# Newton's Interpolation Formula: Prove that $a_2$ is the second divided difference?

Given an interpolating polynomial $P(x) = a_0 + a_1(x-x_0) +a_2(x-x_0)(x-x_1)$, prove that $a_2 = f[x_0,x_1,x_2]$ where $[\cdot]$ denotes the divided difference.

I am having trouble with this problem because the algebra just doesn't work out! Can someone point out my error below?

By definition of interpolation $P(x_2) = f_2$ (where $f_2$ is shorthand for $f(x_2)$). So we have

$$f_2 = a_0 + a_1(x_2-x_0)+a_2(x_2-x_0)(x_2-x_1).$$

Here it is known that $a_0 = f_0$ and $a_1 = f[x_1,x_0] = \frac{f_1 - f_0}{x_1 - x_0}$. So we must show that $a_2 = f[x_0,x_1,x_2]$ by algebraic manipulation. Proceeds below:

$$f_2 = f_0 +\frac{f_1-f_0}{x_1-x_0}(x_2-x_0)+a_2(x_2-x_0)(x_2-x_1)$$

Rearranging: $$f_2 = \frac{f_0(x_1-x_0)+(f_1-f_0)(x_2-x_0)+a_2(x_2-x_0)(x_2-x_1)(x_1-x_0)}{x_1-x_0}$$

Rearranging to solve for $a_2$:

$$a_2 = \frac{(x_1-x_0)f_2 - f_0(x_1-x_0) - (f_1-f_0)(x_2-x_0)}{(x_2-x_0)(x_2-x_1)(x_1-x_0)}.\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(1)$$

Now working out $f[x_0,x_1,x_2]$ gives

$$f[x_0,x_1,x_2] = \frac{(f_2-f_1)(x_1-x_0) - (f_1-f_0)(x_2-x_1)}{(x_2-x_0)(x_2-x_1)(x_1-x_0)}.\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(2)$$

Immediately, I see a problem. The denominators in (1) and (2) are the same, meaning that the numerators must be equal in order to complete the proof. But in the numerator of (1) there is no possibility for an $f_1x_1$ term, which is required in the expansion of (2). So how can they be equal?

If anyone can point out my error, that would be greatly appreciated.

Thanks

You are very close to the answer, only a step or two away. Simplifying your expression for $a_2$ gives
$$a_2 = \frac{(x_1-x_0)f_2 + f_1x_0 - f_1x_2 + f_0(x_2-x_1)}{(x_2-x_0)(x_2-x_1)(x_1-x_0)}$$
Now add $f_1x_1 - f_1x_1$ to the numerator of $a_2$ and you will get the desired result.
• Thanks! Yes, it was the $f_1x_1 - f_1x_1$ trick that I was missing. Nov 22, 2016 at 17:32