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The Real And Complex Number Systems

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have<br />

∫ x<br />

0<br />

∞∑<br />

t 2n − 1 2<br />

n=0<br />

∞∑<br />

t n =<br />

n=0<br />

=<br />

∫ x<br />

0<br />

∫ x<br />

0<br />

1<br />

1 − t − 1<br />

2 2 (1 − t) dt<br />

(<br />

1 1<br />

2 1 − t + 1 )<br />

1 + t<br />

= 1 log (1 + x) .<br />

2<br />

− 1 2<br />

( ) 1<br />

dt<br />

1 − t<br />

<strong>And</strong> as x = 1,<br />

∞∑ (<br />

x 2n+1 / (2n + 1) − x n+1 / (2n + 2) )<br />

n=0<br />

=<br />

=<br />

∞∑<br />

n=0<br />

∞∑<br />

n=0<br />

1<br />

2n + 1 − 1<br />

2n<br />

(−1) n+1<br />

n + 1<br />

by <strong>The</strong>orem8.14.<br />

= log 2 by Abel’s Limit <strong>The</strong>orem.<br />

9.22 Prove that ∑ a n sin nx and ∑ b n cos nx are uniformly convergent on<br />

R if ∑ |a n | converges.<br />

Proof: It is trivial by Weierstrass M-test.<br />

9.23 Let {a n } be a decreasing sequence of positive terms. Prove that<br />

the series ∑ a n sin nx converges uniformly on R if, and only if, na n → 0 as<br />

n → ∞.<br />

Proof: (⇒) Suppose that the series ∑ a n sin nx converges uniformly on<br />

R, then given ε > 0, there exists a positive integer N such that as n ≥ N,<br />

we have<br />

∣ ∣∣∣∣ 2n−1<br />

∑<br />

a k sin kx<br />

< ε. (*)<br />

∣<br />

Choose x = 1<br />

2n , then sin 1 2<br />

k=n<br />

≤ sin kx ≤ sin 1. Hence, as n ≥ N, we always<br />

19

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