BSM510 Numerical Analysis

Transcription

1 BSM510 Numerical Analysis Polynomial Interpolation Prof. Manar Mohaisen Department of EEC Engineering

2 Review of Precedent Lecture Polynomial Regression Multiple Linear Regression Nonlinear Regression

3 Lecture Content Introduction to Interpolation Newton Interpolation Polynomial Lagrange Interpolation Polynomial Extrapolation and Oscillations 3

4 Introduction to Interpolation Interpolation Estimate intermediate values between precise data points Polynomial interpolation Using n data points, find an (n-1)th-order polynomial that passes by all the points Matlab format f() x = a + a x + a x + L + a xn n f x = px + p x + + p x + p () n 1 n L 1 n 1 n 4

5 Introduction to Interpolation Determining Polynomial Coefficients Example nd order polynomial f() x = px + p x + p 1 3 x f(x) p(300) + p (300) + p = p(400) + p (400) + p = p(500) + p (500) + p = , p , p = , p y p p = p x 5

6 Introduction to Interpolation Drawback of the precedent algorithm Vandermonde matrix 90, A = 160, cond( A) = , Matrices of this type are very ill-conditioned! Look for alternative methods! % file: nd order polynomial interpolation format long x =[ ]; y = [ ]; p = polyfir(x, y, ); % find intermediate value >> d = polyval(p, 350) d =

7 Newton Interpolation Polynomial Linear Interpolation f ( x) f( x ) f( x ) f( x ) f( x ) f( x ) = f () x = f( x ) + ( x x ) x x x x 1 1 x x Example (log() = ) Find log() using log(1) and log(6) Find log() using log(1) and log(4) f () = 0 + ( 1) = f () = 0 + ( 1) = y x 7

8 Newton Interpolation Polynomial Quadratic Interpolation f () x = b + b ( x x ) + b ( x x )( x x ) Substitute with x = x 1 f ( x ) = b 1 1 Substitute x = x b = f( x ) f( x ) 1 x x 1 Finally, b 3 = f( x ) f( x ) f( x ) f( x ) 3 1 x x x x 3 1 x x 3 1 8

9 Newton Interpolation Polynomial Quadratic Interpolation Example x f(x) = b 1 b f( x ) f( x ) = 1 x x = b 3 f( x ) f( x ) f( x ) f( x ) 3 1 x x x x = 3 1 x x = Then, f () ( 1) ( ( 1)( 4) x = x x x f () = 9

10 Newton Interpolation Polynomial General Form of Newton s Interpolating Polynomials f ( x) = b + b ( x x ) + L+ b ( x x )( x x ) L( x x ) 1 1 n 1 n 1 With b = f( x ) 1 1 b = f[ x, x ] 1 f( x ) f( x ) fx [, x] = i j i j x x i j b = f[ x, x, x ] 3 1 b 3 = f[ x, x ] f[ x, x ] 3 1 x x 3 1 b = f[ x, x, L, x, x ] n n n

11 Newton Interpolation Polynomial Graphical depiction Finite divided difference Divided difference table x i f(x i ) First Second Third x 1 f(x 1 ) f[x,x 1 ] f[x 3,x,x 1 ] f[x 4, x 3,x,x 1 ] x f(x ) f[x 3,x ] f[x 4, x 3,x ] x 3 f(x 3 ) f[x 4,x 3 ] x 4 f(x 4 ) 11

12 Newton Interpolation Polynomial Example Estimate log() with a third-order Newton s interpolating polynomial f ( x) = b + b ( x x ) + b ( x x )( x x ) + b ( x x )( x x )( x x ) x i f(x i ) First Second Third f 3 () x = ( x 1) ( ( x 1)( x 4) ( x 1)( x 4)( x 6) f () =

13 Lagrange Interpolation Polynomial Linear Lagrange Interpolation Polynomial x x x x f ( x ) = L f ( x ) + L f ( x ) = f ( x ) + f ( x ) x x 1 x x 1 1 nd-order Lagrange Interpolation Polynomial ( x x )( x x ) ( x x )( x x ) ( x x )( x x ) f () x = f( x ) + f( x ) + f( x ) ( x x )( x x ) 1 ( x x )( x x ) ( x x )( x x ) (n-1)th-order Lagrange Interpolation Polynomial n n i i i x = 1 = 1 i x x f () x L()( x f x ), L() x n 1 = = x i j i j j 13

14 Lagrange Interpolation Polynomial Example Estimate f(15) Linear Interpolation x i f(x i ) x x x x fx ( ) = x x fx ( ) + x x fx ( ) = = nd-order interpolation ( x x )( x x ) ( x x )( x x ) ( x x )( x x ) f () x = 3 f( x ) f( x ) + 1 f( x ) ( x x )( x x ) 1 ( x x )( x x ) ( x x )( x x ) (15 0)(15 40) (15 0)(15 40) (15 0)(15 0) = = (0 0)(0 40) (0 0)(0 40) (0 0)(40 0) 14

15 Inverse Interpolation Given f(x), we need to find x f(x) x Solution 1: switch the values of x and f(x) The values of the new x is not guaranteed to be evenly spaced This leads to oscillations in the obtained function Solution : maintain x and f(x) Obtain the polynomial as a function of x Solve the root of the equation Example: f ( x ) = x Find x such that f(x) = = x x = ,

16 Extrapolation and Oscillation Extrapolation Estimating a value of f(x) that lies outside the range of known points x i 16

17 Dangers of Extrapolation Example Interpolate the first 8 points using 7-th order polynomial Find f(x) at 000 using the interpolation function Result: f(000) =

18 Oscillations Interpolation The more is the better? (higher order polynomial is better than lower order polynomials?) Not usually true: higher order polynomials are ill-conditioned (sensitive) Example: Runge s function f() x = x x 18

19 Lecture Summary Introduction to Interpolation Newton Interpolation Polynomial Lagrange Interpolation Polynomial Extrapolation and Oscillations 19