Carefully tear this page off the rest of your exam. UNIVERSITY OF TORONTO SCARBOROUGH MATA37H3 : Calculus for Mathematical Sciences II

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1 Carefully tear this page off the rest of your exam. UNIVERSITY OF TORONTO SCARBOROUGH MATA37H3 : Calculus for Mathematical Sciences II REFERENCE SHEET The (natural logarithm and exponential functions ( The (natural logarithm function is defined: log x = ( For all a, b >, we have log ab = log a + log b. x t dt. (3 For all n N and all x >, we have log x n = n log x. (4 exp = log. (5 exp = exp. (6 e = exp(. (7 For a > and x R, we define the function a x by a x = exp(x log a. Inverse trigonometric functions Let f : [ π/, π/] [, ] by f(x = sin x, g : [, π] [, ] by g(x = cos x, and h : ( π/, π/ R by h(x = tan x. Then we define the inverse trigonometric functions by arcsin = f, arccos = g, arctan = h. Definition of partitions, upper and lower sums ( P is a partition of [a, b] if P is a finite subset of [a, b] which contains both endpoints. ( Given a function f defined on [a, b] and a partition P of [a, b], set L(f, P = j n m j (t j t j where P = {t, t,..., t n } with t i < t i+ for all i, and m j = inf{f(x : x [t j, t j ]}. Similarly, define U(f, P = j n M j (t j t j where M j = sup{f(x : x [t j, t j ]}. (3 For any partitions P and Q of [a, b], and any function f which is defined and bounded on [a, b], we have L(f, P U(f, Q.

2 UNIVERSITY OF TORONTO SCARBOROUGH MATA37H3 : Calculus for Mathematical Sciences II FINAL EXAMINATION April 3rd, Duration 3 hours Aids: none NAME (PRINT: STUDENT NO: Last/Surname First/Given Name KEY TUTORIAL: Tutorial section (Number or Schedule Name of TA Qn. # Value Score COVER PAGE Total TOTAL: Please read the following statement and sign below: I understand that any breach of academic integrity is a violation of The Code of Behaviour on Academic Matters. By signing below, I pledge to abide by the Code. SIGNATURE:

3 Final Exam MATA37H3 page of ( Prove each of the following assertions. You may freely quote the properties listed on the reference page without proof, but any other properties of log or exp must be explicitly proved in your solution. (You are allowed to refer to any theorems from lecture or the text which are not about log or exp. (a (5 points x log x =. Given M, it suffices to show that log x > M for all sufficiently large x. By the Archimedean property, there exists a natural number n > M log. Since log > (as can be seen from the definition of log, property (3 from the reference sheet implies that log n = n log > M. Finally, applying the fundamental theorem of calculus to the definition of the logarithm, we see that d log x = > for all x > ; in particular, log x is dx x an increasing function. It follows that for all x > n, (b (5 points x e x =. log x > log n > M. First, observe that exp is a function: log is increasing, hence is injective, so its inverse is a function. Moreover, since log is increasing, we conclude that exp must be increasing as well. It follows immediately that e >, since exp > exp. Thus we may apply definition (7 to conclude that e x = exp(x for all x, since log e = by property (6. Given M, it suffices to prove that e x > M for all sufficiently large x. If M, then we are immediately done, since e x > for all x >. If M >, then for all x > log M we have since exp is increasing. x (c (5 points x e =. x e x = exp(x > exp(log M = M Since both x and e x tend to as x, we may apply L Hôpital s rule to deduce: x x e = x x x e x (where we have used Property (5 that exp = exp. Since both x and e x tend to with x, we may apply L Hôpital once more to deduce x x e x = x This last quantity clearly tends to, since the numerator is constant and the denominator tends to. More precisely, given any ɛ >, let x = log. For ɛ all x > x, we have e x >, since exp is increasing. Finally, since ɛ ex >, we conclude that e x < ɛ. e x. continued on page 3

4 Final Exam MATA37H3 page 3 of Page intentionally left blank for scratch work. continued on page 4

5 Final Exam MATA37H3 page 4 of ( For this problem, you may freely refer to any of the properties listed on the reference sheet. (a (8 points Suppose f : [a, b] R is a bounded function, and let P denote the set of all partitions of [a, b]. Prove that sup L(f, P inf U(f, P. Fix any partition Q P. By property (3 on the reference sheet, we see that U(f, Q is an upper bound on the set {L(f, P : P P}. It follows that sup L(f, P U(f, Q, since the left side is the least upper bound. In the argument above, the partition Q P was arbitrary. Thus, sup L(f, P is a lower bound on the set {U(f, Q : Q P}. It follows that sup L(f, P inf U(f, Q, Q P since the right hand side is the greatest lower bound. (b (8 points Show by example that it s possible for sup L(f, P < inf U(f, P. (You must prove that your example really is an example! Let f(x = { if x Q if x Q. I claim that for all partitions P P, we have L(f, P = and U(f, P =. Indeed, suppose P P, say P = {t, t,..., t n } with t = a, t n = b, and t i < t j whenever i < j. Then m i = inf f(x = x [t i,t i ] for all i, since Q is dense. Similarly, since the irrationals are dense, we have M i = sup f(x =. x [t i,t i ] It follows that L(f, P = and U(f, P = as claimed. We conclude that sup L(f, P = < = inf U(f, P. continued on page 5

6 Final Exam MATA37H3 page 5 of (3 (8 points Determine the exact numerical value of ( n n + = n= You must justify your answer to receive credit. [Hint: Find the Taylor series of arctan x.] I first claim that the Taylor series for arctan x is given by arctan x = x x3 3 + x5 5 x7 7 + (* One can obtain this in the standard way (write arctan x = a +a x+a x +, plug in x = to find a ; differentiate both sides and plug in x = to find a ; differentiate both sides and plug in x = to find a ; etc.. Instead, I describe a tricky shortcut. First, we find the derivative of arctan x. To do this, we evaluate d tan(arctan x dx in two different ways. On one hand, it simply equals (since tan(arctan x = x. On the other hand, by chain and quotient rules, d dx tan(arctan x = cos (arctan x d arctan x. dx It follows that d dx arctan x = cos (arctan x = + x Recall the expansion + x = x + x x 3 + (note that this is the Taylor series of, and is also the formula for the sum of an +x infinite geometric series. Plugging in x = t yields d dt arctan t = + t = t + t 4 t 6 + Integrating both sides from to x gives the Taylor expansion claimed in (*. Plugging in x = into the Taylor expansion (* gives /3 + /5 /7 + /9 = π 4 continued on page 6

7 Final Exam MATA37H3 page 6 of (4 ( points Suppose that f is integrable on [a, b], and that f = g for some function g. Prove that b a f = g(b g(a. [Warning: f might not be continuous on the interval!] I first claim that L(f, P g(b g(a U(f, P for any partition P of [a, b]. To see this, suppose P = {t, t,..., t n } is a partition of [a, b], with t i < t j whenever i < j. By the Mean Value Theorem, for every i there exists x i [t i, t i ] such that Since f = g, it follows that g (x i = g(t i g(t i t i t i. f(x i ( t i t i = g(ti g(t i for all i. From the definition of L(f, P and infimum, we see that L(f, P = ( inf f (t i t i [t i,t i ] i f(x i ( t i t i i = ( g(t i g(t i i ( = g(b g(a. Similarly, we see that U(f, P g(b g(a. Thus, ( is proved. The definitions of inf and sup immediate imply that sup L(f, P g(b g(a inf U(f, P. P P Since f in integrable by hypothesis, so we conclude that sup L(f, P = P b a b a f = inf P f = g(b g(a. U(f, P, continued on page 7

8 Final Exam MATA37H3 page 7 of (5 Evaluate the integrals below. You must justify your answer to receive credit. (a (8 points arctan x dx. We first integrate by parts. Set u = arctan x and dv = dx. This gives du = dx and v = x, whence +x arctan x dx = x arctan x x dx ( + x Our next task is to evaluate the integral on right hand side of (, which I ll do via u-substitution. Let u = + x. Then du = x dx, so x + x dx = u du = log u = log Plugging this back into ( gives arctan x dx = π 4 log (b (8 points 3 We have x x dx. 3 x 3 x dx = = 3 3 = x x + dx x ( + x + log(x = + log. Alternatively, one can solve this by u-substitution. Letting u = x, we have du = dx, whence 3 x x dx = u + ( du = + du u u which gives the same result as before. 3 dx continued on page 8

9 Final Exam MATA37H3 page 8 of Page intentionally left blank for scratch work. continued on page 9

10 Final Exam MATA37H3 page 9 of (6 Suppose F is a function satisfying F (x = whenever x <, and F (x = x / x/ F x. (In particular, assume that F is integrable on [x/, x /] x. (a (7 points Prove that F ( =. [Careful! There s something to prove here.] I claim that F is integrable on [, ], and that F =. If F ( =, we re done. Thus we may suppose (temporarily that F (. Given ɛ >, let ɛ δ = F ( and consider the partition P = {, δ, } of the interval [, ]. Then ( ( L(F, P = ( δ + min{, F (} ( δ ( = + min{, F (} δ { = + min, ( F ( } δ. Similarly, we find U(F, P = + max {, ( F ( } δ. Thus, if F ( > we have L(F, P = and U(F, P = + ɛ ; otherwise, we have L(F, P = ɛ and U(F, P =. Either way, we conclude that U(F, P L(F, P < ɛ. Since ɛ > was arbitrary, this shows that F is integrable on [, ]. Now that we know the integral exists, computing it is easy. From above, we see that for all ɛ >, ɛ < L(F, P F U(F, P < + ɛ. The squeeze theorem then implies that F ( = F = as claimed. (b (7 points Prove that F is not differentiable at. For F to be differentiable at, the left- and right-hand derivatives would have to agree, i.e. F ( + h F ( h h = h + F ( + h F ( h Since F (x = for all x, the left-hand derivative is. I claim that the right-hand derivative is 3, and hence, that F is not differentiable at. To do this, we apply the fundamental theorem of calculus. Let G be an antiderivative of F, say, G(x = x F. The fundamental theorem implies that x / ( x ( x F (x = F = G G x/ for all x such that F is integrable on [, x]. Differentiating and applying chain rule yields ( x F (x = xg ( x G ( x = xf ( x F. It follows that F ( = F ( F ( = 3 as claimed. continued on page

11 Final Exam MATA37H3 page of (c (6 points Prove that F is continuous at. We must show that F (x = F (. Since F is constant to the left of, we x have F (x =, so it suffices to prove F (x =, i.e. x x + We have (+h / (+h/ F = = h + +4k+k +k +k F + = ( k + (+h / (+h/ F =. F where k = h/ +4k+k F +4k+k Let m = inf { f(x : x [, 3] } and M = sup { f(x : x [, 3] }. (Note that these both exist, since f is integrable, and hence bounded, on [, 3]. Then for all h [, ] we have m f( + h M, whence m(h + h = m(4k + 4k F. +4k+k We conclude by the squeeze theorem that Also, we have It follows that x h + +4k+k F =. ( k = h =. h + h + h F (x = F ( + h + + = h + = h + =. (+h / ( (+h/ F ( k + F M(4k + 4k = M(h + h. +4k+k Thus the right- and left-hand its agree, and we conclude that F is continuous at. F Total Marks = points

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