Prove that the function \(\cos \sqrt {x}\) is not periodic.
Prove that amongst any 7 different numbers it is always possible to choose two of them, \(x\) and \(y\), so that the following inequality was true: \[0 < \frac{x-y}{1+xy} < \frac{1}{\sqrt3}.\]
The Babylonian algorithm for deducing \(\sqrt{2}\). The sequence of numbers \(\{x_n\}\) is given by the following conditions: \(x_1 = 1\), \(x_{n + 1} = \frac 12 (x_n + 2/x_n)\) (\(n \geq 1\)).
Prove that \(\lim\limits_{n\to\infty} x_n = \sqrt{2}\).
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?
The iterative formula of Heron. Prove that the sequence of numbers \(\{x_n\}\) given by the conditions \(x_1 = 1\), \(x_{n + 1} = \frac 12 (x_n + k/x_n)\), converges. Find the limit of this sequence.
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: \(f (x ^*) = x^*\).
The numbers \(a_1, a_2, \dots , a_k\) are such that the equality \(\lim\limits_{n\to\infty} (x_n + a_1x_{n - 1} + \dots + a_kx_{n - k}) = 0\) is possible only for those sequences \(\{x_n\}\) for which \(\lim\limits_{n\to\infty} x_n = 0\). Prove that all the roots of the polynomial P \((\lambda) = \lambda^k + a_1 \lambda^{k-1} + a_2 \lambda^{k -2} + \dots + a_k\) are modulo less than 1.
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 \(a_1 = 2\), \(a_{n + 1} = a_n/2 + a_n^2/8\) (\(n \geq 1\)).