Recursive functions and Gamma functions

Question

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...

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.