Properties of Legendre Polynomials

Chapter C Properties of Legendre Polynomials C Definitions The Legendre Polynomials are the everywhere regular solutions of Legendre s Equation, which are possible only if x 2 )u 2xu + mu = x 2 )u ] + mu =, C.) m = nn + ), n =,,2,. C.2) We write the solution for a particular value of n as P n x). It is a polynomial of degree n. If n is even/odd then the polynomial is even/odd. They are normalised such that P n ) =. P x) =, P x) = x, P 2 x) = 3x 2 )/2, P x) = 5x 3 3x)/2. then, treating x 2 ) n = x ) n x + ) n as a product and using Leibnitz rule to differentiate n times, we have vx) = 2 n n! n!x + )n + terms with x ) as a factor), C2.3 It satisfies the equation Finally v) = n!2n 2 n n! =. x 2 )v 2xv + nn + )v =, since, if hx) = x 2 ) n, then h = 2nx x 2 ) n, x 2 )h + 2nxh =. Now differentiate n + times, using Leibnitz, to get or x 2 )h n+2 2n + )xh n+ 2 n + )n h n + 2nxh n+ + 2nn + )h n =, 2 x 2 )h n+2 2xh n+ + nn + )h n =. As the equation is linear and v h, v satisfies the equation also. C2.4 And that s it Thus vx) is proportional to the regular solution of Legendre s equation and v) = P) =, so vx) = Px). C2 Rodrigue s Formula They can also be represented using Rodrigue s Formula d n P n x) = 2 n n! dx nx2 ) n. This can be demonstrated through the following observations C2. Its a polynomial The right hand side of C.3) is a polynomial. C2.2 It takes the value at. If vx) = d n 2 n n! dx nx2 ) n, 26 C.3) C3 Orthogonality of Legendre Polynomials The differential equation and boundary conditions satisfies by the Legendre Polynomials forms a Sturm-Liouiville system actually a generalised system where the boundary condition amounts to insisting on regularity of the solutions at the boundaries). They should therefore satisfy the orthogonality relation If P n and P m are solutions of Legendre s equation then Integrating the combination P m C.5) P n C.6) gives P n x)p m x) dx =, n m. C.4) x 2 )P n] + nn + )P n =, C.5) x 2 )P m] + mm + )P m =. C.6) P m x 2 )P n ] P n x 2 )P m ] dx + nn + ) mm + )] P n P m dx =. 27

Using integration by parts gives, for the first integral, Pm x 2 )P ] n P n x 2 )P m] P m x2) P n P n x2 )P m dx =, } {{ } = as P m,n and their derivatives are finite at x = ± i.e. they are regular there). Hence, if n m C4 What is P2 n dx? P n P m dx =. We can evaluate this integral using Rodrigue s formula. We have I n = P 2 n dx = 2 2n n!) 2 d n dx nx2 ) n dn dx nx2 ) n dx. Integrating by parts gives ] d 2 2n n!) 2 n )I n = dx nx2 ) n d n dx nx2 ) n dn+ dx n+x2 ) n dx. C.7) Note that differentiating x 2 ) n anything less than n times leaves an expression that has x 2 ) as a factor the first of these two terms vanishes. Similarly, integrating by parts n times gives I n = )n 2 2n n!) 2 x 2 ) n d2n dx 2nx2 ) n dx. The 2n)th derivative of the polynomial x 2 ) n, which has degree 2n is 2n)!. Thus I n = )n 2n)! 2 2n n!) 2 x 2 ) n dx. The completion of this argument is left as an exercise. One way to proceed is to use the transformation s = x + )/2 to transform the integral and then use a reduction formula to show that s n s) n ds = n!)2 2n + )!. The final result is P 2 n dx = 2 2n +. C5 Generalised Fourier Series C.8) Sturm-Liouiville theory does more than guarantee the orthogonality of Legendre polynomials, it also shows that we can represent functions on,] as a sum of Legendre Polynomials. Thus for suitable fx) on,] we have the generalized Fourier series fx) = a n P n x). 28 C.9) To find the coefficients a n, we multiply both sides of this expression by P m x) and integrate to obtain 2 P m x)fx) dx = a n P n x)p m x) dx = a m 2m +, C6 a n = n + ) fx)p n x) dx. 2 A Generating Function for Legendre Polynomials C6. Definition We consider a function of two variables Gx,t) such that Gx,t) = P n x)t n, n= C.) C.) the Legendre Polynomials are the coefficients in the Taylor series of Gx,t) about t =. Our first task is to identify what the function Gx,t) actually is. C6.2 Derivation of the generating function. We know that, in spherical polar coordinates, the function r is harmonic, away from r =, i.e. 2 r = 2 x =. It is a harmonic function independent of φ. Similarly / x x ) is harmonic away from x = x. If x is a unit vector in the z-direction, then So the function x x 2 = x x ).x x ) = x.x 2x.x + x.x = r 2 2r cos θ +. C.2) r 2 2r cos θ + is harmonic, regular at the origin and independent of φ. We should therefore be able to write it in the form, r 2 2r cos θ + = A n r n P n cos θ). 29 n= C.3)

x z x φ θ r x x If we can show that A n =, then the replacement of cos θ by x and r by t gives the required result. To find A n, we evaluate the function along the positive z-axis, putting cos θ =, noting that So r 2 2r + = = r ) 2 r = r = + r + r2 + r 3 +, r <. r n = A n r n P n ) = A n r n, n= n= A n =. Thus, with x = cos θ and t = r, n= Gx,t) = = t n P n x). 2xt + t 2 C6.3 Applications of the Generating Function. n= x y C.4) Generating functions can be applied in many ingenious ways, somethimes best left for examination questions. As an example, we can differentiate Gx,t) with respect to t to show that G t = x t 2xt + t 2 ) 3/2 = 2xt + t 2 ) G = x t)g. t Now write Gx,t) as a sum of Legendre polynomials to get 2xt + t 2 ) n P n x)t n = x t) P n x)t n. n= Now comparing the coefficients of t gives P x) = xp x),, as P =, we have P = x, as expected. Comparing the general coefficient of t n, n >, gives n + )P n+ 2xn P n +n )P n = xp n P n, 3 n= or, rearranging n + )P n+ 2n + )xp n +n P n =, C.5) a recursion relation for Legendre polynomials. Differentiating Gx,t) with respect to x, and proceeding in a similat way yields the result P n+ 2xP n + P n = P n, n. C.6) Combining C.5) and C.6), or obtaining a relationship between G x and G t, shows These need not be learnt. C6.4 Solution of Laplace s equation. P n+ P n = 2n + )P n. n= C.7) Remember where we first came across Lagrange polynomials. If 2 u = and u is regular at θ =,π in spherical polar coordinates and with / φ = then ur,θ) = A n r n + B ) n r n+ P n cos θ). C.8) C7 Example: Temperatures in a Sphere The steady temperature distribution ux) inside the sphere r = a, in spherical polar coordinates, satisfies 2 u =. If we heat the surface of the sphere u = fθ) on r = a for some function fθ), what is the temperature distribution within the sphere? The equation and boundary conditions do not depend on φ so we know that u is of the form C.8). Further more we expect u to be finite at r = B n =. We find the coefficients A n by evaluating this on r = a. We require fθ) = A n a n P n cos θ). n= We can find A n using the orthogonality of the polynomials C.7). However in C.7), the integration is with respect to x and not cos θ. If x = cos θ, then dx = sinθ dθ. The interval x is the interval π θ. Multiply through by sin θ P m θ) and integrate in θ to obtain So π sin θfθ)p m cos θ) dθ = ur,θ) = n= = a n A n ) n= a n A n ) n= π sin θ P n cos θ)p m cos θ) dθ P n x)p m x) dx = 2am A m 2m +. n + ) r ) n π Pn cos θ) fν)p n cos ν)sin ν dν. 2 a C.9) C.2) Let us heat the northern hemisphere and leave the southern half cold, fθ) = for θ π/2 and fθ) = for π/2 < θ π. Then the integral in C.2) is π/2 sin ν P n cos ν) dν = 3 P n x) dx.

An integration of C.7) gives 2n + ) We know that P q) dq = so x P n q) dq = P n+ P n ] x = P nx) P n+ x), n >. ur,θ) = 2 + 2 r ) n Pn cos θ)p n ) P n+ )). a n= Note that the temperature at the centre of the hemisphere r = ) is /2, which might be expected. The Legendre polynomials of odd degree are odd and will be zero at the origin the coefficients in the sum will be zero for even values of n. Hence ur,θ) = 2 + 2 r a) r ) 2m P2m+ cos θ)p 2m ) P a 2m+) )). C.2) m= We need the values at the origin of the polynomials of even degree. Putting x = in C.4) gives Therefore and t n P n ) = n= + t 2 t 4 = 2 t2 +.3 2.2 2!.3.5 2.2.2 3! + = ) m2m )2m 3) 3. 2 m m! m= = ) m 2m)! 2 2m m!) 2t2m. m= P 2m ) P 2m+) ) = ) m 2m)! 2 2m m!) 2 2m + 2)! )m+ 2 2m+2 m + )!) 2 = ) m 2m)! 2 2m m!) 2 + 2m + ), 2m + 2 ur,θ) = 2 + 2 m= t 6 t 2m r a) r ) 2m P2m+ cos θ)) m 2m)! a 2 2m m!) 2 + 2m + ). 2m + 2 C.22) We will evaluate this expression along the axis of the sphere. Here cos θ = ±, depending if we are in the northern or southern hemisphere. The polynomials in appearing are odd so take the value ± at ±. This can be accounted for by allowing r to be negative it measures the directed distance from the centre in a northerly direction. ur) = 2 + 2 r a) r ) 2m ) m 2m)! a 2 2m m!) 2 + 2m + ). 2m + 2 m=.5.25.75.5.25 - -.5.5 We can see that the convergence is not good near the poles. The line that actually attains the values and at the poles is the exact solution ur) = 2 + r/a)2 + + r/a) 2 2r/a) + r/a) 2. This is attained as follows. Equation C.2) tells us ur) = 2 + 2 r a) r ) 2m P2m ) P a 2m+) )). m= Identifying t with r/a), and using C.22),χ recognising that the sum only contains the even powers of t, gives r/a) 2m P 2m ) r/a) 2m 2 P 2m ) = + r/a) 2 r/a) 2 + r/a) 2 m= m= Changing the index in the second sum, using P =, we find And so m= r/a) 2m P 2m ) P 2m+2 )) r/a) 2 = + r/a) 2 r/a) 2 + r/a) 2. ) ur) = 2 + r 2 a + r/a) 2 r/a) 2 + r/a) + 2 r/a) 2, which simplifies to the result above. A graph of the solution obtained by summing this series to,4,7,,3 terms is shown below 32 33

Note that, defining ω 2 = λc 2, we would have ended up with λ rather than λ 2 in D.4), but that would lead to lots of s in what follows. Note now that to find the frequency of oscillation we need to solve the eigenvalue problem D.4), with ua,θ) = from D.2) and u finite at r =. Chapter D Oscillation of a circular membrane D The Problem and the initial steps in its solution D. The problem If we have a circular drum, radius a, and hit it, we will set the drum vibrating. To study this vibration, we need to solve the wave equation c 2 2 ψ = ψ tt, r a D.) with c the speed of wave motion in the drum s material and ψ the displacement of the drum s surface. We have the boundary condition ψa,θ) =, θ < 2π, D.2) corresponding to the drum being fixed at its circular edge. We also expect ψ to be finite at r =, the drum s centre. It we hit the drum,, at t =, ψ =, ψ t r,θ) = fr), then we must also impose this initial condition. D.2 Separating out the time dependence We look for oscilliatory solutions, writing ψr,θ,t) = ur,θ)expiωt). D.3) This is equivalent to looking for solutions of the type ψ = ux)tt) and knowing in advance that the equation for T will have the form T + ω 2 T = where ω is related to the separation constant. This has solutions proportional to cos ωt and sin ωt and we choose to write these in exponential form. The value of ω is the frequency of the disturbance. Doing so leads to c 2 2 u = ω 2 u, or 2 u + λ 2 u =, D.4) after dividing through by the exponential factor). Here ω = λc and we need to solve Helmholtz equation for the spatial part of the solution. We should expand a little on the shorthand, expiωt), that we are using to describe the temporal part of the solution. The solution of the equation for T is T = A ω cosωt) + B ω sinωt). To write this in exponential form we can take the real part of A ω ib ω )cos ωt + isin ωt), i.e. of A ω ib ω )expiωt). We therefore have a complex constant multiplying D.3) which we have omitted. We can write this complex constant in modulus-argument form as R ω expiǫ ω t). 34 D.3 Separating out the θ dependence u rr + r u r + r 2u θθ + λ 2 u =. D.5) We are looking for solitions that are 2π-periodic in θ as we have a circular drum. We know that if we look for solutions of the form ur,θ) = Rr)Θθ) then the θ-dependence will satisfy Θ + p 2 Θ =, where to further ensure periodicity in θ, p 2 and finally to ensure 2π-periodicity p = n for integer n, n =,, 2, 3,... The case n = corresponds to solutions with no θ-dependence. In general the θ-dependence of the solution is like sinnθ) and cosnθ). We choose to write both of these together as expinθ) strictly R n expinθ)expiǫ n )) and look for solutions of the type u = Rr)expinθ). Subsitution into D.5), dividing out the exponential factor and multiply by r 2 gives and if we write z = λr, R = wz), r 2 R + rr + λ 2 r 2 n 2 )R =, z 2 w + zw + z 2 n 2 )w =. D.6) D.7) We have obtained solutions of this equation for integer n through series and found that it has two independent solutions. We can then write wz) = A n J n z) + B n Y n z). D.8) The point z = corresponds to the centre of the membrane and we wish our solution to be analytic here. This means that we must set B n = and we consider only solutions finite at z =, z = A n J n z). At this stage our solution is of the form ψr,θ,t) = λ A nλ J n λr)expinθ)expicλt) n= D.9) where A nλ = R nλ expiǫ λ )expiǫ n ) are complex) constants that we need to find so as to satisfy the initial conditions. In relating this to our definitions above we have used R nλ = R ω R n. We return to this solution later, but for the present we look more closely at the properties of the solutions of Bessel s equation so as to firstly enable us to fix possible values of λ and secondly to express the initial conditions as generalised Fourier Series. D.4 On Bessel Functions D.4a Expression as a series For general p we have the series solution J p x) = ) j x ) 2j+p D.) Γj + )Γj + p + ) 2 j= 35

with Γz) the Gamma function, extending the factorial function to non-integer argument and with n! = Γn + ). The second independent solution is J p x) if p is not an integer. However if p is an integer then as J n = ) n J n, these solutions are linearly dependent. This can be seen as follows. We have the series ) r z ) 2r±ν J ±ν =,ν =,,2,. r!r ± ν)! 2 r=n r= If we look at the solution J ν, and recall that x! = Γ+x) has singularities at x =, 2, 3, then we realise that the first few terms in the sum become zero as ν n an integer. These terms correspond to r n =, 2, 3,, n, corresponding to r = n,n 2,,. Thus, if ν is an integer we need start the series at r = n for J n. This gives ) r z ) 2r n ) r ) n z ) 2r+n J n z) = = = ) n J n z), r!r n)! 2 r + n)!r)! 2 r= using r old = r new +n. To cover this case also a second linearly independent solution Y p x) is taken. This is defined as Y p x) = J p cos pπ J p x) D.) sin pπ and the limit p n considered if p = n. Note Y p is often written N p.) For small x x ) p J p x) D.2) Γp + ) 2 Y x) 2/π)lnx/2) + γ +...) D.3) Y p x) Γp) ) 2 p D.4) π x D.4b Behaviour for large x. For large x all these solutions behave like a damped sinusoidal function with yx) Asinx+ǫ)/ x. This is easy to demonstrate, writing wx) = fx)yx) and substituting into D.7) x 2 f y + 2f y + fy ) + xf y + fy ) + x 2 n 2 )fy = The coeficient of y can be made zero if 2x 2 f + xf = giving f x /2. Choosing f = x /2 and dividing by x 3/2 gives ) y /4 n2 + + x 2 y =. For large x, this is well approximated by y + y = y = Asinx + ǫ) and the result. TablePlotBesselJi, x], {x,, 5}, PlotPoints -> 5], {i,, 5}];Show%].5 -.5 - -.5-2 -2.5-3 2 4 6 8 2 4 TablePlotBesselYi, x], {x,, 5}, PlotPoints -> 5], {i,, 5}]; Show%,PlotRange -> {-3, }] All the solutions have an infinite number of zeros. The zeros of J n are denoted j nm J n j nm ) = and < j n < j n2 < j n3 <... D.5 Determination of λ j n j n2 j n3 j n4 j n5 n = 2.448 5.52 8.6537.795 4.939 n = 3.837 7.56.73 3.323 6.47 n = 2 5.356 8.472.698 4.796 7.96...... We have the boundary condtion ψa,θ) =, i.e. wλa) =. Thus λ is determined λa = j nm or λ = j nm /a, m =,2,3,. Thus the possible frequencies of the drum s vibration are determined as the doubly infinite family For each frequency the vibration is described by ω = ω nm = cj nm /a D.5).8.6.4.2 -.2 -.4 2 4 6 8 2 4 36 J n j nm r/a)expicj nm t/a)expinθ) D.6) The general solution is an arbitray linear combination of all these modes D.9) becomes the double sum ψr,θ,t) = A nm J n j nm r/a)expinθ)expicj nm t/a). D.7) n= m= Here now A nm = R nm expiǫ n )expiǫ m ). The constants R nm can be related to the amplitude of a particular mode, ǫ n to its orientation realtive to the line θ = and ǫ m to its phase where in the sinusoidal temporal cycle it started). 37

n =, m = n = 3, m = 4 << NumericalMath BesselZeros ;n = ; m = 2;jnm = BesselJZerosn, m]m]]; radialr_] = BesselJn, jnm r];azimuththeta_] = ReExpI n theta]]; wrapf_, r_, theta_] = Ifr <, fr, theta], ];polarrx_, y_] = Sqrtx^2 + y^2]; polarthetax_, y_] = Ifx >, ArcTany/x], Ify >, Pi/2 + ArcTan-x/y], -Pi/2 - ArcTanx/y]]];moder_, theta_] = radialr]azimuththeta]; surf = Plot3Dwrapmode, polarrx, y], polarthetax,y]], {x, -, },{y, -, }, PlotRange -> All, PlotPoints -> {,}, Lighting -> False, Mesh ->False, Axes -> False, Boxed -> False];cont = ContourPlotwrapmode, polarrx, y], polarthetax, y]], {x, -,}, {y, -, }, PlotRange -> All, PlotPoints-> {, }, FrameTicks->None];ShowGraphicsArray{surf, cont}]] D.6 Incorporating the Initial Conditions Our initital conditions are that, at t =, ψ = and ψ t = fr). Putting t = in D.7) gives n =, m = 2 = A nm J n j nm r/a)expinθ). n= m= Differentiating and putting t = in D.7) fr) = n= m= icj nm /a)a nm J n j nm r/a)expinθ) where A nm are complex constants as above) and it is understood that we take the real part of the sum. The right hand side has no θ-dependence and we can deduce that we need only the component n = from the first sum. Thus we must find A m = A m = R m expiǫ m ), say, such that, taking real parts, = fr) = R m expiǫ m )J n j m r/a). m= icj m /a)r m expiǫ m )J n j m r/a). m= D.8) D.9) n =, m = 2 38 This can be achieved by setting expiǫ m ) = i, remembering that R m is real. The choice i is driven by a desire for neatness in D.9). 39

Note that we have been a little overgeneral in our presentation here. Right at the start we could have described the time-dependence by the solution sinωt). This is zero but has a non-zero time derivative at t = and is obviously that which is required for our particular initial conditions. This corresponds to our current choice of a purely imaginary value for expiǫ m ). Similarly we could have realised that, as the initial and boundary conditions had no θ-dependence, neither the final solution and included only the n = mode from the start. If there was some θ-dependence in the initial condition, then we would deal with each Fourier component in θ) separately. We are looking to represent fr) as a generalised Fourier Series in r, otherwise known as a Fourier-Bessel Series. This is possible as the differential equation satisfied by the Bessel functions, together with the boundary conditions - finiteness at r = and at r = a are a Sturm-Liouvuille problem. D.7 Orthogonality of Bessel Functions The set of functions J n λr) in r a) with λ = j m /a are eigenfunctions of the Sturm-Liouiville problem r 2 R + rr + λ 2 r 2 n 2 )R =, R) finite, Ra) = D.2) We can rewite this as rr ] ) + λ 2 r n2 R =, r with solution R = J n λr). Let λ = λ i be such that J n λ i a) = and λ be a different value. We have ) r Jn λ i r) ] + λ 2ir n2 J n λ i r) =, D.2) r ) r Jn λr) ] + λ 2 r n2 J n λr) =. D.22) r If we multiply D.2) by J n λr), D.22) by J n λ i r), subtract and integrate we get J n λr) r J n λ i r) ] J n λ i r) r J n λr) ] dr + λ 2 i λ2 ) Use integration by parts on the first integral to get n 2 J n λ i r)j n λr) J n λr)j n λ i r)) } {{ } = r J n λ i r)j n λr) dr dr r =. Jn λr) { r J n λ i r) } J n λ i r) { r J n λr) }] J n λr) r J n λ i r) J n λ i r) r J n λr) dr } {{ } +λ 2 i λ2 ) = r J n λ i r)j n λr) dr =. Now J n λr) = λj n λr) so, using the fact that J nλ i a) =, we have the result D.23) r J n λ i r)j n λr) dr = aλ i J nλ i a) J nλa) λ 2 λ 2. D.24) i Thus if λ λ i and Jλa) =, i.e. λ is another, distinct, eigenvalue, then r J n λ i r)j n λr) dr =. 4 D.25) To see what happens if λ = λ i, let λ λ i and use l Hopital s rule. r J n λ i r) 2 dr = aλ i J n λ λ ia) J nλa) λ=λi λ λ2 λ 2 i ) λ=λi = a2 J 2 n λ i a) ] 2. D.26) We can take λ and λ i as different roots of J, j m. We shall see in the section on the generating function that follows that J n± x) = n x J nx) J nx), = J ) 2 = J ) 2, and we can easily evaluate the values of J at the zeros of J. D.8 The solution We can now consider D.9), multiply by r J rj n /a) and integrate between and a. Only one element of the sum survives due to the orthogonality and we have r Jrj n /a)fr) dr = R n cj n /a)a 2 /2)J j n )] 2, giving R n We make the substitution for A m = R m expiǫ m ) = ir m in D.7) and take the real part of the result to yeild ψr,θ,t) = m= 2 r Jrj m/a)fr) dr cj m aj j m )] 2 J rj m /a)sincj m t/a). D.27) The animation at http://www.ucl.ac.uk/ ucahdrb/mathm242/ illustrates this solution and used the following commands << NumericalMath BesselZeros ;fx_] = Sin2Pi x]; stop = ; c = ;jm = BesselJZeros, stop]; coef = Map2NIntegratex BesselJ, # x] fx], {x,, }]/c# BesselJ, #]^2) &, jm]; solnr_, t_] := Trcoef*MapBesselJ, r#]&, jm]*mapsinc # t] &, jm]]; rest_] := Module{},wrapf_, r_]=ifr <, fr, t], ]; polarrx_, y_] = Sqrtx^2 + y^2]; surf = Plot3Dwrapsoln, polarrx, y]], {x, -, }, {y, -, }, PlotRange -> {-.3,.3}, PlotPoints -> {2, 2},Lighting -> False, Mesh -> False, Axes -> False, Boxed -> False]]; Export"res.gif", Tableresi], {i,, 6,.}]] D.8a Generating Function There is a generating function for the Bessel functions J n. It turns out that n= x J n x)t n = exp t )). D.28) 2 t This can be derived by considering solutions of Helmholtz equation for Gx) in Cartesian and in polar coordinates. 2 G + G =, G xx + G yy + G =, r 2 G rr + rg r + G θθ + r 2 G =. One solution is G = expiy), corresponding to a plane wave. In polar coordinates, this is G = expir sin θ). We have seen that the solution in polar coordinates can be found by separating out 4

the angular dependence expinθ) leaving the radial dependence satisfying Bessel s equation. The plane wave is regular at the origin so we do not need the Y n solutions. We know from section A8 that we should be able to write G as follows G = expir sin θ) = A n J n r)expinθ) D.29) for some constants A n. If we write t = expiθ) sinθ = t t )/2, and replace r by x then we have x exp t )) n= = A n J n x)t n. D.3) 2 t To show that the A n are all equal to one we use our knowledge of the expansions of J n x) for small x, obtained from the series solutions. From D.2) J n n) = 2 n. We start by isolating a particular J m from the sum D.29) by effectively using the orthogonality properties of cosnθ) and sinnθ) which lie behind the idea of Fourier Series. We multiply D.29) by exp imθ) and integrate expir sin θ imθ)dθ = A n J n r) Now differentiate m times with respect to r and put r = to find i m sin m θ)expir sin θ)dθ = 2πA m J m mr), = Using the exponential form for sinθ gives i m sin m θdθ = 2πA m /2 m. expinθ imθ)dθ = 2πA m J m r) 2. If we make the replacement t /t, we see that Gx, /t) = exp t )) 2 + t = exp t )) = Gx,t), 2 t J n x) )n t n = t n J n x). However the left hand side is also tn J n x)) n, using n, rather than n to take us through the sum. Thus and comparing coefficients of t n, Thus from D.3), with n =, t n J n x) )n = J n x) = ) n J n x), J x) = J x), J x) = J x). t n J n x), J 2 x) = J 2 x). 3. If we choose to differentiate G with respect to t, then we can show 2n x J nx) = J n x) + J n+ x). D.32) D.33) D.34) A m = 2m 2π 2 mexp iθ exp iθ)m exp imθ)dθ = 2π exp 2imθ)) m dθ = 2π =, 2π as the only non-zero contribution to the integral comes from integrating the first term in the binomial expansion of the integrand. D.8b Use of the Generating Function. We see that t n J nx) = 2 G x = 2 and comparing coefficients of powers of t, t ) G t t n+ J n x) 2 t n J n x), J nx) = 2 J nx) J n+ x)). D.3) 42 43