Sring 2 Instructions Please write your solutions on your own aer. These roblems should be treated as essay questions. A roblem that says give an examle requires a suorting exlanation. In all roblems, you should exlain your reasoning in comlete sentences. Students in Section 5 should answer questions 6 in Parts A and B, and otionally the extra-credit question in Section D. Students in Section 2 (the honors section) should answer questions 3 in Part A and questions 7 9 in Part C, and otionally the extra-credit question in Section D. Part A, for both Section 2 and Section 5 In this art of the exam, your task is to analyze the following roosition from three different oints of view. Proosition. If n is a natural number, then e n > C n.. Prove the roosition by induction on n. [Recall that e 2:78.] Solution. Evidently e > C, since e 2:78, so the basis ste of the induction holds. Suose it has been shown that e n > C n for a certain natural number n. Multilying both sides by e shows that e nc > e Cen > 2Cen > 2Cn D C.nC/. Thus the validity of the roosition for a certain natural number imlies the validity of the roosition for the next natural number. Since the basis ste holds, and the induction ste holds, the roosition is valid for every natural number. 2. The figure below suggests the more general statement that e x > C x for every nonzero real number x (because the grah of e x is convex). y D e x y D C x May, 2 Page of 6 Dr. Boas
Sring 2 Prove that e x > Cx for every nonzero real number x by writing the Taylor olynomial for e x of degree and examining the remainder term. [Recall Lagrange s theorem about aroximation by Taylor olynomials: f.x/ D P n kd f.k/.x kš /.x x / k C f.nc/.c/.x x.nc/š / nc for some oint c between x and x. Take x equal to and n equal to.] Solution. If f.x/ D e x, then every derivative of f.x/ equals e x, so f./ D and f./ D. Then Taylor s formula says that e x D C x C e c x 2 =2 for some oint c between and x. Since x 2 > when x, and e c > for every oint c, the remainder term is ositive. Thus e x > C x when x. 3. Evidently e x > for every ositive real number x. By integrating on a suitable interval, deduce that e x > C x for every ositive real number x. Solution. The continuous function e t x >, then is ositive when t >, so if < Z x.e t / dt D e t t x D.ex x/.e / D e x x : Adding C x to both sides shows that C x < e x when x >, as claimed. Part B, for Section 5 only 4. State the following theorems: (a) the Bolzano Weierstrass theorem about sequences; (b) the squeeze theorem (sandwich theorem) about its of functions; (c) Rolle s theorem about differentiable functions. Solution. The statements are in the textbook (Theorems 2.4, 5.2, and 7.9). May, 2 Page 2 of 6 Dr. Boas
Sring 2 5. Give an examle of a function f that satisfies all three of the following roerties: the function f is continuous on the oen interval.; 2/, the function f is not differentiable at the oint, and R 2 f.t/ dt is a divergent imroer integral. Solution. One way to make the function continuous but nondifferentiable at the oint is to use a iecewise definition like this: ( =x; if < x < ; f.x/ D ; if x: The same effect can be achieved without exlicitly introducing cases: namely, f.x/ D max.=x; /: A different examle is jx j C. You know from class that the imroer x integral R dx diverges. x There are many other ossible examles. e cos.n/ 6. Prove that n! n D. Solution. Evidently < e cos.n/ < 3 for every n, since the cosine function never exceeds (and e < 3). Therefore < ecos.n/ n < 3 n : But any constant times =n has it when n!. The squeeze theorem imlies that e cos.n/ =n also has it. Alternatively, you could go back to the definition of it: if a ositive " is secified, and N D d3="e, then < e cos.n/ =n < 3=n < " when n > N. Part C, for Section 2 only 7. State the following theorems: (a) the Heine Borel theorem characterizing comact sets of real numbers; May, 2 Page 3 of 6 Dr. Boas
Sring 2 (b) some theorem from this course in which the word countable aears; (c) some (other) theorem from this course named after Cantor. Solution. The Heine Borel theorem states that a set of real numbers is comact if and only if every covering of the set by oen intervals can be reduced to a finite subcover. Some results in the textbook concerning countability are Theorems 4.35, 5.6, and 5.63 and Corollary 8.9. One of Cantor s theorems is the uncountability of the real numbers (Theorem 2.4); another is the theorem on nested closed sets (Corollary 4.25). 8. Give an examle of a function f that satisfies all of the following roerties: the function f is continuous on the oen interval.; /, the uer it su x!c f.x/ equals, the lower it inf x!c f.x/ equals, and the integral R f.t/ dt is a convergent imroer integral. Solution. Since R = x dx converges, so does an integral with a smaller ositive integrand. Here is one examle that fits the secified conditions: f.x/ D sin2.=x/ x : There are many other ossible examles. 9. Prove that n! Z nc n t dt D. C t 3 Solution. In rincile, the integral can be evaluated exlicitly, but you do not want to carry out that comutation. You can, however, make the following estimate: Z nc Z t nc Z < n C t dt < t nc 3 n t dt D 3 n t dt D 2 n n C < n : The squeeze theorem now imlies that the it of the integral equals. Alternatively, observe that R t dt is a convergent imroer integral Ct 3 (the integrand is less than =t 2, and you know from class that R =t 2 dt converges). The integrals R n t dt therefore form a bounded increasing Ct 3 sequence, hence a Cauchy sequence. Since the integral R nc t n dt is the Ct 3 difference of two consecutive terms of this Cauchy sequence, the integral has it. May, 2 Page 4 of 6 Dr. Boas
Sring 2 Part D, otional extra-credit question for both Section 2 and Section 5. Alfie, Beth, Gemma, and Delma are studying the it x2 C 9x x2 C 4x : Alfie says, simlifying the square roots converts the roblem to.x C 3 x /.x C 2 x / D x D : Beth says, the it of a sum is the sum of the its, so the answer is x2 C 9x x2 C 4x D D : Gemma says, by the mean-value theorem, u v D 2.u v/, so the c it equals 2 c Œ.x2 C 9x/.x 2 C 4x/ D 2 5x D : c Delma says, by l Hôital s rule, the it equals 2x C 9 2 2x C 4 x 2 C 9x 2 B D @ 2x q C 9 x 2 C 4x 2x C 9 x D D : 2x C 4 C q A 2x C 4 x Who (if anyone) is right, and why? Identify the mistakes in the erroneous arguments. Solution. Everyone is wrong: the value of the it is actually 5=2. Alfie s algebra is faulty: the square root of a sum is not equal to the sum of the square roots. Beth overlooks that the it law alies only if the its exist as finite values. The exression is undefined, an indication that a subtler argument is needed. May, 2 Page 5 of 6 Dr. Boas
Sring 2 Gemma has a great idea, but the execution falls short. The intermediate oint c is not a constant but deends on x. Indeed, the value of c lies between x 2 C 4x and x 2 C 9x. Therefore x D x 2 < x 2 C 4x < c < x 2 C 9x D x C.9=x/: Accordingly, 5 2 C.9=x/ < 5x 2 c < 5 2 : Since the it of the left-hand side as x! equals 5=2, the squeeze theorem imlies that the original it equals 5=2. Thus Gemma s argument can be corrected to give the right answer. Delma is wrong because l Hôital s rule does not aly to undefined exressions of the form. The standard trick for simlifying a difference of square roots algebraically is the following: x2 C 9x C x x2 C 9x x2 C 4x 2 C 4x x2 C 9x C x 2 C 4x D.x2 C 9x/.x 2 C 4x/ x2 C 9x C x 2 C 4x D 5x x2 C 9x C x 2 C 4x 5 D : C.9=x/ C C.4=x/ Now it is evident that the it as x! is equal to 5=2. May, 2 Page 6 of 6 Dr. Boas