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We know that: $$e^{z+z^{-1}}=\underbrace{\left(\sum_{n=0}^{\infty}\frac{z^n}{n!}\right)\cdot\left(\sum_{m=0}^{\infty}\frac{z^{-m}}{m!}\right)}_{=:\alpha}.$$ Now I am having trouble, applying the following (Cauchy product for series) on $\alpha$: $$\left(\sum_{i=0}^{\infty}a_iz^i\right)\cdot\left(\sum_{j=0}^{\infty}b_jz^j\right)=\sum_{k=0}^{\infty}c_k z^k, \quad c_k=\sum_{l=0}^{k}a_lb_{k-l}. \quad (\ast)$$ My approach was the following:
With ($\ast$) I find that \begin{align} \left(\sum_{n=0}^{\infty}\frac{z^n}{n!}\right)\cdot\left(\sum_{m=0}^{\infty}\frac{z^{-m}}{m!}\right) &= \sum_{k=0}^{\infty}c_k, \\ c_k &= \sum_{l=0}^{k}\left(\frac{1}{l!}\frac{1}{(-(k-l))!}\right) \\ &= \sum_{l=0}^{k}\frac{1}{l!(l-k)!}, \end{align} which results in $$e^{z+z^{-1}}=\sum_{k=0}^{\infty}\left(\sum_{l=0}^{k}\frac{1}{l!(l-k)!}\right)z^k.$$ The result I am looking for is $$e^{z+z^{-1}}=\sum_{k=-\infty}^{\infty}\left(\sum_{l=0}^{\infty}\frac{1}{l!(l+k)!}\right)z^k, \quad \forall k\in\mathbb{N}_0.$$ What am I missing? Also, I suspect that I have to use the following $$\alpha=\left(\sum_{n=0}^{\infty}\frac{z^n}{n!}\right)\cdot\left(\sum_{m=-\infty}^{0}\frac{z^m}{(-m)!}\right)$$ in order to get a sum from $-\infty$ to $\infty$, but I do not know how.

73g0
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  • What have you tried, and where are you getting stuck? – Steven Stadnicki Jan 18 '22 at 18:46
  • As a simple for-instance: can you figure out what the coefficient of $z^0$ should be? – Steven Stadnicki Jan 18 '22 at 18:47
  • Your forgot the $b_j$ on the left side of your Cauchy sum. But since one of your series is a power series in $z^{-1},$ the equivalent Cauchy sums are infinite. For example, the coefficient of $z^0$ in $$\sum_{n=0}^{\infty}\frac{1}{n!\cdot n!}.$$ I doubt that has a closed form. – Thomas Andrews Jan 18 '22 at 18:47
  • @StevenStadnicki I updated the question. – 73g0 Jan 18 '22 at 19:20
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    The result you're 'expecting' to get is clearly wrong; as Thomas Andrews points out, the coefficient of $z^0$ is definitely not 1. You can see that your own result is incorrect because since $e^{z+z^{-1}}$ is symmetric with respect to $z$ and $z^{-1}$, so the Laurent series coefficients must likewise be symmetric; for every positive coefficient $a_kz^k$ there's an equal negative coefficient $a_kz^{-k}$. – Steven Stadnicki Jan 18 '22 at 19:48
  • @ThomasAndrews I believe it's representable as $I_0(2)$ with $I_0$ a modified Bessel function, though of course whether that's a closed form is a matter of opinion. – Steven Stadnicki Jan 18 '22 at 21:41
  • @StevenStadnicki You are right! I edited the question (the upper limit of the sum should not have been $k$, but $\infty$ instead). What am I missing now? – 73g0 Jan 19 '22 at 06:05
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    The problem is that you are forcing the Cauchy product on a problem where it is not applicable. The Cauchy product is for two power series in the same variable. Here you have one power series in $z$ and another in $1/z$. There is a (formal) extension to Laurent series that you can use though: $$ \left( {\sum\limits_{n = - \infty }^\infty {a_n z^n } } \right)\left( {\sum\limits_{n = - \infty }^\infty {b_n z^n } } \right) = \sum\limits_{n = - \infty }^\infty {\left( {\sum\limits_{k = - \infty }^\infty {a_k b_{n - k} } } \right)z^n } . $$ – Gary Jan 19 '22 at 06:13
  • @Gary How do I have to transform the sums in order to make them suitable for the Cauchy product? (How do I get to the form you present in your comment?) – 73g0 Jan 19 '22 at 06:19
  • See the rigorous derivation in the answer to this question. It shows how to multiply two Laurent series in the intersection of their annuli of convergence. – Gary Jan 19 '22 at 06:26

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