Counterexample to the Bounded Orbit Conjecture
Counterexample to the Bounded Orbit Conjecture
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In our eReader you can find the full English version of the book. Read Counterexample to the Bounded Orbit Conjecture Online - link to read the book on full screen. Our eReader also allows you to upload and read Pdf, Txt, ePub and fb2 books. In the Mini eReder on the page below you can quickly view all pages of the book - Read Book Counterexample to the Bounded Orbit Conjecture
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Observe k k that if p is on h (A), then h" (p) is on A; thus, the positive orbit of every point on h (A) is bounded for every k > 0.
To check points between h k_1 (A) and h k (A), for k >_ 1 , we first consider a point p in the region above A k 1/2 and below h k (A). Let q be the point on h (A) which is directly above p (i.e. q is the intersection of h (A) with the vertical line containing p). Since q has bounded positive orbit and h n (p) is directly below h n (q), for all n > 0, the positive ...orbit of p must also be bounded. Next suppose p is between A k)1/ ( n+1 j and \^ /n , where n > 2. By construction there is a positive integer m such that h m (p) is between A, . ,„ and k+m,l/2 h (A). Thus, the above argument shows that p has a bounded positive orbit. In fact, since a point between h" k (A) and h" k+1 (A), k > 0, is between A and h(A) after k applications of h, all po : nts in B have bounded positive orbits.
One can see that the negative orbits of points in B are bounded by recalling that h _1 (p) = -h(-p).
What to read after Counterexample to the Bounded Orbit Conjecture?
You can find similar books in the "Read Also" column, or choose other free books by Boyles, Stephanie Marion to read onlineMoreLess
To quickly assess the difficulty of the text, read a short excerpt:
Observe k k that if p is on h (A), then h" (p) is on A; thus, the positive orbit of every point on h (A) is bounded for every k > 0.
To check points between h k_1 (A) and h k (A), for k >_ 1 , we first consider a point p in the region above A k 1/2 and below h k (A). Let q be the point on h (A) which is directly above p (i.e. q is the intersection of h (A) with the vertical line containing p). Since q has bounded positive orbit and h n (p) is directly below h n (q), for all n > 0, the positive ...orbit of p must also be bounded. Next suppose p is between A k)1/ ( n+1 j and \^ /n , where n > 2. By construction there is a positive integer m such that h m (p) is between A, . ,„ and k+m,l/2 h (A). Thus, the above argument shows that p has a bounded positive orbit. In fact, since a point between h" k (A) and h" k+1 (A), k > 0, is between A and h(A) after k applications of h, all po : nts in B have bounded positive orbits.
One can see that the negative orbits of points in B are bounded by recalling that h _1 (p) = -h(-p).
What to read after Counterexample to the Bounded Orbit Conjecture?
You can find similar books in the "Read Also" column, or choose other free books by Boyles, Stephanie Marion to read onlineMoreLess
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