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: \(f (x ^*) = x^*\).
An iterative polyline serves as a geometric interpretation of the iteration process. To construct it, on the \(Oxy\) plane, the graph of the function \(f (x)\) is drawn and the bisector of the coordinate angle is drawn, as is the straight line \(y = x\). Then on the graph of the function the points \[A_0 (x_0, f (x_0)), A_1 (x_1, f (x_1)), \dots, A_n (x_n, f (x_n)), \dots\] are noted and on the bisector of the coordinate angle – the points \[B_0 (x_0, x_0), B_1 (x_1, x_1), \dots , B_n (x_n, x_n), \dots.\] The polygonal line \(B_0A_0B_1A_1 \dots B_nA_n \dots\) is called iterative.
Construct an iterative polyline from the following information:
a) \(f (x) = 1 + x/2\), \(x_0 = 0\), \(x_0 = 8\);
b) \(f (x) = 1/x\), \(x_0 = 2\);
c) \(f (x) = 2x - 1\), \(x_0 = 0\), \(x_0 = 1{,}125\);
d) \(f (x) = - 3x/2 + 6\), \(x_0 = 5/2\);
e) \(f (x) = x^2 + 3x - 3\), \(x_0 = 1\), \(x_0 = 0{,}99\), \(x_0 = 1{,}01\);
f) \(f (x) = \sqrt{1 + x}\), \(x_0 = 0\), \(x_0 = 8\);
g) \(f (x) = x^3/3 - 5x^2/x + 25x/6 + 3\), \(x_0 = 3\).
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?
The numbers \(a_1, a_2, \dots , a_k\) are such that the equality \(\lim\limits_{n\to\infty} (x_n + a_1x_{n - 1} + \dots + a_kx_{n - k}) = 0\) is possible only for those sequences \(\{x_n\}\) for which \(\lim\limits_{n\to\infty} x_n = 0\). Prove that all the roots of the polynomial P \((\lambda) = \lambda^k + a_1 \lambda^{k-1} + a_2 \lambda^{k -2} + \dots + a_k\) are modulo less than 1.
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 that the tangent to the graph of the function \(f (x)\), constructed at coordinates \((x_0, f (x_0))\) intersects the \(Ox\) axis at the coordinate: \(x_0 -\frac{f(x_0)}{f'(x_0)}\).
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 \(x_2 = x_0\).
The sequence of numbers \(a_1, a_2, a_3, \dots\) is given by the following conditions \(a_1 = 1\), \(a_{n + 1} = a_n + \frac {1} {a_n^2}\) (\(n \geq 0\)).
Prove that
a) this sequence is unbounded;
b) \(a_{9000} > 30\);
c) find the limit \(\lim \limits_ {n \to \infty} \frac {a_n} {\sqrt [3] n}\).
We are given rational positive numbers \(p, q\) where \(1/p + 1/q = 1\). Prove that for positive \(a\) and \(b\), the following inequality holds: \(ab \leq \frac{a^p}{p} + \frac{b^q}{q}\).