Lets go back to the case of a string fixed at
0
0
and
L
L
,
its
n
t
h
n
t
h
harmonic is
y
n
(
x
,
t
)
=
A
n
sin
(
n
π
x
L
)
cos
(
ω
n
t
−
δ
n
)
y
n
(
x
,
t
)
=
A
n
sin
(
n
π
x
L
)
cos
(
ω
n
t
−
δ
n
)
In fact all the modes could be permitted, and so any possible motion of the
string can be completely specified
by:
y
(
x
,
t
)
=
∑
n
=
1
∞
A
n
sin
(
n
π
x
L
)
cos
(
ω
n
t
−
δ
n
)
.
y
(
x
,
t
)
=
∑
n
=
1
∞
A
n
sin
(
n
π
x
L
)
cos
(
ω
n
t
−
δ
n
)
.
This has been rigorously shown by mathematicians but the complete proof is
beyond our scope in this course. Lets accept the mathematicians word on this.
We could take a snapshot of this function at a time
t
=
t
0
t
=
t
0
.
Then we could write
y
(
x
)
=
∑
n
=
1
∞
B
n
sin
(
n
π
x
L
)
y
(
x
)
=
∑
n
=
1
∞
B
n
sin
(
n
π
x
L
)
where
B
n
=
A
n
cos
(
ω
n
t
0
−
δ
n
)
.
B
n
=
A
n
cos
(
ω
n
t
0
−
δ
n
)
.
Likewise we could look at one point at space and look at the oscillations as a
function of time. In that case we would get.
y
(
t
)
=
∑
n
=
1
∞
C
n
cos
(
ω
n
t
−
δ
n
)
y
(
t
)
=
∑
n
=
1
∞
C
n
cos
(
ω
n
t
−
δ
n
)
Lets work with the time
snapshot,
y
(
x
)
=
∑
n
=
1
∞
B
n
sin
(
n
π
x
L
)
y
(
x
)
=
∑
n
=
1
∞
B
n
sin
(
n
π
x
L
)
We need to figure out what the
B
n
B
n
factors are and this is what Fourier figured out. We can multiply both sides
by the
sin
sin
of a particular harmonic
y
(
x
)
s
i
n
(
n
i
π
x
L
)
=
∑
n
=
1
∞
B
n
sin
(
n
π
x
L
)
s
i
n
(
n
i
π
x
L
)
y
(
x
)
s
i
n
(
n
i
π
x
L
)
=
∑
n
=
1
∞
B
n
sin
(
n
π
x
L
)
s
i
n
(
n
i
π
x
L
)
and now we can integrate both sides
Recall
cos
(
θ
−
φ
)
=
cos
θ
cos
φ
+
sin
θ
sin
φ
cos
(
θ
−
φ
)
=
cos
θ
cos
φ
+
sin
θ
sin
φ
cos
(
θ
+
φ
)
=
cos
θ
cos
φ
−
sin
θ
sin
φ
cos
(
θ
+
φ
)
=
cos
θ
cos
φ
−
sin
θ
sin
φ
So
sin
θ
sin
φ
=
1
2
[
c
o
s
(
θ
−
φ
)
−
cos
(
θ
+
φ
)
]
sin
θ
sin
φ
=
1
2
[
c
o
s
(
θ
−
φ
)
−
cos
(
θ
+
φ
)
]
Thus
This is
equal to zero at the limits
0
,
L
0
,
L
except for the particular case when
n
=
n
i
n
=
n
i
.
In that case
∫
sin
(
n
π
x
L
)
sin
(
n
i
π
x
L
)
ⅆ
x
=
∫
sin
2
(
n
π
x
L
)
ⅆ
x
∫
sin
(
n
π
x
L
)
sin
(
n
i
π
x
L
)
ⅆ
x
=
∫
sin
2
(
n
π
x
L
)
ⅆ
x
So you get
After all
that we should see that for
each term in the sum
is zero, except the case where
n
i
=
n
n
i
=
n
.
Thus we can simplify the equation:
∫
0
L
y
(
x
)
sin
(
n
π
x
L
)
ⅆ
x
=
L
2
B
n
.
∫
0
L
y
(
x
)
sin
(
n
π
x
L
)
ⅆ
x
=
L
2
B
n
.
or
B
n
=
2
L
∫
0
L
y
(
x
)
sin
(
n
π
x
L
)
ⅆ
x
B
n
=
2
L
∫
0
L
y
(
x
)
sin
(
n
π
x
L
)
ⅆ
x
The above is a very specific form of the Fourier Series for a function
spanning an interval from
0
0
to
L
L
and passing through zero at
x
=
0
x
=
0
.
One could write a more general case for the Fourier Series which applies to an
interval spanning
−
L
−
L
to
L
L
and not constrained to pass through zero. In that case one can write
y
(
x
)
=
a
0
2
+
∑
n
=
1
∞
[
a
n
cos
(
n
π
x
L
)
+
b
n
sin
(
n
π
x
L
)
]
y
(
x
)
=
a
0
2
+
∑
n
=
1
∞
[
a
n
cos
(
n
π
x
L
)
+
b
n
sin
(
n
π
x
L
)
]
where
A
n
=
1
L
∫
−
L
L
y
(
x
)
cos
(
n
π
x
L
)
ⅆ
x
n
=
0
,
1
,
2
,
3
,
…
A
n
=
1
L
∫
−
L
L
y
(
x
)
cos
(
n
π
x
L
)
ⅆ
x
n
=
0
,
1
,
2
,
3
,
…
and
B
n
=
1
L
∫
−
L
L
y
(
x
)
sin
(
n
π
x
L
)
ⅆ
x
n
=
1
,
2
,
3
,
…
B
n
=
1
L
∫
−
L
L
y
(
x
)
sin
(
n
π
x
L
)
ⅆ
x
n
=
1
,
2
,
3
,
…
You can then look at the symmetry of the problem and see if just
sin
sin
or
cos
cos
can be used. For example if
y
(
−
x
)
=
y
(
x
)
y
(
−
x
)
=
y
(
x
)
then use cosines. If
y
(
−
x
)
=
−
y
(
x
)
y
(
−
x
)
=
−
y
(
x
)
use the sines.
In fact Fourier's theorem can be taken to a next step. This is Fourier's
integral theorem. That is any function (even if it is not periodic) can be
represented by
f
(
x
)
=
1
π
∫
0
∞
[
A
(
k
)
cos
(
k
x
)
+
B
(
k
)
s
i
n
(
k
x
)
]
d
k
f
(
x
)
=
1
π
∫
0
∞
[
A
(
k
)
cos
(
k
x
)
+
B
(
k
)
s
i
n
(
k
x
)
]
d
k
where
A
(
k
)
=
∫
−
∞
∞
f
(
x
)
cos
(
k
x
)
ⅆ
x
A
(
k
)
=
∫
−
∞
∞
f
(
x
)
cos
(
k
x
)
ⅆ
x
B
(
k
)
=
∫
−
∞
∞
f
(
x
)
sin
(
k
x
)
ⅆ
x
B
(
k
)
=
∫
−
∞
∞
f
(
x
)
sin
(
k
x
)
ⅆ
x
A
A
and
B
B
are called the Fourier transforms of
f
(
x
)
f
(
x
)
Lets look at an example.
f
(
x
)
=
E
o
|
x
|
<
L
/
2
f
(
x
)
=
0
|
x
|
>
L
/
2
f
(
x
)
=
E
o
|
x
|
<
L
/
2
f
(
x
)
=
0
|
x
|
>
L
/
2
right
away you can set
B
(
x
)
=
0
B
(
x
)
=
0
from
symmetry arguments
A
(
k
)
=
∫
−
∞
∞
f
(
x
)
cos
(
k
x
)
ⅆ
x
=
∫
−
L
/
2
L
/
2
E
0
cos
(
k
x
)
ⅆ
x
=
E
o
k
sin
(
k
x
)
|
−
L
/
2
L
/
2
=
E
o
k
[
sin
(
k
L
2
)
−
sin
(
−
k
L
2
)
]
=
2
E
o
k
sin
(
k
L
2
)
=
E
0
L
sin
(
k
L
2
)
k
L
2
A
(
k
)
=
∫
−
∞
∞
f
(
x
)
cos
(
k
x
)
ⅆ
x
=
∫
−
L
/
2
L
/
2
E
0
cos
(
k
x
)
ⅆ
x
=
E
o
k
sin
(
k
x
)
|
−
L
/
2
L
/
2
=
E
o
k
[
sin
(
k
L
2
)
−
sin
(
−
k
L
2
)
]
=
2
E
o
k
sin
(
k
L
2
)
=
E
0
L
sin
(
k
L
2
)
k
L
2
Up until now in the course we have been dealing with very simple waves. It
turns out that any complicated wave that can possibly exist can be constructed
from simple harmonic waves. So while it may seem that an harmonic wave is an
over simplification, it can be used in even the most complex cases.
Fourier Synthesis
Listen to Fourier Series
