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factored form as (x x1) ?? ?? ?? (x xn) where the xi are the Chebyshev nodes as described

 

 

 

in Theorem 3.8.

 

Chebyshev’s theorem follows directly from these facts.

 

 

Proof of Theorem 3.6. Let Pn(x) be a monic polynomial with an even smaller absolute

 

maximum on [−1,1]; in other words, |Pn(x)| < 1/2n1 for 1 x 1. This assumption

 

leads to a contradiction. Since Tn(x) alternates between 1 and 1 a total of n + 1 times

 

(Fact 6), at these n + 1 points the difference Pn Tn/2n1 is alternately positive and

 

negative. Therefore, Pn Tn/2n1 must cross zero at least n times; that is, it must have

 

at least n roots. This contradicts the fact that, because Pn and Tn/2n1 are monic, their

 

difference is of degree n 1.

 

3.3.3 Change of interval

 

So far our discussion of Chebyshev interpolation has been restricted to the interval [−1,1],

 

 

 

because Theorem 3.6 is most easily stated for this interval. Next, we will move the whole

 

 

methodology to a general interval [a,b].

 

The base points are moved so that they have the same relative positions in [a,b] that

 

they had in [−1,1]. It is best to think of doing this in two steps: (1) Stretch the points by the

 

factor (b a)/2 (the ratio of the two interval lengths), and (2) Translate the points by

 

(b + a)/2 to move the center of mass from 0 to the midpoint of [a,b]. In other words,

 

 

 

move from the original points

 

cos

 

 

odd π

 

2n

 

 

 

to

 

 

b a

 

 

 

2

 

cos

 

 

odd π

 

2n

 

+ b + a

 

 

 

2

 

 

.

 

 

With the new Chebyshev base points x1, . . . , xn in [a,b], the corresponding upper

 

 

 

bound on the numerator of the interpolation error formula is changed due to the stretch by

 

 

(b a)/2 on each factor x xi . As a result, the minimax value 1/2n1 must be replaced

 

by [(b a)/2]n/2n1.

 

 

 

 

Chebyshev interpolation nodes

 

 

 

On the interval [a,b],

 

 

xi = b + a

 

 

 

2

 

 

+ b a

 

 

 

2

 

cos

 

 

(2i 1

 

2n

 

for i = 1,...,n. The inequality

 

|(x x1) ?? ?? ?? (x xn)| ≤

 

 ba

 

 

 

2

 

 

n

 

2n1 (3.14)

 

holds on [a,b].

 

 

 

The next example illustrates the use of Chebyshev interpolation in a general interval.

 

 

EXAMPLE 3.11 Find the four Chebyshev base points for interpolation on the interval [0,π/2], and find an

 

upper bound for the Chebyshev interpolation error for f (x) = sin x on the interval.

 



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