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

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 $$x=0$$: $$$$\left.\left(\frac{\mathrm d}{\mathrm d x} \right)^n J_0(x)\right|_{x=0}$$$$ Can I compute it? If yes, how?

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

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

• seems to work great, do you have a bibliographic reference on this approach? – Picaud Vincent Oct 2 '18 at 17:19
• 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)? – Winther Oct 2 '18 at 17:44
• @ 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. – skbmoore Oct 2 '18 at 18:04
• @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. – skbmoore Oct 2 '18 at 18:04
• @skbmoore thanks for the reference. I was not aware of this formula. – Picaud Vincent Oct 2 '18 at 21:04

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.

• 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? – Graz Oct 3 '18 at 16:16
• @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 $a_{n+2} = -\frac{a_n}{n^2}$. – Winther Oct 3 '18 at 16:19
• (Sorry, small typo above. It should be $a_{n} = -\frac{a_{n-2}}{n^2}$) – Winther Oct 3 '18 at 16:45