GENERATING FUNCTION. and f( 1) = a m ( 1) m = a 2k a 2k+1 So a2k+1 = 1 (f(1) f( 1)).

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1 GENERATING FUNCTION. Examples Question.. Toss a coin n times and find the probabilit of getting exactl k heads. Represent H b x and T b x 0 and a sequence, sa, HTHHT b (x )(x 0 )(x )(x )(x 0 ). We see that all the possible outcomes of n tosses are represented b the expansion of f(x) (x 0 +x ) n (+x) n. The coefficient of x k in f(x) is the number of outcomes with exactl k heads. Assuming the coin is fair, we see that the probabilit is ( n k) /2 n. Next, let us factor the probabilit into the generating function f(x). B that we mean we want the coefficient of x k gives the probabilit of getting exactl k heads. To make the problem less trivial, let us assume that the coin is not necessaril fair with p for head and q for tail (p + q ). We want to find f(x) a k x k with a k the probabilit of getting exactl k heads. It is not ver hard to come up with f(x) (q + px) n a k x k So a k p k q n k( n k). A more complicated version of the problem is the following. Question.2. You have coins C,...,C n. For each k, C k is biased so that, when tossed, it has probabilit /(2k + ) of falling heads. If the n coins are tossed, what is the probabilit of an odd number of heads? Let p k /(2k+) and q k p k. It is not hard to set up the generating function as in the previous example f(x) (q + p x)(q 2 + p 2 x)...(q n + p n x n ) a m x m with a m the probabilit of getting exactl m heads. We are looking for a + a 3 + a Here is a neat trick to find a +a 3 +a given a series f(x) a m x m. Note that f() a m a 2k + a 2k+ and f( ) a m ( ) m a 2k a 2k+ So a2k+ (f() f( )). 2 In our case, f() and f( ) /(2n+). So the probabil of getting odd number of heads is n/(2n + ).

2 2 GENERATING FUNCTION Given f(x) a m x m. What if we want to find a 0 +a 3 +a a3k? More generall, What if we want to find a pk+r for some fixed p and r. Hint: use roots of unit. Question.3. Find the number of was of changing a 500 dollar bill into $, $2, $5, $0, $20 s. Let k,k 2,k 3,k 4,k 5 be the number of $, $2, $5, $0, $20 s, respectivel. We are looking at the numbers of nonnegative integral solutions of the equation: Set up the correspondence k + 2k 2 + 5k 3 + 0k k (k,k 2,k 3,k 4,k 5 ) (x k )(x 2k 2 )(x 5k 3 )(x 0k 4 )(x 20k 5 ) Then the number of the solutions of the equation is the coefficient of x 500 in f(x) k 0 x k k 2 0 x 2k 2 k 3 0 x 5k 3 k 4 0 x 0k 4 ( + x + x )( + x 2 + x )( + x 5 + x ) ( + x 0 + x )( + x 20 + x ) ( x)( x 2 )( x 5 )( x 0 )( x 20 ) k 5 0 x 20k 5 Of course, to find the Talor expansion of f(x), we have to write f(x) as a sum of partial fractions, which is doable in theor but impossible b hand. But we can still sa a lot about the coefficients of f(x). Let f(x) p(n)x n, i.e., p(n) is the number of the was changing n dollar. Then the following is true: p(n) grows at the order of n 4 as n. I leave this as an exercise. p(20n) is a polnomial in n of degree 4. This is a little bit harder to prove. p(20n) 4 p(20k) k0 0 l 4 l k n l k l This uses interpolation formula. Once we know that p(20n) is a polnomial of degree 4 and the values of p(20n) at five points, sa p(0),p(20),p(40),p(60),p(80), we can find p(20n) b interpolation. More generall, the number of nonnegative integral solutions (x,x 2,...,x m ) of the equation λ x + λ 2 x λ m x m n

3 GENERATING FUNCTION 3 is given b the coefficients of the generating function f(x) ( x λ )( x λ 2)...( x λ m). What if we are looking for integral solutions x i within the range, sa α i x i β i? What is the corresponding generating function? What if we are looking for odd solutions, solutions with x i (mod 3) and etc? What are the corresponding generating functions? Question.4. Let n be a positive integer. Find the number of polnomials P(x) with coefficients in {0,,2,3} such that P(2) n? Let P(x) a 0 + a x + a 2 x a k x k +... We are looking at the solutions of a 0 + 2a + 4a k a k +... n with 0 a k 3. The number of such solutions are given b the coefficient of x n in f(x) ( + x + x 2 + x 3 )( + x 2 + x 4 + x 6 )( + x 4 + x 8 + x 2 )... ( + x 2k + x 2(2k) + x 3(2k) ) k0 x 2k+2 x 2k ( x)( x 2 ) k0 ( ) ( ) ( ) 4 x + 2 ( x) x ( ( )n + ) (n + ) x n 2 n0 Another wa to express the result is n/2 +. Find a purel combinatorial solution to the problem. Question.5. Find the number of subsets of {,..., 2003}, the sum of whose elements is divisible b 5. Set up the correspondence: Then S {,...,2003} S {s,s 2,...,s k } σ(s) x s x s 2...x s k. σ(s) f(x) ( + x)( + x 2 )...( + x 2003 ) The coefficient of x n in f(x) is the number of subsets of {,...,2003}, the number of whose elements is exactl n. So if f(x) a n x n,

4 4 GENERATING FUNCTION we are looking for the sum a 0 + a 5 + a Let r n exp(2nπi/5) be the 5-th roots of unit. Then a 0 + a 5 + a (f(r 0) + f(r ) + f(r 2 ) + f(r 3 ) + f(r 4 )). Obviousl, f(r 0 ) For k 4, f(r k ) ( ( + r 0 k )( + r k )( + r2 k )( + r3 k )( + r4 k )) 400 ( + r k )( + r2 k )( + r3 k ) (2 + 2r k + r 2 k + 2r3 k + r4 k ) and hence 4 f(r k ) (8 2 2 ) k So the answer is (/5)( ). Replace 2003 b n 5k + l and redo the computation. Find a purel combinatorial solution to the problem. Question.6. Suppose that each of n people writes down the numbers,2,3 in random order in one column of a 3 n matrix, with all orders being equall likel and independent. Show that for some n 995, the event that the row sums are consecutive integers is at least four times likel as the event that the row sums are equal. Instead of (,2,3), we use (,0,). Set up the correspondence: a b x a b c Then a 3 n such matrix can be represented b a term in the expansion of ( φ n (x,) x + + x + x + x + ) n Let a n be the constant term in φ n (x,). It is not hard to see that a n is exactl the number of matrices with equal row sums. Let b n,c n,d n,e n,f n,g n be the coefficients of x,,x/,/x,/x,/ in φ n (x,). It is not hard to see that b n + c n + d n + e n + f n + g n are the number of matrices with consecutive row sums. Actuall, from ( φ n+ (x,) x + + x + x + x + ) φ n (x,) we see that a n+ b n + c n + d n + e n + f n + g n

5 GENERATING FUNCTION 5 We are essentiall asked to show that a n /a n+ /4 for n large. In other word, if we put ϕ(t) a n t n, it suffices to show that the radius of convergence of ϕ(t) is less than /4. This suggests us to put all φ n (x,) together into H(x,,t) φ n (x,)t n If we expand H(x,,t) as an infinite series in (x,) (one technical issue here: H(x,,t) expands as a Laurent series), ϕ(t) is the constant term. The rest of the proof is quite technical. You need to know some complex analsis to understand it. I will list the ke steps. First, compute H(x,,t): x H(x,,t) x t(x 2 + x 2 + x x + ) x t( + 2 ) + ( t 2 t)x (t + t)x 2 Second, expand H(x,,t) in x gives us that ϕ(t) is the constant term in the Laurent series of ( t 2 t) 2 4t 2 ( + ) 2 Then ϕ(t) 2πi C ( t 2 t) 2 4t 2 ( + ) 2 d where C { } is the unit circle. Finall, it takes some effort to show that ϕ(t) is a meromorphic function in t and it has singularities when () ( t 2 t) 2 4t 2 ( + ) 2 has a double root on the unit circle. This gives us singularities of ϕ(t) at t /6 and t /2. So the radius of convergence of ϕ(t) is at most /6. 2. Exercise Question 2.. Let p(n) be the number of was of changing n dollars into coins and bills of $, $2, $5, $0, $20 s. Show that p(n) lim n n Question 2.2. Throw a dice n times and find the probabilit of getting a sum which is divisible b 5.