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Newton-Raphson question

  1. Jan 4, 2005 #1
    Let f(x)=x^2 - a, where a>0. The roots of the equation f(x)=0 are +root(a), -root(a).

    Show that if x1 > 0 is any initial estimate for root(a), then the Newton-Raphson method gives the iteration formula

    x n-1 = 1/2( xn + a/xn ),

    n>=1
     
  2. jcsd
  3. Jan 4, 2005 #2
    Sorry, the last post wasn't clear, the iteration formula is..

    Xn-1 = (1/2)(Xn + a/Xn) , n >= 1
     
  4. Jan 4, 2005 #3

    learningphysics

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    Shouldn't this be:
    Xn+1 = (1/2)(Xn + a/Xn) , n >= 1

    Did you apply the newton-raphson method Xn+1=Xn-f(xn)/f'(xn) ?
     
  5. Jan 4, 2005 #4
    yes, it's Xn-1 instead, my fault
    I tried the newton's method, but still have trouble solving the question
     
  6. Jan 4, 2005 #5

    learningphysics

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    Can you show what you did? The given formula arrives almost immediately by applying the method.
     
  7. Jan 4, 2005 #6

    Integral

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    Newtons Method is;

    [tex] x_{n+1} = x_n - \frac {f(x_n)} {f'(x_n)} [/tex]

    You are given

    [tex] f(x)= x^2 - a [/tex]

    Compute [tex] f'(x) = 2x [/tex]

    Now apply Newtons method:
    [tex]x_{n+1}= x_{n}- \frac {f(x_n)} {f'(x_n)} = x_n - \frac {x_n^2 - a } {2x_n} = \frac { 2 x^2_n - x^2_n +a} {2x_n}= \frac {x^2_n + a} {2x_n} = \frac 1 2 ( x_n + \frac a {x_n})[/tex]
     
    Last edited: Jan 4, 2005
  8. Jan 4, 2005 #7
    trap, who is your TA?
     
  9. Jan 4, 2005 #8
    thx for the help guys!!
     
  10. Jan 4, 2005 #9
    ??? Answer me
     
  11. Jan 4, 2005 #10
    kataKonik, hi..you must be taking the same course as i am...
     
  12. Jan 4, 2005 #11
    Yes, that is why I asked you that question.
     
  13. Jan 4, 2005 #12
    well...I'm not even sure if i can use the web to help me with these problem sets..so i just wanna keep my identity as a secret
     
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