Chapter 4 Roots of Polynomials. Objectives Understand the importance of finding polynomial roots in...
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Chapter 4 Roots of Polynomials
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Transcript of Chapter 4 Roots of Polynomials. Objectives Understand the importance of finding polynomial roots in...
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• Conventional method• Muller’s method• Bairstow’s method• Conclusions
Polynomials in… (1)General solutions of linear ODE
Solve for general solution
Change to characteristic equations:
The results can be :-
Polynomials in…(2)General solutions of linear ODE
Polynomials in…(3)Problem :
nnon xaxaxaaxf 2
21)(
Follow these rules:
1.For an nth order equation, there are n real or complex roots.
2.If n is odd, there is at least one real root.
3.If complex roots exist, they will be in conjugate pairs (that is, l+mi and l-mi), where i=sqrt(-1).
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.
Müller method (1) Like Secant, Müller’s method
obtains a root estimate by projecting a parabola to the x axis through three function values.
Müller method (2)
,3,2,1)()(
)(
)()()(
1
11
1
1
ixfxf
xxxfxx
xx
xfxfxf
ii
iiiii
ii
iii
Secant method (linear approximation)
Müller method (3)
Müller method(Parabola or 2nd order
approximation)
Must use three points to approximate function
Müller method (4)Müller methodology derivation
cxxbxxaxf )()()( 22
22
Write the equation in a convenient form at point x2:
We then have three eqns now (from x0, x1, and x2)
cxxbxxaxf
cxxbxxaxf
cxxbxxaxf ooo
)()()(
)()()(
)()()(
222
222
212
211
22
2
Müller method (5)Step III Reduce to two eqns
)()()()(
)()()()(
212
2121
22
22
xxbxxaxfxf
xxbxxaxfxf ooo
Right now u can solve for a and b from
When u know a, b, c you are ready toestimate root from
cxxbxxaxf )()()( 22
22
Müller method (6)Step IV Here the new estimated root is
acbb
cxx
4
2223
Error can be derived from
%1003
23
x
xxa
Two roots, but which one ?
Müller method (7)Summary of algorithm
Start with 3 points [x0,f(x0)] [x1,f(x1)] and [x2,f(x2)]
)(
)()()()(
x-xhx-xh If
2111
1
1
121
1
121o1o
xfcahbhh
a
h
xfxf
h
xfxf
o
o
o
oo
Calculate a, b, and c from
Müller method (8)Summary of algorithm (cont’d)
acbb
cxx
4
2223
Calculate new root from
%1003
23
x
xxa
Calculate error
Check whether
sa
new xi-1 = old xi
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:
0 to11
nitbab
ab
iii
nn
nnon xaxaxaaxf 2
21)(Start with
123211 )(
nnn xbxbxbbxf
Dividing with x-t yields
and a remainder R=b0 The coefficients of polynomial are
Bairstow’s method (2)To permit the evaluation of complex
roots, Bairstow’s method divides the polynomial by a quadratic factor x2-rx-s:
02
)(
)(
21
11
1
231322
ton-isbrbab
rbab
ab
brxbR
xbxbxbbxf
iiii
nn-n-
nn
o
nn
nnn
For the remainder to be zero, bo and 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)Using a similar approach to Newton Raphson method, both bo and b1 can be expanded as function of both r and s in Taylor series.
),(),(
),(),(
22
221111
srOss
br
r
bbssrrb
srOss
br
r
bbssrrb
oooo
ooo bss
br
r
b
bss
br
r
b
111
Neglect higher-order termsWe estimate Δr and Δs from
How can we find these partial derivatives ???
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:
31
21
1
21
11
12
cs
bc
r
b
s
bc
r
b
toniscrcbc
rcbc
bc
oo
iiii
nnn
nn
where
then
obscrc
bscrc
21
132
Bairstow’s method (5)At each step, the error can be estimated as
%100
%100
,
,
s
s
r
r
sa
ra
2
42 srrx
The roots can be determined from
Bairstow’s method (6)• At this point three possibilities exist:
1.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.
2.The quotient is quadratic. The remaining two roots are evaluated directly, using
3.The quotient is a 1st order polynomial. The remaining single root can be evaluated simply as x=-s/r.
2
42 srrx
