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Searching for an asymptotic to exp[0.5]
For multiple reasons there is a intrest to investigate a type of function :

Exp( A ln(x)^3 ).

For instance g" = 0. ( gaussian , Tommy-Sheldon )

Also for the theory of " fake polynomials " :

Fake [ exp( A ln(x)^3 ) ] = exp ( G( ln(x) ) ).

g(x) = A x^3 , G(x) is the fake A x^3 then.

This fake A x^3 gives a g " <> 0 , wich is intresting for our methods.

One could consider

Fakemethod ( exp( a ln^2(x) + A ln^3(x) + ... ) )

= Fake [ exp( a ln^2(x) + polynomialfake [ A x^3 ] + ... ) ].

This way the gaussian AND Tommy-Sheldon come back into play while taking into ACCOUNT g "' (x) !

--

Error term studies are intresting too

---

Slightly off-topic , but there must be a J-like equation for exp( A ln(x)^3 ) right ?

---

Regards

Tommy1729

Reply
(02/16/2016, 03:17 AM)tommy1729 Wrote: I need the ratio's of the real a_n vs fake a_n for the function exp(x) , where the fake is the limit of the Tommy-Sheldon iterations.

From post#141, we have the Gaussian approximation error for a_n for exp(x) is the same as the error in Stirling's approximation for n!, or approximately

Numerical results using the iterated Gaussian, post#150, do not show improved behavior, over Gaussian, for exp(x), for a_n where n>39, although it is an improvement for smaller values of n. I have no idea what the expected behavior is for the iterated Gaussian approach in post#150 is, which Tommy refers to as "Tommy-Sheldon iterations". One interesting note on post#150, if F_n were exactly equal to exp(x), than F_n+1 is the Gaussian approximation...

a_n for n=1-50 error term for Gaussian, vs 20 iterations of post#150, using 5000 term Taylor series approximation; here the error term ratio is printed as , so Gaussian looks like instead of
Code:
n a_n Gaussian error     a_n post#150 error
1 0.0844375514192275 0.0683398515477305
2 0.0844142416333461 0.0607258884762405
3 0.0841935537563679 0.0539730385655355
4 0.0840332149853975 0.0478920432435974
5 0.0839199291390415 0.0422747285213201
6 0.0838370948924301 0.0369976779272284
7 0.0837743008426081 0.0319840348760891
8 0.0837252084955692 0.0271826096143000
9 0.0836858377963707 0.0225573573477991
10 0.0836535913240025 0.0180817574806996
11 0.0836267114154591 0.0137356044170443
12 0.0836039699542637 0.00950306495393020
13 0.0835844845780208 0.00537144303115174
14 0.0835676057991404 0.00133036115853485
15 0.0835528453405350 -0.00262880126422681
16 0.0835398292592345 -0.00651330495603565
17 0.0835282664554722 -0.0103293386111823
18 0.0835179270079519 -0.0140822322763638
19 0.0835086269484465 -0.0177766186034792
20 0.0835002173553064 -0.0214165568027041
21 0.0834925764050155 -0.0250056293731362
22 0.0834856034885243 -0.0285470186139815
23 0.0834792147938839 -0.0320435678797153
24 0.0834733399466525 -0.0354978311600175
25 0.0834679194244035 -0.0389121136095030
26 0.0834629025452544 -0.0422885049797763
27 0.0834582458872422 -0.0456289074255043
28 0.0834539120347256 -0.0489350588073065
29 0.0834498685755973 -0.0522085523576732
30 0.0834460872927128 -0.0554508533850614
31 0.0834425435070680 -0.0586633135474178
32 0.0834392155405340 -0.0618471831168360
33 0.0834360842735316 -0.0650036215728523
34 0.0834331327786455 -0.0681337067965566
35 0.0834303460154027 -0.0712384430865797
36 0.0834277105746333 -0.0743187681777040
37 0.0834252144632697 -0.0773755594108080
38 0.0834228469223184 -0.0804096391772155
39 0.0834205982721904 -0.0834217797398696
40 0.0834184597807136 -0.0864127075170098
41 0.0834164235500404 -0.0893831069003792
42 0.0834144824193701 -0.0923336236687975
43 0.0834126298809658 -0.0952648680487083
44 0.0834108600073941 -0.0981774174656687
45 0.0834091673882766 -0.101071819024386
46 0.0834075470751343 -0.103948591749587
47 0.0834059945331426 -0.106808228615549
48 0.0834045055988069 -0.109651198388331
49 0.0834030764427299 -0.112477947301598
50 0.0834017035367683 -0.115288900584174
- Sheldon
Reply
The integral method can handle small g " much better.
So i Will be focusing more on integral type formula's in the future.

For instance focusing on the truncated integral method

Sup & g " , g "' , ... g ^(n) &

Where n is picked such that we get Sup , and & * & stands for the integral taking into account the first n derivatives of g.

Likewise second to one supremum , Inf , second to one Inf etc are considered.

This might take Some time.

In fact, I do not know An easy way to compute such Sup.

Although approximations seem easy.

---

Not sure about the future for Tommy-Sheldon iterations.
Maybe if we replace iterating the gaussian with iterating integral methods.
Or not.

---

Regards

Tommy1729
Reply
About sheldon's pics for fake exp^[1/2].

I believe fake exp^[1/n] satisfies the functional equation Well in the zone going from positive reals > n to arg(z) = arg( (-1)^(2/n) ).
That is imho intuitive.
And it clearly holds for n = 2 , Well according to the plots at least.

I believe this property carries over to most solutions exp^[1/n] ( no fake , actual ).

This imho suggest that most analytic exp^[1/n] are analytic in the same zone described above.

A strong conjecture is that if the fake holds the functional equation in its zone , the the corresponding actual function ( of which the fake is considered ) is NEC analytic in that zone.

This makes me wonder about iterations of similar functions like sinh ; if they have that property too.

Probably another way to say it is :

" the fake function is not only a good approximation on the positive reals , but also on the region near it ! "

Although that is open for interpretation and debate I assume.

Regards

Tommy1729
Reply
I now use the integral method with g" and g " ".
So the second and fourth derivative.

The Logic is that the odd derivatives do not contribute to the abs of the derivatives of f.

So taking into account g " " Gets me closer to the abs of the n th deriv of f , which is usually An improvement.

Also using g " " " might be too much , too close to the abs , unless of course f is entire and has more than 90 % positive derivatives , and all derivatives strictly decreasing( from n th to n+1 th ).

Regards

Tommy1729
Reply
Consider the rectangle (2,3,2+ 0,0001 i , 3+0,0001 i).

This is zone a.

Exp( zone a) = zone b , exp(zone(b) ) = zone c.

F0 = ln(2sinh^[1/2](exp))

F1 a = ln F0( exp(zone a) ) =< F0(zone a) + g1.

F2 a = ln F1 ( exp ( zone a) ) =< Ln( f0 a + g1) =< f0( a ) + g1 + g1/x^2

By induction

Foo =< f(a) + g1/x^2 + g1/x^4 +... Converges

Qed m- test

Regards

Tommy1729
Reply
If we compute fake f(y^(2^m)) and then substitute y = x^(1/2^m) and let m go to +oo ,

Call the resulting coef g(t) such that g(t) belongs to x^t ,

Then , we get a very good approximation ( for all fake methods ! ) towards

f(x) =~ integral[0,oo] g(t) x^t dt.

The ratio of LHS to RHS as x Goes to oo is conjectured therefore to be - for most f - equal to sqrt(2 pi).

-----

From fake function theory it is logical to conjecture

F(x) = lim_n C sum_{i = n - sqrt(n)}^{n} t_i x^i.

Where t_i are the Taylor coef.

----

I found the Continuüm hypothesis is strongly related to fake function theory and tetration.

Basically I found a way to describe the cardinality of sets ( all sets ?? At least most , another intresting idea ! ) as Taylor series of real-entire functions.

The cardinality of a set is then

Lim x -> w f(x).

Where f is a real-entire Taylor series.

( and w is the card of the integers )

There is no need for AC.

We proceed by noting card( ~ exp(w) ) = card( 2^w ).
And also card( A + w ) = card(A) for card(A) >= w ( Card A nonfinite ).

Then we notice card ( growth C ) = card ( growth D ) IFF growth C = growth D.


( sheldon's growth of f := lim slog( f^[n] ) / n )

Then we use fake-half(x) = t(x).

And note

1) card(w) < f(w) < card(exp(w))

2) card(w) = card( fractions )
Card(exp(w)) = card( reals )

3) from fake function theory we know f(f(w)) = exp(w) and f(w) exists !!

4) the continuüm hypothese is basically 1).

5) the " tail " of f(w) is sufficiënt so arguments that the first Degree V Taylor polynomial bijects to w and thus does not increase card is NO PROBLEM.

6) as for the error terms card(w + w) = card(w).
Card(2^w) = card( w^j k^w + w^l ) for k >= 2 , j,l >= 0.



Qed.

Regards

Tommy1729

Ps : the new conjectures are

1) card of any set = Some Taylor of w.

2) it seems we have an uncountable number of card between card(N) and card® because any growth between 0,1 exists as a distinct card.
Therefore the idea of real Beth Numbers seems to destroy the idea of countable alephs or at least aleph 2.
The consequences are not yet understood ...


Will set theory , calculus and tetration finally be Friends after the " long war " ?


Some names might be pleasant.

Tetrational set theory.
Fake set theory ??

Asymptotic set theory ??

Beth set theory ???


With special thanks to Sheldon as cofounder of the essential fake function theory.


Regards

Tommy1729
The master
Reply
More on that :



Long ago , I was challenged on sci.math to answer my own question.

There cannot be a bijection from exp^[1/2](w) to w nor 2^w.

Here is why :

Let f(x) be exp^[1/2](x) + o(1).

If card f(w) = card w then

card f(f((w)) = card f( card f(w) ) = card f(w) = card(w)

But f(f(w)) = 2^w

Contradiction

If card f(w) = card 2^w then

card f(f(w)) = card f ( card f(w) ) = card f ( 2^w ) = card(2^(2^w))

But f(f(w)) = 2^w

Contradiction

Likewise for exp^[a](w).


Regards

Tommy1729
Reply
To clarity a bit on those taylors.

Consider the set of unreduced positive fractions.

This has card w^2. Or ordinal w^2 if you want.

Now consider those positive fractions a/b.

G(x) = x + 1/x.

Now the cardinality ( resp ordinal ) of g(a/b) is clearly equal to w^2 / 2 or card ( w^2 / 2 ).

Although both are countable , this justifies the Taylor series.

( + is just adding elements or cardinalities )

Regards

Tommy1729

Reply
To conclude the final thing about this set theory CH thing.

Although i have given the impression that CH is false there is a problem with that.

Just one big problem.

If one considers just the last term of the taylor ( t_w w^w ) - because all w^i with i finite are countable , then neither the fake last term nor another last term give a correct solution.

Keep in Mind that an iteration of x^n / a = x^n^2 / a^(n+1).

Also things like ln(w) and 1,3^ w are not consistent. ( ln is too small , 1,3 is not a base for the reals that has uniqueness )

Also results ( for either f or f(f) ) like w^w / e^w , 4^w/2^w , w^w/w^w , w - 1 , etc are undefined.

Hence there are no Nice solutions.

It follows that CH Cannot be constructively be disproven by additions and products of sets/Cardinals/ordinals ( taylors ).

Since additions , products and powersets are the only potential way to increase ordinals / Cardinals / sets ,

... This implies CH must be true !

Consider this

Let M be a Set with card between countable and uncountable.

Let there be a way to describe card M by a metaset of integers.

So , Lets say that M is a subset of the reals.

How many countable or finite sets do we need to describe M ?
So we take either a finite amount of sets of integers , finite sets or an infinite amount of them. And we use Sums and products.

Notice there is nothing between taking a finite amount and an infinite amount.

After taking ( internal ) bijections we can always reduce to
Card M = 2^a n^b + 2^c n^d + e.

Where a , B, C , d , e are non-uncountable.
Notice a part of a countable or finite set is also finite or countable.

So card M must be countable or uncountable !!

Qed.

Tommy1729
Reply


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