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1 MTH 33 Solutions to Exam November th, 08 Name: Section: Recitation Instructor: INSTRUCTIONS Fill in your name, etc. on this first page. Without fully opening the exam, check that you have pages through 3. Show all your work on the standard response questions. Write your answers clearly! Include enough steps for the grader to be able to follow your work. Don t skip limits or equal signs, etc. Include words to clarify your reasoning. Do first all of the problems you know how to do immediately. Do not spend too much time on any particular problem. Return to difficult problems later. If you have any questions please raise your hand. You will be given exactly 90 minutes for this exam. Remove and utilize the formula sheet provided to you at the end of this exam. ACADEMIC HONESTY Do not open the exam booklet until you are instructed to do so. Do not seek or obtain any kind of help from anyone to answer questions on this exam. If you have questions, consult only the proctor(s). Books, notes, calculators, phones, or any other electronic devices are not allowed on the exam. Students should store them in their backpacks. No scratch paper is permitted. If you need more room use the back of a page. You must indicate if you desire work on the back of a page to be graded. Anyone who violates these instructions will have committed an act of academic dishonesty. Penalties for academic dishonesty can be very severe. All cases of academic dishonesty will be reported immediately to the Dean of Undergraduate Studies and added to the student s academic record. I have read and understand the above instructions and statements regarding academic honesty:. SIGNATURE Page of 3

2 MTH 33 Solutions to Exam November th, 08 Standard Response Questions. Show all work to receive credit. Please BOX your final answer.. Determine if the following series converge or diverge. If the series converges, also compute the sum. You must show all of your work and support your conclusions. 3 n (a) (7 points) n Use the ratio test: Since all terms are positive for n, we don t need absolute value a n+ = 3n a n n n+ 3 = 3 n a n+ So L = lim = 3 < and the series converges. a n Alternatively, a n = 3 n 3 = ( ) n 3, so it is a geometric series of ratio 3 < and the n 3 series converges. Since ar n = (b) (7 points) a r, in our case a = and r = 3, so 3 n + 5 3n + Consider the test for divergence: n + 5 Let a n = 3n +. lim a n = lim n + 5 3n + { ( ) } n 3 = 3 3 = 3. = lim + 5 n 3 + n = lim = 3 0 Therefore, by the test for divergence, the series a n diverges. Page of 3

3 MTH 33 Solutions to Exam November th, 08. Determine if the following series converge or diverge. You must show all of your work and justify your use of any series convergence tests. 3n + 5 (a) (7 points) n Use the ratio test: Since all terms are positive for n, we don t need absolute value a n+ = 3(n + ) + 5 n a n n+ 3n + 5 = 3(n + ) + 5 ( 3n + 5 ) 3n + 6n + 8 = 3n = n + 8 n n a n+ So L = lim = and the series converges. a n (b) (7 points) 3 n Consider the limit comparison test: All terms are positive for n, so we may apply the test. Let a n = 3 n and b n = 3 n a n lim = lim b n = lim 3 n 3n 3 n 3 n = lim = 0 = 3 n Therefore, by the limit comparison test, either both a n and b n converge or they both diverge. But b n is a geometric series with ratio = 3 which converges. Therefore the series converges. Page 3 of 3

4 MTH 33 Solutions to Exam November th, For each of the following functions, find its 3rd degree Taylor polynomial centered at the given a. (a) (7 points) f(x) = sin(5x), centered at a = 0. We know that sin(t) = t t3 + HOT (Higher Order Terms, i.e. terms with powers of the variable 6 t higher than 3) is the Taylor series at t = 0 (i.e. MacLaurin series) of sin(t). So, substituting t = 5x, we get sin(5x) = 5x 53 x 3 + HOT. 6 Since this is a power series centered at x = 0 which equals the function sin(5x) for all x in some open interval containing x = 0, it has to be the Taylor series of that function at x = 0. (Recall the derivation of the Taylor series formula.) So the 3rd degree Taylor polynomial of the function sin(5x) is T 3 (x) = 5x 53 x 3 6. (b) (7 points) g(x) = ln(x), centered at a =. g(x) = ln(x) g() = ln() g (x) = x g () = g (x) = x g () = g (x) = x 3 g () = So the 3rd degree Taylor polynomial of the function g(x) = ln(x) centered at a = is T 3 (x) = g() + g ()! (x ) + g ()! (x ) + g () (x ) 3 3! = ln() +! (x )! (x ) + (x )3 3! = ln() + (x ) 8 (x ) + (x )3. I.e. the solution is T 3 (x) = ln() + (x ) 8 (x ) + (x )3. Page of 3

5 MTH 33 Solutions to Exam November th, 08. (7 points) Find the Maclaurin series (Taylor series centered at a = 0) representation of f(x) = 3x + x. Express your answer in sigma notation. 3x + x = a r with a = 3x and r = x. So f(x) can be written as the geometric series. ar n = 3x ( x ) n = ( ) n 3x n+. (7x ) n 5. (7 points) Find the interval of convergence for the power series. (Leave your answer as (n + 3) n=3 an open interval; you do not have to test the end points for convergence.) Since n=3 (7x ) n (n + 3) = ( 7 n (n + 3) x n, the center of the interval of convergence is 7) n=3 the center of the power series, i.e. x = 7. To find the radius of convergence, we apply the ratio test with a n = a n+ ( a n = 7n+ (n + ) x n+ 7) ( ) n + 3 = 7 n + x = n + 7 x 7 n (n + 3) 7 n ( 7 n (n + 3) x n. 7) ( x n 7) a n+ So L = lim a n = 7 x 7. For the series to converge by the ratio test, we want L <, i.e. 7 x 7 <, which holds if and only if x 7 < ( 7, i.e. if x is in 7 7, 7 + ). 7 ( In other words, the interval of convergence (ignoring the convergence at the endpoints) is 7, 3 ). 7 Page 5 of 3

6 MTH 33 Solutions to Exam November th, 08 Multiple Choice. Circle the best answer. No work needed. No partial credit available. 6. ( points) Find the limit of the sequence a n where the n th term is given by a n = A. 0 3n + cos(3n). n B. 3 C. 3 D. E. The sequence does not have a limit. 7. ( points) Which statement about the series n= ln(3n) n is true? A. It diverges by using a comparison test with B. It converges by using a comparison test with n. C. It diverges by using a comparison test with n. D. It converges by using a comparison test with E. It diverges by the ratio test. n. n. 8. ( points) Which statement is true about the series n= n(ln(n)) p? A. The integral test shows that the series converges for all p. B. The integral test shows that the series diverges for p. C. The integral test hypotheses are not met by this series, so it cannot be applied. D. The integral test hypotheses are met by this series, however the test is inconclusive. E. None of the above are true. Page 6 of 3

7 MTH 33 Solutions to Exam November th, ( points) Find the radius of convergence of A. / B. /3 C. 3 D. E. + (x + 3) n+5. (n )! 0. ( points) Determine whether the following series are absolutely convergent, conditionally convergent, or divergent: () n=3 n ln(n) and () ( ) n n A. () is absolutely convergent; () is divergent. B. () is conditionally convergent; () is divergent. C. () is absolutely convergent; () is conditionally convergent. D. () is divergent; () is conditionally convergent. E. Both () and () are conditionally convergent.. ( points) Which statement is true about the series A. By the ratio test, the series converges. B. By the ratio test, the series diverges. C. The ratio test is inconclusive for this series, but the series converges by another test. n? D. The ratio test is inconclusive for this series, but the series diverges by another test. E. None of the above are true. Page 7 of 3

8 MTH 33 Solutions to Exam November th, 08. ( points) By Taylor s Inequality, on the interval [, ], the difference between e x and its degree Taylor polynomial centered at x = 0 is at most: A. B. x. x 3 6. C. e x 3 6. D. E. x. e 3 x. 3. ( points) The Taylor series of the function f(x), centered at a =, is given by What is the value of the third derivative f ()? n + 5 (x ) n. n! A. 5 B. 9/ C. 9 D. /6 E.. ( points) The Taylor series up to order, centered at a = 0, for f(x) = ln( + x) x sin(x) is A. + x 3x + x3 x B. x x + x3 3 x C. x x + x3 6 x D. x 3x + x3 3 x E. x 3x + x3 3 + x! Page 8 of 3

9 MTH 33 Solutions to Exam November th, 08 More Challenging Questions. Show all work to receive credit. Please BOX your final answer. 5. (a) ( points) A ball falls from a height of m and continues bouncing forever. Each time it hits the floor, it bounces back to 3 of the previous height. Write a (numerical) series that represents the total vertical distance traveled by the ball (ignore any horizontal displacement). (Hint: don t forget to include the distance traveled before the first bounce.) m The vertical distance travelled before the first bounce is. Then the first bounce reaches a height of 3, the second bounce reaches a height of 3 ( 3 ) times the previous bounce, i.e., the third ( ) 3 ( ) 3 3 bounce reaches a height of, the fourth bounce reaches a height of, and so on, so that ( ) n 3 the n-th bounce reaches a height of. For each bounce, the ball goes up and down, i.e. it travels twice the height of the bounce. ( ) ( ) ( ) ( ) n 3 So the total vertical distance = = +. (b) ( points) Compute the sum of the series you found in part (a). ( ) n 3 ( ) n 3 3 = = ar n = a r with a = 3 and r = 3, so + ( ) n 3 = = (6 points) By the ratio test, the radius of convergence of the power series If b n = (a n ), does the power series a n x n is found to be R =. b n x n converge at x = 3? Justify your reasoning. The ratio test requires checking L = lim a n+ x n+ a n x n = lim a n+ a n x. If the radius of convergence is R = and it has been found by the ratio test, then this means that L < precisely for Page 9 of 3

10 MTH 33 Solutions to Exam November th, 08 x <, i.e. that L = lim a n+ a n =. Then M = lim b n+ b n = lim a n+ a n whenever M = lim b n+ x n+ b n x n = lim =, so, again by the ratio test, the series b n x n converges b n+ b n x <, i.e. it converges at least in the interval (, ). (Note: The wording at least is because we are not checking the endpoints of the interval (, ), where the series might converge as well.) Thus the series b n x n converges at x = 3. Page 0 of 3

11 MTH 33 Solutions to Exam November th, 08 DO NOT WRITE BELOW THIS LINE. Page Points Score Total: 06 No more than 00 points may be earned on the exam. Page of 3

12 MTH 33 Solutions to Exam November th, 08 Integrals FORMULA SHEET PAGE Derivatives Volume: Suppose A(x) is the cross-sectional area of the solid S perpendicular to the x-axis, then the volume of S is given by d d (sinh x) = cosh x dx Inverse Trigonometric Functions: (cosh x) = sinh x dx V = b a A(x) dx d dx (sin x) = x d dx (csc x) = x x Work: Suppose f(x) is a force function. The work in moving an object from a to b is given by: W = b dx = ln x + C x tan x dx = ln sec x + C a f(x) dx sec x dx = ln sec x + tan x + C a x dx = ax ln a + C for a Integration by Parts: u dv = uv Arc Length Formula: v du d dx (cos x) = x d dx (tan x) = d dx (sec x) = d + x dx (cot x) = x x + x If f is a one-to-one differentiable function with inverse function f and f (f (a)) 0, then the inverse function is differentiable at a and (f ) (a) = f (f (a)) Hyperbolic and Trig Identities Hyperbolic Functions sinh(x) = ex e x cosh(x) = ex + e x csch(x) = sinh x sech(x) = cosh x L = b a + [f (x)] dx tanh(x) = sinh x cosh x cosh x sinh x = sin x = ( cos x) cos x = ( + cos x) sin(x) = sin x cos x coth(x) = cosh x sinh x sin A cos B = [sin(a B) + sin(a + B)] sin A sin B = [cos(a B) cos(a + B)] cos A cos B = [cos(a B) + cos(a + B)] Page of 3

13 MTH 33 Solutions to Exam November th, 08 FORMULA SHEET PAGE Series nth term test for divergence: If lim a n does not exist or if lim a n 0, then the series a n is divergent. The p-series: and divergent if p. Geometric: If r < then is convergent if p > np ar n = a r The Integral Test: Suppose f is a continuous, positive, decreasing function on [, ) and let a n = f(n). Then (i) If (ii) If then then f(x) dx is convergent, a n is convergent. f(x) dx is divergent, a n is divergent. The Comparison Test: Suppose that a n and b n are series with positive terms. (i) If b n is convergent and a n b n for all n, then a n is also convergent. (ii) If b n is divergent and a n b n for all n, then a n is also divergent. The Limit Comparison Test: Suppose that an and b n are series with positive terms. If a n lim = c b n where c is a finite number and c > 0, then either both series converge or both diverge. Alternating Series Test: If the alternating series ( ) n b n satisfies (i) 0 < b n+ b n (ii) lim b n = 0 for all n then the series is convergent. The Ratio Test (i) If lim a n+ a n = L <, then the series a n is absolutely convergent. (ii) If lim a n+ a n = L > or lim a n+ a n =, then the series a n is divergent. (iii) If lim a n+ a n =, the Ratio Test is inconclusive. f (n) (0) Maclaurin Series: f(x) = n! x n Taylor s Inequality If f (n+) (x) M for x a d, then the remainder R n (x) of the Taylor series satisfies the inequality R n (x) M (n + )! x a n+ for x a d Some Power Series e x x n = n! sin x = cos x = ( ) n xn+ (n + )! ln( + x) = ( ) n xn (n)! ( ) n xn n R = R = R = R = x = x n R = Page 3 of 3

Without fully opening the exam, check that you have pages 1 through 12.

MTH 33 Exam 2 November 4th, 208 Name: Section: Recitation Instructor: INSTRUCTIONS Fill in your name, etc. on this first page. Without fully opening the exam, check that you have pages through 2. Show