The derivative of
φt
φ
t
(if it exists!) also satisfies a dilation equation.
To see this, take the derivative of both sides of the dilation
equation.
ddtφt=ddt2∑nhnφ2t−n
t
φ
t
t
2
n
h
n
φ
2
t
n
or
φt=2∑nhndφ2t−ndt
φ
t
2
n
h
n
t
φ
2
t
n
As
ddtφ2t−n=2dφ2t−nd
t
φ
2
t
n
2
φ
2
t
n
we have

φ′t=2∑n2hndφ2t−nd
φ
t
2
n
2
h
n
φ
2
t
n

(1)
Therefore,

φ′t
φ
t
satisfies a dilation equation with coefficients

2hn
2
h
n
. We can find values of

φ′t
φ
t
on the integers

t=n
t
n
using the procedure described already. We will get
the matrix equation

2Pφ′=φ′
2
P
φ
φ
As a result, we see that

11 must be
an eigenvalue of the matrix

2P
2
P
; otherwise, the derivative of

φt
φ
t
does not exist on the integers, meaning that

φt
φ
t
is not differentiable. Equivalently,

1/2
12
is an eigenvalue of

PP.

It turns out that if
hn
h
n
satisfies the sum rule

∑n−1nnhn=0
n
1
n
n
h
n
0

(2)
in addition to this

sum rule, then

1/2
12 is an eigenvalue of

PP.

We will not show that if
hn
h
n
satisfies the sum rules, this equation from the module on
the dilation wavelet equation and Equation 2, then
1/2
12 is an eigenvalue of PP.
(We already know that 11 is an eigenvalue of
PP.) We will do this by showing
that
1/2
12 is an eigenvalue of
PT
P
.

Define
e
0
≔111⋮1
≔
e
0
1
1
1
⋮
1
e
1
≔012⋮N−1
≔
e
1
0
1
2
⋮
N
1
We already saw that
e
0
P=
e
0
e
0
P
e
0
which proves that 1 is an eigenvalue of
PP. Now let's looks at
e
1
P
e
1
P
.
e
1
P=2∑nnh2n∑nnh2n−1∑nnh2n−2⋮∑nnh2n−(N−1)
e
1
P
2
n
n
h
2
n
n
n
h
2
n
1
n
n
h
2
n
2
⋮
n
n
h
2
n
N
1
We now simplify the right hand side. First notice that if
hn
h
n
satisfies this equation, then
∑n−1nnhn=0⇔∑n2nh2n−∑n(2n−1)h2n−1=0
⇔
n
1
n
n
h
n
0
n
2
n
h
2
n
n
2
n
1
h
2
n
1
0
⇔∑n2nh2n−∑n2nh2n−1+∑nh2n−1=0
⇔
n
2
n
h
2
n
n
2
n
h
2
n
1
n
h
2
n
1
0
⇔∑n2nh2n−∑n2nh2n−1+12=0
⇔
n
2
n
h
2
n
n
2
n
h
2
n
1
1
2
0
⇔∑nnh2n−∑nnh2n−1+12×2=0
⇔
n
n
h
2
n
n
n
h
2
n
1
1
2
2
0
⇔∑nnh2n−1=∑nnh2n+12×2
⇔
n
n
h
2
n
1
n
n
h
2
n
1
2
2
The last line says that the second term of
e
1
P
e
1
P
is equal to the first term plus the constant
12×2
1
2
2
. Let's find an expression for the third term
∑nnh2n−2
n
n
h
2
n
2
.

∑nnh2n−2=∑k(k+1)h2k=∑kkh2k+∑kh2k=∑nnh2n+12=∑nnh2n+22×2
n
n
h
2
n
2
k
k
1
h
2
k
k
k
h
2
k
k
h
2
k
n
n
h
2
n
1
2
n
n
h
2
n
2
2
2

(3)
That means that the third term of

e
1
P
e
1
P
is equal to the first term plus twice the previous
constant. In general, it can be similarly shown that

∑n−1nnhn=0 ⇔∑nnh2n−k=∑nnh2n+k2×2
⇔
n
1
n
n
h
n
0
n
n
h
2
n
k
n
n
h
2
n
k
2
2

(4)
Therefore

e
1
P=C
e
0
+12
e
1
e
1
P
C
e
0
1
2
e
1

(5)
with

C=2∑nnh2n
C
2
n
n
h
2
n
We can show that

1/212 is an eigenvalue of

PP by constructing the vector

e
0
+β
e
1
e
0
β
e
1
. For the right choice for

ββ, this vector turns out to
be an eigenvector with eigenvalue

1/212.

(
e
0
+β
e
1
)P=
e
0
P+β
e
1
P=
e
0
+β(C
e
0
+12
e
1
)=(1+βC)
e
0
+β2
e
1
e
0
β
e
1
P
e
0
P
β
e
1
P
e
0
β
C
e
0
1
2
e
1
1
β
C
e
0
β
2
e
1

(6)
Therefore,

(
e
0
+β
e
1
)P=12(
e
0
+β
e
1
)
e
0
β
e
1
P
1
2
e
0
β
e
1
provided

1+βC=12
1
β
C
1
2
. Solving for

ββ
gives

β=−12C
β
1
2
C
Therefore, we have the following result

(∑nhn=2)∧(∑n−1nhn=0)∧(∑n−1nnhn=0)⇒1 and 1/2 are eigenvalues of P
n
h
n
2
n
1
n
h
n
0
n
1
n
n
h
n
0
1
and
12
are eigenvalues of
P
If
hn
h
n
satisfies these sum rules, then we can use the
procedure described before to calculate
φ′t
φ
t
on the dyadic rationals. If
hn
h
n
does not satisfy these sum rules then
φt
φ
t
is not differentiable.

The sequence
hn
h
n
satisfies the sum rules

∑n−1nhn=0
n
1
n
h
n
0

(7)
∑n−1nnhn=0
n
1
n
n
h
n
0

(8)
if and only if

Hz
H
z
can be written as

Hz=Qzz-1+12
H
z
Q
z
z
1
2
To see this, differentiate

Hz
H
z
,

H′z=Qz(-2z-1)(z-1+1)+dQzdz-1+12
H
z
Q
z
-2
z
z
1
Q
z
z
1
2
and set

z=-1
z
-1
. Then

(H-1=0)∧(H′-1=0)
H
-1
0
H
-1
0
We already saw that

H-1=0
H
-1
0
gives

Equation 7. Writing
out

H′-1
H
-1
we get

Hz=∑nhnz−n
H
z
n
h
n
z
n

(9)
H′z=−∑nhnnz−(n+1)
H
z
n
h
n
n
z
n
1

(10)
H′-1=∑nhnn−1n
H
-1
n
h
n
n
1
n

(11)
This gives sum rule

Equation 8. The
result above can therefore be stated as

(H1=2)∧(H-1=0)∧(H′-1=0)⇒1 and 1/2 are eigenvalues of P
H
1
2
H
-1
0
H
-1
0
1
and
12
are eigenvalues of
P
or as

(H1=2)∧(Hz=Qzz-1+12)⇒1 and 1/2 are eigenvalues of P
H
1
2
H
z
Q
z
z
1
2
1
and
12
are eigenvalues of
P
For the existence of higher derivatives, the matrix
PP must have further eigenvalues
of the for
12k
1
2
k
.

In general,

∀0≤k≤K−1:(∑nhn=2)∧(∑n−1nnkhn=0)⇒11/21/4…12K−1 are eigenvalues of P
0
k
K
1
n
h
n
2
n
1
n
n
k
h
n
0
1
12
14
…
1
2
K
1
are eigenvalues of
P

(12)
Or in terms of

Hz
H
z
:

(H1=2)∧(Hz=Qzz-1+1K)⇒11/21/4…12K−1 are eigenvalues of P
H
1
2
H
z
Q
z
z
1
K
1
12
14
…
1
2
K
1
are eigenvalues of
P

(13)
For signal processing applications, it is desirable that

φt
φ
t
be relatively smooth, and differentiability indicates
the smoothness of the function. However,

φt
φ
t
can not be infinitely differentiable when

hn
h
n
is finite in length. When

hn
h
n
is finite in length, then

φt
φ
t
will be finite in length, and the matrix

PP will have a finite number of
eigenvalues. In particular,

PP
will have a finite number of eigenvalues of the form

12k
1
2
k
. Therefore,

φt
φ
t
can not be infinitely differentiable. The scaling
functions that are the most differentiable are the spline
scaling functions. But they are not orthogonal, as we will see
later.

We can write the equation
2Pφ′=φ
2
P
φ
φ
as
(2P−I)φ′=0
2
P
I
φ
0
When we computed samples of
φt
φ
t
we used the normalization
∑nφn=A
n
φ
n
A
in this equation. What normalization should
we use when we compute samples of
φ′t
φ
t
? We will see that
∑ndφnd
n
φ
n
is not the right quantity to normalize, but that
∑nndφnd
n
n
φ
n
is.

Begin with the dilation equation for the derivative of
φt
φ
t
.
φ′t=2∑k2hkdφ2t−kd
φ
t
2
k
2
h
k
φ
2
t
k
Set
t=n∈Z
t
n
,
φ′n=2∑k2hkdφ2n−kd
φ
n
2
k
2
h
k
φ
2
n
k
Sum over nn,

∑ndφnd=∑n2∑k2hkdφ2n−kd=2∑n∑k2hkdφ2n−2kd+2∑n∑k2h2k−1dφ2n−2k+1d=2∑k2h2k∑ndφ2n−2kd+2∑k2h2k−1∑ndφ2n−2k+1d=2∑k2h2k∑ndφ2nd+2∑k2h2k−1∑ndφ2n−1d=2∑k2h2k∑ndφ2nd+2∑k2h2k−1∑ndφ2n−1d=22∑ndφ2nd+22∑ndφ2n−1d=2∑ndφ2nd+2∑ndφ2n−1d=2∑ndφnd
n
φ
n
n
2
k
2
h
k
φ
2
n
k
2
n
k
2
h
k
φ
2
n
2
k
2
n
k
2
h
2
k
1
φ
2
n
2
k
1
2
k
2
h
2
k
n
φ
2
n
2
k
2
k
2
h
2
k
1
n
φ
2
n
2
k
1
2
k
2
h
2
k
n
φ
2
n
2
k
2
h
2
k
1
n
φ
2
n
1
2
k
2
h
2
k
n
φ
2
n
2
k
2
h
2
k
1
n
φ
2
n
1
2
2
n
φ
2
n
2
2
n
φ
2
n
1
2
n
φ
2
n
2
n
φ
2
n
1
2
n
φ
n

(14)
That is,

∑ndφnd=2∑ndφnd
n
φ
n
2
n
φ
n
or

∑ndφnd=0
n
φ
n
0
This assumes of course that

φ′t
φ
t
is defined on the integers, for which we need the
sum rule

∑n−1nnhn=0
n
1
n
n
h
n
0
which we saw before. Similar to what we saw before,
if we define

S
j
≔12j∑ndφn2jd
≔
S
j
1
2
j
n
φ
n
2
j
and use the same procedure used before, we find that

S
j
=2
S
j
−
1
S
j
2
S
j
−
1
Because

S
0
=∑ndφnd=0
S
0
n
φ
n
0
we find that

S
j
=0
S
j
0
for all

j≥0
j
0
. Taking the limit

j→∞
j
gives

∫dφtddt=limit j→
∞
S
j
=0
t
φ
t
j
S
j
0
or

∫dφtddt=∑ndφnd=0
t
φ
t
n
φ
n
0

(15)
Therefore, we can not use the quantity

∑ndφnd
n
φ
n
to normalize

φ′t
φ
t
.

Let us consider now instead the quantity
T
j
≔12j∑kk2jdφk2jd
≔
T
j
1
2
j
k
k
2
j
φ
k
2
j
Notice that
limit j→
∞
T
j
=∫tdφtddt
j
T
j
t
t
φ
t
Set
t=k2j
t
k
2
j
in the dilation equation,
φ′k2j=2∑n2hndφ2k2j−nd
φ
k
2
j
2
n
2
h
n
φ
2
k
2
j
n
Multiply by
k2j
k
2
j
and sum over kk,

∑kk2jdφk2jd=∑kk2j2∑n2hndφ2k2j−nd=2∑nhn∑kk2j−1dφk2j−1−nd=2∑nhn∑l(l2j−1+n)dφl2j−1d=2∑nhn∑ll2j−1dφl2j−1d+2∑nhn∑lndφl2j−1d=2∑ll2j−1dφl2j−1d∑nhn+2∑nnhn∑ldφl2j−1d=2∑ll2j−1dφl2j−1d2+0=2∑ll2j−1dφl2j−1d
k
k
2
j
φ
k
2
j
k
k
2
j
2
n
2
h
n
φ
2
k
2
j
n
2
n
h
n
k
k
2
j
1
φ
k
2
j
1
n
2
n
h
n
l
l
2
j
1
n
φ
l
2
j
1
2
n
h
n
l
l
2
j
1
φ
l
2
j
1
2
n
h
n
l
n
φ
l
2
j
1
2
l
l
2
j
1
φ
l
2
j
1
n
h
n
2
n
n
h
n
l
φ
l
2
j
1
2
l
l
2
j
1
φ
l
2
j
1
2
0
2
l
l
2
j
1
φ
l
2
j
1

(16)
because

∑ldφl2j−1d=0
l
φ
l
2
j
1
0
. Dividing both sides by

2j
2
j
gives

12j∑kk2jdφk2jd=12j−1∑kk2j−1dφk2j−1d
1
2
j
k
k
2
j
φ
k
2
j
1
2
j
1
k
k
2
j
1
φ
k
2
j
1
or

T
j
=
T
j
−
1
T
j
T
j
−
1
Taking the limit

j→∞
j
gives

∫tdφtddt=∑nndφnd
t
t
φ
t
n
n
φ
n

(17)
This is similar to

this equation.

If we choose to normalize
φt
φ
t
such that
∫φtdt=A
t
φ
t
A
, then what value of
∫tdφtddt
t
t
φ
t
does this imply? By the product rule,
tφt′=φt+tdφtd
t
φ
t
φ
t
t
φ
t
Integrating both sides,
∫−∞∞dtφtddt=∫−∞∞φtdt+∫−∞∞tdφtddt
t
t
φ
t
t
φ
t
t
t
φ
t
(This is just integration by parts.) If
φt
φ
t
is of finite support (or decays sufficiently fast),
then
∫−∞∞dtφtddt=tφt|t=−∞∞=0
t
t
φ
t
t
t
φ
t
0
Therefore we get

∫−∞∞tdφtddt=−∫−∞∞φtdt
t
t
φ
t
t
φ
t

(18)
So if we choose to normalize

φt
φ
t
so that

∫φtdt=A
t
φ
t
A
, then we also have

∫tdφtddt=−A
t
t
φ
t
A
. Combining this with

Equation 17 indicates that we should normalize the
eigenvector when we compute samples of

φ′t
φ
t
by

∑nndφnd=−A
n
n
φ
n
A
Appending this normalization to the equations we already have
for the integer samples of

φ′t
φ
t
gives the following matrix system.

(
2P−I
012…N−1
)φ′=(
0
−A
)
2
P
I
0
1
2
…
N
1
φ
0
A

(19)
We can compute samples of

φ′t
φ
t
on a finer grid by the same refinement method used
for

φt
φ
t
.

Develop the procedure for the second derivative
φt′′
2
φ
t
. Give a condition that ensures that
φt′′
2
φ
t
exists on the integers, and show why the condition
works. Show how to compute
φt′′
2
φ
t
on the integers. Show what normalization on
φt′′
2
φ
t
is induced by the normalization
∫φtdt=A
t
φ
t
A
. Using Matlab, make plots of
φ′t
φ
t
and
φt′′
2
φ
t
for some examples
hn
h
n
that satisfy the appropriate conditions.

Given KK, the shortest
hn
h
n
for which
Hz=Qzz-1+1K
H
z
Q
z
z
1
K
has
Qz
Q
z
equal to a constant chosen so that
H1=2
H
1
2
. The result is
Hz=2z-1+12K
H
z
2
z
1
2
K
It turns out that the scaling function
φt
φ
t
corresponding to this is a square pulse repeatedly
convolved with itself. The result is a spline -
a function that is piecewise polynomial.