Prove gamma (n+1/2) = (2npi^1/2)/(n4^n) by induction

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The discussion focuses on proving the equation Gamma(n + 1/2) = (2n√π)/(n4^n) using mathematical induction. The user begins by applying the recursive property of the Gamma function, specifically Gamma(n + 1/2) = (n - 1/2)Gamma(n - 1/2), and establishes the base case with Gamma(1/2) = √π. The user then assumes the formula holds for n and proceeds to express Gamma(n + 1 + 1/2) in terms of Gamma(n + 1/2), indicating a structured approach to the proof.

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Pami
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I tried solving this question this way:
Gamma(n+1/2)
=(n+1/2-1)gamma(n+1/2-1)
=(n-1/2)gamma(n-1/2)
=(2n-1)/2 gamma (2 n-1)/2
Don't know what to do next
 
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To begin with \Gamma (\frac{1}{2})=\sqrt{\pi}. From there: \Gamma (1+\frac{1}{2})= \frac{1}{2}\Gamma(\frac{1}{2})=\frac{1}{2}\sqrt{\pi}. Checks against the formula.
Assume that the formula is correct for n. Then \Gamma(n+1+\frac{1}{2})=(n+\frac{1}{2})\Gamma(n+\frac{1}{2})...
 

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