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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