1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Proof using Rolle's thm or the Mean Value Thm

  1. Jul 20, 2006 #1
    it is known thm of interpolation that:
    [tex]\frac{f[x_1, x_2] - f[x_0, x_1]}{x_2 - x_0}[/tex] = f"(c)}/2
    where c is between the minimum and maximum of [tex]x_0, x_1, x_2[/tex]

    and where
    [tex]f[x_0, x_1] = \frac{f(x_1) - f(x_0)}{x_1 - x_0} = f'(d) [/tex] where d is between [tex] x_0[/tex] and [tex] x_1 [/tex] by the mean value theorem.

    Is it possible and if so, can anyone halp me prove this equality:
    [tex]\frac{f[x_1, x_2] - f[x_0, x_1]}{x_2 - x_0}[/tex] = f"(c)/2
    using possibly Rolle's Thm, the Mean Value Thm, the Taylor series expansion, among others? (i.e., using only elelmentary calculus)

    Last edited: Jul 20, 2006
  2. jcsd
  3. Jul 20, 2006 #2


    User Avatar
    Science Advisor

    Assuming that f is twice differentiable of some interval containing [itex]x_0[/itex], [itex]x_1[/itex], and [itex]x_2[/itex], let [itex]g(x)= \frac{f(x)- f(x_1)}{x- x_1}[/itex]. Apply the mean value theorem to g(x).
  4. Jul 20, 2006 #3
    cheers for the help, HallsofIvy,
    applying the mean value thm to g(x), indeed I would get:
    [tex]\frac{f[x_1, x_2] - f[x_0, x_1]}{x_2 - x_0}[/tex]
    on the left hand side of the equation.

    On the right side, I would have g'(c), but isn't
    [tex]g'(x) = \frac{f'(x)(x-x_1) - f(x)}{(x-x_1)^2}[/tex]
    [tex]g'(c) = \frac{f'(c)(c-x_1) - f(c)}{(c-x_1)^2}[/tex]

    how is this equal to f"(c)/2?


    ETA: I was able to see that [tex]g(x) = f'(c)[/tex] where c is between [tex]x[/tex] and [tex] x_1[/tex]...
    I guess if I let [tex]c = \frac{x-x_1}{2}[/tex], I could get:
    g'(c) = f"(d)/2
    with d between the maximum of [tex]x_0, x_1, x_2[/tex]

    but what if I let
    [tex]c = \frac{x-x_1}{3}[/tex]?
    Wouldn't I get
    g'(c) = f"(d)/3 as a result?
    I'm confused...:(
    Last edited: Jul 20, 2006
  5. Jul 21, 2006 #4
    okay, the best I can come up with is
    [tex]g'(x) = \frac{f'(x)(x-x_1)-[f(x)-f(x_1)]}{(x-x_1)^2} [/tex]
    [tex]=\frac{f'(x)(x-x_1) - f'(d)(x-x_1)}{(x-x_1)^2}[/tex]

    where c and d are between x and x1.

    how is this equal to f"(s)/2???

    I did try an arbtrary f(x) and for some reason, the two really are equivalent
    Last edited: Jul 21, 2006
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook