11/10/2015, 11:19 PM

Alot has already been said about the superroots.

We know that lim n-> oo for x_n^^n = y for y > exp(1/e) [eta] gives x_n = eta.

Also all results about slog and sexp relate.

PROOF SKETCHES

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First i point out that when you have a nonzero radius ,

the eulersum = analytic continuation whenever and wherever both converge.

BUT analytic continuation is USUALLY NOT the correct solution.

For instance

x^x^x^... = x^[oo] = y

Has solution x = y^(1/y)

IFF

Dom y , range x are in the sheltron region.

Clearly x=y^(1/y) is the analytic continuation , but thus false.

Tommy's lemma : for all n -> W•n(0) = 0.

Conjecture 3.1.

Conjecture 3.1 conjectures lim n -> oo

W•n (v) = -v exp(-v).

Clearly the RHS has radius oo.

But the algebra dictates that the radius can be at most :

X^(1/x) = exp(ln(x) / x).

Now v = ln(x) so x = exp(v).

Therefore exp( ln(x) / x) = exp ( v exp(-v) ).

Hence v exp(-v) = W•n(v) = - v exp(-v)

=> contradiction .. Unless for v satisfying

V exp(-v) = v exp(-v)

The solution set is v ={0,oo}.

So the radius of lim W•n = 0.

V= 0 implies x = 1.

X = 1 is within the sheltron region so v = 0 is valid.

Qed

A second proof

Let n = oo

The fractal argument :

W•n(v) = a

<=> v = (b exp ( b exp ( ...(*)) =

(b exp( b exp ( ... (a)).

So v = ( b exp(*))^oo.

A fractal within the sheltron.

V = fixpoint [b exp(*)]

==> solve b exp(A) = A.

==> A = - W(-b).

=> - W(-b) = v. --> b = v exp(-v) = W•n(v).

Similar too previous proof ; v must be 0 ==> radius = 0.

Qed.

Too explain the fractal argument

Notice (b exp(a))^[2] for fixed b and variabele a is NOT equal too

(a exp(a))^[2] .. Even if we set a = b !

For a = b , The difference is in the second case

a exp(a) exp( a exp(a) )

Whereas in the first case

a exp(a exp(a) ).

That is why I use a and b and then set them equal.

That looks confusing but is correct.

---

So we get 2 proofs with W•n(v) = v exp(-v).

However that is only valid within the sheltron.

V = ln(x) -> 0 = ln(1).

Min( | ln(1) - ln(e^(-1/e)) | , | ln(1) - ln(e^(1/e)) | ) = 1/e.

So W•n(v) = v exp(-v) is valid within " radius " 1/e.

And analytic continuation does not help.

Hope that helps.

Regards

Tommy1729

We know that lim n-> oo for x_n^^n = y for y > exp(1/e) [eta] gives x_n = eta.

Also all results about slog and sexp relate.

PROOF SKETCHES

---

First i point out that when you have a nonzero radius ,

the eulersum = analytic continuation whenever and wherever both converge.

BUT analytic continuation is USUALLY NOT the correct solution.

For instance

x^x^x^... = x^[oo] = y

Has solution x = y^(1/y)

IFF

Dom y , range x are in the sheltron region.

Clearly x=y^(1/y) is the analytic continuation , but thus false.

Tommy's lemma : for all n -> W•n(0) = 0.

Conjecture 3.1.

Conjecture 3.1 conjectures lim n -> oo

W•n (v) = -v exp(-v).

Clearly the RHS has radius oo.

But the algebra dictates that the radius can be at most :

X^(1/x) = exp(ln(x) / x).

Now v = ln(x) so x = exp(v).

Therefore exp( ln(x) / x) = exp ( v exp(-v) ).

Hence v exp(-v) = W•n(v) = - v exp(-v)

=> contradiction .. Unless for v satisfying

V exp(-v) = v exp(-v)

The solution set is v ={0,oo}.

So the radius of lim W•n = 0.

V= 0 implies x = 1.

X = 1 is within the sheltron region so v = 0 is valid.

Qed

A second proof

Let n = oo

The fractal argument :

W•n(v) = a

<=> v = (b exp ( b exp ( ...(*)) =

(b exp( b exp ( ... (a)).

So v = ( b exp(*))^oo.

A fractal within the sheltron.

V = fixpoint [b exp(*)]

==> solve b exp(A) = A.

==> A = - W(-b).

=> - W(-b) = v. --> b = v exp(-v) = W•n(v).

Similar too previous proof ; v must be 0 ==> radius = 0.

Qed.

Too explain the fractal argument

Notice (b exp(a))^[2] for fixed b and variabele a is NOT equal too

(a exp(a))^[2] .. Even if we set a = b !

For a = b , The difference is in the second case

a exp(a) exp( a exp(a) )

Whereas in the first case

a exp(a exp(a) ).

That is why I use a and b and then set them equal.

That looks confusing but is correct.

---

So we get 2 proofs with W•n(v) = v exp(-v).

However that is only valid within the sheltron.

V = ln(x) -> 0 = ln(1).

Min( | ln(1) - ln(e^(-1/e)) | , | ln(1) - ln(e^(1/e)) | ) = 1/e.

So W•n(v) = v exp(-v) is valid within " radius " 1/e.

And analytic continuation does not help.

Hope that helps.

Regards

Tommy1729