1 / 20

By Jiaji Pan and Jason Hung

Roots of nonlinear Equations. By Jiaji Pan and Jason Hung. Root for Polynomials. Transfer function of a system Zeros : Poles :. Newton’s method. f(x) given function while and k < N compute , k = k + 1 End Root = . Value of Polynomial . Let

shlomo
Download Presentation

By Jiaji Pan and Jason Hung

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Roots of nonlinear Equations By Jiaji Pan and Jason Hung

  2. Root for Polynomials Transfer function of a system Zeros: Poles:

  3. Newton’s method f(x) given function while and k < N compute , k = k + 1 End Root =

  4. Value of Polynomial Let Rewrite this polynomial with a given z Where = Then at x = z

  5. Obtained the coefficients of and Comparing the coefficient = = for k = n-1,n-2,…0 The coefficients are computed recursively by knowing (the coefficients of ) Then the value of at a given value z is found ----

  6. Value of Polynomial From rewrite the polynomial with z: Where = =

  7. Obtained the coefficients of and Comparing the coefficient = = for k = n-2,n-3,…0 are computed by knowing (the coefficients of ) Then the value ofat a given z is found ----

  8. Combine Horner’s Method with Newton Method When we do the newton method iteration When f(x) = Now with the Horner’s method.

  9. Algorithm Given: The coefficients of polynomial: initial guess , tolerance, iteration index k = 0,j = 0 do { 1. z = , , 2.compute b for j = n-1,n-2,…,1 = = end 3. = 4. } While | > tolerance

  10. Advantage of this method More efficient additions: 2n vs 2n in newton raphson multiplications : 2n − 2 vs in newton raphson :Do not require to find the derivative of polynomial

  11. Solving for Complex Roots • Note that none of the methods: the Newton, the Secant, the Fixed Point iteration Methods, etc. cannot produce an approximation of a complex root starting from real approximations. • Muller’s Method

  12. Muller’s Method: Steps • Approximate x2 from x0 and x1 • Approximation of x2­ is computed as the x-intercept of the secant line passing through (x0,f(x0)) and (x1,f(fx1)) • Approximating x3 from x2, x0 and x1 • Having (x0,f(x0)), (x1,f(x1)) and (x2,f(x2)) at hand, next approximation x3 is computed as the x-intercept of the parabola passing through these three points.

  13. Muller’s Method: Steps • Computer x4 and successive approximations. (ie. xnis approximated using xn-1, xn-2, and xn-3)

  14. Muller’s Method: Math • Let the equation of the parabola P be: P(x) = a(x-x2)2+b(x-x2)+c • Since P(x) passes through (x0,f(x0)), (x1,f(x1)) and (x2,f(x2)), we have: P(x0) = f(x0) = a(x0 – x2)2 + b(x0 – x2) + c P(x1) = f(x1) = a(x1 – x2)2 + b(x1 – x2) + c P(x2) = f(x2) = c

  15. Muller’s Method: Math • Knowing c = f(x2), we can now obtain a and b by solving the first two equations: • From there we can obtain x3 by solving P(x3) = 0. P(x3) = a(x3 – x2)2 + b(x3 – x2) + c = 0

  16. Muller’s Method: Math • To solve for xn-xn-1: • Solving for x3:

  17. Muller’s Method: Example • Given equation: f(x) = x3-2x2-5 • Starting guesses: • x0= -1 • x1 = 0 • x2 = 1

  18. Muller’s Method: Example

  19. Muller’s Method: Example • Solutions: x = -0.345323724017634 +/- 1.318726779571698i x = 2.690647448028386 • Verify: f(x) = (x+0.34532+1.3187i)*(x+0.34532-1.3187i)*(x- 2.6906) = x3-1.99996x2-0.0000203916x+i(4.4E-16)-4.99971 = x3-2x2-5

More Related