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.
Old calculator I.
a) Suppose that we want to find \(\sqrt[3]{x}\) (\(x> 0\)) on a calculator that can find \(\sqrt{x}\) in addition to four ordinary arithmetic operations. Consider the following algorithm. A sequence of numbers \(\{y_n\}\) is constructed, in which \(y_0\) is an arbitrary positive number, for example, \(y_0 = \sqrt{\sqrt{x}}\), and the remaining elements are defined by \(y_{n + 1} = \sqrt{\sqrt{x y_n}}\) (\(n \geq 0\)).
Prove that \(\lim\limits_{n\to\infty} y_n = \sqrt[3]{x}\).
b) Construct a similar algorithm to calculate the fifth root.
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 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 \(\mu (a, b)\).
The Newton method (see Problem 61328) does not always allow us to approach the root of the equation \(f(x) = 0\). Find the initial condition \(x_0\) for the polynomial \(f(x) = x (x - 1)(x + 1)\) such that \(f(x_0) \neq x_0\) and \(x_2 = x_0\).