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AND THE LAMBERT

If one looks at the iteration-

N[n 1]  a N [n] subject to N  a

one finds-

a aa N  a a , N  a a and N  a a

Whether or not this continued of an exponentiation converges or not depends very much on the form of ‘a’. If a=2 we get N=2, N=22=4, N=24=16, N=216=65536. So things clearly blow up as the iteration continues toward n equal to . On the other hand if a=1/2, we get N=1/sqrt(2) , N=(1/2)1/sqrt(2) and N[  ]=0.64118575.. Thus the iteration converges. Note that ‘a’ need not necessarily be real but rather can take on the complex value z=x+iy. The above iteration is commonly referred to as tetration indicating the fourth mathematical after , , and exponentiation. There are several different representations for tetration. These are-

N N N N 4 N  N  4

The arrows are perhaps the clearest notation indicating in this case that there are a total of four appearances of N. Sometimes the stacking of the Ns is also referred to as a power tower. Remember that the iteration process defined above differs from a multiplicative operation such as-

(2^2^2)^2  (4^2)^2  (16)^2  256

Tetration is not associative with this last operation since it reads-

2^(2^(2^2))=2^(2^4)=2^16=65536

and typically has values much larger than the former. The fact that the get very large rapidly with tetration makes this form of representation ideal for very . Many of you will be familiar with the (after which the familiar search engine Google is named). One defines a googol as –

10 1 googol  1010  10100

and the even larger number, the , as-

1010 1 googolplex=  1010  10google Although such numbers are huge they still are in the sense of the WWI soldiers ditty, a long long way to infinity. To reach infinity you need –

10  

Let us next turn our attention to all those numbers for which the iteration given above converges. For the purpose of generalization we will -

N=z=x+iy and a=A+iB

This yields-

x[n  1]  iy[n  1]  (A  iB) x[n]iy[n] subject to z  A  iB

and so has both real and imaginary parts. The first two iterations yield-

( AiB ) z  (A  iB)( AiB) and z  (A  iB)( AiB)

From this we see at once that the nth iteration is simply-

z[n]  (A  iB)  (n  1)

Since we are only interested in those numbers z and a for which the iteration converges, we can quickly calculate this converengce by noting z[n+1]=z[n] there. Thus we have the non-linear equation-

z  (A  iB) z

This equation lends itself to a particularly simple solution for the special case of A=0 and B=1. There we find-

z  exp(iz / 2) which, by Euler’s , is equivalent to-

x  exp(y / 2)cos(x / 2) and y  exp(y / 2)sin(x / 2)

These may be combined to yield the two implicit equations-

x 2  y 2  exp(y) and tan(x / 2)  y / x

Solving these two equations numerically produces the result-

x=0.438282936.. and y=0.36059247.. Thus the iteration convergers to-

z=0.438282936 +i 0.36059247

in the . The precise trajectory to the convergence point is given by-

z[n+1]=exp(iz[n]π/2)

If we make a point plot of all these iterations starting with an initial condition z=1 so that z=i, we find the following interesting three branch inward moving spiral-

The spiral is seen to move nicely to the convergence point. We first discovered this spiral some 15 years ago in our undergraduate complex variable course here at the University of Florida. It does not seem to matter where one starts the iteration as long as ‘a’ is set to i. The convergence point remains the same.

Different spiral patterns will be created when the form of a=A+iB is changed. We have discussed such trajectories in an earlier article.

Going back to the expression-

z  exp(iz / 2 one notes its resemblance to the Lambert function defined as-

W (u)  u exp(W (u))

So if we set iπz/2=-W(u), we get u=-iπ/2. Thus one finds –

 2  i z  W ( )  0.43828293672703211162  i0.36059247187138548596 i 2

The Lambert function may also be used to determine the final convergence point for any other starting point –

a=A+iB= exp(i)

We have the equation-

z  R z exp(iz )  exp{z(ln(R)  i )}

From this it follows that u=-(lnR+i) and-

1 z  W ( ln(R)  i ) ln(R)  i

If we now let a=1+i=sqrt(2)exp(iπ/4), we find the result-

z  0.6410264785 + i0.5236284612I

An iteration beginning with a=1/2 has A=1/2, B=0, R=1/2, and =0 . It produces the real convergence point-

1 z  W (ln(2))  0.64118574450498598451 ln(2)

As a final evaluation consider a=sqrt(2) so that R=sqrt(2) and =0. Here we have-

2 2 2  2 2    2  W ((1/ 2) ln(2))  2 ln(2)

It produces the interesting identity-

1 ln(2)  W (ln( ))  0.69314718055994530943… 2

April 2014