Calculus, Linear Algebra, Fourier Analysis Fermin Viniegra

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Calculus,
Linear Algebra,
Fourier Analysis
Fermin Viniegra
25th January 2004
1 Review of Calculus in one Variable
1.1 Functions
A system (Mechanical, thermodynamical, etc) is defined by its properties.
Each property is defined by its set of possible values.
To each set of possible values we can assign a variable: x, y,....
The values of one variable often depend on the values of another variable
They are called the dependent variable and the independent variable.
The set of values of the independent variable make up the Domain=D
The set of values of the dependent variable make up the Range=R
A function from a set D to a set R is a rule that assigns to each element of D a unique element in R.
In terms of the variables we define a function as:
y
f x
where x is defined in the set D and y is defined in the set R:
x
y
D
R
The graph of a function is defined by the ordered pair x y
.y
x
y
x2
1
1
2
4
3
9
4
16
16
9
4
1
1
2
2
3
4
.x
1 Review of Calculus in one Variable
Not all relations between the set D and the set R are functions:
Problem: You work for a tile manufacturer, you are asked to provide with a tile that loses the most temperature at a specific pressure p 0 .
Which one is it?....and why?
1.2 Derivatives
The Derivative of a function y
f x is defined as the instantaneous rate of change of that function with respect to x:
dy
dx
f x
In terms of the limit, the derivative is defined as
f x
lim
f x
h 0
x
f
function
D
B
values
y=f(x)
System
sets of posible
h f x
h
NUMBERS
R
NUMBERS
NUMBERS
y
y1
x
x1
y2
x2
Function
Not a Function
3
1 Review of Calculus in one Variable
The key calculation linking this two definitions is:
f x
lim
h 0
f x
lim
h 0
x
y2 lim
x1 x2 x 2 ∆y
lim
∆x 0 ∆x
dy
dx
h f x
h
h f x
h x
y1
x1
∆y
y1
y2
x
x1
f x
h
x
h
x+h
x2
∆x
A simple example for limits. Take
f(x)
Example
Notice first that: f x
f(x+h)
f x
x2
2
Temperature
C
B
A
Pressure
p0
4
1 Review of Calculus in one Variable
so from the definition:
h f x
lim
h 0
h
x h 2 x2
lim
h 0
h
2
x 2hx h2 x2
lim
h 0
h
2
2hx h
lim
h 0
h
lim 2x h
f x
f x
h 0
therefore we establish that the rule is
2x
d 2
x
dx
2x
Interpretation of the Derivative
From the graph of the function (curve) we analyse the limit lim h Identify the points f x
0
f x h f x h
:
h and f x on the plot.
draw a line between those points
Taking the Limit:
The left-red-dot approaches the right-red-dot.
The intervals ∆x and ∆y become differentials dx and dy
The straight line crossing the dots is now tangent to the curve
The slope of the tangent line to the curve is now equal to the derivative of the function!
5
1 Review of Calculus in one Variable
g(x)=mx+c1
∆y
m=
∆x
f(x+h)
∆y
f(x)
c1
x
∆x
Limit
h
0
x+h
Straight Line
function
f(x) f(x+h)
dy
g(x)=mx+c2
c2
dx
Slope of the
Straight Line
tangent to
the curve
m=
dy
dx
Derivative of
the function
From the limit definition, it is possible to obtain several rules of differentiation here are a list of the most common:
derivative of a constant c
dc
dx
0
power rule for positive and negative integer powers of x ( if n negative then it is required that x
d n
x
dx
nxn 1
the constant multiple rule
d
cxn
dx
6
cnxn 1
0, we cannot divide by zero!)
1 Review of Calculus in one Variable
the sum and difference rule
d u v
dx
d
u v
dx
du dv
dx dx
du dv
dx dx
the product rule
d
uv
dx
dv du
v
dx
dx
the quotient rule, suppose v
u
0 at a point then
d u
dx v
dv
v du
dx u dx
v2
the derivative of the sine and cosine
d
sin x
dx
d
cos x
dx
cos x
sin x
the derivative of the tan x, cot x, sec x and csc x
d
tan x
dx
d
sec x
dx
d
cot x
dx
d
csc x
dx
sec2 x
sec x tan x
csc2 x
csc x cot x
Another typical example of derivative is the velocity and the acceleration (on one direction only)
v
a
where the notation
dn y
dxn
dr
dt
dv
dt
d2r
dt 2
represent the n-th derivative of the function y
f x
Answer to the problem of tempreatures:
Call the temperature of each tile TA , TB and TC then from the graph of the function we see that the slope of the line tangent to each curve
at p0 satisfies the relation:
mA
mB
mC
0
these slopes are equal to the derivatives so:
dTA
dp
p0
dTB
dp
p0
dTC
dp
p0
the tile that loses the most temperature at a specific pressure p 0 is A.
7
0
1 Review of Calculus in one Variable
1.3 Integrals
Integral: The exact area under the graph of a function
Area
b
F b F a
f x dx
a
this is the Fundamental Theorem of Calculus
Also known by its ability to be the antiderivative.
x
F x
f x dx
a
d
F x
dx
for f continuous on a b , existing f on a b and x
f x
a b
In terms of the graph of the function the integral will be
Problem: look at the following graph of an accelerated particle (curvy line). Using the little squares as reference answer the following
questions:
a) At which instant(s) is the particle reducing its velocity?
b)What is its velocity at t=2s?
c) What is the velocity at t=5s?
Answers:
First of all from the definition of the linear acceleration a
dv
dt
b
vt
a t dt
a
So the velocity will be the area of included by the graph of a t .
Temperature
Temperature
C
B
A
B
A
C
p0
Pressure
p0
8
Pressure
1 Review of Calculus in one Variable
a) Identify the instant(s) in which the particle reduces its velocity?
The value of the integral is always positive that means that the velocity is never diminishing
b)What is its velocity at t=2s?
Its v=2 m/s
c) What is the velocity at t=5s?
Its v=12 m/s
y
A3
f(x)
A1
b
a
A4
A2
b
f(x) dx = A1 − A 2 + A 3 − A 4
a
meters
seconds2
a(t)
4
3
2
1
t
1 2 3
4 5 6 7
9
8 9
seconds
x
2 Multi-variable Calculus
2.1 Variables
x
y
z=f(x,y)
f
System
sets of posible
D
B
values
function
NUMBERS
R
NUMBERS
NUMBERS
2.2 Level Curves
The set of points x y f x y
in space for x y in the domain of f is called the graph of f .
The graph of f is also called the surface z
f xy
The set of points in the plane where a function f x y has a constant value f x y
c is called a level curve
The set of level curves is called the Contour Plot.
2.3 Limits in surfaces
When the value of a real-valued function f x y lie arbitrarily close to a fixed real number L as we approach (but never reach) the point
x0 y0 from a neighbouring point x y in the Domain, then we said that L is the limit of the function f x y as x y approaches x 0 y0 :
x y
lim
f xy
x 0 y0 L
An interesting feature of limits is that they determine if a function is continuous or not.
10
2 Multi-variable Calculus
Continuity
A function is continuous at the point x 0 y0 when:
1. f is defined at x0 y0
2. lim
x y
x 0 y0 f xy
3. lim
x y
x 0 y0 f xy
L
exists
f x 0 y0
Example of Continuity test
2x2 y
x4 y2
Show that there is no limit for the function f x y
consider the lines: y
we substitute y
kx2 for x
0,
kx2 back to the function
f x kx2
f xy
y kx2
2x2 kx2
x4 kx2
2k
2
1 k
2
we calculate the limit along y
kx2
x y
lim
0 0
f xy
x y
lim
f xy
0 0
y kx2
2k
1 k2
Conclusion: the limit depends on which path we chose to approach the point 0 0 . Different k means different level curves means
different paths. Different limit values for different paths means NO defined Limit! No existence of limit means NO continuity of
the function at that point
z
z
z=f(x,y)
f(x,y)=1−x2 −y2
c=1−x2 −y2
y= (1−c)−x2
x
y
x
y
11
2 Multi-variable Calculus
2.4 Partial Derivatives
The partial derivatives are a very simple extension of the definition from the usual one-independent variable case. Consider the function
f x y we can calculate the derivatives with respect to y and x
Partial Derivative with respect to x (take y
constant
y0)
∂f x y
∂x
Partial Derivative with respect to y (take x
constant
f x
lim
h y 0 f x y0
h
h 0
x0)
∂f x y
∂y
f x0 y
lim
h 0
h f x0 y
h
In practice, the way to calculate them is very simple, fix one of the independent variables to a constant, and calculate the derivative with
respect to the other variable:
∂f x y
∂x
and
∂f x y
∂y
d f x y0
dx
Notice the very, very important change on the symbols ∂ d f x0 y
dy
d and the subindex 0 in x 0 and y0 to indicate that the variable is treated as a
constant.
Names of the two kinds of derivatives:
df
dx
T OTAL derivative o f f with respect to x
∂f
∂x
PARTIAL derivative o f f with respect to x
Notation for partial derivatives:
fx
∂f
∂x
∂f
∂y
∂2 f
∂x∂y
∂2 f
∂y∂x
∂2 f
∂x∂x
fy
fxy
fyx
fxx
∂ ∂f
∂x ∂y ∂ ∂f
∂y ∂x ∂ ∂f
∂x ∂x Example
Lets try to find the values of the derivatives
∂f
∂x
and
∂f
∂y
at the point 1 2 if the function is
f xy
x2
5xy
12
y3
2
2 Multi-variable Calculus
we start by considering y
and now take x
x0
y0
constant so
∂f
∂x
constant:
∂f
∂y
2x
5x
5y
3y2
We can now evaluate the value of the partial derivatives of the function at the desired point, 1 2 . We obtain:
∂f
∂x
21
22
2
4
6
1 2
∂f
∂y
21
1 2
2
32
Very simple!
13
2
12
14
3 The Standard Linear Approximation
Until now we have studied how to find the shape of a surface given by the function f x y . We know how to calculate the level curves
of that surface and we know how to find the slope of the lines tangent to the surface at any given point. This technics are a valuable
knowledge.
In this section we will divert a little bit from that line of research and turn to a more subtle problem, the problem of what is the shape of a
surface at a point P when we see it with a magnifying glass or zoom-in around that point.
What we find in most of the cases is that the shape at a close distance, resemblance the one of a flat surface. This means that with a certain
degree of accuracy we can imagine our original surface as been constructed by little pieces of flat surfaces or planes all glued together.
The advantage of this idea is that it is always easier to understand and manipulate information when it is in linear form (that is straight
lines, planes, etc), than in quadratic form (parabola, circle, spheres, ellipsoids,...) than in higher degree forms (very complicated shapes).
This simple but profound realisation has led to incredible technological and scientific developments.
There are three important elements of the linear approximation of a surface f x y at the point P:
1) the size of the region R around the point P for which we want to approximate the surface f x y with the tangent plane at that
point,
2) the slope of that tangent plane and
3) the accuracy A of this approximation for that particular region.
To understand the mathematics of the linearisation process, we will need to introduce some mathematical concepts
Translation of coordinates, we can translate a coordinate system with origin at 0 0 0 to any other point x 0 y0 z0 via the transformations x x x x0 , y y y y0 where:
x x0
∆x
y y0
∆y
∆z
z z0
Polynomials of different degrees
A polynomial of degree n, in two variables seen from a coordinate system centered at 0 0 0 is
z
a1 x
b1 y
1
1
a2 x b 2 y 2
a3 x b 3 y 3
2!
3!
14
1
an x b n y
n
n
3 The Standard Linear Approximation
We can shift the polynomial to a different coordinate system, with origin at x 0 y0 z0 and it’s points will have coordinates x
x x0
∆x, y
y y0
∆y
z
a1 x
b1 y
1
a2 x b2 y
2!
2
1
a3 x b3 y
3!
3
1
an x bn y
n
n
or equivalently:
z z0
a1 ∆x
b1 ∆y
1
1
a2 ∆x b2 ∆y 2
a3 ∆x b3 ∆y 3
2!
3!
1
an ∆x bn ∆y
n
n
So we did two things, we find the form of a general surface, and we learn how to move it around.
How do we express that two surfaces are coincident at the point P in a mathematical language?
Consider the surface given by the function f x y and the plane given by the polynomial of degree 1 z
x 0 y0 then a first observation is that at z
a1 ∆x
z 0 (since ∆x
let say that the point of contact between them is at P
z0
b1 ∆y and
0 ∆y
0),
f xy
f x0 y0 and because they the two surfaces are touching at that point then
f x 0 y0
so z
f x 0 y0
a1 ∆x
z0
b1 ∆y now a1 and b1 are the slopes of the plane, which will coincide with the partial derivatives of the
function at that point:
z
f x 0 y0
f x x0 y0 x x 0
f y x0 y0 y y 0
Standard linear approximation.
All the standard linear approximation implies is that for the points x y relatively close to P
look approximately like a plane, the plane z
f x 0 y0
f x 0 y0
x 0 y0 then the function f x y will
fy x0 y0 y y0 that is
f x x0 y0 x x 0
f x x0 y0 x x 0
f x y f y x0 y0 y y 0
Error estimation
Of course the higher the degree of the polynomial, the better the approximation of z to the function f x y . But to know if that is
needed we calculate first how accurate the linear aproximation is.
The formula for that is
E
1
∆x
2
∆y
2
M
where M is the highest value among the set of numbers { f xx x0 y0
15
fxy x0 y0
fyy x0 y0 }
3 The Standard Linear Approximation
Chain Rule
16
3 The Standard Linear Approximation
Maxima, Minima and Saddle Points
17
3 The Standard Linear Approximation
Homework 1
1. Show that for v x and u x
d u
dx v
2. Show that
dv
v du
dx u dx
v2
d
cos2x
dx
2sin2x
3. Show that
d
sec x
dx
4. Show that
sec x tan x
d
tanx
dx
sec2 x
5. Derive the Integration by Parts formula
f x g x f x g x dx
6. Calculate
xex dx
7. Calculate
xlnx dx
8. Find
dx
a2
9. Find
x2
10. Using Partial Fractions find
x2
3
2
dx
x2 4
2x2 3
dx
x x 1 2
18
f x g x dx
3 The Standard Linear Approximation
Homework 2
Find the function’s a) Domain, b) Range and c) Level Curves of the next functions.
1. f x y
xy.
1
x y .
2. f x y
3. f x y
ex ey
ex ey .
19
3 The Standard Linear Approximation
Homework 3
1) Find: a) Domain, b) level curves, of the function
f xy
x2
x
y2
2) Plot the surface which it represents,
3) Find if the function is continuous at 0 0 by using limits
4) Calculate the partial derivatives of this same function.
5) Find the slope of the tangent lines to the surface at the points 0 0 2 and 0 2 0 . Draw them in your Plot.
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