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 Powerful way to perform continuum sum tommy1729 Ultimate Fellow     Posts: 1,370 Threads: 335 Joined: Feb 2009 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 « Next Oldest | Next Newest »

 Messages In This Thread Powerful way to perform continuum sum - by JmsNxn - 08/10/2013, 09:06 PM RE: Powerful way to perform continuum sum - by tommy1729 - 08/11/2013, 02:00 AM RE: Powerful way to perform continuum sum - by JmsNxn - 08/11/2013, 04:05 PM RE: Powerful way to perform continuum sum - by tommy1729 - 08/11/2013, 07:31 PM RE: Powerful way to perform continuum sum - by JmsNxn - 08/11/2013, 08:18 PM RE: Powerful way to perform continuum sum - by tommy1729 - 08/11/2013, 10:29 PM RE: Powerful way to perform continuum sum - by tommy1729 - 08/11/2013, 11:05 PM RE: Powerful way to perform continuum sum - by JmsNxn - 08/12/2013, 07:17 PM

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