# MacLaurin series of $\ln(1-x^2)$

The MacLaurin series for $\ln(1 + x)$ is obtained from the series for $\frac{1}{1 + x}$ by integration. Use this and appropriate substitutions to obtain the MacLaurin series for $\ln(1-x^2)$.

I did $$\int\frac{2}{x-\frac{1}{x}}$$ but I'm not sure how I'd go about the rest of the substitution in order to find the series.

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If you wanted to use the same type of derivation used to come up with the Maclaurin series of $\ln (1-x)$ this is how it would be done

\begin{align} \frac{d}{dx}(\ln (1-x^2)) &= -\frac{2x}{1-x^2} \\ -\frac{2x}{1-x^2} &= \sum_{n=0}^\infty -2x(x^2)^n \\ &=-2\sum_{n=0}^\infty x^{2n+1} \\ \ln(1-x^2) &= \int -2\sum_{n=0}^\infty x^{2n+1} \mathrm{d}x \\ &=-2\sum_{n=0}^\infty \frac{x^{2n+2}}{2n+2} \\ &= -\sum_{n=0}^\infty \frac{x^{2n+2}}{n+1} \end{align}

By expanding out the first few terms we can see that this is indeed the same as is confirmed above.

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I think what the question is asking is to use $$\log(1+x)=x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\dots\tag{1}$$ then substitute $x\mapsto-x$ $$\log(1-x)=-x-\frac{x^2}2-\frac{x^3}3-\frac{x^4}4-\dots\tag{2}$$ Then using the identity that $\log(ab)=\log(a)+\log(b)$ to get \begin{align} \log(1-x^2) &=-2\frac{x^2}{2}-2\frac{x^4}4-2\frac{x^6}6-\dots\\ &=-x^2-\frac{x^4}2-\frac{x^6}3-\dots\tag{3} \end{align} Note than you could also substitute $x\mapsto-x^2$ into $(1)$ to get $(3)$ directly.

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Well, presumably you already know the series for $\ln(1+x)$. The hint is just telling you to take that series and replace $x$ by $-x^2$.

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So the series would be the integral of each term from $(-x)^{2n}$? – user51462 Apr 10 '13 at 22:20
You think? What's the series for $\ln(1+x)$? – Javier Apr 10 '13 at 22:22
The integral of sigma(-x)^n – user51462 Apr 10 '13 at 22:23
Is ln(1-x^2) = $\int\sum_{n=1}^{\infty} \frac{(x)^{2n+2}}{n+1}$ correct? – user51462 Apr 11 '13 at 0:49
@user51462: As far as I can tell, it's $\ln(1-x^2) = -\sum_{n=1}^{\infty} \frac{x^{2n}}{n}$. How did you get that? You should know the series for $\frac{1}{1-x}$, get $\frac{1}{1+x}$ from that, then get $\ln(1+x)$ from that, and finally get $\ln(1-x^2)$ from that. – Javier Apr 11 '13 at 1:17

Two solutions, both based on $\ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n x^n}$ for $-1 < x < 1$:

1. Put $-x^2$ for $x$. Then $\ln(1-x^2) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n (-x^2)^n} =-\sum_{n=1}^{\infty} \frac{1}{n x^{2n}}$.

2. Put $-x$ for $x$. Then $\ln(1-x) = \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n (-x)^n} =\sum_{n=1}^{\infty} \frac{-1}{n x^{n}}$. Add this to the original series, and

\begin{align} \ln(1-x^2) = \ln(1-x)+\ln(1+x) &=\sum_{n=1}^{\infty} (\frac{(-1)^{n-1}}{n x^n}+\frac{-1}{n x^n})\\ &=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}-1}{n x^n}\\ &=\sum_{n=1, n \text{ even}}^{\infty} \frac{(-1)^{n-1}-1}{n x^n}\\ &=\sum_{n=1}^{\infty} \frac{-2}{2n x^{2n}}\\ &=-\sum_{n=1}^{\infty} \frac{1}{n x^{2n}}\\ \end{align}

Fortunately, the answers agree.

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