Question

question

Answer 1

I made up this, to write a power series representation for x^2/(2-2x).

Answer 2

\[\frac{ x^2 }{ 2-2x }=\frac{ x^2 }{ 2 }\sum_{n=1}^{\infty}x^n\]\[=\frac{ 1 }{ 2 }\sum_{n=1}^{\infty}x^{n+2}\]

Answer 3

is this correct ?

Answer 4

HI!!

Answer 5

yes it is right

Answer 6

just to make sure, I am using the fact that\[\frac{ 1 }{ 1-x}=\sum_{n=1}^{\infty}x^n\]

Answer 7

oh, tnx

Answer 8

since
\[\frac{xx^2}{2-2x}=\frac{1}{2}\frac{x^2}{1-x}\]

Answer 9

so, what if I had to write the series for ln(1-x) ?

Answer 10

\[\frac{1}{1-x}=\sum_{n=1}^{\infty}x^n\]\[-\ln(1-x)=\sum_{n=1}^{\infty}-\frac{x^{n+1}}{n+1}\]

Answer 11

doing good ?

Answer 12

differentiate it, although this series is well known

Answer 13

I am integrating on both sides.

Answer 14

negative sign in sigma is wrong, but..

Answer 15

you have it right, i meant differentiate
\[\ln(1-x)\] to get
\[-\frac{1}{1-x}\]

Answer 16

take the power series and integrate term by term
you have it

Answer 17

oh, so \[\ln(1-x)=-\sum_{n=1}^{\infty}\frac{x^{n+1}}{n+1}\]

Answer 18

because the integral of 1/(1-x) is -ln(1-x)

Answer 19

and then I divide by -1 on both sides

Answer 20

yeah that is it

Answer 21

tnx:)
kind of barely got a hang on this...

Answer 22

lol divide by -1 = change the sign

Answer 23

tnx for helpingme out

Answer 24

\[\color\magenta\heartsuit\] course you did all the work

Answer 25

\[\frac{1}{1-x}=\sum_{n=1}^{\infty}\frac{x^{n}}{1}\]\[\int\limits~\frac{1}{1-x}~dx=\int\limits~\sum_{n=1}^{\infty}\frac{x^{n}}{1}~dx\]\[-\ln(1-x)=\sum_{n=1}^{\infty}\frac{x^{n+1}}{n+1}\]\[\ln(1-x)=-\sum_{n=1}^{\infty}\frac{x^{n+1}}{n+1}\]

Answer 26

got it... !

Answer 27

yay

Answer 28

fyi, the lower limit is n=0 in
\[ \frac{1}{1-x} = \sum_{n=0}^\infty x^n \]