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By Serge Lang

This 5th version of Lang's booklet covers the entire subject matters generally taught within the first-year calculus series. Divided into 5 elements, every one element of a primary direction IN CALCULUS comprises examples and purposes when it comes to the subject lined. additionally, the rear of the ebook comprises distinctive ideas to various the routines, permitting them to be used as worked-out examples -- one of many major advancements over earlier versions.

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Extra resources for A first course in calculus

Example text

19. 20 (p. 57). Let m and n be nonnegative integers. Let ∞ φ(a) := j=0 2 aj q mj +nj (q)j and ψ(a) := φ(aq) . 24) and 1 = ψ(a) + aq m+n ψ(a)ψ(aq)ψ(aq 2 ) · · · ψ(aq 2m−1 ). Proof. We have ∞ φ(a) − φ(aq) = j=0 ∞ = j=1 aj q mj 2 (1 − q j ) (q)j 2 aj q mj +nj (q)j−1 ∞ = aq m+n j=0 = aq +nj m+n 2 aj q mj +2mj+nj (q)j φ(aq 2m ), which is the first assertion. 24) by φ(a) and noting that φ(aq 2m ) = ψ(a)ψ(aq)ψ(aq 2 ) · · · ψ(a2m−1 ), φ(a) we obtain the second assertion. 21 (p. 10). For |xy| < 1 and a and b arbitrary, ∞ ∞ (ab)n (xy)n(n+1)/2 y −n (ab)n (xy)n(n+1)/2 =1−b+b .

40). For |a| < 1, ∞ (a)∞ ∞ 2 an an bn q n = . (q)n (bq)n (q)n (bq)n n=0 n=0 Proof. 10), let both a and b tend to 0. Then replace t by a and c by bq, and lastly multiply both sides by (a)∞ . 2. 2, but additionally we require other formulas that have appeared often in Ramanujan’s work. The Rogers–Fine identity [149, p. 1) is needed in this chapter. 1). 2). 2) = := (q)k ⎩ , otherwise. q (q) (q)k− For any complex numbers a, b, and any nonnegative integer n, n j=0 n (−1)j a−j bj q j(j−1)/2 = (b/a)n .

3) Also, for |z| < 1 and any nonnegative integer N , ∞ n=0 1 n+N n . 1 (p. 5). For any complex number a, ∞ ∞ (−a; q)2n+1 q 2n+1 + (−a)n q n(n+1)/2 2) (q; q n+1 n=0 n=0 = (−aq; q)∞ (q; q 2 )∞ ∞ (−a)n q n(n+1)/2 . (−aq; q)n n=0 Proof. 5), set t = q, b = −q, and a = 0; then replace c by aq. We therefore deduce that ∞ qn (−q; q)∞ = (q; q) (aq; q) (aq; q)∞ (q; q 2 )∞ n n n=0 ∞ (−a; q)2n q 2n (q 2 ; q 2 )n n=0 (−q; q)∞ − (aq; q)∞ (q 2 ; q 2 )∞ ∞ (−a; q)2n+1 q 2n+1 . 1), set h = 1, a = 0, and t = q. Then, replacing c by aq and letting b tend to 0, we find that ∞ qn 1 = (q; q) (aq; q) (q; q) (aq; q)∞ n n ∞ n=0 ∞ (−a)n q n(n+1)/2 .

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