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Feeds
15:02
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A: Find a power series for $\ln(1−x)$. Use the sigma symbol below to write your final answer.

Andrew LiHint: We know that: $${1\over 1-x} = \sum_{k=0}^\infty x^k$$ And integrating both sides: $$\int {1\over 1-x}\,\mathrm dx = \int \sum_{k=0}^\infty x^k\, \mathrm dx$$ Then realize that: $$-\ln |1-x| = \int {1\over 1-x} \,\mathrm dx = \int \sum_{k=0}^\infty x^k\,\mathrm dx = \sum_{k=0}^\infty ...

tien lee
tien lee
Can you please explain the last step
Andrew Li
Andrew Li
@tienlee Did you use the hint?
@JohnDoe Really? I'll edit it then.
tien lee
tien lee
@AndrewLi, so, $ln(1-x)=-\sum _{ k=0 }^{ \infty }{ \frac{x^{k+1}}{k+1} } $
Andrew Li
Andrew Li
@tienlee Yes, that is correct, in the interval of convergence.
tien lee
tien lee
@AndrewLi then for b) the answer will be $ln(1-x)=-x-\sum _{ n=1 }^{ \infty }{ \frac{x^{n+1}}{n+1} } $
Andrew Li
Andrew Li
15:02
@tienlee If you want just one summation, you can reindex to $-\sum_{n = 1}^\infty {x^n\over n}$
tien lee
tien lee
@AndrewLi first I substituted $0$, and then continued with n=1
Andrew Li
Andrew Li
@tienlee Yes, but since we have $n + 1$ in the denominator and a power of $n+1$, we can change that to just $n$ and start from $n=1$ rather than $n=0$.
Hello? I don't have much time to chat @tienlee
tien lee
tien lee
user image
Is this correct
Andrew Li
Andrew Li
No. If you want to reindex you're looking for \sum_{n=1} \frac{x^n}{n}
Also notice that my answer states ln|1-x| not ln(1-x). But in the interval of convergence for the series (|x| < 1), they are the same thus the series is still applicable
tien lee
tien lee
Please, Please help me,
Andrew Li
Andrew Li
15:14
What don't you understand
John Doe
John Doe
@tienlee If you are still confused about something, I would like to suggest asking another question. You're more likely to get help that way, and possibly will get different perspectives in answers that may help you where you're stuck
Andrew Li
Andrew Li
^^ That's a good idea
Frank W.
Frank W.
15:34
If it’s the reindexing issue, one possible idea might be to replace $n$ with $n-1$ so the lower bound is $n=0$ and the rest of the fraction becomes $\frac {x^{(n-1)+1}}{(n-1)+1}$

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