Introduction Linear system Nonlinear equation Interpolation

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1 Interpolation Interpolation is the process of estimating an intermediate value from a set of discrete or tabulated values. Suppose we have the following tabulated values: y y 0 y 1 y 2?? y 3 y 4 y 5 x x 0 x 1 x 2 x 3 x 4 x 5 How do we determine a missing value e.g. y 2? Let us consider a polynomial p(x) that fits to the tabulated values such that y i = p(x i ). Intermediate values of y can be estimated by evaluating this polynomial.

2 Interpolation Example: Develop an interpolation formula for the function e 0.5x using the values at x 0 = 0 and x 1 = 2, and estimate the value of e 0.5x at x = 1. Solution: Let y = f (x) := e 0.5x. A simple approach would be to connect points (x 0, f (x 0 )) and (x 1, f (x 1 )) using a straight line y = mx + c.

3 Interpolation We can evaluate y at x = 1, which is y = m + c and this is the estimate of e 0.5x at x = 1. To find m and c, we use tabulated values to get the following: y 0 = mx 0 + c y 1 = mx 1 + c We solve this linear system using a method of our choice - such as - Gauss elimination, Jacobi etc.

4 Interpolation Using Gauss elimination, we get c = x 0y 1 x 1 y 0 x 0 x 1 m = y 0 y 1 x 0 x 1. Therefore, is the interpolating polynomial. y = x Evaluating this polynomial at x = 1, we get y = Using a calculator, we get the exact value is The error of this linear interpolation is %. What is the degree of the polynomial?

5 Interpolation: how to find the polynomial? We can re-write the straight line such that y = a 0 + a 1 (x x 0 ). It is easy to see that a 0 = y 0 if x = x 0 and a 1 (x 1 x 0 ) = y 1 a 0 if x = x 1 or a 0 = 1.0 and a 1 = y 1 y 0 x 1 x 0 = Similarly, we can write a polynomial of degree 2 y(x) = a 0 + a 1 (x x 0 ) + a 2 (x x 0 )(x x 1 ) and a polynomial of degree 3 y(x) = a 0 +a 1 (x x 0 )+a 2 (x x 0 )(x x 1 )+a 3 (x x 0 )(x x 1 )(x x 2 ) Note that the number of tabulated data = the number of constants for each polynomial.

6 Interpolation Find a general interpolating polynomial of degree n and discuss the limitations. Let us suppose that we have a data set {(x i, y i )} such that y i = f (x i ) for i = 0,..., n. Let the polynomial be of the form f (x) = y = a 0 + a 1 x + a 2 x a n 1 x n 1 + a n x n. To agree with the data set, the polynomial must pass through the point (x i, y i ) for i = 0, 1,..., n. Thus y i = a 0 + a 1 x i + a 2 x 2 i + + a n 1 x n 1 i + a n x n i.

7 Interpolation These equations can be expressed in matrix form as Ba = y, where the Vandermonde matrix is 1 x 0 x0 2 x0 3 x n 0 1 x 1 x1 2 x1 3 x n 1 B = x n xn 2 xn 3 xn n a = a 0 a 1 a 2. a n y = y 0 y 1 y 2. y n

8 Interpolation Home work exercise The Vandermonde matrix is given by Ba = y, where B = [x j i ] for i, j = 0,..., n, a = [a 0,..., a n ] T, and y = [y 0,..., y n ] T. 1. Show that det(b) = Π 0 j<i n (x i x j ) 2. Let y i = f (x i ) for i = 0,..., n represent n + 1 distinct data set. Show that there is a unique interpolating polynomial p(x) of degree n such that p(x i ) = y i.

9 Interpolation: remarks The system of equations Ba = y can be solved using either a direct method (e.g. Gauss elimination) or an iterative method (e.g. Jacobi). A solution exists if and only if the rank of the augmented matrix [B y] is equal to that of the coefficient matrix [B]. The solution will be unique if the rank is n. If rank < n, then one or more equations are redundant. However, note that the system is sensitive to the condition number of the matrix B, and hence determining the polynomial such that a = (B) 1 y is not an efficient approach.

10 Fitting a polynomial to a data set Suppose we have the following tabulated data. Let us estimate f (3.0) by fitting a cubic polynomial to given data. Let p(x) = a 0 + a 1 x + a 2 x 2 + a 3 x 3 be a cubic polynomial. We get x = 3.2 a 0 + a 1 (3.2) + a 2 (3.2) 2 + a 3 (3.2) 3 = 22.0 x = 2.7 a 0 + a 1 (2.7) + a 2 (2.7) 2 + a 3 (2.7) 3 = 17.8 x = 1.0 a 0 + a 1 (1.0) + a 2 (1.0) 2 + a 3 (1.0) 3 = 14.2 x = 4.8 a 0 + a 1 (4.8) + a 2 (4.8) 2 + a 3 (4.8) 3 = 38.3 x y = f (x) Solving these equations, we get a 0 = , a 1 = , a 2 = , a 3 = Hence the polynomial is x x x 3.

11 The derivation of Lagrange polynomial The Lagrange interpolation does not require the solution of a linear system. Let us now derive a quadratic Lagrange interpolation formula. Consider a quadratic polynomial that passes through three points (x i, y i ), i = 0, 1, 2 and express this polynomial in the form l(x) = a 0 (x x 1 )(x x 2 ) + a 1 (x x 0 )(x x 2 ) + a 2 (x x 0 )(x x 1 ), where a 0, a 1, and a 2 are constants that have to be determined such that l(x) passes through given data points (x i, y i ), i = 0, 1, 2 such that l(x i ) = y i.

12 The derivation of Lagrange polynomial Therefore, we get y 0 = a 0 (x 0 x 1 )(x 0 x 2 ) y 1 = a 1 (x 1 x 0 )(x 1 x 2 ) y 2 = a 2 (x 2 x 0 )(x 2 x 1 ) Hence a 0 = a 1 = a 2 = y 0 (x 0 x 1 )(x 0 x 2 ) y 1 (x 1 x 0 )(x 1 x 2 ) y 2 (x 2 x 0 )(x 2 x 1 )

13 Using values of a 0, a 1, and a 2 in l(x), we get, + y 1 (x x 0 )(x x 2 ) (x 1 x 0 )(x 1 x 2 ) We can also write where L i (x) = l(x) = y 0 (x x 1 )(x x 2 ) (x 0 x 1 )(x 0 x 2 ) I (x) = 2 j=0,j i +y 2 (x x 0 )(x x 1 ) (x 2 x 0 )(x 2 x 1 ) 2 y i L i (x), i=0 [ x xj x i x j ], i = 0, 1, 2

14 Example Write out the Lagrangian polynomial from this table: Solution: Let where L i (x) = x y I (x) = 2 j=0,j i n y i L i (x), i=0 ( x xj x i x j ), i = 0, 1, 2

15 Example Therefore L(x) = (x 4.1)(x 7.1) 12.4 ( )( ) + (x 2.1)(x 7.1) 7.3 ( )( ) (x 2.1)(x 4.1) ( )( ) = x x

16 Example Verify that L(x) reproduces the y values for each x values. Solution We find that L(2.1) = 12.4 L(4.1) = 7.3 L(7.1) = 10.1 Example Using L(x) estimate y at x = 3 and at x = 8. Solution We find that L(3) = and L(8) =

17 Example Write out the cubic Lagrangian polynomial from this table: Solution: where Let L i (x) = x y I (x) = 2 j=0,j i n y i L i (x), i=0 ( x xj x i x j ), i = 0, 1, 2 (x 4)(x 3)(x 8) L(x) = 1 (2 4)(2 3)(2 8) (x 2)(x 3)(x 8) (x 2)(x 4)(x 8) (4 2)(4 3)(4 8) (3 2)(3 4)(3 8) (x 2)(x 4)(x 3) + 9 = 37 (8 2)(8 4)(8 3) 60 x x x 199 5