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When Newton Raphson Fails

  1. Jul 16, 2006 #1
    Hello. I've been approached with a problem of explaining why Newton Raphson method fails for some functions. I came across a book in Numerical Analysis (Kellison's book) that the method may fail if

    1.) f'(x)=0
    2.) The initial value is taken at a maximum or minumum point,
    3.) The initial value is taken at a point of inflection,
    4.) The initial value is taken near a maximum point and a minimum point,
    5.) The initial value is taken near a point of inflection.

    Now I can explain (1). Newton's Method will fail since the iteration is given by

    x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}

    Therefore making the whole thing undefined.

    As for (2), since minima and maxima have f'(x)=0, they will fail for the same reasons as (1)

    As for (3), ... I'm totally clueless. I dont know how a point of inflection (f''(x)=0) could be related to the Iteration that uses f'

    As for (4), I also know that choosing near a minima or maxima might make the method "oscillate", but what I'm looking for are more concrete answers; something that can be related to f'(x) or something that resembles a proof.

    As for (5), I have no idea why it may fail for an initial value near a point of inflection..

    All help is appreciated,

    Last edited: Jul 16, 2006
  2. jcsd
  3. Jul 16, 2006 #2
    I'm still in the early stages of calc myself but...

    for 5 I try to explain it to myself like this: imagine you're using Newton on a function that behaves a bit like x^2 - C as it heads for the x axis but with a couple of inflexion points near the area you choose as your guess...If for your first guess the concavity of the curve is up then the tangent at the curve will be steeper than the secant between the root you want and the value of your first guess...so as you follow the tangent you are moving in the same direction as the secant but not as fast (along the x-axis)...such that your guess takes you closer to the root, future guesses under favourable conditions such as this would get you closer still.

    But now suppose that you're unlucky for the second guess. The function has changed it's concavity between the x value of our first guess and the x value of our second guess. Now it's the secant that is steeper than the tangent (unless the tangent changes it's direction), for any point on this tangent you will be above the curve...and you will hit the x axis at the other side of your root or further away from it...from this point onwards you might find yourself attracted to a different root or worse still sent on a wild goose chase

    Last edited: Jul 16, 2006
  4. Jul 16, 2006 #3
    Also if the derivative is not continuous at the root, then convergence may fail to occur.

    Indeed, let f(0) = 0 and f(x) = x + x^2\sin(2/x) elsewhere.
  5. Jul 16, 2006 #4


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    For 4/, when the initial value x0 is near the minima, or maxima, then, it may be possible that: [tex]f'(x_0) \approx 0[/tex]. Hence, it will make x1 considerably large: [tex]x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}[/tex]
    Not really sure about 3/, and 5/...
  6. Jul 16, 2006 #5
    http://www.karlscalculus.org/NRbox.html makes a pretty good analysis of the method. There you will find answers to your questions.
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