Piecewise polynomial interpolation

Transcription

1 Chapter 2 Piecewise polynomial interpolation In ection.6., and in Lab, we learned that it is not a good idea to interpolate unctions by a highorder polynomials at equally spaced points. However, it transpires that it is possible to obtain a very good approximations using a very simple method. The trick is to use a spline: a piecewise polynomial interpolating unction. We ll consider three important example o splines:. linear splines 2. (natural) cubic splines. 3. Hermite piecewise cubics. In this section, we always have N equally spaced points: let h = (b a)/n, then a = x, b = x N and x i = x + ih. Oten these are reerred to as knots points (or simply as knots), and denote the set o knot points by ω N := {x i } N i=. For more details about splines, have a look at [M3, Chap. ], and [98, Lectures and ]. 2.. Construction Deinition 2... Let be a unction that is continuous on [a, b]. The linear spline interpolant to is the continuous unction l such that (i) l(x i ) = (x i ) or each i =,,..., N, (ii) l is a linear unction l i on each interval [x i, x i ]. It is easy to write down a ormula or the l i, based on Lagrange polynomials. et h = (b a)/n. Then, or x [x i, x i ] l i (x) = (x i ) x i x h + (x i ) x x i. (2.) h Example Write down the linear spline interpolant to (x) = e x at the knot points {,, }. 2. Linear Interpolating plines We irst study the piecewise linear interpolant, also called a linear spline. We will see that they have important properties, including (a) they are easy to construct and analyse; (b) the bound on the error decreases as the number o interpolation points increases; (c) the error we get using a linear spline is no more than twice the error using the best possible (piecewise linear) approximation; amd (d) o all the interpolants to at a given set o points, the linear spline is the one with the smallest st derivative Analysis We know that i p N is the polynomial o degree N that interpolates at N equally spaced points, it does not ollow that p N as n. But as we will see, the piecewise linear interpolant to converges to, albeit slowly. This is veriied in the ollowing theorem, which is a direct consequence o Theorem.3.3.

2 2 Theorem uppose that, and are all continuous and deined on the interval [a, b]. Let l be the linear spline interpolant to on the N + equally spaced points a = x < x < x N = b with h = x i x i = (b a)/n. Then l h2 8, (Here, as usual, g is deined as max a x b g(x).) (x) l 4 (x) 5 5 (x).9 l 8 (x) (x).9 l 6 (x) It now ollows directly rom Theorem 2..3 that lim l =. n Example Figure 2. shows linear spline interpolations o Runge s example: (x) = on [ 5, 5]. + x2 These diagrams appear to support our assertion that the error tends to zero as n. Example uppose you are interpolating (x) = e x on N equally spaced intervals between x = and x N =. What value o N would you have to take to ensure that the maximum error is less than 2? 5 5 Fig. 2.: Linear spline interpolants to /( + x 2 ) on [ 5, 5]. Contrast with Figure Best approximation For the next part o the analysis it will help to think o piecewise linear interpolation as an operator. Then we can compare the linear spline to all the other piecewise linear approximations. First, observe that one can deine an ininite number o piecewise linear unctions on a given set o knot points ω N. We ll call the set o these unctions L. Deinition For a ixed set o knot points ω N, let L be the operator that maps the continuous unction to its linear spline interpolant l L. Now suppose that g L. Then L(g) = g. That is L is a projection: L(L()) = L(). It is not hard to see that one could ind a dierent unction ˆl L that is a better approximation o in sense that max (x) ˆl(x) < max (x) l(x). x x x n x x x n However, l is very easy to ind, and the associated error is no worse than twice (x) ˆl(x). Theorem 2..7 (tewart s Aternotes goes to grad school,

3 2 Lecture ). Let l = L(). For all ˆl L, l 2 ˆl. (That L is a projection is key to the proo.) 2..5 Exercises Exercise 2.. Page 28 o the Department o Education s old Mathematics Tables ( The Log Tables ) reports that ln() =, ln(.5) =.455 and ln(2) =.693. (i) Write down the linear spline l that interpolates (x) = ln(x) at the points x =, x =.5 and x 2 = 2. (ii) Use this to estimate ln(x) at x =.2. How does this compare to the value in the tables? (.823) (iii) Give an estimate or the maximum error: max (x) l(x). x 2 (iv) What value o n would you choose to ensure that (x) l(x). or all x [, 2] Minimum Energy The inal interesting property o l that we will study is called the minimum energy property. Deinition Let u be a unction that is continuous and deined on the interval [a, b] except, maybe, at the (countable set) ω N o knot points Then the 2-norm o u is u 2,[a,b] := ( b u 2 (x)dx ) /2. Usually we just write this as u 2. Let H be the set o all unctions u that are continuous on [a, b] and have u 2 <. Note that l H, even though we have not properly deined l at the mesh points ω N. Theorem 2..9 (üli and Mayers, Thm..2). Let w be any unction in H that interpolates the unction at the points in ω N. Let l we the linear spline interpolant o. Then l 2 w 2. a Exercise 2.2. As an alternative to (2.), one can deine the linear spline interpolant to a unction is as a linear combination o a set o piecewise linear basis unctions {ψ i } N i= : { i = j ψ i (x j ) = i j (i) Write down a ormula or the ψ i (x); (ii) derive a ormula or l(x) in terms o the ψ i. This exercise is useul: we ll be using these basis unctions (called hat unctions) in the inal section o the course. ψ i (x) ψ i (x) ψ i+ (x) x i x i x i+ Fig. 2.2: ome hat unctions More precisely, we should say everywhere, except on a set o measure zero. What that means is very important to the areas o measure theory and unctional analysis. However, since some o you are not taking those courses until next year, we ll air-brush over the exact deinition.

4 Cubic plines The Cubic pline 2 is perhaps the most useul and popular interpolating unction used in numerical analysis, automotive and aeronautical engineering, computer and ilm animation, digital photography, inancial modelling, and as many other areas as there are human endeavours where continuous processes are modelled by discrete ones. One could argue that the most undamental shortcoming o the linear spline interpolant to (see, or example, Figure 2.) is that they are not smooth : that is, although l i (x i ) = l i+ (x i ), it is not generally the case that l i (x i) = l i+ (x i). Also, the interpolating unction cannot capture the curvature o. (I you think about this last statement, you ll see that it can be expressed as l (x) = or all x [a, b].) Cubic splines try to balance the simplicity o the linear spline approach by using a low-order polynomial on each interval with the desire or curvature and more smoothness by using cubics, and by orcing adjacent ones, and their irst and second derivatives, to agree at knot points. Deinition Let be a unction that is continuous on [a, b]. The cubic spline interpolant to is the continuous unction such that (i) or i =,..., N, on each interval [x i, x i ] let (x) = s i (x), where each o the s i is a cubic polynomial. (ii) s i (x i ) = (x i ) or i =,..., N, (iii) s i (x i ) = (x i ) or i =,..., N, (iv) s i (x i) = s i+ (x i) or i =,..., N, (v) s i (x i) = s i+ (x i) or i =,..., N. o, we have deined the cubic spline as a unction that interpolates at N + points, has continuous irst and second derivatives on [x, x N ] and is a cubic polynomial on each o the n intervals [x i, x i ]. That is, it is piecewise cubic. We know that one can write a cubic as p 3 (x) = a + a x + a 2 x 2 + a 3 x 3. Thus it takes 4 terms to uniquely deine a single cubic. To uniquely determine N cubics requires 4N terms. They can be ound by solving 4N (linearly independent) equations. But an inspection o (ii) (iv) above reveals only 4N 2 equations. 2 It is generally accepted that (mathematical) splines were irst described by Isaac choenberg (93-99) during WW2. However their physical realisation had been used in the ship-building and aircrat industries prior to that. The missing equations can be chosen in a number o ways: (i) by setting (x ) = and (x N ) =. This is called a natural spline, and is the approach we ll take. (ii) by setting (x ) = and (x N ) =. called a clamped spline. This is (iii) set (x ) = (x N ) and (x ) = (x N ). This is the periodic spline and is used or interpolating, say, trigonometric unctions. (iv) only use N 2 components o the spline: s 2,..., s N. But extend to two end ones so that s 2 (x ) = (x ) and s N = (x N ). This is called the not-aknot condition. In Figure 2.3 we show the natural cubic spline interpolants to (x) = /( + x 2 ) (with a = 5 and b = 5 as usual) on N = 4, 8 and 6 intervals. Compared with Figure 2., we seem to get signiicantly better approximation. Moreover, the rate at which the error decreases with respect to h is much aster..5 Cubic pline on 4 intervals Cubic pline on 8 intervals Cubic pline on 6 intervals Fig. 2.3: Cubic spline interpolants to /( + x 2 ). Compare with Figure 2.

5 Constructing Cubic plines ome notation: ˆ h = x i x i = (x N x )/N or all i. ˆ i := (x i ). To construct the spline, irst observe that i is piecewise cubic, then is piecewise quadratic, and is piecewise linear. (i) Let σ i = (x i ) or i =,..., N. and so, or i =, 2,..., N α i = i h h 6 σ i, Notice that this gives β i = α i. β i = i h h 6 σ i. (2.3) (iv) o, now we just need the equations or σ i. Two o these come rom the act that this is a natural cubic spline, with (x ) = and (x N ) =, thereore σ = and σ N =. The remaining N equations come rom the act that is continuous at x, x 2,..., x N. o we set that s i (x i ) = s i+ (x i ). Ater some work (see Exercise 2.3), we can show this means that the system is: σ =, (2.4a) 6 (σ i + 4σ i + σ i+ ) = h 2 ( i 2 i + i+ ) (2.4b) or i =,..., N, (ii) Integrate twice: σ N = A cubic spline example (2.4c) Example Find the natural cubic spline interpolant to at the points x =, x =, x 2 = 2 and x 3 = 3 where =, = 2, 2 = and 3 =. [We won t do all the details in class, so here they are] olution: From (2.2) we see we are looking or (x)= We get s i (x) = α i (x x i ) + β i (x i x)+ σ i 6h (x i x) 3 + σ i 6h (x x i ) 3, (2.2) or x [x i, x i ]. (iii) The α i and β i arose as the constants o integration. To ind them: α x + β ( x) + σ 6h ( x)3 + σ 6h x3,x [, ], α 2 (x ) + β 2 (2 x) + σ 6h (2 x)3 + σ 2 6h (x )3,x [, 2], α 3 (x x 2 ) + β 3 (3 x) + σ 2 6h (3 x)3 + σ 3 6h (x 2)3,x [2, 3]. We irst solve or the σ i using (2.4): σ 4 σ 8 = 6 This gives 4 σ 2 σ 3 σ =, σ = 24 5, σ 2 = 6 5, σ 3 =. Now use (2.3) to get α = 4 5, α 2 = 4 5, α 3 =. and β = β 2 = 4 5, β 3 = 4 5.

6 24 The answer is 4 x x3, x [, ], 4 4 (x ) + (2 x) 5 5 (x) = 4 5 (2 x)3 + (x 5 )3, x [, 2], 4 (3 x) + (3 5 5 x)3. x [2, 3]. This is shown in Figure 2.4. Example What is the smallest value o N that you must take to ensure that, i interpolating (x) = e x on N equally sized intervals, the error is less that 8? How does this compare with a linear spline interpolation (see Example 2..5)? (x i,y i ) Fig. 2.4: pline interpolant rom Example Error Estimates We state the ollowing error estimates without proo. You don t need to know these, or be able to prove them. But you should be able to use them i required. Theorem I C 4 [a, b] and is its cubic spline interpolant on N + equally spaced points, then where M 4 = (iv) M 4h 4, 24 M 4h 3, 3 8 M 4h 2, Example Give an upper bound on the error or the cubic spline interpolant to = e x on the interval [, ] with N = mesh points. Recall Theorem 2..9 or linear splines. There is an analogous result or natural cubic splines. Theorem Let u be any unction that interpolates at ω N, and is such that u H 2 (x, x N ). Then 2 u 2. The proo is not hard - it is analogous to the proo o Theorem Exercises Exercise 2.3. When deducing the system o equations or the natural cubic spline, we showed how to construct the ormulation given in (2.2) and the relationship between σ i, α i and β i in (2.3). Now careully show to deduce the system (2.4). Exercise 2.4. (For students who did MA385). Write the equations in (2.4) as a linear system Aσ = b, where A is an n n matrix. how that A is nonsingular, and hence that the system as a unique solution. Exercise 2.5. Find the natural cubic spline interpolant to (x) = sin(πx/2) at the nodes {x i } 3 i= = {,, 2, 3}. Calculate value o the interpolant at x = 2.5. What is the error at this point? Exercise 2.6. Take (x) = ln(x), x =, x N = 2. What value o N would you have to take to ensure that ln(x) (x) 4 or all x [, 2]?

7 Piecewise Hermite Interpolation In this section we introduce another type o cubic spline. It is not as smooth as the natural spline o the precious section only it and its irst derivative are continuous on [x, x N ] and it requires that we know (x i ). But its easier to construct (don t have to solve a linear system) and analyse than that natural spline. As beore, or short-hand, we write (x i ) as i and (x i ) as i. And, as ever, we ll simpliy the analysis by taking the points to be equally spaced: x i x i = h or each i) PCHIP Deinition Given a set o interpolation points x < x < < x N, the Piecewise Cubic Hermite pline Interpolant (PCHIP),, to the unction, satisies (i) C [x, x N ], (ii) (x i ) = (x i ) and (x i ) = (x i ) or i =,,..., N. (iii) On each interval [x i, x i ], is a cubic polynomial. We can construct these splines as ollows. On each interval [x i, x i ], let be the cubic polynomial i given by i (x) = c + c (x x i )+ c 2 (x x i ) 2 + c 3 (x x i ) 3, (2.5) Then we can show... Figure 2.5 shows some PCHIP interpolants to (x) = /( + x 2 ) on the interval [ 5, 5]. Compare with Figures 2. and PCHIP pline on 4 intervals PCHIP pline on 8 intervals PCHIP pline on 6 intervals Fig. 2.5: PCHIP interpolants to /( + x 2 ). Compare with Figure 2.3 We now want to prove an error estimate. The norm we use is := max x x x N (x) (x) Theorem Let C 4 [x, x N ] and let be the Hermite Cubic pline interpolant to it at the N equally spaced points a = x < x < < x N = b. Then h4 384 (iv). This gives that c = i, c = i, c 2 = 3 h 2 ( i i ) h ( i + 2 i ), c 3 = h 2 ( i + i ) 2 h 3 ( i i ).

8 Estimating derivatives You can ignore this section. It is included or completeness. It should make sense to anyone who took MA385, given that it relies only on Taylor series. As we deined the PCHIP interpolant one needs to know in order to be able to construct it. However, you might notice that the MATLAB unction >> pchip only requires, not. How does it do this, one might wonder? It turns out that there are important variants on the PHCIP scheme that don t involve knowing,,..., N, but instead uses approximations or. There are several approaches or deriving these estimates, including (i) Geometrically. Recall, all we are really looking or is the slope o the tangent to at x = x i. (ii) By inding a polynomial that interpolates and dierentiating that. (iii) Taylor s Theorem. For more details see [96, Lecture 24]. Or look back at your notes or MA385, Lecture 4. i lope is (x i ) (x i ) x i x i i lope is (x = lim i ) (x i σ) σ σ lope is i+ x i x i x i+ Fig. 2.6: Estimating (x i )dx As suggested by Figure 2.6 one could take (x i ) i i x i x i = h ( i i ) (x i+ ) (x i ) x i+ x i Or (x i ) i+ i = x i+ x i h ( i+ i ). ince one seems to over estimate (x i ), and the other seems to under-estimate, we could take the average: (x i ) i + i+ x i+ x i = 2h ( i + i ) These approximations can also be easily derived rom Taylor s Theorem: or some η (a, b). Using this we can get i+ = i + h (x i ) + h2 2 (x i ) + h3 6 (τ ), i = i h (x i ) + h2 2 (x i ) h3 6 (τ 2 ), or τ (x i, x i+ ) and τ 2 (x i, x i ). Now subtract the st equation rom the second to get the ormula and error estimate Exercises Exercise 2.7. Recall Exercise 2.5. Calculate the value to the PCHIP interpolant to (x) = sin(πx/2) at the nodes {x i } 3 i= = {,, 2, 3} at the point x = 2.5. What is the error at this point? Exercise 2.8. Let (x) = ln(x). Let l and be the piecewise linear and Hermite cubic spline interpolants (respectively) to on n+ equally spaced points = x < x < < x N = 2. What value o n would you have to take to ensure that (i) max x 2 (x) l(x) 4? (ii) max x 2 (x) (x) 4? Also, using the code you developed in Lab 2, compare these theoretical results with the value needed on practice. Exercise 2.9. There are ways o constructing the PCHIP, other than (2.5). For example, let s = x x k, then (x) = h3 3hs 2 + 2s 3 h 3 k + 3hs2 2s 3 h 3 k + s(s h) 2 h 2 k + s2 (s h) h 2 k how that this is the same as the PCHIP. (x) = (a) + (x a) (x a)2 (a) + (a)+ 2 (x a) 3 (x a)n (a) + + (n) (a)+ 3! n! (x a) n+ (n+) (η), (n + )!