Working a bit on
About the inequality conjectured as $x!>\left(\arctan\left(\cosh\left(x\right)\right)\right)^{a}$ for $x>0$ and fixed $a$
I got the inequality:
$$\frac{\sqrt{\pi}}{2}<\left(\pi-e\right)!$$
My first attempt was to translate into an integral reprentation as we have:
$${\displaystyle \int _{0 }^{\infty }e^{-x^{2}}\,dx=\frac{1}{2}{\sqrt {\pi }}.}$$
and
$${\displaystyle \Gamma (z)=\int _{0}^{\infty }x^{z-1}e^{-x}dx}$$
and compare the different integrals.
Some other information :
We have a quite good approximation of the minimum of the Gamma function taking :
$$(\pi-e+0.5)/2\simeq x_{min}=0.4616\cdots$$
Ps: $\Gamma(0.5)=\frac{\sqrt{\pi}}{2}$
Second attempt :
As it's hard to find the right trick to compare the two integrals , I have the idea to express the inequality as :
$$\frac{\sqrt{\pi}}{2}\frac{\sqrt{\pi}}{2}<\frac{\sqrt{\pi}}{2}\left(\pi-e\right)!$$
And use this link where we have :
$$
\Gamma(p)\Gamma(q)=4\int_0^{\infty}\int_0^{\infty}x^{2p-1}y^{2q-1}\operatorname{e}^{-x^2-y^2}\operatorname{d}\!x\operatorname{d}\!y.\tag 1
$$
And use :
$$\int_{0}^{\infty}\int_{0}^{\infty}e^{-\left(x^{2}+y^{2}\right)}dxdy=\frac{\pi}{4}$$
Unfortunetaly I cannot procedd further .
Last attempt :
It seems we have the inequality on $[0,\pi-e]$:
$$f(x)=\frac{\left(\left(\frac{x^{2}+1}{x+1}\right)^{45.9}+\left(\frac{x^{2}+2}{x+2}\right)^{45.9}\right)^{\frac{1}{45.9}}}{\left(2\right)^{\frac{1}{45.9}}}\leq x!$$
Remains to show :
$$f(\pi-e)>\sqrt{\frac{\pi}{4}}$$
Wich is (numerically) true.
I would like to know three things:
- How to finish it by hand ?
- Have you an alternative proof ?
- Is it a well-know result ?
Thanks in advance.
Best Answer
Suppose that you consider the function $$f(x)=(x-e)!-\frac {1} 2\sqrt{x}$$ and you search for its zero.
Let $x=y+e$ and consider that you look for the zero of function $$g(y)=\Gamma(y+1)-\frac 12 \sqrt{y+e}$$ $$g'(y)=\Gamma (y+1)\, \psi^{(0)} (y+1)-\frac{1}{4 \sqrt{y+e}}$$ $$g''(y)=\frac{1}{8 (y+e)^{3/2}}+\Gamma (y+1) \left(\psi ^{(0)}(y+1)^2+\psi ^{(1)}(y+1)\right)$$
By inspection $$g'(0)=-\frac{1}{4 \sqrt{e}}-\gamma$$ $$g'\left(\frac{1}{2}\right)=\frac{\sqrt{\pi }}{2} (2-\gamma -2 \log (2))-\frac{1}{4 \sqrt{\frac{1}{2}+e}}\quad <0$$ $$g'(1)=1-\frac{1}{4 \sqrt{1+e}}-\gamma\quad >0$$
$$\left|\frac{g'(1)}{g'\left(\frac{1}{2}\right)}\right| \sim 2.74$$ So the derivative cancels closer to $\frac{1}{2}$ than to $1$.
For $y_0=\frac 12$, all derivatives involves known numbers; so, we can use one single iteration of Newton-like methods of order $n$ and have explicit formulae for any $y_{(n)}$.
Adding $e$ to the result and computing their decimal representation, the results are $$\left( \begin{array}{ccc} n & x_{(n)} & \text{method} \\ 2 & 3.117817 & \text{Newton} \\ 3 & 3.146499 & \text{Halley} \\ 4 & 3.140573 & \text{Householeder} \\ 5 & 3.141912 & \text{no name} \\ 6 & 3.141614 & \text{no name} \\ \cdots & \cdots & \\ \infty & 3.141668 \end{array} \right)$$