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hw3solutions

# hw3solutions - Math 318 HW#3 Solutions 1 Exercise 6.5.2...

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Math 318 HW #3 Solutions 1. Exercise 6.5.2. Find suitable coeﬃcients ( a n ) so that the resulting power series a n x n (a) converges absolutely for all x [ - 1 , 1] and diverges oﬀ of this set; Answer. Consider the series X n =1 x n n 2 . Then for all x [ - 1 , 1], ± ± x n n 2 ± ± 1 n 2 ; since 1 n 2 converges, the Weierstrass M-test implies that the above series converges absolutely on [ - 1 , 1]. On the other hand, for any x R such that | x | > 1, the sequence of terms ( x n /n ) diverges, so we see that the above series diverges away from [ - 1 , 1]. (b) converges conditionally at x = - 1 and diverges at x = 1; Answer. Consider the series X n =1 x n n . We’ve already seen in class that this series converges conditionally at x = - 1 and diverges at x = 1. (c) converges conditionally at both x = - 1 and x = 1. Answer. Consider the series X n =1 ( - 1) n x 2 n n . (It may look like I’m cheating by ﬁddling with the exponent on x , but the above is really just shorthand for the series a n x n where a n = 0 when n is odd and a n = ( - 1) n/ 2 / ( n/ 2) when n is even.) Notice that this series satisﬁes the hypotheses of the Alternating Series Test for any x [ - 1 , 1], so it converges on this interval. Notice that the series ± ± ± ( - 1) n x 2 n n ± ± ± reduces to the harmonic series when x = ± 1, so the series only converges conditionally at x = ± 1. (d) Is it possible to ﬁnd an example of a power series that converges conditionally at x = - 1 and converges absolutely at x = 1? Answer. No. If there were such a series a n x n , then, when x = - 1, we would have X n =0 | a n x n | = X n =0 | a n || ( - 1) n | = X n =0 | a n | = X n =0 | a n (1) n | , which converges by hypothesis. But this contradicts the fact that a n x n only converges conditionally at x = - 1. 2. (a) Exercise 2.7.12 (Summation by Parts). Let ( x n ) and ( y n ) be sequences, and let s n = x 1 + x 2 + ··· + x n . Use the observation that x j = s j - s j - 1 to verify the formula n X j = m +1 x j y j = s n y n +1 - s m y m +1 + n X j = m +1 s j ( y j - y j +1 ) . 1

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Proof. Using the fact that x j = s j - s j - 1 , we have n X j = m +1 x j y j = n X j = m +1 ( s j - s j - 1 ) y j = s m +1 y m +1 - s m y m +1 + ... + s n y n - s n - 1 y n = s n y n - s m y m +1 + n - 1 X j = m +1 s j ( y j - y j +1 ) .
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hw3solutions - Math 318 HW#3 Solutions 1 Exercise 6.5.2...

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