# taylor series of ln(1+x)?

Compute the taylor series of $ln(1+x)$

I've first computed derivatives (upto the 4th) of ln(1+x)

$f^{'}(x)$ = $\frac{1}{1+x}$
$f^{''}(x) = \frac{-1}{(1+x)^2}$
$f^{'''}(x) = \frac{2}{(1+x)^3}$
$f^{''''}(x) = \frac{-6}{(1+x)^4}$
Therefore the series:
$ln(1+x) = f(a) + \frac{1}{1+a}\frac{x-a}{1!} - \frac{1}{(1+a)^2}\frac{(x-a)^2}{2!} + \frac{2}{(1+a)^3}\frac{(x-a)^3}{3!} - \frac{6}{(1+a)^4}\frac{(x-a)^4}{4!} + ...$

But this doesn't seem to be correct. Can anyone please explain why this doesn't work?

The supposed correct answer is:
As $ln(1+x) = \int (\frac{1}{1+x})dx$
$ln(1+x) = \Sigma_{k=0}^{\infty} \int (-x)^k dx$

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A Taylor series is a Maclaurin series with $a=0$. – aschepler Jul 26 '14 at 0:14
Even then, what i have in terms of signs is different. – Minu Jul 26 '14 at 0:16
@aschepler Other way round. A Maclaurin series is a Taylor series with $a=0.$ – alexqwx Jul 26 '14 at 0:17

You got the general expansion about $x=a$. Here we are intended to take $a=0$. That is, we are finding the Maclaurin series of $\ln(1+x)$. That will simplify your expression considerably. Note also that $\frac{(n-1)!}{n!}=\frac{1}{n}$.

The approach in the suggested solution also works. We note that $$\frac{1}{1+t}=1-t+t^2-t^3+\cdots\tag{1}$$ if $|t|\lt 1$ (infinite geometric series). Then we note that $$\ln(1+x)=\int_0^x \frac{1}{1+t}\,dt.$$ Then we integrate the right-hand side of (1) term by term. We get $$x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots,$$ precisely the same thing as what one gets by putting $a=0$ in your expression.

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If a=0 in my expansion, $ln(1+x) = 0 + x - \frac{x^2}{2!} + \frac{2(x^3)}{3!} - \frac{6(x^4)}{4!}+...$ Which is not the same ! – Minu Jul 26 '14 at 0:20
Why doesn't this method work? – Minu Jul 26 '14 at 0:23
It is the same, I mentioned the cancellation in the answer. For instance, $2!/3!=1/3$, $6/4!=1/4$, and so on. – André Nicolas Jul 26 '14 at 0:24
aren't the signs different? – Minu Jul 26 '14 at 0:27
You are welcome. The question you asked about the $\frac{1}{1+t}=\cdots$ is briefly dealt with above by the mention of geometric series. You are probably familiar with $1+r+r^2+r^3+\cdot=\frac{1}{1-r}$ (sum of infinite geometric series). Put $r=-t$. – André Nicolas Jul 26 '14 at 0:49

Note that $$\frac{1}{1+x}=\sum_{n \ge 0} (-1)^nx^n$$ Integrating both sides gives you \begin{align} \ln(1+x) &=\sum_{n \ge 0}\frac{(-1)^nx^{n+1}}{n+1}\\ &=x-\frac{x^2}{2}+\frac{x^3}{3}-... \end{align} Alternatively, \begin{align} &f^{(1)}(x)=(1+x)^{-1} &\implies \ f^{(1)}(0)=1\\ &f^{(2)}(x)=-(1+x)^{-2} &\implies f^{(2)}(0)=-1\\ &f^{(3)}(x)=2(1+x)^{-3} &\implies \ f^{(3)}(0)=2\\ &f^{(4)}(x)=-6(1+x)^{-4} &\implies \ f^{(4)}(0)=-6\\ \end{align} We deduce that \begin{align} f^{(n)}(0)=(-1)^{n-1}(n-1)! \end{align} Hence, \begin{align} \ln(1+x) &=\sum_{n \ge 1}\frac{f^{(n)}(0)}{n!}x^n\\ &=\sum_{n \ge 1}\frac{(-1)^{n-1}(n-1)!}{n!}x^n\\ &=\sum_{n \ge 1}\frac{(-1)^{n-1}}{n}x^n\\ &=\sum_{n \ge 0}\frac{(-1)^{n}}{n+1}x^{n+1}\\ &=x-\frac{x^2}{2}+\frac{x^3}{3}-... \end{align} Edit: To derive a closed for for the geometric series, let \begin{align} S&=1-x+x^2-x^3+...\\ xS&=x-x^2+x^3-x^4...\\ S+xS&=1\\ S&=\frac{1}{1+x}\\ \end{align} To prove in the other direction, use the binomial theorem or simply compute the series about $0$ manually.

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Can you tell me How you got the summation: $\frac{1}{1+x} = \Sigma_{n=0}^{\infty} (-1)^n x^n$? I understand the rest. – Minu Jul 26 '14 at 0:44