Taylor series of $\ln \left(x+ \sqrt{x^2+1}\right)$

Question

How to compute the taylor series of $\ln \left(x+ \sqrt{x^2+1}\right)$?

All I know is $\ln(1+u) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{u^n}{n}$ but I do not know how to proceed.

Thanks.

Answer 1

Hint: $\dfrac{d}{dx}\ln\left(x+\sqrt{x^2+1}\right)=\dfrac{1}{\sqrt{x^2+1}}$ and use binomial series.

Answer 2

$(\ln (x+\sqrt{1+x^2}))' = (1+x^2)^{-1/2}$

Find the series for $(1+u)^{-1/2}$. Replase $u$ by $x^2$ and then integrate the series.

Answer 3

First we note that the function $g$ defined as

$$g(x)\equiv \frac{d}{dx}log(x+\sqrt{x^2+1})=\frac{1}{\sqrt{1+x^2}}$$

Then, the series for $g$ is straightforward to obtain and is expressed by

$$g(x)=1+\sum_{n=1}^{\infty}(-1)^n\frac{(2n-1)!!x^{2n}}{2^{n}n!}$$

Since we can integrate a power series term by term over its interval of convergence we find that

$$\begin{align} \log(x+\sqrt{x^2+1})&=\int_0^xg(t)\,dt\\\\ &=x+\sum_{n=1}^{\infty}(-1)^n\frac{(2n-1)!!}{2^{n}n!}\int_0^x t^{2n}\,dt\\\\ &=x+\sum_{n=1}^{\infty}(-1)^n\frac{(2n-1)!!x^{2n+1}}{2^{n}n!(2n+1)} \end{align}$$

Answer 4

Most answers seem to give the answer to the first derivative, but it might be good to know how they found that, namely with the chain rule. The chain rule states

$$ \frac{df(g(x))}{dx} = \frac{df(g)}{dg} \frac{dg(x)}{dx}. $$

So in this case for the first derivative $f(g)$ and $g(x)$ would be $ln(g)$ and $x+\sqrt{x^2+1}$ respectively, such that

$$ \frac{df(g)}{dg} \frac{dg(x)}{dx} = \frac{1}{g(x)} \left(1 + \frac{x}{\sqrt{x^2+1}}\right), $$

which after substitution of $g(x)$ can be simplified.

The same can be done for any of the higher derivatives of the resulting expression.