A la recherche de la topologie perdue: I Du côté de chez by Lucien Guillou, Alexis Marin

By Lucien Guillou, Alexis Marin

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Check. If we put f (t) = −2 e−t , then t 0 u f (u) cos(t − u) du = −2 t it = −2 Re e = −2 Re eit · = −2 Re ue t 0 −(1+i)u t u e−u cos(t − u) du = −2 Re du 0 −1 + i −(1+i)t te 2 it = −2 Re e u e−(1+i)u −(1 + i) t u e−u ei(t−u) du 0 + 0 eit 1+i t e−(1+i)u du 0 eit e−(1+i)t − 1 (1 + i)2 1 −t 1 e − eit = t e−y − sin t, 2i 2i + 2 Re e−t · t · (−1 + i) + 2 Re 2 and we have tested our solution. 43 Find a function f ∈ F, such that t 0 f (u) f (t − u) du = 8(sin t − t cos t), t ∈ R+ . We put F (z) = L{f }(z).

We shall demonstrate both methods here. (a) Decomposition and rules of calculations. It follows from z −1 2 + = L 2 e−2t − e−t (z), = z+1 z+2 (z + 1)(z + 2) that the inverse Laplace transform is given by z (z + 1)(z + 2) f (t) = L◦−1 = 2 e−2t − e−t . Residuum formula. Since z = −1 and z = −2 are simple poles, we get by the residuum formula and Rule Ia that f (t) = res z ezt ; −1 + res (z + 1)(z + 2) z ezt ; −2 (z + 1)(z + 2) = −1 −t −2 −2t e + e 1 −1 = 2 e−2t − e−t . (b) Rules of calculation and use of tables.

2n = · (2n)! (2n − 1)! (2n − 1)! (2n − 1)! n=1 (z). (2n − 1)! n=1 t ∈ R. Alternatively we even have the estimate cosh 1 z −1 ≤ C |z|2 for |z| ≥ R. Please click the advert what‘s missing in this equation? You could be one of our future talents MAERSK INTERNATIONAL TECHNOLOGY & SCIENCE PROGRAMME Are you about to graduate as an engineer or geoscientist? Or have you already graduated? P. Moller - Maersk. com 53 Complex Functions Examples c-8 The Laplace transform It follows from the series expansions +∞ 1 z ezt cosh −1 = +∞ +∞ +∞ tm m 1 tm m 1 1 1 1 z · z · · 2n = · 2n−1 , m!

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