asymptotic expansion of $\ln(\cosh(x))$

Question

I am currently looking into some C++ code which approximates the function in the title as follows

$$\ln\cosh(x))\approx x- \ln(2); \qquad x \geq 12$$

The approximation is plausible to me, since plotting the difference between the function results in an $error < 10^{-3}$. However how does one arrive at such an approximation? The usual way of calculating the Taylor expansion at $x_0=12$ does not result in such a nice expression for me.

Answer 1

Start from the definition: $$\cosh(x)=\frac{e^x+e^{-x}}2$$ For large $x$, you have $e^x>>e^{-x}$, so $\cosh x\approx\frac{e^x} 2$. Now take the log, and you get your answer.

Answer 2

Recall that

$$ \cosh x = \frac{e^{x} + e^{-x}}{2} $$

And that for $\color{blue}{x \gg 0}$ the term $e^{-x}$ is very small, so that

$$ \cosh x = \frac{e^{x} + e^{-x}}{2} \approx \frac{e^{x} + 0}{2} ~~~\mbox{for}~~ \color{blue}{x \gg 0} $$

Or equivalently

$$ \ln \cosh x \approx x - \ln 2 ~~~\mbox{for}~~ \color{blue}{x \gg 0} $$

Probably the author of the code decided that 12 was a large number to ignore the contribution from $e^{-x}$, which actually is $e^{-12} \approx 10^{-5}$

Answer 3

To complement the previous answers with an estimate of the error: \begin{align} \log(\cosh x)-(x-\log2)&=\log\frac{e^x+e^{-x}}2-\log\frac{e^x}2\\ &=\log(1+e^{-2x})\\ &\le e^{-2x} \end{align} for $x>0$.

Answer 4

Here is a useful approximation if we expand in both $x$ and $z=e^{-x}$.

\begin{align*} \log(\cosh(x))&=\log(e^x+e^{-x})-\log(2)\\ &=x-\log(2)+\log(e^x+e^{-x})-x\\ &=x-\log(2)+\log(z+z^{-1})+\log(z)\\ &=x-\log(2)+\log(1+z^2)\\ &=x-\log(2)+\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}{z^{2n}}\\ &=x-\log(2)+\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}e^{-2xn} \end{align*}

This came up in Cardy Scaling and Renormalization in Statistical Physics, problem 3.3, where $\log(2\cosh(K))$ is a free energy density and you care only about its non-analytic parts in the infinite sum.