Prove that the function \(\cos \sqrt {x}\) is not periodic.
Prove the formulae: \(\arcsin (- x) = - \arcsin x\), \(\arccos (- x) = \pi - \arccos x\).
Prove that amongst any 7 different numbers it is always possible to choose two of them, \(x\) and \(y\), so that the following inequality was true: \[0 < \frac{x-y}{1+xy} < \frac{1}{\sqrt3}.\]
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.
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?