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I'm glad to share my last (little) discovery concerning the Gamma function or here $x!$

The problem :

Let $x>0$ and $1\leq a \leq \left(\frac{\pi}{e}\right)^{e}$ then it seems we have :

$$f(x)=x!>\left(\arctan\left(\cosh\left(x\right)\right)\right)^{a}$$



As attempt we know that the Gamma function is convex so I think we can use strong convexity to got :

$$x!\geq f'(a)(x-a)+\frac{f''(a)}{2}(x-a)^2$$

On a segment obviously .

Next I'm stuck even using derivatives . One other important inequality on the arctangent function is Shafer- Fink inequality .

As remark the right hand side behaves as a derivatives (graphicaly speaking) .

Have you an a proof (the problem) and an explanation of this fact (the remark) ?

Thanks in advance for all your patience and effort .

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1 Answer 1

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Just of few remarks for a start

Consider $$a(x)=\frac{\log (x!)}{\log \left(\tan ^{-1}(\cosh (x))\right)}$$ The first derivative $$a'(x)=\frac{ \psi ^{(0)}(x+1)}{ \log \left(\tan ^{-1}(\cosh (x))\right)}-\frac{\sinh (x) \log (x!)}{\left(\cosh ^2(x)+1\right) \tan ^{-1}(\cosh (x)) \log ^2\left(\tan ^{-1}(\cosh (x))\right)}$$ shows two zeros on each side of $1$.

The local maximum corresponds to $$x_1= 0.88763805477167701049699323506\cdots$$ $$a(x_1)=0.99560883432768261234933352921\cdots$$

The local minimum corresponds to

$$x_2= 1.13014684117571091749239939997\cdots$$ $$a(x_2)=1.48343672926714924369749424169\cdots$$ which is a bit larger than $\left(\frac{\pi }{e}\right)^e$

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  • $\begingroup$ See also $\sqrt{11/5}$ dear Claude . $\endgroup$ May 20, 2022 at 8:15
  • $\begingroup$ @ErikSatie. Much better indeed. By the way, long time no speak ! Cheers :-) $\endgroup$ May 20, 2022 at 8:21

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