General Power Series James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University March 29, 2018 Outline Power Series Consequences
With all these preliminaries behind us, we can now look at what are called general power series. Definition The power series in t at base point p is an (t p)n where (an) is a sequence of real numbers. The largest interval (p R, p + R) where this series converges is the interval of convergence of the power series and R is called the radius of convergence. The derived series of this series is n=1 n an (t p)n 1 The integrated series of this series is an n+1 (t p)n+1. From our work in the many examples we have done, we suspect the radius of convergence and interval of convergence for the derived and integrated series is the same as the original series. We also can see all power series converge at one point for sure: the base point p. Theorem Given the power series an (t p)n, the radius of convergence of the series is R = 1/ρ where ρ = limn an+1 an as long as this limit exists. If ρ = 0, then R = and the series converges everywhere. If ρ =, R = 0 and the series only converges at the base point p. The radius of convergence of the derived and integrated series are also R. The series, the derived series and the integrated series are continuous on the interval of convergence and these series converge uniformly on compact subsets of this interval. If R = 0, then all three series converge at only the base point p and questions of continuity are moot. Further, on the interval of convergence, the derived series is the same as taking the derivative of the original series term by term and the integrated series is the same as integrating the original series term by term.
Proof Note the last statement in the theorem means ( ) an(t p) n = (an(t p) n ) = b ( an(t p) )dt n = a Using the ratio test, we have = b an+1 t p n+1 an+1 lim n an t p n = lim n an n an(t p) n 1 an(t p) n dt a b an a n + 1 (t p)n+1 dt t p = ρ t p. where we assume this limit exists. The series thus converges when t p < 1/ρ and letting R = 1/ρ, we see the interval of convergence if (p R, p + r). Let the limit function be S. Proof The derived series has a different ratio test calculation: we get (n + 1) an+1 t p n+1 n + 1 an+1 lim n n an t p n = lim n n an t p = ρ t p. The series thus converges when t p < 1/ρ and letting R = 1/ρ, we see the interval of convergence if (p R, p + r). Let the limit function be D. The integrated series is handled similarly. lim n an+1 n+2 t p n+1 n+1 t p n n = lim an n an+1 t p = ρ t p. n + 1 an The series thus converges when t p < 1/ρ and letting R = 1/ρ, we see the interval of convergence if (p R, p + r). Let the limit function be J. Thus, both the derived and integrated series have the same radius of convergence and interval of convergence as the original series.
Proof To examine uniform convergence of these series when R > 0, first consider the case where R is finite; i.e. ρ > 0 and finite. Look at the intervals [p R + r, p + R r] (p R, p + R). We see R + r t p R r implies t p < R r. Then, letting β = 1 r/r < 1. we have an t p n an (R r) n = an β n R n Define Kn = an β n R n. Using the ratio test, we see lim n an+1 β n+1 R n+1 an+1 an β n R n = lim n an βr = ρβr = (1/R)βR = β < 1 and so Kn converges. This tells us the series converges uniformly on [p R + r, p + R r] to a function U. Proof Since the partial sums of the series are polynomials, the limit function U is continuous by the Second Weierstrass Uniform Convergence Theorem for series. Further, by uniquess of limits S = U on [p R + r, p + R r]. Next, given any t (p R, p + R), t is in some [p R + r, p + R r] and so S is continuous at t. Since t is arbitrary, S is continuous on (p R, p + R). If ρ = 0 giving us R =. Look at intervals of the form [p T, p + T ] for any T > 0. Then t p < T and an t p n an T n Define Kn = an T n. Using the ratio test, we see lim n an+1 T n+1 an+1 an T n = lim n an T = ρ T = 0
Proof So Kn converges. This tells us the series converges uniformly on [p T, p + T ] to a function U. Since the partial sums of the series are polynomials, the limit function U is continuous by the Second Weierstrass Uniform Convergence Theorem for series. Further, by uniquess of limits S = U on [p T, p + T ]. Next, given any t R, t is in some [p T, p + T ]and so S is continuous at t. Since t is arbitrary, S is continuous on R. The arguments to show the derived and integrated series converge to a continuous function on (p R, p + R) with the convergence uniform on compact subsets [a, b] (p R, p + R) are similar. Hence, we know S n unif D on [a, b] and S n ptws D on (p R, p + R) with D continuous on (p R, p + R). If we let Jn be the antiderivative of Sn with integration constant unif ptws chosen to be zero, then Jn J on [a, b] and Jn J on (p R, p + R) with J continuous on (p R, p + R). Proof To see that S = D, we need to check the conditions of the Derivative Interchange Theorem: 1. Fix a t in [p, p + R). Sn is differentiable on [p, t]: True. 2. S n is Riemann Integrable on [p, t]: True as each is a polynomial. 3. There is at least one point t0 [0, t] such that the sequence (Sn(t0)) converges. True as the original series converges at t = p. 4. S n unif y on [p, t] and the limit function y is continuous. True as S n unif D on [p, t] and D is continuous. The conditions of derivative interchange theorem are thus satisfied and unif we can say there is a function W on [p, t] so that Sn W on [p, t] and W = D. Since limits are unique, we then have W = S with S = D. We can do a similar argument for t (p R, p]. Hence, on (p R, p + R), S = D.
Proof If we look at the integrated series, we have Jn = unif Sn J on [a, b] (p R, p + R). Since convergence is uniform on [a, b], we have b limn a Sn(t) dt = b (limn Sn(t)) dt or a b Sn(t) dt = b S(t) dt. limn a These last two results are the same as saying ( ) an(t p) n = b ( an(t p) )dt n = a a = (an(t p) n ) = n an(t p) n 1 b an(t p) n dt a b an a n + 1 (t p)n+1 dt Using these ideas we can define a new vector space of functions: the analytic functions. Definition A function f is called analytic at p if there is a power series an(t p)n with a positive radius of convergence R so that f (t) = an(t p)n on the interval of convergence (p R, p + R). We say f has a local power series expansion in this case. Comment The power series converges to a function S which is continuous on (p R, p + R) and so the function f must be continuous there as well. Hence, we are saying the function f has an alternate series form which matches it exactly on the interval of convergence.
Comment We know the series expansion is differentiable and the derived series that results has the same radius of convergence. Hence, we know f (t) = n=1 n an(t p)n 1. This immediately says f (p) = a1. We can find as many higher order derived series as we wish. Thus, we know f must be infinitely differentiable and that the k th derivative must be f (k) = n=k n(n 1)... (n k + 1) an(t p)n k. This tells us f (k) (p) = k(k 1)... (k k + 1)ak = k!ak. Hence, we see ak = f (k) (p)/k! which is the same coefficient we see in Taylor polynomials. Comment It is easy to see sums, differences and scalar multiples of functions analytic at t = p are still analytic. These combinations may have a different radius of convergence but they will still have a finite one and hence are analytic at t = p. The set of functions analytic at t = p is thus a vector space of functions. Not all functions are analytic. f (t) = t is not analytic at t = 0 even though it is continuous on [0, ) because it does not have derivatives at t = 0. The C bump functions discussed earlier are not analytic at their endpoints. Recall the function ha b(x) defined by for any a < b is given by ha b(x) = fa(x) gb(x) = e 0, x a b a (x a)(x b), a < x < b 0, x b We can show the n th derivative h (n) a b (x) is zero at x = a and x = b for all orders n. Hence, if ha b(x) was analytic at x = a, we would have ha b(x) = an(x a) p and by our earlier comments, we know an = h (n) a b (a)/n! = 0. Thus, the power series would sum to identically zero locally at x = a and that contradicts the behavior of the bump function to the right of x = a. We can say similar things about what happens at x = b.
Another example is f (t) = t which is not differentiable at all. It cannot have a local power series expansion at x = 0. What about the Weierstrass Approximation Theorem? The functions f (t) = t is continuous locally at t = 0 and so there is a sequence of unif Bernstein polynomials pn(t) so that pn f on say [ 1, 1]. Then we can write pn(t) = Qn k=0 ak,n t n = a0,n + a1,nt + a2,nt 2 +... + aqn,nt Qn There is no guarantee in the Weierstrass Approxation Theorem that the degree of these polynomials increase monotonically. For example, we could have Q1 = 5, Q2 = 3, Q3 = 7, Q4 = 11 and so forth. To say it differently, suppose we stop at n = 100 and let N = max{q1,..., Q100}. Then all the polynomials can be written as degree N polynomials by just letting the coefficients corresponding to powers of t not is the original polynomial be zero. For example, if N = 200 p1(t) = a0,1 + a1,1t + a2,1t 2 +... + a5,1t 5 This would be a polynomial of degree 200 by just seting the coefficients ak,1 = 0 if k > 5. We would then do a similar thing for p2 and so on. We could have p2(t) = a0,2 + a1,2t + a2,2t 2 +... + a7,2t 7 which is also a polynomial of degree 200 by setting some coefficients to zero. If there was a power series expansion, we would have to have a sequence (bn) so that p1(t) = a0,1 + a1,1t + a2,1t 2 +... + a5,1t 5 = b0 + b1t +... + b5t 5 p2(t) = a0,2 + a1,2t + a2,2t 2 +... + a7,2t 7 = b0 + b1t +... + b5t 5 + b6t 6 + b7t 7 which implies equality in coefficients for the p1 and p2 which we do not know is true. Hence, the polynomials we find invoking the Weierstrass Approximation Theorem do not, in general, come from a power series expansion.
Comment It is easy to see the collection of all continuous functions locally defined on R at the point p is a vector space. From our discussions above, we see the set of function analytic at p is a strict subset of this space. It should be clear now that the vector space of Riemann Integrable functions on the set [a, b] is quite interesting. The vector space of all continuous functions on [a, b] is a strict subset of it as there are many discontinuous functions which are Riemann Integrable. The set of all polynomials is also a vector space on [a, b] and from the Weierstrass Approximation Theorem the metric space (C([a, b]), ) has within it a very special subset. Definition We say E is a dense subset of the metric space (X, d) if given x in E, for all ɛ > 0, there is an y E so that d(x, y) < ɛ. If the set E is countable, we say E is a countable dense subset. A metric space with a countable dense subset is called separable. Comment Q is countably dense in R. By the Weierstrass Approximation Theorem, the vector space of polynomials on [a, b] is dense in (C([a, b]), ). If we restrict our attention to polynomials with rational coefficients, we can show that it is a countably dense subset. Here is a sketch of the proof of this statement. Given ɛ > 0, there is a polynomial p so that f p < ɛ/2 on [a, b]. We have p(x) = N k=0 akx k where N is the degree of the polynomial p. Now the function θ(x) = 1 + x + x 2 +... + x N is continuous on [a, b], so there is a positive constant B so that θ < B on [a, b]. Since Q is dense in R, there are rationals {r0,..., rn} so that ri ai < ɛ/(2b).
Then if q is the polynomial q(x) = N k=0 rkx k, q(x) p(x) N ak rk x k < ɛ/(2b) θ(x) < ɛ/(2) k=0 which implies q p ɛ/2. Combining, we see f q < ɛ/2 + ɛ/2 = ɛ. Hence, the set of polynomials with rational coefficients is dense in C([a, b], ). Why is this set countable? We need a general theorem for this. Theorem A finite union of countable sets is countable and an countably infinite union of countable sets is countable. Proof The first part is an induction. The base case for k = 1 is obvious, so let s assume any union of k countable sets is countable. Let A1 to Ak be countable sets and then by assumption B = k i=1 Ai is countable. Thus there is a 1 1 and onto function φ between B and N. Let C = k+1 i=1 Ai = B Ak+1. We know Ak+1 is countable so there is a function θ that is 1 1 and onto between Ak+1 and N. Write C as the disjoint union C = C1 C2 C3 = (B Ak+1) (B C Ak+1) (B A C k+1). Each of these sets is countable so they can be labeled as C1 = {ri}, C2 = {si} and C3 = {ti}. Define the 1 1 and onto function ψ by ψ(x) = { 3n 2, x = rn C1 3n 1, x = sn C2 3n, x = tn C3 Hence, C is countable. This completes the POMI proof.
Proof To show countably infinite unions of countable sets are countable, we can use a similar argument. Let {Ai} be a countable collection of countable sets. We can show that we can rewrite the union i=1ai as a union of disjoint sets by defining the sequence of sets {Bi} by B1 = A1 and Bi = Ai Bi 1 C. Then i=1 Ai = i=1bi. Each Bi is countable and can be enumerated as Bi = {aj,i} Define the 1 1 and onto function ψ by {1/1, 2/1, 3/1,..., j/1,...}, x = aj,i B1 {1/2, 2/2, 3/2,..., j/2,...}, x = aj,2 B2 ψ(x) =.. {1/n, 2/n, 3/n,..., j/n,...}, x = aj,n Bn.. The range of ψ is the positive rationals with repeats and we already know this is a countable set. Thus, i=1ai is countable. With this theorem, we can show the set of polynomials with rational coefficients is countable. Call this set P and call the polynomials of degree less than or equal to n, with rational coefficients Pn. Then P = Pn. We can show each Pn is countable and so the union of all of them is countable also. Hence the polynomials with rational coefficients are dense in (C([a, b]), ). This says (C([a, b]), ) is a separable metric space as it possesses a countable dense subset.
Homework 27 27.1 Show i=1 Ai = i=1 Bi where B1 = A1 and Bn = An BC n 1. 27.2 Show P3 is countable. 27.3 If f, g and h are analytic at x = p, prove 3f + 5g 4h is analytic at x = p.