Taylor Development of f(x)=ln(1-x)

[Telemachus]

Homework Statement

Hi there. I have this exercise which I'm trying to solve now. It says:

Using that \displaystyle\sum_{n=0}^{\infty}x^n=(1-x)^{-1} find one Taylor development for the function f(x)=\ln(1-x)

So, I've made some derivatives:
f^1(x)=\displaystyle\frac{-1}{(1-x)},f^2(x)=\displaystyle\frac{-1}{(1-x)^2},f^3(x)=\displaystyle\frac{-2}{(1-x)^3},f^4(x)=\displaystyle\frac{-6}{(1-x)^4},f^5(x)=\displaystyle\frac{-24}{(1-x)^5}

And then:

\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{f^b(x_0)(x-x_0)^n}{n!}=-\displaystyle\frac{(x-x_0)}{(1-x_0)}-\displaystyle\frac{(x-x_0)^2}{2(1-x_0)^2}-\displaystyle\frac{2(x-x_0)^3}{6(1-x_0)^3}-\displaystyle\frac{6(x-x_0)^4}{24(1-x_0)^4}-\displaystyle\frac{24(x-x_0)^5}{120(1-x_0)^5}+\ldots+-\displaystyle\frac{(x-x_0)^n}{n(1-x_0)^n}

I have two problems with this. In the first place, the general expression that I've found (which is probably wrong) doesn't work for n=0, it does for the others values of n. I thought of starting the summation at 1, but I'm not sure if this is valid. In the second place I don't know how to use the relation the problem gives at the beginning. I can see that I have (1-x_0) for every term, but I couldn't make it fit inside the summation.

So this is what I got: \displaystyle\sum_{n=1}^{\infty}-\displaystyle\frac{(x-x_0)^n}{n(1-x_0)^n}

Bye there, thanks for your help and suggestions.

 

[Sethric]

What is \int \frac{1}{1-x} dx?

 

[micromass]

The point of the exercise was probably the following. The first derivative gives:

f^\prime(x)=-(1-x)^{-1}.

So, using the relation, we get

f^\prime(x)=\sum_{n=0}^{+\infty}{x^n}.

Now, integrate both sides to get a Taylor expansion of f.

Now, the method outlined above is a lot easier than your method, since I don't have to make all that differentiations. But I still like your method more, since it gives a general answer, i.e. you've actually calculated the Taylor series with arbitrary x₀, while my method only gives a Taylor series around 0...

 

[Telemachus]

f'(x)dx, thanks Sethric.

Thank you micromass, nice point. It actually gives an approximation only on the interval of convergence, which goes from -1<x<1, I think it's because this interval is where the function f'(x) is uniformly convergent, but I'm not sure about this. I mean, in the interval where its not uniformly convergent we can't integrate "inside" the summation, right?