Proving tanh(w) < w < sinh(w)

Question

Problem

I have the following inequality $$ tanh(w) < w < sinh(w) $$ where $w$ is a complex number and also $w > 0$. This suspiciously looks like a mean value theorem problem but I can't figure out how to transform this into an approachable problem. Looks like $w$ corresponds to the $c$ and the boundaries where $f$ is continous and differentiable are $$\Big]tanh(w), sinh(w)\Big[$$

I tried applying Euler's formula: $$ \begin{align} &\frac{e^w-e^{-w}}{e^w+e^{-w}} < w < \frac{e^w-e^{-w}}{2} \\ \implies \; &\frac{2}{e^w+e^{-w}} < \frac{2w}{e^w-e^{-w}} < 1 \end{align} $$

Any hints?

Answer 1

You can restate this as $\int_0^w\operatorname{sech}^2xdx<\int_0^w1dx<\int_0^w\cosh xdx$, which is trivial for $w>0$ since $\cosh x>1$ for $x>0$.

Answer 2

Since $w$ is a complex number and $w>0$, we obtain that $w$ is a real number.

The left inequality.

We need to prove that $$w>\frac{e^{2w}-1}{e^{2w}+1}$$ or $$(w-1)e^{2w}>-1-w,$$ which is obviously true for $w\geq1.$

Let $0<w<1.$

Thus, it's enough to prove that $$e^{2w}<\frac{1+w}{1-w}$$ or $f(w)>0,$ where $$f(w)=\frac{1}{2}(\ln(1+w)-\ln(1-w))-w.$$ Now, $$f'(w)=\frac{1}{2}\left(\frac{1}{1+w}+\frac{1}{1-w}\right)-1=\frac{1}{1-w^2}-1=\frac{w^2}{1-w^2}>0,$$ which says $$f(w)>f(0)=0$$ and we are done.

The right inequality we can prove by the similar way:

We need to prove that $$2w<\frac{e^{2w}-1}{e^w}$$ or $$e^{2w}-2we^w-1>0$$ or $$e^w>w+\sqrt{w^2+1}$$ or $g(w)>0,$ where $$g(w)=w-\ln(w+\sqrt{w^2+1}).$$ Now, $$g'(w)=1-\frac{1+\frac{w}{\sqrt{w^2+1}}}{w+\sqrt{w^2+1}}=1-\frac{1}{\sqrt{w^2+1}}>0,$$ which says $$g(w)>g(0)=0.$$

Answer 3

Refer to the diagram below depicting the unit hyperbola with center $O$, vertex $A$, focus $C$, and general point $P$. $OP$ intersects the tangent of $A$ at $R$.

If the area of hyperbolic sector $OAP$ is $\tfrac{\tau}{2}$, then by definition $$\begin{align} OQ &= \cosh \tau & QP &= \sinh \tau & AR &= \tanh \tau \end{align}$$

so that the areas of triangles $OAR$, $OQP$ are given by

$$\begin{align} \lvert OAR \rvert &= \tfrac{1}{2}\tanh \tau & \lvert OQP \rvert &= \tfrac{1}{2}\cosh\tau \sinh \tau\text{.} \end{align}$$

But $OAR\subset OAP \subset OQP$, whence $$\tfrac{1}{2}\tanh\tau < \tfrac{1}{2}\tau < \tfrac{1}{2}\cosh \tau \sinh \tau\text{.}$$

The result then follows by the duplication formula for $\sinh$.