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Rational operators (a {t} b); a,b > e solved
#8
(06/06/2011, 08:47 AM)JmsNxn Wrote: 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.
Well not on the whole complex plane, but on the real axis, wouldnt that be nice?
I anyway wonder whether thats possible at all.
As is not even differentiable at 1, I wonder whether f is. Did you compare the derivations from left and right?


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

oh I thought }t{ was kinda the inverse in t. But as I see that it is not ... there are already conventions that we use on the forum, even ASCII capable:
a [1] b = a+ b
a [2] b = a*b
a [3] b = a^b

so the operator is [t] (one off from your {t}). But then for the inverse functions
b [t] x = y
y /[t] x = b
b [t]\ y = x

for example
x /[1] y = x - y
x [1]\ y = -x + y


x /[4] p is the p-th superroot (which is not equal to x [4] 1/p)
b [4]\ x is the superlogarithm to base b.

This is also in sync with the convention for quasigroups, i.e. groups with non-associative operation but with left- and right-inverse. There the left- and right inverse are written as / and \.

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Messages In This Thread
RE: Rational operators (a {t} b); a,b > e solved - by bo198214 - 06/06/2011, 09:23 AM

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