$\Gamma(n-1) = \frac{2(n+3)}{n+1}$, what is $n$? I know that the gamma function is an integral and I know that for whole numbers it can be shown as a factorial, but I have no idea how to go about this problem
How to solve $\Gamma(n-1) = 2(n+3)/(n+1)$
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calculus
gamma-function
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1Clearly for large $n$, $\frac{2(n+3)}{n+1} = \frac{2n + 2 + 4}{n + 1} = 2 + \frac{4}{n+1}$ is constant and $\Gamma$ is massively increasing. So for positive $n$, the solution may only lie near $0$. For negative solutions however this might get trickier. – 2017-01-26
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1I would say, numericaly solving: http://www.wolframalpha.com/input/?i=newton%27s+method+%5CGamma(n-1)+-+%7B2(n%2B3)%5Cover(n%2B1)%7D,+x0%3D4 – 2017-01-26