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 $h$.


1 Answer 1


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.

  • $\begingroup$ 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 $\endgroup$
    – Noah16
    Dec 3, 2018 at 14:27
  • $\begingroup$ 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. $\endgroup$ Dec 3, 2018 at 14:45
  • $\begingroup$ ah ya I understand you. could you please add some of the calculations to show how you got $a_{approx}$ above...THNAKS $\endgroup$
    – Noah16
    Dec 3, 2018 at 15:10
  • $\begingroup$ 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 $\endgroup$
    – Noah16
    Dec 19, 2018 at 17:51
  • $\begingroup$ unfortunately, if you considered the term you neglect then the bound is negative and we come back to the same problem ! $\endgroup$
    – Noah16
    Dec 19, 2018 at 18:20

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