2
$\begingroup$

I have been working on a transform (the notes are a little rough and the hosting website has some fraction formatting issues recently) but the idea is there:

Basically, the transform $\mathcal{I}_x[f(x)](s)$, will extract the coefficients of the scaled inverse function, $f^{-1}(x)/x$, in a manner analogous to the Mellin transform.

I have found a pattern that seems to generate Gamma function identities that Mathematica cannot simplify or verify. This was done by acting on simple polynomial expressions for which I previously found that the principal inverse (via series reversion) was a generalised hypergeometric function which led to the following results:

\begin{align} \mathcal{I}[x+x^2](s) = \frac{\Gamma(1-\frac{2s}{1})\Gamma(\frac{s}{1}) }{\Gamma(2-s)}\\ \mathcal{I}[x+x^3](s) = \frac{\Gamma(1-\frac{3s}{2})\Gamma(\frac{s}{2})}{2 \Gamma(2-s)} \\ \mathcal{I}[x+x^4](s) = \frac{8 \cdot 3^{s-\frac{5}{2}}\pi \Gamma(-4s/3)\Gamma(s/3)}{\Gamma(2/3-s/3)\Gamma(4/3-s/3)\Gamma(-s/3)} \stackrel{?}{=}\frac{\Gamma(1-\frac{4s}{3})\Gamma(\frac{s}{3})}{3 \Gamma(2-s)} \end{align} the last one appears to be true, but Mathematica won't simplfy it. In short the generalised conjecture is that the Mellin transform of this generalised hypergeometric function has a very simple result $$ \mathcal{M}_x\left[\;_{(m-1)}F_{(m-2)}\left(\left\{\frac{1}{m},\frac{2}{m},\cdots,\frac{m-1}{m}\right\};\left\{\frac{2}{m-1},\cdots,\frac{m-2}{m-1},\frac{m}{m-1}\right\};-\frac{m^mx^{m-1}}{(m-1)^{m-1}}\right)\right](s) = \frac{\Gamma\left(1-\frac{m s}{m-1}\right)\Gamma\left(\frac{s}{m-1}\right)}{(m-1)\Gamma(2-s)} = \mathcal{I}_x[x+x^m](s) $$ which only makes sense for integer $m$. Any way to prove this general statement? How do we simplify the identity that Mathematica couldn't?

$\endgroup$
2
$\begingroup$

For the Gamma function identity, we use the multiplication formula for the Gamma function, choosing $n=3, z=\frac{2}{3}-\frac{s}{3}$: \begin{align} \Gamma\left(nz\right)&=(2\pi)^{(1-n)/2}n^{nz-(1/2)}\prod_{k=0}^{n-1}\Gamma\left(z+\frac{k}{n}\right)\\ \Gamma\left(2-s\right)&=\frac{1}{2\pi}3^{3/2-s}\Gamma\left( \frac{2}{3}-\frac{s}{3} \right)\Gamma\left(1-\frac{s}{3} \right)\Gamma\left( \frac{4}{3}-\frac{s}{3} \right) \end{align} Now, with the functional relation $ \Gamma\left( 1+u \right)=u\Gamma(-u) $ applied with $u=-s/3$ and $u=-4s/3$ \begin{equation} \frac{\Gamma(1-\frac{4s}{3})\Gamma(\frac{s}{3})}{3 \Gamma(2-s)}=\frac{8\pi\Gamma(-\frac{4s}{3})\Gamma(\frac{s}{3})}{3^{5/2-s}\Gamma\left( \frac{2}{3}-\frac{s}{3} \right)\Gamma\left(-\frac{s}{3} \right)\Gamma\left( \frac{4}{3}-\frac{s}{3} \right)} \end{equation} as expected.

For the Mellin transform question, we evaluate the inverse transform of the function \begin{equation} f(s)=\frac{\Gamma(1-\frac{ms}{m-1})\Gamma(\frac{s}{m-1})}{ \Gamma(2-s)} \end{equation} assuming that the domain of analicity is $0<\Re (s) <\frac{m-1}{m}$. It can be noticed that the poles in the left complex half-plane are $s_p=-p(m-1)$, with $p=0,1,2\ldots$ Their residues are \begin{equation} r_p=\frac{(m-1)\Gamma(mp+1)}{\Gamma\left(2+p(m-1) \right)}\frac{(-1)^p}{p!} \end{equation} In the following we use the multiplication formula above with $z=1/n$, \begin{equation} \prod_{k=1}^{n-1}\Gamma\left(\frac{k}{n}\right)=(2\pi)^{(n-1)/2}n^{-1/2} \end{equation} to write \begin{align} \Gamma\left(np\right)&=(2\pi)^{(1-n)/2}n^{np-(1/2)}\prod_{k=0}^{n-1}\Gamma\left(p+\frac{k}{n}\right)\\ &=(2\pi)^{(1-n)/2}n^{np-(1/2)}\Gamma(p)\prod_{k=1}^{n-1}\Gamma\left(\frac{k}{n}\right)\left( \frac{k}{n} \right)_p\\ &=n^{np-1}\Gamma(p)\prod_{k=1}^{n-1}\left( \frac{k}{n} \right)_p \end{align}

Using the functional relation this residue can be converted into a ratio of Gamma functions: \begin{align*} r_p&=\frac{mp(m-1)}{p(m-1)\left( 1+p(m-1) \right)}\frac{\Gamma(mp)}{\Gamma\left(p(m-1) \right)}\frac{(-1)^p}{p!}\\ &=\frac{m}{\left( m-1 \right)\left(p+\frac{1}{m-1} \right)} \frac{m^{mp-1}\Gamma(p)\prod_{k=1}^{m-1}\left( \frac{k}{m} \right)_p}{(m-1)^{(m-1)p-1}\Gamma(p)\prod_{k=1}^{m-2}\left( \frac{k}{m-1} \right)_p} \frac{(-1)^p}{p!}\\ &=\frac{\Gamma\left(p+\frac{1}{m-1} \right)}{\Gamma\left(p+1+\frac{1}{m-1} \right)} \frac{\prod_{k=1}^{m-1}\left( \frac{k}{m} \right)_p}{\prod_{k=1}^{m-2}\left( \frac{k}{m-1} \right)_p} \left(- \frac{m^m}{(m-1)^{m-1}} \right)^{p}\\ &=\frac{\left(\frac{1}{m-1} \right)_p\Gamma\left( \frac{1}{m-1} \right)}{\left(1+\frac{1}{m-1} \right)_p\Gamma\left(1+ \frac{1}{m-1} \right)} \frac{\prod_{k=1}^{m-1}\left( \frac{k}{m} \right)_p}{\prod_{k=1}^{m-2}\left( \frac{k}{m-1} \right)_p} \left(- \frac{m^m}{(m-1)^{m-1}} \right)^{p}\\ &=\left( m-1 \right)\frac{\left(\frac{1}{m-1} \right)_p}{\left(1+\frac{1}{m-1} \right)_p} \frac{\prod_{k=1}^{m-1}\left( \frac{k}{m} \right)_p}{\prod_{k=1}^{m-2}\left( \frac{k}{m-1} \right)_p} \left(- \frac{m^m}{(m-1)^{m-1}} \right)^{p} \end{align*}

The inverse transform can thus be expressed as a hypergeometric series: \begin{align*} \mathcal{M}_x^{-1}\left[ \frac{1}{m-1}f(s)\right]&=\frac{1}{m-1}\sum_{p=0}^\infty r_px^{p(m-1)}\\ &=\;_{(m-1)}F_{(m-2)}\left(\left\{\frac{1}{m},\frac{2}{m},\cdots,\frac{m-1}{m}\right\}; \left\{\frac{2}{m-1},\cdots,\frac{m-2}{m-1},\frac{m}{m-1}\right\};-\frac{m^mx^{m-1}}{(m-1)^{m-1}}\right)\\ \end{align*} which is the proposed identity.

$\endgroup$
  • $\begingroup$ Thank you so much. That's a very clear explanation that solves both problems directly. $\endgroup$ – Benedict W. J. Irwin Jun 7 '18 at 8:29
  • $\begingroup$ You're welcome. I was interested by your inverse transform. $\endgroup$ – Paul Enta Jun 7 '18 at 13:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.