Solutions to Math 1b Midterm II

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1 Solutions to Math b Midterm II Tuesday, pril 8, 006. (6 points) Suppose that the power series a n(x + ) n converges if x = 7 and diverges if x = 7. ecide which of the following series must converge, must diverge, or may either converge or diverge (inconclusive). ircle your answer. You do not need to justify your answers. (a) If x = 8, the power series onverges iverges Inconclusive (b) If x =, the power series onverges iverges Inconclusive (c) If x = 3, the power series onverges iverges Inconclusive (d) If x =, the power series onverges iverges Inconclusive (e) If x = 5, the power series onverges iverges Inconclusive (f) If x = 5, the power series onverges iverges Inconclusive (a) Inconclusive, (b) onverges, (c) Inconclusive, (d) Inconclusive, (e) Inconclusive, (f) onverges.. (0 points) etermine whether each series below converges absolutely, converges conditionally, or diverges. e careful to justify each of your answers by explicitly referring to the test that you used and explaining how you used it. (a) (b) (c) (d) n 4 + n + ( + n 3 n n= sin(/n) n 3 n n!(ln n) 3 )

2 (e) n= ( ) n n ln n (a) Since n 4 + n + < n, 4 and the series on the right-hand side of the inequality is a convergent p-series, the series converges by the omparison Test. Since all of the terms are positive, the series converges absolutely. (b) Since ( + ) < n 3 n ( + ) = n n ( ) = n and the series on the right-hand side of the inequality is a convergent geometric series, the series converges by the omparison Test. Since all of the terms are positive, the series converges absolutely. (c) Since sin(/n) n 3 < n, 3 6 n and the series on the right-hand side of the inequality is a convergent p-series, the series converges absolutely by the omparison Test. (d) Use the Ratio Test. Since a n+ a n = n+ n!(ln n)3 = (n + )!(ln(n + )) 3 n n we can determine that lim a n+ /a n = 0 <. absolutely. ( ) 3 ln n, ln(n + ) Thus, the series converges (e) Since the terms, a n = ( ) n /(n ln n) are decreasing and lim a n = 0, the series converges by the lternating Series Test. To determine the convergence of n ln n we use the Integral Test, x ln x dx = n= ln u du = ln u ln =. Since the integral diverges, the series diverges. Thus, the series is conditionally convergent.

3 3. (0 points) Let f be a function having derivatives of all orders for all real numbers. The third-degree Taylor polynomial for f about x = is given by (a) ind f( ), f ( ), and f ( ). T 3 (x) = 3 8 (x + ) (x + )3. (b) etermine whether f has a local minimum, a local maximum, or neither at x =. Justify your answer. (c) Use T 3 (x) to find an approximation for f(0). (d) The fourth derivative of f satisfies the inequality f (4) (x) 4 for all x in the closed interval [, 0]. ind an error bound on the approximation for f(0) that you found in part (c). (a) f( ) =, f ( ) = 0, and f ( ) = 3/4 (since f ( )/! = 3/8). (b) Since f ( ) = 0, and f ( ) = 3/4, f has a local a local maximum at x = by the Second erivative Test. (c) T 3 (0) = /6 (d) f(0) T 3 (0) M 4! = (/4) 4 4 = 6 4. (6 points) Suppose we know that a n converges to 0.8. We are given no other information about the infinite series. or each of the following statements circle True if the statement must be true, alse if the statement must be false, and Inconclusive if the statement could be either true or false. You do not need to justify your answers. (a) lim a n = 0.8 True alse Inconclusive (b) lim a n = 0 True alse Inconclusive (c) lim a n+ a n > True alse Inconclusive (d) a n+ < a n for all n True alse Inconclusive 3

4 (e) lim a n = True alse Inconclusive (f) lim S n = 0.8, True alse Inconclusive where S n = a + a + + a n. (a) alse, (b) True, (c) alse, (d) Inconclusive, (e) True, (f) True. 5. (6 points) The graph of y = f(x) is given below. ssume that f is infinitely differentiable everywhere. The Taylor series for f(x) about x = 0 is given by f(x) = c n x n = c 0 + c x + c x + c 3 x 3 +. exam0_graphs.nb y x (a) etermine whether c 0 is positive, negative, or zero. xplain your reasoning. (b) etermine whether c is positive, negative, or zero. xplain your reasoning. (c) etermine whether c is positive, negative, or zero. xplain your reasoning. (a) Negative, c 0 = f(0) < 0 (b) Positive, c = f (0) > 0 (c) Positive, c = f (0)/! > 0 since the function is concave up at x = (4 points) onsider the following power series. (a) x x6 3! + x0 5! 4

5 (b) x4! + x8 4! (c) x! x4 4! + x6 6! (d) x x 3 + x 5 (e) x x4! + x6 4! ill in the letter of the series that corresponds to the given function. You do not need to justify your answer. cos x x cos x sin(x ) x + x cos x is (c); x cos x is (e); sin(x ) is (a); x is (d). + x 7. (0 points) ind the interval of convergence for the power series (x + 5) n. n 3 n If the interval of convergence is finite, make sure that you determine the convergence at each endpoint and justify your conclusions. Using the Ratio Test, a n+ a n = (x + 5) n+ (n + ) 3 n 3 n n+ (x + 5) n = n x + 5, 3(n + ) we can determine that lim a n+ /a n = x + 5 /3. This limit is less than one on the interval x + 5 < 3 or 8 < x <, and the series converges on this interval. We must consider the endpoints separately. t x =, the series 3 n n 3 = n n is a convergent p-series. t x = 8, the series ( 3) n n 3 = ( ) n n n converges by the lternating Series Test. Thus, the power series converges on the interval 8 x. 5

6 8. (0 points) (a) Write down a power series expansion for e x3. (b) Write down a power series expansion for e x3 dx and determine its radius of convergence. (c) Use your answer in part (b) to find a series for / 0 e x3 dx. (d) If you approximate the definite integral / 0 e x3 dx by taking the partial sum consisting of the first four nonzero terms of the series that you obtained in part (c), what is the maximum error for your approximation. (a) Since the power series expansion for e x3 e x = + x + x! + x3 3! + x4 4! + = e x3 = x 3 + x6! x9 3! + x = 4! (b) To find a power series expansion for obtained in (a), is x n n!, ( ) n x 3n. n! e x3 dx, we integrate the power series e x3 dx = c 0 + x x4 4 + x7 7! x0 0 3! + x3 3 4! = c 0 + ( ) n x 3n+ (3n + ) n! where c 0 is an arbitrary constant. The radius of convergence does not change R =. / (c) e x3 dx = ! 0 0 3! ! 0 (d) Since this is an alternating series, the maximum error is bounded by the fifth term of the series, 3 3 4! 9. (8 points) ssume that the square below has sides of length one and that,,, and are the midpoints of the sides. 6

7 exam0_graphs.nb exam0_graphs.nb ê ê exam0_graphs.nb exam0_graphs.nb (a) If the indicated pattern is continued indefinitely, write an infinite series that will give the area of the shaded region? (b) What is the area of the shaded region if the indicated pattern is continued indefinitely? That is, what is the sum of the series that you found in part (a)? (a) = (b) The series in (a) is geometric. Thus, = 7 3 (/) = 4. 7