# UniversitaireWiskundeCompetitie. Problem 2005/4-A We have k=1. Show that for every q Q satisfying 0 < q < 1, there exists a finite subset K N so that

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1 Problemen/UWC NAW 5/7 nr juni Problemen/UWC UniversitireWiskundeCompetitie Edition 005/4 For Session 005/4 we received submissions from Peter Vndendriessche, Vldislv Frnk, Arne Smeets, Jn vn de Lune, en PG Kluit Problem 005/4-A We hve k= = Consequently prtil sums must stisfy k(k+) k(k + ) < Show tht for every q Q stisfying 0 < q <, there eists finite subset K N so tht k(k + ) = q Solution This problem ws solved by Peter Vndendriessche, Vldislv Frnk nd Arne Smeets The solution below is bsed on tht of Vldislv Frnk First note tht k(k+) = k k+ Hence k(k+) + (k+)(k+) + + (k+n )(k+n) = k k+ + k+ + k+n k+n = k k+n Consequently it suffices to represent every rtionl number between 0 nd s + k, where 3 k If two consecutive numbers re equl, they simply cncel out, so we llow equl numbers This will be useful in finl step of proof Let b be our rtionl number There is nturl number n such tht n+ < b n Consider = n b = b n bn The numertor of this frction is non-negtive becuse b n, but less thn, the numertor of b, becuse (n + ) < 0 We hve b = n We now pply the sme lgorithm to Let m be nturl number such tht m+ < m b n The clim is tht m n Nmely, = bn < bn, hence n m n + > n If we continue this lgorithm, we obtin b = ( ( 3 ( ( ) ))) Notice tht the lgorithm cn only be repeted finitely mny times, s the numertor decreses t ech step We now hve b = + 3 ± If is even we re done In other cse we my ssume tht > nd chnge into ( ) Here ( ) nd we re done Of course =, s otherwise b = = which is impossible As generliztion, V Frnk shows tht for ny irrtionl number in the intervl [0,] there eists n infinite sum Problem 005/4-B We consider the progressive rithmetic nd geometric mens of the function sequence f n () = n, n N, > 0, = These re nd A n = A n () = n ( n ) = n n( ) G n = G n () = ( ++ +(n ) ) n = n The Mrtins-property reds A n+ /A n G n+ /G n In our cse this gives n n+ n + n Eindredctie: Mtthijs Coster Redctiedres: UWC/NAW Mthemtisch Instituut Postbus RA Leiden Prove, more generlly, tht + + for >, > 0, = Solution This problem ws solved by Jn vn de Lune, Peter Vndendriessche, Vldislv Frnk nd Arne Smeets The solution below is bsed on tht of Peter Vndendriessche

2 48 NAW 5/7 nr juni 006 Problemen/UWC Let f (t) be (smooth) non-negtive function tht is conve on [, b] nd let [, y] [, b] such tht + y = + b We then hve b f (t)dt f (t)dt y To prove this, consider, for given f (t),, nd y, the function g(t) defined by g(t) = f () + ( f (y) f ())(t ) y g(t) is the line through the points (, f ()) nd (y, f (y)) Notice tht the conveity of f gives g(t)dt f (t)dt Let h(t) = g(t) g(t)dt + f (t)dt, then y h(t)dt = f (t)dt By conveity we hve f (t) g(t) h(t) for t [, ] [y, b] Since h(t) is the eqution of line nd + y = + b, we hve Combining these results we find: b h(t)dt = h(t)dt y b f (t)dt f (t)dt + b y f (t)dt = + h(t)dt + b y h(t)dt + b h(t)dt b f (t)dt = = y y f (t)dt y h(t)dt y Problem B is specil cse of this result For R + 0, =, let f (t) = t Then f (t) = t log () 0 Therefore f (t) is conve We hve to distinguish two cses: (, 0) Then 0 < + < + Apply the lemm to the intervl [ +, ] [0, + ] (0, ) Then 0 < < + < + Apply the lemm to the intervl [, + ] [0, + ] Notice tht in the first cse the sign in both numertor nd denomintor chnges on the right side of the eqution: from which we cn deduce + 0 t dt + + t dt, + + ( ) It is esy to prove the generliztion z y b b z y y, where y + z = + b nd < < y < z < b Problem 005/4-C A finite geometry is geometric system tht hs only finite number of points For n ffine plne geometry, the ioms re s follows: Given ny two distinct points, there is ectly one line tht includes both points The prllel postulte: Given line L nd point P not on L, there eists ectly one line through P tht is prllel to L 3 There eists set of four points, no three colliner

3 Problemen/UWC NAW 5/7 nr juni We denote the set of points by P, nd the set of lines by L Let σ be n utomorphism of (P, L ) (mening tht three colliner points of P re mpped onto three colliner points of P nd three non-colliner points of P re mpped onto three non-colliner points of P ) Prove tht there eists point P P with σ(p) = P or line L L with σ(l) = L or σ(l) L = Solution This problem hs been solved by Leendert Bleijeng nd Peter Vndendriessche The solution below is bsed on their solutions First we will prove the following lemm: Lemm Let M, L L, then M = L Proof Suppose tht M L > then M = L Therefore we my ssume tht M L = Let M = m nd L = l By Aiom 3 we know tht there eists P P such tht P L nd p M Through P we cn construct line prllel to L nd l lines tht intersect L in its l points In the sme wy we cn construct, through P, line prllel to M nd m lines tht intersect M in its m points Let us now determine the number of lines through P; this equls l + nd m + If M L = 0, pick points L nd b M Let N be the line through nd b Then by the previous rgument L = N nd M = N We conclude tht ll lines consist of n equl number of points, sy s Lemm P < L Proof Let P = p nd L = l Every two points define line, nd there re p(p ) pirs of points Ech line hs s points nd is counted s(s ) times Therefore l = p(p ) In order to show tht p < l we hve to prove tht s(s ) < p or p > s(s ) s s + The third iom tells us tht there eist three non-colliner points, b, c P Let L be the line through nd b, M the line through nd c By the prllel postulte, through every point on L there is ectly one line prllel to M Strting with s points on L, we find s lines, ll consisting of s points Therefore p s Suppose tht σ(p) = p, for ll p P, nd tht σ(l) = L nd σ(l) L = for ll L L Consider the function µ : L P given by µ(l) = σ(l) L µ is well defined since σ(l) L is lwys unique point Now suppose tht µ(l) = µ(m) or σ(l) L = σ(m) M = p, nd σ(q) = p Then q L nd q M We know tht q = p Therefore L = M nd µ is injective However, if µ is injective, then P L, which contrdicts the previous lemm Problem 005/4-* stisfy We hve k= /k = (π /6) Consequently prtil sums must k < π 6 Given ny q Q stisfying 0 < q < (π /6), does there eist finite subset K N \{} so tht k = q? Solution This problem ws solved by PG Kluit The solution below is bsed on his solution Let q = ki, where k i re different integers Let m be the lest common multiple of ll k i in the sum For ech such k i number k i eists such tht k ik i = m Then q = m (k i ), tht is, q cn be written s frction with denomintor m nd the numertor sum of squres of different divisors of m This rises the question: given m, which numbers cn be written s sums of squres of different divisors of m? We will show tht for highly composite numbers m, more specificlly m = n!, the nswer will be tht sufficiently mny integers cn be written s sums of squres to prove the problem

4 50 NAW 5/7 nr juni 006 Problemen/UWC Lemm Let n 5 be n integer nd let 3 = d < d < d m = n!/3 be ll divisors of n! between 3 nd n!/3 Then d k > d k+ for k < m Proof Let us prove this by induction For n = 5 the divisors d,, d re 3, 4, 5, 6, 8, 0,, 5, 0, 4, 30, nd 40 It is esy to verify the lemm We ssume the lemm is true for n We hve to prove tht the Lemm holds for the divisors of (n + )! The divisors of (n + )! tht re less thn n!/3 clerly stisfy the lemm Even though there my be more divisors, this cnnot influence the inequlity Suppose tht d k nd d k+ re two successive divisors of (n + )!, with n!/3 d k < d k+ (n + )!/3 Let d k d k = d k+d k+ = (n + )! Then d k+ nd d k re two successive divisors of (n + )! with 3 d k+ < d k 3(n + ) As for n 5 we hve 3(n + ) < n!/3, this suffices to conclude the proof Lemm Let n [9, 56] be n integer Then n cn be represented s sum of different squres d + + d k, where d < < d k 0 Proof The proof cn be found by the enumertion of 8 representtions There is slightly shorter proof which will be left to the reder Lemm Let n N, n Then every integer [9,σ (n!) n! 9] cn be represented s = d k, where the d k re different divisors of n! Here σ m () = d d m Proof Let L kn = [9, t] be the longest intervl in [9, ) whose integers cn ll be represented s sum of different squres of some of the first k divisors of n! Let l kn = L kn, the length of the intervl In the proof the nottion will be bbrevite to l k = L k if it is cler which n is ment In the second lemm we sw tht l 0 = 8 Notice tht l = 49 (= 8 + ) Any 56 is represented by the divisors lesser thn or equl to 0, while the integers re represented using We will show in generl tht l k+ = l k + d k+ by induction, s long s d k+ < n!/ The proof will be given in two steps In the first step we prove tht d k+ < l k+ given d k < l k nd l k+ = l k + d k+ In the second step we will prove tht l k+ = l k + d k+ given d k < l k Using these two steps nd the bsic ssumption (k = ) we cn prove for rbitrry k tht l k+ = l k + d k+ First step Given d k < l k nd l k+ = l k + d k+ we find tht d k+ < l k+ Proof The first lemm tells us tht d k+ < d k Therefore we hve d k+ < d k+ + d k < d k+ + l k < l k+ Second step Given d k < l k we find tht l k+ = l k + d k+ Proof The proof is comprble to the proof bove For L k, it is cler tht L k+ s well, while for the numbers L k+ \L k notice tht l k + 8 < l k + d k+ + 8 If we use the number d k+ to represent the sum, we find for the rest y = d k+ tht l k d k+ + 8 < y l k + 8 Using Lemm gin we hve l k d k+ + 8 > l k d k + 8 > 8 Therefore y L k We cn rewrite the results l k+ = l k + d k+ s for k m, where d m = n!/3 l k = k di 8, i=

5 Problemen/UWC NAW 5/7 nr juni In order to complete the proof of Lemm 3 we need to prove tht l (m+)n = l mn + d n!/, where d m+ = n!/ Notice tht 4 < Therefore for rbitrry L (m+)n, we find either L mn or 4 n! L mn Now we find l k = k di 8, i= for k m +, where d m+ = n!/ This concludes the proof of this lemm Theorem For every q Q such tht 0 < q < π 6 k, finite subset K N eists, such tht Proof Let q Q with 0 < q < π 6 We cn find n n N fulfilling ech of the three following properties by choosing n sufficiently lrge Moreover ech of these properties is monotonic, mening tht if it is true for some n 0, it will be true for ll n > n 0 n, If q = /b, where nd b hve no common divisors, then b divides n!, If q = /(n!), then n is chosen such tht 8 < < σ (n!) (n!) 8 To prove the eistence of 3) notice tht = q σ lim (n!) (n!) 8 n n! = π 6 Now Lemm 3 my be pplied, showing tht cn be represented s sum of squres of different divisors of n! This gives us the sought for representtion of q Remrk A solution of the Str Problem turns out to hve been published in Ron Grhm s On Finite Sums of Unit Frctions, Proc London Mth Soc(4), 964, pp The bsic ides behind the two solutions re similr Grhm strts with multiplictive set S, which in the Str Problem is the set of squres Grhm then defines P(S), the set of sums of elements of S Using the nottion S for the set of inverses of the elements of S, Grhm shows tht if P(S) contins ll positive integers, up to finite number, S is finite, nd s n+ /s n is bounded, then p q P(S ) whenever q s for some s S Moreover, for every ɛ > 0 there is n s P(S ) such tht s p q < ɛ

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### Presentation Problems 5 Presenttion Problems 5 21-355 A For these problems, ssume ll sets re subsets of R unless otherwise specified. 1. Let P nd Q be prtitions of [, b] such tht P Q. Then U(f, P ) U(f, Q) nd L(f, P ) L(f, Q).

### Infinite Geometric Series Infinite Geometric Series Finite Geometric Series ( finite SUM) Let 0 < r < 1, nd let n be positive integer. Consider the finite sum It turns out there is simple lgebric expression tht is equivlent to

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### The usual algebraic operations +,, (or ), on real numbers can then be extended to operations on complex numbers in a natural way: ( 2) i = 1 Mth50 Introduction to Differentil Equtions Brief Review of Complex Numbers Complex Numbers No rel number stisfies the eqution x =, since the squre of ny rel number hs to be non-negtive. By introducing

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### f(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral Improper Integrls Every time tht we hve evluted definite integrl such s f(x) dx, we hve mde two implicit ssumptions bout the integrl:. The intervl [, b] is finite, nd. f(x) is continuous on [, b]. If one

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### The final exam will take place on Friday May 11th from 8am 11am in Evans room 60. Mth 104: finl informtion The finl exm will tke plce on Fridy My 11th from 8m 11m in Evns room 60. The exm will cover ll prts of the course with equl weighting. It will cover Chpters 1 5, 7 15, 17 21, 23

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### p(t) dt + i 1 re it ireit dt = Note: This mteril is contined in Kreyszig, Chpter 13. Complex integrtion We will define integrls of complex functions long curves in C. (This is bit similr to [relvlued] line integrls P dx + Q dy in R2.) CSCI 2400 Models of Computtion, Section 3 Solutions to Homework 4 Problem 1. ll the solutions below refer to the Pumping Lemm of Theorem 4.8, pge 119. () L = f n b l k : k n + lg Let's ssume for contrdiction WHEN IS A FUNCTION NOT FLAT? YIFEI PAN AND MEI WANG Abstrct. In this pper we prove unique continution property for vector vlued functions of one vrible stisfying certin differentil inequlity. Key words: