Prime conditions
for Laplace transforms
No.1
Takao Saito
Thank you for our world
Prime conditions
for Laplace transforms
No.1
Preface
Now, I explain about the prime conditions for the Laplace transforms.
In general, Now Laplace transform (T (0) operation) has been treated as
identities. So, we have been contained the property of prime number.
Textbook
Operator algebras for Laplace transforms (No.1∼4)
Papers
The projection operator for Laplace transforms
Pascal’s triangle matrix for Laplace transforms
The diagonal conditions for Laplace transforms
The ring conditions for Now Laplace transforms
The prime numbers for Laplace transforms
Now, let′s consider with me!
Address
695-52 Chibadera-cho
Chuo-ku Chiba-shi
Postcode 260-0844 Japan
URL: http://opab.web.fc2.com/index.html
(Mon) 5.Feb.2018
Takao Saito
Contents
Preface
§ Chapter 1
D(a) and Y (a) operation
◦ The relation of a0 and eas for D(a) and Y (a) operations · · · 7
F (a) and T (a) operation
◦ The relation of a0 and eas for F (a) and T (a) operations · · · 11
◦ Some results · · · · · · · · · 16
§ Chapter 2
D(a) and Y (a) operation
◦ The relation of unitary and e0s for D(a) and Y (a) operations · · ·17
F (a) and T (a) operation
◦ The relation of unitary and e0s for F (a) and T (a) operations · · ·23
◦ Some results · · · · · · · · · 29
§ Chapter 3
D(a) and Y (a) operation
◦ The relation of D(a) and D(0) operations · · ·30
F (a) and T (a) operation
◦ The relation of T (a) and T (0) operations · · ·33
◦ Some results · · · · · · · · · 37
◦ Conclusion · · · · · · · · · · · · 39
◦ References · · · · · · · · · · · · 41
§ Chapter 1
In this chapter, I explain about the relation of eas and a0 for Now
Laplace transform. In general, we have been obtained e0s = 00. So the
projective conditions of eas and a0 have this condition.
D(a) and Y (a) operations
Now, D(a) operations have defined as following forms.
D(a) = easD(a) = eas
(
a0
a0
)
, O(a) = easO(a) = eas
(
00 − a0
00 − a0
)
and
H(a) = D(a) + O(a) = eas
(
00
00
)
Especially, D(0) operation has the following.
D(0) =
(
00
00
)
On the other hand, Y (a) operations have defined as following forms.
Y (a)f(t) = easY(a)f(t) =
∫ ∞
a
f(t)easdt , N(a)f(t) = easN (a)f(t) =
∫ a
0
f(t)easdt
and
W (a)f(t) = Y (a)f(t) + N(a)f(t) =
∫ ∞
0
f(t)easdt
Especially, Y (0) operation has the following.
Y (0)f(t) =
∫ ∞
0
f(t)e0sdt
© The relation of a0 and eas of D(a) and Y (a) operations
7
Now, a0 has been existed as the element of the matrix of D(a) op-
eration. On the other hand, e0s has been existed as the kernel of Y(a)
operation. In this time, we have the following relation.
a0 = 00 = e0s
So, I am able to represent as follows.
D(a) iso←→ Y(a)
So I also have
D(a) = easD(a) iso←→ easY(a) = Y (a).
In a word,
D(a)
iso←→ Y (a).
Especially, if a = 0 then we have
D(0)
iso←→ Y (0).
In a word, this relation of
00 = e0s
has been lived in D(0) and Y (0) operations. And, we also have the
following by based the relation.
D(a)
iso←→ Y (a)
In this time, this relation is able to extend to as follows.
a0
iso←→ eas
And, if we have eas = e0s then it also has
a0 = eas.
This relation has been applied to explain the jump up structures for
D(a) operations. In a word, a0 has been treated as D(0) operation. On
8
the contrary, eas has been treated as D(a) operation. Now, since a0 = eas
then we are able to have the following relation.
D(0) = D(a)
As a whole, we will have the following relations.
D(a)
P ↓ =
D(0) = Y (0)
= ↑ P
Y (a)
N.B. a
P−→ 0.
P is projection operator.
In a word, D(0) operation is able to represent as Y (0) operation.
By this condition, we are able to also have
D(0) = 00 = e0s = Y (0).
Therefore, D(0) operation has been obtained as no-definition. So,
D(0) operation is able to fix to D(a) operation. In this time, we are able
to also have inclusion monomorphism in this structure. In a word, we
also have the following relation.
D(0)
homo−→ D(a)
and it’s possible.
Aa a whole, we are able to have the following relations.
D(0a)
iso←→ D(a)
So we are able to also have the following.
D(0) = D(a)
9
In a word, the poperty of D(0) operation has been preserved to the
property of D(a) operation. Especially, if the norm of D(0) operation is
one then we have the unital conditions for D(a) operations. In general,
if we have the following condition
‖D(0)‖ = ‖D(a)‖
then D(0) operation does not necessarily to have unital condition because
it has no-defined. However, this D(0) operation has 1 and 0, a.e. So, we
have to obtain as follows.
D(0) = 1, 0.
On the contrary, we are able to also have
D(0) = eas = D(a).
In a word, we do not have always the necessity as D(0) = 1, 0. But if
it’s able to represent as 1 and 0 for all numbers then we are able to treat
only 1 and 0 in D(0) operation.
Similarly, we also have the following relation by taking the inclusion
monomorphism.
Y (0) = Y (a)
And the poperty of Y (0) operation has been preserved to the property
of Y (a) operation. Especially, if the norm of Y (0) operation is one then
we have the unital conditions for Y (a) operations. In general, if we have
the following condition
‖Y (0)‖ = ‖Y (a)‖
then Y (0) operation does not necessarily to have unital condition because
it also has no-definition by e0s. However, this Y (0) operation has 1 and
0, a.e. So, we have to obtain as follows.
Y (0) = 1, 0.
On the contrary, we are able to also have
Y (0) = eas = Y (a).
In a word, we do not have always the necessity as Y (0) = 1, 0. But if
it’s able to represent as 1 and 0 for all numbers then we are able to treat
only 1 and 0 in Y (0) operation.
10
F (a) and T (a) operations
Now, F (a) operations have defined as following forms.
F (a) = eas
(
a0
a a0
)
, G(a) = eas
(
00 − a0
−a 00 − a0
)
and
H(a) = F (a) + G(a) = eas
(
00
00
)
On the other hand, T (a) operations have defined as the following
forms.
T (a)f(t) =
∫ ∞
a
f(t)e−(t−a)sdt , S(a)f(t) =
∫ a
0
f(t)e−(t−a)sdt
and
R(a)f(t) = T (a)f(t) + S(a)f(t) =
∫ ∞
0
f(t)e−(t−a)sdt
© The relation of a0 and eas of F (a) and T (a) operations
Now, a0 has been existed as the element of the matrix of F(a) op-
eration. On the other hand, e0s has been existed as the kernel of T (a)
operation. In this time, we have the following relation.
a0 = 00 = e0s
So, I am able to represent as follows.
F(a) iso←→ T (a)
So I also have
F (a) = easF(a) iso←→ easT (a) = T (a).
In a word,
F (a)
iso←→ T (a).
11
Especially, if a = 0 then we have
F (0)
iso←→ T (0).
In a word, this relation of
00 = e0s
has been lived in F (0) and T (0) operations. And, we also have the
following by based the relation.
F (a)
iso←→ T (a)
In this time, this relation is able to extend to as follows.
a0
iso←→ eas
And, if we have eas = e0s then it also has
a0 = eas.
This relation has been applied to explain the jump up structures for
the Laplace transforms. In a word, a0 has been treated as the kernel of
Now laplace transforms (T (0) operation). On the contrary, eas has been
treated as the kernel of Extended Laplace transforms (T (a) operation).
Now, since a0 = eas then we are able to have the following relation.
T (0) = T (a)
As a whole, we will have the following relations.
F(a)
P ↓ =
F (0) = T (0)
= ↑ P
T (a)
N.B. a
P−→ 0.
P is projection operator.
12
In a word, T (0) operation (Now Laplace transform) is able to rep-
resent as F (0) operation. So we have
T (0) = 00 = e0s.
Therefore, T (0) operation (Now Laplace transform) has been obtained
as no-definition. So, T (0) operation (Now Laplace transform) is able to
fix to T (a) operation (Extended Laplace transform). In this time, we are
able to also have inclusion monomorphism in this structure. In a word,
we also have the following relation.
T (0)
homo−→ T (a)
and it’s possible.
Aa a whole, we are able to have the following relations.
T (0a)
iso←→ T (a
