Problem source

\begin{questionparts}
\item 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 $x^{4}.$ 
\item 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 $x^{4}.$
\end{questionparts}

Solution source

\begin{questionparts}
\item \begin{align*}
&& f(x) &= \ln (1+e^x) \\
&& f'(x) &= \frac{1}{1+e^x} \cdot e^x  \\
&&&= \frac{e^x}{1+e^x} \\
\Rightarrow && \ln [f'(x)] &= x - \ln (1+e^x) \\
&&&= x - f(x) \\
\\
\Rightarrow && \frac{f''(x)}{f'(x)} &= 1 - f'(x) \\
\Rightarrow && f''(x) &= f'(x) - [f'(x)]^2 \\
&& f'''(x) &= f''(x) - 2f'(x) f''(x) \\
&& f^{(4)}(x) &= f'''(x) - 2[f''(x)]^2-2f'(x)f'''(x)
\end{align*}

\begin{align*}
f(0) &= \ln 2 \\
f'(0) &= \tfrac12 \\
f''(0) &= \tfrac12 -\tfrac14 \\
&= \tfrac 14 \\
f'''(0) &= \tfrac14 - 2 \tfrac12 \tfrac 14 \\ 
&= 0 \\
f^{(4)}(0) &= -2 \cdot  \tfrac1{16} \\
&= -\frac18
\end{align*}

Therefore $f(x) = \ln 2 + \tfrac12 x + \tfrac18 x^2 - \frac1{8 \cdot 4!} x^4 + O(x^5)$ 


\item 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*}
\end{questionparts}