Given that
\[
\mathrm{f}(x)=\ln(1+\mathrm{e}^{x}),
\]
prove that \(\ln[\mathrm{f}'(x)]=x-\mathrm{f}(x)\) and that \(\mathrm{f}''(x)=\mathrm{f}'(x)-[\mathrm{f}'(x)]^{2}.\)
Hence, or otherwise, expand \(\mathrm{f}(x)\) as a series in powers of \(x\) up to the term in \(x^{4}.\)
Given that
\[
\mathrm{g}(x)=\frac{1}{\sinh x\cosh2x},
\]
explain why \(\mathrm{g}(x)\) can not be expanded as a series of non-negative powers of \(x\) but that \(x\mathrm{g}(x)\) can be so expanded. Explain also why this latter expansion will consist of even powers of \(x\) only. Expand \(x\mathrm{g}(x)\) as a series as far as the term in \(x^{4}.\)
As \(x \to 0\), \(g(x) \to \infty\) therefore there can be no power series about \(0\). But as \(x \to 0, x g(x) \not \to \infty\) as \(\frac{x}{\sinh x}\) is well behaved.
We can also notice that \(x g(x)\) is an even function, since \(\cosh x\) is even and \(\frac{x}{\sinh x}\) is even, therefore the power series will consist of even powers of \(x\)
\begin{align*}
\lim_{x \to 0} \frac{x}{\sinh x \cosh 2 x} &= \lim_{x \to 0} \frac{x}{\sinh x} \cdot \lim_{x \to 0} \frac{1}{\cosh2 x} \\
&= 1
\end{align*}
Notice that
\begin{align*}
\frac{x}{\sinh x \cosh 2 x} &= \frac{4x}{(e^x - e^{-x})(e^{2x}+e^{-2x})} \\
&= \frac{4x}{(2x + \frac{x^3}{3} + \cdots)(2 + 4x^2 + \frac43 x^4 + \cdots )} \\
&= \frac{1}{1+\frac{x^2}{6}+\frac{x^4}{5!} + \cdots } \frac{1}{1 + 2x^2 + \frac23 x^4 + \cdots } \\
&= \left (1-(\frac{x^2}{6} + \frac{x^4}{5!})+ (\frac{x^2}{6} )^2 + O(x^6)\right) \left (1-(2x^2+\frac23 x^4)+ (2x^2)^2 + O(x^6)\right) \\
&= \left (1 - \frac16 x^2 + \frac{7}{360} x^4 + O(x^6) \right) \left (1 - 2x^2+ \frac{10}3x^4 + O(x^6) \right) \\
&= 1 - \frac{13}{6} x^2 + \frac{1327}{360}x^4 + O(x^6)
\end{align*}