Numerical Analysis/Lagrange exercise

Find the interpolating polynomial passing through the points $(1,2)$, $(3,4)$, $(5,6)$, $(7,8)$, using the Lagrange method.

Solution:

By using the Lagrange method, we need to find the lagrange basis polynominals first.Since we know

$\begin{array}{cccccccc}{x}_0& =1& & & & & f(x_0)& =2\\ x_1& =3& & & & & f(x_1)& =4\\ x_2& =5& & & & & f(x_2)& =6\\ x_3& =7& & & & & f(x_3)& =8.\end{array}$

So we can get the basis polynominals as following:

${\ell }_0(x)=\frac{x-x_1}{x_0-x_1}\cdot \frac{x-x_2}{x_0-x_2}\cdot \frac{x-x_3}{x_0-x_3}=-\frac{1}{48}(x-3)(x-5)(x-7)$

${\ell }_1(x)=\frac{x-x_0}{x_1-x_0}\cdot \frac{x-x_2}{x_1-x_2}\cdot \frac{x-x_3}{x_1-x_3}=\frac{1}{16}(x-1)(x-5)(x-7)$

${\ell }_2(x)=\frac{x-x_0}{x_2-x_0}\cdot \frac{x-x_1}{x_2-x_1}\cdot \frac{x-x_3}{x_2-x_3}=-\frac{1}{16}(x-1)(x-3)(x-7)$

${\ell }_3(x)=\frac{x-x_0}{x_3-x_0}\cdot \frac{x-x_1}{x_3-x_1}\cdot \frac{x-x_2}{x_3-x_2}=\frac{1}{48}(x-1)(x-3)(x-5).$

Thus the interpolating polynomial then is:

$\begin{array}{cc}L(x)& =f(x_0){\ell }_0(x)+f(x_1){\ell }_1(x)+f(x_2){\ell }_2(x)+f(x_3){\ell }_3(x)\\ & =2\cdot \frac{1}{16}(x-1)(x-5)(x-7)+4\cdot \frac{-1}{16}(x-1)(x-3)(x-7)+6\cdot \frac{-1}{16}(x-1)(x-3)(x-7)+8\cdot \frac{1}{48}(x-1)(x-3)(x-5)\\ & =\frac{5}{12}{x}^3-\frac{65}{12}{x}^2+\frac{247}{12}x-\frac{163}{12}.\end{array}$

Therefore, we get the Lagrange form interpolating polynomial:

$L(x)=\frac{5}{12}{x}^3-\frac{65}{12}{x}^2+\frac{247}{12}x-\frac{163}{12}.$