taylor expansion ln(x+1)

Question

turingtest · Answer

about x=0 I assume?

turingtest · Answer

the Taylor series of a function about a point x=a is given by$$f(x)=\sum_{n=0}^\infty{f^{(n)}(a)\over n!}(x-a)^n$$

turingtest · Answer

if as I assume, you just want it about x=0 you get$$f(x)=\sum_{n=0}^\infty{f^{(n)}(0)\over n!}x^n$$

anonymous · Answer

true
this is a well known series, and you will get it using the method above, but it is usually computed by considering the series 
$$\frac{1}{1+x}=\sum_{k=0}^{\infty}(-1)^{k+1}x^k=1-x+x^2-x^3+...$$

anonymous · Answer

since $\frac{d}{dx}\ln(1+x)=\frac{1}{1+x}$ a simple method is to take the taylor series for $\frac{1}{1+x}$ and integrate term by term

turingtest · Answer

so where exactly are you having trouble with this?

turingtest · Answer

this is the expansion around x=0 right?

anonymous · Answer

yes, it is!

turingtest · Answer

$$f(x)=\sum_{n=0}^\infty{f^{(n)}(a)\over n!}(x-a)^n$$you want the expansion about x=0 so this becomes$$f(x)=\sum_{n=0}^\infty{f^{(n)}(0)\over n!}x^n$$ so the first term is$${f(0)\over 0!}x^0$$and what is that equal to?