# Differentiatiang Bessel functions

1. Feb 21, 2008

### Pere Callahan

Hi all,

I am trying to find an expression for the values of the derivates of the Bessel-$$J_1$$ functions at two.

The function is defined by

$$J_1(x)=\sum_{k=0}^\infty{\frac{(-1)^k}{(k+1)!k!}\left(\frac{x}{2}\right)^{2k+1}}$$

this I can differentiate term by term, finding for the n^th derivative at two:

$$J_1^{(n)}(2)=\sum_{k=\left\lfloor\frac{n}{2}\right\rfloor}^\infty{\frac{(-1)^k(2k+1)!}{(k+1)!k!(2k+1-n)!2^n}}$$

Does anybody know a nice expression for this? It can probably be written as the sum of two Besselfunctions evaluated at two, but I have no idea how to find the coefficients...

Thanks

-Pere

Edit:

I found that for n even the result can be written as a linear combination of $$J_1(2)\text{ and } J_2(2)$$. When multiplied by 2^n the coefficients are all integral ... (Experimental results).
I tried the sequence of these integers in that famous integers sequence database, but nothing ...

Last edited: Feb 21, 2008
2. Feb 21, 2008

### Mute

There's an identity

$$J_s'(z) = \frac{1}{2}\left(J_{s-1}(z) - J_{s+1}(z)\right)$$

So the nth derivative would look like

$$J_s^{(n)}(z) = \frac{1}{2}\left(J_{s-1}^{(n-1)}(z) - J_{s+1}^{(n-1)}(z)\right)$$

I guess you might be able to figure out an expression that way. (It might perhaps be best to start with n = 2, 3... to see if you can see a pattern forming, and then use an inductive argument).

(edit: apparently I forgot the forum tag slashes are not the same direction as in LaTeX...)

Last edited: Feb 21, 2008
3. Feb 21, 2008

### Pere Callahan

Thanks for your answer. That's what I actually did first

I got:

$$J_\nu^{(n)}(x)=\frac{1}{2^n}\sum_{k=0}^n{(-1)^k\left(\stackrel{n}{k}\right)J_{\nu-n+2k}(x)}$$

Ok, granted, that's way nicer than the infinte sum I posted above, but still it's a sum of n Bessel functions evaluated at two.. I want to reduce it so that it only involves two Besselfunctions.....any idea how this can systematically be done?

Thanks

-Pere

Last edited: Feb 22, 2008
4. Feb 23, 2008

### Mute

I can't think of anything else to do that seems likely to produce anything nice. You might be able to use the identity

$$J_{s-1}(z) + J_{s+1}(z) = \frac{2s}{z}J_s(z)$$
to move things around, but given that all the terms in that sum have different coefficients it might not work out so nicely (if at all).

You might see if any of the identities in http://www.math.sfu.ca/~cbm/aands/page_358.htm would help (though I didn't see any that looked useful).

5. Feb 23, 2008

### John Creighto

Sense Bessel functions are orthogonal why should should you be able to empress a sum of n Bessel functions as a some of only two Bessel functions?

6. Feb 23, 2008

### Pere Callahan

Only their values evaluated at two .

7. Feb 23, 2008

### John Creighto

So you are trying to evaluate the derivatives at X equal to 2. How about.

$$J_1(x)=\sum_{k=0}^\infty{\frac{(-1)^k}{(k+1)!k!}\left(\frac{x}{2}\right)^{2k+1}}$$
$$J_1(x)=\sum_{k=0}^\infty{\frac{(-1)^k}{(k+1)!k!}\left(\frac{x-2+2}{2}\right)^{2k+1}}$$
Using the binomial theorem.
$$J_1(x)=\sum_{k=0}^\infty{\frac{(-1)^k}{(k+1)!k!}\sum_{m=0}^{2k+1}\left(\stackrel{2k+1}{m}\right)\left(\frac{2}{2}\right)^{2k+1-m}\left(\frac{x-2}{2}\right)^{m}}$$
$$J_1(x)=\sum_{k=0}^\infty{\frac{(-1)^k}{(k+1)!k!}\sum_{m=0}^{2k+1} \left(\stackrel{2k+1}{m}\right)\left(\frac{x-2}{2}\right)^{m}}$$

Now if you can figure out how to reverse these sums you will get the Taylor coefficients at X equal to 2. This will give you your derivatives directly without having to evaluate Bessel functions. I guess this will still give you an infinite series though.

Last edited: Feb 23, 2008
8. Feb 23, 2008

### Pere Callahan

Hi john, this looks like a good idea, thanks. I will try to work it out

9. Feb 23, 2008

### John Creighto

Here's what I get when I change the order of the sum:

$$J_1(x)=\sum_{m=0}^{\infty} \sum_{k=2m-1}^\infty{\frac{(-1)^k}{(k+1)!k!}\left(\stackrel{2k+1}{m}\right)\left(\frac{x-2}{2}\right)^{m}}$$

Or equivalently:

$$J_1(x)=\sum_{m=0}^{\infty} \sum_{k=2m-1}^\infty{\frac{(-1)^k(2k+1)!}{(k+1)!(k!)(2k+1-m)!m!}\left(\frac{x-2}{2}\right)^{m}}$$

Okay, well, duh!. This gives me the exact same thing that you got above by differentiating the series term by term and then setting x=2.

10. Feb 23, 2008

### Pere Callahan

HI John, I just realized that as well - that it gives the same thing.

Hm, maybe it's easier to find a way to avoid calculating these numbers altogether .... after all I'm not interested in them for their own sake but just in order to calculate something else

Thanks for your help anyways.

11. Feb 23, 2008

### John Creighto

What's the problem that you are trying to solve?

12. Feb 23, 2008

### Pere Callahan

I have some nasty integral as functions of their upper integration bound. These integrals involve Bessel functions (and some other stuff as well). Since a closed form for these integrals seems not to exist I tried to find their derivatives at the upper integration bound equal to zero (the lower integration bound is always zero, so this makes some sense). The derivatives of the integral-function at zero involve derivatives of the Bessel function at two (because the Bessel functions only occur as

$$J_1(2e^{-x})$$

inside the integral ... However even it were possible to find the values of the derivatives of the Bessel function, the overall expression I would get looks ugly to say the least ...

My goal was to then write my function as a power series but it seems to be no more handy than the original integral representation ..