JTPIDPS

# JTPIDPS - C&O 330 NOTES ON THE JACOBI TRIPLE PRODUCT...

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IDENTITY PROFESSOR D.M.JACKSON 1. The Jacobi Triple Product Identity These are notes on the Jacobi Triple Product Identity and its use in proving the Euler Pentagonal Number Theorem and the mod 5 and 7 congruences for the partition number. Ihave included in a few more of the details than I included in the lectures. Let P be the set of all partitions and let D n be the set of all partitions of n into distinct parts only. Let T k denote the partition ( k,k 1 ,..., 1) . Lemma 1.1. [Sylvester’s Decomposition] P×{ T k } f [ j 0 D k + j × ( D j ∪D j 1 ) where D 0 ∪D 1 = D 0 . Proof. Append the reverse (1 , 2 ,...,k )o f T k to the top of the Ferrers diagram for π P , and consider the staircase that continues the pro±le of the Ferrers diagram for π. The length of the staircase is k + j. The staircase partitions the diagram into a partition α obtained by summing the columns of ? ’s below the staircase, and a partition β obtained by summing the ? ’s in rows above the staircase. The number of rows in β is j or j 1 . The partitions α and β necessarily have distinct parts, induced by the staircase. The construction is clearly reversible. Theorem 1.2. [Jacobi Triple Product Identity] Y m 1 ( 1 q 2 m )( 1+ yq 2 m 1 )( 1+ y 1 q 2 m 1 ) = X k = −∞ y k q k 2 . Date : October 5, 2002.

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2P R OFESSOR D.M.JACKSON Proof. From Lemma 1.1, by counting partitions with respect to the sum of their parts, marked by q ,wehave q ( k +1 2 ) Y m 1 (1 q m ) 1 = X j 0 £ s k + j ¤ Y a 1 (1 + sq a ) · £ t j ¤ Y b 1 ( 1+ tq b ) + £ t j 1 ¤ Y b 1 ( 1+ tq b ) = X j 0 £ s k + j t j ¤ (1 + t ) Y m 1 (1 + sq m )(1+ tq m ) = X j 0 £ s k + j t j ¤ Y m 1 (1 + sq m ) ( 1+ tq m 1 ) . We now change variables from s and t to s and u through st = u. Then q ( k +1 2 ) Y m 1 (1 q m ) 1 = £ s k ¤ X j 0 £ u j ¤ Y m 1 (1 + sq m ) ( 1+ us 1 q m 1 ) so q ( k +1 2 ) Y m 1 (1 q m ) 1 = £ s k ¤ Y m 1 (1 + sq m ) ( 1+ s 1 q m 1 ) . (1.1) We next sum over k from −∞ to + by making use of the following symmetry in k. Replacing s by s 1 , we have q ( k +1 2 ) Y m 1 (1 q m ) 1 = £ s k ¤ Y m 1 ( 1+ s 1 q m )( 1+ sq m 1 ) . Now replace s by qS, noting that £ s k ¤ = q k £ S k ¤ . Then q ( k +1 2 ) Y m 1 (1 q m ) 1 = q k £ S k ¤ Y m 1 ( 1+ S 1 q m 1 ) (1 + Sq m ) so, replacing S by s, q ( k +1 2 ) Y m 1 (1 q m ) 1 = £ s k ¤ Y m 1 (1 + sq m ) ( 1+ s 1 q m 1 ) since ( k +1 2 ) k = ( k +1 2 ) . Thus (1 . 1) holds with k replaced by k. Thus summing (1 . 1) over k from −∞ to + we have X k = −∞ s k q ( k +1 2 ) Y m 1 (1 q m ) 1 = X k = −∞ s k £ s k ¤ Y m 1 (1 + sq m ) ( 1+ s 1 q m 1 ) = Y
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JTPIDPS - C&O 330 NOTES ON THE JACOBI TRIPLE PRODUCT...

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