Worksheet o Geeratig Fuctios October 26, 205 This worksheet is adapted from otes/exercises by Nat Thiem. Derivatives of Geeratig Fuctios. If the sequece a 0, a, a 2,... has ordiary geeratig fuctio A(x, the what sequece has ordiary geeratig fuctio A (x? By assumptio, A(x = a x = a 0 + a x + a 2 x 2 + =0 Its derivative is (otice how the costat term drops off A (x = = a x = (m + a m+ x m m=0 = a + 2a 2 x + 3a 3 x 2 +... Therefore, the ew sequece is b = ( + a +, i.e. 2. Compute the derivative of questio. Aswer: ( x 2 b 0 = a, b 2 = 2a 2, b 3 = 3a 3,... x with respect to x (this is a pure calculus 3. Now expad the result as a ifiite series i powers of x. This uses Biomial Theorem: ( x 2 = 0 ( 2 ( x = 0 ( 2 ( x. 4. Combie the last three parts to prove that ( 2 ( = ( +. (ote: this ca be prove more directly; the poit is to illustrate the use of geeratig fuctios Let A(x = x which we kow to be the geeratig fuctio of the sequece a = (the sequece of all s.
By the first part, A (x is the geeratig fuctio of the sequece b =, i.e., 2, 3, 4, (the itegers. Therefore it has series expasio A (x = 0( + x. (Note how the x 0 term is etc. By the secod part, we kow that A (x has series expasio A (x = ( 2 ( x. 0 Comparig coefficiets of x, we obtai ( 2 ( = ( +. 5. If the sequece e 0, e, e 2,... has expoetial geeratig fuctio E(x, the what sequece has expoetial geeratig fuctio E (x? By assumptio, E(x = x e. 0 Takig derivatives, E (x = = = = m 0 e x e e x (! x (! m+ xm e m!. Notice how I wet to the trouble to write the result i the form (stuff x 0 agai. Whe we put it i that form, it is the expoetial geeratig fuctio of the stuff. That is, E (x is the expoetial geeratig fuctio of the sequece f = e +. I other words, the sequece is simply shifted. 2
2 Products of Ordiary Geeratig Fuctios. Suppose A(x is the ordiary geeratig fuctio for a 0, a, a 2,... ad B(x is the ordiary geeratig fuctio for b 0, b, b 2,.... Write dow the sequece havig A(xB(x as ordiary geeratig fuctio. Let A(x = 0 a x, B(x = 0 b x. The A(xB(x = a x b m x m. 0 m 0 To deal with this, I have to fid out the coefficiet of each x k i the product. The product will have terms x k for all k 0. To make such a term, we have to combie a a x ad a b m x m such that + m = k. There are various ways to do this. We have A(xB(x = (a 0 b k + a b k + + a k b 0 x k = ( a i b k i x k. k 0 k 0 The stuff i the iterior brackets is the sequece, which is c = a i b k i. 2. Give a ordiary geeratig fuctio A(x for a sequece a 0, a, a 2,..., what sequece has ordiary geeratig fuctio xa(x? We ca do this by appealig to the previous problem. We have B(x = x, i.e. b =. So it is a i. 3 Products of Expoetial Geeratig Fuctios. Suppose E(x is the expoetial geeratig fuctio for e 0, e, e 2,... ad F (x is the expoetial geeratig fuctio for f 0, f, f 2,.... Write dow the sequece havig E(xF (x as expoetial geeratig fuctio. Here, the mai challege is just rememberig to put it i expoetial form at the ed. We are takig a product of E(xF (x = x e x m f m. m! 0 m 0 3
Therefore, as before or by callig o the before (with a = e / ad b = f /, we obtai E(xF (x = k 0 ( e i f k i i! (k i! Now we just have to massage this back ito expoetial form (because the questio asked what sequece has E(xF (x as its expoetial geeratig fuctio. ( E(xF (x = k 0 ( = k 0 ( = k 0 Therefore the fial sequece is e i i! e i i! ( k e i f k i. i f k i (k i! f k i (k i! k! x k. x k x k k! ( k x e k i f k i i k! 2. Suppose E(x is the expoetial geeratig fuctio for a sequece e 0, e, e 2,.... What sequece has geeratig fuctio e x E(x? We ca apply the last part, with F (x = e x which is the expoetial geeratig fuctio of the sequece of all s. Therefore f i = ad we have ( k e i. i e 3. Use the last problem to figure out what sequece has x x as its expoetial geeratig fuctio. This meas settig E(x = x. Sice x = 0 x = 0 x we see that the sequece e = has fuctio. So usig the product formula, x ( k i!. i as its expoetial geeratig 4
This ca be rewritte, if you like, as a sum of fallig factorials: (k i. 4. Show that 2 = m=0 ( m. Hit: Compute e 2x as a series directly ad as a product of kow geeratig fuctios, ad compare. Computig e 2x directly Sice e x = 0 x, we have e 2x = (2x = 2 x. 0 0 Computig it as a product of geeratig fuctios The product is e x e x. We are usig the product formula with e = ad f =, so we obtai a ew series e x e x = ( ( x m. 0 m=0 Comparig coefficiets i the two expressios for the same series, we obtai ( 2 =. m 4 A example we kow m=0. What sequece has ordiary geeratig fuctio ( x k? It is a sequece we have studied i this class. First method. This is a k-fold product of /( x. We have x = + x + x2 + x 3 + ad so takig the product with itself k times: x x = ( + x + x2 + x 3 + ( + x + x 2 + x 3 +. where each product is k copies. To determie the coefficiet of x i the result, we must choose oe term from each of the series o the right. That is, we must choose a iteger (the expoet i each of the factors. For example, suppose k = 3. If we pick ( + x + x 2 + ( + x + x 2 + ( + x + x 2 + this gives x 2 x 2 = x 4. Here, 4 = 2+0+2. Therefore we obtai oe term x for each way of writig as a ordered sum of o-egative itegers. Therefore, this is the geeratig fuctio for weak compositios of 5
2. Prove the last i aother way. Hit: you could use biomial theorem, or you could use the techiques we used to describe the geeratig fuctio for p. Secod Method Now we ca also use biomial theorem. ( k ( k ( x k = ( x k = 0 ( x. = 0 We ca ow try to simplify the resultig coefficiet ( k a = ( ( x. Usig the defiitio of the geeralized biomial coefficiet, this is ( k( k ( k + a = ( k(k + (k + = ( ( k(k + (k + = (k +! = (k! ( k + = Fially, I claim this is the umber of ways of makig a weak compositio of ito k parts. For, such compositios are i bijectio with strigs of s ad k / s. For ay compositio = r + r 2 + + r k, write r s, the a /, the r 2 s, the a / etc., edig with r k s. For example, / / correspods to 3 = 2 + + 0. To cout these, we must arrage the stars ad k lies. That is, from + k symbol positios, choose the positios where we put stars, ad let the rest be lies. 6