Bessel function anarchy

[sambellamy]

I am asked to show that the first order Bessel function

J₁(x) = Ʃ^∞_n=0 (-1)ⁿ(x²ⁿ)(x) / (n)!(n+1)!(2²ⁿ)(2)

is a solution to the differential equation

x² J₁''(x) + xJ₁'(x) + (x² -1) J₁(x) = 0

I am having some trouble. I have found:

J₁'(x) = Ʃ^∞_n=0 (-1)ⁿ(2n+1)(x²ⁿ) / (n)!(n+1)!(2²ⁿ)(2)

and x J₁'(x) = Ʃ^∞_n=0 (-1)ⁿ(2n+1)(x²ⁿ⁺¹) / (n)!(n+1)!(2²ⁿ)(2)

J₁''(x) = Ʃ^∞_n=1 (-1)ⁿ(2n+1)(2n)(x^2n-1) / (n)!(n+1)!(2²ⁿ)(2)

J₁''(x) = Ʃ^∞_n=0 (-1)ⁿ⁺¹(2n+3)(2n+2)(x²ⁿ) / (n=1)!(n+2)!(2²ⁿ)(2²)

and x² J₁''(x) = Ʃ^∞_n=0 (-1)ⁿ⁺¹(2n+3)(2n+2)(x²ⁿ)(x²) / (n=1)!(n+2)!(2²ⁿ)(2²)

and (x² -1) J₁(x) = Ʃ^∞_n=0 [(-1)ⁿ(x²ⁿ)(x³) / (n)!(n+1)!(2²ⁿ)(2) - (-1)ⁿ(x²ⁿ)(x) / (n)!(n+1)!(2²ⁿ)(2)]

with all that work, I get that the whole thing is

Ʃ^∞_n=0 [ (x²ⁿ)(x) / (n+1)!(2²ⁿ)(2) ] times [(-1)ⁿ⁺¹(2n+3)(2n+2)(x) / (n+2)!(2) + (-1)ⁿ(2n+1)/ (n)! + (-1)ⁿ(x²) / (n)! - (-1)ⁿ/ (n)! ]

what have I done? Please help!

 

[Ishuda]

When trying to prove things like this, what you generally want to do is to get the the powers of x to be the same under the summation. For J₁(x), than is x²ⁿ⁺¹ so first get the other summations in that form.

J₁(x) is in that form
J₁(x) = \(\displaystyle \Sigma_0^\infty (-1)^n \frac{x^{2n+1}}{n!\space (n+1)!\space 2^{2n+1}}\)

x J₁'(x) is in that form
x J₁'(x) = \(\displaystyle \Sigma_0^\infty (-1)^n (2n+1)\space \frac{x^{2n+1}}{n!\space (n+1)!\space 2^{2n+1}}\)

x² J₁''(x) is in that form
x² J₁''(x) = \(\displaystyle \Sigma_0^\infty (-1)^n 2n\space (2n+1)\space \frac{x^{2n+1}}{n!\space (n+1)!\space 2^{2n+1}}\)
Note than the n=0 term is zero, so we can start the summation at one, i.e.
x² J₁''(x) = \(\displaystyle \Sigma_1^\infty (-1)^n 2n\space (2n+1)\space \frac{x^{2n+1}}{n!\space (n+1)!\space 2^{2n+1}}\)


To complete the part we need, we still need to calculate x² J₁(x). It is not in the form we want, i.e.
x² J₁(x) = \(\displaystyle \Sigma_0^\infty (-1)^n \frac{x^{2n+3}}{n!\space (n+1)!\space 2^{2n+1}}\)
We can put it in the form we want by substituting m-1 for n. First we note that when n=0, m=1 so that the summation will start at one. Thus we have
x² J₁(x) = \(\displaystyle \Sigma_1^\infty (-1)^{m-1}\space \frac{x^{2(m-1)+3}}{(m-1)!\space (m-1+1)!\space 2^{2(m-1)+1}}\)
= -\(\displaystyle \Sigma_1^\infty (-1)^{n}\space \frac{x^{2n+1}}{(n-1)!\space (n)!\space 2^{2n-1}}\)
=-\(\displaystyle \Sigma_1^\infty (-1)^{n}\space (n\space (n+1)\space 4)\space\frac{x^{2n+1}}{(n-1)!\space n\space (n)!\space (n+1)\space 2^{2n-1}\space 4}\)
=-\(\displaystyle \Sigma_1^\infty (-1)^{n}\space (n\space (n+1)\space 4)\space\frac{x^{2n+1}}{n!\space (n+1)!\space 2^{2n+1}}\)
where I replaced the dummy summation variable m by n.

To make notation easier let
\(\displaystyle \alpha_n = (-1)^n \frac{x^{2n+1}}{n!\space (n+1)!\space 2^{2n+1}}\)
and we can write
x² J₁''(x) = \(\displaystyle \Sigma_1^\infty 2n\space (2n+1)\space \alpha_n \)
x J₁'(x) = \(\displaystyle \Sigma_0^\infty (2n+1)\space \alpha_n\)
(x²-1) J₁(x) = -[\(\displaystyle \Sigma_1^\infty (n\space (n+1)\space 4)\space \alpha_n\) + \(\displaystyle \Sigma_0^\infty \alpha_n\)]

Now pull out all the n=0 terms and add them - they should add to zero. You can then write what is left over as
\(\displaystyle \Sigma_1^\infty a_n \alpha_n\)
where a_n is the sum of the terms under the summation sign and the a_n should be zero. That, of course, is if I haven't made any misteaks.

Edit: Speaking of such, dumb misteaks that is, I had left in the (-1)ⁿ in one of those summations and had to take it out.

 

[sambellamy]

x² J₁''(x) is in that form
x² J₁''(x) = \(\displaystyle \Sigma_0^\infty (-1)^n 2n\space (2n+1)\space \frac{x^{2n+1}}{n!\space (n+1)!\space 2^{2n+1}}\)
Note than the n=0 term is zero, so we can start the summation at one, i.e.
x² J₁''(x) = \(\displaystyle \Sigma_1^\infty (-1)^n 2n\space (2n+1)\space \frac{x^{2n+1}}{n!\space (n+1)!\space 2^{2n+1}}\)

I thought that I had to change the n terms in the function to n+1 when I change the index from n=1 to n=0?

I also thought the indexes had to match if we were to begin adding and subtracting the series?

 

[Ishuda]

I thought that I had to change the n terms in the function to n+1 when I change the index from n=1 to n=0?

If you were changing the index as I did for the x² J₁(x), you would have to change it that way. However what was done here was to just drop the term. That is suppose you have the sum
\(\displaystyle \Sigma_{n=0} n\) = 0 + 1 + 2 + 3 + 4 + ...
We can also write that as
\(\displaystyle \Sigma_{n=1} n\) = 1 + 2 + 3 + 4 + ...

I also thought the indexes had to match if we were to begin adding and subtracting the series?

They do:
\(\displaystyle \Sigma_0^\infty (2n+1)\space \alpha_n\) = \(\displaystyle \alpha_0 + \Sigma_1^\infty (2n+1)\space \alpha_n\)
and
\(\displaystyle \Sigma_1^\infty (n\space (n+1)\space 4)\space \alpha_n + \Sigma_0^\infty \alpha_n = \Sigma_1^\infty (n\space (n+1)\space 4)\space \alpha_n + \alpha_0 + \Sigma_1^\infty \alpha_n\)
= \(\displaystyle \alpha_0 + \Sigma_1^\infty (n\space (n+1)\space 4)\space \alpha_n + \Sigma_1^\infty \alpha_n\)
So "pull out all the n=0 terms" and everything else has a starting index of 1 and, since the indexes match we can add them.