How does the Bessel Function Expansion relate to J_{0}(u+v)?

[Another1]

Bessel function

using $$g(x,t)=g(u+v,t)=g(u,t)g(v,t)$$

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

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my solution

$$g(u+v,t)=e^{\frac{u+v}{2}(t-\frac{1}{t})}$$
$$g(u+v,t)=e^{\frac{u}{2}(t-\frac{1}{t})}\cdot e^{\frac{v}{2}(t-\frac{1}{t})}$$
$$g(u+v,t)=\sum_{n=-\infty}^{\infty}J_{n}(u)t^{n}\sum_{n=-\infty}^{\infty}J_{n}(v)t^{n}$$

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

this is wrong
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please help me to solve this soluion

 

[Another1]

using $$g(x,t)=g(u+v,t)=g(u,t)g(v,t)$$

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

___________________________________________________________________________________________
my solution

$$g(u+v,t)=e^{\frac{u+v}{2}(t-\frac{1}{t})}$$
$$g(u+v,t)=e^{\frac{u}{2}(t-\frac{1}{t})}\cdot e^{\frac{v}{2}(t-\frac{1}{t})}$$
$$g(u+v,t)=\sum_{n=-\infty}^{\infty}J_{n}(u)t^{n}\sum_{n=-\infty}^{\infty}J_{n}(v)t^{n}$$

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

this is wrong
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please help me to solve this soluion

now i can solve it thankkkk !

$$g(u+v,t)=g(u,t)g(v,t)$$

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

$$\sum_{n=-\infty}^{\infty}J_{n}(u+v)t^n=\sum_{l=-\infty}^{\infty}J_{l}(u)t^l\sum_{m=-\infty}^{\infty}J_{m}(v)t^m$$

let n = 0
$$J_{0}(u+v)= \left(...+J_{-1}(u)t^{-1}+J_{0}(u)+J_{1}(u)t^1+...\right)\left(...+J_{-1}(v)t^{-1}+J_{0}(v)+J_{1}(v)t^1+...\right)$$

but $$J_{-n}(u)=(-1)^{n}J_{n}(u)$$
so...
$$J_{0}(u+v)= J_{0}(u)J_{0}(v)+2J_{1}(u)J_{-1}(v)+2J_{2}(u)J_{2}(v)+...$$
$$J_{0}(u+v)= J_{0}(u)J_{0}(v)+2\sum_{s=1}^{\infty}J_{s}(u)J_{-s}(v)$$

 

[Joppy]

Funions!

 

[Another1]

Funions!

oh sorry I mean function n and c missing from word