Bessel function - obtaining a first-order differential equation

[matematicar73]

Hi there,
I'm stuck with an example that probably is very simple but I can't find the way around it.

f(z,t) = sum_(-∞)^∞ [ t^(n) * J_(n) (z)] where J_(n) (z) is a Bessel function of order n.

I need to obtain a first-order differential function relating df/dz and f(z,t) and I have to deduce that f(z,t) = exp [z/2 * (t - 1/t)].

so far I know that df/dz = sum_(-∞)^∞ [t^(n) * J'_(n) (z)] and with the Bessel function property 2J'_(n) (z) = J_(n-1) (z) - J_(n+1) (z)
there is:

df/dz = sum_(-∞)^∞ [1/2(t^n * J_(n-1) (z) - t^n * J_(n+1) (z))

any help going [FONT=Helvetica Neue, Helvetica, Arial, sans-serif]further?

more given properties of Bessel functions are:
J_(n) (-z) = J_(-n) (z) = (-1)^n * J_(n) (z)
J_(n) (0) = 0 for all n greater than 0
J_(0) (0) = 1[/FONT]

 

[Ishuda]

Hi there,
I'm stuck with an example that probably is very simple but I can't find the way around it.

f(z,t) = sum_(-∞)^∞ [ t^(n) * J_(n) (z)] where J_(n) (z) is a Bessel function of order n.

I need to obtain a first-order differential function relating df/dz and f(z,t) and I have to deduce that f(z,t) = exp [z/2 * (t - 1/t)].

so far I know that df/dz = sum_(-∞)^∞ [t^(n) * J'_(n) (z)] and with the Bessel function property 2J'_(n) (z) = J_(n-1) (z) - J_(n+1) (z)
there is:

df/dz = sum_(-∞)^∞ [1/2(t^n * J_(n-1) (z) - t^n * J_(n+1) (z))

any help going further?

more given properties of Bessel functions are:
J_(n) (-z) = J_(-n) (z) = (-1)^n * J_(n) (z)
J_(n) (0) = 0 for all n greater than 0
J_(0) (0) = 1

EDIT:
Break the sum into two parts and rearrange the indices, i.e.

\(\displaystyle f'(z)\, =\, \)\(\displaystyle \displaystyle \sum_{n\, =\, -\infty}^{n\, =\, \infty}\,\) \(\displaystyle \,\dfrac{1}{2}\, \bigg[ t^n\, J_{n-1}(z)\, -\, t^n\, J_{n+1}(z) \bigg]\)

\(\displaystyle \displaystyle =\, \sum_{n\, =\, -\infty}^{n\, =\, \infty}\,\) \(\displaystyle \, \dfrac{1}{2}\, t^n \,J_{n-1}(z)\, -\, ... \)

\(\displaystyle \displaystyle =\, \sum_{n\, =\, -\infty}^{n\, =\, \infty}\,\) \(\displaystyle \, \dfrac{1}{2}\, t^{n+1}\, J_{n}(z)\, -\, ...\)

\(\displaystyle =\, \dfrac{1}{2}\, t\, f(x)\, -\, ...\)

to finally get an equation something like

f'(z) = something * f(z)

and go from there

FORGET THIS PART [part of which I removed]
f'(z) = \(\displaystyle \underset{n=-\infty}{\overset{n=\infty}{\Sigma}}\, \frac{1}{2}\, [t^n J_{n-1}(z)\, -\, t^n\, J_{n+1}(z)]\)
= \(\displaystyle \frac{1}{2}\underset{n=-\infty}{\overset{n=\infty}{\sum}}\, t^n J_{n-1}(z)\, -\, \frac{1}{2}\underset{n=-\infty}{\overset{n=\infty}{\Sigma}}\, t^n\, J_{n+1}(z)\)
= \(\displaystyle \frac{1}{2}\underset{n=-\infty}{\overset{n=-1}{\sum}}\, t^n J_{n-1}(z)\) + ...
= \(\displaystyle \frac{1}{2}\underset{n=\infty}{\overset{n=1}{\sum}}\, t^{-n} J_{-(n+1)}(z)\) + ...
...