advanced engineering mathematics 2014/dr.ryadh-math-3rd/advanced math3.pdfdr. riyadh jabbar advanced...

Advanced Engineering Mathematics

Computer Engineering Department

Dr. Eng. Riyadh J.S. Al-Bahadili

Third Year

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Dr. Riyadh Jabbar Advanced Engineering Mathematics 3rd class computer engineering

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Advanced Engineering Mathematics Chapter (1) Laplace Transformation

Laplace Transformation LT is a method for solving the differential equations DE and

corresponding initial and boundary value problems.

1. Definition

Given a function in time-domain, ( ) defined for all t≥0, multiply by and integrate with

respect to t from zero to ∞. Then if the resulting integral exists, it is defined as F(s)

( ) = ℒ[ ( )] = ∫ ( )

F(S) is called Laplace transform of ( ), while ( ) is the inverse Laplace transform (ILT) of F(S)

( ) = ℒ [ ( )] Note: generally, S is a complex value ( = + )

Example.1

Find F(S) if ( ) =1 for t≥0

Solution: ( ) Here is known as unit-step function u(t)

( ) = ℒ[ ( )] = ∫ ( ) = ∫ 1 = [− ] =

Hence ℒ[ ( ) = 1] =

F(S)

* S

S=0

? Note: for stable system, poles must be located in the left side in the pole-zero figure.




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Example.2

Find F(S) if ( ) = ; > 0, ≥ 0

Solution:

( ) = ℒ[ ] = ∫ = ∫ ( ) = [− ( ) ( ) ] =

Hence ℒ[ ] =

? Exercise.1

Find F(S) if ( ) = ; > 0, ≥ 0

2. Linearity of LT

If ( ) = ℒ[ ( )] and ( ) = ℒ[ ( )] , then

ℒ[ ( ) + ( )] = ℒ[ ( )] + ℒ[ ( )] = ( ) + ( )

Example.3

Find LT for cos( )

Solution:

( ) = cos( ) = = +

( ) = ℒ[cos( )] = ℒ + ℒ = + = Hence ℒ[cos( )] = Example.4

Find LT for ( ) =

Solution:

ℒ[ ] = ∫

To solve this integral, = , ℎ = = /




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ℒ[ ] = ∫ ( ) = ∫

Since ∫ = Γ( + 1) … , ℎ Γ( + 1) = !, : So ℒ[ ] = ( ) or ℒ = ( ) ? Exercise.2

Find LT for sin( ) , sinh( ) , cosh ( )

3. LT of derivatives

If ( ) = [ ( )] , ℎ ℒ[ ( )] = ∫ ( ) = , and = ( ) , so = − , = ( )

Using integration by parts: ∫ = [ ] − ∫

∫ ( ) = [ ( ) ] + ∫ ( ) = − (0) + ( )

So ℒ[ ( )] = ( ) − (0)

In the same way: ℒ[ ( )] = [ℒ[ ( )]] − (0)

So ℒ[ ( )] = ( ) − (0) − (0)

In general: ℒ[ ( )] = ( ) − (0) − (0) … − (0)

Example.5

Using LT of second derivative, find LT for ( ) =

ℒ[ ( )] = ( ) − (0) − (0)

( ) = , ( ) = 2 , ( ) = 2, (0) = 0, (0) = 0

ℒ[ ( )] = ℒ[2] = 2ℒ[1] = Hence = ( ) − 0 − 0,→ ( ) =




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? Exercise.3

Find LT for ( ) = ( ) , ( ) = ( )

(± Hint: sin 2 = 2 . ( ) = (1 + 2 )/2 )

4. Special theories in LT

1. LT of integration: ℒ[∫ ( ) ] = ( )

2. First shifting theorem: ℒ[ ( )] = ( − )

3. Second shifting theorem: ℒ[ ( − ) ( − )] = ( )

4. ℒ[ ( )] = ( )

5. ℒ[ ( )] = (−1) ( )( )

6. Convolution theorem: ℒ[∫ ( ) ( − ) ] = ( ). ( )

7. Periodic function: if ( ) is a periodic with period T

ℒ[ ( )] = ( ∫ ( ) ? Exercise.4

Prove the theories 1, 2, 4, and 7

Example.6

Find ℒ[ 4 ] We have ℒ[ ( )] = ( − ), and ℒ[ 4 ] = Hence ℒ[ 4 ] = ( ) = ? Exercise.5

Find LT for: 2 , 3 , 2 , 4 5 , 3 + 6, 5 2 − 3 2

5. Unit step function

The unit step function ( ) is defined as:




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( ) = 1, ≥ 00, < 0 And ( − ) = 1, ≥ 0, <

0 a

ℒ[ ( )] = ∫ 1 = ℒ[ ( − )] = ∫ 1 = ℒ[ ( − ) ( )] = ( )

? Exercise.6

Find ℒ[8 − 2 ( − 2)] 6. Inverse Laplace Transform using Residue Theorem

Let ( ) = ( )( )…( ) ( ) Can be converted to partial fractions

( ) = ∑ ( ) , where is the mth pole of order n

And [ = ] = = lim → [ ( )! {( − ) ( ) }]

ℒ [ ( )] = ∮ ( ) = ∑ = ∑{ }

Example.7

Find ℒ … ( . ( ) = ℒ [ ( )])

( ) = = ( )( ) , here we have 2 pols (S=-2 and S=-1), both of order n=1

[ = −2] = lim → ( + 2) ( )( ) = lim → ( ) = 3 [ = −1] = lim → ( + 1) ( )( ) = lim → ( ) = −2

1 1




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Hence ( ) = ℒ [ ( )]) = ∑ = 3 − 2 Example.8

Find ℒ [ ( )( ) ] ℒ ( )( ) = ∑ [ = −1 1, = 2 2] [ = −1] = lim → ( + 1) ( )( ) = lim → ( ) = [ = 2] = lim → ( − 2) ( )( ) = lim → ( )

= lim → ( ) ( ) = − Hence ℒ ( )( ) = ∑ = + − ? Exercise.7

Find ILT for the following functions using Residue Theorem:

1. ( ) 2. ( ) ( ) 3. ( ) 4. ( ) 5. ( )( ) 6. 7. ( )

7. Solving DE using LT

LT ( )

, , ℒ[ ],ℒ ′ , ℒ[ ] ( ) = ℒ [ ( )]




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Example.9

Solve the following DE using LT: ( ) + 4 ( ) + 3 ( ) = 0 , ℎ (0) = 3, (0) = 1

Solution:

We have: ℒ[ ( )] = ( ) − (0) And ℒ[ ( )] = ( ) − (0) − (0)

( ) + 4 ( ) + 3 ( ) = 0

[ ( ) − (0) − (0)] + 4[ ( ) − (0)] + 3 ( ) = 0

[ ( ) − 3 − 1] + 4[ ( ) − 3] + 3 ( ) = 0

( ) = = ( )( ) ( ) = ℒ [ ( )] = ℒ ( )( ) = ∑ [ = −3 1, = −1 1]

[ = −3] = lim → ( + 3) ( )( ) = lim → ( ) = −2 [ = −1] = lim → ( + 1) ( )( ) = lim → ( ) = 5 So ( ) = ∑ = 5 − 2 ? Exercise.8

Solve the following DEs using LT:

1. ( ) + ( ) = 1 , ℎ (0) = 1, (0) = 0

2. ( ) − 3 ( ) + 2 ( ) = 2 , ℎ (0) = 2, (0) = −1

? Exercise.9

Find the current value ( ) in the following circuit ( (0) = 0):




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Chapter (2) Vector Analysis

1. 2D & 3D Vectors

Physical quantities like mass, length, density are determined by their magnitude and called scalars.

Other quantities like force, velocity are determined by both magnitude and direction. These are

called vectors. Vectors are represented by their components parallel to the axes of the coordinate

system.

• 2D Vectors

Vector from (0, 0) to (1, 0) is denoted by , and the vector from (0, 0) to (0, 1) is denoted by . Then any vector in x-y plane can be defined in terms and .

Given two points ( , ), ( , ), then 2D vector from → denoted by is

= ( − ) + ( − )

? Note: = −

• Vector Operations

Let = + = + + = ( + ) + ( + ) − = ( − ) + ( − )

• Vector Length | | Let = + , then its length is

| | = +

1,0

0,1 x

y

3

1 3 +




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• Unit vector If = + , then the unit vector in direction is

= | | =

? Exercise.1

Prove that = 1

• Normal Vector If = + then the normal vector is

= | |

Example.1

Find a unit normal vector to the curve = at the point (2,4) and pointing from toward

the concave inside of the curve.

Solution:

To find we should first find (tangent to the curve at ). Needs point and the slope at

= , = 2 …

=

2 1

4




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= 4 … ( = 2)

To form we need another point 1, the equation of a line is = +

= 4 + ; apply p(2,4) to find the constant c, 4 = 8 + c → c = −4

Therefore, the equation of the tangent is = 4 − 4

Apply = 0 to find 1 → = 1, hence 1(1,0)

= 1 = (2 − 1) + (4 − 0) = + 4 = | | | | = √1 + 4 = √17

= = −4 + Hence = | | = √

• 3D vectors

= + +

Given two points ( , , ), ( , , ), then 3D vector from → denoted by is

= ( − ) + ( − ) + ( − )

= ( − ) + ( − ) + ( − )

2. Dot Product .

The scalar product of two vectors and is a scalar and defined by:

. = cos




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Where is the angle between and

e.g.

. = . = . = 1.1 cos 0 = 1

. = . = . = 1.1 cos 90 = 0

= cos = .

Let = + + = + +

. = + +

? Exercise.2

If = − 2 − 2 = 6 + 3 + 2 , find

a) .

b) Angle between

c)

• Orthogonal Vectors

Two vectors and are said to be orthogonal if their dot product is zero

ℎ vectors iff . = 0




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3. Cross Product

The cross product of two vectors and is given by

= sin = −

= sin , ℎ ℎ

Let = + + = + +

=

? Note:

= = = 0 = = −

= = − = − = • Parallel Vectors

Two vectors and are said to be parallel if their cross product is zero

vectors iff = 0




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Example.2

Find a vector that is perpendicular to both = + + = + Solution:

is the vector perpendicular to both = = 11 11 10 = − + ? Exercise.3

Let = 5 − + , = + 3 − 2 , = −15 + 3 − 3

Which pairs of vectors are (1) parallel (2) perpendicular?

4. Lines and Planes in Space

Suppose that L is a line in a space that passes through ( , , ) and is parallel to the vector = + + , then L is the set of all points ( , , ) for which is parallel to . Then the equation of the line L is

= = = Where t is a scalar

The parametric equations of line L are: = + , = + , and = +

? Exercise.4

Find the parametric equations of the line joining the points (1,2,−1) (−1,0,1).

To obtain equation of plane, assume a point ( , , ) on a plane and the normal to that plane

is = + + , the point ( , , ) lies in the plane if the vector is perpendicular

to , i.e. . = 0

= ( − ) + ( − ) + ( − )

Hence the equation of plane is

( − ) + ( − ) + ( − ) = 0

? Note: If we have equation of line or plane + + = ,then the normal vector is = + +




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Example.3

Find a plane that passes through (1,−1,3) and is parallel to the plane 3 + + = 7

Solution:

To find the equation of plane, we need a point on it and a vector normal to it

(1,−1,3) , = 3 + +

3( − 1) + ( + 1) + ( − 3) = 0 3 + + = 5

Example.4

Find the distance between the point (2,−3,4) and the plane + 2 + 2 = 13

Solution:

= + 2 + 2

The line passing through p and parallel to is

= = = … = + 2, = 2 − 3, = 2 + 4

We can find the intersection point Q as follows

( + 2) + 2(2 − 3) + 2(2 + 4) = 13 → = 1

Hence = 1 + 2 = 3, = −1, = 6 → (3,−1,6)

Distance = | | = (3 − 2) + (−1 + 3) + (6 − 4) = 3

? Exercise.5

Find an equation of the plane through the points: (1,1,−1), (2,0,2), (0,−2,1)

[R Check answer:7 − 5 − 4 = 6]

(2,−3,4)




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? Exercise.6

Find the angle between two planes 3 − 6 − 2 = 7 2 + − 2 = 5

[R Check answer: = 79 ]

5. Gradient, Divergence, and Curl

Let the operator ∇ defined as follows

∇= + + If ∅( , , ) ( , , ) have partial derivative, then the gradient (grad) of the function ∅ is

∅ = ∇∅ = ∅ + ∅ + ∅

If ∅( , , ) = is an equation of a surface, then ∇∅ gives the normal at that surface.

? Exercise.7

Find the unit normal vector to the surface 2 + 4 − 5 = −10 at the point (3,−1,2)

Let = + + , the divergence of vector is

= ∇. = + + . ( + + )

= ∇. = + +

The curl of vector is

= ∇X =




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? Exercise.8

Prove that ∇. ∅ = (∇∅). + ∅(∇. )

6. Line Integral

Suppose the curve C is defined by parametric equations:

= ( ), = ( ), = ℎ( )

The line integral ∫ in 3D is defined by

∫ = ∫ [ ( ), ( ),ℎ( )] ( ) + ( ) + ( )

In 2D, ∫ can be written as

∫ = ∫ ( , ) 1 + ( )

Example.5

If = + , C is the line segment between (0,0) (1,1), evaluate ∫

Solution:

The equation of the path (the line C from point A to point B) is = . This line can be

parameterized as: = = , where 0 ≤ ≤ 1 . Hence = 1 = 1

∫ = ∫ [ ( ), ( ),ℎ( )] ( ) + ( ) + ( ) , = + = +

∫ = ∫ ( + )√1 + 1 = √

Another solution using ∫ = ∫ ( , ) 1 + ( )

We have = , = 1, = + = = +

∫ = ∫ ( + )√1 + 1 = √2∫ ( + ) = √




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Example.6

Evaluate the line integral for = + along the path

( ) = + (1 − ) + , ℎ 0 ≤ ≤ 1

Solution:

We have = , = 1 − , = → = 1, = −1, = 1

= + = + (1 − )

∫ = ∫ [ ( ), ( ),ℎ( )] ( ) + ( ) + ( ) = ∫ ( + (1 − ) )√3 = √ ? Exercise.9

Evaluate the line integral for = + − along the path C1: ( ) = + and the path

C2: ( ) = + + , ℎ 0 ≤ ≤ 1. Path C1 from (0, 0, 0) to (1, 1, 0), and Path C1 from

(1, 1, 0) to (1, 1, 1). [R Check answer: √ ]

• Physical meaning for ∫

∫ May be interpreted as:

v If represents the line charge distribution in / , then ∫ will represent the total charge.

v If represents the force , then ∫ is the work done .

Let = + + , where M,N, and P are functions to x, y, z. This force moves the particle

along curve C from point ( , , ) to point ( , , ). The work done is:

= ∫ . , where = + +

Hence = ∫ + +




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Example.7

The force = ( − ) + ( − ) + ( − ) moves a particle from (0, 0, 0) to (1, 1, 1) along

the curve: = , = , = , ℎ 0 ≤ ≤ 1. Find the work done.

Solution:

We have = ( − ), = ( − ), = ( − )

= ∫ ( − ) + ( − ) + ( − )

= , = 2 , = 3 = ∫ ( − ) + 2( − ) + 3( − ) = − ? Exercise.10

For example.7, find the work done by in moving the particle one cycle around the circle: =cos , = sin , = 0, ℎ 0 ≤ ≤ 2 . [R Check answer: = ]

• Independent of the path

Let be a vector field with components M, N, and P that are continuous in the region D, then the

necessary and sufficient condition that ∫ . to be Independent of the path joining A and B in

the region D, is that there exist a differentiable function such that:

= ∇ = + +

Example.8

Find a function such that: = ∇ = 2 + 2 + 2 , then evaluate = ∫ . between (2,2,4) (−1,−1,2) along:

a) The straight line

b) The curve: = − 2, = − 2, = 2 , ℎ 2 > > 1

Solution:

We can guess that if = + + , then = ∇ = 2 + 2 + 2 , so = ∫ . must

be Independent of the path C.




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a) The equation of the line is: = = = So = , = ,→ = , =

∫ . = ∫ 2 + 2 + 2 = 18 >

b) = = − 2 → = = 2 = 2 → = 2

∫ . = ∫ [4( − 2)2 + 8 ] = 18 >

• Conservative Fields

is conservative if = ∇ , i.e. = , = , =

• Exact Differential

The expression: + + is an Exact Differential if:

= , = , = Or ∇ = 0

? Note: → Conservative → Independent of the path

? Exercise.11

Show that ∫ + 2 + 2 is independent of the path between A & B.

? Exercise.12

Let C denote the curve whose vector equation ( ) = ( cos ) + ( sin ) , find ∫ ( ) ⁄

along C from(1, 0) ( , 0). [R Check answer:1 − ]

Example.9

Suppose = ( cos + ) + ( − sin ) + ( + ), Is conservative? If so, find

such that = ∇

Solution:

We have: = cos + , = − sin , = +




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= − sin + =

= =

= =

Hence + + is an Exact Differential.

We can find by integrating the system of equations:

= = cos + = = − sin

= = +

( , , ) = ∫ = ∫ = ∫( cos + ) = cos + + ( , )

~ The problem here is to find ( , )

( , , ) = − sin + + ( , )

Since ( , , ) = = − sin , then ( , ) = 0 → ( , ) = ℎ( )

Hence ( , , ) = cos + + ℎ( )

~ The problem now is to find ℎ( )

( , , ) = + ( )

Since ( , , ) = = + , then ( ) = → ℎ( ) = +

Finally ( , , ) = cos + + +

7. Green’s Theorem

If ℛ is a closed region of the xy-plane bounded by a simple closed curve C and if M, N are

continuous functions of x, y, then

C

ℛ

X

Y




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∮ ( + ) = ∬ − ℛ

Example.10

Let C be the circle = a cos , = sin and = − + , verify Green’s Theorem.

Solution:

We have = − = , = − sin , = a cos

∮ ( + ) = ∮ − +

= ∫ −( sin )(− sin ) + (a cos ) (a cos )

= ∫ (sin + cos ) = ∫ = 2 >

While

∬ − ℛ = ∬ (1 + 1) ℛ = 2∬ ℛ

Recall that = ( → )

2∬ ℛ = 2 ∫ ∫ = 2 [ ] = 2 >

? Exercise.13

Use Green’s Theorem to find the area of ellipse: = a cos , = sin , ℎ 0 ≤ < 2

[? Note: = ∮ ( + ) = ∬ − ℛ ]

? Exercise.14

Evaluate the line integral ∫ 3 + 2 by applying Green’s formula. Where C is the

boundary 0 ≤ ≤ , 0 ≤ ≤ sin . [R Check answer: −2]

? Exercise.15

If C is the circle + = and ℛ is inside C. Verify Green’s Theorem if:

j = , = k = = l = − , =

X

Y C ℛ




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Chapter (3) Special Functions

1. Gamma Function ( )

Gamma function is defined by the integral

Γ( ) = ∫

? Note: Γ(0): , Γ(1) = 1, Γ(1/2) = √

We have two simple expressions to evaluate Gamma function:

• When n is an integer; n= 1, 2, 3…

Γ( + 1) = !

• When (real or integer)

Γ( + 1) = Γ( )

• When n is negative

Γ( ) = ( )

Example.1

Find ( ) ( ) , ( / ) ( / ) , ( / ) ( / ) Solution:

Since Γ( + 1) = ! , ( ) ( ) = ! ( !) = 30

Since Γ( + 1) = Γ( ) , ( / ) ( / ) = ( ) ( ) ( ) = ( )( ) ( ) ( ) =




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( / ) ( / ) = ( )( ) ( ) ( ) =

? Exercise.1

Prove that: Γ(1) = 1, Γ(1/2) = √

Example.2

Use Gamma function to evaluate the following integrals: ∫ , ∫

Solution:

Γ( ) = ∫

• ℎ − 1 = 3 → = 4, ∫ = Γ(4) = 3! = 6 • ∫

Let = 2 → = 2 So ∫ = ∫ ( ) = ∫ = Γ(7) = !

? Exercise.2

1. Find Γ − Γ −

2. Evaluate ∫ 3. Evaluate ∫ / / , [±hint: assume = 2 ], [R Check answer: 2 / Γ(11/2)]

2. Beta Function ( , )

Beta function is defined by the integral:

( , ) = ∫ (1 − )

We can also evaluate Beta function using Gamma function:

( , ) = ( ) ( ) ( ) , ℎ , > 0

? Note: ( , ) = ( , )

Example.3

Use Beta function to evaluate the following integrals: ∫ (1 − ) , ∫ √




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Solution:

( , ) = ∫ (1 − ) = ( ) ( ) ( )

• ∫ (1 − ) = ∫ (1 − )

This is Beta function with = 5 = 4

∫ (1 − ) = (5,4) = ( ) ( ) ( ) = ! ! ! = • Let = 2 → ∫ √ = 4√2 ∫ = 4√2 ∫ (1 − )

This is Beta function with = 3 = Hence ∫ √ = 4√2 3, = √ ( ) ( ) ( ) = √

? Exercise.3

Evaluate ∫ − [±hint: assume = ], [R Check answer: ]

? The advantage of Gamma and Beta functions is to evaluate other integrations:

~ ∫ / . = ( , )

~ ∫ = ( ) (1 − ) , ℎ 0 < < 1

3. Bessel Function J(n) Bessel functions is generated as solution for the DE

+ + ( − ) = 0

Where ≥ 0 . The solution for this equation is

( ) = ( ) + ( )

( ) is called Bessel function of first kind of order n, while ( ) is called Bessel function of

second kind of order n.

( ) = ∑ ( ) ! ( )

The second solution of Bessel equation for all values of n (real and integer) is:




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( ) = ( ) ( )

• When n is an integer, ( ) can be calculated as:

( ) = (−1) ( ) … = 1, 2, 3 …

• When n is not integer, the general solution will be:

= ( ) + − ( )

Where ( ) = ∑ ( ) ! ( )

Example.4

For = 0 → ( ) = 1 − + − + ⋯

For = 1 → ( ) = − ( ) + ( ) − ( ) +⋯

? Exercise.4

Prove that / ( ) = sin / ( ) = cos

Important identities for Bessel functions

A. [ ( )] = ( )

B. [ ( )] = − ( )




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C. ( ) = [ ( ) − ( )] D. ( ) + ( ) = ( ) >

Example.5

Prove identity A.

Proof:

( ) = ∑ ( ) ! ( )

( ) = ∑ ( ) ! ( )

[ ( )] = ∑ (−1) (2 +2 ) 2 +2 −12 +2 ! ( + +1)∞ =0 = ∑ (−1) ( + ) +2 −12 +2 −1 ! ( + ) ( + )∞ =0

= ( )

Example.6

Prove identity B.

Proof:

( ) = ∑ ( ) ! ( )

( ) = ∑ ( ) ! ( )

[ ( )] = ∑ ( ) ( ) ! ( )

Let = + 1 ; = 1,2, …∞ ; = 0,1,2, …∞

[ ( )] = ∑ ( ) ( ) ( ) ( ) ( )! ( )

= − ∑ ( ) ( ) ( ) ! ( ) = − ( )

? Exercise.5

Prove identity C.

? Exercise.6

Prove identity D.




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? Exercise.7

Express ( ) in terms of ( ) ( ). [±Hint: use identity D]

Example.7

Evaluate ∫ ( )

Solution:

Using [ ( )] = − ( ) … identity B

Or ∫ ( ) = − ( ) + . Let = ∫ ( ( ) )

Integration by parts: let = → = 2

= ( ) → = ∫ ( ) = − ( )

= − ( ) + ∫2 ( ( )) = − ( ) + 2 ∫ ( )

Since ∫ ( ) = − ( ) + . = − ( ) − 2 ( ) + . Hence ∫ ( ) = − ( ) − ( ) + . Example.8

Reduce the following DE to Bessel equation using =

+ + 4( − ) = 0

Solution:

Let = , ( ) → ( )

= 2 , = 2

= = = 2

= 2 = 2 + 2 = 2 + 2 2 = 2 + 4

Hence 2 + 4 2 2 + 2 + 4 2 − 2 = 0

+ + ( − ) = 0

This Bessel equation has the solution: ( ) = ( ) + ( )




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4. Legendre Functions

Legendre functions is generated as solution for the DE

(1 − ) − 2 + ( + 1) = 0

The solution for this equation is

( ) = ( ) + ( )

( ) is called Legendre polynomial and ( ) is called Legendre function of second kind.

( ) = ∑ ( ) ( )! ! ( )! ( )!

Where = , : , :

? Note: (1) = 1, (−1) = (−1)

( ) = 1 , ( ) = , ( ) = , ( ) =

Some properties of Legendre function

• Orthogonality

∫ ( ) ( ) = 0 … ≠

∫ ( ) ( ) = … =

• Recurrence Formula

( ) = ( ) − ( )

( ) − ( ) = (2 + 1) ( )




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Example.9

Given that, ( ) = 1 ( ) = , find ( ) ( ) using recurrence formula

Solution:

( ) = ( ) − ( )

Let = 1 → ( ) = ( ) − ( ) = − =

Let = 2 → ( ) = ( ) − ( ) = − =

? Exercise.8

Find: ( ) , ( ) , ∫ ( ) ( ) , ∫ | ( )|




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Chapter (4) Complex Variables

Complex variable = + defines a point in the complex z-plane

1. Functions of Complex Variables

• Polynomial Function

( ) = + + + ⋯+ = ∑

Where , , … , are complex constants, and n is the degree of ( ) . Special case ( ) = + is called linear transformation

• Rational Function

= ( ) ( ) Special case = is called bilinear transformation.

• Exponential Function

= = ± = (cos ± )

? Note: = 1 + + ! + ! +⋯+ ! , → ∞

• Trigonometric & Hyperbolic Functions

sin = , cos =

sinh = , cosh =

?Note: sin( ) = ℎ( ) , ℎ( ) = ( ) cos( ) = ℎ( ) , cosh( ) = ( )

= +




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• Inverse Trigonometric & Hyperbolic Functions

If = sin → = sin

sin = ln( ± √1 − )

cos = ln( ± √ − 1)

tan = ln( )

sinh = ln( + √ + 1)

cosh = ln( + √ − 1)

tanh = ln( )

• Logarithmic Function

= + →

= ∟ →

= ( ), = 0,1,2 … →

= (cos + sin ) →

If = ln → = ln( ( )) = ln + ( + 2 )

In general:

= ( ) >

Where = 0,1, … , ( − 1) = 0, ±1, ±2, …

? Note: +1 = 1∟0 , − 1 = 1∟ , + = 1∟ , − = 1∟ Example.1

Find the roots of √1

Solution:

Let = √1 → = 1 , ( . . ℎ − 1 = 0)

= +1 = 1∟0 → = 1 , = 0 , = 3

n-point




32 | P a g e

We have = ( )

= 1 ( ) , = 0,1,2

= 1 / ( ) → =

= 0 → = = 1 = 1∟0 > = 1 → = = 1∟ > = 2 → = = 1∟ > ℎ ~Check [ ( − ) − − ↔ ( − 1) ]

Example.2

Prove that: sin = ln( ± √1 − )

Solution:

Let w = sin → = sin =

− − 2 = 0

Multiply by

− 2 − 1 = 0 → − 2 − 1 = 0

? Note: + + = 0 → = ±√

Hence

= ±√ = ± √1 −

Take ln for two sided

= ln ± √1 − → = ln ± √1 − But w = sin

So sin = ln ± √1 −

= 1




33 | P a g e

Exercise.1

Find the roots for √8 √−1

Exercise.2

Prove that cos = ln( ± √ − 1)

2. Derivative of ( )

If ( ) is single valued in some region ℛ of z-plane, the derivative of ( ) is defined as:

( ) = ( ) = lim∆ → ( ∆ ) ( )∆ If the limit exist for = then ( ) is called differentiable at . If the limit exist for all z such that | − | < for some > 0 then ( ) is called analytic at . If the limit exist for all z in region ℛ then ( ) is called analytic in ℛ.

• Cauchy-Riemann Equations

Let ( ) = ( , ) + ( , )

The necessary condition that ( ) be analytic in region ℛ is that ( , ) and ( , ) satisfy:

= = −

Example.3

Check the following functions if they are analytic: ( ) = − , ( ) =

Solution:

• ( ) = − → = , = −

= 1 , = −1 → ≠ , ( )

• ( ) = = ( + ) = ( − ) + 2 → = − , = 2

= 2 , = 2 → = = −2 , = 2 → = −

( )




34 | P a g e

Example.4

Show that: ( sin ) = √

Let = → = , =

= sin → = cos . → 1 = cos .

Or = 1cos

Since 2 + 2 = 1 → 2 = 1 − 2 = 1 − 2

Or cos = 1 − 2

Now = = √

But = , ℎ ( ) = √

Exercise.3

Show that: ( ℎ ) =

3. Complex Integration

The integration of ( ) in the complex z-plane along the path C:

∫ ( ) = ∫ ( + ) ( + ) = ∫ ( − ) + ∫ ( + )

Example.5

Evaluate ∫ along the path C, where C is defined by the path from (0,0) to (1,0) then to (1, 2)

Solution:

(1,0)

(1, 2)




35 | P a g e

= ( + ) = ( − ) + 2 ( ) = = + → = ( − ) , = 2

∫ ( ) = ∫ ( − ) + ∫ ( + ) + ∫ ( − ) + ∫ ( + )

~ 0 : = 0, = 0, : 0 → 1

~ : = 1, = 0, : 0 → 2

∫ = ∫ + ∫ 0 + ∫ (−1)2 + ∫ (1 − ) = − (1 + 2)

Example.6

Evaluate ∮ where C is the contour defined by the circle | | =

Solution:

Let = = , 0 ≤ < 2

∮ = ∫ ( ) = ∫

= ∫ = = (1 − 1) = 0

Exercise.4

Express the following functions to the form + : , , , ln

Exercise.5

Evaluate ∫ where C is defined by line = 2 from point (0,0) (1, 2)

Exercise.6

Evaluate ∮ where C is the circle | − 2| = 3

• Simply and multiply connected regions

A region ℛ is called simply connected if any simple closed curve, which lies in, ℛ can be shrunk to a point without leaving ℛ.

A region ℛ that is not simply connected is called multiply connected.

| | =




36 | P a g e

4. Cauchy Integral Theorems

• If ℛ is a region, simply or multiply connected, whose boundary C is sectionally smooth, and if ( ) is analytic and ( ) is continues within the boundary C, then:

∮ ( ) = 0 (?See Example.6)

• If ( ) is analytic in simply connected region ℛ then:

∫ ( ) ℎ ℎ → (?See Example.7)

Simply Connected Region

Multiply Connected Regions

C

C C C1 C1 C2

ℛ

ℛ

ℛ

C

ℛ

P2

P1




37 | P a g e

• If ( ) is analytic within and on the boundary of a region bounded by two closed curves

C1 and C2, then:

∮ ( ) = ∮ ( )

• If ( ) is analytic in a region bounded by non-overlapping simple closed curve 1, 2, … , , then:

∮ ( ) = ∮ ( ) + ∮ ( ) +⋯+ ∮ ( )

• If ( ) is analytic inside and on the boundary C of simply connected region ℛ and if a is any point inside ℛ , then

∮ ( ) = 2 , = 10 , ℎ (? See Example.8)

∮ ( ) ( ) = 2 ( ) (? See Example.9)

∮ ( ) ( ) = ! ( )( ), = 0, 1, … (? See Example.10)

Exercise.7

Use Green’s theorem to prove ∮ ( ) = 0

C1

C2

C1

C2

Cn

C

a

C ℛ




38 | P a g e

Example.7

Evaluate ∫ along:

(1) The curve: = , = , ℎ 1 ≤ ≤ 2 (2) The line from (1 + ) (2 + 4) (3) The lines from (1 + ) → (2 + ) → (2 + 4)

Solution:

We have ∫ = ∫ ( + ) ( + ) = ∫ ( − + 2 ) ( + )

= ∫ ( 2 − 2) − 2 + ∫ 2 + ( 2 − 2)

(1) ∫ ( − ) − 2 2 + ∫ 2 + ( − )2 = − − 6 >

(2) The equation of the line from (1,1) (2,4)

= → = 3 − 2

[ − (3 − 2) ] − 2 (3 − 2)3 + 2 (3 − 2) +

[ − (3 − 2) ]3 = − 863 − 6

(3) From (1,1) (2,1) → = 1 , = 0

∫ ( − 1) + ∫ 2 = + 3

From (2,1) (2,4) → = 2 , = 0

∫ (−4 ) + ∫ (4 − ) = −30 − 9

+ 3 + (−30 − 9) = − − 6 >

(1 + ) (2 + )

(2 + 4 )

(1)

(2) (3)

>




39 | P a g e

Example.8

Evaluate ∮ ( ) if C is: (1) circle | | = 1 , (2) circle | + | = 4

Solution:

(1) The point ( = 3) is outside the circle | | = 1 , so

∮ ( ) = 0

(2) The point ( = 3) is inside the circle | + | = 4 , so

∮ ( ) = 2

Example.9

Evaluate ∮ if C is: (1) circle | − | = 1 , (2) circle | + | = 1

3

| | = 1

| + | = 4

| − | = 1

| + | = 1

−




40 | P a g e

Solution:

+ 1 = ( + )( − ) , ∮ ( ) ( ) = 2 ( )

(1) ∮ ( )( ) = ∮ ( ) ( ) → ( ) = ( ) , = ∮ ( ) ( ) = 2 ( ) = 2 =

(2) ∮ ( )( ) = ∮ ( ) ( ) → ( ) = ( ) , = − ∮ ( ) ( ) = 2 (− ) = 2 = −

Example.10

Evaluate ∮ ( ) , where C is any simple closed curve enclosed the point = 1

Solution:

∮ ( ) ( ) = ! ( )( )

We have = 2, = 1, ( ) = 5 − 3 + 2

∮ ( ) ( ) = ! ( = 1)

( ) = 10 − 3 → ( ) = 10 → (1) = 10

Hence ∮ ( ) ( ) = ! 10 = 10 ?Note: if the point = 1 is outside the region, then, ∮ 5 2−3 +2( −1)3 = 0

Exercise.8

Evaluate the following complex integrals:

(1) ∮ , ℎ : | − 1| = 3




41 | P a g e

(2) ∮ ( ) , ℎ : | − 1| = 3

(3) ∮ ( ) ( ) , ℎ : | − 1| = 1, : | − | = 1

(4) ∮ ( ) , ℎ : | | = 3

(5) ∮ ( ) ( )( ) , ℎ : | | = 3

[R Check answer:j −2 k 2 (1 − ) l 2 , m n 4 ]

5. Residue Theorem

If ( ) is analytic inside and on a simple closed curve except at a finite number of singular points

inside ℂ, (i.e. points a, b, c …) at which the residues are (ℛ [ ],ℛ [ ],ℛ [ ], … ) respectively,

then:

∮ ( ) ℂ = 2 .∑(ℛ [ ],ℛ [ ],ℛ [ ], … )

ℛ [ ] = lim → ( )! [( − ) ( )]

Where p is any point inside C whose order n

Example.11

Using Residue theorem, evaluate ∮ ( ) ( )( ) , ℎ : | | = 3/2

Solution:

We have three simple roots (poles) at = 0, = 1, = 2

? Only = 0 = 1 are inside C, ( ) = (4−3 ) ( −1)( −2) ℛ [ = 0] = lim → [ (4−3 ) ( −1)( −2)] = 2

ℛ [ = 1] = lim → [( − 1) (4−3 ) ( −1)( −2)] = −1

Hence ∮ ( ) ( )( ) = 2 ∑ℛ = 2 (2 − 1) = 2

ℂ

a b

C

| | = 3/2

1 2 0




42 | P a g e

Exercise.9

Try out resolving all previous exercises using Residue theorem

Exercise.10

Using Residue theorem, evaluate

(1) ∮ ( ) , ℎ : | − 1| = 1

(2) ∮ ( ) ( ) , ℎ : | | = 2

[R Check answer:j − k (sin 0.5) ]

• Solving Real Integrals

We can use complex integrals to solve real integrals that contain trigonometric functions.

Example.12

Evaluate ∫ using complex integral

Solution:

= , = , sin = =

∫ = ∮ / ( ) = ∮ = ∮ ( )

= ∮ ( ) ( ) We have two roots: = −3 = − Only = − is inside C

Real Integrals ∫

( , sin , cos , … )

Complex Integrals ∮

→ , →

: | | = 1




43 | P a g e

ℛ = − = lim → [( + ) 1 +3 ( +13 )] = lim → +3 = ∫ = ∮ ( ) = 2 ∑ℛ = 2 = Exercise.11

Evaluate the following real integrals using complex integral

(1) ∫ ( )

(2) ∫ √ (3) ∫

[R Check answer:j /12 k 2 l ]




44 | P a g e

Chapter (5) Numerical Analysis

1. Isolation of the roots

If ( ) is continues function, then any number for which ( ) = 0 is called a root of the

equation ( ) = 0 . The interval of the root [ , ] is found when the sign of ( ) changes. Note

that the smallest possible interval [ , ] contains one and only one root.

Example.1

Isolate the roots (find the intervals) of the equation: ( ) = − 6 + 2 = 0

Solution:

This equation has three roots, either the three roots are equal, or one root is real and the others two

are complex conjugate.

x -5 -3 -2 -1 0 1 2 3 10

Sign of

f(x) - - + + + - - + +

Hence

• First root is located in interval [−3,−2] • Second root is located in interval [0, 1] • Third root is located in interval [2, 3]

Exercise.1

Isolate the roots (find the intervals) of the equation: ( ) = − − 1 = 0

2. Accuracy of approximation root Let

: is the exact root of ( ) = 0

: is the approximated root to Where both are located in the interval [ , ] . Then the absolute error for this

approximation is:




45 | P a g e

− ≤ ( ) ( )

? Note: Types of errors:

• absolute error: = − • relative error: = − 1

• percentage error: % = . 100

Example.2

The approximated root of the equation − − 1 = 0 in the interval [1, 2] is found to be 1.22,

find the accuracy of this approximation.

Solution:

( ) = − − 1 , ( ) = 4 − 1

= (1.22) = −0.00467

= (1.22) = 4.448

= − ≤ ′ ≤ 0.004674.448 ≤ 0.00156 ≤ (1.56)10−3 3. Methods for locating the root

A. Fixed-Point Iteration

The function ( ) = 0 , which has an isolated root in the interval [ , ] , is solved iteratively

by:

= ( )

The iteration process will be convergent for [ , ] if: ′( ) < 1

Where is the initial value.

Note: The function ( ) = 0 is replaced by = ( ) , and then it can be solved iteratively:

= ( )

= ( )

…




46 | P a g e

= ( )

Now = lim → [ ]

x

y y=x

y=g(x)

y=x

y=g(x)

x

y

Convergence

0 < ( ) < 1

Convergence

−1 < ( ) < 0

x x

y y y=x

y=g(x)

y=x y=g(x)

Divergence ( ) > 1

Divergence ( ) < −1




47 | P a g e

Example.3

Find the root of ( ) = + − 1000 in the interval [9,10] using fixed-point method.

Solution:

• Let = 1000 − → ( ) = 1000 − , ( ) = −3 For [9,10] → | ( )| > 1

Hence = 1000 − diverges

• Let = − → ( ) = − , ( ) = +

For [9,10] → | ( )| > 1

Hence = − diverges

• Let = 1000 − → = (1000 − ) /

( ) = (1000 − ) / , ( ) = ( ) / For [9,10] → | ( )| < 1

So = (1000 − ) / can be converged

Start with = 10 = (1000 − ) = 10 = (1000 − 10) = 9.966554934 = 9.966554934 = (1000 − 9.966554934) = 9.966667165 = 9.966667165 = (1000 − 9.966667165) = 9.966666789 = 9.966666789 = (1000 − 9.966666789) = 9.966666790 = 9.966666790 = (1000 − 9.966666790) = 9.966666790

Exercise.2

Find the range of for convergence then find an approximate root for function:

( ) = − 3




48 | P a g e

Exercise.3

Find the root of ( ) = − − 1 in the interval [1,2].

B. Bisection Method

In this method, we must start with an initial interval [ , ], where ( ) ( ) have opposite

signs. Bisection method systematically moves the endpoints of all interval closer and closer

together until we obtain an interval of small width that brackets the exact root . The decision step for this process of interval halving is first to choose the midpoint =

and then to analyze three possibilities:

(1) If ( ) ( ) have opposite signs, then lies in[ , ]. (2) If ( ) ( ) have opposite signs, then lies in[ , ]. (3) If ( ) = 0 , then = .

Accuracy: | − | ≤ ~ ⎯ , ( ) ≈ 0

Example.4

Use Bisection method to find the root of ( ) = + 2 − − 1 in the interval [0,1] Solution:

( ) ≠ ( ) [ , ]

( ) ≠ ( ) [ , ]




49 | P a g e

[ , ] → [0,1] , = = 0.5

( ) = (0) = − , ( ) = (1) = + , ( ) = (0.5) = −

[ , ] → [0.5,1] , = = 0.75

( ) = (0.5) = − , ( ) = (1) = + , ( ) = (0.75) = −

[ , ] → [0.75,1] , = = 0.875

( ) = (0.75) = − , ( ) = (1) = + , ( ) = (0.875) = +

… And so on

New interval [0.8593, 0.875] → = . . = 0.867 ≈ , ( ) = 0.00146

Accuracy ≤

Exercise.4

Use Bisection method to find the root of ( ) = 1 −2 for the intervals [3,7] [1,7]

Exercise.5

Suppose that the bisection method is used to find the root of ( ) in the interval [2,7]. How many

times this interval must be bisected to guarantee that the approximation has an accuracy of 5 10 .

New interval [ , ]

New interval [ , ]

New interval [ , ]




50 | P a g e

C. False Position Method

This method was developed because the bisection method converges at a low speed. A better

approximation is obtained if we find the point ( , 0) where the secant line L joining the points ( , ( )) and ( , ( )) crosses the x-axis.

= − ( ).( ) ( ) ( )

(1) If ( ) ( ) have opposite signs, then lies in[ , ]. (2) If ( ) ( ) have opposite signs, then lies in[ , ]. (3) If ( ) = 0 , then = .

Exercise.6

Use False Position method to find the root of ( ) = + 2 − − 1 in the interval [0,1]

D. Newton-Raphson Method

( )

( )

( , 0) ( , 0)

=

= ( )




51 | P a g e

The equation of the tangent line at = is:

( ) = ( )

At the point ( 1, 0) → ′( 0) = − ( 0) 1− 0 , or = − ( 0) ′( 0) In general: = − ( ) ( ) … ( ) ( ) have the same sign

Example.5

Use Newton method to compute the approximated root for ( ) = − 3 + 75 − 10000 in the interval [−11,−10]. Solution:

( ) = − 3 + 75 − 10000 , [ , ] → [−11,−10]

( ) = 4 − 6 + 75

( ) = 12 − 6

If = = −10 → (−10) = − , (−10) = +

So = = −10 is not suitable choice

Thus = = −11 → (−11) = + , (−11) = +

= − ( ) ( ) → = −

= −11 → = −10.3338 → = −10.2618 → = −10.261 → = −10.261

( ) = (7.27) 10

Exercise.7

Use Newton method to find a root in each of the following functions:

(1) ( ) = tan − , , (2) ( ) = − + 2 , ℎ = −1.5

(3) ( ) = ( − 2) , ℎ = 2.1




52 | P a g e

4. Newton Iteration to Find Square Roots

Assume that > 0 is a real number and let > 0 be an initial value to √ , then

= ( )

Proof:

We want to find = √ , . . = , ( ) = − = 0 , ( ) = 2

= − ( ) ( ) → = − 2− 2 = ( )

Example.6

Use Newton iteration to find √5 , ℎ = 2

Solution:

= ( ) , ℎ = 5 0 = 2

= ( ) = 2.25

= . ( . ) = 2.236111 → = 2.236067 → = 2.236067

Exercise.8

Use Newton iteration to find:

(1) √8 , ℎ = 3 (2) √91 , ℎ = 10

5. Curve Fitting

A. Least-Squares Method

Suppose that we have a table of K data ( , ) . The polynomial to fit this data is:

= + + +⋯+

Special case when = 1 , i.e. = + called straight-line fitting.




53 | P a g e

The main issue is to find the coefficients , , … , so that the error between the function Y

and the data ( , ) is minimum.

The mean square error between Y (fitting) and (table) is

= ∑ [ − ] = ∑ [( + + +⋯+ ) − ]

→ = 0

0 = ∑ 2[( + + +⋯+ ) − ] = 0

→ + ∑ + ∑ +⋯+ ∑ = ∑

1 = ∑ 2[( + + +⋯+ ) − ] = 0

→ ∑ + ∑ +⋯+ ∑ = ∑

…

= ∑ 2[( + + +⋯+ ) − ] = 0

→ ∑ + ∑ +⋯+ ∑ = ∑

, ,

,

= +




54 | P a g e

Alternatively, in the matrix form:

⎣⎢⎢⎢⎢⎡ ∑ ∑

∑ ∑ ∑ ∑ ∑ ∑ ⋯ ∑ ∑ ∑ ⋮ ⋱ ⋮ ∑ ∑ ∑ ⋯ ∑ ⎦⎥⎥⎥⎥⎤ ⎣⎢⎢⎢⎡ ⋮⋮ ⎦⎥⎥

⎥⎤ =⎣⎢⎢⎢⎢⎢⎡ ∑ ∑ ∑ ⋮⋮∑ ⎦⎥⎥

⎥⎥⎥⎤

Example.7

Find a straight-line that fit the following data ( , ) :

Solution:

We have = 8 points, = + , = 1

∑ =1∑ =1 ∑ 2 =1 0 1 = ∑ =1∑ =1 8 2020 92 0 1 = 3725 → 0 1 = 8 2020 92 3725 = 8.6−1.6 Hence = 8.6 − 1.6

Exercise.9

Find the least squares line = + for the following data ( , )

B. Power Fit Method

=

-1 0 1 2 3 4 5 6 10 9 7 5 4 3 0 -1

-2 -1 0 1 2 1 2 3 3 4




55 | P a g e

= ∑ [ − ]

→ = 0

= 2∑ [ − ] = 2∑ [ − ] = 0

= ∑ ∑

• Root mean-square error function

Root mean-square error function ( ) is defined by:

( ) = ∑ [ ( ) − ] /

Note: ? A function ( ) is said to be best fit if it has minimum ( )

Example.8

Find ( ) for = 8.6 − 1.6 in Example.7

Solution:

We have ( ) = 8.6 − 1.6 , = 8 ( ) = 8.6 − 1.6 [ ( ) − ]

-1 10 10.2 0.04 0 9 8.6 0.16 1 7 7.0 0.00 2 5 5.4 0.16 3 4 3.8 0.04 4 3 2.2 0.64 5 0 0.6 0.36 6 -1 -1.0 0.00

( ) = ∑ [ ( ) − ] / = (1.4) / ≈ 0.4183




56 | P a g e

Exercise.10

Find the power fits = 2 = 3for the following data ( , )

Evaluate ( ) to determine which curve fits best.

Exercise.11

Find the power fits = = 2 for the following data ( , )

Evaluate ( ) to determine which curve fits best.

C. Data Linearization Method

The original points ( , ) are transformed into the points ( , ) . This method is used for other functions to generate linear function:

= +

∑ =1∑ =1 ∑ 2 =1 0 1 = ∑ =1∑ =1

• Fit by =

= → ln = ln

ln = ln + ln → ln = ln + (− )

? Transform to = +

= ln , = 0 = ln , 1 = −

2.0 2.3 2.6 2.9 3.2 5.1 7.5 10.6 14.4 19.0

0.5 0.8 1.1 1.8 4.0 7.1 4.4 3.2 1.9 0.9

( , ) → ( , ) ( 0, 1) → ( , )




57 | P a g e

Example.9

Fit the following data ( , ) by exponential form =

Solution:

(1) Transform = into = + = ln , = 0 = ln , 1 = −

(2) Linearize Data ( , ) → ( , )

= ln , =

(3) Find [ 0, 1]

∑ =1∑ =1 ∑ 2 =1 0 1 = ∑ =1∑ =1 3 66 20 0 1 = −3.926−15.858 → 0 1 = 0.693−1

(4) Find [ , ] = ln → 0.693 = ln → = 2

= − → −1 = − → = 1

Hence = 2

Exercise.12

Fit the following data ( , ) by exponential form =

[R Check answer: = 1.579 0.391 ]

0 2 4 2 0.27 0.0365

0 2 4 0.693 -1.309 -3.31

0 1 2 3 4 1.5 2.5 3.5 5 7.5




58 | P a g e

• Fit by = + = +

= , = , = , =

• Fit by = + = +

= +

= , = , = , =

• Fit by =

= +

= +

= , = , = , =

Example.10

Fit the following data ( , ) by =

Solution:

= +

= , = , = 0 , =

Linearize Data ( , ) → ( , )

2 4 8 8 4 2

0.5 0.25 0.125 8 4 2




59 | P a g e

∑ =1∑ =1 ∑ 2 =1 0 1 = ∑ =1∑ =1 3 0.8750.875 0.328 0 1 = 145.25 → 0 1 = 016 = 0 → = 0

= → = 16

Hence =

Exercise.13

Find , , , for the following functions: = , = where L is constant.

Exercise.14

Find the least-squares parabola = 0 + 1 + 2 2 for the following data ( , ):

(-3, 15), (-1, 5), (1, 1), (3, 5)

Exercise.15

You have 3 points (1, 6), (2, 4), and (4, 3). Find the best function to fit these data from the following functions:

(1) = + (2) = (3) = +



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