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Worked Examples · Example 4

Q.Using Lagrange's interpolation formula, find yy at x=2x=2 given the points (1,2),(3,10),(4,17)(1,2),(3,10),(4,17).

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Setting up the three terms (x0=1,y0=2x_0=1,y_0=2; x1=3,y1=10x_1=3,y_1=10; x2=4,y2=17x_2=4,y_2=17; evaluating at x=2x=2)

Term 1: y0(x−x1)(x−x2)(x0−x1)(x0−x2)=2⋅(2−3)(2−4)(1−3)(1−4)=2⋅(−1)(−2)(−2)(−3)=2⋅26=23\text{Term 1: } y_0\frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}=2\cdot\frac{(2-3)(2-4)}{(1-3)(1-4)}=2\cdot\frac{(-1)(-2)}{(-2)(-3)}=2\cdot\frac26=\frac23

Term 2: y1(x−x0)(x−x2)(x1−x0)(x1−x2)=10⋅(2−1)(2−4)(3−1)(3−4)=10⋅(1)(−2)(2)(−1)=10⋅−2−2=10\text{Term 2: } y_1\frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}=10\cdot\frac{(2-1)(2-4)}{(3-1)(3-4)}=10\cdot\frac{(1)(-2)}{(2)(-1)}=10\cdot\frac{-2}{-2}=10

Term 3: y2(x−x0)(x−x1)(x2−x0)(x2−x1)=17⋅(2−1)(2−3)(4−1)(4−3)=17⋅(1)(−1)(3)(1)=−173\text{Term 3: } y_2\frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}=17\cdot\frac{(2-1)(2-3)}{(4-1)(4-3)}=17\cdot\frac{(1)(-1)}{(3)(1)}=-\frac{17}3

Summing …

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