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I want to find the Taylor series of $f(x)=\log(1+x), x \in (-1,+\infty), \xi=0, I=(-1,1)$

$$f'(x)=\frac{1}{1+x}$$

$$\sum_{n=0}^{\infty} (-1)^n x^n=\frac{1}{1+x}, x\in (-1,1)$$

$$f(x)=f(0)+\int_0^x f'(t) dt \Rightarrow f(x)=\int_0^x \frac{1}{1+x}dt= \int_0^x \sum_{n=0}^{\infty} (-1)^n x^n=\sum_{n=0}^{\infty} \int_0^x (-1)^n x^n=\sum_{n=0}^{\infty}(-1)^n \frac{x^{n+1}}{n+1}$$

But..I found that the Taylor series of $f(x)=\log(1+x)$ is equal to $\sum_{n=1}^{\infty} \frac{(-1)^{n-1} (n-1)!}{(1+x)^n}$..

What have I done wrong???

evinda
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    The very last series isn't even a taylor series. – Git Gud May 26 '14 at 17:29
  • So,you mean that my reslt is right? Or are both of them wrong? – evinda May 26 '14 at 17:33
  • What you did is right, but it doesn't say anything about $x\in ]1,+\infty[$. – Git Gud May 26 '14 at 17:33
  • How can I check what happens at these points? – evinda May 26 '14 at 17:34
  • @evinda where did you find that expansion? – S L May 26 '14 at 17:35
  • Which one? The last one? – evinda May 26 '14 at 17:36
  • @evinda my very naive try would be to try to find a pattern for $f^{(n)}(0)$ and then prove it by induction. – Git Gud May 26 '14 at 17:36
  • @AndréNicolas At which relation? – evinda May 26 '14 at 17:46
  • @evinda My naive try works. – Git Gud May 26 '14 at 17:48
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    @evinda: You probably found the general $n$-th derivative. Recall that the coefficient of $x^n$ is $\frac{f^{(n)}(0)}{n!}$. It may be that you forgot to evaluate the derivatives at $0$, and forgot about the division by $n!$. – André Nicolas May 26 '14 at 17:54
  • @AndréNicolas I wanted to find the power serie,without the use of the formula: $\sum_{n=0}^{\infty} \frac{f^{(n)}(0)x^n}{n!}$,but integrating the first derivative and using the geometric series ... Can't I do it in this way??? – evinda May 26 '14 at 17:59
  • Yes, you can, and it is the best way to do it. The comment was about your last sentence, "But I found $\dots$." – André Nicolas May 26 '14 at 18:01
  • @AndréNicolas A ok.. I found this in my notes..But it must be wrong.. Is that what I have done right or have I done something wrong? – evinda May 26 '14 at 18:02
  • Your series evaluation by term by term integration of the series for $\frac{1}{1+t}$ is good. There is a technical issue about justifying the term by term integration. Depending on the level of the course one quotes the appropriate theorem or one does not worry about it. – André Nicolas May 26 '14 at 18:05
  • @AndréNicolas I think that I haven't get taught it... But finding the power series in that way..how can I find the radius of convergence?? – evinda May 26 '14 at 18:07
  • It is (always) the same as the radius of convergence of the series you are integrating term by term. Or if you really want to you can use the Ratio Test or Root Test on the series you ended up with. The radius of convergence is $1$. The only difference is that the integrated series converges at $x=1$, while the the series for $\frac{1}{1+t}$ does not. – André Nicolas May 26 '14 at 18:09
  • Andre Oh sorry!!! I didn't mean the radius of convergence,but the lagrange remainder..how can I find it??? – evinda May 26 '14 at 18:22

1 Answers1

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Here $f'(x)=\frac{1}{1+x}$, and $f''(x)=-\frac{1}{(1+x)^2}$, $f'''(x)=\frac{2}{(x+1)^3}$, and inductively,

$f^{(n)}(x)=(-1)^n\frac{(n-1)!}{(x+1)^n}$.

Thus, by the Taylor expansion centred at $x_0=0$ $\ln(1+x)=\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n=\sum_{n=1}^\infty(-1)^n\frac{(n-1)!}{n!}x^n=\sum_{n=1}^\infty(-1)^n\frac{x^n}{n}$

Let us try to show that $|\frac{x-t}{1+t}|^n\le|x|^n$

We see that for $-1\lt x\lt 0$, and $x\lt t\lt 0$,

$\large(\frac{x-t}{1+t})=f(t)$ (for $x$ fixed).

Here $f'(t)=(\frac{(1+t)(-1)-(x-t)(1)}{(1+t)^2})=(\frac{-1-x}{(1+t)^2})\le 0,\forall t: x\lt t\lt 0$

Thus $f(t)\ge(t+\delta)$, for $\delta$ positive.

Thus $\large(\frac{x-t}{1+t})\ge(\frac{x-0}{1+0})=x$

Thus $|\large(\frac{x-t}{1+t})|\le |x|$

Thus $|\large(\frac{x-t}{1+t})|^n\le |x|^n$

Ellya
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  • I understand...But..to check if we found the right power series,we have to calculate the radius of convergence..It is:

    $$R_{n, \xi}(x, \zeta)=\frac{(-1)^{n+1} n! x^{n+1}}{(\zeta+1)^{n+1}(n+1)!}=\frac{(-1)^{n+1} x^{n+1}}{(\zeta+1)^{n+1}(n+1)}, \zeta \in [0,x] \text{ or } \zeta \in [x,0]$$

    • At the case when 0<x<1:

    $$|R_{n, \xi}(x, \zeta)| \leq \frac{1}{n+1} \to 0$$

    -But,at the case when $-1<x<0$:

    $$|R_{n,0}(x)|=\int_0^x |\frac{(x-t)}{1+t}|^n dt$$

    According to my notes,the relation : $|\frac{(x-t)}{1+t}|^n \leq |x|^n$ stand...But...why is it like that?

     – evinda May 26 '14 at 17:55
  • Would not the same apply for $-1\lt x\lt 0$? as we still have |R_{n,\Xi}|\le\frac{1}{n+1}$ – Ellya May 26 '14 at 18:12
  • According to my notes, we can't conclude anything when $-1<x<0$ for $\frac{|x|^{n+1}}{(1+\zeta)^{n+1}}$ ? Or do you have an idea? – evinda May 26 '14 at 18:21
  • @evinda see my edit. – Ellya May 26 '14 at 18:38