fll)

Consider the four data points given in Example 4.3. Let's interpolate for /(3.44) using linear, quadratic, and cubic interpolation using Neville's algorithm. Rearranging the data in order of closeness to x = 3.44 yields the following set of data:

(3.44 - 3.40)0.285714 - (3.44 - 3.50)0.294118 3.50 - 3.40

(4.55a)

Thus, the result of linear interpolation is /(3.44) =1;(1) = 0.290756. To evaluate 1;(2) ,;;(1) must first be evaluated. Thus,

(0) (0») )Jil) = (x - X2)f3 - (x - X3)fz = (3.44 - 3.500.298507 - (3.44 - 3.350.285714 x3 - x2 3.35 - 3.50

= 0.290831 (4.55b)

Evaluating 1;(2) gives (I) \£(1) ( )fl2) = (x - XI)fz - (x - x311l = 3.44 - 3.400.290831 - (3.44 - 3.35)0.290756 x3 - Xl 3.35 - 3.40

= 0.290696 (4.56)