Math306&307 - Neumerical Analysis - Lec 1 - Bisection Method

Math 307 - Programming & Numerical MethodsNumerical Analysis through C++

Sqn Ldr Athar Kharal

Humanities and Science DepartmentCollege of Aeronautical Engineering

PAF Academy Risalpur

Sqn Ldr Athar Kharal (H and S, CAE) Fundamentals of Scienti�c Computing 1 / 11

Roots of an equation: Bisection Method

De�nitionFor the equation f (x) , a value x = x0 is said to be its root if f (x0) = 0.

Example

Suppose we have equation x2 � 3x + 1.Then x = 2. 62 and x = 0.382 arethe two roots of this equation because�

x2 � 3x + 1�jx=2.62 = �7. 6� 10�5 t 0

and�x2 � 3x + 1

�jx=0.382 = �8. 92� 10�6 t 0.

Questions:

I How we know these two are the roots of this equation?I Answer: Wait!

I What is a root, graphically?I Answer: Value of x for which f (x) intersects the x-axis.


Roots of an equation: Bisection MethodDe�nitionFor the equation f (x) , a value x = x0 is said to be its root if f (x0) = 0.

Example


x2 � 3x + 1�jx=2.62 = �7. 6� 10�5 t 0

and�x2 � 3x + 1

�jx=0.382 = �8. 92� 10�6 t 0.

Questions:





Example


x2 � 3x + 1�jx=2.62 = �7. 6� 10�5 t 0

and�x2 � 3x + 1

�jx=0.382 = �8. 92� 10�6 t 0.

Questions:





Example


x2 � 3x + 1�jx=2.62 = �7. 6� 10�5 t 0

and�x2 � 3x + 1

�jx=0.382 = �8. 92� 10�6 t 0.

Questions:





Example


x2 � 3x + 1�jx=2.62 = �7. 6� 10�5 t 0

and�x2 � 3x + 1

�jx=0.382 = �8. 92� 10�6 t 0.

Questions:I How we know these two are the roots of this equation?

I Answer: Wait!




Example


x2 � 3x + 1�jx=2.62 = �7. 6� 10�5 t 0

and�x2 � 3x + 1

�jx=0.382 = �8. 92� 10�6 t 0.

Questions:I How we know these two are the roots of this equation?I Answer: Wait!




Example


x2 � 3x + 1�jx=2.62 = �7. 6� 10�5 t 0

and�x2 � 3x + 1

�jx=0.382 = �8. 92� 10�6 t 0.


I What is a root, graphically?

I Answer: Value of x for which f (x) intersects the x-axis.



Example


x2 � 3x + 1�jx=2.62 = �7. 6� 10�5 t 0

and�x2 � 3x + 1

�jx=0.382 = �8. 92� 10�6 t 0.




Lets see some examples:

Here is plot ofx2 � 3x + 1

Example

1 1 2 3 4

2

1

1

2

x

y

Roots: 0.38 and 2.62

Here is another plot forx3 � 9x2 + 23x � 15. Weshall call this eq. as CAEequation. Why?

Example

2 4 6

4

2

2

4

x

y

Roots: ?, ?, ?




Example

1 1 2 3 4

2

1

1

2

x

y



Example

2 4 6

4

2

2

4

x

y

Roots: ?, ?, ?




Example

1 1 2 3 4

2

1

1

2

x

y



Example

2 4 6

4

2

2

4

x

y

Roots: ?, ?, ?




Example

1 1 2 3 4

2

1

1

2

x

y



Example

2 4 6

4

2

2

4

x

y

Roots: ?, ?, ?


Example

Here is another plot of x5 � 52x4 + 945x3 � 7094x2 + 20 600x � 14 400

5 10 15 20

5000

10000

15000

20000

x

y

Make the word out of roots?

Announcement:

I Make the word and get a gift of my choice. Guess, what my choice is?


Example

Here is another plot of x5 � 52x4 + 945x3 � 7094x2 + 20 600x � 14 400

5 10 15 20

5000

10000

15000

20000

x

y

Make the word out of roots?

Announcement:I Make the word and get a gift of my choice. Guess, what my choice is?


Sign-jumps and Roots

De�nitionFor two points xi and xj , if f (xi ) and f (xj ) have opposite signs, then wesay that f (x) has a sign-jump in the interval [xi , xj ] .

ProblemIs there some relation between the roots and sign-jumps of f (x) ?

10 200

10000

20000

x

y

Look for sign-jumps

SolutionIf there is a sign-jump in some interval of x-axis, then there always lies aroot in that interval.





10 200

10000

20000

x

y

Look for sign-jumps






10 200

10000

20000

x

y

Look for sign-jumps






10 200

10000

20000

x

y

Look for sign-jumps

SolutionIf there is a sign-jump in some interval of x-axis, then there always lies aroot in that interval.Sqn Ldr Athar Kharal (H and S, CAE) Fundamentals of Scienti�c Computing 5 / 11

Metohd of Bisection

1 Check some values of f (x) to �nd a sign jump and get searchinterval as [a, b] .

2 If f (a) = 0 or f (b) = 0 then root is a or b. Stop. Otherwise proceedforward.

3 Calculate x0 =a+ b2, check if jf (x0)j � δ, if not then proceed

4 Check sign jump in [a, x0] and [x0, b] .


Metohd of Bisection






Metohd of Bisection






Metohd of Bisection






Example

We �nd a real root of the equation f (x) = x3 � x � 1, within j.01j ofexact value.

SolutionFor getting an initial guess of a and b, let us look for some sign-jumps:

x 0 1 2f (x) �1 �1 5

Hence a real root lies in the interval [1, 2] i.e. a = 1, b = 2.

So the �rst approximation to the exact root is x0 =a+ b2

=1+ 22

= 1. 5.

Now we have to check if x0 is itself a root or not.As f (1.5) = 0.875 > .001, hence 1.5 is not a root it is just an improvedapproximation.Hence the root is either in sub-interval : [a, x0] = [1, 1.5] or in[x0, b] = [1.5, 2] .


Example



x 0 1 2f (x) �1 �1 5



=1+ 22

= 1. 5.



Example



x 0 1 2f (x) �1 �1 5



=1+ 22

= 1. 5.



Example



x 0 1 2f (x) �1 �1 5

Hence a real root lies in the interval [1, 2]

i.e. a = 1, b = 2.


=1+ 22

= 1. 5.



Example



x 0 1 2f (x) �1 �1 5



=1+ 22

= 1. 5.



Example



x 0 1 2f (x) �1 �1 5



=1+ 22

= 1. 5.



Example



x 0 1 2f (x) �1 �1 5



=1+ 22

= 1. 5.

Now we have to check if x0 is itself a root or not.

As f (1.5) = 0.875 > .001, hence 1.5 is not a root it is just an improvedapproximation.Hence the root is either in sub-interval : [a, x0] = [1, 1.5] or in[x0, b] = [1.5, 2] .


Example



x 0 1 2f (x) �1 �1 5



=1+ 22

= 1. 5.

Now we have to check if x0 is itself a root or not.As f (1.5) = 0.875 > .001, hence 1.5 is not a root it is just an improvedapproximation.

Hence the root is either in sub-interval : [a, x0] = [1, 1.5] or in[x0, b] = [1.5, 2] .


Example



x 0 1 2f (x) �1 �1 5



=1+ 22

= 1. 5.



Solution

For this sign-jumps are to be checked: f (a) = �1 and f (1.5) = +0.875.So we proceed to compute x1 from [1, 1.5] as: x1 =

a+ x02

=1+ 1.52

=

1. 25,

f (x1) = f (1.25) = �0.2969 < 0.x2 =x0 + x12

=1.5+ 1.25

2= 1.3750

and f (x2) = 0.2246 and jf (x2)j 6� 0.01

x3 =x1 + x22

=1.3750+ 1.25

2= 1.3125 but f (x3) = �0.0515 and

jf (x3)j 6� 0.01

x4 =x2 + x32

= 1.3438 and f (x4) = 0.0826 and jf (x4)j 6� 0.01.

x5 =x3 + x42

= 1.3282 and f (x5) = 0.0146 and jf (x5)j 6� 0.01

x6 =x5 + x42

= 1.3204 and f (x6) = �0.0186 and jf (x6)j 6� 0.01

x7 =x6 + x52

= 1.3243 and f (x7) = �0.0018 and jf (x7)j < 0.01, henceour required root is

x7 = 1.3243. �


SolutionFor this sign-jumps are to be checked: f (a) = �1 and f (1.5) = +0.875.So we proceed to compute x1 from [1, 1.5] as:

x1 =a+ x02

=1+ 1.52

=

1. 25,

f (x1) = f (1.25) = �0.2969 < 0.x2 =x0 + x12

=1.5+ 1.25

2= 1.3750

and f (x2) = 0.2246 and jf (x2)j 6� 0.01

x3 =x1 + x22

=1.3750+ 1.25

2= 1.3125 but f (x3) = �0.0515 and

jf (x3)j 6� 0.01

x4 =x2 + x32

= 1.3438 and f (x4) = 0.0826 and jf (x4)j 6� 0.01.

x5 =x3 + x42

= 1.3282 and f (x5) = 0.0146 and jf (x5)j 6� 0.01

x6 =x5 + x42

= 1.3204 and f (x6) = �0.0186 and jf (x6)j 6� 0.01

x7 =x6 + x52


x7 = 1.3243. �


SolutionFor this sign-jumps are to be checked: f (a) = �1 and f (1.5) = +0.875.So we proceed to compute x1 from [1, 1.5] as: x1 =

a+ x02

=1+ 1.52

=

1. 25,

f (x1) = f (1.25) = �0.2969 < 0.x2 =x0 + x12

=1.5+ 1.25

2= 1.3750

and f (x2) = 0.2246 and jf (x2)j 6� 0.01

x3 =x1 + x22

=1.3750+ 1.25

2= 1.3125 but f (x3) = �0.0515 and

jf (x3)j 6� 0.01

x4 =x2 + x32

= 1.3438 and f (x4) = 0.0826 and jf (x4)j 6� 0.01.

x5 =x3 + x42

= 1.3282 and f (x5) = 0.0146 and jf (x5)j 6� 0.01

x6 =x5 + x42

= 1.3204 and f (x6) = �0.0186 and jf (x6)j 6� 0.01

x7 =x6 + x52


x7 = 1.3243. �



a+ x02

=1+ 1.52

=

1. 25,

f (x1) = f (1.25) = �0.2969 < 0.

x2 =x0 + x12

=1.5+ 1.25

2= 1.3750

and f (x2) = 0.2246 and jf (x2)j 6� 0.01

x3 =x1 + x22

=1.3750+ 1.25

2= 1.3125 but f (x3) = �0.0515 and

jf (x3)j 6� 0.01

x4 =x2 + x32

= 1.3438 and f (x4) = 0.0826 and jf (x4)j 6� 0.01.

x5 =x3 + x42

= 1.3282 and f (x5) = 0.0146 and jf (x5)j 6� 0.01

x6 =x5 + x42

= 1.3204 and f (x6) = �0.0186 and jf (x6)j 6� 0.01

x7 =x6 + x52


x7 = 1.3243. �



a+ x02

=1+ 1.52

=

1. 25,

f (x1) = f (1.25) = �0.2969 < 0.x2 =x0 + x12

=1.5+ 1.25

2= 1.3750

and f (x2) = 0.2246 and jf (x2)j 6� 0.01

x3 =x1 + x22

=1.3750+ 1.25

2= 1.3125 but f (x3) = �0.0515 and

jf (x3)j 6� 0.01

x4 =x2 + x32

= 1.3438 and f (x4) = 0.0826 and jf (x4)j 6� 0.01.

x5 =x3 + x42

= 1.3282 and f (x5) = 0.0146 and jf (x5)j 6� 0.01

x6 =x5 + x42

= 1.3204 and f (x6) = �0.0186 and jf (x6)j 6� 0.01

x7 =x6 + x52


x7 = 1.3243. �



a+ x02

=1+ 1.52

=

1. 25,

f (x1) = f (1.25) = �0.2969 < 0.x2 =x0 + x12

=1.5+ 1.25

2= 1.3750

and f (x2) = 0.2246 and jf (x2)j 6� 0.01

x3 =x1 + x22

=1.3750+ 1.25

2= 1.3125 but f (x3) = �0.0515 and

jf (x3)j 6� 0.01

x4 =x2 + x32

= 1.3438 and f (x4) = 0.0826 and jf (x4)j 6� 0.01.

x5 =x3 + x42

= 1.3282 and f (x5) = 0.0146 and jf (x5)j 6� 0.01

x6 =x5 + x42

= 1.3204 and f (x6) = �0.0186 and jf (x6)j 6� 0.01

x7 =x6 + x52


x7 = 1.3243. �



a+ x02

=1+ 1.52

=

1. 25,

f (x1) = f (1.25) = �0.2969 < 0.x2 =x0 + x12

=1.5+ 1.25

2= 1.3750

and f (x2) = 0.2246 and jf (x2)j 6� 0.01

x3 =x1 + x22

=1.3750+ 1.25

2= 1.3125 but f (x3) = �0.0515 and

jf (x3)j 6� 0.01

x4 =x2 + x32

= 1.3438 and f (x4) = 0.0826 and jf (x4)j 6� 0.01.

x5 =x3 + x42

= 1.3282 and f (x5) = 0.0146 and jf (x5)j 6� 0.01

x6 =x5 + x42

= 1.3204 and f (x6) = �0.0186 and jf (x6)j 6� 0.01

x7 =x6 + x52


x7 = 1.3243. �



a+ x02

=1+ 1.52

=

1. 25,

f (x1) = f (1.25) = �0.2969 < 0.x2 =x0 + x12

=1.5+ 1.25

2= 1.3750

and f (x2) = 0.2246 and jf (x2)j 6� 0.01

x3 =x1 + x22

=1.3750+ 1.25

2= 1.3125 but f (x3) = �0.0515 and

jf (x3)j 6� 0.01

x4 =x2 + x32

= 1.3438 and f (x4) = 0.0826 and jf (x4)j 6� 0.01.

x5 =x3 + x42

= 1.3282 and f (x5) = 0.0146 and jf (x5)j 6� 0.01

x6 =x5 + x42

= 1.3204 and f (x6) = �0.0186 and jf (x6)j 6� 0.01

x7 =x6 + x52


x7 = 1.3243. �



a+ x02

=1+ 1.52

=

1. 25,

f (x1) = f (1.25) = �0.2969 < 0.x2 =x0 + x12

=1.5+ 1.25

2= 1.3750

and f (x2) = 0.2246 and jf (x2)j 6� 0.01

x3 =x1 + x22

=1.3750+ 1.25

2= 1.3125 but f (x3) = �0.0515 and

jf (x3)j 6� 0.01

x4 =x2 + x32

= 1.3438 and f (x4) = 0.0826 and jf (x4)j 6� 0.01.

x5 =x3 + x42

= 1.3282 and f (x5) = 0.0146 and jf (x5)j 6� 0.01

x6 =x5 + x42

= 1.3204 and f (x6) = �0.0186 and jf (x6)j 6� 0.01

x7 =x6 + x52


x7 = 1.3243. �



a+ x02

=1+ 1.52

=

1. 25,

f (x1) = f (1.25) = �0.2969 < 0.x2 =x0 + x12

=1.5+ 1.25

2= 1.3750

and f (x2) = 0.2246 and jf (x2)j 6� 0.01

x3 =x1 + x22

=1.3750+ 1.25

2= 1.3125 but f (x3) = �0.0515 and

jf (x3)j 6� 0.01

x4 =x2 + x32

= 1.3438 and f (x4) = 0.0826 and jf (x4)j 6� 0.01.

x5 =x3 + x42

= 1.3282 and f (x5) = 0.0146 and jf (x5)j 6� 0.01

x6 =x5 + x42

= 1.3204 and f (x6) = �0.0186 and jf (x6)j 6� 0.01

x7 =x6 + x52

= 1.3243 and f (x7) = �0.0018 and jf (x7)j < 0.01,

hence

our required root isx7 = 1.3243. �



a+ x02

=1+ 1.52

=

1. 25,

f (x1) = f (1.25) = �0.2969 < 0.x2 =x0 + x12

=1.5+ 1.25

2= 1.3750

and f (x2) = 0.2246 and jf (x2)j 6� 0.01

x3 =x1 + x22

=1.3750+ 1.25

2= 1.3125 but f (x3) = �0.0515 and

jf (x3)j 6� 0.01

x4 =x2 + x32

= 1.3438 and f (x4) = 0.0826 and jf (x4)j 6� 0.01.

x5 =x3 + x42

= 1.3282 and f (x5) = 0.0146 and jf (x5)j 6� 0.01

x6 =x5 + x42

= 1.3204 and f (x6) = �0.0186 and jf (x6)j 6� 0.01

x7 =x6 + x52


x7 = 1.3243. �


Algorithm1 Choose xl and xu as two guesses for the root such thatf (xl ) f (xu) < 0, or in other words, f (x) changes sign between xland xu .

2 Estimate the root, xm of the equation f (x) = 0 as the mid-pointbetween xl and xm as

xm =xl + xu2

.

3 Now check the following:

1 If f (xl ) f (xm) = 0 then root is xm , stop further computation.

2 If f (xl ) f (xm) < 0 then root lies between xl and xm ; then xl = xl andxu = xm .

3 If f (xl ) f (xm) > 0 then root lies between xm and xu ; then xl = xmand xu = xu .




xm =xl + xu2

.








xm =xl + xu2

.








xm =xl + xu2

.








xm =xl + xu2

.








xm =xl + xu2

.






4. Find the new estimate of root

xm =xl + xu2

.

5. Repeat Step 3-4 untill jf (xm)j < δ (where δ is the user de�ned errortolerance) or the number of iterations m � M (where M is the maximumnumber of iterations allowed by the user).

Thank you


Documents

Math306&307 - Neumerical Analysis - Lec 1 - Bisection Method