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.

Let $\alpha$ be some positive noninteger real constant and $n$ be an arbitrary nonnegative integer. Consider a series $$ S_{n}(x) = \sum\limits_{k=0}^{\infty} {\alpha \choose k} \frac{x^k}{\Gamma(1-\alpha+k)\Gamma(1+n\alpha - k)}\ $$ Is it possible to find it's sum?

share|improve this question
    
Probably not. The reason is that the behaviour for gamma at negative numbers is not so simple. To many poles and some chaotic stuff related to good rational approximations I think (continued fractions for alpha and such , compare to n sec(n) ). This is not an answer of course. But Im not even sure S_n(x) is complex differentiable in all x and alpha. –  mick Oct 24 '12 at 21:19
    
@mick thank you for comment, what did you mean by "compare to $n \sec(n)$"? Can you give me some reference please? –  Nimza Oct 24 '12 at 21:26
    
What happens when $k > \alpha$? –  Jacob Oct 24 '12 at 21:26
1  
@Jacob do you mean ${ \alpha \choose k}$? We define it via Gamma functions, we are not afraid of negative arguments of Gamma. –  Nimza Oct 24 '12 at 21:29
    
Well can you estimate the min value of (n^4 sec(n)^2 - 1) accurately ? Now add some variables and taylor series and binomium and ask for a closed form. You might want to wiki or google Flint Hill series if my questions seems weird or unfamiliar. Since gamma(-ax) behaves equally 'difficult' as sec(an) and your taylor series is quite exotic looking + the fact that its derivative is probably not elementary either ( both with respect to alpha and/or x) , I doubt if there is a solution. And if there is I assume it to be very very general , like generalized hypergeo or such. If differentiable. –  mick Oct 24 '12 at 21:39

1 Answer 1

Let's do the simple case $m=0$. The series is $$ S_0(x)=\sum_{k=0}^\infty\frac{x^k}{\Gamma(1-\alpha+k)\Gamma(1-k)} $$ Use the convention $1/\Gamma(z)=0$ when $z$ is a pole of $\Gamma$. Then all but the first term vanishes and we get $$ S_0(x) = \frac{1}{\Gamma(1-\alpha)} $$

Maple says $$ S_1(x)=-\frac{\operatorname{sin} \bigl(\pi (\alpha + 1)\bigr)}{\pi \alpha} \; {}_2F_1\bigl([-\alpha,-\alpha],[1 - \alpha],x\bigr) $$
I leave $m>1$ to the reader (ha ha).

share|improve this answer
    
Haha, I like your humor. Nevertheless $S_{n}(x)$ are coefficients of some series and they all are important. –  Nimza Oct 24 '12 at 21:47

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.