Transcript Engineering Computation Part 2 E. T. S. I. Caminos, Canales y Puertos Roots of Equations Objective: Solve for x, given that f(x) = 0 -orEquivalently, solve.

Engineering
Computation
Part 2
E. T. S. I. Caminos, Canales y Puertos
1
Roots of Equations
Objective:
Solve for x, given that f(x) = 0
-orEquivalently, solve for x such that
g(x) = h(x) ==> f(x) = g(x) – h(x) = 0
E. T. S. I. Caminos, Canales y Puertos
2
Roots of Equations
Chemical Engineering (C&C 8.1, p. 187):
van der Waals equation; v = V/n (= volume/# moles)
Find the molal volume v such that
f(v) = ( p +
a
v
2
)(v - b) - RT = 0
p = pressure,
T = temperature,
R = universal gas constant,
a & b = empirical constants
E. T. S. I. Caminos, Canales y Puertos
3
Roots of Equations
Civil Engineering (C&C Prob. 8.17, p. 205):
Find horizontal component of tension, H, in a
cable that passes through (0,y0) and (x,y)
H
 wx  
f(H) = cosh 
 -1 + y 0 - y = 0
w
 H  
w = weight per unit length of cable
E. T. S. I. Caminos, Canales y Puertos
4
Roots of Equations
Electrical Engineering (C&C 8.3, p. 194):
Find the resistance, R, of a circuit such that the
charge reaches q at specified time t
f(R) = e-Rt

2
1
R

 
2L
cos 
-
 LC  2L  



q

t =0
 q0

L = inductance,
C = capacitance,
q0 = initial charge
E. T. S. I. Caminos, Canales y Puertos
5
Roots of Equations
Mechanical Engineering (C&C 8.4, p. 196):
Find the value of stiffness k of a vibrating mechanical
system such that the displacement x(t) becomes zero at
t= 0.5sec. The initial displacement is x0 and the initial
velocity is zero. The mass m and damping c are known,
and λ = c/(2m).



x(t) = x0e-t cos  t  + sin  t   = 0



in which
k c2
μ=
- 2
m 4m
E. T. S. I. Caminos, Canales y Puertos
6
Roots of Equations
Determine real roots of :
• Algebraic equations (including polynomials)
• Transcendental equations such as f(x) = sin(x) + e-x
• Combinations thereof
E. T. S. I. Caminos, Canales y Puertos
7
Roots of Equations
Engineering Economics Example:
A municipal bond has an annual payout of
$1,000 for 20 years. It costs $7,500 to purchase
now. What is the implicit interest rate, i ?
1  1  i  n 

Solution: Present-value, PV, is: PV  A 
i


in which:
PV = present value or purchase price = $7,500
A = annual payment = $1,000/yr
n = number of years = 20 yrs
i
= interest rate = ? (as a fraction, e.g., 0.05
= 5%)
E. T. S. I. Caminos, Canales y Puertos
8
Roots of Equations
Engineering Economics Example (cont.):
We need to solve the equation for i:
1  1  i 20 

7,500  1, 000 
i


Equivalently, find the root of:
1  1  i 20 
0
f (i)  7,500  1,000 
i


E. T. S. I. Caminos, Canales y Puertos
9
Roots of Equations
Engineering Economics Example
$10,000
f(i) = PV - g(i)
$5,000
$0
0%
20%
40%
60%
80%
100%
-$5,000
-$10,000
Excel
-$15,000
Interest Rate, i
E. T. S. I. Caminos, Canales y Puertos
10
Roots of Equations
• Graphical methods:
– Determine the friction coefficient c necessary for a
parachutist of mass 68.1 kg to have a speed of 40
m/seg at 10 seconds.
– Reorganizing.
E. T. S. I. Caminos, Canales y Puertos
11
Roots of Equations
Two Fundamental Approaches
1. Bracketing or Closed Methods
- Bisection Method
- False-position Method
2. Open Methods
- One Point Iteration
- Newton-Raphson Iteration
- Secant Method
E. T. S. I. Caminos, Canales y Puertos
12
Bracketing Methods
f(x)
x
a)
f(x)
x
b)
f(x)
x
c)
f(x)
xl
d)
xu
x
In Figure a) we have the case of f(xl)
and f(xu) with the same sign, and there
is no root in the interval (xl,xu).
In Figure b) we have the case of f(xl)
and f(xu) With different sign, and there
is a root in the interval (xl,xu).
In Figure c) we have the case of f(xl)
and f(xu) with the same sign, and there
are two roots.
In Figure d) we have the case of f(xl)
and f(xu) with different sign, and there is
an odd number of roots.
E. T. S. I. Caminos, Canales y Puertos
13
Bracketing Methods
• Though the cases above are generally valid, there are
cases in which they do not hold.
In Figure a) we have the case of f(xl)
and f(xu) with different sign, but there is
a double root.
f(x)
a)
x
b)
x
f(x)
f(x)
xl
c)
In Figure b) We have the case of f(xl)
and f(xu) With different sign, but there
are two discontinuities.
In Figure c) we have the case of f(xl)
and f(xu) with the same sign, but there
is a multiple root.
xu
x
E. T. S. I. Caminos, Canales y Puertos
14
Bracketing Methods (Bisection method)
Bisection Method
f(x)
xl  xu
xr 
2
f(x)
f(xu)
f(xu)
f(x1) f(xr) > 0
(x1)
f(xr)
(xu) x
xr => x1
f(xr)
(x1)
(xr)(xu) x
f(x1)
f(x1)
E. T. S. I. Caminos, Canales y Puertos
15
Bracketing Methods (Bisection method)
Bisection Method:
Given lower and upper bounds, xl and xu, which bracket the
root: f(xl) f(xu) < 0
x1  x u
1) Estimate the Root by midpoint: x r 
2
2) Revise the bracket:
f(xl) f(xr) < 0, xr –> xu,
f(xl) f(xr) > 0, xr –> xl
3) Repeat steps 1-2 until:
old
x new

x
a) |f(xr)| < k,
b) a < s , with a = r new r 100%
xr
c) | xl – xu | < 
E. T. S. I. Caminos, Canales y Puertos
d) maximum # of iterations is
reached. (Always do this in
iteration algorithms.)
16
Bracketing Methods (Bisection method)
Engineering Economics Example:
We need to solve the equation for i:
1  1  i 20 

7,500  1, 000 
i


Equivalently, find the root of:
1  1  i 20 
0
f (i)  7,500  1,000 
i


Make conservative guesses at the upper and lower
bounds:
100% interest rate, f(1.0) = 6,500
0% interest rate, f(0.0) = -12,500
E. T. S. I. Caminos, Canales y Puertos
17
Bracketing Methods (Bisection method)
Engineering Economics Example
$10,000
f(i) = PV - g(i)
$5,000
$0
0%
20%
40%
60%
80%
100%
-$5,000
-$10,000
Excel
-$15,000
Interest Rate, i
E. T. S. I. Caminos, Canales y Puertos
18
Bracketing Methods (Bisection method)
Engineering Economics Example:
Score Sheet for Rootfinding Example:
Method
Bisection
Initial Est(s). s = 2 E-2
(0.00, 1.00)
(0.05, 0.15)
Convergence guaranteed:
E. T. S. I. Caminos, Canales y Puertos
9
6
s = 2 E-7
26
22
1
Eimax
 0.5 Eimax
19
Bracketing Methods (Bisection method)
One important advantage of this method is that one can
calculate the number of required iterations for a given error.
E. T. S. I. Caminos, Canales y Puertos
20
Bracketing Methods (Bisection method)
Parachutist Example:
E. T. S. I. Caminos, Canales y Puertos
21
Bracketing Methods (Bisection method)
Parachutist Example:
E. T. S. I. Caminos, Canales y Puertos
22
Bracketing Methods (Bisection method)
Bisection Method
Advantages:
1. Simple
2. Good estimate of maximum error: E max
3. Convergence guaranteed
xl  xu

2
1
Eimax
 0.5 Eimax
Disadvantages:
1. Slow
2. Requires two good initial estimates which define
an interval around root:
 use graph of function,
 incremental search, or
 trial & error
E. T. S. I. Caminos, Canales y Puertos
23
Bracketing Methods (False-position Method)
False-position Method
f(x)
f(x)
xr  xu 
f (x u )(x1  x u )
f (x1 )  f (x u )
(xr)
f(xu)
f(x1) f(xr) > 0
(x1)
(xu)
x
f(xr)
f(x1)
E. T. S. I. Caminos, Canales y Puertos
x1 = x r
(x1)
f(xu)
(xu)
f(x1)
f(xr)
24
Bracketing Methods (False-position Method)
Similar to bisection. Uses linear interpolation to
approximate the root xr:
f (x u )(x1  x u )
1) x r  x u 
f (x1 )  f (x u )
2) Revise the bracket: f(x1) f(xr) < 0, xr –> xu,
f(x1) f(xr) > 0, xr –> x1
3) Repeat steps 1-2 until:
a) |f(xr)| < k,
b) a < s , with a = new
x r  x old
r
100%
c) |xu – x1 | = d
new
xr
d) maximum # of iterations is reached.
(Always do this in iteration algorithms.)
E. T. S. I. Caminos, Canales y Puertos
25
Bracketing Methods (False-position Method)
Engineering Economics Example
$10,000
f(i) = PV - g(i)
$5,000
$0
0%
20%
40%
60%
80%
100%
-$5,000
-$10,000
Excel
-$15,000
Interest Rate, i
E. T. S. I. Caminos, Canales y Puertos
26
Bracketing Methods (False-position Method)
Engineering Economics Example
$0
7%
9%
11%
13%
15%
f(i) = PV - g(i)
5%
-$5,000
Interest Rate, i
E. T. S. I. Caminos, Canales y Puertos
27
Bracketing Methods (False-position Method)
Score Sheet for False-Position Example:
Method
Initial Est(s).
s = 2 E-2
s = 2 E-7
Bisection
(0.00, 1.00)
(0.05, 0.15)
9
6
26
22
False-pos.
(0.00, 1.00)
(0.05, 0.15)
11
3
28
14
E. T. S. I. Caminos, Canales y Puertos
28
Bracketing Methods (False-position Method)
Parachutist Example:
E. T. S. I. Caminos, Canales y Puertos
29
Bracketing Methods (False-position Method)
Parachutist Example:
E. T. S. I. Caminos, Canales y Puertos
30
Bracketing Methods (False-position Method)
There are some cases in which the false position
method is very slow, and the bisection method gives
a faster solution.
E. T. S. I. Caminos, Canales y Puertos
31
Bracketing Methods (False-position Method)
Summary of False-Position Method:
Advantages:
1. Simple
2. Brackets the Root
Disadvantages:
1. Can be VERY slow
2. Like Bisection, need an initial interval around the
root.
E. T. S. I. Caminos, Canales y Puertos
32
Open Methods
Roots of Equations - Open Methods
Characteristics:
1. Initial estimates need not bracket the root
2. Generally converge faster
3. NOT guaranteed to converge
Open Methods Considered:
- One Point Iteration
- Newton-Raphson Iteration
- Secant Method
E. T. S. I. Caminos, Canales y Puertos
33
Open Methods
• An alternative method
consists of separating the
function into two parts.
E. T. S. I. Caminos, Canales y Puertos
34
Open Methods (Fixed point method)
Fixed point Method
 predict a value of xi+1 as a function of xi.
Convert f(x) = 0
 iteration steps:
to x = g(x)
xi+1 = g(xi )
x(new) = g(x(old) )
E. T. S. I. Caminos, Canales y Puertos
35
Open Methods (Fixed point method)
Example I:
1  1  i 20 

7,500  1, 000 
i


1  1  i 20 

 ii1  1.0 
7.5


Example II:
sin(x)
f (x) 
 1.0  0.0
x
x = sin(x) –> xi+1 = sin(xi)
OR
x = arcsin(x) –> xi+1 = arcsin(xi)
E. T. S. I. Caminos, Canales y Puertos
36
Open Methods (Fixed point method)
Convergence:
Does x move closer to real root (?)
Depends on:
1. nature of the function
2. accuracy of the initial estimate
Interested in:
1. Will it converge or will it diverge?
2. How fast will it converge ?
(rate of convergence)
E. T. S. I. Caminos, Canales y Puertos
37
Open Methods (Fixed point method)
Convergence of the Fixed point Method:
Root satisfies: xr = g(xr)
The Taylor series for function g is:
xi+1 = g(xr) + g'(x)(xi - xr)
xr < x < xi
Subtracting the second equation from the first yields
(xr – xi+1) = g'(x) (xr – xi)
or
E i  1  g' ( )
Ei
1. True error for next iteration is smaller than the true
error in the previous iteration if |g'(x)| < 1.0 (it will
converge).
2. Because g'(x) is almost constant, the new error is
directly proportional to the old error (linear rate of
convergence).
E. T. S. I. Caminos, Canales y Puertos
38
Open Methods (Fixed point method)
Further Considerations:
Convergence depends on how f(x) = 0 is
converted into x = g(x)
So . . .
Convergence may be improved
by recasting the problem.
E. T. S. I. Caminos, Canales y Puertos
39
Open Methods (Fixed point method)
Convergence Problem:
For slowly converging functions
x new  x old
a 
x new
x 100%
can be small, even though xnew is not close to root.
Remedy:
Do not completely rely on a to ensure that the
problem is solved.
Check to make sure |f(xnew) | < k .
E. T. S. I. Caminos, Canales y Puertos
40
Open Methods (Fixed point method)
E. T. S. I. Caminos, Canales y Puertos
41
Open Methods
E. T. S. I. Caminos, Canales y Puertos
42
Roots of Equations
Two Fundamental Approaches
1. Bracketing or Closed Methods
- Bisection Method
- False-position Method
2. Open Methods
- One Point Iteration
- Newton-Raphson Iteration
- Secant Method
E. T. S. I. Caminos, Canales y Puertos
43
Open Methods (Newton-Raphson Method)
Newton-Raphson Method:
Geometrical Derivation:
Slope of tangent at xi is
f (x i )  0
f '(x i ) 
xi  xi1
Solve for xi+1:
f (xi )
xi1  xi 
f '(xi )
[Note that this is the same form as the generalized onepoint iteration, xi+1 = g(xi)]
E. T. S. I. Caminos, Canales y Puertos
44
Open Methods (Newton-Raphson Method)
Newton-Raphson Method
f(x)
Tangent w/slope=f '(xi )
f(x)
f '(x i ) 
f (x i )  0
f '(x i ) 
x i  x i1
f(xi)
f (x i )  0
x i  x i1
f(xi)
f(xi+1)
f(xi+1)
xi+
xi
x
(xi)
x
1
xi1  xi 
f (x i )
f '(x i )
xi+
1
xi = xi+1
E. T. S. I. Caminos, Canales y Puertos
45
Open Methods (Newton-Raphson Method)
First order Taylor Series Derivation:
0 = f(xr)  f(xi) + f '(xi) (xr – xi)
solve for xr to yield next guess xi+1:
f (x i )
x r  xi1  xi 
f '(x i )
This has the form xi+1 = g(xi) with:
g '(x r )  1 
E. T. S. I. Caminos, Canales y Puertos
(f 'f ' f f ")
(f ') 2
46
Open Methods (Newton-Raphson Method)
Newton-Raphson iteration:
f (xi )
xi1  xi 
f '(x i )
This iteration process is repeated until:
1. f(xi+1)  0, i.e., | f(xi+1) | < k,
with k = small number
2.
x i1  x i
a 
100%  s
x i1
3. Maximum number of iterations is reached.
E. T. S. I. Caminos, Canales y Puertos
47
Open Methods
a) Inflection point in the
neighboor of a root.
b) Oscilation in the neighboor
of a maximum or minimum.
c) Jumps in functions with
several roots.
d) Existence of a null
derivative.
E. T. S. I. Caminos, Canales y Puertos
48
Open Methods (Newton-Raphson Method)
Bond Example:
To apply Newton-Raphson method to:
1  1  i 20 
0
f (i)  7,500  1, 000 
i


We need the derivative of the function:
20 



1, 000  1  1  i 
21 
  20 1  i  
f '(i) 

i 
i




E. T. S. I. Caminos, Canales y Puertos
49
Open Methods (Newton-Raphson Method)
Score Sheet for Newton-Raphson Example:
Method
Initial Est(s).
s = 2 E-2
s = 2 E-7
Bisection
(0.00, 1.00)
(0.05, 0.15)
9
6
26
22
False-pos.
(0.00, 1.00)
(0.05, 0.15)
11
3
28
14
N-R
1.0
0. 5
0.25
0.15
0.05
EXCEL
E. T. S. I. Caminos, Canales y Puertos
diverges
diverges
2, but wrong
48
5
7
3
5
4
5
50
Open Methods (Newton-Raphson Method)
Error Analysis for N-R :
Recall that
xi1  xi 
f (xi )
f '(xi )
Taylor Series gives:
f (x r )  f (x i )  f '(x i )(x r  x i ) 
where
f "()
(x r  x i ) 2
2!
xr  x  xi and f(xr) = 0
E. T. S. I. Caminos, Canales y Puertos
51
Open Methods (Newton-Raphson Method)
Dividing through by f '(xi) yields
f (x i )
f "()
0
 (x r  x i ) 
(x r  x i ) 2
f '(x i )
2f '(x i )
f ''()
 (x i1  x i )  (x r  x i ) 
(x r  x i ) 2
2f '(x i )
OR
f ''()
(x r  x i ) 2
2f '(x i )
f "()

Ei 2
2!f '(x i )
(x r  x i1 )  
Ei1
Ei+1 is proportional to Ei2 ==> quadratic rate of convergence.
E. T. S. I. Caminos, Canales y Puertos
52
Open Methods (Newton-Raphson Method)
Summary of Newton-Raphson Method:
Advantages:
1. Can be fast
Disadvantages:
1. May not converge
2. Requires a derivative
E. T. S. I. Caminos, Canales y Puertos
53
Open Methods (Secant Method)
Secant Method
f (x i1 )  f (x i )
f '(x) 
x i1  x i
Approx. f '(x) with backward FDD:
Substitute this into the N-R equation:
to obtain the iterative expression:
E. T. S. I. Caminos, Canales y Puertos
xi1  xi 
f (x i )
f '(x i )
f (x i )(x i1  x i )
x i1  xi 
f (x i1 )  f (x i )
54
Open Methods (Secant Method)
Secant Method
f(x)
f(x)
f '(x i ) 
f(xi-1 ) - f(xi )
xi-1 - xi
f '(x i ) 
f(xi-1 ) - f(xi )
xi-1 - xi
f(xi-1)
f(xi)
f(xi)
xi+
x
xi xi-1
1
x i1  xi 
f (x i )(x i1  x i )
f (x i1 )  f (x i )
xi+ xi
f(xi-1)
x
xi-1
1
xi = xi+1
E. T. S. I. Caminos, Canales y Puertos
55
Open Methods (Secant Method)
1) Requires two initial estimates: xi-1 and xi
These do NOT have to bracket root !
2) Maintains a strict sequence:
x i1  xi 
f (x i )(x i1  x i )
f (x i1 )  f (x i )
Repeated until:
a. | f(xi+1) | < k with k = small number
b.
x i1  x i
a 
100%  s
x i1
c. Max. number of iterations is reached.
3. If xi and xi+1 were to bracket the root, this would be the same as the
False-Position Method. BUT WE DON'T!
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Open Methods (Secant Method)
– In the secant method, the values are replaced in a strict
sequence, xi+1 to xi, and this to xi-1. Thus, the new values
can be on the de same sode of the root, and sometimes
diverge.
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57
Open Methods (Secant Method)
Score Sheet for Secant Example:
Method
Initial Est(s).
s = 2 E-2
s = 2 E-7
Bisection
(0.00, 1.00)
(0.05, 0.15)
9
6
26
22
False-pos.
(0.00, 1.00)
(0.05, 0.15)
11
3
28
14
N-R
1.0
0.5
0.25
0.15
0.05
diverges
2, but wrong
5
3
4
diverges
48
7
5
5
Secant
(0, 1)
diverges
diverges
(0.00, 0.50) 4, but wrong (chaotic) 27
(0.05, 0.15)
3
6
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Open Methods
Why do open methods fail?
Function may not look linear.
Remedy: recast into a linear form. For example,
1  (1  i) 20 
f (i)  7500  
0
i


Is a poorly constrained problem in that there is a
large, nearly flat zone for which the derivative is near
zero. Recast as:
i f(i) = 0 = 7,500 i - 1000 [ 1 - (1+i)-20 ]
E. T. S. I. Caminos, Canales y Puertos
59
Open Methods
1  (1  i) 20 
f (i)  7500  
0
i


Recast as:
i f(i) = 0 = 7,500 i - 1000 [ 1 - (1+i)-20 ]
– The recast function, "i f(i) will have the same roots as f(i)
plus an additional root at i = 0.
– It will not have a large, flat zone, thus:
h(i) = i f(i) = 7,500 i – 1000 [ 1 – (1+ i)–20]
–
To apply N-R we also need the first derivative:
h'(i) = 7,500 - 20,000 (1+ i)-21
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Open Methods
Score Sheet for Open Methods:
Method
Initial Est(s).
s = 2 E-2
s = 2 E-7
N-R
1.0
0. 5
0.25
0.15
0.05
Secant
(0.00, 0.50)
(0.05, 0.15)
4, but wrong
3
27
6
N-R
[as i f(i)]
1.00
0.150
0.050
0.047
0.03
3
2
4
crazy results
converges to i=0
4
4
5
**
**
E. T. S. I. Caminos, Canales y Puertos
diverges
diverges
2, but wrong
48
5
7
3
5
4
5
61
Open Methods
Cases of Multiple Roots
Multiple Roots:
f(x) = (x – 2)2 (x – 4)
3
2
x = 2 represents two of the
three roots.
1
0
-1
-2
-3
1.0
E. T. S. I. Caminos, Canales y Puertos
2.0
3.0
4.0
62
Open Methods
Problems and Approaches:
Cases of Multiple Roots
1.Bracketing Methods fail locating x = 2.
Note that f(x) f(xr) > 0.
2. At x = 2, f(x) = f '(x) = 0.
• Newton-Raphson and Secant methods may experience problems.
• Rate of convergence drops to linear.
• Luckily, f(x)  0 faster than f '(x)  0
3. Other remedies, recasting problem:
Find x such that u(x) = 0 where:
f (x)
u(x) 
0
f '(x)
Note that u(x) and f(x) have same roots.
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Summary -- Rates of Convergence
lim
i
Ei1
Ei
m
A0
m = 1: linear convergence
m = 2: quadratic convergence
Method
Bisection
False Position
Secant, mult. root
NR, mult. root
Secant, single root
NR, single root
Accel. NR, mult. root (f(x)/f'(x)=0)
E. T. S. I. Caminos, Canales y Puertos
m
1
1
1
1
1.618 "super linear"
2
2
64
Multivariate (Multidimensional) Equations
Solve
fi(x1, ..., xn) = 0 for i = 1,...,n
Let X = (x1, ..., xn)T
 Given intial guess Xt, try to solve
df
f (X) 
(Xi1  Xi )  0
dX
where:
df  dfi 



dX  dX j 
 Obtain X = (Xi+1 – Xi) from linear equations:
df
X  f (Xi )
dX
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Alternative Stopping Criteria
1. Always limit number of iterations using an outer DO loop. The
problem may not converge and could try to go on forever.
2. Absolute error criteria for "small" differences: | xt - xt-1 | < 
3. Relative error criteria for "relatively small" changes
| xt – xt-1 | <  | xt |
4. Can combined error criteria 2 & 3 for large and small x-values:
| xt – xt-1 | <  +  | xt |
5. Converge on zero residual
| f(xt) | < k
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Three Performance Criteria
Stopping Criteria:
| xi – xi-1 | <  +  |xi|
or | f(xi) | < k
or Max. iterations
Convergence Criteria:
| xi – xi-1 | <  +  |xi|
and
| f(xi) | < k
N-R and Secant Confirmation of Convergent Behavior:
x in feasible region
and
| f(xi) | ≤ 0.5 | f(xi-1) |
and
| xi – xi-1 | ≤ 0.6 | xi-1 – xi-2 |
otherwise, do Bisection for a while.
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Three Phase Rootfinding Strategy
A real rootfinding problem can be viewed
as having three phases:
1) Opening moves: One needs to find the
region of the parameter space in which
desired root can be found.
Understanding of problem, physical insight, and
common sense are valuable.
2) Middle Game: Use robust algorithm to
reduce initial region of uncertainty.
3) End game: Generate a highly accurate
solution in a few iterations.
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