Today’s class

Today’s class• Roots of Equations• Polynomials

Numerical Methods, Lecture 6 1 Prof. Jinbo Bi

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• Locate the roots of a set of simultaneous equations

• Use extensions of the various open methods

• Fixed-point iteration is not practical for nonlinear systems


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Systems of nonlinear equations

• Using Newton-Raphson method• First order Taylor approximation

• Solve for xi+1 and yi+1


yf

yyxf

xxyxfyxf iii

iiiii

,1

1,1

1,11,1 )()(),(),(

yf

yyxf

xxyxfyxf iii

iiiii

,2

1,2

1,21,2 )()(),(),(


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• Newton-Raphson method can still diverge as with the single equation method

• Can be generalized to more than two equations - requires solving larger linear systems


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Polynomials• A polynomial is of the form:

• 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 n is odd, there is at least one real root• If complex roots exist, they exist in conjugate

pairs (λ+μi and λ-μi)Numerical Methods, Lecture 6 6 Prof. Jinbo Bi

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Polynomials• Uses in electrical engineering

• Solving for poles and zeros in transfer functions

• Solving characteristic equations in linear ordinary differential equations


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Computing with Polynomials


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Computing with Polynomials


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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:

• Root x=4 is known


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Polynomial Deflation


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• Synthetic Division

Polynomial Deflation


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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 method• Bairstow‘s method


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• 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

Muller’s method


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Muller’s method


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• Find a parabola such that it intersects the function at three points x0, x1, and x2

Muller’s method


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• Solve for c

• Substitute back into set of equations

• Define new set of variables

Muller’s method


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• Substitute new variables back into equations

• Solve for a and b

Muller’s method


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• 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

Muller’s method


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• Which of the two quadratic roots do you choose?• Choose the sign that agrees with the sign of

b, making the denominator the largest• That will bring xr closer to x2

Muller’s method


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• How do you pick the three points for the next iteration?• If there are only real roots, choose xr and the

closer two of the orignal three roots to xr.• If there are complex roots, use a sequential

approach: use xr, x2, and x1 to replace x2, x1, and x0

Muller’s method


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• Find roots of f(x)=x3-13x-12• Use x0=4.5, x1=5.5, and x2=5 as the

initial guesses

Example


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Example


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Example


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• 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

Bairstow’s method


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• Factor by x-t

• Factor by x2-rx-s

Bairstow’s method


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• We have to find values of r and s such that the remainder is 0

• The remainder is 0 if b1=b0=0

• b1 and b0 are both functions of r and s, so we can give a Taylor series expansion of both

Bairstow’s method


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Bairstow’s method


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• Solve for Δr and Δs• Use the results to adjust r and s and iterate• Stop when the approximate error in r and s is

sufficient

Bairstow’s method


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• Find roots of f(x)= x5 - 3.5x4 + 2.75x3 + 2.125x5 - 3.875x + 1.25

• Use s= -1, r= -1 as the initial guesses

Example


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• Solve linear system

• New s and r

Example


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• Use s= 0.1381, r= -0.6442

Example


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• Solve linear system

• New s and r

Example


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• Keep going until s= 0.5 and r= -0.5• Then solve the following:

Example


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• The quotient coefficients are as follows

• Now we have to solve:

Example


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

Bairstow’s method


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• Bisection• 2 initial guesses, slow convergence, will not

diverge• False-position

• 2 initial guesses, slow-medium convergence, will not diverge

Roots of equations


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• Newton-Raphson• 1 initial guess, fast convergence, may

diverge• Secant

• 2 initial guesses, medium-fast convergence, may diverge

Roots of equations


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• 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

Roots of equations


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Next class• Linear algebraic equations• Read Chapter 9• HW3, due 09/24


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Today’s class

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