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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)

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

    HallsofIvy

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    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]
    thus
    [tex]g'(c) = \frac{f'(c)(c-x_1) - f(c)}{(c-x_1)^2}[/tex]

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

    thanks

    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]
    [tex]=\frac{f'(x)-f'(d)}{x-x_1}[/tex]

    Thus
    [tex]g'(c)=\frac{f'(c)-f'(d)}{c-x_1}[/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
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