Lecture Notes on
Multivariable Calcul us
by M Niksirat
1
Lecture
Background materials
1 The set of real numbers
Calculus is a branch of mathematics dedicated to exploring the characteristics of “functions.” From a
physical perspective, a function refers to the assignment o f a scalar value to each point within a set,
known as the domain or focal set. The values collectively constitute the range of this assignment or
function This assignment can be represented as f: D
f
!R
f
, where D
f
is the domain of the function,
and R
f
is the range. Generally, the domain of a function is a subset D of R
n
, where R
n
represents
the set of all n-tuples f(x
1
; :::; x
n
)g, with each x
j
denoting a point in R.
To grasp the intricacies of functions, the focal point of Calculus, it is essential to initially comprehend
the properties of a function's domain and range. In this context, we introduce the space R
n
and
delve into its algebraic and topological properties.
The set of real numbers, denoted by R, serves as a mathematical representation of one-dimen-
siona l “continuum” objects, conceptualized as “straight” lines. The term “continuum” implies the
theoretical possibility of inﬁnite divisions within the object. Straightness, a geometric property
easily grasped by all, refers to lines devoid of curvature. This perception gives R a geometric
interpretation as a straight line, commonly referred to as the x-axis or y-axis in the xy-plane or in
thre e-dimensional space.
Similarly, R
2
can be envisioned as an unbounded straight plane without curvature, and R
3
rep-
resents space without curvature, extending this pattern further. The elements o f R
n
have a dual
nature; they can be viewed both as points x=(x; :::; x) and alternatively as vectors x~ =(x; :::; x).
These two interpretations lead to distinct structures on R
n
.
x
2
x
~
x
1
x
n
x
1
x
2
x
n
x
1.1 R
n
as a vector space
As a set of vectors, R
n
b ecomes a vector space through the introduction of two operations: vector
addition and scalar multiplication. For two vectors x~ = (x; :::; x) and y~ =(y ; :::; y), these opera-
tions are deﬁned as follows:
x~ + y~ = (x
1
+ y
1
; : : : x
n
+ y
n
):
Multiplying by a scalar λ, means
λ y
~
= (λx
1
; : : : ; λx
n
):
2
x
~
+ y
~
λx~ ; λ > 0
λ x~ ; λ < 0
x~
x~ ¡ y~
x~
y
~
It is evident that R
n
is closed under these operations, meaning that for any vectors x~ ; y~ in R
n
, and
scalars λ
1
; λ
2
2R, the vectorλ
1
x~ + λ
2
y~ is in R
n
is also in R
n
. Additionally, the fo llowing properties
hold for vectors x~ ; y~ ; z~ 2R
n
, and scalar λ2R:
i. x
~
+ y
~
= y
~
+ x
~
ii. λ (x~ + y~) = λ x~ + λ y~
iii. (x~ + y~) + z~ = x~ + (y~ + z~)
The set fi
^
; j
^
g wh ere i
^
=
1
0
; j
^
=
0
1
is called the standard basis for R
2
. In fact, each vec tor
x~ = (x
1
; x
2
) in R
2
can be uniquely written as
x~ = x
1
i
^
+ x
2
j
^
:
The set fi
^
; j
^
; k
^
g where i
^
= (1; 0; 0); j
^
= (0; 1; 0) and k
^
= (0; 0; 1) is the standard basis for R
3
as any
vector x~ = (x
1
; x
2
; x
3
) can be uniquely expressed as:
x~ = x
1
i
^
+ x
2
j
^
+ x
3
k
^
:
The set fe^
1
; : : : ; e^
n
g where
e^
1
= (1; 0; : : : ; 0); e^
2
= (0; 1; 0; : : : ; 0); : : : ; e^
n
= (0; 0; : : : ; 1)
is the standard basis for R
n
.
As a vector space, R
n
also admits another operation called the dot product. For two vectors
x~ = (x
1
; : : : ; x
n
) and y~ = (y
1
; : : : ; y
n
), the dot product x
~
·y
~
is deﬁned as:
x~ ·y~ =
X
j=1
n
x
j
y
j
:
In particular if y~ = x~ , then
x~ ·x~ =
X
j=1
n
x
j
2
;
3
known as the square magnitude of x~ and is denoted by jx~ j
2
. Accordingly, the magnitude of a vector
x~ is deﬁned as:
jx~ j=
X
j=1
n
x
j
2
v
u
u
u
u
t
:
Deﬁnition 1. Two non-zero vectors x~ and y~ are called orthogonal to each other if x~ · y~ = 0. In
particular the vectors e
^
j
are mutually orthogonal to each other, i.e., e
^
i
·e
^
j
= δ
i; j
where δ
i;j
=
1 i = j
0 i =/ j
:
Exercise 1. Show that the dot product satisﬁes the fo llowing properties:
a) jx~ + y~ jjx~ j+ jy~ j
b) jx~ + y~ j
2
¡jx~ ¡ y~ j
2
= 4 x~ · y~
Exercise 2. Pr ove tha t i f x~
1
; : : : ; x~
m
are mutua lly orthogonal, then
jx~
1
+ ··· + x~
m
j
2
= jx~
1
j+ ··· + jx~
m
j
2
:
An important property of t he dot product is the Cauchy-Schwarz inequality:
x~ · y~ jx~ jjy~ j:
This inequ ality is a consequence of the trivial inequality 0 jλ x~ + y~ j
2
for all scalar λ. Expanding
the right-hand side, we obtain :
λ
2
jx~ j
2
+ 2λ x~ · y~ + jy~ j
2
0:
This inequality holds true, ensuring the validity of the Cauchy-Schwarz inequality. The Cauchy-
Schwarz inequality becomes an equality only if x
~
and y
~
are linearly dependent in the sense x
~
= α y
~
for some scalar α 0. Consequently, we obtain:
¡jx~ jjy~ jx~ ·y~ jx~ jjy~ j:
Through the fact:
¡1
x~ · y~
jx
~
jjy
~
j
1;
we can deﬁne θ, the angle between two vectors x~ and y~ as
cos(θ) =
x~ · y~
jx~ jjy~ j
:
Evidently, the angles θ for 0;
π
2
and π results to the relations x
~
·y
~
=jx
~
jjy
~
j, x
~
·y
~
=0 and x
~
·y
~
=¡jx
~
jjy
~
j.
4
In addition to the dot product, there is another vector multiplication deﬁned in R
3
known as the
cross o r external product. For vect ors x~ =(x
1
; x
2
;x
3
), and y~ = (y
1
; y
2
; y
3
), the cross product is d eﬁned
as:
x~ ×y~ =
i
^
j
^
k
^
x
1
x
2
x
3
y
1
y
2
y
3
= (x
2
y
3
¡x
3
y
2
)i
^
+ (x
3
y
1
¡x
1
y
3
)j
^
+ (x
1
y
3
¡x
3
y
1
)k
^
:
It is evident through the properties of the determinant that x
~
·(x
~
× y
~
) = y
~
·(x
~
× y
~
) = 0 and thus
x
~
×y
~
is orthogonal to the plane spanned by x
~
and y
~
.
θ
x~
y~
A = jx~ × y~ j
x~ × y~
By a simple c alculation, it is seen:
jx~ ×y~ j
2
+ (x~ · y~)
2
= jx~ j
2
jy~ j;
and by the relation (x
~
·y
~
)
2
= jx
~
j
2
jy
~
j
2
cos
2
(θ), we obtain the following equality:
jx~ ×y~ j= jx~ jjy~ jsin(θ):
The ﬁgure below shows the proof by a geometrical argument:
A = ju~ jh = ju~ jjv~ jsin(θ)
θ
v~
u~
h = jv
~
jsin(θ)
This relation in turn leads to:
x
~
×y
~
= jx
~
jjy
~
jsin(θ) n
^
;
where n^ =
x~ × y~
jx~ × y~ j
is the unit ortho gonal vector on the plane spanned by x~ and y~.
Exercise 3. Show the following relati on for any thr ee vectors x~ ; y~ ; z~ in R
3
x~ ·(y~ ×z~) = y~ ·(z~ ×x~ ):
Exercise 4. Pr ove the relation
x~ ×(y~ ×z~) = (x~ ·z~)y~ ¡(x~ · y~)z~
5
Hint : You may wish rst to show the relation when y~ ; z~ are orthogonal, and then generalize this fact for no n-
or thogonal vectors y~ ; z~.
Use the above formula and determine the conditions that the following equality holds
x~ ×(y~ ×z~) = (x~ × y~) ×z~:
For each vector space R
n
, we can deﬁne a vector subspace R
m
for m n . In fact the set of vectors
(
x
1
; : : : ; x
m
; 0; : : : ; 0
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