Taylor series of ln(1+x)

Question

I am trying to find the Taylor series for $f(x)=ln(1+x)$ centered at $x=0$

So I have calculated some of the derivatives,

\begin{align} &f^{(1)}(x)=(1+x)^{-1} , f^{(1)}(0)=1\\ &f^{(2)}(x)=-(1+x)^{-2}, f^{(2)}(0)=-1\\ &f^{(3)}(x)=2(1+x)^{-3}, f^{(3)}(0)=2\\ &f^{(4)}(x)=-6(1+x)^{-4}, f^{(4)}(0)=-6\\ \end{align}

So then, $f^{(n)}(0)=(-1)^{n-1}(n-1)!$

But from here how do I find its Taylor series?

Simply Beautiful Art · Answer 1 · 2017-03-15T22:33:34.060

0

Perhaps this is more enlightening:

$$\ln(1+x)=\int_0^x\frac1{1+t}\ dt=\int_0^x\sum_{n=0}^\infty(-1)^nt^n\ dt=\sum_{n=1}^\infty(-1)^{n+1}\frac{x^n}n$$

where we used the geometric series.

However, if you are looking for the method through Taylor expansions, then recall that

$$\begin{align}\ln(1+x)=f(x)&=f(0)+\sum_{n=1}^\infty\frac{f^{(n)}(0)}{n!}x^n\\&=\sum_{n=1}^\infty\frac{(-1)^{n+1}(n-1)!}{n!}x^n\\&=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}x^n\end{align}$$

edited Mar 15 '17 at 22:33

answered Mar 15 '17 at 22:29

Simply Beautiful Art

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I thought it would give us $\sum_{n \ge 0}\frac{(-1)^{n}}{n+1}x^{n+1}\$ – fr14 Mar 15 '17 at 22:32
That's the same thing. Write out the first few terms and you will see. – Simply Beautiful Art Mar 15 '17 at 22:32
oh, so both answers are correct? – fr14 Mar 15 '17 at 22:33
Yes. I just usually see it written the way it is here (at least when I Google "common Taylor series") – Simply Beautiful Art Mar 15 '17 at 22:34

Taylor series of ln(1+x)

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