What bothered me was the claim that it had to be nowhere analytic.

You have not given proof of that.

It read as if it was a fact.

Besides divergeance on the complex plane is dubious IF you compute the way I DO USE 2sinh.

Im not angry but I cant agree with this without seeing a proof or the words conjecture.

Im not sure how post 61 relates.

Bottomline is every statement should be followed by the words proof or conjecture.

Nowhere analytic is a big claim.

And besides technically you compute things differently .. Although it might be equivalent (or locally equivalent).

Therefore i posted my way of computing again.

Regards

Tommy1729

(12/01/2015, 01:22 PM)tommy1729 Wrote: [ -> ]Im not angry but I cant agree with this without seeing a proof or the words conjecture.

Im not sure how post 61 relates.

Bottomline is every statement should be followed by the words proof or conjecture.

Nowhere analytic is a big claim.

And besides technically you compute things differently .. Although it might be equivalent (or locally equivalent).

Therefore i posted my way of computing again.

It was the sequence leading up to your post#68 (not #61 sorry), "I think the same applies to the base change ... I do not immediately know a way around it." I like my post#65, showing how the Taylor series abruptly changes again somewhere near the somewhere around the 1,352,620th taylor series term. As far as proving the conjecture that the basechange sexp or the TommySexp, or tommy's half exp are nowhere analytic... nope, that hasn't happened. Perhaps someone with a real mathematics education, who is more brilliant than you or I can step in. In the mean time, the calculations I have done, along with posts by Henryk, and Mike are all in agreement.

For now I am done with this conversation. Attempts at proving a non-zero radius of convergence for all counter examples have failed so far, and I don't see anything new. Nor do we have a discussion of how one might proceed on a mathematically rigorous proof. So I see little chance of learning anything useful, or having a productive exchange. So I'm done for now, and I am unlikely to post anything further. Have a good day.

Ok.

Enjoy your day Sheldon.

Sorry I misinterpreted your posts a bit.

Hope my posts are clear.

Regards

Tommy1729

(12/01/2015, 02:58 PM)sheldonison Wrote: [ -> ] (12/01/2015, 01:22 PM)tommy1729 Wrote: [ -> ]Im not angry but I cant agree with this without seeing a proof or the words conjecture.

Im not sure how post 61 relates.

Bottomline is every statement should be followed by the words proof or conjecture.

Nowhere analytic is a big claim.

And besides technically you compute things differently .. Although it might be equivalent (or locally equivalent).

Therefore i posted my way of computing again.

It was the sequence leading up to your post#68 (not #61 sorry), "I think the same applies to the base change ... I do not immediately know a way around it." I like my post#65, showing how the Taylor series abruptly changes again somewhere near the somewhere around the 1,352,620th taylor series term. As far as proving the conjecture that the basechange sexp or the TommySexp, or tommy's half exp are nowhere analytic... nope, that hasn't happened. Perhaps someone with a real mathematics education, who is more brilliant than you or I can step in. In the mean time, the calculations I have done, along with posts by Henryk, and Mike are all in agreement.

For now I am done with this conversation. Attempts at proving a non-zero radius of convergence for all counter examples have failed so far, and I don't see anything new. Nor do we have a discussion of how one might proceed on a mathematically rigorous proof. So I see little chance of learning anything useful, or having a productive exchange. So I'm done for now, and I am unlikely to post anything further. Have a good day.

Im wearing a T-shirt with nr 68 on it.

Is it a sign ? Haha. Just noticed now.

Regards

Tommy1729

Too avoid confusion

R and x real and

R >= 0

X >= 0

A = analytic continuation with respect to x.

L = limit with respect to n going to oo.

Then 2sinh method for base e :

A L A ln^[n]( 2sinh^[R](exp^[n](x)) )

---

Sheldon Ideas are USUALLY focused on variations and making R complex or analytic continuation for R ( or claiming it cant exist ).

I think that is fair to say.

Hope this sketchy summary does not Harm the subject or insult anyone.

But the goal is to clarity.

Regards

Tommy1729

(12/02/2015, 12:28 AM)tommy1729 Wrote: [ -> ]Too avoid confusion

R and x real and

R >= 0

X >= 0

A = analytic continuation with respect to x.

L = limit with respect to n going to oo.

Then 2sinh method for base e :

A L A ln^[n]( 2sinh^[R](exp^[n](x)) )

So then call this function TommySexp(R,x). The work I have done is with TommySexp(R,0) which is a base "e" sexp(z=R) function. While my computation algorithm differs, it gives the same results at the real axis. The resulting conjectures is that all of the Taylor series coefficients of TommySexp(R,0) converge, but the radius of convergence appears to changes in abrupt steps, with the radius of convergence getting arbitrarily small. The math is tricky, so I don't regard this as a completed proof, but its more than enough to convince me that the TommySexp is very likely not defined in the complex plane; and is nowhere analytic.

However, there is another usage, TommySexp(0.5,x) which is the half iterate. No calculations as to the radius of convergence have been done, and I don't think anything has been proven. However, we do know that TommySexp(1,x)=exp(x), so this should be studied more.

If one can show that beyond a given positive real , the m th derivative of the n th iteration ( ln(f_{n-1}(exp(x)) ) converges very fast , then I can complete the proof that it is analytic there.

Most powerful techniques depend on the assumption that things are analytic , such as cauchy's contour type integrals.

But assuming things to prove them is USUALLY bad in math.

This ALMOST forces us to mainly real analysis ...

But that seems harder ...

Regards

Tommy1729

If I had An infinite product for sinh^[r](z) or tommyexp(r,z) that would help.

Maybe a kind of Abelproduct ?

Regards

Tommy1729

In principle a strategy is simple.

Show D^n ln(f(exp(x))) is close to D^n f(x) for all n > -1.

Write f(x) in the product form prod_i (1 + x^i / a_i) and use log(a b) = log a + log b.

So we get ( ignoring radius , continuation , summability ,...)

D^n F(x) ~~ D^n Sum_i ln( 1 + exp( i x) / a_i).

Use Taylor for ln(1 + x^t) and ln(x^3) = 3 ln(x) etc.

Should work.

In principle ... In theory ...

However in practice this is near impossible !?

See here at Gottfried's paper

http://go.helms-net.de/math/musings/drea...quence.pdf
Also posted on MSE by our friend Mick

http://math.stackexchange.com/questions/...of-lnfexpx
Regards

Tommy1729

It seems the method has passed the Weierstrass M-test !!

Regards

Tommy1729