Grrr... This gettin' old is gettin' old. Bessel functions?

Question

I need the power series for the first derivative of the n'th Bessel function of the 1'st kind.

I just moved and all my references are still packed and I can't find it on the 'net.
I think it goes like:
J(sub n)' = 
+ [x^(n-1)] / [(2^n)(0!)((n-1)!)]
- [(n+2)(x^(n+1))] / [(2^(n+2))(1!)((n+1)!)]
+ [(n+4)(x^(n+3))] / [(2^(n+4))(2!)((n+2)!)]
- [(n+6)(x^(n+5))] / [(2^(n+6))(3!)((n+3)!)]
+ ...
yada yada yada.

But I'm not certain. (Guess I must be gettin' old ☺)

Does anybody know off the top of their head or can you dig it out from somewhere?

Thanks.

Doug

Joanna W · Accepted Answer

The Bessel functions of the first kind  are defined as the solutions to the Bessel differential equatation: 
 
Jn = (x/2)n/2nΓ(n+1){1 - (x/2)2/2(2n+2) + (x/2)4/2.4(2n+2)(2n+4) - ...}.

The power series expansion for Bessel functions of the first kind of order n is: nx( )=1( )kx2n+2kk! n + k()!k=0

n " 0, where k !" (k)(k !1)(k ! 2) ... (2)(1) and 0! = 1. For!
n < 0, the integer order functions satisfy Jnx( )= 1( )nJn(x)
!
J"n (x) = ("1)n
Jn (x).
 J(sub n)' = + [x^(n-1)] / [(2^n)(0!)((n-1)!)]
- [(n+2)(x^(n+1))] / [(2^(n+2))(1!)((n+1)!)]
+ [(n+4)(x^(n+3))] / [(2^(n+4))(2!)((n+2)!)]
- [(n+6)(x^(n+5))] / [(2^(n+6))(3!)((n+3)!)]
+
We see that when n is a positive integer, Jn(x) starts off as xn. When n = 0, J0(0) = 1. When n is integral, Jn(0) = 0. In all other cases, Jn is infinite at the origin. In many physical problems, the solution must be defined and well-behaved at the origin, which rules out all solutions except for those with integral n. 

Ratios of Bessel functions of the first kind have continued fraction...
When n = 1/2, we obtain the result that J1/2(x) = (2/πx)1/2sin x. From the recurrence relations we can then find that J-1/2 = (2/πx)1/2cos x. 
In case n is integral, we must search for a second solution linearly independent of Jn. For n = 0, such a function is Neumann's, 
Y0(x) = J0(x) log x + {(x/2)2 - (3/2)(x/2)2/(2!)2 + ...}. This function is tabulated, just like J0(x), and has zero that interlace with those of J0(x). 
Finally, the surprising versatility of Bessel functions is shown in Weber's discontinuous integrals. 

Zeros of Jn(x)
s  n=0 n=1 n=2 n=3 n=4 n=5 
1  2.405 3.832 5.135 6.379 7.586 8.780 
2  5.520 7.016 8.147 9.760 11.064 12.339 
3  8.654 10.173 11.620 13.017 14.373 15.700 
4  11.792 13.323 14.796 16.224 17.616 18.982 
5  14.931 16.470 17.960 19.410 20.827 22.220 
6  18.071 19.616 21.117 22.583 24.018 25.431 
7  21.212 22.760 24.270 25.749 27.200 28.628 
8  24.353 25.903 27.421 28.909 30.371 31.813 
9  27.494 29.047 30.571 32.050 33.512 34.983

MathTutor · Answer

Maybe this helps you?

http://en.wikipedia.org/wiki/Bessel_function

I think that the derivative should be different

The general term of the expansion term would be
 
(-1)^i x^(2n+ i)/n!(n+1)! 2^(2n+i)

So, the derivative would be the sum of terms like this:

(-1)^i (2n+i) x^(2n+i-1)
-------------------------------
n!(n+1)!2^(2n+i)

What you wrote looks a bit different.

DD's wife

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Grrr... This gettin' old is gettin' old. Bessel functions?

2 Answers