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Searching for an asymptotic to exp[0.5]
#31
(05/25/2014, 03:00 PM)sheldonison Wrote: ... Meanwhile, I have a new version I am experimenting with that will greatly reduce the discontinuity at the negative real axis, that may allow approximating both the Weiestrass zeros, and the error term for the convergence to the Kneser half iterate... but there will still be a discontinuity cut point at |L|; still working.

Well, version V works a lot lot better, but it I was hoping for something with more theoretical power. So there's this discontinuity for Kneser's half iterate at the negative real axis. Why not get rid of it, with some sort of mapping that still converges to the half iterate as real(z) increases? Well, I got rid of most of it; not all. Here exp^(0.5) refer's to Kneser's half iterate, with the real axis complex valued, following the cutpoints from above. Inside a circle of radius |L|, we just use the Kneser half iterate. Here is a log10/log10 chart of the relative error of v2,v3,v4,v5 vs the Kneser half iterate. At z=10^5, the error term is already -2.6E-27, where the old algorithm doesn't get that good until z=5.0E11.
   

Here is the algorithm for the version V half iterate approximation.


Now, half_v(-20) ~= -5.48222 - i3.778E-14, so the imaginary discontinuity is starting to get really insignificant. This is because at the negative real axis, we get

so we have cancelled most of the imaginary part, and are left with only the relatively small exp(half(z+3pi i)) term. The function is very nearly 2x the real part of the Kneser half iterate.

Now pretend this is an analytic function, and take the Taylor series of the function. To minimize errors, for a function with a very large magnitude range, use multiple Cauchy integrals as appropriate for different Taylor series terms. Also, it turns out the Taylor series function acts like a Laurent series, where the 1/x and 1/x^2 ... terms quickly decay to irrelevant as real(x) gets larger than 138, and then the oscillating takes over as the dominant error term, continually oscillating from slightly bigger than to slightly less than the Kneser half iterate. For the entire Taylor series, we drop the Laurent terms, and we have the Taylor series function below.
Code:
{halfv= 0.499100407682864282980350196595
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+x^279*  2.00800045291990379681136985289 E-3164
+x^280*  3.88852781952097451271692914060 E-3180
+x^281*  7.20045410377084607877619861129 E-3196
+x^282*  1.27509181724127031046616692488 E-3211
+x^283*  2.15964922735363615422176997757 E-3227
+x^284*  3.49895314766009754591249944461 E-3243
+x^285*  5.42323996238548370698411574285 E-3259
+x^286*  8.04262117412227823541530664764 E-3275
+x^287*  1.14131754981827406752997736453 E-3290
+x^288*  1.55001943773779939813788861284 E-3306
+x^289*  2.01484165993530008162353109238 E-3322
+x^290*  2.50708396515326564914382745069 E-3338
+x^291*  2.98656354216263237756653315717 E-3354
+x^292*  3.40643416590422644964249591711 E-3370
+x^293*  3.72052080393571704625140066739 E-3386
+x^294*  3.89164133408399578097076763129 E-3402
+x^295*  3.89885200481764336374706668619 E-3418
+x^296*  3.74166422420294635127661949707 E-3434
+x^297*  3.44005943228194981602191514912 E-3450
+x^298*  3.03032228244014506693574166427 E-3466
+x^299*  2.55788969580963050575575623475 E-3482
+x^300*  2.06915498791049377955618942408 E-3498
}

- Sheldon
Reply
#32
Thanks for the post sheldon.

My intuition gave me similar ideas , but even my own motivations are not clear to me.

I am both gratefull for your posts but at the same time expected more theoretical motivation.

Well it gets pretty complicated thats almost certain.

I call this " fake function theory " invented by tommy and sheldon Smile

afterall this is getting " bigger " than a 101 calculus textbook.

I like the name " fake function theory " , sounds exciting and not dull or is it just me.

The fact that the name does not reveal what it is immediately might be part of the attraction. Also the " fake " sounds like sherlock thing.

Lots of work left to do.

regards

tommy1729
Reply
#33
(05/29/2014, 11:32 PM)tommy1729 Wrote: Thanks for the post sheldon.

My intuition gave me similar ideas , but even my own motivations are not clear to me.

I am both gratefull for your posts but at the same time expected more theoretical motivation.

Well it gets pretty complicated thats almost certain.

I call this " fake function theory " invented by tommy and sheldon Smile

afterall this is getting " bigger " than a 101 calculus textbook.

I like the name " fake function theory " , sounds exciting and not dull or is it just me.

The fact that the name does not reveal what it is immediately might be part of the attraction. Also the " fake " sounds like sherlock thing.

Lots of work left to do.

regards

tommy1729
I'm imagining one more version, V6. The accuracy won't be any different than V5, but V6 will be exactly real valued at both the positive and negative real axis, but will have a Laurent series representation, and will allow easy understanding of both the error term relative to Kneser, and of the zeros, on the negative real axis. for the Weiestrass representation, such that we can theoretically tie the two representations together, along with Kneser's half iterate. Also, it will have an infinite number of fake zeros near the origin, which we throw away. But version 6 alludes me for many days now .... so I am giving up and version 5 is the best I can do. I like fake function theory.
- Sheldon
Reply
#34
(05/29/2014, 11:54 PM)sheldonison Wrote: I like fake function theory.

Smile
Reply
#35
I wanted to remark that the "fake function" idea is generalizable.
We can find any entire asymptotic with nonnegative derivatives by simply using a rising integer function T(n) :






etc.

This simple idea / equation is very powerfull.

The fundamental theorem of fake function theory.

In combination with the post about the inverse gamma function we could for instance find the asymptotic :



with the method above.

To stay in the spirit of the exp.


regards

tommy1729
Reply
#36
I have another point to add about Weierstrass if you guys are still considering it. I would suggest looking into the stronger, Hadamard factorization theorem.

Since we are dealing with an asymptotic half iterate of exp (I'm going to call it ) I'm going to assume we can say it's order 1?

I.e: Does for some positive M,rho?

If so then Hadamards factorization theorem guarantees that:



for some where is all the zeroes counted with multiplicity.

This also implies, quite adequately that at least for .
I think examining the zeroes is really important. I'm going to rifle through hadamard some more and see if theres anything else we can do. I believe, as tommy does, we can probably prove its order 0, so we can probably cut out and the exponential factors in the product. This would imply a very plain product.

Sorry if you guys are passed this or if it isn't useful but I realize I could've said it before, I just forgot about Hadamard ^_^. Even though its stronger and requires more work than weierstrass we only ever remember weierstrass. Poor hadamard. ^_^
Reply
#37
(06/29/2014, 01:40 PM)JmsNxn Wrote:

for some where is all the zeroes counted with multiplicity.
Your equation is overdetermined I think.



Removing was trivial and the other variable is correct because of the genus theory / theorem in combination with below :


Quote:This also implies, quite adequately that at least for .

Well not from Hadamard alone , but from the asymptotes of the zero's yes.

And in combination with the genus theory we get the result above.

I did assume an upper bound on multiplicity though.

Thanks for the post James.

regards

tommy1729
Reply
#38
Related to mike's conjecture and some of my own about derivatives ,Im intrested in a fake alternating solution.

I use the following strategy :

S(x) = exp^[0.5](x)
For all x > 1 :

- a_n x^n + (a_n + b_n) x^(n+1) < S(x)

n ln(x) + ln( -a_n + (a_n + b_n) x ) < S(ln(x))

(x=exp(X) and b_n/a_n = c_n )

n X + ln(a_n) + ln( - 1 + (1 + c_n) exp(X) ) < S(X)

We could try the same with sexp too.

regards

tommy1729
Reply
#39
(06/30/2014, 12:56 AM)tommy1729 Wrote: Well not from Hadamard alone , but from the asymptotes of the zero's yes.

Maybe I'm misinterpreting your response but a huge lemma that hadamard uses in his proof is that:

if

then NECESSARILY, (by doing some magic with jensen's formula and some other neat complex analysis), if are the zeroes of f

Reply
#40
(06/30/2014, 03:21 PM)JmsNxn Wrote:
(06/30/2014, 12:56 AM)tommy1729 Wrote: Well not from Hadamard alone , but from the asymptotes of the zero's yes.

Maybe I'm misinterpreting your response but a huge lemma that hadamard uses in his proof is that:

if

then NECESSARILY, (by doing some magic with jensen's formula and some other neat complex analysis), if are the zeroes of f


Very intresting.

Well I was simply saying that it did not follow from what your wrote alone.
Hadamard's proof is very intresting.

By the way the constant C is simply f(0).

Let t_n be the absolute value of the zero's.
So if we show that prod (1 + z/t_n) grows slower than exp(x^a) for any real a > 0 then we have

fake exp^[0.5] = f(0) prod (1 + z/t_n).

And I believe that to be the case.

The logic behind that is simply this :

f(0) exp(Az) prod [ (1 + z/t_n) exp(-z/t_n) ]

Now the exp terms in the prod are just there to make the prod convergant IFF that is not yet the case.
But here it already is so we rewrite :

f(0) exp(Az) prod [ (1 + z/t_n) exp(z/t_n) ] =
f(0) exp((A+B)z) prod [ (1 + z/t_n) ]

where B = 1/t_1 + 1/t_2 + ...

prod (1 + z/t_n) grows slower than exp(x^a) for any real a > 0 then we have

A+B = 0

=> f(0) prod [ (1 + z/t_n) ]

And the sky is blue Smile

regards

tommy1729

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