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13 Pages

### FinalSolution

Course: MATH 139, Fall 2010
School: Duke
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Word Count: 3107

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Exam Final - Math 139 Dec 16th Instructor Web page Mauro Maggioni www.math.duke.edu/ mauro/teaching.html You have 3 hours. You may not use books, internet, notes, calculators. The exam should be stapled, written legibly, with your name written at the top of every page, on one side of each page only, and must contain the rearmation of the Duke community standard. You can invoke any Theorems that we discussed in...

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Exam Final - Math 139 Dec 16th Instructor Web page Mauro Maggioni www.math.duke.edu/ mauro/teaching.html You have 3 hours. You may not use books, internet, notes, calculators. The exam should be stapled, written legibly, with your name written at the top of every page, on one side of each page only, and must contain the rearmation of the Duke community standard. You can invoke any Theorems that we discussed in class and are in the sections of the book we studied. No partial credit is given for statements for which proof is not provided. Problems are roughly ordered in increasing diculty, and value (in terms of points). (*) preceeds parts of problems that are for extra credit: most of these are not particularly hard, but do take extra time. 1. Let f C 1 ([a, b]) and f (x) > 0 for all x [a, b]. Prove that f is strictly monotonically increasing, i.e. f (x) < f (y ) for all x < y , x, y [a, b]. [5 pts.] Fix x, y [a, b]. Without loss of generality we may assume that x < y . By the Mean Value Theorem, applied on [x, y ] to the C 1 function f , we have f (y ) f (x) = f ( )(y x) for some point (x, y ). Since f is strictly positive everywhere and y x > 0, the right hand side is strictly positive, proving that f (y ) > f (x). Since x, y were arbitrary, we are done. 2. Show that equation arctan(x) = 1 x has at least one real solution. [5 pts.] Let f (x) = arctan(x) (1 x). f is continuous on the whole real line, f (0) = 1 and f (x) (1 ) =: L > 0 as x + since 1 < 0. Therefore there exists 2 2 2 M > 0 such that f (x) > 0 for x M by denition of limit as x + and the fact that the limit is positive (in fact, there exists M such that |f (x) L| < L/2 and therefore f (x) > L/2 > 0 for x M ). In particular f (M ) > 0, and by the Intermediate Value Theorem applied in [0, M ] to the continuous function f , there exists (0, M ) such that f ( ) = 0, i.e. arctan( ) = 1 . 3. Dene (a) what a metric space (X, ) is; (b) what does it mean for a sequence in (X, ) to converge; (c) what does it mean for a sequence in (X, ) to be Cauchy; (d) what does it mean for (X, ) to be complete. State an interesting Theorem that crucially uses completeness. 1 Finally, consider the set A = +=1 {(x, n ) R2 : x R} in R2 . Show that A is not n complete (with respect to the Euclidean distance in R2 ). (*) Complete A (with respect to the Euclidean distance in R2 ). [10 pts.] (a) A metric space (X, ) is a pair consisting of a set X and a metric , i.e. a function : X X [0, +) satisfying the following conditions: (x, y ) 0 for all x, y X , and (x, y ) = 0 if and only if x = y ; (x, y ) = (y, x) for all x, y X ; (x, z ) (x, y ) + (y, z ) for all x, y, z X . (b) A sequence {xn }n X (i.e. a map N X written as n xn ) is convergent to a point x X if for every > 0 there exists N = N () such that for all n N we have (xn , x) < . (c) A sequence {xn }n X is Cauchy if for every > 0 there exists N = N () such that for all n, m N we have (xn , xm ) < . (d) A metric space is complete if every Cauchy sequence is convergent to some point of X. We state the contraction mapping theorem: let (X, ) be a complete metric space and T a contraction X X , i.e. (T (x), T (y )) (x, y ) for all x, y X , for some < 1 independent of x, y ). Then there exists a unique xed point x X for T , i.e. such that T (x ) = x . Moreover, if x0 is any point in X , the sequence xn = T n (x0 ) converges to x . 1 Consider the sequence {xn }n := {( n , 0)}+=1 , which is contained in A. If the sequence n converges to x A, then it also converges to x R2 since A has the metric of R2 1 restricted to A. But xn (0, 0) in R2 , since ||xn 0||R2 = n 0, therefore it converges to (0, 0) also as a sequence in A. But (0, 0) A. / (*) The completion of A is A := A {(x, 0) : x R|}. Clearly every point in A has to be in the completion of A, since (x, 0) A is the limit of at least one Cauchy sequence, for example of the sequence {(x, 1/n)}n . A is complete: in fact, let {(xn , yn )}n be a Cauchy sequence in A. Such a sequence is also Cauchy in R2 : let (x , y ) R2 be its limit (R2 is complete). Then xn x and yn y (in R). Since the rst coordinate of points in A can be any real number, there is no issue with x . As for y , it can be either 0 or 1/n for some n, since yn {1/n : n N}. To see this, notice that either there exists > 0 such that yn > for all ns, in which case y = 1/n for some n, or for all > 0 there exists n such that yn < , in which case there is a subsequence of {yn }n converging to 0, and since {yn }n is Cauchy, the whole sequence {yn }n converges to 0. We conclude that (x , y ) A, so that A is complete, and it is the smallest complete space including A. 4. Let f C ([0, 1]). Prove or disprove: (a) if 1 0 f (t)dt = 0, then f = 0. (b) if f (x) 0 for every x [0, 1] and (c) if for every x [0, 1], x 0 1 0 f (t)dt = 0, then f = 0. f (t)dt = 0, then f = 0. [10 pts.] (a) False. For example f (x) = sin(2x) is not identically 0 and integrates to 0 on the interval [0, 1]. (b) True. By contradiction, assume that there exists a point s.t. f ( ) > 0. Assume > 0, otherwise we slightly modify the proof to allow for asymmetric intervals around . By continuity, there exists an interval ( , + ) on which f is larger than f ( )/2 > 0. But then 1 0 + f (x)dx f (x)dx f ( )/2 2 > 0 , contradicting our assumptions (here we used the fact that g (x) h(x) for all x [0, 1] implies g h). (c) True. By contradiction, suppose that f = 0, i.e. there exists a point [0, 1] : f ( ) = 0. Without loss of generality we may assume that f ( ) > 0. Proceeding as in (b), assuming > 0 (otherwise we slightly modify the proof to allow for asymmetric intervals around ), f (x) > f ( )/2 on some interval ( , + ). But then + f (t)dt > f ( )/2 2 and at the same time + + f (t)dt = 0 f (t)dt by our assumptions. This is a contradiction. 0 f (t)dt = 0 0 = 0 5. Consider a bounded sequence {xn }n in R. Let x be the set of limit points of {xn }n . (a) Write the denition of x . (b) What can you say about the cardinality of x ? Can it be 0 (i.e. x = )? Finite? Countably innite? Uncountably innite? (c) Is x bounded? Let {yn }n be a second bounded sequence in R, and y the corresponding set of limit points. Let a x and b y . Is a b necessarily a limit point of {xn yn }n ? (*) Prove that x is closed. [10 pts.] (a) x = {y R : there exists a subsequence of {xn }n that converges to y }. (b) x cannot be empty: since {xn }n is bounded, it is contained in an interval [M, M ] for some M > 0, and by Bolzano-Weierstrass {xn }n has a convergent subsequence, whose limit is by denition a point in x . x may be nite: for example if {xn }n converges to a point x R, then x = x . By picking the union of k sequences converging to k dierent points x , . . . , x we obtain a 1 k set of limit points of cardinality exactly k . x may be countable: for example consider the sequence {1, 1/2, 1, 1/2, 1/3, 1, 1/2, 1/3, 1/4, 1, 1/2, 1/3, 1/4, 1/5, 1, 1/2, 1/3, 1/4, 1/5, 1/6, . . . } . Then x = A := {1/n : n N+ } {0}. Indeed, for each n N we can clearly nd a subsequence which is constant and equal to 1/n and therefore 1/n x , so N x . Also there is clearly at least one subsequence converging to 0, so 0 x . On the other hand no other point can be in x : if y R \ A is in x , then no point in sequence has distance to y less than the (positive!) distance of y to the nearest point in A (which is either the point 1/n where n is the nearest integer to 1/y , or 0), so that y cannot be the limit of any subsequence. x may be uncountable: for example consider the set A = Q [0, 1]: since A is countable (it is contained in the countable set Q) we may list the elements of A in a sequence {xn }. Then we claim x = [0, 1]: in fact for every point x [0, 1] there exist innitely many elements of A in the interval (x 1/k, x + 1/k ) [0, 1] (this is a consequence of the Archimedean property, which we discussed in classed and proved in the homework): for each k we may select one such point xnk , making sure that nk+1 > nk (this is possible since there are innitely many points of A in (x 1/k, x + 1/k ) [0, 1], so that they cannot have been exhausted by xn1 , . . . , xnk ). Then this is a subsequence of {xn } that converges to x . (c) x is bounded. In fact, let M > 0 be such that |xn | M for all n (such an M exists because {xn } is bounded). x Let x and {xnk }k a subsequence of {xn } converging to x . Since |xnk | M , necessarily, by taking limits, |x | M . Since x was arbitrary in x , x [M, M ]. No. For example let xn = (1)n and yn = (1)n+1 . Then xn yn = (1)2n+1 = 1 for all ns, so that the set of limit points of {xn yn } is the set {1}. On the other hand 1 x and 1 y , but 1 1 is not a limit point of {xn yn }. Pick a point x which is the limit of a sequence {x }n in x . We want to show that n x x , i.e. that the exists a subsequence of our original sequence {xn } that converges to x . We construct it as follows: for n = 1, let xn1 be an element of {xn } such that |xn1 x | < 1: such an element exists (and in fact there are innitely many) because 1 x x and therefore it is the limit of a subsequence of {xn }n . By induction, having 1 constructed xn1 , . . . , xnk , we let xnk+1 be an element of o{xn }n>nk such that |xnk+1 x +1 | < 1/(k + 1): such element certainly exists since x +1 is a limit point of {xn }. k k Then {xnk }k is a subsequence of {xn } (since {nk }k is constructed to be increasing) and converges to x since |xnk x | |xnk x | + |x x | 1/k + |x x | 0 as k +. k k k 6. Suppose f C (R) and |f (x)| Cx5 for some C > 0 and all x R. Find the domain D of the function n F (x) = lim f (t)dt . n + x Then, not necessarily in the following order, show that: F uniformly continuous on D , F is continuously dierentiable and F is uniformly bounded on D (i.e. M > 0 : |F (x)| < M for all x D ). [Hint: be careful in explaining the manipulations you do with all the limits involved.]. Find F . + (*) Show that F (x) = x f (t)dt. [10 pts.] Since f is continuous, it is Riemann integrable on any bounded interval. |f | is also bounded by, say, M on [1, 1], since f is continuous, so |f | is upper bounded by the function g which is M on [1, 1] and equal to Cx5 for x [1, 1]. We know that g / is Riemann integrable on R, and since |f | is upper bounded by g , |f | is also Riemann integrable on R and therefore f is. In particular f is Riemann integrable on [x, +) for any x R, so that the domain F is R. This also proves (*). We start by showing that F is dierentiable everywhere and its derivative is bounded. n Consider the function Fn = x f (t)dt. Then Fn converges uniformly to F : + |Fn (x) F (x)| = n + f (t)dt n Cx5 cn4 for some constant c, and the righthand side goes to 0 uniformly in x (since it does not depend on x). Similarly, Fn converges uniformly to f (x), since Fn (x) = f (x) (independently of n). We can therefore invoke the Theorem about uniform convergence of a sequence of functions and their derivatives to conclude that F is dierentiable, and F = f (x). But f is bounded on R by the reasoning above (f is dominated by g which is bounded by max{M, C/M 5 } on R). Since F is bounded uniformly on R, we conclude that F is uniformly continuous, since by the mean value theorem we have |F (x) F (y )| = |F ( )| |x y | ||F || |x y | . 7. A set K in a metric space is compact if for every sequence {xn } K there exists a convergent subsequence whose limit is also in K . (a) Give examples of compact and non-compact sets in R (with the usual distance), and describe a large class of compact subsets of R. Is [0, +) compact? (b) Show the the union of two compact sets is a compact set. (*) Generalize this statement to the union of nitely many compact sets. (*) Does the statement hold for the union of innitely many compact sets? (c) Consider C ([0, 1]) as a metric space with the distance induced by || || , i.e. d(f, g ) = ||f g || = sup |f (x) g (x)| , x[0,1] for f, g C ([0, 1]) . Prove that the unit ball {f C ([0, 1]) : ||f || 1} is bounded (i.e. M > 0 such that ||f || M for all f B ), closed (i.e. the limit point of any convergent sequence in B is contained in B ), but not compact. [Hint: nd a sequence in B which has no convergent subsequence.] [15 pts.] (a) By Bolzano-Weierstrass, any bounded closed interval [a, b] is compact. (0, 1] is not compact, since {1/n} tends to 0 but 0 (0, 1]. [0, +) is not compact since {n}n / [0, +) does not have any convergent subsequence, since such subsequence would be Cauchy, but any two elements would be at distance at least 1 from each other. (b) Let K = K1 K2 with K1 , K2 compact. Let {xn }n be a sequence in K . Then {xn }n has innitely many elements at least one of the two sets K1 and K2 . Without loss of generality, let K1 contain innitely many elements of {xn }n , that we may list as a subsequence {xnk }k . This subsequence is completely contained in K1 and therefore has a convergent subsequence to an element of K1 , which is also an element of K . Such a subsequence is a subsequence of the original sequence {xn }n , thereby proving that K is compact. (*)Clearly this may be extended to any nite union of compact sets, since we only used the fact that at least one of them needs to contain countably many elements of the original sequence. (*) No. For example let, for every n N, Kn = {n} (the singleton set containing the integer n). Clearly Kn is compact, +=1 = N, but N is not compact (the sequence n {n}n = {1, 2, 3, . . . } has no convergent subsequence). (c) The unit ball - lets call it B - is clearly bounded by denition, as we may choose M = 1. It is also closed: if {fn } is a sequence in B that converges to f in the metric d, i.e. uniformly (by a Theorem proved in class), then f is in B , since ||f || lim ||f fn || + ||fn || 1 . On the other hand B is not compact: consider the sequence of functions {fn }n with fn (x) = xn . This sequence converges pointwise to the discontinuous function f which is 0 for x [0, 1) and 1 at x = 1. Let fnk be any subsequence: it will also converges pointwise to f , but certainly it cannot converge uniformly to f since fnk is continuous and f is not. We have therefore exhibited a sequence in B which has no convergent subsequence, so B cannot be compact. 8. Let C 1 ([0, 1]) be the set of continuously dierentiable functions on [0, 1]: C 1 ([0, 1]) := {f : [0, 1] R : f is continuously dierentiable } . For f C 1 ([0, 1]) dene ||f ||C 1 := |f (0)| + ||f || = |f (0)| + sup |f (x)| . x[0,1] (a) Prove that ||f ||C 1 is a norm on C 1 ([0, 1]). (b) (*) Prove that C 1 ([0, 1]) is complete with respect to (the distance associated with) this norm. (c) Does + sin(nx)en n=0 converge in this norm? [15 pts.] (a) Let ||h|| = supx[0,1] |h(x)| in what follows. Clearly ||f ||C 1 0, and if ||f ||C 1 = 0 the |f (0)| = 0 and ||f || = 0 so that f (0) = 0 and f is constant, and therefore f is identically 0. Also, ||f ||C 1 = ||(|f (0)| + ||f || ) = || ||f ||C 1 for any R. Finally ||f + g ||C 1 = |f (0) + g (0)| + ||f + g || |f (0)| + ||f || + |g (0)| + ||g || = ||f ||C 1 + ||g ||C 1 (where we used the fact that || || satises the triangle inequality), proving the triangle inequality. (b) (*) Let {fn } be a Cauchy sequence in || ||C 1 . Then |fn (0) fm (0)| ||fn fm ||C 1 implies that the numerical sequence {fn (0)} is Cauchy and therefore convergence to some number that we call f (0); moreover ||fn fm || ||fn fm ||C 1 implies that the sequence of continuous functions {fn } is Cauchy and therefore uniformly convergent to some continuous function g . By applying the Theorem on uniform convergence of derivates, which also needs convergence at one point (0 in this case), we deduce that {fn } converges uniformly to a function f and that f exists and is equal to g ; in particular f C 1 ([0, 1]). In fact, fn converges to f in || ||C 1 , since ||fn f ||C 1 |fn (0) f (0)| + ||fn f || 0 . This proves completeness, since we showed that any Cauchy sequence converges to a point in C 1 ([0, 1]). n k Let fn (x) = k =0 sin(kx)e : by Weierstrass M -test, fn converges uniformly, since k k k | sin(kx)e | e = (1/e) is the k -th term of a geometric sequence with ratio 1/e < 1, therefore convergent. Analogously, fn also converges uniformly, by applying again Weierk k/2 k/2 strass M test: |k cos(kx)e | ke e Cek/2 since kek/2 0 and therefore it is bounded, and the upper bound is the k -th term in geometric series of ratio 1/ e < 1. Let f be the limit of fn : then f is C 1 and moreover f = lim fn (by the same Theorem used above). Then ||f fn ||C 1 = |f (0) fn (0)| + ||f fn || 0 since both terms go to 0 (the rst by pointwise convergence at 0, which is implied the uniform convergence of fn to f , and the second since fn converges to f ). This proves that the series converges in || ||C 1 .
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Duke - MATH - 139
Homework 1 - Math 139Due Sep 7thInstructorOceOce hoursWeb pageMauro Maggioni293 Physics Bldg.Monday 12pm-2pm.www.math.duke.edu/ mauro/teaching.htmlReading: from Reeds textbook: Chapter 1Problems: 1.1: #3, 4, 7 1.4: #1, 10Additional Problems:1
Duke - MATH - 139
Homework 2 - Math 139Due Sep 13thInstructorOceOce hoursWeb pageMauro Maggioni293 Physics Bldg.Monday 1:30pm-3:30pm.www.math.duke.edu/ mauro/teaching.htmlReading: from Reeds textbook: Sections 2.1,2.2Problems:1.1: #8, 10, 11 (in these problems
Duke - MATH - 139
Homework 3 - Math 139Due Sep 22ndInstructorOceOce hoursWeb pageMauro Maggioni293 Physics Bldg.Monday 1:30pm-3:30pm.www.math.duke.edu/ mauro/teaching.htmlReading: from Reeds textbook: Sections 2.4,2.5Problems:1.3: #3(c) , 8. In both exercises,
Duke - MATH - 139
Homework 4 - Math 139Due Sep 29ndInstructorOceOce hoursWeb pageMauro Maggioni293 Physics Bldg.Monday 1:30pm-3:30pm.www.math.duke.edu/ mauro/teaching.htmlReading: from Reeds textbook: Sections 2.6Problems:2.2: #1b, 2a, 4, 52.4: #1, 3, 52.5: #
Duke - MATH - 139
Homework 5 - Math 139Due Oct 6thInstructorOceOce hoursWeb pageMauro Maggioni293 Physics Bldg.Monday 1:30pm-3:30pm.www.math.duke.edu/ mauro/teaching.htmlReading: from Reeds textbook: Sections 6.1,3.2Problems:2.4: #102.6: #1, 3, 96.1: #1(a,c),
Duke - MATH - 139
Homework 6 - Math 139Due Oct 13thInstructorOceOce hoursWeb pageMauro Maggioni293 Physics Bldg.Monday 1:30pm-3:30pm.www.math.duke.edu/ mauro/teaching.htmlReading: from Reeds textbook: Section 3.3Problems: 3.1: #3,7,8,10I suggest you pick and do
Duke - MATH - 139
Homework 7 - Math 139Due Oct 20thInstructorOceOce hoursWeb pageMauro Maggioni293 Physics Bldg.Monday 1:30pm-3:30pm.www.math.duke.edu/ mauro/teaching.htmlReading: from Reeds textbook: Section 3.4Problems:3.2: #7, 8, 9, 113.3: #1, 2, 3
Duke - MATH - 139
Homework 8 - Math 139Due Oct 27thInstructorOceOce hoursWeb pageMauro Maggioni293 Physics Bldg.Monday 1:30pm-3:30pm.www.math.duke.edu/ mauro/teaching.htmlReading: from Reeds textbook: Section 4.1,4.2Problems:3.3: #8, 13, 154.1: #2,3,5,9,10,12
Duke - MATH - 139
Homework 9 - Math 139Due Nov. 3rdInstructorOceOce hoursWeb pageMauro Maggioni293 Physics Bldg.Monday 1:30pm-3:30pm.www.math.duke.edu/ mauro/teaching.htmlReading: from Reeds textbook: Section 4.3,4.5,4.6Problems:4.2: #2, 3, 4, 5, 6, 124.3: #7,
Duke - MATH - 139
Homework 10 - Math 139Due Nov. 10thInstructorOceOce hoursWeb pageMauro Maggioni293 Physics Bldg.Monday 1:30pm-3:30pm.www.math.duke.edu/ mauro/teaching.htmlReading: from Reeds textbook: Section 5.1,5.2Problems:4.5: #6,8 (for (b) use #6)4.6: #2
Duke - MATH - 139
Homework 11 - Math 139Due Nov. 17thInstructorOceOce hoursWeb pageMauro Maggioni293 Physics Bldg.Monday 1:30pm-3:30pm.www.math.duke.edu/ mauro/teaching.htmlReading: from Reeds textbook: Section 5.3,5.6Problems:5.1: #1,4,115.2: #9,15Additional
Duke - MATH - 139
Homework 12 - Math 139Due Thu Dec. 2ndInstructorOceOce hoursWeb pageMauro Maggioni293 Physics Bldg.Monday 1:30pm-3:30pm.www.math.duke.edu/ mauro/teaching.htmlReading: from Reeds textbook: Section 5.7, 6.1 (review), 6.2. I would suggest alsoread
Duke - MATH - 139
Homework 13 - Math 139Due Wed Dec. 8thInstructorOceOce hoursWeb pageMauro Maggioni293 Physics Bldg.Monday 1:30pm-3:30pm.www.math.duke.edu/ mauro/teaching.htmlReading: from Reeds textbook: Section 6.3, 6.4Problems:6.2: #1,5,8,96.3: #3,6
Duke - MATH - 139
Oce Problem 1 - Math 139Due Oct 13thInstructorOceOce hoursWeb pageMauro Maggioni293 Physics Bldg.Monday 1:30pm-3:30pm.www.math.duke.edu/ mauro/teaching.htmlOce consultation: Oct. 7th, 8th, 1pm-3pm, usual oce hours on Oct. 11th. Possiblyother ti
Duke - MATH - 139
Oce Problem 2 - Math 139Due midnight Dec. 10thInstructorOceWeb pageMauro Maggioni293 Physics Bldg.www.math.duke.edu/ mauro/teaching.htmlIn this Oce Problem you will be constructing the Riemann integral for functionsf : Rn R with n 1.(i) You will
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Business-to-BusinessIncentive Marketing BySmall Business ResourceBrought to you in association withWhats in this presentationWhats required in setting up an incentive programPit falls to look out forBrought to you in association withWhy use incent
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Online Advertising Models BySmall Business Resource LtdBrought to you in association withE-Mail AdvertisingUnsolicited commercial e-mail (UCE or spam)Opt-in or solicited e-mailAd supported e-mailHTML mail with banner adsSponsoring e-mail discussio
BYU - BUS M - 201
Lesson 6E-STRATEGIES AND PLANSWhat is a Small BusinessEntrepreneur?Entrepreneur n. a business manor woman of positive disposition whoattempts to make profit from opportunitiesby risk, initiative and guidance from2-small-business.comSESSION OBJECT
BYU - BUS M - 201
BYU - BUS M - 201
BYU - BUS M - 201
Small Business ResourcePower Point SeriesChatMarketingWhat is Chat MarketingChatMarketingistheuseofonlinechatroomstopromoteyourproductorservice.Getting StartedChatroomsareusuallybrokenintocategories.Findtherightchatroomwhereyourtargetedaudiencewo
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ImplementationThe implementation processAn action checklistTotal quality and marketingManaging the organisation/stakeholder interfaceActivities to establish and build customerrelationships Relationship marketing McKinsey 7-S frameworkThe Marketin
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Promotions DecisionsElements in the communication processPromotions mixThe promotions messageExecutions styleMedia choice?Promotional objectivesElements in the Communication ProcessSenderEncodingMessageMediaDecodingNoiseFeedbackResponseRec
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Pricing Decisions Pricing strategies Pricing exercise Ten ways to increase priceswithout increasing price WinklerQualityLowLowPriceHighHighEconomyStrategye.g. TescospaghettiPenetratione.g. Telewestcable phonesSkimminge.g. New film oral
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Channel and Distribution TacticsBucklins definition of distributionTodays system of exchangeChannel intermediariesSix basic channel decisionsSelection considerationPotential Influence Strategies - Frazier and Sheth(1989) Frequencies of use of infl
BYU - BUS M - 201
Products Decisions Product and Service ClassificationSystem The Product Life Cycle Introduction to product matrices Boston Matrix (Growth/Share) Ansoffs Matrix (Product Market)Product and ServiceClassification System Convenience goods - little ef
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Strategic Development Product Life Cycle (Revisited in Product) Bowmans Competitive Strategy Options New Product Development (NPD)Five stages of the PLC Product development - sales are zero, investmentcosts are high Introduction - profits do not ex
BYU - BUS M - 201
Buyer BehaviourDominant Family Purchase - Cozenza 1985Demographic FactorsThe Consumer Buying ProcessMaslows hierarchy of needsUK socioeconomic classification schemeTypes of buyer behaviourThe Buying Decision ProcessOrganisational Buyer BehaviourD
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Why is marketing planningnecessary? Systematic futuristic thinking by management better co-ordination of company efforts development of better performance standards forcontrol sharpening of objectives and policies better prepare for sudden new deve
BYU - BUS M - 201
Definitions of marketingMarketing is the management process thatidentifies, anticipates and satisfies customerrequirements profitablyThe Chartered Institute of MarketingThe right product, in the rightplace, at the right time, and at theright price
BYU - BUS M - 201
Organized LaborWhy Organized Labor? Collective Bargaining Unions bargain for contracts on behalf of its members Strength in numbersthey cant fire us all! Except Air Traffic Controllers (1980s) and Baseball Umpires(1990s) Many were fired for going o
BYU - BUS M - 201
Instructions:1.2.3.4.5.6.Take the act, law, or word you were given in classand find out about it using the Internet!Using Word, create a colorful and creative, one-pageposter.You should include the act, law, or word youresearched in large font
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Sherman ActClean Air Act of Patents1970AgeDiscriminationinEmploymentActClayton Act of Toxic1914SubstancesControl Act of1976TrademarksAmericanswithDisabilitiesActWheeler-LeaAct of 1938SocialresponsibilityCopyrightsOSHAFDANationalE
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Businesses, Workers, and theLawChapter 5Government Regulation of Business Regulations are rules that governmentagencies issue to implement laws. Businesses spend a lot of time and moneymaking sure they comply with laws andregulations and do not fa
BYU - BUS M - 201
GROUP # 1You are working as a cashier at Target, a jobyou love. A customer gave you a \$50 bill butwalked off without receiving his change.GROUP # 2You are in charge of payroll. You enter thehours employees work into the system, andfrom there, paych
BYU - BUS M - 201
Ethics and SocialResponsibilityChapter 4Ethics in Business Ethics set of moral principles or valuesthat govern behavior Managers must face sometimes difficultethical decisions in business situations. Code of Ethics document that outlinesthe princ
BYU - BUS M - 201
Name _B384 Business ManagementChapter 19 Review Guide1. What are three reasons why management controls needed?2. What are the basic principles of a feedback system?3. What are the three basic requirements for management control?1.2.3.4. Explain t
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ManagementControlChapter 19The Importance of Management Control Managers plan under the assumption thateverything will run smoothly However, you anticipate what could go wrong Managers use CONTROLS to identify andcorrect problems Without controls
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Managing Conflict and StressChapter 17What is Conflict? Conflictstrugglebetween people withopposing needs,wishes, or demandsMythTruthConflicts isdysfunctionalConflict is anormal part of lifeAll conflicts canbe resolvedMost conflicts canat
BYU - BUS M - 201
WORKING WITH EMPLOYEESChapter 15Management TalkWhat makes us stand out is our heritage as acooperative and commitment to providing agreat work environment. While we work hard tocreate a challenging and enjoyable workenvironment, it takes great peop