Question

So the exercise I had to do was:
Find the power representation of xln(x-1).
The way to go was finding the power series representation of ln(x-1) and then multiply it with x.

But why can't you first find the derative of xln(x-1) which is x/(x-1) + ln(x-1)

And then make power series of them both. And then find the antiderative of that.

Because when they ask to find power serie represenation of ln(x-1) you have to first find derative of this then convert it into series and then find antiderative of this series.

So why can't i do the same with xln(x-1)..? So apply same method

Answer 1

i mean ln(1-x) so all ln(x-1) are meant to be ln(1-x)

Answer 2

You certainly could use the method you suggested, but it sounds like the question assumes you already know the power series for \(\ln(1-x)\) or \(\ln(x-1)\).

In the case that you didn't, I agree that your method is much more accommodating.
\[f(x)=x\ln(1-x)~~\Rightarrow~~f'(x)=\ln(1-x)-\frac{x}{1-x}\]
The second term is fairly easy, since
\[\frac{1}{1-x}=\sum_{n=0}^\infty x^n~~~\text{for }|x|<1\]
but that first term can be a problem (assuming we don't readily know what the power series is). Taking another derivative - just the first term this time - you get
\[f'(x)=\int\frac{d}{dx}\ln(1-x)~dx-\frac{x}{1-x}=-\int\frac{1}{1-x}~dx-\frac{x}{1-x}\]
Then converting to power series, you have
\[\begin{align*}f'(x)&=-\int\sum_{n=0}^\infty x^n~dx-x\sum_{n=0}^\infty x^n\\
&=-\sum_{n=0}^\infty \frac{x^{n+1}}{n+1}+C_1-\sum_{n=0}^\infty x^{n+1}
\end{align*}\]
Integrating again, you have
\[\begin{align*}f'(x)&=-\int\sum_{n=0}^\infty x^n~dx-x\sum_{n=0}^\infty x^n\\
&=-\sum_{n=0}^\infty \frac{x^{n+2}}{(n+1)(n+2)}+C_1x+C_2-\sum_{n=0}^\infty \frac{x^{n+2}}{n+2}+C_3
\end{align*}\]
From here, you would attempt to condense into one series.

Answer 3

That last \(f'(x)\) should just say \(f(x)\).

Knowing the power series before hand is a great help because you wouldn't have to deal with any of those pesky \(C\)s, not to mention the work is much simpler.
\[\ln(1-x)=-\sum_{n=1}^\infty\frac{x^n}{n}\]
Then
\[x\ln(1-x)=-\sum_{n=1}^\infty\frac{x^{n+1}}{n}\]