3100practice3

# 3100practice3 - PRACTICE PROBLEMS FOR THIRD MATH 3100...

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Unformatted text preview: PRACTICE PROBLEMS FOR THIRD MATH 3100 MIDTERM On Power Series 1) Find all values of x ∈ R at which the following power series converge. a) ∑ ∞ n =1 ( n 2 +5 3 n 3 +4 ) ( x − 1) 2 n 3 n . Solution: This is actually rather tricky, enough for it to be best to apply the Root Test: for any polynomial function P ( n ), one has lim n →∞ | P ( n ) | 1 n = 1, so the first factor in the sum approaches 1. Therefore lim n →∞ ( | x − 1 | 2 n 3 n ) 1 n = | x − 1 | 2 3 . This is less than 1 iff | x − 1 | ≤ √ 3, i.e., if x ∈ (1 − √ 3 , 1 + √ 3). Finally, we check the endpoints: when x = 1 + √ 3, we get the series ∑ ∞ n =1 n 2 +5 3 n 3 +4 , which is divergent by Limit Comparison to the harmonic series. When x = 1 − √ 3 we get the same se- ries, which is still divergent. Therefore the interval of convergence is (1 − √ 3 , 1+ √ 3). b) ∑ ∞ n =2 ( x +3) 2 n +1 n log n . Solution: Since n ≤ n log n ≤ n 2 , ( n log n ) 1 n → 1 and thus lim n →∞ ( | x + 3 | 2 n +1 n log n ) 1 /n = lim n →∞ | x + 3 | 2+ 1 n = | x + 3 | 2 . This quantity is less than 1 iff | x + 3 | < 1 if x ∈ ( − 4 , − 2). Plugging in x = − 2 we get the series ∑ ∞ n =2 1 n log n , which diverges by the Condensation/Integral Test. Plugging in x = − 4, we get the negative of this series (note: not the alternating version: this is because ( − 1) 2 n +1 = − 1 for all n ) which is again divergent. So the interval of convergence is ( − 4 , − 2). 2) Exhibit a power series ∑ n a n ( x − c ) n with interval of convergence [ e,π ). Solution: We may as well choose any real numbers A < B and find a power series with interval of convergence [ A,B ). We need the point c to be the midpoint of this interval, i.e., c = A + B 2 . Moreover, the radius of convergence is half the length of the interval, hence R = B − A 2 . So a good first attempt is ∑ n ( x − c ) n R n , which amounts to a a geometric series with r = x − c R , so is convergent precisely for | x − c | < R . In other words this series converges precisely on the open interval ( A,B ). We also want convergence at the left endpoint, so let’s take instead ∑ n ( x − c ) n nR n : when we plug in the left endpoint x = c − B − A 2 , we get the series ∑ n ( − 1) n n , which converges, and when we plug in the right endpoint we get ∑ n 1 n , which diverges. So a correct c ⃝ Pete L. Clark, 2011. 1 2 PRACTICE PROBLEMS FOR THIRD MATH 3100 MIDTERM answer is ∞ ∑ n =1 ( x − A + B 2 ) n n ( B − A 2 ) n . 3) You may have noticed that many naturally occurring power series have radius of convergence R = 1. This problem gives a kind of justification. a) Let { a n } be a bounded sequence of real numbers. Show that the radius of con- vergence of ∑ n a n x n is at least 1....
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## This note was uploaded on 06/06/2011 for the course MATH 3100 taught by Professor Staff during the Spring '08 term at UGA.

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3100practice3 - PRACTICE PROBLEMS FOR THIRD MATH 3100...

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