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• Lattice structures

for Laplace transforms

No.2

Takao Saito

1

• Thank you for our world

2

• Lattice struactures

for Lapalace transforms

3

• Preface

This paper has been presented about lattice structure for Laplace

transforms. The lattice form is able to represent as matrix conditions. In

general, lattice structure for Laplace transforms includes Pascal’s triangle

matrix. In this time, I want to explain about the two-dimensional matrix

forms. This matrices forms are satisfied the Cauchy problems. As a

whole, lattice form and Pascal’s matrix are isomorphic. Finally, I have

been extended to two-parameters for Laplace transforms.

Papers

About solvers of differential equations of relatively

for Laplace transforms

Operator algebras for Laplace transforms

Rings and ideal structures for Laplace transforms

Extension and contaction for transrated operators

Operator algebras for group conditions

Some matrices rings for Laplace transforms

Pascal’s triangle matrix for Laplace transforms

Now, let′s consider with me!

Chuo-ku Chiba-shi

Postcode 260-0844 Japan

URL: http://opab.web.fc2.com/index.html

(Sun) 24.Jul.2011 Takao Saito

4

• Contents

Preface

§ Chapter 4

◦ About relations of T (a) and T (0) · · · · · · 7 ◦ The matrix condition for lattice structures · · · · · · 15 ◦ Some results · · · · · · 17

§ Chapter 5

◦ The solution of lattice operator in two dimensions · · · · · · 18 ◦ Lattice and matrix operators for 2-dimensions (first form)· · · 21 ◦ Lattice and matrix operators for 2-dimensions (irreducible) 23 ◦ Lattice and matrix operators for 2-dimensions (second form) 25 ◦ Lattice and matrix operators for 2-dimensions (irreducible) 27 ◦ F(a) to L′(a) cobditions · · · · · · 29 ◦ Duality of L′(a) operation · · · · · · 30 ◦ Some results · · · · · · 32

§ Chapter 6

◦ Formulae for T (a, b) operation (particular case) · · · · · · 33 ◦ Formulae for F (a, b) operation · · · · · · 35 ◦ Formulae for L(a, b) operation · · · · · · 37

5

• F (a, b) (Pascal′s triangle matrix)

◦ Figure of basic operator algebras (Pascal’s triangle) · · · 40 ◦ Figure of basic operator algebras (Duality) · · · · · · 40 ◦ Basic operator algebras (Pascal’s triangle) · · · · · · 41 ◦ Basic operator algebras (Duality) · · · · · · 41 ◦ Basic normed operator algebras (Pascal’s triangle) · · · · · · 42 ◦ Basic normed operator algebras (Duality) · · · · · · 42

T (a, b) (Laplace transforms)

◦ Figure of basic operator algebras · · · · · · 44 ◦ Figure of basic operator algebras (Duality) · · · · · · 44 ◦ Basic operator algebras (Laplace transforms) · · · · · · 45 ◦ Basic operator algebras (Duality) · · · · · · 45 ◦ Basic normed operator algebras (Laplace transforms) · · · 46 ◦ Basic normed operator algebras (Duality) · · · · · · 46 ◦ T (a, b) and L(a, b) operations · · · · · · 47 ◦ Some results · · · · · · 49

◦ Conclusion · · · · · · 50

◦ Reference · · · · · · 51

6

• § Chapter 4

In this chapter, I want to explane the lattice strucrture of T (a) op-

erations. As the first step, T (a) operation could extend to seriese and

satisfied following.

N.B. In this case, T (a) is preserved from T (0).

Now, based system of T (a) operation is represented following.

T (a)f(t) = T (a) ∞∑

n=0

f(0)(n)

n! tn =

∞∑

n=0

f(0)(n)

n! T (a)tn

= ∞∑

n=0

f(0)(n)

n! {

n−1∑

k=0

(nCk)a n−k k!

sk+1 +

a0 · n! sn+1

}

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

∫ ∞ a

f(t)e−(t−a)sdt

= ∞∑

n=1

n−1∑

k=0

f(0)(n)

n!

n!

(n− k)!k!a n−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

= ∞∑

n=1

n−1∑

k=0

f(0)(n)

(n− k)! an−k

sk+1 +

∞∑

n=0

f(0)(n)

n!

a0n!

sn+1

On the other hand, we are able to have following formula.

a0 = e0·s for all a.

So we have

= ∞∑

n=1

n−1∑

k=0

f(0)(n)

(n− k)! an−k

sk+1 +

∞∑

n=0

f(0)(n)

n! e0·sT (0)tn

7

• = ∞∑

n=1

n−1∑

k=0

f(0)(n)

(n− k)! an−k

sk+1 + R(0)f(t) = −S1(a)f(t) + R(0)f(t)

N.B. We shall attend that the term of a0 is included in R(0) operation.

Becaurse an−k of first term is changed to a0 as n = k. Furthermore this a0 is changed to 00.

In this time,

R(0) = T (0).

N.B. R(a)f(t) def . =

∫ ∞ 0

f(t)e−(t−a)sdt = easT (0)f(t)

Therefore

T (a)f(t) = −S1(a)f(t) + R(0)f(t) = −S1(a)f(t) + T (0)f(t).

Hence

T (a)f(t) = ∞∑

n=1

n−1∑

k=0

f(0)(n)

(n− k)! an−k

sk+1 + T (0)f(t)

↓ ↓ ↓ ring ideal ring

(see, p.43, Chapter 2, No.7, N o.2) N.B. R(0) = T (0)

Of caurse it’s a compact operators.

8

• T (a) ∫ tν 0 · · ·

∫ t2 0

∫ t1 0

f(t)

t dtdt1 · · · dtν−1 =

∞∑

n=1

n+ν−2∑

k=0

f(0)(n)

n(n + ν − 1− k)! an+ν−1−k

sk+1 +

1

sν T (0)

f(t)

t

T (a) ∫ t2 0

∫ t1 0

f(t)

t dtdt1 =

∞∑

n=1

n∑

k=0

f(0)(n)

n(n + 1− k)! an+1−k

sk+1 +

1

s2 T (0)

f(t)

t

T (a) ∫ t1 0

f(t)

t dt =

∞∑

n=1

n−1∑

k=0

f(0)(n)

n(n− k)! an−k

sk+1 +

1

s T (0)

f(t)

t

T (a) f(t)

t =

∞∑

n=1

n−1∑

k=0

f(0)(n+1)

(n + 1)(n− k)! an−k

sk+1 + T (0)

f(t)

t

T (a)[ f(t)

t ]′ =

∞∑

n=1

n−1∑

k=0

f(0)(n+2)

(n + 2)(n− k)! an−k

sk+1 + sT (0)

f(t)

t − f ′(0)

T (a)[ f(t)

t ]′′ =

∞∑

n=1

n−1∑

k=0

f(0)(n+3)

(n + 3)(n− k)! an−k

sk+1 + s2T (0)

f(t)

t − sf ′(0)− f(0)

′′

2

T (a)[ f(t)

t ](ν) =

∞∑

n=1

n−1∑

k=0

f(0)(n+ν+1)

(n + ν + 1)(n− k)! an−k

sk+1 +sνT (0)

f(t)

t −

ν∑

k=1

sν−k f(0)(k)

k

N.B. Now T (a) operation is preserved with T (0).

9

• Similarly

T (a) ∫ tν 0 · · ·

∫ t2 0

∫ t1 0

f(t)dtdt1 · · · dtν−1 = ∞∑

n=1

n+ν−2∑

k=0

f(0)(n−1)

n(n + ν − 1− k)! an+ν−1−k

sk+1 +

1

sν T (0)f(t)

T (a) ∫ t2 0

∫ t1 0

f(t)dtdt1 = ∞∑

n=1

n∑

k=0

f(0)(n−1)

(n + 1− k)! an+1−k

sk+1 +

1

s2 T (0)f(t)

T (a) ∫ t1 0

f(t)dt = ∞∑

n=1

n−1∑

k=0

f(0)(n−1)

(n− k)! an−k

sk+1 +

1

s T (0)f(t)

based system · · · T (a)f(t) = ∞∑

n=1

n−1∑

k=0

f(0)(n)

(n− k)! an−k

sk+1 + T (0)f(t)

T (a)f ′(t) = ∞∑

n=1

n−1∑

k=0

f(0)(n+1)

(n− k)! an−k

sk+1 + T (0)f ′(t)

T (a)f ′′(t) = ∞∑

n=1

n−1∑

k=0

f(0)(n+2)

(n− k)! an−k

sk+1 + T (0)f ′′(t)

T (a)f (ν)(t) = ∞∑

n=1

n−1∑

k=0

f(0)(n+ν)

(n− k)! an−k

sk+1 + T (0)f (ν)(t)

N.B. Now T (a) operation is preserved with T (0).

10

• And

T (a) ∫ tν 0 · · ·

∫ t2 0

∫ t1 0

tf(t)dtdt1 · · · dtν−1 = ∞∑

n=0

n+ν−1∑

k=0

f(0)(n−1)

(n + ν − k)! an+ν−k

sk+1 +

1

sν T (0)tf(t)

T (a) ∫ t2 0

∫ t1 0

tf(t)dtdt1 = ∞∑

n=0

n+1∑

k=0

nf(0)(n−1)

(n + 2− k)! an+2−k

sk+1 +

1

s2 T (0)tf(t)

T (a) ∫ t1 0

tf(t)dt = ∞∑

n=0

n∑

k=0

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