Physics 116A Solutions to Homework Set #9 Winter 2012

Transcription

1 Physics 116A Solutios to Homework Set #9 Witer 1 1. Boas, problem Simplify Γ( 1 )Γ(4)/Γ( 9 ). Usig xγ(x) Γ(x + 1) repeatedly, oe obtais Γ( 9) 7 Γ( 7) 7 5 Γ( 5 ), etc. util fially obtaiig Γ( 9) Γ( 1 ). Hece, usig Γ(4) 3! 6, it follows that Γ( 1)Γ(4) Γ( ) Boas, problem Express as a Γ fuctio 1 ( x l x) 1 3 dx. Itroduce a ew variable x e u. The dx e u du ad l(1/x) l e u u. Notig that x u ad x 1 u, it follows that 1 ( x l x) 1 3 dx u 3 e 3u du 1 v 3 e v dv Γ(4) I the secod step above, I used the overall mius sig to iterchage the lower ad upper its of itegratio. At the third step, I chaged the itegratio variable oce more to v 3u (the its of itegratio are uchaged). Fially, I used the defiitio of the Gamma fuctio [eq. (3.1) o p. 538 of Boas] followed by Γ(4) 3! 6 to obtai the fial result. 3. Boas, problem Show that the biomial coefficiet ( p ) ca be writte i terms of Gamma fuctios as: ( p ) Γ(p + 1)! Γ(p + 1). (1) The biomial coefficiet is defied i eq. (13.6) o p. 8 of Boas as: ( ) p p(p 1)(p ) (p + 1).! To prove that it ca be writte as i eq. (1), we first multiply umerator ad deomiator by Γ(p + 1), ( ) p p(p 1)(p ) (p + 1) Γ(p + 1). ()! Γ(p + 1) 1

2 By repeatedly employig xγ(x) Γ(x + 1), the umerator of eq. () ca be writte as: p(p 1)(p ) (p + 3)(p + )(p + 1)Γ(p + 1) p(p 1)(p ) (p + 3)(p + )Γ(p + ) p(p 1)(p ) (p + 3)Γ(p + 3) p(p 1)(p )Γ(p ) p(p 1)Γ(p 1) pγ(p) Γ(p + 1). Isertig this last result back ito eq. () yields as requested. ( ) p Γ(p + 1)! Γ(p + 1) 4. Boas, problem , part (a). Use eq. (5.4) o p. 541 of Boas to show that assumig that is a iteger. Γ ( 1 ) Γ ( 1 + ) ( 1), (3) We start from the reflectio formula [eq. (5.4) o p. 541 of Boas]: If we set p 1, the Γ(p)Γ(1 p) Isertig p 1 ito eq. (4) yields: 1 p 1 (1 ) + 1. Γ ( 1 ) Γ ( 1 + ) Fially, we make use of the trigoometric idetity, si p. (4) si [ (. (5) 1 )] si [ ( 1 )] si ( 1 ) si ( 1 ) cos() cos ( 1 ) si() cos(), Boas restricts to be a positive iteger. However, eq. (3) holds for egative itegers as well. I fact, otig that ( 1) ( 1) for ay iteger, it follows that eq. (3) is umodified uder the iterchage of ad. Note that for, eq. (3) also yields a correct result, [Γ( 1 )].

3 after usig si( 1 ) 1 ad cos( 1 ). Usig the fact that cos() ( 1) for ay iteger, it follows that si [ ( 1 )] ( 1), for ay iteger. Isertig this result back ito eq. (5) ad usig ( 1) ( 1) for ay iteger, we obtai: Γ ( 1 ) Γ ( 1 + ) ( 1), for ay iteger, which is the desired result. 5. Boas, problem Express the followig itegral as a Beta fuctio, ad the i terms of Gamma fuctios, y dy (1 + y 3 ). Whe possible, use Gamma fuctio formulae to write the exact aswer for the itegral. Hece, First, we make a chage of variables by defiig x y 3. The y x 1/3 ad dx 3y dy dy dx 3x /3. y dy (1 + y 3 ) 1 x 1/3 3 (1 + x) dx. Usig eqs. (6.5) ad (7.1) o p. 543 of Boas, B(p, q) Γ(p)Γ(q) Γ(p + q) y p 1 dy. (6) (1 + y) p+q Thus, we idetify p ad q 4. Usig eq. (6) ad Γ( 4) 1Γ( 1 ), it follows that y dy Γ(3)Γ(43) 13B(3, 43) Γ(3)Γ(13). (1 + y 3 ) 3Γ() 9Γ() Fially we employ the reflectio formula [cf. eq. (5.4) o p. 541 of Boas], Γ(p)Γ(1 p) with p 3, Γ() 1 ad si(/3) 1 3 to obtai: y dy (1 + y 3 ) Γ(3)Γ(13) 9Γ() 3 si(p), (7) 9 si(/3) 3 7.

4 6. Boas, p. 545, problem Prove B(, ) B(, 1 )/ 1. Use this result to derive the duplicatio formula for the gamma fuctio, Check this formula for the case of 1 4. Γ() 1 Γ()Γ( + 1 ). We begi with eq. (6.4) o p. 543 of Boas: Settig p q, B(p, q) / (si θ) p 1 (cos θ) q 1 dθ. (8) B(, ) / 1 (si θ cos θ) 1 dθ 1 / (si θ) 1 dθ 1 1 / ( si θ cos θ) 1 dθ (si φ) 1 dφ, where we have employed the trigoometric idetity, si θ si θ cos θ. I the last step above, we chaged itegratio variables by defiig φ θ. Note that the correspodig its of itegratio chaged as well. At this poit, oe should observe that: (si φ) 1 dφ / (si φ) 1 dφ + / (si φ) 1 dφ / (si φ) 1 dφ, sice the fuctio si φ is uchaged uder the trasformatio φ φ. Hece, B(, ) 1 1 (si φ) 1 dφ 1 / (si φ) 1 dφ B(, 1 ) 1, after usig eq. (8) to evaluate the fial itegral over φ. Hece, we have prove that: B(, ) B(, 1 ) 1 (9) Usig B(p, q) Γ(p)Γ(q)/Γ(p + q), we ca rewrite eq. (9) as: Γ () Γ() Γ()Γ( 1) 1 Γ( + 1). Solvig for Γ(), ad makig use of Γ( 1 ), we arrive at the duplicatio formula for the gamma fuctio, Γ() 1 Γ()Γ( + 1 ) 4

5 Let us check this formula for the case of 14. Γ( 1 )? 1/ Γ(14)Γ(34). (1) Usig the reflectio formula give i eq. (7), it follows that Γ(14)Γ(34) si(/4). Thus, which cofirms eq. (1). 1/ Γ(14)Γ(34) Γ( 1 ), 7. Boas, problem Use Stirlig s formula to evaluate ()! (!). Thus, Stirlig s formula is give i eq. (11.5) o p. 553 of Boas,! Γ( + 1) e [ ( )] O. (11) Hece, ()! [ ( (!) () e (4) 1/ 1/ 1 + O 1 )] e () [ 1 + O ( 1 ()! (!) 1. )] 1 [1 + O ( )] Boas, p. 558, problem Expad the itegrads of the complete elliptic itegrals, / dθ / K(k) 1 k si θ, E(k) 1 k si θ dθ. i a power series i k si θ (assumig small k), ad itegrate term by term to fid power series approximatios for E(k) ad E(k). I this problem, we shall make use of the followig two power series expasios: 1 ( ) 1/ x ( 1) ( 1)!! x, 1 < x 1, (1) 1 + x ()!! ( ) 1/ 1 + x x ( 1) 1 ( 3)!! ()!! 5 x, 1 < x 1. (13)

6 Eq. (1) was obtaied i problem 1 of homework set #, ad eq. (13) was obtaied i Example o p. 9 of Boas. Settig x k si θ, ad employig eq. (1) i the defiitio of the complete elliptic itegral K(k), we fid: K(k) ( 1)!! ()!! / k (si θ) dθ, (14) after otig that ( 1) ( 1) ( 1) 1 for ay iteger. The itegral over θ ca be evaluated i terms of the Beta fuctio usig eq. (8), / (si θ) dθ 1 B( + 1, 1 ) Γ( + 1 )Γ( 1 ) Γ( + 1) This last result ca be further simplified by employig the expressio for Γ( + 1 ) derived i class. For completeess, I will rederive this result here. Applyig repeatedly the idetity, Γ(x + 1) xγ(x), it follows that for iteger 1, Γ( + 1) ( 1)Γ( 1) ( 1)( 3)Γ( 3) ( 1 )( 3)( 5)Γ( 5) ( 1 1 )( 3)( 5) 3 Γ( 1) ( 1)( 3)( 5) 3 1, where we have used Γ( 1 ). I class, we rewrote this result as: Γ( + 1 ) ( 1)!!, where ( 1)!! ( 1)( 3)( 5) 3 1 is the product of the first positive odd itegers. By defiig ( 1)!! 1, the rage of validity of the above formula is exteded to all o-egative itegers. Hece, usig Γ( + 1)!, it follows that:. / (si θ) dθ Γ( + 1 )Γ( 1 ) Γ( + 1) ( 1)!! +1! ( 1)!! ()!!. (15) I the last step, we used the fact that: ()!! ( ) ( 4) 6 4 ( 1) ( ) 3 1!. Isertig the result of eq. (15) ito eq. (14), K(k) [ ( 1)!! ()!! ] k [ 1 + ( 1 ) k + ( ) k 4 + ( ] ) k 6 + A similar computatio yields the power series expasio of E(k) about k. Settig x k si θ, ad employig eq. (13) i the defiitio of the complete elliptic itegral E(k), we fid: E(k) [ 1 ( 3)!! ()!! 6 ] / k (si θ) dθ, (16)

7 after otig that ( 1) 1 ( 1) ( 1) 1 1 for ay iteger. Isertig the result of eq. (15) ito eq. (16) the yields E(k) [ 1 1 ( 1)!!( 3)!! [()!!] k ] [ 1 ( ) 1 k ( ) 1 4 3k 4 ( ] ) 5k 6 9. Boas, problem Use Stirlig s formula to show that: x B(x, ) Γ(x). Usig eq. (7.1) o p. 543 of Boas for the beta fuctio, it follows that x B(x, ) x Γ()Γ(x) Γ(x + ). (17) It is coveiet to use Γ(+1) Γ() ad Γ(x++1) (x+)γ(x+) to rewrite eq. (17) as x B(x, ) x (x + ) Γ( + 1)Γ(x). (18) Γ(x + + 1) Applyig Stirlig s formula give i eq. (11) to each of the two Gamma fuctios appearig i eq. (18), we obtai: Thus, x B(x, ) ( ) x (x + ) e () 1/ 1 x+ Γ(x) (x + ) x+ e (x+) [(x + )] 1/ ex Γ(x). x + ( ) 1 x+ x B(x, ) e x Γ(x). (19) x + We ow must evaluate the it o the right had side. It is coveiet to write: ( ) 1 x+ ( ) ( ) 1 x ( x + x + x + x + The followig two its are easily obtaied: ) ( x + ( 1 + x ) 1 x 1, (1 + x ) e x. ) 1 x ( 1 + x ) ( 1 + x ) 1 x. The first it above follows from the fact that (x/). The secod it above should be well kow from your calculus course. But just i case you have 7

8 forgotte it, oe ca easily rederive it by takig the logarithm, computig the correspodig it, ad the expoetiatig the result. I this case, for large we have ( l 1 + x ) ( l 1 + x ) [ ( )] x 1 + O x + O It follows that (1 l + x ) x (1 + x ) e x. Hece, oe immediately obtais: ( ) 1 x+ e x. x + Pluggig this result back ito eq. (19) yields: ( ) 1. x B(x, ) Γ(x) () 8