Recurrence Relation Involving the gamma Function

Question

I'm having some doubts about my approach to the following problem. I am given that the function $k(z)$ is defined such that,

$$k(z)=\Gamma\left(\frac{1}{2}+z\right)\Gamma\left(\frac{1}{2}-z\right)\cos{\pi z}$$

I'm required to find the recurrence relation linking $k(z+1)$ and $k(z)$ and to then evaluate $k(z)$ for one specific integer value and thus find $k(z)$ for any real, integer value. My attempt was as follows.

Note that $\Gamma(s+1)=s\Gamma(s)$ and so

\begin{align*} k(z+1)&=\Gamma\left(\frac{1}{2}+z+1\right)\Gamma\left(\frac{1}{2}-z+1\right)\cos{(\pi z + \pi)} \\ &=\left(\frac{1}{2}+z\right)\Gamma\left(\frac{1}{2}+z\right)\left(\frac{1}{2}-z\right)\Gamma\left(\frac{1}{2}-z\right)\cos{(\pi z + \pi)} \end{align*}

Then since $\cos{(\pi z + \pi)}=-\cos{\pi z}$ we have that

$$k(z+1)=\left(z^2-\frac{1}{4}\right)k(z)$$

If we then consider the case $z=0$, we have

$$k(0)=\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)=\pi$$

From this we see that

\begin{align*} &k(1)=-\frac{1}{4}\pi\\ &k(2)=-\frac{3}{16}\pi\\ &k(3)=-\frac{45}{64}\pi\\ &\vdots \end{align*}

I can't spot any pattern here other than the $4^z$ in the denominator which is making me think i've done something wrong, maybe in my use of $\Gamma(s+1)=s\Gamma(s)$? Any advice would be really appreciated.

Answer 1

You made a sign error. \begin{align} \frac{k(n+1)}{k(n)} &=\frac{\Gamma\left(\frac{1}{2}+n+1\right)\Gamma\left(\frac{1}{2}-n\color{red}{-1}\right)(-\cos(\pi n))}{\Gamma\left(\frac{1}{2}+n\right)\Gamma\left(\frac{1}{2}-n\right)\cos(\pi n)}\\ &=\frac{\left(\frac{1}{2}+n\right)\Gamma\left(\frac{1}{2}+n\right)\frac{\Gamma\left(\frac{1}{2}-n\right)}{\left(\frac{1}{2}-n-1\right)}(-1)}{\Gamma\left(\frac{1}{2}+n\right)\Gamma\left(\frac{1}{2}-n\right)}\\ &=-\frac{\frac{1}{2}+n}{-\frac{1}{2}-n}\\ &=1 \end{align} Hence $k(n)=k(0)=\pi$.

Answer 2

Note that using the identity you've developed, $$ \begin{align} k(n) &=k(0)\prod_{j=0}^{n-1}\left(j^2-\frac14\right)\\ &=\frac\pi{4^n}\prod_{j=0}^{n-1}(2j-1)(2j+1)\\ &=\frac\pi{4^n}(-1\cdot1)(1\cdot3)(3\cdot5)\cdots\big((2n-3)\cdot(2n-1)\big)\\ &=\frac\pi{4^n}\frac{-1}{2n-1}(1\cdot3\cdot5\cdots(2n-1))^2\\ &=\frac\pi{4^n}\frac{-1}{2n-1}\left(\frac{(2n)!}{2^nn!}\right)^2 \end{align} $$ This matches the sequence you've listed.

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Euler's Reflection Formula

Euler's Reflection Formula for the Gamma Function is proven in this answer: $$ \Gamma(z)\Gamma(1-z)=\pi\csc(\pi z) $$ Therefore, $$ \begin{align} \Gamma\left(\frac12-z\right)\Gamma\left(\frac12+z\right) &=\pi\csc\left(\frac\pi2-\pi z\right)\\ &=\pi\sec(\pi z) \end{align} $$ This means that $$ \Gamma\left(\frac12-z\right)\Gamma\left(\frac12+z\right)\cos(\pi z)=\pi $$ This indicates there is a problem, which Guest543212345 has pointed out.

Answer 3

Hint: $\quad\Gamma(1+z)~\Gamma(1-z)=\dfrac{\pi z}{\sin\pi z}\qquad$ and $\qquad\Gamma\bigg(\dfrac12+z\bigg)~\Gamma\bigg(\dfrac12-z\bigg)=\dfrac\pi{\cos\pi z}$

See Euler's reflection formula, Legendre's duplication formula, and Gauss' multiplication theorem.