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.
Prove that for a monotonically increasing function \(f (x)\) the equations \(x = f (f (x))\) and \(x = f (x)\) are equivalent.
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 \(\mu (a, b)\).
Prove that the tangent to the graph of the function \(f (x)\), constructed at coordinates \((x_0, f (x_0))\) intersects the \(Ox\) axis at the coordinate: \(x_0 -\frac{f(x_0)}{f'(x_0)}\).
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.
Prove that if the function \(f (x)\) is convex upwards on the line \([a, b]\), then for any distinct points \(x_1, x_2\) in \([a; b]\) and for any positive \(\alpha_{1}, \alpha_{2}\) such that \(\alpha_{1} + \alpha_ {2} = 1\) the following inequality holds: \(f(\alpha_1 x_1 + \alpha_2 x_2 ) > \alpha_1 f (x_1) + \alpha_2 f(x_2)\).
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 \((1, 1, 1, 1)\).