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.
Best Answer
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.