Problem source

Given that $y = \ln ( x + \sqrt{x^2 + 1})$, show that 
$ \displaystyle \frac{\d y}{\d x} = \frac1 {\sqrt{x^2 + 1} }\;$.
Prove by induction that, for $n \ge 0\,$,
\[
\l x^2 + 1 \r y^{\l n + 2 \r} + \l 2n + 1 \r x y^{\l n + 1 \r} + n^2
y^{\l n \r} = 0\;,
\]
where $\displaystyle y^{(n)} = \frac{\d^n y}{\d x^n}$ and $y^{(0)} =y\,$.
Using this result in the case $x = 0\,$, or otherwise, show that the
 Maclaurin series for $y$ begins
\[
x - {x^3 \over 6} +{3 x^5 \over 40}
\]
and find the next non-zero term.

Solution source

\begin{align*}
&& y & = \ln ( x + \sqrt{x^2+1}) \\
\Rightarrow && \frac{\d y}{\d x} &= \frac{1}{x+\sqrt{x^2+1}} \cdot \frac{\d }{\d x} \left ( x + \sqrt{x^2+1} \right) \\
&&&= \frac{1}{x+\sqrt{x^2+1}} \left (1 + \frac12 \frac{2x}{\sqrt{x^2+1}} \right) \\
&&&= \frac{1}{x+\sqrt{x^2+1}} \left ( \frac{\sqrt{x^2+1} + x}{\sqrt{x^2+1}}\right) \\
&&&= \frac{1}{\sqrt{x^2+1}}
\end{align*}

Note that $\displaystyle y^{(2)} = - \frac12 \frac{2x}{(x^2+1)^{3/2}} = - \frac{x}{(x^2+1)^{3/2}}$, and in particular $(x^2+1)y^{(2)} + xy^{(1)} = 0$.

Now applying Leibnitz formula:

\begin{align*}
0 &= \left ( (x^2+1)y^{(2)} + xy^{(1)}  \right )^{(n)} \\
&= \left ( (x^2+1)y^{(2)}\right )^{(n)} + \left (xy^{(1)}  \right )^{(n)} \\
&= (x^2+1)y^{(n+2)} +n2xy^{(n+1)} + \binom{n}{2}2y^{(n)} + xy^{(n+1)} + n y^{(n)} \\
&= (x^2+1)y^{(n+2)} + (2n+1)xy^{(n+1)} + (n^2-n+n)y^{(n)} \\
&= (x^2+1)y^{(n+2)} + (2n+1)xy^{(n+1)} + n^2y^{(n)}
\end{align*}

as required.


When $x = 0$:

\begin{align*}
&& y(0) &= \ln(0 + \sqrt{0^2+1})  \\
&&&= \ln 1 = 0 \\
&& y'(0) &= \frac{1}{\sqrt{0^2+1}} = 1 \\
&& y^{(n+2)} &= -n^2 y^{(n)} \\
&& y^{(2k)} &= 0 \\
&& y^{(3)} &= -1 \\
&& y^{(5)} &= 3^2 \\
&& y^{(7)} &= - 5^2 \cdot 3^2 \\
\end{align*}

Therefore the Maclaurin series about $x = 0$ is 
\begin{align*}
y &= x - \frac{1}{3!} x^3 + \frac{3^2}{5!} x^5 - \frac{3^2 \cdot 5^2}{7!} x^7 + \cdots \\
&= x - \frac{1}{6} x^3 + \frac{3}{1 \cdot 2 \cdot 4 \cdot 5} x^5 - \frac{5}{1 \cdot 2 \cdot 4 \cdot 2 \cdot 7} x^7 + \cdots \\
&= x - \frac{1}{6}x^3 + \frac{3}{40} x^5 - \frac{5}{56} x^7 + \cdots
\end{align*}