Midterm 2

# Midterm 2 - Math 5C Exam#2 Solutions Date July 16th 2010...

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Math 5C: Exam #2 Solutions Date: July 16 th , 2010 Score: out of 60 1. (10) Match each Maclaurin series to the function from the following list it represents by ﬁlling in the blank space below the series. (Note: All listed function are C at x = 0 under the convention f (0) := lim x 0 f ( x ) .) (i) k =0 ( - 1) k x 2 k = k =0 ( - x 2 ) k = 1 1 - ( - x 2 ) = 1 1+ x 2 (ii) k =0 ( - 1) k k ! x 2 k = k =0 ( - x 2 ) k k ! = e - x 2 (iii) k =0 ( - 1) k (2 k )! x 2 k = cos( x ) (iv) k =0 ( - 1) k (2 k +1)! x 2 k = 1 x k =0 ( - 1) k (2 k +1)! x 2 k +1 = 1 x sin( x ) (v) k =0 ( - 1) k 2 k +1 x 2 k = 1 x k =0 ( - 1) k 2 k +1 x 2 k +1 = 1 x arctan( x ) 2. (5+5) Use appropriate convergence tests to discern the convergence (absolute, conditional, etc.) or divergence of the following series. Identify which tests you are using. (2a) Let k =1 u k = k =1 ( - 1) k k + k . As {| u k |} n =1 is a monotone decreasing sequence that converges to zero, by the Alternating Series Test, the series converges. For k =1 | u k | , consider the Limit Comparison Test to the harmonic series k =1 1 k . As lim k →∞ 1 k | u k | = lim k →∞ k + k k = 1 > 0 , the original series converges absolutely if and only if the harmonic series converges. As the harmonic series diverges, k =1 ( - 1) k k + k converges conditionally . Note: Using the direct Comparison Test ( k =1 | u k | ≥ k =1 1 2 k ) also shows absolute divergence. (2b) Let k =1 u k = k =1 sin k ( 1 k ) . As sin k ( 1 k ) > 0 for k N , by the Root Test, as lim k →∞ ( sin k ( 1 k ) ) 1 k = lim k →∞ sin( 1 k ) = sin(0) = 0 = ρ < 1 , the series converges (absolutely).

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3. (5+5) Let f ( x ) = sin (2 x ) .
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Midterm 2 - Math 5C Exam#2 Solutions Date July 16th 2010...

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