What will the sequence from the previous problem 61297 be converging towards if we choose \(x_1\) as equal to \(-1\) as the initial condition?
Old calculator I.
a) Suppose that we want to find \(\sqrt[3]{x}\) (\(x> 0\)) on a calculator that can find \(\sqrt{x}\) in addition to four ordinary arithmetic operations. Consider the following algorithm. A sequence of numbers \(\{y_n\}\) is constructed, in which \(y_0\) is an arbitrary positive number, for example, \(y_0 = \sqrt{\sqrt{x}}\), and the remaining elements are defined by \(y_{n + 1} = \sqrt{\sqrt{x y_n}}\) (\(n \geq 0\)).
Prove that \(\lim\limits_{n\to\infty} y_n = \sqrt[3]{x}\).
b) Construct a similar algorithm to calculate the fifth root.
Method of iterations. In order to approximately solve an equation, it is allowed to write \(f (x) = x\), by using the iteration method. First, some number \(x_0\) is chosen, and then the sequence \(\{x_n\}\) is constructed according to the rule \(x_{n + 1} = f (x_n)\) (\(n \geq 0\)). Prove that if this sequence has the limit \(x * = \lim \limits_ {n \to \infty} x_n\), and the function \(f (x)\) is continuous, then this limit is the root of the original equation: \(f (x ^*) = x^*\).
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 \(\lim \limits_ {n \to \infty} \frac {a_n} {\sqrt [3] n}\).
We are given rational positive numbers \(p, q\) where \(1/p + 1/q = 1\). Prove that for positive \(a\) and \(b\), the following inequality holds: \(ab \leq \frac{a^p}{p} + \frac{b^q}{q}\).
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: \(f (\alpha_{1} x_1 + \dots + \alpha_{n} x_n) > \alpha_{1} f (x_1) + \dots + \alpha_{n} f (x_n)\).
Let \(p\) and \(q\) be positive numbers where \(1 / p + 1 / q = 1\). Prove that \[a_1b_1 + a_2b_2 + \dots + a_nb_n \leq (a_1^p + \dots a_n^p)^{1/p}(b_1^q +\dots + b_n^q)^{1/q}\] The values of the variables are considered positive.