Question

If $f(x) = x^2$, then the second order divided difference for the points $x_0, x_1, x_2$ will be:

• A. $-1$
• B. $\dfrac{-1}{x_1 - x_0}$
• C. $1$
• D. $\dfrac{1}{x_2 - x_1}$

Hint:

For polynomial functions, divided differences of degree $n$ are constant when the polynomial is of degree $n$.

Answer 1

The Correct Option is C

Step 1: Recall the formula for second order divided difference.
For $f(x) = x^2$, \[ f[x_0, x_1, x_2] = \frac{f[x_1, x_2] - f[x_0, x_1]}{x_2 - x_0}. \]

Step 2: Compute first order divided differences.
\[ f[x_0, x_1] = \frac{f(x_1) - f(x_0)}{x_1 - x_0} = \frac{x_1^2 - x_0^2}{x_1 - x_0} = x_1 + x_0. \] \[ f[x_1, x_2] = \frac{f(x_2) - f(x_1)}{x_2 - x_1} = \frac{x_2^2 - x_1^2}{x_2 - x_1} = x_2 + x_1. \]

Step 3: Substitute into second order formula.
\[ f[x_0, x_1, x_2] = \frac{(x_2 + x_1) - (x_1 + x_0)}{x_2 - x_0} = \frac{x_2 - x_0}{x_2 - x_0} = 1. \]

Step 4: Conclusion.
The second order divided difference is always $1$.