Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
add comment

1 Answer

up vote 14 down vote accepted

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

share|improve this answer
3  
formula added... –  Gerald Edgar Apr 15 '12 at 16:53
4  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.