Differentiating Series of Functions

Transcription

1 Differentiating Series of Functions James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University October 30, 017

2 Outline 1 Differentiating Series

3 Differentiating Series Let s look at the problem of differentiating a series of functions. We start with an old friend, the series n=0 tn which we know converges pointwise to S(t) = 1/(1 t) on ( 1, 1). It also is known to converge uniformly on compact subsets of [ 1, 1]. We only showed the argument for sets like [ 1 + r, 1 r] but it is easy enough to go through the same steps for [c, d] in ( 1, 1). We used the uniform convergence to prove the pointwise limit had to continuous on the compact subsets of ( 1, 1) and easily extended the continuity to the whole interval ( 1, 1). This type of argument did not need to know the pointwise limit was 1/(1 t) and so we wokred through some other examples to illustrate how it works with series for which we cannot find the pointwise limit function. Our primary tools were the UCC which was a bit cumbersome to use, and the Second Weierstrass Uniform Convergence Theorem, which was much faster.

4 Differentiating Series Now let s look at what is called the derived series of n=0 tn. This is the series we get by differentiating the original series term by term. Let s call this series D(t) = n tn 1. Note the n = 0 term is gone as the derivative of a constant is zero. This is the series constructed from the partial sums S n. Using the ratio test, we find (n + 1) t n lim n n t n 1 (n + 1) = lim t = t n n Thus the series converges when t < 1 or on the interval ( 1, 1). At t = 1, the series is n which diverges and at t = 1, the series ( 1)n 1 n diverges by oscillation. To find out about uniform convergence, note on the interval [ 1 + r, 1 r] for sufficiently small r, we have t 1 r = ρ < 1. Define the sequence (K n ) by sup n t n 1 nρ n 1 = K n t [ 1+r,1 r]

5 Differentiating Series Then nρn 1 converges by the ratio test. We conclude by the Second Weierstrass Uniform Convergence Theorem that the series n tn 1 converges uniformly on [ 1 + r, 1 r] to a function V. Since each partial sum of n tn 1 is continuous, the uniform limit function V is continuous also and D = V on [ 1 + r, 1 r] since limits are unique. Finally, if t 0 in ( 1, 1) is arbitrary, it is inside some [ 1 + r, 1 r] and so V = D is continuous there also.

6 Differentiating Series Let s check the conditions of the derivative interchange theorem, applied to the sequence of partial sums (S n ). Fix a t in [0, 1). 1 S n is differentiable on [0, t]: True. S n is Riemann Integrable on [0, t]: True as each is a polynomial. 3 There is at least one point t 0 [0, t] such that the sequence (S n (t 0 )) converges. True as the series n=0 tn converges at t = 0. 4 S n unif y on [0, t] and the limit function y is continuous. True as we have just shown S n unif D on [0, t]. The conditions of derivative interchange theorem are satisfied and we can unif say there is a function W on [0, t] so that S n W on [0, t] and W = D. Since limits are unique, we then have W = S with S = D. We can do a similar argument for t ( 1.0]. Since we also know S(t) = 1/(1 t), this proves that on ( 1, 1) ( 1 1 t = ) ( ) 1 t n = nt n 1 1 = = 1 t (1 t) n=0

7 Example Examine convergence of the derived series for The derived series here is lim n 1 n+1 t n 1 n t n 1 = lim 5n ( n ) tn. 5 ( n ) tn 1. Using the ratio test, we find n 1 t = t /. This series thus converges when t / < 1 or t <. at t =, we have it sums to infintity. at t =, we have diverges by oscillation. 5 n 1 = n 10 which diverges since 5 ( ) n 1 = n ( 1)n which

8 Hence, this series converges to a function D pointwise on (, ). Recall the original series also converged at t = but we do not have convergence of the derived series at this endpoint. To find out about uniform convergence, note on the interval [ + r, r] for sufficiently small r, we have t / 1 r = ρ < 1. Define the sequence (K n ) by sup t [ 1+r,1 r] 1 5 n t n 1 = sup t [ 1+r,1 r] 5 ( t /)n 1 ρ n 1 = K n Then ρn 1 converges by the ratio test. We conclude by the Second Weierstrass Uniform Convergence Theorem that the derived series 5 ( n ) tn 1 converges uniformly on [ + r, r] to a function V. Since each partial sum of 5 ( n ) tn 1 is continuous, the uniform limit function V is continuous also and D = V on [ + r, r] since limits are unique.

9 Finally, if t 0 in (, ) is arbitrary, it is inside some [ + r, r] and so V = D is continuous there also. Note we have shown S n unif D on [ + r, r] for all r with [ + r, r] (, ). Let s check the conditions of the derivative interchange theorem, applied to the sequence of partial sums (S n ). Fix a t in [0, ). 1 S n is differentiable on [0, t]: True. S n is Riemann Integrable on [0, t]: True as each is a polynomial. 3 There is at least one point t 0 [0, t] such that the sequence (S n (t 0 )) converges. True as the series 5n ( n ) tn converges at t = 0. 4 S n unif y on [0, t] and the limit function y is continuous. True as we have just shown S n unif D on [0, t].

10 The conditions of the derivative interchange theorem are satisfied and we unif can say there is a function W on [0, t] so that S n W on [0, t] and W = D. Since limits are unique, we then have W = S with S = D. We can do a similar argument for t (, 0]. Hence, on (.), ( S (t) = ) 5n ( n ) tn = 5 ( n ) tn 1 = D(t) This is why we say we can differentiate the series term by term!

11 Example Examine convergence of the derived series for 1 3 n t n. The derived series is n 3 n t n 1 Applying the ratio test, we find lim n n+1 3 n+1 t n n 3 n t n 1 = lim n n + 1 n t /3 = t /3. This series thus converges when t /3 < 1 or t < 3. at t = 3, we have n 3 3 n 1 = n comparison to 1 n. 1 3n which diverges by at t = 3, we have n 3 ( 3) n 1 = n ( 1)n 1 n 3 which diverges by oscillation. Hence, this series converges to a function D pointwise on ( 3, 3)

12 To find out about uniform convergence, note on the interval [ 3 + r, 3 r] for sufficiently small r, we have t /3 1 r = ρ < 1. Define the sequence (K n ) by sup n t/3 n 1 nρ n 1 = K n t [ 1+r,1 r] Then ρn 1 converges by the ratio test. We conclude by the Second Weierstrass Uniform Convergence Theorem that the derived series 5 t n 1 converges uniformly on n [ 3 + r, 3 r] to a function V. Since each partial sum of n(t/3)n 1 is continuous, the uniform limit function V is continuous also and D = V on [ 3 + r, 3 r] since limits are unique. Finally, if t 0 in ( 3, 3) is arbitrary, it is inside some [ 3 + r, 3 r] and so V = D is continuous there also. Note we have shown S n unif D on [ 3 + r, 3 r] for all r with [ 3 + r, 3 r] ( 3, 3).

13 Let s check the conditions of the derivative interchange theorem applied to the sequence of partial sums (S n ). Fix a t in [0, 3). 1 S n is differentiable on [0, t]: True. S n is Riemann Integrable on [0, t]: True as each is a polynomial. 3 There is at least one point t 0 [0, t] such that the sequence (S n (t 0 )) converges. True as the series 1 3 t n converges at n t = 0. 4 S n unif y on [0, t] and the limit function y is continuous. True as we have just shown S n unif D on [0, t]. The conditions of derivative interchange theorem are satisfied and we can unif say there is a function W on [0, t] so that S n W on [0, t] and W = D. Since limits are unique, we then have W = S with S = D. We can do a similar argument for t ( 3, 0].

14 Hence, on ( 3, 3) ( ) S (t) = (t/3) n = n=0 n(t/3) n 1 = D(t) The series n=0 (t/3)n is a geometric series and converges to 1/(1 (t/3)) on ( 3, 3), Hence, we can also say ( ) ( 1 ) = (t/3) n = 1 (t/3) n=0 n(t/3) n 1 1/3 = (1 (t/3))

15 We can take higher derivatives of series. The series obtained by differentiating the derived series term by term is called the second derived series. The third derived series is the series derived from the second derived series and so forth. It is easiest to explain how this works in an example. Example For the series 5 n=0 8(n+1) t n examine the first, second and third derived series for convergence. The first, second and third derived series are (First Derived Series:) This is (Second Derived Series:) This is n= (Third Derived Series:) This is n=3 5n 8(n+1) t n 1. 5(n 1)n 8(n+1) t n. 5(n )(n 1)n 8(n+1) t n 3.

16 We find all of these series converge on ( 1, 1) using the ratio test: we want all of these limits less than 1: (First Derived Series:) lim n 5(n+1) 8(n+) t n (Second Derived Series:) lim n 5n = lim 8(n+1) t n 1 n 5(n)(n+1) 8(n+) t n 1 (Third Derived Series:) lim n = lim 5(n 1)n 8(n+1) t n n 5(n 1)(n)(n+1) 8(n+) t n 5(n )(n 1)n 8(n+1) t n 3 = lim n (n + 1) 3 t = t n(n + ) n(n + 1) 3 t = t (n 1)n(n + ) (n 1)n(n + 1) 3 t = t (n )(n 1)(n + )

17 It is clear from these limits that all three series converge pointwise to a limit function on ( 1, 1). We could test if they converge at the endpoints but we will leave that to you. We check all three series for uniform convergence in the same way. On the interval [ 1 + r, 1 r] for small enough r, we have t 1 r = ρ < 1. We have these estimates: (First Derived Series:) 5n 8(n + 1) t n 1 5(n + 1) 8(n + 1) ρn 1 1 (n + 1) ρn 1 So K n = 1 (n+1) ρn 1. Since K n converges by the ratio test since 0 ρ < 1, we see the first derived series converges uniformly on [ 1 + r, 1 r].

18 (Second Derived Series:) n= 5(n 1)n 8(n + 1) t n n= 5(n + 1) 8(n + 1) ρn n= ρ n So K n = ρ n. Since K n converges by the ratio test since 0 ρ < 1, we see the second derived series converges uniformly on [ 1 + r, 1 r]. (Third Derived Series:) n=3 5(n )(n 1)n 8(n + 1) t n 3 n=3 5(n + 1) 3 8(n + 1) ρn 3 (n + 1)ρ n 3 n=3 So K n = (n + 1)ρ n 3. Since K n converges by the ratio test since 0 ρ < 1, we see the third derived series converges uniformly on [ 1 + r, 1 r].

19 By the Second Weierstrass Uniform Convergence Theorem, we see all these derived series converge uniformly. We have (First Derived Series:) S n (Second Derived Series:) S n (Third Derived Series:) S n unif D on [ 1 + r, 1 r]. unif E on [ 1 + r, 1 r]. unif F on [ 1 + r, 1 r]. Since the partial sums of these derived series are continuous and the convergence is uniform D, E and F are continuous on [ 1 + r, 1 r]. Further given any t ( 1, 1), t is in some [ 1 + r, 1 r] and so we know D, E and F are continuous on ( 1, 1). Now we check the derivative interchange for these derived series

20 (First Derived Series:) 1 Fix a t in [0, 1). S n is differentiable on [0, t]: True. S n is Riemann Integrable on [0, t]: True as each is a polynomial. 3 There is at least one point t 0 [0, t] such that the sequence (S n (t 0 )) converges. True as the original series converges at t = 0. 4 S n unif y on [0, t] and the limit function y is continuous. True as we have just shown S n unif D on [0, t]. The derivative interchange theorem conditions are satisfied and so unif there is a function W on [0, t] with S n W on [0, t] and W = D. Since limits are unique, we then have W = S with S = D. Do a similar argument for t ( 1.0]. So on ( 1, 1) ( S (t) = ) 5 8(n + 1) tn 1 = 5n 8(n + 1) tn 1 = D(t)

21 (Second Derived Series:) 1 Fix a t in [0, 1). S n is differentiable on [0, t]: True. S n is Riemann Integrable on [0, t]: True as each is a polynomial. 3 There is at least one point t 0 [0, t] such that the sequence (S n(t 0 )) converges. True as the derived series converges at t = 0. 4 S n unif y on [0, t] and the limit function y is continuous. True as we have just shown S n unif E on [0, t]. The derivative interchange theorem is satisfied and so there is X on [0, t] so that S n unif X on [0, t] and X = E. Limits are unique, so X = D with D = E. From the first derived series, we have S = D = E. We can do a similar argument for t ( 1.0].

22 Hence, on ( 1, 1) ( D (t) = ( S (t) = n=0 ) 5n 8(n + 1) tn 1 = 5 8(n + 1) tn 1 ) = = n= ( n= 5(n 1)n 8(n + 1) tn = E(t) 5n 8(n + 1) tn 1 ) 5(n 1)n 8(n + 1) tn = E(t)

23 (Third Derived Series:) 1 Fix a t in [0, 1). S n is differentiable on [0, t]: True. S n is Riemann Integrable on [0, t]: True as each is a polynomial. 3 There is at least one point t 0 [0, t] such that the sequence (S n (t 0 )) converges. True as the second derived series converges at t = 0. 4 S n unif y on [0, t] and the limit function y is continuous. True as we have just shown S n unif F on [0, t]. The conditions of derivative interchange theorem are satisfied and we can say there is a function Y on [0, t] so that S n unif Y on [0, t] and Y = F. Since limits are unique, we then have Y = E with E = F. Using the results from the first and second derived series, we have S = E = F. We can do a similar argument for t ( 1, 0].

24 Hence, on ( 1, 1) ( E (t) = n= ( S (t) = ( = n= n=0 ) 5(n 1)n 8(n + 1) tn = 5 8(n + 1) tn 1 5(n 1)n 8(n + 1) tn ) = n=3 ) ( = n=3 5(n )(n 1)n 8(n + 1) t n 3 = F (t) 5n 8(n + 1) tn 1 ) 5(n )(n 1)n 8(n + 1) t n 3 = F (t)

25 Homework 6 You need to follow the full arguments we have done for the examples in this Lecture for these problems. So lots of explaining! 6.1 Examine the convergence of the derived series of n=0 5 n4 n t n. 6. Examine the convergence of the derived series and the second derived series of n=0 6 n t n.