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1 15. Alternating Series and Integral Tests Integral test Here are the conditions under which an integral test is advisable: 0 a n Z n+1 n f (x)dx and Z 1 f (x)dx < = X n=1 a n converges . a n Z n+1 n f (x)dx and Z 1 f (x)dx = = X n=1 a n diverges . Example 1 (i) n=1 1 n = . (ii) n=1 1 n 2 converges . (iii) n=1 1 n p converges iff p> 1. Theorem 1.1 (Alternating Series Theorem) If a 1 a 2 ≥···≥ 0 and lim n→∞ a n =0, then the alternating se- ries s n = k=1 (-1) k+1 a k converges. Moreover, the par- tial sums s n = n k=1 (-1) k+1 a k satisfies |s - s n |≤ a n for all n. 1
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Page 1: Integral testyaozhong/lecture 12_2020.pdf · Integral test Here are the conditions under which an integral test is advisable: 0 a n Z n+1 n f(x)dx and Z 1 1 f(x)dx

1 15. Alternating Series and Integral Tests

Integral testHere are the conditions under which an integral test isadvisable:

0 ≤ an ≤∫ n+1

n

f(x)dx and

∫ ∞

1

f(x)dx <∞

=⇒∞∑n=1

an converges .

an ≥∫ n+1

n

f(x)dx and

∫ ∞

1

f(x)dx =∞

=⇒∞∑n=1

an diverges .

Example 1 (i)∑∞

n=11n =∞.

(ii)∑∞

n=11n2 converges .

(iii)∑∞

n=11np converges iff p > 1.

Theorem 1.1 (Alternating Series Theorem) If a1 ≥a2 ≥ · · · ≥ 0 and limn→∞ an = 0, then the alternating se-ries sn =

∑∞k=1(−1)k+1ak converges. Moreover, the par-

tial sums sn =∑n

k=1(−1)k+1ak satisfies |s− sn| ≤ an forall n.

1

Page 2: Integral testyaozhong/lecture 12_2020.pdf · Integral test Here are the conditions under which an integral test is advisable: 0 a n Z n+1 n f(x)dx and Z 1 1 f(x)dx

2 17. Continuous function

Definition 2.1 Let f be a real-valued function whose do-main is a subset of R. The function f is continuous atx0 in Dom(f) if, for every sequence xn in Dom(f) con-verging to x0, we have limn→∞ f(xn) = f(x0). If f iscontinuous at each point of a set S ⊆ Dom(f), then f issaid to be continuous on S. The function f is said to becontinuous if it is continuous on Dom(f).

Theorem 2.2 Let f be a real-valued function whose do-main is a subset of R. Then f is continuous at x0 inDom(f) if and only iffor each ε > 0 there exists δ > 0 such that x ∈ Dom(f)and |x− x0| < δ imply |f(x)− f(x0)| < ε.

Example 2 Let f(x) = 2x2 + 1 for x ∈ R. Prove f iscontinuous in R by using the definition and the theorem.

Example 3 Let f(x) = x2 sin 1x for x 6= 0 and f(0) = 0.

Prove f is continuous at 0.

Example 4 Let f(x) = 1x sin

1x2 for x 6= 0 and f(0) = 0.

Prove f is discontinuous at 0.

Let xn =1√

2nπ+π2

.

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Page 3: Integral testyaozhong/lecture 12_2020.pdf · Integral test Here are the conditions under which an integral test is advisable: 0 a n Z n+1 n f(x)dx and Z 1 1 f(x)dx

Theorem 2.3 Let f be a real-valued function with Dom(f) ⊆R. If f is continuous at x0 in Dom(f), then |f | andkf, k ∈ R, are continuous at x0.

Proof

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Page 4: Integral testyaozhong/lecture 12_2020.pdf · Integral test Here are the conditions under which an integral test is advisable: 0 a n Z n+1 n f(x)dx and Z 1 1 f(x)dx

Theorem 2.4 Let f and g be real-valued functions thatare continuous at x0 in R. Then

(i) f + g is continuous at x0;

(ii) fg is continuous at x0;

(iii) f/g is continuous at x0 if g(x0) 6= 0.

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Page 5: Integral testyaozhong/lecture 12_2020.pdf · Integral test Here are the conditions under which an integral test is advisable: 0 a n Z n+1 n f(x)dx and Z 1 1 f(x)dx

Theorem 2.5 If f is continuous at x0 and g is contin-uous at f(x0), then the g ◦ f is continuous at x0.

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Page 6: Integral testyaozhong/lecture 12_2020.pdf · Integral test Here are the conditions under which an integral test is advisable: 0 a n Z n+1 n f(x)dx and Z 1 1 f(x)dx

Example 5 e|x| + sinx+ cos(x3) is continuous.

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Page 7: Integral testyaozhong/lecture 12_2020.pdf · Integral test Here are the conditions under which an integral test is advisable: 0 a n Z n+1 n f(x)dx and Z 1 1 f(x)dx

Example 6 if f and g are continuous, then max(f, g)are continuous.

SOLUTION max(f, g) = 12(f + g) + 1

2 |f − g|

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Page 8: Integral testyaozhong/lecture 12_2020.pdf · Integral test Here are the conditions under which an integral test is advisable: 0 a n Z n+1 n f(x)dx and Z 1 1 f(x)dx

Example 7 The Gateway Arch in St. Louis is builtaround a mathematical curve called “catenary”. Theheight of this catenary above the ground at a point x feetfrom the center line is

y = 693.8597− 34.3836(e.0100333x + e−.0100333x) .

Graph this function with window [−299.2239, 299.2239]×[0, 700]. plot(x,693.8597-34.3836*(exp(0.010033*x)+exp(-0.010033*x)))

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Page 9: Integral testyaozhong/lecture 12_2020.pdf · Integral test Here are the conditions under which an integral test is advisable: 0 a n Z n+1 n f(x)dx and Z 1 1 f(x)dx

3 18. Properties of continuous functions

Theorem 3.1 Let f be a continuous real-valued functionon a closed interval [a, b]. Then f is a bounded function.Moreover, f assumes its maximum and minimum valueson [a, b]; that is, there exist x0 and y0 in [a, b] such thatf(x0) ≤ f(x) ≤ f(y0) for all x ∈ [a, b].

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Page 10: Integral testyaozhong/lecture 12_2020.pdf · Integral test Here are the conditions under which an integral test is advisable: 0 a n Z n+1 n f(x)dx and Z 1 1 f(x)dx

Theorem 3.2 If f is a continuous real-valued functionon an interval I, then f has the intermediate value prop-erty on I: Whenever a < b and y lies between f(a) andf(b) [i.e., f(a) < y < f(b) or f(b) < y < f(a)], thereexists at least one x in (a, b) such that f(x) = y.

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Page 11: Integral testyaozhong/lecture 12_2020.pdf · Integral test Here are the conditions under which an integral test is advisable: 0 a n Z n+1 n f(x)dx and Z 1 1 f(x)dx

Corollary 3.3 If f is a continuous real-valued functionon an interval I, then the set f(I) = {f(x) : x ∈ I} isalso an interval or a single point.

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Page 12: Integral testyaozhong/lecture 12_2020.pdf · Integral test Here are the conditions under which an integral test is advisable: 0 a n Z n+1 n f(x)dx and Z 1 1 f(x)dx

Example 8 Let f be a continuous function mapping [0, 1]into [0, 1]. In other words, Dom(f) = [0, 1] and f(x) ∈[0, 1] for all x ∈ [0, 1]. Show f has a fixed point, i.e., apoint x0 ∈ [0, 1] such that f(x0) = x0.

Proof Consider g(x) = f(x)− x.

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Page 13: Integral testyaozhong/lecture 12_2020.pdf · Integral test Here are the conditions under which an integral test is advisable: 0 a n Z n+1 n f(x)dx and Z 1 1 f(x)dx

Figure 1: Traffic Jam to the Peak:Mount Everest

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Page 14: Integral testyaozhong/lecture 12_2020.pdf · Integral test Here are the conditions under which an integral test is advisable: 0 a n Z n+1 n f(x)dx and Z 1 1 f(x)dx

Figure 2: Mount Everest

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Page 15: Integral testyaozhong/lecture 12_2020.pdf · Integral test Here are the conditions under which an integral test is advisable: 0 a n Z n+1 n f(x)dx and Z 1 1 f(x)dx

Example 9 Let f be a continuous function mapping [0, 1]into [0, 1]. In other words, Dom(f) = [0, 1] and f(x) ∈[0, 1] for all x ∈ [0, 1]. Show f has a fixed point, i.e., apoint x0 ∈ [0, 1] such that f(x0) = x0; x0 is left fixed byf .

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Page 16: Integral testyaozhong/lecture 12_2020.pdf · Integral test Here are the conditions under which an integral test is advisable: 0 a n Z n+1 n f(x)dx and Z 1 1 f(x)dx

Example 10 Show that if y > 0 and m ∈ N, then y hasa positive m-th root.

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Page 17: Integral testyaozhong/lecture 12_2020.pdf · Integral test Here are the conditions under which an integral test is advisable: 0 a n Z n+1 n f(x)dx and Z 1 1 f(x)dx

Theorem 3.4 Let f be a continuous strictly increasingfunction on some interval I. Then f(I) is an intervalJ by Corollary 18.3 and the inverse function f−1 repre-sents a function with domain J . The function f−1 is acontinuous strictly increasing function on J .

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Page 18: Integral testyaozhong/lecture 12_2020.pdf · Integral test Here are the conditions under which an integral test is advisable: 0 a n Z n+1 n f(x)dx and Z 1 1 f(x)dx

Theorem 3.5 Let f be a one-to-one continuous functionon an interval I. Then f is strictly increasing or strictlydecreasing.

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