Prove that the polynomial \(x^{2n} - nx^{n + 1} + nx^{n - 1} - 1\) for \(n > 1\) has a triple root of \(x = 1\).
Let \(a, b\) be positive integers and \((a, b) = 1\). Prove that the quantity cannot be a real number except in the following cases \((a, b) = (1, 1)\), \((1,3)\), \((3,1)\).
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) \((x^2 + x + 1)^3\)?
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.
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: \(f (x ^*) = x^*\).