Analytic Theory of Continued Fractions II: Proceedings of a by Christopher Baltus, William B. Jones (auth.), Wolfgang J.

By Christopher Baltus, William B. Jones (auth.), Wolfgang J. Thron (eds.)

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Example text

Entries in on the two the 0 th and (6) Jacobi's from are 0 we we 42 the is i n c o n c l u s i v e M2n,2(z), odd normal we downward can = > for the 1, and conclude from the (-1) 2, the 0 by > (5) second for s o m e st row is get in 0 we ""' The any ... fact that 0 both obtain H 2(-2n)(c) n+1 so we do not n entries ~ ) _ R ( _ 2 n ) ( c )H ( - ( 2 n - 2 ) -2n+2 2n (6), H~n(2n-2))(C) that (z) n = 0 1 2 ' M2n+l,2 . . left e d g e of a s q u a r e block. the argument see identity by M2n+l,2(z ) and M2n+2,2(z) there the row a n d a n y n = O, 1, completely that either satisfies APT-fraction a information at on with entries are (S) information However, that by see is (c) the in most the 2 2 x 2 block × 2 must 2, ...

If this for s u f f i c i e n t l y = 1 for all n. 5] for n=l K(an/l) has a s e q u e n c e ig(n)_~(n) I < (g(n)} 1 2(n+l)qn _ of right or w r o n g 8 n+l dk l~k+l+i~ i ~ tails Such that for all n. k=n+l Moreover, there exists an ~ Re ~(n) + ~ M ~ ~ such that g(n) satisfies n ~ M, b e c a u s e Re g(n) 8 ~ dk 1 12k+l+i~ I ~ - ~ as n ~ k=n+l and Im g(n) < Im ~(n) + ~ 8 ~ dk 12k+l+i~l k=n+l _ 1 • n+l(~ - 8 ~ dk ]2k+l+i~l) < 0 from s o m e n on. 7. The 41 . for a 0, u ~ O. 6 case fraction are may rather complicated.

Stated = shown m=l positive we real be C monotonically points have and again can ~ • intimately sequence problem hence we the (C) M-table Once H ( - ( n - k ) ) (C) n normal. Lo(C) z = 0 and you established ... _ . we +1, for argument 1, does function that for (n = O, there exist ~(t) each ±i, on (-~, n, ± 2 .... )? a real-valued, ~) with bounded, infinitely many 33 In [5], Jones and Thron showed that if C s a t i s f i e s the c o n d i t i o n H ( - 2 n ) ( c ) > 0 and H ( - 2 n ) ( c ) > 0 for n = O, 1 ....

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