On an infinitely long strip of paper, we write an endless row of digits.
We start by writing \(1,2,3,4\). After that, each new digit is chosen like this: add the previous four digits and write down only the last digit of that sum.
So the beginning looks like \(1234096\dots\).
Will the four digits \(8123\) ever appear next to each other somewhere in this endless row?
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 sequence of numbers \(a_n\) is given by the conditions \(a_1 = 1\), \(a_{n + 1} = a_n + 1/a^2_n\) (\(n \geq 1\)).
Is it true that this sequence is limited?
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 \(\mu (a, b)\).