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# ENUMERATIVE COMBINATORIC MA 614

UK

GPA 3.54

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This 2 page Class Notes was uploaded by Kennith Herman on Friday October 23, 2015. The Class Notes belongs to MA 614 at University of Kentucky taught by Staff in Fall. Since its upload, it has received 9 views. For similar materials see /class/228148/ma-614-university-of-kentucky in Mathematics (M) at University of Kentucky.

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Date Created: 10/23/15

Sign Reversing Involutions and Lattice Paths Because I was covering the material a little quickly on Friday I decided to write up some notes on the lattice paths Let ul 01 771 1171 7 k 1 712 10 and 772 k 171 7 The problem is to count the number of ordered pairs P1 P2 of lattice paths using only the steps E and N for which P1 goes from 711 to 771 P2 goes from 712 to 772 and P1 and P2 do not intersect Denote this set of paths by F Draw pictures for yourself We begin by considering a larger set of ordered pairs of lattice paths S P1P2 P1 goes from 711 to 7 and P2 goes from 712 to 77 for 139 77 j We sign the set S by SJr P1P2 P1u1 7 771 and P2 712 7 772 and S P1P2 P1 711 7 772 and P2 712 7 771 So EP1P2 P1P2 6 8 and EP1P2 71P1P2E ST Now de ne an involution 7139 on S as follows If P1 P2 do not intersect then set 7TP1 P2 P1 P2 a xed point If P1 P2 intersect then let q be the rst lattice point of intersection as you follow the paths Set 7TP1 P2 Pf PZ where P is obtained by following P1 from 711 to q and then following P2 from q to the end of P2 and P2 is obtained by following P2 from 712 to q and then following P1 from q to the end of P1 Draw pictures We now make the following observations 1 FCSf P All pairs of paths in SF must intersect in fact paths in S must cross so F is the set of xed points of 7139 C40 The number of paths P ul 7 771 is the same as the number of paths from 0 0 to 1171 7 k namely 4 Similarly the number of paths P 712 7 772 is the same as the number of paths from 00 to 1171 7 k namely U The number of paths P ul 7 772 is the same as the number of paths from 0 0 to k 7 171 7 k 7 1 namely 03 The number of paths P 712 7 771 is the same as the number of paths from 0 0 to k 7171 7 k 1 namely 5 Thus the number of pairs of paths in SJr 00 Similarly the number of pairs of paths in S kn 1 Therefore FZea 165 m SH 7 m 7 1 k f 1 2 1 1 2 As a consequence7 227 2 0 and so S 22 and thus the sequence yields 01 Z k 07 771 is con rmed to be log concave

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