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

${\begin{aligned}x_{0}&=1&&&&&f(x_{0})&=2\\x_{1}&=3&&&&&f(x_{1})&=4\\x_{2}&=5&&&&&f(x_{2})&=6\\x_{3}&=7&&&&&f(x_{3})&=8.\end{aligned}}$

So we can get the basis polynominals as following:

$\ell _{0}(x)={x-x_{1} \over x_{0}-x_{1}}\cdot {x-x_{2} \over x_{0}-x_{2}}\cdot {x-x_{3} \over x_{0}-x_{3}}=-{1 \over 48}(x-3)(x-5)(x-7)$

$\ell _{1}(x)={x-x_{0} \over x_{1}-x_{0}}\cdot {x-x_{2} \over x_{1}-x_{2}}\cdot {x-x_{3} \over x_{1}-x_{3}}={1 \over 16}(x-1)(x-5)(x-7)$

$\ell _{2}(x)={x-x_{0} \over x_{2}-x_{0}}\cdot {x-x_{1} \over x_{2}-x_{1}}\cdot {x-x_{3} \over x_{2}-x_{3}}=-{1 \over 16}(x-1)(x-3)(x-7)$

$\ell _{3}(x)={x-x_{0} \over x_{3}-x_{0}}\cdot {x-x_{1} \over x_{3}-x_{1}}\cdot {x-x_{2} \over x_{3}-x_{2}}={1 \over 48}(x-1)(x-3)(x-5).$

Thus the interpolating polynomial then is:

${\begin{aligned}L(x)&=f(x_{0})\ell _{0}(x)+f(x_{1})\ell _{1}(x)+f(x_{2})\ell _{2}(x)+f(x_{3})\ell _{3}(x)\\[10pt]&=2\cdot {1 \over 16}(x-1)(x-5)(x-7)+4\cdot {-1 \over 16}(x-1)(x-3)(x-7)+6\cdot {-1 \over 16}(x-1)(x-3)(x-7)+8\cdot {1 \over 48}(x-1)(x-3)(x-5)\\[10pt]&={\frac {5}{12}}x^{3}-{\frac {65}{12}}x^{2}+{\frac {247}{12}}x-{\frac {163}{12}}.\end{aligned}}$

Therefore, we get the Lagrange form interpolating polynomial:

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