Chapter 2 solving nonlinear equations

Chapter 2 Solving Nonlinear Equations
Consider a scalar-valued function f (x) and the corresponding equation
f (x)= 0.
- A solution (or a root) to this equation is the value of x that satisﬁes this
equation.
- This solution is the point on the x-axis where f (x) crosses or touches the
x-axis.
- Then only one of the following can occur.
f(x) has a unique solution
f(x) has no solution
f(x) has many (can be inﬁnitely many) solutions

Approaches for finding roots of equations numerically
Consider again the equation f (x)= 0.
Bracketing methods
- Use an initial interval that contains the solution
- Repeatedly decrease the size of this interval to get as close to the exact
solution as possible
- Always converge to a solution
- Examples: Bisection method, Falsi method
Open methods
- Require a “good” initial guess (close enough to the true solution)
- May not converge to a solution (depending on the initial guess).
- If the open methods converges to a solution, it is usually more efficient than
the bracketing methods (i.e. converge faster)
- Example: Newton-Raphson method, secant method, fixed-point iteration
Bisection Method:
Step 1: Find an interval [a, b] such that, for the root 0,
=
)
(x
f 
f (a)f (b) < 0
Step 2: Given [a, b], compute approximate solution by using the midpoint:
2
~ b
a
x


Step 3: Decrease the interval:
If 0
)
~
(
)
( 
x
f
a
f , set x
b ~

If 0
)
~
(
)
( 
x
f
a
f , set x
a ~

Update new interval [a, b] and go back to Step 2.
Given [a, b] and TOL, if n is the number of iterations in Bisection method
TOL
a
b
n

 2
log
Example Let a = 0.1 and b = 1.0. Suppose the equation f (x)= 0 has a root
in the interval [a, b]. How many steps of the bisection method are needed to
determine the root with an error of at most 0.5 x 10-8
if [a, b] is used as an
initial interval?

Example Compute the approximate root of x4
- 2x3
- 10 = 0 by using three
iterations of the bisection method with the initial interval [1, 3]. Calculate the
residual, absolute error and relative error in each iteration

Example Suppose we want to determine where the graphs of y = 3x and y =
ex
intersect on the interval [0, 1].Compute the approximate root from the ﬁrst
three iterations with the initial interval [0, 1] by using the bisection method.
Calculate the residual, absolute error and relative error in the last iteration.

False Position Method
Step 1: Find an interval [a, b] such that, for the root 0,
=
)
(x
f 
f (a)f (b) < 0
Step 2: Given [a, b], compute approximate solution by using:
)
(
)
(
)
(
)
(
~
a
f
b
f
a
bf
b
af
x



Step 3: Decrease the interval:
If 0
)
~
(
)
( 
x
f
a
f , set x
b ~

If 0
)
~
(
)
( 
x
f
a
f , set x
a ~

Update new interval [a, b] and go back to Step 2.
- 2x3
iterations of the false position method with the initial interval [1, 3].
Calculate the absolute error and relative error in each iteration

Newton’s Method
)
(
)
(
1
1
1






j
j
j
j
x
f
x
f
x
x
Example Suppose we want to determine where the graphs of y = 3x and y =
ex
intersect on the interval [0, 1].Compute the approximate root by using
Newton’s Method with the initial value x0 = 1 repeated until the residual error
is less than 10-4
. How many iterations are required to perform?

Secant Method
)
(
)
(
)
(
1
1
1







j
j
j
j
j
j
j
x
f
x
f
x
x
x
f
x
x
- 2x3
iterations of the secant method with the initial values x0 = 1, x1 = 2. Calculate
the absolute error in the last iteration.

Example: The following table shows the approximated root of
[0,1]
x
0
3 

 x
e x
by using Secant method with the initial points x0 = 0, x1 = 1.
- Determine the values of a1, b2, c3 in the table.
- Suppose we want to have the approximation with absolute error less than
5 x10-2
. How many iterations has to be used?

Fixed-Point Iteration Method
Re-write the equation as x = g(x)
• Choose a starting value x0
• While(stopping condition not met)
xi+1 = g(xi)
Example: Apply the ﬁxed-point procedure to approximate a root of f (x)= 0
where f (x)= cos(x) - xex
with
x
xe
x
x
g
e
x
x
g
x
x




)
cos(
)
(
)
cos(
)
(
2
1
and starting with x0 = 1 for both g1 and g2.

Chapter 2 solving nonlinear equations

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