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I have to calculate the following quantity

$$\gamma(8,(R/\alpha)^{1/4})$$

where $\gamma(s,x)$ is the lower incomplete gamma function. Choosing the series representation I'm trying to actually sum the series, but I can't find a compact answer. Let $x=R/\alpha$

$$\gamma(8,x^{1/4})=\Gamma(8)x^{2}e^{-x^{1/4}}\sum_{n=0}^{\infty}\frac{x^{n/4}}{\Gamma(9+n)}$$

I was thinking in manipulations like multiplying the series by the factor of $x^2$ but the point will be in some convenient way of writing

$$\frac{\Gamma(8)}{\Gamma(9+n)} $$

to get the $1/n!$ for the exponential series. So far

$$\frac{\Gamma(8)}{\Gamma(9+n)}=\frac{7!}{(8+n)!}=\frac{7!}{7!8(8+1)...(8+n)} $$

but I can't give the number the way I want. Of course I'm assuming that what I want to do is possible, which may not be the case.

Thanks for your time.

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1 Answer 1

There isn't a "compact" answer.

Since the series is the sum of positive terms, it is quite suitable for a grind-it-out summation.

Letting $y = x^{1/4}$, the series is $\sum_{n=0}^{\infty} \dfrac{y^n}{(n+8)!}$.

The first term is $\dfrac1{8!}$, and the ratio of consecutive terms is $\dfrac{\dfrac{y^{n+1}}{(n+9)!}}{\dfrac{y^n}{(n+8)!}} =\dfrac{y}{n+9} $.

Note that the ratio is greater than $1$ until $n+9 \ge y$, so you have to accumulate at least $y-9$ terms. After you get to this many terms, all subsequent terms decrease, so you can stop when $\dfrac{term}{total} < \epsilon$, where $\epsilon$ determins how accurate you want the answer to be. $\epsilon < 10^{-10}$ is probably good enough.

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Thanks. Actually I was interested in an analytic formula, we're not going to do numerical work as fas as I know. I found that $$y^{8}\sum_{n=0}^{\infty}\frac{y^{n}}{(n+8)!}=e^{y}-\sum_{n=0}^{7}\frac{y^{n}}{‌​n!} $$ –  Jorge Sep 28 '13 at 8:48
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