Question

Find a power series representation for f(x) = ln((1+x)/(1-x)). f(x) = sum n=0 to infinity of ?

Answer 1

might want to split the log into 2 parts

Answer 2

ln(a/b) = ln(a) - ln(b)

now taking derivatives if need be is suitable

Answer 3

ln(1+x)-ln(1-x)
d/dx = (1/1+x)-(1/1-x)

Answer 4

a shortcut now is to do the long division to come up with a series for those derivatives

Answer 5

OR, combine them in that state and do the division

Answer 6

oops (-1/1-x)

Answer 7

ok, ill try.

Answer 8

\[\frac{(1-x)+(1+x)}{1-x^2}\]
\[\frac{2}{1-x^2}\]

Answer 9

I dont understand what im supposed to be doing when you said 'do the division'

Answer 10

umm, divide :/

\[\frac{2}{1-x^2}=2(\frac{1}{1-x^2})\]

so


1+x^2+x^4+x^6+.....
-----------------
1-x^2 | 1
-(1-x^2)
---------
x^2
-(x^2-x^4)
---------
x^4

Answer 11

so the derivative of the stated problem is:\[2(1+x^2+x^4+x^6+...)\]
therefore its integration should be equal to the stated problem ...
\[2(\frac11x^1+\frac13x^3+\frac15x^5+\frac17x^7+...)\]

Answer 12

do we need to go over how to do long division?

Answer 13

No, I know how. I was just confused as to why/what to divide.

Answer 14

I don't believe my professor showed us how to solve these problems like this.

Answer 15

in alot of the cases in a textbook, division will produce an equivalent function in polynomial form

Answer 16

for example\[\frac19\]
lets say 9 = 1-x
\[\frac{1}{1-x}\]
we are still doing the division of:

---------
9 | 1

or written another way

---------
1-x | 1



1+x+x^2+x^3+x^4+....
---------
1-x | 1
-1(1-x)
-------
x
-x(1-x)
-------
x^2
-x^2(1-x)
---------
x^3

Answer 17

9=1-x when x=-8

therefore: \(1-8+8^2-8^3+8^4_...=\frac19\)

Answer 18

Oh, ok. I'm following

Answer 19

thats all i got :)

Answer 20

kids want food ... so bye

Answer 21

Thank you.

Answer 22

When you did the derivative, you're in an okay position to start finding the power series representations.
\[\begin{align*}\frac{d}{dx}f(x)&=\frac{1}{1+x}-\frac{1}{1-x}\\&=\frac{1}{1-(-x)}-\frac{1}{1-x}\\
&=\sum_{n=0}^\infty (-x)^n-\sum_{n=0}^\infty x^n
\end{align*}\]
From here, you could simplify by seeing which terms if any cancel out.