The Runge Phenomenon

4.2.6 The Runge Phenomenon

The remainder term in interpolation is, according to Theorem 4.2.3, equal to

Here ξ x depends on x, but one can say that the error curve behaves for the most part like 2

i =1 (x −x i ) . A similar curve is also typical for error curves arising from least squares approximation. This contrasts sharply with the error curve for

a polynomial curve y = c n

Taylor approximation, whose behavior is described approximatively by y = c(x − x 0 ) n . It is natural to ask what the optimal placing of the interpolation points x 1 ,...,x n

2 n = i =1 (x −x i ) in the

should be in order to minimize the maximum magnitude of M n (x)

interval in which the formula is to be used. For the interval [−1, 1] the answer is given directly by the minimax property (Lemma 3.2.4) of the Chebyshev polynomials—choose

M n (x) =T n (x)/ 2 n −1 .

Thus the interpolation points should be taken as the zeros of T n (x) . (In the case of an interval

1 [a, b] one makes the linear substitution x = 1

2 (a + b) + 2 (b − a)t.)

Example 4.2.8.

Use the same notations as before. For f (x) = x n the interpolation error becomes

f (x) − p(x) = M n (x) , because f (n) (x)/n ! ≡ 1. Figure 4.2.1 shows the interpolation error with n equidistant points on [−1, 1] and with n Chebyshev points on the same interval, i.e.,

2 respectively, for n = 6 and n = 12. The behavior of the error curves as shown in Figure 4.2.1

n −1

is rather typical for functions where f (n) (x) is slowly varying. Also note that the error increases rapidly when x leaves the interval int(x 1 ,x 2 ,...,x n ) . In the equidistant case, the error is also quite large in the outer parts of the interval.

Equidistant interpolation can give rise to convergence difficulties when the number of interpolation points becomes large. This difficulty is often referred to as Runge’s phe- nomenon , and we illustrate it in the following example. A more advanced discussion is given in Sec. 4.3.5, by means of complex analysis.

378 Chapter 4. Interpolation and Approximation

1 x 10

n for f (x) = x , using n = 12: Cheby- shev points (solid line) and equidistant points (dashed line).

Figure 4.2.1. Error of interpolation in P

Example 4.2.9.

The graph of the function

√ where i = −1, is the continuous curve shown in Figure 4.2.2, and is approximated in

two different ways by a polynomial of degree 10 in [−1, 1]. The dashed curve has been

Figure 4.2.2. Polynomial interpolation of 1/(1 + 25x 2 ) in two ways using 11 points: equidistant points (dashed curve), Chebyshev abscissae (dashed-dotted curve).

4.2. Interpolation Formulas and Algorithms 379 determined by interpolation on the equidistant grid with m = 11 points

(4.2.48) The dash-dot curve has been determined by interpolation at the Chebyshev points

x i = −1 + 2(i − 1)/(m − 1), i = 1 : m.

The graph of the polynomial obtained from the equidistant grid has—unlike the graph of

f —a disturbing course between the grid points. The agreement with f near the ends of

5 5 ] the agreement is fairly good. Such behavior is typical of equidistant interpolation of fairly high degree, and

1 , the interval is especially bad, while near the center of the interval [− 1

can be explained theoretically. The polynomial obtained from interpolation at Chebyshev points agrees much better with f , but still is not good. The function f is not at all suited for approximation by one polynomial over the entire interval. One would get a much better result using approximation with piecewise polynomials; see Sec. 4.4.

Notice that the difference between the values of the two polynomials is much smaller at the grid points of the equidistant grid than in certain points between the grid points, especially in the outer parts of the interval. This intimates that the values which one gets by equidistant interpolation with a polynomial of high degree can be very sensitive to disturbances in the given values of the function. Put another way, equidistant interpolation using polynomials of high degree is in some cases an ill-conditioned problem, especially in

the outer parts of the interval [x 1 ,x m ]. The effect is even worse if one extrapolates—i.e., if one computes values of the polynomial outside the grid. But equidistant interpolation

works well near the center of the interval. Even with equidistant data one can often get a more well behaved curve by—instead of interpolating—fitting a polynomial of lower degree using the method of least squares. Generally, if one chooses n < 2√m, then the polynomial fit is quite well-conditioned, but higher values of n should be avoided. 131

If one intends to approximate a function in [−1, 1] and one can choose the points at which the function is computed or measured, then one should choose the Chebyshev

points. Using these points, interpolation is a fairly well-conditioned problem in the entire interval. The risk of disturbing surprises between the grid points is insignificant. One can also conveniently fit a polynomial of lower degree than n−1, if one wishes to smooth errors in measurement; see Sec. 4.5.5.

Example 4.2.9 shows how important it is to study the course of the approximating curve p ∗ (x) between the points which are used in the calculation before one accepts the approximation . When one uses procedures for approximation for which one does not have a complete theoretical analysis, one should make an experimental perturbational calculation. In the above case such a calculation would very probably reveal that the interpolation polynomial reacts quite strongly if the values of the function are disturbed by small amounts, say ±10 −3 . This would give a basis for rejecting the unpleasant dashed curve in the example, even if one knew nothing more about the function than its values at the equidistant grid points.

131 This fact is related to the shape of the so-called Gram polynomials; see Sec. 4.5.5.

380 Chapter 4. Interpolation and Approximation

ReviewQuestions

4.2.1 Prove the theorem which says that the interpolation problem for polynomials has a unique solution.

4.2.2 When is linear interpolation sufficient?

4.2.3 Derive Newton’s interpolation formula.

4.2.4 Derive Newton’s interpolation formula for the equidistant case, starting from New- ton’s general interpolation formula. How is this formula easily remembered?

4.2.5 Derive the Lagrange interpolation formula. Show how it can be rewritten in barycen- tric form. When is the latter form more efficient to use?

4.2.6 Neville’s and Aitken’s interpolation algorithms both perform successive linear inter- polation. What is the difference between these?

4.2.7 The fast algorithm in Sec. 4.2.5 for solving primal Vandermonde systems can give surprisingly accurate results provided that the points x 1 ,x 2 ,...,x n satisfy certain conditions. Which?

4.2.8 (a) Why is it important to study the course of the approximating curve p ∗ (x) between the points which are used in the calculation before one accepts the approximation?

(b) What is a good choice of interpolation points in the interval [a, b], if one wants to get a small error?

Problems and Computer Exercises

4.2.1 (a) Compute f (3) by quadratic interpolation in the following table:

Use the points 1, 2, and 4, and the points 2, 4, and 5, and compare the results. (b) Compute f (3) by cubic interpolation.

4.2.2 Compute f (0) using one of the interpolation formulas treated above on the following table:

f (x) 0.64987 0.62055 0.56074 0.43609 The interpolation formula is here used for extrapolation. Use also Richardson ex-

trapolation and compare the results.

4.2.3 Work out the details of Example 4.2.3 (about divided differences for 1/(z − x)).

4.2.4 (a) Consider the two polynomials p(x) and q(x), both in P n , which interpolate f (x) at the points x

1 n ,...,x

n , and x 2 ,...,x n

+1 , respectively. Assume that {x i } i =1 is an increasing sequence, and that f (x) has constant sign in the interval [x 1 ,x n +1 ].

(n)

4.3. Generalizations and Applications 381 Show that f (x) is contained between p(x) and q(x) for all x ∈ [x 1 ,x n +1 ].

1 (x) and f 2 (x) have the same constant sign in [x 1 ,x n +1 ]. Formulate and prove a kind of generalization of the result in (a).

(b) Suppose that f (x) = f (n) 1 (x) −f 2 (x) , where both f

(n)

4.2.5 Using the barycentric formula (4.2.29) the interpolation polynomial can be written as

p(x) =

Show by taking f (x) ≡ 1 and equating the coefficients for x n −1 on both sides that the support coefficients satisfy n

i =1 w i = 0.