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Prime conditions for Laplace Operator algebras for Laplace transforms (No.1»4) Papers The projection operator for Laplace transforms Pascal’s triangle matrix for Laplace transforms

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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