Combining Taylor expansions How do you taylor expand the function $F(x)={x\over \ln(x+1)}$ using standard results? (I know that WA offers the answer, but I want to know how to get it myself.) I know that $\ln(x+1)=x-{x^2\over 2}+{x^3\over 3}+…$ But I don't know how to take the reciprocal. In general, given a function $g(x)$ with a known Taylor series, how might I find $(g(x))^n$, for some $n\in \mathbb Q$?
Also, how might I evaluate expressions like $\ln(1+g(x))$ where I know the Taylor expansion of $g(x)$ (and $\ln x$). How do I combine them?
Thank you.
 A: You have
$$F(x)=\frac{x}{\sum_{n\ge 1}\frac{(-1)^{n+1}}nx^n}=\frac1{\sum_{n\ge 0}{\frac{(-1)^n}{n+1}}x^n}=\frac1{1-\frac{x}2+\frac{x^2}3-+\dots}\;.$$
Suppose that $F(x)=\sum\limits_{n\ge 0}a_nx^n$; then you want 
$$1=\left(1-\frac{x}2+\frac{x^2}3-+\dots\right)\left(a_0+a_1x+a_2x^2+\dots\right)\;.$$
Multiply out and equate coefficients:
$$\begin{align*}
1&=a_0\,;\\
0&=a_1-\frac{a_0}2=a_1-\frac12,\text{ so }a_1=\frac12\,;\\
0&=a_2-\frac{a_1}2+\frac{a_0}3=a_2-\frac14+\frac13=a_2+\frac1{12},\text{ so }a_2=-\frac1{12}\,;
\end{align*}$$
and so on. In general $$a_n=\frac{a_{n-1}}2-\frac{a_{n-2}}3+\dots+(-1)^{n+1}\frac{a_0}{n+1}$$ for $n>0$, so 
$$\begin{align*}&a_3=-\frac1{24}-\frac16+\frac14=\frac1{24}\;,\\
&a_4=\frac1{48}+\frac1{36}+\frac18-\frac15=-\frac{19}{720}\;,\\
&a_5=-\frac{19}{1440}-\frac1{72}+\frac1{48}-\frac1{10}+\frac16=\frac{29}{480}\;,
\end{align*}$$
and if there’s a pattern, it isn’t an obvious one, but you can get as good an approximation as you want in relatively straightforward fashion;
$$F(x)=1+\frac{x}2-\frac{x^2}{12}+\frac{x^3}{24}-\frac{17x^4}{720}+\frac{29x^5}{480}+\dots$$
already gives two or three decimal places over much of the interval of convergence.
A: You can do this by long division:
$\dfrac{x}{x-{x^2\over 2}+{x^3\over 3}-\cdots} = 1 + \dfrac{{x^2\over 2}-{x^3\over 3}+\cdots}{x-{x^2\over 2}+{x^3\over 3}\cdots} = 1 + \frac{x}{2} +  \dfrac{-{x^3\over 12}+\cdots}{x-{x^2\over 2}+\cdots} = 1 + \frac{x}{2} -  \frac{x^2}{12} + \cdots  $
A: Concerning the multiplicative inverse of $\ln(1+x)$ you may use the classical division (here $x$ is divided by $\ln(1+x)$ ) : 
$\begin{array} {r|l}
x +0 x^2+0 x^3 & x-\frac{x^2}2+\frac{x^3}3+\cdots \\
-x+\frac{x^2}2-\frac{x^3}3 & 1+\frac x2-\frac {x^2}{12}\\
\frac{x^2}2-\frac{x^3}3 & \\
-\frac{x^2}2+\frac{x^3}4 & \\
-\frac{x^3}{12} & \\
\end{array}
$ 
So that $\displaystyle \frac x{\ln(1+x)}=1+\frac x2-\frac {x^2}{12}$

Concerning integer powers ($f(x)^n$ with $n\in \mathbb{N}$) just apply distributivity of multiplication $n-1$ times ($f(x)=\ln(1+x)$ here) :
$\ln(1+x)^2= \left(x-\frac{x^2}2+\frac{x^3}3+\cdots\right)\cdot\left(x-\frac{x^2}2+\frac{x^3}3+\cdots\right)$
$=x\cdot\left(x-\frac{x^2}2+\frac{x^3}3+\cdots\right)-\frac{x^2}2\cdot\left(x-\frac{x^2}2+\frac{x^3}3+\cdots\right)+\frac{x^3}3\cdots$
For higher powers use :
$\ln(1+x)^3=\ln(1+x)^2\cdot \ln(1+x)$  and so on...  
Sometimes it will be more convenient to rewrite the exponent in powers of $2$.
For example $\displaystyle f^9= f^{1+2^3}=f\cdot(((f^2)^2)^2$.

To compute $(1+f(x))^p$ you may use following expansion :
$(1+f(x))^p= 1+ p f(x)+ \frac{p(p-1)}{2!} f(x)^2 + \frac{p(p-1)(p-2)}{3!} f(x)^3+\cdots \binom{p}{k} f(x)^k+\cdots$
Note that you'll sometimes have to divide or multiply by a power of $x$ to apply 
these methods (in the division we replaced $\frac1{\ln(1+x)}$ by $\frac x{\ln(1+x)}$, here we have $1+f(x)$ so that when computing for example $(x+f(x))^p$ we may rewrite this as $x^p\cdot(1+\frac{f(x)}x)^p$).
For any power $p$ (or when $\ln(1+f(x))$ is simple) you may use :
$\displaystyle (1+f(x))^p=e^{p\cdot \ln(1+f(x))}=e^{p\cdot \left(f(x)-\frac{f(x)^2}2+\cdots\right)}$
