Chapter 6 Finding the Roots of Equations The Bisection Method Copyright © The McGraw-Hill...

Chapter 6Finding the Roots of

Equations

The Bisection Method

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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?

Evaluate function at 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)/2Evaluate function at

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.


Chapter 6 Finding the Roots of Equations The Bisection Method Copyright © The McGraw-Hill...

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