Problems

Age
Difficulty
Found: 24

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.

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) \(|a_{1000} - 2| < (3/4)^{1000}\).

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 \(\mu (a, b)\).

For which natural \(n\) does the number \(\frac{n^2}{1.001^n}\) reach its maximum value?

We consider a sequence of words consisting of the letters “A” and “B”. The first word in the sequence is “A”, the \(k\)-th word is obtained from the \((k-1)\)-th by the following operation: each “A” is replaced by “AAB” and each “B” by “A”. It is easy to see that each word is the beginning of the next, thus obtaining an infinite sequence of letters: AABAABAAABAABAAAB...

a) Where in this sequence will the 1000th letter “A” be?

b) Prove that this sequence is non-periodic.