Math 12 Midterm Exam #3 2011

Math 12 Midterm Exam #3 April 27, 2011 Answer Key 1. [15 Points] Find the Interval and Radius of Convergence for each of the following power series. Analyze carefully and with full justification. ∞ X (2n)! x2n+1 (a) (3n)! n=0 Use Ratio Test. lim n→∞ (2(n + 1))! x2(n+1)+1 (3n)! x2n+3 (2n + 2)! (3(n + 1))! · = lim 2n+1 · 2n+1 n→∞ x (2n)! (3n + 3)! (2n)! x (3n)! = lim x2 · n→∞ (3n)! (2n + 2)(2n + 1)(2n)! · (2n)! (3n + 3)(3n + 2)(3n + 1)(3n)! (2n + 2)(2n + 1)|x|2 = 0 < 1 for all x. n→∞ (3n + 3)(3n + 2)(3n + 1) = lim The Ratio Test gives convergence for all x. So Interval of Convergence I = (−∞, ∞) with Radius of Convergence R = ∞ . (b) ∞ X (3x + 5)n n=0 n2 7n Use Ratio Test. (3x + 5)n+1 (3x + 5)n+1 n2 7n |3x + 5| (n + 1)2 7n+1 = lim lim · · = n n 2 n+1 n→∞ n→∞ (3x + 5) (3x + 5) (n + 1) 7 7 n2 7n |3x + 5| < 1 or |3x + 5| < 7. 7 2 That is −7 < 3x + 5 < 7 =⇒ −12 < 3x < 2 =⇒ −4 < x < 3 Endpoints: The Ratio Test gives convergence for x when ∞ •x = ∞ X 7n X 1 2 The original series becomes = which is a convergent p-series with p = 2 > 1. 2 n 3 n 7 n2 n=0 ∞ X n=0 ∞ (−7)n X ∞ X (−1)n (−1)n 7n = which is convergent by n2 7n n2 7n n2 n=0 n=0 n=0 Absolute Convergence Test since the absolute series above converges. (can also use AST here, but not needed if you use ACT) •x = −4 The original series becomes 1   2 7 Finally, Interval of Convergence I = −4, with Radius of Convergence R = . 3 3 2. [10 Points] Consider the function f (x) that satisfies the following f (4) = 2 f ′ (4) = −3 f ′′ (4) = 6 7 f ′′′ (4) = −1 Find the Taylor polynomial of degree 3 for T3 (x) = f (4) + f ′ (4)(x − 2) + = 2 − 3(x − 4) + 6 7 2! f (x) centered at a = 4. f ′′′ (4) f ′′ (4) (x − 4)2 + (x − 4)3 2! 3! (x − 4)2 − 1 (x − 4)3 3! 3 1 = 2 − 3(x − 4) + (x − 4)2 − (x − 4)3 7 6 3. [10 Points] Find the MacLaurin series representation for each of the following functions. ∞ X State the Radius of Convergence for each series. Your answer should be in sigma notation n=0 (a) f (x) = x2 e−3x First ex = ∞ X xn n=0 Then e−3x = n! ∞ X (−3x)n n=0 Finally, x2 e−3x = n! x2 = ∞ X (−1)n 3n xn n=0 ∞ X (−1)n 3n xn n=0 n! n! = ∞ X (−1)n 3n xn+2 n=0 n! 2 (b) f (x) = x arctan(3x) First arctan x = = ∞ X (−1)n n=0 R 1 dx = 1 + x2 n=0 x2n+1 . Here R = 1 or |x| < 1. 2n + 1 Next, arctan(3x) = ∞ X 2n+1 n (3x) (−1) Finally, x arctan(3x) = x ∞ X (−1)n n=0 or R = 31 . = 2n + 1 n=0 4. Z X ∞ 1 dx = 1 − (−x2 ) Z ∞ X (−1)n n=0
2 n −x dx = Z X ∞ (−1)n x2n dx n=0 32n+1 x2n+1 2n + 1 ∞ X 32n+1 x2n+1 32n+1 x2n+2 (−1)n = . Here |3x| < 1 so |x| < 13 , 2n + 1 2n + 1 n=0 Use a Power Series representation for x ln(1 + x3 ) to estimate the given integral 1 within the given error. Justify in words that your error is indeed less than . 10 [10 Points] Estimate Z 1 x ln(1 + x3 ) dx 0 X X 1 1 (−1)n xn . (−x)n = = = 1+x 1 − (−x) n=0 n=0 Then, ln(1 + x) = Next, ln(1 + x3 ) = R ∞ X Finally, x ln(1 + x3 ) (−1)n xn = 1 ∞ =x n=0 ∞ X x ln(1 + x3 ) dx = 0 = (−1)n xn+1 . n+1 (−1)n (x3 )n+1 X (−1)n x3n+3 = . n+1 n+1 n=0 Z ∞ X n=0 n=0 ∞ X n=0 Now, 1 10 ∞ ∞ First with error less than x8 x11 x5 − + + ... 5 8 · 2 11 · 3 Z 1 ∞ (−1)n x3n+3 X (−1)n x3n+4 = n+1 n+1 n=0 ∞ 1X 0 n=0 = 0  (−1)n x3n+4 n+1 dx = ∞ X n=0 (−1)n x3n+5 (3n + 5)(n + 1) 1 0  1 1 1 1 − + + . . . − (0 − 0 + 0 + . . .) ≈ 5 16 33 5 Note this is an alternating series. Use the Alternating Series Estimation Theorem. If we approximate the actual sum with only the first term, the error from the actual sum will be at most the 1 1 1 absolute value of the next term, . Here < as desired. 16 16 10 3 5. (a) [15 Points] Find the sum for each of the following series. ∞ X (−1)n 7n
n ∞ X − 79 7 = = e− 9 n! 9n n! n=0 (b) n=0 ∞ X (−1)n π 2n n=0 62n (2n)! = ∞ X (−1)n n=0
π 2n 6 (2n)! = cos π  6 = √ 3 2
2n ∞ ∞ ∞ ∞ X X X X (−1)n π3 (−1)n π 2n (−1)n π 2n (−1)n π 2n (c) = = = 9n (2n + 1)! (32 )n (2n + 1)! 32n (2n + 1)! (2n + 1)! n=0 n=0 n=0 n=0
2n+1   √  X ∞  π   3  √3 (−1)n π3 (−1)n 3 3 3 3 3
sin = = = = = π π (2n + 1)! π 3 π 2 2π 3 (2n + 1)! n=0 n=0
π 2n+1 ∞ X 6. [20 Points] Volumes of Revolution π . Rotate this 2 region about the horizontal line y = −3 . Set-up, BUT DO NOT EVALUATE!!, the integral to compute the volume of the resulting solid using the Washer Method. Sketch the solid, along with one of the approximating washers. (a) Consider the region bounded by y = cos x, y = ex + 2, x = 0, and x = See me for a sketch. Z Z π
2 V = π (outer radius)2 − (inner radius)2 ) dx = π 0 π 2 0 (ex + 5)2 − (cos x + 3)2 dx (b) Consider the region bounded by y = ex , y = ln x, x = 1 and x = 2. Rotate the region about the y-axis. COMPUTE the volume of the resulting solid using the Cylindrical Shells Method. Sketch the solid, along with one of the approximating shells. See me for a sketch. Z Z 2 2π radius height dx = V = 1 1 for both pieces 2 = 2π xex − 2 ex 1 −  x2 x2 ln x − 2 4 2πx(ex − ln x) dx = 2π  2! 1 = 2π  2e2 − e2 Z − 2 1 (e1 xex dx − − e1 )
− Z  2 1  x ln x dx I.B.P. here 4 4 ln 2 − − 2 4  1 1 ln 1 − 2 4     1 3 2 2 = 2π e − ln 4 + = 2π e − 2 ln 2 + 1 − 4 4 7. [20 Points] Consider the Parametric Curve represented by x = 3 − 2t and y = et + e−t . 4  (a) Compute the arclength of this parametric curve for 0 ≤ t ≤ 1. s Z 1p Z 1q Z 1  2  2 dy dx 2 2 t −t (−2) + (e − e ) dt = 4 + e2t − 2 + e−2t dt + dt = L= dt dt 0 0 0 Z 1q Z 1p e2t + 2 + e−2t dt = (et + e−t )2 dt = 0 = Z 0 1 1 et + e−t dt = et − e−t 0 0

1 = e − e−1 − e0 − e0 = e − e−1 = e − e (b) Compute the surface area obtained by rotating this curve about the x-axis, for 0 ≤ t ≤ 1. s   2 Z 1 Z 1 q dy dx 2 2πy(t) (−2)2 + (et − e−t )2 dt 2πy(t) + dt = S.A. = dt dt 0 0 Z 1 Z 1 q


2π et + e−t (et + e−t dt from part(b) above. (et + e−t )2 dt = 2π et + e−t 0 0      1 2t 1 2 1 0 1 −2t 1 1 −2 1 0 e +2+e dt = 2π = 2π = 2π e + 2t − e e +2− e e +0− e − 2 2 2 2 2 2 0 0




= 2π 21 e2 + 2 − 12 e−2 − 12 + 0 − 21 = 2π 12 e2 + 2 − 21 e−2 = π e2 + 4 − e−2 Z 1 2t  −2t *************************************************************************** OPTIONAL BONUS Do not attempt these unless you are completely done with the rest of the exam. *************************************************************************** OPTIONAL BONUS #1 Compute the sum ∞ X n 7n n=0 First ∞ X n=0 n comes from 7n Recognize ∞ X n nx = =x 1 (1 − x)2 n=0  = x (1 − x)2 1 nxn with x = . 7 n−1 nx n=0 n=0  ∞ X ∞ X x=x ∞ X n−1 nx n=0 d =x dx 1 Finally, substituting x = , we get 7 1 1 1 ∞ X n 7 =  7 2 =  72 = 7 = n 36 7 36 1 6 n=0 1− 49 7 7 5 ∞ X n=0 n x ! d =x dx  1 1−x  OPTIONAL BONUS #2 Compute the sum ∞ X n2 7n n! n=0 First ∞ X n=0 n2 comes from 7n n! ∞ X ∞ X ∞ X n=0 n2 xn 1 with x = n! 7 ∞ ∞ X X n2 xn n2 xn−1 x n · nxn−1 d nxn Recognize = =x =x n! n! n! dx n! n=0 n=0 n=0 n=0 ! !!   ∞ ∞ X X nxn−1 xn d d d d x d x =x x =x x e =x dx n! dx dx n! dx dx n=0 ! d =x dx ∞ X nxn−1 x n=0 n! n=0 d (xex ) = x (xex + ex ) = x2 ex + xex dx 1 Finally, substituting x = , we get 7 !     ∞ 2 2 X n 1 1 1 8 1 1 1 1 e7 = = + e7 + e7 = n 7 n! 7 7 49 7 49 =x n=0 OPTIONAL BONUS #3 Compute the sum ∞ X n3 n=0 First ∞ X n3 n=0 3n Recognize comes from ∞ X =x d dx ∞ X n=0 x n3 xn with x = n=0 n3 xn = ∞ X ! n · nxn−1 x d dx n3 xn−1 x = x ∞ X n=0 d =x dx !! nxn−1 x ∞ X n=0 n=0 n=0 d =x dx ∞ X x =x !!! ∞ X n=0 d dx 1 7 n2 · nxn−1 n · nxn−1 x d dx 7n x ! ∞ X n=0 d =x dx ∞ X n2 xn n=0 d d =x x dx dx !! ∞ X n=0 ! nxn !! nxn−1     ∞ X d 1 d d d d d n =x x =x x x x x dx dx dx dx dx dx 1 − x n=0        d d d 1 x d x x =x x =x dx dx (1 − x)2 dx dx (1 − x)2       (1 − x)2 (1) − x(2(1 − x)(−1)) (1 − x)((1 − x) + 2x d d =x x = x x dx (1 − x)4 dx (1 − x)4        1+x x + x2 (1 − x)3 (1 + 2x) − (x + x2 )3(1 − x)2 (−1) d d x =x =x =x dx (1 − x)3 dx (1 − x)3 (1 − x)6     (1 − x)2 ((1 − x)(1 + 2x) + 3(x + x2 )) (1 − x)(1 + 2x) + 3(x + x2 ) =x =x (1 − x)6 (1 − x)4 6 ! 1 − x + 2x − 2x2 + 3x + 3x2 (1 − x)4  1 + 4x + x2 x + 4x2 + x3 =x =x = (1 − x)4 (1 − x)4  2  3 1 1 1 + + 4 ∞ X n3 91 1 7 7 7 = = Finally, substituting x = , we get 4  n 7 7 216 1 n=0 1− 7    7