How to derive the general term for the Taylor series around $0$ for this function?
I found, using Wolfram Alpha, that:
$$e^{a x} I_0 (|a| x)= \sum_{n=0}^\infty \frac{(2n)!}{n!^3} \frac{(a x)^n}{2^n}$$
I suspect that there should be a nice series for the general function.
$$f(a,b,x)=e^{ax} J_0 (bx)$$
I see several ways to derive it:
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Finding an ODE for $f(a,b,x)$ and solving it with power series. Then maybe a recurrence relation will allow us to find a closed form for the coefficients.
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Using the integral representation of the Bessel function.
I'll try the second way:
$$J_0(x)= \frac{2}{\pi} \int_0^1 \frac{\cos (x u) du}{\sqrt{1-u^2}}$$
$$e^{a x} J_0(x)=\frac{2}{\pi} \int_0^1 e^{a x} \frac{\cos (x u) du}{\sqrt{1-u^2}}$$
Let's try expanding:
$$e^{a x} \cos (u x)=\frac{1}{2} \left(e^{(a+i u) x}+e^{(a-i u) x} \right)= \\ = \frac{1}{2} \sum_{n=0}^\infty \frac{(a+iu)^n+(a-iu)^n}{n!} x^n$$
So we get:
$$e^{a x} J_0 (b x)= \sum_{n=0}^\infty C_n(a,b) \frac{x^n}{n!}$$
Where:
$$C_n(a,b)=\frac{1}{\pi} \int_0^1 \frac{(a+ib u)^n+(a-ib u)^n}{\sqrt{1-u^2}} du$$
Can we simplify this expression somehow?
I suppose we could use Binomial series and then each term becomes a Beta function. On the other hand, we can directly represent this integral as a sum of two hypergeometric functions. I'll see what I can do, but I would welcome other ideas and answers.
Best Answer
The $n$-th derivative of the product is \begin{equation} \frac{d^n}{dx^n}f\left( a,b,x \right)=\sum_{k=0}^n\binom{n}{k}a^{n-k}b^ke^{ax}\left.\frac{d^k}{dz^k}J_0(z)\right|_{z=bx} \end{equation} We want to evaluate it at $x=0$, then \begin{equation} C_n(a,b)=\left.\frac{d^n}{dx^n}f\left( a,b,x \right)\right|_{x=0}=\sum_{k=0}^n\binom{n}{k}a^{n-k}b^k\left.\frac{d^k}{dz^k}J_0(z)\right|_{z=0} \end{equation} Using the formula for the $k$-th derivative of the Bessel function (DLMF), \begin{equation} \left.\frac{d^k}{dz^k}J_0(z)\right|_{z=0}=\frac{1}{2^{k}}\sum_{p=0}^{k}(-1)^{p}\binom{k}{p}% J_{2p-k}\left(0\right) \end{equation} In this sum, the only non-vanishing Bessel functions at the origin have a zero index, i.e. $p=k/2$. Then, for integer $s$, \begin{equation} \left.\frac{d^k}{dz^k}J_0(z)\right|_{z=0}=\begin{cases} (-1)^s2^{-2s}\binom{2s}{s} & \text{for } k=2s \\ 0 & \text{for } k=2s+1 \end{cases} \end{equation} thus \begin{equation} C_n(a,b)=\sum_{s=0}^{\lfloor n/2 \rfloor}(-1)^s2^{-2s}\binom{n}{2s}\binom{2s}{s}a^{n-2s}b^{2s} \end{equation} finally, using the explicit expression of the Legendre polynomials, we obtain \begin{equation} C_n(a,b)=\operatorname{sign}(a)\left( a^2 +b^2\right)^{n/2}P_n\left( \frac{\left|a\right|}{\sqrt{a^2+b^2}} \right) \end{equation} Taking into account the parity properties of the polynomials, this expression can be written as: \begin{equation} C_n(a,b)=\left( a^2 +b^2\right)^{n/2}P_n\left( \frac{a}{\sqrt{a^2+b^2}} \right) \end{equation}
This result can be checked using the exponential generating function of the Legendre polynomials \begin{equation} \sum_{n=0}^\infty P_n(z)\frac{t^n}{n!}=e^{tz}{}_0F_1\left( ;1;\frac{t^2\left( z^2-1\right)}{4} \right) \end{equation} when $-1<x<1$. Using the hypergeometric representation of the Bessel function \begin{equation} J_0(u)={}_0F_1\left( ;1;\frac{-u^2}{4} \right) \end{equation} By choosing $t^2\left( z^2-1 \right)=-b^2x^2$ and $tz=ax$, we obtain \begin{equation} z=\frac{\pm1}{\sqrt{1+\frac{b^2}{a^2}}}\quad t=\pm ax \sqrt{1+\frac{b^2}{a^2}} \end{equation} then the generating function is identical to the Taylor expansion.