Question

Find a power series for ln(x-1).

Answer 1

\[\Large\rm \frac{d}{dx}\ln(x-1)=\frac{1}{x-1}\]
This implies,\[\Large\rm \ln(x-1)=\int\limits \color{orangered}{\frac{1}{x-1}}dx\]Let's mess with this orange part, turn it into geometric series.

Answer 2

Factoring out a -1 from the denominator:\[\Large\rm -\int\limits\limits \color{orangered}{\frac{1}{1-x}}dx\]

Answer 3

Very nice!

Answer 4

Recall that the convergent geometric series is:\[\Large\rm \sum_{n=0}^{\infty} r^n = \frac{1}{1-r}\]

Answer 5

So we want to use that in reverse:\[\Large\rm -\int\limits\limits\limits \color{orangered}{\sum_{n=0}^{\infty}x^n}dx\]Make sense so far? :o

Answer 6

Anyway, from there you can do some fancy business.
Bring the integral into the sum,\[\Large\rm -\sum_{n=0}^{\infty} \int\limits x^n ~dx\]Integrating,\[\Large\rm -\sum_{n=0}^{\infty} \frac{1}{n+1}x^{n+1}\]

Answer 7

I appreciate the help, although i can't really understand it because it's coming through as just about of words and symbols. I can't figure out what you're trying to say! You get an A for effort though :)

Answer 8

*a bunch of words and symbols

Answer 9

Apparently it is an issue with my laptop not letting me view it properly. Used my smartphone and can see it just fine. Thank you so much!

Answer 10

Sometimes the equations do not show up correctly.

ln(x-1) is not in the usual table of series
you could try to create a Taylor series {check] it out, with derivatives of first and higher orders.

#zepdrix showed a clever way to approach the problem
recognizing that ln(x-1) is to the integral of dx/(x-1) and that
1/(x-1) is a well-known power series 1 + x + x^2 + x^3...,
so the integral gives you x + (1/2) x^2 + (1/3)x^3 + ...

Answer 11

Yes, I got that I should have the integral of 1/x-1 dx but didn't know where to go from there. So I just say that the power series is x + (1/2)x^2 etc and that's the answer? It can't be that simple can it?????

Answer 12

I think it is as simple as that: you integrate the 1/(x-1) power series term by trm.