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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.

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I'm not sure there's a way to do it without re-calculating the coefficients. – Alex Becker Feb 19 '12 at 9:36
I have figured this out. Thanks anyways. – tycho Feb 19 '12 at 10:37

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 $

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Thank you, Henry. – tycho Feb 19 '12 at 12:00

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


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;


already gives two or three decimal places over much of the interval of convergence.

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Thank you, Brian. – tycho Feb 19 '12 at 11:59

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) :

$\ln(1+x)^2= \left(x-\frac{x^2}2+\frac{x^3}3+\cdots\right)\cdot\left(x-\frac{x^2}2+\frac{x^3}3+\cdots\right)$


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 :

$\displaystyle (1+f(x))^p=e^{p\cdot \ln(1+f(x))}=e^{p\cdot \left(f(x)-\frac{f(x)^2}2+\cdots\right)}$

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Thank you, Raymond. – tycho Feb 19 '12 at 12:00

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