If y(1) = – 3, y(3) = 9, y(4) = 30, and y(6) = 132, find the four-point
Lagrange interpolation polynomial that takes the same values as the function y at the given points.
"\\def\\arraystretch{1.5}\n \\begin{array}{c:c:c:c:c}\n x & 1 & 3 & 4 & 6 \\\\ \\hline\n y & -3 & 9 & 30 & 132\n\\end{array}"
The Lagrange interpolation polynomial is
"f(x)= \\frac{(x-x_1)(x-x_2)(x-x_3)}{(x_0-x_1)(x_0-x_2)(x_0-x_3)}y_0+ \\frac{(x-x_0)(x-x_2)(x-x_3)}{(x_1-x_0)(x_1-x_2)(x_1-x_3)} y_1+ \\frac{(x-x_0)(x-x_1)(x-x_3)}{(x_2-x_0)(x_2-x_1)(x_2-x_3)}y_2+ \\frac{(x-x_0)(x-x_1)(x-x_2)}{(x_3-x_0)(x_3-x_1)(x_3-x_2)} y_3=\\\\"
"=\\frac{1}{10}(x-3)(x-4)(x-6)+\\frac{3}{2}(x-1)(x-4)(x-6)-5(x-1)(x-3)(x-6)+\\frac{22}{5}(x-1)(x-3)(x-4)=\\\\"
"\\frac{1}{10}(x^3-13x^2+54x-72)+\\frac{3}{2}(x^3-11x^2+34x-24)-5(x^3-10x^2+27x-18)+\\frac{22}{5}(x^3-8x^2+19x-12)=\\\\"
"=x^3-3x^2+5x-6" .
Answer: "f(x)=x^3-3x^2+5x-6."
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