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Math Help - Newton's method convergence

  1. #1
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    Newton's method convergence

    Suppose that f(x) has a zero at x = p and that f'(p) != 0. Show that when f''(p) = 0 Newton's methos converges to p with order at least 3.



    Is this solved using a Taylor series expansion. And finding a delta and an M?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by CarmineCortez View Post
    Suppose that f(x) has a zero at x = p and that f'(p) != 0. Show that when f''(p) = 0 Newton's methos converges to p with order at least 3.



    Is this solved using a Taylor series expansion. And finding a delta and an M?
    Yes it used Taylor series, but I think you need the existance of the third derivative to prove it, and its continuity at p (without these conditions you just get that the convergence is faster than quadratic).

    -Opps.. f''(p)=0, sp yes, with suitable conditions on the existance and continuity of the third derivative convergence is of order 3 or better.

    CB
    Last edited by CaptainBlack; September 29th 2009 at 07:42 PM.
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  3. #3
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    so i have f(x) = f'(p)(x-p) + (f'''(p)*(x-p)^3)/3!

    then letting x = p_n and using the mean value thm.

    f(p_n) = f(p_n)-f(p) + (f'''(p)*(p_n-p)^3)/3!

    So something is not right here....
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  4. #4
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    Quote Originally Posted by CarmineCortez View Post
    so i have f(x) = f'(p)(x-p) + (f'''(p)*(x-p)^3)/3!

    then letting x = p_n and using the mean value thm.

    f(p_n) = f(p_n)-f(p) + (f'''(p)*(p_n-p)^3)/3!

    So something is not right here....
    First I would (without loss of generality) assume p=0 as it makes the algebra easier.

    Then we have:

    \varepsilon_{n+1}=\varepsilon_n-\frac{f(\varepsilon_n)}{f'(\varepsilon_n)}

    Now I would expand the right hand side as a power series in \varepsilon_n about 0, remembering that f(0)=f''(0)=0.

    CB
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