Question

Find the maclaurin series for f(x)
when f(x) = ln(1-x^4)

Answer 1

a good gimmick would be to take the derivative, get that power series, then integrate term by term

Answer 2

that way you don't have to keep taking successive derivatives

Answer 3

I dont know how to do that

Answer 4

just take derivative of ln(1-x^4)?

Answer 5

yes

Answer 6

oh wait, maybe that is not going to work here

Answer 7

damn
ok then you have to do it by hand

Answer 8

\(f(0)=0\) so the constant is 0

Answer 9

\[f'(x)=\frac{4x^3}{1-x^4}\] so \[f'(0)=0\] as well. hmmm

Answer 10

actually it is
\[f'(x)=\frac{-4x^3}{1-x^4}\]

Answer 11

ok I got derivative but im not sure what to do next, the answer has summation in front of it like \[f(x)=\sum_{n=1}^{infinity}\]

Answer 12

and then asks for your answer

Answer 13

well this is going to take a lot of work i think

Answer 14

you are going to get zeros until the \(x^4\) term

Answer 15

I dont mind I just dont know what the steps I need to take to get there

Answer 16

okay i was trying to think of a better way to do it, maybe for example writing it as
\[\log(1-x^4)=\log(1+x^2)+\log(1-x)+\log(1+x)\]

Answer 17

you have to keep taking successive derivatives, and evaluate them at 0
those are your coefficients

Answer 18

well not exactly, it is \(\frac{f^{n}(0)}{n!}x^n\)

Answer 19

i can't come up with a better way of doing it than by hand

Answer 20

lol its ok I gave up, ill just ask my professor tomorrow, I just didnt go to class today

Answer 21

thx anyway

Answer 22

which is going to suck because the derivatives get nastier and nastier
first 4 will give 0, so the first non zero term is the \(x^4\) term

Answer 23

you might try using \[\log(1-x^4)=\log(1+x^2)+\log(1-x)+\log(1+x)\] and expanding each term
the mclauren series for the last two are well known

Answer 24

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

Answer 25

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