Tommy's Gamma trick ?

[tommy1729]

Here I Will explain the Gamma trick.

Tet(x) is computed from interpolating 1/tet(x).

So we need An interpolation for 1/tet(x).

Let Cs be the continuum Sum.

1/tet(x) = Cs ( 1/tet(x) ) - Cs ( 1/ln(tet(x)) )

Obvious.

So we need to interpolate Cs( 1/tet(x) ).

We use Cs( 1/tet(x) + ln(x) ) - Cs( ln(x) ).

Cs ln = lngamma.

Use the generalized formula < replacing product by comm operator > described below to interpolate the Cs f(x) = Cs( 1/tet(x) + ln(x) ).

Lim n -> oo

Gamma(z) = n ! n^Z / [ z (z+1) (z+2) ... (z+n)].

A Well known limit btw.

In this case the comm operator is

A • B = f^[-1](f(A) + f(B)).

So we have a limit form for the Cs(f) and therefore also

Cs(1/tet(x)).

From there we get the continuüm product Cp (1/tet(x)).
And finally Cp(tet(x)).

However to compute f(A) we need the values for tet(x).

But we know tet ' (x) = Cp(tet(x)).

We use that to set Up a recursion equation.

The details are complicated ...

Notice this is different from mike3 method for Cs and the euler formula for Cs.

Also uniqueness is not yet achieved and a few more details need to be added.

Forgive me for the incompleteness Im still considering.

Regards

Tommy1729

[tommy1729]

So we divide the interval (1,2) into K pieces.
Then we define m variabels est(m/k) = sexp(m/K + 1) for 1 =< m =< k.

Also the derivative is estimated (est(m/k) - est((m-1)/k))/k.

And est ' (0) = est ' (1/k) = 1 , est ' (1) = e.

More later.

Regards

Tommy1729

[tommy1729]

By estimating est , we also estimated f.

We set est(1/K) = 1 + 1/k.

And we can use

A • B = f^[-1](f(A) + f(B)) to get the simpler

A * B = f(A) + f(B).

Then when we set lim n -> oo to lim N -> oo ; n = N^k , (so) we (can) compute the

CS(f(x)). [ by the estimate ]

And then exp [ Cs(f(x)) - Cs(ln(x)) ] = Cp(f(x))/ gamma(x) = g(x).

Then g(x)^{-1} = Cp(tet(x)) = (est(x) - est(x-1/k)) ~ sexp'(x)


For x > 1.

What is the desired equation.

----

End of method.

It is assumed tet ' (x) = Cp( tet(x) )

1) extends beyond x E (1,2).

2) the functional equation must follow.

Also

3) we get a C^oo result

In fact

4) we get An analytic function.

All intuitively logical and we could add the functionaliteit equation to the conditions.

5) uniqueness assumed, related to 2) , 4) ofcourse .

Regards

Tommy1729

[tommy1729]

Let f be continu and real and satisfy :

For x = Y


\( f(1) = 1 \) (1)

\( f ' (x) = \Pi_1^Y f(x) \) (2)

\( f(2) = e \)

For real \( 1 < Y < 2 \)

1) does that imply f is C^oo in \( x E [1,Y] \) ?

2) does that imply f is analytic in \( [1,Y] \) ?

3) is the solution Unique ? If not, how many variables are there ?

4) how to prove existance ?

5) does it follow \( f ' (2x) = \Pi_1^{2Y} f(x) \) ?

6) is the equation \( f ' (x) = exp[ \Sigma 1^Y f(x-1) ] \) equivalent ?

7) does it Neccessarily follow \( f(x+1) = exp {f(x)} \) for x>1 ?

8 ) If f is analytic on \( [1,Y] \) does that imply it is analytic in \( [2,1+Y] \) ?

Regards

Tommy1729

[tommy1729]

Let f be continu and real and satisfy :

For x = Y


\( f(1) = 1 \) (1)

\( f ' (x) = \Pi_1^Y f(x) \) (2)

\( f(2) = e \)

For real \( 1 < Y < 2 \)

1) does that imply f is C^oo in \( x E [1,Y] \) ?

2) does that imply f is analytic in \( [1,Y] \) ?

3) is the solution Unique ? If not, how many variables are there ?

4) how to prove existance ?

5) does it follow \( f ' (2x) = \Pi_1^{2Y} f(x) \) ?

6) is the equation \( f ' (x) = exp[ \Sigma 1^Y f(x-1) ] \) equivalent ?

7) does it Neccessarily follow \( f(x+1) = exp {f(x)} \) for x>1 ?

8 ) If f is analytic on \( [1,Y] \) does that imply it is analytic in \( [2,1+Y] \) ?

Regards

Tommy1729

Some answers that I can prove


1) yes !

Proof sketch : C^0 -> C^1 because the product operator is C^0.
C^0^[2] = C^1.

C^1 -> C^oo by induction.

2) maybe. Need to consider the rate of the nth derivative.

3) if f(x) is a solution then f(x) + periodic is not.
The gamma method removes the + periodic part.

Partial answer

5) yes. 7) yes . 8 ) yes.
Easy.

Regards

Tommy1729

[tommy1729]

Due to the nature of the continuum product ( and the previous results) there is a UNIQUE solution that is C^oo in ]1,2[.

Since exp is analytic , the interval where f is C^00 extends to ]1,oo[.

That is close to a new tetration : show existance for An f that is C^00 in ]1,2] ; =
Done !

Regards

Tommy1729

[tommy1729]

Mick did Some effort to put it on MSE.
But the results are not so good.
Poor Mick.
He learned the continuum sum and product but app cant use it without confusing others.
His downvotes and closure are unjustified. Despite many time-consuming debates and edits.

Its painful to watch.
Not only that but the Large distance between tetration and mainstream ...

On the other hand perhaps people here have use for it.

Thanks Mick

http://math.stackexchange.com/questions/...x-pi-1x-fx

http://math.stackexchange.com/questions/...axb-fa-b-x

Regards

Tommy1729

[tommy1729]

I was able to convince mick to rewrite his version of question 6.

It seems to do better.
Not sure if ppl understand , but no downvotes -> suggest it.

Unfortunately no attention yet.

Is MO better for this ?


http://math.stackexchange.com/questions/...l-equation

Regards

Tommy1729