> 4. Interpolation and Approximation Most functions cannot be evaluated exactly: √ x, ex, ln x, trigonometric functions since by using a computer we are limited to the use of elementary arithmetic operations +, −, ×, ÷ With these operations we can only evaluate polynomials and rational functions (polynomial divided by polynomials). 4. Interpolation Math 1070 > 4. Interpolation and Approximation Interpolation Given points x0, x1, . , xn and corresponding values y0, y1, . , yn find a function f(x) such that f(xi) = yi, i = 0, . , n. The interpolation function f is usually taken from a restricted class of functions: polynomials. 4. Interpolation Math 1070 > 4. Interpolation and Approximation > 4.1 Polynomial Interpolation Theory Interpolation of functions f(x) x0, x1, . , xn f(x0), f(x1), . , f(xn) Find a polynomial (or other special function) such that p(xi) = f(xi), i = 0, . , n. What is the error f(x) = p(x)? 4. Interpolation Math 1070 > 4. Interpolation and Approximation > 4.1.1 Linear interpolation Linear interpolation Given two sets of points (x0, y0) and (x1, y1) with x0 6= x1, draw a line through them, i.e., the graph of the linear polynomial x − x1 x − x0 x0 x1 `(x) = y0 + y1 x0 − x1 x1 − x0 y0 y1 (x − x)y + (x − x )y `(x) = 1 0 0 1 (5.1) x1 − x0 We say that `(x) interpolates the value yi at the point xi, i = 0, 1, or `(xi) = yi, i = 0, 1 Figure: Linear interpolation 4. Interpolation Math 1070 > 4. Interpolation and Approximation > 4.1.1 Linear interpolation Example Let the data points be (1, 1) and (4,2). The polynomial P1(x) is given by (4 − x) · 1 + (x − 1) · 2 P (x) = (5.2) 1 3 √ The graph y = P1(x) and y = x, from which the data points were taken. √ Figure: y = x and its linear interpolating polynomial (5.2) 4. Interpolation Math 1070 > 4. Interpolation and Approximation > 4.1.1 Linear interpolation Example Obtain an estimate of e0.826 using the function values e0.82 =· 2.270500, e0.83 =· 2.293319 Denote x0 = 0.82, x1 = 0.83. The interpolating polynomial P1(x) x interpolating e at x0 and x1 is (0.83 − x) · 2.270500 + (x − 0.82) · 2.293319 P (x) = (5.3) 1 0.01 and P1(0.826) = 2.2841914 while the true value s e0.826 =· 2.2841638 to eight significant digits. 4. Interpolation Math 1070 > 4. Interpolation and Approximation > 4.1.2 Quadratic Interpolation Assume three data points (x0, y0), (x1, y1), (x2, y2), with x0, x1, x2 distinct. We construct the quadratic polynomial passing through these points using Lagrange’s folmula P2(x) = y0L0(x) + y1L1(x) + y2L2(x) (5.4) with Lagrange interpolation basis functions for quadratic interpolating polynomial (x−x1)(x−x2) L0(x) = (x0−x1)(x0−x2) (x−x0)(x−x2) L1(x) = (5.5) (x1−x0)(x1−x2) (x−x0)(x−x1) L2(x) = (x2−x0)(x2−x1) Each L (x) has degree 2 ⇒ P (x) has degree ≤ 2. Moreover i 2 Li(xj) = 0, j 6= i 1, i = j for 0 ≤ i, j ≤ 2 i.e., Li(xj) = δi,j = Li(xi) = 1 0, i 6= j the Kronecker delta function. P2(x) interpolates the data P2(x) = yi, i=0,1,2 4. Interpolation Math 1070 > 4. Interpolation and Approximation > 4.1.2 Quadratic Interpolation Example Construct P2(x) for the data points (0, −1), (1, −1), (2, 7). Then (x−1)(x−2) x(x−2) x(x−1) P (x)= · (−1)+ · (−1)+ · 7 (5.6) 2 2 −1 2 Figure: The quadratic interpolating polynomial (5.6) 4. Interpolation Math 1070 > 4. Interpolation and Approximation > 4.1.2 Quadratic Interpolation With linear interpolation: obvious that there is only one straight line passing through two given data points. With three data points: only one quadratic interpolating polynomial whose graph passes through the points. Indeed: assume ∃Q2(x), deg(Q2) ≤ 2 passing through (xi, yi), i = 0, 1, 2, then it is equal to P2(x). The polynomial R(x) = P2(x) − Q2(x) has deg(R) ≤ 2 and R(xi) = P2(xi) − Q2(xi) = yi − yi = 0, for i = 0, 1, 2 So R(x) is a polynomial of degree ≤ 2 with three roots ⇒ R(x) ≡ 0 4. Interpolation Math 1070 > 4. Interpolation and Approximation > 4.1.2 Quadratic Interpolation Example Calculate a quadratic interpolate to e0.826 from the function values e0.82 =· 2.27050 e0.83 =· 2.293319 e0.84 =· 2.231637 0.82 0.83 0.84 With x0 = e , x1 = e , x2 = e , we have · P2(0.826) = 2.2841639 to eight digits, while the true answer e0.826 =· 2.2841638 and · P1(0.826) = 2.2841914. 4. Interpolation Math 1070 > 4. Interpolation and Approximation > 4.1.3 Higher-degree interpolation Lagrange’s Formula Given n + 1 data points (x0, y0), (x1, y1),..., (xn, yn) with all xi’s distinct, ∃ unique Pn, deg(Pn) ≤ n such that Pn(xi) = yi, i = 0, . , n given by Lagrange’s Formula n X Pn(x) = yiLi(x) = y0L0(x) + y1L1(x) + ··· ynLn(x) (5.7) i=0 n Y x−xj (x−x0) ··· (x−xi−1)(x−xi+1) ··· (x−xn) where Li(x)= = , xi −xj (xi −x0) ··· (xi −xi−1)(xi −xi) ··· (xi −xn) j=0,j6=i where Li(xj) = δij 4. Interpolation Math 1070 > 4. Interpolation and Approximation > 4.1.4 Divided differences Remark; The Lagrange’s formula (5.7) is suited for theoretical uses, but is impractical for computing the value of an interpolating polynomial: knowing P2(x) does not lead to a less expensive way to compute P3(x). But for this we need some preliminaries, and we start with a discrete version of the derivative of a function f(x). Definition (First-order divided difference) Let x0 6= x1, we define the first-order divided difference of f(x) by f(x1) − f(x2) f[x0, x1] = (5.8) x1 − x2 If f(x) is differentiable on an interval containing x0 and x1, then the mean value theorem gives 0 f[x0, x1] = f (c), for c between x0 and x1. Also if x0, x1 are close together, then x + x f[x , x ] ≈ f 0 0 1 0 1 2 usually a very good approximation. 4. Interpolation Math 1070 > 4. Interpolation and Approximation > 4.1.4 Divided differences Example Let f(x) = cos(x), x0 = 0.2, x1 = 0.3. Then cos(0.3) − cos(0.2) f[x , x ] = =· −0.2473009 (5.9) 0 1 0.3 − 0.2 while x + x f 0 0 1 = − sin(0.25) =· −0.2474040 2 so f[x0, x1] is a very good approximation of this derivative. 4. Interpolation Math 1070 > 4. Interpolation and Approximation > 4.1.4 Divided differences Higher-order divided differences are defined recursively Let x0, x1, x2 ∈ R distinct. The second-order divided difference f[x1, x2] − f[x0, x1] f[x0, x1, x2] = (5.10) x2 − x0 Let x0, x1, x2, x3 ∈ R distinct. The third-order divided difference f[x1, x2, x3] − f[x0, x1, x2] f[x0, x1, x2, x3] = (5.11) x3 − x0 In general, let x0, x1, . , xn ∈ R, n + 1 distinct numbers. The divided difference of order n f[x1, . , xn] − f[x0, . , xn−1] f[x0, . , xn] = (5.12) xn − x0 or the Newton divided difference. 4. Interpolation Math 1070 > 4. Interpolation and Approximation > 4.1.4 Divided differences Theorem n Let n ≥ 1, f ∈ C [α, β] and x0, x1, . , xn n + 1 distinct numbers in [α, β]. Then 1 (n) f[x0, x1, . , xn] = n! f (c) (5.13) for some unknown point c between the maximum and the minimum of x0, . , xn. Example Let f(x) = cos(x), x0 = 0.2, x1 = 0.3, x2 = 0.4. The f[x0, x1] is given by (5.9), and cos(0.4) − cos(0.3) f[x , x ] = =· −0.3427550 1 2 0.4 − 0.3 hence from (5.11) −0.3427550 − (−0.2473009) f[x , x , x ] =· = −0.4772705 (5.14) 0 1 2 0.4 − 0.2 For n = 2, (5.13) becomes 1 1 1 f[x , x , x ] = f 00(c) = − cos(c) ≈ − cos(0.3) =· −0.4776682 0 1 2 2 2 2 which is nearly equal to the result in (5.14). 4. Interpolation Math 1070 > 4. Interpolation and Approximation > 4.1.5 Properties of divided differences The divided differences (5.12) have special properties that help simplify work with them. (1) Let (i0, i1, . , in) be a permutation (rearrangement) of the integers (0, 1, . , n). It can be shown that f[xi0 , xi1 , . , xin ] = f[x0, x1, . , xn] (5.15) The original definition (5.12) seems to imply that the order of x0, x1, . , xn is important, but (5.15) asserts that it is not true. For n = 1 f(x0) − f(x1) f(x1) − f(x0) f[x0, x1] = = = f[x1, x0] x0 − x1 x1 − x0 For n = 2 we can expand (5.11) to get f(x0) f(x1) f(x2) f[x0, x1, x2] = + + (x0 − x1)(x0 − x2) (x1 − x0)(x1 − x2) (x2 − x0)(x2 − x1) (2) The definitions (5.8), (5.11) and (5.12) extend to the case where some or all of the xi coincide, provided that f(x) is sufficiently differentiable. 4. Interpolation Math 1070 > 4. Interpolation and Approximation > 4.1.5 Properties of divided differences For example, define f(x0) − f(x1) f[x0, x0] = lim f[x0, x1] = lim x1→x0 x1→x0 x1 − x0 0 f[x0, x0] = f (x0) For an arbitrary n ≥ 1, let all xi → x0; this leads to the definition 1 f[x , .