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

  1. May 19, 2010 #1
    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) = [tex]\sum[/tex]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.
    please help.

    --
    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. jcsd
  3. May 19, 2010 #2

    Cyosis

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    Find the Maclaurin series for the derivative of ln(1+x^3), by using the geometric series.
     
  4. May 19, 2010 #3

    vela

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    Education Advisor

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