Roots00 posted

file: roots.ppt p. 1

ROOTS OF EQUATIONS

ENGR 351

Numerical Methods for Engineers

Southern Illinois University Carbondale

College of Engineering

Dr. L.R. Chevalier

Dr. B.A. DeVantier


Quadratic Formula

xb b ac

a

f x ax bx c

2

2

4

2

0( )

This equation gives us the roots of the algebraic functionf(x)

i.e. the value of x that makes f(x) = 0

How can we solve for f(x) = e-x - x?


Roots of Equations

• Plot the function and determine where it crosses the x-axis

• Lacks precision• Trial and error

f(x)=e-x-x

-2

0

2

4

6

8

10

-2 -1 0 1 2

x

f(x)


Overview of Methods

• Bracketing methodsGraphing method

Bisection method

False position

• Open methodsOne point iteration

Newton-Raphson

Secant method


• Understand the graphical interpretation of a root

• Know the graphical interpretation of the false-position method and why it is usually superior to the bisection method

• Understand the difference between bracketing and open methods for root location

Specific Study Objectives


• Understand the concepts of convergence and divergence.

• Know why bracketing methods always converge, whereas open methods may sometimes diverge

• Realize that convergence of open methods is more likely if the initial guess is close to the true root



• Know the fundamental difference between the false position and secant methods and how it relates to convergence

• Understand the problems posed by multiple roots and the modification available to mitigate them

• Use the techniques presented to find the root of an equation

• Solve two nonlinear simultaneous equations



Bracketing Methods

• Graphical

• Bisection method

• False position method


Graphical(limited practical value)

x

f(x)

x

f(x)

x

f(x)

x

f(x)

consider lowerand upper boundsame sign,no roots or even # of roots

opposite sign,odd # of roots


Bisection Method

• Takes advantage of sign changing

• f(xl)f(xu) < 0 where the subscripts refer to lower and upper bounds

• There is at least one real root

x

f(x)

x

f(x)

x

f(x)


Algorithm• Choose xu and xl. Verify sign change

f(xl)f(xu) < 0

• Estimate root

xr = (xl + xu) / 2

• Determine if the estimate is in the lower or upper subinterval

f(xl)f(xr) < 0 then xu = xr RETURN

f(xl)f(xr) >0 then xl = xr RETURN

f(xl)f(xr) =0 then root equals xr - COMPLETE


Error

a

present approx previous approx

present

.100

Let’s consider an example problem:


•f(x) = e-x - x

•xl = -1 xu = 1

EXAMPLE

Use the bisection method to determine the root

3.718282

-0.63212

-2

0

2

4

6

8

10

-2 -1 0 1 2

x

f(x)


SOLUTION

3.718282

-0.63212

1

-2

0

2

4

6

8

10

-2 -1 0 1 2

x

f(x)


-0.63212

1

0.106531

-2

0

2

-1 0 1 2

x

f(x)SOLUTION


False Position Method

• “Brute Force” of bisection method is inefficient

• Join points by a straight line

• Improves the estimate

• Replacing the curve by a straight line gives the “false position”


xl

xu

f(xl)

f(xu)next estimate, xr

f x

x x

f x

x x

x xf x x x

f x f x

l

r l

u

r u

r uu l u

l u

Based on similar triangles


EXAMPLE

Determine the root of the following equation using the false position method starting with an initial estimate of xl=4.55 and xu=4.65

f(x) = x3 - 98

-40

-30

-20

-10

0

10

20

30

4 4.5 5

x

f(x)


Pitfalls of False Position Method

f(x)=x10-1

-5

05

10

15

2025

30

0 0.5 1 1.5

x

f(x)


Open Methods

• Simple one point iteration

• Newton-Raphson method

• Secant method

• Multiple roots

• In the previous bracketing methods, the root is located within an interval prescribed by an upper and lower boundary


Open Methods cont.• Such methods are said to be convergent

solution moves closer to the root as the computation progresses

• Open methodsingle starting value

two starting values that do not necessarily bracket the root

• These solutions may divergesolution moves farther from the root as the

computation progresses


The tangentgives next estimate.xi

f(x)

x

f(xi)

xi+1

f(xi+1 )


Solution can “overshoot”the root and potentiallydiverge

x0

f(x)

x

x1x2


Simple one point iteration

• Open methods employ a formula to predict the root

• In simple one point iteration, rearrange the function f(x) so that x is on the left hand side of the equationi.e. for f(x) = x2 - 2x + 3 = 0

x = (x2 + 3) / 2


Simple one point iteration

• In simple one point iteration, rearrange the function f(x) so that x is on the left hand side of the equationi.e. for f(x) = sin x = 0

x = sin x + x

• Let x = g(x)

• New estimate based onx i+1 = g(xi)


EXAMPLE(solution presented in notes)

• Consider f(x) = e-x -3x

• g(x) = e-x / 3

• Initial guess x = 0

-6-4-202468

10121416

-2 -1 0 1 2

x

f(x)


Initial guess 0.000

g(x) f(x) a

0.333 -0.283

0.239 0.071 39.561

0.263 -0.018 9.016

0.256 0.005 2.395

0.258 -0.001 0.612

0.258 0.000 0.158

0.258 0.000 0.041

-6-4-202468

10121416

-2 -1 0 1 2

x

f(x)


Newton Raphson

tangent

dy

dxf

f xf x

x x

rearrange

x xf x

f x

ii

i i

i ii

i

'

'

'

0

1

1

f(xi)

xi

tangent

xi+1


Newton RaphsonPitfalls


EXAMPLE

Use the Newton Raphson method to determine the root off(x) = x2 - 11 using an initial guess of xi = 3 -20

0

20

40

60

80

100

0 2 4 6 8 10

x

f(x)


In your program code, check for problems of divergence

• Include an upper limit on the number of iterations

• Establish a tolerance, s

• Check to see if a is increasing

What if derivative is difficult to evaluate?

SECANT METHOD


Secant method

f x

f x f x

x xi i

i i

'

1

1

Approximate derivative using a finite divided difference

What is this? HINT: dy / dx = y / x

Substitute this into the formula for Newton Raphson


Secant method

ii

iiiii

i

iii

xfxf

xxxfxx

xf

xfxx

1

11

1 '

Substitute finite difference approximation for thefirst derivative into this equation for Newton Raphson


Secant method

• Requires two initial estimates• f(x) is not required to change signs, therefore this

is not a bracketing method

ii

iiiii xfxf

xxxfxx

1

11


x

f(x)

1

2

new est.

x

f(x)

1

new est.

2

FALSE POSITION

SECANT METHOD

The new estimateis selected from theintersection with thex-axis


Multiple Roots

• Corresponds to a point where a function is tangential to the x-axis

• i.e. double rootf(x) = x3 - 5x2 + 7x -3

f(x) = (x-3)(x-1)(x-1)

i.e. triple root

f(x) = (x-3)(x-1)3

-4

-2

0

2

4

6

8

10

0 1 2 3 4

x

f(x) multiple root


Difficulties

• Bracketing methods won’t work

• Limited to methods that may diverge

-4

-2

0

2

4

6

8

10

0 1 2 3 4

x

f(x) multiple root


• f(x) = 0 at root

• f '(x) = 0 at root

• Hence, zero in the denominator for Newton-Raphson and Secant Methods

• Write a “DO LOOP” to check is f(x) = 0 before continuing

-4

-2

0

2

4

6

8

10

0 1 2 3 4

xf(x) multiple root


Multiple Roots

x xf x f x

f x f x f xi i

i i

i i i

1 2

'

' ' '

-4

-2

0

2

4

6

8

10

0 1 2 3 4

x

f(x) multiple root


Systems of Non-Linear Equations

• We will later consider systems of linear equationsf(x) = a1x1 + a2x2+...... anxn - C = 0

where a1 , a2 .... an and C are constant

• Consider the following equationsy = -x2 + x + 0.5

y + 5xy = x3

• Solve for x and y


Systems of Non-Linear Equations cont.

• Set the equations equal to zeroy = -x2 + x + 0.5

y + 5xy = x3

• u(x,y) = -x2 + x + 0.5 - y = 0• v(x,y) = y + 5xy - x3 = 0• The solution would be the values of x and y

that would make the functions u and v equal to zero


Recall the Taylor Series

f x f x f x hf x

hf x

h

f x

nh R

where h step size x x

i i ii i

ni n

n

i i

12 3

1

2 3'

' '

!

' ' '

!

!. . . . . .


Write a first order Taylor series with respect to u and v

u uu

xx x

u

yy y

v vv

xx x

v

yy y

i ii

i ii

i i

i ii

i ii

i i

1 1 1

1 1 1

The root estimate corresponds to the point whereui+1 = vi+1 = 0


Therefore

x xu

v

yv

u

yu

x

v

y

u

y

v

x

y yu

v

yv

u

yu

x

v

y

u

y

v

x

i i

ii

i

i i i i

i i

ii

i

i i i i

1

1

This is a 2 equation version of Newton-Raphson


Therefore

x xu

v

yv

u

yu

x

v

y

u

y

v

x

y yu

v

yv

u

yu

x

v

y

u

y

v

x

i i

ii

i

i i i i

i i

ii

i

i i i i

1

1

THE DENOMINATOROF EACH OF THESEEQUATIONS ISFORMALLYREFERRED TOAS THE DETERMINANTOF THEJACOBIAN

This is a 2 equation version of Newton-Raphson


EXAMPLE

• Determine the roots of the following nonlinear simultaneous equationsy = -x2 + x + 0.5

y + 5xy = x3

• Use and initial estimate of x=0, y=1


APPLIED PROBLEM

The concentration of pollutant bacteria C in a lakedecreases according to:

Determine the time required for the bacteria to be reduced to 10 using Newton-Raphson method.

C e et t 80 202 0 1.


You buy a $20 K piece of equipment for nothing downand $5K per year for 5 years. What interest rate are you paying? The formula relating present worth (P), annualpayments (A), number of years (n) and the interest rate(i) is:

A Pi i

i

n

n

1

1 1

Use the bisection method

APPLIED PROBLEM


PREVIOUS QUIZGraphically illustrate the Newton Raphson Method and bi-section method for finding the roots of an equation on graphs provided. Only show two iterations. Be sureto select initial guesses which avoid pitfalls (i.e. zeroslope).


PREVIOUS QUIZ

Given the Taylor series approximation, describe thedetail given by a) zero order approximation; b) firstorder approximation; c) second order approximation.

f x f x f x hf x

hf x

h

f x

nh R

where h step size x x

i i ii i

ni n

n

i i

12 3

1

2 3'

' '

!

' ' '

!

!. . . . . .


PREVIOUS EXAM QUESTION

Given the equation:

f(x) = x4 - 3x2 + 6x -2 = 0

a) Indicate on the graph an initial estimate for the Newton Raphson Method where

- the solution will diverge- a reasonable choice

b) Solve to three significant figures

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