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(a) Let 0 < a < 1. Show the series or converges uniformly on [-a, a] to ₁¹. n=0 (b) Does the series on converge uniformly on (-1, 1) to ₁¹ -? Explain. n=0 1

(a) Let 0 < a < 1. Show the series or converges uniformly on [-a, a] to ₁¹. n=0 (b) Does the series on converge uniformly on (-1, 1) to ₁¹ -? Explain. n=0 1

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.2: Arithmetic Sequences
Problem 67E
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use the first one and apply to 25.9 to prove it by M-test

 

Theorem 6.5.2. If a power series o anx" converges absolutely at a point xo, then the power series converges uniformly on the compact interval [-c, c] where c = = |xo|. in=0 Proof. Suppose for some xo we have that Since we have the convergence of converges. With Mn = |an| for each n = 0, 1, 2, 3, . . . , we have for all x satisfying |x|≤|xo| that |anx^| ≤ |anx| = Mn. ∞ n=0 Σ'anno| n=0 [M₁ = 19,2 the Weierstrass M-Test implies that ∞ Σ n=0 n=0 converges uniformly on A = [-c, c] for c = anxen = |xol. We can now prove that when a power series converges on an open interval (-R, R) with R> 0 or R = ∞, the power series is a continuous function on (-R, R). For a fixed x₁ satisfying 0 < x₁ < R we pick xo such that x₁ < x < R. The convergence of Σ an implies by Theorem 6.5.1 that Σ anx solutely on the open interval |x|< xo. n=0 Since 0 < x1 < xo, we have that Σo anx" converges absolutely at x₁. By Theorem 6.5.2, we have that o anx" converges uniformly on [-x1, x1]. n=0 Since each fn(x) = anx" is continuous on [-x₁, x₁], we have by Theorem 6.4.2 that the power series is continuous on [-x₁, x1]. converges ab- Since x₁ satisfying 0 < x₁ < R is arbitrary, we have that anx" is continuous on (-R, R).
Transcribed Image Text:Theorem 6.5.2. If a power series o anx" converges absolutely at a point xo, then the power series converges uniformly on the compact interval [-c, c] where c = = |xo|. in=0 Proof. Suppose for some xo we have that Since we have the convergence of converges. With Mn = |an| for each n = 0, 1, 2, 3, . . . , we have for all x satisfying |x|≤|xo| that |anx^| ≤ |anx| = Mn. ∞ n=0 Σ'anno| n=0 [M₁ = 19,2 the Weierstrass M-Test implies that ∞ Σ n=0 n=0 converges uniformly on A = [-c, c] for c = anxen = |xol. We can now prove that when a power series converges on an open interval (-R, R) with R> 0 or R = ∞, the power series is a continuous function on (-R, R). For a fixed x₁ satisfying 0 < x₁ < R we pick xo such that x₁ < x < R. The convergence of Σ an implies by Theorem 6.5.1 that Σ anx solutely on the open interval |x|< xo. n=0 Since 0 < x1 < xo, we have that Σo anx" converges absolutely at x₁. By Theorem 6.5.2, we have that o anx" converges uniformly on [-x1, x1]. n=0 Since each fn(x) = anx" is continuous on [-x₁, x₁], we have by Theorem 6.4.2 that the power series is continuous on [-x₁, x1]. converges ab- Since x₁ satisfying 0 < x₁ < R is arbitrary, we have that anx" is continuous on (-R, R).
25.9 (a) Let 0 < a < 1. Show the series or converges uniformly 2=0 on [-a, a] to ₁¹ 1-x 1 (b) Does the series on converge uniformly on (-1, 1) to ¹? Explain. in=0
Transcribed Image Text:25.9 (a) Let 0 < a < 1. Show the series or converges uniformly 2=0 on [-a, a] to ₁¹ 1-x 1 (b) Does the series on converge uniformly on (-1, 1) to ¹? Explain. in=0
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