SOLUTIONS TO RAINVILLE S SPECIAL FUNCTIONS (1960)

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1 SOLUTIONS TO RAINVILLE S SPECIAL FUNCTIONS 96 SOLUTIONS BY SYLVESTER J PAGANO AND LEON HALL; EDITED BY TOM CUCHTA Cotets Preface Chapter Solutios 4 Chapter Solutios 9 3 Chapter 3 Solutios 7 4 Chapter 4 Results Cited 3 5 Chapter 4 Solutios 4 6 Results from Chapter 5 used 39 7 Chapter 5 Solutios 4 8 Chapter 6 Solutios 48 9 Chapter 7 Solutios 64 Chapter 8 Solutios 67 Chapter 9 Solutios 76 Chapter Solutios 77 3 Chapter Solutios 98 4 Chapter Solutios 5 Chapter 3 Solutios 6 Chapter 4 Solutios 7 Chapter 5 Solutios 4 8 Chapter 6 Solutios 46 9 Chapter 7 Solutios 53 Chapter 8 Solutios 56 Chapter 9 Solutios 7 Chapter Solutios 7 3 Chapter Solutios 84

2 SOLUTIONS BY SYLVESTER J PAGANO AND LEON HALL; EDITED BY TOM CUCHTA Preface Earl D Raiville bega givig lectures o Special Fuctios at the Uiversity of Michiga i 946 The course was well received, ad his otes became the basis f his book, Special Fuctios, published i 96 Also i 946, Sylvester J Pagao received his BS degree i Electrical Egieerig at the Missouri School of Mies ad Metallurgy MSM That fall, Pagao was appoited Istruct i Mathematics at MSM I 95, Pagao, ow Assistat Profess, spet the summer at Michiga, where he ad Raiville presumably met I the summers of 96, 963, ad 964, Pagao was agai at Michiga, this time as a Natioal Sciece Foudatio Sciece Faculty Fellow Raiville passed away i 966, the same year Pagao was promoted to Profess at the Uiversity of Missouri Rolla UMR, MSM uder its ew ame I the sprig of 966, I was a sophome at UMR ad was takig Elemetary Differetial Equatios; the textbook was the third editio so of Raiville s Elemetary Differetial Equatios This would be a better sty if Pagao had bee the istruct i that class, but he was t I did have a class from Pagao later, whe I was a begiig graduate studet; it was Operatioal Calculus, ad we used the Operatioal Mathematics book by RV Churchill Churchill was Raiville s PhD advis Michiga 939, but I kew othig of ay of these coectios at the time Pagao was a member of both my MS ad PhD committees at UMR Figure Photograph of Earl D Raiville from the Uiversity of Michiga

3 SOLUTIONS TO RAINVILLE S SPECIAL FUNCTIONS 96 3 Figure These photographs are of Sylvester J Pagao as a college sei i 946 Pagao retired i 986, a year after I had retured to Rolla as a faculty member Somehow, i the process of him cleaig out his office, I got his collectios of wked problems from Raiville s Special Fuctios, Raiville s Itermediate Differetial Equatios, ad Churchill s Operatioal Mathematics The way I remember it is that I kew these problem solutios existed, ad whe Pagao retired, I asked him if I could have them, a request to which he graciously agreed As to how I kew he eve had this material, I do t remember f sure These problem solutios had writte almost certaily go back to the three summers Pagao spet at Michiga i the 96s, ad some might eve date back to his earlier 95 visit; Pagao cotiued to wk o them as he taught the courses himself i the 96s 98s Begiig i the mid 99s I bega to teach both Operatioal Calculus ad Special Fuctios from time to time at UMR F Operatioal Calculus, I used Churchill s book, ad Pagao s problem solutios were quite useful I developed my ow otes f Special Fuctios because it was a summer course desiged so our graduate studets who taught i the summer would have somethig to take, ad some of these studets had ot yet studied complex variables The last time I offered Special Fuctios was the summer of, ad I was able to talk Tom Cuchta, oe of the studets i that class, ito wkig o trascribig Pagao s problem solutios from Raiville s Special Fuctios ito electroic fm usig L A TEX To my surprise ad delight, Cuchta fiished the job by the ed of! We leared, however, that Pagao had ot provided solutios to all the problems he wrote up solutios to 96 out of 3 problems i the book That s 85%, so is pretty good, but I decided I would wk o fiishig the job, ad Cuchta agreed to keep addig to the TEXfile as me problems were completed As of ow April 5, there are less tha problems left to fiish The ed is i sight Sylvester Pagao was a good example of a type of mathematics profess that seems to be disappearig these days He did t publish ay papers, but he was a

4 4 SOLUTIONS BY SYLVESTER J PAGANO AND LEON HALL; EDITED BY TOM CUCHTA outstadig teacher ad he kept buildig his kowledge of mathematics throughout his career Whe I talk with alumi from our departmet, ad ofte with egieerig graduates from other departmets, the perso they most frequetly ask about is Profess Pagao; they fodly remember him as oe of the good oes from their studet days These alums are right he was oe of the good oes, ad he should be remembered These problem solutios are mostly his, ad the rest were ispired by him There is a lot of iterestig mathematics here I hope you ejoy it Leo M Hall Profess Emeritus, Mathematics Missouri S&T Chapter Solutios Problem Show that the followig product coverges ad fid its value: [ ] ˆˆ Solutio Solutio by Leo Hall By Theem 3, page 3, this product coverges 6 absolutely because coverges absolutely So, if [ ] 6 P + k + k + 9 k k + 5k + 3 k + k + 9 k [ ][ ] [ 3 + ][ 3 + 9] [7 9][ + + 3] [ 3][ ] the lim P lim Note: The use of Theem 3 is ot eeded because fidig the value of the ifiite product is sufficiet itself to show covergece

5 SOLUTIONS TO RAINVILLE S SPECIAL FUNCTIONS 96 5 Problem Show that Solutio Solutio by Leo Hall Let P + k k k + k k k [3 4 + ][ 3 ] lim P + lim Problem 3 Show that diverges to Solutio 3 Solutio by Leo Hall k P k k 3 k 3 4 Sice lim P lim the product diverges to [The product does ot coverge to because oe of the terms i the product are ] Problem 4 Ivestigate the product + z i z < Solutio 4 Solutio by Leo Hall Let P P + z, + z k The P + z + z P + z + z + z 4 + z + z + z 4 + z 6

6 6 SOLUTIONS BY SYLVESTER J PAGANO AND LEON HALL; EDITED BY TOM CUCHTA Assume P + z z k + z k The P + P + z + P + z + [ P ] + z + z + + z + + z + + z z z z k So we have show by iductio that + P + z z k, which is a geometric series covergig to z + z, f z < + z + z ad same process wks Thus, +z coverges absolutely to z Problem 5 Show that exp diverges Solutio 5 Solutio by Leo Hall Let P S log P exp ad let S is the th partial sum of the harmoic series, which diverges As i the proof of Theem, page 3, P exp S ad k k k lim P lim exp S exp lim S Thus, because {S } diverges, so does {P } Problem 6 Show that exp diverges to Solutio 6 Solutio by Leo Hall Let as i Problem 5 The log P S P exp S exp k k k k exp k k k k

7 Because ad so k k diverges to Problem 7 Test SOLUTIONS TO RAINVILLE S SPECIAL FUNCTIONS 96 7 diverges to we have z lim P exp Solutio 7 Solutio by Leo Hall The product diverges to f ay z such that z ±m, m a positive iteger F all z such that z, we have by Theem 3, page 3, that z is absolutely coverget because z is absolutely coverget I fact, Problem 8 Show that z z π 6 ] [ + + coverges to uity ] Solutio 8 Solutio by Leo Hall Let P [ + k+ k k Case : is eve The P 3 4 Rearragig, we get P f eve Case : is odd The P 3 4 I both cases lim P Problem 9 Test f covergece: p f real p + Solutio 9 Solutio by Leo Hall F p > the series of positive umbers p is kow to be coverget eg, by the Itegral Test Thus, p is absolutely coverget by Theem 3

8 8 SOLUTIONS BY SYLVESTER J PAGANO AND LEON HALL; EDITED BY TOM CUCHTA F < p, > ad so covergece ad absolute covergece are the p same Because the series p diverges f < p, our product diverges by Theem 3 F p <, let p q where q > The p q + q But lim q, so i this case our product diverges by Theem Summary: p diverges whe p a p, ad coverges whe p > Problem Show that the usual covetio at z si z z Solutio Solutio by Leo Hall Let The si z z a z si z z z is absolutely coverget f all fiite z with + a z 3! + 5! 4 7! 6 + [ ] z 6 + O Thus, there exists a costat M such that ad because M z 4 a z < M, coverges, the product + a z si z coverges absolutely ad uifmly f all fiite z by Theems 3 ad 4 If z the product is, with the usual covetio, Problem Show that if c is ot a egative iteger, [ z z exp c + ] is absolutely coverget f all fiite z z z 6

9 SOLUTIONS TO RAINVILLE S SPECIAL FUNCTIONS 96 9 Solutio Let + a z z z exp c + + z + z! + z 3 3! 3 + z + z c + + z 3! + z 4 3! z + c +! z + c + 3! 3! z 3 + c + c + c + z + c c k c + z + k! k c + zk k3 c + c + z c + z + O Thus, f c ot a egative iteger ad f ay fiite z, there is a costat M such that a z M ad so by Theems 3 ad 4 the product coverges absolutely ad uifmly Chapter Solutios ˆˆ Problem Start with Γ z Γz γ z z +, prove that Γ z Γz Γ z Γz Γ z + Γz + log, ad thus derive Legedre s duplicatio fmula, page 4 Solutio Applyig three times ad simplifyig yields Γ z Γz Γ z Γz Γ z + Γz + γ z + γ + z + γ + z + z + lim z + k k z + + lim [ k [ k z + k + H + + z + + lim z + k + k k z + + z + + lim z lim H H + + lim [H log H log + log log ] [γ γ + log ] log Problem Show that Γ γ + log π z + + z + + z lim z + k + lim k z + + k k k k ] z + k H + z + k + H k ] k z + k + H H

10 SOLUTIONS BY SYLVESTER J PAGANO AND LEON HALL; EDITED BY TOM CUCHTA Solutio By Problem, we kow that Now let z to get ad so, algebra yields Γ z Γz Γ z Γz Γ z + Γz + log Γ Γ Γ Γ Γ Γ Γ Γ Γ log Γ But Γ, Γ γ, Γ π, hece ad by rearragemet, Γ γ log, π log, Γ γ + log π Problem 3 Use Euler s itegral fm Γz Solutio 3 From Γz Γz + zγz u t z dv e t dt Γz + e t t z dt to show that e t t z dt, f Rz >, itegratio by parts yields e t t z dt du zt z v e t [ t z e t ] + z e t t z dt where lim t t z e t coverges f Rz > Problem 4 Show that Γz lim z Bz, Solutio 4 From page 8, we kow but Hece Bz, ΓzΓ Γz + + zγz,! z Γz lim, z Γz! z Γz Γz lim z Bz,! z Problem 5 Derive the followig properties of the beta fuctio: a pbp, q + qbp +, q; b Bp, q Bp +, q + Bp, q + ; c p + qbp, q + qbp, q; d Bp, qbp + q, r Bq, rbq + r, p

11 SOLUTIONS TO RAINVILLE S SPECIAL FUNCTIONS 96 Solutio 5 a We kow Bp, q ΓpΓq Γp + q, so b c d pbp, q + pγpγq + Γp + q + Γp + qγq Γp + + q ote: p q ad q p is this the symmetric property? Bp, q ΓpΓq Γp + q ΓpΓq Γp + q + p + q p + qγpγq Γp + q + pγpγq Γp + q + + qγpγq Γp + q + Γp + Γq ΓpΓq + Γp + q + + Γp + q + Bp +, q + Bp, q + p + qγpγq + p + qbp, q + Γp + q + p + qγpγq + p + qγp + q ΓpΓq + Γp + q ΓpqΓq Γp + q qbp, q Bp, qbp + q, qbp +, q ΓpΓq Γp + qγ Γp + q Γp + q + ΓpΓqΓ Γp + q + ΓqΓ ΓpΓq + Γq + Γp + q + Bq, Bq +, p

12 SOLUTIONS BY SYLVESTER J PAGANO AND LEON HALL; EDITED BY TOM CUCHTA Problem 6 Show that f positive itegral, Bp, + p + Solutio 6 F iteger ad usig Theem 9 pg 3, Problem 7 Evaluate ΓpΓ + Bp, + Γp + + ΓpΓ + p + Γp + Γp p + pγp pp + p + + x p x q dx Solutio 7 Let A + x p x q dx Now let y + x, x y, x y y Hece A p y p q y q dy p+q y p y q dy p+q Bp, q Problem 8 Show that f k, α k Note particularly the special case α k α α k Solutio 8 Cosider α k f k The α k αα + α + k αα + α + k [α + kα + k + α + ] α + α + α + k α α + k k α k α k k α α k Note f α, that k! k k

13 SOLUTIONS TO RAINVILLE S SPECIAL FUNCTIONS 96 3 Problem 9 Show that if α is ot a iteger, Γ α Γ α α Solutio 9 Cosider f α ot equal to a iteger as desired Γ α Γ α Γ α αγ α Γ α α α Γ α Γ α α α α + Γ α, α I the followig problems, the fuctio P x : x x Problem Evaluate x Solutio To evaluate P ydy x P ydy whe P y y y iteger so that m If m x < m +, the x m ad x P ydy x P ydy y m dy [ y m ] x m x m Let m be a [ x m [ P x ] 4 P x 8 Problem Use itegratio by parts ad the result of the above exercise to show that P xdx + x 8 + Solutio Cosider dv P xdx u + x v P x 8 m ] P x dx ad use itegratio by parts + x du + x dx P x + x dx P x 4 + x + P x 4 + x dx

14 4 SOLUTIONS BY SYLVESTER J PAGANO AND LEON HALL; EDITED BY TOM CUCHTA { Now max P x } 4 4 ad P 4 implies ad P x + x dx + P x 4 + x dx P x + x dx P x + x dx 8 [ ] 4 + x dx 8 + x [ ] + Problem With the aid of the above problem, prove the covergece of Solutio P x P x dx coverges lim dx but from Exer- + x + x cise, Hece P xdx + x lim P x dx < + x Problem 3 Show that The prove that ad thus coclude that P xdx + x P x + x dx lim + lim P xdx + x N P xdx + x 8 + y dy + + y is coverget y dy + + y Solutio 3 Solutio by Leo Hall Because P x is periodic with period, it is clear that P x + x dx + P x + x dx Let x + y The + P x + x dx P + y + + y dy P y + + y dy y + + y dy

15 SOLUTIONS TO RAINVILLE S SPECIAL FUNCTIONS 96 5 This establishes the first set of equalities y [ y + + dy + 3 ] dy y + + [ y + 3 ] logy log log y To determie the covergece of dy we compare with the kow y + + coverget series usig the limit compariso test Thus, ad y y + + dy coverges k + k k [ [ ] O ] + O [ ] O O + y lim y + + dy, y y + + dy P x + x dx Problem 4 Apply Theem, page 7, to the fuctio fx + x ; let ad thus coclude that γ y P ydy Solutio 4 Solutio by Leo Hall Let fx Theem, page 7 + x gives with p, q, + k + x dx f xb xdx + +

16 6 SOLUTIONS BY SYLVESTER J PAGANO AND LEON HALL; EDITED BY TOM CUCHTA So, + k log y : x Problem 5 Use the relatio ΓzΓ z to prove that π si πz + + si x si y [cosx y cosx + y] ΓcΓ cγc a bγa + b + c Γc aγa + cγc bγb + c Solutio 5 Note that ad c a b a + b + c, c a a + c, c b b c, so we ca use the gamma fuctio relatio four times to get ΓcΓ cγc a bγa + b + c Γc aγa + cγc bγb + c B x + x dx B y + dy y B y y dy Now usig the give trig idetity, we get, cotiuig the equality: ad the elemetary result Γ cγc Γc a bγa + b + c ΓaΓ aγbγ b π si πc a si πc b π siπc si πc a b siπc si πc a b si πc a si πc b siπc si πc a b [cos πa + b cos πc a b] [cos πb a cos πc a b] [cos πa + b cos πc a b] [cos πb a cos πa + b] [cos πa + b cos πc a b] siπa siπb siπc si πc a b siπa siπb si πc si πc a b Cacelig mius sigs ad multiplyig ad dividig by π yields as desired Γc Γ cγc a bγa + b + c ΓaΓ aγbγ b

17 SOLUTIONS TO RAINVILLE S SPECIAL FUNCTIONS Chapter 3 Solutios ˆˆ Problem With the assumptios of Watso s lemma, show, with the aid of the k covergece of the series F t a exp t r i t a + δ, that f k t a, there exists a positive costat c such that F t k + a k exp t r < c exp t r k Solutio We wish to show that there exists c such that f t a see problem f t > a, F t k a k t k r < c t + r uder the coditio of Watso s lemma By the covergece of i t a we write F t a k t k r where c > k+ k a k a k r f t a + δ where δ > k+ a k t k r t + r k+ t + r k+ r, a k t k r a k a k r < c t + Remember from Watso s lemma F t a t r k a t r F t Problem With the assumptios of Watso s lemma, page 4, show that f t > a, there exist positive costats c ad β such that F t k + a k exp t r < c exp t e βt r k Solutio Uder the assumptio of Watso s lemma we wish to show that f t > a, there exist costats c, β such that F t k a k t k r < c e βt t + r

18 8 SOLUTIONS BY SYLVESTER J PAGANO AND LEON HALL; EDITED BY TOM CUCHTA Now f t >, we have give F t < ke bt Hece F t a k t k r < ke bt + t + r a k t k r k k ke bt + a k t k r, but k < ad t > a, so F t a k t k r < ke bt + t + r a k a k r k k but a k a k r coverges Hece, sice t > a, there exist costats M, M such k that F t a k t k a + r < M e bt t + r r + M t + r t k k < M e bt t + r + M t + r < c e βt t + r e bt Problem 3 Derive the asymptotic expasio 6 immediately precedig these exercises by applyig Watso s lemma to the fuctio f x te xt t + t dt ad the itegratig the resultat expasio term by term Solutio 3 Cosider Watso s lemma with F t t + t Hece F t t + t + t + t ; t < Choose r, a, δ 3 Also f t, et ad Hece F t < e t ; t, satisfyig coditio of Watso s Lemma, ad F t t f t Hece by Watso s lemma f f x f x Γ / x / t + t < te xt dt we have + t! x

19 SOLUTIONS TO RAINVILLE S SPECIAL FUNCTIONS 96 9 The x f sds After itegratig, this becomes x! s ds x f sds! x Now d e xt dx + t dt te xt + t dt f x Let A label the itegral i the middle of the last fmula Hece fx e xt + t dt, which we label as B Also ote that itegrals A ad B are uifmly coverget Hece Therefe By, So lim fx lim x x fx x f sds fs e xt dt + t x fx! x ; x, Rx > e xt fx + t dt! x! x + ; x, Rx > Problem 4 Establish 6, page 43, directly, first showig that fx k k!x k + e xt t + + t dt, ad thus obtai ot oly 6 but also a boud o the err made i computig with the series ivolved

20 SOLUTIONS BY SYLVESTER J PAGANO AND LEON HALL; EDITED BY TOM CUCHTA Solutio 4 Solutio by Leo Hall Because So fx + t k t k + k+ + t k t k we have k t k k t k + + t + k t k + e xt + t dt e xt + t+ + t k t k dt + + k e xt t k dt + + k t k e xt t + + t dt e xt t + + t dt Itegratio by parts give the reductio fmula f x as specified This, plus the fact that The fx e xt t k dt kk x e xt dt x, yields k k! + + xk+ fx k k! x k+ < < e xt t k dt e xt t + + t dt e xt t + + t dt e xt t + dt e Rext t + dt +! [Rex] +3 I the regio arg x π, >, if Rex > N, the x > N x +! [Rex] +3 O x, ad so fx k k! x k+ as x i the sect arg x < π, > si ad as

21 SOLUTIONS TO RAINVILLE S SPECIAL FUNCTIONS 96 Problem 5 Use itegratio by parts to establish that f real x, x e t t dt e x e t x Solutio 5 Cosider fx dt as x Usig itegratio by parts, x t u dv e t dt fx [ t t e t t e t dt x x u t v e t [ x e x + t e t + x t 3 e t dt u [ t dv e t dt e x x ] [ x ] t 3 e t 3 x x du t 3 v e t This patter ca clearly cotiue fever, so we ca write fx e x Now sice < x < t, ex fx Hece so e x fx x k k! x k ! k k! x k+ k k! x k+ e x fx t e t dt e x x +!e x! < < x + ex +! x + x x t + e t dt e t dt + t e t dt O x + O x +, x + x + as x Problem 6 Let the Hermite polyomials H x be defied by expxt t H xt t 4 e t dt f all x ad t, as i Chapter Also let the complemetary err fuctio erfc x be defied by erfc x erf x exp β dβ π x

22 SOLUTIONS BY SYLVESTER J PAGANO AND LEON HALL; EDITED BY TOM CUCHTA Apply Watso s lemma to the fuctio F t expxt t ; obtai exp x s exp β dβ H xs, s, s x ad thus arrive at the result [ ] t π exp t x erfc t x H xt, t + Solutio 6 Because exp t t, ad if F t expxt t we ca write F t H x t, H x! exp t so the first coditio i Watso s Lemma is satisfied f ay fixed x with r, a, ad ay δ > Also, F t expxt t e t e xt < e xt f t, so the secod coditio i Watso s Lemma is satisfied with k aythig > ad b x Thus, we get e st H xγ F tdt!s as s, [ exp t + x ] s t dt H xs, as s Completig the square i the expoet leads to exp [x ] [ s exp t + ] s x dt H xs as s, ad if we let β t + s x we get [ exp x ] s e β dβ H xs, s x as s Usig the defiitio of erfc, ad the makig the substitutio t s iverse Laplace trasfm we get [ π exp x ] s as s as t + as desired erfc s x [ π t exp x ] erfc t t x H xs, H xt, ot the

23 SOLUTIONS TO RAINVILLE S SPECIAL FUNCTIONS 96 3 Problem 7 Use itegratio by parts to show that if Rα >, ad if x is real, x e t t α dt x α e x of which Problem 5 is the special case α α x +, x, Solutio 7 Itegratio by parts with u t α ad dv e t dt yields x e t t α dt e x x α α x e t t α+ dt The same itegratio by parts with α replaced by α + applied to the last itegral gives [ e t t α dt e x x α α ] + αα + e t t α+ dt x x Cotiuig, after + itegratios by parts, we have x e t t α dt e x x α+ e x x α e t t α dt x k α k x k+ + + α + k α k x k+ e x x α α + x x x e t t α++ dt e t t α++ dt Thus, we have the desired asymptotic series if the right side of the last equatio is O x + as x This is true because ex x α α + e t t α++ dt x e x x α α + x α++ e t dt x e x α + x + e x α + x + O x + as x By defiitio, 4 Chapter 4 Results Cited F a, b; c; ˆˆ a b c Theem If Rc a b > ad if c is either zero a egative iteger, F a, b; c; ΓcΓc a b Γc aγc b

24 4 SOLUTIONS BY SYLVESTER J PAGANO AND LEON HALL; EDITED BY TOM CUCHTA pg66, : + z a F Theem If z < ad a, b; F c; Theem 3 If z <, a, a + ; 4z + z + a b; z z <, z z a F F a, c b; a, a b + ; x c; b + ; z z F a, b; c; z + z c a b F c a, c b; c; z Theem 4 If b is either zero a egative iteger ad if y < ad y y <, a, a + ; y a F y a, b; y b + F y b; ; Theem 5 Kf a + b + is either zero a egative iteger, ad if x < ad 4x x <, a, b; F 4x x a + b + Theem 6 If c is either zero a egative iteger, ad if both x < ad 4x x <, c a a, a; F x z c F, c + a ; 4x x c; c; Problem Show that a, b; d dx F c; 5 Chapter 4 Solutios x ˆˆ ab c F a +, b + ; c + ; x

25 SOLUTIONS TO RAINVILLE S SPECIAL FUNCTIONS 96 5 Solutio From F a, b; c; a b, c we get Problem Show that F d F a, b; c; x dx a, b; a + b + ; Solutio Wish to evaluate F a b x c c! a + b + x c + ab a + b + x c c + ab F a +, b + ; c + ; x c Γ Γ c + a a, b; a + b + ; a + b + Γ Γ c a + From Theem 5, F a, b; a + b + x F a, b; a + b + 4x x f x <, 4x x < We eed to use x, but ote that Ra+b+ a b > Hece, by Theem, F a, b; a + b + F a, b; a + b + Γ Γ a + b + Γ b + Γ a + Problem 3 Show that F a, a; c; c ΓcΓ c a + Γ Γ c + a

26 6 SOLUTIONS BY SYLVESTER J PAGANO AND LEON HALL; EDITED BY TOM CUCHTA c a, a; Solutio 3 Cosider F x x c F, c + a ; c; c; By Theem 6, f x <, 4x x <, c a a, a; F x x c F, c + a ; 4x x c; c; Sice R c c + a c a + >, we may use Theem to coclude that a, a; c c F F, c + a ; c; c; ΓcΓ, c + a c a + c Γ Γ as desired Problem 4 Obtai the result, b; F c; c b c Solutio 4 Cosider F, b; c; At oce, if Rc b >, F, b; c; ΓcΓc b + Γc + Γc b c b c Actually the coditio Rc b > is ot ecessary because of the termiatio of the series ivolved Problem 5 Obtai the result, a + ; F c; Solutio 5, a + ; F c; + a c c By Exercise 9, Chapter, if c a is oitegral, Hece,, a + ; F a; Γ α Γ α ΓcΓc a Γc + Γc a α c + a c, 4x x

27 SOLUTIONS TO RAINVILLE S SPECIAL FUNCTIONS 96 7 as desired Problem 6 Show that b ; F a; Solutio 6 F, b ; a; Of course a opositive iteger, as usual a + b a a + b ΓaΓa + b + Γa + Γa + b + a + b a a + b Problem 7 Prove that if g F, α; + α ; ad α is ot a iteger, the g f, g Solutio 7 Let g F, α; + α ; The k α k α k α k g k! + α k k!α Hece, compute the series α g t α k α k t k! k! α t α t t α t α Therefe, g ad g f Note: easiest to choose α iteger, ca actually do better tha that probably Problem 8 Show that d dx [ x a + F a, b; c; x ] a x x a F a +, b; c; x Solutio 8 Cosider D [x a + F a, b; c; x] D d We have dx D [x a + F a, b; c; x] D a k b k x +k+a c k k! a k + k + a + k + a k + ax k+ a b k c k k! a k a +k x k+a b k c k a k k! a + k a x k+a b k k!c k a x a F a +, b; c; x

28 8 SOLUTIONS BY SYLVESTER J PAGANO AND LEON HALL; EDITED BY TOM CUCHTA Problem 9 Use equatio, page 66, with z x, b, i which is a o-egative iteger, to coclude that, a; F x x a a, a + ; F 4x + a + ; x + a + ; Solutio 9 From o page 66 we get a + z a, a + ; [ F 4z + z F + a b; a, b; z + a b; Use z x, b to arrive at a x a, a + ; a, ; F 4x x F + a + ; + a + ; as desired Problem I Theem 3, page 65, put b γ, a γ +, 4x + x z ad thus prove that F γ, γ + ; z γ Solutio Theem 4 gives us a, b; + x a 4x F + x ab; Put b γ, a γ + ad The Now x whe z, so z F 4x + x z ] γ + z [ zx + z x + z a, a b + ; x b + ; zx 3 ± z 4z z ± z zx z z z + z x z z z[ z] z + z x ],

29 SOLUTIONS TO RAINVILLE S SPECIAL FUNCTIONS 96 9 Thus ad The we obtai x z + z + x 4x + x 4 z + z + z + z 4 z, a check Now with b γ, a γ +, Theem 4 yields γ γ + + F, γ; z γ; Sice x z z + z ad + x + z, x 4 z + z Thus we have γ, γ + F ; γ+ z + z γ; z + z γ +, ; F x γ + ; ; F x ; x + z γ, as defied Now we use Theem 3 to see that γ, γ + F ; γ, γ z F ; z γ; γ; so that we also get as desired F γ, γ ; z γ; γ +, z z

30 3 SOLUTIONS BY SYLVESTER J PAGANO AND LEON HALL; EDITED BY TOM CUCHTA Problem Use Theem 6 to show that a, a; x c F x x a c F c; Solutio By Theem 6, a, a; x c F x a; which we wished to obtai c a, c a + ; 4xx x c; c a F, c + a ; 4x x c; c a F, c + a ; x c; x a c c a, c c a + ; F c a x a c, c a + ; F Problem I the differetial equatio 3, page 54, f w F a, b; c; z itroduce a ew depedet variable u by w z a u, thus obtaiig c; c; 4xx x z z u + z[c + a b z]u + ac bu Next chage the idepedet variable to x by puttig x z Show that the z equatio f u i terms of x is x x d u + [c a + c b + x]du ac bu, dx dx ad thus derive the solutio a, c b; w z a z F z c; Solutio We kow that w F a, b; c; z is a solutio of the equatio I put w z a u The z zw + [c a + b + z]w abw w z a u + a z a u, w z a u + a z a u + aa + z a u + x x,

31 SOLUTIONS TO RAINVILLE S SPECIAL FUNCTIONS 96 3 Hece the ew equatio is z zu +azu +aa+z z u+cu +ca z u a+b+zu aa+b+z z u abu, z zu +[c b a+z]u + z [a +az+ca a +ab+az ab z]u, z z u + z[c + a b z]u + ac bu Now put x z x The z z x, z, ad we use equatio x o page of IDE f the chage of variable First, dx dz z x : d x dz z 3 x3 The old equatio above may be writte d [ u dz + c z z + a b ] du ac b + z dt z z u, which the leads to the ew equatio { x 4 d u c x [ dx + x 3 x x }] du + a b x dx x x d u du + [ x { c x + a b x}] ac bu, dx dx 3x x d u + [x a b + c + x]du ac bu dx dx Now 3 is a hypergeometric equatio with parameters γ c, α + beta + a b + c +, αβ ac b Hece α a, β c b, γ c Oe solutio of 3 is u F a, c b; c; x, b x3 ac u, x so oe solutio of equatio is W z a F a, c b; c; z z Problem 3 Use the result of Exercise ad the method of Sectio 4 to prove Theem Solutio 3 We kow that i the regio i commo to z < ad z z <, there is a relatio a, c b; z a z F AF a, b; c; z+bz c F a+ c, b+ c; z c; z z c;

32 3 SOLUTIONS BY SYLVESTER J PAGANO AND LEON HALL; EDITED BY TOM CUCHTA Sice c is either zero a egative iteger, the last term is ot aalytic at z Hece B The use z to obtai A, so A Hece a, c b; F a; b; c; z z a z F z c; Problem 4 Prove Theem 3 by the method suggested by Exercises ad 3 Solutio 4 Solutio by Leo Hall Note that the first two parameters i F a, b; c; z are iterchageable, so results ivolvig oe of them also apply to the other By Exercise, Let w z z F a, b; c; z z a F so this becomes a, c b; c; z z F a, b; c; z z a F a, c b; c; w Agai, by Exercise, applied to the secod parameter, F a, b; c; z z a w c b w F c a, c b; c; w But w z, ad w z, so w as desired F a, b; c; z z a z c b F c a, c b; c, z z c a b F c a, c b; c; z Problem 5 Use the method of Sectio 39 to prove that if both z < ad z <, ad if a, b, c are suitably restricted, a, b; a, b; F z ΓcΓc a b Γc aγc b F z c; + b + c; c a, c b; ΓcΓa + b c zc a b + F z ΓaΓb c a b + ; Solutio 5 Solutio by Leo Hall We deote the hypergeometric differetial equatio: z zw z + [c a + b + z]w z abwz by HGDE If we make the chage of variable z y, the HGDE becomes y yw y + [c a + b + y]w y abwy where c a + b + c Thus, two liearly idepedet solutios are F a, b; c ; y ad y c F a + c, b + c ; c ; y These solutios as fuctio of z are valid i z < ad are F a, b; a + b + c; z

33 SOLUTIONS TO RAINVILLE S SPECIAL FUNCTIONS ad z c a b F c a, c b; c a b + ; z Thus, i the regio D where both z < ad z <, F a, b; c; z AF a, b; a+b+ c; z+b z c a b F c a, c b; c a b+; z f some costats A ad B Assume Rec a b > ad c a egative iteger ad let z iside the regio D to get F a, b; c; A + B Thus, by Theem 8, page 49, we get A ΓcΓc a b Γc aγc b Now, let z iside the regio D ad assume Re c > ad either a+b+ c c a b + is zero a egative iteger The ad we get AF a, b; a + b + c; + BF c a, c b; c a b + ; B Agai usig Theem 8, this becomes AF a, b; a + b + c; F c a, c b; c a b + ; B ΓcΓc a bγa+b+ cγ c Γc aγc bγb+ cγa+ c Γc a b+γ c Γ bγ a By Exercise 5, page 3 the umerat is equal to Hece, Γ c Gammac Γc a bγa + b + c ΓaΓ aγbγ b Γ cγc Γc a bγa + b + c B ΓaΓbΓc a b + Γ c Γc cγ c c a + b cγa + b c ΓaΓbc a bγc a bγ c ΓcΓa + b c ΓaΓb This yields the desired fmula f F a, b; c; z i terms of the give hypergeometric fuctios of z Problem 6 I a commo otatio f the Laplace trasfm Show that L L{F t} a, b; s F s + ; e st F tdt fs; L {fs} F t a, b; z F ; z e t

34 34 SOLUTIONS BY SYLVESTER J PAGANO AND LEON HALL; EDITED BY TOM CUCHTA Solutio 6 Let A L s F a, b; z We wish to evaluate A Now + s; a b z { } A L s + s But Hece The Therefe Γ + s ss + sγ + s + Γ + sγ + s Γ + s + Γ, s; s F + s; k s k s k! + s k ss + { } L ss + k k!s + k k e kt k! e t a b z e t A F a, b; ; z e t There are may other ways of doig Exercise 6 Probably the easiest, but most udesirable, is to wk from right to left i the result to be proved It is hard to see ay chace f discoverig the relatio that way Problem 7 With that otatio of Exercise 6 show that +, 3 + ; L{t si at} aγ + s + F 3 ; a s Solutio 7 We wish to obtai the Laplace Trasfm of t si at Now t k a k+ t +k+ si at k +! ad L {t m } Γm + s m+

35 SOLUTIONS TO RAINVILLE S SPECIAL FUNCTIONS Hece L {t k a k+ Γ + k + si at} k +!s +k+ aγ + k + k a k s + k s k aγ + k k s + 3 k! s k k +, + 3 ; Problem 8 Obtai the results aγ + s + F 3 ; log + x xf, ; ; x, arcsi x xf, ; 3 ; x, arcta x xf, ; 3 ; x Solutio 8 Usig +! we kow that x + log + x + x x xf, ; ; x Next, start with y usig to get x y dy Thus we arrive at y x+ + arcsi x 3 xf, ; 3 ; x x+ a s a k k

36 36 SOLUTIONS BY SYLVESTER J PAGANO AND LEON HALL; EDITED BY TOM CUCHTA Fially fm to obtai x + y dy + y x + + y 3 arcta x xf, ; 3 ; x Problem 9 The complete elliptic itegral of the first kid is K Show that K π F, ; ; k Solutio 9 From K π K π dφ k si φ dφ k si φ π we obtai k si φdφ x+ But π Hece si φdφ B +, K π Γ + Γ Γ Γ + k π F, ; ; k Problem The complete elliptic itegral of the secod kid is π E k si θdθ Show that E π F, ; ; k π Solutio From E Hece π E π k si θdθ, we get π k si φdφ k E π F, ; ; k Problem From the cotiguous fuctio relatios -5 obtai the relatios 6-

37 SOLUTIONS TO RAINVILLE S SPECIAL FUNCTIONS a bf af a+ bf b+, a c + F af a+ c F c, 3 [a + b cz]f a zf a+ c c ac bzf c+, 4 zf F a c c bzf c+, 5 zf F b c c azf c+, 6 [a c + b az]f a zf a+ c af a, 7 a + b cf a zf a+ c bf b, 8 c a bf c af a b zf b+, 9 b a zf c af a b zf b+, [ z + c b z]f c af a c zf c, [b c + a bz]f b zf b+ c bf b, [b + a cz]f b zf b+ c c ac bzf c+, 3 b c + F bf b+ c F c, 4 [ b + c a z]f c bf b c zf c, 5 [c + a + b + cz]f c zf c c c ac bzf c+ Solutio From 3 ad 4 we get [a + b cz c a z]f a zf a+ c af a, 6 [a c + b az]f a zf a+ c af a From 3 ad 5 we get [a + b cz c b z]f a zf a+ c bf b, From ad 6 we get 7 [a + b c]f a zf a+ c bf b [a b z a + c b az]f c af a b zf b+, From 6 ad 7 we get 8 [c a b]f c af a b zf b+ 9 b a zf c af a c bf b Use ad 6 to obtai [a c + z a + c b az]f c af a c zf c, [ a + c b z]f c af a c zf c From ad 7 we get [a + b c a b z]f b zf b+ c bf b, [b c + bz]f b zf b+ c bf b, which checks with 6 Easier method: i 6 iterchage a ad b From ad 3 we get [a + b cz a b z]f b zf b+ c c ac bzf c+,

38 38 SOLUTIONS BY SYLVESTER J PAGANO AND LEON HALL; EDITED BY TOM CUCHTA [b + a cz]f b zf b+ c c ac bzf c+, me easily foud by chagig b to a ad a to b i 3 I iterchage a ad b to get 3 b c + F bf b+ c F c I iterchage a ad b to get 4 [ b + c a z]f c bf b c zf c From ad 3 we get [a + b cz a c + z]f c zf c c c ac bzf c+, 5 [c +a+b c+z]f c zf c c c ac bzf c+ Problem The otatio used i Exercise ad i Sectio 33 is ofte exteded as i the examples F a, b+ F a, b + ; c; z, F b+, c+ F a, b + ; c + ; z Use the relatios 4 ad 5 of Exercise to obtai F a F b + c b azf c+ ad from it, by chagig b to b + to obtai the relatio c bf c af a, b+ + a b zf b+, c bf a, b; c; z c af a, b + ; c; z + a b zf a, b + ; c; z, aother relatio we wish to use i Chapter 6 Solutio From Exercise equatio 4 ad 5 we get From the above we get Now replace b by b + to write zf F a c c bzf c+, zf F b c c azf c+ F a F b + c b azf c+ F a, b+ F + c b + azf b+, c+, F a, b; c; z F a, b + ; c; z + c b + azf a, b + ; c + ; z Problem 3 I equatio 9 of Exercise shift b to b + to obtai the relatio c bf c af c, b+ + a b zf b+, c bf a, b; c; z c af a, b + ; c; z + a b zf a, b + ; c; z, aother relatio we wish to use i Chapter 6

39 SOLUTIONS TO RAINVILLE S SPECIAL FUNCTIONS Solutio 3 Equatio 9 of Exercise is from which we may write b a zf c af a c bf b b + a zf b+ c af a, b+ c b F, c b F a, b; c; z c af a, b + ; c; z + a b zf a, b + ; c; z Theem 7 If z is oitegral, 6 Results from Chapter 5 used ˆˆ ΓzΓ z π si πz Theem 8 Dixo s Theem The follow is a idettiy of a, b, ad c are so restricted that each of the fuctios ivolved exists: a, b, c; 3F + a b, + a c; Bα, β α+β p+k+sf q+k+s Γ + aγ + a bγ + a cγ + a b c Γ + aγ + a bγ + a cγ + a b c Theem 9 If Rα >, Rβ >, ad if k ad s are positive itegers, the isdie the regio of covergece of the resultat series t a,, a p ; x α t x β pf q cx k t x s b,, b q ; a,, a p, α k, α +,, α + k, β k k s, β + s Theem If z is oitegral, ΓzΓ z,, β + s ; s b,, b q, α + β k + s, α + β +,, α + β + k + s ; k + s k + s π si πz Theem Dixo s Theem The follow is a idettiy of a, b, ad c are so restricted that each of the fuctios ivolved exists: a, b, c; 3F Γ + aγ + a bγ + a cγ + a b c + a b, + a c; Γ + aγ + a bγ + a cγ + a b c Theem If Rα >, Rβ >, ad if k ad s are positive itegers, the isdie the regio of covergece of the resultat series t a,, a p ; x α t x β pf q cx k t x s b,, b q ; k k s s ct k+s k + s k+s

40 4 SOLUTIONS BY SYLVESTER J PAGANO AND LEON HALL; EDITED BY TOM CUCHTA Bα, β α+β p+k+sf q+k+s Problem Show that ; ; F x F a; b; Solutio Cosider the product F ; a; x F ; b; x a,, a p, α k, α + k,, α + k k, β s, β + s,, β + s ; s b,, b q, α + β k + s, α + β +,, α + β + k + s ; k + s k + s 7 Chapter 5 Solutios x ˆˆ F 3 a + b, a + b ; 4x a, b, a + b ; x +k a k b k!, x a k b k k! k! b k k x a k k! b F, b ; x b a; We the use the result i Exercise 6, page 69, to get a + b x F ; a; x F ; b; x b a a + b a + b a b a + b a + b x a + b F 3 F, a + b ; 4x a, b, a + b ; Problem Show that t x t x [ x t x ] dx πt F Solutio We use theem o the itegral A t x t x [ x t x ] dx 4, 3 4 ; t 4 ; 6 k k s s ct k+s k + s k+s

41 SOLUTIONS TO RAINVILLE S SPECIAL FUNCTIONS F Now A t x t x F ; ; x t x dx, so i Theem we pput alpha 3, β, p, q, a, c, k, s The result is as desired 3 A B, t 5 F 4 3 Γ Γ A t F Γ, 3 4, 5 4, 4, 3 4 ; t + 4, 3 4 ; t 4, ; 4, 3 4, 4 4, 5 4 ; π t F Problem 3 With the aid of Theem, show that ad that Γ + a Γ + a cos πaγ a Γ a Γ + a b Γ + a b si πb aγb a si πb aγb a, 4, 3 4 ; t 4 Thus put Dixo s theem Theem i the fm a, b, c; 3F cos πa si πb a Γ aγb aγ + a cγ + a b c si πb a + a b, + a c; Γ aγb aγ + a cγ + a b c Solutio 3 We first ote that, sice ΓzΓ z π si πz, Γ + a Γ a aγ a γ a Γ + a Γ a aγaγ a si πa π π si πa cos π a si π a si π a cos πa a, b, c; cos πa b si π a Γ aγ b a si πb a + a b, + a c; Γ a Γb aγ ; 6, Γ + a cγ + a b c + a c Γ + a b c,

42 4 SOLUTIONS BY SYLVESTER J PAGANO AND LEON HALL; EDITED BY TOM CUCHTA 3F, α, β ; α, β; so log as a, a, b a, b a are ot itegers ad the gamma fuctios ivolved have o poles But ow we have arrived at a idetity f o-itegral values of certai parameters which has the property that both members are well-behaved if a is a egative iteger zero It follows that the idetity cotiues to be valid f a, a oegative iteger If a +, a odd egativer iteger, cos πa Problem 4 Use the result i Exercise 3 to show that if is a o-egative iteger,, α, β ; 3F α, β;!α β α α β Solutio 4 If a i the idetity of Exercise 3 above, ad if we chose b α, c β, we obtai as desired cos π si πα + Γ + Γα + Γ + β + Γ α + β + si πα + Γ + Γα + Γ + β + Γ α + β + si πα!α ΓβΓβ α + si πα α Γβ + Γβ α!α β α α β, Problem 5 With the aid of the fmula i Exercise 4 prove Ramauja s theem: F α; β; x F α; β; Solutio 5 Cosider the product F α; β; x F α; β; x α, β α; x F 3 β, β, β + ; k α k α k x β k β k k! k! k α k β k α x k!β k α k β, α, β ; 3F β, α ; x 4 α x β Sice the product of the two F s is a eve fuctio of x, we may coclude that, α, β ; 3F β, α ;

43 SOLUTIONS TO RAINVILLE S SPECIAL FUNCTIONS ad that, α, β ; F α; β; x F α; β; x 3F β, α ;!α β α α x, α β!β α x!β by Exercise 4 Hece we get Ramauja s theem α, β α; F α; β; x F α; β; x F 3 β, β, β + ; x 4 Problem 6 Let γ 3 F, a, b ; a, b; Use the result i Exercise 3 to show that γ + ad γ!a + b 3 a b a + b Solutio 6 From Exercise 3, we get α, β, γ; πa cos 3F si πβ α Γ αγβ α Γ + α γγ + α β γ si πβ α Γ + α β, + α γ; α Γβ αγ + α γγ + α β γ Cosider γ 3 F, a, b ; a, b; We wish to use α, but π cos f odd Hece γ + ad γ 3 F, a, b ; a, b; Therefe i the result from Exercise?? we put α, β a, γ b, ad thus obtai cosπ si π a Γ + Γ a ΓbΓ + a + b + 3 γ si π a Γ + Γ aγb + Γ + a + b + cos π si π a! a + b 3, si π a a b a + b so that γ!a + b 3 a b a + b Problem 7 With the aid of the result i Exercise 6 show that 3 a + b, 3 a + b, 3 a + b + ; F ; a, b; t F ; a, b; t 3 F 8 7t 64 a, b, a +, b, b +, a + b, a + b;

44 44 SOLUTIONS BY SYLVESTER J PAGANO AND LEON HALL; EDITED BY TOM CUCHTA Solutio 7 Let us cosider the product ψt F ; a, b; t F ; a, b; t γ k a k b k k!a k b k t a b γ t!a b, t a b i terms of the γ of Exercise 6 above We already kew that γ + which checks with the fact that Ψt is a eve fuctio of t Sice, by Exercise 6, we have ψt γ!a + b 3 a b a + b, a + b 3 t a b a a + b 3 3 a+b a b a a+ F, a, b; x F ; a, b; x 3 F 8 a+b a+b t b b+ a+b a+b, 3 a + b, 3 a + b, 3 a + b + ; 7t 64 a, b, a, a +, b, b +, a + b, a + b ; Problem 8 Prove that k γ b c k γ b k γ c k x k k, b, c; 3F k! k!γ k γ + b k, γ + c k; x γ b γ c x 3F γ, +, γ ; γ + b, γ + c ; 4x x ad ote the special case γ b + c, Whipple s theem

45 SOLUTIONS TO RAINVILLE S SPECIAL FUNCTIONS γ b, γ c; Solutio 8 Let ψ F t γ; x + xt The γ b γ c t [ x + xt] ψ γ γ b γ c x k x k t +k k! k!γ [ ] γ b k γ c k x k x k t k! k!γ k [ ] γ k k x k γ b γ c x t k! γ + c k γ + b k x k, γ ψ 3F,, γ ; γ + b, γ + c ; 4x x γ b γ c x t γ But also, sice t x + xɛ t + xt, b, c; Ψ t b+c γ + xt b+c γ F γ; Hece t x + xt ψ t b+c γ + xt b+c γ b c t [ x t] γ t b+c + xt b+c γ k b c x k t k t, k! k!γ ψ t b+c γ + xt b+c γ k b +k c +k x k t x t +k k!γ +k, b + k, c + k; + xt b+c γ t b+c γ F γ + k; Now F b + k, c + k; γ + k; t t γ b c k F γ b, γ c; γ + k; t

46 46 SOLUTIONS BY SYLVESTER J PAGANO AND LEON HALL; EDITED BY TOM CUCHTA Therefe γ b, γ c; ψ + xt b+c γ F t k b k c k xt k k!γ γ + k; k + xt b+c γ k b k c k γ b γ c x k t +k k!γ +k, + xt b+c γ s b s c s γ b s γ c s x s t s! s!γ s γ b c x t, b, c; 3F γ + b, γ + c ; ψ k, b, c; 3F γ + c b, γ + c x; x x, k γ b c k γ b k γ c k x k t k!γ k k! By equatig coefficiets of t i the two expasios we obtai the desired idetity Exercises 9- below use the otatio of the Laplace trasfm as i Exercise 6, page 7 Problem 9 Show that a,, a p ; L tc pf q b,, b q ; zt Γ + c s +c p+ F q + c, a,, a p ; b,, b q ; z s Solutio 9 We kow that L {t m } L tc pf q a,, a p ; b,, b q ; zt Γm + s m+ The, a a p z b b q L {t+k } a a p z + c Γ + c b b p s +k+ + c, a,, a p ; F + c z s +c p+f q 5 b,, b q ; Problem Show that a L,, a p ; F q+ s p s +, b,, b q ; z p F q+ a,, a p ;, b,, b q ; z e Solutio I Chapter 4, Exercise 6 we foud that { } L e t ss +

47 SOLUTIONS TO RAINVILLE S SPECIAL FUNCTIONS It follows that a L,, a p ; s p F q+ + s, b,, b q ; Problem Show that { L s k } k + ; s z k+ F ; z pf q+ a,, a p ;, b,, b q ; Solutio Cosider s k s z k+ s 4 k + ; 5 k+ s F ; By Exercise 9 with c, k + ; L Γ s F ; Problem Show that d dz p F q a,, a p ; b,, b q ; Solutio a d,, a p ; dz p F q b,, b q ; z z z s p m q j zt z 5, k + ; L Γ s F ; k + ; t F zt ; k + ; F zt ; a m b j a +,, a p + ; pf q b +,, b q + ; a a p z b b q! a + a + z b + b q + a a a +,, a p + ; p pf q b b q b +,, b q + ; z e t z z s Problem 3 I Exercise 9 page 7, we foud that the complete elliptic itegral of the first kid is give by Kk π F, ; ; k z

48 48 SOLUTIONS BY SYLVESTER J PAGANO AND LEON HALL; EDITED BY TOM CUCHTA Show that t K xt xdx π arcsi t Solutio 3 We are give that Kk π F, ; ; k Now cosider A t K xt xdx By the itegral of Sectio 56, ie Theem 57 with α, β, k, s, etc t A π F, ; ; xt x dx π B,,,, ; t 4F 3 t,, 3 ; π ΓΓ t F Γ, ; 3 ; t 4 t π F, ; 3 t ; t π arcsi, by Exercise 8, Chapter 4 8 Chapter 6 Solutios Problem By collectig powers of x i the summatio o the left, show that ˆˆ J + x x J ydy

49 SOLUTIONS TO RAINVILLE S SPECIAL FUNCTIONS Solutio x J k+ x +k+ + k +! k, k x + k! + k +! k x + + k +!, ; F x + +! + ; Γ + Γ + x + Γ + Γ + +! x + + x y dy x J ydy Problem Put the equatio of Theem 39, page 3, ito the fm [ ] A exp zt t J z + J z[t + t ] Use equatio A with t i to coclude that cos z J z + k J k z, Solutio We kow that [ ] exp zt t si z k k J k+ z J zt J zt + J z + Now J z J z Hece [ z exp t ] J ze + J z + J zt t t J z + J z[t + t ] Now use t i The i i i i ad [ π exp i ] expiz cos z + i si z i J zt

50 5 SOLUTIONS BY SYLVESTER J PAGANO AND LEON HALL; EDITED BY TOM CUCHTA Therefe cos z + i si z J z + But J z J z, so we have cos z i si z J z + i J z i J z ote: above argumet ca be doe me easily by equatig eve ad odd fuctio of z Thus we get cos z J z + i d z J z + J z ad i si z si z i + J + z, J + z Problem 3 Use t e iθ i equatio A of Exercise to obtai the results cosz siθ J z + J k z coskθ, siz siθ k J k+ z sik + θ Solutio 3 Put t e iθ The t + t e iθ + N e iθ ad [ z exp t ] expiz siθ cosz siθ + i siz siθ t [ z Also e iθ + e iθ [+ ] cosθ+[ ] siθ From exp t ] t J z + J z[t + 9 t ] we thus obtai cosz si θ+i siz si θ J z+ J z[+ cosθ+ siθ] Now equate eve fuctio of z o the two sides, the odd fuctios of z o the two sides to get cosz si θ J z + J k z coskθ ad siz si θ k J k+ z sik + θ

51 SOLUTIONS TO RAINVILLE S SPECIAL FUNCTIONS 96 5 Problem 4 Use Bessel s itegral, page 4, to obtai f itegral i the relatios B [ + ]J z π π π cosθ cosz siθdθ, C [ ]J z siθ siz siθdθ π With the aid of B ad C show that f itegral k, J k z π J k+ z π π π Solutio 4 We kow that The J z π π Now chage z to z to get J z π We the obtai B π π coskθ cosz sithetaadθ, sik + θ siz siθdθ, cosk + θ cosz siθdθ, J z π sikθ siz siθdθ π cosθ cosz si θdθ + π π cosθ z si θdθ π cosθ cosz si θdθ π [ + ]J z π π π siθ siz si θdθ π siθ siz si θdθ cosθ cosz si θdθ C [ ]J z siθ siz si θdθ π Use B with k ad C with k + to obtai J k z π π π coskθ cosz si θdθ, J k+ z sik + θ siz si θdθ π Use B with k + ad C with k to obtai π π cosk + θ cosz si θdθ, sikθ siz si θdθ Problem 5 Expad cosz siθ ad siz siθ i Fourier series over the iterval π < θ < π Thus use Exercise 4 to obtai i aother way the expasios i Exercise 3

52 5 SOLUTIONS BY SYLVESTER J PAGANO AND LEON HALL; EDITED BY TOM CUCHTA Solutio 5 I the iterval π < θ < π, the Fourier approximatio of fθ is fθ a + a cosθ + b siθ, i which a π π π π fθ cosθdθ, b fθ siθdθ π π Cosider first fθ cosz si θ, a eve fuctio of θ F this fuctio a π π cosθ cosz si θdθ, b By the results i Exercise 4 we obtai a k+, a k J k z Hece cosz si θ J z + J k z coskθ, π < θ < π Next, let fθ siz si θ, a odd fuctio of θ F this fuctio, a, b π k π siθ siz si θdθ From Exercise 4 we get b k, b k+ J k+ z Hece siz si θ J k+ z sik + θ, π < θ < π [ ] Problem 6 I the product of exp xt t the coefficiet of t ad thus show that J x + J x F real x coclude that J x ad J x f Solutio 6 We kow that ad thus that exp [ γ t ] J xt t [ by exp ] xt t, obtai [ exp γ t ] k J k xj xt +k t The coefficiet of t o the right is k J xj x, from which we obtai J x + J xj x

53 SOLUTIONS TO RAINVILLE S SPECIAL FUNCTIONS But J x J x Hece It follows at oce, f real x, that J x + Jx J x, F x, Problem 7 Use Bessel s itegral to show that J x f real x ad itegral Solutio 7 Bessel s itegral is J x π π cosθ x si θdθ F real x ad, cosθ x si θ Hece J x π π dθ Problem 8 By iteratio of equatio 8, page, show that m m dm dz m J z m k C m,k J +m k z, where C m,k is the biomial coefficiet Solutio 8 We have, with D d, from 8, page, dz The Let us use iductio Assume DJ z J z J + z D J z DJ z DJ + z J z J z J z + J + z J z J z + J + z m D m J z m m k C m,k J +m k z, as we k ow is true f m, The m m+ D m+ J z m k C m,k [J +m k z J +m k+ z] m+ m m k+ C m,k J +m+ k z + m+ k C m,k J +m+ k z k m m+ k [C m,k + C m,k ]J +m+ k z + m+ J +m+ z + J m z k

54 54 SOLUTIONS BY SYLVESTER J PAGANO AND LEON HALL; EDITED BY TOM CUCHTA Now C m,k + C m,k C m+,k Pascal s triagle, so that usig the last two terms also, m+ m+ D m+ J z m+ k C m+,k J +m+ k z, which completes the iductio Problem 9 Use the result i Exercise, page 5, to obtai the probduct of two Bessel fuctios of equal argumet Solutio 9 We kow already that a + b F ; a; x F ; b; x F, a + b ; 3 a, b, a + b ; The z z m J zj m z Γ + Γm + F ; + ; z 4 F ; m + ; z 4 z +m + m +, + m + ; Γ + Γm + F 3 z +, m +, + m + ; Problem Start with the power series f J z ad use the fm, page 8, of the Beta fuctio to arrive at the equatio f R > J z z Γ Γ + Solutio We kow that J z π k z k+ k+ k!γk + + si φ cosz cos φdφ, 4x k k!γk + + z k+ k Also k Γk + + Γ k + Γ + Γ Γk + + Γ + B k +, + Γ Γ + π Γ Γ + cos k φ si φdφ

55 SOLUTIONS TO RAINVILLE S SPECIAL FUNCTIONS Therefe J z Γ Γ z Γ z Γ Problem Use the property + + π π si φ cosz cos φdφ si φ cosz cos φdφ d dx F ; a; u du a dx F ; a + ; u to obtai the differetial recurrece relatio 6 of Sectio 6 Solutio We kow that d dx F ; a; u du a dx F ; a + ; u Sice we obtai z J z Γ + F d dz [z J z] which yields 6 of Sectio 6 Problem Expad Γ z z Γ + F z J + z, z ; + ; z 4 F ; + ; z 4 ; + ; z 4 F ; + α; xt t 4 i a series of powers of x ad thus arrive at the result t x t α J α t xt J α+ tx

56 56 SOLUTIONS BY SYLVESTER J PAGANO AND LEON HALL; EDITED BY TOM CUCHTA Solutio Cosider F [ ; + α; ] 4 xt t We obtai F Now ; + α; so we obtai F xt t 4 t x t + α α+ t J α+ t Γα + + F ; + α; xt t 4 k k t +k w +k + α +k k! k t k t x k k! + α + k + α t F ; + α + ; t 4 ; + α + ; t, 4 α t J +α tx Γ + α x Γ + α Γ + α + α t xt α t J +α tx Γ + αj α t xt Γ + α, α t x Jα t t J +α tx xt Problem 3 Use the realtios 3 ad 6 of Sectio 6 to prove that: F real x, betwee ay two cosecutive zeros of x F x, there lies oe ad oly oe zero of x F + x Solutio 3 We are give that d dx [x J x] x J x, d [ x J x ] x J + x dx We kow that J x has exactly zeros at x Let the others we have proved there are ay o the axis of reals be at α,, α,, The curve y x J x has its real zeros oly at the α s By Rolle s theem we see that the zeros β,, β,, of y x J + x are such that a odd umber of them lie betwee each two cosecutive α s The curve y x + J + x

57 SOLUTIONS TO RAINVILLE S SPECIAL FUNCTIONS has its zeros at x ad at the β s But so the α s lie betwee cosecutive β s y x + J x, Problem 4 F the fuctio I z of Sectio 65 obtai the followig properties by usig the methods, but ot the results, of this chapter: Solutio 4 Solutio by Leo Hall zi z zi z I z, zi z zi + z + I z, 3I z I z + I + z, 4I z z[i z I + z] I z i J iz z Γ + F ; + ; z 4 z z Γ + + k k! 4 z k+ k+ k!γk + + k so + k + + k +, So, as i the method of Sectio 6, which is equivalet to which is d dz [z I z] Γ + + k Γk + + z z k+ k+ k!γ + k z z I z, z k+ k+ k!γ + k z k+ k+ k!γk + + z I z + z I z z I z, zi z zi z I z,

58 58 SOLUTIONS BY SYLVESTER J PAGANO AND LEON HALL; EDITED BY TOM CUCHTA ad Similarly, d dz [z I z] which is Addig ad : d dz k z k k+ k!γk + + z k k+ k!γk + + z k+ k++ k!γk z I + z, z I z z I z z I + z, zi z I z + zi + z, zi z zi z + zi + z I z I z + I + z, which is 3 Equatig the right sides of ad : which is 4 zi z I z zi + z + I z, I z z[i z I + z], Problem 5 Show that I z is oe solutio of the equatio z w + zw z + w Solutio 5 Solutio by Leo Hall Because I is a F fuctio times z, we kow from Sectio 46 that u F ; b; y is a solutio of y d y dy + bdu dy u, ad so F ; + ; z is a solutio of 4 z d u + + du zu dz dz Thus is w z u, I z will be a solutio of z + w z w + + z w + + [z w z w] z + w z [zw w [ + z + + z + z]w] z w + zw + z w

59 SOLUTIONS TO RAINVILLE S SPECIAL FUNCTIONS Problem 6 Show that, f Re >, I z Γ Γ + z π si φ coshz cosφdφ Solutio 6 Solutio by Leo Hall F ot a egative iteger, ad usig the result of Problem, I z i iz Γ Γ + I z i J iz, π si φ coshz cos φdφ The powers of i cacel, ad because cosiw cosh w we get I z z Γ Γ + π si φ coshz cos φdφ f Re > as desired Problem 7 F egative itegral defie I z I z, thus completig the defiitio i Sectio 65 Show that I z N I z ad that [ ] exp zt + t I zt Solutio 7 Solutio by Leo Hall We have I zt I zt + I zt + I + zt + I zt Now proceed exactly as i the proof of Theem 39, the oly differece beig that I z ivolves F ; + ; z whereas J z ivolves F ; + ; z, to 4 4 get [ ] I zt exp zt + t Problem 8 Use the itegral evaluated i Sectio 56 to show that t [ xt x] J xt xdx πt + t J+ Solutio 8 t [ xt x] J xt xdx Γ + t x t x F ; + ; xz x 4 dx

60 6 SOLUTIONS BY SYLVESTER J PAGANO AND LEON HALL; EDITED BY TOM CUCHTA t Now use Theem 37 with α +, β +, p, q, b +, c ][4 k, s The result is, +, + ; [ xt x] J xt xdx Γ + B +, + t+ F +, +, + 3 ; ; Γ + Γ + Γ + Γ + t+ F t ; Γ + Γ ; t + t + r t Γ + Γ + 3 F ; 3 Γ + Γ + + t + Γ + z + z ΓzΓ By Legedre s duplicatio fmula, Γz π 3 + Γ + Γ + Γ + π J + t, we get t 4 4 Hece t [ xt x] J xt xdx + t + π t + J + t πt + J + Problem 9 By the method of Exercise 8 show that ad, i geeral, that Solutio 9 Cosider x siα xdx πα J α, x c x J α xdx Γc x siα xdx c J +cα α We kow that si z z F ; 3 ; z 4

61 SOLUTIONS TO RAINVILLE S SPECIAL FUNCTIONS 96 6 Hece x siα xdx α x x F ; 3 ; x α dx 4 We ow use Theem 3 with t, α 3, β 3, p, q, b 3, c α, k, s We get 4 Now let us tur to x siα xdx αb 3, 3 x c x J α xdx α Γ + F 3 3 Γ Γ α Γ3 3 ; α 3, 3 ; F ; 3; α 4 4 x c x F ; + ; α x dx 4 ad use Theem 37 with α +, β c, p, q, b +, t, c α, k, s The result is 4 x t x t J α xdx α Γ + as desired Problem Show that t Solutio Cosider B +, cf + ; +, c + + ; Γ + Γc α Γ + Γ + c + F α c Γc J+c α, exp[ xt x]i [xt x]dx t t exp β dβ exp[ xt x]i [xt x]dx Now ; c + + ; α 4 exp[ xt x]i [xt x] exp[ xt x] F [ ; ; x x ] I Kummer s secod fmula we have F a; a; z e z F ; + ; z 4 α 4 Use a ad z xt x to get exp[ xt x]i [xt x] F ; ; 4xt x

62 6 SOLUTIONS BY SYLVESTER J PAGANO AND LEON HALL; EDITED BY TOM CUCHTA The, t exp[ xt x]i [xt x]dx t F ; ; 4xt x dx, to which we may apply Theem 37 with α, β, p, q, a, b, c 4, k, s We thus get t exp[ xt x]i [xt x]dx Problem Show that t Solutio t B, tf t F ; 3 ; t,, ; 4 t 4,, 3 ; 3 t + + t β dβ t exp β dβ [xt s] exp[4xt x]dx π exp t t + I t t [xt x] exp[4xt x]dx x t x F ; ; 4xt xdx We use Theem 37 with α, β, p q, k, s, c 4 : t [xt x] exp[4xt x]dx B, t F, ; 4t, ; F ; ; t Γ Γ Γ π exp t F ; t; t4 6 t t π exp I 4

63 SOLUTIONS TO RAINVILLE S SPECIAL FUNCTIONS Problem Solutio by Leo Hall Obtai Neuma s expasio Solutio Let z F z The { d z J+k z} dz + k + k!j +k z, k! + k + k! k! z + k z J+k z z J +k z z J+k z z [ J +k z ] z J +kz [ J +k z + kj +k + k z J +k z z Usig 8, Sectio 6, ad, Sectio 6, ad simplifyig gives { d z J+k z} z dz + k [kj +k z + kj +k+ z] So [ z F z k + k! k! J +k z ] + k! J +k+ z k! Note that the k term i the first series is zero, ad so shiftig the idex i the first series yields the secod series Thus, F z, makig F z costat From the structure of the Bessel fuctios, we see that z z F ad so F z, from which Neuma s expasio immediately follows Problem 3 Solutio by Leo Hall Prove Theem 39, page 3, by fmig the product of the series f exp zt ad the series f exp zt Solutio 3 Theem 39 is: f t ad f all fiite z, [ z exp t ] J zt t We kow ad exp z t z exp t m z t m z m m t m m! z! t ]