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Homework Help: Binomial series

  1. May 20, 2010 #1
    1. The problem statement, all variables and given/known data

    Expand [tex](1-3x)^{\frac{1}{3}}[/tex] in ascending power of x , up to the term x^3 . By using an appropriate substitution for x , show that [tex]\sqrt[3]3=\frac{33809}{19683}[/tex]



    2. Relevant equations



    3. The attempt at a solution

    the expansion would be 1-x-x^2-(5/3)x^3 and the range of x whixh make this expansion valid is |-3x|<1/3 .

    My question is how do i find this appropriate substitution for x ? Instead the guessing way , is there a proper way ?
     
  2. jcsd
  3. May 20, 2010 #2

    HallsofIvy

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    Well, since you have [itex](1- 3x)^{1/3}[/itex] and want [itex](3)^{1/3}[/itex] you probably want to make 1- 3x= 3. Solve that for x.
     
  4. May 20, 2010 #3
    i don think so because -1/3<x<1/3 for this expansion to be valid .
     
  5. May 20, 2010 #4
    Hi thereddevils

    Maybe this method can be applied. Since we want to find the cube root and the range for x is -1/3<x<1/3 , we can deduce that (1-3x) should be fraction.

    So, the denominator should be a number that is integer and has a cube root, which is also integer , such as 8, 27, etc

    Now we can try (1-3x) = 3/8 or (1-3x) = 3/27 , etc..

    But cube root of 3 is not 33809 / 19683.
    33809 / 19683 is closer to square root of 3. Maybe you can re-check the question :smile:
     
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