tan 和 tanh⁻¹

hbghlyj

$0<x<\frac\pi2,$ 求证$$\tan x>\frac{\pi}4\cdot\ln\frac{x+\pi/2}{\pi/2-x}$$

hbghlyj

WolframAlpha或Matlab
>> limit(tan(x)/(log(x+pi/2)-log(pi/2-x)),x,pi/2,'left')
ans =
Inf

>> limit(tan(x)/(log(x+pi/2)-log(pi/2-x)),x,pi/2,'right')
ans =
-Inf

hbghlyj

$\frac2\pi x<\sin x⇒\frac{4x^2}{\pi^2}<\sin^2x⇒\cos^2x<1-\frac{4x^2}{\pi^2}⇒\frac1{\cos^2x}>\frac1{1-\frac{4x^2}{\pi^2}}$
从0到$x$积分得$$\tan x>\frac\pi2\arctanh(\frac2\pi x)$$即$$\tan x>\frac{\pi}4\cdot\ln\frac{x+\pi/2}{\pi/2-x}$$