CM2202: Scientific Computing and Multimedia
Applications
Lab Class Week 5
School of Computer Science & Informatics
Vectors
Calculus: Differentiation
Calculus: Integration
Vector operators
Definition (Vector Addition, Subtraction and Scalar Multiplication
in Rn )
Let v and w be two vectors in Rn and k a real number. The
following rules are well-defined:
v + w = (v1 + w1 , v2 + w2 , . . . , vn + wn ).
v − w = (v1 − w1 , v2 − w2 , . . . , vn − wn )
kv = (kv1 , kv2 , . . . , kvn ).
These rules coincide with the geometrical interpretation for
two-dimensional vectors (see previous definitions).
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Vectors
Calculus: Differentiation
Calculus: Integration
Vector Addition, Subtraction and Scalar Multiplication in
MATLAB
MATLAB directly supports vector addition, subtraction and scalar
multiplication:
>> v=[1 2 5];
>> w=[3 -1 1];
>> v+w
ans =
4
1
6
>> v-w
ans =
-2
3
4
>> 3 * v
ans =
3
6
15
>> w*(-1)
ans =
-3
1
-1
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Vectors
Calculus: Differentiation
Calculus: Integration
Scalar product
Definition (Scalar product)
Given two vectors v and w in Rn with components (v1 , v2 , . . . , vn )
and (w1 , w2 , . . . , wn ). We define the scalar product (or (standard)
inner product, dot product) of v and w as
v.w or hv, wi =
n
X
vi wi
i=1
Note what the scalar product does:
It takes two vectors and assigns them a real number.
Problem (Scalar product)
Work out the scalar product of vectors v = (1, 2) and w = (2, 3)
Note the notations v.w and hv, wi are equivalent.
We use the v.w notation.
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Vectors
Calculus: Differentiation
Calculus: Integration
Scalar product using MATLAB
MATLAB provides a vector function dot that computes the dot
product of two vectors (of any, identical dimension).
>> v = [3 2 -1]
>> w = [2 -1 1]
>> dot(v, w)
ans =
3
>> sum(v.*w)
ans =
3
dot(v, w) is equivalent to sum(v.*w) note v.*w is an array
multiplication that returns a vector of the same size.
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Vectors
Calculus: Differentiation
Calculus: Integration
Properties of scalar products
Theorem (Cauchy-Schwarz inequality)
Let v and w be vectors in Rn
Then they satisfy the Cauchy-Schwarz inequality
|v.w| ≤ kvkkwk.
Theorem (Angle Between Two Vectors)
If n = 2, 3 we even have the relation
v.w = kvkkwk cos(θ)
We call θ the angle between v and w .
6 / 24
Vectors
Calculus: Differentiation
Calculus: Integration
Euclidean norm of a vector
Definition (Euclidean norm of a vector)
For a vector v ∈ Rn we define its norm as
√
kvk = v.v
This norm is called the Euclidean norm of the vector v.
The Euclidean norm of a vector coincides with the length of the
vector in R2 and R3 .
v2
y
(v1, v2)
!v!
v
v1
x
By Pythagoras’ Theorem, kvk =
q
√
v12 + v22 = v.v
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Vectors
Calculus: Differentiation
Calculus: Integration
Euclidean norm of a vector in MATLAB
The default behaviour of MATLAB function norm for a given
vector input is to return the Euclidean norm (also called 2-norm):
>> v = [3 4]
>> norm(v)
ans =
5
>> sqrt(dot(v, v))
ans =
5
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Vectors
Calculus: Differentiation
Calculus: Integration
The Vector Cross Product
Besides the scalar product that maps two vectors from Rn to R we
also need a product that maps two vectors from Rn to a vector in
Rn .
Definition (The vector cross product in R2 )
We define the vector cross product of v, w ∈ R2 as a mapping
× : R2 × R2 7→ R with
v × w = v1 w2 − v2 w1
The vector product in R2 is anti-symmetric, i.e.
v × w = −w × v
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Vectors
Calculus: Differentiation
Calculus: Integration
The Vector Cross Product (cont.)
Definition (The vector cross product in R3 )
We define the vector cross product of v, w ∈ R3 as a mapping
× : R3 × R3 7→ R3 with


v2 w3 − v3 w2
v × w = v3 w1 − v1 w3  .
v1 w2 − v2 w1
The vector product in R3 is also anti-symmetric, i.e.
v × w = −w × v
The vector cross product has very useful properties, especially:
for finding orthogonal vectors in R3 .
for area and volume calculations in R2 and R3 respectively.
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Vectors
Calculus: Differentiation
Calculus: Integration
Vector cross products in MATLAB
MATLAB provides a vector function cross to compute the cross
product of two vectors in R3 :
>> v=[1 2 3];
>> w=[-1 1 2];
>> cross(v, w)
ans =
1
-5
3
11 / 24
Vectors
Calculus: Differentiation
Calculus: Integration
Differentiation in MATLAB
Derivative of a poly() structure - polyder(), Poly Diff Eg.m
>> a = [ 3 6 9 ] ; % 3 x ˆ2 + 6 x + 9
>> k = p o l y d e r ( a )
k = 6
6
>>b = [ 1 2 0 ] ;
>> k = p o l y d e r ( b ) % x ˆ2 + 2 x
k =
2
2
>> k = p o l y d e r ( a , b ) %(3x ˆ2 + 6 x + 9 ) ( x ˆ2 + 2 x )
k = 12
36
42
18
MATLAB code for this at Poly Diff Eg.m
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Vectors
Calculus: Differentiation
Calculus: Integration
Symbolic Differentiation in MATLAB
Symbolic Derivative via diff(), Symb Diff Eg.m
>>syms x ;
% Declare f (x)
>>f = x ˆ2 + 2∗ x +1;
% Differentiate f (x)
>>d f = d i f f ( f )
a n s = 2∗ x + 2
MATLAB code for this at Symb Diff Eg.m
13 / 24
Vectors
Calculus: Differentiation
Calculus: Integration
Computing and Plotting Tangent to a Curve
Symb Diff Eg.m continued
% G r a d i a n t a t ( x0 , y0 )
x0 = 1 ; y0= 5 ;
% compute m + c ( y−a x i s I n t e r c e p t )
m = s u b s ( d f , x0 ) ;
c = y0 − m∗ x0 ;
%D e c l a r e
t a n e q n = m∗ x + c ;
%P l o t f ( x ) and t a n g e n t a t ( x0 , y0 )
e z p l o t ( t a n e q n ,[ −50 5 0 ] ) ;
h o l d on ;
e z p l o t ( f ,[ −50 5 0 ] ) ;
MATLAB code for this at Symb Diff Eg.m
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Vectors
Calculus: Differentiation
Calculus: Integration
Symb Diff Eg.m Output
x2 + 2 x + 1
2500
2000
1500
1000
500
0
−50
−40
−30
−20
−10
0
x
10
20
30
40
50
15 / 24
Vectors
Calculus: Differentiation
Calculus: Integration
Stationary Points in MATLAB
MATLAB Example:Stationary point eg.m
syms x
f = x ˆ3 −3∗x ˆ2 −9∗x + 2 ;
df = d i f f ( f ) ; % 1 st d e r i v a t i v e
d f 2 = d i f f ( d f ) ; %2nd d e r i v a t i v e
% f i n d Turning p o i n t s
% n e e d d o u b l e ( ) from c o n v e r t s y m b o l i c v a r i a b l e
t u r n p t s x = double ( s o l v e ( df ) ) ;
t u r n p t s y = double ( subs ( f , t u r n p t s x ) ) ;
max min = d o u b l e ( s u b s ( d f 2 , t u r n p t s x ) ) ;
....
P l o t R e s u l t s ( s e e S t a t i o n a r y p o i n t e g .m f i l e )
Full MATLAB code at: Stationary point eg.m
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Vectors
Calculus: Differentiation
Calculus: Integration
Stationary point eg.m Output
x3 − 3 x2 − 9 x + 2
600
400
200
0
Max
Min
−200
−400
−600
−800
−1000
−1200
−10
−8
−6
−4
−2
0
x
2
4
6
8
10
17 / 24
Vectors
Calculus: Differentiation
Calculus: Integration
MATLAB Integration: poly() Integrals
MATLAB lets you integrate poly() Polynomials:
polyint() and trapz()
>> p = [ 1 0 0 ] % p = x ˆ2
% Indefinite Integral
>> p o l y i n t ( p ) % Ans = x ˆ3/3
ans =
0.3333
0
0
0
% Definite I n t e g r a l ( via numerical integration )
>> t r a p z ( −6:6 , p o l y v a l ( p , − 6 : 6 ) )
ans =
146
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Vectors
Calculus: Differentiation
Calculus: Integration
MATALB Example: Area Under Graph (1)
MATLAB Code to solve and plot above example: integral area.m
% d e c l a r e f u n c t i o n and p l o t
syms x
plot range = [ −10 ,10];
it
f = x ˆ2;
ezplot (f , plot range );
h o l d on ;
% define definite integral
i n t l i m i t s = [−6
6];
limits
% Integrate
intf = int ( f )
% Indefinite
Integral
% Definite Integral via Indefinite result
i n t v a l i n d = int ( f , i n t l i m i t s (1) , i n t l i m i t s (2))
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Vectors
Calculus: Differentiation
Calculus: Integration
MATALB Example: Area Under Graph (2)
integral area.m Cont.
% I d i o t Check ! D e f i n i t e I n t e g r a l v i a I n d e f i n i t e r e s u l t
subs ( i n t f , i n t l i m i t s (2)) − subs ( i n t f , i n t l i m i t s (1))
% not r e a l l y a n e c e s s a r y b i t o f code as
%Def . I n t i s t h e way t o do i t !
% S e t up p l o t
range = i n t l i m i t s ( 1 ) : 0 . 1 : i n t l i m i t s ( 2 ) ;
y = s u b s ( f , r a n g e ) ; %s a m p l e v a l u e s on c u r v e
% Shade a r e a b e l o w t h e c u r v e
area ( range , y , ’ FaceColor ’ , [ 0 , 0 , 1 ] ,
’ LineStyle ’ ,
’ none ’ ) ;
Note: area()- useful plot function to shade areas of given graph
values.
MATLAB code for this at integral area.m
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Vectors
Calculus: Differentiation
Calculus: Integration
Example: Area Between Two Curves (1)
MATLAB Code: integral area between.m
% declare functions
syms x
plot range = [ −3 ,3];
f 1 = −x ˆ2 + 6 ;
f 2 = x ˆ2 − 2∗ x + 2 ;
% Find Points of I n t e r s e c t i o n of cur ves
r o o t s i n t e r s e c t = s o r t ( double ( s o l v e ( f1 − f2 ) ) ) ;
% Definite Integral
area intersect =
i n t ( f 1 − f2 , r o o t s i n t e r s e c t ( 1 ) , r o o t s i n t e r s e c t ( 2 ) )
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Vectors
Calculus: Differentiation
Calculus: Integration
Example: Area Between Two Curves (1)
integral area between.m Cont.
% S e t up p l o t
range = r o o t s i n t e r s e c t ( 1 ) : 0 . 1 : r o o t s i n t e r s e c t ( 2 ) ;
y = s u b s ( f 1 , r a n g e ) ; %s a m p l e v a l u e s on c u r v e f 1
% Shade Are a f 1
area ( range , y , ’ FaceColor ’ , [ 0 , 0 , 1 ] ,
h o l d on ;
’ LineStyle ’ ,
’ none ’ )
% ’ r u b out ’ a r e a a b o v e t h e c u r v e
y = s u b s ( f 2 , r a n g e ) ; %s a m p l e v a l u e s on c u r v e f 2
area ( range , y , ’ FaceColor ’ , [ 1 , 1 , 1 ] , ’ L i n e S t y l e ’ ,
’ none ’ )
% Plot funtions
e z p l o t ( f1 , p l o t
hf2 = e z p l o t ( f2
s e t ( hf2 , ’ Color ’
over p l o t range
range );
, plot range );
, ’ black ’ ) ;
MATLAB code for this at integral area between.m
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Vectors
Calculus: Differentiation
Calculus: Integration
MATLAB Example: Area Under x-axis (1)
MATLAB Code: integral area under.m
% declare function
syms x
plot range = [ −5 ,5];
i n t l i m i t s = [ −3 ,3];
f = x ˆ3;
% Definite Integral via Indefinite result
int area = int ( f , i n t l i m i t s (1) , i n t l i m i t s (2))
% WRONG ( f o r A rea ) !
Answer i s ZERO
% F o r x ˆ3 n e e d t o s p l i t a t x = 0
% E x e r c i s e : W r i t e a g e n e r a l f u n c t i o n t o work t h i s p o i n t o u t .
% Do Area P r o p e r l y
i n t a r e a = abs ( i n t ( f , i n t l i m i t s ( 1 ) , 0 ) ) + abs ( i n t ( f , 0 , i n t l i m i t s ( 2 )
23 / 24
Vectors
Calculus: Differentiation
Calculus: Integration
MATLAB Example: Area Under x-axis (2)
MATLAB Code: integral area under.m
% S e t up p l o t
range = i n t l i m i t s ( 1 ) : 0 . 1 : 0 ;
y = s u b s ( f , r a n g e ) ; %s a m p l e v a l u e s on c u r v e
% Shade a r e a b e l o w t h e c u r v e
area ( range , y , ’ FaceColor ’ , [ 1 , 0 , 0 ] ,
h o l d on ;
’ LineStyle ’ ,
’ none ’ ) ;
range = 0 : 0 . 1 : i n t l i m i t s ( 2 ) ;
y = s u b s ( f , r a n g e ) ; %s a m p l e v a l u e s on c u r v e
% Shade a r e a b e l o w t h e c u r v e
area ( range , y , ’ FaceColor ’ , [ 0 , 0 , 1 ] ,
’ LineStyle ’ ,
’ none ’ ) ;
%p l o t f u n c t i o n
h f =e z p l o t ( f , p l o t r a n g e ) ;
s e t ( hf , ’ Color ’ , ’ black ’ ) ;
MATLAB code for this at integral area under.m
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