sol108c3

# sol108c3 - MA108C HW 3 SOLUTIONS PAUL NELSON(AND SOME...

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MA108C HW 3 SOLUTIONS PAUL NELSON (AND SOME MODIFICATIONS BY FOKKO VAN DE BULT) (1) By the generalized binomial theorem, z = [1 + ( z - 1)] 1 / 2 = X k =0 ± 1 / 2 k ² ( z - 1) k for | z | < 1 , where ( α k ) = α ( α - 1) ··· ( α - k +1) k ! for α C ,k Z . The ratio of successive terms is ( 1 / 2 k +1 ) ( z - 1) k +1 / ( 1 / 2 k ) ( z - 1) k = 1 / 2 - k k +1 ( z - 1) which tends to 1 - z as k → ∞ , so the radius of convergence of our series is 1. Let us now prove this via the general theory of holomorphic functions. Deﬁne f ( z ) = e 1 2 log z for z C \ R 0 , where log( re ) = log r + for θ ( - π,π ), as in the ﬁrst homework. Then f ( z ) = z for z R > 0 , so the power series for f at z = 1 is as above. Since f is holomorphic in { x + iy : x > 0 } ⊃ { z : | z - 1 | < 1 } and the power series for a holomorphic function at a point converges absolutely in any disc centered at that point in which the function is holomorphic, we see that the radius of convergence of f at z = 1 is at least 1. The radius of convergence cannot exceed 1; otherwise, f would extend to a holomorphic function F on { z : | z - 1 | < 1 + ± } for some ± > 0, so that F is holomorphic in a neighborhood of the origin. By continuity, F (0) = 0. For small enough h R > 0 we have F ( h ) = f ( h ) = h , so F ( h ) - F (0) h = h - 1 / 2 . This expression does not have a limit

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sol108c3 - MA108C HW 3 SOLUTIONS PAUL NELSON(AND SOME...

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