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I want to show that the power series around $0$ corresponding to the function $f:x\mapsto \log(1-x)$ is $\sum_{n=1}^{\infty}{-\frac{x^n}{n}}$.

I know that the series $\sum_{n\ge 1}{-\frac{x^n}{n}}$ with radius of convergence $R=1$ hence we can define a map $\forall x\in (-1,1)$ by the sum $g(x)=\sum_{n=1}^{\infty}{-\frac{x^n}{n}}$

Now I want to show that $f=g$ are equal. What should I exactly verify to do this?

palio
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  • You might want to differentiate the series and the function and to compare the results. – Did Nov 23 '13 at 09:45
  • This is the Taylor series of $log(1-x)$ around zero. May be what you are trying to do is to prove Taylor's theorem in this case http://en.wikipedia.org/wiki/Taylor%27s_theorem – hhsaffar Nov 23 '13 at 09:54
  • $\frac{df}{dx}=\frac{-1}{1-x}$ while $\frac{dg}{dx}\sum_{n=0}^{\infty}{-x^n}$. We now go back where we started comparing a function to a sum of a series. Do you take the sum of the geometric series for granted and so we can use the fact that $\frac{1}{1-x}=\sum_{n=0}^{\infty}{x^n}$ – palio Nov 23 '13 at 09:58
  • @hhsaffar do you mean that i should show that $log(1-x)$ is $k$ differentiable for each $k$ and that the $k$th derivative of $log(1-x)$ is exactly $-x^k/k$? is this enough to state the result? – palio Nov 23 '13 at 10:02
  • What determines "what's enough" is the problem's context, what you can use and what you can't. Also have another look at Taylor series what you wrote is not exactly right. I have to go now. Good luck :) – hhsaffar Nov 23 '13 at 10:08
  • The context is clear from my post above, i want to prove that for $x\in (-1,1)$, $\log(1-x)=\sum_{n=0}^{\infty}{\frac{-x^n}{n}}$ – palio Nov 23 '13 at 10:33
  • It seems your first comment means to object to the suggestion I made. If so, note that the expansion of 1/(1-x) can, either be taken for granted, or rederived (and that this derivation is somewhat easier that the expansion of log(1-x)). Is this the nature of your objection? (Unrelated: Please use @.) – Did Nov 23 '13 at 10:49
  • yes i see that the sum of the power series is just the sum of the finite sequence $S_N=\sum_{n=0}^{N}{x^n}=\frac{1-x^{N+1}}{1-x}$ which goes to $\frac{1}{1-x}$ when $N\to \infty$ and $|x|<1$. – palio Nov 23 '13 at 11:18

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If you know (from Taylor theorem) that $$\log(1+x)=\sum_{n=1}^{\infty} (-1)^{n+1}\frac{x^n}{n}$$ for $x \in (-1,1)$, you can simply change $x$ with $-x$. Note that then your $x$ also belong to $(-1,1)$.

user110822
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  • That is exactly my question, under what conditions we can write $$\log(1+x)=\sum_{n=1}^{\infty} (-1)^{n+1}\frac{x^n}{n}$$? I can't find this Taylor theorem that identifies a function with a power series, even the wikipedia article does not seem to state it explicitly. Would you please tell me what are the conditions to verify to state this result – palio Nov 23 '13 at 11:11
  • Well, that series has ratio of convergence $R=1$, and in $x=1$ it converges by Leibniz test, and in $x=-1$ it diverges. You wrote $x\in (-1,1)$, so I used that, but you can say for $\log(1+x)$ that $x \in (-1,1]$. – user110822 Nov 23 '13 at 11:40