We solve problems

Problem #PRU-61297

Problems Calculus Number sequences Limit of a sequence, convergence Algebra and arithmetic Sequences Recurrent relations Recurrent relations (other)

15–16 3.0

The Babylonian algorithm for deducing $\sqrt{2}$ . The sequence of numbers ${x_{n}}$ is given by the following conditions: $x_{1} = 1$ , $x_{n + 1} = \frac{1}{2} (x_{n} + 2 / x_{n})$ ( $n \geq 1$ ).

Prove that $lim_{n \to \infty} x_{n} = \sqrt{2}$ .

Problem #PRU-61298

Problems Calculus Number sequences Limit of a sequence, convergence Algebra and arithmetic Sequences Recurrent relations Recurrent relations (other)

15–16 3.0

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?

Problem #PRU-61299

Problems Calculus Number sequences Limit of a sequence, convergence Algebra and arithmetic Sequences Recurrent relations Recurrent relations (other)

15–16 4.0

The iterative formula of Heron. Prove that the sequence of numbers ${x_{n}}$ given by the conditions $x_{1} = 1$ , $x_{n + 1} = \frac{1}{2} (x_{n} + k / x_{n})$ , converges. Find the limit of this sequence.

Problem #PRU-61304

Problems Calculus Functions of one variable. Continuity Continuous functions (general properties) Number sequences Limit of a sequence, convergence

15–16 3.0

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 $x * = lim_{n \to \infty} x_{n}$ , and the function $f (x)$ is continuous, then this limit is the root of the original equation: $f (x^{*}) = x^{*}$ .

Problem #PRU-61311

Problems Calculus Number sequences Limit of a sequence, convergence Algebra and arithmetic Sequences Recurrent relations Recurrent relations (other)

15–16 4.0

The algorithm of the approximate calculation of $\sqrt[3]{a}$ . The sequence ${a_{n}}$ is defined by the following conditions: $a_{0} = a > 0$ , $a_{n + 1} = 1 / 3 (2 a_{n} + a / a_{n}^{2})$ ( $n \geq 0$ ).

Prove that $lim_{n \to \infty} a_{n} = \sqrt[3]{a}$ .

Problem #PRU-61313

Problems Calculus Number sequences Limit of a sequence, convergence Algebra and arithmetic Sequences Recurrent relations Recurrent relations (other)

15–16 4.0

The sequence of numbers ${a_{n}}$ is given by $a_{1} = 1$ , $a_{n + 1} = 3 a_{n} / 4 + 1 / a_{n}$ ( $n \geq 1$ ). Prove that:

a) the sequence ${a_{n}}$ converges;

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

Problem #PRU-61314

Problems Calculus Number sequences Limit of a sequence, convergence Algebra and arithmetic Sequences Recurrent relations Recurrent relations (other)

15–16 4.0

Find the limit of the sequence that is given by the following conditions $a_{1} = 2$ , $a_{n + 1} = a_{n} / 2 + a_{n}^{2} / 8$ ( $n \geq 1$ ).

Problem #PRU-61324

Problems Calculus Number sequences Limit of a sequence, convergence Algebra and arithmetic Mean values Sequences Recurrent relations Recurrent relations (other)

15–16 3.0

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 $a_{0} = a$ , $b_{0} = b$ , $a_{n + 1} = \frac{2 a_{n} b_{n}}{a_{n} + b_{n}}$ , $b_{n + 1} = \sqrt{a_{n} b_{n}}$ ( $n \geq 0$ ).

We denote it by $ν (a, b)$ . Prove that $ν (a, b)$ is related to $μ (a, b)$ (see problem number 61322) by $ν (a, b) \times μ (1 / a, 1 / b) = 1$ .

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 $μ (a, b)$ .

Problem #PRU-79402

Problems Calculus Number sequences Limit of a sequence, convergence Algebra and arithmetic Trigonometry Trigonometrical inequalities

15–16 4.0

Prove that the sequence $x_{n} = \sin (n^{2})$ does not tend to zero for $n$ that tends to infinity.

Problem #PRU-61308

Problems Calculus Number sequences Limit of a sequence, convergence

15–16 3.0

The numbers $a_{1}, a_{2}, \dots, a_{k}$ are such that the equality $lim_{n \to \infty} (x_{n} + a_{1} x_{n - 1} + \dots + a_{k} x_{n - k}) = 0$ is possible only for those sequences ${x_{n}}$ for which $lim_{n \to \infty} x_{n} = 0$ . Prove that all the roots of the polynomial P $(λ) = λ^{k} + a_{1} λ^{k - 1} + a_{2} λ^{k - 2} + \dots + a_{k}$ are modulo less than 1.