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Tommy's Gamma trick ?
#1
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
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Messages In This Thread
Tommy's Gamma trick ? - by tommy1729 - 10/14/2015, 09:37 PM
RE: Tommy's Gamma trick ? - by tommy1729 - 10/16/2015, 12:28 PM
RE: Tommy's Gamma trick ? - by tommy1729 - 10/17/2015, 12:22 AM
RE: Tommy's Gamma trick ? - by tommy1729 - 10/22/2015, 12:34 PM
RE: Tommy's Gamma trick ? - by tommy1729 - 11/02/2015, 01:43 AM
RE: Tommy's Gamma trick ? - by tommy1729 - 11/04/2015, 12:06 AM
RE: Tommy's Gamma trick ? - by tommy1729 - 11/04/2015, 01:26 PM
RE: Tommy's Gamma trick ? - by tommy1729 - 11/07/2015, 01:02 PM

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