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The first equation must be simplified to achieve the result of the second equation, q is just a constant here. I have no clue about which algebraic manipulations I need to use to achieve this result. This solution was obtainned by MAPLE software, however I couldn´t get the steps from this software.

\begin{equation} \begin{gathered} \frac{\Gamma(\frac{1}{2}+\frac{1}{q-1})}{\Gamma(\frac{1}{q-1})}\frac{\Gamma(\frac{q}{q-1})}{ \Gamma(\frac{q}{q-1}+\frac{1}{2})} = \frac{2}{(q+1)} \\ \end{gathered} \end{equation}

JC94
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1 Answers1

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I suggest you learn how to solve this yourself. Here's a hint:

The LHS simplifies to

$$\frac{\Gamma\left(\frac{1}{2}+\frac{1}{q-1}\right)\Gamma\left(\frac{1}{q-1}+1\right)}{\Gamma\left(\frac{1}{2}+\frac{1}{q-1}+1\right)\Gamma\left(\frac{1}{q-1}\right)}$$

... Now you can use $\Gamma(x+1) = x\Gamma(x)$

Here are the steps:

We can simplify the expressions above by considering $$x = \frac{1}{2}+\frac{1}{q-1}$$ $$y = \frac{1}{q-1}$$

So that we have the following,

$$\frac{\Gamma(x)\Gamma(y+1)}{\Gamma(x+1)\Gamma(y)}$$

Now using the identity $\Gamma(x+1) = x\Gamma(x)$, we can write out

$$ = \frac{y\Gamma(x)\Gamma(y)}{x\Gamma(x)\Gamma(y)}$$

and now simplifying by cancelling $\Gamma(x)$ and $\Gamma(y)$ we get

$$ = \frac{y}{x}$$ which when plugging expressions for $x$ and $y$ back in yield

$$ = \frac{1}{q-1}\frac{1}{\frac{1}{2}+\frac{1}{q-1}}$$ $$ = \frac{2(q-1)}{(q-1)(q+1)}$$ $$ = \frac{2}{q+1}$$

Hope that helps!

Mako
  • 546