SIPCASydney International Piano Competition of Australia
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Check out the self-congratulatory SIPCA website for instance.
Figure 3 and 4 represent the graph of [f.sub.[chi]], where [chi] is a SIPCA and [[phi].sub.1] = Id and [[sigma].sub.R] respectively.
A).sup.Z] be a SIPCA. We say t [is less than or equal to] A is a transition point if [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is a SIPCA.
B).sup.Z] be a SIPCA. The transition points of Id x [phi] are the transition points of [phi] and {B, 2B, ..., AB}.
x [[phi].sub.1] is a SIPCA with [[phi].sub.i] : [[0 ...
[A.sub.i]).sup.Z] be a SIPCA and x a non-transition point with [f.sub.[chi]] (x) = x (or 0).
[A.sub.i]).sup.Z] be a SIPCA. Then x [member of] [[0 ...
Proof: If [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is a SIPCA then it is clearly surjective.
[A.sub.i]).sup.Z] be a SIPCA. If z is the last zero of [f.sub.[chi]] (that is the largest value z in the domain such that [f.sub.[chi]](z) = 0), then there are exactly z pairs (x, y) such that x < y, [f.sub.[chi]] (x) = x, and [f.sub.[chi]] (y) = 0.
[A.sub.i]).sup.Z] be a SIPCA. For all non-transition points such that [f.sub.[chi]](x) = x, there exist points p < q < r < x such that x - r = q - p, [f.sub.[chi]](p) = p, [f.sub.[chi]](q) = 0, and [f.sub.
By Proposition 20 we can use induction to see there exists a SIPCA X : [[0 ...
For bigger neighbourhoods not all surjective NCCA are SIPCA.