By Stephanie Laurens, Kasey Michaels, Nicola Cornick
A woman of Expectations
Jack Lester seeks the proper bride—one who's beautiful and sort, yet who additionally loves him regardless of his wealth. that's why he retains this trifling element hidden. but if Sophie Winterton enters his existence, believing he needs to marry into wealth to run his family's property, how will he persuade her that she is the lady he desires—and that he will be the husband she deserves?
Secrets of a Courtesan
Eve Nightingale concept she had placed her prior as mistress to the Duke of Welburn in the back of her. but if the good-looking duke strolls into her small village, she reveals it difficult to maintain her secrets and techniques concealed…and to maintain her center from stirring for him as soon as again.
How to Woo a Spinster
Still single at twenty-eight, woman Emmaline Daughtry has resigned herself to spinsterhood. Then Captain John Alistair arrives at her door—the very photograph of the right lover of her such a lot deepest goals. yet can a guy with a mystery and a lady who's by no means identified love locate happiness jointly?
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Additional info for A Lady of Expectations and Other Stories
Sin 2θ − 4 sin θ (n − 1)θ 1 15 = =− sin θ + sin nθ. 1 Find all Laurent series from 0, which are a solution of the diﬀerential equation z 3 f (z) + f (z) = 0, and determine its domain of convergence. 1 It is actually possible to solve the equation by inspection. However, since the trick is far from evident, we shall here start with the standard solution. ♦ Laurent series solution. Since z0 = 0 is a singular point of the diﬀerential equation, we can at most expect a Laurent series solution (possibly non at all).
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N=0 If we replace n by −n in (1), then n ∈ Z, a−n+2 = n a−n , so (3) a−n = 1 a−n+2 , n n ∈ Z. If n = 1, then a−1 = a1 = 0. If n = 3, then a−3 = 1 a−1 = 0, 3 hence by induction, a−2n+1 = 0, n ∈ N. The series is now reduced to +∞ f (z) = a−2n z −2n . n=0 We now replace n by 2n in (3). Then we obtain the following recursion formula for a −2n : (4) a−2n = 1 1 a−2n+2 = a−2(n−1) , 2n 2n n ∈ N. Finally, we put cn = a−2n . Then we derive from (4) the following recursion formula, cn = 1 cn−1 , 2n n ∈ N.