06/06/2011, 06:53 AM
(06/06/2011, 02:45 AM)JmsNxn Wrote:
But James, this is not analytic at
We can say:
where
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

PS:
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