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Chapter 4 Roots of Polynomials PowerPoint PPT Presentation


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Chapter 4 Roots of Polynomials. Objectives. Understand the importance of finding polynomial roots in engineering applications Know the conventional method concept Know the Muller’s method concept Know the Bairstow’s method. Content. Polynomials in engineering and science

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Chapter 4 Roots of Polynomials

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Chapter 4 roots of polynomials

Chapter 4

Roots of Polynomials


Chapter 4 roots of polynomials

Objectives

  • Understand the importance of finding polynomial roots in engineering applications

  • Know the conventional method concept

  • Know the Muller’s method concept

  • Know the Bairstow’s method


Chapter 4 roots of polynomials

Content

  • Polynomials in engineering and science

  • Conventional method

  • Muller’s method

  • Bairstow’s method

  • Conclusions


Polynomials in 1

Polynomials in… (1)

General solutions of linear ODE

Solve for general solution

Change to characteristic equations:

The results can be :-


Polynomials in 2

Polynomials in…(2)

General solutions of linear ODE


Polynomials in 3

Polynomials in…(3)

  • Problem :

  • Follow these rules:

  • For an nth order equation, there are n real or complex roots.

  • If n is odd, there is at least one real root.

  • If complex roots exist, they will be in conjugate pairs (that is, l+mi and l-mi), where i=sqrt(-1).


Conventional

Conventional…

  • Only real roots exist

    However,

    • Finding good initial guesses complicates both the open and bracketing methods, also the open methods could be susceptible to divergence.

  • Real and complex roots of polynomials – Müller and Bairstow methods.


M ller method 1

Müller method (1)

  • Like Secant, Müller’s method obtains a root estimate by projecting a parabola to the x axis through three function values.


M ller method 2

Müller method (2)

Secant method

(linear approximation)


M ller method 3

Müller method (3)

Müller method

(Parabola or 2nd order

approximation)

Must use three points to approximate function


M ller method 4

Müller method (4)

Müller methodology derivation

  • Write the equation in a convenient form at point x2:

  • We then have three eqns now (from x0, x1, and x2)


M ller method 5

Müller method (5)

  • Step III Reduce to two eqns

Right now u can solve for a and b from

When u know a, b, c you are ready to

estimate root from


M ller method 6

Müller method (6)

  • Step IV Here the new estimated root is

Two roots, but which one ?

  • Error can be derived from


M ller method 7

Müller method (7)

  • Summary of algorithm

Start with 3 points [x0,f(x0)] [x1,f(x1)] and [x2,f(x2)]

Calculate a, b, and c from


M ller method 8

Müller method (8)

  • Summary of algorithm (cont’d)

Calculate new root from

Calculate error

Check whether

new xi-1 = old xi


Bairstow s method 1

Bairstow’s method (1)

  • An iterative approach loosely related to both Müller and Newton Raphson methods.

  • Based on dividing a polynomial by a factor x-t:

Start with

Dividing with x-t yields

and a remainder R=b0

The coefficients of polynomial are


Bairstow s method 2

Bairstow’s method (2)

  • To permit the evaluation of complex roots, Bairstow’s method divides the polynomial by a quadratic factor x2-rx-s:

For the remainder to be zero, boand b1 must be zero. However, it is unlikely that our initial guesses at the values of r and s will lead to this result, so we do this…


Bairstow s method 3

Bairstow’s method (3)

Using a similar approach to Newton Raphson method, both bo and b1can be expanded as function of both r and s in Taylor series.

Neglect higher-order terms

We estimate Δr and Δs from

How can we find

these partial

derivatives ???


Bairstow s method 4

Bairstow’s method (4)

Partial derivatives can be obtained by a synthetic division of the b’s in a similar fashion the b’s themselves are derived:

where

then


Bairstow s method 5

Bairstow’s method (5)

At each step, the error can be estimated as

The roots can be determined from


Bairstow s method 6

Bairstow’s method (6)

  • At this point three possibilities exist:

    • The quotient is a third-order polynomial or greater. The previous values of r and s serve as initial guesses and Bairstow’s method is applied to the quotient to evaluate new r and s values.

    • The quotient is quadratic. The remaining two roots are evaluated directly, using

    • The quotient is a 1st order polynomial. The remaining single root can be evaluated simply as x=-s/r.


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