Question

Using Newton's divided difference method, the second divided difference for the function values f(1)=2, f(2)=3, f(4)=7 is approximately equal to ____.

• A. 1.5
• B. 1
• C. 0.5
• D. 0.33

Hint:

Build the first divided differences f[x0,x1] and f[x1,x2], then combine them over (x2-x0) to get the second divided difference.

Answer 1

The Correct Option is D

Step 1: The second divided difference can also be computed directly from the symmetric formula, without building a difference table:

\[f[x_0,x_1,x_2] = \frac{f(x_0)}{(x_0-x_1)(x_0-x_2)} + \frac{f(x_1)}{(x_1-x_0)(x_1-x_2)} + \frac{f(x_2)}{(x_2-x_0)(x_2-x_1)}\]

Step 2: Substitute \(x_0=1,x_1=2,x_2=4\) and \(f(x_0)=2,f(x_1)=3,f(x_2)=7\). Compute each denominator:

\[(x_0-x_1)(x_0-x_2)=(1-2)(1-4)=(-1)(-3)=3\] \[(x_1-x_0)(x_1-x_2)=(2-1)(2-4)=(1)(-2)=-2\] \[(x_2-x_0)(x_2-x_1)=(4-1)(4-2)=(3)(2)=6\]

Step 3: Substitute into the formula:

\[f[x_0,x_1,x_2] = \frac{2}{3}+\frac{3}{-2}+\frac{7}{6}\]

Step 4: Convert to a common denominator of 6:

\[f[x_0,x_1,x_2] = \frac{4}{6}-\frac{9}{6}+\frac{7}{6} = \frac{4-9+7}{6} = \frac{2}{6} = \frac{1}{3}\]

Step 5: This confirms the value found from the divided difference table.

\[\boxed{f[x_0,x_1,x_2] = \frac{1}{3} \approx 0.33}\]