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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
file: roots.ppt p. 2
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?
file: roots.ppt p. 3
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)
file: roots.ppt p. 4
Overview of Methods
• Bracketing methodsGraphing method
Bisection method
False position
• Open methodsOne point iteration
Newton-Raphson
Secant method
file: roots.ppt p. 5
• 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
file: roots.ppt p. 6
• 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
Specific Study Objectives
file: roots.ppt p. 7
• 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
Specific Study Objectives
file: roots.ppt p. 8
Bracketing Methods
• Graphical
• Bisection method
• False position method
file: roots.ppt p. 9
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
file: roots.ppt p. 10
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)
file: roots.ppt p. 11
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
file: roots.ppt p. 12
Error
a
present approx previous approx
present
.100
Let’s consider an example problem:
file: roots.ppt p. 13
•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)
file: roots.ppt p. 14
SOLUTION
3.718282
-0.63212
1
-2
0
2
4
6
8
10
-2 -1 0 1 2
x
f(x)
file: roots.ppt p. 15
-0.63212
1
0.106531
-2
0
2
-1 0 1 2
x
f(x)SOLUTION
file: roots.ppt p. 16
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”
file: roots.ppt p. 17
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
file: roots.ppt p. 18
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)
file: roots.ppt p. 19
Pitfalls of False Position Method
f(x)=x10-1
-5
05
10
15
2025
30
0 0.5 1 1.5
x
f(x)
file: roots.ppt p. 20
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
file: roots.ppt p. 21
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
file: roots.ppt p. 22
The tangentgives next estimate.xi
f(x)
x
f(xi)
xi+1
f(xi+1 )
file: roots.ppt p. 23
Solution can “overshoot”the root and potentiallydiverge
x0
f(x)
x
x1x2
file: roots.ppt p. 24
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
file: roots.ppt p. 25
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)
file: roots.ppt p. 26
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)
file: roots.ppt p. 27
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)
file: roots.ppt p. 28
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
file: roots.ppt p. 29
Newton RaphsonPitfalls
file: roots.ppt p. 30
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)
file: roots.ppt p. 31
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
file: roots.ppt p. 32
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
file: roots.ppt p. 33
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
file: roots.ppt p. 34
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
file: roots.ppt p. 35
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
file: roots.ppt p. 36
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
file: roots.ppt p. 37
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
file: roots.ppt p. 38
• 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
file: roots.ppt p. 39
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
file: roots.ppt p. 40
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
file: roots.ppt p. 41
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
file: roots.ppt p. 42
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'
' '
!
' ' '
!
!. . . . . .
file: roots.ppt p. 43
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
file: roots.ppt p. 44
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
file: roots.ppt p. 45
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
file: roots.ppt p. 46
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
file: roots.ppt p. 47
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.
file: roots.ppt p. 48
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
file: roots.ppt p. 49
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).
file: roots.ppt p. 50
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'
' '
!
' ' '
!
!. . . . . .
file: roots.ppt p. 51
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