1 SAMPLE QUESTIONS FOR FINAL EXAM REAL ANALYSIS I FALL Fid the followig usig the defiitio of the Riema itegral: a 0 x + dx 3 Cosider the partitio P x 0 3, x 3 +, x 3 +,......, x of the iterval [ 3, 0]. Observe that for all k,,..., 3}, if x + : x k x x k } x k k + k 5 +, ad similarly sup x + : x k x x k } 5 + k. Thus, lettig f x x +, L f, P 3 k k 3 3, x 3 0} Similarly, U f, P 3 k 5 + k Thus, 6 3 L f, P L f U f U f, P By the Order Limit Theorem ad the algebraic limit theorem with, we have 6 L f U f 6, so L f U f 0 3 x + dx 6.
2 REAL ANALYSIS I FALL 006 b 3 x dx Cosider the partitio P x 0, x +, x +,..., x k + k,..., x 3 + 3, x 3 } of the iterval [, ]. Observe that 3 x x, so the fuctio f x 3 x is decreasig for x 0 ad icreasig for x 0. The, for all k,,..., 3}, sup } 3 x 3 + : x k x x k k if k 3 + k if + k 3, ad if } 3 x 3 + : x k x x k if k k 3 + k if + k 3 Thus, L f, P k k 3 + k k 3 k + 3 k+ 3 k Similarly, U f, P k k 3 + k k + 4 k 3 k k+ + 4 k 3 k Thus, k k 3 k k L f, P L f U f U f, P 6 5. By the Order Limit Theorem ad the algebraic limit theorem with, we have
3 6 L f U f 6, so L f U f c x dx 3 Cosider the partitio SAMPLE QUESTIONS FOR FINAL EXAM 3 3 x dx 6. P x 0 3, x 3 +, x 3 +,......, x , x 4 } of the iterval [ 3, ]. Lettig ψ x x, observe that xk if k 3 because x is decreasig here if ψ x : x [x k, x k ]} x k if k > 3 because x is icreasig here 3 + k if k k. if k > 3 Thus, L ψ, P Note that 3 k 3 k 3 + k + 6 k + 4 k k3+ k 4 k3 4 k3+ 6 k 3 + k k k 4 4 7, k 3 k k k3+ k. so L ψ, P Similarly, 3 + k sup ψ x : x [x k, x k ]} if k k if k > 3,
4 4 REAL ANALYSIS I FALL 006 so ad U ψ, P 4 k3+ 3 k 3 k 3 + k + 6 k k 4 k k 3 k k 4 k3+ 4 k k 6 + k k3+ k, which implies 7 +, U ψ, P Sice L ψ, P L ψ U ψ U ψ, P, we have L ψ U ψ for all N, so that we ca use the order limit theorem as to coclude that the fuctio ψ is itegrable ad that L ψ U ψ 3 x dx 0. 5 Usig the defiitio of Riema itegral, fid ψ x 0 if x 3 7 if x 3 ψ x dx if it exists, if For k N, cosider the partitio P k,, +, } of the iterval [, ]. The 3 k 3 k [ if ψ x : x 3 k, 3 + ]} 0, ad k [ sup ψ x : x 3 k, 3 + ]} 7; k the ifima ad suprema of ψ over the other itervals are zero. Thus, L f, P k U f, P k k k. Sice L f, P k L f U f U f, P k, we have 0 L f U f 4 k
5 SAMPLE QUESTIONS FOR FINAL EXAM 5 for all k N. Usig the order limit theorem as k, we have that L f U f ψ x dx 0. x 6 Prove usig the defiitio that lim x+3. x 3 x Proof: For ay ε > 0, let δ mi ε, } > 0. If 0 < x 3 < δ, the < x < 3 ad x 3 < ε, which implies x x 3 < ε, so that x x 3 x x < ε, or x x + 3 x < ε. 7 Usig the defiitio, prove that each fuctio below is cotiuous o its domai. a g x x+ Proof: For ay ε > 0, if c, let δ mi ε c + 4, c + } > 0. If 4 x c < δ, the 4 c + < x c < c +, so 4 c c + < x < c + c +, or c + c + < x + < c + + c +, so that x + > c + > 0, or c + < x + We also have that x c < δ implies c x < ε 4 c +, which implies c x < ε x + c +, so that c x x + c + < ε, or x + c + < ε. Thus, g is cotiuous at ay c such that c. b h x 4x + 3 Proof: Let c be ay elemet of [ 3,, the domai of h. 4 Case : If c 3, the for ay ε > 0, let δ ε > 0. If x c 4 4 x < δ ad x 3, the 4 4 x < ε, so that 4x + 3 h x h 4 3 < ε. Thus, h is cotiuous at 3 4.
6 6 REAL ANALYSIS I FALL 006 Case : If c > 3, the for ay ε > 0, if δ mi ε c + 3, 3 4c + 3} > 0 ad if 6 x c < δ, the 4x x c c > 4 3 4c c 6 4c + 3, 4 so that 4c + 3 < 4x + 3. Further, agai if x c < δ, The 4 x c < ε 3 4c + 3, so that 4x 4x 4c < ε c + 3, or 4x 4x + 3 4c + 3 < ε c x + 3 4c + 3 4x c + 3 < ε, or 4x + 3 4c + 3 < ε. Thus, h is cotiuous at c. 8 Prove or disprove: If φ : [0, ] R is cotiuous, ad if x is a Cauchy sequece such that x [0, ] for all N, the φ x is Cauchy. True: same proof as 0b below. 9 Usig the defiito, prove that the derivative of x is. x Observe that if x > 0, y x lim y x y x lim y x y + x x by the algebraic limit theorem. 0 Prove that the derivative of is, usig the defiitio. x x Proof: Let f x. If x ad y are i the domai of f, that is x, y R \ 0}, the x y x y x x y yx y x yx. We may ow take the limit as y x of the expressio above, usig the algebraic limit theorem ad the cotiuity of g t algebraic cotiuity theorem. Thus, the limit tx exists ad is f x lim y x y x y x lim y x yx x x x. Assumig Rolle s Theorem, prove the Mea Value Theorem. Hit: Apply Rolle s Theorem to the fuctio g x f x fb fa x. b a Determie if the each series below coverges or diverges, ad prove your aswer.
7 SAMPLE QUESTIONS FOR FINAL EXAM 7 a 3 Observe that for, 4, so that 3 4, or 4 3. Thus, for, Sice 0 3 3/ / 4 3 coverges by the p-series test p 3 >, 3/ 3 coverges by the direct compariso test. b Observe that this series is alteratig. Sice 0 < < for all N, sice +3+ lim 0, the Squeeze Theorem implies that lim 0. Next, observe that +3+ for all 4, , so the absolute values of the terms of the series are decreasig. By the alteratig series test, the series coverges. 3 True or False. Prove your respose you may refer to theorems. a If φ : 0, ] R is cotiuous, ad if x is a Cauchy sequece such that x 0, ] for all N, the φ x is Cauchy. False: For example, let φ x which is cotiuous o 0, ] by the algebraic x properties of cotiuous fuctios, ad cosider the sequece x defied by x 0, ]. The x coverges to 0 ad is thus Cauchy, but φ x, ad so h x diverges ad is therefore ot Cauchy. b If φ : [0, ] R is cotiuous, ad if x is a Cauchy sequece such that x 0, for all N, the φ x is Cauchy. True. Sice [0, ] is closed, x coverges to a poit x [0, ]. The, sice φ is cotiuous, φ x coverges to φ x. Thus, φ x is Cauchy, sice it is a coverget sequece. c Every bouded fuctio o [a, b] is cotiuous o [a, b]. False: For example, let a 0, b, ad cosider the fuctio q : [0, ] R defied by if x q x 0 otherwise The q x for all x [0, ], so q is bouded. However, q is ot cotiuous at ; eve though is a limit poit of [0, ], lim q x 0 q 0. x
8 8 REAL ANALYSIS I FALL 006 d Every bouded fuctio o [a, b] is itegrable o [a, b]. False: For example, if if x Q g x 0 otherwise the g is bouded 0 g x x, but each upper sum is b a, ad each lower sum is zero, so the fuctio g is ot Riema itegrable. e If lim x x is a coverget sequece, the the set x N} x} is compact. True: Sice x coverges, the set x N} is bouded, say x B for all N. The every elemet of S x N} x} is bouded i absolute value by max B, x }. Because lim x x, every eighborhood of x cotais all but fiitely may terms of the sequece x. Thus, if y R x}, the if we let ε y x, V ε y V ε x, so that V ε y y} ca cotai at most a fiite umber of terms x,..., x k } of the sequece. Thus, if ε mi ε, y x,..., y x k } > 0, we have that V eε y y}. Thus, y is ot a limit poit of S. We have show that o elemet other tha x is a limit poit of S, so S is closed. Sice S is closed ad bouded, S is compact, by the Heie-Borel Theorem. f Ay bouded set that cotais all of its limit poits is compact. True: Sice such a set would be closed ad bouded, the Heie-Borel Theorem implies that the set is compact. g Ay set that cotais all of its limit poits is compact. False: For example, N is closed but ot bouded, ad so it cotais all of its limit poits there are oe but is ot compact. h The set x R x 3 3x 6 + 7x 7x 00} is closed. True: Call this set A; observe that A f, 00], where f is the fuctio defied by f x x 3 3x 6 +7x 7x. The fuctio f is cotiuous because it is a polyomial, a sum of products of j x x ad costats, ad the algebraic properties of cotiuous fuctios the imply f is cotiuous. The set, 00] is closed, ad its complemet 00, is ope. Observe that B R \ Af 00, is ope because it is the cotiuous iverse image of a ope set. Thus, A R \ B is closed. i If f is cotiuous o,,, the it is also cotiuous o, ] [,. False: Let if x [, ] f x 0 otherwise The fuctio is costat ad thus cotiuous o,,, but f is ot cotiuous at or at. For istace, the defiitio of cotiuous does ot hold at for ε, because for ay δ > 0, f x f > for x δ, eve though x δ < δ. j If the fuctio φ : R R satisfies the itermediate value property, the it is cotiuous. False: For example, let si φ x x if x 0 0 if x 0 The φ is ot cotiuous at 0, because φ δ, δ [, ] for every δ > 0 so the defiitio of cotiuous caot hold at 0 with ε. O the other had, φ satisfies the itermediate value property IVP. Sice φ is cotiuous away from 0,
9 SAMPLE QUESTIONS FOR FINAL EXAM 9 it automatically satisfies the IVP for itervals cotaied i R \ 0}. Next, if [a, b] is ay iterval cotaiig zero, φ [a, b] φ a, b [, ], so there are a ifiite umber of poits x i a, b such that φ x is ay possible itermediate value. k If f is oegative ad itegrable o [0, ] ad if f 0, the f x 0 0 for every x 0,. False: Let f x if x 0 otherwise The this fuctio is oegative, ad it is ot idetically zero o 0,. However, observe that every lower sum will be zero for this fuctio sice every iterval cotais poits x where f x 0. Next, for k >, cosider the partitio P k defied by P k 0,, + },. k k The U f, P k k. k k k Sice L f, P k L f U f U f, P k, ad sice lim U f, P k 0 ad k L f, P k 0, by the algebraic limit theorem, we have 0 L f U f 0, so that f is itegrable ad f 0. 0 l If f ad g are two itegrable fuctios o [, ], the f g f g. True: f g f g f g f g. m If A is bouded, the if A is a limit poit of A. False: For example, the set A, } has o limit poits, but if A. The fuctio 0 if t is ratioal Bubba t t if t is irratioal is cotiuous at 0. True: Give ε > 0, if t 0 < δ ε, the Bubba t Bubba 0 Bubba t 0 0 if t is ratioal t if t is irratioal < ε. o If b f x dx 0, f is cotiuous, ad f x 0 for all x [a, b], the a f x 0 for all x [a, b]. True: Suppose that f x is ot always 0; that is, suppose that f c < 0 for some c [a, b]. The, sice f is cotiuous, lettig ε fc > 0, there exists δ > 0 such
10 0 REAL ANALYSIS I FALL 006 that x c < δ ad x [a, b] implies that f x f c < fc, so f x f c < fc or f x < fc. Thus, b 0 f x dx f x dx + f x dx f x dx a [a,b] I c I c I c f c dx f c δ < 0, I c where I c [a, b] is a iterval of legth δ that icludes c as a edpoit. This is a cotradictio, so f x 0 for all x [a, b]. p Every real umber is a limit poit of the set of real umbers. True, sice every real umber x is a limit of the sequece a defied by a x + for all N. 4 Suppose that F is a differetiable fuctio o R, ad suppose that F 0, F, ad F 4 0. Prove that there is a real umber x 0, 4 such that F x. 3 Proof: By the Mea Value Theorem, there exists c 0, such that F F F 0 c, ad also there exists b, 4 such that F F 4 F b. By the Darboux Theorem, F satisfies the Itermediate Value Property, so sice c < b ad F c < 3 < F b, there exists x c, b 0, 4 such that F x. 3 5 Let p : [0, ] R be a cotiuous fuctio. Which of the followig must be true? Justify your resposes. a There exists a costat a > 0 such that p x p y < a for all x, y [0, ]. True: By the Extreme Value Theorem, a cotious fuctio achieves its maximum ad miimum o a compact set, so sice [0, ] is compact, i particular p is bouded, say p t M for all t [0, ]. The p x p y p x + p y M. So a M + works. b There exists a costat b > 0 such that p x p y < for all x, y [0, ] that satisfy x y < b. True: Sice a cotiuous fuctio o a compact set is uiformly cotiuous, p is uiformly cotiuous o [0, ], which implies the statemet above with ε > 0. c There exists a costat c > 0 such that p x p y < c x y for all x, y [0, ]. p x p y False: You ca costruct a fuctio that is cotiuous o [0, ] so that x y is ubouded. For example, let x si p x x if 0 < x 0 if x 0 Observe that p is cotiuous o 0, ], sice the fuctios defied by g x si x, g x x are cotiuous, ad by the fact that the compositios ad algebraic combiatios of cotiuous fuctios are cotiuous. Observe that 0 is a limit poit of [0, ], ad sice for x > 0, x x si x x ad lim x 0, the Squeeze + Theorem implies that lim x si 0. x 0 + x x 0
11 SAMPLE QUESTIONS FOR FINAL EXAM Thus, p is cotiuous at 0 as well. O the other had, for ay N, we may let x 4 + π, y, ad so x, y [0, ], but 4 π 4 + π p x p y 4 + π si 4 π 4 π si x y 4 + π 4 π 4 + π + 4 π 4 π 4 + π , which is ubouded as. 6 What facts are impicitly used to make the followig deductios? x + y < 4x + y x + > : assumig x + ad 4x + are ozero, 4x + assumig x + ad 4x + are the same sig, ad usig y the fact that r x is a strictly decreasig fuctio. x y > x + 4x + > : assumig first that 4x + y x + ad are both positive ad usig the fact 4x + that s x x is strictly icreasig o 0, ; the assumig that is greater tha ad usig 4x + the fact that x > x if x >. 7 Cosider the fuctio g defied by 0 if x 0 g x x cos x if x [ π, 0 0, π] Is this fuctio bouded? Is this fuctio cotiuous at zero?... differetiable at zero?... Is g x cotious at zero? Provide justificatio. Aswer: The fuctio g is bouded, because g x x π for all x i the domai of x cos x 0 g. If ε > 0, ad if x < δ ε, the x 0 x < ε. Thus, g is differetiable x at zero with g 0 0, ad as a result g is also cotiuous at zero. Observe that by the algebraic properties of derivatives, if x 0, g is differetiable at x, ad g x x cos x x x si 3 x x cos x x x si.
12 REAL ANALYSIS I FALL 006 Observe that g π + π ad x π π π + π cos + π π + π π /, + π si π + π π + π approaches zero ad π / decreases without boud as icreases, so lim g x. We coclude by the sequetial criterio that g x has o limit as x 0. Thus, g is ot cotious at 0. 8 Cosider the fuctio if t is ratioal G t e t if t is irratioal Fid the set S of real umbers at which G is cotiuous. Justify your respose. Solutio: G is cotiuous oly at 0. Observe that sice 0 G t e t, ad sice lim e t 0, the Squeeze Theorem implies that lim G t 0. Thus t 0 t 0 lim G t 0, so that lim G t G 0, which implies that G is cotiuous at t 0 t 0 0. O the other had, if t 0, the for ε et > 0 ad for ay δ > 0, by the desity of Q i R ad of R Q i R, there are irratioal umbers x ad ratioal umbers y such that x t < δ ad y t < δ so that max G x G t, G y G t } mi e t, e x, e y }. The latter quatity is strictly greater tha ε et if δ is sufficietly small, showig that G is ot cotiuous at t. 9 Suppose that f is a sequece of bouded, real-valued fuctios o R. If 3 f k+ x f k x for all x R ad all k N, prove that f x coverges absolutely ad uiformly o R. Proof: Note: Ratio Test does ot work, because you do ot kow that lim f k+x f k x exists. Plus it oly gives you poitwise covergece. Observe that f x f 3 x, ad i geeral if f k x f 3 k x, the f k+ x f 3 k x f 3 k x. By iductio, f x 3 f x for all, ad the equatio is also valid for. Sice f is bouded o R, there exists M > 0 such that f x M for all x R, so that f x 3 M for all x R. Sice 3 M is a coverget geometric series, the Weierstrass M-test implies that f x coverges absolutely ad uiformly o R. 0 Prove that the Taylor series for f x si x that is cetered at zero coverges to si x at each x R.
13 Proof: If p k x SAMPLE QUESTIONS FOR FINAL EXAM 3 k 0 f 0! x is the k th partial sum of the Taylor series of si x, the Lagrage Remaider Theorem implies that for each x R, si x p x f + c x +, +! where c is a umber betwee 0 ad x. Observe that sice d si x cos x, d cos x dx dx si x, si x, ad cos x, we have that f + t for all t R. The 0 si x p x f + c +! x+ x + +!. Next, for fixed x R, cosider the series x + x + x + +!. Observe that for ay +! +! x +, which has a limit of zero as. By the ratio test, the series coverges, so by the divergece criterio for series, lim iequality above, + x +! 0, so by the Squeeze Theorem ad the lim si x p x 0, so the Taylor series for si x coverges to si x for all x R. Prove usig basic priciples that a sequece of cotiuous fuctios that coverges uiformly coverges to a cotiuous fuctio. Proof: Suppose f is a sequece of cotiuous fuctios that coverges uiformly to a fuctio f o A R. Let c be ay elemet of A. By the defiitio of uiform covergece, give ay ε > 0, we may choose N N such that for all N, f x f x < ε 3 for all x A. Next, for that specific N, the cotiuity of f N implies that we may choose δ > 0 so that if x A ad x c < δ the f N x f N c < ε. The if x c < δ, we 3 have f x f c f x f N x + f N x f N c + f N c f c f x f N x + f N x f N c + f N c f c < ε 3 + ε 3 + ε 3 ε. Thus f is cotiuous at c. Give a example of a sequece of cotiuous fuctios that coverges but does ot coverge to a cotious fuctio. Justify your respose. For each N, let α : [0, ] R be defied by α x x ; all of these fuctios are cotiuous by the algebraic cotiuity theorem. The let β x lim α x 0 if 0 x < if x Observe that lim β x 0 β, so β is ot cotiuous at. x 3 Let B be a bouded, oempty set of real umbers, ad let b be the least upper boud of B. If b is ot a elemet of B, which of the followig is ecessarily true? Justify your resposes. a B is closed. False: for example, if B 0,, b / B. b B is ot ope. False: see above. c b is a limit poit of B.
14 4 REAL ANALYSIS I FALL 006 True. By the sup lemma, give ay ε > 0, there is a x B such that b ε < x b, but sice b / B we also have b ε < x < b. Thus, every ε-eighborhood of b cotais a poit x B b} B; thus, b is a limit poit of B. d No sequece i B coverges to b. False. By part c, b is a limit poit of B, so there exists a sequece of poits i B b} B that coverges to b. e There is a ope iterval cotaiig b that cotais o poits of B. False. By part c, b is a limit poit of B. Every ope iterval cotaiig b is a ope set, so there exists ε > 0 so that V ε b is a subset of that iterval. But every such V ε b cotais a poit of B b} B, so the ope iterval does as well. 4 Prove that every oempty ope set i R is a uio of ope itervals. Proof: If A is a oempty ope subset of R, the for every x A, there is a ε > 0 such that V ε x A. The cosider B V ε x. x A,V εx A The otatio above implies that the uio is take over all possible x A ad all possible ε > 0 such that V ε x A. The B is a uio of ope itervals, ad y B, the y V ε x A for some ε > 0 ad some x A, so y A. Next, if y A, there exists ε > 0 such that y V ε y A, so y B. Thus, A B is a uio of itervals. 5 Assumig that f ad g are real-valued fuctios o R, egate the followig statemet: For each s R, there exists r R such that if f r > 0, the g r > 0. Negatio: There exists s R such that for all r R such that f r > 0, we have g r 0. φx 6 Suppose that φ : R R is a fuctio such that lim x 0 x is a real umber L ad φ 0 0. Which of the followig are ecessarily true? Justify your resposes. a φ is differetiable at 0. True: defiitio of the derivative: φ φ x φ 0 0 lim L. x 0 x 0 b L 0. False: For example, if φ x x for all x R, the φ 0 L. c lim φ x 0. x 0 True: Differetiable implies cotiuous at zero, so sice 0 is a limit poit of the domai, lim φ x φ 0 0. x 0 7 Provide a defiitio for the statemet lim f x. x c lim f x, the lim x c x c f x 0. The prove that if The statemet lim f x meas that give M > 0, δ > 0 such that 0 < x c < x c δ ad x domai of f implies f x M. If this is true, the, give ε > 0, choose N N such that < ε. The choose δ as above with M N. The 0 < x c < δ N ad x domai of f \ 0} implies that 0 fx < ε. Thus, lim N x c f x 0. 8 Give a example of a fuctio α : R R that is uiformly cotiuous ad a example of a fuctio β : R R that is ot uiformly cotiuous. Solutio: For example, let α t t. The, give ay ε > 0, if t x < δ ε, the α t α x t x < ε. Thus α is uiformly cotiuous.
15 SAMPLE QUESTIONS FOR FINAL EXAM 5 Next, let β x x. Observe that for every δ > 0, there exists such that δ >, or δ >. The + δ < δ, but β + δ β + δ δ + 4 δ >, so the defiitio of uiformly cotiuous does ot hold for ay ε <. 9 Be able to prove the product ad quotiet rules. OK, I m able. 30 Prove that if f is differetiable o a iterval with f x o that iterval, the f ca have at most oe fixed poit. Note: a fixed poit of f is a umber x such that f x x. Proof: Cosider the fuctio h x f x x. The by hypothesis, h is the sum of differetiable fuctios ad is thus differetiable o the iterval, ad h x 0 o that iterval. Suppose that h a 0 ad h b 0 for some a ad b i the iterval ie both a ad b are fixed by f. The, by Rolle s Theorem, if a b, the there exists c betwee x ad y such that h c 0, which is a cotradictio. Thus, there is at most oe fixed poit for f o that iterval.
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