1
$\begingroup$

Is there any known formulae for the derivative of the Bessel function with respect to the order of the Bessel function?

$\endgroup$
1
  • $\begingroup$ which Bessel functions are you talking about? There are lots of different kinds. $\endgroup$ Nov 4, 2009 at 0:26

6 Answers 6

3
$\begingroup$

series... from Maple

$${\frac {d}{dr}}{{\rm J}_r\left(2\right)}= \sum _{k=0}^{\infty }{\frac { \left( -1 \right) ^{k+1}\Psi \left( 1+r+k \right) }{\Gamma \left( 1+r+k \right) \Gamma \left( 1+k \right) }}$$

$\endgroup$
1
$\begingroup$

I suggest looking at Landau's paper:http://www.emis.de/journals/EJDE/conf-proc/04/l1/landau.pdf

$\endgroup$
1
$\begingroup$

Abramowitz and Stegun give a couple of special cases but don't give a general result. Starting from some of the integral or series representations and differentiating you can get a corresponding integral or series for the derivative, but I would guess that it's unlikely to simplify to a "known" function in the general case. An example they give is (for the spherical Bessel function $j_\nu(x)$):

$$[ \frac{d}{d\nu} j_\nu(x) ]_{\nu=0} = \frac{\pi}{2x}(\operatorname{Ci}(2x)\sin x - \operatorname{Si}(2x)\cos x)$$

They also give examples evaluated at $\nu=-1$ and similar results for the case of the "other" spherical bessel $y_\nu(x)$.

$\endgroup$
0
$\begingroup$

As I understand it you are looking for $D_\nu(x):=\frac{d}{d\nu} J_\nu(x)$. Perhaps you would like to explain a bit why you are looking at $D_\nu$?

I have worked a bit with the Legendre functions of the first ($P_\nu$) and second kind ($Q_\nu$). Where my primary interest was to find estimates in $\nu$ and both parameters. To find such estimates basically I used relations together with integral representations. At one point I estimated an integral expression for $\frac{d}{d\nu} Q_\nu(x)$ in order to see that for fixed $x>1$ it is decreasing with respect to $\nu$. (The main reason to these studies was to prove a Tauberian theorem for spaces like $L^1_w(G//K)$ - the double coset space of $G=SL(2,R)$ - where $\hat{f}(s)=\int_1^\infty f(x)P_s(x)dx$ is the Fourier transform.)

$\endgroup$
0
$\begingroup$

The following paper might also be interesting:

http://www.tandfonline.com/doi/full/10.1080/10652469.2016.1164156

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.