Question

Find a Lagrange’s interpolating polynomial for the data given below:

x0 = 1, x1 = 2.5, x2 = 4 and x3 = 5.5

f(x0) = 4, f(x1) = 7.5, f(x2) = 13 and f(x3) = 17.5

Also, find the value of f(5).

Answer 1

$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} x & 1 & 2.5 & 4 & 5.5 \\ \hline f(x)& 4& 7.5 & 13 & 17.5 \\ \hdashline \end{array}$

By Lagrange’s interpolation formula we have:

$f(x)= \frac{(x-x_1)(x-x_2)(x-x_3)}{(x_0-x_1)(x_0-x_2)(x_0-x_3)}y_0+ \frac{(x-x_0)(x-x_2)(x-x_3)}{(x_1-x_0)(x_1-x_2)(x_1-x_3)} y_1+ \frac{(x-x_0)(x-x_1)(x-x_3)}{(x_2-x_0)(x_2-x_1)(x_2-x_3)}y_2+ \frac{(x-x_0)(x-x_1)(x-x_2)}{(x_3-x_0)(x_3-x_1)(x_3-x_2)} y_3= \frac{(x-2.5)(x-4)(x-5.5)}{(1-2.5)(1-4)(1-5.5)}4+ \frac{(x-1)(x-4)(x-5.5)}{(2.5-1)(2.5-4)(2.5-5.5)} 7.5+ \frac{(x-1)(x-2.5)(x-5.5)}{(4-1)(4-2.5)(4-5.5)}13+ \frac{(x-1)(x-2.5)(x-4)}{(5.5-1)(5.5-2.5)(5.5-4)} 17.5=-\tfrac{16}{81}(x^3-12x^2+45.75x-55)+\tfrac{10}{9}(x^3-10.5x^2+31.5x-22)-\tfrac{52}{27}(x^3-9x^2+21.75x-13.75)+\tfrac{70}{81}(x^3-7.5x^2+16.5x-10)=-\tfrac{4}{27}x^3+\tfrac{14}{9}x^2-\tfrac{15}{9}x+\tfrac{115}{27}$

$f(5)= -\tfrac{4}{27}\cdot 5^3+\tfrac{14}{9}\cdot5^2-\tfrac{15}{9}\cdot 5+\tfrac{115}{27}=\tfrac{440}{27}$

Answer: $f(x)= -\tfrac{4}{27}x^3+\tfrac{14}{9}x^2-\tfrac{15}{9}x+\tfrac{115}{27}$ and $f(5)=\tfrac{440}{27}$ .