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These definitions are called the inverse branch formalism.
REMARK 4.19. An element (f )1 of 2 C(Us) can be thought of as a function from C
to R which is undefined at some points (i.e. C \ U"y(ii(U)) and takes on multiple values
at others (if p e C, then for each i 1,..., m and for each x e (i-1(p), the function takes
on the value fi(x) at p). Iff fe C(C), a natural way to get an element of EilC(Ui) is to
consider (fo ()'2.
The idea is that A' is the approximation of L' obtained by summing over the good
branches, and that B7 is a summation over the branches thrown out in the (n - j)th step.
This is made explicit in Proposition 4.20 below.
Note: the value - is chosen merely to simplify calculations with Lemma 2.7; the point is
just to choose some exponentially decaying quantity depending only on n j.
For the remainder of this section, we will suppose that , n, m, (Ui) l, (i) l, (Tj)-1
(/j)= , and 0 < c < 1 are as in Definition 4.18, and that for each j 0,..., n, (Zi)),
AJ, and B7 are given by Definition 4.18. We will write r : mult((0i)i.
PROPOSITION 4.20. Fix f E C(C). For all k= 1,..., n,
n-1
(4.2.6) A[f] = (Lk[f] o (i)1 - Bj [L[f]].
j=k
Furthermore, if K C C, then for all j = 0, . . . , n- 1
m
g2UPKT (on)
(4.2.7) sup (B7[f])i < r(2deg2(Tj) + 1 + c-2(n-j))e SUPKj sup(f).
1 /1(K) Kj
where K := T (K). If f > 0, then the left hand side is positive.
PROOF. The proof of (4.2.6) is by backwards induction on k. If k n, the identity is trivial.
If we suppose that the formula is true for k = k + 1 for all f e C(C), then we can substitute
Lk [f] for f, yielding
A"+I[Lk[f]] = (L"[f] o (i)i1 - By[LJ([f]]
j k+1
Subtracting the reverse of (4.2.5) with j = k yields (4.2.6). Thus we are done.
52
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Simmons, David. Random Iteration of Rational Functions, dissertation, May 2012; Denton, Texas. (https://digital.library.unt.edu/ark:/67531/metadc115157/m1/57/: accessed April 25, 2024), University of North Texas Libraries, UNT Digital Library, https://digital.library.unt.edu; .