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TRANSCRIPT
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) =∫ ∞
af(t)easdt , N(a)f(t) = easN (a)f(t) =
∫ a
0f(t)easdt
and
W (a)f(t) = Y (a)f(t) + N(a)f(t) =∫ ∞
0f(t)easdt
Especially, Y (0) operation has the following.
Y (0)f(t) =∫ ∞
0f(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. aP−→ 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) =∫ ∞
af(t)e−(t−a)sdt , S(a)f(t) =
∫ a
0f(t)e−(t−a)sdt
and
R(a)f(t) = T (a)f(t) + S(a)f(t) =∫ ∞
0f(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. aP−→ 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)
So we are able to also have the following.
T (0) = T (a)
In general, since T (a)iso←→ F(a)
iso←→ F (a) then T (a) operation is able
to replace to as follows.
T (a) →(
1
a 1
)← F(a)
iso←→ F (a)
In a word, the poperty of F (0) operation has been preserved to the
property of F (a) operation. Especially, if the norm of F (0) operation is
one then we have the unital conditions for F (a) operations. In general,
if we have the following condition
‖F (0)‖ = ‖F (a)‖
then F (0) operation does not necessarily to have unital condition because
it has no-defined. However, this F (0) operation has 1 and 0, a.e. So, we
have to obtain as follows.
F (0) = 1, 0.
13
On the contrary, we are able to also have
F (0) = eas = F (a).
In a word, we do not have always the necessity as F (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 F (0) operation.
Similarly, we also have the following relation by taking the inclusion
monomorphism.
T (0) = T (a)
And the poperty of T (0) operation has been preserved to the property
of T (a) operation. Especially, if the norm of T (0) operation is one then
we have the unital conditions for Y (a) operations. In general, if we have
the following condition
‖T (0)‖ = ‖T (a)‖then T (0) operation does not necessarily to have unital condition because
it also has no-definition by e0s. However, this T (0) operation has 1 and
0, a.e. So, we have to obtain as follows.
T (0) = 1, 0.
On the contrary, we are able to also have
T (0) = eas = T (a).
For example, we are able to obtain the following relation.
T (a)T (b) = easebs = e(a+b)s = T (a + b) (Semi− group)
In a word, we do not have always the necessity as T (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 T (0) operation.
Since this flow of D(a) to T (a) operations then I have the following
relation,
T (a)homo−→ D(a).
14
In a word, the Extended Laplace transforms (T (a) operation) is able
to represent by D(a) operation. Therefore we are able to have the re-
lation of a0 and eas. Moreover, this D(a) operation is able to also be
D(0) operation. So we have the conception of T (a) operation has been
obtained the property of 00 = e0s. In other words, if we have T (a) opera-
tion (Extended Laplace transform) then we should have the property as
follows.
a0 = 00 = e0s = eas
In a word, these operators has been treated as D(a), D(0), T (0) and
T (a) operations, resp.
a0 ←→ D(a) , 00 ←→ D(0)
e0s ←→ T (0) , eas ←→ T (a)
In general, a0 has been obtained the isomorphic conditions with eas.
However, the norms of a0 and eas are not same. In a word, we should
have the following relations.
a0 6= 00a = e0as = eas as a 6= 0
N.B. ‖00a‖ = ‖e0as‖ = ‖eas‖
In this time, since a0 is able to be 1 then we are able to use to unitary
condition. And a0 and eas are able to treat as Now Laplace transform
(T (0) operation) and Extended Laplace transform (T (a) operation), resp.
On the contrary, if a = 0 of a0 then we should have the following relation.
T (0) = T (a)
15
Some results
◦ We have the following relation.
00 = e0s
◦ The 00 has been obtained as follows.
a0 = 00
◦ Since 00 has no-definition then we also have no-definition of e0s.
◦ So, we are able to be
eas = e0s = 00 = a0.
◦ In this time, we have the following condition.
T (a)iso←→ D(0)
◦ T (a) operation (Extended Laplace transform) is able to represent as
D(0) operation. In a word,
T (a)iso←→
(00
00
).
16
§ Chapter 2
In this chapter, I explain about the prime conditions for Now Laplace
transfors. The Now Laplace transforms are able to treat as unitary con-
ditions. So, we have one element. On the contrary, the one element does
not include in prime numbers.
D(a) and Y (a) operations
Now, D(a) operations have defined as following forms.
D(a) = eas
(a0
a0
), O(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) =∫ ∞
af(t)easdt , N(a)f(t) =
∫ a
0f(t)easdt
and
W (a)f(t) = Y (a)f(t) + N(a)f(t) =∫ ∞
0f(t)easdt
© The relation of unitary and e0s for D(a) and Y (a) operations
The norm of umitary has 1. On the contrary, e0s is able to have any
numbers as e0s = 00. So, e0s also have 1. In a word, e0s also has unitary.
17
In other words, e0s has been extended from unitary. In general, we have
the following relation.
eas = e0s
In a word, D(0) operation is able to be to D(a) operation. In a word,
D(a) = D(0)
as a0 = 00.
And it’s possible.
So, in general, we don’t need to define as e0s = 1. In this time, eas
is able to take the prime number. So e0s dose not have prime number as
1. (N.B. Peas = e0s.) These things will be able to combine each other.
e0s also has prime number because of eas = e0s. However, if e0s = 1 then
it’s not prime number. On the contrary speaking, e0s has any numbers
without 1. In a word, prime numbers are not able to be unitary if it’s
now conditions. In this time, I recommend to decompose 1 as 1a. N.B.
eas is prime number. So, 0a conditions has been also existed each other.
In a word, we have the following relation.
1aiso←→ 0a
iso←→ a
Now, in general, e0s has been obtained as any numbers. So, e0s does
not need only one and zero. So e0s has been extended from unitary
conditions. If D(0) operation has defined on unitary conditions then
I do not need the relation of prime number. However, if it has been
defined on e0s then we should have the conception of prime numbers.
So, e0s also has “1” element. The D(0) operation of e0s has been also
called as “Now Laplace transform”. Similarly, e0s is able to also have
“0” element. And it’s possible. As a whole, e0s has been generated from
1 and 0 elements. In a word, these are identities and characteristics
condition. In this time, e0s has proper. Of course, e0s is able to also
have other numbers. In a word, e0s has been obtained representative
elements as 1 and 0. So, we should consider the relation of 1 of prime
number if 1 is able to be prime number and 1 of representative element
as multiplicative identity. Similarly, additive condition also has same. In
general, this 1 has a property of kernel. In a word, the prime number
18
has been generated itself without kernel. Similarly, if this number has 0
then we have annihilation. So, the prime number has been also generated
itself without this annihilation.
In this time, we have the following relation.
not prime number ≡ prime number
So, since 1 and 0 elements do not have prime number then 1 and 0 ele-
ments are able to also include in prime number. So, 1 and 0 elements are
able to be representative elements of prime numbers. In other words, the
all prime numbers are able to decompose at the representative elements.
In a word,
1aiso←→ 0a
iso←→ eas (A)
eas is prime number.
This relation are able to also explain with the following relation.
KerP = Im⊥P
Precisely speaking,
KerP = Im⊥P ∗
in complex spaces.
In this case, since P has prime numbers then we have P ∗ = P in
complex spaces. In a word, upper relation (A) has been treated in Banach
spaces.
Now D(a) operations have the following.
D(a) = easD(a) = eas
(a0
a0
)
So we have
D(a) =
(a0
a0
)
In this time, the D(a) operation also has
D(a) =
(1
1
)and
(0
0
)
19
as a0 = 1, 0.
In this time, it’s identities of D(0) operation.
In a word, the diagonal of D(a) operation has been extended from 1
with zero. In a word,
a0 = 1, 0 for all a.
In general, since 00 has been no defined then we are able to have all
numbers as a = 0,∞. In fact, we have 00 = ∞0 because it has the
following.
00 = 0−0 = {00}−1 =1
00=
10
00= {1
0}0 = ∞0
In a word, the previous form has special case. In general, we should
have as following relation.
a0 = 00
and it’s possible.
So I am able to represent as follows.
D(a) =
(00
00
)
In this time, the 00 are able to have independently. And this repre-
sentation has been extended from original form of D(a) operation. The
extension of this theory is the following.
a0 → 00
on D(a) operation.
Especially, if a = 0 then we should have the following form.
D(0) =
(00
00
)= F (0)
iso←→ T (0)
In a word, T (0) operation (Now Laplace teransform) is able to also
represent as D(0) = F (0) operation.
20
In this time, I also have the following forms.
(1
1
),
(1
0
),
(0
1
),
(0
0
)
These forms have been obtained the projective conditions. In a word,
D(0) operations are able to extend to the projective conditions.
N.B. In general, I’m able to have all numbers for 00.
On the other hand, Y(a) and Y(0) = Y (0) operations are able to
have the following relation.
Y(a) ≈ Y(0)
Because of
Y(a) =∫ ∞
ae0s1dt ≈
∫ ∞
0e0s2dt = Y(0).
Precisely speaking, it’s able to be
Y(a) = Y(0).
In a word, we are able to have the following.
I(a) = Y(a) =∫ ∞
ae0s1dt =
∫ ∞
0e0s2dt = Y(0) = I(0)
N.B. I(a), I(b) are identities.
And it’s possible.
If we consider on Y(a) operations then we should have the following
relation, at least.
Y(a)iso←→ Y(0a)
In a word, a element has been existed in zero. So we are able to also
have
Y(a) = Y(0a)
and it’s possible.
21
In this time, if we consider the D(a) operations then we have
D(a) = D(0)
The background of this relation has the following form.
Y(a)iso←→ D(a)
Moreover, we have been obtained
Y (a)iso←→ D(a).
22
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) =∫ ∞
af(t)e−(t−a)sdt , S(a)f(t) =
∫ a
0f(t)e−(t−a)sdt
and
R(a)f(t) = T (a)f(t) + S(a)f(t) =∫ ∞
0f(t)e−(t−a)sdt
© The relation of unitary and e0s for F (a) and T (a) operations
The norm of umitary has 1. On the contrary, e0s is able to have any
numbers as e0s = 00. So, e0s also have 1. In a word, e0s also has unitary.
In other words, e0s has been extended from unitary. In general, we have
the following relation.
eas = e0s
In a word, T (0) operation (Now Laplace transforms) is able to be to
T (a) operation (Extended Laplace transforms). In a word,
T (a) = T (0)
and it’s possible.
So, in general, we don’t need to define as e0s = 1. In this time, eas
is able to take the prime number. So e0s dose not have prime number as
23
1. (N.B. Peas = e0s.) These things will be able to combine each other.
e0s also has prime number because of eas = e0s. However, if e0s = 1 then
it’s not prime number. On the contrary speaking, e0s has any numbers
without 1. In a word, prime numbers are not able to be unitary if it’s
now conditions. In this time, I recommend to decompose 1 as 1a. N.B.
eas is prime number. So, 0a conditions has been also existed each other.
In a word, we have the following relation.
1aiso←→ 0a
iso←→ a
Now, in general, e0s has been obtained as any numbers. So, e0s does
not need only one and zero. So e0s has been extended from unitary
conditions. If the Lapalce transform has defined on unitary conditions
then I do not need the relation of prime number. However, if it has been
defined on e0s then we should have the conception of prime numbers.
So, e0s also has “1” element. The Laplace transform of e0s has been
also called as “Now Laplace transform”. Similarly, e0s is able to also
have “0” element. And it’s possible. As a whole, e0s has been generated
from 1 and 0 elements. In a word, these are identities and characteristics
condition. In this time, e0s has proper. Of course, e0s is able to also
have other numbers. In a word, e0s has been obtained representative
elements as 1 and 0. So, we should consider the relation of 1 of prime
number if 1 is able to be prime number and 1 of representative element
as multiplicative identity. Similarly, additive condition also has same. In
general, this 1 has a property of kernel. In a word, the prime number
has been generated itself without kernel. Similarly, if this number has 0
then we have annihilation. So, the prime number has been also generated
itself without this annihilation.
In this time, we have the following relation.
not prime number ≡ prime number
So, since 1 and 0 elements do not have prime number then 1 and 0 ele-
ments are able to also include in prime number. So, 1 and 0 elements are
able to be representative elements of prime numbers. In other words, the
all prime numbers are able to decompose at the representative elements.
24
In a word,
1aiso←→ 0a
iso←→ eas (A)
eas is prime number.
This relation are able to also explain with the following relation.
KerP = Im⊥P
Precisely speaking,
KerP = Im⊥P ∗
in complex spaces.
In this case, since P has prime numbers then we have P ∗ = P in
complex spaces. In a word, upper relation (A) has been treated in Banach
spaces.
Now F (a) operations have the following.
F (a) = easF(a) = eas
(a0
a a0
)
So we have
F(a) =
(a0
a a0
)
In this time, the original form of F(a) operation has
F(a) =
(1
a 1
)
In a word, the diagonal of F(a) operation has been extended from 1
to a0 with zero. In a word,
a0 = 1, 0 for all a.
In general, since 00 has been no defined then we are able to have all
numbers as a = 0,∞. In fact, we have 00 = ∞0 because it has the
following.
00 = 0−0 = {00}−1 =1
00=
10
00= {1
0}0 = ∞0
25
In a word, the previous form has special case. In general, we should
have as following relation.
a0 = 00
and it’s possible.
So I am able to represent as follows.
F(a) =
(00
a 00
)
In this time, the 00 are able to have independently. And this repre-
sentation has been extended from original form of F(a) operation. The
extension of this theory is the following.
1 → a0 → 00
on the diagonal of F(a) operation.
Especially, if a = 0 then we should have the following form.
F (0) =
(00
00
)= D(0)
iso←→ T (0)
In a word, T (0) operation (Now Laplace teransform) is able to also
represent as D(0) = F (0) operation.
In this time, I also have the following forms.
(1
1
),
(1
0
),
(0
1
),
(0
0
)
These forms have been obtained the projective conditions. In a word,
Now Laplace transforms (T (0) operations) are able to extend to the pro-
jective conditions.
If 0 → a then I present as follows.
(1
a 1
),
(1
a 0
),
(0
−a 1
),
(0
−a 0
)
26
These forms have the projective conditions for T (a) operations (Ex-
tended Laplace transforms). And most left form has been obtained Pas-
cal’s triangle matrix. For example, if it’s 3 orderd matrix condition then
we have
F(a) =
1
a 1
a2 2a 1
.
In general, if I cosider the n-th orderd Pascal’s matrix conditions then
we have been obtained as follows.
F(a)
F(a) =
1
a 1
a2 2a 1
a3 3a2 3a 1...
......
.... . .
an−1n−1C1a
n−2n−1C3a
n−3n−1Cka
n−1−k 1
In general, T (a) and T (0) = T (0) operations are able to have the
following relation.
T (a) ≈ T (0)
Because of
T (a) =∫ ∞
ae−(t−0)sdt ≈
∫ ∞
0e−(t−0)sdt = T (0).
Precisely speaking, it’s able to be
T (a) = T (0).
In a word, we are able to have the following.
I(a) = T (a) =∫ ∞
ae−(t−0)sdt =
∫ ∞
0e−(t−0)sdt = T (0) = I(0)
N.B. I(a), I(b) are identities.
27
If we consider on F(a) operations then we should have the following
relation, at least.
F(a)iso←→ F(0a)
In a word,
F(a) =
(1
a 1
)iso←→
(1
0a 1
)= F(0a)
In a word, a element has been existed in zero. So we are able to also
have
F(a) = F(0a)
and it’s possible.
Ref. Since we have this relation then we are able to operate the rank
form. In a word, this content does not change in spite of the form has
been changed.
Especially, if this has been been treated on eigenspaces then we are
able to have equivalence. In this time, this relation has been represented
as follows.
F(a) = F(0)
The background of this relation has the following form.
T (a)iso←→ F(a)
Moreover, we have been obtained
T (a)iso←→ F (a).
28
Some results
◦ Now we have
eas = e0s.
◦ T (a) operation has the following relation.
T (a) = T (0)
◦ Not prime number ≡ Prime number
◦ In general,
aiso←→ 0a
iso←→ 1a
◦ Especially, if eas is prime number then we have
eas iso←→ 0aiso←→ 1a
◦ T (a) operation also has the following relation.
T (a) = T (0)
29
§ Chapter 3
In this chapter, I explain about the relation of T (a) and T (0) opera-
tions. In this time, we have been obtained as D(a) = D(0). So we will
have T (a) = T (0) and it’s possible.
D(a) and Y (a) operations
Now, D(a) operations have defined as following forms.
D(a) = eas
(a0
a0
), O(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) =∫ ∞
af(t)easdt , N(a)f(t) =
∫ a
0f(t)easdt
and
W (a)f(t) = Y (a)f(t) + N(a)f(t) =∫ ∞
0f(t)easdt
© The relation of D(a) and D(0) operations
Now, D(a) operation has been obtained as follows.
D(a) =
(a0
a0
)
30
In this time, if a < 0 and a > 0 in finite numbers then we have as
follows.
a0 = 1.
So we have
a0 = 1.
On the contrary, if a = 0 and a →∞ then we have
a0 = 00 = ∞0
and it’s no-defined. This form has been obtained as follows.
D(0) =
(00
00
)
In a word, D(0) operation has a part of D(a) operation. So we have
the following relation.
D(0) ⊆ D(a)
On the other hand, since 00 has no-definition then 00 is able to adopt
all numbers. So we also have
D(0) ⊇ D(a).
In a word, we have the following relation.
D(0) = D(a)
In a word,
D(a) = D(0).
In this time, the norm of D(a) operations are able to have one. In a
word,
‖D(a)‖ = ‖D(0)‖ = 1.
Moreover, we also have
a0 = 00 (a 6= 0).
And it has the following relation.
D(a) = easD(a) = easD(0) = H(a)
So we have
D(a) = H(a)
31
as a0 = 00.
Moreover, if we have eas = e0s then we also have
D(a) = D(0).
So, the core form for D(a) operations has the following conditions.
D(a) = D(0)
Similarly, since Y(a)iso←→ D(a) then we should have the following
relation.
Y(a) = Y(0)
So we have
Y(a) =∫ ∞
ae0sdt =
∫ ∞
0e0sdt = Y(0).
In a word,
I(a) =∫ ∞
adt =
∫ ∞
0dt = I(0)
Therefore,
I(a) = I(0).
N.B. ‖I(a)‖ = ‖I(0)‖ = 1.
I(a) and I(0) are identities.
In a word, if we consider Y(a) operation then we should have the
following condiion.
I(a) = I(0)
Moreover, this relation has been applied to as follows.
Y (a) = easY(a) = easY(0) = W (a)
So we have
Y (a) = W (a)
as I(a) = I(0).
Moreover, if we have eas = e0s then we also have
Y (a) = Y (0).
So, the core form for Y (a) operations has the following conditions.
Y(a) = Y(0)
32
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) =∫ ∞
af(t)e−(t−a)sdt , S(a)f(t) =
∫ a
0f(t)e−(t−a)sdt
and
R(a)f(t) = T (a)f(t) + S(a)f(t) =∫ ∞
0f(t)e−(t−a)sdt
© The relation of T (a) and T (0) operations
Now, F(a) operation has been obtained as follows.
F(a) =
(a0
a a0
)
In this time, if a < 0 and a > 0 in finite numbers then we have as
follows.
a0 = 1.
So we have
a0 = 1.
On the contrary, if a = 0 and a →∞ then we have
a0 = 00 = ∞0
33
and it’s no-defined. This form has been obtained as follows.
F(0) =
(00
00
)
In a word, F(0) operation has a part of F(a) operation. So we have
the following relation.
F(0) ⊆ F(a)
On the other hand, since 00 has no-definition then 00 is able to adopt
all numbers. So we also have
F(0) ⊇ F(a).
N.B. F (a) operations have been dominated by the diagonal.
In a word, we have the following relation.
F(0) = F(a)
In a word,
F(a) = F(0).
In this time, the norm of F(a) operations are able to have one. In a
word,
‖F(a)‖ = ‖F(0)‖ = 1.
Moreover, we also have
a0 = 00 (a 6= 0).
And it has the following relation.
F (a) = easF(a) = easF(0) = H(a)
So we have
F (a) = H(a)
as a0 = 00.
Moreover, if we have eas = e0s then we also have
F (a) = F (0)
34
N.B. a0 = 00.
So, the core form for F (a) operations has the following conditions.
F(a) = F(0)
This F (a) operation has been originated from Pascal’s triangle matrix.
Similarly, since T (a)iso←→ F(a) then we should have the following
relation.
T (a) = T (0)
In fact, we have as follows.
T (a) =∫ ∞
ae−(t−0)sdt = e−as = e0s =
∫ ∞
0e−(t−0)sdt = T (0)
N.B. eas = e0s.
So we have
T (a) = T (0).
In a word,
I(a) =∫ ∞
adt =
∫ ∞
0dt = I(0)
Therefore,
I(a) = I(0).
N.B. ‖I(a)‖ = ‖I(0)‖ = 1.
I(a) and I(0) are identities.
In a word, if we consider T (a) operation then we should have the
following condiion.
I(a) = I(0)
Moreover, this relation has been applied to as follows.
T (a) = easT (a) = easT (0) = R(a)
So we have
T (a) = R(a)
as I(a) = I(0).
35
Moreover, if we have eas = e0s then we also have
T (a) = T (0).
In a word, we are able to treat as follows.
Extended Laplace transform︸ ︷︷ ︸T (a) operation
= Now Laplace transform︸ ︷︷ ︸T (0) operation
as I(a) = I(0).
And, the core form for T (a) operations has the following conditions.
T (a) = T (0)
36
Some results
◦ The fundamental conception of T (a) operations has the following rela-
tion.
T (a) = T (0)
N.B. It is not necessary to be always a = 0.
◦ In this time, we need the followig relation.
I(a) =∫ ∞
adt =
∫ ∞
0dt = I(0)
N.B. I(a) and I(0) are identities.
◦ This T (a) operation has prime conditions for Extended Laplace trans-
forms (T (a) operation)
◦ This T (0) operation has prime conditions for Now Laplace transforms
(T (0) operation)
◦ T (a) operation (Extended Laplace transform) has the following relation.
T (a) = R(a)
N.B. R(a) = easT (0).
◦ Moreover, if we have eas = e0s then we also have
T (a) = T (0).
And it’s possible.
In a word, we are able to treat as follows.
Extended Laplace transform︸ ︷︷ ︸T (a) operation
= Now Laplace transform︸ ︷︷ ︸T (0) operation
as I(a) = I(0).
37
38
Conclusion
Now, we have the following relation.
eas = e0as
In this time, eas is part of the kernel for Extended Laplace transform
(T (a) operation) and e0as is part of the kernel for Now Laplace transform
(T (0a) operation). In a word, we have the following condition.
T (a)iso←→ T (0a)
The next step, we have the jump up condition from T (0) operation
(Now Laplace transform) to T (a) operations (Extended Laplace trans-
form). In this time, we have the following relation.
T (a) = easT (0)
In this time, since eas has been defined as prime number then we were
able to have the following relation.
eas iso←→ 1aiso←→ 0a
By this means, we should consider that the prime numbers include the
1 and 0 elements.
Finally, we were able to have the following relation.
T (a) = T (0)
Since this fundamental form then we have
T (a) = easT (a) = easT (0) = e0sT (0) = T (0)
N.B. eas = e0s.
So, we are able to treat as follows.
Extended Laplace transform︸ ︷︷ ︸T (a) operation
= Now Laplace transform︸ ︷︷ ︸T (0) operation
as I(a) = I(0).
(Sat) 17.Feb.2018
Now let′s go to the next papers with me!
39
40
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