Part 2 Roots of Equations

1

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Part 2

Roots of Equations

•Why?

•But

a

acbbxcbxax

2

40

22

?0sin

?02345

xxx

xfexdxcxbxax

2


3


Nonlinear Equation Solvers

Bracketing Graphical Open Methods

BisectionFalse Position (Regula-Falsi)

Newton Raphson

Secant

All Iterative

4


Chapter 5 Bracketing Methods

(Or, two point methods for finding roots)• Two initial guesses for the

root are required. These guesses must “bracket” or be on either side of the root.

== > Fig. 5.1

• If one root of a real and continuous function, f(x)=0, is bounded by values x=xl, x

=xu then f(xl) . f(xu) <0. (The function changes sign on opposite sides of the root)

5


Figure 5.2

6


Figure 5.3

7


The Bisection MethodFor the arbitrary equation of one variable, f(x)=0

1. Pick xl and xu such that they bound the root of interest, check if f(xl).f(xu) <0.

2. Estimate the root by evaluating f[(xl+xu)/2].

3. Find the pair • If f(xl). f[(xl+xu)/2]<0, root lies in the lower interval,

then xu=(xl+xu)/2 and go to step 2.

8


• If f(xl). f[(xl+xu)/2]>0, root lies in the upper interval, then xl= [(xl+xu)/2, go to step 2.

• If f(xl). f[(xl+xu)/2]=0, then root is (xl+xu)/2 and terminate.

4. Compare s with a

(5.2, p. 118)

5. If a< s, stop. Otherwise repeat the process.

%100

newr

oldr

newr

ax

xx

Step f (x ) = 2 - e -x x R = -ln(2) = a

t

XL -1 XR -0.5 XU 0

f(XL) -0.71828 f(XR) 0.351279 f(XU) 1

f(XL)*f(XR) -0.25232 f(XL)*f(XR) 0.351279

XL -1 XR -0.75 XU -0.5

f(XL) -0.71828 f(XR) -0.117 f(XU) 0.351279

f(XL)*f(XR) 0.084039 f(XL)*f(XR) -0.0411

XL -0.75 XR -0.625 XU -0.5

f(XL) -0.117 f(XR) 0.131754 f(XU) 0.351279

f(XL)*f(XR) -0.01542 f(XL)*f(XR) 0.046282

XL -0.75 XR -0.6875 XU -0.625

f(XL) -0.117 f(XR) 0.011263 f(XU) 0.131754

f(XL)*f(XR) -0.00132 f(XL)*f(XR) 0.001484

-0.69314718

1

2 0.333333

4 0.090909 0.005647

0.056853

3 0.2 0.068147

10


Evaluation of Method

Pros• Easy• Always find root• Number of iterations

required to attain an absolute error can be computed a priori.

Cons• Slow• Know a and b that

bound root• Multiple roots• No account is taken of

f(xl) and f(xu), if f(xl) is closer to zero, it is likely that root is closer to xl .

11


How Many Iterations will It Take?

• Length of the first Interval Lo=b-a

• After 1 iteration L1=Lo/2

• After 2 iterations L2=Lo/4

• After k iterations Lk=Lo/2k

saR

ka x

L %100

12


• If the absolute magnitude of the error is

and Lo=2, how many iterations will you have to do to get the required accuracy = 10-4 in the solution?

dan

na E

Ln

LE

,

02

0 log2

153.14)2log(

102

log 4

nn

13


The False-Position Method(Regula-Falsi)

• If a real root is bounded by xl and xu of f(x)=0, then we can approximate the solution by doing a linear interpolation between the points [xl, f(xl)] and [xu, f(xu)] to find the xr value such that l(xr)=0, l(x) is the linear approximation of f(x).

== > Fig. 5.12

14


Procedure

1. Find a pair of values of x, xl and xu such that fl=f(xl) <0 and fu=f(xu) >0.

2. Estimate the value of the root from the following formula (Refer to Box 5.1)

and evaluate f(xr).

lu

luulr ff

fxfxx

15


3. Use the new point to replace one of the original points, keeping the two points on opposite sides of the x axis.

Use the same selecting rules as the bisection method.

If f(xr)=0 then you have found the root and need go no further.

16


4. See if the new xl and xu are close enough for convergence to be declared. If they are not go back to step 2.

• Why this method?– Usually faster– Always converges for a single root.

See Sec.5.3.1, Pitfalls of the False-Position Method

Note: Always check by substituting estimated root in the original equation to determine whether f(xr) ≈ 0.

• One-sidedness: Watch out (p. 128)

Step f (x ) = 2 - e -x x R = -ln(2) = a

t

XL -1 XR -0.58198 XU 0

f(XL) -0.71828 f(XR) 0.210428 f(XU) 1

f(XL)*f(XR) -0.15115 f(XL)*f(XR) 0.210428

XL -1 XR -0.67669 XU -0.58198

f(XL) -0.71828 f(XR) 0.03264 f(XU) 0.210428

f(XL)*f(XR) -0.02344 f(XL)*f(XR) 0.006868

XL -1 XR -0.69075 XU -0.67669

f(XL) -0.71828 f(XR) 0.004797 f(XU) 0.03264

f(XL)*f(XR) -0.00345 f(XL)*f(XR) 0.000157

XL -1 XR -0.6928 XU -0.69075

f(XL) -0.71828 f(XR) 0.000699 f(XU) 0.004797

f(XL)*f(XR) -0.0005 f(XL)*f(XR) 3.36E-06

4 0.002962 0.00035

0.016454

3 0.020345 0.002402

-0.69314718

1

2 0.139969

18


Chapter 6

Open Methods• Open methods

are based on formulas that require only a single starting value of x or two starting values that do not necessarily bracket the root.

Figure 6.1

19


Simple Fixed-point Iteration

... 2, 1,k ,given )(

)(0)(

1

okk xxgx

xxgxf

•Bracketing methods are “convergent”.

•Fixed-point methods may sometime “diverge”, depending on the stating point (initial guess) and how the function behaves.

•Good for calculators.

•Rearrange the function so that x is on the left side of the equation:

20


xxg

or

xxg

or

xxg

xxxxf

21)(

2)(

2)(

02)(2

2

Example:

g(x) = 1+2/x g(x) = x2-2 g(x) =

xold xnew xold xnew xold xnew

1 3 1 -1 1 1.7323 1.667 -1 -1 1.732 2.155

1.667 2.2 -1 -1 2.155 1.9282.2 1.909 -1 -1 1.928 2.037

1.909 2.048 -1 -1 2.037 1.9822.048 1.977 -1 -1 1.982 2.0091.977 2.012 -1 -1 2.009 1.9952.012 1.994 -1 -1 1.995 2.0021.994 2.003 -1 -1 2.002 1.9992.003 1.999 -1 -1 1.999 2.0011.999 2.001 -1 -1 2.001 2

(x+2) 0̂.5

22


g(x) = 1+2/x g(x) = x2-2 g(x) =

xold xnew xold xnew xold xnew

0 ##### 0 -2 0 1.414##### ##### -2 0 1.414 2.414##### ##### 0 ##### 2.414 1.828##### ##### ##### ##### 1.828 2.094##### ##### ##### ##### 2.094 1.955##### ##### ##### ##### 1.955 2.023##### ##### ##### ##### 2.023 1.989##### ##### ##### ##### 1.989 2.006##### ##### ##### ##### 2.006 1.997##### ##### ##### ##### 1.997 2.001##### ##### ##### ##### 2.001 1.999

(x+2) 0̂.5

23


Convergence

• x=g(x) can be expressed as a pair of equations:

y1=x

y2=g(x) (component equations)

• Plot them separately.

Figure 6.2

24


Conclusion

• Fixed-point iteration converges if

x)f(x) line theof (slope 1)( xg

•When the method converges, the error is roughly proportional to or less than the error of the previous step, therefore it is called “linearly convergent.”

25


Newton-Raphson Method

• Most widely used method.• Based on Taylor series expansion:

)(

)(

)(0

g,Rearrangin

0)f(x when xof value theisroot The!2

)()()()(

1

1

1i1i

32

1

i

iii

iiii

iiii

xf

xfxx

xx)(xf)f(x

xOx

xfxxfxfxf

Newton-Raphson formula

Solve for

26


• A convenient method for functions whose derivatives can be evaluated analytically. It may not be convenient for functions whose derivatives cannot be evaluated analytically.

Fig. 6.5

27


Fig. 6.6

28


The Secant Method• A slight variation of Newton’s method for

functions whose derivatives are difficult to evaluate. For these cases the derivative can be approximated by a backward finite divided difference.

,3,2,1)()(

)(

)()()(

1

11

1

1

ixfxf

xxxfxx

xx

xfxfxf

ii

iiiii

ii

iii

29


• Requires two initial estimates of x , e.g, xo, x1. However, because f(x) is not required to change signs between estimates, it is not classified as a “bracketing” method.

• The scant method has the same properties as Newton’s method. Convergence is not guaranteed for all xo, f(x).

Fig. 6.7

30


Fig. 6.8

31


Multiple Roots

• None of the methods deal with multiple roots efficiently, however, one way to deal with problems is as follows:

)(

)(1xfindThen

)(

)()(Set

ii

i

i

ii

xu

xu

xf

xfxu

This function has

roots at all the same locations as the original function

32


Fig. 6.10

33


• “Multiple root” corresponds to a point where a function is tangent to the x axis.

• Difficulties– Function does not change sign at the multiple root,

therefore, cannot use bracketing methods.– Both f(x) and f′(x)=0, division by zero with

Newton’s and Secant methods.

34


Systems of Linear Equations

0),,,,(

0),,,,(

0),,,,(

321

3212

3211

nn

n

n

xxxxf

xxxxf

xxxxf

35


• Taylor series expansion of a function of more than one variable

)()(

)()(

11111

11111

iii

iii

ii

iii

iii

ii

yyy

vxx

x

vvv

yyy

uxx

x

uuu

•The root of the equation occurs at the value of x and y where ui+1 and vi+1 equal to zero.

36


y

vy

x

vxvy

y

vx

x

v

y

uy

x

uxuy

y

ux

x

u

ii

iiii

ii

i

ii

iiii

ii

i

11

11

•A set of two linear equations with two unknowns that can be solved for.

37


xv

yu

yv

xu

xu

vxv

uyy

xv

yu

yv

xu

yu

vyv

uxx

iiii

ii

ii

ii

iiii

ii

ii

ii

1

1

Determinant of the Jacobian of the system.

38


Chapter 7

Roots of Polynomials• The roots of polynomials such as

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 root exist in conjugate pairs (that is, +i and -i), where i=sqrt(-1).

39


Conventional Methods

• The efficacy of bracketing and open methods depends on whether the problem being solved involves complex roots. If only real roots exist, these methods could be used. However,– Finding good initial guesses complicates both the

open and bracketing methods, also the open methods could be susceptible to divergence.

• Special methods have been developed to find the real and complex roots of polynomials – Müller and Bairstow methods.

40


Müller Method• Müller’s method obtains a root estimate by

projecting a parabola to the x axis through three function values. Figure 7.3

41


Müller Method

cxxbxxaxf )()()( 22

22

• The method consists of deriving the coefficients of parabola that goes through the three points:

1. Write the equation in a convenient form:

42


2. The parabola should intersect the three points [xo, f(xo)], [x1, f(x1)], [x2, f(x2)]. The coefficients of the polynomial can be estimated by substituting three points to give

3. Three equations can be solved for three unknowns, a, b, c. Since two of the terms in the 3rd equation are zero, it can be immediately solved for c=f(x2).

cxxbxxaxf

cxxbxxaxf

cxxbxxaxf ooo

)()()(

)()()(

)()()(

222

222

212

211

22

2

)()()()(

)()()()(

212

2121

22

22

xxbxxaxfxf

xxbxxaxfxf ooo

43


)(

)()(

)()()()(

x-xhx-xh

If

2111

1

112

11

112

11

12

121

1

1

121o1o

xfcahbhh

a

hahbh

hhahhbhh

xx

xfxf

xx

xfxf

o

o

oooo

o

oo

Solved for a and b

44


• Roots can be found by applying an alternative form of quadratic formula:

• The error can be calculated as

• ±term yields two roots, the sign is chosen to agree with b. This will result in a largest denominator, and will give root estimate that is closest to x2.

acbb

cxx

4

2223

%1003

23

x

xxa

45


• Once x3 is determined, the process is repeated using the following guidelines:

1. If only real roots are being located, choose the two original points that are nearest the new root estimate, x3.

2. If both real and complex roots are estimated, employ a sequential approach just like in secant method, x1, x2, and x3 to replace xo, x1, and x2.

46


Bairstow’s Method• Bairstow’s method is an iterative approach loosely

related to both Müller and Newton Raphson methods.• It is based on dividing a polynomial by a factor x-t:

2 to1

iprelationsh recurrenceby calculated

are tscoefficien the,bRreminder awith

)(

)(

1

o

23211

221

nitbab

ab

xbxbxbbxf

xaxaxaaxf

iii

nn

nnn

nnon

47


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

02

iprelationsh recurrence simple a Using

)(

)(

21

11

1

231322

ton-isbrbab

rbab

ab

brxbR

xbxbxbbxf

iiii

nn-n-

nn

o

nn

nnn

48


• 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, a systematic approach can be used to modify our guesses so that bo and b1 approach to zero.

• 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.

49


ooo

oooo

bss

br

r

b

bss

br

r

b

ss

br

r

bbssrrb

ss

br

r

bbssrrb

111

1111

as estimated

be willguessesour improve toneededr and sin

changes The roots.at s andr of values the toclose

adequately are guesses initial that theassuming

),(

),(

50


• If partial derivatives of the b’s can be determined, then the two equations can be solved simultaneously for the two unknowns r and b.

• 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

22

cs

bc

r

b

s

bc

r

b

where

toniscrcbc

rcbc

bc

oo

iiii

nnn

nn

51


• Then

• At each step the error can be estimated as

obscrc

bscrc

21

132 Solved for r and s, in turn are employed to improve the initial guesses.

%100

%100

,

,

r

r

r

r

sa

ra

52


2

42 srrx

• The values of the roots are determined by

• 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 the above eqn.

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

Refer to Tables pt2.3 and pt2.4

Documents

Part 2 Roots of Equations