# Complicated bound after using Stirling's approximation

I have this inequality $$\frac{1}{a}\exp\bigl\{-\frac{4}{h^2}\bigr\} \geq \frac{1}{f}$$ where $$a \leq \Bigl(\pi^{d/2}\Gamma(\frac{1}{2}d+1)^{-1} + 1\Bigr) \left(\frac{h^{d+1}}{2} \Gamma \left(\frac{d+1}{2} \right) + h^d \left(\frac{d}{2}\right)^\frac{d}{2} \exp\Bigl\{-\frac{d}{2}\Bigr\} \right)$$ I tried to lowerbound $$h$$ in the inequality using the properties of the Gamma function and Stirling's approximation but I still see it complicated to be used in order to bound $$h$$.

Here is my work

Again \begin{align} a & \leq \Bigl(\pi^{d/2}\Gamma(\frac{1}{2}d+1)^{-1} + 1\Bigr) \left(\frac{h^{d+1}}{2} \Gamma \left(\frac{d+1}{2} \right) + h^d \left(\frac{d}{2}\right)^\frac{d}{2} \exp\Bigl\{-\frac{d}{2}\Bigr\} \right) \nonumber \\ \text{let \frac{d}{2}=n, then} \nonumber \\ a & \leq \Bigl(\pi^{n}\Gamma(n+1)^{-1} + 1\Bigr)\left( \frac{h^{2n+1}}{2} \Gamma \left( n+\frac{1}{2} \right) + h^{2n} n^n \exp\Bigl\{-n\Bigr\} \right) \nonumber \end{align} If $$\Gamma(n+1) = n\Gamma(n)=n(n-1)!=n!$$ and $$\Gamma(n+\frac{1}{2})=\frac{(2n)!}{4^n n!}\sqrt(\pi)=\frac{(2n-1)!!}{2^n}\sqrt(\pi)=\binom {n-\frac{1}{2}}{n}n! \sqrt(\pi)$$ then

\begin{align} a & \leq \Bigl(\frac{\pi^{n}}{n!} + 1\Bigr)\left( \frac{h^{2n+1}}{2} \frac{(2n)!}{4^n n!}\sqrt(\pi) + h^{2n} n^n \exp\Bigl\{-n\Bigr\} \right)\nonumber \\ & \leq h^{2n+2} \Bigl( \Bigl(\frac{\pi^{n}}{n!} + 1\Bigr) \left( \frac{(2n)!}{2^{2n }n!}\sqrt(\pi) + n^n \exp\Bigl\{-n\Bigr\} \right) \Bigr) \nonumber \end{align} Using stirling's approximation \begin{align} a & \leq h^{2n+2} \Bigl( \Bigl( \frac{1}{\sqrt{2}} \frac{\pi^{n-(1/2)}e^n}{n^{n+(1/2)}} + 1\Bigr)\left(\frac{1}{\sqrt{2}} \frac{e^n}{n^{n+(1/2)}} + \left(\frac{n}{e}\right)^n\right) \Bigr) \nonumber \\ & \leq h^{2n+2}\Bigl( \Bigl( \pi^{n-\frac{1}{2}} \left( \frac{e}{n}\right)^n +1 \Bigr) \Bigl(\left( \frac{e}{n}\right)^n + \left( \frac{n}{e}\right)^n \Bigr) \Bigr) \nonumber \\ & = h^{2n+2}\Bigl( \pi^{n-\frac{1}{2}} \Bigl( \left( \frac{e}{n}\right)^{2n} +1 \Bigr) + \left( \frac{e}{n}\right)^n + \left( \frac{n}{e}\right)^n \Bigr) \nonumber \end{align} What I can do more to simplify $$a$$ in order to use it to bound $$h$$ or is there a better way to bound $$h$$.

I understand from the reference to Stirling that you are looking for a large-$$d$$ approximation of

$$a_{\rm max}= \Bigl(\pi^{d/2}\Gamma(\frac{1}{2}d+1)^{-1} + 1\Bigr) \left(\frac{h^{d+1}}{2} \Gamma \left(\frac{d+1}{2} \right) + h^d \left(\frac{d}{2}\right)^\frac{d}{2} \exp\left(-\frac{d}{2}\right)\right).$$ With some algebra I arrived at

$$a_{\rm max}\approx\left(h^{d+1}\sqrt{\frac{\pi}{2} } +h^d\right) \exp\left[\frac{d}{2} \left(\ln \left(\frac{d}{2}\right)-1\right)\right]\equiv a_{\rm approx},\;\;d\gg 1.$$

This is already quite accurate for moderately small $$d$$, here is a plot for $$d=10$$. Shown are $$a_{\rm max}/f$$ (blue), $$a_{\rm approx}/f$$ (green), and $$\exp(-4/h^2)$$ (orange) as a function of $$h$$ for $$f=10^5$$.

The condition $$\exp(-4/h^2)\geq a_{\rm max}/f$$ is reached in an interval $$(h_{\rm min},h_{\rm max})$$ around $$h_0$$ given by

$$4/h_0^2=\ln f-\frac{d}{2} \left(\ln \left(\frac{d}{2}\right)-1\right)$$

This interval only exists if $$f$$ is large enough, you need $$f\gtrsim\exp\left[\frac{d}{2} \left(\ln \left(\frac{d}{2}\right)-1\right)\right]\equiv f_{\rm min}.$$

Irrespective of these large-$$d$$ approximations, the inequality in the OP is always violated for large enough $$h$$, because $$a_{\rm max}$$ grows as $$h^{d+1}$$ for large $$h$$, while $$\exp(-4/h^2)$$ tends to unity. So $$h_{\rm max}<\infty$$. Moreover, $$h_{\rm min}>0$$ because $$a_{\rm max}$$ vanishes as $$h^d$$ for small $$h$$, while $$\exp(-4/h^2)$$ vanishes more rapidly.

• Thanks for your help and cooperation. Concerning $d$ it is not necessarily too large but I can say greater than $10$. Maybe I wan't clear enough but I am trying to show that (using a lowerbound) $h$ is greater than a positive number! but according to your work it's still negative bound Dec 3, 2018 at 14:27
• why negative bound? $h_0$ is a real positive number if $f$ is large enough and the interval in $h$ in which your inequality holds is a narrow interval centered on $h_0$; so you can conclude that $h$ is near $h_0>0$ provided $f>f_{\rm min}$. If $f<f_{\rm min}$ you cannot satisfy your inequality. Dec 3, 2018 at 14:45
• ah ya I understand you. could you please add some of the calculations to show how you got $a_{approx}$ above...THNAKS Dec 3, 2018 at 15:10
• Thanks for your cooperation and sorry for this late response. It seems that there is something wrong in the calculations. I think you just expanded $$\left(\frac{h^{d+1}}{2} \Gamma \left(\frac{d+1}{2} \right) + h^d \left(\frac{d}{2}\right)^\frac{d}{2} \exp\left(-\frac{d}{2}\right)\right)$$ and neglect this amount $$\Bigl(\pi^{d/2}\Gamma(\frac{1}{2}d+1)^{-1} + 1\Bigr)$$ which is much more than one Dec 19, 2018 at 17:51
• unfortunately, if you considered the term you neglect then the bound is negative and we come back to the same problem ! Dec 19, 2018 at 18:20