(x/p, z) + x + O x log z .
This completes the proof of (ii).
□ Now we argue that Lemma 3.9 (ii) is analogous to the recurrence (3.6). Let Vn,m = Φ(en, em) en .
Assuming that Φ(x, z)/x is slowly varying in x and in z, we get from Lemma 3.9 (ii) nVn,m ≈ n−m X k=m+1 Vn−k,m X ek−1 x1/2, thus we may assume that z0 ⩽z ⩽x1/2.
We iterate the recurrence in Lemma 3.9 (ii) in a manner similar to the way we analyzed (3.6), however there are more delicate error terms to analyze. Define ∆(x, z) by log z x Φ(x, z) = ω(u) −z x + ∆(x, z), where u = log x log z . With z fixed, define ∆∗ N = max z⩽x⩽zN |∆(x, z)| (N = 2, 3, . . .).
By Lemma 3.9 (i), we have ∆∗ 2 ≪ 1 log z .
Now suppose that N ⩾3 and zN−1 < x ⩽zN. Divide Lemma 3.9 (ii) by ux, obtaining log z x Φ(x, z) = 1 u + O 1 u log z + 1 u log z X z<p⩽x/z log p p log z x/p Φ(x/p, z) .
Since z/x ⩽1/√x < 1 u log z by our assumptions on z and u, we get (3.9) ω(u) + ∆(x, z) = 1 u + O 1 u log z + 1 u log z X z<p⩽x/z log p p ω log(x/p) log z −zp x + ∆ x p , z .
In (3.9), the summands −zp/x contribute, by the Prime Number Theorem, ≪ z/x u log z X z<p⩽x/z log p ≪ 1 u log z .
Since x/p < x/z < zN−1, the summands ∆(x/p, z) contribute an amount which in absolute value does not exceed ∆∗ N−1 u log z X z<p⩽x/z log p p = ∆∗ N−1 log x (log x −2 log z + O(1)) ⩽(1 −1/N)∆∗ N−1 44 3. INTEGERS WITHOUT SMALL PRIME FACTORS AND PERMUTATIONS WITHOUT SMALL CYCLES if z is large enough, where we used Mertens’ estimate (0.6). Using (0.6), partial summation and the rapid decay of ω′(u) (Theorem 3.3), we obtain X z<p⩽x/z log p p ω log(x/p) log z = Z x/z z 1 t ω log(x/t) log z dt + + Z x/z z O(e−√log t)ω′ log(x/t) log z dt t log z + O(1) = (log z) Z u−1 1 ω(v) dv + O(1).
Inserting all of these estimates into (3.9), we find that |∆(x, z)| ⩽ −ω(u) + 1 u + 1 u Z u−1 1 ω(v) dv + 1 −1 N ∆∗ N−1 + O 1 u log z = 1 −1 N ∆∗ N−1 + O 1 N log z using the recurrence (3.3) and that N −1 ⩽u ⩽N. Taking a maximum over all zN−1 < x ⩽zN we get ∆∗ N ⩽max ∆∗ N−1, 1 −1 N ∆∗ N−1 + O 1 N log z !
.
Iterating this gives ∆∗ N ≪ 1 log z (N = 2, 3, . . .).
From the definition of ∆∗ N we conclude that ∆(x, z) ≪ 1 log z (u ⩾2).
This completes the proof.
□ An unusual application of these theorems is the evaluation of the constant in the Mertens’ product formula (0.7).
It is rather straightforward to obtain a version of (0.7) where e−γ is replaced by some unspecified constant, and evaluation of the constant is not that easy.
Corollary 3.10. Assume a weak form of Mertens’ product formula (3.10) Y p⩽z 1 −1 p ≍ 1 log z (z ⩾2).
Then Y p⩽z 1 −1 p ∼e−γ log z (z →∞).
Proof. This proof was suggested by Granville. It is easy to check that in the proof of Theorems 3.7 and 3.8, we did not use (0.7) anywhere in the strong form. Fix z and let x = z10 log2 z. Then u = log x log z = 10 log2 z > 5 log2 x if z is large enough. Using (3.10) and Theorem 3.7, we have Φ(x, z) = x Y p⩽z 1 −1 p 1 + O(e−u/2 log z) = x Y p⩽z 1 −1 p 1 + O(1/ log4 z) .
On the other hand, using Theorem 3.8, followed by Theorem 3.5, Φ(x, z) = xω(u) log z 1 + O 1 log z = xe−γ log z 1 + O 1 log z .
The claim follows.
□ 3. Exercises 45 Theorem 3.11. Uniformly for 2 ⩽z ⩽x we have Φ(x, z) = x Y p⩽z 1 −1 p 1 + O(e−u/5) .
Proof. When u ⩾4.8 log2 x, equivalently, z ⩽x1/4.8 log2 x, we use Theorem 3.7 and get Φ(x, z) = x Y p⩽z 1 −1 p 1 + O(e−u/2 log z) = x Y p⩽z 1 −1 p 1 + O(e−u/5) .
When 1 ⩽u ⩽2, the claim follows from (3.8). Now suppose that 2 ⩽u ⩽4.8 log2 x. By Theorem 3.8, followed by Mertens’ estimate (0.7), we have (3.11) Φ(x, z) = xω(u) log z 1 + O 1 log z = xeγω(u) Y p⩽z 1 −1 p 1 + O 1 log z .
Now 1 log z = u log x ≪ u eu/4.8 ≪e−u/5 and the claim follows from Theorem 3.5.
□ 3. Exercises Exercise 3.1. Show that Φ(x, z) ≍x/ log z uniformly for x ⩾2z ⩾4. Be careful with the case of small x.
Exercise 3.2. Show that Ψ(x, y) ≪xe−u uniformly for all x ⩾y ⩾3, where u = log x log y .
Exercise 3.3 (,Exercise 03). (Integers with a large smooth part). For an integer n and y ⩾2 let ny be the product of all prime powers dividing n with the prime ⩽y. Define, for y ⩽z ⩽x the function Θ(x, y, z) = #{n ⩽x : ny > z}.
(a) Show that Θ(x, y, z) ⩽ X z k.
(b) Combine (a) with Theorems 2.6 and 3.4 to deduce that ρ(u) + Z u−1 0 ρ(v)ω(u −v) dv = 1 (u ⩾1).
Remark: This provides a ‘combinatorial proof’ of a purely analytic statement.
CHAPTER 4 Poisson approximation of small cycle lengths and small prime divisors 1. Small cycles of permutations Let 1 ⩽k ⩽n and consider the problem of modeling Ck = (C1(σ), . . . , Ck(σ)) by the random vector Zk = (Z1, . . . , Zk), Zj d = Pois(1/j).
We especially desire a good approximation when k is large, as opposed to bounded (ref. Theorem 1.9). We express our results in terms of the Total Variational Distance dT V (X, Y ) between two random variables X and Y taking values in a discrete space Ω, defined by (4.1) dT V (X, Y ) := sup U⊂Ω P(X ∈U) −P(Y ∈U).
The supremum occurs when U = {ω ∈Ω: P(X = ω) > P(Y = ω)}, hence (4.2) dT V (X, Y ) = X ω∈Ω max 0, P(X = ω) −P(Y = ω) .
Replacing U by Ω\ U, we see that dT V (X, Y ) = dT V (Y, X).
In comparing Ck and Zk, the space of values is Ω= Nk 0.
Lemma 4.1. We have dT V (Ck, Zk) = X h∈Nk 0 k Y j=1 1 jhjhj! max 0, e−Hk −Un′,k , where n′ = n′(h) = n −Pk j=1 jhj.
Proof. From (4.2) we have dT V (Ck, Zk) = X h∈Nk 0 max 0, P(Zk = h) −P(Ck = h) .
Clearly, P(Zk = h) = e−Hk k Y j=1 (1/j)hj hj!
.
Now fix h, write g = h1 + 2h2 + · · · + khk and consider P(Ck = h). If g > n, then P(Ck = h) = 0. Now suppose that g ⩽n. Write σ = σ1σ2, where σ1 is the product of the cycles of length at most k and permutes a subset I of [n] of size g, and σ2 is the product of the cycles of length greater than k and permutes [n] \ I of size n′ = n −g. By Cauchy’s formula (Theorem 1.2), applied to σ1, it follows that P(Ck = h) = Un′,k k Y j=1 (1/j)hj hj!
, 46 2. The Kubilius model of small prime factors of integers 47 and the lemma follows.
□ Theorem 4.2 (Poisson distribution of small cycles). Let 1 ⩽k ⩽n. Then dT V (Ck, Zk) ≪e−n/(5k).
Proof. Consider a generic vector h = (h1, · · · , hk) ∈Nk 0 and let n′ = n −(h1 + 2h2 + · · · + khk). If n′ > k then By Theorem 3.6, Un′,k = e−Hk 1 + O e−n′/(5k) .
If n′ ⩽k we’ll use the trivial bound max(0, e−Hk −Un′,k) ⩽e−Hk, and thus for all h we have max 0, e−Hk −Un′,k ≪e−Hk−n′/(5k).
Therefore, and thus X h∈H1 k Y j=1 (1/j)hj hj!
max 0, e−Hk −Un′,k ≪e−Hk−n/(5k) X h∈Nk 0 k Y j=1 (1/j)hjejhj/(5k) hj!
= exp ( −Hk −n 5k + k X j=1 ej/(5k) j ) ≪e−n/(5k), using (2.1) in the last step with u = e1/5 and w = e1/(5k). The theorem now follows from Lemma 4.1.
□ Remarks. By a more sophisticated sieve method than that used to prove Theorem (2.1), see , it is possible to prove that dT V (Ck, Zk) ≪e−f(n/k), where f(x) ∼x log x as x →∞. This is the true order (the asymptotics of the logarithm of the left side), and a result of Arratia and Tavaré . Sharper bounds are known, and are expressed in terms of the Dickman and Buhstab functions (see ).
We almost immediately obtain the following corollary, by grouping together integers into sets.
Corollary 4.3. Let k ⩽n. Then, for any subset T ⊆[k] and A ⊆N0, we have P CT (σ) ∈A = P ZT ∈A + O(e−n/(5k)).
where ZT d = Pois(H(T)).
As long as k = o(n) as n →∞, the error term is o(1) and this establishes, in a very strong form, the validity of the Poisson model for Ck.
2. The Kubilius model of small prime factors of integers We will make formal a probabilistic interpretation of various results about the distribution of integers which have been stated in earlier sections. Consider a randomly chosen integers n ∈[1, x]. Such an integer n has a canonical prime factorization as n = Y p⩽x pvp.
We regard each of the exponents vp as random variables (they depend on p and also on x). We compute exactly Px(vp = k) = 1 ⌊x⌋ x pk − x pk+1 = 1 pk − 1 pk+1 + O 1 x , the error term being relatively small when pk is small. Moreover, the variables vp are quasi-independent; that is, the correlations are small, again provided that the primes are small. The variables vp corresponding to large p are very dependent on each other, for example the event (vp > 0, vq > 0) is impossible if pq > x.
48 4. POISSON APPROXIMATION OF SMALL CYCLE LENGTHS AND SMALL PRIME DIVISORS The model of Kubilius is a sequence of idealized random variables which remove the error terms above, and is thus easier to compute with. For each prime p, define the random variable Xp that has domain N0 and such that P(Xp = k) = 1 pk − 1 pk+1 = 1 pk 1 −1 p (k = 0, 1, 2, . . .).
Furthermore, the variables Xp are all independent. If y is small compared with x, we expect that the random vector Xy = (Xp : p ⩽y) has distribution close to that of the random vector Vy = (vp : p ⩽y).
Again, Vy depends on x as well.
Recall the definition (4.1) of the total variation distance and the basic identity (4.2).
Lemma 4.4. We have dT V (Xy, Vy) = X P +(m)⩽y max 0, ζy m − 1 ⌊x⌋Φ x m, y , ζy = Y p⩽y 1 −1 p .
Proof. (cf. Tenenbaum ). Fix u = (up : p ⩽y) and write m = Q p⩽y pup. Then P(Xy = u) = Y p⩽y P(Xp = up) = Y p⩽y 1 pup 1 −1 p = ζy m and Px (Vy = u) = 1 ⌊x⌋#{ℓ∈N : mℓ⩽x, P −(ℓ) > y} = 1 ⌊x⌋Φ x m, y .
□ The lemma follows from (4.2).
Theorem 4.5 (Kubilius model approximation). Let 2 ⩽y ⩽x. Then dT V (Xy, Vy) ≪exp n −log x 5 log y o .
Proof. Let δ = 1/(5 log y), so that 0 < δ ⩽1/3. Let m satisfy P +(m) ⩽y. If m ⩽x/y2 then Theorem 3.11 implies that Φ x m, y = x mζy 1 + O(e−δ log(x/m)) .
For m > x/y2 we’ll just use the trivial bound max 0, ζy m − 1 ⌊x⌋Φ x m, y ⩽ζy m.
Thus, for all m we have max 0, ζy m − 1 ⌊x⌋Φ x m, y ≪ζy me−δ log(x/m) = ζy xδm1−δ .
By Lemma 4.4, dT V (Xy, Vy) ≪ζyx−δ X P +(m)⩽y m−1+δ = ζyx−δ Y p⩽y 1 + 1 p1−δ + 1 p2−2δ + · · · ≪x−δ log y exp ( X p⩽y pδ p ) .
2. The Kubilius model of small prime factors of integers 49 Since pδ = 1 + O(δ log p), Mertens’ estimates (0.5) and (0.6) imply that the final sum on p is log2 y + O(1).
The theorem follows.
□ We next use the Kubilius model to show that prime factors have an approximate Poisson distribution.
There are two complications. First, as with permutations, large prime factors (those > xc for some fixed c > 0) cannot be Poisson distributed because they are highly dependent on each other, and the number of such factors is limited (trivially bounded by 1/c). Secondly, and unlike the case of permutations, the small prime factors also cannot be Poisson distributed (that is, as x →∞). Take the case ω(n, 2), which equals 0 or 1, each with probability tending to 1 2 as x →∞. Likewise, for fixed t, ω(n, t) takes only finitely many values and thus cannot approach a Poisson limit as x →∞. Hence, in the result stated below, the Poisson approximation reveals itself only when “intermediate prime factors” of n are dominant, that is, those in an interval (y, z] where y →∞and log z log x →0 as x →∞. For a set T of primes, denote UT = X p∈T 1(Xp ⩾1), WT = X p∈T Xp which, in the Kubilius model, are model for ω(n; T) and Ω(n; T), respectively. Since E 1(Xp ⩾1) = 1/p and E Xp = 1/(p −1) we have (4.3) E UT = H(T), E WT = H′(T) := X p∈T 1 p −1.
Define also H′′(T) := X p∈T 1 p2 .
Theorem 4.6. Let T be a finite subset of the primes, and suppose either Y = UT or Y = WT . Let H = E Y , using the formulas (4.3). Let Z d = Pois(H). Then P (Y = k) −P(Z = k) ≪ ( H′′(T) Hk k! e−H 1 k+1 + k−H H 2 0 ⩽k ⩽1.9H H′′(T)e0.9H(1.9)−k k > 1.9H.
Proof. Write H′′ = H′′(T). When k = 0, P(Z = 0) = e−H and P(Y = 0) = P(∀p ∈T : Xp = 0) = Y p∈T 1 −1 p = e−H+O(H′′) = e−H(1 + O(H′′)), and the desired inequality follows.
For k ⩾1, we work with moment generating functions. For any complex s, (0.11) implies (4.4) E sZ = e(s−1)H.
If Y = UT , then H = H(T), and uniformly for complex s with |s| ⩽2 we have E sUT = Y p∈T E s1(Xp⩾1) = Y p∈T 1 + s −1 p = e(s−1)H+O(|s−1|2H′′) = e(s−1)H 1 + O |s −1|2H′′(T) (4.5) If Y = WT then H = H′(T) and uniformly for |s| ⩽1.9 we have E sWT = Y p∈T E sXp = Y p∈T 1 + s −1 p −s = Y p∈T 1 + s −1 p −1 + (s −1)2 (p −1)(p −s) = e(s−1)H+O(|s−1|2H′′) = e(s−1)H 1 + O(|s −1|2H′′(T)) .
(4.6) 50 4. POISSON APPROXIMATION OF SMALL CYCLE LENGTHS AND SMALL PRIME DIVISORS Then, for any 0 < r ⩽1.9, (4.4), (4.5) and (4.6) imply P(Y = k) −P(Z = k) = 1 2πi I |s|=r E sY −E sZ sk+1 dz = 1 rk Z 1 0 e−2πikθh E (re2πiθ)Y −E (re2πiθ)Zi dθ ≪H′′ rk Z 1/2 0 re2πiθ −1 2e(r cos(2πθ)−1)H dθ.
Now, for 0 ⩽θ ⩽1 2, r cos(2πθ) −1 = r −1 −2r sin2(πθ) ⩽r −1 −8rθ2 and re2πiθ −1 2 = (r −1 −2r sin2(πθ))2 + sin2(2πθ) ≪(r −1)2 + θ2, so we obtain P(Y = k) −P(Z = k) ≪H′′ e(r−1)H rk Z 1/2 0 (|r −1|2 + θ2)e−8rθ2H dθ ≪H′′ e(r−1)H rk |r −1|2 √ 1 + rH + 1 (1 + rH)3/2 .
(4.7) When 1 ⩽k ⩽1.9H, we take r = k/H in (4.7) and obtain, using Stirling’s formula, P(Y = k) −P(Z = k) ≪H′′ Hkek−H kk |k/H −1|2 k1/2 + 1 k3/2 ≪H′′ e−HHk k! k −H H 2 + 1 k !
.
This completes the proof when k ⩽1.9H.
When k > 1.9H we take r = 1.9 and the result follows in this case as well from (4.7).
□ Theorem 4.7. Let T be a finite subset of the primes. Then dT V UT , Pois(H(T)) ≪ H′′(T) 1 + H(T) and dT V WT , Pois(H′(T)) ≪ H′′(T) 1 + H(T), Proof. Let Y ∈{UT , WT }. If Y = UT , let H = H(T) and if Y = WT , let H = H′(T). Let Z d = Pois(H).
Again, write H′′ = H′′(T). From (4.2), dT V (Y, Z) ⩽ ∞ X k=0 P(Z = k) −P(Y = k) .
Consider two cases. First, if H ⩽2, we have by Theorem 4.6, dT V (Y, Z) ≪H′′ + X k>1.9H H′′(1.9)−k ≪H′′.
If H > 2, Theorem 4.6 likewise implies that X k>1.9H |P(Y = k) −P(Z = k)| ≪H′′ X k>1.9H e0.9H (1.9)k ≪H′′e−0.3H 3. Exercises 51 and also X k⩽1.9H |P(Y = k) −P(Z = k)| ≪H′′e−H ∞ X k=0 Hk k!
" 1 k + 1 + k(k −1) + k −2kH + H2 H2 # = H′′e−H ∞ X k=0 Hk (k + 1)! + ∞ X k=2 Hk−2 (k −2)! + 1 H ∞ X k=1 Hk−1 (k −1)!+ −2 ∞ X k=1 Hk−1 (k −1)! + ∞ X k=0 Hk k!
= H′′e−H eH −1 H + eH + eH H −2eH + eH ≪H′′ H .
This proves the bound when H > 2, upon noting that H′(T) ≍H(T).
□ We can use Theorem 4.6 to deal with prime factors in an arbitrary collection of subsets, by a simple com-binatorial device. The following is a consequence of Exercise 4.1. Here ZTi d = Pois(H(Ti)), and ZT1, . . . , ZTm are independent.
Corollary 4.8. Let T1, . . . , Tm be disjoint sets of primes. Then dT V ((UT1, . . . , UTm), (ZT1, . . . , ZTm)) ≪ m X j=1 H′′(Tj) max(1, H(Tj)).
Combining this Corollary with the Kubilius model (Thm. 4.5), we conclude that the “intermediate” (not too small and not too large) prime factors of an integer are “Poisson distributed”.
Theorem 4.9. Let 2 ⩽y ⩽x, and let T1, . . . , Tm be disjoint sets of primes ⩽y. Then dT V ((ω(n; T1), . . . , ω(n; Tm)), (ZT1, . . . , ZTm)) ≪ m X j=1 H′′(Tj) max(1, H(Tj)) + O e−log x 5 log y .
Proof. Let ω = (ω(n; T1), . . . , ω(n; Tm)), U = (UT1, . . . , UTm) and Z = (ZT1, . . . , ZTm) By the triangle inequality for dT V , which follows easily from the definition (4.1), we have dT V (ω, Z) ⩽dT V (ω, U) + dT V (U, Z) ⩽dT V (Vy, Xy) + dT V (U, Z).
The theorem now follows by combining Theorem 4.5 with Corollary 4.8.
□ 3. Exercises Exercise 4.1. (a) Prove that if X1, . . . , Xm are independent discrete random variables, and Y1, . . . , Ym are independent discrete random variables (with Yj having the same domain as Xj), then dT V ((X1, . . . , Xm), (Y1, . . . , Ym)) ⩽ m X j=1 dT V (Xj, Yj).
(b) Let Xj d = Pois(λj) for 1 ⩽j ⩽m, where 0 < λj ⩽1 for each j. Also suppose that Yj is a Bernouilli random variable, with P(Yj = 0) = 1 −λj, P(Yj = 1) = λj for each j. Show that dT V ((X1, . . . , Xm), (Y1, . . . , Ym)) ≪ m X j=1 λ2 j.
52 4. POISSON APPROXIMATION OF SMALL CYCLE LENGTHS AND SMALL PRIME DIVISORS Exercise 4.2. For each j ∈N, let Zj be Poisson with parameter 1/j, and Z1, Z2, . . . independent.
(a) Show that P(Zj ⩽1 ∀j) = e−γ.
(b) Let An be the probability that a random σ ∈Sn has distinct cycle sizes. Prove that lim n→∞An = e−γ.
(This is a result of Lehmer from 1972).
Exercise 4.3. Let 2 ⩽y ⩽x, and let T1, . . . , Tm be disjoint sets of primes ⩽y. Then dT V ((Ω(n; T1), . . . , Ω(n; Tm)), (ZT1, . . . , ZTm)) ≪ m X j=1 H′′(Tj) max(1, H(Tj)) + O e−log x 5 log y .
CHAPTER 5 Central Limit Theorems 1. Gaussian approximation of Poisson variables It is well-known that, as λ →∞that Pois(λ) approaches a Gaussian distribution. This is a special case of the Central Limit Theorem. Below we record a quantitative version with explicit error term, and provide an elementary proof.
Proposition 5.1 (Poisson CLT). Uniformly for real λ ⩾1, X d = Pois(λ), and real z, we have P X ⩽λ + z √ λ = Φ(z) + O λ−1/2 , where Φ(z) = 1 √ 2π Z z −∞ e−1 2 t2 dt is the distribution function of the standard Gaussian distribution.
Proof. We may assume that λ is sufficiently large. Let h∗= 3 p λ log(1 + λ). First observe that by the Poisson Tails Proposition 0.3 and the crude bounds for Q(x) (0.16), we have P(|X −λ| > h∗) ⩽2e−3 log(1+λ) = 2 (1 + λ)3 .
Likewise, (5.1) Z |t|>3√ log(1+λ) e−1 2 t2 dt ≪ 1 (1 + λ)3 .
Consequently, we may assume that |z| ⩽h∗, and deduce P X ⩽λ + z √ λ = e−λ X λ−h∗⩽k⩽λ+z √ λ λk k! + O 1 λ3 .
For |k −λ| ⩽h∗, Stirling’s formula implies that k! = k e k √ 2πλ 1 + O |k −λ| + 1 λ .
Write k = λ + u. Then, for |u| ⩽h∗, we have e−λ λk k! = 1 + O |u|+1 λ √ 2πλ e−λ eλ λ + u λ+u = 1 + O |u|+1 λ √ 2πλ eu (1 + u/λ)λ+u = 1 + O |u|+1 λ √ 2πλ exp u −(λ + u) u λ −1 2 u λ 2 + O u λ 3 = 1 + O 1 + |u| λ + |u|3 λ2 e−u2 2λ √ 2πλ .
53 54 5. CENTRAL LIMIT THEOREMS It follows that e−λ X λ−h∗⩽k⩽λ+z √ λ λk k! = M + E, where M = 1 √ 2πλ X λ−h∗⩽k⩽λ+z √ λ e−(k−λ)2 2λ and E ≪ 1 √ λ X k 1 + |k −λ| λ + |k −λ|3 λ2 e−|k−λ|2 2λ .
For some integer a ⩾1 we have (a −1) √ λ ⩽|k −λ| ⩽a √ λ. Summing over all a gives E ≪ ∞ X a=1 a + a3 √ λ e−(a−1)2/2 ≪ 1 √ λ .
By Euler summation, M = 1 √ 2πλ " Z λ+z √ λ λ−h∗ e−(t−λ)2 2λ dt − Z λ+z √ λ λ−h∗ {t} t −λ λ e−(t−λ)2 2λ dt + O(1) # .
The integral involving {t} is O(1). The first equals, by (5.1), √ λ Z z −3√ log(1+λ) e−1 2 u2 du = √ λ Z z −∞ e−1 2 u2 du + O(λ−5/2), and hence M = 1 √ 2π Z z −∞ e−1 2 u2 du + O 1 √ λ = Φ(z) + O 1 √ λ .
The proof is complete.
□ 2. Central Limit Theorems for cycles Combining Theorem 4.3 with the Central Limit Theorem for Poisson variables (Theorem 5.1 below) establishes a Central Limit Theorem for the count of cycles whose lengths lie in an arbitrary set I ⊂[n].
Theorem 5.2. Let I ⊂[n]. Uniformly for all I and any real w, Pσ∈Sn CI(σ) ⩽H(I) + w p H(I) = Φ(w) + O log(2H(I)) p H(I) !
.
Proof. Let H = H(I). We may assume that H ⩾100, otherwise the claim is trivial. If |w| ⩾√3 log H then the result follows from Corollary 1.12, since the left side is thus O(1/H) = Φ(w) + O(1/H) if w ⩽ −√3 log H and is 1 −O(1/H) = Φ(w) + O(1/H) if w ⩾√3 log H. Suppose now that |w| < √3 log H, let A = H + w √ H, m = n 5 log H , J = I ∩[m].
Because H(I \ J) = X m<k⩽n k∈I 1 k ⩽H((m, n] ∩N) ⩽log log H + O(1) we have H(J) = H + O(log log H). Thus, A = H(J) + w′p H(J), w′ = w + O log log H √ H .
2. Central Limit Theorems for cycles 55 Let Y be a Poisson random variable with parameter H(J). Thus, by Theorem 4.3 and Proposition 5.1, Pσ(CI(σ) ⩽A) ⩽Pσ(CJ(σ) ⩽A) = P(Y ⩽A) + O(e−n/5m) = Φ(w′) + O H(J)−1/2 + e−n/5m = Φ(w′) + O 1 √ H = Φ(w) + O log log H √ H .
We also have A −5 log H = H(J) + w′′p H(J), w′′ = w + O log H √ H and it follows that Pσ(CI(σ) ⩽A) ⩾Pσ CJ(σ) ⩽A −5 log H and CI\J(σ) ⩽5 log H = Pσ (CJ(σ) ⩽A −5 log H) , since min(I \ J) ⩾m ⩾n/(5 log H) implies that CI\J(σ) ⩽5 log H always. Hence, by Theorem 4.3 and Lemma 5.1, Pσ(CI(σ) ⩽A) ⩾Φ(w′′) + O(1/ √ H) = Φ(w) + O log H √ H .
The theorem follows by combining the upper and lower bounds for P(CI(σ) ⩽A).
□ The special case I = [n] was established by Goncharov , without a specific rate of convergence.
Goncharov analyzed carefully the asymptotics of the Stirling number of the first kind, s(n, m), the absolute value of which counts the number of permutations σ ∈Sn with C(σ) = m. Since Hn = log n + O(1) and Φ has bounded derivative, we quickly arrive at the following.
Theorem 5.3. Let n ⩾100 and w be real. Then Pσ∈Sn C(σ) ⩽log n + w p log n = Φ(w) + O log2 n √log n .
Proof. Letting log n + w√log n = Hn + w′√Hn, we have w −w′ ≪ 1 √log n.
Hence, Pσ C(σ) ⩽log n + w p log n = Φ(w′) + O log Hn √Hn = Φ(w) + O log2 n √log n .
□ The big-O term in Theorem 5.2 cannot be made smaller than 1/ p H(I) since CI(σ) is integer valued, and thus the left side is constant in intervals of w of length 1/ p H(I), while Φ′(w) ≫1 if w is bounded. We remark that when H(I) is bounded, CI(σ) is expected to have Poisson distribution with small parameter, and this cannot be approximated by a Gaussian.
We also derive that the j-th smallest cycle of σ, denoted Dj(σ) (with ties allowed), also obeys the Gaussian law, refining Theorem 1.22.
56 5. CENTRAL LIMIT THEOREMS Theorem 5.4. Uniformly for j in the range 1 ⩽j ⩽log n − p (log n) log log n and for any real w, Pσ∈Sn log Dj(σ) ⩽j + w p j = Φ(w) + O log(2j) √j .
Proof. We may assume that j ⩾10 and that n is sufficiently large, the statement being trivial otherwise.
We may also assume that |w| ⩽√log j, since the statement for w outside this range follows from the monotonicity of P(log Dj(σ) ⩽j + w√j), as a function of w, the statement for the two points w = ±√log j and the fact that Φ(−√log j) ≪1/j1/2 and Φ(√log j) = 1 −O(1/j1/2).
Let k = j ej+w√jk , so by hypothesis, log k ⩽j + p j log j ⩽j + p (log n) log log n ⩽log n.
Then Dj(σ) ⩽k is equivalent to Ck ⩾j. As Hk = log k + O(1) and √Hk = √j + O(|w| + 1), we have j −1 = Hk −u p Hk, where u = w + O w2 + 1 √j .
By Theorem 5.2, Pσ(Dj(σ) ⩽k) = Pσ(Ck ⩾j) = 1 −Pσ(Ck ⩽j −1) = 1 −Φ(−u) + O log Hk √Hk = Φ(u) + O log(2j) √j .
Also, Φ(u) = Φ(w) + O w2 + 1 √j = Φ(w) + O log(2j) √j and the proof is complete.
□ 3. Central Limit theorems for prime factors Theorem 5.5 (Prime factors CLT). Suppose that T is a subset of the primes in [2, x]. For any real w, Px ω(n; T) ⩽H(T) + w p H(T) = Φ(w) + O log(2H(T)) p H(T) !
.
Proof. Let H = H(T). Assume H ⩾100, else the conclusion is trivial. As in the proof of Theorem 5.2, the conclusion in the case |w| ⩾√3 log H follows from Theorem 1.13 and Proposition 0.3.
Suppose now that |w| < √3 log H, let A = H + w √ H, y = x1/(5 log H), J = T ∩[2, y].
Because H(T \ J) ⩽ X y 0 and z ∈R. Show that if j = j(x) →∞as x →∞ and j(x) ⩽(1 −ε) log2 x, then Px log2 pj+1(n) −log2 pj(n) ⩽z →1 −e−z (x →∞).
Exercise 5.5. Fix ε > 0 and z ∈R. Show that if j = j(n) →∞as n →∞and j(n) ⩽(1 −ε) log n, then Pσ∈Sn log Dj+1(σ) −log Dj(σ) ⩽z →1 −e−z (n →∞).
This matches the spacing distribution of points in a Poisson process.
CHAPTER 6 The concentration of divisors of integers and permutations 1. Concentration of divisors Erdős conjectured in 1948 that almost all integers have two divisors d and d′ with d < d′ < 2d.
This may seem counterintuitive, given that we have already shown (cf. Theorem 1.21) that a typical integer has about 2j divisors less than eej, and hence the k-th smallest divisor of a typical integer, for k large, is about exp{k1/ log 2}. Hence, one may be led to believe that dk+1/dk →∞for a typical n, as long as k →∞ (and τ(n) −k →∞by symmetry). This, however, is faulty reasoning, as (i) as we see in Theorem 5.7, there is a very large “normal” range of deviation for each prime pj(n), on a log log-scale; however having two close prime factors is genuinely rare (see Exercise 1.5). Also, (ii) divisors form from combinations of prime factors in complicated ways, which lack the “independence” of prime divisors. Erdős proved that the set {n ∈N : ∃d, d′|n with d < d′ < 2d} has asymptotic density, say δ. Note that δ ⩾1/6 because 2 < 3 < 2 · 2.
The question whether δ = 1 would remain open for another 36 years.
Working heuristic: Assume that the set {log(d′/d) : dd′|n, (d, d′) = 1} ⊂[−log n, log n] is well-distributed for a typical integer n. Since #{(d, d′) : dd′|n, (d, d′) = 1} ⩾3ω(n), we expect that |{dd′|n : (d, d′) = 1, | log(d′/d)| ⩽σ}| ≈3ω(n) σ log n.
The right hand side above is at least 1 for σ ≳(log n)3−ω(n) ≈(log n)1−log 3 = o(1).
Maier and Tenenbaum proved that this heuristic is in fact close to the truth.
Theorem 6.1 (Maier, Tenenbaum, 1984 ). Fix ε > 0. Then almost all integers n have two divisors d and d′ such that d < d′ < d 1 + (log n)1−log 3+ε .
It is possible to show more, that for almost all n there are intervals (y, ey] containing many divisors of n. We define the Erdős-Hooley ∆-function ∆(n) := max t #{d|n, log d ∈(t, t + 1]}, that is to say the maximum number of divisors n has in any interval of logarithmic length 1. Its normal order (almost sure behavior) has proven quite mysterious. Work on the distibution of ∆(n) began with Erdős , Erdős and Nicolas [20, 21] and Hooley in the 1970s. Further work on the normal and average behavior of ∆(n) can be found in works of Tenenbaum [62, 63], Hall and Tenenbaum [42, 43, 44], Maier and Tenenbaum [52, 53, 54], and most recently Ford, Green and Koukoulopoulos . See also [45, Ch.
5,6,7]. Tenenbaum’s survey paper [65, p. 652–658] includes a history of the function ∆(n) and description of many applications in number theory.
The best bounds for ∆(n) for “normal” n currently known were obtained in papers of Maier and Tenen-baum (upper bound) and Ford, Green and Koukoulopoulos (lower bound). For almost all n we have (6.1) (log2 n)η−o(1) ⩽∆(n) ⩽(log2 n)log 2+o(1), where η = 0.35332277270132346711 . . . is a specific constant.
In this section, we prove a weaker lower bound, recovering the bound proved by Maier and Tenenbaum .
59 60 6. THE CONCENTRATION OF DIVISORS OF INTEGERS AND PERMUTATIONS Theorem 6.2. Fix ε > 0. For almost all n we have ∆(n) ⩾(log log n)η1−ε, where η1 = log 2 log log 3 log 3−1 = 0.2875404895 . . .
The same argument will provide a measure of the concentration of divisors of random permutations.
Theorem 6.3. For a permutation σ on Sn, denote by ∆(σ) := max r #{τ|σ : |τ| = r}.
Then, for all but o(n!) of the permutations σ ∈Sn, we have ∆(σ) ⩾(log n)η1−o(1).
Here the terms o(1) refer to functions that →0 as n →∞.
2. A random model of prime factors and cycle lengths As we have seen by Theorem 4.2, the number of cycles of size i in a random permutation is well-approximated by a Pois(1/i) variable. Also, by Theorem 4.9, ω(n; T) is well approximated by Pois(H(T)) for a set T of primes which are ⩽xo(1) and not ”too thin”. In particular, if i is much larger than K and T is the set of primes in (ei/K, e(i+1)/K] then by Mertens’ Theorem (0.5), H(T) ≈1/i and hence ω(n; T) is well-approximate by Pois(1/i). In turn, when i is large, by Exercise 4.1 (b), Pois(1/i) is well-approximated by a Bernouilli random variable Y with P(Y = 1) = 1/i. This motivates the following random model.
Definition 6.4. We define A to be the random set of positive integers such that P(i ∈A) = 1/i for each i, and the events i ∈A are independent for different values of i. That is, if Y1, Y2, . . . are independent Bernouilli random variables with P(Yi = 1) = 1/i for each i, then A = {i : Yi = 1}.
The property that an integer has k close divisors can be modeled by the event that A has k equal subset sums.
Definition 6.5. Let k ⩾2 be an integer. Let βk be the supremum of all exponents c < 1 for which the following is true with probability →1 as D →∞: there are distinct sets A1, · · · , Ak ⊂A ∩[Dc, D] with X a∈A1 a = · · · = X a∈Ak a.
It is not a priori obvious that βk exists. We will establish this later in Theorem 6.9. In particular, we have 0 < βk < 1/10 for all k.
Define (6.2) ζk = log k log(1/βk).
Theorem 6.6 (). Fix k ⩾2 and ε > 0. For almost every positive integer n, we have ∆(n) ≫(log log n)ζk−ε.
Theorem 6.7 (). Fix k ⩾2 and ε > 0. Then, for all but on→∞(n!) of the permutations σ ∈Sn, we have ∆(σ) ⩾(log n)ζk−ε.
We also prove a more general version of Theorem 6.1. Define αk be the supremum of all real numbers α such that for almost every n ∈N, n has k divisors d1 < · · · < dk with dk ⩽d1(1 + (log n)−α). In 1964, Erdős conjectured that α2 = log 3 −1, and this was confirmed by Erdős and Hall (upper bound) and Maier and Tenenbaum (lower bound). Maier and Tenenbaum showed that αk ⩽log 2 k + 1 (k ⩾3) 2. A random model of prime factors and cycle lengths 61 and (this is not stated explicitly in ) (6.3) αk ⩾(log 3 −1)m3m−1 (3 log 3 −1)m−1 (2m−1 < k ⩽2m, m ∈N).
See also [65, p. 655–656]1. In particular, it is not known if α3 > α4, although Tenenbaum conjectures that the sequence (αk)k⩾2 is strictly decreasing.
Ford, Green and Koukoulopoulos connected αk to the constant βk from the random model.
It is easy to see that for any k ⩾2, αk ⩽α2 ⩽1. Indeed, αk ⩽α2 is obvious. Let δ = (log x)−1−ε for arbitrary fixed ε > 0. If n has two divisors d, d′ with d < d′ ⩽d(1 + δ), then WLOG there exist two such divisors with (d, d′) = 1, hence dd′|n. Thus, the number of n ⩽x divisible by two such divisors is at most X 1/δ⩽d⩽√x X d<d′⩽d(1+δ) x dd′ ⩽x X 1/δ<d⩽√x dδ d2 ≪δx log x = o(x) as x →∞. This proves α2 ⩽1.
Theorem 6.8 (). For all k ⩾2, αk ⩾βk/(1 −βk).
The authors in conjecture the corresponding upper bound αk ⩽βk/(1 −βk). The methods of allow one to compute βk for any k using a finite procedure. In particular, we have β3 = log 3 −1 log 3 + 1 ξ = 0.02616218797316965133 . . .
and β4 = log 3 −1 log 3 + 1 ξ + 1 ξλ = 0.01295186091360511918 . . .
where ξ = log 2 −log(e −1) log(3/2) , λ = log 2 −log(e −1) 1 + log 2 −log(e −1) −log(1 + 21−ξ).
In it is also proved that sup k⩾2 ζk ⩾η = 0.3533227727 . . .
which, combined with Theorem 6.6, gives the lower bound in (6.1). The authors of conjecture that lim supk→∞ζk = η. Most of the paper is devoted to relating βk to a certain combinatorial optimization problem, which grows very complex as k increases.
In these notes we will concentrate on the easiest case k = 2 and prove Theorem 6.9. We have β2 = 1 − 1 log 3 = 0.0897607 . . . and, for any k ⩾2, βk ⩾β⌈log k log 2⌉ 2 .
Trivially, βk ⩽β2 for all k. Thus, we see that βk exists for all k ⩾2 and 0 < βk ⩽β2 < 1/11 for all k.
Combining Theorem 6.9 with Theorems 6.6 and 6.7 gives Theorems 6.2 and 6.3.
Also, combining Theorems 6.8 and 6.9 gives Theorem 6.1.
It remains to prove Theorems 6.6, 6.7, 6.8, and 6.9.
Further Remarks. The average order of ∆(n) is also rather mysterious, the best known bounds being log log x ⩽1 x X n⩽x ∆(n) ⩽ec√log log x, for some c > 0. The lower bound is due to Hall and Tenenbaum (see [45, Theorem 60]) and the upper bound is from an unpublished manuscript of Koukoulopoulos which slightly refines a bound of Tenenbaum (see also [45, Theorem 70]).
1The factor 3m−1 is missing in the stated lower bounds for αk in .
62 6. THE CONCENTRATION OF DIVISORS OF INTEGERS AND PERMUTATIONS 3. Proof of Theorems 6.6, 6.7, and 6.8 In this section we assume Theorem 6.9 and deduce Theorems 6.6, 6.7, and 6.8. In particular, 0 < βk < 1/10 for all k.
We begin with a simple combinatorial argument, first used in a related context in the work of Maier-Tenenbaum , which shows how to use equal subsums in multiple intervals (uc, u] to create many more common subsums in A. For any finite subset S of positive integers, denote by ΣS the sum of the elements of S.
Lemma 6.10. Let k ⩾1 be an integer and fix ε > 0. Let C, D be parameters with log u ⩽(log v)o(1) and u →∞as v →∞. Then, with probability →1 as v →∞, there are distinct A1, . . . , AM ⊂A ∩(u, v] with ΣA1 = · · · = ΣAM and M ⩾(log v)ζk−ε.
Proof. Fix δ ∈(0, βk), depending on ε, set α := βk −δ and assume that v is sufficiently large in terms of ε, δ. Set m := jlog log v −log log u −log α k and consider the intervals Ji := (vαi+1, vαi], i = 0, 1, . . . , m−1. Due to the choice of m, these all lie in (u, v].
For each i ∈{0, 1, . . . , m−1}, let Ei be the event that there are distinct Ai,1, . . . , Ai,k ⊂Ji with equal sums.
Then, by the definition of βk, if v is large enough then P(Ei) ⩾1 −δ, uniformly in i. These events Ei are all independent. The Law of Large Numbers then implies that, with probability at least 1−δ, at least (1−2δ)m of them occur. Suppose we are in this event, suppose that Ei occurs for i ∈I, where |I| ⩾(1 −2δ)m. Then there are numbers si, i ∈I, and sets Ai,1, . . . , Ai,k ⊂Ji all with sum si.
It is now clear that for each j = {ji : i ∈I} ∈[k]I, the set Bj = [ i∈I Ai,ji lies in (u, v] and has sum P i∈I si. Moreover, these sets are distinct. Let M = k|I|, then M ⩾k(1−2δ)m ⩾k−1 log v log u (1−2δ)(log k)/(−log α) .
Recall the definition (6.2) of ζk.
By our assumption on u, and if δ is small enough, the right side is ⩾(log v)ζk−ε, as required.
□ Lemma 6.11. Let x be a large parameter, suppose that (6.4) 1 ⩽K ⩽(log x)1/2, and let I = (u, v] ∩N, where (6.5) u = 100K(log2 x)2 , v = K log x 5 log3 x −1.
For i ∈I, let Ti be the set of primes in (ei/K, e(i+1)/K], and define the random set B = {i ∈(u, v] : ω(n; Ti) ⩾1}.
Uniformly for any collection I of subsets of I, we have P(A ∩I ∈I ) = Px(B ∈I ) + O(1/ log2 x).
Proof. For u < i ⩽v, let ωi = 1(ω(n; Ti) ⩾1), Pi d = Pois(H(Ti)) (these being independent for different i) and let Zi = 1(Pi ⩾1). Each Ti is contained in [log x, y], where y = x1/5 log3 x. Hence, by Theorem 4.9 and (6.5), dT V (ω, Z) ≪ 1 log2 x + X i∈I H′′(Ti) ≪ 1 log2 x.
3. Proof of Theorems 6.6, 6.7, and 6.8 63 By the strong form of Mertens’ estimate (0.5), H(Ti) = log i + 1 K −log i K + O(e−(i/K)1/2) = 1 i + O 1 i2 .
Hence, if Yi is a Bernouilli variable with P(Yi = 1) = 1/i, then by Exercise 4.1 (b), dT V (Z, Y) ⩽ X i∈I dT V (Zi, Yi) ≪ X i∈I 1 i2 ≪ 1 log2 x.
By the triangle inequality, dT V (ω, Y) ⩽dT V (ω, Z) + dT V (Z, Y) ≪ 1 log2 x and the claim follows.
□ Proof of Theorem 6.6. Fix ε > 0, let x be large, let K = (log2 x)2 and define u, v by (6.5). Define B and sets Ti as in Lemma 6.11.
Throughout the proof, o(1) means a function →0 as x →∞.
By Lemma 6.10, with probability 1 −o(1), there are distinct sets A1, . . . , AM of A ∩(u, v] with equal sums and M ⩾(log2 x)ζk−ε. We also have that with probability 1 −o(1), |A ∩(u, v]| ⩽2 log v. Indeed, by Markov’s inequality, P(|A ∩(u, v]| > 2 log v) = P(2|A∩(u,v]| > 22 log v) ⩽v−log 4E 2|A∩(u,v]| = v−log 4 D Y j=u+1 1 −1 i + 2 i = (v/u)v−log 4 = o(1) (v →∞).
Let F be the event that A ∩(u, v] has at most 2 log v elements and has M distinct subsets with equal sums.
By the above discussion, P(F) = 1 + o(1). By Lemma 6.11, the corresponding event F ′ for the random set B also holds with probability 1 −o(1); that is, F ′ is the event that |B ∩(u, v]| ⩽2 log v and that there are distinct subsets B1, . . . , BM of B with equal sums. If we are in the event F ′ and n is divisible by Q b∈B pb, where pb ∈Tb for each b ∈B, then let di = Q b∈Bi pb for each i ⩽M. For 1 ⩽i < j ⩽M we have | log di −log dj| = X b∈Bi log pb − X b∈Bj log pb ⩽|Bi| + |Bj| K + 1 K X b∈Bi b − X b∈Bj b = |Bi| + |Bj| K ⩽4 log v K ≪ 1 log2 x.
Thus, there are M divisors di of n whose logarithms all lie in a single interval of length O(1/ log2 x) < 1. It follows that Px(∆(n) ⩾M) = 1 −o(1), as required for Theorem 6.6.
□ Proof of Theorem 6.7. Fix ε > 0. Let u = log n and v = n/ log n. For each j, let Zj d = Pois(1/j), with Z1, Z2, . . . independent. For a random permutation σ ∈Sn, let C = {j : Cj(σ) ⩾1}, and define the random set e A = {j : Zj ⩾1}. By Theorem 4.2, Exercise 4.1 (b), and the triangle inequality, we have dT V (A ∩(u, v], C ∩(u, v]) ⩽dT V (A ∩(u, v], e A ∩(u, v]) + dT V ( e A ∩(u, v], C ∩(u, v]) = o(1) as n →∞. By Lemma 6.10, with probability →1 as n →∞, A ∩(u, v] has M distinct subsets A1, . . . , AM with equal sums, where M ⩾(log v)ζk−ε. Hence, C has distinct subsets S1, . . . , SM with equal sums with probability →1 as n →∞. Each subset Sj corresponds to a distinct divisor of σ, the size of the divisor being the sum of elements of Sj. As ε > 0 is arbitrary, the result follows.
□ 64 6. THE CONCENTRATION OF DIVISORS OF INTEGERS AND PERMUTATIONS Proof of Theorem 6.8. Fix 0 < c < βk 1−βk , so that c < 1/10. Let x be large and set K = (log x)c.
Define u, v by (6.5), let D = v and define c′ by Dc′ = u. By assumption, c′ ∼ c c + 1 (x →∞) and therefore there is a δ > 0 so that c′ ⩽βk −δ for sufficiently large x. Let n be a random integer chosen uniformly in [1, x]. With probability →1 as x →∞, ω(n) ⩽2 log2 x (e.g., Theorem 1.8). By the definition of βk and Lemma 6.11, with probability 1 −o(1), the set B defined in that Lemma has k distinct subsets B1, . . . , Bk with equal sums. For each b ∈B, let pb be a prime factor of n lying in Tb. Let di = Q b∈Bi pb, for 1 ⩽i ⩽k. We have X b∈Bi log pb − X b∈Bj log pb ⩽|Bi| + |Bj| K ⩽4 log2 x (log x)c .
Thus, max(dj) ⩽min(dj) exp O log2 x (log x)c = min(dj) 1 + O log2 x (log x)c .
Since c is arbitrary subject to c < βk/(1 −βk), we conclude that αk ⩾βk/(1 −βk).
□ It remains to prove Theorem 6.9, which we accomplish in the next section.
4. Proof of Theorem 6.9 Our proof use the method introduced by Maier and Tenenbaum in , adapted to the random set A.
This can be thought of as a ‘global-to-local’ principle, where the local behavior of divisor ratios is deduced from a result about the global distribution of ratios. A similar principle will be utilized in the study of integers with a divisor in a given interval.
It is easy to see the claimed lower bound on βk given that β2 exists, following the idea in Lemma 6.10.
Indeed, fix very small ε > 0 and m ∈N, and set α = β2 −ε. Let D be large and set Ii = (Dαi, Dαi−1] for 1 ⩽i ⩽m. By the definition of β2, with probability →1 as D →∞there are distinct sets Ai,j ∈Ii, 1 ⩽i ⩽m, 1 ⩽j ⩽2 such that ΣAi,1 = ΣAi,2 (1 ⩽i ⩽m).
It follows that the 2m sets Bj = Sm i=1 Ai,ji, where j = (j1, . . . , jm) ∈{1, 2}m, all have equal sums. It follows that β2m ⩾αm. As ε is arbitrary, β2m ⩾βm 2 . The claim now follows from the obvious fact that (βk)∞ k=2 is a decreasing sequence (k + 1 equal subset sums implies k equal subset sums).
It thus remains to establish the value of β2. First, we show that |A ∩I| ≈H(I) with high probability.
Lemma 6.12. Let I be a finite set of positive integers, and 1 ⩽θ ⩽ p H(I). Then P |A ∩I| −H(I) ⩾θ p H(I) ⩽2e−1 3 θ2.
Proof. Let H = H(I). By hypothesis, H ⩾1. For any λ > 0 we have E λ|A∩I| = Y j∈I 1 + λ −1 j ⩽e(λ−1)H.
Take λ = 1 + θ/ √ H. Then P |A ∩I| ⩾H + θ √ H ⩽E λ|A∩I|−H−θ √ H 1 ⩽λ−H−θ √ H 1 e(λ1−1)H = e−Q(λ1) ⩽e−1 3 θ2 using (0.4) at the last step. Similarly, if λ2 = 1 −θ/ √ H then P |A ∩I| ⩽H −θ √ H ⩽E λ|A∩I|−H+ψ √ H 2 ⩽e−Q(λ2) ⩽e−1 3 θ2.
□ 4. Proof of Theorem 6.9 65 Lemma 6.13. Given 1 ⩽C < D, ψ ⩾1 and 0 < ε ⩽1, the probability that both # A ∩(C, v] −log(v/C) ⩽ε log(v/C)) (Ceψ ⩽v ⩽D), # A ∩(u, D] −log(D/u) ⩽ε log(D/u) (C ⩽u ⩽De−ψ) is ⩾1 −Oε(e−(1/3)ε2ψ).
Proof. The probability in question is at least 1 minus the probability that (6.6) #A ∩(ek, el] −(l −k) ⩾ε(l −k) −4 for some k, l ∈N with either k = ⌊log C⌋and k + ψ −1 ⩽l ⩽log D + 1 or l = ⌊log D⌋+ 1 and log C − 1 ⩽k ⩽l −ψ + 2.
This comes from the monotonicity of #(A ∩(C, v]) as a function of v, and the monotonicity of #(A ∩(u, D]) as a function of u (cf., the proof of Theorem 1.18). By Lemma 6.12 with θ = ε(l −k)1/2 + O((l −k)−1/2), the probability that (6.6) holds is O(e−1 3 ε2(k−l)). Summing on k, l gives the lemma.
□ To establish the lower bound on β2, we first analyze the global distribution of divisor ratios.
For 1 ⩽C < D define the random set λ(C, D) = n ΣA1 −ΣA2 : A1 ⊆A ∩(C, D], A2 ⊆A ∩(C, D], A1 ∩A2 = ∅, A1 ̸= ∅, A2 ̸= ∅ o , where ΣA is the sum of the elements of A.
Lemma 6.14. Fix ε satisfying 0 < ε < 1 200.
Suppose that 1 ⩽C < D and C ⩽D1− 1 log 3 −ε.
For any 10 ⩽ξ ⩽log D, with probability 1 −Oε(1/ log ξ) we have #λ(C, D) ⩾D/ξ and Σ(A ∩[1, D]) ⩽ξD.
Proof. We may assume that ξ ⩾e20, else the lemma is trivial. Firstly, E Σ(A ∩[1, D]) = D, hence by Markov’s inequality, (6.7) P Σ(A ∩[1, D]) > ξD ⩽1 ξ .
We will use a second moment argument. Let V denote the number of quadruples (A1, A2, A3, A4) with each Ai a nonempty subset of A′, A1 ∩A2 = A3 ∩A4 = ∅and (6.8) ΣA1 −ΣA2 = ΣA3 −ΣA4.
Our main task is to show that with probability 1 −Oε(1/ log ξ) we have (6.9) V ⩽ξ · 32|A′| 2D and |A′| ⩾log D log 3 .
Assuming (6.9), let gm = #{(A1, A2) : ∅̸= A1 ⊆A′, ∅̸= A2 ⊆A′, A1 ∩A2 = ∅, ΣA1 −ΣA2 = m}.
In this notation, V = P m g2 m and λ(C, D) = {m : gm > 0}. By Cauchy’s inequality, (3|A′| −2|A′|+1)2 = X m gm 2 ⩽#λ(C, D) X m g2 m = #λ(C, D) V.
The lemma now follows from (6.9), since (3|A′| −2|A′|+1)2 ⩾1 232|A′| for large enough D.
66 6. THE CONCENTRATION OF DIVISORS OF INTEGERS AND PERMUTATIONS To prove (6.9), we first separate offthose solutions of (6.8) with A1 = A3 and A2 = A4. Thus, V ⩽3|A′| + V ∗, where V ∗counts solutions with A1 ̸= A3 or A2 ̸= A4. Let ψ = log ξ 3 . Let E be the event that |A ∩(B, D)| ⩾(1 −ε) log(D/B) (C ⩽B ⩽De−ψ).
This implies that (6.10) |A ∩(B, D)| ⩾(1 −ε) log(D/B) −ψ (C ⩽B ⩽D).
By Lemma 6.13, P(E) ≪e−(1/3)ε2ψ ≪ε 1/ log ξ. Assume now that we are in event E. In particular, since log(D/C) ⩾( 1 log 3 + ε) log D, we have 3|A′| ⩾3(1−ε) log(D/C)−ψ ⩾D if D is large enough, giving the second part of (6.9). Hence, (6.11) V ⩽ξ 4 · 32|A′| D + V ∗.
Our next task is to show that (6.12) T := E V ∗1(E) 32|A′| ≪ξ1/2 D .
Assuming (6.12), Markov’s inequality gives P V ∗⩾ξ 4 · 32|A′| D and E ⩽ T (ξ/4)D−1 ≪ξ−1/2, which implies that (6.9) holds with probability 1 −Oε(1/ log ξ), as required.
It remains to prove (6.12). Assuming E, consider solutions of (6.8) with A1 ̸= A3 or A2 ̸= A4. Elements in A1 ∩A3 and in A2 ∩A4 cancel each other. Let n∗be the largest uncancelled element, and write A′ = Q ∪{n∗} ∪Q′, max Q < n∗< min Q′.
In particular, all elements of Q′ cancel out in (6.8), that is, (6.13) ΣA1 ∩(Q ∪{n∗}) −ΣA2 ∩(Q ∪{n∗}) = ΣA3 ∩(Q ∪{n∗}) −ΣA4 ∩(Q ∪{n∗}).
Also, by (6.10), if M = max Q then |Q′| = |A ∩(M, D]| −1 ⩾(1 −ε) log(D/M) −ψ −1 =: k(M).
(6.14) Given Q, the intersections Ai ∩Q for 1 ⩽i ⩽4, and one of the O(1) possibilities for which sets Ai contain n∗(there are 6 possibilities, namely n∗may lie in a single set Ai, or in A1 and A4 or in A2 and A3), the element n∗is uniquely determined by (6.13). We will compute T as follows: • Fix M ∈(C, D] and Q ⊂(C, M] with M ∈Q; compute P(A ∩(C, M] = Q; • Fix the sets Ai ∩Q, i = 1, 2, 3, 4; • Fix one of the possibilities for n∗; compute P(A ∩(M, n∗] = {n∗}; • Fix Q′ ⊆(n∗, D] for which (6.14) holds; compute P(A ∩(n∗, D] = Q′; • there are then 3|Q′| ways to form the sets Ai ∩Q′, 1 ⩽i ⩽4, namely each element of Q′ lies in A1 ∩A3, or in A2 ∩A4 or neither (it cannot be in both since A1, A2 are disjoint).
• By (6.14), we have 3|Q′| 32|A′| = 1 32+2|Q|+|Q′| ⩽ 1 32+k(M)+2|Q| .
4. Proof of Theorem 6.9 67 Following this process, we have T ⩽ X C<M⩽D 1 32+k(M) X Q⊆(C,M] M∈Q P(A ∩(C, M] = Q) 32|Q| X A1∩Q,...,A4∩Q X n∗ P A ∩(M, n∗] = {n∗} × × X Q′⊆(n∗,D] |Q′|⩾k(M) P A ∩(n∗, D] = Q′ .
The inner sum over Q′ is clearly ⩽1. For each n∗, P A ∩(M, n∗] = {n∗} ⩽1 n∗⩽1 M .
Also, with Q fixed there are 32|Q| ways to choose A1 ∩Q, . . . , A4 ∩Q. Thus, T ≪ X C<M⩽D 1 M · 3k(M) X Q⊆(C,M] M∈Q P A ∩(C, M] = Q .
The inner sum on Q equals P(M ∈A) = 1/M. Recalling (6.14), we see that T ≪ε 3ψ X C<M⩽D 1 M 2(D/M)(1−ε) log 3 = 3ψD−(1−ε) log 3 X C<M⩽D M (1−ε) log 3−2 ≪3ψ D , as required for (6.12). This completes the proof of the lemma.
□ Proof that β2 ⩾1 − 1 log 3. We use a device from to build up a solution of ΣA1 = ΣA2, starting with Lemma 6.14.
Fix ε ∈(0, 1 200), let N be large and let ξ = ξ(N) = log2 N, let D0 = N 1−ε and Dj = D0(3ξ)j for j ⩾1. Let C = D 1− 1 log 3 −ε 0 and put λj = λ(C, Dj) for j ⩾0. Let Ej be the event that 0 ̸∈λj, and let E′ j be the event that Ej holds and also that |λj| ⩾Dj/ξ and Σ(A ∩(C, Dj]) ⩽ξDj. By Lemma 6.14, 0 ⩽P Ej −P E′ j ≪1/ log ξ.
Let Hj be the event that there are integers n, n′ ∈A with ξDj < n < n′ ⩽3ξDj and n −n′ ∈λj, and define Fj = E′ j ∧Hj. Since Fj implies that 0 ∈λj+1 we have P Ej+1 ⩽P Ej −P Fj.
With A ∩(C, Dj] fixed such that E′ j holds we have P Hj ⩾ X m∈λj m>0 X ξDj0 X ξDj0 X ξDj\ 0 and c′ > 0, which depend only on ε, P Ej+1 ⩽P Ej −cξ−2P E′ j = (1 −c/ξ2)P Ej + c′ log ξ c ξ2 .
Iterating this, starting with PE0 ⩽1, gives PEj ⩽(1 −c/ξ2)j + c′c ξ2 log ξ j−1 X h=0 (1 −c/ξ2)h.
Taking J = ξ3 , we conclude that P EJ ≪1/ log ξ.
Since DJ < N, it follows that with probability 1 −O(1/ log ξ), there are distinct sets A1, A2 ⊆A ∩(C, N] with ΣA1 = ΣA2. As ε > 0 is arbitrary, this proves that β2 ⩾1 − 1 log 3.
□ Proof of the upper bound β2 ⩽1 − 1 log 3. Fix ε > 0 small, let N be large and put C = N 1− 1 log 3 +ε.
It suffices to show that with probability →0 as N →∞, A ∩(C, N] has two disjoint, nomempty subsets with equal sums. Let E be the event that (6.15) |A ∩(C, B]| ⩽log(B/C) + ε 2 log N (C ⩽B ⩽N).
If N is large enough, this occurs if we have |A ∩(C, B]| ⩽(1 + ε/4) log(B/C) (C log N ⩽B ⩽N).
By Lemma 6.13 with ψ = log log N, P E = o(1) as N →∞.
The probability that there exist distinct nonempty sets A1, A2 ∈A ∩(C, N] with ΣA1 = ΣA2 is at most P(E) + E S · 1(E), where S is the number of pairs A1, A2 of distinct subsets of A ∩(C, N] with equal sums. For counting S, WLOG let M = max A1 > max A2, A′ 1 = A1 \ {M} and M ′ = max(A′ 1 ∪A2). Given A′ 1 and A2, M is unique determined (if it exists; it must also satisfy M > M ′). Also, with M ′ fixed, (6.15) implies that |A ∩(C, M ′]| ⩽log(M ′/C) + ε 2 log N =: k(M ′).
Then E S · 1(E) ⩽2 X C<M ′⩽N X A′⊆(C,M ′] M ′∈A′ |A′|⩽k(M ′) P(A ∩(C, M ′] = A′) X A′ 1,A2⊆A′ A′ 1∩A2=∅ P M ∈A ∩(M ′, N] .
The innermost probability is 1 M ⩽ 1 M ′ , while the number of pairs (A′ 1, A2) equals 3|A′| ⩽3k(M ′). Thus, E S · 1(E) ⩽2 X C<M ′⩽N 3k(M ′) M ′ X A′⊆(C,M ′] M ′∈A′ P(A ∩(C, M ′] = A′) ⩽2 X C<M ′⩽N 3k(M ′) (M ′)2 ≪3(ε/2) log N Clog 3 X C<M ′⩽N (M ′)log 3−2 ≪ε 3(ε/2) log NN −ε log 3 ≪N −ε/2.
□ 5. Exercises Exercise 6.1. Let 2 ⩽ℓ⩽k. Show that βk ⩾β⌈log k log ℓ⌉ ℓ .
CHAPTER 7 Integers with a divisor in a given interval 1. Exact formulas For 0 < y < z, let τ(n; y, z) be the number of divisors d of n which satisfy y < d ⩽z. Define H(x, y, z) to be the number of positive integers n ⩽x with τ(n; y, z) > 0, and Hr(x, y, z), the number of n ⩽x with τ(n; y, z) = r. By inclusion-exclusion, H(x, y, z) = X k⩾1 (−1)k−1 X y<d1<···<dk⩽z x lcm[d1, · · · , dk] , but this is not useful for estimating H(x, y, z) unless z −y is small. With y and z fixed, however, this formula implies that the set of positive integers having at least one divisor in (y, z] has an asymptotic density, i.e.
ε(y, z) := lim x→∞ H(x, y, z) x = X k⩾1 (−1)k−1 X y<d1<···<dk⩽z 1 lcm[d1, · · · , dk].
In these notes we primarily focus on the case y ⩽x3/4. Since d|n if and only if (n/d)|n, if n ≈x then we expect that H(x, y, z) ≈H(x, x/z, x/y).
This has been proved in , although the details are rather messy due to the fact that many n are significantly smaller than x, and we actually require short interval versions, that is, estimating H(x, y, z) − H(x′, y, z) from below.
2. Easy bounds when z is small or large When z −y is small compared with y, it is very rare to have more than one divisor in (y, z].
Theorem 7.1. For 2 ⩽y + 1 ⩽x3/4 with z = (1 + η)y ∈[y + 1, 2y]. Then H(x, y, z) = x X y 0, and the constant log 4 −1 cannot be replaced by a smaller number. See also [45, p. 38–39].
Proof. We start with a simple truncated form of inclusion-exclusion, which implies that X y<d⩽z jx d k − X y<d1<d2⩽z x [d1, d2] ⩽H(x, y, z) ⩽ X y<d⩽z jx d k .
The first sum over d equals O(ηy) + x X y<d⩽z 1 d.
69 70 7. INTEGERS WITH A DIVISOR IN A GIVEN INTERVAL In the sum over d1, d2, let m = (d1, d2), ti = di/m for i = 1, 2. We also have m ⩽d2 −d1 ⩽z −y = ηy.
Therefore, X y<d1<d2⩽z x [d1, d2] ⩽x X m⩽ηy 1 m X y/m<t1 0. For some y0(c), if y ⩾y0(c) and y1+c ⩽z ⩽x then H(x, y, z) ≫c x.
(b) For all 2 ⩽y ⩽z ⩽x, we have H(x, y, z) = x 1 + O log y log z .
Proof. For (a), we will show more, that there are ≫c x integers n ⩽x with exactly one prime divisor in (y, z]. Let T denote the set of primes in (y, z]. We may assume without loss of generality that c ⩽1 and that z = y1+c. We have, by Mertens’ estimate (0.5), H(T) = log(1 + c) + O(1/ log y) ∈[ 1 2 log(1 + c), 1] if y0(c) is large enough. By Exercise 1.5, Px(ω(n, T) ⩾1) ⩾E x ω(n, T) − ω(n, T) 2 ⩾1 x X y\ z. If z ⩽√x then by Exercise 3.1, #{n ⩽x : τ(n, y, z) = 0} ⩾ X m⩽y1/2 Φ x m, z ≫ X m⩽y1/2 x m log z ≍xlog y log z .
3. The critical case z = 2y Besicovitch showed in 1934 that (7.1) lim inf y→∞ε(y, 2y) = 0, and used this to construct an infinite set A of positive integers such that its set of multiples B(A ) = {am : a ∈A , m ⩾1} does not possess asymptotic density. Erdős in 1935 showed lim y→∞ε(y, 2y) = 0 and in 1960 gave the further refinement ε(y, 2y) = (log y)−E+o(1) (y →∞), where E = 1 −1 + log log 2 log 2 = 0.086071 . . . .
In 1984, Tenenbaum refined the bounds to x (log y)E exp{c p log2 y log3 y} ≪H(x, y, 2y) ≪ x (log y)E(log2 y)1/2 , valid for 100 ⩽y ⩽√x, where c > 0 is a constant. Hall and Tenenbaum’s book Divisors [45, Ch. 2] gives a simpler proof of Tenenbaum’s theorem.
Theorem 7.3 (Ford ). Uniformly for 4 ⩽y ⩽x1/2 we have H(x, y, 2y) ≍ x (log y)E(log2 y)3/2 .
Remarks. Theorem 1 of establishes the order of H(x, y, z) for all x, y, z, improving upon cruder estimates of Tenenbaum .
Our proof of the lower bound implicit in Theorem 7.3 gives a somewhat stronger conclusion, where we restrict to integers in an interval which are squarefree and free of small prime factors. We will leave the details as an exercise, Exercise 7.1 below.
Theorem 7.4. Fix c′ < 1 < c, and w ⩾1. We have, for 4 ⩽y ⩽x1/2, #{c′x < n ⩽x : µ2(n) = 1, τ(n, y, cy) ⩾1} ≫c,c′ H(x, y, z).
Remarks. In Theorem 7.4, we consider c, c′, w all fixed. In , the order of magnitude of #{n ⩽x : P −(n) > w, τ(n, y, 2y) ⩾1} was determined for all x, y, w. The results change behavior depending on the relative size of y, w.
4. Some applications of Theorem 7.3 1. Distinct products in a multiplication table, a problem of Erdős from 1955 (, ). Let A(x) be the number of positive integers n ⩽x which can be written as n = m1m2 with each mi ⩽√x. Earlier, see Theorem 1.28, we showed that A(x) ≪x/(log x)E.
Theorem 7.5. We have A(x) ≍ x (log x)E(log log x)3/2 .
72 7. INTEGERS WITH A DIVISOR IN A GIVEN INTERVAL Proof. Evidently A(x) is at least the number of n ⩽x/4 with a divisor in ( 1 4 √x, 1 2 √x].
Also, if n = m1m2 with m1 ⩽√x, m2 ⩽√x then for some integer k ⩾0, 2−k−1√x < m1 ⩽2−k√x, and hence n ⩽2−kx. Thus, H x 4 , √x 4 , √x 2 ⩽A(x) ⩽ X k⩾0 H x 2k , √x 2k+1 , √x 2k .
The theorem now follows quickly from Theorem 7.3. For the right side, use Theorem 7.3 when k ⩽10 log2 x and the trivial upper bound x/2k when k > 10 log2 x.
□ 2. Distribution of Farey gaps (Cobeli, Ford, Zaharescu ).
Theorem 7.6. Let ( 0 1, 1 Q, . . . , Q−1 Q , 1 1) denote the sequence of Farey fractions of order Q, and let N(Q) denote the number of distinct gaps between successive terms of the sequence. Then N(Q) ≍ Q2 (log Q)δ(log log Q)3/2 .
Proof. The distinct gaps are precisely those products qq′ with 1 ⩽q, q′ ⩽Q, (q, q′) = 1 and q+q′ > Q.
Thus, max(q, q′) > Q/2, so N(Q) ⩽H(Q2, Q/2, Q) and the upper bound follows from Theorem 7.3. For the lower bound, consider squarefree 0.3Q2 < n ⩽0.36Q2 with a divisor in (0.5Q, 0.6Q]. The complementary divisor then lies in (0.5Q, 0.72Q]. Hence N(Q) ⩾1 2 #{0.3Q2 < n ⩽0.36Q2 : µ2(n) = 1, τ(n, 0.5Q, 0.6Q) ⩾1} and the lower bound follows from Theorem 7.4.
□ 3. Density of unions of residue classes. Given moduli m1, . . . , mk, let δ0(m1, . . . , mk) be the minimum, over all possible residue classes a1 mod m1, . . . , ak mod mk, of the density of integers which lie in at least one of the classes.
By a theorem of Rogers (see [41, p. 242–244]), the minimum is achieved by taking a1 = · · · = ak = 0 and thus δ0(m1, . . . , mk) is the density of integers possessing a divisor among the numbers m1, . . . , mk. When m1, . . . , mk consist of the integers in an interval (y, z], then δ0(m1, . . . , mk) = ε(y, z).
4. Partial Möbius divisor sums, which was first studied by Erdős and Hall . Define M(n, y) = X d|n d⩽y µ(d).
Theorem 7.7 (K. Ford, unpublished). Let 10 ⩽y ⩽√x. The number of integers n ⩽x with M(n, y) ̸= 0 is ≍ x (log y)E(log2 y)3/2 .
This requires versions of H(x, y, 2y) which count integers free of prime factors ⩽w, uniformly in w, as well as a version counting integers with exactly one divisor in (y, 2y]. The former is dealt with in and the latter in .
Here we argue more crudely and show that (7.2) #{n ⩽x : M(n, y) ̸= 0} ≪ x (log y)E/2 (y ⩽x1/3).
Let w = exp{(log y)E/2}. By Theorem 3.8, #{n ⩽x : P −(n) > w} = Φ(x, w) ≪ x (log y)E/2 .
Now consider n ⩽x with p = P −(n) ⩽w. If pa|n with pa > w2 then for some d ∈N with d > w, d2|n. The number of such n is at most ≪ X d>w x d2 ≪x w ≪ x log y .
5. A heuristic for H(x, y, 2y) 73 Now consider n ⩽x where p = P −(n) ⩽w, pa∥n, pa ⩽w2, n = pam. If τ(n, y, py) = 0 then for any squarefree d|m, pd ⩽y and hence M(n, y) = X d|m d⩽y µ(d) + µ(pd) = 0.
The number of such n is at most X p⩽w pa⩽w2 H x pa , y, py ≪ X p⩽w pa⩽w2 X 0⩽k⩽log p log 2 H x pa , 2ky, 2k+1y ≪ X p⩽w a⩾1 x log p pa(log y)E ≪ x (log y)E/2 by Mertens’ estimate (0.6). This proves (7.2).
5. A heuristic for H(x, y, 2y) Write n = n′n′′, where n′ is composed only of primes ⩽2y and n′′ is composed only of primes > 2y. For simplicity, assume n′ is squarefree and n′ ⩽y100. Assume for the moment that the set D(n′) = {log d : d|n′} is uniformly distributed in [0, log n′]. If n′ has k prime factors, then the expected value of τ(n′, y, 2y) should be about 2k log 2 log n′ ≍ 2k log y. This is ≫1 precisely when k ⩾k0 + O(1), where k0 := j log log y log 2 k . Using the fact that, e.g. Theorem 1.13 for the upper bound, the number of n ⩽x with n′ having k prime factors is of order x log y (log log y)k k!
, we obtain a heuristic estimate for H(x, y, 2y) of order x log y X k⩾k0+O(1) (log log y)k k!
≍x(log log y)k0 k0! log y ≍ x (log y)δ(log log y)1/2 .
This is slightly too big, and the reason stems from the uniformity assumption about D(n′). In fact, for most n′ with about k0 prime factors, the set D(n′) is far from uniform, possessing many clusters of divisors and large gaps between clusters. This substantially decreases the likelihood that τ(n′, y, 2y) ⩾1. If we write n′ = p1 · · · pk, where p1 < p2 < . . . < pk, then we expect log log pj ≈j log log y k0 = j log 2 + O(1) for each j.
The Central Limit Theorem for prime factors ⩽2y (e.g. Theorem 5.7) tell us that with high probability there is a j for which log log pj ⩽j log 2 −c√log log y, where c is a small positive constant. Thus, the 2j divisors of p1 · · · pj will be clustered in an interval of logarithmic length about ≪log pj ⩽2je−c√ log2 y. On a logarithmic scale, the divisors of n′ will then lie in 2k−j translates of this cluster, the total length of the clusters being ≪2ke−c√ log2 y. A measure of the degree of clustering of the divisors of an integer a is given by (7.3) L (a) = [ d|a [−log 2 + log d, log d) L(a) = meas L (a).
The probability that τ(n′, y, 2y) ⩾1 should then be about L(n′)/ log y. Making this precise leads to the upper and lower bounds for H(x, y, 2y) given below in Proposition 7.8. The upper bound for L(a) given in Lemma 7.9 (iii) below quantifies how small L(a) must be when there is a j with log log pj considerably smaller than j log 2.
What we really need to count is n for which n′ has about k0 prime factors and L(n′) ≫log n′. This roughly corresponds to asking for log log pj ⩾j log 2 −O(1) for all j. The analogous problem from statistics theory is to ask for the likelihood than given k0 random numbers in [0, 1], there are ⩽k0x + O(1) of them 74 7. INTEGERS WITH A DIVISOR IN A GIVEN INTERVAL which are ⩽x, uniformly in 0 ⩽x ⩽1. Later, we will see that this probability is about 1/k0 ≍1/ log log y and this leads to the correct order of H(x, y, 2y) given in Theorem 7.3.
6. A global-to-local principle In this section, we estimate H(x, y, 2y) in terms of an average over L(a), as defined in (7.3). As L(a) captures the global distribution of divisors of a, we call this a ’global-to-local’ principle.
Introduce the notation P(x) = {n ∈N : µ2(n) = 1, P +(n) ⩽x}.
Proposition 7.8. If y0 is sufficiently large and y0 ⩽y ⩽√x, then H(x, y, 2y) ≍ x log2 y X a∈P(y) L(a) a .
We first show some basic inequalities for L(a) and then relate sums of the type on the RHS in Proposition 7.8.
Lemma 7.9. We have (i) L(a) ⩽min(τ(a) log 2, log(2a)); (ii) If (a, b) = 1, then L(ab) ⩽τ(b)L(a); (iii) If p1 < · · · < pk, then L(p1 · · · pk) ⩽min 0⩽j⩽k 2k−j(log(2p1 · · · pj)).
Proof. Part (i) is immediate, since L (a) is the union of τ(a) intervals of length log 2, all contained in [−log 2, log a). Part (ii) follows from L (ab) = [ d|b {u + log d : u ∈L (a)}.
Combining parts (i) and (ii) with a = p1 · · · pj and b = pj+1 · · · pk yields (iii).
□ Lemma 7.10. Let w2 ⩾w1 ⩾2. Then X a∈P(w2) L(a) a ≪ log w2 log w1 2 X a∈P(w1) L(a) a .
Proof. Given a ∈L (w2), write a uniquely as a = a1a2 where P +(a1) ⩽w1 < P −(a2). By Lemma 7.9 (ii), L(a) ⩽τ(a2)L(a1). Thus, X a∈P(w2) L(a) a = X a1∈P(w1) L(a1) a1 X p|a2⇒w1 y1/2}. Then clearly X a∈P(2y) L(a) a log2(P +(a) + y2/3/a) ≪ X a∈P1 L(a) a log2 P +(a) + 1 (log y)2 X a∈P(2y) L(a) a .
For a ∈P1, let p = P +(a) and a = pb, so b > y1/4. By Lemma 7.9 (ii), L(a) ⩽2L(b) and thus X a∈P1 L(a) a log2 P +(a) ⩽2 X p⩽y1/4 1 p log2 p X b∈P(p) b>y1/4 L(b) b ⩽2 X p⩽y1/4 1 p log2 p 43 log3 y X b∈P(p) L(b) log3 b b .
Next, X b∈P(p) L(b) log3 b b = X b∈P(p) L(b) b X p1|b,p2|b,p3|b (log p1)(log p2)(log p3) ⩽8 X p1,p2,p3⩽p (log p1)(log p2)(log p3) [p1, p2, p3] X t∈P(p) L(t) t , where we have written b = [p1, p2, p3]t and used Lemma 7.9 (ii) again. Considering separately the three cases (p1 = p2 = p3, two of the pi equal, all pi distinct), we find that X p1,p2,p3⩽p (log p1)(log p2)(log p3) [p1, p2, p3] ≪(log p)3 by Mertens’ estimate. Extending the range of t to t ∈P(2y), we get X a∈P1 L(a) a log2(P +(a) + y2/3/a) ≪ X p⩽y1/4 1 p log2 p (log p)3 (log y)3 X t∈P(2y) L(t) t .
A final application of Mertens’ estimate concludes the proof.
□ We are now ready to embark on the proof of Proposition 7.8.
We begin with the lower bound, which is easier. Consider integers n = ap1p2b ⩽x with p1 and p2 prime, a ⩽y1/5 < p1 < p2 ⩽1 4y4/5 < P −(b), and with log(y/p1p2) ∈L (a). The last condition implies that there is a divisor d|a with d 2 ⩽ y p1p2 < d, which is equivalent to y < dp1p2 ⩽2y. Thus, for such n, τ(n, y, 2y) ⩾τ(ap1p2, y, 2y) ⩾1. In particular, y4/5 ⩽y/a < p1p2 ⩽2y, so that x/ap1p2 ⩾x/(2y6/5) ⩾ 1 2y4/5. Thus, by Exercise 3.1, for each triple a, p1, p2, the number of possible b is ⩾Φ x ap1p2 , 1 4y4/5 ≫ x ap1p2 log y .
Now L (a) is the disjoint union of intervals of length ⩾log 2, all contained in [−log 2, 1 5 log y]. For each such interval [u, v] we have by Mertens’ estimae (0.5) X u⩽log(y/p1p2)⩽v y1/5<p1<p2< 1 4 y4/5 1 p1p2 ⩾ X 8y1/5 yδ is ⩽x X n′′>yδ 1 n′′ ≪ x yδ/2 , since the number of squarefull integers below w is O(√w). If n′′ ⩽yδ, then for some f|n′′, n′ has a divisor in (y/f, 2y/f], hence (7.5) H(x, y, 2y) ⩽ X n′′⩽yδ X f|n′′ H∗ x n′′ , y f , 2y f + O x yδ/2 .
6. A global-to-local principle 77 For any n′′ and f, y/f ⩾y1−δ. Next, we show that (7.6) H∗(x1, y1, 2y1) −H∗( 1 2x1, y1, 2y1) ≪x1 X a∈P(2y1) L(a) a log2 y3/4 1 /a + P +(a) (y1−δ ⩽y1 ⩽x 1 2 +δ 1 ).
Consider squarefree n ∈( 1 2x1, x1] with a divisor in (y1, 2y1]. Put z1 = 2y1, y2 = x1 4y1 , z2 = x1 y1 . Then n = m1m2 with yi < mi ⩽zi (i = 1, 2). For some j ∈{1, 2} we have p = P +(mj) < P +(m3−j). Write n = abp, P +(a) < p < P −(b).
Then b > p and this is crucial to our argument. Since τ(ap, yj, zj) ⩾1 and y2 ⩾1 4x 1 2 −δ 1 ⩾1 4 y 1/2−δ 1/2+δ 1 ⩾y3/4 1 if δ is small enough, we have p ⩾yj/a ⩾y3/4 1 /a.
Thus, p ⩾Q(a) := max(P +(a), y3/4 1 /a). We also have p ⩽min(z1, z2) ⩽2y1. As noted earlier, b > p and so ap ⩽x1/p. Hence, by (3.8), given a and p, the number of possible b is ⩽Φ x1 ap, p ≪ x1 ap log p ⩽ x1 ap log Q(a), Since a has a divisor in (yj/p, zj/p], we have log(yj/p) ∈L (a) or log(2yj/p) ∈L (a) (the latter case is needed if j = 2 since z2 = 4y2). Since L (a) is the disjoint union of intervals of length ⩾log 2 with total measure L(a), by repeated use of Mertens’ sum estimate (0.5), we obtain X log(cyj/p)∈L (a) p⩾P +(a) 1 p ≪ L(a) log Q(a) (c = 1, 2), and (7.6) follows upon summing over all possible a and over j = 1, 2.
Write x2 = x/n′′, y1 = y/f. Each n ∈(x2/yδ 1, x2] lies in an interval (2−r−1x2, 2−rx2] for some integer 0 ⩽r ⩽δ log y1 log 2 . We note that y1 ⩾y1−δ since f ⩽n′′ ⩽yδ and also x1 = 2−rx2 ⩾x2y−δ 1 ⩾xy−2δ ⩾y2−2δ ⩾y2−2δ 1 , which implies y1 ⩽x 1 2(1−δ) 1 ⩽x 1 2 +δ 1 . Applying (7.6) with x1 = 2−rx2 and summing over r we find that H∗(x2, y1, 2y1) ≪x2 yδ 1 + x2 X a∈P(2y1) L(a) a log2 y3/4 1 /a + P +(a) .
The first term x2/yδ 1 may be ignored because L(1) = log 2 and thus the term a = 1 is ≫1/ log2 y1. Thus, by (7.5), H(x, y, 2y) ≪ x yδ/2 + x X n′′⩽yδ 1 n′′ X f|n′′ X a∈P(2y/f) L(a) a log2 (y/f)3/4/a + P +(a) .
Again, the term x/yδ/2 is negligible and may be omitted. We have (y/f)3/4 ⩾(y1−δ)3/4 ⩾y2/3 for any pair (n′′, f) and X n′′ τ(n′′) n′′ = Y p 1 + 3 p2 + 4 p3 + · · · ≪1.
This completes the proof of (7.4).
Combining (7.4) with Lemma 7.11, we find that H(x, y, 2y) ≪ 1 log2 y X a∈P(2y) L(a) a .
78 7. INTEGERS WITH A DIVISOR IN A GIVEN INTERVAL Finally, applying Lemma 7.10 with w1 = y and w2 = 2y completes the proof of the upper bound in Proposition 7.8.
7. Completion of the lower bound in Theorem 7.3 Lemma 7.12. For any finite set A of positive integers, X a∈A L(a) a ⩾(log 2) P a∈A τ(a) a 2 P a∈A W (a) a , where W(a) = |{(d, d′) : d|a, d′|a, | log d/d′| ⩽log 2}|.
Proof. Since τ(a) log 2 = Z τ(a, eu, 2eu) du = Z 1(u ∈L (a))τ(a, eu, 2eu) du and R 1(u ∈L (a)) du = L(a), by the Cauchy-Schwarz inequality, X a∈A τ(a) a !2 (log 2)2 = X a∈A 1 a Z τ(a, eu, 2eu) du !2 ⩽ X a∈A L(a) a ! X a∈A 1 a Z τ 2(a, eu, 2eu) du !
.
Now Z τ 2(a, eu, 2eu) du = Z #{d|a, d′|a, eu < d, d′ ⩽2eu} du = X d|a,d′|a max 0, log 2 −| log(d′/d)| ⩽(log 2)W(a) and the proof is complete.
□ We apply Lemma 7.12 with sets A of integers whose prime factors are localized.
To simplify later analysis, partition the primes into sets D1, D2, . . ., where each Dj consists of the primes in an interval (λj−1, λj], with λj ≈λ2 j−1. More precisely, let λ0 = 1.9 and define inductively λj for j ⩾1 as the largest prime so that (7.7) X λj−1<p⩽λj 1 p ⩽log 2.
For example, λ1 = 2, D1 = {2}, λ2 = 7 and D2 = {3, 5, 7}. The left side of (7.7) is = log 2 + O(1/λj), hence by Mertens’ estimate (0.5), (7.8) log log λj −log log λj−1 = log 2 + O(1/ log λj−1).
We claim that this implies (7.9) log λj = 2j+O(1) (j ⩾0).
To see (7.9), let wj = log log λj. Since each Dj contains at least one prime, wj →∞as j →∞and hence, for some j0, if j ⩾j0 then the big-O term in (7.8) is ⩽0.1 and wj −wj−1 ⩾log 2−0.1 ⩾1/2. It then follows that the big-O term is ≪e−wj ≪e−j/2, and thus (7.8) implies (wj −j log 2) −(wj−1 −(j −1) log 2) ≪e−j/2.
By Cauchy’s criterion, limj→∞(wj −j log 2) exists and (7.9) follows.
For a vector b = (b1, . . . , bJ) of non-negative integers, let A (b) be the set of square-free integers a composed of exactly bj prime factors from Dj for each j, and having no other prime factors.
7. Completion of the lower bound in Theorem 7.3 79 Lemma 7.13. Assume b = (b1, . . . , bJ). Then X a∈A (b) W(a) a ≪(2 log 2)b1+···+bJ b1! · · · bJ!
J X j=1 2−j+b1+···+bj.
Proof. Let B = b1 + · · · + bJ and for j ⩾0 let Bj = P i⩽j bj. Let a = p1 · · · pB, where (7.10) pBj−1+1, . . . , pBj ∈Dj (1 ⩽j ⩽J) and the primes in each interval Dj are unordered. Since W(p1 · · · pB) is the number of pairs Y, Z ⊆{1, . . . , B} with (7.11) X i∈Y log pi − X i∈Z log pi ⩽log 2, we have (7.12) X a∈A (b) W(a) a ⩽ 1 b1! · · · bJ!
X Y,Z⊆{1,...,B} X p1,...,pB (7.10),(7.11) 1 p1 · · · pB .
By (7.7), we have X pi 1 pi ⩽log 2.
When Y = Z, the inner sum on the right side of (7.12) is ⩽(log 2)B, and there are 2B such pairs Y, Z. When Y ̸= Z, let I = max[(Y ∪Z) −(Y ∩Z)]. With all the pi fixed except for pI, (7.11) implies that U ⩽pI ⩽4U for some number U. Let E(I) be defined by BE(I)−1 < I ⩽BE(I), i.e. pI ∈DE(I). By Mertens’ bound (0.5) and (7.9), X U⩽pI⩽4U pI∈DE(I) 1 pI ≪ 1 max(log U, log λE(I)−1) ≪2−E(I).
Thus, by (7.7) the inner sum in (7.12) is ≪2−E(I)(log 2)B. For each I, there are 2B−I+14I−1 = 2B+I−1 corresponding pairs Y, Z. Hence, by (7.12), X a∈A (b) W(a) a ≪(2 log 2)B b1! · · · bJ!
" 1 + B X I=1 2I−E(I) # ≪(2 log 2)B b1! · · · bJ!
1 + J X j=1 2−j X Bj−1<I⩽Bj 2I .
The sum on I on the right side is ⩽2Bj+1 = 21+b1+···+bj. When j = 1, 2−j+b1+···+bj ⩾2−1 and thus we may remove the additive term 1 on the right side above. The claimed bound follows.
□ Now suppose that y is sufficiently large, M is a sufficiently large, fixed positive integer, bi = 0 for i < M, and bj ⩽Mj for each j (this last constraint is very mild, as we expect that each Dj contains only 1 prime on average). Let k = bM + · · · + bJ. By (7.9), X a∈A (b) τ(a) a = 2k J Y j=M 1 bj!
X p1∈Dj 1 p1 X p2∈Dj p2̸=p1 1 p2 · · · X pbj ∈Dj pbj ̸∈{p1,...,pbj −1} 1 pbj ⩾2k J Y j=M 1 bj!
log 2 − bj λj−1 bj ⩾ (2 log 2)k 2bM! · · · bJ!.
(7.13) Let k = log log y log 2 −2M , J = M + k −1.
80 7. INTEGERS WITH A DIVISOR IN A GIVEN INTERVAL Let B be the set of vectors (b1, . . . , bJ) with bi = 0 for i < M and b1 + · · · + bJ = k. Let B∗be the set of b ∈B with bj ⩽Mj for each j ⩾M. If b ∈B∗and a ∈A (b), then by (7.9), P +(a) ⩽λJ ⩽exp 2J+O(1) ⩽y if M is large enough. Put (7.14) f(b) := J X h=M 2M−1−h+bM+···+bh = J X h=M 2(bM−1)+···+(bh−1).
Since (7.15) (bM −1) + · · · + (bJ −1) = k −(J −M + 1) = 0, we have f(b) ⩾1 and hence, by Lemma 7.13, X a∈A (b) W(a) a ≪(2 log 2)k bM! · · · bJ! M−1 X j=1 2−j + 21−Mf(b) !
≪(2 log 2)k bM! · · · bJ!f(b).
By Proposition 7.8, Lemma 7.12, plus (7.13), we have for large y (7.16) H(x, y, 2y) ≫x(2 log 2)k log2 y X b∈B∗ 1 bM! · · · bJ!f(b).
Roughly speaking, f(b) ≈g(b) := max j 2(bM−1)+···+(bj−1).
Observe that the product of factorials is unchanged under permutation of bM, . . . , bJ. Given real numbers z1, · · · , zk with zero sum, there is a cyclic permutation z′ of the vector z = (z1, . . . , zk) all of whose partial sums are ⩾0: let i be the index minimizing z1 + · · · + zi and take z′ = (zi+1, . . . , zk, z1, . . . , zi).
In combinatorics, this fact is know as the cycle lemma. Thus, there is a a cyclic permutation b′ of b with g(b′) = 1. Thus, we expect that 1/f(b′) will be ≫1/k on average over b′ and that 1/f(b) ≫1/k on average over b ∈B. This is essentially what we prove next; see (7.18) below.
Lemma 7.14. For positive real numbers x1, . . . , xr with product X, let xr+i = xi for i ⩾1. Then r−1 X j=0 r X h=1 x1+j · · · xh+j !−1 ∈ 1 max(1, X), 1 min(1, X) .
Proof. Put y0 = 1 and yj = x1 · · · xj for j ⩾1. The sum in question is r−1 X j=0 r X h=1 yh+j yj !−1 = r−1 X j=0 yj y1+j + · · · + yr+j .
Since yr = X, y1+j + · · · + yr+j = X(y0 + · · · + yj) + y1+j + · · · + yr−1 ∈[min(1, X)(y0 + · · · + yr−1), max(1, X)(y0 + · · · + yr−1)].
□ We have (7.17) X b∈B∗ 1 bM! · · · bJ!f(b) ⩾S0 − X M⩽jMj 1 bM! · · · bJ!f(b).
7. Completion of the lower bound in Theorem 7.3 81 Let xi = 2bM−1+i−1 for 1 ⩽i ⩽k, and xi = xi−k for i > k. By (7.14) and (7.15), x1 · · · xk = 1 and f(b) = x1 + x1x2 + · · · + x1x2 · · · xk.
By Lemma 7.14 and the multinomial theorem, (7.18) S0 = X b∈B 1 bM! · · · bJ!
1 k k−1 X j=0 k X h=1 x1+j · · · xh+j !−1 = kk−1 k! .
To bound S1(j), apply Lemma 7.14 with xi = 2bj+i−1 for 1 ⩽i ⩽J −j. By (7.15) and bj > jM we have X := x1 · · · xJ−j = 2(bj+1−1)+···+(bJ−1) = 2−(bM−1)−···−(bj−1) ⩽2j−M+1−Mj < 1.
From the definition of J and our assumption j ⩽k/M, we have J −j ⩾k/2. Write b′ = (bM, . . . , bj−1, bj+1, . . . , bJ), whose sum of components is k −bj. We will sum over cyclic permutations of (bj+1, . . . , bJ) using Lemma 7.14. Ignoring the terms with h ⩽j in (7.14), we have f(b) ⩾ J X h=j+1 2M−1−h+bM+···+bh = J X h=j+1 2M−1−j+bM+···+bj+(bj+1−1)+···+(bJ−1) = 2M−1−j+bM+···+bj(x1 + x1x2 + · · · x1 · · · xJ−j).
Since the variables bi are unrestricted for i ̸= j, and have sum k −bj, we get S1(j) ⩽ X bj>Mj 1 bj!
X b′ 1 Q i̸=j bi! 2M−1−j+bM+·+bj · 1 J −j J−j−1 X i=0 J−j X h=1 x1+i · · · xh+i !−1 ⩽ X bj>Mj 1 bj!
X b′ 1 Q i̸=j bi! 2M−1−j+bM+·+bj · 1 (J −j)X = 2k J −j X bj>Mj 1 2bjbj!
X b′ Y i̸=j 2−bi bi!
= 2k J −j X bj>Mj 1 2bjbj!
k−1 2 k−bj (k −bj)! , 82 7. INTEGERS WITH A DIVISOR IN A GIVEN INTERVAL using the multinomial theorem in the last step. We conclude that S1(j) ⩽2 k X bj>Mj (k −1)k−bj bj!(k −bj)!
= 2(k −1)k−1 k · k!
X bj>Mj k(k −1) · · · (k −bj + 1) bj!(k −1)bj−1 ⩽2(k −1)k−1 k!
X bj>Mj 1 bj!
⩽kk−1 k!
2 (Mj)!.
Hence, if M ⩾2 then (7.19) X M⩽j k0 and use the bound L(a) ⩽log(2a) ≪log(a) (since a ⩾2). We have Tk ≪ X a∈P(y) ω(a)=k 1 a X p|a log p ⩽ X p⩽y log p p X b∈P(y) ω(b)=k−1 1 b ≪(log y)(log2 y + O(1))k−1 (k −1)!
.
Since log2 y+O(1) k−1 ⩾1/ log 2 −o(1) ⩾1.3 for large y, the sum over k is dominated by the smallest term k = k1 := k0 + ⌈10 log3 y⌉. Hence X k⩾k0+10 log3 y Tk ≪(log y)(log2 y + O(1))k1−1 (k1 −1)!
≪(log y)(log2 y)k1−1 (k1 −1)!
≪(log y)(log2 y)k0 k0!
1 log 2 k1−1−k0 ≪(log y)2−E p log2 y (log2 y)10 log log 2 ≪(log y)2−E (log2 y)3 .
□ Remarks. By the same argument applied to all k we deduce X k Tk ≪(log y)2−E p log2 y , which is too big by a factor log2 y. The correct order is achieved by using the more sophisticated bound for L(a) given by Lemma 7.9 (iii). In particular, this captures when there are an unusually large number of small prime factors of a, this forcing L(a) to be small.
For k near k0, we bound Tk in terms of a mutivariate integral. Since P p⩽z 1/p = log log z + O(1), by partial summation we expect for “nice” functions f that X p1<···<pk⩽y f log2 p1 log2 y , · · · , log2 pk log2 y p1 · · · pk ≈(log2 y)k Z · · · Z 0⩽ξ1⩽···⩽ξk⩽1 f(x) dx.
84 7. INTEGERS WITH A DIVISOR IN A GIVEN INTERVAL For a = p1 . . . pk, the function L(a) is not very regular as a function of p1, . . . , pk. However, the most common way for L(a) to be small is for a to have many small prime factors, and the bound in Lemma 7.9 (iii) captures this nicely. Moreover, this bound has the useful property of being monotone in each variable pi Lemma 7.16. Suppose y is large and k0/2 ⩽k ⩽2k0. Then Tk ≪(2 log2 y)kUk(k0), Uk(v) = Z · · · Z 0⩽ξ1⩽···⩽ξk⩽1 min 0⩽j⩽k 2−j(1 + 2vξ1 + · · · + 2vξj) dx.
Proof. Recall the definition of λi, Di from the previous section. Consider a = p1 · · · pk, where p1 < · · · < pk ⩽y, and define ji by pi ∈Dji (1 ⩽i ⩽k). By (7.9), pi ∈Dj implies that log pi ≪2j. By (7.9), 0 ⩽ji ⩽k1 for each i, where k1 = k0 + O(1). By Lemma 7.9 (iii), L(a) ⩽2k+1 min 0⩽g⩽k 2−g log(2p1 · · · pg) ≪2kF(j), where F(j) = min 0⩽g⩽k 2−g(1 + 2j1 + · · · + 2jg).
For each 1 ⩽j ⩽k1, let bj = #{i : pi ∈Dj} Then Tk ≪2k X 1⩽j1⩽···⩽jk⩽k1 F(j) k1 Y j=1 (log 2)bj b1!
≪(2 log 2)k b1! · · · bk1!
X j F(j).
Extend the domain of F to include k-tuples of non-negative real numbers. It is clear that if ji −1 ⩽ti ⩽ji for each i, then F(j) ⩽2F(t). Therefore, writing Bi = b1 + · · · + bi as before, 1 b1! · · · bk1!
X j F(j) = X j F(j) k1 Y j=1 Z · · · Z j−1⩽tBj−1+1⩽···⩽tBj ⩽j 1dt ⩽2 Z · · · Z 0⩽t1⩽···⩽tk⩽k1 F(t) dt.
Making the change of variables ti = k1ξi for each i, we see that the multiple integral on the right side equals kk 1 Z · · · Z 0⩽ξ1⩽···⩽ξk⩽1 min 0⩽g⩽k 2−g 1 + 2k1ξ1 + · · · + 2k1ξg dξ.
Recalling that k1 = k0 + O(1), we conclude that Tk ≪(2k0 log 2 + O(1))kUk(k0).
Lastly, (2k0 log 2 + O(1))k ≪(2 log log y)k since k ⩽2k0, and the lemma follows.
□ Estimating Uk(v) is the most complex part of the argument. For comparison purposes, observe that the region of integration has volume 1/k! and that the integrand is roughly ≪min(1, 2v−k). This leads to an upper bound roughly like Uk(v) ≪ 1 k!(2k−v + 1).
The next lemma will be proved in the next section. In the case |k −v| small, this improves upon the trivial bound by a factor 1/k.
Lemma 7.17. Suppose k, v are integers with 0 ⩽k ⩽2v. Then Uk(v) ≪ 1 + |v −k|2 (k + 1)!(2k−v + 1).
9. Upper bound, part II 85 Proof of Theorem 7.3, upper bound, assuming Lemma 7.17. By Lemmas 7.16 and 7.17, X k0⩽k⩽k0+10 log3 y Tk ≪ X k0⩽k⩽2k0 2k0((k −k0)2 + 1) (log2 y)k (k + 1)! ≪(2 log2 y)k0 (k0 + 1)!
≍(log y)2−E (log2 y)3/2 .
and X k0−10 log3 y⩽k⩽k0 Tk ≪ X k⩽k0 ((k0 −k)2 + 1)(2 log2 y)k (k + 1)!
≪(2 log2 y)k0 (k0 + 1)!
≍(log y)2−E (log2 y)3/2 .
Combining these with Lemma 7.15 and Proposition 7.8 completes the proof.
□ 9. Upper bound, part II The goal of this section is to prove Lemma 7.17, and thus complete the proof of the upper bound in Theorem 7.3.
Let Y1, . . . , Yk be independent, uniformly distributed random variables in [0, 1]. Let ξ1 be the smallest of the numbers Yi, let ξ2 be the next smallest, etc., so that 0 ⩽ξ1 ⩽· · · ⩽ξk ⩽1. The numbers ξi are the order statistics for Y1, . . . , Yk. Then k!Uk(v) is the expectation of the random variable X = min 0⩽j⩽k 2−j(1 + 2vξ1 + · · · + 2vξj).
Heuristically, we expect that ξi ≈i/k, and so when k ≈v we guess that (7.20) E X ≪E min 1⩽j⩽k 2−j+vξj.
It remains to understand the distribution of min1⩽j⩽k vξj −j. Let Qk(u, v) be the probability that ξi ⩾i−u v for every i, that is, vξi −i ⩾−u for all i. In the special case v = k, Smirnov in 1939 showed that Qk(x √ k, k) ∼1 −e−2x2 for each fixed x. The corresponding probability estimate for two-sided bounds on the ξi was established by Kolmogorov in 1933 and together these limit theorems are the basis of the Kolmogorov-Smirnov goodness-of-fit statistical tests.
In the next lemma, we prove a uniform estimate for Qk(u, v) from . Stronger asymptotics are known, e.g. . The remainder of the section is essentially devoted to proving (7.20). The details are complicated, but the basic idea is that if 2−j(2vξ1 +· · ·+2vξj) is much large than 2vξj−j, then for some large l, the numbers ξj−l, . . . , ξj are all very close to one another. As shown below in Lemmas 7.21 and 7.22, this is quite rare.
Lemma 7.18. Let w = u + v −k. Uniformly in u ⩾0 and w ⩾0, we have Qk(u, v) ≪(u + 1)(w + 1)2 k .
Proof. Without loss of generality, suppose k ⩾100, u ⩽k/10 and w ⩽ √ k. If min1⩽i⩽k(ξi −i−u v ) < 0, let l be the smallest index with ξl < l−u v and write ξl = l−u−λ v , so that 0 ⩽λ ⩽1. Let Rl(λ) = Vol 0 ⩽ξ1 ⩽· · · ⩽ξl−1 ⩽l −u −λ v : ξi ⩾i −u v (1 ⩽i ⩽l −1) .
Then we have Qk(u, v) = 1 −k!
v Z 1 0 X u+λ⩽l⩽k Rl(λ) Vol l −u −λ v ⩽ξl+1 ⩽· · · ⩽ξk ⩽1 dλ = 1 −k!
v Z 1 0 X u+λ⩽l⩽k Rl(λ) (k −l)!
k + w + λ −l v k−l dλ.
86 7. INTEGERS WITH A DIVISOR IN A GIVEN INTERVAL Now suppose that ξk ⩽1−2w+2 v = k−u−w−2 v . Then min1⩽i⩽k(ξi −i−u v ) < 0. Defining l and λ as before, we have 1 −2w + 2 v k = k! Vol 0 ⩽ξ1 ⩽· · · ⩽ξk ⩽1 −2w + 2 v = k!
v Z 1 0 X u+λ⩽l⩽k−w−2+λ Rl(λ) (k −l)!
k −l −w −2 + λ v k−l dλ.
Thus, for any A > 0, we have Qk(u, v) = 1 −A 1 −2w + 2 v k −k!
v Z 1 0 X k−w−2+λ 0 and b > 0. Then n X k=0 n k (a + k)k−1(b + n −k)n−k−1 = 1 a + 1 b (n + a + b)n−1.
Proof. From Riordan, p. 18–20. Define An(x, y; p, q) = n X k=0 n k (x + k)k+p(y + n −k)n−k+q where p, q ∈Z, x > 0 and y > 0. The formula in Lemma 7.19 is the case p = q = −1. We first observe that replacing k with n −k yields (7.21) An(x, y; p, q) = An(y, x; q, p) 9. Upper bound, part II 87 Next, by the Pascal relation n k = n−1 k−1 + n−1 k , we get (setting h = k −1 in the first sum) An(x, y; p, q) = n−1 X h=0 n −1 h (x + 1 + h)h+p+1(y + (n −1) −h)n−1−h+q + n−1 X k=0 n −1 k (x + k)k+p(y + 1 + (n −1) −k)(n−1)−k+q+1 = An−1(x + 1, y; p + 1, q) + An−1(x, y + 1; p, q + 1).
(7.22) Another identity is obtained by splitting offone factor x + k, thus An(x, y; p, q) = n X k=0 n k (x + k)(x + k)k−1+p(y + n −k)n−k+q = xAn(x, y; p −1, q) + n n X k=1 n −1 k −1 (x + k)k−1+p(y + n −k)n−k+q = xAn(x, y; p −1, q) + nAn−1(x + 1, y; p, q).
Applying (7.22) to the first term we get (7.23) An(x, y; p, q) = xAn−1(x, y + 1; p −1, q + 1) + (x + n)An−1(x + 1, y; p, q).
For brevity, write Bn(x, y) = An(x, y; −1, 0). Taking p = 0, q = −1 in (7.23) and applying (7.21) we deduce that Bn(x, y) = An(y, x; 0, −1) = yAn−1(y, x + 1; −1, 0) + (y + n)An(y + 1, x; 0, −1) = yBn−1(y, x + 1) + (y + n)Bn−1(x, y + 1).
(7.24) It follows from (7.24) by an easy induction on n ∈N that Bn(x, y) = x−1(x + y + n)n (n ∈N, x > 0, y > 0).
Inserting this into (7.22) and using (7.21) we deduce that An(x, y; −1, −1) = An−1(x + 1, y; 0, −1) + An−1(x, y + 1; −1, 0) = Bn−1(y, x + 1) + Bn−1(x, y + 1) = (1/x + 1/y)(x + y + n)n−1, and this completes the proof of the lemma.
□ Lemma 7.20. If t ⩾2, b ⩾0 and t + a + b > 0, then X 1⩽j⩽t−1 j+a>0 t j (a + j)j−1(b + t −j)t−j−1 ⩽e4(t + a + b)t−1.
Proof. Let Ct(a, b) denote the sum in the lemma. We may assume that a > 1−t, otherwise Ct(a, b) = 0.
If a ⩾−1, put A = max(1, a) and B = max(1, b). By Lemma 7.19, Ct(a, b) ⩽Ct(A, B) ⩽ 1 A + 1 B (t + A + B)t−1 ⩽2(t + a + b + 3)t−1 ⩽2e 3(t−1) t+a+b (t + a + b)t−1 < e4(t + a + b)t−1.
(7.25) Next assume a < −1. For c > 0, (1 −c/x)x is an increasing function for x > c, thus we have (a + j)j−1 = (j −1)j−1 1 + a + 1 j −1 j−1 ⩽(j −1)j−1 1 + a + 1 t −1 t−1 .
88 7. INTEGERS WITH A DIVISOR IN A GIVEN INTERVAL Thus, by (7.25), Ct(a, b) ⩽ t + a t −1 t−1 Ct(−1, b) ⩽e4 (t + a)(t + b −1) t −1 t−1 = e4 t + a + b + (a + 1)b t −1 t−1 ⩽e4(t + a + b)t−1.
□ We now complete the proof of the upper bound for H(x, y, 2y) in Theorem 7.3. Recall our heuristic that min 1⩽j⩽k 2−j 2vξ1 + · · · + 2vξj ≈min 1⩽j⩽k 2−j+vξj.
This is violated if the minimum occurs at j = j∗and there are many ξi clustered near ξj∗. The next lemma captures the likelihood of such an event.
Lemma 7.21. Suppose g, k, s, u, v ∈Z satisfy 1 ⩽g ⩽k −1, s ⩾0, v ⩾k/2, u ⩾0, u + v ⩾k + 1.
Let R be the set of (ξ1, . . . , ξk) satisfying (7.26) 0 ⩽ξ1 ⩽· · · ⩽ξk ⩽1, ξi ⩾i −u v (1 ⩽i ⩽k) and such that, for some l ⩾max(g + 1, u), we have (7.27) l −u v ⩽ξl ⩽l −u + 1 v , ξl−g ⩾l −u −s v .
Then Vol(R) ≪g2(2(s + 1))g g!
(u + 1)(u + v −k)2 (k + 1)!
.
Proof. Fix l satisfying max(u, g + 1) ⩽l ⩽k. Let Rl be the subset of ξ satisfying (7.26) and (7.27) for this particular l. We have Vol(Rl) ⩽V1V2V3V4, where, by Lemma 7.18, V1 = Vol{0 ⩽ξ1 ⩽· · · ⩽ξl−g−1 ⩽l−u+1 v : ξi ⩾i−u v ∀i} = l −u + 1 v l−g−1 Vol{0 ⩽θ1 ⩽· · · ⩽θl−g−1 ⩽1 : θi ⩾ i−u l−u+1 ∀i} = l −u + 1 v l−g−1 Ql−g−1(u, l −u + 1) (l −g −1)!
≪ l −u + 1 v l−g−1 (u + 1)g2 (l −g)! , V2 = Vol{ l−u−s v ⩽ξl−g ⩽· · · ⩽ξl−1 ⩽l−u+1 v } = 1 g!
s + 1 v g , V3 = Vol{ l−u v ⩽ξl ⩽l−u+1 v } = 1 v , V4 = Vol{ξl+1 ⩽· · · ⩽ξk ⩽1 : ξi ⩾i−u v ∀i} (note that ξl+1 ⩾l+1−u v ) = u + v −l −1 v k−l Qk−l(0, u + v −l −1) (k −l)!
≪ u + v −l v k−l (u + v −k)2 (k −l + 1)! .
9. Upper bound, part II 89 Thus Vol(R) ≪(s + 1)g(u + 1)g2(u + v −k)2 g!vk(k + 1 −g)!
X l⩾g+1 l⩾u k + 1 −g l −g (l −u + 1)l−g−1(u + v −l)k−l.
By Lemma 7.20 (with t = k + 1 −g, a = g + 1 −u, b = u + v −k −1, j = l −g), the sum on l is ≪(v + 1)k−g ≪vk−g = vk (k + 1)g k + 1 v g ≪vk2g (k −g + 1)!
(k + 1)!
and the lemma follows.
□ To bound Uk(v), we will bound the volume of the set T (k, v, m) = {ξ ∈Rk : 0 ⩽ξ1 ⩽· · · ⩽ξk ⩽1, 2vξ1 + · · · + 2vξj ⩾2j−m (1 ⩽j ⩽k)}.
Lemma 7.22. Suppose k, v, m are integers with 1 ⩽k ⩽2v and m ⩾0. Set b = k −v. Then Vol(T (k, v, m)) ≪ Y 22b−m(k + 1)!, Y = ( b if b ⩾m + 5 (m + 5 −b)2(m + 1) if b ⩽m + 4 .
Proof. Let r = max(5, b −m) and ξ ∈T (k, v, m). Then either (7.28) ξj > j−m−r v (1 ⩽j ⩽k) or (7.29) min 1⩽j⩽k(ξj −j−m v ) = ξl −l−m v ∈[ −h v , 1−h v ] for some integers h ⩾r + 1, 1 ⩽l ⩽k.
Let V1 be the volume of ξ ∈T (k, v, m) satisfying (7.28). If b ⩾m + 5, then (7.28) is not possible since (7.28) implies that ξk > k −m −r v = k −b v = 1.
Thus, if (7.28) holds then b ⩽m + 4 and r = 5. By Lemma 7.18, V1 ⩽Qk(m + 5, v) k!
≪(m + 6)(m + 5 −b)2 (k + 1)!
≪ Y 22b−m(k + 1)!.
If (7.29) holds, then there is an integer t satisfying (7.30) t ⩾h −3, ξl−2t ⩾l−m−2t v .
To see (7.30), suppose such an t does not exist. Then 2vξ1 + · · · + 2vξl ⩽2h−32vξl + X t⩾h−3 2t · 2vξl−2t ⩽2h−32l−m−h+1 + X t⩾h−3 2t2l−m−2t ⩽2l−m−1, a contradiction. Let V2 be the volume of ξ ∈T (k, v, m) satisfying (7.29). Fix h and t satisfying (7.30) and use Lemma 7.21 with u = m + h, g = 2t, s = 2t. The volume of such ξ is ≪(m + h + 1)(m + h −b)2 (k + 1)!
(4t + 2)2t22t (2t)!
≪(m + h + 1)(m + h −b)2 22t+3(k + 1)!
.
The sum of 2−2t+3 over t ⩾h −3 is ≪2−2h. Summing over h ⩾r + 1 gives V2 ≪(m + r + 2)(m −b + r + 2)2 22r+1(k + 1)!
≪ Y 22b−m(k + 1)!.
□ 90 7. INTEGERS WITH A DIVISOR IN A GIVEN INTERVAL Proof of Lemma 7.17. Assume k ⩾1, since the lemma is trivial when k = 0. Put b = k −v. For integers m ⩾1, consider ξ satisfying 2−m+1 ⩽min 0⩽j⩽k 2−j 1 + 2vξ1 + · · · + 2vξj < 2−m+2.
For such ξ and for 1 ⩽j ⩽k we have 2−j 2vξ1 + · · · + 2vξj ⩾max(2−j, 21−m −2−j) ⩾2−m, so that ξ ∈T (k, v, m). By Lemma 7.22, Uk(v) ⩽ X m⩾1 22−m Vol(T (k, v, m)) ≪ 1 (k + 1)!
X m⩾1 2−mYm 22b−m , Ym = ( b if m ⩽b −5 (m + 5 −b)2(m + 1) if m ⩾b −4 .
Next, X m⩾1 2−mYm 22b−m ≪ X 1⩽m⩽b−5 b 2m22b−m + X m⩾max(1,b−4) (m + 5 −b)2(m + 1) 2m .
The proof is completed by noting that if b ⩾5, each sum on the right side is ≪b2−b and if b ⩽4, the first sum is empty and the second is ≪(5 −b)2 ≪1 + b2. This completes the proof of Lemma 7.17.
□ 10. Counting integers with a given number of divisors in an interval Recall τ(n, y, z) = #{d|n : y < d ⩽z} and define Hr(x, y, z) = {n ⩽x : τ(n, y, z) = r}.
Of particular interest is the case r = 1. Similarly, the exact formula Hr(x, y, z) = X k⩾r (−1)k−r k r X y<d1<···<dk⩽z x lcm[d1, · · · , dk] implies the existence of εr(y, z) = lim x→∞ Hr(x, y, z) x .
for every fixed pair y, z.
In , it is proved that for every r ⩾1, Hr(x, y, 2y) ≫r H(x, y, 2y), that is, a positive proportion of all integers n ⩽x with a divisor in (y, 2y] have exactly r such divisors. In particular, this disproved a 1960 conjecture of Erdős.
11. Exercises Exercise 7.1. Fix c′ < 1 < c. Suppose that 3 ⩽y ⩽√x. (a) Show that #{c′x < n ⩽x : µ2(n) = 1, τ(n, y, cy) ⩾1} ≫c,c′ 1 log2 y X a∈P(y) L(a) a .
(b) Use (a) to prove Theorem 7.4.
Exercise 7.2. If 1 ⩽y ⩽x1/ log2 x, show that H(x, y, 2y) x ∼ε(y, 2y) (x →∞).
CHAPTER 8 Permutations with a fixed set of a given size 1. Introduction and notation Let k, n be integers with 1 ⩽k ⩽n/2. What is i(n, k), the probability that a random σ ∈Sn fixes some set of size k? Equivalently, what is the probability that the cycle decomposition of σ contains disjoint cycles with lengths summing to k? This is analogous to the problem of bounding H(x, y, 2y) from the previous Chapter, and we will develop a parallel theory based on the same ideas.
The size of i(n, k) has only recently been at all well understood. The lower bound limn→∞i(n, k) ≫ log k/k is contained in a paper of Diaconis, Fulman and Guralnick in 2008, while the upper bound i(n, k) ≪k−1/100 is due to Łuczak and Pyber in 1993 (These authors did not make any special effort to optimise the constant 1/100, but their method does not lead to a sharp bound.) Pemantle, Peres, and Rivin [57, Theorem 1.7] proved that limn→∞i(n, k) = k−E+o(1), where as before, E = 1 −1 + log log 2 log 2 ≈0.08607.
Theorem 8.1 (Eberhard–Ford–Green, 2016). For 1 ⩽k ⩽n/2, i(n, k) ≍ 1 kE(1 + log k)−3/2 .
Since i(n, n −k) = i(n, k), Theorem 8.1 establishes the order of i(n, k) for all n, k.
Theorem 8.1 has implications for a conjecture of Cameron related to random generation of the sym-metric group. Cameron conjectured that the proportion of σ ∈Sn contained in a transitive subgroup not containing An tends to zero: this was proved by Łuczak and Pyber using their bound i(n, k) ≪k−1/100.
Cameron further guessed that this proportion might decay as fast as n−1/2+o(1) (see [51, Section 5]). However Theorem 8.1 has the following corollary.
Corollary 8.2. The proportion of σ ∈Sn contained in a transitive subgroup not containing An is ≫n−E(log n)−3/2, provided that n is even and greater than 2.
Proof. By Theorem 8.1 the proportion of σ ∈Sn fixing a set B1 of size n/2 is ≍n−E(log n)−3/2. Such a permutation σ must also fix the set B2 = {1, . . . , n} \ B1, and thus preserve the partition {B1, B2} of {1, . . . , n}. Since |B1| = |B2|, the set of all τ preserving this partition is a transitive subgroup not containing An.
□ In by Eberhard, Ford and Koukoulopoulos, it is shown (among other things) that the proportion of σ ∈Sn contained in a transitive subgroup not containing An is ≍n−E(log n)−3/2, provided that n is even and greater than 2.
Whether or not a permutation σ has a fixed set of size k depends only on the vector C = (C1(σ), C2(σ), . . . , Ck(σ)) listing the number of cycles of length 1, 2, . . . , k, respectively, in σ. Crucial to our argument is the fact (e.g., Theorem 4.2) that C has limiting distribution (as n →∞) equal to Xk = (X1, X2, . . . , Xk), where the Xi are independent and Xi has Poisson distribution with parameter 1/i (for short, Xi d = Pois(1/i)).
91 92 8. PERMUTATIONS WITH A FIXED SET OF A GIVEN SIZE A simple corollary is that the limit i(∞, k) = limn→∞i(n, k) exists for every k. Define, for any finite list c = (c1, c2, . . . , ck) of non-negative integers, the quantity (8.1) L (c) = {m1 + 2m2 + · · · + kmk : 0 ⩽mj ⩽cj for j = 1, 2, . . . , k .
We immediately obtain that (8.2) i(∞, k) = P(k ∈L (Xk)).
This makes it easy to compute i(∞, k) for small values of k. For example we have the extremely well known result (derangements) that i(∞, 1) = P(X1 ⩾1) = 1 −1 e ≈0.6321, and the less well known fact that i(∞, 2) = 1 −P(X1 = X2 = 0) −P(X1 = 1, X2 = 0) = 1 −2e−3/2 ≈0.5537.
When k is allowed to grow with n, the vector C is still close to being distributed as Xk, the total variation distance between the two distributions decaying rapidly as n/k →∞(Theorem 4.2). This fact is, however, not strong enough for our application. We must establish an approximate analog of (8.2), showing that i(n, k) has about the same order as P(k ∈L (Xk)), uniformly in k ⩽n/2.
Instead of directly estimating the probability of a single number lying in L (Xk), however, we apply a global-to-local principle analogous to Proposition 7.8.
Proposition 8.3. i(n, k) ≍1 kE |L (Xk)| uniformly for 40 ⩽k ⩽n/2.
Theorem 8.1 follows immediately from this and the next proposition, when k ⩾40.
Proposition 8.4. E |L (Xk)| ≍k1−E(1 + log k)−3/2.
When 1 ⩽k ⩽39, we argue more directly. If n = 2k then Cauchy’s formula (Lemma 1.2) implies that i(2k, k) ⩾Pσ∈Sn(Ck(σ) = 2) = 1 2k2 ≫1.
If n > 2k then Lemma 8.5 gives i(n, k) ⩾Pσ∈Sn Ck(σ) = 1, Cj(σ) = 0 (j < k) ⩾ 1 k(2k + 1) ≫1.
Thus, Theorem 8.1 follows when k ⩽39 as well.
Question 1. Is there some constant C such that i(∞, k) ∼Ck−E(log k)−3/2?
Question 2. Is i(∞, k) monotonically decreasing in k?
Data collected by Britnell and Wildon shows that this is so at least as far as i(∞, 30), and of course a positive answer is plausible just from the fact that i(∞, k) →0.
2. The global-to-local principle In this section we prove Proposition 8.3.
Lemma 8.5. Let 1 ⩽m < n and c1, . . . , cm be non-negative integers satisfying c1 + 2c2 + · · · + mcm ⩽n −m −1.
Then 1 2m + 1 m Y i=1 (1/i)ci ci!
⩽Pσ(C1(σ) = c1, . . . , Cm(σ) = cm) ⩽ 1 m + 1 m Y i=1 (1/i)ci ci!
.
Proof. Let t = c1 + 2c2 + · · · + mcm. If Cj(σ) = cj for 1 ⩽j ⩽m, write σ = σ1σ2, where all the cycles in σ1 have length ⩽m and all the cycles in σ2 have length > m. Applying Cauchy’s Theorem (Lemma 1.2) for σ1 and (3.5) for σ2 completes the proof.
□ 2. The global-to-local principle 93 As in the introduction, let X1, X2, . . . be independent random variables with distribution Xj d = Pois(1/j).
We record here that (8.3) E |L (Xk)| = X c1,...,ck⩾0 |L (c)|P(X1 = c1) · · · P(Xk = ck) = e−Hk X c1,...,ck⩾0 |L (c)| Qk i=1 ci!ici .
Lemma 8.6. Let k ∈N, c1, . . . , ck ⩾0, I ⊂[k] and c′ i = ci for i ̸∈I, c′ i = 0 for i ∈I. then |L (c)| ⩽|L (c′)| Y i∈I (ci + 1).
Proof. Clearly, L (c) is the union of Q i∈I(ci + 1) translates of L (c′).
□ Lemma 8.7. Suppose that ℓ′ ⩽ℓ. Then E |L (Xℓ)| ⩽ℓ+ 1 ℓ′ + 1E |L (Xℓ′)|.
Proof. By Lemma 8.6, |L (Xℓ)| ⩽(1 + Xℓ′+1) · · · (1 + Xℓ)|L (Xℓ′)|. Thus by independence, E |L (Xℓ)| ⩽ ℓ Y i=ℓ′+1 E (1 + Xi) E |L (Xℓ′)| = ℓ+ 1 ℓ′ + 1E |L (Xℓ′)|.
□ Lemma 8.8. For any j ⩽k we have E |L (Xk)|Xj ⩽3 j E |L (Xk)|.
Suppose that j1, . . . , jh ⩽k are distinct integers and that a1, . . . , ah are positive integers. Then E |L (Xk)|Xa1 j1 · · · Xah jh ≪a1,...,ah 1 j1 . . . jh E |L (Xk)|.
Proof. Define X′ k by putting X′ j1 = · · · = X′ jh = 0 and X′ j = Xj for all other j. By Lemma 8.6, we have |L (Xk)| ⩽|L (X′ k)|(1 + Xj1) · · · (1 + Xjh) ⩽|L (Xk)|(1 + Xj1) · · · (1 + Xjh).
Thus by independence E |L (Xk)|Xa1 j1 · · · Xah jh ⩽E |L (Xk)| h Y i=1 (E Xai ji + E Xai+1 ji ).
When h = a1 = 1 and j1 = j, we have E Xj + X2 j = 2E X 2 + 2E X = 1 j2 + 2 j ⩽3 j .
In general, for X d = Pois(λ) with λ ⩽1 we have E Xm ≪m λ and the lemma follows.
□ We turn now to the proof of Proposition 8.3. In what follows write S(Xℓ) = X1 + 2X2 + · · · + ℓXℓ= max L (Xℓ).
We will treat the lower bound and upper bound in Proposition 8.3 separately, the former being somewhat more straightforward than the latter.
Proof of Proposition 8.3 (Lower bound). When k ⩾40, let r = ⌊k/20⌋, so that r ⩾2, and consider the permutations σ = ασ1σ2β ∈Sn, where σ1 and σ2 are cycles, |α| ⩽4r < |σ1| < |σ2| < 16r, all cycles in α have length ⩽r, all cycles in β have length at least 16r, and α has a fixed set of size k−|σ1|−|σ2|.
If Ci(α) = ci for 1 ⩽i ⩽r, then the last condition is equivalent to k −|σ1| −|σ2| ∈L (c). In particular |σ1| + |σ2| ⩽k, and hence n −|α| −|σ1| −|σ2| ⩾4 5k ⩾k −4r ⩾16r.
94 8. PERMUTATIONS WITH A FIXED SET OF A GIVEN SIZE Fix c and ℓ1, ℓ2 with 4r < ℓ1 < ℓ2 < 16r such that k −ℓ1 −ℓ2 ∈L (c). By Proposition 8.5, the probability that a random σ ∈Sn has ci cycles of length i (1 ⩽i ⩽r), one cycle each of length ℓ1, ℓ2 and no other cycles of length < 16r is at least ≫ 1 rℓ1ℓ2 Qr i=1 ci!ici ≫ 1 r3 Qr i=1 ci!ici .
Now L (c) ⊂[0, 4r]. Hence, for any ℓ1 satisfying 4r + 1 ⩽ℓ1 ⩽8r −1, there are |L (c)| admissible values of ℓ2 > ℓ1 for which k −ℓ1 −ℓ2 ∈L (c). We conclude that i(n, k) ≫1 r2 X c1,··· ,cr⩾0 S(c)⩽4r |L (c)| r Y i=1 (1/i)ci ci!
.
As in (8.3), the sum above equals eHrE |L (Xr)|1(S(Xr) ⩽4r). Hence, by , we see that i(n, k) ≫1 r E |L (Xr)|1(S(Xr) ⩽4r).
To estimate this, we use the inequality 1(S(Xr) ⩽4r) ⩾1 −S(Xr) 4r .
By Lemma 8.8 (first part) we have E |L (Xr)|S(Xr) = r X j=1 E |L (Xr)|jXj ⩽3rE |L (Xr)|.
It follows that i(n, k) ≫1 r E |L (Xr)|.
Finally, the lower bound in Proposition 8.3 is a consequence of this and Lemma 8.7.
□ Proof of Proposition 8.3 (Upper bound). Temporarily impose a total ordering on the set of all cycles fromed from subsets of [n], first ordering them by length, then imposing an arbitrary ordering of the cycles of a given length. Let σ ∈Sn have a divisor of size k. Let k1 = k and k2 = n −k. Then σ = σ1σ2, where |σ1 = k1 and |σ2| = k2. For some j ∈{1, 2}, the largest cycle in σ, with respect to our total ordering, lies in σ3−j. Let δ be the largest cycle in σj, and note that |δ| ⩽min(k1, k2) = k. Write σ = αδβ, where α is the product of all cycles dividing σ which are smaller than δ and β is the product of all cycles which are larger than σ. In particular |β| ⩾|δ| since β contains the largest cycle in σ, and thus (8.4) |δ| ⩽|β| = n −|δ| −|α|.
By definition of δ and α, α has a divisor of size kj −|δ|. Suppose |δ| = ℓ, Cj(α) = cj for j ⩽ℓand c = (c1, c2, . . . , cℓ). Then kj −ℓ∈L (c). For ℓand c satisfying this last condition, we count the number of possible pairs α, δ using Lemma 1.2 to first count the number of products αδ (noting that Cℓ(αδ) = cℓ+ 1)) and then counting the number of possible ways to choose δ (which equals cℓ+ 1). The total count is n n −|α| −ℓ (|α| + ℓ)!
Y i<ℓ (1/i)ci ci!
× 1 (cℓ+ 1)!ℓcℓ+1 × (cℓ+ 1) = n!
ℓ(n −|α| −ℓ)!
Y i⩽ℓ (1/i)ci ci!
.
Given α and δ, (8.4) and (3.5) imply that the number of choices for β is at most (n −|α| −ℓ)!/ℓ. Thus i(n, k) ⩽ 2 X j=1 k X ℓ=1 1 ℓ2 X c1,...,cℓ⩾0 kj−ℓ∈L (c) Y i⩽ℓ (1/i)ci ci!
= 2 X j=1 X c1,...,ck⩾0 Y i⩽k (1/i)ci ci!
X m(c)⩽ℓ⩽k kj−ℓ∈L (c) 1 ℓ2 , 2. The global-to-local principle 95 where m(c) = max{i : ci > 0} ∪{1}. With c fixed, note that ℓ⩾max(m(c), kj −S(c)). Also, the number of ℓsuch that kj −ℓ∈L (c) is at most |L (c)|. Thus, the innermost sum on the right side above is at most |L (c)| max m(c), kj −S(c) 2 .
Like (8.3), using we thus see that (8.5) i(n, k) ⩽2ek E |L (Xk)| max(m(Xk), k −S(Xk))2 .
To bound this we use the inequality 1 max(m, k −S)2 ⩽4 k2 1 + S2 m2 , which can be checked in the cases S ⩾k/2 and S ⩽k/2 separately. It follows from this and (8.5) that (8.6) i(n, k) ⩽8e1 k E |L (Xk)| + 8e1 k E |L (Xk)|S(Xk)2 m(Xk)2 .
The first of these two terms is what we want, but the second requires a keener analysis. By conditioning on m = m(Xk) we have E |L (Xk)|S(Xk)2 m(Xk)2 = k X m=1 1 m2 X c1,...,cm⩾0 cm⩾1 |L (c)|S(c)2P(Xm = c)P(Xm+1 = · · · = Xkj = 0) = kj X m=1 1 m2 E |L (Xm)|S(Xm)21(Xm ⩾1)eHm−Hk ⩽e k k X m=1 1 mE |L (Xm)|S(Xm)2Xm.
In the last step we used the crude inequality 1(Xm ⩾1) ⩽Xm.
Letting Ym = |L (Xm)|, expanding S(Xm)2 = (X1 + 2X2 + · · · + mXm)2 and using (8.6), we arrive at (8.7) i(n, k) ≪1 k E |L (Xk)| + 1 k k X m=1 1 m m X i,i′=1 ii′E YmXiXi′Xm.
The innermost sum is estimated using Lemma 8.8, splitting into various cases depending on the set of distinct values among i, i′, m.
Case 1: i, i′, m all distinct. Then ii′E YmXiXi′Xm ≪1 mE Ym.
Case 2: i = i′ ̸= m. Then ii′E YmXiXi′Xm ≪ i mE Ym ≪E Ym.
Case 3: i = i′ = m. Then ii′E YmXiXi′Xm ≪mE Ym.
Case 4: i ̸= i′ = m or i′ ̸= i = m. In both cases ii′E YmXiXi′Xm ≪E Ym.
Summing over all cases, it follows that m X i,i′=1 ii′E YmXiXi′Xm ≪mE Ym.
Since clearly E Ym ⩽E Yk for every m ⩽k the result follows from this and (8.7).
□ 96 8. PERMUTATIONS WITH A FIXED SET OF A GIVEN SIZE 3. The lower bound in Proposition 8.4 We begin by noting that from (8.3) and the inequality Hk ⩽1 + log k, it follows (8.8) E |L (Xk)| ⩾1 ek X c1,...,ck⩾0 |L (c)| k Y i=1 (1/i)ci ci!
.
Fix r = c1 + · · · + ck. We claim that (8.9) X c1+···+ck=r |L (c)| k Y i=1 (1/i)ci ci!
= 1 r!
k X a1,...,ar=1 |L ∗(a)| a1 · · · ar , where (8.10) L ∗(a) = n X i∈I ai : I ⊆[r] o .
To see (8.9), we start from the right side and set ci = |{j : aj = i}| for each i. Then L (c) = L ∗(a), Qk i=1 ici = a1 · · · ak, and each c = (c1, . . . , ck) comes from r!
c1!···ck! different choices of a1, . . . , ak.
Now let J = j log k log 2 k and suppose that b1, . . . , bJ are arbitrary non-negative integers with sum r. Consider the part of the sum on the right side of (8.9) in which bi = #{j : 2i−1 ⩽aj ⩽2i −1} (i = 1, 2, . . . , J), aj ⩽2J −1 (∀j).
Writing D(b) = J Y i=1 {2i−1, . . . , 2i −1}bi, we have (8.11) 1 r!
2J−1 X a1,...,ar=1 |L ∗(a)| a1 · · · ar = X b1+···+bJ=J 1 b1! · · · bJ!
X d∈D(b) |L ∗(d)| d1 · · · dr .
To see this, fix b1, . . . , bJ and observe that there are r!
b1!···bJ! ways to choose which bi of the variables a1, . . . , ar lie in [2i−1, 2i −1] for 1 ⩽i ⩽J.
We now take only the terms with r = J. Combining (8.8), (8.9) and (8.11) gives (8.12) E |L (Xk)| ≫1 k X b1+···+bJ=J 1 b1! · · · bJ!
X d∈D(b) |L ∗(d)| d1 · · · dJ .
Lemma 8.9. For any b = (b1, . . . , bJ) with b1 + · · · + bJ = J we have X d∈D(b) |L ∗(d)| d1 · · · dJ ≫ (2 log 2)J PJ i=1 2b1+···+bi−i .
Proof. Given ℓ⩾0, let R(d, ℓ) be the number of I ⊆[J] with ℓ= P i∈I di. Also, define λi = 2i−1 X j=2i−1 1 j .
Since 1 j ⩾ 1 2j + 1 2j+1 for all j, λi ⩾λi+1. Also, limi→∞λi = log 2. Thus we conclude that log 2 ⩽λi ⩽1 (1 ⩽i ⩽J).
3. The lower bound in Proposition 8.4 97 Since P ℓR(d, ℓ) = 2J, By Cauchy-Schwarz, 22J J Y j=1 λ2bj j = X d∈D(b) 1 d1 · · · dJ X ℓ R(d, ℓ) 2 = X d∈D(b) 1 d1 · · · dJ X ℓ∈L ∗(d) R(d, ℓ) 2 ⩽ X d∈D(b),ℓ R(d, ℓ)2 d1 · · · dJ X d∈D(b) |L ∗(d)| d1 · · · dJ .
(8.13) Our next aim is to establish an upper bound for the first sum on the right side. We have (8.14) X d∈D(b),ℓ R(d, ℓ)2 d1 · · · dJ = X Y,Z⊂[J] S(Y, Z), S(Y, Z) = X d∈D(b) P i∈Y di=P i∈Z di.
1 d1 · · · dJ .
If Y = Z, then evidently S(Y, Z) = λb1 1 · · · λbJ J . If Y and Z are distinct, let j = max(Y △Z) be the largest coordinate at which Y and Z differ. With all of the quantities di fixed except for dj, we see that dj is uniquely determined by the relation P i∈Y di = P i∈Z di. If we define e(j) ∈[J] uniquely by b1 + · · · + be(j)−1 + 1 ⩽j ⩽b1 + · · · + be(j), then dj ⩾2e(j)−1, regardless of the choice of d1, . . . , dj−1, dj+1, . . . , dJ and thus S(Y, Z) ⩽ J Y i=1 i̸=j X di 1 di · 1 2e(j)−1 = λb1 1 · · · λbJ J λ−1 e(j) 2e(j)−1 ≪λb1 1 · · · λbJ J 2e(j) .
(Here, the sums over di are over the appropriate dyadic intervals required so that d ∈D(b).) Since the number of pairs of subsets Y, Z ⊂[J] with max(Y △Z) = j is exactly 2J+j−1, we get from this and (8.14) that J Y j=1 λ−bj j X d∈D(b),ℓ R(d, ℓ)2 d1 · · · dJ ≪2J + 2J J X j=1 2j−e(j) = 2J + 2J J X i=1 2−i X j:e(j)=i 2j ≪2J + 2J J X i=1 2b1+···+bi−i ≪2J J X i=1 2b1+···+bi−i.
Comparing with (8.13), and using again that λi ⩾log 2, completes the proof.
□ Combining Lemma 8.9 and (8.12), we obtain E |L (Xk)| ≫(2 log 2)J k X b1+···+bJ=J 1 b1! · · · bJ! PJ i=1 2b1+···+bi−i .
Applying the Lemma 7.14 with xi = 2bi−1, 1 ⩽i ⩽J, the multiple sum over b1, . . . , bJ equals 1 J X b1+···+bJ=J 1 b1! · · · bJ! = 1 J · JJ J! ≍ eJ J3/2 , using the multinomial theorem and Stirling’s formula. Recalling that J = log k log 2 + O(1), the lower bound in Proposition 8.4 now follows.
98 8. PERMUTATIONS WITH A FIXED SET OF A GIVEN SIZE 4. The upper bound in Proposition 8.4 Recall from (8.10) the definition of L ∗(a). From (8.3), the bound Hk ⩾log k and (8.9) we obtain (8.15) E |L (Xk)| ⩽1 k X r 1 r!
k X a1,...,ar=1 |L ∗(a)| a1 · · · ar .
Let ˜ a1, ˜ a2, . . . be the increasing rearrangement of the sequence a, so that ˜ a1 ⩽˜ a2 ⩽· · · . For 0 ⩽j ⩽r, L ∗(a) ⊂ ( m + X i∈I ˜ ai : 0 ⩽m ⩽ j X i=1 ˜ ai, I ⊂{j + 1, . . . , r} ) , from which it follows immediately that |L ∗(a)| ⩽G(e a), where, for any real t1, . . . , tr, we define (8.16) G(t) = min 0⩽j⩽r 2r−j (t1 + · · · + tj + 1) .
As in the previous chapter, we replace the sum in (8.15) with an integral, using that G(a) is increasing in each coordinate.
Lemma 8.10. For any r ⩾1, we have 1 r!
k X a1,...,ar=1 |L ∗(a)| a1 · · · ar ≪(2Hk)r Z Ωr min 0⩽j⩽r 2−j(kξ1 + · · · + kξj + 1)dξ, where Ωr = {(ξ1, . . . , ξr) : 0 ⩽ξ1 ⩽ξ2 ⩽. . . ⩽ξr ⩽1}.
Proof. Motivated by the fact that 1 a = Z exp(Ha) exp(Ha−1) dt t , for each a define the product sets R(a) = r Y i=1 [exp (Hai−1) , exp (Hai)] .
By (8.16), we have k X a1,...,ar=1 |L ∗(a)| a1 · · · ar ⩽ k X a1,...,ar=1 G(e a) a1 · · · ar = k X a1,...,ar=1 G(e a) Z R(a) dt t1 · · · tr .
Consider some t ∈R(a), so that exp{Hai−1} ⩽ti ⩽exp{Hai} for 1 ⩽i ⩽r. Let ˜ t1 ⩽· · · ⩽˜ tr be the increasing rearrangement of t. From Hm ⩾log(m + 1) we have ˜ ti ⩾˜ ai for all i. Hence G(e a) ⩽min 0⩽j⩽r 2r−j(˜ t1 + · · · + ˜ tj + 1) = G(˜ t) for all t ∈R(a).
This yields k X a1,...,ar=1 G(˜ a) Z R(a) dt t1 · · · tr ⩽ k X a1,...,ar=1 Z R(a) G(˜ t) t1 · · · tr dt = Z exp(Hk) 1 · · · Z exp(Hk) 1 G(˜ t) t1 · · · tr dt.
The integrand on the right is symmetric in t1, . . . , tr. Making the change of variables ti = eξiHk yields k X a1,...,ar=1 |L ∗(a)| a1 · · · ar ⩽(2Hk)rr!
Z Ωr min 0⩽j⩽r 2−j eξ1Hk + · · · + eξjHk + 1 dξ.
The lemma follows from the upper bound Hk ⩽1 + log k.
□ 5. Exercises 99 In the notation of Lemma 7.17, we have by Lemma 8.10 the bound (8.17) E |L (Xk)| ≪1 k X r (2Hk)rUr(v), v = log k log 2.
Now Lemma 7.17 provides the bound Ur(v) ≪ 1 + |v −r|2 (r + 1)!(2r−v + 1), uniformly for 0 ⩽r ⩽2v. Set r∗= ⌊v⌋.
In what follows, we will use the observation that an/(n + 1)! is increasing for n ⩽a −2 and decreasing thereafter. If r = r∗+ m with 0 ⩽m ⩽v, then we have (2Hk)rUr(v) ≪(2Hk)r (r + 1)! · 1 + m2 2m ≪(2Hk)r∗ (r∗+ 1)! · 2Hk r∗ m · 1 + m2 2m ≪k2−E(log k)−3/2 · (1 + m2)(log 2 + 0.1)m, the log 2 + 0.1 coming from the assumption that k ⩾40.
Summed over m, this gives a total of ≪ k2−E(log k)−3/2.
Next suppose that r = r∗−m, m ∈N. Then we have (2Hk)rUr(v) ≪(2Hk)r (r + 1)! · (1 + m2) ≪(2Hk)r∗ (r∗+ 1)! · r∗ 2Hk m · (1 + m2) ≪k2−E(log k)−3/2 · 1 + m2 (2 log 2)m .
Summed over m, we again get a total of ≪k2−E(log k)−3/2.
There remains the range r > 2v. Here, we use the trivial bound Ur(v) ⩽1/r! and thus X r>2v (2Hk)rUr(v) ≪ X r>2v (2Hk)r r!
≪e2Hk(1−Q(1/ log 2)) ≪k2−2E.
We conclude that X r (2Hk)rUv(r) ≪k2−E(log k)−3/2.
Combined with (8.17), this proves the upper bound in Proposition 8.4.
5. Exercises Exercise 8.1. If 1 ⩽k ⩽n/ log n, show that i(n, k) ∼i(∞, k) (n →∞).
CHAPTER 9 Sets of permutations with equal sized divisors 1. Equal sized divisors of several permutations Let σ1, . . . , σr be random permutations in Sn, chosen independently. By Theorem 8.1, the probability that they all have a divisor of size k is i(n, k)r ≍r 1 (kE(1 + log k))r (1 ⩽k ⩽n/2).
For σ ∈Sn define D(σ) = {|β| : β|σ}, the set of sizes of divisors of σ.
We note that {0, n} ⊆D(σ) always. We wish to also bound the probability that D(σ1)∩· · ·∩D(σr) contains an element in [k, n/2]. Crudely, from the above and the fact that 12E > 1 this is at most (9.1) X k⩽m⩽n/2 i(n, k)r ≪r k1−rE (1 + log k)(3/2)r (r ⩾12).
This estimate is wasteful, since the events m1 ∈D(σ1) ∩· · · ∩D(σr) and m2 ∈D(σ1) ∩· · · ∩D(σr) are not independent. In fact, if r ⩾4 then it is rare for D(σ1) ∩· · · ∩D(σr) to have a large element, but this is not true for r = 3. This ultimately depends on the inequalities 3(1 −log 2) < 1 < 4(1 −log 2).
Theorem 9.1 (Pemantle-Peres-Rivin ; 2016). Let σ1, . . . , σ4 be random permutations in Sn, chosen independently. There is a real number α > 0 so that P D(σ1) ∩· · · ∩D(σ4) = {0, n} ⩾α.
The authors if gave a proof of 9.1 which is much simpler than the original proof in , and we given an even simpler proof below.
Theorem 9.2 (Eberhard-Ford-Green ; 2017). With probability →1 as n →∞, we have |D(σ1) ∩D(σ2) ∩D(σ3)| →∞ as n →∞.
Lemma 9.3. (a) Suppose that h, m are integers with h < m < n −h. Then Pσ m ∈D(σ), Ch = 0 ≪h−2.
(b) If 1 ⩽λ ⩽2, 1 ⩽h ⩽k ⩽n/2 and k < m < n −k then Pσ m ∈D(σ), Ch = 0, C(h,k](σ) ⩽λ log(k/h) ≪h−2(k/h)−2Q(λ/2).
Proof. For (a), factor each such σ as σ = σ1σ2 where |σ1| = m and |σ2| = n −m. By By Theorem 1.9, the number of such σ is ≪ n m · m!
h · (n −m)!
h ≪n!
h2 , 100 1. Equal sized divisors of several permutations 101 For (b), WLOG k ⩾10. Fix ℓ⩽λ log(k/h) and consider permutations σ with Ch = 0, C(h,k](σ) = ℓ and such that m ∈D(σ).
Write σ = σ1σ2 with |σ1| = m, |σ2| = n −m.
Then C(h,k](σ1) = ℓ1 and C(h,k](σ2) = ℓ2, where ℓ1 + ℓ2 = ℓ. By Theorem 1.9, the number of such σ, for a given choice of ℓ1, ℓ2, is ≪ n m (Hk −Hh)ℓ1 h · (k/h)ℓ1! m! · (Hk −Hh)ℓ2 h · (k/h)ℓ2! (n −m)! = n!(1 + log(k/h))ℓ k2ℓ1!ℓ2!
.
Summing over ℓ1 + ℓ2 = ℓ, we see that Pσ m ∈D(σ), Ch = 0, C(h,k](σ) = ℓ) ≪(2 log(k/h) + 2)ℓ k2ℓ!
.
We sum over ℓ⩽λ log(k/h), using the bound for Poisson tails (Prop. 0.3) and noting that (2 log(k/h)+2)ℓ≪ (2 log(k/h))ℓ. We get Pσ m ∈D(σ), Ck ⩽(1 + ε) log k ≪h−2e−(2 log(k/h))Q(λ/2) = h−2(k/h)−2Q(λ/2).
□ Lemma 9.4. Let 1 ⩽h ⩽n/10. There is a real number c > 0 so that with probability ≪h−4−c, Ch = 0 for 1 ⩽i ⩽4 and D(σ1) ∩· · · ∩D(σ4) contains an element other than 0 and n.
Proof. Fix 0 < ε ⩽1 2, to be determined later. WLOG h ⩾10. Let k = 2rh ⩽n/2 for a non-negative integer r. Let σ1, σ2, σ3, σ4 be random permutations of Sn.
Let Fk be the event that Ch = 0 (1 ⩽i ⩽4), and for some i, C(h,k](σi) > (1 + ε) log(k/h).
Let Gk be the event that Ch = 0 (1 ⩽i ⩽4), and that for some m ∈(k, 2k] ∩(h, n/2], we have m ∈D(σ1) ∩· · · ∩D(σ4). Let Hk be the event that Gk holds and that C(h,k](σi) ⩽(1 + ε) log(k/h) for each i. Evidently, the probability in the lemma is at most (9.2) X h⩽k⩽n/2 P Gk ⩽ X h⩽k⩽h3 P Gk + X h3 h3 then P Fk ≪ 1 h4(k/h)Q(1+ε) .
By Lemma 9.3 (a), we have P Gk ≪k h8 .
Now suppose that h3 < k ⩽n/2 and consider the event Hk. By Lemma 9.3 (b), we get that P Hk ≪ k h8(k/h)8Q( 1+ε 2 ) .
Choose ε so 1 −8Q( 1+ε 2 ) = −Q(1 + ε); the solution is ε = 0.08895343 · · · . Inserting our bounds for Fk, Gk and Hk into (9.2), we see that the probability in the lemma is ≪h3 h8 + 1 h4+2Q(1+ε) + 1 h5+16Q( 1+ε 2 ) ≪ 1 h4+2Q(1+ε) .
□ Proof of Theorem 9.1. Let h be a sufficiently large constant and n0 = 3h. If n ⩽n0, the probability in question is at least the probability that σ1 is an n-cycle, which is 1/n ⩾1/n0. Now suppose that n > n0, in particular n > 3h. By (3.5) and Lemma 9.4, P Ch = 0 (1 ⩽i ⩽4); D(σ1) ∩· · · ∩D(σ4) = {0, n} ⩾ 1 (2h + 1)4 −O 1 h4+c .
Taking h large enough proves the theorem.
□ 102 9. SETS OF PERMUTATIONS WITH EQUAL SIZED DIVISORS 2. Application: Invariable generation of Sn By Dixon’s theorem , two random elements σ1, σ2 of Sn generate at least the whole alternating group An with probability tending to 1 as n →∞. It is less clear how large the group generated by σ′ 1, σ′ 2 must be when σ′ 1 and σ′ 2 are allowed to be arbitrary conjugates of σ1 and σ2. Following Dixon we say that a list σ1, . . . , σr ∈Sn has a property P invariably if σ′ 1, . . . , σ′ r has property P whenever σ′ i is conjugate to σi for every i. How many random elements of Sn must we take before we expect them to invariably generate Sn?
This problem is connected with computational Galois theory. Given a polynomial f ∈Z[x] of degree n with no repeated factors, information about the Galois group can be gained by reducing f modulo various primes p and factorizing the reduced polynomial fp over Z/pZ. By classical Galois theory, if fp has irreducible factors of degrees n1, . . . , nr then the Galois group G of f over Q has an element with cycle lengths n1, . . . , nr.
Moreover by Frobenius’s density theorem, if G = Sn then the frequency with which a given cycle type arises is equal to the proportion of elements in Sn with that cycle type. Thus if we suspect that G = Sn then the number of times we expect to have to iterate this procedure before proving that G = Sn is controlled by the expected number of random elements required to invariably generate Sn.
Łuczak and Pyber were the first to prove the existence of a constant C such that C random permutations σ1, . . . , σC ∈Sn invariably generate Sn with probability bounded away from zero.
Their method does not directly yield a reasonable value of C, but recently Pemantle, Peres, and Rivin proved that we may take C = 4. The key to their proof is Theorem 9.1.
Theorem 9.5 (Pemantle-Peres-Rivin ; 2016). For some α > 0, the probability that random σ1, . . . , σ4 invariably generate Sn is at least α, for any n.
Theorem 9.6 (Eberhard-Ford-Green ; 2017). With probability →1 as n →∞, random σ1, σ2, σ3 do not invarably generate Sn.
The connection between common size fixed sets of σ1, . . . , σr and invariable generation is clear: if m ∈ D(σ1) ∩· · · ∩D(σr), 0 < m < n, then there are conjugates σ′ 1, . . . , σ′ r each mapping {1, 2, · · · , m} to itself (a fixed set). In this case, it is clear that σ′ 1, . . . , σ′ r do not generate Sn.
What if D(σ1) ∩· · · ∩D(σr) = {0, n}?
It is simple to see that σ′ 1, . . . , σ′ r generate some transitive subgroup of Sn; a subgroup H of Sn is transitive if for every pair i, j ∈[n] there is some element of H which maps i to j. Examples of transitive subgroups include Sn, An as well as imprimitive transitive subgroups such as this example when n = 2k is even: Let I1 = {1, 2, . . . , k} and I2 = {k + 1, . . . , 2k} amd let G = {σ : either σ(I1) = I1, σ(I2) = I2 or σ(I1) = I2, σ(I2) = I1}.
Lemma 9.7. Suppose that D(σ1)∩· · ·∩D(σr) = {0, n}. Then σ1, . . . , σr generate a transitive subgroup of Sn.
Proof. Let G be the group generated by σ1, . . . , σr . Define a relation ∼on [n] by a ∼b if there is a σ ∈G with σ(a) = b. This is clearly an equivalence relation, and thus partitions [n]. If G is not a transitive subgroup, then there is a non-trivial equivalence class I, ∅̸= I ̸= [n], and clearly I is a fixed set of each σj.
□ From Lemma 9.7, Theorem 9.6 now follows from Theorem 9.2. Also, from Theorem 9.1, it follows that for some α′ > 0, with probability at least α′ four random permutations σ1, . . . , σ4 generate a transitive subgroup of Sn. It is known (a Theorem of Łuczak-Pyber , which also uses the theory of permutations) that the probability that a random σ ∈Sn lies in a transitive subgroup of Sn other than Sn or An is o(1) as n →∞. We will not prove this here. For the latest estimates, see . Note that if σ does not lie in a transitive subgroup, then neither do any of its conjugates ˜ σ, since σ and ˜ σ are indistinguishable in a group-theoretic sense. It follows then that for some α′′ > 0, with probability at least α′′ four random permutations σ1, . . . , σ4 generate either An or Sn.
To distinguish An from Sn, it suffices to choose σ1 to be an odd permutation (a permutation which is the product of an odd number of transpositions (ab)) in order to conclude that σ′ 1, . . . , σ′ 4 generates Sn.
2. Application: Invariable generation of Sn 103 Recalling the proof of Theorem 9.1 from the previous section, it suffices to show that if σ is a random odd permutation, then Pσ(Ch = 0) ≫1/h, i.e., an analog of the lower bound in (3.5).
Lemma 9.8. Let n be sufficiently large and 1 ⩽h ⩽n1/3. Then P(Ch = 0, σ odd) ≫1 h.
We first establish a general result about random odd permutations.
Lemma 9.9 (Random odd permutations). Let τ ∈Sn be a random odd permutation, and 1 ⩽k ⩽n. Then dT V (C1(τ), . . . , Ck(τ), Zk) ≪k/n, where Zk = (Z1, . . . , Zk), Zi d = Pois(1/i) (1 ⩽i ⩽k), Zi independent.
Proof. WLOG k ⩽n/10. Choose σ ∈Sn uniformly at random, and define ρ by putting ρ = 1 if σ is odd and ρ = (12) if σ is even. Then τ = σρ is uniformly distributed over odd permutations. We will show that with high probability, (9.3) (C1(σ), . . . , Ck(σ)) = (C1(σρ), . . . , Ck(σρ)).
In particular, (9.3) holds provided that neither of the following holds: (i) 1 and 2 are in different cycles, each of length ⩽k; or (ii) 1 and 2 are in the same cycle of σ and there are fewer than k elements in between them (in either direction).
To see this, observe that for any strings of numbers A, B (including the empty strings), if σ has cycles (1A) and (2B) and ρ = (12) then σρ has a cycle (1A2B). Also, if σ has a cycle (1A2B) and ρ = (12), then σρ has cycles (1A) and (2B).
We will show that the probability that (9.3) fails is O( k n). Then, by the Poisson distribution of Small Cycles (Theorem 4.3), dT V (C1(σ), . . . , Ck(σ), Zk) ≪e−n/5k ≪k/n and the lemma follows.
Consider (i). Given positive a, b ⩽k, the probability that (1A) and (2B) are cycles in σ with |A| = a, |B| = b, equals 1 n! · (n −2)!
a!b!(n −a −b −2)!a!b!(n −a −b −2)! = 1 n(n −1).
Thus, the probability that (i) holds is O(k2/n2).
Now consider the probability that 1 and 2 are both contained in the same cycle of σ and are close together. Let this cycle be (1A2B), where |A| = a and |B| = b with min(a, b) ⩽k −1. The probability of having such a cycle with a, b fixed equals 1 n(n−1) by the same logic. Hence, the probability that (ii) holds is O(k/n).
□ Proof of Lemma 9.8. By Lemma 9.3, Pσ(Ch = 0, σ odd) = 1 2Pτ∈Sn\An(Ch = 0) = 1 2P(Z1 = · · · = Zh = 0) + O(h/n) = (1/2)e−Hh + O(1/n2/3) ≫1/h if n is large enough.
□ 104 9. SETS OF PERMUTATIONS WITH EQUAL SIZED DIVISORS 3. Application: Irreducibility of polynomials over Q A famous conjecture of Odlyzko and Poonen states that a random polynomial (9.4) xn + an−1xn−1 + · · · + a1x + 1, where each ai ∈{0, 1} is randomly chosen, is irreducible with probability →1 as n →∞. Variations on this problem put the coefficients aj in other finite sets, e.g. {−1, 1}. Here are some highlights of what’s known.
• Almost all polynomials of shape (9.4) with an ∈{0, 1} have no irreducible factor ⩽cn/ log n, for some c > 0 (Konyagin , 1999). This was improved in Bary-Soroker, Koukoulopoulos and Kozma in 2022, where the conclusion is that with probability →1 as n →∞, the polynomial has no irreducible factor ⩽cn, for some c > 0.
• Almost all polynomials of shape (9.4) with an ∈{1, 2, . . . , 210} are irreducible (Bary-Soroker and Kozma , 2020). The methods were refined in work of Bary-Soroker, Koukoulopoulos and Kozma in 2022, where they show the same conclusion with an ∈{1, 2, . . . , 35} and other finite sets.
• Assuming the GRH for Dedekind zeta-functions, almost all polynomials of shape (9.4) with an ∈ {0, 1} are irreducible (Bruillard, Varju , 2019).
Here we discuss the proof of Bary-Soroker and Kozma’s “210” theorem, because of its connection to random permutations. The authors consider polynomials f(x) = xn + an−1xn−1 + · · · + a1x + a0 with each ai ∈{1, 2, . . . , 210} chosen uniformly at random. We sketch the proof of Theorem 9.10. Let ξ(n) →∞as n →∞. With probability →1 as n →∞, f(x) has no factor of degree in [ξ(n), n −1].
The key idea is that 210 = 2 · 3 · 5 · 7, and there is a natural bijection between such f and the vector (f2, f3, f5, f7) of its reduction modulo 2, 3, 5 and 7. Moreover, each fp is a random polynomial over Fp and (this is crucial!) the Chinese Remainder Theorem implies that the random polynomials f2, f3, f5, f7 are independent. Also, if f has a factor or degree k, then so do f2, f3, f5, f7 (these need not be irreducible!).
The second key idea is the well-known fact that the distribution of the sizes of the irreducible factors of a random polynomial of degree n over Fp is roughly the same as the distribution of the sizes of cycles of a random permutation σ ∈Sn. This is relatively easy to establish since we have exact formulas in both cases.
Lemma 9.11. Let f be a random monic polynomial of degree n over Fq, where q is a prime power. Let If(r) be the number of irreducible factors of f of degree r. Then, if m1 + 2m2 + · · · + nmn = n, P If(1) = m1, . . . , If(n) = mn = n Y i=1 α(i, mi), where α(i, 1) = 1 i X j|i µ(i/j)qj−i, α(i, m) = (α(i, 1) + q−i(m −1))(α(i, 1) + q−i(m −2)) · · · α(i, 1) m!
Proof. The main task is to show that α(i, m)qim is the number of m-tuples (duplicates allowed) of monic, irreducible polynomials of degree i over Fq. Once this is established, we follow the idea of Lemma 1.1, getting E If(1) m1 · · · If(n) mn = n Y i=1 α(i, mi) and then noticing that if m1 +2m2 +· · ·+nmn = n, the product of binomials on the left equals 1 if and only if If(r) = mr for all r. We begin with the interpretation of α(i, 1). Let πq(d) be the number of irreducible 4. Exercises 105 polynomials of degree d (all polynomials will be monic and over Fq). Let I denote a generic irreducible polynomial of any degree. Then nqn = X deg F =n X Iv|F deg(I) = X v·deg I⩽n deg(I) · |{F : deg(F) = n, Iv|F}| = X v·deg I⩽n deg(I) · qn−v·deg(I) = qn n X m=1 q−m X d|m dπq(d).
Divide by qn, then apply this at n and n −1 and take the difference. This gives qn = X d|n dπq(d).
By Möbius inversion, nπq(n) = X d|n µ(d)qn/d = X d|n µ(n/d)qd, and therefore α(d, 1) = q−dπq(d), as desired.
Finally, the number of ways one can choose m irreducible polynomials of degree d is πq(d) + m −1 m = qdmα(d, m).
□ For large i we have α(i, 1) = 1 i + O(q−i/2) ≈1 i and α(i, m) ≈α(i, m)i m!
≈(1/i)m m!
.
Thus, the distribution of (If(1), . . . , If(k)) is about the same as the distribution of (C1(σ), . . . , Ck(σ)). For precise statements, see .
Heuristically, the probability that f2, . . . , f7 all have a factor of the same degree in [ξ(n), n −1] should be approximately the same as the probability that four random permutations each have a divisor of size k for some k ∈[ξ(n), n −1]. This latter probability is →0 as n →∞, essentially the proof of Theorem 9.4.
See Exercise 9.1 below.
4. Exercises Exercise 9.1. Show that for some constant c > 0, the probability that D(σ1) ∩· · · ∩D(σ4) contains an element in [K, n −K] is O(K−c).
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11293 | https://www.biochemtuition.com/wp-content/uploads/2015/07/C2-Exponentials-Logarithms-Answers.pdf | Solomon Press EXPONENTIALS AND LOGARITHMS C2 Answers - Worksheet A 1 a log10 1000 = 3 b log3 81 = 4 c log2 256 = 8 d log7 1 = 0 e log3 1 27 = −3 f log32 1 2 = 1 5 − g log19 19 = 1 h log36 216 = 3 2 2 a 53 = 125 b 24 = 16 c 105 = 100 000 d 230 = 1 e 1 2 9 = 3 f 10−2 = 0.01 g 2−3 = 1 8 h 61 = 6 3 a = log7 72 b = log4 43 c = log2 27 d = log3 33 = 2 = 3 = 7 = 3 e = log5 54 f = log8 81 g = log7 70 h = log15 15−1 = 4 = 1 = 0 = −1 i = log3 3−2 j = lg 10−3 k = log16 1 4 16 l = log4 3 2 4 = −2 = −3 = 1 4 = 3 2 m = log9 5 2 9 n = log100 3 2 100 − o = log25 3 2 25 p = log27 2 3 27 − = 5 2 = 3 2 − = 3 2 = 2 3 − 4 a 5x = 25 b 26 = x c x3 = 64 d 10−3 = x x = 2 x = 64 x = 4 x = 1 1000 e 2 3 x = 16 f 5x = 1 g x1 = 9 h 10x = 1012 x = 64 x = 0 x = 9 x = 12 i logx 7 = 1 2 j 41.5 = x k 1 3 x − = 0.1 l log8 x = 1 3 − 1 2 x = 7 x = 8 x = 1000 1 3 8 − = x x = 49 x = 1 2 5 a = loga (4 × 7) b = loga (10 ÷ 5) c = loga 62 = loga 28 = loga 2 = loga 36 d = loga (9 ÷ 1 3 ) e = loga 1 2 25 + loga 32 f = loga 48 − loga 23 − loga 1 2 9 = loga 27 = loga 5 + loga 9 = loga 48 − loga 8 − loga 3 = loga (5 × 9) = loga [48 ÷ (8 × 3)] = loga 45 = loga 2 6 a = 5 logq x b = 15 2 logq x c = logq x−1 d = logq 1 3 x = −logq x = 1 3 logq x e = 4 logq 1 2 x − f = 2 logq x + 5 logq x g = logq x−2 + logq x−3 h = 6 logq x − 2 logq x = −2 logq x = 7 logq x = −2 logq x − 3 logq x = 4 logq x = −5 logq x Dr. Faisal Rana www.biochemtuition.com faisal.rana@biochemtuition.com C2 EXPONENTIALS AND LOGARITHMS Answers - Worksheet A page 2 Solomon Press 7 a = lg (5 × 4) b = lg (12 ÷ 6) c = lg 23 d = lg 34 − lg 9 = lg 20 = lg 2 = lg 8 = lg 81 − lg 9 = lg (81 ÷ 9) = lg 9 e = lg 1 2 16 − lg 1 5 32 f = lg 10 + lg 11 g = lg 1 50 + lg 102 h = lg 103 − lg 40 = lg 4 − lg 2 = lg (10 × 11) = lg 1 50 + lg 100 = lg 1000 − lg 40 = lg (4 ÷ 2) = lg 110 = lg ( 1 50 × 100) = lg (1000 ÷ 40) = lg 2 = lg 2 = lg 25 8 a = log3 (54 ÷ 2) b = log5 (20 × 1.25) c = log2 24 + log3 33 = log3 27 = log5 25 = 4 + 3 = log3 33 = log5 52 = 7 = 3 = 2 d = log6 (24 × 9) e = log3 (12 ÷ 4) f = log4 (18 ÷ 9) = log6 216 = log3 3 = log4 2 = log6 63 = 1 = log4 1 2 4 = 3 = 1 2 g = log9 (4 × 0.25) h = lg 22 + lg 25 i = log3 1 3 8 − log3 18 = log9 1 = lg 4 + lg 25 = log3 2 − log3 18 = 0 = lg (4 × 25) = log3 (2 ÷ 18) = lg 100 = log3 1 9 = lg 102 = log3 3−2 = 2 = −2 j = log4 1 3 64 + (2 × log5 52) k = 1 2 log5 25 16 + log5 102 l = log3 5 − log3 62 − log3 15 4 = log4 4 + (2 × 2) = log5 1 2 25 16 ( ) + log5 100 = log3 [5 ÷ (36 × 15 4 )] = 1 + 4 = log5 5 4 + log5 100 = log3 1 27 = 5 = log5 ( 5 4 × 100) = log3 3−3 = log5 125 = −3 = log5 53 = 3 Dr. Faisal Rana www.biochemtuition.com faisal.rana@biochemtuition.com Solomon Press EXPONENTIALS AND LOGARITHMS C2 Answers - Worksheet B 1 a = log10 a + log10 b b = log10 a + log10 b7 c = log10 a3 − log10 b d = log10 a + log10 1 2 b = log10 a + 7 log10 b = 3 log10 a − log10 b = log10 a + 1 2 log10 b e = 2 log10 ab f = −log10 ab g = log10 3 2 a + log10 5 2 b h = 3(log10 a2 − log10 1 3 b ) = 2log10 a + 2log10 b = −log10 a − log10 b = 3 2 log10 a + 5 2 log10 b = 6 log10 a − log10 b 2 a = logq 82 b = logq 1 3 8 c = logq 16 − logq q d = logq 4 + logq q3 = 2y = 1 3 y = logq 4 3 8 − 1 = logq 2 3 8 + 3 = 4 3 y − 1 = 2 3 y + 3 3 a = lg (2 × 32) b = lg (25 × 3) c = lg 9 − lg 16 d = lg (2 × 3) − lg 23 = lg 2 + 2 lg 3 = 5 lg 2 + lg 3 = lg 32 − lg 24 = lg 2 + lg 3 − 3 lg 2 = a + 2b = 5a + b = 2 lg 3 − 4 lg 2 = lg 3 − 2 lg 2 = 2b − 4a = b − 2a e = 1 2 lg 6 f = 3 2 lg 24 + 1 2 lg 34 g = 4lg3 − 3(lg2+lg3) h =lg(6×10)+lg(2×10)−2 = 1 2 (lg 2 + lg 3) = 6 lg 2 + 2 lg 3 = lg 3 − 3 lg 2 = lg 6 + 1 + lg 2 + 1 − 2 = 1 2 (a + b) = 6a + 2b = b − 3a = lg 2 + lg 3 + lg 2 = 2a + b 4 a = log5 10 − log5 2 b = log12 16 + log12 9 c = log4 8 = log5 5 = log12 144 = log4 3 2 4 = 1 = 2 = 3 2 d = 4 7 7 log 3 log 3 e = log27 3 2 12 72 f = 2 11 11 log 5 log 5 − = 7 7 4log 3 log 3 = log27 12 12 12 6 12 6 12 × × × × × = 11 11 2log 5 log 5 − = 4 = log27 1 3 = 1 3 − = −2 5 a x = 31.8 b x = 5−0.3 c x − 3 = 82.1 x = 7.22 x = 0.617 x = 3 + 82.1 x = 81.8 d 1 2 x + 1 = 43.2 e log2 3y = 5.3 f log6 (1 − 5t) = −0.6 x = 2(43.2 − 1) 3y = 25.3 1 − 5t = 6−0.6 x = 167 y = 1 3 × 25.3 t = 1 5 (1 − 6−0.6) y = 13.1 t = 0.132 6 a = log2 x5 b = log2 (x2 + 4x) c = log2 x2 + log2 x = log2 x3 d = log2 (x − 2)3 − log2 x4 e = log2 2 1 1 x x − + f = log2 x − 2 log2 x + 2 3 log2 x = log2 3 4 ( 2) x x − = log2 ( 1)( 1) 1 x x x + − + = 1 3 −log2 x = log2 (x − 1) = log2 1 3 x − Dr. Faisal Rana www.biochemtuition.com faisal.rana@biochemtuition.com C2 EXPONENTIALS AND LOGARITHMS Answers - Worksheet B page 2 Solomon Press 7 a log3 5x = log3 (2x + 3) b log9 10x = 3 2 5x = 2x + 3 10x = 3 2 9 = 27 x = 1 x = 2.7 c log4 1 x x − = log4 3 + log4 2 = log4 6 d log5 5 2 x x + = log5 6 x x + 1 x x − = 6 5 2 x x + = 6 x x + x = 6x − 6 5x2 = (x + 2)(x + 6) = x2 + 8x + 12 x = 6 5 x2 − 2x − 3 = 0 (x + 1)(x − 3) = 0 x = −1, 3 log5 x not real for x = −1 ∴ x = 3 e log6 x2 = log6 5(2x − 5) f log7 4x − log7 1 6 x − = 1 x2 = 5(2x − 5) log7 4x(x − 6) = 1 x2 − 10x + 25 = 0 4x(x − 6) = 7 (x − 5)2 = 0 4x2 − 24x − 7 = 0 x = 5 x = 24 576 112 8 ± + = 3 ± 1 2 43 log7 4x not real for x = 3 − 1 2 43 ∴ x = 3 + 1 2 43 [ = 6.28 (3sf)] 8 a logx y = 2 ⇒ y = x2 b log5 x − 2 log5 y = log5 2 ⇒ 2 x y = 2 sub. x3 = 27 ⇒ x = 2y2 x = 3 sub. 3y2 = 12 ∴ x = 3, y = 9 y2 = 4 for real log5 y, y > 0 ∴ y = 2 ∴ x = 8, y = 2 c logy 32 = 5 2 − ⇒ 5 2 y − = 32 d logy x = 3 2 ⇒ 3 2 y = x ⇒ y = 2 5 32 − = 1 4 ⇒ 1 2 y = 1 3 x sub. log2 x = 3 − 2 log2 1 4 sub. 4 1 3 x = 20 log2 x = 3 − (−4) = 7 1 3 x = 5 x = 27 = 128 x = 53 = 125 ∴ x = 128, y = 1 4 ∴ x = 125, y = 25 e loga x + loga 3 = 1 2 loga y ⇒ 3x = 1 2 y f log10 y + 2 log10 x = 3 ⇒ x2y = 103 ⇒ y = 9x2 log2 y − log2 x = 3 ⇒ y x = 23 sub. 3x + 9x2 = 20 ⇒ y = 8x 9x2 + 3x − 20 = 0 sub. 8x3 = 1000 (3x + 5)(3x − 4) = 0 x3 = 125 for real loga x, x > 0 ∴ x = 4 3 x = 5 ∴ x = 4 3 , y = 16 ∴ x = 5, y = 40 Dr. Faisal Rana www.biochemtuition.com faisal.rana@biochemtuition.com Solomon Press EXPONENTIALS AND LOGARITHMS C2 Answers - Worksheet C 1 a 1.78 b 0.778 c 2.40 d −0.398 2 a x = lg 14 = 1.15 b 10x = 4 c 3x = lg 49 x = lg 4 = 0.60 x = 1 3 lg 49 = 0.56 d x − 4 = lg 23 e 2x + 1 = lg 130 f (102)x = 102x = 5 x = 4 + lg 23 = 5.36 x = 1 2 (lg 130 − 1) = 0.56 2x = lg 5 x = 1 2 lg 5 = 0.35 3 let y = loga b ⇒ ay = b y logc a = logc b y = log log c c b a ∴ loga b = log log c c b a 4 a = lg7 lg2 = 2.81 b = lg172 lg20 = 1.72 c = lg49 lg5 = 2.42 d = lg4 lg9 = 0.631 5 a x lg 3 = lg 12 b x lg 2 = lg 0.7 c −y lg 8 = lg 3 d 1 2 x lg 4 = lg 0.3 x = lg12 lg3 x = lg0.7 lg2 y = −lg3 lg8 x = 2lg0.3 lg4 x = 2.26 x = −0.515 y = −0.528 x = −1.74 e (t + 3) lg 5 = lg 24 f (4 + x) lg 3 = lg 16 g (2x + 4)lg 7 = lg 12 h 23x + 1 = 12.4 t = lg24 lg5 − 3 x = lg16 lg3 − 4 x = 1 2 ( lg12 lg7 − 4) (3x + 1) lg 2 = lg 12.4 t = −1.03 x = −1.48 x = −1.36 x = 1 3 ( lg12.4 lg2 − 1) x = 0.877 i (2 − 3x) lg 4 = lg 32.7 j x lg 5 = (x − 1) lg 6 x = 1 3 (2 − lg32.7 lg4 ) x (lg 6 − lg 5) = lg 6 x = −0.172 x = lg6 lg6 lg5 − = 9.83 k (y + 2) lg 7 = (y + 1) lg 9 l (5 − x) lg 4 = (2x − 1) lg 11 y (lg 9 − lg 7) = 2 lg 7 − lg 9 x (2 lg 11 + lg 4) = 5 lg 4 + lg 11 y = 2lg7 lg9 lg9 lg7 − − = 6.74 x = 5lg4 lg11 2lg11 lg4 + + = 1.51 m ( 1 2 x + 3) lg 4 = (1 − 2x) lg 5 n (3y − 2) lg 2 = (2y + 5) lg 3 x ( 1 2 lg 4 + 2 lg 5) = lg 5 − 3 lg 4 y (3 lg 2 − 2 lg 3) = 5 lg 3 + 2 lg 2 x = 1 2 lg5 3lg4 lg4 2lg5 − + = −0.652 y = 5lg3 2lg2 3lg2 2lg3 + − = −58.4 o 72x + 4 = 113x − 4 p 3x + 1 = 24 + x (2x + 4) lg 7 = (3x − 4) lg 11 (x + 1) lg 3 = (4 + x) lg 2 x (3 lg 11 − 2 lg 7) = 4 lg 7 + 4 lg 11 x (lg 3 − lg 2) = 4 lg 2 − lg 3 x = 4lg7 4lg11 3lg11 2lg7 + − = 5.26 x = 4lg2 lg3 lg3 lg2 − − = 4.13 Dr. Faisal Rana www.biochemtuition.com faisal.rana@biochemtuition.com C2 EXPONENTIALS AND LOGARITHMS Answers - Worksheet C page 2 Solomon Press 6 a (2x + 3)(2x − 2) = 0 b (3x − 1)(3x − 4) = 0 c 52x − 8(5x) + 12 = 0 2x = −3 [no sols], 2 3x = 1, 4 (5x − 2)(5x − 6) = 0 x = 1 x = 0, lg4 lg3 = 0, 1.26 5x = 2, 6 x = lg2 lg5 , lg6 lg5 = 0.43, 1.11 d 2(42x) − 7(4x) + 3 = 0 e 2(22y) + 7(2y) − 15 = 0 f 3(32x) − 17(3x) + 10 = 0 (2(4x) − 1)(4x − 3) = 0 (2(2y) − 3)(2y + 5) = 0 (3(3x) − 2)(3x − 5) = 0 4x = 1 2 , 3 2y = −5 [no sols], 3 2 3x = 2 3 , 5 x = 1 2 − , lg3 lg4 = 1 2 − , 0.79 y = 3 2 lg lg2 = 0.58 x = 2 3 lg lg3 , lg5 lg3 = −0.37, 1.46 g 52t + 5(5t) − 24 = 0 h 3(32x) − 18(3x) + 15 = 0 i 3(42x) − 16(4x) + 5 = 0 (5t + 8)(5t − 3) = 0 3(3x − 1)(3x − 5) = 0 (3(4x) − 1)(4x − 5) = 0 5t = −8 [no sols], 3 3x = 1, 5 4x = 1 3 , 5 t = lg3 lg5 = 0.68 x = 0, lg5 lg3 = 0, 1.46 x = 1 3 lg lg4 , lg5 lg4 = −0.79, 1.16 7 a y b y y = 5x y = 2x y = ( 1 3 )x y = 3x (0, 1) (0, 1) O x O x c y d y y = 4x (0, 8) y = 4x − − − − 1 y = 2x + 3 y = 2x (0, 1) x O x 8 a y 9 x = 0 ⇒ y = −4 y = 0 ⇒ 2x = 5 x = lg5 lg2 AB2 = 42 + ( lg5 lg2 )2 = 21.391 AB = 4.63 O x b (3, 29) ⇒ 29 = 2 + a3 a3 = 27 a = 3 (0, 1) (0, 0) (0, 3) y = 2 Dr. Faisal Rana www.biochemtuition.com faisal.rana@biochemtuition.com Solomon Press EXPONENTIALS AND LOGARITHMS C2 Answers - Worksheet D 1 a = log10 3 2 2 a log3 x = 5 4 = log10 3 − log10 2 x = 5 4 3 = 3.95 (3sf) = b − a b 3 log3 x − 5 log3 x = 4 b = log10 (23 × 3) log3 x = −2 = 3 log10 2 + log10 3 x = 3−2 = 1 9 = 3a + b c = log10 (1.5 × 100) = log10 1.5 + log10 100 = b − a + 2 3 a i = log2 1 2 q = 1 2 log2 q = 1 2 p 4 2000 = 1000 × 1.0224t ii = log2 8 + log2 q = 3 + p 2 = 1.0224t b 3 + p − 1 2 p = 2 4t lg 1.022 = lg 2 p = log2 q = −2 t = lg2 4lg1.022 = 7.96 ∴ q = 2−2 = 1 4 ∴ 8 years 5 a (0, −3) 6 a log3 1 2 x x + − = 1 b k = −4 1 2 x x + − = 3 c ( 1 3 )x − 4 = 0 x + 1 = 3x − 6 ( 1 3 )x = 4 x = 7 2 x = 1 3 lg4 lg = −1.26 (3sf) b (2x + 1) lg 3 = (x − 4) lg 2 x (lg 2 − 2 lg 3) = lg 3 + 4 lg 2 x = lg3 4lg2 lg2 2lg3 + − 7 a i = 2−1(2x) = 1 2 t 8 a log2 (3x + 5) + 3 = 7 ii = 2(22x) = 2(2x)2 = 2t2 3x + 5 = 24 = 16 b 2t2 − 7t + 6 = 0 x = 11 3 (2t − 3)(t − 2) = 0 b log2 (x + 1) + log2 (3x − 1) = 5 t = 2x = 3 2 , 2 (x + 1)(3x − 1) = 25 = 32 x = 3 2 lg lg2 , 1 = 0.585 (3sf), 1 3x2 + 2x − 33 = 0 (3x + 11)(x − 3) = 0 x = 11 3 − , 3 for real log2 (3x − 1), x > 1 3 ∴ x = 3 Dr. Faisal Rana www.biochemtuition.com faisal.rana@biochemtuition.com C2 EXPONENTIALS AND LOGARITHMS Answers - Worksheet D page 2 Solomon Press 9 a x + 4 = 5 4 x 10 a t = 0 ⇒ n = 2000 x = 16 b 3600 = 3 18000 1 8c− + b y + 2 = 12 1 y + 1 + 8c−3 = 5 (y + 2)(y + 1) = 12 c−3 = 1 2 y2 + 3y − 10 = 0 c3 = 2 (y + 5)(y − 2) = 0 c = 3 2 y > 0 ∴ y = 2 c 4000 = 18000 1 8 t c− + c logy x = log2 16 = 4 1 + 8c−t = 9 2 c−t = 7 16 −t = 7 16 3 lg lg 2 t = 3.578 weeks = 25 days 11 a i log8 x2 = 2 log8 x = 2y 12 log2 y − log2 (3 − 2x) = 1 ⇒ 3 2 y x − = 2 ii y = log8 x ⇒ x = 8y = 23y ⇒ y = 6 − 4x ∴ log2 x = 3y log4 xy = 1 2 ⇒ xy = 1 2 4 = 2 b 3(2y) + 3y = 6 sub. x(6 − 4x) = 2 y = log8 x = 2 3 2x2 − 3x + 1 = 0 ∴ x = 2 3 8 = 4 (2x − 1)(x − 1) = 0 x = 1 2 , 1 ∴ x = 1 2 , y = 4 or x = 1, y = 2 13 a y 14 a when x = 1, LHS = 8 − 4(4) + 2 + 6 = 0 ∴ x = 1 is a solution b 23x = (2x)3 = u3 y = ( 1 2 )x y = 2x + 1 22x = (2x)2 = u2 ∴ (I) ⇒ u3 − 4u2 + u + 6 = 0 c x = 1 ⇒ u = 2 ∴ (u − 2) is a factor O x b at A, 2x + 1 = ( 1 2 )x (2x)2 + 2x = 1 22x + 2x − 1 = 0 c 2x = 1 1 4 2 −± + 2x = 1 5 2 −− [no sols] or 1 5 2 −+ (u − 2)(u2 − 2u − 3) = 0 ∴ 2x = 1 2 5 − 1 2 (u − 2)(u − 3)(u + 1) = 0 ∴ y = ( 1 2 5 − 1 2 ) + 1 = 1 2 ( 5 + 1) u = 2x = −1 [no sols], 2 or 3 x = 1 (given) or lg3 lg2 = 1.58 u2 − 2u − 3 u − 2 u3 − 4u2 + u + 6 u3 − 2u2 − 2u2 + u − 2u2 + 4u − 3u + 6 − 3u + 6 (0, 2) (0, 1) Dr. Faisal Rana www.biochemtuition.com faisal.rana@biochemtuition.com |
11294 | https://www.nejm.org/doi/abs/10.1056/NEJMcp1001124 | Gout | New England Journal of Medicine
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Author: Tuhina Neogi, M.D., Ph.D.Author Info & Affiliations
Published February 3, 2011
N Engl J Med 2011;364:443-452
DOI: 10.1056/NEJMcp1001124
VOL. 364 NO. 5
Copyright © 2011
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This article reviews the use of NSAIDs, colchicine, glucocorticoids, and corticotropin for acute flares of gout and the use of long-term urate-lowering therapies. Both well-established therapies and newer urate-lowering therapies are discussed.
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An audio version of this article is available at NEJM.org.
Dr. Neogi reports serving as a core expert panel leader for the American College of Rheumatology Gout Treatment Guidelines.
Disclosure forms provided by the author are available with the full text of this article at NEJM.org.
No other potential conflict of interest relevant to this article was reported.
I thank Drs. Saralynn Allaire, Hyon Choi, and Yuqing Zhang for their review of the first draft of the manuscript and Dr. Robert Terkeltaub for his review of an earlier version of Figure 1.
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New England Journal of Medicine
Volume 364 • Number 5 • February 3, 2011
Pages: 443-452
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Copyright © 2011 Massachusetts Medical Society. All rights reserved.
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Published online: February 3, 2011
Published in issue: February 3, 2011
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Clinical Medicine General
Emergency Medicine General
Rheumatoid Arthritis
Urology/Prostate Disease General
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Tuhina Neogi, M.D., Ph.D.
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From the Section of Clinical Epidemiology Research and Training Unit, Boston University School of Medicine; and the Department of Epidemiology, Boston University School of Public Health — both in Boston.
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Address reprint requests to Dr. Neogi at the Clinical Epidemiology Unit, Boston University School of Medicine, 650 Albany St., Suite X-200, Boston, MA 02118, or at tneogi@bu.edu.
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11295 | https://api.pageplace.de/preview/DT0400.9781292034003_A24620400/preview-9781292034003_A24620400.pdf | Inorganic Chemistry Gary L. Miessler Paul J. Fischer Donald A. T arr Fifth Edition Pearson Education Limited Edinburgh Gate Harlow Essex CM20 2JE England and Associated Companies throughout the world Visit us on the World Wide Web at: www.pearsoned.co.uk © Pearson Education Limited 2014 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without either the prior written permission of the publisher or a licence permitting restricted copying in the United Kingdom issued by the Copyright Licensing Agency Ltd, Saffron House, 6–10 Kirby Street, London EC1N 8TS.
All trademarks used herein are the property of their respective owners. The use of any trademark in this text does not vest in the author or publisher any trademark ownership rights in such trademarks, nor does the use of such trademarks imply any affi liation with or endorsement of this book by such owners. British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library Printed in the United States of America ISBN 10: 1-292-02075-X ISBN 13: 978-1-292-02075-4 Table of Contents P E A R S O N C U S T O M L I B R A R Y I 1. Introduction to Inorganic Chemistry 1 Gary L. Miessler/Paul J. Fischer/Donald A. Tarr 2. Atomic Structure 9 Gary L. Miessler/Paul J. Fischer/Donald A. Tarr 3. Simple Bonding Theory 47 Gary L. Miessler/Paul J. Fischer/Donald A. Tarr 4. Symmetry and Group Theory 79 Gary L. Miessler/Paul J. Fischer/Donald A. Tarr 5. Molecular Orbitals 129 Gary L. Miessler/Paul J. Fischer/Donald A. Tarr 6. Acid–Base and Donor–Acceptor Chemistry 187 Gary L. Miessler/Paul J. Fischer/Donald A. Tarr 7. The Crystalline Solid State 237 Gary L. Miessler/Paul J. Fischer/Donald A. Tarr 8. Chemistry of the Main Group Elements 273 Gary L. Miessler/Paul J. Fischer/Donald A. Tarr 9. Coordination Chemistry I: Structures and Isomers 341 Gary L. Miessler/Paul J. Fischer/Donald A. Tarr 10. Coordination Chemistry II: Bonding 389 Gary L. Miessler/Paul J. Fischer/Donald A. Tarr 11. Coordination Chemistry III: Electronic Spectra 441 Gary L. Miessler/Paul J. Fischer/Donald A. Tarr 12. Coordination Chemistry IV: Reactions and Mechanisms 477 Gary L. Miessler/Paul J. Fischer/Donald A. Tarr 13. Organometallic Chemistry 517 Gary L. Miessler/Paul J. Fischer/Donald A. Tarr II 14. Organometallic Reactions and Catalysis 589 Gary L. Miessler/Paul J. Fischer/Donald A. Tarr Greek Alphabet and Names and Symbols for the Elements 629 Gary L. Miessler/Paul J. Fischer/Donald A. Tarr Appendix: Character Tables 633 Gary L. Miessler/Paul J. Fischer/Donald A. Tarr Electron Configurations of the Elements, Physical Constants, and Conversion Factors 645 Gary L. Miessler/Paul J. Fischer/Donald A. Tarr Appendix: Useful Data 649 Gary L. Miessler/Paul J. Fischer/Donald A. Tarr 667 Index Introduction to Inorganic Chemistry 1 What Is Inorganic Chemistry?
If organic chemistry is defined as the chemistry of hydrocarbon compounds and their derivatives, inorganic chemistry can be described broadly as the chemistry of “everything else.” This includes all the remaining elements in the periodic table, as well as carbon, which plays a major and growing role in inorganic chemistry. The large field of organo-metallic chemistry bridges both areas by considering compounds containing metal–carbon bonds; it also includes catalysis of many organic reactions. Bioinorganic chemistry bridges biochemistry and inorganic chemistry and has an important focus on medical applications. Environmental chemistry includes the study of both inorganic and organic compounds. In short, the inorganic realm is vast, providing essentially limitless areas for investigation and potential practical applications.
2 Contrasts with Organic Chemistry Some comparisons between organic and inorganic compounds are in order. In both areas, single, double, and triple covalent bonds are found (Figure 1); for inorganic compounds, these include direct metal—metal bonds and metal—carbon bonds. Although the maxi-mum number of bonds between two carbon atoms is three, there are many compounds that contain quadruple bonds between metal atoms. In addition to the sigma and pi bonds common in organic chemistry, quadruply bonded metal atoms contain a delta (d) bond (Figure 2); a combination of one sigma bond, two pi bonds, and one delta bond makes up the quadruple bond. The delta bond is possible in these cases because the metal atoms have d orbitals to use in bonding, whereas carbon has only s and p orbitals energetically accessible for bonding.
Compounds with “fivefold” bonds between transition metals have been reported ( Figure 3), accompanied by debate as to whether these bonds merit the designation “quin-tuple.” In organic compounds, hydrogen is nearly always bonded to a single carbon. In inor-ganic compounds, hydrogen is frequently encountered as a bridging atom between two or more other atoms. Bridging hydrogen atoms can also occur in metal cluster compounds, in which hydrogen atoms form bridges across edges or faces of polyhedra of metal atoms. Alkyl groups may also act as bridges in inorganic compounds, a function rarely encoun-tered in organic chemistry except in reaction intermediates. Examples of terminal and bridging hydrogen atoms and alkyl groups in inorganic compounds are in Figure 4.
Some of the most striking differences between the chemistry of carbon and that of many other elements are in coordination number and geometry. Although carbon is usually limited to a maximum coordination number of four (a maximum of four atoms bonded H H H H H H B B From Chapter 1 of Inorganic Chemistry, Fifth Edition. Gary L Miessler, Paul J. Fischer, Donald A. Tarr. Copyright © 2014 by Pearson Education, Inc. All rights reserved.
1 Introduction to Inorganic Chemistry H H H H H H C C F F H H H H C C C H H C N N O O W S S S S S S S S S S W Organic Inorganic Organometallic 3Hg Hg42+ OC CH3 Mn C C C O C O C O O C O C O CO CO CO O C O C O NR2 NR2 C CH3 OC C Cr CH3 OC6H5 I Cr NR2 R2N Cl Cl Cl Cl Os Os Cl ClCl Cl 2-Cl Cl ClCl Cl Cl Cl Cl Re Re 2-FIGURE 1 Single and Multiple Bonds in Organic and Inorganic Molecules.
+ + + Sigma Pi s s s p p p Delta d d d FIGURE 2 Examples of Bonding Interactions.
to carbon, as in CH4), numerous inorganic compounds have central atoms with coordina-tion numbers of five, six, seven, and higher; the most common coordination geometry for transition metals is an octahedral arrangement around a central atom, as shown for [TiF6]3- (Figure 5). Furthermore, inorganic compounds present coordination geometries different from those found for carbon. For example, although 4-coordinate carbon is nearly always tetrahedral, both tetrahedral and square-planar shapes occur for 4-coordinate com-pounds of both metals and nonmetals. When metals are in the center, with anions or neu-tral molecules (ligands) bonded to them (frequently through N, O, or S), these are called coordination complexes; when carbon is the element directly bonded to metal atoms or ions, they are also classified as organometallic complexes.
i-Pr i-Pr i-Pr Cr Cr i-Pr i-Pr i-Pr i-Pr i-Pr FIGURE 3 Example of Fivefold Bonding.
OC CO H H3 C H3 C Cr H3C H3C CH3 CH3 Al Al H H H H H H B B CO CO O C Cr O C O O O C C C Li Li Li Li = CH3 Each CH3 bridges a face of the Li4 tetrahedron.
O C -FIGURE 4 Examples of Inorganic Compounds Containing Terminal and Bridging Hydrogens and Alkyl Groups.
2 Introduction to Inorganic Chemistry The tetrahedral geometry usually found in 4-coordinate compounds of carbon also occurs in a different form in some inorganic molecules. Methane contains four hydrogens in a regular tetrahedron around carbon. Elemental phosphorus is tetratomic (P4) and tet-rahedral, but with no central atom. Other elements can also form molecules in which outer atoms surround a central cavity; an example is boron, which forms numerous structures containing icosahedral B12 units. Examples of some of the geometries found for inorganic compounds are in Figure 5.
Aromatic rings are common in organic chemistry, and aryl groups can also form sigma bonds to metals. However, aromatic rings can also bond to metals in a dramatically different fashion using their pi orbitals, as shown in Figure 6. The result is a metal atom bonded above the center of the ring, almost as if suspended in space. In many cases, metal atoms are sandwiched between two aromatic rings. Multiple-decker sandwiches of metals and aromatic rings are also known.
Carbon plays an unusual role in a number of metal cluster compounds in which a carbon atom is at the center of a polyhedron of metal atoms. Examples of carbon-centered clusters with five, six, or more surrounding metals are known (Figure 7). The striking role that carbon plays in these clusters has provided a challenge to theoretical inorganic chemists.
In addition, since the mid-1980s the chemistry of elemental carbon has flourished. This phenomenon began with the discovery of fullerenes, most notably the cluster C60, dubbed “buckminsterfullerene” after the developer of the geodesic dome. Many other fullerenes (buckyballs) are now known and serve as cores of a variety of derivatives. In CF3 CF3 F3C F3C Fe Ni Ni Mo S S Mo Zn Zn Cr S S + FIGURE 6 Inorganic Compounds Containing Pi-Bonded Aromatic Rings.
Fe1CO23 Fe1CO23 C 1CO23Fe Fe1CO23 1CO23Fe 1CO23Ru 1CO22 Fe1CO23 Fe1CO23 C 1CO23Fe Fe1CO23 1CO23Fe Ru1CO22 Ru1CO23 C Ru1CO23 1CO23Ru Ru1CO23 1CO22Ru OC FIGURE 7 Carbon-Centered Metal Clusters.
FIGURE 5 Examples of Geometries of Inorganic Compounds.
F F Ti Xe F F H F F F F F Pt P P P P N N N N B H H H H H Cl H H Cl H H B N B F B12H12 2- (not shown: one hydrogen on each boron) 3-B B B B B B B B B B B 2-B I F F F F F F F F F Ti Xe F F H F F F F F Pt P P P P N N N N B H H H H H Cl H H Cl H H B N B F B12H12 2- (not shown: one hydrogen on each boron) 3-B B B B B B B B B B B 2-B I F F F F F F F 3 Introduction to Inorganic Chemistry addition, numerous other forms of carbon (for example, carbon nanotubes, nanoribbons, graphene, and carbon wires) have attracted much interest and show potential for applica-tions in fields as diverse as nanoelectronics, body armor, and drug delivery. Figure 8 provides examples of these newer forms of carbon.
The era of sharp dividing lines between subfields in chemistry has long been obsolete. Many of the subjects in this text, such as acid–base chemistry and organometallic reactions, are of vital interest to organic chemists. Other topics such as oxidation–reduction reactions, spectra, and solubility relations interest analytical chemists. Subjects related to structure determination, spectra, conductivity, and theories of bonding appeal to physical chemists. Finally, the use of organometallic catalysts provides a connection to petroleum and poly-mer chemistry, and coordination compounds such as hemoglobin and metal-containing enzymes provide a similar tie to biochemistry. Many inorganic chemists work with profes-sionals in other fields to apply chemical discoveries to addressing modern challenges in medicine, energy, the environment, materials science, and other fields. In brief, modern inorganic chemistry is not a fragmented field of study, but has numerous interconnections with other fields of science, medicine, technology, and other disciplines.
The remainder of this chapter is devoted to a short history of the origins of inorganic chemistry and perspective on more recent developments, intended to provide a sense of connection to the past and to place some aspects of inorganic chemistry within the context of larger historical events.
3 The History of Inorganic Chemistry Even before alchemy became a subject of study, many chemical reactions were used and their products applied to daily life. The first metals used were probably gold and copper, which can be found in the metallic state in nature. Copper can also be readily formed by the reduction of malachite—basic copper carbonate, Cu2(CO3)(OH)2—in charcoal fires. Silver, tin, antimony, and lead were also known as early as 3000 bce. Iron appeared in FIGURE 8 The Fullerene C60, a Fullerene Compound, a Carbon Nanotube, Graphene, a Carbon Peapod, and a Polyyne “Wire” Connecting Platinum Atoms.
4 Introduction to Inorganic Chemistry classical Greece and in other areas around the Mediterranean Sea by 1500 bce. At about the same time, colored glasses and ceramic glazes were introduced, largely composed of silicon dioxide (SiO2, the major component of sand) and other metallic oxides, which had been melted and allowed to cool to amorphous solids.
Alchemists were active in China, Egypt, and other centers of civilization early in the first centuries ce. Although much effort went into attempts to “transmute” base metals into gold, alchemists also described many other chemical reactions and operations. Distillation, sublimation, crystallization, and other techniques were developed and used in their stud-ies. Because of the political and social changes of the time, alchemy shifted into the Arab world and later—about 1000 to 1500 ce—reappeared in Europe. Gunpowder was used in Chinese fireworks as early as 1150, and alchemy was also widespread in China and India at that time. Alchemists appeared in art, literature, and science until at least 1600, by which time chemistry was beginning to take shape as a science. Roger Bacon (1214–1294), recog-nized as one of the first great experimental scientists, also wrote extensively about alchemy.
By the seventeenth century, the common strong acids—nitric, sulfuric, and hydro-chloric—were known, and systematic descriptions of common salts and their reactions were being accumulated. As experimental techniques improved, the quantitative study of chemical reactions and the properties of gases became more common, atomic and molecu-lar weights were determined more accurately, and the groundwork was laid for what later became the periodic table of the elements. By 1869, the concepts of atoms and molecules were well established, and it was possible for Mendeleev and Meyer to propose different forms of the periodic table. Figure 9 illustrates Mendeleev’s original periodic table. The chemical industry, which had been in existence since very early times in the form of factories for purifying salts and for smelting and refining metals, expanded as methods for preparing relatively pure materials became common. In 1896, Becquerel discovered radioactivity, and another area of study was opened. Studies of subatomic particles, spectra, and electricity led to the atomic theory of Bohr in 1913, which was soon modified by the quantum mechanics of Schrödinger and Heisenberg in 1926 and 1927.
Inorganic chemistry as a field of study was extremely important during the early years of the exploration and development of mineral resources. Qualitative analysis methods were The original table was published in Zeitschrift für Chemie, 1869, 12, 405. It can be found in English translation, together with a page from the German article, at web.lemoyne.edu/~giunta/mendeleev.html. See M. Laing, J. Chem. Educ., 2008, 85, 63 for illustrations of Mendeleev’s various versions of the periodic table, including his handwritten draft of the 1869 table.
Ti = 50 V = 51 Cr = 52 Mn = 53 Fe = 56 Ni = Co = 59 Cu = 63.4 Zn = 65.2 ? = 68 ? = 70 As = 75 Se = 79.4 Br = 80 Rb = 85.4 Sr = 87.6 Ce = 92 La = 94 Di = 95 Th = 118 ?
Zr = 90 Nb = 94 Mo = 96 Rh = 104.4 Ru = 104.2 Pd = 106.6 Ag = 108 Cd = 112 Ur = 116 Sn = 118 Sb = 122 Te = 128?
J = 127 Cs = 133 Ba = 137 ? = 180 Ta = 182 W = 186 Pt = 197.4 Ir = 198 Os = 199 Hg = 200 Au = 197? Bi = 210? Tl = 204 Pb = 207 Mg = 24 Al = 27.4 Si = 28 P = 31 S = 32 Cl = 35.5 K = 39 Ca = 40 ? = 45 ?Er = 56 ?Yt = 60 ?In = 75.6 H = 1 Li = 7 Be = 9.4 B = 11 C = 12 N = 14 O = 16 F = 19 Na = 23 FIGURE 9 Mendeleev’s 1869 Periodic Table. Two years later, Mendeleev revised his table into a form similar to a modern short-form periodic table, with eight groups across.
5 Introduction to Inorganic Chemistry developed to help identify minerals and, combined with quantitative methods, to assess their purity and value. As the Industrial Revolution progressed, so did the chemical industry. By the early twentieth century, plants for the high volume production of ammonia, nitric acid, sulfuric acid, sodium hydroxide, and many other inorganic chemicals were common.
Early in the twentieth century, Werner and Jørgensen made considerable progress on understanding the coordination chemistry of transition metals and also discovered a number of organometallic compounds. Nevertheless, the popularity of inorganic chem-istry as a field of study gradually declined during most of the first half of the century. The need for inorganic chemists to work on military projects during World War II rejuve-nated interest in the field. As work was done on many projects (not least of which was the Manhattan Project, in which scientists developed the fission bomb), new areas of research appeared, and new theories were proposed that prompted further experimental work. A great expansion of inorganic chemistry began in the 1940s, sparked by the enthusiasm and ideas generated during World War II.
In the 1950s, an earlier method used to describe the spectra of metal ions surrounded by negatively charged ions in crystals (crystal field theory)1 was extended by the use of molecular orbital theory2 to develop ligand field theory for use in coordination compounds, in which metal ions are surrounded by ions or molecules that donate electron pairs. This theory gave a more complete picture of the bonding in these compounds. The field devel-oped rapidly as a result of this theoretical framework, availability of new instruments, and the generally reawakened interest in inorganic chemistry.
In 1955, Ziegler3 and Natta4 discovered organometallic compounds that could cata-lyze the polymerization of ethylene at lower temperatures and pressures than the common industrial method at that time. In addition, the polyethylene formed was more likely to be made up of linear, rather than branched, molecules and, as a consequence, was stronger and more durable. Other catalysts were soon developed, and their study contributed to the rapid expansion of organometallic chemistry, still a rapidly growing area.
The study of biological materials containing metal atoms has also progressed rapidly. The development of new experimental methods allowed more thorough study of these compounds, and the related theoretical work provided connections to other areas of study. Attempts to make model compounds that have chemical and biological activity similar to the natural compounds have also led to many new synthetic techniques. Two of the many biological molecules that contain metals are in Figure 10. Although these molecules have very different roles, they share similar ring systems.
One current area that bridges organometallic chemistry and bioinorganic chemistry is the conversion of nitrogen to ammonia: N2 + 3 H2 h 2 NH3 This reaction is one of the most important industrial processes, with over 100 million tons of ammonia produced annually worldwide, primarily for fertilizer. However, in spite of metal oxide catalysts introduced in the Haber–Bosch process in 1913, and improved since then, it is also a reaction that requires temperatures between 350 and 550 °C and from 150–350 atm pressure and that still results in a yield of only 15 percent ammonia. Bacteria, however, manage to fix nitrogen (convert it to ammonia and then to nitrite and nitrate) at 0.8 atm at room temperature in nodules on the roots of legumes. The nitrogenase enzyme that catalyzes this reaction is a complex iron–molybdenum–sulfur protein. The structure of its active sites has been determined by X-ray crystallography.5 A vigorous area of modern inorganic research is to design reactions that could be carried out on an industrial scale that model the reaction of nitrogenase to generate ammonia under mild conditions. It is estimated that as much as 1 percent of the world’s total energy consumption is currently used for the Haber–Bosch process.
Inorganic chemistry also has medical applications. Notable among these is the development of platinum-containing antitumor agents, the first of which was the cis isomer of Pt(NH3)2Cl2, 6 Introduction to Inorganic Chemistry cisplatin. First approved for clinical use approximately 30 years ago, cisplatin has served as the prototype for a variety of anticancer agents; for example, satraplatin, the first orally avail-able platinum anticancer drug to reach clinical trials. These two compounds are in Figure 11.
4 Perspective The premier issue of the journal Inorganic Chemistry was published in February 1962. Much of the focus of that issue was on classic coordination chemistry, with more than half its research papers on synthesis of coordination complexes and their structures and proper-ties. A few papers were on compounds of nonmetals and on organometallic chemistry, then a relatively new field; several were on thermodynamics or spectroscopy. All of these topics have developed considerably in the subsequent half-century, but much of the evolution of inorganic chemistry has been into realms unforeseen in 1962.
The 1962 publication of the first edition of F. A. Cotton and G. Wilkinson’s landmark text Advanced Inorganic Chemistry6 provides a convenient reference point for the status of inorganic chemistry at that time. For example, this text cited only the two long-known forms of carbon, diamond and graphite, although it did mention “amorphous forms” attrib-uted to microcrystalline graphite. It would not be until more than two decades later that carbon chemistry would explode with the seminal discovery of C60 in 1985 by Kroto, Curl, Smalley, and colleagues,7 followed by other fullerenes, nanotubes, graphene, and other forms of carbon (Figure 8) with the potential to have major impacts on electronics, materials science, medicine, and other realms of science and technology.
As another example, at the beginning of 1962 the elements helium through radon were commonly dubbed “inert” gases, believed to “form no chemically bound compounds” because of the stability of their electron configurations. Later that same year, Bartlett CH3 CH3 CH2 H2NOC CH2 H3C CH2 CONH2 CH2 CH2 CH2 CH3 NH2 OH OH CH2CH2CONH2 CH2CONH2 CH2CH2CONH2 CH3 CH3 CH3 CH3 CH3 H H H H O H H3C CH2OH CH CH CO CH2 CH2 NH CH3 N N N N Co N N N N N H H N O O O O-O P HO H CH2 CH2 COOC20H39 (b) (a) CH2 Mg CH3 CH3 C O CH2CH3 H3C H3C CH HC C H C COOCH3 HC H H H H CONH2 H H N N N N FIGURE 10 Biological Molecules Containing Metal Ions. (a) Chlorophyll a, the active agent in photosynthesis. (b) Vitamin B12 coenzyme, a naturally occurring organome-tallic compound.
Cl O O Cl NH3 NH3 P Cl Cl Pt Cl O O C C O O Cl NH3 NH3 NH3 Pt N H2 CH3 CH3 Cl Cl Pt FIGURE 11 Cisplatin and Satraplatin.
The authors of this issue of Inorganic Chemistry were a distinguished group, including five recipients of the Priestley Medal, the highest honor conferred by the American Chemical Society, and 1983 Nobel Laureate Henry Taube.
For reviews of modes of interaction of cisplatin and related drugs, see P. C. A. Bruijnincx, P. J. Sadler, Curr. Opin. Chem. Bio., 2008, 12, 197 and F. Arnesano, G. Natile, Coord. Chem. Rev., 2009, 253, 2070.
7 Introduction to Inorganic Chemistry reported the first chemical reactions of xenon with PtF6, launching the synthetic chemistry of the now-renamed “noble” gas elements, especially xenon and krypton;8 numerous compounds of these elements have been prepared in succeeding decades.
Numerous square planar platinum complexes were known by 1962; the chemistry of platinum compounds had been underway for more than a century. However, it was not known until Rosenberg’s work in the latter part of the 1960s that one of these, cis@Pt(NH3)2Cl2 (cisplatin, Figure 11), had anticancer activity.9 Antitumor agents containing platinum and other transition metals have subsequently become major tools in treatment regimens for many types of cancer.10 That first issue of Inorganic Chemistry contained only 188 pages, and the journal was published quarterly, exclusively in hardcopy. Researchers from only four countries were represented, more than 90 percent from the United States, the others from Europe. Inorganic Chemistry now averages approximately 550 pages per issue, is published 24 times annually, and publishes (electronically) research conducted broadly around the globe. The growth and diversity of research published in Inorganic Chemistry has been paralleled in a wide variety of other journals that publish articles on inorganic and related fields.
In the preface to the first edition of Advanced Inorganic Chemistry, Cotton and Wilkinson stated, “in recent years, inorganic chemistry has experienced an impressive renaissance.” This renaissance shows no sign of diminishing.
With this brief survey of the marvelously complex field of inorganic chemistry, we now turn to the details in the remainder of this text. The topics included provide a broad introduction to the field. However, even a cursory examination of a chemical library or one of the many inorganic journals shows some important aspects of inorganic chemistry that must be omitted in a textbook of moderate length. The references cited in this text suggest resources for further study, including historical sources, texts, and reference works that provide useful additional material.
References 1. H. A. Bethe, Ann. Physik, 1929, 3, 133. 2. J. S. Griffith, L. E. Orgel, Q. Rev. Chem. Soc., 1957, XI, 381. 3. K. Ziegler, E. Holzkamp, H. Breil, H. Martin, Angew. Chem., 1955, 67, 541. 4. G. Natta, J. Polym. Sci., 1955, 16, 143. 5. M. K. Chan, J. Kin, D. C. Rees, Science, 1993, 260, 792. 6. F. A. Cotton, G. Wilkinson, Advanced Inorganic Chemistry, Interscience, John Wiley & Sons, 1962. 7. H. W, Kroto, J. R. Heath, S. C. O’Brien, R. F. Curl, R. E. Smalley, Nature (London), 1985, 318, 162. 8. N. Bartlett, D. H. Lohmann, Proc. Chem. Soc., 1962, 115; N. Bartlett, Proc. Chem. Soc., 1962, 218. 9. B. Rosenberg, L. VanCamp, J. E. Trosko, V. H. Mansour, Nature, 1969, 222, 385. 10. C. G. Hartinger, N. Metzler-Nolte, P. J. Dyson, Organometallics, 2012, 31, 5677 and P. C. A. Bruijnincx, P. J. Sadler, Adv. Inorg. Chem., 2009, 61, 1; G. N. Kaluderovi ´ c, R. Paschke, Curr. Med. Chem., 2011, 18, 4738.
General References For those who are interested in the historical development of inorganic chemistry focused on metal coordination compounds during the period 1798–1935, copies of key research papers, including translations, are provided in the three-volume set Classics in Coordination Chemistry, G. B. Kauffman, ed., Dover Publications, N.Y. 1968, 1976, 1978. Among the many general reference works available, three of the most useful and complete are N. N. Greenwood and A. Earnshaw’s Chemistry of the Elements, 2nd ed., Butterworth-Heinemann, Oxford, 1997; F. A. Cotton, G. Wilkinson, C. A. Murillo, and M. Bochman’s Advanced Inorganic Chemistry, 6th ed., John Wiley & Sons, New York, 1999; and A. F. Wells’s Structural Inorganic Chem-istry, 5th ed., Oxford University Press, New York, 1984. An interesting study of inorganic reactions from a different perspec-tive can be found in G. Wulfsberg’s Principles of Descriptive Inorganic Chemistry, Brooks/Cole, Belmont, CA, 1987.
8 Understanding the structure of the atom has been a fundamental challenge for centuries. It is possible to gain a practical understanding of atomic and molecular structure using only a moderate amount of mathematics rather than the mathematical sophistication of quantum mechanics. This chapter introduces the fundamentals needed to explain atomic structure in qualitative and semiquantitative terms.
1 Historical Development of Atomic Theory Although the Greek philosophers Democritus (460–370 bce) and Epicurus (341–270 bce) presented views of nature that included atoms, many centuries passed before experimental studies could establish the quantitative relationships needed for a coherent atomic theory. In 1808, John Dalton published A New System of Chemical Philosophy,1 in which he proposed that … the ultimate particles of all homogeneous bodies are perfectly alike in weight, figure, etc. In other words, every particle of water is like every other particle of water; every particle of hydrogen is like every other particle of hydrogen, etc.2 and that atoms combine in simple numerical ratios to form compounds. The terminology he used has since been modified, but he clearly presented the concepts of atoms and molecules, and made quantitative observations of the masses and volumes of substances as they combined to form new substances. For example, in describing the reaction between the gases hydrogen and oxygen to form water Dalton said that When two measures of hydrogen and one of oxygen gas are mixed, and fired by the electric spark, the whole is converted into steam, and if the pressure be great, this steam becomes water. It is most probable then that there is the same number of particles in two measures of hydrogen as in one of oxygen.3 Because Dalton was not aware of the diatomic nature of the molecules H2 and O2, which he assumed to be monatomic H and O, he did not find the correct formula of water, and therefore his surmise about the relative numbers of particles in “measures” of the gases is inconsistent with the modern concept of the mole and the chemical equation 2H2 + O2 S 2H2O.
Only a few years later, Avogadro used data from Gay-Lussac to argue that equal volumes of gas at equal temperatures and pressures contain the same number of mole-cules, but uncertainties about the nature of sulfur, phosphorus, arsenic, and mercury vapors delayed acceptance of this idea. Widespread confusion about atomic weights and molecular formulas contributed to the delay; in 1861, Kekulé gave 19 different possible formulas for acetic acid!4 In the 1850s, Cannizzaro revived the argument of Avogadro and argued that From Chapter 2 of Inorganic Chemistry, Fifth Edition. Gary L Miessler, Paul J. Fischer, Donald A. Tarr. Copyright © 2014 by Pearson Education, Inc. All rights reserved.
Atomic Structure 9 Atomic Structure everyone should use the same set of atomic weights rather than the many different sets then being used. At a meeting in Karlsruhe in 1860, Cannizzaro distributed a pamphlet describing his views.5 His proposal was eventually accepted, and a consistent set of atomic weights and formulas evolved. In 1869, Mendeleev6 and Meyer7 independently proposed periodic tables nearly like those used today, and from that time the development of atomic theory progressed rapidly.
1.1 The Periodic Table The idea of arranging the elements into a periodic table had been considered by many chemists, but either data to support the idea were insufficient or the classification schemes were incomplete. Mendeleev and Meyer organized the elements in order of atomic weight and then identified groups of elements with similar properties. By arranging these groups in rows and columns, and by considering similarities in chemical behavior as well as atomic weight, Mendeleev found vacancies in the table and was able to predict the proper-ties of several elements—gallium, scandium, germanium, and polonium—that had not yet been discovered. When his predictions proved accurate, the concept of a periodic table was quickly accepted. The discovery of additional elements not known in Mendeleev’s time and the synthesis of heavy elements have led to the modern periodic table.
In the modern periodic table, a horizontal row of elements is called a period and a vertical column is a group. The traditional designations of groups in the United States differ from those used in Europe. The International Union of Pure and Applied Chem-istry (IUPAC) has recommended that the groups be numbered 1 through 18. In this text, we will use primarily the IUPAC group numbers. Some sections of the periodic table have traditional names, as shown in Figure 1.
1 3 55 Alkali Metals Coinage Metals Chalcogens Halogens Noble Gases Alkaline Earth Metals 72 87 89 104 22 40 80 112 30 48 57 21 39 81 5 49 13 31 86 71 58 Lanthanides 103 90 Actinides 10 Transition metals 2 1 Groups (IUPAC) 3 6 13 2 5 8 7 10 9 12 11 18 15 14 17 4 16 IA Groups (European tradition) IIIA VIA IIIB IIA VA VIIA VIII IIB IB 0 VB IVB VIIB IVA VIB IA Groups (American tradition) IIIB VIB IIIA IIA VB VIIB VIIIB IIB IB VIIIA VA IVA VIIA IVB VIA FIGURE 1 Numbering Schemes and Names for Parts of the Periodic Table.
10 Atomic Structure 1.2 Discovery of Subatomic Particles and the Bohr Atom During the 50 years after the periodic tables of Mendeleev and Meyer were proposed, experimental advances came rapidly. Some of these discoveries are listed in Table 1.
Parallel discoveries in atomic spectra showed that each element emits light of specific energies when excited by an electric discharge or heat. In 1885, Balmer showed that the energies of visible light emitted by the hydrogen atom are given by the equation E = RHa 1 22 - 1 n2 h b where nh = integer, with nh 7 2 RH = Rydberg constant for hydrogen = 1.097 107 m-1 = 2.179 10-18 J = 13.61 eV and the energy of the light emitted is related to the wavelength, frequency, and wavenumber of the light, as given by the equation E = hv = hc l = hcv where h = Planck constant = 6.626 10-34 J s v = frequency of the light, in s-1 c = speed of light = 2.998 108 m s-1 l = wavelength of the light, frequently in nm v = wavenumber of the light, usually in cm-1 In addition to emission of visible light, as described by the Balmer equation, infrared and ultraviolet emissions were also discovered in the spectrum of the hydrogen atom. The energies of these emissions could be described by replacing 22 by integers nl 2 in Balmer’s original equation, with the condition that nl 6 nh (l for lower level, h for higher level). These quantities, n, are called quantum numbers. (These are the principal quantum numbers; other quantum numbers are discussed in Section 2.2.) The origin of this energy was unknown until Niels Bohr’s quantum theory of the atom,8 first published in 1913 and refined over the following decade. This theory assumed that negatively charged electrons in atoms move in stable circular orbits around the positively charged nucleus with no absorp-tion or emission of energy. However, electrons may absorb light of certain specific energies TABLE 1 Discoveries in Atomic Structure 1896 A. H. Becquerel Discovered radioactivity of uranium 1897 J. J. Thomson Showed that electrons have a negative charge, with charge/mass = 1.76 1011 C/kg 1909 R. A. Millikan Measured the electronic charge as 1.60 10-19 C; therefore, mass of electron = 9.11 10-31 kg 1911 E. Rutherford Established the nuclear model of the atom: a very small, heavy nucleus surrounded by mostly empty space 1913 H. G. J. Moseley Determined nuclear charges by X-ray emission, establishing atomic numbers as more fundamental than atomic masses 11 Atomic Structure and be excited to orbits of higher energy; they may also emit light of specific energies and fall to orbits of lower energy. The energy of the light emitted or absorbed can be found, according to the Bohr model of the hydrogen atom, from the equation E = R a 1 n 2 l - 1 n 2 h b where R = 2p2mZ2e4 (4pe0)2h2 m = reduced mass of the electron/nucleus combination: 1 m = 1 me + 1 mnucleus me = mass of the electron mnucleus = mass of the nucleus Z = charge of the nucleus e = electronic charge h = Planck constant nh = quantum number describing the higher energy state nl = quantum number describing the lower energy state 4pe0 = permittivity of a vacuum This equation shows that the Rydberg constant depends on the mass of the nucleus and on various fundamental constants. If the atom is hydrogen, the subscript H is commonly appended to the Rydberg constant (RH).
Examples of the transitions observed for the hydrogen atom and the energy levels responsible are shown in Figure 2. As the electrons drop from level nh to nl, energy is released in the form of electromagnetic radiation. Conversely, if radiation of the correct energy is absorbed by an atom, electrons are raised from level nl to level nh. The inverse-square dependence of energy on n results in energy levels that are far apart in energy at small n and become much closer in energy at larger n. In the upper limit, as n approaches infinity, the energy approaches a limit of zero. Individual electrons can have more energy, but above this point, they are no longer part of the atom; an infinite quantum number means that the nucleus and the electron are separate entities.
E XE RCISE 1 Determine the energy of the transition from nh = 3 to nl = 2 for the hydrogen atom, in both joules and cm-1 (a common unit in spectroscopy, often used as an energy unit, since v is proportional to E). This transition results in a red line in the visible emission spectrum of hydrogen.
When applied to the hydrogen atom, Bohr’s theory worked well; however, the theory failed when atoms with two or more electrons were considered. Modifications such as ellip-tical rather than circular orbits were unsuccessfully introduced in attempts to fit the data to Bohr’s theory.9 The developing experimental science of atomic spectroscopy provided extensive data for testing Bohr’s theory and its modifications. In spite of the efforts to “fix” the Bohr theory, the theory ultimately proved unsatisfactory; the energy levels predicted by the Bohr equation above and shown in Figure 2 are valid only for the hydrogen atom and 12 Atomic Structure other one-electron situations such as He+, Li2+, and Be3+. A fundamental characteristic of the electron—its wave nature—needed to be considered.
The de Broglie equation, proposed in the 1920s,10 accounted for the electron’s wave nature. According to de Broglie, all moving particles have wave properties described by the equation l = h mu l = wavelength of the particle h = Planck constant m = mass of the particle u = velocity of the particle Balmer series (visible transitions shown) Paschen series (IR) Quantum Number n Energy 0 RH RH -RH RH 6 5 4 3 36 9 -RH 2 1 Lyman series (UV) RH 1 -1 16 -1 1 4 -1 25 -q FIGURE 2 Hydrogen Atom Energy Levels.
Multiplying RH by Z2, the square of the nuclear charge, and adjusting the reduced mass accordingly provides an equation that describes these more exotic one-electron situations.
13 Atomic Structure Particles massive enough to be visible have very short wavelengths, too small to be measured. Electrons, on the other hand, have observable wave properties because of their very small mass.
Electrons moving in circles around the nucleus, as in Bohr’s theory, can be thought of as standing waves that can be described by the de Broglie equation. However, we no longer believe that it is possible to describe the motion of an electron in an atom so precisely. This is a consequence of another fundamental principle of modern physics, Heisenberg’s uncertainty principle,11 which states that there is a relationship between the inherent uncertainties in the location and momentum of an electron. The x component of this uncertainty is described as x px Ú h 4p x = uncertainty in the position of the electron px = uncertainty in the momentum of the electron The energy of spectral lines can be measured with high precision (as an example, recent emission spectral data of hydrogen atoms in the solar corona indicated a difference between nh = 2 and nl = 1 of 82258.9543992821(23) cm-1)!12 This in turn allows precise deter-mination of the energy of electrons in atoms. This precision in energy also implies preci-sion in momentum (px is small); therefore, according to Heisenberg, there is a large uncertainty in the location of the electron (x is large). This means that we cannot treat electrons as simple particles with their motion described precisely, but we must instead consider the wave properties of electrons, characterized by a degree of uncertainty in their location. In other words, instead of being able to describe precise orbits of electrons, as in the Bohr theory, we can only describe orbitals, regions that describe the probable location of electrons. The probability of finding the electron at a particular point in space, also called the electron density, can be calculated—at least in principle.
2 The Schrödinger Equation In 1926 and 1927, Schrödinger13 and Heisenberg11 published papers on wave mechan-ics, descriptions of the wave properties of electrons in atoms, that used very different mathematical techniques. In spite of the different approaches, it was soon shown that their theories were equivalent. Schrödinger’s differential equations are more commonly used to introduce the theory, and we will follow that practice.
The Schrödinger equation describes the wave properties of an electron in terms of its position, mass, total energy, and potential energy. The equation is based on the wave function, , which describes an electron wave in space; in other words, it describes an atomic orbital. In its simplest notation, the equation is H= E H = Hamiltonian operator E = energy of the electron = wave function The Hamiltonian operator, frequently called simply the Hamiltonian, includes deriva-tives that operate on the wave function. When the Hamiltonian is carried out, the result is a constant (the energy) times . The operation can be performed on any wave function An operator is an instruction or set of instructions that states what to do with the function that follows it. It may be a simple instruction such as “multiply the following function by 6,” or it may be much more complicated than the Hamiltonian. The Hamiltonian operator is sometimes written H n with the n (hat) symbol designating an operator.
14 Atomic Structure describing an atomic orbital. Different orbitals have different wave functions and different values of E. This is another way of describing quantization in that each orbital, character-ized by its own function , has a characteristic energy. In the form used for calculating energy levels, the Hamiltonian operator for one-electron systems is H = -h2 8p2m a 02 0x2 + 02 0y2 + 02 0z2 b - Ze2 4p e02x2 + y2 + z2 This part of the operator describes the kinetic energy of the electron, its energy of motion.
This part of the operator describes the potential energy of the electron, the result of electrostatic attraction between the electron and the nucleus. It is commonly designated as V.
where h = Planck constant m = mass of the electron e = charge of the electron 2x2 + y2 + z2 = r = distance from the nucleus Z = charge of the nucleus 4pe0 = permittivity of a vacuum This operator can be applied to a wave function , c -h2 8p2m a 02 0x2 + 02 0y2 + 02 0z2 b + V(x, y, z)d (x, y, z) = E (x, y, z) where V = -Ze2 4pe0 r = -Ze2 4pe02x2 + y2 + z2 The potential energy V is a result of electrostatic attraction between the electron and the nucleus. Attractive forces, such as those between a positive nucleus and a negative electron, are defined by convention to have a negative potential energy. An electron near the nucleus (small r) is strongly attracted to the nucleus and has a large negative potential energy. Electrons farther from the nucleus have potential energies that are small and negative. For an electron at infinite distance from the nucleus (r = ), the attraction between the nucleus and the electron is zero, and the potential energy is zero. The hydrogen atom energy level diagram in Figure 2 illustrates these concepts.
Because n varies from 1 to , and every atomic orbital is described by a unique , there is no limit to the number of solutions of the Schrödinger equation for an atom. Each describes the wave properties of a given electron in a particular orbital. The probability of finding an electron at a given point in space is proportional to 2. A number of condi-tions are required for a physically realistic solution for : 1. The wave function must be single-valued. 2. The wave function and its first derivatives must be continuous.
There cannot be two probabilities for an electron at any position in space.
The probability must be defined at all posi-tions in space and cannot change abruptly from one point to the next. 3. The wave function must approach zero as r approaches infinity.
For large distances from the nucleus, the probability must grow smaller and smaller (the atom must be finite).
15 Atomic Structure 2.1 The Particle in a Box A simple example of the wave equation, the particle in a one-dimensional box, shows how these conditions are used. We will give an outline of the method; details are available else-where. The “box” is shown in Figure 3. The potential energy V(x) inside the box, between x = 0 and x = a, is defined to be zero. Outside the box, the potential energy is infinite. This means that the particle is completely trapped in the box and would require an infinite amount of energy to leave the box. However, there are no forces acting on it within the box. The wave equation for locations within the box is -h2 8p2m a 02 (x) 0x2 b = E (x), because V(x) = 0 Sine and cosine functions have the properties we associate with waves—a well-defined wavelength and amplitude—and we may therefore propose that the wave characteristics of our particle may be described by a combination of sine and cosine functions. A general solution to describe the possible waves in the box would then be = A sin rx + B cos sx where A, B, r, and s are constants. Substitution into the wave equation allows solution for r and s (see Problem 8a at the end of the chapter): r = s = 22mE 2p h Because must be continuous and must equal zero at x 6 0 and x 7 a (because the particle is confined to the box), must go to zero at x = 0 and x = a. Because cos sx = 1 for x = 0, can equal zero in the general solution above only if B = 0. This reduces the expression for to = A sin rx At x = a, must also equal zero; therefore, sin ra = 0, which is possible only if ra is an integral multiple of p: ra = { np or r = {np a G. M. Barrow, Physical Chemistry, 6th ed., McGraw-Hill, New York, 1996, pp. 65, 430, calls this the “particle on a line” problem. Other physical chemistry texts also include solutions to this problem.
V = q V = q V a 0 x V = 0 FIGURE 3 Potential Energy Well for the Particle in a Box. 4. The integral L all space AA dt = 1 5. The integral L all space AB dt = 0 The total probability of an electron being somewhere in space = 1. This is called normalizing the wave function. A and B are wave functions for electrons in different orbitals within the same atom. All orbitals in an atom must be orthogonal to each other. In some cases, this means that the axes of orbitals must be perpendicular, as with the px, py, and pz orbitals.
Because the wave functions may have imaginary values (containing 2-1), (where designates the complex conjugate of ) is used to make the integral real. In many cases, the wave functions themselves are real, and this integral becomes L all space A 2 dt.
16 Atomic Structure where n = any integer 0. Because both positive and negative values yield the same results, substituting the positive value for r into the solution for r gives r = np a = 22mE 2p h This expression may be solved for E: E = n2h2 8ma2 These are the energy levels predicted by the particle-in-a-box model for any particle in a one-dimensional box of length a. The energy levels are quantized according to quantum numbers n = 1, 2, 3, c Substituting r = np/a into the wave function gives = A sin npx a And applying the normalizing requirement L dt = 1 gives A = A 2 a The total solution is then = A 2 a sin npx a The resulting wave functions and their squares for the first three states—the ground state (n = 1) and first two excited states (n = 2 and n = 3)—are plotted in Figure 4.
The squared wave functions are the probability densities; they show one difference between classical and quantum mechanical behavior of an electron in such a box. Classi-cal mechanics predicts that the electron has equal probability of being at any point in the box. The wave nature of the electron gives it varied probabilities at different locations in the box. The greater the square of the electron wave amplitude, the greater the probability of the electron being located at the specified coordinate when at the quantized energy defined by the .
If n = 0, then r = 0 and = 0 at all points. The probability of finding the particle is L dx = 0; if the particle is an electron, there is then no electron at all.
2 0 .2 .4 .6 .8 1 -.5 1.5 1 .5 0 2 0 .2 .4 .6 x/a .8 1 1 0 -1 -2 Particle in a box n = 3 2 0 .2 .4 .6 x/a .8 1 1 0 -1 -2 Particle in a box n = 2 ±2 Wave function ± ±2 Wave function ± ±2 Wave function ± x/a Particle in a box n = 1 FIGURE 4 Wave Functions and Their Squares for the Particle in a Box with n = 1, 2, and 3.
17 Atomic Structure 2.2 Quantum Numbers and Atomic Wave Functions The particle-in-a-box example shows how a wave function operates in one dimension. Mathematically, atomic orbitals are discrete solutions of the three-dimensional Schrödinger equations. The same methods used for the one-dimensional box can be expanded to three dimensions for atoms. These orbital equations include three quantum numbers, n, l, and ml. A fourth quantum number, ms, a result of relativistic corrections to the Schrödinger equation, completes the description by accounting for the magnetic moment of the electron. The quantum numbers are summarized in Table 2. Tables 3 and 4 describe wave functions.
The quantum number n is primarily responsible for determining the overall energy of an atomic orbital; the other quantum numbers have smaller effects on the energy. The quantum number l determines the angular momentum and shape of an orbital. The quantum number ml determines the orientation of the angular momentum vector in a magnetic field, or the position of the orbital in space, as shown in Table 3. The quantum number ms determines the orientation of the electron’s magnetic moment in a magnetic field, either in the direction of the field 1+1 22 or opposed to it 1-1 22. When no field is present, all ml values associated with a given n—all three p orbitals or all five d orbitals—have the same energy, and both ms values have the same energy. Together, the quantum numbers n, l, and ml define an atomic orbital.
The quantum number ms describes the electron spin within the orbital. This fourth quantum number is consistent with a famous experimental observation. When a beam of alkali metal atoms (each with a single valence electron) is passed through a magnetic field, the beam splits into two parts; half the atoms are attracted by one magnet pole, and half are attracted by the opposite pole. Because in classical physics spinning charged particles generate magnetic moments, it is common to attribute an electron’s magnetic moment to its spin—as if an electron were a tiny bar magnet—with the orientation of the magnetic field vector a function of the spin direction (counterclockwise vs. clockwise). However, the spin of an electron is a purely quantum mechanical property; application of classical mechanics to an electron is inaccurate.
One feature that should be mentioned is the appearance of i( = 2-1) in the p and d orbital wave equations in Table 3. Because it is much more convenient to work with Also called the azimuthal quantum number.
TABLE 2 Quantum Numbers and Their Properties Symbol Name Values Role n Principal 1, 2, 3, . . . Determines the major part of the energy l Angular momentum 0, 1, 2, . . ., n - 1 Describes angular dependence and contributes to the energy ml Magnetic 0, {1, {2, c, {l Describes orientation in space (angular momentum in the z direction) ms Spin { 1 2 Describes orientation of the electron spin (magnetic moment) in space Orbitals with different l values are known by the following labels, derived from early terms for different families of spectroscopic lines: l 0 1 2 3 4 5, … Label s p d f g continuing alphabetically 18 Atomic Structure real functions than complex functions, we usually take advantage of another property of the wave equation. For differential equations of this type, any linear combination of solu-tions to the equation—sums or differences of the functions, with each multiplied by any coefficient—is also a solution to the equation. The combinations usually chosen for the p orbitals are the sum and difference of the p orbitals having ml = + 1 and –1, normalized by multiplying by the constants 1 22 and i 22 , respectively: 2px = 1 22 (+1 + -1) = 1 2 A 3 p 3R(r)4 sin u cos f 2py = i 22 (+1 - -1) = 1 2 A 3 p 3R(r)4 sin u sin f TABLE 3 Hydrogen Atom Wave Functions: Angular Functions Angular Factors Real Wave Functions Related to Angular Momentum Functions of u In Polar Coordinates In Cartesian Coordinates Shapes Label l ml z (u, f) (x, y, z) 0(s) 0 1 22p 1 22 1 22p 1 22p z y x z y x s 1(p) 0 1 22p 26 2 cos u 1 2 A 3 p cos u 1 2 A 3 p z r pz +1 1 22p eif 23 2 sin u w z g 1 2 A 3 p sin u cos f 1 2 A 3 p x r z y x z y x px -1 1 22p e-if 23 2 sin u 1 2 A 3 p sin u sin f 1 2 A 3 p y r py 2(d) 0 1 22p 1 2A 5 2 (3 cos 2 u-1) z 1 4A 5 p (3 cos2 u - 1) 1 4A 5 p (2z2 - x2 - y2) r2 z y x z y x dz2 +1 1 22p eif 215 2 cos u sin u w z g 1 2 A 15 p cos u sin u cos f 1 2 A 15 p xz r2 dxz -1 1 22p e-if 215 2 cos u sin u 1 2 A 15 p cos u sin u sin f 1 2A 15 p yz r2 z y x dyz +2 1 22p e2if 215 4 sin2 u w z g 1 4 A 15 p sin2 u cos 2f 1 4A 15 p (x2 - y2) r2 z y x dx2-y2 -2 1 22p e-2if 215 4 sin2 u 1 4 A 15 p sin2 u sin 2f 1 4A 15 p xy r2 z y x dxy Source: Hydrogen Atom Wave Functions: Angular Functions, Physical Chemistry, 5th ed.,Gordon Barrow (c) 1988. McGraw-Hill Companies, Inc.
NOTE: The relations (eif - e-if)/(2i) = sin f and (eif + e-if)/2 = cos f can be used to convert the exponential imaginary functions to real trigonometric functions, combining the two orbitals with ml = { 1 to give two orbitals with sin f and cos f. In a similar fashion, the orbitals with ml = { 2 result in real functions with cos2 f and sin2 f. These functions have then been converted to Cartesian form by using the functions x = r sin u cos f, y = r sin u sin f, and z = r cos u.
19 Atomic Structure TABLE 4 Hydrogen Atom Wave Functions: Radial Functions Radial Functions R(r), with s = Zr/a0 Orbital n l R(r) 1s 1 0 R 1s = 2c Z a0 d 3/2 e -s 2s 2 0 R 2s = 2c Z 2a0 d 3/2 (2 - s)e -s/2 2p 1 R 2p = 1 23 c Z 2a0 d 3/2 se -s/2 3s 3 0 R 3s = 2 27c Z 3a0 d 3/2 (27 - 18s + 2s2)e -s/3 3p 1 R 3p = 1 8123 c 2Z a0 d 3/2 (6 - s)s e -s/3 3d 2 R 3d = 1 81215 c 2Z a0 d 3/2 s2 e -s/3 We should really call this the d2z2-x2-y2 orbital!
The same procedure used on the d orbital functions for ml = { 1 and {2 gives the functions in the column headed (u, f) in Table 3, which are the familiar d orbitals. The dz2 orbital (ml = 0) actually uses the function 2z2 - x2 - y2, which we shorten to z2 for convenience. These functions are now real functions, so = and = 2.
A more detailed look at the Schrödinger equation shows the mathematical origin of atomic orbitals. In three dimensions, may be expressed in terms of Cartesian coordinates (x, y, z) or in terms of spherical coordinates (r, u, f). Spherical coordinates, as shown in Figure 5, are especially useful in that r represents the distance from the nucleus. The spheri-cal coordinate u is the angle from the z axis, varying from 0 to p, and f is the angle from the x axis, varying from 0 to 2p. Conversion between Cartesian and spherical coordinates is carried out with the following expressions: x = r sin u cos f y = r sin u sin f z = r cos u In spherical coordinates, the three sides of the volume element are r du, r sin u df, and dr. The product of the three sides is r2 sin u du df dr, equivalent to dx dy dz. The volume of the thin shell between r and r + dr is 4pr2 dr, which is the integral over f from 0 to p and over u from 0 to 2p. This integral is useful in describing the electron density as a function of distance from the nucleus.
can be factored into a radial component and two angular components. The radial function R describes electron density at different distances from the nucleus; the angular functions and describe the shape of the orbital and its orientation in space. The two angular factors are sometimes combined into one factor, called Y: (r, u, f) = R(r)(u)(f) = R(r)Y(u, f) u Spherical coordinates Volume element u r r f x y z x y du r sin u df r sin u df rdu dr f z FIGURE 5 Spherical Coordinates and Volume Element for a Spherical Shell in Spherical Coordinates.
20 Atomic Structure R is a function only of r; Y is a function of u and f, and it gives the distinctive shapes of s, p, d, and other orbitals. R, and are shown separately in Tables 3 and 4.
Angular Functions The angular functions and determine how the probability changes from point to point at a given distance from the center of the atom; in other words, they give the shape of the orbitals and their orientation in space. The angular functions and are determined by the quantum numbers l and ml. The shapes of s, p, and d orbitals are shown in Table 3 and Figure 6.
In the center of Table 3 are the shapes for the portion; when the portion is included, with values of f = 0 to 2p, the three-dimensional shapes in the far-right col-umn are formed. In the three-dimensional diagrams of orbitals in Table 3, the orbital lobes are shaded where the wave function is negative. The different shadings of the lobes repre-sent different signs of the wave function . It is useful to distinguish regions of opposite signs for bonding purposes.
Radial Functions The radial factor R(r) (Table 4) is determined by the quantum numbers n and l, the principal and angular momentum quantum numbers.
The radial probability function is 4pr2R2. This function describes the probability of finding the electron at a given distance from the nucleus, summed over all angles, with the 4pr2 factor the result of integrating over all angles. The radial wave functions and radial probability functions are plotted for the n = 1, 2, and 3 orbitals in Figure 7. Both R(r) and 4pr2R2 are scaled with a0, the Bohr radius, to give reasonable units on the axes of the FIGURE 6 Selected Atomic Orbitals.
(Selected Atomic Orbitals by Gary O. Spessard and Gary L. Miessler. Reprinted by permission.) 21 Atomic Structure 3p 15 Radial function a0R(r) 0 30 -.1 0 .1 .2 2 3 .3 .4 5 25 20 10 3d Radial function a0R(r) 0 30 0 .1 .2 .3 2 3 .4 .5 5 25 20 10 1s r/a0 15 Radial function a0R(r) 0 30 Radial Wave Functions Radial Probability Functions 0 .4 .8 1.2 2 3 1.6 2 5 25 20 10 3s r/a0 15 Probability a0r2R2 Probability a0r2R2 Probability a0r2R2 Probability a0r2R2 Probability a0r2R2 Probabilitya0r2R2 0 30 0 .1 .2 .3 .4 .5 .6 5 25 20 10 0 .1 .2 .3 .4 .5 .6 0 .1 .2 .3 .4 .5 .6 0 .1 .2 .3 .4 .5 .6 0 .1 .2 .3 .4 .5 .6 0 .1 .2 .3 .4 .5 .6 3p r/a0 15 0 30 5 25 20 10 3d r/a0 15 0 30 5 25 20 10 2s r/a0 15 0 30 5 25 20 10 2p r/a0 15 0 30 5 25 20 10 1s r/a0 15 0 30 5 25 20 10 2p r/a0 15 Radial function a0R(r) 0 30 0 .2 .4 .6 2 3 .8 1 5 25 20 10 3s 15 Radial function a0R(r) 0 30 -.1 0 .1 .2 2 3 .3 .4 5 25 20 10 2s r/a0 15 r/a0 r/a0 15 r/a0 Radial function a0R(r) 0 30 -.2 0 .2 .4 2 3 .6 .8 5 25 20 10 FIGURE 7 Radial Wave Functions and Radial Probability Functions.
22 Atomic Structure graphs. The Bohr radius, a0 = 52.9 pm, is a common unit in quantum mechanics. It is the value of r at the maximum of 2 for a hydrogen 1s orbital (the most probable distance from the hydrogen nucleus for the 1s electron), and it is also the radius of the n = 1 orbit according to the Bohr model.
In all the radial probability plots, the electron density, or probability of finding the electron, falls off rapidly beyond its maximum as the distance from the nucleus increases. It falls off most quickly for the 1s orbital; by r = 5a0, the probability is approaching zero. By contrast, the 3d orbital has a maximum at r = 9a0 and does not approach zero until approximately r = 20a0. All the orbitals, including the s orbitals, have zero probability at the center of the nucleus, because 4pr2R2 = 0 at r = 0. The radial probability functions are a combination of 4pr2, which increases rapidly with r, and R2, which may have maxima and minima, but generally decreases exponentially with r. The product of these two factors gives the characteristic probabilities seen in the plots. Because chemical reactions depend on the shape and extent of orbitals at large distances from the nucleus, the radial probability functions help show which orbitals are most likely to be involved in reactions.
Nodal Surfaces At large distances from the nucleus, the electron density, or probability of finding the electron, falls off rapidly. The 2s orbital also has a nodal surface, a surface with zero electron density, in this case a sphere with r = 2a0 where the probability is zero. Nodes appear naturally as a result of the wave nature of the electron. A node is a surface where the wave function is zero as it changes sign (as at r = 2a0 in the 2s orbital); this requires that = 0, and the probability of finding the electron at any point on the surface is also zero.
If the probability of finding an electron is zero (2 = 0), must also be equal to zero. Because (r, u, f) = R (r)Y(u, f) in order for = 0, either R(r) = 0 or Y(u, f) = 0. We can therefore determine nodal surfaces by determining under what conditions R = 0 or Y = 0.
Table 5 summarizes the nodes for several orbitals. Note that the total number of nodes in any orbital is n – 1 if the conical nodes of some d and f orbitals count as two nodes. Mathematically, the nodal surface for the dz2 orbital is one surface, but in this instance, it fits the pattern better if thought of as two nodes.
TABLE 5 Nodal Surfaces Angular Nodes [Y(u, f) = 0] Examples (number of angular nodes) s orbitals 0 p orbitals 1 plane for each orbital d orbitals 2 planes for each orbital except dz2 1 conical surface for dz2 Radial Nodes [R(r) 0] Examples (number of radial nodes) 1s 0 2p 0 3d 0 2s 1 3p 1 4d 1 3s 2 4p 2 5d 2 pz dx2 - y2 y y x x z x = -y x = y 23 Atomic Structure Angular nodes result when Y = 0, and are planar or conical. Angular nodes can be determined in terms of u and f but may be easier to visualize if Y is expressed in Cartesian (x, y, z) coordinates (see Table 3). In addition, the regions where the wave function is posi-tive and where it is negative can be found. This information will be useful in working with molecular orbitals. There are l angular nodes in any orbital, with the conical surface in the dz2 orbitals—and other orbitals having conical nodes—counted as two nodes.
Radial nodes (spherical nodes) result when R = 0. They give the atom a layered appear-ance, shown in Figure 8 for the 3s and 3pz orbitals. These nodes occur when the radial func-tion changes sign; they are depicted in the radial function graphs by R(r) = 0 and in the radial probability graphs by 4pr2R2 = 0. The lowest energy orbitals of each classifica-tion (1s, 2p, 3d, 4f, etc.) have no radial nodes. The number of radial nodes increases as n increases; the number of radial nodes for a given orbital is always equal to n - l - 1. Nodal surfaces can be puzzling. For example, a p orbital has a nodal plane through the nucleus. How can an electron be on both sides of a node at the same time without ever having been at the node, at which the probability is zero? One explanation is that the prob-ability does not go quite to zero on the basis of relativistic arguments.
Again, counting a conical nodal surface, such as for a dz2 orbital, as two nodes.
0.0316 0.0316 0.10 0.10 0.10 0.10 0.10 0.316 Node Node 0.10 0.316 0.316 Node Node Node 0.316 0.1 0.2 0.1 Node Node Node (a) Cl:3s (c) Cl:3pz (d) Ti3+:3dz2 (e) Ti3+:3dx2 - y2 (f) Ti3+:3dx2 - y2 1 Å 1 Å 1 Å 1 Å 1 Å 0.0316 0.10 0.316 Node (b) C:2pz 1 Å z z x y y z FIGURE 8 Constant Electron Density Surfaces for Selected Atomic Orbitals. (a)–(d) The cross-sectional plane is any plane containing the z axis. (e) The cross section is taken through the xz or yz plane. (f) The cross section is taken through the xy plane.
(Figures (b)–(f) Reproduced with permission from E. A. Orgyzlo and G.B. Porter, in J. Chem. Educ., 40, 258. Copyright 1963. American Chemical Society.) A. Szabo, J. Chem. Educ., 1969, 46, 678 explains that the electron probability at a nodal surface has a very small but finite value.
24 Atomic Structure Another explanation is that such a question really has no meaning for an electron behav-ing as a wave. Recall the particle-in-a-box example. Figure 4 shows nodes at x/a = 0.5 for n = 2 and at x/a = 0.33 and 0.67 for n = 3. The same diagrams could represent the amplitudes of the motion of vibrating strings at the fundamental frequency (n = 1) and multiples of 2 and 3. A plucked violin string vibrates at a specific frequency, and nodes at which the amplitude of vibration is zero are a natural result. Zero amplitude does not mean that the string does not exist at these points but simply that the magnitude of the vibration is zero. An electron wave exists at the node as well as on both sides of a nodal surface, just as a violin string exists at the nodes and on both sides of points having zero amplitude.
Still another explanation, in a lighter vein, was suggested by R. M. Fuoss to one of the authors in a class on bonding. Paraphrased from St. Thomas Aquinas, “Angels are not material beings. Therefore, they can be first in one place and later in another without ever having been in between.” If the word “electrons” replaces the word “angels,” a semitheo-logical interpretation of nodes would result.
E X A M P L E 1 Nodal structure of pz The angular factor Y is given in Table 3 in terms of Cartesian coordinates: Y = 1 2 A 3 p z r This orbital is designated pz because z appears in the Y expression. For an angular node, Y must equal zero, which is true only if z = 0. Therefore, z = 0 (the xy plane) is an angular nodal surface for the pz orbital, as shown in Table 5 and Figure 8. The wave function is positive where z 7 0 and negative where z 6 0. In addition, a 2pz orbital has no radial (spherical) nodes, a 3pz orbital has one radial node, and so on.
Nodal structure of dx2-y2 Y = 1 4A 15 p (x 2 - y 2) r2 Here, the expression x 2 - y 2 appears in the equation, so the designation is dx2-y2.
Because there are two solutions to the equation Y = 0 (setting x 2 - y 2 = 0, the solutions are x = y and x = - y), the planes defined by these equations are the angular nodal surfaces. They are planes containing the z axis and making 45° angles with the x and y axes (see Table 5). The function is positive where x 7 y and negative where x 6 y. In addition, a 3dx2-y2 orbital has no radial nodes, a 4dx2-y2 has one radial node, and so on.
EXERCISE 2 Describe the angular nodal surfaces for a dz2 orbital, whose angular wave function is Y = 1 4 A 5 p (2z 2 - x 2 - y 2) r2 EXERCISE 3 Describe the angular nodal surfaces for a dxz orbital, whose angular wave function is Y = 1 2 A 15 p xz r2 The result of the calculations is the set of atomic orbitals familiar to chemists. Figure 6 shows diagrams of s, p, and d orbitals, and Figure 8 shows lines of constant electron density 25 Atomic Structure in several orbitals. Different shadings of the orbital lobes in Figure 6 indicate different signs of the electron wave amplitude, and the outer surfaces shown enclose 90% of the total electron density of the orbitals. The orbitals we use are the common ones used by chemists; others that are also solutions of the Schrödinger equation can be chosen for special purposes.14 The reader is encouraged to make use of Internet resources that display a wide range of atomic orbitals—including f, g, and higher orbitals—show radial and angular nodes, and provide additional information. 2.3 The Aufbau Principle Limitations on the values of the quantum numbers lead to the aufbau (German, Aufbau, building up) principle, where the buildup of electrons in atoms results from continually increasing the quantum numbers. The energy level pattern in Figure 2 describes electron behavior in a hydrogen atom, where there is only one electron. However, interactions between electrons in polyelectronic atoms require that the order of filling orbitals be specified when more than one electron is in the same atom. In this process, we start with the lowest n, l, and ml values (1, 0, and 0, respectively) and either of the ms values (we will arbitrarily use+ 1 2 first). Three rules will then give us the proper order for the remaining electrons, as we increase the quantum numbers in the order ml, ms, l, and n. 1. Electrons are placed in orbitals to give the lowest total electronic energy to the atom. This means that the lowest values of n and l are filled first. Because the orbitals within each subshell (p, d, etc.) have the same energy, the orders for values of ml and ms are indeterminate. 2. The Pauli exclusion principle15 requires that each electron in an atom have a unique set of quantum numbers. At least one quantum number must be different from those of every other electron. This principle does not come from the Schrödinger equation, but from experimental determination of electronic structures. 3. Hund’s rule of maximum multiplicity16 requires that electrons be placed in orbitals to give the maximum total spin possible (the maximum number of parallel spins). Two electrons in the same orbital have a higher energy than two electrons in different orbitals because of electrostatic repulsion (see below); electrons in the same orbital repel each other more than electrons in separate orbitals. Therefore, this rule is a consequence of the lowest possible energy rule (Rule 1). When there are one to six electrons in a p subshell, the required arrangements are those given in Table 6. (The spin multiplicity is the number of unpaired electrons plus 1, or n + 1). Any other arrangement of electrons results in fewer unpaired electrons. Two examples are and
This is only one of Hund’s rules.
TABLE 6 Hund’s Rule and Multiplicity Number of Electrons Arrangement Unpaired e-Multiplicity 1 1 2 2 2 3 3 3 4 4 2 3 5 1 2 6 0 1 26 Atomic Structure Hund’s rule is a consequence of the energy required for pairing electrons in the same orbital. When two negatively charged electrons occupy the same region of space (same orbital) in an atom, they repel each other, with a Coulombic energy of repulsion, Pc, per pair of electrons. As a result, this repulsive force favors electrons in different orbitals ( different regions of space) over electrons in the same orbitals.
In addition, there is an exchange energy, Pe, which arises from purely quantum mechani-cal considerations. This energy depends on the number of possible exchanges between two electrons with the same energy and the same spin. For example, the electron configuration of a carbon atom is 1s2 2s2 2p2. The 2p electrons can be placed in the p orbitals in three ways: (1) (2) (3) Each of these corresponds to a state having a particular energy. State (1) involves Coulombic energy of repulsion, Pc, because it is the only one that pairs electrons in the same orbital. The energy of this state is higher than that of the other two by Pc as a result of electron–electron repulsion.
In the first two states, there is only one possible way to arrange the electrons to give the same diagram, because there is only a single electron in each having + or – spin; these electrons can be distinguished from each other on this basis. However, in the third state, the electrons have the same spin and are therefore indistinguishable from each other. Therefore, there are two possible ways in which the electrons can be arranged: 1 2 2 (one exchange of electrons) 1 Because there are two possible ways in which the electrons in state (3) can be arranged, we can say that there is one pair of possible exchanges between these arrangements, described as one exchange of parallel electrons. The energy involved in such an exchange of parallel electrons is designated Pe; each exchange stabilizes (lowers the energy of) an electronic state, favoring states with more parallel spins (Hund’s rule). Therefore, state (3), which is stabilized by one exchange of parallel electrons, is lower in energy than state (2) by Pe.
The results of considering the effects of Coulombic and exchange energies for the p2 configuration may be summarized in an energy diagram: ∑e ∑c Energy (1) (2) (3) State (3) is the most stable; its electrons are in separate orbitals and have parallel spin; because state (3) has one possible exchange of electrons with parallel spin, it is lower in energy than state (2) by Pe. State (1) is highest in energy because it has two electrons in the same orbital and is therefore higher in energy than state (2) by Pc. Neither state (1) nor state (2) is stabilized by exchange interactions (zero Pe).
27 Atomic Structure In summary: Coulombic energy of repulsion Pc is a consequence of repulsion between electrons in the same orbital; the greater the number of such paired electrons, the higher the energy of the state. Exchange energy Pe is a consequence of parallel electron spins in separate orbitals; the greater the number of such parallel spins (and consequently the greater the number of exchanges), the lower the energy of the state.
Both Coulombic and exchange energies must be taken into account when comparing the energies of different electronic states.
E X A M P L E 2 Oxygen With four p electrons, oxygen could have two unpaired electrons ( ), or it could have no unpaired electrons ( ).
a. Determine the number of electrons that could be exchanged in each case, and find the Coulombic and exchange energies.
1 2 3 2 1 3 3 2 1 1 3 2 This configuration has one pair, energy contribution Pc.
1 2 3 2 1 3 3 2 1 1 3 2 One electron with T spin and no possibility of exchange.
1 2 3 2 1 3 3 2 1 1 3 2 Four possible arrangements for electrons with c spin; three exchange possibilities (1–2, 1–3, 2–3), shown below; energy contribution 3Pe.
1 2 3 2 1 3 3 2 1 1 3 2 Overall, 3Pe + Pc. has two pairs in the same orbitals and one exchange possibility for each spin pair.
Overall, 2Pe + 2Pc.
b. Which state, , or , is lower in energy?
The state is lower in energy because it has less Coulombic energy of repulsion (Pc in comparison with 2Pc) and is stabilized by a greater number of exchanges (3Pe in comparison with 2Pe).
EXERCISE 4 A third possible state for the p4 configuration would be (1) (2) (3) . Determine the Coulombic and exchange energies of this state, and compare its energy with the energies of the states determined in the preceding example. Draw a sketch showing the relative energies of these three states for oxygen’s p4 configuration.
EXERCISE 5 A nitrogen atom, with three 2p electrons, could have three unpaired electrons ( ), or it could have one unpaired electron ( ).
a. Determine the number of electrons that could be exchanged in each case and the Coulombic and exchange energies. Which state would be lower in energy?
In atoms with more than one electron (polyelectronic atoms), all electrons are subject to some Coulombic repulsion energy, but this contribution is significantly higher for electrons that are paired within atomic orbitals.
28 Atomic Structure b. A third possible state for a 2p3 configuration would be (1) (2) (3) . Determine its Coulombic and exchange energies, and compare the energy of this state with the energies determined in part a.
When the orbitals are degenerate (have the same energy), both Coulombic and exchange energies favor unpaired configurations over paired configurations. However, if there is a difference in energy between the levels involved, this difference, together with the Coulombic and exchange energies, determines the final configuration, with the configu-ration of lowest energy expected as the ground state; energy minimization is the driving force. For atoms, this usually means that one subshell (s, p, d) is filled before another has any electrons. However, this approach is insufficient in some transition elements, because the 4s and 3d (or the higher corresponding levels) are so close in energy that the sum of the Coulombic and exchange terms is nearly the same as the difference in energy between the 4s and 3d. Section 2.4 considers these cases.
Many schemes have been used to predict the order of filling of atomic orbitals. Klechkowsky’s rule states that the order of filling of the orbitals proceeds from the low-est available value for the sum n + l. When two combinations have the same value, the one with the smaller value of n is filled first; thus, 4s1n + l = 4 + 02 fills before 3d1n + l = 3 + 22. Combined with the other rules, this gives the order of filling of most of the orbitals. One of the simplest methods that fits most atoms uses the periodic table organized as in Figure 9. The electron configurations of hydrogen and helium are clearly 1s1 and 1s2. After that, the elements in the first two columns on the left (Groups 1 and 2) are filling s orbitals, with l = 0; those in the six columns on the right (Groups 13 to 18) are filling p orbitals, with l = 1; and the ten in the middle (the transition elements, Groups 3 to 12) are filling 1s 2s 3s 4s 5s 6s 5d 7s s block p block d block f block 2s 3s 4s 5s 6s 7s 6d 6d 3d 4d 5d 6d 3d 4d 5d 6d 3d 4d 5d 3d 4d 6p 2p 5p 3p 4p 6p 2p 5p 3p 4p 6p 2p 5p 3p 4p 6p 2p 5p 3p 4p 6p 2p 5p 3p 4p 5p 3p 4p 6p 4f 4f 5f 5f 4f 5f 5d 6d 3d 4d 4f 5f 5d 6d 3d 4d 4f 5f 5d 6d 3d 4d 4f 5f 5d 6d 3d 4d 4f 5f 5d 6d 3d 4d 4f 5f 5d 6d 3d 4d 4f 5f 4f 5f 4f 5f 4f 5f 4f 5f 4f 5f 2p 1s 1 Groups (IUPAC) 3 6 13 2 5 8 7 10 9 12 11 18 15 14 17 4 16 IA (US traditional) IIIB VIB IIIA IIA VB VIIB VIIIB IIB IB VIIIA VA IVA VIIA IVB VIA FIGURE 9 Atomic Orbital Filling in the Periodic Table.
For recent perspective on electron configurations, energies of atomic orbitals, the periodic system, and related topics, see S-G. Wang and W. H. E. Schwarz, Angew. Chem. Int. Ed., 2009, 48, 3404.
29 Atomic Structure d orbitals, with l = 2. The lanthanide and actinide series (numbers 58 to 71 and 90 to 103) are filling f orbitals, with l = 3. These two methods are oversimplifications, as shown in the following paragraphs, but they do fit most atoms and provide starting points for the others.
2.4 Shielding In polyelectronic atoms, energies of specific levels are difficult to predict quantitatively. A useful approach to such predictions uses the concept of shielding: each electron acts as a shield for electrons farther from the nucleus, reducing the attraction between the nucleus and the more distant electrons.
Although the quantum number n is most important in determining the energy, quantum number l must also be included in calculating the energy in atoms having more than one electron. As the atomic number increases, electrons are drawn toward the nucleus, and the orbital energies become more negative. Although the energies decrease with increasing Z, the changes are somewhat irregular because of the shielding of outer electrons by inner electrons. The electron configurations of atoms from the resulting order of orbital filling are shown in Table 7.
As a result of shielding and other subtle interactions between electrons, exclusive reliance on n to rank orbital energies (higher energy with higher quantum number n), which works for one-electron species, holds only for orbitals with lowest values of n (see Figure 10) in polyelectronic species. In multielectron atoms (and ions), for higher values of n, as the split in orbital energies with different values of quantum number l becomes comparable in magnitude to the differences in energy caused by n, the simplest order does not hold.
For example, consider the n = 3 and n = 4 sets in Figure 10. For many atoms the 4s orbital is lower in energy than the 3d orbitals; consequently the order of filling is …3s, 3p, 4s, 3d, 4p… rather than the order based strictly on increasing n …3s, 3p, 3d, 4s, 4p… Similarly, 5s begins to fill before 4d, and 6s before 5d. Other examples can be found in Figure 10.
Slater17 formulated rules that serve as an approximate guide to this effect. These rules define the effective nuclear charge Z as a measure of the attraction of the nucleus for a particular electron: Effective nuclear charge Z = Z - S, where Z = nuclear charge S = shielding constant Slater’s rules for determining S for a specific electron: 1. The atom’s electronic structure is written in order of increasing quantum numbers n and l, grouped as follows: (1s) (2s, 2p) (3s, 3p) (3d) (4s, 4p) (4d) (4f ) (5s, 5p) (5d) (and so on) 2. Electrons in groups to the right in this list do not shield electrons to their left. 3. The shielding constant S for electrons in these groups can now be determined. For ns and np valence electrons: a. Each electron in the same group contributes 0.35 to the value of S for each other electron in the group. Exception: A 1s electron contributes 0.30 to S for another 1s electron. Example: For a configuration 2s2 2p5, a particular 2p electron has six other elec-trons in the (2s, 2p) group. Each of these contributes 0.35 to the value of S, for a total contribution to S of 6 0.35 = 2.10.
Slater’s original numbering scheme has been changed for convenience.
30 Atomic Structure TABLE 7 Electron Configurations of the Elements Element Z Configuration Element Z Configuration H 1 1s1 Cs 55 3Xe46s1 He 2 1s2 Ba 56 3Xe46s2 Li 3 3He42s1 La 57 3Xe46s2 5d1 Be 4 3He42s2 Ce 58 3Xe46s2 4f 1 5d1 B 5 3He42s2 2p1 Pr 59 3Xe46s2 4f 3 C 6 3He42s2 2p2 Nd 60 3Xe46s2 4f 4 N 7 3He42s2 2p3 Pm 61 3Xe46s2 4f 5 O 8 3He42s2 2p4 Sm 62 3Xe46s2 4f 6 F 9 3He42s2 2p5 Eu 63 3Xe46s2 4f 7 Ne 10 3He42s2 2p6 Gd 64 3Xe46s2 4f 7 5d1 Tb 65 3Xe46s2 4f 9 Na 11 3Ne43s1 Dy 66 3Xe46s2 4f 10 Mg 12 3Ne43s2 Ho 67 3Xe46s2 4f 11 Al 13 3Ne43s2 3p1 Er 68 3Xe46s2 4f 12 Si 14 3Ne43s2 3p2 Tm 69 3Xe46s2 4f 13 P 15 3Ne43s2 3p3 Yb 70 3Xe46s2 4f 14 S 16 3Ne43s2 3p4 Lu 71 3Xe46s2 4f 14 5d1 Cl 17 3Ne43s2 3p5 Hf 72 3Xe46s2 4f 14 5d2 Ar 18 3Ne43s2 3p6 Ta 73 3Xe46s2 4f 14 5d3 W 74 3Xe46s2 4f 14 5d4 K 19 3Ar44s1 Re 75 3Xe46s2 4f 14 5d5 Ca 20 3Ar44s2 Os 76 3Xe46s2 4f 14 5d6 Sc 21 3Ar44s2 3d1 Ir 77 3Xe46s2 4f 14 5d7 Ti 22 3Ar44s2 3d2 Pt 78 3Xe46s1 4f 14 5d9 V 23 3Ar44s2 3d3 Au 79 3Xe46s1 4f 14 5d10 Cr 24 3Ar44s1 3d5 Hg 80 3Xe46s2 4f 14 5d10 Mn 25 3Ar44s2 3d5 Tl 81 3Xe46s2 4f 14 5d10 6p1 Fe 26 3Ar44s2 3d6 Pb 82 3Xe46s2 4f 14 5d10 6p2 Co 27 3Ar44s2 3d7 Bi 83 3Xe46s2 4f 14 5d10 6p3 Ni 28 3Ar44s2 3d8 Po 84 3Xe46s2 4f 14 5d10 6p4 Cu 29 3Ar44s1 3d10 At 85 3Xe46s2 4f 14 5d10 6p5 Zn 30 3Ar44s2 3d10 Rn 86 3Xe46s2 4f 14 5d10 6p6 Ga 31 3Ar44s2 3d10 4p1 Ge 32 3Ar44s2 3d10 4p2 Fr 87 3Rn47s1 As 33 3Ar44s2 3d10 4p3 Ra 88 3Rn47s2 Se 34 3Ar44s2 3d10 4p4 Ac 89 3Rn47s2 6d1 Br 35 3Ar44s2 3d10 4p5 Th 90 3Rn47s2 6d2 Kr 36 3Ar44s2 3d10 4p6 Pa 91 3Rn47s2 5f 2 6d1 U 92 3Rn47s2 5f 3 6d1 Rb 37 3Kr45s1 Np 93 3Rn47s2 5f 4 6d1 Sr 38 3Kr45s2 Pu 94 3Rn47s2 5f 6 Am 95 3Rn47s2 5f 7 Y 39 3Kr45s2 4d1 Cm 96 3Rn47s2 5f 7 6d1 Zr 40 3Kr45s2 4d2 Bk 97 3Rn47s2 5f 9 Nb 41 3Kr45s1 4d4 Cf 98 3Rn47s2 5f 9 6d1 Mo 42 3Kr45s1 4d5 Es 99 3Rn47s2 5f 11 Tc 43 3Kr45s2 4d5 Fm 100 3Rn47s2 5f 12 Ru 44 3Kr45s1 4d7 Md 101 3Rn47s2 5f 13 Rh 45 3Kr45s1 4d8 No 102 3Rn47s2 5f 14 Pd 46 3Kr44d10 Lr 103 3Rn47s2 5f 14 6d1 Ag 47 3Kr45s1 4d10 Rf 104 3Rn47s2 5f 14 6d2 Cd 48 3Kr45s2 4d10 Db 105 3Rn47s2 5f 14 6d3 In 49 3Kr45s2 4d10 5p1 Sg 106 3Rn47s2 5f 14 6d4 Sn 50 3Kr45s2 4d10 5p2 Bh 107 3Rn47s2 5f 14 6d5 Sb 51 3Kr45s2 4d10 5p3 Hs 108 3Rn47s2 5f 14 6d6 Te 52 3Kr45s2 4d10 5p4 Mt 109 3Rn47s2 5f 14 6d7 I 53 3Kr45s2 4d10 5p5 Ds 110 3Rn47s1 5f 14 6d9 Xe 54 3Kr45s2 4d10 5p6 Rg 111 3Rn47s1 5f14 6d10 Cna 112 3Rn47s2 5f14 6d10 Elements with configurations that do not follow the simple order of orbital filling.
a Evidence for elements 113–118 has been reviewed by IUPAC; see R. C. Barber, P. J. Karol, H. Nakahara, E. Vardaci, E. W. Vogt, Pure Appl. Chem., 2011, 83, 1485. In May 2012, IUPAC officially named element 114 (flerovium, symbol Fl) and element 116 (livermorium, Lv).
Source: Actinide configurations are from J. J. Katz, G. T. Seaborg, and L. R. Morss, The Chemistry of the Actinide Elements, 2nd ed., Chapman and Hall, New York and London, 1986. Configurations for elements 100 to 112 are predicted, not experimental.
31 Atomic Structure b. Each electron in n – 1 groups contribute 0.85 to S. Example: For the 3s electron of sodium, there are eight electrons in the (2s, 2p) group. Each of these electrons contributes 0.85 to the value of S, a total contribu-tion of 8 0.85 = 6.80. c. Each electron in n – 2 or lower groups contributes 1.00 to S. 4. For nd and nf valence electrons: a. Each electron in the same group contributes 0.35 to the value of S for each other electron in the group. (Same rule as 3a.) b. Each electron in groups to the left contributes 1.00 to S.
These rules are used to calculate the shielding constant S for valence electrons. Subtracting S from the total nuclear charge Z gives the effective nuclear charge Z on the selected electron: Z = Z - S Calculations of S and Z follow.
E X A M P L E 3 Oxygen Use Slater’s rules to calculate the shielding constant and effective nuclear charge of a 2p electron.
Rule 1: The electron configuration is written using Slater’s groupings, in order: (1s2)(2s2, 2p4) To calculate S for a valence 2p electron: Rule 3a: Each other electron in the (2s2, 2p4) group contributes 0.35 to S. Total contribution = 5 0.35 = 1.75 Rule 3b: Each 1s electron contributes 0.85 to S. Total contribution = 2 0.85 = 1.70 5g 5f 4f 5d 5p 4d 5s 4p 3d 4s 3p 3s 2p 2s 1s n = 1 n = 2 n = 3 n = 4 n = 5 FIGURE 10 Energy Level Splitting and Overlap. The differences between the upper levels are exaggerated for easier visualization. This diagram provides unambiguous electron configurations for elements hydrogen to vanadium.
32 Atomic Structure Total S = 1.75 + 1.70 = 3.45 Effective nuclear charge Z = 8 - 3.45 = 4.55 So rather than feeling the full +8 nuclear charge, a 2p electron is calculated to feel a charge of +4.55, or about 57% of the full nuclear charge.
Nickel Use Slater’s rules to calculate the shielding constant and effective nuclear charge of a 3d and 4s electron.
Rule 1: The electron configuration is written (1s2)(2s2, 2p6)(3s2, 3p6)(3d8)(4s2) For a 3d electron: Rule 4a: Each other electron in the (3d8) group contributes 0.35 to S. Total contribution = 7 0.35 = 2.45 Rule 4b: Each electron in groups to the left of (3d8) contributes 1.00 to S. Total contribution = 18 1.00 = 18.00 Total S = 2.45 + 18.00 = 20.45 Effective nuclear charge Z = 28 - 20.45 = 7.55 For a 4s electron: Rule 3a: The other electron in the (4s2) group contributes 0.35 to S.
Rule 3b: Each electron in the (3s2, 3p6)(3d8) groups (n – 1) contributes 0.85. Total contribution = 16 0.85 = 13.60 Rule 3c: Each other electron to the left contributes 1.00. Total contribution = 10 1.00 = 10.00 Total S = 0.35 + 13.60 + 10.00 = 23.95 Effective nuclear charge Z = 28 - 23.95 = 4.05 The effective nuclear charge for the 4s electron is considerably smaller than the value for the 3d electron. This is equivalent to stating that the 4s electron is held less tightly than the 3d and should therefore be the first removed in ionization. This is consistent with experimental observations on nickel compounds. Ni2+, the most common oxida-tion state of nickel, has a configuration of [Ar] 3d8, rather than [Ar] 3d6 4s2, corre-sponding to loss of the 4s electrons from nickel atoms. All the transition metal atoms follow this same pattern of losing ns electrons more readily than (n – 1)d electrons.
EXERCISE 6 Calculate the effective nuclear charge on a 5s, 5p, and 4d electron in a tin atom.
EXERCISE 7 Calculate the effective nuclear charge on a 7s, 5f, and 6d electron in a uranium atom.
Justification for Slater’s rules comes from the electron probability curves for the orbit-als; Slater devised these rules semiempirically using equations modeled after wavefunc-tion equations to fit experimental data for atoms. Slater’s approach results in rules that provide useful approximations for the effective nuclear charge an electron in an atom actually experiences after shielding is taken into account. The s and p orbitals have higher probabilities near the nucleus than do d orbitals of the same n, as shown earlier in Figure 7. Therefore, the shielding of 3d electrons by (3s, 3p) electrons is calculated as 100% effective, a contribution of 1.00. At the same time, shielding of 3s or 3p electrons by (2s, 2p) electrons is estimated as 85% effective, a contribution of 0.85, because the 3s and 3p orbitals have regions of significant probability close to the nucleus. Therefore, electrons in these orbitals are not completely shielded by (2s, 2p) electrons.
33 Atomic Structure A complication arises at Cr (Z = 24) and Cu (Z = 29) in the first transition series and in an increasing number of atoms with higher atomic numbers in the second and third transition series. This effect places an extra electron in the 3d level and removes one electron from the 4s level. Cr, for example, has a configuration of [Ar] 4s1 3d5 rather than [Ar] 4s2 3d4. Traditionally, this phenomenon has often been explained as a consequence of the “special stability of half-filled subshells.” Half-filled and filled d and f subshells are, in fact, fairly common, as shown in Figure 11. A more complete explanation considers both the effects of increasing nuclear charge on the energies of the 4s and 3d levels and the interactions between electrons sharing the same orbital.18 This approach requires totaling all contributions to the energy of the configuration of electrons, including the Coulombic and exchange energies; results of the complete calculations are consistent with the configurations determined by experimental data.
Slater’s rules have been refined to improve their match with experimental data. One relatively simple refinement is based on the ionization energies for the elements hydrogen through xenon, and it provides a calculation procedure similar to that proposed by Slater.19 A more elaborate method incorporates exponential screening and provides energies that are in closer agreement with experimental values.20 Another explanation that is more pictorial and considers electron–electron interactions was proposed by Rich.21 He explained electronic structures of atoms by considering the differ-ence in energy between the energy of one electron in an orbital and two electrons in the same orbital. Although the orbital itself is usually assumed to have only one energy, the electrostatic repulsion of the two electrons in one orbital adds the electron-pairing energy described in Section 2.3 as part of Hund’s rule. We can visualize two parallel energy levels, each with electrons of only one spin, separated by the electron-pairing energy, as shown in Figure 12.
For example, an Sc atom has the valence configuration 4s2 3d1. By Rich’s approach, the first electron is arbitrarily considered to have ms = - 1 2. The second electron, with ms = + 1 2, completes the 4s2 configuration—but the total energy of these two electrons is greater than twice the energy of the first electron, because of the Coulombic energy of repulsion, c. In Figure 12(a) Sc is shown as having three electrons: in ascending order these are 4s (ms = - 1 2), 4s (ms = + 1 2), and 3d (ms = - 1 2). The next element, Ti, also Ar Cl S P Si Al Ge As Se Br Kr Ga Zn Cu Ni Co Fe Mn Cr 4s2 4s1 4s2 4s1 V Ti Sc 3d8 3d10 3d10 3d7 3d6 3d5 3d5 3d3 3d2 3d1 Sn Sb Te I Xe In Cd Ag Pd Rh Ru Tc Mo Nb 4d10 Zr Y 5s2 5d10 6s2 6d10 6d10 6d 9 5f 14 5f 14 5f 14 5f 14 5f 14 5f 14 5f 14 5f 13 5f 12 5f 11 5f 9 5f 9 5f 7 5f 7 5f 6 5f 4 5f 3 5f 2 6d2 6d1 6d1 6d1 6d1 6d1 6d2 6d3 6d 4 6d 5 6d6 6d7 7s1 7s2 6d1 6s1 6s1 5d10 5d9 5d7 5d6 5d5 6s2 4f 14 4f 14 4f 14 5d 4 5d3 5d2 4f 14 4f 13 4f 12 4f 11 4f 10 4f 9 4f 7 4f 7 4f 6 4f 5 4f 4 4f 3 4f 1 5d1 5d1 5d1 5s1 5s2 5s1 4d10 4d10 4d8 4d7 5s1 5s1 4d5 4d5 4d4 5s1 4d2 4d1 Pb Uuq Cn Rg Ds Mt Hs Bh Sg Db Rf No Md Fm Es Cf Bk Cm Am Pu Np U Pa Th Ac Ra Fr Uuh Uuo Bi Po At Rn TI Hg Au Pt Ir Os Re W Ta Hf 5f 14 6d1 5d1 4f 14 Lr Lu Yb Tm Er Ho Dy Tb Gd Eu Sm Pm Nd Pr Ce La Ba Cs Sr Rb Ca K Mg Na Filled d Half-flled d Half-flled f Filled f FIGURE 11 Electron Configu-rations of Transition Metals, Including Lanthanides and Actinides. Solid lines sur-rounding elements designate filled (d10 or f14) or half-filled (d5 or f7) subshells. Dashed lines surrounding elements desig-nate irregularities in sequential orbital filling, also found within some of the solid lines.
34 Atomic Structure has one 4s electron with each spin, then two 3d electrons, each with ms = - 1 2. The two 3d electrons, by Hund’s rule, have parallel spin.
As the number of protons in the nucleus increases, the effective nuclear charge for all electrons increases and the energy levels decrease in energy; their electrons become more stable. Figure 12 illustrates that the energy of the 3d subshell decreases more dramatically relative to 4s as one moves across the first transition series; this trend generally holds for (n - 1)d and ns orbitals. A rationalization for this trend is that orbitals with shorter most probable distances to the nucleus are stabilized more as Z increases relative to orbitals with greater most probable distances. Because the 3d orbitals have shorter most probable distances from the nucleus than the 4s orbital, the 3d orbitals are stabilized more than the 4s as the nuclear charge increases.
The effective nuclear charge that an electron experiences generally increases as the most probable distance of the electron from the nucleus decreases; these electrons are less susceptible to shielding by electrons farther from the nucleus (for example, in Slater’s rules electrons with greater most probable distances to the electron in question do not contribute at all to S). Since the most probable distance from the nucleus increases as n increases (Figure 7), the 3d subshell ultimately stabilizes its electrons more than the 4s orbital once Z gets sufficiently high. Regardless of the relative orbital energies, the observed electronic configuration is always the one of lowest energy. Electrons fill the lowest available orbitals in order up to their capacity, with the results shown in Figure 12 and in Table 7.
The schematic diagram in Figure 12(a) shows the order in which the levels fill, from bottom to top in energy. For example, Ti has two 4s electrons, one in each spin level, and two 3d electrons, both with the same spin. Fe has two 4s electrons, one in each spin level, five 3d electrons with spin-1 2, and one 3d electron with spin + 1 2. For vanadium, the first two electrons enter the 4s, -1 2 and 4s, +1 2 levels; the next three are all in the 3d, -1 2 level, and vanadium has the configuration 4s2 3d3. The 3d, -1 2 line crosses the 4s, +1 2 line between V and Cr. When the six electrons of chromium are filled in from the lowest level, Number of electrons in half subshell 3d 3d 4s 4s 1 Electron spin 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 5 5 5 5 5 5 5 5 5 5 4 5 5 5 5 5 5 Sc Ti V Cr Mn Fe Co Ni Cu Zn Sc+ Ti+ V+ Cr+ Mn+ Fe+ Co+ Ni+ Cu+ Zn+ (a) E (b) 5 5 2 2 2 3 3 3 4 2 1 + 2 1 + 2 1 + 2 1 + 2 1 -2 1 -2 1 -2 1 -FIGURE 12 Schematic Energy Levels for Transition Elements. (a) Schematic interpretation of electron configurations for transition elements in terms of intraorbital repulsion and trends in subshell energies. (b) A similar diagram for ions, showing the shift in the crossover points on removal of an electron. The shift is even more pronounced for metal ions having 2+ or greater charges. As a consequence, transition-metal ions with 2+ or greater charges have no s electrons, only d electrons in their outer levels. Similar diagrams, although more complex, can be drawn for the heavier transition elements and the lanthanides.
(Rich, R. L., Periodic Coorelate, 1st Ed., (c) 1965. Reprinted and Electroni-cally reproduced by permission of Pearson Education Inc, Upper Saddle River, NJ 07458.) 35 Atomic Structure chromium has the configuration 4s1 3d5. A similar crossing gives copper its 4s1 3d10 structure. This approach to electron configurations of transition metals does not depend on the stability of half-filled shells or other additional factors.
Formation of a positive ion by removal of an electron reduces shielding; the effective nuclear charge for all electrons increases dramatically. On the basis of the most probable distance effect discussed previously, (n - 1)d orbitals will be lower in energy than ns orbitals in the cation, as shown in Figure 12(b). As a result, the remaining electrons occupy the d orbitals. A common rule in introductory chemistry is that electrons with highest n—in this case, those in the s orbitals—are always removed first when ions are formed from the transition elements. A perhaps more mature perspective on this idea is that regardless of which electron is lost to form a transition metal ion, the lowest energy electron configura-tion of the resulting ion will always exhibit the vacancy in the ns orbital. This effect is even stronger for 2+ ions, where the effective nuclear charge is even higher. Transition metal cations have no s electrons, only d electrons in their outer levels.
A similar, but more complex, crossing of levels appears in the lanthanide and actinide series. The simple explanation would have these elements start filling f orbitals at lanthanum (57) and actinium (89), but these atoms have one d electron instead. Other elements in these series also show deviations from the “normal” sequence. Rich has explained these situations using similar diagrams.21 3 Periodic Properties of Atoms A valuable aspect of the arrangment of atoms on the basis of similar electronic configurations within the periodic table is that an atom’s position provides information about its properties. Some of these properties, and how they vary across periods and groups, are now discussed.
3.1 Ionization Energy The ionization energy, also known as the ionization potential, is the energy required to remove an electron from a gaseous atom or ion: A n+(g) h A (n+1)+(g) + e - ionization energy (IE) = U where n = 0 (first ionization energy), n = 1 (second ionization energy), and so on.
As would be expected from the effects of shielding, the ionization energy varies with different nuclei and different numbers of electrons. Trends for the first ionization energies of the early elements in the periodic table are shown in Figure 13. The general trend across a period is an increase in ionization energy as the nuclear charge increases. However, the experimental values show a break in the trend in the second period at boron and again at oxygen. Because boron is the first atom to have an electron in a higher energy 2p orbital that is shielded somewhat by the 2s electrons, boron’s 2p electron is more easily lost than the 2s electrons of beryllium; boron has the lower ionization energy.
Energy Be B 2p 2s 2s At the fourth 2p electron, at oxygen, a similar decrease in ionization energy occurs. Here, the fourth electron shares an orbital with one of the three previous 2p electrons 36 Atomic Structure ( ), and the repulsion between the paired electrons (Pc) reduces the energy necessary to remove an electron from oxygen; oxygen has a lower ionization energy than nitrogen, which has the 2p configuration .
Similar patterns appear in the other periods, for example Na through Ar and K through Kr, omitting the transition metals. The transition metals have less dramatic differences in ionization energies, with the effects of shielding and increasing nuclear charge more nearly in balance.
Much larger decreases in ionization energy occur at the start of each new period, because the change to the next major quantum number requires that the new s electron have a much higher energy. The maxima at the noble gases decrease with increasing Z, because the outer electrons are farther from the nucleus in the heavier elements. Overall, the trends are toward higher ionization energy from left to right in the periodic table (the major change) and lower ionization energy from top to bottom (a minor change). The differences described in the previous paragraph are superimposed on these more general changes.
3.2 Electron Affinity Electron affinity can be defined as the energy required to remove an electron from a negative ion: A-(g) h A(g) + e - electron affinity (EA) = U Because of the similarity of this reaction to the ionization for an atom, electron affin-ity is sometimes described as the zeroth ionization energy. This reaction is endothermic (positive U) except for the noble gases and the alkaline earth elements. The pattern of electron affinities with changing Z, shown in Figure 13, is similar to that of the ionization energies, but for one larger Z value (one more electron for each species) and with much smaller absolute numbers. For either of the reactions, removal of the first electron past a noble gas configuration is easy, so the noble gases have the lowest electron affinities. The electron affinities are all much smaller than the corresponding ionization energies, because electron removal from a negative ion (that features more shielding of the nuclear charge) is easier than removal from a neutral atom.
Comparison of the ionization and electron affinity graphs in Figure 13 shows similar zigzag patterns, but with the two graphs displaced by one element: for example, electron affinity shows a peak at F and valley at Ne, and ionization energy a peak at Ne and valley at Ionization energy Electron affnity Atomic number kJ/mol 10 20 Ne F N O C B Be Li H He Ne He Be N Ar Cl S P Mg Si Se Ge Ga Cu Zn Cr Ca Pd Cd In Sn Sb I Te K As Al Na F Kr Kr Xe Xe Ba Ba I Ar Zn Br Br Rb Sr Cd Cs Cl -500 0 500 1000 1500 2000 2500 30 40 50 60 FIGURE 13 Ionization Energies and Electron Affinities. Ionization energy = U for M(g) h M+(g) + e-(Data from C. E. Moore, Ionization Potentials and Ionization Limits, National Standards Reference Data Series, U.S. National Bureau of Stand-ards, Washington, DC, 1970, NSRDS-NBS 34) Electron affinity = U for M-(g) h M(g) + e- Data from H. Hotop and W. C. Lineberger, J. Phys. Chem. Ref. Data, 1985, 14, 731). Numerical values are in Appendices B-2 and B-3.
Historically, the definition has been - U for the reverse reaction, adding an electron to the neutral atom. The definition we use avoids the sign change.
37 Atomic Structure Na. The patterns in these two quantities can more easily be seen by plotting energy against the number of electrons in each reactant, as shown in Figure 14 for electron affinity and first and second ionization energy.
The peaks and valleys match for all three graphs because the electron configura-tions match—for example, there are peaks at 10 electrons and valleys at 11 electrons. At 10 electrons, all three reactant species (F–, Ne, and Na+) have identical 1s2 2s2 2p6 configura-tions; these are by definition isoelectronic species. The relatively high energy necessary to remove an electron from these configurations is typical for configurations in which electron shells are complete. The next electron, in an 11-electron configuration, is the first to occupy a higher energy 3s orbital and is much more easily lost, providing a valley in each graph, corresponding to removal of an electron from the 11-electron species Ne–, Na, and Mg+.
E XE RCISE 8 Explain why all three graphs in Figure 14 have maxima at 4 electrons and minima at 5 electrons.
3.3 Covalent and Ionic Radii The sizes of atoms and ions are also related to the ionization energies and electron affinities. As the nuclear charge increases, the electrons are pulled in toward the center of the atom, and the size of any particular orbital decreases. On the other hand, as the nuclear charge increases, more electrons are added to the atom, and their mutual repulsion keeps the outer orbitals large. The interaction of these two effects, increasing nuclear charge and increas-ing number of electrons, results in a gradual decrease in atomic size across each period. Table 8 gives nonpolar covalent radii, based on bond distances in nonpolar molecules. There are other measures of atomic size, such as the van der Waals radius, in which collisions with other atoms are used to define the size. It is difficult to obtain consistent data for any such measure, because the polarity, chemical structure, and physical state of molecules change drastically from one compound to another. The numbers shown here are sufficient for a general comparison of different elements.
There are similar challenges in determining the size of ions. Because the stable ions of the different elements have different charges and different numbers of electrons, as well as different crystal structures for their compounds, it is difficult to find a suitable set of num-bers for comparison. Earlier data were based on Pauling’s approach, in which the ratio of the radii of isoelectronic ions was assumed to be equal to the ratio of their effective 4 12 5 10 11 8 9 7 6 3 2 1 Number of electrons in reactant Second ionization energy First ionization energy Electron affnity Energy (kJ/mol) 0 1000 2000 3000 4000 5000 7000 6000 Be+ He+ B+ C+ N+ O+ F+ Ne+ Na+ Al+ Mg+ B-C-N-O-F-Ne- Na-Be-He- Li-H-H Li B Be He C N F Ne Na Mg O Li+ FIGURE 14 First and Second Ionization Energies and Electron Affinities (©2003 Beth Abdella and Gary Miessler, reproduced with permission) 38 Atomic Structure TABLE 8 Nonpolar Covalent Radii (pm) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 H He 32 31 Li Be B C N O F Ne 123 89 82 77 75 73 71 69 Na Mg Al Si P S Cl Ar 154 136 118 111 106 102 99 98 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 203 174 144 132 122 118 117 117 116 115 117 125 126 122 120 117 114 111 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe 216 191 162 145 134 130 127 125 125 128 134 148 144 140 140 136 133 126 Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Ra 235 198 169 144 134 130 128 126 127 130 134 149 148 147 146 (146) (145) Source: Data from R. T. Sanderson, Inorganic Chemistry, Reinhold, New York, 1967, p. 74; and E. C. M. Chen, J. G. Dojahn, W. E. Wentworth, J. Phys. Chem. A, 1997, 101, 3088.
TABLE 9 Crystal Radii for Selected Ions Z Element Radius (pm) Alkali metal ions 3 Li+ 90 11 Na+ 116 19 K+ 152 37 Rb+ 166 55 Cs+ 181 Alkaline earth ions 4 Be2+ 59 12 Mg2+ 86 20 Ca 2+ 114 38 Sr 2+ 132 56 Ba 2+ 149 Other cations 13 Al3+ 68 30 Zn2+ 88 Halide ions 9 F -119 17 Cl-167 35 Br -182 53 I -206 Other anions 8 O 2-126 16 S2-170 Source: Data from R. D. Shannon, Acta Crystallogr. 1976, A32, 751 for six-coordinate ions. nuclear charges. More recent calculations are based on a number of considerations, including electron density maps from X-ray data that show larger cations and smaller anions than those previously found. Those in Table 9 were called “crystal radii” 39 Atomic Structure TABLE 10 Crystal Radius and Nuclear Charge Ion Protons Electrons Radius (pm) O2-8 10 126 F -9 10 119 Na+ 11 10 116 Mg2+ 12 10 86 Al3+ 13 10 68 TABLE 11 Crystal Radius and Total Number of Electrons Ion Protons Electrons Radius (pm) O2-8 10 126 S2-16 18 170 Se2-34 36 184 Te2-52 54 207 TABLE 12 Crystal Radius and Ionic Charge Ion Protons Electrons Radius (pm) Ti2+ 22 20 100 Ti3+ 22 19 81 Ti4+ 22 18 75 by Shannon22 and are generally different from the older values of “ionic radii” by +14 pm for cations and –14 pm for anions, as well as being revised to accommodate more recent measurements. The radii in Table 9 can be used for rough estimation of the packing of ions in crystals and other calculations, as long as the “fuzzy” nature of atoms and ions is kept in mind.
Factors that influence ionic size include the coordination number of the ion, the cova-lent character of the bonding, distortions of regular crystal geometries, and delocalization of electrons (metallic or semiconducting character. The radius of the anion is also influ-enced by the size and charge of the cation. Conversely, the anion exerts a smaller influence on the radius of the cation.23 The values in Table 10 show that anions are generally larger than cations with similar numbers of electrons. The radius decreases as nuclear charge increases for ions with the same electronic structure, with the charge on cations having a strong effect, for example in the series Na+, Mg2+, Al3+. Within a group, the ionic radius increases as Z increases because of the larger number of electrons in the ions and, for the same element, the radius decreases with increasing charge on the cation. Examples of these trends are shown in Tables 10, 11, and 12.
40 General References Additional information on the history of atomic theory can be found in J. R. Partington, A Short History of Chemistry, 3rd ed., Macmillan, London, 1957, reprinted by Harper & Row, New York, 1960, and in the Journal of Chemical Education. For an introduction to atomic theory and orbitals, see V. M. S. Gil, Orbitals in Chemistry: A Modern Guide for Students, Cambridge University Press, Cambridge, 2000, UK, pp. 1–69. A more thorough treatment of the electronic structure of atoms is in M. Gerloch, Orbitals, Terms, and States, John Wiley & Sons, New York, 1986. Many Internet sites provide images of atomic orbitals, their wave equations, nodal behavior, and other characteristics. Two examples are and 1 Determine the de Broglie wavelength of a. an electron moving at 1/10 the speed of light.
b. a 400 g Frisbee moving at 10 km/h.
c. an 8.0-pound bowling ball rolling down the lane with a velocity of 2.0 meters per second.
d. a 13.7 g hummingbird flying at a speed of 30.0 miles per hour. 2 Using the equation E = R Ha 1 22 - 1 n2 h b, determine the energies and wavelengths of the visible emission bands in the atomic spectrum of hydrogen arising from nh = 4, 5, and 6. (The red line, corresponding to nh = 3, was calculated in Exercise 1.) 3 The transition from the n = 7 to the n = 2 level of the hydrogen atom is accompanied by the emission of radia-tion slightly beyond the range of human perception, in the ultraviolet region. Determine the energy and wavelength. 4 Emissions are observed at wavelengths of 383.65 and 379.90 nm for transitions from excited states of the hydro-gen atom to the n = 2 state. Determine the quantum num-bers nh for these emissions. 5 What is the least amount of energy that can be emitted by an excited electron in a hydrogen atom falling from an excited state directly to the n = 3 state? What is the quantum num-ber n for the excited state? Humans cannot visually observe the photons emitted in this process. Why not? 6 Hydrogen atom emission spectra measured from the solar corona indicated that the 4s orbital was 102823.8530211 cm-1, and 3s orbital 97492.221701 cm-1, respectively, above the 1s ground state. (These energies have tiny uncertainties, and can be treated as exact numbers for the sake of this problem.) Calculate the difference in energy (J) between these levels on the basis of these data, and compare this difference to that 1. John Dalton, A New System of Chemical Philosophy, 1808; reprinted with an introduction by Alexander Joseph, Peter Owen Limited, London, 1965. 2. Ibid., p. 113. 3. Ibid., p. 133. 4. J. R. Partington, A Short History of Chemistry, 3rd ed., Macmillan, London, 1957; reprinted, 1960, Harper & Row, New York, p. 255. 5. Ibid., pp. 256–258. 6. D. I. Mendeleev, J. Russ. Phys. Chem. Soc., 1869, i, 60. 7. L. Meyer, Justus Liebigs Ann. Chem., 1870, Suppl. vii, 354. 8. N. Bohr, Philos. Mag., 1913, 26, 1. 9. G. Herzberg, Atomic Spectra and Atomic Structure, 2nd ed., Dover Publications, New York, 1994, p. 18. 10. L. de Broglie, Philos. Mag., 1924, 47, 446; Ann. Phys. Paris, 1925, 3, 22. 11. W. Heisenberg, Z. Phys., 1927, 43, 172. 12. Ralchenko, Yu., Kramida, A.E., Reader, J., and NIST ASD Team (2011). NIST Atomic Spectra Database (ver. 4.1.0), [Online]. Available: [2012, August 21]. National Institute of Standards and Technol-ogy, Gaithersburg, MD. 13. E. Schrödinger, Ann. Phys. (Leipzig), 1926, 79, 361, 489, 734; 1926, 80, 437; 1926, 81, 109; Naturwissenshaften, 1926, 14, 664; Phys. Rev., 1926. 28, 1049. 14. R. E. Powell, J. Chem. Educ., 1968, 45, 45. 15. W. Pauli, Z. Physik, 1925, 31, 765. 16. F. Hund, Z. Physik, 1925, 33, 345. 17. J.C. Slater. Phys. Rev., 1930, 36, 57. 18. L. G. Vanquickenborne, K. Pierloot, D. Devoghel, J. Chem. Ed., 1994, 71, 468 19. J. L. Reed, J. Chem. Educ., 1999, 76, 802. 20. W. Eek, S. Nordholm, G. B. Bacskay, Chem. Educator, 2006, 11, 235. 21. R. L. Rich, Periodic Correlations, W. A. Benjamin, Menlo Park, CA, 1965, pp. 9–11. 22. R. D. Shannon, Acta Crystallogr., 1976, A32, 751. 23. O. Johnson, Inorg. Chem., 1973, 12, 780.
Problems References Atomic Structure 41 Atomic Structure obtained by the Balmer equation in Section 1.2. How well does the Balmer equation work for hydrogen? (Data from Y. Ralchenko, A. E. Kramida, J. Reader, and NIST ASD Team (2011). NIST Atomic Spectra Database (ver. 4.1.0), [Online]. Available: http:// physics.nist.gov/asd [2012, January 18]. National Institute of Standards and Technol-ogy, Gaithersburg, MD.) 7 The Rydberg constant equation has two terms that vary depending on the species under consideration, the reduced mass of the electron/nucleus combination and the charge of the nucleus (Z). a. Determine the approximate ratio between the Rydberg constants for isoelectronic He+ (consider the most abun-dant helium-4 isotope) and H.
b. Use this ratio to calculate an approximate Rydberg constant (J) for He+.
c. The difference between the He+ 2s and 1s orbitals was reported as 329179.76197(20) cm-1. Calculate the He+ Rydberg constant from this spectral line for comparison to your value from b. (Data from the same reference as Problem 6.) 8 The details of several steps in the particle-in-a-box model in this chapter have been omitted. Work out the details of the following steps: a. Show that if = A sin rx + B cos sx (A, B, r, and s are constants) is a solution to the wave equation for the one-dimensional box, then r = s = 22mEa2p h b b. Show that if = A sin rx, the boundary condi-tions ( = 0 when x = 0 and x = a) require that r = { np a , where n = any integer other than zero.
c. Show that if r = { np a , the energy levels of the particle are given by E = n2h2 8ma2 d. Show that substituting the value of r given in part c into = A sin rx and applying the normalizing requirement gives A = 22/a. 9 For the 3pz and 4dxz hydrogen-like atomic orbitals, sketch the following: a. The radial function R b. The radial probability function a0 r2R 2 c. Contour maps of electron density. 10 Repeat the exercise in Problem 9, for the 4s and 5dx2-y2 orbitals. 11 Repeat the exercise in Problem 9, for the 5s and 4dz2 orbitals. 12 The 4fz(x2-y2) orbital has the angular function Y = (constant) z(x 2 - y 2)/r3.
a. How many radial nodes does this orbital have?
b. How many angular nodes does it have?
c. Write equations to define the angular nodal surfaces. What shapes are these surfaces?
d. Sketch the shape of the orbital, and show all radial and angular nodes. 13 Repeat the exercise in Problem 12, for the 5fxyz orbital, which has Y = (constant) xyz/r3. 14 The label for an fz3 orbital, like that for a dz2 orbital, is an abbreviation. The actual angular function for this orbital is Y = (constant) z(5z 2 - 3r2)/r3. Repeat the exercise in Problem 12, for a 4fz3 orbital. (Note: recall that r2 = x 2 + y 2 + z 2). 15 a. Determine the possible values for the l and ml quan-tum numbers for a 5d electron, a 4f electron, and a 7g electron.
b. Determine the possible values for all four quantum numbers for a 3d electron.
c. What values of ml are possible for f orbitals?
d. At most, how many electrons can occupy a 4d orbital? 16 a. What are the values of quantum numbers l and n for a 5d electron?
b. At most, how many 4d electrons can an atom have? Of these electrons how many, at most, can have ms = - 1 2?
c. A 5f electron has what value of quantum number l? What values of ml may it have?
d. What values of the quantum number ml are possible for a subshell having l = 4? 17 a. At most, how many electrons in an atom can have both n = 5 and l = 3?
b. A 5d electron has what possible values of the quantum number ml?
c. What value of quantum number l do p orbitals have? For what values of n do p orbitals occur?
d. What is the quantum number l for g orbitals? How many orbitals are in a g subshell? 18 Determine the Coulombic and exchange energies for the following states, and determine which state is favored (has lower energy): a. b. 19 Two excited states for a d4 configuration are shown. Which is likely to have lower energy? Explain your choice in terms of Coulombic and exchange energies.
W: X: W: X: 20 Two excited states for a d5 configuration are shown. Which is likely to have lower energy? Why? Explain your choice in terms of Coulombic and exchange energies.
Y: Z: Y: Z: and and and and 42 21 What states are possible for a d3 configuration? Determine the Coulombic and exchange energies for each, and rank the states in terms of relative energy. 22 Provide explanations of the following phenomena: a. The electron configuration of Cr is [Ar] 4s1 3d5 rather than [Ar] 4s2 3d4.
b. The electron configuration of Ti is [Ar] 4s2 3d2, but that of Cr2+ is [Ar] 3d4. 23 Give electron configurations for the following: a. V b. Br c. Ru3+ d. Hg2+ e. Sb 24 Predict the electron configurations of the following metal anions: a. Rb-b. Pt2- (See: A. Karbov, J. Nuss, U. Weding, M. Jansen, Angew. Chem. Int. Ed., 2003, 42, 4818.) 25 Radial probability plots shed insight on issues of shield-ing and effective nuclear charge. Interpret the radial prob-ability functions in Figure 7 to explain why the general order of orbital filling is n = 1, followed by n = 2, followed by n = 3. Interpret the graphs for 3s, 3p, and 3d to rationalize the filling order for these orbitals. 26 Briefly explain the following on the basis of electron configurations: a. Fluorine forms an ion having a charge of 1–.
b. The most common ion formed by zinc has a 2+ charge.
c. The electron configuration of the molybdenum atom is [Kr] 5s1 4d5 rather than [Kr] 5s2 4d4. 27 Briefly explain the following on the basis of electron configurations: a. The most common ion formed by silver has a 1+ charge.
b. Cm has the outer electron configuration s2d1 f 7 rather than s2 f 8.
c. Sn often forms an ion having a charge of 2+ (the stannous ion). 28 a. Which 2+ ion has two 3d electrons? Which has eight 3d electrons?
b. Which is the more likely configuration for Mn2+: [Ar] 4s2 3d3 or [Ar] 3d5? 29 Using Slater’s rules, determine Z for a. a 3p electron in P, S, Cl, and Ar. Is the calculated value of Z consistent with the relative sizes of these atoms?
b. a 2p electron in O2-, F-, Na+ and Mg2+. Is the cal-culated value of Z consistent with the relative sizes of these ions?
c. a 4s and a 3d electron of Cu. Which type of electron is more likely to be lost when copper forms a positive ion?
d. a 4f electron in Ce, Pr, and Nd. There is a decrease in size, commonly known as the lanthanide contraction, with increasing atomic number in the lanthanides. Are your values of Z consistent with this trend? 30 A sample calculation in this chapter showed that, according to Slater’s rules, a 3d electron of nickel has a higher effective nuclear charge than a 4s electron. Is the same true for early first-row transition metals? Using Slater’s rules, calculate S and Z for 4s and 3d electrons of Sc and Ti, and comment on the similarities or differ-ences with Ni. 31 Ionization energies should depend on the effective nuclear charge that holds the electrons in the atom. Calculate Z (Slater’s rules) for N, P, and As. Do their ionization energies seem to match these effective nuclear charges? If not, what other factors influence the ioniza-tion energies? 32 Prepare a diagram such as the one in Figure 12(a) for the fifth period in the periodic table, elements Zr through Pd. The configurations in Table 7 can be used to determine the crossover points of the lines. Can a diagram be drawn that is completely consistent with the configurations in the table? 33 Why are the ionization energies of the alkali metals in the order Li >, Na > K > Rb? 34 The second ionization of carbon (C + h C2+ + e-) and the first ionization of boron (B h B + + e+) both fit the reaction 1s2 2s2 2p1 h 1s2 2s2 + e-. Compare the two ionization energies (24.383 eV and 8.298 eV, respectively) and the effective nuclear charge Z. Is this an adequate explanation of the difference in ionization energies? If not, suggest other factors. 35 Explain why all three graphs in Figure 14 have maxima at 4 electrons and minima at 5 electrons. 36 a. For a graph of third ionization energy against atomic number, predict the positions of peaks and valleys for elements through atomic number 12. Compare the positions of these peaks and valleys with those for first ionization energies shown in Figure 13.
b. How would a graph of third ionization energies against the number of electrons in reactant compare with the other graphs shown in Figure 14? Explain briefly. 37 The second ionization energy involves removing an elec-tron from a positively charged ion in the gas phase (see preceding problem). How would a graph of second ion-ization energy vs. atomic number for the elements helium through neon compare with the graph of first ionization energy in Figure 13? Be specific in comparing the posi-tions of peaks and valleys. 38 In each of the following pairs, pick the element with the higher ionization energy and explain your choice.
a. Fe, Ru b. P, S c. K, Br d. C, N e. Cd, In f. Cl, F Atomic Structure 43 Atomic Structure 39 On the basis of electron configurations, explain why a. sulfur has a lower electron affinity than chlorine.
b. iodine has a lower electron affinity than bromine.
c. boron has a lower ionization energy than beryllium.
d. sulfur has a lower ionization energy than phosphorus. 40 a. The graph of ionization energy versus atomic num-ber for the elements Na through Ar (Figure 13) shows maxima at Mg and P and minima at Al and S. Explain these maxima and minima.
b. The graph of electron affinity versus atomic number for the elements Na through Ar (Figure 13) also shows maxima and minima, but shifted by one element in comparison with the ionization energy graph. Why are the maxima and minima shifted in this way? 41 The second ionization energy of He is almost exactly four times the ionization energy of H, and the third ionization energy of Li is almost exactly nine times the ionization energy of H: IE (MJ mol-1) H(g) h H +(g) + e-1.3120 He+(g) h He2+(g) + e-5.2504 Li2+(g) h Li3+(g) + e-11.8149 Explain this trend on the basis of the Bohr equation for energy levels of single-electron systems. 42 The size of the transition-metal atoms decreases slightly from left to right in the periodic table. What factors must be con-sidered in explaining this decrease? In particular, why does the size decrease at all, and why is the decrease so gradual? 43 Predict the largest and smallest radius in each series, and account for your choices: a. Se2- Br - Rb+ Sr2+ b. Y3+ Zr4+ Nb5+ c. Co4+ Co3+ Co2+ Co 44 Select the best choice, and briefly indicate the reason for each choice: a. Largest radius: Na+ Ne F -b. Greatest volume: S2- Se2- Te2-c. Highest ionization energy: Na Mg Al d. Most energy necessary to remove an electron: Fe Fe2+ Fe3+ e. Highest electron affinity: O F Ne 45 Select the best choice, and briefly indicate the reason for your choice: a. Smallest radius: Sc Ti V b. Greatest volume: S2- Ar Ca2+ c. Lowest ionization energy: K Rb Cs d. Highest electron affinity: Cl Br I e. Most energy necessary to remove an electron: Cu Cu+ Cu2+ 46 There are a number of Web sites that display atomic orbitals. Use a search engine to find a complete set of the f orbitals.
a. How many orbitals are there in one set (for example, a set of 4f orbitals)?
b. Describe the angular nodes of the orbitals.
c. Observe what happens to the number of radial nodes as the principal quantum number is increased.
d. Include the URL for the site you used for each, along with sketches or printouts of the orbitals. (Two useful Web sites at this writing are orbitals.com and winter.group.shef.ac.uk/orbitron.) 47 Repeat the exercise in Problem 46, this time for a set of g orbitals.
44 Atomic Structure s 1s 1s + 1s sn s 1s - 1s 1 E = RHa 1 22 - 1 32b = RHa 5 36b = 2.179 10-18 Ja 5 36b = 3.026 10-19 J = 1.097 107 m-1a 5 36b = 1.524 106 m-1 m 100 cm = 1.524 104 cm-1 2 The nodal surfaces require 2z2 - x2 - y2 = 0, so the angular nodal surface for a dz2 orbital is the conical surface where 2z2 = x2 + y2.
3 The angular nodal surfaces for a dxz orbital are the planes where xz = 0, which means that either x or z must be zero. The yz and xy planes satisfy this requirement.
4 One pair with c spin, one pair with T spin, one exchange possibility for each; energy contribution 2Pe. One pair (first orbital), energy contribution Pc. Total: 2Pe + Pc ∑e ∑c 2∑e + ∑c 3∑e + ∑c 2∑e + 2∑c 5 a. If the three 2p electrons all have the same spin, as in 3 1 2 , there are three exchange possibilities (1 and 2, 1 and 3, or 2 and 3) and no pairs. Overall, the total energy is 3Pe. If there is one unpaired electron, as in , there is one electron with T spin, and no possibility of exchange; two electrons with c spin, with one exchange possibility; and one pair. Overall; the total energy is Pe + Pc. Because Pe is negative and Pc is positive, the configuration with three unpaired electrons has a much lower energy. b. If the three 2p electrons avoid pairing, but do not all have mutually parallel spins, as in c c T , only the two c electrons can exchange. The total energy is Pe. The energy of this state is intermediate between those in part a. It is 2Pe higher than c c c and lower than c T c by Pc.
Answers to Exercises 45 Atomic Structure 6 Tin Total 5p 5s 4d Z 50 50 50 50 (1s2) 2 2 2 2 (2s22p6) 8 8 8 8 (3s23p6) 8 8 8 8 (3d10) 10 10 10 10 (4s24p6) 8 8 0.85 8 0.85 8 (4d 10) 10 10 0.85 10 0.85 9 0.35 (5s25p2) 4 3 0.35 3 0.35 Z 5.65 5.65 10.85 7 Uranium Total 7s 5f 6d Z 92 92 92 92 (1s2) 2 2 2 2 (2s22p6) 8 8 8 8 (3s23p6) 8 8 8 8 (3d 10) 10 10 10 10 (4s24p6) 8 8 8 8 (4d 10) 10 10 10 10 (4f 14) 14 14 14 14 (5s25p6) 8 8 8 8 (5d 10) 10 10 10 10 (5f 3) 3 3 2 0.35 3 (6s26p6) 8 8 0.85 8 (6d 1) 1 1 0.85 (7s2) 2 1 0.35 Z 3.00 13.30 3.00 8 With 4 electrons, the electron configurations of B+, Be, and Li– are all 1s2 2s2. Because the effective nuclear charge is greater for each 1s2 2s2 configuration than the 1s2 2s1 configuration of the preceding element, more energy is necessary to remove an electron from the species with 1s2 2s2 configurations. For 5 electrons (C+, B, and Be-) the configurations are all 1s2 2s2 2p1. Because the 2p orbitals are significantly higher in energy than the 2s orbitals, in each case it is much easier to remove an electron from a 1s2 2s2 2p1 configuration than from the 1s2 2s2 configuration of the preceding element. When a sixth electron is present (a 1s2 2s2 2p2 configuration), more energy is required to remove an electron because the electron being removed must overcome greater effective nuclear charge than for the preceding 5-electron species.
46 Simple Bonding Theory We now turn from the use of quantum mechanics and its description of the atom to an el-ementary description of molecules. Although most of our discussion of chemical bonding uses the molecular orbital approach, less rigorous methods that provide approximate pictures of the shapes and polarities of molecules are also useful. This chapter provides an overview of Lewis dot structures, valence shell electron-pair repulsion (VSEPR), and related topics. The ideas of this chapter provide a starting point for molecular orbital descriptions.
Ultimately, any description of bonding must be consistent with experimental data on bond lengths, bond angles, and bond strengths. Angles and distances are most frequently determined by diffraction (X-ray crystallography, electron diffraction, neutron diffraction) or spectroscopic (microwave, infrared) methods. For many molecules, there is general agreement on the nature of the bonding, although there are alternative ways to describe it. For others, there is considerable difference of opinion on the best way to describe the bonding. In this chapter we describe some useful qualitative approaches, including some of the opposing views.
1 Lewis Electron-Dot Diagrams Lewis electron-dot diagrams, although oversimplified, provide a good starting point for analyzing the bonding in molecules. Credit for their initial use goes to G. N. Lewis,1 an American chemist who contributed much to the understanding of thermodynamics and chemical bonding in the early twentieth century. In Lewis diagrams, bonds between two atoms exist when they share one or more pairs of electrons. In addition, some molecules have nonbonding pairs, also called lone pairs, of electrons on atoms. These electrons contribute to the shape and reactivity of the molecule but do not directly bond the atoms together. Most Lewis structures are based on the concept that eight valence electrons, corre-sponding to s and p electrons outside the noble gas core, form a particularly stable arrange-ment, as in the noble gases with s2 p6 configurations. An exception is hydrogen, which is stable with two valence electrons. Also, some molecules require more than eight electrons around a given central atom, and some molecules require fewer than eight electrons.
Simple molecules such as water follow the octet rule, in which eight electrons surround the central atom. Each hydrogen atom shares two electrons with the oxygen, forming the famil-iar structure with two bonds; the O atom accommodates two bonding pairs and two lone pairs: O H H (a) The treatment of water via molecular orbital theory results in an electronic structure in which each of these elec-tron pairs has a unique energy. This model is supported by spectroscopic evidence, and indicates one limitation of the Lewis model.
From Chapter 3 of Inorganic Chemistry, Fifth Edition. Gary L Miessler, Paul J. Fischer, Donald A. Tarr. Copyright © 2014 by Pearson Education, Inc. All rights reserved.
47 Simple Bonding Theory Shared electrons are considered to contribute to the electronic requirements of both atoms involved; thus, the electron pairs shared by H and O in the water molecule are counted toward both the 8-electron requirement of oxygen and the 2-electron requirement of hydrogen.
The Lewis model defines double bonds as containing four electrons and triple bonds as containing six electrons: O O C H H C C 1.1 Resonance In many Lewis structures, the choice of which atoms are connected by multiple bonds is arbitrary. When alternate locations for single bonds and multiple bonds are possible that all afford valid Lewis structures, a structure demonstrating each option should be drawn. For example, three drawings (resonance structures) of CO3 2- are needed (Figure 1) to show the double bond in each of the three possible CiO positions. In fact, experimental evidence shows that all three CiO bonds are equivalent, with bond lengths (129 pm) between typical CiO double-bond and single-bond distances (116 pm and 143 pm, respectively). All three drawings are necessary to describe the structure, with each drawing contributing equally to describe the bonding in the actual ion. This is called resonance; there is more than one possible way in which the valence electrons can be placed in a Lewis structure. Note that in resonance structures, such as those shown for CO3 2- in Figure 1, the electrons are arranged differently, but the nuclei remain in fixed positions.
The species CO3 2- and NO3 - have the same number of electrons (i.e., they are isoelectronic) and use the same orbitals for bonding. Their Lewis diagrams are identical except for the identity and formal charge (Section 1.3) of the central atom.
When a molecule has several resonance structures, its overall electronic energy is lowered, making it more stable. Just as the energy levels of a particle in a box are lowered by making the box larger, the electronic energy levels of the bonding electrons are lowered when the electrons can occupy a larger space.
1.2 Higher Electron Counts When it is impossible to draw a structure consistent with the octet rule because additional valence electrons remain to be assigned after the octet rule is satisfied on all atoms, it is necessary to increase the number of electrons around the central atom. An option limited to elements of the third and higher periods is to use d orbitals for this expansion, although theoretical work suggests that expansion beyond the s and p orbitals is unnecessary for most main group molecules.2 In most cases, two or four added electrons will complete the bonding, but more can be added if necessary. For example, 10 electrons are required around chlorine in ClF3 and 12 around sulfur in SF6 (Figure 2). The increased number of electrons is often described as an expanded shell or an expanded electron count. The term hypervalent is used to describe central atoms that have electron counts greater than the atom’s usual requirement.
There are examples with even more electrons around the central atom, such as IF7 (14 electrons), [TaF8]3- (16 electrons), and [XeF8]2- (18 electrons). There are rarely more than 18 electrons (2 for s, 6 for p, and 10 for d orbitals) around a single atom in the top half of the periodic table, and crowding of the outer atoms usually keeps the number below this, even for much heavier atoms that have f orbitals energetically available.
C O O O 2-C O O O 2-C O O 2-O FIGURE 1 Lewis Diagrams for CO3 2-.
48 Simple Bonding Theory 1.3 Formal Charge Formal charge is the apparent electronic charge of each atom in a molecule, based on the electron-dot structure. Formal charges help assess resonance structures and molecular topology, and they are presented here as a simplified method of describing structures, just as the Bohr model is a simple method of describing electronic configurations in atoms. Both of these methods have limitations, and other approaches are more accurate, but they can be useful as long as their imperfections are kept in mind.
Formal charges can help in eliminating resonance structures expected to contribute very little to the electronic ground state of the molecule, and, in some cases, suggesting multiple bonds beyond those required by the octet rule. It is essential, however, to remem-ber that formal charge is only a tool for assessing Lewis structures, not a measure of any actual charge on the atoms. The number of valence electrons available in a free atom of an element minus the total for that atom in the molecule—determined by counting lone pairs as two electrons and bonding pairs as one electron assigned to each atom—is the formal charge on the atom: = F F F F F F F F F F F F F F F Cl S S FIGURE 2 Structures of ClF3 and SF6.
Formal charge = ° number of valence electrons in a free atom of the element ¢ - a number of unshared electrons on the atomb - anumber of bonds to the atom b S A B C N 1-C S N 2-C S N 1-1+ C FIGURE 3 Resonance Structures of Thiocyanate, SCN-.
In addition, Charge on molecule or ion = sum of formal charges Resonance structures that contribute more to the electronic ground state of the species generally (a) have smaller magnitudes of formal charges, (b) place negative formal charges on more electronegative elements (in the upper right-hand part of the periodic table), and (c) have smaller separation of charges. Three examples—SCN -, OCN -, and CNO-—will illustrate the use of formal charges in describing electronic structures.
E X AMPLE 1 SCN− In the thiocyanate ion, SCN -, three resonance structures are consistent with the elec-tron-dot method, as shown in Figure 3. Structure A has only one negative formal charge on the nitrogen atom, the most electronegative atom in the ion. Structure B has a single negative charge on the S, which is less electronegative than N. Structure C has charges of 2- on N and 1+ on S, consistent with the relative electronegativities of these atoms but also has a large magnitude 2- charge and greater charge separation than the other structures. Therefore these structures lead to the prediction that structure A contributes the most to the electronic ground state of SCN−, structure B contributes an intermediate amount, and any contribution from C is minor in describing the electronic ground state of SCN -.
The bond lengths in Table 1 are somewhat consistent with this conclusion, with SCN - bond lengths between those of structures A and B. Protonation of the ion forms HNCS, consistent with a negative charge on N in SCN-. The bond lengths in HNCS are close to those of double bonds, consistent with the structure HiN “C“S.
49 Simple Bonding Theory E X AMPLE 2 OCN− The isoelectronic cyanate ion, OCN - (Figure 4), has the same possibilities, but the larger electronegativity of O is expected to make structure B contribute more to the electronic ground state in cyanate relative the contribution of B in thiocyanate. The protonation of cyanate results in two isomers, 97% HNCO and 3% HOCN, consistent with a major contribution of structure A and a small, but significant, contribution from B. The bond lengths in OCN - and HNCO in Table 2 are reasonably consistent with this analysis. Formal charge arguments provide a good starting point to assess Lewis structures, and reactivity patterns are also useful to gain experimental insight about electron distributions.
O N 1-C O N 2-C O N 1-1+ C A B C FIGURE 4 Resonance Structures of Cyanate, OCN-.
TABLE 2 Table of O—C and C—N Bond Lengths (pm) O—C C—N OCN -126 117 HNCO 118 120 Single bond 143 147 Double bond 116 (CO2) 128 (approximate) Triple bond 113 (CO) 116 Data from A. F. Wells, Structural Inorganic Chemistry, 5th ed., Oxford University Press, New York, 1984, pp. 807, 926, 933–934; S. E. Bradforth, E. H. Kim, E. W. Arnold, D. M. Neumark, J. Chem. Phys., 1993, 98, 800.
TABLE 1 Table of S—C and C—N Bond Lengths (pm) S—C C—N SCN - (in NaSCN) 165 118 HNCS 156 122 Single bond 181 147 Double bond 155 128 (approximate) Triple bond 116 Data from A. F. Wells, Structural Inorganic Chemistry, 5th ed., Oxford University Press, New York, 1984, pp. 807, 926, 934–936.
E X AMPLE 3 CNO− The isomeric fulminate ion, CNO- (Figure 5), can be drawn with three similar struc-tures, but the resulting formal charges have larger magnitudes than in OCN -. Because the order of electronegativities is C < N < O, none of these are ideal structures, and it is not surprising that this ion is unstable. The only common fulminate salts are of mercury and silver; both are explosive. Fulminic acid is linear HCNO in the vapor phase, consistent with the greatest contribution from structure C; coordination complexes of CNO-with transition-metal ions are known with MCNO structures.3 50 Simple Bonding Theory EXERCISE 1 Use electron-dot diagrams and formal charges to predict the bond order for each bond in POF3, SOF4, and SO3F -.
Some molecules have satisfactory electron-dot structures with octets but have more reasonable formal charge distributions in their structures with expanded electron counts. In each of the cases in Figure 6, the actual molecules and ions are consistent with electron counts greater than 8 on the central atom and with a large contribution from the resonance structure that uses multiple bonds to minimize formal charges. The multiple bonds may also influence the shapes of the molecules.
1.4 Multiple Bonds in Be and B Compounds A few molecules—such as BeF2, BeCl2, and BF3—seem to require multiple bonds to sat-isfy the octet rule for Be and B, even though multiple bonds for F and Cl are not generally expected on the basis of the high electronegativities of these halogens. Structures minimiz-ing formal charges for these molecules have only four electrons in the valence shell of Be and six electrons in the valence shell of B, in both cases fewer than the usual octet. The alternative, requiring eight electrons on the central atom, predicts multiple bonds, with BeF2 analogous to CO2 and BF3 analogous to SO3 (Figure 7). These structures, however, result in nonideal formal charges (2- on Be and 1+ on F in BeF2, and 1- on B and 1+ on the double-bonded F in BF3) on the basis of the usual rules.
A B C C O 1+ 1+ 2-N C O 1-N C O 3-1+ 1+ 1-N FIGURE 5 Resonance Structures of Fulminate, CNO-.
F F F N SNF3 SO2Cl2 XeO3 SO3 2-S N S O S O Xe O S N 0 0 2+ 2-2+ 1-3+ 1-1+ 1-12 S O 0 0 12 Xe O 0 0 14 S O 0 0,1-10 S Cl O Cl O O O O Xe O O O S F F F N S Cl O Cl O S O O O S O O O Xe Molecule Atom Formal Charge Formal Charge Atom Expanded to: Octet Expanded S FIGURE 6 Formal Charge and Expanded Electron Counts on Central Atom.
51 Simple Bonding Theory In solid BeF2, a complex network is formed with a Be atom coordination number of 4 (see Figure 7). BeCl2 dimerizes to a 3-coordinate structure in the vapor phase, but the lin-ear monomer is formed at high temperatures. This monomeric structure is unstable due to the electronic deficiency at Be; in the dimer and the network formed in the solid-state, the halogen atoms share lone pairs with the Be atom in an attempt to fill beryllium’s valence shell. The monomer is still frequently drawn as a singly bonded structure, with only four electrons around the beryllium and the ability to accept lone pairs of other molecules to relieve its electronic deficiency (Lewis acid behavior).
Bond lengths in all the boron trihalides are shorter than expected for single bonds, so the partial double-bond character predicted seems reasonable despite the nonideal formal charges of these resonance forms. While a small amount of double bonding is possible in these molecules, the strong polarity of the B–halogen bonds and the ligand close-packing (LCP) model (Section 2.4) have been used to account for the short bonds without the need to invoke multiple bonding. The boron trihalides combine readily with other molecules that can contribute a lone pair of electrons (Lewis bases), forming a roughly tetrahedral structure with four bonds: F F F H H H H H N H N B F F F B + Because of this tendency, boron trihalides are frequently drawn with only six electrons around the boron.
Other boron compounds that cannot be adequately described via simple electron-dot structures include hydrides such as B2H6, and many more complex molecules.
Be Be Be Be Be Predicted Actual solid Solid Predicted F F F F F F F Be B Cl Cl F F Cl Cl Be F F F F Be F F F F F F B F B F Be Cl Cl Cl Cl Be Be Vapor Cl Cl Cl Cl F bond length is 131 pm; The B the calculated single-bond length is 152 pm.
Predicted FIGURE 7 Structures of BeF2, BeCl2, and BF3 . (Data from A. F. Wells, Structural Inor-ganic Chemistry, 5th ed., Oxford University Press, Oxford, Eng-land, 1984, pp. 412, 1047.) 52 Simple Bonding Theory 2 Valence Shell Electron-Pair Repulsion Valence shell electron-pair repulsion (VSEPR) is an approach that provides a method for predicting the shape of molecules based on the electron-pair electrostatic repulsion described by Sidgwick and Powell4 in 1940 and further developed by Gillespie and Nyholm5 in 1957 and in the succeeding decades. Despite this method’s simple approach, based on Lewis electron-dot structures, the VSEPR method in most cases predicts shapes that compare favorably with those determined experimentally. However, this approach at best provides approximate shapes for molecules. The most common method of determining the actual structures is X-ray diffraction, although electron diffraction, neutron diffraction, and many spectroscopic methods are also used.6 The basis of the VSEPR approach is that electrons repel each other because they are negatively charged. Quantum mechanical rules dictate that electrons can be accommodated in the same region of space as bonding pairs or lone pairs, but each pair repels all other pairs. According to the VSEPR model, therefore, molecules adopt geometries such that valence electron pairs position themselves as far from each other as possible to minimize electron–electron repulsions. A molecule can be described by the generic formula AXmEn, where A is the central atom, X stands for any atom or group of atoms surrounding the central atom, and E represents a lone pair of electrons. The steric number (SN m n) is the total number of positions occupied by atoms or lone pairs around a central atom; lone pairs and bonding pairs both influence the molecular shape.
Carbon dioxide is a molecule with two atoms attached (SN = 2) to the central atom via double bonds. The electrons in each double bond must be between C and O, and the repulsion between these electron groups forces a linear structure on the molecule. Sulfur trioxide has three atoms bound to the sulfur (SN = 3), with equivalent partial double-bond character between sulfur and each oxygen, a conclusion rendered by analysis of its resonance forms. The best positions for the oxygens to minimize electron–electron repul-sions in this molecule are at the corners of an equilateral triangle, with OiSiO bond angles of 120°. The multiple bonding does not affect the geometry, because all three bonds are equivalent in terms of bond order.
The same pattern of finding the Lewis structure and then matching it to a geometry that minimizes the repulsive energy of bonding electrons is followed through steric numbers 4, 5, 6, 7, and 8 where the outer atoms are identical in each molecule, as shown in Figure 8.
Bond angles and distances are uniform in each of these structures with two, three, four, and six electron pairs. Neither the corresponding 5- nor 7-coordinate structures can have uniform angles and distances, because there are no regular polyhedra with these num-bers of vertices. The 5-coordinate molecules have a trigonal bipyramidal structure, with a central triangular plane of three positions plus two other positions above and below the center of the plane. The 7-coordinate molecules have a pentagonal bipyramidal structure, with a pentagonal plane of five positions and positions above and below the center of the plane. The regular square antiprism structure (SN = 8) is like a cube that has had the top face twisted 45° into the antiprism arrangement, as shown in Figure 9. It has three differ-ent bond angles for adjacent fluorines. [TaF8]3- has square antiprismatic geometry but is distorted from this ideal in the solid.7 The steric number is also called the number of electron pair domains.
O O C O O O S 53 Simple Bonding Theory F F X X X X X X X X 109.5° 45° 45° 70.5° 99.6° F F F F Ta M F F FIGURE 9 Conversion of a Cube into a Square Antiprism.
Steric Number Geometry Examples Calculated Bond Angles 2 Linear CO2 180° 3 Trigonal (triangular) SO3 120° 4 Tetrahedral CH4 109.5° 5 Trigonal bipyramidal PCl5 120°, 90° 6 Octahedral SF6 90° 7 Pentagonal bipyramidal IF7 72°, 90° 8 Square antiprismatic [TaF8]3− 70.5°, 99.6°, 109.5° FIGURE 8 VSEPR Predictions.
F F F F F F F F Cl P C C O O H H H H O S O O Cl Cl Cl Cl Ta I F F F F F F F S F F F F F F 54 Simple Bonding Theory 2.1 Lone-Pair Repulsion Bonding models are useful only if their explanations are consistent with experimental data. New theories are continually being suggested and tested. Because we are working with such a wide variety of atoms and molecular structures, a single approach will unlikely work for all of them. Although the fundamental ideas of atomic and molecular structures are relatively simple, their application to complex molecules is not. To a first approximation, lone pairs, single bonds, double bonds, and triple bonds can all be treated similarly when predicting molecular shapes. However, better predictions of overall shapes can be made by considering some important differences between lone pairs and bonding pairs. These methods are sufficient to show the trends and explain the bonding, as in rationalizing why the HiNiH angle in ammonia is smaller than the tetrahedral angle in methane and larger than the HiOiH angle in water.
As a general guideline, the VSEPR model predicts that electron-pair repulsions involv-ing lone pairs (lp) are stronger than those involving bonding pairs (bp) in the order lp-lp repulsions 7 lp-bp repulsions 7 bp-bp repulsions Steric Number = 4 The isoelectronic molecules CH4, NH3, and H2O (Figure 10) illustrate the effect of lone pairs on molecular shape. Methane has four identical bonds between carbon and each of the hydrogens. When the four pairs of electrons are arranged as far from each other as possible, the result is the familiar tetrahedral shape. The tetrahedron, with all HiCiH angles measuring 109.5°, has four identical bonds.
Ammonia also has four pairs of electrons around the central atom, but three are bond-ing pairs between N and H, and the fourth is a lone pair on the nitrogen. The nuclei form a trigonal pyramid with the three bonding pairs; the lone pair occupies the fourth region in space resulting in a tetrahedral arrangement of the four electron groups. Because each of the three bonding pairs is attracted by two positively charged nuclei (H and N), these pairs are largely confined to the regions between the H and N atoms. The lone pair, on the other hand, is attracted solely by the nitrogen nucleus; it has no second nucleus to confine it to a small region of space. Consequently, the lone pair tends to spread out and to occupy more space around the nitrogen than the bonding pairs. As a result, the HiNiH angles are 106.6°, nearly 3° smaller than the angles in methane.
The same principles apply to the water molecule, in which two lone pairs and two bonding pairs repel each other. Again, the electron pairs adopt a nearly tetrahedral arrange-ment, with the atoms arranged in a V shape. The angle of largest repulsion, between the two lone pairs, cannot be measured. However, the lone pair–bonding pair (lp–bp) repulsion is greater than the bonding pair–bonding pair (bp–bp) repulsion; as a result, the HiOiH bond angle is only 104.5°, another 2.1° decrease from the ammonia angles. The net result is that we can predict approximate molecular shapes by assigning more space to lone electron pairs; lone pairs are able to spread out and occupy more space since they are attracted to one nucleus rather than two.
Steric Number = 5 For the trigonal bipyramidal geometry, there are two unique locations for electron pairs, axial and equatorial. If there is a single lone pair, for example in SF4, the lone pair occu-pies an equatorial position. This position provides the lone pair with the most space and minimizes the interactions between the lone pair and bonding pairs. If the lone pair were axial, it would have three 90° interactions with bonding pairs; in an equatorial position, it has only two such interactions, as shown in Figure 11. The actual structure is distorted by the lone pair as it spreads out in space and effectively squeezes the rest of the molecule together.
C C H H H H H H 109.5° H H O O H H 104.5° H H N N H H H H 106.6° H H FIGURE 10 Shapes of Methane, Ammonia, and Water.
S F F F F Equatorial lone pair (observed structure) Axial lone pair F F F F S FIGURE 11 Possible Structures of SF4.
55 Simple Bonding Theory ClF3 provides a second example of the influence of lone pairs in molecules having a steric number of 5. There are three possible structures for ClF3, as shown in Figure 12.
In determining the feasibility of different structures, lone pair–lone pair interactions should be considered first, followed by lone pair–bonding pair interactions. These interac-tions at angles of 90° or less are generally considered destabilizing; larger angles gener-ally render structures more feasible. For example, in ClF3, structure B can be eliminated quickly because of the 90° lp-lp angle. The lp-lp angles are large for A and C, so the choice must come from the lp-bp and bp-bp angles. Because the lp-bp angles are more important, C, which has only four 90° lp-bp interactions, is favored over A, which has six such interactions. Experiments have confirmed that the structure is based on C, with slight distortions due to the lone pairs. The lone pair–bonding pair repulsion causes the lp–bp angles to be larger than 90° and the bp-bp angles to be less than 90° (actually, 87.5°). The CliF bond distances show the repulsive effects as well, with the axial fluorines (approxi-mately 90° lp-bp angles) at 169.8 pm and the equatorial fluorine (in the plane with two lone pairs) at 159.8 pm.8 Angles involving lone pairs cannot be determined experimentally.
Angles in Possible Structures Experimental Interaction A B C lp–lp 180° 90° 120° Cannot be determined lp–bp 6 at 90° 3 at 90° 4 at 90° Cannot be determined 2 at 120° 2 at 120° bp–bp 3 at 120° 2 at 90° 2 at 90° 2 at 87.5° 1 at 120° Axial Cl—F 169.8 pm Equatorial Cli F 159.8 pm Additional examples of structures with lone pairs are illustrated in Figure 13. The structures based on a trigonal bipyramidal arrangement of electron pairs around a central atom always place any lone pairs in the equatorial plane, as in SF4, BrF3, and XeF2. The resulting shapes minimize both lone pair–lone pair and lone pair–bonding pair repulsions. The shapes are called seesaw (SF4), distorted T (BrF3), and linear (XeF2).
Steric Numbers = 6 and 7 In octahedral structures, all six positions are equivalent. When a single lone pair is present, it typically repels adjacent bonding pairs, reducing bond angles accordingly, as for IF5 in Figure 13. In octahedron-based structures with two lone pairs, lone pair–lone pair repulsion is minimized if these pairs are trans, and this is the shape that is adopted. Square planar XeF4, also shown in Figure 13, is an example. Recently XeF3 -, which would be expected to have a steric number of 6 and three lone pairs, has been reported in the gas phase, but attempts to prepare salts of this ion have been unsuccessful.9 The shape that minimizes electron-pair repulsions for a steric number of 7 is the pen-tagonal bipyramid, shown in Figure 8. IF7 (in the margin) and TeF7 2- exhibit this shape, with both axial and equatorial fluorines. If a single lone pair is present, in some cases the lone pair causes distortion. The nature of this distortion is not always easy to ascertain; XeF6 is a classic example.10 In other cases the structure is octahedral (see Problem 26) with the lone pair not stereochemically active. Two lone pairs minimize their repulsions by adopting axial (trans) positions, with the atoms all in the equatorial plane. Two known examples are XeF5 - (in the margin) and IF5 2-.
Cl F F F F F F Cl F F F Cl F F F A B C Experimental Cl FIGURE 12 Possible Structures of ClF3.
A lone pair that appears in the Lewis-dot structure but has no apparent effect on the molecular geometry is classified as not stereo chemically active. The VSEPR model assumes that all lone pairs are stereochemically active and therefore do affect the molecular geometry.
I F F F F F F F Xe F F F F F -56 Simple Bonding Theory E X AMPLE 4 SbF4 has a single lone pair on Sb. Its structure is therefore similar to SF4, with a lone pair occupying an equatorial position. This lone pair causes considerable distortion, giving an FiSbiF (axial positions) angle of 155° and an FiSbiF (equatorial) angle of 90°.
SF5 has a single lone pair. Its structure is based on an octahedron, with the ion distorted away from the lone pair, as in IF5 .
SeF3 has a single lone pair. This lone pair reduces the SeiF bond angle significantly, to 94°.
EXERCISE 2 Predict the structures of the following ions. Include a description of distor-tions from the ideal angles (for example, less than 109.5° because…).
NH2 - NH4 + I3 - PCl6 -2.2 Multiple Bonds The VSEPR model considers double and triple bonds to have slightly greater repulsive effects than single bonds because of the repulsive effect of p electrons that increase the electron density between the bonded atoms beyond that present in a s bond. For example, the H3CiCiCH3 angle in (CH3)2C“CH2 is smaller, and the H3CiC“CH2 angle is larger than the trigonal 120° (Figure 14).11 Cl Cl Be Steric Number None 1 2 3 Number of Lone Pairs on Central Atom 2 3 4 H H H Cl Cl 95° Sn B F F F C H H H H 106.6° N H H 104.5° O P Cl Cl Cl Cl Cl S F F F F F F F F F Br F F F 173° 86.2° 81.9° Xe F F F F Xe F F 101.6° F F S 5 6 F F F F I FIGURE 13 Structures Containing Lone Pairs.
Sb F F F F -S F F F F F -Se F F F + H H 122.2° 115.6° C C H3C H3C FIGURE 14 Bond Angles in (CH3)2C=CH2.
57 Simple Bonding Theory Additional examples of the effect of multiple bonds on molecular geometry are shown in Figure 15. Comparing Figures 13 and 15, we see that multiple bonds tend to occupy the same positions as lone pairs. For example, the double bonds to oxygen in SOF4, ClO2F3, and XeO3F2 are all equatorial, as are the lone pairs in the matching compounds of steric number 5, SF4, BrF3, and XeF2. Multiple bonds, like lone pairs, also tend to occupy more space than single bonds, causing distortions that squeeze the rest of the molecule together. In molecules that have both lone pairs and multiple bonds, these features may compete for space; examples are shown in Figure 16. As a generalization, lone pairs often have a greater influence than multiple bonds in dictating molecular geometry.
E X AMPLE 5 HCP, like HCN, is linear, with a triple bond: HiC‚P: IOF4 has a single lone pair on the side opposite the oxygen. The lone pair has a slightly greater repulsive effect than the double bond to oxygen, as shown by the average OiIiF angle of 89°. (The extra repulsive character of the I “O bond places it opposite the lone pair.) SeOCl2 has both a lone pair and a selenium–oxygen double bond. The lone pair has a greater effect than the double bond; the CliSeiCl angle is reduced to 97° by this effect, and the CliSeiO angle is 106°.
EXERCISE 3 Predict the structures of the following. Indicate the direction of distortions from the regular structures.
XeOF2 ClOF3 SOCl2 S O O O 108° 126° F F C O 115° O O N O O C Steric Number 1 2 3 4 Number of Bonds with Multiple Bond Character 2 3 4 5 6 Xe O O O 94° F F F F F F F F F F S S N 111° 110° 125° 90.7° 103° 120° Cl Cl O O O S Cl O O O O O O S F Xe F O O O F I O F F F F 2- The bond angles of these molecules have not been determined accurately. However, spectroscopic measurements are consistent with the structures shown.
-FIGURE 15 Structures Containing Multiple Bonds.
O O F F 102° ~180° Xe O F F F F I O F F F 98° 168° 91° -I FIGURE 16 Structures Contain-ing Both Lone Pairs and Multiple Bonds.
Se Cl Cl O I O F F F F -58 Simple Bonding Theory 2.3 Electronegativity and Atomic Size Effects Electronegativity is a measure of an atom’s ability to attract electrons from a neighboring atom to which it is bonded; it can be viewed as the ability of an atom to win the competi-tion to attract shared electrons. Electronegativity was mentioned earlier as a guide in the use of formal charges. It also can play an important role in determining the arrangement of outer atoms around a central atom and in rationalizing bond angles. The effects of elec-tronegativity and atomic size frequently parallel each other, but in some cases, the sizes of outer atoms and groups may play the more important role.
Electronegativity Scales Linus Pauling introduced the concept of electronegativity in the 1930s as a means of describing bond energies. Pauling recognized that polar bonds have higher bond energies than nonpolar bonds formed from the same elements. For example, he observed that the bond energy of HCl, 432 kJ/mol, was much higher than the average of the bond energies of H2 (436 kJ/mol) and Cl2 (243 kJ/mol). He related the difference between actual and average bond energies to the difference in electronegativity between the elements involved. He also made adjustments for the sake of convenience, most notably to give the elements C through F equally spaced values of 2.5 through 4.0. Some early Pauling electronegativity values are in Table 3. The value of 4.0 for fluorine is still commonly used as a reference point for other electronegativity scales.
More recent values have been derived from other molecular and atomic properties, such as ionization energies and electron affinities. Table 4 summarizes approaches used for a variety of electronegativity scales; examining differences among these is beyond the scope of this text. In most cases the different methods give similar electronegativity val-ues, sometimes with the exception of the transition metals.12 We choose to use the values reported by Mann, Meek, and Allen (Table 5) based on configuration energies (CE), the average ionization energies of valence electrons in ground state free atoms. For s- and p-block elements the configuration energies are defined as follows:13 CE = nes + mep n + m where n = number of s electrons m = number of p electrons es, es = experimental 1-electron s and p energies† TABLE 3 Early Values of Pauling Electronegativities H 2.1 C N O F 2.5 3.0 3.5 4.0 Si P S Cl 1.8 2.1 2.5 3.0 Ge As Se Br 1.8 2.0 2.4 2.8 Values used by Pauling, converted to kJ/mol. L. Pauling, The Nature of the Chemical Bond, 3rd ed., 1960, Cornell University Press, Ithaca, NY, p. 81.
Earlier, Pauling had assigned fluorine an electronegativity of 2.00; see L. Pauling, J. Am. Chem. Soc., 1932, 54, 3570.
†Multiplet averaged values from C. E. Moore, Ionization Potentials and Ionization Limits Derived From the Analyses of Optical Spectra, NSRDS-NBS-34, Washington, D.C., 1971; Atomic Energy Levels, NSRDS-35, Washington, D.C., 1971, Vol. III.
59 Simple Bonding Theory The configuration energies are multiplied by a constant to give values comparable to the Pauling scale to enable convenient comparison between the scales. Pauling’s calculation of electronegativities from bond energies requires averaging over a number of compounds in an attempt to minimize experimental uncertainties and other minor effects. Methods that use ionization energies and other atomic properties can be calculated more directly. The electronegativities reported are suitable for most uses, but the actual values for atoms in different molecules can differ depending on the specific electronic environment of the atoms. The concept of electronegativity varying for a given atom on the basis of its specific bonds within a molecule is usually not intro duced in introductory chemistry, but is a consequence of modern electronegativity scales.
It is important to emphasize that all electronegativities are measures of an atom’s ability to attract electrons from a neighboring atom to which it is bonded. A critique of all electronegativity scales, and particularly Pauling’s, is that each scale cannot be successfully applied to all situations; all of these scales have deficiencies on the basis of the specific assumptions used in their development.21 TABLE 4 Electronegativity Scales Principal Authors Method of Calculation or Description Pauling14 Bond energies Mulliken15 Average of electron affinity and ionization energy Allred & Rochow16 Electrostatic attraction proportional to Z/r 2 Sanderson17 Electron densities of atoms Pearson18 Average of electron affinity and ionization energy Allen19 Average energy of valence shell electrons, configuration energies Jaffé20 Orbital electronegativities TABLE 5 Electronegativity (Pauling Units) 1 2 12 13 14 15 16 17 18 H He 2.300 4.160 Li 0.912 Be 1.576 B 2.051 C 2.544 N 3.066 O 3.610 F 4.193 Ne 4.787 Na 0.869 Mg 1.293 Al 1.613 Si 1.916 P 2.253 S 2.589 Cl 2.869 Ar 3.242 K 0.734 Ca 1.034 Zn 1.588 Ga 1.756 Ge 1.994 As 2.211 Se 2.424 Br 2.685 Kr 2.966 Rb 0.706 Sr 0.963 Cd 1.521 In 1.656 Sn 1.824 Sb 1.984 Te 2.158 l 2.359 Xe 2.582 Cs 0.659 Ba 0.881 Hg 1.765 Tl 1.789 Pb 1.854 Bi (2.01) Po (2.19) At (2.39) Rn (2.60) Source: J. B. Mann, T. L. Meek, L. C. Allen, J. Am. Chem. Soc., 2000, 122, 2780, Table 2.
For a recent approach that addresses some of the limitations of the Allen method, see P. Politzer, Z. P. Shields, F. A. Bulat, J. S. Murray, J. Chem. Theory Comput., 2011, 7, 377.
60 Simple Bonding Theory With the exception of helium and neon, which have large calculated electronegativi-ties and no known stable compounds, fluorine has the largest value, and electronegativity decreases toward the lower left corner of the periodic table. Although usually classified with Group 1 (IA), hydrogen is quite dissimilar from the alkali metals in its electronega-tivity, as well as in many other chemical and physical properties. Hydrogen’s chemistry is distinctive from all the groups.
Electronegativities of the noble gases can be calculated more easily from ionization energies than from bond energies. Because the noble gases have higher ionization energies than the halogens, calculations suggest that the electronegativities of the noble gases may exceed those of the halogens (Table 5).22 The noble gas atoms are somewhat smaller than the neighboring halogen atoms—for example, Ne is smaller than F—as a consequence of a greater effective nuclear charge. This charge, which is able to attract noble gas electrons strongly toward the nucleus, is also likely to exert a strong attraction on electrons of neighbor-ing atoms; hence, the high electronegativities predicted for the noble gases are reasonable.
Electronegativity and Bond Angles By the VSEPR approach, trends in many bond angles can be explained by electronegativity. Consider the bond angles in the following molecules: Molecule X–P–X Angle (°) Molecule X–S–X Angle (°) PF3 97.8 OSF2 92.3 PCl3 100.3 OSCl2 96.2 PBr3 101.0 OSBr2 98.2 As the electronegativity of the halogen increases, the halogen exerts a stronger pull on electron pairs it shares with the central atom. This effect reduces the concentration of electrons near the central atom, decreasing somewhat the repulsion between the bonding pairs near the central atom, and allows the lone pair to have more impact in compress-ing the halogen–central atom–halogen angles. Consequently, the molecules with the most electronegative outer atoms, PF3 and OSF2, have the smallest angles.
If the central atom remains the same, molecules that have a larger difference in electronegativity values between their central and outer atoms have smaller bond angles. The atom with larger electronegativity draws the shared electrons toward itself and away from the central atom, reducing the repulsive effect of these electrons. The compounds of the halogens in Table 6 show this effect; the compounds containing fluorine have smaller angles than those containing chlorine, which in turn have smaller angles than those con-taining bromine. The lone pair exerts a relatively larger effect, and forces smaller bond angles, as the electronegativity of the outer atom increases. An alternative explanation for this trend is size: as the size of the outer atom increases in the order F 6 Cl 6 Br, the bond angle increases. Additional compounds showing the effects of electronegativity on bond angles are also given in Table 6.
Similar considerations can be made in situations where the outer atoms remain the same, but the central atom is changed, for example, Molecule Bond Angle (°) Molecule Bond Angle (°) H2O 104.5 NCl3 106.8 H2S 92.1 PCl3 100.3 H2Se 90.6 AsCl3 98.9 In these cases, as the central atom becomes more electronegative, it pulls electrons in bonding pairs more strongly toward itself, increasing the concentration of electrons near the central atom. 61 Simple Bonding Theory The net effect is that an increase in bonding pair–bonding pair repulsions near the central atom increases the bond angles. In these situations the molecule with the most electronegative central atom has the largest bond angles. Additional examples can be found in Table 6, where mol-ecules having the same outer atoms, but different central atoms, are shown in the same column.
E XE RCISE 4 Which molecule has the smallest bond angle in each series?
a. OSeF2 OSeCl2 OSeBr2 (halogen–Se–halogen angle) b. SbCl3 SbBr3 SbI3 c. PI3 AsI3 SbI3 Effects of Size In the examples considered so far, the most electronegative atoms have also been the smallest. For example, the smallest halogen, fluorine, is also the most electronegative. Consequently, we could have predicted the trends in bond angles on the basis of atomic size, with the smallest atoms capable of being crowded together most closely. It is impor-tant to also consider situations in which size and electronegativity might have opposite effects, where a smaller outer group is less electronegative than a larger group attached to a central atom. For example, Molecule C i N i C Angle (°) N(CH3)3 110.9 N(CF3)3 117.9 In this case VSEPR would predict that the more electronegative CF3 groups would lead to a smaller bond angle because they would withdraw electrons more strongly than CH3 groups. That the bond angle in N(CF3)3 is actually 7° larger than in N(CH3)3 suggests that in this case, size is the more important factor, with the larger CF3 groups requiring more space. The point at which the size of outer atoms and groups becomes more important TABLE 6 Bond Angles and Lengths Molecule Bond Angle (°) Bond Length (pm) Molecule Bond Angle (°) Bond Length (pm) Molecule Bond Angle (°) Bond Length (pm) Molecule Bond Angle (°) Bond Length (pm) H2O 104.5 97 OF2 103.3 96 OCl2 110.9 170 H2S 92.1 135 SF2 98.0 159 SCl2 102.7 201 H2Se 90.6 146 SeCl2 99.6 216 H2Te 90.2 169 TeCl2 97.0 233 NH3 106.6 101.5 NF3 102.2 137 NCl3 106.8 175 PH3 93.2 142 PF3 97.8 157 PCl3 100.3 204 PBr3 101.0 220 AsH3 92.1 151.9 AsF3 95.8 170.6 AsCl3 98.9 217 AsBr3 99.8 236 SbH3 91.6 170.7 SbF3 87.3 192 SbCl3 97.2 233 SbBr3 98.2 249 Source: N. N. Greenwood and A. Earnshaw, Chemistry of the Elements, 2nd ed., Butterworth-Heinemann, Oxford, 1997, pp. 557, 767; A. F. Wells, Structural Inorganic Chemistry, 5th ed., Oxford University Press, Oxford, 1987, pp. 705, 793, 846, and 879; R. J. Gillespie and I. Hargittai, The VSEPR Model of Molecular Geometry, Allyn and Bacon, Needham Heights, MA, 1991.
62 Simple Bonding Theory than electronegativity can be difficult to predict, but the potential of large outer atoms and groups to affect molecular shape should not be dismissed.
Molecules Having Steric Number = 5 For main group atoms having a steric number of 5, it is instructive to consider the relative bond lengths for axial and equatorial positions. For example, in PCl5, SF4, and ClF3, the central atom–axial distances are longer than the distances to equatorial atoms, as shown in Figure 17. This effect has been attributed to the greater repulsion of lone and bonding pairs with atoms in axial positions (three 90° interactions) than with atoms in equatorial positions (two 90° interactions).
In addition, there is a tendency for less electronegative groups to occupy equatorial posi-tions, similar to lone pairs and multiply bonded atoms. For example, in phosphorus compounds having both fluorine and chlorine atoms, in each case the chlorines occupy equatorial positions (Figure 18). The same tendency is shown in compounds having formulas PF4CH3, PF3(CH3)2, and PF2(CH3)3, with the less electronegative CH3 groups also equatorial (Figure 19). One can envision the electron density of the P—A bond, where A is the less electronegative atom, being concentrated closer to the phosphorus in such cases, leading to a preference for equatorial positions by similar reasoning applied to lone pairs and multiple bonds.
The relative effects on bond angles by less electronegative atoms are, however, typi-cally less than for lone pairs and multiple bonds. For example, the bond angle to equatorial positions opposite the Cl atom in PF4Cl is only slightly less than 120°, in contrast to the greater reduction in comparable angles in SF4 and SOF4 (Figure 20).
Predicting structures in some cases is challenging. Phosphorus compounds contain-ing both fluorine atoms and CF3 groups provide an intriguing example. CF3 is an electron withdrawing group whose electronegativity has been calculated to be comparable to the more electronegative halogen atoms. Does CF3 favor equatorial positions more strongly than F? Trigonal bipyramidal phosphorus compounds containing varying numbers of F and CF3 groups with both axial and equatorial CF3 groups are known (Figure 21). When two or three CF3 groups are present, the orientations are truly a challenge to explain: these groups are axial in PF3(CF3)2 but equatorial in PF2(CF3)3! In both cases the more sym-metrical structure, with identical equatorial groups, is preferred. Cl Cl Cl Cl Cl P 202 213 165 154 160 170 F F F F S F F Cl F FIGURE 17 Bond Distances in PCl5, SF4, and ClF3.
F F Cl F F P 200 158 159 160 154 200 154 F Cl Cl F P F 200 F Cl Cl F P Cl FIGURE 18 PClF4, PCl2F3, and PCl3F2.
F F CH3 F F P 178 161 164 168 154 155 180 F H3C F F P CH3 181 F H3C H3C F P CH3 FIGURE 19 PF4CH3, PF3(CH3)2, and PF2(CH3)3.
See H. Oberhammer, J. Grobe, D. Le Van, Inorg. Chem., 1982, 21, 275 for a discussion of these structures.
For an analysis of different approaches to determining the electronegativity of CF3, see J. E. True, T. D. Thomas, R. W. Winter, G. L. Gard, Inorg. Chem., 2003, 42, 4437.
63 |
11296 | https://www.youtube.com/watch?v=eJnS-PQVvk0 | Integral of x^(-1/3)
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00:00 Integrate x^(-1/3)
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Integrate x^(-1/3) Integrate x^(-1/3) Convert 1 to 3/3 Convert 1 to 3/3 Simplify expression Simplify expression Rewrite expression Rewrite expression Add integration constant +C Add integration constant +C Final answer! Final answer! |
11297 | https://www.rcet.org.in/uploads/academics/rohini_78114304250.pdf | ROHINI COLLEGE OF ENGINEERING AND TECHNOLOGY EC8451 ELECTROMAGNETICFIELDS Poynting Vector and Power Flow in Electromagnetic Fields Electromagnetic waves can transport energy from one point to another point. The electric and magnetic field intensities asscociated with a travelling electromagnetic wave can be related to the rate of such energy transfer. Let us consider Maxwell's Curl Equations: Using vector identity the above curl equations we can write .............................................(5.35) In simple medium where and are constant, we can write and Applying Divergence theorem we can write, ...........................(5.35) ROHINI COLLEGE OF ENGINEERING AND TECHNOLOGY EC8451 ELECTROMAGNETICFIELDS The term represents the rate of change of energy stored in the electric and magnetic fields and the term represents the power dissipation within the volume. Hence right hand side of the equation (5.35) represents the total decrease in power within the volume under consideration. The left hand side of equation (5.35) can be written as where (W/mt2) is called the Poynting vector and it represents the power density vector associated with the electromagnetic field. The integration of the Poynting vector over any closed surface gives the net power flowing out of the surface. Equation (5.35) is referred to as Poynting theorem and it states that the net power flowing out of a given volume is equal to the time rate of decrease in the energy stored within the volume minus the conduction losses. Poynting vector for the time harmonic case: For time harmonic case, the time variation is of the form , and we have seen that instantaneous value of a quantity is the real part of the product of a phasor quantity and when is used as reference. For example, if we consider the phasor then we can write the instanteneous field as .................................(5.37) when E0 is real. Let us consider two instanteneous quantities A and B such that where A and B are the phasor quantities. i. e, ROHINI COLLEGE OF ENGINEERING AND TECHNOLOGY EC8451 ELECTROMAGNETICFIELDS Therefore, ..............................(5.39) Since A and B are periodic with period , the time average value of the product form AB, denoted by can be written as .....................................(5.40) Further, considering the phasor quantities A and B, we find that and , where denotes complex conjugate. ..............................................(5.41) The poynting vector can be expressed as ...................................(5.4 2) If we consider a plane electromagnetic wave propagating in +z direction and has only component, from (5.42) we can write: Using (5.41) ROHINI COLLEGE OF ENGINEERING AND TECHNOLOGY EC8451 ELECTROMAGNETICFIELDS ........................................(5.43) where and , for the plane wave under consideration. For a general case, we can write .....................(5.44) We can define a complex Poynting vector and time average of the instantaneous Poynting vector is given by . |
11298 | https://opencw.aprende.org/courses/physics/8-02-electricity-and-magnetism-spring-2002/video-lectures/lecture-9-currents-resistivity-and-ohms-law/ | Lecture 9: Currents, Resistivity and Ohm's Law | Video Lectures | Electricity and Magnetism | Physics | MIT OpenCourseWare
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Home » Courses » Physics » Electricity and Magnetism » Video Lectures » Lecture 9: Currents, Resistivity and Ohm's Law
Lecture 9: Currents, Resistivity and Ohm's Law
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{'English - US': '/courses/physics/8-02-electricity-and-magnetism-spring-2002/video-lectures/lecture-9-currents-resistivity-and-ohms-law/8.02_L09.srt'}
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Topics covered:
Currents
Resistivity
Ohm's Law
Instructor/speaker: Prof. Walter Lewin
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Download this transcript - PDF (English - US)
When positive charges move in this direction, then per definition, we say the current goes in this direction.
When negative charges go in this direction, we also say the current goes in that direction, that's just our convention.
If I apply a potential difference over a conductor, then I'm going to create an electric field in that conductor.
And the electrons -- there are free electrons in a conductor -- they can move, but the ions cannot move, because they are frozen into the solid, into the crystal.
And so when a current flows in a conductor, it's always the electrons that are responsible for the current.
The electrons fuel the electric fields, and then the electrons try to make the electric field 0, but they can't succeed, because we keep the potential difference over the conductor.
Often, there is a linear relationship between current and the potential, in which case, we talk about Ohm's Law.
Now, I will try to derive Ohm's Law in a very crude way, a poor man's version, and not really 100 percent kosher, it requires quantum mechanics, which is beyond the course -- beyond this course -- but I will do a job that still gives us some interesting insight into Ohm's Law.
If I start off with a conductor, for instance, copper, at room temperature, 300 degrees Kelvin, the free electrons in copper have a speed, an average speed of about a million meters per second.
So this is the average speed of those free electrons, about a million meters per second.
This in all directions.
It's a chaotic motion.
It's a thermal motion, it's due to the temperature.
The time between collisions -- time between the collisions -- and this is a collision of the free electron with the atoms -- is approximately -- I call it tau -- is about 3 times 10 to the -14 seconds.
No surprise, because the speed is enormously high.
And the number of free electrons in copper per cubic meter, I call that number N, is about 10 to the 29.
There's about one free electron for every atom.
So we get twen- 10 to the 29 free electrons per cubic meter.
So now imagine that I apply a potential difference -- piece of copper -- or any conductor, for that matter -- then the electrons will experience a force which is the charge of the electron, that's my little e times the electric field that I'm creating, because I apply a potential difference.
I realize that the force and the electric field are in opposite directions for electrons, but that's a detail, I'm interested in the magnitudes only.
And so now these electrons will experience an acceleration, which is the force divided by the mass of the electron, and so they will pick up, a speed, between these collisions, which we call the drift velocity, which is A times tau, it's just 8.01.
And so A equals F divided by Me.
F is e E, so we get e times E divided by the mass of the electrons, times tau.
And that is the the drift velocity.
When the electric field goes up, the drift velocity goes up, so the electrons move faster in the direction opposite to the current.
If the time between collisions gets larger, they -- the acceleration lasts longer, so also, they pick up a larger speed, so that's intuitively pleasing.
If we take a specific case, and I take, for instance, copper, and I apply over the -- over a wire -- let's say the wire has a length of 10 meters -- I apply a potential difference I call delta V, but I could have said just V -- I apply there a potential difference of 10 volts, then the electric field -- inside the conductor, now -- is about 1 volt per meter.
And so I can calculate, now, for that specific case, I can calculate what the drift velocity would be.
So the drift velocity of those free electrons would be the charge of the electron, which is 1.6 times 10 to the -19 Coulombs.
The E field is 1, so I can forget about that.
Tau is 3 times 10 to the -14, as long as I'm room temperature, and the mass of the electron is about 10 to the -30 kilograms.
And so, if I didn't slip up, I found that this is 5 times 10 to the -3 meters per second, which is half a centimeter per second.
So imagine, due to the thermal motion, these free electrons move with a million meters per second.
But due to this electric field, they only advance along the wire slowly, like a snail, with a speed on average of half a centimeter per second.
And that goes very much against your and my own intuition, but this is the way it is.
I mean, a turtle would go faster than these electrons.
To go along a 10-meter wire would take half hour.
Something that you never thought of.
That it would take a half hour for these electrons to go along the wire if you apply potential difference of 10 volts, copper 10 meters long.
Now, I want to massage this further, and see whether we can somehow squeeze out Ohm's Law, which is the linear relation between the potential and the current.
So let me start off with a wire which has a cross-section A, and it has a length L, and I put a potential difference over the wire, plus here, and minus there, potential V, so I would get a current in this direction, that's our definition of current, going from plus to minus.
The electrons, of course, are moving in this direction, with the drift velocity.
And so the electric field in here, which is in this direction, that electric field is approximately V divided by L, potential difference divided by distance.
In 1 second, these free electrons will move from left to right over a distance Vd meters.
So if I make any cross-section through this wire, anywhere, I can calculate how many electrons pass through that cross-section in 1 second.
In 1 second, the volume that passes through here, the volume is Vd times A but the number of free electrons per cubic meter is called N, so this is now the number of free electrons that passes, per second, through any cross-section.
And each electron has a charge E, and so this is the current that will flow.
The current, of course, is in this direction, but that's a detail.
If I now substitute the drift velocity, which we have here, I substitute that in there, but then I find that the current -- I get a e squared, the charge squared, I get N, I get tau, I get downstairs, the mass of the electron, and then I get A times the electric field E.
Because I have here, is electric field E.
When you look at this here, that really depends only on the properties of my substance, for a given temperature.
And we give that a name.
We call this sigma, which is called conductivity.
Conductivity.
If I calculate, for copper, the conductivity, at room temperature, that's very easy, because I've given you what N is, on the blackboard there, 10 to the 29, you know what tau is at room temperature, 3 times 10 to the -14, so for copper, at room temperature, you will find about 10 to the 8.
You will see more values for sigma later on during this course.
This is in SI units.
I can massage this a little further, because E is V divided by L, and so I can write now that the current is that sigma times A times V divided by L.
I can write it down a little bit differently, I can say V, therefore, equals L divided by sigma A, times I.
And now, you're staring at Ohm's Law, whether you like it or not, because this is what we call the resistance, capital R.
We often write down rho for 1 over sigma, and rho is called the resistivity.
So either one will do.
So you can also write down -- you can write down V equals I R, and this R, then, is either L divided by sigma A, or L times rho -- let me make it a nicer rho -- divided by A.
That's the same thing.
The units for resistance R is volts per ampere, but we call that ohm.
And so the unit for R is ohm.
And so if you want to know what the unit for rho and sigma is, that follows immediately from the equations.
The unit for rho is then ohm-meters.
So we have derived the resistance here in terms of the dimensions -- namely, the length and the cross-section -- but also in terms of the physics on an atomic scale, which, all by itself, is interesting.
If you look at the resistance, you see it is proportional with the length of your wire through which you drive a current.
Think of this as water trying to go through a pipe.
If you make the pipe longer, the resistance goes up, so that's very intuitively pleasing.
Notice that you have A downstairs.
That means if the pipe is wider, larger cross-section, it's also easier for the current to flow, it's easier for the water to flow.
So that's also quite pleasing.
Ohm's Law, also, often holds for insulators, which are not conductors, even though I have derived it here for conductors, which have these free electrons.
And so now, I want to make a comparison between very good conductors, and very good insulators.
So I'll start off with a -- a chunk of material , cross-sectional area A -- let's take it 1 millimeter by 1 millimeter -- so A is 10 to the -6 square meters.
So here I have a chunk of material, and the length of that material, L, is 1 meter.
Put a potential difference over there, plus here, and minus here.
Current will start to flow in this direction, electrons will flow in this direction.
The question now is, what is the resistance of this chunk of material?
Well, very easy.
You take these equations, you know L and A, so if I tell you what sigma is, then you can immediately calculate what the resistance is.
So let's take, first, a good conductor.
Silver and gold and copper are very good conductors.
They would have values for sigma, 10 to the 8, we just calculated for copper, you've seen in front of your own eyes.
So that means rho would be 10 to the minus 8, it's 1 over sigma.
And so in this particular case, since A is 10 to the -6 the resistance R is simply 10 to the 6th times rho.
Because L is 1 meter.
So it's very easy -- resistance here, R, is 10 to the -2 ohms.
1/100 of an ohm.
For this material if it were copper.
Let's now take a very good insulator.
Glass is an example.
Quartz, porcelain, very good insulators.
Now, sigma, the conductivity, is extremely low.
They vary somewhere from 10 to the -12 through 10 to the -16.
So rho, now, the resistivity, is something like 10 to the 12 to 12 to the +16, and if I take 10 to the 14, just I grab -- I have to grab a number -- then you'll find that R, now, is 10 to the 20 ohms.
A 1 with 20 zeros.
That's an enormous resistance.
So you see the difference -- 22 orders of magnitude difference between a good conductor and a good insulator.
And if I make this potential difference over the wire, if I make that 1 volt, and if I apply Ohm's Law, V equals I R, then I can also calculate the current that is going to flow.
If I R is 1, then the current here is 100 amperes, and the current here is 10 to the -20 amperes, an insignificant current, 10 to the -20 amperes.
I first want to demonstrate to you that Ohm's Law sometimes holds, I will do a demonstration, whereby you have a voltage supply -- put a V in here -- and we change the voltage in a matter of a few seconds from 0 to 4 volts.
This is the plus side, this is the minus side, I have connected it here to a resistor which is 50 ohms -- we use this symbol for a resistor -- and here is a current meter.
And the current meter has negligible resistance, so you can ignore that.
And I'm going to show you on an oscilloscope -- we've never discussed an oscilloscope, but maybe we will in the future -- I'm going to show you, they are projected -- the voltage is go from 0 to 4, versus the current.
And so it will start here, and by the time we reach 4 volts, then we would have reached a current of 4 divided by 50, according to Ohm's Law, I will write down just 4 divided by 50 amperes, which is 0.08 amperes.
And if Ohm's Law holds, then you would find a straight line.
That's the whole idea about Ohm's Law, that the potential difference, linearly proportional to the current.
You double the potential difference, your current doubles.
So let's do that, let's take a look at that, you're going to see that there -- and I have to change my lights so that you get a good shot at it -- oh, it's already going.
So you see, horizontally, we have the current, and vertically, we have the voltage.
And so it takes about a second to go from 0 to 4 -- so this goes from 0 to 4 volts -- and you'll see that the current is beautifully linear.
Yes, I'm blocking it -- oh, no, it's my reflection, that's interesting.
Ohm's Law doesn't allow for that.
So you see how beautifully linear it is.
So now, you may have great confidence in Ohm's Law.
Don't have any confidence in Ohm's Law.
The conductivity sigma is a strong function of the temperature.
If you increase the temperature, then the time tau between collisions goes down, because the speed of these free electrons goes up.
It's a very strong function of temperature.
And so if tau goes down, then clearly, what will happen is that the conductivity will go down.
And that means rho will go up.
And so you get more resistance.
And so when you heat up a substance, the resistance goes up.
A higher temperature, higher resistance.
So the moment that the resistance R becomes a function of the temperature, I call that a total breakdown of V equals I R, a total breakdown of Ohm's Law.
If you look in your book, they say, "Oh, no, no, no, that's not a breakdown.
You just have to adjust the re- the resistance for a different temperature." Well, yes, that's an incredible poor man's way of saving a law that is a very bad law.
Because the temperature itself is a function of current, the higher the current the higher the temperature.
And so now, you get a ratio, V divided by I, which is no longer constant.
It becomes a function of the current.
That's the end of Ohm's Law.
And so I want to show you that if I do the same experiment that I did here, but if I replace this by a light bulb of 50 ohms -- it's a very small light bulb, resistance when it is hot is 50 ohms, when it is cold, it is 7 ohms.
So R cold of the light bulb is roughly 7 ohms, I believe, but I know that when it is hot, it's very close to the 50 ohms.
Think it's a little lower.
What do you expect now?
Well, you expect now, that when the resistance is low in the beginning, you get this, and then when the resistance goes up, you're going to get this.
I may end up a little higher current, because I think the resistance is a little lower than 50 ohms.
And if you see a curve like this, that's not linear anymore.
So that's the end of Ohm's Law.
And that's what I want to show you now.
So, all I do is, here I have this little light bulb -- for those of you who sit close, they can actually see that light bulb start glowing, but that's not important, I really want you to see that V versus I is no longer linear, there you go.
And you see, every time you see this light bulb go on, it heats up, and during the heating up, it, um, the resistance increases.
And it's the end of Ohm's Law, for this light bulb, at least.
It was fine for the other resistor, but it was not fine for this light bulb.
There is another way that I can show you that Ohm's Law is not always doing so well.
I have a 125 volt power supply, so V is 125 volts -- this is the potential difference -- and I have a light bulb, you see it here, that's the light bulb -- the resistance of the light bulb, cold, I believe, is 25 ohms, and hot, is about 250 ohms.
A huge difference.
So if the resistance -- if I take the cold resistance, then I would get 5 amperes, but by the time that the bulb is hot, I would only get half an ampere.
It's a huge difference.
And what I want to show you, again with the oscilloscope, is the current as a function of time.
When you switch on a light bulb, you would expect, if Ohm's Law holds, that when you switch on the current -- or switch on the voltage, I should say -- that you see this.
This is then your 5 amperes.
And that it would stay there.
That's the whole idea.
Namely, that the voltage divided by the current remains a constant.
However, what you're going to see is like this.
Current goes up, but then the resistance goes down, then the resistance goes up, when the current goes up, the resistance goes up, and then therefore the current will go down, and will level off at a level which is substantially below this.
So you're looking there -- you're staring at the breakdown of Ohm's Law.
And so that's what I want to show you now.
So, here we need 125 volts -- and there is the light bulb, and when I throw this switch, you will see the pattern of the current versus time -- you will only see it once, and then we freeze it with the oscilloscope -- turn this off -- so look closely, now.
There it is.
Forget these little ripples that you see on it, it has to do with the way that we produce the 125 volts.
And so you see here, horizontally, time, the time between two adjacent vertical lines is 20 milliseconds.
And so, indeed, very early on, the current surged toward -- to a very high value, and then the filament heats up, and so the resistance goes up, the light bulb, and the current just goes back again.
From the far left to the far right on the screen is about 200 milliseconds.
That's about 2/10 of a second.
And here you get a current level which is way lower than what you get there.
That's a breakdown of Ohm's Law.
It is actually very nice that resistances go up with light bulbs when the temperature goes up.
Because, suppose it were the other way around.
Suppose you turn on a light bulb, and the resistance would go down.
Light bulb got hot, resistance goes down, that means the current goes up.
Instead of down, the current goes up.
That means it gets hotter.
That means the resistance goes even further down.
That means the current goes even further up.
And so what it would mean is that every time you turn on a light bulb, it would, right in front of your eyes, destruct itself.
That's not happening.
It's the other way around.
So, in a way, it's fortunate that the resistance goes up when the light bulbs get hot.
All right.
Let's now be a little bit more qualitative on some networks of resistors, and we'll have you do a few problems like that, whereby we just will assume, naively, that Ohm's Law holds.
In other words, we will always assume that the values for the resistances that we give you will not change.
So we will assume that the heat that is produced will not play any important role.
So we will just use Ohm's Law, for now, and if you can't use it, we will be very specific about that.
So suppose I have here, between point A and point B, suppose I have two resistors, R1 and R2.
And suppose I apply a potential difference between A and B, that this be plus, and this be minus, and the potential difference is V.
And you know V, this is known, I give you V, I give you this resistance, and I give you that one.
So I could ask you now, what is the current that is going to flow?
I could also ask you, then, what is the potential difference over this resistor alone -- which I will call V1 -- and what is the potential difference over the second resistor, which I call V2?
Very straightforward question.
Well, you apply, now, Ohm's Law, and so between A and B, there are two resistors, in series.
So the current has to go through both, and so the potential difference V, in Ohm's Law, is now the total current times R1 plus R2.
Suppose these two resistors were the same, they had the same length, same cross-sectional area.
If you put two in series, you have twice the length.
Well, so, twice the length, remember, resistance is linearly proportional with the length of a wire, and so you add them up.
So now you know R1 and you know R2, you know V, so you already know the current, very simple.
You can also apply Ohm's Law, as long as it holds, for this resistor alone.
So then you get that V1 equals I times R1, so now you have the voltage over this resistor, and of course, V2 must be the current I times R2.
And so you have solved your problem.
All the questions that I asked you, you have the answers to.
We could now have a slightly different problem, whereby point A is here, but now we have a resistor here, which is R1, and we have here, R2.
This is point B, and this is R2.
And the potential difference is V, that is, again, given, and now I could ask you, what, now, is the current that will flow here?
And then I can also ask you, what is the current that would go through one -- resistor one, and what is the current that could go through resistor two?
And I would allow you to use Ohm's Law.
So now you say, "Aha! The potential difference from A to B going this route, that potential difference, is V, that's a given." So V must now be I1 times R1.
That's Ohm's Law, for this upper branch.
But, of course, you can also go the lower branch.
So the same V is also I2 times R2.
But whatever current comes in here must split up between these two, think of it as water.
You cannot get rid of charges.
The number of charges per second that flow into this juncture continue on, and so I, the total current, is I1 plus I2.
And so now, you see, you have all the ingredients that you need to solve for the current I -- for the current I1, and for the current I2.
And you can turn this into an industry, you can make extremely complicated networks of resistors -- and if you were in course 6, you should love it -- I don't like it at all, so you don't have to worry about it, you're not going to get very complicated resistor networks from me -- but in course 6, you're going to see a lot of them.
They're going to throw them -- stuff them down your throat.
The conductivity of a substan- substance goes up if I can increase the number of charge carriers.
If we have dry air, and it is cold, then the resistivity of cold, dry air at 1 atmosphere -- so rho for air, cold, dry, 1 atmosphere -- cold means temperature that we have outside -- it's about 4 times 10 to the 13.
That is the resistivity of air.
It is about what it is in this room, maybe a little lower, because the temperature is a little higher.
If I heat it up -- the air -- then the conductivity will go up.
Resistivity will go down, because now, I create oxygen and nitrogen ions by heating up the air.
Remember when we had this lightning, the step leader came down, and we created a channel full of ions and electrons, that had a very low resistivity, a very high conductivity.
And so what I want to demonstrate to you, that when I create ions in this room, that I can actually make the conductivity of air go up tremendously.
Not only will the electrons move, but also the ions, now, will start to move.
And the way I'm going to do that is, I'm going to put charge on the electroscope -- oh, that is not so good -- no harm done.
I'm going to put charge on the electroscope, and you will see that the conductivity of air is so poor that it will stay there for hours.
And then what I will do, I will create ions in the vicinity of the electroscope.
But let's first put some charge on the electroscope.
I have here a glass rod and I'll put some charge on it.
OK, that's a lot of charge.
And, uh, the r- the air is quite dry, conductivity is very, very small, and so the charge cannot go off through the air to the surroundings, to the earth.
But now I'm going to create ions there by heating it up, and I decided to do that with a candle, because a candle is very romantic, as we all know.
So here I have this candle -- look how well the charge is holding, eh?
-- and here's my candle.
And I will bring the candle -- oh, maybe 20 centimeters from the electroscope.
Look at it, look at it, already going.
It's about fifteen centimeters away.
I'll take my candle away, and it stops again.
So it's all due to the fact that I'm ionizing the air there, creating free electrons as well as ions, and they both participate now in the current, and the charge can flow away from the electroscope through the earth, because the conductivity now is so much higher.
I stop again, and it stops.
You see in front of your eyes how important the temperature is, in this case, the presence of the ions in the air.
If I have clean, distilled water -- I mean, clean water.
I don't mean the stuff that you get in Cambridge, let alone did I mean the stuff that is in the Charles River, I mean clean water, that has a pH of 7.
That means 1 out of 10 to the 7 of the water molecules is ionized, H+ and OH-.
The conductivity, by the way, is not the result of the free electrons, but is really the result of these H+ and OH- ions.
It's one of the cases whereby not the -- the electrons are maj- the major responsibility for the current.
If I have add 3% of salt, in terms of weight, then all that salt will ionize, so you get sodium plus and Cl- ions, you increase the number of ions by an enormous factor.
And so the conductivity will soar up by a factor of 300000, or up to a million, because you increase the ions by that amount.
And so it's no surprise then, for you, that the conductivity of seawater is a million times higher -- think about it, a million times higher -- than the conductivity of distilled water.
And I would like to give you the number for water -- so this is distilled water -- that is about 2 times 10 to the 5 ohm-meters.
.
That is the resistivity, 2 times 10 to the 5 ohm-meters.
I have here, a bucket of distilled water.
I'll make a drawing for you on the blackboard there.
So here is a bucket of distilled water, and in there, is a copper plate, and another copper plate, and here is a light bulb, and this will go straight to the outlet [wssshhht], stick it in, 110 volts.
This light bulb has 800 ohm resistance when it is hot.
You see the light bulb here.
You can calculate what this resistance is between the two plates, that's easy, you have all the tools now.
If you know the distance, it's about 20 centimeters, and you know the surface area of the plates, because remember, the resistance is inversely proportional with A, so you have to take that into account -- and you take the resistivity of water into account, it's a trivial calculation, you can calculate what the resistance is of this portion here.
And I found that this resistance here is about 2 megaohms.
2 million ohms.
So, when I plug this into the wall, the current that will flow is extremely low, because it has to go through the 800 ohms, and through the 2 megaohms.
So you won't see anything, the light bulb will not show any light.
But now, if I put salt in here, if I really manage to put 3% in weight salt in here, then this 2 megaohm will go down to 2 ohms, a million times less.
So now, the light bulb will be happy like a clam at high tide, because 2 ohms here, plus the 800, the 2 is insignificant.
And this is what I want to -- to demonstrate to you now, the enormous importance of increasing ions.
I increased ions here by heating the air, now I'm going to increase the ions by adding salt.
And so the first thing that I will do is, I will stick this in here.
There's the light bulb.
And I make a daring prediction that you will see nothing.
There we go.
Nothing.
Isn't that amazing?
You didn't expect that, right?
Physics works.
You see nothing.
If I take the plates out, and touch them with each other, what will happen?
There you go.
But this water has such a huge resistance that the current is too low.
Well, let's add some -- not pepper -- add some salt.
Yes, there's salt in there.
It's about as much as I would put on my eggs in the morning -- stir a little -- ah, hey, look at that.
Isn't that amazing?
And when I bring them closer together, it will become even brighter, because L is now smaller, the distance is smaller.
I bring them farther apart, it's amazing.
Just a teeny, weeny little bit of salt, about as much as I use on my egg, let alone -- what the hell, let's put everything in there -- that's a [unintelligible] I put everything, then, of course, you go almost down to the 2 ohms, and the light bulb will be just burning normally.
But even with that little bit of salt, you saw the huge difference.
My body is a fairly good conductor -- yours too, we all came out of the sea -- so we are almost all water -- and therefore, when we do experiments with little charge, like the Van de Graaff, being a student, then we have to insulate ourselves very carefully, putting glass plates under us, or plastic stools, to prevent that the charge runs down to the earth.
In fact, the resistance, my resistance between my body and the earth is largely dictated by the soles of my shoe, not by my body, not by my skin.
But if you look at my soles, then you get something like this, and it has a certain thickness, and this, maybe 1 centimeter.
This, now, is L in my calculation for the resistance, because current may flow in this direction, so that's L.
Well, how large is my foot?
Let's say it's 1 foot long -- no pun implied -- and let's say it's about 10 centimeters wide.
So you can calculate what the surface area A is, you know what L is, and if you know, now what the resistivity is for my sole, I can make a rough guess, I looked up the material, and I found that the resistivity is about 10 to the 10.
So I can now calculate what the resistance is in this direction.
And I found that that resistance then, putting in the numbers, is about 10 billion ohm.
And you will say, "Wow!" Oh, it's 4, actually.
Well, big deal.
4 billion ohm.
So you will say, "That's enormous resistance!" Well, first of all, I'm walking on two feet, not on one, so if I would be standing one the whole lecture, it would probably be 4 billion, but if I have two feet on the ground, it's really 2 billion, you will say, "Well, that's still extremely large!" Well, it may look large, but it really isn't, because all the experiments that we are doing here in 26-100, you're dealing with very small amounts of charge.
Even if you take the Van de Graaff -- the Van de Graaff, say, has 200,000 volts -- and let's assume that my resistance is 2 times 10 to the 9 ohms, two feet on the ground.
So when I touch the Van de Graaff, the current that would flow, according to Ohm's Law, would be 100 microamperes.
That means, in 1 second, I can take 100 microcoulombs of the Van de Graaff, but the Van de Graaff has only 10 microcoulombs on it.
So the resistance of 4 billion or 2 billion ohms is way too low for these experiments that we have been doing in 26-100, and that's why we use these plastic stools, and we use these glass plates in order to make sure that the current is not draining off the charge that we need for the experiments.
I want to demonstrate that to you, that, indeed, even with my shoes on -- that means, even with my 2 billion ohm resistance to the ground -- that it will be very difficult for me, for instance, to keep charge on an electroscope.
I'm going to put charge on this electroscope by scuffing my feet.
But, since I keep my -- I have my shoes on, I'm not standing on the glass plate, the charge will flow through me.
You can apply Ohm's Law.
And you will see that as I do this -- I'm scuffing my feet now -- that I can only keep that electroscope charged as long as I keep scuffing.
But the moment that I stop scuffing, it's gone.
Start scuffing again, that's fine, but the moment that I stop scuffing, it goes off again.
Even though this resistance is something like 2 billion ohms.
Let alone if I take my shoes off.
I apologize for that.
If now I scuff, I can't even get any charge on the electroscope, because now, the resistance is so ridiculously low, I don't even have the 2 billion ohms, I can't even put any charge on the electroscope.
It's always very difficult for us to do these experiments unless we insulate ourselves very well.
And if, somehow, the weather is a little damp, we can very thin films of water onto our tools, and then the current can flow off just through these very thin layers of water.
That's why we always like to do these experiments in winter, so that the conductivity of the air is very low, no water anywhere.
Here you see a slide of a robbery.
I have scuffed my feet across the rug, and I am armed with a static charge.
Hand over all your money, or I'll touch your nose.
This person either never took 8.02 or he is wearing very, very special shoes.
See you on Wednesday.
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Recursion: The Pros and Cons
William Dale
6 min readMay 9, 2018
Ah, recursion. How many nights have I poured over your hows and whys? Alas, no longer! For I have conquered your enigmatic conviction. Your wretched desires shall haunt the recesses of my conscious ne’er more.
Ok whew, moving on. Recursion. I’ve spent a lot of time trying to get to the bottom of what recursion is and what the benefits and faults are of using the method. On the surface it seems like a difficult concept to grasp, but after a little thought, seeing examples and making analogies, the concept becomes a bit more clear. So what is recursion?
Recursion by definition is “when a thing is defined in terms of itself.” In this case we are referring to mathematical or programatic functions. With respect to a programming function, recursion happens when a function calls itself within its own definition. It calls itself over and over again until a base condition is met that breaks the loop.
There are 2 main parts of a recursive function; the base case and the recursive call. The base case is important because without it, the function would theoretically repeat forever (in application there would be what is referred to as a “stack overflow” to stop the repetition which we will touch on a little later). Below is an example of a simple recursive function.
So what is happening in that picture above? That is a simple recursive function to calculate the value of n! (n factorial). Factorial means the product of an integer and each subsequent integer below it up to and including 1. In the above example we are calculating the factorial for n = 3 (3 2 1 = 6). The function starts at the uppermost box in the diagram. The function is
def factorial(n) if n > 1 return factorial(n - 1) n else return 1 end end
Our base case (the point at which the repetition stops) is when n is no longer greater than 1. At this point the function will return 1 as the value and we will move back up the “stack” of boxes until we have our final answer. The method above repeatedly calls factorial on n-1 (it is also necessary to change the input value so that it moves closer to the base case with each recursive call, otherwise we will never reach the base case and we will be stuck in RECURSIVE PURGATORY) until it reaches the base case, which is 1. When the base case is reached, the function returns 1. 1 is then the value that is passed back up so that the previous call of factorial(n-1) = 1. n here is equal to 2 so we get 1 2 = 2. 2 is then passed up, n is equal to 3 so we have 3 2 = 6 for the final value. (If we would have gone up one more, we would have returned 6, n would be equal to 4 so 6 4 = 24, which is the correct value for 4!) WOOHOO you did recursion!
Ok, so we generally know the basics on how recursion works. But why is any of this important? When and why would we choose recursion over any other algorithmic method, such as say, iteration? Well there are several pros and cons to recursion.
PROS:
Recursion can reduce time complexity. This was somewhat counter-intuitive to me since in my experience, recursion sometimes increased the time it took for a function to complete the task. An example of this is calculating fibonacci numbers. If you calculate the fibonacci sequence up to a number n using recursion rather than iteration, the time to complete the task when compared to that of the iterative approach was much greater. However, if you memoize the result (aka save the value of each calculation for further use in the recursive call) you can in fact reduce the time complexity (read a great answer response for more information about memoization here).
Recursion adds clarity and reduces the time needed to write and debug code. This one is valid to a point. If you know your input into a function is going to be small, then recursion is certainly a good choice if you want to de-clutter your code. If your input is sufficiently large however, the sacrifice of speed and memory for the sake of clarity becomes much less attractive and functional.
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Recursion is better at tree traversal. This one is a little more advanced. An extremely simplified version of what this means is as follows: A tree is a collection objects that are linked to one another (imagine leaves on a tree connected by branches that are in turn connected to other branches all the way to the roots). One of the more efficient ways to traverse these trees when looking for a specific leaf (or node) is by recursively following a single branch until the end of that branch until you find the value you are looking for. Again, this is extremely abstracted and simplified for what is actually happening and I urge you to look further into what is actually happening in tree traversal.
Recursion in the above tree diagram would be beneficial when used on preorder tree traversal.
CONS:
Recursion uses more memory. Because the function has to add to the stack with each recursive call and keep the values there until the call is finished, the memory allocation is greater than that of an iterative function.
Recursion can be slow. If not implemented correctly (as stated above with memoization) it can be much slower than iteration. It is actually pretty difficult to write a recursive function where the speed and memory will be less than that of an iterative function completing the same task. The reason that recursion is slow is that it requires the allocation of a new stack frame.
I know I mentioned a lot about recursion vs iteration above, so lets look more into that.
Both iteration and recursion are repetitive processes that repeat a certain process until a certain condition is met. They are both used in programming to complete tasks where a task has to be repeated in order to solve the problem.
Iteration: A function repeats a defined process until a condition fails. This is usually done through a loop, such as a for or while loop with a counter and comparative statement making up the condition that will fail. An infinite loop for iteration occurs when the condition never fails.
Recursion: Instead of executing a specific process within the function, the function calls itself repeatedly until a certain condition is met (this condition being the base case). The base case is explicitly stated to return a specific value when a certain condition is met. An infinite recursive loop occurs when the function does not reduce its input in a way that will converge on the base case.
The Stack
As stated above, recursion is memory intensive because it requires an allocated stack frame, which can be shown by the above columns/buckets. For every call of the function, another element is added to the stack and once the base case is reached (at the top of the stack, or the last entry), the element is “popped” off of the top and that value is passed to the value below it. The stack is another interesting topic to look into, and I would suggest checking it out as there is too much information to go into here. I hope I have provided a basic view of how recursion uses the stack.
In conclusion, there is a great article written about the importance of knowing about recursion here that is definitely worth the read. Obviously there is A LOT more information on recursion but I hope that I have at least touched on some major areas to give you a direction in which to explore great topics on recursion a little further.
Recursion
## Written by William Dale
49 followers
·4 following
Responses (2)
Write a response
What are your thoughts?
Giulio Mecocci
Sep 11, 2019
```
I think this article can be summarized as: recursive functions are easier to write and think about once you’ve understood the paradigm, especially when operating on datastructures like trees (the reason being that they introduce a stack implicitly —…
```
6
Lior Malka, PhD
Sep 4, 2021
Recursion uses more memory. Because the function has to add to the stack with each recursive call and keep the values there until the call is finished, the memory allocation is greater ...
```
I’m not sure. For example, say you want to do limited depth DFS, where you only want to visit vertices up to distance k from the starting vertex. If you use the non recursive version of DFS you’ll need to track the depth using a map or some other data structure and this would cost more memory compared to recursion
```
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