# Tetration Forum

Full Version: Searching for an asymptotic to exp[0.5]
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Notice how both S9 and the gaussian do Well for f(x^a).

Regards

Tommy1729
Let f(x) satisfy the conditions.

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Let g(x) satisfy the conditions.
Let m be a positive integer.
Let s be a nonnegative real.
Let r be a positive real.

Notice min [ (f(x) + s x^m) / x^m ] = min[ f(x)/x^m ] + s.

Therefore using S9

Fake [ f(x) + s x^m ] = fake [ f(x) ] + s x^m.

Let ++ be close to addition.

Then by the above and

Fake( f(x) ) => a_n
==>
fake ( f(x) x ) => b_n = a_(n-1)

And also

Fake ( r f(x) ) = r fake ( f(x) )

We get

Fake [ f(x) + g(x) ] = Fake [ f(x) ] ++ Fake [ g(x) ]

This is a simple but important fact.

Notice this is not immediately clear from - with respect - the S9 formulation with all the ln's , exp and g and h.

It might be good homework to try and find these results with the g h formulation of S9.
Ofcourse the g h formulations has its OWN benefits.
But imho this simple notation of min ( f(x) / x^n ) i still use has been put back on the map.

Guess this should be called

Another road of investigation should be how other methods like the gaussian or Tommy-sheldon iterations deal with this.

By now the idea of distributive must have come to your mind.

A wild conjecture would be

Tfdl =

Tommy's fake distributive law.

Law , conjecture or theorem ...

This refers to tommy's generalized distributive law.

See elsewhere on the forum.

For S9 :

Notice fake [ fake( f(x) ) + fake( g(x) ) ] = fake^( f(x) ) + fake^( g(x) ).

Clearly related. Fake^ means fake fake.

I call that tommy's fake fake formula.

I think these results Will be useful.

For none S9 methods we can probably replace + on the RHS with ++.
This needs more investigation.

Using the squeeze theorem and comparison theorem ( from " limit calculus " ) Will be useful to know.

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Coffee break :p

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For the second half of this post I Will post An idea that is in my head since post 9.

In post 16 Sheldon improved the overestimate S9 with a division.

It also seems Natural to Try with a substraction.

For methods that are general and do not use zero's it seems unnatural to not use the S9 part.

( e.g. Gaussian = S9 / sqrt .. )

In particular after the properties given in the first half of this post, and in general after rereading this thread.

So this idea occured :

Tommy's tommynaci sequence

And use f n) = f n-1) + f n-2).

Already known as the Evangelist series.

F n-1) = ((3*sqrt(5)+1)*(((1+sqrt(5))/2)^n)+(3*sqrt(5)-1)*(((1-sqrt(5))/2)^n))/(2*sqrt(5)).

For those intrested.

Anyway

a_n = min [ ( f(x) - ... - a_(n_19) x^(n-19) - a_(n-12) x^(n-12) - a_(n-7) x^(n-7) -

a_(n-5) x^(n-5) - a_(n-2) x^(n-2) ) / x^n ]

Tommy's Evangelist recursion.
Or the tommynaci method.

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Notice for exp(x) we get the Exact solution

By using a_n = min [ ( f(x) - a0 - a1 x - a2 x^2 - ... - a_(n-1) x^(n-1) ) / x^n ]

But that does not work for nonentire f or f with Some negative derivatives.

Hence Tommy's Evangelist recursion is used instead for such.
This implies the method does not perform well for exp.

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But way too many questions ...

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How Well does the tommynacci method work ?

Time for medicine.

But as you probably can tell im feeling better and no longer confused.

Unfortunaly not at full potential. So i probably missed something trivial too add.

Take care

Regards

Tommy1729

I designed them for nonnegative g ' '.

I suggest - with doubt - that in case of a negative , we remove the minus sign.

In other words the absolute value.

Although counterintuit , notice the second deriv already influences the first
Derivative.

Regards

Tommy1729
Im considering to replace g'' in the gaussian method or the Tommy-Sheldon iterations by ln( e + exp(g'')).

Notice the condition f '' > 0 implies

For x > 1 :
g '' + g' (g'-1) > 0.

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How good the Tommy-Sheldon iterations work are depending alot on how good the gaussian is.

Lets say the gaussian is off by a factor 5.

Then the next iteration gives

New g = ln( 5 f (*) ) = ln(5) + old g.
This change seems too weak.

Funny because if we consider gaussian( 5 f) then this weak change is exactly what we need ( to get S9 close to gaussian and both to correct )
, since min( 5 f/ x^m) = 5 min ( f/ x^m).

Reconsidering things seems necessary.

Fake function theory is tricky.

However , i assumed here the gaussian was off by a factor 5.
It is not certain this is possible.

What then gives back confidence to the Tommy-Sheldon iterations.

Or by the replacement suggested Above ,

" The exponential Tommy-Sheldon iterations ".

Shorthand ets.

Regards

Tommy1729

In context of the previous post , i assume the gaussian can be off by about
( inspired by the ets and the binary partition )
ln(e + exp(0)) / ln(e) = ln(e+1) ~ 1,313

Regards

Tommy1729
Solving g '' + g ' ( g ' - 1) = C is very insightfull !

Regards

Tommy1729
Not sure if its a good idea.
I have not considered it much.

I call it the egyptian method

Min(a) + min(b) =< min(a+b)

And

1/x = 1/(x+1) + 1/(x^2 + x).

So instead of min( f(x)/x^n )

Try min ( f(x) / (x^n + 1) ) + min ( f(x) / (x^2n + x^n) ).

Regards

Tommy1729
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.

Also , related and usefull ,

1 / ( 1 - a_n/fake a_n ).

[ i assumed the limit of the ratio going to 1 indeed ].

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Identical question for exp^.

Im close to insights.

Regards

Tommy1729
Intresting is the fake of exp( (2 pi)^{-1} * 0,5 * ln(x)^2 ).

This makes the gaussian agree on the usual S9.

Methods agreeing could indicate good estimates.

So how good is it ? And how true is the indication ?

Will we get the searched O(1/n^2) ??

Need to study this.

Regards

Tommy1729
Notice how the second iteration of the Tommy-Sheldon iterations gives a nonnegative coëfficiënt for g '' , therefore being better than the gaussian in " most " cases.

Naturally in all cases where g " < 0 , but more investigation is needed for g " > 0.

Regards

Tommy1729
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