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Homework Help: Newtons Method - A little help

  1. Aug 2, 2010 #1
    1. The problem statement, all variables and given/known data

    Solve each equation using newtons method. The problem I am working on right now is:

    [tex] x^{5}+x^{3}+x=1 [/tex]

    which is the same as:

    [tex] x^{5}+x^{3}+x-1=0 [/tex]


    2. Relevant equations

    [tex] x_{2} = x_{1}-\frac {f(x)}{f'(x)} [/tex]

    with x1 being the first guess to what an f(x)=0 would be.

    3. The attempt at a solution

    So I put the equation into my calculator to graph it and It is very clear from the graph that f(x)=0 on the x>0 side so I choose x1=1 and continue like so:

    [tex] x_{2} = 1 - \frac {f(x)}{f'(x)} = 1- \frac {2}{9} = \frac {7}{9} [/tex]

    and after 5 iterations I get:

    [tex] x_{5} = .6368843716 [/tex]

    which looks right to me but when I check the answer in the back of the book it says that it should be -.63688, so just the opposite of what I got. The graph of the equation clearly shows that f(x)=0 on x>0 so did I do something wrong or does the book have another typo?

    thanks
     
  2. jcsd
  3. Aug 2, 2010 #2

    Char. Limit

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    Gold Member

    Another typo.
     
  4. Aug 3, 2010 #3

    hunt_mat

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    Homework Helper

    I've just spent the last 5 months play with Newton's method... The general method is
    [tex]
    x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}
    [/tex]
    Where in your case:
    [tex]
    f(x)=x^{5}+x^{3}+x-1
    [/tex]
    The derivative of this is:
    [tex]
    f'(x)=5x^{4}+3x^{2}+1
    [/tex]
    And in your case Newton's method is
    [tex]
    x_{n+1}=x_{n}-\frac{x_{n}^{5}+x_{n}^{3}+x_{n}-1}{5x_{n}^{4}+3x_{n}^{2}+1}
    [/tex]
    So all you have to do is plug numbers in. Newtons method converges very quickly in my experience, 2 or 3 iterations will usually do the trick.
     
  5. Aug 3, 2010 #4

    hunt_mat

    User Avatar
    Homework Helper

    As a check, plug your solution back into the equation and check.
     
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