Applied Calculus - download pdf or read online

By Deborah Hughes-Hallett

ISBN-10: 0471681210

ISBN-13: 9780471681212

The ebook arrived in a couple of days and used to be the 1st of my textbooks to reach. My basically grievance is that the ebook used to be indexed as being in "very sturdy" , yet i'd examine it in "good" or perhaps even "fair" . the canopy was once worn to the purpose that it sort of feels this e-book has been round the block greater than a pair occasions. That acknowledged, i might say spending $10 for a $120 e-book is well worth the additional put on and tear.

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Considering the series ∞ � 1 = an , n(n + 1) n=1 n=1 ∞ � it is seen that an = 1 . com 50 Calculus 3b General series; tricks and methods in solutions of problems a) We get by a decomposition an = 1 1 1 . = − n n+1 n(n + 1) b) The sequence of segments sn is written down and reduced, = a1 + a2 + · · · + a n � � � � � � � � 1 1 1 1 1 1 1 − + ··· + − + − + = 1− n n+1 2 4 2 3 2 1 . = 1− n+1 It is seen that the series is reduced like an old-fashioned telescope, therefore the name. c) Finally, we get by taking the limit � � ∞ � 1 1 = 1.

Xn . With only one obvious exception, they n=0 xn . They are all convergent in the whole n! a) Quotient like series, � = 1. ) �∞ 1 = n=0 xn , 1−x n=0 �∞ 1 = n=0 (−1)n xn , 1+x � � �∞ α (1 + x)α = n=0 xn , n ln(1 + x) = arctan x = �∞ n=1 �∞ n=0 �∞ (−1)n−1 n x , n �∞ xn = n=0 (−1) �∞ n=0 �∞ n=1 � α n 1 , 1−x |x| < 1, x = 1 , 1+x |x| < 1, xn = (1 + x)α , |x| < 1, n n � (−1)n−1 n x = ln(1 + x), n α∈R |x| < 1, (−1)n 2n+1 �∞ (−1)n 2n+1 x = arctan x, |x| < 1. x , n=0 2n + 1 2n + 1 If in the third line α = n ∈ N0 , then (1 + x)n is a polynomial.

1) Leibniz’s criterion is obtained when an = (−1)n−1 . In fact, in this case we get from condition c) that � � � � n � n � �� � �� 1 for n odd, � � � k−1 � ak � = � (−1) |sn | = � �= 0 for n even, � � � � k=1 k=1 hence the sequence of segments is bounded, which is sufficient. The sequence of segments is of course not convergent. 2) The Fourier series �∞ n=1 1 sin nx is pointwise convergent. n For x = pπ, p ∈ Z, there is nothing to prove, because the zero series is convergent. Choose a fixed x0 �= pπ, p ∈ Z.

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Applied Calculus by Deborah Hughes-Hallett


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