The sequence of numbers \(a_1, a_2, \dots\) is given by the conditions \(a_1 = 1\), \(a_2 = 143\) and
for all \(n \geq 2\).
Prove that all members of the sequence are integers.
Which term in the expansion \((1 + \sqrt 3)^{100}\) will be the largest by the Newton binomial formula?
Prove that in any infinite decimal fraction you can rearrange the numbers so that the resulting fraction becomes a rational number.
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?
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 \(\mu (a, b)\).