By Mark Fellowes, Nicholas Battey

Adapt or die: it’s nature’s most famed crucial. yet how does evolution truly take place? It’s too gradual to determine, yet it’s happening throughout you, for all time. no matter if you’re on best of the major phrases - edition? normal choice? Parent-offspring clash? - you continue to want a few context to place them in. From populations to speciation and polymorphism to evolutionary psychology, here’s the one-stop resource for all you want to recognize. Evolution unlocks the laboratory of existence, dissecting it into the 50 most important subject matters that supply the lacking hyperlinks to appreciate the common world’s four-billion-year ancestry and the method of ordinary choice within which species both adapt in myriad methods - mutation, ingenuity, and intelligence - to satisfy the demanding situations of a altering atmosphere, or die. resolve the improvement of dwelling organisms, at micro and macro point - from genes to geniuses.

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C1 , . . , cq , 0) · TS∪{˜s} = ˜ 0, а из леммы 1 следует, что (c , . . , c , 0)т ∈ L⊥ (T ) = L⊥ (P ), 1 q S∪{˜ s} S∪{˜ s} Проблема Шёнберга: критерий вырожденности матрицы l1 -расстояний 43 поэтому (c1 , . . , cq , 0) · PS∪{˜s} = ˜0 и равенство (1) выполняется для точки s˜ (достаточно заметить, что скалярное произведение вектора (c1 , . . , cq , 0) и последнего столбца матрицы PS∪{˜s} равно нулю). Если равенство (1) выполнено для всех точек пространства, тогда, в частности, оно выполнено и для точек системы S, откуда следует сингулярность системы.

Vol(P), где vol(P) — объём многогранника P = P(P) = { (x1 , . . , xn ) ∈ Rn | 0 xi 1, vi vj ⇒ x i xj } . Понятно, что P(P) — выпуклый многогранник, заключённый в единичный куб E n ⊂ Rn и ограниченный гиперплоскостями xi = xj для всех сравнимых в P элементов vi и vj . Из вышеприведённого следуют способы нахождения чисел e(P ) : 1) подсчёт некоторых сохраняющих порядок отображений (или перестановок) — “прямой” метод; 2) подсчёт некоторых цепей; 3) оценка величины vol(P); 4) использование рекуррентных соотношений и производящих функций — стандартный подход в перечислительной комбинаторике.

109] и M = M (замыкание в сильной топологии), получим, что пара {a, a1 } ∈ M × M . Поэтому a1 , g = a, g1 для всех пар {g, g1 } ∈ A∗ . Равенство A = A = J(⊥ (A∗ )) влечет {a, a1 } ∈ A. 109] {a, a1 } ∈ AM . Таким образом, J(⊥ B) ⊂ AM . Отсюда и из равенства (⊥ B)⊥ = B ∗ ∗ ˜ ˜ получаем AM ⊂ B. Равенство AM = B доказано. О слабых решениях дифференциальных включений 29 Приступаем непосредственно к доказательству теоремы. Пусть y – слабое решение (1). Так как A ⊂ A, для y справедлива теорема 4. По теореме 2 функция v в равенстве (9) является слабым решением (1).

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