Searching for an asymptotic to exp[0.5]

[tommy1729]

About that perpendicular thing...

I think I was able to show that the first occurance - between the first 2 zero's starting from 0 going to -oo - is true if all zero's are on the negative real line.
Basicly it uses hadamard and says that adding a linear term (by product) does not change the property. I started with a polynomial of degree 2 and then used induction.

Probably this can be extended to all occurances.

regards

tommy1729

[tommy1729]

Let f(z) be the entire fake half-iterate of exp(z).

Now we do not have a sexp and hence also no theta(z) function.

However at least at the positive real line there must other entire functions that satisfy the functional equation very well.

Let u(z),v(z),w(z) give these other solutions f_i(z) :

f_1(z) = f(z+u(z)) = f(z)(1+v(z)) = f(z) + w(z).

(Here I assumed u,v,w are Always related , that is another subject)

Now it seems logical to assume u,v,w also grow much slower then exp , likely like exp^[1/2].
This gives then a similar hadamard product for these functions.

Let m be an integer > -1.
Now U := u,v,w must satisfy U(f^[m](0)) = 0.
Which btw is consistant with carlson and hadamard.

So U is completely determined by the other zero's apart from U(f^[m](0)) = 0 (and a constant C of course).
And that determines how well the functional equation is satisfied off the real line !!

It would be amazing if one of u,v,w is somehow optimal WITHOUT other zero's then U(f^[m](0)) = 0.

Of particular intrest is the cases where u,v,w have no zero's in the region where our Original f(z) satisfies the functional equation well.

Because if it has a zero there (point "Q") then it must also have a zero at point f(Q) which complicates matters.

regards

tommy1729

[tommy1729]

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

To combine 2 ideas of me we can use this to find an alternative asymptotic half-iterate F(z) of exp(z) such that F(z) = F(-z).
Or even the additional F^[2](z) = F^[2](-z) if we take F(0) = 0.

We do that by defining F(z) = a0 + a1 x + a3 x^3 + a5 x^5 + ...

In other words we take T(n) = 2n + 1 and then solve like described above.

The main difference with 2sinh^[0.5](z) is that we get a different region of almost agreement with exp(z) or even 2sinh(z) after 2 iterations.
Also , though related , notice the derivatives of 2sinh^[0.5](z) are not all positive.
Btw we should also get the similar hadamard product for this F(z) as well as for 2sinh^[0.5].
But how do the plots look like ? And where are the zero's ?
This requires some additional work.

An intresting lemma for the 2sinh if true would be this :

A := 2 pi
2sinh^[0.5](z) = a0 + a1 z + a2 z^2 + ...
g(z) := |a0| + |a1| z + |a2| z^2 + ...

Conjecture : |g(z)| < 1 + A 2sinh^[0.5](|z|)^A

Stronger and weaker variants are also of intrest.

regards

tommy1729

[tommy1729]

Let f(z) be the entire fake half-iterate of exp(z).

Now we do not have a sexp and hence also no theta(z) function.

However at least at the positive real line there must other entire functions that satisfy the functional equation very well.

Let u(z),v(z),w(z) give these other solutions f_i(z) :

f_1(z) = f(z+u(z)) = f(z)(1+v(z)) = f(z) + w(z).

(Here I assumed u,v,w are Always related , that is another subject)

Now it seems logical to assume u,v,w also grow much slower then exp , likely like exp^[1/2].
This gives then a similar hadamard product for these functions.

Let m be an integer > -1.
Now U := u,v,w must satisfy U(f^[m](0)) = 0.
Which btw is consistant with carlson and hadamard.

So U is completely determined by the other zero's apart from U(f^[m](0)) = 0 (and a constant C of course).
And that determines how well the functional equation is satisfied off the real line !!

It would be amazing if one of u,v,w is somehow optimal WITHOUT other zero's then U(f^[m](0)) = 0.

Of particular intrest is the cases where u,v,w have no zero's in the region where our Original f(z) satisfies the functional equation well.

Because if it has a zero there (point "Q") then it must also have a zero at point f(Q) which complicates matters.

regards

tommy1729

It turns out 1 + v(z) cannot both be entire and grow slower then O(exp(|z|/|a|) and still satisfy the functional equation well.
Hence there is no (nice) v(z) !!
I can prove that if you want, its trivial. It follows from considering the zero's of both 1+v(z) and f(z).

hence the attention goes to f_1(z) = f(z+u(z)).

It seems u(z) changes the function in the sense that the region where f_1^[2](z) - exp(z) << 1 is different from f(f(z)) - exp(z) << 1.

This can be explained by induction starting from the real line.
However that does assume the conjectures made about f(z).

It seems u(z) makes a " wobble " just like we saw with the theta functions.

Im a bit puzzled by u(z).

If u(z) has so many zero's this suggests (carlson , hadamard) u(z) grows at least like exp^[0.5](|z|/|a|) for z with negative real part.

But then |f(z+u(z))| grows like O(exp(|z|/|a|)) ?

This implies that f(f(z)) grows like exp(z) near the positive real line and like exp(exp(z)) near (but not on) the negative real line for sufficiently large z ?!?

That would be one weird function.
Like the half-iterate of both exp(z) and exp(exp(z)).
Thus the half-iterate of a twice fake/asymptotic function itself.

Fake function theory seems like a neverending story.

regards

tommy1729

[sheldonison]

Let f(z) be the entire fake half-iterate of exp(z).

Now we do not have a sexp and hence also no theta(z) function.

However at least at the positive real line there must other entire functions that satisfy the functional equation very well.
....
It seems u(z) makes a " wobble " just like we saw with the theta functions.

We could also look at

Still on vacation; enjoyed reading your posts. I generated a quick graph of . The first zero crossing is at z=137.8052295808. For smaller value of z, the error term is dominated by the 1/z laurent term I mentioned earlier; for larger values of z, it is dominated by Here is the graph. The logarithmic spikes are the zero crossings of f(f(z))-exp(z). Notice how it wobbles around exp(z). The error term is dominated by the error term for f(z), which I graphed in an earlier post. f(z) has nearly the same first error term. f(f(z))/exp(z) has a larger error than f(z)/exp^0.5(z), but they similarly. The error term for f(f(z))/exp(z) at z=1000 is f(f(z))/exp(z)-1=-1.138369235934 E-12.
   

[tommy1729]

Let f(x) = a_0 + a_1 x + a_2 x^2 + ...
Let a_0 , a_1 , a_2 > 0
Assume a_2 > a_3 , a_3 > a_4 , ... > 0

Then the goal is to find a_n for n > 2.

Assume a_n > (n-1) a_(n+1)

Some motivation.

First we try to solve for a single variable.
Tetration is a difficult subject so multivariable ideas/equations are complicated in particular
when we are not working with matrices.

Another thing. We prefer not to solve for an equation that contains both a_(n-1) and a_n.

The reason is that we get disagreement.

example : F(a_10,a_11) = 0 F(a_11,a12) = 0 Now we have 2 conflicting values for a_11.
Besides solving for a_(n-1) when we already had its value is unlogical.

Solving for a_n and a_(n+1) makes a bit more sense since we did not yet have the value of a_(n+1).
example : given a_5 it makes sense to solve for a_6,a_7.

Keeping that in the back of our mind , what type of equation should we solve ?

Logical would be a truncated taylor series.

Sheldon approximately solved a_n x^n = exp^[0.5](x).

However the truncation is extreme here.

It is true that for every x , there must be an n such that a_n x^n is the most dominant term.
But a_n is independant of the value of x. Hence the equation a_n x^n = exp^[0.5](x) makes some sense.

However a more dominant approximation of a taylor series is a_q x^q ,
where q is between n and n+1 and a_q = a_n.

This shows that sheldons equation is probably valid within a ratio of x^(q-n).


On average that is x^(1/2).
So sheldon's solution S(x) probably satisfies S(x)/x^(3/4) - C < f(x) < S(x) x^(3/4) + C for some constant C.
In fact sheldon mentioned the correcting factor x^(1/2) himself.

A less extreme truncation would lead to better results.

How to get more dominant terms ?

We need to consider the contributions of a_m x^m for a general a_m from a random taylor series with positive a_i.

The contributions with respect to m look like a gaussian curve g where g(m - t) is smaller than g(q)
and g(m + t2) is smaller than g(q) for suff large t and assuming m = q + o(2).

the top of this curve at g(m) has growing m where m grows with x.

but a gauss curve is symmetric.

so a_n x^n + a_(n+1) x^(n+1) are the most dominant terms and both terms can have similar contribution !

Now perhaps you are thinking : why not solve a_n x^n + a_(n+1) x^(n+1) + a_(n+2) x^(n+2) + a_(n+3) x^(n+3) ?

Its a reasonable idea ... but It violates all logic above.
First there are too many variables.
Second we get a lot disagreement.

But most importantly its no longer likely that a_n x^n is the dominant term !!
Its quite more likely to be a_(n+1) x^(n+1) + a_(n+2) x^(n+2) which then is basicly the same as what is proposed now.
Notice that if a_n x^n is not one of the dominant terms , its more like the term a_(n-1) x^(n-1).
But then we violate another logic of above ; we solve for values we already have.


Notice the assumption "Assume a_n > (n-1) a_(n+1)" only makes this argument stronger.

These consideration lead me to

a_n x^n + a_(n+1) x^(n+1) = exp^[0.5](x).

2 unknowns is a bit problematic and I do not want to get disagreement.
So we use the assumption "Assume a_n > (n-1) a_(n+1)".
Notice this assumption makes sense since f grows more like a polynomial then an exponential.

The assumption is also consistant with sheldon's equation , plots and all results sofar.

a_n x^n + a_n/(n-1) x^(n+1) = exp^[0.5](x)

Now we need approximations for exp^[0.5](x) without having our f(x).
I use my 2sinh method for that when x is large.

We continue :

a_n x^n (1 + 1/(n-1) x) = exp^[0.5](x)

exp(X) = x , take ln on both sides :

ln(a_n) + n ln(x) + ln(1 + 1/(n-1) x) = ln(exp^[0.5](x))

replace x with exp(X)

Now the strenght of my 2sinh shows : ln(exp^[0.5](exp(X))) = exp^[0.5](X)

we get :

ln(a_n) + n X + ln(1 + 1/(n-1) exp(X)) = exp^[0.5](X)

We know from that ln(1 + "Large") is about ln("large") + 1/"large".

SO we can further reduce :


ln(a_n) + n X + ln(1/(n-1) exp(X)) + (n-1)/exp(X) = exp^[0.5](X)

Slightly less formal but it seems the term (n-1)/exp(X) is so small we can neglet it and remove it.

we then get :


ln(a_n) + n X + ln(1/(n-1) exp(X)) = exp^[0.5](X)

Simplify

ln(a_n) + n X - ln(n-1) + X = exp^[0.5](X)

Simplify even Further :


ln(a_n) + (n+1) X - ln(n-1) = exp^[0.5](X)

ln(a_n) = Max [ exp^[0.5](X) - (n+1) X + ln(n-1) ]
= Max [exp^[0.5](X) - (n+1) X] + ln(n-1)

exp^[0.5](X) - (n+1) X has one minimum value, where the derivative is zero.


At the minimum, the derivative will be equal 0, so defining
Now define




And then finally the improved solution is :


I considered further improvements but things got dubious and I stumbled upon problems with the principles used here.
In other words be careful.

regards

tommy1729

" truth is that what does not go away when you stop believing in it "
tommy1729

[tommy1729]

Some further thought :



There is good news and bad news.
Basicly both are that this is in a way very similar to the Original estimate for a_n combined with our estimate for a_(n+1).





Just as our estimate for a_(n+1) from a_n and the Original estimate of a_n.

But maybe a refinement of the previous post helps.

Am I wrong in assuming we prefer to solve for one variable ?

regards

tommy1729

[tommy1729]

EDIT

(removed nonsense)

[sheldonison]

Some further thought :

Hey Tommy,

I'm not able to understand the part of your equation; these are big numbers. For n=50, , perhaps you intended a ln somewhere?

[sheldonison]

...to understand the negative real axis, you mostly have to understand how Kneser's half iterate behaves for positive , due to the equation . And the definition I'm using for negative axis comes down to plus stuff, where stuff can be shown too small to create zeros anywhere else, since halfk(z) can be shown to have a large absolute value at the negative axis, so stuff winds up acting like a small perturbation of the zeros of the approximation.

Before we can understand "stuff", we need an exact converging equation for a_n. Here is the equation, as a converging integral. Of course, using h_n is a pretty good approximation, but the limit converges, and gets more accurate as y goes to infinity. First, we need an equation for the trough of , at a particular real value of y, where we want the local minimum of , which is where the derivative has a zero crossing.



We note that z=0 has a derivative of zero, since the function has a local maximum at zi=0, but we are interested in the local minimum, where the real part at the minimum gets arbitrarily large negative as y gets bigger.

Then we get this equation for all of the Taylor series coefficients of the asymptotic half iterate, f(z). Of course, you can use y=h(n), for our previous "best behaved" radius, but the conjecture is that this integral converges as y gets arbitrarily large, for all values of a_n. This conjecture only depends on the first real minimum getting arbitrarily negative for exp^{0.5}. In this equation, exp(y) is the radius of a circle, where we are wrapping the approximation around the circle multiple times until we get to the local minimum. Each z=pi i corresponds to half way around the circle. For example, for y=2, we get mi(2)~=5.65pi, so we wrap the approximation around the circle 5.65 times, for a radius of exp(2). This limit converges rapidly as y increases.



A more difficult conjecture is that this equation also converges for all negative values of n, where we are talking about a Laurent Series! Then, here are the golden values of a_n, for the asymptotic half iterate, accurate to 32 decimal digits.

Code:

{half=
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+x^225*  1.9519651728833711120704211891455 E-2346
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+x^227*  1.3259100688231246355692990305644 E-2375
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+x^229*  7.2959763189818210857434787158098 E-2405
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+x^237*  8.3717096964008922909601293473603 E-2523
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+x^242*  3.0818343021308110175424084637158 E-2597
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+x^264*  4.2297484953859892645978946228906 E-2931
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+x^267*  2.0973678925508310140708454839134 E-2977
+x^268*  7.0204513860124048128929361569859 E-2993
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+x^270*  6.8462320075794491042780947822524 E-3024
+x^271*  1.9950930497586676299541608886840 E-3039
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+x^273*  1.4764149510058814271782272044002 E-3070
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+x^277*  4.6791248960258165813941772872856 E-3133
+x^278*  9.9138356530228591442513442058349 E-3149
+x^279*  2.0080004529199037968113698528820 E-3164
+x^280*  3.8885278195209745127169291405925 E-3180
+x^281*  7.2004541037708460787761986112735 E-3196
+x^282*  1.2750918172412703104661669248805 E-3211
+x^283*  2.1596492273536361542217699775661 E-3227
+x^284*  3.4989531476600975459124994446056 E-3243
+x^285*  5.4232399623854837069841157428345 E-3259
+x^286*  8.0426211741222782354153066476251 E-3275
+x^287*  1.1413175498182740675299773645312 E-3290
+x^288*  1.5500194377377993981378886128387 E-3306
+x^289*  2.0148416599353000816235310923698 E-3322
+x^290*  2.5070839651532656491438274506800 E-3338
+x^291*  2.9865635421626323775665331571620 E-3354
+x^292*  3.4064341659042264496424959171075 E-3370
+x^293*  3.7205208039357170462514006673813 E-3386
+x^294*  3.8916413340839957809707676312809 E-3402
+x^295*  3.8988520048176433637470666861845 E-3418
+x^296*  3.7416642242029463512766194970626 E-3434
+x^297*  3.4400594322819498160219151491178 E-3450
+x^298*  3.0303222824401450669357416642614 E-3466
+x^299*  2.5578896958096305057557562347395 E-3482
+x^300*  2.0691549879104937795561894240776 E-3498
}

I mentioned "stuff" in a previous post. The 1/x Laurent series is an excellent approximation for -stuff, where here we use a 5 term approximation. If the |z| is fairly large, then we might want to use more terms in the approximation, there is always some stuff left over.


And here are the first 20 terms of the Laurent series, also generated and printed to 32 decimal digits accuracy. Generating the Laurent series is much much more difficult computationally, since there is no optimal h_n value for negative values of n, so extremely high precision calculations are required for the Kneser half iterate. I call this function the half_error term, since it is the dominant term for stuff, for small radius's. By the way, I also conjecture that even though all of these Laurent series terms converge, they don't behave very well as n gets arbitrarily large negative, and the function itself probably doesn't converge anywhere. But the truncated Laurent (1/z) series is very useful for understanding the "stuff" terms of the half iterate. A 20 term truncated Laurent series seems to work pretty well.

"Stuff" in the equation above, can be approximated by -half_error(z), since at any given radius, any function can be approximated arbitrarily well by a Laurent series. We throw away the negative terms for our entire half iterate. So naturally, there is an error term generated by the negative terms we threw away. I will give a more complete thorough analysis of all of the "stuff" near some of the zeros of f(z), and at the positive real axis, in a later post, by using the mi function, the the exp^{0.5}(y+mi(y)) function, and the Taylor/Laurent series.

Code:

{half_error=
+x^ -1*  0.0038981927072789568905822639730044
+x^ -2* -0.0017214881666811347076719510576548
+x^ -3* -0.00013712753070231088880268844304361
+x^ -4*  0.00013907417275118396879544895662740
+x^ -5*  0.000051320769626162064593356837293471
+x^ -6* -0.000022366390692218719076651284271858
+x^ -7* -0.000023771055718461525133758346306702
+x^ -8*  0.0000013731519550483503617729218008457
+x^ -9*  0.000012027038063261925914628822119819
+x^-10*  0.0000039087491754042668374627189270986
+x^-11* -0.0000057892790783328757123356666460684
+x^-12* -0.0000051218889509626574137447251389938
+x^-13*  0.0000019098976291086407044050882321080
+x^-14*  0.0000048547491203337398558987063953054
+x^-15*  0.00000074114812516008642387127350420916
+x^-16* -0.0000038309228524031537802224892859947
+x^-17* -0.0000025748868434074919176037441576404
+x^-18*  0.0000022951530139076283185894597305398
+x^-19*  0.0000036906301942910425788903290199039
+x^-20* -0.00000035885452386419832591366749556753}