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ddd [48]
4 years ago
14

Use lagrange interpolation to find a polynomial that passes through the points. (a) (0, 1), (2, 3), (3, 0) (b) (−1, 0), (2, 1),

(3, 1), (5, 2)

Mathematics
1 answer:
Alexus [3.1K]4 years ago
6 0
The Lagrange interpolation polynomial is a sum of products.

(a)
f(x)=1\times \frac{(x-2)(x-3)}{(0-2)(0-3)}+3\times \frac{(x-0)(x-3)}{(2-0)(2-3)}+0\times \frac{(x-0)(x-2)}{(3-0)(3-2)}
This can be simplified to
f(x)=\frac{1}{3}(-4x^{2}+11x+3)

(b)
g(x)=0+\frac{1(x+1)(x-3)(x-5)}{(2+1)(2-3)(2-5)}+\frac{1(x+1)(x-2)(x-5)}{(3+1)(3-2)(3-5)}+\frac{2(x+1)(x-2)(x-3)}{(5+1)(5-2)(5-3)}
This can be simplified to
g(x)=\frac{1}{24}(x^{3}-6x^{2}+11x+18)

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