Unformatted text preview: B + n M X k =0 a k ( s n k B ) AB fl fl fl fl fl + n X k = n M +1  a k  s n k  ≤ fl fl fl fl fl ˆ n M X k =0 a k A !fl fl fl fl fl  B  + n M X k =0  a k  ² 3 α + C ² 3 C = fl fl fl fl fl ˆ n M X k =0 a k A !fl fl fl fl fl  B  + α ² 3 α + C ² 3 C < e 3 C  B  + 2 ² 3 < ² 3 + 2 ² 3 = ². Notice: There was no need to use the absolute convergence of the series ∑ ∞ n =1 b n for this part. This is known as Mertens’ Theorem: If ∑ a n converges absolutely, ∑ b n converges , then ∑ c n converges and equals ∑ a n · ∑ b n . 2. Problem 20 of Chapter VII of Rosenlicht (pp. 16263). Solution. The radii are 1 , 1 , 1 , ∞ , 1 /e . 3. Problem 22 of Chapter VII of Rosenlicht (p. 163). Solution. Let f ( z ) = ∑ ∞ n =0 c n ( z a ) n . The assumption is that not all coefficients are zero; we have to prove there exists δ > 0 such that f ( z ) 6 = 0 for 0 <  z a  < δ . The exercise is stated for the real case, but the same result holds in the complex case. Let n be the first index with c n 6 = 0; that is n ∈ N is such that c k = 0 if k < n , c n 6 = 0. Then f ( z ) = c n ( z a ) n + c n +1 ( z a ) n +1 + ··· = ( z a ) n g ( z ) where g ( z ) = ∞ X k = n c k ( z a ) k n = ∞ X k =0 c k + n ( z a ) k . It is easy to see that the radius of convergence of the series defining g is the same as that of f . So g is a continuous function in  z a  < r and g ( a ) = c n 6 = 0. By continuity, there is δ > 0 such that g ( z ) 6 = 0 if  z a  < δ . Since  z a  n 6 = 0 if z 6 = a , it follows that f ( z ) 6 = 0 for <  z a  < δ . 4. C ( K ) is a Banach space. Solution. The Theorem on page 90 of Rosenlicht has a much more general result of which the completeness of C ( K ) is an immediate conse quence. 3...
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 Spring '11
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 CN, k=0, ak sn−k

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