GENERATING FUNCTIONS

Size: px

Start display at page:

Download "GENERATING FUNCTIONS"

Charlene Snow
5 years ago
Views:

1 GENERATING FUNCTIONS XI CHEN. Exapes Questio.. Toss a coi ties ad fid the probabiity of gettig exacty k heads. Represet H by x ad T by x 0 ad a sequece, say, HTHHT by (x (x 0 (x (x (x 0. We see that a the possibe outcoes of tosses are represeted by the expasio of f(x (x 0 +x (+x. The coefficiet of x k i f(x is the uber of outcoes with exacty k heads. Assuig the coi is fair, we see that the probabiity is ( k /2. Next, et us factor the probabiity ito the geeratig fuctio f(x. By that we ea we wat the coefficiet of x k gives the probabiity of gettig exacty k heads. To ake the probe ess trivia, et us assue that the coi is ot ecessariy fair with p for head ad q for tai (p + q. We wat to fid f(x a k x k with a k the probabiity of gettig exacty k heads. It is ot very hard to coe up with f(x (q + px a k x k So a k p k q k( k. A ore copicated versio of the probe is the foowig. Questio.2. You have cois C,..., C. For each k, C k is biased so that, whe tossed, it has probabiity /(2k + of faig heads. If the cois are tossed, what is the probabiity of a odd uber of heads? Let p k /(2k+ ad q k p k. It is ot hard to set up the geeratig fuctio as i the previous exape f(x (q + p x(q 2 + p 2 x...(q + p x a x with a the probabiity of gettig exacty heads. We are ookig for a + a 3 + a Here is a eat trick to fid a +a 3 +a give a series f(x a x. Note that f( a a 2k + a 2k+ ad So f( a ( a 2k a 2k+ a2k+ 2 (f( f(.

2 2 XI CHEN I our case, f( ad f( /(2+. So the probabiy of gettig odd uber of heads is /(2 +. Give f(x a x. What if we wat to fid a 0 +a 3 +a a3k? More geeray, What if we wat to fid a pk+r for soe fixed p ad r. Hit: use roots of uity. Questio.3. Fid the uber of ways of chagig a 500 doar bi ito $, $2, $5, $0, $20 s. Let k, k 2, k 3, k 4, k 5 be the uber of $, $2, $5, $0, $20 s, respectivey. We are ookig at the ubers of oegative itegra soutios of the equatio: k + 2k 2 + 5k 3 + 0k k Set up the correspodece (k, k 2, k 3, k 4, k 5 (x k (x 2k 2 (x 5k 3 (x 0k 4 (x 20k 5 The the uber of the soutios of the equatio is the coefficiet of x 500 i f(x x k k 0 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 fid the Tayor expasio of f(x, we have to write f(x as a su of partia fractios, which is doabe i theory but ipossibe by had. But we ca sti say a ot about the coefficiets of f(x. Let f(x p(x, i.e., p( is the uber of the ways chagig doar. The the foowig is true: p( grows at the order of 4 as. I eave this as a exercise. p(20 is a poyoia i of degree 4. This is a itte bit harder to prove. p(20 4 p(20k k0 0 4 k k This uses iterpoatio forua. Oce we kow that p(20 is a poyoia of degree 4 ad the vaues of p(20 at five poits, say p(0, p(20, p(40, p(60, p(80, we ca fid p(20 by iterpoatio. More geeray, the uber of oegative itegra soutios (x, x 2,..., x of the equatio λ x + λ 2 x λ x

3 GENERATING FUNCTIONS 3 is give by the coefficiets of the geeratig fuctio f(x ( x λ ( x λ 2...( x λ. What if we are ookig for itegra soutios x i withi the rage, say α i x i β i? What is the correspodig geeratig fuctio? What if we are ookig for odd soutios, soutios with x i (od 3 ad etc? What are the correspodig geeratig fuctios? Questio.4. Let be a positive iteger. Fid the uber of poyoias P (x with coefficiets i {0,, 2, 3} such that P (2? Let P (x a 0 + a x + a 2 x a k x k +... We are ookig at the soutios of a 0 + 2a + 4a k a k +... with 0 a k 3. The uber of such soutios are give by the coefficiet of x i 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 ( ( + ( + x 2 0 Aother way to express the resut is /2 +. Fid a purey cobiatoria soutio to the probe. Questio.5. Fid the uber of subsets of {,..., 2003}, the su of whose eeets is divisibe by 5. Set up the correspodece: The 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 coefficiet of x i f(x is the uber of subsets of {,..., 2003}, the uber of whose eeets is exacty. So if f(x a x, we are ookig for the su a 0 + a 5 + a Let r exp(2πi/5 be the 5-th roots of uity. The a 0 + a 5 + a (f(r 0 + f(r + f(r 2 + f(r 3 + f(r 4.

4 4 XI CHEN Obviousy, f(r For k 4, ad hece 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 4 f(r k ( k So the aswer is (/5( Repace 2003 by 5k + ad redo the coputatio. Fid a purey cobiatoria soutio to the probe. Questio.6. Suppose that each of peope writes dow the ubers, 2, 3 i rado order i oe cou of a 3 atrix, with a orders beig equay ikey ad idepedet. Show that for soe 995, the evet that the row sus are cosecutive itegers is at east four ties ikey as the evet that the row sus are equa. Istead of (, 2, 3, we use (, 0,. Set up the correspodece: a b x a y b c The a 3 such atrix ca be represeted by a ter i the expasio of ( φ (x, y x + y + x y + y x + x + y Let a be the costat ter i φ (x, y. It is ot hard to see that a is exacty the uber of atrices with equa row sus. Let b, c, d, e, f, g be the coefficiets of x, y, x/y, y/x, /x, /y i φ (x, y. It is ot hard to see that b + c + d + e + f + g are the uber of atrices with cosecutive row sus. Actuay, fro ( φ + (x, y x + y + x y + y x + x + φ (x, y y we see that a + b + c + d + e + f + g We are essetiay asked to show that a /a + /4 for arge. I other word, if we put ϕ(t a t, it suffices to show that the radius of covergece of ϕ(t is ess tha /4. This suggests us to put a φ (x, y together ito H(x, y, t φ (x, yt If we expad H(x, y, t as a ifiite series i (x, y (oe techica issue here: H(x, y, t expads as a Lauret series, ϕ(t is the costat ter.

5 GENERATING FUNCTIONS 5 The rest of the proof is quite techica. You eed to kow soe copex aaysis to uderstad it. I wi ist the key steps. First, copute H(x, y, t: xy H(x, y, t xy t(x 2 y + xy 2 + x 2 + y 2 + x + y xy t(y + y 2 + (y ty 2 tx (ty + tx 2 Secod, expad H(x, y, t i x gives us that ϕ(t is the costat ter i the Lauret series of y (y ty 2 t 2 4t 2 (y + 2 y The ϕ(t 2πi C y (y ty 2 t 2 4t 2 (y + 2 y dy where C { y } is the uit circe. Fiay, it takes soe effort to show that ϕ(t is a eroorphic fuctio i t ad it has siguarities whe (y (y ty 2 t 2 4t 2 (y + 2 y has a doube root o the uit circe. This gives us siguarities of ϕ(t at t /6 ad t /2. So the radius of covergece of ϕ(t is at ost /6. 2. Bioia Theore The cassica Newto s Bioia Theore tes us that ( ( ( ( (2. ( + x + x x 0 for a positive itegers, where ( (2.2!!(!. 0 Questio 2.. Why is (, defied by (2.2, aways a iteger? Proof. It is easy to verify ( ( (2.3 + ( which is actuay the reatio i Pasca triage. The we ca prove ( Z by iductio o. Proof 2. Let! p a pa pa be the factorizatio of! with prie factors p i. The every a i is give by (2.4 a i p i p 2 i µ p µ i x

6 6 XI CHEN where x is the argest iteger x. It is easy to check (2.5 x + y x + y for a x, y R. Therefore, (2.6 p µ p µ + for a p ad µ. If we factorize!,! ad (! as (2.7 we aways have p µ! p a pa pa! p b p b p b (! p c pc pc (2.8 a i b i + c i for a i. Therefore, ( (2.9 is a iteger.!!(! pa b c p a 2 b 2 c p a b c Questio 2.2. Fid the uber of traiig 0 s i 000!. Questio 2.3. Show that ( is eve for a eve ad odd. (2.3 is a specia case of Questio 2.4. For a N, ( + 2 i0 ( i ( 2. i Actuay, Bioia Theore ca be geeraized to ( r (2.0 ( + x r x 0 for x <, where r ca be ay rea uber ad ( r r(r...(r + (2..! This foows fro Tayor expasio of f(x ( + x r : (2.2 f ( (0 r(r...(r +. Questio 2.5. Verify that ( r for a r R ad N. ( + r (

7 Questio 2.6. Prove that for x < ad a Z +. GENERATING FUNCTIONS 7 ( x Questio 2.7. Prove that ( /2 0 ( + x for a Z +. ( 2 2 ( 2 3. Exercise Questio 3.. Cosider the geera hoogeeous poyoia of degree i the p variabes x, x 2,..., x p. How ay ters does it have? Questio 3.2. Fid the geeratig fuctio for the uber of itegra soutios to x + x 2 + x 3 + x 4 k, where 0 x, 3 x 2, 2 x 3 5, x 4 5. Questio 3.3. Let p( be the uber of ways of chagig doars ito cois ad bis of $, $2, $5, $0, $20 s. Show that p( i Questio 3.4. Throw a dice ties ad fid the probabiity of gettig a su which is divisibe by 5. Questio 3.5. I how ay ways ca you put the ecessary staps i oe row o a airai etter set iside the U.S., usig 2, 4, 6, 8 cets staps? The postage is 0 cets. (Differet arrageets of the sae vaues are regarded as differet ways. Questio 3.6. Suppose we have a arge suppy of red, white, ad bue baoos. How ay differet buches of 0 baoos are there, if each buch ust have at east oe baoo of each coor ad the uber of white baoos ust be eve? Questio 3.7. Prove that the expressio gcd(, ( is a iteger for a pairs of itegers. Questio 3.8. Copute for Z +. 0 ( 4 +

8 8 XI CHEN Questio 3.9. Prove that for a Z +. Questio 3.0. Prove that for a Z +. 0 Questio 3.. Copute for Z +. 0 ( 2 ( 2 2 ( 2 ( 2 ( ( Questio 3.2. Prove that ( ( for Z +. Refereces 2 2 [G] David Guichard. A Itroductio to Cobiatorics ad Graph Theory. oie/cgt.pdf. [P-S] George Póya ad Gabor Szegö. Probes ad Theores i Aaysis I. Spriger- Verag, Beri Heideberg, Cetra Acadeic Buidig, Uiversity of Aberta, Edoto, Aberta T6G 2G, CANADA E-ai address: xiche@ath.uaberta.ca

CS 70 Second Midterm 7 April NAME (1 pt): SID (1 pt): TA (1 pt): Name of Neighbor to your left (1 pt): Name of Neighbor to your right (1 pt):

CS 70 Second Midterm 7 April NAME (1 pt): SID (1 pt): TA (1 pt): Name of Neighbor to your left (1 pt): Name of Neighbor to your right (1 pt): CS 70 Secod Midter 7 April 2011 NAME (1 pt): SID (1 pt): TA (1 pt): Nae of Neighbor to your left (1 pt): Nae of Neighbor to your right (1 pt): Istructios: This is a closed book, closed calculator, closed