Find the Interpolating Polynomial?

Question

Find the Interpolating Polynomial?

 

a) Standard form
b) Lagrange
c) Newton

 

x0 = 2    f(x) = 1
x1 = 4    f(x) = 0
x2 = 7    f(x) = -21/4

 

 

Answer 1

To find the interpolating polynomial, we can use different methods like the standard form, Lagrange interpolation, or Newton interpolation. Let's calculate the interpolating polynomial using all three methods for the given data points.

a) Standard Form:

The standard form of an interpolating polynomial can be found using the following equation:

P(x) = a0 + a1(x - x0) + a2(x - x0)(x - x1) + ...

Given the data points:

x0 = 2, f(x0) = 1

x1 = 4, f(x1) = 0

x2 = 7, f(x2) = -21/4

Let's substitute these values into the standard form equation:

P(x) = a0 + a1(x - 2) + a2(x - 2)(x - 4)

To find the coefficients a0, a1, and a2, we can use the given data points:

When x = x0 = 2:

1 = a0 + a1(2 - 2) + a2(2 - 2)(2 - 4)

1 = a0

When x = x1 = 4:

0 = a0 + a1(4 - 2) + a2(4 - 2)(4 - 4)

0 = a0 + 2a1

When x = x2 = 7:

-21/4 = a0 + a1(7 - 2) + a2(7 - 2)(7 - 4)

-21/4 = a0 + 5a1 + 15a2

Solving the above system of equations, we get:

a0 = 1

a1 = -1/2

a2 = -3/20

Therefore, the interpolating polynomial in standard form is:

P(x) = 1 - (1/2)(x - 2) - (3/20)(x - 2)(x - 4)

b) Lagrange Interpolation:

The Lagrange interpolation formula for three points is:

P(x) = f(x0) * L0(x) + f(x1) * L1(x) + f(x2) * L2(x)

Where L0(x), L1(x), and L2(x) are the Lagrange basis polynomials.

L0(x) = ((x - x1)(x - x2))/((x0 - x1)(x0 - x2))

L1(x) = ((x - x0)(x - x2))/((x1 - x0)(x1 - x2))

L2(x) = ((x - x0)(x - x1))/((x2 - x0)(x2 - x1))

Substituting the given values into the Lagrange interpolation formula:

P(x) = 1 * L0(x) + 0 * L1(x) + (-21/4) * L2(x)

Simplifying the Lagrange basis polynomials:

L0(x) = ((x - 4)(x - 7))/((2 - 4)(2 - 7)) = -(1/10)(x^2 - 11x + 28)

L1(x) = ((x - 2)(x - 7))/((4 - 2)(4 - 7)) = (1/6)(x^2 - 9x + 14)

L2(x) = ((x - 2)(x - 4))/((7 - 2)(7 - 4)) = -(7/20)(x^2 - 6x + 8)

Substituting the Lagrange basis polynomials into the Lagrange interpolation formula:

P(x) = 1 * (-(1/10)(x^2 - 11x + 28)) + 0 * (1/6)(x^2 - 9x + 14) + (-21/4) * (-(7/20)(x^2 - 6x + 8))

Simplifying the equation:

P(x) = (1/10)x^2 - (11/10)x + (7/10) + (21/4)x^2 - (63/10)x + (147/20)

P(x) = (55/4)x^2 - (168/10)x + (7/10) + (147/20)

P(x) = (55/4)x^2 - (84/5)x + (7/10) + (147/20)

P(x) = (55/4)x^2 - (84/5)x + (77/20)

Therefore, the interpolating polynomial in Lagrange form is:

P(x) = (55/4)x^2 - (84/5)x + (77/20)

c) Newton Interpolation:

The Newton interpolation formula uses divided differences. We can calculate the divided differences using the given data points:

f[x0] = f(x0) = 1

f[x1] = f(x1) = 0

f[x2] = f(x2) = -21/4

f[x0, x1] = (f[x1] - f[x0]) / (x1 - x0) = (0 - 1) / (4 - 2) = -1/2

f[x1, x2] = (f[x2] - f[x1]) / (x2 - x1) = ((-21/4) - 0) / (7 - 4) = -7/12

f[x0, x1, x2] = (f[x1, x2] - f[x0, x1]) / (x2 - x0) = ((-7/12) - (-1/2)) / (7 - 2) = -1/12

The Newton interpolation formula is:

P(x) = f[x0] + f[x0, x1](x - x0) + f[x0, x1, x2](x - x0)(x - x1)

Substituting the calculated divided differences:

P(x) = 1 + (-1/2)(x - 2) + (-1/12)(x - 2)(x - 4)

Simplifying the equation:

P(x) = 1 - (1/2)(x - 2) - (1/12)(x - 2)(x - 4)

Therefore, the interpolating polynomial in Newton form is:

P(x) = 1 - (1/2)(x - 2) - (1/12)(x - 2)(x - 4)

To summarize:

a) Standard form: P(x) = 1 - (1/2)(x - 2) - (3/20)(x - 2)(x - 4)

b) Lagrange form: P(x) = (55/4)x^2 - (84/5)x + (77/20)

c) Newton form: P(x) = 1 - (1/2)(x - 2) - (1/12)(x - 2)(x - 4)