1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor

    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:
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook