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I found the first four derivatives of $ f(x) = ln(1+x) $

Then for all n > 1, $$ f^n(x) = \frac{(-1^{n+1})(n-1)!}{(1+x)^n } $$

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

By definition Maclaurin Series are defined as:

$$\sum_{n=0}^{\infty} \frac{f^n(0)}{n!}x^n$$

*Since the $ f^n(0) $ is only true when n > 1 then,

$$\sum_{n=1}^{\infty} \frac{f^{n+1}(0)}{n+1!}x^{n+1}$$

I continue by replacing each term and I get:

$$ ln(1+x) = - \sum_{n=1}^{\infty} \frac{(-1)^n}{n*(n+1)}x^{n+1}$$

I know that the answer should be :

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

Where did I go wrong?

"*" : Unsure about the step.

I'm really bad at taylor series so if you have any good sites for resources, it would be very much appreciated. Thank you for any help and answers.

A. Port
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  • I think at "Since the $f^n$..." you turned the sum from $a_0+a_1x+\ldots$ to $a_2x^2+\ldots$ so you're missing the two first terms – Jam Oct 02 '17 at 18:37
  • The * step is indeed wrong. Expand the formulas with concrete values of $n$. –  Oct 02 '17 at 18:37
  • I think I know why your sum went wrong: if you increase the first index of a sum, you'd need to decrease the other indices. So you've turned $\sum_0 a_n$ into $\sum_1 a_{n+1}$ when it really should have been $\sum_1 a_{n-1}$ – Jam Oct 02 '17 at 18:41

2 Answers2

2

$$f(0)=\ln(1+0)=0,\\f'(0)=\frac1{1+0}=1,\\f''(0)=-\frac1{(1+0)^2}=-1,\\f'''(0)=\frac2{(1+0)^3}=2,\\f''''(0)=-\frac{3!}{(1+0)^4}=-3!,\\\cdots$$

then

$$\ln(1+x)=0+x-\frac{1x^2}2+\frac{2x^3}{3!}-\frac{3!x^3}{4!}+\cdots$$

2

If $f(x)=\log(1+x)$, then we have

$$\begin{align} f^{(1)}(x)&=(1+x)^{-1}\\\\ f^{(2)}(x)&=-(1)(1+x)^{-2}\\\\ f^{(3)}(x)&=(-1)(-2)(1+x)^{-3}\\\\ \vdots f^{(n)}(x)&=(-1)(-2)\cdots (-(n-1))(1+x)^{-n}\\\\ \end{align}$$

Therefore, we can write

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

Mark Viola
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