Kvant 1971, No. 4 · Pages

Popular science physics and mathematics
quantum
4
1971
magazine
Academy
Sciences of the USSR
and the Academy of Pedagogical
Sciences of the USSR

In the room:

Economics and linear inequalities 1 A. B. Katok
Ohm’s law 8 Ya. A. Smorodinsky
Logical problems and propositional algebra 14 L. L. Tsinman
Fundamentals of vortex theory 21 N. E. Zhukovsky
Doppler effect 30 L. G. Aslamazov

Problem «Kvant» 33
Solutions to problems M28—M30, F41—F47 35
Student’s workshop Trigonometric inequalities 43
Entrance exams in mathematics at the Moscow Institute of Electronic Engineering 51
V. B. Demyanov
Information Correspondence mathematical tournament 56
Olympics, Olympics 57
Reviews. Bibliography On both sides of the mirror 58
Collector’s Corner USSR - the birthplace of astronautics 60
Answers, directions, solutions 62
«Kvant» for younger schoolchildren, 3rd cover page
Crossword 4th cover page

inequality $x_2 \geq 0$ form a half-space lying “behind” the plane $x_1Ox_3$, and points satisfying inequality $x_3 \geq 0$ form a half-space lying above the plane $x_1Ox_2$.

Thus, the points $(x_1, x_2, x_3)$, for which all three inequalities $x_1 \geq 0, x_2 \geq 0, x_3 \geq 0$ are simultaneously satisfied, lie in the intersection of the three described half-spaces, that is, in the positive edge (“1/8 part”) of the space (see Fig. 1).

The points $(x_1, x_2, x_3)$, for which $a_{11}x_1 + a_{12}x_2 + a_{13}x_3 = p_1$, form a plane passing through the points $P_1 = \left(\frac{p_1}{a_{11}}, 0, 0\right)$ on the $Ox_1$ axis, $Q_1 = \left(0, \frac{p_1}{a_{12}}, 0\right)$ on the $Ox_2$ axis, and $S_1 = \left(0, 0, \frac{p_1}{a_{13}}\right)$ on the $Ox_3$ axis. Points for which $a_{11}x_1 + a_{12}x_2 + a_{13}x_3 \leq p_1$, (3)

lie in one of the half-spaces into which this plane divides space. Since $p_1 > 0$, then the coordinates of the point $O = (0, 0, 0)$ satisfy inequality (3). Therefore, we need to choose the half-space $L_1$ that contains this point. Points $(x_1, x_2, x_3)$ whose coordinates satisfy the system of inequalities

$$\begin{cases} x_1 \geq 0, x_2 \geq 0, x_3 \geq 0, \\ a_{11}x_1 + a_{12}x_2 + a_{13}x_3 \leq p_1, \end{cases}$$

lie at the intersection of the half-space $L_1$ with a positive edge. This intersection is represented by an irregular tetrahedron $OP_1Q_1S_1$ (Fig. 2).

Points whose coordinates satisfy the inequality $a_{21}x_1 + a_{22}x_2 + a_{23}x_3 \leq p_2$ lie in the half-space $L_2$, which contains the point $O$ and is bounded by the plane passing through the points

$$P_2 = \left(\frac{p_2}{a_{21}}, 0, 0\right),$$

$$Q_2 = \left(0, \frac{p_2}{a_{22}}, 0\right),$$

$$S_2 = \left(0, 0, \frac{p_2}{a_{23}}\right).$$

Solutions to the system of inequalities

$$\begin{cases} x_1 \geq 0, x_2 \geq 0, x_3 \geq 0, \\ a_{11}x_1 + a_{12}x_2 + a_{13}x_3 \leq p_1, \\ a_{21}x_1 + a_{22}x_2 + a_{23}x_3 \leq p_2 \end{cases}$$

there will be an intersection of the tetrahedron $OP_1Q_1S_1$ with the half-space $L_2$ or, what is the same, the intersection of the tetrahedra $O_1P_1Q_1S_1$ and $OP_2Q_2S_2$. The following cases are possible:

a) All three points $P_2Q_2S_2$ lie at the intersection of the tetrahedron $OP_1Q_1S_1$ (this means $\frac{p_1}{a_{11}} \leq \frac{p_2}{a_{21}}$; $\frac{p_1}{a_{12}} \leq \frac{p_2}{a_{22}}$, $\frac{p_1}{a_{13}} \leq \frac{p_2}{a_{23}}$).

Then the intersection of the tetrahedra $OP_1Q_1S_1$ and $OP_2Q_2S_2$ coincides with the tetrahedron $OP_1Q_1S_1$ (Fig. 3, a).

b) Some of the points $P_2, Q_2, S_2$ lie inside, and non-tortic ones lie outside

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tetrahedron $OP_1Q_1S_1$ (this means that among the numbers

$$\frac{p_1}{a_{11}} - \frac{p_2}{a_{21}}, \frac{p_1}{a_{12}} - \frac{p_2}{a_{22}}, \frac{p_1}{a_{13}} - \frac{p_2}{a_{23}}$$

there is at least one positive and one negative). Then the plane $P_2Q_2S_2$ “cuts off the corner” of the tetrahedron $OP_1Q_1S_1$; the intersection of the tetrahedra $OP_1Q_1S_1$ and $OP_2Q_2S_2$ will be a polyhedron with six vertices (on Fig. 3, b the polyhedron $OP_2Q_2D_1D_2S_1$).

c) All three points $P_2$, $Q_2$, $S_2$ lie inside the tetrahedron $OP_1Q_1S_1$, that is

$$\frac{p_1}{a_{11}} \geq \frac{p_2}{a_{21}}, \frac{p_1}{a_{12}} \geq \frac{p_2}{a_{22}}, \frac{p_1}{a_{13}} \geq \frac{p_2}{a_{32}}.$$

In this case, the intersection of the tetrahedra $OP_1Q_1S_1$ and $OP_2Q_2S_2$ coincides with $OP_2Q_2S_2$.

By adding subsequent inequalities of system (2) to system (4), we see that each subsequent inequality either does not change in any way the set of solutions to the system of previous inequalities (as in case a)), or the constructed plane “cuts off” part of the polyhedron corresponding to the solution of the system (as in cases b) and c)).

Just like on a plane, a set of a point in space is called convex if, together with each property of two points, it entirely contains the segment connecting them. Since a half-space is a convex set and the general part is convex

convex sets, a very polyset, then the set of the point corresponding to the solution to system (2) is represented by the convex polyhedron $A$. However, from the vertices of this polyhedron there will be a point $O$, and the edges emanating from this vertex will be non-tortic segments of the coordinate axes. Note that, in contrast to what we have when solving the problem “about chicken feed,” the set of admissible points in this case turned out to be limited.

Let us call a plane supporting a convex set in space if this set lies entirely in one of the half-spaces on which this plane divides the space, and at the same time at least one point of the polyset lies on the plane.

Consider a family of parallel planes

$$x_1 + x_2 + x_3 = K.$$

For $K < 0$ these planes do not intersect the positive octane and, therefore, the polyhedron $A$. Plane

$$x_1 + x_2 + x_3 = 0$$

is the support to the polyhedron $A$ and intersects with it at a single point (0, 0, 0). The entire polyhedron $A$ lies in the half-space $x_1 + x_2 + x_3 \geq 0$. The point (0, 0, 0) corresponds to the minimum of the linear function

5

$f(x) = x_1 + x_2 + x_3$ defined on the polyhedron $A$. As $K$ increases further, the plane $x_1 + x_2 + x_3 = K$ will intersect the polyhedron $A$ along the polygon.

There is a value (Fig. 4) $K = K_0$ such that the plane $x_1 + x_2 + x_3 = K_0$ will again support the polyhedron $A$, but the polyhedron will lie in the half-space $x_1 + x_2 + x_3 \leq K_0$.

Obviously, the linear function $f(x) = x_1 + x_2 + x_3$ reaches a maximum on the polyhedron $A$ at all points of intersection of the polyhedron $A$ with the plane $x_1 + x_2 + x_3 = K_0$. As you can see, in the family of parallel planes given by equations (5), there are two planes supporting the polyhedron $A$, one of which corresponds to the minimum, and the other to the desired maximum of the function $x_1 + x_2 + x_3$ on $A$. The intersection of a convex polyhedron with a reference plane can consist of either a single vertex of the polygon, or points of some edge of the polyhedron, or points of some face of the polyhedron. Each of these three cases can happen in our situation.

The faces of the polyhedron $A$ lie in the plane
$$x_i = 0 \quad (i = 1, 2, 3),$$
$$a_{j1}x_1 + a_{j2}x_2 + a_{j3}x_3 = p_j$$
$(j = 1, \dots, n)$

and not necessarily in all of them. Each edge lies at the intersection of two

Рис. 4.

of these planes, and each vertex is at the intersection of several (three or more) planes. Further, more than three planes intersect at a vertex; you can select three of these planes, the intersection of which consists only of this vertex. The coordinates of the intersection point of three planes can be found by solving the system of equations given by these planes.

So, you can search for the vertices of the polyhedron $A$ as follows. It is possible to compose all body triples of equations from $(n+3)$ equations (6). (The number of such tracks is $\frac{(n+3)(n+2)(n+1)}{6}$).

Each triple can be considered as a system of three linear equations with the third unknown. If a solution $(x_1, x_2, x_3)$ of this system exists and is unique, then it is necessary to check whether the remaining $n$ inequalities of system (2) are satisfied for the numbers $x_1, x_2, x_3$. If the inequalities are satisfied, then the point $(x_1, x_2, x_3)$ is the vertex of the polyhedron $A$. After this, to solve the linear programming problem, it is enough to compare the values ​​of the functions $f(x) = x_1 + x_2 + x_3$ in all the resulting vertices and select those in which this value is the most accurate. This procedure involves quite a lot of computational work, especially if $n$ is large enough.

In fact, the solutions to the majority of the system will not be the vertices of the polyhedron $A$, and, in addition, to solve a linear programming problem it is not necessary to know all the vertices of the polyhedron. Therefore, such an indefinite search is not used when solving a linear programming problem.

A nanomore widespread and economical way of solving the linear programming problem was thought up in 1947 by American mathematicians J. Von Neumann and J. Danzg, who did not then know about the work of L.V. Kantorovich and independently came to the method of linear programming. They

6

They called their method the simplex method. We will now describe the geometric idea of ​​the simplex method as applied to our three-dimensional problem. In fact, when using the simplex method, the problem is transformed by introducing some additional unknowns, which makes it possible to simplify the solution process from an algebraic point of view. However, to preserve geometric clarity, we will not do this.

So let’s start from the top
$$O = (0, 0, 0),$$

of whose existence we are sure, and consider the edges of the polyhedron $A$ emerging from this vertex. Let $P$ be the second end of one of these edges. Function value
$$f(x) = x_1 + x_2 + x_3$$

at point $P$ is greater than at O. Let us now consider all the edges of the polyhedron $A$ extending from point $P$. When moving from point $P$ along some of these edges, the value of function $x_1 + x_2 + x_3$ may kill or remain constant. We will immediately discard these edges and select one of the edges, when moving along which the value of the function $x_1 + x_2 + x_3$ increases. Let $P'$ be the second end of this edge. The value of the function $x_1 + x_2 + x_3$ at vertex $P'$ is greater than at vertex $P$. Let’s consider all the edges leaving the point $P'$ and select some edge $P'P''$, when moving along which the value of the function $x_1 + x_2 + x_3$ increases, etc.

At each step we will obtain a vertex of the polyhedron $A$, in which the value of the function $x_1 + x_2 + x_3$ is greater than in the previous ones. In the end we will come to such a peak
$$M = (x_1^0, x_2^0, x_3^0),$$
that when moving along any edge emerging from this vertex, the values ​​of the function $x_1 + x_2 + x_3$ will not arise. This means that the plane
$$x_1 + x_2 + x_3 = x_1^0 + x_2^0 + x_3^0,$$

support to the polyhedron $A$ and $M$ is one of the solutions to the linear programming problem. To find other solutions, you need to find edges coming out of $M$ such that when moving along which the value of the function $x_1 + x_2 + x_3$ remains constant.

If there are no such edges, then $M$ is the only solution to the problem.

If there is one such edge, then the solutions to the problem are all points of this edge. If there are two edges, then all points of the face in which these edges lie serve as solutions.

Most often, three edges emanate from each vertex of the polyhedron $A$. In this case, at each step it is easier to check only two edges (since the edge on which the previous vertex lies can be immediately discarded). If along these common edges the function values ​​are at two vertex positions and select the one in which this value is greater. If there are more edges emanating from a vertex, along which the function values ​​are in two positions of the vertices, then select the one in which this value is greater. If there are more edges emanating from a vertex, along which the function values ​​are in two positions of the vertices, then select the one in which this value is greater. This method may turn out to be more economical from a computational point of view, since it does not require first finding the ends of all edges emanating from a given vertex.

All geometric concepts that turn out to be useful when considering linear programming problems with two and three variables (line, plane, parallelism, convex set, support straightness, etc.) can be formulated purely algebraically in the language of equations and inequalities. When using this language, unlike geometric, as a rule, there will be no significant differences between the cases of two, three and a large number of transformations.

variables, and the corresponding definitions are transferred by analogy to the general case.

On the other hand, the value of a geometric language lies in the fact that intuition can suggest some constructions and methods for solving problems that are difficult to figure out if you use only algebraic language. The Simpleskmethod provides a good example of such a technique. Because of this, it turns out to be useful to introduce geometric language in the case of an arbitrary number of variables.

Exercises

1. Prove that if the planes
   $$x_1 + x_2 + x_3 = K$$
   are parallel to some face of the polyhedron $A$, then the plane
   $$x_1 + x_2 + x_3 = K_0$$
   coincides with the plane of this face.

2. Is it true that if the planes
   $$x_1 + x_2 + x_3 = K$$
   are parallel to some edge of the polyhedron $A$, then the plane
   $$x_1 + x_2 + x_3 = K_0$$
   goes through this rib?

3. a) Let the problem of choosing technological resources set the task of maximizing not the output, but the approximation, that is, the difference between the price of the produced product and the cost of using the resource in production, and it is known that the price of a unit of produced product is equal to $q$ rubles, the cost of using units of the $j$ resource is equal to $q_j$ rubles ($j=1,\dots,m$).

6. Let the choice problem be technological, so that the output is no less than $S$ units and at the same time that the costs are minimal.

Formulate each of these problems as a linear programming problem. Prove that there is exactly one value $S$ for which the solutions to problem 6) coincide with the solutions to problem a).

Literature

A. S. Solodovnikov, Introduction to linear algebra and linear programming. M., “Enlightenment”, 1966.

8

Ohm’s law
Y.A.SMORODINSKY

Simple at first glance

Typically Ohm’s law is formulated this way. Let’s take a conductor with current (Fig. 1). Let a potential difference $V_A - V_B$ be applied to its ends, and $V_A > V_B$. then the strength of the current flowing through the conductor will be determined by the formula

$$I = \frac{V_A - V_B}{R},$$

where $R$ is the conductor resistance. In addition, they add that if a conductor with a constant cross-section $S$ is uniform along the entire length $l$, then $R = \frac{1}{\sigma} \frac{l}{S}$, where $\sigma$ is called specific conductivity, and $\frac{1}{\sigma} = \rho$ is resistivity.

The graph of the dependence of the current on the applied potential difference is expressed as a straight line (Fig. 2).

Ohm’s law looks very simple and very powerful. After all, it is clear that the greater the potential difference,

$$V_A \quad V_B \quad I$$

Fig. 1.

tem more and current. But why is there a simple proportionality between these two quantities, why does the quantity $R$ itself not change with increasing current? Experience has shown that Ohm’s law is true for both very small and very large currents in metals. Such a large potential difference was applied to the wiring that over a length of 1 cm the potential changes by a million volts. And yet the current could be determined using Ohm’s law. True, in fact, in such experiments very small deviations from this law were observed, but they could be explained by the fact that the conductor, despite all precautions, heated up, and its resistance changed due to changes in temperature. This seemingly simple law raises interesting questions.

In which direction does the current flow?

Why does current flow in the direction of decreasing potential, and not in the direction of increasing it? Let’s take a battery and connect it to some resistance, for example, turn on a light bulb. According to Ohm’s law, the circuit will consume current from the positive terminal of the battery through the light bulb to the imprinted terminal, reducing the potential difference between the terminals, which of course will be restored by the battery, but this consumes energy all the time (this will be discussed later).

What would happen if the current went in the opposite direction? Terminal charges

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is expressed temporally by an exponential function *).

Knowing the properties of the exponential function, we can write that the charge changes as follows:

$$Q(t) = Qe^{-\frac{2t}{CR}}$$

Thus, the charge will decrease according to an exponential law. Technically speaking, the charges will disappear only after an infinitely long time, but in practice they will become invisible very quickly.

Typically, the time for establishing equilibrium is called the time during which the charges of the balls (and therefore the current in the circuit) decreases by $e$ times. For our example, this time is $\frac{CR}{2}$.

History of discovery

We still consider Ohm’s law to be one of the most important laws of physics. When Georg Simon Ohm (1787-1854) discovered this law, they simply did not believe him. Ohm’s main work was published in 1927 in Berlin and was called “Galvanic goal, mathematical developed by the doctor.”

*) The exhibitor is described in «Kvant» No. 1 for this year. Only there α is chosen equal to one.

**) Let’s check the last formula. Let’s consider two moments of time $t$ and $t+\Delta t$.

$$Q(t+\Delta t) = Qe^{-\frac{2}{CR}(t+\Delta t)}$$

$$Q(t) = Qe^{-\frac{2}{CR}t}$$

The difference between these two functions can be calculated (for small $\Delta t$). Note that we are talking about changing the function $e^x$, and $x$ changes from $\left(-\frac{2}{CR}t\right)$ to $\left[-\frac{2}{CR}(t+\Delta t)\right]$, that is, by values ​​$\Delta x = -\frac{2}{CR}\Delta t$.

Thus, $\Delta Q = -\frac{2}{CR}\Delta t \cdot Q$, since the rate of change of an exponential function is proportional to itself.

G.S. Ohm." But his first research began earlier; their results were published in 1825. To understand how difficult it was to achieve recognition of these works, you need to know what tools an experimental physicist had at his disposal 150 years ago.

In 1820, Ampere discovered that an electric current deflects the magnetic needle of a compass. This allowed us to basically change the “current strength” and understand that there is a difference between current strength and voltage. By this time, two current sources were known: the electric battery “voltaic” (invented in 1799 by A. Volta) and the thermodynamic element invented by Seebeck in 1821. However, he did not foresee that it was possible to establish mathematical laws of battery operation. Physicists were surprised at the change in the action of the battery with different connecting elements, but they could not see any clear patterns here. In addition, galvanic batteries at that time were flat, the direction at their ends changed during the experiment (the batteries “sat down”), which greatly confused the results of the experiments. Ohm abandoned these batteries and conducted his first experiments using a thermodynamic element as a current source, which gave a significantly lower current, but the voltage was stable. The potential difference of such an element is about 0.0045 volts. To change the current obtained from thermal action, Ohm came up with a new method. The current-carrying conductor was positioned along the direction of a freely suspended magnetic needle (i.e., along the meridian), and the angle at which the thread had to be twisted changed so that the needle did not deviate from this direction. The angle of twist of the thread characterized the current strength.

The first formula that Ohm received was of the form

$$X = \frac{a}{b+x},$$

In the place where the current source is located, there is a sharp jump in the potential difference, which determines the electromotive force in the circuit. If you are only interested in the external circuit, then the last formula can be rewritten as follows:

$$IR = \varepsilon - Ir.$$

We see that the potential difference at the terminals of the current source is less than the electromotive force by values ​​$Ir$.

Looking now at Ohm’s formula

$$X = \frac{am}{bm+x},$$

we see how well Ohm understood the role of the galvanic elements in an electrical circuit. He not only established that e.d. s. battery (composed of series-connected elements) is proportional to the number of elements, but also introduced the number of elements into the denominator of the formula. This means that he well understood the role of the internal resistance of the battery, which (as is very convenient for us now) is also proportional to the number of elements connected in series.

Ohm’s local law

Let’s write another formula for Ohm’s law. To do this, we use the current density - the ratio of the current flowing through a very small area $S$ (located perpendicularly directed to the current) to the magnitude of this area

$$j = \frac{I}{S}.$$

Then from Ohm’s law it follows that

$$j = \frac{V_A - V_B}{RS} = \sigma \frac{V_A - V_B}{l}.$$

If points $A$ and $B$ are located very close to each other, then the values

$$\frac{V_A - V_B}{l} = E$$

called the electric field intensity. Voltage $E$ is a vector directed towards the highest fast-musical potential. Since $j$ is also a vector and it is directed in the same direction, we can write Ohm’s law in vector form

$$\vec{j} = \sigma \vec{E}.$$

In this form, Ohm’s law can be applied to any body at any point. This expression does not include any geometric dimensions of the body, but only the conductivity $\sigma$, determined by the substance in the place where we measure the current density. In this form, Ohm’s law is local in nature, that is, it includes quantities that relate only to one selected point (and not to the entire chain, as was the case in the formulas that we wrote above).

Elementary theory of Ohm’s formula

We can now explain Ohm’s law very simply. We carry current in metals using electrons that move under the influence of an electric field. Random thermal movement occurs in the conductor. Under the influence of an electric field, a more ordered movement is superimposed on it. (Usually its speeds are less than thermal ones.) The electric field acts on electrons with a force

$$\vec{F} = e\vec{E}.$$

Why don’t electrons move at the same speed? Because they collide with atoms and lose their speed. This is explained in detail in the article by D. A. Frank-Kamenetsky (see «Kvant» No. 9, 1970).

The electron gains additional speed only during that short time $\tau$, which it moves almost freely in the interval between two collisions; this time is called “free run time”. In time $\tau$ the electron will gain additional speed equal to

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acceleration $\frac{e}{m} \vec{E}$ multiplied by τ:
$$\vec{u} = \frac{e}{m} \vec{E} \tau.$$

If the number of electrons (“free electrons,” that is, those that are not “attached” to atoms) per unit volume is denoted by $N$, then an electric current will flow through $1 \, cm^2$ equal to the charge of the electron multiplied by the number of electrons passing through $1 \, cm^2$ in 1 second

$$\vec{j} = \frac{1}{2} \frac{e^2}{m} N \tau \vec{E}.$$

In this formula, together with the speed $u$ we wrote half of it. It is simply the arithmetic average of the initial and final velocities between two collisions.

From here we can write an expression for conductivity (as a proportionality coefficient between $\vec{j}$ and $\vec{E}$):

$$\sigma = \frac{1}{2} \frac{e^2 N \tau}{m}.$$

Joule heat and current source work

Thus, behind the simple formula of Ohm’s law there are many complex phenomena hidden. When current flows, electrons constantly receive energy from the electric field and always lose it when interacting with atoms. As a result, as is well known, heat is released

*) Sometimes the same formula is written slightly differently, replacing τ by the formula

$$\tau = \frac{\text{свободный пробег}}{\text{средняя тепл. скорость}} = \frac{\lambda}{v_{\text{cp}}}.$$

The mean free path is the average path taken by an electron between two collisions. Then

$$\sigma = \frac{1}{2} \frac{e^2 N \lambda}{m v_{\text{cp}}}.$$

This formula is called the Drude formula.

(for example, an electric stove heats up, the wire in which has a high resistance).

The amount of heat that is released in a section of a conductor with resistance is determined by the Joule-Lenz formula

$$Q = \frac{1}{J} I^2 R,$$

where $J$ is the mechanical equivalent of heat:

$$J = 4,184 \frac{\partial \chi}{\kappa \cdot \kappa a l}.$$

What resources generate heat? Answer: just as the current source does this work, it maintains the potential at its terminals.

Returning to Ohm’s formula for a closed circuit, we can say that the energy expended in the form of heat in the entire circuit, including the interior of the current source itself, is equal to $E I$, that is, the product of E. d. c. on current strength

$$I^2 (R + r) = E I.$$

The work done by a current source occurs either due to chemical energy (in batteries) or due to mechanical energy (in generators). The last formula expresses the law of conservation of energy for a circuit consisting of a current source closed by some resistance.

The formula of Ohm’s law allows you to calculate any direct current circuit, and in a slightly modified form it can be generalized to alternating currents.

Exercises

1. Find by what law the potentials of two balls similar to those drawn on figure 3 will change if the charge of one is $Q=0,01$ coulomb and the other is 0.02 coulomb. Set the product $CR$ equal to $10^{-12}$ sec.

2. After how long will the difference between the charges of the balls decrease by 20 times?

14

Logic problems and propositional algebra

L. L. Tsinman

Let’s start with the formulation of several logical problems.

Task 1. After the parent-teacher meeting, one of the parents approached the teacher.

“You didn’t name my son among the good students,” he said. “But my Fedya is an excellent student and, moreover, the best athlete in the class.”

“Yes, you’re right,” answered the teacher. “But we consider a good student to be a student who studies well, is disciplined, helps those who are lagging behind in their studies, and, in addition, participates in the work of a scientific circle or engages in argument.” And your Fedya…

What else was the teacher going to say to Fyodor Palais?

Problem 2. Once, therefore, it was necessary to simultaneously interrogate three witnesses: Claude, Jacques and Dick. Their testimony was contradicted by others, and each of them was accused of lying. Claude said that Jacques was lying, Jacques accused Dick of lying, and Dick insisted, therefore, not to believe either Claude or Jacques. But therefore he quickly brought them out into the open without asking them a single question. Which of the witnesses answered the truth?

Problem 3. One man was captured by savages and was imprisoned in a dungeon with two exits. The leader of the savages offered the captive the following chance of salvation: “One exit leads to freedom, and the other leads to certain death. You can choose anyone. We help my witness make a choice. He’ll stay here

would answer one question - any one we wish to ask him. But I warn myself: if this guard is in a good mood, then he is telling the truth, but if he is in a bad mood, then he is lying!

After a moment of reflection, the assembled loyik asked one question, after which he unmistakably chose the exit that led to freedom. What was that question?

Logical problems similar to those listed often appear in popular science publications. Solving them does not require any special knowledge from the field of logic, only fairly well-developed logical thinking is required. But imagine a person who takes on the solution of some numerical problem (movement, pouring) and does not know how to freely do algebraic transformations and solve equations. You find yourself in approximately the same position when solving logical problems without using an auxiliary apparatus, which is called propositional algebra or algebra of logic (as opposed to the algebra of numbers studied at school).

The purpose of the article is to introduce the reader to the initial information from this necessity of algebra and, using the knowledge gained, to solve the problems formulated above. Note, however, that propositional algebra is not only used to solve logical problems (this is just one of its applications), but it is also important

an integral part of one of the modern branches of mathematics is mathematical logic.

Proposition algebra elements are statements, for example: “Fedya is an excellent student,” “Fedya is a good student,” “Dick is lying.” When considering statements, we will be interested in only one question - whether a given statement is true or false. So, the first of the given statements is true, the second is false (see the condition of problem 1), the third statement is also either true or false, but to clarify this issue it is necessary to solve problem 2. To denote statements in the algebra of logic, we will use capital Latin letters A, B, C,… Record

zn. {A}=1

means that the value of statement A is true, and the notation

zn. {A}=0

means that the meaning of statement A is false.

Thus, any statement can take one of two values: 0 or 1.

In offensive speech, from simple statements with the help of various conjunctions, parts, connectives, we form compound statements. So from statement A and statement B one can make the following statements: “A and B”; “A or B”; “not A” (“it is not true that A”); “if A, then B”; “either A or B” and others.

Examples of measures

“(Claude lies) u (Jacques lies)”;
“it is not true that (Fedya is a good student)”;
“if (Claude is telling the truth), then (Jacques is lying).”

Let us now introduce operations on statements in our algebra, and we will define these operations so that they correspond to some of the most common ways of processing compound statements in ordinary language.

Logical product of statements (conjunction)

The logical product of a statement A and a statement B is denoted as follows: A ∧ B. This operation must correspond to the compound statement “A and B”. We consider such a compound statement to be true if and only if the given statements are true. For example, as a logical product operation it is natural to take the following truth table:

| A | B | A ∧ B |
| :— | :— | :— |
| 1 | 1 | 1 |
| 1 | 0 | 0 |
| 0 | 1 | 0 |
| 0 | 0 | 0 |

Logical sum of propositions (disjunction)

Logical sum notation: A ∨ B. This operation must correspond to the compound statement “A or B”. We consider the following statement to be true if and only if at least one of these statements is true. It is in this sense that in task 1 the teacher used the word or in the definition of a good student.

The truth table for the operation or looks like this:

| A | B | A ∨ B |
| :— | :— | :— |
| 1 | 1 | 1 |
| 1 | 0 | 1 |
| 0 | 1 | 1 |
| 0 | 0 | 0 |

16

Negation of a statement
The notation for the negation of a statement is $A : \bar{A}$. The negation of the statement $A$ corresponds to the statement “not $A$” (“it is not true that $A$”). Therefore, it is determined by the following truth table:

| $A$ | $\bar{A}$ |
| :— | :— |
| 1 | 0 |
| 0 | 1 |

Using the introduced operations in statement algebra, complex algebraic expressions can be constructed from individual statements $A, B, C$…, for example: $\bar{A} \vee B; \bar{A} \wedge B; (A \vee \bar{B}) \wedge C$.

As in numerical algebra, the condition is that the operation of multiplication is older than the operation of addition. This allows you to avoid writing some parentheses. Now the expression

$((A \wedge B) \vee ((\bar{A} \vee B) \wedge C))$

can be rewritten in the form

$A \wedge B \vee (\bar{A} \vee B) \wedge C$.

If you know the meaning of each statement included in an algebraic expression, then using truth tables you can find the value of this algebraic expression. Let us find, for example, the values ​​of the algebraic expression $\bar{A} \vee \bar{B}$ for all possible values ​​of its components $A$ and $B$:

| $A$ | $B$ | $\bar{A}$ | $\bar{B}$ | $\bar{A} \vee \bar{B}$ |
| :— | :— | :— | :— | :— |
| 1 | 1 | 0 | 0 | 0 |
| 1 | 0 | 0 | 1 | 1 |
| 0 | 1 | 1 | 0 | 1 |
| 0 | 0 | 1 | 1 | 1 |

First, the third and fourth columns are filled in using the values ​​in the first two columns, and then the last column is filled in according to the definition of addition. This table is called the truth table for the expression $\bar{A} \vee \bar{B}$.

Now let’s build a truth table for $\bar{A} \wedge \bar{B}$:

| $A$ | $B$ | $\bar{A} \wedge \bar{B}$ |
| :— | :— | :— |
| 1 | 1 | 0 |
| 1 | 0 | 1 |
| 0 | 1 | 0 |
| 0 | 0 | 1 |

The following columns in the above tables turned out to be the same. This means that for any values ​​of the statements $A$ and $B$ the expressions $\bar{A} \vee \bar{B}$ and $\bar{A} \wedge \bar{B}$ take on the same meanings. We will call such algebraic expressions equivalent. To record equivalence, the identity sign is used:

$$\bar{A} \wedge \bar{B} \equiv \bar{A} \vee \bar{B}$$

These two equivalent expressions have the same meaningful meaning, but expressed in different words.

Indeed, if $A$ and $B$ are some statements, then $\bar{A} \wedge \bar{B}$ expresses the complex statement “it is not true that both given statements are true”, and $\bar{A} \vee \bar{B}$ expresses the complex statement “at least one of the given statements is false”.

Algebraic expressions in which the last column in the truth table consists of ones (of zeros) will also be called truly true (equivalently false) and denoted by the letter $H$ (letter $J$).

Using a truth table, it is always easy to determine whether two given algebraic expressions are equivalent or not.

Let us present the simplest equivalences, some of which we

17

In what follows we will use them as rules.

1. $A \wedge B \equiv B \wedge A$.
2. $A \vee B \equiv B \vee A$.
3. $(A \wedge B) \wedge C \equiv A \wedge (B \wedge C)$.
4. $(A \vee B) \vee C \equiv A \vee (B \vee C)$.
5. $A \wedge (B \vee C) \equiv (A \wedge B) \vee (A \wedge C)$.
6. $A \vee (B \wedge C) \equiv (A \vee B) \wedge (A \vee C)$.
7. $A \wedge A \equiv A$.
8. $A \vee A \equiv A$.
9. $\overline{A} \wedge \overline{B} \equiv \overline{A} \vee \overline{B}$.
10. $\overline{A} \vee \overline{B} \equiv \overline{A} \wedge \overline{B}$.
11. $\overline{A} \equiv A$.
12. $A \wedge \overline{A} \equiv \overline{J}$.
13. $A \vee \overline{A} \equiv \overline{H}$.
14. $A \wedge \overline{J} \equiv \overline{J}$.
15. $A \vee \overline{H} \equiv \overline{H}$.
16. $A \wedge \overline{H} \equiv A$.
17. $A \vee \overline{J} \equiv A$.

If we imagine the substantive meaning of rules 1-17 as we did above with rule 9, then these equivalences will become obvious and easy to remember.

Some of the listed rules, namely 1-5, are similar to the well-known properties of the operations of addition and multiplication in numerical algebra. Therefore, in propositional algebra we can perform algebraic transformations similar to those we do in numerical algebra. In particular, we can “multiply” two algebraic statements using the rule for multiplying a polynomial by a polynomial.

But propositional algebra also has its own specific rules, for example 6.

Solution to problem 1

Let us introduce notation for the statement:

$A$: “Fedya is a good student”;
$B$: “Fedya is disciplined”;
$C$: “Fedya helps those who are behind in their studies”;
$D$: “Fedya participates in the work of a scientific circle”;
$E$: “Fedya plays sports.”

Now the statement “Fedya is a good student,” according to the “defined” good student, this teaching,

body, can be written like this:

$A \cup B \cup C$ ($D \cup E$).

In the language of propositional algebra, this complex statement will take the form of an algebraic expression

$A \wedge B \wedge C \wedge (D \vee E)$.

The teacher claims that “it is not true that (Fedya is a good student)”, that is

$$\{A \wedge B \wedge C \wedge (D \vee E)\} = 1.$$

Let us now transform the algebraic expression on the left side, first according to rule 9 (though in this case the negation is above the logical product of the four components, but it is easy to see that the equivalence of 9 remains valid in this more general form), and then according to rule 10:

$$\begin{align*} \{A \vee \overline{B} \vee \overline{C} \vee (D \vee E)\} &= 1; \\ \{A \vee \overline{B} \vee \overline{C} \vee \overline{D} \wedge \overline{E}\} &= 1. \end{align*}$$

Fedin Popau will claim (and the teacher agreed with him) that

$$\{A\} = 1$$

(Fedya is an excellent student!), and also

$$\{E\} = 1$$

(Fedya is the best skier in the class!). Then in (1) the first and last two terms are false and they can be discarded according to rules 17 and 14.

We get that

$$\{B \vee \overline{C}\} = 1.$$

This means that the teacher wanted to complain with Fedya either because Fedya is undisciplined, or because he does not help those who are behind in their studies (or maybe Fedya has become both of these shortcomings at once).

Solution to problem 2

Let us denote the testimony of witnesses Claude, Jacques and Dick correspondingly by the letters $K$, $J$, $D$. We do not know which of these readings are true and which are false. But we know the following:

1. either Claude told the truth, and then Jacques lied, or Claude lied, and then Jacques told the truth;

2. One student came to take an exam with an automatic examiner. Five questions appeared on the machine’s screen, each of which required a “yes” or “no” answer.
   After answering all the questions (to do this, you need to press the appropriate button), the machine, giving each correct answer 1 point, gives the overall score for the exam. The student carefully went through all the questions and was saddened to admit that he did not know the answer to any of them.
   All his knowledge boiled down to the following:

1. the first and last questions require opposite answers;
2. on the contrary, the second and fourth questions should have the same answers;
3. at least one of the first two questions requires the answer “yes”, then the fifth requires the answer “no”.
   In addition, student experience tells him that there are always more questions requiring a “yes” answer than questions requiring a “no” answer.
   The picture turned out to be sad. But the student did not despair. He knew this subject poorly, but he was very good at solving logical problems. The leaf quickly made some calculations, he was happy to discover that he was guaranteed 4 points, and if he was lucky, he would get all 5 points. What calculations did the student make?

3. Writers A, B, C, D write with subpseudonyms X, Y, Z, W. It is required to find out which pseudonym each writer uses if the following is known:

1. if D is not X, then B is X;
2. if B is neither X nor W, then A is X;
3. if C is not W, then B is Z;
4. if D is Y, then B is not X;
5. if A is not X, then C is not Y.

For a more complete acquaintance with propositional algebra, we can also recommend the following literature:

L. A. Kaluzhin, What is mathematical logic. M., “Science”, 1964, pp. 1 and 2.
I. M. Yaglom, Extraordinary algebra. Series “Popular lectures on mathematics”, M., “Science”, 1968.
R. R. Stoll, Sets, logic, axiomatic theories. M., “Enlightenment”, 1968, 11, §§ 2.1-2.4.
P. S. Novikov, Elements of mathematical logic. M., “Fizmatgiz”, 1959
A. Grzegorczyk, Popular logic. M., “Science”, 1965.
A. B. Trakhtenbrot, Algorithms and machine problem solving. M., “Fizmatgiz”, 1960.
H. Freudenthal, The Language of Logic. M., “Science”, 1969.

HOW IS IT EASIER TO SHARE?

What is the easiest way to divide a smaller number by a larger one if the larger number ends in 9?

Let you need to divide the number a by the number b, and a < b. Let b = 10c + 9. Now check that

$$\frac{a}{10c+9} = 10^{-1} \frac{a-p_1}{c+1} + \frac{a-p_1}{c+1} + 10p_1$$

where $p_1$ is the remainder when $a$ is divided by $c+1$, so $\frac{a-p_1}{c+1}$ is an integer.

Let $\frac{a}{b} = 0, \alpha_1 \alpha_2 \alpha_3 \dots$ be the decimal notation of the quotient. Then $\alpha_1 = \frac{a-p_1}{c+1}$ and $\alpha_1 + 10p_1 = 0, \alpha_2 \alpha_3 \dots$.

Denoting $\alpha_1 + 10p_1$ by $a_1$, we write the equality $\frac{a_1}{10c+9} = 0, \alpha_2 \alpha_3 \dots$, from which we similarly find $\alpha_2$ and so on.

Examples of measures

1. We need to divide 37 by 89. We do this: we divide 37 not by 89, but by 9, since the number of ones in the divisor is 9, and the number of tens is 8, 8+1=9. Then the remainder is added before the quotient and again divided by 9 and so on. Partial decimal digits of the fraction $\frac{37}{89}$.

$$-\frac{37}{36} \left| \frac{9}{4} \right| - \frac{14}{9} \left| \frac{9}{1} \right|$$

$$-\frac{51}{45} \left| \frac{9}{5} \right| - \frac{65}{63} \left| \frac{9}{7} \right|$$

So, $\frac{37}{89} = 0,4157\dots$

(For the ending, see page 29.)

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the solution was first obtained by Euler analytically, but remained lost among the mass of formulas, and only thanks to simple and visual interpretations of Poinsot was presented before the eyes of scientists with all clarity.

What role fell to Poinsot in explaining the question of the motion of a solid body, the same role belongs to Helmholtz in explaining the question of the motion of a liquid.

Almost all of Helmholtz’s works on mechanics are related to fluid mechanics. It can be said that modern hydrodynamics owes its development mainly to Helmholtz. Meanwhile, the most important work of the scientist in this area appeared in 1858*), 43 years after the formulas containing the reasons for the preservation of wines were found by Cauchy. But Cauchy considered the result he obtained only from the analytical side and did not foresee the mass of questions that could be solved with proper geometric illumination of the conclusions.

I will now try to explain to you as simply as possible the concept of wine established by Helmholtz.

Let us imagine a cylindrical vessel of finite height (Fig. 1) with a scale larger

*) This refers to Helmholtz’s work on the equations of hydrodynamics corresponding to linear motion.

Fig 1

with a solid base, filled with liquid (or gas), and suppose that this liquid moves like this: a central cylindrical column of some thickness rotates like a solid body around its axis, and the mass of liquid always spins around this column in circles with speeds inversely proportional to the distance from the axis of the column, and these speeds, increasing as they approach the central column, transform on its surface into the speed of the column.

This movement of liquid is called a vine, and the cylindrical column that characterizes it is called a vine cord. Let’s say that the direction of the cord is equal to half the product of the velocity of the liquid on the surface of the cord and the perimeter of the normal section of the cord.

Twice the value of this product is called the circulation velocity.

In general, the circulation of speed along some closed contour within the movement of a fluid is equal to the product of the length of the contour and the average of all the velocities of the points of the contour in the direction tangent to the contour.

Since in the movement of a fluid, let us depict on figure 1, the velocities are inversely proportional to the radii, the velocity circulation along all horizontal circles having a center on the axis of the column and enclosing it are equal to each other and, therefore, equal to the double direction of the cord, and the velocity circulation along contours consisting of segments of two circles between segments of two radii, and lying outside the cord (contour ABCD on figure 2) are equal to zero. In addition, it can be proven that the circulation of speed along any closed contour that encloses the direction is equal to twice the direction of the direction, and the circulation of speed along any closed contour that does not cover the direction is equal to zero.

Fig. 5.

equal, then this center would recede into an infinite distance, and both cords of bezaline would move forward in a direction perpendicular to the straight line connecting the centers, as can be seen from figure 5.

figure 6 shows the trajectories (paths) of three vinre cords, of which 1 and 2 rotate counterclockwise, and 3 rotate clockwise.

The concept we have established of a straight cord enclosed in a very wide cylindrical vessel extends to cords that originate in a liquid mass. In this case, vinre cords can be searched by drawing up circulations

Fig. 6.

Fig. 7.

25

During the movement, either its ends will lie on the edges of the liquid (on the walls of the vessel or on the free surface), or it will remain closed.

In fact, in order to leave the walls of the vessel, the base of the vortex must be reduced in size to zero; and since the circulation of speed along the contour of the base must remain unchanged, the convergence would require that the speed of the round fluid at the bottom of the cord increase to infinity.

The hydrodynamic pressure of the fluid decreases as the speed increases. Therefore, when the vortex pressure on the wall of the vessel decreases, the pressure in this place will quickly decrease, and the remaining mass of the liquid will press on the particles of the end of the vortex cord and prevent them from moving away from the wall. The vortex cord, so to speak, sticks with its ends to the walls of the vessel. If the end of the cord lies on a free surface, then such suction can be seen by the funnel that forms on the free surface at the “bottom” of the cord.

If the ends of the vortex cord do not lie on the edges of the liquid, then they must be closed together, and thus a closed vortex cord is obtained - one in which, so to speak, both ends are sucked to each other.

The simplest type of closed vortex filament appears to be a vortex ring, shown at figure 7.

All parts of the liquid lying outside the ring swirl in a closed curve passing through the ring so that the circulation of velocity along all these curves is the same and equal to the circulation of velocity on the contour of the cross section of the ring. By crossing the vortex of the ring, we will receive various circulations for the trajectory of its part. Liquid point velocities

the largest on the surface of the ring. They decrease with distance from this surface inside the ring and are equal to zero on a certain center line. They also decrease as they move away from the ring in the mass of liquid surrounding it. For fluid points at a significant distance from the ring, the velocities are inversely proportional to the cubes of the distance from the ring.

We have seen that two straight parallel cords originating in a liquid mass, around which the liquid is round with equal directions of the vortex in opposite directions, will move in a direction perpedicular to the plane drawn through them. For the same reason, the vortex ring will not remain motionless, but will move in a direction perpedicular to the plane of the ring, in the direction in which the liquid flows out of the ring.

We see on figure 7 that particles of the liquid mass moving in the upper closed trajectories will move to the lower edge of the ring and move it to the right; in the same way, particles of the liquid mass moving along the lower closed trajectories will move to the upper edge of the ring and also move it to the right. All rings will move evenly to the right side, carrying behind them the round liquid around them. This movement will be faster, the greater the direction of the vortex and the smaller the size of the ring.

We said that inside an ideal liquid there should be persistence, and new cords cannot be formed. Meanwhile, in nature we often see and fade vortices. This is due to the fact that our waters and air have a certain degree of viscosity, due to which the above theoretical results are somewhat modified. On the one hand, vortices can become quenched (mainly

26

in those places in which two layers of liquid slide over each other at different speeds), on the other hand, the vinrys that stick together are not preserved, but gradually disappear.

Helmogoly demonstrated the formation of direct vortices with one excellent experiment, described in his speech on vinre storms. We will repeat this experience here.

A small hole is made in the denne of the cylindrical vessel (Fig. 8), plugged with a stopper. The vessel is filled with water. By means of air jets directed by a tube to one edge of the free surface of the water, we cause the liquid to slowly rotate. The liquid begins to flow out of the hole, approaching from the edges of the vessel to its axis. Since the circulation speed along a circle drawn from a point on the cylinder axis through the same particles of liquid should not change with time, then with a decrease in the radii of these circles the speed of part of the liquid will increase. The rotation of the liquid as it approaches the axis will become faster and faster, and we will notice a sharply formed winry, above which a funnel will appear, deeper and deeper.

I will also demonstrate the formation of winry using a rapidly rotating disk.

At figure 9 the device of Prof. F. N. Shvedova. Through the bottom of a glass cylindrical vessel, a vertical axis is sold in the stuffing box, ending with a small disk. This axis can be brought into rapid rotation through endless repairs. Water and oil are poured into the vessel, which floats on top of the water. Rotating the disk, we will notice that the water gradually begins to rotate and forms a winding cord above the disk, which is replaced along the funnel at the interface between water and oil. This funnel is filled with oil, which descends to the disk in the form of a descending tornado. At the moment when the oil comes into contact with the disk, its entire mass is dispersed through the water.

Even more interesting is the way direct winrays are formed in the air. The air above the surface of the water causes rotation with the help of a special rapidly rotating impeller placed at a certain height above the water (Fig. 10). The air vinry captures water along its axis and lifts it in the form of an ascending tornado all the way to the lionfish.

Vortex rings in the air are demonstrated using the Theta instrument. It consists of a box (Fig. 11), the back side of which is covered with leather, and a hole with sharp edges is made in the front. Form from-

28

Until now, we have produced the rings from a round hole. Let’s now try to form them from an elliptical and square hole. We see that in this case the ring does not retain the shape of the hole, but oscillates, trying to transform into a round ring, which is thus the only stable form of a closed cord.

Let us now consider the influence of foreign objects on the output loop. Bringing solid bodies from the side to the moving ring, we see that they push the ring away. But if the ring runs onto a stationary plane parallel to its plane, then, approaching it, it increases more and more in size, so to speak, spreads over the plane. If we let the peg run onto a knife, the plane of which passes through the axis of the peg, then the latter will be cut by the knife into two half-rings, the ends of which will slide along the surface of the knife; but, having passed this surface, the ends come together again, and the ring is restored.

In addition to smoke rings in the air, you can also observe air rings in the water. This interesting phenomenon, which at first glance seems paradoxical, is simply explained by the fact that, due to the centrifugal force, significant pressure is applied to the axes of the output ring. If, when forming a ring ring, we introduce several air bubbles into the water, they will immediately climb into the place of the liquid where the pressure is lowest, that is, on the axis of the ring, and will remain there all the time while the ring moves along the mass of water, despite the fact that air is 800 times lighter than water.

I will show here a device for the formation of air rings in water, which is a modification of the device of the professional Osborne Reynolds. There is (Fig.12) a large glass bathtub filled with water. A glass bent at a right angle is immersed in it.

Fig. 12.

naya tube. A sleeve made of a rubber ball is put on the upper end of the tube coming out of the water, through which air can be driven into the tube and water can be pushed out of it. Quickly squeezing the ball, we push a container of water out of the horizontal bend of the tube and do this so that the air almost reaches the lower end of the tube, but does not come out of it in large quantities. The container of water, having flowed out into the surrounding liquid, is wrapped in a glass ring. At the same time, several bubbles will be pushed out along with the water; the wine ring will be clearly visible: it will be formed as if from shiny grains of beads. Running along the entire bathtub, the wells will hit the opposite wall of it and here, expanding, disappear.

Using our device, we can clearly demonstrate the reflection of rings from the free surface of water. To do this, just turn the tube so that its lower end points slightly upward. The ring, running up to the free surface of the liquid, reflected from it

29

shrinks, and the angle of incidence is equal to the angle of reflection.

Since real liquids have viscosity and rub against the walls of the vessels in which they move, during their movement they are constantly filled with vortex cords. Helmholtz showed that a liquid mass in any imaginable movement can be considered as continuously filled with vortex cords, and as a means of studying the movement of these cords.

Brief information about the scientists mentioned in N. E. Zhukovsky’s lecture.

Euler is the greatest mathematician of the 18th century, a St. Petersburg academician, one of the creators of analytical mechanics. Laplace called Euler the father of modern mathematical analysis.

Poinsot is a French scientist, mathematician and mechanic. In 1834 he obtained an elegant geometric solution to some problems in the dynamics of rotation.

Helmholtz was an outstanding German physicist and mathematician of the 19th century. He became famous, in particular, for his work on acoustics (Helmholtz resonators) and thermodynamics. He is also the founder of the vortex theory.

Cauchy is a French scientist of the late 18th - first half of the 19th century, who was rightfully considered one of the creators of modern mathematics. He left over 800 memoirs on various branches of mathematics.

Shvedov V.F. is a professor at Odessa University, a well-known specialist in the hydrodynamics of viscous fluids (second half of the 19th century).

Tag is a professor of physics in Edinburgh (England, second half of the 19th century).

Reynolds is a professor at Owen College in Manchester (England, second half of the 19th century). He became famous for working on hydrodynamics.

2. 28 must be divided by 59. 5+1=6.

$$\frac{28}{24} \cdot \frac{6}{4} - \frac{44}{42} \cdot \frac{6}{7}$$

$$\frac{27}{24} \cdot \frac{6}{4} - \frac{34}{30} \cdot \frac{6}{5}$$

So, $$\frac{28}{59} = 0.4745$$

This method of division is simpler than the generally accepted one and easier to perform. It was proposed by the German mathematician G. Spitzer in 1847. But soon they forgot about him.

G. D. Novinsky

PROBLEMS ABOUT THE WHEEL

1. The wheel rolls down slightly on an inclined plane. Since the movement of the wheel is accompanied by energy losses at the point of contact with the plane, even in the absence of air resistance, the wheel, starting from a certain moment, will roll evenly. It follows that the force R with which the plane acts on the wheel must be directed upward and equal to the gravitational force P. But then the force R will be created by a turning moment relative to the center O, and therefore the wheel will not rotate uniformly, but rather accelerated.

Deal with the contradiction.

(Continued on page 34.)

30

DOPPER effect
L. G. Aslamazov

In a famous anecdote about the sign of physicist Robert Wood, it is said that when Bud’s car passed an intersection on a red light and was stopped, Bud explained his mistake to the effective Doppler: for a moving person, the frequency of the light source changes and the traffic light signal seemed green to him. They say that the policeman turned out to be no layman. He knew that a noticeable change in frequency could only occur at a very high speed of the car, close to the speed

31

Sveta. Pozny the policeman did not argue with the meaning of the physicist and simply fined him for speeding. And although this venerable story may discourage us from ever trying to explain anything in our lives by the Doppler effect, nevertheless, apparently everyone has encountered this effect. When a buzzing train passes us or an airplane flies over us, we hear a sudden change in the tone of the sound. This is also Doppler effective.

Why does the frequency of the sound of a moving source change? Let’s try to figure it out. Let’s draw a picture of waves emitted by a moving source. The speed of sound propagation in the medium does not depend on the speed of the source, and the wave pattern will be a set of circles, the centers of which will be mixed in the direction of movement of the source (picture at the beginning of the article). Let now the observed stand, for example, behind the source, so that the source moves away from it with speed $v$. Let us assume that initially the distance between the source and the observed was equal to $u$, ($u$ is the speed of sound). then after 1 second the distance between them will be $u + v$ and at this distance all $\omega$ maxima created by the source per unit of time decrease ($\omega$ - sound frequency). Before observing for 1 second only those maxima that were initially at a distance of $u$ from it. Therefore, the frequency of the sound $\omega_1$ perceived by the observable will be less, and

$$\frac{\omega_1}{\omega} = \frac{u}{u + v},$$

from where $\omega_1 = \omega \frac{1}{1 + \frac{v}{u}}$

If the source approaches the application, then similarly we obtain

$$\omega_1 = \omega \frac{1}{1 - \frac{v}{u}}$$

If it is not the source, but the receiver of the sound that moves, the same change in frequency occurs qualitatively, but quantitatively the result will be somewhat different. If the applicator moves towards the source with a speed of $v$, then in 1 second it will pass not mismo $\omega$, but a mismo of a larger number of maxima $\omega_2$, and $\frac{\omega_2}{\omega} = \frac{u + v}{u}$, hence $\omega_2 = \omega \left(1 + \frac{v}{u}\right)$.

If the receiver moves away from the source, it will note a lower frequency

$$\omega_2 = \omega \left(1 - \frac{v}{u}\right).$$

This change in frequency when a source or application moves is called the Doppler effect. Considering $v$ positive in both cases when the distance increases, we can write

$$\omega_1 = \frac{1 \omega}{1 + \frac{v}{u}}, \quad \omega_2 = \omega \left(1 - \frac{v}{u}\right).$$

If the speed of the source is small compared to the speed of sound, then, using formulas for approximate calculations, $\frac{1}{1 + \frac{v}{u}}$ can be approximately replaced by $1 - \frac{v}{u}^*$, and then $\omega_1 \approx \omega_2$. However, in modern

*) Check this for example for $\frac{u}{v} = 0.01$. 0.05: 0.1; 0.2

32

The technical speed of the source or receiver is not at all small compared to the speed of sound (for example, the speed of aircraft), and then the Doppler efficiency in both cases is significantly different.

Now that we have understood the Doppler efficiency for sound waves, it would seem that there is nothing simpler than writing similar formulas for light - it is enough to put the speed of sound $u$ together with the speed of light $c$ everywhere, and the answer is ready. However, here we are faced with an amazing obstruction: such formulas cannot be correct, since they contradict Galileo’s principle of relative.

According to this principle, it is impossible to distinguish uniform rectilinear motion from a state of rest, in other words, any physical phenomena must proceed in the same way and any physical experiments must give the same result in all coordinate systems moving relative to each other at constant speeds (that is, in inertial systems). We can consider, for example, the case of movement of a source and a stationary observable in a reporting system moving with the source, and then, according to our formulas, the change in the frequency of light in both systems would be different.

In the propagation of sound, this generality should not confuse us. After all, here, in addition to the source and receiver, the medium in which the sound propagates plays a role. The movement of the source to the receiver and the receiver to the source gives different results precisely because in the first case the relative movement

33

M76. In a company of $n$ people, every two strangers had exactly two mutual acquaintances, and every two acquaintances had no more common acquaintances. Prove that everyone in this company knows the same number of people.

M77. The lengths of the two sides of the triangle are 10 and 15. Prove that the bicsectiras of the angle between them is not greater than 12.

N. B. Vasiliev

M78. Prove that every non-negative integer can be represented as $\frac{(x+y)^2+3x+y}{2}$, where $x$ and $y$ are non-negative integers, and what is the singular representation.

M79. Two points $P$ and $Q$ move along two intersecting straight lines with the same constant speed $v$. Prove that there is a fixed point $A$ on the plane, the distances from which to the points $P$ and $Q$ at any time are equal.

I. F. Sharyein

M80. Table $m \times n$ contains arbitrary numbers. It is allowed to simultaneously change the sign of all numbers in one column or all numbers in one row. Prove that by repeating this operation several times, you can obtain a table in which the sum of the numbers in any column and the sum of the numbers in any row are non-negative.

A. S. Schwartz

F88. A vessel, partially filled with a structure, moves with horizontal acceleration, resulting in the surface of the path being inclined to the horizon at a certain constant angle.

The top of the vessel is closed with a lid.

How will this angle change if the vessel is filled to the top with water?

F89. The electric field voltage in an electromagnetic wave of frequency $\omega=2 \cdot 10^{16} \text{ сек}^{-1}$, amplitude modulated with frequency $\Omega = 2 \cdot 10^{15} \text{ сек}^{-1}$, changes with time according to the law $E=a (1+\cos \Omega t) \cos \omega t$, where $a$ is a constant. Determine the energy of the electrons knocked out by this wave from the atoms of gaseous hydrogen with an induction energy of $W=13,5 \text{ эв}$.

A. D. Gladun, L. G. Aslamaozov

F90. An ideal gas of mass $m$, located at a temperature $T$, is cooled isochorically so that the pressure drops $n$ times. The gas then expands at constant pressure. In the final state its temperature is equal to the initial one. Determine the work done by the gas. The molecular weight of the gas is $\mu$.

F91. To power the device, the voltage at its input must be set as accurately as possible. For this

34

Fig. 1.

It uses two rheostats connected as shown in figure 1. The lengths of the rheostats are the same, and the resistance of one of them is 10 times greater than the resistance of the other. What should I do to set the voltage as accurately as possible? How many times will the accuracy of voltage setting be greater than in the case when only one rheostat is used? How to turn on rheostats if to power the devices you need to set more precisely not the voltage, but the current?

F92. The mass hoop $m$ stands on the mass board $M$ (Fig. 2). The coefficient of friction between the dock and the hoop is $k$. The board lies on a smooth table. With what acceleration will the board move if the hoop is pulled with a force of $F$?

S. N. Domozhakov

Fig. 2.

PROBLEMS ABOUT THE WHEEL

(Continued. See page 29 for the beginning.)

2. A wheel rolls along a horizontal plane. Since the friction force $F$ is directed to the left, the speed of the wheel will decrease. But the moment of this force relative to the center $O$ is directed clockwise, and therefore, the speed of rotation of the wheel should increase.

How will the speed of the wheel change?

3. When you press the brake pedal, the brake pads are pressed against the wheels and their rotation slows down. But the interaction forces between wheels and brake pads are internal and, therefore, cannot reduce the speed of automation. Why does it stop automatically?

4. As is well known, the force driving the train is the friction force of the locomotive wheels on the rails, and the friction force between the rails and the wheels of the cars is the braking force. But the wheels of the locomotive and the carriages are made of the same materials, and all the carriages are much larger than the weight of the locomotive.

Why is the locomotive in a state of motion?

(For ending see page 42)

35

SOLUTIONS

In this issue we publish the solution to problems M28–M30

M28

(Continuation of the solution, see “Quant” No. 3, p. 33). Let us recall that through $f_1(n)$ we denote the smallest integer greater than or equal to $\log_2 n$ (it is equal to the smallest number of checks, after which it is possible to check, for which one can choose one radioactive ball from $n$), through $f_2(n)$ - the smallest number that $f_2(11) \geq 7$. Below we will very briefly prove that $f_2(6) \geq 5$;

Fig 1.

choose one radioactive ball from $n$), through $f_2(n)$ - the smallest number that $f_2(11) \geq 7$. Below we will very briefly prove that $f_2(6) \geq 5$;

36

$f_2(8) \geq 6; \quad f_2(11) \geq 7; \quad f_2(10) \leq 6;$
$f_2(15) \leq 7; \quad f_2(21) \leq 8.$

Readers who wish to verify that all the numbers in the bottom row of the table on page 35 are correct will have to prove, in addition, that $f_2(16) \geq 8;$
$f_2(22) \leq 8; \quad f_2(23) \geq 9; \quad f_2(25) \leq 9$ (indeed, it is clear that the function $f_2$ is non-decreasing, that is, $m > n,$ then $f_2(m) \geq f_2(n)$, therefore it is enough to correctly indicate those values ​​$n,$ for which $f_2(n+1)$ is strictly greater than $f_2(n).$

Before proving the promised inequalities, note that determining the exact value of $f_2(n)$ for any $n$ in this problem; as in many other tasks, it is not very interesting. The point is that the appropriate method for selecting balls is apparently very complicated. On the other hand, it is easy to indicate very simple algorithms for selecting balls, requiring only a slightly larger number of checks than $f_2(n),$. For example, if they first produce one radioactive ball, then grow it and choose another from the remaining ones, then this will take no more than $f_1(n)+f_1(n-1)$ checks, and, as is not difficult to show, the difference $f_1(n)+f_1(n-1) - f_1(C_n^2)$ is always no more than 2, therefore, the difference $f_1(n)+f_1(n-1)-f_2(n)$ is also no more than 2. In practical problems like this, if the number of checks $k$ is not too small, one or two extra checks will most likely play a role special role, if only the verification method was simpler (say, for $n=15$ we can no longer reduce a diagram similar to figure 1 on the magazine page, and it becomes difficult to understand the description of the verification method).

Below we use the following notation: the “-” sign means that there are no radioactive balls in the pile being checked, and the “+” sign means that there are some.

In 4 checks it is impossible to select 2 balls out of 6

Let the first group of 6-q balls be checked, where $q=2, 3, 4$ or 5. Then the outcome “-” corresponds to $C_q^2$ options (both radioactive balls among $q$ others), the outcome “+” corresponds to $(C_6^2-C_q^2)$ options. However, of these numbers, $2^3=8$ is greater.

In 5 checks it is impossible to select 2 balls out of 8

If
$C_q^2 < 2^4 = 16$ and $C_8^2 - C_q^2 < 16,$ then $q$ can only be equal to 6. If the outcome is “-”, then it remains to select 2 balls out of 6, which is impossible in 4 checks.

In 6 checks it is impossible to select 2 balls out of 11

If $C_q^2 < 2^5 = 32$ and $C_{11}^2 - C_q^2 < 32,$ then $q=8,$ but $f_2(8) \geq 6$.

For 6 checks you can select 2 balls out of 10

The first check is a pile of 4 balls {1-4}. “-” for 5 checks of 2 balls out of 6 {5-10}. “+” check {4-6}. “+-” for 4 checks of 2 balls out of 7 {1-3, 7-10}, if it is known that {1-3} give “+” (see left half Fig. 1). “++” check {5, 6}. “++-” for 3 checks 1 ball from {1-3, 7-10}. “+++” for 1 check 1 ball from {5, 6} and for 2-1 from {1-4}.

For 7 checks you can select 2 balls out of 15

First check {1-5}. “+” check {5-9}. “+-” for 5 checks 2 balls out of 10 {1-4, 10-15}, among which {1-4} give “+” (see 10 balls, “+”). “++” check {5}. “++-” for 4 checks 1 of {1—4, 6—15}. “++-” for 2 checks 1 from {1-4} and for 2 checks 1 from {6-9}.

After 8 checks you can select 2 balls out of 21

First check {1-6}. In addition to those indicated on figure and very important checks, “+-” is also indicated for 6 checks 2 out of 15 {1-5, 12-21}, among which {1-5} give “+” (see 15 balls, “+”). “++-” for 2 checks 1 from {1-4} and for 2 checks 1 from {8-11}.

37

Fig. 2.

M29

n identical coins lie on the table, forming a closed chain (see Fig. 2). How many revolutions will a coin of the same size make before it is crimped once around the outside of the entire chain, as shown in figure?

How would the answer change if coin $M$ had a radius $k$ times different from the radius of each coin in the chain?

Taking the radius of the coins that make up the chain as one. From figure 3 it is clear that during the time that a coin of radius $k$ rolls along the arc $\alpha$ of a stationary circle of radius 1, it will rotate through an angle $\alpha (1+1/k)$: at this point figure the radii $MA$ and $M'A''$ are parallel, $\angle A''M'B = \angle AOB = \alpha$

Fig. 3.

and, since the arcs $\widehat{A'B}$ and $\widehat{AB}$ are equal in length, $\angle BMA' = \alpha/k$, therefore the entire angle $\angle A''MA'$ at which to turn the coin $M$ is equal to $\alpha + \alpha/k$ (in particular, at $k=1$ this angle is $2\alpha$).

Now let’s find the sum of arcs consisting of points of fixed coins that the coin $M$ touched while rolling along the chain. Let $O_1, O_2, \ldots, O_n$ be the centers of the coins in the chain. The sum of the arcs lying inside the polygon $O_1O_2 \ldots O_n$ is equal to the sum of its internal angles, that is, $\pi (n-2)$. The sum of the arcs lying outside the polygon is therefore equal to $2\pi n - \pi (n-2) = \pi (n+2)$. You also need to subtract from it

Fig. 4, a.

Fig. 4, b.

38

Fig. 5.
the sum of arcs lying in the depressions between two adjacent coins in which $M$ does not fall. In each of the $n$ recesses, the sum of two such arcs is equal to $2\pi/3$ for $k=1$ (Fig. 4a) and $2\arccos \frac{1}{k+1}$ in the general case (Fig. 4, 6). So, the sum of the arcs along which the coin $M$ will roll is equal to $\pi(n+2)-2\pi n/3$ (in the general case $\pi(n+2)-2n\arccos \frac{1}{k+1}$). To find out the required number of revolutions, you need to multiply this value by 2 (in general, by $1+1/k$) and divide by 2π.

Answer. $\left(\frac{n}{3} + 2\right)$
revolution at $k=1$;
$\frac{k+1}{2k}\left(n - \frac{2}{\pi}n\arccos \frac{1}{k+1} + 2\right)$
turnover in the general case ($n \geqslant 3$).

Fig. 6.

In this solution we took advantage (by counting the “losses” in the indentations between coins) that the $M$ coin is rolled over all $n$ coins in a row, without exception. As a rule, reader A. Virovyansky from Gorky, who sent us a solution, noted that this condition is necessary for the problem to have a definite answer. For a chain with “narrow gaps” (see Fig. 5), of course, it is impossible to find out how much the coin $M$ will turn just by the number $n$. A sufficient but not necessary condition for the problem to be solvable is the following: multilevel $O_1$, $O_2...O_n$ vipucles.

Think about how to get the answer if, together with a closed chain, there is simply one coin, two coins touching each other, and generally an open chain of $n$ coin $O_1,...,O_n$, in which $O_k$ touches $O_{k+1}$, arranged so that $M$ can roll back and forth over all the coins in this order:

$O_1$, $O_2,...,O_{n-1}$, $O_n$, $O_{n-1},...,O_1$ (Fig. 6) (Indication: In the formulas written above, it is enough to close $n$ to $2n-2$.).

M30

Prove that $N$ points in a plane can always be covered by several non-intersecting circles whose sum of diameters is less than $N$ and the distance between any two of which is greater than 1. (Under consideration between two search circles is the distance between their closest points.)

We will need the following obvious lemma.

If two circles of diameters $a'_1$ and $a'_2$ intersect (have a common point), then they can be folded into one circle of diameter no more than $a'_1+a'_2$ (Fig. 7, a, 6).

Let’s construct a circle with a center at each of the given $N$ points, having a radius $a$ (and slightly larger than $1/2$; more precisely, the value $a$ will be chosen below). If among these circles they turn out to be intersecting, then, using my lem, we replace any two intersecting circles (no matter which ones) with one circle covering them. If among the obtained circles there are still intersecting ones, we will again use lemmy, etc. In general, let there be some system of circles that: 1) cover all given points together with circles of radius $a$ with centers at these points and 2) have a sum of diameters no more than $N \cdot 2a$. Then if among them there are intersecting ones, then we can

39

Let us use the lemma and construct a new system of fewer circles that satisfies the same conditions 1), 2), and so on until we obtain a system of $k$ circles, no two of which intersect.

Let us now reduce the radius of each of these $k$ circles by amounts $b$, leaving their centers in place ($b$ is greater than $1/2$ and less than $a$; the exact value of $b$ is indicated below). Then the resulting $k$ circles:

1. contain all point data,
2. have a sum of diameters no more than $N \cdot 2a - k \cdot 2b \leq 2Na - 2b$,
3. are separated from each other by no less than $2b$.

It is clear that if we choose $a$ and $b$ so that the inequalities $b < a$; $2Na - 2b < N$, $2b > 1$, then all the requirements of the task will be satisfied. It was enough, for example, to take $a = \frac{1}{2} + \frac{1}{2N}$ and $b = \frac{1}{2} + \frac{1}{4N}$ from the very beginning.

Of course, such $a$ and $b$ were needed only because the conditions required satisfying strict inequalities: the sum of the diameters is less than $N$, the distance between the circles is greater than 1. If we took a simple $a = b = 1/2$, then it would be proven that our $N$ points can be covered (for some $k$) by circles so that the distance between there were at least 1 circles, and the sum of their diameters was no more than $N - k$. Note that since this statement can be proven for any unified dimension, then (by choosing this united one a little less) from it we can

In this issue we publish the solution problem F41—F47 *)

F41

How many times less is the security on a moonlit night at half a beam than on a sunny day? The height of the Moon and the Sun above the horizon is the same. Assume that the illuminated hemisphere of the Moon uniformly scatters light into space.

The radius of the Moon is taken to be 2000 km, and the distance from the Moon to the Sun is 400,000 km.

Since the distances from the Sun to the Earth and the Moon are large compared to the diameter of the Sun, then in calculations we can assume that the Sun is an accurate source of light, uniformly emitting light energy throughout space. Let us assume that the luminous intensity of this source, that is, the energy emitted by the Sun into a unit solid angle in 1 second, is equal to $I$. Then the illumination of the Earth’s surface on a bright sunny day will be equal to $E_c = \frac{1}{L^2}$, where $L$ is the distance from the Sun to the Earth.

The Moon illuminates the Earth with reflected sunlight. Since the distance from the Sun to the Moon can be taken equal to the distance from the Sun

*) Solution problem F43 will be published in «Kvant» No. 5

40

to the Earth, then the illumination of the lunar surface at half moon is also equal to $\frac{I}{L^2}$. Light energy reaches the entire surface of the Moon

$$W = \frac{I}{L^2} \pi r^2.$$

Since this energy then dissipates evenly in all directions along the “hemisphere”, energy is emitted into a unit solid angle

$$I_\pi = \frac{I \pi r^2}{L^2 \cdot 2\pi}$$

(full solid angle is 4 π, half is 2π). Now it is easy to find the illumination of the Earth’s surface during the full moon. Considering the Moon to be an exact source with luminous intensity $I_\pi$, we obtain

$$E_\pi = \frac{I r^2}{2 L^2 l^2},$$

where $l$ is the distance from the Moon to the Earth.

The ratio of the illuminance of the Earth at half moon and on a sunny day is equal to

$$\frac{E_\pi}{E_c} = \frac{1}{2} \left( \frac{r}{l} \right)^2 = \frac{1}{80 000}.$$

F42

The rope will be thrown over the block, with part of the rope lying on the table and part on the floor. After the rope was released, it began to move. Find the speed of steady uniform motion of the rope. The height of the table is $h$ (Fig. 8).

Fig. 8.

The easiest way to solve the problem is to use Newton’s second law.

During time $\Delta t$, a piece of rope of length $\Delta l = v \Delta t$ is involved in motion. If the mass of a unit length of rope is denoted by $\mu$, then the mass of a piece of rope $\Delta l$ is equal to $\Delta m = \mu \Delta l = \mu v \Delta t$. During the time $\Delta t$, the mass $\Delta m$ of the rope is given an impulse (amount of motion) equal to $\Delta mv = \mu v^2 \Delta t$. It is clear that this change in momentum appears due to the difference in gravitational forces acting on the left and right parts of the rope. This difference is equal to $mgh$.

According to Newton’s second law, we can write

$$\mu v^2 \Delta t = \mu gh \Delta t.$$

From here

$$v = \sqrt{gh}.$$

The correct solution was sent by N. Ezhov from Odessa and A. Viktorov from Tula.

F44

Two dolphins swim towards each other. However, they produce a sound of frequency $\omega$. What frequency does the other dolphin hear if the speeds of the dolphins relative to the water are the same and equal to $v$? The speed of sound in water is $u$.

Using the result obtained in the “Doppler Effect” table, we find that the second dolphin will hear sound with a frequency

$$\omega_2 = \omega \frac{u + v}{u - v}.$$

Indeed, since the first dolphin is moving, the frequency of the sound waves that are received by the motionless receiver in front is equal to

$$\omega_1 = \frac{\omega}{1 - \frac{v}{u}} = \frac{\omega u}{u - v}.$$

41

Fig. 9.

Thanks to the movement of the second dolphin, it will pour out the sound with a frequency

$$\omega_2 = \omega_1 \left( 1 + \frac{v}{u} \right) = \omega_1 \frac{u + v}{u} = \omega \frac{u + v}{u - v}.$$

F45

A spherical capacitor filled with a dielectric and charged to a certain potential difference is discharged through its dielectric. What will be the magnetic field of the discharge currents in the space between the spheres?

Let us select a narrow end “conductor” in the insulator. It is clear that for any point $A$ (Fig. 9) we can always find a “conductor” with another concept, such that the current flowing through it creates a field at point $A$, the direction of which is equal to the directionality of the current field of the “conductor” we have selected and has the opposite direction. This means that the magnetic field of the discharge currents between the spheres is zero.

F46

A cube falls onto a ball from a height of 1 m, and then jumps almost 1 m. To what height does the ball jump?

To what height will the ball jump if exactly the same ball falls on it from a height of 1 m?

At the moment when the cube leaves the ball, the speed of the cube is equal to the speed of the top point of the ball. Let’s denote this speed $v$. Then the cube moves freely. Therefore, having written down the law of conservation of energy for it

$$\frac{Mv^2}{2} = Mgh$$

($h — высота подъема$), we find that $v = \sqrt{2gh}$.

The speed of the bottom point of the ball at the moment when the cube comes off it is zero, which means the speed of the center of the ball is $\frac{v}{2}$. Having written down the law of conservation of energy for the ball

$$\frac{m \left( \frac{v}{2} \right)^2}{2} = mgh_1$$

where $h_1$ is the height of the cube, we find

$$h_1 = \frac{h}{4} \approx 25 \text{ cm}$$

You will figure out for yourself how to create the flow of balls.

42

If the ABC process (Fig. 10) is performed on an ideal gas, then the amount of heat Q is imparted to it. What amount of heat is imparted to the gas during the ADC process?

The heat that is imparted to the gas goes to increase its internal energy W and to the work done during the expansion of the gas. Since the initial and final states, and therefore the temperatures of the gas, are the same for both processes, the changes in the internal energy of the gas are also the same. As for the work done during gas expansion, in the first case it is equal to $P_2(V_2-V_1)$, and in the second: $P_1(V_2-V_1)$.

From the law of conservation of energy it follows that

$$Q = W + P_2(V_2-V_1)$$

And

$$Q_1 = W + P_1(V_2-V_1),$$

where $Q_1$ is the amount of heat during the process

$$A \rightarrow D \rightarrow C.$$

From these equations it is easy to find that

$$Q_1 = Q - P_2(V_2-V_1) + P_1(V_2-V_1) = Q - (P_2-P_1)(V_2-V_1).$$

The value $(P_2-P_1)(V_2-V_1)$ is numerically equal to the area of ​​rectangle ABCD. Therefore, our result is quite natural.

The difference $Q-Q_1$ should obviously be equal to the difference in the work done during the expansion of the gas, and the work during the expansion of the gas is numerically equal to the area of ​​the figure formed by the graph of the dependence $P$ on $V$ and the explanation $V$.

The correct solution was sent by A. Rozozin from Izhevsk.

I. Sh. Slobodetsky

43

We will call trigonometric inequalities those containing the unknown $x$, which is part of the arguments of trigonometric functions. Solving a trigonometric inequality means, as always, finding all the values ​​of the unknown that satisfy this inequality.

Solving the simplest trigonometric inequalities does not require any special knowledge beyond the school curriculum. Namely, in the examples discussed below, in addition to the properties of trigonometric functions, only the concept of the absolute value of a number (example 4), the rules for solving quadratic inequalities (examples 5-8) and the properties of a logarithmic function (examples 9, 10) are used. This information is sufficient to cope with the examples given at the end of the article for independent solutions.

From the outset, however, one fundamental point must be noted. In trigonometric inequalities, the arguments of trigonometric functions are treated not as angles or arcs, but as real numbers. Since experience shows that many high school graduates do not have complete clarity on this issue, let us dwell on it in a little more detail. The current applicant is advised to carefully consider what is said below in this regard.

Generally trigonometric functions are introduced first as functions of an angle (more precisely, even just an acute angle). Then the concept of an argument of a trigonometric function is expanded - they begin to consider arc functions. It turns out that it is convenient not to limit the arcs contained within one revolution, that is, having values ​​from $0^\circ$ to $360^\circ$, but to consider arcs whose magnitudes are expressed by any number of degrees (both positive and negative). With this understanding of the argument, trigonometric functions turn out to be periodic. The next step is to move from the hail measurement of arcs to the radian measurement. And finally, he comes to the concept of a trigonometric function in 6th numbers. Namely, the value of the trigonometric function of a number is understood as the value of a given function for that arc, the value of which in radians is expressed by this number.

In physics, trigonometric functions of a numerical argument naturally appear when describing various periodic processes, that is, processes whose character is completely repeated at certain intervals. That is why in mathematics it is advisable to consider trigonometric functions of a numerical argument. In fact, we do this already when we build graphs of trigonometric functions. In fact, we plot the values ​​of real numbers along the abscissa axis.

Let us now consider some trigonometric inequalities

44

STV, let’s start with the simplest example.

Example 1. Solve inequality $tg x < 1$.

Solution. First, we will show how not to “solve” this inequality, that is, we will analyze some of the incorrect answers to the question.

Another applicant, noticing that $tg \frac{\pi}{4} = 1$ and that near the value $x = \frac{\pi}{4}$ the function $tg x$ increases with increasing $x$, writes the answer in the form: $x < \frac{\pi}{4}$. Already by the appearance of the answer, we can say that it is incorrect, since it does not reflect the periodicity of the function $y = tg x$. It is easy to indicate specific values ​​of $x$ that satisfy the condition $x < \frac{\pi}{4}$, but do not satisfy the given inequality. For example, with $x = -\frac{2\pi}{3}$ we have: $-\frac{2\pi}{3} < \frac{\pi}{4}$, but $tg\left(-\frac{2\pi}{3}\right) = \sqrt{3} > 1$; at $x = -\frac{\pi}{2}$ the lying part of the inequality does not make sense, therefore, this value does not satisfy the inequality either, although again $-\frac{\pi}{2} < \frac{\pi}{4}$.

It also happens that an applicant tries to take into account the frequency, but at the same time shows helplessness. Namely, often the “answer” is given in the form: $x < \frac{\pi}{4} + k\pi$, where $k = 0, \pm 1, \pm 2, \dots$.

This “answer” is even worse because it makes no sense at all. It is easy to verify that there are simply no numbers $x$ satisfying the condition $x < \frac{\pi}{4} + k\pi$ for all $k = 0, \pm 1, \pm 2, \dots$.

To get the correct answer, it is useful to consider the graph of the function $y = tg x$ (Fig. 1), since this will make the reasoning more geometrically clear.

The inequality $tg x < 1$ is satisfied by those and only those values ​​$x$ for which the corresponding points of the graph lie below the dotted horizontal line $y = 1$. Further, from figure 1 it is clear that if we find all the solutions belonging to some specific interval of length $x$, then all other solutions will differ from those found by shifting to the right or left by $\pi, 2\pi, 3\pi$, etc. The reason for this is that the number $x$ is the period of the function $y = tg x$.