Math 409-502Harold P. [email protected] of the third examinationGood job!Here are the scores:95 94 94 92 92 91 90 90 89 88 8885 85 80 79 78 78 77 76 75 66 34Reminder: the comprehensive final examination will be held in this room on Tuesday, Decem-ber 14, from 8:00–10:00 AM.Math 409-502 December 3, 2004 — slide #2Things can go wrong in the limitContinuity. On the interval [0, 1], let fn(x) = x1/n.Then limn→∞fn(x) =(1, if x 6= 0,0, if x = 0,so the limit of these continuous functions is discontinuous.Derivatives. Let gn(x) =1nxn. Then³limn→∞gn(x)´0= 0 when 0 ≤ x ≤ 1, but limn→∞g0n(x) =limn→∞xn−1=(0, if x 6= 1,1, if x = 1.Here (derivative of the limit) 6= (limit of the derivatives).Integrals. Let hn(x) =(2n+1, if12n+1≤ x ≤12n,0, otherwise.ThenR10limn→∞hn(x) dx = 0, but limn→∞R10hn(x) dx = 1.Here (integral of the limit) 6= (limit of the integrals).Math 409-502 December 3, 2004 — slide #3Uniform convergenceA sequence of functions { fn}∞n=1converges uniformly on an interval to a function f if for ev-ery e > 0 there is an N such that | fn(x) − f (x)| < e for all x whenever n > N.“Uniform” means that N can be chosen to be independent of x.Example. Let fn(x) = sin(x +1n). Then the sequence of functions fn(x) converges uniformly tothe function f (x) = sin(x) on the unbounded interval (−∞, ∞).For suppose e > 0 is given. By the mean-value theorem, sin(x +1n) − sin(x) =1ncos(c) forsome c depending on x, so | fn(x) − f (x)| ≤1n. Therefore | fn(x) − f (x)| < e when n > 1/e. Sowe can take N = 1/e, which is independent of x.Math 409-502 December 3, 2004 — slide #4Theorems about uniform convergence1. If a sequence of continuous functions fnconverges uniformly on an interval to a function f ,then the limit function f is continuous.2. If a sequence of differentiable functions fnconverges (uniformly) on an interval to a func-tion f , and if the sequence of derivatives f0nconverges uniformly to a function g, then the limitfunction f is differentiable, and f0= g.3. If a sequence of integrable functions fnconverges uniformly on an interval [a, b] to a func-tion f , then f is integrable andRbaf (x) dx = limn→∞Rbafn(x) dx.(This is a stronger theorem than the one in the book.)In all three cases, the word “sequence” can be replaced by the word “series” (because conver-gence of a series means convergence of the sequence of partial sums).Math 409-502 December 3, 2004 — slide #5HomeworkRead sections 22.1–22.5, pages 305–318.Math 409-502 December 3, 2004 — slide