The function F is given on the whole real axis, and for each x the equality holds: F $(x + 1)$ F $(x)$ + F $(x + 1)$ + 1 = 0.
Prove that the function F can not be continuous.
Does there exist a number h such that for any natural number n the number [$h \times 1969^n$] is not divisible by [$h \times 1969^{n-1}$]?
Gary drew an empty table of $50 \times 50$ and wrote on top of each column and to the left of each row a number. It turned out that all 100 written numbers are different, and 50 of them are rational, and the remaining 50 are irrational. Then, in each cell of the table, he wrote down a product of numbers written at the top of its column and to the left of the row $($ the “multiplication table” $)$. What is the largest number of products in this table which could be rational numbers?
In the Republic of mathematicians, the number α $>$ 2 was chosen and coins were issued with denominations of 1 pound, as well as in $α^k$ pounds for every natural k. In this case α was chosen so that the value of all the coins, except for the smallest, was irrational. Could it be that any amount of a natural number of pounds can be made with these coins, using coins of each denomination no more than 6 times?
Let us show that the mathematicians could choose a number
– the root of the equation $α^2$ + α = 7. It is clear that α $>$ 2. It is easy to see that for a natural m,
where $a_m$ and $b_m$ are integers, and $a_m$ $<$ 0 $<$ $b_m$ for odd m and $a_m$ $>$ 0 $>$ $b_m$ for even m. Hence, the number $α_m$ is irrational.
It remains to show that for any natural number n the sum of n pounds can be made in the required way. Consider all the ways to collect n pounds in coins issued $($at least one such method exists: you can take n pound coins$)$. Choose from them the way in which the smallest number of coins is used. Suppose that some coin of value $α^k$ $(k ≥ 0)$ occurs using this method at least 7 times. Then you can replace 7 $α^k$ coins with coins of the values of $α^{k + 1}$ and $α^{k + 2}$. The total value of the coins will not change $(α^{k + 1} + α^{k + 2} = 7α^k)$, and their number will decrease. This contradicts the result of our method.
As is not hard to see, for any desired α the sum of 7 pounds can be collected only in the following way where 7 = α + α². Hence, the presented value of α is unique.
It could.
The equations $ax^2$ + bx + c = 0 $(1)$ and – $ax^2$ + bx + c $(2)$ are given. Prove that if $x_1$ and $x_2$ are, respectively, any roots of the equations $(1)$ and $(2)$, then there is a root $x_3$ of the equation ½ $ax^2$ + bx + c such that either $x_1$ $≤$ $x_3$ $≤$ $x_2$ or $x_1$ $≥$ $x_3$ $≥$ $x_2$.
Let f $(x)$ = 1/2 $ax^2$ + bx + c.
These values have different signs, so one of the roots of the trinomial is located between $x_1$ and $x_2$.
The triangle $C_1C_2O$ is given. Within it the bisector $C_2C_3$ is drawn, then in the triangle $C_2C_3O$ – bisector $C_3C_4$ and so on. Prove that the sequence of angles $γ_n$ = $C_{n + 1}C_nO$ tends to a limit, and find this limit if $C_1OC_2$ = α.
It follows from the condition that $C_{n + 1}C_{n + 2}$ is the bisector of the angle of the triangle $C_nC_{n + 1}O$ $($ see Fig. $)$, So $2γ_{n + 1}$ + $γ_n$ + α = π. $(1)$
Let us prove that $γ_n$ tends to the limit β = $(π-α)$ / 3. We rewrite $(1)$ thus: 2 $(γ_{n + 1} – β)$ = β – $γ_n$. It follows that | $γ_{n + 1}$ – β | = ½ | $γ_n$ – β |, that is, the difference between $γ_n$ and β in the transition from n to n + 1 is halved. Therefore, | $γ_n$ – β | = $2^{1-n}$ | $γ_1$ – β | tends to zero.
$(π-α)$ / 3.
Author: V.A. Popov
On the interval [0; 1] a function f is given. This function is non-negative at all points, f $(1)$ = 1 and, finally, for any two non-negative numbers $x_1$ and $x_2$ whose sum does not exceed 1, the quantity $f (x_1 + x_2)$ does not exceed the sum of $f (x_1)$ and $f (x_2 )$.
a) Prove that for any number x on the interval [0; 1], the inequality $f (x_2) ≤ 2x$ holds.
b) Prove that for any number x on the interval [0; 1], the $f (x_2) ≤ 1.9x$ must be true?
a) A function satisfying the condition of the problem increases monotonically. In fact, if $x_2 \geq x_1$ and $x_2 \geq 1$, then $f (x_2) \geq f (x_1) + f (x_2-x_1)$ and $f (x_2-x_1) \geq 0$; consequently, $f (x_2) \geq f (x_1)$. Therefore, if $½ $<$x \geq 1$, since $f (1) = 1$, we have $f (x) \leq 1 \leq 2x$.
Now let $0 $<$x \leq ½$. We take a natural number n such that $½ $<$nx \leq 1$ $($it is clear that this can always be done$)$. It is easy to prove $($by induction$)$ that if $x \geq 0$ and $nx \leq 1$, then $f (nx) \geq nf (x)$. For $½ $<$nx \leq 1$, it is already proved that $f (nx) \leq 2 \times nx$.
Therefore, $nf (x) \leq f (nx) \leq 2 \times nx$, whence $f (x) \leq 2x$ and for $0 $<$x \leq ½$. Now, it only remains to find out what will happen at zero. We take $0 $<$x_1 \leq 1$; then $f (x_1) = f (0 + x_1) \geq f (0) + f (x_1)$.
But by definition $f (0) \geq 0$, therefore, $f (0) = 0$.
b) The answer: no, it is not true.
Let us give an example. We define the function as follows: f $($x$)$ = 
If $1 \geq x_1 \geq x_2 \geq 0$ and $x_1 + x_2 \leq 1$, then $x_2 \leq ½$ and $f (x_2) = 0$. Consequently, $f (x_1 + x_2) \geq f (x_1) + f (x_2)$ and the function in question satisfies the requirements of the problem. At the same time, $f (\frac{51}{100}) = 1> 1.9 \times \frac{51}{100}$.
b) No, it is not true.
The Newton method $($see Problem 61328$)$ does not always allow us to approach the root of the equation f $($x$)$ = 0. Find the initial condition $x_0$ for the polynomial f $(x)$ = x $(x – 1)$ $(x + 1)$ such that f $(x_0)$ $\neq$
$x_0$ and $x_2$ = $x_0$.
Let p and q be nonzero real numbers and $p^2$ – 4q $>$ 0. Prove that the following sequences converge:
a$)$ $y_0$ = 0, $y_{n + 1} = \frac{q}{p-y_n} (n \geq 0)$;
b$)$ $z_0$ = 0, $z_{n + 1}$ = p – q/$z_n$ $( n \geq 0 )$.
Establish a connection between the limiting values of these sequences y *, z * and the roots of the equation $x^2$ – px + q = 0.
Cut the interval [-1, 1] into black and white segments so that the integrals of any a$)$ linear function; b$)$ a square trinomial in white and black segments are equal.
Old calculator II. The derivative of the function ln x for x = 1 is 1. Hence
$\lim\limits_{x\to0}$
= $\lim\limits_{x\to0}$
= 1.
Use this fact to approximate the natural logarithm of the number N. As in Problem 61302, the standard arithmetic operations and the square root extraction operation are allowed.
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 $\sqrt{2}$.
Let f $(x)$ be a polynomial of degree n with roots $α_1$, …, $α_n$. We define the polygon M as the convex hull of the points $α_1$, …, $α_n$ on the complex plane. Prove that the roots of the derivative of this polynomial lie inside the polygon M.
Let f $(x)$ = $(x – a)$ $(x – b)$ $(x – c)$ be a polynomial of the third degree with complex roots a, b, c.
Prove that the roots of the derivative of this polynomial lie inside the triangle with vertices at the points a, b, c.
Use the equality
and the result of task 61134.
Use the equality
and the result of task 61134.
Using the theorem on rational roots of a polynomial $($ see Problem 61013 $)$, prove that if p / q is rational and cos $(p / q)$ $^{\circ}$ ≠ 0, ± 1, ± 1, then
cos $(p / q)$ $^{\circ}$ is irrational.
Prove that there are infinitely many composite numbers among the numbers [$2^k$ $\sqrt{2}$] $(k = 0, 1, …)$.
Prove that for any infinite continued fraction [$a_0$; $a_1$, …, $a_n$, …] there exists a limit of its suitable fractions – an irrational number α. Explain why if this number α is decomposed into an infinite continued fraction by the algorithm of task 60606, then an infinite continued fraction is obtained.
$x_1$ is the real root of the equation $x^2$ + ax + b = 0, $x_2$ is the real root of the equation $x^2$ – ax – b = 0.
Prove that the equation $x^2$ + 2ax + 2b = 0 has a real root, enclosed between $x_1$ and $x_2$. $($ a and b are real numbers $)$.
Prove that for any natural number $a_1> 1$ there exists an increasing sequence of natural numbers $a_1, a_2, a_3$, …, for which $a_1^2+ a_2^2 +…+ a_k^2$ is divisible by $a_1+ a_2+…+ a_k$ for all k ≥ 1.
Let us prove that for any numbers $a_1, …, a_n$ satisfying the condition of the problem, one can find a number $a_{n + 1}$ such that $S_{n+1} = a_1^2+ …+ a_{n+1}^2$
is divisible by $s_{n + 1} = a_1 + … + a_n + a_{n + 1}$. It follows from the equality 
that $S_{n + 1}$ is divisible by $s_{n + 1}$ if $S_n+ s_n^2$ is divisible by $s_{n + 1}$, since $a_{n + 1} + s_n = s_{n + 1}$. Thus, it suffices to take 
– in this case $S_n+ s_n^2 = s_{n+1}$. In this case 
See the solution above.
