0
$\begingroup$

I am not sure whether this is rather an MO or MSE question but it results from my research, so I put it here.

In my effort to find (or to disprove the existence of) $k,l,h\in\mathbb{N}$ such that $2^{2(k+h)+l}-3^{h+l}|-(2^{2k}+2)3^{l-1}+2^{2k+l}$ and the quotient of the two being negative, I found that this is equivalent to finding an integer $n\in(-\mathbb{N})$ such that $A_{h,k}\left(-\dfrac{3^{l-1}}{2^{l}}\right)=n$ with $A_{h,k}$ being the Mobius transformation given by $$A_{h,k}=\begin{pmatrix}2^{2k}+2 & 2^{2k} \\ 3^{h+1} & 2^{2(k+h)}\end{pmatrix}.$$ Just for completeness: the matrix $A_{h,k}$ has positive determinant for any $k,h,l$. Since the Gamma function $\Gamma$ has poles on the negative integers, it could be possible to disprove the existence of such an $n$ by choosing a family of closed curves $\lbrace C_{\varepsilon}\left(-\frac{3^{l-1}}{2^{l}}\right)\rbrace$ where $C_{\varepsilon}\left(-\frac{3^{l-1}}{2^{l}}\right)$ contracts to $-\dfrac{3^{l-1}}{2^{l}}$ as $\varepsilon \to 0$ and proving that $$\lim_{\varepsilon\to 0}\oint_{C_{\varepsilon}\left(-\frac{3^{l-1}}{2^{l}}\right)}\Gamma(A_{h,k}(z))\mathrm{d}z=0$$ meaning that there is no pole at the point $-\dfrac{3^{l-1}}{2^{l}}$. And since there is no pole of $\Gamma\circ A_{h,k}$ at $-\dfrac{3^{l-1}}{2^{l}}$ the transformation $A_{h,k}$ does not map $-\dfrac{3^{l-1}}{2^{l}}$ to a negative integer. My problem is now the evaluation of the integral above. A change of variables does not help, since it asks exactly the same question as the initial problem.

Resolving the integral is related to calculating the residues of $\Gamma$. But this is never done using the integral definition of residues but rather the identity $z\Gamma(z)= \Gamma(z+1)$ and some limit considerations when we let $z$ tend to one of the poles of $\Gamma$. See for example the proofwiki proof of the residues. This is based on the simple poles of $\Gamma$ and the Laurent series expansion around one of them, e.g., $n$ as defined above. If $\Gamma\circ A_{h,k}$ has no pole at $-\frac{3^{l-1}}{2^{l}}$, then it is holomorphic in an environment of $-\frac{3^{l-1}}{2^{l}}$ and so the coefficients $a_i$ of its Laurent expansion around $-\frac{3^{l-1}}{2^{l}}$ satisfies $a_{-1}=0$. Therefore, I should obtain

$$\lim_{z\to -\frac{3^{l-1}}{2^{l}}} \left(z+\frac{3^{l-1}}{2^{l}}\right)(\Gamma\circ A_{h,k})(z)=0.$$

Is the reasoning until this point correct? Here I get stuck, since the technique, which is used for the $\Gamma$ function to obtain the residue at $n$ does not work here, since $\frac{3^{l-1}}{2^{l}}$ is not an integer. How could I proceed here?

This seems like a general problem/approach

Whenever a quotient of can be expressed as the action of some "reasonable" Mobius transform $A$ on a rational number $q$ one could either look at $\Gamma\circ A$ or $\Gamma\circ (-A)$ to check for negative or positive solutions, respectively, to the initial problem using the approach discussed above IF it even works out already in my example. Thoughts on this idea or references where this is already discussed are highly appreciated as well.

$\endgroup$

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.