08/11/2013, 10:29 PM

I was looking for posts about the continuum product and uniqueness related to tetration (and in general the construction of superfunctions).

But I was a bit dissapointed. Not much has been said or concluded compared to other topics here (including myself). Im sure we know more than we said.

But to settle this here - which I feel is about time after all those years ! - I will reformulate the problem.

( base e is used here , analogues must exist for other bases )

Continuum product = exp( Continuum sum ln ) by def.

In analogue with integrals

To do a numeric integral we need to know a point where integral f(x) = 0. And without that point , the symbolic integral is only " generalized nonsense " in the sense that we do not understand the symbolic integral very well.

I think this clarifies what I mean : http://mathworld.wolfram.com/SoldnersConstant.html

Now back to the continuum sum.

we need a point where Continuum sum f(x) = 1. Also very well known as the " empty product ".

***

It is often problematic to take the Continuum product before this " empty product point ", let alone solve equations involving Continuum product and some derivatives (expecially with the restriction of being COMPLEX ANALYTIC )

( this has already been discussed by mike3 )

***

It is also desired that (super)functions have at most 1 real fixpoint for x > 0.

Now tetration should satisfy

sexp ' (x) dx = Continuum product [sexp(x)]

Now comes something that is often done wrong so pay attention

When ppl consider the product equation they do

sexp ' (3,2) = sexp(3,2)*sexp(2,2)*sexp(1,2)*sexp ' (0,2).

However when we by def take sexp(0)=0.

Ppl think its ok to go to "0,2" but its NOT OK IF they think sexp '(0,2) has to be Continuum product sexp(0,2).

Since sexp(1)=1 the CONTINUUM PRODUCT IS AT BEST VALID FOR x > 1.

( after the *empty product* as discussed above )

1) Now this ALSO requires ofcourse that sexp has only 2 fixpoints for x>=0 : {0,1}

2) It is also required that sexp ' (1) = 1 for the product to work. VERY convenient that this works well with the fixpoint condition 1).

Now a sexp(x) analytic for x>=1 and continu for x>=0

should then satisfy :

for x >= 1 : sexp ' (x) = Continuum product [sexp(x)]

More specific we need to include the remarks about the "numerical computation" and hence arrive at

for x >= 1 :

sexp ' (x) - sexp ' (1) = Continuum product[sexp(x)] - Continuum product[sexp(1)]

We know that sexp '(1) = Continuum product[sexp(1)] = 1 =empty product hence :

for x >= 1 :

sexp ' (x) = Continuum product (from 1 to x) [sexp(x)]

( since x is also > 0 we can write by using the q-derivative ...)

(for x >=1)

=> [sexp(q x) - sexp(x)]/[(q-1)x] = exp(Continuum sum (from 1 to x) [sexp(x-1)] )

( Notice we could use l'hopital here because we used q-derivative , this would lead to adding more continuum products in the equations , but its not certain that this could be helpfull )

So far the equation and first step to computations.

It is not clear to me that this would yield a solution that is also analytic FOR 0 < x =< 1 ??

However we are not finished.

How about uniqueness ? We wonder about uniqueness and might even desire it for computation of a solution to the equation. ( or for a proof of the consistancy of a method to solve it )

In order not to retype everything lets use f(x) and g(x).

f(x) = sexp(x) satisfying the equation (and properties) described above.

f(x) + g(x) is also "a sexp(x)" satisfying the equation and properties.

Now we can build the "complex" equation :

for x >= 1 and g(x) =/= 0

f ' (x) + g ' (x) = CP [ f(x) + g(x) ]

=> g ' (x) = CP [ f(x) + g(x) ] - CP [ f(x) ]

Notice if analytic continuation applies to g(x) we can show that from the above equation we must have g(0) = 0 however we already have that property. It might however play a role in non-tetration-dynamics.

Notice g(1) = 0 and g ' (1) = 0.

IF lim x -> 1 : g(x) / g ' (x) = 1

g ' (1) = CP[ f(1) + g(1) ] - CP [ f(1) ] = 0

As expected.

Note that IF Q > 1 : g ' (Q) = 0 then g ' (Q) = CP [ f(Q) + G(Q) ] - CP [ f(Q) ] = 0 HENCE

g ' (Q) = 0 => g(Q) = 0 WHICH contradicts the required properties !

Hence there is no such g(x) (apart from g(x) = 0) !!!

This means f(x) + g(x) is strictly between f(x) and x or strictly above f(x).

This restricts the " second sexp " that f(x) + g(x) can be !!

Lets call this " Lemma1729 " , " theorem1729 " or " property1729 "

( Im narcistic and did not use tex , forgive me. Naming this however might be usefull for the future talks about tetration )

NOW COMES THE FINAL KEY ARGUMENT :

f(1) + g(1) = 1

f(2) + g(2) = e

f(3) + g(3) = e^e

This must be true if g(x) is not identically 0 , BUT it contradicts "theorem1729" HENCE IT MUST BE THAT A C^1 G DOES NOT EXIST !!

**Tommy's continuum product theorem**

Q.E.D.

Now just a few simple integral transforms from James and we have a new method of tetration

Many thanks to the tetration forum and its members.

regards

tommy1729

But I was a bit dissapointed. Not much has been said or concluded compared to other topics here (including myself). Im sure we know more than we said.

But to settle this here - which I feel is about time after all those years ! - I will reformulate the problem.

( base e is used here , analogues must exist for other bases )

Continuum product = exp( Continuum sum ln ) by def.

In analogue with integrals

To do a numeric integral we need to know a point where integral f(x) = 0. And without that point , the symbolic integral is only " generalized nonsense " in the sense that we do not understand the symbolic integral very well.

I think this clarifies what I mean : http://mathworld.wolfram.com/SoldnersConstant.html

Now back to the continuum sum.

we need a point where Continuum sum f(x) = 1. Also very well known as the " empty product ".

***

It is often problematic to take the Continuum product before this " empty product point ", let alone solve equations involving Continuum product and some derivatives (expecially with the restriction of being COMPLEX ANALYTIC )

( this has already been discussed by mike3 )

***

It is also desired that (super)functions have at most 1 real fixpoint for x > 0.

Now tetration should satisfy

sexp ' (x) dx = Continuum product [sexp(x)]

Now comes something that is often done wrong so pay attention

When ppl consider the product equation they do

sexp ' (3,2) = sexp(3,2)*sexp(2,2)*sexp(1,2)*sexp ' (0,2).

However when we by def take sexp(0)=0.

Ppl think its ok to go to "0,2" but its NOT OK IF they think sexp '(0,2) has to be Continuum product sexp(0,2).

Since sexp(1)=1 the CONTINUUM PRODUCT IS AT BEST VALID FOR x > 1.

( after the *empty product* as discussed above )

1) Now this ALSO requires ofcourse that sexp has only 2 fixpoints for x>=0 : {0,1}

2) It is also required that sexp ' (1) = 1 for the product to work. VERY convenient that this works well with the fixpoint condition 1).

Now a sexp(x) analytic for x>=1 and continu for x>=0

should then satisfy :

for x >= 1 : sexp ' (x) = Continuum product [sexp(x)]

More specific we need to include the remarks about the "numerical computation" and hence arrive at

for x >= 1 :

sexp ' (x) - sexp ' (1) = Continuum product[sexp(x)] - Continuum product[sexp(1)]

We know that sexp '(1) = Continuum product[sexp(1)] = 1 =empty product hence :

for x >= 1 :

sexp ' (x) = Continuum product (from 1 to x) [sexp(x)]

( since x is also > 0 we can write by using the q-derivative ...)

(for x >=1)

=> [sexp(q x) - sexp(x)]/[(q-1)x] = exp(Continuum sum (from 1 to x) [sexp(x-1)] )

( Notice we could use l'hopital here because we used q-derivative , this would lead to adding more continuum products in the equations , but its not certain that this could be helpfull )

So far the equation and first step to computations.

It is not clear to me that this would yield a solution that is also analytic FOR 0 < x =< 1 ??

However we are not finished.

How about uniqueness ? We wonder about uniqueness and might even desire it for computation of a solution to the equation. ( or for a proof of the consistancy of a method to solve it )

In order not to retype everything lets use f(x) and g(x).

f(x) = sexp(x) satisfying the equation (and properties) described above.

f(x) + g(x) is also "a sexp(x)" satisfying the equation and properties.

Now we can build the "complex" equation :

for x >= 1 and g(x) =/= 0

f ' (x) + g ' (x) = CP [ f(x) + g(x) ]

=> g ' (x) = CP [ f(x) + g(x) ] - CP [ f(x) ]

Notice if analytic continuation applies to g(x) we can show that from the above equation we must have g(0) = 0 however we already have that property. It might however play a role in non-tetration-dynamics.

Notice g(1) = 0 and g ' (1) = 0.

IF lim x -> 1 : g(x) / g ' (x) = 1

g ' (1) = CP[ f(1) + g(1) ] - CP [ f(1) ] = 0

As expected.

Note that IF Q > 1 : g ' (Q) = 0 then g ' (Q) = CP [ f(Q) + G(Q) ] - CP [ f(Q) ] = 0 HENCE

g ' (Q) = 0 => g(Q) = 0 WHICH contradicts the required properties !

Hence there is no such g(x) (apart from g(x) = 0) !!!

This means f(x) + g(x) is strictly between f(x) and x or strictly above f(x).

This restricts the " second sexp " that f(x) + g(x) can be !!

Lets call this " Lemma1729 " , " theorem1729 " or " property1729 "

( Im narcistic and did not use tex , forgive me. Naming this however might be usefull for the future talks about tetration )

NOW COMES THE FINAL KEY ARGUMENT :

f(1) + g(1) = 1

f(2) + g(2) = e

f(3) + g(3) = e^e

This must be true if g(x) is not identically 0 , BUT it contradicts "theorem1729" HENCE IT MUST BE THAT A C^1 G DOES NOT EXIST !!

**Tommy's continuum product theorem**

Q.E.D.

Now just a few simple integral transforms from James and we have a new method of tetration

Many thanks to the tetration forum and its members.

regards

tommy1729