APPM 1360 Exam 3 Fall 2018

1. [36 pts] Determine if the following series converge absolutely, converge conditionally or diverge. Be sure to fully justify your answer and state any test you use. (2n−1)π √ ∞ 3n ∞ 2 ∞ sin 3 n X e X 28n + 5 2 2 · 4 2 · 4 · 6 X 2 (a) (b) √ (c) − + − + ··· (d) √ nn 7 3 2 2 · 5 2 · 5 · 8 n n=1 n=1 16n + 3n + 2 n=1 SOLUTION:

(a) s 3n 3 n e e lim = lim = 0 < 1 n→∞ nn n→∞ n

Series converges absolutely by the Root Test. (b) Each term of the series is positive and as n gets large, the terms begin to look like 28n2/4n7/2 = 7n−3/2.

2 √ 28n +5 7/2 3/2 7/2 ! 5 16n7+3n3+2 28n + 5n 1/n 28 + n2 lim 1 = lim √ √ = lim n→∞ n→∞ 7 2 7 n→∞ q 16n7+3n2+2 n3/2 16n + 3n + 2 1/ n n7

5 28 + n2 28 = lim q = √ = 7 (positive and ﬁnite) n→∞ 3 2 16 16 + n5 + n7

∞ X 1 Since is a p–series with p = 3/2 > 1 it converges, and the original series converges by the Limit n3/2 n=0 Comparison Test. Since all of the terms of the series are positive, the series converges absolutely. 2 · 4 · 6 ····· (2n) (c) Note that the terms of the series are a = (−1)n and we consider n 2 · 5 · 8 ····· (3n − 1)

n+1 2·4·6·····(2n)(2(n+1)) a (−1) lim n+1 = lim 2·5·8·····(3n−1)(3(n+1)−1) 2·4·6·····(2n) n→∞ an n→∞ (−1)n 2·5·8·····(3n−1) ( hh (((( hhh (2 ·(4 (· 6 ····· (2n)(2n + 2) 2 · 5 · 8 ·····h(3hnh−h1) = lim hh · (((h n→∞ hhhh 2 · 4 (· 6(·····( (2n) 2 · 5 · 8 ····· (3hnh−hh1)(3n + 2) (( 2n + 2 1/n 2 + 2/n 2 = lim = lim = < 1 n→∞ 3n + 2 1/n n→∞ 3 + 2/n 3

so the series converges absolutely by the Ratio Test. (−1)n−1n1/3 (−1)n−1 (d) The terms of the series can be written as = showing that the original series is simply an n1/2 n1/6 1 alternating series with b = . Since n n1/6 1 lim bn = lim = 0 n→∞ n→∞ n1/6 and 1 1 n + 1 > n =⇒ (n + 1)1/6 > n1/6 =⇒ b = > = b for n ≥ 1 n n1/6 (n + 1)1/6 n+1 the series converges by the Alternating Series Test. Now consider √ ∞ sin (2n−1)π 3 n ∞ X 2 X 1 √ = n n1/6 n=1 n=1

which is a p–series with p = 1/6 < 1 and thus diverges. Consequently the original series converges conditionally.

2. [32 pts] The following problems are not related.

∞ X (−1)n−1 (a) Find the sum of . n! n=1 ∞ 1 + x X (b) Find the power series for . Write your answer in the form c xn for an appropriate c . (1 − x)2 n n n=0 ∞ X n (c) Answer True or False, justifying your answer. If the power series cnx converges at x = 3, then n=0 ∞ X n 5 n (−1) cn 2 converges. n=0 (d) Find the first four terms of the binomial series representation for (1 + 5x)1/5. Be sure to compute the binomial coefficients and completely simplify your final answer.

SOLUTION:

(a)

∞ ∞ ∞ ∞ ! X (−1)n−1 X (−1)n(−1)−1 X (−1)n X (−1)n (−1)0 = = − = − − n! n! n! n! 0! n=1 n=1 n=1 n=0 1 e − 1 = − e−1 − 1 = 1 − e−1 = 1 − = e e

(b)

∞ ∞ ∞ ∞ ∞ 1 X d 1 1 d X X d X X = xn =⇒ = = xn = xn = nxn−1 = (n + 1)xn 1 − x dx 1 − x (1 − x)2 dx dx n=1 n=0 n=0 n=1 n=0 Then

∞ ∞ 1 + x 1 x X X = + = (n + 1)xn + x (n + 1)xn (1 − x)2 (1 − x2) (1 − x)2 n=0 n=0

∞ ∞ ∞ ∞ X X X X = (n + 1)xn + (n + 1)xn+1 = (n + 1)xn + nxn n=0 n=0 n=0 n=1

∞ ∞ ∞ ∞ X X X X = (n + 1)xn + nxn = [(n + 1) + n] xn = (2n + 1)xn n=0 n=0 n=0 n=0 Another way to look at this is

∞ 1 + x 1 X = (1 + x) = (1 + x) nxn−1 (1 − x)2 (1 − x)2 n=1 = (1 + x) 1 + 2x + 3x2 + 4x3 + 5x4 + ···

= 1 + 2x + 3x2 + 4x3 + 5x4 + ··· + x + 2x2 + 3x3 + 4x4 + 5x5 + ···

∞ X = 1 + 3x + 5x2 + 7x3 + 9x4 + ··· = (2n + 1)xn n=0

Alternatively, the power series for 1/(1 − x)2 can be obtained from the binomial series representation and then manipulated as above. Or, one can ﬁnd the derivatives of the original function and build the Taylor series from scratch. (c) True. Since the power series converges at x = 3 and it is centered at 0, we know that the radius of convergence is n 5 n 5 n 5 at least 3, implying that the interval of convergence is at least (−3, 3]. Since (−1) 2 = − 2 and − 2 is in ∞ X n 5 n this minimum interval of convergence, (−1) cn 2 converges. n=0 (d)

∞ ∞ ∞ X 1 X 1 X 1 (1 + x)1/5 = 5 xn =⇒ (1 + 5x)1/5 = 5 (5x)n = 5 5nxn n n n n=0 n=0 n=0 1 1 1 1 = 5 + 5 5x + 5 52x2 + 5 53x3 + ··· 0 1 2 3

Now 1 5 = 1 0

1 1 1 5 = 5 = (−1)0 1 1! 511!

1 1 − 4 1 · 4 5 = 5 5 = (−1)1 2 2! 522!

1 1 − 4 − 9 1 · 4 · 9 5 = 5 5 5 = (−1)2 3 3! 533! 1 1 · 4 1 · 4 · 9 =⇒ (1 + 5x)1/5 = 1 + (−1)0 5x + (−1)1 52x2 + (−1)2 53x3 + ··· 511! 522! 533! = 1 + x − 2x2 + 6x3 − · · ·

3. [12 pts] Suppose f (n)(−4) = (−5)11−(n+1)2nn! for n ≥ 0. (a) Find the Taylor series for f centered at −4, writing your answer using sigma notation. (b) Find f, the sum of the Taylor series in part (a). Simplify your answer.

SOLUTION: ∞ X n (a) The Taylor series we seek is cn(x + 4) where n=0

f (n)(−4) (−5)11−(n+1)2nn! 5 2 n c = = = (−5)11−111−n2n = − n n! n! 11 11

∞ n ∞ n X 5 2 5 X 2 =⇒ − (x + 4)n = − (x + 4)n 11 11 11 11 n=0 n=0

(b) From part (a)

∞ n ∞ n 5 X 2 5 X 2(x + 4) 5 1 − (x + 4)n = − = − 11 11 11 11 11 2(x+4) n=0 n=0 1 − 11 −5 5 = = 11 − (2x + 8) 2x − 3

√ 4. [12 pts] Let g(x) = x. (a) Find the second degree Taylor polynomial of g, T (x), centered at 4. √ 2 (b) Use T2(x) to approximate 5. Write your final answer as a single fraction. (c) Use Taylor’s formula to find an error bound for your approximation in (b), simplifying your final answer.

SOLUTION: (a)

f(x) = x1/2 =⇒ f(4) = 2 1 1 1 1 f 0(x) = x−1/2 =⇒ f 0(4) = = 2 2 41/2 4 1 1 1 1 1 f 00(x) = − x−3/2 =⇒ f 00(4) = − = − 2 2 4 43/2 32 1 1 3 3 f 000(x) = − − x−5/2 = 2 2 2 8x5/2

f 00(4) 1 1 T (x) = f(4) + f 0(4)(x − 4) + (x − 4)2 = 2 + (x − 4) − (x − 4)2 2 2! 4 64 (b) √ 1 1 1 1 128 + 16 − 1 143 5 ≈ T (5) = 2 + (5 − 4) − (5 − 4)2 = 2 + − = = 2 4 64 4 64 64 64 (c)

000 f (z) 3 |R2(x)| = (x − 4) 4 < z < 5 3! Since f 000(x) is decreasing on [4, 5], it attains its largest value at the left endpoint so that 3 3 3 |f 000(z)| < = = 8 45/2 8(32) 256

Thus 3 1 |R (5)| < (5 − 4)3 = 2 3!256 512

5. [8 pts] Let x(t) = 3 sin(t/2) and y(t) = 3 cos(t/2) with 0 ≤ t ≤ π. (a) Sketch the parametric curve and indicate with an arrow the direction in which the curve is traced as the parameter increases. (b) Eliminate the parameter to ﬁnd a Cartesian equation of the curve. Be sure to include any restrictions, if any, on the Cartesian variables.

SOLUTION:

t x y y 3 0 0 3 √ π/3 3/2 3 3/2 (a) √ √ π/2 3 2/2 3 2/2 √ 2π/3 3 3/2 3/2 x π 3 0 3 (b) x x2 x = 3 sin(t/2) =⇒ = sin(t/2) =⇒ = sin2(t/2) (1) 3 3 y y 2 y = 3 cos(t/2) =⇒ = cos(t/2) =⇒ = cos2(t/2) (2) 3 3 Adding (1) and (2) yields

x2 y 2 + = sin2(t/2) + cos2(t/2) = 1 =⇒ x2 + y2 = 9 3 3 with the restriction that x ≥ 0 and y ≥ 0.