solns12 - Mathematics 185 Intro to Complex Analysis Fall...

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Mathematics 185 – Intro to Complex Analysis Fall 2009 – M. Christ Solutions to Selecta from Problem Set 12 4.4.4 Show that π 2 sin 2 ( πz ) = n = -∞ ( z - n ) - 2 for all z C \ Z . Solution. We deduce this from the formula π cot( πz ) = z - 1 + X n 6 =0 ± ( z - n ) - 1 + n - 1 ² , (1) where the sum is over all integers n 6 = 0, positive and negative. We know the latter formula to be valid for all z C \ Z . The connection between the two formulas is that π 2 sin 2 ( πz ) is the derivative of π cot( πz ), while n = -∞ ( z - n ) - 2 is obtained, at least formally, by differentiating the right-hand side n = -∞ ( z - n ) - 1 term by term. Thus our task is to justify this term by term differentiation. Let F N ( z ) = N n = - N ( z - n ) - 1 . It was implicitly shown in the course of the proof of (1) that F N ( z ) π cot( πz ) uniformly on D , as N → ∞ . (The difference F N ( z ) - π cot( πz ) is expressed as an integral over C N , and the bound established for this integral tends to zero as N → ∞ , uniformly for z D .) Therefore the sequence of derivatives also converges uniformly on any disk centered at w of strictly smaller radius; F 0 N d dz ( π cot( πz )) uniformly for all z B ( w,r/ 2), by one of our general theorems. Since F 0 N ( z ) = N n = - N ( z - n ) - 2 , this is the desired conclusion. 4.4.8 Prove that n =0 ( - 1) n - 1 n - 2 = π 2 12 . Solution. Apply the residue theorem to f ( z ) = π z 2 sin( πz ) = z - 2 π csc( πz ), using the familiar contours C N . Calculate Res( f,n ) = n - 2 1 cos( πn ) = ( - 1) n n - 2 for 0 6 = n Z ; using csc instead of cot is natural, because it gives the alternating ± signs which we’re aiming for. Use the expansion sin( πz ) = πz (1 - π 2 z 2 / 6 + O ( z 4 )) to calculate π π sin( πz ) = π πz (1 - π 2 z 2 / 6 + O ( z 4 )) - 1 = z - 1 (1 + π 2 z 2 / 6 + O ( z 4 )) to obtain z - 2 π π sin( πz ) = z - 3 (1 + π 2 z 2 / 6 + O ( z 4 )) = z - 3 + π 2 6 z - 1 + O ( z ) 1
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and therefore Res( f, 0) = π 2 6 . By the Residue Theorem, for each integer N 1, (2 πi ) - 1 Z C N f ( z ) dz = π 2 6 + - 1 X n = - N ( - 1) n n - 2 + N X n =1 ( - 1) n n - 2 . If we can show that R C N f ( z ) dz 0 as N → ∞ then we obtain the desired formula. Now | Z C N f ( z ) dz | ≤ ( C N ) max z C N h | z | - 2 π | sin( πz ) | - 1 i . Since ( C N ) = 8 N + 4 and | z | - 2 N - 2 for every z C N , it suffices to prove that | sin( πz ) | - 1 1000 for every z C N , for every integer N 1. This is very similar to a corresponding bound established in class and in our text for
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solns12 - Mathematics 185 Intro to Complex Analysis Fall...

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