Univalent functions by Christian Pommerenke; Gerd Jensen

By Christian Pommerenke; Gerd Jensen

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Next, we have a single domino covering the third and fourth tile. This indicates that we must create the adjacency 1 2 by placing 2 next to 1 in the permutation 2 1 3, shifting 2 and 3 accordingly. Then we get the permutation 3 1 2 4, and we add 5 to break up the adjacency 1 2. This results in the permutation 3 1 5 2 4. Next, we have a stack of four squares, which means that we must place 6 in the fourth rightmost position. This gives us the permutation 3 6 1 5 2 4. The last tile is a stack of two dominoes, which indicates that we must create the adjacency 2 3 and break it up with 8.

This result will be essential in proving infinite continued fraction expansions. 13. (Loya, 2006). It turns out that many irrational numbers have surprisingly beautiful continued fraction expansions. For example, 6 = 02 + 12 − π2 14 (5) 24 12 + 22 − 34 22 + 32 − 32 + 42 − 44 . 42 + 52 − . 12 to Euler’s sum, we have π2 = 6 1 (12 )2 12 + (6) (−22 )2 (−22 ) − (12 ) + (32 )2 (32 ) − (−22 ) + (−42 ) − (32 ) + (−42 )2 .. 1 = (7) 14 12 + 24 −(12 + 22 ) + 34 22 + 32 + −(32 + 42 ) + 44 ... 1 = (8) 14 12 − 24 12 + 22 − 34 22 + 32 − 32 + 42 − 44 ..

Proof. The proof will proceed by strong induction. For n = 0, 1! + 0! = 2, which is consistent with the fact that there are 2 ways to tile a 1-board with height condition 2. For n = 1, 2! + 1! = 3, which is consistent with the fact that there are 3 ways to tile a 2-board with height conditions 2, (1, 1). Suppose we have (n + 1)! + n! ways to tile a (n + 1)-board for all 1 ≤ n ≤ k. Then consider a (k + 2)-board. The (k + 2)nd tile has height conditions (k, k + 1). We condition on whether the board ends in a stack of squares or dominoes.

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