Mathwords: Inverse of a Matrix
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Inverse of a Matrix
Matrix Inverse
Multiplic
Splitting the Dierence.
Figure 12.4.
are easily computed:
1
1
ak =
a0 =
(x) dx =
1
dx = 1,
0
1
(x) cos k x dx =
cos k x dx = 0,
2 ,
1
1
k
(x) sin k x dx =
sin k x dx =
bk =
0
0,
0
k = 2 l + 1 odd,
k = 2 l even.
Therefore, the Fourier series for the st
144
Lecture 8: Fourier Methods
Although the preceding remarks have been made for one-dimension they appropriately carry over to higher
dimensions.
It follows from (8.32) that if we encounter a Fourier transform which can be factored then the inverse
tra
Lecture 8: Fourier Methods
147
that is illuminated all others receiving no light one nds that frequency of ring is proportional to the
logarithm of the light intensity, Weber-Fechner law (2.5). To avoid carrying the logarithm, we measure
illumination by t
150
Lecture 8: Fourier Methods
Inspection of Figure 8.6 shows that the response to the right of the discontinuity and is greatest and least
to the left of it. Clearly this follows from the ommatidium at the right getting less inhibition, and the one
on th
146
Lecture 8: Fourier Methods
is purely a notational change, (8.38), and is introduced in order to conform with Matlab usage. It then follows
from (8.42) that
Fk =
1
N
N
An e2ink /N ,
(8.45)
n=1
so that (8.43) and (8.45) exactly represents the sampled kn
Lecture 8: Fourier Methods
149
=
n
N
1
N
2ink/N
ke
(8.54)
k=1
with
N
k
2ink/N
ne
=
(8.55)
n=1
known. Next the unknown rn is written as
1
rn =
N
N
rk e2ink/N .
(8.56)
k=1
Substitution into (8.47) yields
rk =
k
1 + 2 cos(2k/N )
(8.57)
and the problem is sol
Lecture 8: Fourier Methods
141
an =
1
T
T /2
e2int/T F (t)dt =
T /2
T
1
T
e2int/T F (t)dt.
(8.10)
0
The analogue of (8.2) is
1
T
T /2
T /2
|F (t)|2 dt =
|an |2 .
(8.11)
n
= 1/T has the dimensions of reciprocal time, and thus is a frequency, and (8.9) in
148
Lecture 8: Fourier Methods
inhibitory network is basically the same for all ommatidia. Putting the pieces together, the response of the
nth receptor is governed by the Hartline-Ratli equation
rn =
n
(rn1 + rn+1 ),
(8.47)
where we make the further ass
Lecture 8: Fourier Methods
139
converges very nicely and quickly everywhere in the interval as seen in Figure 8.1. Note that a Taylor
expansion is local, and by contrast, a Fourier expansion uses global functions, viz., sinusoids.
Exercise 8.1. (a) Obtain
Lecture 8: Fourier Methods
145
It then follows from (8.9) that
Fk = F (k) =
an e2ink/N .
(8.36)
An important observation is that
Fk+N = Fk
(8.37)
so that the original function has become an N -periodic sequence. Next if we use the Euler rule to approximat
140
Lecture 8: Fourier Methods
Fig. 8.2:
The square wave, sgn(x) (8.6) for 1 < x < 1, is shown. The number of
terms in (8.6) is indicated. Note the over and undershoot at the discontinuities, Gibbs
phenomenon.
Exercise 8.3. Find the Fourier series for (a)
Lecture 8: Fourier Methods
143
Exercise 8.5 Suppose the temperature T in a bar of unit length 0 x 0 is initially T o = x. We have
shown T satises
T
2T
=
t
x2
(8.25)
T
= 0 at x = 0, 1.
x
(8.26)
Both ends are insulated at
Solve for T = T (x, t).
As touched
152
Lecture 8: Fourier Methods
s =
1
t
(8.65)
is the sampling frequency. It then follows that the sampled version of (8.64) is
Sk = S(kt) = e2i
Fig. 8.8:
kt
T
= e2ik s .
(8.66)
S(t), (8.64), for = 10/9 is sampled uniform at s = 1, resulting in the
aliased
Lecture 8: Fourier Methods
153
In Figure 8.8 we plot the sampled version of (8.64). These plots conrm the informal discussion given
above. We obtain an aliased version of (8.64) which gives an incorrect depiction of the original signal. Observe
that if
S(
Lecture 8: Fourier Methods
151
produces two solutions for . Can you put these together so that = no ? You will need a pedestal to avoid
negative ring rates. Graph and interpret your answer.
Sampling & Aliasing
At this point the relation between the exact
a
Lemma 12.11. If f (x) is odd and integrable on the symmetric interval [ a, a ], then
a
f (x) dx = 0. If f (x) is even and integrable, then
a
f (x) dx = 2
a
f (x) dx.
a
0
The next result is an immediate consequence of applying Lemmas 12.10 and 12.11 to
t
Example 12.2. Consider the function f (x) = x. We may compute its Fourier
coecients directly, employing integration by parts to evaluate the integrals:
1
a0 =
1
bk =
x dx = 0,
1
ak =
1
x sin k x dx =
1
x cos k x dx =
x cos k x sin k x
+
k
k2
x sin k x cos
1
0.5
-1
1
2
4
3
-0.5
-1
Figure 12.2.
Piecewise Continuous Function.
The simplest example of a piecewise continuous function is the step function
(x) =
1,
0,
x > 0,
x < 0.
(12.38)
It has a single jump discontinuity at x = 0 of magnitude 1, and is continuo
1
0.5
-1
1
2
3
4
-0.5
-1
Figure 12.3.
Piecewise C1 Function.
Thus, at each point, including jump discontinuities, the graph of f (x) has well-dened
right and left tangent lines. For example, the function f (x) = | x | is piecewise C1 since it
is continuou
Even this very simple example has remarkable and nontrivial consequences. For instance, if we substitute x = 1 in (12.30) and divide by 2, we obtain Gregorys series
2
1
1
1
1
= 1
+
+
.
(12.36)
4
3
5
7
9
While this striking formula predates Fourier theor
the derivatives of the function at the origin. The partial sums
n
sn (x) = c0 + c1 x + + cn xn =
ck xk
k=0
of a power series are ordinary polynomials, and the same convergence issues arise.
Although supercially similar, in actuality the two theories are p
with cos l x for l > 0, and invoking the underlying linearity of the inner product, yields
a
f ; cos l x = 0 1 ; cos l x
2
+
[ ak cos k x ; cos l x + bk sin k x ; cos l x ]
k=1
= al cos l x ; cos l x = al ,
since, by the orthogonality relations (12.26), a
Such partial dierential equations model diusion processes in which a quadratic energy
functional is decreasing as rapidly as possible. A good physical example is the ow of
heat in a body; the heat disperses throughout the body so as to decrease the therma
in which represents the vibrational frequency. Substituting into the wave equation
and the associated boundary conditions, we deduce that v(x) must be a solution to the
eigenvalue problem
d2 v
(12.21)
+ 2 v = 0,
v(0) = 0 = v(),
dx2
in which 2 = plays the
equations that govern the dynamics of continuous mechanical systems. In our reconstrucdtion of Fouriers thought processes, let us start by reviewing what we have learned.
In Chapter 6, we characterized the equilibrium equations of discrete mechanical and
to represent the general solution to the periodic heat equation. As in the discrete version, the coecients ak , bk are found by substituting the solution formula into the initial
condition (12.10), whence
u(0, x) =
a0
+
2
ak cos k x + bk sin k x = f (x).
154
Lecture 8: Fourier Methods
is a signal of period T = 1/. Then the highest frequency in the signal (8.72) is N /2. If we now sample at
t = T /N
1
Fk = F (kt) =
N
N/2
An e2ink/N ,
(8.73)
N/2+1
then we can exactly determine cfw_An from
N
An =
e2ink/N Fk
138
Lecture 8: Fourier Methods
f
2
1
=
f (s)f (s)ds
0
|an |2 = a 2 ,
=
(8.2)
n
so the transformation between function and fourier series is unitary! A rational approximation to a function
might be based on
an e2is = fN
f
(8.3)
|n|<N
with fN
2
being for ex
142
Lecture 8: Fourier Methods
and
F (t) =
F ()e2it d,
(8.19)
i.e. (8.18) and (8.19) are a Fourier pair that correspond to (7.98). Just as we showed that the nite Fourier
transform is unitary, (8.2), the same informal argument given above shows that
|F (t