By Vladimir I. Arnol'd
В книге написанной на основе лекции для студентов, рассказывается о рождении современной математики и теоретической физики в трудах великих ученых XVII века. Некоторые идеи Ньютона и Гюйгенса опередили свое время на несколько столетий и получили развитие только в последние годы. Об этих идеях, включая несколько новых результатов, также расказано в книге.
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Extra resources for Huygens and Barrow, Newton and Hooke: pioneers in mathematical analysis and catastrophe theory from evolvents to quasicrystals
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21 for an application. 24. 1. Construct an analytical proof of Young’s inequality. 2. Assume a, b, c, d > 0 and prove the following: (a) a2 + b2 ≥ 2ab. (b) a4 + b4 ≥ 2a2 b2 . (c) a2 + b2 + c2 ≥ ab + bc + ca. (d) a4 + b4 + c4 + d4 ≥ 4abcd. (e) (a + b)(b + c)(c + a) ≥ 8abc. 3. Use AM–GM to show that n! < n+1 2 n . Prove also that if n > 1 is a natural number, then [45] (2n − 1)!! , where (2n − 1)!! = (2n − 1) · (2n − 3) · (2n − 5) · · · · · 5 · 3 · 1 (2n)!! = (2n) · (2n − 2) · (2n − 4) · · · · · 4 · 2.
3. If f (x) and g(x) are integrable on [a, b] with f (x) ≤ g(x), then b a b f (x) dx ≤ g(x) dx. a 22 2. Methods from the Calculus Proof. With the notation described above, we form Riemann sums: n n f (xi )∆x ≤ i=1 g(xi )∆x. 1. 1 (Simple Estimate). If f (x) is integrable on [a, b] with m ≤ f (x) ≤ M , then b m(b − a) ≤ f (x) dx ≤ M (b − a). 2 (Modulus Inequality). If f (x) is integrable on [a, b], then b b f (x) dx ≤ a |f (x)| dx. a The second corollary follows from the inequalities −|f (x)| ≤ f (x) ≤ |f (x)| and plays the role of the triangle inequality for integrals.
Underestimating an integral. By the quotient rule for differentiation, d dx 1 x x f (u) du 0 ≥0 as required. It is possible to obtain other inequalities involving integrals through an ad hoc consideration of areas bounded by various plane curves. This simple process, reminiscent of the integral test from calculus, is probably best described in an example. 12. The function f (x) = xp , where −1 < p < 0, is strictly decreasing. 4 it is apparent that n+1 1 n xp dx < n kp < xp dx. 0 k=1 Hence, after carrying out the integrations, (n + 1)p+1 − 1 < p+1 n kp < k=1 np+1 .