0
$\begingroup$

Stirling’s approximation can be extended to a very well known inequality -

$$\sqrt{2\pi n}\left(\frac{n}{e}\right)^n \leq n! \leq e\sqrt{n}\left(\frac{n}{e}\right)^n$$

How can we use this to prove,

$$\ln \frac{(h+f)!}{(h-g)!} = (f+g)\ln h \pm O\left(\frac{(f+g)^2}{h}\right) $$

when $f+g=o(h)$

My Attempt

$$ \begin{align} \ln \frac{(h+f)!}{(h-g)!} &= \ln \frac{\left(h+f\right)^{h+f+1/2}}{\left(h-g\right)^{h-g+1/2}} + \ln \frac{e^{h-g}}{e^{h+f}}\\ &= (h+f+1/2)\ln(h+f) - (h-g+1/2)\ln(h-g) - (f+g) \end{align} $$

At this point I can use property of logarithms $\ln(h+f)=\ln h+\ln(1+f/h)$ and $\ln(h-g)=\ln h+\ln(1-g/h)$ to extract $\ln h$ terms. But that would leave a lot of other terms with itself which I'm not able to collate.

How should I move ahead from hear? If this is not the right approach then I hope to be pointed towards right direction.

$\endgroup$
  • $\begingroup$ I never saw the very well known inequality :-) $\endgroup$ – Yves Daoust Mar 22 at 10:41
  • $\begingroup$ I'm sorry, I should not have presumed too much. But the inequality is there on wiki, and quite straight forward to prove as well. $\endgroup$ – Prateek Mar 22 at 11:05
1
$\begingroup$

To continue, you can use the Taylor expansion of logarithm, just the beginning: $$ \ln(1+x) = x + \mathcal{O}(x^2)$$ That gives you $$ (h+f+1/2)\ln(1+f/h) = f + \mathcal{O}\left(\frac{f^2}{h}\right) $$ and $$-(h-g+1/2)\ln(1+g/h) = g + \mathcal{O}\left(\frac{g^2}{h}\right) $$

However, note that Stirling's approxiamtion won't give you the exact equality as you're writing it. What you can get from them is $$ \ln (n!) = (n+1/2)\ln n - n + \xi(n)$$ where $\xi(n) \in [\ln\sqrt{2\pi},1]$. This last term is negligible compared to the leading term, but it doesn't vanish. In the end, the best you can get this with this method is $$ \ln\frac{(h+f)!}{(h-g)!} = (f+g)\ln h + \xi + \mathcal{O}\left(\frac{f^2}{h},\frac{g^2}{h}\right) $$ where $\xi = \xi(h+f)-\xi(h-g)$, $|\xi| \le 1-\ln\sqrt{2\pi}$

$\endgroup$

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.