= + + + + + + + + + ...

Let's solve the example problem presented in Section 5.1 by the three procedures presented above. Consider the following three data points:

First, fit the quadratic polynomial, Pz(x) = ao + alx + azxz, to the three data points: 0.294118 = ao + al (3.4) + az(3.4)z 0.285714 = ao + al (3.5) + az(3.5)z 0.277778 = ao + a)(3.6) + az(3.6)z

(5.12a) (5.12b) (5.12c)

Solving for ao, al> and az by Gauss elimination gives ao = 0.858314, al = -0.245500, and az = 0.023400. Substituting these values into Eqs. (5.7a) and (5.7b) and evaluating at X = 3.5 yields the solution for the direct fit polynomial:

P;(3.5) = -0.245500 + (0.04680)(3.5) = -0.081700 P'2(x) = 0.046800

(5.12d) (5.12e)

Substituting the tabular values into Eqs. (5.9a) and (5.9b) and evaluating at X = 3.5 yields the solution for the Lagrange polynomial:

I 2(3.5) - (3.5 + 3.6) 2(3.5) - (3.4 + 3.6) Pz(3.5) = (3.4 _ 3.5)(3.4 _ 3.6) (0.294118) + (3.5 _ 3.4)(3.5 _ 3.6) (0.285714)

2(3.5) - (3.4 + 3.5) + (3.6 _ 3.4)(3.6 _ 3.5) (0.277778) = -0.081700 (5.13a)

"() 2(0.294118) 2(0.285714) 2(0.277778)Pz 3.5 = + + --c::--:---'-::-.,.-:-:-:,------,---'--:,........"..(3.4 - 3.5)(3.4 - 3.6) (3.5 - 3.4)(3.5 - 3.6) (3.6 - 3.4)(3.6 - 3.5) =0.046800 (5.13b)

A divided difference table must be constructed for the tabular data to use the divided difference polynomial. Thus,