1. (60 pts True/False. Indicate whether the statement is true or false. If the statement is false, present an explanation or counterexample. (No explanation necessary for each true statement.) (2 points for each true statement, 5 points for each false statement with explanation/counterexample. Since there are 18 parts, there must be 10 true statements and 8 false statements.) ________(a) The negation of \"x, (~p(x) q(x))\" is \" x such that (~q(x) p(x))\". ________(b) Suppose S is a nonempty subset of the real numbers. The negation of "S is a closed set" is "S is an open set." Page 1 of 12 S= n N ________(c) is an accumulation point (i.e., limit point) of the set [ ) +1 ,5 n . ________(d) If S is a nonempty subset of the real numbers and S contains an open interval, then the set S is uncountable. ________(e) > 0, number N such that positive integer n, if n > N, then n4 s = n ________(f) Let 7n /2 . Then sequence (sn) converges to 7 7 |91 /n 1|< . since s n+ 1 (n+1)4 7n /2 1 n+1 4 1 1 4 1 7 lim =lim (n+1)/ 2 4 =lim 1 /2 =lim 1+ = = sn n 7 n 7 7 n 7 7 . ( ) ( ) Page 2 of 12 ________(g) If lim s n=1 n , then lim [ ( sn )n ] =1 n . ________(h) Every Cauchy sequence of real numbers is a monotone sequence. _______(i) If (an) is any sequence of increasing negative real numbers and (bn) is any sequence an of nonzero real numbers converging to 0, then the quotient sequence bn diverges. ( ) ________(j) Let D be an interval of the form [a, b]. Given any function f with domain D, if f is continuous on D, then f (D), the range of f, contains a maximum and a minimum. Page 3 of 12 ________(k) Let D be a nonempty subset of R, let f : D R, and let c be an accumulation point of D. Suppose there exists a sequence (sn) in D with each sn c such that (sn) converges to c, but (f (sn)) diverges. Then f does not have a finite limit at c. ________(l) The series (1 )n n=1 ________(m) The series 1 1 12 n n 3 ( )( converges absolutely. n (1+ 1n ) n=1 ) converges. ________(n) Let D be a nonempty subset of R. If function f: D R is continuous on D and D is compact, then f is uniformly continuous on D. Page 4 of 12 ________(o) 2 Let f ( x )=x 3 x . If |x 0| 1, then |f ( x )f (0)| 2 . ________(p) If a function f is differentiable at c, then f is continuous at c. ________(q) lim x 0 tan ( 3 x ) 3 = sin ( 4 x ) 4 ________(r) Suppose f ( x )=x sin(x) on the interval [0, ]. Then there exists at least one value c (0, ) such that the derivative f (c) = 0. Page 5 of 12 2. (24 pts) Complete the following table. For each set, determine the infimum, minimum, supremum, and maximum (1 pt each). (If it does not exist, say \"X\"). Indicate whether the set is countable and whether it is compact (2 pts each). No explanations required. Infimum Set S Minimum Maximum Supremum Countable? (Yes / No) Compact? (Yes / No) {x R : x is irrational and x 2 } { 1 :n N 3 n+1 } 2 1 n N 3 , 6 2 n n ( 3. (6 + 4 + 4 = 14 pts) ) cos ( n) Let s n=n (a) List the first six numbers in the sequence. (b) State lim sup sn and lim inf sn. s (c) Does the sequence ( n ) converge, or diverge to or diverge to -, or just diverge? Explain. Page 6 of 12 4. (18 pts) Prove by induction: 3 + 7 + 11 + 15 + ... + (4n 1) = n(2n + 1) 5. (10 pts) Find the radius of convergence of the power series n=1 for all n N. 1 xn n . n(0.4) Show work. Page 7 of 12 6. (12 + 8 = 20 pts) (a) Determine whether the limit exists. If so, find the limit and prove it, using the - definition (Limits Notes, p.1 or Def. 3.1.3 in Lebl). If not, explain carefully why it does not exist. 1 3 lim 6 x cos 3 x 0 6 x ( ) 3 (b) Is f ( x )= 6 x cos 1 6 x ( ) 3 uniformly continuous on the interval (0, 1) ? Explain. Page 8 of 12 7. (6 + 6 = 12 pts) Let f ( x )= Since 2+cos ( x) . 1e x f (2 )= we have Consider the following discussion: 2+cos (2 ) 3 3 2+cos ( ) 1 1 = >0 and f ( 1 )= = <0 , 2 0.86 1e 1.72 f (2 )> 0 > f ( 1 ) . Thus, by the Intermediate Value Theorem, there exists a value c in the interval (2,1 ) such that f ( c )=0. (a) Critique the discussion. Is it valid or not? What are the flaws, if any? Be specific. (b) Is it true that there exists a value c in the interval (2,1 ) such that f ( c )=0 ? Why? Explain. Page 9 of 12 8. (7 + 16 + 4 = 27 pts) Consider: Let f ( x )= { 2 3 x + 5, x 1 86 ln ( x+2 ) , x>1 (a) Is the function f continuous at x = 1? Explain. (b) Find each of the following limits, if they exist (the left-hand and right-hand limits for the derivative at 1). Show work. f ( x ) f (1) x(1) lim x 1 f ( x )f (1) x(1) lim x 1+ Page 10 of 12 (c) Is the function f differentiable at x = 1? Explain. If yes, what is the value of the derivative f (1) ? Page 11 of 12 9. (4 + 6 + 5 = 15 pts) For x [3, 7] and n N, define f n ( x )= x3 n . 4 ( ) (a) What is f (3)? ____ What is f (4)? ____ (no explanation required) Let f (x) = lim fn(x). What is f (6.5)? ____ What is f (7)? ____ (b) Determine the numerical value of f (x) for each x [3, 7] ----- you should find a relatively simple multi-part formula for f. Just state your formula. (c) Does (fn) converge uniformly to f on [3, 7] ? Explain. Page 12 of 12 Differentiation (notes by S. Sands) The Derivative (Def. 4.1.1, Lebl, p. 131) Definition: Let f: I R be a real-valued function defined on an interval I containing the point c. (We allow the possibility that c is an endpoint of the interval.) We say that f is differentiable at c (or has a derivative at c) if the limit \u0007\b \u0007\b lim \u0004\u0006 exists and is finite. We call the limit the derivative of f at c, denoted f (c), so \u0007\b = \u0007\b \u0007\b whenever the limit exists and is finite. If the function f is differentiable at each point of the set S I, then f is said to be differentiable on S, and the function f : S R is called the derivative of f on S. Examples: For relatively simple functions, such as low degree polynomials, some powers of x, and simple piecewise-defined functions, it is relatively easy to find the derivative using the limit definition. \u0012 sin \u0004 , 0\u0016 Example: Let \u0007\b = \u000f . f is not differentiable at x = 0 because the limit 0, = 0 lim\u0004\u0017 \u0018\b\u0004 \u0019\u0018\b\u0017 \u0004\u0019\u0017 = lim\u0004\u0017 \u001d \u001e \u0004 \u001a\u001b\u001c \u0019\u0017 \u0004 \u0012 = lim\u0004\u0017 sin \u0004 does not exist. Geometrically, the difference quotient \u0018\b\u0004 \u0019\u0018\b\u0006 \u0004\u0019\u0006 represents the slope of the secant line through the points (c, f(c)) and (x, f(x)). As x approaches c, this ratio approaches the slope of the tangent line at c. Page 1 of 7 Sequential Condition for Derivatives: Let I be an interval containing the point c and f: I R. Then f is differentiable at c iff for every sequence (xn) in I that converges to c with xn c for all n, the sequence \u0018\b\u0004! \u0019\u0018\b\u0006 \u0004! \u0019\u0006 " converges. Furthermore, if f is differentiable at c, then the sequence of quotients converges to f (c). Example: Using the sequential condition to show that a function is not differentiable at a point: Let f(x) = |x| and let xn = (-1)n/n for positive integers n. Then the sequence (xn) converges to 0. The corresponding sequence of quotients has terms \u0018\b\u0004! \u0019\u0018\b\u0006 \u0004! \u0019\u0006 = # \b1 n #\u0019\u0017 ! n \b1 \u0019\u0017 ! . For n even, the quotient is 1 but for n odd, the quotient is 1, so the sequence of quotients oscillates between 1 and 1, and so does not converge. REMARK: f(x) = |x| is continuous on R but not differentiable at 0. (Prop. 4.1.4, Lebl) Theorem. If f: I R is differentiable at a point c in I, then f is continuous at c. Differentiation Rules (Prop. 4.1.5, 4.16, 4.17, Lebl) Theorem. Suppose that f: I R and f: I R are differentiable at c in I. Then (a) if k R, then the function kf is differentiable at c and (kf)(c) = k f (c). (b) Sum Rule: The function f + g is differentiable at c and (f + g) (c) = f (c) + g (c). (c) Product Rule: The function f g is differentiable at c and (f g) (c) = f (c) g (c) + g(c) f (c). (d) Quotient Rule: If g(x) 0, then the function f /g is differentiable at c and \u0007 ( &\b \u0007 ( \b \u0007\b &( \b % ' \b = & [&\b\f+ Page 2 of 7 Power Rules. (Integer exponents) For any positive integer n, if f(x) = xn for all x in R, then f (x) = nxn - 1 for all x in R. For any negative integer n, if f(x) = xn for all x 0, then f (x) = nxn - 1 for all x 0. (Prop. 4.1.8, Lebl, p. 132) Theorem. Chain Rule. Let I and J be intervals in R, let f: I R and g: I R, where f(I) J, and let c in I. If f is differentiable at c and g is differentiable at f(c), then the composite function g o f is differentiable at c and (g o f) (c) = g (f(c)) f (c). Examples: (1) Differentiate h(x) = sin (1/x) for x 0. h(x) = sin (1/x) = (g o f)(x) where f(x) = 1/x = x1 and g(x) = sin x. Since f (x) = x2 = 1/x2 (power rule) and g (x) = cos x (assumed), applying the chain rule, h(x) = g (f(x)) f (x) = [cos (1/x)] (1/x2) = (1/x2) cos (1/x). (2) Differentiate r(x) = x sin (1/x) for x 0. Note that r(x) = x h(x). Applying the product rule, r (x) = x h (x) + 1 h(x) = x (1/x2) cos (1/x) + sin (1/x) = (1/x) cos (1/x) + sin (1/x) . (3) There is a continuous function on R that has a derivative at all but one real number. f(x) = |x| qualifies (see earlier example on previous page). Another example: \u0012 sin \u0004 , 0\u0016 Let ,\b = \u000f This is an extension of the function r(x) in part (2). 0, = 0 The derivative has been found for all x 0. When x = 0, the derivative does not exist. (See the second example on the first page.) Page 3 of 7 Calculus Theorems (Th. 4.2.2, Lebl) Theorem: If f is differentiable on an open interval (a, b) and if f assumes its maximum or minimum at a point c (a, b), then f (c) = 0. Remark: Recall that in calculus when f continuous on closed interval [a, b], to find the maximum and minimum of the function, we look at the y -values corresponding to: (1) points c where f (c) = 0 , (2) endpoints a and b, and (3) points c where f (c) does not exist. (Th. 4.2.3, Lebl) Rolle's Theorem. Let f be a function which is continuous on [a, b] and differentiable on (a, b), and suppose that f(a) = f(b). Then there exists at least one point c (a, b) such that f (c) = 0. The Mean Value Theorem is an extension of Rolle's Theorem where the endpoints do not have the same y-values. (Th. 4.2.4, Lebl) Mean Value Theorem. Let f be a function which is continuous on [a, b] and differentiable on (a, b). Then there exists at least one point c (a, b) such that \u0007\b- \u0007\b. \u0007\b = -. Page 4 of 7 Examples: The Mean Value Theorem can be used to prove Bernoulli's inequality: (1 + x)n 1 + nx, for x > -1, and all n in N. The Mean Value Theorem can be used to estimate the value of a function near a point. For instance, 40 can be approximated using the Mean Value Theorem and the knowledge that 40 is relatively close to 36, a perfect square. Additional familiar calculus concepts can be established. (Prop. 4.2.5, Lebl) Theorem. Let f be a function which is continuous on [a, b] and differentiable on (a, b). If f (x) = 0 for all x (a, b), then f is constant on (a, b). Corollary. Let f and g be continuous on [a, b] and differentiable on (a, b). Suppose that f (x) = g (x) for all x (a, b). Then there exists a constant C such that f = g + C on [a, b]. (Prop. 4.2.6, Lebl) Theorem. Let f be differentiable on an interval I. Then (a) if f (x) > 0 for all x I, then f is strictly increasing on I, and (b) if f (x) < 0 for all x I, then f is strictly decreasing on I. (Darboux Th. 4.2.9, Lebl) Intermediate Value Theorem for Derivatives. Let f be differentiable on [a, b] and suppose that k is a number between f (a) and f (b). Then there exists a point c (a, b) such that f (c) = k. (Th. 4.4.2, Lebl) Inverse Function Theorem. Suppose f is differentiable on an interval I and f (x) 0 for all x I. Then f is injective, f 1 is differentiable on f(I), and \b\u0007 \u0019\u0012 ( \b1 = \u0012 \u0018\t2 \b\u0004 \u0012 or, equivalently, \u0007 ( \b = \b\u00183\u001d 2 \b4 or, equivalently, \u0007 ( \b = 52 \b4 \u0012 where y = f(x). where y = f(x). where y = f(x) and g = f1. Page 5 of 7 (Cor. 4.4.3, Lebl) Power Rule (Power of form 1/n) For any positive integer n, if f(x) = x1/n for all x > 0, then f (x) = (1/n) x1/n 1 for x > 0. Proof: Since f is injective, the inverse function exists. To apply the Inverse Function Theorem, we need to know f 1, the inverse function. y = f(x) = x1/n Now raise both sides to the nth power. so yn = x, and so f 1 (y) = yn . g(y) = yn is the inverse function. g (y) = n yn 1 by the earlier power rule for positive integer powers. Applying the Inverse Function Theorem, \u0007 ( \b = \u0012 52 \b4 Substituting for y, = \u0012 64 !3\u001d \u0012 6 \u0012 = 1 \u001967\u0012 where y = f(x) = x1/n 6 \u0012 1 \u001967\u0012 = 8 \u0012/6 : 6 \u001967\u0012 \u0012 = 6 \u00196 \u001d "7\u0012/6 ! \u0012 \u0012 \u001d = \u0019\u00127\u0012/6 = !\t\u0019\t\u0012 6 6 So, f (x) = (1/n) x1/n 1 Power Rule (Rational Exponents) For any nonzero integers m and n, if f(x) = xm/n for all x > 0, then f (x) = (m/n) xm/n - 1 for x > 0. To establish this, use the fact that f(x) = xm/n = (x1/n )m and apply the Chain Rule and the previous example. Page 6 of 7 L'Hospital's Rule Cauchy Mean Value Theorem: Let f and g be functions that are continuous on [a, b] and differentiable (a, b). Then there exists at least one point c (a, b) such that [f (b) f (a)] g(c) = [g(b) g(a)] f (c). Indeterminate forms: 0/0, / , 0 , 1, 0, 00, . (Th. 2.4.1, TRENCH) L'Hospital's Rule for the indeterminate form 0/0. Let f and g be functions that are continuous on [a, b] and differentiable (a, b). Suppose that c [a, b] and that f (c) = g (c) = 0. Suppose also that g(c) 0 for x U, where U is the intersection of (a, b) and some deleted neighborhood of c. If lim\u0004\u0006 \u0018(\b\u0004 5(\b\u0004 = ;, with L R, then lim\u0004\u0006 Examples (indeterminate form 0/0): lim\u0004\u0017 \u001a\u001b\u001c \u0004 \u0004 = 1, lim\u0004\u0012 +\u0004 < \u0019\t=\u0004\t7\t\u0012 \u0004\t\u0019\t\u0012 \u0018\b\u0004 5\b\u0004 = ;. = 1, lim\u0004\u0017 \u0012\u0019>?\u001a \u0004 \u0004< \u0012 = , lim\u0004\u0017 + @ <\u001e \u0019\u0012 \u0004 =2 Limits at Infinity Definition. Let f: (b, ) R where b R. We say that L R is the limit of f as x , written lim \u0007\b = ; \u0004 provided that for each > 0 there exists a real number N > b such that x > N implies that | f(x) - L | < . Definition. Let f: (b, ) R where b R. We say that f tends to as x , written lim \u0007\b = \u0004 provided that for each R there exists a real number N > b such that x > N implies that f(x) > . (Th. 2.4.1, TRENCH) L'Hospital's Rule for the indeterminate form / . Let f and g be differentiable on (b, ). Suppose that lim\u0004B \u0007\b = lim\u0004B &\b = , and that g(x) 0 for x (b, ). If lim\u0004B \u0018(\b\u0004 5(\b\u0004 = ;, with L R, then lim\u0004B Examples (indeterminate forms / , 0 , 00) lim\u0004 D\u001c \u0004 \u0004 = 0, \u0018\b\u0004 5\b\u0004 lim\u0004\u00177 \b \b ln = 0, =;. lim\u0004\u00177 \u0004 = 1 Page 7 of 7 Infinite Series (Week 7) Definition. If |\u0003\u0004 |\tconverges, then the series \u0003\u0004 converges is said to converge absolutely (or to be absolutely convergent). If \u0003\u0004 converges but |\u0003\u0004 | diverges, then \u0003\u0004 is said to converge conditionally (or be conditionally convergent). (Def. 2.5.12, Lebl) Theorem (Alternating Series Test) If (an) is a decreasing sequence of positive numbers and lim an = 0, then the alternating series \u00061\t\u0004 \u0003\u0004 converges. Example: The alternating harmonic series \u00061\t\u0004 \u0004 converges conditionally. Theorem (Root Test) Given a series \u0003\u0004 , let \u000f = lim sup|\u0003\u0004 |\u000b/\u0004 . (a) If < 1, then the series converges absolutely. (b) If > 1, the series diverges. (c) Otherwise = 1, and the test gives no information about convergence or divergence. Definition. Let \u0006\u0003\u0004 \u001a \u0004\u0018\u0019 be a sequence of real numbers. \u0004 \u001d \u001e The series \u001a \u0004\u0018\u0019 \u0003\u0004 \u001b = \u0003\u0019 + \u0003 \u001b + \u0003\u001d \u001b + \u0003\u001e \u001b + ... (Section 2.6.5, Lebl) is called a power series (centered at 0). The number an is called the nth coefficient of the series. Definition. More generally, a power series centered at x0 has the form \u0003\u0004 \u0006\u001b \u001b\u0019 \u0004 . Example: \u001b \u0004 is a geometric power series which converges iff | x | < 1. Root Criterion for Power Series. Let \u0003\u0004 \u001b \u0004 be a power series centered at 0 and let R = lim sup |an|1/n. Define by 1 , "#\t0 < ! < + ' = ! +, "#\t! = 0 0, "#\t! = + Then the series converges absolutely whenever | x | < and diverges whenever | x | > . (When = , we take this to mean that the series converges absolutely for all real x. When = 0, then the series converges only at 0.) (Prop. 2.6.9 and 2.6.10, Lebl) is called the radius of convergence. When = 0, the power series converges only at 0. When 0 < < , the power series converges for x in the interval (-, ). (Convergence/divergence at the endpoints and must be determined separately. The theorem does not tell us anything about the behavior of the series for x = or x = .) When = , the power series converges for all real numbers x, that is, for any x in (-,). Page 1 of 2 The set of values for which the power series converges is called the interval of convergence. Note that this interval has one of the following categorizations: open, or closed, or half-open (including one of the two endpoints), depending upon the power series in question. Example: For \u001b \u0004 , the radius of convergence is 1. The interval of convergence is (-1, 1). lim ( )*+, ( exists, set ! = lim ( )*+, (. Ratio Criterion for Power Series. Given power series If )* The radius of convergence is )* a x n n centered at 0, (The limit could be infinite.) 1 , "#\t0 < ! < + ' = ! +, "#\t! = 0 0, "#\t! = + Example: For \u001b \u0004 , the radius of convergence is 1. The interval of convergence is [-1, 1). Example: For \u0004 \u0004- \u001b \u0004 , the radius of convergence is 1. The interval of convergence is [-1, 1]. Example: For \u001b \u0004 , the radius of convergence is . The interval of convergence is (-,). \u0004! Example: For /\u0004 \u001b \u0004 , the radius of convergence is 0. The interval of convergence is {0}. REMARK: For a power series centered at x0, the same convergence tests apply, with the interval of convergence centered at x0. Example: For \u0006\u001b 1\t\u0004 , x0 = 1, the radius of convergence is 1, and the interval of convergence is (x0 - 1, x0 + 1) = (1 - 1, 1 + 1) = (0, 2). Page 2 of 2 Infinite Series (Week 6) Convergence of Infinite Series Summation notation \u0005\u0003\u0006\u0007 \u0002\u0003 = \u0002\u0007 + \u0002\u0007 + ... + \u0002\u0005 Def. 2.5.1 Lebl, p. 72 Let (sn) be the sequence of partial sums defined by \u0005 = \u0005\u0003\u0006 \u0002\u0003 = \u0002 + \u0002\u000e + ... + \u0002\u0005 . If (sn) converges to a real number s, we say that the infinite series \u000f \u0005\u0006 \u0002\u0005 is convergent and we write \u000f \u0002 = . \u0005\u0006 \u0005 s is called the sum of the series. A series that is not convergent is called divergent. If lim sn = + we say that the series diverges to + and we write \u000f \u0005\u0006 \u0002\u0005 = + . Example: Consider the infinite series \u000f \u0005\u0006 \u0005(\u0005 ) . Partial sum: \u0005 = \u000e + \u000e\u0014 + + \u0005(\u0005 ) + \u000e\u0014 + \u000e By induction, it can be shown that \u0005 Therefore, lim \u0005 = lim \u0005 = 1. \u0005 + \u0005(\u0005 ) = \u0005 for every n N. So, the infinite series converges to 1, and we write \u000f \u0005\u0006 \u0005(\u0005 ) = 1. An alternative approach: Thinking of partial fractions from calculus, note that \u0005(\u0005 ) = \u0005 \u0005 . So, \u000f \u0005\u0006 \u0005(\u0005 ) 1 1 = \u000f \u0005\u0006 \u001c\u001d \u001d+1\u001e 1 and \u0005 = \u001f \u000e + \u001f 2 1 3 1 +\u001f 3 1 4 1 ++\u001f \u001d 1 \u001d+1 (called a telescoping sum) = 1 \u0005 Therefore, lim \u0005 = lim \u001f1 \u0005 = 1, just as with our other approach. Example: The harmonic series 1 + \u000e + \u0014 + $ + = \u000f \u0005\u0006 \u0005 diverges to +. (Shown in Sequence notes, week 4, page 7) Page 1 of 4 Theorem. Suppose that \u0002\u0005 = and %\u0005 = &. (Prop. 2.5.10, Lebl, page 75) Then (\u0002\u0005 + %\u0005 ) = + &, and ('\u0002\u0005 ) = ' for every k R. Theorem. If \u0002\u0005 is a convergent series, then lim an = 0. (Prop. 2.5.8, Lebl) Proof: If \u0002\u0005 converges, then the sequence of partial sums (sn) must have a finite limit. Call it s. Note that sn - sn - 1 = an. So, lim (sn - sn - 1) = lim an. lim sn - lim sn - 1 = lim an. s - s = lim an. 0 = lim an. I have specifically stated the theorem's related result which is familiar from calculus: Corollary (nth Term Test): If lim an 0, then \u0002\u0005 diverges. Note that the corollary is the contrapositive of the theorem and thus follows directly from the theorem. Example: \u000f \u0005\u0006 \u0005 (\u0005 $ diverges, because lim \u0005 (\u0005 $ = ( \t0. Theorem. (Cauchy Criterion for Series) (Prop. 2.5.7, Lebl) The infinite series \u0002\u0005 converges iff for each > 0 there exists a number N such that, if n m > N, then|\t\u0002\u0007 + \u0002\u0007 + ... + \u0002\u0005 | < . Proof: Suppose that \u0002\u0005 converges. Then the sequence (sn) of partial sums converges, and so (sn) must be a Cauchy sequence. Given > 0, there exists N such that m, n > N implies that | sn - sm | < . So if n m > N + 1, then m - 1 > N, so that | sn - sm - 1 | < . But |sn - sm - 1| = |(\u0002 + \u0002\u000e + ... + \u0002\u0005 ) (\u0002 + \u0002\u000e + ... + \u0002\u0007+ )| = |\t\u0002\u0007 + \u0002\u0007 + ... + \u0002\u0005 | so |\t\u0002\u0007 + \u0002\u0007 + ... + \u0002\u0005 | < as desired. Conversely, suppose that for each > 0 there exists a number N such that n m > N implies that|\t\u0002\u0007 + \u0002\u0007 + ... + \u0002\u0005 | < . Then for n > m > N, we have m + 1 > N so that |sn - sm | = |\t\u0002\u0007 + \u0002\u0007 \u000e + ... + \u0002\u0005 | < . Therefore, the sequence of partial sums (sn) must be a Cauchy sequence and therefore converges. Since the sequence of partial sums converges, \u0002\u0005 converges. Page 2 of 4 \u0005 2 3 Geometric Series \u000f \u0005\u0006( , = 1 + , + , + , + Partial sum: \u0005 = 1 + , + , \u000e + + , \u0005 It can be proven by induction that 1 + , + ,\u000e + + ,\u0005 = +- ./0 +- for every n N, r 1. Another result from our week 3 study of sequences showed that lim\u0005\u000f 2 \u0005 = 0 iff |x| < 1. So, lim\u0005\u000f , \u0005 = 0 provided |r| < 1. If |r| < 1, lim \u0005 = lim +- ./0 +- = +345 - ./0 +- = +\t-\t345 - . +- = +-(() +- = +- \u0005 If r = 1, then \u000f \u0005\u0006( 1 = 1 + 1 + 1 + , which diverges to +. \u0005 If r = -1, then \u000f \u0005\u0006((1) = 1 1 + 1 1 + 1 . Note that if n is even, \u0005 = 1 but if n is odd, \u0005 = 0. Therefore the sequence (sn) of partial sums does n \u0005 not converge, and so \u000f \u0005\u0006((1) diverges. Alternatively, lim (1) does not exist, so the series diverges. If |r| > 1, lim an = lim rn 0, so the geometric series diverges. Summary: \u0005 2 3 The geometric series \u000f \u0005\u0006( , = 1 + , + , + , + = For |r| 1, the series diverges. p-Series Example: \u00056 . 1 1, only if |r| < 1. The p-series converges if p > 1 and diverges if p 1. (Prop. 2.5.15, Lebl) Convergence Tests Theorem (Comparison Test) (Def. 2.5.14, Lebl) Let \u0002\u0005 and %\u0005 be infinite series of nonnegative terms. Then (a) If \u0002\u0005 converges and 0 bn an for all n, then %\u0005 converges. (b) If \u0002\u0005 = + and 0 an bn for all n, then %\u0005 = +. Examples: (\u0005 )7 . Since 0 < (\u0005 )7 < \u00057 , and \u00057 is a convergent p-series with p = 2, the given series converges by the comparison test. \u0005+\u000e . Since 0 < \u0005 < \u0005+ series diverges by the comparison test. \u000e (for n > 2) and the harmonic series \u0005 diverges, the given Page 3 of 4 Definition. If |\u0002\u0005 |\tconverges, then the series \u0002\u0005 converges is said to converge absolutely (or to be absolutely convergent). If \u0002\u0005 converges but |\u0002\u0005 | diverges, then \u0002\u0005 is said to converge (Def. 2.5.12, Lebl) conditionally (or be conditionally convergent). Theorem: If a series converges absolutely, then it converges. Example: (+\u000b). \u00057 (Prop. 2.5.13, Lebl) converges absolutely. (Series of absolute values is a p-series with p = 2). Theorem (Ratio Test) Let \u0002\u0005 be an infinite series of nonzero terms. ;./0 Suppose lim : ;. (Prop. 2.5.17, Lebl) : exists and is equal to L. (a) If < < 1, then the series converges absolutely. (b) If < > 1, the series diverges. (c) Otherwise, the test gives no information about convergence or divergence. Theorem (Root Test) Given a series \u0002\u0005 , let ? = lim sup|\u0002\u0005 |\u000b/\u0005 . (Prop. 2.6.1, Lebl) (a) If < 1, then the series converges absolutely. (b) If > 1, the series diverges. (c) Otherwise = 1, and the test gives no information about convergence or divergence. When applying the root test, it can be handy to recall that lim \u001d\u000b/\u0005 = 1. Examples \u0005 \u000e. converges (can apply the ratio test or the root test). ;./0 lim : ;. : = lim (\u0005 ) \u000e. \u000e./0 \u0005 \u00050/. lim sup|\u0002\u0005 |\u000b/\u0005 = lim sup (\u000e. )0/. = lim sup \u000e = lim \u000e \u001f1 + \u0005 = \u000e < 1. \u00050/. \u000e = \u000e lim \u001d\u000b/\u0005 = \u000e < 1 + \u00147 + \u000eD + \u0014E + \u000eF + \u0014G + converges (compare to the convergent geometric series with r = 1/2.) \u000e. \u0005! converges (can apply the ratio test). Page 4 of 4 Sequences and Series of Functions Pointwise and Uniform Convergence Definition. Let (fn) be a sequence of functions defined on a subset S of R. Then (fn) converges pointwise on S if for each x in S, the sequence of numbers (fn(x)) converges. If (fn) converges pointwise on S, then we define f: S R by \u0001\u0002\u0003\u0004 = lim\t \u0001 \u0002\u0003\u0004 for each x in S and we say that (fn)converges to f pointwise on S. (Def. 6.1.1, Lebl) Example. Let fn : [0, 1] R where fn (x) = xn for each positive integer n. For each x in [0, 1), lim\t \u0001 \u0002\u0003\u0004 = lim\t \u0003 = 0, and lim\t \u0001 \u00021\u0004 = lim\t 1 = 1. So, (fn) converges pointwise to the function \u0001\u0002\u0003\u0004 = \u000f 0, \u0011\u0001\t0 \u0003 < 1\u0014 1, \u0011\u0001\t\u0003 = 1 Note that although each fn is continuous (and differentiable) on [0, 1], the limit function f is discontinuous (and not differentiable) at x = 1. SEE VIDEO and Interactive Version of this example at http://sandsduchon.org/sands/MATH301/Videos/Chapter9/MATH301Chapter9Videos.html Page 1 of 5 Example. For x in [0, 1] and n 2, let 1 \u0019\u001a \u0003, \u0011\u0001\t0 \u0003 \u0019 2 1 2 \u0001 \u0002\u0003\u0004 = \u0019\u001a \u001c\u0003 \u001e , \u0011\u0001 \u0003 \u0014 \u0019 \u0019 \u0019 \u0017 2 \u0016 0, \u0011\u0001 < \u0003 1 \u0015 \u0019 \u0018 \u0016 SEE Interactive Version of this example at http://sandsduchon.org/sands/MATH301/Videos/Chapter9/MATH301Chapter9Videos.html Then lim\t \u0001 \u0002\u0003\u0004 = 0 so the limit function is f(x) = 0. Note that \u001f" \u0001 \u0002\u0003\u0004 \u0003 = 1 but \u001f" lim\t \u0001 \u0002\u0003\u0004 \u0003 = \u001f" \u0001\u0002\u0003\u0004 \u0003 = 0 so we have ! ! ! ! ! lim # \u0001 \u0002\u0003\u0004 \u0003 # lim \u0001 \u0002\u0003\u0004 \u0003 " " Example. For x in [0, 2] and positive integers n, let \u0001 \u0002\u0003\u0004 = $%& ' . Then (fn) converges to f (x) = 0. SEE VIDEO and Interactive Version of this example at http://sandsduchon.org/sands/MATH301/Videos/Chapter9/MATH301Chapter9Videos.html However, it turns out that the sequence of derivatives (fn) does not converge to f . In fact, the sequence (fn) does not converge for any x. Example. For x in [0, 1] and positive integers n, let \u0001 \u0002\u0003\u0004 = 2\u0003 + . Then (fn) converges to f (x) = 2x. ' It turns out that lim\t \u001f" \u0001 \u0002\u0003\u0004 \u0003 = \u001f" lim\t \u0001 \u0002\u0003\u0004 \u0003 and also it turns out that the sequence of derivatives (fn) converges to f . ! ! Page 2 of 5 (Def. 6.1.6, Lebl) Definition. let (fn) be a sequence of functions defined on a subset S of R. Then (fn) converges uniformly on S to a function f defined on S if for each > 0 there exists a number N such that for all x in S and all n > N , | fn (x) f(x) | < . To say that a sequence (fn) converges uniformly on S is to say that there exists a function f to which (fn) converges uniformly on S. Example. Let fn : [0, 1] R where fn (x) = xn for each positive integer n. (same sequence as in earlier example) It turns out that although (fn) converges pointwise to f, the sequence (fn) does not converge uniformly to f. Discussion: (fn) converges uniformly on [0,1] if there is a function f satisfying: For each > 0 there exists a number N such that for all x in [0,1], n > N implies that | fn (x) - f(x) | < . NOT converging uniformly on [0, 1] to f means: > 0 such that for all N , x in [0,1] and n > N for which | xn - f(x) | . Proof that the sequence (fn) does not converge uniformly to f: Pick = 1/2 and say that x in [0, 1), so f(x) = 0 and | xn - f(x) | = xn. We want xn 1/2. So, \u0003 +\u001a = \u00021/2\u0004!/ = 2.!/ , ! Thus for = 1/2 , given any N, for n > N, let \u0003 = 2.!/ . Then \u0003 > 02.!/ 1 = 1/2 SEE VIDEO and Interactive Version of this example at http://sandsduchon.org/sands/MATH301/Videos/Chapter9/MATH301Chapter9Videos.html Page 3 of 5 Example. The earlier example with the sequence (fn) depicted in figure 35.2, the convergence to f is not uniform. Example. For x in [0, 2] and positive integers n, let \u0001 \u0002\u0003\u0004 = $%& ' . Then (fn) converges uniformly to f (x) = 0 on the interval [0, 2]. SEE VIDEO of proof at http://sandsduchon.org/sands/MATH301/Videos/Chapter9/MATH301Chapter9Videos.html Cauchy criterion for uniform convergence of sequences of functions (Prop. 6.1.13, Lebl) Theorem. let (fn) be a sequence of functions defined on a subset S of R. There exists a function f such that (fn) converges to f uniformly on S iff the Cauchy criterion is satisfied: for every > 0, there exists a number N such that | fn (x) - fm(x) | < for all x in S and all m, n > N. *Theorem. Let (fn) be a sequence of continuous functions defined on a set S and suppose that (fn) converges uniformly on S to a function f: S R. Then f is continuous on S. (Prop. 6.2.2, Lebl) Example: Let fn : [0, 1] R where fn (x) = xn for each positive integer n. The sequence (fn) converges to \u0001\u0002\u0003\u0004 = \u000f 0, \u0011\u0001\t0 \u0003 < 1\u0014 1, \u0011\u0001\t\u0003 = 1 but the sequence (fn) cannot converge uniformly to f. Otherwise, by the theorem*, the limit f must be continuous on [0, 1], but clearly in this example, f is discontinuous at x = 1. Theorem. Let (fn) be a sequence of continuous functions defined on an interval [a, b] and suppose that (fn) converges uniformly on [a, b] to a function f. Then 2 2 lim # \u0001 \u0002\u0003\u0004 \u0003 = # \u0001\u0002\u0003\u0004 \u0003 3 3 Theorem. Suppose that (fn) converges to f on an interval [a, b]. Suppose also that each (fn) exists and is continuous on [a, b], and the sequence (fn) converges uniformly on [a, b]. Then lim\t \u0001 4 \u0002\u0003\u0004 = \u0001 4 \u0002\u0003\u0004 for each x in [a, b]. Page 4 of 5 Definition. If \u0002\u0001 \u0004 5" is a sequence of functions defined on a set S, the series \t5" \u0001 is said to converge pointwise on S iff the sequence of partial sums given by 7 \u0002\u0003\u0004 = 85" \u00018 \u0002\u0003\u0004 converges pointwise on S. (A similar definition applies to uniform convergence.) Theorem. Weierstrass M-Test. Suppose that (fn) is a sequence of functions defined on S and (Mn) is a sequence of nonnegative numbers such that| fn (x) | Mn for all x in S and all positive integers n. If 9 converges, then \u0001 converges uniformly on S. (Th. 3, Chapter 4, Sec. 13, Zakon, p.240) $%& ' Example. Consider the series : $%& ' <: \u001a so : ; <= $%& ' \u001a |$%& '|, \u001a, ! \u001a, ; \u001a ! =: ; \u001a ! for x in R. and : ; converges (geometric series with r =1/2) \u001a ! ; converges by the Weierstrass M-Test, with 9 = :\u001a; Example. Consider the series 5" \u0001 where \u0001 \u0002\u0003\u0004 = ! . It turns out that the series converges pointwise on R but does not converge uniformly on R. However, the series converges uniformly on any closed interval [-t, t]. ', Theorem. Let 5" \u0001 be a series of functions defined on a set S. Suppose that each fn is continuous on S and that the series converges uniformly to a function f on S. Then \u0001 = \u0019=0 \u0001\u0019 is continuous on S. Theorem. Let 5" \u0001 be a series of functions defined on an interval [a, b]. Suppose that each fn is continuous on [a, b] and that the series converges uniformly to a function f on [a, b]. 2 2 Then \u001f3 \u0001\u0002\u0003\u0004 \u0003 = \u0019=0 \u001f3 \u0001 \u0002\u0003\u0004 \u0003 . Example. The geometric series 5" \u0002@\u0004 = ! !\tA\tB for t in the interval (-1, 1). It can be shown by the Weierstrass M-test that the series converges uniformly on any interval [-r, r] contained in (-1, 1). According to the theorem above, if x is in (-1, 1), we can integrate term by term and get so, \u0019=0 \u0019=0 \u0019=0 ' ' \u0003 A! @ # = C # \u0002@\u0004 @ = C\u00021\u0004 # @ @ = C\u00021\u0004 \u0019+1 " 1+@ " " ' \u0003 A! \u0003\u001a \u0003E \u0003G ln\u00021 + \u0003\u0004 = C\u00021\u0004 = \u0003 + +\t... \u0019+1 2 3 4 \u0019=0 Theorem. Let 5" \u0001 be a series of functions that converges to a function f on an interval [a, b]. Suppose that for each n, fn exists and is continuous on [a, b] and that the series of derivatives 5" \u0001 is uniformly convergent on [a, b]. Then \u0001 4 \u0002\u0003\u0004 = \u0001 \u0002\u0003\u0004 for all x in [a, b]. \u0019=0 \u0019 Page 5 of 5 The union ofall closed intervals oftheform U\" = [L Illl]. J'I ThiE union equal5 (U: I}: which i5 not a timed Eat. The union ofall closed intervals oftheform U\" = [L Illl]. J'I ThiE union equal5 (U: I}: which i5 not a timed Eat