Brewster H.D. Mathematical Physics
Подождите немного. Документ загружается.
142
Linear
Algebra
00
<A.,j>
=
I>n
<~j'~n
>
J
n=l
00
=
LCnN/Dij
n=l
=Cj0.
So, the expansion coefficient
is
Nj
<A,
A>:·
Let's determine
if
the set J
of
functions
<l>n(x)
= sin
nx
for n =
1,
2,
...
is
orthogonal on the interval [= 1t,
<1>],
We need to show that <
<l>n'
<l>m
> = 0 for n
-:f.
m.
Thus, we have for n
-:f.
m
<
<I>
<I>
> = r sinnxsinmxdx
n' m
7t
=
.!.
r [cos(n - m)x - cos(n + m)x ]dx
2
7t
=.!.[sin(n
- m)x
2
n-m
sin(n + m)x]7t =
O.
n+m
-7t
Here we have made use
of
a trigonometric identity for the product
of
two
sines.
So, we have determined that the set
<l>n(x)
= sin nx for n =
1,
2,
... is
an
orthogonal set
of
functions on the interval [= 1t, 1t]. Just as with vectors in
three dimensions, we can normalize our basis functions to arrive
at
an
orthonormal basis. This
is
simply done by dividing by the length
of
the vector.
Recall that the length
of
a vector was obtained as v =
~
.
In the same way, we define the norm
of
our functions by
11/11
=~<
l,j
>.
Note, there are many types
of
norms, but this will be sufficient for us.
For the above basis
of
sine functions, we want to first compute the norm
of
each function. Then we would like to find a new basis from this one such that
each basis eigenfunction has unit length and
is
therefore an orthonormal basis.
We first compute
11<1>
112
= [ sin
2
nxdx
n 7t
Linear
Algebra
=
~
r [1-cos2nx]dx
2 1t
=
~[x
_ sin
2nX]1t
= 1t.
2
2n_1t
We have found from this computation that
<
<P
l
,
<Pn
> =
nOij
143
1
and that
lI<Pnll
=J;..
Defining
'Vnex)
=
~<Pnex),
we have normalized the
<Pn's
and have obtained an orthonormal basis
of
functions on
[-
n,
<P]'
Expansions offunctions in trigonometric bases occur often and originally
resulted from the study
of
partial differential equations named Fourier series.
Chapter 5
Complex
Representations
of
Functions
COMPLEX
REPRESENTATIONS
OF
WAVES
We have seen that we can seek the frequency content
of
a functionf(t)
defined on an interval [0,
11
by looking for the Fourier coefficients
in
the
Fourier series expansion
ao
~
21tnt
.
21tnt
f(t)=
2+
~ancosT+bnsmT
n=l
The coefficients take forms like
2£1'
21tnt
a = -
f(t)cos--dt
n T T
However, trigonometric functions can be written in a complex exponential
form. This is based on Euler's formula (or, Euler's identity).
e
i9
= cos 9 + i sin
9.
The complex conjugate is found by replacing i with i to obtain
e--9
= cos 9 - isin
9.
Adding these expressions, we have
2 cos
9 = e
i9
+
e-
i
9.
Subtracting the exponentials leads to an expression for the sine function.
Thus, we have the important result that sines and cosines can be written as
complex exponentials.
So, we can write
e
i9
+ e-
i9
cos
9=---
2
e
i9
_e-
i9
sin
9=---
2i
21t
nt 1
(21tint
_
21tint
)
cos---y:-
=
2"
e T + e T .
Complex
Representations
of Functions
145
We can use this information to rewrite our series as a sum over complex
exponentials in the form
00
f(t)
= L
21tint
e e T
n
n=-oo
where the Fourier coefficients now take the form
21tint
en
= (
f(t)e-
T
In fact, in order to connect our analysis to ideal signals over an infinite
interval and containing a continuum
of
frequencies, we will see the above
sum become an integral and we will naturally find ourselves needing to
work with functions
of
complex variables and perform complex integrals.
We can extend these ideas to develop a complex representation for waves.
We obtained the solution
1[00
00
]
u(x,
t)=
2"
~AnSinkn(x+et)+
~AnSinkn(X-et)
We can replace the sines with their complex forms as
u(x,
t)
~
~j
[
~
A,
(/;,,(n,,)
- e
-ik.
(n,t)
) ]
+
[f
An
(eikn(x-ct) - e
-ikn(X+ct»)]
n=l
Now, defining
k_n
=
-kn'
we can rewrite this solution in the form
00
"
[e
eikn(:x.+ct)
+ d eikn(x-ct) ]
u(x,
t)
=
~
n n
n=-oo
Such representations are also possible for waves propagating over the
entire real line.
In such cases we are not restricted to discrete frequencies and wave
numbers. The sum
of
the harmonics will then be a sum over a continuous
range, which means that our sums become integrals.
So, we are then lead to
the complex representation
u(x, t) =
[00
[e(k)eik(X+ct) + d(k)eikll(X-ct)
Jdk.
The forms
eikex+
ct
) and eikex-ct) are complex representations
of
what
146 Complex Representations
of
Functirms
are called plane waves in one dimension. The integral represents a general
wave form consisting
of
a sum over plane waves, typically representing
wave packets. The Fourier coefficients in the representation can be complex
valued functions and the evaluation
of
the integral may be done using
methods from complex analysis. We would like to
be
able to compute
such integrals.
With the above ideas in mind,
we
will now take a tour
of
complex
analysis. We will first review some facts about complex numbers and then
introduce complex functions. This will lead us to the calculus
of
functions
of
a complex variable, including differentiation and complex integration.
COMPLEX
NUMBERS
Complex numbers were first introduced in order to solve some simple
problems.
The
history
of
complex
numbers
only
extends
about
two
equations
such as x
2
+ 1 =
O.
The
solution
is x =
±H.
Due
to
the
usefulness
of
this concept, which was not realized at first, a special symbol
was introduced - the imaginary unit,
i = H .
A complex number
is
a number
of
the form z = x + iy, where x and y
are real numbers. x is called the real part
of
z and y is the imaginary part
of
z.
Examples
of
such numbers are 3 + 3i, - Ii, 4i and
5.
Note that 5 = 5
+
Oi
and 4i = 0 + 4i.
There
is a geometric representation
of
complex numbers in a two
dimensional plane, known as the complex plane
C.
This is given
by
the
Argand
diagram,
here
we
can
think
of
the complex
number
z = x + iy as a
point
(x,
y)
in the
complex
plane
or
as a vector.
The
magnitude,
or
length,
of
this
vector
is
called
the
complex
modulus
oflzl =
~x2
+
y2.
We
can
also
use
the
geometric
picture
to
develop
a
polar
representation
of
complex numbers. We can see that in terms
of
rand
e
we have that
Thus,
x = r cos
a,
y=
r sin
a.
z
=x
+
iy=
r(cos a + i sin
a)
= re
i9
•
Here we have used Euler's formula.
So, given
r and a we have z =
ria.
However, given the Cartesian form,
z = x +
iy,
we can also determine the polar form, since
r
=
Jx
2
+
y2
tan
a=
Y
x
Complex
Representations
of
Functions
iy
•
x+iy
x
147
Fig. The Argand Diagram for Plotting Complex Numbers in the Complex z-plane
Note that r =
Izl.
Example: Write 1 + i
in
polar form.
If
one locates 1 + i
in
the complex
plane, then it might be possible to immediately determine the polar form
from the angle and length
of
the "complex vector".
If
one did not see the
polar form from the plot in the z-plane, then one can systematically
determine the results.
We
want to write 1 + i
in
polar form. 1 + i =
ria
for
some
rand
8.
Using the above relations, we have r
=~x2
+ y2
=J2
and
tan 8 =
~
=
1.
This gives 8 =
~.
So, we have found that
1 + i =
J2e
i1C
/
4
We
also
have
the
usual
operations.
We can
add
two
complex
numbers and obtain another complex number. This is simply done by
adding the real parts and the imaginary parts. So,
;y
2;
1 + ;
1
2
x
Fig. Locating 1 + i
in
the Complex z-plane
148
Complex
Representations of
Functions
(3
+
20
+
(1
- i) = 4 + i.
We can also multiply two complex numbers
just
like we multiply any
binomials, though we now can use the fact
that;2 =
-1.
For example,
we
have
(3
+ 2i)(1 - i) = 3 +
2i
- 3i +
2i(-
i) = 5 -
i.
We can even divide one complex number into another one and get a
complex number as the quotient. Before we do this, we need to introduce the
complex conjugate,
Z,
of
a complex number. The complex conjugate
of
z = x
+
iy,
where x and
yare
real numbers, is given as
z
=x-
iy.
Complex conjugates satisfy the following relations for complex numbers
z and
wand
real number x.
z+w
=
z+w
zw
=
zw
z
=z
x
=x.
One consequence is that the complex conjugate
of
z =
riB
is
z = re
i6
= cos 8 + i sin 8 = cos 8 - i sin 8 =
re-
6
Another consequence is that
zz
= re
i6
re-
ie
=
r2
..
Thus, the product
of
a complex number with its complex conjugate is a
real number. We can also write this result in the form
zz
= (x + iy)(x - iy) =
x2
+ r =
JzJ2.
Therefore,
we
have
JzJ2
=
zz.
Now
we are in a position to write the quotient
of
two complex numbers
in the standard form
of
a real plus
an
imaginary number. As an example,
3+2i
we-
1
--·
.
-l
This is accomplished by multiplying the numerator and denominator
of
this expression
by
the complex conjugate
of
the denominator:
3+2i
3+2i(1+i)_1+5i
--=--
--
---
1-i
1-i
l+i
2
3+
2i 1
5.
Therefore, we have the quotient is H = 2 + 2
'
·
Complex
Representations of Functions
149
We can also look at powers
of
complex numbers. For example,
(1
+ i)2 =
2i,
(1
+ i)3 =
(1
+
i)(2i)
=
2i
-
2.
But, what is
(1
+ i)1/2 =
JI+i?
In general, we want to find the nth root
of
a complex number. Let
t
=
zl=n.
To find t
in
this case
is
the same as asking for the solution
of
t"-z=O
given
z.
But, this
is
the root
of
an nth degree equation, for which we expect
n roots. We can answer our question
if
we write z
in
polar form, z = ,eia.
Then,
rllne
ieln
rlln
[cos~
+
iSin~]
If
we use this to obtain an answer to our problem, we get
(I
+ i)I12 =
(J2e
irr/4
t2
= 21/4 eirrel7t/8
But this
is
only one solution. We expected two
solutions~
The problem is that
the
polar representation for z is not unique. We
note that
e
2krri
=
1,
k =
0,
±
1,
±2, ....
So, we can rewrite
z as z = ,eia e
2krri
= ,ei(e+2krr). Now, we have that
zIln
=
rllnei(e+2krr)ln
lin
[
(8
+
2kTt)
.'
(8
+
2kTt)]
r cos + 1sm
n n
We note that we only get different values for
k =
0,
1,
... , n -
1.
Now, we can finish our example.
(
.fie
irr/4
e
2krri
)112
(1
+ i)1/2 =
'---v--'
insert 1 = e
2kltl
= 2114 e
i
(rr/8+krr)
= {2114 e
17t/8
,
2114
e9rri 18}
Finally, what
is
'il?
Our first guess would
be'il
=
1.
But, we know
that there should be n roots. These roots are called the nth roots
of
unity.
150 Complex Representations
of
Functions
and
Using the above result in equation with r = 1 and e = 0, we have that
[
21tk
.'
21tk]
11
=
Cos-;-+lsm-;-
,
For example, we have
[
21tk
.'
21tk]
VI
=
Cos-;-+lsm-;-
,
These three roots can be written out as
3r.
1
=1
-!+
fj
i
-!-
fj
i
'\/1 , 2
2'
2
2'
iy
k=O,
...
,n-l.
k=0,1,2.
-1
1 x
-i
Fig. Locating the Cube roots
of
Unity in the Complex z-plane
Note, the reader can verify that these are indeed the cube roots
of
unity.
We note that
We can locate these cube roots
of
unity in the complex plane. We see
that these points lie on the unit circle and are at the vertices
of
an equilateral
triangle. In fact, all nth roots
of
unity lie on the unit circle and are the vertices
of
a regular n-gon.
Complex
Representations of
Functions
COMPLEX
VALUED
FUNCTIONS
151
We would like to next explore complex functions and the calculus
of
complex functions. We begin by defining a function that takes complex
numbers into complex numbers,
f C
~
C.
It is difficult to visualize such
functions.
One typically uses two copies
of
the complex plane to indicate how
such functions behave. We will call the domain the z-plane and the image
will lie in the w-plane.
Iy
21
z-plane
2 x
iy
21
w-plane
w
2 u
Fig. Defining a Complex Valued Function on C
We let z = x + iy and w = u + iv. Then we can define our function as
w =
j(z)
= f (x + iy) = u(x,
y)
+ iv(x, y).
We see that one can view this function as a function
of
z or a function
of
x and y.
Often,
we have an
interest
in
writing
out
the
real
and
imaginary parts
of
the function, which can be viewed as functions
of
two
variables.
Example:
j(z)
= z2.
For example, we can look at the simple functionj(z) =
z2.
It
is
a simple
matter to determine the real and imaginary parts
of
this function. Namely,
we have
Z2
= (x + iy)2 = x
2
- T + 2ixy.
Therefore, we have that
u(x,
y)
= x
2
-
T,
vex,
y)
= 2xy.
Example:
j(z)
= e
Z
•
For this case, we make use
of
Euler's Formula.
j(z)
=
e=
=
tr+
iy
=
tre
iy
= tr(cos y + i sin y).
Thus, u(x.
y)
=
tr
cos y and
vex.
y)
=
tr
sin
y.
Example:
j(z)
=
In
z.