Lecture 9. Curve fitting. Interpolation. Lecture in Numerical Methods from 28. April 2015 UVT. Lecture 9. Numerical. Interpolation his o

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1 Curve fitting. Lecture in Methods from 28. April 2015 to ity Interpolation FIGURE A S Splines Piecewise relat UVT

2 Agenda of today s lecture 1 Interpolation Idea Splines Piecewise Interpolation to ity Interpolation FIGURE A S Splines Piecewise relat 1.2

3 Interpolation Idea Interpolation is stimation of values between well-known (or precisely-known) discrete points. to ity Interpolation FIGURE A S Splines Piecewise relat 1.3

4 Interpolation Idea Interpolation is stimation of values between well-known (or precisely-known) discrete points. We later discuss also extrapolation which is estimating values beyond the limits e observed data. to ity Interpolation FIGURE A S Splines Piecewise relat 1.3

5 Interpolation Idea Interpolation is stimation of values between well-known (or precisely-known) discrete points. We later discuss also extrapolation which is estimating values beyond the limits e observed data. What is the use of trend analysis? to ity Interpolation FIGURE A S Splines Piecewise relat 1.3

6 Example Say, for a better prediction e free-fall velocity e bungee jumper, we want to take into account also air density 406 POLYNOMIAL INTERPOLATION viscosity, which we can take from a well-known fluid mechanics book as: TABLE 17.1 Density (ρ), dynamic viscosity (μ), kintic viscosity (v) as a function of temperature (T ) at 1 atm as reported by White (1999). T, C ρ, kg/m 3 μ, N s/m 2 v, m 2 /s Can we estimate air density viscosity at 10 C? desired temperature based on the densities that bracket it. The simplest approach is to determine quation for the straight line connecting the two adjacent values use this equation to estimate the density at the desired intermediate temperature. Although such linear interpolation is perfectly adequate in many cases, error can be introduced n the data exhibit significant curvature. In this chapter, we explore a ber of different approaches for obtaining adequate estimates for such situations. to ity Interpolation FIGURE A S Splines Piecewise relat 17.1 INTRODUCTION TO INTERPOLATION 1.4

7 interpolation For n data points, there is only one polynomial of order (n 1) that passes through all the points f (x) = a 0 + a 1 x + a 2 x a n 1 x n 1. interpolation consists in determining the unique polynomial that fits the data. to ity Interpolation FIGURE A S Splines Piecewise relat 1.5

8 Interpolation: Example Rrks Determine the coefficients e parabola that passes through the last 3 density values e previous Table: x 1 = 300 y 1 = x 2 = 400 y 2 = x 3 = 500 y 3 = to ity Interpolation FIGURE A S Splines Piecewise relat 1.6

9 Interpolation: Example Rrks Determine the coefficients e parabola that passes through the last 3 density values e previous Table: x 1 = 300 y 1 = x 2 = 400 y 2 = x 3 = 500 y 3 = We should solve p 1 p 2 p 3 = Condition ber e matrix is in the l 2 norm. (So about 6 digits e solution are questionable!) As the ber of equations becomes larger, the condition ber increases we need alternative approaches!. to ity Interpolation FIGURE A S Splines Piecewise relat 1.6

10 Newton interpolation polynomial: linear version Simplest form of interpolation: 410 POLYNOMIAL connect INTERPOLATION 2 points with a straight line. Idea: determine the line connecting the 2 points; write it out as function of x: f 1 (x). f(x) f(x 2 ) f 1(x) f(x 1) x 1 x x 2 to ity Interpolation FIGURE A S FIGURE 17.2 Graphical depiction of linear interpolation. The shaded areas indicate the similar triangles used to derive the Newton linear-interpolation formula [Eq. (17.5)]. f 1 (x) = f (x 1 ) + f (x 2) f (x 1 ) (x x 1 ). x 2 x 1 Newton linear-interpolation formula. x approximation e first derivative [recall Eq. (4.20)]. In gen, the smaller the interval Splines Piecewise relat between the data points, the better the approximation. This is due to the fact that, as Interpolation the his o interval decreases, a continuous function be better approximated by a straight line. This characteristic is demonstd in the following example. EXAMPLE 17.2 Linear Interpolation Problem Statement. Estimate the natural logarithm of 2 using linear interpolation. First, perform the computation by interpolating between ln 1 = 0 ln 6 = Then, repeat the procedure, but use a smaller interval from ln 1 to ln 4 ( ). Note that the true value of ln 2 is

11 Example Estimate ln 2. Interpolate first between ln 1 ln 6 = Then use a smaller interval: from ln 1 to ln 4 = to ity Interpolation FIGURE A S Splines Piecewise relat 1.8

12 Newton interpolation polynomial: linear version we denote it f 1 (x), since it is a first order interpolation; to ity Interpolation FIGURE A S Splines Piecewise relat 1.9

13 Newton interpolation polynomial: linear version we denote it f 1 (x), since it is a first order interpolation; observe the finite difference fraction the formula f 1 (x) = f (x 1 ) + f (x 2) f (x 1 ) x 2 x 1 (x x 1 ); to ity Interpolation FIGURE A S Splines Piecewise relat 1.9

14 Newton interpolation polynomial: linear version we denote it f 1 (x), since it is a first order interpolation; observe the finite difference fraction the formula f 1 (x) = f (x 1 ) + f (x 2) f (x 1 ) x 2 x 1 (x x 1 ); the smaller the interval the more accu the approximation to ity Interpolation FIGURE A S Splines Piecewise relat 1.9

15 Newton interpolation polynomial: quadratic version Given 3 points, (x 1, f (x 1 )), (x 2, f (x 2 )), (x 3, f (x 3 )), the second-order interpolating polynomial f 2 (x) = b 1 + b 2 (x x 1 ) + b 3 (x x 1 )(x x 2 ). Fitting the known points, we find to ity Interpolation FIGURE A S Splines Piecewise relat 0

16 Newton interpolation polynomial: quadratic version Given 3 points, (x 1, f (x 1 )), (x 2, f (x 2 )), (x 3, f (x 3 )), the second-order interpolating polynomial f 2 (x) = b 1 + b 2 (x x 1 ) + b 3 (x x 1 )(x x 2 ). Fitting the known points, we find b 1 = f (x 1 ) b 2 = f (x 2) f (x 1 ) x 2 x 1 f (x 3 ) f (x 2 ) x b 3 = 3 x 2 f (x 2) f (x 1 ) x 2 x 1. x 3 x 1 to ity Interpolation FIGURE A S Splines Piecewise relat 0

17 Rrks first 2 terms in f 2 (x) are the same as in f 1 (x); to ity Interpolation FIGURE A S Splines Piecewise relat 1

18 Rrks first 2 terms in f 2 (x) are the same as in f 1 (x); check the finite differences fractions f 2 (x) = f (x 1 ) + f (x 2) f (x 1 ) x 2 x 1 (x x 1 )+ f (x 3 ) f (x 2 ) x 3 x 2 f (x 2) f (x 1 ) x 2 x 1 x 3 x 1 (x x 1 )(x x 2 ). to ity Interpolation FIGURE A S Splines Piecewise relat 1

19 Rrks x 1 = 1 f (x 1 ) = 0 x 2 = 4 f (x 2 ) = x 3 = 6 f (x 3 ) = first 2 terms in f 2 (x) are the same as in f 1 (x); check the finite Solution. differences Applying Eq. (17.7) fractions yields b 1 = 0 Equation (17.8) f 2 (x) = f (x 1 ) + f (x gives 2) f (x 1 ) (x x 1 )+ x 2 x 1 b 2 = = f 4(x 3 ) f 1 (x 2 ) x 3 x 2 f (x 2) f (x 1 ) x 2 x 1 (x x 1 )(x x 2 ). FIGURE 17.4 x 3 x 1 to ity Interpolation FIGURE acting on a The use of quadratic interpolation to estimate ln 2. The linear interpolation from x = 1 to 4Idea isforces also included for comparison. bungee free-falling A S f (x) 2 f(x) ln x True value Quadratic estimate Linear estimate 5 f 2 (x) x Splines Piecewise relat 1

20 for Newton s interpolating polynomial We can genize to fit an (n 1)th order polynomial to n data: f n 1 (x) = b 1 + b 2 (x x 1 ) b n (x x 1 )(x x 2 )... (x x n 1 ), re b 1, b 2,..., b n are given by b 1 = f (x 1 ) b 2 = f [x 2, x 1 ] b 3 = f [x 3, x 2, x 1 ]... b n = f [x n,..., x 2, x 1 ]. to ity Interpolation FIGURE A S Splines Piecewise relat 2

21 for Newton s interpolating polynomial We can genize to fit an (n 1)th order polynomial to n data: f n 1 (x) = b 1 + b 2 (x x 1 ) b n (x x 1 )(x x 2 )... (x x n 1 ), re b 1, b 2,..., b n are given by b 1 = f (x 1 ) b 2 = f [x 2, x 1 ] b 3 = f [x 3, x 2, x 1 ]... b n = f [x n,..., x 2, x 1 ]. to ity Interpolation FIGURE A S Splines Piecewise relat 2

22 Finite divided differences f [x i, x j ] = f (x i) f (x j ) x i x j f [x i, x j, x k ] = f [x i, x j ] f [x j, x k ] x i x k... f [x n,..., x 2, x 1 ] = f [x n,..., x 2 ] f [x n 1,..., x 1 ] POLYNOMIAL INTERPOLATION. x n x 1 x i x 1 x 2 x 3 x 4 f(x i ) f(x 1 ) f(x 2 ) f(x 3 ) f(x 4 ) First f [x 2, x 1 ] f [x 3, x 2 ] f [x 4, x 3 ] Second f [x 3, x 2, x 1 ] f [x 4, x 3, x 2 ] Third f [x 4, x 3, x 2, x 1 ] FIGURE 17.5 Graphical depiction e recursive nature of finite divided differences. This representation is referred to as a divided difference table. to ity Interpolation FIGURE A S Splines Piecewise relat 3

23 Linear Lagrange Interpolating Suppose the interpolating polynomial is a weighted sum e known values f (x 1 ) f (x 2 ): f (x) = L 1 f (x 1 ) + L 2 f (x 2 ), with L 1 = 1 at x 1, L 1 = 0 at x 2, L 2 = 1 at x 2, 0 at x 1. to ity Interpolation FIGURE A S Splines Piecewise relat 4

24 Linear Lagrange Interpolating Suppose the interpolating polynomial is a weighted sum e known values f (x 1 ) f (x 2 ): f (x) = L 1 f (x 1 ) + L 2 f (x 2 ), with L 1 = 1 at x 1, L 1 = 0 at x 2, L 2 = 1 at x 2, 0 at x 1. Then L 1 = x x 2 L 2 = x x 1, x 1 x 2 x 2 x 1 f 1 (x) = x x 2 x 1 x 2 f (x 1 ) + x x 1 x 2 x 1 f (x 2 ) is the linear Lagrange interpolating polynomial. to ity Interpolation FIGURE A S Splines Piecewise relat 4

25 Lagrange Interpolating Second order: f 2 (x) = (x x 2)(x x 3 ) (x 1 x 2 )(x 1 x 3 ) f (x 1) + (x x 1)(x x 3 ) (x 2 x 1 )(x 2 x 3 ) f (x 2)+ Genly re f n 1 (x) = L i (x) = (x x 1 )(x x 2 ) (x 3 x 1 )(x 3 x 2 ) f (x 3). n L i f (x i ), i=1 n j=1,j i re n is the ber of data points. x x j x i x j, to ity Interpolation FIGURE A S Splines Piecewise relat 5

26 on fitting a parabola through the first three known points. f (x) x 1 Interpolation x 2 True curve of interpolating polynomial t is a step in the unknown so divergence is expected! x 3 x to ity Interpolation FIGURE A S Splines Piecewise relat 6

27 on fitting a parabola through the first three known points. f (x) x 1 Interpolation x 2 True curve of interpolating polynomial it is a step in the unknown so divergence is expected! x 3 x to ity Interpolation FIGURE A S Splines Piecewise relat 6

28 Possible dangers of interpolation using high-order polynomials may be a trap, due to their oscillatory behavior; to ity Interpolation FIGURE A S Splines Piecewise relat 7

29 Possible dangers of interpolation using high-order polynomials may be a trap, due to their oscillatory behavior; in addition, high-order polynomials are ill-conditioned thus highly sensitive to round-off errors. to ity Interpolation FIGURE A S Splines Piecewise relat 7

30 Possible dangers of interpolation using high-order polynomials may be a trap, due to their oscillatory behavior; in addition, high-order polynomials are ill-conditioned thus highly sensitive to round-off errors. Runge s function f (x) = x 2. Interpolate the function on [ 1, 1] with polynomials of various order. to ity Interpolation FIGURE A S Splines Piecewise relat 7

31 Idea 430 SPLINES AND PIECEWISE INTERPOLATION f (x) Bypass the oscillatory behavior by employing low-order polynomials in a piecewise fashion, to subsets of data points. Such polynomials are called splines. Best use: n the function undergoes an abrupt change. 0 (a) f (x) 0 (b) f (x) 0 (c) f (x) x x x to ity Interpolation FIGURE A S 0 (d) x Splines Piecewise relat A visual representation of a situation re splines are superior to higher-ord polynomials. The function to be fit undergoes an abrupt increase at x = 0. P indicate that the abrupt change induces oscillations in interpolating polynom because it is limited to straight-line connections, a linear spline (d) provides FIGURE 18.1 acceptable approximation. 8