ERROR IN LINEAR INTERPOLATION

Transcription

1 ERROR IN LINEAR INTERPOLATION Let P 1 (x) denote the linear polynomial interpolating f (x) at x 0 and x 1, with f (x) a given function (e.g. f (x) = cos x). What is the error f (x) P 1 (x)? Let f (x) be twice continuously differentiable on an interval [a, b] which contains the points {x 0, x 1 }. Then for a x b, f (x) P 1 (x) = (x x 0) (x x 1 ) f (c x ) 2 for some c x between the minimum and maximum of x 0, x 1, and x. If x 1 and x are close to x 0, then f (x) P 1 (x) (x x 0) (x x 1 ) f (x 0 ) 2 Thus the error acts like a quadratic polynomial, with zeros at x 0 and x 1.

2 EXAMPLE Let f (x) = log 10 x; and in line with typical tables of log 10 x, we take 1 x, x 0, x For definiteness, let x 0 < x 1 with h = x 1 x 0. Then f (x) = log 10 e x 2 [ log ] 10 e cx [ 2 ] log10 e = (x x 0 ) (x 1 x) log 10 x P 1 (x) = (x x 0) (x x 1 ) 2 2c 2 x We usually interpolate with x 0 x x 1 ; in that case, (x x 0 ) (x 1 x) 0, x 0 c x x 1

3 (x x 0 ) (x 1 x) 0, x 0 c x x 1 and therefore [ ] log10 e (x x 0 ) (x 1 x) 2x1 2 log 10 x P 1 (x) [ ] log10 e (x x 0 ) (x 1 x) For h = x 1 x 0 small, we have for x 0 x x 1 [ ] log10 e log 10 x P 1 (x) (x x 0 ) (x 1 x) Typical high school algebra textbooks contain tables of log 10 x with a spacing of h =.01. What is the error in this case? To look at this, we use [ ] log10 e 0 log 10 x P 1 (x) (x x 0 ) (x 1 x) 2x 2 0 2x 2 0 2x 2 0

4 By a simple calculation, Therefore, max (x x 0 ) (x 1 x) = h2 x 0 x x log 10 x P 1 (x) h2 4 [ ] log10 e.= h x 2 0 If we want a uniform bound for all points 1 x 0 10, we have 0 log 10 x P 1 (x) h2 log 10 e 8 0 log 10 x P 1 (x).0543h 2 x 2 0. =.0543h 2 For h =.01, as is typical of the high school text book tables of log 10 x, 0 log 10 x P 1 (x)

5 In most tables, a typical entry is given to only four decimal places to the right of the decimal point, e.g. log =.7332 Therefore the entries are in error by as much as Comparing this with the interpolation error, we see the latter is less important than the rounding errors in the table entries. From the bound 0 log 10 x P 1 (x) h2 log 10 e 8x 2 0. =.0543 h2 x 2 0 we see the error decreases as x 0 increases, and it is about 100 times smaller for points near 10 than for points near 1.

6 THE QUADRATIC CASE For n = 2, we have f (x) P 2 (x) = (x x 0) (x x 1 ) (x x 2 ) f (3) (c x ) (*) 3! with c x some point between the minimum and maximum of the points in {x, x 0, x 1, x 2 }. To illustrate the use of this formula, consider the case of evenly spaced nodes: x 1 = x 0 + h, x 2 = x 1 + h Further suppose we have x 0 x x 2, as we would usually have when interpolating in a table of given function values (e.g. log 10 x). The quantity Ψ 2 (x) = (x x 0 ) (x x 1 ) (x x 2 ) can be evaluated directly for a particular x.

7 Graph of using (x 0, x 1, x 2 ) = ( h, 0, h): Ψ 2 (x) = (x + h) x (x h) y h h x

8 In the formula ( ), however, we do not know c x, and therefore we replace f (3) (c x ) with a maximum of f (3) (x) as x varies over x 0 x x 2. This yields f (x) P 2 (x) Ψ 2(x) 3! f max (3) (x) (**) x 0 x x 2 If we want a uniform bound for x 0 x x 2, we must compute max Ψ 2 (x) = max (x x 0 ) (x x 1 ) (x x 2 ) x 0 x x 2 x 0 x x 2 With a change of variables x = x 1 + t h, x x 0 = (1 + t) h, x x 1 = t h, x x 2 = (t 1) h max Ψ 2 (x) = h 3 x 0 x x 2 max 1 t 1 t t 3 = 2h3 3 3 Combined with ( ), this yields f (x) P 2 (x) h3 f 9 3 max (3) (x) x 0 x x 2 for x 0 x x 2.

9 For f (x) = log 10 x, with 1 x 0 x x 2 10, this leads to log 10 x P 2 (x) = h3 9 3 max 2 log 10 e x 0 x x 2 x h3 x0 3 For the case of h =.01, we have For comparison, log 10 x P 2 (x) x log 10 x P 1 (x)

10 Question: How much larger could we make h so that quadratic interpolation would have an error comparable to that of linear interpolation of log 10 x with h =.01? The error bound for the linear interpolation was , and therefore we want the same to be true of quadratic interpolation. Using a simpler bound, we want to find h so that log 10 x P 2 (x) h This is true if h = Therefore a spacing of h =.04 would be sufficient. A table with this spacing and quadratic interpolation would have an error comparable to a table with h =.01 and linear interpolation.

11 AN ERROR FORMULA: THE GENERAL CASE Recall the general interpolation problem: find a polynomial P n (x) for which deg(p n ) n P n (x i ) = f (x i ), i = 0, 1,, n with distinct node points {x 0,..., x n } and a given function f (x). Let [a, b] be a given interval on which f (x) is (n + 1)-times continuously differentiable; and assume the points x 0,..., x n, and x are contained in [a, b]. Then f (x) P n (x) = (x x 0) (x x 1 ) (x x n ) f (n+1) (c x ) (n + 1)! with c x some point between the minimum and maximum of the points in {x, x 0,..., x n }.

12 As shorthand, introduce Ψ n (x) = (x x 0 ) (x x 1 ) (x x n ) a polynomial of degree n + 1 with roots {x 0,..., x n }. Then f (x) P n (x) = Ψ n(x) (n + 1)! f (n+1) (c x ) When bounding the error we replace f (n+1) (c x ) with its maximum over the interval containing {x, x 0,..., x n }, as we have illustrated earlier in the linear and quadratic cases. We concentrate on understanding the size of Ψ n (x) or Ψ n (x) (n + 1)!

13 ERROR FOR EVENLY SPACED NODES We consider first the case in which the node points are evenly spaced, as this seems the natural way to define the points at which interpolation is carried out. Moreover, using evenly spaced nodes is the case to consider for table interpolation. For evenly spaced node points on [0, 1], For this case, h = 1 n, x 0 = 0, x 1 = h, x 2 = 2h,..., x n = nh = 1 Ψ n (x) = x (x h) (x 2h) (x 1)

14 Using the following table n M n n M n E E E E E E E E E E 12 we can observe that the maximum M n max x 0 x x n Ψ n (x) (n + 1)! becomes smaller with increasing n. Graphs of Ψ n (x) are shown next for the cases of n = 2,..., 9.

15 y n = 2 y n = 3 1 x 1 x y n = 4 y n = 5 1 x 1 x Figure: Graphs of Ψ n (x) on [0, 1] for n = 2, 3, 4, 5

16 y n = 6 y n = 7 1 x 1 x y n = 8 y n = 9 1 x 1 x Figure: Graphs of Ψ n (x) on [0, 1] for n = 6, 7, 8, 9

17 More detailed view the graph of with evenly spaced nodes: Ψ 6 (x) = (x x 0 ) (x x 1 ) (x x 6 ) x 0 x 1 x 2 x 3 x 4 x 5 x 6 x

18 What can we learn from the given graphs? Observe that there is enormous variation in the size of Ψ n (x) as x varies over [0, 1]; and thus there is also enormous variation in the error as x so varies. For example, in the n = 9 case, max x 0 x x 1 max x 4 x x 5 Ψ n (x) = (n + 1)! Ψ n (x) = (n + 1)! and the ratio of these two errors is approximately 49. Thus the interpolation error is likely to be around 49 times larger when x 0 x x 1 as compared to the case when x 4 x x 5. When doing table interpolation, the point x at which we interpolate should be centrally located with respect to the interpolation nodes {x 0,..., x n } being used to define the interpolation, if possible.

19 Discussion on Convergence Consider now the problem of using an interpolation polynomial to approximate a given function f (x) on a given interval [a, b]. In particular, take interpolation nodes a x 0 < x 1 < < x n 1 < x n b and produce the interpolation polynomial P n (x) that interpolates f (x) at the given node points. We would like to have Does it happen? Recall the error bound max f (x) P n(x) 0 as n a x b max f (x) P n(x) max a x b a x b Ψ n (x) (n + 1)! max a x b f (n+1) (x) We begin with an example using evenly spaced node points.

20 RUNGE S EXAMPLE Use evenly spaced node points: h = b a n, x i = a + ih for i = 0,..., n For some functions, such as f (x) = e x, the maximum error goes to zero quite rapidly. But the size of the derivative term f (n+1) (x) in max f (x) P n(x) max a x b a x b Ψ n (x) (n + 1)! max a x b f (n+1) (x) can badly hurt or destroy the convergence of other cases. In particular, we show the graph of f (x) = 1/ ( 1 + x 2) and P n (x) on [ 5, 5] for the cases n = 8 and n = 12. The case n = 10 is in the text on page 127. It can be proven that for this function, the maximum error on [ 5, 5] does not converge to zero. Thus the use of evenly spaced nodes is not necessarily a good approach to approximating a function f (x) by interpolation.

21 y y=p 10 (x) y=1/(1+x 2 ) x Figure: Runge s example with n = 10

22 OTHER CHOICES OF NODES Recall the general error bound max f (x) P n(x) max a x b a x b Ψ n (x) (n + 1)! max a x b f (n+1) (x) There is nothing we really do with the derivative term for f ; but we can examine the way of defining the nodes {x 0,..., x n } within the interval [a, b]. We ask how these nodes can be chosen so that the maximum of Ψ n (x) over [a, b] is made as small as possible.

23 There is an elegant solution to this problem, and it is taken up in 4.6. The node points {x 0,..., x n } turn out to be the zeros of a particular polynomial T n+1 (x) of degree n + 1, called a Chebyshev polynomial. These zeros are known explicitly, and with them max Ψ n(x) = a x b ( ) b a n+1 2 n This turns out to be smaller than for evenly spaced cases; and although this polynomial interpolation does not work for all functions f (x), it works for all differentiable functions and more. 2

24 ANOTHER ERROR FORMULA Recall the error formula f (x) P n (x) = Ψ n(x) (n + 1)! f (n+1) (c) Ψ n (x) = (x x 0 ) (x x 1 ) (x x n ) with c between the minimum and maximum of {x 0,..., x n, x}. A second formula is given by f (x) P n (x) = Ψ n (x) f [x 0,..., x n, x] To show this is a simple, but somewhat subtle argument. Let P n+1 (x) denote the polynomial of degree n + 1 which interpolates f (x) at the points {x 0,..., x n, x n+1 }. Then P n+1 (x) = P n (x) + f [x 0,..., x n, x n+1 ] (x x 0 ) (x x n )

25 Substituting x = x n+1, and using the fact that P n+1 (x) interpolates f (x) at x n+1, we have f (x n+1 ) = P n (x n+1 ) + f [x 0,..., x n, x n+1 ] (x n+1 x 0 ) (x n+1 x n ) In this formula, the number x n+1 is completely arbitrary, other than being distinct from the points in {x 0,..., x n }. To emphasize this fact, replace x n+1 by x throughout the formula, obtaining f (x) = P n (x) + f [x 0,..., x n, x] (x x 0 ) (x x n ) = P n (x) + Ψ n (x) f [x 0,..., x n, x] provided x x 0,..., x n. This formula is also easily true for x a node point provided f (x) is sufficiently differentiable (to insure the existence of the divided difference f [x 0,..., x n, x]). This shows f (x) P n (x) = Ψ n (x) f [x 0,..., x n, x] (1)

26 Compare the two error formulas: Then f (x) P n (x) = Ψ n (x) f [x 0,..., x n, x] = f [x 0,..., x n, x] = f (n+1) (c) (n + 1)! Ψ n(x) (n + 1)! f (n+1) (c) for some c between the smallest and largest of the numbers in {x 0,..., x n, x}. To make this symmetric in its arguments, let x = x n+1, m = n + 1. Then f [x 0,..., x m 1, x m ] = f (m) (c) (2) m! with c an unknown number between the smallest and largest of the numbers in {x 0,..., x m }. This was given in the preceding section where divided differences were introduced.