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Phys 3750
Complex Fourier Series
Overview and Motivation : We continue with our discussion of Fourier series,
which is all about representing a function as a linear combination of harmonic
functions. The new wrinkle is that we now use complex forms of the harmonic
functions.
Key Mathematics : More Fourier Series! And a cute trick that often comes in handy
when calculating integrals.
I. The Complex Fourier Series
Last time we introduced Fourier Series and discussed writing a function f ( ) x defined
on the interval − L ≤ x ≤ L as
( ) (^) ∑[ ( ) ( )]
∞
=
1
0 cos sin n
L
nx L n
nx f x n
π π
where the Fourier coefficients are given as
( ) ∫ −
L
L
f x dx 2 L
α 0 , (2a)
( ) ( ) ∫ −
L
L
L
nx n f x dx L
π α cos
, (2b)
( ) ( ) ∫ −
L
L
L
nx n f x dx L
β 1 sin^ π. (2c)
While there is nothing wrong with this description of Fourier Series, it is often
advantageous to use the complex representations of the sine and cosine functions,
( nL^ π x^ ) = ( ein π xL + e − in^ π xL ) 2
cos , (3a)
( nL^ x ) ( e inxL einxL ) i
π (^) = π − − π 2
sin. (3a)
If we insert these expressions into Eq. (1) we obtain
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( ) [ ( ) ( )]
∞
=
= + + −^ − − −
1
0 2
n
inxL inxL n
inxL inxL f x n e e i e e
α α π π β π^ π , (4)
which can be rearranged as
( ) ∑[( ) ( ) ]
∞
=
1
0 2
n
inxL n n
inxL f x n i n e i e
α α β π α β^ π. (5)
This doesn't look any simpler, but notice what happens if we define a new set of
coefficients (which are simply linear combinations of the of the current coefficients
a n and β n ),
c 0 = α 0 (6a)
cn = 12 ( α (^) n − i β n ) (6b)
c − n = 21 ( α (^) n + i β n ) (6c)
Then we can write Eq. (5) as
( ) [ ]
∞
=
− = + + −
1
0 n
inxL n
inxL f x c cn e c e
or even more simply as
( )
∞
=−∞
n
inxL f x cn e
Using Eq. (2) it is not hard to show that the coefficients cn in Eq. (6) are given by
( )
−
L
L
inxL n f x e dx L
c^ π 2
Equations (8) and (9) are known as the complex Fourier series representation of the
function f ( ) x. Notice that with the complex representation there is only one
expression needed for all of the Fourier coefficients.
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Not this is OK, as long as we do not use it for more than it is worth: for any nonzero
value of n Eq. (14) is perfectly fine. But what about the case of n = 0? Then this
expression is undefined. What this means is that we must explicitly set n = 0 in Eq.
(13) and reevaluate it. But for n = 0 , we see from Eq. (9) that c 0 is just the average
value of the function, which is A 2. Putting this all together we can represent the
function f 1 ( ) x as
( )
( )
( )
∞
≠
=−∞
0
(^122)
1 cos
2
n
n
einxL n
n A
A
f x^ π
Now this is a valid representation of the function f 1 ( x ), but you may be wondering
about something. We know that the function f 1 ( x )is real, but the rhs of Eq. (15)
appears to have an imaginary part, because ei^ n π x^ L = cos ( nL π x ) + i sin( nL^ π x ). So what is the
deal? Well, it is not too difficult to see that the imaginary part of each positive- n term
is exactly cancelled by the imaginary part of the corresponding negative- n term. So,
Eq. (15) is indeed real.
III. The Gaussian Function
Let's take another look at the Gaussian function and think a bit about representing it
as a Fourier Series. You should recall that the Gaussian function is defined as
( )
(^2) σ 2 σ
x G x e
−
where σ is known as the width parameter. Let's assume in this example that L >> σ.
Then we have something like the following picture, where we have set σ = 0. 2 and
L = 2.
1 0.5 0 0.5 1
0
1
f 1 ( )x
A
x
L
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2 1 0 1 2
0
1
GaussianGaussian
x
G(x)
Let's calculate the coefficients cn. Using Eq. (9) we have
∫ −
L
L
x inxL n e e dx L
c σ^ π
2 2
2
Now this integral can be expressed in terms of the error function, which is the integral
of the Gaussian function, but the expression is pretty messy. However, there is an
approximate solution to the integral in Eq. (17) that is quite simple and very accurate,
as we now show. For the conditions that we assumed, namely L >> σ, the Gaussian
function is nearly zero for x ≥ L. Because of this we can extend the limits of
integration in Eq. (17) to m ∞, and with very little loss of accuracy we can write
∫
∞
−∞
= e − e − dx L
c (^) n x σ^ in^ π xL
2 2
2
We can also take advantage of the properties of integrals of odd and even functions if
we write e ( n xL ) i ( n x L )
inxL
π =cos − sin
−
, which turns Eq. (18) into
[ (^ )^ (^ )] ∫
∞
−∞
− = e n x L − i n xL dx L
c
x
n π^ π
σ cos sin 2
Now the Gaussian function is even, so the integral of e x σ^ [ i sin( n π xL )]
2 2
− − is zero, and
so we are left with
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approximately 3× 2 L (π σ). On the scale of this graph, the truncated Fourier series
certainly does a good job of representing the Gaussian function. (As above, we have
again set σ = 0. 2 and L = 2 ).
2 1 0 1 2
0
1
Gaussian Fourier Representation
Gaussian Fourier Representation
x
G(x), Four. Rep.
However, there is an inherent limitation to using Fourier series to represent a
nonperiodic function such as a Gaussian. That limitation is illustrated in the next
figure, which plots the Gaussian and its Fourier series over an interval larger than
− L ≤ x ≤ L. Within the interval the match is very good (as we saw in the last graph),
but outside the interval the match is pretty lousy. Why? Well, that is because the
harmonic functions that make up the Fourier series all repeat on any interval with
length 2 L. Thus, the Fourier representation of the Gaussian function has periodicity
2 L.
Well, you might say that there is no problem here. I'll just pick a value of L that is
10 6 2 2 6 10
0
1
Gaussian Fourier Representation
Gaussian Fourier Representation
x
G(x), Four. Rep.
− L L
Phys 3750
larger than any value of x where I might want to evaluate the original function. That
might work in practice, but we might also ask the question: is there a Fourier-series
representation that will work for all x? The answer is yes, and we will discuss that in
a few lectures after this one.
Right now I just want to point out that getting to such a representation is not at all
trivial. Consider the following. We have represented the Gaussian as a linear
combination of harmonic functions
inxL e
π
. If we want to use a Fourier representation
for all x , then somehow we must take the limit where L →∞. What does that mean
for the harmonic functions? It looks like all of the harmonic functions will simply
become equal to 1 (which seems pretty bad!). There is a resolution to this dilemma,
but this illustrates that taking the L →∞limit of the Fourier-series representation is
somewhat nontrivial.
Exercises
* 12.1 Calculate the integral on the rhs of Eq. (10) and show that it is nonzero only if
n = m.
* 12.2 Consider the result for the coefficient for the triangle function,
( ) 2 2
1 cos
n
n c (^) n A
= , which is undefined for n = 0. Use l'Hôspital's rule to show that as
n → 0 , cn → A 2 , the result for c 0.
* 12.3 Using Eq. (2) in Eq. (6b), show that cn is given by Eq. (9).
** 12.4 Fourier series example. Consider the function ( )
− 0
e x
e x f x x
x
(a) Plot this function. Explain why this function is even?
(b) Find a real analytic expression for the Fourier coefficients cn for this function.
(Hint: You can use the fact that f ( x )is even to simplify your determination of the
coefficients.)
(c) Let L =5. Plot the function and its truncated Fourier representation for several
values of M. What is the minimum reasonable value for M necessary to represent
f ( ) x on this interval?
