Expanding logarithm into series How to expand $f(x)=\ln\left( x+\sqrt{1+x^2} \right)$ into series at $x_0=0$? I've tried using Taylor's formula but counting consecutive derivatives was inconvenient and I couldn't find the general formula.
 A: This is a somewhat laborious expansion, but it has a nice closed formula. 
We begin with
$$f(x) = \frac{1}{\sqrt{1+x^2}}$$
and note that you function $F$ is such that $F'(x) = f(x)$. We can make use of the Generalized Binomial Theorem, namely
$$(1+\mathrm x)^\alpha = \sum_{n=0}^\infty {\alpha \choose n}\mathrm x^n$$
In your case, set $\mathrm x= x^2$ and $\alpha = -\dfrac{1}{2}$.
$$(1+x^2)^{-1/2}=\frac{1}{\sqrt{1+x^2}}=\sum_{n=0}^\infty {-1/2 \choose n}x^{2n}$$
It is clear the most important calculation will be that of $$ c_n={-1/2 \choose n}$$
Writing this explicitly, it gives
$$\eqalign{
  & {c_n} = \frac{{\left( { - \frac{1}{2}} \right)\left( { - \frac{1}{2} - 1} \right)\left( { - \frac{1}{2} - 2} \right) \cdots \left( { - \frac{1}{2} - n + 1} \right)}}{{n!}}  \cr 
  & {c_n} = \frac{1}{{n!}}\prod\limits_{k = 0}^{n - 1} {\left( { - \frac{1}{2} - k} \right)}   \cr 
  & {c_n} = \frac{1}{{n!}}{\left( { - 1} \right)^n}\prod\limits_{k = 0}^{n - 1} {\left( {\frac{1}{2} + k} \right)}   \cr 
  & {c_n} = \frac{1}{{n!}}{\left( { - 1} \right)^n}\prod\limits_{k = 0}^{n - 1} {\left( {\frac{{2k + 1}}{2}} \right)}   \cr 
  & {c_n} = \frac{1}{{n!}}{\left( { - \frac{1}{2}} \right)^n}\prod\limits_{k = 0}^{n - 1} {\left( {2k + 1} \right)}  \cr} $$
Note that for $n=1$ the product is empty, thati is, it is $1$. 
If we write the product explicitly, we get
$$\prod\limits_{k = 0}^{n - 1} {\left( {2k + 1} \right)}  = 1 \cdot 3 \cdots \left( {2n - 3} \right)\left( {2n - 1} \right)$$
We can "complete" it by adjoining the even numers:
$$\prod\limits_{k = 0}^{n - 1} {\left( {2k + 1} \right)}  = {2^n}n!\frac{{1 \cdot 3 \cdots \left( {2n - 3} \right)\left( {2n - 1} \right)}}{{{2^n}n!}} = \frac{{1 \cdot 2 \cdot 3 \cdot 4 \cdots \left( {2n - 3} \right)\left( {2n - 2} \right)\left( {2n - 1} \right)2n}}{{{2^n}n!}}$$
and get
$$\prod\limits_{k = 0}^{n - 1} {\left( {2k + 1} \right)}  = \frac{{\left( {2n} \right)!}}{{{2^n}n!}} = \frac{1}{{{2^n}}}\frac{{\left( {2n} \right)!}}{{\left( {2n - n} \right)!}}$$
so that
$${c_n} = {\left( { - 1} \right)^n}\frac{1}{{{2^n}n!}}\prod\limits_{k = 0}^{n - 1} {\left( {2k + 1} \right)}  = {\left( { - 1} \right)^n}\frac{1}{{{4^n}}}\frac{{\left( {2n} \right)!}}{{n!\left( {2n - n} \right)!}} = {\left( { - 1} \right)^n}\frac{1}{{{4^n}}}{2n \choose n}$$
Thus we have that
$$\frac{1}{{\sqrt {1 + {x^2}} }} = \sum\limits_{n = 0}^\infty  {{c_n}} {x^{2n}} = \sum\limits_{n = 0}^\infty  {{{\left( { - \frac{1}{4}} \right)}^n}}{2n \choose n} {x^{2n}}$$
Thus, from our previous considerations, we get
$$\log \left( {x + \sqrt {1 + {x^2}} } \right) = \int\limits_0^x {\frac{{dt}}{{\sqrt {1 + {t^2}} }}}  = \sum\limits_{n = 0}^\infty  {{{\left( { - \frac{1}{4}} \right)}^n}}{2n \choose n}  \frac{{{x^{2n + 1}}}}{{2n + 1}}$$
A: Check that your function is the inverse hyperbolic sine: $$f(x)=\operatorname{arsinh}x\text{, where }\sinh x=\frac{e^x-e^{-x}}{2}$$ and you'll be able to get the derivative using the theorem for the inverse function...:)
