Answer to Question #110428 in Quantitative Methods for Gayatri Yadav

Question

Given the following data, estimate the value of f(5) using
i) Lagrange’s interpolation
ii) Newton’s divided difference interpolation
x 1 4 6 7 10
f(x) -3 9 10 9 8

Answer 1

$\begin{matrix} i&0&1&2&3&4\\ x & 1&4&6&7&10 \\ f(x)&-3&9&10&9&8 \end{matrix}$

i) Lagrange’s interpolation 

$f(x)=\frac{(x-4)(x-6)(x-7)(x-10)}{(1-4)(1-6)(1-7)(1-10)}\cdot(-3)+\\ +\frac{(x-1)(x-6)(x-7)(x-10)}{(4-1)(4-6)(4-7)(4-10)}\cdot9+\\ +\frac{(x-1)(x-4)(x-7)(x-10)}{(6-1)(6-4)(6-7)(6-10)}\cdot10+\\ +\frac{(x-1)(x-4)(x-6)(x-10)}{(7-1)(7-4)(7-6)(7-10)}\cdot9+\\ +\frac{(x-1)(x-4)(x-6)(x-7)}{(10-1)(10-4)(10-6)(10-7)}\cdot8$

$f(5)=\frac{(5-4)(5-6)(5-7)(5-10)}{(-270)}+\\ +\frac{(5-1)(5-6)(5-7)(5-10)}{(-12)}+\\ +\frac{(5-1)(5-4)(5-7)(5-10)}{(4)}+\\ +\frac{(5-1)(5-4)(5-6)(5-10)}{(-6)}+\\ +\frac{(5-1)(5-4)(5-6)(5-7)}{81}=\\ =\frac{1}{27}+\frac{10}{3}+10-\frac{10}{3}+\frac{8}{81}=10.14$

ii) Newton’s divided difference interpolation  

$\begin{matrix} 1 & -3 \\ &&\frac{9+3}{4-1}=4\\ 4 & 9&&\frac{0.5-4}{6-1}=-0.7\\ &&\frac{10-9}{6-4}=0.5&&\\ 6&10&&\frac{-1-0.5}{1-7}=0.25&&\\ &&\frac{9-10}{7-6}=-1&&\\ 7&9&&\frac{-0.3+1}{4-10}=-0.12\\ &&\frac{8-9}{10-7}=-0.3\\ 10&8 \end{matrix}$

$\begin{matrix} \frac{0.25+0.7}{7-1}=0.16\\ &&\frac{-0.06-0.16}{10-1}=-0.02\\ \frac{-0.12-0.25}{10-4}=-0.06\\ \end{matrix}$

Resulting coefficients are:[-3,4,-0.7,0.16,-0.02]

Create Target Polynomials

$f(x)=-3+4(x-1)-0.7(x-1)(x-4)+\\+0.16(x-1)(x-4)(x-6)-\\ -0.02(x-1)(x-4)(x-6)(x-7)\\ f(5)=-3+4(5-1)-0.7(5-1)(5-4)+\\+0.16(5-1)(5-4)(5-6)-\\ -0.02(5-1)(5-4)(5-6)(5-7)=9.4$