So the exercise I had to do was: Find the power representation of $x\ln(1-x)$. The way to go was finding the power series representation of $\ln(1-x)$ and then multiply it with $x$. But why can't you first find the derative of $x\ln(1-x)$ which is $x/(1-x) + \ln(1-x)$ and then make power series of them both. And then find the antiderative of that. Because when they ask to find power series representation of $\ln(1-x)$ you have to first find derivative of this then convert it into series and then find antiderivative of this series. So why can't i do the same with $x\ln(1-x)$..?
So why can't I apply the same method? Why can't I solve this exercise in same way, by first finding the derivative then turning it into a power series and then integrating it?
I did $f(x)= x\ln (1-x)$
$f^\prime(x) = -x/(1-x) + \ln(1-x)$
I already knew $\ln(1-x)= Σ-(1/n) x^{n+1}$ (with n=0)
This in power series gives
$-x\sum x^n + \sum -1/n x^{n+1}$ (both n=0)
This gives $\sum-x^{n+1} + \sum-(1/n) x^{n+1}$
Now I integrate both and get
$\sum-(1/(n+2)) \cdot x^{n+2} + \sum-1/n \cdot 1/(n+2) x^{n+2} $
Best Answer
Hint
Start with $$\frac{1}{1-x}=\sum_{i=0}^{\infty}x^i$$ and integrate with respect to $x$. You then have $$\log(1-x)=-\sum_{i=1}^{\infty}\frac{x^i}{i}$$ So, $$x^n \log(1-x)=-\sum_{i=1}^{\infty}\frac{x^{n+i}}{i}$$
Added later to this answer
If you want to start with the derivative and then integrate, there is no problem. If $$f(x)=x\log(1-x)$$ $$f'(x)=\log (1-x)-\frac{x}{1-x}$$ So, use the expansion of $\frac{1}{1-x}$, muliply each term by $x$ and substract all of that from the expansion of $\log (1-x)$. You end with $$f'(x)=-2 x-\frac{3 x^2}{2}-\frac{4 x^3}{3}-\frac{5 x^4}{4}-\frac{6 x^5}{5}-\frac{7 x^6}{6}+O\left(x^7\right)$$ Now, integrate wrt $x$ for the same result.