Lambert w-function

When playing around with my TI-83+ I found out that the
the sequence a(n) defined by the recursive relation a(n+1) = x*e^-a(n) where a(1) can be anything and x>=-1/e has the limit W(x) when n-->inf and W is the Lambert w-function.

I could get it to converge with -1/e < e < 7/2. With x=-1/e it seems to simply oscillate and never converge and with numbers larger than 7/2 it would converge incredibly slowly. The calculators accuracy probably isn't enough to make it converge with larger values.

I conjecture that the sequence does converge to W(x) for every value x>-1/e, perhaps even for -1/e.

We need a proof or disproof of this.

Wait a second.

a(n+1)=x*e^-a(n) <=> a(n+1)e^a(n)=x

So, if a(n) does converge then for the limit, a, applies that ae^a=x <=> W(x)=a.

We just have to prove that a(n) converges for all values x>=-1/e.

I suspect this may be related
http://en.wikipedia.org/wiki/Tetration#Extension_to_infinite_heights

Indeed it obviously is related but Wikipedia doesn't have proper proofs.

All that's necessary is to prove that for every x>=-1/e the sequence a(n) is a Cauchy sequence.

It boils down to evaluating |a(n+1)-a(n)|=|xe^-a(n)-a(n)|.

The radius of convergence is e you retards not 7/2.

You are using an ancient and horrible algorithm for finding a fix point of function (point where x=f(x)), it is slow and has horrible radius.

It will diverge when |f'(x)| > 1, solve for x or see first line.

Make a ti-basic program like this to play with

x = 'initial value'  # Choose whatever you want
while xe^x-c
x = ce^-x            # This is fixpoint alg for ce^-x
disp x               # Move this to after end if you want
end

It gives you x = W(c) for c in (-exp(-1),exp(1)).

Now I shall return to my cool life of totally having lots of sex and having many friends, fuck yeah.

Uh, yes that's exactly what I did with my calculator some years ago.

And I'm not interested in the practical value of this sequence, that's completely irrelevant. I'm interested in the proof for the fact that the radius of convergence is e (as you say it is).

So, you completely missed the point.

|f'(x)| > 1 as I said, solve for x.

http://en.wikipedia.org/wiki/Fixed_point_iteration

So for a fixed x, W(x) is a fixed point of the function f(a) = x e^-a, so as described in that article, repeatedly applying f should converge to W(x).  It doesn't say anything about convergence there, but it might somewhere else.

So, you use the Banach fixed point theorem. That requires f to be a contraction, which it ought to be since the iteration does converge to where f(a)=a <=> x = ae^a.

And the iteration only converges for x in [-1/e,e], thus the fixed point a=W(x) is in [-1,1].

To prove convergence we need to show that the Lipschitz constant (I assume it is Lipschitz continuous) of f is <1 when x is in [-1/e,e]. Once shown, the Banach fixed point theorem says that for f:R-->R there exists an unique fixed point a=f(a) that is the limit of the iterative sequence.

But how do we prove that?

The lipschitz constant is the max of the absolute value of the derivative on [-1,1] (basically because of MVT).  So you can calculate it from that, I think.

But that maximum depends on x: it's |x|*e which is >1 only when x is in ]-1/e,1/e[. So x=-1/e and x in [1/e,e] have to considered separately? Looking at the derivative on [-1,1] is no longer useful, the domain has to be altered to lower the max. This is getting too complicated.

Also, the Banach FPT has the assumption that the domain and codomain of the contraction are the same perfect set. With real numbers that means it has to be a closed and bounded set.