Bernoulli Polynomial BesselJ expansion

Question

I have been reading a paper on classes of polynomials and it gives the following series: $$J_{\nu }(x)=\sum _{n=0}^{\infty } \frac{\left(x^{\nu } B_n\left(x^2\right)\right) \left(\frac{(-1)^{n+2} 2^{-\nu -2 n+2}}{\Gamma (n+\nu )}+(-1)^{n+1} 2^{1-n} J_{n+\nu -1}(1)\right)}{n!}$$ $$J_{\nu }(x)=\sum _{n=0}^{\infty } \frac{\left((-1)^n 2^{-\nu -2 n-1} \left(\, _0F_1\left(;n+\nu +1;-\frac{1}{4}\right)+1\right)\right) \left(x^{\nu } E_n\left(x^2\right)\right)}{n! \Gamma (n+\nu +1)}$$ where $B_n$ is Bernoulli Polynomial and $E_n$ is the Euler Polynomial but I try Taylor series but it is different it is seem convergent and fast for the most values it possible get it from Euler Naclaurin series

Answer 1

The first expression follows from theorem 1 in the paper linked in the comment. It states that, under suitable growth properties, \begin{equation} f(X)=\int_0^1f(t)\,dt+\sum_{n=1}^\infty\left[ f^{(n-1)}(1)-f^{(n-1)}(0)\right]\frac{B_n(X)}{n!} \tag{1}\label{eq1} \end{equation} Starting from the series expansion for the Bessel function \begin{equation} J_\nu(x)=\sum_{m=0}^\infty\frac{(-1)^m2^{-2m-\nu}}{\Gamma(m+1+\nu)}\frac{x^{2m+\nu}}{m!} \end{equation} denoting $X=x^2$ and \begin{equation} f(X)=x^{-\nu} J_\nu(x) \end{equation} we have directly \begin{equation} f(X)=\sum_{m=0}^\infty\frac{(-1)^m2^{-2m-\nu}}{\Gamma(m+1+\nu)}\frac{X^m}{m!} \end{equation} so that \begin{equation} f^{(n-1)}(0)=\frac{(-1)^{n-1}2^{-2n-\nu+2}}{\Gamma(n+\nu)} \end{equation} Now, since \begin{equation} \frac{d}{dX}=\frac{1}{2x}\frac{d}{dx} \end{equation} we have \begin{equation} \left[\frac{d}{dX} \right]^{n-1}\left( f(X) \right)=2^{-n+1}\left[\frac{1}{x}\frac{d}{dx} \right]^{n-1}\left( \frac{J_\nu(x)}{ x^{\nu}} \right) \end{equation} Using the derivation property for the Bessel function, we deduce \begin{equation} f^{(n-1)}(X)=(-1)^{n-1}2^{-n+1}\frac{J_{n+\nu-1}(X)}{X^{n+\nu-1}} \end{equation} and thus \begin{equation} f^{(n-1)}(1)=(-1)^{n-1}2^{-n+1}J_{n+\nu-1}(1) \end{equation} We have also from G\&R (6.561.9) \begin{align} \int_0^1f(t)\,dt&=2\int_0^1 x^{-\nu+1}J_\nu(x)\,dx\\ &=\frac{2^{2-\nu}}{\Gamma(\nu)}-2J_{\nu-1}(1) \end{align} Plugging all these results into expression \eqref{eq1}, one obtains \begin{equation} x^{-\nu} J_\nu(x)=\frac{2^{2-\nu}}{\Gamma(\nu)}-2J_{\nu-1}(1) +\sum_{n=1}^\infty\left[ (-1)^{n-1}2^{-n+1}J_{n+\nu-1}(1)-\frac{(-1)^{n-1}2^{-2n-\nu+2}}{\Gamma(n+\nu)}\right]\frac{B_n(x^2)}{n!} \end{equation} Remarking that the first term corresponds to a term $n=0$ in the summation, the result can be written as \begin{equation} J_\nu(x)=\sum_{n=0}^\infty(-1)^{n}\left[\frac{2^{-2n-\nu+2}}{\Gamma(n+\nu)}- 2^{-n+1}J_{n+\nu-1}(1)\right]\frac{ x^{\nu} B_n(x^2)}{n!} \end{equation}