It is known that \(\cos \alpha^{\circ} = 1/3\). Is \(\alpha\) a rational number?
Let \(f (x)\) be a polynomial of degree \(n\) with roots \(\alpha_1, \dots , \alpha_n\). We define the polygon \(M\) as the convex hull of the points \(\alpha_1, \dots , \alpha_n\) on the complex plane. Prove that the roots of the derivative of this polynomial lie inside the polygon \(M\).
a) Using geometric considerations, prove that the base and the side of an isosceles triangle with an angle of \(36^{\circ}\) at the vertex are incommensurable.
b) Invent a geometric proof of the irrationality of \(\sqrt{2}\).
Let \(z_1\) and \(z_2\) be fixed points of a complex plane. Give a geometric description of the sets of all points \(z\) that satisfy the conditions:
a) \(\operatorname{arg} \frac{z - z_1}{z - z_2} = 0\);
b) \(\operatorname{arg} \frac{z_1 - z}{z - z_2} = 0\).
Prove that the function \(\cos \sqrt {x}\) is not periodic.
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.
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: \(f (x ^*) = x^*\).