Week02_Bracketing Methods.pptx
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CHE 555NUMERICAL METHODS AND
OPTIMIZATION
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WEEK 2
Roots of Equations: Bracketing method Graphical method Bisection method False Position method
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At the end of this topic, the students will be able:
To identify and apply the bracketing methods to solve roots of equations
LESSON OUTCOMES
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• Root of equation is obtained at value of x when the f(x)=0• Before digital computers, root of algebraic and transcendental
equations could be found by direct method that can be solved analytically, but not easy for complex
function plot function and determine where it crosses the x axis, but it lacks of
precision
• Why?
• But
aacbbxcbxax
240
22
?0sin?02345
xxxxfexdxcxbxax
Roots of Equation
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Roots of Equations
Bracketing methods Open methods
Methods start with two initial guesses that bracket (or contain) the root and then systematically reduce the width of the bracket.
Methods involve systematic trial-and-error iterations but no need for the initial guesses to bracket the root.
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Graphical Methods
• This is a simple method to get an estimate of the root of the equation f(x) =0. • The function f(x) is plotted against x and the location(s) where it crosses the x axis (i.e. f(x) = 0) provides a rough approximation of the root(s).
• Graphical techniques are limited practical value (not precise).
• It can be utilized to obtain rough estimates of roots.
• These estimates can be used as initial guesses for advanced numerical methods.
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Example
Use the graphical approach to determine the drag coefficient c needed for a parachutist of mass m = 68.1kg to have a velocity of 40m/s after free falling for time t=10s. Note: acceleration due to gravity is 9.8 m/s2.
SolutionFrom equation that derived from Newton’s second law for the parachutist’s velocity:
It can be expressed in the form of f(c) = 0.
vec
gmcf tmc )1()( )/(
Inserting all the known parameters:
40)1(38.667)( 146843.0 cec
cf
Various values of c can be inserted into the right hand side of eq. to calculate f(c)
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The resulting curve crosses the c axis between 12 and 16.
Visual inspection of the plot provides a rough estimate of the root at 14.75.
c f(c)4 34.115
8 17.653
12 6.06716 -2.26920 -8.401
f(xl) f(xu) < 0
f(xl) and f(xu) have different signs then,
Lower bound, f(xl)
Upper bound, f(xu)
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f(x)
x
* *
f(x)
x
* *
f(x)
x
*
*
f(x)
x
*
*
a. f(xl) and f(xu) have the same sign, no roots or even number of roots in interval
c. f(xl) and f(xu) have the same sign, there are roots or even number of roots in interval
b. Function has different signs at the end points, there will be an odd number of roots in the interval
d. Function has different signs at the end points, there will be an odd number of roots in the interval
Xl Xu
Xl
Xu
Xl Xu
Xl
Xu
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Bisection MethodFrom the graphical method, we found that when a function f(x) is continuous and real in the interval from xl to xu , and f(xl) and f(xu) have opposite signs,
then there is at least one real root between xl and xu.
f(xl) f(xu) < 0
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Choose xl and xu guesses point, and ensure f(xl)f(xu) < 0
Estimate the root, xr by
• If f(xl) f(xr) <0 the root lies in the lower subinterval, then set xr as xu, and repeat to find new xr
If f(xl) f(xr) >0 the root lies in the upper subinterval, then set xr as xl, and repeat to find new xr
If f(xl)f(xr)=0 the root equals to xr, STOP!!
Compare ɛs with ɛa.If ɛa< ɛs, stop.
Otherwise repeat the process.
Procedure for bisection method
% 100εa newr
oldr
newr
xxx
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ExampleUse the bisection method to determine the drag coefficient c needed for a parachutist of mass m = 68.1kg to have a velocity of 40m/s after free falling for time t=10s. Note: acceleration due to gravity is 9.8 m/s2. Given εs=0.5% and true value = 14.7802.
Solution Guess two values of the unknown that give values for f(c) with different signs. From previous example, f(c) changes sign between 12 and 16. Therefore, set xl = 12 and xu = 16.
Estimate the root,
Compute the product of function value at lower bound and at midpoint,
No sign change occurs between lower bound and midpoint. Thus, root must be located between 14 and 16. Set xl = 14 and xu = 16.
142
16122
ulr
xxx
0 519.9)569.1(067.6)14()12()()( ffxfxf rl
40)1(38.667)( 146843.0 cec
cf
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Calculate new xr,
Compare εa with εs.
Compute the product of function value at lower bound and at midpoint,
The root is between 14 and 15. Set xl = 14 and xu = 15.
Calculate new xr,
Compare εa with εs.
152
16142
ulr
xxx
0 667.0)425.0(569.1)15()14()()( ffxfxf rl
5.142
15142
ulr
xxx
sa ε%667.6% 10015
1415% 100ε
newr
oldr
newr
xxx
sa ε%448.3% 1005.14155.14% 100ε
new
r
oldr
newr
xxx
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Repeat the calculation until termination criteria is met.
Iteration xl xu xr εa(%) εt(%)
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
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Exercise
Find the root of the 3rd-order polynomial using bisection
in the interval [3.75,5]. The tolerance εs is 0.5%.
0810)( 23 xxxxf
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False Position Method The false-position method takes into account the magnitudes of the function at the lower and upper bounds, f(xl) and f(xu).
For example, if f(xl) is closer to zero than f(xu) then xl is likely to be closer to the root xr than xu is.
This results in an improved estimate of the root.
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)()(
))((
uxflxf
uxlxuxfuxrx
If a real root is bounded by xl and xu of f(x) = 0, then we can approximate the solution by doing a linear interpolation between the points [xl, f(xl)] and [xu, f(xu)] to find the xr value such that l(xr) = 0, l(x) is the linear approximation of f(x).Thus, using similar triangle we can write,
Solving for xr ,
l(xr)
Derivation of method: page 125
ur
u
lr
l
xxxf
xxxf
)()(
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Choose xl and xu guesses point, , and ensure f(xl)f(xu) < 0
Estimate the root, xr from
If f(xr) = 0 the root equals to xr, terminate the computation
Compare εs with εa If εa< εs, stop. Otherwise repeat the process.
Procedure for false position method
)()())((
ul
uluur xfxf
xxxfxx
Evaluate f(xr), replace xr whichever of xl or xu yields a function value with the same sign as f(xr).
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ExampleUse the false position method with guess of xl = 12 and xu = 16 to determine the drag coefficient c needed for a parachutist of mass m = 68.1kg to have a velocity of 40m/s after free falling for time t=10s. Note: acceleration due to gravity is 9.8 m/s2. Given εs=0.5% and true value = 14.7802.
40)1(38.667)( 146843.0 cec
cf
First iteration
9113.14)2688.2(0669.6
)1612(2688.216
2688.2)( 16
0669.6)( 12
rx
uxfux
lxflx
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Second iteration
%79.0%1007942.14
9113.147942.14
7942.14)2543.0(0669.6
)9113.1412(2543.09113.14
]by replace [ 2543.0)( 9113.14
0669.6)( 12
5428.1)2543.0)(0669.6()()(
a
r
r
rx
xuxuxfux
lxflx
xflxf
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Third iteration
%0846.0%1007817.14
7942.147817.14
7817.14)0273.0(0669.6
)7942.1412(0273.07942.14
]by replace [ 0273.0)( 7942.14
0669.6)( 12
1656.0)0273.0)(0669.6()()(
a
r
r
rx
xuxuxfux
lxflx
xflxf
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22
Exercise
Find the root of the 3rd-order polynomial using FP
in the interval [3.75,5]. The tolerance εs is 0.5%.
0810)( 23 xxxxf
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23
Use both the bisection and false-position method to find the root of
in the interval [0, 2] until the approximate percent relative error is less than 5%.
02)( 4 xxf
Lets try
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Exercise(5.4) Determine the roots of f(x) = -12 – 21x +18x2 – 2.75x3 with
a) Bisection b) False position
Using initial guess of xl = -1 and xu = 0 and stopping criterion of 1%.
(5.6) Determine the positive real root of ln(x4)=0.7 a) Using three iterations of the bisection method with
initial guesses of xl = 0.5 and xu = 2, b) Using three iterations of the false position method with
same initial guesses as in (a).