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Newtons Method and Finding the 5th root

  1. Oct 11, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the 5th root of 36 accurate to four decimal places



    2. Relevant equations

    xn+1 = xn - f(xn)/f'(xn)



    3. The attempt at a solution

    First I attempted to write the fifth root of 36 in exponential form as show below:

    Let the 5th root of 36 = x
    Let f(x) = x^1/5 - 36
    So, f'(x) = 1/5x^-4/5 Is this right so far?
     
  2. jcsd
  3. Oct 11, 2011 #2

    LCKurtz

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    No. You need to solve x5 - 36 = 0.
     
  4. Oct 11, 2011 #3
    oh okay, but i am confused because i thought that the fifth root of a number in exponent
    form was that number raised to the 1/5. I don't understand how it is x raised to the fifth.
     
  5. Oct 11, 2011 #4

    LCKurtz

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    If you solved that equation by radicals wouldn't you get [itex]x=36^{\frac 1 5}[/itex]? And isn't that what you are asked to find? So you want the root of that equation.
     
  6. Oct 11, 2011 #5
    oh okay. So then it is like below?

    f(2) = 2^5 - 36
    = - 4
    and

    f'(2) = 5(2^4)
    = 80
     
  7. Oct 11, 2011 #6

    LCKurtz

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    No. Don't you have any worked examples in your text? You mentioned above the recursion formula for xn+1. What do you get for that? You need to use it.
     
  8. Oct 11, 2011 #7
    it's okay, it says that in the text...got it from here. thx.
     
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