# MacLaurin Series for f(x)=ln|1+x^3|

1. May 19, 2010

### Forknknife

1. The problem statement, all variables and given/known data
Find the MacLaurin series representation for f(x)=ln|1+x^3|

2. Relevant equations
1/(1-x) = $$\sum$$x^n = 1+x+x^2+x^3+......... |x|<1

3. The attempt at a solution
right.
so maclaurin series by default means it expands as a taylor series where x=0
f(0)= ln|1+x^3| = 0
f'(0)= 3(0)^2/(1+0^3)^1 = 0/1 = 0
f''(0)= -3(0)^2/(1+0^3)^2 = -0/1 = 0
f'''(0)= 6(0)^6-42(0)^3+6/(1+0^3)^3 = 6
so on and so forth

it's taken me a few hours but so far, i can't seem to use that relevant equation to find the maclaurin series.

i may have gotten kinda close, b/c i've found so far that
fn(0)=(very complex numerator that kind of looks like a quadratic or polynomial formula with switching signs)/(1+x3)n

i've done upto 9th derivative(and further more) and found that this goes something like:
0,0,0,6,0,0,-360,0,0,120960,0,0,-119750400,0,0,261534973600
and that the series the pretty much follows (-1)^(n-1)*3*(n-1)!, where n starts at 0, with 2 blanks in between.

i know i'm missing something, and this isn't that hard.

--
then i'm supposed to find radius of convergence, which i think i can probably get using by ratio test..
then need to find out how many terms are needed to appx it within 0.0001...
then find out why appx'ing integral of ln|1+x^3| from 0 to 2 by using the series representation is wrong.
but i think i can figure those out once the first part is done.

Last edited: May 19, 2010
2. May 19, 2010

### Cyosis

Find the Maclaurin series for the derivative of ln(1+x^3), by using the geometric series.

3. May 19, 2010

### vela

Staff Emeritus
Or find the Maclaurin series for ln(1+t) and then let t=x^3.