Polynomial Interpolation: Inverse Interpolation

Problem 17.9.

Employ inverse interpolation to determine the value of x that corresponds to f(x) = 0.93 for the following tabulated data: 

Note that the values in the table were generated with the function f(x) = x²/(1+x²).

(a) Determine the correct value analytically. 

(b) Use quadratic interpolation and the quadratic formula to determine the value numerically. 

(c) Use cubic interpolation and bisection to determine the value numerically. 

Solution:

(a) Using the function available, manipulate it to determine corresponding x value:

(b) A quadratic interpolating polynomial can be fit to the last three points: (3, 0.9), (4, 0.941176), and (5, 0.961538) using polyfit function:

>> format long

>> x = [3 4 5];

>> fx = [0.9 0.941176 0.961538];

>> a = polyfit(x,fx,2)

a =

  -0.010407000000000   0.114024999999999   0.651588000000002

Thus, the best-fit parabola that passes through the last three points is f₂(x) = -0.010407x² + 0.114025x + 0.651588. 

We must therefore find the root of 

0.93 = -0.010407x² + 0.114025x + 0.651588

or

0 = -0.010407x² + 0.114025x - 0.278412

The quadratic formula can be used to calculate:

Thus, the second root, 3.67295, is a good estimate of the true value. 

(c) A cubic interpolating polynomial can be fit to the last four points: (2, 0.8), (3, 0.9), (4, 0.941176), and (5, 0.961538) using polyfit function:

>> format long

>> x = [2 3 4 5];

>> fx = [0.8 0.9 0.941176 0.961538];

>> a = polyfit(x,fx,3)

a =

   0.006335000000000  -0.086427000000002   0.411770000000007   0.271487999999994

Thus, the best-fit cubic that passes through the last four points is f₃(x) = 0.006335x³ - 0.086427x² + 0.41177x + 0.271488. 

We must therefore find the root of 

0.93 = 0.006335x³ - 0.086427x² + 0.41177x + 0.271488

or

0 = 0.006335x³ - 0.086427x² + 0.41177x - 0.658512

Bisection method (bisect.m) can be employed to find the root of this polynomial. Using initial guesses of x_l = 3 and x_u = 4 (that bracket the root), a value of 3.61884 is obtained.

>> fx = @(x) 0.006335*x^3 - 0.086427*x^2 + 0.41177*x - 0.658512;

>> [root,fx,ea,iter]=bisect(fx,3,4)

root =

   3.618841171264648

fx =

     1.082876788238707e-08

ea =

     5.270606093347705e-05

iter =

    19