Chapter 1 Solving Nonlinear Equations
Introduction:
Solving Nonlinear Equations
Interval Halving (Bisection) Method
Linear Interpolation Method
Newton’s Method
Muller’s Method
Fixed-Point Iteration: x= g(x) Method
Newton’s Method for polynomials
Bairstow’s Method for Quadratic Factors
Other Methods for Polynomials
Multiple Roots
Solve f(x) = 0 Find the roots (zeros) of the equation
Algebraic functions
3x + 5 = 0
linear eq.
2
19x + 5x –3 = 0
quadratic eq.
45x3 + x2 - 5x +3 = 0 cubic eq.
:
anxn + an-1xn-1 + ……+ a1x1 + a0 = nth degree poly
Transcendental functions
4ex lnx + sin x + x2 - 5 = 0
curve
Find the root(s) of the eq. Graphically -- find where the curve crosses x-axis.
1. Bisection Method:
Intermediate Value Thm
If f(x)  C[a, b] and k is any number between f(a) and f(b), then there exits a number
c in (a, b) for which f(c) = k.
A. Algorithm f(x) = 0
Assume that f(x) is continuous [x1 , x2]
Try to find two points x1 and x2  f(x1) * f(x2) < 0
(x1 and x2 are on the opposite side of x-axis)
Repeat
Set x3 = (x1 + x2) / 2
If f(x1) * f(x3) < 0
Set x2 = x3
Else
Set x1 = x3
Endif
Until (x1 - x2  / 2)<  (tolerance value) or f(x3) ~ 0
1
A. Examples :
Page 42
f(x) = x3 + x2 - 3x - 3 = 0
Max. error is at most half the width of the last interval
Error after n iterations < (x1 - x2  / 2n ) < 
:
How many iterations are required to obtain a max error of ?
Pick  = 0.01 and x1 = 1, x2 = 2, then n =?
accurate up to one decimal place.
Note : the value of n is completely independent of the characteristic of the function f(x).
B. Convergence
When do we stop the iteration?
 The problem is solved i.e. f(xk) = 0 for some xk
 The iteration has converged xk  xk-1 or xk - xk-1 
 The iteration has reached the tolerance value 
0
C. Conclusion:
Adv and Disadv -- Can not handle multiple roots (Tangent to x-axis)
 Slow, but completely reliable and good for all problems
(Algebraic or transcendental fcts)
Guaranteed to work -number of iterations can be calculated to achieve a specified accuracy
2
2. Linear Interpolation Method
● The Secant Method
● Linear Interpolation (False Position)
** The
Secant Method **
A. Algorithm:
To determine a root of f(x)
pick two points x0 and x1 that are near the root.
(not need to be opposite side of the f(x) along x-axis)
If f(x0)  < f(x1)
swap x0 with x1
Repeat
Set x2 = x1 – f(x1) * {(x0 - x1) / [f(x0) - f(x1)} ----------- (1/m)
Set x0 = x1
Set x1 = x2
Until (f(x2) <  (tolerance value) or f(x2) ~ 0
Derive the formula -Notation y = f(x), y0 = f(x0) , and y1 = f(x1) …..
y – y1 = m(x-x1) , where m = (y1 - y0) /(x1 - x0)
If (x2, 0 ) on the line  0 - y1 = m (x2 - x1)
Solve for x2
x2 - x1 = - y1 * (1/ m)
x3 = x2 - y2 * (1/ m)
x4 = x3 - y3 * (1/ m)
………
xi = xi-1 - yi-1 * (1/ m)
B. Examples
Page 46

x2 = x1 - y1 * (1/ m)
f(x) = x3 + x2 - 3x - 3
f(x) = 3x + sin (x) – e x
f(x) = x2 - 2
3
C. Convergence
In general it is faster than Bisection Method
i+1 = k(i)p
__
Secant Method ----- p = (1 + √ 5 ) / 2 ~ 1.62
Newton Method ----- p = 2
Muller's Method ----- p = 1.8
D. Conclusion
Advantage and disadv ---- pick any two points
--- conv pretty fast
--- Always uses the most recent infor and thus can't freeze at one point
--- Secant method does not always conv, but if it does it conv faster than
False Position method
If the slope of the secant line is close to 0… Div
One way to get around apply Bisection method to reach within a safe region then
Apply Secant method.
** Linear Interpolation (False Position) **
A. Algorithm:
Find two points x0 and x1  {f(x0) * f(x1)}<0
(x1 and x0 are on the opposite side of x-axis)
Repeat
Set x2 = x1 – f(x1) * {(x0 - x1) / [f(x0) - f(x1)]} ----------- (1/m)
If f(x2) * f(x0) < 0
Set x1 = x2
Else
Set x0 = x2
EndIf
Until (f(x2) <  (tolerance value) or f(x2) ~ 0
4
B. Examples
f(x) = 3x + sin(x) - ex = 0 , x0 = 0, x1 = 1
f(x) = x3 + x2 - 3x - 3 = 0, x0 = 1, x1 = 2
iteration
1
2
3
4
5
interval
[1,
2]
[1.57142, 2]
[1.70542, 2]
[1.72788, 2]
[ 1.7324, 2]
x2
1.57142
1.70540
f(x0)
-4.0
-1.364
f(x1)
3.0
3.0
1.72788 -0.24784 3.0
1.73140 -0.03936 3.0
………….
f(x2)
-1.36449
-0.24784
-0.03936
-0.00615
C. Convergence
In general it is faster than Bisection Method
Check –
The iteration has reach the tolerance value
(f(xr) <  (tolerance value) or f(xr) ~ 0
The iteration has converged (f(xk) - f(xk-1) < 
D. Conclusion:
** Conver faster than Bisection and more accurate
(False position Method –uses straight line to predict the new point x2
The amount of computation per step requires more time than Bisection)
** If f(x) has significant curvature between x0 and x1, then it damages the
speed of the conv.
*** Newton’s Method ***
A. Algorithm: find xr  f(xr) = 0
Find a point x0 such that x0 is reasonably close to the root.
Compute f(x0) and f (x0)
If f(x0)  0 and f (x0)  0
Repeat
Set x1 = x0
Set (x0 - x1 < tolerance value 1) OR (f(x0 ) < tolerance value 2)
5
B. Examples
1. f(x) = 3x + sin(x) - ex = 0 , x0 = 0.0
Page 49
iteration
1
2
x0
0.0
0.33333
x1
0.333333
0.36017
3
0. 36017
0.3604217
The root is correct up to seven significant digits
2. Find nth root of a function --------___
 N
___
N
3
,
____
N
n
,
……
x2 = k where k = 1,2, 3, …
example -___
 28
___
2
___
2
3
___
2
3K
____
 540
5
or
___
3
3___
3
3K___
3
….
__
5
__
6
___
7
___
5
__
6
7
N
3K__
3K___
3
___
5
3K
3
__
6
3K
3___
7
___
N
3___
N
3. Complex Roots
f(x) = x2 + x +1 = 0, and f (x0) 2X + 1
let x0 = 1.0 + 1.0i
…….
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