Prove by Induction (fraction with pi)

[Seeker555]

By induction prove
\(\displaystyle \Gamma (n+0.5) = \frac{1.3.5.\cdots . (2n-1)}{2^n}\sqrt{\pi}\)


Recall \(\displaystyle \Gamma(0.5)=\sqrt{\pi}\)


That's the exact wording of the question. I know induction is something to do with inputting a n+1 but im not sure hwo to go about proving. Thanks.

 

[daon2]

Do you know the definition of the Gamma function?

For induction, choose the smallest value for which the proposition asserts the equation is true. You haven't given that, though I will assume n=0. Show that P(0) is true, where P(n) is the statement "The given equation holds true for n."

Then you may assume P(n) is true (or P(k) is true for all 0<= k <= n), and use this to show that P(n+1) is true. That is, show:

\(\displaystyle \Gamma(n+ 0.5) = \frac{1\cdot 3\cdots (2n-1)}{2^n}\sqrt{\pi} \implies \Gamma((n+1) + 0.5) = \frac{1\cdot 3\cdots (2n+1)}{2^{n+1}}\sqrt{\pi}\)

Note a direct observation reduces this to proving:

\(\displaystyle \Gamma(n+1.5) = \frac{(2n+1)}{2}\Gamma(n+0.5)\)