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I was wondering if anyone can help me with the procedure of simplifying the following formula: $$ e^{ - \beta x} \mathcal{T}_b \sum\limits_m {i^m J_m (kr)e^{im(\theta - \alpha )} } $$ using the following identity: $$ e^{ - \beta x} = e^{ - \beta b} \sum\limits_s {( - 1)^s I_s (\beta r)e^{is\theta } } $$ to the following formula in which the index of the modified Bessel function is modified to $m-s$: $$ e^{ - \beta b} \sum\limits_m {( - 1)^m U_m (r)e^{im\theta } } $$ where $U_m$ is equal to $$ \mathcal{T}_b \sum\limits_s {( - i)^s I_{m - s} (\beta r)J_s (kr)e^{ - is\alpha } } . $$ In these formulas $\theta$, $\alpha$, and $x$ are arbitrary real variables and $m$ and $s$ are integers varying from $-\infty$ to $+\infty$ and other variables are constant. Also $x = b + r \cos(\theta)$ and $y = r \sin(\theta)$.

Gary
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math student
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  • Next time please use MathJax to format your equations. The answer to your question: multiply the first and the second equations and collect the coefficient of the different powers of $e^{i\theta}$. If you call $z=e^{i\theta}$, it is just a multiplication of Laurent series. – Gary Jan 10 '22 at 09:33
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    How I_(m-s) term is created? @Gary – math student Jan 10 '22 at 10:19
  • Think about power series first. Do you know the Cauchy product of two power series? It is $$ \left( {\sum\limits_{m = 0}^\infty {a_m x^m } } \right)\left( {\sum\limits_{m = 0}^\infty {b_m x^m } } \right) = \sum\limits_{m = 0}^\infty {\left( {\sum\limits_{s = 0}^m {a_{m - s} b_s } } \right)x^m } . $$ Can you see that $m-s$? – Gary Jan 10 '22 at 10:31
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    Thanks a lot @Gary – math student Jan 11 '22 at 04:48

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