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Question Number 204468 by MathedUp last updated on 18/Feb/24

$$\mathrm{How}\mathrm{Can}\mathrm{we}\mathrm{prove}\underset{h=-\mathrm{\infty }}{\sum ^{\mathrm{\infty }}}{J}_h\left(z\right)=1$$

Answered by Peace last updated on 19/Feb/24

$$J_{n-1}\left(x\right)+j_{n+1}\left(x\right)=\frac{2n}{x}{j}_n\left(x\right).....\left(1\right)$$$$let{f}_x\left(y\right)=\underset{-\mathrm{\infty }}{\sum ^{\mathrm{\infty }}}{j}_n\left(x\right)y^n$$$$\Rightarrow {yf}_x\left(y\right)+\frac{f\left(y\right)}{y}=\frac{2}{xy}\mathrm{\Sigma }{nj}_n\left(x\right)y^{n-1}=\frac{2y}{x}\left(f_x^{\prime }\left(y\right)\right)$$$$\Rightarrow f_x^{\prime }\left(y\right)=f_x\left(y\right)(\frac{x}{2}+\frac{x}{2y^2})$$$$f_x\left(y\right)=e^{\frac{x}{2}(y-\frac{1}{y})}=\underset{-\mathrm{\infty }}{\sum ^{\mathrm{\infty }}}{j}_n\left(x\right)y^n;y=1\Rightarrow 1=\underset{-\mathrm{\infty }}{\sum ^{+\mathrm{\infty }}}{j}_n\left(x\right)$$$$$$$$$$

Commented by MathedUp last updated on 19/Feb/24

$$thx!$$

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