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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 \(a_1 = 2\), \(a_{n + 1} = a_n/2 + a_n^2/8\) (\(n \geq 1\)).

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

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 the polynomial \(P (x)\) is divisible by its derivative if and only if \(P (x)\) has the form \(P(x) = a_n(x - x_0)^n\).

Find the largest value of the expression \(a + b + c + d - ab - bc - cd - da\), if each of the numbers \(a\), \(b\), \(c\) and \(d\) belongs to the interval \([0, 1]\).