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Numerical Methods Course Notes Version 0.1 (UCSD Math 174, Fall ...

Numerical Methods Course Notes Version 0.1 (UCSD Math 174, Fall ...

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5.2. ERRORS IN POLYNOMIAL INTERPOLATION 67<br />

You should verify that along the top line of this pyramid you can read off the coefficients for<br />

Newton’s form, as found in Example Problem 5.4.<br />

⊣<br />

5.2 Errors in Polynomial Interpolation<br />

We now consider two questions:<br />

1. If we want to interpolate some function f(x) at n + 1 nodes over some closed interval, how<br />

should we pick the nodes?<br />

2. How accurate can we make a polynomial interpolant over a closed interval?<br />

You may be surprised to find that the answer to the first question is not that we should make<br />

the x i equally spaced over the closed interval, as the following example illustrates.<br />

Example 5.7 (Runge Function). Let<br />

f(x) = ( 1 + x 2) −1<br />

,<br />

(known as the Runge Function), and let p n (x) interpolate f on n equally spaced nodes, including<br />

the endpoints, on [−5, 5]. Then<br />

lim<br />

n→∞<br />

This behaviour is shown in Figure 5.3.<br />

max |p n(x) − f(x)| = ∞.<br />

x∈[−5,5]<br />

It turns out that a much better choice is related to the Chebyshev Polynomials (“of the first<br />

kind”). If our closed interval is [−1, 1], then we want to define our nodes as<br />

[( ) ] 2i + 1<br />

x i = cos<br />

π , 0 ≤ i ≤ n.<br />

2n + 2<br />

Literally interpreted, these Chebyshev Nodes are the projections of points uniformly spaced on<br />

a semi circle; see Figure 5.4. By using the Chebyshev nodes, a good polynomial interpolant of the<br />

Runge function of Example 5.7 can be found, as shown in Figure 5.5.<br />

5.2.1 Interpolation Error Theorem<br />

We did not just invent the Chebyshev nodes.<br />

follows from the following theorem:<br />

The fact that they are “good” for interpolation<br />

Theorem 5.8 (Interpolation Error Theorem). Let p be the polynomial of degree at most n<br />

interpolating function f at the n+1 distinct nodes x 0 , x 1 , . . . , x n on [a, b]. Let f (n+1) be continuous.<br />

Then for each x ∈ [a, b] there is some ξ ∈ [a, b] such that<br />

f(x) − p(x) =<br />

1<br />

(n + 1)! f (n+1) (ξ)<br />

n∏<br />

(x − x i ) .<br />

You should be thinking that the term on the right hand side resembles the error term in Taylor’s<br />

Theorem.<br />

i=0

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