Scientific Computing. Roots of Equations

Transcription

1 ECE257 Numerical Methods and Scientific Computing Roots of Equations

2 Today s s class: Roots of Equations Polynomials

3 Polynomials A polynomial is of the form: ( x) = a 0 + a 1 x + a 2 x 2 +L+ a n x n f n The roots of a polynomial follow these rules: There will be n roots to an n-th order polynomial. The roots may be real or complex and need not be distinct If complex roots exist, they exist in conjugate pairs (λ+µi i and λ-µi) If n is odd, there is at least one real root

4 Polynomials Uses in electrical engineering Solving for poles and zeros in transfer functions Solving characteristic equations in linear ordinary differential equations

5 Polynomials f n ( x) = a 0 + a 1 x + a 2 x 2 +L+ a n x n F=A[0] FOR I=1 to N K = A[I] FOR J=1 to I K = K * X ENDFOR F = F + K ENDFOR n( n +1) 2 n additions multiplications

6 Polynomials f n ( ( )) ( x) = a 0 + x a 1 + x a 2 +L+ xa n F=A[N] FOR I=N-1 to 0 F = F*X + A[I] ENDFOR n additions n multiplications

7 Polynomial Deflation Given a polynomial and a single known root, deflating the polynomial can reduce the order of the polynomial and remove possible redundant roots Example: x 4 8x 3 20x x 576 Root x=4 is known

8 Polynomial Deflation x 4 x 3 4 x 2 36x +144 x 4 8x 3 20x x 576 x 4 4x 3 4x 3 20x 2 4 x 3 +16x 2 36x x 36x x 144 x x 576 0

9 Polynomial Deflation Synthetic Division x 4 8x 3 20x x

10 Polynomial Deflation Synthetic Division x 3 4x 2 36x

11 Polynomial Deflation Synthetic Division a 0 + a 1 x + a 2 x 2 +L+ a n x n = ( b 0 + b 1 x + b 2 x 2 +L+ b n 1 x n 1 )( x t) + r r=a[n] b[n]=0 for i=n-1 to 0 b[i]=r r = r*t+a[i] endfor n additions n multiplications

12 Polynomial Deflation If the known root is a calculated root, approximation error in that root may be compounded during the deflation process With forward deflation, i.e. new polynomial coefficients are calculated in descending order, it is better to divide by roots of smallest absolute value first Round-off errors can also be reduced by using root polishing.

13 Roots of polynomials Previous methods (Newton-Raphson, bracketing, etc.) have a few problems when applied to polynomials Must determine an initial guess of the root Can not find complex roots May not converge Polynomial-specific methods Müller s s method Bairstow s s method

14 Müller s s method Similar idea to secant method Instead of projecting a straight line through two points to estimate the root, project a parabola through three points to estimate the root

15 Müller s method From Numerical Methods for Engineers, Chapra and Canale, Copyright The McGraw-Hill Companies, Inc.

16 Müller s s method Find a parabola f 2 ( x) = a( x x 2 ) 2 + b( x x 2 ) + c such that it intersects the function at three points x 0, x 1, and x 2 f ( x 0 ) = f 2 ( x 0 ) = a( x 0 x 2 ) 2 + b( x 0 x 2 ) + c f ( x 1 ) = f 2 ( x 1 ) = a( x 1 x 2 ) 2 + b( x 1 x 2 ) + c f ( x 2 ) = f 2 ( x 2 ) = a( x 2 x 2 ) 2 + b( x 2 x 2 ) + c

17 Müller s s method Solve for c c = f ( x 2 ) Substitute back into set of equations f ( x 0 ) f ( x 2 ) = a( x 0 x 2 ) 2 + b( x 0 x 2 ) f ( x 1 ) f ( x 2 ) = a( x 1 x 2 ) 2 + b( x 1 x 2 ) Define new set of variables h 0 = ( x 1 x 0 ) h 1 = ( x 2 x 1 ) ( ) f ( x 0 ) ( ) f ( x 1 ) δ 0 = f x 1 x 1 x 0 δ 1 = f x 2 x 2 x 1

18 Müller s s method Substitute new variables back into equations Solve for a and b h 0 δ 0 + h 1 δ 1 = b( h 0 + h 1 ) a( h 0 + h 1 ) 2 h 1 δ = bh 1 ah 1 2 a = δ 1 δ 0 h 0 + h 1 b = ah 1 + δ 1

19 Müller s s method Now that we have found the coefficients of the parabola, find where it intersects the x-axis to get the next root estimate The intersection point with the x-axis is simply the root of the parabola which we can find using the quadratic formula f 2 ( x r ) = a x r x 2 ( ) 2 + b( x r x 2 ) + c = 0 x r x 2 = b ± b2 4ac 2a = b ± 2c b 2 4ac

20 Müller s s method x r = x 2 + b ± 2c b 2 4ac Which of the two quadratic roots do you choose? Choose the sign that makes the denominator the largest That will bring x r closer to x 2

21 Müller s s method How do you pick the three points for the next iteration? If there are only real roots, choose x r and the closer two of the orignal three roots to x r. If there are complex roots, use a sequential approach: use x r, x 2, and x 1.

22 Example Find roots of f(x)=x 3-13x-12 Use x 0 =4.5, x 1 =5.5, and x 2 =5 as the initial guesses ( ) = f ( 4.5) = ( ) = f ( 5.5) = ( ) = f ( 5) = 48 f x 0 f x 1 f x 2 ( ) = =1.0 ( ) f ( x 0 ) h 0 = x 1 x 0 δ 0 = f x 1 = x 1 x = ( ) = = 0.5 ( ) f ( x 1 ) h 1 = x 2 x 1 δ 1 = f x 2 = x 2 x = 69.75

23 Example x r = x 2 + = a = δ 1 δ 0 h 0 + h 1 = b = ah 1 + δ 1 = c = f x =15 ( ) = ( ) = 48 2c b ± b 2 4ac = ( ) ε a = x x r 2 = x r = 25.74%

24 Example i x r ε a (%)

25 Bairstow s s method Based on dividing the polynomial by (x-t) where t is the root estimate If there is no remainder, we found the root If not, we adjust the guess

26 Bairstow s s method f ( x) = a 0 + a 1 x + a 2 x 2 +L+ a n x n Factor by x-t f ( x) = ( x t) ( b 1 + b 2 x + b 3 x 2 +L+ b n x n 1 ) + b 0 Factor by x 2 -rx-s f ( x) = x 2 rx s b n = a n b i = a i + b i+1 t for i = n 1 to 0 ( )( b 2 + b 3 x + b 4 x 2 +L+ b n x n 2 ) + b 1 (x r) + b 0

27 Bairstow s s method b n = a n We have to find values of r and s such that the remainder is 0 The remainder is 0 if b 1 =b 0 =0 b 1 and b 0 are both functions of r and s, so we can give a Taylor series expansion of both b 1 ( r + Δr,s + Δs) b 1 ( r,s) + b 1( r,s) Δr + b 1( r,s) Δs r s b 0 ( r + Δr,s + Δs) b 0 ( r,s) + b 0( r,s) Δr + b 0( r,s) Δs r s b n 1 = a n 1 + rb n b i = a i + rb i+1 + sb i+2 for i = n 2 to 0

28 Bairstow s s method 0 = b 1 r,s 0 = b 0 r,s c n = b n ( ) + b 1( r,s) r ( ) + b 0( r,s) r Δr + b 1( r,s) c 2 Δr + c 3 Δs = b 1 r,s c 1 Δr + c 2 Δs = b 0 Δs s Δr + b 0( r,s) Δs s ( ) ( r,s) c n 1 = b n 1 + rc n c i = b i + rc i+1 + sc i+2 for i = n 2 to 1

29 Bairstow s s method Solve for r r and s ( ) ( r,s) c 2 Δr + c 3 Δs = b 1 r,s c 1 Δr + c 2 Δs = b 0 Use the results to adjust r and s and iterate Stop when the approximate error in r and s is sufficient ε a,r = Δr r and ε a,s = Δs s

30 Example Find roots of f(x)=x 5-3.5x x x x+1.25 Use s=-1, r=-1 as the initial guesses b 5 = a 5 =1 b 4 = a 4 + rb 5 = 4.5 b 3 = a 3 + rb 4 + sb 5 = 6.25 b 2 = a 2 + rb 3 + sb 4 = b 1 = a 1 + rb 2 + sb 3 = b 0 = a 0 + rb 1 + sb 2 = c 5 = b 5 =1 c 4 = b 4 + rc 5 = 5.5 c 3 = b 3 + rc 4 + sc 5 =10.75 c 2 = b 2 + rc 3 + sc 4 = c 1 = b 1 + rc 2 + sc 3 =

31 Example Solve linear system ( ) ( r,s) c 2 Δr + c 3 Δs = b 1 r,s c 1 Δr + c 2 Δs = b Δr Δs = Δr 4.875Δs = Δr = Δs = New s and r r = 1+ Δr = s = 1+ Δs =

32 Example Use s=0.1381, r= b 5 = a 5 =1 b 4 = a 4 + rb 5 = b 3 = a 3 + rb 4 + sb 5 = b 2 = a 2 + rb 3 + sb 4 = b 1 = a 1 + rb 2 + sb 3 = b 0 = a 0 + rb 1 + sb 2 = c 5 = b 5 =1 c 4 = b 4 + rc 5 = c 3 = b 3 + rc 4 + sc 5 = c 2 = b 2 + rc 3 + sc 4 = c 1 = b 1 + rc 2 + sc 3 =

33 Example Solve linear system ( ) ( r,s) c 2 Δr + c 3 Δs = b 1 r,s c 1 Δr + c 2 Δs = b Δr Δs = Δr Δs = Δr = Δs = New s and r r = Δr = s = Δs =

34 Example Keep going until s=0.5 and r=-0.5 Then solve the following: x 2 rx s = 0 x 2 + x = 0 x = 0.5 ± ( 0.5) ( 0.5) 2 = 0.5,1.0

35 Example The quotient coefficients are as follows b 5 = a 5 =1 b 4 = a 4 + rb 5 = 4 b 3 = a 3 + rb 4 + sb 5 = 5.25 b 2 = a 2 + rb 3 + sb 4 = -2.5 b 1 = a 1 + rb 2 + sb 3 = 0 b 0 = a 0 + rb 1 + sb 2 = 0 Now we have to solve: f ( x) = x 3 4 x x 2.5 = 0

36 Bairstow s s method If no s or r are known, start with s=0, r=0 Converges fast May diverge

37 Roots of equations Bisection 2 initial guesses, slow convergence, will not diverge False-position 2 initial guesses, slow-medium convergence, will not diverge

38 Roots of equations Newton-Raphson 1 initial guess, fast convergence, may diverge Secant 2 initial guesses, medium-fast convergence, may diverge

39 Roots of equations Muller 3 initial guesses, medium-fast convergence, may diverge, only polynomials, complex roots Bairstow 2 initial guesses, medium-fast convergence, may diverge, only polynomials, complex roots

40 Next Lecture Linear algebraic equations Read Chapter PT3, 9 HW3 due 9/30 Exam 1 on Tuesday 10/04