Exam 3 Practice Problems-1

# Exam 3 Practice Problems-1 - Math 142 F: Exam 3 Practice...

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Unformatted text preview: Math 142 F: Exam 3 Practice Problems 1. Find a power series representation for f ( x ) = ln(1+ x 2 ). Find the corresponding radius of convergence. 2. Find the interval and radius of convergence for the power series. ∞ X n =1 (4 x + 1) n n 2 3. Find the interval and radius of convergence for the power series. ∞ X n =1 n ( x + 1) n 4 n 4. Determine whether the given series converges or diverges. ∞ X n =1 cos( n ) n 2 5. Determine whether the given series converges or diverges. ∞ X n =1 n n n ! 6. Determine whether the given series converges or diverges. ∞ X n =1 2 n + 3 3 n + 2 n 7. Find the area enclosed by each curve. (a) r = sin 3 θ (b) r = sin 4 θ 8. Consider the curve with polar equation r = 1 + sin θ . Complete the following. (a) Find dy/dx . (b) Find the equation of the tangent line to the curve when θ = π/ 3. (c) Find all points on the curve where there is either a horizontal or vertical tangent. 9. Find the area of the described region. inside: r = 3 sin θ ; outside: r = 1 + sin θ 1 1 Solutions Problem 1 : We begin by noting that d dx [ln(1 + x 2 )] = 2 x 1 + x 2 . We will use the following steps to find the power series (this is the Section 11.9 approach). Steps : (1) Find a power series representation for 2 x 1 + x 2 . (2) Use the fact the ln(1 + x 2 ) = Z 2 x 1 + x 2 dx (plus a constant C ). Let’s do this : (1) We will use 1 1- x = ∞ X n =0 x n , | x | < 1 (geometric expansion). 2 x 1 + x 2 = 2 x · 1 1- (- x 2 ) = 2 x · ∞ X n =0 (- x 2 ) n , | - x 2 | < 1 = 2 x · ∞ X n =0 (- 1) n x 2 n , | - x 2 | < 1 = ∞ X n =0 2(- 1) n x 2 n +1 , | - x 2 | < 1 So 2 x 1 + x 2 dx = ∞ X n =0 2(- 1) n x 2 n +1 , | x | < 1. Note that | - x 2 | < 1 ⇐⇒ | x | < 1. 2 (2) Use ln(1 + x 2 ) = Z 2 x 1 + x 2 dx (plus a constant C ). ln(1 + x 2 ) = Z 2 x 1 + x 2 dx = Z ∞ X n =0 2(- 1) n x 2 n +1 ! dx, | x | < 1 = ∞ X n =0 Z 2(- 1) n x 2 n +1 dx , | x | < 1 = ∞ X n =0 2(- 1) n Z x 2 n +1 dx , | x | < 1 = ∞ X n =0 2(- 1) n x 2 n +2 2 n + 2 + C, | x | < 1 = ∞ X n =0 (- 1) n x 2 n +2 n + 1 + C, | x | < 1 So we have ln(1+ x 2 ) = ∞ X n =0 (- 1) n x 2 n +2 n + 1 + C, | x | < 1. We can actually find the constant C . Let x = 0 . Then 0 = ln 1 = ∞ X n =0 (- 1) n (0) 2 n +2 n + 1 + C = 0 + C We see that C = 0. Here is the final answer (radius of convergence R = 1). ln(1 + x 2 ) = ∞ X n =0 (- 1) n x 2 n +2 n + 1 , | x | < 1 Problem 2 : Apply the Ratio test: a n = (4 x + 1) n n 2 . a n +1 a n = (4 x + 1) n +1 ( n + 1) 2 · n 2 (4 x + 1) n = (4 x + 1) n 2 ( n + 1) 2 → | 4 x + 1 | as n → ∞ ....
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## This note was uploaded on 11/18/2011 for the course PSY 122 taught by Professor Kim during the Spring '11 term at SUNY Buffalo.

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Exam 3 Practice Problems-1 - Math 142 F: Exam 3 Practice...

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