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Rolle's Theorem

  1. Jul 15, 2012 #1
    I am curious if there is a way of using Rolle's theorem to determine the zero of a function? is there any numerical method for it?

    I am familiar with the basics, When finding the zeros of a function the immediate thought it to set the function equal to zero and solve for x. If the function has more than one root, we know by Rolle's Theorem that the derivative of the function between the two roots must be 0.

    However, does anyone have any ideas on this?
     
  2. jcsd
  3. Jul 15, 2012 #2

    chiro

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    Assuming differentiability of a known function, we can find how many roots exist by finding the turning points as well as the second derivative of those turning points.

    Using the mean-value theorem, we know that if we have two successive turning points with one below the x-axis and one above, then there exists a root between these turning points, (which is what this theorem is getting at).

    At most for a non-constant function, we can only have at maximum n roots where n is the degree of the polynomial.

    The reason to consider the second derivative is for inflection points: If we have a point of inflection like we get in say f(x) = x3 then it means that the function doesn't turn around but keeps going in the same direction in terms of the sign of the first derivative (i.e doesn't decrease from an increase or vice-versa, but instead keeps increasing or decreasing).

    Using all of this information, you can say where the roots must lie if they exist (some functions don't have real roots, but everything does have at least one complex root) and then use the numerical schemes to get a good approximation for the actual root.
     
  4. Jul 15, 2012 #3
    I've probably misinterpreted you, but surely you can't say that.
     
  5. Jul 15, 2012 #4

    chiro

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    No it's true. Consider the function f(x) = x^2 + 1. This function never crosses the x-axis and does not have any real roots. But it has two complex roots of i and -i, but no purely real roots.
     
  6. Jul 15, 2012 #5
    Yes, I appreciate that. But y=1/x has neither real nor complex roots. As I said, I've obviously missed some proviso in what you're saying. Maybe I should read what you said again... it's getting very late where I am! :smile:
     
  7. Jul 15, 2012 #6

    chiro

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    That's not a polynomial and doesn't have a solution for the derivative to be zero.

    Of course you have to check if you get any solutions with the derivatives being equal to zero and in this case you do not.

    The other thing is that this function is not a polynomial: it has a series expansion with a radii of convergence but it is not a polynomial.

    We can differentiate this function from first principles and we can show that it will never have a root for the first derivative.
     
  8. Jul 15, 2012 #7
    Absolutely agreed. I think I was being a bit too pedantic in what I was actually commenting on. :smile:
     
  9. Jul 16, 2012 #8

    haruspex

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    But that's true regardless of whether they're turning points. Also, the OP says nothing about polynomials or the function being twice differentiable..
    What we do know from Rolle is that each pair of roots is separated by at least one root of f', so if you know the roots of f' then you know you only need to look for one root of f between consecutive pairs of roots of f'. Could do this by binary chop, e.g.
    Also need to check for two more: one lower than the lowest root of f' and one higher than the highest. For those, can look at the asymptotic behaviour of f. E.g. if the highest root of f' is f'(a) = 0, and f(a)≠0, then f has a root > a iff f(a) and f(+∞) have opposite sign.
     
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