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While trying to calculate the following infinite sum:

$$ \sum_{k=0}^{\infty} \frac {4^k}{(k!)^{2}}$$

I got the result: $I_{0}(4) = 11.301...$

I've never encountered this function before ($ I_{0}(x) $), can someone please describe it and explain why the above infinite sum converges to an output of this function?

I expected something having to do with the exponential function since $$ \sum_{k=0}^{\infty} \frac {\mu^k}{k!} = e^\mu $$

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The modified Bessel function of the first kind has a power series expansion $$ I_{\alpha}(x)=\sum_{k=0}^{\infty}\frac{1}{k!\Gamma(k+\alpha+1)}\Big(\frac{x}{2}\Big)^{2k+\alpha} $$

Taking $\alpha=0$ and using $\Gamma(k+1)=k!$, and then setting $x=4$, we get $$ I_0(4)=\sum_{k=0}^{\infty}\frac{4^k}{(k!)^2} $$ which is your sum.

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