2
$\begingroup$

I am reading a paper in which the authors arrive at the quotient of Gamma functions $\frac{\Gamma(s+k-1)}{\Gamma(k-1)}$ where we are thinking of $s$ as the variable and $k$ is just some other parameter. They then say

By Stirling's formula, for any vertical strip $0<a\leq \Re(s)\leq b$, we have that $$\frac{\Gamma(s+k-1)}{\Gamma(k-1)}=(k-1)^s\left(1+O_{a,b}((\vert s\vert+1)^2k^{-1})\right)$$

Now I am curious as to derive this from Stirling's formula (since using Stirling's formula is something I have always struggled with).

Since we are in a vertical strip, the version of Stirling's formula that I plan to use is $$ \Gamma(s)=\sqrt{2\pi}s^{s-\frac{1}{2}}e^{-s}\left(1+O\left(\frac{1}{\vert s\vert}\right)\right) $$

If I use this for our quotient, we have $$ \frac{\Gamma(s+k-1)}{\Gamma(k-1)}=\frac{\sqrt{2\pi}(s+k-1)^{s+k-\frac{3}{2}}e^{-s-k+1}\left(1+O\left(\frac{1}{\vert s+k-1\vert}\right)\right)}{\sqrt{2\pi}(k-1)^{k-\frac{3}{2}}e^{1-k}\left(1+O\left(\frac{1}{\vert k-1\vert}\right)\right)}=\frac{(s+k-1)^{s+k-\frac{3}{2}}e^{-s}\left(1+O\left(\frac{1}{\vert s+k-1\vert}\right)\right)}{(k-1)^{k-\frac{3}{2}}\left(1+O\left(\frac{1}{\vert k-1\vert}\right)\right)} $$

I don't know how to conclude from here. It seems as though heuristically since we are in a vertical strip we will have that the real part of $s$ is bounded so if I just say for simplicity that $s=0$, then I see how the $(k-1)^s$ term comes out, but all the other parts with the big $O$ still confuse me. I don't know if there is an easy way to get this or if I'm doing something wrong. Any help would be greatly appreciated.

$\endgroup$
1
  • 1
    $\begingroup$ Considering the case $k\gg1$ and $k\gg s$ $$\frac{\sqrt{2\pi}(s+k-1)^{s+k-\frac{3}{2}}e^{-s-k+1}}{\sqrt{2\pi}(k-1)^{k-\frac{3}{2}}e^{1-k}}=\sqrt{1+\dfrac s{k-1}}e^{-s}\frac{(k-1)^{s+k-1}\big(1+\frac s{k-1}\big)^{s+k-1}}{(k-1)^{k-1}}$$ $$=\sqrt{1+\dfrac s{k-1}}e^{-s}e^{(s+k-1)\ln(1+\frac s{k-1})}(k-1)^s=\sqrt{1+\dfrac s{k-1}}e^{-s}e^{(s+k-1)(\frac s{k-1}-\frac{s^2}{2(k-1)^2}+...)}(k-1)^s$$ $$=(k-1)^2\sqrt{1+\dfrac s{k-1}}e^{-s}e^se^{\frac {s^2}{2(k-1)}+O\big(\frac1{(k-1)^2}\big)}=(k-1)^s\left(1+O\Big(\frac1{k-1}\Big)\right)$$ $\endgroup$
    – Svyatoslav
    Commented Oct 13, 2023 at 3:28

1 Answer 1

Reset to default
4
$\begingroup$

I think that it is easier to start with logarithms $$A=\log \left(\frac{\Gamma (k+s-1)}{\Gamma (k-1)}\right)$$ Assuming that $k$ is large and larger than $s$, using twice Stirling approximation $$\log (\Gamma (x))=x (\log (x)-1)+\log \left(\sqrt{\frac{2\pi}{x}}\right)+\frac{1}{12 x}-\frac{1}{360 x^3}+O\left(\frac{1}{x^5}\right)$$ and continuing with Taylor series gives (with $\color{red}{p=k-1}$) $$A=s \log (p)+\frac{(s-1) s}{2 p}-\frac{(s-1) s (2 s-1)}{12 p^2}+\frac{(s-1)^2 s^2}{12 p^3}+O\left(\frac{1}{p^4}\right)$$ Exponentiating and continuing with Taylor $$e^A=(k-1)^s \left(1+\frac{(s-1) s}{2 (k-1)}+\frac{(s-2) (s-1) s (3 s-1)}{24(k-1)^2}+O\left(\frac{1}{k^3}\right) \right)$$

$\endgroup$
1
  • $\begingroup$ I understand using the twice Stirling approximation, but I am confused when you "continue with Taylor series" to get your formula for $A$, since when trying to work it out on my own, I just wrote $A=\log(\Gamma(s+p))-\log(\Gamma(p))$, and used the twice Stirling, but I didn't get that it simplified as nicely as you have, if you don't mind adding a bit to help explain it more. After that, I see how exponentiating gives me what I want, but now I'm just a little hung up over that one equality for $A$. $\endgroup$
    – Steven Creech
    Commented Oct 13, 2023 at 14:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .