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Rational operators (a {t} b); a,b > e solved
#7
(06/06/2011, 06:53 AM)bo198214 Wrote: But James, this is not analytic at , if we reformulate:

We can say:

where


is addition and composition of analytic functions, except this one function . The whole function can not be analytic. I wonder why it looks so smooth.

I like your definition better--it seems sleeker Tongue. I was sort of aware that there was no way I was gonna produce an analytic function over the whole complex domain, I'm happy with analytic in a few regions.
Quote:But I see you gracefully avoided that problem by just defining it for a,b > e Smile
well hopefully I'll be having to tackle that problem soon.

Quote:PS:
1. , This notation is ambiguous, compare . Please invent a better one!

Alright, from henceforth I shall refer to logarithmic semi operators with the following notation:

And the inverse is given by:


therefore:
etc etc..

Quote:2. , not \alpha but \circ belongs in the exponent: . This notation is derived from the symbol for function composition .

I knew there was something off about my equations. lol

(06/06/2011, 06:02 AM)sheldonison Wrote: Hey James, try my code snippet, which I updated while you were posting. It will work for values of a and b<e, seamlessly.
- Shel

I'm wary about using for defining bases less than e. My complaints are explained by the following points:

fatb(e+0.0001, 2, pi*I) = -0.999999 - 0.00115551*I
fatb(e+0.0001, 1.8, pi*I) = -1.883265702 - 0.00194696*I
fatb(e+0.0001, 1.5, pi*I) = -5.707515375 - 0.011242371*I
fatb(e+0.0001, 1.3, pi*I) = -4.091499848 - 8.531525563*I
fatb(e+0.0001, 1.1, pi*I) = -1.002757644 - 8.536029475*I
fatb(e+0.0001, 1, pi*I) = -8.53659263001*I

Ignoring the drastic jumps in values, observe the hump that occurs in the real transformation. For no reason the values just spike to -5 randomly. This happens with all regular superfunctions of the logarithm. That's what makes the cheta function unique.

But I think, is it possible to create an upper superfunction for ?, perhaps it will give similar smooth results. Except it will be defined for a,b > 2... at least I think so.
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Messages In This Thread
RE: Rational operators (a {t} b); a,b > e solved - by JmsNxn - 06/06/2011, 08:47 AM

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