5th May 2004
Notes by Omar Shouman
Calculus Lecture
Dr.Kuenzer
Define (Hyperbolic functions)
1
1 z0 z1 z 2
(z )0 (z )1 (z ) 2
sinh z  .(e z  e  z )  ((  
 )  (


 ))
2
2 0! 1! 2!
0!
1!
2!
z z3 z5
  
 
1! 3! 5!
Similarly,
1 z
z0 z2 z4
z
cosh z  (e  e ) 


 
2
0! 2! 4!
Fourier Series
Problem
Given a periodic function f(x) with period 2 , we want to represent it in the form:
f (x ) 

a0 
  an cos(nx )   b n sin(nx )
2 n 1
n 1
We need to find the coefficients an , bn .
Suppose given a function f :

x
f (x ) such that:
1) f is periodic with a period 2 , that is, f ( x  2 )  f ( x) for all x 
2) f is piecewise twofold differentiable on (2 , 2 ) , that is, except at a finite number of
points in (2 , 2 ) , f is twofold differentiable (i.e. 2nd derivative exists).
If it is NOT twofold differentiable in x0  (2 , 2 ) , then:
lim f (x )  f (x 0  ) , lim f (x )  f (x 0  ) , f ( x0  ) and f ( x0  ) exist .
x x 0
x x 0
1
We will use the 2 -periodic function e imz for m
Note that: eim( x2 )  eimx .e2 im  eimx (cos(2 m)  i sin(2 m))  eimx .
Thus, eimx is, in fact, 2 -periodic.
Let us assume that it is possible to write:
f (x ) 



n 
n 1
n 1
 c ne inx  c 0   c ne inx  c  ne inx with certain coefficients cn 
Can we get the cn starting from f ?
2
 imx
 f (x )e dx 
0
2

inx
 imx
 (  c ne )(e )dx 
0
n 
 2
 c e
n  0
i ( n m ) x
n
dx
if m  n
2 c m


2
 
1
1
i ( n m ) x 
 c n 
e
 cn
(e i ( n m )2  e 0 )

i
(
n

m
)
i
(
n

m
)
 
0

1

 cn
(cos  2  n  m    i sin  2  n  m    1)  0
i (n  m )

if m  n
Thus,
2
 f (x )e
 imx
dx  2 c m
0
Hence, we can get the coefficients:
cm 
1
2
2
 f (x )e
 imx
dx
0
Theorem
If f :

def
Let c m 
1
2
,
f (x ) , is 2 periodic and piecewise twofold differentiable.
x
2
 f (x ).e
 imx
dx
where cm 
0
Then,

 c .e
n
n
inx
 f ( x),if f is twofold differentiable in x

 1


 2 ( f ( x )  f ( x )), if not
2
ess.
We write: f ( x) 

c
n 
n
.e inx (Complex Fourier series)
Corollary
bn  i(cn  c n ) for n  1, and let a0  2c 0
Let an  cn  c n ,
Then,


n 
n 1
ess.

 c n .e inx  c 0   c n .e inx   c  n .e i (  n )x
f (x ) 
n 1


 c0   cn (cos(nx)  i sin(nx))   c n (cos(nx)  i sin(nx))
n1
n1


 c 0   (c n  c  n ) cos(nx )   i (c n  c  n ) sin(nx )
a0 2
n 1
n 1
an
ess.
f (x ) 
bn

a0 
  an cos(nx )   b n sin(nx )
2 n 1
n 1
So, we get the real Fourier series.
Note that:
Up to this point, we have investigated functions with period 2 . What if the function that we
are interested in is T-periodic? Can we find the Fourier series for it?
If f ( x) is 2 -periodic for some T  0 , that is, f ( x  T )  f ( x) for all x 
consider:
g (x )  f (
, then we can
T
x ) which satisfies:
2
T

T

T

g (x  2 )  f 
(x  2 )   f 
x T   f 
x   g (x )
 2

 2

 2 
ess.
Now, g (x ) 

c
n 
n
.e inx
Then,
2
inx
 2  ess. 
f (x )  g 
x    c ne T
 T  n 
Here, we used another function to get a 2 periodicity again which we can deal with.
3
Lemma (Parseval)
This is lemma is just an application on the Fourier series. It gives us the value to which and
infinite series converges.
In the notation above, we have:


n 
cn 
2
1
2
2
 f (x ) dx
2
0
Proof
1
2



2
1
0 [f (x )] dx  2
2
1
2
1
2
2
 f (x )f (x )dx
0
2

 
ikx 
 ilx
c
e
k
  c l e
0  k

 l 


 c c
k  l 

 ck ck 
k 
k
l
2

0


k 
ck
1

 dx  2




 ck cl
k  l 
2
e
i ( k l ) x
dx
0
if k  l 

if k  l 
2
Example
Let f ( x)  x for x [0, 2  and continued 2 -periodically to
.
First, we calculate:
1
c0 
2
2
 f (x )e
0
2
 i 0x
1 x 2 
1  4 2 
dx 

 .
2  2  0
2  2 
f (0 )  2 , f (0 )  0 .
Now, we calculate the general term for the coefficients of the complex Fourier series for
m  0.
cm 
1
2
2
 imx
 f (x )e dx 
0
1
2
2
 xe
 imx
dx
0
Integrating by parts to evaluate this integral:
1
cm 
2
2
2

1  imx 
1 imx
e
  1
e dx
  x

im
0
0
  im
 1 
1
1
 imx 2 

 

e
 
 2
2 
0
2


im
(

im
)



4
2
Note that: the term  e  imx  is equal to ZERO because:
0
2
e imx   e 2 im  e 0  cos(2 m )  i sin(2 m )  1  1  0  1  0.
0
Therefore,
cm 
1 i
 .
im m
5