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How do we solve recursive functions using gamma function? Basically I'm not able to understand the connection between the two. I found following reduction while reading on web $(i+b)I_{i}=(i-1-\lambda)I_{i-1}+p(1+p\lambda)\delta_{i0}$ to $I_i$=$I_{0}\frac{\Gamma(i+\lambda)\Gamma(b+1)}{\Gamma(i+b+1)\Gamma(\lambda)}$ I'm not able to get this reduction...

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Could you give more information? What are some examples of the kind of things you wish to solve? –  Matthew Conroy Apr 10 '12 at 17:25
    
$Matthew ConroyI want to learn connection between these two things so that I can use it for my equations can you please some pointers? –  user997704 Apr 10 '12 at 17:27
    
What makes you think there is a connection to learn at all? –  Henning Makholm Apr 10 '12 at 17:56
    
want to learn that connection....CAN WE SOLVE RECURSIVE FUNCTION USING GAMMMA FUNCTION? –  user997704 Apr 10 '12 at 17:58
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Writing in capital letters is a terrific way to convince people to stay away from you. If, instead, you want people to help you, you might try to meet them halfway. Let's start with vocabulary. You ask about solving functions, but that makes no sense. Do you mean solving equations? You've used the tag "recurrence relations", so perhaps what you really want to do is solve recurrence relations? Do you have an example of a recurrence relation that somebody solved using the gamma function? or an example of a recurrence relation that you would like to solve using the gamma function? –  Gerry Myerson Apr 11 '12 at 5:17
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1 Answer 1

The gamma function is defined such that $z\Gamma(z)=\Gamma(z+1)$. Multiplying inductively, we have

$$\big((k-1)+z\big)\cdots(2+z)(1+z)z\,\Gamma(z)= \Gamma(k+z).$$

Divide by $\Gamma(z)$ to obtain our major lemma $(k-1+z)\cdots (1+z)z=\Gamma(k+z)/\Gamma(z)$.

We can now solve the following generalized recursive sequence:

$$\begin{array}{c l} I_k & =\frac{a(k-1)+b}{c(k-1)+d}I_{k-1} \\[2pt] & = \frac{a}{c} \frac{(k-1)+\alpha}{(k-1)+\beta} I_{k-1}, \quad \alpha=b/a,~\gamma=d/c \\[2pt] & = \frac{a}{c} \frac{(k-1)+\alpha}{(k-1)+\beta} \frac{a}{c} \frac{(k-2)+\alpha}{(k-2)+\beta}\cdots \frac{a}{c} \frac{2+\alpha}{2+\beta} \frac{a}{c} \frac{1+\alpha}{1+\beta} \frac{a}{c} \frac{0+\alpha}{0+\beta} I_0 \\[2pt] & = \left(\frac{a}{c}\right)^k \frac{\big((k-1)+\alpha\big)\cdots(1+\alpha)\alpha}{\big((k-1)+\beta\big)\cdots(1+\beta)\beta} I_0 \\[2pt] & = (a/c)^k \frac{\Gamma(k+\alpha)/\Gamma(\alpha)}{\Gamma(k+\beta)/\Gamma(\beta)}I_0 \\[2pt] & = (a/c)^k\frac{\Gamma(k+b/a)\Gamma(d/c)}{\Gamma(k+d/c)\Gamma(b/a)} I_0. \end{array}$$

The cases with either $a$ or $c$ equal to zero work just the same as above.

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If I'm not missing $\Gamma(z)$=$\int^\infty_0e^{-t}t^{z-1}dt$..right? –  user997704 Apr 11 '12 at 6:42
    
@user997704: That is indeed the actual definition. It is defined in such a way that the recurrence $z\Gamma(z)=\Gamma(z+1)$ is true, which can be proved with integration by parts. See here. –  anon Apr 11 '12 at 6:46
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