2
$\begingroup$

I am interested in determining a closed expression for the n-th derivative of the Bessel function of the first kind $J_0(x)$, centered in $x=0$: \begin{equation} \left.\left(\frac{\mathrm d}{\mathrm d x} \right)^n J_0(x)\right|_{x=0} \end{equation} Can I compute it? If yes, how?

Thanks in advance!

$\endgroup$
  • 1
    $\begingroup$ Isn't it trivial from the series definition of $J_0$? What is your definition of $J_0$? $\endgroup$ – Jack D'Aurizio Oct 2 '18 at 21:33
3
$\begingroup$

This proof uses operator methods. $$\sum_{n=0}^\infty \frac{t^n}{n!} \frac{d^n}{dx^n} J_0(x)\Big|_{x=0} = \exp{\big(t\,\frac{d}{dx} \big)} J_0(x) \Big|_{x=0} = J_0(x+t) \Big|_{x=0} = J_0(t)= $$ $$= \sum_{n=0}^\infty \frac{(-1/4)^n}{n!^2} t^{2n}$$ Equate coefficients of $t.$ Odd $n$ will yields zero coefficients. Even coefficients imply $$ \frac{d^{2n}}{dx^{2n}} J_0(x)\Big|_{x=0} = (-1/4)^n\frac{(2n)!}{n!^2}=(-1)^n2^{-2n}\binom{2n}{n}$$ Combining we have $$ \frac{d^{n}}{dx^{n}} J_0(x)\Big|_{x=0} = \frac{1+(-1)^n}{2} \, i^n\,2^{-n}\binom{n}{n/2}$$

$\endgroup$
  • $\begingroup$ seems to work great, do you have a bibliographic reference on this approach? $\endgroup$ – Picaud Vincent Oct 2 '18 at 17:19
  • $\begingroup$ Isn't this circular: you use the power-series of $J_0(x)$ to derive the power-series of $J_0(x)$ (which the coefficients you compute really are)? $\endgroup$ – Winther Oct 2 '18 at 17:44
  • $\begingroup$ @ Vincent. exp(hd/dx)f(x)=f(x+h) is a very old formula. It can be thought of as a way to express the McLaurin expansion. I see it in my F.B. Hildebrand's Intro to Numerical Analysis, Ch.5, Dover ed., pg 181, (first ed. was 1956) but this cannot be the first time it was in a popular textbook. $\endgroup$ – skbmoore Oct 2 '18 at 18:04
  • $\begingroup$ @Winther Not circular because I say nothing about $J_0(x)$ in the beginning step, only that I want to find $d^n/dx^n\,J_0(x)|_{x=0}.$ You can use the same trick for other functions, say, $\sin(x),$ and that first step would be the same, just $J_0(x) \to \sin(x).$ The information resides in the Taylor series for whatever function you decide to use, $J_0(x),\sin(x),$ etc., and that's in the last step. $\endgroup$ – skbmoore Oct 2 '18 at 18:04
  • $\begingroup$ @skbmoore thanks for the reference. I was not aware of this formula. $\endgroup$ – Picaud Vincent Oct 2 '18 at 21:04
2
$\begingroup$

The defining ODE of the zeroth order Bessel function is

$$x^2J_0''(x) + xJ_0'(x) + x^2J_0(x) = 0.$$

Solve this using the power-series method (Frobenius method): take the ansatz $J_0(x) = \sum_{n=0}^\infty a_n x^n$ and insert it into the ODE to get a recurrence relation for the $a_n$'s and solve this. With this solution in hand note that $$\left.\frac{d^n}{dx^n}J_0(x)\right|_{x=0} = n! a_n$$ which gives you all the numbers you seek.

$\endgroup$
  • $\begingroup$ Thanks for all your responses and discussions. Regarding the answer above, I guess the recursion relations shall take as a seed the values of the lowest derivative(s) of J_0 in x=0? $\endgroup$ – Graz Oct 3 '18 at 16:16
  • $\begingroup$ @Graz Yes. The initial conditions are taken to be $J_0(0) = 1$ and $J_0'(0) = 0$ so $a_0 = 1$ and $a_1 = 0$ are the initial conditions to apply to the recurrence $a_{n+2} = -\frac{a_n}{n^2}$. $\endgroup$ – Winther Oct 3 '18 at 16:19
  • 1
    $\begingroup$ (Sorry, small typo above. It should be $a_{n} = -\frac{a_{n-2}}{n^2}$) $\endgroup$ – Winther Oct 3 '18 at 16:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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