Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

We have

$$\Gamma(n+1)=n!,\ \ \ \ \ \Gamma(n+2)=(n+1)!$$

for integers, so if $\Delta$ is some real value with

$$0<\Delta<1,$$

then

$$n!\ <\ \Gamma(n+1+\Delta)\ <\ (n+1)!,$$

because $\Gamma$ is monotone there and so there is another number $f$ with

$$0<f<1,$$

such that

$$\Gamma(n+1+\Delta)=(1-f)\times n!+f\times(n+1)!.$$

How can we make this more precise? Can we find $f(\Delta)$?

Or if we know the value $\Delta$, which will usually be the case, what $f$ will be a good approximation?

share|improve this question
1  
I think your $f$ depends not just on $\Delta$, but on $n$ as well. –  froggie May 17 '12 at 15:58

2 Answers 2

up vote 1 down vote accepted

Asymptotically, as $n \to \infty$ with fixed $\Delta$, $$ f(n,\Delta) = \dfrac{\Gamma(n+1+\Delta)-\Gamma(n+1)}{\Gamma(n+2)-\Gamma(n+1)} = n^\Delta \left( \dfrac{1}{n} + \dfrac{\Delta(1+\Delta)}{2n^2} + \dfrac{\Delta(-1+\Delta)(3\Delta+2)(1+\Delta)}{24n^3} + \ldots \right) $$

share|improve this answer
    
Can't one just pull out the $n$'s of the first expression on all four Gammas and then it cancels? –  NikolajK May 17 '12 at 16:32
    
@NickKidman: I don't understand that comment. How do you "pull out the $n$'s" from $\Gamma(n+1+\Delta)$? –  Robert Israel May 17 '12 at 20:29
    
I ment with the gamma identity $\Gamma(z+n)=\Gamma(z)\times...$, but while the argument of $\Gamma$ would become smaller, the whole expression wouldn't look much simpler. –  NikolajK May 17 '12 at 20:47

Well, $\Gamma(1) = \Gamma(2) = 1$, but $\Gamma(\frac{3}{2}) = \frac{\sqrt{\pi}}{2} <1$, so presumably you need $n>1$.

And $f(n, \Delta) = \frac{\Gamma(n+1+\Delta)-\Gamma(n+1)}{\Gamma(n+2)-\Gamma(n+1)}$.

share|improve this answer
    
Well yeah, okay this involves $\Gamma(n+1+\Delta)$ itself. What would be a good practical approximation? I can't seem to expand this expression for $f(n,\Delta)$ in a series. (That is, I can use Mathematica, but it doesn't tell me the thing in terms of $n$, even if there seems to be a pattern. It'll probably come down to a series expanstion of the polygamma function.). $\textbf{edit}$: Wait, you can factor out $n$ in this thing, can't you. –  NikolajK May 17 '12 at 16:25

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.