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Rational operators (a {t} b); a,b > e solved
#5
(06/06/2011, 02:45 AM)JmsNxn 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.

But then on the other hand there is a general problem with semioperators (note that your operator is 1 off the standard notation, i.e. {t}=[t+1]):
(a [0] 0 = 1)
a [1] 0 = a
a [2] 0 = 0
a [3] 0 = 1
a [n] 0 = 1 for n>3

or

(a [0] 1 = 2)
a [1] 1 = a+1
a [2] 1 = a
a [3] 1 = a for n>2

As soon as one defines a [t] 1 = a for t > 2, then the whole analytic function is already determined to be a, i.e. it must also be a for t=1 which is wrong.
Hence an analytic t |-> a[t]1 will not be constant but somehow meandering between the a's, which is somehow really strange.

But I see you gracefully avoided that problem by just defining it for a,b > e Smile

PS:
1. , This notation is ambiguous, compare . Please invent a better one!
2. , not \alpha but \circ belongs in the exponent: . This notation is derived from the symbol for function composition .

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Messages In This Thread
RE: Rational operators (a {t} b); a,b > e solved - by bo198214 - 06/06/2011, 06:53 AM

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