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\over12}}\over {7-9}} = {5\over 12}\\ 3 & 7 &                            & {{-3\over 2} - 0\over {7-5}} = {-3\over 4} \\ &  & {1-7\over 7-3} = {-3\over2} & \\ 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\over2}                   &             & {{{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


 * $$p_4(x) = p_3 + [y_0,y_1,y_2,y_3,y_4](x-x_0)(x-x_1)(x-x_2)(x-x_3).$$
 * $$p_3(x) = \frac{5}{12}x^{3} - 7x^{2} + \frac{427}{12}x -48 + \frac{17}{15}(x-9)(x-5)(x-3)(x-7).$$
 * $$p_3(x) = \frac{7}{15}x^{4} - \frac{647}{60}x^{3} + \frac{1337}{15}x^{2}-\frac{18697}{60}x + 393$$.