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I need to show that $$ \sin(ar\sin\zeta)\cos\zeta+ \sin(ar\cos\zeta)\sin\zeta=\sum_{m=0}^{+\infty}\cfrac{8(2m+1)}{ar}J_{2(2m+1)}(ar)\sin[2(2m+1)\zeta] $$

I think I have to use the relationships $$ \sin(ar\sin\zeta)=2\sum_{n=0}^{+\infty}J_{2n+1}(ar)\sin[(2n+1)\zeta]\\ \sin(ar\cos\zeta)=2\sum_{n=0}^{+\infty}(-1)^nJ_{2n+1}(ar)\cos[(2n+1)\zeta] $$

Substituting these relationships in the left side of the first equality, I see that I come with the two products $\sin[(2n+1)\zeta]\cos\zeta$ and $\cos[(2n+1)\zeta]\sin\zeta$ which can be reduced to $\sin[(2n+2)\zeta]+\sin(2n\zeta)$. Hence I can consider the sum

$$ \sum_{n=0}^{+\infty}[J_{2n+1}(ar)+(-1)^nJ_{2n+1}(ar)] $$ which is zero for odd $n$. So i can put $n=2m$ and get $$ 2\sum_{m=0}^{+\infty}[J_{4m+1}(ar)] $$ but now I don't know

Marco81
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1 Answers1

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Something went wrong in the proposed attempt. As mentioned in the OP, using the given identities, we obtain \begin{align} S&=\sin(ar\sin\zeta)\cos\zeta+ \sin(ar\cos\zeta)\sin\zeta\\ &=2\sum_{n=0}^{+\infty}J_{2n+1}(ar)\left\lbrace\sin[(2n+1)\zeta]\cos\zeta+(-1)^n\cos[(2n+1)\zeta]\sin\zeta\right\rbrace \end{align} Decomposing the summation using the parity of the index $n$ gives \begin{align} S&=2\sum_{p=0}^{+\infty}J_{4p+1}(ar)\sin[(4p+2)\zeta]+2\sum_{p=0}^{+\infty}J_{4p+3}(ar)\sin[(4p+2)\zeta]\\ &=2\sum_{p=0}^{+\infty}\sin[(4p+2)\zeta]\left[J_{4p+1}(ar)+J_{4p+3}(ar)\right] \end{align} Now, using the recurrence relation \begin{equation} J_{\nu-1}\left(z\right)+J_{\nu+1}\left(z\right)=(2\nu/z)J_{\nu}\left(z\right) \end{equation} with $\nu=4p+2$, we obtain \begin{equation} S=\frac{8}{ar}\sum_{p=0}^{\infty}(2p+1)\sin[(4p+2)\zeta]J_{4p+2}(ar) \end{equation}

Paul Enta
  • 14,113