Question

What are the steps for solving a Maclaurin series for f(x): ln((1+x)/(1-x))

Answer 1

i think i would start with
\[\ln(1+x)-\ln(1-x)\]

Answer 2

series for these are well known, although i guess you should derive them

Answer 3

we could do them by hand, or use a trick
either way works

Answer 4

so then for example f'(x) = -2/(x^2-1), f"(x)=4x/(x^2-1)^2, etc, then don't you insert a value determined by x_o then simplify?

Answer 5

whoa way way too much

Answer 6

it will get very ugly very fast doing it that way
break it apart first

Answer 7

that is, start with
\[\ln(1+x)-\ln(1-x)\] compute separately and subtract, it will be way simpler

Answer 8

roger that, ultimately i'm trying to show that the original equation is equal to 2(x+(x^3/3)+(x^5/5)+(x^(2k-1)/(2k-1)+...), I probably should have clarified that in the beginning

Answer 9

here is the very simplest way
put
\[f(x)=\ln(1+x)\]and then
\[f'(x)=\frac{1}{1+x}\]

Answer 10

\[\frac{1}{1+x}\] expands as a geometric series, namely
\[1-x+x^2-x^3+x^4-...\] take the antiderivative term by term to get the expansion for
\[\ln(1+x)\]

Answer 11

alright that makes more sense

Answer 12

you get \[\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...\]

Answer 13

this one is not even bad to do by hand
\[f(x)=\ln(1+x)\]
\[f'(x)=(1+x)^{-1}\]
\[f''(x)=-(1+x)^{-2}\]
\[f^{(3)}(x)=2(1+x)^{-3}\] and so on
then replace \(x\) by 0 and divide by \(k!\) to the get expansion, but the first method is slicker

Answer 14

repeat the process with
\[\ln(1-x)\] take the derivative, recognize the geometric series, then integrate term by term
it is quick

Answer 15

ahhh alright, that makes a lot more sense

Answer 16

have fun

Answer 17

thanks a lot