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Math 360: Series of Constants and Functions

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Math 360: Series of Constants and Functions Math 360: Series of constants and functions D. DeTurck University of Pennsylvania November 26, 2017 D. DeTurck Math 360 001 2017C: 1 / 26 Series A series is an infinite sum, i,e, the sum of a sequence. To make sense of this, given the sequence fang = a1; a2; a3;::: we form a new sequence { the sequence of partial sums fsng where s1 = a1 s2 = a1 + a2 s3 = a1 + a2 + a3 . n X sn = a1 + a2 + ··· + an = ak k=1 D. DeTurck Math 360 001 2017C: 2 / 26 Sum/limit of a series If the sequence of partial sums sn has a limit, say lim sn = S, n!1 then we say that \the series converges to S" or \the sum of the series is S, and write 1 n X X ak = lim ak = S: n!1 k=1 k=1 Otherwise, the series diverges. 1 1 X 1 X 1 For instance, = 1 and = e − 1. 2k n! k=1 k=1 D. DeTurck Math 360 001 2017C: 3 / 26 Cauchy criterion We can bring to bear everything we already know about sequences to series via the sequence of partial sums. For instance, a series is Cauchy if the sequence of partial sums is, in other words given any " > 0 there is an N > 0 such that for all m; n > N we have n X jsn − smj = ak = jam+1 + am+2 + ··· + anj < " k=m+1 A series is convergent if and only if it is Cauchy. In particular (with n = m + 1): If a series is convergent, then lim an = 0. (nth term test). The harmonic series shows that the n!1 converse is not true. D. DeTurck Math 360 001 2017C: 4 / 26 Series with positive terms If ak > 0 for each k, then the sequence of partial sums sn is monotonically increasing. By the fundamental axiom, the sequence 1 X fsng and hence the series ak , will converge if there is an upper k=0 bound to the sequence of partial sums. This observation is at the root of many of the familiar \tests" for convergence: Comparison test Suppose bk ≥ ak ≥ 0 for every k. X X (a) If bk converges then so does ak X X (b) If ak diverges then so does bk In both cases the partial sums for one series bound those of the other, either from above or below. D. DeTurck Math 360 001 2017C: 5 / 26 Integral test Integral test Suppose f (x) ≥ 0 is a decreasing function from [0; 1) to R. Then 1 X the series f (n) converges if and only if n=1 M lim f (x) dx M!1 ˆ1 exists and is finite. 1 We usually write that limit as the \improper integral" f (x) dx. ˆ1 1 1 X X n Proof: Compare f (n) to f (x) dx (for convergence) or ˆ n=1 n=1 n−1 1 X n+1 to f (x) dx (for divergence). ˆ n=1 n D. DeTurck Math 360 001 2017C: 6 / 26 Ratio test Ratio test 1 X For the series ak (assume ak > 0 for the time being), suppose k=1 a lim k+1 = L: k!1 ak (a) If L < 1 then the series converges. (b) If L > 1 then the series diverges. (c) If L = 1 then no conclusion can be drawn X Proof: Comparison with the geometric series r n for r between L and 1. Some other tests (limit comparison test, root test) will be discussed in the homework. D. DeTurck Math 360 001 2017C: 7 / 26 Alternating series 1 X k+1 A series of the form ± (−1) ak (with ak > 0) is called k=1 alternating. Alternating series test If ak is a positive monotonically decreasing sequence then 1 X k+1 (−1) ak converges if and only if ak ! 0. k=1 Proof: Consider the subsequence s2n { it's monotonically increasing and bounded above by a1. Since an ! 0, the subsequence s2n+1 converges to the same limit. But then so does all of sn. D. DeTurck Math 360 001 2017C: 8 / 26 Absolute convergence X Say that a series an of arbitrary numbers converges X absolutely if janj converges. Using the Cauchy criterion, an absolutely convergent series is itself convergent. So we can extend the ratio test by considering lim jan+1=anj. D. DeTurck Math 360 001 2017C: 9 / 26 Pointwise convergence Now we'll begin consideration of sequences and series of functions (pay attention to whether it's a sequence or a series!). Definition: (pointwise convergence) X A sequence ffng or series fn of functions converges pointwise to F in the interval from a to b (open or closed or otherwise) if for every x in the interval the numerical sequence ffn(x)g or series X fn(x) converges to F (x). −nx Think about fe g on [0; 1) or farctan nxg on all of R. (Pointwise limits of continuous functions need not be continuous). Or the sequence f(n + 1)xng on [0; 1). (The integral of the limit is not the limit of the integrals). D. DeTurck Math 360 001 2017C: 10 / 26 Uniform convergence We need a stronger notion of convergence to preserve desirable properties: Definition: (uniform convergence) A sequence fn of functions on a domain D converges uniformly to F if for any " > 0 there is an N > 0 (depending only on " and not on x 2 D) such that jF (x) − fn(x)j < " for all n ≥ N and all x 2 D Define the notion of \uniformly Cauchy" and show that a uniformly Cauchy sequence is uniformly convergent. As we shall see, the limit of a uniformly convergent sequence of continuous functions is continuous. Likewise for integrable, and the integral of the limit is equal to the limit of the integrals. D. DeTurck Math 360 001 2017C: 11 / 26 The Weierstrass M-test A result that is often helpful for showing that a series of functions is uniformly convergent is Theorem (Weierstrass M-test) 1 X Consider the series fn(x) on some domain D. If there are n=1 constants Mn > 0 such that jfn(x)j < Mn for all x 2 D and all n, 1 X and if the series of numbers Mn converges, then the series n=1 1 X fn(x) converges uniformly on D n=1 To prove this, note that the Mn series is Cauchy, so the fn series is uniformly Cauchy. D. DeTurck Math 360 001 2017C: 12 / 26 Three fundamental theorems We'll look at the relationship between uniform convergence and continuity, integrability and differentiability. Theorem (uniform convergence and continuity) Let ffn(x)g be a uniformly convergent sequence of continuous functions on the domain D. Then the limit function is continuous. Proof: Let f (x) be the limit function. To show that f is continuous at x 2 D, let " > 0 be given. Then there is an N > 0 so that 1 jf (x) − fn(x)j < 3 " for all x 2 D and all n > N. For such an n, we 1 have that there is a δ > 0 so that jfn(x) − fn(y)j < 3 " for all y 2 D with jx − yj < δ. Therefore for y 2 D with jy − xj < " we have " jf (x)−f (y)j ≤ jf (x)−f (x)j+jf (x)−f (y)j+jf (y)−f (y)j < 3 = ": n n n n 3 D. DeTurck Math 360 001 2017C: 13 / 26 Three fundamental theorems Theorem (uniform convergence and integrability) Let ffn(x)g be a uniformly convergent sequence of integrable functions on [a; b]. Then the limit function f :[a; b] ! R is also integrable, and b b lim fn(x) dx = f (x) dx: n!1 ˆa ˆa To show: Given " > 0, U(f ; P) < L(f ; P) + " for some partition P. By uniform convergence, there is N > 0 such that for n > N and 1 some partition P, U(fn; P) < L(fn; P) + 3 " and " " f (x) − < f (x) < f (x) + for all x 2 [a; b] n 3(b − a) n 3(b − a) 1 2 But then U(f ; P) < U(fn; P) + 3 " < L(fn; P) + 3 " < L(f ; P) + ". D. DeTurck Math 360 001 2017C: 14 / 26 Three fundamental theorems Theorem (uniform continuity and differentiability) Let ffn(x)g be a sequence of continuously differentiable functions on an interval I that converges pointless to f (x) and such that the 0 sequence of derivatives ffn(x)g converges uniformly on I . Then f (x) is continuously differentiable and f f 0(x) = lim f 0(x) on I . n!1 n Choose x0 2 I and use the preceding theorem to get x x 0 0 f (x)−f (x0) = lim fn(x)−fn(x0) = lim f (t) dt = lim f (t) dt n!1 n!1 n n!1 n ˆx0 ˆx0 which shows that f 0(x) = lim f 0(x), and f 0(x) is continuous by n!1 n 0 uniform convergence of the fn. (There is a version of this theorem that does not assume continuity of the derivatives.) D. DeTurck Math 360 001 2017C: 15 / 26 Corollaries for series Each of the three preceding theorems has a version for series: Corollary 1 X Suppose the series fn(x) converges uniformly on [a; b] Then n=1 1 If the fn are continuous, then the sum is continuous. 2 If the fn are integrable, then the sum is integrable and the series may be integrated term-by-term.
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