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Searching for an asymptotic to exp[0.5]
Let f(x) satisfy the conditions.


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

Fake addition theorem

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^[2]( f(x) ) + fake^[2]( g(x) ).

Clearly related. Fake^[2] 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.


Coffee break :p


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

Start with 2 , 5 , ..

And use fSadn) = fSadn-1) + fSadn-2).

Already known as the Evangelist series.

FSadn-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.


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.


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.


I had Some gaussian variant Ideas about the above.
But way too many questions ...


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




Messages In This Thread
RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 10/11/2015, 07:17 PM

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