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

edited Apr 16 2012 at 0:27

Gerald Edgar
15.1k●1●28●67

asked Apr 15 2012 at 14:44

xmchenhit
53●3

Gerald Edgar · Accepted Answer · 2012-04-15 16:53:38Z UTC

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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 $$ \sum_{n = 0}^{\infty} \frac{x^{n}}{(n!)^{2}} = \mathrm{I}_0 \left(2 \sqrt{x}\right) $$

edited Apr 15 2012 at 16:53

Gerald Edgar

answered Apr 15 2012 at 15:09

Woett
1,021●4●16

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formula added... – Gerald Edgar Apr 15 2012 at 16:53

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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 2012 at 0:41

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

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