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Bracketing Methods
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Chapters 5
Numerical
Methods
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Week 2
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:
() ( )
(Numerical Methods)
().
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Graphical
Method
Bisection
Method
False Position
MethodSimple Fixed
Point Method
Newton Raphson
Method
Secant
Method
Numerical Methods
Roots of Equations
Bracketing Methods
Open Methods
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This chapter on roots of equations deals with methods that use
the fact that a function changes sign around its root.
These techniques are called Bracketing Methods because two guesses
for the root are required, these guesses must bracket the root. These
methods always reduce the width of the bracket.
5.1 Graphical Methods
A simple method for obtaining an estimate of the root of the
equation() is to make a plot of the function and observe whereit crosses the axis. This point, which represents the value for which
() , provides a rough approximation of the root.
x xroot root
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Bracketing Methods
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EXAMPLE 5.1 The Graphical Approach
Problem Statement:
Use the graphical approach to determine the drag coefficient c needed
for a parachutist of mass to have a velocity of 40 m/s afterfree-falling for time .() ( ()) ( )
Solution:using the parameters , , , and :() () ( ())
or
()
(
)
Various values of c can be substituted into
the right-hand side of this equation to
compute the values of().
These points can be plotted as shown in the
figure.
212. . ..
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Special Cases
Graphical techniques are of limited practical value because they are
not precise. However, they can be used to obtain rough estimates ofroots. These estimates can be employed as starting guesses for the
other numerical methods discussed in this and next chapter.
They are also important tools for understanding the properties of the
functions and guessing the pitfalls of the other numerical methods.
Figure (b) shows the case where a single root is
bracketed by negative and positive values of().
However, fig(d) where () and () are also onopposite sides of the axis, shows three rootsoccurring within the interval.
In general, if () and () have opposite signs,there are an odd number of roots in the interval (fig b).
As indicated by fig (a) and (c), if () and ()have the same sign, there are either no roots or even
number of roots between the values.
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Although previous generalizations are usually true,
there are some cases where they do not hold.
For example, functions that are tangential to the axis as in figure(a) and discontinuous functions as infig(b) can violate these principles.
An example of a function that is tangential to the axis
is the cubic equation () ( )( )( ).Notice that makes two terms in thispolynomial equal to zero. Mathematically, iscalled a multiple root (fig a).
.
The existence of such cases makes it difficult to develop general
computer algorithms guaranteed to locate all the roots in an interval.
( (.
()(
.)
( (,().( (..
( )..
Graphical Method Summary
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Problem 5.1
Determine the real roots of() :(a)
Graphically.(b)Using the quadratic formula.
Solution
(a)A plot indicates that roots occur at about and 6.4.
( ( )6, 7 ).
(b)
() ()()
()
( .)
( ) () () () ( )
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5.2 The Bisection Method
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When applying the graphical technique in example 5.1, you have
observed that()changed sign on the opposite sides of the root. In general, if()is realand continuousin the interval from to and
()and()have opposite signs, that is:() () then, there is at least one real root between and .
The bisection method, alternatively called (binary chopping), is one type of
incremental search method in which the interval is always divided in half.
If a function changes sign over an interval, the function value at the
midpoint is evaluated, and the location of the root is then suggested to lie
at the midpoint of that new subinterval.
The process is repeated to obtain refined estimates.
Step 1:Choose lower and upper guesses for the root such that:(ensure that() () )
Step2:Find an estimate of the root
using:
Step3:Determine in which subinterval the root lies:
(a)If() () , the root lies in the lower subinterval.set and return to step 2.
(b)If() () , set and return to step 2.(c)If
() () , the root equals
, terminate.
(iteration).
Steps for the bisection method
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Example 5.3: Bisection Method
Use bisection method to solve the same problem approached
graphically in example 5.1 (page 4).
Solution:
The first step in bisection is to guess two values of the unknown ()that give values for()with different signs.
We can see that the function changes sign between values of 12 and
16. Therefore, the initial estimate of the root () lies at the midpointof the interval
This estimate represents a true percent relative error of
(note that the true value of the root is (14.7802).
Next, we compute the product of the function value at the lower
()and at the midpoint ():() () ()
which is greater than zero.
Since no sign change occurs between the lower
() and the
midpoint (), the root must be located between 14 and 16. Therefore, we create a new interval by redefining the lower ()
to 14 and determining a revised root estimate
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which represents a true percent error of .
()() () )) Therefore, the root is between 14 and 15. The upper () isredefined to 15, and the root estimate for the third iteration is
calculated as:
which represents a percent relative error of . The method can be repeated until the result is accurate enough to
satisfy some standard error (). The following figure shows a graphical description of the bisection
method for the first three iterations.
( )(iterations) ))( .)
.
5.3
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In the previous example the true error does not decrease with each
iteration. However, the interval is halved with each step in the process.
The interval width provides an exact estimate of the upper bound ofthe error for the bisection method.
5.2.1 Termination Criteria and Error Estimates
We must develop an objective criterion for deciding when to terminate
the method. An initial suggestion might be that we end the calculation
when the true error falls below some pre-specified level.
For instance, in Example 5.3, the true percent relative error dropped from
5.3% to 1.9% during the computation. We might decide that we should
terminate when the percent error drops below, say, 0.1% percent.
This strategy is not usefulbecause the error estimates in the example
were based on knowledge of the true root of the function. This would
not be the case in an actual situation. Therefore, we require an error
estimate that does not depend on the knowledge of the root. An
approximate percent relative error can be calculated as||
()
where is the root for the present iteration and is the rootfrom the previous iteration.
The absolute value is used because we are usually concerned with the
magnitude of rather than its sign. When || becomes less than apre-specified stopping criterion , the computation terminates.
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EXAMPLE 5.4 Error Estimates for Bisection
Continue Example 5.3 until the approximate error falls below a stopping
criterion of
. Use the error approximate equation.
Solution:
The result of the first two iterations for
example 5.3 were 14 and 15. Substituting
these values into Eq.(5.2) yields
Recall that the true percent relative true
error for the root estimate of 15 was 1.5%.
Therefore, is greater than . This behavioris clear for the other iterations:
Iteration () ()1 12 16 14 - 5.279
2 14 16 15 6.667 1.487
3 14 15 14.5 3.448 1.896
4 14.5 15 14.75 1.695 0.204
5 14.75 15 14.875 0.840 0.641
6 14.75 14.875 14.8125 0.422 0.219
Thus, after six iterations and the computation can be terminated.( ( ) .)
The plot shows that . thus, when falls below , thecomputation could be terminated with a confidence that the root is
known to be at least as accurate as the pre-specified acceptable level.
( )
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EXAMPLE 5.2
Determine the real root of() (a)
Graphically.(b)Using bisection to locate the root, Employ initial guesses of
and and iterate until the estimated error
Solution
(a)A plot indicates that a single real root occurs at about .(b)First iteration:
() () () (
)Therefore, the new bracket is and
The process can be repeated until () falls below 10%. As summarizedbelow, this occurs after 5 iterations yielding a root estimate of 0.40625.
iteration () () ()() 1 0 1 0.5 -2 0.375 -0.75
2 0 0.5 0.25 -2 -0.73438 1.46875 100.00%
3 0.25 0.5 0.375 -0.73438 -0.18945 0.13913 33.33%
4 0.375 0.5 0.4375 -0.18945 0.08667 -0.01642 14.29%
5 0.375 0.4375 0.40625 -0.18945 -0.05246 0.009939 7.69%
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Problem 5.5
Locate the first non-zero root of sin ( ) , where is in radians.Use a graphical technique and bisection with the initial intervals from 0.5 to1. Perform the computation until is less than of = 2%. Also, perform anerror check by substituting your final answer into the original equation.
Solution
A graph indicates that a nontrivial root (nonzero) is located at about 0.93.
Using bisection,
() () () ()Therefore, the root is in the second interval and the lower guess is
redefined as .All the iterations are displayed in the following table:
() () () 1 0.5 0.354426 1 -0.158529 0.75 0.2597638
2 0.75 0.259764 1 -0.158529 0.875 0.0976216 14.29%
3 0.875 0.097622 1 -0.158529 0.9375 -0.0178935 6.67%
4 0.875 0.097622 0.9375 -0.0178935 0.90625 0.0429034 3.45%
5 0.90625 0.042903 0.9375 -0.0178935 0.921875 0.0132774 1.69%
After five iterations we obtain a root estimate of 0.921875 with an
approximate error of 1.69%, which is below the stopping criterion of 2%.
The result can be checked
() () ()
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5.3 The False-Position Method
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Although bisection is a perfectly valid technique for determining
roots, it's rate is relatively slow and inefficient.
False position is an alternative based on graphical knowledge.
A disadvantage of the bisection method is that, in dividing the
interval from to into equal halves, no consider is taken of themagnitudes of()and().
For example, if ()is much closer to zero than (), it is likelythat the root is closer to than to .
The bisection method uses the graphical knowledge to join ()and()by a straight line. The intersection of this line with the x-axis represents an improved estimate of the root.
The fact that the replacement of the curve by a straight line gives a
"false position" of the root is the origin of its name.
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Using similar triangles,
()
()
()
Which can be solved for
()( ) () () () This is the false position formula. The value of computed with
the previous equation then replaces one of the two initial guesses or , same as in bisection method.
The process is repeated until the root is estimated adequately. The
algorithm is identical to the one for bisection with the exception that
the previous equation is used for step 2.
The same stopping criterion is used to terminate the computation.
()( ) () ()
(()() ) ( )
( ) ( ) ( ) ( )
( (.
Summary for Bisection and False Position Methods
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EXAMPLE 5.5 False Position
Use the false position method to determine the root of the same
equation in Example 5.1.
Solution:
Initiate the computation with guesses of and First iteration:
() () ( )() which has a true relative error of 0.89percent.
Second iteration:
() ()
Therefore, the root lies in the first subinterval, and becomes theupper limit for the next iteration, : ()
()
( ()
which has a true and approximate relatives errors of 0.09 and
0.79percent.
Additional iterations can be performed to improve the estimate of roots.
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EXAMPLE 5.3
Determine the real root of
() (a)Graphically.(b)Using bisection to determine the root to .
Employ initial guesses of and .(c)Perform the same computation as in (b) but use the false-position
method and
.
(a)A plot indicates that a single real root occurs at about .(b) Bisection:
First iteration:
( ) ()() () The new bracket is and . The process is repeateduntil . This occurs after 4 iterations yielding a root estimate of0.59375.
iteration () () () () 1 0.50000 1.00000 0.75000 -1.47813 2.07236 -3.06321
2 0.50000 0.75000 0.62500 -1.47813 0.68199 -1.00807 20.00%
3 0.50000 0.62500 0.56250 -1.47813 -0.28199 0.41682 11.11%
4 0.56250 0.62500 0.59375 -0.28199 0.22645 -0.06386 5.26%
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(c)False position:
First iteration:
()
() ( )
() () () Therefore, the bracket is and .Second iteration: () () ( )
The process can be repeated until the approximate error falls below
0.2%. This occurs after 4 iterations yielding a root estimate of 0.57956.
iteration () () () ()() 1 0.5 1.00000 -1.47813 3.70000 0.64273 0.91879 -1.35808 -
2 0.5 0.64273 -1.47813 0.91879 0.58802 0.13729 -0.20293 9.304%
3 0.5 0.58802 -1.47813 0.13729 0.58054 0.01822 -0.02693 1.289%
4 0.5 0.58054 -1.47813 0.01822 0.57956 0.00238 -0.00351 0.169%
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Problem 5.8
Find the positive square root of 18 using the false-position method
within
. Employ initial guesses of
and
.
Solution
The square root of 18 can be set up as a roots' problem by determining
the positive root of the function:
()
Using false position, the first iteration is
( ) () () () Therefore, the root is in the second interval and the lower guess is
redefined as . The second iteration is ( )
Thus, the computation can be stopped after just two iterations because
0.442% < 0.5%.