Upper bounds on $f(x)=\frac{\cosh(a+x)}{\cosh(b+x)}$

Question

What would be some good upper bounds on the following ratio \begin{align} f(x)=\frac{\cosh(a+x)}{\cosh(b+x)} \end{align} whre $a,b \in \mathbb{R}$.

I tried using $1+\frac{x^2}{2} \le\cosh(x)\le e^{x^2/2}$, but the bounds are not very good.

Also, from the plots that I did the function $f(x)$ seems to be bounded. Thanks.

Answer 1

$$f(x)=\frac{\cosh(a+x)}{\cosh(b+x)}\implies \log f(x)=\log \cosh(a+x) - \log \cosh(b+x) $$

$$\frac{d}{ds}\log \cosh s=\tanh s\implies \log f(x) =-\int_a^b \tanh(x+t)\,dt$$

Then, as $|\tanh(u)| \leqslant 1$ for all $u\in\mathbb{R}$, the triangle inequality for integrals gives immediately that

$$|\log f(x)| \le \left| \int_a^bdt\right|=|b-a|$$

which can be shown to be a sharp bound by looking at $x\to\pm\infty$. Thus:

$$|\log f(x)| \le |b-a| \implies f(x) \in [e^{-|b-a|},e^{|b-a|}] \,\, \forall x\in\mathbb{R}$$

Answer 2

Just from the definition of $\mathrm{cosh}(x)$ you can see that $$\frac{\mathrm{cosh}(a+x)}{\mathrm{cosh}(b+x)} = \frac{e^{a + x} + e^{-a -x}}{e^{b + x} + e^{-b -x}} = \frac{e^{a + 2x} + e^{-a}}{e^{b + 2x} + e^{-b}}$$ and so as $x \to \infty$ this will very quickly approach $e^{a - b}$, and as $x \to -\infty$ it will approach $e^{b - a}$. So all you have to worry about are the turning points in the middle, which you should be able to find and analyse using the first and second derivatives.

Answer 3

It is apparent that the answer will only depend on $\lvert b-a \rvert$, because the hyperbolic cosine is even and we can shift $x$. So let's work with $$ f_c(x) = \frac{\cosh{(x-c)}}{\cosh{x}}. $$ Using the exponential expressions, $$ f_c(x) = \frac{e^{x}e^{-c}+e^{-x}e^{c}}{e^x+e^{-x}} = \frac{e^{-c}e^{2x}+e^{c}}{e^{2x}+1} = e^{-c} + \frac{e^{c}-e^{-c}}{e^{2x}+1} = e^{-c} +\frac{2\sinh{c}}{e^{2x}+1}. $$ This expression is useful because it explicitly describes a function that is increasing or decreasing, depending on the sign of $c$: $e^{2x}$ is increasing, so $1/(e^{2x}+1)$ is decreasing. Hence the extrema occur as $x \to \pm \infty$.

• If $c>0$, the function is decreasing and so the maximum is $e^c$ and the minimum is $e^{-c}$.
• If $c<0$ the function is increasing, so the maximum is $e^{-c}$ and the minimum is $e^{c}$.

In both cases, the extrema are not achieved. Obviously if $c=0$ the function is constant. Hence we obtain $$ e^{|c|} > f_c(x) > e^{-|c|}. $$