Question

Given the table of values

x : 150 152 154 156

y = √x : 12.247 12.329 12.410 12.490

Evaluate √155 using Lagrange’s interpolation formula.

Answer 1

$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} x: & 150 & 152 & 154 & 156 \\ \hline y=\sqrt{x}: & 12.247 & 12.329 & 12.410 & 12.490 \\ \hdashline \end{array}$

By Lagrange’s interpolation formula we have:

$y=f(x)= \frac{(x-x_1)(x-x_2)(x-x_3)}{(x_0-x_1)(x_0-x_2)(x_0-x_3)}y_0+ \frac{(x-x_0)(x-x_2)(x-x_3)}{(x_1-x_0)(x_1-x_2)(x_1-x_3)} y_1+ \frac{(x-x_0)(x-x_1)(x-x_3)}{(x_2-x_0)(x_2-x_1)(x_2-x_3)}y_2+ \frac{(x-x_0)(x-x_1)(x-x_2)}{(x_3-x_0)(x_3-x_1)(x_3-x_2)} y_3$

We put $x=155$ :

$y(155)=f(155)= \frac{(155-152)(155-154)(155-156)}{(150-152)(150-154)(150-156)}12.247+ \frac{(155-150)(155-154)(155-156)}{(152-150)(152-154)(152-156)} 12.329+ \frac{(155-150)(155-152)(155-156)}{(154-150)(154-152)(154-156)}12.410+ \frac{(155-150)(155-152)(155-154)}{(156-150)(156-152)(156-154)} 12.490=0.0625\cdot 12.247-0.3125\cdot 12.329 +0.9375\cdot 12.410+0.3125\cdot 12.490\approx 12.450$

Answer: $\sqrt{155}\approx 12.450.$