The algorithm of the approximate calculation of \(\sqrt[3]{a}\). The sequence \(\{a_n\}\) is defined by the following conditions: \(a_0 = a > 0\), \(a_{n + 1} = 1/3 (2a_n + a/a^2_n)\) (\(n \geq 0\)).
Prove that \(\lim\limits_{n\to\infty} a_n = \sqrt[3]{a}\).
The sequence of numbers \(\{a_n\}\) is given by \(a_1 = 1\), \(a_{n + 1} = 3a_n/4 + 1/a_n\) (\(n \geq 1\)). Prove that:
a) the sequence \(\{a_n\}\) converges;
b) \(|a_{1000} - 2| < (3/4)^{1000}\).
Find the limit of the sequence that is given by the following conditions \(a_1 = 2\), \(a_{n + 1} = a_n/2 + a_n^2/8\) (\(n \geq 1\)).
The sequence of numbers \(\{x_n\}\) is given by the following conditions: \(x_1 \geq - a\), \(x_{n + 1} = \sqrt{a + x_n}\). Prove that the sequence \(x_n\) is monotonic and bounded. Find its limit.
We call the geometric-harmonic mean of numbers \(a\) and \(b\) the general limit of the sequences \(\{a_n\}\) and \(\{b_n\}\) constructed according to the rule \(a_0 = a\), \(b_0 = b\), \(a_{n + 1} = \frac{2a_nb_n}{a_n + b_n}\), \(b_{n + 1} = \sqrt{a_nb_n}\) (\(n \geq 0\)).
We denote it by \(\nu (a, b)\). Prove that \(\nu (a, b)\) is related to \(\mu (a, b)\) (see problem number 61322) by \(\nu (a, b) \times \mu (1/a, 1/b) = 1\).
Problem number 61322 says that both of these sequences have the same limit.
This limit is called the arithmetic-geometric mean of the numbers \(a, b\) and is denoted by \(\mu (a, b)\).
The Newton method (see Problem 61328) does not always allow us to approach the root of the equation \(f(x) = 0\). Find the initial condition \(x_0\) for the polynomial \(f(x) = x (x - 1)(x + 1)\) such that \(f(x_0) \neq x_0\) and \(x_2 = x_0\).
The sequence of numbers \(a_1, a_2, a_3, \dots\) is given by the following conditions \(a_1 = 1\), \(a_{n + 1} = a_n + \frac {1} {a_n^2}\) (\(n \geq 0\)).
Prove that
a) this sequence is unbounded;
b) \(a_{9000} > 30\);
c) find the limit \(\lim \limits_ {n \to \infty} \frac {a_n} {\sqrt [3] n}\).
Prove the inequality: \[\frac{(b_1 + \dots b_n)^{b_1 + \dots b_n}}{(a_1 + \dots a_n)^{b_1 + \dots + b_n}}\leq \left(\frac{b_1}{a_1}\right)^{b_1}\dots \left( \frac{b_n}{a_n}\right)^{b_n}\] where all variables are considered positive.
Inequality of Jensen. Prove that if the function \(f (x)\) is convex upward on \([a, b]\), then for any distinct points \(x_1, x_2, \dots , x_n\) (\(n \geq 2\)) from \([a; b]\) and any positive \(\alpha_{1}, \alpha_{2}, \dots , \alpha_{n}\) such that \(\alpha_ {1} + \alpha_{2} + \dots + \alpha_{n} = 1\), the following inequality holds: \(f (\alpha_{1} x_1 + \dots + \alpha_{n} x_n) > \alpha_{1} f (x_1) + \dots + \alpha_{n} f (x_n)\).
Draw all of the stairs made from four bricks in descending order, starting with the steepest \((4, 0, 0, 0)\) and ending with the shallowest \((1, 1, 1, 1)\).