Problems

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Found: 16

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.

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) \(|a_{1000} - 2| < (3/4)^{1000}\).

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.

Author: I.I. Bogdanov

Peter wants to write down all of the possible sequences of 100 natural numbers, in each of which there is at least one 3, and any two neighbouring terms differ by no more than 1. How many sequences will he have to write out?

Author: I.I. Bogdanov

Peter wants to write down all of the possible sequences of 100 natural numbers, in each of which there is at least one 4 or 5, and any two neighbouring terms differ by no more than 2. How many sequences will he have to write out?

At a contest named “Ah well, monsters!”, 15 dragons stand in a row. Between neighbouring dragons the number of heads differs by 1. If the dragon has more heads than both of his two neighbors, he is considered cunning, if he has less than both of his neighbors – strong, the rest (including those standing at the edges) are considered ordinary. In the row there are exactly four cunning dragons – with 4, 6, 7 and 7 heads and exactly three strong ones – with 3, 3 and 6 heads. The first and last dragons have the same number of heads.

a) Give an example of how this could occur.

b) Prove that the number of heads of the first dragon in all potential examples is the same.

Author: G. Zhukov

The square trinomial \(f (x) = ax^2 + bx + c\) that does not have roots is such that the coefficient \(b\) is rational, and among the numbers \(c\) and \(f (c)\) there is exactly one irrational.

Can the discriminant of the trinomial \(f (x)\) be rational?

At what value of \(k\) is the quantity \(A_k = (19^k + 66^k)/k!\) at its maximum?