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Another1

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**Bessel function**

using \(\displaystyle g(x,t)=g(u+v,t)=g(u,t)g(v,t)\)

to show that \(\displaystyle J_{0}(u+v)=J_{0}(u)J_{0}(v)+2\sum_{s=1}^{\infty}J_{s}(u)J_{-s}(v)\)

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my solution

\(\displaystyle g(u+v,t)=e^{\frac{u+v}{2}(t-\frac{1}{t})}\)

\(\displaystyle g(u+v,t)=e^{\frac{u}{2}(t-\frac{1}{t})}\cdot e^{\frac{v}{2}(t-\frac{1}{t})}\)

\(\displaystyle g(u+v,t)=\sum_{n=-\infty}^{\infty}J_{n}(u)t^{n}\sum_{n=-\infty}^{\infty}J_{n}(v)t^{n}\)

\(\displaystyle J_{n}(u+v)=\sum_{s=0}^{\infty}\frac{(-1)^{s}}{s!(n+s)!}(\frac{u+v}{2})^{n+2s}\)

\(\displaystyle J_{0}(u+v)=\sum_{s=0}^{\infty}\frac{(-1)^{s}}{s!s!}(\frac{u}{2}+\frac{v}{2})^{2s}\)

\(\displaystyle J_{0}(u+v)=\sum_{s=0}^{\infty}\frac{(-1)^{s}}{s!s!}\left\{(\frac{u}{2}+\frac{v}{2})^{2s} \right\}\)

\(\displaystyle J_{0}(u+v)=\sum_{s=0}^{\infty}\frac{(-1)^{s}}{s!s!}\left\{ \sum_{k=0}^{2s}{2s\choose k}\left(\frac{u}{2} \right)^{2s -k}\left(\frac{v}{2} \right)^{k} \right\}\)

\(\displaystyle J_{0}(u+v)=\sum_{s=0}^{\infty}\frac{(-1)^{s}}{s!s!}\left\{ \left(\frac{u}{2}\right)^{2s}+\left(\frac{v}{2}\right)^{2s}+ \sum_{k=1}^{2s-1}{2s\choose k}\left(\frac{u}{2} \right)^{2s -k}\left(\frac{v}{2} \right)^{k} \right\}\)

\(\displaystyle J_{0}(u+v)=J_{0}(u)+J_{0}(v)+\sum_{s=0}^{\infty}\frac{(-1)^{s}}{s!s!}\left\{\sum_{k=1}^{2s-1}{2s\choose k}\left(\frac{u}{2} \right)^{2s -k}\left(\frac{v}{2} \right)^{k} \right\}\)

this is wrong

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please help me to solve this soluion

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