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Newton-Rhapson's Cube Root.

  1. May 7, 2013 #1
    1. The problem statement, all variables and given/known data

    Derive cube root formula using Newton-Rhapson's method. x - y^3 = 0.

    2. Relevant equations
    xn + 1 = xn - f(xn)/f'(xn)


    3. The attempt at a solution
    I know that the solution is (2y + (x/y^2))/3
    I tried using implicit differentiation and stuff but I can't get this out. Any tips?
     
  2. jcsd
  3. May 7, 2013 #2

    LCKurtz

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    I think your use of x and y is confusing you. If you want the cube root of ##n## you might solve ##f(x) = x^3 - n## for its roots using your above formula. No implicit differentiation needed and you can switch the names of the variables at the end if you want to.
     
  4. May 7, 2013 #3
    So, differentiate "n" as a constant and "x^3" as normal? Thanks for that.
     
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