Question

5. Let f(x)=x~ln(x) be defined at x_{0}=2 x_{1}=2.5 , x_{2}=3.3 , ,x_{3}=4
b). Use Newton's divided difference interpolation formula to find f(3.5)

Answer 1

To find f(3.5) using Newton's divided difference interpolation formula, we need to calculate the divided difference table.

The divided difference table is as follows:

x0 | f[x0] = f(2) = 2 * ln(2) = 1.38629
--------------------------------------------------------------
x1 | f[x1] = f(2.5) = 2.5 * ln(2.5) = 2.09861 | f[x0,x1] = (f[x1] - f[x0]) / (x1 - x0) = (2.09861 - 1.38629) / (2.5 - 2) = 0.35430
--------------------------------------------------------------
x2 | f[x2] = f(3.3) = 3.3 * ln(3.3) = 2.42494 | f[x1,x2] = (f[x2] - f[x1]) / (x2 - x1) = (2.42494 - 2.09861) / (3.3 - 2.5) = 0.710125
| f[x0,x1,x2] = (f[x1,x2] - f[x0,x1]) / (x2 - x0) = (0.710125 - 0.35430) / (3.3 - 2) = 0.178625
--------------------------------------------------------------
x3 | f[x3] = f(4) = 4 * ln(4) = 3.17805 | f[x2,x3] = (f[x3] - f[x2]) / (x3 - x2) = (3.17805 - 2.42494) / (4 - 3.3) = 1.5054
| f[x1,x2,x3] = (f[x2,x3] - f[x1,x2]) / (x3 - x1) = (1.5054 - 0.710125) / (4 - 2.5) = 0.79545
| f[x0,x1,x2,x3] = (f[x1,x2,x3] - f[x0,x1,x2]) / (x3 - x0) = (0.79545 - 0.178625) / (4 - 2) = 0.3089125

Now, using Newton's divided difference interpolation formula, we can find f(3.5):

f(3.5) = f(x0) + f[x0,x1](x - x0) + f[x0,x1,x2](x - x0)(x - x1) + f[x0,x1,x2,x3](x - x0)(x - x1)(x - x2)

Plugging in the given values:

f(3.5) = 1.38629 + 0.35430(3.5 - 2) + 0.178625(3.5 - 2)(3.5 - 2.5) + 0.3089125(3.5 - 2)(3.5 - 2.5)(3.5 - 3.3)

Simplifying:

f(3.5) ≈ 1.38629 + 0.35430(1.5) + 0.178625(1.5)(0.5) + 0.3089125(1.5)(0.5)(0.2)

f(3.5) ≈ 1.38629 + 0.53145 + 0.0670056 + 0.0155445

f(3.5) ≈ 1.99928

Therefore, f(3.5) ≈ 1.99928.