Written by MIDN 1/C
Leif Eric Walroth
Understanding Parseval’s Identity
Final Report for Capstone Project
Presented to
Professor Joyner on
26 April 2007
for SM472
The goal of this paper is to state, prove, and
show applications for Parseval’s Identity
PARSEVAL’S IDENTITY
1. INTRODUCTION
I am writing this paper to learn and understand more about Fourier series and
Fourier analysis. My focus will be on Parseval’s Identity.
1.1 THESIS
This paper will introduce, state, and prove Parseval’s Identity and show how it
relates to applications.
1.2. BACKGROUND ON MARC-ANTOINE PARSEVAL
Born into a family of high standing in France, Parseval grew up in a family of
wealthy land-owners. Parseval, who considered himself a squire in his younger
years, lived from 1755-1836. Not much was known about Parseval’s personal
life. It is known that he once fled France from Napoleon when he published
poetry about Napoleon’s regime and Napoleon ordered his arrest. Little is known
about Parseval’s personal life. There is no record of his own family in his older
years or other personal information such as hobbies and interests. Parseval’s
mathematic contributions consisted of only five publications which were
presented to the French Academy of Science. The second of these publications is
where he proved the result known today as Parseval’s Identity or Parseval’s
Theorem.1
Parseval’s theorem occurs in many different contexts. In many problems,
Parseval’s theorem breaks down a given function into linear combinations of
orthogonal eigenfunctions. Parseval’s theorem says that the total energy of the
original function is the sum of the different eigenfunctions’ energies. Although
Parseval’s theorem may be applied to general orthogonal basis of eigenfunctions,
it is mainly discussed with Fourier series and Fourier transform.2
1
1.3 UNDERSTANDING PARSEVAL’S THEOREM
Real and Imaginary Numbers
Let a ∈
on the x-axis and b ∈
on the y-axis.
Let z =a + bi. Let the complex conjugate of z be z.
So z = a − bi
The magnitude of z is equal to the square root of a 2 + b 2
This is written as z = a 2 + b 2 . We have:
z z = (a + bi )(a − bi )
= a 2 + abi − abi + b 2
= a 2 + b2
= z
2
c=z for the graph below.
2
Inner Product on L2 ( I )
Suppose f , g are two square integrable functions on the interval I=[a,b].
b
ie.
∫
2
f ( x) dx exists and is finite. (We say that such a function belongs to L2 ( I ).)
a
Their inner product is defined by
b
( f , g ) = ∫ f ( x) g ( x)dx
a
Properties of the Inner Product
1. (c1 f1 + c2 f 2 , g ) = c1 ( f1 , g ) + c2 ( f 2 , g ) where c1 , c2 ∈
2. ( f , f ) ≥ 0
3. (f , g ) = ( g , f )
Follows that for α ∈
b
(f , α g ) = ∫ f ( x)α g ( x)dx
a
b
(f , α g ) = ∫ f ( x)α g ( x)dx
a
b
(f , α g ) = α ∫ f ( x) g ( x)dx
a
(f , α g ) = α ( f , g )
Using the inner product, we define
f
1/ 2
=( f, f )
to be the "magnitude" or norm of a function.
(This is most commonly denoted by
Then
f −g
f
2
)
is defined to be the distance between f and g .
Orthonormal Family of Functions
Definition: A family of functions {ϕk :k = 0,1, 2...} is orthonormal
if the following properties are satisfied
(1) (ϕk , ϕl ) = 0 if k ≠ l
(2) (ϕk , ϕ k ) = 1 (i.e. ϕk = 1) ∀ k
3
Suppose {ϕ0 , ϕ1,...ϕ n } is an orthonormal set of functions and f is a linear combination of the ϕ m .
Then it is easy to compute the coefficients cm .
n
Suppose f = ∑ ckϕk
k =0
= c0ϕ0 + c1ϕ1 + ... + cnϕn
To find c0 : take the inner product of both sides with ϕ0
( f , ϕ0 ) = (c0ϕ0 + c1ϕ1 + ... + cnϕn , ϕ0 )
= c0 (ϕ0 , ϕ0 ) + c1 (ϕ1 , ϕ0 ) + ... + cn (ϕ n , ϕ0 )
Remember from above that (ϕ k , ϕl ) = 0 if k ≠ l and
(ϕk , ϕk ) = 1 (i.e. ϕk = 1)
So every inner product is 0 except for (ϕ0 , ϕ0 ) which is equal to 1.
( f , ϕ0 ) = (c0 + 0 + ... + 0)
Therefore c0 = ( f , ϕ0 )
In general we get cm = ( f , ϕ m )
Definition: The numbers cm are called the Fourier coefficients of f with respect to the
orthonormal set {ϕ0 , ϕ1 ,..., ϕ n }
Next suppose {ϕ0 , ϕ1 ,..., ϕ n } is an orthonormal set and f is a function in L2 ( I ).
The following theorem is the crucial step in proving Parseval's theorem. It says that
among all linear combinations b0ϕ0 + b1ϕ1 + ... + bnϕn the one that best approximates
f is obtained by letting bi be the Fourier coefficients of f with respect to {ϕ0 , ϕ1 ,..., ϕ n }.
Theorem
Suppose f belongs to L2 ( I ).
i.e. if sn = c0ϕ0 + c1ϕ1 + ... + cnϕn where cm = ( f , ϕm )
and
tn = b0ϕ0 + b1ϕ1 + ... + bnϕn
where bm ∈
then f − sn ≤ f − tn
Note: sn is the closest approximation to f than any other generic or random
linear combination of the ϕn .
Note: tn is a generic or random combination of the ϕn .
4
PROOF
f − tn
2
= ( f − tn , f − t n )
Use the definition of the inner product to solve the above equation.
b
= ∫ ( f ( x) −tn ( x))( f ( x) − tn ( x))dx
a
b
= ∫ ( f ( x)( f ( x) − tn ( x)) − tn ( x)( f ( x) − tn ( x))dx
a
b
b
= ∫ ( f ( x)( f ( x) − tn ( x))dx − ∫ tn ( x)( f ( x) − tn ( x))dx
a
a
= ( f , f − tn ) − (tn , f − tn )
= ( f , f ) − ( f , tn ) − (tn , f ) + (tn , tn )
Each term is now individually expanded.
• (tn , tn ) = (b0ϕ0 + b1ϕ1 + ... + bnϕn , b0ϕ0 + b1ϕ1 + ... + bnϕn )
= (b0ϕ0 , b0ϕ0 ) + (b1ϕ1 , b1ϕ1 ) + ... + (bnϕ n , bnϕn )
There should be n 2 terms, however there are only n terms since
all other terms cancel out where (bnϕn , bmϕm ) and n ≠ m.
So (tn , tn ) = b0 b0 (ϕ0 , ϕ0 ) + b1 b1 (ϕ1 , ϕ1 ) + ... + bn bn (ϕn , ϕn )
= b0 b0 + b1 b1 + ... + bn bn
n
= b0 + b1 + ... + bn = ∑ bk
2
2
2
2
k =0
• ( f , tn ) = ( f , b0ϕ0 + b1ϕ1 + ... + bnϕ n )
= ( f , b0ϕ0 ) + ( f , b1ϕ1 ) + ... + ( f , bnϕn )
= b0 ( f , ϕ0 ) + b1 ( f , ϕ1 ) + ... + bn ( f , ϕ n )
Remember that ( f , ϕm ) = cm .
So ( f , tn ) = b0c0 + b1c1 + ... + bn cn
= ∑ k =0 bk ck
n
5
• (tn , f ) = ( f , tn )
n
= ∑ bk ck
k =0
= ∑ k =0 ck bk
n
2
So f − tn
2
f − tn
2
= f
− ( f , tn ) − (tn , f ) + (tn , tn )
n
− ∑ k =0 bk ck − ∑ k =0 ck bk + ∑ bk
n
2
= f
n
2
k =0
Now we complete the square by adding and subtracting ∑ k =0 ck
n
2
f − tn
n
n
n
(2)
− ∑ ck + ∑ ck − ∑ k =0 bk ck − ∑ k =0 ck bk + ∑ bk
k =0
k =0
k =0
14444444244444443
2
= f
2
2
n
2
n
2
"completing the square"
We next show that
n
n
n
∑ bk − ck = ∑ ck − ∑ k =0 bk ck − ∑ k =0 ck bk + ∑ bk
2
k =0
n
2
n
k =0
n
∑b
k
− ck
2
2
k =0
can be factored as follows
k =0
n
∑b
k
n
− ck = ∑ (bk − ck )(bk − ck )
2
k =0
k =0
n
∑b
k
n
− ck = ∑ (bk bk − ck bk − bk ck +ck ck )
2
k =0
k =0
2
Remember bk = bk bk .
n
∑b
k
n
− ck = ∑ ( bk − ck bk − bk ck + ck )
2
k =0
2
2
k =0
n
n
n
n
n
k =0
k =0
k=0
∑ bk − ck = ∑ bk − ∑ ck bk − ∑ bk ck +∑ ck
2
k =0
2
k =0
2
So we can reduce the equation (2) to
f − tn
2
= f
2
n
n
− ∑ ck + ∑ bk − ck
k =0
2
2
k =0
6
For which choice of bk does this become a minimum?
The minimum is attained where all the numbers bk − ck = 0
i.e. when bk = ck
Theorem: Let {ϕ0 , ϕ1 , ϕ2 ,...} be an orthonormal set in L2 ( I ) where I = [a, b] and
L2 ( I ) = { f : I →
b
∫
such that
2
f ( x) dx < ∞}. Recall the definitions for the inner
a
product and the norm:
1
2
b
( f , g ) = ∫ f ( x) g ( x)dx,
b
f = ( f , f ) = [ ∫ f ( x) dx]
2
1
2
a
a
Suppose f ∈ L2 ( I ) and let ck = ( f , ϕ k )
∞
(1)∑ ck ≤ f
2
2
(Bessel's Inequality)
k =0
→0
(2)Assume that furthermore f − sn
n →∞
n
where sn ( x) = ∑ ckϕ k ( x)
k =0
∞
then ∑ ck = f . (Parseval's Identity)
2
2
k=0
PROOF OF (1):
n
From previous proof, if tn = ∑ bkϕk ( x)
k =0
f − tn = f
2
∞
n
− ∑ ck + ∑ bk − ck
2
k=0
2
k =0
So by replacing tn by sn (which means bk = ck )
we get f − sn
2
= f
2
n
− ∑ ck
2
k =0
which can be rearranged into
n
∑c
k
2
= f
2
− f − sn
2
≤ f
2
k =0
n
∞
Since this is true for ∀n, we get lim∑ ck ≤ f , i.e., ∑ ck ≤ f
n→∞
k =0
2
2
2
2
(Bessel's Inequality)
k =0
7
PROOF OF (2) :
→0.
Now assume that f − sn
Then we get
n →∞
n
∑ ck = f
2
2
n
⇒∑ c
2
− f − sn
2
k
2
= f
(Parseval's Identity)
n →∞ k=0
k=0
We present below an example as well as some special cases of Parseval's Identity.
First we present an application of Parseval's Identity. Consider the set of functions
{ϕ0 , ϕ1 , ϕ2 ,...} in L2 ([0, 2π ]) given by
ϕ 0 ( x) =
1
2π
ϕ 2 n −1 ( x) =
cos(nx)
ϕ2 n ( x) =
π
sin(nx)
π
.
We show below that this set is orthonormal.
∫
2π
∫
2π
0
0
1
dx
2π
2π
x 2π
=∫
=
|0
0
4π 2 2π
2π
0
=
−
=1
2π 2π
ϕ0 ( x)ϕ0 ( x) = ∫
1
2π
1
2π
0
ϕ2 n −1 ( x)ϕ2 n −1 ( x)dx = ∫
2π
cos(nx) cos(nx)
dx
π
π
sin(2nπ ) cos(2nπ ) + 2nπ 2π
|0
=
2nπ
2nπ
= 0+
=1
2nπ
0
Similarly, one shows the following:
∫
2π
∫
2π
∫
2π
0
0
0
ϕ0 ( x)ϕ2 n −1 ( x)dx = ∫
2π
0
ϕ0 ( x)ϕ2 n ( x)dx ∫
1 cos(nx)
dx = 0
π
2π
1 sin(nx)
dx = 0
π
2π
2π
0
ϕ2 n ( x)ϕ 2 n ( x)dx = ∫
2π
0
sin(nx) sin(nx)
π
π
dx = 1
8
∫
2π
∫
2π
∫
2π
0
0
0
ϕ2 n −1 ( x)ϕ2 m −1 ( x)dx = ∫
2π
cos(nx) cos(mx)
π
0
ϕ2 n ( x)ϕ 2 m ( x)dx = ∫
2π
π
sin(nx) sin(mx)
π
0
ϕ2 n ( x)ϕ 2 m −1 ( x)dx = ∫
2π
π
dx = 0 where m ≠ n
sin(nx) cos(mx)
π
0
dx = 0 where m ≠ n
π
dx = 0 where m ≠ n
Theorem: Suppose f ∈ L2 ([0, 2π ])
n
and let sn ( x) = ∑ ck ϕk ( x) where ck = ( f , ϕk ).
k =0
Then f - sn
⇒ 0.
n →∞
The proof is omitted. See W. Rudin, Theorem 8.16.
Let f ( x) = x on [0,2π ]
Compute ck
c0 = ( f , ϕ0 ) = ∫
2π
0
=
2π
f ( x)ϕ0 ( x)dx = ∫ x
0
1
dx
2π
1 x 2 2π
4π 2
2π 2
=
|0 =
2π 2
2 2π
2π
c2 n −1 = ( f , ϕ 2 n−1 ) = ∫
2π
0
=
1
π
cos(nx)
x
π
∫
2π
0
dx
x cos(nx)dx
Now we are going to use integration by parts to compute the above integral.
u=x
du = dx
dv = cos(nx)dx
1
v = sin(nx)
n
c2 n −1 = ( f , ϕ 2 n−1 ) =
=
1
2π 1
1
[ x sin(nx) |02π − ∫
sin(nx)dx]
0 n
π n
1 1
cos(nx) |02π = 0
2
n
π
9
c2 n = ( f , ϕ 2 n ) = ∫
2π
0
=
1
∫
π
2π
x
sin(nx)
π
dx
x sin(nx)dx
0
Once again we use integration by parts to compute the above integral.
dv = sin(nx)dx
1
v = cos(nx)
n
u=x
du = dx
2π 1
1
[− x cos(nx) |02π + ∫
cos(nx)dx]
0 n
π n
1
1
1
2π 1
(− 2π + 2 sin(nx) |02π = −
=
n
π n
π n
c2 n = ( f , ϕ 2 n ) =
1
∞
2π 2 1
2π 1 sin(nx)
−∑
f ( x) = x ~
2π 2π n=1 {
π n
π
{
c0
c2 n
Combining the above two theorems, we obtain Parseval's Identity:
∞
∑
2
2
ck = f . Since
k=0
∞
∑
2
ck =
k =0
f
2
=∫
2π
0
∞
∞
4π 4
1
1
+ 4π ∑ 2 = 2π 3 + 4π ∑ 2 and
2π
n =1 n
n =1 n
2π
f ( x) f ( x)dx = ∫ x 2 dx =
0
x3 2π 8π 3
|0 =
we get
3
3
1 8π 3
=
2
3
n =1 n
∞
2π 3 + 4π ∑
Rearrange with algebra to get
∞
1 2π 3
1 π2
=
⇒
=
∑
2
2
3
6
n =1 n
n =1 n
∞
4π ∑
10
Generally consider the space L2 [−T , T ]
The functions
nπ x
nπ x
1
1
1
ϕ 0 ( x) =
, ϕ2n-1 ( x) =
cos(
), ϕ2n ( x) =
sin(
)
T
T
T
T
2T
form a complete orthonormal set. (Complete means the hypothesis to the
previous theorem is satisfied.)
a0
nπ x
nπ x
) + bn sin(
) be the Fourier series of f
+ ∑ an cos(
T
T
2
1 T
where a 0 = ∫ f ( x)dx
T −T
nπ x
1 T
an = ∫ f ( x) cos(
)dx
T −T
T
nπ x
1 T
bn = ∫ f ( x) sin(
)dx.
T −T
T
Let f ( x) ~
In general, Parseval's Identity says that
∫
T
-T
∞
f(x) dx = ∑ cn
2
2
*
n =0
where cn = ( f , ϕn ). We compute:
c0 = ( f , ϕ0 ) = ∫
T
−T
f ( x)ϕ0 ( x)dx
1 T
T
f ( x)dx =
a0
∫
−T
2T
2T
T 1
1 T T
f ( x)dx =
=
∫
2T T −T
2T T
=
∫
T
−T
f ( x)dx =
T
a0
2T
c2 n −1 = ( f , ϕ2 n −1 )
=∫
T
−T
=
1
T
nπ x
1
cos(
)dx
T
T
nπ x
f ( x) cos(
)dx = T an
T
f ( x)
∫
T
−T
Similarly
c2 n = ( f , ϕ 2 n ) = T bn
11
∞
So
∞
T
a0 ) 2 + ∑ ( T an ) 2 + ( T bn )2
2T
n =1
∑ cn = (
2
n=0
=
∞
T
2
2
2
a0 + ∑ T ( an + bn )
2
n =1
1 2 ∞
2
2
= T ( a0 + ∑ ( an + bn )
2
n =1
So * gives
1 T
1 2 ∞
2
2
2
f
x
dx
a0 + ∑ an + bn
=
(
)
∫
T
−
T
2
n =1
This is Parseval's Identity for real Fourier series.
Parseval's Identity for complex fourier series
Consider the set of functions
1 inTπ x
e
n∈ .
2T
This is a complete orthonormal set in L2 ([−T , T ]).
ϕ n ( x) =
∞
∑ c ϕ ( x) is the Fourier series expansion of f ( x) with respect to
If f ( x) =
n
n
n =−∞
this set, Parseval's Identity gives that
∞
f ( x) =
∑
cnϕn ( x),
n =−∞
∫
T
-T
∞
∑
2
f ( x) dx =
cn
2
*
n =−∞
Recall that the complex Fourier series of f ( x) over [−T , T ] is given by
∞
f ( x)~ ∑ d n e
inπ x
T
n =−∞
− inπ x
1 T
T
(
)
f
x
e
dx
2T ∫−T
T
1
1 − inTπ x
2T ∫ f ( x)
e
dx
=
−T
2T
2T243
14
where d n =
ϕn ( x )
=
1
1
( f ,ϕn ) =
cn
2T
2T
12
∫
So * gives
T
-T
∞
∑
2
f ( x) dx =
2
cn
n =−∞
∞
= 2T
∑
dn
2
n =−∞
∞
1 T
2
f ( x) dx
∫
2T -T
n =−∞
This is Parseval's Identity for complex Fourier Series.
⇒
∑
2
dn =
Finally we present Parseval's Identity for the discrete Fourier Transfrom.
Discrete Fourier Transform
h = (h0 , h1 ,..., hn −1 ) n-tuple of complex numbers
Then their Discrete Fourier Transform is the n-tuple
N −1
( H 0 , H1 ,..., H n-1 ) where H k = ∑ h j e
− i 2π j
N
for k = 0,1,...N -1.
j =0
Parseval's Identity for DFT:
N-1
2
1
hj =
∑
N
j=0
123
h
N −1
∑H
2
k
k =0
2
Proof
N −1
∑
Hk = ∑ Hk Hk
k =0
k =0
2
N −1
N −1 N −1 N −1
= ∑ (∑∑ h j hl e
k =0
− i 2π k
N
Note(2)
H k = h0 + h1e
N −1
N −1 N −1 N −1
∑
k =0
e
i 2π lk
N
)
j =0 l =0
Note(1) H k = h0 + h1e
i 2π k
N
H k = ∑∑∑ h j hl e
2
− i 2π jk
N
+ h2e
+ h2 e
− i 2π jk
N
e
− i 4π k
N
i 4π k
N
+ ... + hN −1e
+ ... + hN −1e
− i 2π ( N −1) k
N
i 2π ( N −1) k
N
i 2π lk
N
j = 0 l =0 k =0
N −1 N −1
N −1
= ∑∑ h j hl ∑ e
j =0 l = 0
k =0
N −1 N −1
N −1
j =0 l = 0
k =0
= ∑∑ h j hl ∑ e
− i 2π jk
N
e
i 2π lk
N
i 2π ( l − j ) k
N
(changing the order of summation)
∗
13
Now consider the sum
N-1
∑e
i 2π ( l − j ) k
N
.
k=0
If l = j , the sum is equal to N .
If l ≠ j , let w = e
i 2π ( l − j )
N
.
Note that w ≠ 1 is an N th root of 1 (w N = 1).
The sum is equal to 1 + w + w2 + ... + w N −1.
(1 − w)(1 + w + w2 + ... + w N −1 ) =
1 + w + w2 + ... + w N −1 -(w + w2 + ... + w N −1 − w N ) = 1 − w N .
Therefore 1 + w + w2 + ... + w N −1 =
N -1
N −1
∑h h N = N∑ h
j
j
j =0
1 − wN
= 0, since w N = 1.
1− w
2
j
j =0
It follows that
N −1 N −1
N −1
j = 0 l =0
k =0
∑∑ h j hl ∑ e
i 2π ( l − j ) k
N
N −1
N −1
j =0
j =0
= ∑ h j hl N = N ∑ h j
2
and equation * implies that
N -1
∑H
k =0
2
k
N −1
= N ∑ hj
2
q.e.d.
j =0
14
References
Walker, James S. Fast Fourier Transforms: Second Edition. CRC Press.
1996. New York, Boca Raton, Washington, D.C., London.
Rudin, Walter. Principles of Mathematical Analysis, 3rd Edition, 1984
McGraw-Hill.
Websites
1. http://www-history.mcs.st-andrews.ac.uk/Biographies/Parseval.html
2. http://www.everything2.com/index.pl?node_id=991847
3. http://www.iun.edu/~mathiho/genealogy/IztokHozoGenealogy.htm
Professor Guidance
Professor Alevras
15