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(06/05/2014, 01:17 AM)JmsNxn Wrote: (06/05/2014, 12:45 AM)mike3 Wrote: I decided to test your integral plugging the in from the Kneser super-logarithm as calculated via other methods. If it works you should get that super-logarithm back out.

However, I wasn't able to quite get it to converge on a numerical test for . The infinite-sum-defined function exhibits what is probably a very slow and very rapidly (tetrationally, I bet! )-increasing period oscillation, and so I'm not sure how well the negative-power factor in the integrand damps it out, i.e. if it damps it out enough to converge.

I want to point out that the Kneser super-logarithm has singularities at the fixed points of the logarithm and . Therefore, on the boundary of the half-plane , there are two singularities. These are logarithmic singularities and so the function is exponentially unbounded on that half-plane. If you need a tight (and not just asymptotic) bound, try a half-plane for some . Then there are no singularities on the boundary and the function is exponentially-bounded on the whole half-plane.

Yep that should fix it, makes a lot more sense.

And as to tommy's qblunt statement. lol, you're wrong.

and as well:

Similarly:

where is an int

Now are you going to doubt the one to one nature of the fourier transform?

SO L is the fixpoint.

that solves 1).

I assume you agree on 2).

So i was wrong with 3) ?

I assume that is what you meant.

Are these integrals showing 3) is wrong ?

Sorry I might be wrong. I wasnt convinced of 3) in the first place.

regards

tommy1729

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It seems you are once again trying to use Ramanujan's master theorem.

As for the remark 3) :

Let N,M,i,j be integers.

we have slog(z+ 2 pi j i) = slog(z).

if tet (a_i) = b_i and tet ' (a_i) = 0

then slog(z) is not analytic at b_i.

it follows slog(z) is also not analytic at b_i + N 2pi i.

also tet ' (a_i + M) = 0

tet(a_i + M) = exp^[M](b_i) which is chaotic most of the time !!

SO your halfplane cannot contain all those points nor the fixpoints L.

Also none of the complex conjugates of those !!!

hence your half-plane is parallel to Re(z) > Q if it exists at all.

That was the reason for my comment about 3).

Maybe a link to H trapmann's paper is usefull.

I hope I have not offended you and you understand my viewpoint (now).

regards

tommy1729

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06/05/2014, 01:25 PM
(This post was last modified: 06/05/2014, 01:34 PM by JmsNxn.)
(06/05/2014, 12:25 PM)tommy1729 Wrote: It seems you are once again trying to use Ramanujan's master theorem.

As for the remark 3) :

Let N,M,i,j be integers.

we have slog(z+ 2 pi j i) = slog(z).

if tet (a_i) = b_i and tet ' (a_i) = 0

then slog(z) is not analytic at b_i.

it follows slog(z) is also not analytic at b_i + N 2pi i.

also tet ' (a_i + M) = 0

tet(a_i + M) = exp^[M](b_i) which is chaotic most of the time !!

SO your halfplane cannot contain all those points nor the fixpoints L.

Also none of the complex conjugates of those !!!

hence your half-plane is parallel to Re(z) > Q if it exists at all.

That was the reason for my comment about 3).

Maybe a link to H trapmann's paper is usefull.

I hope I have not offended you and you understand my viewpoint (now).

regards

tommy1729

Really? Hmm. I was under the impression it would be analytic in a half plane. If it's not then it fails completely. Sorry if I was rude ^_^, I was just excited. The power of using the weyl differintegral for recursion just gets me giddy.

I am using ramanujan's master theorem, but I've expanded on it more and reworded the result as a result in fractional calculus, I also proved it using a different method.

Luckily, if slog is non zero in a half plane but with poles, we can do the exact same procedure with

instead, if it has no poles and is exponentially bounded.

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Well I dont know if you are wrong , Im just suggesting it might not be correct.

Im waiting for other experts to join the discussion. I merely expressed my doubts with arguments in the last few posts.

I dont know what Trappmann said about it and if for instance Trappmann , mike3 and sheldon support your claim that makes quite a strong case.

Im just in the state of " mathematical and general discussion ".

Im under the impression you use stronger versions of some theorems like carlson etc.

Im no expert but I have doubts on such things , in particular combined with uniqueness conditions.

For instance , I believe you said a few times that a function is determined completely by the values at integers if |f(z)| is bounded by an exponential.

That looks like a stronger version of carlson.

Again Im no expert in complex analysis , but another thing is Ramanujan's master theorem.

I have Always been told to use it with great caution.

This is mainly because the sequence f(1) , f(2) , f(3) , ... does not easily define f(-1) uniquely. For instance there many function that interpolate the factorials , but there is only one gamma function.

Then again , you might know more about this master theorem than I do. But the feeling remains with these theorems , IN PARTICULAR because not all the necc conditions are proven yet.

Maybe You use a theorem that generalizes both Carlson and Ramanujan ?

However , and forgive me if this is identical to one of your ideas , but if in the master theorem the Taylor coefficients are an interpolation of exp^[1/2](n) that is entire and has all derivatives at 0 > 0...

Then carlson's theorem applies and we have uniqueness ... therefore we have from the master theorem the values of the fake ( but correct at positive integers) exp^[1/2](-n).

At least thats what I think.

That way It relates to recent ideas of myself and sheldon and the development of " fake function theory ".

But I dont want to force you to think in that way.

You seem to want the correct tet(z),slog(z) etc with its singularities and stuff.

However its also those singularities and stuff that prevents you from being able to use FORMALLY many theorems about analytic functions or entire functions ... So that seems to make things hard.

NOW if exp^[M](b_i) is NOT chaotic , for instance cyclic that might be controllable.

But statistically speaking its chaotic , so the measure of your potential solutions seems 0 in the complex plane.

But that is no disproof.

My understanding of slog is not complete.

I believe that in the master theorem with f(x) the function considered for Taylor development and a(z) its Taylor coefficients ( a(n) ) then

f(z) is analytic for Re(z) >= 0 and a(z) is entire and unique by carlson then the master theorem ALWAYS applies.

SO i assume these conditions to be sufficient.

But I wonder if they can be relaxed.

Maybe f(z) only needs to be analytic in a strip near the real line.

Or maybe a(z) only needs to be analytic in a half-plane containing the real line.

I guess that is in the books.

This also might relate to the continuum sums ... but I think I have said enough.

Just want to add that it makes me wonder how the entire function f(z) given by the Taylor series below looks like.

f(z) = SUM fake exp^[1/2](n)/n! z^n

regards

tommy1729