week 10 - numerical root-finding

7/24/2019 Week 10 - Numerical Root-Finding

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Numerical Root-Finding:Bracketing Methods


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Non-linear Equations

Linear Equations Non-linear Equations

10 + 5 = 15 + 2 = 2 + + 1 = 0 = cos

= 1

=1

+1

= =

= ln = +

= = ( 1 )

What are Nonlinear Equations?


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Root-finding

What is Root-finding?

Example: Find the roots of + 6 + 5 = 0

Example: Find the roots of 2 + 2 = 0 .

Note:

Solve for x:

= ln

Find the roots of:

ln = 0 is the

same as


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Outline

Numerical Root-Finding Methods

1. Bisection Method

2. Regula-Falsi Method

Example: Solve for x in

+ 2 = 2


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Non-linear Equations

Example: Solve for x in + 2 = 2

Lets do the Bisection Method!

1. Turn your equation into a function in x .

= 2 2

*Again, solving for x means finding the rootof f(x), i.e. f(x) = 0 .


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Bisection Method

2. Plot . Find an interval [ a , b] where 1 root exists.

= Choose theinterval such that is monotonic in[a , b] and < 0.

= = = =

Lets say: [1, 3]

NO GOOD!


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Bisection Method



= = = =

Lets say: [4, 5]

Exactly 1root exists!


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Bisection Method

3. Find the midpoint of [ a , b] and label it as c. Evaluate , and ( ).

= . = .

=

=

Bounds: 4, 5

= 4

= 5

= +2

= .


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Bisection Method

4. If < 0, then set the new bounds to [ a , c].If < 0, then set the new bounds to [ c, b].

= . = .

=

=

= .

= 1.887


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Bisection Method

Repeat step 3. Find the midpoint of [ a , b] and label itas c. Evaluate , and ( ).

= . = . . = .

=

Bounds: 4, 4.5 = 4

= 4.5

= .


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Bisection Method


= . = . . = .

=

= 2.070

= .


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Bisection Method

How do you know when to stop?

Ans. Set a tolerance!Lets say, = 0.01


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Bisection Method

Lets look at what really happens

= 2 2


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Bisection Method

Lets look at what really happens Iter a b c f(a) f(b) f(c)

0 4 5 4.5 -2 5 0.377417

1 4 4.5 4.25 -2 0.377417 -1.035192 4.25 4.5 4.375 -1.03519 0.377417 -0.39119

3 4.375 4.5 4.4375 -0.39119 0.377417 -0.02332

4 4.4375 4.5 4.46875 -0.02332 0.377417 0.172832

5 4.4375 4.46875 4.453125 -0.02332 0.172832 0.073717

6 4.4375 4.453125 4.445313 -0.02332 0.073717 0.024941

= 2 2

At Iter = 6 , since 4.453125 4.445313 < 0.01 , then we stop!


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Bisection Method

Lets look at what really happens Iter A B C f(A) f(B) f(C)

14 4.441284 4.441345 4.441314697 -6.6E-06 0.00037 0.00018215 4.441284 4.441315 4.441299438 -6.6E-06 0.000182 8.77E-05

16 4.441284 4.441299 4.441291809 -6.6E-06 8.77E-05 4.06E-05

17 4.441284 4.441292 4.441287994 -6.6E-06 4.06E-05 1.7E-05

18 4.441284 4.441288 4.441286087 -6.6E-06 1.7E-05 5.23E-06

19 4.441284 4.441286 4.441285133 -6.6E-06 5.23E-06 -6.6E-07

= 2 2

At Iter = 19 , since 4.44128513 4.44128609 < 0.000001 , then we stop!

For a tolerance of 0.000001


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Bisection Method

1. Turn your equation into a function of x. Set a tolerance value

(typically 0.0001).2. Plot . Find an interval [ a , b] where 1 root exists.3. Find the midpoint of [ a , b] and label it as c. Evaluate

, and ( ).

4. If < 0, then set the new bounds to [ a , c]. If < 0, then set the new bounds to [ c, b].

5. If < , then STOP. Report = as youranswer. Otherwise, go to step 3.

Summary of steps:


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Bisection Method

Alternatively,instead ofhaving atolerance, youcan also set the

maximumnumber ofiterations .

N = 4


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Bisection Method

YOUR TASK:

Develop a program that carries out the bisectionmethod. Prompt the user for the initial guessinterval [a, b] , and the desired tolerance .

Output the final root , and the number ofiterations needed to reach that final value.


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Regula-Falsi Method

Example: Solve for x in + 2 = 2

Lets do the Regula-falsi Method!

1. Turn your equation into a function in x .

= 2 2

*Again, solving for x means finding the rootof f(x), i.e. f(x) = 0 .


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Regula-Falsi Method



From before:[a, b] = [4, 5]

Exactly 1root exists!


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Regula-Falsi Method

3. Find c using the following formula. Evaluate ( ).

=

= ( )( ) ( )

= =

. = .

= 5 (5)(5 4)

5 (4)

= 557

=307

= 4.286

= +2


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Regula-Falsi Method

Why = ( )( ) ( )

?

f(b)

f(b) f(a)

a b

From similar triangles:

( ) ( )

=

b ac

b c


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Regula-Falsi Method


=

=

=

= 1.726

= .

. = .


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Regula-Falsi Method

1. Turn your equation into a function of x. Set a tolerance value(typically 0.0001).


3. Find = ( )( ) ( )

. Evaluate ( ).

4. If < 0, then set the new bounds to [ a , c]. If < 0, then set the new bounds to [ c, b].

5. If < , then STOP. Report = as youranswer. Otherwise, go to step 3.

Summary of steps:


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Regula-Falsi Method

YOUR TASK:

Develop a program that carries out the regula-falsi(false-position) method. Prompt the user for theinitial guess [a, b] , and the desired tolerance .

Output the final root , and the number ofiterations needed to reach that final value.


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Bracketing Methods

Problems in theBracketing Methods:

- Knowing where to start- Slow convergence


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Solve for the only positive value of x that satisfies:

Use the following methods. Which converges fastest?

1. Bisection Method2. Regula-Falsi Method

Use a tolerance of 0.00001 .

Practice

1

1 12

x x x e

week 10 - numerical root-finding

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