The basics of Bessel functions

The basics of Bessel functions Peter Haggstrom mathsatbondibeach@gmail.com https://gotohaggstrom.com February 15, 2023 1 Historical background to Bessel functions Bessel functions are one of the many pinnacles of 19th century mathematical physics, especially astronomy. Friedrich Bessel was clearly a giant of his time and according to Wikipedia, Gauss recommended him for an honorary PhD from the University of Gottingen [1]. Apparently he was largely self-taught. As you get into this article you will quickly see that it is steeped in essentially cutting edge 19th century mathematics that modern courses have taken the edge off. A complete study involves complex analysis. asymptotic theory and more and so I will only give glimpses of these dimensions. Many of the techniques used in current undergraduate mathematical physics are artefacts of the 18th and 19th centuries which have been simplified, improved and then commoditised. The tentacles of Bessel’s work, like that of Fourier, extend into vast areas of mathematics and science. Apart from astronomy, you will find Bessel functions in quantum mechanics and the structure of DNA (see the Appendix) as well as many areas of engineering. This article does not pretend to cover the full scope of issues relating to Bessel functions and their applications. It is an enormous field requiring exposure to the ”art” of the subject. But you have to start somewhere... I first came across Bessel functions in an undergraduate Fourier theory course in the early 1970s. The textbook we used was by Churchill [2] and was a very good mix of mathematical physics/engineering and analysis. It was an influential textbook in the 1970s and Fischer Black and Myron Scholes used it in their famous paper on the option pricing model that bears their name. They simply quoted one of Churchill’s results to get the solution to the critical partial differential equation they had constructed. Churchill’s book introduced Bessel functions by referring to 1 how the Laplacian transforms in cylindrical coordinates so that when the process of separation of variables is applied you get an equation of this form: dy d2 y + ρ + (λ2 ρ2 − ν 2 )y = 0 2 dρ dρ After the substitution x = λρ is made Bessel’s equation becomes: ρ2 x2 dy(x) d2 y(x) +x + (x2 − n2 )y(x) = 0 2 dx dx (1) (2) where attention has been limited to ν = n where n = 0, 1, 2, . . . because there will be a solution which is analytic for all values of x including the origin ie a power series that is convergent for all x. I have set out in detail in the Appendix how this is actually done. It is a basic building block of all mathematical physics courses. The method of Frobenius is then used. This involves assuming a power series solution of the form: y=x p ∞ X j aj x = j=0 ∞ X aj xp+j (3) j=0 After some algebra (see the Appendix) one arrives at the following form for Bessel’s function of the first kind of order n: Jn (x) = ∞ X j=0 (−1)j  x n+2j j!(n + j)! 2 (4) The series representation in (4) then forms the basis for working out a whole range of properties. It may come as a surprise but Bessel didn’t arrive at (4) via the above route or by working out differential equations of hanging chains etc. The “old school” bible on Bessel functions is a 19th century book by Gray and Mathews [3] which explains that Bessel was originally led to the discovery of his functions by the investigation of a problem connected with elliptical motion. In detail Bessel found a relationship between the eccentric anomaly of a point on an ellipse and a quantity called the mean anomaly. The other “bible” of Bessel functions is Watson [5] which systematically plumbs the depths of the subject. It is not a book for a struggling 2 undergraduate. It is mathematically dense. The relationship Bessel was interested in ultimately had this form: 2 Ar = rπ Z π 0 cos[r(φ − e sin φ)] dφ (5) where e is the eccentricity of the ellipse. As will be shown, Bessel used (5) to get (2) (see the Appendix). If you pick up Eli Stein and Rami Shakarchi’s book on Fourier analysis [4], you will find that they define the Bessel function of order n by the integral (Churchill covers this form as well ): 1 Jn (ρ) = 2π Z 2π eiρ sin θ e−inθ dθ (6) 0 Note that this looks like a Fourier transform of the function eiρ sin θ . Once you view it as suchPit ”morally” (to use one of Eli Stein’s turns of phrase) follows inθ that eiρ sin θ = ∞ and that is why Stein chose that representation. n=−∞ Jn (ρ)e You then have to extract all the same properties from (6) as you do for the series representation (4). This is an interesting and revealing process in itself. Proving that you get (4) from (6) is set out in detail and involves classical analysis. . 1.1 How Bessel arrived at the formula for the eccentric anomaly Gray and Mathews [3] (pages 3-6) set out what Bessel did and I will fill in the gaps in what follows. I actually tried to find out how Bessel did the business by looking at the German papers in which his works were published and I couldn’t find the precise approach they develop and which I reproduce below. Extracts from Bessel’s 1824 paper are set out in the Appendix. Figure 1 provides the framework for the calculations. 3 P is a point on the ellipse, of which AA′ is the major axis, S is a focus and C is the centre. The ordinate NPQ meets the auxiliary circle at Q. In astronomical parlance, the eccentric anomaly of P is the number of radians in the angle ACQ, or φ, which can also be represented as: φ=π area of sector ACQ area of semicircle AQA′ (7) The mean anomaly is defined by this relationship: µ=π area of elliptic sector ASP area of semi- ellipse AP A′ (8) The authors note that by Kepler’s second law of planetary motion, µ is proportional to the time of passage from A to P, supposing that S is the centre of attraction. They then go on to argue that by “orthogonal projection” the following relationship holds: area of ASP : area of AP A′ =area of ASQ : area of AQA′ =(ACQ − CSQ) : area of AQA′ 1 1 1 =( a2 φ − ea2 sin φ) : πa2 2 2 2 =(φ − e sin φ) : π (9) where e is the eccentricity. The principle of orthogonal projection in the context of ellipses is based on the idea of a principal circle associated with the ellipse as indicated in Figure 2. It is part of the geometric thinking set out in Newton’s “Principia” - the principle is undoubtedly there if you look hard enough. To see 4 why the relationship in (9) holds consider Figure 2 below: An ellipse can be considered as the orthogonal projection of its principal circle. In Figure 2 HP is the orthogonal projection of HQ. To see this we have to rotate the circle through an angle θ such that cos θ = ab and when we do that the projection of the circle onto the x − y plane is the ellipse. You can actually demonstrate this by getting two sheets paper one of which has the circle drawn on it. You simply use a needle to makes holes along the circumference at regular intervals and then you rotate that page above the other page. Now turn off the lights and shine a torch perpendicularly onto the circle with holes and you will see an ellipse projected on the other sheet of paper. Figure 3 shows the relationship: 5 The ellipse relationship is derived analytically from these principles in this reference [6]. We also know that the projection of an area is that area multiplied by the cosine of the angle between the two planes involved. Thus the projection of the sector ASQ on the circle is the area ASQ × cos θ = area ASQ × ab . But this equals the area of the elliptical sector ASP. Thus we have: Area sector ASQ b = Area elliptical sector ASP a (10) Area semi-circle AQA′ b ′ = Area semi- ellipse AP A a (11) Similarly we get: The relationship (9) then follows from (10) and (11). To work ou the area of ASQ it is simply the difference between the area of sector ACQ (ie 12 a2 θ ) and the area of the triangle CSQ. From Figure 1 the eccen. This means that the area of triangle tricity e is defined to be the ratio e = SC a 6 CSQ is 21 (ea)a sin φ = 12 ea2 sin φ. From (8) and (9) it follows that this relationship holds: µ = φ − e sin φ (12) Now if µ and φ vary while e is held constant, the difference φ − µ is a periodic function of µ which vanishes at A and A′ where φ = 0, µ ie a multiple of π. On this basis Bessel assumed the following relationship, which is a Fourier sine series: φ−µ= ∞ X Ar sin rµ (13) r=1 Physical conditions would have allowed Bessel to assume that φ − µ is a continuous function of µ so that Dirichlet’s conditions would have allowed him to conclude that the series converges. I am not sure of the relative timing here of Dirichlet’s work on Fourier series and this aspect of Bessel’s work. This then means that we have to find the Ar which are functions of the eccentricity e. What Gray and Mathews do next is to differentiate (13) with respect to µ giving: ∞ X rAr cos rµ = r=1 dφ −1 dµ (14) Remember this is long before Weierstrass and concepts of uniform continuity and uniform convergence which were nevertheless implicit in the works of Cauchy circa 1820. They then multiply by cos rµ and integrate as follows: 1 πrAr = 2 = Z π Z0 π 0  dφ  − 1 cos rµ dµ dµ dφ cos rµ dµ dµ (15) If you mechnically do the following (exchanging summation and integration of course) you will not get what Gray and Mathews get: 7 Z 0 ∞ ∞X 2 rAr cos rµ dµ = r=1 Z 0 ∞ ∞X rAr r=1 ∞ X 1 = rAr 2 r=1 ∞ X Z
1 cos 2rµ + 1 dµ 2 ∞ 0 "
cos 2rµ + 1 dµ 1 1 sin 2rµ + µ = rAr 2 2r r=1 #π (16) 0 ∞ πX rAr = 2 r=1 This differs substantially from the LHS of (15). Are the authors wrong? No, but they have left out some steps and explanation. Let’s write (14) as follows: ∞ X jAj cos jµ = j=1 dφ −1 dµ (17) 1 2 Now multiply (17) by cos rµ and note that cos jµ cos rµ =
 cos (j − r)µ . Then the integral of the LHS of (17) becomes: Z 0 ∞ πX jAj cos jµ dµ = j=1 = Z 0 ∞ πX ∞ X j=1 jAj j=1 jAj Z π 0 
cos (j + r)µ +

 1 cos (j + r)µ + cos (j − r)µ dµ 2

 1 cos (j + r)µ + cos (j − r)µ dµ 2 (18) Rπ Now when j = r the first integral on RHS of (18) is simply zero since 0 cos 2rµ dµ = 0 for r = 1, 2, . . . . The second integral is simply π2 and so the term is π2 rAr . For the remaining terms in the sum ie when j 6= r we have that the RHS of (18) becomes: 8 ∞ X Z π

 1 cos (j + r)µ + cos (j − r)µ dµ 0 2 j=1 #π " ∞ X 1 1 j = Aj sin(j + r)µ + sin(j − r)µ 2 j + r j − r j=1 jAj 0 = ∞ X j j=1 2 (19) Aj [0 + 0 − (0 + 0)] =0 So the LHS of (15) is indeed correct. Next that authors note that φ = 0 when µ = 0 and φ = π when µ = π so they are emboldened to change the independent variable from µ to φ. If we write dφ dφ = f ′ (µ) and so dφ = f ′ (µ) dµ = dµ dµ. On this basis we have: φ = f (µ) then dµ 1 πrAr = 2 Z Z π cos rµ dφ 0 π cos[r(φ − e sin φ)] dφ using (12) Z π 2 cos[r(φ − e sin φ)] dφ =⇒ Ar = rπ 0 = (20) 0 2 Proofs of fundamental properties Some basic properties are set out in Problem 7.1 of Chapter 6 of Stein [4] in relation to the integral representation in equation (6). The proofs of these properties depend on the the way Jn is presented and I will demonstrate two types of proof based on these two representations: 1 Jn (r) = 2π and Z 2π eir sin θ e−inθ dθ 0 9 (21) Jn (r) = ∞ X j=0 Properties for n ∈ Z : (−1)j  r n+2j j!(n + j)! 2 (22) Property 1 Jn (r) is real for all real r (23) J−n (r) = (−1)n Jn (r) (24) 2Jn (r) = Jn−1 (r) − Jn+1 (r) (25) 2n Jn (r) = Jn−1 (r) + Jn+1 (r) r (26) Property 2 Property 3 ′ Property 4 Property 5 r−n Jn (r)
′ = −r−n Jn+1 (r) (27) Property 6 rn Jn (r)
′ = rn Jn−1 (r) (28) Property 7 Jn (r) satisfies: ′′ ′ Jn (r) + r−1 Jn (r) + (1 − 10 n2 )Jn (r) = 0 r2 (29) Property 8 Jn (r) = ∞ X j=0 Property 9 For all real a, b: (−1)j  r n+2j j!(n + j)! 2 Jn (a + b) = X Jl (a) Jn−l (b) (30) (31) l∈Z 2.1 Proofs based on the integral form in equation (6) In what follows n ∈ Z. Property 1 Show that Jn (r) is real for all real r. We can show that Jn (r) = Jn (r) or, equivalently, that the imaginary part of the integral is zero. Z 2π 1 e−ir sin θ+inθ dθ Jn (r) = 2π 0 Z 2π 1 Jn (r) − Jn (r) = (ei(r sin θ−nθ) − e−i(r sin θ−nθ) ) dθ 2π 0 (32) Z 2π
1 = 2i sin r sin θ − nθ dθ 2π 0 =0
To see why this is the case note that f (θ) = sin r sin θ − nθ is an odd function since f (−θ) = −f (θ) and sin θ is 2π periodic. Hence the integral over [0, 2π] is zero. The oddness properties of sin give rise to integrals which look complicated but are easy to evaluate without any calculations and they abound in the study of Bessel functions. For example: Z 2π 1 sin(r sin θ) dθ = 0 (33) 2π 0 1 π Z π 2 sin θ cos(r sin θ) dθ = 0 − π2 11 (34) 1 π Z π 2 Z π 2 cos θ sin(r sin θ) dθ = 0 (35) − π2 cos(r sin θ) dθ = 0 Z π 2 cos(r cos θ) dθ (36) 0 Oddness explains the first three but the last one is explained by the fact that cos θ = sin( π2 − θ) and indeed one can make the substitution u = π2 − θ to convince oneself without any real calculation that the two integrals are equal but this diagram says it all but for the sake of completeness here it is: Rπ 2 0 cos(r sin θ) dθ = − R0 π 2 cos(r sin( π2 − u)) du = Rπ 2 0 cos(r cos u) du
Expanding on the 2π periodicity point, note that f (θ) = sin r sin θ − nθ is an odd function since f (−θ) = −f (θ) and sin θ is 2π periodic. To prove that f (θ) is 2π periodic we need to show that f (θ ± 2π) = f (θ). For the f (θ + 2π) case we have:
f (θ + 2π) = sin r sin(θ + 2π) − n(θ + 2π)
= sin r sin θ − nθ − 2π (37)
= sin r sin θ − nθ =f (θ) A similar process estlablishes that f (θ − 2π) = f (θ). We can replciate this for
cos r sin θ − nθ . 12 Hence the integral over [0, 2π] is zero. Note that if you have a 2π periodic function f (θ), for any real a, the following will be true: Z 2π Z a+2π f (θ) dθ (38) f (θ) dθ = 0 a For instance a cos θ dθ = R 2π 0 cos θ dθ = 0 . This is because: R a+2π Ra R a+2π Ra f (θ) dθ = 0 f (θ) dθ − 0 f (θ) dθ = 0 f (θ) dθ − 0 f (θ + 2π) dθ = R a+2π R a+2π R 2π R 2π f (θ) dθ − 2π f (θ) dθ = 0 f (θ) dθ + a+2π f (θ) dθ = 0 f (θ) dθ. R a+2π R a+2πa 0 R a+2π Property 2 J−n (r) = (−1)n Jn (r) To prove this we proceed as follows: Z 2π 1 eir sin θ einθ dθ J−n (r) = 2π 0 Z 3π 1 = eir sin(u−π) ein(u−π) dθ with u = θ + π 2π π Z 3π 1 = e−ir sin u einu e|−nπi {z } du 2π π =(−1)n Z (−1)n 3π −ir sin u inu = e| {z e } du 2π π this is 2π periodic n Z 2π (−1) = e−ir sin u einu du 2π 0 (39) =(−1)n Jn (r) =(−1)n Jn (r) using Property 1 Property 3 ′ 2Jn (r) = Jn−1 (r) − Jn+1 (r) We start with: 13 (40) 1 Jn (r) = 2π Differentiating: Z 2π eir sin θ e−inθ dθ (41) 0 Z 2π  1 ∂ ir sin θ −inθ  e e dθ Jn (r) = 2π 0 ∂r Z 2π 1 i sin θ eir sin θ e−inθ dθ = 2π 0 (42) Z 2π   1 Jn−1 (r) − Jn+1 (r) = eir sin θ e−i(n−1)θ − e−i(n+1)θ dθ 2π 0 Z 2π   1 iθ −iθ ir sin θ −inθ = dθ e −e e e 2π 0 Z 2π 1 = 2i sin θ eir sin θ e−inθ dθ 2π 0 ′ =2Jn (r) (43) ′ Also: Property 4 2n Jn (r) = Jn−1 (r) + Jn+1 (r) r On the RHS we have, on using integration by parts: 14 (44) Z 2π   1 −i(n−1)θ −i(n+1)θ ir sin θ e +e e Jn−1 (r) + Jn+1 (r) = dθ 2π 0 Z 2π 1 = 2 cos θ eir sin θ e−inθ dθ 2π 0 Z 1 2π e−inθ ir sin θ = e d(ir sin θ) π 0 ir (" ) #2π Z 2π 1 e−inθ ir sin θ in −inθ ir sin θ = e e e dθ + π ir ir 0 0 ) ( Z 2π 1 n 1 1 eir sin θ e−inθ dθ ( − )+ = π ir ir r 0 Z 2π n = eir sin θ e−inθ dθ πr 0 Z 2π 2n eir sin θ e−inθ dθ = 2πr 0 2n = Jn (r) r Property 5 r−n Jn (r)
′ = −r−n Jn+1 (r) (45) (46) From Properties 4 and 3 we have that: 2n ′ Jn (r) − 2Jn (r) r n ′ =⇒ Jn (r) = Jn (r) − Jn+1 (r) r 2Jn+1 (r) = (47) Therefore:  r −n Jn (r) ′ ′ =r−n Jn (r) − nr−n−1 Jn (r) =r−n−1 nJn (r) − r−n Jn+1 (r) − nr−n−1 Jn (r) = − r−n Jn+1 (r) 15 (48) Property 6 rn Jn (r)
′ = rn Jn−1 (r) (49) =rn Jn (r) + nrn−1 Jn (r) i h ′ n =rn Jn (r) + Jn (r) r i hJ n − J n−1 n+1 + Jn (r) =rn 2 r i 2n rn h Jn−1 (r) − Jn+1 (r) + Jn (r) = 2 r i rn h = Jn−1 (r) − Jn+1 (r) + Jn−1 (r) + Jn+1 (r) 2 =rn Jn−1 (r) (50) Starting with the LHS: rn Jn (r)
′ ′ Property 7 Jn (r) satisfies: n2 Jn (r) + r Jn (r) + (1 − 2 )Jn (r) = 0 r ′′ −1 ′ (51) From Property 6 we have:  ′ Jn−1 (r) =r−n rn Jn (r)   n−1 n ′ −n r Jn (r) + nr Jn (r) =r n ′ =Jn (r) + Jn (r) r Therefore: ′ r−(n−1) Jn−1 (r) = r−(n−1) Jn (r) + nr−n Jn (r) 16 (52) (53) Differentiating:  r −(n−1) Jn−1 (r) ′  = r −(n−1) Jn′ (r) + nr −n Jn (r) ′ ′ ′ ′′ =r−(n−1) Jn (r) − (n − 1)r−n Jn + nr−n Jn (r) − n2 r−n−1 Jn (r) ′′ ′ =r−(n−1) Jn (r) + r−n (n − n + 1)Jn (r) − n2 r−n−1 Jn (r) ′′ ′ =r−(n−1) Jn (r) + r−n Jn (r) − n2 r−n−1 Jn (r) (54) But from Property 5, substituting n − 1 for n, we have that: r−(n−1) Jn−1 (r)
′ = −r−(n−1) Jn (r) (55) So from (55) we have that: ′ ′′ ′′ r−(n−1) Jn (r) + r−n Jn (r) − n2 r−n−1 Jn (r) = − r−(n−1) Jn (r) ′ r−(n−1) Jn (r) + r−n Jn (r) + (r−(n−1) − n2 r−n−1 )Jn (r) =0 (56) Multiplying through this last line by rn−1 we have: n2 1 ′ Jn (r) + Jn (r) + (1 − 2 )Jn (r) = 0 r r ′′ (57) Property 8 Jn (r) = ∞ X j=0 (−1)j  r n+2j j!(n + j)! 2 (58) Given the integral representation of Jn (r) this statement may appear improbable at first blush. To prove it involves understanding that a series representation of ez where z is complex is absolutely convergent and when you multiply two such series the Cauchy product of them is also absolutely convergent with a sum equal to the product of the sums (thanks to a theorem by Dirichlet) (see [11], pages 374386). In order to interchange summation and integration the Weierstrass M-test is then needed. We start by writing Jn (r) as follows: 17 Z 2π 1 eir sin θ e−inθ dθ Jn (r) = 2π 0 Z 2π r iθ 1 −iθ = e 2 (e −e ) e−inθ dθ 2π 0     Z 2π −r −iθ r iθ e e 1 e 2 = e 2 e−inθ dθ 2π 0 (59) We now express the components as appropriate exponential series as follows: e  −r −iθ e 2  = = Similarly: e  r iθ e 2  e −inθ ∞ X ( −r e−iθ )k 2 k=0 ∞ X k! rk (−1)k k e−ikθ 2{zk! } k=0 | = =ak (θ) ∞ X ( r eiθ )k 2 k=0 k! e−inθ ∞ X rk ikθ −inθ e e = 2k k! k=0 = (60) (61) ∞ X rk i(k−n)θ e k k! 2 {z } k=0 | =bk (θ) Let dk = 2k1k! . When we multiply (60) and (61) we get the Cauchy product k coefficient of follows (recall that the Cauchy Pm product term for a product of Pr as P two series ( k ak )( k bk ) has the form cm = k=1 ak bm−k ): cm (θ) = = m X k=0 m X (−1)k dk e−ikθ dm−k ei(m−k−n)θ (62) k (−1) dk dm−k e k=0 18 i(m−2k−n)θ Thus we are essentially asserting this (see (59 ): eir sin θ e−inθ = ∞ X cm (θ)rm (63) m=0 so that we ultimately have to show that: 1 Jn (r) = 2π Z 0 ∞ 2π X m cm (θ)r dθ = m=0 ∞ X j=0 r
n+2j (−1)j j!(n + j)! 2 (64) and we will need to prove that we can interchange the integral and sum which will involve the Weierstrass M-test. This is proved below. Now going back to (62) we have that, since: ( Z 2π 0 if p = ±1, ±2, . . . eipθ dθ = 2π if p = 0 0 (65) we must also have that: Z 2π ei(m−2k−n)θ dθ = 0 (66) 0 unless m − 2k − n = 0 ie 2k = m − n. But this means that m − n = 0 or 2k is a positive even integer ie 2j = m − n for j = 0, 1, 2, . . . (67) (which covers the case m ≥ n). Also we have: m − n =2k m − k =n + k m − k =n + j using the fact that j = k on the RHS (68) For j = 0, 1, 2 . . . we have the following (see (62) ) noting that in the sum in cm (θ) when we exchange integration with the sum the terms which are zero drop out and we are left with a single term for each m: 19 Z Z 2π cm (θ) dθ = 0 Z = 2π cn+2j (θ) dθ 0 2π (−1)j dk dn+2j−k dθ 0 (69) j =(−1) 2πdj dn+j because j = k 2π =(−1)j j n+j 2 j!2 (n + j)! but if m 6= n + 2j we have that: Z 2π cm (θ) dθ = 0 (70) 0 We can now finalise the calculation. From (64) we have that (noting that m = n + 2j and exchanging summation and integration): 1 Jn (r) = 2π = = 1 2π Z 0 ∞ 2π X ∞ X j=0 ∞ X j=0 cm (θ)rm dθ m=0 (−1)j 2π rn+2j 2j j!2n+j (n + j)! (71) (−1)j  r n+2j j!(n + j)! 2 2.2 Uniform convergence proof P m To establish that ∞ is uniformly convergent on (0, 2π) in θ we make m=0 cm (θ)r the following estimate and recall that dk = 2k1k! : 20 |cm (θ)rm | = m X dk dm−k ei(m−2k−n)θ rm k=0 ≤|r| m m X k=0 =|r|m = r 2 m = r 2 m = r 2 m |dk | |dm−k | m X 1 1 k m−k 2 k! 2 (m − k)! k=0 m X 1 1 k! (m − k)! k=0 m X 1 m  m! k k=0 m   1 X m m! k=0 k (72) r m 1 m 2 2 m! |r|m = m! which is finite for any fixed r, m and is independent of θ. Thus the Weierstrass M-test establishes the uniform convergence with respect to θ. = Property 9 For all real a, b we have to prove that, using the integral form of the Bessel function of order n (6), we have that: X Jn (a + b) = Jl (a) Jn−l (b) (73) l∈Z As Stein and Shakarchi note ( [4], pages 197 -198) the Bessel function or order n ∈ Z ,Jn (r), is defined to be the nth Fourier coefficient of the function eir sin θ . In other words: Z 2π 1 eir sin θ e−inθ dθ (74) Jn (r) = 2π 0 With this Fourier theoretic perspective it then follows that: 21 e ir sin θ = ∞ X Jn (r)einθ (75) n=−∞ We have that: 1 Jn (a + b) = 2π 1 = 2π Z Z 2π ei(a+b) sin θ e−inθ dθ 0 2π 0 ! Z 2π ∞ X 1 ia sin θ −ilθ ilθ eib sin θ e−inθ dθ e e dθ e 2π 0 l=−∞ {z } | see (75) = = ∞ X l=−∞ ∞ X 1 2π Z 2π 0 eia sin θ e−ilθ dθ × 1 2π Z 2π eib sin θ e−i(n−l)θ dθ 0 ! (76) Jl (a) Jn−l (b) l=−∞ 3 An equivalent representation So far we have been dealing with Bessel functions Jn (r) for integral n. For nonintegral values of n > − 12 there is another definition which is as follows: Z 1 1 rn eirt (1 − t2 )n− 2 dt Jn (r) = n (77) 1 √ 2 Γ(n + 2 ) π −1 A proof of this result is given on page 38 of Gray [3]. In fact there are many equivalent representations and these are given in Watson [5] for R(ν) > − 21 : 22 Problem 7(2) in [4] ( page 213) asks you to show that this formula agrees with integral version given earlier for integral n ≥ 0. In other words one has to show that (77) agrees with: Z 2π 1 Jn (r) = eir sin θ e−inθ dθ (78) 2π 0 A hint is given: Verify (77) for n = 0 and then check that both sides satisfy this recursion formula: ′ (r−n Jn (r)) = −r−n Jn+1 (r) (79) When n√ = 0 in (77) we get with the substitution t = sin θ and noting that Γ( 12 ) = π : Z 1 1 1 J0 (r) = 1 √ eirt (1 − t2 )− 2 dt Γ( 2 ) π −1 Z π 2 1 =√ √ eir sin θ dθ π π − π2 Z π 1 2 ir sin θ e dθ = π − π2 (80) From (74) with n = 0 we have that: 1 J0 (r) = 2π Z 2π eir sin θ dθ (81) 0 and so we have to show that (80) and (78) with n = 0 are equal. We previously showed that for any R 2π real r, Jn (r) is real for all n ∈ Z. This 1 was Property 1. This means that 2π 0 i sin(r sin θ)) dθ = 0 and indeed the same Rπ Rπ can be said for the imaginary part of π1 −2π eir sin θ dθ = π1 −2π i sin(r sin θ) dθ. This 2 2 can be seen without any calculation. So what we really have to show is that the following real parts are equal: 1 π Z π 2 − π2 1 cos(r sin θ)dθ = 2π ? Z 2π cos(r sin θ)dθ (82) 0 We know that for 2π periodic even trigonometric functions that the integral over [0, 2π] will be twice that over [0, π]. In more detail, since cos is even we know that: 23 1 2π Z 2π 0 Z π 2 cos(r sin θ)dθ = cos(r sin θ)dθ 2π 0 Z π 1 cos(r sin θ)dθ = π 0 Z π Z π  1 2 cos(r sin θ)dθ + = cos(r sin θ)dθ π π 0 2 Z Z π  2 1 0 π π = cos(r sin(u + ))du cos(r sin(u + ))du + π − π2 2 2 0 π Z Z  2 1 0 cos(r cos u)du cos(r cos u)du + = π − π2 0 Z 0 Z π   2 1 cos(r sin u)du + = cos(r sin u)du π − π2 0 Z π 1 2 = cos(r sin u) du π − π2 (83) Note that we jhave used the results of the discussion surrounding (36) in finalising the above integral. Thus (78) is correct. Thus the alternative defintion does equal the definition given in equation (6) for n = 0. Going back to the hint Eli Stein gave in his problem mentioned above, we have to show that with n = 0, (73) satisfies both sides of (75) which was proved using (74). Note that with n = 0 in (75) we get the well known result: J0 (r) = −J1 (r) (84) Proving that the integral representation in (73) satisfies Property 5 involves a straightforward integration by parts. We have to show that:  ′ r−n Jn (r) = −r−n Jn+1 (r) (85) We have the following, noting that v below is zero at t = ±1 for the relevant step in integration by parts: 24  r −n Jn (r) ′ Z 1 1 ∂ 1 eirt (1 − t2 )n− 2 dt = n 1 √ 2 Γ(n + 2 ) π ∂r −1 Z 1 1 2 n− 12 irt √ t (1 − t ) e i dt 1 {z } 2n Γ(n + 2 ) π −1 |{z} | =u =dv Z 1 1 1 1 2 eirt (1 − t2 )n+ 2 dt = n 1 √ ×i r 1 2 Γ(n + 2 ) π 2(n + 2 ) −1 Z 1 1 −r = n+1 eirt (1 − t2 )n+ 2 dt 3 √ 2 Γ(n + 2 ) π −1  n+1 −n r Z 1 −r 2 1 = eirt (1 − t2 )n+ 2 dt 3 √ Γ(n + 2 ) π −1 (86) = − r−n Jn+1 (r) In Problem 7.2(b) ( [4], page 213 ) Stein and Shakarchi invite you to establish that: r 2 −1 J 1 (r) = (87) r 2 sin r 2 π Putting n = 12 into (77) we see that: √ Z 1 r J 1 (r) = √ eirt dt √ 2 2 Γ(1) π −1 |{z} =1 √ h irt i r e 1 =√ √ 2 π ir −1 2i sin r =√ √ √ 2 π ri r 2 −1 r 2 sin r = π Problem 7.2 (c) ( [4], page 213 ) asks you to prove that: r 2 −1 lim1 Jn (r) = r 2 cos r π n→− 2 (88) (89) This is trickier since (77) only makes sense for n > − 21 so we clearly have to engage in a limiting argument. It is to be noted that there are other ways to prove 25 that the integral form in (77) gives rise to this result and as Watson demonstrates, the process is very intricate. However, proving the property from the series formulation is not hard and is set out later on. To prove (89), let n = − 21 + ǫ where ǫ > 0. We now examine the behaviour of the LHS of (85) as ǫ → 0. We can use (77 ) because n + 21 = ǫ > 0 and so the integral form is kosher. The LHS of (85 ) using (77) is (noting that the differentiation is with respect to r):  r 1 −ǫ 2 J− 1 +ǫ (r) 2 ′ !′ Z 1 1 1 r 2 −ǫ r− 2 +ǫ eirt (1 − t2 )ǫ−1 dt = − 1 +ǫ √ 2 2 Γ(ǫ) π −1 Z 1 1 = − 1 +ǫ it eirt (1 − t2 )ǫ−1 dt √ 2 2 Γ(ǫ) π −1 Z 1 i = − 1 +ǫ eirt t(1 − t2 )ǫ−1 dt √ |{z} {z } 2 2 Γ(ǫ) π −1 =u | =dv Z i ir 1 irt = − 1 +ǫ e (1 − t2 )ǫ dt √ × 2 2 Γ(ǫ) π 2ǫ −1 Z 1 −r = 1 +ǫ eirt (1 − t2 )ǫ dt √ 2 2 ǫ Γ(ǫ) π −1 Z 1 −r = 1 +ǫ eirt (1 − t2 )ǫ dt √ 2 2 Γ(ǫ + 1) π −1 (90) Now as ǫ → 0 we have that: 2 1 +ǫ 2 −r √ Γ(ǫ + 1) π Z 1 −1 eirt (1 − t2 )ǫ dt → 1 2 −r √ 2 Γ(1) π Z 1 −1 eirt dt = − r 2 sin r (91) π The RHS of ( 85) will be, using (87), as ǫ → 0 : r r 1 1 1 2 −1 2 −ǫ r 2 sin r = − sin r (92) − r 2 J 1 +ǫ (r) → −r 2 J 1 (r) = −r 2 2 2 π π q  1 ′ 2 Note that the LHS of (90 ) in the limit is r J− 1 (r) and this equals − π2 sin r. 2 Thus we integrate with respect to r as follows: 26  ǫ→0 ′ r 2 =− sin r 2 π Z r 1 2 r 2 J− 1 (r) = − sin r dr 2 π r 2 cos r = π r 2 −1 r 2 cos r =⇒ lim1 Jn (r) = π n→− 2 lim r 1 −ǫ 2 J− 1 +ǫ (r) (93) These limit steps work because of continuity, dominated convergence of the integral (or an appeal to uniform convergence of the relevant functions). For a more detailed discussion see [12]. It is somewhat easier to get (87) and (89) using of Property 8 which is the series form of Jn (r) ie Jn (r) = ∞ X j=0 Thus: J 1 (r) = 2 = = ∞  r n+2j (−1)j (−1)j  r n+2j X = j!(n + j)! 2 j!Γ(n + j + 1) 2 j=0 (94) ∞ X (−1)j  r  12 +2j j!Γ( 23 + j) 2 j=0   52   29   21 x 2 x 2 x 2 − + − ... Γ( 23 ) 1! Γ( 25 ) 2! Γ( 72 )   12   52 x 2 x 2   92 x 2 − + − ... (1/2)Γ( 21 ) 1! (3/2)(1/2)Γ( 12 ) 2! (5/2)(3/2)(1/2)Γ( 12 )   12 x o n 2 x2 x4 √ 1− + − ... = 3! 5! (1/2) π   21 x n o 2 x3 x5 √ x− + − ... 3! 5! x(1/2) π r 2 sin x = πx 27 (95) J− 1 (r) is found by an identical process: 2 ∞ X (−1)j  r − 12 +2j J− 1 (r) = 2 j!Γ( 21 + j) 2 j=0  − 12   23   72 = = x 2 Γ( 12 )  − 12 x 2 Γ( 12 )  − 12 x 2 − − x 2 x 2 + − ... 1! Γ( 32 ) 2! Γ( 52 )   23   72 x 2 1! (1/2)Γ( 21 ) + x 2 2! (3/2)(1/2)Γ( 12 ) − ... (96) n o x2 x4 1− = √ + − ... 2! 4! π r 2 cos x = πx 3.1 A series expression for the integral form of Jn+ 21 (r) It is possible to get a series expression for Jn+ 1 (r) from the integral representation 2 above (77 ) and then J 1 (r) falls out. J− 1 (r) can also be found from a similar 2 2 process, but with some complications. Watson does this ( [5], pages 53-4) but leaves out a lot of steps. The process is intricate and basically starts by recursively integrating (77 ) by parts. This gives a summation of derivatives which are amenable to Leibnitz’s rule and after some careful calculations one finally gets to the goal of Jn+ 1 (r). 2 For n > − 12 we know that the integral form is kosher and that it is certainly kosher for positive integers and zero. Thus we have that: 1 Jn+ 1 (r) = 2 rn+ 2 2 n+ 21 √ Z 1 eirt (1 − t2 )n dt Γ(n + 1) π −1 Z 1 r = n+ 1 √ eirt (1 − t2 )n dt 2 2 n! π −1 n+ 21 (97) We now focus on integrating the integral by parts noting that (1 − t2 )n is a polynomial of degree 2n so that when we differentiate it 2n + 1 times it is zero. 28 The crux of the integration by parts is based on this since if we integrate by parts 2n + 1 times the process will terminate. In what follows we will be taking k th derivatives of (1 − t2 )n and we use the convention that when k = 0 we get the 0 2 )n function itself ie d (1−t = (1 − t2 )n . dt0 What we get when we perform the integration by parts is this: #t=1 " Z 1 2n k+1 k 2 n X d (1 − t ) i eirt (1 − t2 )n dt = − eirt rk+1 dtk −1 k=0 (98) t=−1 To see this we need to perform a few integrations by parts to identify the structure. In what follows the basic structure that is iterated is that our u equals the polynomial term and our dv is the exponential term: 29 Z 1 irt −1 2 n Z 1 (1 − t2 )n e|irt {zdt} −1 | {z } =dv =u #t=1 " Z 1 irt e d(1 − t2 )n eirt 2 n (1 − t ) dt − = ir dt −1 ir t=−1 " #t=1 Z 1 irt 1 eirt d0 (1 − t2 )n e d (1 − t2 )n = dt − ir dt0 dt1 −1 ir t=−1 " #t=1 Z 2 n irt 0 e d (1 − t ) 1 1 d1 (1 − t2 )n irt = − |e {zdt} 1 ir dt0 ir −1 | dt {z } e (1 − t ) dt = t=−1 =u =dv #t=1 #t=1 ) (" Z 1 1 1 irt d2 (1 − t2 )n eirt d1 (1 − t2 )n eirt d0 (1 − t2 )n e dt – − = ir dt0 ir ir dt1 ir −1 dt2 t=−1 t=−1 " #t=1 " #t=1 Z 1 2 2 n eirt d0 (1 − t2 )n eirt d1 (1 − t2 )n 1 1 irt d (1 − t ) = dt – + e ir dt0 (ir)2 ir dt1 (ir)2 −1 dt2 t=−1 t=−1 " #t=1 " #t=1 eirt d0 (1 − t2 )n 1 eirt d1 (1 − t2 )n = – ir dt0 (ir)2 ir dt1 t=−1 t=−1 (" #t=1 ) Z 1 3 2 n 1 1 eirt d2 (1 − t2 )n irt d (1 − t ) + e dt − (ir)2 ir dt2 ir −1 dt3 t=−1 #t=1 " #t=1 #t=1 " " eirt d1 (1 − t2 )n eirt d2 (1 − t2 )n eirt d0 (1 − t2 )n – + = ir dt0 (ir)2 dt1 (ir)3 dt2 t=−1 t=−1 t=−1 Z 1 3 2 n d (1 − t ) 1 eirt dt − 3 (ir) −1 dt3 (99) " Bearing in mind that the integral in (99) disappears after 2n + 1 iterations we are left with: 30 Z 1 −1 " 1 d1 (1 − t2 )n 1 d2 (1 − t2 )n 1 d0 (1 − t2 )n − + + ···+ ir dt0 (ir)2 dt1 (ir)3 dt2 #t=1 d2n (1 − t2 )n 1 (ir)2n+1 dt2n t=−1 " #t=1 2n k 2 n k+1 Xi d (1 − t ) = − eirt rk+1 dtk k=0 eirt (1 − t2 )n dt =eirt t=−1 (100) Now we can use Leibnitz’s theorem to work out derivatives at the upper and lower limits. Recall that Lebnitz’s theorem says this: k   dk (1 − t2 )n X k dj (1 − t)n dk−j (1 + t)n = dtk dtj dtk−j j j=0 (101) We have to evaluate this at t = 1 and t = −1. The crux of the use of Leibnitz’s theorem is to note that (1−t2 )n = (1−t)n (1+ t)n . Note that for j < n we have: dj (1 − t)n dtj #t=1 = (−1)j n(n − 1) . . . (n − j + 1)(1 − t)n−j j n #t=1 =0 (102) = 0. This leaves the case when j = n for the term Clearly when j > n , d (1−t) dtj j n k−j n d (1−t) = (−1)n n! . For the term d dt(1+t) note that we only therefore need to k−j dtj consider j ≥ n . Of course 0 ≤ k ≤ 2n . For this second term we differentiate k − n times having differentiated the first component n times, thus we have, noting
k that the binomial coefficient in (101) is n : 31 dk−n (1 + t)n dtk−n #t=1 =n(n − 1)...(n − (k − n) + 1)(1 + t)n−(k−n) =n(n − 1)...(2n − k + 1)(1 + t) = 2n−k #t=1 #t=1 (103) n!22n−k (2n − k)! Thus the Leibnitz rule in (101) gives us the following: #t=1   " k n!22n−k dk (1 − t2 )n n = (−1) n! dtk (2n − k)! n (104) We now have to evaluate the derivative at t = −1 and for this purpose it is easier to swap the components ie: " # " k   # k 2 n n j n k−j X d (1 − t ) k d (1 + t) d (1 − t) = (105) k dt dtj dtk−j j j=0 t=−1 j t=−1 n Note that for j > n, d (1+t) = 0 (because it is a degree n polynomial) and for dtj dj (1+t)n j < n, dtj = 0 because t = −1. So we only need consider j = n which gives us n! for that term. As before we differentiate the second term k − n times which gives us: dk−n (1 − t)n dtk−n # =(−1) k−n t=−1 n(n − 1)...(n − (k − n) + 1)(1 − t) =(−1)k−n n(n − 1)...(2n − k + 1)(1 − t)2n−k =(−1)k−n # # t=−1 t=−1 2n−k n!2 (2n − k)! Finally we have: # " dk (1 − t2 )n dtk n−(k−n) = (−1) t=−1 32 (106) k−n   n!22n−k k n! n (2n − k)! (107) Going back to (100) and using (101), (104) and (107) what we have is this: Z #t=1 2n k+1 k 2 n X d (1 − t ) i eirt (1 − t2 )n dt = − eirt rk+1 dtk −1 k=0 t=−1   2n X ik+1 n!22n−k n k = − eir (−1) + n! rk+1 (2n − k)! n k=n   2n k+1 X n!22n−k i −ir k−n k (−1) n! e rk+1 n (2n − k)! k=n " 1 (108) Using (97) we get the final form of Jn+ 1 (r): 2 1 rn+ 2 Jn+ 1 (r) = n+ 1 √ 2 2 2 n! π Z " 1 −1 eirt (1 − t2 )n dt #     2n k+1 2n k+1 2n−k 2n−k X X k k n!2 n!2 i i (−1)n n! (−1)k−n n! + e−ir = n+ 1 √ − eir k+1 k+1 r (2n − k)! r (2n − k)! n n 2 2 n! π k=n k=n " # 1 2n 2n k+1 2n−k k+1 2n−k X X i k! 2 rn+ 2 (−i) k! 2 + (−1)n+1 e−ir = n+ 1 √ (−1)n+1 eir k+1 (k − n)!(2n − k)! r rk+1 (k − n)!(2n − k)! 2 2 π k=n k=n " # 2n 2n n+1 k+1 n−k n+1 k+1 n−k X X rn (−1) i k! 2 (−1) (−i) k! 2 eir =√ + e−ir k (k − n)!(2n − k)! r rk (k − n)!(2n − k)! 2πr k=n k=n # " 2n 2n n+1 k+1 n−k n+1 k+1 n−k X X (−1) (−i) k! (2r) 1 (−1) i k! (2r) =√ + e−ir eir (k − n)!(2n − k)! (k − n)!(2n − k)! 2πr k=n # " k=n n n n+1 n+j+1 −j n+1 n+j+1 −j X X 1 (−1) (−i) (n + j)! (2r) (−1) i (n + j)! (2r) + e−ir =√ eir j!(n − j)! j!(n − j)! 2πr j=0 j=0 # " n n X X (−1)n+1 (−i)n+j+1 (n + j)! (−1)n+1 in+j+1 (n + j)! 1 −ir ir +e =√ e (2r)j j!(n − j)! (2r)j j!(n − j)! 2πr j=0 j=0 # " n n j−n−1 j−n−1 X X (−i) (n + j)! 1 i (n + j)! + e−ir =√ eir j j!(n − j)! (2r) (2r)j j!(n − j)! 2πr j=0 j=0 r n+ 12 (109) Note that we have made the index change j = k − n and for the factors 33 ij−n−1 and (−i)j−n−1 we have used the following manipulations, observing that (−1)n+1 = (−1)−(n+1) : (−1)n+1 in+j+1 = (i2 )−(n+1) in+j+1 = i−2n−2 in+j+1 = ij−n−1 (110) and (−1)n+1 (−i)n+j+1 = ((−i)2 ))−(n+1) (−i)n+j+1 = (−i)−2n−2 (−i)n+j+1 = (−i)j−n−1 (111) The test of (105) is when we put in n = 0 we should get (87): " # 0 0 X X 1 J 1 (r) = √ (−i)−1 i−1 + e−ir eir 2 2πr j=0 j=0 " # 1 eir − e−ir =√ i 2πr 1 2i sin r =√ 2πr i r 2 sin r = πr (112) Watson ( [5), 53-54) goes on qto use a similar but even more complicated ap2 proach to show that J− 1 (r) = πr cos r. According to Watson ([5], page 53) de 2 la Vallee Poussin developed the asterisked formula set out below: 4 The generating function representation of Jn(x) The generating function that gives us Jn (x) is this: 34 e x 2 t− 1t
∞ X = Jn (x)tn (113) n=−∞
1 x xt x Two things to note at the outset is that because e 2 t− t = e 2 × e− 2t we have a product of an absolutely converging series of ascending powers of t and an absolutely convergent series of descending powers of t. Thus as long as we keep away from t = 0 we should have an expansion that is valid for all values of x, t with t = 0 excluded of course. To see that the generating function (113) actually gives Jn (x) as the coefficients of tn we note that (using the Cauchy product): e x 2 t− 1t
= ∞ nX r=0 =  ∞ ∞ X X r=0 k=0 = = r!  r − ∞ on X  k xt 2 k! k=0 k  r−k xt x 2t 2 k! ∞ X ∞ X (−1)k n=−∞ ∞ X k! o (r − k)!  r x 2 k=0 (114) tr−2k (r − k)!  n+2k ) ( ∞ k x ∞ (−1) X X 2 r=0 k=0 = −x 2t k!(n + k)! tn Jn (x) tn n=−∞ Note that we have re-indexed with r−2k = n so that r−k = n+k. This has the effect of causing n to run from −∞ to +∞ as r, k independently range from 0 to ∞. P  P  P ∞ ∞ ∞ Recall that the product of two infinite series m=0 cm j=0 aj k=0 bk = Pm where cm = r=0 ar bm−r , is called the Cauchy product term. 35 5 Bounds on Jn(x) We can get some bounds on Jn (x) by manipulating the series representation as follows: |Jn (x)| = ≤ ∞ (−1)k X  n+2k x 2 k!(n + k)!  n+2k ∞ (−1)k x X 2 k=0 k=0 x ≤ 2 x = 2 ≤ = n n k!(n + k)! ∞ X k=0 ∞ X k=0 ∞ x n X 2 n! x n 2 n! k=0 ∞ X x 2k 2 k!(n + k)! (115) x 2k 2 k!(n + k)(n + k − 1) . . . (n + 1)n! x 2k 2 k!(n + 1)k !k k=0 | x2 |2 n+1 k! So for n ≥ 0 we have: |Jn (x)| ≤ x n 2 n! e |x|2 4 n+1 ≤ x n 2 n! e |x|2 4 This derivation holds for x real or complex. 5.1 Proofs of the properties using the series approach We now derive the basic properties using the series formulation. Property 1 36 (116) Jn (r) = P∞ (−1)j j=0 j!(n+j)!  n+2j r 2 Clearly Jn (r) is real for all real r. Property 2 J−n (r) = (−1)n Jn (r) J−n (r) = = ∞ X j=0 ∞ X j=0 = n−1 X j=0 = ∞ X j=n ∞ X  r −n+2j (−1)j j!(−n + j)! 2  r −n+2j (−1)j j!Γ(−n + j + 1) 2 j! (−1)j Γ(−n + j + 1) | {z } →∞ when j=0,1,2,...n−1  r −n+2j 2 + ∞ X j=n  r −n+2j (−1)j j!Γ(−n + j + 1) 2  r −n+2j (−1)j let j = n + k j!Γ(−n + j + 1) 2  r n+2k (−1)n+k (n + k)!Γ(k + 1) 2 k=0 ∞ X (−1)k  r n+2k n =(−1) (n + k)!k! 2 k=0 = =(−1)n Jn (r) (117) Because Γ(−n + j + 1) → ∞ for j = 0, 1, 2, . . . n − 1 the reciprocal approaches 0 and we can ignore those terms in the sum. The following graph of the Gamma function demonstrates the relevant behaviour. 37 Property 3 To prove 2 Jn′ (r) = Jn−1 (r) − Jn+1 (r) using the series representation for Jn (r) the standard textbook approach is to show first that: which is Property 6 and which is Property 5.  d n r Jn (r) = rn Jn−1 (r) dr (118)  d  −n r Jn (r) = −r−n Jn+1 (r) dr (119) These will be proved later. Using these two properties we have:  d n r Jn (r) = rn Jn′ (r) + nrn−1 Jn (r) = rn Jn−1 (r) dr (120) rJn′ (r) + nJn (r) = rJn−1 (r) (121)  d  −n r Jn (r) = r−n Jn′ (r) − nr−n−1 Jn (r) = −r−n Jn+1 (r) dr (122) or Similarly: or 38 rJn′ (r) − nJn (r) = −rJn+1 (r) (123) Adding (121) and (123) and dividing by r we have that: 2Jn′ (r) = Jn−1 (r) − Jn+1 (r) (124) Is there a direct “bare hands” proof of this relationship starting with the difference of the two series? Jn−1 (r) − Jn+1 (r) = = ∞ X j=0 ∞ X j=0 ∞  r n−1+2j X  r n+1+2j (−1)j (−1)j − (n − 1 + j)!j! 2 (n + 1 + j)!j! 2 j=0 ∞  r n−1+2k (−1)j (n + j)  r n−1+2j X (−1)k−1 − (n + j)!j! 2 (n + k)!(k − 1)! 2 {z } |k=1 k=j+1 = = ∞ X (−1)j (n + j)  r n−1+2j j=0 ∞ X j=0 = = = 2 j=0 ∞ X ∞ X j=0 ∞ X ∞ (−1)j (n + j)  r n−1+2j X (−1)k−1 k  r n−1+2k − (n + j)!j! 2 (n + k)!k! 2 k=0 ∞ X (−1)j (n + j)  r n−1+2j j=0 = (n + j)!j! ∞ X (−1)k−1 k  r n−1+2k − (n + k)!k! 2 k=1 (n + j)!j! 2 + ∞ X (−1)k k  r n−1+2k (n + k)!k! 2 k=0 ∞ (−1)j (n + j)  r n−1+2j X (−1)j j  r n−1+2j + (n + j)!j! 2 (n + j)!j! 2 j=0 (−1)j  r n−1+2j (n + j + j) j!(n + j)! 2 (−1)j (n + 2j)  r n−1+2j j!(n + j)! 2 j=0 =2Jn′ (r) (125) Note that we can get away with the re-indexing from k = 0 because of the multiplicative factor k which has the effect of adding nothing to the sum. Property 4 39 We have to prove that 2n J (r) r n = Jn−1 (r) + Jn+1 (r) To do this use Properties 5 and 6 (see below). From Property 6 we have that: rJn′ (r) + nJn (r) = rJn−1 (r) (126) and from Property 5 we have that: rJn′ (r) − nJn (r) = −rJn+1 (r) (127) Subtracting (127) from (126 ) we have that:   2nJn (r) =r Jn−1 (r) + Jn+1 (r) 2n =Jn−1 (r) + Jn+1 (r) r (128) Property 5 ∞  d X (−1)j  r n+2j d  −n r Jn (r) = r−n dr dr j=0 j!(n + j)! 2 ∞ r2j d X (−1)j = dr j=0 j!(n + j)! 2n+2j ∞ X (−1)j 2j r2j−1 = j!(n + j)! 2n+2j j=1 =r −n =r−n ∞ X j=1 ∞ X |k=0 = − r−n (−1)j rn−1+2j (j − 1)!(n + j)! 2n−1+2j  r n+1+2k (−1)k+1 k!(n + 1 + k)! 2 {z } k=j−1 ∞ X k=0  r n+1+2k (−1)k k!(n + 1 + k)! 2 = − r−n Jn+1 (r) 40 (129) Property 6 ∞  d X (−1)j  r n+2j d n r Jn (r) = rn dr dr j=0 j!(n + j)! 2 ∞ d X (−1)j r2n+2j = dr j=0 j!(n + j)! 2n+2j = = ∞ X j=0 ∞ X r2n+2j−1 (−1)j (2n + 2j) n+2j j!(n + j)! 2  r 2n+2j−1 (−1)j j!(n − 1 + j)! 2 j=0 ∞ X n =r j=0 n (130)  r n−1+2j (−1)j j!(n − 1 + j)! 2 =r Jn−1 (r) Property 7 We have to show that the series expression for Jn (r) satisfies Bessel’s differential equation:  n2  ′′ −1 ′ (131) Jn (r) + r Jn (r) + 1 − 2 Jn (r) = 0 r We work out the derivatives as follows and then add up the expressions: Jn′ (r) = Jn′′ (r) = ∞ X (−1)j (n + 2j) n+2j−1 r n+2j j!(n + j)! 2 j=0 ∞ X (−1)j (n + 2j)(n + 2j − 1) j=0 j!(n + j)! 2n+2j rn+2j−2 Hence we have that (on multiplying (131) through by r2 ): 41 (132) (133) r 2 Jn′′ (r)+rJn′ (r)+ ∞ X j=0 = ∞ X j=0 2 2
r −n Jn (r) = ∞ X (−1)j (n + 2j)(n + 2j − 1) j!(n + j)! 2n+2j j=0 r n+2j ∞ X (−1)j (−1)j n2 n+2j+2 r − rn+2j n+2j n+2j j!(n + j)! 2 j!(n + j)!2 j=0 ∞ X (−1)j (n + 2j) n+2j + r + j!(n + j)! 2n+2j j=0 ∞ i X (−1) rn+2j h (−1)j rn+2j+2 2 (n + 2j)(n + 2j − 1) + n + 2j − n + j!(n + j)! 2n+2j j!(n + j)! 2n+2j j=0 {z } | j k=j+1 = ∞ X i rn+2k (−1)j rn+2j h (−1)k−1 k 2 2 + (n + 2j) − n j!(n + j)! 2n+2j k!(n + k − 1)! 2n+2k−2 k=1 j=0 ∞ X = = ∞ X j=0 ∞ X ∞ i X (−1)k−1 k(n + k) rn+2k (−1)j rn+2j h 2 4jn + 4j + j!(n + j)! 2n+2j k!(n + k)! 2n+2k−2 k=1 ∞ i X (−1)j−1 j(n + j) rn+2j (−1)j rn+2j h 2 + 4jn + 4j j!(n + j)! 2n+2j j!(n + j)! 2n+2j−2 j=0 j=0 ∞ X = j=0 ∞ i X (−1)j 4j(n + j) rn+2j (−1)j rn+2j h 2 − 4jn + 4j j!(n + j)! 2n+2j j!(n + j)! 2n+2j j=0 = ∞ X j=0 i (−1)j rn+2j h 2 4jn + 4j − 4j(n + j) j!(n + j)! 2n+2j | {z } (134) =0 This establishes that Jn (r) satisfies Bessel’s differential equation. According to Watson ( [5], page 24) , “shortly before the appearance of Bessel’s memoir on planetary perturbations, Poisson had published an important work on the Conduction of Heat, in the course integrals of the R π of which he investigated Rπ 2n 2n+1 θ dθ and 0 cos(z cos θ) sin θ dθ where n is a positive types 0 cos(z cos θ) sin integer or zero. Watson then provides a 3 line proof of the following: Z π zn Jn (z) = cos(z cos θ) sin2n θ dθ 1.3.5. . . . (2n − 1)π 0  n z Z π 2 cos(z cos θ) sin2n θ dθ = 1 1 Γ(n + 2 )Γ( 2 ) 0 42 (135) Watson says it is easy to prove this because all one has to do is expand the integrand in powers of z and then integrate term by term (because the series is uniformly convergent ). Let’s see how “easy” this is. Here is Watson’s 3 line proof: 1 π Z π Z ∞ 1 X (−1)m z 2m π cos(z cos θ) sin θ dθ = cos2m θ sin2n θ dθ π m=0 (2m)! 0 2n 0 = ∞ X (−1)m z 2m 1.3.5 . . . (2n − 1).1.3.5 . . . (2m − 1) (2m)! 2.4.6. . . . (2n + 2m) m=0 =1.3.5 . . . (2n − 1) ∞ X (−1)m z 2m 2n+2m m!(n + m)! m=0 (136) from which the result follows. To flesh this proof out we use the series representation of cos(z cos θ) = in the integral as follows: 1 π Z π 0 Z ∞ π X (z cos θ)2m sin2n θ dθ (2m)! 0 m=0 ∞ 2m Z π 1X m z = cos2m θ sin2n θ dθ (−1) π m=0 (2m)! 0 1 cos(z cos θ) sin θ dθ = π 2n P∞ m=0 (−1) 2m m (z cos θ) (2m)! (−1)m (137) Rπ Rπ Rπ To integrate 0 cos2m θ sin2n θ dθ we first note that 0 cos2m θ sin2n θ dθ = 2 02 cos2m θ sin2n θ dθ. This can be seen by noting that the integrand is an even function and then splitting the domain of integration into [0, π2 ] and [ π2 , π]. We then make the substitution 1 u = sin θ so that we have du = cos θ dθ and cos θ = (1 − u2 ) 2 . Thus we have that: Z π cos 2m 2n θ sin θ dθ =2 0 =2 π 2 Z Z0 1 0 =2 Z 1 0 cos2m θ sin2n θ dθ (1 − u2 )m u2n 1 (1 − u2 ) 2 du 1 (1 − u2 )m− 2 u2n du We now let x = u2 so that dx = 2u du and our integral becomes: 43 (138) 2 Z 1 0 2 m− 12 (1 − u ) 2n u du = Z 1 0 1 1 xn− 2 (1 − x)m− 2 dx 1 1 =B(n + , m + ) 2 2 Γ(n + 21 ) Γ(m + 21 ) = Γ(m + n + 1) (139) In this context it will be recalled that the beta function B(m, n) is defined as follows for m > 0, n > 0: Z 1 xm−1 (1 − x)n−1 dx (140) B(m, n) = 0 and it has the property that: Γ(m)Γ(n) Γ(m + n) At this stage we need a preliminary but useful result which is this: B(m, n) = B(m, n) = 2 Z π 2 sin2m−1 θ cos2n−1 θ dθ (141) (142) 0 To prove this we let x = sin2 θ so that dx = 2 sin θ cos θ dθ. Hence B(m, n) = = Z Z =2 1 xm−1 (1 − x)n−1 dx 0 π 2 (sin2 θ)m−1 (cos2 θ)n−1 2 sin θ cos θ dθ (143) 0 Z π 2 sin2m−1 θ cos2n−1 θ dθ 0 we proceed as follows (noting that m, n > 0 ). We To prove B(m, n) = Γ(m)Γ(n) Γ(m+n) 2 let z = x so that dz = 2x dx and hence Z ∞ Z ∞ 2 m−1 −z x2m−1 e−x dx (144) z e dz = 2 Γ(m) = 0 0 Similarly, Γ(n) = 2 Z ∞ 2 y 2n−1 e−y dy 0 44 (145) Using polar coordinates x = r cos φ and y = r sin φ and remembering that the differential area element is dx dy = rdr dφ or working out the Jacobian we have that: Γ(m) Γ(n) =4 =4 =4 =2 Z Z Z ∞ 0 ∞ Z 0 π 2 φ=0 Z x2m−1 e ∞ −x2 dx ! Z x2m−1 y 2n−1 e−(x ∞ y 2n−1 e −y 2 dy 0 2 +y 2 ) ! dx dy 0 Z ∞ r=0 ∞ 2 r2(m+n)−1 e−r cos2m−1 φ sin2n−1 φ drdφ r=0 ! Z π 2 2 r2(m+n)−1 e−r dr 2 cos2m−1 φ sin2n−1 φ dφ φ=0 (146) ! =Γ(m + n) B(n, m) =Γ(m + n) B(m, n) whwere (142) and (144) have been used. Note that B(m, n) = B(n, m) because if we make the change of variable x = 1 − y we have: Z 1 xm−1 (1 − x)n−1 dx 0 Z 0 (1 − y)m−1 y n−1 dy =− Z 11 y n−1 (1 − y)m−1 dy = B(m, n) = (147) 0 =B(n, m) We need some further building blocks based on the fact that: Z π cos 0 2m 2n θ sin θ dθ = 2 Z 1 0 2 m− 12 (1 − u ) Γ(n + 12 ) Γ(m + 21 ) u du = Γ(m + n + 1) 2n Also we need expressions for Γ(n + 21 ) and Γ(m + 12 ): 45 (148) 1 1 Γ(n + ) =Γ(n − + 1) 2 2 1 1 =(n − ) Γ(n − ) 2 2 (2n − 1) 3 = Γ(n − + 1) 2 2 5 (2n − 1) (2n − 3) Γ(n − + 1) = 2 2 2 (2n − 1) (2n − 3) (2n − 5) 31 1 = ... Γ( ) 2 2 2 22 2 3 1√ (2n − 1) (2n − 3) (2n − 5) ... π = 2 2 2 22 (2n − 1)(2n − 3)(2n − 5) . . . 3.1 √ = π 2n (149) A quick inductive proof of this is as follows. √ π. √ Let P (n) = Γ(n + 12 ) = (2n−1)(2n−3)(2n−5)...3.1 2n The base case P (1) is true since Γ( 23 ) = 12 Γ( 21 ) = 12 π. Using the usual inductive hypothesis we have: P (n + 1) =Γ(n + 1 + 1) 2 1 1 =(n + )Γ(n + ) 2 2 (2n + 1) (2n − 1)(2n − 3) . . . 3.1 √ = π 2 2n (2n + 1)(2n − 1)(2n − 3) . . . 3.1 √ π = 2n+1 (150) Thus we have: Z π cos 0 2m Γ(n + 12 ) Γ(m + 21 ) θ sin θ dθ = Γ(m + n + 1) (151) (2n − 1)(2n − 3) . . . 3.1 (2m − 1)(2m − 3) . . . 3.1π = (m + n)!2m+n 2n 46 We now plug this into the series expression: 1 π Z π Z ∞ 1 X (−1)m z 2m π cos(z cos θ) sin θ dθ = cos2m θ sin2n θ dθ π m=0 (2m)! 0 2n 0 ∞ 1 X (−1)m z 2m (2n − 1)(2n − 3)..3.1 (2m − 1)(2m − 3)..3.1π = π m=0 (2m)! (m + n)!2m+n ∞ X (−1)m z 2m (2m − 1)(2m − 3)..3.1 =(2n − 1)(2n − 3)..3.1 (2m)! (m + n)!2m+n m=0 =(2n − 1)(2n − 3)..3.1 =(2n − 1)(2n − 3)..3.1 ∞ X (−1)m z 2m (2m − 1)(2m − 3)..3.1 2m m!(2m − 1)(2m − 3)...3.1 (m + n)!2m+n m=0 ∞ X (−1)m z 2m this is Watson’s (136) m!(m + n)!2n+2m m=0 ∞ (2n − 1)(2n − 3)..3.1 X (−1)m z n+2m = zn 2n+2m m!(m + n)! m=0  n+2m m z ∞ (−1) 2 (2n − 1)(2n − 3)..3.1 X = zn m!(m + n)! m=0 = (2n − 1)(2n − 3)..3.1 Jn (z) zn (152) Note that in the above derivation we used the fact that: (2m)! = 2m m!(2m − 1)(2m − 3) . . . 5.3.1 (153) This can be established inductively. The base case m = 1 is clearly true. Using the obvious inductive hypothesis the (m + 1)th case is: (2m + 2)! =(2m + 2)(2m + 1)(2m)! =2(m + 1)(2m + 1)2m m!(2m − 1)(2m − 3) . . . 5.3.1 =2m+1 (m + 1)!(2m + 1)(2m − 1( 2m − 3) . . . 5.3.1 Therefore: 47 (154) Z 1 π zn cos(z cos θ) sin2n θ dθ Jn (z) = (2n − 1)(2n − 3)..3.1 π 0 Z π zn = n cos(z cos θ) sin2n θ dθ 2 (n − 12 )(n − 23 ) . . . 52 32 21 π 0 Z π zn cos(z cos θ) sin2n θ dθ = 2n Γ(n+ 1 )π 2 Γ( 12 ) = (155) 0  n z 2 Γ(n + 12 ) Γ( 12 ) Z π cos(z cos θ) sin2n θ dθ 0 Note√that Γ(n + 12 ) = (n − 21 )Γ(n − 12 ) = (n − 12 )(n − 23 ) . . . 32 12 Γ( 12 ) and that = π. Γ( 12 ) Another bound that figures in Watson’s treatise is the following ( [5], page 17) where z can be complex:  n Jn (z) = z 2 n! (1 + θ) where |θ| ≤ e |z|2 4(n+1) −1≤ e |z|2 4 −1 n+1 As usual Watson does not prove this bound but the details are as follows: 48 (156) (157) Jn (z) = = ∞ X (−1)m  z n+2m m!(n + m)! 2 m=0 ∞  z n X (−1)m  z 2m m!(n + m)! 2 "m=0 #  z 2  z 4  z 6  z n 1 1 1 1 − + − + ... = 2 n! 1!(n + 1)! 2 2!(n + 2)! 2 3!(n + 3)! 2  n " # z    z 2  z 4  z 6 2 1 1 1 = − + − ... 1− n! 1!(n + 1) 2 2!(n + 2)(n + 1) 2 3!(n + 3)(n + 2)(n + 1) 2  n = 2 z 2 n! (1 + θ) (158)   2  4  6  1 1 z z z 1 − + − . . . . where θ = − 1!(n+1) 2 2!(n+2)(n+1) 2 3!(n+3)(n+2)(n+1) 2 Hence: |θ| = −   z 2  z 4  z 6  1 1 1 − + − ... 1!(n + 1) 2 2!(n + 2)(n + 1) 2 3!(n + 3)(n + 2)(n + 1) 2 =  z 2  z 4  z 6 1 1 1 − + − ... 1!(n + 1) 2 2!(n + 2)(n + 1) 2 3!(n + 3)(n + 2)(n + 1) 2 ≤  z 2  z 4  z 6 1 1 1 + + + ... 1!(n + 1) 2 2!(n + 2)(n + 1) 2 3!(n + 3)(n + 2)(n + 1) 2 ≤  z 4  z 6  z 2 1 1 1 + + + ... 1!(n + 1) 2 2!(n + 1)2 2 3!(n + 1)3 2  z 2 2 n+1 ≤e z 2 −1 e 2 −1 ≤ n+1 (159) To see why the last line above is true note that: 49 e u2 n+1 !2 !3 1 u2 u2 u2 1 + + + ··· − 1 − 1 =1 + n + 1 2! n + 1 3! n + 1 !2 !3 u2 u2 u2 1 1 = + + ... + n + 1 2! n + 1 3! n + 1 u6 u4 u2 + + ... + n + 1 2!(n + 1)2 3!(n + 1)3 u4 u6 u2 + + + ... ≤ n + 1 2!(n + 1) 3!(n + 1) = (160) 2 eu − 1 = n+1 6 Analyticity of Bessel functions Throughout this paper infinite sums have been differentiated term by term and various other manipulations have been performed which necessarily involved concepts convergence or, better still, analyticity. A complex power series P∞ of uniform n a z with a disc of convergence |z| < R where 0 ≤ R < ∞ converges abn=0 n solutely and uniformly within that disc. (see [8], pages 14-18] . This is such an important result it is worth quickly proving. Watson proves the analyticity of Jν (z) ( [5] , page 44) in a way that basically reconstructs the Weierstrass M-test whereas one can prove generally that a power series will converge uniformly within its disc of convergence and moreover because the power series is holomorphic (note every holomorphioc function is analytic) in its disc of convergence, the derivative series is also a power series obtained by term by term differentiation and that series has the same radius of convergence as the original series. I will show both Watson’s approach and the second using the general theorems since the latter is much quicker. P∞ P n n Let’s suppose that ∞ n=1 an w n=1 an z converges for complex w, that is, n converges. This means that an w → 0 as n → ∞. These terms must also be bounded so there exists an M > 0 such that |an wn | < M for all n. Now if we choose z in the disc |z| < R where 0 < R < |w| the following holds: 50 |an z n | =|an | |z n | ≤|an | |Rn | =|an | |wn | ≤M But R w < 1 so the series P∞ n=0 M R w R w Rn wn (161) n n is geometric and converges. So if we P n take Mn = M R in the Weierstrass M-test we see that ∞ n=1 an z converges w uniformly on the disc |z| < R. n Watson proves analyticity as follows:  ν+2m P∞ (−1)m (ν,z) z = where J (z) = Watson uses the ratio test in the form am+1 ν m=0 m!(m+ν)! 2 am (ν,z)   ν+2m P∞ z . Thus with this identification and for |ν| ≤ N and m=0 am (ν, m) 2 |z| ≤ ∆: 51 am+1 (ν, z) −z 2 = am (ν, z) 4(m + 1)(ν + m + 1) −z 2 ≤ 4m(ν + m) ∆2 ≤ 4m(m − N ) (162) 1 1 since for −N ≤ ν ≤ N , m − N ≤ m + ν ≤ N + m and hence m+ν ≤ m−N and 1 ∆2 1 so m(m+ν) ≤ m(m−N ) for m > 1. So all we need is 4m(m−N ) < 1 for the ratio test to ensure convergence and solving the √quadratic 4m2 − 4mN − ∆2 = yields √ 2 2 n± n2 +∆2 m = and so if we take m > N + N2 +∆ it follows that for m > M = 2 √ ∆2 N + N 2 +∆2 , bm = 4m(m−N < 1. Note that this choice of M is independent of ν and 2 ) 1 1 z. Because (m+1)(m+1−N ) < m(m−N it follows that bm is a decreasing sequence of ) positive terms (since m − N > 0 ) bounded below by zero, and hence bm converges to zero as m → ∞. This means that the ratio at issue must itself be bounded, that is, there exists an r such that 0 < r < 1 and there exists a positive integer P with P ≥ M (all of which are independent of ν and z ) such that for all m ≥ P we have that: am+1 (ν, z) ≤r<1 am (ν, z) (163) Note that the fact that r < 1 ensures that we can ultimately use a convergent geometric series - indeed that is actually the crux of one form of the proof of the Ratio Test. Note also that when m = 0, a0 (ν, z) is bounded . That this is so when ν is non-negative integer is clear but when ν is a negative integer the reciprocal of the Gamma function goes to zero. We have that for m > P ie m ≥ P + k where k = 1, 2, 3 . . . : |am (ν, z)| =|aP +k (ν, z)| ≤r|aP +k−1 (ν, z)| ≤r2 |aP +k−2 (ν, z)| .. . ≤rm−P |aP (ν, z)| |aP (ν, z)| m = r rP 52 (164) is bounded for (ν, z) ∈ [−N, N ] × [−∆, ∆] so there But whatever P is, |aPr(ν,z)| P is some constant C > 0 such that: |am (ν, z)| < Crm (165) Hence the Weierstrass M-test applies to bound the original Bessel series by a convergent geometric series and so Jν (z) converges uniformly: ∞ X m=0 |am (ν, z)| < C ∞ X m=0 rm < ∞ (166) Note that when ν is negative the origin is a problem which Watson recognises in his preamble. The theory of power series is very developed and a basic theorem (see [8] Pwell ∞ page 15 ) is that given a power series n=0 an z n , there exists 0 ≤ R ≤ ∞ such that: (1) If |z| < R the series converges absolutely (2) If |z| > R the series diverges Moreover, if we use the convention that by Hadamard’s formula: 1 0 = ∞ and 1 ∞ = 0 , then R is given 1 1 = lim sup|an | n (167) R R is called the radius of convergence and if you have an infinite radius of convergence you can get all the good features mentioned above, eg uniform convergence. For fixed ν in Jν (z) we can write Jν (z) = identification: |am | =
z ν 2 P∞ (−1)m m!(m + ν)!22m (−1)m 2m . m=0 m!(m+ν)!22m z With this (168) A high level observation is that since, for fixed non-negative ν, m!(m+ν)!22m → ∞ as m → ∞ and the mth root is also unbounded, hence the reciprocal appoaches zero. Similarly, if m + ν + 1 is 0 or a negative integer we know that Γ(m + ν + 53 1) = (m + ν)! in unbounded as is its mth root, and hence |m!(m + ν)!22m | is also unbounded and its reciprocal also approaches zero. So it is no surprise that 1 lim sup|am | m m = 0 so that R = ∞. To make this more rigorous we note that for m = 2k even: m! =(2k)(2k − 1)(2k − 2)(2k − 3) . . . 4.3.2.1  2k + 1   2k − 1 + 2   2k − 2 + 3   2k − (k − 1) + k  ... ≥ 2 2 {z2 } | 2 k terms = =  2k + 1 k (169) 2  m + 1  m2 2 = m+1 and When m = 2k + 1 is odd the above pairing yields k pairs of 2k+2 2 2 2k+2 m = = k + 1 and thus (since k + 1 > ): leaves another term which equals m+1 2 2 2 m! ≥ So m! ≥  m+1 2  m + 1 k  2k + 2   m + 1 k+1  m + 1  m2 ≥ = 2 2 2 2  m2 (170) whether m is odd or even. When ν = 0, 1, 2, . . . it is clear that (m + ν)! ≥ m! ≥ ν non-negative: |am | 1 m 1 (−1)m m = m!(m + ν)!22m   m1 1 = m!(m + ν)!22m ≤  = m+1 2 1  m2  m+1 2 1 2(m + 1) 1  m2 22m ! m1  m+1 2  m2 and hence for (171) So as m → ∞, lim sup|am | m = 0 and hence the radius of convergence R is ∞. 54 As noted above, when ν +m+1 is a negative integer Γ(ν +m+1) is unbounded 1 and in those cases lim sup|am | m = 0 and in the cases when it is non-negative the 1 analysis above applies, so in both cases , lim sup|am | m = 0 . This means that the series has an infinite radius of convergence. 7 APPENDIX 7.1 Showing that Bessel’s integral form (5) solves (2) Recall that: 2 Ar = rπ Z π 0 cos[r(φ − e sin φ)] dφ In what follows we let x = e and u = have: du = dx = Z π Z0 π 0 h rπ Ar 2 = Rπ 0 (172) cos[r(φ − x sin φ)] dφ. We thus  ∂  cos[r(φ − x sin φ)] dφ ∂x sin[r(φ − x sin φ)] r sin φ dφ {z } | {z } | dv u iπ Z π r cos φ r(1 − x cos φ) cos(rφ − rx sin φ) dφ = − r cos φ sin(rφ − rx sin φ) + 0 0 {z } | =0 Z π =r2 cos φ (1 − x cos φ) cos(rφ − rx sin φ) dφ 0 Z π Z π 2 2 cos2 φ cos(rφ − rx sin φ) dφ cos φ cos(rφ − rx sin φ) dφ − r x =r Z0 Z 0π 1 du r2 π ∴ = cos φ cos(rφ − rx sin φ) dφ − r2 cos2 φ cos(rφ − rx sin φ) dφ x dx x 0 0 (173) Noting that: du = dx we have: Z π 0 r sin φ sin[r(φ − x sin φ)] dφ 55 (174) d2 u = dx2 Z π  ∂  r sin φ sin[r(φ − x sin φ)] dφ 0 ∂x Z π 2 =−r sin2 φ cos[r(φ − x sin φ)] dφ Z0 π (1 − cos2 φ) cos[r(φ − x sin φ)] dφ = − r2 Z π Z0 π 2 2 cos2 φ cos[r(φ − x sin φ)] dφ cos[r(φ − x sin φ)] dφ +r =−r 0 {z } |0 (175) =u Therefore: Z π Z d2 u 1 du r2 π 2 2 2 + =−r u+r cos φ cos[r(φ − x sin φ)] dφ + cos φ cos(rφ − rx sin φ) dφ dx2 x dx x 0 0 Z π − r2 cos2 φ cos(rφ − rx sin φ) dφ 0 Z r2 π 2 cos φ cos(rφ − rx sin φ) dφ =−r u+ x 0 ! 2 Z π r (1 − x cos φ) − 1 = − r2 u − cos(rφ − rx sin φ) dφ x 0 x Z r2 π (1 − x cos φ) r2 2 cos(rφ − rx sin φ) dφ = − r u + 2u − x x 0 x Z r2 π r2 2 (1 − x cos φ) cos(rφ − rx sin φ) dφ = − r u + 2u − 2 x x 0 Z  r2 r2 π 1  2 = − r u + 2u − 2 d sin(rφ − rx sin φ) x x 0 r " #π r2 2 = − r u + 2 u − sin(rφ − rx sin φ) x 0 | {z } =0 1 = − r2 (1 − 2 )u x 2 1 d u 1 du + r2 (1 − 2 )u = 0 ∴ 2+ dx x dx x If we put rx = z in (176) we get: 56 (176) r2 d2 u 1 du + )=0 (177) + (1 − dz 2 z dz z2 which is seen to be of the same form as (2) after multiplying through by z 2 . The details are as follows: dz du du dz du = r and dx  dx  = dzdx =r dz   d2 u dx2 d d du = dx r du = dz r du dx dz dz 1 du 1 d2 u 2 Therefore: dx2 + x dx + r (1 − x2 )u 2 2 2 r2 ddzu2 + rz du + r2 (1 − zr2 )u = 0 and dz = d dx dz dx 2 = r2 ddzu2 = 0 implies 2 + (1 − so ddzu2 + z1 du dz r2 )u z2 = 0. 7.2 Separation of variables The technique of separation of variables is a fundamental technique for solving problems in mathematical physics which typically involve partial differential equations of varying degress of complexity. Bessel’s equation arises from solutions to the Laplacian in cylindrical coordinates and we start with that (for the various ways of deriving it see [8] ). ∇2 u = 1 ∂ 2u ∂ 2u ∂ 2 u 1 ∂u + + + ∂r2 r ∂r r2 ∂ϕ2 ∂z 2 (178) We let u(r, ϕ, z) = R(r)Φ(ϕ)Z(z) ie u is an independent product of the radial, angular and height variables. Plugging u = RΦZ into (178) we get: 1 1 ′′ ′′ ′′ ′ ΦZR + ΦZR + 2 RZΦ + ΦRZ =0 r r 1 ′ 1 ′′ 1 ′′ 1 ′′ R + R + 2 Φ + Z =0 on dividing by RΦZ R rR r Φ Z (179) At this stage we have that: 1 ′ 1 ′′ 1 ′′ 1 ′′ R + R + 2 Φ =− Z R rR r Φ Z 57 (180) The LHS of (180) has no z dependence but equals something that does have z dependence. If we differentiated the LHS with respect to z we would get zero since there is no z dependence. But the RHS which does have z dependence equals the LHS so its derivative with respect to z is zero and the only way this can happen is ′′ if − Z1 Z equals a constant. We will replicate this argument below. Now for present purposes it doesn’t really matter whether the constant is positive or negative. If it is negative, ie −k 2 , say, then Z(z) = Aekz + Be−kz . If it is positive, ie k 2 , then Z(z) = A cos kz + B sin kz. For the sake of argument, and because I know the ultimate trajectory of this, let’s just take the constant as −k 2 . So what we have now is this: 1 ′ 1 ′′ 1 ′′ R + R + 2 Φ = −k 2 R rR r Φ (181) 1 ′ 1 ′′ 1 ′′ R + R + k2 = − 2 Φ R rR r Φ (182) Thus we have: Replicating the same logic as above we have that both sides of (182) are equal to some constant n2 : This leads to: 1 ′′ 1 ′ R + R + k 2 =n2 R rR ′′ ′ r2 R + rR + (r2 k 2 − n2 )R =0 (183) which is equation (1). 7.3 Extracts from Bessel’s 1824 paper titled in English ”Investigation of the part of the planetary disturbances which arise from the movement of the sun” Here is the integral form of the Bessel function: 58 Here is Bessel’s differential equation - compare (48) in the extract with with (2) noting that Iki is the equivalent symbol for the modern Ji (k): 59 7.4 The series solution using the Frobenius method As indicated we are looking for a solution to: ′′ ′ x2 y + xy + (x2 − n2 )y = 0 (184) which has the form: y=x ∞ X p j aj x = j=0 ∞ X aj xp+j (185) j=0 This process is intricate but for those used to Fourier theory it should hold no horrors of principle! First we work out the derivatives: X ′ (p + j)aj xp+j−1 (186) (p + j)(p + j − 1)aj xp+j−2 (187) y = j=0 ′′ y = X j=0 Next we make the relevant substitutions into (184) to get: ∞  X j=0  aj (p + j)(p + j − 1)xp+j + aj (p + j)xp+j + (x2 − n2 )aj xp+j =0 ∞  X j=0 ∞ X j=0  aj [(p + j)2 − n2 ]xp+j + aj xp+j+2 =0 aj [(p + j)2 − n2 ]xp+j + 60 ∞ X j=0 aj xp+j+2 =0 (188) Now we divide through by xp to get: ∞ X j=0 ∞ X j=0 ∞ X aj [(p + j)2 − n2 ]xj + 2 2 aj xj+2 =0 j=0 ∞ X j aj [(p + j) − n ]x + (189) j aj−2 x =0 j=2 This becomes: (p−n)(p+n)a0 +(p+1−n)(p+1+n)a1 x+ ∞ X j=2 [(p+j −n)(p+j +n)aj +aj−2 ]xj = 0 (190) This last equation is an identity in x if the coefficient of each power of x vanishes and this will happen if: (a) p = n; or (b) p = −n; and (c) a1 = 0; and (d) (p+j−n)(p+j+n)aj +aj−2 = 0 for j = 2, 3, . . . If (a) or (b) then the constant term involving a0 vanishes. If we choose p = n we get a viable recurrence relation: j(2n + j)aj = − aj−2 =⇒ aj = − 1 aj−2 j(2n + j) (191) If, however, we choose p = −n then we get: (2n − j)jaj = aj−2 (192) Clearly when j = 2n, then 2n − j = 0 and we can’t get a proper recurrence relation so we stick with p = n. 61 Since a1 = 0 it follows from (191) that a3 = 0 and so a5 = 0 and so on. Thus: a2j−1 = 0 for j = 1, 2, . . . (193) For the even indices we have: 1 a2j−2 2j(2n + 2j) 1 a2j−2 =− 2 2 (n + j) 1 a2j−4 a2j−2 = − 2 2 (j − 1)(n + j − 1) 1 (−1)2 a2j−4 =⇒ a2j = j(j − 1)(n + j)(n + j − 1) 24 a2j = − (194) If we perform this process j − 2 times we will arrive at a0 since each iteration drops the index by 2. We therefore arrive at: a2j = 1 a0 (−1)j j! (n + j)(n + j − 1) . . . (n + 1) 22j (195) As a check we try: (−1)3 1 a0 a6 = 3! (n + 3)(n + 2) . . . (n + 1) 26 (196) From (194) we see that: a6 = 1 a2 (−1)2 3 × 2 (n + 3)(n + 2) 24 and from (191) we have that: 62 (197) a2 = −1 a0 + 1) 22 (n (198) Putting (195) and (198) together we can see that (195) is correct. Now since a0 is a common factor in all the terms we let it be equal to the following quantity which leads to a nice final expression: 1 2n n! (199) (−1)j 1 n+2j j!(n + j)! 2 (200) a0 = Hence: a2j = for j = 0, 1, 2, . . . . We are nearly there. Recall that: y= ∞ X aj xp+j j=0 = = ∞ X j=0 ∞ X j=0 a2j xn+2j (201) (−1)j  x n+2j j!(n + j)! 2 =Jn (x) 8 Bessel functions in DNA In the 1950s when the basic research into DNA was being done, the British researchers used X-ray diffraction on sodium thymonucleate fibres to obtain the structure of DNA. Thus in one paper Rosalind Franklin and R G Gosling said the following [9]: 63 ”For a smooth single-strand helix the structure factor on the n-the layer is given by: π Fn = Jn (2πrR)ein(χ+ 2 ) (202) where Jn (u) is the nth -order Bessel function of u , r is the radius of the helix, and R and χ are the radial and azimuthal co-ordinates in the reciprocal space.” The transform of a uniform helix was obtained in the paper by Cochran, Crick and Vann [10]. The “wire” is assumed to be of infinitesimal thickness and of infinite length with radius r and axial spacing P. In what follows I use their symbolism. and essentially repeat their exposition: The helix is defined by these equations: x =r cos(2πz/P ) y =r sin(2πz/P ) z =z (203) With this set up the Fourier transform at a point (ξ, η, ζ) is given by: T (ξ, η, ζ) = Z e2πi(xξ+yη+zζ) dV (204) where dV is a volume element of the helix. Now using ( 203 ) and the fact that dV is proportional to dz we have (subject to a constant of proportionality): T (ξ, η, ζ) = This can be rewritten as : Z P e 0 T (R, ψ, ζ) = Z  2πi rξ cos 2π Pz +rη sin 2π Pz +zζ P e 

2πi rR cos 2π Pz −ψ +zζ 0   dz dz (205) (206) where R2 = ξ 2 + η 2 and tan ψ = ηξ . This last integral ( 206) vanishes unless ζ = Pn , where n is an integer which corresponds to the fact that the X-ray scattering from a helix with has an exact repeat after a vertical distance P, is confined to layer lines at heights ζ = Pn in reciprocal space. Accordingly (206) can be written as: n T (R, ψ, ) = P Z P e 0 2πi rR cos 2π Pz −ψ + nz P 64  dz (207) This integral can be evaluated using the identity: n 2πi Jn (X) = after taking X = 2πRr and φ = Z 2π eiX cos φ einφ dφ (208) 0 2πz . P The final result is: T (R, ψ, π n ) = Jn (2πRr)ein(ψ+ 2 ) P (209) 9 References [1] https://en.wikipedia.org/wiki/Friedrich Bessel [2] Ruel V Churchill Fourier Series and Boundary Value Problems, Second Edition, McGraw-Hill, 1963 [3] Andrew Gray, G B Mathews, , A Treatise on Bessel Functions and their Applications to Phys Merchant Books, 2007 [4] Elias M Stein and Rami Shakarchi, Fourier Analysis: An Introduction, Princeton University Press, 2003. [5] G N Watson, A Treatise on Bessel Functions , Second Edition, Cambridge University Press, 1966 [6] Ali R Amir-Moéz, , Ellipses as Projections of Circles , Pi Mu Epsilon Journal, Vol. p, No.9 (Fall 1993), pp. 573-575, https://www.jstor.org/stable/ 24339931 [7] Peter Haggstrom, Basic Fourier Integrals , https://gotohaggstrom.com/ Basic%20Fourier%20integrals.pdf 65 [8] Peter Haggstrom, The Laplacian in curvilinear coordinates, https://www.gotohaggstrom.com The%20Laplacian%20in%20curvilinear%20coordinates%20-%20the%20full%20story.pdf [9] Rosalind Franklin, R G Gosling, Molecular Configuration in Sodium Thymonucleate, Nature, April 1953, Vol 171, pages 740-741 // [10] W Cochran, F H C Crick, V Vand, The Structure of Synthetic Polypeptides I , Acta Cryst (1952) 5, 581 [11] G H Hardy, A Course of Pure Mathematics, Tenth Edition, Cambridge University Press, 2006 [12] Peter Haggstrom, ”An application of Bernoulli’s inequality to uniform convergence” , https://gotohaggstrom.com/An%20application%20of%20Bernoullis%20inequality% 20to%20uniform%20convergence.pdf [13] Elias M Stein and Rami Shakarchi, “Complex Analysis , Princeton Lectures in Aanalysis II, Princeton University Press, 2003 10 History Created 15 February 2023 66