Uniform Convergence and Series of Functions

Transcription

1 Uniform Convergence and Series of Functions James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University October 7, 017 Outline Uniform Convergence Tests for Series of Functions Examples Integrated Series

2 We have already used the UCC in the context of series multiple times. It is time to state it in a form that is explicitly useful for series and to state and prove a variant of the Weiestrass Test for Uniform Convergence. Definition Let (un) be a sequence of functions defined on S and let (Sn) be the associated sequence of partial sums. For convenience, assume all indexing starts at n = 1. We say (Sn) satisfies the The Uniform Cauchy Criterion For Series of and only if n ɛ > 0 N uk(t) < ɛ, for n > m > N, t S k=m=1 or (Sn) satisfies the The Uniform Cauchy Criterion For Series if and only if ɛ > 0 N n k=m=1 uk(t) < ɛ, for n > m > N where it is understood the sup norm is computed over S. Using the UCC for series, we can prove another test for uniform convergence just like we did for a sequence of functions (xn). The difference is we are specializing to the partial sum sequence. Theorem Let (xn) be a sequence of functions on the set Ω with associated partials sums (Sn). Then ( ) ( ) unif S : Ω R Sn S (Sn) satisfies the UCC. Proof ( :) unif We assume there is an S : Ω R so that Sn S. Then, given ɛ > 0, there is N so that Sn(t) S(t) < ɛ/ for all t Ω. Thus, if n > m > N, we have

3 Proof Sn(t) Sm(t) = Sn(t) S(t) + S(t) Sm(t) Sn(t) S(t) + S(t) Sm(t) < ɛ/ + ɛ/ = ɛ Thus, (Sn) satisfies the UCC. ( :) If (Sn) satisfies the UCC, then given ɛ > 0, there is N so that Sn(t) Sm(t) < ɛ for n > m > N. This says the seq (Sn(ˆt)) is a Cauchy Sequence for each ˆt Ω. Since R is complete, there is a number aˆt so that Sn(ˆt) aˆt. The number aˆt defines a function S : Ω R by S(ˆt) = aˆt. ptws Clearly, Sn S on Ω. From the UCC, we can therefore say for the given ɛ, Sn(t) Sm(t) ɛ/, if n > m > N, t Ω Proof But the absolute function is continuous and so Sn(t) Sm(t) ɛ/, if n > m > N, t Ω unif or S(t) Sm(t) < ɛ when m > N. This shows Sn S as the choice of index m is not important.

4 This allows us to state a new test for uniform convergence specialized to series. Theorem Second Weierstrass Test For Uniform Convergence Let (un) be a sequence of functions defined on the set S and let (Sn) be the associated sequence of partial sums. We assume all indexing starts at n = 1 for convenience. Let the sequence (Kn) be defined by sup t S un(t) Kn. Then Kn converges = (Sn) converges uniformly on S Proof If Kn converges, the sequence of partial sums σn = n converges and so it is a Cauchy Sequence of real numbers. j=1 Kj Proof Thus, given ɛ > 0, there is N so that n > m > N = σn σm = n Kj < ɛ j=m+1 Then if n > m > N, for all t S n n n uj(t) uj(t) Kj < ɛ j=m+1 j=m+1 j=m+1 Thus, (Sn) satisfies the UCC and so there is a function U so that unif Sn U on S where we usually denote this uniform it by U(t) = un(t). Comment Note this says if (Sn) does not converge uniformly on S, the associated series sup t S un must diverge to infinity. Let s go back and do our original problems using these new tools.

5 Example Discuss the convergence of the series n=0 tn. We apply the ratio test here. t n+1 t n = t = t This series converges absolutely if t < 1, diverges if t > 1 and at t = ±1. At t = 1, the partial sums oscillate between 1 and 0 and so the sequence (Sn) does not converge. At t = 1, the partials sum diverge to and so the sequence (Sn) does not converge. Thus, n=0 tn converges pointwise to a function S(t) on ( 1, 1). To determine uniform convergence, let un(t) = t n and for suitably small r, let Kn = sup t [ 1+r,1 r] t n = 1 r = ρ < 1. Since Kn = ρn is a geometric series with 0 ρ < 1, it converges. Hence by the Second Weierstrass Uniform Convergence Theorem (SWUCT), the convergence of the series n=0 tn is uniform on [ 1 + r, 1 r] and so it converges to a function U on [ 1 + r, 1 r]. Since its are unique, S U on [ 1 + r, 1 r]. Further, each Sn is continuous as it is a polynomial in t, and so the it function U = S is continuous on [ r, r]. Note the sequence does not converge uniformly on the whole interval ( 1, 1) as Kn = sup t ( 1,1) t n = 1 and so Kn diverges. Finally if you choose any arbitrary t0 in ( 1, 1), you can find a r so that t0 [ 1 + r, 1 r] and hence S is continuous at t0 for arbitrary t0. This shows S is actually continuous on the full interval ( 1, 1).

6 Example Examine convergence of the series From the ratio test, we find 5(n+1) n+1 t n+1 5n n t n = 5n n tn. n n t = t /. This series thus converges when t / < 1 or t <. at t =, we have 5n n n = comparison to 1/n. 5n which diverges by at t =, we have 5n n ( )n = ( 1)n 5n which converges by the alternating series test. Hence, this series converges to a function S pointwise on [, ). On the interval [ + r, r], we have t/ < 1 r = ρ < 1 and we can defined the sequence (Kn) by sup t [ +r, r] 5n t/ n sup t [ +r, r] 5 t/ n sup t [ +r, r] t/ n = ρ n = Kn. Since Kn is a geometric series with 0 ρ < 1, it converges which implies 5n n tn converges uniformly on [ + r, r] to a function U. Since its are unique, we have S = U on [ + r, r]. Also, since each Sn is continuous as it is a polynomial, the it U = S is continuous on [ + r, r]. Then given any t0 in (, ), t0 is in some [ + r, r] and so S must be continuous on (, ). Note at the endpoint t0 =, we have t Sn(t) = Sn( ) for each n which then implies Sn(t) = Sn( ) = S( ) t

7 Hence, the question of whether or not t S(t) = S( ) is essentially a it interchange issue: is Sn(t) = t t and we don t know how to answer this! Sn(t)? Note in this example, we also really do not know what the pointwise it function is! Example Examine convergence of the series 1 3 n tn. This is the same as the series (t/3)n. Applying the ratio test, we find the series converges when t /3 < 1. at t = 3, we have (1)n =. which diverges. at t = 3, we have ( 1)n which diverges by oscillation between 0 and 1. Hence, this series converges to a function S pointwise on ( 3, 3). We ptws have shown Sn S on ( 3, 3). Is the convergence uniform? Restrict attention to the interval [ 3 + r, 3 r] for suitable r. On this interval, t/3 < 1 r/3 = ρ < 1, we define the sequence (Kn) as follows: sup t/3 n ρ n = Kn. t [ 3+r,3 r]

8 Since Kn is a geometric series with 0 ρ < 1, it converges which implies (t/3)n converges uniformly on [ 3 + r, 3 r] to a function U. Since its are unique, we have S = U on [ 3 + r, 3 r]. Also, since each Sn is continuous as it is a polynomial, the it U = S is continuous on [ 3 + r, 3 r]. Then given any t0 in ( 3, 3), t0 is in some [ 3 + r, 3 r] and so S must be continuous on ( 3, 3). The series we obtain by integrating a series term by term is called the integrated series. Let s look at some examples. Example Examine convergence of the integrated series from n 6 n tn. The integrated series is (n+1) (n+)6n+1 t n+ n (n+1)6 n t n+1 n (n+1)6 n tn+1. So this series converges on ( 6, 6). Also, at t = 6, we have 1n n+1 (n + 1) 1 = t = t /6. n(n + ) 6 =. which diverges. at t = 6, we have 1n n+1 ( 1)n+1 which diverges by oscillation.

9 ptws We have shown Sn S on ( 6, 6). To determine if the convergence is uniform, restrict attention to the interval [ 6 + r, 6 r] for suitable r. On this interval, t/6 < 1 r/6 = ρ < 1, we define the sequence (Kn) as follows: 1n sup t [ 6+r,6 r] n + 1 ( t /6)n+1 1n n+1 ρn+1 = Kn. Using the ratio test, we see Kn converges as 0 ρ < 1. which implies n (n+1)6 n tn+1 converges uniformly on [ 6 + r, 6 r] to a function U. Since its are unique, we have S = U on [ 6 + r, 6 r]. Also, since each partial sum of the integrated series is continuous as it is a polynomial, the it U = S is continuous on [ 6 + r, 6 r]. Then given any t0 in ( 6, 6), t0 is in some [ 6 + r, 6 r] and so S must be continuous on ( 6, 6). Example Examine convergence of the integrated series from The integrated series is find 5n 3 +4 (n +8) 7 n tn. 5n 3 +4 (n+1)(n +8) 7 n tn+1. Using the ratio test, we 5(n+1) 3 +4 (n+)((n+1) +8) 7n+1 t n+ = 5n 3 +4) (n+1)(n +8) 7 n t n+1 5(n + 1) (n + 1)(n + 8) (n + )((n + 1) + 8) 5n 3 t /7 = + 4 t /7. Thus, this series converges on ( 7, 7).

10 at t = 7, we have at t = 7, we have oscillation. ptws We have shown Sn S on ( 7, 7). 7(5n 3 +4) (n+1)(n +8) =. which diverges. 7(5n 3 +4) (n+1)(n +8) ( 1)n+1 which diverges by To determine if the convergence is uniform, restrict attention to the interval [ 7 + r, 7 r] for suitable r. On this interval, t/7 < 1 r/7 = ρ < 1, we define the sequence (Kn) as follows: 7(5n 3 + 4) sup t [ 7+r,7 r] (n + 1)(n + 8) ( t /7)n+1 7(5n3 + 4) (n + 1)(n + 8) ρn+1 = Kn. Using the ratio test, we see Kn converges as 0 ρ < 1. which implies 5n 3 +4 (n+1)(n +8) 7 n tn+1 converges uniformly on [ 7 + r, 7 r] to a function U. Since its are unique, we have S = U on [ 7 + r, 7 r]. Also, since each partial sum of the integrated series is continuous as it is a polynomial, the it U = S is continuous on [ 7 + r, 7 r]. Then given any t0 in ( 7, 7), t0 is in some [ 7 + r, 7 r] and so S must be continuous on ( 7, 7). Comment In our examples, we have seen that the ratio test is an easy way to determine where our series converge and to determine continuity of the pointwise it function we simply use the Weierstrass Uniform Convergence Theorem for Series. Comment Note that the integrated series has the same interval of convergence as the original series in our examples. The next lecture looks at differentiating series of functions and see what happens. This will require that we use the derivative interchange theorem.

11 Homework 5 You need to follow the full arguments we have done for the examples in this Lecture for these problems. Use the Weiestrass Uniform Convergence Theorem for Series here. 5.1 Examine convergence of the series 4 7 n tn and its integrated series. 5. Examine convergence of the series for integrated series. 6 8n (4 n ) tn and its