It is known that \(\cos \alpha^{\circ} = 1/3\). Is \(\alpha\) a rational number?
Let \(a, b\) be positive integers and \((a, b) = 1\). Prove that the quantity cannot be a real number except in the following cases \((a, b) = (1, 1)\), \((1,3)\), \((3,1)\).
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 \(M\).
For what values of \(n\) does the polynomial \((x+1)^n - x^n - 1\) divide by:
a) \(x^2 + x + 1\); b) \((x^2 + x + 1)^2\); c) \((x^2 + x + 1)^3\)?
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 \(\sqrt{2}\).
Find the largest and smallest values of the functions
a) \(f_1 (x) = a \cos x + b \sin x\); b) \(f_2 (x) = a \cos^2x + b \cos x \sin x + c \sin^2x\).
Prove that the function \(\cos \sqrt {x}\) is not periodic.
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.
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: \(ab \leq \frac{a^p}{p} + \frac{b^q}{q}\).