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.
The algorithm of the approximate calculation of \(\sqrt[3]{a}\). The sequence \(\{a_n\}\) is defined by the following conditions: \(a_0 = a > 0\), \(a_{n + 1} = 1/3 (2a_n + a/a^2_n)\) (\(n \geq 0\)).
Prove that \(\lim\limits_{n\to\infty} a_n = \sqrt[3]{a}\).
The sequence of numbers \(\{a_n\}\) is given by \(a_1 = 1\), \(a_{n + 1} = 3a_n/4 + 1/a_n\) (\(n \geq 1\)). Prove that:
a) the sequence \(\{a_n\}\) converges;
b) \(|a_{1000} - 2| < (3/4)^{1000}\).
Find the limit of the sequence that is given by the following conditions \(a_1 = 2\), \(a_{n + 1} = a_n/2 + a_n^2/8\) (\(n \geq 1\)).
The sequence of numbers \(\{x_n\}\) is given by the following conditions: \(x_1 \geq - a\), \(x_{n + 1} = \sqrt{a + x_n}\). Prove that the sequence \(x_n\) is monotonic and bounded. Find its limit.
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 \(\mu (a, b)\).
Hannah placed 101 counters in a row which had values of 1, 2 and 3 points. It turned out that there was at least one counter between every two one point counters, at least two counters lie between every two two point counters, and at least three counters lie between every two three point counters. How many three point counters could Hannah have?
In a row there are 20 different natural numbers. The product of every two of them standing next to one another is the square of a natural number. The first number is 42. Prove that at least one of the numbers is greater than 16,000.