Question

Find a power series representation for f(x)=ln((1+x)/(1−x)).

Answer 1

http://home.scarlet.be/math/taylor.htm#ln%281+x%29-and-its-Macl

Answer 2

The formula for ln(1+x)= (-1)^n-1* x^n/n
The form. for ln(1-x)= x??

I know I need to combine them. Just not sure how.

Answer 3

just add (1-x) at the bottom,
if you expand (1-x) binomially, you will end up having another series at each term.

Answer 4

i would start with the fact that you have
\[\ln(1+x)-\ln(1-x)\]

Answer 5

Oo ... didn't see that ... both was inside log.

Answer 6

ln(1-x)=\[\sum_{n=0}^{\infty} (-1)^{n-1}\frac{x^n}{n}\]

Answer 7

Sorry, thats ln(1+x)

Answer 8

Ooooh, you're saying with that subtraction sign, i can distribute and it'll be posi. and be in the same form?!?

Answer 9

n = 1

Answer 10

My denominator is 1?

Answer 11

start summation from n=1

Answer 12

The issue is that I have to make a formula first to put in the summation.

Answer 13

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

Answer 14

\[ \sum_{n=2i+1, i=0}^{\infty} \frac{-2x^n}{n} \]

Answer 15

Im not allowed to start the sum at 1, lol.

Answer 16

Oo.. there was an error, above
make that - in summation to + at the middle,
I am sure if you can start at 0, since 1/n = undefined at n=0

Answer 17

@sam30317 you got this?

Answer 18

Nooo, = (
I got it wrong & still a bit lost. I just ned to know how to fuse them

Answer 19

ok first off well know series for \(\ln(1+x)\) at x = 0 is
\[x-\frac{x^2}{2}+\frac{x^3}{3}-...=\sum (-1)^{k+1}x^k\]

Answer 20

so far so good?

Answer 21

yes

Answer 22

now to find series for \(\ln(1-x)\) replace \(x\) by \(-x\) in the above series. the even powers (which are all negative above) obviously can't tell the diffreence between \(x\) and \(-x\) so they stay the same, and the odd powers (which are all positive above) become negative

Answer 23

in other words is t is identical to the one above except that ALL the terms are negative,
\[-x-\frac{x^2}{2}-\frac{x^3}{3}-...=-\sum\frac{x^k}{k}\]

Answer 24

ahh

Answer 25

now subtract the second from the first, which means of course get rid of the minus sign and add the terms

Answer 26

http://www.wolframalpha.com/input/?i=ln%28%281%2Bx%29%2F%281-x%29%29

Answer 27

the even powers will add up to zero

Answer 28

and you will get double all the odd ones

Answer 29

\[2x+\frac{2x^3}{3}+\frac{2x^5}{5}+...\]

Answer 30

yay just was the wolf got it! so we know it is right

Answer 31

∑(−1)^k+1 x^k - (x^k/k) ??
I didnt have to solve anything. I just needed to find the formula representation of it. Secondly, Wolf isnt showing what I was looking for. Where do you giys see it?

Answer 32

the formula representation would be
\[2\sum\frac{x^{2k-1}}{2k-1}\]

Answer 33

>____>
Thanks everyone! Ihave to review what you giys told me b/c im a bit brain dead. I have to relearn EVERYTHING taught by Monday.

Answer 34

good luck! stay up late and drink lots of coffee