1 Interpolation: What is it?

Transcription

1 Polynomial Interpolation KEY WORDS interpolation, polynomial interpolation, Lagrange from of interpolation, Newton form of interpolation GOAL To understand the significance of interpolation To understand various forms of polynomial interpolation Interpolation: What is it? In the problem of data approximation, we are given some discrete points (x 0, y 0 ), (x, y ),, (x N, y N ) () and are asked to find a function φ(x) that characterizes the trend of the data points For example, for the following data set: x x 0 = π x = 0 x = π y y 0 = 0 y = y = 0 the function φ(x) = π x satisfies the conditions φ(x i ) = y i for i = 0,,, and therefore may tell something about the trend of the data The trigonometric function φ = cos(x) also satises the conditions φ (x i ) = y i for i = 0,, and may also represent a certain trend of the data Obviously, these two functions behave quite differently for large values of x You may ask which one of the two curves is acceptable It is not clear what the correct answer is to this question because we know nothing else about the data It is possible that both are acceptable, or it might be the case that neither is Notice that one of the functions is a polynomial and the other is a trigonometric function Our objective in this unit is to illustrate how polynomials can be used to predict the trend of the data set This does not mean that polynomials will always do a better job than other functions such as trigonometric functions The procedure that we develop can be generalized to other functions without too much difficulty

2 The Interpolation Problem Suppose we wish to approximate a complicated function f(x) by a simpler function p(x) such that p(x) goes through the data {(x i, f i )} N i=0 (Throughout this discussion, we assume that the data are obtained by evaluating the function f(x); that is, we assume that f i = f(x i ), i = 0,,, N) This means that p(x i ) = f i, i = 0,,, N, and we say that p(x) interpolates the data This is a system of N + equations comprising the interpolating conditions In addition to replacing a complicated function by a simpler one, other purposes for interpolation include: plotting a smooth curve through discrete data points reading between the lines of a table approximating a mathematical function by a simpler one differentiating or integrating tabular data A common choice of the interpolating function p(x) is a polynomial because polynomials are simple in form can be calculated by elementary operations are free from singular points are unrestricted as to the range of values may be differentiated and integrated without difficulty and the coefficients to be determined enter linearly For special cases, however, other types of interpolants may be advantageous For example, if the underlying function is periodic, it is natural to employ trigonometric interpolation The discussion of polynomial interpolation in the following sections revolves around how an interpolating polynomial can be represented, computed, and evaluated Polynomial Interpolation The Power Series Form of the Interpolating Polynomial Consider the power series form of the polynomial p M (x) of degree M: p M (x) = a 0 + a x + + a M x M

3 If the polynomial interpolates the given data {(x i, f i )} N i=0 then the interpolating conditions form a system of N + linear equations p M (x 0 ) a 0 + a x a M x M 0 = f 0 p M (x ) a 0 + a x + + a M x M = f p M (x N ) a 0 + a x N + + a M x M N = f N for the M + coefficients a 0, a,, a M How should the degree M be chosen? Since the polynomial must satisfy N+ interpolation equations, we might expect to need N + parameters to satisfy the conditions If the interpolating polynomial is allowed to be of degree higher than N, there are many polynomials satisfying the interpolation conditions It is easy to show by example that if the degree is less than N, it may not be possible to satisfy all of the conditions Degree N is just right; then there is always a unique solution of the interpolation problem which we now demonstrate When M = N, the coefficient matrix of the linear system of interpolating conditions is the Vandermonde matrix x 0 x 0 x N 0 x x x N V N x N x N xn N It can be shown that the determinant of V N is det(v N ) = i>j(x i x j ) Note that det(v N ) 0, that is, V N is nonsingular, if and only if the nodes {x i } N i=0 are distinct; that is, if and only if x i x j whenever i j Thus when the nodes are distinct, there exist unique coefficients a 0, a,, a N (An alternative approach to showing that V N is nonsingular is the following Suppose that there is a vector c = [c 0, c,, c N ] T such that V N c = 0 It follows that the polynomial q(x) = c 0 + c x + c x + + c N x N is zero at x = x 0,, x N This says that we have a polynomial of degree N with N + roots The only way this can happen is if q is the zero polynomial, that is, when c = 0 Thus V N is nonsingular) EXAMPLE : Determine the power series form of the quadratic interpolating polynomial p (x) = a 0 + a x + a x interpolating the data (, 0), (0, ) and (, )

4 Solution: The interpolating conditions, in the order of the data, are p ( ) a 0 + ( )a + ()a = 0 p (0) a 0 = p () a 0 + ()a + ()a = On solving this linear system of equations, we obtain a 0 =, a = and a = So, in power series form, the interpolating polynomial is p (x) = + x + x Of course, polynomials can be written in many different ways, some more appropriate to a given task than others For example, when interpolating the data {(x i, f i )} N i=0, a Vandermonde system determines the values of the coefficients a 0, a,, a N in the power series form The number of floating point computations needed by Gauss elimination to solve this system grows like N with the degree N of the polynomial However, this cost can be significantly reduced by exploiting properties of the Vandermonde coefficient matrix Another way to reduce the cost of determining the interpolating polynomial is to change the way the interpolating polynomial is represented, as we shall see subsequently The Lagrange Form of the Interpolating Polynomial An interpolating polynomial can be written in various forms, the most common being the Lagrange form and the Newton form The Newton form is probably the most convenient and efficient; however, conceptually, the Lagrange form has several advantages We begin with the Lagrange form since it is easier to understand To illustrate the idea behind the Lagrange form, we consider the two points (x 0, f 0 ), (x, f ) with x 0 x The equation of the straight line through these points is or where p (x) = (x x)f 0 + (x x 0 )f (x x 0 ) (x x ) p (x) = f 0 (x 0 x ) + f (x x 0 ) (x x 0 ) l() 0 (x)f 0 + l () (x)f Note that each of the polynomials is a linear function, and l () 0 (x) = (x x ) (x 0 x ), l() (x) = (x x 0) (x x 0 ) l () 0 (x), l() (x) { l (), i = j i (x j ) = 0, i j 4

5 Consider now the data {(x i, f i )} i=0 The Lagrange form of the quadratic polynomial interpolating this data may be written: where p (x) = l () 0 (x)f 0 + l () (x)f + l () (x)f l () 0 = (x x )(x x ) (x 0 x )(x 0 x ), l() = (x x 0)(x x ) (x x 0 )(x x ), l() = (x x 0)(x x ) (x x 0 )(x x ) Clearly, l () 0, l(), l() are quadratic and l () i (x j ) = {, i = j 0, i j More generally, consider the polynomial p N (x) of degree N interpolating the data {(x i, f i )} N i=0 : N p N (x) = f i l (N) i (x) i=0 where the polynomials l (N) k (x) are polynomials of degree N and have the property that { l (N), k = j k (x j ) = 0, k j so that clearly In this case, p N (x i ) = f i, i = 0,,, N N l (N) (x x 0 )(x x ) (x x k )(x x k+ ) (x x N ) k (x) (x k x 0 )(x k x ) (x k x k )(x k x k+ ) (x k x N ) = j=0,j k The polynomial l (N) k is called a Lagrange basis polynomial The Lagrange basis polynomials are of exact degree N, but the interpolant can be of lower degree Indeed, it is important to appreciate that when f(x) is a polynomial of degree N, the interpolant must be f(x) itself After all, the polynomial f(x) interpolates itself and its degree is N, so, by uniqueness, p N (x) = f(x) ( x xj x k x j ) EXAMPLE : Find the cubic polynomial which interpolates the data (, ), (, ), (, ), (4, 4) Solution: First we calculate the Lagrange basis polynomials: Then we form the interpolating polynomial, p (x) = 6 (x )(x + )(x 4) (x )(x + )(x 4) 0 (x )(x )(x 4) + (x )(x )(x + ) 5 5

6 Unlike the power series form of the interpolating polynomial, the Lagrange form has no coefficients whose values must be determined In a sense, Lagrange interpolation provides an explicit solution of the interpolating conditions In summary, the Lagrange form of the interpolating polynomial is useful theoretically because it does not require solving a linear system explicitly shows how each data value f i affects the overall interpolating polynomial The Newton Form of the Interpolating Polynomial Consider the problem of determining the quadratic polynomial interpolating the data (x 0, f 0 ), (x, f ), (x, f ) Using the data written in this order, the Newton form of the quadratic interpolating polynomial is p (x) = b 0 + b (x x 0 ) + b (x x 0 )(x x ) In this case, the interpolating conditions are or p (x 0 ) b 0 = f 0 p (x ) b 0 + b (x x 0 ) = f p (x ) b 0 + b (x x 0 ) + b (x x 0 )(x x ) = f 0 0 (x x 0 ) 0 (x x 0 ) (x x 0 )(x x ) b 0 b b = Note that the coefficient matrix of this system is lower triangular When the nodes {x i } i=0 are distinct, the diagonal elements of this lower triangular matrix are nonzero Consequently, the linear system has a unique solution which may be determined by forward substitution In fact, f 0 f f b 0 = f 0, b = f b 0 (x x 0 ), b = f b 0 b (x x 0 ) (x x 0 )(x x ) () EXAMPLE : Determine the Newton form of the quadratic interpolating polynomial: p (x) = b 0 + b (x x 0 ) + b (x x 0 )(x x ) of the data (, 0), (0, ) and (, ) 6

7 Solution: Taking the data points in their order in the data, we have x 0 =, x = 0 and x = The interpolating conditions are p ( ) b 0 = 0 p (0) b 0 + (0 ( ))b = p () b 0 + ( ( ))b + ( ( ))( 0)b = and this lower triangular system may be solved to give b 0 = 0, b = and b = So, the Newton form of the quadratic interpolating polynomial is p (x) = 0 + (x + ) + (x + )(x 0) After rearrangement, we observe that this is the same polynomial as the power series form in Example (You may wish to verify this) EXAMPLE : For the data (, ), (, ) and (5, 4), using the data points in the order given, the Newton form of the interpolating polynomial is and the interpolating conditions are p (x) = b 0 + b (x ) + b (x )(x ) p () b 0 = p () b 0 + b ( ) = p (5) b 0 + b (5 ) + b (5 )(5 ) = 4 This lower triangular linear system may be solved by forward substitution giving b 0 = b = b 0 ( ) b = 4 b 0 (5 )b (5 )(5 ) = = 0 Consequently, the Newton form of the interpolating polynomial is p (x) = + (x ) + 0(x )(x ) Generally, this interpolating polynomial is of degree, but its exact degree may be less Here, rearranging into power series form, we have p (x) = + x and the exact degree is ; this happens because the data (, ), (, ) and (5, 4) are collinear 7

8 4 The Newton Divided Difference Interpolating Polynomial In this section, we derive the Newton form of the interpolating polynomial of degree N given the data {(x i, f i )} N i=0, and develop an efficient procedure for the determination of its coefficients Let p k (x) denote the polynomial of degree k interpolating the data {(x i, f i )} k i=0 We now try to determine a polynomial of degree k, P k, such that Since then () implies that p k (x) = p k (x) + P k (x) () p k (x i ) = f i, and p k (x i ) = f i, i = 0,,, k, P k (x i ) = p k (x i ) p k (x i ) = 0, i = 0,,, k Thus, for some constant b k, P k (x) = b k (x x 0 )(x x )(x x ) (x x k ) (4) Again using (), and adding these equations gives p k (x) p 0 (x) = P (x) + P (x) + + P k (x), where p 0 (x) = b 0 Thus, from (4), p k (x) = b 0 +b (x x 0 )+b (x x 0 )(x x )+ +b N (x x 0 )(x x )(x x ) (x x k ) To find the coefficients b i, we see that the coefficient of x k in p k (x) is b k, and since the interpolating polynomial is unique, b k must also be the coefficient of x k in the Lagrange form of the interpolating polynomial, p k (x) which is the coefficient of x k in that is, k i=0 b k = f i (x x 0 ) (x x i )(x x i+ ) (x x k ) (x i x 0 ) (x i x i )(x i x i+ ) (x i x k ) ; k i=0 f i (x i x 0 ) (x i x i )(x i x i+ ) (x i x k ) (5) We denote the right hand side of (5) by f[x 0, x,, x k ], the k-th divided difference of f at x 0, x,, x k Thus: b k = f[x 0, x,, x k ], k = 0,,, N (6) 8

9 Accordingly we have p N (x) = b 0 +b (x x 0 )+b (x x 0 )(x x )+ +b N (x x 0 )(x x )(x x ) (x x N ) (7) This is Newton s divided difference interpolating polynomial Note that by (5), f[x 0, x,, x k ] is symmetric in its arguments, that is f[x 0, x,, x i,, x j,, x k ] = f[x 0, x,, x j,, x i,, x k ] We now derive a recurrence relationship between the divided differences of f which is useful in computing high order differences Note that we could equally well have written p k (x) in the form p N (x) = c 0 +c (x x N )+c (x x N )(x x N )+ +c k (x x N )(x x N )(x x N ) (x x ), (8) where c i = f[x N, x N,, x N i ], i = 0,,, N (9) In particular, the coefficients of x N in (7) and(8) are identical, that is, b N = c N Equating (7) and(8), and taking all the terms to one side yields That is, or b N (x x )(x x ) (x x N ){(x x 0 ) (x x N )} + x N (b N c N ) + a polynomial of degree at most N 0 (0) b N (x x )(x x ) (x x N )(x N x 0 ) + (b N c N )x N + a polynomial of degree at most N 0, () {b N (x N x 0 )+(b N c N )}x N + a polynomial of degree at most N 0 Since the coefficient of x N is zero, we have from which it follows that b N (x N x 0 ) + b N c N = 0, that is, using (6) and (9) in (), we have b N = c N b N x N x 0 ; () f[x 0, x,, x N ] = f[x N, x N,, x ] f[x 0, x,, x N ] x N x 0, and using the symmetry property of divided differences, f[x N, x N,, x ] = f[x, x,, x N ] 9

10 and hence f[x 0, x,, x N ] = f[x, x,, x N ] f[x 0, x,, x N ] x N x 0 () In practice, (7) is used in conjunction with a divided difference table based on () x f[ ] f[, ] f[,, ] f[,,, ] x 0 f[x 0 ] f[x 0, x ] x f[x ] f[x 0, x, x ] f[x, x ] f[x 0, x, x, x ] x f[x ] f[x, x, x ] f[x, x ] x f[x ] The divided differences on the upper diagonal are the coefficients in (7) The first three divided differences are f[x i ] = f(x i ) (4) f[x i, x i+ ] = f[x i+] f[x i ] x i+ x i (5) f[x i, x i+, x i+ ] = f[x i+, x i+ ] f[x i, x i+ ] x i+ x i (6) EXAMPLE 4: Construct a divided difference table for the function f given in the following table, and determine the Newton divided difference interpolating polynomial x 0 5 f(x) 4 : Solution: The divided difference table is: x f[ ] f[, ] f[,, ] f[,,, ] From this table, we obtain 4 p (x) = + (x ) + (x )(x ) (x )(x )x 0

11 to In summary, for the Newton form of the interpolating polynomial, it is easy determine the coefficients evaluate the polynomial at specified values via nested multiplication extend the polynomial to incorporate additional interpolation points and data If the interpolating polynomial is to be evaluated at many points, generally it is best first to determine its Newton form and then to use the nested multiplication scheme to evaluate the interpolating polynomial at each desired point 5 Reading Assignments/References: About Lagrangian interpolation: follow the link org/wiki/lagrange_polynomial About Newton s interpolation: follow the link org/wiki/newton_polynomial