pascal’s triangle matrix for laplace transformsopab.web.fc2.com/no.7.pdfpreface this paper has...

Pascal’s triangle matrix

for Laplace transforms

No.1

Takao Saito

1

Thank you for our world

2

Pascal’s triangle matrix

for Lapalace transforms

3

Preface

This paper has been presented about Pascal’s triangle matrix op-

eration. Previously, I have been represented by F (a) operation. This

F (a) operation is generated from the Pascal’s triangle matrix operation.

As the introduction for this paper I want to treat the group theorem.

Furthermore, how to treat this operation is same with Lalace transforms.

Finally, I want to explain about this application for Pascal’s triangle ma-

trix operation.

Papers

About solvers of differential equations of relatively

for Laplace transforms

Operator algebras for Laplace transforms

Rings and ideal structures for Laplace transforms

Groups and matrix operator for Laplace transforms

Extension and contaction for transrated operators

Operator algebras for group conditions

Some matrices rings 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

(Fri) 15.Apr.2011 Takao Saito

4

Contents

Preface

§ Chapter 1

◦ About eas and a0 · · · · · · 7◦ Definition od e−0·s in group conditions · · · · · · 9

◦ Some results · · · · · · 18

§ Chapter 2

◦ Pascal’s triangle in matrix · · · · · · 19

◦ About isomorphism of F(a) and T (a) operations. · · · · · · 29

◦ Extended Pascal’s operation FL(a) and irreducibility · · · 32

◦ Decomposition form for FL(a) operation · · · · · · 34

◦ Relation of F (a) and T (a) in dual spaces · · · · · · 38

◦ Projection operator for T (a) operation · · · · · · 39

◦ Relation of D(a) and Y (a) in dual spaces · · · · · · 40

◦ Some results · · · · · · 47

§ Chapter 3

◦ Ideal structure and certain ring · · · · · · 48

◦ Geometrical projection · · · · · · 53

◦ Some results · · · · · · 58

5

◦ Conclusion · · · · · · 60

◦ Reference · · · · · · 61

6

§ Chapter 1

In this chapter, I explain that a is finite to infinite of T (a)1 opera-

tion. T (a)1 is very important condition for generating F(a) (Pascal’s)

operation. T (a) operation is able to extend from group condition to ring

condition when a is infinite number.

©About eas and a0.

Now, based system is following.

T (a)f(t) = T (a){∞∑

n=1

f(0)(n)

n!tn + f(0)} = T (a)

∞∑

n=1

f(0)(n)

n!tn + T (a)f(0)

=∞∑

n=1

f(0)(n)

n!T (a)tn+

f(0) · a0

s=

∞∑

n=1

f(0)(n)

n!{

n−1∑

k=0

(nCk)an−k k!

sk+1+

a0 · n!

sn+1}+f(0) · a0

s

=∞∑

n=1

n−1∑

k=0

f(0)(n)

n!

n!

(n− k)!k!an−k k!

sk+1+

∞∑

n=1

f(0)(n) · a0

sn+1+

f(0) · a0

s

=∞∑

n=1

n−1∑

k=0

f(0)(n)

(n− k)!

an−k

sk+1+

∞∑

n=0

f(0)(n) · a0

sn+1

= −S1(a)f(t) + a0∞∑

n=0

f(0)(n)

n!T (0)tn

= −S1(a)f(t) + a0T (0)∞∑

n=0

f(0)(n)

n!tn

= −S1(a)f(t) + a0T (0)f(t)

　 So

T (a)f(t) = −S1(a)f(t) + a0T (0)f(t) − A

On the other hand, since the definition of T (a), we have

T (a)f(t) = −S(a)f(t) + easT (0)f(t) −B

7

Since B-A then we have

{T (a)− T (a)}f(t) = −S(a)f(t) + S1(a)f(t) + (eas − a0)T (0)f(t)

{T (a)− T (a)}f(t) + {S(a)− S1(a)}f(t) = (eas − a0)T (0)f(t)

In this time, if a = 0 then we have

{T (0)− T (0)}f(t) + {S(0)− S1(0)}f(t) = 0 + 0 = 0

Hence

eas = a0 iff a = 0.

So we have following

e0·s = 00.

mPeas = Pa0

N.B. P is projection operator.

Moreover we are able to have very important formula.

e0·s = Pa0 = a0

Hence

e0·s = a0 for any s and a.

This is very important formula for generating F (a) (Pascal’s) opera-

tions.

If T (a) is to Y (a) operations then we should have following

N.B. T (a)f(t) = Y (a)e−stf(t) and S(a)f(t) = N(a)e−stf(t)

{Y (a)e−st−Y (a)e−st}f(t)+{N(a)e−st−N1(a)e−st}f(t) = (eas−a0)Y (0)e−stf(t)

Moreover, let e−stf(t) = g(t). So we have

{Y (a)− Y (a)}g(t) + {N(a)−N1(a)}g(t) = (eas − a0)Y (0)g(t)

8

In this time, if a = 0 then we also hve

e0·s = 00 for all s.

Similarly this condition is able to extend following.

e0·s = a0 for any s and a.

©Definition of e−0·s in group conditions

T (a)1 is calculated following.

T (a)1 =∫ ∞

a1 · e−(t−a)sdt = [−e−(t−a)s

s]∞a =

e−0·s

s

In this time, we are considering on group condition. So a is finite

condition (see, p.27, Chapter 16, No.6, N o.1). On the other hand, we

have following form.

e0·s = a0

In this case, e−0·s = e0·s = a0. So we have

T (a)1 =e−0·s

s=

a0

s= a0 · 1

s= a0T (0)1

This condition is very important to generate the F(a) (Pascal’s matrix)

operation.

For example,

T (a)1 = a0T (0)1

T (a)t = aT (0)1 + a0T (0)t

So this representation for F(a) operation (matrix condition of 2×2) is

following.

F(a)

(T (a)1

T (a)t

)=

(a0

a a0

) (T (0)1

T (0)t

)(1)

9

In this case,

e−0·s = a0

Precisely speaking, a0 is to 00 (ring condition). Since, a is finite

number then we have a0 = 1, only.

So we have

e−0·s = a0 = 1.

Therefore we are able to adopt following form.

e−0·s = 1 as lims→∞−0 · s = 0.

Conseqentry, we have

T (a)1 =e−0·s

s=

1

son group conditions.

Hence

T (a)1 =1

s= T (0)1 (a is finite).

in L1-spaces.

Reference

T (a)t0 =a0

s= a0T (0)t0. P recisely, PT (a) = T (0)

So we have following relations

a0

s= a0 · T (0)1

and other is P↓ ↑Pa0

s= 00 · T (a)1

In general, this group condition is able to extend that a is infinite

10

number in L1-spaces.

If a is infinite number in L1-spaces then we have following condition.

e−0·s = ∞0 = 1

(see, p.43, Chapter 1, No.1, N o.1)

So we have

e−0·s = 1 as lims→∞−0 · s = 0.

Conseqentry, we have

T (a)1 =e−0·s

s=

1

son extended group conditions.

Hence

T (a)1 =1

s= T (0)1 (a is infinite).

in L1-spaces.

N.B. T (∞) ≤ +∞

As a whole, we have following important property.

T (a)1 =1

s= T (0)1

This condition has a joint with T (a) and T (0) operations, respectively.

T (a)1 does not change with T (0)1. Furthermore, T (a)1 is ring condition

with ideal structures. And T (0) is also satisfied ring conditions without

ideal structures. So T (a)1 is same with T (0)1 on Now Laplace transforms.

In general, the relation of T (a) and T (0) operations are represented fol-

lowing.

PT (a) = T (0).

In this time, P is a projection operator as aP→ 0.

If a is to ∞ in L1-space then I am considering as

‖T (∞)‖ = 1 by Now Laplace transforms.

11

N.B. T (∞) = 1.

If this operation is Pascal’s matrix then we have following conditions.

F(∞) =

∞0

∞ ∞0

.... . .

∞n · · · ∞0

(2)

This diagonal is generated as 1 in L1-space. As a whole, we are able

to have following.

‖F(∞)‖ = 1.

Now, T (a) operation is defined as following.

T (a) = easT (a)

If a is infinite condition then we have

‖T (∞)‖ = ‖e∞·sT (∞)‖ ≤ ‖e∞·s‖

In this time, s is able to obtain negative infinite condition. So we have

‖T (∞)‖ = 0 iff (s ≤ 0) in L1 − spaces.

So, T (a) is able to be converged to zero as a → ∞ iff s ≤ 0 in L1-

spaces. In this time, The group structure have been extended to infinite

conditions from finite conditions.

Reference

Now, I could have been arranged following on L1-spaces.


Group + semi− group︸︷︷︸T (a)

− semi− group︸︷︷︸−S(a)

= Group︸︷︷︸R(a)

.

12

for all a.

Group + semi− group︸︷︷︸F (a)

− semi− group︸︷︷︸−G(a)

= Group︸︷︷︸H(a)

.

for all a.

If T (a), F (a) operations are to T (a),F(a) ,resp then we have following



= Group︸︷︷︸T (0)

.

for all a.

Group + semi− group︸︷︷︸F(a)

− semi− group︸︷︷︸−G(a)

= Group︸︷︷︸F(0)

.

for all a.

In this time, T (a) operation is satisfied the unitary condition. So

T (∞) is able to be 1 and S(∞) is able to be 0 in L1-spaces. Similarly,

F(∞) is able to be 1 and G(∞) is able to be 0.

If a is infinite conditions in L2-space then we have following.

e−0·s = ∞0 = 0.

So we should have

lims→∞−0 · s = −∞.

Therefore

s → +∞In this time, we will have

T (∞)1 = 0

Hence

T (∞) = 0 = S⊥(0).

13

If this operation is Pascal’s matrix then we have following conditions.

F (∞) = e∞s

∞0

∞ ∞0

.... . .

∞n · · · ∞0

(3)

This diagonal is generated as zero in L2-space. So it’s kind of ideal.

As a whole, we are able to have following.

F (∞) = 0 = G⊥(0).

Hence

T (∞) = 0 = F (∞) in L2 − spaces.

In this time, S(a) is 1 as a →∞. So

S(∞) = 1 in L2 − spaces.

N.B. S(∞) = T (0)


On the other hand, this structure of these operators has

Group + subgroup︸︷︷︸T (a)

− subgroup︸︷︷︸−S(a)


.

for all a.

Group + subgroup︸︷︷︸F (a)

− subgroup︸︷︷︸−G(a)

= Group︸︷︷︸H(a)

.

for all a.

14

Especially, if T (a), F (a) operations are to T (a),F(a) ,resp in L2-spaces

then we have



= Group︸︷︷︸T (0)

.

for all a.

Group + subgroup︸︷︷︸F(a)

− subgroup︸︷︷︸−G(a)

= Group︸︷︷︸F(0)

.

for all a.

In fact, F(a) operation is generated as Pascal’s matrix form. So we

had better to consider as ring condition. And the ideal structure is rep-

resented as G(a) operation. Now this condition is considering on group

as infinite condition of a. So ideal structure is able to be subgroup as

infinite condition in L2-spaces.


Similarly, if a is infinite conditions in Banach spaces then we have

e−0·s = ∞0 = 0 in L∞ − space.

So we should have

lims→∞−0 · s = −∞.

Therefore

s → +∞In this time, we will have

T (∞)1 = 0 in Banach spaces.

Hence

T (∞) = 0.

Similarly, if this operation is Pascal’s matrix in Banach spaces then we

15

are able to define as following.

F (∞) = e∞s

∞0

∞ ∞0

.... . .

∞n · · · ∞0

on Banach spaces. (4)

This diagonal is generated as zero in L∞-space. So it’s kind of ideal.

Moreover, in this time, e∞·s is lower order than matrix condition. As a

whole, we are able to have following.

F (∞) = 0 as s → +∞.

Hence

T (∞) = 0 = F (∞) in L∞ − spaces.

These fundamental forms are used with completion of T (a) and F (a)

operations on Hilbert or Banach algebras.

16

Reference

If T (a) operation is to Y (a) operation on L1-space then we will have

following with some conditions.

Y (a)f(t) = easY(a)f(t) =∫ ∞

af(t)easdt

if f(t) = 0 as 0 ≤ t ≤ a.

In this time, Y (a) operation is isomorphic with D(a) operation. D(a)

operation defines following.

D(a) = easD(a) = eas

a0

a0

. . .

a0

(5)

So

Y (a)iso←→ D(a)


If a is infinite condition in L1-space then we will have following.

Y(∞) = 1 ←→ I = D(∞).

N (∞) = 0 = O(∞).

On the contrary, if a is infinite condition in L2-space then we should

have following. In this case, some conditions of Y (a) operation has no

necessity. As a whole,

Y⊥(0) = N (∞) = 1 = O(∞) = D⊥(0) as s → +∞.

N⊥(0) = Y(∞) = 0 = D(∞) = O⊥(0) as s → +∞.

N.B. Of caurse, the ideal structures for N (0) and O(0) are converged

to zero.

Similarly, if it’s on Banach space then we have

Y (∞) = 0 = D(∞) as s → +∞.

N.B. (see, p.7 - p.11, Chapter 10, No.4, N o.1)

17

Some results

◦ Precisely speaking, we have following formula.

e0·s = 00.

This formula is able to replace following and it’s very important for gen-

erating F(a) operations.

e0·s = a0.

◦ We have forllowing formula for all a

T (a)1 =1

s= T (0)1

in L1-spaces.

◦ If a is infinite on L2-spaces then we obtain following conditions.

T (∞) = 0 = S⊥(0) in L2 − spaces.

This condition is treated on completion in Hilbert spaces.

◦ These operator algebra have




.

for all a.

(L1-spaces)

◦ If it’s in L2-space then we have




.

for all a.

◦ If this is on Banach space then we also have

T (∞) = 0 in L∞ − spaces.

Similarly, this condition is treated on completion in Banach spaces.

18

§ Chapter 2

In this chapter, I want to explain about F (a) operation (Pascal’s

triangle matrix). In general, matrix condition has a property of ring.

The F (a) operation is isomorphic with T (a) operation. Therefore the

properties of F (a) operations have the properties of T (a) operations, too.

©Pascal′s triangle in matrix.

Now, the original formula is given as following

L(a)f(t) =∫ ∞

af(t)e−(t−a)sdt = T (a)f(t)

This L(a) operation means the Laplace transforms from a condition.

So if a = 0 then we have Now Laplace transforms. So we have L(0) =

L. T (a) operation is also My Laplace transforms. This T (a) operation

is replaced for operator algebras with S(a) operation (ideal) (see, p.21,

Chapter 4, No.2, N o.1). In this time, we consider the power of t in L(a)

operation and tried to compare with “Now Laplace transforms”.

L(a)1 =1

s= L1

L(a)t =a

s+

1

s2= aL1 + Lt

L(a)t2 =a2

s+

2a

s2+

2

s3= a2L1 + 2aLt + Lt2

L(a)t3 =a3

s+

3a2

s2+

3a

s3+

3!

s4= a3L1 + 3a2Lt + 3aLt2 + Lt3

...

N.B. e−0·s = 1 for all s, L = L(0) (Now Laplace transforms).

19

So I could obtain following.

F(a)

L(a)1

L(a)t

L(a)t2

L(a)t3

...

L(a)tn

=

1

a 1

a2 2a 1

a3 3a2 3a 1

......

.... . .

annC1a

n−1nC2a

n−2nCka

n−k 1

L1

Lt

Lt2

Lt3

...

Ltn

(6)

This matrix is distanced from Pascal’s triangle only part of a. How-

ever if a = 1 , it’s described to “Pascal’s triangle. In other words, this

matrix have been extended to “Pascal’s triangle”.

If a = 0 then F(0) is identity and ‖F(a)‖ = 1. Moreover, F(a) have

a following properties.

F(−a) = F(a)−1 (translation operators)

‖F(a)‖ = detF(a) = 1 (identical)

F(0) = I (identity) as 00 def .= 1

F(0) = F∗(0) (Hermitian)

F(a)F(b) = F(a + b) (semi− groups), especially Fn(a) = F(na)

(7)

If a = 0, this condition has related with T (a) operation, then we have

Hermite forms. So “Now Laplace transforms” is generated on Hermite

forms.

Precisely speaking, F(a) has following properties.

F(a)F(b) = F(a + b)

F(0) = I(0) as 00 def .= 1

s− lima→0F(a)v = v

(8)

So we have that

F(a) have a property of transration-operators and semi-groups.

20

Precisely speaking, we should be e−0·s = 00. However, in this time, it is

discussing on a0. So we have following.

e−0·s = a0

For example

T (a)1 =1

sa0 = a0T (0)1

T (a)t =a

sa0+

1

s2a0 = aT (0)1+a0T (0)t

T (a)t2 =a2

sa0+

2a

s2a0+

2

s3a0 = a2T (0)1+2aT (0)t+a0T (0)t2

T (a)t3 =a3

sa0+

3a2

s2a0+

3a

s3a0+

3!

s4a0 = a3T (0)1+3a2T (0)t+3aT (0)t2+a0T (0)t3

...

So we are able to analogize following

T (a)tn =n∑

k=0

nCkan−kT (0)tk − A.

When n = 0, we have following

T (a)1 = a0T (0)1

So we have

T (a)1 =a0

s= a0T (0)1.

If n is to n + 1 then we should have

T (a)tn+1 =∫ ∞

atn+1e−(t−a)sdt = [tn+1·(−e−(t−a)s

s)]∞a +

n + 1

s

∫ ∞

atne−(t−a)sdt

=an+1

s+

n + 1

s

n∑

k=0

nCk · an−k · T (0)tk − by A

21

=an+1

s+

n + 1

s

n∑

k=0

n!

k!(n− k)!·an−k· k!

sk+1=

an+1

s+

n∑

k=0

(n + 1)!

(k + 1)!(n− k)!·an−k·(k + 1)!

sk+2

= an+1T (0)1+n∑

k=0

n+1Ck+1 ·an−k ·T (0)tk+1 =n∑

k=−1

n+1Ck+1 ·an−k ·T (0)tk+1

So we have

T (a)tn+1 =n+1∑

k=0

n+1Ck · an+1−k · T (0)tk

Therefore this hypothesis (A) was correct.

Hence

T (a)tn =n∑

k=0

nCkan−kT (0)tk.

This condition is able to be represented by matrix form. This matrix

form is called by F(a) operation.

So I could generate following.

F(a)

T (a)1

T (a)t

T (a)t2

T (a)t3

...

T (a)tn

=

a0

a a0

a2 2a a0

a3 3a2 3a a0

......

.... . .

annC1a

n−1nC2a

n−2nCka

n−k a0

T (0)1

T (0)t

T (0)t2

T (0)t3

...

T (0)tn

(9)

In this case, we have to attention to treat the element a0 on diagonal.

In general, I have been defined this forms. This F(a) operation is used

on previous papers.

22

The dual form or involution of Pascal’s triangle matrix is following form.

F∗(a)

L(a)1

L(a)t

...

L(a)t(n−2)

L(a)t(n−1)

L(a)tn

∗

=

L1

Lt

...

Lt(n−2)

Lt(n−1)

Ltn

∗

1 a a2 · · · an−1 an

1 2a · · · · · · nC1an−1

. . ....

1 (n− 1)a nCn−2a2

1 na

1

(10)

N.B. Lf(t) = L(0)f(t)

F∗(−a) = F∗(a)−1 , F∗(0) = I∗

‖F∗(a)‖ = detF∗(a) = 1 , F∗(0) = F(0)

F∗(a)F∗(b) = F∗(a + b) , F∗n(a) = F∗(na)

(11)

Similarly, it is able to extend to following as a0.

F∗(a)

T (a)1

T (a)t

...

T (a)t(n−2)

T (a)t(n−1)

T (a)tn

∗

=

T (0)1

T (0)t

...

T (0)t(n−2)

T (0)t(n−1)

T (0)tn

∗

a0 a a2 · · · an−1 an

a0 2a · · · · · · nC1an−1

. . ....

a0 (n− 1)a nCn−2a2

a0 na

a0

(12)

N.B. L(a)f(t) = T (a)f(t)

23

Similarly

F(−a) = F(a)−1

L(− a)1

L(− a)t

L(− a)t2

L(− a)t3

...

L(− a)tn

=

1

−a 1

a2 −2a 1

−a3 3a2 −3a 1

......

. . .

(−1)nan (−1)n−1nC1a

n−1 (−1)n−knCka

n−k 1

L1

Lt

Lt2

Lt3

...

Ltn

(13)

This matrix condition is able to extend to following. We have to attend

to treat the diagonal in matrix.

F(−a) = F(a)−1

T (− a)1

T (− a)t

T (− a)t2

T (− a)t3

...

T (− a)tn

=

a0

−a a0

a2 −2a a0

−a3 3a2 −3a a0

......

. . .

(−1)nan (−1)n−1nC1a

n−1 (−1)n−knCka

n−k a0

T (0)1

T (0)t

T (0)t2

T (0)t3

...

T (0)tn

(14)

Now Laplace transforms is used as a = 0, however this condition is 00.

It’s treated as no-defined. Therefore , we should correct the consideration

of “Now Laplace transforms”. Of caurse, it’s said to be unitary operator.

Especially

Now, F(1) have a property of Pascal’s triangle.

24

F(1)

L(1)1

L(1)t

L(1)t2

L(1)t3

...

L(1)tn

=

1

1 1

1 2 1

1 3 3 1

......

.... . .

1 nC1 nC2 nC3 nCk 1

L1

Lt

Lt2

Lt3

...

Ltn

(15)

F(1)

L1

Lt

Lt2

Lt3

...

Ltn

=

1

1 1

1 2 1

1 3 3 1

......

.... . .

1 nC1 nC2 nC3 nCk 1

L(− 1)1

L(− 1)t

L(− 1)t2

L(− 1)t3

...

L(− 1)tn

(16)

This form is proved by semi-groups.

T (1) = F(1)T (0)

T (1)T (−1) = F(1)T (0)T (−1)

T (0) = F(1)T (−1)

So we have above.

N.B. L = T (0) , L(a) = T (a)


25

In this time, F(1) (Pascal’ triangle) operation does not change for

T (a) operation.

Similarly if f(t) = (ebt, tebt, t2ebt, · · · , tnebt)T then following.

F(a)

L(a)ebt

L(a)tebt

L(a)t2ebt

L(a)t3ebt

...

L(a)tnebt

= eab

1

a 1

a2 2a 1

a3 3a2 3a 1

......

.... . .

annC1a

n−1nC2a

n−2nCka

n−k 1

Lebt

Ltebt

Lt2ebt

Lt3ebt

...

Ltnebt

(17)

F(−a) = F(a)−1

L(− a)ebt

L(− a)tebt

L(− a)t2ebt

L(− a)t3ebt

...

L(− a)tnebt

= e−ab

1

−a 1

a2 −2a 1

−a3 3a2 −3a 1

......

.... . .

(−1)nan (−1)n−knCka

n−k 1

Lebt

Ltebt

Lt2ebt

Lt3ebt

...

Ltnebt

(18)

Consequentry, we will extend to following.

26

F(a)

T (a)ebt

T (a)tebt

T (a)t2ebt

T (a)t3ebt

...

T (a)tnebt

= eab

a0

a a0

a2 2a a0

a3 3a2 3a a0

......

.... . .

annC1a

n−1nC2a

n−2nCka

n−k a0

T (0)ebt

T (0)tebt

T (0)t2ebt

T (0)t3ebt

...

T (0)tnebt

(19)

F(−a) = F(a)−1

T (− a)ebt

T (− a)tebt

T (− a)t2ebt

T (− a)t3ebt

...

T (− a)tnebt

= e−ab

a0

−a a0

a2 −2a a0

−a3 3a2 −3a a0

......

.... . .

(−1)nan (−1)n−knCka

n−k a0

T (0)ebt

T (0)tebt

T (0)t2ebt

T (0)t3ebt

...

T (0)tnebt

(20)

Of course F(a) is Hermitian iff a = 0.

For example, we are able to fix that the norm of T (a) operation is

to F (a) operation. This condition is treated with extended the norm

by Hahn Banach theorem. In this case, since this conditions, we have

following form.

T (a)est = easF(a)T (0)est as b = s.

This means that∫ ∞

aeste−(t−a)sdt = easF(a)

∫ ∞

0este−stdt

∫ ∞

aeasdt = easF(a)

∫ ∞

0dt

eas∫ ∞

adt = easF(a)

∫ ∞

0dt

Therefore

Y(a)1 = F(a)Y(0)1 , I(a) = F(a).

27


Furthermore I can write following.

T (a) = αF(a)T (0)

α is kind of norm of T (a). This property is very important to concider

irreducibility. This normed condition is following.

‖T (a)‖ = ‖αF(a)‖‖T (0)‖ = |eas|‖T (0)‖ iff let α = eas

Therefore

‖T (a)‖ = ‖easF(a)‖‖T (0)‖ = |eas|‖T (0)‖ = |eas|On the other hand

‖T (a)‖ = ‖F (a)‖‖T (0)‖ = |eas|


Therefore

‖F (a)‖ = ‖T (a)‖ = |eas|.

Now, we want to consider relation of the operation F(a) with T (a).

‖F(a)‖ = detF(a) = 1. On the other hand, ‖T (a)‖ is also 1. There-

fore ‖F(a)‖ = ‖T (a)‖ = 1 and it’s unitary operators. Therefore this

operation has two kinds forms. First condition (FP (a) operator) is that

diagonal is generated all 1. Sencond condition (FL(a) operator) is that

diagonal is represented as a0. In general, FP (a) is embedded in FL(a).

However , interesting case is a = 0. This condition has no-defined.

Ideal structure of this matrix has same conditions. Moreover I want

to consider the fundamental properties and extend to F (a) and T (a).

N.B. F(a) is matrix operators.

28

In this case, we are treating FP (a) operation. So

FP (a) = easFP (a) = eas

1

a 1

a2 2a 1...

.... . .

annC1a

n−1nCka

n−k 1

(21)

and

T (a)f(t) = easT (a)f(t) =∫ ∞

af(t)e−(t−a)sdt

respectively.

Hence, I define following.

‖FP (a)‖ = ‖T (a)‖ = |eas|.

©About isomorphism of F(a) and T (a). (s is finite.)

Since Pascal’s triangle (FP (a)), we have following.　

FP (a)　　　　　　　　　　　

T (a)1

T (a)t

T (a)t2

T (a)t3

...

T (a)tn

=

1

a 1

a2 2a 1

a3 3a2 3a 1

......

.... . .

annC1a

n−1nC2a

n−2nCka

n−k 1

T (0)1

T (0)t

T (0)t2

T (0)t3

...

T (0)tn

(22)

N.B. a 6= 0

Since this mtrix conditions then we have

T (a)tn = anT (0)1 + · · ·+n Ckan−kT (0)tk + · · ·+ T (0)tn

29

　　　　 = an 1

s+ · · ·+n Cka

n−k · k!

sk+1+ · · ·+ n!

sn+1

　　　　 =n−1∑

k=0

nCkan−k · k!

sk+1+

n!

sn+1=

n−1∑

k=0

nPkan−k

sk+1+

n!

sn+1　�．

Similarly

FP (b)

T (b)1

T (b)t

T (b)t2

T (b)t3

...

T (b)tn

=

1

b 1

b2 2b 1

b3 3b2 3b 1

......

.... . .

bnnC1b

n−1nC2b

n−2nCkb

n−k 1

T (0)1

T (0)t

T (0)t2

T (0)t3

...

T (0)tn

(23)

N.B. b 6= 0

T (b)tn = bnT (0)1 + · · ·+n Ckbn−kT (0)tk + · · ·+ T (0)tn

　　　　 = bn 1

s+ · · ·+n Ckb

n−k · k!

sk+1+ · · ·+ n!

sn+1

　　　　 =n−1∑

k=0

nCkbn−k · k!

sk+1+

n!

sn+1=

n∑

k=0

nPkbn−k

sk+1+

n!

sn+1�．

　 Since �,� then we have

T (a)tn =n−1∑

k=0

nPkan−k

sk+1+

n!

sn+1=

n−1∑

k=0

nPkbn−k

sk+1+

n!

sn+1= T (b)tn　 iff　T (a) = T (b).

n−1∑

k=0

nPk

sk+1{an−k − bn−k} = 0

　　 So　 an−k − bn−k = 0　 as　 s 6= ∞.

(a− b)(ap−1 + ap−2b + · · ·+ abp−2 + bp−1) = 0

30

　　　　　　　　　　　　　　　　　　　　　　　N.B.　 p = n− k

In this time, a 6= 0 and b 6= 0.

Therefore　 a = b　 iff 　 T (a) = T (b).　　　　　

　 So

aiso←→ T (a).　�

Similarly, if this operation is represented as FP (a) then we have following.

　 an − bn = 0.　

(a− b)(an−1 + an−2b + · · ·+ abn−2 + bn−1) = 0

Therefore

a = b　 iff 　FP (a) = FP (b).

So

aiso←→ FP (a)　�

Since �,� then we have

T (a)iso←→ a

iso←→ FP (a).

Hence

T (a)iso←→ FP (a)

　　　　　　　　　　　　　　　　　 if　f(t) = (1　 t　 t2　 · · ·　 tn)T

N.B. FP (a) is finite rank.

Furthermore we will find that T (a) and FP (a) are compact operators.

31

Therefore T (a) operation also have a following properties.

T (−a) = T (a)−1 (translation operators)

‖T (a)‖ = 1 (identical)

T (0) = I (identity) as 00 def .= 1

T (0) = T ∗(0) (Hermitian)

T (a)T (b) = T (a + b) (semi− groups), especially T n(a) = T (na)

(24)

©Extended Pascal′s operation FL(a) and irreducibility

Now, I want to refer the matrix FL(a) and it’s irreducibility. FL(a) is

the matrix that the diagonal has a0.

We have following.

T (a)1

T (a)t

T (a)t2

T (a)t3

...

T (a)tn

=

a0

a a0

a2 2a a0

a3 3a2 3a a0

......

.... . .

annC1a

n−1nC2a

n−2nCka

n−k a0

T (0)1

T (0)t

T (0)t2

T (0)t3

...

T (0)tn

(25)

On the other hand, in general, T (a) operation is able to represent as

following.

T (a) = a0T (a)

So the irreducible form is genaerated only the diagonal.

FE(a) = D(a)

T (a)1

T (a)t

T (a)t2

T (a)t3

...

T (a)tn

=

a0

a0

a0

a0

. . .

a0

T (a)1

T (a)t

T (a)t2

T (a)t3

...

T (a)tn

(26)

32

In this time, D(a) operations are defined following.


a0

a0

a0

a0

. . .

a0

(27)


Especially, the characteristic form for T (a) operation is represented

following.

T (a) = 00T (a)

So the irreducible form is genaerated only the diagonal.

FE(0)

T (a)1

T (a)t

T (a)t2

T (a)t3

...

T (a)tn

=

00

00

00

00

. . .

00

T (a)1

T (a)t

T (a)t2

T (a)t3

...

T (a)tn

(28)

This T (0) operation have difference for T (a) operation only the part of

S(a). In general, S(a) is satisfied ideal structure. Therefore the property

of T (0) is same with T (a). Similarly FP (a) is same for FE(0) with the

part of G(a). Of caurse, G(a) operation is ideal. The projection operator

has following forms.

PFp(a) = FE(0) and T (0) = PT (a) , simultaneously.

N.B. In this case, ‖T (a)‖ = 1.

33

Reference

G(a) = −easG(a) = −eas

a

a2 2a

a3 3a2 3a...

.... . .

annC1a

n−1 · · · na

(29)

and

S(a)f(t) = easS(a)f(t) =∫ a

0f(t)e−(t−a)sdt = (0).

©Decomposition form for FL(a) operation

Now, let decompose the matrix operator to diagonal ring and ideal.

So we have following.

T (a)1

T (a)t

T (a)t2

T (a)t3

...

T (a)tn

=

a0

a a0

a2 2a a0

a3 3a2 3a a0

......

.... . .

annC1a

n−1nC2a

n−2nCka

n−k a0

T (0)1

T (0)t

T (0)t2

T (0)t3

...

T (0)tn

(30)

↙ decomposed ↘

a0

a0

a0

a0

. . .

a0

+

a

a2 2a

a3 3a2 3a

......

. . .

annC1a

n−1nC2a

n−2nCka

n−k na

(31)

34

ring ideal

This form is very important to commute with following operator algebras.

General form · · · F(a) = H(a) + {−G(a)}

F (a) = easH(a) + {−G(a)} · · · adjunction operators

Especially, these forms are to following conditions. (irreducible forms)

F (a) = easH(0) + {−G(a)} · · · adjunction operators

In general, we should that eas is not a0. However if there operations

are unitary conditions then we have following form.

F(a) = a0H(0) + {−G(a)} = FL(a)

Therefore we are able to have following properties.

Since T (a)iso↔ F (a) and T (a)

iso↔ F(a) then we have

General form · · · T (a) = R(a) + {−S(a)}

T (a) = easR(a) + {−S(a)} · · · adjunction operators


T (a) = easR(0) + {−S(a)} · · · adjunction operators

35

In general, we should that eas is not a0. However if there operations

are unitary conditions then we have following form.

T (a) = a0R(0) + {−S(a)}

In this time, G(a), G(a) operations are satisfied the property of ideal

structures. So S(a), S(a) operations also generate the property of idial

structures.

Similarly, the dual condition is following.

T (a)1

T (a)t

T (a)t2

...

T (a)t(n−1)

T (a)tn

∗

=

T (0)1

T (0)t

T (a)t2

...

T (0)t(n−1)

T (0)tn

∗

a0 a a2 · · · an

a0 2a · · · an−1

a0 3a......

a0nCka

n−k

a0

(32)

↙ decomposed ↘

a0

a0

a0

. . .

a0

+

a a2 · · · an

2a · · · an−1

3a...

na

(33)

ring ideal

This form is very important to commute with following operator algebras.

General form · · · F∗(a) = H∗(a) + {−G∗(a)}

F (a) = easH(a) + {−G(a)} · · · adjunction operators

36


F ∗(a) = easH∗(0) + {−G∗(a)} · · · adjunction operators

Similarly, we have this form iff there operations are unitary conditions.

F∗(a) = a0H∗(0)︸︷︷︸Hermitian

+ {−G∗(a)} = F∗L(a)

Therefore we are able to also have following properties.

Since T ∗(a)iso↔ F ∗(a) and T ∗(a)

iso↔ F∗(a) then we have

General form · · · T ∗(a) = R∗(a) + {−S∗(a)}

T (a) = easR(a) + {−S(a)} · · · adjunction operators


T ∗(a) = easR∗(0) + {−S∗(a)} · · · adjunction operators

Similarly, we have this form iff there operations are unitary conditions.

T ∗(a) = a0R∗(0)︸︷︷︸Hermitian

+ {−S∗(a)}

In this time, G∗(a), G∗(a) operations are satisfied the property of ideal

structures. So S∗(a), S∗(a) operations also generate the property of idial

structures.

37

©Relation of F (a) and T (a) in dual space.

'

&

$

%

F−1(a) = F (−a)

F ∗−1(a) = F ∗(−a)

←→

¾

½

»

¼

F (a)iso←→ F ∗(a)

e−as l l eas

'

&

$

%

F−1(a) = F(−a)

F∗−1(a) = F∗(−a)

←→

¾

½

»

¼

F(a)iso←→ F∗(a)

‖F(a)‖ = 1

iso m unitary m iso

‖T (a)‖ = 1'

&

$

%

T −1(a) = T (−a)

T ∗−1(a) = T ∗(−a)

←→

¾

½

»

¼

T (a)iso←→ T ∗(a)

e−as l l eas

'

&

$

%

T−1(a) = T (−a)

T ∗−1(a) = T ∗(−a)

←→

¾

½

»

¼

T (a)iso←→ T ∗(a)

38

　 Now the operation T (a) have two kinds of representations. First

form is

T (a) = F(a)T (0)

and other is P↓ ↑PT (a) = D(0)T (a) (characteristic form)

　 In this case, D(0) = PF(a) = F(0) and T (0) = PT (a). The norm of

lower equation is represented as eas. In this time, upper T (a) is also ex-

tended as T (a) = αF(a)T (0). Especially, in this time, α = 1. Therefore,

the norm of T (a) operation is able to be 1. So we have

‖T (a)‖ = 1.

In fact, the norm of T (a) has variable condition. In this case, I have

been treated that the norm of T (a) operation is 1 (unitary operator)

with Pascal’s triangle operation. Of caurse, the norm of Pascal’s triangle

operation is 1, too.

On the contrary, lower type of T (a) operation does not need to be

1. So, in general, we are able to consider on eas. This condition is

concentrated with Hahn-Banach theorem. From 1 to extend to eas is

using this theorem.


This F (a) operation is following.

F(a)

F (a) = easF(a) = eas

a0

a a0

a2 2a a0

a3 3a2 3a a0

......

. . .

annC1a

n−1 · · · na a0

(34)

Especially, F(a) is Pascal’s triangle matrix operator.

39

©Relation of D(a) and Y (a) in dual space.

'

&

$

%

D−1(a) = D(−a)

D∗−1(a) = D∗(−a)

←→

¾

½

»

¼

D(a)iso←→ D∗(a)

e−as l l eas

'

&

$

%

D−1(a) = D(−a)

D∗−1(a) = D∗(−a)

←→

#

"

Ã

!

D(a) = D∗(a)

(Hermitian)

‖D(a)‖ = 1

iso m unitary m iso

‖Y(a)‖ = 1'

&

$

%

Y−1(a) = Y(−a)

Y∗−1(a) = Y∗(−a)

←→

#

"

Ã

!

Y(a) = Y∗(a)

(Hermitian)

e−as l l eas

'

&

$

%

Y −1(a) = Y (−a)

Y ∗−1(a) = Y ∗(−a)

←→

¾

½

»

¼

Y (a)iso←→ Y ∗(a)

40

The normed form of F (a), T (a) operations in dual space is following.

'

&

$

%

‖F−1(a)‖ = ‖F (−a)‖ = |e−as|

‖F ∗−1(a)‖ = ‖F ∗(−a)‖ = |e−as|←→

¾

½

»

¼

‖F (a)‖ = ‖F ∗(a)‖ = |eas|

e−as l l eas

'

&

$

%

‖F−1(a)‖ = ‖F(−a)‖ = 1

‖F∗−1(a)‖ = ‖F∗(−a)‖ = 1

←→

¾

½

»

¼

‖F(a)‖ = ‖F∗(a)‖ = 1

‖F(a)‖ = 1

iso m unitary m iso

‖T (a)‖ = 1'

&

$

%

‖T −1(a)‖ = ‖T (−a)‖ = 1

‖T ∗−1(a)‖ = ‖T ∗(−a)‖ = 1

←→

¾

½

»

¼

‖T (a)‖ = ‖T ∗(a)‖ = 1

e−as l l eas

'

&

$

%

‖T−1(a)‖ = ‖T (−a)‖ = |e−as|

‖T ∗−1(a)‖ = ‖T ∗(−a)‖ = |e−as|←→

¾

½

»

¼

‖T (a)‖ = ‖T ∗(a)‖ = |eas|

41

Similarly, the normed form of D(a), Y (a) operations in dual space

is following.

'

&

$

%

‖D−1(a)‖ = ‖D(−a)‖ = |e−as|

‖D∗−1(a)‖ = ‖D∗(−a)‖ = |e−as|←→

¾

½

»

¼

‖D(a)‖ = ‖D∗(a)‖ = |eas|

e−as l l eas

'

&

$

%

‖D−1(a)‖ = ‖D(−a)‖ = 1

‖D∗−1(a)‖ = ‖D∗(−a)‖ = 1

←→

¾

½

»

¼

‖D(a)‖ = ‖D∗(a)‖ = 1

‖D(a)‖ = 1

iso m unitary m iso

‖Y(a)‖ = 1'

&

$

%

‖Y−1(a)‖ = ‖Y(−a)‖ = 1

‖Y∗−1(a)‖ = ‖Y∗(−a)‖ = 1

←→

¾

½

»

¼

‖Y(a)‖ = ‖Y∗(a)‖ = 1

e−as l l eas

'

&

$

%

‖Y −1(a)‖ = ‖Y (−a)‖ = |e−as|

‖Y ∗−1(a)‖ = ‖Y ∗(−a)‖ = |e−as|←→

¾

½

»

¼

‖Y (a)‖ = ‖Y ∗(a)‖ = |eas|

42

In this time, if s is finite number then we should treat as 00 = 1.

On the contrary, if s is infinite number then we are not able to define the

value of e0·∞.

F(a) F (a)

a0

a a0

.... . .

an · · · nCkan−k a0

, eas

a0

a a0

.... . .

an · · · nCkan−k a0

(35)

l l(A) (B)

On the contrary, T (a) have two forms.

T (a) + S1(a) = 00T (0) = R(0) − (A)

T (a) + S(a) = easT (0) = R(a) − (B)

(see, p.7, -A,-B, Chapter 1, No.1, N o.2)

The norm condition of T (a) of upper form (A) is represented fol-

lowing.

T (a)f(t) =∫ ∞

af(t)e−(t−a)sdt

‖T (a)‖ = ‖ker(s, t)‖ = ‖e−(t−a)s‖{a,∞} = a0 = ‖F(a)‖Therefore

‖F(a)‖ = ‖T (a)‖ = a0 = 00 = ‖T (0)‖ = ‖F(0)‖.

On the contrary, the T (a) operation of (B) is extended from T (0)

operation. R(a) operation to T (a) operation is preserved. In this time,

the norm of T (a) operation is represented following.

T (a)f(t) =∫ ∞

af(t)e−(t−a)sdt as 0 ≤ t ≤ ∞.

43

Since Hahn Banach theorem, ‖T (a)‖ is extended in 0 ≤ t ≤ a as

f(t) 6= 0. Therefore

‖T (a)‖ = ‖ker(s, t)‖ = ‖e−(t−a)s‖{0,∞} = ‖eas‖ = ‖R(a)‖.

So

‖T (a)‖ = ‖R(a)‖ = ‖eas‖ (extended)

N.B. R(a)f(t) =∫∞0 f(t)e−(t−a)sdt


Since S1(a) is ideal, T (a) is preserved from R(0) operations. In this

case, ‖T (a)‖ = 00 = 1. And, PT (a) = T (0) = R(0). On the other

hand, (B) is preserved to ‖R(a)‖ operation (R(0) → R(a) form compare

(A) form. Essentially, the property of R(a) = easT (0) and R(0), T (0)

are same. R(a) and R(0), T (0) generated ring boundary. However, in

general, eas 6= 00. So R(a) 6= 00T (0). On this time, T (a) is able to extend

to ‖eas‖ by Hahn-Banach theorem → (B). Therefore, the norm of (B) is

preserved to R(a) operation. And S(a) is able to be ideal condition. So

the concept for the (A) form has same for the (B) form. Slightly different

is that the norm on ring boundary is extended. This norm condition will

be treated in C∗-algebras. Of caurse, T (a) operation that it’s extended

by Hahn-Banach theorem is also treated on C∗-algebras. ((B) form)

Similarly, D(a) and D(a) operations are treated with Y(a) and Y (a)

operations. Slight different is that the ideal structure does not have. So

the ring has same conditions.

D(a) D(a)

a0

a0

. . .

a0

, eas

a0

a0

. . .

a0

(36)

l l(C) (D)

44

Now, Y (a) have two forms.

Y (0) + N1(0) = 00Y (0) = W (0) − (C)

Y (a) + N(a) = easY (0) = W (a) − (D)

The norm condition of Y (a) of upper form (C) is represented following.

Y (a)f(t) =∫ ∞

af(t)easdt as (a ≤ t ≤ ∞)

‖Y (0)‖ = ‖ker(s, t)‖ = ‖e0·s‖{0,∞} = 00 = ‖D(0)‖Therefore

‖Y (0)‖ = 00 = ‖D(0)‖.

On the contrary, the Y (a) operation of (D) is extended from Y (0)

operation. W (a) operation to Y (a) operation is preserved. In this time,

the norm of Y (a) operation is represented following.

Y (a)f(t) =∫ ∞

af(t)easdt as 0 ≤ t ≤ ∞.

Since Hahn Banach theorem, ‖Y (a)‖ is constant in 0 ≤ t ≤ a as

f(t) 6= 0. Therefore

‖Y (a)‖ = ‖ker(s, t)‖ = ‖eas‖{0,∞} = ‖eas‖ = ‖W (a)‖.

So

‖Y (a)‖ = ‖W (a)‖ = ‖eas‖ (extended from Y (0))

N.B. W (a)f(t) =∫∞0 f(t)easdt


Clearly, since N1(a) is ideal, Y (a) is preserved from Y (0) oper-

ations with characteristic condition. So we have, ‖Y (a)‖ = 00 = 1.

And, PY (a) = Y (0) = W (0). On the other hand, (D) is preserved to

‖W (a)‖ operation (W (0) → W (a) form compare (C) form). Essentially,

the property of W (a) = easY (0) and W (0), Y (0) are same. W (a) and

W (0), Y (0) generated ring boundary. However, in general, eas 6= 00. So

W (a) 6= 00Y (0). On this time, Y (a) is able to extend to ‖eas‖ by Hahn-

Banach theorem → (D). Therefore, the norm of (D) is preserved to W (a)

45

operation. And N(a) is able to be ideal condition. So the concept for the

(C) form has same for the (D) form. Slightly different is that the norm

on ring boundary is extended. This norm condition will be treated in C∗-algebras. Of caurse, Y (a) operation that it’s extended by Hahn-Banach

theorem is also treated on C∗-algebras. ((D) form)

My Laplace transform is represented as T (a) operation. Since the

Pascal’s operation then we have following form.

T (a) = αF(a)T (0)

In this time, F(a) is identical. Therefore T (a) is able to have various

forms by the coefficients α.

On the contrary, Y (a) operation has a property of FE(a) conditions.

So Y (a)iso←→ FE(a) = D(a) ((C) form).

46

Some results

◦ F(a) (Pascal’s triangle matrix) operation is generated from power func-

tion of T (a) operation (My Laplace transforms).

F(a) =

a0

a a0

a2 2a a0

a3 3a2 3a a0

......

.... . .

annC1a

n−1nC2a

n−2nCka

n−k a0

(37)

Especially, if a = 1 then we have Pascal’s triangle.

◦ Now, we have following form.

T (a)1 =1

sa0 = a0T (0)1

This form is very important condition to generate F(a) (Pascal’s tri-

angle matrix) operation.

◦ F(a) operation has following properties.

F(−a) = F(a)−1 , F(0) = I(0)

‖F(a)‖ = detF(a) = 1 , F(0) = F∗(0)

F(a)F(b) = F(a + b) , Fn(a) = F(na)

(38)

◦ Pascal’s matrix operation F(a) is isomorphic with T (a) operation

(My Laplace transforms).

F(a)iso←→ T (a).

◦ T (a) operation has same properties with F (a) operation. So we have

T (−a) = T (a)−1 , T (0) = I(0)

In this case ‖T (a)‖ = 1 , T (0) = T ∗(0)

T (a)T (b) = T (a + b) , T n(a) = T (na)

(39)

The -∗ algebras are same, too.

◦ F (a) operation is also isomorphic with T (a) operation (extended

form).

47

§ Chapter 3

In this chapter, I want to explain for extended definition of ideal

structure for F (a) operation (Pascal’s triangle matrix). Previously type

of G(a) operation is not extended to L2-space. Now, this G(a) operation

was able to extend to all condition of a.

©Ideal structure and certain ring

Now, we have been treated that the ideal G(a) operation is following

in L1-spaces.

G(a) = −

0

a 0

a2 2a 0

a3 3a2 3a 0

......

.... . .

annC1a

n−1nC2a

n−2nCka

n−k 0

(40)

iff a is finite.

In fact, this ideal structure has generated from following operator

algebras.

F(a) + G(a) = H(a)

In this time, let H(a) is identical identity. So

H(a) =

00

00

. . .

00

=

1

1. . .

1

= I(0). (41)

48


Essentially H(a) should adopt as 00. However, now H(a) operation is

satisfied the following condition in L1, L2-spaces.

e0·s = 00 = a0 = 1

Therefore, precisely speaking, the ideal structure G(a) is formed as

G(a) = −

a0 − 1

a a0 − 1

a2 2a a0 − 1

a3 3a2 3a a0 − 1

......

.... . .

annC1a

n−1nC2a

n−2nCka

n−k a0 − 1

(42)

for all a.

This condition is able to also treat on L2-spaces.

N.B.

F(a) =

a0

a a0

a2 2a a0

a3 3a2 3a a0

......

.... . .

annC1a

n−1nC2a

n−2nCka

n−k a0

(43)

for all a.

For example, if a →∞ in L1-space then we have

∞0 = 1

49

So

G(∞) = −(

0

∞ 0

)=

(0

−∞ 0

)= ideal. (44)

On the contrary, a →∞ in L2-space then we have

∞0 = 0

So

G(∞) = −( −1

∞ −1

)=

(1

−∞ 1

)= indentical. (45)

This condition has maximal ideal to ring condition. It is property of

orthogonal spaces (Hilbert spaces). In this time,

G(∞) = I = F⊥(0).

Therefore(

1

−∞ 1

)=

(1

1

)⊥. (46)

The identity of orthogonal space is also moved to identity. These

identities are separated to four kinds. It’s following.

(1

−∞ 1

) (1

∞ 1

) (1 −∞

1

) (1 ∞

1

), resp. (47)

So I defined following.

G(a) = easG(a) = −eas

a0 − 1

a a0 − 1

a2 2a a0 − 1

a3 3a2 3a a0 − 1

......

.... . .

annC1a

n−1nC2a

n−2nCka

n−k a0 − 1

(48)

50

for all a.

Similarly,

H(a) = easH(a) = eas

1

1

1

1. . .

1

(49)

for all a.

As a whole, we have following operator algebras with T (a) operations.

Pascal′s matrix operation

F (a) + G(a) = H(a).

mT (a) + S(a) = R(a).

Laplace transforms


Similarly,

Diagonal matrix operation

D(a) + O(a) = H(a).

mY (a) + N(a) = W (a).

Integral operation


51

N.B.


a0

a0

. . .

a0

(50)

for all a.

and

O(a) = easO(a) = eas

1− a0

1− a0

. . .

1− a0

(51)

for all a.

These operator algebras are extended from privious definition. There

are treated on L1, L2-spaces. And I want to pay attention to satisfy

following certain condition.

e0·s = 00 = a0 = 1 (L1 − space)

If a is ∞ in L2-space then we should have following.

e0·s = 00 = a0 = ∞0 = 0 (L2 − space)

as lims→∞ 0 · s = −∞.

This condition is very important form for these paperes.

(see, p.16,32, Chapter 13,14, resp, No.5, N o.1, etc)

Since this conditions, 00 have variable number. This situation has

no defined. Especially, this {1}, {0} in L1-space or {1, 0} in L2-space

generate the projection operator.


52

For example, I want to suggest following.

©Geometrical projection

Geometrical projection operator is defined as following.

P =

1

1

0

(52)

Now, I define the following projection operator

P (0) =

00

00

00

(53)

00 are independed on the diagonal. So we are able to have many

forms for projection operator.

P (0) = F(0) = D(0) ⊇ identity.

If P (0) is identity then we have following property.

P (0) = P ∗(0) = P 2(0) = P−1(0).

This property is able to pass for F(a),D(a) operations. Q(0) is defined

as following.

Q(0) = I(0)− P (0)

In general, P (a) operation is defined

F(a) ⊇ P (a) =

a0

a0

a0

⊇ identity. (54)

Reference

The kind of F(a) operation is following.

F(a) =

a0

a a0

a2 2a a0

. (55)

53

So F(a) has possibility to be projection operator. Moreover, since this

conditions, I am able to have identity in projection operator.

P (a) = D(a).

Let’s consider that the identity exists in projection operators. There-

fore the identity also has following property.

P (a) = P ∗(a) = P 2(a) = P−1(a) = P (2a)

N.B. P (2a) originates in left-invariant on group-rings.

Similarly, D(a) also has same property. So

D(a) = D∗(a) = D2(a) = D−1(a) = D(2a).

In this case, there forms are satisfied with field condition. If the

operation is F(a) then we should the normed form. So

‖F(a)‖ = ‖F∗(a)‖ = ‖F2(a)‖ = ‖F−1(a)‖ = ‖F(2a)‖ = {1, 0}

‖P (a)‖ = ‖P ∗(a)‖ = ‖P 2(a)‖ = ‖P−1(a)‖ = ‖P (2a)‖ = {1, 0}

On the other hand, we have

P (0) = Q⊥(∞) = {1} , Q(0) = P⊥(∞) = {0} . . . L2 − spaces.

P (0) = P (∞) = {1} , Q(0) = Q(∞) = {0} . . . L1 − spaces.

{P (a) + Q(a) = {1}P (a)Q(a) = {0} for all a.

(56)

For example,

If P (0) is given as

P (0) =

1

1

0

= Q⊥(∞) (57)

54

then Q(∞) is following.

Q(∞) =

0

0

1

= P⊥(0). (58)

Similarly

Let P (∞) is given as

P (∞) =

1

1

0

= Q⊥(0) (59)

then Q(0) is following.

Q(0) =

0

0

1

= P⊥(∞) (60)

So we have

P (0) = P (∞) , Q(0) = Q(∞)

in L1-spaces.

If P (∞) is given as

P (∞) =

0

0

1

= Q⊥(0) (61)

then Q(0) is following.

55

Q(0) =

1

1

0

= P⊥(∞) (62)

In this time, we have P (0) = Q(0) and P (∞) = Q(∞). It contradicts

for the properties of projection.

So we should consider

P (0) = {1, 0} = P (∞) , Q(0) = {0, 1} = Q(∞) , respectively.

and

P (0) = Q⊥(∞) = {1, 0} , Q(0) = P⊥(∞) = {0, 1} . . . L2 − spaces.

P (0) = P (∞) = {1, 1} , Q(0) = Q(∞) = {0, 0} . . . L1 − spaces.

{P (a) + Q(a) = {1, 1}P (a)Q(a) = {0, 0} for all a.

(63)

< P (a), Q(a) >= 0

P (a) ⊥ Q(a).

Precisely, the orthogonal condition of P (a) and Q(a) have to represent

on no defined. So the based condition for orthogonal system is following.

P (a) =

a0

a0

a0

= P⊥(a) (64)

Therefore

P (a)P⊥(a) =

a0

a0

a0

a0

a0

a0

=

a0

a0

a0

= I(a) = identitical.(65)

56

Simirarly

Q(a) =

1− a0

1− a0

1− a0

= Q⊥(a) (66)

Therefore

Q(a)Q⊥(a) =

1− a0

1− a0

1− a0

1− a0

1− a0

1− a0

(67)

=

1− a0

1− a0

1− a0

= I(a) = identitical. (68)

So we have

P (a)Q(a) =

a0

a0

a0

1− a0

1− a0

1− a0

=

0

0

0

= 0(69)

Hence

P (a) ⊥ Q(a)

lF (a) ⊥ G(a)

lT (a) ⊥ S(a).


57

Some results

◦G(a) operation is ideal that Pascal’s triangle operation F (a). In general,

G(a) = easG(a) = −eas

a0 − 1

a a0 − 1

a2 2a a0 − 1...

.... . .

annC1a

n−1nCka

n−k a0 − 1

(70)

for all a.

This G(a) operation is able to treat on L1 and L2-spaces.

◦ Similarly with T (a)iso←→ F (a), we have following form.

G(a)iso←→ S(a) , H(a)

iso←→ R(a)

◦ We are able to have following form in L2-spaces.

F(∞) = 0 = G⊥(0) ↔ T (∞) = 0 = S⊥(0)

F(0) = I = G⊥(∞) ↔ T (0) = I = S⊥(∞)

◦ The projection operator is able to define following.

P (a) =

a0

a0

a0

Q(a) =

1− a0

1− a0

1− a0

(71)

For example,

P (∞) =

∞0

∞0

∞0

=

1

1

0

= Q⊥(0) (72)

◦ Projection operator includes identity. In this cace, the projection oper-

ator generates identical identity that is D(a) and H(a) operations.

◦ Since F(0) = D(0) then we are able to represent as

F(0) = P (0) = Q⊥(∞) = G⊥(∞).

58

Conclusion

Now, Pascal’s triangle matrix operation is genereated from power func-

tion of My Laplace transforms. At the first time, T (a)1 is most important

condition for generating the matrix form. In general,

T (a)1 =1

s= T (0)1 for all a.

in L1-spaces.

As extension for this T (a)1, we have following condition on this matrix

operation (F(a) operation). In this case,

T (a)1 =a0

s= a0T (0)1 for all a.

in L1,L2-spaces.

The core formula of this situation is following condition.

(Integral operation) e0·s = a0 (matrix operation)

for all a.

Pricisely speaking, we have to be to 00 from a0. However, the Pas-

cal’s triangle matrix operation is treating as a0. And it’s proper condition

for other theorems. T (a) operation is able to explain from this formula.

So we were able to have that T (a) operation is isomorphic with F (a)

operation. Since using the property then T (a) operation has following

conditions.

T (−a) = T (a)−1 (translation operators) , ‖T (a)‖ = 1 (identical)

T (0) = I (identity) as 00 def .= 1 , T (0) = T ∗(0) (Hermitian)

T (a)T (b) = T (a + b) (semi− groups), especially T n(a) = T (na), etc

(73)

Finally, if a = 0 for F (a) operation then we have projection oprator

as P (0). Identity (F (0) operation) is included in the projection operator.

So we are able to extend to other operations by using this projection

operators.

(Fri) 15.Apr.2011

Now, let′s go to next pepers with me!

59

References

[1] Bryan P.Rynne and Martin A.Youngson, Linear functional anal-

ysis, Springer, SUMS, 2001.

[2] P.M.Cohn, Springer, SUMS, An Introduction to Ring Theory.

SUMS, 2002.

[3] M.A.Naimark, Normed Rings, P.Noordhoff,Ltd, 1959.

[4] Iain T.Adamson, Introduction to Field Theory, Cambridge Uni-

versity Press, 1982.

[5] Charles E.Rickart, General theory of Banach algebras, D.Van

Nostrand Company,Inc,1960.

[6] Shoichiro Sakai, C∗−algebras and W ∗−algebras, Springer, 1971.

[7] Hille and Philips, Functional analysis and semi-Groups, AMS,

1957.

[8] J.L.Kelley · Isaac Namioka Linear topological spaces, D.Van

Nostrand Company,Inc, 1961.

[9] Takao Saito, Operator algebras for Laplace transforms,(reprint)

(Mon)1.Dec.2010 (the latest).

[10] Takao Saito, Rings and ideal structures for Laplace transforms,

(reprint) (Mon)10.Dec.2007.

Furthermore

[11] Micheal O Searcoid, Elements of Abstract Analysis, Springer,

SUMS, 2002.

[12] Israel Gohberg and Seymour Goldberg, Basic operator theory,

Birkhauser, 1980.

[13] Harry Hochtadt, Integral Equations, John & Sons,Inc, 1973.

[14] Emil G.Milewski, Rea’s Problem Solvers Topology, 1998.

[15] Irina V.Melnikova, Abstract Cauchy problems, Chapman, 2001.

[16] Paul L.Butzer Hubert Berens, Semi-groups of Operators and

Approximation, Springer, 1967

[17] Kosaku Yoshida, Functional Analysis, Springer, 1980.

60

[18] Richard V. Kadison John R.Ringrose Fundamentals of the the-

ory of Operator Algebras, AMS

[19] Dunford & Schwartz Linear operators I,II,III, Wiley.

61

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