# Laurent Series, c-5 by Mejlbro L.

By Mejlbro L.

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**Example text**

In the latter case, however, we have the following equivalent equations, dg 1 1 − w exp − w2 · g(w) 0 = exp − w2 dw 2 2 d 1 2 dg 1 + · g(w) = exp − w exp − w2 dw dw 2 2 d 1 = exp − w2 g(w) . dw 2 An integration gives with an arbitrary constant c ∈ C, 1 exp − w2 2 g(w) = c, w ∈ C \ {0}. Finally, it follows from g(w) = f (z) and w = f (z) = g(w) = c · exp 1 2 w 2 = c · exp 1 that z 1 2z 2 z ∈ C \ {0}. 2 Find all Laurent series from 0, which are solutions of the diﬀerential equation z 2 − z f (z) = (1 − 2z) f (z), and determine the domain of convergence.

And 1) Find the Maclaurin series of f . Find also the Maclaurin series of g and the coeﬃcients of this series. 2) Find the Laurent series of f in the annulus |z| > 1. Also, ﬁnd the Laurent series of g in the same annulus and the coeﬃcients of this series. 3) Finally, ﬁnd the Laurent series of g in the set deﬁned by |z − 1| > 0. 1) The Maclaurin series of f is given by +∞ f (z) = zn, |z| < 1. n=0 Hence by termwise diﬀerentiation and insertion, g(z) = z 2 +∞ n(n − 1)z n−2 + z n=2 +∞ = +∞ n=1 n2 z 2 = n=0 +∞ n2 z n , +∞ n z n−1 = n(n − 1)z n + n=2 +∞ n zn n=1 (n=0) (n=0) for |z| < 1.

B2(n−1)+1 = · · · = 1! b1 = a−1 , hence a−2n = b2n = 1 a0 (2n)! and a−2n−1 = b2n+1 = 1 a−1 , (2n + 1)! n ∈ N0 . Thus the formal Laurent series solution is given by +∞ f (z) = a0 +∞ 1 1 1 1 · 2n + a−1 · 2n+1 = a0 cosh (2n)! z (2n + 1)! z n=0 n=0 1 z + a−1 sinh 1 z . The determination of the domain of convergence C \ {0} is trivial, because we only consider known 1 < +∞. series which are convergent for z Only the zero solution can be extended to all of C. com 64 Complex Funktions Examples c-5 Laurent series solution of differential equations Alternatively, the coeﬃcient a0 (z) = z 4 leads one the the idea of transforming the diﬀerential 1 equation into a diﬀerential equation in the new variable w = , z = 0, w = 0.