Proof using Rolle's thm or the Mean Value Thm

[island-boy]

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

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

Is it possible and if so, can anyone help me prove this equality:
\frac{f[x_1, x_2] - f[x_0, x_1]}{x_2 - x_0} = 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

 

[HallsofIvy]

Assuming that f is twice differentiable of some interval containing x_0, x_1, and x_2, let g(x)= \frac{f(x)- f(x_1)}{x- x_1}. Apply the mean value theorem to g(x).

 

[island-boy]

cheers for the help, HallsofIvy,
applying the mean value thm to g(x), indeed I would get:
\frac{f[x_1, x_2] - f[x_0, x_1]}{x_2 - x_0}
on the left hand side of the equation.

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

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

thanks

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

but what if I let
c = \frac{x-x_1}{3}?
Wouldn't I get
g'(c) = f"(d)/3 as a result?
I'm confused...:(

 

[island-boy]

okay, the best I can come up with is
g'(x) = \frac{f'(x)(x-x_1)-[f(x)-f(x_1)]}{(x-x_1)^2}
=\frac{f'(x)(x-x_1) - f'(d)(x-x_1)}{(x-x_1)^2}
=\frac{f'(x)-f'(d)}{x-x_1}

Thus
g'(c)=\frac{f'(c)-f'(d)}{c-x_1}
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