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Newtons method

  1. Dec 8, 2006 #1
    so I was helping my friend today and ran into a problem.

    the problem was to use newtons method to approx. the intersection points of two graphs.


    so I subtracted f-g, found the derivative and pluged in some guesses.

    Except all of my guesses just blew upwards in values instead of zooming in on a point. whats the deal?

  2. jcsd
  3. Dec 8, 2006 #2
    Your guesses were too far from the root. To ensure convergence, the distance between your initial guess and the root has to be less than [itex] 2 \delta \over M [/itex] where

    [tex]|f'(x)| \geq \delta > 0[/tex]

    [tex]|f''(x)| \leq M[/tex]

    for all [itex]x \epsilon [a,b][/itex] in the interval [itex][a,b][/itex] you're considering
    Last edited: Dec 8, 2006
  4. Dec 8, 2006 #3
    Your totally right. Thats one tight interval to make a guess for though. I was only about .2 away from a working guess. Thanks
  5. Dec 20, 2006 #4
    Hi Robierob,

    When u compute f(x)-g(x), u should get x² - cos x on which u can apply Newton's method. Actually u can ensure that your guess is not far from the actual root by finding when your function changes from positive to negative or vice versa. There will be a root when the function changes sign provided that it is continuous for the interval where the root lies. For example, there is a root between x=a and x=b when a and b are consecutive and f(a) f(b) < 0. U can deduce the values of a and b by substitution and use a starting value between a and b for your approximation. That should help u arrive at your root quickly.
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