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A Note: Playing With Elementary & Contours + Imagery

John Gill Summer 2017

Abstract: Starting from a rudimentary definition of functional and basic complex variable theory several examples of functional integrals and their values are derived. In addition, images of the complex plane in which each point is associated with a functional integral are presented. This is not the functional integral used in physics, nor are these examples related to applications.

From Wikipedia: “ In an ordinary integral there is a to be integrated (the integrand) and a region of space over which to integrate the function (the domain of integration). The process of integration consists of adding up the values of the integrand for each point of the domain of integration . . . where the domain of integration is divided into smaller and smaller regions. For each small region, the value of the integrand cannot vary much, so it may be replaced by a single value. In a functional integral the domain of integration is a space of functions. For each function, the integrand returns a value to add up.”

b n

The common integral fxdx()= Lim∑ fc ()kn, ∆ x kn , , where x is the (real) variable of ∫ n→∞ a k=1 n integration and f( x ) is the integrand. Π[a , b ] = x is a partition of [a , b ] and n{ k, n }k=1

∆xkn, = x k+ 1, n − x kn , with xkn,≤ c kn , ≤ x k+ 1, n . All the subintervals shrink to 0 as n → ∞ .

On the other hand, the integral ∫F[ f ] dµ T has as its variable of integration “points” in a S S , i.e., functions f∈ S . And the “subintervals” are sets (σ − algebras ) of functions requiring a non-negative definition of “size” or “”, µT , paralleling the concept. F[ f ] is a functional operating on a function f∈ S and returning a real or for each such function.

I start from this very rudimentary and simple definition of functional integral and not from the usual interpretation seen in path integrals. Also, when measures are discussed I include pre- measures and cylinder measures under this rubric. The space S will be a space of defined contours in the complex plane.

In order to establish a functional integral in the context of sets of contours, S , we need a way of quantifying or measuring sets, T , within S , and a designation of a collection of such sets dependent upon n; also, a reasonable way of extracting an individual contour, z( t ) , within each set and assigning a numerical value to it. We want the “size” of each of these sets to diminish to 0 as n → ∞ . Thus the “functional integral” here is a hybrid concept.

Example : S :{ztzttit():()=+ρ , 0 ≤≤ ρ 1, t ∈ [0,1] } . Think of these contours as vectors in the first quadrant. Assume the unit square is divided into n× n squares. Focus on the right-most k k +1 vertical line, divided into equal subintervals. Let Tk, n ={(): z t n ≤ρ < n } . The simplest measure

1 2 k is µT = . Now, define a functional as Fzt[ ( )]= 1 + i ρ , ρ = , where the contour k, n n kn, kn , kn , n k k +1 terminates at lowest point of the subinterval [n , n ) . Then

n k2  1 F[ z ]µ T= Lim∑ 1 +⋅≈  .5417, n = 10,000 ∫ n→∞ 2 S k=1 n  n

This particular contour from Tk, n is representative of all contours in this set since

n n n 2 2k − 1 11+−ρ 2 1 1 +≤k →→∞ 0, n n∑()k, n n ∑()n2 ∑ 3 k=1 k = 1 k = 1 n

Of course, this is a in disguise.

Let the set Sz be a space of contours ζ s (t ) with t∈[0,1], s ∈ [0,1] . s denotes a family of individual contours originating at the point z , and t ranging from 0 to 1 describes a contour. The first choice of a contour space is

Sz:{ζ s ( t ): ζ s ( t )=+ z sSin (5 t ) − itCos (5 s ), z =+ 1 i } A collection of sets in this space is given by Tab(,)={ζ s ():0 t ≤≤<< asb 1 } . These cylinder sets generate a σ −algebra . The measure we apply is the traversed by the points

ζ s(1) over the s -intervals (the red and green contour in the image below). The next step is to designate a particular collection of sets

k k +1 Tk, n={ζ s (): tsn ≤< n } , kn:0 → , n → ∞ .

k+1 n   2 Then µTk, n= ∫ d ζ s (1) . For a functional on Sz , set Fζk= z − ζ k (1) for that particular n  n k n ⇒ contour from each subset. Then λ(z )= ∫ FdT [ ζµs ] λ (1+ i ) = ∫ FdT[ ζµs ] ≈ 2.2078 Sz S 1+i

An accurate representation would not show individual contours as s is a continuous variable.

z(1+ t ( e s − 1) ) Example : Let ζ ()t= , zxiy = + . In the image below z = 1+i. The measure s 1 −t2 () xs + iy and functional are the same.

Image over a region: Dark = small moduli , light = large. λ(z ) = ∫ F[ ζs ] dT µ : Sz

[-10,10] z Example : ζ (t )= , z = x + iy . The same details as in the previous s 1+t() Cos () xs − iSin () ys example. z=1+I in the first image. λ(z ) = ∫ F[ ζs ] dT µ : The image is over [-10,10]. Sz

z(1+ iSin () t Cos ( sxy ) ) Example : ζ s (t )= , z = x + iy z=1+I λ(z ) = ∫ F[ ζs ] dT µ : 1+t() Cos () xs − iSin () ys Sz

[-10,10] z Example : ζ s (t )= , z = x + iy , z=1+I, λ(z ) = ∫ F[ ζs ] dT µ : 1+t() sCos (5 x ) + iSin (5 y ) Sz

[-10,10] z( Cos(2 t )+ iSin (2 t ) ) Example : ζ s (t )= , z = x + iy , z=1+I λ(z ) = ∫ d µ T : 1+ts() Cos () xy + iSin () xy Sz

[-10,10]

A contour measure :

Example : ζ s()2t= t + is ( + 1) Sin (2) st . Let Tab(,)={ζ s ():0 t ≤≤<< asb 1 } . This cylinder set generates a sigma-algebra. For a measure, set µTab( , )= Area between ζa ( t ) and ζ b ( t ) . And k k +1 for a functional, set F[ζb ( t )]= ζ b (1) . Specify Tk, n={ζ s ( t ): n ≤ s < n } . Then

n

F[ζµs ] dT = LimF∑ [ ζµ kn/ (1)] T kn , ≈ 3.2455 . ∫ n→∞ k=0 S0

nk+1  n k With µTk, n =+1() 1 − Cos (2) −+− 1() 1 Cos (2) k+ 1 n  k n

Contours are related to complex vector fields in the following way:

Given a vector f( z ) in C , to determine a formula for a contour, z( t ) , write dz =ϕ()z = fz () − z and solve for z( t ) . For instance, ϕ()zz= 2 ⇒ fzzz() = 2 + , leading to dt z z( t ) = 0 . 1− z0 t

This approach is taken in the following example where the focus is upon an attractor in a complex :

Example : Consider the set of contours in the f( z ) = z 2 . There are two fixed points in the space, α = 0 (attractor) and β = 1 (repulsor), that govern flow. Each circle centered at α = 0 is a locus of terminating points of contours z( t ), t ∈ [0,1] . Corresponding to * each such circle of radius rj (in red in figure) is a circle of radius rj (in blue) of initial points of

1 these contours. Let S be the space of contours of the form {z( t ): z (1) ≤ 2} .

k Now set ε =, 1 ≤k ≤ n . For the rings R about the origin, formed by consecutive radii k, n 2n k, n

εkn,, ε k+ 1, n , set Tkn,={ ztz ( ): (1) ∈ R kn , } . We have dz z 1  ⇒ 0 =ϕ(z ) = fzz ( ) − zt ( ) = t t . Partition D= z ≤  into diminishing rings dt zee0(1− ) + 2 

n

D= Lim∪ Rknkn,,, R =≤ zε kn ,∩ z > ε kn+ 1, . Now observe that n→∞ () () k=1

Tkn,={ ztz( ): (1) ∈ R kn , } corresponds in a one-to-one manner to Zkn,={ zz 0: (1) ∈ R kn , } . The two blue circles in the figure define the boundary of Zk, n .

An obvious choice for µTk, n would be the area between the blue circles. However, the algebra involved is unpleasant, so, having no application in mind, we default to the much simpler difference between the two circumferences: ck22+ ck 2 − 4 n 2 ∆C = 4 eπ n , c= e − 1 k, n ()()ck22−4 nck 22 ( + 1) 2 − 4 n 2

The remaining decisions are to designate a zkn,( t ) ∈ T kn , as n → ∞ and F[ z ( t )] in a way that

n

∑Fz[kn, ( t )] ⋅ µ T kn , is a proper mathematical object and is computable. Toward this end k=1 observe that very large values of n produce z0 ' s virtually on the circumference of a (blue)

circle Ck, n . So that all zkn,( t ) ∈ T kn , roughly correspond to points on this one circle. Hence, we choose

4eπ nk Fz[ ( t )] = C = kn, kn , ()c2 k 2− 4 n 2

Define

n n 22 2 2  2 2 2 kck(+ ck − 4 n )  FzdT[ ]µ= Lim∑ Fz [kn, ( t )] ⋅ µ T kn , = 16 e π nLim ∑ 2 . ∫ n→∞ n →∞ 22 2 2 2 2  S k=1 k = 1 ()()ck−4 n ck ( + 1) − 4 n 

And ∫FzdT[ ]µ ≈ 44.324, n = 100,000 . S

1 Example : The same as the one above, except ε =, 2 ≤ k →∞ . In this context the integrand k k

is a step-functional and the integral collapses to a common . Tk={ ztz( ): (1) ∈ R k } where ∩ Rk=( z ≤ε k) ( z > ε k +1 ) . Again, Tk corresponds to Zk={ zz0 : (1) ∈ R k } , the region between two blue circles in the figure. The choice of the area between the two circles as a measure of

Tk is reasonable as the algebra is fairly simple: 2 2 4 2 (kk (+ 1) − c )(2 k + 1) µTAek=∆= k π 2 2 , ce =− 1 . ()()kc22−( k + 1) 22 − c

2eπ k And Fzt[ ( )] = C = . k k ()k2− c 2

32 2 24  ∞ ∞ 2ekkkπ ( (+ 1) − c )(2 k + 1)  Hence FzdT[ ]µ=∑ Fzt [k ( )] ⋅ µ T k = ∑ 3 2 ≈396.47, n = 100 . ∫ 22 22  S k=1 k = 1 ()()kc−( k + 1) − c 