$\left.\left(\frac{\mathrm d}{\mathrm d x}\right)^n J_0(x)\right|_{x=0}={}$?

Question

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!

Answer 1

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.

Answer 2

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 $$ \frac{d^{n}}{dx^{n}} J_0(x)\Big|_{x=0} = \frac{1+(-1)^n}{2} \, i^n\,2^{-n}\binom{n}{n/2}$$