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