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Math Help - Fi nd polynom roots using Newton-Raphson method

  1. #1
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    Fi nd polynom roots using Newton-Raphson method

    Hi all

    I try to devellop an algorithm to find roots of any polynom using the Newton Raphson method
    x = x - fx/f'x
    for ex f(x) = x^5 -3^2 + 4

    x = x - (x^5 -3^2 + 4) / (5*x^4 -6*x)
    and repeat for each result of x until you get f(x) = 0

    the problem is the method get only one solution according to the first x we choose
    questions
    which x to choose for xo ?
    and how to know if the are another solution ?
    Thanks in advance
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Fi nd polynom roots using Newton-Raphson method

    I would probably begin here:

    Descartes' rule of signs - Wikipedia, the free encyclopedia

    The article gives links to other methods as well.
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  3. #3
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    Re: Fi nd polynom roots using Newton-Raphson method

    Usually, start by sketching a graph or, often better, graphs.

    For this example, you could start by splitting f(x)=x^{5}-3x^{2}+4 into f(x)=g(x)-h(x), where g(x)=x^{5} and h(x)=3x^{2}-4. The graphs of  y=g(x) and y=h(x) are both easy to sketch and the x coordinates of their points of intersection will be the the values of x for which  f(x)=0. (If you don't like this, your alternative is to sketch the graph of  y=f(x) and look for intersections with the x-axis.)

    A further point is that complex roots occur as conjugate pairs. That means that a fifth order equation like this will have either 1,3 or 5 real roots to go with either 4,2 or zero complex roots.

    The graphs should show you that there is a single intersection to the left of the origin. You might have some doubt as to intersections to the right of the origin but substituting x = 1 and 2 should convince you that there aren't any.

    Next is to try to come up with a suitable first approximation. Do this by substituting some (negative in this case) values for x and look for a pair of values for x where  g(x) and  h(x) cross over each other, or, oops in this case, they are equal to each other. Newton -Raphson is not needed for this example !
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