2
$\begingroup$

Working on one of other question Show the inequality $\frac{\sqrt{\pi}}{2}<\left(\pi-e\right)!$ I found :

$$\frac{\left(\pi-e+\frac{1}{2}\right)}{2}\simeq x_{min}=0.4616\cdots$$

Where $x_{min}$ verify $f'(x_{min})=0$ of $f(x)=x!$ for $x>0$

Now I refine it to get :

$$\frac{\left(\pi-e+\frac{1}{2}\right)}{2}-\frac{2\left(\pi^{2}-e^{2}\right)^{e+\pi}}{\left(\pi^{2}+e^{2}\right)^{e+\pi}}\simeq x_{min}$$

Edit after Tyma Gaidash comment :

We have a better result with :

$$x_{min}\simeq\frac{\left(\pi-e+0.5\right)}{2}-2\frac{\left(\pi^{2}-e^{2}\right)^{\left(\pi+e\right)}}{\left(\pi^{2}+e^{2}\right)^{\left(\pi+e\right)}}+\frac{\left(\pi^{3}-e^{3}\right)^{\frac{7}{4}\left(\pi+e\right)}}{\left(\pi^{3}+e^{3}\right)^{\frac{7}{4}\left(\pi+e\right)}}-\frac{\left(\pi^{4}-e^{4}\right)^{2\left(\pi+e\right)}}{\left(\pi^{4}+e^{4}\right)^{2\left(\pi+e\right)}}+\frac{\left(\pi^{5}-e^{5}\right)^{\frac{13}{5}\left(\pi+e\right)}}{\left(\pi^{5}+e^{5}\right)^{\frac{13}{5}\left(\pi+e\right)}}-\frac{\left(\pi^{6}-e^{6}\right)^{\frac{18}{5}\left(\pi+e\right)}}{\left(\pi^{6}+e^{6}\right)^{\frac{18}{5}\left(\pi+e\right)}}$$

We can continue but it converge very slowly .

Is it the beginning of something like a power series or can we find something like $$\frac{\left(\pi-e+\frac{1}{2}\right)}{2}+\sum_{n=2}^{\infty}\frac{a_{n\ }\left(\pi^{n}-e^{n}\right)^{b_{n}\left(e+\pi\right)}}{\left(\pi^{n}+e^{n}\right)^{b_{n}\left(e+\pi\right)}}=x_{min}$$ where $a_n,b_n$ are rational. How to explain that ?

$\endgroup$
5
  • $\begingroup$ See also the value $$\left(\pi-e+0.5\right)\cdot\frac{1}{2}-2\left(\frac{\left(\pi^{2}-e^{2}\right)}{\pi^{2}+e^{2}}\right)^{\left(\pi+e\right)}-\left(\frac{\left(\pi^{3}-e^{3}\right)}{\pi^{3}+e^{3}}\right)^{\left(\frac{7}{4}\left(\pi+e\right)\right)}$$ $\endgroup$ Jun 7, 2022 at 12:02
  • $\begingroup$ I want to says that $a_n,b_n$ can be negative . $\endgroup$ Jun 10, 2022 at 7:55
  • $\begingroup$ Can you give a few more terms, 4 maybe, than the part directly after “Now I redefine to get:”? $\endgroup$ Jun 10, 2022 at 13:28
  • $\begingroup$ @TymaGaidash Thanks for the edit ,let me the time to do that . $\endgroup$ Jun 10, 2022 at 14:55
  • $\begingroup$ It might be just a coincidence, however, I would advise you to look at link. Inverting the digamma function might give you interesting series expansions for $x_{min}$. $\endgroup$ Jun 13, 2022 at 9:44

2 Answers 2

1
+50
$\begingroup$

Let $$f_n(a_n)=\left(\frac{\pi ^n-e^n}{\pi ^n+e^n}\right)^{a_n(e+\pi ) }$$

$$\frac{1}{2}\left(\pi-e+\frac{1}{2} \right)-2 f_2(1)-f_3\left(\frac{7}{4}\right)-f_4\left(\frac{5}{2}\right)+f_5\left(3\right)+$$ $$f_6\left(\frac{7}{2}\right)-f_7\left(\frac{17}{4}\right)+f_8\left(6\right)+f_9\left(7\right)-f_{10}\left(\frac{35}{4}\right)-f_{11}\left(\frac{23}{2}\right)+f_{12}\left(14\right)$$ gives $\color{red}{0.4616321449}82$ to be compared to $\color{red}{0.461632144968}$

$\endgroup$
2
  • $\begingroup$ Perhaps you mean $f_{n}(a)$ in the first line? $\endgroup$ Jun 13, 2022 at 10:28
  • $\begingroup$ @TianVlašić. Thanks for pointing it ! $\endgroup$ Jun 13, 2022 at 10:30
1
$\begingroup$

Like Claude Leibovici's answer we have :

Let :

$$f\left(x\right)=x!$$

And :

$$h(x)=f'(x)$$

Then set :

$$A=\frac{\left(\pi-e+0.5\right)}{2}-2\frac{\left(\pi^{2}-e^{2}\right)^{\left(\pi+e\right)}}{\left(\pi^{2}+e^{2}\right)^{\left(\pi+e\right)}}+\frac{\left(\pi^{3}-e^{3}\right)^{\frac{7}{4}\left(\pi+e\right)}}{\left(\pi^{3}+e^{3}\right)^{\frac{7}{4}\left(\pi+e\right)}}-\frac{\left(\pi^{4}-e^{4}\right)^{2\left(\pi+e\right)}}{\left(\pi^{4}+e^{4}\right)^{2\left(\pi+e\right)}}+\frac{\left(\pi^{5}-e^{5}\right)^{\frac{13}{5}\left(\pi+e\right)}}{\left(\pi^{5}+e^{5}\right)^{\frac{13}{5}\left(\pi+e\right)}}-\frac{5}{4}\frac{\left(\pi^{6}-e^{6}\right)^{\frac{18}{5}\left(\pi+e\right)}}{\left(\pi^{6}+e^{6}\right)^{\frac{18}{5}\left(\pi+e\right)}}+\frac{6}{5}\frac{\left(\pi^{7}-e^{7}\right)^{\frac{19}{4}\left(\pi+e\right)}}{\left(\pi^{7}+e^{7}\right)^{\frac{19}{4}\left(\pi+e\right)}}-\frac{251}{200}\frac{\left(\pi^{8}-e^{8}\right)^{\frac{31}{5}\left(\pi+e\right)}}{\left(\pi^{8}+e^{8}\right)^{\frac{31}{5}\left(\pi+e\right)}}+\frac{\left(\pi^{9}-e^{9}\right)^{\frac{48}{5}\left(\pi+e\right)}}{\left(\pi^{9}+e^{9}\right)^{\frac{48}{5}\left(\pi+e\right)}}-\frac{\left(\pi^{10}-e^{10}\right)^{\frac{52}{5}\left(\pi+e\right)}}{\left(\pi^{10}+e^{10}\right)^{\frac{52}{5}\left(\pi+e\right)}}$$

Then a numerical approximation gives :

$$|h(A)|<4* 10^{-15}$$

We have $A=0.461632144968\cdots$

$\endgroup$

You must log in to answer this question.

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