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As the title say I try to better understand the approximation due to Ramanujan :

$$\frac{\sqrt{\pi}\left(\frac{x}{e}\right)^{x}\left(8x^{3}+4x^{2}+x+\frac{1}{30}\right)^{\frac{1}{6}}}{x!}>1,\forall x>0$$

What's the intuition behind this ?

I managed myself to find some other formula like :

Let $x>0$ then we have :

$$\left(\frac{-\frac{1}{30}+\sqrt{1+x\sqrt{1+x\sqrt{1+x\sqrt{1+x\sqrt{...}}}}}}{x}\right)^{\frac{1}{6}}<\frac{x!}{\sqrt{2\pi x}\left(\frac{x}{e}\right)^{x}}$$

A more spectacular example is :

Let $x>0$ then we have :

$$f\left(x+1\right)-\left(\left(\sqrt{\pi}\left(\frac{x}{e}\right)^{x}\left(8x^{3}+4x^{2}+x+\frac{1}{100}\right)^{\frac{1}{6}}\right)^{-1}x!\right)>0$$

Where : $$f\left(x\right)=\left(\frac{\left(\sqrt{1+x\sqrt{1+x\sqrt{1+x\sqrt{\cdot\cdot\cdot}}}}\right)}{x}\tanh\left(x\right)\right)^{\frac{1}{x}}$$

So starting from this example we see that the inspiration of Ramanujan is not composite but is for sure at this point a nice thinking .

Second remark :

We can also say that the Stirling's approximation is not unique see :

$$\lim_{x\to\infty}q\left(x\right)=\lim_{x\to\infty}\left(\frac{x!}{f\left(x\right)}\right)^{-1}\cdot\frac{\left(h\left(x\right)\right)^{-1}}{\left(x-\frac{1}{12}\right)}=1$$

Where :

$$h\left(x\right)=\exp(13/(99x^{\frac{9}{2}})+5/(63x^{\frac{7}{2}})-12/(35x^{\frac{5}{2}})+7/(15x^{1.5})-1/(15x^{5})+9/(40x^{4})-5/(24x^{3})-1/(12x^{2})-\sqrt{x}-1/(3\sqrt{x})+3/(4x)+1/2),f\left(x\right)=\sqrt{2\pi x}\left(1-\sqrt{x}+x\right)^{1+x}e^{-x}$$

Now let $h(x)$ as above and introduce :

$$F\left(x\right)=\sqrt{\pi}\left(1-\sqrt{x}+x\right)^{1+x}e^{-x}\cdot(8x^{3}+4x^{2}+x+1/30)^{\frac{1}{6}}$$

We have :

$$\lim_{x\to\infty}\left(\frac{x!}{F\left(x\right)}\right)^{-1}\cdot\frac{\left(h\left(x\right)\right)^{-1}}{x}=1$$

So it seems under a slight modification of the Ramaanujan's approximation there is some invariant ($h(x)$ doesn't change).

Is it purely a calculus trick ? Or in other words Does Ramanujan have a single idea in mind ? If so have you some example where Ramanujan draws perfectly ?

Any helps is greatly welcome .

Miss and Mister cassoulet char
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  • You can also use the tag [intuition] in this case:) – Nicolás A. Jun 05 '23 at 08:58
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    One can write power series of $\frac{\Gamma(x+1)}{\sqrt{2\pi x}\left(\frac{x}{e}\right)^{x}}$ for $x\to\infty$. Then one can consider expression of types $(1+a/x)^{1/2}$, $(1+b/x+c/x^2)^{1/4}$, $(1+d/x+e/x^2+f/x^3)^{1/6}$ and choose coefficients $a$, $b$, $c$, $d$, $e$, $f$ to get the same power series starting coefficients. The simplest formula (with $a$) looks like $\frac{x!}{\sqrt{2\pi x}\left(\frac{x}{e}\right)^{x}}>\sqrt{1+\frac{1}{6x}}$. Ramanujan formula is 3rd order formula with $d$, $e$, $f$. – Ivan Kaznacheyeu Jun 05 '23 at 09:44
  • @IvanKaznacheyeu Good point ! Perhaps we need to see a proof ? – Miss and Mister cassoulet char Jun 05 '23 at 09:56
  • Not the subject : $$\left(\frac{\sqrt{1+x\sqrt{1+x\sqrt{1+x\sqrt{1+x\sqrt{1+x\sqrt{1+x\sqrt{1+x\sqrt{1+x\sqrt{1+x}}}}}}}}}}{x}\tanh\left(x\right)\right)^{\frac{1}{x}},g\left(x\right)=1+\frac{\left(f\left(x+1\right)-\left(\frac{\sqrt{\pi}\left(\frac{x}{e}\right)^{x}\left(8x^{3}+4x^{2}+x+\frac{1}{100}\right)^{\frac{1}{6}}}{x!}\right)^{-1}\right)}{12},h\left(x\right)=\frac{\left(x^{\frac{x}{x+1}}+\sqrt{x}-x\right)1}{10}+\frac{9}{10},|g\left(3\right)-h\left(3\right)|<6*10^{-7}$$ So it seems there is someting with Shannon's entropy – Miss and Mister cassoulet char Jun 07 '23 at 08:14
  • You could be interested by https://math.stackexchange.com/questions/3629388/on-ramanujans-approximation-n-sim-sqrt-pi-big-frac-ne-bign-sqrt-62/3629529#3629529 – Claude Leibovici Jun 07 '23 at 12:22
  • @ClaudeLeibovici Thanks ! You're a master at it ! – Miss and Mister cassoulet char Jun 08 '23 at 07:57
  • @ClaudeLeibovici See the graph of $$g\left(x\right)=\frac{\left(\int_{0}^{x}\frac{\sqrt{\pi}\left(\frac{y}{e}\right)^{y}\left(8y^{3}+4y^{2}+y+\frac{1}{30}\right)^{\frac{1}{6}}}{y!}dy\right)}{x-\frac{1}{12}},\left(\sqrt{2\pi x}\left(\frac{x}{e}\right)^{x}\right)^{-1}\cdot x!,g\left(x+0.11\right)$$ – Miss and Mister cassoulet char Jun 08 '23 at 11:44
  • @ClaudeLeibovici And $$\frac{\left(\frac{\left(x\right)!}{\sqrt{2\pi x}\left(\frac{x}{e}\right)^{x}}+\frac{x+0.11}{x-\frac{1}{12}+0.11}\right)}{2}$$ – Miss and Mister cassoulet char Jun 08 '23 at 12:32
  • @ClaudeLeibovici see also https://mathworld.wolfram.com/One-NinthConstant.html – Miss and Mister cassoulet char Jun 08 '23 at 13:55

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