Example One

Topology Algebra Differential Equations Integrals Complex example

Topology

Multiplication by an axis.

Let us note the following formulas:

E xy ϕ ( y ) = X × E xy ϕ ( y )

E xy [ ϕ ( x ) ⩔ ϕ ( y ) ] = [ E x ϕ ( x ) × Y ] ⋃ [ X × E y ϕ ( y ) ] E xy [ ϕ ( x ) ⩓ ϕ ( y ) ] = [ E x ϕ ( x ) × Y ] ⋂ [ X × E y ϕ ( y ) ] = E x ϕ ( x ) × E y ϕ ( y )

Algebra

Example

f ( x ) = ( x + y ) 3

= ∑ n = 0 3 ( 3 n ) x 3 - n y n

by the Binomial theorem = x 3 + 3 x 2 y + 3 x y 2 + y 3

Example

Let A = ( 1 2 3 4 5 6 7 8 9 ) .

The characteristic polynomial is equivalent to the determinant of the matrix B = A - λ I .

det B = det ( 1 - λ 2 3 4 5 - λ 6 7 8 9 - λ ) = 0

We can compute det B by using cofactors:

( 1 - λ ) ‖ 5 - λ 6 8 9 - λ ‖ - 2 ‖ 4 6 7 9 - λ ‖ + 3 ‖ 4 5 - λ 7 8 ‖ = 0

The roots of this cubic polynomial are thethree eigenvalues of A: λ 1 = 0 , λ 2 = 3 2 ( 5 - 33 ) , λ 3 = 3 2 ( 5 + 33 )

Differential Equations

Suppose v ≥ 2 and ζ 1 , ⃛ , ζ v - 1 ∈ C 2 ( G ) . Then

∑ α = 1 v ( - 1 ) α ∂ ∂ x α [ ∂ ( ζ 1 , ⃛ , ζ α - 1 , ζ α , ⃛ , ζ v - 1 ) ∂ ( x 1 ,⃛ , x α - 1 , x α +1 , ⃛ , x v - 1 ) ] = 0 .

Integration

We are concerned with the equation

d 2 x d t 2 + ( a + ϵcos t ) x = 0

We shall carry out a complete study only for a = ω 2 > 0 The equation is then equivalent to dX dt =AX+εP(t)X where

A= ( 0 1 - a ω 2 0 ) and P ( t ) = ( 0 0 - cos t 0 )

We shall carry calculate an approximation to Φ ϵ ( t , 0 ) by the method of section 1.1 . Thus let

Φ ϵ ( t , 0 )= exp (At)+ε Φ 1 (t)+ ϵ 2 Φ 2 (t)+…

We shall calculate exp (A2Π)+ε Φ 1 (2Π)+ ϵ 2 Φ 2 (2Π), the beginning of the expansion of Φ ϵ ( 2Π , 0 ), and then an approximation to the second order in ε of tr Φ ϵ ( 2 Π , 0 ). We know that

exp [A(t-s)]= ( cosω ( t - s ) 1 ω sinω ( t - s ) - ωsinω ( t - s ) cosw(t-s) )

and

tr expΦ ϵ ( A 2 Π)=2cos 2ωΠ

Let us calculate

Φ 1 (t)= ∫ 0 t exp [A(t-s)]P(s)exp(As)ds

= ∫ 0 t ( - 1 ω sinω ( t - s ) cosωs coss - 1 ω 2 sinω ( t - s ) sinωs coss - cosω ( t - s ) cosωs coss - 1 ω cosω ( t - s ) sinωs coss ) ds

Thus tr Φ 1 (2Π)= - 1 ω ∫ 0 2 Π sin 2 Πωcoss ds

Complex Example

Exampl1

( ∑ i ∈ n a l i x i ) det K ( t = 1 , x 1 , ⃛ , x n ; l ‖ l )

= ( Π i ∈ n x i ⩓ ) ∑ I ⊆ n - { l } ( - 1 ) ‖ I ‖ A ( λ ) ( I ‖ I ) det A ( λ ) ( I _ ⋃ { l } ‖ I _ ⋃ { l }).

Example2

d 2 P d Ω s d ω s = r e 2 ∫ V < S i > d 3 r ∫ ‖ e ⩓ . Π ↔ . e ⩓ ‖ 2 k 2 fδ ( k . v - ω ) d 3 v

= r e 2 ∫ V < S i > d 3 r ∫ ‖ 1 - ( 1 - s ⩓ . i ⩓ ) ( 1 - β i ) ( 1 - β s ) β e 2 ‖ 2 ‖ 1 - β i 1 - β s ‖ 2

× ( 1 - β 2 ) fδ( k . v - ω ) d 3 v