bisection method

Chapter 6
Finding the Roots of
Equations
The Bisection Method
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Finding Roots of Equations
• In this chapter, we are examining equations with
one independent variable.
• These equations may be linear or non-linear
• Non-linear equations may be polynomials or
generally non-linear equations
• A root of the equation is simply a value of the
independent variable that satisfies the equation
Engineering Computation: An Introduction Using MATLAB and Excel

Classification of Equations
• Linear: independent variable appears to the first
power only, either alone or multiplied by a
constant
• Nonlinear:
– Polynomial: independent variable appears
raised to powers of positive integers only
– General non-linear: all other equations

Finding Roots of Equations
• As with our method for solving simultaneous non-
linear equations, we often set the equation to be
equal to zero when the equation is satisfied
• Example:
• If we say that
then when f(y) =0, the equation is satisfied

Solution Methods
• Linear: Easily solved analytically
• Polynomials: Some can be solved analytically
(such as by quadratic formula), but most will
require numerical solution
• General non-linear: unless very simple, will
require numerical solution

The Bisection Method
• In the bisection method, we start with an interval
(initial low and high guesses) and halve its width
until the interval is sufficiently small
• As long as the initial guesses are such that the
function has opposite signs at the two ends of the
interval, this method will converge to a solution
• Example: Consider the function

Bisection Method Example
• Consider an initial interval of ylower = -10 to yupper = 10
• Since the signs are opposite, we know that the
method will converge to a root of the equation
• The value of the function at the midpoint of the
interval is:

• The method can be better understood by looking
at a graph of the function:
Interval

• Now we eliminate half of the interval, keeping the
half where the sign of f(midpoint) is opposite the
sign of f(endpoint)
• In this case, since f(ymid) = -6 and f(yupper) = 64, we
keep the upper half of the interval, since the
function crosses zero in this interval

• The interval has now been bisected, or halved:
New Interval

• New interval: ylower = 0, yupper = 10, ymid = 5
• Function values:
• Since f(ylower) and f(ymid) have opposite signs, the
lower half of the interval is kept

• At each step, the difference between the high and
low values of y is compared to 2*(allowable error)
• If the difference is greater, than the procedure
continues
• Suppose we set the allowable error at 0.0005. As
long as the width of the interval is greater than
0.001, we will continue to halve the interval
• When the width is less than 0.001, then the
midpoint of the range becomes our answer

• Excel solution:
Initial
Guesses
Is interval width
narrow enough to stop?
Evaluate function at
lower and mid values.
If signs are same (+ product),
eliminate lower half of interval.

• Next iteration:
New Interval (if statements
based on product at the end
of previous row)
Is interval width narrow
enough to stop?
lower and mid values.
If signs are different (- product),
eliminate upper half of interval.

• Continue until interval width < 2*error (16
iterations)
Answer:
y = 0.857

• Or course, we know that the exact answer is 6/7
(0.857143)
• If we wanted our answer accurate to 5 decimal
places, we could set the allowable error to
0.000005
• This increases the number of iterations only from
16 to 22 – the halving process quickly reduces the
interval to very small values
• Even if the initial guesses are set to -10,000 and
10000, only 32 iterations are required to get a
solution accurate to 5 decimal places

Bisection Method Example - Polynomial
• Now consider this example:
• Use the bisection method, with allowed error of
0.0001

• If limits of -10 to 0
are selected, the
solution converges
to x = -2

• If limits of 0 to 10
are selected, the
solution converges
to x = 4

• If limits of -10 to 10 are selected, which root is
found?
• In this case f(-10) and f(10) are both positive, and
f(0) is negative

• Which half of the interval is kept?
• Depends on the algorithm used – in our example,
if the function values for the lower limit and
midpoint are of opposite signs, we keep the lower
half of the interval

• Therefore, we converge to the negative root

In-Class Exercise
• Draw a flow chart of the algorithm used to find a
root of an equation using the bisection method
• Write the MATLAB code to determine a root of
within the interval x = 0 to 10

Input lower and upper
limits low and high
Define tolerance tol
while high-low > 2*tol
mid = (high+low)/2
lower limit and
midpoint:
fl = f(low), fm = f(mid)
fl*fm > 0?
Keep upper half of
range:
low = mid
Keep lower half
of range:
high = mid
Display root (mid)
YESNO

MATLAB Solution
• Consider defining the function
as a MATLAB function “fun1”
• This will allow our bisection program to be used
on other functions without editing the program –
only the MATLAB function needs to be modified

MATLAB Function
function y = fun1(x)
y = exp(x) - 15*x -10;
Check values at x = 0 and x = 10:
>> fun1(0)
ans =
-9
>> fun1(10)
ans =
2.1866e+004
Different signs, so a root exists within this range

Set tolerance to 0.00001;
answer will be accurate to 5
decimal places

Find Root
>> bisect
Enter the lower limit 0
Enter the upper limit 10
Root found: 4.3135

What if No Root Exists?
• Try interval of 0 to 3:
>> bisect
Root found: 3.0000
• This value is not a root – we might want to add a
check to see if the converged value is a root

Modified Code
• Add “solution tolerance” (usually looser than
convergence tolerance):
• Add check at end of program:

Check Revised Code
>> bisect
No root found

Numerical Tools
• Of course, Excel and MATLAB have built-in tools
for finding roots of equations
• However, the examples we have considered
illustrate an important concept about non-linear
solutions:
Remember that there may be many roots to a
non-linear equation. Even when specifying an
interval to be searched, keep in mind that there
may be multiple solutions (or no solution) within
the interval.

bisection method

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