# Is there an explicit formula for $\sum x^n/(n!)^2$?

Let $f(x)=\sum_{n\geq 0}\frac{1}{n!n!}x^{n}$, is there an explicit formula for $f(x)$?

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## 1 Answer

This is, for $x \ge 0$, a special case of the modified Bessel function of the first kind. Have a look here: http://mathworld.wolfram.com/ModifiedBesselFunctionoftheFirstKind.html $$\sum_{n = 0}^{\infty} \frac{x^{n}}{(n!)^{2}} = \mathrm{I}_0 \left(2 \sqrt{x}\right)$$

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formula added... –  Gerald Edgar Apr 15 '12 at 16:53
Here is a probabilistic interpretation of this equation: Let $X_1$ and $X_2$ be iid Poisson random variables with rate $\sqrt{x}$. Then, $\mathbb P(X_1 = X_2) = e^{-2 \sqrt{x}} I_0(2 x)$. –  cardinal Apr 16 '12 at 0:41