Show $\left | \cosh{x}-1 \right |\leq 3 \left | x \right |$ for $\left | x \right |<1/2$

Question

I have to show that $\left | \cosh{x}-1 \right |\leq 3 \left | x \right |$ for $\left | x \right |<1/2$. I can not use derivatives, series of the trigonometric functions etc. I have to use generel inequalities i.e like $$1+x\le e^x\le\frac{1}{1-x}$$

How would I do this. I am completely lost. I know the definition of hyperbolic cosine but I can not work this one out.

Answer 1

First proof We know that $\text{ cosh} \left(x\right)-1 = 2 {\text{ sinh}}^{2} \left(\frac{x}{2}\right)$. Let $t = \left|x\right|/2$, it remains to prove that

\begin{equation} 0 \leqslant t \leqslant \frac{1}{4} \Longrightarrow {\text{ sinh}}^{2} \left(t\right) \leqslant 3 t \end{equation}

For such $t$, we have

\begin{equation} \text{ sinh} \left(t\right) \leqslant \frac{1}{2} \left({e}^{t}-1\right) \leqslant \frac{1}{2} \left(\frac{1}{\left(1-t\right)}-1\right) \leqslant \frac{t}{2 \left(1-t\right)} \end{equation}

It remains to prove that $\frac{t^2}{4 \left(1-t\right)^2} \leqslant 3 t$ for $t \in I = \left[0 , 1/4\right]$, but this is equivalent to $P(t) = 12 t^2 - 25 t + 12\ge 0$, which is true because the roots of $P$ are $3/4$ and $4/3$.

Second proof It suffices to prove the result for $0\le x\le 1/2$. For such $x$ we have \begin{equation} \cosh x - 1 = \frac{e^x + e^{-x}}{2}- 1 \le e^x - 1 \le \frac{1}{1-x} - 1 = \frac{x}{1-x} \end{equation} It remains to show that $\frac{x}{1-x}\le 3 x$ but this is equivalent to $1\le 3 (1-x)$, that is to say $x\le \frac{2}{3}$, which is true. QED

Answer 2

$\cosh(x) =\sum_{n=0}^{\infty} \dfrac{x^{2n}}{(2n)!} $ so, if $0 \le x \le a < 1$,

$\begin{array}\\ \cosh(x)-1 &=\sum_{n=1}^{\infty} \dfrac{x^{2n}}{(2n)!}\\ &=x^2\sum_{n=1}^{\infty} \dfrac{x^{2n-2}}{(2n)!}\\ &=x^2\sum_{n=0}^{\infty} \dfrac{x^{2n}}{(2n+2)!}\\ &=\dfrac{x^2}{2}\sum_{n=0}^{\infty} \dfrac{x^{2n}}{(2n+2)!/2}\\ &\le\dfrac{x^2}{2}\sum_{n=0}^{\infty}x^{2n}\\ &\le\dfrac{x^2}{2(1-x^2)}\\ &=x\dfrac{x}{2(1-x^2)}\\ &\le x\dfrac{a}{2(1-a^2)} \qquad\text{since } \dfrac{x}{1-x^2} \text{ in increasing for }0 < x < 1\\ \end{array} $

If $a = \dfrac12$, $\dfrac{a}{2(1-a^2)} =\dfrac{1/2}{2(1-1/4)} =\dfrac{1}{4(3/4)} =\dfrac13 $

so $|\cosh(x)-1| \le x/3$ for $|x| \le \dfrac12$ (actually, according to Wolfy, for $|x| \lt 0.644$).

Answer 3

Focusing on positive $x$ (symmetry takes care of the rest), we need to show that, for $x\in[0,\frac 12]$, we have $\cosh x \leq 3x + 1$.

Assuming that you can use the fact that $\cosh x$ is a convex function, you know that $\cosh x \leq 2 \cosh \frac 12 \cdot x$, when $x \in [0,\frac 12]$. So, we just need to show that $2\cosh \frac 12 \cdot x \leq 3x+1$.