Making Change of n cents

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1 A Complete High School Proof of Schur s Theorem on Making Change of n cents William Gasarch Univ. of MD at College Park Introduction Let a, a 2,..., a L be coin denominations. Assume you have an unlimited number of each coin. How many ways can you make n cents with these coins? Schur s theorem gives the answer asymptotically and also yields the coefficient of the dominant term. We state it: Theorem. Let a < < a L N be relatively prime. Think of them as coin denominations. As n goes to infinity the number of ways to make change of n cents is n L (L!a a 2 a L + O(n L 2. I volunteered to present a proof of Schur s theorem to high school students taking pre-calculus. I then realized that the proof of Schur s theorem I knew, from Wilf s wonderful book on generating functions [], uses generating function, roots of unity, partial fractions, and the Taylor series for. ( x L The material on generating functions and roots of unity is such that I could prove and use just what was needed. Partial fractions could certainly be taught to high school students; however, I wanted to prove they worked, not show them a cook book way to decompose. I then realized that University of Maryland, College Park, MD 20742, gasarch@cs.umd.edu

2 I had never seen a proof that they worked; however, I doubt this would be hard to obtain. Getting a Taylor series without calculus for ( x L seemed like an interesting challenge. I succeeded in my goals. We present a complete high school proof of Schur s theorem, essentially Wilf s proof, with the following features:. We prove the partial fractions decomposition that we need. The result is standard; however, our proof is short and makes our treatment self contained. We doubt our proof is new though we have not been able to find a reference. 2. We obtain the Taylor series of ( x L we have not been able to find a reference. without calculus. We doubt our proof is new though 3. We obtain that for all r M where M LCM(a,..., a L there is a polynomial h r of degree L such that if n r (mod M then h r (n is the number of ways to make change of n. We doubt the result is new though we have not been able to find a reference. 2 Induction Proof of Partial Fractions Decomposition Lemma 2.. For all n N, for all c, d C, c d, there exists A, A,..., A n such that ( cx( dx A n n cx + A k ( dx. k k 2. For all n,..., n L N, for all c,..., c L C, distinct complex numbers, there exists A i,j such that L ( c i x n i i A i,j ( c i x. j i j 2

3 Proof: We prove this by induction on n. Base Case: n. We need to solve for A, A in this equation: cx dx A cx + A dx. A( dx + A ( cx, so (A + A (da + ca x. Hence A + A and da + ca 0. These can be easily solved to yield A Note that we are using c d. Induction Hypothesis (IH: We assume the lemms true for n. c and A c d d. c d By the IH there exists A, A 2,..., A n (we purposely make these off by one so that later they will be what we want such that Hence A ( cx( dx n n cx + k A k+ ( dx k. [ ( cx( dx n dx A + n cx k ] A k+ A + n ( dx k ( cx( dx k2 A k ( dx k A cx + A dx + n k2 A cx + n k A k ( dx j A k ( dx k (by the n case 2 We prove this by induction on L i n i. Base Case: L i n i. This only happens when L and n which is trivial. Induction Hypothesis (IH: Assume the lemms true for all (n,..., n L with L i n i < L i n i. Clearly L ( c i x n i i [ c x ( c x n L ( c i n i ] 3

4 We rewrite what is in the square brackets as ( c x n ( c 2 n 2 ( c 3 n 3 ( c L n L This is in the exact form of the lemma we are proving though note that the sum of the exponents is ( L i n i < L i n i. Hence by the IH there exists A,j+ (we purposely make these off by one so that later they will be what we want and A i,j such that the expression in square brackets is the following (we seperate out the first term for notationaly convinence. n j A,j+ ( c x + L j Hence our original product, L i, is ( c i x n i j A i,j ( c i x j [ n c x j A,j+ ( c x + L j j We can re-index the first summation to get: A ] [ n i,j ( c i x j j A,j+ ( c x + L j+ j A ] i,j ( c i x j n j2 A,j ( c x + L j j By Part there exists constants A i,j and A i,j,k such that A i,j ( c x( c i x j j A i,j ( c x( c i x L j Let L n j A i,j A,. Then j A L i,j c x + n A i,j c x A, c x. j j j k A i,j,k ( c i x k 4

5 For 2 i L and j n i let A i,j n i ni k jk A i,j,k. Then j j k A i,j,k ( c i x L k n i k jk Hence our original product, L i, is ( c i x n i A i,j,k ( c i x L k k A i,j ( c i x k n j2 A,j ( c x j + A, c x + L k A i,j ( c i x L k i j A i,j ( c i x j. The usual theorem about partial fraction decomposition that is used in calculus starts with a polynomial over the reals and factors it into linear and quadratic polynomials over the reals. This version can easily be derived from Lemma 2. 3 Non Calculus Proof of the Taylor Series for ( x n We obtain the Taylor expansion for ( x L Def 3. via combinatorics, not calculus.. If n N then [n] is the set {,..., n}. 2. An L-set of X is a subset of X of size L. Lemma 3.2 For all n, L, n i0 ( L +i L ( L+n L. Proof: The term L-set will mean L-set of {,..., L + n} throughout. ( L+n L. We solve the following problem two ways: How many L-sets are there? Clearly the answer is Another way to solve this problem is to partition the L-sets based on the set s largest element. The largest element in any L-set is of the form L + i where 0 i n. The number of L-sets with 5

6 largest element L + i is the number of (L -sets of {,..., L + i}, namely ( L +i L. Hence the number of L-sets is n i0. This yields our result. ( L +i L Lemma 3.3 For all L, ( x L n0 Proof: We prove this by induction on L. ( L +n L x n. Base Case: L. This is the well known geometric series x i0 xi. Induction Hypothesis (IH: Assume the lemms true for L : From the IH we obtain: ( x L i0 ( L 2 + i L 2 x i ( x L ( x L x ( ( ( L 2 + i x i x j L 2 i0 j0 ( L 2 + i ( n ( L 2 + i ( L + n x i+j x n x n L 2 L 2 L i0 j0 n0 i0 n0 This last equality used Lemma Lemma about Roots of Unity Lemma 4. If 0 x y 2π, cos(x cos(y, and sin(x sin(y then x y. Proof: Since cos(x cos(y either x y or x + y 2π. Since sin(x sin(y either x y or x + y {π, 3π}. Since 2π / {π, 3π}, x y. 6

7 Lemma 4.2 Let a < < a L N be relatively prime. Let g(x (x a (x a L. When g(x is factored completely into linear terms the factor (x occurs L times and all of the other linear factors occur L times. Proof: Clearly x occurs in all L of the polynomials (x and hence occurs L times. Each polynomial (x has distinct roots, so if another linear term occurs L times it has to occur as a factor in each (x. Assume that there exists ω such that (x ω divides each (x. We will show that a,..., a L have a nontrivial common factor and hence are not relatively prime. For all i L let ω i be the primitive ith root of unity. For all i, since x ω divides x a, ω is an th root of unity. In particular there exists A a such that ω A ω. Since A a, a does not divide A. Hence there is some prime power p c that divides a but does not divide A. Let 2 i L. We show that p divides. Since ω is an th root of unity there exists B such that ω A ω ω B i. Hence ω A ω B i cos 2πA a + sin 2πA a cos 2πB + sin 2πB Hence cos 2πA a sin 2πA a cos 2πB sin 2πB By Lemma 4. A/a B/. Therefore A Ba, hence a must divide A. Since p c divides a but not A, p must divide. 5 Schur s Theorem Theorem 5. Let a < < a L N be relatively prime. Think of them as coin denominations.. Let M LCM(a,..., a L. Let 0 r M. There is a polynomial h r of degree L 7

8 such that if n r (mod M then h r (n is the number of ways to make change of n cents. 2. For all 0 r M the coefficient of of n L in h r is (L!a a 2 a L. Note that the coefficient does not depend on r. 3. (This follows from Parts and 2. The number of ways to make change of n cents is n L (L!a a 2 a L + O(n L 2. Proof: We get a formula for the number of ways to make change of n cents and then prove Parts and 2. Part 3 follows from Parts and 2. The number of ways to make change of n cents is the coefficient of x n in f(x ( + x a + x 2a + ( + x a 2 + x 2a 2 + ( + x a L + x 2a L + ( x a ( x a 2 ( x a L For all i L, j, let α i,j be the jth th roots of unity (we think of as being the 0th root of unity. Formally α i,j cos 2πj + sin 2πj. Let be the number of times the factor ( α i,j x appears in ( x a ( x a 2 ( x a L. By Lemma 4.2 L. We rewrite f(x and use Lemma 2. and 3.3 to obtain f(x ( x L L i ai j ( α i,jx i A i ( x + L i i a i j k A i,j,k ( α i,j x k. i n0 ( i + n A i x n + i i a i j k n0 ( k + n A i,j,k α n k i,jx n. 8

9 ( L ( i + n A i + i n0 i i a i j k ( k + n A i,j,k αi,j n x n. k Since α i,j is an ith root of unity, αi,j n αi,j n mod M. Hence if n r (mod M then the coefficient of x n in f(x, which is the answer we seek, is h r (n i ( i + n A i + i i a i j k ( k + n A i,j,k αi,j r k Clearly this is a polynomial in n of degree L. 2 We need to find A L. ( x a ( x a 2 ( x a L Multiply both sides by ( x L i A i ( x + L i i a i j k A i,j,k ( α i,j x k. ( x L L ( x a ( x a 2 A ( x a L L + A i ( x L i + i i a i j k A i,j,k ( x L ( α i,j x k. The left hand side can be rewritten as ( + x + x x a ( + x + x x a 2 ( + x + x x a L As x approaches (from the left the LHS approaches a a 2 a L and the RHS approaches A L. Hence A L a a 2 a L. Therefore 9

10 h r (n (n + (n + 2 (n + L L ( i + n + A i + (L!a a 2 a L i Since all L i i a i j k ( k + n A i,j,k αi,j r k h r (n n L (L!a a 2 a L + O(n L 2. 6 Acknowledgment We would like to thank Daniel Smolyak, Larry Washington, Sam Zbarsky for proofreading and discussion. References [] H. Wilf. Generatingfunctionology. Academic Press, Waltham, MA,

Making Change for n Cents: Two Approaches

Making Change for n Cents: Two Approaches Maing Change for n Cents: Two Approaches William Gasarch arxiv:406523v4 [mathho] 26 Jun 204 Univ of MD at College Par Abstract Given a dollar, how many ways are there to mae change using pennies, nicels,