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Math Help - Newton's method

  1. #1
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    Newton's method

    Let f(x) = (cos(x))^2 - 3*x

    Show that abs(f''(x)/f'(y))<=B, for x,y exist in [0,1], B is constant>0
    Hence or otherwise deduce that Newton's method will converge.

    I found that
    abs(f''(x)/f'(y)) = abs( (2(sin(x)^2) - 2*(cos(x)^2)) / (-2cosy*siny - 3))

    I can deduce that abs(2(sin(x)^2) - 2*(cos(x)^2)) is <=2

    and that abs(-2cosy*siny - 3)>=3

    so abs( (2(sin(x)^2) - 2*(cos(x)^2)) / (-2cosy*siny - 3))<=2/3

    and so B=1? Am I correct??

    Thank you very much!!!
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  2. #2
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    Re: Newton's method

    Couldn't edit my post, but I can now see that B>0 as a positive constant can be a fraction so is B=2/3?

    And after that how can I show that Newton's method converges? If the condition abs(f''(x)/f'(y))<=B, for x,y exist in [0,1] is satisfied then don't we automatically know that the method converges?

    Thank you very much!!!
    Last edited by Darkprince; November 9th 2011 at 03:03 PM.
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  3. #3
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    Re: Newton's method

    Can someone help me please? Thanks again!
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