By Mejlbro L.

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Additional resources for Power series, c-4

Example text

19 1) Find the radius of convergence R for the power series +∞ 1 n z . 2 n n=1 Prove that the series is absolutely and uniformly convergent in the closed disc {z ∈ C | |z| ≤ R}. 2) Find the set K ⊂ C, for which the series +∞ 1 nz e n2 n=1 is convergent for z ∈ K and divergent for z ∈ / K. 1 n z → +∞ for n → +∞ due to the order of magnitudes. Thus the necessary n2 condition of convergence is not satisﬁes. We therefore conclude that R ≤ 1. If instead |z| ≤ 1, then we get the estimate 1) If |z| > 1, then +∞ +∞ 1 n 1 π2 .

Z| |z| n=1 n=0 hence in C. e. 4 This example indicates a method of determining the Laplace transformed of a piecewise continuous function, which is only = 0 on a closed bounded interval. 14 Assume that g(z) is analytic in |z| < R, and that g(0) = 0. Apply Weierstraß’s 1 . double series theorem in order to ﬁnd a power series of 1 − g(z) 1 Find in particular the ﬁrst three terms in the power series expansion (from 0) for . cos z Since +∞ +∞ 1 = {g(z)}n = gn (z) 1 − g(z) n=0 n=0 for |g(z)| < 1, and since g(0) = 0 implies that |g(z)| < 1 in a neighbourhood of 0 (where the convergence is uniform for |g(z)| ≤ k < 1), and c0 = 0, it follows that +∞ 1 = 1 − g(z) n=0 n +∞ cp z p=1 p +∞ = z n=0 n n +∞ cp+1 z p .

Determine the radius of convergence of the series. Since ez − 1 = z + 1 2 z + ··· 2! is diﬀerent from zero for z = 2i p π, p ∈ Z, we conclude that lim z→0 z = lim ez − 1 z→0 z + 1 2 z = lim z 2 + o (z 2 ) z→0 1 + 1 2 1 = 1 = B0 . z + o(z) It follows that B0 = 1. Furthermore, the power series is convergent in the largest open disc of centrum at 0 which does not contain any number of the form 2i p π, p ∈ Z \ {0}. The two closest singularities of the point of expansion z0 = 0 are ±2i π, so we conclude that the radius of convergence is |±2i π| = 2π.