x 0 2 3 4
y 7 11 28 63
a)_Using these values, use the Lagrange interpolation process to obtain a polynomial of least degree.
b)_rearrange the points in the table of a) and find the newton form of the interpolating polynomial. Show that they are the same even though their forms maybe different.
To find the polynomial of least degree using the Lagrange interpolation process, we can use the following steps:
a) Lagrange interpolation process:
Step 1: Write down the given x and y values in a table.
x 0 2 3 4
-------------------------
y 7 11 28 63
Step 2: Determine the number of given data points, denoted by n, which is equal to the number of x and y values.
In this case, n = 4.
Step 3: Define the Lagrange basis polynomials, denoted by L_i(x), for each data point using the formula:
L_i(x) = ∏[(x - x_j) / (x_i - x_j)] for i ≠ j,
where i and j are indices of the data points, and i ≠ j denotes that i and j are not equal.
L_0(x) = [(x - x_1)(x - x_2)(x - x_3)] / [(x_0 - x_1)(x_0 - x_2)(x_0 - x_3)]
L_1(x) = [(x - x_0)(x - x_2)(x - x_3)] / [(x_1 - x_0)(x_1 - x_2)(x_1 - x_3)]
L_2(x) = [(x - x_0)(x - x_1)(x - x_3)] / [(x_2 - x_0)(x_2 - x_1)(x_2 - x_3)]
L_3(x) = [(x - x_0)(x - x_1)(x - x_2)] / [(x_3 - x_0)(x_3 - x_1)(x_3 - x_2)]
Step 4: Substitute the given x values into each Lagrange basis polynomial and multiply with the corresponding y values.
P(x) = Σ[y_i * L_i(x)] for i from 0 to n-1,
where Σ denotes the sum of terms.
P(x) = [7 * L_0(x)] + [11 * L_1(x)] + [28 * L_2(x)] + [63 * L_3(x)]
Simplifying the above expression will give us the polynomial of least degree.
b) Newton form of the interpolating polynomial:
Step 1: Rearrange the given table of points in ascending order of x values.
x 0 2 3 4
-------------------------
y 7 11 28 63
Step 2: Calculate the divided differences using the Newton's divided difference formula:
f[x_0] = y_0 = 7
f[x_1, x_0] = (y_1 - y_0) / (x_1 - x_0) = (11 - 7) / (2 - 0) = 2
f[x_2, x_1, x_0] = ((y_2 - y_1) / (x_2 - x_1) - f[x_1, x_0]) / (x_2 - x_0) = ((28 - 11) / (3 - 2) - 2) / (3 - 0) = 5
f[x_3, x_2, x_1, x_0] = ((y_3 - y_2) / (x_3 - x_2) - f[x_2, x_1, x_0]) / (x_3 - x_0) = ((63 - 28) / (4 - 3) - 5) / (4 - 0) = 6
Step 3: Express the interpolating polynomial in Newton form:
P(x) = f[x_0] + f[x_1, x_0] * (x - x_0) + f[x_2, x_1, x_0] * (x - x_0) * (x - x_1) + f[x_3, x_2, x_1, x_0] * (x - x_0) * (x - x_1) * (x - x_2)
Simplifying the above expression will give us the polynomial in Newton form.
Comparing the polynomials obtained using the Lagrange interpolation process and the Newton form, we will see that they are equivalent even though their forms may be different.