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)
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
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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
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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
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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.