Question

In Newton's form of the interpolation polynomial we need to compute the coefficients, \\ $c_0 = f[x_0], c_1 = f[x_0, x_1], \dots, c_n = f[x_0, x_1, \dots, x_n]$. In the table of divided differences \\ we proceed column by column and the needed coefficients are in the uppermost di-\\ agonals. A simple 1D array, $c$ of size $n+1$, can be used to store and compute these \\ values. We just have to compute them from bottom to top to avoid losing values we \\ have already computed. The following pseudocode does precisely this:\\ \\ for $j = 0, 1, \dots, n$\\ $\qquad c_j = f_j;$\\ end\\ for $k = 1, \dots, n$\\ $\qquad$ for $j = n, n-1, \dots, k$\\ $\qquad \qquad c_j = (c_j - c_{j-1})/(x_j - x_{j-k});$\\ $\qquad$ end\\ end\\ The evaluation of the interpolation polynomial in Newton's form can be then done \\ with the Horner-like scheme seen in class:\\ \\ $p = c_n$\\ for $j = n-1, n-2, \dots, 0$\\ $\qquad p = c_j + (x - x_j) \cdot p;$\\ end\\ (a) Write computer codes to compute the coefficients $c_0, c_1, \dots, c_n$ and to evaluate the \\ corresponding interpolation polynomial at an arbitrary point $x$. Test your codes and \\ turn in a run of your test.\\ (b) Consider the function $f(x) = e^{-x^2}$ for $x \in [-1, 1]$ and the nodes $x_j = -1 + j(2/10)$, \\ $j = 0, 1, \dots, 10$. Use your code(s) in (a) to evaluate $P_{10}(x)$ at the points $x_j = -1 + $\\$j(2/100)$, $j = 0, 1, \dots, 100$ and plot the error $f(x) - P_{10}(x)$.

          In Newton's form of the interpolation polynomial we need to compute the coefficients, \\
$c_0 = f[x_0], c_1 = f[x_0, x_1], \dots, c_n = f[x_0, x_1, \dots, x_n]$. In the table of divided differences \\
we proceed column by column and the needed coefficients are in the uppermost di-\\
agonals. A simple 1D array, $c$ of size $n+1$, can be used to store and compute these \\
values. We just have to compute them from bottom to top to avoid losing values we \\
have already computed. The following pseudocode does precisely this:\\
\\
for $j = 0, 1, \dots, n$\\
$\qquad c_j = f_j;$\\
end\\
for $k = 1, \dots, n$\\
$\qquad$ for $j = n, n-1, \dots, k$\\
$\qquad \qquad c_j = (c_j - c_{j-1})/(x_j - x_{j-k});$\\
$\qquad$ end\\
end\\
The evaluation of the interpolation polynomial in Newton's form can be then done \\
with the Horner-like scheme seen in class:\\
\\
$p = c_n$\\
for $j = n-1, n-2, \dots, 0$\\
$\qquad p = c_j + (x - x_j) \cdot p;$\\
end\\
(a) Write computer codes to compute the coefficients $c_0, c_1, \dots, c_n$ and to evaluate the \\
corresponding interpolation polynomial at an arbitrary point $x$. Test your codes and \\
turn in a run of your test.\\
(b) Consider the function $f(x) = e^{-x^2}$ for $x \in [-1, 1]$ and the nodes $x_j = -1 + j(2/10)$, \\
$j = 0, 1, \dots, 10$. Use your code(s) in (a) to evaluate $P_{10}(x)$ at the points $x_j = -1 + $\\$j(2/100)$, $j = 0, 1, \dots, 100$ and plot the error $f(x) - P_{10}(x)$.

In Newton's form of the interpolation polynomial we need to compute the coefficients,

c0 = f[x0], c1 = f[x0, x1], …, cn = f[x0, x1, …, xn]. In the table of divided differences

we proceed column by column and the needed coefficients are in the uppermost di-

agonals. A simple 1D array, c of size n+1, can be used to store and compute these

values. We just have to compute them from bottom to top to avoid losing values we

have already computed. The following pseudocode does precisely this:

for j = 0, 1, …, n

cj = fj;

end

for k = 1, …, n

for j = n, n-1, …, k

cj = (cj - cj-1)/(xj - xj-k);

end

end

The evaluation of the interpolation polynomial in Newton's form can be then done

with the Horner-like scheme seen in class:

p = cn

for j = n-1, n-2, …, 0

p = cj + (x - xj) · p;

end

(a) Write computer codes to compute the coefficients c0, c1, …, cn and to evaluate the

corresponding interpolation polynomial at an arbitrary point x. Test your codes and

turn in a run of your test.

(b) Consider the function f(x) = e^-x^2 for x ∈ [-1, 1] and the nodes xj = -1 + j(2/10),

j = 0, 1, …, 10. Use your code(s) in (a) to evaluate P10(x) at the points xj = -1 +
j(2/100), j = 0, 1, …, 100 and plot the error f(x) - P10(x).

Added by Danny B.

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