By Peter Walters

The first a part of this advent to ergodic conception addresses measure-preserving modifications of likelihood areas and covers such subject matters as recurrence homes and the Birkhoff ergodic theorem. the second one half specializes in the ergodic conception of continuing modifications of compact metrizable areas. a number of examples are specified, and the ultimate bankruptcy outlines effects and functions of ergodic concept to different branches of mathematics.

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**Extra resources for An Introduction to Ergodic Theory (Graduate Texts in Mathematics)**

**Example text**

10). This completes the proof. The next entry can be found in Slater’s compendium [262, equation (35)]. 6 (p. 35). 8). Then ∞ (−q; q 2 )n q (n+1)(n+2)/2 qf (−q, −q 7 ) . 2 and Auxiliary Results 33 Proof. 1), set h = 2, a = c = −q 2 , t = −q, and b = 0. Multiplying both sides of the resulting identity by q/(1 + q), we ﬁnd that ∞ ∞ (−q; q 2 )n q (n+1)(n+2)/2 (−q)n q(−q; q 2 )∞ = . 8) three times altogether, we ﬁnd that ∞ ∞ qn 1 q (n−1)/2 = (1 − (−1)n ) (q; q)2n+1 2 n=0 (q; q)n n=0 1 = √ 2 q 1 1 − √ √ ( q; q)∞ (− q; q)∞ 1 (−q 1/2 ; q 2 )∞ (−q 3/2 ; q 2 )∞ (q 2 ; q 2 )∞ = √ 2 q(q; q)∞ −(q 1/2 ; q 2 )∞ (q 3/2 ; q 2 )∞ (q 2 ; q 2 )∞ 1 = √ 2 q(q; q)∞ 1 =√ q(q; q)∞ = = 1 (q; q)∞ ∞ 2 qn −n/2 ∞ − 2 (−1)n q n −n/2 n=−∞ n=−∞ ∞ (2n+1)2 −(2n+1)/2 q n=−∞ ∞ 4n2 +3n q n=−∞ f (q, q 7 ) .

14), we ﬁnd that ∞ ∞ ∞ (−1)n q (n+1)(n+2)/2 (−1)n q (n+1)(n+2)/2+m(2n+1) = 2n+1 (q)n (1 − q ) (q)n n=0 n=0 m=0 ∞ ∞ q m+1 = m=0 ∞ (−1)n q n(n−1)/2+n(2m+2) (q)n n=0 q m+1 (q 2m+2 )∞ = m=0 ∞ = q(q)∞ qm (q)2m+1 m=0 = qf (q, q 7 ), which completes the proof. 8 can be found in Slater’s paper [262, equation (37)]. 8 (p. 35). 8), then ∞ (−q; q 2 )n q n(n+1)/2 f (−q 3 , −q 5 ) . 15) Proof. 1), we set h = 2, a = −q 2 , b = 0, and c = t = −q. Upon using Euler’s identity, we ﬁnd that ∞ ∞ (−q; q 2 )n q n(n+1)/2 (−q)n 2 2 = (−q; q) (−q; q ) .

8), we ﬁnd that S= (iq 2 ; q 2 )∞ (−i1/2 q; q)∞ (−i−1/2 q; q)∞ 2(−q; q 2 )∞ (iq 2 ; q 2 )∞ +(i1/2 q; q)∞ (i−1/2 q; q)∞ = 1 2(q; q)∞ (−q; q 2 )∞ (−i1/2 ; q)∞ (−i−1/2 q; q)∞ (q; q)∞ 1 + i1/2 + = 1 2(q; q)∞ (−q; q 2 )∞ 1 1 + i1/2 + = (i1/2 ; q)∞ (i−1/2 q; q)∞ (q; q)∞ 1 − i1/2 1 1 − i1/2 1 2(1 − i)(q; q)∞ (−q; q 2 )∞ ∞ in/2 q n(n−1)/2 n=−∞ ∞ (−1)n in/2 q n(n−1)/2 n=−∞ ∞ (1 − i1/2 ) in/2 q n(n−1)/2 n=−∞ ∞ +(1 + i1/2 ) (−1)n in/2 q n(n−1)/2 n=−∞ = 1 (1 − i)(q; q)∞ (−q; q 2 )∞ ∞ ∞ in q n(2n−1) − n=−∞ in+1 q n(2n+1) n=−∞ 50 2 The Sears–Thomae Transformation = ∞ 1 (1 − i)(q; q)∞ (−q; q 2 )∞ ∞ in q n(2n−1) − i (−i)n q n(2n−1) n=−∞ , n=−∞ where we replaced n by −n in the latter sum.