Math 2, Fall 27 Schaeffer/Kemeny Final Exam (December th, 27) Last/Family Name First/Given Name Seat # Failure to follow the instructions below will constitute a breach of the Stanford Honor Code: You may not write anything on this page other than your name, seat #, and signature. You may not use a calculator or any notes or book during the exam.* You may not access your phone or any other electronics during the exam for any reason. * You must sit in your assigned seat during the exam.* You may not communicate with anyone other than the course staff during the exam, or look at anyone else s solutions during the exam.* Additionally, you may not discuss the contents of this exam with ANYONE other than the course staff until : PM tonight. You have 8 minutes to complete this exam. If the course staff must ask you to stop writing or to turn in your exam more than once after time is called, you may receive a score of zero. *During the exam is defined as before you have handed in your exam and left the testing site. I understand and accept these instructions. Signature: Remember to show your work and justify your answer if required. Present all solutions in as organized a manner as possible. The last page of the exam is blank, and can be used for extra work. If you think it would help for us to look at this work, you should indicate that CLEARLY on the problem s page. Do not detach the last page of the exam. GOOD LUCK!
Here are some tips: If you have time, it s always a good idea to check your work when possible. If you get the wrong answer but show your work, you have a better chance of receiving partial credit. DO NOT attempt to estimate any of your answers as decimals. For example, than.682, because it is exact. is a much better answer The last page of the exam are blank, and can be used for extra work. If you think it would help for us to look at this work, you should indicate that CLEARLY on the problem s page. Integral table entries you may need Z du u 2 a 2 = 2a ln u a u + a + C Z du (u r)(u s) = r s ln u r u s + C Z du u 2 + a 2 = u a arctan + C a Z du p a 2 u 2 = arcsin u a In the entries above, a, r, s are constants such that a 6= and r 6= s. Values of arcsine and arctangent + C The table below gives important values of the arcsine and arctangent functions: x 2 p p p2 3 3 2 p 3 arcsin x /6 /4 /3 /2 undef. undef. arctan x /6 /4 /3 /2 For negative values: arcsin( x) = arcsin(x) and arctan( x) = arctan(x). Blank entries in the table are not nice multiples of. Bounds from the integral and alternating series tests X If f(n) converges by the integral test, then If n=a Z a f(x) dx apple X f(n) apple f(a)+ n=a X ( ) n a n converges by the alternating series test, then Z a f(x) dx error in the Nth partial sum apple the (N + )th term in the series 2
Section Improper Integrals, Sequences, Geometric Series. Fill in the blanks below with the appropriate asymptotic relation:,, or (as x or n!). a. e x2 (e x ) 2 b. 3 x (/2) x c. n! n d. x p x 2 + x 2 + e. x 2 e x < 3 x f. x 2 + e x p x + e x g. x 2 + e x p x + e x h. x 2 + px e x + e x i. x. ln x j. n! n n.2 a. For which values of p does All p>. b. For which values of p does All p<. Z Z dx converge? Describe all such values. xp dx converge? Describe all such values. xp 3
Z e /x.3 Partario is attempting to evaluate the integral x 2 dx, illustrated below: By substituting u = Z e /x x, he correctly finds x 2 dx = e/x + C. Because the integrand is not defined at x =, Partario continues as follows: Z e /x x 2 dx =lim b! Z b e /x x 2 h dx =lim b! e /xi b h = e lim e /bi = e e = b! So the integral diverges! Is Partario correct or incorrect? If you believe he is correct, write PARTARIO IS CORRECT. Otherwise, explain what went wrong in Partario s solution and write down the correct answer. Partario is incorrect. The integral is the limit of R b as b! (that is, as b approaches from the left). This makes a difference because h e /bi = e lim b! (/b) = e = lim b! while the limit from the other direction is h lim e /bi = e lim b! + (/b) = e = b! + This is because the asymptote of x heads to when approaching x from the left, but it heads to + when approaching from the right. In any case, replacing Partario s limits with lim, we arrive at the answer b! e e =/e. 4
In problems.4.6, evaluate the improper integral, showing all work. If you believe the integral diverges, write DIVERGES. Some integration formulas and values of inverse trigonometric functions are available for your use on pg 2. Draw boxes around your final answers..4.5.6 Z dx 4x 2 + Break this into simple integrals R + R. By the integration formula on page 2, we have (with u =2x, du =2dx, and a =) Z dx 4x 2 + = Z du 2 u 2 + a 2 = 2 arctan(2x)+c The integral from to infinity is therefore 2 [arctan() we have R = R. Thus, the final answer is /2. Z 4 4 x 3 dx arctan()] = /4, and since the function is even, We have to break this up as R 4 + R 4. The integral R 4 dx diverges by p-test, so the integral DIVERGES. x 3 Z x 2 e x3 dx By substitution u = x 3 and du = 3x 2 dx, we have Z x 2 e x3 dx = e x3 3 + C The limit of e x3 3 as x!is, so the answer is e = 3 3e.7 Does the integral Z It converges (to.8 Does the integral ln xdxconverge or diverge? You do not need to justify your answer. ). This was Problem D (parts a and b) on HW2. Z ln(x + ) dx converge or diverge? Carefully justify your answer. x 7/6 Hint: If 2 apple p apple, then apple ln(x + ) apple xp for all real numbers x. Break it up as R + R. On [, ] we use apple ln(x + ) apple x to find apple ln(x + ) x 7/6 apple x x 7/6 = x /6 5
so the integral R converges by comparison with R x /6 dx. For the integral on [, ), we use the fact that ln(x + ) x q for any power q>. Then, for example, ln(x + ) x /2, which means that for some a and some positive constant c, we have ln(x + ) apple cx /2 whenever x a. Thus for x a, ln(x + ) x 7/6 apple x/2 x 7/6 = x 3/2 and the integral of x 3/2 from a to infinity converges by p-test. This completes the argument that the integral converges..9 a. Evaluate the sum 6 3 +6 4 +6 5 + +6. You may leave your answer unsimplified, but it may not contain more than one + or sign, and it may not contain P. 98 =6 3 ( + 6 + 6 2 + +6 97 )=6 3 6 97 5. b. Evaluate the geometric series 4 4 2 + 4 3 +. (Note: The series is alternating.) 44 = 4 X ( /4) n = 4 + 4 = 4 4 5 =/5 c. Express the repeating decimal.222 as a fraction of whole numbers in lowest terms. = 2 + 2 2 + = 2 = 2 999 = 7 333. Section 2 Sequences, Series, Power Series, and Taylor Series 2. a,b. Write down an example of a sequence that is bounded but does not converge, or state that no such sequence exists. ( ) n c. Write down an example of a sequence that is bounded and monotone but does not converge, or state that no such sequence exists. No such example exists: All bounded monotone sequences converge. 6
X 2.2 Suppose that the series a n converges conditionally (and not absolutely). Which of the following statements must be true? Circle all true statements. Here highlighted in blue. a n+ i. If the limit L = lim exists, then L =. n! a n ii. The series P a n is alternating: Consecutive nonzero terms always have opposite signs. iii. The sequence {a n } contains both infinitely many positive terms as well as infinitely many negative terms. iv. The limit of the sequence {a n } is. v. The limit of the sequence { a n } is. /n 2 vi. The limit lim, provided it exists and is finite, is equal to zero. n! a n NX vii. The limit lim a n exists and is finite. N! viii. None of the above. 2.3 Determine whether each series (a e) below converges or diverges. If the series converges (either absolutely or conditionally), circle it. Here highlighed in blue. If the series diverges, draw an X through it. Below each series, fill in the blank with the roman numeral I VII corresponding to the reason you chose convergence/divergence. a. X n= sin n n 4 b. X n= sin(n ) n c. X n= cos(n ) n d. X + (sin n) 2 n= n e. X n= + (sin n) 2 V. IV. or V. or VI. IV. II. III. Under each of the five series, make sure to include one of the following reasons: I. The series converges by comparison with the harmonic series. II. The series diverges by comparison with the harmonic series. III. The series diverges by the divergence test (terms do not tend to zero). IV. The series converges by the alternating series test. P V. an converges, so P a n converges too (absolute convergence guarantees convergence). VI. The terms of the series are all zero, so the series converges. VII. The series converges by the integral test. VIII. None of the above. 7
2.4 Below are ten infinite series (a j). If the series converges (either absolutely or conditionally), circle it. Here highlighted in blue. If the series diverges, draw an X through it. a. X 3n + b. X sin(n) c. X n 2 p 4n 9 + d. X n=2 n ln n e. X ( ) n n. f. X n! 3 (3n)! g. X 2 n n 2 3 n + h. X n= +2 ( ) n n 3 i. X n= ln n n 4 j. X n! n You do not need to show your work. SPACE BELOW FOR SCRATCH WORK (MORE AT END OF EXAM) 8
X 2.5 Let F (x) be the power series Show all work and draw a box around your final answer. The center is at x =. The ratio test yields ( ) n (x + ) 3n+ 8 n. What is F (x) s interval of convergence? (3n + ) L(x) = lim n! ( )(x + ) 3 (3n + 4) 8 (3n + ) = x + 3 8 Solving for L(x) =, we find the endpoints of the IoC: 3 and. Plugging in x = 3 yields the series F ( 3) = X which diverges. At the other endpoint, ( ) n ( 2) 3n+ 8 n (3n + ) = X ( ) n ( 8) n ( 2) 8 n (3n + ) = 2 X 3n + F () = X ( ) n (2) 3n+ 8 n (3n + ) = X ( ) n (8) n (2) 8 n (3n + ) =2 X ( ) n 3n + which converges conditionally. The IoC is therefore ( 3, ]. 2.6 In (a d) write down the Taylor series for the given function at x =in P form. a. cos x X ( ) n x 2n (2n)! b. ln( + x) X ( ) n x n+ n + c. x arctan x X ( ) n+ x 2n+ d. n= 2n + x +x 3 X ( ) n x 3n+ 9
2.7 Below are four equations involving infinite series, each with one unknown quantity: X X n n! =3 X ( ) n Y 2n (2n)! =2 X p 3 ( ) n 3 n (2n + ) = Z X W n = In the spaces below, write down values for X, Y, Z, and W that make each of the equations above true, or write DNE if no such value exists. Your answers may not contain any unresolved inverse trig functions. A table of inverse trig values is available for your use on pg 2. a. X =ln3 b. Y = DNE c. Z = arctan(/ p 3) = /6 d. W = DNE
Section 3 Applications of Taylor Series 3. Quintana is studying a mystery function G(t), which she knows to be smooth (infinitely differentiable) at t =. She is attempting to approximate the value of G(.4) using the quadratic approximation (2nd degree Taylor polynomial), P 2 (x) at x =. To make sure her estimate is not too far off from the actual value, she uses the Lagrange error bounds, which take the form G(.4) P 2 (.4) apple M?? a. Help Quintana out by complete the Lagrange error bounds she would use in this case, (that is, rewrite the above, filling in the?s). You may leave M as is. The right hand side should be M 3! (.4)3. b. Explain what M is in the Lagrange error bounds, as written above. A number larger than G (3) (the magnitude of G s third derivative) on the interval between and.4. 3.2 The Taylor series for the tangent function at x =has a rather complicated closed form, which is why we never had you learn it (unlike those for sine and cosine). The first few terms of the series are tan x = x + x3 3 + 2x5 5 + 7x7 35 + and the series converges on ( 2, 2 ). Using this information, or whatever method you prefer, answer the questions below. a. Using the 3rd degree Taylor polynomial of tan x at x =, write down an estimate for Z tan xdx. You do not need to simplify your answer, which should be a sum of fractions. 2 + 2 b. If you instead use the 4th degree Taylor polynomial of tan x at x =to estimate the integral, will your estimate be better, worse, or the same? The same. c. Let f(x) = tan x sin x. What is the value of f (5) ()? The coefficient of x 5 in f(x) s Taylor series at x =is 2 answer is 5!/8, which is 5. d. Evaluate the limit lim x! apple tan x x 3 sin x 5. 2 =/8. This is equal to f () /5!, so the The first nonzero term in the Taylor series for tan x sin x is the term containing x 3, which is x3 3 x ( 3 x3 6 )= 2. Thus, replacing the numerator with x3 /2, we see that the limit is /2.
3.3 Consider the definite (proper) integral Z sin(x 2 ) dx. a. Express the value of the integral above as an infinite series (using P ). X ( ) n x 4n+3 (4n + 3) (2n + )! b. The series you obtained in the previous part should be alternating (if it s not, you should check your work!). Using the bounds from the alternating series test (on pg 2), write down a finite sum of fractions that is guaranteed to estimate the value of the integral to within 3 =. Your answer may not contain P or. Two terms is enough. 3.4 Suppose y = f(x) is a solution to the differential equation y +4xy + y = satisfying f() = and f () = can find its Taylor series there.. Assume that f(x) is smooth (infinitely differentiable) at x =, so we a. Writing f(x) =c + c x + c 2 x 2 + for the Taylor series of f(x) at x =, what is the Taylor series for f (x) at x =? Express your answer as an infinite (power) series. X The derivative is c +2c 2 x +3c 3 x 2 + = (n + )c n+ x n. b. What is the Taylor series for f (x) at x =? You may express your answer either as an infinite (power) series, or as (the first three nonzero terms)+. X The second derivative is 2c 2 +6c 3 x + 2c 4 x 2 + = (n + 2)(n + )c n+2 x n. c. What are c and c equal to? c =and c =. d. Find P 4 (x), the 4th degree Taylor polynomial for f(x) at x =. x 2 x2 + 5 6 x3 + 3 8 x4 2