Numerical Analysis/Newton form exercise

1. Using the Newton form, find the interpolating polynomial passing through the points $(x_{0},y_{0})=(9,9)$, $(x_{1},y_{1})=(5,7)$, $(x_{2},y_{2})=(3,7)$, and $(x_{3},y_{3})=(7,1)$.

Solution:

To find the four coefficients, we can use the divided differences.

${\begin{matrix}x_{i}&y_{i}&[y_{0},y_{1}]&[y_{0},y_{1},y_{2}]&[y_{0},y_{1},y_{2},y_{3}]\\9&9&&\\&&{7-9 \over 5-9}={1 \over 2}&\\5&7&&{{0-1 \over 2} \over {3-9}}={1 \over 12}\\&&{7-7 \over 3-5}={0}&&{{{-3 \over 4}-{1 \over 12}} \over {7-9}}={5 \over 12}\\3&7&&{{-3 \over 2}-0 \over {7-5}}={-3 \over 4}\\&&{1-7 \over 7-3}={-3 \over 2}&\\7&1&&\\\end{matrix}}$

Thus, the four coefficients are $x_{0}$ and the upper diagonal of the calculated divided differences.

Using the Newton Forward Divided Difference formula, we find the polynomial is

$p_{3}(x)=[y_{0}]+[y_{0},y_{1}](x-x_{0})+[y_{0},y_{1},y_{2}](x-x_{0})(x-x_{1})+[y_{0},y_{1},y_{2},y_{3}](x-x_{0})(x-x_{1})(x-x_{2}).$

$p_{3}(x)=9+{\frac {1}{2}}(x-9)+{\frac {1}{12}}(x-9)(x-5)+{\frac {5}{12}}(x-9)(x-5)(x-3).$

$p_{3}(x)={\frac {5}{12}}x^{3}-7x^{2}+{\frac {427}{12}}x-48$.

 

2. Find the interpolating polynomial passing through the four points given in the exercise above as well as through the point $(x_{4},y_{4})=(4,2)$

Solution:

To find the new coefficient, we can simply add the new point to the end and calculate $[y_{0},y_{1},y_{2},y_{3},y_{4}]$.

${\begin{matrix}x_{i}&y_{i}&[y_{0},y_{1}]&[y_{0},y_{1},y_{2}]&[y_{0},y_{1},y_{2},y_{3}]&[y_{0},y_{1},y_{2},y_{3},y_{4}]\\9&9&&\\&&{1 \over 2}&\\5&7&&{1 \over 12}\\&&{0}&&{5 \over 12}\\3&7&&{-3 \over 4}&&{{{-23 \over 12}-{5 \over 12}} \over {4-9}}={17 \over 15}\\&&{-3 \over 2}&&{{{7 \over 6}-{-3 \over 4}} \over {4-5}}={-23 \over 12}\\7&1&&{{-1 \over 3}-0 \over {-3 \over 2}}={7 \over 6}\\&&{2-1 \over 4-7}={-1 \over 3}&\\4&2&&\\\end{matrix}}$

Thus, we find the polynomial $p_{4}$ is

$p_{4}(x)=p_{3}+[y_{0},y_{1},y_{2},y_{3},y_{4}](x-x_{0})(x-x_{1})(x-x_{2})(x-x_{3}).$

$p_{3}(x)={\frac {5}{12}}x^{3}-7x^{2}+{\frac {427}{12}}x-48+{\frac {17}{15}}(x-9)(x-5)(x-3)(x-7).$

$p_{3}(x)={\frac {7}{15}}x^{4}-{\frac {647}{60}}x^{3}+{\frac {1337}{15}}x^{2}-{\frac {18697}{60}}x+393$.